Datasets:

Infi-MM

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InfiMM-WebMath-40B

like 26

[ "Small utility functions\n\nmyouter(x, y, fun)\nshiftleft(x, k = 1)\nshiftright(x, k = 1)\n\n## Arguments\n\nx a vector. a vector. a non-negative integer. a function, see Details'.\n\n## Details\n\nmyouter(x,y,fun) computes the outer product of x and y using the function fun. The result is a matrix with $$(i,j)$$th element equal to fun(x[i],y[j]). It is not required for fun to be able to work with vector arguments. The function does the computations in R using a simple double loop. So, it is a convenience function, not a speed improving one.\n\nshiftright(x,k) rotates the vector x k positions to the right.\n\nshiftleft(x,k) rotates the vector x k positions to the left.\n\n## Value\n\nfor myouter, a matrix, as described in Details'\n\nfor shiftleft and shiftright, a vector" ]

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[ "# #0148a6 Color Information\n\nIn a RGB color space, hex #0148a6 is composed of 0.4% red, 28.2% green and 65.1% blue. Whereas in a CMYK color space, it is composed of 99.4% cyan, 56.6% magenta, 0% yellow and 34.9% black. It has a hue angle of 214.2 degrees, a saturation of 98.8% and a lightness of 32.7%. #0148a6 color hex could be obtained by blending #0290ff with #00004d. Closest websafe color is: #003399.\n\n• R 0\n• G 28\n• B 65\nRGB color chart\n• C 99\n• M 57\n• Y 0\n• K 35\nCMYK color chart\n\n#0148a6 color description : Dark blue.\n\n# #0148a6 Color Conversion\n\nThe hexadecimal color #0148a6 has RGB values of R:1, G:72, B:166 and CMYK values of C:0.99, M:0.57, Y:0, K:0.35. Its decimal value is 84134.\n\nHex triplet RGB Decimal 0148a6 `#0148a6` 1, 72, 166 `rgb(1,72,166)` 0.4, 28.2, 65.1 `rgb(0.4%,28.2%,65.1%)` 99, 57, 0, 35 214.2°, 98.8, 32.7 `hsl(214.2,98.8%,32.7%)` 214.2°, 99.4, 65.1 003399 `#003399`\nCIE-LAB 32.687, 19.811, -55.643 9.211, 7.394, 37.016 0.172, 0.138, 7.394 32.687, 59.064, 289.597 32.687, -16.338, -76.69 27.191, 12.885, -61.679 00000001, 01001000, 10100110\n\n# Color Schemes with #0148a6\n\n• #0148a6\n``#0148a6` `rgb(1,72,166)``\n• #a65f01\n``#a65f01` `rgb(166,95,1)``\nComplementary Color\n• #019ba6\n``#019ba6` `rgb(1,155,166)``\n• #0148a6\n``#0148a6` `rgb(1,72,166)``\n• #0d01a6\n``#0d01a6` `rgb(13,1,166)``\nAnalogous Color\n• #9ba601\n``#9ba601` `rgb(155,166,1)``\n• #0148a6\n``#0148a6` `rgb(1,72,166)``\n• #a60d01\n``#a60d01` `rgb(166,13,1)``\nSplit Complementary Color\n• #48a601\n``#48a601` `rgb(72,166,1)``\n• #0148a6\n``#0148a6` `rgb(1,72,166)``\n• #a60148\n``#a60148` `rgb(166,1,72)``\n• #01a65f\n``#01a65f` `rgb(1,166,95)``\n• #0148a6\n``#0148a6` `rgb(1,72,166)``\n• #a60148\n``#a60148` `rgb(166,1,72)``\n• #a65f01\n``#a65f01` `rgb(166,95,1)``\n• #01275a\n``#01275a` `rgb(1,39,90)``\n• #013273\n``#013273` `rgb(1,50,115)``\n• #013d8d\n``#013d8d` `rgb(1,61,141)``\n• #0148a6\n``#0148a6` `rgb(1,72,166)``\n• #0153bf\n``#0153bf` `rgb(1,83,191)``\n• #015ed9\n``#015ed9` `rgb(1,94,217)``\n• #0169f2\n``#0169f2` `rgb(1,105,242)``\nMonochromatic Color\n\n# Alternatives to #0148a6\n\nBelow, you can see some colors close to #0148a6. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0171a6\n``#0171a6` `rgb(1,113,166)``\n• #0163a6\n``#0163a6` `rgb(1,99,166)``\n• #0156a6\n``#0156a6` `rgb(1,86,166)``\n• #0148a6\n``#0148a6` `rgb(1,72,166)``\n• #013aa6\n``#013aa6` `rgb(1,58,166)``\n• #012ca6\n``#012ca6` `rgb(1,44,166)``\n• #011fa6\n``#011fa6` `rgb(1,31,166)``\nSimilar Colors\n\n# #0148a6 Preview\n\nThis text has a font color of #0148a6.\n\n``<span style=\"color:#0148a6;\">Text here</span>``\n#0148a6 background color\n\nThis paragraph has a background color of #0148a6.\n\n``<p style=\"background-color:#0148a6;\">Content here</p>``\n#0148a6 border color\n\nThis element has a border color of #0148a6.\n\n``<div style=\"border:1px solid #0148a6;\">Content here</div>``\nCSS codes\n``.text {color:#0148a6;}``\n``.background {background-color:#0148a6;}``\n``.border {border:1px solid #0148a6;}``\n\n# Shades and Tints of #0148a6\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #00040a is the darkest color, while #f6faff is the lightest one.\n\n• #00040a\n``#00040a` `rgb(0,4,10)``\n• #000d1e\n``#000d1e` `rgb(0,13,30)``\n• #001531\n``#001531` `rgb(0,21,49)``\n• #001e45\n``#001e45` `rgb(0,30,69)``\n• #012658\n``#012658` `rgb(1,38,88)``\n• #012f6c\n``#012f6c` `rgb(1,47,108)``\n• #01377f\n``#01377f` `rgb(1,55,127)``\n• #014093\n``#014093` `rgb(1,64,147)``\n• #0148a6\n``#0148a6` `rgb(1,72,166)``\n• #0150b9\n``#0150b9` `rgb(1,80,185)``\n• #0159cd\n``#0159cd` `rgb(1,89,205)``\n• #0161e0\n``#0161e0` `rgb(1,97,224)``\n• #016af4\n``#016af4` `rgb(1,106,244)``\n• #0c74fe\n``#0c74fe` `rgb(12,116,254)``\n• #1f7ffe\n``#1f7ffe` `rgb(31,127,254)``\n• #338afe\n``#338afe` `rgb(51,138,254)``\n• #4695fe\n``#4695fe` `rgb(70,149,254)``\n• #5aa0fe\n``#5aa0fe` `rgb(90,160,254)``\n• #6dabfe\n``#6dabfe` `rgb(109,171,254)``\n• #81b7fe\n``#81b7fe` `rgb(129,183,254)``\n• #94c2fe\n``#94c2fe` `rgb(148,194,254)``\n• #a8cdfe\n``#a8cdfe` `rgb(168,205,254)``\n• #bbd8ff\n``#bbd8ff` `rgb(187,216,255)``\n• #cfe3ff\n``#cfe3ff` `rgb(207,227,255)``\n• #e2eeff\n``#e2eeff` `rgb(226,238,255)``\n• #f6faff\n``#f6faff` `rgb(246,250,255)``\nTint Color Variation\n\n# Tones of #0148a6\n\nA tone is produced by adding gray to any pure hue. In this case, #4e5359 is the less saturated color, while #0148a6 is the most saturated one.\n\n• #4e5359\n``#4e5359` `rgb(78,83,89)``\n• #48525f\n``#48525f` `rgb(72,82,95)``\n• #415166\n``#415166` `rgb(65,81,102)``\n• #3b506c\n``#3b506c` `rgb(59,80,108)``\n• #344f73\n``#344f73` `rgb(52,79,115)``\n• #2e4e79\n``#2e4e79` `rgb(46,78,121)``\n• #284d7f\n``#284d7f` `rgb(40,77,127)``\n• #214c86\n``#214c86` `rgb(33,76,134)``\n• #1b4c8c\n``#1b4c8c` `rgb(27,76,140)``\n• #144b93\n``#144b93` `rgb(20,75,147)``\n• #0e4a99\n``#0e4a99` `rgb(14,74,153)``\n• #0749a0\n``#0749a0` `rgb(7,73,160)``\n• #0148a6\n``#0148a6` `rgb(1,72,166)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0148a6 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "## Mathematics MCQ on Types of Relations for Class 12, JEE and Engineering Exams\n\nMCQ on Types of Relations:  To be an expert in JEE Mathematics, it is absolutely necessary to practice and be familiar will all the concepts as well as the questions of different types. This is essential to gain mastery over the subject. We have also often heard the common saying, “Practice Makes a Man Perfect”, hence students have to practice, practice and practice till they master the subject.", null, "In this post we are providing you MCQ on Types of Relations, which will be beneficial for you in upcoming JEE and Engineering Exams.\n\n## MCQ on Types of Relations\n\nQ1.  Which of the following relations is symmetric and transitive but not reflexive for the set I = {4, 5}?\na) R = {(4, 4), (5, 5)}\nb) R = {(4, 5), (5, 4)}\nc) R = {(4, 5), (5, 4), (4, 4)}\nd) R = {(4, 4), (5, 4), (5, 5)}\n\nd) R = {(4, 4), (5, 4), (5, 5)}\n\nR= {(4, 5), (5, 4), (4, 4)} is symmetric since (4, 5) and (5, 4) are converse of each other thus satisfying the condition for a symmetric relation and it is transitive as (4, 5)∈R and (5, 4)∈R implies that (4, 4) ∈R. It is not reflexive as every element in the set I is not related to itself.\n\nQ2. Which of these is not a type of relation?\na) Reflexive\nb) Symmetric\nc) Surjective\nd) Transitive\n\nc) Surjective\n\nQ3. Which of the following relations is reflexive but not transitive for the set T = {7, 8, 9}?\na) R = {0}\nb) R = {(7, 7), (8, 8), (9, 9)}\nc) R = {(7, 8), (8, 7), (8, 9)}\nd) R = {(7, 8), (8, 8), (8, 9)}\n\nb) R = {(7, 7), (8, 8), (9, 9)}\n\nThe relation R= {(7, 7), (8, 8), (9, 9)} is reflexive as every element is related to itself i.e. (a,a) ∈ R, for every a∈A. and it is not transitive as it does not satisfy the condition that for a relation R in a set A if (a1, a2)∈R and (a2, a3)∈R implies that (a1, a3) ∈ R for every a1, a2, a3 ∈ R.\n\nQ4. Let R be a relation in the set N given by R={(a,b): a+b=5, b>1}. Which of the following will satisfy the given relation?\na) (2,1) ∈ R\nb) (2,3) ∈ R\nc) (4,2) ∈ R\nd) (5,0) ∈ R\n\nb) (2,3) ∈ R\n\n(2,3) ∈ R as 2+3 = 5, 3>1, thus satisfying the given condition.\n(4,2) doesn’t belong to R as 4+2 ≠ 5.\n(2,1) doesn’t belong to R as 2+1 ≠ 5.\n(5,0) doesn’tbelong to R as 0⊁1\n\nQ5. Let I be a set of all lines in a XY plane and R be a relation in I defined as R = {(I1, I2):I1 is parallel to I2}. What is the type of given relation?\na) Equivalence relation\nb) Reflexive relation\nc) Symmetric relation\nd) Transitive relation\n\na) Equivalence relation\n\nThis is an equivalence relation. A relation R is said to be an equivalence relation when it is reflexive, transitive and symmetric.\nReflexive: We know that a line is always parallel to itself. This implies that I1 is parallel to I1 i.e. (I1, I2)∈R. Hence, it is a reflexive relation.\nSymmetric: Now if a line I1 || I2 then the line I2 || I1. Therefore, (I1, I2)∈R implies that (I2, I1)∈R. Hence, it is a symmetric relation.\nTransitive: If two lines (I1, I3) are parallel to a third line (I2) then they will be parallel to each other i.e. if (I1, I2) ∈R and (I2, I3) ∈R implies that (I1, I3) ∈R.\n\nQ6. (a,a) ∈ R, for every a ∈ A. This condition is for which of the following relations?\na) Equivalence relation\nb) Reflexive relation\nc) Symmetric relation\nd) Transitive relation\n\nb) Reflexive relation\n\nQ7. Which of the following relations is transitive but not reflexive for the set S={3, 4, 6}?\na) R = {(1, 1), (2, 2), (3, 3)}\nb) R = {(1, 2), (1, 3), (1, 4)}\nc) R = {(3, 3), (4, 4), (6, 6)}\nd) R = {(3, 4), (4, 6), (3, 6)}\n\nd) R = {(3, 4), (4, 6), (3, 6)}\n\nQ8. Which of the following relations is symmetric but neither reflexive nor transitive for a set A = {1, 2, 3}.\na) R = {(1, 2), (2, 1)}\nb) R = {(1, 1), (2, 2), (3, 3)}\nc) R = {(1, 1), (1, 2), (2, 3)}\nd) R = {(1, 2), (1, 3), (1, 4)}\n\na) R = {(1, 2), (2, 1)}\n\nQ9. (a1, a2) ∈R implies that (a2, a1) ∈ R, for all a1, a2∈A. This condition is for which of the following relations?\na) Equivalence relation\nb) Reflexive relation\nc) Symmetric relation\nd) Transitive relation\n\nc) Symmetric relation\n\nQ10. An Equivalence relation is always symmetric.\na) True\nb) False\n\na) True", null, "", null, "" ]

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[ "# C# API Reference¶\n\nThe C# API Reference provides details on all the classes which make up the Leap Motion API.\n\nNote that any coordinates, directions, and transformations reported by these classes are expressed relative to the Leap Motion coordinate system, not your Unity game world. To convert position vectors to Unity coordinates, use the Vector class extension ToUnityScaled(). To convert direction vectors to Unity coordinates, use ToUnity(). These scripts are defined in a C# script included with our Unity core asset package, LeapUnityExtensions.\n\nYou can convert a transform expressed as a matrix from the Leap Motion API to a Unity Quaternion rotation with the Matrix class extension Rotation() and a Vector3 translation with Translation() (also defined in the script LeapUnityExtensions).\n\nWhen you get Vectors from the Hand scripts, these vectors have already been transformed into Unity world space relative to the HandController object. In other words, if you rotate the controller object in the scene, the hand positions are also rotated. You can transform points from Leap Motion coordinates to Unity world coordinates in the same way using the Unity Transform.TransformPoint() function after calling ToUnityScaled():\n\nLeap.Vector position = finger.TipPosition;\nVector3 unityPosition = position.ToUnityScaled(false);\nVector3 worldPosition = handController.transform.TransformPoint(unityPosition);\n\n\nYou can do the same for direction vectors with TransformDirection():\n\nLeap.Vector direction = finger.Direction;\nVector3 unityDirection = direction.ToUnity(false)\nVector3 worldDirection = handController.transform.TransformDirection(unityDirection);\n\n\nTo convert a rotation relative to the controller object in the scene, multiply the the rotation by the rotation of the controller:\n\nQuaternion leapRotation = arm.Basis.Rotation(false);\nQuaternion finalRotation = controller.transform.rotation * leapRotation;" ]

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[ "# Multiple panes\n\nChart area can be divided into multiple panes using the `rowDefinitions` and `columnDefinitions` properties.\n\n### Row Definitions\n\nTo split the chart area vertically into a number of rows, use `rowDefinitions` of the chart.\n\n• You can allocate space for each row by using the `unit` option that determines whether the chart area should be split by percentage or pixels for the given `rowHeight` value of the rowDefinitions.\n\n• To associate a vertical axis to a row, specify the rowDefinitions index value to the `rowIndex` property of the chart axis.\n\n• To customize each row’s horizontal line, use `lineColor` and `lineWidth` property.\n\n• ts\n• ``````this.rowDefinitions = [{\n// Split first row of the chart area\nunit: 'percentage',\nlineColor: 'Gray',\nrowHeight: 50,\nlinewidth: 0\n}, {\n// Split second row of the chart area\nunit: 'percentage',\nlineColor: 'green',\nrowHeight: 50,\nlinewidth: 0\n}];\nthis.axes = [{\n//Create secondary axis and bind it to second row of chart area\nname: \"yAxis1\",\nrowIndex: 1\n}];``````\n• html\n• ``````<ej-chart id=\"chartcontainer\" [rowDefinitions]=\"rowDefinitions\" [axes]=\"axes\">\n<e-seriescollection>\n<!--Binding vertical axis name-->\n<e-series yAxisName=\"yAxis1\">\n</e-series>\n</e-seriescollection>\n</ej-chart>``````", null, "Row Span\n\nFor spanning the vertical axis along multiple panes vertically, you can use `rowSpan` property of axis.\n\n• html\n• ``````<ej-chart id=\"chartcontainer\" [rowDefinitions]=\"rowDefinitions\" [axes]=\"axes\" [primaryYAxis.rows]=2>\n<e-seriescollection>\n<!--Binding vertical axis name-->\n<e-series yAxisName=\"yAxis1\">\n</e-series>\n</e-seriescollection>\n</ej-chart>``````", null, "## Column Definitions\n\nTo split the chart area horizontally into a number of columns, use `columnDefinitions` of the chart.\n\n• You can allocate space for each column by using the `unit` option that determines whether the chart area should be split by percentage or pixels for the given `columnWidth` value of the columnDefinitions.\n\n• To associate a horizontal axis to a column, specify the columnDefinitions index value to the `columnIndex` property of the chart axis.\n\n• ts\n• ``````// Splitting chart area into multiple columns\nthis.columnDefinitions = [{\n// Split first column of the chart area\nunit: 'percentage',\ncolumnWidth: 50,\n}, {\n// Split second column of the chart area\nunit: 'percentage',\ncolumnWidth: 50,\n}];\n\nthis.axes = [{\n//Create secondary axis and bind it to second column of chart area\nname: \"xAxis1\",\ncolumnIndex: 1\n}];``````\n• html\n• ``````<ej-chart id=\"chartcontainer\" [columnDefinitions]=\"columnDefinitions\" [axes]=\"axes\">\n<e-seriescollection>\n<!--Binding horizontal axis name-->\n<e-series xAxisName=\"xAxis1\">\n</e-series>\n</e-seriescollection>\n</ej-chart>``````", null, "Column Span\n\nFor spanning the horizontal axis along multiple panes horizontally, you can use `columnSpan` property of axis.\n\n• html\n• ``````<ej-chart id=\"chartcontainer\" [columnDefinitions]=\"columnDefinitions\" [axes]=\"axes\" [primaryXAxis.columnSpan]=2>\n<e-seriescollection>\n<!--Binding horizontal axis name-->\n<e-series xAxisName=\"xAxis1\">\n</e-series>\n</e-seriescollection>\n</ej-chart>``````", null, "" ]

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[ "", null, "The Evolution of the\n\nR E A L  N U M B E R S\n\n2\n\n# THE RATIO OF TWONATURAL NUMBERS\n\nThe definition\n\nTHE STUDENT WHO HAS DONE THE previous lesson can begin to appreciate that there is always a name for how any two natural numbers are related.  7 is half of 14.  24 is three times 8.  10 is two thirds of 15.  Those names are called their ratio.  Ratio is the spoken language of arithmetic.\n\nDEFINITION.  The ratio of two natural numbers is their relationship with respect to relative size, which we can always name. Specifically, we can say that one number is a multiple of the other (so many times it), a part of it, or parts of it.\n\n(Topic 1.   Cf. Euclid, VII. Def. 20.)\n\nExample 1.  Multiple   What ratio has 15 to 5?\n\nAnswer.  15 is three times 5.\n\nThat is the ratio -- the relationship -- of 15 to 5.\n\nWe do not answer \"3 to 1,\" because we want to name the ratio of 15 to 5 explicitly.  It is true that 15 has the same ratio to 5 that 3 has to 1.  3 is three times 1, just as 15 is three times 5.\n\nThe two numbers in a ratio are called the terms; the first term and the second.\n\nNotice that we answer with a complete sentence beginning with the first term and ending with the second:  \"15 is three times 5.\"  For, a ratio is a relationship.\n\nExample 2.  Part   What ratio has 5 to 15?\n\nAnswer.  5 is the third part of 15.\n\nThat is called inverse ratio of 15 to 5.  The terms are exchanged.\n\nExample 3.  Parts   What ratio has 10 to 15?\n\nAnswer.  10 is two thirds of 15.", null, "Those are the three types of ratio:  One number is a multiple of the other (so many times it), a part of it, or parts of it.\n\nProblem 10.   What ratio have the following?  Answer with a complete sentence beginning with the first term.\n\na)   2 to 10?   2 is the fifth part of 10.\n\n b)  10 to 2? 10 is five times 2.  A larger number is always so many times a smaller.\n\nc)   7 to 1?   7 is seven times 1.\n\nd)   1 to 7?   1 is the seventh part of 7.\n\ne)   25 to 100?   25 is the fourth part of 100.\n\nf)   75 to 100?   75 is three fourths of 100.\n\ng)   12 to 6?   12 is two times 6, or twice as much as 6, or double 6.\n\nh)   6 to 12?   6 is half of 12.\n\ni)   40 to 8?   40 is five times 8.\n\nj)   24 to 6?   24 is four times 6.\n\nk)   6 to 24?   6 is the fourth part of 24.\n\nl)   10 to 1?   10 is ten times 1.\n\nm)   1 to 10?   1 is the tenth part of 10.\n\nThe ratio of a smaller number to a larger\n\n2 to 3\n\nWe can always name the ratio of any smaller number to a larger simply by letting each number say its name.  Let the smaller number say its cardinal name -- One, two, three, . . .  Let the larger number say its ordinal name -- third, fourth, fifth . . . .\n\nExample 4.   What ratio has 2 to 3?\n\nAnswer.  \"2 is two thirds of 3.\"\n\n2 says its cardinal name, \"two.\"  3 says its ordinal name, \"third.\"\n\nExample 5.   What ratio has 4 to 5?\n\nAnswer.  \"4 is four fifths of 5.\"  Each number says its name.\n\nTo see this, consider that 1 is one fifth of 5:", null, "2 is two fifths of 5.\n\n3 is three fifths of 5.\n\n4 is four fifths of 5.\n\nEach number says its name.\n\nProblem 11.   What ratio has\n\na)   5 to 8?   5 is five eighths of 8.\n\nb)   3 to 4?   3 is three fourths of 4.\n\nc)   2 to 9?   2 is two ninths of 9.\n\nc)   99 to 100?   99 is ninety-nine hundredths of 100.\n\nWhat ratio has 8 to 12?  While it is correct to say that 8 is eight twelfths of 12, we will see how to express that ratio with the smallest numbers that have that ratio, that is, with the lowest terms. (Lesson 3: The theorem of the common divisor.)\n\nA mixed number of times\n\nBy a mixed number of times, we mean a whole number of times plus a part.\n\nExample 6.   How much is two and a half times 8?\n\nAnswer.  \"Two and a half times 8\" means\n\nTwo times 8 plus half of 8.\n\nTwo times 8 is 16.  Half of 8 is 4.  16 plus 4 is 20.\n\nExample 7.   A cheese sells for \\$6 a pound, and you buy three and a half pounds.  How much do you pay?\n\n Answer. Three pounds cost \\$18. Half a pound costs \\$3. You pay \\$21.\n\nThat is, \"Three and a half times 6\" means\n\nThree times 6 plus half of 6.\n\n18 + 3 = 21.\n\nThis is a mixed number of times:  A whole number of times plus a part.\n\nExample 8.   How much is five and a quarter times 8?\n\n Answer. \"Five times 8 is 40. \"A quarter (or a fourth) of 8 is 2. \"40 + 2 = 42.\"\n\nProblem 12.   How much is\n\na)   Two and a half times 40?   80 + 20 = 100\n\nb)   One and a half times 12?   12 + 6 = 18\n\nc)   One and a quarter times 20?   20 + 5 = 25\n\nd)   Two and a quarter times 8?   16 + 2 = 18\n\ne)   Three and a half times 10?   30 + 5 = 35\n\nf)   Five and a third times 6?   30 + 2 = 32\n\ng)   Two and a quarter times 100?   200 + 25 = 225\n\nh)   Two and three quarters times 100?   200 + 75 = 275\n\nMixed ratio\n\nRatio and division\n\nWe just saw that \"20 is two and a half times 8.\" That statement expresses the ratio of 20 to 8.  It is called a mixed ratio.  In a mixed ratio, the larger number is a multiple of the smaller number, plus a part or parts of the smaller number.\n\nExample 9.   What ratio has 25 to 10?\n\nAnswer.  We can decompose 25 into a multiple of 10 plus a remainder:\n\n25 = 20 + 5.\n\n25 is made up of two 10's, plus a remainder of 5.  The remainder 5 is a part of 10, namely half.  Therefore we say,\n\n\"25 is two and a half times 10.\"\n\nTwo times 10 is 20; half of 10 is 5; 20 plus 5 is 25.\n\nWe always say that a larger number is so many times a smaller number.  25 is two and a half times 10.\n\nExample 10.   What ratio has 13 to 3? That is, 13 is how many times 3?\n\nTo answer, we can divide 13 by 3.\n\n13 ÷ 3 = 4 R 1.\n\n13 is made up of four 3's with remainder 1.\n\nThe remainder 1 is a part of 3 -- it is the third part.  We say,\n\n\"13 is four and a third times 3.\"\n\nNotice again:  We always say that a larger number is so many times a smaller.\n\nExample 11.   What ratio has 50 to 40?\n\nAnswer.   50 is one and a quarter times 40.\n\nFor, 50 contains 40 one time with remainder 10.\n\nThe remainder 10 is a quarter of 40.  Therefore,\n\n50 is one and a quarter times 40.\n\nWe now see that we can always express in words the relationship -- the ratio -- of any two natural numbers.\n\nWe also see the relationship between ratio and division.  The quotient of two numbers indicates the ratio of those numbers.  The ratio of 15 to 5, for example, is indicated by 15 ÷ 5 = 3.  This implies:\n\n15 = 3 × 5.\n\n\"15 is three times 5.\"\n\nThe traditional notation for ratio is 15 : 3, which is the divison sign ÷ but without the bar.\n\nProblem 13.   Express each ratio.  (The larger number is how many times the smaller number?)\n\na)   45 to 10?   45 is four and a half times 10.\n\nb)   20 to 8?   20 is two and a half times 8.\n\nc)   22 to 4?   22 is five and a half times 4.\n\nd)   5 to 2?   5 is two and a half times 2.\n\ne)   7 to 2?   7 is three and a half times 2.\n\nf)   13 to 2?   13 is six and a half times 2.\n\ng)   5 to 4?   5 is one and a quarter times 4.\n\nh)   9 to 4?   9 is two and a quarter times 4.\n\ni)   11 to 4?   11 is two and three quarters times 4.\n\nj)   11 to 3?   11 is three and two thirds times 3.\n\nk)   44 to 6?   44 is seven and a third times 6.\n\nFinally, then, we see that we can always express in words the ratio of any two natural numbers.", null, "Next Topic:  Proportions\n\nPlease make a donation to keep TheMathPage online.\nEven \\$1 will help." ]

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[ "# ColorHistogramFilter¶", null, "Filter point indices using color histogram by comparing with reference histogram\n\nMethods for histogram comparison is configurable from multiple methods. (See parameter ~compare_policy) After computing distance between input histograms and reference, filter by thresholding (See parameter ~distance_threshold) Reference histogram can be set as ~reference topic or as a parameter ~reference_histogram.\n\n## Subscribing Topics¶\n\n• ~input (jsk_recognition_msgs/ColorHistogramArray)\n\nInput color histogram arrayThe order of each histograms must be the same as the order of input cluster point indices.\n\n• ~input/indices (jsk_recognition_msgs/ClusterPointIndices)\n\nInput point indices\n\n• ~input/reference (jsk_recognition_msgs/ColorHistogram)\n\nReference histogram\n\nIt can be set as a parameter. See parameter ~reference_histogram.\n\n## Publishing Topics¶\n\n• ~output (jsk_recognition_msgs/ColorHistogramArray)\n\nFiltered color histogram array\n\n• ~output/indices (jsk_recognition_msgs/ClusterPointIndices)\n\nFiltered cluster point indices\n\n## Parameters¶\n\n• ~queue_size (Int, default: 100)\n\nQueue size for message synchronization\n\n• ~bin_size (Int, default: 100)\n\nBin size for histogram\n\n• ~compare_policy (Enum[Int], default: CORRELATION)\n\nPolicy for histogram values to compare\n\n• 0: CORRELATION\n• Use correlation\n• 1: BHATTACHARYYA\n• Use bhattacharyya distance\n• 2: INTERSECTION\n• Use vector intersection\n• 3: CHISQUARE\n• Use chi-square between two vectors\n• 4: KL_DIVERGENCE\n• Use Kullback-Leibler divergence for comparing two vectors\n• ~distance_threshold (Double, default: 0.6)\n\nColor histograms and point cloud indices whose similarities are above this value are published as filtered topics.\n\n• ~flip_threshold (Bool, default: false)\n\nPublish indices whose distance from reference is higher than ~distance_threshold if this value is false. If this value is true, publish indices whose is lower than threshold.\n\n• ~reference_histogram (Float[])\n\nReference histogram\n\nIt can also be set as topic. See ~input/reference topic." ]

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[ "## 56737\n\n56,737 (fifty-six thousand seven hundred thirty-seven) is an odd five-digits prime number following 56736 and preceding 56738. In scientific notation, it is written as 5.6737 × 104. The sum of its digits is 28. It has a total of 1 prime factor and 2 positive divisors. There are 56,736 positive integers (up to 56737) that are relatively prime to 56737.\n\n## Basic properties\n\n• Is Prime? Yes\n• Number parity Odd\n• Number length 5\n• Sum of Digits 28\n• Digital Root 1\n\n## Name\n\nShort name 56 thousand 737 fifty-six thousand seven hundred thirty-seven\n\n## Notation\n\nScientific notation 5.6737 × 104 56.737 × 103\n\n## Prime Factorization of 56737\n\nPrime Factorization 56737\n\nPrime number\nDistinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 56737 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 10.9462 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 56,737 is 56737. Since it has a total of 1 prime factor, 56,737 is a prime number.\n\n## Divisors of 56737\n\n2 divisors\n\n Even divisors 0 2 2 0\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 56738 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 28369 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 238.195 Returns the nth root of the product of n divisors H(n) 1.99996 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 56,737 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 56,737) is 56,738, the average is 28,369.\n\n## Other Arithmetic Functions (n = 56737)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 56736 Total number of positive integers not greater than n that are coprime to n λ(n) 56736 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5747 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares\n\nThere are 56,736 positive integers (less than 56,737) that are coprime with 56,737. And there are approximately 5,747 prime numbers less than or equal to 56,737.\n\n## Divisibility of 56737\n\n m n mod m 2 3 4 5 6 7 8 9 1 1 1 2 1 2 1 1\n\n56,737 is not divisible by any number less than or equal to 9.\n\n• Arithmetic\n• Prime\n• Deficient\n\n• Polite\n\n• Prime Power\n• Square Free\n\n## Base conversion (56737)\n\nBase System Value\n2 Binary 1101110110100001\n3 Ternary 2212211101\n4 Quaternary 31312201\n5 Quinary 3303422\n6 Senary 1114401\n8 Octal 156641\n10 Decimal 56737\n12 Duodecimal 28a01\n20 Vigesimal 71gh\n36 Base36 17s1\n\n## Basic calculations (n = 56737)\n\n### Multiplication\n\nn×i\n n×2 113474 170211 226948 283685\n\n### Division\n\nni\n n⁄2 28368.5 18912.3 14184.2 11347.4\n\n### Exponentiation\n\nni\n n2 3219087169 182641348707553 10362522201620434561 587938422153338595687457\n\n### Nth Root\n\ni√n\n 2√n 238.195 38.4257 15.4336 8.92839\n\n## 56737 as geometric shapes\n\n### Circle\n\n Diameter 113474 356489 1.01131e+10\n\n### Sphere\n\n Volume 7.65046e+14 4.04522e+10 356489\n\n### Square\n\nLength = n\n Perimeter 226948 3.21909e+09 80238.2\n\n### Cube\n\nLength = n\n Surface area 1.93145e+10 1.82641e+14 98271.4\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 170211 1.39391e+09 49135.7\n\n### Triangular Pyramid\n\nLength = n\n Surface area 5.57562e+09 2.15245e+13 46325.6\n\n## Cryptographic Hash Functions\n\nmd5 bd4cc228a6a753ba2b024997ee4a626d 6c88de424673e06818581ebd580a5da9d18eef1a 5d9aee823b1ce11cbae324a74bacbc630d358ac6bd17c0d031653409c9595cb4 d21f5b616a9929123fa7f0eb7deeda0d1d9df1d18e610b9770859654d372f4364aed383a885a910fa702e470c3640f9676be0bfb176c5018d06bec0d983f64b5 f902ba68b4134d13ebc30b8a31ea06570369da8c" ]

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[ "# Accelaration of the chain as a function of x\n\n## Homework Statement\n\nA uniform flexible chain of length L ,with weight per unit length λ , passes over a small frictionless peg..It is released from a rest position with a length of chain x hanging from one side and a length L-x from the other side .Find the accelaration a as a function of x.\n\n## The Attempt at a Solution\n\nI am not sure how to approach the problem.I feel there can be three ways to approach .\n\n1. Work energy method\n2 Newtons law F=Ma considering dx element .\n3. Centre of Mass\n\n## The Attempt at a Solution\n\n#### Attachments\n\n•", null, "chain.jpg\n17.4 KB · Views: 507\n\nSince the question talks about acceleration, Newton's force law would be the right track to follow. You will need the center of mass of either part of the chain. Good luck!\n\nCan u elaborate??\n\nCentre of mass will be half the distance on either part..then?\n\nCan u elaborate??\nSince the question talks about acceleration, Newton's force law would be the right track to follow.\nSince Newton's law has acceleration explicitly (the a of F = ma) and conservation of energy usually has velocities, Newton's laws will lead you to the result more conveniently (There is some intuition also involved here, which I cannot really explain).\n\nYou will need the center of mass of either part of the chain. Good luck!\n\nThe two segments of the chain, one on each side of the peg, can be treated as blocks of mass λs1 and λs2 (where s1 and s2 are the lengths of each segment) hanging from a massless string. This reduces your problem to a standard 'two blocks on a pulley' scenario, which I'm sure you have come across before.\n\nAs the chain slides, s1 and s2 change, thus changing the acceleration.\n\nIn the standard 'two blocks on a pulley' scenario the masses on the two sides are constant, here the masses are constantly changing ??\n\nYes. The scenarios are not the same, but similar.\n\nOnce you do the calculation as I outline and put values for s1 and s2 as x and (l-x), your acceleration will be a function of x.\n\nIn a normal two blocks on a pulley, the mass of blocks on either side is a constant, hence the acceleration is constant.\n\nSorry ...i m not able to get it....Can i get more insight ??\n\nehild\nHomework Helper\nIn the standard 'two blocks on a pulley' scenario the masses on the two sides are constant, here the masses are constantly changing ??\n\nThey are constantly changing. The two pieces interact at the peg, with some tension. You can imagine that, for an instant, you have two rods, length x and L-x, and both with linear density λ, connected with a piece of string wrapped around the pulley. This string provides the tension. Can you find the acceleration of such system?\n\nehild\n\n#### Attachments\n\nYes i can find the accelaration in this case ... so that means the accelaration is continuously changing since the masses are changing ....increasing continuously...\n\nWhat is the force acting on this system ? Both the parts are being pulled down by gravity...How is tension acting ?? What is the motion of Centre of mass ??\n\nCan we approach the problem in some other way?\n\nehild\nHomework Helper\nYes i can find the accelaration in this case ... so that means the accelaration is continuously changing since the masses are changing ....increasing continuously...\n\nWhat is the result you got? What does it mean?\n\nehild\n\na={(2x-L)/L}g is the answer . since L continuously decreases 'a' increases .\n\nehild\nHomework Helper\nL is the length of the chain, it is constant. x increases, so a increases, but how long? And what happens if x=L? Can the acceleration increase any further?\n\nWhat happens if x=L/2 at the beginning?\n\nehild\n\nIf x=L then chain undergoes free fall i.e accelaration is 'g' .If x=L/2 at the beginning then system remains at rest . am i correct ?\n\nehild\nHomework Helper\nRight. In case of x=L/2 at the beginning, but it has some velocity, x will change, and it will accelerate with time.\n\nehild" ]

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[ "## Infix expression to postfix form using the stack function, Data Structure & Algorithms\n\nAssignment Help:\n\nQ. Convert the following given Infix expression to Postfix form using the stack function: x + y * z + (p * q + r ) * s, Follow general precedence rule and suppose that the expression is legal.\n\nAns.\n\nThe given infix expression is as follows :-\n\nx + y * z + (p * q + r) * s\n\nSymb                Stack                    Postfix Expression x empty  x\n\n+                       +                           x\n\ny                       +                           xy\n\n*                       +*                         xy\n\nz                       +*                         xyz\n\n+                       +                           xyz*+ (          +(         xyz*+\n\np                       +(                         xyz*+p\n\n*                       +(*                       xyz*+p q       +(*       xyz*+p\n\n+                       +(+                       xyz*+pq*\n\nr                        +(+                       xyz*+pq*r\n\n)                        +                           xyz*+pq*r+\n\n*                       +*                         xyz*+pq*r+\n\ns                        +*                         xyz*+pq*r+s\n\n= xyz*+pq*r+s*+ The postfix expression obtained  is :- xyz*+pq*r+s*+\n\n#### Efficient algorithms.., implementation of fast fourier transforms for non p...\n\nimplementation of fast fourier transforms for non power of 2\n\n#### Algorithm for multiplication of two sparse matrices using li, algorithm for...\n\nalgorithm for multiplication of two sparse matrices using linked lists..\n\n#### Functions for inserting and deleting at either of the end, Q. Develop a rep...\n\nQ. Develop a representation for a list where insertions and deletions can be done at either end. Such a structure is known as a Deque (Double ended queue). Write functions for inse\n\n#### Techniques of representing polynomials using arrays, Q. Explain any three m...\n\nQ. Explain any three methods or techniques of representing polynomials using arrays. Write which method is most efficient or effective for representing the following polynomials.\n\n#### Postfix expression algorithm, Write an algorithm to calculate a postfix exp...\n\nWrite an algorithm to calculate a postfix expression.  Execute your algorithm using the given postfix expression as your input : a b + c d +*f ↑ . T o evaluate a postfix expr\n\n#### The two famous methods for traversing, The two famous methods for traversin...\n\nThe two famous methods for traversing are:- a) Depth first traversal b) Breadth first\n\n#### Abstract data type-stack, Conceptually, the stack abstract data type mimics...\n\nConceptually, the stack abstract data type mimics the information kept into a pile on a desk. Informally, first we consider a material on a desk, where we might keep separate stack\n\n#### C++ function, Write c++ function to traverse the threaded binary tree in in...\n\nWrite c++ function to traverse the threaded binary tree in inorder traversal\n\n#### Linked list, write an algorithm for multiplication of two sparse matrices u...\n\nwrite an algorithm for multiplication of two sparse matrices using Linked Lists\n\n#### Multiple Queues in a single dimension array, Implement multiple queues in a...\n\nImplement multiple queues in a single dimensional array. Write algorithms for various queue operations for them.", null, "", null, "" ]

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[ "", null, "# Digital Logic OR Gate\n\n## Digital Logic OR Gate – Digital Gates\n\n### What is Logic OR Gate\n\nA logic gate which gives low state “0” only and only when all of the inputs are low state “0”. Otherwise, provide high state “1” as output.\n\nLogical OR gate can have 2 or more than 2 input lines but it has one output line. It can be either true or false, true being High state”1” and false being Low state”0”.", null, "OR Gate works on submission bases. i.e.\n\nIf either “A” or “B” is “1”, Then “C” is 1.\n\nGood to know:\n\nThese terms also used in binary logic gates and circuits\n\n0 or 1\n\nLow or High\n\nTrue or False\n\nON or OFF\n\n#### OR Gate Logic Symbol, Boolean Expression & Truth Table\n\n##### OR Gate Symbol\n\nThere are 3 types of symbols used for OR gate all over the world.\n\nAmerican National Standards Institute (ANSI) MILITARY", null, "International Electrotechnical Commission (IEC) EUROPEAN", null, "Deutsches Institut für Normung (DIN) GERMANY", null, "##### Boolean Expression\n\nIn Boolean algebra (+) sign is used for OR operation, which gives high state “1” if either one or both of the operands are a high state. In most of the programming languages (|) is used for Bit wise OR operation while (||) is used for logical OR operation.\n\nC = A+B,         C=A|B,            C=A || B,\n\n##### Truth Table\n\nA mathematical table used to specify input to output logic combination of a digital circuit is known as a truth table, the truth table of OR Gate is given below.", null, "### Construction and Working Mechanism of OR Gate\n\n#### OR Gate using Resistor – Diode Logic\n\nIn resistor-diode logic (RDL), the diode is used as a switching unit. In the schematic given below RDL logic OR gate is given, in which 2 diodes and 1 pull down resistor are used. Pull down resistor means that whenever the diodes are in reversed bias it will pull output voltages down to 0 volts (LOW state “0”).\n\nDiodes are used in a parallel configuration. When both of the input A and B are “0”, diodes will be reverse bias meaning no current will flow; as a result, output volts will be pulled down i.e. LOW STATE “0”. And if there is even a single high state input, then its corresponding diode will become forward bias resulting in high state “1” flow out as output. Hence resulting in the truth table given below.\n\nThere is also 0.7 volts drop in the diode.", null, "", null, "#### OR Gate using Resistor – Transistor Logic\n\nIn Resistor-Transistor logic (RTL), Bipolar Junction Transistor (BJT) is used as a switching unit. BJTs are high power consumption devices as it operates on base current. RTL logic schematic of OR gate is given in the figure given below. In this schematic two NPN transistor is used in parallel configuration and 2 input resistors and a pull-down resistor.\n\nThere are two discrete input lines connected with separate NPN transistor’s base. And an output line is taken between resistor and NPN Transistors.\n\nNPN transistors switch on when there is a HIGH input at its base and current starts flowing through the transistor.\n\nWhen both of the inputs are LOW state “0”, BJTs will be switched off, as a result, there will be no current flow and OUTPUT “C” will be pulled down to “0 LOW state”.\n\nWhen there is a single HIGH state input, one of the BJTs connected in parallel will turn on and start conducting current and eventually producing voltage Vcc “high state 1” across load resistor “R” which is taken as OUTPUT C.\n\n#### OR Gate using MOS Logic\n\nIn MOS logic, the main switching units are MOSFETs, MOSFETs are fast switching, low power consumption devices as it operates on gate voltages, MOS logic schematic of OR gate is given in the figure below.\n\nMOS logic uses the same working principle as RTL logic, the only difference is use MOSFETs instead of BJTs. As you can see in the figure given below inputs are fed through the input resistor to the gates of two parallel connected NMOSFETs and output is taken across load resistor “R1”.\n\nNMOSFETs switch on when there is high state “1” on its gate and starts conducting current through it. And turns off when there is LOW state and become open(no flow of current).\n\nWhen both input lines are LOW state “0”, both NMOSFETs will be switched off and there will be no current flow. So the only path connected to output will be GND (LOW state 0).\n\nIf there is a single high state input or both of the inputs are a high state then one or both of the NMOSFETs will be switched on respectively. It will give a closed path for current to flow through Vdd to GND. As a result, there will be a voltage drop across LOAD resistor resulting in HIGH state “1” at the output.", null, "", null, "#### OR Gate From Other Logic Gates\n\nOR gate functionality can be achieved with the help different combinations of the different logic gate. Some of them are given below;\n\nNAND and NOR gates are Universal gates which can be used to implement any kind of Boolean function. It’s very easy to implement,\n\n##### OR Gate constructed from NOR Gate", null, "##### OR Gate from NOR, NOT Gate\n\nOR gate constructed from NOR GATE is given above. NOR is actually INVERT of OR. So placing an INVERT/NOT gate to the output of NOR gate will provide OR gate’s result. Or placing a NOR gate with common input line in front of NOR gate will also do the job as it will act as an INVERTER.\n\nNAND gates operation is same as AND gate but with an inverted output. So feeding it inverted input will provide the Operation OR as explained in below Boolean expressions\n\nC = (A + B)              Taking complement on both side\n\nC’= (A + B)’             De Morgan’s law\n\nC’= (A’ . B’)             Taking complement again on both side\n\nC= ( A’ . B’)’\n\n### OR Gate with Multiple Input\n\nOR gate can have more than 2 inputs and can be used according to the requirements of the schematic design.\n\nOR gate with “n” number of input is given below.", null, "#### Truth Table\n\nNOTE: In the table given below “X” means “don’t care”. It can be “1”and it can be “0”. Which means, as long as there is a single input carrying “1”, the output will be always “1” so there is no reason to put a check on other inputs, that’s why it’s called “don’t care X”.", null, "#### Resistor-Diode Logic\n\nIN RDL (Resistor-Diode logic) logic, a diode is used on every input line, thus adding diodes can potentially increase the number of input lines as shown in the figure below. Each input is fed to separate diode. Place as many diodes as you would like, to create a Multi-Input RDL OR GATE.", null, "#### Resistor-Transistor Logic\n\nIn RTL (Resistor-Transistor Logic), Transistors are used as switching unit. To increase the input lines we have to increase the number of transistors connected in parallel as shown in the figure below.", null, "#### MOS Logic\n\nThe idea to create a Multi-Input OR gate in MOS logic is exactly same as in RTL logic. Increasing the number of MOSFETs in parallel can potentially increase the number of input lines. The figure given below shows MOS logic OR gate schematic.", null, "Multi-input OR gate can be made from cascading setup of 2-input OR gates as shown in the figure given below.\n\nOUT                =             IN1 + IN2 + IN3\n\nOUT                =          (IN1 + IN2) + IN3", null, "OUT              =            IN1 + IN2 + IN3 + IN4\n\nOUT              =          (IN1 + IN2) + (IN3 + IN4)", null, "### TTL and CMOS Logic OR Gate IC’s\n\nCommercially up to 4-Input OR gate IC’s are available in the market, having two OR gates in a single package.\n\nSuch as:\n\nTTL Logic OR Gates\n\n• 74LS32 Quad 2-input OR Gate\n• 744075 TTL 3-Input OR Gate\n\nCMOS Logic OR Gate\n\n• CD4071 Quad 2-input OR Gate\n• CD4075 Triple 3-input CMOS OR Gate\n• CD4072 Dual 4-input CMOS OR Gate\n\nHex 2-input OR drivers\n\n• 74832\n\n####", null, "", null, "##### Pinout for 7432 TTL OR Gate IC\n PIN Number Description 1 Input Gate 1 2 Input Gate 1 3 Output  Gate 1 4 Input Gate 2 5 Input Gate 2 6 Output Gate 2 7 Ground 8 Output Gate 3 9 Input Gate 3 10 Input Gate 3 11 Output Gate 4 12 Input Gate 4 13 Input Gate 4 14 Positive Supply\n\n#### 4072 CMOS OR Gate IC (4-Inout)", null, "", null, "Commonly available OR Logic Gate IC are given in the table below:\n\n 4071 Quad 2-Input OR Gate 4072 Dual 4-Input OR Gate 4073 Triple 3-Input AND Gate 4075 Triple 3-Input OR Gate 7432 Quad 2-Input OR Gate 741G32 Single 2-Input OR Gate 741G86 Single 2-Inout Exclusive OR Gate 74832 Hex 2-Input OR Drivers 741G3208 Single 3-Input OR-AND Gate 744075 Single 3-Inpout OR Gate 744078 8-input OR / NOR Gate\n\n### OR Gate Applications\n\nOR Gate has been widely used in practical life applications such as:\n\n• To detect exceed of specific quantities / parameters (such as temperature or pressure) and produce command signal for the system to take required action.\n• Alarm circuit for car door system\n• In calculators, computers and digital logic circuits." ]

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[ "## 概念\n\n\nhello world!\n\n\n\nstr1 = 'hello world!'\nstr2 = 'hello! How are you doing?'\n\n\n## 思路\n\n\ns1 = 'abcdefg'\ns2 = 'defgcde'\n\n\ns2的每一个字符去比对s1的字符，如果相同,对应s1字符索引值为1不同为0，于是有了下面这个矩阵\n\ns1abcdefg\n-x1x2x3x4x5x6x7\ny10001000\ny20000100\ny30000010\ny40000001\ny50010000\ny60001000\ny70000100\n\ns1abcdefg\n-x1x2x3x4x5x6x7\ny10001000\ny20000200\ny30000030\ny40000004\ny50010000\ny60002000\ny70000300\n\n## 实现代码\n\ndef findit(str1, str2):\n\"\"\"\n求两个字符串的最长公共子串\n:param str1: 字符串一\n:param str2: 字符串二\n:return: 最长公共子串\n\"\"\"\nmatrix = [ * len(str1) for i in range(len(str2))]\n# 初始化阵列表，一次性开辟空间\nxmax = 0\nxindex = 0\n\nfor y, i in enumerate(str2):\n# y轴\nfor x, j in enumerate(str1):\n# x轴\nif x == 0 or y == 0:\n# 如果 x,y轴处于边界状态，那么不需要加上层，否则超出边界\n# 相等直接等于1\nif i == j:\nmatrix[y][x] = 1\nelse:\nif i == j:\nmatrix[y][x] = matrix[y - 1][x - 1] + 1\n\nif matrix[y][x] > xmax:\nxmax = matrix[y][x]\nxindex = x + 1\n# xindex = x\n# xindex += 1 # 切片后不包，所以需要+1\n# 记录阵列最大值及其索引\n\nreturn str1[xindex - xmax:xindex]\n\ns1 = 'abcdefg'\ns2 = 'defgcde'\n\nprint(findit(s1, s2))" ]

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[ "/JavaScript\n\n# label\n\nThe labeled statement can be used with `break` or `continue` statements. It is prefixing a statement with an identifier which you can refer to.\n\n## Syntax\n\n```label:\nstatement\n```\n`label`\n\nAny JavaScript identifier that is not a reserved word.\n\n`statement`\n\nA JavaScript statement. `break` can be used with any labeled statement, and `continue` can be used with looping labeled statements.\n\n## Description\n\nYou can use a label to identify a loop, and then use the `break` or `continue` statements to indicate whether a program should interrupt the loop or continue its execution.\n\nNote that JavaScript has no `goto` statement, you can only use labels with `break` or `continue`.\n\nIn strict mode code, you can't use `let` as a label name. It will throw a `SyntaxError` (let is a reserved identifier).\n\n## Examples\n\n### Using a labeled continue with for loops\n\n```let i, j;\n\nloop1:\nfor (i = 0; i < 3; i++) { //The first for statement is labeled \"loop1\"\nloop2:\nfor (j = 0; j < 3; j++) { //The second for statement is labeled \"loop2\"\nif (i === 1 && j === 1) {\ncontinue loop1;\n}\nconsole.log(`i = \\${i}, j = \\${j}`);\n}\n}\n\n// Output is:\n// \"i = 0, j = 0\"\n// \"i = 0, j = 1\"\n// \"i = 0, j = 2\"\n// \"i = 1, j = 0\"\n// \"i = 2, j = 0\"\n// \"i = 2, j = 1\"\n// \"i = 2, j = 2\"\n// Notice how it skips both \"i = 1, j = 1\" and \"i = 1, j = 2\"\n```\n\n### Using a labeled continue statement\n\nGiven an array of items and an array of tests, this example counts the number of items that passes all the tests.\n\n```let itemsPassed = 0;\nlet i, j;\n\ntop:\nfor (i = 0; i < items.length; i++) {\nfor (j = 0; j < tests.length; j++) {\nif (!tests[j].pass(items[i])) {\ncontinue top;\n}\n}\n\nitemsPassed++;\n}\n```\n\n### Using a labeled break with for loops\n\n```let i, j;\n\nloop1:\nfor (i = 0; i < 3; i++) { //The first for statement is labeled \"loop1\"\nloop2:\nfor (j = 0; j < 3; j++) { //The second for statement is labeled \"loop2\"\nif (i === 1 && j === 1) {\nbreak loop1;\n}\nconsole.log(`i = \\${i}, j = \\${j}`);\n}\n}\n\n// Output is:\n// \"i = 0, j = 0\"\n// \"i = 0, j = 1\"\n// \"i = 0, j = 2\"\n// \"i = 1, j = 0\"\n// Notice the difference with the previous continue example\n```\n\n### Using a labeled break statement\n\nGiven an array of items and an array of tests, this example determines whether all items pass all tests.\n\n```let allPass = true;\nlet i, j;\n\ntop:\nfor (i = 0; i < items.length; i++) {\nfor (j = 0; j < tests.length; j++) {\nif (!tests[j].pass(items[i])) {\nallPass = false;\nbreak top;\n}\n}\n}\n```\n\n### Using a labeled block with break\n\nYou can use labels within simple blocks, but only `break` statements can make use of non-loop labels.\n\n```foo: {\nconsole.log('face');\nbreak foo;\nconsole.log('this will not be executed');\n}\nconsole.log('swap');\n\n// this will log:\n\n// \"face\"\n// \"swap\"\n```\n\n### Labeled function declarations\n\nLabels can only be applied to statements, not declarations. Still, the Annex B: Additional ECMAScript Features for Web Browsers section defines a legacy grammar to standardize labeled function declarations in non-strict code.\n\n```L: function F() {}\n```\n\nIn strict mode code, however, this will throw a `SyntaxError`:\n\n```'use strict';\nL: function F() {}\n// SyntaxError: functions cannot be labelled\n```\n\nGenerator functions can neither be labeled in strict code, nor in non-strict code:\n\n```L: function* F() {}\n// SyntaxError: generator functions cannot be labelled\n```\n\n## Browser compatibility\n\nDesktop Mobile Server\nChrome Edge Firefox Internet Explorer Opera Safari WebView Android Chrome Android Firefox for Android Opera Android Safari on IOS Samsung Internet Deno Node.js\n`label`\n1\n12\n1\n4\n4\n1\n4.4\n18\n4\n10.1\n1\n1.0\n1.0\n0.10.0" ]

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[ "# [NOI2018] 归程\n\n## 输入输出样例\n\n### 输入样例 #1\n\n1\n4 3\n1 2 50 1\n2 3 100 2\n3 4 50 1\n5 0 2\n3 0\n2 1\n4 1\n3 1\n3 2\n\n### 输出样例 #1\n\n0\n50\n200\n50\n150\n\n### 输入样例 #2\n\n1\n5 5\n1 2 1 2\n2 3 1 2\n4 3 1 2\n5 3 1 2\n1 5 2 1\n4 1 3\n5 1\n5 2\n2 0\n4 0\n\n### 输出样例 #2\n\n0\n2\n3\n1\n\n## 说明\n\n### 更多样例 更多样例请在附加文件中下载。 #### 样例 3 见附加文件中的 return3.in 与 return3.ans。 该样例满足海拔为一种，且不强制在线。 #### 样例 4 见附加文件中的 return4.in 与 return4.ans。 该样例满足图形态为一条链，且强制在线。 #### 样例 5 见附加文件中的 return5.in 与 return5.ans。 该样例满足不强制在线。 ### 样例 1 解释 第一天没有降水，Yazid 可以坐车直接回到家中。 第二天、第三天、第四天的积水情况相同，均为连接 1，2 号节点的边、连接 3，4 号点的边有积水。 对于第二天，Yazid 从 2 号点出发坐车只能去往 3 号节点，对回家没有帮助。因此 Yazid 只能纯靠徒步回家。 对于第三天，从 4 号节点出发的唯一一条边是有积水的，车也就变得无用了。Yazid 只能纯靠徒步回家。 对于第四天，Yazid 可以坐车先到达 2 号节点，再步行回家。 第五天所有的边都积水了，因此 Yazid 只能纯靠徒步回家。 ### 样例 2 解释 本组数据强制在线。 第一天的答案是 $0$，因此第二天的 $v=\\left( 5+0-1\\right)\\bmod 5+1=5$，$p=\\left(2+0\\right)\\bmod\\left(3+1\\right)=2$。 第二天的答案是 $2$，因此第三天的 $v=\\left( 2+2-1\\right)\\bmod 5+1=4$，$p=\\left(0+2\\right)\\bmod\\left(3+1\\right)=2$。 第三天的答案是 $3$，因此第四天的 $v=\\left( 4+3-1\\right)\\bmod 5+1=2$，$p=\\left(0+3\\right)\\bmod\\left(3+1\\right)=3$。 ### 数据范围与约定 所有测试点均保证 $T\\leq 3$，所有测试点中的所有数据均满足如下限制： - $n\\leq 2\\times 10^5$，$m\\leq 4\\times 10^5$，$Q\\leq 4\\times 10^5$，$K\\in\\left\\{0,1\\right\\}$，$1\\leq S\\leq 10^9$。 - 对于所有边：$l\\leq 10^4$，$a\\leq 10^9$。 - 任意两点之间都直接或间接通过边相连。 **为了方便你快速理解，我们在表格中使用了一些简单易懂的表述。在此，我们对这些内容作形式化的说明：** - 图形态：对于表格中该项为 “一棵树” 或 “一条链” 的测试点，保证 $m = n-1$。除此之外，这两类测试点分别满足如下限制： - 一棵树：保证输入的图是一棵树，即保证边不会构成回路。 - 一条链：保证所有边满足 $u + 1 = v$。 - 海拔：对于表格中该项为 “一种” 的测试点，保证对于所有边有 $a = 1$。 - 强制在线：对于表格中该项为 “是” 的测试点，保证 $K = 1$；如果该项为 “否”，则有 $K = 0$。 - 对于所有测试点，如果上述对应项为 “不保证”，则对该项内容不作任何保证。 $n$|$m$|$Q=$|测试点|形态|海拔|强制在线 -|-|-|-|-|-|- $\\leq 1$|$\\leq 0$|$0$|1|不保证|一种|否 $\\leq 6$|$\\leq 10$|$10$|2|不保证|一种|否 $\\leq 50$|$\\leq 150$|$100$|3|不保证|一种|否 $\\leq 100$|$\\leq 300$|$200$|4|不保证|一种|否 $\\leq 1500$|$\\leq 4000$|$2000$|5|不保证|一种|否 $\\leq 200000$|$\\leq 400000$|$100000$|6|不保证|一种|否 $\\leq 1500$|$=n-1$|$2000$|7|一条链|不保证|否 $\\leq 1500$|$=n-1$|$2000$|8|一条链|不保证|否 $\\leq 1500$|$=n-1$|$2000$|9|一条链|不保证|否 $\\leq 200000$|$=n-1$|$100000$|10|一棵树|不保证|否 $\\leq 200000$|$=n-1$|$100000$|11|一棵树|不保证|是 $\\leq 200000$|$\\leq 400000$|$100000$|12|不保证|不保证|否 $\\leq 200000$|$\\leq 400000$|$100000$|13|不保证|不保证|否 $\\leq 200000$|$\\leq 400000$|$100000$|14|不保证|不保证|否 $\\leq 1500$|$\\leq 4000$|$2000$|15|不保证|不保证|是 $\\leq 1500$|$\\leq 4000$|$2000$|16|不保证|不保证|是 $\\leq 200000$|$\\leq 400000$|$100000$|17|不保证|不保证|是 $\\leq 200000$|$\\leq 400000$|$100000$|18|不保证|不保证|是 $\\leq 200000$|$\\leq 400000$|$400000$|19|不保证|不保证|是 $\\leq 200000$|$\\leq 400000$|$400000$|20|不保证|不保证|是" ]

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[ "# Open Collections\n\n## UBC Theses and Dissertations", null, "## UBC Theses and Dissertations\n\n### Speech genre and temporal conceptual metaphor use in the discourse of speakers with autism spectrum disorders… Sun, Peter 2007\n\nMedia\n831-ubc_2007-0601.pdf [ 6.29MB ]\nJSON: 831-1.0100980.json\nJSON-LD: 831-1.0100980-ld.json\nRDF/XML (Pretty): 831-1.0100980-rdf.xml\nRDF/JSON: 831-1.0100980-rdf.json\nTurtle: 831-1.0100980-turtle.txt\nN-Triples: 831-1.0100980-rdf-ntriples.txt\nOriginal Record: 831-1.0100980-source.json\nFull Text\n831-1.0100980-fulltext.txt\nCitation\n831-1.0100980.ris\n\n#### Full Text\n\n`SPEECH GENRE AND TEMPORAL CONCEPTUAL METAPHOR USE IN THE DISCOURSE OF SPEAKERS WITH AUTISM SPECTRUM DISORDERS (ASDs) by P E T E R S U N B.A. , The University of British Columbia, 2005 A THESIS S U B M I T T E D IN P A R T I A L F U L F I L L M E N T OF T H E R E Q U I R E M E N T S F O R T H E D E G R E E OF M A S T E R OF A R T S in T H E F A C U L T Y OF G R A D U A T E STUDIES (English) T H E U N I V E R S I T Y OF BRIT ISH C O L U M B I A August 2007 © Peter Sun, 2007 A B S T R A C T The thesis explores how temporal spoken text and metaphors of time are used in semi-structured conversational discourse by speakers with Aut ism Spectrum Disorders. The focus on time, its structure (through genre) and metaphoric representation, is a potentially revealing line of research for better understanding communication difficulties as well as patterns of conceptualization in A S D . Metaphors, in general, are difficult for people with A S D (Happe, 1993, 1995). Time in this context is an interesting concept to examine as it is largely expressed using metaphor (Lakoff & Johnson, 1999). The thesis also provides an opportunity to explore how temporal metaphors, as ingrained concepts of nature, are used by individuals who traditionally struggle with figurative metaphor. Individuals with A S D use temporal metaphors and the findings here may not only serve as a contribution to our knowledge about A S D , but also to the understanding of semantics and philosophy of time. i i TABLE OF CONTENTS Abstract i i Table of Contents i i i List of Tables '. v List of Figures vi i List of Abbreviations v i i i Acknowledgements ix Chapter 1 - Introduction 1 Chapter 2 - Background 7 1. Autism Spectrum Disorders 7 2. Data 7 3. Speech Genre 8 4 Cognitive Linguistics 9 Chapter 3 - Methods 12 A . Cohesion Analysis 12 B. Speech Genre Analysis 15 C. Conceptual Metaphor Analysis 18 Chapter 4 - Speech Genre Analysis 27 Part 1: Individual Text Analysis 28 Part 2: Analysis of A l l Transcripts 51 Conclusion 60 Chapter 5 - Metaphor Analysis 62 Part 1: Individual Text Analysis 64 i i i Part 2: Analysis of A l l Texts 95 Conclusion 109 Chapter 6 - Conclusion 112 Works Cited 120 Appendix 1 - Temporal Stretches Extracted from 9 Transcripts 122 Appendix 2 - Summary of Genre Types and Generic Stages Used in 9 Texts 144 Appendix 3 - Utterances that Use Units of Time Extracted from 9 Texts 146 Appendix 4 - Analysis of Metaphors, Metonymies, and Senses of Time of Utterances that Use Temporal Units 151 iv LIST OF TABLES Table 3.1 - External conjunctive relations of the temporal type 14 Table 3.2 - Storytelling genres and their structure 15 Table 3.3 - Description of generic stages used in storytelling genres 16 Table 3.4 - Generic structure of Procedure 17 Table 3.5 - Genre types and associated generic structure used for analysis 18 Table 4.1 - Genre types used for analysis 27 Table 4.2 - Summary of genre types and generic stages used by speakers in text 1 30 Table 4.3 - Summary of genre types and generic stages used by speakers in text 2 31 Table 4.4 - Generic structure of the Anecdote and the Specific Recount 32 Table 4.5 - Summary of genre types and generic stages used by speakers in text 3 38 Table 4.6 - Summary of genre types and generic stages used by speakers in text 4 39 Table 4.7 - Summary of genre types and generic stages used by speakers in text 5 41 Table 4.8 - Summary of genre types and generic stages used by speakers in text 6 42 Table 4.9 - Summary of genre types and generic stages used by speakers in text 7 46 Table 4.10 - Summary of genre types and generic stages used by speakers in text 8 50 Table 4.11 - Summary of genre types used by speakers in 9 texts 51 Table 4.12 - Summary of generic stages used by speakers in 9 texts 52 Table 5.1 - Lines 140, 141, 146, 147, 151, and 153 from text 1 64 Table 5.2 - Summary of metaphors and metonymies used in text 1 66 Table 5.3 - Lines 1, 3, 4, 6, and 8 from text 2 67 Table 5.4 - Summary of metaphors and metonymies used in text 2 71 Table 5.5- Lines 45, 70, and 115 from text 3 72 v Table 5.6 - Lines 21, 43, 127, 176, 177, 205, 206, 209, 210, and 222 from text 3 73 Table 5.7 - Lines 82 and 107 from text 3 74 Table 5.8 - Lines 94, 96, 105, and 191 from text 3 75 Table 5.9 - Summary of metaphors and metonymies used in text 3 81 Table 5 . 1 0 - L i n e s 134, 151, 153 from text 4 82 Table 5.11 - Lines 140 and 144 from text 4 . 82 Table 5.12 - Summary of metaphors and metonymies used in text 4 84 Table 5 . 1 3 - Lines 9 and 16 from text 5 85 Table 5.14 - Summary of metaphors and metonymies used in text 5 86 Table 5 . 1 6 - L i n e s 14, 25 ,26 ,71 ,72 , 74, 76, 113,114, 115, 120, and 150 from text 7 87 Table 5.17 - Lines 24, 27, and 53 from text 7 88 Table 5.18 - Summary of metaphors and metonymies used in text 7 90 Table 5.19 - Summary of metaphors and metonymies used in text 8 92 Table 5.20 - Summary of metaphors and metonymies used in text 9 94 Table 5.21 - Number of utterances that use basic metaphors of time in texts 1 through 9 96 Table 5.22 - Summary of metaphors and metonymies used in texts 1 through 9 97 Table 5.23 - Number of utterances that use metonymies of time in texts 1 through 9.... 98 Table 5.24 - Number of utterances that use other basic metaphors of time in texts 1 through 9 99 Table 5.25 - Number of utterances that use Evans' senses of time 99 vi LIST OF FIGURES Figure 5.1 - Researcher's frame of the Time Orientation and Moving Observer metaphors 67 Figure 5.2 - Chi ld 's frame of the Time Orientation and the Moving Observer Metaphors 67 Figure 5.3 - Construal of \"weekends\" from line 219 of text 2 69 Figure 5.4 - \"Twelve rides\" construal of time in line 192 of text 3 76 V l l L I S T O F A B B R E V I A T I O N S Transcr ip t ion Conventions R E S : = Researcher CHI : = Chi ld (Research Participant) A C K N O W L E D G E M E N T S Time is a valuable commodity. Thank you Dr. Jessica de Vi l l iers and Dr. Barbara gier for being generous with your time while supervising this thesis. C H A P T E R 1 - INTRODUCTION This thesis investigates how temporal spoken text and metaphors of time are produced and understood in the discourse of people with Aut ism Spectrum Disorders (ASDs). The approach taken integrates speech genre analysis (Martin & Plum, 1997; Halliday & Hasan, 1989; Eggins & Slade, 1997; Plum, 2004) from the Systemic Functional Tradition in Linguistics with an analysis of conceptual approaches to metaphor from Cognitive Linguistics (Lee 2001; Lakoff & Johnson 1999; Evans 2003) in a complementary way. Speech genre analysis proposes that speakers organize and structure texts in predictable stages to achieve an overall purpose. Conceptual metaphor is the process of understanding one conceptual domain in terms of another domain. In the thesis, I identify and examine stretches of temporally sequenced events in a corpus of 9 semi-structured conversational texts produced by children and adolescents diagnosed with A S D . O f particular interest in the examination is the marked use of speech genre types in the discourse of A S D . Contextually appropriate discourse is particularly difficult for speakers with A S D . While the discourse of individuals with A S D typically does not conform to expected patterns, problems with the contextual use and variation of language are equally detrimental to social success but are also not well-understood in A S D . There are numerous theories of A S D ; however, few deal with the difficulties in communication. deVill iers and Szatmari (2004) found that individuals with A S D have difficulties with chronological organization in the quantity of information and linearity of serial events. Speech genre analysis is an appropriate approach to examining the discourse of speakers with A S D as it looks at the structure of text in a functional way. Specifically, speech 1 genre theory is sensitive to a configuration of contextual factors, including the nature of the text, the participants and their relationship to one another, and the role language plays in the experience. In this thesis, speech genre types and their associated generic structure wi l l provide a means to better understand how individuals with A S D structure time and temporal concepts in conversational texts. In this thesis, the term genre refers to speech genre. Also of interest is the conceptualization of time among individuals with A S D , as temporal concepts are largely communicated using conceptual metaphor (Lakoff & Johnson, 1999). Conceptual metaphor is used when one conceptual domain is understood in terms of another conceptual domain. Meaning is achieved in conceptual metaphor by recognizing a set of systematic correspondences between the two domains. Time is unconsciously conceptualized and so deeply entrenched in our cognitive system that typical speakers often do not realize that they are using conceptual metaphor. Classic examples include spending, saving, or measuring time. The T I M E IS M O N E Y conceptual metaphor, among other conceptual metaphors of time, are relatively stable, are not arbitrary, and are widespread throughout cultures (Lakoff & Johnson, 1999, 134). In this thesis, unless otherwise noted, the term metaphor refers to conceptual metaphor. It is widely reported that even the most verbally able individuals with A S D fail to understand nonliteral speech (Happe, 1993, 1995). The problems that people with A S D experience in such metaphoric language use can significantly encumber communication and social success. Yet, despite a growing literature on problems with metaphor and other figurative language use in A S D , these difficulties are still not well-understood. The analysis of conceptual metaphors, metonymies, and senses of time is appropriate to 2 approach temporal texts produced by speakers with A S D as these individuals must have an understanding of conceptual metaphors of time and temporal concepts. This thesis aims to investigate the degree to which individual use and comprehend conceptual metaphors of time. The findings may not only serve as a contribution to the knowledge of the disorder, but may also comment on the degree of the conventionality of temporal metaphor and contribute to the philosophy of time. The thesis analyzes nine spoken texts produced by children and adolescents diagnosed with A S D . It applies two approaches: Systemic Functional Discourse Analysis and Cognitive Linguistics. First, a speech genre analysis of stretches of spoken discourse that involve temporally sequenced events is undertaken to examine the extent to which the research participants under investigation predictably engage in discourse based on contextual factors. Fol lowing this speech genre analysis, stretches of temporally sequenced events are examined for the use of temporal conceptual metaphors. The result is a well-articulated evaluation of the use of speech genre and conceptual metaphor types in 9 conversational texts of speakers with A S D . The speech genre analysis follows a framework developed by Eggins and Slade ' (1997) who found ten types of speech genres used in casual conversations. O f these ten types, they label four types as storytelling genres. The four types of storytelling genres are Narrative, Anecdotes, Recounts, and Exemplum. Bui lding on Eggins and Slade's work, further subcategorization of the Recount speech genre into Specific Recounts and General Recounts was made. (Specific Recounts involve a Record of Events that is unique and non-repetitive while General Recounts involve a Record of Events that is habitual or that occurs on a recurring basis.) A n emergent speech genre type that uses 3 external temporal conjunctions but does not adhere to any of the speech genres described by Eggins and Slade (1997) is Procedure (Plum, 2004). Fol lowing Plum, Procedure is incorporated in the classification. To approach temporal metaphors produced by speakers with A S D , Lakoff and Johnson's (1999) conceptual metaphors of time and Evans' (2003) senses of time are applied. In the analysis of temporal conceptual metaphors, Lakoff and Johnson (1999) were followed and the approach to the analysis of the lexical item, \"t ime\" followed Evans (2003). Lakoff and Johnson's conceptual metaphors of time have been described in Philosophy in the Flesh (1999). Although time as a concept is chiefly metaphorical, it is realized in many ways by different conceptual metaphors. The selected conceptual metaphors of time used for analysis of metaphor are: 1) Time Orientation metaphor; 2) Moving Time metaphor; 3) Moving Observer metaphor; 4) Event-for-Time metonymy; 5) Distance-Time metonymy; 6) Time as a Resource metaphor; and 7) Time as Money metaphor. In all stretches of temporally sequential texts from the 9 transcripts, all instances were noted where the word \"t ime\" or units of time (i.e. \"tomorrow\", \"two minutes\", \"June the fifth\", \"1985\", etc.) were uttered. These uses of time were categorized according to Lakoff and Johnson's conceptual metaphors using Lakoff & Johnson's (1999) criteria. Where the lexical item \"t ime\" explicitly appeared, these instances were categorized according to Evans' (2003) distinct senses of time: 1. Duration Sense 2. Moment Sense 3. Instance Sense 4. Event Sense 5. Matrix Sense 6. Agentive Sense 4 7. Measurement-system Sense The analysis of speech genre in the nine texts found that the individuals with A S D used certain speech genre types and not others. They also used some speech genre types more frequently than others. The Recount genre (Specific and General) was used the most. The thesis examines closely the speech genre types used and considers the frequency of particular stages in the genres. Some children appeared to have stronger conversational ability marked by longer generic stages and longer turns. These speakers did use a greater variety of generic stages but appeared to favour a particular generic stage. The analysis of conceptual metaphor, metonymies, and senses of time showed that individuals with A S D appeared to use some metaphors more often than others. A s with genres, some individuals appeared to be more able than others to use temporal metaphors, metonymies and senses of time marked by a more varied and a greater quantity of these expressions. The speakers that appeared to be more competent with temporal expressions showed unconventional uses related to the specificity of conceptual metaphors, metonymies, and senses of time. Fol lowing the speech genre and metaphor analyses, the results of the two investigations are compared and discussed. Both approaches to analyzing the same 9 texts showed a relationship where the research participants with the most conversational engagement also used the greatest variety of speech genre types and the most temporal conceptual metaphors. The findings in this thesis are important because speech genre analysis and the analysis of conceptual metaphors are appropriate to examine the pragmatic language use of time and temporal concepts that occur in the conversations of 5 ind iv iduals w i t h A S D . Individuals w i t h A S D struggle w i t h pragmatic and figurative language use and few studies examine the conversational diff icul t ies o f ind iv iduals w i t h A S D . Th i s thesis w i l l serve as a contr ibut ion to the knowledge o f the conversational diff icul t ies o f the disorder and comment on the degree o f the convent ional i ty o f temporal conceptual metaphors that occur i n spoken discourse. Chapter 1 has introduced the thesis that w i l l examine the use o f speech genres and temporal conceptual metaphors, metonymies , and senses o f t ime i n spoken texts produced by speakers w i t h A S D . Chapter 2, Background, describes A S D , speech genre analysis, and Cogn i t i ve L ingu i s t i c s . The second chapter also identifies the source o f the data used i n this thesis. Chapter 3, Methods, outlines cohes ion analysis, speech genre analysis, and the analysis o f temporal conceptual metaphors, metonymies , and senses o f t ime used to approach the data. Chapter 4, Speech Genre Analysis, describes the use o f speech genre that occur i n each o f the 9 texts. Patterns that occur i n the use o f speech genres are also identif ied across the 9 texts. Chapter 5, Metaphor Analysis, describes the use o f temporal conceptual metaphors, metonymies , and senses o f t ime that occur i n the same 9 texts. Patterns o f temporal metaphors, metonymies , and senses o f t ime that occur across the 9 texts are then identif ied and discussed. Chapter 5, Conclusion, discusses observations and f indings that emerged f rom both the analysis o f speech genre and the analysis o f temporal conceptual metaphors, metonymies , and senses o f t ime. 6 C H A P T E R 2 - B A C K G R O U N D 1. Aut ism Spectrum Disorders A u t i s m Spectrum Disorder ( A S D ) is an umbre l la term for a spectrum o f neurocogni t ive disorders, inc lud ing autism, high-funct ioning aut ism ( H F A ) , Asperge r ' s Syndrome ( A S ) and Pervasive Deve lopmenta l Disorder N o t Otherwise Speci f ied ( P D D -N O S ) that is characterized by severe impairments i n communica t ion and social rec iproci ty . It is w e l l k n o w n that communica t ion i n A S D is associated w i t h diff icult ies i n pragmatics, i nc lud ing the product ion and interpretation o f f igurative metaphor (Happe, 1995). Since t ime and temporal concepts are v i r tual ly imposs ible to conceptual ize wi thout conceptual metaphor ( L a k o f f & Johnson, 1999, 139), an invest igat ion o f the use o f temporal concepts by speakers w i t h A S D may help to better understand the nature o f the disorder. People w i t h A S D are also k n o w n to have di f f icul ty us ing discourse i n contextually appropriate ways , i nc lud ing problems w i t h relevance (Happe, 1993), organizat ion (de V i l l i e r s & Szatmari , 2004) and cohesion (Fine , 1994). A n analysis o f temporal ly sequenced stretches o f discourse us ing a contextually sensitive descript ive approach such as functional speech genre analysis may also in fo rm our understanding o f the diff icul t ies w i t h language associated w i t h A S D . 2. Data Data for this thesis came f rom a fo l low-up study o f ch i ld ren diagnosed w i t h A S D at C h e d o k e - M c M a s t e r Hosp i t a l i n H a m i l t o n , Ontario. A u d i o recordings o f semi-structured conversations between a research technician and 9 research participants diagnosed w i t h A S D were col lected and transcribed pr ior to the analysis undertaken i n the present research. F o r the present examinat ion, temporal stretches o f discourse were isolated f rom 7 the 9 transcripts of semi-structured conversational texts. For a full description of the data collection procedures see de Vi l l iers et al. 2007. 3. Speech Genre In the early 20 t h century, genre study was limited to comparing the differences among texts in the disciplines of rhetoric and literature. In 1953, Bakhtin claimed that texts (both spoken and written) were heterogeneous in nature and suggested that a broader perspective was necessary in communication to account for similarities among texts. For example, Bakhtin noted the fact that utterances involve a finalization of a turn that can be sensed by participants in a conversation. Bakhtin explained that each text can be extremely varied, for example, daily dialogue as opposed to business documents; but, their forms are in fact combinations of speech genres: Each separate utterance is individual, of course, but each sphere in which language is used develops its own relative stable types of these utterances. This we may call speech genres (1986, 60). In the Systemic Functional Tradition, Halliday provides a similar account on the nature of texts using the concept of register. \" . . . every text is in some sense like other texts; and for any given text there wi l l be some that it resembles more closely\" (Halliday & Hasan, 1989, 42). Register analysis claims that participants can predict textual features based on 3 situational variables: the nature of the text, the participants and their relationship to one another, and the role language plays in the experience. These variables are used to account for contextual aspects of text. Speech genre theory extends upon these variables and attempts to understand why the text was produced; speech genre analysis considers the cultural purposes of text both in content and form. It claims that texts are composed 8 of stages that are culturally imperative. The generic structure of a text is composed of a sequence of intermediate stages that serve a final goal. For example, Hasan (Halliday & Hasan, 1989) compared conversational transactions at a fruit stand. Hasan notes that these \"service encounter\" texts were generic to the extent that they are made up of obligatory stages in the following order— Sales Request (SR), Sales Compliance (SC), Sale (S), Purchase (P), and Purchase Closure (PC): SR= Can I have ten oranges and a kilo of bananas please? SC= Yes, anything else? No thanks. S= That'll be dollar forty. P= Two dollars. PC= Sixty, eighty, two dollars. Thank you. Example 2.1 - Service Encounter (Halliday & Hasan, 1989, 59) According to Hasan, a successful transaction at a fruit stand of the Service Encounter genre type requires a SR, SC, S, P, and PC in serial order. The use of speech genre analysis is a revealing way to describe discourse in this thesis as it provides a means to examine how individuals with A S D structure time and temporal concepts in spoken text. Speech genre analysis enables a discussion about pragmatic language use; this thesis is therefore able to examine the degree to which text produced by speakers with ASD conforms to culturally accepted norms in content and form. 4. Cognitive Linguistics The thesis uses Lakoff and Johnson (1999) and Evans' (2003) cognitive approaches to time. Both stem from the discipline of Cognitive Linguistics. Cognitive Linguistics assumes that linguistic expression is associated with specific construal (Lee, 2001, 2). It assumes that thought operates at a level below cognitive awareness (Lakoff & 9 Johnson, 1980, 10) and that although there may be formal rules that govern language use, these rules are rooted in cognition. Lakof f and Johnson's approach to time uses the concept of conceptual metaphor. Conceptual metaphor allows us to think about a phenomenon in different ways (Lee, 2001, 6). For example, \"I spend too much time at the mal l , \" uses the conceptual metaphor T I M E IS M O N E Y , while, \"time passes quickly at the mal l , \" uses the conceptual metaphor T I M E IS M O T I O N 1 . Both linguistic expressions of conceptual metaphor present different construal of the same phenomenon using a source and target domain. In the case of T I M E IS M O N E Y , \"money\" is the source and \"t ime\" is the target. The source domain is relatively more experientially concrete while the target is more abstract; therefore, our conception of money (more concrete) helps us structure our experience of time (more abstract). Conceptual metaphors differ from figurative metaphors in their conventionality. Linguistic expressions of conceptual metaphors of time are so deeply entrenched in thought and conventionalized to the point that they may no longer be recognized as metaphorical expressions. Although figurative metaphors may be based on conceptual metaphors, they require a higher level of cognitive function. Figurative metaphors are more literary and often less clear but richer in meaning (Kovecses, 2002, 45). The distinction between conceptual metaphors and figurative metaphors is relevant to this thesis as individuals with A S D are known to struggle with non-literal language. These difficulties are most obvious with expressions using figurative metaphors. Since individuals with A S D do conceptualize time and temporal events, they must use conceptual metaphors of time. This thesis wi l l examine the degree to which individuals 1 Or PASSAGE OF TIME IS MOTION 10 with A S D use and comprehend temporal metaphors as temporal discourse cannot exist without conceptual metaphor. Evans' senses of time approaches time and temporal concepts from a framework that classifies a range of distinct lexical concepts for time. This approach considers instances where lexical concepts are paired with the lexeme \"t ime\". Although Evans' approach differs from the analysis of time using conceptual metaphors, the investigation of the ways in which individuals with A S D use the lexeme \"t ime\" is an interesting area of investigation. Analyzing spoken discourse using both cognitive approaches to time wi l l reveal the ways in which these speakers use and comprehend temporal concepts. 11 C H A P T E R 3 - M E T H O D S This thesis examines semi-structured conversational texts of nine speakers diagnosed with ASD using a combined approach of cohesion and genre analysis from Systemic Functional Linguistics (Halliday, 2004) and analysis of metaphor from Cognitive Linguistics (Lakoff & Johnson, 1999). The focus of analysis is on temporal discourse, both in generic structures and linguistic expressions of metaphor. This combined approach is appropriate for several reasons. First, temporal patterns in the discourse of speakers with ASD are underdescribed in both cognitive linguistic and functional linguistic literature. Both approaches use microanalytic techniques to uncover patterns that would otherwise go unobserved. Both approaches are descriptive of discourses situated in context. Moreover, by using functional cohesion analysis, I was able to extract, in a systematic way, stretches of temporally sequenced events from the nine original transcripts that served as the data for the cognitive linguistic analysis of metaphor. The results of the genre and metaphor analyses of different speakers' texts were also compared. Thus, the two theoretical descriptive approaches were applied in complementary ways. A . Cohesion Analysis Before beginning the analyses of temporal genres and metaphors in the nine texts, temporal stretches in the semi-structured conversations had to be identified. To isolate these, stretches of text were first identified with temporally sequenced events using Halliday and Hasan's (1976) cohesive subtypes. Cohesion can be defined as meaning relations within text (Halliday & Hasan, 1976, 29) and instances of cohesion are realized 12 where the interpretation of one element in discourse is dependent upon another (Halliday & Hasan, 1976, 4). Cohesion concerns the way in which the meaning of the elements is interpreted. \"Where the interpretation of any item in the discourse requires making reference to some other item in the discourse, there is cohesion\" (Halliday & Hasan, 1976, 11). For example: la . John is nice. lb . He likes to help people. Example 3.1 - Cohesive tie In example 3.1, both John and he refer to the same entity; or, restated, the proper name John and the personal pronoun he, form a cohesive tie. In Halliday and Hasan's framework (1976), cohesion in text can be realized by reference (as in example 3.1), substitution, ellipsis, conjunction, or lexical collocation. Relevant to temporally sequenced events are temporal conjunctions, a subtype of Halliday and Hasan's conjunction group. Halliday and Hasan (1976) differentiate instances of internal temporal relations from external temporal relations (263). The former express successivity in the communication process while the latter concern successivity in the events talked about (Halliday & Hassan, 1976, 263): Internal temporal relation: First, I w i l l talk about the car's engine, and then I wi l l discuss its colour. After, I conclude with comments about its manufacturer. External temporal relation: First, John opened the car door then sat in the seat. Next, he started the engine then backed out of the garage. Example 3.2 - Internal and external temporal relations. In this thesis, external conjunctions of the temporal type were used to locate and extract stretches of temporal discourse (see Appendix 1) because they are a feature that identifies the temporal sequencing of events as \"There is always some feature of which 13 we can say, 'This is typically associated with this or that use of language'\" (Halliday & Hasan, 1989, 40). External temporal conjunctions are typically associated, but not exclusive, to texts that communicate events as they happen(ed). Below, Table 3.1 provides a summary of external conjunctive relations of the temporal type: Table 3.1 is a list of external conjunctive relations of the temporal type. Table 3.1 has been removed due to copyright restrictions. Please see Hall iday & Hasan, 1976, 266. Table 3.1 - External conjunctive relations of the temporal type (Halliday & Hasan, 1976, 266) Table 3.1 identifies the external conjunctive relations of the temporal time that were used to extract stretches of temporal text from 9 conversational transcripts. The use of these conjunctions identifies stretches of texts that describe sequential events. These texts form the data that was used for the analysis of speech genres and conceptual metaphors. 14 B. Speech Genre Analysis In Chapter 4, a full speech genre analysis was conducted of the stretches of text extracted from the 9 transcripts and they were grouped according to a framework developed by Eggins and Slade (1997). Eggins and Slade (1997) found ten types of genre used in casual conversations. Of these 10 types, they label 4 types as storytelling genre. They are Narrative, Anecdotes, Recounts, and Exemplum. Table 3.2 summarizes their generic structure. The present paper classifies the stretches of text from the 9 transcripts of speakers with ASD according to Eggins and Slade's 4 storytelling genre types. Genre (Types) Generic structure Narrative (Abstract) A (Orientation) A Complication A Evaluation A Resolution A (Coda) Anecdote (Abstract) A (Orientation) A Remarkable Event A Reaction A (Coda) Exemplum (Abstract) A (Orientation) A Incident A Interpretation A (Coda) Recount (Abstract) A (Orientation) A Record of Events A (Coda) Table 3.2 - Storytelling genres and their structure (Eggins & Slade, 1997) Martin notes these storytelling genres are based on narrative clauses that share basic generic stages at their beginnings and ends (1992, 564). Table 3.3 provides a description of the stages in generic structure. 15 Generic Stage Description Abstract Establishes the point of the text and signals that a story is about to be told Orientation Orients listeners to what is to follow in terms of people, actions, time and place Complication Temporally orders actions leading to a crisis Remarkable Event Temporally orders actions outlining a remarkable event which the narrator wants to share her reaction to Reaction The evaluation of the events establishes the significance of the story Incident Outlines temporally sequenced events in order to elucidate interpretative comments or moral judgement Interpretation A moral interpretation or judgement of incident is relayed Record of Events Provides a sequence of events with ongoing appraisal Resolution Actions resolve crisis Evaluation Evaluates or presents appraisal Coda Makes point about text as a whole Returns text to present Table 3.3 - Description of generic stages used in storytelling genres In Table 3.3, the shared stages in generic structure are Abstract, Orientation and Coda and are identified as optional with parentheses; the remaining generic stages are obligatory to the genre type. For example, a Narrative must contain a Complication stage, followed by an Evaluation and a Resolution stage; it may or may not contain an Abstract, Orientation, or Coda. The Abstract stage provides thematic information for the stretch of discourse and the Orientation stage describes the setting while the Coda may complete the stretch of discourse by glossing the entire sequence. Descriptions of the storytelling genre types follow: Narratives have a Complication, Evaluation, and Resolution because these texts build in tension and excitement to reach a crisis that is then resolved. Anecdotes are texts that are similar to narrative as they also focus on crisis. Anecdotes contain a Remarkable Event; a sequence of events builds towards a crisis but unlike Narratives, the crisis is not resolved. 16 Alternatively, the string of events is told to highlight event(s) that are bizarre or unusual for the purpose of sharing a reaction to these events. Exemplums are texts that have a prescriptive nature and suggest how the world should or should not be. Exemplums contain an Incident, a set of temporally sequenced events that are told in order to focus on the significance of these events (rather than their problematic nature). The Interpretation stage of the Exemplum text relates the story to the larger context of culture and a moral point of some kinds is then made. Finally, Recounts are about a temporal sequence experienced by the narrator. In these texts, the Record of Events tells how one event leads to another; the goal is simply to relay succession of events. I further subcategorize the Recount genre into Specific Recounts and General Recounts. I define Specific Recounts to involve a Record of Events that is unique and non-repetitive while General Recounts involve a Record of Events that is habitual or that occurs on a recurring basis. A n additional genre type emerged that used external temporal conjunctions but did not adhere to any of the genres described by Eggins and Slade (1997). I found this genre type to be Procedure (Plum,2004). Procedures are texts that involve temporally sequenced events or steps in a How To stage that outlines the means to achieve a desired state or result. A recipe is a classical example of a Procedure text. Table 3.4 details the generic structure of Procedure. Genre Generic structure Procedure (Abstract) A Orientation A How To A (Coda) Table 3.4 - Generic structure of Procedure (Plum, 2004) The 6 genre types, including my 2 subtypes of Recount genre, are used for the analysis of temporally sequenced text from conversations of speakers with A S D . Table 17 3.5 generalizes the generic structure of these genre types used to code spoken discourse in this project. Genre Generic structure Narrative (Abstract) A (Orientation) A Complication A Evaluation A Resolution A (Coda) Anecdote (Abstract) A (Orientation) A Remarkable E v e n t A Reaction A (Coda) Exemplum (Abstract) A (Orientation) A Incident A Interpretation A (Coda) Specific Recount (Abstract) A (Orientation) A Specific Record of Events A (Coda) General Recount (Abstract) A (Orientation) A General Record of Events A (Coda) Procedure (Abstract) A Orientation A How To A (Coda) Table 3.5 - Genre types and associated generic structure used for analysis As in Eggins and Slade (1997), the stages in the six genre types include optional stages in parentheses and required stages without parentheses. Evaluation is an additional optional stage that may occur throughout the generic structure in the 6 genre types at any point. The purpose of Evaluation is to sustain the story and establish its contextual significance (Eggins & Slade, 1997, 238). Analyzing conversational texts produced by speakers with A S D using speech genre and their associated generic stages wi l l provide insight into how these individual produce temporal discourse. C . Conceptual Metaphor Analysis In Chapter 5, an analysis of metaphors in the Cognitive Linguistics tradition was performed on each of the temporal stretches identified with the cohesion analysis. Both Lakoff and Johnson's conceptual metaphors of time and Evans' senses of time were used, with Evans' approach being applied where further semantic specification for the lexeme \"t ime\" helped to show unconventional uses of the lexeme \"time\". 18 Lakoff and Johnson's Conceptual Metaphors of Time (1999) This approach to the analysis of metaphors of time in discourse is developed from Lakof f and Johnson's Philosophy in the Flesh (1999). Lakoff and Johnson (1999) describe a variety of conceptual metaphors of time and metonymies for time. Metonymy, similar to metaphor, is defined as using one entity to refer to another (Lakoff & Johnson, 1980, 35.) For example, \"the rock concert was long,\" uses the Event-for-Time metonymy where the event, \"the rock concert,\" stands metonymically for a specific period of time. This approach to analyzing temporal discourse classifies metaphors and metonymies according to the following categories described by Lakof f & Johnson (1999): A . Time Orientation metaphor B. Moving Time metaphor C. Moving Observer metaphor D. Event-for-Time metonymy E. Distance-for-Time metonymy F. Time as a Resource metaphor G. Time as Money metaphor The following outlines and provides examples of these metaphors and metonymies: Time Orientation Metaphor The Time Orientation metaphor is the most basic metaphor. Its construal situates an observer who is at the present who faces the future. Lakof f and Johnson provide the following examples: That's all behind us now. We're looking ahead to the future. He has a great future in front of him. (1999, 140) 19 Moving Time Metaphor In this metaphor, Lakoff and Johnson (1999) describe time using a lone, stationary observer who faces a fixed direction. A n indefinite sequence of objects (times) move past the observer from front to back and these objects are also conceptualized as having fronts that face their direction of motion (Lakoff & Johnson, 1999, 141). The motion of the objects (time) moving past the observer represents the \"passage\" of time. The Moving Time metaphor can be combined with the Time Orientation metaphor to create the following composite mapping: The location of the observer -> The present The space in front of the observer -> The future The space behind the observer -> The past Objects -> Times The motion of objects past the observer -> The \"passage\" of time (Lakoff & Johnson, 1999, 142) Here are some examples of the Moving Time metaphor: The deadline is approaching. The time for action has arrived. The summer just zoomed by. The time for end-of-summer sales has passed In the weeks following next Tuesday, there wi l l be very little to do. (Lakoff & Johnson, 1999, 143) Moving Observer Metaphor The Moving Observer metaphor differs from the Moving Time metaphor because the observer is no longer fixed in one location. Alternatively, the locations (times) are fixed on a path upon which the observer moves. The motion of the observer represents the \"passage\" of time while the distance moved by the observer represents the amount of time \"passed.\" 20 When we combine the Moving Observer metaphor with the Time Orientation metaphor, we have the following composite mapping: The location of the observer -> The present The space in front of the observer -> The future The space behind the observer -> The past Locations on the observer's path of motion -> Times The motion of the observer -> The \"passage\" of time The distance moved by the observer -> The amount of time \"passed\" (Lakoff & Johnson, 1999, 146) Lakof f & Johnson provide the following examples: There's going to be trouble down the road. His visit to Russia extended over many years. She arrived on time. We're coming up on Christmas. We passed the deadline. (1999, 146) Event-for-Time Metonymy The Event-for-Time metonymy can co-occur with other metaphors of time. Through metaphor, temporal moments are represented as locations, substances, or motion and events are bounded and realized with respect to these temporal moments. For example, in the sentence, \"The rock concert is approaching,\" the event (rock concert) metonymically stands for the time (duration) of the concert. Distance-for-Time Metonymy Distance can also stand metonymically for time as in \"I slept for fifty miles while she drove\" (Lakoff & Johnson, 1999, 152). In this example, the time it took to drive \"fifty miles\" is the amount of time the speaker slept. Time as a Resource Metaphor The Time as a Resource metaphor is a characteristic way of conceptualizing time in Western culture (Lakoff & Johnson, 1999, 161). This metaphor maps information from 21 a resource domain (source) onto the time domain (target). Lakoff & Johnson describe this mapping: The resource The user of the resource The purpose that requires the resource The value of the resource The value of the purpose -> Time -> The agent (the user of time) -> The purpose that requires time -> The value of the time The value of the purpose (1999, 162-3) Linguistic expressions of this metaphor include: You've used up all of your time. The job took up three hours. He used his time efficiently. Time as Money Money is a type of resource; therefore, the Time as Money metaphor is similar to the Time as a Resource metaphor. They are constituents of the same system. In the Time as Money metaphor, words like \"budget,\" \"spend,\" \"invest,\" \"profit,\" and \"loss\" prompt for this metaphor. This mapping is as follows: Money The user of the money The purpose that requires the money The value of the money The value of the purpose -> Time -> The user of time (the agent) -> The purpose that requires time -> The value of the time -> The value of the purpose (Lakoff & Johnson, 1999, 163-4) Lakoff and Johnson provide the following linguistic expressions of this metaphor: I have to budget my time. I spent too much time on that. I've invested a lot of time on this project. (1999, 164) The above 7 types of conceptual metaphors form the framework for the analysis of temporal metaphors in this thesis. 22 Evans ' Senses of T ime (2003) This cognitive approach to the analysis of time is developed by Evans (2003). Evans presents a framework that determines a range of distinct lexical concepts for time. This approach considers instances where lexical concepts are paired with the lexeme \"time\". It suggests that there are 8 distinct senses of time: 1. The Duration Sense 2. The Moment Sense 3. The Instance Sense 4. The Event Sense 5. The Matrix Sense 6. The Agentive Sense 7. The Measurement-system Sense 8. The Commodity Sense I outline and provide examples for each of Evans' senses of time that wi l l analyze the use of the lexeme \"t ime\" in the conversational texts produced by speakers with A S D . The Durat ion Sense The Duration Sense is a lexical concept that constitutes an interval bounded by two \"boundary\" events. There is an onset and an offset that creates a temporal interval. The duration sense can be elaborated in terms of physical length, quality of experience, and temporal compression and protracted duration. Evans (2003) provides the following examples: The relationship lasted a long/short time. During their ill-fated marriage they fought a lot/some/much of the time. He returned to Germany for good in 1857, moving for a time to Berl in. Time flies (by) when you're having fun. Time crawls (by) when you're bored. (110-115) 23 Moment Sense The Moment Sense prompts for a conceptualisation of a discrete of punctual point or moment without reference to its duration. Evans provides the following examples: The time for a decision has arrived / come. Doctors had warned that Daniel, five, of Sinfin, Derby, could die at any time. What size was she the time of change? What time is it? (2003,123) Instance Sense The Instance Sense prompts for a reading in which an instance of a particular event, activity, process or state is being referenced. Evans provides the following examples: Devine improved for the fourth time this winter when he reached 64.40 meters at a meeting in Melbourne. This time, it was a bit more serious because I got a registered letter. He did it 50 times in a row. Once it was clear that the room could not be held, he would order its evacuation, men leaving two at a time by the far window. (2003, 131) Event Sense The Event Sense prompts for a conceptualisation in which a specific event is referenced. A n event constitutes an occurrence of some type, characterized by certain features or characteristics which mark the occurrence as is by being temporally discreet. This sense references an experiential point in an event-sequence (event embedded in ongoing experience/event-sequences). The following linguistic expressions are examples of the Event Sense of time: The young woman's time [=labour] approached. The man had every caution given him not a minute before to be careful with the gun, but his time was coming as his poor shipmates say and with that they console themselves. 24 The barman called time. (Evans, 2003,135) Matrix Sense In the Matrix Sense, \" t ime\" prompts an entity that is unbounded. Evans provides the following examples: Time flows/runs/goes on forever. Time has no end. Those mountains have stood for all time. Nothing can outlast time. (Evans, 2003,142) Agentive Sense The Agentive Sense prompts time as an entity that serves to bring about change. Here are some examples: Time, the subtle thief of youth. Time has aged me. Time has left its scars Time has yellowed the pages Time transformed her Time reveals all (Evans, 2003, 159) Measurement-System Sense In the Measurement-System Sense, physical symbols can be used to represent or measure time. The temporal measurements arise due to the correlation between periodic behaviour in the external world and our experience of duration. Evans provides the following examples: Clock—serves to divide day into hours, minutes, seconds... In quick time (dance), 108 paces, or 270 feet, are taken in a minute; and in slow time, seventy-five paces, or 187 feet... To play music out of time. In the 1850s Railway time was introduced as standard. We get paid double time on public holidays (2003, 169) 25 Commodi ty Sense The Commodity Sense prompts time as an entity that is valuable. It therefore can be for exchanged, traded, acquired, and possessed: Remember that time is money. Time has become a scarce commodity. Everyone wants more of it. They are selling time-shares on the Costa Blanca. The psychiatrist charges a lot for her time. A few techniques to create more time in your day. (Evans, 2003, 177) Fol lowing Evans, the above 8 senses of time form the framework to analyze the use of the lexeme \"t ime\" in conversational texts produced by speakers with A S D . This chapter has described the methods used in this thesis. Cohesion analysis was used to isolate stretches of temporal text from 9 conversational transcripts produced by speakers with A S D . Speech genre analysis and conceptual metaphor analysis was used to describe observations in each of the 9 texts and the patters across all 9 texts. 26 CHAPTER 4 - SPEECH GENRE ANALYSIS In this chapter I examine generic structure in temporal discourse in semi-structured conversations involving speakers with Autism Spectrum Disorders (ASDs). My approach uses genre analysis (Eggins & Slade, 1997; Plum, 2004; Martin, 1997) from functional discourse analysis to examine a corpus of nine texts. The chapter examines how speakers with ASD produce, structure, and comprehend temporally sequenced events in spoken text. Genre analysis from the discipline of discourse analysis is used to classify stretches of spoken text that involve sequences of events. This chapter proceeds in two parts. Part 1 describes each speaker's use of genre in the 9 individual texts. Part 2 extends upon this description by analyzing patterns of genre use across the 9 texts. To identify speakers' use of genre, stretches of spoken text that involve sequences of events were isolated from 9 transcripts of transcribed conversations based on the speakers' use of Halliday & Hasan's (1976) external conjunctive relations of the temporal type. These stretches of text were classified into 6 genre types that follow the work of Eggins and Slade (1997) and Plum (2004). Table 1.1 identifies the 6 genre types used for analysis in this chapter and their generic structure. Appendix 1 presents the full set of extracted data coded for genre types and generic stages. Genre Generic structure Narrative (Abstract) A (Orientation) A Complication A Evaluation A Resolution A (Coda) Anecdote (Abstract) A (Orientation) A Remarkable Event A Reaction A (Coda) Exemplum (Abstract) A (Orientation) A Incident A Interpretation A (Coda) Specific Recount (Abstract) A (Orientation) A Specific Record of Events A (Coda) General Recount (Abstract) A (Orientation) A General Record of Events A (Coda) Procedure (Abstract) A Orientation A How To A (Coda) Table 4 . 1 - Genre types used for analysis 27 PART 1: Individual Text Analysis This section begins the analysis of genre by describing the individual speaker's use of genre in the 9 texts. In these texts, both the researcher (RES) and the child (CHI) contribute to the generic structure of the text because turn-taking is a characteristic of conversational discourse. Textl Text 1 contains 2 stretches of temporally sequenced text. The first was a General Recount and the second was a Specific Recount. The General Recount text proceeds through an Orientation stage followed by a Record of Events: ORIENTATION RES: what do you do when you go home from s c h o o l W i l l ? [RECORD OF EVENTS CHI: ah -: I p l a y games. Example 4.1 - General Recount 1, transcript 1, lines 79-80 As can be seen from Example 1.1, the General Recount identified is not produced entirely by the research participant. Rather, the researcher utters the Orientation as a question and the child utters the Record of Events in response to the researcher's question. The participant that contributes each generic stage is identified because conversational texts, especially clinical texts, involve a semi-structured turn-taking pattern where both participants share the construction of the genre. The child's Record of Events involves a single event and the researcher solicits it. The Specific Recount proceeds through the generic stages of OrientationA Record of EventsA OrientationA Record of EventsA Record of EventsA OrientationA Record of Events: ORIENTATION RES: so do you l i k e p i z z a W i l l ? CHI: mmhm. 28 RECORD OF EVENTS RES: I had p i z z a l a s t n i g h t f o r supper. ORIENTATION RES: what d i d you have f o r supper l a s t n i g h t ? RECORD OF EVENTS CHI RES CHI RES I had p i z z a t o o . d i d you? yeah. oh -: . RECORD OF EVENTS RES: and the n i g h t b e f o r e I had po r k chops. |ORIENTATION| can you remember what you had t h e n i g h t b e f o r e ? RES : CHI RES RES RECORD OF EVENTS p o r k chops t o o . you had p o r k chops t o o . hm. Example 4.2 - Specific Recount 1, transcript 1, Lines 138-150 The second text's sequence repeats the Orientation A Record of Events sequence 3 times with an additional Record of Events between the second and third sequence. In this, each Orientation stage following the initial Orientation functions to re-establish the temporal setting when supper was eaten. In both the General Recount and Specific Recount, all Orientation stages were uttered by the researcher in the form of questions. The Records of Events represent the child's responses to these questions. In the 2 stretches of text, the researcher utters all 4 of the Orientation stages. The child only utters solicited Records of Events and each describes a single event. A turn-taking pattern is observed where the researcher poses a question and the child answers it. It appears that the child requires structure from the researcher's questions to sustain turn-taking in the conversation. Table 4.2 summarizes the genres types and generic stages that speakers in text 1 use. 29 CU X s 3 z • M ' u u c ets u H Z o •o u cu e < a. E CU U s o u c 10 a © u 3 •a cu u o >- X < cu O CA S GO o VI X U X < B o e •a cu c TS cu o GO X I B e e > Ui C M O cu OS •a cu 5 •o cu o CO X s cu W C M o •a cu 3 S o o EE •o cu o CA B 5 cu o to XI U X o o e CU > tit _cu X eg -4 « E cu a! T3 cu '3 \"o CA B 5 •a cu o CO X U x cu > U _cu X « -4 cu 05 B e u es cu Qi •o cu 5 cu o co 2 Is U >> x B _© U C3 CU o U •a cu • M \"o CA B 5 cu o to X CO T3 O U B .© et s 13 > B 5 •O cu o Vi X B .© CO s \"et > Table 4.2 - Summary of genre types and generic stages used by speakers in text 1 Text 2 Text 2 contains a single General Recount (see Appendix 1). This text uses only Orientation and Record of Events stages in its generic structure with the exception of an evaluative comment that also serves as a Coda uttered by the researcher. The General Recount contains 3 Orientations, 3 Record of Events, and a Coda. Two of the 3 Orientations were uttered by the researcher and the remaining Orientation was uttered by the child and solicited by the researcher. A l l 3 Records of Events were uttered by the child and solicited by the researcher. The generic structure of the text repeats the OrientationA Record of Events sequence 3 times. The participants' turn-taking followed a pattern where the researcher posed a question and the child answered the question. The conversation relies heavily on this turn-taking pattern to sustain conversation as all of the 30 child's responses were solicited by the researcher. Table 4.3 summarizes the genre type and generic stages that speakers in text 2 use. 1. tu X E 3 E 3 \"a. E x c «3 0 0 0 0 1 0 0 0 X B o S 3 XI X o o SB 0 0 0 ja U >^ s > Ed J£ x cs ±£ u a a as x 0 0 0 1 0 1 •a o U e/3 X! U >-> x Table 4.3 - Summary of genre types and generic stages used by speakers in text 2 Text 3 Text 3 contains a Specific Recount and 3 General Recounts. The Specific Recount is particularly interesting because it contains a Reaction stage that is normally found in Anecdotes. It appears that the child and researcher have a misunderstanding in the conversation's genre (see Example 4.3). The Specific Recount is also discontinuously realized. The Specific Recount proceeds through the following genre stages: • Abstract • OrientationA • EvaluationA • Record of EventsA 31 • React ion A • Orientation A • Record of Events A • Orientation A • Record of Events A • Orientation A • Record of Events A • Orientation A • Record of Events A • Orientation A • Evaluat ion A • Orientation A • Record of Events A • Evaluat ion A • Evaluation/Coda In the Specific Recount, the child engages in a Recount while the researcher appears to engage in an Anecdote. Table 4.4 compares the generic structure of an Anecdote with the generic structure of a Specific Recount. Genre Generic structure Anecdote (Abstract) A (Orientation) A Remarkable E v e n t A Reaction A (Coda) Specific Recount (Abstract) A (Orientation) A Specific Record of Events A (Coda) Table 4.4 - Generic structure of the Anecdote and the Specific Recount In Chapter 3, Methods, it is noted that the beginnings and endings of genre types may use some of the same generic stages. Both the Anecdote and Specific Recount may begin with Abstracts and Orientations and they may conclude with Codas. These shared stages and the nature of the Remarkable Event and Specific Record of Events stages in both genre types appear to cause the confusion between the participants. The researcher begins the Specific Recount with an Abstract. The child then utters an Orientation that is solicited by the researcher. The researcher evaluates and then asks a question. The child responds to the researcher's question with a Record of Events. This Record of Events is solicited and it describes 6 events in serial order. The researcher 32 appears to mistake the child's Record of Events for a Remarkable Event and interrupts the child with a Reaction, \"wow.\" The researcher then attempts to reorient the conversation with a question; however, the child answers this question with a polar response and returns to continue his Record of Events: RECORD OF EVENTS CHI: CHI: CHI : CHI : Sophie and I went t o Recordman and a t Recordnan Recordman we found a S t a n Rogers tape c a l l e d P o e t i c J u s t i c e [!] w i t h two r a d i o p l a y e r s c a l l e d H a r r i s and t h e major and the s i s t e r s . um J Sophie found a Sophie found a t h i n g t h a t wasn't Tom W a i t s wasn't the Wake i t wasn't R i c k W a k e f i e l d b u t i t was t h e W a l l f l o w e r s # c a l l e d b r i n g i n g down the h o r s e . i t was the o n l y W a l l f l o w e r s album t h e y e v e r r e c o r d e d . ANECDOTE RES : REACTION ORIENTATION RES RES CHI RES had you h e a r d about them b e f o r e ? t h i s group? no! RECORD OF EVENTS CHI: CHI: Example 4.3 uh the so t h e n we went t o an a n t i q u e shop, and t h e n back t o V i n y l R e c o r d s . Excerpt from Specific Recount 1, transcript 3, lines 26-36 The misinterpretation of genre types may be better understood by considering the context of situation. In this text, what is happening ideationally is a casual conversation between a researcher and a child. Both participants appear to understand that the conversation genre belongs to a storytelling genre type. The level of specificity of language appears to be the source of confusion. The child retells events with great detail; for example, the naming of each music album. This child's use of detail is more typical of a Remarkable Event stage found in the Anecdote genre type. The researcher appears to draw this conclusion based on the child's use of detail. Typically in a Recount, information in a Record of Events would be kept more general compared to a 33 Remarkable Event in an Anecdote. Compare the level of specificity and detail in the following invented examples: Recount: RECORD OF EVENTS Y e s t e r d a y n i g h t I made ma c a r o n i and cheese. A f t e r d i n n e r , I watched some T.V. Example 4.4 - Typical Record of Events stage in a Recount Anecdote: REMARKABLE EVENT Y e s t e r d a y n i g h t I made macar o n i and cheese. I t o o k a cup o f d r y m a c a r o n i and put i t i n a p o t . Then I went o v e r t o the s i n k , and when I h e l d the pot under the f a u c e t and tuned on the hot water t a p , s m e l l y brown water came out o f the t a p ! REACTION Yuck! Example 4.5 - Typical Remarkable Event and Reaction Stages in an Anecdote In the Remarkable Event stage, information is typically retold with more detail compared to the Record of Events stage. Overly specific information when relating text to context would hint to the audience that that the detailed retelling of events is intentional. In a Recount or an Anecdote, audiences may not establish the text's genre type until after the optional elements have been uttered because both may begin with the same optional elements. The child uses high level of detail in the Record of Events and the researcher believes that the child is retelling a Remarkable Event. This example suggests that speakers with A S D may have difficulty with the coordination of contextual information producing utterances in conversations that can mislead audiences. In the Specific Recount, the child continues in an extended Record of Events following the researcher's Reaction and (re)Orientation. The child's Record of Events describes 12 events in serial order. The researcher attempts to reorient the child during his Record of Events 4 times, but the child always returns to his Record of Events. These attempts to reorient, including the child's return to a Record of Events, are shaded in Example 4.6: 34 I R E C O R D O F E V E N T S | R E S : w h a t d i d y o u s e e i n T o r o n t o ? C H I : < I > [>] . R E S : < o r i n > [<] T o r o n t o ? C H I : f i r s t I w e n t t o K i n g s b u r y . C H I : a n d v i s i t e d m y a u n t E d d y a n d J u d y W i n s t o n C H I : o n o n T u e s d a y a n d W e d n e s d a y J u d y W i n s t o n a n d J we b o t h w e n t t o t h e a i r p o r t t o p i c k u p S o p h i e . C H I : ( a ) n ( d ) S o p h i e i n t r o d u c e d u s t o a b a n d c a l l e d t h e W a l l f l o w e r s w i t h J a c o b D i l l o n i n i t . R E S : o h w h e r e d o e s t h e b a n d p l a y ? C H I : f r o m A m e r i c a . R E S : o h f r o m A m e r i c a . C H I : S o p h i e a n d I w e n t t o R e c o r d m a n . C H I : a n d a t R e c o r d n a n R e c o r d m a n w e f o u n d a S t a n R o g e r s t a p e c a l l e d P o e t i c J u s t i c e [ ! ] w i t h t w o r a d i o p l a y e r s c a l l e d H a r r i s a n d t h e m a j o r a n d t h e s i s t e r s . C H I : u m J S o p h i e f o u n d a S o p h i e f o u n d a t h i n g t h a t w a s n ' t T o m W a i t s w a s n ' t t h e W a k e i t w a s n ' t R i c k W a k e f i e l d b u t i t w a s t h e W a l l f l o w e r s # c a l l e d b r i n g i n g d o w n t h e h o r s e . C H I : i t w a s t h e o n l y W a l l f l o w e r s a l b u m t h e y e v e r r e c o r d e d . ANECDOTE R E A C T I O N R E S : w o w . _ _ _ _ _ _ _ 0 R'l- E-NTi A;T-;-l :0N R E S - r j {.'.riaffi^ R E S ; \" , t h i s , g r o u p j ? [ C H I : . ;>•:'\"'• n o f j R E S : • no?[~\"~~\"~ C H I Y • ;y\\}:.;^ n.i;jah,, t-hV-'.;so^then:,we/\\$e\"n._'i t d ^ a j ^ a r f t & L q u e s h o p ,| C H I : a n d t h e n b a c k t o V i n y l R e c o r d s . C H I : w h e n I f l i p p e d t h r o u g h t h e t h e l e t t e r b i n t h e B o w i e [ ! ] s e c t i o n I f o u n d C h a n g e s [ ! ] O n e B o w i e . C H I : a n d t h e n i n the C o h e n [ ! ] s e c t i o n I r i f l e d t h r o u g h t i l l I w a s s t o p p e d d e a d b y t h e b l u e r a i n c o a t s t a r i n g out f r o m t h e n e w J e n n i f e r W a r n e s a l b u m c a l l e d F a m o u s B l u e R a i n c o a t . C H I : a n d I g o t i t o n L P a n d c a s s e t t e . C H I : a n d t h e n w e a n d t h e n we w e n t h o m e t o a n d t h e n a n d t h e n w e f o u h o m e a n d t h e n we d r o v e b a c k t o K i n g s b u r y a n d K i n g s b u r y . C H I : a n d l i s t e n e d t o i t o n t h e w a y - : a y t h e r e . C H I : a n d a t K i n g s b u r y I a t e s o m e l a s a g n a a n d b u t t e r s c o t c h i c e c r e a m . C H I : a n d w e w e n t o u t s i d e a t n i g h t a n d p l a y e d i n t h e s n o w u m < a n d > [>] . R E S : < d o y o u > [<] . C H I : t h e n e x t d a y I w e I w e n t t o t h e b o o k s h o p . J C H I : ; \" a r i d , gbtf,\"trie\" Eng l i sh^ 'Ve f sYo lTro f irth^e'rJudiFli , . B e o r i s ' book' •Alexander .And^The^e*^ a o v e d > [>;/•; '/; | 0R1 E N T A T 3 O N R E S T \\ < e x c u s e me Jwill>;\"[<l .] R E S : , ' d i d y o u * H a v e - c o u s i n s t he re , T t p T p T a ^ JCHI: ' ' n o - : i f\"~ R E S . : . , : - n o . j u s t < a d u l t s . e h . hm> . [>]; ?j 35 RECORD ••OF'. EVENTS (CHI::. . < i . t ' s dr.> <] ;,• t h'e ri* ^ ^ e n t l l t ;o B i l l y / a h d ^ N a n c y i t s ^ f a r m . C H I : <and> [>] ORIENTATION RES JCHI RES [CHI RES' JCHI RES' RES (CHI RES RES: they- ihadi-a l o t ' [ ! ' l i k e about ',twentyvor ! f i f ty'\\pr.-'whatiitwould, you think?! t' tiwJenty____b\\Tj • twent-y^fewo.?P** <yeah>„ [>] ' <and w> [-<] what what ' c o l o u r s »were they<.?j were t h e y i H b l s t i e n s b l a c k a n d - w h i t e 'Jones, or^ 'were . t, f f—• u • t h e y .were - b l a c k -and, w h i t e - H o l s t - i e n s . ] • were , they_?j uhhuh. | RECORD. OF EVENTS CHI CHI CHI CHI CHI CHI CHI CHI CHI CHI CHI CHI CHI CHI excuse me ' and ^excuse/ 'mevand'-^ and I went o u t s i d e a g a i n . and I t o o k a b a t h . and t h e n [!] I went t o s l e e p . and t h e n I went t o b e d . and t u r n e d o f f t he l i g h t . t h e n e x t m o r n i n g I p a c k e d my s u i t c a s e up . and we went back t o T o r o n t o , bu t my f a t h e r was gone . I s a t down and r e a d - : . and s u d d e n l y my f a t h e r went my f a t h e r came b a c k , he w a l k e d me o v e r t o t he Sam's on Yonge s t r e e t , and B l o o r s t r e e t - : . and I . PRIENTATION RES: ; w h a t ' s . ; , that p l a c e - W i l l ; ? L (CHI: \"\"' ' ' \" y . i t • ' s; '3vit ' 'S;;-a:rri;ey<?Sam^s;^<wBe%fe>[^ : . ' ::j£he^thave;jiis^ey RES:: jfe;<: whe'rie < j] \"M RECORD OF\" EVENTS C H I : , '\"but I am'Sflrry they , d i d no.t^ h a v e a r t y ' ; W i n c h e s t e r LPs there..) Example 4.6 - Excerpt from Specific Recount 1, transcript 3, Lines 16-81 The shaded areas in the above example show a rigid adherence to the Record of Events stage of the Recount genre. The Specific Recount in Text 3 includes 7 Orientations, 6 Records of Events, a Reaction, 3 Evaluations, and a Coda. Six of the 7 Orientations and all the Records of Events were uttered by the child. O f these 6 Orientations, 5 were solicited by the researcher while only 1 of the 6 Record of Events was solicited by the researcher. The ratio 5:1 of unsolicited Record of Events to 36 unsolicited Orientations points again to preferred or rigid use of the Record of Events generic stages in genre. In the Specific Recount, the researcher uttered the Reaction, and 2 of the 3 Evaluations. The child uttered a single Evaluation and it was solicited by the researcher. Text 3 also contains 3 General Recounts. The first General Recount proceeds through Orientation A Record of Events A Orientation A Evaluation/Coda. A l l stages are uttered by the researcher. The second General Recount proceeds through Abstract^ Orientation A Evaluat ion A Orientation A Record of Events A Orientation A Record of Events A Evaluation/Coda. The Abstract is uttered by the child and solicited by the researcher. Two of the 3 Orientation stages were uttered by the child and solicited by the researcher and both Records of Events were uttered by the child without solicitation. The first Record of Events describes a single event while the second Record of Events describes 2 events in serial order. Both the Evaluation and the Evaluation/Coda are uttered by the researcher. The third General Recount proceeds through Orientation A Record of Events A Evaluat ion A Record of Events A Coda. Two Records of Events are uttered by the child and are solicited by the researcher. Both Records of Events describe 2 events in serial order. The Orientation, Evaluation, and Coda are uttered by the researcher. Text 3 contains 13 Orientations (4 by the researcher and 9 by the child). Seven of the child's Orientations are solicited by the researcher. The child utters 10 Records of Events, 8 of which are unsolicited. The researcher utters a single Reaction, all Codas (4) and 6 of the 7 Evaluations. (The child's Evaluation is solicited by the researcher.) Interestingly, in this text, the child engages in Records of Events that often describe 37 multiple events in serial order. Table 4.5 summarizes the genre types and generic stages that the speakers in text 3 use. tu X E 3 Z es u H o> > CS u es Z o •a w V B < _3 \"5. E V X Ed B 3 O u O S a. Vi 3 O u 02 o> B tu a o> >-3 •a U o < 0 0 o> \"o B w •a 01 o Vi £1 B •O o> \"o t» B •a tu o XI B O B o> 7 S 2 U Ed •a u o u tu tu B 13 tU o (/I X U >> x B 01 >• Ed •a u o u 0> 2 S 8 U o to X \"o B 5 •a 0> o 05 X o H o X B tu > Ed 0> X a it cs E 01 B S •a 01 o CA B t/3 o •a x o> > Ed — X cs E 01 ai B O es tu P S 01 CA B 3 o> o a) x es o> o U •a V \"o CA 8 w •a o Vi X es •o o U B cs _3 > Ed 0 0 o i o •a 0> -*-t '3 \"o CA B 3 VI © VJ x B cs a \"5 >• Ed Table 4.5 - Summary of genre types and generic stages used by speakers in text 3 Text 4 Text 4 contains a General Recount, and a Procedure. The General Recount proceeds through Orientation A Record of Events A Evaluation. The researcher utters the Orientation in the form of a question. The child responds with a Record of Events that describes a single event. The child then utters an unsolicited Evaluation. The Procedure describes how to make Kraft Dinner. The Procedure proceeds through Abstract^ Orientation A Record of Events A Coda. It follows the Procedure genre described by Eggins and Slade perfectly because it uses all mandatory and optional stages of the Procedure genre type. The researcher utters the Abstract in the form of a question: 38 \"Do you do any cooking at home Wi l l ? \" The researcher then orients. The child utters a How To stage that involves 3 steps. The child provides an unsolicited Coda. Text 4 contains 2 stretches of spoken discourse that use temporally sequenced events. The researcher uses a single Abstract. The speakers use 2 Orientations; both speakers utter a single Orientation. The child's Orientation is unsolicited. The child utters a single Record of Events and a single How To stage that are both solicited by the researcher. The child utters a single unsolicited Evaluation. In text 4, the child demonstrates competency in using the Procedure genre type in addition to the General Recount genre type. The child is also able utter an unsolicited Coda and evaluates without solicitation once. Table 4.6 summarizes the genre types and generic stages that the speakers in text 4 use. X Q. E tu x W C v> x u CD u S CD o o x U >> x e o c CD o CD \"o CA e o Vi X Vi X o H o X U x CD Vi X y X Pi cs O U o X es •O O U 73 CD '5 \"o CA S CD O t/J X c o es _s > 0 0 1 1 0 0 0 u 1 u Table 4.6 - Summary of genre types and generic stages used by speakers in text 4 39 Text 5 Text 5 contains a Specific Recount and a Procedure. The Specific Recount proceeds through Orientation A Record of Events A Record of Events A Evaluation/Coda. The researcher utters the Orientation, the first Record of Events, and the Evaluation/Coda. The child utters the second Record of Events and it is solicited by the researcher. The child's Record of Events describes a single event. The Procedure from Text 5 proceeds through Abstract A Orientation A Evaluat ion A How to A Evaluat ion A Evaluation/Coda. The child utters the Abstract that is solicited by the researcher. The researcher utters the Orientation and the child follows with an unsolicited Evaluation. The child then utters a How To solicited by the researcher and then evaluates without solicitation. The How To describes 3 steps. Finally, the researcher utters the Evaluation/Coda. Text 5 contains a single solicited Abstract uttered by the child, 2 Orientations uttered by the researcher, 2 Record of Events (one solicited and uttered by the child and the other uttered by the researcher), a solicited How To uttered by the child, 2 Codas uttered by the researcher, and 4 Evaluations. Two of the evaluations are unsolicited and are uttered by the child and 2 are uttered by the researcher. The child's utterances are all solicited by the researcher with the exception of 2 unsolicited Evaluations. Table 4.7 summarizes the genre types and generic stages that the speakers in text 5 use. 40 X es o •o u C D S < 0 0 E 3 x 0 0) O S a, 75 o u CD OS B tu 0 3 T3 tu |3 \"3 CA e o CO U X X 1 S es B o> •c o •o tu [3 \"© CA B to\" •a o X B O S . o 0 B > W C M O •a OS 0) B 3 o> o JS U x B 0> > W o T3 © o S3 T5 3 ST •a o t/3 X © s S3 B C D > u C D X es 0 CD B CO 1—' •a C D O to x -*-» B CD > X es es E CD a; CD ai 73 C D '3 \"© CA B CO* 73 CD O CO X! B u es s> O S es T3 © T 3 CD -*-» [3 \"o CA B CO~ © co X es 73 o U es 73 CD 3 CO o CO X B _o CS _3 \"es > H 2 U Table 4.7 - Summary of genre types and generic stages used by speakers in text 5 Text 6 Text 6 contains a General Recount and 2 Procedures. The General Recount proceeds through an Orientation A Record of Events. The researcher solicits both stages and they are uttered by the child. The child's Record of Events describes 2 events in serial order. Both Procedures proceed through Orientation A How T o A Evaluation/Coda. In both cases, the child utters all elements. A l l stages in the Procedures are unsolicited except for the Orientation in the first Procedure. Both How To stages describe 2 steps in serial order. Text 6 contains 3 Orientations, one uttered by the researcher, and 2 uttered by the child. O f the 2 Orientations produced by the child, one is solicited by the researcher and the other is not. The text also contains 2 How To stages, 2 Codas, and an Evaluation. 41 These stages are all unsolicited and uttered by the child. Table 4.8 summarizes the genre types and generic stages that the speakers in text 5 use. X E z *u cu CA c es u H o •o u CD B E _a \"E. E 41 X w B 3 O eu tu Qi u tu a. CO B 3 O cu <u OS tu B tu o •a tu u o o CS CA XI < tu -*-» \"3 CA B *— 73 tu o CO >> XI u 0! X < B tu o 0 0 s •o tu o CO X U >> X B 0) o 1 s 1 u B CU > o u tu ai B 3 w •O CD O CO X B > W O •d u o cu o H o SB 73 CU •«-» ]3 \"o Cfl B 3 o X o H o SB 2 U CU > X ! S3 1-es E at 73 '3 \"o CA B 3 73 CU o CO X i B cu >• W X CS cs E CU OS cu es 73 eu o CA B 3 CO* © CO XI B _© CU CS es 73 o U 73 eu '3 \"o CA s 3 73 cu o CO X i es 73 O u 2 U es 73 > 3 co~ o CO 73 >> X B O CS > 1 u Table 4.8 - Summary of genre types and generic stages used by speakers in text 6 Text 7 Text 7 contains 4 General Recounts. The first proceeds through Abstract A Orientation A Evaluat ion A Record of Events A Orientation A Record of Events A Record of Events A Record of Events A Orientation A Evaluation/Coda. The child in this text demonstrates advanced use of generic stages of the General Recount. The Abstract is solicited by the researcher and uttered by the child. The first Record of Events describes a single event. The Orientation and first Record of Events, produced by the child, are unsolicited. The researcher reorients the conversation and solicits a second Record of Events from the child. The second Record of Events describes 5 events in serial order. 42 The second, third, and fourth Records of Events in the first General Recount are interesting because the fourth Record of Events is discontinuously realized from the second Record of Events, while the third Record of Events refers to a different temporal setting: |RECQRD OF E V E N T S ( 2 ) | C H I : w e l l we t a k e t h e um f o u r + f o r t y + f i v e b u s . C H I : um - : a n d t h a t i s t h o s e um t h o . RECORD OF E V E N T S (3) C H I : w e l l i t ' s n o t F u n t r a c k . C H I : i t u s e d t o b e F u n t r a c k . C H I : i t ' s T r e n t w a y <now> [ > ] . R E S : <mmhm> [<] mmhm? RECORD OF E V E N T S (4) C H I : a n d um # a n d we t a k e t h e we t a k e t h e f o u r + f o r t y + f i v e b u s . C H I : t h e n um i t g o e s t o a l l t h e s e l i t t l e t o w n s b e f o r e i t s t o p s a t um i n H a m i l t o n . Example 1.7 - General Recount 1, transcript 7, lines 30-37 The child discusses the bus ride he generally takes to work in the second Record of Events, then shifts to the third Record of Events where he describes the before and after facts of the bus' name/type. In the fourth Record of Events, the child returns to continue the second Record of Events by repeating it and then proceeds through additional events in serial order. That is, the child suspends the temporal sequences in the second Record of Events, shifts to another temporal sequence (the third Record of Events) and then continues the second Record of Events in the fourth Record of Events. The third and fourth Records of Events both describe 2 events in serial order. The final Orientation is an unsolicited list of towns (possibly in serial order) uttered by the child with an Evaluation/Coda uttered by the researcher. The second General Recount in Text 7 has a simple structure of Orientation A Record of Events. The Orientation is uttered by the researcher and the Record of Events is solicited by the researcher and uttered by the child. This Record of Events describes 2 43 events in serial order. The third and fourth General Recounts in Text 7 followed a similar simple generic structure. The third General Recount proceeds through Orientation A Record of Events A Evaluat ion A Record of Events. The Orientation is uttered by the researcher and both Records of Events are uttered by the child. The first Record of Events is solicited by the researcher and describes a single event while the second Record of Events is unsolicited and describes 2 events in serial order. The first and second Records of Events are discontinuously realized and they are separated by a brief Evaluation uttered by the researcher. The child utters the entire fourth General Recount. The generic structure proceeds through Orientation A Record of Events A Orientation A Record of Events. The initial Orientation is solicited by the researcher while the remaining stages are unsolicited. Both Records of Events are a result of a discontinuously realized single Record of Events and the Orientation that separates them functions to further specify the situation under which the first Record of Events occurs. Here the child describes his physical activity, reorients the situation, and then continues describing the serial events in his workout: [ R E C O R D O F E V E N T S | C H I : < w e l l > [<] w e u I u s e d t o d o t h a t f o r s p o r t s n i g h t . C H I : I I d o n ' t v e r y m u c h n o w . C H I : b u t I d o o n e # b u t um o n S u n d a y s I g o t o u h m y f i t n e s s c l a s s e s . C H I : a n d I d o l i f t i n g l i f t i n g o n e u p . O R I E N T A T I O N C H I : w e l l t h o s e t h o s e w e i g h t t h i n g s . C H I : e x c e p t t h e y ' r e n o t t h e b i g s o r t o f w e i g h t s . C H I : b e c a u s e I t r i e d t o l i f t t h o s e . C H I : < a n d c o u l d n ' t d o t h a t > [% c h u c k l i n g ] . C H I : s o i t ' s m o r e o f t h e m a c h i n e s o r t o f w e i g h t s . R E S : u h h u h ? R E C O R D O F E V E N T S C H I : y o u d o t h e . C H I : e x c u s e m e . C H I : < x x x > [ > ] . R E S : <0 [ = ! c h u c k l e s ] > [ < ] . C H I : um # I d o u m . C H I : w h a t e l s e d o I d o ? 44 C H I : I d o t h i s u m # t h e o n e s w h e r e y o u g o l i k e t h i s . R E S : o h r i g h t . C H I : t h e n y o u # y o u d o t h a t . R E S : y e a h . C H I : I d o um s o m e t h i n g w i t h # t h a t . C H I : I d o # . Example 4.8 - General Recount 4, transcript 7, lines 148-169 Text 7 contains 7 Orientations; 3 are unsolicited and uttered by the child while 4 are uttered by the researcher. Text 7 also contains 9 Records of Events all uttered by the child; 4 Records of Events are solicited by the researcher while 5 are not. Several of the Record of Events stages describe multiple events in serial order. The researcher utters a Coda and 3 Evaluations. The child in Text 7 is flexible with generic structure because he is able to divide Records of Events with an optional stage (an Orientation). He is also able and to produce discontinuously realized spoken discourse. Table 4.9 summarizes the genre types and generic stages that the speakers in text 7 use. 45 «U Si E s Z e cs H 0 0 a D. E cu X W u CU D. CO 0 1 u s o o CO XI u >> X •a i_ o cu cu SL 4 S 5 U x o H o X 0 0 0 0 0 1 0 c 3 tO~ X U x X cs •o o U \"O cu 'u \"o s 3 T3 CU o CO X ! cs > bi Table 4.9 - Summary of genre types and generic stages used by speakers in text 7 Text 8 Text 8 contains 2 Specific Recounts, a General Recount, and a Procedure. The 2 Specific Recounts are simple in generic structure; they proceed through Orientation A Record of Events. In the first Specific Recount, the child utters both the Orientation and the Record of Events. In the Record of Events, the researcher attempts to solicit a How To with the questions, \"How does it get f i l led up?\" but the child responds with a Record of Events that contains a single event. If the child had responded to the researcher's question with a How To, the text would be classified as a Procedure rather than a Specific Recount. 46 p R I E N T A T I O N | R E S : < a n d > [<] y o u w e r e t e l l i n g me a b o u t a c e m e n t t r u c k b e f o r e . C H I : mm - : y e s ! C H I : i t h a d a w e t h e a v y l o a d . R E S : a w e t h e a v y l o a d - : . C H I : o f c o n c r e t e . R E S : f r o m w h a t ? C H I : f o r t h e m f o r f o r t h e m i g h t y m i x e r . C H I : t h a t x x x . R E C O R D O F E V E N T S R E S : h o w d o e s i t g e t f i l l e d u p ? C H I : j u s t g o x x x . Example 4.9 - Recount 1, transcript 8, lines 9-18 In the second Specific Recount, the researcher asks the child \"what does it make?\" In this question, \"it\" refers to wet cement mix when it dries. Instead of responding to the researcher's question, again, the child utters a Record of Events that describes a single event: p R I E N T A T I O N | R E S : a n d w h a t d o e s i t m a k e ? R E C O R D O F E V E N T S C H I : um # a n d i t t u r n s t h a t w a y i n s t e a d o f c o m i n g t h i s w a y . i t = t r u c k C H I : t h a t ' s b e c a u s e t h e y a r e m a n y m a c h i n e s i n t h e c o n s t r u c t i o n s i t e . R E S : t h e r e a r e m a n y w h i c h ? C H I : t h e m t h e r e a r e m a n y m a c h i n e s i n t h e c o n s t r u c t i o n s i t e . Example 4.10 - Recount 2, transcript 8, lines 38-42 The child violates the conventions of generic stages in both Specific Recounts because he neglects the solicited question and responds with Records of Events. In example, 4.10, the researcher solicits a Record of Events and the child responds with a Record of Events; however, the child's Record of Events does not answer the researcher's question. The General Recount proceeds through Orientation A Record of Events A Orientation A Record of Events. The child utters the Orientation that is solicited by the researcher. A n interesting aspect of the Orientation is that it contains what Hasan (Halliday and Hasan, 1989) term a probe. In Hasan's analysis of sales encounters, i f a person were to enter into a store and \"hang about\" the vendor may attempt to provoke a 47 Sales Request (SR). The vendor could attempt to do so by asking \"Can I help you?\" or \"Are you alright?\" (Halliday & Hasan, 1989, 66). Hasan explains, \"It [a probe] consists of some device that is calculated to bring about the kind of behaviour on the part of some(one) participant...\" (Halliday & Hasan, 1989, 66). In the General Recount the child asks the researcher \"what what did Sophia do?\" This question functions as a probe that attempts to solicit a question from the researcher. It appears that the child knows that the researcher wi l l not be able to answer this question and wi l l respond accordingly with another question. The researcher responds with the anticipated question in the following example: O R I E N T A T I O N R E S C H I R E S C H I R E S C H I C H I C H I C H I R E S R E S C H I w h a t ' s a d o g < g a m e > [ > ] ? I h a v e [<] s o m u c h I h a v e s o m u c h f u n . w h a t ' s a d o g g a m e l i k e ? i t ' s i t ' s x x x t r y i n g t o g e t t h e o t h e r d o g s o u t . o h w h o d o y o u p l a y w i t h ? u m - : s o m e t i m e I p l a y w i t h S o p h i a w h a t w h a t d i d S o p h i a ? w e l l # w h a t d i d s h e d o ? w h a t d i d s h e d o t o m e ? I d o n ' t k n o w . d i d s h e d o s o m e t h i n g t o y o u ? y e s . R E C O R D O F E V E N T S C H I : n o w s h e w a s s t u c k i n i n b e t w e e n i n b e t w e e n m a y b e s h e g o t s t u c k i n b e t w e e n . O R I E N T A T I O N R E S C H I . R E S R E S i n b e t w e e n w h a t ? i n b e t w e e n # t h e t r e e s , o h i n b e t w e e n t h e t r e e s ? < w h e r e i n > [ > ] ? R E C O R D O F E V E N T S C H I : n o w S o p h i a w a s g o t t e n s t u c k i n b e t w e e n # o n a s u m m e r d a y . Example 4.11 - General Recount 1, transcript 8, line 119-136 The child appears so determined to probe for a specific question that he probes the researcher with 3 consecutive questions. When the researcher responds to the probe with the question, \"D id she do something to you?\" the child begins a Record of Events. The researcher then attempts to clarify the setting by soliciting the child to reorient. The child 48 responds and continues with an unsolicited Record of Events. Both of the child's Records of Events in the General Recount describe a single event. In total, Text 8 contains 5 Orientations; the researcher utters 2 Orientations while the child utters 3 Orientations. The child's Orientations are solicited by the researcher. This text also contains 5 Records of Events that are all uttered by the child; 4 are solicited by the researcher while 1 is not. The majority of the conversation follows an Orientation A Record of Events pattern. The child violates the genre initiated by the researcher and utters a Record of Events where another generic stage would typically be used or responds with a Record of Events that does not answer the researcher's question. The child also probes the researcher 2 times in this text. In the first probe, the child attempts to probe the researcher by repeating the same question 3 times. In the second probe, the child is successful at probing the researcher with his first attempt. Table 4.10 summarizes the genre types and generic stages that the speakers in text 8 use. 49 H oo Transcript Number u o Narrative 1.10 - Summary of genre types and generic stages used 1 o Anecdote 1.10 - Summary of genre types and generic stages used 1 o Exemplum 1.10 - Summary of genre types and generic stages used 1 t o Specific Recount 1.10 - Summary of genre types and generic stages used 1 - General Recount 1.10 - Summary of genre types and generic stages used 1 - Procedure 1.10 - Summary of genre types and generic stages used 1 o Abstract 1.10 - Summary of genre types and generic stages used 1 o Abstract by Child Solicited (S)/Unsolicited (U) 1.10 - Summary of genre types and generic stages used 1 t_/l Orientation 1.10 - Summary of genre types and generic stages used 1 Orientation by Child Solicited (S)/Unsolicited (U) 1.10 - Summary of genre types and generic stages used 1 Record of Events 1.10 - Summary of genre types and generic stages used 1 C — </> 4^ Record of Events by Child Solicited (S)AJnsolicited (U) 1.10 - Summary of genre types and generic stages used 1 O How To 1.10 - Summary of genre types and generic stages used 1 o How To by Child Solicited (S)/Unsolicited (U) 1.10 - Summary of genre types and generic stages used 1 o Remarkable Event oy speakers in text 8 o Remarkable Event by Child Solicited (S)/Unsolicited (U) oy speakers in text 8 © Reaction oy speakers in text 8 o Reaction by Child Solicited (S)/Unsolicited (U) oy speakers in text 8 o Coda oy speakers in text 8 o Coda by Child Solicited (S)/Unsolicited (U) oy speakers in text 8 o Evaluation oy speakers in text 8 o Evaluation by Child Solicited (S)AJnsolicited (U) P A R T 2: Analysis of A l l Transcr ipts In Part 1, stretches of text that use temporal events were categorized according to 6 genre types. The generic structure of each stretch was described and patterns were observed within individual texts. In this section, patterns across the 9 text are observed and discussed. In the 9 texts, speakers use Specific Recounts, General Recounts, and Procedures but do not use Narratives, Anecdotes, or Exemplums. In total, the speakers use 5 Specific Recounts, 12 General Recounts, and 5 Procedures. General Recounts are the most frequently used genre type in the 9 texts. Table 4.11 details these findings. Transcript Specific General Number Narrative Anecdote Exemplum Recount Recount Procedure 1 0 0 0 1 1 0 2 0 0 0 0 1 0 3 0 0 0 1 3 0 4 0 0 0 0 1 1 5 0 0 0 1 0 1 6 0 0 0 0 1 2 7 0 0 0 0 4 0 8 0 0 0 2 1 1 9 0 0 0 0 0 0 Total 0 0 0 5 12 5 Table 4 . 1 1 - Summary of genre types used by speakers in 9 texts The generic stages that speakers use in the 9 texts correspond with the genre types found in the texts (see Appendix 2). There is an exception where a Reaction stage appears within a Specific Recount. Table 1.12 shows that the speakers (researcher and research participants) use Orientations most frequently (39 times), followed by Records of Events (37 times) and Evaluations (17 times). Although Orientations are an optional stage among the 6 genre types, speakers use Orientations more than other mandatory stages. 51 The speakers in the 9 texts use Records of Events frequently as it is a mandatory stage of the Recount genres but less often than Orientations. The Recount genres include both General Recounts and Specific Recounts. Together, these genre types were used the most by the speakers in the texts. Speakers also use Evaluations frequently. Evaluations are another optional stage found in all 6 of the genre types. Speakers use Reactions the least among the generic stages. A single Reaction appears in the 9 texts. This observation is not surprising as no Anecdotes are found in the 9 texts. Appendix 2 summarizes all genre types and generic stages used by speakers in the 9 texts. Number of Number of Number of Times used by Child Generic Times Times Used Stage Used by Speakers by Researcher Abstract 5 3 2(2S) Orientation 39 20 19(13S6U) Record of Events 37 4 33 (19S 14U) How To 4 0 4 (2S 2U) Remarkable Event 0 0 0 Reaction 1 1 0 Coda 11 8 3 (3U) Evaluation 17 12 5 (1S4U) Table 4.12 - Summary of generic stages used by speakers in 9 texts The child uses Records of Events the most frequently (33 times), followed by Orientations (19 times) and Evaluations (5 times). The child uses 33 Records of Events while the researcher uses 4 Records of Events. This finding may be explained by the semi-structured nature of the conversational texts. The researcher orients Recount genres and solicits Records of Events from the child; therefore, 19 of the child's total 33 Records of Events are solicited by the researcher. Yet a remaining 14 Records of Events uttered by the child are not solicited and these warrant further examination. They can, perhaps, 52 be better understood from the analysis of individual texts. Earlier observations suggest that children with A S D have a tendency to use Records of Events in several ways. On 4 occasions in text 3, the child ignores or responds to Orientations that interrupt a Records of Events, but then returns to the Record of Events without solicitation. It can also be observed that the children sometimes respond using Records of Events from the Recount genre when other stages and genres are solicited by the researcher. This happens, for example, in text 8, when the researcher solicits a Procedure and the child responds using a Specific Recount. The children's use of probes to cause the researcher to solicit Records of Events is a third observation that supports that children prefer to use Records of Events of the Recount genres. The use of probes is found twice in text 9. The first probe appears in a General Recount while the second probe appears in a Procedure. Although the second probe is found in a Procedure, the utterances that fol low the probe appear to describe a Record of Events. These three observations suggest that children with A S D may have a preference for using Records of Events in spoken discourse. The children use a total of 19 Orientation in the 9 texts. Those who use the most Orientations are the speakers in texts 3, 7, and 8. The child in text 3 uses 9 Orientations and the children in text 7 and 8 each use 3 Orientations. In each of the remaining texts (1, 2, 4, 5, 6, and 9) research participants use 2 or less Orientation stages. Stil l a further distinction should be made between those who produce solicited and unsolicited Orientations. The children who use unsolicited Orientation stages are found in texts 3, 6, and 7 (producing 2, 1 and 3 instances respectively). As noted earlier, speakers use Orientation stages to establish what is to fol low in terms of people, actions, times and places in text; therefore, it can be suggested that children who use unsolicited 53 Orientations may have stronger conversational abilities. Another observation helps confirm that the children in texts 3, 6, and 7 who use unsolicited Orientations have stronger conversational abilities. These children are the only speakers with A S D among the 9 texts who describe multiple events in serial order while engaged in Record of Events stages. Evaluations, the third most used generic stage by the children, may serve as another indication of conversational abilities. Evaluations offer appraisal and serve as running commentary between generic stages. The research participants in texts 3, 4, 5, and 6 use Evaluations. Those in texts 1 ,2 ,7 ,8 and 9 do not. O f the Evaluations, only the children in texts 3, 4, 5 and 6 offer unsolicited Evaluations. (Child 3 utters a solicited Evaluation, child 5 utters 2 unsolicited Evaluations, and children 4 and 6 each utter a single unsolicited Evaluation.) One observation that can be made is that the children in texts 3 and 6 are the ones who use both unsolicited Orientations and Evaluations. This may suggest that they have a greater command over optional stages in generic structure. The research participants in the texts use Abstracts the least among all the generic stages examined. Abstracts establish the point of the text and signals that a story is about to be told. The children use Abstracts only twice, and both times these stages were solicited by the researcher. In contrast, the extensive use of Records of Events by the children points to their significant contributions to the Recount genres and a relative ease in this genre. A s discussed in Chapter 3, Methods, the Recount genres were subdivided into 2 sub-types; Specific Recounts and General Recounts. Including both types, 17 Recounts are identified in the 9 texts: 12 General Recounts and 5 Specific Recounts. Together, these 17 54 instances of Recount genres form the majority of the genre types of the stretches of discourse uttered by the speakers in the 9 texts. Thus patterning in the Recount genres emerges as an interesting area for description. Several patterns can be observed related to Recount genre use. The children in texts 3 and 7 appear to have greater conversational ability. This is suggested by longer dialogue between turns Chi ld 3 utters 130 lines within 60 turns. This child utters an average of 2.17 lines per turn. Chi ld 7 utters 61 lines within 19 turns. This child utters an average of 3.21 lines per turn. These averages can be compared to less conversationally engaged children who utter a single line per turn. The research participants in texts 3 and 7 are the only ones to use both unsolicited Orientations and Records of Events. They are also among the children who describe multiple events in serial order when engaged in Records of Events. The children in texts 3 and 7 also produce Records of Events that were discontinuously realized. The analysis of unsolicited generic stages is interesting because it demonstrates that usually where they occur, the child has consistent strength to utter them across multiple generic stages. Children 4, 5, 6, and 8 also use unsolicited generic stages but they are infrequent. Texts 1, 5, and 9, do not contain any Orientations produced by research participants. The child in text 1 only uses 3 solicited Records of Events. In text 5 the child only uses a single solicited Record of Events and 2 unsolicited Evaluations. The child in text 9 does not use any genre types or generic stages. This pattern provides further evidence that weak conversational ability among the children may result in solicited stages and it is consistent across multiple generic stages. It is also important to note the possibility that weaker conversational abilities result in closer adherence to 55 mandatory stages of generic structure. The child in text 5 does use 2 unsolicited Evaluations, however this is an exception. Some of the children who are weaker conversationalists appear to have coping techniques. Such techniques are evident in Texts 1, 2, and 8. In Text 1, the child provides solicited answers to questions posed by the researcher that mirror the researcher (and appear coincidental): |ORIENTATION| . RES: so do you l i k e p i z z a W i l l ? C H I : mmhm. RECORD OF EVENTS RES: I had p i z z a l a s t n i g h t f o r s u p p e r . ORIENTATION RES: what d i d you have f o r suppe r l a s t n i g h t ? |RECORD OF EVENTS C H I : I had p i z z a t o o . RES: d i d you? C H I : y e a h . RES: oh - : . RECORD OF EVENTS RES: and the n i g h t b e f o r e I had p o r k c h o p s . ORIENTATION RES: can you remember what you had the n i g h t b e f o r e ? |RECORD OF EVENTS C H I : p o r k chops t o o . RES: you had p o r k chops t o o . RES: hm. Example 4.12 - Recount 1, transcript 1, lines 138-150 Also in Text 1, the child provides a solicited response that might be considered too general as answer to fully address the researcher's question: |ORIENTATION| RES: what do you do when you go home from s c h o o l W i l l ? RECORD OF EVENTS C H I : ah -: I p l a y games. Example 4.13 - General Recount 1, transcript 1, lines 79-80 In Text 2, the child uses minimal and polar responses that do not advance the conversation: 56 O R I E N T A T I O N R E S : s o w h a t d o y o u d o o n t h e w e e k e n d s W i l l ? C H I : n o t h i n g . R E C O R D OF E V E N T ' S R E S : d o y o u h a v e a w o r k e r t h a t c o m e s ? C H I : n o p e . R E S : a n d s p e n d s s o m e t i m e w i t h y o u ? C H I : y e s . | O R I E N T A T I O N R E S : hm w h a t ' s h i s n a m e ? C H I : T o m . I R E C O R D O F E V E N T S ] R E S : a n d w h a t d o y o u d o w i t h T o m C H I : g o o u t w i t h h i m . R E S : mmhm? O R I E N T A T I O N R E S : w h a t d o y o u d o w h e n y o u ' r e o u t ? |RECORD O F E V E N T S C H I : p l a y g o l f . R E S : g o l f ? C H I : y e s . Example 4.14 - General Recount 1, transcript 2, lines 219-233 In Text 8, the child uses the elaborate technique of probing the researcher to solicit specific questions while demonstrating difficulties in other areas of the conversation. Another pattern is the unvaried use of conjunctions of the temporal type. In the 9 texts, the children largely link temporal events in sequence by using the conjunctive (not coordiating) and and the temporal simple conjunction of the sequential type (and) then. It is true that these are among the most common ways that people generally communicate temporally sequenced events. However, in Texts 3 and 7, there are stretches of discourse where the conjunctions are unvaried and almost exclusive to \"and\" and \"(and) then\", reflecting a pattern of serial ordered relations that may be characteristic in A S D (de Vi l l iers & Szatmari, 2004): IRECORD OF EVENTS| CHI: e x c u s e me a n d e x c u s e me a n d t h e n w e w e n t h o m e . CHI: a n d I w e n t o u t s i d e a g a i n . CHI: a n d I t o o k a b a t h . CHI: a n d t h e n [ ! ] I w e n t t o s l e e p . CHI: a n d t h e n I w e n t t o b e d . CHI: a n d t u r n e d o f f t h e l i g h t . CHI: t h e n e x t m o r n i n g I p a c k e d m y s u i t c a s e u p . 57 C H I : and we went back t o T o r o n t o . Example 4.15 - Excerpt from Recount 1, transcript 3, lines 64-71 |RECQRD OF EVENTS| 2 3 . C H I : we go on the v a n f i r s t . 24 . C H I : and t h e n and t h e n we t h e n we work # t h e n we work um s o m e t h i n g l i k e uh n i n e + t h i r t y t o n i n e + t h i r t y t o t w e l v e [ ! ] . 2 5 . C H I : t h e n we have l u n c h a t t w e l v e a t t h a t . 2 6 . C H I : t h e n we s t a r t back a t work a t one o ' c l o c k . 2 7 . C H I : and t h e n we go r i g h t a l l t he way t h r o u g h t o f o u r + t h i r t y . 2 8 . RES: oh - : . Example 4.16 - Excerpt from Recount 1, transcript 7, lines 23-28 Inflexible use of conjunctions in Text 7 only occurs once while it is a consistent pattern throughout Text 3. Excerpts that fol low a rigid pattern of question by the researcher and response by the child (Texts 1 and 2) did not contain any conjunctions uttered by the child because turn-taking proceeded through a question from the researcher and response from the child. That is, responses tended to be without elaboration. The researcher typically led the conversations in all texts by soliciting responses; however, this varies in degree and is less prevalent in texts where the child engages in extended discourse. The children with the less strong conversational abilities used little or no temporal conjunctions. Interestingly, children with stronger conversational abilities (Texts 3 and 7) are both children who present patterns of inflexible temporal conjunction use in their Recounts. Further discourse produced by speakers with A S D who are able to sustain longer turns and are engaged in extended Records of Events is needed to determine i f inflexible temporal conjunction use is a characteristic of speakers with A S D . A final pattern concerning temporal discourse that describes sequences of events in the future emerged. In these \"Future Projections\", the children use external conjunctive relations of the temporal type to sequence unrealized events in serial order. Future 58 Projections are not included in the genre analysis, but because of their temporal dimension, the stretches are used in Chapter 5, Metaphor. C O N C L U S I O N Genre analysis provides insight into how individuals with ASD use the contextual configuration of relevant situational information in producing spoken text. This chapter used genre analysis to approach the investigation of how individuals with ASD produce temporal texts that involve sequences of events. Specifically, the analysis of temporal stretches of text found that the research participants used the Record of Events generic stage the most frequently, followed by Orientations. These research participants may favour the Records of Events stage in temporal texts: evidence suggests that children ignore or respond to interruptions during Records of Events and then continue on with the the Record of Events. Also, children with ASD may use the Record of Events stage when other genre types are solicited by the researcher. Furthermore, individuals with ASD may also probe for questions that solicit Records of Events. This chapter identified conversational difficulties found in the spoken discourse of individuals with ASD that are illuminated with genre analysis. The level of detail used when engaging in Records of Events can be responsible for confusion between a speaker and audience. Individuals with ASD may ignore context in some cases, (for example in the use of a Recount when a Procedure is solicited); however, it is also apparent that speakers with ASD do use genre, but in different ways. The speakers looked at in this thesis largely engage in Recounts and for the most part have a mastery of the Record of Events generic stage. Those who are most competent and use both unsolicited Orientations and Records of Events (the children in texts 3 and 7) appear to be the children with stronger conversational abilities. These children engage in longer turns and use discontinuously realized Records of Events. These same children were among the 60 research participants who described multiple events in serial order when engaged in Records of Events. Probing also provides further evidence that context is important to speakers with ASD. The mechanism seeks to establish a specific contextual configuration so that the individual may utter spoken discourse in context. Individuals with weaker conversational abilities appear to rely on their interlocutor's probes to establish specific contexts. It may be that speakers with ASD have more knowledge of context than they are able to make use of in conversation. This project applied an established genre framework to discourses of speakers with ASD. While the use of an established descriptive framework was warranted, indeed necessary in a field with recognized, replicable methodologies, it does not focus on speakers' individual variation. A study replicating these findings in a larger group of speakers with ASD would provide more conclusive findings and could also be used to investigate the possibility that speakers with ASD may have their own varieties that may involve specific generic structures. 61 C H A P T E R 5 - M E T A P H O R A N A L Y S I S This chapter examines the use of metaphors of time, metonymies of time, and senses of time in stretches of temporal discourse found in 9 semi-structured conversational texts produced by speakers with Autism Spectrum Disorders (ASDs). The chapter proceeds in two parts. Part 1 describes the use of metaphors, metonymies, and senses of time in each individual text. Patterns within each text are identified. Part 2 extends upon the observations from part 1 by examining patterns in the use of metaphors, metonymies, and senses of time across the 9 texts. Utterances that use units of time found within stretches of temporal discourse are identified. These utterances are then analyzed for the use of temporal metaphors, metonymies and sense of time. The methods for analysis and their theoretical frameworks are previously outlined in Chapter 3, Methods. Metaphors and metonymies of time are classified according to Lakoff and Johnson's (1999) framework. The categories selected from Lakoff and Johnson (1999) are: Time Orientation metaphor Moving Observer metaphor Moving Time metaphor Event-for-Time metonymy Distance-for-Time metonymy Time-for-Distance metonymy Time is a Resource metaphor Time is Money metaphor Where the lexeme \"t ime\" appears, it is classified according to Evans' (2003) senses of time. Evans' (2003) senses of time are: Duration Sense Moment Sense Instance Sense Event Sense Matrix Sense 62 Agentive Sense Measurement-system Sense Commodity Sense In this chapter, utterances produced by both the child and researcher are discussed. It is important to note that although these speakers' utterances are compared in form and number, they are not produced equally due partly to the semi-structured nature of these conversational texts. 63 P A R T 1: Indiv idual Text Analysis Text 1 There are 6 utterances in text 1 that use temporal units. The researcher utters 5 of these 6 utterances. The child utters a single solicited response that uses a single temporal unit. A l l utterances use the Time Orientation metaphor where the future is construed as the space in front of the observer, the present is the observer's current location, and the past is the space behind the observer. This metaphor is prompted by the prepositions \"last\" and \"before\" and the lexeme \"tonight.\" These lexical items motivate spatial deixis where locations establish meaning in relation to each other. Table 5.1 shows this data. The temporal units are in bold and lexemes that prompt for temporal metaphors and metonymies are italicized. Line Speaker Utterance 140 RES I had p i z z a l a s t night f o r s u p p e r . 141 RES what d i d you have f o r suppe r last night? 146 RES and t he night before I had p o r k c h o p s . 147 RES can you remember what you had t h e night before? 151 RES what a r e you h a v i n g tonight # f o r s u p p e r ? 153 CHI I ' m h a v i n g r i c e f o r s u p p e r tonight. Table 5.1 - Lines 140, 141, 146, 147, 151, and 153 from text 1 In text 1, the researcher asks the child what he had for supper last night, the night before last, and what he wi l l have tonight for supper. The first of these questions establishes the temporal setting in the past (line 141). The researcher then reorients the temporal setting by moving further into the past (line 147). The researcher's final question shifts the temporal setting ahead into the future (line 151). The child's responses indicate that the child comprehends the Time Orientation metaphor. In addition, the child also comprehends the shifts in temporal setting prescribed by the researcher's questions. 64 The child uses spatial deixis to form temporal relations when the researcher shifts the temporal setting. In line 141, the researcher asks, \"what did you have for supper last night?\" The child replies, \"I had pizza too.\" Later in line 147, the researcher asks, \"can you remember what you had the night before?\" Here the researcher is implicitly asking the child what he had for supper the night before last. The child replies \"pork chops too.\" The child's responses suggest that the child first comprehends the initial shift into the past (\"last night\"). With reference to this point, the child is able to use this information to answer the researcher's next question that shifts the setting one night further into the past to \"the night before.\" Line 147 further demonstrates that the child is able to comprehend questions with temporal units in a pragmatic way. Literally, the researcher's question \"can you remember what you had the night before?\" solicits a polar yes/no response; however, the child correctly assumes that the researcher is inquiring about what the child actually ate for dinner and not i f the child is able to remember what he ate. In one instance, the child utters a response that includes the temporal unit \"tonight.\" \"Tonight\" is the only unit of time uttered by the child in text 1. The temporal unit, \"tonight\", is a bounded event in the sense that it uses deixis to reference a specific period of time. There are two possible reasons for the child's limited use of temporal units. First, the child provides minimal non-elaborative and polar responses to the researcher's questions. The conversation therefore follows a turn-taking pattern where the researcher poses questions and the child answers these questions. Second, the conversational turn-taking pattern and the child's limited use of temporal units indicates that the child may be more able to comprehend temporal metaphors compared to his ability to produce temporal metaphors. Although these two reasons may explain the lack 65 of temporal units uttered by the child, the child does demonstrate that he is able to comprehend the researcher's utterances that use temporal units in a pragmatic way. Table 5.2 summarizes the types of metaphors and metonymies that the speakers use in text 1. The number corresponding to each metaphor or metonymy represents the number of utterances that use that type of metaphor or metonymy. B •o cu icit sol c 2 2 2 S 2 2 Is !s is is U U U U U •a _: r j >. ;•> •o . _ 'c — i by >> JS Chi w >> -3 >. rb 2 Vi M> _ L. _ u Chi p P = P o o ' - ^ f Utterance ces esearcher Child Solic f Temporal by Child n Metapho n Metapho :r Metapho jr Metapho etaphor etaphor by Metonymy Metonymy ne Metonyi -Time Metonyi ice Metonyi ice Metonyi rce Metaph rce Metaph )y Child Time Time by Cl — iber o tteran OS o _ CU nits —> —' ervi ervi 4* me me -Tir -Time Metonyi stan stan sou sou >> ca CM o nse of be iber o tteran q s _\"S -3 rient rient Vi (A Ui] Ulj H H \"or or E ir-Dii cu Qi on on cn _ nse of E c 3 CU s E 3 rient rient o o H H _ o _ i o> • cu i ir-Dii cs cd Sei cu 1/1 3 z Z •a CU an z o O O BC C BC B BC B BC B _. t-fi nc1 nc • Vi [A \"c« \"cn Text i Total Ignor Utter Utt( Total Temf Time Time Movi Movi Movi Movi Even Even Dista Dista Time Time Time Time Time Time Evan B « > 1 6 0 5 1 s 6 1 6 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Table 5.2 - Summary of metaphors and metonymies used in text 1 Text 2 Text 2 contains 7 utterances that use 10 temporal units. The child utters 2 unsolicited lines that use 3 temporal units and the researcher utters 5 lines that use 7 temporal units. Lines 1, 3, 4, 6, and 8 use the Time Orientation and the Mov ing Observer metaphors. The Time Orientation metaphor is prompted by the lexemes \"today\" and \"tomorrow.\" A s seen in text 1, lexical items such as \"today\" and \"tomorrow\" motivate spatial deixis where temporal locations become relative to one another. The Moving Observer metaphor is prompted indirectly by an intermediate metaphor. Lakoff and Johnson describe this intermediate metaphor as one where numbers are points on a line 66 (1999, 155). These numerical values prompt for the Moving Observer metaphor because \"the numbers pick out points on a line that metaphorically represent instants of t ime\" (Lakoff & Johnson, 1999,155). When we combine the Time Orientation metaphor with the Moving Observer metaphor, we arrive at a composite mapping where locations on the observer's path of motion (numbers on a line) represent times. Line Speaker Utterance 1 RES today i s June the s i x t h . 3 CHI < f i f t h > [<] 4 RES i s i t the f i f t h ? 6 CHI tomorrow's the s i x t h . 8 RES t o d a y i s June the f i f t h . Table 5.3 - Lines 1, 3, 4, 6, and 8 from text 2 Lines 1 through 8 reflect the researcher mistaking the date as \"June the sixth.\" The child corrects the researcher by stating that it is June 5 t h . Lines 1 through 3 indicate that the child and researcher have different frames (or versions) of the Time Orientation metaphor: 1. RES: t o d a y i s June the s i x t h . 2 . RES: and <I> [>]. 3. C H I : < f i f t h > [<] Figures 5.1 and 5.2 compare the researcher's initial frame of the Time Orientation and Moving Observer metaphors with the child's frame of the same metaphors. 5 6 7 - » June: past present future Figure 5.1 - Researcher's frame of the Time Orientation and Moving Observer metaphors 4 5 6 -> June: past present future Figure 5.2 - Chi ld 's frame of the Time Orientation and the Moving Observer metaphors 67 In Figure 5.1, the researcher construes the present date as \"June the sixth.\" The space behind the researcher (the past) includes June the 5th, 4 t h , 3 r d . . . and the space ahead of the researcher (the future) includes June the 7 t h , 8 t h , 9 t h... The child's frame construes the values of these points differently. The child construes the present as June the \"fifth\", the past as June 4 t h , 3 r d , 2 n d . . . , and the future is June 6 t h , 7 t h , 8 t h... Both speakers' frames depend on spatial deixis whereby their present location provides contextual information that determines the values of the spaces ahead and behind the speakers. The fact that the child is able to correct the researcher demonstrates three competencies. First, the child is able to construe his own frame of the composite mapping as shown in Figure 5.2. Second, the child is able to access the researcher's frame of the composite mapping as shown in Figure 5.1. Third, the child is able to compare both frames and conclude that the researcher has assigned incorrect dates, or values, to the points of his composite mapping. The child's process of prompting two frames and comparing them occurs in lines 1 through 3. Lines 4 through 8 reflect a process where the researcher adopts the child's frame of the composite mapping: 4 . R E S i s i t t h e f i f t h ? 5 . C H I y e s . 6 . C H I t o m o r r o w ' s t h e s i x t h 7 . R E S o h o k a y . 8 . R E S t o d a y i s J u n e t h e f i f t h . In line 6, the child makes a significant contribution. The child provides the researcher with an additional access point to the child's frame of his composite metaphor. In line 4, the researcher asks \" is it the fifth?\" The child responds \"yes\" and then states \"tomorrow's the sixth.\" The child's utterances appear to serve two purposes. First, the child attempts to convince the researcher that it is June the fifth by negotiating spatial 68 deixis. Second, the child elaborates by providing additional information that serves as an access point for the researcher to the child's frame. In line 8, the researcher adopts the child's frame and agrees that \"today is June the fifth.\" Line 219 is uttered by the researcher and also uses the Time Orientation and the Moving Observer metaphors to create a composite mapping. The preposition \" o n \" in the utterance \"so what do you do on the weekends W i l l ? \" prompts for a composite mapping because the locations on the observer's path of motion represent times. Furthermore, the plural temporal unit, \"weekends\" elicits multiple points upon the observer's path of motion. Some examples of these points are marked with a \" * \" in Figure 5.3. * * * < - I I \" > past week present week next week M T W T F S S M T W T F S S M T W T F S S Figure 5.3 - Construal of \"weekends\" from line 219 of text 2 The researcher's question, \"so what do you do on the weekends W i l l ? \" asks the child to generalize the weekends along the child's path of motion. Restated, the researcher is inquiring about the habitual activities that the child does on the weekends. Among these locations are the points included in Figure 5.3; however, the researcher's question may prompt for other weekends that extend further into both the future and past. The child replies with the minimal and evasive response \"nothing\": 2 19 . RES: so what do you do on the weekends W i l l ? 220 . C H I : n o t h i n g . 2 2 1 . RES: do you have a w o r k e r t h a t comes? 222 . C H I : nope . 223 . RES: and spends some t i m e w i t h you? 224 . C H I : y e s . The researcher rephrases the question twice before the child provides an appropriate answer. The child's initial response to the researcher's question in line 219 suggests that the child may experience difficulty with temporal construal that prompt for 69 a generic understanding of temporal points. General terms, like \"weekends\" found in line 219, require a more global cognitive approach to process the temporal construal. They require the child to elicit all weekends and then select general events that occur on the majority of the weekends. Although, the child is able to entertain two specific frames as show in lines 1-8 of this text, he may experience difficulty with generalizations that elicit a multitude of unspecified points. \"T ime\" is the final temporal unit used in line 223 of text 2. The researcher asks \"and spends some time with you?\" According to Lakoff & Johnson, the verb \"to spend\" and the quantifier \"some\" coupled with the lexeme \"t ime\" prompts for the Time is Money metaphor. It is based on the Resource Schema where a purpose requires an amount of a resource (Lakoff & Johnson, 1999,162). In the Time as Money metaphor, \"spends\" prompts for a purpose that requires money. When mapped onto the time domain, this purpose is one that requires time. This metaphor had achieved a level of conventionality whereby it has been lexicalized as a literal expression in English. Alternatively, fol lowing Evans' (2003) the lexeme \"t ime\" in this case prompts for the Commodity Sense of time. This sense construes time as a valuable commodity. The child comprehends the Time is Money metaphor (Lakoff & Johnson, 1999) or the Commodity Sense (Evans, 2003) of time and replies with the minimal, polar response \"yes.\" The utterances in text 2 provide insight into this child's comprehension and use of temporal metaphors. In lines 1 through 8, the child contributes utterances that are unsolicited and he elaborates upon these voluntary utterances. These lines use the most basic metaphors of time that include the Time Orientation and Mov ing Observer metaphors to create composite mappings. The child's voluntary and elaborative 70 utterances suggest that the child is competent with these metaphors of time. Again, in line 219, the same metaphors of time are used; however, they construe time in a generalized way. The child appears to have difficulty when these metaphors elicit an indefinite number of temporal points and requires the child to generalize. Line 223 uses the Time is Money metaphor or the Commodity Sense of time. The child appears to comprehend this metaphor/sense of time but responds with a non-elaborative, polar response. The child's minimal, polar response to this metaphor/sense of time may suggest that he has a stronger ability to comprehend this metaphor than to produce it. Table 5.4 summarize the metaphors, metonymies, and senses of time of time that the speakers use in text 2. a isolicited 2 ted (S) / Units by Chi by Chi Child by Chile >> iy by CI iy by CI >r by Cl 2 Text Number Total Number of Utterances Ignored Utterances Utterances by Researcher Utterances by Child Solici Total Number of Temporal Temporal Units by Child Time Orientation Metaphor Time Orientation Metaphor Moving Observer Metaphoi Moving Observer Metaphoi Moving Time Metaphor Moving Time Metaphor by Event-for-Time Metonymy Event-for-Time Metonymy 1 Distance-for-Time Metonyn Distance-for-Time Metonyn Time-for-Distance Metonyn Time-for-Distance Metonyn Time is a Resource Metaphi Time is a Resource Metaph< Time is Money Time is Money by Child Evans' Sense of Time Evans' Sense of Time by Ch 2 1 2 7 0 5 U 0 3 6 2 6 2 0 0 0 0 0 0 0 0 0 0 1 0 1 0 Table 5.4 - Summary of metaphors and metonymies used in text 2 71 Text 3 There are 26 utterances that use 35 temporal units in text 3. The child utters 16 lines; 12 of these lines are unsolicited by the researcher and 4 are solicited. The researcher utters 10 lines that use temporal units. The child utters lines 45, 70, and 115 (see Table 5.5). These utterances use the Time Orientation metaphor. In these lines, the adjective \"next\" that precedes each temporal unit prompts for the Time Orientation metaphor as it situates the present at the observer's current position with the future located ahead of the observer. These expressions motivate spatial deixis. In the following tables, temporal units are in bold while lexemes that prompt for metaphors and metonymies of time are italicized. Line Speaker Utterance 4 5 C H I t h e n e x t day I w e I w e n t t o t h e b o o k s h o p . 7 0 C H I t h e next morning I p a c k e d m y s u i t c a s e u p . 1 1 5 C H I a n d t h e n e x t morning w h e n I w o k e u p I h a d s o m e b r e a d Table 5.5- Lines 45, 70, and 115 from text 3 Lines 21, 43, 127, 176, 177, 205, 206, 209, 210, and 222 also use the Time Orientation metaphor. 72 Line Speaker Utterance 2 1 C H I on on Tuesday a n d Wednesday J u d y W i n s t o n a n d J w e b o t h w e n t 4 3 C H I a n d w e w e n t o u t s i d e at night a n d p l a y e d i n t h e s n o w u m < a n d > [>] . 1 1 2 7 C H I < t h e y ' r e c o m i n g > [>] a n ( d ) v i s i t a n d v i s i t i n g me in t h e summer in July [!] 1 7 6 R E S t e l l me a b o u t y o u r u h y o u d o s o m e t h i n g on Saturday ' s u s u a l l y . 1 7 7 R E S w h a t d o y o u u s u a l l y d o on Saturdays? 2 0 5 C H I a n d h a d a N i e l s e n L P w h i c h i s c a l l e d T h e P o i n t r e l e a s e d in 1985 l i k e l i k e o u r v i d e l i k e o u r v i d e o T h e P o i n t w h i c h w a s a w h i c h w a s r e p l a c e d a t t h e v i d e o b o x . 2 0 6 C H I a n d b y a n d o n e a n d o n c e m y m o t h e r a n d I g o t i t in u m July a n d August # b o u g h t i t i n June a n d August. 2 0 9 C H I b u t t h e v i d e o w a s n o t r e l e a s e d i n 1985. 2 1 0 C H I i t w a s u h r e l e a s e d b e f o r e t h e L P in 1984 - : ## u h 2 2 2 R E S t e l l me w h a t y o u r f a v o u r i t e - : t h i n g i s t o d o in t h e summer time. Table 5.6 - Lines 21, 43, 127, 176, 177, 205, 206, 209, 210, and 222 from text 3 The utterances in Table 2.6 combine the Time Orientation metaphor with the Moving Observer metaphor to form composite mappings. Both the Time Orientation metaphor and the Moving Observer metaphors are prompted by the prepositions \" i n , \" \"at,\" or, \" o n \" that precede the temporal units. These prepositions prompt for a construal where the observer arrives at temporal points along a path of motion. The child utters 7 of the 11 lines that use the Time Orientation and Moving Observer metaphors while the researcher utters 3 lines. Lines 176 and 177 uttered by the researcher require a generalized construal of time similar to line 219 in text 2. The researcher repeats the question twice before the child provides an appropriate response in line 178. This observation suggests that the child in text 3, like the child in text 2, may experience difficulties with utterances that elicit multiple temporal points and require the child to generalize across these points. 73 In line 222, the researcher uses the lexeme \"summer time.\" Although this unit of time is underdefined, the child replies to the researcher's question with an elaborative response. The child utters line 82 and the researcher utters line 107. Both lines 82 and 107 use the Time Orientation metaphor plus additional metaphors. Line Speaker Utterance 8 2 C H I m y f a t h e r w e n t t h e r e w a n m y f a t h e r a n d I w e n t t h e r e o n c e b e f o r e g r a d e # w e n t t h e r e o n c e before grade five started. 1 0 7 R E S d i d y o u s l e e p during the car trip? Table 5.7 - Lines 82 and 107 from text 3 In line 107, the researcher uses the Time Orientation and Mov ing Observer metaphors. These metaphors are prompted by the mapping where the motion of the observer (car trip) is understood as the \"passage of time\" (Lakoff & Johnson, 1999, 146). This expression also uses the Event-for-Time metonymy where the \"car trip\" also stands for an amount of time during which the observer could sleep. The child appears to comprehend the researcher's question and replies to it with a minimal, polar response. Line 82, uttered by the child uses the Time Orientation metaphor. The Time Orientation metaphor is prompted by the preposition \"before.\" The preposition suggests a temporal point located in the past. Line 82 also uses the Event-for-Time metonymy. The event, when \"grade five started,\" references a temporal moment. In addition to these metaphors and this metonymy, the child's utterance expresses Evans' Instance Sense of time. Recall that the Instance Sense \"prompts for a reading in which an instance of a particular event, activity, or state is being referenced...\" (Evans, 131). The lexeme \"once\" prompts the Instance Sense because it is implicitly understood as \"one time.\" The child is therefore not only able to construe the Instance Sense of time, but also is able to 74 articulate a contextually appropriate form of this sense of time. Line 206 shown in Table 2.6 also uses the lexeme \"once.\" Lines 94, 96, 105, and 191 use the Time Orientation and Moving Observer metaphors. In addition to these metaphors, the utterances also use the Event-for-Time metonymy and the Distance-for-Time metonymy. L i n e Speaker Utterance 94 RES was i t a long d r i v e ? 96 CHI i t was a long d r i v e . 105 CHI and t h e c a r t r i p w a s long [!] . 191 RES hm how long have you been d o i n g t h a t ? Table 5.8 - Lines 94, 96, 105, and 191 from text 3 In these utterances, the events, \"drive,\" \"trip,\" and \"horseback riding\" (the pronoun \"that\" in line 191 refers to \"horseback riding\" found in line 182) metonymically stand for stretches of time during which these events occurred. In all cases, \"long\" modifies the event and prompts the Distance-for-Time metonymy. These metonymies presuppose the Time Orientation and the Moving Observer metaphors as the events are construed in terms of physical distance along the observer's path of motion. The child's utterance in line 96 is in response to the researcher's question in line 94. The child utters line 105 as a continuing turn and an elaborative statement. In line 192, the child comprehends and responds to the researcher's question from line 191: 191. RES: hm how l o n g have you been d o i n g t h a t ? 192. CHI: I have been d o i n g i t f o r t w e l ( v e ) I have d I have done i t f o r t w e l v e r i d e s -: . In line 192, the child combines the Time Orientation metaphor with the Moving Observer metaphor. The child's utterance also uses the Event-for-Time metonymy where a single ride stands for a specific amount of time. The child construes the length of time he has participated in the activity of horseback in terms of 12 occasions of horseback riding. 75 These 12 occasions of horseback riding are located in the past and they exist as points along the child's temporal path. These points prompt for Time Orientation and the Mov ing Observer metaphors. Figure 5.4 shows this construal. *1 *2 *3 *4 *5 *6 *7 *8 *9 *10 *11 *12 past present future * denotes \" a r ide\" Figure 5.4 - \"Twelve rides\" construal of time in line 192 of text 3 \"Twelve rides\" expresses a length of time in an unconventional way. A typical speaker would likely describe the same experience using the term \"twelve times.\" The lexeme \"t ime\" in this context would be considered an example of Evans' Instance Sense of time. Line 192 suggests that the child chooses to construe time using the Event-for-Time metonymy, in terms of a sequence events, as opposed to simply the number of times the same event reoccurred. The child's construal suggests that the \"twelve rides\" are heterogeneous where as a typical speaker's response would use \"twelve times\" that construes the rides as a set of homogeneous events. This observation is closely related to the children's difficulties with generalizations seen already in texts 2 and earlier in text 3. Generalizations require a construal where each instance is construed as a reoccurrence of the same event. It is possible that speakers with A S D do not view these events as reoccurrences; rather, they are construed as discreet and dissimilar events. The child's response in line 192 does not clearly answer the researcher's question from line 191. It can be inferred that the child has been going horseback riding for 12 weeks when lines 176 through 178 are taken into account: 1 7 6 . R E S : t e l l me a b o u t y o u r u h y o u d o s o m e t h i n g o n S a t u r d a y ' s u s u a l l y . 1 7 7 . R E S : w h a t d o y o u u s u a l l y d o o n S a t u r d a y s ? 1 7 8 . C H I : I g o h o r s e b a c k r i d i n g . 76 Although lines 176 to 178 establish that the child usually goes horseback riding on Saturdays, the length of time the child has been horseback riding remain uncertain. Line 139 and line 144 again use metonymy and the Time Orientation/Moving Observer composite mapping. The researcher uses the Time-for-Distance metonymy. These metaphors and this metonymy are prompted by the proposition \"from\" or the verb \"to drive\" coupled with the temporal unit \"half an hour\": 139. RES: when you were t h e r e you were o n l y about a h a l f an hour from a farm where I yu I grew up. 140. CHI 141. RES 142. RES 143. CHI 14 4. RES 145. RES 146. RES 147. CHI what farm i s t h a t ? w e l l t h a t ' s where my mom and dad owned a farm, and I l i v e d when I was a l i t t l e g i r l . yes [ ! ] . and my mom used t o d r i v e a h a l f an hour, and t e a c h i n K i n g s b u r y , she t a u g h t h i g h s c h o o l . <did she> [>] ? The Time-for-Distance metonymy reverses the source and target domains of the Distance-for-Time metonymy where \"time duration can stand metonymically for distance\" (Lakoff and Johnson, 1999, 152). Lakoff and Johnson explain \"half an hour, the time it takes to travel the distance, stands for the distance\" (1999, 152). In line 139 and 144, the researcher uses time to metonymically stand for a distance travelled. The child appears to comprehend these metaphors as he prompts the researcher to elaborate in line 140 and 147. Lines 98, 101, 103, and 170 use the Time as a Resource metaphor. Line 170, uttered by the researcher, is a classic example of the Time is a Resource metaphor: 170. RES: and we need t o do about # t h r e e o r f o u r more m i n u t e s . The adjective \"more\" prompts for the Resource Schema where a purpose requires an amount of a resource (Lakoff & Johnson, 1999, 161). In the Time is a Resource 77 metaphor, the purpose that requires the resource is mapped onto the time domain where it becomes a purpose that requires time. The child appears to comprehend this metaphor as he responds, \"yes Dor is\" in line 174. Lines 98, 101, and 103 also use the Time is a Resource metaphor along with other metaphors and metonymies. The researcher asks a question in line 98 and the child responds to this question in line 101. The child repeats his utterance from line 101 in line 103: 9 8 . R E S : h o w l o n g d i d i t t a k e y o u W i l l ? 9 9 . R E S : < j u s t a b o u t > [>] . 1 0 0 . C H I : < i t > [<] . 1 0 1 . C H I : i t t o o k i t t o o k w o i t t o o k f i v e h u n d r e d a n d s i x t y f o u r m i n u t e s . 1 0 2 . R E S : d i d i t r e a l l y ? 1 0 3 . C H I : i t t o o k f i v e h u n d r e d a n d s i x t y f o u r m i n u t e s . Lines 98, 101, and 103 prompt for the Time is a Resource metaphor because they use the verb \"to take.\" This verb uses multiple elements and scenarios from the Resource Schema. Specifically, time is a resource and the child is the user of this resource. The purpose is a trip and it requires (\"takes\") an amount of the time resource. The child must \"use up\" an amount of time to make this trip. The result is a portion of time that has been used/taken away and is no longer available to the child. Line 98 uses the Time Orientation metaphor, the Moving Observer metaphor, and the Distance-for-Time metonymy in addition to the Time is a Resource metaphor. \" L o n g \" prompts for the Distance-for-Time metonymy while \"it\" refers to a drive and prompts for the Time Orientation and the Moving Observer metaphors. Line 101 uses the Time Orientation metaphor and the Moving Observer metaphor again prompted by \"it\" in addition to the Time is a Resource Metaphor. 78 Lines 101 and 103 uttered by the child use temporal units in an unconventional way. The child uses \"f ive hundred and sixty four minutes\" to describe the amount of time a trip takes. This construal divides the trip into 564 increments of one minute. We would expect a typical speaker to use larger increments of time; for example, a speaker may use \"9 hours and 24 minutes\" or alternatively approximate the time of the trip as \"9 and a half hours.\" Yet again, the child's discourse points to possible difficulties with generalization. A pattern emerges from the child's utterances in lines 101, 103, and 192. Although lines 101, 103, and 192 are construed using different metaphors and senses of time, they demonstrate the child in text 3 has the tendency to divided units of time or events that metonymically stand for time into smaller, than usual increments. In lines 101 and 103 the child construes 9 hours and 24 minutes as \"f ive hundred and sixty four minutes.\" In line 192, the child construes a length of time in terms of 12 rides. It appears that the child has the tendency a to construe continuous experiences or larger units in term of smaller increments. A second observation points to a discrepancy between the construals presented in lines 82 and 192. In line 192, the child uses \"rides\" as opposed to the typical Instance Sense construal that would use the lexeme \"time.\" In lines 82 and 206, the child not only uses the Instance Sense, but also uses it in a contextually appropriate way. The child must comprehend Evans' Instance Sense of time because he uses the variant \"once\" in lines 82 and 206 but avoids its use in line 192.1 also note that in both cases, the construals that involve the Instance Sense of time are preceded by false starts. It is possible that the child in text 3 comprehends Evans' Instance Sense but may experience difficulties in articulating the sense using the actual lexical item \"t ime.\" A possible 79 explanation is the fact that the child's construal of \"r ides\" suggests a heterogeneous sequence of events. This construal of time is unlike Evans' Instance Sense of time that motivates a homogeneous understanding of events. In general, the spoken discourse produced by the child in text 3 demonstrates an extensive ability to comprehend and articulate multiple construals of time. They include Lakof f and Johnson's Time Orientation, Moving Observer, and Time is a Resource Metaphors. Also the child is also able to understand and use the Event-for-Time Metonymy, the Distance-Time Metonymy, and the Instance Sense of time. In the case of the Instance Sense of time, the child does not explicitly use the lexeme \"time.\" In addition to the child's strength in the use of these temporal expressions in spoken discourse, the child is also able to engage in extended conversation using unsolicited, elaborative, and continuing turns that involve temporal units; however, there remain several instances where the child still provides minimal or polar responses in response to utterances that contain temporal units. Table 5.9 provides a summary of the metaphors, metonymies, and senses of time used in text 3. The child's utterances that use the variant \"once\" o f the Instance Sense of time are included in the Table 5.9. 80 § solicited Un •a Event-for-Time Metonymy by Child 2 2 m 2 Text Number Total Number of Utterances Ignored Utterances Utterances by Researcher Utterances by Child Solicited (S) / Total Number of Temporal Units Temporal Units by Child Time Orientation Metaphor Time Orientation Metaphor by Chil Moving Observer Metaphor Moving Observer Metaphor by Chi Moving Time Metaphor Moving Time Metaphor by Child Event-for-Time Metonymy Event-for-Time Metonymy by Child Distance-for-Time Metonymy Distance-for-Time Metonymy by Cl Time-for-Distance Metonymy Time-for-Distance Metonymy by Cl Time is a Resource Metaphor Time is a Resource Metaphor by Cl Time is Money Time is Money by Child Evans'Sense of Time Evans' Sense of Time by Child 2 1 4 S 12 3 2 2 1 2 1 3 6 0 0 U 5 3 4 5 0 2 0 0 7 4 5 2 2 0 4 2 0 0 3 2 Table 5.9 - Summary of metaphors and metonymies used in text 3 Text 4 Text 4 contains 10 temporal units found in 7 utterances. The child utters 5 lines; 4 of these 5 lines are unsolicited while a single line is solicited by the researcher. The researcher utters 2 lines that use temporal units in text 4. Lines 134, 151, and 153 contain 4 temporal units and use the Time Orientation metaphor. In line 134, the child's utterance construes \"last year and the year before\" as increments along a linear path located in the past. This utterance is elaborative and relies on spatial deixis. The child's construal of time in line 134 divides the period of the past 2 years into one-year increments. The construal of time is unconventional as it would be more typical to state \"last two years\" as opposed to \"last year and the year before.\" L ike the example in line 192 in text 3, this construal of time suggests that \"last year and the year before\" are heterogeneous periods of time. The more typical expression \"two years\" construes the period as homogeneous. In the following tables, temporal units are in bold and lexemes that prompt for metaphors and metonymies are italicized. 81 Line Speaker Utterance 1 3 4 C H I b u t s h e h a s n ' t v i s i t me l a s t year a n d t h e year b e f o r e . 1 5 1 C H I a n d I ' m w i t ' s s o m e t h i n g f o r me t o look forward to. 1 5 3 R E S f o r e v e r y o n e t o look forward to e h ? Table 5.10- Lines 134, 151, 153 from text 4 The child utters line 151 and the researcher repeats the same information in line 153. In line 151, the child construes time using the Time Orientation metaphor because he talks about a point ahead of him as \"something for me to look forward to\" in the future. Line 151 is elaborative and is produced by the child without solicitation. The researcher mirrors the child's construal in line 153. Line 140 and line 144 use a composite mapping that combines the Time Orientation metaphor with the Moving Observer metaphor. Line Speaker Utterance 1 4 0 C H I < s h e ' s > [<] v i s i t i n g me at Christmas. 1 4 4 R E S i s s h e c o m i n g Christmas day? Table 5.11- Lines 140 and 144 from text 4 The child utters line 140 and the preposition \"at\" that precedes the temporal unit \"Christmas\" prompts for the composite mapping. In line 144, the researcher utters \"is she coming Christmas day?\" The conversational nature of this spoken text allows the researcher to skip the preposition \"on\" that would likely precede the temporal unit \"Christmas day.\" In line 144, the \"implied\" preposition prompts for the Time Orientation and Moving Observer metaphors. The child appears to comprehend the researcher's question in line 144 and replies with an elaborative response in line 145: 1 4 4 . R E S : i s s h e c o m i n g C h r i s t m a s d a y ? 1 4 5 . C H I : s h e ' s s p e n d i n g t h e n i g h t s s p e n d i n g n i g h t s t h e r e . Line 145 uses the Time is Money metaphor. This metaphor is prompted by the verb \"to spend.\" \"She,\" (the child's cousin) is visiting the child during Christmas and therefore will be spending time with the child. Two observations provide insight into the child's 82 . unconvent ional construal o f t ime i n this utterance. Firs t , the c h i l d is over ly general stating that \"she's spending . . . nights there.\" A l t h o u g h this may be true, typ ica l ly we w o u l d expect a speaker to quantify the number o f nights dur ing w h i c h the cous in w i l l vis i t . T y p i c a l speakers w o u l d use this convent ional ized expression o f t ime i n an incremental way ; however , the c h i l d does not do this. The utterance involves a false start that may suggest d i f f icul ty i n p roduc ing the T i m e is M o n e y metaphor. Second, the c h i l d uses the pronoun \"there\" to express where his cous in w i l l spend her t ime. T y p i c a l l y , i f someone were to v is i t over Chris tmas, we w o u l d use the p rox ima l p ronoun \"here\" as opposed to the distal p ronoun \"there.\" In this expression, the c h i l d appears to have diff icul t ies w i t h dietic reference. These diff icult ies w i t h both spatial deixis and the T i m e is M o n e y metaphor may be rooted i n the l ex ica l i zed expression associated w i t h \"spending\" t ime. T h i s expression requires contextual information for it to be used appropriately. A l t h o u g h we observed that the c h i l d i n text 2 was able to comprehend an expression that uses the verb \"to spend\" w i t h concepts o f t ime, this is the first occas ion where a c h i l d attempts to produce this k i n d o f metaphorical expression. Moreove r , the response the c h i l d provides i n l ine 145 does not answer the researcher's question i n l ine 144. The f inal temporal unit is found i n l ine 190. The c h i l d utters, \"but first y o u have to wai t t i l l the w a # for the water to b o i l [!]\" Th i s utterance combines the T i m e Orientat ion metaphor w i t h the M o v i n g T i m e metaphor. The point where \"the water b o i l s \" is construed as being ahead o f the observer. The observer remains stationary and waits for the point to pass the observer. T h i s utterance is not sol ic i ted by the researcher. In text 4, the c h i l d demonstrates the abi l i ty to use and comprehend a variety o f metaphors. The c h i l d is able to use and comprehend the most basic metaphors o f t ime. 83 They include the Time Orientation, the Moving Observer, and the Moving Time metaphors. In addition, the child uses the Time is Money metaphor; however, it is construed in an unconventional way. The child is too general in expressing the time his cousin wi l l spend visiting him and demonstrates difficulty with deixis. Table 5.12 summarizes the temporal metaphors and metonymies the speakers use in text 4. u CD E H Table 5.12 - Summary of metaphors and metonymies used in text 4 Text 5 In text 5, the researcher utters 2 lines that use temporal units. These temporal units are found in lines 9 and 16. Line 9 uses the temporal unit \"night\" in a question. The adjective \"last\" that precedes \"night\" prompts for the Time Orientation Metaphor as it locates the past as the space behind the observer. \"Last night\" becomes a point behind the observer. The researcher uses a check, \"didn't I?\" The child comprehends and replies to the researcher's question and check with a polar \"yeah\" in line 10. 84 Line Speaker Utterance 9 R E S I t a l k e d t o y o u o n t h e t e l e p h o n e l a s t night t o o d i d n ' t I ? 1 6 R E S t h a t s o u n d s l i k e q u i t e a long drive. Table 5.13 - Lines 9 and 16 from text 5 The second utterance uses the temporal unit \" long drive.\" The utterance prompts for a construal that uses the Event-for-Time metonymy where the \"dr ive\" metonymically represents a length of time. This length of time is modified by the preceding adjective \" long\" that prompts for the Distance-for-Time metonymy. The \" long\" distance travelled during the ride metonymically stands for an extended duration of time. This utterance also uses the Time Orientation and Moving Observer metaphors because the distance travelled by the observer is an amount of time that has passed. Line 16 is followed by a continuing line uttered by the researcher. It cannot be determined i f the child comprehends this metaphor. The child does not utter any temporal units in text 5. The child demonstrates that he comprehends the construal of time in line 9 that uses the Time Orientation metaphor. Although the child elaborates and provides unsolicited utterances, the child replies with minimal responses to questions that contain temporal units posed by the researcher. Table 5.13 summarizes the metaphors and metonymies speakers use in text 5. 85 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 u Table 5.14 - Summary o f metaphors and metonymies used i n text 5 Text 6 Text 6 does not contain any utterances that used temporal units. Text 7 Text 7 contains 23 utterances. S i x o f these 23 utterances are ignored. L i n e s 20 and 22 are ignored because they use the id iomat ic expression \"quite a t r ip . \" L i n e s 30 and 36 are ignored because they use the term \"four+forty+five bus\" where t ime modif ies the bus. A l s o , l ines 121 and 148 are ignored because they use the contract ion \"sportsnight.\" The remain ing 17 utterances contain 22 temporal units. The c h i l d utters 11 o f the 17 l ines; 2 lines are sol ic i ted by the researcher w h i l e 9 are not. The researcher utters 6 lines. L i n e 12 is uttered by the researcher and uses the temporal unit \" n o w . \" T h i s lexeme prompts for the T i m e Orientat ion metaphor because it situates the researcher and the c h i l d at a point o n the observer 's path that is the present t ime. The lexeme \" n o w \" also motivates spatial deixis . The ch i ld comprehends this construal o f t ime as he provides a polar response, \"yes .\" The c h i l d then elaborates: 86 1 2 . R E S : a n d a r e y o u w o r k i n g n o w ? 1 3 . C H I : u m - : y e s . Lines 14, 25, 26, 71, 72, 74, 76, 113, 114, 115, 120, and 150 (12 utterances) prompt for the Time Orientation and the Moving Observer metaphors. The child utters 9 of these 12 utterances. The researcher utters 3 lines. These utterances use temporal units that are preceded or followed by prepositions. The prepositions function to orient the observer either \"on , \" \"at,\" \" i n , \" or \"after\" the temporal unit that is located along the observer's path. The following table summarizes the utterances that prompt for the Time Orientation and Moving Observer metaphors. In the following tables, the temporal units are in bold and prepositions are italicized. Line Speaker Utterance 1 4 C H I I w o r k a t u m o n Tuesdays a n d Thursdays a t a t u m M a i l b o x e s . 2 5 C H I t h e n we h a v e l u n c h at twelve a t t h a t . 2 6 C H I t h e n we s t a r t b a c k a t w o r k at one o'clock. 7 1 R E S w h a t t i m e d o y o u l e a v e h e r e in t h e morning? 7 2 C H I um - i t ' s u s u a l l y after eight. 7 4 C H I between eight+fifteen a n d eight+thirty w e u s u a l l y l e a v e . 7 6 C H I a n d w e g e t u h t o M a i l b o x e s a t nine+thirty. 1 1 3 R E S s o w h a t d o y o u u s u a l l y d o in t h e evenings w h e n y o u g e t b a c k h e r e ? 1 1 4 R E S w h a t d o y o u d o a f t e r supper? 1 1 5 C H I w e l l Tuesdays after w o r k i n g w e h a v e t o d o a w o r k o u t 1 2 0 C H I a n d on thurs w e l l # w e l l on Thursdays I u s e d t o g o t o um t o 1 5 0 C H I b u t I d o o n e # b u t um on Sundays I g o t o u h m y f i t n e s s Table 5 . 1 6 - L i n e s 14, 25 ,26 ,71 ,72 , 74, 76, 113, 114, 115, 120, and 150 from text 7 Lines 71, 114, and 115 that use the preposition \"after\" are cases of the Moving Observer metaphor because of a prior preposition that precedes these utterances. For example, in line 71, the researcher asks, \"What time do you leave here in the morning?\" The child responds in line 72, \"um-: it's usually after eight.\" The child's response uses the Moving Observer metaphor because \" i n \" from line 72 contextual establishes the same construal in line 73. 87 In lines 24, 27, and 53, the speaker construes time using the Time Orientation metaphor, the Moving Observer metaphor, and the Distance-for-Time metonymy. Line Speaker Utterance 24 CHI and t h e n and then we then we work # th e n we work um something l i k e uh nine+thirty to nine+thirty t o twelve [ ! ] • 27 CHI and t h e n we go r i g h t a l l t h e way t h r o u g h t o four+thirty. 53 RES t h a t makes q u i t e a long day doesn't i t ? Table 5 . 1 7 - Lines 24, 27, and 53 from text 7 In line 24, the child utters \"and then and then we then we work # then we work um something like uh nine+thirty to nine+thirty to twelve [!]\" The child construes himself as an observer upon a temporal path and faces forward as described by the Time Orientation metaphor. The child travels from the point \"nine+thirty\" to \"twelve\" and the distance that the child travels between these points metonymically represent a length of time. Line 27 is also uttered by the child and prompts for the same construal. The child states, \"and then we go right all the way through to four+thirty.\" This utterance construes the passing of time as movement along a temporal path and the distance travelled metonymically represents a length of time. The temporal unit \"day\" is modified by the adjective \" long\" in line 53. Again, in this expression the speaker metonymically uses distance to stand for a length of time. In lines 29 and 71, the researcher uses the lexeme \"t ime.\" According to Evans (2003), both utterances use the Moment Sense of time where a discrete punctual point is conceptualized without reference to its duration. The child appears to comprehend both cases of the Moment Sense as the child provides elaborative responses: 29. RES: and what time does t h a t g et you back h e r e ? 30. CHI: w e l l we t a k e t h e um f o u r + f o r t y + f i v e bus. 31. CHI: um -: and t h a t i s t h o s e um t h o . 32. CHI: w e l l i t ' s not F u n t r a c k . 88 7 1 . R E S : w h a t t i m e d o y o u l e a v e h e r e i n t h e m o r n i n g ? 7 2 . C H I : , um - : i t ' s u s u a l l y a f t e r e i g h t . 7 3 . R E S : u h h u h ? 7 4 . C H I : b e t w e e n e i g h t + f i f t e e n a n d e i g h t + t h i r t y w e u s u a l l y l e a v e . 7 5 . R E S : m m h m . 7 6 . C H I : a n d w e g e t u h t o M a i l b o x e s a t n i n e + t h i r t y . A particularly interesting observation that differentiates this child from others is this child's ability to generalize using metaphor, metonymies, and senses of time. In this text, many of the utterances that contain metaphors, metonymies and senses of time discuss events that occur on a regular basis. Line 113 and line 114, however, do point out a recurring difficulty that the children with A S D appear to have with generalizations. In line 113, the researcher asks, \"so what do you usually do in the evenings when you get back here?\" The child does not respond and the researcher rephrases the question as \"what do you do after supper?\" This pattern of rephrasing questions that involve generalization was also observed in lines 176 and 177 in text 3 and lines 219 through 223 in text 2. Nonetheless, the child in this text produces and comprehends a variety of temporal metaphors and metonymies in text 7. They include the Time Orientation metaphor, the Moving Observer metaphor, and the Distance-for-Time metonymy. The child is particularly competent with the composite mapping that combines the Time Orientation and Moving Observer metaphors. These metaphors were used together in all of the child's utterances. The child appears to comprehend Evans' Moment Sense of time. He provides elaborative and extended responses to questions posed by the researcher. Table 5.18 provides a summary of the metaphors, metonymies, and senses of time the speakers use in text 7. 89 X E 3 z o H US o c 6 D 3 c 3 s •a X I 9 U 'E \"3 _ o a. E o 3 Z o H c 3 o C L E c o E H X I 2 s o E H Xt O ex c \"> o X o X I a. X o or B > O o X a. E H en B X i o a. E H OD E > •o >> E B O > S 5 U >-> XI >> E x x X Q _ OS E H c o 2 E H B E H E H —• o '0 Table 5.18 - Summary o f metaphors and metonymies used i n text 7 Text 8 Text 8 contains 4 utterances that use temporal units. L i n e 58 is ignored because the c h i l d utters the temporal unit \"midn igh t \" that appears to be part o f the proper name \"midn igh t express.\" O f the remain ing 3 utterances, the c h i l d utters a single sol ic i ted l ine w h i l e the researcher utters 2 l ines. The c h i l d uses 2 temporal units i n l ine 136 and the researcher utters a single temporal unit i n l ine 137. In both lines 136 and 137, the speakers use the temporal unit \"a summer day.\" In both l ines, this unit is preceded by the preposi t ion \" o n . \" T h i s propos i t ion construes \" a summer day\" as a temporal point upon an observer 's path and prompts for the T i m e Orientat ion and M o v i n g Observer metaphors: 1 3 6 . C H I : n o w C h r i s t i n a w a s g o t t e n s t u c k i n b e t w e e n # o n a s u m m e r d a y . 1 3 7 . R E S : o n a s u m m e r d a y . In l ine 136, the c h i l d appears to have dif f icul ty expressing when Chr i s t ina was \"stuck i n between.\" First , the c h i l d uses the lexeme \" n o w \" w h i c h uses the T i m e Orientat ion 90 metaphor to express the current temporal moment; however, the child then uses both the past-tense form of the verb \"to be\" and the verb phrase \"get stuck\" in the form \"gotten stuck.\" The child struggles with spatial deixis as the lexemes \"now\" and the past-tense clash instead of establishing relative temporal meaning. The child's utterance appears to be unsolicited, elaborative, and continuing in turn. The researcher's utterance in line 137 appears to repeats the child's words, \"on a summer day,\" to acknowledge the child's statement. Line 62 contains the temporal unit, \"during that train ride\" and is a question uttered by the researcher: 6 2 . R E S : I g u e s s y o u w o u l d s l e e p d u r i n g t h a t t r a i n r i d e e h ? 6 3 . C H I : y e s I w o u l d . This question construes a hypothetical situation because of the conjunction \" i f uttered by the child in line 58 and the verb \" w i l l \" in the form \"would\" in line 62 that expresses the conditional. This utterance uses the Time Orientation metaphor because the situation is unrealized and located in the future. The Time Orientation metaphor is coupled with the Moving Observer metaphor to create a composite mapping where the distance travelled by the researcher represents an amount of time that passes. Line 62 also uses the Event-for-Time metonymy as the train ride stands metonymically for an amount of time that that the child could sleep. The child comprehends the researcher's question and responds \"yes I would,\" in the following line. This response is not elaborative. In text 8, the child is able to use and comprehend both the Time Orientation and Moving Observer metaphors. The child also appears to comprehend and use the Event-for-Time metonymy. The child does produce solicited and unsolicited utterances in text 91 to H 03 a r/j e 3 3 CD <-+ 03 73 cr o >-t CO g o. 3 <D o 3 fD X O O oo Text Number Total Number of Utterances - Ignored Utterances Utterances by Researcher o Utterances by Child Solicited (S) / Unsolicited (U) 4> Total Number of Temporal Units to Temporal Units by Child U> Time Orientation Metaphor o Time Orientation Metaphor by Child u> Moving Observer Metaphor o Moving Observer Metaphor by Child o Moving Time Metaphor o Moving Time Metaphor by Child - Event-for-Time Metonymy o Event-for-Time Metonymy by Child o Distance-for-Time Metonymy o Distance-for-Time Metonymy by Child o Time-for-Distance Metonymy o Time-for-Distance Metonymy by Child o Time is a Resource Metaphor o Time is a Resource Metaphor by Child o Time is Money o Time is Money by Child o Evans' Sense of Time o Evans' Sense of Time by Child Text 9 Text 9 contains a single temporal unit uttered by the researcher in line 254: 2 5 4 . R E S : w h a t d o y o u d o a f t e r s c h o o l ? 2 5 5 . C H I : s u p p e r . 2 5 6 . R E S : s u p p e r ? 2 5 7 . R E S : y e a h ? This utterance uses the Time Orientation metaphor and the Event-for-Time metonymy. The researcher's utterances prior to line 254 construes \"after school\" as a temporal stretch of time located in the future. The Event-for-Time metonymy is prompted because the utterance uses the event of children finishing school to represent a specific temporal moment. Line 254 therefore motivates spatial deixis where a temporal relationship is established between the point that the child finishes school and the events that occur after this point. In this line, the researcher asks a question that requires the child to generalize about what he usually does after school. Although the child responds stating \"supper,\" this response is peculiar. Supper is typically located much further in the evening in relation to the time a child finishes school. Typical speakers would likely describe events that occur on a regular basis closer to the point when school finishes. A possible explanation for the child's response is that supper is the consistent event that reoccurs. It is possible that the child may recall the events that occur after school on a daily basis but is unable to generalize across the days. Supper is an event that is consistent and occurs everyday after school. It is also possible the child participates in preparing supper (shopping, cooking, etc.). Similar problems with generalizations were observed in texts 2 and 3. The child's response is minimal and he does not elaborate. The conversational turn-taking follows a question-response pattern, with questions posed by the researcher. The child provides minimal responses, does not 93 H sa CT\" ft to o I 3 o 3 CD tr o i-i CO 3 o CO C CO CD CL Text Number - Total Number of Utterances o Ignored Utterances - Utterances by Researcher o Utterances by Child Solicited (S) / Unsolicited (U) - Total Number of Temporal Units o Temporal Units by Child - Time Orientation Metaphor o Time Orientation Metaphor by Child o Moving Observer Metaphor o Moving Observer Metaphor by Child o Moving Time Metaphor o Moving Time Metaphor by Child - Event-for-Time Metonymy o Event-for-Time Metonymy by Child o Distance-for-Time Metonymy o Distance-for-Time Metonymy by Child o Time-for-Distance Metonymy o Time-for-Distance Metonymy by Child © Time is a Resource Metaphor o Time is a Resource Metaphor by Child o Time is Money o Time is Money by Child o Evans' Sense of Time o Evans' Sense of Time by Child PART 2: Analysis of All Texts This section summarizes the use of temporal metaphors, temporal metonymies, and senses of time produced by speakers across all 9 texts. General and specific patterns in the use of these devices are identified and discussed. Texts 1 through 9 contain 76 utterances that use temporal units. Seven of these 76 utterances are ignored. O f the remaining 69 lines, the child utters 35 lines that use 52 temporal units; 8 of the child's utterances are solicited while 27 are not. The researcher utters 34 lines that use 38 temporal units. Table 5.22 summarizes the utterances that use temporal units in texts 1 through 9. For a case-by-case analysis of each instance of metaphor, metonymy, or sense of time, refer to Appendix 2. Four general observations are made from Table 5.22. First, the majority of temporal units uttered by both the researcher and the child construe time using Lakoff and Johnson's (1999) Time Orientation metaphor and Moving Observer metaphor. The Time Orientation metaphor occurs in 65 of the total 69 utterances. Thirty-four of the 65 utterances that use the Time Orientation metaphor are uttered by the child. Forty-eight of the 69 total utterances use the Moving Observer metaphor. The child uses the Moving Observer metaphor 27 times. According to Lakoff and Johnson (1999) another basic metaphor of time is the Moving Time metaphor; however, the speakers in texts 1 through 9 seldom use the Moving Time metaphor. The child uses the Moving Time metaphor once in text 4. In general, the child used more of each metaphor type than the researcher. In total there are 114 times that the Time Orientation metaphor, Mov ing Observer metaphor or the Moving Time metaphor occur. Table 5.21 summarizes these observations. Table 5.22 suggests that in general the children with A S D comprehend and 95 use the Time Orientation and Moving Observer metaphors most frequently among the three most basic temporal metaphors types. Speaker Total Time Moving Moving Utterances Orientation Observer Time Metaphor Metaphor Metaphor Chi ld 35 34 27 1 Researcher 34 31 21 0 Total 69 65 48 1 Table 5.21 - Number of utterances that use basic metaphors of time in texts 1 through 9 96 Total SO oo ^4 c/1 Ul to - Text Number -i o\\ - to U) o to 1^ to os 1^ OS Total Number of Utterances ~4 O - os o o o o o o Ignored Utterances - OS o to to o vyi Utterances by Researcher to OO G GO so to C oo o o •u — C oo — to GO c to c 00 Utterances by Child Solicited (S) / Unsolicited (U) 00 so - *• to to o to o Ul *. o Os Total Number of Temporal Units u> to o o to U) Ul - Temporal Units by Child o\\ <J1 - u> o to OS to OS Os Time Orientation Metaphor t*i o - - o o to ~ Time Orientation Metaphor by Child A. oo o Ul OS o - to to o OS o Moving Observer Metaphor to -J o - - o o - to to o Moving Observer Metaphor by Child o o o o o - o o o Moving Time Metaphor - o o o o o - o o o Moving Time Metaphor by Child O - - o o - o o o Event-for-Time Metonymy •U o o o o © © o o Event-for-Time Metonymy by Child 00 o o to o - o o o Distance-for-Time Metonymy Ul o o - o o o to © o Distance-for-Time Metonymy by Child to o o © o o o to o o Time-for-Distance Metonymy o o o o o o o o o o Time-for-Distance Metonymy by Child o o o o o o o © Time is a Resource Metaphor to o o o o o o to o o Time is a Resource Metaphor by Child tO o o o o o - o - o Time is Money _ o o o o o - o o o Time is Money by Child Os o o to o o o U) - o Evans' Sense of Time N> o o o o o o to o © Evans' Sense of Time by Child A second general observation indicates that the speakers in texts 1 through 9 produced a fewer number of utterances that use metonymies of time. The Event-for-Time metonymy occurs in 10 of the total 69 utterances; the child utters 4 of these 10 lines. Eight of the total 69 utterances use the Distance-for-Time metonymy; the child utters 3 of these 8 lines. Two utterances use the Time-Distance-Metonymy and none of these lines are uttered by the child. In total, the researcher utters more lines that use each type of metonymy than the child. In total, 20 utterances use one or more types of metonymy. Table 5.23 summarizes these observations. Speaker Total Event-for Distance-for- Time-for-Utterances Time Time Distance Metonymy Metonymy Metonymy Chi ld 35 4 3 0 Researcher 34 6 5 2 Total 69 10 8 2 Table 5.23 - Number of utterances that use metonymies of time in texts 1 through 9 A third general observation indicates that speakers produce utterances that use even fewer of Lakoff and Johnson's (1999) other metaphors of time. These metaphors are the Time is a Resource Metaphor and the Time is Money metaphor. Four of the total 69 utterances use the Time is a Resource metaphor; 2 of these lines are uttered by the child. Two of the total 69 utterances use the Time is Money metaphor; the child utters 1 of these lines. In general, the child uses as many or fewer of these metaphors of time compared to the researcher. In total, the Time is a Resource or the Time is Money metaphor occur in 6 lines of text. Table 5.24 summarizes these observations. 98 Speaker Total Utterances Time is a Resource Time is Money Metaphor Chi ld 35 2 1 Researcher 34 2 2 Total 69 4 2 Table 5.24 - Number of utterances that use other basic metaphors of time in texts 1 through 9 A final general observation shows that the speakers in texts 1 through 9 use Evans' senses of time the least among the expressions examined. Six of the 69 total utterances use one of Evans' senses of time; the researcher utters 4 of these lines while the child utters 2 lines. Table 5.25 summarizes these observations. Speaker Total Utterances Evans' Senses of Time Chi ld 35 2 Researcher 34 4 Total 69 6 Table 5.25 - Number of utterances that use Evans' senses of time General findings of the use of temporal metaphors and metonymies in spoken discourse suggest that the speakers examined use the Time Orientation metaphor and the Moving Observer metaphor the most frequently among the most general metaphors of time. In the 9 texts, the child uses the Time Orientation metaphor, the Mov ing Observer metaphor and the Moving Time metaphor in utterances more frequently than the researcher. Closer examination indicates that although general temporal metaphors are overall used more often by the child than the researcher, it is not the case when each child is examined on an individual text basis. The children in texts 3, 4, and 7 are the only speakers that use the Time Orientation metaphor and the Mov ing Observer metaphor as often as or more often than the researcher. The remaining children (in texts 1, 2, 5, 6, 8, and 9) use these metaphors less often in utterances than the researcher. The child in text 4 99 is the only child that uses the Moving Time metaphor. A l l speakers examined use the Time Orientation metaphor and the Moving Observer metaphor. Preliminary findings suggest that both the researcher and child use less temporal metonymies in utterances than they use general temporal metaphors. Closer examination indicates that temporal metonymies only occur in texts 3, 5, 7, 8, and 9; however, the children in texts 3 and 7 are the only children who use temporal metonymies in their utterances. Chi ld 3 uses the Event-for-Time metonymy 4 times and the Distance-for-Time metonymy twice. Chi ld 7 uses the Distance-for-Time metonymy once. In the remaining texts where metonymies of time appear, they are uttered by the researcher. Speakers in texts 1 through 9 use the Time is a Resource metaphor and the Time is Money metaphor even less than they use temporal metonymies. These lexicalized metaphors occur in texts 2, 3, and 4. Texts 3 and 4 are the only texts where the child uses the Time is a Resource metaphor or the Time is Money metaphor. Although the children appear to comprehend the Time is a Resource metaphor when uttered by the researcher, the cases where the children express these metaphors appear to be unconventional. A possible reason for this observation is that both the Time is a Resource and Time as Money metaphors have achieved a level of literalness where they have become lexicalized expression. Lexicalized expression use is subject to contextual conventions and children with A S D do not use the Time as a Resource and Time as Money metaphors in contextually appropriate ways. The child in text 3 uses the Time is a Resource metaphor twice and is overly specific with the amount of time a trip took. The child in text 4 uses the Time is Money metaphor once and underspecifies the number of night his cousin wi l l spend visiting him. 100 F i n a l l y , E v a n s ' senses o f t ime occur the least among the types o f expressions examined. E v a n s ' senses o f t ime are used w i t h i n utterances i n texts 2, 3, and 7. The c h i l d i n text 3 is the on ly c h i l d among the 9 ch i ldren that uses E v a n s ' sense o f t ime. M o r e o v e r , recal l that this c h i l d does not use the lexeme \" t ime,\" rather, the c h i l d uses the term \"once\" that is a variant o f E v a n s ' Instance Sense o f t ime. Therefore, throughout the texts examined, none o f the chi ldren actually use the lexeme \" t ime\" i n spoken discourse. O v e r a l l , a patter emerges where chi ldren i n texts 3, 4, and 7 use the most and the greatest variety o f metaphors, metonymies, and senses o f t ime. The chi ldren that are most competent w i t h the basic metaphors o f t ime appear to be the same ch i ld ren who use temporal metonymies , the T i m e is a Resource /Time is M o n e y metaphor, and E v a n s ' senses o f t ime. The remaining chi ldren utter very l imi ted or no utterances that use expression other than the T i m e Orientat ion and M o v i n g Observer metaphors. Several more specific f indings emerge f rom the analysis o f the 9 texts. These f indings suggest 4 patterns: the chi ldren i n these texts demonstrate a greater abi l i ty to interpret as opposed to produce metaphors, metonymies , and senses o f t ime; some chi ldren experience dif f icul ty w i t h generalization, unspecif ic construal , or construal that prompt for mul t ip le reference points; t ime and temporal events are construed and expressed by the ch i ldren i n smaller than typica l units; and where typ ica l speakers w o u l d l i ke ly use the lexeme \" t ime\" and senses o f t ime the c h i l d construes t ime and senses o f t ime i n unconvent ional ways . Firs t , the majority o f ch i ldren are more able to interpret metaphors, metonymies , and senses o f t ime compared to their abi l i ty to produce these same expressions. Table 5.22 suggests that the ch i ldren w i t h A S D are more l i ke ly to produce the most general 101 temporal metaphors more often than the researcher; however , we f ind upon closer observation that this is not the case. A few chi ldren do use the most general metaphors o f t ime more often than the researcher but the majority o f ch i ldren do not. The total score for general metaphors o f t ime is being dr iven by a few ch i ld ren w h o frequently use them. The conversational abil i t ies o f ch i ldren may be a large determinate o f temporal metaphor use by chi ldren w i t h A S D . The imbalance between the product ion and interpretation o f temporal expressions among the majori ty o f the ch i ldren i n the 9 texts may be caused by the fact that chi ldren w i t h less strong conversat ional abil i t ies require more structured conversations. These more structured conversations entail a turn-taking pattern where the researcher asks questions and the c h i l d responds to these questions. Chapter 4, Genre notes this reoccurr ing turn-taking pattern. In addi t ion to more structured conversations, ch i ldren w i t h less strong conversational abil i t ies sometimes provide m i n i m a l and non-elaborative responses that consequently are un l ike ly to include expressions that contain either temporal metaphors or units o f t ime. In the majori ty o f the texts, the researcher uses more temporal expressions and the c h i l d appears to understand the researcher's expressions. T h i s fact raises the poss ibi l i ty that the ch i ldren w i t h A S D i n the 9 texts are more l i ke ly to comprehend temporal metaphors than they are able to use them. It may, o f course, be the case that the 'terseness' noted ( m i n i m a l responsiveness) affords fewer opportunities for product ion. The question warrants further consideration. It is possible that diff icult ies w i t h temporal metaphor use may be rooted i n the ch i ld ren ' s inabi l i ty to articulate more compl ica ted construals o f t ime. Several ch i ldren demonstrated diff icul t ies w i t h temporal specif ici ty and generalizations i n terms o f 102 temporal points prompted in spoken discourse. In text 2, the child is asked a question and he provides an unlikely response: 2 1 9 . R E S : s o w h a t d o y o u d o o n t h e w e e k e n d s W i l l ? 2 2 0 . C H I : n o t h i n g . The child uses a minimal response and does not answer the researcher's question. The researcher rephrases the question two additional times before the child provides an appropriate response: 2 2 1 . R E S : d o y o u h a v e a w o r k e r t h a t c o m e s ? 2 2 2 . C H I : n o p e . 2 2 3 . R E S : a n d s p e n d s s o m e t i m e w i t h y o u ? 2 2 4 . C H I : y e s . Line 219 uses the temporal unit, \"weekends,\" that prompts for multiple points upon the observer's path. Although we cannot determine the degree to which the child comprehends this metaphor, again, it appears that metaphors that involve generalized or unspecific events pose problems with individuals with A S D . These construals of time elicit a number of temporal points and require the child to generalize across these points. Lines 176 through 178 in text 3 provide another example: 1 7 6 . R E S : t e l l me a b o u t y o u r u h y o u d o s o m e t h i n g o n S a t u r d a y ' s u s u a l l y . ' 111. R E S : w h a t d o y o u u s u a l l y d o o n S a t u r d a y s ? 1 7 8 . C H I : I g o h o r s e b a c k r i d i n g . In line 178, the child answers the researcher's question; however, before doing so, the question has to be rephrased. The researcher uses the temporal unit, \"Saturdays,\" that again prompt for multiple points upon the observer's path. The need for the researcher to rephrase the question may indicate that the child has processing difficulties. The need to rephrase a generalized question in the context of generalizations comes up again in text 7: 1 1 3 . R E S : s o w h a t d o y o u u s u a l l y d o i n t h e e v e n i n g s w h e n y o u g e t b a c k h e r e ? 1 1 4 . R E S : w h a t d o y o u d o a f t e r s u p p e r ? 103 1 1 5 . C H I : w e l l T u e s d a y s a f t e r w o r k i n g w e h a v e t o d o a w o r k o u t d o w n s t a i r s . Text 9 provides a final example of the children's difficulties with questions that generalize and elicit multiple temporal points: 2 5 4 . R E S : w h a t d o y o u d o a f t e r . s c h o o l ? 2 5 5 . C H I : s u p p e r . Although the child responds to the researcher's question with a correct answer, the child's response is unconventional. Cases where children with A S D are required to generalize seem to pose difficulties as they require a homogeneous perspective of time and events. It is possible that children with A S D may elicit multiple temporal points but it is the task that requires them to see similarities and generalize across these points that is the source of difficulty. The child in text 7 appears to be significantly more capable at generalizing compared to the other children. He appears to be the only child that is more able to generalize in his utterances; however, he still requires the researcher to rephrase a question. In text 3, the child is asked to generalize again; however, the child demonstrates difficulty in a different way, providing a more specific answer than that requested by the researcher: 2 2 2 . R E S : t e l l me w h a t y o u r f a v o u r i t e - : t h i n g i s t o d o i n t h e s u m m e r t i m e . 2 2 3 . C H I : I I g o o n v a c a t i o n t o P r i n c e E d w a r d I s l a n d . 2 2 4 . C H I : a n ( d ) w h e n K a t h y i s a w a y i n M o n ( t ) r e a l I g o c a m p i n g . 2 2 5 . C H I : a n d a n d a f t e r I g o c a m p i n g I p a c k u p m y s u i t c a s e . . . Although the child answers the researcher's question that uses the temporal unit \"summer time,\" it appears that the child replies referencing a specific occurrence and not a general one. Again, this example suggests that the child prefers a heterogeneous construal of 104 events as opposed to a homogenous one. Moreover, the child uses a false start which may again indicate processing difficulties. This chapter has shown that children with A S D demonstrate competence with some of the most basic metaphors of time; however, when these metaphors elicit multiple temporal points or motivate a homogeneous construal of time, the children with A S D appear to struggle with these expressions. Discoursal patterns coinciding include minimal and non-elaborative responses, false starts, and the need for the question to be rephrased. A second pattern that concerns temporal specificity emerged among the texts analyzed. Several children showed the tendency to construe time and temporal events in smaller than typical units. This observation somewhat differs from the previous pattern. In this pattern, the children show competence is expressing time but construe time in overly specific units. Three examples are provided. Minutes are used as opposed to hours and a two-year period may be divided into 2 single-year segments. Text 3 provides the first example: 9 8 . R E S : h o w l o n g d i d i t t a k e y o u W i l l ? 9 9 . R E S : < j u s t a b o u t > [>] . 1 0 0 . C H I : < i t > [<] . 1 0 1 . C H I : i t t o o k i t t o o k w o i t t o o k f i v e h u n d r e d a n d s i x t y f o u r m i n u t e s . 1 0 2 . R E S : d i d i t r e a l l y ? 1 0 3 . C H I : i t t o o k f i v e h u n d r e d a n d s i x t y f o u r m i n u t e s . In lines 98 through 103, \"it\" refers to a (car) drive. We would expect typical speakers to be more general and to describe the length of time using hours, not minutes. Note that the child's response begins with a false start. A second example shows that in text 4 the child construes a length of time using smaller than typical units: 105 1 3 1 . R E S h o w d o e s s h e l i k e T o r o n t o ? 1 3 2 . C H I s h e l i k e s i t < w o n d e r b a r > [! ] 1 3 3 . R E S mmhm? 1 3 4 . C H I b u t s h e h a s n t v i s i t me l a s t 1 3 5 . R E S u m . In lines 131 through 135, \"she\" refers to the child's little cousin. The child describes the period of time that his cousin has not visited Toronto in terms of 2 1 -year increments. We would expect typical speakers to simply state that Ingrid has not been to Toronto in the past 2 years. A third example in text 3 shows that a child construes a length of time in terms of a series of shorter events as opposed to a longer continuous event: 1 9 1 . R E S : h m h o w l o n g h a v e y o u b e e n d o i n g t h a t ? 1 9 2 . C H I : I h a v e b e e n d o i n g i t f o r t w e l ( v e ) I h a v e d I h a v e d o n e i t f o r t w e l v e r i d e s - : . 1 9 3 . R E S : t w e l v e r i d e s o h . 1 9 4 . C H I : y e s - : . In lines 191 through 194, \"that\" and \"it\" refers to the activity of horseback riding. The child does not answer the researcher's question. Instead of stating a length of time that he has done the activity, the child expresses the number of times as a sequence of the same event. The length of time is therefore construed in terms of a series of smaller temporal events as opposed to a single, continuous duration of time. Note again that the child begins his response with a false start. The above examples from texts 3 and 4 provide further evidence that children with A S D experience difficulties with temporal specificity. These difficulties are apparent where utterances prompt for multiple temporal points in terms of generalizations and where utterances prompt for events or ongoing events where smaller units can be used. Again, the observations in texts 3 and 4 suggest that the child uses a heterogeneous construal of time. Two 1-year increments heterogeneously differentiate one year for the 106 other whereas the more typical expression \"two years\" construes the period of time as homogeneous. Twelve rides also construes the 12 events as heterogeneous entities while the more typical \"12 times\" expression assumes an ongoing homogeneous recurrence of the same event. The majority of these findings are observed in texts 3, 4, and 7 where children demonstrate the ability to use a greater variety of temporal metaphors, metonymies, and sense of time. The data examined suggest that children with A S D may construe and communicate time differently. Where typical speakers would likely use the lexeme \"t ime\" and senses of time, the child construed time and senses of time in unconventional ways. There is comprehension of the researcher's utterances that used the Commodity, Duration, and Moment Senses of time; however, the Instance Sense is the only sense of time used by child 3 in the 9 texts. Chi ld 3 uses the lexeme \"once\" on two occasions that prompt for the Instance Sense of time. \"Once\" is a contextually appropriate variant of the Instance Sense and is therefore construed as \"one time\" in text 3: 8 2 . C H I : m y f a t h e r w e n t t h e r e w a n m y f a t h e r a n d I w e n t t h e r e o n c e b e f o r e g r a d e # w e n t t h e r e o n c e b e f o r e g r a d e f i v e s t a r t e d . 2 0 6 . C H I : a n d b y a n d o n e a n d o n c e m y m o t h e r a n d I g o t i t i n um J u l y a n d A u g u s t # b o u g h t i t i n J u n e a n d A u g u s t . Yet none of the children used the lexeme \"t ime\" itself, and in several instances, time is construed unconventionally. In text 4, we see that the child construed the length of time he has being going horseback riding as \"twelve rides\": . 1 9 2 . C H I : I h a v e b e e n d o i n g i t f o r t w e l ( v e ) I h a v e d I h a v e d o n e i t f o r t w e l v e r i d e s - : . This construal uses 12 consecutive events to describe a length of time. It is unusual as we would expect typical speakers to use the Instance Sense of time. For example, \"I have 107 done it twelve t imes.\" H o w indiv iduals w i th A S D use senses o f t ime warrants further invest igat ion as these two examples show both competency and diff icul t ies w i t h the Instance Sense o f t ime. Further analysis o f texts where speakers w i t h A S D use the lexeme \" t ime\" w o u l d provide more evidence for patterns i n the use o f senses o f t ime. 108 CONCLUSION T h i s chapter has identif ied that speakers w i t h A S D do comprehend and produce temporal metaphors, temporal metonymies , and senses o f t ime. The analysis found that i n general, the chi ldren i n the 9 texts use the T i m e Orientat ion and M o v i n g Observer metaphors the most often. Me tonymies o f t ime and the less basic metaphors o f t ime are used less by the ch i ld ren w i t h A S D . C h i l d r e n w i t h A S D used E v a n s ' senses o f t ime the least. In addi t ion to these general observations, it was identif ied that the majori ty o f ch i ldren w i t h A S D used fewer temporal metaphors, metonymies , and senses o f t ime than the researcher. U p o n closer examinat ion, it was found that ch i ldren i n text 3, 4, and 7, are exceptions who used the T i m e Orientat ion and M o v i n g Observer metaphors more frequently than the researcher. The ch i ld ren i n text 1, 3, 4, and 7 are i n fact the only ch i ld ren who used other metaphors, metonymies , and sense o f t ime, besides the T i m e Orienta t ion metaphor and the M o v i n g Observer metaphor. Moreove r , patterns are ident i f ied among speakers that suggest d i f f icul ty w i t h t ime i n general iz ing events. General izat ions prompt for mul t ip le temporal points and motivate homogeneous construals o f events and t ime. Other specific patterns that were identif ied include difficult ies w i t h specif ici ty i n terms o f temporal units and diff icul t ies w i t h senses o f t ime and the lexeme \" t ime .\" The most obvious examples o f these patterns are found i n texts 3, 4, and 7. The ch i ld ren wi th A S D i n texts 3, 4, and 7 are the ch i ld ren that used the greatest variety o f metaphors, metonymies , and senses o f t ime; yet, these ch i ld ren are among the ones that demonstrated the greatest diff icult ies w i th generalizations, specif ic i ty , and the lexeme \" t ime .\" C h i l d r e n responded to generalizations or questions that required 109 construals involving multiple temporal points with responses that were unlikely or that used false starts. They required the question to be restated or replied with utterances that did not answer the question. Lengths of time were observed to be construed in smaller than usual units. The lexeme \"t ime\" was never used and where the lexeme \"t ime\" would typically be expected, the child replaced \"t ime\" with a series of events. The child did comprehend the researcher's utterances that used senses of time and also used the lexeme, \"once.\" These observations suggest that children with A S D have the tendency to express heterogeneous construals of time and struggle with expressions that motivate homogeneous construals. Further analysis of the lexeme \"t ime\" with a specific focus on temporal cycles and recurrence may provide interesting and more conclusive findings that would help better understand how individuals with A S D use the lexeme \"time.\" The fact that the research participants did not use the lexeme \"t ime\" suggests that Evans' approach to time has limited applications in the context of the discourse of individuals with A S D . This limit provides further insight into the analysis of time as the research participants were able to express time and temporality without using the lexeme. This observation suggests that conceptual metaphor, in general, may be a more effective way to discuss the cognitive aspects of time. Although Evans' framework enables a discussion in this thesis, it precedes the theory of conceptual metaphor in the context of speakers with A S D . Throughout texts 1 through 9, many of the child's utterances demonstrate unusual temporal construals that we would not expect in typical conversations. These utterances violate or ignore contextual factors and protocol of conversational discourse. Unusual utterances among the children examined were largely found to be uttered by the children 110 who used the greatest variety of metaphors, metonymies, and senses of time. Context and genre establish the generally accepted conventions of conversation. A definite relationship exists between the expressions analyzed and story telling genres. A variety of general and specific observations concerning temporal expressions are noted in this chapter. Ill CHAPTER 6 - CONCLUSION This thesis examined temporal discourse produced by 9 speakers with A S D from two linguistic disciplines in a complementary way. First, using speech genre analysis, this thesis explored the ways in which individuals with A S D structured stretches of texts that involve sequences of events. These stretches of text were described following a framework from Eggins and Slade (1997) and Plum (2004) and patterns in the use of speech genre types and generic stages were identified across all 9 texts. Second, these same stretches of text were examined for temporal conceptual metaphors, metonymies and senses of time following Lakoff and Johnson (1999) and Evans (2003). The speakers' uses of conceptual metaphors, metonymies, and senses of time were described and patterns in the linguistic expressions of these devices were identified across the 9 texts. The examination of semi-structured conversational texts is an interesting line of inquiry as individuals with A S D struggle with contextually appropriate discourse in conversation (de Vi l l iers et al., 2006). The \"storytelling genres\" (Eggins & Slade, 1997) in such texts also place relatively high demands on semantic and episodic memory (Asp & de Vi l l iers, forthcoming). Thus the data was well-suited to this inquiry. The speech genre analysis was found to be an effective means of better understanding the ways in which these individuals contextually use and vary language in spoken discourse. Time and temporal concepts were another engaging area to examine as these concepts are largely communicated using conceptual metaphor. It is widely acknowledged, individuals with A S D struggle the interpretation and use of figurative metaphor, tending to interpret metaphors in an overly literal way. By combining these two approaches, it was found that 112 the individuals with A S D investigated have problems with temporal specificity in the context of storytelling conversations. Speech genre analysis of the stretches of temporal texts found that children with A S D engaged exclusively in Specific Recounts, General Recounts, and Procedures. The research participants did not use Narratives, Anecdotes, or Exemplums as speech genre types in the stretches extracted from the 9 texts. Among the generic stages of the speech genre types, children used Records of Events the most, followed by Orientations, and Evaluations. A pattern also emerged where children with A S D appeared to have a tendency to use Records of Events. This finding was based on the observation that a large number of Records of Events were found, that children ignored utterances that interrupted Records of Events, they returned to Records of Events i f they were interrupted, and they provided Records of Events when other generic stages were solicited. The research participants in texts 3 and 7 appeared to have stronger conversational abilities marked by longer dialogue between turns and the ability to use a variety of unsolicited generic stages. These same children were the participants who also described multiple events in serial order and produced discontinuously realized Records of Events. Interestingly, although the participants from texts 3 and 7 may have had stronger conversational abilities, they were inflexible with temporal conjunction use. They largely sequenced events using \"then\" or \"and (then).\" In contrast, the participants in texts 1,5, and 9 showed weaker conversational abilities marked by fewer or no unsolicited and optional generic stages. Children in texts 2 and 8 also had weaker conversational abilities. They exhibited what may have been coping techniques that 113 included mirroring the researcher's responses, using minimal responses, and probing for specific questions. The analysis of temporal conceptual metaphors, metonymies, and senses of time found several general patterns. The research participants with ASD used the Time Orientation and Moving Observer metaphors the most frequently. They used metonymies of time, the Time is a Resource metaphor, and the Time is Money metaphor less frequently and Evans' senses of time the least. The research participants in texts 3, 4, and 7 used the most and the greatest variety of temporal conceptual metaphors, metonymies, and senses of time. The remaining participants used the Time Orientation metaphor and the Moving Observer metaphor, but very few or none of the other conceptual metaphors, metonymies, or senses of time. Other patterns in the use of conceptual metaphors, metonymies, and senses of time suggest that some children with ASD were more able to interpret the above temporal expressions as opposed to producing them. The research participants appeared to be more able to comprehend some temporal conceptual metaphors uttered by the researcher, but lacked the resources to use theses same metaphors in contextually appropriate ways. A number of expressions of conceptual metaphors, metonymies, and senses of time uttered by the children demonstrated an attempt to use the expression but difficulties with specificity resulted in utterances that were unexpected or unconventional, Some children further experienced difficulties with generalizations, unspecific construals of time, or consturals that prompted for multiple temporal points. The research participants also expressed time in smaller than typical units, tended to construe events heterogeneously, and where the lexeme \"time\" would normally occur, expressed the lexical item in 114 unconventional ways. The most obvious examples were found in texts 3, 4, and 7, where children were the most able at using conceptual metaphors, metonymies, and senses of time. Although time is typically expressed using figurative and conceptual metaphors, individuals who struggle with figurative expressions showed general competency with some conceptual metaphors of time; specifically, these conceptual metaphors were the Time Orientation and Moving Observer metaphors. Conceptual metaphors differ from figurative metaphors in their intensity of conventionality. Individuals with A S D are knows to have difficulties with figurative expressions but the findings of this thesis indicate that the research participants interpreted and produced linguistic expressions that used conventional (conceptual) metaphors of time. Overall, patterns from the two analytic approaches employed suggest interesting possibilities. Specifically, a relationship was found between stronger conversational ability and a stronger ability to use speech genre, conceptual metaphors, metonymies, and senses of time. In addition, unconventional patterns in the use of speech genre, metaphors, metonymies, and senses of time seemed to occur among the most conversationally able speakers with A S D . It appears that there is a relationship between conversational success and conceptual metaphor use in A S D and further studies in this area are warranted. The findings also suggest that unconventional use of speech genre types and metaphor are not necessarily indicative of conversational ability in A S D . Further studies would be needed to explore this hypothesis. The children in texts 3 and 7 were the children that used the most and the greatest variety of generic stages and speech genre types in the stretches of texts extracted from 115 the 9 transcripts. Children in texts 3, 4, and 7 used the most and greatest variety of temporal metaphors, metonymies, and sense of time in the same in the same 9 texts. These observations suggest that speakers who are more able to use speech genre types and stages may also be more able to use temporal conceptual metaphors, metonymies, and senses of time. Both the analyses show specific difficulties with speech genre use and conceptual metaphor use. Speech genre analysis revealed that although the children in texts 3 and 7 had stronger conversational abilities marked by their varied use of generic stages, longer turns, and the ability to produce discontinuously realized Records of Events and Records of Events that described multiple events in serial order, their use of temporal conjunctions was unvaried. They largely sequenced events in serial order using \"then\" and \"and (then).\" The child in text 3 was overly specific with his description of events, causing the researcher to mistake the child's Recount for an Anecdote. This child also provided many obvious examples where he either ignored utterances by the researcher that interrupted Records of Events or responded to the researcher's utterances and then returned to the Record of Events. It may be that individuals with A S D who are more conversationally able favour the Recount genre. The unvaried use of conjunctions, the specific description of events, and the tendency to return to Records of Events are all unconventional conversational qualities that support and may even encourage children with A S D to use the Record of Events stage of the Recount genre. Metaphor analysis revealed that these same children were among the most apparent to construe time and temporal events in unconventional ways. Chi ld 3 was overly specific and described 9 hours and 24 minutes as \"f ive hundred and sixty four 116 minutes.\" This child also described the fact that he had gone horseback riding 12 times as \"twelve rides,\" an example that suggests the child may replace the more conventional expression using the lexeme \"t ime\" with the lexeme \"ride.\" Furthermore, as discussed, \"twelve rides\" suggests a heterogeneous construal of temporal events where each ride may be different, whereas \"12 times\" would suggest 12 occasions of the same event. Both the children in texts 3 and 7 appeared to have difficulties with generalizations. Questions that solicited multiple temporal points and then required the child to generalize, (e.g. \"What do you usually do in the evenings?\", were ignored by the child or needed to be rephrased by the researcher. A recurring pattern that was apparent among the most conversationally engaged children with A S D was the inappropriate use of detail. From speech genre analysis, difficulties with specificity appeared in the detailed description of events. From conceptual metaphor analysis, this same difficulty appeared in the use of smaller temporal units, generalized events, and the recurrence of similar events. It is l ikely that the less conversationally engaged individuals with A S D may experience these same difficulties; however, these patters may be more difficult to detect because of the ways these individuals participate in conversation. Speech genre analysis showed that the less conversationally engaged individuals required more structure from the researcher, did not elaborate, and tended to use minimal and polar responses. Individuals with A S D who were less conversationally engaged appeared, at times, to compensate for their difficulties. As just noted, speech genre analysis showed that these individuals participated in a turn-taking pattern where the researcher would ask a question and the child would respond to the question. On occasion, the child did not 117 respond to questions, provided unlikely answers, or copied the researcher's responses. The speech genre analysis also showed that the less conversationally engaged children appeared not to elaborate and one child used probes to solicit specific questions from the researcher. The analysis of conceptual metaphor showed that these same children used fewer conceptual metaphors, metonymies, and senses of time and largely used only the Time Orientation metaphor and the Moving Observer metaphor. Again, the limited use of metaphors can be linked back to patterns found from speech genre analysis. These individuals often provided responses to questions that were minimal and unelaborative; therefore, limited use of metaphor, metonymies, and senses of time could be expected. All the research participants showed unconventional patterns in their use of storytelling speech genres, conceptual metaphors, metonymies, and senses of time. The more conversationally engaged individuals showed specific difficulties with both speech genre and conceptual metaphor that related to specificity of time and temporal events in different ways. The less conversationally engaged individuals also showed difficulties and sometimes techniques that helped compensate for their difficulties with speech genre and conceptual metaphor. Conversational specificity is highly dependent on context and in this thesis specificity in relation to speech genre and conceptual metaphors of time was identified as a particular area of difficulty for individuals with ASD. Although individuals with ASD may experience difficulties with speech genre, they do appear use genre and generic stages in conversation, but in unconventional ways. This finding suggests that speakers with ASD do rely on context and attempt to make shared and meaningful contributions in conversational discourse. 118 Individuals with ASD also used metaphors, metonymies, and senses of time in unconventional ways. Difficulties with conceptual metaphors are rooted in specificity. The fact that the speakers with ASD used temporal conceptual metaphors, metonymies, and senses of time indicates that they attempted to make shared temporal contributions in spoken discourse. Their success in using some conceptual metaphors also serves as a contribution to the philosophy of time, since metaphor use, especially in spoken discourse, is particularly sensitive to context. People with ASD who struggle with figurative metaphor comprehend and produce some conceptual metaphors, metonymies, and senses of time. Most interestingly, the research participants showed competency with some of the most conventionalized metaphors of time. Conceptual metaphors explored in this thesis are among the most conventionalized temporal metaphors. As figurative metaphors are extensions of conceptual metaphors, temporal metaphors exist on a continuum that ranges from conventional (conceptual) metaphors to poetic (figurative) metaphors. Individuals with ASD must reach a point on the continuum of the conventionality of metaphors where they begin to struggle with metaphorical expression. The tendency for individuals to use a limited number of conceptual metaphors in this thesis provides evidence that individuals with ASD begin to show difficulties with figurative language somewhere at the conceptual level. 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Philosophy in the Flesh: The Embodied Mind and Its Challenge to Western Thought. New York: Basic Books. 120 Lakoff, G. & Johnson, M . (1980). Metaphors We Live By. Chicago: University of Chicago Press. Lee, D. (2001). Cognitive Linguistics: An Introduction. Victoria, Australia: Oxford University Press. Mart in, J. R. & Plum, G. A . (1997). Construing Experience: Some Story Genres. Journal of Narrative and Life History, 7 (1-4), 299-308. Plum, G.A. (2004). Text and Contextual Condition in Spoken English: A Genre-based Approach. [Online]. Available U R L http://ses.librarv.usvd.edu.aU/bitstream/2123/608/2/adt-NU20040629.09514001 front.pdf 121 APPENDIX 1 Temporal Stretches Extracted from 9 Transcripts TEMPORAL STRETCHES: TRANSCRIPT 1 G E N E R A L R E C O U N T 1 O R I E N T A T I O N 7 9 . R E S : w h a t d o y o u d o w h e n y o u g o h o m e f r o m s c h o o l W i l l | R E C 0 R D O F E V E N T S 8 0 . C H I : a h - : I p l a y g a m e s . S P E C I F I C R E C O U N T 1 1 3 8 . R E S : 1 3 9 . C H I : 1 4 0 . R E S : 1 4 1 . R E S : 1 4 2 . C H I : 1 4 3 . R E S : 1 4 4 . C H I : 1 4 5 . R E S : 1 4 6 . R E S : 1 4 7 . R E S : 1 4 8 . C H I : 1 4 9 . R E S : 1 5 0 . R E S : O R I E N T A T I O N s o d o y o u l i k e p i z z a W i l l ? rarahra. R E C O R D O F E V E N T S I h a d p i z z a l a s t n i g h t f o r s u p p e r . O R I E N T A T I O N w h a t d i d y o u h a v e f o r s u p p e r l a s t n i g h t ? R E C O R D O F E V E N T S I h a d p i z z a t o o . d i d y o u ? y e a h . o h - : . R E C O R D O F E V E N T S a n d t h e n i g h t b e f o r e I h a d p o r k c h o p s . O R I E N T A T I O N c a n y o u r e m e m b e r w h a t y o u h a d t h e n i g h t b e f o r e ? R E C O R D O F E V E N T S p o r k c h o p s t o o . y o u h a d p o r k c h o p s t o o . h m . F U T U R E P R O J E C T I O N 1 1 5 1 . R E S w h a t a r e y o u h a v i n g t o n i g h t # f o r s u p p e r ? 1 5 2 . R E S d o y o u k n o w ? 1 5 3 . C H I I ' m h a v i n g r i c e f o r s u p p e r t o n i g h t . 1 5 4 . R E S o h w h a t d o y o u e a t w i t h t h e r i c e ? 1 5 5 . C H I t u r k e y . 1 5 6 . R E S t u r k e y a n d r i c e . 1 5 7 . R E S t h a t s o u n d s d e l i c i o u s . 1 5 8 . R E S m m h m . TEMPORAL STRETCHES: TRANSCRIPT 2 F U T U R E P R O J E C T I O N 1 1 . R E S t o d a y i s J u n e t h e s i x t h . 2 . R E S a n d < I > [>] . 3 . C H I < f i f t h > [<] . 4 . R E S i s i t t h e f i f t h ? 5 . C H I y e s . 6 . C H I t o m o r r o w ' s t h e s i x t h 7 . R E S o h o k a y . 8 . R E S t o d a y i s J u n e t h e f i f t h . G E N E R A L R E C O U N T 1 O R I E N T A T I O N 2 1 9 . R E S : 2 2 0 . C H I : 2 2 1 . R E S : 2 2 2 . C H I : 2 2 3 . R E S : 2 2 4 . C H I : 2 2 5 . R E S : 2 2 6 . C H I : 2 2 7 . R E S : 2 2 8 . C H I : 2 2 9 . R E S : 2 3 0 . R E S : 2 3 1 . C H I : 2 3 2 . R E S : 2 3 3 . C H I : 2 3 4 . R E S : 2 3 5 . C H I : s o w h a t d o y o u d o o n t h e w e e k e n d s W i l l ? n o t h i n g . R E C O R D O F E V E N T S d o y o u h a v e a w o r k e r t h a t c o m e s ? n o p e . a n d s p e n d s s o m e t i m e w i t h y o u ? y e s . O R I E N T A T I O N hm w h a t ' s h i s n a m e ? T o m . R E C O R D O F E V E N T S ] a n d w h a t d o y o u d o w i t h T o m ? g o o u t w i t h h i m . mmhm? O R I E N T A T I O N ! w h a t d o y o u d o w h e n y o u ' r e o u t ? [ p l a y g o l f . R E C O R D O F E V E N T S 1 g o l f ? y e s , E V A L U A T I O N / C O D A ^ o h t h a t ' s e x c i t i n g , y e s . TEMPORAL STRETCHES: TRANSCRIPT 3 S P E C I F I C R E C O U N T 1 - D I S C O U N T I N U O U S L Y R E A L I Z E D ABSTRACT 9. 10 . 11. 12. 13. 14 . 15. 16. 1.7 . 18 . 19. 20. 21. RES : CHI: RES: CHI: RES : RES CHI CHI CHI CHI RES RES CHI RES CHI CHI CHI I hear you went on a h o l i d a y r e c e n t l y . I [>] • ORIENTATION <where d i d you go?> [<] I went t o T o r o n t o and K i n g s b u r y . EVALUATION wow. I l i k e your a c c e n t , yes . yes . you l o v e my a c c e n t . I d you do. mmhm. RECORD OF EVENTS what d i d you see i n T o r o n t o ? <I> [>] • <or in> [<] T o r o n t o ? f i r s t I went t o K i n g s b u r y . and v i s i t e d my aunt Eddy and Judy W i n s t o n . on on Tuesday and Wednesday Judy W i n s t o n and J we b o t h went t o t h e a i r p o r t t o p i c k up Sophie. 22. CHI: (a)n(d) Sophie i n t r o d u c e d us t o a band c a l l e d t h e W a l l f l o w e r s w i t h Jacob D i l l o n i n i t . 23. 24 . 25. 26. 27 . RES CHI RES CHI CHI oh where does t h e band p l a y ? from A m e r i c a , oh from A m e r i c a . Sophie and I went t o Recordman. and a t Recordnan Recordman we found a S t a n Rogers tape !] w i t h two r a d i o p l a y e r s c a l l e d H a r r i s and the c a l l e d P o e t i c J u s t i c e major and the s i s t e r s . 28. CHI: um J Sophie found a Sophie found a t h i n g t h a t wasn't Tom W a i t s wasn't t h e Wake i t wasn't R i c k W a k e f i e l d but i t was t h e W a l l f l o w e r s # c a l l e d b r i n g i n g down the h o r s e . 29. CHI: i t was the o n l y W a l l f l o w e r s album t h e y e v e r r e c o r d e d A N E C D O T E REACTION 30. RES: ORIENTATION 31. 32. 33. 34 . RES RES CHI RES had you h e a r d about them b e f o r e ? t h i s group? no! no? RECORD OF EVENTS uh t h e . s o t h e n we went t o an a n t i q u e shop, and t h e n back t o V i n y l Records.. when I f l i p p e d t h r o u g h the the l e t t e r b i n the Bowie 35. CHI 36. CHI 37. CHI [!] s e c t i o n I found Changes [!] One Bowie. 38. CHI: t i l l I was s t o p p e d dead by t h e b l u e r a i n c o a t s t a r i n g out from t h e new J e n n i f e r Warnes album c a l l e d Famous B l u e R a i n c o a t . 39. CHI: and I got i t on LP and c a s s e t t e . and t h e n i n t h e Cohen [!] s e c t i o n I r i f l e d t h r o u g h 125 4 0 . C H I : a n d t h e n w e a n d t h e n w e w e n t h o m e t o a n d t h e n a n d t h e n w e f o u h o m e a n d t h e n w e d r o v e b a c k t o K i n g s b u r y a n d K i n g s b u r y . 4 1 . C H I : a n d l i s t e n e d t o i t o n t h e w a y - : a y t h e r e . 4 2 . C H I : a n d a t K i n g s b u r y I a t e s o m e l a s a g n a a n d b u t t e r s c o t c h i c e c r e a m . 4 3 . C H I : a n d w e w e n t o u t s i d e a t n i g h t a n d p l a y e d i n t h e s n o w um < a n d > [>] . 4 4 . R E S : < d o y o u > [<] . 4 5 . C H I : t h e n e x t d a y I w e I w e n t t o t h e b o o k s h o p . 4 6 . C H I : a n d g o t t h e E n g l i s h v e r s i o n o f t h e J u d i t h B e o r i s b o o k A l e x a n d e r A n d T h e T e r r i b l e H o r r i b l e N o G o o d V e r y B a d D a y < w h i c h I l o v e d > [>] • 4 7 . 4 8 . 4 9 . 5 0 . R E S R E S C H I R E S C H I : 5 1 . f a r m . 5 2 . C H I : O R I E N T A T I O N < e x c u s e me W i l l > [<] . d i d y o u h a v e c o u s i n s t h e r e t o p l a y w i t h t o o ? n o - : ! n o j u s t < a d u l t s e h hm> [>] ? R E C O R D O F E V E N T S ! < i t ' s i n > [<] t h e n I w e n t t o B i l l y a n d N a n c y ' s d a i r y < a n d > [>] . O R I E N T A T I O N 5 3 . R E S < h o w m a n y u h > [<] c o w s w o u l d t h e y h a v e o n t h a t d a i r y f a r m ? 54 . C H I t h e y h a d a l o t [ ! ] . 5 5 . R E S l i k e a b o u t t w e n t y o r f i f t y o r w h a t w o u l d y o u t h i n k ? 5 6 . C H I t t w e n t y t w o . 5 7 . R E S t w e n t y t w o ? 5 8 . C H I < y e a h > [>] 5 9 . R E S < a n d w> [<] w h a t w h a t c o l o u r s w e r e t h e y ? 6 0 . R E S w e r e t h e y H o l s t i e n s b l a c k a n d w h i t e o n e s o r w e r e . 6 1 . C H I t h e y w e r e b l a c k a n d w h i t e H o l s t i e n s . 6 2 . R E S w e r e t h e y ? 6 3 . R E S u h h u h . R E C O R D O F E V E N T S 64 . C H I e x c u s e me a n d e x c u s e me a n d t h e n w e w e n t h o m e . 6 5 . C H I a n d I w e n t o u t s i d e a g a i n . 6 6 . C H I a n d I t o o k a b a t h . 6 7 . C H I a n d t h e n [ ! ] I w e n t t o s l e e p . 6 8 . C H I a n d t h e n I w e n t t o b e d . 6 9 C H I a n d t u r n e d o f f t h e l i g h t . 7 0 C H I t h e n e x t m o r n i n g I p a c k e d m y s u i t c a s e u p . 7 1 C H I a n d w e w e n t b a c k t o T o r o n t o . 7 2 C H I b u t m y f a t h e r w a s g o n e . 7 3 C H I I s a t d o w n a n d r e a d - : 7 4 C H I a n d s u d d e n l y m y f a t h e r w e n t m y f a t h e r c a m e b a c k . 7 5 C H I h e w a l k e d me o v e r t o t h e S a m ' s o n Y o n g e s t r e e t . 7 6 C H I a n d B l o o r s t r e e t - : . 7 7 C H I a n d I . O R I E N T A T I O N 7 8 R E S w h a t ' s t h a t p l a c e W i l l ? 7 9 C H I i t ' s a i t ' s a n e w S a m ' s < w h e r e > [>] t h e y h a v e u s e d L P s . 8 0 R E S < w h e r e > [<] . R E C O R D O F E V E N T S 8 1 C H I b u t I a m s o r r y t h e y d i d n o t h a v e a n y J e s s y [ ! ] W i n c h e s t e r L P s t h e r e . 126 8 2 . C H I : m y f a t h e r w e n t t h e r e w a n m y f a t h e r a n d I w e n t t h e r e o n c e b e f o r e g r a d e # w e n t t h e r e o n c e b e f o r e g r a d e f i v e s t a r t e d . 8 3 . C H I : a n d w e # f o u n d t w o n e w o n e s # L e a r n t o L o v e I t a n d T a l k M e m p h i s F o r M e . 8 4 . C H I : a n d t h ( e n ) a n d t h e n [ ! ] a s w e w e n t t o t h e S a m ' s o n B l o o r s t r e e t I f o u n d J e s s y W i n c h e s t e r ' s f i r s t a l b u m [ ! ] a n d t h e n a l b u m [ ! ] a n d h i s e i g h t h i s s i x t h o n e c a l l e d T o u c h O n T h e R a i n y S i d e a n ( d ) a n d a n d w e p a y a n d a t a p e o f M e n d l e s o n J o e c a l l e d B o r n T o C u d d l e s o C u d d l e . h o m e 8 5 . C H I a n d w e p a i d f o r t h e m . 8 6 . C H I a n d t h e n we w a l k e d o v e r t o V a n h o u t s . 8 7 . C H I a n d I a t e a n a n a i m o b a r a n d d r a n k s o m e w a t e r . 8 8 . C H I a n d a t a x i [ ! ] p i c k e d u s u p . 8 9 . C H I a n d a n d a t a x i c a m e a n d t o o k u s b a c k a n d t o o k t o T o r o n t o # < a n d > [>] O R I E N T A T I O N 9 0 . R E S 9 1 . R E S 9 2 . C H I 9 3 . R E S 94 . R E S 9 5 . R E S 9 6 . C H I 9 7 . R E S 9 8 . C H I 9 9 . R E S 1 0 0 C H I : 1 0 1 C H I : m i n u t e s 1 0 2 R E S : 1 0 3 C H I : 1 0 4 . C H I : 1 0 5 . C H I : 1 0 6 . R E S : <um > [<] h o w d i d y o u g e t t o T o r o n t o ? d i d y o u f l y o r d i d y o u g o b y c a r ? s h w e w e n t b y c a r - : . y e a h ? E V A L U A T I O N w a s i t a l o n g d r i v e ? i t m u s t h a v e b e e n e h ? i t w a s a l o n g d r i v e . m m h m . h o w l o n g d i d i t t a k e y o u W i l l ? < j u s t a b o u t > [>] . < i t > [<] . i t t o o k i t t o o k w o i t t o o k f i v e h u n d r e d a n d s i x t y f o u r d i d i t r e a l l y ? i t t o o k f i v e h u n d r e d a n d s i x t y f o u r m i n u t e s . e n o u g h t o m a k e me s w e a t . a n d t h e c a r t r i p w a s l o n g [ ! ] . m m h m . O R I E N T A T I O N 1 0 7 R E S d i d y o u s l e e p d u r i n g t h e c a r t r i p ? 1 0 8 C H I n o - : ! 1 0 9 R E S n o ? 1 1 0 C H I I w a s a w a k e d u r i n g t h e t r i p . 1 1 1 R E S mmhm? R E C O R D O F E V E N T S 1 1 2 C H I a n d I s l e p a n d I s l i p a n d I s l e p t o n a m a t t r e s s a t h o m e . 1 1 3 R E S mmhm? 1 1 4 C H I a t t r e s s a f t e r t h e m a t t r e s s i n m y g r a n d m o t h e r a n d g r a n d f a t h e r ' s a p a r t m e n t . 1 1 5 C H I a n d t h e n e x t m o r n i n g w h e n I w o k e u p I h a d s o m e b r e a d a n d p e a n u t b u t t e r f o r b r e a k f a s t . 1 1 6 C H I a n d t h e n . 1 1 7 C H I e k e a k f a s t 1 1 8 C H I a n d t h e n I h a d t o p a c k u p . 1 1 9 C H I a n d d e p a r t e d f o r h o m e . 1 2 0 C H I a n d I c r i e d a b o u t l e a v i n g T o r o n t o . 1 2 1 R E S d i d y o u ? 1 2 2 C H I y e s . E V A L U A T I O N 1 2 3 . R E S : I s u p p o s e i t w a s s a d l e a v i n g y o u r g r a n d p a r e n t s e h ? 127 1 2 4 . C H I : y e s i t w a s s a d -... ( C O N T I N U E D 3 4 L I N E S L A T E R ) E V A L U A T I O N / C O D A ^ 1 5 8 . R E S : s o y o u h a d a p r e t t y g o o d t r i p . 1 5 9 . C H I : < y e s - : > [>] . 1 6 0 . R E S : < s o u n d s l i k e > [<] . 1 6 1 . R E S : y e a h . FUTURE PROJECTION 1 1 2 5 . R E S : d o t h e y c o m e a n d v i s i t 1 2 6 . C H I : y e s . 1 2 7 . C H I : < t h e y ' r e c o m i n g > t h e s u m m e r i n J u l y [ ! ] . 1 2 8 . R E S : < o h t h a t ' s n i c e > [<] . 1 2 9 . R E S : . o h t h a t ' l l b e n i c e . 1 3 0 . C H I : i n J u . 1 3 1 . C H I : y e s . [ N o t e : i n F u t u r e P r o j e c t i o n 1 t h e y y o u h e r e t o o ? [>] a n ( d ) v i s i t a n d v i s i t i n g me i n = g r a n d p a r e n t s ] GENERAL RECOUNT 1 O R I E N T A T I O N 1 3 2 . R E S 1 3 3 . C H I 1 3 4 . R E S 1 3 5 . C H I 1 3 6 . R E S 1 3 7 . R E S 1 3 8 . C H I 1 3 9 . R E S a f a r m w h e r e d o y o u k n o w w h e n y o u w e r e a t K i n g s b u r y W i l l , w h e r e ? y o u r e m e m b e r w h e n y o u w e r e a t K i n g s b u r y ? w h a t ? w h e n y o u v i s i t e d K i n g s b u r y , d o y o u r e m e m b e r t h a t s t h a t t o w n ? y e s . w h e n y o u w e r e t h e r e y o u w e r e o n l y a b o u t a h a l f a n h o u r f r o m I y u I g r e w u p . 1 4 0 . C H I w h a t f a r m i s t h a t ? 1 4 1 . R E S w e l l t h a t ' s w h e r e m y mom a n d d a d o w n e d a f a r m . R E C O R D O F E V E N T S 1 4 2 . R E S a n d I l i v e d w h e n I w a s a l i t t l e g i r l . 1 4 3 . C H I y e s [ ! ] . 1 4 4 . R E S a n d m y mom u s e d t o d r i v e a h a l f a n h o u r . 1 4 5 . R E S a n d t e a c h i n K i n g s b u r y . 1 4 6 . R E S s h e t a u g h t h i g h s c h o o l . O R I E N T A T I O N 1 4 7 . C H I < d i d s h e > [> ? 1 4 8 . R E S < d o y o u k n o w > [<] d o y o u k n o w d o y o u k n o w w h a t s u b j e c t s h e t a u g h t ? 1 4 9 . C H I w h a t [ ! ] d i d s h e t e a c h ? 1 5 0 . R E S s h e t a u g h t - : m a t h e m a t i c s # t o h i g h s c h o o l s t u d e n t s . 1 5 1 . C H I m y t m y t m y t e a m y t e a c h e r t e a c h e s g e o g r a p h y [ ! ] . 1 5 2 . R E S d o e s s h e ? 1 5 3 . C H I g e o g r a p h y f r o m C a n a d a -1 5 4 . R E S mmhm? E V A L U A T I O N / C O D A ^ 128 1 5 5 . R E S : i t ' s i n t e r e s t i n g t o l e a r n a b o u t y o u r o w n c o u n t r y i n g e o g r a p h y i s n ' t i t ? 1 5 6 . C H I : y e s - : . 1 5 7 . R E S : mmhm - : . F U T U R E P R O J E C T I O N 2 1 6 6 . C H I : o h f o h f o h D o r i s p l e a s e D o r i s p l e a s e s t o p t h e t a p e . 1 6 7 . C H I : a n d a h a n d r e w i n d i t t o p l a y . 1 6 8 . C H I : a n d l e t ' s r e r e c o r d o u r v o i c e . 1 6 9 . R E S : w e l l we h a v e b e e n r e c o r d i n g o u r v o i c e . 1 7 0 . R E S : a n d we n e e d t o d o a b o u t # t h r e e o r f o u r m o r e m i n u t e s . 1 7 1 . R E S : a n d t h e n I ' l l s t o p a n d r e w i n d . 1 7 2 . R E S : a n d y o u c a n h e a r y o u r s e l f . 1 7 3 . R E S : o k a y ? 1 7 4 . C H I : y e s D o r i s . 1 7 5 . C H I : I c a n . G E N E R A L R E C O U N T 2 PLBSRTACT 1 7 6 . R E S • t e l l me a b o u t y o u r u h y o u d o s o m e t h i n g o n S a t u r d a y ' u s u a l l y . 1 7 7 . R E S w h a t d o y o u u s u a l l y d o o n S a t u r d a y s ? 1 7 8 . C H I I g o h o r s e b a c k r i d i n g . O R I E N T A T I O N 1 7 9 . R E S w h e r e ' s t h a t W i l l ? 1 8 0 . C H I i t ' s i n N o r t h Y o r k a t K i n g s w a y . 1 8 1 . R E S mmhm? E V A L U A T I O N 1 8 2 . R E S t h a t ' s a l o n g w a y s t o g o t o g o h o r s e b a c k r i d i n g e h ? 1 8 3 . C H I y e a h - : . 1 8 4 . R E S m m . 1 8 5 . C H I y e s . O R I E N T A T I O I \\ 1 8 6 . R E S i s i t o u t s i d e o r i n s i d e ? 1 8 7 . C H I i t ' s i n d o o r s 1 8 8 . R E S w o w . 1 8 9 . C H I s o m e t i m e s I r i d e o u t [! ] • 1 9 0 . R E S mmhm? 1 9 1 . R E S hm h o w l o n g h a v e y o u b e e n d o i n g t h a t ? 1 9 2 . C H I I h a v e b e e n d o i n g i t f o r t w e l ( v e ) I h a v e d I i t f o r t w e l v e r i d e s - : . 1 9 3 . R E S t w e l v e r i d e s o h . 1 9 4 . C H I y e s - : . 1 9 5 . R E S d o y o u r mom a n d d a d r i d e h o r s e b a c k t o o ? 1 9 6 . C H I n o [ ! ] . 1 9 7 . ^ R E S n o . 1 9 8 . C H I n o . 1 9 9 . C H I m y f r i e n d C r a i g H o l l y t a k e s m e . 2 0 0 . R E S u m . 2 0 1 . R E S d o e s h e l i v e i n D u n d a s ? 2 0 2 . C H I y e y e s . 2 0 3 . C H I h e w e n h e l i v e s i n B u r l i n g t o n . 1 2 9 R E C O R D O F E V E N T S 2 0 4 . C H I : 2 0 5 . C H I : r e l e a s e d w h i c h w a s 2 0 6 . C H I : J u l y a n d 2 0 7 . C H I 2 0 8 . C H I 2 0 9 . C H I 2 1 0 . C H I a n d 1 9 8 9 2 1 1 . R E S 2 1 2 . C H I 2 1 3 . R E S 2 1 4 . R E S h e w e n t t o s c h o o l i n T o r o n t o . O R I E N T A T I O N a n d h a d a N i e l s e n L P w h i c h i s c a l l e d T h e P o i n t i n 1 9 8 5 l i k e l i k e o u r v i d e l i k e o u r v i d e o T h e P o i n t w h i c h w a s a r e p l a c e d a t t h e v i d e o b o x . R E C O R D O F E V E N T S g o t i t i n u m a n d b y a n d o n e a n d o n c e m y m o t h e r a n d I A u g u s t # b o u g h t i t i n J u n e a n d A u g u s t , b u t . A u g u s t u m . b u t t h e v i d e o w a s n o t r e l e a s e d i n 1 9 8 5 . i t w a s u h r e l e a s e d b e f o r e t h e L P i n 1 9 8 4 - : ## u h 1 9 8 4 E V A L U A T I O N / C O D A ^ y o u k n o w s o m u c h a b o u t m u s i c W i l l . I d ( o ) < I d o > [>] D o r i s . < y o u d o > [<] . < y o u k n o w s o m u c h > [>] . GENERAL RECOUNT 3 O R I E N T A T I O N 2 2 2 . R E S : t i m e . 2 3 . C H I : 2 2 4 . C H I : 2 2 5 . C H I : a g a i n . 2 2 6 . C H I : 2 2 7 . R E S : 2 2 8 . C H I : 2 2 9 . C H I : 2 3 0 . C H I : t h e o t h e r 2 3 1 . C H I : 2 3 2 . R E S 2 3 3 . C H I 2 3 4 . C H I t e l l me w h a t y o u r f a v o u r i t e t h i n g i s t o d o i n t h e s u m m e r R E C O R D O F E V E N T S I I g o o n v a c a t i o n t o P r i n c e E d w a r d I s l a n d . a n ( d ) w h e n K a t h y i s a w a y i n T o r o n t o I g o c a m p i n g . a n d a n d a f t e r I g o c a m p i n g I p a c k u p m y s u i t c a s e a n d d a n d l e a v e f o r T o r o n t o . E V A L U A T I O N t h a t s o u n d s l i k e f u n . y e s . R E C O R D O F E V E N T S I ' l l ( h ) a f t a I ' l l ( h ) a f t a p i c k u p E s t h e r a n d X . a n d t h e n g o t o t h e t w o S a m ' s # t h e f i r s t o n Y o n g e a n d o n e u h B l o o r . um u h we f i n d a l o t o f g r e a t t r e a s u r e s - : t h e r e u m . CODA^ I c a n i m a g i n e , y e a h . y o u c a n i m a g i n e t h a t D o r i s ! 130 TEMPORAL STRETCHES: TRANSCRIPT 4 F U T U R E P R O J E C T I O N 1 2 0 . R E S 2 1 . C H I 2 2 . C H I 2 3 . C H I 2 4 . C H I 2 5 . R E S 2 6 . 2 7 . 2 8 . 2 9 . 3 0 . 3 1 . 3 2 . 3 3 . C H I R E S R E S C H I C H I C H I C H I C H I a n d w h a t a r e g o n n a h a g o i n g t o h a v e f o r l u n c h ? J J J a m e s . I ' m g o n n a a c t u a l l y n o t r e a l l y . a P i l l s b u r y + P i z z a + p o p . a n d I ' m g o n n a p r e t e n d i t ' s J a m e s . o h . < a n d e a t h i m > [% c h u c k l e s ] . h m . < w h y ' s > [>] t h a t ? [ b e c a u s e ] [<] b e c a u s e J a m e s ' s n o t n i c e t o m e . a n d I ' l l p r e t e n d m y d e s s e r t i s W a l t e r , a n d I e a t W a l t , a n d e a t h i m . ( b e ) c a u s e W a l t e r i s n ' t n i c e t o me e i t h e r . G E N E R A L R E C O U N T 1 1 1 1 . C H I : O R I E N T A T I O N 1 0 8 . R E S : s o w h a t d o y o u d o w h e n y o u g o t o B o w e n + I s l a n d ? R E C O R D O F E V E N T S 1 0 9 . C H I : p l a y w i t h I n g r i d . 1 1 0 . R E S : m m h m . E V A L U A T I O N s h e ' s o u r f a v o u r i t e c o u s h e ' s o u r f a v o u r i t e l i t t l e c o u s i n . F U T U R E P R O J E C T I O N 2 1 3 1 . R E S h o w d o e s s h e l i k e T o r o n t o ? 1 3 2 . C H I s h e l i k e s i t < w o n d e r b a r > [ ! ] . 1 3 3 . R E S mmhm? 1 3 4 . C H I b u t s h e h a s n ' t v i s i t me l a s t y e a r a n d t h e y e a r b e f o r e 1 3 5 . R E S u m . 1 3 6 . C H I i t ' s < a s h a m e > [ > ] . 1 3 7 . R E S < s o y o u h a v e > [ < ] . 1 3 8 . R E S m m h m . 1 3 9 . R E S s o y o u h a v e a < b r o t h e r x x > [ > ] . 1 4 0 . C H I < s h e ' s > [<] v i s i t i n g me a t C h r i s t m a s . 1 4 1 . R E S t h a t ' l l b e e x c i t i n g w o n ' t < i t > [ > ] ? 1 4 2 . C H I <mmhm> [ < ] . 1 4 3 . R E S m m h m . 1 4 4 . R E S i s s h e c o m i n g C h r i s t m a s d a y ? 1 4 5 . C H I s h e ' s s p e n d i n g t h e n i g h t s s p e n d i n g n i g h t s t h e r e . 1 4 6 . R E S m m h m . 1 4 7 . R E S a n d y o u r a u n t a n d u n c l e a s w e l l ? 1 4 8 . C H I y e a h . 1 4 9 . R E S m m h m . 1 5 0 . C H I U n c l e J a k e a n d A u n t M e l i s s a . 1 5 1 . C H I a n d I ' m w i t ' s s o m e t h i n g f o r me t o l o o k f o r w a r d t o . 1 5 2 . R E S t h a t ' s r i g h t . 131 1 5 3 . R E S : f o r e v e r y o n e t o l o o k f o r w a r d t o e h ? 1 5 4 . C H I : m m h m . [ N o t e : i n F u t u r e P r o j e c t i o n 2 s h e = I n g r i d ] p r o c e d u r e 1 1 7 9 . R E S : 1 8 0 . C H I : 1 8 1 . R E S : 1 8 2 . C H I : 1 8 3 . 1 8 4 . 1 8 5 . 1 8 6 . 1 8 7 . 1 8 8 . 1 8 9 . 1 9 0 . b o i l R E S R E S C H I R E S C H I C H I R E S C H I [ ! ] 1 9 1 . C H I : 1 9 2 . C H I : A B S T R A C T d o y o u d o a n y c o o k i n g a t h o m e # < W i l l > [ > ] ? <um y e a h > [ < ] . O R I E N T A T I O N w h a t d o y o u c o o k ? I d i d u n c o o k i n g f o r x x b a t c h K r a f t + D i n n e r a n d o a t m e a l , HOW T O c a n y o u t e l l me h o w t o m a k e K r a f t + D i n n e r ? h o w ' d y o u < d o i t > [ > ] ? < y o u j u s t > [<] a d d c h c h e e s e a n d . mmhm? a n ( d ) c h e e s e . t h e n y o u b o i l i t i n t h e p o t . mmhm? b u t f i r s t y o u h a v e t o w a i t t i l l t h e w a # f o r t h e w a t e r CODA^ t h a t ' s a l l . y h o w y o u m a k e K r a f t + D i n n e r . TEMPORAL STRETCHES: TRANSCRIPT 5 S P E C I F I C RECOUNT 1 9 . R E S : 1 0 . C H I : 1 1 . R E S : 1 2 . R E S : O R I E N T A T I O N I t a l k e d t o y o u o n t h e t e l e p h o n e l a s t n i g h t t o o d i d n ' t y e a h . m m h m . R E C O R D O F E V E N T S y o u a n d y o u r mom h a d j u s t g o n e f o r a d r i v e . R E C O R D O F E V E N T S 1 3 . R E S : w h e r e d i d y o u g o ? 1 4 . C H I : w e l l w e w e r e w e j u s t c a m e b a c k f w e w e r e b a c k f r o m m y c o u s i n ' s h o u s e # u p i n W o o d b r i d g e . 1 5 . R E S : m m . E V A L U A T I O N / C O D A ^ 1 6 . R E S : t h a t s o u n d s l i k e q u i t e a l o n g d r i v e . PROCEDURE 1 A B S T R A C T 1 6 3 . R E S : 1 6 4 . C H I : 1 6 5 . R E S : 1 6 6 . C H I : 1 6 7 . R E S : p u d d i n g t 1 6 8 . C H I : 1 6 9 . R E S : 1 7 0 . R E S : 1 7 1 . C H I : 1 7 2 . C H I : 1 7 3 . C H I : 1 7 4 . R E S : 1 7 5 . R E S : 1 7 6 . C H I : 1 7 7 . C H I : 1 7 8 . C H I • a n d d o y o u h a v e a f a v o u r i t e f o o d ? y e s . w h a t ? u h c h e e s e p i z z a a n d p i n e a p p l e p u d d i n g . o h r i g h t I r e m e m b e r y o u t e l l i n g me a b o u t t h e p i n e a p p l e i f o r e . y e a h . m m . O R I E N T A T I O N d o y o u e v e r m a k e c h e e s e p i z z a ? w e l l # y e s . E V A L U A T I O N b u t m y mom d o e s n ' t m y p a r e n t s d o n ' t l i k e c h e e s e , w e l l m y d a d l i k e s h i s c h e e s e o n h i s p i z z a , mmhm? HOW T O h o w d o y o u m a k e t h e c h e e s e p i z z a ? w e l l we s p r i n k l e a b i t o f m o z z a r e l l a . a n d t h e n we p u t o n t h e t o p p i n g s . t w o l a y e r s o f c h e e s e . E V A L U A T I O N , 1 7 9 . C H I 1 8 0 . R E S 1 8 1 . C H I 1 8 2 . C H I m i g h t t a s t 1 8 3 . R E S 1 8 4 . R E S 1 8 5 . C H I 1 8 6 . R E S 1 8 7 . R E S : 1 8 8 . C H I : b u t s o m e t i m e s c h e e s e t a s t e a w f u l d o n ' t y o u t h i n k ? . t a s t e s a w f u l ? y e a h . b e c a u s e i t s h o w s i f y o u i f i t d o e s n ' t g o i n t h e f r i d g e e s o u r , y e a h . o r a b i t m o u l d y ? y e a h . h m . E V A L U A T I O N / C O D A ^ b u t i t ' s p r e t t y g o o d w h e n i t ' s f r e s h i s n ' t i t ? y e a h . 1 8 9 . R E S : m m h m . 1 9 0 . R E S : I e s p e c i a l l y l i k e i t o n p i z z a s , 1 9 1 . C H I : m m h m . 1 9 2 . C H I : me t o o . 1 9 3 . R E S : mmm. TEMPORAL STRETCHES: TRANSCRIPT 6 GENERAL RECOUNT 1 1 4 7 . R E S 1 4 8 . R E S 1 4 9 . R E S 1 5 0 . C H I 1 5 1 . C H I 1 5 2 . R E S 1 5 3 . C H I 1 5 4 . R E S O R I E N T A T I O N w h a t d o y o u d o f o r g r o c e r i e s a n d f o o d a n d t h a t ? i s t h e r e a s t o r e n e a r b y ? d o y o u h a v e t o g o b y b o a t ? R E C O R D O F E V E N T S w e l l w e h a v e t o g e t o f f t h e b o a t . a n d u h t r a v e l t o t h e s t o r e < i n > [>] <mmhm> [ < ] ? I G A . mmhm? t o w n . PROCEDURE 1 O R I E N T A T I O N 1 6 5 . R E S : a n d w h a t d o y o u d o o u t s i d e a t t h e c o t t a g e ? 1 6 6 . C H I : w e l l # w e l l a t t h e c o t t a g e w e w e # w e h a v e w e h a v e t h i s w e h a v e t h i s l u m b e r m i l l u p h e r e w h i c h i s u s e d f o r s a t i n y s h e d l u m b e r s h e d o r l u m b e r m i l l c a l l e d . 1 6 7 . C H I : a n d t h e r e ' s t h i s t h e r e ' s t h i s m o t h i s a l l t h e s e t o o l s a n d l o t s o f s k i n n y w o o d a n d t h i n g s s t o r e d i n h e r e . 1 6 8 . C H I : t h e r e w a s a l s o t h i s m o t o r . 1 6 9 . C H I : t h e r e w a s a l s o t h i s m o t o r a p l u g a n d s a n a n d s o m e s a n d s t o n e w h e e l s t o s h a r p e n t h e a x e s . 1 7 0 . R E S : 1 7 1 . C H I : w o o d . 1 7 2 . C H I : 1 7 3 . C H I : e n d . 1 7 4 . R E S : 1 7 5 . C H I : s a n d s a n d . m m h m . HOW T O p l u s i f y o u g e t l i k e a r e c t a n g a s k i n n y r e c t a n g l e d p i e c e o f a n d p u t t h e e n d o n t h e e n d o f t h e s a n d s t o n e w h e e l . i t i t i t s m o o t h l y s a n d s r i g h t t h r o u g h i t l e a v i n g a s m o o t h mmhm? I t h i n k w h e n I g e t u p t h e r e y o u c a n g o s a n d s a n d s a n d s a n d E V A L U A T I O N / C O D A J 1 7 6 . 1 7 7 . C H I : R E S : m a k e s a p e r f e c t f a n c y s t i c k , d o e s i t ? PROCEDURE 2 | O R I E N T A T I O N | a n d u h # I s h a r I c a n s h I ' d g e t a n a x e a n d u h s h k i n d o f t h e r e ' s s p a r k s f l y i n g i n t h e a i r . b u t I k n o w b u t I ' v e l e a r n e d t o s h a r p e n , a n d k n o w h o w t o h a n d l e i t . u h h u h ? I k n o w h o w t o h a n d l e t h e s a n d # g r i n d e r m a c h i n e , p l u s t h e a x e s . |HOW T O 1 8 5 . C H I : p l u s I p u p u s h g e t t h e b o d y o f t h e a x e . 1 7 8 . C H I : s h a r p e n i t . 1 7 9 . C H I : 1 8 0 . C H I : 1 8 1 . C H I : 1 8 2 . R E S : 1 8 3 . C H I : 1 8 4 . C H I : 135 1 8 6 . C H I : a n d p u t i t a l o n g t h e s a n d # s t o n e t o l i k e k i n d o f c l e a n i t t o t a k e t h e r u s t o f f . 1 8 7 . C H I : a n d I i t r e v e a l e d p r i n t i n g s a y i n g t h a t i t w a s m a d e i n S w e d e n E V A L U A T I O N / C O D A ^ 1 8 8 . C H I : t h e y m a k e g o o d s t e e l t h e r e . 1 8 9 . R E S : m m h m . 136 TEMPORAL STRETCHES: TRANSCRIPT 7 GENEREAL RECOUNT 1 ABSTRACT .12 . 13 . RES : CHI: and a r e you w o r k i n g now? um -: yes. ORIENTATION 14 . CHI I work a t um on Tuesdays and Thursdays 15. CHI and t h a t ' s um i n S t r a t f o r d 16. RES uhhuh? 17 . RES so you t r a v e l from here t o t h e r e ? 18 . CHI yes . 19. RES oh r e a l l y ? EVALUATION 20. RES <that's> [>] q u i t e a t r i p . 21. CHI <yeah> [<]. 22 . CHI uh ye yeah i t i s q u i t e a t r i p . RECORD OF EVENTS 23. 24 . CHI: CHI: we go on the van f i r s t . and t h e n and th e n we t h e n we work # t h e n we work um something l i k e uh n i n e + t h i r t y t o n i n e + t h i r t y t o t w e l v e [!] 25. 26. 27 . 28 . CHI CHI CHI RES t h e n we have l u n c h a t t w e l v e a t t h a t . t h e n we s t a r t back at work a t one o ' c l o c k . and t h e n we go r i g h t a l l the way t h r o u g h t o f o u r + t h i r t y . oh -: . ORIENTATION 29. RES: and what time does t h a t get you back he r e ? RECORD OF EVENTS 30. 31. CHI: CHI : w e l l we t a k e the um f o u r + f o r t y + f i v e bus. um -: and t h a t i s t h o s e um t h o . RECORD OF EVENTS 32 . CHI: 33. CHI: 34 . CHI: 35. RES : 36. CHI: 37 . CHI: um i n Ha 38. CHI: 39. RES : 40. CHI: 41. CHI: 42 . RES : 43. CHI: 44 . CHI: 45. CHI: 46. CHI: 47 . CHI: 48. CHI: 49. CHI: 50. CHI: 51 RES : w e l l i t ' s not F u n t r a c k . i t used t o be F u n t r a c k . i t ' s Trentway <now> [>] . <mmhm> [<] mmhm? RECORD OF EVENTS and um # and we t a k e the we t a k e t h e f o u r + f o r t y + f i v e bus. t h e n um i t goes t o a l l t h e s e l i t t l e towns b e f o r e i t s t o p s a t ORIENTATION < l i k e > [>] um what i s i t ? <oh -: > [<] . uh B e a m s v i l l e V i n e l a n d um # um t h a t F o o d l a n d . not F o o d l a n d . P o r t P e r r y <maybe> [>]? <yeah> [<]. w e l l w yes yes. P o r t P e r r y um and Foo d l a n d . yeah. t h a t ' s what i t ' s c a l l e d . i t ' s a l i t t l e town ne a r . i t ' s n e ar i t ' s a t sc a Scugog. we s t o p by t h e r e t o o . mmhm? 137 5 2 . R E S : 5 3 . R E S : 5 4 . C H I : 5 5 . C H I : G E N E R A L 7 1 . R E S : 7 2 . C H I : 7 3 . R E S : 7 4 . C H I : 7 5 . R E S : 7 6 . C H I : 7 7 . R E S : G E N E R A L 1 1 3 . R E S h e r e ? 1 1 4 . R E S 1 1 5 . C H I d o w n s t a i 1 1 6 . C H I 1 1 7 . C H I 1 1 8 . R E S 1 1 9 C H I 1 2 0 C H I u m . 1 2 1 C H I 1 2 2 C H I 1 2 3 C H I m m h m . E V A L U A T I O N / C O D A ^ t h a t m a k e s q u i t e a l o n g d a y d o e s n ' t i t ? i t d o e s . y e a h . O R I E N T A T I O N w h a t t i m e d o y o u l e a v e h e r e i n t h e m o r n i n g ? R E C O R D O F E V E N T S um - : i t ' s u s u a l l y a f t e r e i g h t , u h h u h ? b e t w e e n e i g h t + f i f t e e n a n d e i g h t + t h i r t y w e u s u a l l y l e a v e , m m h m . a n d w e g e t u h t o M a i l b o x e s a t n i n e + t h i r t y . u h h u h . O R I E N T A T I O N s o w h a t d o y o u u s u a l l y d o i n t h e e v e n i n g s w h e n y o u g e t b a c k w h a t d o y o u d o a f t e r s u p p e r ? I R E C O R D O F E V E N T S l w e l l T u e s d a y s a f t e r w o r k i n g w e h a v e t o d o a w o r k o u t a n d y o u h a v e t o l i f t w e i g h t s . 0 [ = ! c h u c k l e s ] . E V A L U A T I O N o h g o o d n e s s , y e a h r i g h t . R E C O R D O F E V E N T S a n d o n t h u r s w e l l # w e l l o n T h u r s d a y s I u s e d t o g o t o um t o w h a t w a s i t a s s p o r t s + n i g h t i n u m i n G r i m s b y . I d o n ' t n o w . u h I I d o b o w l i n g n o w . G E N E R A L R E C O U N T 4 1 4 4 . R E S : 1 4 5 . C H I : 1 4 6 . R E S : 1 4 7 . R E S : 1 4 8 . C H I : 1 4 9 . C H I : 1 5 0 . C H I : c l a s s e s . 1 5 1 . C H I : O R I E N T A T I O N < d o y o u > [<] d o y o u d o y o u t a k e p a r t i n o t h e r s p o r t s W i l l ? u h - : . a r e t h e r e a n y o t h e r s p o r t s t h a t y o u d o ? b a s e b a l l o r < s o c c e r - : > [ > ] . R E C O R D O F E V E N T S < w e l l > [<] we u I u s e d t o d o t h a t f o r s p o r t s n i g h t . I I d o n ' t v e r y m u c h n o w . b u t I d o o n e # b u t u m o n S u n d a y s I g o t o u h m y f i t n e s s a n d I d o l i f t i n g l i f t i n g o n e u p . 138 ORIENTATION 152 . CHI w e l l t h o s e t h o s e w e i g h t t h i n g s . 153 . CHI e x c e p t t h e y ' r e no t t he b i g s o r t o f w e i g h t s . 154 . CHI because I t r i e d t o l i f t t h o s e . 155 . CHI <and c o u l d n ' t do t h a t > [% c h u c k l i n g ] . 156 . CHI so i t ' s more o f t h e mach ine s o r t o f w e i g h t s . 157 . RES uhhuh? RECORD OF EVENTS 158 . CHI you do t h e . 159 . CHI e x c u s e me. 160 . CHI <xxx> [>] . 1 6 1 . RES <0 [=! c h u c k l e s ] > [<] . 162 . CHI um # I do um. 163 . CHI what e l s e do I do? 164 . CHI I do t h i s um # t h e ones where you go l i k e t h i s 165 . RES oh r i g h t . 166 . CHI t h e n you # you do t h a t . 167 . RES y e a h . 168 . CHI I do um s o m e t h i n g w i t h # t h a t . 169 . CHI I do #. 139 TEMPORAL STRETCHES: TRANSCRIPT 8 S P E C I F I C R E C O U N T 1 O R I E N T A T I O N 9 . R E S : < a n d > [<] y o u w e r e t e l l i n g me a b o u t a c e m e n t t r u c k b e f o r e 1 0 C H I mm - : y e s ! 1 1 C H I i t h a d a w e t h e a v y l o a d . 1 2 R E S a w e t h e a v y l o a d - : . 1 3 C H I o f c o n c r e t e . 14 R E S f r o m w h a t ? 1-5 C H I f o r t h e m f o r f o r t h e m i g h t y m i x e r . 1 6 C H I t h a t x x x . R E C O R D O F E V E N T S 1 7 . R E S h o w d o e s i t g e t f i l l e d u p ? 1 8 . C H I j u s t g o x x x . F U T U R E P R O J E C T I O N 1 2 4 . R E S a n d t h e n w h a t w o u l d i t d o ? 2 5 . C H I i t w o u l d m o v e a l o n g . 2 6 . R E S w h e r e w o u l d i t m o v e a l o n g t o ? 2 7 . C H I t h e c o n s t r u c t i o n s i t e . 2 8 . R E S t h e c o n s t r u c t i o n s i t e . 2 9 . R E S a n d w h a t w o u l d i t d o t h e r e ? 3 0 . C H I p o u r c e m e n t . 3 1 . R E S p o u r c e m e n t ? 3 2 . C H I y e s . 3 3 . C H I a n d t h e n i t h a r d e n s . 3 4 . R E S a n d w h e n i t h a r d e n s w h a t h a p p e n s ? 3 5 . C H I t h e n i t d r i e s [ ! ] . 3 6 . R E S i t d r i e s . 3 7 . R E S u h h u h . [ N o t e : i n F u t u r e P r o j e c t i o n 1 i t = c e m e n t t r u c k ] S P E C I F I C R E C O U N T 2 p R I E N T A T I O N | 3 8 . R E S : a n d w h a t d o e s i t m a k e ? R E C O R D O F E V E N T S 3 9 . C H I : um # a n d i t t u r n s t h a t w a y i n s t e a d o f c o m i n g t h i s w a y . i t = t r u c k 4 0 . C H I : t h a t ' s b e c a u s e t h e y a r e m a n y m a c h i n e s i n t h e c o n s t r u c t i o n s i t e . 4 1 . R E S : t h e r e a r e m a n y w h i c h ? 4 2 . C H I : t h e m t h e r e a r e m a n y m a c h i n e s i n t h e c o n s t r u c t i o n s i t e . [ N o t e : i n S p e c i f i c R e c o u n t 2 i t = c e m e n t ] 140 F U T U R E P R O J E C T I O N 2 w h a t h a p p e n s i f y o u g o o n t h e t h e m i d n i g h t e x p r e s s ? I d o n ' t k n o w , w h a t h a p p e n s ? w h e n y o u g e t w h e n u m # t h e u h y a n d t h e n y o u w a k e u p a t t h e 5 8 . . C H I : 5 9 . R E S : 6 0 . R E S : 6 1 . C H I : w h i s t l e < w h e n y o u g e t o n > [ ? ] 6 2 . R E S 6 3 . C H I 6 4 . R E S 6 5 . R E S I g u e s s y o u w o u l d s l e e p d u r i n g t h a t t r a i n r i d e e h ? y e s I w o u l d . I s e e . a n d t h e w h i s t l e w o u l d w w a k e y o u u p y o u m e a n ? F U T U R E P R O J E C T I O N 3 1 0 3 . C H I 1 0 4 . R E S 1 0 5 . C H I 1 0 6 . C H I 1 0 7 . R E S 1 0 8 . R E S 1 0 9 . C H I 1 1 0 . R E S w h a t h a p p e n s u h w h e n u h t h e t r a i n c r a s h ? w h a t d o y o u t h i n k h a p p e n s ? t h e n t h e m e n w i l l d i e . t h e n t h e y w i l l m a k e t h e m [ ? ] g o t o t h e h o s p i t a l . w e l l t h e y w o u l d i f t h e y w e r e i l l u h s i c k w o u l d n ' t t h e y ? o r i f t h e y w e r e i n j u r e d . i f t h e y w e r e i n j t h e y w e r e i n j u r e d # w o u l d n ' t t h e y ? m m h m . G E N E R A L R E C O U N T 1 O R I E N T A T I O N 1 1 9 . R E S : 1 2 0 . C H I : 1 2 1 . R E S : 1 2 2 . C H I : 1 2 3 . R E S : 1 2 4 . C H I : 1 2 5 . C H I : 1 2 6 . C H I : 1 2 7 . C H I : 1 2 8 . R E S : 1 2 9 . R E S : 1 3 0 . C H I : 1 3 1 . C H I : s t u c k i n 1 3 2 . R E S : 1 3 3 . C H I : 1 3 4 . R E S : 1 3 . 5 . R E S : 1 3 6 . C H I : 1 3 7 . R E S : w h a t ' s a d o g < g a m e > [ > ] ? I h a v e [<] s o m u c h I h a v e s o m u c h f u n . w h a t ' s a d o g g a m e l i k e ? i t ' s i t ' s x x x t r y i n g t o g e t t h e o t h e r d o g s o u t . o h w h o d o y o u p l a y w i t h ? u m - : s o m e t i m e I p l a y w i t h S o p h i a w h a t w h a t d i d S o p h i a d o ? w e l l # w h a t d i d s h e d o ? w h a t d i d s h e d o t o m e ? I d o n ' t k n o w . R E C O R D O F E V E N T S d i d s h e d o s o m e t h i n g t o y o u ? y e s . n o w s h e w a s s t u c k i n i n b e t w e e n i n b e t w e e n m a y b e s h e g o t b e t w e e n . O R I E N T A T I O N i n b e t w e e n w h a t ? i n b e t w e e n # t h e t r e e s , o h i n b e t w e e n t h e t r e e s ? < w h e r e i n > [ > ] ? R E C O R D O F E V E N T S n o w S o p h i a w a s g o t t e n s t u c k i n b e t w e e n # o n a s u m m e r d a y . o n a s u m m e r d a y . 141 F U T U R E P R O J E C T I O N 4 1 5 7 . C H I w h a t w o u l d s c r a t c h y o u . 1 5 8 . C H I w o u l d n ' t t h a t b e t e r r i b l e ? 1 5 9 . R E S m m h m . 1 6 0 . C H I a w f u l ! 1 6 1 . C H I s t i n k . 1 6 2 . C H I a n d t h e n y o u h a t e l i o n s . 1 6 3 . C H I < s t i n k h a i r y > [ ? ] . 1 6 4 . C H I c a u s e s h e ' s s t i n k y . 1 6 5 . C H I a n d t h e n < I l i I - : l e a v e > [ ? ] F U T U R E P R O J E C T I O N 5 2 6 3 . C H I : u m - : w h a t d o y o u [ ! ] d o # i f y o u ' r e s t u c k i n b e t w e e n i n -# w h a t h a p p e n s i f y o u ' r e w h a t h a p p e n s i f y o u ' r e f e e l i n g a l i t t l e d o w n 2 6 4 . R E S 2 6 5 . R E S 2 6 6 . C H I 2 6 7 . C H I i f y o u ' r e f e e l i n g a l i t t l e d o w n ? h m . y o u r h e a d w i l l g e t r e d . t h e n y o u ' l l s c r e a m . P R O C E D U R E 1 p R I E N T A T I O N | 3 2 8 . R E S : t h a t i s a t u r n t a b l e . 3 2 9 . R E S : h o w d o y o u w o r k i t ? |HOW TOl 3 3 0 . C H I d o i t l i k e t h a t . 3 3 1 . R E S o h I s e e . 3 3 2 . C H I , t o g e t t o t h e o u t s i d e . 3 3 3 . R E S u m . 3 3 4 . C H I s o y o u g o o n t h e t o t o g e t o n t h e t u r n t r a y 3 3 5 . R E S g e t o n t h e t u r n t a b l e . 3 3 6 . C H I s a y \" l o o k o u t y o u d o n ' t f a l l \" . 3 3 7 . R E S l o o k o u t y o u d o n ' t f a l l . 3 3 8 . C H I s 1 ## h e ' s g o i n g o n t h e t a b l e . 3 3 9 . C H I w h a t h a p p e n s i f y o u g o o n t h e t u r n t a b l e . 3 4 0 . R E S y o u t u r n a r o u n d . 142 TEMPORAL STRETCHES: TRANSCRIPT 9 F U T U R E P R O J E C T I O N 1 252. RES: <um> [<] what are, you gonna do when <you go> [>] home.today? 253. CHI: <0 [=! t h r o a t n o i s e s ] > [<] . 254. RES: what do you do a f t e r s c h o o l ? 255. CHI: supper. 256. RES: supper? 257. RES: yeah? 143 APPENDIX 2 Summary of Genre Types and Generic Stages Used in 9 Texts Total SO 00 - J NJ Transcript Number e o o © © © © © © © Narrative o o © © © © © © © © Anecdote e o © © © © © © © © Exemplum in o NJ © © - © © - Specific Recount o - - © - - - General Recount cn o - © NJ © © © Procedure >j\\ o © - © - - NJ © © Abstract NI o © © © 00 © 00 © © Abstract by Child Solicited (S)/Unsolicited (U) Ul o NJ NJ Orientation 13S 6U o OJ 00 C IS 1U © 00 7S 2U 00 © Orientation by Child Solicited (S)/UnsoIicited (U) Ul - J o - NJ Ul Record of Events 19S 1- o 00 4^ 00 00 00 00 NJ 00 00 # 00 OJ 00 e G c G Record of Events by Child Solicited (S)/Unsolicited (U) o © © NJ - © © o How To co c o © © NJ C 00 00 © © © How To by Child Solicited (S)/Unsolicited (U) o o © © © © © © © © Remarkable Event o o © © © © © © © © Remarkable Event by Child Solicited (S)/Unsolicited (U) o © © © © © - © © Reaction o © © © © © © © © © Reaction by Child Solicited (S)/Unsolicited (U) t—* o © - NJ NJ - - © Coda e © © © NJ G © c © © © Coda by Child Solicited (S)/Unsolicited (U) - J o © — - © Evaluation 1S4U o © © C NJ G C 00 © © Evaluation by Child Solicited (S)/Unsolicited (U) - Ul © © © NJ NJ - Future Projection APPENDIX 3 Utterances that Use Units of Time Extracted from 9 Texts Text 1 140. RES: I had p i z z a l a s t night f o r supper. 141. RES: what d i d you have f o r supper l a s t night? 14 6. RES: and th e night b e f o r e I had por k chops. 147. RES: can you remember what you had th e night b e f o r e ? 151. RES: what are you h a v i n g tonight # f o r supper? 153. CHI: I'm h a v i n g r i c e f o r supper tonight. Text 2 1. RES: today i s June the sixth. 3. CHI: <fifth> [<] . 4. RES: i s i t the f i fth? 6. CHI: tomorrow's th e sixth. 8. RES: today i s June the f i f t h . 219. RES: so what do you do on th e weekends W i l l ? 223. RES: and spends some time w i t h you? Text 3 21. CHI: on on Tuesday and Wednesday Judy Winston and J we b o t h went t o t h e a i r p o r t t o p i c k up So p h i e . 43. CHI: and we went o u t s i d e a t night and p l a y e d i n t h e snow um <and> [>] • 45. CHI: the nex t day I we I went t o the bookshop. 70. CHI: t h e n e x t morning I packed my s u i t c a s e up. 82. CHI: my f a t h e r went t h e r e w an my f a t h e r and I went t h e r e once b e f o r e grade # went t h e r e once b e f o r e grade five started. 94. RES: was i t a long drive? 96. CHI: i t was a long drive. 98. RES: how long d i d i t t a k e you W i l l ? ( i t = th e d r i v e ) 101. CHI: i t too k i t to o k wo i t too k five hundred and sixty four minutes. ( i t = t h e d r i v e ) 147 1 0 3 . C H I : i t t o o k f i v e hundred and s i x t y four minutes. 1 0 5 . C H I : a n d t h e c a r t r i p was long [!] . 1 0 7 . R E S : d i d y o u s l e e p during the car t r i p ? 1 1 5 . C H I : a n d t h e n e x t morning w h e n I w o k e u p I h a d s o m e b r e a d 1 2 7 . C H I : < t h e y ' r e c o m i n g > [>] a n ( d ) v i s i t a n d v i s i t i n g me i n t h e summer i n July [!] . 1 3 9 . R E S : w h e n y o u w e r e t h e r e y o u w e r e o n l y a b o u t a h a l f an hour f r o m a f a r m w h e r e I y u I g r e w u p 1 4 4 . R E S : a n d m y mom u s e d t o d r i v e a h a l f an hour. 1 7 0 . R E S : a n d w e n e e d t o d o a b o u t # three or four more minutes. 1 7 6 . R E S : t e l l me a b o u t y o u r u h y o u d o s o m e t h i n g o n Saturday ' s u s u a l l y . 1 7 7 . R E S : w h a t d o y o u u s u a l l y d o o n Saturdays? 1 9 1 . R E S : h m h o w long h a v e y o u b e e n d o i n g t h a t ? 1 9 2 . C H I : I h a v e b e e n d o i n g i t f o r t w e l ( v e ) I h a v e d I h a v e d o n e i t f o r twelve rides -: . 2 0 5 . C H I : a n d h a d a N i e l s e n L P w h i c h i s c a l l e d T h e P o i n t r e l e a s e d i n 1985 l i k e l i k e o u r v i d e l i k e o u r v i d e o T h e P o i n t w h i c h w a s a w h i c h w a s r e p l a c e d a t t h e v i d e o b o x . 2 0 6 . C H I : a n d b y a n d o n e a n d once m y m o t h e r a n d I g o t i t i n u m July a n d August # b o u g h t i t i n June a n d August. 2 0 9 . C H I : b u t t h e v i d e o w a s n o t r e l e a s e d i n 1985. 2 1 0 . C H I : i t w a s u h r e l e a s e d b e f o r e t h e L P i n 1984 - : ## u h 1 9 8 4 a n d 1 9 8 9 - : . 2 2 2 . R E S : t e l l me w h a t y o u r f a v o u r i t e - : t h i n g i s t o d o i n t h e summer time. Text 4 1 3 4 . C H I : b u t s h e h a s n ' t v i s i t me l a s t year a n d t h e year b e f o r e . 1 4 0 . C H I : < s h e ' s > [<] v i s i t i n g me a t Christmas. 1 4 4 . R E S : i s s h e c o m i n g Christmas day? 1 4 5 . C H I : s h e ' s s p e n d i n g t h e night s s p e n d i n g nights t h e r e . 1 5 1 . C H I : a n d I ' m w i t ' s s o m e t h i n g f o r me t o look forward to. 1 5 3 . R E S : f o r e v e r y o n e t o look forward to e h ? 148 190. CHI: but f i r s t you have t o wait t i l l the wa # f o r the water t o b o i l [!]. Text 5 9. RES: I t a l k e d t o you on the t e l e p h o n e l a s t night t o o d i d n ' t I ? 16. RES: t h a t sounds l i k e q u i t e a long drive. Text 6 [none] Text 7 12. RES: and ar e you w o r k i n g now? 14. CHI: I work a t um on Tuesdays and Thursdays a t a t um M a i l b o x e s . 20. RES: <that's> [>] quite a t r i p . 22. CHI: uh ye yeah i t i s quite a t r i p . 24. CHI: and t h e n and then we then we work # th e n we work um something l i k e uh nine+thirty t o nine+thirty t o twelve [!]. 25. CHI: t h e n we have l u n c h a t twelve a t t h a t . 26. CHI: t h e n we s t a r t back a t work a t one o'clock. 27. CHI: and then we go r i g h t a l l t h e way t h r o u g h t o four+thirty. 29. RES: and what time does t h a t get you back he r e ? 30. CHI: w e l l we t a k e t he um four+forty+five bus. 36. CHI: and um # and we t a k e t he we t a k e t he fo 149 ur+forty+five b u s . 5 3 . R E S : t h a t m a k e s q u i t e a long day d o e s n ' t i t ? 7 1 . R E S : w h a t t i m e d o y o u l e a v e h e r e i n t h e morning? 7 2 . C H I : um - : i t ' s u s u a l l y a f t e r eight. 7 4 . C H I : b e t w e e n eight+fifteen a n d eight+thirty w e u s u a l l y l e a v e . 7 6 . C H I : a n d w e g e t u h t o M a i l b o x e s a t nine+thirty. 1 1 3 . R E S : s o w h a t d o y o u u s u a l l y d o i n t h e evenings w h e n y o u g e t b a c k h e r e ? 1 1 4 . R E S : w h a t d o y o u d o after supper? 1 1 5 . C H I : w e l l Tuesdays a f t e r w o r k i n g w e h a v e t o d o a w o r k o u t d o w n s t a i r s . 1 2 0 . C H I : a n d o n thurs w e l l # w e l l o n Thursdays I u s e d t o g o t o u m t o u m . 1 2 1 . C H I : w h a t w a s i t a s s p o r t s + n i g h t i n u m i n G r i m s b y . 1 4 8 . C H I : < w e l l > [<] w e u I u s e d t o d o t h a t f o r s p o r t s n i g h t . 1 5 0 . C H I : b u t I d o o n e # b u t um o n Sundays I g o t o u h m y f i t n e s s c l a s s e s . Text 8 5 8 . C H I : w h a t h a p p e n s i f y o u g o o n t h e t h e midnight e x p r e s s ? 6 2 . R E S : I g u e s s y o u w o u l d s l e e p during that train ride e h ? 1 3 6 . C H I : now C h r i s t i n a w a s g o t t e n s t u c k i n b e t w e e n # o n a summer day. 1 3 7 . R E S : o n a summer day. Text 9 2 5 4 . R E S : w h a t d o y o u d o after school? 150 APPENDIX 4 Analysis of Metaphors, Metonymies, and Senses of Time of Utterances that Use Temporal Units 1 5 Text Os o Line Number CHI RES RES RES RES RES Speaker I'm having rice for supper tonight. what are you having tonight # for supper? ,, can you remember what you had the night before? and the night before I had pork chops. what did you have for- supper last night? I had pizza last night for supper. Utterance w Solicited (S) / Unsolicited (U) Number of Temporal Units Ignored Utterance Time Orientation Metaphor Moving Observer Metaphor Moving Time Metaphor Event-for-Time Metonymy Distance-for-Time Metonymy Time-for-Distance Metonymy Time is a Resource Metaphor Time is Money Metaphor Evans's Sense of Time Evans' Sense of Time t o to to to to to to Text t o to to 00 - Line Number RES RES RES CHI RES CHI RES Speaker and spends some time with you? so what do you do on the weekends Will? today is June the fifth. tomorrow's the sixth. is it the fifth? A Hi H-Hi V \"A\" today is June the sixth. Utterance C C Solicited (S) / Unsolicited (U) - t o to - t o Number of Temporal Units Ignored Utterance Time Orientation Metaphor Moving Observer Metaphor Moving Time Metaphor Event-for-Time Metonymy Distance-for-Time Metonymy Time-for-Distance Metonymy Time is a Resource Metaphor - Time is Money Metaphor - Evans's Sense of Time Commodity Evans' Sense of Time 3 to Text Line Number n x Speaker BJ A 3 rt CD < < P - \" W 1-1 3 O P -g 3 CD ua v p-3 — V r r — cr CD Bi 3 T J 3 a cr cr 0) ro 3 « CD B cr O rt a M H -O 3 H-a o s- p-r^ \" 6-cT 0 O a w t-h 3 0J >< cr cr CD CD Hi Mi o o 1-1 1-1 Hi ID h rt cr CD a> H i-t cr m CD 3 l-l a a a 0 CD M ff rt \" cr O CD 3 cr Hi * p -(D CD cr CD |a CD O 3 O CD s CD 3 r t cr CD r t M cr CD CO P - z 3 CD 3 \"rt r t CT ro o A ai 3 3 P a iQ V rt v n> — 3 a 3 3 3 (0 S ttJ CD 3 0-r t a> crra E 0 CD 0) 3 a r t | r t O C i C a re 65 Solicited (S) / Unsolicited (U) Number of Temporal Units Ignored Utterance Time Orientation Metaphor Moving Observer Metaphor Moving Time Metaphor Event-for-Time Metonymy Distance-for-Time Metonymy Time-for-Distance Metonymy Time is a Resource Metaphor Time is Money Metaphor Evans's Sense of Time Evans' Sense of Time 3 139 RES w h e n y o u w e r e t h e r e y o u w e r e o n l y a b o u t a h a l f an hour f r o m a f a r m w h e r e I y u I g r e w u p 1 1 1 3 144 RES a n d my mom u s e d t o d r i v e a h a l f an hour. 1 1 1 1 3 170 RES a n d we n e e d t o d o a b o u t # three or four more minutes. 2 1 3 176 RES t e l l me a b o u t y o u r u h y o u d o s o m e t h i n g o n Saturday ' s u s u a l l y . 1 1 1 3 177 RES w h a t d o y o u u s u a l l y d o o n Saturdays ? 1 1 1 3 191 RES hm h o w long h a v e y o u b e e n d o i n g t h a t ? 1 1 1 1 1 3 192 CHI I h a v e b e e n d o i n g i t f o r t w e l ( v e ) I h a v e d I h a v e d o n e i t f o r t w e l v e r i d e s - : . S 1 1 1 1 * Instance 3 205 CHI a n d h a d a . N i e l s e n L P w h i c h i s c a l l e d T h e P o i n t r e l e a s e d i n 1985 l i k e l i k e o u r v i d e l i k e o u r v i d e o T h e P o i n t w h i c h w a s a w h i c h w a s r e p l a c e d a t t h e v i d e o b o x . u 1 1 1 3. 206 CHI a n d b y a n d o n e a n d o n c e my m o t h e r a n d I g o t i t i n um J u l y a n d August # b o u g h t i t i n June a n d August. u 4 1 1 3 209 CHI b u t t h e v i d e o w a s n o t r e l e a s e d „ i n 1985. u 1 1 . 1 3 210 CHI i t w a s u h r e l e a s e d b e f o r e t h e L P i n 1984 - : ## u h 1984 a n d 1989 -u 3 1 1 3 222 RES t e l l me w h a t y o u r f a v o u r i t e - : t h i n g i s t o d o i n t h e summer time. 1 1 1 1 Duration 4*. 4*. 4*. 4^ Text SO o U l U) 4* U i 4^ 4^ O U) 4^ Line Number CHI RES CHI CHI RES CHI CHI Speaker but first you have to wait till the wa # for the water to boil for everyone to look forward to eh? and I'm w it's something for me to look forward to. she's spending the night s spending nights there. is she coming Christmas day? <she's> [<] visiting me at Christmas. but she hasn't visit me last year and the year before. Utterance C C CO G G Solicited (S) / Unsolicited (U) - - to Number of Temporal Units Ignored Utterance - - - - - Time Orientation Metaphor - Moving Observer Metaphor Moving Time Metaphor Event-for-Time Metonymy Distance-for-Time Metonymy Time-for-Distance Metonymy Time is a Resource Metaphor - Time is Money Metaphor Evans's Sense of Time Evans' Sense of Time Ul ui Text -o Line Number RES RES Speaker that sounds like quite drive. I talked to you on the last night too didn't I Utterance a long telephone Solicited (S) / Unsolicited (U) — Number of Temporal Units Ignored Utterance - - Time Orientation Metaphor - Moving Observer Metaphor Moving Time Metaphor Event-for-Time Metonymy - Distance-for-Time Metonymy Time-for-Distance Metonymy Time is a Resource Metaphor Time is Money Metaphor Evans's Sense of Time Evans' Sense of Time 8SI -0 -o -4 -0 •o - J Text - o OS ~ J - J t o -o u> u> U ) Os o t o v© t o - J t o Os t o U l t o t o to to © IO Line Number CHI CHI CHI RES RES CHI CHI 1 RES CHI CHI CHI CHI CHI RES CHI RES Speaker and we get uh to Mailboxes at nine+thirty. between eight+fifteen and eight+thirty we usually leave. um - it's usually after eight. what time do you leave here in the morning? that makes quite a long day doesn't it? and um # and we take the we take the four+forty+five bus. well we take the um four+forty+five bus. and what time does that get you back here? and then we go right all the way through to four+thirty. then, we start back at work at one o'clock. then we have lunch at twelve at that. and then and then we then we work # then we work um something like uh nine+thirty to nine+thirty to twelve [!]. uh ye yeah it is quite a trip. <that's> [>] quite a trip. I work at um on Tuesdays and Thursdays at at um Mailboxes. and are you working now? Utterance C G t o G t o G G G G G G Solicited (S) / Unsolicited (U) t o Number of Temporal Units - Ignored Utterance - - - - - - - - Time Orientation Metaphor - - - - 1—» - - - Moving Observer Metaphor Moving Time Metaphor Event-for-Time Metonymy Distance-for-Time Metonymy Ti m e-fo r-Dista n ce Metonymy Time is a Resource Metaphor Time is Money Metaphor - Evans's Sense of Time Moment Moment Evans' Sense of Time 7 113 RES s o w h a t d o y o u u s u a l l y d o i n t h e e v e n i n g s w h e n y o u g e t b a c k h e r e ? 1 1 1 7 114 RES w h a t d o y o u d o a f t e r s u p p e r ? 1 1 1 7 115 CHI w e l l T u e s d a y s a f t e r w o r k i n g we h a v e t o d o a w o r k o u t S 1 1 1 7 120 CHI a n d o n t h u r s w e l l # w e l l o n T h u r s d a y s I u s e d t o g o t o um t o U 2 1 1 7 121 CHI w h a t w a s i t a s s p o r t s + n i g h t i n um i n G r i m s b y . U 1 1 7 148 CHI < w e l l > [<] we u I u s e d t o d o t h a t f o r s p o r t s n i g h t . s 1 1 7 150 CHI b u t I d o o n e # b u t um o n S u n d a y s I g o t o u h my f i t n e s s u 1 1 1 00 00 OO o o Text U l - J U ) Os ON U l 00 Line Number RES RES RES CHI Speaker on a summer day. now Christina was gotten stuck in between # on a summer day! I guess you would sleep during that train, ride eh? what happens if you go on the the midnight express? Utterance G Solicited (S) / Unsolicited (U) - Number of Temporal Units Ignored Utterance - Time Orientation Metaphor - Moving Observer Metaphor Moving Time Metaphor Event-for-Time Metonymy Distance-for-Time Metonymy Time-for-Distance Metonymy Time is a Resource Metaphor Time is Money Metaphor Evans's Sense of Time Evans' Sense of Time SO Text to 4^ Line Number RES Speaker what do you do after sc] Utterance hool? Solicited (S) / Unsolicited (U) - Number of Temporal Units Ignored Utterance - Time Orientation Metaphor Moving Observer Metaphor Moving Time Metaphor Event-for-Time Metonymy Distance-for-Time Metonymy Time-for-Distance Metonymy Time is a Resource Metaphor Time is Money Metaphor Evans's Sense of Time Evans' Sense of Time `\n\n#### Cite\n\nCitation Scheme:\n\nCitations by CSL (citeproc-js)\n\n#### Embed\n\nCustomize your widget with the following options, then copy and paste the code below into the HTML of your page to embed this item in your website.\n``` ```\n<div id=\"ubcOpenCollectionsWidgetDisplay\">\n<script id=\"ubcOpenCollectionsWidget\"\nsrc=\"{[{embed.src}]}\"\ndata-item=\"{[{embed.item}]}\"\ndata-collection=\"{[{embed.collection}]}\"", null, "Our image viewer uses the IIIF 2.0 standard. To load this item in other compatible viewers, use this url:\n`https://iiif.library.ubc.ca/presentation/dsp.831.1-0100980/manifest`" ]

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[ "# Suppose a consumer has the following utility function: U(x1,x2) = alnx1 + (1-a)lnx2\n\n##### Description\n\nFind the Marginal Rate of Substitution (MRS) for the following utility functions:\n\na. U(X,Y) = X0.3Y0.4\n\nb. U(X,Y) = XaY1-a\n\n2. Suppose a consumer has the following utility function: U(x1,x2) = alnx1 + (1-a)lnx2\n\na. Use the Lagrange method to find the demand functions for x1 and x2\n\nb. Suppose the price of good 1 is \\$10 and the price of good 2 is \\$20 and the consumer’s\n\nincome is \\$500. Find the quantities of x1 and x2 if a=1/2.\n\n3. A consumer has the following utility function: U(X1,X2) = X1X2 + 3X1 + X2\n\nWhere X1 is her consumption of pastries with a price of \\$8 and X2 is her consumption of books\n\nwith a price of \\$12. Her income is \\$212. Determine the number of pastries and books that will\n\nmaximize the consumer’s utility.\n\n4. A consumer has the following demand function for good 1: x1 = (3/4)(m/p1)\n\nThe original price of the good is \\$2 and the consumer’s income is \\$200. Calculate the substitution\n\neffect, the income effect and the total effect for this consumer, when the price of good changes to\n\n\\$1.\n\nBasic features\n• Free title page and bibliography\n• Unlimited revisions\n• Plagiarism-free guarantee\n• Money-back guarantee" ]

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[ "# sxlib.5alc - Man Page\n\na portable CMOS Standard Cell Library\n\n## Description\n\nsxlib library contains standard cells that have been developed at UPMC-ASIM/LIP6. This manual gives the list of available cells, with their behavior, width, maximum delay and input fan-in. This manual gives also few thumb rules to help the user to well use the cells. The given delay are the maximum (that means worst case for a generic .35 micron process). More precise delay can be found in ALLIANCE VHDL behavior files (.vbe file). Cell-name is built that way <behavior>_<output drive> (see explanations below).\n\n```Four files are attached to each cell:-\n- ALLIANCE Layout ............... cell-name.ap\n- ALLIANCE Transistor net-list .. cell-name.al\n- ALLIANCE VHDL behavior ........ cell-name.vbe\n- Compiled HILO behavior ........ 0000000xx.dat\n\nAnd few files more:-\n- CATAL ......................... ALLIANCE catalog file\n- sxlib.cct ..................... Cell definition for HILO CAD tools\n- CIRCUIT.idx ................... HILO catalog file\n- sxlib.lib ..................... Cell definition for Synopsys CAD tools\n- sxlib.db ...................... Compiled cell definition for Synopsys\n- sxlib.sdb ..................... Icon definition for Synopys```\n\n## Physical Outline\n\nsxlib uses the symbolic layout promoted by Alliance in order to provide process independence. All dimensions are in lambda units. The mapping to a specific process CIF or GDS2 layout must be performed by the s2r tool (symbolic to real), which uses a value for the lambda (e.g. 1 lambda=0.3um).\n\n``` _________________\n50 | VDD |\n45 |_________________| x : place of virtual connector.\n40 | x |\n35 | x x | they are named : name_<y>\n30 | x x |\n25 | x x | for example : i0_20\n20 | x | i0_25\n15 | x | i0_30\n10 |_________________|\n5 | VSS |\n0 |_________________|\n0 5 10 15 20 25 30```\n\nAll cells are 50 lambdas high and N times 5 lambdas wide, where N is the number of pitches. That is the only physical information given in the cell list below.  Power supplies are in horizontal ALU1 and are 6 lambdas wide.  Connectors are inside the cells, placed on a 5x5 grid. Half layout design rules are a warranty for any layer on any face, except for the power supply and NWELL.  Cells can be abutted in all directions whenever the supply is well connected and connectors are always placed on the 5x5 grid.\n\n## Delay Model\n\nCells have been extracted and simulated by using a generic 0.35um process in order to give realistic values for the delays and capacitances.  We chose to give only the worst delay for each output signal, though it is not very realistic (since delay depends on each input, an input can be easily up to twice faster than another). However, we just wanted to give an idea of the relative delay.\n\nFurthermore, we added 0.6ns to each output delay in order to take into account the delay due to the signal commutation. We have supposed the output drives the maximum capacitance. This capacitance have been computed as follow. We considered that a good slope signal for this process was 0.8ns.  Then we searched for the capacitance required to obtain the same input and output slope (0.8ns) for the smaller inverter (inv_x1). That was 125fF. We simulated the same inverter without output capacitance. The delay difference was about 0.6ns. This result is not exactly the same for all cells, but 0.6ns is a good approximation.\n\nThe given delay is then a worst case (70degree, 2.7Volt, slow process, worst input), an idea of the typical delay can be obtain by dividing worst delay by 1.5, and best delay by dividing by 2.  More detailed data can be found in GENERIC data included in the VHDL files (.vbe). Examples can be found at the end of this manual.\n\nAt last, to get a very better idea about the real delay without simulating the spice transistor netlist, it is required to use the TAS (1) tool, which is a timing static analyzer able to give the longer and the shorter path for a given process.\n\n## Output Drive\n\nThe output drive of a cell gives an information on the faculty for the cell to drive a big capacitance. This faculty depends on the rising and falling output resistance. The smaller the resistance, the bigger can be the capacitance.  Minimum drive is x1. This corresponds to the smallest available inverter (inv_x1). x2 means the cell is equivalent (from the driving point of view) at two smaller inverters in parallel, and so on.\n\nThe maximum output drive is x8. It is limited because of the maximum output slope and the maximum authorized instantaneous current. If it was bigger the output slope could be very tight and the current too big.\n\nWith the 0.35um process, an x1 is able to drive about 125fF, x2 -> 250fF, x4 -> 500fF,x8 -> 1000fF. This is just an indication since if a cell is overloaded, the only consequence is to increase the propagation time. On the other hand, it is not very good to under-load a cell because this leads to a signal overshoot. Actually, for big gate, such as noa3ao322_x1, x1 means maximal driving strength reachable with a single logic layer, that can be much less than an inv_x1. That is why is the cell list below contains more precise drive strengh. As you can see noa3ao322_x1 as a output drive strengh of 0.6, that means 0.6 time an inverter, so say it can drive about 0.6*125fF=75fF.\n\nWith the 0.35um process, a 1 lambda interconnect wire is about 0.15fF, an average cell fan-in is 10fF. Then, if it needs about 50 lambdas to connect 2 cells, an x1 cell is able to drive about 7 cells (125/(10+50*.15)=7). With 100 lambdas, 5 cells, with 750 lambdas only 2 cells. Note that 50 lambdas means cells are very close one from each other, nearly abutted, 100 lambdas is an average value.\n\nAll this are indications.  Only a timing analysis on the extracted transistor net-list from layout can tell if a cell is well used or not (see tas(1) for informations about static timing analysis).\n\n## Behavior\n\nFor most of cells, the user can deduce the cell behavior just by reading its name.  That is very intuitive for inverter and more complex for and/or cells.  For the last, the name gives the and/or tree structure.  The input order for the VHDL interface component is always the alphabetic order.\n\n```inv : inversor buffer\nbuf : buffer\n[n]ts : [not] tree-state\n[n]xr<i> : [not] xor <i> inputs\n[n]mx<i> : [not] multiplexor <i> inputs with coded command\n[n][sd]ff<i> : [not] [static|dynamic] flip-flop <i> inputs\n[n]oa... : [not] and/or function (see below)\n\nand_or cell (YACC (1) grammar):-\n\nNAME : n OA_CELL -> not OA_CELL\n| OA_CELL -> OA_CELL\n\nOA_CELL : OPERATOR INPUTS -> function with INPUTS inputs\n| OPERATOR OA_CELLS INPUTS -> function with INPUTS inputs\nwhere some inputs are OA_CELL\n\nOPERATOR : a -> and\n| o -> or\n| n -> not\n\nOA_CELLS : OA_CELLS OA_CELL -> list of OA_CELL\n| OA_CELL -> last OA_CELL of the list\n\nINPUTS : integer -> number of inputs\n\nThe input names are implicit and formed that way i<number>.\nThey are attributed in order beginning by i0.\n\nnx where x is a number means there are x inverters in parallel. For\nexample an23 is an and with 3 inputs of which two are inverted, that\nis and( not(i0), not(i1), i2).\n\nExamples:- (some are not in sxlib)\n\nna2 : not( and(i0,i1))\non12 : or( not(i0), i1)\nnoa2a22 : not( or( and(i0,i1), and(i2,i3)))\nnoa23 : not( or( and(i0,i1), i3))\nnoao22a34 : not( or( and( or(i0,i1), i2), and(i3,i4,i5), i6, i7))\n\nNote that xr2 could not be expressed with an and/or formulea even if\nxr2 = or( and( not(i0), i1), and( not(i1), i0)) = oan12an122\nbut the input names are not well distributed.```\n\n## Cell List\n\nAll available cells are listed below. The first column is the pitch width. The pitch value is 5 lambdas. The height is 50. Area is then <number>*5*50.\n\nThe second column is the output drive strenght compared with the inv_x1 output drive strenght (see explanation above in section Output Drive).\n\nThe following column is the delay in nano-seconds. Remember this delay corresponds to the slower input+0.6ns (see explanation above in section Delay Model).\n\nThe last column gives the function behavior with input capacitance. / means not, + means or, . means and, ^ means xor. Each input is followed by fan-in capacitance in fF, (e.g. i0<11> means i0 pin capacitance is 11fF).\n\nFor some cells, such as fulladder, it was not possible to internally connect all inputs. That means there are several inputs that must be externally connected. In the following list, these inputs are followed by a star (*) character in the equation.\n\nFor example, fulladder equation is sout <= (a* . b* . cin*). a* replaces a0, a1, a2, a3 that must be explicitly connected by the user. Note also few cells have more than one output. In that case there are several lines in the list, one by output.\n\n```=================================================================\nWIDTH NAME DRIVE DELAY BEHAVIOR with cin\n-------------------------------------------------------- INVERSOR\n3 inv_x1 1.0 0.7 nq <= /i<8>\n3 inv_x2 1.6 0.7 nq <= /i<12>\n4 inv_x4 3.6 0.7 nq <= /i<26>\n7 inv_x8 8.4 0.7 nq <= /i<54>\n---------------------------------------------------------- BUFFER\n4 buf_x2 2.1 1.0 q <= i<6>\n5 buf_x4 4.3 1.0 q <= i<9>\n8 buf_x8 8.4 1.0 q <= i<15>\n------------------------------------------------------ THREE STATE\n6 nts_x1 1.2 0.8 IF (cmd<14>) nq <= /i<14>\n8 nts_x2 2.4 0.9 IF (cmd<18>) nq <= /i<28>\n10 ts_x4 4.3 1.1 IF (cmd<19>) q <= i<8>\n13 ts_x8 8.4 1.2 IF (cmd<19>) q <= i<8>\n-------------------------------------------------------------- AND\n4 na2_x1 1.0 0.9 nq <= /(i0<11>.i1<11>)\n7 na2_x4 4.3 1.2 nq <= /(i0<10>.i1<10>)\n5 na3_x1 0.9 1.0 nq <= /(i0<11>.i1<11>.i2<11>)\n8 na3_x4 4.3 1.3 nq <= /(i0<10>.i1<10>.i2<10>)\n6 na4_x1 0.7 1.0 nq <= /(i0<10>.i1<11>.i2<11>.i3<11>)\n10 na4_x4 4.3 1.4 nq <= /(i0<10>.i1<11>.i2<11>.i3<11>)\n5 a2_x2 2.1 1.0 q <= (i0<9>.i1<11>)\n6 a2_x4 4.3 1.1 q <= (i0<9>.i1<11>)\n6 a3_x2 2.1 1.1 q <= (i0<10>.i1<10>.i2<10>)\n7 a3_x4 4.3 1.2 q <= (i0<10>.i1<10>.i2<10>)\n7 a4_x2 2.1 1.2 q <= (i0<10>.i1<10>.i2<10>.i3<10>)\n8 a4_x4 4.3 1.3 q <= (i0<10>.i1<10>.i2<10>.i3<10>)\n5 an12_x1 1.0 1.0 q <= (/i0<12>).i1<9>\n8 an12_x4 4.3 1.1 q <= (/i0<9>).i1<11>\n--------------------------------------------------------------- OR\n4 no2_x1 1.0 0.9 nq <= /(i0<12>+i1<12>)\n8 no2_x4 4.3 1.2 nq <= /(i0<12>+i1<11>)\n5 no3_x1 0.8 1.0 nq <= /(i0<12>+i1<12>+i2<12>)\n8 no3_x4 4.3 1.3 nq <= /(i0<12>+i1<12>+i2<11>)\n6 no4_x1 0.6 1.1 nq <= /(i0<12>+i1<12>+i2<12>+i3<12>)\n10 no4_x4 4.3 1.4 nq <= /(i0<12>+i1<12>+i2<12>+i3<12>)\n5 o2_x2 2.1 1.0 q <= (i0<10>+i1<10>)\n6 o2_x4 4.3 1.1 q <= (i0<10>+i1<10>)\n6 o3_x2 2.1 1.1 q <= (i0<10>+i1<10>+i2<9>)\n10 o3_x4 4.3 1.2 q <= (i0<10>+i1<10>+i2<9>)\n7 o4_x2 2.1 1.2 q <= (i0<10>+i1<10>+i2<10>+i3<9>)\n8 o4_x4 4.3 1.3 q <= (i0<12>+i1<12>+i2<12>+i3<12>)\n5 on12_x1 1.0 0.9 q <= (/i0<11>)+i1<9>\n8 on12_x4 4.3 1.1 q <= (/i0<9>)+i1<10>\n--------------------------------------------------------- AND/OR 3\n6 nao22_x1 1.2 0.9 nq <= /((i0<14>+i1<14>).i2<14>)\n10 nao22_x4 4.3 1.3 nq <= /((i0<8> +i1<8>) .i2<9>)\n6 noa22_x1 1.2 0.9 nq <= /((i0<14>.i1<14>)+i2<14>)\n10 noa22_x4 4.3 1.3 nq <= /((i0<8> .i1<8>) +i2<9>)\n6 ao22_x2 2.1 1.2 q <= ((i0<8>+i1<8>).i2<9>)\n8 ao22_x4 4.3 1.3 q <= ((i0<8>+i1<8>).i2<9>)\n6 oa22_x2 2.1 1.2 q <= ((i0<8>.i1<8>)+i2<9>)\n8 oa22_x4 4.3 1.3 q <= ((i0<8>.i1<8>)+i2<9>)\n--------------------------------------------------------- AND/OR 4\n7 nao2o22_x1 1.2 1.0 nq <= /((i0<14>+i1<14>).(i2<14>+i3<14>))\n11 nao2o22_x4 4.3 1.4 nq <= /((i0<8> +i1<8>) .(i2<8> +i3<8>))\n7 noa2a22_x1 1.2 1.0 nq <= /((i0<14>.i1<14>)+(i2<14>.i3<14>))\n11 noa2a22_x4 4.3 1.4 nq <= /((i0<8> .i1<8>) +(i2<8> .i3<8>))\n9 ao2o22_x2 2.1 1.2 q <= ((i0<8>+i1<8>).(i2<8>+i3<8>))\n10 ao2o22_x4 4.3 1.3 q <= ((i0<8>+i1<8>).(i2<8>+i3<8>))\n9 oa2a22_x2 2.1 1.2 q <= ((i0<8>.i1<8>)+(i2<8>.i3<8>))\n10 oa2a22_x4 4.3 1.4 q <= ((i0<8>.i1<8>)+(i2<8>.i3<8>))\n--------------------------------------------------------- AND/OR 5\n7 noa2ao222_x1 0.7 1.1 nq <= /((i0<11>.i1<11>)+((i2<13>+i3<13>).i4<13>))\n11 noa2ao222_x4 4.3 1.4 nq <= /((i0<11>.i1<11>)+((i2<11>+i3<11>).i4<11>))\n10 oa2ao222_x2 2.1 1.2 q <= ((i0<8> .i1<8>) +((i2<8> +i3<8>) .i4<8>))\n11 oa2ao222_x4 4.3 1.3 q <= ((i0<8> .i1<8>) +((i2<8> +i3<8>) .i4<8>))\n--------------------------------------------------------- AND/OR 6\n10 noa2a2a23_x1 0.8 1.2 nq <= /((i0<13>.i1<14>) +(i2<14>.i3<14>)\n+(i4<14>.i5<14>))\n13 noa2a2a23_x4 4.3 1.3 nq <= /((i0<13>.i1<14>) +(i2<14>.i3<14>)\n+(i4<14>.i5<14>))\n12 oa2a2a23_x2 2.1 1.4 q <= ((i0<13>.i1<14>) +(i2<14>.i3<14>)\n+(i4<14>.i5<14>))\n13 oa2a2a23_x4 4.3 1.4 q <= ((i0<13>.i1<14>) +(i2<14>.i3<14>)\n+(i4<14>.i5<14>))\n--------------------------------------------------------- AND/OR 7\n9 noa3ao322_x1 0.6 1.2 nq <= /((i0<13>.i1<13>.i2<12>)\n+((i3<13>+i4<13>+i5<13>).i6<13>))\n11 noa3ao322_x4 4.3 1.4 nq <= /((i0<10>.i1<9>.i2<9>)\n+((i3<9>+i4<9>+i5<9>).i6<9>))\n10 oa3ao322_x2 2.1 1.2 q <= /((i0<10>.i1<9>.i2<9>)\n+((i3<9>+i4<9>+i5<9>).i6<9>))\n11 oa3ao322_x4 4.3 1.3 q <= /((i0<10>.i1<9>.i2<9>)\n+((i3<9>+i4<9>+i5<9>).i6<9>))\n--------------------------------------------------------- AND/OR 8\n14 noa2a2a2a24_x1 0.6 1.4 nq <= /((i0<14>.i1<14>)+(i2<13>.i3<13>)\n+(i4<13>.i5<13>)+(i6<14>.i7<14>))\n17 noa2a2a2a24_x4 4.3 1.7 nq <= /((i0<14>.i1<14>)+(i2<14>.i3<13>)\n+(i4<13>.i5<13>)+(i6<14>.i7<14>))\n15 oa2a2a2a24_x2 2.1 1.5 q <= ((i0<14>.i1<14>)+(i2<14>.i3<13>)\n+(i4<13>.i5<13>)+(i6<14>.i7<14>))\n16 oa2a2a2a24_x4 4.3 1.6 q <= ((i0<14>.i1<14>)+(i2<14>.i3<13>)\n+(i4<13>.i5<13>)+(i6<14>.i7<14>))\n------------------------------------------------------ MULTIPLEXER\n7 nmx2_x1 1.2 1.0 nq <= /((i0<14>./cmd<21>)+(i1<14>.cmd))\n12 nmx2_x4 4.3 1.3 nq <= /((i0<8>./cmd<14>)+(i1<9>.cmd))\n9 mx2_x2 2.1 1.1 q <= (i0<8>./cmd<17>)+(i1<9>.cmd)\n10 mx2_x4 4.3 1.3 q <= (i0<8>./cmd<17>)+(i1<9>.cmd)\n12 nmx3_x1 0.4 1.2 nq <= /((i0<9>./cmd0<15>)\n+(((i1<8>.cmd1<15>)+(i2<8>./cmd1)).cmd0))\n15 nmx3_x4 4.3 1.7 nq <= /((i0<9>./cmd0<15>)\n+(((i1<8>.cmd1<15>)+(i2<8>./cmd1)).cmd0))\n13 mx3_x2 2.1 1.4 q <= ((i0<9>./cmd0<15>)\n+(((i1<8>.cmd1<15>)+(i2<8>./cmd1)).cmd0))\n14 mx3_x4 4.3 1.6 q <= ((i0<9>./cmd0<15>)\n+(((i1<8>.cmd1<15>)+(i2<8>./cmd1)).cmd0))\n-------------------------------------------------------------- XOR\n9 nxr2_x1 1.2 1.1 nq <= /(i0<21>^i1<22>)\n11 nxr2_x4 4.3 1.2 nq <= /(i0<20>^i1<21>)\n9 xr2_x1 1.2 1.0 q <= (i0<21>^i1<22>)\n12 xr2_x4 4.3 1.2 q <= (i0<20>^i1<21>)\n-------------------------------------------------------- FLIP-FLOP\nnq <=/((i0<11>./cmd<13>)+(i1<7>.cmd))\n18 sff1_x4 4.3 1.7 IF RISE(ck<8>)\nq <= i<8>\n24 sff2_x4 4.3 1.9 IF RISE(ck<8>)\nq <= ((i0<8>./cmd<16>)+(i1<7>.cmd))\n28 sff3_x4 4.3 2.4 IF RISE(ck<8>)\nq <= (i0<9>./cmd0<15>)\n+(((i1<8>.cmd1<15>)+(i2<8>./cmd1)).cmd0)\n16 halfadder_x2 2.1 1.2 sout <= (a<27>^b<22>)\n2.1 1.0 cout <= (a.b)\n18 halfadder_x4 4.3 1.3 sout <= (a<27>^b<22>)\n4.3 1.1 cout <= (a.b)\n20 fulladder_x2 2.1 1.8 sout <= (a*<28>^b*<28>^cin*<19>)\n2.1 1.4 cout <= (a*.b*+a*.cin*+b*.cin*)\n21 fulladder_x4 4.3 2.2 sout <= (a*<28>^b*<28>^cin*<19>)\n4.3 1.5 cout <= (a*.b*+a*.cin*+b*.cin*)\n---------------------------------------------------------- SPECIAL\n3 zero_x0 0 0 nq <= '0'\n3 one_x0 0 0 q <= '1'\n2 tie_x0 0 0 Body tie cell\n1 rowend_x0 0 0 Empty cell\n==================================================================```\n\n## New Cells\n\nIt is possible to add new cells in the library just by providing the 3 files .ap, .al and .vbe in the standard cell directory.  The layout view can be created with the symbolic editor graal.  The physical outline is given above.  The net-list view can be automatically generated with the lynx extractor.  The behavioral view must be written by the designer and checked with the yagle functional abstractor.  The file must contain the generic fields in order to be used by the logic synthesis tools and the I/Os terminals must be in the same order (alphabetic) in the .vbe and .al files.\n\nIf you develop new cells, please send the corresponding files to [email protected]\n\n## VHDL Files\n\nYou can find below the commented VHDL GENERIC for the na2_x4 cell.\n\n```ENTITY na2_x4 IS\nGENERIC (\nCONSTANT area : NATURAL := 1750; -- lamba * lambda\nCONSTANT transistors : NATURAL := 10; -- number of\nCONSTANT cin_i0 : NATURAL := 10; -- femto Farad for i0\nCONSTANT cin_i1 : NATURAL := 10; -- femto Farad for i1\nCONSTANT tplh_i1_nq : NATURAL := 606; -- propag. time in pico-sec\n-- from i1 falling\n-- to nq rizing\nCONSTANT rup_i1_nq : NATURAL := 890; -- resitance in Ohms when nq\n-- rizing due to i1 change\nCONSTANT tphl_i1_nq : NATURAL := 349; -- propag time when nq falls\nCONSTANT rdown_i1_nq : NATURAL := 800; -- resist when nq falls\nCONSTANT tplh_i0_nq : NATURAL := 557; -- idem for i0\nCONSTANT rup_i0_nq : NATURAL := 890;\nCONSTANT tphl_i0_nq : NATURAL := 408;\nCONSTANT rdown_i0_nq : NATURAL := 800\n);\nPORT (\ni0 : in BIT;\ni1 : in BIT;\nnq : out BIT;\nvdd : in BIT;\nvss : in BIT\n);```" ]

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[ "# Critical Formulas You Need For The FAR Test\n\nThe financial accounting and reporting (FAR) test covers topics related to accounting transactions and generating financial statements. This test is all about the day-to-day work of accounting, and the amount of information you need to know can be overwhelming. Understanding some key formulas can help you answer a number of questions correctly. Use these formulas as you study for the FAR test.\n\n## Inventory formula\n\nFor any companies, inventory is the largest asset on the balance sheet, and it’s important to understand how the inventory balance is calculated. The formula is:\n\n(Beginning inventory) + (purchases) = (goods available for sale) – (cost of goods sold) = ending inventory\n\nNow, let’s assume that you own a hardware store, and that you’re inventory calculation is:\n\n\\$1,000,000 beg inventory + \\$500,000 purchases = \\$1,500,000 goods available for sale\n\nOnly two things can happen to goods available for sale: You either sell the goods, or they stay on your shelf as inventory. If your cost of goods sold is \\$800,000, for example, you have ending inventory of (\\$1,500,000 – \\$800,000 = \\$700,000).\n\nI hope that bit of logic can help you understand this formula. If you can visualize products moving in and out of a retail shop, you’ll have an easier time with the inventory formula.\n\n## Pensions", null, "Many CPA candidates find the pension section of the test so difficult to understand that they just pass on trying these questions. I get it — the language can be difficult. However, I’ve found that using a “bucket analogy” to explain pensions can help students. Here’s a pension liability formula for a defined benefit plan, a plan in which the employer pays a specific dollar amount as a pension benefit:\n\n(Contributions) + (expected rate of return) – (service costs) – (interest costs)\n\nIf the number is positive, you have more assets than you need to pay your pensions. If, on the other hand, the number is negative, you have an underfunded liability.\n\nYou can view pensions this way: A company that offers a pension is obligated to invest funds to make pension payments. The business contributes money into the pension bucket, and the dollars in the bucket earn a rate of return. The firm estimates that rate of return, which is why it’s called an expected rate.\n\nWhen a worker retires, the employee receives a pension, and the cost of that pension is based on the number of years the employee worked, and the ending salary amount. That service cost comes out of the bucket, along with interest payments that workers earn on the dollar amount in the bucket. The pension questions can be more complicated than this basic formula, but the bucket analogy will get you on the right track.\n\n## Diluted earnings per share (EPS)\n\nI want to make the distinction between basic EPS and diluted EPS, because both topics are on the FAR test. Basic EPS is (Earnings per common share) / (weighted average common stock shares outstanding). Weighted average is simply [(beginning balance) + (ending balance)] / 2. EPS tells you the dollar amount of earnings that a firm generates per common stock share.\n\nThe diluted earnings concept assumes that any security that can be converted into additional shares of common stock is converted. Now, think about that for a minute: The dollar amount of earnings has not changed — only the number of common stock shares. Assume, for example, that a company’s EPS is (\\$1,000,000 / 200,000 shares), or \\$5 per share. If the number of common stock shares increases to 250,000, EPS declines to \\$4 per share. In other words, the additional shares have diluted EPS.\n\nSeveral types of securities can be converted into additional common stock shares, including stock options, rights, and warrants. Convertible securities, such as convertible bonds and convertible preferred stock, also dilute EPS. To understand dilution, remember putting chocolate powder into milk. The powder dilutes and spreads out into the milk — that visual tool might help you understand the concept.\n\nThe FAR test requires the CPA candidate to digest a great deal of information. Use these formulas to answer more test questions correctly and pass the test.\n\nBy the way, sign up for our 1 Week Free Trial to try out Magoosh GMAT Prep!" ]

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[ "Related Articles\n\n# Queries to find the minimum array sum possible by removing elements from either end\n\n• Last Updated : 24 May, 2021\n\nGiven an array arr[] consisting of N distinct integers and an array Q[] representing queries, the task for every query Q[i] is to find the minimum sum possible by removing the array elements from either end until Q[i] is obtained.\n\nExamples:\n\nInput: arr[] = {2, 3, 6, 7, 4, 5, 1}, Q[] = {7, 6}\nOutput: 17 11\nExplanation:\nQuery 1: By popping elements from the end, sum = 1 + 5 + 4 + 7 = 17.\nQuery 2: Popping elements from the front, sum = 2 + 3 + 6 = 11.\n\nInput: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, Q[] = {4, 6, 3}\nOutput: 10 21 6\n\nNaive Approach: The simplest approach to solve the given problem is to traverse the given array from both the ends for each query Q[i] and print the minimum sum obtained from both the traversals till the element with value Q[i] is obtained.\n\nBelow is the implementation of the above approach:\n\n## C++14\n\n `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find the minimum sum for``// each query after removing elements``// from either ends``void` `minSum(``int` `arr[], ``int` `N, ``int` `Q[],``            ``int` `M)``{``    ``// Traverse the query array``    ``for` `(``int` `i = 0; i < M; i++) {``        ``int` `val = Q[i];` `        ``int` `front = 0, rear = 0;` `        ``// Traverse the array from``        ``// the front``        ``for` `(``int` `j = 0; j < N; j++) {``            ``front += arr[j];` `            ``// If element equals val,``            ``// then break out of loop``            ``if` `(arr[j] == val) {``                ``break``;``            ``}``        ``}` `        ``// Traverse the array from rear``        ``for` `(``int` `j = N - 1; j >= 0; j--) {``            ``rear += arr[j];` `            ``// If element equals val, break``            ``if` `(arr[j] == val) {``                ``break``;``            ``}``        ``}` `        ``// Print the minimum of the``        ``// two as the answer``        ``cout << min(front, rear) << ``\" \"``;``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 3, 6, 7, 4, 5, 1 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `Q[] = { 7, 6 };``    ``int` `M = ``sizeof``(Q) / ``sizeof``(Q);` `    ``// Function Call``    ``minSum(arr, N, Q, M);` `    ``return` `0;``}`\n\n## Java\n\n `// Java program for the above approach``import` `java.util.*;``class` `GFG``{` `// Function to find the minimum sum for``// each query after removing elements``// from either ends``static` `void` `minSum(``int` `arr[], ``int` `N, ``int` `Q[],``            ``int` `M)``{``  ` `    ``// Traverse the query array``    ``for` `(``int` `i = ``0``; i < M; i++)``    ``{``        ``int` `val = Q[i];``        ``int` `front = ``0``, rear = ``0``;` `        ``// Traverse the array from``        ``// the front``        ``for` `(``int` `j = ``0``; j < N; j++)``        ``{``            ``front += arr[j];` `            ``// If element equals val,``            ``// then break out of loop``            ``if` `(arr[j] == val)``            ``{``                ``break``;``            ``}``        ``}` `        ``// Traverse the array from rear``        ``for` `(``int` `j = N - ``1``; j >= ``0``; j--)``        ``{``            ``rear += arr[j];` `            ``// If element equals val, break``            ``if` `(arr[j] == val)``            ``{``                ``break``;``            ``}``        ``}` `        ``// Print the minimum of the``        ``// two as the answer``        ``System.out.print(Math.min(front, rear) + ``\" \"``);``    ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``2``, ``3``, ``6``, ``7``, ``4``, ``5``, ``1` `};``    ``int` `N = arr.length;``    ``int` `Q[] = { ``7``, ``6` `};``    ``int` `M = Q.length;` `    ``// Function Call``    ``minSum(arr, N, Q, M);``}``}` `// This code is contributed by shikhasingrajput`\n\n## Python3\n\n `# Python3 program for the above approach` `# Function to find the minimum sum for``# each query after removing elements``# from either ends``def` `minSum(arr, N, Q, M):``  ` `    ``# Traverse the query array``    ``for` `i ``in` `range``(M):``        ``val ``=` `Q[i]` `        ``front, rear ``=` `0``, ``0` `        ``# Traverse the array from``        ``# the front``        ``for` `j ``in` `range``(N):``            ``front ``+``=` `arr[j]` `            ``# If element equals val,``            ``# then break out of loop``            ``if` `(arr[j] ``=``=` `val):``                ``break` `        ``# Traverse the array from rear``        ``for` `j ``in` `range``(N ``-` `1``, ``-``1``, ``-``1``):``            ``rear ``+``=` `arr[j]` `            ``# If element equals val, break``            ``if` `(arr[j] ``=``=` `val):``                ``break` `        ``# Prthe minimum of the``        ``# two as the answer``        ``print``(``min``(front, rear), end ``=` `\" \"``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``2``, ``3``, ``6``, ``7``, ``4``, ``5``, ``1``]``    ``N ``=` `len``(arr)``    ``Q ``=` `[``7``, ``6``]``    ``M ``=` `len``(Q)` `    ``# Function Call``    ``minSum(arr, N, Q, M)` `    ``# This code is contributed by mohit kumar 29.`\n\n## C#\n\n `// C# program for the above approach``using` `System;` `class` `GFG{``    ` `// Function to find the minimum sum for``// each query after removing elements``// from either ends``static` `void` `minSum(``int``[] arr, ``int` `N, ``int``[] Q,``                   ``int` `M)``{``    ` `    ``// Traverse the query array``    ``for``(``int` `i = 0; i < M; i++)``    ``{``        ``int` `val = Q[i];``        ``int` `front = 0, rear = 0;`` ` `        ``// Traverse the array from``        ``// the front``        ``for``(``int` `j = 0; j < N; j++)``        ``{``            ``front += arr[j];`` ` `            ``// If element equals val,``            ``// then break out of loop``            ``if` `(arr[j] == val)``            ``{``                ``break``;``            ``}``        ``}`` ` `        ``// Traverse the array from rear``        ``for``(``int` `j = N - 1; j >= 0; j--)``        ``{``            ``rear += arr[j];`` ` `            ``// If element equals val, break``            ``if` `(arr[j] == val)``            ``{``                ``break``;``            ``}``        ``}`` ` `        ``// Print the minimum of the``        ``// two as the answer``        ``Console.Write(Math.Min(front, rear) + ``\" \"``);``    ``}``}`` ` `// Driver Code``static` `public` `void` `Main()``{``    ``int``[] arr = { 2, 3, 6, 7, 4, 5, 1 };``    ``int` `N = arr.Length;``    ``int``[] Q = { 7, 6 };``    ``int` `M = Q.Length;`` ` `    ``// Function Call``    ``minSum(arr, N, Q, M);``}``}` `// This code is contributed by rag2127`\n\n## Javascript\n\n ``\nOutput:\n`17 11`\n\nTime Complexity: O(N*M)\nAuxiliary Space: O(1)\n\nEfficient Approach: To optimize the above approach, the idea is to use the Prefix Sum technique to solve this problem. Follow the steps below to solve the problem:\n\n• Create two auxiliary Maps, say M1 and M2.\n• Traverse the array from the front and insert the current sum calculated till each index in the Map M1 along with the element.\n• Similarly, traverse the array from the back and insert the current sum calculated till each index in the map M2 along with the element.\n• Traverse the array Q[] and for each element Q[i], print minimum of M1[Q[i]] and M2[Q[i]] as the minimum possible sum.\n\nBelow is the implementation of the above approach:\n\n## C++14\n\n `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find the minimum sum``// for each query after removing``// element from either ends till each``// value Q[i]``void` `minOperations(``int` `arr[], ``int` `N,``                   ``int` `Q[], ``int` `M)``{``    ``// Stores the prefix sum from``    ``// both the ends of the array``    ``map<``int``, ``int``> m1, m2;` `    ``int` `front = 0, rear = 0;` `    ``// Traverse the array from front``    ``for` `(``int` `i = 0; i < N; i++) {``        ``front += arr[i];` `        ``// Insert it into the map m1``        ``m1.insert({ arr[i], front });``    ``}` `    ``// Traverse the array in reverse``    ``for` `(``int` `i = N - 1; i >= 0; i--) {``        ``rear += arr[i];` `        ``// Insert it into the map m2``        ``m2.insert({ arr[i], rear });``    ``}` `    ``// Traverse the query array``    ``for` `(``int` `i = 0; i < M; i++) {` `        ``// Print the minimum of the``        ``// two values as the answer``        ``cout << min(m1[Q[i]], m2[Q[i]])``             ``<< ``\" \"``;``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 3, 6, 7, 4, 5, 1 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `Q[] = { 7, 6 };``    ``int` `M = ``sizeof``(Q) / ``sizeof``(Q);` `    ``// Function Call``    ``minOperations(arr, N, Q, M);` `    ``return` `0;``}`\n\n## Java\n\n `// Java program for the above approach``import` `java.util.*;` `class` `GFG{``    ` `// Function to find the minimum sum``// for each query after removing``// element from either ends till each``// value Q[i]``static` `void` `minOperations(``int``[] arr, ``int` `N,``                          ``int``[] Q, ``int` `M)``{``    ` `    ``// Stores the prefix sum from``    ``// both the ends of the array``    ``Map m1 = ``new` `HashMap();``    ``Map m2 = ``new` `HashMap();` `    ``int` `front = ``0``, rear = ``0``;` `    ``// Traverse the array from front``    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ``front += arr[i];` `        ``// Insert it into the map m1``        ``m1.put(arr[i], front);``    ``}` `    ``// Traverse the array in reverse``    ``for``(``int` `i = N - ``1``; i >= ``0``; i--)``    ``{``        ``rear += arr[i];` `        ``// Insert it into the map m2``        ``m2.put(arr[i], rear);``    ``}` `    ``// Traverse the query array``    ``for``(``int` `i = ``0``; i < M; i++)``    ``{``        ` `        ``// Print the minimum of the``        ``// two values as the answer``        ``System.out.print(Math.min(m1.get(Q[i]),``                                  ``m2.get(Q[i])) + ``\" \"``);``    ``}``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int``[] arr = { ``2``, ``3``, ``6``, ``7``, ``4``, ``5``, ``1` `};``    ``int` `N = arr.length;``    ``int``[] Q = { ``7``, ``6` `};``    ``int` `M = Q.length;``    ` `    ``// Function Call``    ``minOperations(arr, N, Q, M);``}``}` `// This code is contributed by offbeat`\n\n## Python3\n\n `# Python3 program for the above approach` `# Function to find the minimum sum``# for each query after removing``# element from either ends till each``# value Q[i]``def` `minOperations(arr, N, Q, M):``  ` `    ``# Stores the prefix sum from``    ``# both the ends of the array``    ``m1 ``=` `{}``    ``m2 ``=` `{}``    ``front ``=` `0``    ``rear ``=` `0``    ` `    ``# Traverse the array from front``    ``for` `i ``in` `range``(N):``        ``front ``+``=` `arr[i]``        ` `        ``# Insert it into the map m1``        ``m1[arr[i]] ``=` `front``    ` `    ``# Traverse the array in reverse``    ``for` `i ``in` `range``(N ``-` `1``, ``-``1``, ``-``1``):``        ``rear ``+``=` `arr[i]``        ` `        ``# Insert it into the map m2``        ``m2[arr[i]] ``=` `rear``    ` `    ``# Traverse the query array``    ``for` `i ``in` `range``(M):``      ` `        ``# Print the minimum of the``        ``# two values as the answer``        ``print``(``min``(m1[Q[i]], m2[Q[i]]),end``=``\" \"``)` `# Driver Code``arr ``=` `[``2``, ``3``, ``6``, ``7``, ``4``, ``5``, ``1` `]``N ``=` `len``(arr)``Q ``=` `[``7``,``6``]``M ``=` `len``(Q)` `# Function Call``minOperations(arr, N, Q, M)` `# This code is contributed by avanitrachhadiya2155`\n\n## C#\n\n `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{``    ` `// Function to find the minimum sum``// for each query after removing``// element from either ends till each``// value Q[i]``static` `void` `minOperations(``int``[] arr, ``int` `N,``                          ``int``[] Q, ``int` `M)``{``    ` `    ``// Stores the prefix sum from``    ``// both the ends of the array``    ``Dictionary<``int``,``               ``int``> m1 = ``new` `Dictionary<``int``,``                                        ``int``>();``    ``Dictionary<``int``,``               ``int``> m2 = ``new` `Dictionary<``int``,``                                        ``int``>();` `    ``int` `front = 0, rear = 0;` `    ``// Traverse the array from front``    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ``front += arr[i];` `        ``// Insert it into the map m1``        ``m1[arr[i]] = front;``    ``}` `    ``// Traverse the array in reverse``    ``for``(``int` `i = N - 1; i >= 0; i--)``    ``{``        ``rear += arr[i];` `        ``// Insert it into the map m2``        ``m2[arr[i]] = rear;``    ``}` `    ``// Traverse the query array``    ``for``(``int` `i = 0; i < M; i++)``    ``{``        ` `        ``// Print the minimum of the``        ``// two values as the answer``        ``Console.Write(Math.Min(m1[Q[i]],``                               ``m2[Q[i]]) + ``\" \"``);``    ``}``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int``[] arr = { 2, 3, 6, 7, 4, 5, 1 };``    ``int` `N = arr.Length;``    ``int``[] Q = { 7, 6 };``    ``int` `M = Q.Length;` `    ``// Function Call``    ``minOperations(arr, N, Q, M);``}``}` `// This code is contributed by ukasp`\n\n## Javascript\n\n ``\nOutput:\n`17 11`\n\nTime Complexity: O(N + M)\nAuxiliary Space: O(N)\n\nAttention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.\n\nIn case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.\n\nMy Personal Notes arrow_drop_up" ]

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[ "# Measurement of CP-Violating and Mixing-Induced Observables in B_{s}^{0}→ϕγ Decays\n\nLHCb Collaboration\n\nРезультат исследований: Материалы для журналаСтатья\n\n3 Цитирования (Scopus)\n\n### Аннотация\n\nA time-dependent analysis of the B_{s}^{0}→ϕγ decay rate is performed to determine the CP -violating observables S_{ϕγ} and C_{ϕγ} and the mixing-induced observable A_{ϕγ}^{Δ}. The measurement is based on a sample of pp collision data recorded with the LHCb detector, corresponding to an integrated luminosity of 3  fb^{-1} at center-of-mass energies of 7 and 8 TeV. The measured values are S_{ϕγ}=0.43±0.30±0.11, C_{ϕγ}=0.11±0.29±0.11, and A_{ϕγ}^{Δ}=-0.67_{-0.41}^{+0.37}±0.17, where the first uncertainty is statistical and the second systematic. This is the first measurement of the observables S and C in radiative B_{s}^{0} decays. The results are consistent with the standard model predictions.\n\nЯзык оригинала Английский 1 Physical Review Letters 123 8 https://doi.org/10.1103/PhysRevLett.123.081802 Опубликовано - 23 авг 2019\n\n### ASJC Scopus subject areas\n\n• Physics and Astronomy(all)" ]

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[ "# right diagonal matrix in c\n\nFor instance, the following matrix is a Toeplitz matrix: Any N×N matrix A of the form is a Toeplitz matrix if A (i, j) = A (i+1, j+1) = A (i+2, j+2) and so on.. Matrices that are similar to triangular matrices are called triangularisable. C Program to find the sum of all diagonal elements of a given matrix. A square matrix D = [d ij] n x n will be called a diagonal matrix if d ij = 0, whenever i is not equal to j. Program to find sum of main diagonal elements of a matrix /** * C program to find sum of main diagonal elements of a matrix */ #include #define SIZE 3 // Matrix size int main() { int A[SIZE][SIZE]; int row, col, sum = 0; /* Input elements in matrix from user */ printf(\"Enter elements in matrix of size %dx%d: \\n\", SIZE, SIZE); for(row=0; row Write a program in C to find transpose of a given matrix. To force diag to build a matrix from variable-size inputs that are not 1-by-: or :-by-1, use: A square matrix in which every element except the principal diagonal elements is zero is called a Diagonal Matrix. 2. elements satisfying i+j\n\nPosted in 게시판." ]

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[ "# Lesson 12 Homework Answer Key 5.3\n\nLesson 12 Homework Answer Key 5.3. P = (2 × 2 cm) + (2 × 4 cm) + 3 cm + 5 cm + 6 cm + 8 cm = 34 cm (equations. Lesson 3 homework answer key.\n\n[get] lesson 15 homework 5.3 answer key. Www.youtube.com eureka math grade 5 module 1 lesson 3 homework reply key, for instance, a. P = (2 × 2 cm) + (2 × 4 cm) + 3 cm + 5 cm + 6 cm + 8 cm = 34 cm (equations.\n\n### The Mad Is 5.3 Degrees For City B, Which Means That, On Average, The Monthly Temperatures Differ By 5.3 Degrees From The Mean Of 63 Degrees.\n\n60 x 5 = 300 300 + 12 = 312. A scale of 2 would be more appropriate for a bar graph. P = (2 × 7 m) + 3 m + 2 m + 9 m + 4 m = 32 m (equations may vary) b.\n\n### Eureka Math Lesson 21 Homework 5.2.\n\n[get] lesson 15 homework 5.3 answer key. Lesson 12 homework 5.3 answer key five key features of mavericks, apple’s new operating system for macs. Labeled 3 m, 7 m;\n\n### Www.instructables.com 60 X 5 = 300 300 + 12 = 312.\n\nLesson 3 homework answer key. Slope = = = 15000 uncover the. This image representes lesson 3 homework 4.2 reply key.\n\n### The Mad Is 5.3 Levels For Metropolis B, Which Signifies That, On Common, The.\n\nWww.youtube.com eureka math grade 5 module 1 lesson 3 homework reply key, for instance, a. P = (2 × 2 cm) + (2 × 4 cm) + 3 cm + 5 cm + 6 cm + 8 cm = 34 cm (equations. Eureka math grade 5 module 5 lesson 12 problem set answer key question 1.\n\n### Lesson 12 Homework 5.3 Answer Key Green Hornet Original Vintage Movie Poster, 1940 Key Luke :\n\nThe biggest unit in 51 is tens. Shown division using an array. We will match you with an good and we form 5 module 5 lesson 3 homework answer key testament supervise your cooperation from start to finish.\n\n## Comment", null, "" ]

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[ "The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.", null, "Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)\n A291786 a(n) = number of iterations of k -> (psi(k)+phi(k))/2 (A291784) needed to reach a prime or a power of a prime or 1, or -1 if that doesn't happen. 5\n 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 3, 0, 2, 1, 0, 0, 3, 0, 2, 2, 1, 0, 6, 0, 1, 0, 5, 0, 4, 0, 0, 9, 8, 7, 6, 0, 5, 4, 3, 0, 5, 0, 2, -1, 1, 0, -1, 0, -1, 6, 5, 0, 4, -1, -1, 2, 1, 0, -1, 0, 4, 3, 0, 3, 2, 0, -1, -1, -1, 0, -1, 0 (list; graph; refs; listen; history; text; internal format)\n OFFSET 1,12 COMMENTS Primes and prime powers are fixed points under the map f(k) = (psi(k)+phi(k))/2, so in that case we take a(n)=0. (If n = p^k, then psi(n) = p^k(1+1/p), phi(n) = p^k(1-1/p), and their average is p^k, so n is a fixed point under the map.) Since f(n)>n if n is not a prime power, there can be no nontrivial cycles. Wall (1985) observes that the trajectories of 45 and 50 are unbounded, so a(45) = a(50) = -1. See A291787, A291788. REFERENCES Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004. See Section B41, p. 147. LINKS C. R. Wall, Unbounded sequences of Euler-Dedekind means, Amer. Math. Monthly, 92 (1985), 587. FORMULA a(n) = 0 iff n is in A000961. - M. F. Hasler, Sep 03 2017 PROG (PARI) A291786(n, L=n)=n>1&&for(i=0, L, isprimepower(n)&&return(i); n=A291784(n)); -(n>1) \\\\ The suggested search limit L=n is only empirical and might require revision. The code also currently assumes that the prime powers are the only cycles. - M. F. Hasler, Sep 03 2017 CROSSREFS Cf. A000010, A001615, A291784, A291785, A291787, A291788. Sequence in context: A022898 A072780 A124452 * A004603 A174951 A275326 Adjacent sequences:  A291783 A291784 A291785 * A291787 A291788 A291789 KEYWORD sign,more AUTHOR N. J. A. Sloane, Sep 02 2017 EXTENSIONS Initial terms corrected and more terms from M. F. Hasler, Sep 03 2017 STATUS approved\n\nLookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam\nContribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent\nThe OEIS Community | Maintained by The OEIS Foundation Inc.\n\nLast modified April 17 09:29 EDT 2021. Contains 343064 sequences. (Running on oeis4.)" ]

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[ "", null, "# Python filter()\n\nThe filter() function is used for returning an iterator, here the elements are filtered through the function.it helps to test each element in the sequence to be true or not.\n\n``````\nfilter(function, iterable) #Where iterable can be a list, string, tuple, dictionary , set etc ```\n```\n\n## filter() Parameters:\n\nIt takes functions and iterable as parameters.\n\nParameter Description Required / Optional\nfunction A function that tests if the element is accepted or not Required\niterable  Iterable may be a set, lists, tuples, etc... Required\n\n## filter() Return Value\n\nInput Return Value\niterable(if element) iterator(filtered list)\niterable(if not element) False\n\n## Examples of filter() method in Python\n\n### Example 1: How filter() works for iterable list?\n\n``````\n# list of letters\nletters = ['a', 'b', 'd', 'e', 'i', 'j', 'o']\n\n# function that filters vowels\ndef filter_vowels(letter):\nvowels = ['a', 'e', 'i', 'o', 'u']\n\nif(letter in vowels):\nreturn True\nelse:\nreturn False\n\nfiltered_vowels = filter(filter_vowels, letters)\n\nprint('The filtered vowels are:')\nfor vowel in filtered_vowels:\nprint(vowel)\n```\n```\n\nOutput:\n\n```The filtered vowels are:\na\ne\ni\no\n```\n\n### Example 2: How filter() works using a pre-defined function?\n\n``````\n# Returns the elements which are multiples of 5\ndef multipleOf5(n):\nif(n % 5 == 0):\nreturn n\nmyList = [10, 25, 17, 9, 30, -5]\nmyList2 = list(filter(multipleOf5, myList))\nprint(myList2)\n```\n```\n\nOutput:\n\n```[10, 25, 30, -5]\n```\n\n### Example 3: How filter() method works without the filter function?\n\n``````\n# random list\nrandom_list = [1, 'a', 0, False, True, '0']\n\nfiltered_list = filter(None, random_list)\n\nprint('The filtered elements are:')\nfor element in filtered_list:\nprint(element)\n```\n```\n\nOutput:\n\n```The filtered elements are:\n1\na\nTrue\n0\n```\nVIEW ALL\nVIEW ALL\n##### OtherTutorials\nVIEW ALL", null, "" ]

[ null, "https://559987-1802630-raikfcquaxqncofqfm.stackpathdns.com/assets/uploads/course/python.jpg", null, "https://559987-1802630-raikfcquaxqncofqfm.stackpathdns.com/assets/images/sub2.png", null ]

[ " Can you solve these 6th-grade math questions? | The notebook\nDecember 1 — 1:45 pm, 2015\n\n# Can you solve these 6th-grade math questions?\n\nThe following multiple-choice questions are from a selection of sample test items provided by the Pennsylvania Department of Education for the 6th-grade PSSA exam in math. Calculator use is permitted. The exam also includes open-ended questions where students are expected to explain their answers.\n\n1. Emily is making bows using ribbon. She has two pieces of ribbon to use. One is 23 yards long. The other is 4 1/4 yards long. She needs 1 5/6 yards of ribbon to make each bow. What is the greatest number of bows Emily can make?\n\nA. 12\n\nB. 14\n\nC. 15\n\nD. 19\n\n2. Gracie is rewriting the expression (24 + 40) as an integer times the sum of two integers. By factoring out a 2, she knows she can rewrite the expression as 2 times the sum of two integers. What are all the other numbers greater than 2 that Gracie can factor out of (24 + 40) to rewrite the expression as an integer times the sum of two integers?\n\nA. 4, 8\n\nB. 4, 6, 8\n\nC. 3, 5, 6, 10, 12, 20\n\nD. 3, 4, 5, 6, 8, 10, 12, 20\n\n3. Quincy and Ray keep track of their scores in a game. The person with the greater score is winning the game. Quincy has a score of –70, and Ray has a score of –60. Which statement best explains who is winning and how many points away from 0 that person is?\n\nA. Ray is winning and needs to lose 60 points to get to 0.\n\nB. Ray is winning and needs to gain 60 points to get to 0.\n\nC. Quincy is winning and needs to lose 70 points to get to 0.\n\nD. Quincy is winning and needs to gain 70 points to get to 0.\n\n4. At a factory, a machine tests 1 out of every 75 items produced for quality. The machine requires a safety check after testing 450 items. The factory produces 303,750 items each month. How many safety checks does the machine require each month?\n\nA. 6\n\nB. 9\n\nC. 50\n\nD. 54\n\n5. When a farmer harvests chicken eggs, he expects 2% of the eggs to be cracked. How many eggs would the farmer expect to be cracked when harvesting 350 eggs?\n\nA. 3\n\nB. 7\n\nC. 18\n\nD. 70\n\n6. Alex and Payton each have a favorite pancake recipe.\n\n• Alex’s recipe uses 7 1/2 cups of flour for 5 batches.\n\n• Payton’s recipe uses 3/4 cup of flour more per batch than Alex’s recipe uses per batch.\n\nWhich expression can be used to determine the number of cups of flour used to make x batches of Payton’s pancake recipe?\n\nA. 2 1/4 x\n\nB. 8 1/4 x\n\nC. 1 1/2 x + 3/4\n\nD. 7 1/2 x + 3/4\n\n7. This soccer season, Gavin scored 9 fewer than 3 times the number of goals that Rico scored. Rico scored 12 goals. The value of which expression is equivalent to the number of goals Gavin scored this soccer season?\n\nA. 3(4 – 3)\n\nB. 3(12 – 9)\n\nC. 9(4 – 1)\n\nD. 9(36 – 1)\n\n8. Which expression uses exactly three terms and is equivalent to 6(2 + x + x + y)?\n\nA. 8 + 8x + 7y\n\nB. 12 + 12x + 6y\n\nC. 8 + 6x + 6x + 6y\n\nD. 12 + 6x + 6x + 6y\n\n9. Michael has \\$68. Craig has \\$24 less than Michael has. Michael spends \\$20 on a new hat. The solution of which equation represents the amount of money (x), in dollars, Craig has after Michael buys the hat?\n\nA. x + 4 = 48\n\nB. x + 4 = 68\n\nC. x + 20 = 44\n\nD. x + 24 = 48\n\n10. The heights, rounded to the nearest foot, of the trees in a park are listed below.\n\n23 13 8 52 26 42 48 52\n\nWhat is the median of the tree heights?\n\nA. 33 feet\n\nB. 34 feet\n\nC. 39 feet\n\nD. 44 feet" ]

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[ "# #129e6c Color Information\n\nIn a RGB color space, hex #129e6c is composed of 7.1% red, 62% green and 42.4% blue. Whereas in a CMYK color space, it is composed of 88.6% cyan, 0% magenta, 31.6% yellow and 38% black. It has a hue angle of 158.6 degrees, a saturation of 79.5% and a lightness of 34.5%. #129e6c color hex could be obtained by blending #24ffd8 with #003d00. Closest websafe color is: #009966.\n\n• R 7\n• G 62\n• B 42\nRGB color chart\n• C 89\n• M 0\n• Y 32\n• K 38\nCMYK color chart\n\n#129e6c color description : Dark cyan - lime green.\n\n# #129e6c Color Conversion\n\nThe hexadecimal color #129e6c has RGB values of R:18, G:158, B:108 and CMYK values of C:0.89, M:0, Y:0.32, K:0.38. Its decimal value is 1220204.\n\nHex triplet RGB Decimal 129e6c `#129e6c` 18, 158, 108 `rgb(18,158,108)` 7.1, 62, 42.4 `rgb(7.1%,62%,42.4%)` 89, 0, 32, 38 158.6°, 79.5, 34.5 `hsl(158.6,79.5%,34.5%)` 158.6°, 88.6, 62 009966 `#009966`\nCIE-LAB 57.716, -46.453, 16.644 15.182, 25.663, 18.34 0.257, 0.434, 25.663 57.716, 49.345, 160.287 57.716, -48.333, 29.353 50.659, -35.159, 13.997 00010010, 10011110, 01101100\n\n# Color Schemes with #129e6c\n\n• #129e6c\n``#129e6c` `rgb(18,158,108)``\n• #9e1244\n``#9e1244` `rgb(158,18,68)``\nComplementary Color\n• #129e26\n``#129e26` `rgb(18,158,38)``\n• #129e6c\n``#129e6c` `rgb(18,158,108)``\n• #128a9e\n``#128a9e` `rgb(18,138,158)``\nAnalogous Color\n• #9e2612\n``#9e2612` `rgb(158,38,18)``\n• #129e6c\n``#129e6c` `rgb(18,158,108)``\n• #9e128a\n``#9e128a` `rgb(158,18,138)``\nSplit Complementary Color\n• #9e6c12\n``#9e6c12` `rgb(158,108,18)``\n• #129e6c\n``#129e6c` `rgb(18,158,108)``\n• #6c129e\n``#6c129e` `rgb(108,18,158)``\n• #449e12\n``#449e12` `rgb(68,158,18)``\n• #129e6c\n``#129e6c` `rgb(18,158,108)``\n• #6c129e\n``#6c129e` `rgb(108,18,158)``\n• #9e1244\n``#9e1244` `rgb(158,18,68)``\n• #0a593d\n``#0a593d` `rgb(10,89,61)``\n• #0d704d\n``#0d704d` `rgb(13,112,77)``\n• #0f875c\n``#0f875c` `rgb(15,135,92)``\n• #129e6c\n``#129e6c` `rgb(18,158,108)``\n• #15b57c\n``#15b57c` `rgb(21,181,124)``\n• #17cc8b\n``#17cc8b` `rgb(23,204,139)``\n• #1ae39b\n``#1ae39b` `rgb(26,227,155)``\nMonochromatic Color\n\n# Alternatives to #129e6c\n\nBelow, you can see some colors close to #129e6c. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #129e49\n``#129e49` `rgb(18,158,73)``\n• #129e55\n``#129e55` `rgb(18,158,85)``\n• #129e60\n``#129e60` `rgb(18,158,96)``\n• #129e6c\n``#129e6c` `rgb(18,158,108)``\n• #129e78\n``#129e78` `rgb(18,158,120)``\n• #129e83\n``#129e83` `rgb(18,158,131)``\n• #129e8f\n``#129e8f` `rgb(18,158,143)``\nSimilar Colors\n\n# #129e6c Preview\n\nThis text has a font color of #129e6c.\n\n``<span style=\"color:#129e6c;\">Text here</span>``\n#129e6c background color\n\nThis paragraph has a background color of #129e6c.\n\n``<p style=\"background-color:#129e6c;\">Content here</p>``\n#129e6c border color\n\nThis element has a border color of #129e6c.\n\n``<div style=\"border:1px solid #129e6c;\">Content here</div>``\nCSS codes\n``.text {color:#129e6c;}``\n``.background {background-color:#129e6c;}``\n``.border {border:1px solid #129e6c;}``\n\n# Shades and Tints of #129e6c\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #02110c is the darkest color, while #ffffff is the lightest one.\n\n• #02110c\n``#02110c` `rgb(2,17,12)``\n• #042318\n``#042318` `rgb(4,35,24)``\n• #063424\n``#063424` `rgb(6,52,36)``\n• #084630\n``#084630` `rgb(8,70,48)``\n• #0a583c\n``#0a583c` `rgb(10,88,60)``\n• #0c6948\n``#0c6948` `rgb(12,105,72)``\n• #0e7b54\n``#0e7b54` `rgb(14,123,84)``\n• #108c60\n``#108c60` `rgb(16,140,96)``\n• #129e6c\n``#129e6c` `rgb(18,158,108)``\n• #14b078\n``#14b078` `rgb(20,176,120)``\n• #16c184\n``#16c184` `rgb(22,193,132)``\n• #18d390\n``#18d390` `rgb(24,211,144)``\n• #1ae49c\n``#1ae49c` `rgb(26,228,156)``\n• #2be7a4\n``#2be7a4` `rgb(43,231,164)``\n• #3de9ab\n``#3de9ab` `rgb(61,233,171)``\n• #4eebb3\n``#4eebb3` `rgb(78,235,179)``\n• #60edbb\n``#60edbb` `rgb(96,237,187)``\n• #72efc2\n``#72efc2` `rgb(114,239,194)``\n• #83f1ca\n``#83f1ca` `rgb(131,241,202)``\n• #95f3d1\n``#95f3d1` `rgb(149,243,209)``\n• #a6f5d9\n``#a6f5d9` `rgb(166,245,217)``\n• #b8f7e0\n``#b8f7e0` `rgb(184,247,224)``\n• #caf9e8\n``#caf9e8` `rgb(202,249,232)``\n• #dbfbf0\n``#dbfbf0` `rgb(219,251,240)``\n• #edfdf7\n``#edfdf7` `rgb(237,253,247)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\nTint Color Variation\n\n# Tones of #129e6c\n\nA tone is produced by adding gray to any pure hue. In this case, #565a59 is the less saturated color, while #04ac70 is the most saturated one.\n\n• #565a59\n``#565a59` `rgb(86,90,89)``\n• #4f615b\n``#4f615b` `rgb(79,97,91)``\n• #48685d\n``#48685d` `rgb(72,104,93)``\n• #416f5e\n``#416f5e` `rgb(65,111,94)``\n• #3b7560\n``#3b7560` `rgb(59,117,96)``\n• #347c62\n``#347c62` `rgb(52,124,98)``\n• #2d8364\n``#2d8364` `rgb(45,131,100)``\n• #268a66\n``#268a66` `rgb(38,138,102)``\n• #209068\n``#209068` `rgb(32,144,104)``\n• #19976a\n``#19976a` `rgb(25,151,106)``\n• #129e6c\n``#129e6c` `rgb(18,158,108)``\n• #0ba56e\n``#0ba56e` `rgb(11,165,110)``\n• #04ac70\n``#04ac70` `rgb(4,172,112)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #129e6c is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "### Sn = [n(n+1)(2n+1)]/ 6\n\n iii. Sn = n(n+1)(2n+1) 6 Sol. Sn = n(n+1)(2n+1) 6 ∴ S1 = 1(1+1)[2(1)+1] = 1(2)(3) = 6 = 1 6 6 6 ∴ S2 = 2(2+1)[2(2)+1] = 2(3)(5) = 30 = 5 6 6 6 ∴ S3 = 3(3+1)[2(3)+1] = 3(4)(7) = 2× 7 = 14 6 6\n\nWe know that,\nt1 = S1 = 1\nt2 = S2 – S1 = 5 – 1 = 4\nt3 = S3 – S2 = 14 – 5 = 9\nThe first three terms of the sequence are 1, 4 and 9." ]

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[ "# polyphase sample rate conversion with non-integer factor\n\nI want to do sample rate conversion by subsequently upsampling with factor I=5, and then downsampling with factor D=9.\n\nI have designed a nyquist sample rate conversion filter h() of length M, with matlab's filterbuilder tool for interpolation factor I=5, and decimation factor D=9. Since D > I, I have chosen the normalized cutoff frequency of the filter to be at 1/9, by setting \"Band\" under filter specifications to 9.\n\nThen I compute I polyphase filters pk(n), which have length K=M/I by sampling h() according to:\n\npk(n) = h(k + n*I), for k=0,...,I-1 and n=0,...,K-1.\n\nThen I only compute the output samples for each polyphase filter, which are sampled by the downsampling operation. In other words, I compute the output y[m], by sampling the outputs of the polyphase filters yk[] according to:\n\ny[m] = yk[ floor( m * D / I ) ]\nk = ( m * D ) modulo I\n\n\nThus, I do not compute samples of yk[] that are not used in the output.\n\nI apply these polyphase filter's first to do sample rate conversion for the rows of the image. Then I apply the same filters to do sample rate conversion for the columns.\n\nHowever, I get a distorted output image which still, clearly contains alias:", null, "Does anyone know If I'm conceptually doing anything wrong?\n\n• What do mean that you set \"Band\" to \"9\". What exactly did you do when you designed the filter? – Jim Clay Aug 15 '13 at 18:02\n• In the matlab command line I typed \"filterbuilder\", then I selected \"Nyquist\" then a menu pops up in which you can specify filter options: \"Band\": 9, \"Impulse response\": FIR, \"Filter order mode\": minimum, \"Filter Type\": Sample-rate converter, \"Interpolation Factor\": 5, \"Decimation Factor\": 9. I leave the rest of the options to their default settings. As far as I understood \"Band\" is the inverse of the cutoff frequency. – Luc Aug 15 '13 at 20:52\n\nYour approach seems conceptually sound. I suspect that the problem is one of two things: 1) you have implemented it incorrectly, 2) Your filter bandwidth is too wide.\n\nYou are implementing the \"smart\" approach in that you are doing it the computationally efficient way. I would try doing it the dumb way (insert I-1 zeroes between every sample, filtering with the entire filter, get rid of the extra samples) to verify your implementation and to see if the filter itself is the problem. You can verify your implementation by comparing your resulting samples against the samples produced by the dumb approach. After you have filtered the upsampled data one of the sample \"phases\" (there are \"D\" phases whose start samples are samples 0...D-1) should be the same as the smart approach's results. If not, something is wrong in your implementation of either the smart or dumb approach.\n\nIf the results do match, then the problem is your filter. You need to look closely at the bandwidth to make sure that it isn't too wide. Since D > I the ideal filter bandwidth, assuming a perfect brick wall filter, is $\\frac{I}{2D}$, where the factor of two comes from the Nyquist rate.\n\n• Hi Jim, I thought that the normalized cutoff frequency of the lowpass filter for sample rate conversion should be min( 1/D, 1/I )? When I set \"Band\" to 9, I can see from the frequency response plot that the cutoff frequency (6dB) of the lowpass filter lies at 1/9 – Luc Aug 15 '13 at 21:26\n• You're right, looking at the sample rate of the input image normalized by the nyquist frequency the brick wall filter should have a cutoff frequency of I/D. Looking at the sample rate of the upsampled image, in which I-1 zeros are added in between each original sample, the normalized cutoff frequency should be min( 1/D, i/I ) – Luc Aug 18 '13 at 22:18\n\nI solved my problem. The design methodology as described above is sound. However, there was an error in my implementation of the FIR filter. For symmetrical filters you can replace convolution by correlation, and that's what I did. However, since the polyphase filters are asymmetrical this led to incorrect results. I implemented the FIR filter by convolution and now my sample rate converter works like a charm.\n\nBelow you can see the correctly resampled image:", null, "And here's the original input image:", null, "• Can you add the correctly filtered image? It'd be nice for comparison. – Peter K. Aug 18 '13 at 21:53\n• Hi, Peter. I've added the correctly resampled image and the original image. Could you accept my answer and upvote it please? Thank you. – Luc Aug 18 '13 at 22:19" ]

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[ "# How To Fill Percentage In Upsc Form In Cgpa Format?\n\nContents\n\n## How to fill CGPA in UPSC form?\n\nIt all depends on the college from which you are getting your degree. But if you college doesn’t has any conversion criteria then you can simply multiply CGPA by 10 and get the required percentage.\n\n## How to fill out CGPA marks?\n\nStep 1: Add the grade points i.e 9+8+7+8+8 = 40. Step 2: Divide the sum by 5 i.e 40/5 = 8. Thus, your CGPA is 8.0.\n\n## How do I convert my CGPA to percentage?\n\nSimply multiply your CGPA by 9.5, and you’ll have your percentage. For instance, suppose you have a 9.4 CGPA and need to convert it to a percentage. Multiply it by 9.5, for example, (9.4 X 9.5) = 89.3. So your percentage is 89.3 percent.\n\n## What percentage is 7.6 CGPA?\n\nHere the percentage from CGPA will be obtained by multiplying the CGPA by 10. For example, if your CGPA is 7.6, then the percentage will be 76%.\n\n## What CGPA is 70%?\n\nThe Percentage you received is 70. To find the CGPA, divide 70 by 9.5. The CGPA you will receive is 7.4.\n\n## What CGPA is 75 %?\n\nFor example, to convert 75 percent to CGPA, we divide it by 9.5 and the resulting number 7.9 is the CGPA out of 10.\n\n## What CGPA is 73 %?\n\nWhat is the CGPA of 73%? To calculate CGPA of 73%, we can simply multiply it by 9.5, which will be 7.68 CGPA.\n\n## What CGPA is 83%?\n\nA 3.0 GPA, or Grade Point Average, is equivalent to a B letter grade on a 4.0 GPA scale, and a percentage grade of 83–86.\n\n## What CGPA is 82%?\n\n2.7 GPA is a B- Letter Grade or 80–82% – GPA Calculator.\n\n## What CGPA is 97 %?\n\nA+ GPA. An A+ letter grade is equivalent to a 4.0 GPA, or Grade Point Average, on a 4.0 GPA scale, and a percentage grade of 97–100." ]

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[ "# zbMATH — the first resource for mathematics\n\nElliptic regularity with continuous and branching edge asymptotics. (English) Zbl 0718.35012\nMultivariate approximation and interpolation, Proc. Int. Workshop, Duisburg/FRG 1989, ISNM Int. Ser. Numer. Math. 94, 275-284 (1990).\n[For the entire collection see Zbl 0703.00022.]\nThe paper gives a description of the asymptotics of solutions of elliptic equations in non-smooth domains where the singular set is of the type of an edge, outside of which the structure is $$C^{\\infty}$$. The difficulty is that the data of the asymptotics $\\sum^{\\infty}_{j=0}\\sum^{m_ j}_{k=0}\\zeta_{jk}r^{-p_ j} \\log^ kr\\quad as\\quad r\\to 0$ with r being the ‘distance’ to the edge, such as the $$p_ j\\in {\\mathbb{C}}$$, $$m_ j\\in {\\mathbb{N}}$$, and the coefficients $$\\zeta_{jk}\\in {\\mathbb{C}}$$ may depend on the edge variable y, and there are in general jumps or more chaotic changes with varying y. This is to be expected even for solutions which are $$C^{\\infty}$$ outside the edge. In the author’s work [Integral Equations Oper. Theory 11, No.4, 557-602 (1988; Zbl 0671.58040)] the singular functions have been described in this $$C^{\\infty}$$ case, by using $$C^{\\infty}$$ functions of analytic functionals in the complex Mellin plane which are pointwise discrete and of finite order. The present article extends that result to solutions in weighted Sobolev spaces. Background are the more general continuous asymptotics, earlier introduced by the author [Math. Nachr. 136, 7-57 (1988; Zbl 0664.58041)].\nReviewer: B.-W.Schulze\n##### MSC:\n 35B40 Asymptotic behavior of solutions to PDEs 35B65 Smoothness and regularity of solutions to PDEs 35J25 Boundary value problems for second-order elliptic equations 35C20 Asymptotic expansions of solutions to PDEs\n##### Keywords:\nsingular set; edge; Mellin plane" ]

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[ "# #19200b Color Information\n\nIn a RGB color space, hex #19200b is composed of 9.8% red, 12.5% green and 4.3% blue. Whereas in a CMYK color space, it is composed of 21.9% cyan, 0% magenta, 65.6% yellow and 87.5% black. It has a hue angle of 80 degrees, a saturation of 48.8% and a lightness of 8.4%. #19200b color hex could be obtained by blending #324016 with #000000. Closest websafe color is: #003300.\n\n• R 10\n• G 13\n• B 4\nRGB color chart\n• C 22\n• M 0\n• Y 66\n• K 87\nCMYK color chart\n\n#19200b color description : Very dark (mostly black) green.\n\n# #19200b Color Conversion\n\nThe hexadecimal color #19200b has RGB values of R:25, G:32, B:11 and CMYK values of C:0.22, M:0, Y:0.66, K:0.87. Its decimal value is 1646603.\n\nHex triplet RGB Decimal 19200b `#19200b` 25, 32, 11 `rgb(25,32,11)` 9.8, 12.5, 4.3 `rgb(9.8%,12.5%,4.3%)` 22, 0, 66, 87 80°, 48.8, 8.4 `hsl(80,48.8%,8.4%)` 80°, 65.6, 12.5 003300 `#003300`\nCIE-LAB 11.02, -7.722, 11.72 0.978, 1.264, 0.509 0.355, 0.459, 1.264 11.02, 14.035, 123.381 11.02, -2.236, 8.831 11.242, -4.148, 5.185 00011001, 00100000, 00001011\n\n# Color Schemes with #19200b\n\n• #19200b\n``#19200b` `rgb(25,32,11)``\n• #120b20\n``#120b20` `rgb(18,11,32)``\nComplementary Color\n• #201d0b\n``#201d0b` `rgb(32,29,11)``\n• #19200b\n``#19200b` `rgb(25,32,11)``\n• #0f200b\n``#0f200b` `rgb(15,32,11)``\nAnalogous Color\n• #1d0b20\n``#1d0b20` `rgb(29,11,32)``\n• #19200b\n``#19200b` `rgb(25,32,11)``\n• #0b0f20\n``#0b0f20` `rgb(11,15,32)``\nSplit Complementary Color\n• #200b19\n``#200b19` `rgb(32,11,25)``\n• #19200b\n``#19200b` `rgb(25,32,11)``\n• #0b1920\n``#0b1920` `rgb(11,25,32)``\n• #20120b\n``#20120b` `rgb(32,18,11)``\n• #19200b\n``#19200b` `rgb(25,32,11)``\n• #0b1920\n``#0b1920` `rgb(11,25,32)``\n• #120b20\n``#120b20` `rgb(18,11,32)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #0a0d04\n``#0a0d04` `rgb(10,13,4)``\n• #19200b\n``#19200b` `rgb(25,32,11)``\n• #283312\n``#283312` `rgb(40,51,18)``\n• #374618\n``#374618` `rgb(55,70,24)``\n• #45591f\n``#45591f` `rgb(69,89,31)``\nMonochromatic Color\n\n# Alternatives to #19200b\n\nBelow, you can see some colors close to #19200b. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #1e200b\n``#1e200b` `rgb(30,32,11)``\n• #1d200b\n``#1d200b` `rgb(29,32,11)``\n• #1b200b\n``#1b200b` `rgb(27,32,11)``\n• #19200b\n``#19200b` `rgb(25,32,11)``\n• #17200b\n``#17200b` `rgb(23,32,11)``\n• #16200b\n``#16200b` `rgb(22,32,11)``\n• #14200b\n``#14200b` `rgb(20,32,11)``\nSimilar Colors\n\n# #19200b Preview\n\nThis text has a font color of #19200b.\n\n``<span style=\"color:#19200b;\">Text here</span>``\n#19200b background color\n\nThis paragraph has a background color of #19200b.\n\n``<p style=\"background-color:#19200b;\">Content here</p>``\n#19200b border color\n\nThis element has a border color of #19200b.\n\n``<div style=\"border:1px solid #19200b;\">Content here</div>``\nCSS codes\n``.text {color:#19200b;}``\n``.background {background-color:#19200b;}``\n``.border {border:1px solid #19200b;}``\n\n# Shades and Tints of #19200b\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #020301 is the darkest color, while #f8fbf3 is the lightest one.\n\n• #020301\n``#020301` `rgb(2,3,1)``\n• #0e1106\n``#0e1106` `rgb(14,17,6)``\n• #19200b\n``#19200b` `rgb(25,32,11)``\n• #242f10\n``#242f10` `rgb(36,47,16)``\n• #303d15\n``#303d15` `rgb(48,61,21)``\n• #3b4c1a\n``#3b4c1a` `rgb(59,76,26)``\n• #475a1f\n``#475a1f` `rgb(71,90,31)``\n• #526924\n``#526924` `rgb(82,105,36)``\n• #5d7829\n``#5d7829` `rgb(93,120,41)``\n• #69862e\n``#69862e` `rgb(105,134,46)``\n• #749533\n``#749533` `rgb(116,149,51)``\n• #80a338\n``#80a338` `rgb(128,163,56)``\n• #8bb23d\n``#8bb23d` `rgb(139,178,61)``\n• #96bf44\n``#96bf44` `rgb(150,191,68)``\n• #9ec453\n``#9ec453` `rgb(158,196,83)``\n• #a6c961\n``#a6c961` `rgb(166,201,97)``\n• #aece70\n``#aece70` `rgb(174,206,112)``\n• #b7d37e\n``#b7d37e` `rgb(183,211,126)``\n• #bfd88d\n``#bfd88d` `rgb(191,216,141)``\n• #c7dd9c\n``#c7dd9c` `rgb(199,221,156)``\n• #cfe2aa\n``#cfe2aa` `rgb(207,226,170)``\n• #d8e7b9\n``#d8e7b9` `rgb(216,231,185)``\n• #e0ecc7\n``#e0ecc7` `rgb(224,236,199)``\n• #e8f1d6\n``#e8f1d6` `rgb(232,241,214)``\n• #f0f6e5\n``#f0f6e5` `rgb(240,246,229)``\n• #f8fbf3\n``#f8fbf3` `rgb(248,251,243)``\nTint Color Variation\n\n# Tones of #19200b\n\nA tone is produced by adding gray to any pure hue. In this case, #161615 is the less saturated color, while #1c2a01 is the most saturated one.\n\n• #161615\n``#161615` `rgb(22,22,21)``\n• #161813\n``#161813` `rgb(22,24,19)``\n• #171912\n``#171912` `rgb(23,25,18)``\n• #171b10\n``#171b10` `rgb(23,27,16)``\n• #181d0e\n``#181d0e` `rgb(24,29,14)``\n• #181e0d\n``#181e0d` `rgb(24,30,13)``\n• #19200b\n``#19200b` `rgb(25,32,11)``\n• #1a2209\n``#1a2209` `rgb(26,34,9)``\n• #1a2308\n``#1a2308` `rgb(26,35,8)``\n• #1b2506\n``#1b2506` `rgb(27,37,6)``\n• #1b2704\n``#1b2704` `rgb(27,39,4)``\n• #1c2803\n``#1c2803` `rgb(28,40,3)``\n• #1c2a01\n``#1c2a01` `rgb(28,42,1)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #19200b is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# How to Iterate Through a List in Python\n\nPython is a versatile programming language that offers multiple ways to work with lists. In this article, we will dive deep into various techniques for iterating through a list in Python. By the end of this article, you will have a solid understanding of different iteration methods and when to use them.\n\n## 1. Using For Loop to Iterate Through a List\n\nA `for` loop is the most common and straightforward way to iterate through a list in Python. Here’s a simple example:\n\n```my_list = [1, 2, 3, 4, 5] for item in my_list: print(item)```\n\nIn this example, the `for` loop iterates over each element in the `my_list` list and prints it.\n\n## 2. Using While Loop for List Iteration\n\nA `while` loop can also be used to iterate through a list in Python. You need to use an index variable to access the elements of the list:\n\n```my_list = [1, 2, 3, 4, 5] index = 0 while index < len(my_list): print(my_list[index]) index += 1```\n\nIn this example, the `while` loop iterates over each element in the `my_list` list using an index variable and prints it.\n\n## 3. Using the Enumerate Function\n\nThe `enumerate()` function is a built-in Python function that allows you to iterate through a list while keeping track of the index. Here’s how to use it:\n\n```my_list = ['apple', 'banana', 'cherry'] for index, element in enumerate(my_list): print(f\"Index: {index}, Element: {element}\")```\n\nIn this example, the `enumerate()` function returns both the index and the element for each iteration, allowing you to perform operations that require the index.\n\n## 4. Using List Comprehensions\n\nList comprehensions are a concise way to create a new list by iterating through an existing list and applying a transformation to each element. Here’s a simple example:\n\n```my_list = [1, 2, 3, 4, 5] squared_list = [item ** 2 for item in my_list] print(squared_list) # Output: [1, 4, 9, 16, 25]```\n\nIn this example, we create a new list containing the square of each element in `my_list` using a list comprehension.\n\n## 5. Using the Map Function\n\nThe `map()` function is a higher-order function that applies a given function to each item in an iterable, such as a list, and returns an iterator. Here’s an example:\n\n```def square(item): return item ** 2 my_list = [1, 2, 3, 4, 5] squared_list = list(map(square, my_list)) print(squared_list) # Output: [1, 4, 9, 16, 25]```\n\nIn this example, the `square()` function is applied to each element in `my_list`, and the result is converted back to a list using the `list()` function.\n\n## 6. Using the Filter Function\n\nThe `filter()` function is another higher-order function that filters the elements of an iterable based on a given function. Here’s a simple example:\n\n```def is_even(item): return item % 2 == 0 my_list = [1, 2, 3, 4, 5] even_list = list(filter(is_even, my_list)) print (even_list) # Output: [2, 4]```\n\nIn this example, the `is_even()` function is used to filter the even elements in `my_list`, and the result is converted back to a list using the `list()` function.\n\n## 7. Using the Zip Function\n\nThe `zip()` function is a built-in Python function that allows you to iterate through multiple lists in parallel. Here’s how to use it:\n\n```list1 = [1, 2, 3] list2 = [4, 5, 6] for item1, item2 in zip(list1, list2): print(f\"Item1: {item1}, Item2: {item2}\")```\n\nIn this example, the `zip()` function combines the elements from `list1` and `list2` in pairs, allowing you to perform operations that require elements from both lists.\n\n## 8. Using Nested Loops for Multidimensional Lists\n\nWhen working with multidimensional lists, such as matrices, nested loops can be used to iterate through the elements:\n\n```matrix = [ [1, 2, 3], [4, 5, 6], [7, 8, 9]] for row in matrix: for item in row: print(item, end=' ') print()```\n\nIn this example, a nested `for` loop is used to iterate through the rows and columns of the `matrix` list.\n\n## 9. Using List Methods to Iterate Through Elements\n\nPython lists have built-in methods, such as `pop()` and `remove()`, which can be used to iterate through the elements of a list while modifying it. Here’s an example:\n\n```my_list = [1, 2, 3, 4, 5] while my_list: item = my_list.pop(0) print(item) print(my_list) # Output: []```\n\nIn this example, the `pop()` method is used to remove and return the first element of the list until the list is empty.\n\n## 10. Using Itertools Module for Advanced Iteration\n\nThe `itertools` module provides a collection of tools for handling iterators, including advanced iteration techniques. Here’s an example using the `itertools.cycle()` function:\n\n```import itertools my_list = [1, 2, 3] cycled_list = itertools.cycle(my_list) for i in range(10): print(next(cycled_list))```\n\nIn this example, the `cycle()` function creates an iterator that cycles through the elements of the `my_list` indefinitely, allowing you to perform repeated iterations.\n\n## 11. Conclusion\n\nPython offers multiple ways to iterate through a list, each with its own advantages and use cases. By understanding these techniques, you can write more efficient and readable code. From basic `for` loops to advanced techniques using the `itertools` module, this comprehensive guide has covered various methods to help you master list iteration in Python.\n\n## 12. FAQ\n\n##### Q1: Can I iterate through multiple lists at once?\n\nYes, you can use the `zip()` function to iterate through multiple lists in parallel.\n\n##### Q2: How can I iterate through a list in reverse order?\n\nYou can use the `reversed()` function or slice notation to iterate through a list in reverse order.\n\n##### Q3: Can I use a list comprehension to iterate through a list and filter its elements?\n\nYes, you can use a list comprehension with a conditional expression to filter the elements of a list.\n\n##### Q4: How can I apply a function to every element in a list?\n\nYou can use the `map()` function or a list comprehension to apply a function to every element in a list.\n\n##### Q5: What is the difference between `map()` and a list comprehension?\n\n`map()` is a higher-order function that applies a given function to each item in an iterable and returns an iterator. List comprehensions, on the other hand, are a concise way to create a new list by iterating through an existing list and applying a transformation to each element. List comprehensions can also include an optional condition for filtering elements. While both methods can be used to perform similar operations, list comprehensions are often considered more Pythonic and easier to read.\n\nIt’s important to remember that the `map()` function returns an iterator, so you need to convert it back to a list using the `list()` function if you want a list as the output. List comprehensions, on the other hand, directly create a new list." ]

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[ "## December 01, 2019\n\nThis just keeps coming up, so let's explore the dynamic a bit.\n\nYou have a paid search program. Last year in November you spent \\$100,000, and you obtained the following outcome:\n• Spend = \\$100,000.\n• Clicks = 200,000.\n• Cost per Click = \\$0.50.\n• Conversion Rate = 1.8%.\n• Orders = 3,600.\n• Average Order Value = \\$100.\n• Profit Factor = 30%.\n• Profit = (3,600*100)*0.30 - \\$100,000 = \\$8,000.\n• Profit per Order = (8,000 / 3,600) = \\$2.22.\nThis year, however, your merchandise productivity is -10%. Your metrics, as a consequence, look different.\n\n• Spend = \\$100,000.\n• Clicks = 200,000.\n• Cost per Click = \\$0.50.\n• Conversion Rate = (1.8%*0.90) = 1.62%.\n• Orders = 3,240.\n• Average Order Value = \\$100.\n• Profit Factor = 30%.\n• Profit = (3,240*100)*0.30 - \\$100,000 = (\\$2,800).\n• Profit per Order = (-2,800 / 3,240) = (\\$0.86).\nBecause of a merchandise productivity issue, marketing metrics look worse ... you generated \\$2.22 of profit per order last year, you generate a loss of \\$0.86 per order this year.\n\nAs a marketer, you're likely to \"optimize\" performance. You'll cut back on marketing spend, and your metrics change as a result.\n• Spend = \\$80,000.\n• Clicks = 179,000.\n• Cost per Click = \\$0.45.\n• Conversion Rate = (1.8%*0.90) = 1.62%.\n• Orders = 2,900.\n• Average Order Value = \\$100.\n• Profit Factor = 30%.\n• Profit = (2,900*100)*0.30 - \\$100,000 = \\$7,000.\n• Profit per Order = (7,000 / 2,900) = \\$2.41.\nThe marketer \"optimized\" performance. But there is a two-stage outcome that must be understood, an outcome caused by a 10% drop in merchandise productivity.\n• Orders dropped from 3,600 to 3,240 because of merchandise productivity declines.\n• Orders then drop from 3,240 to 2,900 because of how marketing responded to merchandise productivity declines.\nIn other words, a 10% drop in merchandise productivity results in a 19% drop in orders because marketing optimizes performance.\n\nNow you likely have a new customer issue.\n\nIt's important to work carefully with your Chief Financial Officer on these issues. It may be more profitable long-term to lose money today and acquire enough new customers to protect the future of your business. Just be sure to do the math and figure out what makes the most sense for your business, ok?", null, "" ]

[ null, "https://1.bp.blogspot.com/-ytcsRkCyf-c/YWnGfjebkCI/AAAAAAAAKTQ/upunC2CXiAAIbRJIZMZIMO2emGzlZT8sQCLcBGAsYHQ/w400-h225/mtd_20211015c.jpg", null ]

[ "# Single and double angle plots interpretation for Astigmatism Analysis\n\nThere are two ways of representing the results of astigmatism analysis The first way is the most common way that everyone knows because it is the way That is more common or more familiar with the real location around the eye It is the Single Angle representation. In this kind of Single Angle representation We have a polar plot from 0 to 360 degrees and we know that the zero is the same location that the 360 degrees and that are coincident with the locations in the eye. Anyway when we work with astigmatism analysis the most common and the most adequate could be to work in a double angle even though we can show all the results in a single angle plot. All our calculations are going to be performed in a double angle plot and in the next lesson We will explain what is the reason of this kind of analysis based on the double angle. The most important here in order to interpret correctly the result is that if we are working with a single angle if our results are representing in a 360 degrees plot, our results are going to be classified in with-the-rule astigmatism in the vertical orientation. You can see here that the blue area is the area that contains all the cases that are with-the-rule cases with-the-rule astigmatism and we have the against-the-rule in the horizontal location Also we have this kind of oblique locations in yellow. Anyway, This is very familiar for the surgeon because you see exactly what are the locations of the astigmatism in the same way that are located in the real eye. Anyway if you have a double angle plot you have to see first What is the number of angles you have in the plot You can see here that you have from 0 to 180 So the data are going to be represented in a different way. All your with-the-rule cases are going to be located at the left side of the plot and all there against-the-rule cases are going to be located at the right side. So in the vertical locations you are going to have the oblique positions So you have to know before to interpret the results If you are working with that single angle or if you are working in a double angle plot. Even though you can represent in both sides you can see here exactly what is the corneal topography for the different kinds of astigmatism. E ven though you can use both plots. Always the calculations are going to be conducted in a double angle plot and later you have to transform the location of the points to a single angle if you want to represent a single angle and usually If you are going to represent in a single angle you are going to represent only the upper side of the plot. Not all the plot like in the double angle because here you have one hundred and eighty degrees and here You have also one hundred and eighty degrees." ]

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[ "Home » Programming » Data Structure and Algorithm » Stack Data Structure and its Operations in C\n\n# Stack Data Structure and its Operations in C", null, "A stack is non-primitive linear data structure with predefined capacity. Additionally, it is an ordered list which allows the insertion of new data items and deletion of existing data items from one end only, known as top of stack (TOS). A stack is said to be empty or underflow if there are not any data items in the stack, whereas if the stack is full it is said to be overflow.\n\n## Basic Feature of Stack Data Structure\n\n• Stack is an Abstract Data Type (ADT).\n• It is also called Last-In-First-Out (LIFO) since the last item added in the stack (which is in the top) is removed at first from the top.\n• To insert a data item in stack we use the “push” function. Likewise, to delete  a data item we use the “pop” function.\n• We cannot access data items randomly in stack due to restriction.\n\n## Basic Operations\n\n• PUSH operation: The “PUSH” operation inserts new data items in stack. In this case, the data item most recently pushed/added is at the Top.\n• POP operation: The “POP” operation deletes the data item which is at the top of the stack.\n• createStack operation: This operation creates a new empty stack.\n• isFull Operation: This  operation checks if the stack is full or not (i.e. stack overflow).\n• isEmpty Operation: This operation checks if the stack is empty or not (i.e. stack underflow).\n• Peek operation: This operation returns the top data item of the stack without removing it.\n\n## Applications of Stack Data Structure\n\nStack data structure are used in computer in various ways. Some of them are:\n\n• String Reversal: Stack can be used in reversal of string. Firstly, we push every character of the string in the stack and then pop them one by one to reverse a string.\n• Expression Evaluation and Conversion: We can evaluate the prefix, postfix and infix expressions using stacks. Also, they can be helpful to convert one form of expression to another form (like infix to postfix, postfix to prefix, etc.).\n• Syntax Parsing: Parsing of program blocks, syntax of expressions, etc. can be done using stack. If you want to read more about syntax parsing and parsers you can read it in Wikipedia.\n• Parenthesis Checking: Compilers can use the stack to check the correct opening and closing of parenthesis.\n• Recursion: Stack is used to store and restore the recursive function and their arguments.\n\n## Implementation of Stack\n\nThere are two ways to implement stack. They are:\n\n1. Array Implementation of stack: This is also called static implementation because they are limited in size. They are quick because there is no overhead to allocate, link, unlink and deallocate the storage blocks.\n2. Linked List implementation of stack: This is also called dynamic implementation because the size of stack is not limited. However, there is storage block allocation, linking, unlinking and deallocation overhead in this implementation. Due to which linked lists are inefficient than array implementation.\n\n### Static (Array) implementation of stack data structure in C\n\nStatic or array implementation of stack data structure in C uses a one dimensional array to store the data. Also, we use a variable “top” to keep track of the top position of the stack. We increment top by 1 every time we add any data and decrement by 1 every time we delete any item. Generally, in this implementation we set the value of top to -1 (top = -1) to indicate the empty stack.\n\nWe will be using structure to store data here. However, we can use array and “top” variables individually.\n\n```#define size 50\nstruct Stack {\nint items[size];\nint top;\n};\ntypedef struct Stack STACK;```\n\n#### Creating empty stack:\n\nIn C implementation, the value of top = -1 shows the stack is empty.\n\n```void createStack(STACK *s) {\ns->top = -1;\n}```\n\n#### Stack Underflow:\n\nStack Underflow means that there aren’t any data items in the stack. That means the value of top will be -1 (top = -1). The following function returns 1 if the stack is empty else returns 0.\n\n```int isEmpty(STACK *s) {\nif(s->top == -1)\nreturn 1;\nelse\nreturn 0;\n}```\n\n#### Stack Overflow:\n\nStack Overflow means that we cannot push data items any more in the stack. In this case, the top lies in the highest location (top = size – 1) of the stack. The following function returns 1 if the stack is full else returns 0.\n\n```int isFull(STACK *s) {\nif(s->top == size - 10)\nreturn 1;\nelse\nreturn 0;\n}```\n\n#### Algorithm for PUSH operation\n\nThis algorithm is used for insertion of an item at the top of the stack.\n\n```Step 1: Check for the stack overflow (i.e. if the stack is full or not).\nStep 2: If there is overflow, then print the error and exit.\nStep 3: If there isn't overflow, then increment the value of top by 1 and add the item.```\n\n#### Algorithm for POP operation\n\nThis algorithm deletes or removes an item from top of the stack and assigns it to a variable.\n\n```Step 1: Check for the stack underflow (i.e if the stack is empty or not).\nStep 2: If there is underflow, then print the error and exit.\nStep 3: If there isn’t underflow, then assign the top value to variable and decrement the top by 1.```\n\n#### Program to implement Stack Data Structure in C\n\n```#include<stdio.h>\n#include<conio.h>\n\n/* size gives the maximum size of the stack */\n#define size 50\n\nstruct Stack {\nint items[size];\nint top;\n};\n/* defines \"struct Stack\" as \"STACK\" */\ntypedef struct Stack STACK;\n\n/* This function creates a empty stack */\nvoid createStack(STACK *s) {\ns->top = -1;\n}\n\n/* This function returns 1 if the stack is empty else returns 0 */\nint isEmpty(STACK *s) {\nif(s->top == -1)\nreturn 1;\nelse\nreturn 0;\n}\n\n/* This function returns 1 if the stack is full else returns 0 */\nint isFull(STACK *s) {\nif(s->top == size - 1)\nreturn 1;\nelse\nreturn 0;\n}\n\n/* This function push the data item in the stack */\nvoid push(STACK *s, int data) {\nif(isFull(s) == 1) {\nprintf(\"Stack overflow\");\nexit(1);\n} else {\ns->top++;\ns->items[s->top] = data;\n}\n}\n\n/* This function removes the top data item and returns it. */\nint pop(STACK *s) {\nint value;\nif(isEmpty(s) == 1) {\nprintf(\"Stack Underflow\");\nexit(1);\nreturn 0;\n} else {\nvalue = s->items[s->top];\ns->top--;\nreturn value;\n}\n}\n\n/* This function returns the top data item of the stack. */\nint peek(STACK *s) {\nif(isEmpty(s) == 1) {\nprintf(\"Underflow\");\nexit(1);\nreturn 0;\n} else {\nreturn s->items[s->top];\n}\n}\n\n/* This function prints all the data items in the stack. */\nvoid printStack(STACK *s) {\nint i;\nif(isEmpty(s) == 1) {\nprintf(\"\\nStack is Empty.\");\n} else {\nfor(i=0; i<=s->top; i++) {\nprintf(\"\\n%d\", s->items[i]);\n}\n}\n}\n\nint main(){\nint item, choice;\nint flag=1;\nSTACK st;\nSTACK *s;\ns = &st;\nclrscr();\ncreateStack(s);\ndo {\n/* Showing the menu items */\nprintf(\"\\n1. Push the data item.\");\nprintf(\"\\n2. Pop the data item.\");\nprintf(\"\\n3. Check the top item.\");\nprintf(\"\\n4. Display all data items.\");\nprintf(\"\\n5. Exit.\");\n\n/* Taking choice */\nscanf(\"%d\", &choice);\n\n/* Switching according to the choice */\nswitch(choice) {\ncase 1:\nprintf(\"\\nEnter the number to be pushed: \");\nscanf(\"%d\", &item);\npush(s, item);\nbreak;\ncase 2:\nitem = pop(s);\nprintf(\"\\nPopped item is %d.\", item);\nbreak;\ncase 3:\nitem = peek(s);\nprintf(\"\\nTop item is %d\", item);\nbreak;\ncase 4:\nprintf(\"All the data items in stack are: \");\nprintStack(s);\nbreak;\ncase 5:\nflag = 0;\nbreak;\ndefault:\nprintf(\"Invalid Choice!!!\");\n}\n} while(flag == 1);\nreturn 0;\n}```\n##### Output\n```_______ Menu _______\n1. Push the data item.\n2. Pop the data item.\n3. Check the top item.\n4. Display all data items.\n5. Exit.\n\nEnter the number to be pushed: 5" ]

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[ "", null, "Search:\n\n# Finding duplicates in a range\n\nAsked By: Howard    Date: Nov 18    Category: MS Office    Views: 1874\n\nI have this code which finds duplicates from a range and gives red\ncolor to the duplicates. I want someone to modify this code so that\nboth the exact match cases should get red color.\n\neg... i have william, henry, suzanne, william, henry in a range of\ncells. Both the william's should get red color and both the henry's\ntoo. With the code i have only one william and one henry becomes red\ncolor.\n\nHere's the code.\n\nPrivate Sub CommandButton1_Click()\nDim i, j, k\nDim objdict As Dictionary\nDim objsheet As Worksheet\nSet objdict = New Dictionary\nSet objsheet = Sheets(\"Sheet1\")\nk = 1\nFor j = 2 To 5\nFor i = 1 To 250\nIf objdict.exists(objsheet.Cells(i, j).Value) Then\nobjsheet.Cells(i, j).Font.Color = 255\nElse\nk = k + 1\nEnd If\nNext\nNext\n\nEnd Sub\n\nShare:\n\nYou don't need VBA for this. Conditional formatting will do it. Base\nyour conditional formula on a COUNTIF statement, highlighting wheere the\nCOUNTIF gives a value  of >1 when the individual cell is compared to the\nwhole range  of names.\n\nI designed today a formula how does just that and even uses a dynamic range\nfor itself to function...\n\n=AANTAL.ALS(VERSCHUIVING(\\$A\\$2;0;0;AANTALARG(\\$A:\\$A)-1;AANTALARG(\\$A\\$2:\\$B\\$2));A\n2)>1\n\nwhere:\nAANTAL.ALS = COUNTIF\nVERSCHUIVING = OFFSET\nAANTALARG = COUNTA\n\nBe sure to be on A1 of the range  (= A2 in my example because row 1 contains\nlabels). Notice that it takes all the rows in account and only two columns\nso\n\n=AANTAL.ALS(VERSCHUIVING(\\$A\\$2;0;0;AANTALARG(\\$A:\\$A)-1;2);A2)>1\n\ndoes the same as the formula above.\n\nThis worked for me:\n\nPrivate Sub CommandButton1_Click()\nDim i, j, k, j1, i1, limit\nDim objdict As Dictionary\nDim objsheet As Worksheet\nSet objdict = New Dictionary\nSet objsheet = Sheets(\"Sheet1\")\nk = 1\nFor j = 2 To 5\nFor i = 1 To 250\nIf objdict.exists(objsheet.Cells(i, j).Value) Then\nobjsheet.Cells(i, j).Font.Color = 255\nIf Not IsEmpty(objsheet.Cells(i, j)) Then\nFor j1 = 2 To j\nIf j1 = j Then limit = i - 1 Else limit = 250\nFor i1 = 1 To limit\nIf objsheet.Cells(i, j).Value = _\nobjsheet.Cells(i1, j1).Value _\nThen objsheet.Cells(i1, j1).Font.Color = 255\nNext i1\nNext j1\nEnd If\nElse\nIf Not IsEmpty(objsheet.Cells(i, j)) Then objdict.Add objsheet.Cells\n(i, j).Value, k\nk = k + 1\nEnd If\nNext\nNext\nEnd Sub\n\nThere were a few problems with it adding the first empty cell it came\nacross as an item in the dictionary which then slowed the rest down\nas it scanned for repeats of empty cells and changed the empty cell's\nfont to red  but that is now catered for.\n\nDidn't find what you were looking for? Find more on Finding duplicates in a range Or get search suggestion and latest updates." ]

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[ "#### You may also like", null, "### Understanding Hypotheses", null, "Can you work out the means of these distributions using numerical methods?\n\n# Stats Statements\n\n##### Age 16 to 18 Challenge Level:\n\nRussell from Willenhall School Sports College gave answers to five of the parts of this problem using a good mix of examples and results from distributions. Other contributions came from anonymous solution submitters and from teachers attending the Goldman Sachs Teacher Inspiration Day .\n\n1) This doesn't have to be true. For example, in the set of results $0,0,58,72,51,63,60,56$ only $2$ out of $8$ got less than the average mark of $45$ because of the two extreme cases of the two people that put their name on the paper and then left! It is true if the results are normally (or symmetrically) distributed. The less symmetrical the distribution, the less likely that half the students will be under average.\n\nThis is usually true when lots of people take a test and the result is symmetrically distributed about the mean (like the normal distribution). It is not usually true when the results are skewed with large outliers for some reason\n\n2) This is always false unless everyone gets exactly the same mark\n\n3) Because the population is large, the question only says 'about half' and weights of adults are likely to be normally distributed, the result is likely to be true.\n\n4) The total score over N games will be an even number. But the average might be even or odd. For example, scoring $10$ and $20$ over $2$ games gives an average of $15$. Scoring $10$, $20$ and $30$ over $3$ games gives an average of $20$.\n\n5) This is sometimes true. For example, when rolling a fair die the standard deviation is $\\sqrt{\\frac{35}{12}} \\approx 1.71$. I could roll the die three times and get $3, 4, 4$. This has a range of $1$, which is less than $1.71$. It can also obviously be false. For the example of the roll of a die you are very likely to observe a range larger than the standard deviation.\n\nFor a normal $N(0,1)$ distribution, the probability of a random variable $X$being within half a standard deviation of the mean is\n$$P(-0.5< X< 0.5) = \\Phi(0.5) -\\Phi(-0.5) =0.69-0.31=0.38$$\nThe chance of 3 results occurring in this range is $0.388^3 = 0.05$. From this we can see that there is a small chance that 3 or more results will lie within 1 standard deviation of each other. (although this does not show it directly, because we could in a very unlikely set of results draw 3 numbers far from the mean which just happen to be close to each other)\n\nWe think that this helps to show that in almost all situations it is very unlikely that 3 or more randomly generated numbers are within 1 standard deviation of each other.\n\n6) This is definitely true for distributions like normal where the range of possible values is infinite. Let's look at a different distribution. For a binomial distribution $B(N, p)$ the variance is $Np(1-p)$. With a binomial distribution the smallest possible outcome is $0$ and the largest is $N$. So the theoretical maximum range is $N$. The result is true for a binomial $B(N, p)$ if\n$$\\sqrt{Np(1-p)}\\leq \\frac{1}{2}N$$\nThis is only true in the case that $p(1-p)\\leq \\frac{N}{4}$ which is only false in the special case when $N=1$ and $p=0.5$. For a dice, half the range is 3 which is bigger then the standard deviation of $1.8$. So it seems that the result can be false, but only under very special circumstances.\n\n7) Chebyshev's inequality says that the probability that a random number is more then $k$ standard deviations from the mean is not more than $\\frac{1}{k^2}$. So, in this case the probability would be $\\frac{1}{9}$. This means that the result is sometimes false. For the special case of a normal distribution, the chance of being within $3$ standard deviations of the mean is $0.0027$. So, the result is true for normal distributions.\n\n8) This is always true by the law of large numbers, assuming that the average outcome is defined.  (The precise statement of the law of large numbers is somewhat technical, but in most everyday cases this is true.)\n\n9) This is always the case, using Chebyshev's inequality. For a normal distribution, the probability of being within 10 standard deviations is about $1.5\\times 10^{-23}$. So, for most distributions it is really, really, really likely that the sample is within 10 standard deviations of the mean.\n\n10) Although this sounds like it ought to be true, it is not. This counter example shows why. The correlation between two random variables $X$ and $Y$ with standard deviations $\\sigma_X$ and $\\sigma_Y$ is\n$$\\frac{E(XY)-E(X)E(Y)}{\\sigma_X\\sigma_Y}$$\nSo, this is zero if and only if $E(XY) = E(X)E(Y)$.\nConsider rolling a die twice. Let $A$ and $B$ be the result in each case. The make two new random variables $X=A+B$ and $Y=A-B$. Then $E(XY) = E((A+B)(A-B)) = E(A^2-B^2) = E(A^2) - E(B^2)$. Since $A$ and $B$ are identically distributed, we see that $E(XY)=0$. Also, it is easy to see that $E(Y)=0$. So, the two random variables $X$ and $Y$ have correlation zero. However, they are clearly dependent\n\nSo we have shown that correlation zero does not imply independence, although independence zero DOES imply zero correlation." ]

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[ "January 2022 enrollments are closed and the course commences on 8th of January, 2022. For future cohorts\n\nFAIR-RATIONS Solution\n\n/*\n* Author: Arpit Bhayani\n* https://arpitbhayani.me\n*/\n#include <cmath>\n#include <cstdio>\n#include <cstdlib>\n#include <climits>\n#include <deque>\n#include <iostream>\n#include <list>\n#include <limits>\n#include <map>\n#include <queue>\n#include <set>\n#include <stack>\n#include <vector>\n\n#define ll long long\n\n#define MIN(a, b) a < b ? a : b\n#define MAX(a, b) a > b ? a : b\n\nusing namespace std;\n\nint i = 0;\nchar ch;\nwhile((ch = getchar()) != '\\n') {\nstr[i++] = ch;\n}\nstr[i] = '\\0';\nreturn i;\n}\n\nint arr;\nint main(int argc, char *argv[]) {\nint n;\nscanf(\"%d\", &n);\n\nfor(int i = 0 ; i < n ; i++) {\nscanf(\"%d\", &arr[i]);\n}\n\nint sum = 0;\nfor(int i = 0 ; i < n ;i++) {\nsum += arr[i];\n}\n\nif(sum & 1) {\nprintf(\"NO\\n\");\n}\nelse {\nint count = 0;\nfor(int i = 0 ; i < (n-1) ; i++) {\nif(arr[i] & 1) {\narr[i]++;\narr[i+1]++;\ncount += 2;\n}\n}\ncout << count << endl;\n}\n\nreturn 0;\n}" ]

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[ "# Module Bigarray.Genarray\n\n`module Genarray: `sig` .. `end``\n\n`type `('a, 'b, 'c)` t `\n\nThe type `Genarray.t` is the type of Bigarrays with variable numbers of dimensions. Any number of dimensions between 0 and 16 is supported.\n\nThe three type parameters to `Genarray.t` identify the array element kind and layout, as follows:\n\n• the first parameter, `'a`, is the OCaml type for accessing array elements (`float`, `int`, `int32`, `int64`, `nativeint`);\n• the second parameter, `'b`, is the actual kind of array elements (`float32_elt`, `float64_elt`, `int8_signed_elt`, `int8_unsigned_elt`, etc);\n• the third parameter, `'c`, identifies the array layout (`c_layout` or `fortran_layout`).\n\nFor instance, `(float, float32_elt, fortran_layout) Genarray.t` is the type of generic Bigarrays containing 32-bit floats in Fortran layout; reads and writes in this array use the OCaml type `float`.\n\n`val create : `('a, 'b) Bigarray.kind -> 'c Bigarray.layout -> int array -> ('a, 'b, 'c) t``\n\n`Genarray.create kind layout dimensions` returns a new Bigarray whose element kind is determined by the parameter `kind` (one of `float32`, `float64`, `int8_signed`, etc) and whose layout is determined by the parameter `layout` (one of `c_layout` or `fortran_layout`). The `dimensions` parameter is an array of integers that indicate the size of the Bigarray in each dimension. The length of `dimensions` determines the number of dimensions of the Bigarray.\n\nFor instance, `Genarray.create int32 c_layout [|4;6;8|]` returns a fresh Bigarray of 32-bit integers, in C layout, having three dimensions, the three dimensions being 4, 6 and 8 respectively.\n\nBigarrays returned by `Genarray.create` are not initialized: the initial values of array elements is unspecified.\n\n`Genarray.create` raises `Invalid_argument` if the number of dimensions is not in the range 0 to 16 inclusive, or if one of the dimensions is negative.\n\n`val init : `('a, 'b) Bigarray.kind -> 'c Bigarray.layout -> int array -> (int array -> 'a) -> ('a, 'b, 'c) t``\n\n`Genarray.init kind layout dimensions f` returns a new Bigarray `b` whose element kind is determined by the parameter `kind` (one of `float32`, `float64`, `int8_signed`, etc) and whose layout is determined by the parameter `layout` (one of `c_layout` or `fortran_layout`). The `dimensions` parameter is an array of integers that indicate the size of the Bigarray in each dimension. The length of `dimensions` determines the number of dimensions of the Bigarray.\n\nEach element `Genarray.get b i` is initialized to the result of `f i`. In other words, `Genarray.init kind layout dimensions f` tabulates the results of `f` applied to the indices of a new Bigarray whose layout is described by `kind`, `layout` and `dimensions`. The index array `i` may be shared and mutated between calls to f.\n\nFor instance, ```Genarray.init int c_layout [|2; 1; 3|]       (Array.fold_left (+) 0)``` returns a fresh Bigarray of integers, in C layout, having three dimensions (2, 1, 3, respectively), with the element values 0, 1, 2, 1, 2, 3.\n\n`Genarray.init` raises `Invalid_argument` if the number of dimensions is not in the range 0 to 16 inclusive, or if one of the dimensions is negative.\n\n• Since 4.12.0\n`val num_dims : `('a, 'b, 'c) t -> int``\n\nReturn the number of dimensions of the given Bigarray.\n\n`val dims : `('a, 'b, 'c) t -> int array``\n\n`Genarray.dims a` returns all dimensions of the Bigarray `a`, as an array of integers of length `Genarray.num_dims a`.\n\n`val nth_dim : `('a, 'b, 'c) t -> int -> int``\n\n`Genarray.nth_dim a n` returns the `n`-th dimension of the Bigarray `a`. The first dimension corresponds to `n = 0`; the second dimension corresponds to `n = 1`; the last dimension, to `n = Genarray.num_dims a - 1`.\n\n• Raises `Invalid_argument` if `n` is less than 0 or greater or equal than `Genarray.num_dims a`.\n`val kind : `('a, 'b, 'c) t -> ('a, 'b) Bigarray.kind``\n\nReturn the kind of the given Bigarray.\n\n`val layout : `('a, 'b, 'c) t -> 'c Bigarray.layout``\n\nReturn the layout of the given Bigarray.\n\n`val change_layout : `('a, 'b, 'c) t -> 'd Bigarray.layout -> ('a, 'b, 'd) t``\n\n`Genarray.change_layout a layout` returns a Bigarray with the specified `layout`, sharing the data with `a` (and hence having the same dimensions as `a`). No copying of elements is involved: the new array and the original array share the same storage space. The dimensions are reversed, such that `get v [| a; b |]` in C layout becomes `get v [| b+1; a+1 |]` in Fortran layout.\n\n• Since 4.04.0\n`val size_in_bytes : `('a, 'b, 'c) t -> int``\n\n`size_in_bytes a` is the number of elements in `a` multiplied by `a`'s `Bigarray.kind_size_in_bytes`.\n\n• Since 4.03.0\n`val get : `('a, 'b, 'c) t -> int array -> 'a``\n\nRead an element of a generic Bigarray. `Genarray.get a [|i1; ...; iN|]` returns the element of `a` whose coordinates are `i1` in the first dimension, `i2` in the second dimension, ..., `iN` in the `N`-th dimension.\n\nIf `a` has C layout, the coordinates must be greater or equal than 0 and strictly less than the corresponding dimensions of `a`. If `a` has Fortran layout, the coordinates must be greater or equal than 1 and less or equal than the corresponding dimensions of `a`.\n\nIf `N > 3`, alternate syntax is provided: you can write `a.{i1, i2, ..., iN}` instead of `Genarray.get a [|i1; ...; iN|]`. (The syntax `a.{...}` with one, two or three coordinates is reserved for accessing one-, two- and three-dimensional arrays as described below.)\n\n• Raises `Invalid_argument` if the array `a` does not have exactly `N` dimensions, or if the coordinates are outside the array bounds.\n`val set : `('a, 'b, 'c) t -> int array -> 'a -> unit``\n\nAssign an element of a generic Bigarray. `Genarray.set a [|i1; ...; iN|] v` stores the value `v` in the element of `a` whose coordinates are `i1` in the first dimension, `i2` in the second dimension, ..., `iN` in the `N`-th dimension.\n\nThe array `a` must have exactly `N` dimensions, and all coordinates must lie inside the array bounds, as described for `Genarray.get`; otherwise, `Invalid_argument` is raised.\n\nIf `N > 3`, alternate syntax is provided: you can write `a.{i1, i2, ..., iN} <- v` instead of `Genarray.set a [|i1; ...; iN|] v`. (The syntax `a.{...} <- v` with one, two or three coordinates is reserved for updating one-, two- and three-dimensional arrays as described below.)\n\n`val sub_left : `('a, 'b, Bigarray.c_layout) t -> int -> int -> ('a, 'b, Bigarray.c_layout) t``\n\nExtract a sub-array of the given Bigarray by restricting the first (left-most) dimension. `Genarray.sub_left a ofs len` returns a Bigarray with the same number of dimensions as `a`, and the same dimensions as `a`, except the first dimension, which corresponds to the interval `[ofs ... ofs + len - 1]` of the first dimension of `a`. No copying of elements is involved: the sub-array and the original array share the same storage space. In other terms, the element at coordinates `[|i1; ...; iN|]` of the sub-array is identical to the element at coordinates `[|i1+ofs; ...; iN|]` of the original array `a`.\n\n`Genarray.sub_left` applies only to Bigarrays in C layout.\n\n• Raises `Invalid_argument` if `ofs` and `len` do not designate a valid sub-array of `a`, that is, if `ofs < 0`, or `len < 0`, or `ofs + len > Genarray.nth_dim a 0`.\n`val sub_right : `('a, 'b, Bigarray.fortran_layout) t -> int -> int -> ('a, 'b, Bigarray.fortran_layout) t``\n\nExtract a sub-array of the given Bigarray by restricting the last (right-most) dimension. `Genarray.sub_right a ofs len` returns a Bigarray with the same number of dimensions as `a`, and the same dimensions as `a`, except the last dimension, which corresponds to the interval `[ofs ... ofs + len - 1]` of the last dimension of `a`. No copying of elements is involved: the sub-array and the original array share the same storage space. In other terms, the element at coordinates `[|i1; ...; iN|]` of the sub-array is identical to the element at coordinates `[|i1; ...; iN+ofs|]` of the original array `a`.\n\n`Genarray.sub_right` applies only to Bigarrays in Fortran layout.\n\n• Raises `Invalid_argument` if `ofs` and `len` do not designate a valid sub-array of `a`, that is, if `ofs < 1`, or `len < 0`, or `ofs + len > Genarray.nth_dim a (Genarray.num_dims a - 1)`.\n`val slice_left : `('a, 'b, Bigarray.c_layout) t -> int array -> ('a, 'b, Bigarray.c_layout) t``\n\nExtract a sub-array of lower dimension from the given Bigarray by fixing one or several of the first (left-most) coordinates. `Genarray.slice_left a [|i1; ... ; iM|]` returns the 'slice' of `a` obtained by setting the first `M` coordinates to `i1`, ..., `iM`. If `a` has `N` dimensions, the slice has dimension `N - M`, and the element at coordinates `[|j1; ...; j(N-M)|]` in the slice is identical to the element at coordinates `[|i1; ...; iM; j1; ...; j(N-M)|]` in the original array `a`. No copying of elements is involved: the slice and the original array share the same storage space.\n\n`Genarray.slice_left` applies only to Bigarrays in C layout.\n\n• Raises `Invalid_argument` if `M >= N`, or if `[|i1; ... ; iM|]` is outside the bounds of `a`.\n`val slice_right : `('a, 'b, Bigarray.fortran_layout) t -> int array -> ('a, 'b, Bigarray.fortran_layout) t``\n\nExtract a sub-array of lower dimension from the given Bigarray by fixing one or several of the last (right-most) coordinates. `Genarray.slice_right a [|i1; ... ; iM|]` returns the 'slice' of `a` obtained by setting the last `M` coordinates to `i1`, ..., `iM`. If `a` has `N` dimensions, the slice has dimension `N - M`, and the element at coordinates `[|j1; ...; j(N-M)|]` in the slice is identical to the element at coordinates `[|j1; ...; j(N-M); i1; ...; iM|]` in the original array `a`. No copying of elements is involved: the slice and the original array share the same storage space.\n\n`Genarray.slice_right` applies only to Bigarrays in Fortran layout.\n\n• Raises `Invalid_argument` if `M >= N`, or if `[|i1; ... ; iM|]` is outside the bounds of `a`.\n`val blit : `('a, 'b, 'c) t -> ('a, 'b, 'c) t -> unit``\n\nCopy all elements of a Bigarray in another Bigarray. `Genarray.blit src dst` copies all elements of `src` into `dst`. Both arrays `src` and `dst` must have the same number of dimensions and equal dimensions. Copying a sub-array of `src` to a sub-array of `dst` can be achieved by applying `Genarray.blit` to sub-array or slices of `src` and `dst`.\n\n`val fill : `('a, 'b, 'c) t -> 'a -> unit``\n\nSet all elements of a Bigarray to a given value. `Genarray.fill a v` stores the value `v` in all elements of the Bigarray `a`. Setting only some elements of `a` to `v` can be achieved by applying `Genarray.fill` to a sub-array or a slice of `a`." ]

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[ "", null, "• 393K Members\n• 3,245 Online\n• 424K Conversations\n\nNew Contributor\n\n# I need a (complicated?) formula\n\nHi All,\n\nI have been using Excel for a few years but am new to formulas (aside from the basic addition, subtraction, multiplication, etc.).  I have a worksheet which contains dollar amounts in one column and dates in another column.  The column that has the dates is completed only when the policy is issued so it has many empty rows.  I need a formula that will determine which rows in the date column (column E) contain a date and take the corresponding dollar amount from the premium column (column C) and add the totals for written business.  I have tried using several \"IF\" functions and \"LOOKUP\" functions but I am missing something because none of the formulas are pulling just the premiums that correlate to the dates in the written column.  Can anyone please tell me what I am doing wrong.  Thanks so much!\n\n2 Replies\n\n# Re: I need a (complicated?) formula\n\nLinda,\n\nsounds like a job for SUMIF() or SUMIFS().\n\n# RE: I need a (complicated?) formula\n\nThanks. I will keep trying!\nRelated Conversations" ]

[ null, "https://www.facebook.com/tr", null ]

[ "Create random combinations in Excel\n\nImagine that you need to create random combinations of letters in Excel. In this example I want to create a 3 letters random combinations list. I want the result to be like this:\n\n TEX JYY QCX CDH NTW\n\nTo get this kind of combinations I used the following formula:\n\n=CHAR(RANDBETWEEN(65,90))&CHAR(RANDBETWEEN(65,90))&CHAR(RANDBETWEEN(65,90))\n\nI’ve used the RANDBETWEEN() function to generate values between 65 and 90 because they are the ANSI codes of letters A to Z. The RANDBETWEEM() function has the following arguments:\n\nRANDBETWEEN(bottom,top)\n\nbottom is the smallest integer that the function will return and top is the highest.\n\nAfter the generation of the random number, I use CHAR() function to return the corresponding character from the ANSI table of characters. This function has the following syntax:\n\nCHAR(number)\n\nnumber is a number between 1 and 255 that specifies which character we want to return.\n\nOn my formula, combining 3 times the formula CHAR(RANDBETWEEN(65,90) I get a combination of 3 letters.\n\nIf I wanted to get a random list of numbers between 100 and 1000, I could use the following formula:\n\n=RANDBETWEEN(100,1000)\n\nThis will give me a list of number like the one below:\n\n 941 486 970 952 376\n\n1 comentários:\n\nAnonymous said...\n\nThat's not a combination. It's a permutation." ]

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[ "Home | Menu | Get Involved | Contact webmaster", null, "", null, "", null, "", null, "", null, "0 / 12\n\n# Number 14281\n\nfourteen thousand two hundred eighty one\n\n### Properties of the number 14281\n\n Factorization 14281 Divisors 1, 14281 Count of divisors 2 Sum of divisors 14282 Previous integer 14280 Next integer 14282 Is prime? YES (1676th prime) Previous prime 14251 Next prime 14293 14281st prime 155087 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 11011111001001 Octal 33711 Duodecimal 8321 Hexadecimal 37c9 Square 203946961 Square root 119.50313803411 Natural logarithm 9.5666852614529 Decimal logarithm 4.1547586191542 Sine -0.62895102873931 Cosine 0.77744491988035 Tangent -0.8089975413771\nNumber 14281 is pronounced fourteen thousand two hundred eighty one. Number 14281 is a prime number. The prime number before 14281 is 14251. The prime number after 14281 is 14293. Number 14281 has 2 divisors: 1, 14281. Sum of the divisors is 14282. Number 14281 is not a Fibonacci number. It is not a Bell number. Number 14281 is not a Catalan number. Number 14281 is not a regular number (Hamming number). It is a not factorial of any number. Number 14281 is a deficient number and therefore is not a perfect number. Binary numeral for number 14281 is 11011111001001. Octal numeral is 33711. Duodecimal value is 8321. Hexadecimal representation is 37c9. Square of the number 14281 is 203946961. Square root of the number 14281 is 119.50313803411. Natural logarithm of 14281 is 9.5666852614529 Decimal logarithm of the number 14281 is 4.1547586191542 Sine of 14281 is -0.62895102873931. Cosine of the number 14281 is 0.77744491988035. Tangent of the number 14281 is -0.8089975413771\n\n### Number properties\n\n0 / 12\nExamples: 3628800, 9876543211, 12586269025" ]

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[ "# Time-scale calculus\n\n(Redirected from Time scale calculus)\nJump to navigation Jump to search\n\nIn mathematics, time-scale calculus is a unification of the theory of difference equations with that of differential equations, unifying integral and differential calculus with the calculus of finite differences, offering a formalism for studying hybrid discrete–continuous dynamical systems. It has applications in any field that requires simultaneous modelling of discrete and continuous data. It gives a new definition of a derivative such that if one differentiates a function which acts on the real numbers then the definition is equivalent to standard differentiation, but if one uses a function acting on the integers then it is equivalent to the forward difference operator.\n\n## History\n\nTime-scale calculus was introduced in 1988 by the German mathematician Stefan Hilger. However, similar ideas have been used before and go back at least to the introduction of the Riemann–Stieltjes integral which unifies sums and integrals.\n\n## Dynamic equations\n\nMany results concerning differential equations carry over quite easily to corresponding results for difference equations, while other results seem to be completely different from their continuous counterparts. The study of dynamic equations on time scales reveals such discrepancies, and helps avoid proving results twice — once for differential equations and once again for difference equations. The general idea is to prove a result for a dynamic equation where the domain of the unknown function is a so-called time scale (also known as a time-set), which may be an arbitrary closed subset of the reals. In this way, results apply not only to the set of real numbers or set of integers but to more general time scales such as a Cantor set.\n\nThe three most popular examples of calculus on time scales are differential calculus, difference calculus, and quantum calculus. Dynamic equations on a time scale have a potential for applications, such as in population dynamics. For example, they can model insect populations that evolve continuously while in season, die out in winter while their eggs are incubating or dormant, and then hatch in a new season, giving rise to a non–overlapping population.\n\n## Formal definitions\n\nA time scale (or measure chain) is a closed subset of the real line $\\mathbb {R}$", null, ". The common notation for a general time scale is $\\mathbb {T}$", null, ".\n\nThe two most commonly encountered examples of time scales are the real numbers $\\mathbb {R}$", null, "and the discrete time scale $h\\mathbb {Z}$", null, ".\n\nA single point in a time scale is defined as:\n\n$t:t\\in \\mathbb {T}$", null, "### Operations on time scales", null, "The forward jump, backward jump, and graininess operators on a discrete time scale\n\nThe forward jump and backward jump operators represent the closest point in the time scale on the right and left of a given point $t$", null, ", respectively. Formally:\n\n$\\sigma (t)=\\inf\\{s\\in \\mathbb {T} :s>t\\}$", null, "(forward shift operator / forward jump operator)\n$\\rho (t)=\\sup\\{s\\in \\mathbb {T} :s", null, "(backward shift operator / backward jump operator)\n\nThe graininess $\\mu$", null, "is the distance from a point to the closest point on the right and is given by:\n\n$\\mu (t)=\\sigma (t)-t.$", null, "For a right-dense $t$", null, ", $\\sigma (t)=t$", null, "and $\\mu (t)=0$", null, ".\nFor a left-dense $t$", null, ", $\\rho (t)=t.$", null, "### Classification of points\n\nFor any $t\\in \\mathbb {T}$", null, ", $t$", null, "is:\n\n• left dense if $\\rho (t)=t$", null, "• right dense if $\\sigma (t)=t$", null, "• left scattered if $\\rho (t)", null, "• right scattered if $\\sigma (t)>t$", null, "• dense if both left dense and right dense\n• isolated if both left scattered and right scattered\n\nAs illustrated by the figure at right:\n\n• Point $t_{1}$", null, "is dense\n• Point $t_{2}$", null, "is left dense and right scattered\n• Point $t_{3}$", null, "is isolated\n• Point $t_{4}$", null, "is left scattered and right dense\n\n### Continuity\n\nContinuity of a time scale is redefined as equivalent to density. A time scale is said to be right-continuous at point $t$", null, "if it is right dense at point $t$", null, ". Similarly, a time scale is said to be left-continuous at point $t$", null, "if it is left dense at point $t$", null, ".\n\n## Derivative\n\nTake a function:\n\n$f:\\mathbb {T} \\rightarrow \\mathbb {R}$", null, ",\n\n(where ℝ could be any Banach space, but is set to the real line for simplicity).\n\nDefinition: The delta derivative (also Hilger derivative) $f^{\\Delta }(t)$", null, "exists if and only if:\n\nFor every $\\epsilon >0$", null, "there exists a neighborhood $U$", null, "of $t$", null, "such that:\n\n$|f(\\sigma (t))-f(s)-f^{\\Delta }(t)(\\sigma (t)-s)|\\leq \\varepsilon |\\sigma (t)-s|$", null, "for all $s$", null, "in $U$", null, ".\n\nTake $\\mathbb {T} =\\mathbb {R} .$", null, "Then $\\sigma (t)=t$", null, ", $\\mu (t)=0$", null, ", $f^{\\Delta }=f'$", null, "; is the derivative used in standard calculus. If $\\mathbb {T} =\\mathbb {Z}$", null, "(the integers), $\\sigma (t)=t+1$", null, ", $\\mu (t)=1$", null, ", $f^{\\Delta }=\\Delta f$", null, "is the forward difference operator used in difference equations.\n\n## Integration\n\nThe delta integral is defined as the antiderivative with respect to the delta derivative. If $F(t)$", null, "has a continuous derivative $f(t)=F^{\\Delta }(t)$", null, "one sets\n\n$\\int _{r}^{s}f(t)\\Delta (t)=F(s)-F(r).$", null, "## Laplace transform and z-transform\n\nA Laplace transform can be defined for functions on time scales, which uses the same table of transforms for any arbitrary time scale. This transform can be used to solve dynamic equations on time scales. If the time scale is the non-negative integers then the transform is equal to a modified Z-transform:\n\n${\\mathcal {Z}}'\\{x[z]\\}={\\frac {{\\mathcal {Z}}\\{x[z+1]\\}}{z+1}}$", null, "## Partial differentiation\n\nPartial differential equations and partial difference equations are unified as partial dynamic equations on time scales.\n\n## Multiple integration\n\nMultiple integration on time scales is treated in Bohner (2005).\n\n## Stochastic dynamic equations on time scales\n\nStochastic differential equations and stochastic difference equations can be generalized to stochastic dynamic equations on time scales.\n\n## Measure theory on time scales\n\nAssociated with every time scale is a natural measure defined via\n\n$\\mu ^{\\Delta }(A)=\\lambda (\\rho ^{-1}(A)),$", null, "where $\\lambda$", null, "denotes Lebesgue measure and $\\rho$", null, "is the backward shift operator defined on $\\mathbb {R}$", null, ". The delta integral turns out to be the usual Lebesgue–Stieltjes integral with respect to this measure\n\n$\\int _{r}^{s}f(t)\\Delta t=\\int _{[r,s)}f(t)d\\mu ^{\\Delta }(t)$", null, "and the delta derivative turns out to be the Radon–Nikodym derivative with respect to this measure\n\n$f^{\\Delta }(t)={\\frac {df}{d\\mu ^{\\Delta }}}(t).$", null, "## Distributions on time scales\n\nThe Dirac delta and Kronecker delta are unified on time scales as the Hilger delta:\n\n$\\delta _{a}^{\\mathbb {H} }(t)={\\begin{cases}{\\frac {1}{\\mu (a)}},&t=a\\\\0,&t\\neq a\\end{cases}}$", null, "## Integral equations on time scales\n\nIntegral equations and summation equations are unified as integral equations on time scales.\n\n## Fractional calculus on time scales\n\nFractional calculus on time scales is treated in Bastos, Mozyrska, and Torres." ]

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[ "#### Background knowledge\n\n• ##### 1.Basic concepts\n\nMotor is an energy conversion device that converts electric energy into mechanical energy or mechanical energy into electric energy, using magnetic field as media.\nPMDC motor is an energy conversion device that converts electric energy into mechanical energy, usingpermanent magnetic field as media provided by permenant magnets like ferrite magnets and neodymium magnets.\nEvery motor needs two basic conditions to function: magnetic field and current.\n\n• ##### 2.Classification of motor\n\nThere are many ways to classify the motors.\nThe motors Kinmore makes belong to brush type\nstrontium ferrite permanent magnet DC motor.", null, "• ##### 3.Basic theories\n\nResearch to the motors is based on the following five scientific laws. In order to have a preliminary acquaintance to motor principles, we need to known these laws first.", null, "##### (1) Law of electromagnetic induction (Faraday 1831)\n\nConductors (of finite dimensions) moving through a uniform magnetic field\nwill have currents induced within them.\nThe direction of the current is judged by right hand rule and follows the\nequation: E=B*L*V\nE: Electromotive force (Unit: V)\nB: Magnetic flux density of magnetic field (1 Tesla=104 Gauss)\nL: Effective length of conductor (Unit: m)\nV: Velocity of the conductor (Unit: m/s)\nSee figure 1 to the right, if we connect a lead wire to the conductor,induced\ncurrent will be generated.\n\n• ##### (2) Biot-Savart Law\n\nConductors with current within them will generate electro\nmagnetic force in a magnetic field. The direction is judged\nby left hand rule, (see figure 2) and follows the equation:\nF=B*I*L\nF: Electromagnetic force (Unit: N)\nI: Current in the inductor (Unit: A)\nB: Magnetic flux density of the magnetic field (Unit: Tesla)\nL: Effective length of the conductor (Unit: m)\nLeft hand rule is also called as motor rule.\nRight hand rule isalso called as generator rule.", null, "•", null, "##### (3) Kirchhoff's circuit laws (See figure 3)\n\nKCL ΣI=0: At any node (junction) in an electrical circuit, the sum of\ncurrents flowing into that node is equal to the sum of currents flowing\nout of that node\nKVL ΣU=0: The directed sum of the electrical potential differences\n(voltage) around any closed network is zero.\n\n• ##### (4) Law of conservation of energy\n\nThe total amount of energy in an isolated system\nremains constant over time.\n\n##### (5) Ampère's circuital law\n\nIn short, conductors with current within them generate magnetic\nfield around them. The direction of the magnetic filed is judged by\nright hand thrumb rule and follows the equation. (See figure 4)\n∮H×dL=∑I=IA+IB+IC+…\nH: magnetic field intensity (Unit: A/M)\nL: Length of conductor (Unit: M)\nI: Current (Unit: A)", null, "• ##### 4.Basic principles", null, "2-pole PMDC motor\n2-bar commutator\n2-conductors (1-loop coil) simple armature.\nAccording to Biot-Savart Law and left-hand rule,armature\nruns in CCW direction.\nIt is a simple but unpractical motor.(Figure 5)\n\n##### (1) Electric potential (Figure 6)\n\nFrom V=E+2△U+I*r we get E=V-2△U-I*r\nMeanwhile E=KE*Φ*n(armature back EMF)\nV: power supply voltage (Unit: V)\n2△U: brush voltage drop (Unit: V)\nI: armature current (Unit: A)\nR: rotor resistance (Unit: Ω)\nKE: EMF constant = Z/60 (for a 2-pole motor.\nZ: number of conductors)\nΦ: magnetic flux (Unit: Weber) = average magnetic\nflux density B * width of magnetic pole *effective\nlength of rotor\nN: speed (Unit: rpm)\n\n##### (2) Torque\n\nTE=KTΦ*I（electromagnetic torque: N.M）\nKT: torque constant = Z/2π\nΦ: magnetic flux (unit: Weber)\nI: armature current (unit: A)\n\n##### (3) Relationship between power and torque：\n\nP=T*n/97500 P: power（unit: W）\nT: torque (unit: g.cm)\nn: speed (unit: rpm)\nWhen the unit of T is “N?m”, P=T*n/9.55（unit: W）", null, "•", null, "##### (4) Energy equation(Figure 7)\n\nP1=2△U*I+I2r+PE\nPE=P2+PFe+Pmec\nPE: electromagnetic power P2: output power\nPmec: mechanical loss PFe: iron loss\nP2=P1-2△U*I-I2r-PFe-Pmec (unit: W)\nEfficiency: η=P2/P1*100%\nPFe+Pmec is also called no load power\nP0=PFe+Pmec\nPE=P2+P0 and TE=T2+T0\n\n• ##### (5) Energy transmission graph: (Figure 8)", null, "• ##### 6.Performance characteristic (Figure 9)\n\nn=f(T2) relationship between speed & torque.\nI=f(T2) relationship between current & output power\nη=f(T2) relationship between efficiency & torque\nP2=f(T2) relationship between output power & torque", null, "•", null, "##### (1) I=f(T2)\n\nI=TE/KT*Φ=(T0+T2)/KT*Φ=T0/KT*Φ+T2/KT*Φ=I0+[1/KT*Φ]*T2 (liner equation)\nI0: no load current Φ: constant\nAt stall, n=0, E=0, according to Figure 6, current Ist=(U-2△U)/r\n\n##### (2) n=f(T2)\n\nE=V-2△U-I*r=KEΦ*n\nn=(V-2△U-I*r)/KE*Φ={U-2△U-[(I0+T2)/KT*Φ]*r}/KE*Φ\n=(U-2△U-I0*r)/KE*Φ-r/KE*KT*Φ2*T2\n= n0-[r/KE*KT*Φ2]*T2（equation of lines）\n\n##### (3) P2=f(T2)\n\nP2=T2*n/9.55=[n0-(V/KE*KT*Φ2)*T2]/9.55=[n0*T2-(r/KE*KT*Φ2)*(T2)2]/9.55\n\nP2 is a second-degree parabola (Figure 10)\n\n• (5) Energy transmission graph: (Figure 8)\n(Equation iscomplicated thus is omitted here.)", null, "##### (1) Turns of coil and magnet wire diameter (other parameters remain unchanged)\n\nWe know from 5.1 that the potential constant KE increases when the turns of coil increase. Motor speed n is therefore lowered. On the contrary, when the turns of coil decrease, the motor speed increases. When the diameter of the magnet wire increases, the rotor resistance r reduces. Back EMF of the rotor increases (E=V-2△U-I*r). The motor speed n therefore increases. On the contrary, when the diameter of the magnet wire decreases, the motor speed n decreases. The current at stall is in inverse proportion to the resistance r.Turns of the coil and diameter of the magnet wire restrict each other under the space limit of the lamination slot. We should clearly understand such relationship when we try to adjust the motor parameters.\n\n##### (2) Magnetic flux (other parameters remain unchanged)\n\nMagnets with higher magnetic flux density and longer lamination sheets will both increase the magnetic flux Φ. From 5.1 and 6.2 we know that speed n decreases. At the same time, load (T2) has less influence over speed n. The characteristic of the motor is thus called hard. On the contrary, if we use magnets with lower magnetic flux density and shorter lamination sheets, the characteristic of the motor is called soft.\n\n##### (3) Air gap\n\nSee figure 12, the magnetization curve of the air gap\nΦδ=-μ0*(Sδ/δ)*Fδ\nΦδ: Air gap flux\nSδ: Air gap area\nΔ: Air gap length\nFδ: Air gap magnetomotive force(magnetic EMF)\nPermeance angle: α=tg-1[μ0*(Sδ/δ)].\nWe can see that when δ is longer, α is smaller, air gap flux Φδ is smaller. Motor speed will increase if the other parameters remain unchanged. On the contrary, when δ is shorter, α is larger, air gap flux Φδ is larger. Motor speed will decrease. We will see the same result as we see in 7.2. We usually pursue the maximum possible value of (Φδ*Fδ) in motor design.\n\n##### (4) Effective volume D2*L\n\nMotor torque is proportional to D2*L.\n[D: diameter of the rotor L: length of the rotor]\nMotor power is proportional to D2*L *n." ]

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[ "Case Based Reasoning\n\n# Case-Based Reasoning\n\nCase-Based Reasoning uses the first two principles of Instance Based Learning:\n\n1. Lazy Learning\n2. Classify new instances based on those similar to them\n\nBut not the third\n3. CBR doesn't represent instances as real-valued points in an n-dimensional Euclidean space.\n\nIn CBR, instances are typically represented using more rich symbolic descriptions, and the methods used to retrieve similar instances are correspondingly more elaborate.\n\nFor instance, we might represent an instance as the following (learner to create new mechanical designs):", null, "As such they're more 'cases' than instances - quite different from each other, but with the same purpose.\n\n# CBR Cycle\n\n• RETRIEVE the most similar case or cases\n• REUSE the information and knowledge useful to solve the problem from the case\n• REVISE the proposed solution\n• RETAIN the parts of this experience likely to be useful for future problem solving\n\n# Similarity\n\nThe similarity metric different for Cases is quite different than the Euclidean distance for instances, due to their rich symbolic nature.\n\nFor instance we can use a match and mismatch function (counting the number of variables that match or don't as a proportion) to come up with something like:\n\n(1)\n\\begin{align} score(Q,C) = \\frac{match(Q,C) - mismatch(Q,C)}{|C|} \\end{align}" ]

[ null, "http://wiki.bethanycrane.com/local--files/case-based-reasoning/Screen%20Shot%202012-04-15%20at%2010.32.20%20PM.png", null ]

[ "# What sort of operations can be applied on a Hilbert spaces?\n\nI was reading the paper No Universal Flipper for Quantum States. In this paper they have tried to prove by contradiction that a universal flipping machine cannot exist. By flipping I mean if I have a qubit $\\alpha|0 \\rangle +\\beta|1\\rangle$, then the flipped qubit will be $\\beta^*|0 \\rangle -\\alpha^*|1\\rangle$. In the paper they take an entangled system between two parties Alice and Bob, and apply an operation on Bob's side. But they say that the operation is not necessarily unitary ( its just linear ) but it does preserve the trace of Alice's reduced density operator before and after the operation.\n\nBut I thought on closed quantum mechanical systems only unitary operations could be applied. Am I missing something?\n\n• Evolution operators have to be unitary because you want the scalar product to be conserved by evolution. This is not true for e.g. observables, that are surely admitted in QM as operators (they are the \"building blocks\" of the theory), but are self-adjoint and not unitary. – yuggib Mar 25 '15 at 9:13\n• by observables do yo mean measurement operators ? – sashas Mar 25 '15 at 9:40\n• I mean e.g. energy, momentum, angular momentum, spin operators. Experimentally, when you perform a measurement of one of these quantities and obtain a result, you are projecting on a spectral subspace of those operators, and that again is not a unitary operation. – yuggib Mar 25 '15 at 9:49\n• the example of flipping you gave is implemented by an antiunitary operator, so my feeling is that somehow this is related to the question whether antilinear operator should be considered or not. Unfortunately I'm not an expert of quantum computing/information so I don't really know the answer to this last question. – Phoenix87 Mar 25 '15 at 14:51\n\nCorrect, the evolution of a quantum system does not necessarily need to be unitary. However, if the system is closed, it needs to be unitary. The latter is due to the fact that we assume the system evolves due to the Schrödinger equation with a self-adjoint Hamiltonian, which induces a unitary evolution operator via Stone's theorem.\n\nFor non-closed systems, the idea is the following: Let's suppose you have your system (denote it by a state $\\rho$) and maybe some other system, the environment (denoted by a state $\\rho_E$). Then you start out with the state $\\rho\\otimes \\rho_E$ and you can perform any unitary operation you want, i.e. $\\tilde{\\rho}:=U(\\rho\\otimes \\rho_E)U^*$. Now so far, we have dealt with what you learn in your first quantum theory course. The idea is that in our case, we are simply not interested in (or can't keep track of) the evolution of the environment. Hence the state we have at the end is the reduced density matrix of $\\tilde{\\rho}$ restricted to our system, i.e:\n\n$$\\rho_{final}=\\operatorname{tr}_E(U(\\rho\\otimes \\rho_E)U^*)$$ with the partial trace over the environment. Now it turns out that this is (more or less) the most general form of what is called a completely positive map and hence we usually say that those are the maps governing quantum mechanics in closed as well as open systems. In order to keep the normalization of a state, these need to be trace-preserving, hence we say that a general quantum evolution is governed by a completely positive trace-preserving map (CPTP) or quantum channel.\n\nNote that the only point to argue is about $\\rho\\otimes \\rho_E$: It is not so clear that we should have to start with an uncorrelated systems, but there are good reasons to do so.\n\nNow, maybe you don't even know how the environment looks like. These systems are called \"open quantum systems\" and one way to describe their evolution is by as master equation as in statistical equation (for quantum mechanics, the corresponding type is also often known as Lindblad-equation). However, the corresponding map on density matrices is also a quantum channel and thus of the form outlined above.\n\nAnother characterization of quantum channels might be helpful for your understanding: They can be axiomatized (in finite dimensions for simplicity) via:\n\n• A quantum channel is a linear map from matrices to matrices (we need linearity physically; see below).\n• It is a positive map, i.e. it sends positive semidefinite matrices to positive semidefinite matrices (states are positive semidefinite and our map better send states to states).\n• Moreover, if $T$ is the map, then also $T\\otimes \\operatorname{id}_n$, where $\\operatorname{id}_n$ is the identity is also a positive map for all $n$ (this is what is meant by \"complete\" positivity: If our map acts just on a subsystem, the natural extension to the whole system should definitely be a valid map).\n• It preserves the trace of the matrices.\n\nExcept for maybe the third property, it should be immediately clear that the other properties should be true for any quantum evolution starting of any system - and these should also be the properties the authors use for their proof.\n\nLinearity: Finally, let's talk about antiunitary evolutions: It turns out that a map as proposed is not linear (it's antilinear), so we have to exclude antiunitary maps, too, because they would violate the linearity of quantum mechanics. This linearity however is established by many experiments and lies at the heart of quantum theory - there is no good reason why we should abandon it." ]

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[ "# Convert 7 ounces to metric tablespoons (7 oz to metric tblsp conversion)\n\n## 7 ounces is equal to how many metric tablespoons?\n\n7 ounces is equal to 13.79 metric tablespoons.\n\n## How many metric tablespoons are in 7 ounces?\n\nThere are 13.79 metric tablespoons in 7 ounces.\n\n• 7 ounces = 13.79 metric tablespoons\n• 7.1 ounces = 13.987 metric tablespoons\n• 7.2 ounces = 14.184 metric tablespoons\n• 7.3 ounces = 14.381 metric tablespoons\n• 7.4 ounces = 14.578 metric tablespoons\n• 7.5 ounces = 14.775 metric tablespoons\n• 7.6 ounces = 14.972 metric tablespoons\n• 7.7 ounces = 15.169 metric tablespoons\n• 7.8 ounces = 15.366 metric tablespoons\n• 7.9 ounces = 15.563 metric tablespoons\n\n## How to convert 7 ounces to metric tablespoons?\n\nTo convert 7 ounces to metric tablespoons, multiply the value in ounces by 1.97.\n\n### What is the formula to convert 7 ounces to metric tablespoons?\n\nThe conversion formula to convert 7 ounces to metric tablespoons is :\nmetric tablespoons = ounces × 1.97\n\n### What is the conversion factor to convert 7 ounces to metric tablespoons?\n\nThe conversion factor to convert 7 ounces to metric tablespoons is 1.97\n\n### Examples to convert oz to metric tblsp\n\n#### Example 1\n\nConvert 7.2 oz to metric tblsp.\n\nSolution:\nConverting from ounces to metric tablespoons is very easy.\nWe know that 1 oz = 1.97 metric tblsp.\n\nSo, to convert 7.2 oz to metric tblsp, multiply 7.2 oz by 1.97 metric tblsp.\n\n7.2 oz = 7.2 × 1.97 metric tblsp\n7.2 oz = 14.184 metric tblsp\n\nTherefore, 7.2 ounces converted to metric tablespoons is equal to 14.184 metric tblsp.\n\n#### Example 2\n\nConvert 7.8 oz to metric tblsp.\n\nSolution:\n1 oz = 1.97 metric tblsp\n\nSo, 7.8 oz = 7.8 × 1.97 metric tblsp\n7.8 oz = 15.366 metric tblsp\n\nTherefore, 7.8 oz converted to metric tblsp is equal to 15.366 metric tblsp.\n\nIf you don't want to do the calculation from 7 ounces to metric tablespoons manually, you can simply use our 7 ounces to metric tablespoons calculator.\n\n### How to use the 7 ounces to metric tablespoons converter?\n\nTo use the 7 ounces to metric tablespoons converter, follow these steps:\n\n1. Enter the value in ounces that you want to convert.\n2. Click \"Convert\".\n3. To copy the conversion steps, click \"Copy\".\n4. To report an incorrect conversion, click \"Report incorrect conversion\".\n5. To reset the converter, click \"Reset\".\n6. To convert metric tablespoons to ounces, click \"Swap\".\n7. To convert between other units of weight, select the units from the drop-down menus.\n\n## Ounces to metric tablespoons conversion table\n\nThe ounces to metric tablespoons conversion chart below shows a list of various ounces values converted to metric tablespoons\n\nOunces (oz) Metric tablespoons (metric tblsp)\n7 oz 13.79 metric tblsp\n7.05 oz 13.8885 metric tblsp\n7.1 oz 13.987 metric tblsp\n7.15 oz 14.0855 metric tblsp\n7.2 oz 14.184 metric tblsp\n7.25 oz 14.2825 metric tblsp\n7.3 oz 14.381 metric tblsp\n7.35 oz 14.4795 metric tblsp\n7.4 oz 14.578 metric tblsp\n7.45 oz 14.6765 metric tblsp\n7.5 oz 14.775 metric tblsp\n7.55 oz 14.8735 metric tblsp\n7.6 oz 14.972 metric tblsp\n7.65 oz 15.0705 metric tblsp\n7.7 oz 15.169 metric tblsp\n7.75 oz 15.2675 metric tblsp\n7.8 oz 15.366 metric tblsp\n7.85 oz 15.4645 metric tblsp\n7.9 oz 15.563 metric tblsp\n7.95 oz 15.6615 metric tblsp\n\n## Ounces (oz)\n\n### What is an ounce?\n\nAn ounce is a unit of weight in the US customary system and imperial system of measurement that is equal to one-sixteenth of a pound.\n\n### What is the symbol/abbreviation of the ounce?\n\nThe symbol/abbreviation of the ounce is oz.\n\n### What is the history/origin of the ounce?\n\nThe ounce, derived from the Latin word \"uncia\" meaning one twelfth, traces its origins to ancient Rome where it denoted 1/12th of a Roman pound (libra). The British later adopted and refined this system. The avoirdupois ounce, now widely used, was standardized in the UK in the 14th century and is prevalent in English-speaking countries.\n\n### What is the ounce used for?\n\nThe ounce is mainly used to measure the weight of smaller items or quantities of substances, like for example food products, medicine, postal items, jewelries, precious metals and many more.\n\n### How to measure weight in ounces?\n\nTo measure weight in ounces, you can use a scale (e.g kitchen scales, postal scales or jewelry scales) that is calibrated in ounces. Place the item you want to weigh on the scale and it will show you the weight in ounces.\n\n### 7 ounces equivalents in other weight units\n\n• 7 ounces = 992.25 carats\n• 7 ounces = 3062.5 grains\n• 7 ounces = 198.45 grams\n• 7 ounces = 0.1984465 kilograms\n• 7 ounces = 0.003906252 long hundredweights (UK)\n• 7 ounces = 0.0001953126 long tons (UK)\n• 7 ounces = 0.0001984465 metric tons (or tonnes)\n• 7 ounces = 198446661.91 micrograms\n• 7 ounces = 198446.5 milligrams\n• 7 ounces = 127.61 pennyweights\n• 7 ounces = 0.4375 pounds\n• 7 ounces = 0.004375 short hundredweights (US)\n• 7 ounces = 0.00021875 short tons (US)\n• 7 ounces = 0.03125003 stones\n• 7 ounces = 6.380206 troy ounces\n• 7 ounces = 0.5316843 troy pounds\n\n## Metric tablespoons (metric tblsp)\n\n### What is a metric tablespoon?\n\nA metric tablespoon is a unit of volume measurement in the metric system and is equal to 15 milliliters (ml).\n\n### What is the symbol/abbreviation of the metric tablespoon?\n\nThe symbol/abbreviation of the metric tablespoon is tbsp or Tbsp.\n\n### What is the metric tablespoon used for?\n\nThe metric tablespoon is primarily used for measuring ingredients in cooking and baking, especially for small quantities of liquids or dry ingredients like spices. It is commonly used in recipes in countries that have adopted the metric system.\n\n### How to measure volume in metric tablespoons?\n\nTo measure volume in metric tablespoons, follow these steps:\n\n• Select the correct measuring spoon. Metric measuring spoons are typically available in 1 tbsp and 1/2 tbsp sizes. They are labeled in milliliters (ml). One metric tablespoon is equal to 15 ml and half metric tablespoon is equal to 7.5 ml.\n• Fill the spoon to the brim. If you are measuring a dry ingredient, fluff it up with a fork before measuring to ensure accuracy.\n• Level off the spoon with a knife or other straight edge. This will remove any excess ingredient and ensure that you are measuring the correct volume.\n\nAlternatively, you can use a graduated measuring cup that has tablespoon markings. Simply pour the liquid or dry ingredient into the cup until it reaches the desired tablespoon marking.\n\n### 7 metric tablespoons to ounces conversion (metric tblsp to oz)\n\n• 7 metric tablespoons = 3.550295 ounces\n• 7.1 metric tablespoons = 3.6010135 ounces\n• 7.2 metric tablespoons = 3.651732 ounces\n• 7.3 metric tablespoons = 3.7024505 ounces\n• 7.4 metric tablespoons = 3.753169 ounces\n• 7.5 metric tablespoons = 3.8038875 ounces\n• 7.6 metric tablespoons = 3.854606 ounces\n• 7.7 metric tablespoons = 3.9053245 ounces\n• 7.8 metric tablespoons = 3.956043 ounces\n• 7.9 metric tablespoons = 4.0067615 ounces\n\n### 7 metric tablespoons equivalents in other volume units\n\n• 7 metric tablespoons = 0.4619356 canadian cups\n• 7 metric tablespoons = 10.5 centiliters\n• 7 metric tablespoons = 105 cubic centimeters\n• 7 metric tablespoons = 0.00370804 cubic feet\n• 7 metric tablespoons = 6.407492 cubic inches\n• 7 metric tablespoons = 0.000105 cubic meters\n• 7 metric tablespoons = 105000 cubic millimeters\n• 7 metric tablespoons = 0.0001373351 cubic yards\n• 7 metric tablespoons = 0.000105 kiloliters\n• 7 metric tablespoons = 0.105 liters\n• 7 metric tablespoons = 0.42 metric cups\n• 7 metric tablespoons = 21 metric teaspoons\n• 7 metric tablespoons = 105000 microliters\n• 7 metric tablespoons = 105 milliliters\n• 7 metric tablespoons = 0.0006415773 UK barrels\n• 7 metric tablespoons = 3.695482 UK fluid ounces\n• 7 metric tablespoons = 0.02309678 UK gallons\n• 7 metric tablespoons = 0.739095 UK gills\n• 7 metric tablespoons = 0.1847741 UK pints\n• 7 metric tablespoons = 0.0923874 UK quarts\n• 7 metric tablespoons = 7.42 UK tablespoons\n• 7 metric tablespoons = 29.54 UK teaspoons\n• 7 metric tablespoons = 0.000908089 US barrels (dry)\n• 7 metric tablespoons = 0.00089474 US barrels (federal)\n• 7 metric tablespoons = 0.0008806 US barrels (liquid)\n• 7 metric tablespoons = 0.0006604304 US barrels (oil)\n• 7 metric tablespoons = 0.4438091 US cups\n• 7 metric tablespoons = 3.55047 US fluid ounces\n• 7 metric tablespoons = 0.02383717 US gallons (dry)\n• 7 metric tablespoons = 0.02773806 US gallons (liquid)\n• 7 metric tablespoons = 0.887621 US gills\n• 7 metric tablespoons = 0.1906975 US pints (dry)\n• 7 metric tablespoons = 0.2219042 US pints (liquid)\n• 7 metric tablespoons = 0.0953484 US quarts (dry)\n• 7 metric tablespoons = 0.1109521 US quarts (liquid)\n• 7 metric tablespoons = 7.07 US tablespoons\n• 7 metric tablespoons = 21.28 US teaspoons" ]

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[ "# Definition of a Minimal Set\n\nA few times while studying math I have encountered the notion of a \"minimal set\". For example, given some set of subsets, what is the \"minimal\" sigma algebra generated by that set of subsets? Or, in the example I am currently studying: one of the ZFC axioms ensures the existence of an inductive set. Therefore, we define the natural numbers to be the minimal inductive set. I am trying to think of some other examples, but I can't remember off the top of my head. Feel free to comment if you think of some other examples of minimal sets.\n\nHere is my question. I have seen the \"minimal\" set defined different ways, and I am not sure which statement is definition and which is an implication of the definition.\n\nThe first definition I have seen is to call a set $M$ satisfying some property minimal if, for every other set $A$ satisfying the same property as the set $M$, we have $M \\subseteq A$: \\begin{equation*} M \\textrm{ is the minimal set with property } P \\iff (\\forall A \\textrm{ satisfying property P }) M \\subseteq A \\end{equation*}\n\nThe second definition I have seen is to define the minimal set $M$ satisfying some property to be the intersection of all other sets satisfying the same property: \\begin{equation*} M \\textrm{ is the minimal set in some class } C \\iff M=\\bigcap C \\end{equation*}\n\nThe second definition seems more concise, however in the example I am studying right now (defining the set of natural numbers to be the minimal inductive set), I don't know if such a set $C$ (the set of all inductive sets) exists, so I am not even sure if the right hand side of definition 2 even makes any formal logical sense.\n\n• The two are equivalent (assuming closure under intersections), so you can treat either as the definition as you prefer. – Qudit Mar 15 '15 at 20:45\n• Thanks for your comment. I will try to prove that these two are equivalent. Typically, however, which statement is usually treated as the primary definition? – Mathemanic Mar 15 '15 at 20:47\n• The first definition is generalizable to orders other than $\\subseteq$, so it's better. – Git Gud Mar 15 '15 at 20:47\n• @EthanAlvaree I'm not sure. I've definitely seen both as well. – Qudit Mar 15 '15 at 20:49\n• @GitGud, I'm not sure what you mean by generalizable to orders other than $\\subseteq$. Can you please explain more? – Mathemanic Mar 15 '15 at 20:50\n\nThe first definition is the more general one (and, as has been said in the comment, can be generalized to arbitrary partial orders besides $\\subseteq$). The second definition is not always correct. However, if the class $C$ is nonempty has the property that an arbitrary intersection of members of $C$ is again in $C$, then the second definition is equivalent to the first definition.\nRegarding your final problem about a minimal inductive set, note that $C$ need only be a nonempty class, not necessarily a set. In case you worry that $\\bigcap$ is only definied for sets, not for (proper) classes of sets: No, the definition $$\\tag1\\bigcap C:=\\left\\{\\,x\\mid \\forall c\\in C\\colon x\\in c\\,\\right\\}$$ is perfectly fine and defines a set for any nonempty(!) class $C$ though admittedly $(1)$ uses class builder, not set builder notation. But let $S\\in C$ be an arbitrary set and define $$\\tag2\\bigcap C:=\\left\\{\\,x\\in S\\mid \\forall c\\in C\\colon x\\in c\\,\\right\\},$$ then the result does not depend on the choice of $S$ (why?) and as $(2)$ is an instance of the Axiom Schema of Comprehension, this shows that $\\bigcap C$ is a set. (Then finally, as $C$ is closed under arbitrary intersection, we see that $\\bigcap C$ is again an element of $C$ and surely the mnimal element in the sense of the first definition; If $C$ denotes the class of inductoive sets, then the usual formulation of the Axiom of Infnity can be rephrased as simply: $C$ is not empty - which is precisely what we need)\n• Yes, I was worried about $\\bigcap C$ being defined in set theory when $C$ is not a set. I think I understand, but please tell me if this is right. So $\\bigcap C$ in both definitions $(1)$ and $(2)$ are equal to each other? But definition $(2)$ is preferred because it represents how we actually use the ZFC axioms to ensure existence of the minimal inductive set, correct? For instance, your set $S$ is the set whose existence is provided by the axiom of infinity, correct? I think I understand, but please let me know if I got this right. Thanks for your great answer! – Mathemanic Mar 15 '15 at 21:15\nGiven a set $S$ of sets you can consider the containment relation between sets. This relation is what is called a partial order, and makes $S$ into a poset. A minimal element $s\\in S$ is an element that is not properly contained by another element in $S$. Of course when $S$ is closed under arbitrary intersections the minimal element of $S$ is the intersection of all the elements of $S$." ]

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[ "## Markets Risks\n\n#### Topical Issues on Risks in Financial Markets.", null, "Last updated on 13 September 2016\n\nThe method for calculating Vega risk charge and Delta risk charge is exactly same i.e. in the first step, risk weighted sensitivity is calculated for each risk factor. Next, risk weighted sensitivities are aggregated within a bucket to calculate risk position and finally risk positions are aggregated across buckets. However, there are differences in the calculation of individual components like risk sensitivity and correlations.\n\nStep 1 – Calculate net sensitivity", null, "$s_k$ for each risk factor across all the portfolios on a trading desk. It is important to note that sensitivity in this case is not simply vega as most trading and risk professionals know it. Vega risk sensitivity is a multiplication of an option vega and the implied option volatility. So portfolio level vega risk sensitivity to a given risk factor is the sum of  option level vega risk sensitivities to that risk factor. Mathematically,", null, "$s_k =\\sum_i \\nu_{ik} * \\sigma_{ik}$ where i represents the position and k represents the risk factor.\n\nIn case of an interest rate portfolio,", null, "$k$ has two dimensions – option maturity and the residual maturity of the underlying", null, "$\\nu_k$ is required to be mapped to these maturity tenors along both the dimensions – 0.5y, 1y, 3y, 5y and 10y. Most likely,", null, "$k$ will have to be mapped to one of these tenors using linear or an internally approved interpolation method. For rest of the risk classes(FX, Equity etc.)", null, "$k$ will be defined in only one dimension – option maturity which are mapped to the same tenors as for an interest rate portfolio.\n\nAs mentioned in my previous post, I will cover sensitivity definitions for all the risk classes separately.\n\nStep 2 – Calculate risk weighted sensitivity", null, "$WS_k$ which is the multiplication of net sensitivity", null, "$s_k$ and the corresponding risk weight", null, "$RW_k$. Mathematically, it is represented as", null, "$WS_k = RW_ks_k$\n\nVega risk weights are the function of risk class and its liquidity horizon which are capped at 100%. Following equation to be used to calculate vega risk weight of each risk class:", null, "$RW_k = min[RW_\\sigma*\\sqrt{\\frac{LH_{Risk Class}}{10}};1]$\n\nHere,", null, "$RW_\\sigma$ is set at 55% and", null, "$LH_{Risk Class}$ is the regulatory liquidity horizon as per the table below.\n\n Risk Class", null, "$LH_{Risk Class}$", null, "$RW_{Risk Class}$ GIRR 60 100% FX 40 100% Commodity 120 100% Equity-Large Cap 20 78% Equity-Small Cap 60 100% CSR (all risk class variants) 120 100%\n\nStep 3 – Risk position,", null, "$s_k$, for each bucket b (definition of bucket is different for each risk class) is calculated by aggregating risk weighted sensitivities within each bucket. Aggregation requires use of prescribed correlations", null, "$\\rho_{kl}$. Given below formula to be used for the aggregation within each bucket.", null, "$K_b=\\sqrt{max(\\sum_kWS^2_k + \\sum_k\\sum_{k\\neq l}\\rho_{kl}WS_kWS_l,0)}$\n\nThis equation can easily be implemented in Excel using MMULT function or perhaps by implementing the logic in VBA. However the critical part of this equation is the correlation matrix.\n\n### Same Bucket Correlations for GIRR", null, "$\\rho_{kl}$", null, "$\\rho_{kl}$ between", null, "$WS_k$ and", null, "$WS_l$ within same bucket (eg. USD or EUR/USD) and same underlying curve (eg. USD Libor) with different tenors is set at", null, "$\\rho_{kl} = min[\\rho_{kl}^{(option maturity)} * \\rho_{kl}^{(underlying maturity)};1]$\n\n•", null, "$\\rho_{kl}^{(option maturity)} = e^{-\\alpha.\\frac{|T_k-T_l|}{min(T_k;T_l)}}$ Here", null, "$T_k$ refers to the option maturity in years related to the weighted vega sensitivity", null, "$WS_k$; and", null, "$\\alpha$ is set at 1%;\n•", null, "$\\rho_{kl}^{(underlying maturity)} = e^{-\\alpha.\\frac{|T_k^U-T_l^U|}{min(T_k^U;T_l^U)}}$ Here", null, "$T_k^U$ refers to the maturity of the underlying in years related to the weighted vega sensitivity", null, "$WS_k$; and", null, "$\\alpha$ is set at 1%.\n\nExample of a partial correlation matrix", null, "$\\rho_{kl}$ is provided in a picture below. This shows a correlation between a 1y1y weighted vega risk sensitivity and 1y10y weighted vega risk sensitivity is 91%.", null, "### Same Bucket Correlations for Other Risk Classes", null, "$\\rho_{kl}$\n\nFor other risk classes", null, "$\\rho_{kl}$ is set at", null, "$\\rho_{kl} = min[\\rho_{kl}^{(Delta)} * \\rho_{kl}^{(option maturity)};1]$ where\n\n•", null, "$\\rho_{kl}^{(option maturity)} = e^{-\\alpha.\\frac{|T_k-T_l|}{min(T_k;T_l)}}$ Here", null, "$T_k$ refers to the option maturity in years related to the weighted vega sensitivity", null, "$WS_k$; and", null, "$\\alpha$ is set at 1%. Refer to sub-correlation matrix inside a red box in the above picture for", null, "$\\rho_{kl}^{(option maturity)}$ for other risk classes;\n•", null, "$\\rho_{kl}^{(Delta)}$ is the correlation that is applicable to equivalent Delta risk factors. For example, $latex \\rho_{kl}^{(Delta)} for the equity options in Advanced Economy Industrials stocks (Bucket 6) is 25%. Step 4 – Calculate Vega risk charge by aggregating vega risk positions across all the buckets. Following formula to be used for this aggregation", null, "$Vega Risk Charge = \\sqrt{\\sum_bK^2_b + \\sum_b\\sum_{c \\neq b}\\gamma_{bc}S_bS_c}$", null, "$S_b = \\sum_kWS_k$ for all the risk factors in a bucket b", null, "$S_b = max(min(\\sum_kWS_k,K_b),-K_b)$ when a number under square root is negative using first approximation of", null, "$S_b$", null, "$\\gamma_{bc}$ = 50% for aggregating interest rate vega risk positions between buckets. This correlation is different for each risk class. ### Example for an FX Portfolio Pictures below show steps of calculating a vega risk charge for a hypothetical FX portfolio (first picture). For the simplicity it is assumed that vegas are perfectly matched to the prescribed option maturity tenors.", null, "Second picture shows at-the-money implied volatilities at the pre-specified tenors. The smile is ignored in these calculations. Note that these volatilities are hypothetical and do not represent current or any other market condition.", null, "Picture below has steps to calculate", null, "$K_b$ and", null, "$S_b$.", null, "Next a formula in Step 4 above is used to calculate vega risk charge. For FX risk class,", null, "$\\gamma_{bc}$ is 60% which is same as was used for calculating delta risk charge. This formula can be implemented in Excel using MMULT resulting in a vega risk charge of$5,980,059 which is approximately $6 million for the aggregate portfolio vega of$210k.\n\nNext I will present curvature risk charge.\n\n1.", null, "#### Carme\n\nHi Anshu,\nDo you happen to have an excel example of how you calculate the correlation values in the matrix and the risk position for each bucket when interest rate derivatives?\nBesides, I understand this works for swaptions as we have both maturities as inputs for valuation. But i don´t see this for caps (caplets) – do you maybe have qn example for both?\nMany thanks in advanced and congrats for the contents, it´s been really helpful so far. Best,\nCarme\n\n•", null, "Carme, sorry I didn’t see your comment earlier. Thanks for dropping by. Correlation calcs are relatively straight and as per the formula provided in the standard. I don’t have example for swaptions and caps but I can prepare one. For caps, I would assume that correlation matrix can be created for 1m, 3m and 6m. Anyways there is not much liquidity in most currencies for maturities other than 3m.\n\n2.", null, "#### Laurent\n\nHi Anshu,\n\nThanks for the different articles on FRTB, which I found very useful.\nI would also be interested in any Excel file you have created through those articles. I would like to try to get the implementation right and to play a bit with the inputs to see how the results change. Is there any chance you could send them over?\n\nKind regards,\nLaurent\n\n•", null, "•", null, "" ]

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[ "", null, "Browse", null, "Search", null, "Login", null, "", null, "Home : SROUND", null, "", null, "", null, "Q11285 - INFO: SROUND\nSROUND A Spread formula that allows you to scientifically round values.\n\nSYNTAX:\n\nSROUND(Cell Reference/Value, Decimal Places)\n\nCell Reference/Value: The specifies the value to round.\n\nDecimal Places:  Number indicating how many places to the right of the decimal are included in the rounding.\n\nNOTES:\n\nScientific Rounding:  When the fractional part is exactly 0.5, the function always rounds it to the nearest even number.  For example, 0.5 rounds to 0, and 1.5 rounds to 2.  Normal rounding rounds digits 1,2,3, and 4 down.  Rounds digits 5,6,7,8, and 9 up.\n\nYou can only display up to 10 significant digits using this function.\n\nIn Spread Reports if you want to use normal rounding, simply set the decimal places by using the Increase/Decrease decimal buttons or use Format, Cell.\n\nSee Scientific vs Normal rounding for a table showing the key differences between the rounding techinques.\n\nSee SROUNDZ for rounding of values with qualifiers (i.e. <2).\n\nEXAMPLES:\n\n=SROUND(1.2250,2) returns 1.22\n\n=SROUND(G8,1) returns 1.6 (where the cell G8 = 1.65)\n\n=SROUND(1.45,1) returns 1.4\n\n=sround(GAVG(1,1,\"M\"),1) 3.1 (Where the GAVG Function returned 3.106)\n\n=SROUNDZ(\"<2.5\", 0) <2\n\n=SROUNDZ(GAVGZ(1,1,\"M\",1, 2),1) <3.1 (where GAVGZ returns \"<3.11\")", null, "Related Articles No Related Articles Available.\n\n Article Attachments No Attachments Available.", null, "", null, "", null, "Created on 6/2/2008 12:10 PM. Last Modified on 8/13/2018 11:01 AM. Last Modified by Scott Dorner. Article has been viewed 3458 times. Rated 0 out of 10 based on 0 votes.", null, "Print Article", null, "Email Article" ]

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[ "# wheels¶\n\nFunctions and dsp model to model the mechanic of the wheels.\n\nFunctions\n\n calculate_r_dynamic Calculates the dynamic radius of the wheels [m]. calculate_r_wheels Calculates the radius of the wheels [m] from the tyre dimensions. calculate_tyre_dimensions Calculates the tyre dimensions from the tyre code. calculate_wheel_power Calculates the wheel power [kW]. calculate_wheel_powers Calculates power at the wheels [kW]. calculate_wheel_speeds Calculates rotating speed of the wheels [RPM]. calculate_wheel_torques Calculates torque at the wheels [N*m]. default_tyre_code Return one of the most popular tyre code according to the r dynamic. define_tyre_code Returns the tyre code from the tyre dimensions. identify_r_dynamic Identifies the dynamic radius of the wheels [m]. identify_r_dynamic_v1 Identifies the dynamic radius of the wheels [m]. identify_r_dynamic_v2 Identifies the dynamic radius of the wheels [m]. identify_r_dynamic_v3 Identifies the dynamic radius of the wheels [m]. identify_r_dynamic_v4 Identifies the dynamic radius of the wheels [m]. identify_tyre_dynamic_rolling_coefficient Identifies the dynamic rolling coefficient [-].\n\nDispatchers\n\n dsp It models the wheel dynamics." ]

[ null ]

[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: Simplify problems for checking easy equalities...\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg53860] Re: Simplify problems for checking easy equalities...\n• From: \"Drago Ganic\" <drago.ganic at in2.hr>\n• Date: Tue, 1 Feb 2005 04:08:10 -0500 (EST)\n• References: <cti5nd\\$8n2\\[email protected]>\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```Hi Cyrus,\nuse assumptions! The simplification qou expect is not generally valid.\nSimplify or Refine are enough for the elementary functions!\n\nFullSimplify[Log[x^n] - n Log[x], n \\[Element] Integers && x > 0]\n0\n\nSimplify[Log[x^n] - n*Log[x], n \\[Element] Integers && x > 0]\n0\n\nRefine[Log[x^n] - n*Log[x], n \\[Element] Integers && x > 0]\n0\n\nThose assumptions are used implicity in the unary function PowerExpand:\nLog[x^n] - n*Log[x] // PowerExpand\n0\n\nGreetings,\nDrago\n\"Cyrus Erik Eierud\" <cyruserik at tele2.se> wrote in message\nnews:cti5nd\\$8n2\\$1 at smc.vnet.net...\n>\n> My problem is that I can not simplify what to me seems as a very\n> simple equality task. This is what I want Mathematica to return zero\n> for:\n>\n> in:= FullSimplify[Log[x^n] - n*Log[x]]\n>\n> out= -n Log[x] + Log[x^n])\n>\n> I have used Simplify to check equalities, but the one above (and many\n> other equations similar to the one above) just don't simplify. Am I\n> doing anything wrong or does anyone know of a better way to check\n> equalities?\n>\n> Appreciate any help,\n> Cyrus Eierud, Student\n> cyruserik at tele2.se\n>\n\n```\n\n• Prev by Date: Re: Simplify problems for checking easy equalities...\n• Next by Date: New Web Site for Mathematica Users using WikiMedia\n• Previous by thread: Re: Simplify problems for checking easy equalities...\n• Next by thread: Re: Simplify problems for checking easy equalities..." ]

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[ "## Factorials\n\n.\n\n#### Definition\n\nFor any natural number", null, "", null, "This means that the factorial of any natural number (also called the counting numbers or positive integers; meaning all integers from 1 to infinity) is equal to the product of all of the integers from 1 up to that number.\n\nExamples                Explanation", null, "", null, "", null, "", null, "Calculating Factorials\n\nThe factorial operation is only defined for positive integers and 0. 0! is defined to be equal to 1. This means that 0! never needs to be calculated. The factorial of will never appear when you are simply calculating the factorial of a larger number, but it may appear when working more complicated problems.\n\nIt can be easier to calculate factorials of large numbers if you know the factorial of a smaller number.\n\nExamples                Explanation", null, "", null, "The example immediately above was solved using the value of 5! which was calculated in an earlier example.\n\nBecause any number multiplied by 1 does not change in value, this step can be skipped when calculating factorials.\n\nExamples                Explanation", null, "", null, "#### Try these exercises:\n\nInstructions.\n\n1.", null, "2.", null, "3.", null, "4.", null, "5.", null, "6. Find , and express the difference as a factorial.", null, "7.", null, "8.", null, "9. Why doesn’t the factorial of a negative number make sense?\n10.", null, "1.", null, "2.", null, "3.", null, "4.", null, "5.", null, "6.", null, "7.", null, "8.", null, "9. The factorial of a number is calculated by multiplying integers from 1 up to the number. A similar process for negative numbers would require the operands to get smaller instead of larger.\n\n10.", null, "" ]

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[ "", null, "", null, "", null, "Practice Problems for  the Lander Math Contest\nModule 7, Problems (19-21)\n\n19) What is the probability of not getting 3 in rolling a fair die\nfive times?\n\n20) A basketball player has 65% chance to throw the ball into\nthe basket. What is the probability of at most 6 successful\nthrows out of 7?\n\n21) In a bag with 12 coins 5 are normal, 2 are two-sided\nheads and the remaining coins are two-sided tails. What is\nthe probability that a coin taken at random from the bag and\nbeing tossed 4 times, turns out to be exactly 3 heads and one\ntail?\n\n22) For additional practice, try some or all of the Practice\nThis Concept problems in the upper right hand corner green", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "" ]

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[ "``````pragma solidity ^0.4.25;\n\n/**\n* Math operations with safety checks\n*/\nlibrary SafeMath {\nfunction mul(uint a, uint b) internal pure returns (uint) {\nuint c = a * b;\nassert(a == 0 || c / a == b);\nreturn c;\n}\n\nfunction div(uint a, uint b) internal pure returns (uint) {\nassert(b > 0);\nuint c = a / b;\nassert(a == b * c + a % b);\nreturn c;\n}\n\nfunction sub(uint a, uint b) internal pure returns (uint) {\nassert(b <= a);\nreturn a - b;\n}\n\nfunction add(uint a, uint b) internal pure returns (uint) {\nuint c = a + b;\nassert(c >= a);\nreturn c;\n}\n\nfunction max64(uint64 a, uint64 b) internal pure returns (uint64) {\nreturn a >= b ? a : b;\n}\n\nfunction min64(uint64 a, uint64 b) internal pure returns (uint64) {\nreturn a < b ? a : b;\n}\n\nfunction max256(uint256 a, uint256 b) internal pure returns (uint256) {\nreturn a >= b ? a : b;\n}\n\nfunction min256(uint256 a, uint256 b) internal pure returns (uint256) {\nreturn a < b ? a : b;\n}\n}\n\ncontract ERC20Protocol {\n/* This is a slight change to the ERC20 base standard.\nfunction totalSupply() constant returns (uint supply);\nis replaced with:\nuint public totalSupply;\nThis automatically creates a getter function for the totalSupply.\nThis is moved to the base contract since public getter functions are not\ncurrently recognised as an implementation of the matching abstract\nfunction by the compiler.\n*/\n/// total amount of tokens\nuint public totalSupply;\n\n/// @param _owner The address from which the balance will be retrieved\n/// @return The balance\nfunction balanceOf(address _owner) public constant returns (uint balance);\n\n/// @notice send `_value` token to `_to` from `msg.sender`\n/// @param _to The address of the recipient\n/// @param _value The amount of token to be transferred\n/// @return Whether the transfer was successful or not\nfunction transfer(address _to, uint _value) public returns (bool success);\n\n/// @notice send `_value` token to `_to` from `_from` on the condition it is approved by `_from`\n/// @param _from The address of the sender\n/// @param _to The address of the recipient\n/// @param _value The amount of token to be transferred\n/// @return Whether the transfer was successful or not\n\n///if you want to use privacy transaction,you need to implement this function in your contract\n/// @notice send `_value` token to `_to` from `msg.sender`\n/// @param _to The address of the recipient\n/// @param _toKey the ota pubkey\n/// @param _value The amount of token to be transferred\n/// @return Whether the transfer was successful or not\nfunction otatransfer(address _to, bytes _toKey, uint256 _value) public returns (string);\n\n///check privacy transaction\n/// @param _owner The address from which the ota balance will be retrieved\n/// @return The balance\nfunction otabalanceOf(address _owner) public constant returns (uint256 balance);\n\n/// @notice `msg.sender` approves `_spender` to spend `_value` tokens\n/// @param _spender The address of the account able to transfer the tokens\n/// @param _value The amount of tokens to be approved for transfer\n/// @return Whether the approval was successful or not\nfunction approve(address _spender, uint _value) public returns (bool success);\n\n/// @param _owner The address of the account owning tokens\n/// @param _spender The address of the account able to transfer the tokens\n/// @return Amount of remaining tokens allowed to spent\n\n}\n\n//the contract implements ERC20Protocol interface with privacy transaction\ncontract StandardToken is ERC20Protocol {\n\nusing SafeMath for uint;\nstring public constant name = \"WanToken-Beta\";\nstring public constant symbol = \"WanToken\";\nuint public constant decimals = 18;\n\nfunction transfer(address _to, uint _value) public returns (bool success) {\n\nif (balances[msg.sender] >= _value) {\nbalances[msg.sender] -= _value;\nbalances[_to] += _value;\nTransfer(msg.sender, _to, _value);\nreturn true;\n} else { return false; }\n}\n\nfunction transferFrom(address _from, address _to, uint _value) public returns (bool success) {\n\nif (balances[_from] >= _value && allowed[_from][msg.sender] >= _value) {\nbalances[_to] += _value;\nbalances[_from] -= _value;\nallowed[_from][msg.sender] -= _value;\nTransfer(_from, _to, _value);\nreturn true;\n} else { return false; }\n}\n\nfunction balanceOf(address _owner) public constant returns (uint balance) {\nreturn balances[_owner];\n}\n\nfunction approve(address _spender, uint _value) public returns (bool success) {\n\nassert((_value == 0) || (allowed[msg.sender][_spender] == 0));\n\nallowed[msg.sender][_spender] = _value;\nApproval(msg.sender, _spender, _value);\nreturn true;\n}\n\nreturn allowed[_owner][_spender];\n}\n\n// privacy balance, bytes for public key\nmapping (address => uint256) public privacyBalance;\nmapping (address => bytes) public otaKey;\n\n//this only for initialize, only for test to mint token to one wan address\nfunction initPrivacyAsset(address initialBase, bytes baseKeyBytes, uint256 value) public {\nprivacyBalance[initialBase] = value;\notaKey[initialBase] = baseKeyBytes;\n}\n\n// return string just for debug\nfunction otatransfer(address _to, bytes _toKey, uint256 _value) public returns (string) {\nif(privacyBalance[msg.sender] < _value) return \"sender token too low\";\n\nprivacyBalance[msg.sender] -= _value;\nprivacyBalance[_to] += _value;\notaKey[_to] = _toKey;\nreturn \"success\";\n}\n\n//check privacy balance\nfunction otabalanceOf(address _owner) public view returns (uint256 balance) {\nreturn privacyBalance[_owner];\n}\n}\n``````" ]

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[ "# Introduction⌗\n\nIn this post I plan to give an introduction to reversible programming languages, and some introduction to the programming language Rust. Then I will introduce my amalgamation of those two things, RRust. The overview I will give here is not going to be too deep, In the end, I will provide some articles which go deeper into it all. RRust was produced for my Master’s Thesis, so the recounting here goes into some details which may already be known to Rust users. Finally, any comments or questions are welcome and can be sent to [email protected].\n\n# Reversible Computing⌗\n\n## Introduction⌗\n\nReversible computing is the idea that you can for a given function `f(x) := y` you can generate the inverse function that takes the output and generates the input `f^-1(y) := x`. This of course could be done by hand, but we want to make a machine, here donoted by I, that can make the transformation of the code. That is that `I(f) := f^-1`. To generalize this a bit we can look at `x` and `y` as states and `f` as a function that goes from state `x` to state `y`, this is how we will look at it in the following chapters since we are going to model our functions as mutations to variables. So we want to generate the inverse function `f^-1` that takes state y and mutates it to state x.\n\n## Which Functions Can Be Reversible?⌗\n\nA good question to ask here is if all functions can be reversible or if there are some limitations here.\n\nSo the first thing we can single out here is that we need the functions to be deterministic, that is that for any given input the function will always produce the same output, but we have to go a bit further than that here. We also need backward determinism, that is that any output can only be produced from a single input, which can also be seen as that the inverse function is deterministic.\n\nFrom this, we can see that a given function will have to be bijective.", null, "Bijective function from the set X to the set Y\n\nThere is a small caveat to this, we cannot always know before a computation if it will succeed, so we need to include a third possibility, in the form of an `error`. So any function is reversible if it is bijective or results in an `error`.\n\n## Historical Background⌗\n\nReversible computing has long been a topic that has been worked on to some degree for a couple of reasons. Early [Bennet] argued that a computer only had to dissipate heat when erasing information, since you have to keep all information to keep a given computation reversible this was given as an example of a future, more efficient system. Even if it had to do more actual computations to compute the same information.\n\nMore recently reversible programming languages have shown up in the field of quantum computers, again because of the need to keep all information at any given point in a computation. Though they are often a lot lower level than the languages I will present here, in theory these languages would be able to run on a powerful enough quantum computer.\n\n## Reversible Programming Languages⌗\n\nThe reversible languages we will look at in this post are not the ones used for quantum computers, we will look at some that are higher level.\n\n### Janus⌗\n\nThe first one I have here is Janus, it is made for general invertibility and I will go further into how it works in a later section.\n\n### Reversible Erlang⌗\n\nReversible Erlang is a subset of Erlang that introduces the concept of rollbacks. By using that the subset is reversible it is possible to rollback to a checkpoint by running in reverse until you hit it. Lanese2018.\n\n### Hermes⌗\n\nHermes is a small reversible programming language constructed to be used to implement encryption algorithms. It uses the reversibility and type system to defend against certain types of side-channel attacks. MOGENSEN2022102746\n\n## Janus⌗\n\nJanus is a reversible programming language first described at Caltech around 1982 and formalized in 2007. [Lutz86, Yokoyama2007]\n\nThe main reversible language I will focus on is Janus, this is because I am using the general idea of that to implement the reversibility in my language.\n\n### Example⌗\n\nWe start with a small example of Janus, just to give a bit of taste of the syntax:\n\n``````procedure copy(int arr[], int payload[])\nlocal int i = 0 // Create a local variable\nfrom i = 0 loop // Loop from i == 0\narr[i] += payload[i] // Add the value\ni += 1 // Increment counter\nuntil i = 2048 // until i == 2048\ndelocal i = 2048 // Uninitialize local variable\n``````\n\n### How does Janus ensure reversibility?⌗\n\nJanus uses specialized structures to ensure reversibility the first we will look at the constructs it uses for control flow. Janus has two types of control-flow: Conditionals and for-until loops. Furthermore, it uses the syntax to ensure reversibility, so we will look at that as well.\n\n#### Conditionals⌗\n\nTo try to understand how conditionals work we will look at a small example and try to explain it. So we want to look at the following example:\n\n``````if c₁ then\ns₁\nelse\ns₂\nfi c₂\n``````\n\nSo what does this mean, we can look at it as `c₁` as a pre-condition and `c₂` as a post-condition. So if `c₁` is true then `s₁` is run and then `c₂` must be true, and conversely, if `c₁` is false then `s₂` is run and `c₂` must be false afterward. Another way we can look at it is by looking at a flow-chart:\n\nIn the flow-chart representation, we can see what we described above. Here we can also see how the inverse is going to look, we can swap `c₁` with `c₂`, and then get the inverse of `s₁` and `s₂`.\n\n#### Loops⌗\n\nThe loops work similarly to the conditionals, though the conditions are slightly different compared to them. We again look at an example:\n\n``````from c₁\ndo s₁\nloop s₂\nuntil c₂\n``````\n\nThis construct will probably seem a bit more alien to programmers, it is not clear what `do` and `loop` does in this context. To reverse the loop we swap `c₁` and `c₂` then we take the inverse of `s₁` and `s₂`:\n\n``````from c₂\ndo I[s₁]\nloop I[s₂]\nuntil c₁\n``````\n\nNow we have an overview of how both directions will look, so we can try and follow the flow of it. For the forward we start with `c₁` being true, and then it must be false every time following that until `c₂` is true and the loop is terminated. The reason for this is that when we reverse the loop it will terminate too early if `c₁` is ever true before what is the end of it when going forward.\n\nWe can also try and understand it by looking at a flow-chart as before:\n\nThe reason for the last part here is to ensure reversibility. To create the inverse version of this loop we swap `c₁` and `c₂`, then take the inverse of `s₁` and `s₂`.\n\nThen we can see that we start with `c₂` being true only once similarly to when it was the exit condition. And `c₁` is true only when the loop is finished.\n\n#### Syntax⌗\n\nThe Janus syntax is pretty barebones and not too complicated, so we will go over it here. It should be noted that the Janus version that we target has a few more features notable function arguments and locals, but we will come to them when we present RRust.\n\n``````\n# Syntax Domains:\np ∈ Progs[Janus]\nx ∈ Vars[Janus]\nc ∈ Cons[Janus]\nid ∈ Idens[Janus]\ns ∈ Stmts[Janus]\ne ∈ Exps[Janus]\n⨁ ∈ ModOps[Janus]\n⨀ ∈ Ops[Janus]\n\n# Grammar\np ::= d* (procedure id s)⁺\nd ::= x | x[c]\ns ::= x ⨁= e | x[e] ⨁= e |\nif e then s else s fi e |\nfrom e do s loop s until e |\ncall id | uncall id | skip | s s\ne ::= c | x | x[e] | e ⨀ e\nc ::= 0 | 1 | ... | 4294967295\n⨁ ::= + | - | ^\n⨀ ::= ⨁ | * | / | % | */ | & | '|' | && | \"||\" |\n< | > | = | != | <= | >=\n``````\n\nThe most interesting part of the syntax for me is the statements s, the first two forms it can take both cause mutation of a variable or an array. We can see that the only methods that are allowed to cause mutations are `⊕=`, which again expands into `+=`, `-=` and `^=`. These are the only operators that are allowed because they are the only ones that can be generally reversed. For expressions, all operators are allowed this is because they are not inversed when reversing the procedure. In s we also have `call` and `uncall` which allows us to both call a function and call the reverse of that function. Another interesting thing to note here is that we only allow the use of integers, this is because floating point numbers do not have reversible semantics.\n\n# Rust⌗\n\nI will not dive too deep into Rust or its syntax in this section as there already exists good walkthroughs of the language which cover it well. Though I will get into the main parts of the language that enables its safety and explain them briefly. If you are an intermediate user of Rust you can probably skip or skim this section.\n\n## What Is Rust?⌗\n\nThe official site explains rust with this short motto:\n\nA language empowering everyone to build reliable and efficient software.\n\nAlthough this does not say too much about the language. Rust is a programming language first designed by Graydon Hoare and later incubated at Mozilla Research to be used in their products. The main feature of the language is to be safer than other languages in the same class such as C and C++. Today it is used widely in loads of companies such as Amazon and Discord.\n\n## Why Rust?⌗\n\nSo why should a company use Rust instead of the tested and stable languages like C and C++? One of the major reasons to use Rust is that it can rule out a class of issues related to memory safety, which is the cause of a lot of exploits. Microsoft for example has said that around 70% of their vulnerabilities are some form of memory corruption msrc_proactive_2019. Another reason to move to Rust is that it is a more “modern” language compared to C and C++ it has a bunch of features from functional languages such as pattern matching. It also has a more strict type system which again allows one to catch issues earlier. Furthermore, it has an integrated package manager and build system that makes it easier to pull in dependencies.\n\n## How does Rust ensure memory-safety?⌗\n\nRust uses a few different constructs to ensure safety here we will go through the most prominent.\n\n### Ownership⌗\n\nOwnership is a concept that is central in Rust, it can be seen as a variable that every value has, and it follows a few rules:\n\n• Each value in Rust has a variable that’s called its /owner/.\n• There can be only one owner at a time.\n• When the owner goes out of scope, the value will be dropped. So what do these rules tell us? Firstly every value has some /owner/ which must be unique for every value, and when it leaves a scope it is dropped. Dropped here means that a possible deconstructor is run, and it is deallocated.\n\nLet us have a look at a small example of ownership.\n\n``````let v = vec![1, 2, 3];\n{ // start a new scope\nlet w = v; // move v into the scope\n} // w goes out of scope and is dropped\n\nv // Error\n``````\n\nSo in this example, we create a new value `v` which here is a Vec, it is a non-`copy`1 type, so ownership is transferred into the new scope. There it is assigned to `w` which means that `w` now is the owner of the value. It is then dropped after it goes out of scope. Attempting to compile this code will end out with a compiler error:\n\n``````error[E0382]: borrow of moved value: `v`\n--> src/main.rs:7:1\n|\n2 | let v = vec![1, 2, 3];\n| - move occurs because `v` has type `Vec<i32>`, which does not implement the `Copy` trait\n3 | { // start a new scope\n4 | let w = v; // move v into the scope\n| - value moved here\n...\n7 | v // Error\n| ^ value borrowed here after move\n``````\n\nIt is important that the error here happens at compile time as otherwise, this could cause a use-after-free memory corruption.\n\n### Borrowing⌗\n\nTo use the value of a non-`copy` type in a new scope without moving it you can use borrowing:\n\n``````let v = vec![1, 2, 3];\n{ // start a new scope\nlet w = &v; // borrow v in the scope\n} // drop the reference to v\n\nv // No error\n``````\n\nHere instead of moving the whole value into the new scope you only move a reference. A reference is similar to a pointer from other languages though it gives better guarantees at compile-time.\n\nThere are two types of borrows in Rust:\n\n• `&T` Immutable references\n• `&mut T` Mutable references There are a few rules related to those, immutable references may exist in any number as long as there are no mutable references. Mutable reference must be the only borrow of a value, no other borrows may exist at the same time.\n\nTo ensure that these rules are upheld Rust uses a system called the Borrow Checker. It will construct a set of regions called lifetimes in which a borrow is valid, it can then check if the rules are upheld there.\n\nFor example, if we try to use both a mutable and an immutable borrow at the same time we will run into an error:\n\n``````let mut v0 = Vec::new();\nv0.push(1);\nlet b0 = &v0;\nlet b1 = &mut v0;\nprintln!(\"{}, {}\", b0, b1);\n``````\n\nHere both `b0` and `b1` are used at the same time leading to the following error:\n\n``````error[E0502]: cannot borrow `v0` as mutable because it is also borrowed as immutable\n--> src/lib.rs:4:15\n|\n3 | let b0 = &v0;\n| -- immutable borrow occurs here\n4 | let b1 = &mut v0;\n| ^^ mutable borrow occurs here\n5 | println!(\"{}, {}\", b0, b1);\n| -- immutable borrow later used here\n``````\n\nIf either of `b0` or `b1` is removed from line 5 it will result in a successful compilation since the borrow checker can see that none of the borrowing rules are breached.\n\n### Unsafe⌗\n\nIn any case, where there is a possibility of breaking memory safety it is necessary to use the `unsafe` keyword which allows one to use functions and methods that are marked in that way. For example, it is unsafe to read from a raw pointer.\n\n### Bound checking⌗\n\nThe last thing I will highlight that Rust does to ensure safety is that array indexing is bound checked and will cause a run-time panic if it fails, thus aborting the program2.\n\n## Meta-Programming⌗\n\nThe last feature of Rust I will highlight here does not have much to do with safety, it instead concerns the powerful meta-programming Rust allows. Rust has two major types of meta-programming:\n\n### `macro_rules!`⌗\n\n• Can match on tokens and types of tokens.\n• Can produce code following that.\n• Is built into the compiler.\n• Needs to be hygenic\n\n### Procedural Macros⌗\n\n• Can do arbitrary transformations of a stream of tokens.\n• Works as a compiler plugin.\n• Often uses a library like `syn` to parse the tokens into a token tree.\n• Can be unhygienic\n\n# RRust⌗\n\nSo now we have looked at reversible computing and the language Janus. We also had a detour into Rust and a few features for how it ensures safety. Now I will give some reasoning as to why I made RRust and how it solves issues with Janus, then I will finally present RRust.\n\n## Issues With The Safety Of Janus⌗\n\nJanus has two modes of evaluation a safe interpreter that runs the code and a translation into C++. The translation into C++ gives us some issues when it comes to memory safety. If we for example have a function with the following signature:\n\n``````procedure copy(int arr[], int payload[])\n``````\n\nThe C++ will generate the following functions:\n\n``````void copy_forward(int *arr, int *payload);\nvoid copy_reverse(int *arr, int *payload);\n``````\n\nHere we see a few issues, C and C++ allow for mutable aliasing, so there is nothing to stop `arr` and `payload` from referencing the same memory, either in full or partially. In both cases, it could cause operations that damage the reversibility of the program. The other main issue is that it allows out-of-bound writing since it uses direct array access.\n\n## The Idea⌗\n\nThe idea I have is to use the safety features of Rust to ensure that it is memory-safe. Rust checks for out-of-bound access and mutable aliasing is not possible3.\n\nTo implement the language I will use the meta-programming features of Rust to transform the code into a reversible version.\n\n## How could a translation look?⌗\n\nWe will start by looking at the example from before:\n\n``````procedure copy(int arr[], int payload[])\nlocal int i = 0\nfrom i = 0 loop\ni += 1\nuntil i = 2048\ndelocal int i = 2048\n``````\n\nThis function copies by addition from payload into arr. With the Rust version, we want to make it similar to how Janus translates it. We use a zero-sized type as a kind of namespace for the functions instead of using a prefix:\n\n``````struct Copy; // Zero Sized Type\nimpl Copy {\nfn forward(arr: &mut [i32], payload: &mut [i32]) { /* forward code */ }\nfn backwards(arr: &mut [i32], payload: &mut [i32]) { /* backwards code */ }\n}\n``````\n\nTo have it only mutate through a mutable reference instead of taking ownership and returning a value simplifies the code. This also means that it is not possible to alias `arr` with `payload` since it would in that case not compile.\n\nI will also show how the `forward` and `backwards` code could look, it is very similar to how the same program would look in C++, The biggest difference is that arrays are bound checked4:\n\n``````struct Copy;\nimpl Copy {\nfn forward(arr: &mut [i32], payload: &mut [i32]) {\nlet mut i = 0;\nassert!(i == 0);\nwhile !(i == 2048) {\ni += 1;\nassert!(!(i == 0));\n}\nif i != 2048 {\npanic!(\"Delocal failed {} != {}\", i, 2048);\n}\ndrop(i);\n}\nfn backwards(arr: &mut [i32], payload: &mut [i32]) {\nlet mut i = 2048;\nassert!(i == 2048);\nwhile !(i == 0) {\ni -= 1;\nassert!(!(i == 2048));\n}\nif i != 0 {\npanic!(\"Delocal failed {} != {}\", i, 0);\n}\ndrop(i);\n}\n}\n``````\n\n## How would this program look in RRust?⌗\n\nNow we have a good idea about how the finished program should look, but the bigger question now is how it should look before being transformed into that. We can base the structure upon Janus with a flavor of Rust, so here I present the example from above in RRust:\n\n``````rfn!(Copy, (arr: &mut [i32], payload: &mut [i32]), {\nlet mut i = 0;\nrloop!(i == 0,\n{\ni += 1;\n},\ni == 2048);\ndelocal!(i, 2048);\n});\n``````\n\nWe use a series of macros to construct each part of it, `rfn!` creates the structure of the code, `rloop!` creates a reversible loop with a pre-condition and a post-condition and `delocal!` creates checks if the value is correct and makes it impossible to use the local value afterward.\n\nTo ensure that the program was reversible I made a secondary transpiler that compiled the Rust code into Janus which could then also be checked.\n\n## Defining a syntax⌗\n\nI created a syntax of the part of Rust I deemed to be compatible with Janus since I was translating to that as well:\n\n``````\nprogram ::= rfn!( name, (args), body);\nargs ::= arg | args, arg\narg ::= id: &mut type\nlist ::= [ scalar ]\nscalar ::= i8 | i16 | i32 | i64 | u8 | u16 | u32 | u64 | isize | usize\nnumber ::= scalar :: MIN | scalar :: MIN + 1 | ... |\nscalar :: MAX -1 | scalar :: MAX\nname ::= [A-Z|a-z][A-Z|a-z|0-9|_]*\nbody ::= { stmt }\nstmt ::= stmt ; | stmt ; stmt | { stmt } | * stmt | let id = expr |\nid ⊕= expr | l[expr] ⊕= expr |\nrif | rloop | delocal | name::forward(fargs) |\nname::backwards(fargs) | swap(id, id)\nexpr ::= number | *expr | l[expr] | expr ⊙ expr\nfargs ::= farg | fargs, farg\nfarg ::= expr | id\n⊕ ::= + | - | ^\n⊙ ::= ⊕ | * | / | & | '|'\n⊗ ::= == | >= | <= | < | > | !=\nrif ::= rif!(bool, body, bool) | rif!(bool, body, body, bool)\nrloop ::= rloop!(bool,body,bool) | rloop!(bool,body,body,bool)\ndelocal ::= delocal!(id, expr)\nbool ::= expr ⊗ expr | true | false\n``````\n\nHere the most important parts to observe are the two statements that allow mutation both uses the `⊕=` that we saw with Janus as well. As well as only allowing integer types for the same reason as Janus.\n\n## Transformation⌗\n\nSo to facilitate the transformation of the code we use a few different features of the way that the language is set up. Local Inversibility, local invertibility here means that any syntax segment can be inversed in isolation, and there is no need for full program analysis to figure out how to translate a part, this also makes the second feature fit right in. I use recursive decent to go through the program and transform it. So in each part we do the local inversion, and then we recursively invert every part of that statement.\n\nThese two parts make it possible to transform the code. The only thing missing now is the rules we apply to perform these transformations. I want to define two sets of transformations one to Rust that follows the semantics of the translation into C++ and one that translates into Janus.\n\n## RRust to Rust⌗\n\nIn this section, I will present the transformation rules that go from RRust into Rust. I will be using `F[stmt]` for forward translation rules and `R[stmt]` for reverse translation.\n\n### Reversible If⌗\n\nThe first construct we will look at is reversible if statements, as with Janus, they have a pre and a post condition, here the pre-condition is `c₁` and the post-condition is `c₂`.\n\n``````rif!(c₁, b₁, b₂, c₂);\n``````\n\nThis then gets expanded into the following Rust code:\n\n``````if c₁ {\nF[b₁]\nassert!(c₂);\n} else {\nF[b₂]\nassert!(!(c₂));\n}\n``````\n\nWe can see here that it is as with the graph we looked at earlier, if `c₁` then `c₂` and if `!c₁` then `!c₂`.\n\nThis was the forward translation, the reverse translation is similar, but we exchange `c₁` and `c₂` and recursivly reverse:\n\n``````if c₂ {\nR[b₁]\nassert!(c₁);\n} else {\nR[b₂]\nassert!(!(c₁));\n}\n``````\n\nHere we can see that the same rules applies as well, e.g. if `c₂` then `c₁`.\n\n### Reversible loop⌗\n\nThe reversible loop again works as the one in Janus, it has a condition that is only true when entering and a post-condition that similarily is only when exiting the loop.\n\n``````rloop!(c₁, b₁, b₂, c₂);\n``````\n\nThis expands in the forward direction to:\n\n``````assert!(c₁);\nF[b₁];\nwhile !(c₂) {\nF[b₂];\nassert!(!(c₁));\nF[b₁];\n}\n``````\n\nHere we have the form of loop we saw on the graph earlier, `c₁` must be true entering and `c₂` must be true when exiting. Again the reverse version is very similar. We only swap `c₁` and `c₂` around and recursively reverse.\n\n``````assert!(c₂);\nR[b₁];\nwhile !(c₁) {\nR[b₂];\nassert!(!(c₂));\nR[b₁];\n}\n``````\n\n### Other constructs⌗\n\nNow we have looked at the biggest constructs, now we look at all the smaller ones and try to put a few words to each of why they are constructed in that way. We only need to go through the reverse translation as the forward translation has no changes. I will start with arguable one of the most important ones:\n\n• `R[s₁; s₂]` We need to split them apart and then swap the positions since we need to do all operations in the reverse order: `R[s₂]; R[s₁]`.\n• `R[s₁;]` The semicolon is moved out and then recursively translated: `R[s₁];`.\n• `R[{ s₁ }]` Here we also move the brackets out: `{ R[s₁] }`.\n• `R[* s₁]` We also do that with dereferences: `* R[s₁]`.\n• `R[let id₁ = e₁]` let statements will be switched to be after the delocal statement, so we need to swap them around: `delocal!(id₁, e₁)`.\n• `R[delocal!(id₁, e₁)]` and the reverse is the same: `let id₁ = e₁`.\n• `R[id₁ ⊕= e₁]` Mutating statements gets the operators transformed: `id R[⊕] e₁`.\n• `R[l[e₁] ⊕= e₂]` And similarly with list mutation: `l[e₁] R[⊕]= e₂`.\n• `R[name::forward(f₁)]` Forward calls gets transformed into reverse calls: `name::backwards(f₁)`.\n• `R[name::backwards(f₁)]` and reverse calls gets transformed into forward calls: `name::forward(f₁)`.\n• `R[swap(id₁, id₂)]` Swap is its own self-inverse so no changes with it: `swap(id₁, id₂)`.\n\n### Operators⌗\n\nThe 3 operators we skipped over are not translated in any complicated way, `+` becomes `-`, and the reverse is the same. For xor `^` it is similar to `swap` since it is self-inverse, so it does not change.\n\n### Aliasing⌗\n\nThe transformations done in the previous sections is enough to write reversible programs, but it does not remove the possibility to write programs that will not be reversible. To ensure this we need to insert checks for aliasing as we discussed in the Janus section. The Janus-generated C++ code will not check for it at any point, so it is possible to give inputs that will cause issues. Rust ensures that we cannot give aliased arguments, but we also need mutation operations not to be aliased. The specific forms that we need to ensure are the following:\n\n``````// Also the case for +, - and ^.\n\na += b; // where a == b\n\nl[x] += l[y]; // where x == y\n``````\n\nTo ensure this we insert a check if they point to the same value:\n\n``````if core::ptr::eq(&(e₁), &(e₂)) {\npanic!();\n}\ne₁ += e₂;\n``````\n\nHere the references will be implicit cast to pointers which we then can check for equality, and fail if they are equal.\n\n## RRust to Janus⌗\n\nAs said further up, I also implemented a transformation of RRust into Janus. The transformation was rather mechanical and was mostly done by removing all the extra notation Rust uses. The most complicated part of that was the parser needed for the macros. I am not going to fully present it here, but if you are interested in seeing more about it feel free to contact me about it.\n\n# Example RRust programs⌗\n\nHere are a few examples of RRust code, they are reworked examples of Janus code.\n\n``````\nrfn!(Fib, (x1: &mut i32, x2: &mut i32, n: &mut i32), {\nrif!(\n*n == 0,\n{\n*x1 += 1;\n*x2 += 1;\n},\n{\n*n -= 1;\nFib::forward(x1, x2, n);\n*x1 += *x2;\nstd::mem::swap(x1, x2);\n},\n*x1 == *x2\n);\n});\n\nlet mut x1 = 0;\nlet mut x2 = 0;\nlet mut n = 10;\n\nFib::forward(&mut x1, &mut x2, &mut n);\n\nassert_eq!(x1, 89);\nassert_eq!(x2, 144);\nassert_eq!(n, 0);\n\nFib::backwards(&mut x1, &mut x2, &mut n);\n\nassert_eq!(x1, 0);\nassert_eq!(x2, 0);\nassert_eq!(n, 10);\n``````\n``````\nrfn!(Factor, (num: &mut usize, fact: &mut [usize; 20]), {\nlet mut tryf = 0;\nlet mut i = 0;\nrloop!(\ntryf == 0 && *num > 1,\n{\nNextTry::forward(&mut tryf);\nrloop!(\nfact[i] != tryf,\n{\ni += 1;\nfact[i] += tryf;\nlet mut z = *num / tryf;\nstd::mem::swap(&mut z, num);\ndelocal!(z, *num * tryf);\n},\n*num % tryf != 0\n);\n},\ntryf * tryf > *num\n);\n\nrif!(\n*num != 1,\n{\ni += 1;\nfact[i] ^= *num;\n*num ^= fact[i];\nfact[i] ^= *num;\n},\n{\n*num -= 1;\n},\nfact[i] != fact[i - 1]\n);\n\nrif!(\n(fact[i - 1] * fact[i - 1]) < fact[i],\n{\nrloop!(\ntryf * tryf > fact[i],\n{\nNextTry::backwards(&mut tryf);\n},\ntryf == 0\n);\n},\n{\ntryf -= fact[i - 1];\n},\n(fact[i - 1] * fact[i - 1]) < fact[i]\n);\n\nZeroI::forward(&mut i, fact);\ndelocal!(i, 0);\ndelocal!(tryf, 0);\n});\n\nrfn!(ZeroI, (i: &mut usize, fact: &mut [usize; 20]), {\nrloop!(\nfact[*i + 1] == 0,\n{\n*i -= 1;\n},\n*i == 0\n);\n});\n\nrfn!(NextTry, (tryf: &mut usize), {\n*tryf += 2;\nrif!(\n*tryf == 4,\n{\n*tryf -= 1;\n},\n*tryf == 3\n);\n});\n\nlet mut num = 840;\nlet mut fact = [0; 20];\n\nFactor::forward(&mut num, &mut fact);\nprint!(\"Num: {}, Factors: \", num);\nfor i in 1u64..=6 {\nprint!(\"{}: {}\", i, fact[i as usize]);\nif i != 6 {\nprint!(\", \");\n} else {\nprintln!(\".\");\n}\n}\n\n// Num: 0, Factors: 1: 2, 2: 2, 3: 2, 4: 3, 5: 5, 6: 7.\n\nFactor::backwards(&mut num, &mut fact);\nprint!(\"Num: {}, Factors: \", num);\nfor i in 1u64..=6 {\nprint!(\"{}: {}\", i, fact[i as usize]);\nif i != 6 {\nprint!(\", \");\n} else {\nprintln!(\".\");\n}\n}\n\n// Num: 840, Factors: 1: 0, 2: 0, 3: 0, 4: 0, 5: 0, 6: 0.\n``````\n\n# Conclusion⌗\n\nSo what did I show with this whole blog? And did it reach the goals I had set for myself?\n\nI have shown that it is possible to use the meta-programming of Rust to construct a safe reversible programming language, or at least one safer than the code generated by the language Janus. Following that I have shown that it is possible to use meta-programming in Rust for more advanced program transformations.\n\nBut some parts that I wanted to have shown are still missing, I believe it is possible to use the type system of Rust to ensure reversibility better than what I have done, specifically the need to ensure that a type is not `Copy` since that will break the rules. I have tried various things, but I never found a way that did not get in the way when writing code or was in some sense trivial to misuse.\n\n# Looking forward⌗\n\nI have put the code up on GitHub at https://github.com/erk-/rrust, and I intend to work on it when I have some time. I have a few things that I want to have done with it including finding a way to make better bounds with the help of the type system. Any contributions are appreciated.\n\nThe main things that I want to do looking forward are, in no particular order:\n\n• Use the type system of Rust to ensure types are not copied.\n• Return errors instead of panicking when ending up in an error state.\n• A side effect of this will be the need to be able to mark input values as poisoned, since the value cannot be used after it enters a fail\n• Allow functions and methods that do not mutate, for example there is no reason that it should be impossible to call `len` on a slice since it will be the same forward and in reverse. I believe that this is correct for any function that does not mutate, barring internal mutability.\n\n## The thermodynamics of computation — a review⌗\n\n[web]\n\nComputers may be thought of as engines for transforming free energy into waste heat and mathematical work. Existing electronic computers dissipate energy vastly in excess of the mean thermal energykT, for purposes such as maintaining volatile storage devices in a bistable condition, synchronizing and standardizing signals, and maximizing switching speed. On the other hand, recent models due to Fredkin and Toffoli show that in principle a computer could compute at finite speed with zero energy dissipation and zero error. In these models, a simple assemblage of simple but idealized mechanical parts (e.g., hard spheres and flat plates) determines a ballistic trajectory isomorphic with the desired computation, a trajectory therefore not foreseen in detail by the builder of the computer. In a classical or semiclassical setting, ballistic models are unrealistic because they require the parts to be assembled with perfect precision and isolated from thermal noise, which would eventually randomize the trajectory and lead to errors. Possibly quantum effects could be exploited to prevent this undesired equipartition of the kinetic energy. Another family of models may be called Brownian computers, because they allow thermal noise to influence the trajectory so strongly that it becomes a random walk through the entire accessible (low-potential-energy) portion of the computer’s configuration space. In these computers, a simple assemblage of simple parts determines a low-energy labyrinth isomorphic to the desired computation, through which the system executes its random walk, with a slight drift velocity due to a weak driving force in the direction of forward computation. In return for their greater realism, Brownian models are more dissipative than ballistic ones: the drift velocity is proportional to the driving force, and hence the energy dissipated approaches zero only in the limit of zero speed. In this regard Brownian models resemble the traditional apparatus of thermodynamic thought experiments, where reversibility is also typically only attainable in the limit of zero speed. The enzymatic apparatus of DNA replication, transcription, and translation appear to be nature’s closest approach to a Brownian computer, dissipating 20–100kT per step. Both the ballistic and Brownian computers require a change in programming style: computations must be renderedlogically reversible, so that no machine state has more than one logical predecessor. In a ballistic computer, the merging of two trajectories clearly cannot be brought about by purely conservative forces; in a Brownian computer, any extensive amount of merging of computation paths would cause the Brownian computer to spend most of its time bogged down in extraneous predecessors of states on the intended path, unless an extra driving force ofkTln2 were applied (and dissipated) at each merge point. The mathematical means of rendering a computation logically reversible (e.g., creation and annihilation of a history file) will be discussed. The old Maxwell’s demon problem is discussed in the light of the relation between logical and thermodynamic reversibility: the essential irreversible step, which prevents the demon from breaking the second law, is not the making of a measurement (which in principle can be done reversibly) but rather the logically irreversible act of erasing the record of one measurement to make room for the next. Converse to the rule that logically irreversible operations on data require an entropy increase elsewhere in the computer is the fact that a tape full of zeros, or one containing some computable pseudorandom sequence such as pi, has fuel value and can be made to do useful thermodynamic work as it randomizes itself. A tape containing an algorithmically random sequence lacks this ability.\n\n## Microsoft Security Response Center. A proactive approach to more secure code.⌗\n\nhttps://msrc-blog.microsoft.com/2019/07/16/a-proactive-approach-to-more-secure-code/\n\n## Ivan Lanese, Naoki Nishida, Adrián Palacios, and Germán Vidal. A theory of reversibility for erlang. CoRR, abs/1806.07100, 2018.⌗\n\narXiv\n\nIn a reversible language, any forward computation can be undone by a finite sequence of backward steps. Reversible computing has been studied in the context of different programming languages and formalisms, where it has been used for testing and verification, among others. In this paper, we consider a subset of Erlang, a functional and concurrent programming language based on the actor model. We present a formal semantics for reversible computation in this language and prove its main properties, including its causal consistency. We also build on top of it a rollback operator that can be used to undo the actions of a process up to a given checkpoint.\n\n## Christopher Lutz. Janus: a time-reversible language. 1986. Letter to R. Landauer.⌗\n\n.pdf\n\nJANUS is a compiler and interpreter for the time-reversible language JANUS. The JANUS compiler is written in SLIMEULA, and compiles the code into an internal SLIMEULA Class structure which can be interpreted directly. ‘SLIMEULA’ means SIMULA running on a DECSYSTEM-20. JANUS is considered to be a throw-away piece of code. It will not be maintained and is not purported to be robust. The compiler consists of four major parts: A lexical analyzer which tokenizes the input stream and generates a symbol table; a recursive descent parser made of the Init Code of SLIMEULA Classes; an interpreter which consists of the procedure ’exec’ com- mon to all of the Classes created in the parsing; and the runtime command scanner\n\n## Tetsuo Yokoyama and Robert Glück. A reversible programming language and its invertible self-interpreter. In Proceedings of the 2007 ACM SIGPLAN Symposium on Partial Evaluation and Semantics-Based Program Manipulation, PEPM ‘07, page 144–153, New York, NY, USA, 2007. Association for Computing Machinery.⌗\n\n[DOI, web]\n\nA reversible programming language supports deterministic forward and backward computation. We formalize the programming language Janus and prove its reversibility. We provide a program inverter for the language and implement a self-interpreter that achieves deterministic forward and backward interpretation of Janus programs without using a computation history. As the self-interpreter is implemented in a reversible language, it is invertible using local program inversion. Many physical phenomena are reversible and we demonstrate the power of Janus by implementing a reversible program for discrete simulation of the Schrödinger wave equation that can be inverted as well as run forward and backward.\n\nKeywords: reversible computing, program inversion, non-standard interpreter hierarchy, self-interpreter, Janus, reversible programming language\n\n## Torben Ægidius Mogensen. Hermes: A reversible language for lightweight encryption. Science of Computer Programming, 215:102746, 2022.⌗\n\n[DOI, web]\n\nHermes is a domain-specific language for writing lightweight encryption algorithms: It is reversible, so it is not necessary to write separate encryption and decryption procedures. Hermes uses a type system that avoids several types of side-channel attacks, by ensuring no secret values are left in memory and that operations on secret data spend time independent of the value of this data, thus preventing timing-based attacks. We show a complete formal specification of Hermes, argue absence of timing-based attacks (under reasonable assumptions), and compare implementations of well-known lightweight encryption algorithms in Hermes and C.\n\nKeywords: Lightweight encryption, Side-channel attacks, Reversible programming languages, Domain-specific languages\n\n1. `Copy` types include small types such as integers that efficiently can be copied. Rust will implicitly copy the value into a new, therefore it does not transfer ownership. https://doc.rust-lang.org/std/marker/trait.Copy.html ↩︎\n\n2. It is possible to opt out of such bound-checks with the use of `unsafe` and specifically the `get_unchecked` function. ↩︎\n\n3. Without the use of `unsafe`↩︎\n\n4. Note that there is an important part missing here, I will get back to that later. ↩︎" ]

[ null, "https://blog.erk.dev/posts/rrust/images/bijection.png", null ]

[ "# Document (#28667)\n\nAuthor\nSkulschus, M.\nWiederstein, M.\nTitle\nXSLT und XPath\nImprint\nBonn : MITP-Verlag\nYear\n2005\nPages\n303 S\nIsbn\n3-8266-1532-8\nAbstract\nMit XSLT können XML-Dokumente in Formate wie HTML oder Text transformiert werden. Das Buch zeigt, wie Sie die Technik nutzen, um XML-Daten flexibel einzusetzen. Die Syntax und die praktische Arbeit mit XSLT und der Basistechnik für Transformationen, XPath, werden von den Autoren verständlich vorgestellt. Es wird gezeigt, wie Stylesheets erzeugt werden und XPath zur Adressierung benutzt wird. Dabei wird neben XPath 1.0 auch die neue Version 2.0 samt der Änderungen ausführlich beschrieben. Die Autoren gehen auf die unterschiedlichen Anforderungen ein, die bestimmte Ein- und Ausgabeformate mit sich bringen, darunter HTML, Text, CSS und XML. Das Buch wird mit einer XSLT-Referenz abgerundet, die die möglichen Elemente auflistet und ihre Funktion und Syntax kompakt erklärt. Zahlreiche Tipps, Hinweise und Beispiele lockern das Werk auf und helfen, die neu erlernten Kenntnisse schnell praktisch umzusetzen und dabei typische Einsteigerfehler zu vermeiden.\nXSLT (eXtensible Stylesheet Language forTransformations) ist eine W3C-Syntax, die speziell für die Transformation von XML-Dokumenten geschaffen wurde. In diesem Buch werden die Versionen 1.0 und 2.0 behandelt. Mit XSLT können XML-Dokumente in Formate wie HTML, Text und andere XML-Formate transformiert werden. Diese Technologie lässt sich in (fast) allen Programmiersprachen und in vielen Datenbanken für die XMLVerarbeitung nutzen und stellt die beste Möglichkeit dar, aus mehreren Anwendungen heraus die gleiche Transformation aufzurufen. Dieses Buch stellt Ihnen die Syntax vor und erläutert Ihnen umfassend die Arbeitsweise mit XSLT und XPath 1.0 und 2.0. Nach einer Einführung in die Grundkonzepte gehen die Autoren direkt detailliert auf das Arbeiten mit Vorlagen und XPath ein. Im Folgenden werden alle Techniken ausführlich behandelt, die Sie für professionelle XML-Transformationen brauchen: Kontrollstrukturen, Parameter und Variablen, Sortieren und Gruppieren und Sonderarbeiten für verschiedene Ein-und Ausgabeformate. Zahlreiche Beispiele ermöglichen es Ihnen, die vorgestellten Techniken sofort nachzuvollziehen und anhand des im Internet verfügbaren Quellcodes selber zu bearbeiten. Zusätzlich werden alle neuen Techniken in Schema-Zeichnungen umgesetzt. Neben XSLT stellt dieses Werk auch die zweite Basistechnologie für Transformationen dar: XPath in der neuen Version 2.0. Mit dieserTechnik hat man die Möglichkeit, XML-Strukturen zu lokalisieren und auszuwählen. Beide Technologien gehören eng zusammen. Aus dem Inhalt: - Definition und Aufruf von Vorlagen, Vorlagen-Typen und -Alternativen - XPath: Adressierung, Lokalisierung, Filtern - Algorithmen: Kontrollstrukturen in XSLT und XPath - Auslagerung und Wiederverwendung mit globalen Parametern und eingebetteten Dateien - Einsatz von Parametern und Variablen - Sortieren, Nummerieren und Gruppieren - Unterschiedliche Algorithmen für unterschiedliche Daten-Modellierungen\nContent\nDas Buch wird ergänzt um 2 weitere Titel: Der zweite Band stellt die Syntax von XSL-FO 1.0 bzw. XSL 1.0 mit dem Titel Extensible Stylesheet Language (XSL), Version 1.0, W3C Recommendation 15 October 2001 unter http: //www. w3.org/TR/xs1/ dar. Sie lernen in diesem Band, wie Druckerzeugnisse wie z.B. PDF-Dateien aus XML-Daten erzeugt werden. Dies erfordert einen eigenen Standard mit einer Vielzahl an unterschiedlichen Elementen, um Seitenbereiche, Seitenverläufe, Text- und Absatzformate sowie zusätzliche Dokumenteigenschaften anzugeben wie Inhaltsverzeichnisse. Der dritte Band stellt keine besondere Syntax dar, sondern kombiniert die einzelnen vorgestellten Standards mit allgemeinen Techniken, die nur die Transformation betreffen, mit Ideen, wie XSLT-Anwendungen aufgebaut werden können, und solchen Techniken wie die Verwendung einer Datenbank, um XML- und sogar XSLT Daten abzuspeichern und dynamisch für Transformationsabläufe zusammenzusetzen.\nObject\nXSLT\nXPath\n\n## Similar documents (content)\n\n1. Stein, M.: Workshop XML (2001) 0.19\n```0.19415097 = sum of:\n0.19415097 = product of:\n0.97075486 = sum of:\n0.012784275 = weight(abstract_txt:wird in 2464) [ClassicSimilarity], result of:\n0.012784275 = score(doc=2464,freq=1.0), product of:\n0.04305564 = queryWeight, product of:\n1.1421081 = boost\n3.800634 = idf(docFreq=2570, maxDocs=42306)\n0.009918976 = queryNorm\n0.29692453 = fieldWeight in 2464, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.800634 = idf(docFreq=2570, maxDocs=42306)\n0.078125 = fieldNorm(doc=2464)\n0.042731833 = weight(abstract_txt:ihnen in 2464) [ClassicSimilarity], result of:\n0.042731833 = score(doc=2464,freq=1.0), product of:\n0.08745153 = queryWeight, product of:\n1.4096345 = boost\n6.2545214 = idf(docFreq=220, maxDocs=42306)\n0.009918976 = queryNorm\n0.48863447 = fieldWeight in 2464, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.2545214 = idf(docFreq=220, maxDocs=42306)\n0.078125 = fieldNorm(doc=2464)\n0.051823653 = weight(abstract_txt:buch in 2464) [ClassicSimilarity], result of:\n0.051823653 = score(doc=2464,freq=2.0), product of:\n0.086880304 = queryWeight, product of:\n1.622381 = boost\n5.398855 = idf(docFreq=519, maxDocs=42306)\n0.009918976 = queryNorm\n0.59649485 = fieldWeight in 2464, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n5.398855 = idf(docFreq=519, maxDocs=42306)\n0.078125 = fieldNorm(doc=2464)\n0.40372235 = weight(abstract_txt:xslt in 2464) [ClassicSimilarity], result of:\n0.40372235 = score(doc=2464,freq=1.0), product of:\n0.56367487 = queryWeight, product of:\n6.198659 = boost\n9.167778 = idf(docFreq=11, maxDocs=42306)\n0.009918976 = queryNorm\n0.71623266 = fieldWeight in 2464, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.167778 = idf(docFreq=11, maxDocs=42306)\n0.078125 = fieldNorm(doc=2464)\n0.45969275 = weight(abstract_txt:xpath in 2464) [ClassicSimilarity], result of:\n0.45969275 = score(doc=2464,freq=1.0), product of:\n0.61463684 = queryWeight, product of:\n6.472808 = boost\n9.573242 = idf(docFreq=7, maxDocs=42306)\n0.009918976 = queryNorm\n0.74790955 = fieldWeight in 2464, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.573242 = idf(docFreq=7, maxDocs=42306)\n0.078125 = fieldNorm(doc=2464)\n0.2 = coord(5/25)\n```\n2. 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Becker, H.-G.: MODS2FRBRoo : Ein Tool zur Anbindung von bibliografischen Daten an eine Ontologie für Begriffe und Informationen (2010) 0.08\n```0.075122744 = sum of:\n0.075122744 = product of:\n0.6260229 = sum of:\n0.025311572 = weight(abstract_txt:wird in 1266) [ClassicSimilarity], result of:\n0.025311572 = score(doc=1266,freq=2.0), product of:\n0.04305564 = queryWeight, product of:\n1.1421081 = boost\n3.800634 = idf(docFreq=2570, maxDocs=42306)\n0.009918976 = queryNorm\n0.58788055 = fieldWeight in 1266, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n3.800634 = idf(docFreq=2570, maxDocs=42306)\n0.109375 = fieldNorm(doc=1266)\n0.03550003 = weight(abstract_txt:werden in 1266) [ClassicSimilarity], result of:\n0.03550003 = score(doc=1266,freq=2.0), product of:\n0.06501034 = queryWeight, product of:\n1.8565319 = boost\n3.530313 = idf(docFreq=3368, maxDocs=42306)\n0.009918976 = queryNorm\n0.5460675 = fieldWeight in 1266, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n3.530313 = idf(docFreq=3368, maxDocs=42306)\n0.109375 = fieldNorm(doc=1266)\n0.5652113 = weight(abstract_txt:xslt in 1266) [ClassicSimilarity], result of:\n0.5652113 = score(doc=1266,freq=1.0), product of:\n0.56367487 = queryWeight, product of:\n6.198659 = boost\n9.167778 = idf(docFreq=11, maxDocs=42306)\n0.009918976 = queryNorm\n1.0027257 = fieldWeight in 1266, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.167778 = idf(docFreq=11, maxDocs=42306)\n0.109375 = fieldNorm(doc=1266)\n0.12 = coord(3/25)\n```\n5. Öttl, S.; Streiff, D.; Stettler, N.; Studer, M.: Aufbau einer Testumgebung zur Ermittlung signifikanter Parameter bei der Ontologieabfrage (2010) 0.07\n```0.07155728 = sum of:\n0.07155728 = product of:\n0.44723302 = sum of:\n0.01022742 = weight(abstract_txt:wird in 1258) [ClassicSimilarity], result of:\n0.01022742 = score(doc=1258,freq=1.0), product of:\n0.04305564 = queryWeight, product of:\n1.1421081 = boost\n3.800634 = idf(docFreq=2570, maxDocs=42306)\n0.009918976 = queryNorm\n0.23753962 = fieldWeight in 1258, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.800634 = idf(docFreq=2570, maxDocs=42306)\n0.0625 = fieldNorm(doc=1258)\n0.04440655 = weight(abstract_txt:techniken in 1258) [ClassicSimilarity], result of:\n0.04440655 = score(doc=1258,freq=1.0), product of:\n0.10411286 = queryWeight, product of:\n1.538066 = boost\n6.8243704 = idf(docFreq=124, maxDocs=42306)\n0.009918976 = queryNorm\n0.42652315 = fieldWeight in 1258, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.8243704 = idf(docFreq=124, maxDocs=42306)\n0.0625 = fieldNorm(doc=1258)\n0.024844844 = weight(abstract_txt:werden in 1258) [ClassicSimilarity], result of:\n0.024844844 = score(doc=1258,freq=3.0), product of:\n0.06501034 = queryWeight, product of:\n1.8565319 = boost\n3.530313 = idf(docFreq=3368, maxDocs=42306)\n0.009918976 = queryNorm\n0.38216758 = fieldWeight in 1258, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n3.530313 = idf(docFreq=3368, maxDocs=42306)\n0.0625 = fieldNorm(doc=1258)\n0.36775422 = weight(abstract_txt:xpath in 1258) [ClassicSimilarity], result of:\n0.36775422 = score(doc=1258,freq=1.0), product of:\n0.61463684 = queryWeight, product of:\n6.472808 = boost\n9.573242 = idf(docFreq=7, maxDocs=42306)\n0.009918976 = queryNorm\n0.59832764 = fieldWeight in 1258, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.573242 = idf(docFreq=7, maxDocs=42306)\n0.0625 = fieldNorm(doc=1258)\n0.16 = coord(4/25)\n```" ]

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[ "#### Volume 10, issue 1 (2006)\n\n Download this article", null, "For screen For printing", null, "", null, "Recent Issues", null, "", null, "The Journal About the Journal Editorial Board Subscriptions Editorial Interests Editorial Procedure Submission Guidelines Submission Page Ethics Statement ISSN (electronic): 1364-0380 ISSN (print): 1465-3060 Author Index To Appear Other MSP Journals\nAutomorphic forms and rational homology 3–spheres\n\n### Frank Calegari and Nathan M Dunfield\n\nGeometry & Topology 10 (2006) 295–329\n arXiv: math.GT/0508271\n##### Abstract\n\nWe investigate a question of Cooper adjacent to the Virtual Haken Conjecture. Assuming certain conjectures in number theory, we show that there exist hyperbolic rational homology 3–spheres with arbitrarily large injectivity radius. These examples come from a tower of abelian covers of an explicit arithmetic 3–manifold. The conjectures we must assume are the Generalized Riemann Hypothesis and a mild strengthening of results of Taylor et al on part of the Langlands Program for ${GL}_{2}$ of an imaginary quadratic field.\n\nThe proof of this theorem involves ruling out the existence of an irreducible two dimensional Galois representation $\\rho$ of $Gal\\left(\\overline{ℚ}∕ℚ\\left(\\sqrt{-2}\\right)\\right)$ satisfying certain prescribed ramification conditions. In contrast to similar questions of this form, $\\rho$ is allowed to have arbitrary ramification at some prime $\\pi$ of $ℤ\\left[\\sqrt{-2}\\right]$.\n\nIn the next paper in this volume, Boston and Ellenberg apply pro–$\\phantom{\\rule{0.3em}{0ex}}p$ techniques to our examples and show that our result is true unconditionally. Here, we give additional examples where their techniques apply, including some non-arithmetic examples.\n\nFinally, we investigate the congruence covers of twist-knot orbifolds. Our experimental evidence suggests that these topologically similar orbifolds have rather different behavior depending on whether or not they are arithmetic. In particular, the congruence covers of the non-arithmetic orbifolds have a paucity of homology.\n\n##### Keywords\nvirtual Haken Conjecture, Cooper's question, rational homology sphere, injectivity radius, automorphic forms, Galois representations\n##### Mathematical Subject Classification 2000\nPrimary: 57M27, 11F80, 11F75\n##### Publication", null, "" ]

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[ "# Is there a term for the standard deviation of a sample as a percentage of the mean?\n\nI’m trying to compare the variability within two samples which are quite different in scale (pupil capacity of different types of educational establishment (nursery, primary and secondary schools). My approach is to calculate the standard deviation for each sample, then divide that by the sample mean.\n\nI’ve not seen this metric used, so I’m wondering is there an accepted name for it? Or is there is a better metric to use which would explain why I haven’t seen it?\n\n$$c_{\\rm v} = \\frac{\\sigma}{\\mu}$$" ]

[ null ]

[ "# RoyaltyStat Blog\n\n### Posts by Topic\n\nTransfer pricing tax compliance is devoid of external CUP (comparable uncontrolled prices). Therefore, we must select (under the TNMM and under the Profit Split Method) the most appropriate “net” profit indicator (NPI) from comparable uncontrolled enterprises. Most appropriate is the most reliable among competing profit indicators. In economics and statistics, reliability is measured by the coefficient of variation (standard deviation / mean) of the selected variable.\n\nHere is the OECD definition of NPI: “The ratio of net profit to an appropriate base (e.g. costs, sales, assets). The transactional net margin method [TNMM] relies on a comparison of an appropriate net profit indicator for the controlled transaction with the same net profit indicator in comparable uncontrolled transactions.” OECD (2017), § B.3.3 (Determination of net profit). See OECD Transfer Pricing Guidelines for Multinational Enterprises and Tax Administrations 2017, OECD Publishing, Paris, located at: http://dx.doi.org/10.1787/tpg-2017-en\n\nIn transfer pricing analysis, one important issue of contention is if, prima facie, asset return (OECD lingo: “return on assets”) is more reliable than profit margin. Lovers of asset return argue that because this NPI includes two multiplicative components, profit margin and asset turnover, asset return provides more financial information than profit margin about comparable enterprises. We show that this purported à priori advantage of asset return over profit margin is not supported by a comparison of their coefficients of variation (measure of reliability).\n\nFrom the DuPont profit identity, define R = [M ∙ T] = asset return, where M = profit margin (operating profit / revenue) and T = asset turnover (revenue / asset). Define random variable R with mean µR and standard deviation σR, M with mean µM and standard deviation σM, and T with mean µT and standard deviation σT.\n\nHere is a major problem in selecting the most appropriate NPI: If M and T are independent variables, find E[M ∙ T] and Var[M ∙ T]. These two statistical measures are important to determine the coefficient of variation.\n\nWe must select the NPI with the smallest coefficient of variation, i.e., we must select the most reliable NPI computed from uncontrolled comparable enterprises:\n\n(1)  If M and T are independent variables, E[R] = E[M ∙ T] = µ= µM µT . However, the variance of R is complex to evaluate because it shows compound errors.\n\n(2)  Var[M ∙ T] = E[M2 T2] – {E[M]E[T]}2 = σR2\n\n= (µM2 + σM2) (µT2 + σT2) − µM2 µT2\n\n= σM2 σT2 + µM2 µT2 + σM2 µT2 + σT2 µM2 − µM2 µT2\n\nσR2 = σM2 σT2 + σM2 µT2 + σT2 µM2\n\nThe standard deviation is the square root of (2). See product of independent variables (without our statistical details): https://en.wikipedia.org/wiki/Variance\n\nWe can show that the coefficient of variation of M can’t exceed that of R = M ∙ T; therefore, if M and T are independent variables, profit margin is more reliable than asset return.\n\nWe define the coefficient of variation of random variable X (which can be M, T or R) as:\n\n(3)  KX = √(σX2 / µX2)\n\nwhere, like (1) and (2), we have these two statistical parameters defined as:\n\n(4)  µR = µM µT and\n\n(5)  σR2 = σM2 σT2 + σM2 µT2 + σT2 µM2\n\nIt follows that the coefficient of variation of asset return is computed with formula:\n\n(6)  KR2 = (σR2 / µR2) = {σM2 σT2 + σM2 µT2 + σT2 µM2} / µM2 µT2\n\nFrom (6), and the definition of the coefficient of variation in (3), we obtain:\n\n(7)  KR2 = {KM2 (KT2 + 1) +  KT2} > KM2\n\nTherefore, we find our basic result that the coefficient of variation of R is greater than of M:\n\n(8)  KR > KM because KT2 > 0.\n\nIn selecting NPI, inequality (8) shows that if M and T are independent variables, profit margin is more reliable than asset return.\n\nProfit margin has several additional advantages over asset return as a more reliable NPI to determine arm's length taxable profits. First, profit margin is a pure number (it has no dimension). Second, asset return suffers from differences in asset vintages or bases (different historical costs of asset acquisition among different comparable enterprises). Third, asset (denominator of asset return) suffers from differences in depreciation rates. See OECD (2017), ¶ 2.104.\n\nTakeaway: If profit margin (M) and asset turnover (T) are independent variables, we have a strong statistical reason to use profit margin instead of asset return as the most reliable (most appropriate) NPI in transfer pricing compliance.\n\nPublished on Oct 22, 2018 7:31:38 AM\n\nEdnaldo Silva (Ph.D.) is founder and managing director of RoyaltyStat. He helped draft the US transfer pricing regulations and developed the comparable profits method called TNNM by the OECD. He can be contacted at: [email protected]\n\nRoyaltyStat provides premier online databases of royalty rates extracted from unredacted license agreements\nand normalized company financials (income statement, balance sheet, cash flow). We provide high-quality data, built-in analytical tools, customer training and attentive technical support.", null, "Topics: Net Profit Indicator" ]

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[ "You are on page 1of 43\n\n# Chapter 9: Monoprotic Acid-Base Equilibria Chapter 11: Acid-Base Titrations\n\n## I. Solutions and Indicators for Neutralization Titrations\n\nA. Standard Solutions: The standards solutions used as titrants for unknown w eak acids or bases are alw ays strong bases or acids, respectively. Standard titrant acids: dilute solutions of HCl, HClO 4 , or H 2 SO 4 . Standard titrant bases: dilute solutions of NaOH, KOH. The primary standards may not be strong acids or bases (e.g., potassium acid phthalate, sodium oxalate, sodium bicarbonate).\n\nB. The Theory of Indicator Behavior 1. pH-sensitive dyes have long been used as indicators. Normally, the basic form (In) on the dye has a color different from the acid form, HIn: HIn + H2O <====> H3O+ + InIn + H2O <====> OH- + HIn+\nKa = [H 3O + ][In ] [HIn] [OH ][HIn + ] [In]\n\n[Eq.1]\n\nKb =\n\n[Eq. 2]\n\nWe can see that both equilibrium constant expressions above can be written:\n[ H 3O + ] = Ka[HIn] [In ]\nKa[HIn + ] [In]\n\n[Eq. 1]\n\n[ H 3O ] =\n\n## {Note: Kw =KaKb = [OH-][H+]}\n\n[Eq. 2]\n\nTherefore, the [H3O+] determines the ratio of the acid/conjugate base form of the indicator. To see the color of a particular form (acid or base) of the indicator, that form must be present at tenfold higher concentration.\n\nFor example, To see the color of acid form, the ratio [HIn]/[In] must be equal to or greater than 10. To see the color of base form, the ratio [HIn]/[In] must be equal to or less than 0.1. This means that: For acid color [H3O+] > Ka (10/1) For base color [H3O+] < Ka (1/10) Hence: indicator range = pKa 1, and the pH change in the area of the equivalence point must match this range or at least overlap it significantly to use this indicator for endpoint detection. (note: how to select the indicator)\n\n9-1\n\nC.\n\nTitration Curves may be linear-segment curve or a sigmoidal curve depending on what is plotted on the y-axis. The X-axis units are always reagent or titrant volume. The Y-axis may be in increments of analyte reacted or product formed (linear-segment curve) or a p-function such as pH (s-curve). The equivalence point is characterized by large changes in the relative concentrations of the reagent and analyte. (See Table 10-2)\n\nTitration Curves\n\n9-2\n\n## D. The Titration of a Strong Acid with a Strong Base\n\nExample: Determination of HCl concentration by titration with NaOH\n\nNaOH + HCl\n\nNaCl + H2O\n\n## moles = CNaOHVNaOH = CHClVHCl\n\n1. H3O+ in the titration medium has two sources a. From the H2O solvent b. From the acid solute - usually this is in great excess relative contribution from water because the Kw is so small c. The mass balance equation describing this situation is:\n[ H 3 O + ] = C H C l + [ O H -] = C HCl\n\nd . T h e s a m e is tr u e fo r a s tr o n g b a s e a n d w e c a n w r ite : [ O H -] = C N a O H + [ H 3 O + ] = C N a O H\n\n2. Before the equivalence point We calculate the pH of the titration medium from the concentration of unreacted strong acid:\nMoles of [ H 3O + ] Moles of [OH ] [ H 3O ] = Vacid + Vbase\n+\n\npH = - log [H3O+] 3. At the equivalence point All of the acid has reacted with the titrant base. For a strong acid titrated with a strong base, the salt is a strong electrolyte and therefore completely dissociated. It does not react with H2O. The resulting solution is neutral (pH = 7.00) because: HCl + NaOH <====> H2O + Na+ + Cl2 H2O <====> H3O+ + OH4. After the equivalence point we calculate the pH of the titration medium from the concentration of unreacted strong base:\nKw Moles of [OH ]added Moles of [ H 3O + ] = [OH ] = [ H 3O + ] Vacid + Vbase\n\n9-1\n\n9-1\n\n9-2\n\nAny Questions?\n\n5. The Effect of Concentration on the shape of the curve: With a very dilute solution of strong acid which is titrated with a very dilute solution of strong base, there will be a smaller relative change in the pH immediately before and after the equivalence point\n\n6.\n\nThe selection of an indicator The indicator range (detectable color change) should occur in the area of the equivalence point. Concentration effect i. titration of 0.0500 M HCl with 0.1000 M NaOH three indicators (phenolphthalein, bromothymol blue, Bromocresol green) have color changes in this range. ii. for the more dilute titration medium (0.000500 M HCl with 0.001000 M NaOH), only one of the indicators (bromothymol blue) is now suitable. iii. This is because the relative pH change for the second curve is so small that two of the indicators change color before or after the equivalence point.\n\nAny Questions?\n\nE.\n\nBuffer Solutions: A buffer solution resists changes in pH. Buffers usually consist of a weak acid/conjugate base pair mixture in solution. Since the titration of a weak acid (or base) with a strong base (or acid) will form a buffer solution, the curves constructed for these titration systems will appear quite different from those where all the reactants are completely\n\nE1. The Calculation of the pH of Buffer Solutions: Example: We are preparing a buffer in which the acid, HA, and its salt, NaA, are being added to the solution to give CHA and CNaA For the species HA and A- we can write: a. Pertinent Equilibria HA + H2O <====> A- + H3O+ A-+ H2O <====> HA + OH-\n\n## 2H2O <====> H3O+ + OHb. Equilibrium Expressions\n\n[H 3 O + ][A - ] Ka = [HA] [OH ][HA] Kb = [A - ]\n\n## [Eq. 1] [Eq. 2] [Eq. 3]\n\nKw =[H3O+] [OH-] c. Mass balance equations: [HA] = CHA - [H3O+] + [OH-] [A-] = CNaA + [H3O+] - [OH-]\n\n[Eq. 4] [Eq. 5]\n\nd. Approximations: Since the concentrations of these species are likely to be negligible relative the CHA and CNaA, we can approximate: [HA] = CHA [A-] = CNaA This assumption is true only when Ka < 10-3 and the relative concentrations of the acid or its conjugate base are relatively high. e. Solving equations If we rearrange Eq.1 and solve for [H3O+] then\n[H 3O + ] = K a [HA] [A - ]\n\n[Eq.1]\n\n## Taking -log of both sides of Eq.1\n\npH = pK a log C [HA] [A - ] = pK a + log = pK a + log NaA [A - ] [HA] C HA\n\nThis equation implies that the pH of a buffer solution is independent of the dilution of the solution since the relative concentrations of conjugate base/acid do not change upon dilution\n\n9-3\n\n## HCOO- + H2O < ==> HCOOH + H3O+\n\npH = pK a + log C NaA C HA\n\n9-4\n\npH = pK a + log\n\nC NH 3 C NH +\n4\n\npH = pK a + log\n\nC NH 3 C NH +\n4\n\nAny Questions?\n\nF. Properties of Buffer Solutions 1. Effect of Dilution: Theoretically, pH does not change with dilution. However, ionic strength changes with dilution (and therefore so will Ka). For buffers whose Ka values are strongly influenced by ionic strength, we may see a pH change over large changes in concentration. 2. Effect of Temperature: Since Ka changes as a function of temperature, we can expect buffer pH to change with changes in temperature. 3. Effect of Added Acids or Bases: Buffer solutions tend to resist pH change, although the ratio of base/acid changes depending on the amount of acid or base added 4. Buffer Capacity: The number of moles of strong acid or strong base that causes the pH of 1.00 L of buffer to change by 1.00 pH unit. The buffering capacity of the system for acid or base falls off as the concentration ratio of weak acid to conjugate base in the solution becomes larger or smaller than 1.\n\nIf ([A-]/[HA])<<1, the system will not buffer acid effectively; If ([A-]/[HA])>>1, the system will not buffer base effectively. If ([A-]/[HA])= 1, buffering capacity to both acids and bases is considered most effective and the pKa for the system is within 1 unit of the desired\n\nG.\n\n## Titration Curves for Weak Acids: There are 4 areas to consider\n\n1) 2) 3) 4)\n\nbefore the addition of base before the equivalence point (buffer region 1) at the equivalence point after the equivalence point\n\n1. Before the addition of base: Calculated from the concentration and Ka of the weak acid. 2. After the addition of strong base but before the equivalence point:\nMoles of [OH ]added Vtotal Moles of [OH ]added = pK a + log pH = pK a + log Moles of [H 3O + ] Moles of [H 3O + ] Vtotal\n\nNote: The half-neutralized point pH = pKa and hence you can measure pKa from titration curve. The half-neutralized point means pH at\nVb = 1 Ve 2\n\nNote: Vb : volume of titrant Ve: volume of titrant needed to reach the equivalence point Note: You will need these questions and concepts for the calculation in Chem 322 (potentiometric titration experiments).\n\n3. At the equivalence point: The predominant equilibrium is the hydrolysis of H2O by the salt of the weak acid: A- + H2O <====> HA + OH[OH ][HA] Kb = [A - ]\n\nGenerally, you can solve the equilibrium constant equation for [OH-] and assume that\n[A ] =\n-\n\nMoles of [H 3O + ] Vtotal\n\n[OH ] = K b C b Thus,\n\nFor weak acids that are titrated with strong bases, the pH at the equivalence point will be basic. 4. Beyond the equivalence point: Both the anion of the weak acid and the excess base are sources of [OH-]. However, due to LeChatelier's Principle the addition of [OH-] in the form of a strong base will suppress the hydrolysis by the weak acid anion so that:\nMoles of [OH ]added Moles of [H 3O + ] [OH ] = Vtotal\n\n## 9-5 9-4 HOAC + H2O H3O+ + OAC-\n\n[ H 3O + ] = K a C a\n\n[OH ] = K b C b\n\nH.\n\n## Titration Curves for Weak Bases: There are 4 areas to consider\n\n1) 2) 3) 4)\n\nbefore the addition of acid before the equivalence point (buffer region 1) at the equivalence point after the equivalence point\n\n9-6\n\n[OH ] = K b C b\n\n## Moles of [OH ] pH = pK a + log Moles of [H 3O + ]\n\n[ H 3O + ] = K a C a\n\n## Moles of [H 3O ]added Moles of [OH ] [ H 3O ] = Vtotal\n\n+\n\nI. The Effect of Concentration: Again, as with the titration of strong acids with strong bases, as the concentration of the weak acid or base becomes more dilute, the relative change in pH at the equivalence point decreases making the endpoint less sharp.\n\nA: 0.1000 M acid with 0.1000 M base B: 0.001000 M acid with 0.001000 M base\n\nI. The Effect of Reaction Completeness: The smaller the Ka, the less sharp the endpoint when a weak acid is titrated with a strong base (Figure 10-11). This depends also on concentration, so that weaker acids can be titrated if concentrated solutions are used.\n\nAny Questions?\n\nSummary\nHow to select standard solutions and indicators for neutralization titrations Theory of Indicator Behavior Titration Curves Theories: Plots and shapes of plots\nParameters: pH, Volume of titrants, equivalence point Calculations: Ka, pH, indicator choices, half-neutralized point Effects: Concentration, reaction, composition, temperature, equilibrium constants The titration of a strong acid with a strong base The titration of a week acid with a strong base The titration of a strong base with a strong acid The titration of a week base with a strong acid\n\nBuffer solutions:\nDefinition and properties (e.g., buffer capacity) Calculation of pH of the buffer solution Applications\n\nHomework\n9-B, D, E, 2, 5, 6, 22, 23, 27, 28, 33 11-A, B, F, 3, 6, 14\n\n## Before working on Homework,\n\nPractice with all examples that we discussed in the class and examples in the textbook!!" ]

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[ "ADDITION AND SUBTRACTION OF ALGEBRAIC EXPRESSIONS\n\nThe addition and subtraction of algebraic expressions are almost similar to the addition and subtraction of numbers.\n\nHow to add and subtract algebraic expressions ?\n\nWhen adding and subtracting algebraic expressions, we first categorize the terms into two types.\n\nThey are like terms and unlike terms.\n\nLike terms :\n\nThe terms whose variables and exponents are the same are known as like terms.\n\nThat means,\n\n3x2 and 4x2 are like terms and they can be added or subtracted from each other.\n\nWhen adding or subtracting like terms, the coefficients are added or subtracted and the variables remain unchanged.\n\nWe can only combine like terms by adding or subtracting them with one another.\n\nUnlike terms :\n\nThe terms having different variables are unlike terms.\n\nThat means,\n\n2x2 and 4y2 are unlike terms and they cannot be added or subtracted from each other.\n\nUnlike terms cannot be combined by adding or subtracting them with one another.\n\nProblem 1 :\n\nFind the sum of the following expressions.\n\n(i)  7p + 6q, 5p – q, q + 16p\n\n(ii)  a + 5b + 7c, 2a + 10b + 9c\n\n(iii)  mn + t, 2mn – 2t, -3t + 3mn\n\n(iv)  u + v, u – v, 2u + 5v, 2u – 5v\n\n(v)  5xyz – 3xy, 3zxy – 5yx\n\nSolution :\n\n(i)\n\nGiven, 7p + 6q, 5p – q, q + 16p\n\n= 7p + 6q + 5p – q + q + 16p\n\nBy combining like terms,\n\n=  (7p + 5p + 16p) + (6q – q + q)\n\n=  (7 + 5 + 16)p + (6 – 1 + 1)q\n\n=  28p + 6q\n\n(ii)\n\nGiven, a + 5b + 7c, 2a + 10b + 9c\n\n=  a + 5b+ 7c + 2a + 10b + 9c\n\nBy combining like terms,\n\n=  (a + 2a) + (5b + 10b) + (7c + 9c)\n\n=  (1 + 2)a + (5 + 10)b + (7 + 9)c\n\n=  3a + 15b + 16c\n\n(iii)\n\nGiven, mn + t, 2mn – 2t, -3t + 3mn\n\n= mn + t + 2mn – 2t - 3t + 3mn\n\nBy combining like terms,\n\n=  (mn + 2mn + 3mn) + (t – 2t – 3t)\n\n=  (1 + 2 + 3)mn + (1 – 2 – 3)t\n\n=  6mn – 4t\n\n(iv)\n\nGiven, u + v, u – v, 2u + 5v, 2u – 5v\n\n=  u + v + u – v + 2u + 5v + 2u – 5v\n\nBy combining like terms,\n\n=  (u + u + 2u + 2u) + (v – v + 5v – 5v)\n\n= (1 + 1 + 2 + 2)u + (1 – 1 + 5 – 5)v\n\n=  6u\n\n(v)\n\nGiven, 5xyz – 3xy, 3zxy – 5yx\n\n=  5xyz – 3xy + 3zxy – 5yx\n\nBy combining like terms,\n\n=  (5xyz + 3zxy) – (3xy + 5yx)\n\n=  (5 + 3)xyz – (3 + 5)xy\n\n=  8xyz – 8xy\n\nProblem 2 :\n\nSubtract :\n\n(i)  13x + 12y – 5 from 27x + 5y – 43\n\n(ii)  3p + 5 from p – 2q + 7\n\n(iii)  m + n from 3m – 7n\n\n(iv)  2y + z from 6z – 5y\n\nSolution :\n\n(i)\n\nGiven,  13x + 12y – 5 from 27x + 5y – 43\n\n=  (27x + 5y – 43) – (13x + 12y – 5)\n\n=  27x + 5y – 43 – 13x – 12y + 5\n\nBy combining like terms,\n\n=  (27x – 13x) + (5y – 12y) – (43 – 5)\n\n=  14x – 7y - 38\n\n(ii)\n\nGiven, 3p + 5 from p – 2q + 7\n\n=  (p – 2q + 7) - (3p + 5)\n\n=  p - 2q + 7 - 3p - 5\n\n=  -2p - 2q + 2\n\n(iii)\n\nGiven, m + n from 3m – 7n\n\n=  (3m – 7n) – (m + n)\n\n=  3m – 7n - m - n\n\nBy combining like terms,\n\n=  (3m – m) + (-7n - n)\n\n=  (3 – 1)m + (-7 - 1)n\n\n=  2m – 8n\n\n(iv)\n\nGiven, 2y + z from 6z – 5y\n\n=  (6z – 5y) – (2y + z)\n\n=  (6z – 5y) + (-2y - z)\n\n=  (6z - z) + (-5y – 2y)\n\n=  (6 - 1)z + (-5 – 2)y\n\n=  5z – 7y\n\nProblem 3 :\n\nSimplify :\n\n(i)  (x + y – z) + (3x – 5y + 7z) – (14x + 7y – 6z)\n\n(ii)  p + p + 2 + p + 3 – p – 4 – p – 5 + p + 10\n\n(iii)  n + (m + 1) + (n + 2) + (m + 3) + (n + 4) + (m + 5)\n\nSolution :\n\n(i)\n\nGiven, (x + y – z) + (3x – 5y + 7z) – (14x + 7y – 6z)\n\nBy combining like terms,\n\n=  (x + 3x – 14x) + (y – 5y – 7y) + (-z + 7z + 6z)\n\n=  (1 + 3 – 14)x + (1 – 5 – 7)y + (-1 + 7 + 6)z\n\n=  -10x – 11y + 12z\n\n(ii)\n\nGiven, p + p + 2 + p + 3 – p – 4 – p – 5 + p + 10\n\nBy combining like terms,\n\n=  (p + p + p – p – p + p) + (2 + 3 – 4 – 5 + 10)\n\n=  (1 + 1 + 1 – 1 – 1 + 1)p + (6)\n\n=  2p + 6\n\n(iii)\n\nGiven, n + (m + 1) + (n + 2) + (m + 3) + (n + 4) + (m + 5)\n\nBy combining like terms,\n\n=  (m + 1) + (m + 3) + (m + 5) + n + (n + 2) + (n + 4)\n\n=  (m + m + m + 1 + 3 + 5 + n + n + n + 2 + 4)\n\n=   3m + 3n + 15", null, "Apart from the stuff given above if you need any other stuff in math, please use our google custom search here.\n\nKindly mail your feedback to [email protected]\n\nRecent Articles", null, "1. Multiplicative Inverse Worksheet\n\nJan 19, 22 09:34 AM\n\nMultiplicative Inverse Worksheet\n\n2. Multiplicative Inverse\n\nJan 19, 22 09:25 AM\n\nMultiplicative Inverse - Concept - Examples" ]

[ null, "https://www.onlinemath4all.com/images/onlinemath4all1.png.pagespeed.ce.ae60VyIxFs.png", null, "https://www.onlinemath4all.com/objects/xrss.png.pagespeed.ic.nVmUyyfqP7.png", null ]

[ "# physics\n\nan ice skater starts out traveling with a velocity of (-3,-7) m/s. he performs a 3 second maneuver and ends with a velocity of (0, 5) m/s. (a) what is his average acceleration over this period? (b) a different ice skater starts with the same initial velocity, accelerates at (1.5, 3.5) m/s2 for 2 seconds, and then at (0, 5 m.s2 for 1 seconds. what is his final velocity?\n\n(a)\n(b)\n\n1. 👍 0\n2. 👎 0\n3. 👁 206\n1. (a) The components of the average acceleration and the changes in the velocity components,(3, 12) divided by the time interval. You do the numbers\n\n(b) Add the two different velocity vector changes during the two time intervals PLUS the initial velocity vector.\n\nYou have only specified one component of the acceleration during the last 1 second. Perhaps you did not copy the question correctly.\n\n1. 👍 0\n2. 👎 0\n\n## Similar Questions\n\n1. ### Physics\n\nan ice skater with a mass of 80kg pushes off against a second skater with a mass of 32kg. both skaters are initially at rest. a. what is the total momentum of the system after the push off? b. if the larger skater moves off with a\n\nasked by Titi on November 10, 2009\n2. ### physics\n\nSkater 1 has a mass of 45 kg and is at rest. Skater 2 has a mass of 50 kg and is moving slowly at a constant velocity of 3.2 m/s [E]. Skater 3 has a mass of 75 kg and is moving quickly at a constant velocity of 9.6 m/s [E]. Which\n\nasked by shan on April 16, 2012\n3. ### physics\n\ntwo ice skaters initially at rest push each other.if one skater whose mass is 60kg has a velocity of 2m/s.find the velocity of other skater whose mass is 40kg\n\nasked by kavya on August 3, 2014\n4. ### Physics\n\na 65-kg ice skater moving to the right with a velocity of 2.50m/s throws a .150kg snowball to the right with a velocity of 32.0m/s relative to the ground a) what is the velocity of the ice skater after throwing the snowball? b) a\n\nasked by Sarah on April 26, 2012\n1. ### physics\n\nTwo fi gure skaters are moving east together during a performance. Skater 1 (78 kg) is behind skater 2 (56 kg) when skater 2 pushes on skater 1 with a force of 64 N [W]. Assume that no friction acts on either skater. T / I (a)\n\nasked by shan on April 16, 2012\n2. ### physics\n\nAn ice skater is traveling in a straight, horizontal line on the ice with a velocity of v = 6.5 m/s in the positive x-direction. The coefficient of kinetic friction between the skates and the ice is μk = 0.23. v = 6.5 m/s μk =\n\nasked by TJ on October 9, 2014\n3. ### 12th Physics\n\nA 55 kg ice skater is at rest on a flat skating rink. A 198 N horizontal forceis needed to set the skater in motion.However, after the skater is inmotion, a horizontal force of 175 N keeps the skater moving at a constantvelocity.\n\nasked by Melissa on December 8, 2009\n4. ### Physics\n\nA 74.5 kg ice skater moving to the right with a velocity of 2.62 m/s throws a 0.21 kg snowball to the right with a velocity of 23.8 m/s relative to the ground. (a) What is the velocity of the ice skater after throwing the\n\nasked by Elizabeth on April 24, 2011\n1. ### Physic HELP!!!!\n\nAn ice skater moving at 10.0 m/s coast to a halt in 1.0 x 102 m on a smooth ice surface. What is the coefficient of friction between the ice and the skates?\n\nasked by @-@ on October 26, 2008\n2. ### physics\n\nA 67.3 kg ice skater moving to the right with a velocity of 2.52 m/s throws a 0.151 kg snowball to the right with a velocity of 33.7 m/s relative to the ground. What is the velocity of the ice skater after throwing the snowball?\n\nasked by Emily on December 2, 2012\n3. ### Physics\n\nIn an ice show a 40.0 kg skater leaps into the air and is caught by an initially stationary 65.0 kg skater. (a) What is their final velocity assuming negligible friction and that the leaper's original horizontal velocity was 4.00\n\nasked by Amy on July 14, 2013\n4. ### college\n\nAt the center of a 50 m diameter circular ice rink, a 75 kg skater traveling north at 2.5 m/s collides with and holds onto a 60 kg skater who had been heading west at 3.5 m/s. How long will it take them to get to the edge of the\n\nasked by charlotte on December 1, 2010" ]

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[ "Semilinear parabolic equation in ${𝐑}^{N}$ associated with critical Sobolev exponent\nAnnales de l'I.H.P. Analyse non linéaire, Volume 27 (2010) no. 3, p. 877-900\n\nWe consider the semilinear parabolic equation ${u}_{t}-\\Delta u={|u|}^{p-1}u$ on the whole space ${𝐑}^{N}$, $N⩾3$, where the exponent $p=\\left(N+2\\right)/\\left(N-2\\right)$ is associated with the Sobolev imbedding ${H}^{1}\\left({𝐑}^{N}\\right)\\subset {L}^{p+1}\\left({𝐑}^{N}\\right)$. First, we study the decay and blow-up of the solution by means of the potential-well and forward self-similar transformation. Then, we discuss blow-up in infinite time and classify the orbit.\n\nDOI : https://doi.org/10.1016/j.anihpc.2010.01.002\nClassification:  35K55\nKeywords: Parabolic equation, Critical Sobolev exponent, Cauchy problem, Stable and unstable sets, Self-similarity\n@article{AIHPC_2010__27_3_877_0,\nauthor = {Ikehata, Ryo and Ishiwata, Michinori and Suzuki, Takashi},\ntitle = {Semilinear parabolic equation in ${\\mathbf{R}}^{N}$ associated with critical Sobolev exponent},\njournal = {Annales de l'I.H.P. Analyse non lin\\'eaire},\npublisher = {Elsevier},\nvolume = {27},\nnumber = {3},\nyear = {2010},\npages = {877-900},\ndoi = {10.1016/j.anihpc.2010.01.002},\nzbl = {1192.35099},\nmrnumber = {2629884},\nlanguage = {en},\nurl = {http://www.numdam.org/item/AIHPC_2010__27_3_877_0}\n}\n\nIkehata, Ryo; Ishiwata, Michinori; Suzuki, Takashi. Semilinear parabolic equation in ${\\mathbf{R}}^{N}$ associated with critical Sobolev exponent. Annales de l'I.H.P. Analyse non linéaire, Volume 27 (2010) no. 3, pp. 877-900. doi : 10.1016/j.anihpc.2010.01.002. http://www.numdam.org/item/AIHPC_2010__27_3_877_0/\n\n L.A. Caffarelli, B. Gidas, J. Spruck, Asymptotic symmetry and local behavior of semilinear elliptic equations with critical Sobolev growth, Comm. Pure Appl. Math. 42 (1989), 271-297 | MR 982351 | Zbl 0702.35085\n\n T. Cazenave, P.L. Lions, Solutions globales d'equations de la shaleur semi lineaires, Comm. Partial Differential Equations 9 (1984), 955-978 | MR 755928 | Zbl 0555.35067\n\n W. Chen, C. Li, Classification of solutions of some nonlinear elliptic equations, Duke Math. J. 63 (1991), 615-622 | MR 1121147 | Zbl 0768.35025\n\n E.B. Davis, Heat Kernels and Spectral Theory, Cambridge University Press, Cambridge (1989) | MR 990239 | Zbl 0699.35006\n\n M. Escobedo, O. Kavian, Variational problems related to self-similar solutions of the heat equation, Nonlinear Anal. 11 (1987), 1103-1133 | MR 913672 | Zbl 0639.35038\n\n M. Escobedo, E. Zuazua, Large time behavior for convection–diffusion equations in ${R}^{N}$, J. Funct. Anal. 100 (1991), 119-161 | MR 1124296 | Zbl 0762.35011\n\n V. Georgiev, Semilinear Hyperbolic Equations, MSJ Mem. vol. 7, Math. Soc. Japan (2000) | MR 1807081 | Zbl 0959.35002\n\n R. Ikehata, The Palais–Smale condition for the energy of some semilinear parabolic equations, Hiroshima Math. J. 30 (2000), 117-127 | MR 1753386 | Zbl 0953.35067\n\n R. Ikehata, T. Suzuki, Semilinear parabolic equations involving critical Sobolev exponent: Local and asymptotic behavior of solutions, Differential Integral Equations 13 (2000), 437-477 | MR 1775238 | Zbl 1016.35005\n\n K. Ishige, T. Kawakami, Asymptotic behavior of solutions for some semilinear heat equations in ${𝐑}^{N}$, preprint | MR 2505375\n\n M. Ishiwata, Existence of a stable set for some nonlinear parabolic equation involving critical Sobolev exponent, Discrete Contin. Dyn. Syst. (2005), 443-452 | MR 2192702 | Zbl 1173.35344\n\n M. Ishiwata, On the asymptotic behavior of radial positive solutions for semilinear parabolic problem involving critical Sobolev exponent, in preparation\n\n T. Kato, Perturbation Theory for Linear Operators, Springer-Verlag, New York (1976) | MR 407617\n\n O. Kavian, Remarks on the large time behaviour of a nonlinear diffusion equation, Ann. Inst. H. Poincaré 4 (1987), 423-452 | MR 921547 | Zbl 0653.35036\n\n T. Kawanago, Asymptotic behavior of solutions of a semilinear heat equation with subcritical nonlinearity, Ann. Inst. H. Poincaré 13 (1996), 1-15 | Numdam | MR 1373470 | Zbl 0847.35060\n\n T. Kawanago, Existence and behavior of solutions for ${u}_{t}=\\Delta {u}^{m}+{u}^{\\ell }$, Adv. Math. Sci. Appl. 7 (1997), 367-400 | MR 1454672 | Zbl 0876.35061\n\n V. Komornik, Exact Controllability and Stabilization, Multiplier Method, Masson, Paris (1994) | MR 1359765 | Zbl 0937.93003\n\n M. Otani, Existence and asymptotic stability of strong solutions of nonlinear evolution equations with a difference term of subdifferentials, Colloq. Math. Soc. Janos Bolyai, Qualitative Theory of Differential Equations, vol. 30, North-Holland, Amsterdam (1980) | Zbl 0506.35075\n\n M. Otani, ${L}^{\\infty }$-energy method and its applications, nonlinear partial differential equations and their applications, GAKUTO Internat. Ser. Math. Sci. Appl. 20 (2004), 505-516 | MR 2087494 | Zbl 1061.35035\n\n L.E. Payne, D.H. Sattinger, Saddle points and unstability of nonlinear hyperbolic equations, Israel J. Math. 22 (1975), 273-303 | MR 402291 | Zbl 0317.35059\n\n P. Poláčik, private communication\n\n D.H. Sattinger, On global solution of nonlinear hyperbolic equations, Arch. Ration. Mech. Anal. 30 (1968), 148-172 | MR 227616 | Zbl 0159.39102\n\n T. Suzuki, Semilinear parabolic equation on bounded domain with critical Sobolev exponent, Indiana Univ. Math. J. 57 no. 7 (2008), 3365-3396 | MR 2492236 | Zbl 1201.35116\n\n H. Tanabe, Equations of Evolution, Pitman, London (1979) | MR 533824" ]

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[ "## Sunday, March 1, 2009\n\n### Three Dimensional Poisson's Equation\n\nThis is a follow up to the post I wrote a while back about implementing numerical methods. The goal is to demonstrate a work-flow using tools that come included in the Fedora Linux (and many other) distributions.\n1. In Maxima\n• Define governing equations\n• Substitute finite difference", null, "expressions for differential expressions in the governing equations\n• Output appropriate difference expressions to Fortran", null, "using f90()\n2. In Fortran: wrap the expression generated by Maxima in appropriate functions, subroutines and modules\n\n3. In Python\n• Compile modules with f2py, which comes packaged along with numpy\n• Write the high-level application code, parsing input decks, performing optimization, grid convergence studies, or visualizations that use the compiled Fortran modules\n\n### Maxima\n\nThe governing equation is the three-dimensional Poisson's equation.", null, "In Cartesian coordinates", null, "The Maxima code to define this equation is straightforward:\ndepends(u,x); depends(u,y); depends(u,z); /* Poisson's equation: */ p : 'diff(u,x,2) + 'diff(u,y,2) + 'diff(u,z,2) = -f;\nNow that we have the continuous differential equation defined we can decide on what sort of discrete approximation we are going to solve. The simple thing to do with second derivatives is to use central differences of second order accuracy, which require only information from closest neighbours in the finite difference grid. Replacing the differential expressions with difference expressions is accomplished in Maxima with the ratsubst() function.\n/* substitue difference expressions for differential ones: */\np : ratsubst((u[i-1,j,k] - 2*u[i,j,k] + u[i+1,j,k])/dx**2 , 'diff(u,x,2), p);\np : ratsubst((u[i,j-1,k] - 2*u[i,j,k] + u[i,j+1,k])/dy**2 , 'diff(u,y,2), p);\np : ratsubst((u[i,j,k-1] - 2*u[i,j,k] + u[i,j,k+1])/dz**2 , 'diff(u,z,2), p);\n/* substitute the correct array value of f:*/\np : ratsubst(f[i,j,k], f, p);\nThis assumes that the solution u and the forcing function f are stored in three dimensional arrays, if not then the indexing in the difference expressions becomes more complicated. If we want to apply a Gauss-Seidel method to solve our elliptic problem we just need to solve p for u[i,j,k], and then dump that expression to Fortran format.\ngs : solve(p, u[i,j,k]);\ndel : part(gs,2) - u[i,j,k];\n/* output for fortran: */\nwith_stdout(\"gs.f90\", f90(d = expand(del)));\nThis gives the expression for the update to the solution at each iteration of the solver.\n\n### Fortran\n\nThe Maxima expressions developed above need control flow added so they get applied properly to our solution array.\ndo k = 1, nz\ndo j = 1, ny\ndo i = 1, nx\nd = (dy**2*dz**2*u(i+1,j,k)+dx**2*dz**2*u(i,j+1,k)+dx**2*dy**2* & u(i,j,k+1)+dx**2*dy**2*dz**2*f(i,j,k)+dx**2*dy**2* & u(i,j,k-1)+dx**2*dz**2*u(i,j-1,k)+dy**2*dz**2*u(i-1,j,k))/ & ((2*dy**2+2*dx**2)*dz**2+2*dx**2*dy**2)-u(i,j,k) u(i,j,k) = u(i,j,k) + w*d\nend do\nend do\nend do\nThe expression for the update was generated from our governing equations in Maxima, one of the things to notice is the w that multiplies our update, this is so we can do successive over relaxation. Also, we'll go ahead and package several of these iterative schemes into a Fortran module\nmodule iter_schemes\nimplicit none\ncontains\n\n...bunches of related subroutines and functions ...\n\nend module iter_schemes\n\n### Python\n\nNow we want to be able to call our Fortran solution schemes from Python. Using F2py makes this quite simple:\n[command-line]\\$ f2py -m iter_schemes -c iter_schemes.f90\nThis should result in a python module iter_schemes.so that can be imported just as any other module.\nimport numpy\nfrom iter_schemes import *\n\n... do some cool stuff with our new module in Python ...\n\n### Conclusion\n\nThis approach might seem like over-kill for the fairly simple scalar equation we used, but think about the complicated update expressions that can be generated for large vector partial differential equations like the Navier-Stokes", null, "or Maxwell's equations", null, ". Having these expressions automatically generated and ready to be wrapped in loops can save lots of initial development and subsequent debugging time. It also makes generating unit tests easier, and so encourages good development practice", null, ". But does it work? Of course, here's the convergence of the Gauss-Seidel and SOR schemes discussed above along with a Jacobi scheme and a conjugate gradient scheme applied to a 50 by 50 by 50 grid with a manufactured solution", null, ".", null, "The plot was generated with the Python module matplotlib.\n\n1.", null, "2.", null, "Dear jstults,\n\nHi,\n\nHope all is fine with you.\n\nIn the first I'm very thankfull for your really valuable materials about solving Poisson's equation in three dimensions, numeerically.\n\nWould you mind Mail me via [email protected] for more discussions, please?\n\nBests,\n\n3.", null, "Just post your question; that way other folks can benefit from the discussion. There's probably plenty of folks who have the same or similar questions.\n\n2-D simplifies things a bit so they are good example problems to talk about.\n\nFire away...\n\n4.", null, "What sort of discretization did you use? What does the discrete system matrix (or expression) that you came up with look like?\n\nThe Thomas algorithm (Gaussian elimination for a tridiagonal system) only works for a 1-D problem. You'll have to use relaxation or some other iterative method; or if the problem is small enough just get Matlab to solve the system for you (this is easiest when you are learning).\n\nOnce you have inverted the small problem directly with Matlab, then you can check that your iterative code is giving (roughly) the same answer.\n\nThe other tip I would give you is to break up your solution into components that can be tested individually.\n\nWrite a routine that calculates your forcing function, and test that it is operating correctly.\n\nWrite a routine that calculates the derivatives, and test that it is operating correctly.\n\nWrite a routine that takes the forcing function and derivatives and performs your chosen iterative update, and test that it is operating correctly. For example, you can check to see that it converges to the \"direct\" solution that Matlab gave you.\n\nIf you follow this sort of methodical approach you should be able to locate which part of your solution method (derivative calculation, forcing function, iterative update) is incorrect.\n\n5.", null, "I don't have time to debug your code. If you break the solution up into chunks and unit test those chunks, then you probably won't need me to look at your code because you will have located the problem yourself.\n\nI'd be happy to help once you identify a specific \"part\" that is broken, and you aren't sure how to fix it. Finding the broken part is a useful experience for you, and I wouldn't want to rob you of it.\n\n6.", null, "It is simply a collaboration if you are willing.\n\nWe can help and write a paper (ISI), including Numerical parts and also analytical.\nIn effect it seems that solving of this PDE and relevanted initial conditions is too hard for me. I tried but couldn't....\n\nPlease let me know, what do you think about collaboration in this project?\n\n7.", null, "Thank you, I appreciate your offer, but I don't think this lines up with my current research interests.\n\nI think it is good to get these posts up though, perhaps one of the other readers will be interested (this is the most read post on my site, ~150 views/month), and they can contact you through your profile.\n\n8.", null, "O.K. Grate.\n\nReally thanks for your interesting discussion. I just wait for a professional person in this Field.\n9.", null, "10.", null, "" ]

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[ "", null, "Mathematics Magazine Home Online Math Tests Math Book Subscribe Contact Us", null, "Advertise Here. E-mail us your request for an advertising quote!\n\nToroid\n\nby Liliana Usvat\n\nIn mathematics, a toroid is a doughnut-shaped object, such as an O-ring. It is a ring form of a solenoid.\n\nA toroid is used as an inductor in electronic circuits, especially at low frequencies where comparatively large inductances are necessary.\n\nTorus\n\nIn geometry, a torus (pl. tori) is a surface of revolution generated by revolving a circle in three-dimensional space about an axis coplanar with the circle. If the axis of revolution does not touch the circle, the surface has a ring shape and is called a ring torus or simply torus if the ring shape is implicit.\n\nA torus can be defined parametrically by:", null, "", null, "", null, "where\n\nθ, φ are angles which make a full circle, starting at 0 and ending at 2π, so that their values start and end at the same point,\nR is the distance from the center of the tube to the center of the torus,\nr is the radius of the tube.\n\nR and r are also known as the \"major radius\" and \"minor radius\", respectively. The ratio of the two is known as the \"aspect ratio\". A doughnut has an aspect ratio of about 2 to 3.\n\nAn implicit equation in Cartesian coordinates for a torus radially symmetric about the z-axis is", null, "or the solution of f(x, y, z) = 0, where", null, "Algebraically eliminating the square root gives a quartic equation,", null, "The three different classes of standard tori correspond to the three possible relative sizes of r and R. When R > r, the surface will be the familiar ring torus. The case R = r corresponds to the horn torus, which in effect is a torus with no \"hole\". The case R < r describes the self-intersecting spindle torus. When R = 0, the torus degenerates to the sphere.\n\nWhen R ≥ r, the interior", null, "of this torus is diffeomorphic (and, hence, homeomorphic) to a product of an Euclidean open disc and a circle.\n\nThe surface area and interior volume of this torus are easily computed using Pappus's centroid theorem giving", null, "", null, "These formulas are the same as for a cylinder of length 2πR and radius r, created by cutting the tube and unrolling it by straightening out the line running around the center of the tube. The losses in surface area and volume on the inner side of the tube exactly cancel out the gains on the outer side.\n\nAs a torus is the product of two circles, a modified version of the spherical coordinate system is sometimes used. In traditional spherical coordinates there are three measures, R, the distance from the center of the coordinate system, and θ and φ, angles measured from the center point. As a torus has, effectively, two center points, the centerpoints of the angles are moved; φ measures the same angle as it does in the spherical system, but is known as the \"toroidal\" direction. The center point of θ is moved to the center of r, and is known as the \"poloidal\" direction. These terms were first used in a discussion of the Earth's magnetic field, where \"poloidal\" was used to denote \"the direction toward the poles\".In modern use these terms are more commonly used to discuss magnetic confinement fusion devices.\n\n## n-dimensional torus\n\nThe torus has a generalization to higher dimensions, the n-dimensional torus, often called the n-torus or hypertorus for short. (This is one of two different meanings of the term \"n-torus\".) Recalling that the torus is the product space of two circles, the n-dimensional torus is the product of n circles. That is:", null, "The 1-torus is just the circle: T1 = S1. The torus discussed above is the 2-torus, T2. And similar to the 2-torus, the n-torus, Tn can be described as a quotient of Rn under integral shifts in any coordinate. That is, the n-torus is Rn modulo the action of the integer lattice Zn (with the action being taken as vector addition). Equivalently, the n-torus is obtained from the n-dimensional hypercube by gluing the opposite faces together.\n\nAn n-torus in this sense is an example of an n-dimensional compact manifold. It is also an example of a compact abelian Lie group. This follows from the fact that the unit circle is a compact abelian Lie group (when identified with the unit complex numbers with multiplication). Group multiplication on the torus is then defined by coordinate-wise multiplication.\n\nToroidal groups play an important part in the theory of compact Lie groups. This is due in part to the fact that in any compact Lie group G one can always find a maximal torus; that is, a closed subgroup which is a torus of the largest possible dimension. Such maximal tori T have a controlling role to play in theory of connected G. Toroidal groups are examples of protori, which (like tori) are compact connected abelian groups, which are not required to be manifolds.\n\n# Magnetic Field of Toroid", null, "Finding the magnetic field inside a toroid is a good example of the power of Ampere's law. The current enclosed by the dashed line is just the number of loops times the current in each loop. Amperes law then gives the magnetic field by", null, "The toroid is a useful device used in everything from tape heads to tokamaks.\n\n### Free energy\n\nThere are dedicated independent scientists around the world that claim that we can generate unlimited clean energy by just tapping into the ‘torus’, a shape that supposedly pervades the universe, and which could yield endless free energy.\n\nToroid and Nature\n\nThe Sun has a large toroidal field surrounding it — the heliosphere — that is itself embedded inside a vastly larger toroidal field encompassing the Milky Way galaxy. Our Earth’s magnetic field is surrounding us and is inside the Sun’s field, buffering us from the direct impact of solar electromagnetic radiation. Earth’s atmosphere and ocean dynamics are toroidal and are influenced by the surrounding magnetic field. Ecosystems, plants, animals, etc all exhibit torus flow dynamics and reside within and are directly influenced by (and directly influence) the Earth’s atmospheric and oceanic systems. And on it goes inward into the ecosystems and organs of our bodies, the cells they’re made of, and the molecules, atoms and sub-atomic particles.\n\nContinuing our exploration of the torus as a form and flow process, one of the key characteristics of it is that at its very center, the entire system comes to a point of ultimate balance and stillness — in other words, perfect centeredness.\n\nPhylosophy\n\nThe torus is the oldest structure in existence and without it nothing could exist. The toroidal shape is similar to a donut but rather than having an empty central “hole”, the topology of a torus folds in upon itself and all points along its surface converge together into a zero-dimensional point at the center called the Vertex.\n\nIt has even been suggested that the torus can be used to define the workings of consciousness itself.  In other words…consciousness has a geometry! The geometric shape used to describe the self-reflexive nature of consciousness is the torus. The torus allows a vortex of energy to form which bends back along itself and re-enters itself. It ‘inside-outs’, continuously flowing back into itself. Thus the energy of a torus is continually refreshing itself, continually influencing itself.\n\nAll Toroids have a black hole at one end and a white hole at the other. Black holes suck in energy and white holes emit it.  So in our human body toroids we have black (negatively charged) and white (positively charged) holes.\n\nWhen the torus is in balance and the energy is flowing we are in a perfect state to clear ourselves of anything that is ‘not self’ anything that prevents us being our authentic selves.\n\nThe infinity symbol an ancient two dimensional representation of the 3D double toroidal energy flow – self generating, continual, never ending.\n\nTheorists in astrophysics now argue that each electron holds at its core “zero point energy,” as does the universe at large.\n\nZero point energy is a place where there is no sound or light. This nothingness, issuing the essence of everything, exists at the heart of creation. It is the place wherein miraculous manifestation from nothing to something happens.\n\n \"Chance favors the prepared mind.\" - Louis Pasteur" ]

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[ "# Triangle calculator SAS\n\nPlease enter two sides of the triangle and the included angle\n°\n\n### Obtuse scalene triangle.\n\nSides: a = 49   b = 31   c = 60.65993560088\n\nArea: T = 755.3399379532\nPerimeter: p = 140.6599356009\nSemiperimeter: s = 70.33296780044\n\nAngle ∠ A = α = 53.45325997546° = 53°27'9″ = 0.93329238595 rad\nAngle ∠ B = β = 30.54774002454° = 30°32'51″ = 0.53331527122 rad\nAngle ∠ C = γ = 96° = 1.67655160819 rad\n\nHeight: ha = 30.83301787564\nHeight: hb = 48.7321572873\nHeight: hc = 24.90442993276\n\nMedian: ma = 41.47332291449\nMedian: mb = 52.91552977475\nMedian: mc = 27.58882335815\n\nInradius: r = 10.74399806307\nCircumradius: R = 30.49767423499\n\nVertex coordinates: A[60.65993560088; 0] B[0; 0] C[42.19992402183; 24.90442993276]\nCentroid: CG[34.28661987424; 8.30114331092]\nCoordinates of the circumscribed circle: U[30.33296780044; -3.18877776125]\nCoordinates of the inscribed circle: I[39.33296780044; 10.74399806307]\n\nExterior(or external, outer) angles of the triangle:\n∠ A' = α' = 126.5477400245° = 126°32'51″ = 0.93329238595 rad\n∠ B' = β' = 149.4532599755° = 149°27'9″ = 0.53331527122 rad\n∠ C' = γ' = 84° = 1.67655160819 rad\n\n# How did we calculate this triangle?\n\n### 1. Calculation of the third side c of the triangle using a Law of Cosines", null, "Now we know the lengths of all three sides of the triangle and the triangle is uniquely determined. Next we calculate another its characteristics - same procedure as calculation of the triangle from the known three sides SSS.", null, "### 2. The triangle circumference is the sum of the lengths of its three sides", null, "### 3. Semiperimeter of the triangle", null, "### 4. The triangle area using Heron's formula", null, "### 5. Calculate the heights of the triangle from its area.", null, "### 6. Calculation of the inner angles of the triangle using a Law of Cosines", null, "", null, "", null, "", null, "" ]

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[ "", null, "Address 5519 53rd St, Lubbock, TX 79414 (806) 791-8200 http://www.tylertech.com\n\n# calculate probability of type i error Dickens, Texas\n\nOne cannot evaluate the probability of a type II error when the alternative hypothesis is of the form µ > 180, but often the alternative hypothesis is a competing hypothesis of A medical researcher wants to compare the effectiveness of two medications. Trying to avoid the issue by always choosing the same significance level is itself a value judgment. Remarks If there is a diagnostic value demarcating the choice of two means, moving it to decrease type I error will increase type II error (and vice-versa).\n\nwhat fraction of the population are predisposed and diagnosed as healthy? The probability of rejecting the null hypothesis when it is false is equal to 1–β. Compute the probability of committing a type I error. Thank you,,for signing up!\n\nIn other words, the probability of Type I error is α.1 Rephrasing using the definition of Type I error: The significance level αis the probability of making the wrong decision when They are also each equally affordable. The t statistic for the average ERA before and after is approximately .95. What is the probability that a randomly chosen genuine coin weighs more than 475 grains?\n\nProbabilities of type I and II error refer to the conditional probabilities. I hope you be so nice to tell me what I did wrong for b. $$\\frac{1.9^2}{2}-\\frac{0.1^2}{2} = \\frac{9}{5}$$ –Danique Jun 23 '15 at 17:44 @Danique In b ConclusionThe calculated p-value of .35153 is the probability of committing a Type I Error (chance of getting it wrong). Where y with a small bar over the top (read \"y bar\") is the average for each dataset, Sp is the pooled standard deviation, n1 and n2 are the sample sizes\n\nType I error When the null hypothesis is true and you reject it, you make a type I error. So setting a large significance level is appropriate. For P(D|B) we calculate the z-score (225-300)/30 = -2.5, the relevant tail area is .9938 for the heavier people; .9938 × .1 = .09938. Another good reason for reporting p-values is that different people may have different standards of evidence; see the section\"Deciding what significance level to use\" on this page. 3.\n\nBecause the applet uses the z-score rather than the raw data, it may be confusing to you. Let A designate healthy, B designate predisposed, C designate cholesterol level below 225, D designate cholesterol level above 225. Suppose that the standard deviation of the population of all such bags of chips is 0.6 ounces. Alternative hypothesis (H1): μ1≠ μ2 The two medications are not equally effective.\n\nIn this classic case, the two possibilities are the defendant is not guilty (innocent of the crime) or the defendant is guilty. Most people would not consider the improvement practically significant. If the cholesterol level of healthy men is normally distributed with a mean of 180 and a standard deviation of 20, at what level (in excess of 180) should men be A t-Test provides the probability of making a Type I error (getting it wrong).\n\nHow do I debug an emoticon-based URL? The table below has all four possibilities. When the null hypothesis states µ1= µ2, it is a statistical way of stating that the averages of dataset 1 and dataset 2 are the same. We get a sample mean that is way out here.\n\nProbabilities of type I and II error refer to the conditional probabilities. Type II error A type II error occurs when one rejects the alternative hypothesis (fails to reject the null hypothesis) when the alternative hypothesis is true. Find Iteration of Day of Week in Month Safety of using images found through Google image search What does Billy Beane mean by \"Yankees are paying half your salary\"? The following examines an example of a hypothesis test, and calculates the probability of type I and type II errors.We will assume that the simple conditions hold.\n\nSometimes there may be serious consequences of each alternative, so some compromises or weighing priorities may be necessary. One decides to test H0 : θ = 2 against H1 : θ = 2 by rejecting H0 if x ≤0.1 or x ≥ 1.9. Then we have some statistic and we're seeing if the null hypothesis is true, what is the probability of getting that statistic, or getting a result that extreme or more extreme In this case there would be much more evidence that this average ERA changed in the before and after years.\n\nThe probability of a type II error is denoted by *beta*. Sorry for being not clear. There is always a possibility of a Type I error; the sample in the study might have been one of the small percentage of samples giving an unusually extreme test statistic. Formula: Example : Suppose the mean weight of King Penguins found in an Antarctic colony last year was 5.2 kg.\n\nI should note one very important concept that many experimenters do incorrectly. This is an instance of the common mistake of expecting too much certainty. P(C|B) = .0062, the probability of a type II error calculated above. See Sample size calculations to plan an experiment, GraphPad.com, for more examples.\n\nThe null hypothesis is \"defendant is not guilty;\" the alternate is \"defendant is guilty.\"4 A Type I error would correspond to convicting an innocent person; a Type II error would correspond However, there is some suspicion that Drug 2 causes a serious side-effect in some patients, whereas Drug 1 has been used for decades with no reports of the side effect. Created by Sal Khan.ShareTweetEmailThe idea of significance testsSimple hypothesis testingIdea behind hypothesis testingPractice: Simple hypothesis testingType 1 errorsNext tutorialTests about a population proportionTagsType 1 and type 2 errorsVideo transcriptI want to To have p-value less thanα , a t-value for this test must be to the right oftα.\n\nSo you should have $\\int_{0.1}^{1.9} \\frac{2}{5} dx = \\frac{3.6}{5}=0.72$. –Ian Jun 23 '15 at 17:46 Thanks! more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed C.K.Taylor By Courtney Taylor Statistics Expert Share Pin Tweet Submit Stumble Post Share By Courtney Taylor An important part of inferential statistics is hypothesis testing. How will the z-buffers have the same values even if polygons are sent in different order?\n\nIf the cholesterol level of healthy men is normally distributed with a mean of 180 and a standard deviation of 20, at what level (in excess of 180) should men be Last updated May 12, 2011 English Español Français Deutschland 中国 Português Pусский 日本語 Türk Sign in Calculators Tutorials Converters Unit Conversion Currency Conversion Answers Formulas Facts Code Dictionary Download Others Excel This is why replicating experiments (i.e., repeating the experiment with another sample) is important. The generally accepted position of society is that a Type I Error or putting an innocent person in jail is far worse than a Type II error or letting a guilty\n\nLooking at his data closely, you can see that in the before years his ERA varied from 1.02 to 4.78 which is a difference (or Range) of 3.76 (4.78 - 1.02 Statistics Statistics Help and Tutorials Statistics Formulas Probability Help & Tutorials Practice Problems Lesson Plans Classroom Activities Applications of Statistics Books, Software & Resources Careers Notable Statisticians Mathematical Statistics About Education Common mistake: Neglecting to think adequately about possible consequences of Type I and Type II errors (and deciding acceptable levels of Type I and II errors based on these consequences) before The trial analogy illustrates this well: Which is better or worse, imprisoning an innocent person or letting a guilty person go free?6 This is a value judgment; value judgments are often\n\nPlease enter a valid email address. Because if the null hypothesis is true there's a 0.5% chance that this could still happen. The greater the signal, the more likely there is a shift in the mean. Hence P(AD)=P(D|A)P(A)=.0122 × .9 = .0110.\n\nWhat is the probability that a randomly chosen coin weighs more than 475 grains and is counterfeit? The vertical red line shows the cut-off for rejection of the null hypothesis: the null hypothesis is rejected for values of the test statistic to the right of the red line" ]

[ null, "http://digitalhaunt.net/maps/Dickens_TX.png", null ]

[ "# The Great Mathematical Problems\n\n(US Title: Visions of Infinity: The Great Mathematical Problems)\n\nSynopsis: There are some mathematical problems whose significance goes beyond the ordinary – like Fermat’s Last Theorem or Goldbach’s Conjecture – they are the enigmas which define mathematics. The Great Mathematical Problems explains why these problems exist, why they matter, what drives mathematicians to incredible lengths to solve them and where they stand in the context of mathematics and science as a whole. It contains solved problems – like the Poincaré Conjecture, cracked by the eccentric genius Grigori Perelman, who refused academic honours and a million-dollar prize for his work, and ones which, like the Riemann Hypothesis, remain baffling after centuries.Stewart is the guide to this mysterious and exciting world, showing how modern mathematicians constantly rise to the challenges set by their predecessors, as the great mathematical problems of the past succumb to the new techniques and ideas of the present.\n\nPublished: March 2013 | ISBN-13: 978-1846681998\n\nAuthor’s Homepage: http://freespace.virgin.net/ianstewart.joat" ]

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[ "", null, "TypeScript 2.0 Beta 版已经发布一段时间了, 相比于 1.8 增加了很多新的特性. 本文译自 TypeScript Wiki, 由于官方新特性一览多数是直接修改自对应特性 issue 内容, 部分内容和之前的译文有重叠.\n\n### 编译器理解 null 和 undefined 类型\n\nTypeScript 有两个特殊的类型, Null 和 Undefined, 他们分别对应了值 null 和 undefined. 过去这些类型没有明确的名称, 但 null 和 undefined 现在可以在任意类型检查模式下作为类型名称使用.\n\n#### --strictNullChecks\n\n--strictNullChecks 选项会启用新的严格空值检查模式.\n\n``````// 使用 --strictNullChecks 选项编译\nlet x: number;\nlet y: number | undefined;\nlet z: number | null | undefined;\nx = 1; // 正确\ny = 1; // 正确\nz = 1; // 正确\nx = undefined; // 错误\ny = undefined; // 正确\nz = undefined; // 正确\nx = null; // 错误\ny = null; // 错误\nz = null; // 正确\nx = y; // 错误\nx = z; // 错误\ny = x; // 正确\ny = z; // 错误\nz = x; // 正确\nz = y; // 正确\n\n``````\n\n#### 使用前赋值检查\n\n``````// 使用 --strictNullChecks 选项编译\nlet x: number;\nlet y: number | null;\nlet z: number | undefined;\nx; // 错误, 引用使用前没有被赋值\ny; // 错误, 引用使用前没有被赋值\nz; // 正确\nx = 1;\ny = null;\nx; // 正确\ny; // 正确\n\n``````\n\n#### 可选参数和属性\n\n``````// 使用 --strictNullChecks 选项编译\ntype T1 = (x?: number) => string; // x 的类型为 number | undefined\ntype T2 = (x?: number | undefined) => string; // x 的类型为 number | undefined\n\n``````\n\n#### 非 null 和非 undefined 类型收窄\n\n``````// 使用 --strictNullChecks 选项编译\ndeclare function f(x: number): string;\nlet x: number | null | undefined;\nif (x) {\nf(x); // 正确, x 的类型在这里是 number\n}\nelse {\nf(x); // 错误, x 的类型在这里是 number?\n}\nlet a = x != null ? f(x) : \"\"; // a 的类型为 string\nlet b = x && f(x); // b 的类型为 string?\n\n``````\n\n#### 类型收窄中带点的名称\n\n``````interface Options {\nlocation?: {\nx?: number;\ny?: number;\n};\n}\n\nfunction foo(options?: Options) {\nif (options && options.location && options.location.x) {\nconst x = options.location.x; // x 的类型为 number\n}\n}\n\n``````\n\n#### 表达式运算符\n\n``````// 使用 --strictNullChecks 选项编译\nfunction sum(a: number | null, b: number | null) {\nreturn a + b; // 产生类型为 number 的值\n}\n\n``````\n\n&& 运算符会根据左边操作数的类型将 null 和/或 undefined 添加到右边被操作数的类型上, 而 || 运算符会在结果的联合类型中同时移除左边被操作数类型中的 null 和 undefined.\n\n``````// 使用 --strictNullChecks 选项编译\ninterface Entity {\nname: string;\n}\nlet x: Entity | null;\nlet s = x && x.name; // s 的类型为 string | null\nlet y = x || { name: \"test\" }; // y 的类型为 Entity\n\n``````\n\n#### 类型拓宽\n\nnull 和 undefined 类型在严格空值检查模式中不会被拓宽为 any.\n\n``````let z = null; // z 的类型为 null\n\n``````\n\n#### 非空断言运算符\n\n``````// 使用 --strictNullChecks 选项编译\nfunction validateEntity(e: Entity?) {\n// 当 e 是 null 或非法 entity 时抛出异常\n}\n\nfunction processEntity(e: Entity?) {\nvalidateEntity(e);\nlet s = e!.name; // 声明 e 非 null 并且访问 name\n}\n\n``````\n\n### 基于控制流的类型分析\n\nTypeScript 2.0 实现了对于本地变量和参数基于控制流的类型分析. 过去对类型收窄进行的分析仅限于 if 语句和 ?: 条件表达式, 并没有包含赋值和控制流结构带来的影响, 比如 return 和 break 语句. 在 TypeScript 2.0 中, 类型检查器会分析语句和表达式中所有可能的控制流, 以尽可能得出包含联合类型的本地变量及参数在指定位置最准确的类型 (收窄的类型).\n\n``````function foo(x: string | number | boolean) {\nif (typeof x === \"string\") {\nx; // x 的类型在这里是 string\nx = 1;\nx; // x 的类型在这里是 number\n}\nx; // x 的类型在这里是 number | boolean\n}\n\nfunction bar(x: string | number) {\nif (typeof x === \"number\") {\nreturn;\n}\nx; // x 的类型在这里是 string\n}\n\n``````\n\n``````function test(x: string | null) {\nif (x === null) {\nreturn;\n}\nx; // x 在这之后的类型为 string\n}\n\n``````\n\n``````function mumble(check: boolean) {\nlet x: number; // 类型不允许 undefined\nx; // 错误, x 是 undefined\nif (check) {\nx = 1;\nx; // 正确\n}\nx; // 错误, x 可能是 undefined\nx = 2;\nx; // 正确\n}\n\n``````\n\n### 带标记的联合类型\n\nTypeScript 2.0 实现了对带标记的 (或被区别的) 联合类型. 特别的, TS 编译器现在支持通过对可识别的属性进行判断来收窄联合类型, 并且支持 switch 语句.\n\n``````interface Square {\nkind: \"square\";\nsize: number;\n}\n\ninterface Rectangle {\nkind: \"rectangle\";\nwidth: number;\nheight: number;\n}\n\ninterface Circle {\nkind: \"circle\";\n}\n\ntype Shape = Square | Rectangle | Circle;\n\nfunction area(s: Shape) {\n// 在下面的语句中, s 的类型在每一个分支中都被收窄了\n// 根据可识别属性的值, 允许变量的其他属性在没有类型断言的情况下被访问\nswitch (s.kind) {\ncase \"square\": return s.size * s.size;\ncase \"rectangle\": return s.width * s.height;\n}\n}\n\nfunction test1(s: Shape) {\nif (s.kind === \"square\") {\ns; // Square\n}\nelse {\ns; // Rectangle | Circle\n}\n}\n\nfunction test2(s: Shape) {\nif (s.kind === \"square\" || s.kind === \"rectangle\") {\nreturn;\n}\ns; // Circle\n}\n\n``````\n\n### never 类型\n\nTypeScript 2.0 引入了新的原始类型 never. never 类型代表从来不会出现的值的类型. 特别的, never 可以是永远不返回的函数的返回值类型, 也可以是变量在类型收窄中不可能为真的类型.\n\nnever 类型有以下特征:\n\n• never 是任何类型的子类型, 并且可以赋值给任何类型.\n• 没有类型是 never 的子类型或者可以复制给 never (除了 never 本身).\n• 在一个没有返回值标注的函数表达式或箭头函数中, 如果函数没有 return 语句, 或者仅有表达式类型为 never 的 return 语句, 并且函数的终止点无法被执行到 (按照控制流分析), 则推导出的函数返回值类型是 never.\n• 在一个明确指定了 never 返回值类型的函数中, 所有 return 语句 (如果有) 表达式的值必须为 never 类型, 且函数不应能执行到终止点.\n\n``````// 返回 never 的函数必须有无法被执行到的终止点\nfunction error(message: string): never {\nthrow new Error(message);\n}\n\n// 推断的返回值是 never\nfunction fail() {\nreturn error(\"一些东西失败了\");\n}\n\n// 返回 never 的函数必须有无法被执行到的终止点\nfunction infiniteLoop(): never {\nwhile (true) {\n}\n}\n\n``````\n\n``````// 推断的返回值类型为 number\nfunction move1(direction: \"up\" | \"down\") {\nswitch (direction) {\ncase \"up\":\nreturn 1;\ncase \"down\":\nreturn -1;\n}\nreturn error(\"永远不应该到这里\");\n}\n\n// 推断的返回值类型为 number\nfunction move2(direction: \"up\" | \"down\") {\nreturn direction === \"up\" ? 1 :\ndirection === \"down\" ? -1 :\nerror(\"永远不应该到这里\");\n}\n\n// 推断的返回值类型为 T\nfunction check<T>(x: T | undefined) {\nreturn x || error(\"未定义的值\");\n}\n\n``````\n\n``````function test(cb: () => string) {\nlet s = cb();\nreturn s;\n}\n\ntest(() => \"hello\");\ntest(() => fail());\ntest(() => { throw new Error(); })\n\n``````\n\n### 只读属性和索引签名\n\n• 一个只有 get 访问符没有 set 访问符声明的属性被认为是只读的.\n• 在枚举对象的类型中, 枚举成员被认为是只读属性.\n• 在模块对象的类型中, 导出的 const 变量被认为是只读属性.\n• 在 import 语句中声明的实体被认为是只读的.\n• 通过 ES2015 命名空间导入来访问的实体被认为是只读的 (比如当 foo 为 import * as foo from \"foo\" 声明时, foo.x 是只读的).\n\n``````interface Point {\n}\n\nvar p1: Point = { x: 10, y: 20 };\np1.x = 5; // 错误, p1.x 是只读的\n\nvar p2 = { x: 1, y: 1 };\nvar p3: Point = p2; // 正确, p2 的只读别名\np3.x = 5; // 错误, p3.x 是只读的\np2.x = 5; // 正确, 但由于是别名也会改变 p3.x\n\n``````\n``````class Foo {\nconstructor() {\nthis.b = \"hello\"; // 在构造函数中允许赋值\n}\n}\n\n``````\n``````let a: Array<number> = [0, 1, 2, 3, 4];\nb = 5; // 错误, 元素是只读的\nb.push(5); // 错误, 没有 push 方法 (因为它会改变数组)\nb.length = 3; // 错误, length 是只读的\na = b; // 错误, 缺少会改变数组值的方法\n\n``````\n\n### 指定函数的 this 类型\n\n``````function f(this: void) {\n// 确保 `this` 在这个单独的函数中不可用\n}\n\n``````\n\n#### 回调函数中的 this 参数\n\n``````interface UIElement {\naddClickListener(onclick: (this: void, e: Event) => void): void;\n}\n\n``````\n\nthis: void 说明 addClickListener 期望 onclick 是一个不要求 this 类型的函数.\n\n``````class Handler {\ninfo: string;\n// 啊哦, 这里使用了 this. 把它用作回调函数可能会在运行时崩掉\nthis.info = e.message;\n};\n}\nlet h = new Handler();\n\n``````\n\n### tsconfig.json 对 glob 的支持\n\n``````{\n\"compilerOptions\": {\n\"module\": \"commonjs\",\n\"noImplicitAny\": true,\n\"preserveConstEnums\": true,\n\"outFile\": \"../../built/local/tsc.js\",\n\"sourceMap\": true\n},\n\"include\": [\n\"src/**/*\"\n],\n\"exclude\": [\n\"node_modules\",\n\"**/*.spec.ts\"\n]\n}\n\n``````\n\n• * 匹配零个或多个字符 (不包括目录分隔符)\n• ? 匹配任意一个字符 (不包括目录分隔符)\n• **/ 递归匹配子目录\n\n### 模块解析增强: 基准 URL, 路径映射, 多个根目录及追踪\n\nTypeScript 2.0 提供了一系列模块解析配置来告知编译器从哪里找到给定模块的声明.\n\n#### 基准 URL\n\n``````{\n\"compilerOptions\": {\n\"baseUrl\": \"./modules\"\n}\n}\n\n``````\n\n``````import A from \"moduleA\";\n\n``````\n\n#### 路径映射\n\nTypeScript 编译器支持在 tsconfig.json 里使用 \"path\" 属性来声明这样的映射.\n\n``````{\n\"compilerOptions\": {\n\"paths\": {\n\"jquery\": [\"node_modules/jquery/dist/jquery.d.ts\"]\n}\n}\n}\n\n``````\n\n#### 使用 rootDirs 配置虚拟路径\n\n`````` src\n└── views\n└── view1.ts (imports './template1')\n└── view2.ts\n\ngenerated\n└── templates\n└── views\n└── template1.ts (imports './view2')\n\n``````\n\n\"rootDirs\" 指定了将在运行时合并的多个根目录的列表. 所以对于我们的例子, tsconfig.json 文件应该像这样:\n\n``````{\n\"compilerOptions\": {\n\"rootDirs\": [\n\"src/views\",\n\"generated/templates/views\"\n]\n}\n}\n\n``````\n\n#### 追踪模块解析\n\n--traceResolution 提供了一个方便的途径帮助理解模块是如何被编译器解析的.\n\n``````tsc --traceResolution\n\n``````\n\n### 简易外围模块声明\n\ndeclarations.d.ts\n``````declare module \"hot-new-module\";\n\n``````\n\n``````import x, {y} from \"hot-new-module\";\nx(y);\n\n``````\n\n### 模块名中的通配符\n\nTypeScript 2.0 支持使用通配符 (*) 来声明一组模块名称; 这样, 声明一次就可以对应一组扩展, 而不是单个资源.\n\n``````declare module \"*!text\" {\nconst content: string;\nexport default content;\n}\n\n// 有的使用另一种方式\ndeclare module \"json!*\" {\nconst value: any;\nexport default value;\n}\n\n``````\n\n``````import fileContent from \"./xyz.txt!text\";\nimport data from \"json!http://example.com/data.json\";\nconsole.log(data, fileContent);\n\n``````\n\n``````declare module \"myLibrary\\*\";\n\n``````\n\n``````import { readFile } from \"myLibrary\\fileSystem\\readFile\";\n\n``````\n\n### 支持 UMD 模块定义\n\nmath-lib.d.ts\n``````export const isPrime(x: number): boolean;\nexport as namespace mathLib;\n\n``````\n\n``````import { isPrime } from \"math-lib\";\nisPrime(2);\nmathLib.isPrime(2); // 错误: 在模块中不能使用全局定义\n\n``````\n\n``````mathLib.isPrime(2);\n\n``````\n\n### 可选的类属性\n\n``````class Bar {\na: number;\nb?: number;\nf() {\nreturn 1;\n}\ng?(): number; // 可选方法的函数体可以被省略\nh?() {\nreturn 2;\n}\n}\n\n``````\n\n``````function test(x: Bar) {\nx.a; // number\nx.b; // number | undefined\nx.f; // () => number\nx.g; // (() => number) | undefined\nlet f1 = x.f(); // number\nlet g1 = x.g && x.g(); // number | undefined\nlet g2 = x.g ? x.g() : 0; // number\n}\n\n``````\n\n### 私有和受保护的构造函数\n\n``````class Singleton {\nprivate static instance: Singleton;\n\nprivate constructor() { }\n\nstatic getInstance() {\nif (!Singleton.instance) {\nSingleton.instance = new Singleton();\n}\nreturn Singleton.instance;\n}\n}\n\nlet e = new Singleton(); // 错误: 'Singleton' 的构造函数是私有的.\nlet v = Singleton.getInstance();\n\n``````\n\n### 抽象属性和访问器\n\n``````abstract class Base {\nabstract name: string;\nabstract get value();\nabstract set value(v: number);\n}\n\nclass Derived extends Base {\nname = \"derived\";\nvalue = 1;\n}\n\n``````\n\n### 隐式索引签名\n\n``````function httpService(path: string, headers: { [x: string]: string }) { }\n\n\"Content-Type\": \"application/x-www-form-urlencoded\"\n};\n\nhttpService(\"\", { \"Content-Type\": \"application/x-www-form-urlencoded\" }); // 正确\n\n``````\n\n### 使用 --lib 加入内建类型声明\n\n• dom\n• webworker\n• es5\n• es6 / es2015\n• es2015.core\n• es2015.collection\n• es2015.iterable\n• es2015.promise\n• es2015.proxy\n• es2015.reflect\n• es2015.generator\n• es2015.symbol\n• es2015.symbol.wellknown\n• es2016\n• es2016.array.include\n• es2017\n• es2017.object\n• es2017.sharedmemory\n• scripthost\n\n``````tsc --target es5 --lib es5,es6.promise\n\n``````\n``````\"compilerOptions\": {\n\"lib\": [\"es5\", \"es6.promise\"]\n}\n\n``````\n\n### 使用 --noUnusedParameters 和 --noUnusedLocals 标记未使用的声明\n\nTypeScript 2.0 提供了两个新的选项来帮助你保持代码整洁. --noUnusedParameters 可以标记所有未使用的函数或方法参数为错误. --noUnusedLocals 可以标记所有未使用的本地 (未导出的) 声明, 比如变量, 函数, 类, 导入项等等. 另外, 类中未使用的私有成员也会被 --noUnusedLocals 标记为错误.\n\n``````import B, { readFile } from \"./b\";\n// ^ 错误: `B` 被声明但从未使用\n\nexport function write(message: string, args: string[]) {\n// ^^^^ 错误: `arg` 被声明但从未使用\nconsole.log(message);\n}\n\n``````\n\n``````function returnNull(_a) { // 正确\nreturn null;\n}\n\n``````\n\n### 函数形参和实参列表结尾处的逗号\n\n``````function foo(\nbar: Bar,\nbaz: Baz, // 形参列表可以以逗号结尾\n) {\n// 实现...\n}\n\nfoo(\nbar,\nbaz, // 实参列表也可以\n);\n\n``````\n\n### 新的 --skipLibCheck\n\nTypeScript 2.0 添加了一个新的 --skipLibCheck 编译器选项来跳过对声明文件 (扩展名为 .d.ts 的文件) 的类型检查. 当程序包含了大的声明文件时, 编译器会花掉很多时间对这些已知没有错误的声明进行类型检查, 跳过这些声明文件的类型检查能够显著缩短编译时间.\n\n### 支持多个声明中出现重复的标示符\n\nTypeScript 2.0 放宽了相关约束并允许不同代码块中出现重复的标示符, 只要它们的类型等价.\n\n``````interface Error {\nstack?: string;\n}\n\ninterface Error {\ncode?: string;\npath?: string;\nstack?: string; // 正确\n}\n``````\n\n### 新的 --declarationDir\n\n--declarationDir 允许将声明文件生成到和 JavaScript 文件不同的位置.\n\n 原文为 \"Null- and undefined-aware types\"" ]

[ null, "https://pic2.zhimg.com/f3cc17489d01b2a533bbc73c5208f36c_1440w.jpg", null ]

[ "# INTRODUCTION TO THE LIFE TABLES\n\nTable 10.1 is the 2004 Life Table for all males in the United States. The first column to the left is the age of the man at his last birthday. In the rest of the table entry defi­nitions, this single number is just referred to by the letter x.\n\nThe second column is called q(x) . This notation is referred to as functional notation. It’s saying that the variable q “is a function of x.” Don’t worry about this. Column q is the probability of dying sometime between your xth and (x + 1)st birthday. For example, the probability of dying between your twentieth and twenty – first birthday, that is, when you’re 20 years old, is 0.001266. This definition, as stated, is correct—but should be elaborated on. This probability that you ’ tl die during your twentieth year is 0.001266 assumes that you made it to your twentieth birthday. It is not the same as the probability that, when you’re born, you’ll live to be 20 years old. I’ll get to that calculation in a few pages.\n\nNote that the probability that a man will die at age 0 (before his first birthday) is higher than the probability of his dying at any other age up to age 54. Getting born and living your first year are relatively risky activities.\n\nThe bottom entry in the leftmost column is “100 or over.” In other words, all the probabilities of living to 100, 101, 102, and so on are lumped together into one catchall entry. This is because there simply aren’t enough people living past 100 in the United States today to make probability calculations meaningful. This is analo­gous to the 100-person life insurance group example above; relatively small changes in the number of people dying in a small group swing calculations so much that you can’t draw very reliable conclusions.\n\nThe value of q associated with age = 100 or more is 1.000. A probability of 1.00 is called the “certain event.” The table is reflecting the fact that eventually, everyone will die. If you make it to age 100, you’ll die during your 100th year, your 101st year, and so on, but it’s a certainty that you will die.\n\nThe entries in the third and fourth columns of the table are calculated in an interactive manner, so they must be described together. As I’ve discussed, to do meaningful probability calculations on a relatively unlikely event (an event that has a probability much less than 1.00), we need a large group of people. It is conven­tional to use a group of 100,000 as a “standard starting point.” I don’t know if this is the best size to start with, or even how to calculate if it’s the best size to start with, but it is the standard starting point, so that’s where this table starts the third column: for x = 0 (age = 0-1), l = 100,000.", null, "0 20 40 60 80 100 Age Figure 10.3 Probable of number of men dying each year after birth out of the original 100,000 group based on the 2004 Life Table.\n\nIf there are 100,000 men starting out and the probability of each of them dying during their first year of life is 0.007475, then during this year (0.007475) (100,000) = 747 men will die.\n\nIf I start with 100,000 men and during the year 747 of them die, then going into the next year of life (x = 1), there are 100,000 – 747 = 99,253 men alive. This is the second entry in the l column.\n\nNow the work gets repetitive: Looking at the x = 1 line in the table, there are\n\n99.253 men starting out the year, and there’s a probability of 0.000508 that each of them will die. The number of men most likely to die during this year is therefore (99,253)(0.000508) = 50. This is the second entry in the d column.\n\nGoing to the third entry of column 1, 99,253 men started the year and 50 of them died during the year. The column l entry for the x = 2 line is therefore\n\n99.253 – 50 = 99,202 (again, you can’t see the rounding that actually occurred). Following this pattern all the way down to the x = 100 or more line, the column\n\nl entry is 1,261 men reaching their one hundredth birthday. Since the probability of each of them dying during their one hundredth year or any year thereafter is 1.000, the column d entry is (1,261)(1.00) = 1,261. At this point, there’s nobody left of the original 100,000 men and we stop.\n\nFigure 10.3 is a graph of column d, This graph peaks at about age 83. This means that the most likely age for a man to die is 83 years old. It does not mean\nthat the average life expectancy of all men, looking ahead from when they’re born, is 83 years. To find this average life expectancy, you first need to understand how to calculate expected, or average, values from a graph." ]

[ null, "http://cacasa2.info/img/1143/image082.jpg", null ]

[ "", null, "# Volume Profile Indicator and POCs\n\n#### dusty\n\n##### New member\nDoes anyone have a VPOC study for ThinkorSwim that closely resembles the one on TTD MarketWebs Christian Fromhertz uses it? Is there any way to take the volume profile VAH/VAL/POC plots and move them forward, so the previous days plots are projected onto current day?\n\nSee attached for reference.", null, "Last edited by a moderator:\n•", null, "#### gilda\n\n##### New member\n@dusty This is the one I'm using.\n\nCode:\n``````# Profile TPO & Volume\ninput profileType = {default Time, Volume};\ninput pricePerRowHeightMode = {default Ticksize, Automatic, Custom};\ninput customRowHeight = 1.0;\ninput timePerProfile = {default Day, Week, Month, Year, Hour, Chart, \"Opt Exp\"};\ninput multiplier = 1;\ninput OnExpansionProfile = No;\ninput OnExpansionValueArea = No;\ninput profiles = 2;\ninput showPointOfControl = Yes;\ninput showValueArea = Yes;\ninput showValueAreaCloud = Yes;\ninput ValueAreaPercent = 70;\ninput ShowHighLow = Yes;\ninput opacity = 5;\ninput PaintBars = No;\ninput ShowExtensions = No;\ninput DynamicHideExtensions = Yes;\ninput ShowLabel = No;\ndef FibExt1 = 1.618;\ndef FibExt2 = 2.618;\ndef FibExt3 = 4.236;\ndef SE = ShowExtensions;\ndef ShowProfileValueAreaCloud = no;\n\ndef period;\ndef yyyymmdd = GetYYYYMMDD();\ndef seconds = SecondsFromTime(0);\ndef month = GetYear() * 12 + GetMonth();\ndef year = GetYear();\ndef day_number = DaysFromDate(First(yyyymmdd)) + GetDayOfWeek(First(yyyymmdd));\ndef dom = GetDayOfMonth(yyyymmdd);\ndef dow = GetDayOfWeek(yyyymmdd - dom + 1);\ndef expthismonth = (if dow > 5 then 27 else 20) - dow;\ndef exp_opt = month + (dom > expthismonth);\nswitch (timePerProfile) {\ncase Chart:\nperiod = 0;\ncase Hour:\nperiod = Floor(seconds / 3600 + day_number * 24);\ncase Day:\nperiod = CountTradingDays(Min(First(yyyymmdd), yyyymmdd), yyyymmdd) - 1;\ncase Week:\nperiod = Floor(day_number / 7);\ncase Month:\nperiod = Floor(month - First(month));\ncase Year:\nperiod = Floor(year - First(year));\ncase \"Opt Exp\":\nperiod = exp_opt - First(exp_opt);\n}\n\ndef CloseByPeriod = close(Period = timePerProfile)[-1];\ndef Openbyperiod = open(Period = timePerProfile)[-1];\ndef NewDay = if !IsNaN(CloseByPeriod) then 0 else 1;\n\nrec Count = if period != period then (Count + period - period) % 1 else Count;\ndef Cond = Count < Count + period - period;\ndef height;\nswitch (pricePerRowHeightMode) {\ncase Automatic:\nheight = PricePerRow.AUTOMATIC;\ncase Ticksize:\nheight = PricePerRow.TICKSIZE;\ncase Custom:\nheight = customRowHeight;\n}\n\nprofile VOL = if profileType == profileType.Volume then VolumeProfile(\"startNewProfile\" = Cond, \"onExpansion\" = onExpansionProfile, \"NumberOfProfiles\" = profiles, \"PricePerRow\" = height, \"Value Area Percent\" = ValueAreaPercent) else TimeProfile(\"startNewProfile\" = Cond, \"OnExpansion\" = onExpansionProfile, \"NumberOfProfiles\" = profiles, \"PricePerRow\" = height, \"Value Area Percent\" = ValueAreaPercent);\n\ndef con = CompoundValue(1, onExpansionProfile, no);\nrec pc = if IsNaN(VOL.GetPointOfControl()) and con then pc else VOL.GetPointOfControl();\nrec hVA = if IsNaN(VOL.GetHighestValueArea()) and con then hVA else VOL.GetHighestValueArea();\nrec lVA = if IsNaN(VOL.GetLowestValueArea()) and con then lVA else VOL.GetLowestValueArea();\nrec HVA_Last = if period == period then HVA_Last else hVA;\nrec PC_Last = if period == period then PC_Last else pc;\nrec LVA_Last = if period == period then LVA_Last else lVA;\nrec hProfile = if IsNaN(VOL.GetHighest()) and con then hProfile else VOL.GetHighest();\nrec lProfile = if IsNaN(VOL.GetLowest()) and con then lProfile else VOL.GetLowest();\ndef plotsDomain = IsNaN(close) == onExpansionProfile;\n\nrec hP_Last = if period == period then hP_Last else hProfile;\nrec lP_Last = if period == period then lP_Last else lProfile;\n\nplot VAH = if !showValueArea then Double.NaN else if IsNaN(close) then HVA_Last else if !OnExpansionValueArea then HVA_Last else Double.NaN;\nplot POC = if IsNaN(close) then PC_Last else if !OnExpansionValueArea then PC_Last else Double.NaN;\nplot VAL = if !showValueArea then Double.NaN else if IsNaN(close) then LVA_Last else if !OnExpansionValueArea then LVA_Last else Double.NaN;\nplot High = if !ShowHighLow then Double.NaN else if IsNaN(close) then hP_Last else if !OnExpansionValueArea then hP_Last else Double.NaN;\nplot Low = if !ShowHighLow then Double.NaN else if IsNaN(close) then lP_Last else if !OnExpansionValueArea then lP_Last else Double.NaN;\n\nDefineGlobalColor(\"Profile\", GetColor(7));\nDefineGlobalColor(\"Point Of Control\", GetColor(5));\nDefineGlobalColor(\"Value Area\", GetColor(8));\n\nVOL.Show(GlobalColor(\"Profile\"), if showPointOfControl then GlobalColor(\"Point Of Control\") else Color.CURRENT, if ShowProfileValueAreaCloud then GlobalColor(\"Value Area\") else Color.CURRENT, opacity);\n\nPOC.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nPOC.SetDefaultColor(Color.DARK_GRAY);\n#POC.SetDefaultColor(CreateColor(32,49,57));\nPOC.SetLineWeight(1);\nPOC.HideTitle();\n\nVAH.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nVAH.SetDefaultColor(Color.DARK_GREEN);\nVAH.SetLineWeight(1);\nVAH.HideBubble();\nVAH.HideTitle();\n\nVAL.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nVAL.SetDefaultColor(Color.DARK_RED);\nVAL.SetLineWeight(1);\nVAL.HideBubble();\nVAL.HideTitle();\n\nHigh.SetPaintingStrategy(PaintingStrategy.DASHES);\nHigh.SetDefaultColor(CreateColor(38, 38, 8));\nHigh.SetLineWeight(1);\nHigh.HideBubble();\nHigh.HideTitle();\n\nLow.SetPaintingStrategy(PaintingStrategy.DASHES);\nLow.SetDefaultColor(CreateColor(38, 38, 8));\nLow.SetLineWeight(1);\nLow.HideBubble();\nLow.HideTitle();\n#Paint Bars\n\n#AssignPriceColor(if !PaintBars then Color.CURRENT else if open >= VAH and close >= VAH then CreateColor(0, 204, 0) else if open <= VAL and close <= VAL then CreateColor(204, 0, 0) else Color.Light_GRAY);\nAssignPriceColor(if !PaintBars then Color.CURRENT else if open >= VAH and close >= VAH then Color.Green else if open <= VAL and close <= VAL then Color.Red else Color.Light_GRAY);\n#Value Area Cloud\n\n#DefineGlobalColor(\"Value Area Cloud\", (CreateColor(20, 20, 20)));\nDefineGlobalColor(\"Value Area Cloud\", Color.DARK_GRAY);\ndef cloudhigh = if showValueAreaCloud and IsNaN(close) then HVA_Last else if !OnExpansionValueArea then HVA_Last else Double.NaN;\ndef cloudlow = if showValueAreaCloud and IsNaN(close) then LVA_Last else if !OnExpansionValueArea then LVA_Last else Double.NaN;\nAddCloud (cloudhigh, cloudlow, GlobalColor(\"Value Area Cloud\"));\n#Chart Label\n\ndef InsideValueArea = close < HVA_Last and close > LVA_Last;\ndef BelowValue = close < LVA_Last;\n\nAddLabel(ShowLabel, close, if InsideValueArea then Color.GRAY else if BelowValue then Color.RED else Color.GREEN);\n#Fibonacci Extensions\n\ndef VAWidth = VAH - VAL;\n\nplot E1H = if SE < 1 then Double.NaN else VAH + VAWidth * (FibExt1 - 1);\nE1H.SetHiding(DynamicHideExtensions and close < VAH);\nE1H.SetDefaultColor(CreateColor(0, 51, 0));\nE1H.SetPaintingStrategy(PaintingStrategy.DASHES);\nE1H.SetLineWeight(1);\nE1H.HideBubble();\nE1H.HideTitle();\nplot E1L = if SE < 1 then Double.NaN else VAL - VAWidth * (FibExt1 - 1);\nE1L.SetHiding(DynamicHideExtensions and close > VAL);\nE1L.SetDefaultColor(CreateColor(51, 0, 0));\nE1L.SetPaintingStrategy(PaintingStrategy.DASHES);\nE1L.SetLineWeight(1);\nE1L.HideBubble();\nE1L.HideTitle();\n\nplot E2H = if SE < 1 then Double.NaN else VAH + VAWidth * (FibExt2 - 1);\nE2H.SetHiding(DynamicHideExtensions and close < E1H);\nE2H.SetDefaultColor(CreateColor(0, 51, 0));\nE2H.SetPaintingStrategy(PaintingStrategy.DASHES);\nE2H.SetLineWeight(1);\nE2H.HideBubble();\nE2H.HideTitle();\nplot E2L = if SE < 1 then Double.NaN else VAL - VAWidth * (FibExt2 - 1);\nE2L.SetHiding(DynamicHideExtensions and close > E1L);\nE2L.SetDefaultColor(CreateColor(51, 0, 0));\nE2L.SetPaintingStrategy(PaintingStrategy.DASHES);\nE2L.SetLineWeight(1);\nE2L.HideBubble();\nE2L.HideTitle();\n\nplot E3H = if SE < 1 then Double.NaN else VAH + VAWidth * (FibExt3 - 1);\nE3H.SetHiding(DynamicHideExtensions and close < E2H);\nE3H.SetDefaultColor(CreateColor(0, 51, 0));\nE3H.SetPaintingStrategy(PaintingStrategy.DASHES);\nE3H.SetLineWeight(1);\nE3H.HideBubble();\nE3H.HideTitle();\nplot E3L = if SE < 1 then Double.NaN else VAL - VAWidth * (FibExt3 - 1);\nE3L.SetHiding(DynamicHideExtensions and close > E2L);\nE3L.SetDefaultColor(CreateColor(51, 0, 0));\nE3L.SetPaintingStrategy(PaintingStrategy.DASHES);\nE3L.SetLineWeight(1);\nE3L.HideBubble();\nE3L.HideTitle();``````\n\nLast edited by a moderator:\n•", null, "#### john3\n\n##### Active member\n2019 Donor\nThis one has everything you asked for and more.\n\nI have never used it with equities, only with futures. Be careful with the volume or TPO profile in TOS, it uses aggregated values, not a true tick, so it might lag the real profile, especially intraday. In addition, the values for different time frames often do not match.\n\nCode:\n``````#v1.1 - 5/10/18\n#- Able to get Implied Volatity at any specified time. Default set to 9PM est\n#- Modified Settlement to take closing price of the last bar of the day. Previous version used \"Closing Price\" at 16:15 which was usually nowhere near ending settlement\n#- Added option to substitute symbol used in IV/Deviation calculation for instruments with #no option chains such as /NKD\n#- Added customizable deviation line (hidden as default)\n\n#===========================\n#Steps if you want your VA to match u/uberbotman's:\n#1) Set your chart to 30M, and show extended hours\n#2) Go to options, make sure ShowLabels is set to \"Yes\" (hit apply if needed)\n#3) Copy VAH, POC and VAL from the labels in the top left corner of your chart into Manual input locations\n#4) Set Value Area Area Mode to Manual\n\ndeclare upper;\ndeclare once_per_bar;\n#============================\n#Inputs\n#============================\n\ninput SetMode = {default Auto, Manual};#Hint SetMode: Select Auto to update Settlement automatically in input manually. \\n\\n NOTE: ToS doesn't support Settlement so LAST closing price is used. This is usually pretty close to within a single tick, but Manually entering Settlement from CME website or UBM's nightly post is more accurate\ninput Settlement = 2444.50;#Hint Settlement: Enter Settlement value when SetMode is set to Manual\ninput IVMode = {default Auto, Manual};#Hint IVMode: Select Auto to update Implied Volatility at time chosen on IVSettleTime\ninput IVSettleTime = 2100;#Hint IVSettleTime: Enter time_value of desired Implied Volatility \"settlement\" \\n\\n NOTE: If time selected is not visible on chart (i.e. 2100 on a chart not showing extended trading hours), IV will not calculate\ninput Volatility = 11.2;#Hint Volatility: Enter Implied Volatility value when IVMode is set to Manual\ninput IVSymbolMode = {default Auto, Manual};#Hint IVSymbolMode: Select Manual if you wish to use a different instrument's IV instead of what is on the chart\ninput Symbol = \"EWJ\";#Hint Symbol: Enter symbol of Implied Volatility you wish to substitute with. For use with products with no option chains (i.e. /NKD)\ninput ValueAreaMode = {default Auto, Manual};#Hint ValueAreaMode: Select Auto to update Value Areas automatically and is rounded to the nearest tick. Manual selection changes Value Area for today only \\n\\n NOTE: Value Area is typically calculated on a 30min chart, if the timeframe changes, then calculated value area may also change. To lock in the values from the 30min chart follow these steps: \\n#1) Set your chart to 30M, and show extended hours \\n#2) Go to options, make sure ShowLabels is set to \"Yes\" (hit apply if needed) \\n#3) Copy VAH, POC and VAL from the labels showing in the top left corner of the chart into the Manual input locations \\n#4) Set Value Area Area Mode to Manual and hit Apply\ninput ValueAreaHigh = 2445.00;#Hint ValueAreaHigh: Enter Value Area High when ValueAreaMode is set to Manual\ninput PointOfControl = 2442.00;#Hint PointOfControl: Enter Point of Control when ValueAreaMode is set to Manual\ninput ValueAreaLow = 2440.00;#Hint ValueAreaLow: Enter Value Area Low when ValueAreaMode is set to Manual\ninput ShowTodayOnly = {default \"No\", \"Yes\"};#Hint ShowTodayOnly: Show/Hide chart plots for previous days\ninput ShowBubbles = {\"No\", default \"Yes\"};#Hint ShowBubbles: Show/Hide chart bubbles\ninput ShowCloud = {\"No\", default \"Yes\"};#Hint ShowCloud: Show/Hide Value Area cloud\ninput ShowLabels = {\"No\", default \"Yes\"};#Hint ShowLabels: Show/Hide Labels with Value Area data and IV used for Std Deviation calculations in Auto setting\ninput ProfileType = {default Volume, Time};#Hint ProfileType: Switch between Volume Profile and Time Profile\ninput valueAreaPercent = 70;#Hint valueAreaPercent: Set the size of the Value Area\ninput ShowVWAP = {\"No\", default \"Yes\"};#Hint ShowVWAP: Show daily VWAP\ninput CustomDev = 3.5;#Hint CustomDev: Enter custom deviation line\n\n#============================\n#Std Dev Calc / Plot\n#============================\n\ndef CloseTime2 = SecondsTillTime(0000) >= 0;\ndef OpenTime2 = SecondsFromTime(1700) >= 0;\ndef MarketOpen = OpenTime2 and CloseTime2;\ndef NewDay = IsNaN(close(period = “Day”)[-1]);\ndef Chart = MarketOpen and NewDay;\ndef bar = BarNumber();\n\ndef SettleTime = 1800;\nrec SetTimeValue = if(secondstilltime(SettleTime)== 0,close(),SetTimeValue);\ndef SetAtTime = if(SetTimeValue == 0, double.nan,SetTimeValue);\n\ndef Set = if SetMode == SetMode.\"Manual\" and NewDay then Settlement else SetAtTime;\n\nrec IVTimeValue = if(secondstilltime(IVSettleTime)== 0,imp_volatility(symbol = if IVSymbolMode == IVSymbolMode.\"Auto\" then GetSymbol() else Symbol),IVTimeValue);\ndef IVSet = if(IVTimeValue==0, double.nan,IVTimeValue);\ndef Vol = if IVMode == IVMode.\"Manual\" and NewDay then Volatility else IVSet;\n\ndef a = Vol;\ndef b = a / Sqrt(252);\ndef SD = b * Set;\n\nplot Settle = Round(If (MarketOpen, Set, If (ShowTodayOnly, Double.NaN, Set)) / TickSize(), 0) * TickSize();\nSettle.SetDefaultColor(Color.DARK_GREEN);\nSettle.SetStyle(Curve.FIRM);\nSettle.SetLineWeight(2);\nAddChartBubble(bar == HighestAll(bar) and ShowBubbles, Settle, \"Set\", Color.DARK_GREEN, no);\n\nplot StdDev025H = Round(If (Chart, Set + (SD * 0.25), If (ShowTodayOnly, Double.NaN, Set + (SD * 0.25))) / TickSize(), 0) * TickSize();\nStdDev025H.SetDefaultColor(Color.GRAY);\nStdDev025H.SetStyle(Curve.SHORT_DASH);\nAddChartBubble(bar == HighestAll(bar) and ShowBubbles, StdDev025H, \"0.25 Std Dev \", Color.GRAY, no);\n\nplot StdDev05H = Round(If (Chart, Set + (SD * 0.5), If (ShowTodayOnly, Double.NaN, Set + (SD * 0.5))) / TickSize(), 0) * TickSize();\nStdDev05H.SetDefaultColor(Color.CYAN);\nStdDev05H.SetStyle(Curve.SHORT_DASH);\nAddChartBubble(bar == HighestAll(bar) and ShowBubbles, StdDev05H, \"0.5 Std Dev \", Color.CYAN, no);\n\nplot StdDev1H = Round(If (Chart, Set + (SD * 1), If (ShowTodayOnly, Double.NaN, Set + (SD * 1))) / TickSize(), 0) * TickSize();\nStdDev1H.SetDefaultColor(Color.VIOLET);\nStdDev1H.SetStyle(Curve.LONG_DASH);\nAddChartBubble(bar == HighestAll(bar) and ShowBubbles, StdDev1H, \"1 Std Dev\", Color.VIOLET, no);\n\nplot StdDev1_5H = Round(If (Chart, Set + (SD * 1.5), If (ShowTodayOnly, Double.NaN, Set + (SD * 1.5))) / TickSize(), 0) * TickSize();\nStdDev1_5H.SetDefaultColor(Color.PLUM);\nStdDev1_5H.SetStyle(Curve.LONG_DASH);\n#AddChartBubble(bar == HighestAll(bar) and ShowBubbles, StdDev1_5H, \"1.5 Std Dev\", Color.PLUM, no);\n\nplot StdDev2H = Round(If (Chart, Set + (SD * 2), If (ShowTodayOnly, Double.NaN, Set + (SD * 2))) / TickSize(), 0) * TickSize();\nStdDev2H.SetDefaultColor(Color.PLUM);\nStdDev2H.SetStyle(Curve.FIRM);\nStdDev2H.SetLineWeight(3);\nAddChartBubble(bar == HighestAll(bar) and ShowBubbles, StdDev2H, \"2 Std Dev\", Color.PLUM, no);\n\nplot StdDev3H = Round(If (Chart, Set + (SD * 3), If (ShowTodayOnly, Double.NaN, Set + (SD * 3))) / TickSize(), 0) * TickSize();\nStdDev3H.SetDefaultColor(Color.DARK_RED);\nStdDev3H.SetStyle(Curve.FIRM);\nStdDev3H.SetLineWeight(3);\nAddChartBubble(bar == HighestAll(bar) and ShowBubbles, StdDev3H, \"3 Std Dev\", Color.DARK_RED, no);\n\nplot StdDev025L = Round(If (Chart, Set + (SD * -0.25), If (ShowTodayOnly, Double.NaN, Set + (SD * -0.25))) / TickSize(), 0) * TickSize();\nStdDev025L.SetDefaultColor(Color.GRAY);\nStdDev025L.SetStyle(Curve.SHORT_DASH);\nAddChartBubble(bar == HighestAll(bar) and ShowBubbles, StdDev025L, \"-0.25 Std Dev\", Color.GRAY, no);\n\nplot StdDev05L = Round(If (Chart, Set + (SD * -0.5), If (ShowTodayOnly, Double.NaN, Set + (SD * -0.5))) / TickSize(), 0) * TickSize();\nStdDev05L.SetDefaultColor(Color.CYAN);\nStdDev05L.SetStyle(Curve.SHORT_DASH);\nAddChartBubble(bar == HighestAll(bar) and ShowBubbles, StdDev05L, \"-0.5 Std Dev\", Color.CYAN, no);\n\nplot StdDev1L = Round(If (Chart, Set + (SD * -1), If (ShowTodayOnly, Double.NaN, Set + (SD * -1))) / TickSize(), 0) * TickSize();\nStdDev1L.SetDefaultColor(Color.VIOLET);\nStdDev1L.SetStyle(Curve.LONG_DASH);\nAddChartBubble(bar == HighestAll(bar) and ShowBubbles, StdDev1L, \"1 Std Dev\", Color.VIOLET, no);\n\nplot StdDev1_5L = Round(If (Chart, Set + (SD * -1.5), If (ShowTodayOnly, Double.NaN, Set + (SD * -1.5))) / TickSize(), 0) * TickSize();\nStdDev1_5L.SetDefaultColor(Color.PLUM);\nStdDev1_5L.SetStyle(Curve.LONG_DASH);\n#AddChartBubble(bar == HighestAll(bar) and ShowBubbles, StdDev1_5L, \"-1.5 Std Dev\", Color.PLUM, no);\n\nplot StdDev2L = Round(If (Chart, Set + (SD * -2), If (ShowTodayOnly, Double.NaN, Set + (SD * -2))) / TickSize(), 0) * TickSize();\nStdDev2L.SetDefaultColor(Color.PLUM);\nStdDev2L.SetStyle(Curve.FIRM);\nStdDev2L.SetLineWeight(3);\nAddChartBubble(bar == HighestAll(bar) and ShowBubbles, StdDev2L, \"-2 Std Dev\", Color.PLUM, no);\n\nplot StdDev3L = Round(If (Chart, Set + (SD * -3), If (ShowTodayOnly, Double.NaN, Set + (SD * -3))) / TickSize(), 0) * TickSize();\nStdDev3L.SetDefaultColor(Color.DARK_RED);\nStdDev3L.SetStyle(Curve.FIRM);\nStdDev3L.SetLineWeight(3);\nAddChartBubble(bar == HighestAll(bar) and ShowBubbles, StdDev3L, \"-3 Std Dev\", Color.DARK_RED, no);\n\nplot CustDev = Round(If (Chart, Set + (SD * CustomDev), If (ShowTodayOnly, Double.NaN, Set + (SD * CustomDev))) / TickSize(), 0) * TickSize();\nCustDev.Hide();\nCustDev.SetDefaultColor(Color.YELLOW);\nCustDev.SetStyle(Curve.FIRM);\nCustDev.SetLineWeight(2);\n\n# =================================\n# Volume Profile Definition Section\n# =================================\n\ndef profiles = 50;\ndef customRowHeight = 1.0;\ndef multiplier = 1;\ndef onExpansion = ShowTodayOnly;\ndef yyyymmdd = GetYYYYMMDD();\ndef seconds = SecondsFromTime(0);\ndef period = CountTradingDays(Min(First(yyyymmdd), yyyymmdd), yyyymmdd) - 1;\ndef month = GetYear() * 12 + GetMonth();\ndef day_number = DaysFromDate(First(yyyymmdd)) + GetDayOfWeek(First(yyyymmdd));\ndef dom = GetDayOfMonth(yyyymmdd);\ndef dow = GetDayOfWeek(yyyymmdd - dom + 1);\ndef expthismonth = (if dow > 5 then 27 else 20) - dow;\ndef exp_opt = month + (dom > expthismonth);\ndef height = PricePerRow.TICKSIZE;\n\nrec count = CompoundValue(1, if period != period then (count + period - period) % multiplier else count, 0);\ndef cond = count < count + period - period;\n\n#============================\n# Plot POC VAH VAL Section\n#============================\n\nprofile tpo = if ProfileType == ProfileType.Volume then VolumeProfile(\"startNewProfile\" = cond, \"onExpansion\" = onExpansion, \"numberOfProfiles\" = profiles, \"pricePerRow\" = height, \"value area percent\" = valueAreaPercent)\nelse TimeProfile(\"startNewProfile\" = cond, \"onExpansion\" = onExpansion, \"numberOfProfiles\" = profiles, \"pricePerRow\" = height, \"value area percent\" = valueAreaPercent);\n\nrec PC = if cond == 1 then tpo.GetPointOfControl() else PC;\nplot POC = Round(If(ValueAreaMode == ValueAreaMode.\"Auto\", PC, if NewDay then PointOfControl else PC) / TickSize(), 0) * TickSize();\nPOC.SetDefaultColor(Color.DARK_RED);\nPOC.SetStyle(Curve.FIRM);\nPOC.SetLineWeight(2);\n\nrec hVA = if cond == 1 then tpo.GetHighestValueArea() else hVA;\nplot VAH = Round(If(ValueAreaMode == ValueAreaMode.\"Auto\", hVA, if NewDay then ValueAreaHigh else hVA) / TickSize(), 0) * TickSize();\nVAH.SetDefaultColor(Color.RED);\nVAH.SetStyle(Curve.FIRM);\n\nrec lVA = if cond == 1 then tpo.GetLowestValueArea() else lVA;\nplot VAL = Round(If(ValueAreaMode == ValueAreaMode.\"Auto\", lVA, if NewDay then ValueAreaLow else lVA) / TickSize(), 0) * TickSize();\n;\nVAL.SetDefaultColor(Color.GREEN);\nVAL.SetStyle(Curve.FIRM);\n\n#============================\n# VWAP Plot\n#============================\ndef VWAP1 = Round(vwap(period = AggregationPeriod.DAY) / TickSize(), 0) * TickSize();\nplot VWAP = if ShowVWAP > 0 then VWAP1 else Double.NaN;\nVWAP.SetPaintingStrategy(PaintingStrategy.DASHES);\nVWAP.SetDefaultColor(Color.DARK_GRAY);\n\n#============================\n#Value Area Cloud & Labels\n#============================\n\ndef VArea = Between(close, VAL, VAH);\ndef VAreaAbove = close > VAH;\ndef VAreaBelow = close < VAL;\n\ndef Cloudhigh = if ShowCloud then VAH else Double.NaN;\ndef Cloudlow = if ShowCloud then VAL else Double.NaN;\nDefineGlobalColor(\"cloud\", Color.DARK_GRAY);\n\nAddLabel(ShowLabels, Concat(\"VAH:\", Round(VAH)), if close > VAH then Color.GREEN else Color.RED) ;\nAddLabel(ShowLabels, Concat(\"POC:\", Round(POC)), if close > POC then Color.GREEN else Color.RED);\nAddLabel(ShowLabels, Concat(\"VAL:\", Round(VAL)), if close > VAL then Color.GREEN else Color.RED);\nAddLabel(ShowLabels, \"IV: \" + AsPercent(IVSet), if IsAscending(imp_volatility(period = AggregationPeriod.DAY, priceType = PriceType.LAST)) then Color.RED else if IsDescending(imp_volatility(period = AggregationPeriod.DAY, priceType = PriceType.LAST)) then Color.GREEN else Color.WHITE);\nAddLabel(ShowLabels, if VArea then \"Inside Value Area\" else if VAreaAbove then \"Above Value Area\" else \"Below Value Area\", if VArea then Color.ORANGE else if VAreaAbove then Color.GREEN else Color.RED);\n\n#============================\n#============================\n# BLOCK CODE BELOW\ninput UseAlerts = {false, default true};\ninput AlertType = {default \"BAR\", \"ONCE\", \"TICK\"};\ninput AlertSound = {default \"Bell\", \"Chimes\", \"Ding\", \"NoSound\", \"Ring\"};\ndef Signal = (close crosses POC) or (close crosses VAH) or (close crosses VAL) or (close crosses StdDev05H) or (close crosses StdDev1H) or (close crosses StdDev2H) or (close crosses StdDev05L) or (close crosses StdDev1L) or (close crosses StdDev2L);\n\nLast edited:\n\n#### dusty\n\n##### New member\nYeah I am aware. I never use current day profile anyways, hence why I asked for an indicator that plots prev. day onto current day. TOS is not great for VPRO/TPO with the aggregate tick, but I have noticed with equities its close enough to get the job done. All I look for is price to break, and build above or below VAH/VAL and use previous VPOC as targets. There's no way I'd use it for futures trading.\n\nLast edited:\n\n#### lgol\n\n##### New member\nThis is working well for me however I am finding that i cannot change the # of Bars, Hours etc. length. Basically, it's stuck on 1 bar, 1Hour, 1 Day etc. and I'd like to have this as a variable length (the way it works on the standard version of VP). Thanks!\n\nLast edited:\n\n#### Shinthus\n\n##### Member\n2019 Donor\nHey all. Someone mentioned that ThinkorSwim's volume profile study was basically Center of Gravity (COG). I forget who and where this was said - can someone please confirm or refute?\n\nAlso, I noticed the way Volume Profile is calculated on TOS is different than how it's calculated on Tradingview but I don't use Tradingview because I would rather not pay for indicators.... I also took a liking to Tradingview's version and find it more accurate and easier to interpret. Does anybody know the calculation are different and whether or not we can convert that from pinescript to TOS?\n\nThis Volume Profile script is for 24 hours BEGINNING with the RTH OPEN.\n\nCode:\n``````# Volume Profile for RTH and GlobeX\n# Mobius\n# Chat Room Discussion 03.26.2018\n\ninput pricePerRowHeightMode = {AUTOMATIC, TICKSIZE, default CUSTOM};\ninput customRowHeight = 1.0;\ninput onExpansion = no;\ninput profiles = 5; #Hint profiles: for just RTH 1 for GlobeX and RTH 2\ninput showPointOfControl = yes;\ninput showValueArea = yes;\ninput valueAreaPercent = 68;\ninput opacity = 5;\ninput RthBegin = 0930;\ninput RthEnd = 1600;\n\ndef TS = TickSize();\ndef Active = getTime() >= RegularTradingStart(getYYYYMMDD());\ndef cond = getTime() crosses above RegularTradingStart(getYYYYMMDD());\ndef height;\nswitch (pricePerRowHeightMode) {\ncase AUTOMATIC:\nheight = PricePerRow.AUTOMATIC;\ncase TICKSIZE:\nheight = PricePerRow.TICKSIZE;\ncase CUSTOM:\nheight = customRowHeight;\n}\nprofile vol = VolumeProfile(\"startNewProfile\" = cond, \"onExpansion\" = onExpansion, \"numberOfProfiles\" = profiles, \"pricePerRow\" = height, \"value area percent\" = valueAreaPercent);\ndef con = CompoundValue(1, onExpansion, no);\ndef pc = if IsNaN(vol.GetPointOfControl()) and con\nthen pc\nelse vol.GetPointOfControl();\ndef hVA = if IsNaN(vol.GetHighestValueArea()) and con\nthen hVA\nelse Round(vol.GetHighestValueArea(), 0);\ndef lVA = if IsNaN(vol.GetLowestValueArea()) and con\nthen lVA\nelse vol.GetLowestValueArea();\ndef hProfile = if IsNaN(vol.GetHighest()) and con\nthen hProfile\nelse vol.GetHighest();\ndef lProfile = if IsNaN(vol.GetLowest()) and con\nthen lProfile\nelse vol.GetLowest();\ndef plotsDomain = IsNaN(close) == onExpansion;\n\nplot POC = if plotsDomain\nthen Round(pc / TS, 0) * TS\nelse Double.NaN;\nplot ProfileHigh = if plotsDomain\nthen Round(hProfile / TS, 0) * TS\nelse Double.NaN;\nplot ProfileLow = if plotsDomain\nthen Round(lProfile / TS, 0) * TS\nelse Double.NaN;\nplot VAHigh = if plotsDomain\nthen Round(hVA / TS, 0) * TS\nelse Double.NaN;\nplot VALow = if plotsDomain\nthen Round(lVA / TS, 0) * TS\nelse Double.NaN;\n\nDefineGlobalColor(\"Profile\", GetColor(1));\nDefineGlobalColor(\"Point Of Control\", GetColor(5));\nDefineGlobalColor(\"Value Area\", GetColor(8));\n\nvol.Show(GlobalColor(\"Profile\"), if showPointOfControl then GlobalColor(\"Point Of Control\") else Color.CURRENT, if showValueArea then GlobalColor(\"Value Area\") else Color.CURRENT, opacity);\n\nPOC.SetDefaultColor(GlobalColor(\"Point Of Control\"));\nPOC.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nVAHigh.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nVAHigh.HideBubble();\nVAHigh.HideTitle();\nVALow.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nVALow.HideBubble();\nVALow.HideTitle();\nVAHigh.SetDefaultColor(GlobalColor(\"Value Area\"));\nVALow.SetDefaultColor(GlobalColor(\"Value Area\"));\nProfileHigh.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nProfileLow.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nProfileHigh.SetDefaultColor(GetColor(3));\nProfileLow.SetDefaultColor(GetColor(3));\nProfileHigh.Hide();\nProfileLow.Hide();\ndef bubble = isNaN(close) and !isNaN(close);\nAddChartBubble(bubble, VAHigh, \"VAH\", color = Color.YELLOW, yes);\n# End Code Volume Profile RTH and GlobeX``````\n\n•", null, "#### markos\n\n##### Well-known member\nVIP\nToS's built in Volume Profile is correct according to Professor Jeff Bierman, CTA. He worked as the chief market technician for ThinkorSwim for 5 years before the buyout by TDA. Keep in mind, just like the Fractal Energy Indicator is similar to the Chop Indicator even though the math constructs are different.\n\nCode:\n``````#VolumeProfile_RTHvOvernight\n#JoeBone87 in TSL May 2019 (No Guarantee it will plot but math should be good)\n\ninput pricePerRowHeightMode = {default AUTOMATIC, TICKSIZE, CUSTOM};\ninput customRowHeight = 1.0;\ninput onExpansion = no;\ninput profiles = 1000;\ninput showPointOfControl = yes;\ninput showValueArea = yes;\ninput valueAreaPercent = 70;\ninput opacity = 50;\n\ninput rthbegin = 0930;\ninput rthend = 1600;\ndef count = secondsfromTime(rthbegin)>0 and secondstillTime(rthend)>0;\ndef cond = count != count;\ndef height;\nswitch (pricePerRowHeightMode) {\ncase AUTOMATIC:\nheight = PricePerRow.AUTOMATIC;\ncase TICKSIZE:\nheight = PricePerRow.TICKSIZE;\ncase CUSTOM:\nheight = customRowHeight;\n}\n\nprofile vol = VolumeProfile(\"startNewProfile\" = cond, \"onExpansion\" = onExpansion, \"numberOfProfiles\" = profiles, \"pricePerRow\" = height, \"value area percent\" = valueAreaPercent);\ndef con = CompoundValue(1, onExpansion, no);\ndef pc = if IsNaN(vol.GetPointOfControl()) and con then pc else vol.GetPointOfControl();\ndef hVA = if IsNaN(vol.GetHighestValueArea()) and con then hVA else vol.GetHighestValueArea();\ndef lVA = if IsNaN(vol.GetLowestValueArea()) and con then lVA else vol.GetLowestValueArea();\n\ndef hProfile = if IsNaN(vol.GetHighest()) and con then hProfile else vol.GetHighest();\ndef lProfile = if IsNaN(vol.GetLowest()) and con then lProfile else vol.GetLowest();\ndef plotsDomain = IsNaN(close) == onExpansion;\n\nplot POC = if plotsDomain then pc else Double.NaN;\nplot ProfileHigh = if plotsDomain then hProfile else Double.NaN;\nplot ProfileLow = if plotsDomain then lProfile else Double.NaN;\nplot VAHigh = if plotsDomain then hVA else Double.NaN;\nplot VALow = if plotsDomain then lVA else Double.NaN;\n\nDefineGlobalColor(\"Profile\", GetColor(1));\nDefineGlobalColor(\"Point Of Control\", GetColor(5));\nDefineGlobalColor(\"Value Area\", GetColor(8));\n\nvol.Show(GlobalColor(\"Profile\"), if showPointOfControl then GlobalColor(\"Point Of Control\") else Color.CURRENT, if showValueArea then GlobalColor(\"Value Area\") else Color.CURRENT, opacity);\nPOC.SetDefaultColor(GlobalColor(\"Point Of Control\"));\nPOC.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nVAHigh.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nVALow.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nVAHigh.SetDefaultColor(GlobalColor(\"Value Area\"));\nVALow.SetDefaultColor(GlobalColor(\"Value Area\"));\nProfileHigh.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nProfileLow.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nProfileHigh.SetDefaultColor(GetColor(3));\nProfileLow.SetDefaultColor(GetColor(3));\nProfileHigh.Hide();\nProfileLow.Hide();\n\ninput bubbles = yes;\ninput n = 2;\ndef n1 = n + 1;\nAddChartBubble(bubbles and !IsNaN(close[n1]) and IsNaN(close[n]), VAHigh[n1], \"V-VAH\", color = Color.YELLOW, yes);\nAddChartBubble(bubbles and !IsNaN(close[n1]) and IsNaN(close[n]), VALow[n1], \"V-VAL\", Color.YELLOW, no);\nAddChartBubble(bubbles and !IsNaN(close[n1]) and IsNaN(close[n]), POC[n1], \"V-POC\", Color.RED, no);``````\n\nThe Volume Profile script below lets you set the starting time.\n\nCode:\n``````# Volume Profile for User Set Time\n# Mobius\n# Chat Room Discussion 03.26.2018\n\ninput pricePerRowHeightMode = {AUTOMATIC, TICKSIZE, default CUSTOM};\ninput customRowHeight = 1.0;\ninput onExpansion = no;\ninput profiles = 5; #Hint profiles: for just RTH 1 for GlobeX and RTH 2\ninput showPointOfControl = yes;\ninput showValueArea = yes;\ninput valueAreaPercent = 68;\ninput opacity = 5;\ninput Begin = 0930;\n\ndef TS = TickSize();\ndef Active = SecondsTillTime(Begin) == 0;\ndef cond = Active;\ndef height;\nswitch (pricePerRowHeightMode) {\ncase AUTOMATIC:\nheight = PricePerRow.AUTOMATIC;\ncase TICKSIZE:\nheight = PricePerRow.TICKSIZE;\ncase CUSTOM:\nheight = customRowHeight;\n}\nprofile vol = VolumeProfile(\"startNewProfile\" = cond, \"onExpansion\" = onExpansion, \"numberOfProfiles\" = profiles, \"pricePerRow\" = height, \"value area percent\" = valueAreaPercent);\ndef con = CompoundValue(1, onExpansion, no);\ndef pc = if IsNaN(vol.GetPointOfControl()) and con\nthen pc\nelse vol.GetPointOfControl();\ndef hVA = if IsNaN(vol.GetHighestValueArea()) and con\nthen hVA\nelse Round(vol.GetHighestValueArea(), 0);\ndef lVA = if IsNaN(vol.GetLowestValueArea()) and con\nthen lVA\nelse vol.GetLowestValueArea();\ndef hProfile = if IsNaN(vol.GetHighest()) and con\nthen hProfile\nelse vol.GetHighest();\ndef lProfile = if IsNaN(vol.GetLowest()) and con\nthen lProfile\nelse vol.GetLowest();\ndef plotsDomain = IsNaN(close) == onExpansion;\n\nplot POC = if plotsDomain\nthen Round(pc / TS, 0) * TS\nelse Double.NaN;\nplot ProfileHigh = if plotsDomain\nthen Round(hProfile / TS, 0) * TS\nelse Double.NaN;\nplot ProfileLow = if plotsDomain\nthen Round(lProfile / TS, 0) * TS\nelse Double.NaN;\nplot VAHigh = if plotsDomain\nthen Round(hVA / TS, 0) * TS\nelse Double.NaN;\nplot VALow = if plotsDomain\nthen Round(lVA / TS, 0) * TS\nelse Double.NaN;\n\nDefineGlobalColor(\"Profile\", GetColor(1));\nDefineGlobalColor(\"Point Of Control\", GetColor(5));\nDefineGlobalColor(\"Value Area\", GetColor(8));\n\nvol.Show(GlobalColor(\"Profile\"), if showPointOfControl then GlobalColor(\"Point Of Control\") else Color.CURRENT, if showValueArea then GlobalColor(\"Value Area\") else Color.CURRENT, opacity);\n\nPOC.SetDefaultColor(GlobalColor(\"Point Of Control\"));\nPOC.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nVAHigh.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nVAHigh.HideBubble();\nVAHigh.HideTitle();\nVALow.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nVALow.HideBubble();\nVALow.HideTitle();\nVAHigh.SetDefaultColor(GlobalColor(\"Value Area\"));\nVALow.SetDefaultColor(GlobalColor(\"Value Area\"));\nProfileHigh.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nProfileLow.SetPaintingStrategy(PaintingStrategy.HORIZONTAL);\nProfileHigh.SetDefaultColor(GetColor(3));\nProfileLow.SetDefaultColor(GetColor(3));\nProfileHigh.Hide();\nProfileLow.Hide();\ndef bubble = isNaN(close) and !isNaN(close);\nAddChartBubble(bubble, VAHigh, \"VAH\", color = Color.YELLOW, yes);\n# End Code Volume Profile User Set Time``````\n\nVolume Profile Scanner\n\nCode:\n``````# Mr. Script\ndef yyyymmdd = GetYYYYMMDD();\ndef day_number = DaysFromDate(First(yyyymmdd)) + GetDayOfWeek(First(yyyymmdd));\ndef period = Floor(day_number / 7);\ndef cond = 0 < period - period;\nprofile vol = VolumeProfile(\"startNewProfile\" = cond, \"onExpansion\" = no);\nvol.Show(\"va color\" = Color.YELLOW);\ndef b = vol.GetLowestValueArea();\nplot c = close <= b and close >= (b*.9);\nc.setPaintingStrategy(paintingStrategy.BOOLEAN_ARROW_UP);``````\n\n•", null, "#### John808\n\n##### New member\n2019 Donor\nHi @Shinthus, here is another cog indicator https://usethinkscript.com/threads/center-of-gravity-cog-indicator-for-thinkorswim.138/ I think this might be the cog indicator you're looking for if you were in bluesgirl's chatroom. (I'm in the same room", null, "That's similar to what she uses. From what I understand with the cog indicator is you want to play the edges and also look at the curves of the waves. So if the waves are starting to curl up price may have found a bottom and start to move up. Maybe similar to boillinger bands.\n\nAs for volume profile here is a picture of the study I use on TOS. It looks at the volume profile for every individual day, so the levels are pretty much the same across all time frames. I can post the settings I have if you're interested", null, "", null, "I like to mark off the Value area Highs/Lows (the yellow lines) and the point of control (red line). I use these lines as support/resistance and targets for the next day.", null, "Oh, all of those lines are hand-drawn lol. With Volume profile, there's a concept called VPOC or virgin point of control, which means if a point of control(the red line) hasn't been tested the next day after it's been made, the price will naturally trend towards it. The \"put\" target level was the value area high from 8/9/19, the same day the VPOC level is from. I'm still practicing with volume profile, but I like it so far.\n\nHopefully this helps.\n\nLast edited by a moderator:\n\n#### Jmp2626\n\n##### New member\n2019 Donor\nVIP\nAlso would like to post historical POC (Point of Control) to follow average POC.\n\nCode:\n``````# plots POC, VAH and VAL for profiles starting at RTH Open and ending 1,2,3 etc bars later\n# Not a realistic solution to the problem\n# Nube\n\ndef bn = BarNumber();\ndef na = Double.NaN;\n\nscript TestPro {\ninput Start = 1;\ninput nBars = 3;\ndef bn = BarNumber();\ndef na = Double.NaN;\ndef EndBar = if bn == start + nBars\nthen bn else endBar;\ndef NewProfile = bn == start or\nbn crosses above endBar;\n\nprofile testProfile = VolumeProfile(\"startNewProfile\" = NewProfile\n, \"onExpansion\" = no, \"numberOfProfiles\" = 1000, \"pricePerRow\" = TickSize(), \"value area percent\" = 68);\n\nplot hva = if bn == EndBar\nthen testProfile.GetHighestValueArea()\nelse na;\nplot lva = if bn == EndBar\nthen testProfile.GetLowestValueArea()\nelse na;\nplot poc = if bn == EndBar\nthen testProfile.GetPointOfControl()\nelse na;\n}\ndef RTHOpen = GetDay() == GetLastDay() &&\nGetTime() crosses above\ndef RTHBar = CompoundValue(1,\nif RTHOpen\nthen bn\nelse RTHBar,0);\n\ndef hva =\nif bn == RTHBar + 1 then TestPro(RTHBar,1).hva else\nif bn == RTHBar + 2 then TestPro(RTHBar,2).hva else\nif bn == RTHBar + 3 then TestPro(RTHBar,3).hva else\nif bn == RTHBar + 4 then TestPro(RTHBar,4).hva else\nif bn == RTHBar + 5 then TestPro(RTHBar,5).hva else\nif bn == RTHBar + 6 then TestPro(RTHBar,6).hva else\nif bn == RTHBar + 7 then TestPro(RTHBar,7).hva else\nif bn == RTHBar + 8 then TestPro(RTHBar,8).hva else\nif bn == RTHBar + 9 then TestPro(RTHBar,9).hva else\nif bn == RTHBar + 10 then TestPro(RTHBar,10).hva else\nif bn == RTHBar + 11 then TestPro(RTHBar,11).hva else\nif bn == RTHBar + 12 then TestPro(RTHBar,12).hva else\nif bn == RTHBar + 13 then TestPro(RTHBar,13).hva else\nif bn == RTHBar + 14 then TestPro(RTHBar,14).hva else\nna;\n\ndef lva =\nif bn == RTHBar + 1 then TestPro(RTHBar,1).lva else\nif bn == RTHBar + 2 then TestPro(RTHBar,2).lva else\nif bn == RTHBar + 3 then TestPro(RTHBar,3).lva else\nif bn == RTHBar + 4 then TestPro(RTHBar,4).lva else\nif bn == RTHBar + 5 then TestPro(RTHBar,5).lva else\nif bn == RTHBar + 6 then TestPro(RTHBar,6).lva else\nif bn == RTHBar + 7 then TestPro(RTHBar,7).lva else\nif bn == RTHBar + 8 then TestPro(RTHBar,8).lva else\nif bn == RTHBar + 9 then TestPro(RTHBar,9).lva else\nif bn == RTHBar + 10 then TestPro(RTHBar,10).lva else\nif bn == RTHBar + 11 then TestPro(RTHBar,11).lva else\nif bn == RTHBar + 12 then TestPro(RTHBar,12).lva else\nif bn == RTHBar + 13 then TestPro(RTHBar,13).lva else\nif bn == RTHBar + 14 then TestPro(RTHBar,14).lva else\nna;\n\ndef poc =\nif bn == RTHBar + 1 then TestPro(RTHBar,1).poc else\nif bn == RTHBar + 2 then TestPro(RTHBar,2).poc else\nif bn == RTHBar + 3 then TestPro(RTHBar,3).poc else\nif bn == RTHBar + 4 then TestPro(RTHBar,4).poc else\nif bn == RTHBar + 5 then TestPro(RTHBar,5).poc else\nif bn == RTHBar + 6 then TestPro(RTHBar,6).poc else\nif bn == RTHBar + 7 then TestPro(RTHBar,7).poc else\nif bn == RTHBar + 8 then TestPro(RTHBar,8).poc else\nif bn == RTHBar + 9 then TestPro(RTHBar,9).poc else\nif bn == RTHBar + 10 then TestPro(RTHBar,10).poc else\nif bn == RTHBar + 11 then TestPro(RTHBar,11).poc else\nif bn == RTHBar + 12 then TestPro(RTHBar,12).poc else\nif bn == RTHBar + 13 then TestPro(RTHBar,13).poc else\nif bn == RTHBar + 14 then TestPro(RTHBar,14).poc else\nna;\n\nplot\nLoVA = if bn > HighestAll(RTHBar)\nthen lva else na;\nplot\nHiVA = if bn > HighestAll(RTHBar)\nthen hva else na;\nplot\nPointOfControl = if bn > HighestAll(RTHBar)\nthen poc else na;\n\nLoVA.SetDefaultColor(Color.Violet);\nLoVA.SetStyle(Curve.Long_Dash);\nLoVA.EnableApproximation();\nHiVA.SetStyle(Curve.Long_Dash);\nHiVA.SetDefaultColor(Color.Violet);\nHiVA.EnableApproximation();\nPointOfControl.SetDefaultColor(Color.Yellow);\nPointOfControl.EnableApproximation();``````\n\n#### horserider\n\n##### Well-known member\nVIP\n•", null, "markos\n\n#### skynetgen\n\n##### Well-known member\nHere is a script which shows if price is above or below prevday POC (also it has elderimpulse scan there as well)\nhttps://tos.mx/Sf1HpE\n\n•", null, "markos\n\n#### Jmp2626\n\n##### New member\n2019 Donor\nVIP\nPlots the average POC for all Chart Data. Can be used as boundaries for likely price range intraday.\n\nCode:\n``````# Average Point OF Control (POC) with Standard Deviation Bands\n# Mobius\n# V01.09.2018\n# Plots the average point of control value and standard deviation for high and low price as bands around the average POC.\n\ninput SD_Multiplier = 2;\n\ndef h = high;\ndef l = low;\ndef x = BarNumber();\ndef RTHBar1 = if getTime() crosses above RegularTradingStart(getYYYYMMDD())\nthen x\nelse Double.NaN;\ndef cond = x == (RTHBar1);\nprofile vol = VolumeProfile(\"startNewProfile\" = cond, \"onExpansion\" = no, \"numberOfProfiles\" = 100, \"pricePerRow\" = PricePerRow.TICKSIZE, \"value area percent\" = 68);\ndef pc = if IsNaN(vol.GetPointOfControl())\nthen pc\nelse vol.GetPointOfControl();\ndef sumPc = if !isNaN(RTHbar1)\nthen sumPC + pc\nelse sumPc;\ndef sumD = if !isNaN(RTHbar1)\nthen sumD + 1\nelse sumD;\ndef SDH = if isNaN(StDevAll(h))\nthen SDH\nelse StDevAll(h);\ndef SDL = if isNaN(StDevAll(l))\nthen SDL\nelse StDevAll(l);\nplot Avgpc = (sumPc + pc) / (sumD + 1);\nAvgpc.SetDefaultColor(Color.Cyan);\nplot upper = Avgpc + (SD_Multiplier*SDH);\nupper.SetDefaultColor(Color.Gray);\nplot lower = Avgpc - (SD_Multiplier*SDL);\nlower.SetDefaultColor(Color.Gray);\n# End Code``````\n\n#### horserider\n\n##### Well-known member\nVIP", null, "Something like that ? @Jmp2626\n\n#### markos\n\n##### Well-known member\nVIP\n@horserider That's great! Please post the code and a share. Thanks", null, "#### horserider\n\n##### Well-known member\nVIP\nYou people really want this? Just a moving average that can be duplicated by using the already existing SMA or EMA in ToS. Here is the POC avg compared to 100 length SMA and EMA.", null, "I doubt it is worth the trouble. If you have a POC that is plotted each day just add something like this :\n\nCode:\n``````input length =20;\nplot smapoc = simpleMovingAvg (poc, length);``````\n\nAdjust length to fit what you want.\n\n•", null, "XeoNoX\n\n#### lindosskier\n\n##### Member\nNew here to the forum, but had a few questions regarding thinkscript for VolumeProfile in TOS. I know there is a study there already, but I need to change it a bit to fit my needs.\n\n1)How can I be showing ALL the Weekly VOLUMEPROFILES On the chart (For all the Weeks on the chart)? When I click \"No\" in the Expansion input, it shows me all the VolumeProfiles, but it plots them on their respective Weeks' charts, when I need them to show next week's one (for each one of them - One period forward, so it can serve as support or resistance for next week). - You can take a look at Market Webs and/or Christian Fromhertz from Tribeca Trade Group (active on twitter)\n2)How can I be having the Monthly Volume Profile shown when I have the Weekly Chart Up, the Weekly Volume profile shown when I have the Daily Chart up and the Daily Volume profile shown when I have the 60min. Chart up? and\n3)On each of the respective charts above, I would need to see the POC plotted, UNTIL the price hits it again in the future (Virgin Point of Control).\nSo, for the Weekly chart the Monthly POC, for the Daily Chart the Weekly POC and for the 60min. chart the Daily POC to be carrying forward, until price hits them in the future. How can I do that?\n\n#### lindosskier\n\n##### Member\nTOS is giving me trouble with this one. Some invalid statements etc. Has anyone run this lately? I would appreciate it, if one could clean it up a bit. From the \"Paint Bars\" down there are a few \"invalid statements\" as it says. It doesn't bring up anything. Blank. It has an exclamation mark within an orange triangle at the top of the script. Am I doing something wrong or is it the script again? thx\n\n#### lindosskier\n\n##### Member\nI did thank you. I wish we could get the code to work, from the original post. @BenTen your help would be much appreciated, if you could let me know what can possibly be wrong with the code. It does not bring up any \"Invalid statements\", but it has the orange triangle with the exclamation mark in it. don't know what that means. thx\n\nAnyone know how to make the original Volume Profile Study on TOS, be offset by 1 period?\nThat way, yesterday's, or last week's range will be showing on today's, or this week's chart?\nSince I can seem to make the script above show on my charts and I do not know how to code, I could use your help.\nThx in advance for everyone's help\n\n#### XeoNoX\n\n##### Well-known member\nVIP\ni hand typed the COG Center of Gravity code for TOS ( thinkorswim ) thinkscript indicator study that was posted above, since i cant copy and paste the picture. Here it is in easier copy and paste format. Please keep entire code intact.\n\nCode:\n``````# Ehlers_COG (Center Of Gravity)\n# by growex\n# Posted by: Mobius\n# 12.18.2017\n# Mobius: Someone was asking for a COG study with bands. Cog is an\n# unbounded oscillartor as designed by Ehlers. It would need to be\n# normalized to the price chart to have bands. Otherwise a 1 bar\n# lag on the oscillator gives very good signals. COG is 1 weighted\n# average shifted backwards to their center of lag or balance thus\n# the name. I was picturing the lagged second line or signal line\n# as the second of two but that's actually the same WAVG just\n# lagged one bar\n#\n# Mobius: This study is an accurate translation of Ehlers Center\n# Of Gravity.\n\ndeclare lower;\n\ninput price = hl2;\ninput length = 10;\n\ndef Num = fold i = 0 to length -1\nwith n\ndo n + (1 + i) * getvalue(price, i , length - 1);\ndef denom = fold j = 0 to length -1\nwith k\ndo k + getvalue(price, j, length -1);\ndef CG = if denom <> 0 then -Num/Denom else double.NaN;\n\nplot cgline = cg;\ncgline.SetDefaultColor(color.green);\nplot trigger = cgline;\ntrigger.setdefaultcolor(color.red);\n•", null, "" ]

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[ "I wrote two solutions. The first one finds all solutions, but it is\nslow (takes about 12 seconds on my machine). I got around this by\nwriting the results to a yaml file and using that for subsequent runs.\nMy second solution just picks a random permutation until if finds one\nthat meets the requirements.\n\n# chess960.rb\nrequire 'permutation'\nrequire 'yaml'\n\nfile = 'positions.yml'\n\ndef create_positions_file(file, pieces)\npositions = Permutation.for(pieces).map{|p| p.project}.uniq\npositions = positions.select do |p|\nare_bishops_on_opposite_colors?(p) && is_king_between_rooks?(p)\nend\n\nFile.open(file ,'w+'){|f| f.write(YAML::dump(positions))}\nend\n\ndef are_bishops_on_opposite_colors?(a)\n(a.index('B') + a.rindex('B')).modulo(2) != 0\nend\n\ndef is_king_between_rooks?(a)\n(a.index('R') < a.index('K')) && (a.index('K') < a.rindex('R'))\nend\n\ncreate_positions_file(file, %w{R N B K Q B N R}) unless\nFile.exist?(file)\n\nrandom = rand(positions.size)\nputs \"Starting position #{random}:\"\n\n{1 => 'White', 8 => 'Black'}.sort.each do |k, color|\nplace = ('a'..'h').to_a.join(k.to_s + ' ') + k.to_s\nputs \"\\n#{color}\\n\\n#{place}\\n #{positions[random].join(' ')}\"\nend\n\n#eof\n\n#chess960_2.rb\nrequire 'permutation'\n\ndef bishops_on_opposite_colors?(a)\n(a.index('B') + a.rindex('B')).modulo(2) != 0\nend\n\ndef king_between_rooks?(a)\n(a.index('R') < a.index('K')) && (a.index('K') < a.rindex('R'))\nend\n\npieces = %w{R N B K Q B N R}\n\nloop do\n@positions = Permutation.for(pieces).random.project(pieces)\nbreak if bishops_on_opposite_colors?(@positions) &&\nking_between_rooks?(@positions)\nend\n\n{1 => 'White', 8 => 'Black'}.sort.each do |k, color|\nplace = ('a'..'h').to_a.join(k.to_s + ' ') + k.to_s\nputs \"\\n#{color}\\n\\n#{place}\\n #{@positions.join(' ')}\"\nend" ]

[ null ]

[ "Anda di halaman 1dari 36\n\nSolutions\n\n1. Write a C program that receives 10 float numbers from the console and sort them in nonascending\norder, and prints the result\n\n#include <stdio.h>\n\n//print message of the numbers the user enter\n\nprintf(\"Numbers that you enter are:\\n %f\\n %f\\n %f\\n %f\\n %f\\n %f\\n %f\\n %f\\n %f\\n\n%f\\n\",numbers,numbers,numbers,numbers,numbers,numbers,numbers,numbers,\nnumbers,numbers);\n\n//print a message of the numbers in decreasing order\n\nprintf(\"decreasing order:\\n\");\n\nfor(i=10;i>=0;i--) //loop for compare the numbers enter\n\n{ for(k=10;k>=0;k--) //loop for compare the numbers with the array numbers[k]\n\n{ if(numbers[i]<numbers[k]) //compare\n\n{\nresult=numbers[i]; //save number\n\nnumbers[i]=numbers[k]; //save the greater number one after another\n\nnumbers[k]=result; } } }\n\n//print the numbers in decreasing number\n\nprintf(\" %f\\n %f\\n %f\\n %f\\n %f\\n %f\\n %f\\n %f\\n %f\\n\n%f\\n\",numbers,numbers,numbers,numbers,numbers,numbers,numbers,numbers,\nnumbers,numbers);\n\nscanf (\"%x\",&p);\n\n} //end of the program\n\n2. Write a C program that receives an integer number and determines if it is a prime number\n\nor not.\n\nand determines if it is a prime number or not*/\n\n#include <stdio.h>\n\n#include <stdlib.h>\n\n#define FALSE 0\n\n#define TRUE 1\n\nint main(void)\n\nprintf(\"Please enter an integer number that is desired to be prime-tested:\\n\"); //User Input\n\nscanf(\"%d\", &num1); //Stores Inputted value in num1 location\n\nprintf(\"\\nThe number you just entered was %d\\n\\n\",num1); //Reminds the user the\nnumber they just entered\n\nprintf(\"1 determines that the inputted number is Prime.\\n\");\n\nprintf(\"0 determines that the inputted number is NOT Prime.\\n\\n %d\", prime_test(num1));\n//Outputs the result of the called function\n\nint prime_test(int n) //Starts next function\n\nlong i; //Initializes i\n\nif (n < 2) return FALSE; //Prime numbers are being assumed greater than 2 (3,7,11,etc. 1\nand zero are excluded.)\n\nfor (i = 2; i < n; i++) //Starts loop\n\nif ((n % i) == 0) return FALSE; //If remainder of division is zero, then it is not a prime number\n\nreturn TRUE; //If conditions are not met, then the number has to be prime,\n\n//as its only divisible by 1 and by itself.\n\n3. Write a C program that receives 11 float numbers of three significant figures from the\n\nconsole. Sort it in non-ascending order and print the midpoint of the list\n#include <stdio.h>\n\n#include <stdlib.h>\n\nmain (void)\n\nint index,x;\n\nprintf(\"Write a number \\n\",index);\n\nscanf(\"%f\",&Number [index]);\n\nfor (x=0; x<11; x++) //This part of the code is to sort the 11 numbers in a non ascending form\n\nif (Number [index] >Number[index+1])\n\nz1=Number [index];\n\nz2=Number [index+1];\n\nNumber [index]=z1;\n\nNumber [index+1]=z2;\n}\n\nprintf (\"Middle Point is %f\\n\",Number); // In this instruction it show you the middle point\n\nsystem(\"PAUSE\");\n\nreturn 0;\n\n4. Write a C program that takes as input two memory addresses between 0 and ((2^8) – 1) and\n\nprints the displacement address (e.g. the distance between two points).\n\n/* This code can be executed using a Dev-C++ compiler.\n\nQuestion 4: Write a C program that takes as input two memory addresses between 0 and ((2^8) – 1)\n\nand prints the displacement address (e.g. the distance between two point).\n\n*/\n\n#include <stdio.h>\n\n#include <stdlib.h>\n\nint main(int argc, char *argv[])\n\nint address1 = -1; // It is initialized to -1 so we can use it as a condition for the while loop.\n\nint address2 = -1;\n\nint displacement; // This wil hold the distance between the addresses.\n\n// This while loop will assure the program to get the correct input we desire for the first address.\n\nwhile(address1 < 0 || address1 > 255) // The input must be bigger than 0 and less than 256 so\nthat we can get out the while loop.\n\n{\nprintf(\"Enter first address: \");\n\n// This while loop will assure the program to get the correct input we desire for the second address.\n\nwhile(address2 < 0 || address2 > 255)// The input must be bigger than 0 and less than 256 so\nthat we can get out the while loop.\n\ndisplacement = fabs(address1 - address2); // This computes the distance of the addresses by\n\nsubtracting them and using absolute value.\n\nprintf(\"Displacement: %d\\n\", displacement);\n\nsystem(\"PAUSE\");\n\nreturn 0;\n\n5. Write a C program that receives as input 4 parameters named x1 ,x2, y1 ,y2 representing\n\ncontinuous points of a linear equation(f(x)=x*m+b). Calculate the rate of change and also indicate if it\nis\n\nincreasing or decreasing.\n\n#include<stdio.h>\n\n#include<conio.h>\n\nvoid main()\n\n{\nint x1,x2,y1,y2;\n\nscanf(\"%d\",&x1);\n\nscanf(\"%d\",&y1);\n\nscanf(\"%d\",&x2);\n\nscanf(\"%d\",&y2);\n\nprintf(\"The equation is decreasing. \\n\");\n\nelse if (m == 0) {\nprintf(\"The equation is not changing. \\n\");\n\nsystem(\"PAUSE\");\n\nreturn 0;\n\n6. Write a C program that reads from the console a random number and determine if it is a\n\npower of 2.\n\n#include <stdio.h>\n\nint main () {\n\nint x;\n\nscanf(\"%d\", &x);\n\n//This function checks whether the number is a power\n\n//of 2 or not\n\nif((x&(x-1))==0){\n\nelse {\n\nprintf(\"The number you selected is not a power of 2\\n\");\n\nsystem(\"PAUSE\");\n\nreturn 0;\n\n}\n7. Write a C program that receives a positive float number and divides it by two until the result\n\nis less than or equal to zero. Print the number of iterations required to meet the specification. Explain\nin\n\nthe comments the behavior of your code.\n\n#include <stdio.h>\n\n#include <stdlib.h>\n\n#include <math.h>\n\nfloat x;\n\nprintf(\"Enter a floating point number: \");\n\nscanf(\"%f\" ,&x); //Asks the user for a positive floating point number.\n\nif(x<=0)\n\n//If the input is negative, throws an exception.\n\nprintf(\"Invalid response!\\n\");\n\nsystem(\"PAUSE\");\n\nreturn 0;\n\nelse{\n\nfloat t;\n\nwhile(!(x<=0)){ //While the answer is not equal or less than zero.\n\nt=x/2;\nx=t; //Updates by the value of the division by 2.\n\nprintf(\"Number of iterations it took:%d\\n \", count); //Number of iterations.\n\nsystem(\"PAUSE\");\n\nreturn 0;}\n\n8. Write a C program that takes as input an 8-bit number (e.g. 11011011). Perform and print\n\nthe logical NOT of the number. (Use the bitwise operations).\n\n/*Question 8: Write a C program that takes as input an 8-bit number (e.g. 11011011). Perform and print\n\nthe logical NOT of the number. (Use the bitwise operations.)*/\n\n#include <stdio.h>\n\n#include <stdlib.h>\n\nint main()\n\n{ //Initializing Variables:\n\nint binary_array;\n\nint notbinaryarray;\n\nint bincountr = 0;\n\nint bincountr2 = 0;\n\nint r,i;\nprintf(\"Enter desired binary number (from Most Significant Bit\\nto Least Significant Bit):\\n\");\n\nfor ( r = 0; bincountr < 8; r++) //Binary number loop\n\nscanf(\"%d\",&binary_array[r]);\n\nbincountr++;\n\nprintf(\"\\nThe inputted binary number was: \\n\");\n\nprintf(\"%d%d%d%d%d%d%d%d\\n\", binary_array,binary_array,binary_array,\n//Displays inputted binary value\n\nbinary_array,binary_array,binary_array,binary_array,binary_array);\n\nprintf(\"\\nThe logical NOT operation of the inputted number is:\\n\");\n\nfor ( i = 0; bincountr2 < 8; i++) //Calculates logical NOT bit by bit using bitwise operation XOR\n\nnotbinaryarray[i]=(binary_array[i] ^ 1 ) ;\n\nbincountr2++;\n\nprintf(\"%d%d%d%d%d%d%d%d\\n\", notbinaryarray,notbinaryarray,notbinaryarray,\n//Displays logical NOT of inputted value.\n\nnotbinaryarray,notbinaryarray,notbinaryarray,notbinaryarray,notbinaryarray);\n\nprintf(\"\\n\");\n\ngetch(); //Another method of writing System Pause in dev c++\n\nreturn 0;\n\n9. Write a C program that takes as input two 8-bit numbers and calculate the AND, OR, and\nXOR logical representations. (Use the bitwise operations).\n\n#include <stdio.h>\n\n#include <stdlib.h>\n\n//Question 9: Write a C program that takes as input two 8-bit numbers and calculate the AND,OR\n\nprintf (\"Enter the first binary number of 8-bits:\\n\");\n\nfor (i=0; i<8; i++) //Asking for the first binary number\n\nprintf (\"Enter the second binary number of 8-bits:\\n\");\n\nfor (i=0; i<8; i++) //Asking for the second binary number\n\nscanf (\"%d\", &number2 [i]);\n\nprintf (\"The AND logical representation of the two entered binary numbers is:\\n\");\n\nfor (i=0; i<8; i++) //AND logical representation using \"&\" like AND operator\n\nprintf (\"%d\\n\", number1 [i] & number2 [i]);\n\nprintf (\"The OR logical representation of the two entered binary numbers is:\\n\");\n\nfor (i=0; i<8; i++) //OR logical representation using \"|\" like OR operator\n\nprintf (\"%d\\n\", number1 [i] | number2 [i]);\n\nprintf (\"The XOR logical representation of the two entered binary numbers is:\\n\");\n\nfor (i=0; i<8; i++) //XOR logical representation using \"^\" like XOR operator\n\n}\n\nsystem(\"PAUSE\"); //Screen pause after the program runs\n\n} //Program End\n\n10. Write a C program that takes as input an 8-bit number. Perform and print the logical NEG\n\n/*\n\nQuestion 10: Write a C program that takes as input an 8-bit number.\n\nPerform and print the logical NEG of this number. (Use the bitwise operations.)\n\n*/\n\n#include <stdio.h>\n\n#include <stdlib.h>\n\nprintf(\"Enter an 8-bit (Only O and 1 are allowed as input)\\n\");\n\n// This for loop will run 8 times to get as input each bit\n\nfor(i=0; i<8; i++)\n\n{\nint bit = -1; // Holds the input from the user. I initialized to -1 to use it also as a flag to check that\nthe input is only 0 or 1.\n\nwhile(bit == -1)// This while loop only permits 0 and 1 as inputs for the array.\n\nscanf(\"%d\", &bit);\n\nbit = -1;\n\nbinary[i] = bit;\n\nfor(i = 0; i<8; i++)\n\nprintf(\"%d\", binary[i]);\n\n/*\n\nThis for loop actually does the real job of negating the binary number.\n\nEach iteration will check what is the bit in the specified index (i), if\n\n*/\n\n{\nif(binary[i]==0)\n\nbinary[i] = 1;\n\nelse\n\nbinary[i]=0;\n\nprintf(\"\\nApplying the NEG operator to the binary number results: \");\n\n// This for loop prints the negated 8-bit binary number received.\n\nfor(i = 0; i<8; i++)\n\nprintf(\"%d\", binary[i]);\n\nprintf(\"\\n\");\n\nsystem(\"PAUSE\");\n\nreturn 0;\n\n11. Write a C program that takes as input two 8-bit binary numbers and perform Binary ADD\n\nand print the result. (Use the bitwise operations).\n\n#include <stdio.h>\n\n#include <stdlib.h>\n\n#define SIZE 8\nint main(int argc, char *argv[])\n\nint f[SIZE];\n\nint s[SIZE];\n\nprintf(\"Enter the first 8-digit binary number\\n\\n\");\n\nfor(i=SIZE-1; i>=0; i--)//this loop assigns the input to a location in the array f\n\nscanf(\"%d\", &key);\n\nf[i] = key;\n\nfor(i=SIZE-1;i>=0;i--)\n\nprintf(\"%d\", f[i]);\n\nprintf(\"\\n\\nEnter the second 8-digit binary number\\n\");\n\nfor (i=SIZE-1; i>=0; i--)//this loop assigns the input to a location in the array s\n\n{\nscanf(\"%d\",&key);\n\ns[i] = key;\n\nfor(i=SIZE-1; i>=0; i--)\n\nprintf(\"%d\", s[i]);\n\nint sum[SIZE+1];\n\nint a=0;\n\nint j, sum1;\n\nfor(j=0; j<SIZE; j++)//This loop excecutes the sum of the binary numbers and assigns the carry to a\nvariable a\n\nsum1 = f[j]+s[j] + a; //when the som is 2 the carry a is equal to 1 and this location im the array is 0\n\nif(sum1 == 2)\n\na = 1;\n\nsum[j] = 0;\n\nelse if(sum1 == 3)\n\nsum[j] = 1;\n\na = 1;\n}\n\na = 0;\n\nsum[j] = sum1;\n\nif(j==SIZE-1)\n\nsum[SIZE]= a;\n\nfor(i=SIZE; i>=0; i--)\n\nprintf(\"%d\", sum[i]);\n\nsystem(\"PAUSE\");\n\nreturn 0;\n\n12. Write a C program that takes as input an 8-bit number and calculates the 2’s complement\n\nlogical representation. (Use the bitwise operations).\n\n#include <stdio.h>\nint main (int argc, char * const argv[]) {\n\n//This is an array of 8 integers called digits\n\nint digits;\n\nint i;\n\nint n;\n\nprintf(\"Please enter an 8 bit number, the first number being 2^0 and so on until 2^7:\\n\");\n\n//In this part the scanned numbers received by the user are\n\n//arranged into each space in the array from 2^0 to 2^7\n\nscanf(\"%d\\n\", &digits);\n\nscanf(\"%d\\n\", &digits);\n\nscanf(\"%d\\n\", &digits);\n\nscanf(\"%d\\n\", &digits);\n\nscanf(\"%d\\n\", &digits);\n\nscanf(\"%d\\n\", &digits);\n\nscanf(\"%d\\n\", &digits);\n\nscanf(\"%d\", &digits);\n\nif(digits==1 && digits[i]==0){\n\ndigits[i]=1;\n\n}\nelse if(digits==1 && digits[i]==1){\n\ndigits[i]=0;\n\nfor(n=1; n<8; n++){\n\nif(digits[i+n]==1){\n\ndigits[i+n]=0;\n\nelse if(digits[i+n]==0){\n\ndigits[i+n]=1;\n\nprintf(\"The numbers 2's compliment is:\\n\");\n\nprintf(\"%d\\n\", digits);\n\nprintf(\"%d\\n\", digits);\n\nprintf(\"%d\\n\", digits);\n\nprintf(\"%d\\n\", digits);\n\nprintf(\"%d\\n\", digits);\n\nprintf(\"%d\\n\", digits);\n\nprintf(\"%d\\n\", digits);\n\nprintf(\"%d\\n\", digits);\nsystem(\"PAUSE\");\n\nreturn 0;\n\n13. Write a C program that generates a random positive float number in a fixed interval with a\n\nmaximum distance between endpoints of 100 (e.g. [0,100] or [101,201]). Both interval points are\ngiven\n\nby the user through console. The generated random number inside the given interval must be less\nthan\n\nor equal to 10975. Calculate the factorial (n!) of the resulting random number.\n\n/*\n\nThis code can be executed with a Dev-C++ compiler.\n\nQuestion 13: Write a C program that generates a random positive float number\n\n100 (e.g. [0,100] or [101,201]). Both interval points are given by\n\nthe user through console. The generated random number inside the given\n\ninterval must be less than or equal to 10975. Calculate the factorial (n!)\n\nof the resulting random number.\n\n*/\n\n#include <stdio.h>\n\n#include <stdlib.h>\n\n#include <time.h>\n\nint main(int argc, char *argv[])\n\n{\n\nint flag = 0; // This flag is used to control the while loops to get the correct input from the user.\n\nint i; // This is the counter for the for loops.\n\ndouble factorial = 1; // This will accumulate the factorial for our random number.\n\nint endpoint1, endpoint2; // This will be the endpoints for the interval that our random number will be\ninside of.\n\nint distance; // This will be the distance between the two endpoints\n\nsrand(time(NULL)); // This will help generate better random numbers by using the current time as the\nseed.\n\n// This while loop will assure the program that the user only inputs a number less than 10875.\n\nwhile(flag == 0)\n\nprintf(\"Enter first endpoint of interval: \");\n\nscanf(\"%d\", &endpoint1);\n\nif(endpoint1 <= 10875 && endpoint1 >= 1) // It makes sure that the number is smaller than 10876\nand bigger than 0.\n\nflag = 1;// When this flag changes to 1, the user's input was correct.\n\nflag = 0; // Since we will used it again for our second input\n\n// This while loop will assure the program that the user only inputs a number that will have a\nmaximum distance of 100 from the first endpoint.\n\nwhile(flag == 0)\n{\n\nprintf(\"Enter second endpoint of interval: \");\n\nscanf(\"%d\", &endpoint2);\n\ndistance = endpoint2 - endpoint1;\n\nif(distance <= 100 && distance >= 1) // It makes sure that the distance between endpoints is smaller\nthan 101 and bihher than 0.\n\nflag = 1;// When this flag changes to 1, the user's input was correct.\n\nprintf(\"The inputs for the interval are [ %d , %d ]\\n\", endpoint1, endpoint2);\n\nint random = endpoint1 + (rand()%distance)+1; // This generates our random number between our two\nendpoints.\n\nprintf(\"Random number: %d\\n\", random);\n\n/*\n\nThis for loop accumulates the factorial of each number until we reach our random number.\n\nUnfortunately I couldn't find a data type that could hold up such big numbers when it passes over 170.\n\nThe unsigned long int can only hold up to 4294967295. The one I used was double which holds up to\n1.8x10^308.\n\nThe more appropiate datatype would be long double which can hold up to 1.2x10^4932.\n\n*/\nfor( i = 1; i <= random; i++)\n\nfactorial *= i;\n\nprintf(\"The factorial for the random number %d is: %f\\n\", random, factorial);\n\nsystem(\"PAUSE\");\n\nreturn 0;\n\n14. Assume an ideal voltage divider circuit. Write a C program that reads from the console a\n\nparameter float (Vin) and a second parameter float (Vout ) and print the load resistor needed for this\n\ndesign with its unit. Fix the other resistor to 1kOhm.\n\n#include <stdio.h>\n\n#include <stdlib.h>\n\n//Question 14: Assume an ideal voltage divider circuit.\n\n//Write a C program that reads from the console a parameter float (Vin)\n\n//and a second parameter float (Vout ) and print the load resistor needed\n\n//for this design with its unit. Fix the other resistor to 1kOhm.\n\nfloat Vin;\n\nfloat R1 = 1000;//asigning the fixed resistor to 1 Kohm\n\nprintf(\"\\nPlease enter the input voltage\\n\");\n\nscanf(\"%f\",&Vin);\n\nprintf(\"\\nEnter the output voltage\\n\");\n\nscanf(\"%f\",&Vout);\n\nprintf(\"The Load resistor needed for this design is %f\\n\",Rload);\n\nsystem(\"PAUSE\");\n\nreturn 0;\n\n15. Write a C program that takes as input a positive fractional number and converts it to\n\nbinary.\n\n#include<stdio.h>\n\n#include<stdlib.h>\n\nmain (void) //Program Start\n\nfloat Num; //Variables initialization. In this code \"float\" is used fot the\n\nfloat Denom; //fraction division and \"int\" is used for the other variables and\n\nint i; //parameters.\n\nint Number1;\nint n=0;\n\nint Numb[n];\n\nint j;\n\nint x=0;\n\nprintf(\"Please enter the positive fraction numerator:\\n\");\n\nscanf(\"%f\", &Num); //Asking for the numerator for the fraction division\n\nprintf(\"Please enter the positive fraction denominator:\\n\");\n\nscanf(\"%f\", &Denom); //Asking for the denominator for the fraction division\n\ndo\n\nprintf(\"\\n\");\n\nfor(i=n-1; i>=0; i--) //Reversing the numbers order\n\nprintf(\"%d\", Numb[i]);\n\nprintf(\".\"); //Point between the integer and decimal part of the number\n\ndo\n\nNum = Num*2; //Multiplpy by 2 for reached the 15 values\n\nj=Num; //Comparation\n\nNum = Num-j;\n\nx++;\n}\n\nwhile(x!=15); //The maximun quantity of values for a binary number is 15\n\nprintf(\"\\n\");\n\nsystem (\"PAUSE\"); //Pausing the result screen after the program run\n\n} //Program End\n\n16. Write a C program that takes a non- negative integer number as input and prints its binary\n\nrepresentation.\n\n/*Assume the output number is an 8 bit binary number.*/\n\n#include <stdio.h>\n\n#include <stdlib.h>\n\nint main()\n\nint inputnumber;\n\nint i;\n\nint n;\n\nprintf(\"Enter the desired integer to be converted to binary (enter -1 to finish)\\n\");\n\nscanf(\"%d\", &inputnumber); //User input\n\nwhile( inputnumber != -1 ) //While loop to input different numbers at one's own convenience\n\nif(inputnumber < 256 && inputnumber >= 0) //restrictions taken from the Range of values\npossible\n\nfor( i = 0; i < 8; i++ )\n\nn = inputnumber % 2;\n\nif( inputnumber == 0 )\n\nbinaryoutput[i]=0;\n\nelse\n\nbinaryoutput[i]=n;\n\ninputnumber = inputnumber / 2;\n\nelse\n\n{\nprintf(\"\\nNumber out of boundary please re-enter to end -1\\n\"); //Wrong input gives a\nchance to repeat.\n\nprintf(\"\\nThe binary representation\n\nis:\\n%d%d%d%d%d%d%d%d\\n\",binaryoutput,binaryoutput,binaryoutput,\n\nbinaryoutput,binaryoutput,binaryoutput,binaryoutput,binaryoutput );\n\nprintf(\"\\nEnter the desired integer to be converted to binary\\n\");\n\nscanf(\"%d\", &inputnumber);\n\nsystem(\"PAUSE\");\n\nreturn 0;\n\n17. Write a C program that takes a 2 digit hexadecimal number from the user through console\n\nand display it in binary format.\n\n#include <stdio.h>\n\nchar number;\n\nint i;\n\n// Here it scans the number given and stores it in the array\n\nscanf(\"%s\", number);\n\nfor(i=0; i<2; i++){\n\n//If the number is equal to one number in hexadecimal it print in\n\n//the binary format\n\nif(number[i] == '0'){\n\nprintf(\"0000\");\n\nprintf(\"0001\");\n\nprintf(\"0010\");\n\nprintf(\"0011\");\n\nprintf(\"0100\");\n\nprintf(\"0101\");\n\nprintf(\"0110\");\n\nprintf(\"0111\");\n}\n\nprintf(\"1000\");\n\nprintf(\"1001\");\n\nprintf(\"1010\");\n\nprintf(\"1011\");\n\nprintf(\"1100\");\n\nprintf(\"1101\");\n\nprintf(\"1110\");\n\nprintf(\"1111\");\n\n}\nelse {\n\nprintf( \"You did not follow the rules\\n\" );\n\nprintf( \"\\n\" );\n\nsystem(\"PAUSE\");\n\nreturn 0;\n\n18. Write a C program that takes an 8-bit number through console as input. Convert it to\n\nhexadecimal format and display it as output.\n\n#include <stdio.h>\n\n#include <string.h>\n\nwhile(strcmp(bin,\"0\")) // Start While Loop With String Compare\n\nCondition\n{\n\nprintf(\"\\nPlease Enter an 8-Bit Binary Number (Press 0 To EXIT): \"); // Prompts For User Input\n\nfgets(bin, sizeof(bin), stdin); // Reads Characters From Stream And Stores Them\nAs A C String\n\nComes First\n\nCondition...\n\n*p = '\\0'; // If False Condition Declare Null\n\ndec = bin2dec(bin);\n\n}\n\nint len, sum = 0; // States Variable Into Memory\n\nlen = strlen(bin) - 1;\n\nb = 1;\n\nif ((n > 1) || (n < 0)) // Makes Sure It Is A Binary Input\n\nputs(\"\\n\\n ERROR! BINARY has only 1 and 0!\\n\"); // Displays Error Message If\nCondition Is Not Met\n\nreturn (0);\n\nb = b<<(len-k);" ]

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[ "Home   Package List   Routine Alphabetical List   Global Alphabetical List   FileMan Files List   FileMan Sub-Files List   Package Component Lists   Package-Namespace Mapping\nRoutine: SDWLE3\n\n# SDWLE3.m\n\nGo to the documentation of this file.\n```SDWLE3 ;IOFO BAY PINES/TEH - WAITING LIST-ENTER/EDIT;06/12/2002 ; 12/14/05 1:28pm ; Compiled April 25, 2006 10:42:02\n```\n``` ;;5.3;scheduling;**263,417,446**;AUG 13 1993;Build 77\n```\n``` ;\n```\n``` ;\n```\n``` ;******************************************************************\n```\n``` ; CHANGE LOG\n```\n``` ;\n```\n``` ; DATE PATCH DESCRIPTION\n```\n``` ; ---- ----- -----------\n```\n``` ; 08/01/2005 SD*5.3*417 Permit multiple teams\n```\n``` ; 04/21/2006 SD*5.3*446 Inter-Facility Transfer\n```\n``` ;\n```\n``` ;\n```\n```EN ;\n```\n``` ;ASK FOR SPECIFIC TEAM (404.51)\n```\n``` K DIR,DIC,DR,DIE,SDTMENT S (DA,SDTMENT)=SDWLDA K SDWLTH,SDWLMAX\n```\n``` S SDWLYN=5,SDWLTYE=1,SDWLVBR=\"SDWLST\"\n```\n``` I \\$D(SDWLST),'SDWLST K SDWLST\n```\n``` I \\$G(SDWLCP3)'=\"\" D\n```\n``` .W !,\"This patient is already on the \",SDWLCP3,\".\" S DIR(0)=\"Y^A0\",DIR(\"B\")=\"NO\",DIR(\"A\")=\"Are you sure you want to continue\" D ^DIR\n```\n``` .I 'Y!(Y[\"^\") S DUOUT=1 Q\n```\n``` I \\$D(DUOUT),DUOUT G END\n```\n``` D GETLIST\n```\n``` S SDWLERR=0,SDWLY=\"Team\",SDWLVAR=\\$S(\\$D(SDWLST):SDWLST,1:0),SDWLSCR=\"\"\n```\n``` S SDWLVBR=\"SDWLST\"\n```\n```EN1 W ! S SDWLS=SDWLY,SDWLX=\\$S(SDWLTYE=1:\"T\",SDWLTYE=2:\"P\",1:\"\"),SDWLSX=\" \"_SDWLS\n```\n``` S SDWLF=\"SCTM(404.51,\"\n```\n``` S SDWLA=0 F S SDWLA=\\$O(^SCTM(404.58,\"B\",SDWLA)) Q:SDWLA=\"\" D\n```\n``` .I \\$D(SDWLCT),SDWLCT=SDWLA Q\n```\n``` .I \\$P(\\$G(^SCTM(404.51,SDWLA,0)),U,7)'=SDWLIN Q\n```\n``` .I +\\$\\$ACTTM^SCMCTMU(SDWLA)=0 S SDWLTH(SDWLA)=\"\"\n```\n``` .S SDWLMAX=0,X=\\$\\$TEAMCNT^SCAPMCU1(SDWLA,DT),SDWLMAX(SDWLA)=\"\" D\n```\n``` ..I X<\\$P(\\$G(^SCTM(404.51,SDWLA,0)),U,8) K SDWLMAX(SDWLA)\n```\n``` N SDWLT S SDWLT=0 F S SDWLT=\\$O(SDWLPLST(1,SDWLT)) Q:SDWLT<1 K SDWLMAX(SDWLT)\n```\n``` S SDWLSCR=\"I \\$P(^(0),U,7)=SDWLIN,'\\$D(SDWLTH(+Y)),\\$D(SDWLMAX(+Y)),'\\$D(SDWLPLST(SDWLTYE,+Y,SDWLIN))\"\n```\n``` D EN2 G END:\\$D(DUOUT)\n```\n``` ;DA=SDWLDA, see EN\n```\n``` S DR=\"5////^S X=SDWLVAR\",DIE=409.3 D ^DIE\n```\n``` N FLG D FLAGS(.FLG,DFN,SDWLVAR)\n```\n``` I 'FLG S DA=SDTMENT,DIE=409.3 D\n```\n``` .S SDINTR=FLG(1),SDREJ=FLG(2),SDMTM=FLG(3)\n```\n``` .S DR=\"32////^S X=SDREJ;34////^S X=SDINTR;38////^S X=SDMTM\" D ^DIE\n```\n``` ;\n```\n``` S @SDWLVBR=SDWLVAR\n```\n``` I \\$D(SDWLVARO),SDWLVARO,SDWLVAR'=SDWLVARO D DELPOS\n```\n``` G END\n```\n```EN2 ;-DIR READ\n```\n``` I '\\$D(SDWLDATA),\\$D(SDWLMAX)'=11 W !,\"No TEAMS are available for this INSTITUTION.\",! S DUOUT=\"\" Q\n```\n``` K DIR,DR,DIE,DIC,DUOUT\n```\n``` S DIR(\"?\")=\"^S X=\"\"?\"\",DIC(\"\"S\"\")=\"\"I \\$P(^SCTM(404.51,+Y,0),U,7)=SDWLIN,'\\$D(SDWLTH(+Y)),\\$D(SDWLMAX(+Y)),'\\$D(SDWLPLST(1,+Y,SDWLINE))\"\" S DIC=404.51,DIC(0)=\"\"EQMNZ\"\" D ^DIC\"\n```\n``` I \\$D(SDWLVAR),SDWLVAR S X=SDWLVAR,SDWLMPX=\\$\\$EXTERNAL^DILFD(409.3,SDWLYN,,SDWLVAR),DIR(\"B\")=SDWLMPX,SDWLVARO=SDWLVAR K X\n```\n``` S DIR(0)=\"FAO\",DIR(\"A\")=\"Select \"_SDWLY_\": \"\n```\n``` D ^DIR\n```\n``` I X[\"^\" S DUOUT=1 Q\n```\n``` S DUOUT=\\$S(X=0:1,X=\"@\":1,\\$D(DTOUT):1,1:0) I 'DUOUT K DUOUT\n```\n``` I X=\"@\" W *7,\" No deleting allowing.\" G EN2\n```\n``` S DIC(\"S\")=SDWLSCR\n```\n``` S DIC(0)=\"EMNZ\",DIC=404.51 D ^DIC I \\$D(DTOUT) S DUOUT=1\n```\n``` I \\$D(DUOUT) Q\n```\n``` I Y<0 W \"??\" G EN2\n```\n``` S SDWLVAR=+Y\n```\n``` Q\n```\n``` ;identify flags\n```\n```FLAGS(FLG,DFN,TEAM) ;\n```\n``` N SDTEAM S SDTEAM=\\$G(TEAM)\n```\n``` ; check if transfer and if multiple teams in institution\n```\n``` S SDCNT=0,SDINTR=0,SDREJ=0,SDMTM=0 D\n```\n``` .S SDWLIN=\\$P(\\$G(^SCTM(404.51,TEAM,0)),U,7)\n```\n``` .I \\$P(^SCTM(404.51,TEAM,0),U,5)'=1 Q ; cannot be primary care provider team\n```\n``` .;identify INTRA-transfer\n```\n``` .;- is patient assigned to PC provider?\n```\n``` .I \\$\\$GETALL^SCAPMCA(DFN) D\n```\n``` ..I \\$G(^TMP(\"SC\",\\$J,DFN,\"PCPOS\",0)) S SDTM=\\$P(^(1),U,3) I SDTM>0 D\n```\n``` ...I \\$P(\\$G(^SCTM(404.51,SDTM,0)),U,7)'=SDWLIN S SDINTR=1 D ; inter transfer ; different institution\n```\n``` ..I '\\$G(^TMP(\"SC\",\\$J,DFN,\"PCPOS\",0)) D\n```\n``` ...;check available PCMM teams in other institutions and if so set up rejection flag\n```\n``` ...S SDINS=\"\"\n```\n``` ...F S SDINS=\\$O(^SCTM(404.51,\"AINST\",SDINS)) Q:SDINS=\"\" I SDINS'=SDWLIN D Q:SDREJ\n```\n``` ....S SDCNT=0,SDT=\"\"\n```\n``` ....F S SDT=\\$O(^SCTM(404.51,\"AINST\",SDINS,SDT)) Q:SDT=\"\" D Q:SDREJ\n```\n``` .....I \\$\\$ACTTM^SCMCTMU(SDT,DT)&(\\$P(\\$G(^SCTM(404.51,SDT,0)),U,5))&'\\$P(\\$G(^SCTM(404.51,SDT,0)),U,10) D\n```\n``` ......S SCTMCT=\\$\\$TEAMCNT^SCAPMCU1(SDT) ;currently assigned\n```\n``` ......S SCTMMAX=\\$P(\\$\\$GETEAM^SCAPMCU3(SDT),\"^\",8) ;maximum set\n```\n``` ......I SCTMCT<SCTMMAX S SDREJ=1\n```\n``` ..;find all teams from institution SDWLIN\n```\n``` ..I SDINTR S SDCNT=0,SDT=\"\" D\n```\n``` ...F S SDT=\\$O(^SCTM(404.51,\"AINST\",SDWLIN,SDT)) Q:SDT=\"\" I \\$P(^SCTM(404.51,SDT,0),U,5)=1 S TEAM(SDT)=\"\",SDCNT=SDCNT+1\n```\n``` S FLG(1)=SDINTR,FLG(2)=SDREJ,FLG(3)=SDMTM\n```\n``` I SDCNT>1 S SDMTM=1,FLG(3)=SDMTM,FLG=1 S SDCC=\"\" F S SDCC=\\$O(TEAM(SDCC)) Q:SDCC=\"\" S TEAM=SDCC N DR,Y D WMT\n```\n``` I SDCNT>1 S TEAM=\\$G(SDTEAM) Q\n```\n``` I SDCNT'>1 N DR,Y S FLG=0 S TEAM=\\$G(SDTEAM) Q\n```\n```WMT D INPUT^SDWLRP1(.RES,DFN_U_1_U_TEAM_U_U_DUZ_\"^^\"_U_SDINTR_U_SDREJ_U_SDMTM)\n```\n``` ;I \\$G(RES) S OK=0,DA=+\\$P(RES,U,2),DIE=\"^SDWL(409.3,\",DR=\"25;S OK=1\" D ^DIE I '\\$G(OK) S DIK=DIE D ^DIK W !,\"Wait list entry deleted\"\n```\n``` Q\n```\n```GETLIST ;GET LIST OF TEAM ASSIGNMENTS - SD*5.3*417\n```\n``` N SDWLDAX,X,Z,SDWLIN K SDWLPLST S SDWLPLST=\"\"\n```\n``` S SDWLDAX=0 F S SDWLDAX=\\$O(^SDWL(409.3,\"B\",SDWLDFN,SDWLDAX)) Q:SDWLDAX=\"\" D\n```\n``` .S Z=\\$G(^SDWL(409.3,SDWLDAX,0)),X=\\$P(Z,U,5),SDWLINE=+\\$P(Z,U,3) Q:X'=1&(X'=2) D\n```\n``` ..S Y=+\\$S(X=1:\\$P(Z,U,6),X=2:\\$P(Z,U,7),1:0) Q:'Y D\n```\n``` ...I \\$P(Z,U,17)[\"O\" S SDWLPLST(X,Y,SDWLINE)=\"\" I \\$D(SDWLST),SDWLST=+Y K SDWLPLST(X,Y,SDWLINE)\n```\n``` S Y=0 F S Y=\\$O(SDWLCPT(Y)) Q:Y=\"\" D\n```\n``` .S SDWLPLST(1,Y,SDWLINE)=\"\" I \\$D(SDWLST),SDWLST=+Y K SDWLPLST(1,Y,SDWLINE)\n```\n``` Q\n```\n```DELPOS ;DELETE POSITIONS FOR OLD TEAM\n```\n``` S SDWLA=0,CNT=0 F S SDWLA=\\$O(^SDWL(409.3,\"B\",SDWLDFN,SDWLA)) Q:SDWLA<1 D\n```\n``` .S X=\\$G(^SDWL(409.3,SDWLA,0)) Q:\\$P(X,U,7)=\"\"\n```\n``` .I \\$P(X,U,5)'=2 Q\n```\n``` .I \\$P(X,U,17)[\"C\" Q\n```\n``` .S SDWLPX=+\\$P(X,U,7) I \\$P(\\$G(^SCTM(404.57,SDWLPX,0)),U,2)'=SDWLVARO Q\n```\n``` .S CNT=CNT+1,^XTMP(\"SDWLE3\",\\$J,CNT)=SDWLA_\";\"_X W !\n```\n``` I 'CNT Q\n```\n``` W !,\"This patient has one or more Wait List entries for PCMM Positions\",!\n```\n``` W !,\"Wait List Type\",?30,\"Waiting For\",?45,\"Institution\",?60,\"Date Entered\",!\n```\n``` S Y=0 F S Y=\\$O(^XTMP(\"SDWLE3\",\\$J,Y)) Q:Y<1 S X=\\$G(^XTMP(\"SDWLE3\",\\$J,Y)),SDWLIEN=\\$P(X,\";\",1) D\n```\n``` .W !,\\$\\$GET1^DIQ(409.3,SDWLIEN,4),?30,\\$\\$GET1^DIQ(409.3,SDWLIEN,6),?45,\\$\\$GET1^DIQ(409.3,SDWLIEN,2),?60,\\$\\$GET1^DIQ(409.3,SDWLIEN,1)\n```\n``` W ! S SDWLET=\\$\\$EXTERNAL^DILFD(409.3,SDWLYN,,SDWLVARO)\n```\n``` K DIR S DIR(\"?\",1)=\"This patient has one or more Wait List entries for PCMM positions.\"\n```\n``` S DIR(\"?\",2)=\"By answering 'YES' you will close the Wait List entries which were listed.\"\n```\n``` S DIR(\"?\")=\"Answer 'NO' to keep those Wait List entries open.\"\n```\n``` S DIR(\"A\")=\"Do you wish to close these POSITION(S) entries? \",DIR(0)=\"Y\",DIR(\"B\")=\"YES\" D ^DIR\n```\n``` I 'Y W *7,\" No POSITIONS closed.\" Q\n```\n``` N DA S SDWLA=0 F S SDWLA=\\$O(^SDWL(409.3,\"B\",SDWLDFN,SDWLA)) Q:SDWLA<1 D\n```\n``` .S X=\\$G(^SDWL(409.3,SDWLA,0)) Q:\\$P(X,U,7)=\"\" D\n```\n``` ..S SDWLP=\\$P(X,U,7) I \\$P(^SCTM(404.57,SDWLP,0),U,2)=SDWLVARO D\n```\n``` ...K DIE,DIC,DR,DICR,DIR S DA=SDWLA,SDWLDISP=\"NN\"\n```\n``` ...S DIE=\"^SDWL(409.3,\",DR=\"21////^S X=SDWLDISP\" D ^DIE\n```\n``` ...S DR=\"19////^S X=DT\" D ^DIE\n```\n``` ...S DR=\"20////^S X=SDWLDUZ\" D ^DIE\n```\n``` ...S DR=\"23////\"\"C\"\"\" D ^DIE\n```\n``` Q\n```\n```END K SDWLA,SDWLMAX,SDWLTH,SDWLSCR,DIR,DIC,DIE,DR,SDWLPLST,SDWLDAX,DTOUT,SDWLCP3,SDWLINE\n```\n``` K X,Y,Z,SDWLPLST,SDWLB,SDWLA,SDWLSX,SDWLS,SDWLVBR,SDWLVAR,SDWLSCR,SDWLF,SDWLYN,SDWLMPX\n```\n``` Q\n```" ]

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[ "• ### Measurement of the beam asymmetry $\\Sigma$ for $\\pi^0$ and $\\eta$ photoproduction on the proton at $E_\\gamma = 9$ GeV(1701.08123)\n\nMay 15, 2017 nucl-ex\nWe report measurements of the photon beam asymmetry $\\Sigma$ for the reactions $\\vec{\\gamma}p\\to p\\pi^0$ and $\\vec{\\gamma}p\\to p\\eta$ from the GlueX experiment using a 9 GeV linearly-polarized, tagged photon beam incident on a liquid hydrogen target in Jefferson Lab's Hall D. The asymmetries, measured as a function of the proton momentum transfer, possess greater precision than previous $\\pi^0$ measurements and are the first $\\eta$ measurements in this energy regime. The results are compared with theoretical predictions based on $t$-channel, quasi-particle exchange and constrain the axial-vector component of the neutral meson production mechanism in these models.\n• ### First Results from The GlueX Experiment(1512.03699)\n\nJan. 14, 2016 hep-ex, nucl-ex, physics.ins-det\nThe GlueX experiment at Jefferson Lab ran with its first commissioning beam in late 2014 and the spring of 2015. Data were collected on both plastic and liquid hydrogen targets, and much of the detector has been commissioned. All of the detector systems are now performing at or near design specifications and events are being fully reconstructed, including exclusive production of $\\pi^{0}$, $\\eta$ and $\\omega$ mesons. Linearly-polarized photons were successfully produced through coherent bremsstrahlung and polarization transfer to the $\\rho$ has been observed.\n• ### A study of decays to strange final states with GlueX in Hall D using components of the BaBar DIRC(1408.0215)\n\nAug. 1, 2014 hep-ex, physics.ins-det\nWe propose to enhance the kaon identification capabilities of the GlueX detector by constructing an FDIRC (Focusing Detection of Internally Reflected Cherenkov) detector utilizing the decommissioned BaBar DIRC components. The GlueX FDIRC would significantly enhance the GlueX physics program by allowing one to search for and study hybrid mesons decaying into kaon final states. Such systematic studies of kaon final states are essential for inferring the quark flavor content of hybrid and conventional mesons. The GlueX FDIRC would reuse one-third of the synthetic fused silica bars that were utilized in the BaBar DIRC. A new focussing photon camera, read out with large area photodetectors, would be developed. We propose operating the enhanced GlueX detector in Hall D for a total of 220 days at an average intensity of 5x10^7 {\\gamma}/s, a program that was conditionally approved by PAC39\n• ### An initial study of mesons and baryons containing strange quarks with GlueX(1305.1523)\n\nMay 7, 2013 nucl-ex\nThe primary motivation of the GlueX experiment is to search for and ultimately study the pattern of gluonic excitations in the meson spectrum produced in $\\gamma p$ collisions. Recent lattice QCD calculations predict a rich spectrum of hybrid mesons that have both exotic and non-exotic $J^{PC}$, corresponding to $q\\bar{q}$ states ($q=u,$ $d,$ or $s$) coupled with a gluonic field. A thorough study of the hybrid spectrum, including the identification of the isovector triplet, with charges 0 and $\\pm1$, and both isoscalar members, $|s\\bar{s}\\ >$ and $|u\\bar{u}\\ > + |d\\bar{d}\\ >$, for each predicted hybrid combination of $J^{PC}$, may only be achieved by conducting a systematic amplitude analysis of many different hadronic final states. Detailed studies of the performance of the \\gx detector have indicated that identification of particular final states with kaons is possible using the baseline detector configuration. The efficiency of kaon detection coupled with the relatively lower production cross section for particles containing hidden strangeness will require a high intensity run in order for analyses of such states to be feasible. We propose to collect a total of 200 days of physics analysis data at an average intensity of $5\\times 10^7$ tagged photons on target per second. This data sample will provide an order of magnitude statistical improvement over the initial GlueX running, which will allow us to begin a program of studying mesons and baryons containing strange quarks. In addition, the increased intensity will permit us to study reactions that may have been statistically limited in the initial phases of GlueX. Overall, this will lead to a significant increase in the potential for \\gx to make key experimental advances in our knowledge of hybrid mesons and excited $\\Xi$ baryons.\n• ### Analytic approximations, perturbation methods, and their applications(0710.5658)\n\nNov. 2, 2007 gr-qc\nThe paper summarizes the parallel session B3 {\\em Analytic approximations, perturbation methods, and their applications} of the GR18 conference. The talks in the session reported notably recent advances in black hole perturbations and post-Newtonian approximations as applied to sources of gravitational waves." ]

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[ "# Benford law and lognormal distributions\n\nBenford’s law is nowadays extremely popular (see e.g. http://en.wikipedia.org/…). It is usually claimed that, for a given set data set, changing units does not affect the distribution of the first digit. Thus, it should be related to scale invariant distributions. Heuristically, scale (or unit) invariance means that the density of the measure  (or probability function) should be proportional to . Thus, because densities integrate to 1, the proportionality coefficient has to be , and therefore, should satisfy the following functional equation, , for all in  and in . The solution of this functional equation is , I guess this can be proved easily solving ordinary differential equation\n\nNow if denotes the first digit of , in base 10, then\n\nWhich is the so-called Benford’s law. So, this distribution looks like that\n\n> (benford=log(1+1/(1:9))/log(10))\n 0.30103000 0.17609126 0.12493874 0.09691001 0.07918125\n 0.06694679 0.05799195 0.05115252 0.04575749\n> names(benford)=1:9\n> sum(benford)\n 1\n> barplot(benford,col=\"white\",ylim=c(-.045,.3))\n> abline(h=0)", null, "To compute the empirical distribution from a sample, use the following function\n\n> firstdigit=function(x){\n+ if(x>=1){x=as.numeric(substr(as.character(x),1,1)); zero=FALSE}\n+ if(x<1){zero=TRUE}\n+ while(zero==TRUE){\n+ x=x*10; zero=FALSE\n+ if(trunc(x)==0){zero=TRUE}\n+ }\n+ return(trunc(x))\n+ }\n\nand then\n\n> Xd=sapply(X,firstdigit)\n> table(Xd)/1000", null, "It is not a mathematical article, so do not expect any formal proof in this paper. At least, we can run monte carlo simulation, and see what’s going on if we generate samples from a lognormal distribution with variance . For instance, with a unit variance,\n\n> set.seed(1)\n> s=1\n> X=rlnorm(n=1000,0,s)\n> Xd=sapply(X,firstdigit)\n> table(Xd)/1000\nXd\n1 2 3 4 5 6 7 8 9\n0.288 0.172 0.121 0.086 0.075 0.072 0.073 0.053 0.060\n> T=rbind(benford,-table(Xd)/1000)\n> barplot(T,col=c(\"red\",\"white\"),ylim=c(-.045,.3))\n> abline(h=0)", null, "Clearly, it not far away from Benford’s law. Perhaps a more formal test can be considered, for instance Pearson’s  (goodness of fit) test.\n\n> chisq.test(T,p=benford)\n\nChi-squared test for given probabilities\n\ndata: T\nX-squared = 10.9976, df = 8, p-value = 0.2018\n\nSo yes, Benford’s law is admissible ! Now, if we consider the case where is smaller (say 0.9), it is a rather different story,", null, "compared with the case where is larger (say 1.1)", null, "It is possible to generate several samples (always the same size, here 1,000 observations), just change the variance parameter and compute the -value of the test. There might be one tricky part: when generating samples from lognormal distributions with small variance, it might be possible that some digits do not appear at all. On that case, there is a problem with the test. So we just use here\n\n> T=table(Xd)\n> T=T[as.character(1:9)]\n> T[is.na(T)]=0\n> PVAL[i]=chisq.test(T,p=benford)$p.value Boxplots of the -value of the test are the following,", null, "When is too small, it is clearly not Benford’s distribution: for half (or more) of our samples, the -value is lower than 5%. On the other hand, when is large (enough), Benford’s distribution is the distribution of the first digit of lognormal samples, since 95% of our samples have -values higher than 5% (and the distribution of the -value is almost uniform on the unit interval). Here is the proportion of samples where the -value was lower than 5% (on 5,000 generations each time)", null, "Note that it is also possible to compute the -value of Komogorov-Smirnov test, testing if the -value has a uniform distribution, > ks.test(PVAL[,s], \"punif\")$p.value", null, "Indeed, if is larger than 1.15 (around that value), it looks like Benford’s law is a suitable distribution for the first digit.\n\n## 11 thoughts on “Benford law and lognormal distributions”\n\n1.", null, "Govind S. Mudholkar says:\n\nBENFORD’S LAW FOR COEFFICIENTS OF MODULAR FORMS …\nhttp://www.ams.org/proc/2011-139-05/…/S0002-9939-2010-10577-4.pdf‎\nby T Anderson – ‎2011 – ‎Cited by 1 – ‎Related articles\nOct 5, 2010 – PROCEEDINGS OF THE. AMERICAN MATHEMATICAL SOCIETY. Volume … Here we prove that Benford’s law holds for coefficients of an in-.\nPrevious article – Proceedings of the American Mathematical Society\nhttp://www.ams.org/proc/2011-139-05/S0002-9939-2010-10577-4/‎\nby T Anderson – ‎2011 – ‎Cited by 1 – ‎Related articles\nBenford’s law for coefficients of modular forms and partition functions … Abstract: Here we prove that Benford’s law holds for coefficients of an infinite class of …\nSimon Newcomb and “Natural Numbers” (Benford’s Law) – American …\nhttp://www.ams.org/samplings/feature-column/fcarc-newcomb‎\nby S Newcomb – ‎Related articles\ntony at math.sunysb.edu … This law was rediscovered by Frank Benford (“The law of anomalous numbers,” …. Frank Benford, The law of anomalous numbers, Proceedings of the American Philosophical Society 78 (1938) 551-572. Theodore P. Hill, Base-invariance implies Benford’s law, Proceedings of the A. M. S. 123 …\nBase-invariance implies Benford’s law – American Mathematical …\nhttp://www.ams.org/proc/1995-123-03/S0002-9939-1995-1233974-8/‎\nby TP Hill – ‎1995 – ‎Cited by 150 – ‎Related articles\nProceedings of the American Mathematical Society. Journals … Abstract: A derivation of Benford’s Law or the First-Digit Phenomenon is given assuming only …\nWorks that this work references – Benford Online Bibliog\n\n2.", null, "Govind S. Mudholkar says:\n\nPlease look for “Scale invariance implies Benford’s law” in the Proceedings of American Mathematical Society. I do not remember the author or year.\n\nGM\n\n3.", null, "NicolasGauvrit says:\n\nThanks Peter, this is exactly the book I should have read before !\n\n1.", null, "Peter Vijn says:\n\nNicholas,\nThe order in which we digest doesn’t matter, as long as the full picture emerges. For Benford the full picture is that it naturally arises from any wide distribution, defined as sigma > 0.45 on the (10)log scale, which reduces it to merely an epiphenomenon.\nWith this knowledge, the popular game of computing and publishing isolated Benford distributions is silly and should best be avoided. If Benford is of interest for a particular dataset, a band of distributions (consisting as the overlay of many Benford distributions) should be plotted instead, obtained by multiplying the original dataset with numbers between 1 and 10 in small increments (e.g. by iterative multiplication with say 1.05) and computing Benford with every step. Essentially, the ‘ones scaling test’ is applied to all nine first digits simultaneously which is a much better basis for further analysis and discussion. But even when using this ‘advanced’ method, all that is accomplished is a inefficient test for the width of the distribution of the original data.\nCheers, Peter\n\n4.", null, "Peter Vijn says:\n\nThanks for this! Very nice. I strongly recommend Steven Smith’s DSP chapter on Benford’s law (http://www.dspguide.com/ch34.htm 1997). After reading this the mystery is fully solved and explained. Your observations in these simulations are exactly in line with this, and there are also no surprises in Nicolas’ paper mentioned above.\n\n5.", null, "NicolasGauvrit says:\n\nYou should be interested in two papers we wrote (in french) some time ago. One proves a theorem from which Benford may be derived, and the second shows that the “generalized” Benford’s law stating that Frac(f(X))~Unif mod 1 is actually more often true than the classical Benford’s law — for some f.\nhttp://www.ehess.fr/revue-msh/pdf/N186R1370.pdf\nhttp://msh.revues.org/10363?file=1\n\nWe also wrote a chapter (in English) about the same subject: http://arxiv.org/abs/0910.1359\n\nCheers,\n\n6.", null, "Carlos J. Gil Bellosta says:\n\nI wrote a critical review of Benford’s Law some time ago in my blog pointing precisely at the magical way in which the first digit probabilities are presented. You can see it at\n\nhttp://www.datanalytics.com/blog/2011/09/15/la-ley-de-benford/\n\nIn fact, there does not seem to be a “single” Benford Law but several depending on the distribution of your data. As you point out, it may be the case that for some of those other distributions Benford’s probabilities are a good proxy, though.\n\n1.", null, "Arthur Charpentier says:\n\nExactly ! I’d be glad to see such a “generalized Benford law” which might work when data have regularly varying (Pareto type) tails, and a more uniform distribution. Benford is one possible distribution, which works well with the power decay. But alternative distribution should be possible for other underlying distributions, as you mention.\n\n7.", null, "DM says:\n\nI can’t really understand what the fuss is over Benford’s law personally. At least things like the Pareto law let you focus on smaller subsets and concentrate your efforts.\n\nThis aside, is there more power in reviewing the link between the coefficient of variation of the lognormal vs Benford’s law – ie the CoV encompasses all of the lognormal variability in a scale invariant metric?\n\n8.", null, "Christopher D. Long says:\n\nFix f(1) in your functional equation, then setting x=1 we get k*f(k) = f(1) for all k, thus f(k) = f(1)/k for all k.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]

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[ "# How is a quintillion written?\n\n## How is a quintillion written?\n\nA quintillion is represented as 1,000,000,000,000,000,000. The prefix “exa” is associated with quintillion. 10 exameters = 105.7 light years.\n\n## What is this number 1000000000000000000000000?\n\nseptillion\nSome Very Big, and Very Small Numbers\n\nName The Number Symbol\nseptillion 1,000,000,000,000,000,000,000,000 Y\nsextillion 1,000,000,000,000,000,000,000 Z\nquintillion 1,000,000,000,000,000,000 E\n\nHow much is a Quatillion?\n\nA million million millions. We could also think of it as a million trillion or a billion billion. The prefix “exa-” means a quintillion.\n\n### What is a quintillion number?\n\nDefinition of quintillion US : a number equal to 1 followed by 18 zeros — see Table of Numbers also, British : a number equal to 1 followed by 30 zeros — see Table of Numbers. Other Words from quintillion Example Sentences Learn More About quintillion.\n\n### How many trillions are in a googol?\n\nTrillion, the next number, is a 1 with twelve zeros after it, or: 1,000,000,000,000. This pattern continues until you get to Ten-duotrigintillion, more commonly known as a Googol (yes, this is where search engine Google got their name from)….Names of Large Numbers.\n\nName Number\nCentillion 1 x 10 303\nGoogolplex 1 x 10 10 100\nSkewes’ Number\n\nHow many zeros are in a decillion?\n\n33\nDecillion/Number of zeros\n\n## How do you write 100000000 in words?\n\n100,000,000 (one hundred million) is the natural number following 99,999,999 and preceding 100,000,001.\n\n## How many zeros are in a jillion?\n\n6\n1,000,000/Number of zeros\n\nHow many zeros is quintillion?\n\n18 zeros\na cardinal number represented in the U.S. by 1 followed by 18 zeros, and in Great Britain by 1 followed by 30 zeros.\n\n### How many trillions make a quintillion?\n\nWhat Exactly Is a Billion?\n\nNumber Short Scale Long Scale\n1012 one trillion one billion\n1015 one quadrillion one thousand billion\n1018 one quintillion one trillion\n1021 one sextillion one thousand trillion\n\n### What is an Octillion equal?\n\nDefinition of octillion US : a number equal to 1 followed by 27 zeros — see Table of Numbers also, British : a number equal to 1 followed by 48 zeros — see Table of Numbers.\n\nWhat is a Duotrigintillion?\n\nDuotrigintillion. A unit of quantity equal to 1099 (1 followed by 99 zeros).\n\n## What is the difference between 1 terabyte and 1 quintillion?\n\n1 Terabyte is 1000 Gigabytes 1 Gigabyte is 1000 Megabytes 1 Megabyte is 1000 Kilobytes 1 Kilobyte is 1000 Bytes. 1 Quintillion is 10 18. 2.5 Quintillion is 2.5 x 10 18. 1 Terabyte is 10 12 bytes. How many terabytes is it then?\n\n## How many zeros are there in a quintillion?\n\nAccording to short scale notation, one quintillion is equal to 1018 or 1 followed by 18 zeroes.According to long scale notation, one quintillion is equal to 1030 or 1 followed by 30 zeroes. How many zeros are in 13 quintillion?\n\nWhat is the meaning of the word quintillion?\n\nDefinition of quintillion. US : a number equal to 1 followed by 18 zeros — see Table of Numbers also, British : a number equal to 1 followed by 30 zeros — see Table of Numbers. Other Words from quintillion Example Sentences Learn More about quintillion.\n\n### How many bytes are there in 1 exabyte?\n\nHow many bytes are in an exabyte? 1 exabyte (EB) = 1 quintillion bytes quintillion = 1,000,000,000,000,000,000 = 10006 = 1018 An exabyte is 1000 petabytes, which is in turn equal to 1000 terabytes, which is equal to 1000 gigabytes, which is equal to 1000 megabytes…. and so on. What is 1 billion x 1 billion equal?" ]

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[ "# On the difference equation yx+a=A + yn/yn-k with A\n\n@article{Saleh2006OnTD,\ntitle={On the difference equation yx+a=A + yn/yn-k with A},\nauthor={M. Saleh and M. Aloqeili},\njournal={Appl. Math. Comput.},\nyear={2006},\nvolume={176},\npages={359-363}\n}\n• Published 2006\n• Computer Science, Mathematics\n• Appl. Math. Comput.\nAbstract We find conditions for the global asymptotic stability of the unique negative equilibrium y ¯ = 1 + A of the equation (0.1) y n + 1 = A + y n y n - k , where y−k, y−k+1, … , y0 ∈ (0, ∞), A\n22 Citations\n\n#### Topics from this paper\n\nOn the rational recursive sequence yn = A + yn-1/yn-m for small A\n• Mathematics, Computer Science\n• Appl. Math. Lett.\n• 2008\nQualitative Behavior of a Rational Difference Equation x_{n+1}=ax_n^2/(cx_n+bx_{n-1})\n• Mathematics\n• 2011 10th International Symposium on Distributed Computing and Applications to Business, Engineering and Science\n• 2011\nQualitative behavior of difference equation of order two\n• E. Elsayed\n• Computer Science, Mathematics\n• Math. Comput. Model.\n• 2009" ]

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[ "base (version 3.6.2)\n\n# prop.table: Express Table Entries as Fraction of Marginal Table\n\n## Description\n\nThis is really `sweep(x, margin, margin.table(x, margin), \"/\")` for newbies, except that if `margin` has length zero, then one gets `x/sum(x)`.\n\n## Usage\n\n`prop.table(x, margin = NULL)`\n\n## Arguments\n\nx\n\ntable\n\nmargin\n\nindex, or vector of indices to generate margin for\n\n## Value\n\nTable like `x` expressed relative to `margin`\n\n`margin.table`\n\n## Examples\n\nRun this code\n``````# NOT RUN {\nm <- matrix(1:4, 2)\nm\nprop.table(m, 1)\n# }\n``````\n\nRun the code above in your browser using DataCamp Workspace" ]

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[ "Cody\n\n# Problem 2927. Matlab Basics II - Max & Index of Max\n\nSolution 1072607\n\nSubmitted on 4 Dec 2016 by SeWoong Lee\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\nx = [1 4 5 2 3]; y_correct = [5,3]; assert(isequal(max_ind_max(x),y_correct))\n\na = 5 b = 3 o = 5 3\n\n2   Pass\nx = [3.2 4.3 -9.8 4.7 -10.9 3.7 -2.5]; y_correct = [4.7,4]; assert(isequal(max_ind_max(x),y_correct))\n\na = 4.7000 b = 4 o = 4.7000 4.0000" ]

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[ "# Managing a Crisis in the New World of Social Media\n\n• Published on\n20-Nov-2014\n\n• View\n382\n\n• Download\n4\n\nDESCRIPTION\n\nA presentation on how to handle PR issues in the new world of social media with personal examples from NGO attacks to severity 2 airline incidents.\n\nTranscript\n\n• 1. Crisis Control: Containing Widespread of Outrage Online & Minimizing Reputation Damage Presentation by: Ali Bullock\n\n2. I H E L P C O M PA N I E S F I G U R E OUT SOCIAL ENGAGEMENT 3. S O C I A L C H A N G E S T H E O L D VA L U E S , T H E O L D N A R AT I V E S , T H E O L D WAY S 4. I S TA R T E D T H E C X FA N PA G E O N FACEBOOK WITH ONE FAN. ME. In 5 years we built a base of over 4 million fans of fans in 20 countries. 5. PEOPLE. ITS ABOUT ENGAGEMENT WITH PEOPLE.. 6. I N S I G H T S :1 9 1 2 7. I N S I G H T S : -W H I S K E YW E I G H T-1 9 1 2L O S SS U P E R F L O U S H A I RL O S S 8. I N S I G H T S :2 0 1 3 9. I N S I G H T S : -W H I S K E YW E I G H T-1 9 1 2L O S SS U P E R F L O U S H A I RL O S S 10. MANAGING A CRISIS IN SOCIAL MEDIA. 11. DONT HAVE A CRISIS IN THE FIRST PACE B u i l d s o c i a ly o u rr e p u t a t i o nm e d i a :M e a n i n g f u la n d B u i l d n e e dy o u r t h o s e U s et r u ec o m m u n i t y a d v o c a t e si n t e r n a lo nC S R( y o u l l l a t e r )e n g a g e m e n ts t r a t e g i e sw i t hy o u re m p l o y e e s H a v es o c i a lm e d i ag u i d e l i n e s 12. 1 .E v e r y t h i n gh a p p e n sa tl i g h t n i n g - s p e e d5 TRENDS IN DIGITAL CRISISMAN AGEMENT 2 .P e o p l ed e m a n d h y p e r -t r a n s p a r e n c y 3 .D i a l o g u ea si m p o r t a n tm e s s a g e 4 .R e p u t a t i o ni sa sd e l i v e r y i ns o c i a l m e d i a 5 .B r a n dd e t r a c t o r s t h es a m eh a v e t o o l s 13. E V E RY O N E M A K E S M I S TA K E S . B E C A U S E T H E Y A R E H U M A N Stupid post to make on a corporate channel. 14. E V E NT H EB E S TO FU S G R E AT D AY, G R E AT D AY,SHOOTINGPHOTOGRAPHINGELEPHANTS FOR 8ELEPHANTS FOR 8HOURS- ALIHOURS- ALIBULLOCK, HEAD OFBULLOCK, HEAD OFCOMMS WWFCOMMS WWF. 15. SOCIAL MEDIA & SPEED 16. # 1 B a d-n e w st h a ne v e rT w i t t e r ,S p e e dH o u r sj u s t W h e nt om a t t e r H a v e a n dg e n e r a t e dac r i s i s ,r e a c t o ft oy o uf a s th o u r s ,m e d i ai nn o ts t r e a m l i n e dm a y ad a y s a p p r o a c ht e a mi np l a c e E x p e r i e n c ei ns o c i a lm e d i ar e s p o n df a s tw i l lam e d i a ,t r a d i t i o n a le m e r g i n go u rr e l e v a n tr e s p o n d i n gn e e da n d l i f e s t r e a m s a l lc o n s u m e r n o tv i aF a c e b o o k M o n i t o r2 4f a s t e rb e f o r ec o l l e c t i v eT h eF i r s ts p r e a d sh e l py o u 17. # 1-S p e e dHour 12 Hour 6T h eMicromediaF i r s t H o u r sHour 182 40 HourCRISIS HITSMainstreamBlogsSharingHour 24Editorial 18. IN 30 SECONDS THIS PHOTO HAD B E E N TA K E N A N D UPLOADED TO SOCIAL MEDIA 19. JOURNALISTS WORKING IN REAL TIME# a l i b u l l o c k# c r i s i s c o m m s 20. I T SA B O U TS Y M P A T H Y,C O M M U N I T Y&E N G A G E M E N TA S I A N AAP R E S SP U TR E L E A S E8 HOURSL AT E R . T H ATW O R K E DW O R L D I NT H EO FC O M M U N I C A T I O N S . O L D 21. I T S E R E S O V OT O O K A S I A N A 8 H O U R S T O N D O U T A P R E S S L E A S E . N O U P D A T E S O N C I A L M E D I A T O F I L L T H E I D 22. T H I SP H O T O W A S N T T A K E N J O U R N A L I S T, I T W A SB Y A P H O T O T A K E N B Y A P A S S E N G E R . 23. H O W S H O U L D H AV E A S I A N A RESPONDED? 24. I T S N O T T H E M I S TA K E S M A D E U S I N G S O C I A L M E D I A T H AT C A U S E B R A N D S T H E B I G G E S T P R O B L E M S , I T S H O W P O O R LY O R S L O W LY THEY DEAL WITH THEM. 25. SAYING NOT HING OR TAKING TOO MUCH TIME TO RESPOND IS AS BAD AS GETTING YOUR FACTS WRONG. 26. O n e - w a y# 3 D i a l o g u e G e t R e a d y f o r 2 - Wa y D i a l o g u ew o r k w h e r em e s s a g i n g a n y m o r ep e o p l e I n v i t i n gd o e s n ti nc r a v ead i a l o g u ec u s t o m e r sc o n v e r s a t i o n e f f e c t i v e w h o y o up r e s ss c r i p t e dm o s tw a yt ob u i l da n dw i l l i fs y s t e mf o rc r i t i c a lb r a n ds u p p o r tc r i s i sh i t ss o l e l yr e l e a s e sa n di n t e r a c t i o n sd o e s n t Aat h e C o m m u n i c a t i n g t h r o u g hi n t oi sg o o d w i l l a d v o c a t e sw o r l ds a t i s f yl i s t e n i n g t oi sr e m a i n i n g r e s p o n s i v e 27. # 4 S e a r c h R e p u t a t i o n s a r e B u i l t o r B r o k e n o n S o c i a l M e d i a 8 0 %o fI n t e r n e tt h e i r O r g a n i c t os e s s i o ns e a r c hs o c i a lt h ea ti sm e d i at o G o o g l eu s e r ss t a r ts e a r c hs e n s i t i v ec o n t e n td u ec r o s s - l i n k i n gd e l i v e r s u n i v e r s a lm a k i n gm u l t i m e d i as e a r c h c r i t i c a l D i f f i c u l t o n c ei tt o i sd i s l o d g e i ns e a r c hc o n t e n t r e s u l t s 28. OLD MEDIA APPROACH 29. WELCOME TO THE NEW WORLD OF SOCIAL (EVERYONE HAS A CAMERA) 30. SOCIAL MEDIA PUTS YOUR BRAND IN THE HAND OF THE CONSUMER. 31. A n# 5 D e t r a c t o r s Yo u r D e t r a c t o r s A r e R e s o u r c e f u li n d i v i d u a lt r a v e la r o u n d S m a l lg e n e r a t e do w na l o n ei sa n dt om e d i a a n n oc a nn i m b l e m e d i ac o n s u m e r i sc r i t i c a li n f l u e n c e rc i r c l e s ,c a no n l yt o d a ys o c i a l L i s t e n i n gt h e i rw o r l df a s tw i t h E v e r y o n et h eo r g a n i z a t i o n s b ec a ne a s i l ym o r eo f t e nv o i c es ol o n g e rm e t r i cf o ri nt r a f f i c b et h ej u d g i n gi n f l u e n c e 32. ANTON CASEY 33. ANTON CASEY; #5 - Detractors REMEMBER. People see EVERYTHING! 34. #5 - Detractors H E H A D N O S O C I A L C A P I TA L TO U S E . 35. M YE X P E R I E N C E S 36. BUT SOMETIMES YOU NEED COURAGE ON SOCIAL MEDIA 37. C AT H AY PA C I F I C S H E PA R DV SS E ACATHAY PACIFICSEVERAL OF WHOM WEREPARTICIPATES IN THEVERBALLY ASSAULTED AT OURDOLPHIN SLAVE TRADE.AIRPORTS BY PEOPLE ANGRY- SEA SHEPERDOVER THE STORY.FROM PICKING UP THE STORY- OUR POSTS ON SOCIALTO POSTING ON FACEBOOKMEDIA TOOK HOURS TO GETOVER 24 HOURS PASSED APPROVALS RATHER THAN MINUTES.THIS WAS TOO LONG - WE DIDNT INFORM STAFF AROUND THE NETWORK, 38. One can imagine what Cathay Pacific officials would say about their companys participation in thisactivity. Perhaps they would tell us that it is only business, and moving dolphins from Japan to China is legal. Let us remember Cathay Pacifics role in this. Let us also be informed. U.S. based American Airlines just entered into a codesharing arrangement with Cathay Pacific. British Airways and Qantas Airlines are also in that elite group. Perhaps those who care about the plight of dolphins should remember this when they make business decisions about which airline to patronize.. - SEA SHEPERD JUNE 8 2011 39. R I C O B A R RY G O O N C N N WA SA B O U TWE KNEW THE STORY WASNTTHEM TO HAND. I WAS HONESTACCURATE. BUT HOW DO YOUAND 100% TRANSPARENT.FACE SOMEONE AS RESPECTED AS RIC OBARRY, ONE OF THEI SAID THAT IF IT TURNED OUTWORLDS LEADING DOLPHINTHE STORY WAS TRUE I WOULDADVOCATES.RESIGN FROM CX. I WAS THAT CONFIDENT IN MY BELIEF THATI WAS CONTACTED BY RICCX WOULD NOT HAVEOBARRYS FOUNDATIONTRANSPORTED THESE(THANKS TO MY VOLENTEERDOLPHINS.WORK PHOTOGRAPHING ANIMALS FOR CHARITIES SUCHRIC DIDNT GO ON CNN.AS PETA, AAF, SPCA, HKDR ACRES AND WWF.)1 YEAR LATER WE BROUGHT OUT A NEW SET OF GUIDELINESI SPOKE TO RICSTHAT INCLUDED A BAN ONREPRESENTATIVES AND GAVESHIPMENTS OF SHARKFINTHEM THE FACTS AS I HADTO 40. Cathay Pacific holds a 32% stake in CKTS but has no direct involvement in its day-to-day management. We will be contacting the handling agent to find out more about this incident. - CATHAY PACIFIC JUNE 8 2011 41. S O C I A L &M E D I AA C C U R A T E T H A T1 4 0N E E D SAQ U I C KR E S P O N S E I N C H A R A C T E R SL E S S. 42. S E V E R I T Y2C R I S I SATC X 43. 9.32am CALL: ALI, WE NEED YOU IN THE CRISIS COMMS CENTER NOW.9.40am COMMS BRIEFING ON A SEVERITY 2 INCIDENT WITH A PLANE MAKING AN EMERGENCY LANDING.10.12am PEOPLE EVACUATED FROM THE PLANE AND IMAGES / POSTS START TO APPEAR ON SOCIAL MEDIA CHANNELS. 44. W H AT W E L E A R N T T H AT D AY WA S I N VA L U A B L EOUR CRISIS CENTER NEEDED UPDATING!WE NEEDED MORE THAN JUST THE CORPORATE COMMSTEAM IN THE ROOM. WE NEEDED CUSTOMER SERVICE,OPS AND PERHAPS CRITICALLY, THE SOCIAL MEDIA TEAM SOCIAL MEDIA UPDATES WOULD NEED TO HAPPEN AT THE SAME TIME AS TRADITIONAL MEDIABRIEFINGSSOCIAL MEDIA CREATED 100 TIMES TH NOISE OTHER CHANNELS DID MOVING FORWARD WE WOULD NEED 24/7 MANNING OF SOCIAL MEDIA ACROSS 20 DIFFERENT COUNTRIESCUSTOMERS AND PASSENGERS USED SOCIAL MEDIA FOR ENQUIRES AND COMMENTSCUSTOMER SERVICE AND LISTENING FOR PASSENGER OR RELATED ENQUIRIES ON SOCIAL AS CUSTOMERS COULD NOT GET THROUGH ON THE PHONE LINES 45. L I S T E N I N G E N A B L E D U S TO M A K E I N F O M R E D D E C I S I O N S A N D C O M M U N I C AT I O N S . 46. 5 K E Y S TO D I G I TA L C R I S I S M A N A G E M E N T Y O U C A N A P P LY TO Y O U R B U S I N E S SS E TU PAL I S T E N I N GP O S TP R O G R A M T O D A YG E TC - S U I T EI M P O R T A N C EO FI D E N T I F YI N F L U E N C E R SF O R( A N DB U Y - I N :S O C I A LT H ET O PT H EM E D I A O N L I N EY O U RB U S I N E S SB E G I NB U I L D I N GR E L A T I O N S H I P S )K N O W O N L I N EH O WY O U( E . G .W I L L S P E A K T W I T T E R ,B L O G ,Y O U T U B E )E S T A B L I S HE N G A G E M E N TS O C I A LG U I D E L I N E S Y O U RM E D I AA C R O S SM A R C O MT E A M 47. 5 K E Y T R A I T S Y O U M U S T H AV E I N C R I S I S1 .C O M P E T E N C E :HANDLE A TOUGH PRESS CONFERENCE WITH DEXTERITY? YOURE DEEMED COMPETENT. LOOK UNEASY BEFORE CAMERAS? YOURE NOT. 2 .C R E D I B I L I T Y :THERES ONE QUESTION THAT DRIVES THE PUBLICS PERCEPTION OF AN EXECUTIVE OR SPOKESPERSON MORE THAN ANY OTHER: DOES HE OR SHE GET IT? A3 .C O M M I T M E N TSET THE RIGHT TONE, WITH ACTIONS AND WORDS TO THE MEDIA, AFFECTEDSTAKEHOKDERS AND EMPLOYEES CONSISTENT MESSAGE OF ADDRESSING AND FIXING THE ISSUE. 4 .C A R I N G :LITTLE MAKES THE PUBLIC TURN ON AN EXECUTIVE OR PUBLIC FIGURE IN CRISIS MORETHAN SOMEONE WHOS CAVALIER TOWARD ANY VICTIMS. 5 . C A P A B I L I T Y:SHOW THAT YOU OR YOUR CEO SOLVE THE PROBLEM AT HAND 48. I NAC R I S I S( B O T HC O M M U N I C A T I O N SI N T E R N A L L YE X T E R N A L L Y )I SA N DE S S E N C I A L . 49. M O V I N GF O RWA R D 50. S O C I A LC H A N G E SV A L U E S , N A R A T I V E S ,T H ET H ET H EO L DO L DO L DW A Y S . 51. T H EL I N E SB E T W E E NC O M M U N I C AT I O N S B L U R R I N GA N DE N C O U R A G E O FA N YD O N TS H O U L DH A S T E NR E M A I N I N G K N O WO RA R EW E L C O M E ,T H EW A L L S .W H E T H E RA D V E R T I S I N G ,T H EP R O F E S S I O N SW EA N DA L LT E A R I N GG R E ATT H E Y R EI D E A SP R ,S O C I A L . E D W A R D B O C H E S , C R E AT I V E D I R E C T O RD O W N 52. THANK YOU FOR YOUR TIME. YOU CAN FIND A COPY OF THIS PRESENTATION HERE: WWW.SLIDESHARE.NET/ALIBULLOCK/P RESENTATIONS LINKEDIN: ALI [email protected] WWW.ALIBULLOCK.COM" ]

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[ "Z80 Assembler\n\n# Z80 Assembler » Statements\n\nStatements are SpectNetIDE specific control flow constructs — thanks again for the inspiration by Simon Brattel — that instruct the compiler about loop-like and conditional compilation.\n\nWhile directives help you to organize your code and include code files optionally according to the compilation context, statements provide you more useful tools to shorten the way you can declare Z80 assembly code.\n\nEach statement can be written with a leading dot, or without it, and the compiler accepts both lowercase and uppercase versions. For example all of these version are valid: `.if`, `if`, `.IF`, and `IF`.\n\n## The LOOP Block\n\nWith LOOP block, you can organize a cycle to emit code. Here is a sample that tells the gist:\n\n``````.loop 6\n.endl\n``````\n\nThis is a shorter way to multiply HL with 64. It is equivalent with the following code:\n\n`````` add hl,hl\n``````\n\nThe `.loop` statement accepts an expression. The compiler repeats the instructions within the loop’s body according to the value of the expression. The `.endl` statement marks the end of the loop.\n\nYou can use many flavors for the `.endl` block closing statement. `.endl`, `endl`, `.lend`, `lend` are all accepted — with fully uppercase letters, too.\n\nLook at this code:\n\n``````counter .equ 2\n; do something (code omitted)\n.loop counter + 1\n.db #80, #00\n.endl\n``````\n\nThis is as if you wrote this:\n\n`````` .db #80, #00\n.db #80, #00\n.db #80, #00\n``````\n\n## The LOOP Scope\n\nThe `.loop` statement declares a scope for all labels, symbols, and variables declared in the loop’s body. Every iteration has its separate local scope. When the assembler resolves symbols, it starts from the scope of the loop, and tries to resolve the value of a symbol. If it fails, steps out to the outer scope, and goes on with the resolution.\n\nCheck this code:\n\n``````value .equ 2\n; do something (code omitted)\n.loop 2\nvalue .equ 5\nld a,value\n.endl\n``````\n\nThe compiler takes it into account as if you wrote this:\n\n`````` ld a,5\nld a,5\n``````\n\nThe `value` symbol declared within the loop, overrides `value` in the outer scope, and thus 5 is used instead of 5.\n\nNonetheless, you when you utilize a different construct, it seems a bit strange at first:\n\n``````value .equ 2\n; do something (code omitted)\n.loop 2\nld a,value\nvalue .equ 5\nld b,value\n.endl\n``````\n\nThe strangeness is that the compiler creates this:\n\n`````` ld a,2\nld b,5\nld a,2\nld b,5\n``````\n\nWhen the assembler resolves `value` in the `ld a,value` instruction, if finds `value` in the outer scope only, as it is not declared yet within the loop’s scope. In the `ld b,value` instruction `value` gets resolved from the inner scope, so it takes 5.\n\n## Variables and Scopes\n\nUnlike symbols that work as constant values, variables (declared with the `.var` pragma, or its syntactical equivalents, the `=` or `:=` tokens) can change their values.\n\nTake a look at this code:\n\n``````counter = 4\n.loop 3\ninnercounter = 4\nld a,counter + innercounter\ncounter = counter + 1\n.endl\n``````\n\nHere, the `counter` variable is defined in the global scope (out of the loop’s scope), while `innercounter` in the local scope of the loop. When evaluating the `counter = counter + 1` statement, the compiler finds `counter` in the outer scope, so it uses that variable to increment its value. This code emits machine code for this source:\n\n``````ld a,#08\nld a,#09\nld a,#0A\n``````\n\nNow, add a single line to the loop’s code:\n\n``````counter = 4\n.loop 3\ninnercounter = 4\nld a,counter + innercounter\ncounter = counter + 1\n.endl\nld b,innercounter\n``````\n\nThe compiler will not compile this code, as it cannot find the value for `innercounter` in the `ld b,innercount` instruction. Because `innercounter` is defined in the local scope of the loop, this scope is immediately disposed as the loop is completed. When the compiler processes the `ld b,innercounter` instruction, the local scope is not available.\n\n## Labels and Scopes\n\nLabels behave like symbols, and they work similarly. When you create a label within a loop, that label is created in the local scope of the loop. The following code helps you understand which labels are the part of the global scope, and which are created in the loop’s scope:\n\n``````.org #8000\nMyLoop: .loop 2\nld bc,MyLoop\nInner:\nld de,MyEnd\nld hl,Inner\nld ix,Outer\nMyEnd: .endl\nOuter: nop\n``````\n\nThe label of the `.loop` statement is part of the outer (global) scope, just like the label that follows the `.endl` statement. However, all labels declared within the loop’s body, including the label of the `.endl` statement belongs to the local scope of the loop.\n\nThus, the compiler translates the code above into this one:\n\n`````` (#8000): ld bc,#8000 (MyLoop)\nInner_1 (#8003): ld de,#800D (MyEnd_1)\n(#8006): ld hl,#8003 (Inner_1)\n(#8009): ld ix,#801A (Outer)\nMyEnd_1 (#800D): ld bc,#8000 (MyLoop)\nInner_2 (#8010): ld de,#801A (MyEnd_2)\n(#8013): ld hl,#8010 (Inner_2)\n(#8016): ld ix,#801A (Outer)\nMyEnd_2\nOuter (#801A): nop\n``````\n\nHere, `Inner_1`, `Inner_2`, `MyEnd_1`, and `MyEnd_2` represents the labels created in the local scope of the loop. The `_1` and `_2` suffixes indicate that each loop iteration has a separate local scope. As you can see, the last iteration of `MyLabel` points to the first outer address (`Outer` label).\n\n## Nesting LOOPs\n\nOf course, you can nest loops, such as in this code:\n\n``````.loop 3\nnop\n.loop 2\nld a,b\n.endl\ninc b\n.endl\n``````\n\nThis code snippet translates to this:\n\n``````nop\nld a,b\nld a,b\ninc b\nnop\nld a,b\nld a,b\ninc b\nnop\nld a,b\nld a,b\ninc b\n``````\n\nWhen you nest loops, each loop has its separate scope.\n\n## The \\$CNT value\n\nIt is very useful to use the `\\$cnt` value that represents the current loop counter. It starts from 1 and increments to the maximum number of loops. This sample demonstrates how you can use it:\n\n``````.loop 2\nouterCount = \\$cnt\n.loop 3\n.db #10 * outerCount + \\$cnt\n.endl\n.endl\n``````\n\nThis code translates to this:\n\n``````.db #11\n.db #12\n.db #13\n.db #21\n.db #22\n.db #23\n``````\n\nYou can observe that each loop has its spearate `\\$cnt` value.\n\nThe `\\$ctn` value has several syntax versions that the compiler accepts: `\\$CNT`, `.cnt`, and `.CNT`.\n\n## The PROC..ENDP Block\n\nIn the previous section you could understand how labels and scopes work for the `.loop` statement. You can utilize this scoping mechanism with the help of the `.proc`..`.endp` statement. This sample code demonstrates the concepts (just as you learned earlier):\n\n``````.org #8000\nMyLabel:\nld de,Outer\nld hl,Mylabel\ncall MyProc\nhalt\n\nMyProc:\n.proc\nld bc,MyProc\nMyLabel:\nld de,MyEnd\nld hl,MyLabel\nld ix,Outer\nret\nMyEnd:\n.endp\nOuter: nop\n``````\n\nThe first `MyLabel` label belongs to the global scope, while the second (within `MyProc`) to the local scope of the procedure wrapped between `.proc` and `endp`. `MyProc` belongs to the global scope too, however, `MyEnd` is the part of the `MyProc` scope, so it is visible only from within the procedure.\n\nThe assembler emits this code:\n\n``````MyLabel (#8000): ld de,#8018 (Outer)\n(#8003): ld hl,#8000 (MyLabel)\n(#8006): call #800A (MyProc)\n(#8009): halt\nMyProc (#800A): ld bc,#800A (MyProc)\nMyLabel_ (#800D): ld de,#8018 (MyEnd)\n(#8010): ld hl,#800D (MyLabel_)\n(#8013): ld ix,#8018 (Outer)\n(#8017): ret\nMyEnd\nOuter (#8018): nop\n``````\n\nYou can nest `PROC` bloks just as `LOOP` blocks. Each `PROC` block has its private scope. When the compiler sees a `PROC` block, it works just as if you wrote `.loop 1`.\n\nNOTE: `PROC` is different than a loop. You cannot use the `\\$cnt` value. Similarly, the `break` and `continue` instructions are unavailable within a `PROC` block.\n\nThe assembler accepts these aliases for `PROC` and `ENDP`: `.proc`, `proc`, `.PROC` , `PROC`, `.endp`, `.ENDP`, `endp`, `ENDP`, `.pend`, `.PEND`, `pend`, `PEND`.\n\n## The REPEAT..UNTIL Block\n\nWhile the `.loop` statement works with an expression that specified the loop counter, the `.repeat`..`.until` block uses an exit condition to create more flexible loops. Here is a sample:\n\n``````counter = 0\n.repeat\n.db counter\ncounter = counter + 3\n.until counter % 7 == 0\n``````\n\nObserve, the `counter % 7 == 0` condition specifies when to exit the loop. Because the exit condition is examined only at the end of the loop, the `.repeat` blocks executes at least once.\n\nThe sample above translates to this:\n\n``````.db 0\n.db 3\n.db 6\n.db 9\n.db 12\n.db 15\n.db 18\n``````\n\nThe `.repeat` block uses the same approach to handle its local scope, symbols, labels, and variables as the `.loop` block. The block also provides the `\\$cnt` loop counter that starts from 1 and increments in every loop cycle.\n\nThis sample demontrates the `.repeat` block in action:\n\n``````.org #8000\ncounter = 0\n.repeat\n.db low(EndLabel), high(Endlabel), \\$cnt\ncounter = counter + 3\nEndLabel: .until counter % 7 == 0\n``````\n\nThe compiler translates the code to this:\n\n``````.db #03, #80, #01\n.db #06, #80, #02\n.db #09, #80, #03\n.db #0C, #80, #04\n.db #0F, #80, #05\n.db #12, #80, #06\n.db #15, #80, #07\n``````\n\n## The WHILE..ENDW Block\n\nWith `.while` loop, you can create another kind of block, which uses entry condition. For example, the following code snippet generates instructions to create the sum of numbers from 1 to 9:\n\n``````counter = 1\nld a,0\n.while counter < 10\ncounter = counter + 1\n.endw\n``````\n\nThe `.while`..`.endw` block uses an entry condition declared in the `.while` statement. Provided, this condition is true, the compiler enters into the body of the loop, and compiles all instructions and statements until it reaches the `.endw` statement. Observe, it may happen that the body of the loop is never reached.\n\nThe compiler translates the code snippet above to the following:\n\n``````ld a,0\n``````\n\nJust like the `.loop` and the `.repeat` blocks, `.while` uses the same approach to handle its local scope, symbols, labels, and variables. This block also provides the `\\$cnt` loop counter that starts from 1 and increments in every loop cycle.\n\nThis code demonstrates the `.while` block with labels and using `\\$cnt` value:\n\n``````counter = 0\n.while counter < 21\n.db low(EndLabel), high(Endlabel), \\$cnt\ncounter = counter + 3\nEndLabel: .endw\n``````\n\nThe compiler translates the code to this:\n\n``````.db #03, #80, #01\n.db #06, #80, #02\n.db #09, #80, #03\n.db #0C, #80, #04\n.db #0F, #80, #05\n.db #12, #80, #06\n.db #15, #80, #07\n``````\n\nYou can use many flavors for the `.endw` block closing statement. `.endw`, `endw`, `.wend`, `wend` are all accepted — with fully uppercase letters, too.\n\n## The FOR..NEXT Loop\n\nTou can use the traditional `.for`..`.next` loop to create a loop:\n\n``````.for myVar = 2 .to 5\n.db 1 << int(myVar)\n.next\n``````\n\nThis loop uses the `myVar` variable as its iteration variable, which iterates from 1 to 4. As you expect, the compiler translates the for-loop into this:\n\n``````.db #04\n.db #08\n.db #10\n.db #20\n``````\n\nYou can specify a `.step` close to change the loop increment value:\n\n``````.for myVar = 1 .to 7 .step 2\n.db 1 << int(myVar)\n.next\n``````\n\nNow, the code translates to this:\n\n``````.db #02\n.db #08\n.db #20\n.db #80\n``````\n\nYou can create a loop with decrementing iteration variable value:\n\n``````.for myVar = 7 .to 1 .step -2\n.db 1 << int(myVar)\n.next\n``````\n\nAs you expect, now you get this translation:\n\n``````.db #80\n.db #20\n.db #08\n.db #02\n``````\n\nJust as with the other statements, you can use the `.for`, `.to`, and `.step` keywords without the `.` prefix, so `for`, `to`, and `step` are also valid.\n\nThe for-loop can do the same stunts as the other kind of loops; it handles labels, symbols, and variables exactly the same way. There’s only one exception, the loop iteration variable. If this variable is found in an outer scope, instead of using that value, the compiler raises an error. You can us the for-loop only with a freshly created variable.\n\nSo both cases in this code raise an error:\n\n``````myVar = 0\n.for myVar = 1 .to 4 ; ERROR: Variable myVar is already declared\n; ...\n.next\n\n.for _i = 1 .to 3\n.for _i = 3 .to 8 ; ; ERROR: Variable _i is already declared\n; ...\n.next\n.next\n``````\n\nAs `i` is a reserved token (it represents the `I` register), you cannot use `i` as a variable name. Nonetheless, `_i` is a valid variable name.\n\nThe for-loop works with both integer and float variables. If any of the initial value, the last value (the one after `.to`), or the increment value (the one after `.step`) is a float value, the for-loop uses float operations; otherwise it uses integer operations.\n\nThis code snippet demonstrates the difference:\n\n``````.for myVar = 1 .to 4 .step 1\n.db 1 << myVar\n.next\n\n.for myVar = 1 .to 4 .step 1.4\n.db 1 << myVar ; ERROR: Right operand of the shift left operator must be integral\n.next\n``````\n\nNonetheless, you can solve this issue with applying the `int()` function:\n\n``````.for myVar = 1 .to 4 .step 1.4\n.db 1 << int(myVar) ; Now, it's OK.\n.next\n``````\n\nYou can still use the `\\$cnt` value in for loops. Just like with other loop, it indicates the count of cycles strating from one and incremented by one in each iteration.\n\n## Maximum Loop Count\n\nIt’s pretty easy to create an infinite (or at least a very long) loop. For example, these loops are obviously infinite ones:\n\n``````.repeat\n.until false\n\n.while true\n.wend\n``````\n\nThe assembler checks the loop counter during compilation. Whenever it exceeds #FFFF (65535), it raises an error.\n\n## The IF..ELIF..ELSE..ENDIF Statement\n\nYou can use the `.if` statement to create branches with conditions. For example, this code emits `inc b` or `inc c` statement depending on whether the value of `branch` is even or odd:\n\n``````.if branch % 2 == 0\ninc b\n.else\ninc c\n.endif\n``````\n\nYou do not have to specify an `.else` branch, so this statement is entirely valid:\n\n``````.if branch % 2 == 0\ninc b\n.endif\n``````\n\nYou can nest if statements like this to manage four different code branches according to the value of `branch`:\n\n``````.if branch == 1\ninc b\n.else\n.if branch == 2\ninc c\n.else\n.if branch == 3\ninc d\n.else\ninc e\n.endif\n.endif\n.endif\n``````\n\nNonetheless, you can use the `.elif` statement to create the code snippet above in clearer way:\n\n``````.if branch == 1\ninc b\n.elif branch == 2\ninc c\n.elif branch == 3\ninc d\n.else\ninc e\n.endif\n``````\n\n## IF and Scopes\n\nUnlike the loop statements, `.if` does not provide its local scope. Whenever you create a symbol, a label or a variable, those get into the current scope. This code defines a label with the same name in each branches. Because the compiler evaluates the `.if` branches from top to down, it either compiles one of the `.elif` branches — the first with a matching condition — or the else branch. Thus, this code does not define `MyLabel` twice:\n\n``````branch = 4 ; Try to set up a different value\n; Do something (omitted from code)\nld hl,MyLabel\n.if branch == 1\ninc b\nMyLabel ld a,20\n.elif branch > 2\nMyLabel ld a,30\ninc c\n.elif branch < 6\ninc d\nMyLabel ld a,40\n.else\nMyLabel ld a,50\ninc e\n.endif\n``````\n\nGenerally, you can decorate any statement with labels. The `.elif` and `.else` statements are exception. If you do so, the compiler raises an error:\n\n``````.if branch == 1\ninc b\nMyLabel ld a,20\n.elif branch > 2\nMyLabel ld a,30\ninc c\nOther .elif branch < 6 ; ERROR: ELIF section cannot have a label\ninc d\nMyLabel ld a,40\nAnother .else ; ERROR: ELSE section cannot have a label\nMyLabel ld a,50\ninc e\n.endif\n``````\n\n## IF Nesting\n\nWhen you nest `.if` statements, take care that each of them has a corresponding `.endif`. Whenever the compiler finds an `.endif`, is associates it with the closest `.if` statement before `.endif`. I suggest you use indentation to make the structure more straightforward, as the following code snippet shows:\n\n``````row = 2\ncol = 2\n; Change row and col (omitted from code)\n.if row == 0\n.if col == 0\n.db #00\n.elif col == 1\n.db #01\n.else\n.db #02\n.endif\n.elif row == 1\n.if col == 0\n.db #03\n.elif col == 1\n.db #04\n.else\n.db #05\n.endif\n.elif row == 2\n.if col == 0\n.db #06\n.elif col == 1\n.db #07\n.else\n.db #08\n.endif\n.else\n.if col == 0\n.db #09\n.elif col == 1\n.db #0A\n.else\n.db #0B\n.endif\n.endif\n``````\n\n## The IFUSED/IFNUSED Statements\n\nSpectNetIDE offers a similar construct to IF..ELIF..ELSE..ENDIF, using the IFUSED or IFNUSED statement instead of IF. These new statements are specialized forms of IF. You can use these statements to emit code depending on whether a symbol (label, `.EQU`, `.VAR`, structure, or structure field) exists and has already been used by the code preceding the IFUSED/IFNUSED statement.\n\nHere are a few examples:\n\n``````MyProc:\nld hl,#5800\nld (hl),a\nret\n; some other code\n\n.ifused MyProc\nMyMsg: .defn \"MyProc is used\"\n.else\nMyMsg: .defn \"MyProc is not used\"\n.endif\n\nMain:\nld hl,MyMsg\n``````\n\nHere, the `.ifused` statement will set the string the `MyMsg` label point to according to whether the `MyProc` label is used, or not. As in this case `MyProc` is defined but not invoked before the `.ifused` statement, HL will point to the “MyProc is not used” message.\n\nShould you call `MyProc` before `.ifused`, HL would point to the other message, “MyProc is used”:\n\n``````MyProc:\nld hl,#5800\nld (hl),a\nret\n; some other code\ncall MyProc\n; some other code\n\n.ifused MyProc\nMyMsg: .defn \"MyProc is used\"\n.else\nMyMsg: .defn \"MyProc is not used\"\n.endif\n\nMain:\nld hl,MyMsg\n``````\n\nThe `.ifnused` statement is the complement of `.ifused`. It is evaluated to a true condition value only if the symbol following `.ifnused` is not defined, or, if defined, is not used.\n\n### IFUSED/IFNUSED Syntax\n\nYou need to specify a symbol after the `.ifused` or `.ifnused` keywords. These symbols must follow the syntax of identifiers. They can be compound names used for modules and structures. So, all of these symbol names are correct:\n\n``````MyLabel\nMyStruct\nMyStruct.FieldX\nMyModule.Main\n::NestedModule.Start.MyProc\n``````\n\nNote: You can use these aliases for `.ifused`: `.IFUSED`, `ifused`, `IFUSED`. Similarly, `.ifnused` accept alternative tokens: `.IFNUSED`, `ifnused`, `IFNUSED`.\n\n### IFUSED/IFNUSED Semantics\n\nThe SpectNetIDE compiler accepts any `.ifused` and `.ifnused` statements until they are syntactically correct. When the assembler tests their condition, it works this way:\n\n• If the specified symbol does not exists, `.ifused` evaluates to false, while `.ifnused` evaluates to true.\n• If the particular symbol exists and it is used in the code section preceding the `.ifused` or `.ifnused` statement, `.ifused` evaluates to true, `.ifnused` to false.\n• If the particular symbol exists and it is not used in the code section preceding the `.ifused` or `.ifnused` statement, `.ifused` evaluates to false, `.ifnused` to true.\n\nThese statements do not support look-ahead in the code. This behavior could lead to paradox situations, like in this example:\n\n``````MyFlag = true\nMyValue: .equ #1234\n; some other code that does not use MyValue\n\n.ifused MyValue\nMyFlag = false;\n.endif\n\n; some other code that does not change MyFlag\n\n.if MyFlag\nld a,MyValue\n.endif\n``````\n\nShould `.ifused` work with look-ahead, this code would make the compiler scratch its virtual head. Because `MyFlag` is set to true, the `.if` statement at the bottom of the code would emit an `ld a,MyValue` instruction. Knowing this fact, the compiler would say that `.ifused MyValue` should be evaluated to true. However, in this case, the body `.ifused` would set `MyFlag` to true, and that would prevent the bottom `.if` to emit `ld a,MyValue`, and then `MyValue` would not be used at all.\n\n## Block Statements without a Closing Statement\n\nThe compiler automatically recognizes if a block does not have a closing statement, and provides an error message accordingly.\n\n## Orphan Closing Statements\n\nWhen the compiler finds a closing statement (such as `.endw`, `.endl`, `.until`, `.endif`, etc.) it will issue an error.\n\n## The BREAK statement\n\nYou can exit the loop — independently of the loop’s exit condition — with the `.break` statement:\n\n``````; LOOP sample\n.loop 5\n.if \\$cnt == 4\n.break\n.endif\n.db \\$cnt\n.endl\n\n; REPEAT sample\n.repeat\n.if \\$cnt == 4\n.break\n.endif\n.db \\$cnt\n.until \\$cnt == 5\n\n; WHILE sample\n.while \\$cnt < 5\n.if \\$cnt == 4\n.break\n.endif\n.db \\$cnt\n.endw\n\n; FOR-loop sample\n.for value = 1 to 5\n.if value == 4\n.break\n.endif\n.db value\n.next\n``````\n\nBecause all these loops are exited at the beginning of the 4th iteration, they produce this output:\n\n``````.db #01\n.db #02\n.db #03\n``````\n\nYou cannot use the `.break` statement outside of a loop construct. If you do so, the compiler raises an error.\n\n## The CONTINUE Statement\n\nYou can interrupt the current iteration of the loop and carry on the next iteration with the `.continue` statement:\n\n``````; LOOP sample\n.loop 5\n.if \\$cnt == 4\n.continue\n.endif\n.db \\$cnt\n.endl\n\n; REPEAT sample\n.repeat\n.if \\$cnt == 4\n.continue\n.endif\n.db \\$cnt\n.until \\$cnt == 5\n\n; WHILE sample\n.while \\$cnt <= 5\n.if \\$cnt == 4\n.continue\n.endif\n.db \\$cnt\n.endw\n\n; FOR-loop sample\n.for value = 1 to 5\n.if value == 4\n.continue\n.endif\n.db value\n.next\n``````\n\nBecause all these loops skip the 4th iteration, they produce this output:\n\n``````.db #01\n.db #02\n.db #03\n; #04 is skipped\n.db #05\n``````\n\nYou cannot use the `.continue` statement outside of a loop construct. If you do so, the compiler raises an error." ]

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