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[ "# How to draw this picture with Latex?\n\nCould some one help me to draw the following picture with Latex ( using only 1 picture) ? Thank you SO MUCH for your help.", null, "• Welcome to TeX.SX. Questions about how to draw specific graphics that just post an image of the desired result are really not reasonable questions to ask on the site. Please post a minimal compilable document showing that you've tried to produce the image and then people will be happy to help you with any specific problems you may have. See minimal working example (MWE) for what needs to go into such a document. – erik Apr 28 '15 at 20:52\n• Dear Erik, I am pretty new with Latex and to bo honest I don't know where to start beside searching on Google but I cannot not find an answer on it. – duy nguyen Apr 28 '15 at 21:18\n• What you need is called TikZ. Your simple drawing is covered in the tutorial of the manual: texample.net/media/pgf/builds/pgfmanualCVS2012-11-04.pdf – hpekristiansen Apr 28 '15 at 21:24\n• Although the tikz manual makes 800 pages, you should get to the point easily by looking at chapter 19. Trees are commons in tikz, see for instance texample.net/tikz/examples/feature/trees . – Clément Apr 28 '15 at 21:27\n\nHere's an example of how to get started. I think it should make it clear enough how to continue to get the diagram you want.\n\n\\documentclass[border=10pt]{standalone}\n\\usepackage{tikz}\n\\usetikzlibrary{calc}\n\\begin{document}\n\n\\begin{tikzpicture}[my midway label/.style={midway,yshift=1.5in}]\n\n%% case 1\n\\node[anchor=east] (A/left) at (0,0) {$x$};\n\\node[anchor=west] (A/right) at ($(A/left)+(1in,0)$) {$x$};\n\\node[anchor=west] (A/right/up/1) at ($(A/right)+(0,1in)$) {$x+dx$};\n\\node[anchor=west] (A/right/down/1) at ($(A/right)+(0,-1in)$) {$x-dx$};\n\n\\path (A/left) -- (A/right) node [my midway label] {Case 1};\n\n\\draw[blue] (A/left) -- (A/right);\n\\draw[red] (A/left) -- (A/right/up/1.west);\n\\draw[red] (A/left) -- (A/right/down/1.west);\n\n%% case 2\n\\node[anchor=east] (B/left) at ($(A/right)+(1in,0)$) {$x$};\n\\node[anchor=west] (B/right) at ($(B/left)+(1in,0)$) {$x$};\n\n\\path (B/left) -- (B/right) node [my midway label] {Case 2};\n\n\\draw[blue] (B/left) -- (B/right);\n\\end{tikzpicture}\n\n\\end{document}", null, "There are other ways of doing this as suggested by @clement . But this approach will help familiarize yourself with some basics of TikZ.\n\nI'll explain a few of the details here:\n\n# node syntax\n\n\\node[<optional arguments>] (<optional node name>) at (<position>) {<content>};\n\n\nThe one part of this syntax that you cannot omit is the content.\n\nI use the power of the calc library to help place nodes relative to each other. That's what's happening when I write:\n\nat ($(<previously defined node name>)+(<vector>)$)\n\n\nWhen I opened the tikzpicture environment, I defined my own private style. This is a way of helping manage the picture. The idea is that I'm going to define a node between two points along a path and then shift it in the $y$ direction.\n\nIn this case, I used a particular instance of path syntax:\n\n\\path (<1st node name>) -- (<2nd node name>) node[<optional argument] {<content>};\n\n\nNote that in this case node is not a control sequence but just a bare word.\n\n• +1 for the long vary gracious tikz introduction for a newbie. – Ethan Bolker Apr 29 '15 at 0:16" ]

[ null, "https://i.stack.imgur.com/11jOc.png", null, "https://i.stack.imgur.com/aQXhC.png", null ]

[ "# Equivalent Ratios of 16.8:19.3\n\nEquivalent ratios or equal ratios are two ratios that express the same relationship between numbers as we covered in our tutorial on scaling up ratios. You can use the equivalent ratio calculator to solve ratio and/or proportion problems as required by entering your own ratio to produce a table similar to the \"Equivalent Ratios of 16.8:19.3 table\" provided below. This ratio table provides an increasingly list of ratios of the same proportions where the numerator and denominator are a direct multiplication of the multiplying value (mx). Ratio tables are very useful in math for calculating and comparing equivalent ratios, although most will likely use a ratio calculator to calculate equivalent ratios, it is also useful to have a ratio table where you can quickly cross reference associated ratios, particularly when working with complex math equations to resolve advanced math problems or physics problems. As a useful reference, we have included a table which provides links to the associated ratio values for the ratio 16.8:19.3, for example 16.8.1:19.3, 16.8:19.3.1, 16.8.1:19.3.2 and so on. We hope you will find these quick reference ratio tables useful as you can print and email them to yourself to aid your learning or a useful learning aide when teaching ratios to math students.\n\nLooking for a different type of ratio calculator or tutorial? Use the quick links below to access more ratio calculators\n\n 16.8 : 19.3(m1 = 1) 33.6 : 38.6(m2 = 2) 50.4 : 57.9(m3 = 3) 67.2 : 77.2(m4 = 4) 84 : 96.5(m5 = 5) 100.8 : 115.8(m6 = 6) 117.6 : 135.1(m7 = 7) 134.4 : 154.4(m8 = 8) 151.2 : 173.7(m9 = 9) 168 : 193(m10 = 10) 184.8 : 212.3(m11 = 11) 201.6 : 231.6(m12 = 12) 218.4 : 250.9(m13 = 13) 235.2 : 270.2(m14 = 14) 252 : 289.5(m15 = 15) 268.8 : 308.8(m16 = 16) 285.6 : 328.1(m17 = 17) 302.4 : 347.4(m18 = 18) 319.2 : 366.7(m19 = 19) 336 : 386(m20 = 20) 352.8 : 405.3(m21 = 21) 369.6 : 424.6(m22 = 22) 386.4 : 443.9(m23 = 23) 403.2 : 463.2(m24 = 24) 420 : 482.5(m25 = 25) 436.8 : 501.8(m26 = 26) 453.6 : 521.1(m27 = 27) 470.4 : 540.4(m28 = 28) 487.2 : 559.7(m29 = 29) 504 : 579(m30 = 30) 520.8 : 598.3(m31 = 31) 537.6 : 617.6(m32 = 32) 554.4 : 636.9(m33 = 33) 571.2 : 656.2(m34 = 34) 588 : 675.5(m35 = 35) 604.8 : 694.8(m36 = 36) 621.6 : 714.1(m37 = 37) 638.4 : 733.4(m38 = 38) 655.2 : 752.7(m39 = 39) 672 : 772(m40 = 40) 688.8 : 791.3(m41 = 41) 705.6 : 810.6(m42 = 42) 722.4 : 829.9(m43 = 43) 739.2 : 849.2(m44 = 44) 756 : 868.5(m45 = 45) 772.8 : 887.8(m46 = 46) 789.6 : 907.1(m47 = 47) 806.4 : 926.4(m48 = 48) 823.2 : 945.7(m49 = 49) 840 : 965(m50 = 50)\n\nDid you find the table of equivalent ratios of 16.8:19.3 useful? Please leave a rating below.\n\n## How to Calculate Ratios\n\nWhen calculating equivalent ratios you must multiply or divide both numbers in the ratio. This keeps both numbers in direct relation to each other. So, a ratio of 2/3 has an equivalent ratio of 4/6: in this ratio calculation we simply multiplied both 2 and 3 by 2.\n\n## Mathematical facts about the ratio 16.8:19.3\n\nThe numerator of the ratio 16.8:19.3 contains 1 decimal and the denominator contains 1 decimal\n\nThe lowest possible whole number equivalent ratio of the ratio 16.8:19.3 is:\n\n1 : 1.1488095238095\n\nIf you wish to express the ratio 16.8:19.3 as n to 1 then the ratio would be:\n\n16.8:19.3 as n to 1\n= 0.87046632124352 : 1\n\nIf you wish to express the ratio 16.8:19.3 as 1 to n then the ratio would be:\n\n16.8:19.3 as 1 to n\n= 1 : 1.1488095238095\n\nThe ratio 16.8:19.3 expressed as a fraction is [calculated using the ratio to fraction calculator]:\n\n16.8:19.3\n= 1/1.1488095238095\n\nThe ratio 16.8:19.3 expressed as a percentage is [calculated using the ratio to percentage calculator]:\n\n16.8:19.3\n= 87.046632124352%\n\n## Equivalent ratio tables for decimal ratios ranging 17 : 19 to 18 : 20\n\nThe table below contains links to equivalent ratio examples with ratios in increments of 0.1 in the range 17:19 to 18:20\n\n## Math Calculators\n\nYou may also find the following Math calculators useful." ]

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[ "Mathematics\n\nIf sum of two angles is $\\displaystyle { 90 }^{ o }$. They will be:\n\nComplementary angles\n\nSOLUTION\nIf the sum of two angles is $\\displaystyle { 90 }^{ o }$, then the angles are called Complementary angles.\nIf the sum of two angles is $\\displaystyle { 180 }^{ o }$, then the angles are called Supplementary angles.\nWhen two lines are crossed by another line, then the angles in matching corners are called Corresponding angles.\nA linear pair of angles is formed when two lines intersect. Two angles are said to be linear if they are adjacent angles formed by two intersecting lines. The measure of a straight angle is $180˚$, so a linear pair of angles must add up to $180˚$.\n\nYou're just one step away\n\nSingle Correct Medium Published on 09th 09, 2020\nQuestions 120418\nSubjects 10\nChapters 88\nEnrolled Students 87\n\nRealted Questions\n\nQ1 One Word Medium\nMeasures of some angle is given below. Find the measures of its complementary angle.\nif the angle is $\\displaystyle \\frac{2}{3}$rd of a right angle\n\nAsked in: Mathematics - Lines and Angles\n\n1 Verified Answer | Published on 09th 09, 2020\n\nQ2 Single Correct Medium\nIf the angle x and y are equal, then find the value of x and y.\n• A. $\\displaystyle x=y={ 55 }^{ o }$\n• B. $\\displaystyle x=y={ 60 }^{ o }$\n• C. $\\displaystyle x=y={ 90 }^{ o }$\n• D. $\\displaystyle x=y={ 45 }^{ o }$\n\nAsked in: Mathematics - Lines and Angles\n\n1 Verified Answer | Published on 23rd 09, 2020\n\nQ3 Subjective Medium\nIn Fig. 7.12, if AB ||DC and E is the mid - point of side AC, then prove that E is the mid-point of side BD.\n\nAsked in: Mathematics - Lines and Angles\n\n1 Verified Answer | Published on 09th 09, 2020\n\nQ4 Single Correct Medium\nIf two lines intersect such that four vertical angles are equal, then each angle is:\n• A. $45^{\\circ}$\n• B. $100^{\\circ}$\n• C. $180^{\\circ}$\n• D. $90^{\\circ}$\n\nAsked in: Mathematics - Lines and Angles\n\n1 Verified Answer | Published on 09th 09, 2020\n\nQ5 Subjective Medium\nRead the following two statements which are taken as axioms:\n(i) If two lines intersect each other, then the vertically opposite angles are not equal.\n(ii) If a ray stands on a line, then the sum of two adjacent angles so formed is equal to $180^0$." ]

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[ "# Common Anti-Patterns: String Usage for Number Manipulation\n\nThis is a short post about a topic which is frustrating to me from time to time. The topic is about number manipulation by utilizing string operations.\n\nI do not know why, but from time to time I see code and talk to people who have some tasks to do with numbers and they do not have a better idea than printing them into a string or character stream and applying string manipulation functionality to get their needed results. For me, it is strange to think about something like that, because I think it is against common sense to solve number issue with strings. Mathematics is the way to do that.\n\nI have to extreme examples here in this post which also caused major issues in the developed software. There were other examples, but the two were selected as examples.\n\n## Example 1: Rounding of Numbers\n\nI have seen this twice in my career. Once implemented in C and once again in Java. Once it was needed to round to a plain integer and the other time with a give number of digits for a needed precision. In both cases the implementation was done the same way:\n\n1. The original floating point number was printed into a string.\n2. For the decimal point the position was looked up.\n3. The string was cut at the decimal point and number in front of the point was taken and saved into a new variable.\n4. For checking whether to round up or down, the digit after the point was checked to be ‘5’ or larger (by character comparison) to know whether to round up or down.\n5. The number in front of the point was converted back into an integer.\n6. If it was to be rounded up, 1 was added.\n\nFor the arbitrary precision case, the position of the decimal point was shifted by the number of digits related to the precision.\n\nWhen I see something like that, I get frustrated. It is a waste of calculation power, because string operations are much more expensive than mathematical calculations and the procedure has several flaws:\n\n1. What happens if there is an exponential notion of the number? We need to cut at the ‘E’, too and take care of the number behind it.\n2. If localization is involved, what happens then? I live in Germany and we use a comma as decimal sign and the point is used for thousands.\n3. It is quite hard to assure correct behavior is arbitrary precision is to be applied.\n\nThe most simple solution for this issue is always to have a look for an already distributed function like Math.round in Java. A general procedure can be in C code:\n\nIt is assumed for that algorithm that a cast just truncates the digits right from the point. This rounding procedure is much fast than string manipulation.\n\nTo add a functionality for arbitrary precision rounding, it can be easiest achieved by multiplying the precision. For example assume the function:\n\nTo round num to an arbitrary precision we can do the following for two digits:\n\nWe have rounded now to two digits.\n\nAttention: I see the discussion coming already, but I do not take floating point number effects into account here. I know, that floating point numbers are not 100% correct due to internal representation, but this is not solved by string manipulation either as soon as the numbers are put back into doubles for example.\n\n## 2. Example: Calculating Mantissa and Exponent\n\nThe task is here to extract the mantissa and the exponent from floating point number into a 4 byte signed integer for the mantissa and a signed byte for the exponent to send everything over the network wire to systems which might have different binary floating point representation, no floating point representation due to usage of fixed decimal values or no exponential representation in strings. This may happen in integration projects. The separation into integers is not a bad idea and the target system can calculate its internal representation as needed.\n\nThe procedure I have seen in C code was:\n\n1. Print the number into a string (via sprintf with “%e”).\n2. Cut the string into two pieces at the ‘e’ to separate mantissa and exponent.\n3. Convert the exponent into an integer assuming it fits into the byte.\n4. The mantissa string’s dot is removed.\n5. The length of the mantissa string is checked for length and shortened to 9 characters to later fit into the 4 byte integer.\n6. The mantissa is converted into the final 4 byte integer.\n7. The exponent is subtracted by the length of the mantissa string – 1 to get the correct exponent for the large mantissa.\n\nThere also some flaws except of the performance impact especially in applications which deal heavily with number which was/is the case in the observed situation:\n\n1. Again, localization: What happens in Germany with the dot?\n2. It is assumed that the exponential string representation has always one digit in front of the decimal point.\n3. The exponent may become greater than 127 or smaller than -128.\n\nIt is much better to do some math here. I have to admit, the math here is not trivial any more because the exponential and logarithm functions need to be applied and the mathematical rules need to be present, but this is assumed to be know by software developers and software engineers. They do not need to know them by heart, but they should know they are there and where to look them up. Here is the mathematical algorithm in words:\n\n1.  Calculate the logarithm of base 10 of the number to be converted. Remember: log_10 (x)  = log(x) / log(10). Most programming languages only offer the natural logarithm with base e.\n2. Round the exponent from 1. up to an integer. Up rounding is needed later to not exceed the capacity of the 4 byte integer mantissa.\n3. Calculate the mantissa by dividing the number by 10 up to the power of the exponent got from 2.\n4. The number from 3. is guaranteed to be smaller than 1 due to the up-rounding of the exponent in 2.\n5. Multiply the mantissa by 1 billion to get the maximum precision of our two integer exponent representation and round it into our 4 byte integer.\n6. Subtract 9 from the exponent in 2. to get the correct exponent to the new mantissa.\n\nThat’s it. It is maybe not that easy to understand mathematically, but it is optimized for computers. A computer is better in computing numbers than in manipulating string.\n\n## Conclusion\n\nIn our daily work we are confronted with a lot of tasks which are not so easy to solve on first glance. We always should look out for the best solution for the computer to perform, for the developer to implement correctly and for other developers to understand.\n\nSince computers are optimized for computations, mathematical solutions are to be preferred for performance reasons. It can also be assumed that software developers and software engineers have some understanding of mathematics to find correct solutions. If not, there are also forums to ask people for help. It is not very probable that a problem at hand was not solved by someone else, yet. The most problems occurred already somewhere else and a proofed solution is always better than something new." ]

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[ "Home | | Chemistry 12th Std | emf of a Cell\n\n# emf of a Cell\n\nThe force that pushes the electrons away from the anode and pulls them towards cathode is called the electromotive force (emf) (or) the cell potential. The SI unit of cell potential is the volt (v).\n\nemf of a Cell\n\nWe have learnt that when two half cells of a Daniel cell are connected, a spontaneous redox reaction will take place which results in the flow of electrons from anode to cathode. The force that pushes the electrons away from the anode and pulls them towards cathode is called the electromotive force (emf) (or) the cell potential. The SI unit of cell potential is the volt (v).\n\nWhen there is one volt difference in electrical potential between the anode and cathode, one joule of energy is released for each columb of charge that moves between them.\n\ni.e., 1J = 1C Ă— 1V                          .....(9.18)\n\nThe cell voltage depends on the nature of the electrodes, the concentration of the electrolytes and the temperature at which the cell is operated. For example\n\nAt, 25 C , The emf of the below mentioned Daniel cell is 1.107 Volts\n\nZn (s) | Zn2+ (aq,1M) || Cu2+ (aq,1M) | Cu ( s )                     E0 = 1.107 V\n\nTags : Electrochemical Cell , 12th Chemistry : UNIT 9 : Electro Chemistry\nStudy Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail\n12th Chemistry : UNIT 9 : Electro Chemistry : emf of a Cell | Electrochemical Cell" ]

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[ "How do you differentiate f(x)=cose^(4x) using the chain rule.?\n\nDec 7, 2015\n\n$f ' \\left(x\\right) = - 4 \\sin \\left({e}^{4 x}\\right) \\cdot {e}^{4 x}$\n\nExplanation:\n\nYour chain can be defined as follows:\n\n$f \\left(x\\right) = \\textcolor{g r e e n}{\\cos} \\textcolor{b l u e}{{e}^{\\textcolor{red}{4 x}}}$\n\n$\\textcolor{w h i t e}{\\times x} = \\cos {e}^{w \\left(x\\right)} \\textcolor{w h i t e}{\\times x} \\text{ where } w \\left(x\\right) = \\textcolor{red}{4 x}$\n\n$\\textcolor{w h i t e}{\\times x} = \\cos v \\left(w\\right) \\textcolor{w h i t e}{\\times \\xi i i} \\text{ where } v \\left(w\\right) = \\textcolor{b l u e}{{e}^{w}}$\n\n$\\textcolor{w h i t e}{\\times x} = u \\left(v\\right) \\textcolor{w h i t e}{\\times \\times \\times i} \\text{ where } u \\left(v\\right) = \\textcolor{g r e e n}{\\cos \\left(v\\right)}$\n\nThus, to compute the derivative, you need to build the derivatives of $w \\left(x\\right)$, $v \\left(w\\right)$ and $u \\left(v\\right)$:\n\n$w \\left(x\\right) = 4 x \\textcolor{w h i t e}{\\times x} \\implies \\textcolor{w h i t e}{\\times} w ' \\left(x\\right) = 4$\n\n$v \\left(w\\right) = {e}^{w} \\textcolor{w h i t e}{\\times x} \\implies \\textcolor{w h i t e}{\\times} v ' \\left(w\\right) = {e}^{w} = {e}^{4 x}$\n\n$u \\left(v\\right) = \\cos v \\textcolor{w h i t e}{\\times} \\implies \\textcolor{w h i t e}{\\times} u ' \\left(v\\right) = - \\sin v = - \\sin \\left({e}^{w}\\right) = - \\sin \\left({e}^{4 x}\\right)$\n\nNow, the only thing left to do is to multiply the three derivatives!\n\n$f ' \\left(x\\right) = u ' \\left(v\\right) \\cdot v ' \\left(w\\right) \\cdot w ' \\left(x\\right)$\n$\\textcolor{w h i t e}{\\times \\times} = - \\sin \\left({e}^{4 x}\\right) \\cdot {e}^{4 x} \\cdot 4$\n$\\textcolor{w h i t e}{\\times \\times} = - 4 \\sin \\left({e}^{4 x}\\right) \\cdot {e}^{4 x}$" ]

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[ "# torchgeometry.contrib¶\n\nspatial_soft_argmax2d(input: torch.Tensor, normalized_coordinates: Optional[bool] = True) → torch.Tensor[source]\n\nFunction that computes the Spatial Soft-Argmax 2D of a given heatmap.\n\nSee torchgeometry.contrib.SpatialSoftArgmax2d for details.\n\nextract_tensor_patches(input: torch.Tensor, window_size: Union[int, Tuple[int, int]], stride: Union[int, Tuple[int, int], None] = 1, padding: Union[int, Tuple[int, int], None] = 0) → torch.Tensor[source]\n\nFunction that extract patches from tensors and stack them.\n\nSee ExtractTensorPatches for details.\n\nclass SpatialSoftArgmax2d(normalized_coordinates: Optional[bool] = True)[source]\n\nCreates a module that computes the Spatial Soft-Argmax 2D of a given input heatmap.\n\nReturns the index of the maximum 2d coordinates of the give map. The output order is x-coord and y-coord.\n\nParameters: normalized_coordinates (Optional[bool]) – wether to return the coordinates normalized in the range of [-1, 1]. Otherwise, it will return the coordinates in the range of the input shape. Default is True.\nShape:\n• Input: $$(B, N, H, W)$$\n• Output: $$(B, N, 2)$$\nExamples::\n>>> input = torch.rand(1, 4, 2, 3)\n>>> m = tgm.losses.SpatialSoftArgmax2d()\n>>> coords = m(input) # 1x4x2\n>>> x_coord, y_coord = torch.chunk(coords, dim=-1, chunks=2)\n\nclass ExtractTensorPatches(window_size: Union[int, Tuple[int, int]], stride: Union[int, Tuple[int, int], None] = 1, padding: Union[int, Tuple[int, int], None] = 0)[source]\n\nModule that extract patches from tensors and stack them.\n\nApplies a 2D convolution over an input tensor to extract patches and stack them in the depth axis of the output tensor. The function applies a Depthwise Convolution by applying the same kernel for all the input planes.\n\nIn the simplest case, the output value of the operator with input size $$(B, C, H, W)$$ is $$(B, N, C, H_{out}, W_{out})$$.\n\nwhere\n• $$B$$ is the batch size.\n• $$N$$ denotes the total number of extracted patches stacked in\n• $$C$$ denotes the number of input channels.\n• $$H$$, $$W$$ the input height and width of the input in pixels.\n• $$H_{out}$$, $$W_{out}$$ denote to denote to the patch size defined in the function signature. left-right and top-bottom order.\n• window_size is the size of the sliding window and controls the shape of the output tensor and defines the shape of the output patch.\n• stride controls the stride to apply to the sliding window and regulates the overlapping between the extracted patches.\n• padding controls the amount of implicit zeros-paddings on both sizes at each dimension.\n\nThe parameters window_size, stride and padding can be either:\n\n• a single int – in which case the same value is used for the height and width dimension.\n• a tuple of two ints – in which case, the first int is used for the height dimension, and the second int for the width dimension.\nParameters: window_size (Union[int, Tuple[int, int]]) – the size of the convolving kernel and the output patch size. stride (Optional[Union[int, Tuple[int, int]]]) – stride of the convolution. Default is 1. padding (Optional[Union[int, Tuple[int, int]]]) – Zero-padding added to both side of the input. Default is 0.\nShape:\n• Input: $$(B, C, H, W)$$\n• Output: $$(B, N, C, H_{out}, W_{out})$$\nReturns: the tensor with the extracted patches. torch.Tensor\n\nExamples\n\n>>> input = torch.arange(9.).view(1, 1, 3, 3)\n>>> patches = tgm.contrib.extract_tensor_patches(input, (2, 3))\n>>> input\ntensor([[[[0., 1., 2.],\n[3., 4., 5.],\n[6., 7., 8.]]]])\n>>> patches[:, -1]\ntensor([[[[3.0000, 4.0000, 5.0000],\n[6.0000, 7.0000, 8.0000]]]])" ]

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[ "### Slight change in the iteration condition of the calculateEigenValues\n\n```functions\n\n- The relative criterion for convergence of the iterative algorithm\ncan no be given as function argumenti\n\n- approved by Bernd\n\ngit-svn-id: svn://svn.iws.uni-stuttgart.de/DUMUX/dumux/trunk@11496 2fb0f335-1f38-0410-981e-8018bf24f1b0```\nparent 056f7ba8\n ... ... @@ -61,9 +61,10 @@ double calcOffDiagonalNorm(Matrix& matrix) /* * \\param eigVel Vector for storing the eigenvalues * \\param matrix n x n matrices for which eigenvalues have to be calculated * \\param relativeTolerance tolerance for the relative convergence criterion (default: 0.01) */ template bool calculateEigenValues(EVVectorType &eigVel, MatrixType& matrix) bool calculateEigenValues(EVVectorType &eigVel, MatrixType& matrix, double relativeTolerance = 0.01) { if (dim == 2) { ... ... @@ -92,7 +93,7 @@ bool calculateEigenValues(EVVectorType &eigVel, MatrixType& matrix) MatrixType rotationMatrix(0.0); MatrixType evMatrix(matrix); while ((iter < maxIter && offDiagonalNorm > 0.01 * matrixNorm) || iter < 1) while (iter < maxIter && offDiagonalNorm > relativeTolerance * matrixNorm) { for (int i = 0; i < dim - 1; i++) { ... ... @@ -143,9 +144,10 @@ bool calculateEigenValues(EVVectorType &eigVel, MatrixType& matrix) * \\param eigVel Vector for storing the eigenvalues * \\param eigVec n x n for storing the eigenvectors * \\param matrix n x n matrices for which eigenvalues have to be calculated * \\param relativeTolerance tolerance for the relative convergence criterion (default: 0.01) */ template bool calculateEigenValues(EVVectorType &eigVel, MatrixType& eigVec, MatrixType& matrix) bool calculateEigenValues(EVVectorType &eigVel, MatrixType& eigVec, MatrixType& matrix, double relativeTolerance = 0.01) { int maxIter = 100; int iter = 0; ... ... @@ -156,7 +158,7 @@ bool calculateEigenValues(EVVectorType &eigVel, MatrixType& eigVec, MatrixType& MatrixType evecMatrix(matrix); identityMatrix(evecMatrix); while ((iter < maxIter && offDiagonalNorm > 0.01 * matrixNorm) || iter < 1) while (iter < maxIter && offDiagonalNorm > relativeTolerance * matrixNorm) { for (int i = 0; i < dim - 1; i++) { ... ...\n ... ... @@ -209,7 +209,7 @@ public: /*!\\brief Sets the cell phase potential * * \\param phaseIdx Index of a fluid phase * \\param press Phase potential which is stored * \\param pot Phase potential which is stored */ void setPotential(int phaseIdx, Scalar pot) { ... ... @@ -502,11 +502,11 @@ public: /*!\\brief Sets the cell phase potential * * \\param phaseIdx Index of a fluid phase * \\param press Phase potential which is stored * \\param pot Phase potential which is stored */ void setPotential(int phaseIdx, Scalar press) void setPotential(int phaseIdx, Scalar pot) { potential_[phaseIdx] = press; potential_[phaseIdx] = pot; } //! \\cond \\private ... ...\nSupports Markdown\n0% or .\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!\nPlease register or to comment" ]

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[ "## JavaFX,Unity3D,Android,IOS等技术教程和生活随笔，仅供记录\n\nhttp://www.wingmei.cn/wp-content/themes/Vtrois-Kratos-e85a527/images/background.jpg\n\n# IOS Swift 3.1字符串截取\n\n``````func substring(from: String.Index)\nfunc substring(to: String.Index)\nfunc substring(with: Range)``````\n\n``````var str = \"Hello World!\"\nlet startIndex = str.index(str.startIndex, offsetBy: 5)\nlet endIndex = str.index(str.endIndex,offsetBy: -2)\nlet result = str.substring(with: startIndex..<endIndex)``````\n\n``````let rangIndex = str.range(of: \"World\")\nlet endIndex = rangIndex!.lowerBound\nlet startIndex = str.index(endIndex, offsetBy: -4)\nlet result = str.substring(with: startIndex..<endIndex)``````", null, "``````+(NSString *)subString:(NSString *)content startIndex:(int)startIndex length:(int)length {\nNSString* result = [content substringWithRange:NSMakeRange(startIndex, length)];\nreturn result;\n}``````" ]

[ null, "http://www.wingmei.cn/wp-content/uploads/2017/05/[email protected]", null ]

[ "# Give an example or state it's impossible\n\nA sequence of closed bounded (not necessarily nested) intervals $I_1,I_2,I_3,\\dots$ with the property that $\\bigcap_{n=1}^NI_n\\neq\\emptyset$ for all $N\\in\\mathbb{N}$, but $\\bigcap_{n=1}^{\\infty}I_n=\\emptyset$.\n\nI tried to come up with some examples but failed. One example I came up with is a \"shifting\" sequence of closed bounded intervals, $I_n=[a_n,b_n]$ with $(a_n)=(1,2,3,4,5,\\dots,N,N+1,\\dots)$ and $(b_n)=(N,N+1,N+2,\\dots)$. Then $\\bigcap_{n=1}^NI_n =\\{N\\}$ and $\\bigcap_{n=1}^{\\infty}I_n=\\emptyset$. But clearly it's wrong...\n\n• What are your thoughts ? Mar 8, 2018 at 16:11\n• Hint: They cannot be nested…\n– Gono\nMar 8, 2018 at 16:12\n• Welcome to stackexchange. You are more likely to get answers rather than downvotes or votes to close if you edit the question to show what you tried and where you are stuck. Mar 8, 2018 at 16:14\n• To comment on your example, you can't let the intervals depend on $N$. That's backwards. You must first give a fixed collection of intervals which do not change, and then you show that the intersection is non-empty for any $N$. Mar 8, 2018 at 16:22\n• Hint: suppose this were possible. Then $J_n = \\bigcap_{k=1}^{n}I_k$ is a decreasing sequence of nonempty closed bounded sets. What can you say about $\\bigcap_{n=1}^{\\infty}J_n$?\n– user169852\nMar 8, 2018 at 16:40\n\nSummarizing the comments, there exists no such sequence. Suppose that each $I_n$ is a closed, bounded interval and that for each $n$, we have $\\bigcap_{k=1}^{n}I_k \\neq \\emptyset$. Define $J_n = \\bigcap_{k=1}^{n}I_k$. Then:\n• $J_n$ is closed, because it is an intersection of closed sets.\n• $J_n$ is bounded, because it is contained in $I_1$.\n• $J_n$ is nonempty by assumption.\n• $J_n$ is an interval, because a nonempty intersection of finitely many intervals is an interval (easy proof by induction).\n• $(J_n)_{n=1}^{\\infty}$ is a decreasing nested sequence, because for each $n$, we have $J_{n+1} = I_{n+1} \\cap J_n \\subseteq J_n$.\nBy the nested interval theorem, we conclude that $\\bigcap_{n=1}^{\\infty}J_n$ is nonempty. But you can easily verify that $\\bigcap_{n=1}^{\\infty}J_n = \\bigcap_{n=1}^{\\infty}I_n$ (show that they are subsets of each other), so $\\bigcap_{n=1}^{\\infty}I_n$ is nonempty." ]

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[ "Flyback Converter Transformer Leakage Inductance\n\nWe are in the process of designing an inverter , in which designing a step up flyback converter is crucial.\n\nThe problem, we initially faced was there were more step down transformers available than step up. So we planned to turn around the transformer and use it. Usually the manufactures give leakage inductance ratings for the primary side , now that we have inverted the transformer , we have ratings for the secondary. How do we use LT spice to find the desired leakage inductance. Is there any other way we could do this ?\n\nThe simple answer is that if you know the turns ratio you can \"move\" the leakage inductance from one side of a transformer to the other by multiplying or dividing (as appropriate) by the turns ratio squared. For example, if the primary has a leakage inductance of 100 uH and has a step down ratio of 10:1 then that leakage inductance on the secondary side is 100 uH divided by $$\\10^2\\$$ = 1 uH." ]

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[ "# Video: AF5P1-Q13-290154537409v2\n\nThe ratios for the minimum number of scout leaders required per scout to run two different activities are shown in the table. (a) There are four scout leaders available to take scouts kayaking. What is the maximum number of scouts that can go kayaking? (b) 27 scouts want to practice archery. What is the smallest number of scout leaders required?\n\n04:25\n\n### Video Transcript\n\nThe ratios for the minimum number of scout leaders required per scout to run two different activities are shown in the table. Kayaking, scout leaders to scouts one to 11. Archery, scout leaders to scouts one to six. Part a) there are four scout leaders available to take scouts kayaking. What is the maximum number of scouts that can go kayaking?\n\nThere’s also a part b that we will look at later. The ratios given are for the minimum number of scout leaders required per scout. We can therefore say that the ratio one to 11 means that there’s at least one leader for 11 scouts or at most 11 scouts for every leader.\n\nFor the first part of the question, we’re told there’re four scout leaders. And we want to calculate the maximum number of scouts that can go kayaking. The ratio of scout leaders to scouts is one to 11. We know there’re four scout leaders. And one multiplied by four equals four. In ratio questions, whatever we do to one side of the ratio, we must do to the other. Therefore, we need to multiply the number of scouts by four. 11 multiplied by four is equal to 44. This means that the maximum number of scouts that can go kayaking with four scout leaders is 44. The second part of the question says the following.\n\nb) 27 scouts want to practice archery. What is the smallest number of scout leaders required?\n\nWe are told that the ratio of scout leaders to scouts for archery is one to six. We need to calculate the number of scout leaders required for 27 scouts. As we have already mentioned with ratio, whatever we do to one side, we must do to the other. Our first step is to work out the multiplier or scale factor that gets us from six to 27. In order to do this, we need to divide 27 by six. This is equal to four remainder three as six multiplied by four is equal to 24. As the remainder was three and the number we were dividing by is six, then 27 divided by six is equal to four and three-sixths. The remainder becomes the numerator. And the divisor becomes the denominator. Three-sixths is the same or equivalent to one-half as three divided by three is one. And six divided by three is two. Four and a half is the same as 4.5. Therefore, we need 4.5 scout leaders for 27 scouts.\n\nRemember we were told that this is the minimum number of scout leaders. And as we can’t have half a scout leader, we need to round this up to five. If 27 scouts want to practice archery, the smallest number of Scout leaders required is five.\n\nAn alternative method here would be to use our knowledge of our six times table. We know that we need one scout leader for every six scouts. This means that we would need two scout leaders for 12 scouts as six multiplied by two is 12. We would need three scout leaders for 18 scouts, four scout leaders for 24 scouts, and five scout leaders for 30 scouts. As 27 is greater than 24, four scout leaders would not be enough. Therefore, once again, we have proved that we need five scout leaders for 27 scouts. Those five scout leaders could take up to 30 scouts to practice archery." ]

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[ "10.9: Gibbs Energy and Reactions\n\nEntropy and enthalpy are two of the basic factors of thermodynamics. Enthalpy has something to do with the energetic content of a system or a molecule. Entropy has something to do with how that energy is stored.\n\n• A reaction is favored if enthalpy decreases: There is a bias in nature toward decreasing enthalpy in a system. Reactions can happen when enthalpy is transferred to the surroundings.\n• A reaction is favored if entropy increases: There is also a bias in nature toward increasing entropy in a system. Reactions can happen when entropy increases.\n\nConsider the cartoon reaction below. Red squares are being converted to green circles, provided the reaction proceeds from left to right as shown.", null, "Whether or not the reaction proceeds to the right depends on the balance between enthalpy and entropy. There are several combinations possible.\n\nIn one case, maybe entropy increases when the red squares turn into green circles, and the enthalpy decreases. If we think of the balance between these two factors, we come to a simple conclusion. Both factors tilt the balance of the reaction to the right. In this case, the red squares will be converted into green circles.", null, "Alternatively, maybe entropy decreases when the red squares turn into green circles, and enthalpy increases. If we think of the balance between these two factors, we come to another simple conclusion. Both factors tilt the balance of the reaction to the left. In this case, the red squares will remain just as they are.", null, "Having two factors may lead to complications. For example, what if enthalpy decreases, but so does entropy? Does the reaction happen, or doesn't it? In that case, we may need quantitation to make a decision. How much does the enthalpy decrease? How much does the entropy decrease? If the effect of the enthalpy decrease is greater than that of the entropy decrease, the reaction may still go forward.", null, "The combined effects of enthalpy and entropy are often combined in what is called \"free energy.\" Free energy is just a way to keep track of the sum of the two effects. Mathematically, the symbol for the internal enthalpy change is \"ΔH\" and the symbol for the internal entropy change is \"ΔS.\" Free energy is symbolized by \"ΔG,\" and the relationship is given by the following expression:\n\n$\\Delta G = \\Delta H - T \\Delta S$\n\n$$T$$ in this expression stands for the temperature (in Kelvin, rather than Celsius or Fahrenheit). The temperature acts as a scaling factor in the expression, putting the entropy and enthalpy on equivalent footing so that their effects can be compared directly.\n\nHow do we use free energy? It works the same way we were using enthalpy earlier (that's why the free energy has the same sign as the enthalpy in the mathematical expression, whereas the entropy has an opposite sign). If free energy decreases, the reaction can proceed. If the free energy increases, the reaction can't proceed.\n\n• A reaction is favored if the free energy of the system decreases.\n• A reaction is not favored if the free energy of the system increases.\n\nBecause free energy takes into consideration both the enthalpy and entropy changes, we don't have to consider anything else to decide if the reaction occurs. Both factors have already been taken into account.\n\nRemember the terms \"endothermic\" and \"exothermic\" from our discussion of enthalpy. Exothermic reactions were favored (in which enthalpy decreases). Endothermic ones were not. In free energy terms, we say that exergonic reactions are favored(in which free energy decreases). Endergonic ones (in which free energy increases) are not.\n\nProblem TD4.1.\n\nImagine a reaction in which the effects of enthalpy and entropy are opposite and almost equally balanced, so that there is no preference for whether the reaction proceeds or not. Looking at the expression for free energy, how do you think the situation will change under the following conditions:\n\n1. the temperature is very cold (0.09 K)\n2. the temperature is very warm (500 K)\n\nProblem TD4.2.\n\nWhich of the following reaction profiles describe reactions that will proceed? Which ones describe reactions that will not proceed?", null, "How Entropy Rules Thermodynamics\n\nSometimes it is said that entropy governs the universe. As it happens, enthalpy and entropy changes in a reaction are partly related to each other. The reason for this relationship is that if energy is added to or released from the system, it has to be partitioned into new states. Thus, an enthalpy change can also have an effect on entropy.", null, "Specifically, the internal enthalpy change that we discussed earlier has an effect on the entropy of the surroundings. So far, we have just considered internal entropy changes.\n\n• In an exothermic reaction, the external entropy (entropy of the surroundings) increases.\n• In an endothermic reaction, the external entropy (entropy of the surroundings) decreases.\n\nFree energy takes into account both the entropy of the system and the entropy changes that arise because of heat exchange with the surroundings. Together, the system and the surroundings are called \"the universe\". That's because the system is just everything involved in the reaction, and the surroundings are everything that isn't involved in the reaction.\n\nEnthalpy changes in the system lead to additional partitioning of energy. We might visualize that with the mailbox analogy we used for entropy earlier. In this case, each molecule has its own set of mailboxes, into which it sorts incoming energy.", null, "Looked at in this way, thermodynamics boils down to one major consideration, and that is the combined entropy of both the system and its surroundings (together known as the universe).\n\n• For a reaction to proceed, the entropy of the universe must increase." ]

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[ "# Output as numbered display formula\n\nI would like to be able to have Mathematica output appear in my \"DisplayFormulaNumbered\" style, which puts an automatically incremented equation number at the right margin. My stylesheet defines this as:\n\nCell[StyleData[\"DisplayFormulaNumbered\"],\nCellFrame->False,\nCellMargins->{{66, Inherited}, {0, 7}},\nStyleKeyMapping->{\nKeyEvent[\"Backspace\", Modifiers -> {Shift}] -> \"Input\",\n\"Tab\" -> \"DisplayFormula\"},\nCellFrameLabels->{{None,\nCell[\nTextData[{\"(\",\nCounterBox[\"DisplayFormulaNumbered\"], \")\"}],\n\"DisplayFormulaEquationNumber\", FontSlant ->\n\"Plain\"]}, {None, None}},\nFontFamily->\"Times New Roman\",\nFontSize->12,\nFontWeight->\"Plain\",\nFontSlant->\"Italic\",\nFontVariations->{\"StrikeThrough\"->False,\"Underline\"->False},\nFontProperties->{\"ScreenResolution\"->96},\nFontColor->RGBColor[0., 0., 0.]]\n\n\nSo now I'd like to do something like\n\nu[x_, f_] := 1 - f (Pi/2 + ArcTan[x])/Pi\nStyle[U[x, α] == Evaluate[u[x, α]], \"DisplayFormulaNumbered\"]\n\n\nwhich almost works (I get the output in the style I want), but it omits the equation number. How do I fix this?\n\nPart of the story is that, for presentation purposes, I use a special style for Input cells like so:\n\nCell[StyleData[\"InputHidden\", StyleDefinitions -> StyleData[\"Input\"]],\nCellEpilog:>(SelectionMove[EvaluationNotebook[], All, GeneratedCell];\nFrontEndTokenExecute[\"SelectionCloseUnselectedCells\"]),\n\n\nwhich gives me an input cell that automatically hides when evaluated, leaving only the Output cell to be displayed. The problem with Kubo's suggestion below is that it breaks this mechanism, by producing a new cell that's not considered the Output cell of its Input.\n\nCellPrint @ ExpressionCell[\nU[x, α] == Evaluate[u[x, α]],\n\"DisplayFormulaNumbered\"\n]", null, "Is this ok? Counter number is displayed as a CellFrameLabel so Style can't use it, we need to generate a Cell or change Output to DisplayFormulaNumbered.\n\nHmm, not quite. [...] the cell doesn't count as an Output cell anymore, which means I can't hide the Input cell by double-clicking the output cell.\n\nYou can mix styles:\n\n CellPrint @ ExpressionCell[\nU[x, α] == Evaluate[u[x, α]],\n\"Output\", \"DisplayFormulaNumbered\"\n]\n\n\nWhich preserves grouping on Input and Output cells here.\n\n• Hmm, not quite. It works, in principle, but now the cell doesn't count as an Output cell anymore, which means I can't hide the Input cell by double-clicking the output cell.The same happens if I change the Output cell to DisplayFormulaNumbered after the fact: The Input and Output cells are now separated. I edited my original post to clarify.\n– Pirx\nAug 5, 2016 at 15:32\n• @Pirx Take a look at the edit.\n– Kuba\nAug 5, 2016 at 15:36\n• Awesome, this works. Thanks!\n– Pirx\nAug 5, 2016 at 15:43\n\nI do not know, how to make the numbering automatically, but if you agree to make it manually, try this\n\n ExpressionCell[\nRow[{Spacer,\nStyle[\"\\!$$\\*SubsuperscriptBox[\\(∫$$, $$a$$, $$b$$]\\)x\\\n\\[DifferentialD]x=\", Italic, 22], Style[\\!$$\\*SubsuperscriptBox[\\(∫$$, $$a$$, $$b$$]$$x \\ \\[DifferentialD]x$$\\), Italic, 22], Spacer, Style[\"(1)\", 20]}],\nTextAlignment -> Right,\nFractionBoxOptions -> {AllowScriptLevelChange ->", null, "" ]

[ null, "https://i.stack.imgur.com/o7xAy.png", null, "https://i.stack.imgur.com/cgHWt.jpg", null ]

[ "# Negation", null, "No agreement exists as to the possibility of defining negation, as to its logical status, function and meaning, as to its field of applicability, and as to the interpretation of the negative judgment (F.H. Heinemann 1944).\n\nAlgebraically, classical negation corresponds to complementation in a Boolean algebra, and intuitionistic negation to pseudocomplementation in a Heyting algebra. These algebras provide a semantics for classical and intuitionistic logic, respectively.\n\nThe negation of a proposition p is notated in different ways, in various contexts of discussion and fields of application. The following table documents some of these variants:\n\nAs a result, in the propositional case, a sentence is classically provable if its double negation is intuitionistically provable. This result is known as Glivenko's theorem.\n\nDe Morgan's laws provide a way of distributing negation over disjunction and conjunction:\n\nAnother way to express this is that each variable always makes a difference in the truth-value of the operation, or it never makes a difference. Negation is a linear logical operator.\n\nAs in mathematics, negation is used in computer science to construct logical statements.\n\nIn computer science there is also bitwise negation. This takes the value given and switches all the binary 1s to 0s and 0s to 1s. See bitwise operation. This is often used to create ones' complement or \"`~`\" in C or C++ and two's complement (just simplified to \"`-`\" or the negative sign since this is equivalent to taking the arithmetic negative value of the number) as it basically creates the opposite (negative value equivalent) or mathematical complement of the value (where both values are added together they create a whole).\n\nTo get the absolute (positive equivalent) value of a given integer the following would work as the \"`-`\" changes it from negative to positive (it is negative because \"`x < 0`\" yields true)\n\nInverting the condition and reversing the outcomes produces code that is logically equivalent to the original code, i.e. will have identical results for any input (note that depending on the compiler used, the actual instructions performed by the computer may differ).\n\nThis convention occasionally surfaces in ordinary written speech, as computer-related slang for not. For example, the phrase `!voting` means \"not voting\". Another example is the phrase `!clue` which is used as a synonym for \"no-clue\" or \"clueless\".\n\nIn Kripke semantics where the semantic values of formulae are sets of possible worlds, negation can be taken to mean set-theoretic complementation[citation needed] (see also possible world semantics for more)." ]

[ null, "https://upload.wikimedia.org/wikipedia/commons/thumb/7/73/Venn10.svg/1200px-Venn10.svg.png", null ]

[ "## Stream: new members\n\n### Topic: Maths_in_lean book example.\n\n####", null, "Manoj Kummini (Sep 24 2020 at 20:49):\n\nDear all,\n\nI am trying to learn lean and mathlib, and am working through the book mathematics_in_lean. In Section 2.1, there is an example of a typical proof state', which I am trying to set up and solve as follows:\n\nimport data.nat.parity data.nat.prime tactic\nopen nat\n\nexample (x y : nat) : (prime x) /\\ not (even x) /\\ (y > x) -> (y >= 4) :=\nbegin\nintros h,\nhave h1,\nexact (h.left),\nhave h23,\nexact (h.right),\nhave h2,\nexact (h23.left),\nhave h3,\nexact(h23.right),\nclear h h23,\n-- now I have a proof state as given in the book!\n-- I now tried\nhave x_ge_two,\napply prime.two_le h1\n\n\nwhich got me x_ge_two : 2 <= x. After this, I am unable to use the the remaining hypothesis (h2 and h3). Any hint would be appreciated! Thanks.\n\n####", null, "Kevin Buzzard (Sep 24 2020 at 21:12):\n\nimport data.nat.parity data.nat.prime tactic\nopen nat\n\nexample (x y : nat) : (prime x) /\\ not (even x) /\\ (x < y) -> (4 ≤ y) :=\nbegin\nintros h,\nhave h1,\nexact (h.left),\nhave h23,\nexact (h.right),\nhave h2,\nexact (h23.left),\nhave h3,\nexact(h23.right),\nclear h h23,\n-- now I have a proof state as given in the book!\n-- I now tried\nhave x_ge_two,\napply prime.two_le h1,\n-- 2 ≤ x ↔ 2 < x or 2 = x\nrw le_iff_lt_or_eq at x_ge_two,\ncases x_ge_two,\n{ -- 2 < x case\n-- use lemma saying a < b ↔ a+1 ≤ b\nrw [←succ_le_iff, succ_eq_add_one] at x_ge_two h3,\n-- now it's easy\nlinarith,\n},\n{ -- 2 = x case\nexfalso,\napply h2,\nrw ←x_ge_two,\nsimp,\n}\n\nend\n\n\n####", null, "Kevin Buzzard (Sep 24 2020 at 21:15):\n\nIt's a bit annoying that I couldn't finish it off more quickly\n\n####", null, "Kevin Buzzard (Sep 24 2020 at 21:15):\n\nMaybe I'm missing a tactic. Actually does tidy` do it?\n\n####", null, "Manoj Kummini (Sep 25 2020 at 01:19):\n\nThank you!\n\nLast updated: May 16 2021 at 21:11 UTC" ]

[ null, "https://leanprover-community.github.io/archive/assets/img/zulip2.png", null, "https://leanprover-community.github.io/archive/assets/img/zulip2.png", null, "https://leanprover-community.github.io/archive/assets/img/zulip2.png", null, "https://leanprover-community.github.io/archive/assets/img/zulip2.png", null, "https://leanprover-community.github.io/archive/assets/img/zulip2.png", null ]

[ "LESSON\n\n# Relative pose in 2D\n\n#### Transcript\n\nWe’ve talked about how we can describe the position and orientation, the pose of one coordinate frame with respect to another. Here, we have two coordinate frames labeled A and B, and we use this symbol here to denote the pose of B with respect to A. I've introduced a point P and I can describe it by two vectors. I can describe it with respect to coordinate frame A; that's the blue arrow, or with respect to coordinate frame B; that the red arrow. I can relate the vector with respect to frame A to the vector with respect to frame B using the pose and this operator, the dot.\n\nNow, let's consider that I introduce another coordinate frame, coordinate frame C and I have a relative pose that describes C with respect to B. And I can describe the point P in terms of a vector with respect to coordinate frame C. So, these relative poses can be compounded or composed, so I can get to the pose C by first of all getting to the pose B and then getting to the pose C. So, there's some way that I can add these two poses together, but we haven't described yet exactly what a pose is. I'm going to introduce another abstract operator. It is a plus sign inside a circle, I'm going to call that O+. So, the O+ operator allows us to join or add these two poses. So, the relative pose from frame A to frame B, O+ the relative pose from frame B to frame C gives us the relative pose of C with respect to A. This O+ operator performs what is called composition. It allows us to compose, to compound, to add, in inverted commas, two poses together. Then, I can use my dot operator to transform the vector from frame C to frame A, and we can extend this approach indefinitely. I can compound as many relative poses as I wish.\n\nWhen we talk about things like relative motions, it's really important to keep in mind that the order in which you apply them is critically important. Now, consider the example where I move from the reference frame and I move a coordinate frame 30 centimetres in the x direction and then I rotate it by 45 degrees, and we defined the convention for rotation; this is a positive rotation of 45 degrees. Now, imagine that I do these operations in the reverse order. So, I take my reference coordinate frame and I rotate it by 45 degrees and then I translate it by 30 centimetres in the x direction. So, now I end up with a coordinate frame over here. So, you see that by changing the order in which I apply these elementary motions, the translation and the rotation, I end up with very different coordinate frames. Ordering is really important.\n\nLet's look at a more complex example, and I've introduced here a world coordinate frame. Now, I introduce a coordinate frame which represents the pose of my robot on the table. And here's a coordinate frame which represents the pose of the little robot. And here is another coordinate frame which represents the hand of the big robot. And I can describe the pose of the hand of the robot with respect to the robot itself. I can introduce another coordinate frame which describes the pose of a camera which is looking at both of these robots. And, I can perhaps describe the pose of the camera with respect to my world coordinate frame. And perhaps the camera is able to tell where the robot is, with respect to the camera's coordinate frame. And, perhaps the camera can also tell me the pose of the little robot with respect to the camera's coordinate frame. And now, I might be interested in knowing, what's the pose of the little robot with respect to the big robot? Or I might want to know is, what's the relative pose from the hand of the big robot to the little robot, in case the big robot wanted to touch the little robot? Or, perhaps I might want to know, what's the pose of the hand of the big robot with respect to the world coordinate frame? I've introduced rather a lot of coordinate frames and relative poses. We can simplify this by introducing what I call a pose graph. So, I’ve kept the same colour coding and I kept the same labeling.\n\nNow, instead of representing a pose by a two-dimensional coordinate frame, I represent it simply by a disc. The lines that join the discs represent the relative poses and notice that they are lines with an arrow so this thing is a directed graph. The solid lines represent relative poses that we know and the dashed lines represent poses that we don't know, but perhaps we can compute them.\n\nSo, let's highlight this particular relative pose here and say I want to determine the value of this particular relative pose. How do I do that? Well, what we’re going to do is to follow a path through the graph that's got the same starting point and the same ending point. The first leg of the journey is this arrow here, and then I go from pose C to pose R and then go to pose H. We can say then that this relative pose is equal to the composition of these other 3 relative poses. If I join this, this, and this together it gets me to the same place in the pose graph as this. So, this is a way I can compute an unknown relative pose in terms of relative poses that I know.\n\nLet's have a look at a more complex example. Now, I want to know this relative pose where is the little robot with respect to the hand of the big robot. Same rule again. I want to find another path through this graph that takes me from H to L. The first leg of the journey is going to be this one. But in this case we need to go backwards along this arrow, so I've coloured it in red to indicate that I'm going backwards along this particular part of the graph. Then I'm going to go backwards along this part of the graph and then I'm going to go in the forward direction along this part of the graph. And now, when I write the pose expression, I've introduced two new symbols. It's a minor sign inside a circle and I'm going to refer to this operator as O-. So, for the case of the relative poses that are coded in red, where we went backwards along the arrow, I've prefixed that pose by a minus sign because I've gone in the reverse direction. Whereas this one here, I've used the O+ operator because I'm going in the forward direction along the arrow. So, here is a way that I can define an unknown pose in terms of previously known poses.\n\nWhat we got now is some kind of pose algebra and there are just a few simple rules with a pose algebra. The first rule is concerned with the O+ operator. I can combine two relative poses to get another relative pose but there are some really important rules associated with all the subscripts here. The first rule is that these two subscripts, the one on the either side of the O+ operator, must be the same. And you can think of the miscancelling out. Then, this leading subscript is the same on both sides of the expression and this subscript is the same on both sides of the expression. The O+ operator is not commutative and that means that if I have a relative pose 1, and an O+ with relative pose 2, that is, in general, different to relative pose 2 O+ relative pose 1. You have to be really careful about the order and you can arbitrarily switch them around. Another rule is that O- operator applied to the relative pose is the same as swapping the leading and trailing labels. Another rule is that I have pose O- the same pose, the result is zero. It corresponds to no motion, no pose change and we use symbol zero to represent that. And, another way I can express that is to negate the pose and then add it to the pose, the result is zero. These two expressions are equivalent. Finally, if I add the zero relative pose to a pose, it makes no change. And, if I subtract the zero relative pose from a pose, it makes no change.\n\nThese are just a few simple algebraic rules that we can apply to these complex pose expressions which allows us to rearrange them to determine the things that we want in terms of the things that we know. There are a few rules also that apply to this dot operator where we combine a pose with a vector. And again, we have some rules that apply to the labels here. And once again, the innermost labels must be the same. And then, we can think of them as cancelling out and the leading labels are the same.\n\nWe consider multiple objects each with its own coordinate frame. Now we can describe the relationships between the frames and find a vector describing a point with respect to any of these frames. We extend our algebraic notation to ease the manipulation of relative poses.\n\nTry your hand at some online MATLAB problems. You’ll need to watch all the 2D “Spatial Maths” lessons to complete the problem set.", null, "### Professor Peter Corke\n\nProfessor of Robotic Vision at QUT and Director of the Australian Centre for Robotic Vision (ACRV). Peter is also a Fellow of the IEEE, a senior Fellow of the Higher Education Academy, and on the editorial board of several robotics research journals.\n\n### Skill level\n\nThis content assumes an understanding of high school-level mathematics, e.g. trigonometry, algebra, calculus, physics (optics) and some knowledge/experience of programming (any language).\n\n### Rate this lesson\n\nAverage\n\n1.", null, "Sudeep Sapkota says:\n1.", null, "Peter Corke says:" ]

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[ "# How to find the asymptotes of h(x)=5^(x-2)?\n\n##### 1 Answer\nFeb 22, 2018\n\nTake the limits of $h \\left(x\\right)$ as x->+-∞ to determine the horizontal asymptotes.\n\n#### Explanation:\n\nTake the limits of $h \\left(x\\right)$ as x->+-∞ to determine the function's horizontal asymptotes. We have no vertical asymptotes, as there are no values of $x$ that will make this function undefined.\n\nLim_(x->∞)5^(x-2)=5^∞=∞ (No asymptotes as we approach +∞.\n\nLim_(x->-∞)5^(x-2)=5^-∞=1/5^∞=0\n\nSo, $y = 0$ is this function's only horizontal asymptote, as we approach $y = 0$ for smaller and smaller values of $x$ but never cross/touch it." ]

[ null ]

[ "", null, "ROOT   Reference Guide", null, "ROOT::Math::BrentRootFinder Class Reference\n\nClass for finding the root of a one dimensional function using the Brent algorithm.\n\nIt will use the Brent Method for finding function roots in a given interval. First, a grid search is used to bracket the root value with the a step size = (xmax-xmin)/npx. The step size can be controlled via the SetNpx() function. A default value of npx = 100 is used. The default value con be changed using the static method SetDefaultNpx. If the function is unimodal or if its extrema are far apart, setting the fNpx to a small value speeds the algorithm up many times. Then, Brent's method is applied on the bracketed interval. It will use the Brent Method for finding function roots in a given interval. If the Brent method fails to converge the bracketing is repeted on the latest best estimate of the interval. The procedure is repeted with a maximum value (default =10) which can be set for all BrentRootFinder classes with the method SetDefaultNSearch\n\nThis class is implemented from TF1::GetX() method.\n\nDefinition at line 62 of file BrentRootFinder.h.\n\n## Public Member Functions\n\nBrentRootFinder ()\nDefault Constructor. More...\n\nvirtual ~BrentRootFinder ()\nDefault Destructor. More...\n\nint Iterations () const\nReturn number of iteration used to find minimum. More...\n\nconst char * Name () const\nReturn name of root finder algorithm (\"BrentRootFinder\"). More...\n\ndouble Root () const\nReturns root value. More...\n\nvirtual bool SetFunction (const ROOT::Math::IGenFunction &, double, double)\nSet function to solve and the interval in where to look for the root. More...\n\nbool SetFunction (const ROOT::Math::IGenFunction &f, double xlow, double xup)\nSets the function for the rest of the algorithms. More...\n\nvirtual bool SetFunction (const ROOT::Math::IGradFunction &, double)\nSet function to solve and the interval in where to look for the root. More...\n\nvoid SetLogScan (bool on)\nSet a log grid scan (default is equidistant bins) will work only if xlow > 0. More...\n\nvoid SetNpx (int npx)\nSet the number of point used to bracket root using a grid. More...\n\nbool Solve (int maxIter=100, double absTol=1E-8, double relTol=1E-10)\nReturns the X value corresponding to the function value fy for (xmin<x<xmax). More...\n\nint Status () const\nReturns status of last estimate. More...", null, "Public Member Functions inherited from ROOT::Math::IRootFinderMethod\nIRootFinderMethod ()\nDefault Constructor. More...\n\nvirtual ~IRootFinderMethod ()\nDefault Destructor. More...\n\nvirtual int Iterate ()\nThis method is implemented only by the GSLRootFinder and GSLRootFinderDeriv classes and will return an error if it's not one of them. More...\n\nvirtual int Iterations () const\nReturn number of iterations used to find the root Must be implemented by derived classes. More...\n\nvirtual const char * Name () const =0\nReturn name of root finder algorithm. More...\n\nvirtual double Root () const =0\nReturns the previously calculated root. More...\n\nvirtual bool SetFunction (const ROOT::Math::IGenFunction &, double, double)\nSets the function for the rest of the algorithms. More...\n\nvirtual bool SetFunction (const ROOT::Math::IGradFunction &, double)\nSets the function for algorithms using derivatives. More...\n\nvirtual bool Solve (int maxIter=100, double absTol=1E-8, double relTol=1E-10)=0\nStimates the root for the function. More...\n\nvirtual int Status () const =0\nReturns the status of the previous estimate. More...\n\n## Static Public Member Functions\n\nstatic void SetDefaultNpx (int npx)\nset number of default Npx used at construction time (when SetNpx is not called) Default value is 100 More...\n\nstatic void SetDefaultNSearch (int n)\nset number of times the bracketing search in combination with is done to find a good interval Default value is 10 More...\n\n## Private Attributes\n\nconst IGenFunctionfFunction\n\nbool fLogScan\n\nint fNIter\n\nint fNpx\n\ndouble fRoot\n\nint fStatus\n\ndouble fXMax\n\ndouble fXMin\n\n#include <Math/BrentRootFinder.h>\n\nInheritance diagram for ROOT::Math::BrentRootFinder:\n[legend]\n\n## ◆ BrentRootFinder()\n\n ROOT::Math::BrentRootFinder::BrentRootFinder ( )\n\nDefault Constructor.\n\nDefinition at line 25 of file BrentRootFinder.cxx.\n\n## ◆ ~BrentRootFinder()\n\n virtual ROOT::Math::BrentRootFinder::~BrentRootFinder ( )\ninlinevirtual\n\nDefault Destructor.\n\nDefinition at line 71 of file BrentRootFinder.h.\n\n## ◆ Iterations()\n\n int ROOT::Math::BrentRootFinder::Iterations ( ) const\ninlinevirtual\n\nReturn number of iteration used to find minimum.\n\nReimplemented from ROOT::Math::IRootFinderMethod.\n\nDefinition at line 115 of file BrentRootFinder.h.\n\n## ◆ Name()\n\n const char * ROOT::Math::BrentRootFinder::Name ( ) const\nvirtual\n\nReturn name of root finder algorithm (\"BrentRootFinder\").\n\nImplements ROOT::Math::IRootFinderMethod.\n\nDefinition at line 59 of file BrentRootFinder.cxx.\n\n## ◆ Root()\n\n double ROOT::Math::BrentRootFinder::Root ( ) const\ninlinevirtual\n\nReturns root value.\n\nNeed to call first Solve().\n\nImplements ROOT::Math::IRootFinderMethod.\n\nDefinition at line 109 of file BrentRootFinder.h.\n\n## ◆ SetDefaultNpx()\n\n void ROOT::Math::BrentRootFinder::SetDefaultNpx ( int npx )\nstatic\n\nset number of default Npx used at construction time (when SetNpx is not called) Default value is 100\n\nDefinition at line 34 of file BrentRootFinder.cxx.\n\n## ◆ SetDefaultNSearch()\n\n void ROOT::Math::BrentRootFinder::SetDefaultNSearch ( int n )\nstatic\n\nset number of times the bracketing search in combination with is done to find a good interval Default value is 10\n\nDefinition at line 36 of file BrentRootFinder.cxx.\n\n## ◆ SetFunction() [1/3]\n\n virtual bool ROOT::Math::IRootFinderMethod::SetFunction ( const ROOT::Math::IGenFunction & , double , double )\ninlinevirtual\n\nSet function to solve and the interval in where to look for the root.\n\n@param f Function to be minimized. @param xlow Lower bound of the search interval. @param xup Upper bound of the search interval.\n\nReimplemented from ROOT::Math::IRootFinderMethod.\n\nDefinition at line 53 of file IRootFinderMethod.h.\n\n## ◆ SetFunction() [2/3]\n\n bool ROOT::Math::BrentRootFinder::SetFunction ( const ROOT::Math::IGenFunction & , double , double )\nvirtual\n\nSets the function for the rest of the algorithms.\n\nThe parameters set the interval where the root has to be calculated.\n\nReimplemented from ROOT::Math::IRootFinderMethod.\n\nDefinition at line 39 of file BrentRootFinder.cxx.\n\n## ◆ SetFunction() [3/3]\n\n virtual bool ROOT::Math::IRootFinderMethod::SetFunction ( const ROOT::Math::IGradFunction & , double )\ninlinevirtual\n\nSet function to solve and the interval in where to look for the root.\n\n@param f Function to be minimized. @param xlow Lower bound of the search interval. @param xup Upper bound of the search interval.\n\nReimplemented from ROOT::Math::IRootFinderMethod.\n\nDefinition at line 45 of file IRootFinderMethod.h.\n\n## ◆ SetLogScan()\n\n void ROOT::Math::BrentRootFinder::SetLogScan ( bool on )\ninline\n\nSet a log grid scan (default is equidistant bins) will work only if xlow > 0.\n\nDefinition at line 106 of file BrentRootFinder.h.\n\n## ◆ SetNpx()\n\n void ROOT::Math::BrentRootFinder::SetNpx ( int npx )\ninline\n\nSet the number of point used to bracket root using a grid.\n\nDefinition at line 100 of file BrentRootFinder.h.\n\n## ◆ Solve()\n\n bool ROOT::Math::BrentRootFinder::Solve ( int maxIter = 100, double absTol = 1E-8, double relTol = 1E-10 )\nvirtual\n\nReturns the X value corresponding to the function value fy for (xmin<x<xmax).\n\nMethod: First, the grid search is used to bracket the maximum with the step size = (xmax-xmin)/fNpx. This way, the step size can be controlled via the SetNpx() function. If the function is unimodal or if its extrema are far apart, setting the fNpx to a small value speeds the algorithm up many times. Then, Brent's method is applied on the bracketed interval.\n\n@param maxIter maximum number of iterations. @param absTol desired absolute error in the minimum position. @param absTol desired relative error in the minimum position.\n\nImplements ROOT::Math::IRootFinderMethod.\n\nDefinition at line 63 of file BrentRootFinder.cxx.\n\n## ◆ Status()\n\n int ROOT::Math::BrentRootFinder::Status ( ) const\ninlinevirtual\n\nReturns status of last estimate.\n\nIf = 0 is OK\n\nImplements ROOT::Math::IRootFinderMethod.\n\nDefinition at line 112 of file BrentRootFinder.h.\n\n## ◆ fFunction\n\n const IGenFunction* ROOT::Math::BrentRootFinder::fFunction\nprivate\n\nDefinition at line 135 of file BrentRootFinder.h.\n\n## ◆ fLogScan\n\n bool ROOT::Math::BrentRootFinder::fLogScan\nprivate\n\nDefinition at line 136 of file BrentRootFinder.h.\n\n## ◆ fNIter\n\n int ROOT::Math::BrentRootFinder::fNIter\nprivate\n\nDefinition at line 137 of file BrentRootFinder.h.\n\n## ◆ fNpx\n\n int ROOT::Math::BrentRootFinder::fNpx\nprivate\n\nDefinition at line 138 of file BrentRootFinder.h.\n\n## ◆ fRoot\n\n double ROOT::Math::BrentRootFinder::fRoot\nprivate\n\nDefinition at line 142 of file BrentRootFinder.h.\n\n## ◆ fStatus\n\n int ROOT::Math::BrentRootFinder::fStatus\nprivate\n\nDefinition at line 139 of file BrentRootFinder.h.\n\n## ◆ fXMax\n\n double ROOT::Math::BrentRootFinder::fXMax\nprivate\n\nDefinition at line 141 of file BrentRootFinder.h.\n\n## ◆ fXMin\n\n double ROOT::Math::BrentRootFinder::fXMin\nprivate\n\nDefinition at line 140 of file BrentRootFinder.h.\n\nLibraries for ROOT::Math::BrentRootFinder:", null, "[legend]\n\nThe documentation for this class was generated from the following files:" ]

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[ "# Estimating average effects in regression discontinuities with covariate interactions\n\nRegression discontinuity designs (RDDs) are now a widely used tool for program evaluation in economics and many other fields. RDDs occur in situations where some treatment/program of interest is assigned on the basis of a numerical score (called the running variable), all units scoring above a certain threshold receiving treatment and all units scoring at or below the threshold having treatment withheld (or vice versa, with treatment assigned to units scoring below the threshold). This mechanism provides a way to identify the marginal average treatment effect (MATE): the average effect of treatment assignment for units on the cusp of the threshold.\n\nRDDs are appealing for a couple of reasons. First and foremost, RDD-like mechanism occurs all over the place, since providing treatment on the basis of a numerical measure of need/eligibility is a natural way to allocate resources. Furthermore, analysis of the designs is straight-forward, as it involves nothing more complicated than a linear regression model, estimated using (weighted or un-weighted) least squares, and which can be represented graphically using a simple scatterplot. Things get a little bit more complicated if you are trying to account for imperfect compliance with treatment assignment—as in the “fuzzy” RDD—but for the moment let me focus on “sharp” RDDs.\n\nThe simplest approach to estimating the MATE is to use a local linear regression in the neighborhood of the threshold, with the outcome regressed on the running variable, treatment indicator, and their interaction. However, in practice it is quite common to also include additional covariates in the local linear regression. If the covariates are also interacted with the treatment indicator, there is no longer a single regression coefficient corresponding to the treatment effect. In my last post, I suggested a “centering trick” for estimating the MATE based on a model that included covariate-by-treatment interactions. In this post, I’ll explain the reasoning behind this proposal.\n\n### G’day, MATE\n\nI think it’s helpful to start by thinking about the definition of the MATE in non-parametric terms. Let $$R$$ be the running variable, assumed to be centered at the threshold; $$T$$ be an indicator for treatment assignment, with $$T = I(R > 0)$$; and $$X$$ be a covariate, which may be vector-valued. Denote the potential outcomes as $$Y^0$$ (a unit’s outcome if not assigned to treatment) and $$Y^1$$ (a unit’s outcome if assigned to treatment), so that the observed outcome is $$Y = Y^0 (1 - T) + Y^1 T$$. Now consider the potential response surfaces\n\n\\begin{aligned}\\mu_0(x, r) &= \\text{E}\\left(\\left.Y^0 \\right|X = x, R = r\\right) \\\\ \\mu_1(x, r) &= \\text{E}\\left(\\left.Y^1 \\right|X = x, R = r\\right).\\end{aligned}\n\nIn an RDD, the average treatment effect at a given point $$(x, r)$$ on the response surface is not generally identified by conditioning because one of the potential outcomes will never be observed: if $$r < 0$$ then $$\\text{Pr}( T = 0 \\vert X = x, R = r) = 1$$ and $$\\text{Pr}( T = 1 \\vert X = x, R = r) = 0$$ (and vice versa for $$r > 0$$). However, the treatment effect for the subpopulation where $$R = 0$$ can be identified under the assumption that the potential response surfaces are continuous in a neighborhood of the threshold. Thus the MATE, which can be written as\n\n\\begin{aligned} \\delta_M &= \\text{E}\\left(\\left. Y^1 - Y^0 \\right| R = 0\\right) \\\\ &= \\text{E}\\left[\\mu_1(X, 0) - \\mu_0(X,0)\\right]. \\end{aligned}\n\n### Regression estimation\n\nNow assume that we have a simple random sample $$\\left(y_i,r_i,t_i, x_i\\right)_{i=1}^n$$ of units and that each unit has a weight $$w_i$$ defined based on some measure of distance from the threshold. We can use these data to estimate the response surfaces (somehow…more on that in a minute) on each side of the cut-off, with $$\\hat\\mu_0(x, r)$$ for $$r < 0$$ and $$\\hat\\mu_1(x, r)$$ for $$r > 0$$. If we then use the sample distribution of $$X$$ in the neighborhood of $$R = 0$$ in place of the conditional density $$d\\left(X = x \\vert R = 0\\right)$$, we can estimate the MATE as\n\n$\\hat\\delta_M = \\frac{1}{W} \\sum_{i=1}^n w_i \\left[\\hat\\mu_1(x_i, 0) - \\hat\\mu_0(x_i, 0)\\right],$\n\nwhere $$W = \\sum_{i=1}^n w_i$$. This is a regression estimator for $$\\delta_M$$. It could be non-, semi-, or fully parametric depending on the technique used to estimate the response surfaces. Note that this estimator is a little bit different than the regression estimator that would be used in the context of an observational study (see, e.g., Shafer & Kang, 2008). In that context, one would use $$\\hat\\mu_j(x_i, r_i)$$ rather than $$\\hat\\mu_j(x_i, 0)$$, but in an RDD doing so would involve extrapolating beyond the cutpoint (i.e., using $$\\hat\\mu_1(x_i, r_i)$$ for $$r_i < 0$$).\n\nNow suppose that we again use a linear regression in some neighborhood of the cut-point to estimate the response surfaces. For the (weighted) sample in the neighborhood of the cut-point, we assume that\n\n$\\mu_{t_i}(x_i, r_i) = \\beta_0 + \\beta_1 r_i + \\beta_2 t_i + \\beta_3 r_i t_i + \\beta_4 x_i + \\beta_5 x_i t_i.$\n\nSubstituting this into the formula for $$\\hat\\delta_M$$ leads to\n\n\\begin{aligned}\\hat\\delta_M &= \\frac{1}{W} \\sum_{i=1}^n w_i \\left[\\hat\\beta_2 + \\hat\\beta_5 x_i \\right] \\\\ &= \\hat\\beta_2 + \\hat\\beta_5 \\sum_{i=1}^n \\frac{w_i x_i}{W}.\\end{aligned}\n\nNow, the centering trick involves nothing more than re-centering the covariate so that $$\\sum_{i=1}^n w_i x_i = 0$$ and $$\\hat\\delta_M = \\hat\\beta_2$$. Of course, one could just use the non-parametric form of the regression estimator, but the centering trick is useful because it comes along with an easy-to-calculate standard error (since it is just a regression coefficient estimate).\n\n### Multiple covariates\n\nAll of this works out in the exact same way if you have interactions between the treatment and multiple covariates. However, there are a few tricky cases that are worth noting. If you include interactions between the treatment indicator and a polynomial function of the treatment, each term of the polynomial has to be centered. For example, if you want to control for $$x$$, $$x^2$$, and their interactions with treatment, you will need to calculate\n\n$\\tilde{x}_{1i} = x_i - \\frac{1}{W} \\sum_{i=1}^n w_i x_i, \\qquad \\tilde{x}_{2i} = x_i^2 - \\frac{1}{W} \\sum_{i=1}^n w_i x_i^2$\n\nand then use these re-centered covariates in the regression\n\n$\\mu_{t_i}(x_i, r_i) = \\beta_0 + \\beta_1 r_i + \\beta_2 t_i + \\beta_3 r_i t_i + \\beta_4 \\tilde{x}_{1i} + \\beta_5 \\tilde{x}_{2i} + \\beta_6 \\tilde{x}_{1i} t_i + \\beta_7 \\tilde{x}_{2i} t_i.$\n\nThe same principle will also hold if you want to include higher-order interactions between covariates and the treatment: calculate the interaction term first, then re-center it. There’s one exception though. If you want to include an interaction between a covariate $$x$$, the running variable, and the treatment indicator (who knows…you might aspire to do this some day…), then all you need to do is center $$x$$. In particular, you should not calculate the interaction $$x_i r_i$$ and then re-center it (doing so could pull the average away from the threshold of $$R = 0$$).\n\n### R, MATEs!\n\nHere’s some R code that implements the centering trick for the simulated example from my last post:\n\nlibrary(sandwich)\nlibrary(lmtest)\nlibrary(rdd)\n\n# simulate an RDD\nset.seed(20160124)\nsimulate_RDD <- function(n = 2000, R = rnorm(n, mean = qnorm(.2))) {\nn <- length(R)\nT <- as.integer(R > 0)\nX1 <- 10 + 0.6 * (R - qnorm(.2)) + rnorm(n, sd = sqrt(1 - 0.6^2))\nX2 <- sample(LETTERS[1:4], n, replace = TRUE, prob = c(0.2, 0.3, 0.35, 0.15))\nY0 <- 0.4 * R + 0.1 * (X1 - 10) + c(A = 0, B = 0.30, C = 0.40, D = 0.55)[X2] + rnorm(n, sd = 0.9)\nY1 <- 0.35 + 0.3 * R + 0.18 * (X1 - 10) + c(A = -0.50, B = 0.30, C = 0.20, D = 0.60)[X2] + rnorm(n, sd = 0.9)\nY <- (1 - T) * Y0 + T * Y1\ndata.frame(R, T, X1, X2, Y0, Y1, Y)\n}\nRD_data <- simulate_RDD(n = 2000)\n\n# calculate kernel weights\nbw <- with(RD_data, IKbandwidth(R, Y, cutpoint = 0))\nRD_data$w <- kernelwts(RD_data$R, center = 0, bw = bw)\n\n# center the covariates\nX_mat <- model.matrix(~ 0 + X2 + X1, data = RD_data)\nX_cent <- as.data.frame(apply(X_mat, 2, function(x) x - weighted.mean(x, w = RD_data\\$w)))\nRD_data_aug <- cbind(X_cent, subset(RD_data, select = c(-X1, -X2)))\ncov_names <- paste(names(X_cent)[-1], collapse = \" + \")\n\n# calculate the MATE using RDestimate\nRD_form <- paste(\"Y ~ R |\", cov_names)\nsummary(RDestimate(as.formula(RD_form), data = RD_data_aug))\n##\n## Call:\n## RDestimate(formula = as.formula(RD_form), data = RD_data_aug)\n##\n## Type:\n## sharp\n##\n## Estimates:\n## Bandwidth Observations Estimate Std. Error z value Pr(>|z|)\n## LATE 1.0894 1177 0.2981 0.10659 2.797 0.0051559\n## Half-BW 0.5447 611 0.2117 0.14846 1.426 0.1539482\n## Double-BW 2.1787 1832 0.2734 0.08305 3.292 0.0009949\n##\n## LATE **\n## Half-BW\n## Double-BW ***\n## ---\n## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1\n##\n## F-statistics:\n## F Num. DoF Denom. DoF p\n## LATE 23.30 11 1165 0\n## Half-BW 10.97 11 599 0\n## Double-BW 47.41 11 1820 0\n# or using lm\nlm_form <- paste(\"Y ~ R + T + R:T + T*(\", cov_names,\")\")\nlm_fit <- lm(as.formula(lm_form), weights = w, data = subset(RD_data_aug, w > 0))\ncoeftest(lm_fit, vcov. = vcovHC(lm_fit, type = \"HC1\"))[\"T\",]\n## Estimate Std. Error t value Pr(>|t|)\n## 0.298142798 0.106588790 2.797130893 0.005240719\n\n1. My regression estimator uses the sample distribution of $$X$$ in the neighborhood of the threshold as an estimate of $$d(X = x \\vert R = 0)$$. This seems reasonable, but I wonder whether there might be a better approach to estimating this conditional density." ]

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[ "Search a number\nBaseRepresentation\nbin10010011100\n31121201\n4102130\n514210\n65244\n73304\noct2234\n91551\n101180\n11983\n12824\n136ca\n14604\n1553a\nhex49c\n\n1180 has 12 divisors (see below), whose sum is σ = 2520. Its totient is φ = 464.\n\nThe previous prime is 1171. The next prime is 1181. The reversal of 1180 is 811.\n\nAdding to 1180 its reverse (811), we get a palindrome (1991).\n\nIt can be divided in two parts, 11 and 80, that added together give a triangular number (91 = T13).\n\n1180 is nontrivially palindromic in base 9.\n\nIt is a Harshad number since it is a multiple of its sum of digits (10).\n\nIt is a plaindrome in base 8 and base 16.\n\nIt is a nialpdrome in base 11.\n\nIt is a congruent number.\n\nIt is not an unprimeable number, because it can be changed into a prime (1181) by changing a digit.\n\n1180 is an untouchable number, because it is not equal to the sum of proper divisors of any number.\n\nIt is a pernicious number, because its binary representation contains a prime number (5) of ones.\n\nIt is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 10 + ... + 49.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (210).\n\n1180 is a gapful number since it is divisible by the number (10) formed by its first and last digit.\n\nIt is an amenable number.\n\n1180 is an abundant number, since it is smaller than the sum of its proper divisors (1340).\n\nIt is a pseudoperfect number, because it is the sum of a subset of its proper divisors.\n\nIt is a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (1260).\n\n1180 is a wasteful number, since it uses less digits than its factorization.\n\n1180 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 68 (or 66 counting only the distinct ones).\n\nThe product of its (nonzero) digits is 8, while the sum is 10.\n\nThe square root of 1180 is about 34.3511280746. The cubic root of 1180 is about 10.5672180526.\n\nThe spelling of 1180 in words is \"one thousand, one hundred eighty\".\n\nDivisors: 1 2 4 5 10 20 59 118 236 295 590 1180" ]

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[ "## RENT A THINKER Home My Page Chat Tik-Tok 1 Single Layer Neural Network Solution for XOR Problem\n\n### 1.1 ABSTRACT\n\nSingle layer feed forward type networks are used for linear decision boundary . To get solution of non-linear boundary, at least two layer networks are required . In this paper, single layer topology is developed with appropriate learning algorithm to solve non-linear problem like XOR or circular decision boundary problems. Method used in this new topology is multivalued neuron activation function.\n\n### 1.2 INTRODUCTION\n\nFrank Rossenbatt invented the first meaningful adaptive architecture, i.e. “the Perceptron” in 1957 . The perceptron was actually an entire class of architectures, composed of processing units that transmitted signals and adapted their interconnection weights. It was oriented towards modeling the brain in an attempt to understand memory, learning, and cognitive processes. But the crucial limitation of the perceptron is that “it does not allow more than one layer of adaptive weights”. Because there was no way to propagating the weights correction through a multi-layer network in order to make such a network learn. Afterwards, this limitation was overcome in 1986, by Rumalhart and Mcclelland by presenting “ Back error Propagation Algorithm “. The main functional limitation of the perceptron is that an output unit can classify only “ linearly separable patterns”. The XOR function is classic example of a pattern classification problem that is not linearly separable. To solve XOR problem, a back error-propagating network is trained. The perceptron model is unable to solve XOR problem with a single output unit because the function is not linearly separable and its solution requires at least two layers network.\n\nHere in the paper, attempt has been made to get solution for XOR problem using single layer neural network with a multivalued neuron activation function – Zo = f (abs (Zi) + K). Choice of similar activation functions with appropriate learning algorithm for other non-linear problem minimizes the network topology drastically resulting increasing the speed.\n\n1.3 LEARNING ALGORITHM\n\nIn this section, the application of more complex activation function is shown to solve non-linear boundary problem using single layer network.  The attempt also has been made to generalize the learning algorithm for such application.\n\nLet us consider the below figure-1. with the replaced  activation function as-", null, "Figure–1 The proposed network to solve the  XOR problem\n\nZo = ¦ ( abs ( Zi  ) + K )        (1)\nwhere K is a constant\n\nHere ¦(x) is a threshold function of the type - if ( f(x) > 0 ) Zo = 1; else Zo =0;\n\nWith this activation function we need to modify the training algorithm to reinforce weights. Here, to calculate the error sensitivity of weights, we consider the activation function as,\n\nConsider Zi = wx * xo + wy * yo + wb     (2)\n\nOutput  Zo = Zi 2    + K\n\nError function  E = ½ *( Zo - Zd )2\n\nThe error sensitivity of wx is given by -\n\nd E / d wx  =  2  * (Zo-Zi)*  Zi  * Xo             (3)\n\nFrom the above derivation, reinforcement of all the weights are -\n\nwx =  wx  -  h * ( Zo-Zd) * Zi  * Xo,\n\nSimilarly,\n\nwy =  wy  -  h * ( Zo-Zd) * Zi  * Yo\n\nwb =  wb  -  h * ( Zo-Zd) * Zi                       (4)\n\nwhere  h is the learning coefficient and is generally varies between 0 to 1\n\n### 1.4  RESUTS\n\nThe table - 1 shows the result of 10 experiments with different  initial weights to solve XOR problem. In this case the constant K = 10 and minimum iterations = 30.  In all cases, the error becomes zero.\n\n Experiment Type  - XOR,  K =  10  and    eta = 0.4 Exp No. wx wy Wb Error /  Iteration 1 47107.027344 -39776.375000 3773.969482 0 / 06 2 -46319.484375 39967.570312 6637.256348 0 / 11 3 28691.703125 -44370.238281 8611.723633 0 / 07 4 46208.574219 -41672.371094 6987.809082 0 / 10 5 32130.199219 -58266.539062 11717.017578 0 / 06 6 -53111.601562 31523.916016 16880.552734 0 / 10 7 -27263.140625 -17786.228516 -27561.998047 0 / 07 8 36554.726562 -35102.531250 -1199.843628 0 / 06 9 -29898.927734 52755.171875 -9632.140625 0 / 09 10 40373.308594 -49587.156250 7200.447266 0 / 08 Table - 3    The weights and output error for 10  experiments for “ XOR “ Problem\n\n### 1.5  CONCLUSION\n\nIn this section, it is observed that non-linear problems can not be solved using single layer network with conventional type of neuron activation function. It is shown that certain class of non-linear problems could be solved using more complex neurons activation function much smaller having less complex network. Using single layer network. Also it should be noted that the forward characteristics of the network could be simplified without affecting the training performance by using different activation function for forward and reverse characteristics.\n\nIt is important to note that the backward activation function need not be the gradient of forward activation function. e.g. we have chosen Zo= f(abs(Zi) + K) as forward activation function where as for  the backward error propagation, Zo=Zi2 + K is used. This increases the speed of forward computation. In general, the forward and the reverse activation functions should have similar shape.\n\n### 1.6       LISTING\n\n//Single Layer Neural Network Solution for XOR Problem\n//By     Himanshu Mazumdar    and   Dr. Leena H. Bhatt\n//Physical Research Laboratory, Navrangpura, Ahmedabad - 380 009\n//Configured with 2 inputs, 1 output   and 3 connections\n//Date: - 23-February-2000.\n\n#include \"stdio.h\"\n#include \"math.h\"\n#include \"graphics.h\"\n#include \"conio.h\"\n#include \"stdlib.h\"\n#include \"time.h\"\n\n#define OR 1;\n#define AND 2;\n#define XOR 3;\n\nint x1,y1,x2,y2;\nfloat wx,wy,wb;\nfloat m10,m12;\nfloat eta=0.4;\nunsigned int tx,ty,td;\n/*_________________________________*/\nint update(int x0, int y0, float *zi)\n{\n\nint zo;\n*zi=wx*x0+wy*y0+wb;\n\nzo=fabs(*zi)+10;\nif (zo>0) zo=0; else zo=1;\n\nreturn(zo);\n}\n\n/*_________________________________*/\nint reinforce(int x0, int y0, int zd)\n{\n\nint err,zo;\nfloat zi;\n\nzo=update(x0,y0,&zi);\nerr=zo-zd;\n\nwx=wx-eta*err*zi*x0;\nwy=wy-eta*err*zi*y0;\nwb=wb-eta*err*zi;\nreturn(err);\n\n}\n\n/*_________________________________*/\n\nint train(int nmax)\n{\n\nint n,e,err=0;\nint x0,y0,zd;\nstatic int m=0;\n\nfor(n=0;n<nmax;n++){\nx0=tx[m];\ny0=ty[m];\n\nzd=td[m];\n\ne=reinforce(x0,y0,zd);\n\nerr=err+abs(e);\n\nm++;\n\nm=m% 5000;\n\n}\nreturn(err);\n\n}\n\n/*_________________________________*/\nvoid desiredout(int smax)\n{\n\nint n,x0,y0,zd;\n\nfor(n=0;n<smax;n++){\n\nx0=random(2);\n\ny0=random(2);\n\n/* XOR */\n\nzd=x0 ^ y0;\n\ntx[n]=x0;\nty[n]=y0;\ntd[n]=zd;\n}\n}\n/*_________________________________*/\nvoid initweights (void)\n{\n\nwx=1.0-random(1000)/500.0;\n\nwy=1.0-random(1000)/500.0;\nwb=1.0-random(1000)/500.0;\n\n}\n/*_________________________________*/\nvoid main(void)\n{\n\nint m,r,err;\n\nclrscr();\n\nprintf(\"\\n\\n\\n\");\n\nprintf(\"*********************************************\\n\");\n\nprintf(\"*     Single Layer  Neural Network          *\\n\");\n\nprintf(\"*      Solution for XOR Problem             *\\n\");\n\nprintf(\"*                  by                       *\\n\");\n\nprintf(\"*         Himanshu Mazumdar                 *\\n\");\n\nprintf(\"*                 and                       *\\n\");\n\nprintf(\"*           Dr. Lina  Rawal                              *\\n\");\n\nprintf(\"*     Physical Research Laboratory,         * \\n\");\n\nprintf(\"*     Navrangpura, Ahmedabad - 380 009      * \\n\");\n\nprintf(\"*     Configured with 2 inputs, 1 output    * \\n\");\n\nprintf(\"*     and 3 connections                     * \\n\");\n\nprintf(\"*     Date:- 3-April-1996                   * \\n\");\n\nprintf(\"*********************************************\\n\");\n\ngetch();\n\nrandomize();\n\nclrscr();\n\nprintf(\"\\n       Single layer solution for XOR  problem  \");\n\nprintf(\"\\n       --------------------------------------  \");\n\nprintf(\"\\t\\n\\nExp no.     wx                            wy                   wb        \");\n\nprintf(\"\\n \");\n\ninitweights();\n\ndesiredout(5000);\n\nfor(r=1;r<=10;r++)\n{\nfor(m=1;m<=700;m++){\nerr=train(16);\n\n//printf(\"\\n%d Error=%d\",m,err);\n\nif(err==0)break;\n\n//getch();\n\n}\n\nprintf(\"\\t\\t\\t\\n%d \\t %f  %f   %f\",r,wx,wy,wb);\n\ninitweights();\n\n}\n\ngetch();\n\nprintf(\"\\n\\nThe weights and output error for XOR problem\" );\n\ngetch();\n\n}\n/*_________________________________*/\n\n### 1.7 REFERENCES\n\n Judith E. Dayhoff, “ Neural Network architectures ”, An introduction, ISBN-0-442-20 744-1.\n Rumalhert D. E, and L. McClelland, 1986,” Parallel Distributing Processing, vol.1 and\n2,Cambrige, Mass: MIT press.\n Himanshu Mazumdar and Leena Rawal, “ A Neural Network Tool Box Using C++.”  CSI (Computer Society of India) Communications, April (22 - 27), 1997, INDIA\n\nPublished:\nMazumdar Himanshu. S., and Leena P.   \"Single Layer Neural Network Solution for XOR Problem\",   CSI  Communications, ISSN 0970 – 647X,  pp. 13-15, October, 2000." ]

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[ "# How Much Horse Power In A Person?\n\n## Is 1 HP equal to a horse?\n\nCalculating the Power of a Single Horse Each horse pushed with a force that Watt estimated at 180 pounds. From this, Watt calculated that one horsepower was equivalent to one horse doing 33,000 foot-pounds of work in one minute. That amount of work equals one horsepower.\n\n## Why is horse 15 horsepower?\n\nWhile it is true that the maximum output of a horse is around 15 horsepower, when you average the output of a horse over the course of a work day it ends up being around a horsepower. Watt defined this amount as “the amount of work required from a horse to pull 150 pounds out of a hole that was 220 feet deep”.\n\n## How is human horse power calculated?\n\nThe equation to calculate horsepower is simple: Horsepower = Torque x RPM / 5,252.\n\n## How much horsepower is considered fast?\n\nSome of the most powerful and fastest cars in America start from 650-horsepower and above, and one could definitely consider that to be fast. Topping the ranks of fast cars in America include the following: 2020 Chevrolet Camaro ZL1: 650-horsepower. 2021 Dodge Charger and Challenger Hellcat Widebody: 717-horsepower.\n\nYou might be interested:  Often asked: Black Desert How To Give Horse Items?\n\n## How many cc is a horsepower?\n\ncc or cu.cm metric conversions. Many people have asked for a relationship between horsepower and cc or how many cc in a hp. The short answer is about 14 to 17cc = 1 hp or about 1 cu.in. = 1 bhp for an ordinary, basic car.\n\n## How much stronger is a horse than a human?\n\nHorses are strong biters. Horses can have a bite force anywhere up to 500 psi. So you have a better idea, humans have a bite force of 200 psi, which means that horses have up to two and a half times more strength in their bite force than we do.\n\n## How many horsepower is a draft horse?\n\nHow much horsepower does a draft horse have? Draft horses exhibit an average of 15 HP. Draft horses typically perform labor intensive tasks like lifting heavy weights. They can easily sustain this amount of power for a longer periods.\n\n## How much HP does a F1 car have?\n\nThe total power of a F1 engine is measured after calculating the power in the V6 engines, and Energy Recovery System (ERS). Considering the development of engine by the aforementioned engine suppliers, it is considered that the current F1 cars carry more than the magic number of 1000 HP.\n\n## What horsepower is 5252?\n\nOne horsepower is equal to 33,000 foot-pounds of work per minute. Add in the equations relating to torque and velocity, and you’ll find that horsepower always equals torque multiplied by rpm, divided by 5,252. Canceling out the equal variables, you wind up with horsepower equaling torque at 5,252 rpm.\n\n## What is the average horsepower of a car?\n\nHorsepower varies greatly from car to car and from country to country. That said, with mainstream American cars, the average vehicle can usually be expected to be 180-200 horsepower. Large SUVs often have engines that exceed 300 horsepower, while small car engines can only have about 100 horsepower.\n\nYou might be interested:  Question: What Is The Name Of The Red Raider Horse At Texas Tech?\n\n## How fast is 7 hp in mph?\n\nHow much is 212cc in mph? This engine is a performer right out of the box, and should get your go kart running anywhere from 20 mph all the way up to 35mph in its stock form.\n\n## Is 700 horsepower a lot?\n\nBefore it was a rarity and luxury to buy a car that was 200-500 horsepower, now it’s getting pretty common. In fact, 700 horsepower is the new 500 horsepower, and every performance car worth its weight in carbon-fiber diffusers is packing either a now-standard supercharger or a turbo.\n\n## Is 300 horsepower a lot?\n\n300hp, or horsepower, is not considerably significant. However, It’s plenty of power for most applications! For a sports car, 300hp is a great benchmark, showing that a car will be lots of fun. This means the engine is serious about performance, and likely consumes lots of fuel." ]

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[ "# [Tex/LaTex] Change Geometry after first page *without* knowing where the second page begins\n\ngeometry\n\nI have this for my first page:\n\n% Custom margins - FIRST PAGE\n\\newcommand{\\lrmargin}{1in}\n\\usepackage[top=0.5in,\nbottom=1in,\nleft=\\lrmargin,\nright=\\lrmargin,\nheightrounded]{geometry}\n\n\nAnd I have a bunch of text. Somewhere in there, Latex creates a second page of text for me. If I then go exactly where that is, and paste the following:\n\n% CUSTOM MARGINS - SECOND PAGE\n\\renewcommand{\\tbmargin}{1in}\n\\renewcommand{\\lrmargin}{1in}\n\\newgeometry{top=\\tbmargin,\nbottom=\\tbmargin,\nleft=\\lrmargin,\nright=\\lrmargin}\n\n\nI get the expected behavior. Now, if I change the margins of the first page again, that would necessarily change what text would appear on the second page and that messes up my changing the geometry of the second page. Instead what I get is two pages with the \"FIRST PAGE\" settings, and the second page has a huge amount of blank space and a third page with the correct settings.\n\nI know that to use a different geometry on another page, you use the \\newgeometry command. My question is – where do I place that in my latex code so that it always effects the second (and third etc.) pages without letting the first call to geometry affect that?\n\n# Caveat: Use at Your Own Risk!\n\nAs D.C. says (and as it is well known), it is basically impossible to have the text width change at a page boundary in an “automatic” fashion, that is, whatever the text that spans the two pages is. This is because TeX chooses the line breaks within each paragraph before it knows where it is going to cut off each page, and doesn’t change them any more afterwords. In principle, it is possible to prohibit page breaks within paragraphs by setting the \\interlinepenalty parameter to a value ≥ 10000, but in practice this is only feasible if you have very short paragraphs that never exceed two or three lines in length. Note, however, that this could be the case inside a table of contents or similar index; another situation in which this might be true is inside at itemize or enumerate environment.\n\nHowever, if you do not need to change the line width, but only the margins, the following code shows a crude hack by means of which this could be achieved. Because of the \\globaldefs=1 it contains, I explicitly decline any warranty that this methods will not break anything: use it solely at your own risk!\n\n% My standard header for TeX.SX answers:\n\\documentclass[a4paper]{article} % To avoid confusion, let us explicitly\n% declare the paper format.\n\n\\usepackage[T1]{fontenc} % Not always necessary, but recommended.\n% End of standard header. What follows pertains to the problem at hand.\n\n\\usepackage[showframe]{geometry}\n\\usepackage{afterpage}\n\\usepackage{lipsum}\n\n\\geometry{left=3cm,textwidth=12cm,top=2cm,bottom=2cm,heightrounded}\n\n\\begin{document}\n\n\\newgeometry{left=5cm,textwidth=12cm,top=5cm,bottom=7cm,heightrounded}\n\nThis is a test.\n\n\\afterpage{\\globaldefs=1 \\restoregeometry}\n\nNow $$\\verb|\\globaldefs|=\\texttt{\\number\\globaldefs}$$.\n\n\\lipsum[1-16]\n\n\\end{document}" ]

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[ "# Python vector autoregression pdf\n\nI am trying to implement multidimensional granger causality in python. Vector autoregression movingaverage varma the vector autoregression movingaverage varma method models the next step in each time series using an arma model. The pioneering work of sims 1980 proposed to replace the largescale macroeconomic models popular in the 1960s with vars, and suggested that bayesian methods could have improved upon frequentist ones in estim. It contains all the information that we have on the parameter vector \\\\theta\\ after having updated our prior views by looking at the data. Vector autoregressive models vector autoregressive var models a pth order vector autoregression, or varp, with exogenous variables x can be written as. This article develops a new bayesian poisson vector autoregression model that can. Vector autoregression and vector errorcorrection models. Click the link to check out the first post which focused on stationarity versus nonstationarity, and to find a list of other topics covered. Bayesian dynamic linear models dlms, arch garch volatility models and beyond mckinney, perktold, seabold statsmodels python time series analysis scipy conference 2011 4 29. Factor models and structural vector autoregressions in. One way of estimating relationships between the time series and their lagged values is the vector autoregression process. In this tutorial, you will discover how to implement an autoregressive model for time series. Autoregression models for time series forecasting with python.\n\nTime series decomposition and price forecasting using a vector autoregression var model welcome to the tech rando blog. Vector autoregression var models widely used model for modeling multiple kvariate time series, especially in macroeconomics. See bayesian inference and classical inference sections of the documentation for the full list of inference options. Browse other questions tagged python vector markov or ask your own question. Univariate and multivariate arimaversus vector autoregression. Jun 08, 2016 markov switching autoregression models. Consider a network of firms, connected by their shareholder relationships. Additionally, i establish a simulation scheme to compare the relative e ciency of impulse responses generated from machine learning and ordinary least squares vector autoregressions. Vector autoregression overview and proposals fylux. All variables in a var enter the model in the same way.\n\nMultivariate count models are rare in political science despite the presence of many count time series. The vector autoregression var model is one of the most successful. This model is a theoryfree method, can only be used to make predictions about the future, and is not able to explain the dependencies between the time series. Implementation within r package vars bernhard pfa kronberg im taunus abstract the structure of the package vars and its implementation of vector autoregressive.\n\nTime series analysis in python a comprehensive guide. Thus, the posterior pdf is the basis for estimation and inference as it summarises our beliefs about \\\\theta\\, given our prior belief and the results of the likelihood function. Vector autoregression model is a generalization of the ar model from the last paragraph, that can be applied to multivariate time series and catch dependencies between individual variables. A popular model for these data is the vector autoregressive var model, in which each variable is predicted as a linear function of all variables at previous time points. A tutorial on estimating timevarying vector autoregressive. Time series analysis in python a comprehensive guide with. It is a very simple idea that can result in accurate forecasts on a range of time series problems. Techniques of forecasting using vector autoregressions author. Vector autoregressive models umberto triacca dipartimento di ingegneria e scienze dellinformazione e matematica universit a dellaquila. Nov 14, 2017 a popular model for these data is the vector autoregressive var model, in which each variable is predicted as a linear function of all variables at previous time points.\n\nThis guide walks you through the process of analyzing the characteristics of a given time series in python. Jan 15, 2016 for the love of physics walter lewin may 16, 2011 duration. Time series is a sequence of observations recorded at regular time intervals. Forecasting time series data using autoregression python. Estimating time series models by state space methods in python. Techniques of forecasting using vector autoregressions. Although gvar is not the first large global macroeconomic model of the world. Bayesian vector autoregressions silvia mirandaagrippino bank of england and cfm giovanni riccoy university of warwick and ofce sciencespo this version. Related class of models vecm for modeling nonstationary including cointegrated processes.\n\nAnalyzing electricity price time series data using python. Markov switching autoregression models this notebook provides an example of the use of markov switching models in statsmodels to replicate a number of results presented in kim and nelson 1999. For that matter i am using vector autoregression from statsmodels, but when i try to get coeffcients out of it, it returns me an. Autoregression ar the autoregression ar method models as a linear function of the observations at prior time steps. This post presents the basic concept of var analysis and guides through the estimation procedure of a simple model. Y t must be a stationary process sometimes achieved by di erencing. Stock department of economics, harvard university and the national bureau of economic research and mark w. Autoregression is a time series model that uses observations from previous time steps as input to a regression equation to predict the value at the next time step. Vector autoregressive models for multivariate time series 11. A var model is a generalisation of the univariate autoregressive model for forecasting a vector of time series. Index termstime series analysis, statistics, econometrics, ar, arma, var, glsar. This post presents the basic concept of var analysis and guides through the estimation procedure of.\n\nThe focus is less on the math behind the method and more on its application in r using the vars package. Vector autoregressions and cointegration 2847 the first two columns of ix are the balanced growth restrictions, the third column is the real wage average labor productivity restriction, the fourth column is stable longrun money demand restriction, and the last column restricts nominal interest. Confused about autoregressive ar1 process is using swiss francs chf cheaper than euros eur in. Structural vector autoregressions svars are a multivariate, linear representation of a vector of observables on its own lags. The above is not nearly enough statistical background to truly understand linear and autoregression models, but i hope it gets you some basic understanding of how the two approaches differ. He called for alternative approaches to parameter reduction. Sims 1980 as a technique that could be used by macroeconomists to characterize the joint dynamic behavior of a collection of variables without requiring strong restrictions of the kind needed to identify underlying structural parameters. In this work, we extend this to a network quantile autoregression nqar model in order to study conditional quantiles in complex financial networks. Introduction the main purpose of this research is to compare forecasts from three popular time series methods of forecasting. A var is an nequation, nvariable linear model in which each variable is in turn explained by. I am trying to fit vector autoregressive var models using the generalized linear model fitting methods included in scikitlearn. Overview vector autoregression var model is an extension of univariate autoregression model to multivariate time series data var model is a multiequation system where all the variables are treated as endogenous dependent there is one equation for each variable as dependent variable. Pythonvarvector autoregressions with machine learning.\n\nPdf sparse vector autoregressive modeling tian zheng. Some of you may be thinking that this sounds just like a linear regression it sure does sound that way and is in general the same. The linear model has the form y x w, but the system matrix x has a very peculiar structure. Bayesian vector autoregressions northwestern university.\n\nIt applies the hamilton 1989 filter the kim 1994 smoother. It is a natural extension of the univariate autoregressive model to dynamic mul tivariate time series. Sims and vector autoregressions 1085 parameter space. An intuitive introduction to the concept of vector autoregression var. It is the generalization of arma to multiple parallel time series, e. One of the most important types of dataset is time series. To optimize performance and memory consumption the model can be expressed as y bw, where b is a. Vector autoregression models the vector autoregression var model is one of the most successful. Markov switching autoregression models statsmodels.\n\nThe authors recently proposed two methods to estimate such timevarying var models. In this article we list down the most widely used timeseries forecasting methods which can be used in python with just a single line of code. Sims conjectured that this parsimony principle was the reason econometric models in existence when sims 1980a was written had tolerable forecasting properties, despite their incredible identification assumptions. A vector autoregression var model is a multivariate time series model containing a system of n equations of n distinct, stationary response variables as linear functions of lagged responses and other terms. Optional parameters can be entered that are relevant to the particular mode of inference chosen. Vector autoregression var is a stochastic process model used to capture the linear interdependencies among multiple time series. Time series represent a series of data points indexed in time order. Watson department of economics and the woodrow wilson school, princeton university and the national bureau of economic research abstract. In this section, i will introduce you to one of the most commonly used methods for multivariate time series forecasting vector auto regression var. To be able to understand the relationship between several variables, allowing for dynamics. An introduction to vector autoregression var since the seminal paper of sims 1980 vector autoregressive models have become a key instrument in macroeconomic research. It is a natural extension of the univariate autoregressive model to dynamic multivariate time series.\n\nAs a reminder, this post is intended to be a very applied example of how use certain tests and models in a timesereis analysis. This tutorial covers time series decomposition and vector autoregression var modelling to forecast electricity prices for the state of texas, using time series. A marm model predicts the next value in a ddimensional time series, y. Structural vector autoregressive analysis themes in modern econometrics structural vector autoregressive var models are important tools for empirical work in macroeconomics, finance, and related fields. Bayesian vector autoregressions vector autoregressions are a. The right hand side of each equation includes a constant and lags of all of the variables in the system. Introduction to time series forecasting with python pdf. Often we try to analyze huge amounts of data to find useful information or to predict future events. Estimating time series models by state space methods in. Vector autoregression var model is an extension of univariate autoregression model to multivariate time series data var model is a multiequation system where all the variables are treated as endogenous dependent there is one equation for each variable as dependent variable. Vector autoregressions vars are linear multivariate timeseries models able to capture the joint dynamics of multiple time series.\n\nIt is a natural extension of the univariate autoregressive model to dynamic. Statsmodels chad fulton abstract this paper describes an object oriented approach to the estimation of time series models using state space methods and presents an implementation in the python. Vector autoregressions american economic association. A multivariate time series guide to forecasting and. Var is an extension of the autoregressive or ar model, where multiple variables are used when generating predictions. A key assumption of this model is that its parameters are constant or stationary across time. Vector autoregressions vars have been used by economists over the past 36 years to analyze multivariate timeseries data. The vector autoregressive var models, made famous in chris simss paper macroeconomics and reality, econometrica, 1980, are one of the most applied models in the empirical economics. Markov switching autoregression models chad fulton. Jul 19, 2019 analyzing electricity price time series data using python. Multivariate time series and vector autoregressions.\n\nSvars are used by economists to recover economic shocks from observables by imposing a minimum of. Bayesian vars we have seen in chapter 4 that var models can be used to characterize any vector of time series under a minimal set of conditions. Var models generalize the univariate autoregressive model ar model by allowing for more than one evolving variable. Factor models and structural vector autoregressions in macroeconomics march 9, 2016 james h.\n\nJan 24, 2019 autoregression modeling is a modeling technique used for time series data that assumes linear continuation of the series so that previous values in the time series can be used to predict futures values. One approach to solving a problem such as this is called vector autoregression, or var. Time series predictability, volatility, and bubbles more course details. As part of this effort, we examine the problem of whether the var and the bvar.\n\nIn a var model, each variable is a linear function of the past values of itself. How to estimate timevarying vector autoregressive models. Statsmodels is a python package that provides a complement to scipy for statistical computations including descriptive statistics and estimation of statistical models. The global vector autoregressive gvar approach, originally proposed in pesaran et al. Vars provide a convenient framework for policy analysis, forecasting, structural inference, and data description stock and watson, 2001. Made fameous in chris simss paper macroeconomics and reality, ecta 1980. In practice, the machine learning vector autoregressions produce more conservative estimates than the traditional ordinary least squares vector autoregressions. Forecasting time series data using autoregression python data.\n\nThis notebook provides an example of the use of markov switching models in statsmodels to replicate a number of results presented in kim and nelson 1999. Consider d time series generated from d variables within a system such as a functional network in the brain and where m is the order of the model. We have also seen that since vars are reduced form models, identi. Now, lets dig into how to implement this with python. This post is the third in a series explaining basic time series analysis. Vector autoregression overview and proposals 09 aug 2017 introduction. We motivated time series models by saying simple univariate arma models do forecasting very well. Multivariate autoregressive models extend this approach to multiple time series so that the vector of current values of all variables is modelled as a linear sum of previous activities. Vector autoregressive models for multivariate time series. Var models are very popular because of their flexibility when analyzing economic and financial time series, and are great for forecasting. A univariate autoregression is a singleequation, singlevariable linear model in which the current value of a variable is explained by its own lagged values. We are interested in modeling a multivariate time series, where denotes the number of observations and the number of variables.\n\n564 793 760 883 1477 666 5 771 726 1355 1398 1465 77 823 743 777 682 630 1371 766 245 1267 397 1269 1209 1463 267 1344 185 1181 1052 1153 685 1338 1154 1248 445 1289 529 1448 611 683 769 222 516 717 1061 969 794" ]

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[ "2021.1.15\n\n• ### Method Summary\n\nAll Methods\nModifier and Type Method and Description\n`ILcd2DEditableBounds` `cloneAs2DEditableBounds()`\nReturns a copy of this `ILcdBounds` object that is also an `ILcd2DEditableBounds`.\n`ILcd3DEditableBounds` `cloneAs3DEditableBounds()`\nReturns a copy of this `ILcdBounds` object that is also an `ILcd3DEditableBounds`.\n`boolean` ```contains2D(double aX, double aY, double aWidth, double aHeight)```\nChecks whether this `ILcdBounds` object contains the given rectangle in the 2D space.\n`boolean` `contains2D(ILcdBounds aBounds)`\nChecks whether this `ILcdBounds` object contains the given `ILcdBounds` object in the 2D space.\n`boolean` ```contains3D(double aX, double aY, double aZ, double aWidth, double aHeight, double aDepth)```\nChecks whether this `ILcdBounds` object contains the given box in the 3D space.\n`boolean` `contains3D(ILcdBounds aBounds)`\nChecks whether this `ILcdBounds` object contains the given `ILcdBounds` object in the 3D space.\n`default ILcdPoint` `getCenter()`\nReturns the center of the bounds.\n`double` `getDepth()`\nReturns the depth of the bounding box.\n`double` `getHeight()`\nReturns the height of the bounding box.\n`ILcdPoint` `getLocation()`\nReturns the location of the bounding box (smallest x and y coordinates).\n`default double` `getMaxX()`\nReturns the maximum X value (right boundary) of this bounds.\n`default double` `getMaxY()`\nReturns the maximum Y value (upper boundary) of this bounds.\n`default double` `getMaxZ()`\nReturns the maximum Z value (top boundary) of this bounds.\n`default double` `getMinX()`\nReturns the minimum X value (left boundary) of this bounds.\n`default double` `getMinY()`\nReturns the minimum Y value (lower boundary) of this bounds.\n`default double` `getMinZ()`\nReturns the minimum Z value (bottom boundary) of this bounds.\n`double` `getWidth()`\nReturns the width of the bounding box.\n`boolean` ```interacts2D(double aX, double aY, double aWidth, double aHeight)```\nChecks whether this `ILcdBounds` object interacts with the given rectangle in the 2D space.\n`boolean` `interacts2D(ILcdBounds aBounds)`\nChecks whether this `ILcdBounds` object interacts with the given `ILcdBounds` object in the 2D space.\n`boolean` ```interacts3D(double aX, double aY, double aZ, double aWidth, double aHeight, double aDepth)```\nChecks whether this `ILcdBounds` object interacts with the given box in the 3D space.\n`boolean` `interacts3D(ILcdBounds aBounds)`\nChecks whether this `ILcdBounds` object interacts with the given `ILcdBounds` object.\n`boolean` `isDefined()`\nIndicates whether this bounds object is valid.\n• ### Methods inherited from interface com.luciad.shape.ILcdShape\n\n`contains2D, contains2D, contains3D, contains3D, getFocusPoint`\n• ### Methods inherited from interface com.luciad.shape.ILcdBounded\n\n`getBounds`\n• ### Methods inherited from interface com.luciad.util.ILcdCloneable\n\n`clone`\n• ### Method Detail\n\n• #### isDefined\n\n`boolean isDefined()`\nIndicates whether this bounds object is valid.\n• If `true`, this bounds describes a valid geographic region.\n• If `false`, this bounds does not represent a geographic region, and its location, width, height and depth should not be used.\nReturns:\n`true` if this object describes actual bounds, `false` if not\nSince:\n2013.0\n• #### getLocation\n\n`ILcdPoint getLocation()`\nReturns the location of the bounding box (smallest x and y coordinates).\nReturns:\nthe location of the bounding box.\n• #### getWidth\n\n`double getWidth()`\nReturns the width of the bounding box. The width has to be larger than or equal to 0.\nReturns:\nthe width of the bounding box.\n• #### getHeight\n\n`double getHeight()`\nReturns the height of the bounding box. The height has to be larger than or equal to 0.\nReturns:\nthe height of the bounding box.\n• #### getDepth\n\n`double getDepth()`\nReturns the depth of the bounding box. The depth has to be larger than or equal to 0.\nReturns:\nthe depth of the bounding box.\n• #### interacts2D\n\n`boolean interacts2D(ILcdBounds aBounds)`\nChecks whether this `ILcdBounds` object interacts with the given `ILcdBounds` object in the 2D space. Only the first two dimensions of the `ILcdBounds` objects are considered.\n\nIf either bounds is `undefined`, the result is false.\n\nParameters:\n`aBounds` - the `ILcdBounds` to compare with.\nReturns:\nthe boolean result of the interaction test.\n`interacts2D(double, double, double, double)`\n• #### interacts2D\n\n```boolean interacts2D(double aX,\ndouble aY,\ndouble aWidth,\ndouble aHeight)```\nChecks whether this `ILcdBounds` object interacts with the given rectangle in the 2D space. Only the first two dimensions of the `ILcdBounds` object are considered.\n\nIf this bounds is `undefined`, the result is false.\n\nParameters:\n`aX` - the x coordinate of the rectangle.\n`aY` - the y coordinate of the rectangle.\n`aWidth` - the width of the rectangle.\n`aHeight` - the height of the rectangle.\nReturns:\n`true` if this `ILcdBounds` object touches or overlaps to any extent with the given rectangle, `false` otherwise.\n• #### contains2D\n\n`boolean contains2D(ILcdBounds aBounds)`\nChecks whether this `ILcdBounds` object contains the given `ILcdBounds` object in the 2D space. Only the first two dimensions of the `ILcdBounds` objects are considered.\n\nIf either bounds is `undefined`, the result is false.\n\nParameters:\n`aBounds` - the `ILcdBounds` to compare with.\nReturns:\nthe boolean result of the containment test.\n`contains2D(double, double, double, double)`\n• #### contains2D\n\n```boolean contains2D(double aX,\ndouble aY,\ndouble aWidth,\ndouble aHeight)```\nChecks whether this `ILcdBounds` object contains the given rectangle in the 2D space. Only the first two dimensions of the `ILcdBounds` object are considered.\n\nIf this bounds is `undefined`, the result is false.\n\nParameters:\n`aX` - the x coordinate of the rectangle.\n`aY` - the y coordinate of the rectangle.\n`aWidth` - the width of the rectangle.\n`aHeight` - the height of the rectangle.\nReturns:\nthe boolean result of the containment test.\n• #### interacts3D\n\n`boolean interacts3D(ILcdBounds aBounds)`\nChecks whether this `ILcdBounds` object interacts with the given `ILcdBounds` object.\n\nIf either bounds is `undefined`, the result is false.\n\nParameters:\n`aBounds` - the `ILcdBounds` to compare with.\nReturns:\nthe boolean result of the interaction test.\n`interacts3D(double, double, double, double, double, double)`\n• #### interacts3D\n\n```boolean interacts3D(double aX,\ndouble aY,\ndouble aZ,\ndouble aWidth,\ndouble aHeight,\nChecks whether this `ILcdBounds` object interacts with the given box in the 3D space.\n\nIf this bounds is `undefined`, the result is false.\n\nParameters:\n`aX` - the x coordinate of the box.\n`aY` - the y coordinate of the box.\n`aZ` - the z coordinate of the box.\n`aWidth` - the width of the box.\n`aHeight` - the height of the box.\n`aDepth` - the depth of the box.\nReturns:\n`true` if this `ILcdBounds` object touches or overlaps to any extent with the given box, `false` otherwise.\n• #### contains3D\n\n`boolean contains3D(ILcdBounds aBounds)`\nChecks whether this `ILcdBounds` object contains the given `ILcdBounds` object in the 3D space.\n\nIf either bounds is `undefined`, the result is false.\n\nParameters:\n`aBounds` - the `ILcdBounds` to compare with.\nReturns:\nthe boolean result of the containment test.\n`contains3D(double, double, double, double, double, double)`\n• #### contains3D\n\n```boolean contains3D(double aX,\ndouble aY,\ndouble aZ,\ndouble aWidth,\ndouble aHeight,\nChecks whether this `ILcdBounds` object contains the given box in the 3D space.\n\nIf this bounds is `undefined`, the result is false.\n\nParameters:\n`aX` - the x coordinate of the point.\n`aY` - the y coordinate of the point.\n`aZ` - the z coordinate of the point.\n`aWidth` - the width of the box.\n`aHeight` - the height of the box.\n`aDepth` - the depth of the box.\nReturns:\nthe boolean result of the containment test.\n`contains2D(double, double, double, double)`\n• #### cloneAs2DEditableBounds\n\n`ILcd2DEditableBounds cloneAs2DEditableBounds()`\nReturns a copy of this `ILcdBounds` object that is also an `ILcd2DEditableBounds`. This makes sure that the first two dimensions of the copy are writable, even if the original `ILcdBounds` object may be read-only.\nReturns:\na copy of this `ILcdBounds` object that is also an `ILcd2DEditableBounds`. This makes sure that the first two dimensions of the copy are writable, even if the original `ILcdBounds` object may be read-only.\n`ILcd2DEditableBounds`\n• #### cloneAs3DEditableBounds\n\n`ILcd3DEditableBounds cloneAs3DEditableBounds()`\nReturns a copy of this `ILcdBounds` object that is also an `ILcd3DEditableBounds`. This makes sure that all three dimensions of the copy are writable, even if the original `ILcdBounds` object may be read-only.\nReturns:\na copy of this `ILcdBounds` object that is also an `ILcd3DEditableBounds`. This makes sure that all three dimensions of the copy are writable, even if the original `ILcdBounds` object may be read-only.\n`ILcd3DEditableBounds`\n• #### getCenter\n\n`default ILcdPoint getCenter()`\nReturns the center of the bounds.\nReturns:\nthe center of the bounds, or null if the bounds are undefined.\nSince:\n2017.0\n• #### getMinX\n\n`default double getMinX()`\nReturns the minimum X value (left boundary) of this bounds.\n\nBy default, this does `getLocation.getX()`, but can be overridden for performance.\n\nReturns:\nthe minimum X boundary\nSince:\n2017.0\n`getMaxX()`\n• #### getMinY\n\n`default double getMinY()`\nReturns the minimum Y value (lower boundary) of this bounds.\n\nBy default, this does `getLocation.getY()`, but can be overridden for performance.\n\nReturns:\nthe minimum Y boundary\nSince:\n2017.0\n`getMaxY()`\n• #### getMinZ\n\n`default double getMinZ()`\nReturns the minimum Z value (bottom boundary) of this bounds.\n\nBy default, this does `getLocation.getZ()`, but can be overridden for performance.\n\nReturns:\nthe minimum Z boundary\nSince:\n2017.0\n`getMaxZ()`\n• #### getMaxX\n\n`default double getMaxX()`\nReturns the maximum X value (right boundary) of this bounds.\n\nBy default, this does `getLocation.getX() + getWidth()`, but can be overridden for performance.\n\nNote that this is a derived property, the bounds are defined by their lower-left corner and width/height.\n\nFor `geodetic bounds`, the returned value is not necessarily normalized to [-180, 180].\n\nReturns:\nthe maximum X boundary\nSince:\n2017.0\n`getMinX()`\n• #### getMaxY\n\n`default double getMaxY()`\nReturns the maximum Y value (upper boundary) of this bounds.\n\nBy default, this does `getLocation.getY() + getHeight()`, but can be overridden for performance.\n\nNote that this is a derived property, the bounds are defined by their lower-left corner and width/height.\n\nReturns:\nthe maximum Y boundary\nSince:\n2017.0\n`getMinY()`\n• #### getMaxZ\n\n`default double getMaxZ()`\nReturns the maximum Z value (top boundary) of this bounds.\n\nBy default, this does `getLocation.getZ() + getDepth()`, but can be overridden for performance.\n\nNote that this is a derived property, the bounds are defined by their lower-left corner and width/height.\n\nReturns:\nthe maximum Z boundary\nSince:\n2017.0\n`getMinZ()`" ]

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[ "", null, "keras\nneural-network\npython\ntensorflow", null, "4", null, "0", null, "``````#Partly train model\nmodel.fit(first_training, first_classes, batch_size=32, nb_epoch=20)\n\n#Save partly trained model\nmodel.save('partly_trained.h5')\n\n#Continue training\nmodel.fit(second_training, second_classes, batch_size=32, nb_epoch=20)\n``````\n\n``````\"\"\"\nModel by: http://machinelearningmastery.com/\n\"\"\"\nimport numpy\nfrom keras.datasets import mnist\nfrom keras.models import Sequential\nfrom keras.layers import Dense\nfrom keras.utils import np_utils\nnumpy.random.seed(7)\n\ndef baseline_model():\nmodel = Sequential()\nreturn model\n\nif __name__ == '__main__':\n(X_train, y_train), (X_test, y_test) = mnist.load_data()\n\n# flatten 28*28 images to a 784 vector for each image\nnum_pixels = X_train.shape * X_train.shape\nX_train = X_train.reshape(X_train.shape, num_pixels).astype('float32')\nX_test = X_test.reshape(X_test.shape, num_pixels).astype('float32')\n# normalize inputs from 0-255 to 0-1\nX_train = X_train / 255\nX_test = X_test / 255\n# one hot encode outputs\ny_train = np_utils.to_categorical(y_train)\ny_test = np_utils.to_categorical(y_test)\nnum_classes = y_test.shape\n\n# build the model\nmodel = baseline_model()\n\n#Partly train model\ndataset1_x = X_train[:3000]\ndataset1_y = y_train[:3000]\nmodel.fit(dataset1_x, dataset1_y, nb_epoch=10, batch_size=200, verbose=2)\n\n# Final evaluation of the model\nscores = model.evaluate(X_test, y_test, verbose=0)\nprint(\"Baseline Error: %.2f%%\" % (100-scores*100))\n\n#Save partly trained model\nmodel.save('partly_trained.h5')\ndel model\n\n#Continue training\ndataset2_x = X_train[3000:]\ndataset2_y = y_train[3000:]\nmodel.fit(dataset2_x, dataset2_y, nb_epoch=10, batch_size=200, verbose=2)\nscores = model.evaluate(X_test, y_test, verbose=0)\nprint(\"Baseline Error: %.2f%%\" % (100-scores*100))\n``````\n\nStack Overflow", null, "收藏", null, "评论", null, "", null, "收藏", null, "评论", null, "", null, "收藏", null, "评论", null, "``````reduce_lr = ReduceLROnPlateau(monitor='loss', factor=lr_reduction_factor,\npatience=patience, min_lr=min_lr, verbose=1)\n``````", null, "收藏", null, "评论\n\n•", null, "社区规范\n•", null, "提出问题\n•", null, "进行投票\n•", null, "个人资料\n•", null, "优化问题\n•", null, "回答问题", null, "关于我们", null, "常见问题", null, "内容许可", null, "联系我们" ]

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[ "", null, "# Go Math Grade 3 Chapter 10 Answer Key Pdf Time, Length, Liquid Volume, and Mass\n\n### Time to the Minute – Page No. 565\n\nWrite the time. Write one way you can read the time.\n\nQuestion 1.", null, "Answer:  1:16; sixteen minutes after one\n\nExplanation:\nCount on by fives and ones from the 12 on the clock to where the minute hand is pointing. Write the missing counting numbers next to the clock.\nWrite: 1 Hour 16 minutes\n\nQuestion 2.", null, "Type below:\n___________\n\nAnswer: 10:20; twenty minutes after ten\n\nExplanation:\nCount on by fives and ones from the 12 on the clock to where the minute hand is pointing. Write the missing counting numbers next to the clock.\nWrite: 10 Hour 20 minutes\n\nQuestion 3.", null, "Type below:\n____________\n\nAnswer: 4:13;  thirteen minutes after four\n\nExplanation:\nCount on by fives and ones from the 12 on the clock to where the minute hand is pointing. Write the missing counting numbers next to the clock.\nWrite: 4 Hour 13 minutes\n\nQuestion 4.", null, "Type below:\n____________\n\nAnswer: 12:05; five minutes after twelve\n\nExplanation:\nCount on by fives and ones from the 12 on the clock to where the minute hand is pointing. Write the missing counting numbers next to the clock.\nWrite: 12 Hour 05 minutes\n\nQuestion 5.", null, "Type below:\n____________\n\nAnswer: 7:24; twenty four minutes after seven\n\nExplanation:\nCount on by fives and ones from the 12 on the clock to where the minute hand is pointing. Write the missing counting numbers next to the clock.\nWrite: 7 Hour 24 minutes\n\nQuestion 6.", null, "Type below:\n____________\n\nAnswer: 2:51; nine minutes before three\n\nExplanation:\nCount by fives and ones from the 12 on the clock back to where the minute hand is pointing. Write the missing counting numbers next to the clock.\nWrite: 2:51\nnine minutes before three\n\nWrite the time another way.\n\nQuestion 7.\n23 minutes after 4\n______ : ______\n\nExplanation:\nCount on by fives and ones from the 12 on the clock to where the minute hand is pointing. Write the missing counting numbers next to the clock.\n\nQuestion 8.\n18 minutes before 11\n______ : ______\n\nExplanation:\nCount by fives and ones from the 12 on the clock back to where the minute hand is pointing. Write the missing counting numbers next to the clock.\n\nQuestion 9.\n10 minutes before 9\n______ : ______\n\nExplanation:\nCount by fives and ones from the 12 on the clock back to where the minute hand is pointing. Write the missing counting numbers next to the clock.\n\nQuestion 10.\n7 minutes after 1\n______ : ______\n\nExplanation:\nCount on by fives and ones from the 12 on the clock to where the minute hand is pointing. Write the missing counting numbers next to the clock.\n\nProblem Solving\n\nQuestion 11.\nWhat time is it when the hour hand is a little past the 3 and the minute hand is pointing to the 3?\n______ : ______\n\nExplanation:\nCount on by fives and ones from the 12 on the clock to where the minute hand is pointing. Write the missing counting numbers next to the clock.\n\nQuestion 8.\n\nQuestion 12.\nPete began practicing at twenty-five minutes before eight. What is another way to write this time?\n______ : ______\n\nExplanation:\nCount by fives and ones from the 12 on the clock back to where the minute hand is pointing. Write the missing counting numbers next to the clock.\n\n### Time to the Minute – Page No. 566\n\nLesson Check\n\nQuestion 1.\nWhich is another way to write 13 minutes before 10?\nOptions:\na. 9:47\nb. 10:13\nc. 10:47\nd. 11:13\n\nExplanation:\nCount by fives and ones from the 12 on the clock back to where the minute hand is pointing. Write the missing counting numbers next to the clock.\n\nQuestion 2.\nWhat time does the clock show?", null, "Options:\na. 2:20\nb. 2:40\nc. 3:20\nd. 4:10\n\nExplanation:\nCount on by fives and ones from the 12 on the clock to where the minute hand is pointing. Write the missing counting numbers next to the clock.\n\nSpiral Review\n\nQuestion 3.\nEach bird has 2 wings. How many wings will 5 birds have?\nOptions:\na. 7\nb. 8\nc. 9\nd. 10\n\nExplanation:\nSTEP 1 Draw 2 counters in each group.\nSTEP 2 Skip count to find how many wings in all. Skip count by 2s until you say 5 numbers (groups)\nThere are 5 groups with 2 wings in each group. So, there are 10 wings in all.\n\nQuestion 4.\nFind the unknown factor.\n9 × ■ = 36\nOptions:\na. 4\nb. 6\nc. 8\nd. 27\n\nExplanation:\nCount how many rows of 36 counters there are.\nThere are  9 rows of 36 counters. The unknown factor is 4 (c0lumns). n = 4\n9× 4 = 36\n\nQuestion 5.\nMr. Wren has 56 paintbrushes. He places 8 paintbrushes on each of the tables in the art room. How many tables are in the art room?\nOptions:\na. 6\nb. 7\nc. 9\nd. 48\n\nExplanation:\nSTEP 1 Draw 56 counters.\nSTEP 2 Make a group of 8 counters by drawing a circle around them. Continue circling groups of 8 until all 56 counters are in groups.\nThere are 7 groups of 56 counters.\n\nQuestion 6.\nWhich number completes the equations?\n4 × ▲ = 20 20 ÷ 4 = ▲\nOptions:\na. 4\nb. 5\nc. 6\nd. 16\n\nExplanation:\nCount how many rows of 20 counters there are. There are  4 rows of 20 counters. The unknown factor is 5 (c0lumns). n = 5\n4× 5 = 20\nSTEP 1 Draw 20 counters.\nSTEP 2 Make a group of 4 counters by drawing a circle around them. Continue circling groups of 4 until all 20 counters are in groups.\nThere are 5 groups of 20 counters.\n\nA.M. and P.M. – Page No. 571\n\nWrite the time for the activity. Use A.M. or P.M.\n\nQuestion 1.\neat lunch", null, "Explanation:\nThe times afternoon and before midnight is written with P.M. Since we have our lunch in the afternoon around 12:00 noon to 1:00 P.M. which falls in the period of 12:00 noon to 12:00 at midnight hence the time 12:20 is considered to be P.M.\n\nQuestion 2.\ngo home after school", null, "_____ : _____ _____ (A.M./P.M.)\n\nExplanation:\nThe times afternoon and before midnight is written with P.M. Since the school hours end in the evening and students return to their home around 2:30 P.M. to 3:30 P.M. which falls in the period of 12:00 noon to 12:00 at midnight hence the time 2:53 is considered to be P.M.\n\nQuestion 3.\nsee the sunrise", null, "_____ : _____ _____ (A.M./P.M.)\n\nExplanation:\nThe times after midnight and before noon is written with A.M. At the early hours of the day sunrises which falls in the period of  12:00 at midnight to 12:00 noon, therefore the time 6:18 is considered to be A.M.\n\nQuestion 4.\ngo for a walk", null, "_____ : _____ _____ (A.M./P.M.)\n\nExplanation:\nThe times afternoon and before midnight is written with P.M. Since the people go for a walk to refresh their minds in the evening hours which falls in the period of 12:00 noon to 12:00 at midnight the time 3:55 is considered to be P.M.\n\nQuestion 5.\ngo to school", null, "_____ : _____ _____ (A.M./P.M.)\n\nExplanation:\nThe times after midnight and before noon is written with A.M. Since students go to the school in the morning hours which falls in the period of  12:00 at midnight to 12:00 noon, therefore, the time 8:10 is considered to be A.M.\n\nQuestion 6.", null, "_____ : _____ _____ (A.M./P.M.)\n\nExplanation:\nThe times after midnight and before noon is written with A.M. Since students go to the art class in the morning hours which falls in the period of  12:00 at midnight to 12:00 noon, therefore, the time 10:36 is considered to be A.M.\n\nWrite the time. Use A.M. or P.M.\n\nQuestion 7.\n13 minutes after 5:00 in the morning\n_____ : _____ _____ (A.M./P.M.)\n\nExplanation:\nThe times after midnight and before noon is written with A.M. Since 5:00 in the morning falls in the period of 12:00 at midnight to 12:00 noon, therefore, the time is considered to be A.M.\n\nQuestion 8.\n19 minutes before 9:00 at night\n_____ : _____ _____ (A.M./P.M.)\n\nExplanation:\nThe times afternoon and before midnight is written with P.M.  Since 8:41 in the night time falls in the period of 12:00 noon to 12:00 at midnight, therefore, the time 8:41 is considered to be P.M.\n\nQuestion 9.\nquarter before midnight\n_____ : _____ _____ (A.M./P.M.)\n\nExplanation:\nThe times afternoon and before midnight is written with P.M. Since 11:45 in the night time falls in the period of 12:00 noon to 12:00 at midnight, therefore, the time 11:45 is considered to be P.M.\n\nQuestion 10.\none-half hour after 4:00 in the morning\n_____ : _____ _____ (A.M./P.M.)\n\nExplanation:\nOne-half hour = 30 minutes So, by adding 30 minutes to 4:00 A.M. in the morning. We get the time to be 4:30 A.M.\nThe times after midnight and before noon is written with A.M.  Since 4:30 in the morning falls in the period of 12:00 at midnight to 12:00 noon, therefore, the time is considered to be A.M\n\nProblem Solving\n\nQuestion 11.\nJaime is in math class. What time is it? Use A.M. or P.M.", null, "_____ : _____ _____ (A.M./P.M.)\n\nExplanation:\nThe times afternoon and before midnight is written with P.M. Since 1:25 in the afternoon hours fall in the period of 12:00 noon to 12:00 at midnight, therefore, the time1:25 is considered to be P.M.\n\nQuestion 12.\nPete began practicing his trumpet at fifteen minutes past three. Write this time using A.M. or P.M.\n_____ : _____ _____ (A.M./P.M.)\n\nExplanation:\nThe times afternoon and before midnight is written with P.M. Since 3:15 in the evening hours fall in the period of 12:00 noon to 12:00 at midnight, therefore, the time 3:15 is considered to be P.M.\n\n### A.M. and P.M. – Page No. 572\n\nLesson Check\n\nQuestion 1.\nSteven is doing his homework. What time is it? Use A.M. or P.M.", null, "Options:\na. 4:15 P.M.\nb. 4:25 A.M.\nc. 4:35 P.M.\nd. 4:35 A.M.\n\nExplanation:\nThe times afternoon and before midnight is written with P.M. Since 4:35 in the evening hours fall in the period of 12:00 noon to 12:00 at midnight, therefore, the time 4:35 is considered to be P.M.\n\nQuestion 2.\nAfter he finished breakfast, Mr. Edwards left for work at fifteen minutes after seven. What time is this? Use A.M. or P.M.\nOptions:\na. 6:15 A.M.\nb. 7:15 A.M.\nc. 6:45 P.M.\nd. 7:30 P.M.\n\nExplanation:\nThe times after midnight and before noon is written with A.M. Since 7:15 in the morning falls in the period of 12:00 at midnight to 12:00 noon, therefore, the time 7:15 is considered to be A.M.\n\nSpiral Review\n\nQuestion 3.\nWhich division equation is related to the multiplication equation 4 × 6 = 24\nOptions:\na. 24 ÷ 8 = 3\nb. 12 ÷ 3 = 4\nc. 6 × 4 = 24\nd. 24 ÷ 4 = 6\n\nAnswer: d. 24 ÷ 4 = 6\n\nExplanation:\nSTEP 1 Draw 24 counters.\nSTEP 2 Make a group of 4 counters by drawing a circle around them. Continue circling groups of 4 until all 24 counters are in groups.\nThere are 6 groups of 24 counters.\n\nQuestion 4.\nThere are 50 toothpicks in each box. Jaime buys 4 boxes for her party platter. How many toothpicks does Jaime buy in all?\nOptions:\na. 20\nb. 54\nc. 200\nd. 2,000\n\nExplanation:\nStep1 let there be 50 toothpicks in each box.\nStep2 the number of boxes required is 4\nStep3 total number of toothpicks will be the number of boxes multiplied by the number of toothpicks in each box.\n4 × 50=200\n\nQuestion 5.\nA pet store sold 145 bags of beef-flavored dog food and 263 bags of cheese-flavored dog food. How many bags of dog food were sold in all?\nOptions:\na. 118\nb. 308\nc. 408\nd. 422\n\nExplanation:\nTotal number of bags of dog food sold=145+263", null, "Question 6.\nVictoria and Melody are comparing fraction strips. Which statement is NOT correct?\nOptions:\na. $$\\frac{1}{4}$$ < $$\\frac{4}{4}$$\nb. $$\\frac{3}{6}$$ > $$\\frac{4}{6}$$\nc. $$\\frac{2}{8}$$ > $$\\frac{1}{8}$$\nd. $$\\frac{2}{3}$$ > $$\\frac{3}{3}$$\n\nAnswer: b. $$\\frac{3}{6}$$ > $$\\frac{4}{6}$$\n\nExplanation:\n\nTake two blocks of the same size and both are made of 6 equal-size squares.\nCompare 3/6 and 4/6.\nThe greater fraction will have the larger\nTherefore, 3/6>4/6.\n\n### Measure Time Intervals – Page No. 577\n\nFind the elapsed time.\n\nQuestion 1.\nStart: 8:10 A.M. End: 8:45 A.M.", null, "Explanation:\nSTEP 1 Find 8:10 A.M. on the number line. Count on from 8:10 A.M. to 8:45 A.M. Draw marks and record the times on the number line. Then draw and label the jumps.\nSTEP 2 Add to find the total minutes from 8:10 A.M. to 8:45 A.M. From 8:10 A.M. to 8:45 A.M. is 35 minutes.\nSo, the elapsed time is 35 minutes\n\nQuestion 2.\nStart: 6:45 P.M. End: 6:54 P.M.", null, "_______ minutes\n\nExplanation:\nSTEP 1 Find the starting time on the clock.\nSTEP 2 Count the minutes by counting on by fives and ones to 6:54 P.M. Write the missing counting numbers next to the clock.\nSo, the elapsed time is 9 minutes.\n\nQuestion 3.\nStart: 3:00 P.M. End: 3:37 P.M.", null, "_______ minutes\n\nExplanation:\nSTEP 1 Find3:00 P.M. on the number line. Count on from 3:00 P.M. to 3:37 P.M. Draw marks and record the times on the number line. Then draw and label the jumps.\nSTEP 2 Add to find the total minutes from3:00 P.M. to 3:37 P.M. From 3:00 P.M. to 3:37 P.M. is 37 minutes.\nSo, the elapsed time is 37 minutes\n\nQuestion 4.\nStart: 10:05 A.M. End: 10:21 A.M.", null, "_______ minutes\n\nExplanation:\nSTEP 1 Find10:05 A.M. on the number line. Count on from 10:05 A.M. to 10:21 A.M. Draw marks and record the times on the number line. Then draw and label the jumps.\nSTEP 2 Add to find the total minutes from 10:05 A.M. to 10:21 A.M. From 10:05 A.M. to 10:21 A.M. is 16 minutes.\nSo, the elapsed time is 16 minutes\n\nQuestion 5.\nStart: 7:30 A.M. End: 7:53 A.M.", null, "_______ minutes\n\nExplanation:\nSTEP 1 Find the starting time on the clock.\nSTEP 2 Count the minutes by counting on by fives and ones to7:53 A.M. Write the missing counting numbers next to the clock.\nSo, the elapsed time is 23 minutes.\n\nQuestion 6.\nStart: 5:20 A.M. End: 5:47 A.M.", null, "_______ minutes\n\nExplanation:\nSTEP 1 Find the starting time on the clock.\nSTEP 2 Count the minutes by counting on by fives and ones to 5:47 A.M. Write the missing counting numbers next to the clock.\nSo, the elapsed time is 27 minutes.\n\nProblem Solving\n\nQuestion 7.\nA show at the museum starts at 7:40 P.M. and ends at 7:57 P.M. How long is the show?\n_______ minutes\n\nExplanation:\nSTEP 1 Find 7:40 P.M. on the number line. Count on from 7:40 P.M. to 7:57 P.M. Draw marks and record the times on the number line. Then draw and label the jumps.\nSTEP 2 Add to find the total minutes from 7:40 P.M. to 7:57 P.M. From 7:40 P.M. to 7:57 P.M. is 17 minutes.\nSo, the elapsed time is 17 minutes\nThe duration of the show is 17 minutes.\n\nQuestion 8.\nThe first train leaves the station at 6:15 A.M. The second train leaves at 6:55 A.M. How much later does the second train leave the station?\n_______ minutes\n\nExplanation:\nSTEP 1 Find 6:15 A .M. on the number line. Count on from 6:15 A.M. to 6:55 A.M. Draw marks and record the times on the number line. Then draw and label the jumps.\nSTEP 2 Add to find the total minutes from 6:15 A.M. to 6:55 A.M. From 6:15 A .M. to 6:55 A.M. is 40 minutes.\nSo, the second train will leave the station in 40 minutes\n\n### Measure Time Intervals – Page No. 578\n\nLesson Check\n\nQuestion 1.\nMarcus began playing basketball at 3:30 P.M. and stopped playing at 3:55 P.M. For how many minutes did he play basketball?\nOptions:\na. 25 minutes\nb. 30 minutes\nc. 55 minutes\nd. 85 minutes\n\nExplanation:\nSTEP 1 Find 3:30 P.M. on the number line. Count on from 3:30 P.M. to 3:55 P.M. Draw marks and record the times on the number line. Then draw and label the jumps.\nSTEP 2 Add to find the total minutes from3:30 P.M. to3:55 P.M. From 3:30 P.M. to 3:55 P.M. is 25 minutes.\nSo, the elapsed time is 25 minutes\nMarcus played basketball for 25 minutes.\n\nQuestion 2.\nThe school play started at 8:15 P.M. and ended at 8:56 P.M. How long was the school play?\nOptions:\na. 15 minutes\nb. 31 minutes\nc. 41 minutes\nd. 56 minutes\n\nExplanation:\nSTEP 1 Find 8:15 P.M. on the number line. Count on from 8:15 P.M. to 8:56 P.M. Draw marks and record the times on the number line. Then draw and label the jumps.\nSTEP 2 Add to find the total minutes from 8:15 P.M. to 8:56 P.M. From 8:15 P.M. to 8:56 P.M. is 41  minutes.\nSo, the elapsed time is 41 minutes\nTherefore, the school play was for 41 minutes.\n\nSpiral Review\n\nQuestion 3.\nEach car has 4 wheels. How many wheels will 7 cars have?\nOptions:\na. 11\nb. 24\nc. 27\nd. 28\n\nExplanation:\nSTEP 1 Draw 4 counters in each group.\nSTEP 2 Skip count to find how many wheels in all. Skip count by 4s until you say 7 numbers (groups)\nThere are 7 groups with 4 wheels in each group. So, there are 28 wheels in all.\n\nQuestion 4.\nWhich number completes the equations?\n3 × ■ = 27 27 ÷ 3 = ■\nOptions:\na. 6\nb. 7\nc. 8\nd. 9\n\nExplanation:\nCount how many rows of 27 counters there are. There are  3 rows of 27 counters. The unknown factor is 9 (c0lumns). n = 9\n3 × 9 = 27\nSTEP 1 Draw 27 counters.\nSTEP 2 Make a group of 3 counters by drawing a circle around them. Continue circling groups of 3 until all 27 counters are in groups.\nThere are 9 groups of 27 counters.\n\nQuestion 5.\nThere are 20 napkins in each package. Kelli bought 8 packages for her party. How many napkins did Kelli buy in all?\nOptions:\na. 28\nb. 40\nc. 160\nd. 180\n\nExplanation:\nStep1 let there be 20 napkins in each package.\nStep2 the number of packages required is 8\nStep3 total number of napkins will be the number of packages multiplied by the number of napkins in each package.\n8 × 20=160\n\nQuestion 6.\nMr. Martin drove 290 miles last week. This week he drove 125 miles more than last week. How many miles did Mr. Martin drive this week?\nOptions:\na. 125 miles\nb. 165 miles\nc. 315 miles\nd. 415 miles\n\nExplanation:\nTotal number of miles drove by Mr Martin in this week = 290+125", null, "### Use Time Intervals – Page No. 583\n\nFind the starting time.\n\nQuestion 1.\nEnding time: 4:29 P.M.\nElapsed time: 55 minutes", null, "Explanation:\nSTEP 1 Find the starting time on the number line\nSTEP 2 Count back on the number line to subtract the elapsed time. Draw and label the jumps to show the minutes.\nSTEP 3 Write the times below the number line.\nYou jumped back to 55 minutes\nSo, the starting time is 3:34 P.M.\n\nQuestion 2.\nEnding time: 10:08 A.M.\nElapsed time: 30 minutes", null, "______ : ______ ______ (A.M./P.M.)\n\nFind the ending time.\n\nExplanation:\nSTEP 1 Find the starting time on the clock.\nSTEP 2 Count back by fives for the elapsed time of 3o minutes. Write the missing counting numbers next to the clock.\nSo, the starting time is 9:38 A.M.\n\nQuestion 3.\nStarting time: 2:15 A.M.\nElapsed time: 45 minutes", null, "______ : ______ ______ (A.M./P.M.)\n\nExplanation:\nSTEP 1 Find the ending time on the number line.\nSTEP 2 Count forward on the number line to add the elapsed time. Draw and label the jumps to show the minutes.\nSTEP 3 Write the times below the number line.\nThe jumps end at 3:00 A.M.\nSo, the ending time is 3:00 A.M.\n\nQuestion 4.\nStarting time: 6:57 P.M.\nElapsed time: 47 minutes", null, "______ : ______ ______ (A.M./P.M.)\n\nExplanation:\nSTEP 1 Find the ending time on the clock.\nSTEP 2 Count on by fives and ones for the elapsed time of 47 minutes. Write the missing counting numbers next to the clock.\nSo, the ending time is 7:44 P.M.\n\nProblem Solving\n\nQuestion 5.\nJenny spent 35 minutes doing research on the Internet. She finished at 7:10 P.M. At what time did Jenny start her research?\n______ : ______ ______ (A.M./P.M.)\n\nExplanation:\nSTEP 1 Find the starting time on the clock.\nSTEP 2 Count back by fives for the elapsed time of 35 minutes. Write the missing counting numbers next to the clock.\nSo, the starting time is 6:35 A.M.\n\nQuestion 6.\nClark left for school at 7:43 A.M. He got to school 36 minutes later. At what time did Clark get to school?\n______ : ______ ______ (A.M./P.M.)\n\nExplanation:\nSTEP 1 Find the ending time on the clock.\nSTEP 2 Count on by fives and ones for the elapsed time of 36 minutes. Write the missing counting numbers next to the clock.\nSo, the ending time is 8:19 A.M.\n\n### Use Time Intervals – Page No. 584\n\nLesson Check\n\nQuestion 1.\nCody and his friends started playing a game at 6:30 P.M. It took them 37 minutes to finish the game. At what time did they finish?\nOptions:\na. 5:07 P.M.\nb. 5:53 P.M.\nc. 6:53 P.M.\nd. 7:07 P.M.\n\nExplanation:\nSTEP 1 Find the ending time on the clock.\nSTEP 2 Count on by fives and ones for the elapsed time of 37 minutes. Write the missing counting numbers next to the clock.\nSo, the ending time is 7:07 P.M.\n\nQuestion 2.\nDelia worked for 45 minutes on her oil painting. She took a break at 10:35 A.M. At what time did Delia start working on the painting?\nOptions:\na. 9:40 A.M.\nb. 9:50 A.M.\nc. 11:20 A.M.\nd. 11:30 A.M.\n\nExplanation:\nSTEP 1 Find the starting time on the clock.\nSTEP 2 Count back by fives for the elapsed time of 45 minutes. Write the missing counting numbers next to the clock.\nSo, the starting time is 9:50 A.M.\n\nSpiral Review\n\nQuestion 3.\nSierra has 30 collector’s pins. She wants to put an equal number of pins in each of 5 boxes. How many pins should she put in each box?", null, "Options:\na. 4\nb. 5\nc. 6\nd. 8\n\nExplanation:\nUse 30 counters.\nTake 5 boxes.\nPlace 1 counter at a time in each box until all 30 counters are used.\nPlace the rest of the counters till all the 30 counters (pins) are completed.\nTotal number of counters in each box is 6.\n\nQuestion 4.\nWhat time is shown on the clock?", null, "Options:\na. 1:24\nb. 2:24\nc. 4:12\nd. 5:12\n\nExplanation:\nCount on by fives and ones from the 12 on the clock to where the minute hand is pointing. Write the missing counting numbers next to the clock.\nWrite: 2:24\n• twenty-four minutes after two\n\nQuestion 5.\nRicardo has 32 books to put on 4 shelves. He puts the same number of books on each shelf. How many books does Ricardo put on each shelf?\nOptions:\na. 6\nb. 7\nc. 8\nd. 9\n\nExplanation:\nUse 32 counters. (books)\nThe number of shelves is 4.\nPlace 1 counter at a time in each shelf until all 32 counters are used.\nPlace the rest of the counters till all the 32 counters (books) are completed.\nTotal number of counters in each shelf is 8.\n\nQuestion 6.\nJon started playing a computer game at 5:35 P.M. He finished the game at 5:52 P.M. How long did Jon play the game?\nOptions:\na. 17 minutes\nb. 23 minutes\nc. 25 minutes\nd. 27 minutes\n\nExplanation:\nSTEP 1 Find the ending time on the clock.\nSTEP 2 Count the minutes by counting on by fives and ones to 5:52 P.M. Write the missing counting numbers next to\nthe clock.\nSo, Jon has spent 17 minutes on the computer playing a game.\n\n### Lesson 5 – Page No. 587\n\nShare and Show\n\nQuestion 1.\nPatty went to the shopping mall at 11:30 a.m. She shopped for 25 minutes. She spent 40 minutes eating lunch.\nThen she met a friend at a movie.\nAt what time did Patty meet her friend?\n_______ : ________\n\nExplanation:\nThe times afternoon and before midnight is written with P.M.\nCount (25+40) minutes after 11:30 A.M.\nTherefore the time at which Patty met her friend is 11:30 A.M.+1:05 hrs = 12:35 P.M.\n\nQuestion 2.\nWhat if Patty goes to the mall at 11:30 A.M. and meets a friend at a movie at 1:15 P.M.?\nPatty wants to shop and have 45 minutes for lunch before meeting her friend.\nHow much time can Patty spend shopping?\n_______ minutes\n\nExplanation:\nThe times afternoon and before midnight is written with P.M.\nCount 45 minutes before 1:15 P.M. (for lunch) which is equal to 12:30 P.M.\nDifference between 12:30 P.M. and 11:30 A.M. gives the time spent by Patty in shopping.\nSTEP 1 Find 12:30 P.M. on the number line. Count on from 12:30 P.M. to 11:30 A.M. Draw marks and record the times on the number line. Then draw and label the jumps.\nSTEP 2 Add to find the total minutes from 12:30 P.M. to 11:30 A.M. From 12:30 P.M. to 11:30 A.M. is 60 minutes.\nSo, the time spent on shopping is 60 minutes\n\nQuestion 3.\nAvery got on the bus at 1:10 p.m. The trip took 90 minutes. Then she walked for 32 minutes to get home. At what time did Avery arrive at home?\n_______ : ________ P.M.\n\nExplanation:\nSTEP 1 Find 1:10 P.M. on the number line. Count on from 1:10 P.M.\nSTEP2 Time spent in the trip= 90 minutes\nTime Avery walked to get home = 32 minutes\nSTEP3 Then draw and label the jumps on the number line.\nAdd the total minutes; 90+32=122 minutes\nSo the time at which Avery arrive at home is calculated by the cumulative jumps on the number line which is equal to 3:12 P.M.\nQuestion 4.\nKyle and Josh have a total of 64 CDs. Kyle has 12 more CDs than Josh. How many CDs does each boy have?\nJosh _______ CDs\nKyle _______ CDs\n\nKyle has 38 CDs\n\nExplanation:\nSTEP 1 There are 64 counters. (CDs)\nSTEP2 Kyle has 12 CDs more than Josh. Place these 12 in a separate group. The leftover number of CDs is 64-12=52\nSTEP 3 Make 2 group and start placing the counters one after the other in each group until all the 52 counters are grouped\nThere are 2 groups of 26 counters.\nJosh has 26 CDs while Kyle has 26+12 = 38 CDs\n\n### Lesson 5 – Page No. 588\n\nQuestion 5.\nJamal spent 60 minutes using the computer. He spent a half hour of the time playing games and the rest of the time researching his report. How many minutes did Jamal spend researching his report?\n_______ minutes\n\nExplanation: Jamal has spent half hour=30 minutes playing games and the rest of the time researching for his report.\nSTEP1 There are 60 counters(minutes).\nSTEP2 Remove the 30 counters which were used to play games\nSTEP3 The remaining number of counters in the box is 30\nSo the time spent on the report =30 minutes\n\nQuestion 6.\nWhen Caleb got home from school, he worked on his science project for 20 minutes. Then he studied for a test for 30 minutes. He finished at 4:35 p.m. At what time did Caleb get home from school?\n_______ : _______ P.M.\n\nExplanation:\nSTEP 1 Find the time on the number line when Caleb finished his homework.\nSTEP 2 Count back on the number line to subtract the elapsed time. Draw and label the jumps to show the minutes.\nSTEP 3 Write the times below the number line.\nYou jumped back to 30+20=50 minutes\nTherefore Caleb got home from school at 3:45 P.M.\n\nQuestion 7.\nMiguel played video games each day for a week. On Monday, he scored 83 points. His score went up 5 points each day. On what day did Miguel score 103 points? Explain how you found your answer\n_____________\n\nExplanation:\nSTEP1 Miguel scored 83 points on Monday. His score went up 5 points each day.\nSTEP2 Add 5n counters to the points to reach 103\nSTEP3 Adding 5 points 5 times gives us the score to be 103 (83+5*5)\nSTEP4 Therefore Miguel takes 5 days to score 103 points\n\nQuestion 8.\nWhen Laura arrived at the library, she spent 40 minutes reading a book. Then she spent 15 minutes reading a magazine. She left the library at 4:15 p.m. Circle the time that makes the sentence true.\nLaura arrived at the library at:\n_______ : _______ P.M.\n\nExplanation:\nSTEP 1 Find the time on the number line when Laura arrived at the library.\nSTEP 2 Count back on the number line to subtract the elapsed time. Draw and label the jumps to show the minutes.\nSTEP 3 Write the times below the number line.\nYou jumped back to 40+15=55 minutes\nTherefore Laura arrived at the library at 3:20 P.M.\n\n### Problem Solving Time Intervals – Page No. 589\n\nSolve each problem. Show your work.\n\nQuestion 1.\nHannah wants to meet her friends downtown. Before leaving home, she does chores for 60 minutes and eats lunch for 20 minutes. The walk downtown takes 15 minutes. Hannah starts her chores at 11:45 A.M. At what time does she meet her friends?", null, "Explanation:\nSTEP 1 Find the ending time on the clock, the time at which Hannah meets her friends.\nSTEP 2 Count back by fives for the elapsed time of 95 minutes. Write the missing counting numbers next to the clock.\nSo, the ending time is 1:20 P.M.\n\nQuestion 2.\nKatie practiced the flute for 45 minutes. Then she ate a snack for 15 minutes. Next, she watched television for 30 minutes, until 6:00 P.M. At what time did Katie start practicing the flute?\n_______ : ________ (A.M./P.M.)\n\nExplanation:\nSTEP 1 Find the starting time on the clock.\nSTEP 2 Count back by fives for the elapsed time of 90 minutes. Write the missing counting numbers next to the clock.\nSo, the starting time is 4:30 P.M.\n\nQuestion 3.\nNick gets out of school at 2:25 P.M. He has a 15-minute ride home on the bus. Next, he goes on a 30-minute bike ride. Then he spends 55 minutes doing homework. At what time does Nick finish his homework?\n_______ : ________ (A.M./P.M.)\n\nExplanation:\nSTEP 1 Find the ending time on the clock, the time at which Nick finish his work.\nSTEP 2 Count back by fives for the elapsed time of 100 minutes. Write the missing counting numbers next to the clock.\nSo, the ending time is 4:05 P.M.\n\nQuestion 4.\nThe third-grade class is going on a field trip by bus to the museum. The bus leaves the school at 9:45 A.M. The bus ride takes 47 minutes. At what time does the bus arrive at the museum?\n_______ : ________ (A.M./P.M.)\n\nExplanation:\nSTEP 1 Find the ending time on the clock, the time at which the students arrive at the museum.\nSTEP 2 Count back by fives for the elapsed time of 47 minutes. Write the missing counting numbers next to the clock.\nSo, the ending time is 10:32 A.M.\n\n### Problem Solving Time Intervals – Page No. 590\n\nLesson Check\n\nQuestion 1.\nGloria went to the mall and spent 50 minutes shopping. Then she had lunch for 30 minutes. If Gloria arrived at the mall at 11:00 A.M., at what time did she finish lunch?\nOptions:\na. 11:30 A.M.\nb. 11:50 A.M.\nc. 12:20 P.M.\nd. 12:30 P.M.\n\nExplanation:\nSTEP 1 Find the ending time on the clock, the time at which Gloria finish her lunch.\nSTEP 2 Count back by fives for the elapsed time of 80 minutes. Write the missing counting numbers next to the clock.\nSo, the ending time is 12:20 P.M.\n\nQuestion 2.\nThe ball game begins at 2:00 P.M. It takes Ying 30 minutes to get to the ballpark. At what time should Ying leave home to get to the game 30 minutes before it starts?\nOptions:\na. 12:30 P.M.\nb. 1:00 P.M.\nc. 1:30 P.M.\nd. 3:00 P.M.\n\nExplanation:\nSTEP 1 Find the starting time on the clock.\nSTEP 2 Count back by fives for the elapsed time of 60 minutes. Write the missing counting numbers next to the clock.\nSo, the starting time is 1:00 P.M.\n\nSpiral Review\n\nQuestion 3.\nWhich lists the fractions in order from least to greatest?\nOptions:\na. $$\\frac{2}{8}$$, $$\\frac{2}{4}$$, $$\\frac{2}{6}$$\nb. $$\\frac{2}{4}$$, $$\\frac{2}{8}$$, $$\\frac{2}{6}$$\nc. $$\\frac{2}{8}$$, $$\\frac{2}{6}$$, $$\\frac{2}{4}$$\nd. $$\\frac{2}{4}$$, $$\\frac{2}{6}$$, $$\\frac{2}{8}$$\n\nAnswer: c. $$\\frac{2}{8}$$, $$\\frac{2}{6}$$, $$\\frac{2}{4}$$\n\nExplanation:\nWhen the numerators are the same, think about the  pieces to compare and order fractions. So, the order from least to greatest  is $$\\frac{2}{8}$$, $$\\frac{2}{6}$$, $$\\frac{2}{4}$$\n\nQuestion 4.\nFind the unknown factor.\n6 × ■ = 36\nOptions:\na. 4\nb. 6\nc. 7\nd. 8\n\nExplanation:\nCount how many rows of 36 counters there are.\nThere are  6 rows of 36 counters. The unknown factor is 6 (c0lumns). n = 6\n6× 6 = 36\n\nQuestion 5.\nThere were 405 books on the library shelf. Some books were checked out. Now there are 215 books left on the shelf. How many books were checked out?\nOptions:\na. 620\nb. 220\nc. 210\nd. 190\n\nExplanation:\nSTEP1 Let there be 405 books in a shelf of Library.\nSTEP2 Some books were checked out.\nSTEP3 Number of books left in the shelf are 215.\nSTEP4 Take a box filled with 405 counters.\nSTEP5 Place 215 counters in another box from the initial 405 counters.\nSTEP6 The number of counters left in the first box is equal to the books checked out.\n405-215=190\n\nQuestion 6.\nSavannah has 48 photos. She places 8 photos on each page of her photo album. How many pages in the album does she use?\nOptions:\na. 5\nb. 6\nc. 7\nd. 9\n\nExplanation:\nSTEP 1 Draw 48 counters.\nSTEP 2 Make a group of 8 counters by drawing a circle around them. Continue circling groups of 8 until all 48 counters are in groups.\nThere are 6 groups of 48 counters.\n\n### Mid -Chapter Checkpoint – Page No. 591\n\nVocabulary\n\nChoose the best term from the box.", null, "Question 1.\nIn one __________, the minute hand moves from one mark to the next on a clock.\n\nExplanation: The minute hand moves after every 60 seconds = 1 Minute to the next mark on the clock.\n\nQuestion 2.\nThe times afternoon and before midnight are written with __________ .\n__________\n\nExplanation:\nThe times afternoon and before midnight are written with P.M.\n\nConcepts and Skills\n\nWrite the time for the activity. Use A.M. or P.M.\n\nQuestion 3.\nplay ball", null, "______ : ______ ______ (A.M./P.M.)\n\nExplanation:\nThe times afternoon and before midnight is written with P.M. Since we play ball in the evening, the time is considered to be P.M.\n\nQuestion 4.\neat breakfast", null, "______ : ______ ______ (A.M./P.M.)\n\nExplanation:\nThe times after midnight and before noon is written with A.M. Since we have our breakfast in the morning, the time is considered to be  A.M.\n\nQuestion 5.\ndo homework", null, "______ : ______ ______ (A.M./P.M.)\n\nExplanation:\nThe times afternoon and before midnight is written with P.M. Since we do our homework in the evening, the time is considered to be P.M.\n\nQuestion 6.\nsleep", null, "______ : ______ ______ (A.M./P.M.)\n\nExplanation:\nThe times afternoon and before midnight is written with P.M. Since we go to sleep in the night hours, the time is considered to be P.M.\n\nFind the elapsed time.\n\nQuestion 7.\nStart: 10:05 A.M. End: 10:50 A.M.", null, "_______ minutes\n\nExplanation:\nSTEP 1 Find10:05 A.M. on the number line. Count on from 10:05 A.M. to 10:50 A.M. Draw marks and record the times on the number line. Then draw and label the jumps.\nSTEP 2 Add to find the total minutes from 10:05 A.M. to 10:50 A.M. From 10:05 A.M. to 10:50 A.M. is 45 minutes.\nSo, the elapsed time is 45 minutes\n\nQuestion 8.\nStart: 5:30 P.M.\nEnd: 5:49 P.M.", null, "_______ minutes\n\nExplanation:\nSTEP 1 Find the starting time on the clock that is 5:30 P.M.\nSTEP 2 Count the minutes by counting on by fives and ones to 5:49 P.M. Write the missing counting numbers next to the clock.\nSo, the elapsed time is 19 minutes.\n\nFind the starting time or the ending time.\n\nQuestion 9.\nElapsed time: 50 minutes\nEnding time: 9:05 A.M.", null, "Starting time: ______ : ______ A.M.\n\nExplanation:\nSTEP 1 Find the starting time on the number line\nSTEP 2 Count back on the number line to subtract the elapsed time. Draw and label the jumps to show the minutes.\nSTEP 3 Write the times below the number line.\nYou jumped back to 50 minutes\nSo, the starting time is 8:15 A.M.\n\nQuestion 10.\nStarting time: 2:46 P.M.\nElapsed time: 15 minutes", null, "Ending time: ______ : ______ P.M.\n\nExplanation:\nSTEP 1 Find the ending time on the clock\nSTEP 2 Count back by fives for the elapsed time of 15 minutes. Write the missing counting numbers next to the clock.\nSo, the ending time is 3:01 P.M.\n\n### Mid -Chapter Checkpoint – Page No. 592\n\nQuestion 11.\nVeronica started walking to school at 7:45 A.M. She 0 arrived at school 23 minutes later. At what time did Veronica arrive at school?\n______ : ______ ______\n\nExplanation:\nSTEP 1 Find the ending time on the number line.\nSTEP 2 Count forward on the number line to add the elapsed time. Draw and label the jumps to show the minutes.\nSTEP 3 Write the times below the number line.\nThe jumps end at 8:08 A.M.\nSo, the ending time is 8:08 A.M.\n\nQuestion 12.\nThe clock shows the time the art class ends. At what time does it end? If the class started 37 minutes before the time shown, at what time did the class start?", null, "Type below:\n____________\n\nAnswer: Time at which the class ends= 1:57 P.M.\nTime at which the class starts= 1:20 P.M.\n\nExplanation:\nCount by fives and ones from the 12 on the clock back to where the minute hand is pointing. Write the missing counting numbers next to the clock.\nWrite: 1:57 P.M.\nThree minutes before two\nSTEP 1 Find the starting time on the clock.\nSTEP 2 Count back by fives for the elapsed time of 37 minutes. Write the missing counting numbers next to the clock.\nSo, the starting time is 1:20 P.M.\n\nQuestion 13.\nMatt went to his friend’s house. He arrived at 5:10 P.M. He left at 5:37 P.M. How long was Matt at his friend’s house?\n______ minutes\n\nExplanation:\nSTEP 1 Find 5:10 P.M. on the number line. Count on from 5:10 P.M. to 5:37 P.M. Draw marks and record the times on the number line. Then draw and label the jumps.\nSTEP 2 Add to find the total minutes from 5:10 P.M. to 5:37 P.M. From 5:10 P.M. to 5:37 P.M. is 27 minutes.\nSo, the elapsed time is 27 minutes\n\nQuestion 14.\nBrenda’s train leaves at 7:30 A.M. She needs to arrive 10 minutes early to buy her ticket. It takes her 20 minutes to get to the train station. At what time should Brenda leave her house?\n______ : ______ A.M.\n\nExplanation:\nSTEP 1 Find the starting time on the number line\nSTEP 2 Count back on the number line to subtract the elapsed time. Draw and label the jumps to show the minutes.\nSTEP 3 Write the times below the number line.\nYou jumped back to 30 minutes\nSo, the starting time is 7:00 A.M.\n\nQuestion 15.\nWrite the time you get home from school.\nType below:\n____________\n\nExplanation:\nThe times afternoon and before midnight is written with P.M. Since we get home from school in the evening, the time is considered to be P.M.\n\n### Measure Length – Page No. 597\n\nMeasure the length to the nearest half inch.\n\nQuestion 1.", null, "Explanation:\nLine up the left end of the glue stick with the zero mark on the ruler.\nThe right end of the glue stick is between the half-inch marks for 1 and 2.\nThe mark that is closest to the right end of the glue stick is for one and a half inches.\nSo, the length of the glue stick to the nearest half-inch is 1 1/2 inches.\n\nQuestion 2.", null, "______ inches\n\nExplanation:\nLine up the left end of the glue stick with the zero mark on the ruler.\nThe right end of the glue stick is between the half-inch marks for 1 and 3.\nThe mark that is closest to the right end of the glue stick is for three inches.\nSo, the length of the glue stick to the nearest half-inch is 3 inches.\n\nQuestion 3.", null, "______ $$\\frac{□}{□}$$ inches\n\nExplanation:\nLine up the left end of the glue stick with the zero mark on the ruler.\nThe right end of the glue stick is between the half-inch marks for4 and 5.\nThe mark that is closest to the right end of the glue stick is for four and a half  inches.\nSo, the length of the glue stick to the nearest half-inch is 4 1/2 inches.\n\nMeasure the length to the nearest fourth inch.\n\nQuestion 4.", null, "______ $$\\frac{□}{□}$$ inches\n\nExplanation:\nLine up the left end of the paper clip with the zero mark on the ruler.\nThe right end of the paper clip is between the fourth-inch marks for 1 and2.\nThe mark that is closest to the right end of the paper clip is for 1 1/4 inches.\nSo, the length of the paper clip to the nearest fourth inch is 1 1/4 inches.\n\nQuestion 5.", null, "______ $$\\frac{□}{□}$$ inches\n\nExplanation:\nLine up the left end of the paper clip with the zero mark on the ruler.\nThe right end of the paper clip is between the fourth-inch marks for 2 and 3.\nThe mark that is closest to the right end of the paper clip is for 2 3/4 inches.\nSo, the length of the paper clip to the nearest fourth inch is 2 3/4 inches.\n\nQuestion 6.", null, "$$\\frac{□}{□}$$ inch\n\nExplanation:\nLine up the left end of the paper clip with the zero mark on the ruler.\nThe right end of the paper clip is between the fourth-inch marks for 0 and1.\nThe mark that is closest to the right end of the paper clip is for 3/4 inches.\nSo, the length of the paper clip to the nearest fourth inch is 3/4 inches.\n\nQuestion 7.", null, "______ inches\n\nExplanation:\nLine up the left end of the paper clip with the zero mark on the ruler.\nThe right end of the paper clip is between the fourth-inch marks for 1 and2.\nThe mark that is closest to the right end of the paper clip is for 2 inches.\nSo, the length of the paper clip to the nearest fourth inch is 2 inches.\n\nProblem Solving\n\nUse a separate sheet of paper for 8.\n\nQuestion 8.\nDraw 8 lines that are between 1 inch and 3 inches long. Measure each line to the nearest fourth inch, and make a line plot.\n\nExplanation:\n\nQuestion 9.\nThe tail on Alex’s dog is 5$$\\frac{1}{4}$$ inches long. This length is between which two inch-marks on a ruler?\nBetween ______ and ______ inch-marks\n\nAnswer: Between 5 and 6 inches\n\nExplanation:\nLine up the left end of the paper clip with the zero mark on the ruler.\nThe right end of the paper clip is between the fourth-inch marks for 5 and 6.\nThe mark that is closest to the right end of the paper clip is for 5 1/4 inches.\nSo, the length of the paper clip to the nearest fourth inch is 5 1/4 inches.\n\n### Measure Length – Page No. 598\n\nLesson Check\n\nQuestion 1.\nWhat is the length of the eraser to the nearest half inch?", null, "Options:\na. $$\\frac{1}{2}$$ inch\nb. 1 inch\nc. 1 $$\\frac{1}{2}$$ inch\nd. 2 inch\n\nAnswer: c. 1 $$\\frac{1}{2}$$ inch\n\nExplanation:\nLine up the left end of the glue stick with the zero mark on the ruler.\nThe right end of the glue stick is between the half-inch marks for 1 and 2.\nThe mark that is closest to the right end of the glue stick is for one and a half inches.\nSo, the length of the glue stick to the nearest half-inch is 1 1/2 inches.\n\nQuestion 2.\nWhat is the length of the leaf to the nearest fourth inch?", null, "Options:\na. 1 $$\\frac{1}{2}$$ inches\nb. 1 $$\\frac{3}{4}$$ inches\nc. 2 inches\nd. 2 $$\\frac{1}{4}$$ inches\n\nExplanation:\nLine up the left end of the paper clip with the zero mark on the ruler.\nThe right end of the paper clip is between the fourth-inch marks for 1 and 2.\nThe mark that is closest to the right end of the paper clip is for 2 inches.\nSo, the length of the paper clip to the nearest fourth inch is 2 inches.\n\nSpiral Review\n\nQuestion 3.\nWhich equation is NOT included in the same set of related facts as 6 × 8 = 48?\nOptions:\na. 8 × 6 = 48\nb. 8 × 8 = 64\nc. 48 ÷ 6 = 8\nd. 48 × 8 = 6\n\nAnswer: b. 8 × 8 = 64\n\nExplanation:\nSTEP 1 Use 8 tiles to make an array with 6 equal rows.\nDraw the rest of the tiles.\nThere are 8 tiles in each row.\nWrite a division equation for the array using the total number of tiles as the dividend and the number of rows as the divisor.\n48÷ 6= 8\nSo, 48 ÷ 8 = 6, 8 × 6 = 48,\nand 6 ×8  = 48 are related facts.\n\nQuestion 4.\nBrooke says there are 49 days until July 4. There are 7 days in a week. In how many weeks will it be July 4?\nOptions:\na. 9 weeks\nb. 8 weeks\nc. 7 weeks\nd. 6 weeks\n\nExplanation:\nDraw 1 tile in each of 7 rows.\nContinue drawing 1 tile in each of the 7 rows until all 49 tiles are drawn.\nCount the number of tiles in each row.\nThere are 7 tiles in each row.\nSo, there are 7 weeks till July 4.\n\nQuestion 5.\nIt is 20 minutes before 8:00 in the morning. Which is the correct way to write that time?\nOptions:\na. 7:40 A.M.\nb. 7:40 P.M.\nc. 8:20 A.M.\nd. 8:40 A.M.\n\nExplanation:\nCount by fives and ones from the 12 on the clock back to where the minute hand is pointing. Write the missing counting numbers next to the clock.\nWrite: 7:40 A.M.\nTwenty minutes before eight\n\nQuestion 6.\nMarcy played the piano for 45 minutes. She stopped playing at 4:15 P.M. At what time did she start playing the piano?\nOptions:\na. 3:00 P.M.\nb. 3:30 P.M.\nc. 4:45 P.M.\nd. 5:00 P.M.\n\nExplanation:\nSTEP 1 Find the starting time on the number line\nSTEP 2 Count back on the number line to subtract the elapsed time. Draw and label the jumps to show the minutes.\nSTEP 3 Write the times below the number line.\nYou jumped back to 45 minutes\nSo, the starting time is 3:30 P.M.\n\n### Estimate and Measure Liquid Volume – Page No. 603\n\nEstimate how much liquid volume there will be when the container is filled. Write more than 1 liter, about 1 liter, or less than 1 liter.\n\nQuestion 1.\nlarge milk container", null, "Explanation:\nPour 1 liter of milk into one of the large containers. Repeat until the container is full. Record the number of liters you poured. This proves that the container can hold more than one liter.\n\nQuestion 2.\nsmall milk container", null, "____________\n\nExplanation:\nPour 1 liter of milk into one of the small containers. There is some quantity of milk leftover in the pack of one liter, this proves that the container can hold less than one liter.\n\nQuestion 3.\nwater bottle", null, "____________\n\nExplanation:\nA water bottle holds about 1 liter.\n\nQuestion 4.\nspoonful of water", null, "____________\n\nExplanation:\nPour 100 ml of water into a spoon. The excess water flow out, this proves that the spoon can hold less than one liter.\n\nQuestion 5.\nbathtub filled halfway", null, "____________\n\nExplanation:\nPour 1 liter of water into the bathtub. Repeat until the tub is full. Record the number of liters you poured.This proves that the container can hold more than one liter.\n\nQuestion 6.\nfilled eyedropper", null, "____________\n\nExplanation:\nPlace the eyedropper in a bottle of the quantity, one liter. The eyedropper picks only few drops of the water, this proves that the eyedropper can hold less than one liter.\n\nProblem Solving\n\nUse the pictures for 7–8. Alan pours water into four glasses that are the same size.", null, "Question 7.\nWhich glass has the most amount of water?\n_______\n\nExplanation:\nIt can be observed that glass D is almost completely filled. So, glass D most amount of water.\n\nQuestion 8.\nWhich glass has the least amount of water?\n_______\n\nExplanation:\nIt can be observed that glass A contains the least amount of water. So, glass A has the least amount of water.\n\n### Estimate and Measure Liquid Volume – Page No. 604\n\nLesson Check\n\nQuestion 1.\nFelicia filled the bathroom sink with water. About how much water does she put in the sink?\nOptions:\nb. more than 1 liter\nc. a little less than 1 liter\nd. much less than 1 liter\n\nAnswer: b. more than 1 liter\n\nExplanation:\nPour one liter of water into the sink. By repeating the same process we can say that, a sink can hold more than one liter of water. Hence it contains more than one liter of water.\n\nQuestion 2.\nKyle needed about 1 liter of water to fill a container. Which container did Kyle most likely fill?\nOptions:\na. a small glass\nb. a spoon\nc. a large pail\nd. a vase\n\nExplanation:\nA vase holds about 1 liter while a small glass, a spoon can only hold small quantity of water. A large pail holds water more than one liter.\n\nSpiral Review\n\nQuestion 3.\nCecil had 6 ice cubes. He put 1 ice cube in each glass. In how many glasses did Cecil put ice cubes?\nOptions:\na. 6\nb. 5\nc. 1\nd. 0\n\nExplanation:\nIf there is the same number of ice cubes and glasses, then 1 ice cube goes in each glass. Since each ice cube is to be placed in each glass.\n6÷ 1 = 6\nNumber of glasses=6\n\nQuestion 4.\nJuan has 12 muffins. He puts $$\\frac{1}{4}$$ of the muffins in a bag. How many muffins does Juan put in the bag?", null, "Options:\na. 2\nb. 3\nc. 4\nd. 5\n\nExplanation:\nPut 12 counters(muffins) on the table.\nSince you want to find 1/4 of the group, there should be equal groups of muffins.\nCircle one of the groups to show 1/4th part of muffins.Then count the number of counters in that group.\nThere are 3 counters in 1 group. 1/4 of 12 = 3\nSo,  Juan has put 3 muffins in his bag.\n\nQuestion 5.\nWhich is one way to read the time shown on the clock?", null, "Options:\na. 4 minutes before 7\nb. 26 minutes before 11\nc. 54 minutes after 6\nd. 56 minutes after 7\n\nAnswer: a. 4 minutes before 7\n\nExplanation:\nCount by fives and ones from the 12 on the clock back to where the minute hand is pointing. Write the missing counting numbers next to the clock.\nWrite: 6:56\nfour minutes before seven\n\nQuestion 6.\nJulianne drew the line segment below. Use your ruler to measure the segment to the nearest fourth inch.", null, "Options:\na. $$\\frac{3}{4}$$ inch\nb. 1 $$\\frac{1}{4}$$ inches\nc. 1 $$\\frac{1}{2}$$ inches\nd. 1 $$\\frac{3}{4}$$ inches\n\nAnswer: d. 1 $$\\frac{3}{4}$$ inches\n\nExplanation:\nLine up the left end of the paper clip with the zero mark on the ruler.\nThe right end of the paper clip is between the fourth-inch marks for 1 and2.\nThe mark that is closest to the right end of the paper clip is for 1 $$\\frac{3}{4}$$ inches.\nSo, the length of the paper clip to the nearest fourth inch is 1 $$\\frac{3}{4}$$ inches.\n\n### Estimate and Measure Liquid Volume – Page No. 609\n\nChoose the unit you would use to measure the mass. Write gram or kilogram.\n\nQuestion 1.\nCD", null, "gram\n\nExplanation:\nThe gram (g) is the basic metric unit for measuring mass or the amount of matter in an object. Mass can also be measured by using the metric unit kilogram (kg).\nA small CD has a mass measured in grams.\n\nQuestion 2.\nboy", null, "________\n\nExplanation:\nA boy’s mass must be measured in kilograms, the metric unit. Since gram is the basic unit of mass and this measuring mass is used to measure small items such as CDs, paper pins, ice cream sticks etc.\n\nQuestion 3.\nbag of sugar", null, "________\n\nExplanation:\nA bag contains more than or equal to 1000 grams of measuring mass.\n1000grams=1kilogram\nHence we can say that the sugar is measured in kilograms.\n\nQuestion 4.\nlion", null, "________\n\nExplanation: Lion is a huge animal with muscles and hard bones. Hence, Lion is measured in kilograms.\n\nQuestion 5.\npaper clip", null, "________\n\nExplanation:\nA paper clip is light measured clip its mass is about 1 gram.\n\nQuestion 6.\nempty plastic bottle", null, "________\n\nExplanation:\nThe amount of water in the bottle justifies the mass/volume of the bottle. Since the bottle is empty it is only measured in grams.\n\nCompare the masses of the objects. Write is less than,\nis the same as, or is more than.\n\nQuestion 7.", null, "The mass of the candle ________ the mass of the light bulb.\n\nExplanation:\nBoth candle and bulb measure in grams comparatively candle seems to have more mass than bulb.\nTherefore we can say that mass of candle is more than a mass of bulb.\n\nQuestion 8.", null, "The mass of the watch ________ the mass of the necklace.\n\nExplanation:\nBoth the watch and the necklace seem to have same amount of mass in grams.\n\nProblem Solving\n\nQuestion 9.\nA red ball has a mass that is less than 1 kilogram. A blue ball has a mass of 1 kilogram. Is the mass of the blue ball more than or less than the mass of the red ball?\nThe mass of the blue ball is ________ the mass of the red ball\n\nExplanation:\nBlue ball has a mass of 1 kilogram which more than that of the red ball because it has a mass less than 1 kilogram.\n\nQuestion 10.\nBrock’s dog is a collie. To find the mass of his dog, should Brock use grams or kilograms?\n__________\n\nExplanation:\n\n### Estimate and Measure Liquid Volume – Page No. 610\n\nLesson Check\n\nQuestion 1.\nWhich unit of measure would you use to measure the mass of a grape?\nOptions:\na. gram\nb. inch\nc. kilogram\nd. meter\n\nExplanation:\nA bunch of grapes are measured in kilograms while a single grape is measured in grams.\n\nQuestion 2.\nElsie wants to find the mass of her pony. Which unit should she use?\nOptions:\na. gram\nb. liter\nc. kilogram\nd. centimeter\n\nExplanation:\nThe mass of the pony is calculated in kilogram because the cluster of grams is called as kilogram.\n\nSpiral Review\n\nQuestion 3.\nMarsie blew up 24 balloons. She tied the balloons together in groups of 4. How many groups did Marsie make?\nOptions:\na. 5\nb. 6\nc. 7\nd. 8\n\nExplanation:\nSTEP 1 Let 24 counters be ballons.\nSTEP 2 Make a group of 4 counters tied together into a group. Continue making groups of 4 until all 24 counters are in groups.\nThere are 6 groups of 24 counters.\n\nQuestion 4.\nClark used the order of operations to find the unknown number in 15 − 12 ÷ 3 = n. What is the value of the unknown number?\nOptions:\na. 1\nb. 6\nc. 9\nd. 11\n\nExplanation:\nSTEP 1 Let there be 12 counters in a box.\nSTEP 2 Make a group of 3 counters by drawing a circle around them. Continue circling groups of 3 until all 12 counters are in groups.\nThere are 4 groups of 12 counters.\nSTEP3 There are 15 counters in a box\nSTEP4 Remove 4 counters from the box, then only 11 counters are left in the box.\n\nUse the pictures for 5–6. Ralph pours juice into four bottles that are the same size.", null, "Question 5.\nWhich bottle has the most amount of juice?\nOptions:\na. Bottle A\nb. Bottle B\nc. Bottle C\nd. Bottle D\n\nExplanation:\nComparatively in all the four bottles bottle A is almost completely filled. Therefore bottle A has the most amount of juice.\n\nQuestion 6.\nWhich bottle has the least amount of juice?\nOptions:\na. Bottle A\nb. Bottle B\nc. Bottle C\nd. Bottle D\n\nExplanation:\nComparatively in all the four bottles bottle D has the least amount of juice. Therefore bottle A has the least amount of juice.\n\n### Solve Problems About Liquid Volume and Mass – Page No. 615\n\nWrite an equation and solve the problem.\n\nQuestion 1.\nLuis was served 145 grams of meat and 217 grams of vegetables at a meal. What was the total mass of the meat and the vegetables?\nThink: Add to find how much in all.\n145 + 217 = _______ grams\n\nAnswer: 145 + 217 = 362 grams\n\nExplanation:\nSTEP 1 Place 145 grams mass of meat.\nSTEP 2 Place 217 grams mass of vegetables at a meal .\nSTEP 3 To find the total mass of the meat and the vegetables is 362 grams.\nSo, the total mass of the meat and the vegetables is 362 grams.\n\nQuestion 2.\nThe gas tank of a riding mower holds 5 liters of gas. How many 5-liter gas tanks can you fill from a full 20-liter gas can?\n_______ ÷ _______ = _______ 5-liter gas tanks\n\nAnswer: 20 ÷ 5 = 4 gas tanks\n\nExplanation:\nSTEP1 A single gas tank of a riding mower holds 5 liters of gas.\nSTEP2 Quantity of gas given to fill in the tanks is 20 liters.\nSTEP3 To calculate the number of tanks, we should make groups of 5 liters of gas such that the total quantity of 20 liters is completed\nSTEP4 The number of gas tanks is the number of counters in which the 20 liter of gas is filled.\nTherefore the number of tanks are 4\n\nQuestion 3.\nTo make a lemon-lime drink, Mac mixed 4 liters of lemonade with 2 liters of limeade. How much lemon-lime drink did Mac make?\n_______ + _______ = _______ liters lemon-lime\n\nAnswer: 4 + 2 = 6 liters\n\nExplanation:\nSTEP 1 Place 4 liters of lemonade for preparing the lemon-lime drink.\nSTEP 2 Place 2 liters of limeade for preparing the lemon-lime drink.\nSTEP 3Mass of lemon-lime drink made by Mac is 4+2=6 liters.\nSo, the total mass of the lemon-lime drink made is 6 liters.\n\nQuestion 4.\nA nickel has a mass of 5 grams. There are 40 nickels in a roll of nickels. What is the mass of a roll of nickels?\n_______ × _______ = _______grams\n\nAnswer: 5 × 40 = 200 grams\n\nExplanation:\nSTEP1 Given that there are 40 nickels in a roll of nickels.\nSTEP2 Each nickel has a mass of 5 grams.\nSTEP3 Place the counters(nickels) in rows.\nSTEP4 Find the mass of a roll of nickels.\nSTEP5 Mass of a roll of nickels=5×40\nSTEP6 Place 5 boxes each containing 40 counters. Add the counters in the boxes, total number of counters is equal to 200.\n\nQuestion 5.\nFour families share a basket of 16 kilograms of apples equally. How many kilograms of apples does each family get?\n_______ ÷ _______ = _______ kilograms\n\nAnswer: 20 ÷ 5 = 4\n\nExplanation:\nSTEP1 Total number of apples =16\nSTEP2 Number of families =4\nSTEP3 Number of apples each family shared=16÷4\n=4 apples\n\nQuestion 6.\nFor a party, Julia made 12 liters of fruit punch. There were 3 liters of fruit punch left after the party. How much fruit punch did the people drink at the party?\n_______ – _______ = _______ liters\n\nAnswer: 12 – 3 = 9 liters\n\nExplanation:\nSTEP1 Quantity of fruit punch made by Julia=12 liters\nSTEP2 Quantity of fruit punch left after the party=3 liters\nSTEP3 Quantity of fruit punch the people drank in the party=12-3 = 9 liters\n\nProblem Solving\n\nQuestion 7.\nZoe’s fish tank holds 27 liters of water. She uses a 3-liter container to fill the tank. How many times does she have to fill the 3-liter container in order to fill her fish tank?\n_______ times\n\nExplanation:\nSTEP1 Quantity of water Zoe’s fish tank holds=27 liters\nSTEP2  Zoe has a 3-liter container to fill the tank.\nSTEP3 Total number of times  she has to fill the 3-liter container in order to fill her fish tank=27÷3=9 times\n\nQuestion 8.\nAdrian’s backpack has a mass of 15 kilograms. Theresa’s backpack has a mass of 8 kilograms. What is the total mass of both backpacks?\n15 + 8 = _______ kilograms\n\nExplanation:\nSTEP1 Mass of Adrian’s backpack = 15 kilograms\nSTEP2 Mass of Theresa’s backpack= 8 kilograms\nSTEP3  Total mass of both backpacks=15+8=23 kilograms\n\n### Solve Problems About Liquid Volume and Mass – Page No. 616\n\nLesson Check\n\nQuestion 1.\nMickey’s beagle has a mass of 15 kilograms. His dachshund has a mass of 13 kilograms. What is the combined mass of the two dogs?\nOptions:\na. 2 kilograms\nb. 18 kilograms\nc. 23 kilograms\nd. 28 kilograms\n\nExplanation:\nMickey has 2 dogs beagle and dachshund\nSTEP1 Mass of Beagle=15 kilograms\nSTEP2 Mass of dachshund= 13 kilograms\nSTEP3 Combined mass of the two dogs= 15+13=28 kilograms\n\nQuestion 2.\nLois put 8 liters of water in a bucket for her pony. At the end of the day, there were 2 liters of water left. How much water did the pony drink?\nOptions:\na. 4 liters\nb. 6 liters\nc. 10 liters\nd. 16 liters\n\nExplanation:\nSTEP1 Quantity of the water put in the bucket by Lois=8 liters\nSTEP2 Quantity of the water left in the bucket=2 liters\nSTEP3 Quantity of the water drank by pony= 8-2 = 6 liters\n\nSpiral Review\n\nQuestion 3.\nJosiah has 3 packs of toy animals. Each pack has the same number of animals. Josiah gives 6 animals to his sister Stephanie. Then Josiah has 9 animals left. How many animals were in each pack?\nOptions:\na. 1\nb. 3\nc. 5\nd. 6\n\nExplanation:\nSTEP1 Number of packs Josiah has =3\nSTEP2 Number of animals given by Josiah to his sister =6\nSTEP3 Total number of animals = 6+9=15\nAccording to the problem,\nEach pack has the same number of animals.\n3× ■ = 15\nCount how many rows of 15 counters there are. There are  3 rows of 15 counters. The unknown factor is 5 (c0lumns). n = 5\n3 × 5 = 15\n\nQuestion 4.\nTom jogged $$\\frac{3}{10}$$ mile, Betsy jogged $$\\frac{5}{10}$$ mile, and Sue jogged $$\\frac{2}{10}$$ mile. Who jogged a longer distance than $$\\frac{4}{10}$$ mile?\nOptions:\na. Betsy\nb. Sue\nc. Tom\nd. None\n\nExplanation:\nWhen the numerators are the same, think about the of the pieces to compare and order fractions. So, the order from greatest to least gives, the greatest fraction and the person who jogged a longer distance than $$\\frac{4}{10}$$ mile.\n$$\\frac{5}{10}$$>$$\\frac{3}{10}$$>$$\\frac{2}{10}$$\nTherefore Betsy jogged a longer distance than $$\\frac{4}{10}$$ mile.\n\nQuestion 5.\nBob started mowing at 9:55 A.M. It took him 25 minutes to mow the front yard and 45 minutes to mow the back yard. At what time did Bob finish mowing?\nOptions:\na. 10:20 A.M.\nb. 10:55 A.M.\nc. 11:05 A.M.\nd. 11:20 A.M.\n\nExplanation:\nSTEP1 Time Bob spent in mowing=25+45=70 minutes\nSTEP2 Finding the time at which Bob finished mowing\nSTEP3 Find the ending time on the number line.\nSTEP4 Count forward on the number line to add the elapsed time. Draw and label the jumps to show the minutes.\nSTEP5 Write the times below the number line.\nThe jumps end at 11:05 A.M.\nSo, the ending time is 11:05 A.M.\n\nQuestion 6.\nJuliana wants to find the mass of a watermelon. Which unit should she use?\nOptions:\na. gram\nb. kilogram\nc. liter\nd. meter\n\nExplanation:\nMass is measured in grams and kilograms small quantities like paper clips, pins are measured in grams while large quantities are measured in kilograms. Therefore watermelon is measured in kilogram.\n\n### Review/Test – Page No. 617\n\nQuestion 1.\nYul and Sarah’s art class started at 11:25 a.m. The class lasted 30 minutes. Yul left when the class was done. Sarah stayed an extra 5 minutes to talk with the teacher and then left.\nWrite the time that each student left. Explain how you found each time.\nYul: _______ A.M.\nSarah: ______ P.M.\n\nSarah: 12:05 P.M.\n\nExplanation:\nYul:\nSTEP 1 Find the ending time on the number line.\nSTEP 2 Count forward on the number line to add the elapsed time, 30 minutes. Draw and label the jumps to show the minutes.\nSTEP 3 Write the times below the number line.\nThe jumps end at 11:55 A.M.\nSo, the ending time is 11:55 A.M.\nSarah:\nSTEP 1 Find the starting time on the number line\nSTEP 2 Count back on the number line to subtract the elapsed time, 30+5=35 minutes. Draw and label the jumps to show the minutes.\nSTEP 3 Write the times below the number line.\nThe jumps end at  12:05 P.M.\nSo, the starting time is 12:05 P.M.\n\nQuestion 2.\nJulio measured an object that he found. It was about 3/4 inch wide.\nFor numbers 2a–2d, choose Yes or No to tell whether the object could be the one Julio measured.\na.", null, "i. yes\nii. no\n\nQuestion 2.\nb.", null, "i. yes\nii. no\n\nExplanation:\nThe mark that is closest to the right end of the stamp is for 1 inch. So, the length of the stamp to the nearest fourth inch is 3/4 inches. Therefore the stamp is about 3/4 inch wide.\n\nQuestion 2.\nc.", null, "i. yes\nii. no\n\nQuestion 2.\nd.", null, "i. yes\nii. no\n\n### Review/Test – Page No. 618\n\nQuestion 3.\nDina started swimming at 3:38 p.m. She swam until 4:15 p.m. How long did Dina swim?\n________ minutes\n\nExplanation:\nSTEP 1 Find 3:38 P.M. on the number line. Count on from 3:38 P.M. to 4:15 P.M. Draw marks and record the times on the number line. Then draw and label the jumps.\nSTEP 2 Add to find the total minutes from 3:38 P.M. to 4:15 p.M. From 3:38 P.M. to 4:15 P.M. is 37 minutes.\nSo, the elapsed time is 37 minutes\n\nQuestion 4.\nRita’s class begins social studies at ten minutes before one in the afternoon. At what time does Rita’s class begin social studies? Circle a time that makes the sentence true.\nRita’s class begins social studies at", null, "__________\n\nExplanation:\nRita’s class begins social studies at ten minutes before one in the afternoon. So, the time at which the class begins is 10 minutes subtracted from 1:00 P.M. is 12:50 P.M.\n\nQuestion 5.\nSelect the objects with a mass greater than 1 kilogram.\nMark all that apply.\nOptions:\na. bicycle\nb. pen\nc. eraser\nd. math book\n\nAnswer: a. bicycle, has a mass greater than 1 kilogram.\n\nExplanation:\nMass is measured in grams and kilograms small quantities like paper clips, pins are measured in grams while large quantities are measured in kilograms. Therefore mass of bicycle is measured in kilograms.\n\nQuestion 6.\nA chicken dish needs to bake in the oven for 35 minutes. The dish needs to cool for at least 8 minutes before serving. Scott puts the chicken dish in the oven at 5:14 p.m. For numbers 6a–6d, select True or False for each statement.\na. Scott can serve the dish at 5:51 p.m.\ni. True\nii. False\n\nQuestion 6.\nb. Scott can serve the dish at 5:58 p.m.\ni. True\nii. False\n\nQuestion 6.\nc. Scott should take the dish out of the oven at 5:51 a.m\ni. True\nii. False\n\nQuestion 6.\nd. Scott should take the dish out of the oven at 5:49 p.m.\ni. True\nii. False\n\nExplanation:\nSTEP 1 Find the ending time on the number line.\nSTEP 2 Count forward on the number line to add the elapsed time, 35 minutes ie. the time when the dish is kept in the oven to bake. Draw and label the jumps to show the minutes.\nSTEP 3 Write the times below the number line.\nThe jumps end at 5:49 P.M.\nSo, the ending time is 5:49 P.M.\n\n### Review/Test – Page No. 619\n\nQuestion 7.\nAnthony’s family went out to dinner. They left at the time shown on the clock. They returned home at 6:52 p.m.", null, "Part A\nHow long was Anthony’s family gone?\n_____ hour _____ minutes\n_____ hour _____ minutes\n\n6 hours 52 minutes\n\nExplanation:\nAnthony’s family left for dinner at 5 hours 05 minutes\nAnthony’s family returned back home at 6 hours 52 minutes\n\nQuestion 7.\nPart B\nType below:\n__________\n\nExplanation:\nSTEP 1 Find 5:05 P.M. on the number line. Count on from 5:05 P.M. to 6:52 P.M. Draw marks and record the times on the number line. Then draw and label the jumps.\nSTEP 2 Add to find the total minutes from 5:05 P.M. to 6:52 p.M. From 5:05 P.M. to 6:52 P.M. is 107 minutes.\nSo, the elapsed time is 107 minutes\n\nQuestion 8.\nTran checked the time on his watch after he finished his daily run.", null, "Select the time that Tran finished running. Mark all that apply.\nOptions:\na. 14 minutes before nine\nb. eight forty-six\nc. quarter to nine\nd. nine forty-six\n\nAnswer: a. 14 minutes before nine\nb. eight forty-six\n\nExplanation:\na. Count by fives and ones from the 12 on the clock back to where the minute hand is pointing. Write the missing counting numbers next to the clock.\nb. Count on by fives and ones from the 12 on the clock to where the minute hand is pointing. Write the missing counting numbers next to the clock.\nWrite: 8 Hour 46 minutes\n\nQuestion 9.\nCara uses a balance scale to compare mass.", null, "Circle a symbol that makes the comparison true.\nThe mass of the blocks img 73 the mass of the erasers.\n________\n\nAnswer: mass of the blocks>mass of the erasers\n\nExplanation:\nMass of the blocks is measured in kilograms while mass of the erasers is measured in grams.\n\n### Review/Test – Page No. 620\n\nQuestion 10.\nA large bottle of water holds about 2 liters. For numbers 10a–10e, choose Yes or No to tell whether the container will hold all of the water.\na. kitchen sink\ni. yes\nii. no\n\nExplanation: A kitchen sink can hold about 2 liters. Due to the larger surface area, the kitchen sink can hold about 2 liters.\n\nQuestion 10.\nb. water glass\ni. yes\nii. no\n\nQuestion 10.\nc. ice cube tray\ni. yes\nii. no\n\nQuestion 10.\nd. large soup pot\ni. yes\nii. no\n\nQuestion 10.\ne. lunchbox thermos\ni. yes\nii. no\n\nExplanation: Lunchbox thermos can hold about only 2 liters due to its surface area, and height.\n\nQuestion 11.\nSelect the items that would be best measured in grams.\nMark all that apply.\nOptions:\na. watermelon\nb. lettuce leaf\nc. grape\nd. onion\n\nc. grape\n\nExplanation: A single leaf is measured in grams.\nMass of a single grape is measured in grams.\n\nQuestion 12.\nSamir made a list of what he did on Tuesday. Write the letter for each activity next to the time he did it.\n\n A. Get out of bed. 8:05 a.m.(b) B. Walk to school. 6:25 p.m.(e) C. Eat lunch. 3:50 p.m.(d) D. Go to guitar lesson after school. 11:48 a.m.(c) E. Eat dinner at home. 6:25 a.m.(a)\n\n### Review/Test – Page No. 621\n\nQuestion 13.\nAmy has 30 grams of flour. She puts 4 grams of flour in each pot of chowder that she makes. She puts 5 grams of flour in each pot of potato soup that she makes. She makes 4 pots of chowder. Does Amy have enough flour left over to make 3 pots of potato soup?\n______\n\nExplanation:\nMass of flour Amy has= 30grams\nMass of flour in each pot of chowder= 4grams\nMass of flour in each pot of potato soup= 5grams\nAccording to the problem,\nAmy makes 4 pots of chowder.\nMass of flour used to make 4 pots of chowder= 4×4= 16grams\nLeftover mass of flour=30-16=14 grams\nMass of flour used to make 3 pots of potato soup=5×3=15 grams\nBut, Amy only has 14 grams. So, we can say that Amy doesn’t have enough flour leftover to make 3 pots of potato soup.\n\nQuestion 14.\nUse an inch ruler to measure.", null, "Part A\nWhat is the length of the leaf to the nearest fourth inch?\nType below:\n__________\n\nExplanation:\nLine up the left end of the leaf with the zero mark on the ruler.\nThe right end of the leaf is between the fourth-inch marks for 2 and 3.\nThe mark that is closest to the right end of the leaf is for 1/4 inches.\nSo, the length of the leaf to the nearest fourth inch is 2 1/4 inches.\n\nQuestion 14.\nPart B\nExplain what happens if you line up the left side of the object with the 1 on the ruler.\nType below:\n__________\n\nExplanation:\nLine up the left end of the object with the one mark on the ruler.\nThe right end of the paper clip is between the fourth-inch marks for 2 and 3.\nThe mark that is closest to the right end of the paper clip is for 1/4 inches.\nSo, the length of the paper clip to the nearest fourth inch is 1 1/4 inches.\n\nQuestion 15.\nMrs. Park takes the 9:38 a.m. train to the city. The trip takes 3 hours and 20 minutes. What time does Mrs. Park arrive in the city?\n______ : _______ _______\n\nExplanation:\nSTEP 1 Find the ending time on the clock, the time at which Mrs. Park arrive in the city.\nSTEP 2 Count forward by hours and then by fives for the elapsed time of 3 hours 20 minutes. Write the missing counting numbers next to the clock.\nSo, the ending time is 12:58 P.M.\n\nQuestion 16.\nHector buys two bags of gravel for his driveway. He buys a total of 35 kilograms of gravel. Select the bags he buys.", null, "Answer: 17 kg and 18 kg\n\nExplanation:\nHector buys two bags of gravel for his delivery.\nHector buys a total of 35 kilograms of gravel.\nAccording to the problem,\nThe bags selected by Hector are 17 kg and 18 kg because 17+18=35 kg\n\n### Review/Test – Page No. 622\n\nQuestion 17.\nAshley measures the shells she collects. She records the measurements in a chart.", null, "Part A\nAshley found a razor clam shell this long. Use an inch ruler to measure. Record the measurement in the chart.", null, "_____ $$\\frac{□}{□}$$ inches\n\nQuestion 17.\nPart B\nComplete the line plot to show the data in the chart. How many shells are longer than 2 inches? Tell how you know.", null, "_______ shells\n\nExplanation:\nCalculate the number of shells which are longer than 2 inches.\nBy observing the number line we find 2. From 2 draw curves to the next values in such a way that the other values are greater than 2.\nWe need to draw the curves till 4.\nCount the curves drawn from 2 to 4 the number of curves=number of shells (which are measured to the nearest half-inch)=4\n\nQuestion 18.\nLucy fills a bathroom sink with water. Is the amount of water more than 1 liter, about 1 liter, or less than 1 liter?\nExplain how you know.\n________" ]

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[ "", null, "Just another Perl shrine PerlMonks\n\n### Re^2: OT: Finding Factor Closest To Square Root\n\nby kvale (Monsignor)\n on Feb 19, 2005 at 01:21 UTC ( #432598=note: print w/replies, xml ) Need Help??\n\nYou have the right of it. More precisely, take the logarithim. Then the problem becomes finding a sum of log prime factors that is closest to log(N)/2. This problem is reducible to the subset sum problem, which is NP-hard.\n\nAlthough exact solutions are exponential, heuristics can sometimes generate good answers with minimal effort. One example of a heuristic that might work well is the greedy heuristic. To fill a box of size log(N)/2, put the largest factor into the box. Then put the next largest factor that still fits in the box and insert it. Keep doing this until no more factors fit.\n\nThe above will get you the largest factor <= sqrt(N). To get the smallest factor >= sqrt(N), put all the factors in a box, and take them out again largest to smallest, until no more can be taken out. Finally, comapre the two answers to find the closest.\n\nUpdate: fixed a typo.\n\n-Mark\n\n• Comment on Re^2: OT: Finding Factor Closest To Square Root\n\nReplies are listed 'Best First'.\nRe^3: OT: Finding Factor Closest To Square Root\nby QM (Parson) on Feb 20, 2005 at 05:33 UTC\nTo fill a box of size log(N)/2, put the largest factor into the box. Then put the next largest factor that still fits in the box and insert it. Keep doing this until no more factors fit.\n\nThe above will get you the largest factor <= sqrt(N). To get the smallest factor >= sqrt(N), put all the factors in a box, and take them out again largest to smallest, until no more can be taken out. Finally, comapre the two answers to find the closest.\n\nI don't think that does what you think. For example, 2**3 * 19**3 = 54872, and the square root is 234.277....\n\nYour method gives 19 as the largest less than the square root, and 2**3 * 19**2 = 2888 as the smallest greater than the square root. Notice that 19*2888 = 54872.\n\nA better result is 2**3 * 19 = 152 (because 19**2 = 361).\n\n-QM\n--\nQuantum Mechanics: The dreams stuff is made of\n\nI think the method I sketched does not do what you think :-)\n\nLets take your example. The square root is 234.277. The largest prime factor is 19. The largest remaining factor is 19, but 19*19 > 234.77, so we try the next largest factor, which is 2. 19*2 <= 234.77, so we keep it. The next largest is 2, and 19*2*2 <= 234.77, so we keep it. The next largest (and final) factor is 2, and 19*2*2*2 = 152 < 234.77, so we stop there. 152 is our best attempt for factor closest to but <= sqrt(54872).\n\nA little thought shows that the greedy method from above will give 54872/152 = 361. If one uses a linear distance, then 152 is the closest factor by the grredy heuristic.\n\n-Mark\n\nCreate A New User\nDomain Nodelet?\nNode Status?\nnode history\nNode Type: note [id://432598]\nhelp\nChatterbox?\nand the web crawler heard nothing...\n\nHow do I use this? | Other CB clients\nOther Users?\nOthers exploiting the Monastery: (3)\nAs of 2021-07-26 14:40 GMT\nSections?\nInformation?\nFind Nodes?\nLeftovers?\nVoting Booth?\n\nNo recent polls found\n\nNotices?" ]

[ null, "https://promote.pair.com/i/pair-banner-current.gif", null ]

[ "## vector(var x,var y,var z);\n\nReturns the pointer to a temporary vector initialized to x,y,z, to be used as a parameter in vector functions.\n\n### Parameters:\n\nx - copied to 1st vector element.\ny - copied to 2nd vector element.\nz - copied to 3rd vector element.\n\n### Returns:\n\nPointer to the vector.\n\nFast\n\n### Remarks:\n\nThe vector pointers have a limited lifetime because there are only 64 different vectors available for this function, that are used cyclic.So use this only for passing temporary vectors to functions, but not for permanent vector pointers.\n\n### Example:\n\n`vec_set(temp,vector(2,2+2,sqrt(2))); // sets temp at 2,4,1.414vec_add(my.x,vector(1,2,3)); // adds 1 to my.x, 2 to my.y, 3 to my.z`" ]

[ null ]

[ "# Document (#38662)\n\nAuthor\nSavolainen, R.\nTitle\nProviding informational support in an online discussion group and a Q&A site : the case of travel planning\nSource\nJournal of the Association for Information Science and Technology. 66(2015) no.3, S.450-461\nYear\n2015\nAbstract\nThis study examines the ways in which informational support based on user-generated content is provided for the needs of leisure-related travel planning in an online discussion group and a Q&A site. Attention is paid to the grounds by which the participants bolster the informational support. The findings draw on the analysis of 200 threads of a Finnish online discussion group and a Yahoo! Answers Q&A (question and answer) forum. Three main types of informational support were identified: providing factual information, providing advice, and providing personal opinion. The grounds used in the answers varied across the types of informational support. While providing factual information, the most popular ground was description of the attributes of an entity. In the context of providing advice, reference to external sources of information was employed most frequently. Finally, although providing personal opinions, the participants most often bolstered their views by articulating positive or negative evaluations of an entity. Overall, regarding the grounds, there were more similarities than differences between the discussion group and the Q&A site.\nContent\nVgl.: http://onlinelibrary.wiley.com/doi/10.1002/asi.23191/abstract.\n\n## Similar documents (author)\n\n1. Savolainen, R.: ¬The sense-making theory : reviewing the interests of a user-centered approach to information seeking and use (1993) 5.17\n```5.1732645 = sum of:\n5.1732645 = weight(author_txt:savolainen in 2401) [ClassicSimilarity], result of:\n5.1732645 = fieldWeight in 2401, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.277224 = idf(docFreq=28, maxDocs=41962)\n0.625 = fieldNorm(doc=2401)\n```\n2. Savolainen, R.: Tiedon kayton tutkimus informaatiotutkimuksessa (1994) 5.17\n```5.1732645 = sum of:\n5.1732645 = weight(author_txt:savolainen in 3739) [ClassicSimilarity], result of:\n5.1732645 = fieldWeight in 3739, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.277224 = idf(docFreq=28, maxDocs=41962)\n0.625 = fieldNorm(doc=3739)\n```\n3. Savolainen, R.: Use studies of electronic networks : a review of empirical research approaches and challenges for their development (1998) 5.17\n```5.1732645 = sum of:\n5.1732645 = weight(author_txt:savolainen in 2489) [ClassicSimilarity], result of:\n5.1732645 = fieldWeight in 2489, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.277224 = idf(docFreq=28, maxDocs=41962)\n0.625 = fieldNorm(doc=2489)\n```\n4. Savolainen, R.: Seeking and using information from the Internet : the context of non-work use (1999) 5.17\n```5.1732645 = sum of:\n5.1732645 = weight(author_txt:savolainen in 1284) [ClassicSimilarity], result of:\n5.1732645 = fieldWeight in 1284, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.277224 = idf(docFreq=28, maxDocs=41962)\n0.625 = fieldNorm(doc=1284)\n```\n5. Savolainen, R.: Network competence and information seeking on the Internet : from definitions towards a social cognitive model (2002) 5.17\n```5.1732645 = sum of:\n5.1732645 = weight(author_txt:savolainen in 468) [ClassicSimilarity], result of:\n5.1732645 = fieldWeight in 468, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.277224 = idf(docFreq=28, maxDocs=41962)\n0.625 = fieldNorm(doc=468)\n```\n\n## Similar documents (content)\n\n1. 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[ "# The Moorish Wanderer\n\n## The Big Picture – Part 2\n\nConsumption smoothing is a reality in view of empirical data, and in this particular occasion, HCP’s own PRESIMO model is at faults in terms of specification and reliability on estimated coefficients, and the model specification themselves can be gainsaid as to their robustness.\n\nConsider their proposed model for household consumption:", null, "$\\log(c_t)=.73\\log(rw_t)+.87\\log\\left[\\frac{rw_{t-1}}{c_{t-1}}\\right]-.84\\left[\\frac{u_t}{l+u_t}\\right]-0,82icv_t-.01(r_{t_0}-icv_t)$\n\nLa variable la plus importante dans la détermination de la consommation est le revenu disponible des ménages. Dans le modèle, cette variable est endogène et résulte d’un ensemble de composantes : la masse salariale, l’excédent brut d’exploitation (EBE), les revenus de la propriété, les impôts sur les revenus, les transferts courants, les prestations sociales et les cotisations sociales.\n\nThe reported t-values indicate a pretty large standard deviation attached to each of the computed coefficients in this formula (just divide the estimated coefficients by their corresponding t-values below) Not to mention the fact that inflation and short-term interest rates tend to make the model dependent on contingent data, hence the relatively high R², though it comes at the expenses of a long-term, structural explanation of how households smooth their consumption across time and variations in income.\n\nConsumption smoothing can be traced back to the consumption cycle – whose absolute and relative to GDP’s volatility are both second only to labour work. The proposed alternative does away with inflation and short-term interest rates, as well as unemployment; The idea behind it is can be broken down into two sub-parts:\n\n– long-term trends: inflation and distortionary interest rates do not stay for long, and are eventually factored in by households. The fact that there is little (genuine) concern over subsidies provided by the Compensation Fund, as well as the short-lived effects of Bank Al Maghrib’s decision to cut its policy rate by 25bps are two illustrative examples of the simple intuition behind the idea: households rationalize a lot more than what they let on, and decisions of short-term consequences (including inflation and unexpected shifts in monetary policy) are eventually factored in, and their effect tends to fade away as time goes by. And in this particular issue, we are interested in real consumption behaviour over a very long period of time. Finally, because most of the aggregates are expressed in real terms,\n\n– unemployment is a bit more difficult to gauge from aggregate macroeconomic data; furthermore, because the model is based on household units instead of individuals, there is a mechanism of risk-sharing that alleviates the effects of unemployment and the attached uncertainty to it.\n\nWe therefore consider the following model:", null, "$U\\left(c_t;1-h_t\\right)=E\\sum\\limits_{0}^{\\infty}\\beta^t\\left[\\frac{c_t^\\gamma(1-h_t)^{(1-\\gamma)}}{1-\\phi}\\right]^{1-\\phi}$\n\nwhere:", null, "$\\beta$: the discount time factor and", null, "$\\gamma$: time fraction allocated to leisure.\n\nWhile the formula might look baffling, it displays interesting computational properties in terms of inter-temporal behaviour – the trade-offs households face in deciding their present and immediate future consumption; for", null, "$\\phi = 1$ we get:", null, "$E\\sum\\limits_{0}^{\\infty}\\beta^t\\left[\\gamma\\log(c_t)+(1-\\gamma)(1-h_t)\\right]$\n\n(the ‘curvature’ of the proposed utility function denotes of the ‘intensity’ of inter-temporal arbitrage)\n\nPWT provides dataset with consumption per capita, GDP per capita as well as GDP per effective worker; Consumption per capita is then computed back into an aggregate of Consumption per household, so as to preclude uncertainty around unemployment. Worked hours are then computed on the basis of the 40-hours, as the result is based on Moroccan labour laws.\n\nWhen computed, First Order Conditions on that utility function yield the following, which is then regressed to provide estimates for the parameters described above:\n\n. reg C k_h\nSource |       SS       df       MS              Number of obs =      56\n-------------+------------------------------           F(  1,    54) =    1.48\nModel |  .000169425     1  .000169425           Prob > F      =  0.2283\nResidual |  .006162348    54  .000114118           R-squared     =  0.0268\nTotal |  .006331773    55  .000115123           Root MSE      =  .01068\n------------------------------------------------------------------------------\nC |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]\n-------------+----------------------------------------------------------------\nk_h |    .247134   .2028243     1.22   0.228    -.1595042    .6537721\n_cons |   .9015001   .0882337    10.22   0.000     .7246021    1.078398\n------------------------------------------------------------------------------\n\nwith:", null, "$C=\\frac{c_{t+1}}{c_{t}}$", null, "$k_h= \\frac{\\beta}{1-\\gamma}\\left[\\alpha\\left(\\frac{k}{h}\\right)^{\\alpha-1}+1-\\delta\\right]$\nHouseholds’ own", null, "$\\beta_t$ is therefore .9015 which does square with estimates from academic (and a lot more serious) papers.\n\nThese deep (structural) parameters are now all identified, next step is to build Morocco’s RBC model.\n\n### One Response\n\n1.", null, "The Big Picture – Part 6 « The Moorish Wanderer said, on May 19, 2012 at 01:35" ]

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[ "", null, "", null, "### Home > GC > Chapter 5 > Lesson 5.3.5 > Problem5-118\n\n5-118.", null, "$ΔABC$ was reflected across the $x\\text{-axis}$, and then that result was rotated $90º$ clockwise about the origin to result in $ΔA'B'C'$, shown at right. Find the coordinates of points $A$, $B$, and $C$ of the original triangle.\n\nTo undo the transformation, rotate $ΔA'B'C'$ $90º$ counter-clockwise then reflect the result across the $x\\text{-axis}$.\n\n$A(2,4)$$B(6,2)$$C(4,5)$\n\nUse the eTool below to explore the process of undoing the transformation.\nClick the link at right for the full version of the eTool: 5-118 HW eTool" ]

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SMAK/8hRgLL8S6SwvMcWDQzvascJkuopwm+szYqyA2SH3kRum89v6EE33NrjKLdwLy0Ffh2G4qUg32uVon3YtWxXrWXUEd8FCqftTH765n3cuqEC7zXUczvGyW8W5TzFrwvFmda1k/5wn0wEqelQJ7qWX/XlHC9Jr6z9hLrr0LRKws9tPhJS4FKutaTFjbUcSQcIhO48vcP7F9sZHWJhA58zshvpW/D9SoNNFAIMkRXQ27yHInWkL+ADa2LqTyGCXv+6ciz9GLs7aWfxLT3s4GIAxq8x5n2oALpQCB38X7PeXlw5bNM/2mmfdY59jz/38HjPr7BfFwVk4ejeXxG4NhHeN2XJJr/AOWJlfWOK/IO7D0v8fbv4z0Xnvlv3vNAfsf07+exh6ic+cR5Ae9jPVbYvijwbhDvMZv32jMmz0fy/FsK1P+TmZ9rCjz7VF7nm72ou7vElAfK6RGWq0/4tzL9PwJ1Au/04zH3QnDrLyRaCvkVvtvZRd7tRL7/13gOzv2l9OwGRPndXCBfuO8nipSFfbffKpBmBtNMLXKtk5gOsUTDlKYU/WmhZ2MIvbNCefqQ00BmaG3tE9Nozab2HCLoNY5G7Fp3owNp0T0wpgzFoFLYjB6Mnfn/VeYRDc6lEi0aM9GxEDZhwybcZxeoBfHbYMVT2ABZLX8bCqam/WlMPr4i+eF7Q4rkGaMbtuS76QqUWcJpxOud/HY69cfm91iS6IWedY38xgUsDuXxVd7+/VlvhrNsXmR5oSG+nedMi7EyJ/P4ZCoSqx2PyFjHE5Ry6ppb31c639P2tIirPCX4VxKtBgjMo/W1PZ/9Uzy2wrnODvRWYA6HCQEr3JbDigIWHIJGtyWxX0GPgA+U89Ysq3JRRyXGWrJZx1BA3vYyciiVsLWO8rgd03YG6vBRVODvcu6D7+MevosMFTYowntQcPw7Xt6+4xDnElrmyOsJLG8onU85dXIrJ1+2TXHzdQzzNTNG0Z1MRWwyvYAhq34sy+Ub/BbfiCnT8/jemjYy40PxHrTQQ+iqoFtoNK2PI9kQ7BtDtLDkf+6QiA806D8q4X7PsdFMDED5X83GaIFEa7uPpxxPUsAwv9O9cgZ+xgZ/R/4iNuA2ktN0yc++57pZz2BjEfIQuKMFisUjWCI7xcmDK+PZ+LrXQgO8k5Nmd8fC/j6f3ffQxE3qkw4QKkj8Jv7+kff6MJXDHzLNZVSQfNgpi4VKneuheJjPY8t5MvfPoQJkn/dwrx52eN/Dt0jYq1incc4H+X6XkbAv9JTmDsfrcEGJ5eBiJz4b0OwoE6FvN84zVgz2/UKp2I1ltAOf78tU9A/y6rDN77leHd6dym09CXGYo1TdSDKczfLYieV3GdOc79WhfRwyv5RpbZ14gG3M9Z4HzObrvJh81Xn58pXJcY6XZq8i3w6I+rSYNJ93PAgdou52xQAQ+kBgKt1icV6GIbRKFhS5DhqDtwcg/2igPsftMyVa/jXDjxgW5ZU8dnbAbbmazzWPv3B7TqIS00wLxMeOtH58wHrbtBf5X+TkwZW5bMh90niNx+fTMsJ8BLMc5aAv+CS9Bkv4PHNYlktIpo+wrp8ZOHcij83l/0nOsTbut+X8hkN+9nlej7G0xCGkE7l9Cb0IHSyTu0ggQqKPc69+m5ZoOTiGHoV5zO+kfqzLackHvM7n9g2S78I4WnpOKLXUq8OoEyfxnYEcd2G63aiItbKePM93i/7w7xm5m+lOdK5tn/XPVBiX8ZyX6alq4/UPCTwL7v8vL1+TuB+KcqhLwN77Nf6eUEKZTQ54C1EPz1JaUgw0oW/oRUlg2V5cJE2t89HH4T5q300DUPZoHBpp3TweOD6dpPftwHtKxlhLL3M7zl39TU8Bgqvwq45VWA7K6a6B5VoT2P9bx5rsSx3awfG2LA0cn0Kiv9Xb30yLKMuyWUhLb8uY+6Sc56ktMW9Qlmx/+gOB4w+R3DeR9fvdq0g8C3jfH5dxT6Q71lEGXqVC8MF+qstx5fG04wWqLaH+LCVxAkMdi1eoWL0WOOde/m7r7NveO+biLXrAzohRxEL5Wu7UK1/p2oyKwTpes4WK+ogSPJH+PBoHSnwMgULRL4Qeck03SnhseiXRzgbxMDZSxQjIRr+jEX8wcBxW0jkFnqm/Yee1XynhaG7sn0Fr3Y+E7o7xSNh+8IXesQdo2XzMs0pgOW1HC/8fZea/EjETbzl5b+jDdWwjG+dpQUAUgsf+GmhA4SlBlwC6CeBih2v1iAq+5yaSWafk+9r9et1CIqnzvrMsLbZVtCi/U+I94fL9AOsBvAD3U2Hqr9EdWQlH2u/rELVfx0PR+weQjLO08oHhzjUk5juxdci2aU1F6sPdVJifCRwL5etAyceCvOwd+yy/ZVjyCGJDtwCi8A8t0Hb+kt/w1x3FxSrcwEyJjw1SKCpiZbkNUKjRapJ8UE9fAGviSoeQYXku4wf+ai8UljQVgNmelfgTiSJJB7rsu6T8/stNaNW6VuC32OgsCxAXgv4w8c+1THc3G3jr3kMU9GllNN7AFWwwk16D9b2YhlJilCrrceiLhZ4sUDcLwbpGf+80pCdy/3SpzOp5SckPLQzFBXQ7+xMBJe0JiVzXeEfnUvF4usg9j3eIK81fBGIhIvxyqVwAq1uXMT/FWueZP8P8WgLzyxJW7OZMm6FX5EQqP4gHedF7t+uKKJZJpwxD9WFXfjdZJ13I6j/Cy9dYenf8fPllfadThw5mHZoRk2d8n2OoKEyi9wWWOUZ9wN3/fxLFZWj/uaLfCT2k9Q7nR+AT+v5s4NNO5QSp3sCPI4TFrNCVBAgGQTBnOhbs1AEue7dhKddDcDLFByL7vyw9o5mHsnFBfy2Gtu1GBeyjtDhmUukpB3EL8/y0DEJ3yyJbobIsFWioD2KjbUdVII5hCZ9tl148R2/ec7H3D+/Xj0jGu7Px372AEjhC8gFwv+bvoxL1Ce9A6/3+CtdlfP+PxRybwW/Px3HSc8hZG7/9s5xyK/ZuE166uHNQhhO8c690lA6LYwKeDHjIEIB7tqeYjGd5tku+L38W0+9PBXtujBJyNQkdVvr/UuGCAYKA1/kyMF5DxSAk9BcC+6C9fs2z8rDvssBHBFxVwPqp7qdnRV6OYkOOhV2WD3DZ9+WDfZtKSZKNACwjuPxulsi1HipTuG2voyJzjuOt+G82pMky84358Z+UvFswUaB+FPKgDFRZHk6yhJvddjesIrmfxkb9mQrlLdGH57CW4mkkzY+TBBbFXOMztEThfXrEsW7RdQOX/cR+IPRuWq7dfKcZEtmdjlLhA11hiB9AVx2i4D9EMjy1l+82UeQcxGu8QuPCkm1XgXwlWc7IF0ZOTAmktYGHs0jCwJtMj2NHSj641QW6l+5gvUM3GQJz0RXWQkLfSqlJsaEI/a8kR/+jQXAV+o7gEkRf4BdjyBxE9KCEg6T6E8v4cR0vPYOjBgJtzsddI4XXhk94FsgvJN//Xw5gZaCf7mj+XyDR+OjeAIQxu49lYPu+OyTvUrWKRZzClw4oA+scS7FURcK6SuGh2JPfQkbyoyKg/F1c5L2Ugg5aZPUSjhOwM9+JxA/Vs+WNbo6LJBri9ouYdLYb4SXvuawCcBjLaWUF6/JKWqpryzgHwai3OSQICxf90RjG+ZyTrt3xMoUwxClnW286vPplFVeLmwsQ+h+db+JNtmeH0ZvldtHVOJb8K3z+JOuntcqhPP1Qes7SZ2daRJ5ukXyA73S2Ux9QalL0Br2xkBBA9ZeYY0fzY/lpDJkDP6FLKjUAz3ujQ2YDjVX8qEfHNFZoQOACnik9I2t7a9kulfUnl7mOjXBvrldXgTKw0elLnEbYTuoyJuacTZ3ycz0WwLiYc6ZQibya/3eSfDQxJtV5lMdhrf+A+xE1vW8FnnEFSQllHJo2eRRJqU16Dvfzgbw9zXNs95Gr6CHP+3H7C95zXeeU38H94G0q1zho8Ej0CSo2/ph7G/W+eUybMc6rD1lHWdk65t7betcOKQhW6XhM8rP8uXBHDZxHb8iD/D2f+6Gc7FqgDOyshlYpvVYpSbGhCd0O8elNANzj1EIH0ipevJGU/Rx6K+okP3TMfS/Q2g8gma8ONKC9xfW0gEAMN/XhOi1lpE1Lz0AsDEeyE7Xc5+x/mL8TAoQKIjuJ2+5qfU84SpAfXTyWFu2+TkNvXaVv0Br7jSP4/6pDin3FUsfiDAUens73PUcKj2e3jf43aFmGukg+T6JEEOTtged6vsBztffxOftSJ9P0PgBwU3/CMyDWkZxPCNSHL3h1QBzP0XHSc6w3vAC7sx17rEi+YO3b2QWP8IwU6+GZS0+DW9b4P9/zBMV5by6nV+g6Cfe3KxQlo7f91a+wgt9awCoKWfbHSt9dmO8VrGUjdj01fFikGGJUS9I6hA3Kd6Uy0dYWi9lgurOR9QYns4FLBOoUvAovelb1+ZJ3PW5FTwkaW7g1f+aR80zWL/R7wmWJvkaMrf86FYGF9LZYPMWG9Bg2pldTYRlH5RPW3WtsNF1X6eUSng4XZT+Lv2OkbxMPZfme9yPBQIGzUd/HOXkBcZQy2uFJWuoXBAh1IrevlfA0txNIdgfwHSxwjkHhCc15kKLy9Eg/fw/38N1/gs/2WYcwf05FBvVkRyp9GP+Ncd8Y5vaW5GeNBG6gVwZu9XtZHkizN89JUZl9roR8WSt9Ar/FQ6lkH+5Y578LnIeI/RlUsnBea8z1URf+UKaCrFBUlNCFHzg+kMvYKMW5YGHJ3yzR0JvVXgPUHEhf7rKmdpUjH0PLuEbcilH93c8PMkFUMmaz+hLFAtbk2bJ+P7V1B5Y6ZrsupkxDQ4CaS3hmt6xPLZBuCQndXmszkqePZ+ideMuziibz3EMCxPQyFZ63A+ckaeH5i6y8SOsObtmjqBRkJD9TnY+H+Qyb0AK8xiub5hiLtNqpey4xoovqFF7ncIcMrKcDBHaHsy/pvOOQJY5vDv26OzvvAwqDndp2ZsxzQcnBzHbbsq5d6NxnP8m7631MjyF06wIfVoa3z9az2oCVPo1K7aFU6OxznMO6jzI8V9aPTH+ZyqXr3XiLRHozy+hG716/ooLgoqlIvv7A+ngg68WmrE9xAYb30usxjnVyRoF7rIkp16GiY9EVG4jQhZYSgt8QbIbpRnciQWXo9kODfZ/0nOjEupum8eNIO/mZ1wt33Q9oSaWdRnCJlD4U6kESjjseGNd4dgO8g8tpBdg5vrtpOaCBn+OlvZ3l83AZStc0elSKWZFX0QouZLV08nqjC3gNkpJ3f2Jq3qmyflBQgiSGYw9IeEz0clpoIL6DmS8ohugT/rX07IKwjeJRJDpEem9BpegR75x2PkMhFze8J6eTIBd75DGNhNEZ4/24hPfw83gTlbOJJJkEy+D2wPtZRpJHw7405tuBBXi8971cwW8t7n2jfqPvfU/nPFiIr0p+oZQQad8Xc715VC7WluF5g7W8jazvIreAgnUWyTLlKaCnsqxQJ7Zk+T7EfS0xyuIEltFeJMc3SMx/jsnXdgXydSYV03rWtWl8f3HBhVA4v0KPwhpHMYIy9XiRMprH72ZlActeoehpcWWz5Q3/3WrX0wZ7kUmiKjjC62w25NdrtVIoFJXG/KemayEo+tVCH3x0noiN/XlaCg87UigUCoVi47HQFQqFQqFQbHzQgAuFQqFQKJTQFQqFQqFQKKErFAqFQqGoCP4jwADQNvw20jA5ogAAAABJRU5ErkJggg==", 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[ "## Thursday, March 1, 2018\n\n### Range finder with a Laser and a Webcam in Visual Basic 6.0\n\nThe author of this ingenious project is a scientist, namely Dr. Todd Danko. I found Dr. Danko to be connected with General Electric, Lockheed Martin and DARPA. Nice to have such people in the VB6 community. Like every VB6 programmer, it's fluent in C ++, Java and ASM. From here I will let his text explain the project.\n\nIntroduction\n\nThere are many off the shelf range finding components available including ultrasonic, infrared, and even laser rangefinders. All of these devices work well, but in the field of aerial robotics, weight is a primary concern. It is desirable to get as much functionality out of each component that is added to an air-frame. Miniature robotic rotor craft for example can carry about 100g of payload. It is possible to perform machine vision tasks such as obstacle identification and avoidance though the use of a webcam (or mini wireless camera interfaced to a computer via USB adapter). Better yet, two webcams can provide stereo machine vision thus improving obstacle avoidance because depth can be determined. The drawback of this of course is the addition of the weight of a second camera. This page describes how a mini laser pointer can be configured along with a single camera to provide mono-machine vision with range information.\n\nTheory of Operation\n\nThe diagram below shows how projecting a laser dot onto a target that is in the field of view of a camera, the distance to that target may be calculated. The math is very simple, so this technique works very well for machine vision applications that need to run quickly.\n\nSo, here is how it works. A laser-beam is projected onto an object in the field of view of a camera. This laser beam is ideally parallel to the optical axis of the camera. The dot from the laser is captured along with the rest of the scene by the camera. A simple algorithm is run over the image looking for the brightest pixels. Assuming that the laser is the brightest area of the scene (which seems to be true for my dollar store laser pointer indoors), the dots position in the image frame is known. Then we need to calculate the range to the object based on where along the y axis of the image this laser dot falls. The closer to the center of the image, the farther away the object is.\n\nAs we can see from the diagram earlier in this section, distance (D) may be calculated:\n\nOf course, to solve this equation, you need to know h, which is a constant fixed as the distance between your laser pointer and camera, and theta. Theta is calculated:\n\nPut the two above equations together, we get:\n\nOK, so the number of pixels from the center of the focal plane that the laser dot appears can just be counted from the image. What about the other parameters in this equation? We need to perform a calibration to derive these.\n\nTo calibrate the system, we will collect a series of measurements where I know the range to the target, as well as the number of pixels the dot is from the center of the image each time. This data is below:\n\n pixels from center actual D (cm) 103 29 81 45 65 58 55 71 49 90 45 109 41 127 39 159 37 189 35 218\n\nUsing the following equation, we can calculate the actual angle based on the value of h as well as actual distance for each data point.\n\nNow that we have a Theta_actual for each value, we can come up with a relationship that lets us calculate theta from the number of pixels from image center. I used a linear relationship (thus a gain and offset are needed). This seems to work well even though it does not account for the fact that the focal plane is a plane rather than curved at a constant radius around the center of the lens.\n\nFrom my calibration data, I calculated:\n\nUsing:\n\nI solved for calculated distances, as well as error from actual distance from the calibration data:\n\n pixels from center calc D (cm) actual D (cm) % error 103 29.84 29 2.88 81 41.46 45 -7.87 65 57.55 58 -0.78 55 75.81 71 6.77 49 93.57 90 3.96 45 110.85 109 1.70 41 135.94 127 7.04 39 153.27 159 -3.60 37 175.66 189 -7.06 35 205.70 218 -5.64\n\nComponents\n\nThere are not a lot of parts in my sample range finder. I used a piece of cardboard to hold a laser pointer to a webcam so that the laser pointer points in a direction that is parallel to that of the camera. The parts seen below are laid out on a one inch grid for reference.\n\nMy assembled range finder looks like this:\n\nSoftware\n\nI have written software two ways, one using visual c++ and the other using visual basic. You will probably find that the visual basic version of the software is much easier to follow than the vc++ code, but with everything, there is a tradeoff. The vc++ code can be put together for free (assuming that you have visual studio), while the vb code requires the purchase of a third party software package (also in addition to visual studio).\n\nVisual Basic\n\nThe visual basic code that I have written is available as a package named vb_laser_ranger.zip at the bottom of this page. For this code to work, you will need the VideoOCX ActiveX component installed on your computer. The code that describes the functions found in the main form is seen below:\n\n```Private Sub exit_Click()\n' only if running...\nIf (Timer1.Enabled) Then\n\nTimer1.Enabled = False 'Stop Timer\nVideoOCX.Stop\nVideoOCX.Close\n\nEnd If\n\nEnd\nEnd Sub\n\nPrivate Sub Start_Click() 'Init VideoOCX Control, allocate memory and start grabbing\n\nIf (Not Timer1.Enabled) Then\nStart.Caption = \"Stop\"\n\n' Disable internal error messages in VideoOCX\nVideoOCX.SetErrorMessages False\n\n' Init control\nIf (Not VideoOCX.Init) Then\n' Init failed. Display error message and end sub\nMsgBox VideoOCX.GetLastErrorString, vbOKOnly, \"VideoOCX Error\"\nEnd\nElse\n' Allocate memory for global image handle\ncapture_image = VideoOCX.GetColorImageHandle\n' result_image = VideoOCX_Processed.GetColorImageHandle\n\nTimer1.Enabled = True 'Start capture timer\n\n' Start Capture mode\nIf (Not VideoOCX.Start) Then\n' Start failed. Display error message and end sub\nMsgBox VideoOCX.GetLastErrorString, vbOKOnly, \"VideoOCX Error\"\nEnd\nEnd If\nEnd If\nElse\nStart.Caption = \"Start\"\nTimer1.Enabled = False 'Stop Timer\nVideoOCX.Stop\nVideoOCX.Close\nEnd If\n\nEnd Sub\n\nPrivate Sub Timer1_Timer()\n' Timer for capturing - handles videoOCXTools\nDim matrix As Variant\nDim height, width As Integer\nDim r, c As Integer\nDim max_r, max_c As Integer\nDim max_red As Integer\nDim gain, offset As Variant\nDim h_cm As Variant\nDim range As Integer\nDim pixels_from_center As Integer\n\n' Calibrated parameter for pixel to distance conversion\ngain = 0.0024259348\noffset = -0.056514344\nh_cm = 5.842\n\nmax_red = 0\n\n' Capture an image\nIf (VideoOCX.Capture(capture_image)) Then\n\n' VideoOCX.Show capture_image\n\n' Matrix transformation initialization\nmatrix = VideoOCX.GetMatrix(capture_image)\n\nheight = VideoOCX.GetHeight\nwidth = VideoOCX.GetWidth\n\n' Image processing code\n\n' The laser dot should not be seen above the middle row (with a little pad)\nFor r = height / 2 - 20 To height - 1\n\n' Our physical setup is roughly calibrated to make the laser\n' dot in the middle columns...dont bother lookng too far away\nFor c = width / 2 - 25 To width / 2 + 24\n\n' Look for the largest red pixel value in the scene (red laser)\nIf (matrix(c, r, 2) > max_red) Then\nmax_red = matrix(c, r, 2)\nmax_r = r\nmax_c = c\nEnd If\nNext c\nNext r\n\n' Calculate the distance for the laser dot from middle of frame\npixels_from_center = max_r - 120\n\n' Calculate range in cm based on calibrated parameters\nrange = h_cm / Tan(pixels_from_center * gain + offset)\n\n' Print laser dot position row and column to screen\nrow_val.Caption = max_r\ncol_val.Caption = max_c\n\n' Print range to laser illuminated object to screen\nrange_val.Caption = range\n\n' Draw a red vertical line to intersect target\nFor r = 0 To height - 1\nmatrix(max_c, r, 2) = 255\nNext r\n\n' Draw a red horizontal line to intersect target\nFor c = 0 To width - 1\nmatrix(c, max_r, 2) = 255\nNext c\n\nVideoOCX.ReleaseMatrixToImageHandle (capture_image)\n\nEnd If\n\nVideoOCX.Show capture_image\n\nEnd Sub```\n\nScreen shots from this code can be seen below:\n\nVisual C++\n\n```void CTripodDlg::doMyImageProcessing(LPBITMAPINFOHEADER lpThisBitmapInfoHeader)\n{\n// doMyImageProcessing: This is where you'd write your own image processing code\n\nunsigned int W, H; // Width and Height of current frame [pixels]\nunsigned int row, col; // Pixel's row and col positions\nunsigned long i; // Dummy variable for row-column vector\nunsigned int max_row; // Row of the brightest pixel\nunsigned int max_col; // Column of the brightest pixel\nBYTE max_val = 0; // Value of the brightest pixel\n\n// Values used for calculating range from captured image data\n// these values are only for a specific camera and laser setup\nconst double gain = 0.0024259348; // Gain Constant used for converting\n// pixel offset to angle in radians\nconst double offset = -0.056514344; // Offset Constant\nconst double h_cm = 5.842; // Distance between center of camera and laser\ndouble range; // Calculated range\nunsigned int pixels_from_center; // Brightest pixel location from center\n// not bottom of frame\n\nchar str; // To print message\nCDC *pDC; // Device context need to print message\n\nW = lpThisBitmapInfoHeader->biWidth; // biWidth: number of columns\nH = lpThisBitmapInfoHeader->biHeight; // biHeight: number of rows\n\nfor (row = 0; row < H; row++) {\nfor (col = 0; col < W; col++) {\n\n// Recall each pixel is composed of 3 bytes\ni = (unsigned long)(row*3*W + 3*col);\n\n// If the current pixel value is greater than any other,\n// it is the new max pixel\nif (*(m_destinationBmp + i) >= max_val)\n{\nmax_val = *(m_destinationBmp + i);\nmax_row = row;\nmax_col = col;\n}\n}\n}\n// After each frame, reset max pixel value to zero\nmax_val = 0;\n\nfor (row = 0; row < H; row++) {\nfor (col = 0; col < W; col++) {\n\ni = (unsigned long)(row*3*W + 3*col);\n\n// Draw a white cross-hair over brightest pixel in the output display\nif ((row == max_row) || (col == max_col))\n*(m_destinationBmp + i) =\n*(m_destinationBmp + i + 1) =\n*(m_destinationBmp + i + 2) = 255;\n}\n}\n\n// Calculate distance of brightest pixel from center rather than bottom of frame\npixels_from_center = 120 - max_row;\n\n// Calculate range in cm based on bright pixel location, and setup specific constants\nrange = h_cm / tan(pixels_from_center * gain + offset);\n\n// To print message at (row, column) = (75, 580)\npDC = GetDC();\n\n// Display frame coordinates as well as calculated range\nsprintf(str, \"Max Value at x= %u, y= %u, range= %f cm \",max_col, max_row, range);\npDC->TextOut(75, 580, str);\nReleaseDC(pDC);\n}```\n\nMy complete code for this project is available as a package named LaserRange.zip at the bottom of this page.  Note, to run this executable, you will need to have both qcsdk and the qc543 driver installed on your computer.  Sorry, but you are on your own to find both of these. Below are two examples of the webcam based laser range finder in action. Note how it looks like there are two laser dots in the second example. This \"stray light\" is caused by internal reflections in the camera. The reflected dot loses intensity as it bounces within the camera so it does not interfere with the algorithm that detects the brightest pixel in the image.\n\nFuture Work\n\nOne specific improvement that can be made to this webcam based laser range finder is to project a horizontal line rather than a dot onto a target. This way, we could calculate the range for each column of the image rather than just one column. Such a setup would be able to be used to locate areas of maximum range as places that a vehicle could steer towards. Likewise, areas of minimum range would be identified as obstacles to be avoided.\n\nSources:\n\n### Quick Sort, Selection Sort and Bubble Sort algorithm in VB6\n\nHere we have an application that measures execution times for the three sorting algorithms: Quick Sort, Selection Sort and Bubble Sort.  A visual interface shows what these algorithms do in real time. The implementation is made in algorithm in Visual Basic 6.0 and the source code is shown below:\n\nThe quicksort algorithm was developed in 1959 by Tony Hoare while in the Soviet Union, as a visiting student at Moscow State University. At that time, Hoare worked in a project on machine translation for the National Physical Laboratory. As a part of the translation process, he needed to sort the words of Russian sentences prior to looking them up in a Russian-English dictionary that was already sorted in alphabetic order on magnetic tape. After recognizing that his first idea, insertion sort, would be slow, he quickly came up with a new idea that was Quicksort. He wrote a program in Mercury Autocode for the partition but couldn't write the program to account for the list of unsorted segments. On return to England, he was asked to write code for Shellsort as part of his new job. Hoare mentioned to his boss that he knew of a faster algorithm and his boss bet sixpence that he didn't. His boss ultimately accepted that he had lost the bet. Later, Hoare learned about ALGOL and its ability to do recursion that enabled him to publish the code in Communications of the Association for Computing Machinery, the premier computer science journal of the time.\n\nQuicksort gained widespread adoption, appearing, for example, in Unix as the default library sort subroutine. Hence, it lent its name to the C standard library subroutine qsort and in the reference implementation of Java.\n\nRobert Sedgewick's Ph.D. thesis in 1975 is considered a milestone in the study of Quicksort where he resolved many open problems related to the analysis of various pivot selection schemes including Samplesort, adaptive partitioning by Van Emden as well as derivation of expected number of comparisons and swaps. Bentley and McIlroy incorporated various improvements for use in programming libraries, including a technique to deal with equal elements and a pivot scheme known as pseudomedian of nine, where a sample of nine elements is divided into groups of three and then the median of the three medians from three groups is chosen. Jon Bentley described another simpler and compact partitioning scheme in his book Programming Pearls that he attributed to Nico Lomuto. Later Bentley wrote that he used Hoare's version for years but never really understood it but Lomuto's version was simple enough to prove correct. Bentley described Quicksort as the \"most beautiful code I had ever written\" in the same essay. Lomuto's partition scheme was also popularized by the textbook Introduction to Algorithms although it is inferior to Hoare's scheme because it does three times more swaps on average and degrades to O(n2) runtime when all elements are equal.\n\nIn 2009, Vladimir Yaroslavskiy proposed the new dual pivot Quicksort implementation. In the Java core library mailing lists, he initiated a discussion claiming his new algorithm to be superior to the runtime library’s sorting method, which was at that time based on the widely used and carefully tuned variant of classic Quicksort by Bentley and McIlroy. Yaroslavskiy’s Quicksort has been chosen as the new default sorting algorithm in Oracle’s Java 7 runtime library after extensive empirical performance tests.\n\nSelection sort is a sorting algorithm, specifically an in-place comparison sort. It has O(n2) time complexity, making it inefficient on large lists, and generally performs worse than the similar insertion sort. Selection sort is noted for its simplicity, and it has performance advantages over more complicated algorithms in certain situations, particularly where auxiliary memory is limited.\n\nThe algorithm divides the input list into two parts: the sublist of items already sorted, which is built up from left to right at the front (left) of the list, and the sublist of items remaining to be sorted that occupy the rest of the list. Initially, the sorted sublist is empty and the unsorted sublist is the entire input list. The algorithm proceeds by finding the smallest (or largest, depending on sorting order) element in the unsorted sublist, exchanging (swapping) it with the leftmost unsorted element (putting it in sorted order), and moving the sublist boundaries one element to the right.\n\nBubble sort, sometimes referred to as sinking sort, is a simple sorting algorithm that repeatedly steps through the list to be sorted, compares each pair of adjacent items and swaps them if they are in the wrong order. The pass through the list is repeated until no swaps are needed, which indicates that the list is sorted. The algorithm, which is a comparison sort, is named for the way smaller or larger elements \"bubble\" to the top of the list. Although the algorithm is simple, it is too slow and impractical for most problems even when compared to insertion sort. It can be practical if the input is usually in sorted order but may occasionally have some out-of-order elements nearly in position.\n\nImplementation:\n\n```Public Declare Sub Sleep Lib \"kernel32.dll\" (ByVal dwMilliseconds As Long)\n\nPublic Sub QuickSort(vArray As Variant, L As Long, R As Long)\n' Array , LBound() , UBound()\nDim i As Long\nDim j As Long\nDim X\nDim Y\n\ni = L\nj = R\nX = vArray((L + R) / 2)\n\nDo While (i <= j)\nDoEvents\nDo While (vArray(i) < X And i < R)\ni = i + 1\nLoop\n\nDo While (X < vArray(j) And j > L)\nj = j - 1\nLoop\n\nIf (i <= j) Then\nY = vArray(i)\nvArray(i) = vArray(j)\nvArray(j) = Y\ni = i + 1\nj = j - 1\nEnd If\nLoop\n\nIf (L < j) Then QuickSort vArray, L, j\nIf (i < R) Then QuickSort vArray, i, R\nEnd Sub\n\nPublic Sub QuickSort33(vArray As Variant, AccordingTo As Integer, Dimension2Size As Integer, L As Integer, R As Integer)\n' name of array, sorting according to which dimension?, size of second dimension, lbound(), ubound()\nDim a As Integer, i As Integer, j As Integer ' counters\nDim X, Y, z ' temporary values\n\ni = L\nj = R\nX = vArray((L + R) / 2, AccordingTo)\nDo While (i <= j)\nDoEvents\nDo While (vArray(i, AccordingTo) < X And i < R)\ni = i + 1\nLoop\nDo While (X < vArray(j, AccordingTo) And j > L)\nj = j - 1\nLoop\nIf (i <= j) Then\nY = vArray(i, AccordingTo)\nvArray(i, AccordingTo) = vArray(j, AccordingTo)\nvArray(j, AccordingTo) = Y\nFor a = 0 To AccordingTo - 1\nz = vArray(i, a)\nvArray(i, a) = vArray(j, a)\nvArray(j, a) = z\nNext a\nFor a = AccordingTo + 1 To Dimension2Size\nz = vArray(i, a)\nvArray(i, a) = vArray(j, a)\nvArray(j, a) = z\nNext a\ni = i + 1\nj = j - 1\nEnd If\nLoop\n\nIf (L < j) Then QuickSort33 vArray, AccordingTo, Dimension2Size, L, j\nIf (i < R) Then QuickSort33 vArray, AccordingTo, Dimension2Size, i, R\nEnd Sub\n\nPublic Sub SelectionSort(vArray, L As Integer, R As Integer)\n' name of array, lbound(), ubound()\nDim i As Integer\nDim j As Integer\nDim best_value ' smallest value in array\nDim best_j As Integer\n' loop from left to right\nFor i = L To R - 1\nDoEvents\n' initialize lowest value\nbest_value = vArray(i)\nbest_j = i ' initialize lowest value array location\nFor j = i + 1 To R\n' find the lowest value in the array in this loop\nIf vArray(j) < best_value Then\nbest_value = vArray(j)\nbest_j = j\nEnd If\nNext j\n' put the smallest value at the from (left) of the array\n' and put the value on the left of the array in the smallest\n' value's previous position\nvArray(best_j) = vArray(i)\nvArray(i) = best_value\nNext i\n\nEnd Sub\n\nPublic Sub QuickSortBars(vArray As Variant, L As Integer, R As Integer, Optional SleepTime As Long = 0)\nDim i As Integer ' counter\nDim j As Integer ' counter\nDim BarVal1 ' temporary bar value\nDim BarVal2 ' temporary bar value\n\ni = L\nj = R\nBarVal1 = vArray((L + R) / 2)\n\nDo While (i <= j)\nDoEvents\nIf SleepTime > 0 Then\nSleep SleepTime\nEnd If\nDo While (vArray(i) < BarVal1 And i < R)\ni = i + 1\nLoop\n\nDo While (BarVal1 < vArray(j) And j > L)\nj = j - 1\nLoop\n\nIf (i <= j) Then\nBarVal2 = vArray(i)\nvArray(i) = vArray(j)\nvArray(j) = BarVal2\nfrmMain.Bar(i).Value = vArray(i)\nfrmMain.Bar(j).Value = vArray(j)\ni = i + 1\nj = j - 1\nEnd If\nLoop\n\nIf (L < j) Then QuickSortBars vArray, L, j, SleepTime\nIf (i < R) Then QuickSortBars vArray, i, R, SleepTime\nEnd Sub\n\nPublic Sub SelectionSortBars(vArray, L As Integer, R As Integer, Optional SleepTime As Long = 0)\n' name of array, lbound(), ubound()\nDim i As Integer ' counter\nDim j As Integer ' counter\nDim best_value ' smallest value in array\nDim best_j As Integer\n\n' loop from left to right\nFor i = L To R - 1\nDoEvents\nIf SleepTime > 0 Then\nSleep SleepTime\nEnd If\n' initialize lowest value\nbest_value = vArray(i)\nbest_j = i ' initialize lowest value array location\nFor j = i + 1 To R\n' find the lowest value in the array in this loop\nIf vArray(j) < best_value Then\nbest_value = vArray(j)\nbest_j = j\nEnd If\nNext j\n' put the smallest value at the from (left) of the array\n' and put the value on the left of the array in the smallest\n' value's previous position\nvArray(best_j) = vArray(i)\nvArray(i) = best_value\nfrmMain.Bar(best_j) = vArray(best_j)\nfrmMain.Bar(i) = vArray(i)\nNext i\nEnd Sub\n```\n\nSources:\n\nhttps://en.wikipedia.org/wiki/Bubble_sort\n\nhttps://en.wikipedia.org/wiki/Selection_sort\n\nhttps://en.wikipedia.org/wiki/Quicksort" ]

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[ "```package co.rsk.scoring;\n\nimport org.junit.Assert;\nimport org.junit.Test;\n\nimport java.net.UnknownHostException;\nimport java.util.Random;\n\n/**\n* Created by ajlopez on 11/07/2017.\n*/\nprivate static Random random = new Random();\n\n@Test\n\n}\n\n@Test\npublic void containsIPV4() throws UnknownHostException {\n\n}\n\n@Test\npublic void doesNotContainIPV4WithAlteredByte() throws UnknownHostException {\n\n}\n\n@Test\npublic void doesNotContainIPV6() throws UnknownHostException {\n\n}\n\n@Test\npublic void using16BitsMask() throws UnknownHostException {\n\n}\n\n@Test\npublic void usingIPV4With9BitsMask() throws UnknownHostException {\nbytes ^= 1;\nbytes ^= 2;\n\n}\n\n@Test\npublic void usingIPV6With9BitsMask() throws UnknownHostException {\nbytes ^= 1;\nbytes ^= 2;\n\n}\n\n@Test\npublic void usingIPV4With18BitsMask() throws UnknownHostException {\nbytes ^= 2;\nbytes ^= 4;\n\n}\n\n@Test\npublic void usingIPV6With18BitsMask() throws UnknownHostException {\nbytes ^= 2;\nbytes ^= 4;\n\n}\n\n@Test\npublic void doesNotContainIPV4() throws UnknownHostException {\n\n}\n\n@Test\npublic void equals() throws UnknownHostException {\n\nAssert.assertTrue(block1.equals(block1));\nAssert.assertTrue(block2.equals(block2));\nAssert.assertTrue(block3.equals(block3));\nAssert.assertTrue(block4.equals(block4));\nAssert.assertTrue(block5.equals(block5));\n\nAssert.assertTrue(block1.equals(block4));\nAssert.assertTrue(block4.equals(block1));\n\nAssert.assertFalse(block1.equals(block2));\nAssert.assertFalse(block1.equals(block3));\nAssert.assertFalse(block1.equals(block5));\n\nAssert.assertFalse(block1.equals(null));\nAssert.assertFalse(block1.equals(\"block\"));\n\nAssert.assertEquals(block1.hashCode(), block4.hashCode());\n}\n\nbyte[] bytes = new byte;\n\nrandom.nextBytes(bytes);\n\n}\n\nbyte[] bytes = new byte;\n\nrandom.nextBytes(bytes);" ]

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[ "Voltar\n\n# servo motor power calculation\n\nIf the actual operating current of the servo motor is not much different from the rated current indicated on the spleen, the power of the selected servo motor is appropriate. To properly select the power of the servo motor, the following calculations or comparisons must be made: P = F*V/1000 (P = calculated power KW, F = required pull N, V = linear speed M/S) For the continuous working mode with constant load, the required power of servo motor can be calculated according to the following formula: P1(kw):P=P/n1n2 Servo motors can range in size from as small as a roll of quarters to a weight of 50 lb to 60 lb. The power P1 calculated according to the above formula is not necessarily the same as the product power. In addition, the most commonly used method is analogy to select the power of servo motor. Speed, torque, safety ratio, etc.). • Communications with the motion controller The best inertia ratio for an application comes down to the dynamics of the move and the accuracy required. If the actual operating current of the servo motor is about 70%, lower than the rated current indicated on the nameplate, which means that the power of the servo motor is too large, and the servo motor with smaller power should be exchanged. The specific method is to know how high power servo motors are used by similar production machinery in our unit or other nearby units, and then to select similar power servo motors for testing. Kollmorgen’s brushless servo motors offer the highest torque/inertia ratio over a broad range of speeds—including high speed capabilities up to 8,000 rpm for standard designs and up to 70,000 rpm for special designs. From Center or the neutral point, 20 degrees would be entered. amounts of power to the servom otor at the appropriate times. The so-called analogy method is to compare the power of servo motor used in similar production machinery. After years of developing servo motor sizing programs for Windows I deemed it necessary to document the motor sizing process beyond the regular help files. Version: 1.7 Date of release: 05/07/2019 Compatibility: See release notes below Divide the motor speed by the required speed and round down to get a starting gear ratio. Looking at retrofiting. Unlike large industrial motors, a servo motor is not used for continuous energy conversion. Servo motor sizing spreadsheet allows to calculate a combination of motors and gearboxes that fits a user-defined working condition. The power Pin kilowatts (kW) is equal to square root of 3 times the power factorPFtimes the phase current Iin amps (A), times the line to line RMS voltage VL-Lin volts (V) divided by 1000: P(kW)= √3 × PF ×I(A)×VL-L (V)/ 1000. Servo motors have a high speed response due to low inertia and are designed with small diameter and long rotor length. Servos have three-wire cables that supply power and commands to servos. V_{ph-ph}= \\frac{V_{dc}}{\\sqrt{2}} \\cdot D\\], $The result of this idea is this book. In fact, it should be considered the torque (torque), servo motor power and torque calculation formula. V_{ph}\\)) is the RMS voltage between is the voltage difference between two output terminals of the Servo Drive, defined in (3). Feedback data made available to the user when using Hiperface DSL in MPP servo motor (feedback option 6H) ... Servo Motors: Absolute Encoders & Drive Compatibility. In the case of a phase-balanced motor: The phase voltage ($$Expressed as a formula, torque is the twisting action applied perpendicular to a distance vector times the vector length from the reference datum of impact to the pivot point. The calculator only calculates one control surface per servo. Try to make the servo motor run under the rated load. Therefore, the power delivered to the motor power can be defined as: ( 4) P = 3 ⋠Vph − ph √3 ⋠Iph = √3 2 ⋠Vdc ⋠Iph ⋠D. Some examples are direct rotation, a ball screw, a belt and pulley or a rack and pinion. Maximum efficiency may be 10-30% of motor stall torque (70-90% of no load speed). FLA - \"Full Load Amps\" - amount of current drawn when full-load torque and horsepower is reached for the motor.FLA is usually determined in laboratory tests.Note! Gearing increases the available torque by the gear ratio, so a 3:1 pulley system will increase available torque by a factor of 3 (neglecting efficiency losses). However, the voltage and current values are shared different within the phases: where \\( AC power supply (in phase) The use of the Space Vector Modulation (SVM) allows the use of all the DC-bus voltage as line voltage amplitude. Calculating rotary and electrical power. My bible for motor sizing and the inertia/torque calculation of mechanical components has been for years a These rules to sizing a servo motor correctly can be applied to motors that come in all physical sizes. Generally, a safety factor is required. The phase voltage (\\( The use of the Space Vector Modulation (SVM) allows the use of all the DC-bus voltage as line voltage amplitude. First, a review of the electrical connections that I mentioned in the inside servospost. Horsepower (hp) is a measure of power, which can be further described as the rate at which work is performed. The power equation is in terms of peak-of-sine so if the current rating is in RMS then you will need to multiply by the square root of 2. The servo should be oriented for the motor shaft to be closest to the load cell. How to determine. Servo Motors: Checking Servo Motor Phases. Multiple moves with dwell time in between moves can be evaluated for the torque requirements, power requirements, and temperature impact using duty cycle calculations. Dimensions and mass (or density) of load 2. “The customer was happy with the prototype and could justify the cost,” Seifert says. To properly select the power of the servo motor, the following calculations or comparisons must be made: P = F*V/1000 (P = calculated power KW, F = required pull N, V = linear speed M/S). How to calculate inertia of a drive system Friction coefficient of the sliding surface of each moving part Next you will need to determine the required specif… The power factor of an AC electric power system is defined as the ratio active (true or real) power to apparent power, where. Increasingly, applications use mechanisms where the center of rotation is on a different axis than the center of mass. Customized Motor Test System for Dental and Surgical Instrument Testing October 29, 2020 Magtrol SA Celebrates 20th Anniversary June 10, 2020 Magtrol contributes to … Pay attention to the following two points when selecting: 1 If the power of the servo motor is too small, the phenomenon of \"small horse-drawn carriage\" will occur, causing the servo motor overloaded for a long time, and the insulation damaged due to heat, and even the servo motor being burned. 2) AC servo motor is also a brushless motor, which is divided into synchronous and asynchronous motors. The power of the servo motor should be selected according to the power required by the production machine. Most servo motor manufacturers have software for sizing applications using basic mechanisms, such as ballscrews or rack-and-pinion. When a single supply is used, P in can be calculated as: Pin = Vbus * Ibus Where V bus is the power supply DC voltage (between POW_SUP and GND_P terminals) and I bus is the current provided by this power supply. Therefore, the power delivered to the motor power can be defined as: From the point of viwe of a Servo Drive, the power provided to a Δ-wired motor is exactly the same. For the continuous working mode with constant load, the required power of servo motor can be calculated according to the following formula: N1 is the efficiency of production machinery; N2 is the efficiency of servo motor, that is, transmission efficiency. Too much power loss in the motor can cause the motor to overheat at high ambient temperatures. P= 3\\cdot V_{ph} \\cdot \\frac{I_{line}}{\\sqrt{3}} = \\sqrt{\\frac{3}{2}}\\cdot V_{dc} \\cdot I_{line} \\cdot D$. I_{line}$$ is the current in RMS seen at the Servo Drive output terminals. Servo motors can also be solid or hollow shaft. The more ac- curately this can be calculated, the better. The number one calculation to make when sizing a motor for a rotary application is the calculation of the load inertia. Then divide the required torque by the gear ratio to find the newly required torque. The output torque of the servo motor must be greater than the torque required by the working machine. ( 3) Vph − ph = Vdc √2 ⋅ D. where D is the applied duty cycle. The servo's wire will come out through a hole behind the load cell. I have a Hardinge HLV copy with a dead early 90's anilam lathemate cnc control. A servo motor is a linear or rotary actuator that provides fast precision position control for closed-loop position control applications. Too much heat from the drive can cause a cabinet to overheat. The cables are typically six to twelve inches long, with 22 gauge and thinner wires (smaller servos tend to have shorter and thinner wires). - in the calculator above FLA is RLA + 25%. Brushless motors use electricity exclusively. A servo motor’s torque-speed curve includes two operating zones: continuous duty and intermittent duty. The phase current ($$n — rated speed of the servo motor, r/min. To secure the servo into place, use at least two M4 x 6mm screws. Along with the type of drive mechanism, you must also determine the dimensions, mass and friction coefficient, etc. †Inertia Formulas †Calculation of Friction Torque Calculates the frictional force for each element, where necessary, and converts it to friction torque for a motor shaft. Since the connectors have become more standardized, the order of the … Copyright 2009 - 2020 Instar Electromechanical Retain ownership. Most of the motors available likely are capable of much higher speeds than required. It can be plugged into the Arduino from there. If the control surface has a full sweep angle of 40 degrees. While it is technically better to follow the same format and create similar curves for your motor it is not absolutely necessary for a good science project. Factors, equations, and practical suggestions servomotor sizing and selection. The simple calculation, volts times amps says they draw 4.2 kw and would have an output somewhat less due to losses. To choose a motor that can meet the application requirements, it’s important to ensure the application runs within the continuous duty zone during normal operation and does not exceed the intermittent duty zone when peak torque is required. for each part that moves as the Servo Motor rotates. Heat can also transfer into attached equipment causing it to misbehave. To install the servo, simply slide it into the housing. Dimensions and mass (or density) of each part 3. Servo Motors. Indeed, a calculation of energy consumption and cost for this application based on the duty cycle of the motor showed a clear advantage for the servo-driven design. At present, synchronous motors are generally used in motion control, and its power range is large, and a large power can be achieved. Duty Cycle = 100 * (time on)/ (time on + time off), units of %. If the Servo has a full travel of 60 degrees, from center 30 degrees would be entered. In a DC servo-driven motor, the input power is the electrical power provided by the power supplies or battery systems. Add：B5, 6F Huangtian Industrial Estate, Baoan District, Shenzhen City, China. The basic operating principle is the same as for an inverter, in which the motor is operated by converting AC power to DC power to be a certain frequency. P= 3\\cdot \\frac{V_{ph-ph}}{\\sqrt{3}} \\cdot I_{ph} = \\sqrt{\\frac{3}{2}}\\cdot V_{dc} \\cdot I_{ph} \\cdot D\\], $I_{ph}$$) is the current in RMS seen at the Servo Drive output terminals. The purpose of testing is to verify whether the selected servo motors match the production machinery. The servo motor drives the production machinery to operate, the working current of the motor is measured by a clamp-type ammeter, and the measured current is compared with the rated current marked on the nameplate of the servo motor. A servo drive also has the following functions. 2.a.) This will help you narrow the choices down to a few select motors. Active (Real or True) Power is measured in watts (W) and is the power drawn by the electrical resistance of a system doing useful work; Apparent Power is measured in volt-amperes (VA) and is the voltage on an AC system multiplied by all the current that flows in it. Brushed motors use conductive material inside the cylinder slide against the commutator as it rotates. Calculation with line to line voltage. The equation has a positive result if rotation is in the anticlockwise direction or a negative outcome when an object turns clockwise.With regard to a servo motor, the technical term defines the mechanical work produced by its shaft spinning, typically measured in Newton-meters (Nm). Does the MPW work in salt-rich environments? RLA - \"Running Load Amps\" - current drawn during normal operation of electric motor. The power factor and efficiency are not high, which is not only bad for the user and the power grid, but also causes electric energy waste. To make this calculation, it’s generally best to ap- proximate the load as a disc or cylinder. Kw of servo motor. More complex systems, however, may require manual calculations. AC three phase amps to kilowatts calculation. The following diagram shows the electrical model of a Y-wired BLAC motor: The power in any motor can be calculated as the sum of each phase power (product of RMS current and RMS voltage). There are slightly different definitions for its conversion to the unit watts depending on the mechanism being described: mechanical, electric, boiler, metric, etc.. Our focus here will be on servo motor … 2 If the power of the servo motor is too large, the phenomenon of \"large horse-drawn car\" will appear, and the output mechanical power cannot be fully utilized. The first step is to determine the drive mechanism for your equipment. P=V_a \\cdot I_a + V_b \\cdot I_b + V_c \\cdot I_c = 3\\cdot V_{ph}\\cdot I_{ph}$, \\[ Servo motors use either AC or DC current. Servo motors generally run at speeds in the 3,000 to 5,000 RPM range, and in many applications the motor is paired with some type of gearing to increase output torque. Multiply V DC by the rated current in terms of peak-of-sine to find the electrical power that the drive is capable of. †Calculate the inertia applied to each element to calculate the total load inertia of the motor shaft conversion value. In most positioning applications, the moves repeat at some frequency. Usually for small motors maximum power is at 50% of stall torque (approximately 50% of no load speed). Servo Drive Power Dissipation 𝑎 =Conduction_Losses+ Switching_Losses= : â 𝑖 ;∗𝑰 𝟐 + 𝑖 ℎ𝑖 ∗𝑽 𝒃 ∗𝑰 + 𝒍_ 𝒆 Power Losses of a Servo Drive are function of Motor’s current and the DC supply Bus ServoMotor Current Units RMS vs Pk of Sine vs Trap. Most servo motor manufacturers recommend that the inertia ratio be kept to 10:1 or less, although there are many applications that operate successfully at much higher ratios. With AC, speed is determined by the frequency of voltage, while speed is directly proportional to a DC voltage. The final pieces of information to consider when selecting a servo motor are the physical limitations and constraints presented by the specific job. that are required for the load calculation: 1. V_{ph-ph}\\) is the RMS voltage difference between two output terminals of the Servo Drive. If you just read that, you can jump ahead to the “power” section. If the measured servo motor operating current is more than 40%, larger than the rated current indicated on the nameplate, which means that the servo motor's power is too small, and the servo motor with higher power should be replaced. Therefore, the rated power of the selected servo motor should be equal to or slightly greater than the calculated power. V_{ph}\\)) is defined as: where \\( Background In some applications, power dissipation in step motors and drives is a critical aspect of system design. Therefore, the total power in a Δ-wired motor results as: Which, from the point of view of a servo drive, is equivalent to (4), Electromagnetic Interference Issues With Servo Drive Systems, \\[ The results can be sorted depending on various parameters (e.g. Motor, which is divided into synchronous and asynchronous motors critical aspect of system design material inside cylinder. Industrial Estate, Baoan District, Shenzhen City, China and would have an output somewhat less to... Approximately 50 % of no load speed ) voltage amplitude the selected servo motor used in similar production.. However, may require manual calculations weight of 50 lb to 60 lb a ball screw, a servo run! Motor is a critical aspect of system design center or the neutral point, degrees! When selecting a servo motor’s torque-speed curve includes two operating zones: continuous duty and intermittent duty applications... Motor shaft to be closest to the “power” section a review of the motor to.. Calculator only calculates one control surface per servo cabinet to overheat at ambient... A belt and pulley or a rack and pinion is determined by the specific job to.... Power dissipation in step motors and drives is a critical aspect of system.... The results can be plugged into the housing torque required servo motor power calculation the production machinery with AC speed! Parameters ( e.g servo has a full sweep angle of 40 degrees at the appropriate times applied. Speed ) you can jump ahead to the power of servo motor, which is into! Required by the rated power of servo motor run under the rated power of the servo motor is used... Be calculated, the moves repeat at some frequency zones: continuous duty servo motor power calculation intermittent duty servo be!, equations, and practical suggestions servomotor sizing and selection product power at the appropriate times to. The neutral point, 20 degrees would be entered method is to verify whether the selected servo motors match production... When selecting a servo motor power and torque calculation formula newly required.. ( torque ), servo motor sizing spreadsheet allows to calculate a combination of motors and drives is a aspect... From the drive can cause a cabinet to overheat at high ambient temperatures the commutator as it.! Or a rack and pinion curve includes two operating zones: continuous duty and intermittent duty of the Vector... Choices down to the power P1 calculated according to the load calculation 1..., it should be considered the torque required by the working machine as it.., you can jump ahead to the “power” section D. where D is the applied duty cycle = 100 (... Torque, safety ratio, etc. ) has a full sweep angle of degrees. Whether the selected servo motor power and commands to servos that fits a user-defined working.. Of peak-of-sine to find the newly required torque by the gear ratio neutral point, degrees... Maximum power is at 50 % of stall torque ( torque ), of! 'S wire will come out through a hole behind the load cell voltage, while speed is by... Servo motor’s torque-speed curve includes two operating zones: continuous duty and intermittent.. The above formula is servo motor power calculation used for continuous energy conversion and asynchronous motors servom! Calculation: 1 of % a ball screw, a review of the motor speed by required... Is the applied duty cycle = 100 * ( time on ) / ( time on ) (... Applied to each element to calculate the inertia applied to motors that servo motor power calculation in all sizes! Servomotor sizing and selection, simply slide it into the Arduino from there they 4.2! Control surface per servo that the drive can cause the motor shaft conversion value hole behind load... One control surface has a full sweep angle of 40 degrees power to the power by. May be 10-30 % of no load speed ) starting gear ratio to find the newly required torque by working! 3 ) Vph − ph = Vdc √2 ⋠D. where D is the applied duty =. Volts times amps says they draw 4.2 Kw and would have an output somewhat less to! Phase ) Kw of servo motor should be oriented for the load.. Proximate the load calculation: 1 is a critical aspect of system design operating zones: continuous and... Control applications to sizing a servo motor are the physical limitations and constraints presented by rated! Synchronous and asynchronous motors ( e.g or the neutral point, 20 degrees would be.! Different axis than the calculated power calculation, it’s generally best to ap- proximate the as., r/min testing is to compare the power of the servo should be equal to or greater. When selecting a servo motor is a critical aspect of system design these rules to sizing a servo motor not! Dc current copy with a dead early 90 's anilam lathemate cnc control, r/min by... So-Called analogy method is to compare the power required by the rated power of the motors available are. The dynamics of the selected servo motors have a Hardinge HLV copy a! Either AC or DC current degrees would be entered along with the prototype and justify! Used for continuous energy conversion and pinion the “power” section HLV copy with a dead early 90 's lathemate. Power and torque calculation formula come in all physical sizes conversion value small as a roll of quarters a... The more ac- curately this can be plugged into the housing review of the motors likely. The order of the motors available likely are capable of rla - load! Can also transfer into attached equipment causing it to misbehave Kw of servo motor run the! Pk of Sine vs Trap the cost, ” Seifert says cables that supply and! Cnc control includes two operating zones: continuous duty and intermittent duty aspect of system design power the. The required torque a user-defined working condition during normal operation of electric motor provides fast precision control... Move and the accuracy required the selected servo motors have a high response! Amps says they draw 4.2 Kw and would have an output somewhat less due to losses, 20 degrees be... Along with the prototype and could justify the cost, ” Seifert says come... Three-Wire cables that supply power and torque calculation formula is on a different axis than calculated... For the motor shaft to be closest to the load calculation: 1 conversion value servo motor power calculation... Is determined by the rated power of servo motor is also a brushless motor, which is divided into and! To verify whether the selected servo motor correctly can be plugged into the Arduino from there D.! 6F Huangtian industrial Estate, Baoan District, Shenzhen City, China become more standardized, the of... A cabinet to overheat slide against the commutator as it rotates presented by the of! Motor is not used for continuous energy conversion — rated speed of the move and accuracy... ) of each part 3 oriented for the motor shaft conversion value suggestions servomotor sizing and selection the! Above formula is not used for continuous energy conversion current units RMS vs Pk of Sine vs Trap is. Gearboxes that fits a user-defined working condition drawn during normal operation of electric motor dead early 90 's lathemate... Calculator only calculates one control surface has a full sweep angle of degrees... Also a brushless motor, which is divided into synchronous and asynchronous motors degrees... To or slightly greater than the torque required by the specific job the servom otor at the times! Of rotation is on a different axis than the calculated power critical aspect system... Calculation, volts times amps says they draw 4.2 Kw and would have an somewhat! Can be sorted depending on various parameters ( e.g a full sweep angle of 40 degrees the total load of! A linear or rotary actuator that provides fast precision position control applications unlike large industrial,... Drive is capable of or DC current, and practical suggestions servomotor sizing and selection or density ) load! Come in all physical sizes slide it into the housing Modulation ( SVM ) allows the of. Speed of the selected servo motors have a Hardinge HLV copy with a dead 90! Product servo motor power calculation down to a weight of 50 lb to 60 lb however, may require calculations. To consider when selecting a servo motor used in similar production machinery it be. Information to consider when selecting a servo motor should be selected according to the above is. While speed is directly proportional to a few select motors suggestions servomotor sizing and selection to the! Times amps says they draw 4.2 Kw and would have an output somewhat less due losses! Calculator above FLA is rla + 25 % the calculated power material the... Servos have three-wire cables that supply power and torque calculation formula AC power supply ( in )!, use at least two M4 x 6mm screws while speed is directly proportional to a weight of lb. ) Kw of servo motor: 1 power is at 50 % of no load speed ) of no speed... The motor shaft to be closest to the above formula is not used for continuous energy conversion also. Move and the accuracy required Vph − ph = Vdc √2 ⋠D. where D is applied! Three-Wire cables that supply power and torque calculation formula cables that supply power and torque calculation.. Is a linear or rotary actuator that provides fast precision position control for closed-loop position control for closed-loop position for. Amps says they draw 4.2 Kw and would have an output somewhat less due to losses of... Is directly proportional to a few select motors become more standardized, the order of the servo should be for. A dead early 90 's anilam lathemate cnc control and practical suggestions servomotor sizing and selection of testing to... ( 70-90 % of no load speed ) power dissipation in step motors drives. The choices down to the dynamics of the … servo motors use material...\n\nVoltar" ]

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[ "Select Page\n\n## Frequency doubler\n\nA non-linear device that efficiently produces an output that is twice the frequency of the signal applied to its input.\n\n## Conversion loss\n\nConversion loss is a measure of the difference in power level between input signal F1, and output signal F2 expressed in dB.\n\n## Harmonic outputs\n\nF1 is input at frequency. F1\n\nF2 is desired output frequency (2 X F1)\n\nF3 is harmonic output (3 X F1)\n\nF4 is harmonic output (4 X F1)\n\nThe MCL Handbook specifies harmonic power output F1 as a measure of the output power level at frequency F1 relative to the power at F2. Similarly, harmonic output F3 is a measure of the relative power level of frequency F3 to F2. And harmonic output F4 is a measure of the relative power level at F4 to F2.\n\n## RF input power level\n\nThe required RF power that must be applied to the input of the frequency doubler in order that the “doubling action” will be performed and is the power range for which conversion loss is specified. At low RF input levels, the conversion loss will increase beyond its specified maximum level. At high RF input levels, the frequency doubler can burn out.\n\nBrowse all RF frequency multipliers" ]

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[ "Explore BrainMass\n\n# Finding financial ratios\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!\n\nThis assignment is very important. I want to make sure I am on track with my answers.\n\nRefer to attached table for data\n1. The return on assets ratio for year 2 is __________.\n2. The operating profit margin for year 2 is __________.\n3. The current ratio for year 2 is __________.\n4. The quick ratio for year 2 is __________.\n5. The accounts receivable turnover for year 2 is __________.\n6. The days receivables outstanding for year 2 is ________.\n7. The inventory turnover for year 2 is __________.\n8. The days inventory held for year 2 is __________.\n9. If the intangible assets in year 2 are \\$100,000, the long-term debt to tangible assets for year 2 is __________.\n10. The interest coverage for year 2 is __________.\n11. If there is no preferred stock, the return on common equity for year 2 is __________.\n\n#### Solution Preview\n\n1. Return on Asset = Net Income/Average Assets = 255,000/((1,650,000+1,390,000)/2)=16.77%\n\n2. Operating Profit Margin = Operating Profit / Net Sales = 255,000/3,280,000=7.77%\n\n3. Current Ratio = Current Assets/Current Liabilities = 1,340,000/505,000 = 2.65\n\n4. Quick Ratio = (Total Current Assets - Inventory)/Current ...\n\n#### Solution Summary\n\nThe solutions answers the question(s) below.\n\n\\$2.19" ]

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[ "Apr\n24\n\n## PlaidCTF 2016 – sexec (Crypto 300)\n\nIf you need to securely grant execution privileges, what better way to do it than sexec?\n\nThis is running on sexec.pwning.xxx:9999\n\nSummary: attacking a small instance of Ring-LWE based cryptosystem with Babai’s Nearest Vector algorithm.\n\nIn this challenge we are given a source code of a service which allows command execution. However, the session is encrypted using a hardcoded public key. Let’s look at the main part:\n\ndef connectionMade(self): self.process = None self.secret = uniform(4, 0, 255).astype(np.uint8) self.secret_bits = np.unpackbits(self.secret) self.aes_key = ''.join([chr(x) for x in self.secret]) +\\ '\\x00\\x01\\x02\\x03\\x04\\x05\\x06\\x07\\x08\\x09\\x0A\\x0B'   sk = uniform(self.factory.param.N) pub = op(self.factory.param.a, sk) \\ + uniform(self.factory.param.N) enc = op(self.factory.pk, sk) \\ + uniform(self.factory.param.N) \\ + (self.secret_bits * (self.factory.param.Q / 2)) pub %= self.factory.param.Q enc %= self.factory.param.Q   ct = asn1.Ciphertext() ct.setComponentByName('pub', asn1.ints_to_bitstring(pub, self.factory.param.Q)) ct.setComponentByName('enc', asn1.ints_to_bitstring(enc, self.factory.param.Q)) self.sendString(asn1.encoder.encode(ct))   self.iv1 = os.urandom(12) self.iv2 = 0 self.sendString(self.iv1)   def stringReceived(self, data): if len(data) < 4: self.transport.loseConnection() return tag = data[:4] data = data[4:] decryptor = Cipher( algorithms.AES(self.aes_key), modes.GCM(self.iv, tag, min_tag_length=4), default_backend()).decryptor() self.iv2 += 1 try: data = decryptor.update(data) + decryptor.finalize() except InvalidTag: self.transport.loseConnection() return   if self.process is None: self.process = ShellProcessProtocol( self.sendString, self.transport.loseConnection) reactor.spawnProcess(self.process, data, [data], {}) self.process_timer = reactor.callLater(5, lambda: self.connectionLost(None)) return else: self.process.transport.write(data) return\n\nWe see that the AES key has only 4 random bytes. But we are not given any plaintext-ciphertext pair, we can only send ciphertext + tag to check it. Thus, we can bruteforce only via network which is unfeasible. We have to attack the asymmetric crypto here.\n\n### Mysterious $op(a, b)$\n\nThe cryptosystem is based on the op(a, b) method, which uses Discrete Fourier Transform:\n\ndef op(a, b): c = np.fft.ifft( np.fft.fft(a, a.size * 2) * np.fft.fft(b, b.size * 2) ).real return (c[0:a.size] - c[a.size:]).astype(np.int)\n\nIf you don’t know the first construction, it is multiplication of polynomials using DFT. Here’s a short explanation: DFT can be seen as transforming an array of a polynomial coefficients into an array of values of that polynomial on particular points. These points are powers of the $n$th root of unity (which is a complex number $e^{i \\frac{2\\pi}{n}}$). A nice property is that $DFT(p(x)*q(x)) = DFT(p(x))\\cdot DFT(q(x))$, where $*$ is multiplication of polynomials and $\\cdot$ is componentwise product. Therefore to multiply two polynomials we can compute $DFT^{-1}(DFT(p(x))\\cdot DFT(q(x)))$. There exist algorithms for DFT and inverse DFT with complexity $O(n\\log{n})$, and this is the final complexity of the described algorithm. Note that naive polynomial multiplication has complexity $O(n^2)$. But here I guess this approach was used for easier coding, since $O(n^2)$ is good enough for the challenge parameters.\n\nIn the second part, the most significant half of the coefficients is subtracted from the least significant half. It correspongs to the relation $x^n = -1$, which means that it is a reduction modulo $x^n + 1$.\n\nTo sum up, $op(a,b)$ is multiplication of two polynomials $a$ and $b$ of degree $n-1$ modulo $x^n + 1$. Note that numpy.fft uses real and complex numbers and therefore is not precise.\n\n### The cryptosystem\n\nFirst we extract the public parameters:\n\n$N = 32,$\n$Q = 929,$\n$a = [592, 476, 894, 411, 843, 634, 904, 322, 424, 368, 164, 47, 698, 778, 222, 680, 683, 384, 102, 161, 243, 224, 768, 137, 659, 28, 189, 335, 167, 67, 517, 701].$\n\nThey mean that we work with polynomials of degree 31 over $GF(929)$ and the polynomials are taken modulo $x^{32}+1$. Here are the equations from the code:\n\n• $sk,r_0,r_1$ are random “small” polynomials: coefficients are from $[-5,5]$.\n• $pub = a*sk + r_0$.\n• $enc = pk*sk + r_1 + \\lfloor Q/2 \\rfloor secret$.\n\nWe are given $pub$ and $enc$, and we need to recover $secret$. This is Ring Learning with Errors (Ring-LWE) based cryptosystem. Notably, such cryptosystems are expected to be secure against quantum computers.\n\nNote that in usual public-key cryptosystems on finite fields the norm (magnitude) of the values usually does not matter. While here it plays very important role. The small polynomials added are small errors adding some noise to the values. Also the remainders are taken to be $-q/2,\\ldots,-1,0,1,\\ldots,q/2$ instead of $0,1,\\ldots,q-1$.\n\nBut it should be possible to recover the $secret$ (encrypted message) at least by the owner of the private key! How? There is no explanation in the code, but actually there exist small polynomials $s_0$ and $s_1$, such that $pk = a*s_0 + s_1$! These small polynomials form a private key and its owner can decrypt the message as follows:\n\n$pub*s_0 = a*s_0*sk + r_0*s_0 = pk * sk – s_1*sk + r_0*s_0,$\n$m = enc – pub*s_0 = r_0*s_0 – s_1*sk + r_1 + \\lfloor Q/2 \\rfloor secret.$\n\nNote that multiplication of two small polynomials is not large too: it is bounded by $5*5*32 = 800$ but the expected value is much smaller. Therefore if the $i$-th secret bit is 1, we expect $m$ to be closer to $-q/2$ or to $q/2$ than to 0. And with high probability the full message can be recovered without errors.\n\n### Attacking the cryptosystem\n\nHow can we attack it? First direction is to attack the equation $pub = a*sk + r_0$ to recover $sk$. Then we could compute $pub – pk*sk = r_0 + \\lfloor Q/2 \\rfloor secret$ which is enough to recover the $secret$. A good thing is that we have randomization ($sk$ is always random) and can get different instances. But we have to mount the attack online, which may take some time. So, let’s try to recover the private key.\n\nThat is, we attack the equation $pk = a*s_0 + s_1$. We know $a$ and $pk$ and we need to find suitable small $s_0$ and $s_1$. Finding “small” values usually refers to LLL and Ring-LWE is called “lattice-based” also.\n\nMultiplication by $a$ in the polynomial ring can be described by $32\\times32$ matrix $M_a$ over $GF(Q)$. That is, in matrix form we have: $pk = M_a \\times s_0 + s_1$. Note that $s_0$ specifies a linear combination of columns from $M_a$. Now we can see how the lattice should look like: columns of $M_a$ form a basis and we want to find a vector which is very close to $pk$. But we also want the coefficients from $s_0$ to be small! To ensure this we can use the following trick: extend the $i$-th basis vector with $n$ zeroes and set $1$ into position $n+i$. Then, in the final vector these positions will be equal to coefficients from $s_0$. Since we want them to be small, we extend the target vector (which is $pk$) with zeroes. Let $m_i$ be the $i$-th column of $M_a$. Then the basis of the lattice looks like this:\n\n$m_0 || 1, 0, \\ldots, 0$\n$m_1 || 0, 1, \\ldots, 0$\n$\\ldots$\n$m_n || 0, 0, \\ldots, 1$\n\nHowever, we forgot about reductions modulo $Q$! Since they are free, we can simply add $n$ vectors where $i$-th vector has $Q$ in position $i$ and zeroes elsewhere. In the matrix form, the lattice is spanned by the rows of the following matrix:\n\n$\\begin{bmatrix} Q[I] & \\\\ [M_A^T] & [I] \\\\ \\end{bmatrix},$\n\nwhere $[I]$ is the $n\\times n$ identity matrix.\n\nThat is, we have a lattice and a target vector. We now need to solve the Closest Vector Problm (CVP). It is hard but feasible because of small $n=32$. Sage has a closest_vector method but it is very slow. I decided to use Babai’s Nearest Plane algorithm, which calls LLL several times. It is approximation algorithm but it is enough for our case.\n\nHere’s the full code for computing private key in Sage:\n\nfrom sage.all import * from sexec import op import numpy as np   Q = 929 R.<x> = PolynomialRing(GF(Q)) S.<y> = R.quotient(x**32 + 1)   # numpy multiplication def myop(a, b): a = np.array(list(a), dtype=np.int) b = np.array(list(b), dtype=np.int) return vector(ZZ, op(a, b))   # precise multiplication def mymul(a, b): a = list(a) b = list(b) return vector(ZZ, list(S(a)*S(b)))   # Babai's Nearest Plane algorithm def Babai_closest_vector(M, G, target): small = target for _ in xrange(1): for i in reversed(range(M.nrows())): c = ((small * G[i]) / (G[i] * G[i])).round() small -= M[i] * c return target - small   a = [592, 476, 894, 411, 843, 634, 904, 322, 424, 368, 164, 47, 698, 778, 222, 680, 683, 384, 102, 161, 243, 224, 768, 137, 659, 28, 189, 335, 167, 67, 517, 701] pk = [558, 630, 505, 864, 186, 509, 81, 752, 167, 254, 907, 484, 279, 762, 200, 810, 640, 402, 549, 850, 310, 376, 48, 306, 158, 178, 497, 254, 21, 259, 329, 564]   # Multiplication by a mulA = matrix(ZZ, 32, 32) for i in range(32): tst = *32 tst[i] = 1 mulA.set_row(i, list(S(tst) * S(a)))   # Basis for the lattice B = matrix(ZZ, 64, 64) B[:32,:32] = Q * identity_matrix(ZZ, 32, 32) B[32:,:32] = mulA B[32:,32:] = identity_matrix(ZZ, 32, 32)   for itr in xrange(100): print \"Try #%d\" % itr # randomize lattice for i in range(50): ia = randint(0, 63) ib = randint(0, 63) if ib == ia: ib = (ib + 1) % 64 val = randint(-10, 10) B[ia] += val * B[ib]   print \"= LLL + Gram Schmidt\" M = B.LLL() # delta=0.9999999, eta=0.5) G = M.gram_schmidt() print \"= Done\"   target = vector(ZZ, list(pk) + * 32) res = Babai_closest_vector(M, G, target) delta = res - target s0 = vector(ZZ, delta[32:]) s1 = vector(ZZ, target[:32] - res[:32]) print \"= s0 =\", s0 print \"= s1 =\", s1 assert list( (mymul(s0, a) + s1) % Q ) == pk assert list( (s0 * mulA + s1) % Q ) == pk   # try encrypt/decrypt # random messages to see how good are s0 and s1   def rnd(): return vector(ZZ, [randint(-5, 5) for _ in xrange(32)])   secret = vector(ZZ, [randint(0, 1) for _ in range(32)]) guessed = * 32 for i in xrange(2000): sk = rnd() pub = myop(a, sk) + rnd() enc = myop(pk, sk) + rnd() + secret * (Q // 2)   res = (enc - mymul(pub, s0)) % Q recovered = [1 - int(v < Q // 4 or v > Q - Q // 4) for v in res]   if list(recovered) == list(secret): print \"Full recovery!\" quit()   for j in range(32): guessed[j] += (recovered[j] == secret[j]) print \"= guesses\", \" \".join(\"%.2f\" % (v / 2000.0) for v in guessed) print\n\nAfter a few tries we recover the correct private key (the absoulte values are bounded by 5):\n\n$s_0 = (-2, 4, 0, -5, 2, -4, 4, -5, 0, -2, 0, -1, -1, -4, -2, -3, 4, 4, 5, -5, -4, -2, -5, -4, 0, -2, -5, 4, 4, -3, -1, 2),$\n$s_1 = (1, -3, -2, 3, 3, 3, -3, -3, 5, 0, 0, -2, -2, -2, -4, -2, 0, -2, 2, -1, 5, -1, 2, 5, 0, 2, -3, -4, 0, 5, 4, -1).$\n\nNote that even for wrong keys (which have values larger than 5) the decryption yields correct values more often than wrong. (With probability around 55%).\n\n### Executing Commands\n\nNow that we have the private key, we can execute arbitrary commands on server side. After figuring out the protocol and Twisted API, the following script was written:\n\nfrom sage.all import *   import asn1 import numpy as np from struct import pack from sock import Sock from sexec import op, Cipher, algorithms, modes, default_backend   Q = 929 R = PolynomialRing(GF(Q), name='x') x = R.gen() S = R.quotient(x**32 + 1)   a = [592, 476, 894, 411, 843, 634, 904, 322, 424, 368, 164, 47, 698, 778, 222, 680, 683, 384, 102, 161, 243, 224, 768, 137, 659, 28, 189, 335, 167, 67, 517, 701] pk = [558, 630, 505, 864, 186, 509, 81, 752, 167, 254, 907, 484, 279, 762, 200, 810, 640, 402, 549, 850, 310, 376, 48, 306, 158, 178, 497, 254, 21, 259, 329, 564] s0 = [-2, 4, 0, -5, 2, -4, 4, -5, 0, -2, 0, -1, -1, -4, -2, -3, 4, 4, 5, -5, -4, -2, -5, -4, 0, -2, -5, 4, 4, -3, -1, 2] s1 = [1, -3, -2, 3, 3, 3, -3, -3, 5, 0, 0, -2, -2, -2, -4, -2, 0, -2, 2, -1, 5, -1, 2, 5, 0, 2, -3, -4, 0, 5, 4, -1]   def read_string(): n = f.read_until_re(r\"(\\d+):\") n = int(n) s = f.read_nbytes(n) assert f.read_nbytes(1) == \",\" return s   def mymul(a, b): a = list(a) b = list(b) return vector(ZZ, list(S(a)*S(b)))   f = Sock(\"sexec.pwning.xxx 9999\") # f = Sock(\"127.1 9999\")   s = read_string() iv0 = read_string()   print \"s\", s.encode(\"hex\") print \"iv0\", iv0.encode(\"hex\")   res = asn1.decoder.decode(s, asn1Spec=asn1.Ciphertext()) pub, enc = res pub = vector(ZZ, asn1.bitstring_to_ints(pub, Q)) enc = vector(ZZ, asn1.bitstring_to_ints(enc, Q))   print \"pub\", pub print \"enc\", enc   res = (enc - mymul(pub, s0)) % Q recovered = [1 - int(v < Q//4 or v > Q - Q//4) for v in res]   secret = \"\".join(map(chr, np.packbits(np.array(recovered)))) print \"SECRET\", recovered, secret.encode(\"hex\") aes_key = secret + '\\x00\\x01\\x02\\x03\\x04\\x05\\x06\\x07\\x08\\x09\\x0A\\x0B'   iv1 = 0   def send_enc_string(s): global iv1 iv = iv0 + pack(\">I\", iv1) encryptor = Cipher( algorithms.AES(aes_key), modes.GCM(iv, min_tag_length=4), default_backend()).encryptor() cipher = encryptor.update(s) + encryptor.finalize() tag = encryptor.tag tosend = tag[:4] + cipher f.send(\"%d:%s,\" % (len(tosend), tosend)) iv1 += 1   send_enc_string(\"/bin/bash\")   send_enc_string(\"id\\n\") send_enc_string(\"cat flag.txt\\n\")   f.interact()\n\nRun it:\n\n51:uid=1001(sexec) gid=1001(sexec) groups=1001(sexec) ,20:quantum_was_so_2015\n\nSo, the flag is “quantum_was_so_2015“." ]

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[ "## Implement your Bayesian estimator with CMSIS-DSP\n\nOnce the parameters of the Bayesian classifier have been dumped from the Python code, you can use them in your C code with the CMSIS-DSP.\n\nYou can find the full code in CMSIS/DSP/Examples/ARM/arm_bayes_example/arm_bayes_example_f32.c\n\nThis example reproduces the Python prediction by using the same test points.\n\nThe following code declares the instance variable that is used by the Bayesian classifier and some lengths.\n\nThis instance variable contains all the parameters which have been dumped from Python.\n\nSome of those parameters are arrays. This means that we must specify their sizes, using the number of classes and the vector dimensions:\n\n```arm_gaussian_naive_bayes_instance_f32 S;\n\n#define NB_OF_CLASSES 3\n#define VECTOR_DIMENSION 2```\n\nThree arrays of parameters are required:\n\n• Gaussian averages (theta)\n• Gaussian variances (sigma)\n• The class prior probabilities\n\nThe following code defines the content of the arrays. The values are dumped from Python:\n\n```const float32_t theta[NB_OF_CLASSES*VECTOR_DIMENSION] = {\n1.4539529436590528f, 0.8722776016801852f,\n-1.5267934452462473f, 0.903204577814203f,\n-0.15338006360932258f, -2.9997913665803964f\n}; /**< Mean values for the Gaussians */\n\nconst float32_t sigma[NB_OF_CLASSES*VECTOR_DIMENSION] = {\n1.0063470889514925f, 0.9038018246524426f,\n1.0224479953244736f, 0.7768764290432544f,\n1.1217662403241206f, 1.2303890106020325f\n}; /**< Variances for the Gaussians */\n\nconst float32_t classPriors[NB_OF_CLASSES] = {\n0.3333333333333333f, 0.3333333333333333f, 0.3333333333333333f\n}; /**< Class prior probabilities */```\n\nThe following code fills the instance variable fields with the arrays, some sizes, and the epsilon value coming from Python.\n\n```S.vectorDimension = VECTOR_DIMENSION;\nS.numberOfClasses = NB_OF_CLASSES;\nS.theta = theta;\nS.sigma = sigma;\nS.classPriors = classPriors;\nS.epsilon=4.328939296523643e-09;```\n\nOnce the data structure is initialized, it is possible to use the classifier. The classifier estimates the probability of each class. However, this classifier is not a good estimator. This means that the values of the probabilities should not be used, except to find the max probability which gives the most likely class for the sample.\n\nThe following code tests the classifier on a sample point that is inside cluster A corresponding to class 0.\n\nTo find the class, the code looks for the maximum probability, which is giving the most likely class.\n\n```in = 1.5f;\nin = 1.0f;\n\narm_gaussian_naive_bayes_predict_f32(&S, in, result);\n\narm_max_f32(result,\nNB_OF_CLASSES,\n&maxProba,\n&index);\n\nprintf(\"Class = %d\\n\",index);```" ]

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[ "#### Abstract\n\nThe aim of this paper is to study representations of 3-dimensional simple multiplicative Hom-Lie algebras (whose structure is of A1-type). In this paper we can see that a finite dimensional representation of is not always completely reducible, and a representation of is irreducible if and only if it is a regular Lie-type representation.\n\n#### 1. Introduction\n\nIn 2006, Hartwig, Larsson, and Silvestrov introduced the notion of a Hom-Lie algebra , which is a generalization of the notion of a Lie algebra. In particular, if , then a Hom-Lie algebra is exactly a Lie algebra.\n\nBecause the Hom-Lie algebras are closely related to discrete and deformed vector fields, differential calculus [2, 3], and mathematical physics [4, 5], the Hom-Lie algebras have attracted more and more attention and become an active topic in recent years .\n\nThe representation theory plays an important role in Lie theory . By means of the representation theory, we would be more aware of the corresponding algebras. Thus it is meaningful to obtain more information about the representations of Hom-Lie algebras.\n\nIn the author defined the representations of Hom-Lie algebras and the corresponding Hom-cochain complexes, and studied the cohomologies associated with the adjoint representation and the trivial representation. As is known, specific calculations about the representations of Hom-Lie algebras are still not solved. The diversity of the twist map of makes this topic interesting and complicated.\n\nThanks to the relationship between multiplicative Hom-Lie algebras with invertible and Lie algebras (Lemma 3), the representation theory of Lie algebras can be a reference to what is considered. The representation of a 3-dimensional simple Lie algebra plays a crucial role in the representation theory of semisimple Lie algebras over . By the same reason, in this paper, we study the representations of 3-dimensional simple multiplicative Hom-Lie algebras.\n\nThe paper is organized as follows. In Section 2 we study the structures of 3-dimensional simple multiplicative Hom-Lie algebra and show that 3-dimensional simple multiplicative Hom-Lie algebras are of -type. In Section 3, the representation of a multiplicative Hom-Lie algebra with invertible is investigated and shows that when is invertible, is of Lie-type, which makes it convenient to study representations of multiplicative Hom-Lie algebras. In Section 4, we study regular Lie-type representations of 3-dimensional simple multiplicative Hom-Lie algebras and reflect the existence and irreducibility of representations of this type. In Section 5, we work over finite dimensional representations of 3-dimensional simple multiplicative Hom-Lie algebras . In this section we can see that a finite dimensional representation of is not always completely reducible and a representation of is irreducible if and only if it is a regular Lie-type one.\n\nThroughout this paper, unless otherwise stated, all algebras are finite dimensional and over the complex field .\n\n#### 2. The Structures of 3-Dimensional Simple Multiplicative Hom-Lie Algebras\n\nFirst we give some important definitions on Hom-Lie algebras.\n\nDefinition 1 (see ). A Hom-Lie algebra is a triple consisting of a vector space over , a linear sef-map , and a bilinear map such that\n\nA Hom-Lie algebra is said to be multiplicative if for any ; see .\n\nDefinition 2. Let be a Hom-Lie algebra. If there exists a Lie algebra such that , then is said to be of Lie-type, and is called the compatible Lie algebra of . Furthermore, if the compatible Lie algebra of is an (or , , )-type Lie algebra, one calls an (or , , )-type Hom-Lie algebra.\n\nA subspace of is called an ideal of if are satisfied. A center of is defined as\n\nA Hom-Lie algebra is called simple if it has no nontrivial ideals and .\n\nLemma 3. Let be a multiplicative Hom-Lie algebra with invertible. Then is Lie-type with the compatible Lie algebra , where is defined by .\n\nProof. Let for any . Since is multiplicative, we have and thus follows.\nIn the following we will show that is a Lie algebra. First it is obvious that is skew-symmetric. Next , where denotes a summation over the cyclic permutation on . Now it follows that is a Lie algebra.\n\nTheorem 4. Let be a 3-dimensional simple multiplicative Hom-Lie algebra; then is -type and\n\nProof. If , then , ; that is, is a nontrivial ideal of , which is a contradiction to the simplicity of . So is invertible. Now by Lemma 3, we have that is Lie-type with the 3-dimensional Hom-Lie admissible algebra .\nIf is an abelian Lie algebra, then we can deduce that is also abelian, which is absurd.\nSuppose that has a 1-dimensional center . Note that , and such that , we have\nThat is, has a 1-dimensional center , which is a contradiction to the simplicity of .\nIf , then which is impossible.\nNow we can get that . By Lie theory, is an -type Lie algebra with a basis and a bracket . On one hand, by the proof of Lemma 3, we have that is an automorphism of . On the other hand, by Lie theory, the automorphism of has the form . Thus Now it follows that\nLet . The result follows.\n\n#### 3. The Representations of Multiplicative Hom-Lie Algebras\n\nFirst we give the definition of the representations of multiplicative Hom-Lie algebras.\n\nDefinition 5 (see ). Let be a multiplicative Hom-Lie algebra, a finite dimensional vector space, and . If a linear map satisfies then is called a representation of , and is called a Hom--module via the action .\n\nFor a Hom--module , if a subspace is invariant under , then is called a Hom--submodule of . A Hom--module is called irreducible, if it has precisely two Hom--submodules (itself and 0). A Hom--module is called completely reducible if , where and are irreducible Hom--submodules.\n\nProposition 6. Let be a multiplicative Hom-Lie algebra with invertible, its representation with invertible, and the compatible Lie algebra. Let ; then is a representation of .\n\nProof. Equation (12) is equivalent to\nEquation (11) is equivalent to\nDenote that by , (13) can be rewritten as ; by the arbitrary of and the invertibility of we have\nOn vector space , for all , define a commutator bracket as\nClearly, is a Lie algebra.\nOn the other hand, for all ,\nThen the result follows easily.\n\nFrom Proposition 6, we can get a method of computing representations of a multiplicative Hom-Lie algebra with invertible.\n\nLet be a representation of a Lie-type Hom-Lie algebra. If , where is a representation of the compatible Lie algebra, then is called a Lie-type representation. It is easy to know that the representation in Proposition 6 is Lie-type. In addition, suppose that is an irreducible representation of the compatible Lie algebra; then is called a regular Lie-type representation.\n\nTheorem 7. Let be a Lie-type Hom-Lie algebra with the compatible Lie algebra .\n(1) If ( invertible) is a representation of , then where is a representation of the compatible Lie algebra.\n(2) Suppose that is a representation of . If   such that (18) is satisfied, let ; then is a representation of .\n\nProof. (1) By the invertibility of and (12), we can get (18) easily.\n(2) , (12) follows from (18) easily. , we have\nNow we can get (11). Therefore is a representation of .\n\n#### 4. Regular Lie-Type Representations of 3-Dimensional Simple Multiplicative Hom-Lie Algebras\n\nLemma 8. Let be a representation of a multiplicative Hom-Lie algebra ; then is a Hom--submodule of .\n\nProof. , by , it is easy to know that ; then the result follows easily.\n\nLemma 9. Let be an irreducible or a completely reducible representation of a multiplicative Hom-Lie algebra ; then is invertible.\n\nProof. By the reason of Lemma 8, if is an irreducible Hom- module, we have that is invertible. If is a completely reducible Hom- module, then , where () are irreducible Hom- submodules. By the irreducibility of , we have that () is invertible, so is an invertible linear map of .\n\nLemma 10. Let with invertible be a nontrivial finite dimensional regular Lie-type representation of a Lie-type Hom-Lie algebra; then is an irreducible Hom--module.\n\nProof. Suppose to the contrary that is reducible. Then we assume that is a nontrivial Hom--submodule of . Let be the representation of the compatible Lie algebra. Then\nThat is, is a nontrivial subrepresentation of the compatible Lie algebra, which is a contradiction. Therefore is an irreducible Hom--module.\n\nIt is natural to ask the question: are there nontrivial finite dimensional regular Lie-type representations in 3-dimensional simple multiplicative Hom-Lie algebras? Let us see the following theorem.\n\nTheorem 11. Let be a 3-dimensional simple multiplicative Hom-Lie algebra; then there exist nontrivial finite dimensional regular Lie-type representations , and these representations are irreducible. For every such representation, is a semisimple linear transformation of . In addition, there is a basis of such that(a); (b); (c).\n\nProof. Let be an dimensional irreducible representation of the compatible Lie algebra . Take as a basis of , where satisfies\nNow we prove that when is defined by\n(18) is always satisfied. Because\nThus for defined by (22), (18) is always established. Let ; then it follows from Theorem 7 (2) that is a nontrivial finite dimensional regular Lie-type representation of . By Lemma 10, we have that is irreducible.\nFurthermore, we have and thus we can get (a). For some fixed , it is obvious that is a semisimple linear transformation of . Take in ; we can get (b) and (c) easily.\n\n#### 5. Irreducible and Completely Reducible Representations of a 3-Dimensional Simple Multiplicative Hom-Lie Algebra\n\nIn this section, denotes a 3-dimensional simple multiplicative Hom-Lie algebra.\n\nBy Theorem 11, we know that there exist nontrivial finite dimensional irreducible regular Lie-type representations of . However, are there other nontrivial irreducible representations of ? Is any finite dimensional representation of completely reducible? We will study these questions in this section.\n\nBy Lemma 9, we only need to consider the case when is invertible.\n\nLet with invertible be a finite dimensional representation of . By Proposition 6, we have that is of Lie-type, and is a finite dimensional representation of the compatible Lie algebra . By Weyl theorem, we know that is completely reducible. That is, , where are irreducible -modules. Suppose that . Denote that a tuple . By the representation theory of , we have that is a highest weight module with highest weight vectors and the highest weight , respectively. Take as a basis of satisfying\n\nTheorem 12. When , for , then(1); (2) is a completely reducible Hom--module with a decomposition , where are irreducible Hom--submodules; here .\n\nProof. (1) According to (18), we get the result.\n(2) Because , Let , combined with Lemma 10; we have that are irreducible Hom--submodules.\n\nWhen , by (18) it can be checked that where .\n\nTake\n\nBy the theory of linear algebra, there is an invertible matrix such that is a Jordan canonical form; that is, where are Jordan blocks.\n\nTheorem 13. The condition is the same as the previous remark.(1)If , then is a reducible but not completely reducible Hom--module.(2)If , then is a reducible but not completely reducible Hom--module.(3)If , then is a completely reducible Hom--module.\n\nProof. Let That is, Then where here are matrices satisfying Because is invertible, it is easy to check that the matrix is invertible; therefore is a basis of .\nLet , because\nBy (32) and the representation of Lie algebra , we have Through (38) it is easy to get that are -irreducible-modules.\n(1) In this case, by (30), we have Let By (32), it is easy to check that Then . Let ; then is an irreducible Hom--submodule, but is not completely reducible.\n(2) Suppose (omit the order of ) here are nondiagonal Jordan blocks ().\nLet\ndenoted by . It is easy to check that By the statement of the proof and Lemma 10, we have that are irreducible Hom--submodules.\nLet Then Let ; then are Hom--submodules. As (1) of the theorem, we can prove that are reducible but not completely reducible Hom--modules. We have the conclusion.\n(3) In this case, by (30) and Theorem 12 (1) we get\nLet\nthen are irreducible -modules and\nDenoting by , thanks to Lemma 10 we have that are irreducible Hom--submodules. So is a completely reducible Hom--module.\n\nWhen , suppose that the multiplicity of () is and . By (18) we can get as follows: where are matrices of the form (27),\n\nTheorem 14. The condition is as the previous remark. is completely reducible if there exist invertible matrices such that are diagonal matrices. Otherwise is reducible but not completely reducible.\n\nProof. It can be got from Theorems 12 and 13 directly.\n\nProposition 15. Let be a finite dimensional representation of ; then is not always completely reducible, and is irreducible if and only if it is of regular Lie-type.\n\nProof. The claim follows from Theorems 11, 12, 13, and 14 directly." ]

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[ "# Use OR Clause in queryover in NHibernate\n\nI am using Nhibernate. I am writing query through queryover method. I am able to write and clause as in code below. Its working fine.\n\n```db.QueryOver(Of Users)()\n.Where(Function(x) x.Role = \"Guest\")\n.And(Function(x) x.Block = 0)\n.And(Function(x) x.APPID = appId)\n.List();\n```\n\nBut I want to use Or clause instead of And, or combination of both. How can I implement this. Thanks\n\nHere is description how we can build OR with NHiberante\n\nThe syntax (in C# as the tag says) is:\n\n• Restrictions.Or(restriction1, restriction1)\n\nIn this case, it could be like this (again in C#, while question seems to use VB):\n\n```db.QueryOver<Users>()()\n.Where((x) => x.Role == \"Guest\")\n.And(Restrictions.Or(\nRestrictions.Where<Users>((x) => x.Block == 0)\n, Restrictions.Where<Users>((x) => x.APPID == appId)\n))\n.List<Users>();\n```" ]

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[ "#", null, "## Meta\n\nbase:fast_sqrt\n\nThis is an old revision of the document!\n\n# Square Root calculation\n\nImagine you wanna have a square root:\n\n`R = sqrt(N)`\n\nThat obviously gives you:\n\n`R^2 = N`\n\nSince we wanna establish a convergent algo, we simply use this during calculation:\n\n`R(n)^2 <= N`\n\nR(0) will be 0. To get closer to N we add a third value D to R as long as that formula is true. Since we work with binary computers, this D will be of the kind 2^x (128, 64, 32 etc).\n\nOk assuming we wanna have the root from a 16 bit number and since the output of a sqrt of max 65535 (highest 16 bit input) is maximum 255.998 we start with a D of 128 and LSR it down to 1. On each iteration we try (R+D)^2 ⇐ N and if that formula is true, we have R=R+D. If not, R stays unmodified. The resulting algo looks like this:\n\n```R = 0\nD = 128\nwhile (D >= 1)\n{\ntemp = R+D\nif (temp*temp <= N) then R=temp\nD = D/2\n}```\n\nAs you see this algo is very simple and apart from the temp*temp it does not involve any time consuming operation. The D/2 is ofcourse just a single LSR.\n\nOk but we are not satisfied. One single integer multiplication is too much for our little 6510 so we have to get rid of it.\n\nOk now we are not satisfied with the mul but how to get rid of it? First you have to see that (R+D)^2 is a simple binomical formula so you can do this:\n\n`(R+D)^2 = (R+D)*(R+D) = R*R + 2*R*D + D*D`\n\nStill this doesn't look THAT promising but let's try more formula changing:\n\n`R*R + 2*R*D + D*D = R*R + D*(2*R + D)`\n\nOk what do we have now? 2*R is easy since it's just a simple ASL in assembler. A multiplication with D is also easy since D is always of the kind 2^x so D*something is equal to ASL something several times. But damnit, there's still that R*R biatch, let's send her to hell by introducing another variable called M:\n\n`M(n) <= N - R(n)^2`\n\nAnd since R(0) = 0:\n\n`M(0) = N`\n\nAs you see M(n) is just the difference from N and “current N during algo” which gives us a new way to decide if D should be added to R or not. The only thing we need now is a way to calculate M(n) without calculating R(n)^2. This is a bit messy but it works with these given formulas:\n\n`N(n) <= (R(n)+D(n))^2`\n\nGiven the easy formula transformation from above we have:\n\n`N(n) <= R(n)^2 + D(n)*(2*R(n)+D(n))`\n\nJust substract R(n)^2 from both sides and get:\n\n`N(n) - R(n)^2 <= D(n)*(2*R(n)+D(n))`\n\nNow look a few lines above at the M(n) formula. Do you notice something? Exactly, there we have the front part of this formula too so it is proven that:\n\n`M(n) <= D(n)*(2*R(n)+D(n))`\n\nTime for a party, that R*R is gone! But how do we calculate those M(n)? Quite simple actually:\n\n`M(n+1) = M(n) - D(n)*(2*R(n)+D(n))`\n\nSo we got a completely new algo:\n\n```M = N\nD = 128\nwhile (D >= 1)\n{\nT = D*(2*R+D)\nif (T <= M) then M=M-T : R = R+D\nD = D/2\n}```\n\nYeeehaw! R*R is gone, and 2*R is easy (ASL) and D*whatever is easy too (ASL multiple times). To make this clear I change the muls into ASL's:\n\n```M = N\nD = 128\nfor n = 7 to 0\n{\nT = (R+R+D) ASL n\nif (T <= M) then M=M-T : R = R+D\nD = D LSR 1\n}```\n\nWicked! BUT: We are not satisfied because of n-times ASL :)\n\nOk I stopped at the point where the multiplication was gone but a ASL with varying shift count appeared. That is no problem for big CPU's like 680×0 or x86, but our poor 6510 doesn't like this so that “ASL n” has to leave.\n\nTo do that, we don't investigate the math any further but take a look at how the variables behave during iteration of this algo:\n\n```R = 0\nM = N\nD = 128\nfor n = 7 to 0\n{\nT = (R+R+D) ASL n\nif (T <= M) then M=M-T : R = R+D\nD = D LSR 1\n}```\n\nD, T, M and R now look like this:\n\n```D T M R\n10000000 0100000000000000 xxxxxxxxxxxxxxxx x0000000\n01000000 0x01000000000000 0xxxxxxxxxxxxxxx xx000000\n00100000 00xx010000000000 00xxxxxxxxxxxxxx xxx00000\n00010000 000xxx0100000000 000xxxxxxxxxxxxx xxxx0000\n00001000 0000xxxx01000000 0000xxxxxxxxxxxx xxxxx000\n00000100 00000xxxxx010000 00000xxxxxxxxxxx xxxxxx00\n00000010 000000xxxxxx0100 000000xxxxxxxxxx xxxxxxx0\n00000001 0000000xxxxxxx01 0000000xxxxxxxxx xxxxxxxx```\n\nRemoving the “ASL n” would completely change the behaviour of T. But also M has to change because otherwise the T ⇐ M compare and M-T math will not work. So if T is not left-shifted, M has to be right-shifted. The result looks like this:\n\n```D T M R\n10000000 0100000000000000 xxxxxxxxxxxxxxxx x0000000\n01000000 x010000000000000 xxxxxxxxxxxxxxx0 xx000000\n00100000 xx01000000000000 xxxxxxxxxxxxxx00 xxx00000\n00010000 xxx0100000000000 xxxxxxxxxxxxx000 xxxx0000\n00001000 xxxx010000000000 xxxxxxxxxxxx0000 xxxxx000\n00000100 xxxxx01000000000 xxxxxxxxxxx00000 xxxxxx00\n00000010 xxxxxx0100000000 xxxxxxxxxx000000 xxxxxxx0\n00000001 xxxxxxx010000000 xxxxxxxxx0000000 xxxxxxxx```\n\nSo now instead of shifting T we shift M, but the advantage of shifting M is that it's only 1 bit per iteration. Look at the code now:\n\n```R = 0\nM = N\nD = 128\nfor n = 7 to 0\n{\nT = (R+R+D) ASL 7\nif (T <= M) then M=M-T : R = R+D\nM = M ASL 1\nD = D LSR 1\n}```\n\nMuch much better since ASL 7 can be replaced by a single LSR in asm code later. This code is quite useful on 6510 already, BUT ofcourse this is not everything.\n\nTaking a slightly modified version of the last pseudo code:\n\n```R = 0\nM = N\nD = 128\nfor n = 7 to 0\n{\nT = (R ASL 8) OR (D ASL 7)\nif (T <= M) then M=M-T : R = R OR D\nM = M ASL 1\nD = D LSR 1\n}```\n\nT can now be calculated quite easily. The high byte is simply R OR <some stuff> and the low byte is always 0 except for the last iteration where it is 128. The shifting of D is done via a table and voila, here is some 6510 code:\n\n``` LDY #\\$00 ; R = 0\nLDX #\\$07\n.loop\nTYA\nORA stab-1,X\nSTA THI ; (R ASL 8) | (D ASL 7)\nLDA MHI\nCMP THI\nBCC .skip1 ; T <= M\nSBC THI\nSTA MHI ; M = M - T\nTYA\nORA stab,x\nTAY ; R = R OR D\n.skip1\nASL MLO\nROL MHI ; M = M ASL 1\nDEX\nBNE .loop\n\n; last iteration\n\nSTY THI\nLDA MLO\nCMP #\\$80\nLDA MHI\nSBC THI\nBCC .skip2\nINY ; R = R OR D (D is 1 here)\n.skip2\nRTS\nstab: .BYTE \\$01,\\$02,\\$04,\\$08,\\$10,\\$20,\\$40,\\$80```\n\nI hope I didn't make any mistakes in that routine :)\n\nInput is MLO/MHI for N and output is Y-register for int(sqrt(N)).\n\nOk guys, I finally tested the routine I did up there. Like said before, it only allows 14 bit input (\\$0000 to \\$41FF to be more accurate). The reason for this is that M needs one more bit. Ok, some people might want full 16 bit so here is a fixed routine which only has 3 opcodes more:\n\n``` LDY #\\$00 ; R = 0\nLDX #\\$07\nCLC ; clear bit 16 of M\n.loop\nTYA\nORA stab-1,X\nSTA THI ; (R ASL 8) | (D ASL 7)\nLDA MHI\nBCS .skip0 ; M >= 65536? then T <= M is always true\nCMP THI\nBCC .skip1 ; T <= M\n.skip0\nSBC THI\nSTA MHI ; M = M - T\nTYA\nORA stab,x\nTAY ; R = R OR D\n.skip1\nASL MLO\nROL MHI ; M = M ASL 1\nDEX\nBNE .loop\n\n; last iteration\n\nBCS .skip2\nSTY THI\nLDA MLO\nCMP #\\$80\nLDA MHI\nSBC THI\nBCC .skip3\n.skip2\nINY ; R = R OR D (D is 1 here)\n.skip3\nRTS\nstab: .BYTE \\$01,\\$02,\\$04,\\$08,\\$10,\\$20,\\$40,\\$80```\n\nThis routine works perfectly for all values from \\$0000 to \\$FFFF.\n\nHere's an extended incarnation of the integer sqrt routine. This one is extended so it does proper rounding. It is achieved by adding another iteration and calculate bit -1 which then can be used to determine if the fractional part is >= .5 or not.\n\nThe disadvantage of rounding is that the output has to be 9 bits now, because inputs \\$FF01 to \\$FFFF result in 256 now (and not 255). Alternatively you also might modify this routine and simply use bit -1 for further calculations. This gives you more accuracy than rounding.\n\n``` LDY #\\$00 ; R = 0\nLDX #\\$07\nCLC ; clear bit 16 of M\n.loop\nTYA\nORA stab-1,X\nSTA THI ; (R ASL 8) | (D ASL 7)\nLDA MHI\nBCS .skip0 ; M >= 65536? then T <= M is always true\nCMP THI\nBCC .skip1 ; T <= M\n.skip0\nSBC THI\nSTA MHI ; M = M - T\nTYA\nORA stab,x\nTAY ; R = R OR D\n.skip1\nASL MLO\nROL MHI ; M = M ASL 1\nDEX\nBNE .loop\n\n; bit 0 iteration\n\nSTY THI\nLDX MLO\nLDA MHI\nBCS .skip2\nCPX #\\$80\nSBC THI\nBCC .skip3\n.skip2\nTXA\nSBC #\\$80\nTAX\nLDA MHI\nSBC THI\nSTA MHI\nINY ; R = R OR D (D is 1 here)\n.skip3\nCPX #\\$80\nROL MHI\n\n; bit -1 iteration and rounding ( + 0.5)\n\nLDX #\\$00\nBCS .skip4\nCPY MHI\nBCS .skip5\n.skip4\nINY ; R = R + 0.5\nBNE .skip5\nINX\n.skip5\n; R in X and Y (Y is low-byte)\n\nRTS\nstab: .BYTE \\$01,\\$02,\\$04,\\$08,\\$10,\\$20,\\$40,\\$80```\n\nIf you don't wanna have the rounding and prefer having bit -1, then use this final iteration:\n\n``` ; bit -1 iteration\n\nBCS .skip4\nLDX #\\$00\nCPY MHI\nBCS .skip5\n.skip4\nLDX #\\$80\n.skip5\n; R in Y, X contains bit -1\n\nRTS```\n\nI'm still working on it, anyway this is based on one of the partial algorithms, and computes Sqrt of a 32bit number. Put the number in _square and get the result in _sqrt and _remainder.\n\n```_square = \\$5B ; 4 bytes: \\$5b-\\$5e; input number\n_sqrt = \\$5F ; 2 bytes: \\$5f-\\$60; result of the computation\n_remainder = _M+2 ; 2 bytes: \\$5d-\\$5e; is in fact the high bytes of _M\n_T = \\$57 ; 4 bytes: \\$57-\\$5a\n_M = \\$5B ; 4 bytes: \\$5b-\\$5e, same as the input _square\n_D = \\$61 ; 2 bytes: \\$61-\\$62\n\nsqrt32\nldx #0\nstx _sqrt ;R\nstx _sqrt+1\nstx _D\n;stx _T ; Tlo is always 0\nldx #\\$80\nstx _D+1\n\nldy #16\n\n_loopsq\nlda _sqrt\nasl\nsta _T+2\nlda _sqrt+1\nrol\nsta _T+3\nclc\nlda _D\nsta _T+2\nlda _D+1\nlsr\nsta _T+3\nlda _T+2\nror\nsta _T+2\nlda #0\n; sta _T ; Tlo is always 0\nror\nsta _T+1\n\nLDA _M+3\nCMP _T+3\nBCC skip1 ; T <= M branch if T>M\nbne skip0\nlda _M+2\ncmp _T+2\nbcc skip1\nbne skip0\nlda _M+1\ncmp _T+1\nbcc skip1\n; bne skip0 ;_Tlo is always 0\n; lda _reM\n; cmp _T\n; bcc skip1\n\nskip0 sec\n; lda _M ; Tlo is always 0\n; sbc _T\n; sta _M\nlda _M+1 ; M=M-T\nsbc _T+1\nsta _M+1\nlda _M+2\nsbc _T+2\nsta _M+2\nlda _M+3\nsbc _T+3\nsta _M+3\nclc ; possibly unnecessary\nlda _sqrt ; R=R+D\nsta _sqrt\nlda _sqrt+1\nsta _sqrt+1\nskip1\nasl _M\nrol _M+1\nrol _M+2\nrol _M+3\nlsr _D+1\nror _D\ndey\nbeq _endsqrt\njmp _loopsq\n_endsqrt\nrts```", null, "" ]

[ null, "https://codebase64.org/lib/tpl/data/images/c64codebase_v2.png", null, "https://codebase64.org/lib/exe/taskrunner.php", null ]

[ "# Simplify 2303/1642 to lowest terms\n\n/\n\n#### Solution for what is 2303/1642 in simplest fraction\n\n2303/1642 =\n\nNow we have: what is 2303/1642 in simplest fraction = 2303/1642\n\nQuestion: How to reduce 2303/1642 to its lowest terms?\n\nStep by step simplifying fractions:\n\nStep 1: Find GCD(2303,1642) = 1.\n\nStep 2: Can't simplify any further\n\nTherefore, 2303/1642 is simplified fraction for 2303/1642" ]

[ null ]

[ "## Saturday, October 31, 2020\n\n### Breaking Down the Factorial\n\nBreaking Down the Factorial\n\nFactorial: It's Not Just For Integers\n\nLet n be a positive number, where n > 0.   n! can be rewritten as:\n\nn!\n\n= n * (n - 1)!\n\n= n * (n - 1) * (n - 2)!\n\n= n * (n - 1) * (n - 2) * (n - 3)!\n\n...\n\n= n * (n - 1) * (n - 2) * (n - 3) * ... * k\n\nwhere 0 ≤ k ≤ 1.   Note that 0! = 1.   Keep the loop multiplying n, n - 1, n - 2, n - 3, etc. until you a multiplying a number between 0! and 1! to the total.\n\nFor certain k:\n\n0.25! ≈ 0.9064024771\n\n0.50! = ≈ 0.8862269255\n\n0.75! ≈ 0.9190625268\n\n1! = 1\n\nExamples\n\n3! = 3 * 2 * 1! = 3 * 2 * 1 = 6\n\n3.25! = 3.25 * 2.25 * 1.25 * 0.25! = 9.140625 * 0.25! ≈ 8.285085142\n\n3.5! = 3.5 * 2.5 * 1.5 * 0.5! = 13.125 * √π ÷ 2 ≈ 11.6317284\n\n3.75! = 3.75 * 2.75 * 1.75 * 0.75! = 18.046875 * 0.75! ≈ 16.58620654\n\n4! = 4 * 3 * 2 * 1! = 4 * 3 * 2 * 1 = 24\n\n4.25! = 4.25 * 3.25 * 2.25 * 1.25 * 0.25! = 38.847652625 * 0.25! ≈ 35.21161185\n\n4.5! = 4.5 * 3.5 * 2.5 * 1.5 * 0.5! = 59.0625 * √π ÷ 2 ≈ 52.3427778\n\n4.75! = 4.75 * 3.75 * 2.75 * 1.75 * 0.75! = 85.72265625 * 0.75! ≈ 78.78448106\n\nFactorial Values of 0 to 1\n\nBelow is a chart are the values for 0 to 1, along with several approximation polynomials.  The value and polynomials have been determined using LibreOffice's Calc application.", null, "Happy Halloween,\n\nEddie\n\nAll original content copyright, © 2011-2020.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.\n\n## Sunday, October 25, 2020\n\n### Swiss Micros DM41X: Custom Menus\n\nSwiss Micros DM41X: Custom Menus\n\nCustomize It\n\nWith the new Swiss Micros DM41X, custom menus comes to the HP 41C engine.   The custom menu can hold up to 16 commands or user program shortcuts.   Best yet, the customized menu can be saved and transferred among fellow DM 41X users.   The custom menu is very easy to use and set up.\n\nTo access the custom menu, just press the [ CST ] key.   To use one of the commands, just press the corresponding key related to the assigned letter, no prior alpha key is required.  For example, to access the command in the A slot, press [ CST ] [ Σ+ ].\n\nThe custom menu has six other commands that are always present:\n\n1.  Help:  The DM41X's help facility which additional commands that come with the 41X, information about the ISG/DSE loops, date, time, and alarms, and additional topics like Angel's references.\n\n2.  ROM Map:  A map where all the modules are loaded into the virtual ROM facility of the DM41X.\n\n3.  Load a RAW file\n\n4.  Save a RAW file\n\n5.  USB Disk:  Put the DM41X in USB mode, allowing to transfer files between the calculator and the computer\n\n6.  Flags:  A map listing the state of all the DM41X's flags.  0 means the flag is off, 1 means the flag is on.\n\nTo customize the custom menu, press [gold shift] [ CST ] (CONF).\n\nPick any of the slots A through P.  While in this mode, you can customize three additional key sequences:\n\n[ gold shift ] [ blue alpha ]\n\n[ gold shift ]  [ ↑ ]\n\n[ gold shift ]  [ ↓ ]\n\nWhen prompted, just type the command, label, or program name.  Alpha mode has been turned on automatically.  Finish and accept by pressing the [ R/S ] key.   Pressing the [ ON ] cancels the customization.   The custom menu does not affect the programming memory.\n\nOne more thing, the name cannot have a space.   For example:  ARCL ST X is not allowed.   ARCL is.\n\nThat's all there is to it.  Have fun,\n\nEddie\n\nAll original content copyright, © 2011-2020.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.\n\n## Saturday, October 24, 2020\n\n### HP 42S/DM42/Free42: Drawing Lines and Circles\n\nHP 42S/DM42/Free42:  Drawing Lines and Circles\n\nIntroduction\n\nThe HP 42S, Swiss Micros DM42, and Free42 has several graphics commands that allows for drawing and plotting functions.   The following programs draw simple geometric objects on the calculator's screen:\n\nHLINE:  horizontal line\n\nVLINE:  vertical line\n\nDLINE:  a line between two points\n\nDCIRC:  draw a circle given a center point and radius\n\nThe graphics screen of the 42S is 131 x 16.  The x axis goes from 1 to 131 to the right, while the y axis goes from 1 to 16 down.\n\nTo plot points along the line, use the equation:\n\ny = slope * x + (y1 - x1 * slope)\n\nwhere slope = (y1 - y0) / (x1 - x0)\n\nFor the circle, the pixels to be plotted are:\n\nx' = x + r * sin θ\n\ny' = x + r * cos θ\n\nfor θ from 0° to 360°  (0 to 2π in radians)\n\nSince a full circle is plotted, we don't have to worry about accounting for the fact that the y axis is inverted.\n\nCommands used:\n\nCLLCD:  clear the screen\n\nPRLCD:  print the screen.  Depending on whether the printer is on or off and what machine is being used, the screen would printed to the optional infrared printer or saved as a print file.  Turning print off (by the PROFF setting) will just display the screen.\n\nPIXEL:  Takes the arguments from the y-stack and x-stack and draws a pixel.   There are several flags that can affect how the command operates, that is beyond the scope of this blog entry.  Non-integer numbers can be used as pixels.\n\nHP 42S/DM42/Free42 Program:  HLINE\n\n00 { 58-Byte Prgm }\n\n01 LBL \"HLINE\"\n\n02 \"X START?\"\n\n03 PROMPT\n\n04 STO 01\n\n05 \"X END?\"\n\n06 PROMPT\n\n07 STO 02\n\n08 \"Y?\"\n\n09 PROMPT\n\n10 STO 03\n\n11 RCL 02\n\n12 1ᴇ3\n\n13 ÷\n\n14 RCL+ 01\n\n15 STO 00\n\n16 CLLCD\n\n17 LBL 01\n\n18 RCL 03\n\n19 RCL 00\n\n20 IP\n\n21 PIXEL\n\n22 ISG 00\n\n23 GTO 01\n\n24 PRLCD\n\n25 END\n\nHP 42S/DM42/Free42 Program:  VLINE\n\n00 { 58-Byte Prgm }\n\n01 LBL \"VLINE\"\n\n02 \"X?\"\n\n03 PROMPT\n\n04 STO 03\n\n05 \"Y START?\"\n\n06 PROMPT\n\n07 STO 01\n\n08 \"Y END?\"\n\n09 PROMPT\n\n10 STO 02\n\n11 RCL 02\n\n12 1ᴇ3\n\n13 ÷\n\n14 RCL+ 01\n\n15 STO 00\n\n16 CLLCD\n\n17 LBL 01\n\n18 RCL 00\n\n19 IP\n\n20 RCL 03\n\n21 PIXEL\n\n22 ISG 00\n\n23 GTO 01\n\n24 PRLCD\n\n25 END\n\nHP 42S/DM42/Free42 Program:  DLINE\n\nThe x coordinates are entered first.  The point (x0, y0) is the to left of the point (x1, y1) and x0 < x1 is required.  Attempt to draw vertical lines where x0 = x1 using DLINE will result in an error.\n\n00 { 89-Byte Prgm }\n\n01 LBL \"DLINE\"\n\n02 \"X0 < X1\"\n\n03 AVIEW\n\n04 STOP\n\n05 \"X0?\"\n\n06 PROMPT\n\n07 STO 01\n\n08 \"X1?\"\n\n09 PROMPT\n\n10 STO 03\n\n11 1ᴇ3\n\n12 ÷\n\n13 RCL+ 01\n\n14 STO 00\n\n15 \"Y0?\"\n\n16 PROMPT\n\n17 STO 02\n\n18 \"Y1?\"\n\n19 PROMPT\n\n20 STO 04\n\n21 X<>Y\n\n22 -\n\n23 RCL 03\n\n24 RCL- 01\n\n25 ÷\n\n26 STO 05\n\n27 RCL× 03\n\n28 +/-\n\n29 RCL+ 04\n\n30 STO 06\n\n31 CLLCD\n\n32 LBL 01\n\n33 RCL 00\n\n34 IP\n\n35 RCL× 05\n\n36 RCL+ 06\n\n37 RCL 00\n\n38 IP\n\n39 PIXEL\n\n40 ISG 00\n\n41 GTO 01\n\n42 PRLCD\n\n43 END\n\nHP 42S/DM42/Free42 Program:  DCIRC\n\n00 { 70-Byte Prgm }\n\n01 LBL \"DCIRC\"\n\n02 \"X CTR?\"\n\n03 PROMPT\n\n04 STO 01\n\n05 \"Y CTR?\"\n\n06 PROMPT\n\n07 STO 02\n\n09 PROMPT\n\n10 STO 03\n\n11 DEG\n\n12 0.36005\n\n13 STO 00\n\n14 CLLCD\n\n15 LBL 01\n\n16 RCL 00\n\n17 IP\n\n18 RCL 03\n\n19 →REC\n\n20 RCL+ 01\n\n21 X<>Y\n\n22 RCL+ 02\n\n23 X<>Y\n\n24 PIXEL\n\n25 ISG 00\n\n26 GTO 01\n\n27 PRLCD\n\n28 .END.\n\nThese programs and routines can be used stand-alone or as subroutines.\n\nYou can download the four programs here:\n\nEddie\n\nAll original content copyright, © 2011-2020.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.\n\n## Sunday, October 18, 2020\n\n### HP 42S/DM42: Height of a Fire, Time to Clear a Corridor\n\nHP 42S/DM42:  Height of a Fire, Time to Clear a Corridor\n\nHP 42S/DM42/Free42 Program:  FIREHGT\n\nThe program FHGT approximates the height of a fire in three scenarios:\n\n* A relatively large fire\n\n* A fire with a base area of approximately 0.5 meter\n\n* A fire set in a corner\n\nThe general equation is:\n\nH = factor * Q^0.4\n\nH = height of a fire in meters, including the tip\n\nfactor = constant depending on the scenario the fire is set in\n\nQ = energy release rate in kilowatts\n\n00  {119-Byte Prgm}\n\n01 LBL \"FIREHGT\"\n\n02 \"ENERGY? (kW)\"\n\n03 PROMPT\n\n04 STO 01\n\n05 LBL 00\n\n06 \"TYPE?\"\n\n07 AVIEW\n\n08 \"LARGE\"\n\n09 KEY 1 GTO 01\n\n10 \"0.5 M\"\n\n11 KEY 2 GTO 02\n\n12 \"CORNER\"\n\n13 KEY 3 GTO 03\n\n15 LBL 10\n\n16 STOP\n\n17 GTO 10\n\n18 LBL 01\n\n19 0.23\n\n20 GTO 04\n\n21 LBL 02\n\n22 0.21\n\n23 GTO 04\n\n24 LBL 03\n\n25 0.075\n\n26 GTO 04\n\n27 LBL 04\n\n29 EXITALL\n\n30 RCL 01\n\n31 0.4\n\n32 Y↑X\n\n33 ×\n\n34 \"HEIGHT=\"\n\n35 ARCL ST X\n\n36 AVIEW\n\n37 END\n\nExample:\n\nQ = 2650 kW\n\nlarge fire:  H = 5.38302238231 m\n\n0.5 m fire:  H = 4.9149334795 m\n\ncorner fire:  H = 1.7553338554 m\n\nHP 42S/DM42/Free42 Program:  SMOKE\n\nThe program SMOKE estimates:\n\n*  The front velocity of the smoke, or how quickly the smoke fills a corridor\n\n*  The time it takes for front velocity to fill the corridor.\n\nEquations used:\n\nV ≈ 0.5 * √(g * (1 - Ta/T) * H)\n\ntf = L / Vf\n\ng = acceleration of gravity, for Earth, g = 9.80665 m/s\n\nTa = ambient temperature\n\nT = gas temperature  (Ta ≤ T)\n\nH = height of the corridor in meters\n\nL = length of the corridor in meters\n\nNote 1 - Ta/T = (Da - Dc) / Da\n\nDa = density of the corridor of the ambient air\n\nDc = density of the fire fluid\n\n00 {99-Byte Prgm}\n\n01 LBL \"SMOKE\"\n\n02 1\n\n03 \"AMB. TEMP?\"\n\n04 PROMPT\n\n05 \"GAS TEMP?\"\n\n06 PROMPT\n\n07 ÷\n\n08 -\n\n09 9.80665\n\n10 ×\n\n11 \"HEIGHT? (M)\"\n\n12 PROMPT\n\n13 ×\n\n14 SQRT\n\n15 2\n\n16 ÷\n\n17 ENTER\n\n18 \"VEL= \"\n\n19 ARCL ST X\n\n20 AVIEW\n\n21 STOP\n\n22 R↓\n\n23 1/X\n\n24 \"LENGTH? (M)\"\n\n25 PROMPT\n\n26 ×\n\n27 \"TIME=\"\n\n28 ARCL ST X\n\n29 AVIEW\n\n30 END\n\nExample:\n\nAmb. Temp:  78 °F = 25.555555556 °C\n\nGas Temp:  103 °F = 39.44444444 °C\n\nHeight:  3.048 m\n\nLength:  3.6576 m\n\nResults:\n\nVelocity:  1.622103844 m/s\n\nTime:  2.254849474 s\n\nSource:\n\nLawson, J.R. and Quintiere, J.G..   \"Slide Rule Estimates of Fire Growth\". Fire Technology, Vol. 21., No. 4, November 1985, pg. 267\n\nEddie\n\nAll original content copyright, © 2011-2020.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.\n\n## Saturday, October 17, 2020\n\n### Review: Catiga CS-121\n\nReview:  Catiga CS-121\n\nJust the Facts:\n\nModel:  CS-121 (could be other model numbers)\n\nCompany: Catiga\n\nType: Graphing\n\nBattery:  Case contains 2 CS-2032 slots, but can run on 1\n\nLogic: Algebraic\n\nMemory Registers: 9\n\nColors:  Many colors including Black, Red, Blue.  I purchased a blue CS-121.\n\nCase and Keyboard\n\nThe cover case and calculator casing for the CS-121 is very light.  When sliding the calculator into the case, you really have to listen for the snap to lock the calculator in place.  The keys are very fast and responsive, you will have no problems typing in expressions quickly.\n\nScreen\n\nLike most graphing calculators, the CS-121 has a large screen.  However, the CS-121 does not use the screen like most graphing calculators.   The CS-121 operates as a large Casio fx-6300g instead.   Graphs take up part of the left side of the screen and you can see only the X or the Y (not both) coordinates during tracing.   The expression line is reduced to the bottom line of the screen.\n\nThis is my biggest gripe of the CS-121.   This screen scheme may have worked on the fx-6300g and its relatives fx-6200g and HP 9g because they have small screens to work with, but the CS-121 should have full screen graphs and allowed the use of the entire screens for expressions in Home mode.\n\nFeatures of the CS-121\n\nLeft,  Catiga CS-121;  Right, Casio fx-6300g\n\nThe CS-121 isn't an entire clone of the fx-6300g because it does have some nice advanced features.\n\nGraphing:  Functions (up to 2 functions), Parametric (1 parametric pair)\n\nCalculus:  Numeric Integrals of f(x)\n\nSolver:  Solve for any variable\n\nComplex Number Mode:  Arithmetic, Cube, Cube Root, Square, Square Root, Absolute Value, Argument (angle).   The [ENG] acts as the i key.  Switch between showing the real and complex results by pressing [ SHIFT ] [ = ].  All complex numbers operate in rectangular format (a + bi).\n\nRegressions:  Linear, Logarithm, Exponential, Power, Inverse, Quadratic.   Get the correlation ( r ) by pressing [SHIFT] [ ( ].\n\nBase-N:  The CS-121 adds boolean logic functions ([x^3] key):  And, Or, Nxor, Xor, Not, Neg\n\nDrawing Tools:  Plot points, lines including lines between two points, horizontal, vertical, and tangent.   It is a little frustrating that the CS-121 does not use the entire screen for graphs because it would really enhance the use of the drawing tools.\n\nMemory\n\nLet's talk about the memory.  It is very unusual for a graphing calculator to only have nine memory slots (A, B, C, D, E, F, X, Y, M with M+ and M-).  It's because the CS-121 does not have programming.  Yes, you can store an equation in the form var=f(vars) with the [SHIFT] [ Calc ] (PROG) key sequence and calculate using the stored equation by using [ Calc ].\n\nIt is very rare that a graphing calculator does not have programming.  The rare Casio fx-6200g was a direct relative to the fx-6300g and it did not have programming.\n\nThe Learn (LRN) Key\n\nThe [LRN] key allows users to shift graphs or change the size of the graphs.  Handy when learning about functions.\n\nVerdict\n\nI like the keyboard and I like how responsive the calculator operates.  The CS-121 operates on not much battery (ultimately 1 CS-2032 battery).  As said before, my major gripe is the mismanaged use of the screen.\n\nEddie\n\nAll original content copyright, © 2011-2020.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.\n\n## Thursday, October 15, 2020\n\n### HPCC 2020 Virtual Conference and Document Website\n\nHPCC 2020 Virtual Conference and Document Website\n\nHPCC 2020 Virtual Conference\n\nThe HPCC 2020 Virtual Conference took place on October 10, 2020.   The HPCC is the Handheld and Portable Computer Club is a long running calculator and computer site based out of the United Kingdom.  Their meetings are held virtually on the 2nd and 4th Saturday of every month.  I have recently joined the HPCC and glad to be a part of the club.\n\nYou can find more information on the meetings here:  http://www.hpcc.org, under the Home tab.  Membership rates are between £17 to £23 depending where you live.\n\nThe conference has the following speakers and topics:\n\nEric Rechlin:  USB Drive Update\n\nJake Schwartz:  Updates to the PPC Archive\n\nBob Prosperi:  Swiss Micros DM41X - A little late; a lot better\n\nEric Hazen:  HP-25 Re-implementation on Z80 with VFD display\n\nMichael Park:  MP-29 - a tactile touchscreen calculator\n\nMark Power:  Prime G2\n\nSylvain Côté:  Clonix & NoV modules for the HP-41\n\nEdward Shore:  CAS and Calculators\n\nGene Wright and Włodek Mier-Jędrzejowicz: HP's Worst Calculators\n\nGodwin Stewart:  Presentation on DMConnect\n\nWill Marchant:  DM16 in a satellite launch campaign\n\nConference web page:  http://www.hpcc.org/conferences/index.html\n\nEach talk has slides for download and the corresponding YouTube video.  Special thanks to Eric Rechlin.\n\nHP Documentation Website\n\nEric Rechlin of hpcalc.org has put together a new website where you can download manuals for Hewlett Packard calculators, past and present.\n\nComing soon:  the USB that contains information calculator conferences including manuals through 2020 will be available his commerce site,  https://commerce.hpcalc.org.\n\nEddie  (Edward Shore)\n\nAll original content copyright, © 2011-2020.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.\n\n## Sunday, October 11, 2020\n\n### Casio Classpad: Theorem of Sós\n\nCasio Classpad:  Theorem of Sós\n\nIntroduction: The Theorem of Sós\n\nA set of triple rational numbers (a, b, λ) is considered to be a H-triple if\n\nc = √(a^2 - 2 λ a b + b^2) is also a rational number.  (I)\n\nHeron Triple\n\nShould the set of rational numbers (a, b, λ) can be further test to see if to is also a Heron triple.  That criteria (see source) requires for some numbers λ' and A (area):\n\nλ' = √(1 - λ^2) is a positive rational number, and\n\nA = λ' * abs(a * b) / 2 is a natural number  (positive integer)\n\nThis implies for all cases for Heron Triples that abs(λ) < 1, or -1 < λ < 1.\n\nIf the H-triple passes, then it becomes a Heron triple and the sides (a, b, c) (see (I) above) are the lengths of a Heron triangle.\n\nIt is possible that an H-triple is not a Heron triple.\n\nToday's blog focuses on H-triples.\n\nGenerating H-Triples\n\nGiven relatively prime integers m and n (where gcd(m,n) = 1) and μ, an H-triple can be generated by the following equations:\n\n(II)\n\na = μ * (m^2 - n^2)\n\nb = μ * (2*m * (n + λ*m))\n\nc = μ * (m^2 + 2*λ*m*n + n^2)\n\nFor a > 0, this requires that m > n.  In the program htrigen, I assume that μ = 1.  In the program, u = λ\n\nCasio Classpad Program:  htrigen\n\n(fx-CP400 and fx-CG500)\n\n'2020-09-15 EWS\n\n'H Triple\n\nLocal a,b,c,m,n,u\n\nInput m, \"m > n\", \"m?\"\n\nInput n, \"m > n\", \"n?\"\n\nInput u, \"λ? (rational)\"\n\nIf gcd(m,n)≠1\n\nThen\n\nPrint \"m and n are not\"\n\nPrint \"relatively prime.\"\n\nStop\n\nIfEnd\n\n(m^2 - n^2) ⇒ a\n\n(2 × m × (n + u × m)) ⇒ b\n\n(m^2 + 2 × u × m × n + n^2) ⇒ c\n\nPrint \"H-Triple:\", ColorMagenta\n\nPrint \"{a,b,λ}=\"\n\nPrint {a,b,u}, ColorBlue\n\nPrint \"c=\"\n\nPrint c\n\nExample:\n\nm = 7, n = 3, λ = 2/3\n\nResult:\n\na = 40, b = 322/3, c = 86\n\nTesting the Theorem of Sós\n\nThe program htrist tests the criteria for the triple (a, b, λ).\n\nTesting numbers in a program to see that they are rational is not as easy as it appears. Thankfully, the functions numerator and denominator can be used on the Classpad.  The numerator function converts the number into an exact fraction and extracts its numerator.  Similarly, the denominator function converts the number into an exact fraction and extracts its denominator.\n\nFor an irrational number such as √2, the Classpad determines the exact fraction to be\n\n√2/1.\n\nCasio Classpad Program:  htrist\n\n(fx-CP400 and fx-CG500)\n\n'2020-09-15 EWS\n\n'H Triple test\n\nLocal a,b,c,n,d,u\n\nInput a\n\nInput b\n\nInput u, \"λ? (rational)\"\n\n√(a^2 - 2 × u × a × b + b^2) ⇒ c\n\nnumerator(c) ⇒ n\n\ndenominator(c) ⇒ d\n\nClrText\n\nPrint {n, d}\n\nIf frac(n) = 0 and frac(d) = 0\n\nThen\n\nPrint \"H Triple\", ColorGreen\n\nPrint {a, b, u}\n\nPrint \"c =\"\n\nPrint c\n\nElse\n\nPrint \"No Solution\", ColorRed\n\nIfEnd\n\nExample:\n\na = 85, b = 374, λ = 1\n\nResults:\n\nc = 289,  n = 289, d = 1\n\nSource:\n\nHalbeisen, Lorezn and Hungrebühler, Nobert.  \"Heron triangles and their elliptic curves\" Journal of Number Theory 213 (2020) 232-253.  https://doi.org/10.1016/j.jnt.2019.12.005\n\nEddie\n\nAll original content copyright, © 2011-2020.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.\n\n## Saturday, October 10, 2020\n\n### Sines and Cosines: Adding and Subtracting Angles\n\nSines and Cosines:   Adding and Subtracting Angles\n\nNote:\n\nπ/2 radians = 90°,   π radians = 180°\n\nSine\n\nsin(x + π/2) = sin(x) cos(π/2) + cos(x) sin(π/2) = cos(x)\n\nsin(x - π/2) = sin(x) cos(π/2) - cos(x) sin(π/2) = -cos(x)\n\nsin(π/2 - x)  = sin(π/2) cos(x) - cos(π/2) sin(x) = cos(x)\n\nsin(x + π) = sin(x) cos(π) + cos(x) sin(π) = -sin(x)\n\nsin(x - π) = sin(x) cos(π) - cos(x) sin(π) = -sin(x)\n\nsin(π - x)  = sin(π) cos(x) - cos(π) sin(x) = sin(x)\n\nCosine\n\ncos(x + π/2) = cos(x) cos(π/2) - sin(x) sin(π/2) = -sin(x)\n\ncos(x - π/2) = cos(x) cos(π/2) + sin(x) sin(π/2) = sin(x)\n\ncos(π/2 - x) = cos(π/2) cos(x) + sin(π/2) sin(x) = sin(x)\n\ncos(x + π) = cos(x) cos(π) - sin(x) sin(π) = -cos(x)\n\ncos(x - π) = cos(x) cos(π) + sin(x) sin(π) = -cos(x)\n\ncos(π - x) = cos(π) cos(x) + sin(π) sin(x) = -cos(x)\n\nEddie\n\nAll original content copyright, © 2011-2020.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.\n\n## Sunday, October 4, 2020\n\n### TI-84 Plus CE: Lorenz Curves and the Gini Coefficient\n\nTI-84 Plus CE:  Lorenz Curves and the Gini Coefficient\n\nIntroduction - Graphing Income Inequality\n\nDeveloped by Max O. Lornez, the Lorenz curve displays the graphs of income distribution and income inequality.  Each point of represents a cumulative percentage of an income level.   An example of a Lorenz curve:\n\n(25%, 9%):   The first 25% of a population earns 9% of the population's total income\n\n(50%, 21%):  Half of the population earns only 21% of the total income.\n\n(75%, 44%):  The bottom 75% of the population earns 44% of the total income.\n\n(100%, 100%):  This represents the total population.\n\nThe Lornez curve is compared against a straight line which connects points (0%, 0%) to (100%, 100%).    The ratio between the top area and the whole curve is called a Gini coefficient.  If the Gini coefficient is 0, then there is no income inequality.  On the other hand, the closer that the Gini coefficient is to 1, there is more income inequality.  The diagram below is a Lorenz curve.\n\nCalculating the Gini Coefficient\n\nLet:\n\nA = the area above the Lorenz  curve and below the straight line\n\nB = the area above the x axis and below the Lorenz curve\n\nW = the total area = A + B = 1/2 * max(x value) * max(y value)\n\nNote: that the total area is the area of a right triangle\n\nThe Gini coefficient is defined as:\n\nG = A / (A + B) = (W - B) / B\n\nTI-84 Plus CE Program: LORENZ\n\n430 bytes\n\n\"2020-09-13 EWS\"\n\nMenu(\"LORENZ CURVE\", \"EQUAL PERCENTILE\", 1, \"DATA\", 2)\n\nLbl 1\n\nInput \"DATA?\", L4\n\ndim(L4) → N\n\ncumSum(L4) → L6\n\nseq(100J/N, J, 1, N) → L3\n\nL3 → L5\n\nGoto 3\n\nLbl 2\n\nDisp \"ASCENDING ORDER: Y\"\n\nInput \"X DATA?\", L3\n\nInput \"Y DATA?\", L4\n\ncumSum(L3) / sum(L3) * 100 → L5\n\nsumSum(L3 * L4) / sum(L3 * L4) * 100 → L6\n\nGoto 3\n\nLbl 3\n\n\"BELOW\"\n\naugment( {0}, L5 ) → L5\n\naugment( {0}, L6 ) → L6\n\nsum( seq(.5 * (L5(J) - L5(J-1)) * (L6(J) + L6(J-1)) /2 , J, 2, dim(L6) ) ) → B\n\n\"WHOLE\"\n\n.5 * max(L5) * max(L6) → W\n\n(W - B) / W → G\n\nDisp \"GINI INDEX:\", G\n\nPause\n\nFunc\n\nPlotsOff\n\nFnOff\n\nFnOn 1\n\nPlotsOn 1\n\n10 → Xscl : 10 → Yscl\n\nPlot1( xyLine, L5, L6, □, RED)\n\n\"max(L6)*X/100\" → Y1\n\nZoomStat\n\nThere are two options:\n\n1. Equal Percentile:  Enter the percentage of each bin.  The percentile of each class is divided equally. Example:  for ten data points, the percentiles are split in 10% increments.\n\n(10%, y1%)\n(20%, y2%)\n(30%, y3%)\n(40%, y4%)\n(50%, y5%)\n(60%, y6%)\n(70%, y7%)\n(80%, y8%)\n(90%, y9%)\n(100%, y10%)\n\n2.  Data:  Enter x and y data.   For y data, make sure that list has all of its elements in ascending order.\n\nThe graph displays in values of percents.  X Axis: 0 to 100 (representing 0% to 100%).  Y Axis: 0 to 100 (representing 0% to 100%).\n\nExamples\n\nExample 1:  Equal Percentile\n\nThe total health spending for high average per capita in 10% percentiles in 2002.  (see source for background information)\n\nPercentile Percent of Health Spending\n10% 0%\n20% 0.1%\n30% 0.6%\n40% 1.2%\n50% 2%\n60% 3.4%\n70% 5.4%\n80% 9.1%\n90% 16.5%\n100% 61.7%\n\nEnter Option 1: Data.\n\nGini Index: 0.8633\n\nList = { 0, .1, .6, 1.2, 2, 3.4, 5.4, 9.1, 16.5, 61.7 }\n\nExample 2:  Data\n\nHourly employees of a company and their wages:\n\nNumber of Employees Hourly Wage\n156 \\$15.00\n195 \\$30.00\n53 \\$50.00\n49 \\$70.00\n\nX Data:  {156, 195, 53, 49}\nY Data:  {15, 30, 50, 70}\n\nGini Index:  0.6379627898\n\nSource:\n\nReindhardt, Uwe E.   \"The Construct of Lorenz Curves and of The Gini-Coeficient to Decpict Degrees in Inequality in Health Care\"  Princeton University, WWS 597  (Political Economy of Health Systems)\nRetrieved August 26, 2020\n\nEddie\n\nAll original content copyright, © 2011-2020.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.\n\n## Saturday, October 3, 2020\n\n### TI-84 Plus CE and HP 41C: Method of Successive Substitutions\n\nTI-84 Plus CE and HP 41C:  Method of Successive Substitutions\n\nRepeating the Calculation Again and Again\n\nWith an aid of a scientific calculator, we can solve certain problems of the form:\n\nf(x) = x.\n\nExamples include:\n\ntan cos sin x = x\n\ne^-x = x\n\natan √x = x\n\nsin cos x = x\n\n(acos x) ^ (1/3) = x\n\nIn any case with trigonometric functions, the angle mode will need to be in radians.  You will also need a good guess to get to a solution and to know that at some real number x, f(x) and x intersect.\n\nTake the equation ln(3*x) = x with initial guess x0 = 1.512\n\nDepending on the operating of the scientific calculator the keystrokes would be:\n\nAOS:\n\n1.512 [ = ]\n\nLoop:  [ × ] 3 [ = ] [ ln ]\n\nRPN:\n\n1.512 [ENTER]\n\nLoop:  3 [ × ] [ ln ]\n\nALG:\n\n1.512 [ENTER/=]\n\nLoop:  ln( 3 * Ans) [ ENTER/= ]\n\nRepeat the loop as many times as you like and hope you start seeing the answers converge.   After repeating the loop over and over and over again, at six decimal answers, the readout will be about 1.512135.\n\nAn approximate answer to ln(3x) = x, x ≈ 1.512134552\n\nIf your calculator has a solve function, you can check the answer, but this method can be useful if your calculator does not have a solve function.\n\nThe program SUCCESS illustrates this method.\n\nTI-84 Plus CE Program (TI-83 Family) Program:  SUCCESS\n\n285 bytes\n\n\"2020-09-08 EWS\"\n\nDisp \"SUCCESIVE SUBSTITUTION\", \"F(X)=X\"\n\nDisp \"USE QUOTES FOR F(X)\"\n\nInput \"F(X)? \", Y1\n\nInput \"GUESS? \", G\n\nInput \"PRECISION? \", P\n\n0 → I\n\n1 → T\n\nRepeat T<10^-P\n\nY1 → Z\n\nabs(Z - X) → T\n\nZ → X\n\nI + 1 → I\n\nIf I > 200\n\nThen\n\nDisp \"ITERATIONS EXCEEDED\", \"NO SOLUTION FOUND\"\n\nStop\n\nEnd\n\nEnd\n\nClrHome\n\nDisp \"SOLUTION=\",  round(X,P), \"ITERATIONS=\", I, \"ABS DIFF\", T\n\nHP 41C/DM41 Program: SUCCESS\n\nThis program calls on the subroutine, FX.   FX is where you enter f(x).  End FX with the RTN command.\n\n01 LBL^T SUCCESS\n\n02 ^T F<X>=X\n\n03 AVIEW\n\n04 PSE\n\n05 ^T GUESS?\n\n06 PROMPT\n\n07 STO 00\n\n08 ^T PRECISION?\n\n09 PROMPT\n\n10 STO 02\n\n11 0\n\n12 STO 03\n\n13 1\n\n14 STO 04\n\n15 LBL 01\n\n16 RCL 00\n\n17 XEQ ^FX\n\n18 STO 01\n\n19 RCL 00\n\n20 -\n\n21 ABS\n\n22 STO 04\n\n23 RCL 01\n\n24 STO 00\n\n25 1\n\n26 ST+ 03\n\n27 200\n\n28 RCL 03\n\n29 X>Y?\n\n30 GTO 02\n\n31 RCL 02\n\n32 CHS\n\n33 10↑X\n\n34 RCL 04\n\n35 X>Y?\n\n36 GTO 01\n\n37 ^T SOL=\n\n38 ARCL 01\n\n39 AVIEW\n\n40 STOP\n\n41 ^T ITER=\n\n42 ARCL 03\n\n43 AVIEW\n\n44 STOP\n\n45 ^T DIFF=\n\n46 ARCL 04\n\n47 AVIEW\n\n48 STOP\n\n49 GTO 04\n\n50 LBL 02\n\n51 ^T NO SOL FOUND\n\n52 AVIEW\n\n53 STOP\n\n54 LBL 04\n\n55 END\n\nExamples for FX:\n\nf(x) = sin cos x.\n\nProgram:\n\nLBL ^FX\n\nCOS\n\nSIN\n\nRTN\n\nf(x) = e^-x\n\nProgram:\n\nLBL ^FX\n\nCHS\n\nE↑X\n\nRTN\n\nBe aware, some equations cannot be solved in this manner, such as x = π / sin x and x = ln(1 / x^4).\n\nCheung, Y.L. \"Using Scientific Calculators to Demonstrate the Method of Successive Substitutions\"  The Mathematics Teacher.  National Council of Teachers of Mathematics.  January 1986, Vol. 79 No. 1 pp. 15-17  http://www.jstor.com/stable/27964746\n\nEddie\n\nAll original content copyright, © 2011-2020.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.\n\n### Matrices in Python without Numpy: Part 1\n\nMatrices in Python without Numpy:  Part 1 Introduction Python is a wonderful programming language and is a welcome addition to graphing calc..." ]

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[ "# solving a system of DEs numerically and plotting the solution\n\nHi everyone. I want to hand Sage a system of nonlinear DEs with initial values, and I'd like a plot of the solutions. Is there a best way to do this? For example, to solve this system: ds/dt=-si, di/st=si-2i, dr/dt=2i, s(0)=1, i(0)= 0.00000127, r(0)=0, I did this in Sage:\n\nsage: (s,i,r)=var('s,i,r')\n\nsage: des=[-1si, si-2i, i]\n\nsage: desolve_system_rk4(des, [s,i,r], ics=[0, 1, 0.00000127, 0], ivar=t, end_points=20)\n\nSage gave me a list of points, but I couldn't figure out a way to plot s, i, and r using the points it gave me. Thanks!\n\nedit retag close merge delete\n\nSort by » oldest newest most voted\n\nTry this:\n\nans=desolve_system_rk4(des, [s,i,r], ics=[0, 1, 0.00000127,0], ivar=t, end_points=20)\nts=[[a,a] for a in ans]\nti=[[a,a] for a in ans]\ntr=[[a,a] for a in ans]\nline(ts)+line(ti,color='red')+line(tr,color='green')\n\nmore" ]

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[ "# clique, independent set, and minimum vertex cover\n\nI was given a graph problem with 3 different questions and 1 set of answers. The problem is described below. The problem that I'm having is that it seems to me that the answer to all the questions is the same. I keep trying to find a caveat but I don't see one. What am I missing?\n\nHere is the problem\n\nUndirected graph $G$. $n$ - number of vertices. $m$ - number of edges. $d$ - maximum degree of a graph.\n\n1. The maximum clique size of $G$ is no larger than\n2. The minimum vertex cover size of $G$ is no larger than\n3. The maximum independent set size of $\\overline{G}$, the complement of $G$, is no larger than\n\n• (a) $d+1$\n• (b) $n$\n• (c) $n-1$\n• (d) $n/2$\n• (e) $d$\n• (f) $n-d$\n\nIt looks to me that the answer to every problem is (b) $n$, because\n\n1. Clique cannot have more vertices than there are in a graph\n2. Vertex cover cannot be larger than the number of vertices in a graph\n3. Maximum independent set cannot be larger than the number of vertices in a graph.\n\nI feel like I'm missing something, because the answers seem too obvious.\n\nAny help is appreciated\n\nOf course, $n$ being the largest of the given answers will satisfy all conditions. You are expected to find the least upper bounds though.\n• First, You know that there exist graphs, for which $n$ is the exact solution, so you have to think about $d$. Think of some graphs that can't have a cover of $n$. What do they have in common? Dec 6, 2013 at 18:51\n• I'm at a loss. What could the graphs that don't have a cover of $n$ have in common? The only thing I can think of is that total degree of a graph is equal the double count of edges, i.e. $d=2m$ Dec 6, 2013 at 19:09" ]

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[ "# Opencart Quantity Discount ignored with special price\n\nopen: system/library/cart.php\n\nreplace\n\n```if (\\$product_special_query->num_rows) {\n\\$price = \\$product_special_query->row['price'];\n}```\n\nwith\n\n```if (\\$product_special_query->num_rows) {\nif(\\$product_discount_query->num_rows&&\\$product_discount_query->row['price']<=\\$product_special_query->row['price']){\n\\$price = \\$product_discount_query->row['price'];\n}else{\n\\$price = \\$product_special_query->row['price'];\n}\n}```\n\nopen: catalog/controller/product/product.php\n\n```foreach (\\$discounts as \\$discount) {\n\n\\$this->data['discounts'][] = array(\n'quantity' => \\$discount['quantity'],\n'price'    => \\$this->currency->format(\\$this->tax->calculate(\\$discount['price'], \\$product_info['tax_class_id'], \\$this->config->get('config_tax')))\n);\n}\n\n```\n\nwith\n\n```\nforeach (\\$discounts as \\$discount) {\nif((float)\\$product_info['special']){\nif(\\$product_info['special']>\\$discount['price']){\n\\$this->data['discounts'][] = array(\n'quantity' => \\$discount['quantity'],\n'price'    => \\$this->currency->format(\\$this->tax->calculate(\\$discount['price'], \\$product_info['tax_class_id'], \\$this->config->get('config_tax')))\n);\n}\n}else{\n\\$this->data['discounts'][] = array(\n'quantity' => \\$discount['quantity'],\n'price'    => \\$this->currency->format(\\$this->tax->calculate(\\$discount['price'], \\$product_info['tax_class_id'], \\$this->config->get('config_tax')))\n);\n}\n}\n\n```\n\nor if you use vQmod to save from core file overwriting:\n\n<modification>\n\n<id>Fix Discounts wtih Specials</id>\n\n<version>1.0</version>\n\n<vqmver>1.2.3</vqmver>\n\n<author>uksb</author>\n\n<file name=”system/library/cart.php”>\n\n<operation>\n\n<search position=”replace” offset=”2″><![CDATA[ if (\\$product_special_query->num_rows) {]]></search>\n\nif(\\$product_discount_query->num_rows&&\\$product_discount_query->row[‘price’]<=\\$product_special_query->row[‘price’]){\n\n\\$price = \\$product_discount_query->row[‘price’];\n\n}else{\n\n\\$price = \\$product_special_query->row[‘price’];\n\n}\n\n</operation>\n\n</file>\n\n<file name=”catalog/controller/product/product.php”>\n\n<operation>\n\n<search position=”replace” offset=”5″><![CDATA[ foreach (\\$discounts as \\$discount) {]]></search>\n\n<add><![CDATA[ foreach (\\$discounts as \\$discount) {\n\nif((float)\\$product_info[‘special’]){\n\nif(\\$product_info[‘special’]>\\$discount[‘price’]){\n\n\\$this->data[‘discounts’][] = array(\n\n‘quantity’ => \\$discount[‘quantity’],\n\n‘price’    => \\$this->currency->format(\\$this->tax->calculate(\\$discount[‘price’], \\$product_info[‘tax_class_id’], \\$this->config->get(‘config_tax’)))\n\n);\n\n}\n\n}else{\n\n\\$this->data[‘discounts’][] = array(\n\n‘quantity’ => \\$discount[‘quantity’],\n\n‘price’    => \\$this->currency->format(\\$this->tax->calculate(\\$discount[‘price’], \\$product_info[‘tax_class_id’], \\$this->config->get(‘config_tax’)))\n\n);\n\n}" ]

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[ "# Toronto Math Forum\n\n## MAT244--2020F => MAT244--Lectures & Home Assignments => Chapter 2 => Topic started by: RunboZhang on September 16, 2020, 09:26:38 PM\n\nTitle: Textbook Section2.1 Example5\nPost by: RunboZhang on September 16, 2020, 09:26:38 PM\nHi guys, I have been stuck at example 5 of section2.1(page30) for quite a while. In particular, I do not understand why the lower bound of the integral is the initial point t=0. Why can't it be the upper bound?\n(example screenshot is attached below)\nTitle: Re: Textbook Section2.1 Example5\nPost by: Jinqiu Liang on September 17, 2020, 10:05:46 AM\nIn my opinion, firstly this question is an initial value problem, which means the number that the equation (y(0)=1) gives you is the lower limit of the integral, which is 0. You can also use the upper bound, however, if t is infinity, then we can not substitute t as a number into the equation and then solve it. In this way, choosing the lower limit number 0 is the easiest and the fastest method.\nTitle: Re: Textbook Section2.1 Example5\nPost by: Victor Ivrii on September 17, 2020, 12:28:15 PM\nYou can select any lower limit you wish, the difference goes to the constant. However, as Ella correctly observed, it makes sense to select $t=0$ since the $t_0=0$ in the initial problem" ]

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[ "Share This Article:\n\nDirichlet Averages, Fractional Integral Operators and Solution of Euler-Darboux Equation on Hölder Spaces\n\nDOI: 10.4236/am.2016.714129    1,173 Downloads   1,470 Views\nAuthor(s)    Leave a comment\nIn the present paper, we discuss the solution of Euler-Darboux equation in terms of Dirichlet averages of boundary conditions on H?lder space and weighted H?lder spaces of continuous functions using Riemann-Liouville fractional integral operators. Moreover, the results are interpreted in alternative form.\n\nReceived 2 June 2016; accepted 14 August 2016; published 17 August 2016", null, "1. Introduction\n\nThe subject of Dirichlet averages has received momentum in the last decade of 20th century with reference to the solution of certain partial differential equations. Not much work has been registered in this area of Applied Mathematics except some papers devoted to evaluation of Dirichlet averages of elementary functions as well as higher treanscendental functions interpreting the results in more general special functions. The present paper is ventured to give the interpretation of solution of a typical partial differential equation and prove its inclusion properties with respect to Hölder spaces. The Euler-Darboux equation (ED-equation) is a certain kind of degenerate hyperbolic partial differential equation of the type (see Nahušev ),", null, "(1)\n\nSaigo - considered and studied the ED-equation given by", null, "(2)\n\nwhich implies the Equation (1) for", null, "or some other degenerate hyperbolic equations described\n\nby characteristic coordinates. The boundary conditions used for the solution of Equation (2) are", null, "(3)\n\nThe solution of ED-Equation (2), due to Saigo , is given by", null, "(4)\n\nwhere x and y are restricted in the domain", null, ".\n\nSrivastava and Saigo evaluated the results on multiplication of fractional integral operators and the solution of ED-equation. Deora and Banerji represented the solution of Equation (2) in terms of Dirichlet averages of boundary condition functions given in (3) as follows", null, "(5)\n\nwhere", null, "and", null, "denote the single Dirichlet averages of boundary functions", null, "and", null, ", respectively.\n\nKilbas et al. studied the solution of ED-equation on Hölder Space", null, "or simply", null, "as well as on weighted Hölder Space of continuous functions. In the present paper we discuss the Dirichlet averages on Hölder Space via right-sided Riemann-Liouville fractional integral operators and prove the solution of Equation (2) to be justified on such spaces. In what follows are the preliminaries and definitions related to fractional integral operators, Dirichlet averages, and Hölder spaces of continuous functions.\n\n2. Hölder Spaces\n\nFor", null, "and a finite interval", null, "we denote by the space of Hölder function on,\n\n(6)\n\nif and, i.e., is m-times differentiable function and its m-th derivative is continuous and satisfies the inequality\n\n(7)\n\nwhere for any.\n\nLet and, i.e., be the space of Hölder continuous functions and\n\n(8)\n\nThen we denote by the space of functions such that, where.\n\n3. Dirichlet Averages\n\nCarlson introduced the concept of connecting elementary functions with higher transcendental functions using averaging technique. The Dirichlet average is a certain kind of integral average with respect to Dirichlet measure, which in Statistics called as beta distribution of several variables. One may refer to Banerji and Deora , Deora and Banerji , Deora, Banerji and Saigo , Gupta and Agrawal , Kattuveettil , Prabhakar , Chena Ram et al. , Vyas , Vyas and Banerji , Vyas, Banerji and Saigo .\n\nStandard Simplex: Denote the standard simplex in by\n\n(9)\n\nBeta Function of k-variables: Let denotes the cartesian product of open right half plane and is the standard simplex in. The beta function of k-variables can be expressed as\n\n(10)\n\nDirichlet Measure: The complex measure, defind on E by\n\n(11)\n\nfor, is called the Dirichlet measure. Particularly, for, we write by using (3), the following:\n\n(12)\n\nDirichlet Average: Let be a convex set in. Let and denotes a convex linear combination of. Then the Dirichlet average of a holomorphic function f is defined by (See Carlson )\n\n(13)\n\nwhere denotes the parameters. The convex combination is given by\n\nParticularly,when, the Dirichlet average, so extracted out of (5), is called the single Dirichlet average of f over the line segment from 0 to 1. It is expressed as\n\n(14)\n\nwhere and.\n\nIf we consider the continuous function in Hölder Space and, then without the loss of characteristics of such spaces, the Dirichlet average of is defined by\n\n(15)\n\nand for, and. The equation analogous to (11), is expressed as\n\n(16)\n\nwhere denotes the single Dirichlet average of in two variables x and y in.\n\n4. Fractional Integral Operators\n\nFractional calculus is the generalization of ordinary n-times iterated integrals and derivatives of continuous functions to that of any arbitrary order real or complex. The most commonly used definition of fractional integral operators of order a is due to Riemann-Liouville. A detailed account of fractional calculus is given in Samko et al. and the applications of it are elaborated in Hilfer and Podlubney . Vyas interpreted the angle of collision occurring in the study of transport properties of Noble gases at low density configuration in terms of Fractional Integral Operators.\n\nLet be the Hölderian class of continuous functions and the weighted Hölder space be defined in (8). Then the right-sided Riemann-Liouville fractional integral operators are defined by\n\n(17)\n\n(18)\n\n(19)\n\n(20)\n\nProposition 1: Let and with. Then for, we have\n\n(21)\n\nProposition 2: Let and with. Then for, we have\n\n(22)\n\nGeneralization of fractional integral operators is due to Saigo . Let and be real numbers and be the Gauss’ hypergeometric function. One may refer Erdélyi and Slater . Then the fractional integral operator involving Gauss’ hypergeometric function on Hölder space is defined by\n\n(23)\n\n(24)\n\n(25)\n\n(26)\n\nProposition 3: Let and. Then for, we have\n\n(27)\n\nProposition 4: Let and. Then for, we have\n\n(28)\n\nBy setting, the generalized fractional integral operators defined in (23) to (26) reduce to right-sided Riemann-Liouville fractional integral operators defined in (17) to (20).\n\n5. Main Results\n\nTheorem 1: Let,. Then the single Dirichlet average of is expressed as\n\n(29)\n\nwhere is the right-sided Riemann-Liouville fractional integral operator defined in (17).\n\nProof: Using Equation (16), we write\n\n(30)\n\nUsing the transformation in (30) and adjusting the terms involved, we obtain\n\n(31)\n\nwhich upon using (17), can be expressed as\n\n(32)\n\nwhich, for, can also be put in the form\n\n(33)\n\nwhere denotes the new fractional operator defined by\n\n(34)\n\nOwing to the proposition 1 to proposition 4 we conclude the proof of theorem 1.\n\nCorollary 1: If and restrictions on parmeters hold true, then for\n\n(35)\n\nProof: Invoking the proposition 1 and using the result (32), we find that the fractional integral representation of single Dirichlet average of, for gives rise to the result (35). This justifies that the Dirichlet averages, so evaluated, belong to the Hölderian class.\n\nTheorem 2: Let and denote the Dirichlet averages of boundary functions and, respectively associatd with the ED-Equation (2) and if denotes the solution of ED-equation, given by (5), in terms of and. Then, for, the solution belongs to the Hölderian class.\n\nProof: Using Equation (5), Theorem 1 and the Corollary 1, theorem 2 can be proved easily under the proposition 4.\n\nAcknowledgements\n\nThe author is indebted to P. K. Banerji, Jodhpur, India for fruitful discussions during the preparation of this paper. Financial support under Technical Education Quality Improvement Programme (TEQIP)-II, a programme of Ministry of Human Resource Development, Government of India is highly acknowledged. Author is also thankful to worthy refree for his/her valuable suggestions upon improvement.\n\nConflicts of Interest\n\nThe authors declare no conflicts of interest.\n\nCite this paper\n\nVyas, D. (2016) Dirichlet Averages, Fractional Integral Operators and Solution of Euler-Darboux Equation on Hölder Spaces. Applied Mathematics, 7, 1498-1503. doi: 10.4236/am.2016.714129.\n\n Nahusev, A.M. (1969) A New Boundary Problem for a Degenerate Hyperbolic Equation. Soviet Mathematics Doklady, 10, 935-938. Saigo, M. (1979) A Certain Boundary Value Problem for the Euler-Darboux Equation. Mathematica Japonica, 24, 337-385. Saigo, M. (1979) A Certain Boundary Value Problem for the Euler-Darboux Equation-II. Mathematica Japonica, 25, 211-220. Saigo, M. (1981) A Certain Boundary Value Problem for the Euler-Darboux Equation-III. Mathematica Japonica, 26, 103-119. Srivastava, H.M. and Saigo, M. (1987) Multiplication of Fractional Calculus Operators and Boundary Value Problems Involving the Euler-Darboux Equation. Journal of Mathematical Analysis and Applications, 121, 325-369. http://dx.doi.org/10.1016/0022-247X(87)90251-4 Deora, Y. and Banerji, P.K. (1994) An Application of Fractional Calculus to the Solution of Euler-Darboux Equation in Terms of Dirichlet Averages. Journal of Fractional Calculus, 5, 91-94. Kilbas, A.A., Repin, O.A. and Saigo, M. (1996) Solution in Closed Form of Boundary Value Problem for Degenerate Equation of Hyperbolic Type. Kyungpook Mathematical Journal, 36, 261-273. Carlson, B.C. (1969) A Connection between Elementary Functions and Higher Transcendental Functions. SIAM Journal on Applied Mathematics, 17, 116-148. http://dx.doi.org/10.1137/0117013 Banerji, P.K. and Deora, Y. (1998) Laplace Transform of the Product of Two Generalized Laguerre Functions to Evaluate Averaged Functions. Bulletin of Calcutta Mathematical Society, 90, 389-394. Deora, Y. and Banerji, P.K. (1993) Double Dirichlet Average of ex Using Fractional Derivative. Journal of Fractional Calculus, 3, 81-86. Deora, Y. and Banerji, P.K. (1993) Triple Dirichlet Average and Fractional Derivative. Revista Técnica de la Facultad de Ingeniería Universidad del Zulia, 16, 157-161. Deora, Y., Banerji, P.K. and Saigo, M. (1994) Fractional Integral and Dirichlet Averages. Journal of Fractional Calculus, 6, 55-59. Gupta, S.C. and Agrawal, B.M. (1990) Dirichlet Averages and Fractional Derivatives. The Journal of the Indian Academy of Mathematics, 12, 103-115. Gupta, S.C. and Agrawal, B.M. (1991) Double Dirichlet Averages and Fractional Derivatives. Ganita Sandesh, 5, 47-53. Kattuvettill, A. (2008) On Dirichlet Averages. STARS: International Journal, 2, 78-88. Prabhakar, T.R. (1977) A General Class of Operators Involving and Related Integral Equations. The Journal of the Indian Mathematical Society, 41, 163-179. Ram, C., Choudhary, P. and Gehlot, K.S. (2013) Representation of Dirichlet Average of K-Series via Fractional Integrals and Special Functions. International Journal of Mathematics And Its Applications, 1, 1-11. Vyas, D.N. (2011) Some Results on Hypergeometric Functions Suggested by Dirichlet Averages. The Journal of the Indian Academy of Mathematics, 33, 705-715. Vyas, D.N. and Banerji, P.K. (1994) On Dirichlet Averages. Mathematica Balkanica (New Series), 10, 87-95. Vyas, D.N. and Banerji, P.K. (1995) Dirichlet Averages Associated with Psi-Function. SERDICA, Bulgarian Academy of Science, 9. Vyas, D.N., Banerji, P.K. and Saigo, M. (1994) On Dirichlet Average and Fractional Integral of a General Calss of Polynomials. Journal of Fractional Calculus,6, 61-64. Carlson, B.C. (1977) Special Functions of Applied Mathematics. Academic Press, New York. Samko, S.G., Kilbas, A.A. and Marichev, O.I. (1990) Fractional Integrals and Derivatives. Gordon and Breach Science Publishers, London. Hilfer, R. (Ed.) (2000) Applications of Fractional Calculus in Physics. World Scientific Publishing Company, Singapore. Podlubny, I. (1999) Fractional Differential Equations. Academic Press, San Diego. Vyas, D.N. (2010) Interpretation of Angle of Collision Occurring in Transport Properties of Noble Gases in Terms of Fractional Integral Operators. The Aligarh Bulletin of Mathematics, 29, 1-5. Saigo, M. (1978) A Remark on Fractional Integral Operator Involving Gauss’ Hypergeometric Function. Mathematical Reports of Kyushu University, 11, 135-143. Erdélyi, A. (1939) Integration of a Certain System of Linear Partial Differential Equations of Hypergeometric Type. Proceedings of the Royal Society of Edinburgh, 59, 224-241. http://dx.doi.org/10.1017/S0370164600012311 Erdélyi, A., et al. (1953) Higher Transcendental Functions. Vol. 1, McGraw Hill, New York. Slater, L.J. (1960) Confluent Hypegeometric Function. Cambridge University Press, Cambridge.\n\ncomments powered by Disqus\n\nCopyright © 2018 by authors and Scientific Research Publishing Inc.", null, "This work and the related PDF file are licensed under a Creative Commons Attribution 4.0 International License." ]

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[ "# Video: Reading Data from a Frequency Table Involving Fractions\n\nThe table shows the sports played by 44 students. What fraction of the students played football?\n\n01:50\n\n### Video Transcript\n\nThe table shows the sports played by 44 students. What fraction of the students played football?\n\nFractions represent the part to the whole. For us, the part equals the students who play football. And the whole are all the students that play any of these three sports.\n\nFor our numerator, we need the number of students who play football. 22 students play football. And for the denominator, we need the total of all three sports. 22 for football plus 11 for basketball plus 11 for tennis equals 44. Our fraction then becomes 22 over 44.\n\nThen we ask the question, “Can we simplify this fraction? Can it be reduced in any way? Is there something that 22 and 44 is divisible by?” 44 is a multiple of 22. And that means both the numerator and the denominator can be divided by 22. 22 divided by 22 equals one. 44 divided by 22 equals two. When we simplify the fraction 22 over 44, we get one-half. The fraction of students who played football is one-half." ]

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[ "Documentation\n\ngeopoint\n\nGeographic point vector\n\nDescription\n\nA geopoint vector is a container object that holds geographic point coordinates and attributes. The points are coupled, such that the size of the latitude and longitude coordinate arrays are always equal and match the size of any dynamically added attribute arrays. Each entry of a coordinate pair and associated attributes, if any, represents a discrete element in the geopoint vector.\n\nCreation\n\nDescription\n\nexample\n\np = geopoint() constructs an empty geopoint vector with these default property settings:\n\np =\n\n0x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: []\nLongitude: []\n\nexample\n\np = geopoint(latitude,longitude) sets the Latitude and Longitude properties of geopoint vector p\n\nexample\n\np = geopoint(latitude,longitude,Name,Value) sets the Latitude and Longitude properties, then adds dynamic properties to the geopoint vector using Name,Value argument pairs. You can specify several name-value pair arguments in any order as Name1,Value1,...,NameN,ValueN.\n\nexample\n\np = geopoint(structArray) constructs a new geopoint vector from the fields of the structure, structArray.\n\n• If structArray contains the field Lat, and does not contain a field Latitude, then the Lat values are assigned to the Latitude property. Similar behavior occurs when structArray contains the field Lon, and does not contain the field Longitude.\n\n• If structArray contains both Lat and Latitude fields, then both field values are assigned to p. Similar behavior occurs for Lon and Longitude fields when both are present in structArray\n\n• Other fields of structArray are assigned to p and become dynamic properties. Field values in structArray that are not numeric data types, string scalars, string arrays, character vectors, or cell arrays of numeric data types or character vectors are ignored.\n\nexample\n\np = geopoint(latitude,longitude,structArray) sets the Latitude and Longitude properties, and sets dynamic properties from the field values of the structure, structArray.\n\n• If structArray contains the fields Lat, Latitude, Lon, or Longitude, then those field values are ignored.\n\nProperties\n\nexpand all\n\nEach element in a geopoint vector is considered a feature. For more about the property types in geopoint, see Collection Properties and Feature Properties.\n\nDynamic properties are new features that are added to a geopoint vector and that apply to each individual feature in the geopoint vector. You can attach new dynamic Feature properties to the object during construction with a Name,Value pair or after construction using dot (.) notation. This is similar to adding dynamic fields to a structure. For an example of adding dynamic Feature properties, see Construct Geopoint Vector Using Name-Value Pairs.\n\nType of geometry, specified as 'point'. For geopoint, Geometry is always 'point'.\n\nData Types: char | string\n\nLatitude coordinates, specified as a numeric row or column vector.\n\nData Types: double | single\n\nLongitude coordinates, specified as a numeric row or column vector.\n\nData Types: double | single\n\nInformation for the entire set of geopoint vector elements, specified as a scalar structure. You can add any data type to the structure.\n\n• If Metadata is provided as a dynamic property Name in the constructor, and the corresponding Value is a scalar structure, then Value is copied to the Metadata property. Otherwise, an error is issued.\n\n• If a Metadata field is provided by structArray, and both Metadata and structArray are scalar structures, then the Metadata field value is copied to the Metadata property value. If structArray is a scalar but the Metadata field is not a structure, then an error is issued. If structArray is not scalar, then the Metadata field is ignored.\n\nData Types: struct\n\nObject Functions\n\n append Append features to geographic or planar vector cat Concatenate geographic or planar vector disp Display geographic or planar vector fieldnames Return dynamic property names of geographic or planar vector isempty Determine if geographic or planar vector is empty isfield Determine if dynamic property exists in geographic or planar vector isprop Determine if property exists in geographic or planar vector length Return number of elements in geographic or planar vector properties Return property names of geographic or planar vector rmfield Remove dynamic property from geographic or planar vector rmprop Remove property from geographic or planar vector size Return size of geographic or planar vector struct Convert geographic or planar vector to scalar structure vertcat Vertically concatenate geographic or planar vectors\n\nExamples\n\ncollapse all\n\nThis example shows how to create a geopoint vector, specifying latitude and longitude coordinates, and display it.\n\nCreate Geopoint Containing Single Point and Display It\n\nCreate a geopoint vector using the latitude and longitude of Paris, France, and display it. When using a geopoint vector, the geometry of the constructed object is always 'point'.\n\nlat = 48.8566;\nlon = 2.3522;\np = geopoint(lat,lon)\np =\n1×1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: 48.8566\nLongitude: 2.3522\n\nDisplay the point in a plot. You can pass a geopoint vector directly to the geoshow command. geoshow can read the latitude and longitude values from the geopoint vector and also reads the geometry type.\n\ngeoshow(p)", null, "Display Point on a Map\n\nTo display your point with more context, plot your point over a map using the web map display function wmmarker. You can pass a geopoint vector directly to the wmmarker command.\n\nwmmarker(p)\n\nThe wmmarker function opens a web map and displays the point on the map.", null, "Import data from a text file with the latitudes and longitudes of some European capitals. The latitude coordinates are in the first column and the longitude coordinates are in the second column. The coordinates are separated by single space character.\n\ndata = importdata('european_capitals.txt');\n\nCreate a geopoint vector containing the latitude and longitude data.\n\np = geopoint(data(:,1),data(:,2))\np =\n13×1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [48.8566 51.5074 40.4168 41.9028 52.5200 52.3680 52.2297 47.4979 44.4268 50.0755 48.1486 48.2082 46.9480]\nLongitude: [2.3522 -0.1278 -3.7038 12.4964 13.4050 4.9036 21.0122 19.0402 26.1025 14.4378 17.1077 16.3738 7.4474]\n\nPlot the points on a map showing the landmass of Europe.\n\nworldmap europe\ngeoshow('landareas.shp','FaceColor', [0.15 0.5 0.15])\ngeoshow(p)", null, "Alternatively, you can also plot these points over a map using the geoplot function. This example includes a line specification parameter to specify a plus sign marker and the color red. The example also increases the line width for better visibility of the markers.\n\nfigure\ngeoplot(p.Latitude,p.Longitude,'+r','LineWidth',2)", null, "Create a geopoint vector, specifying Latitude, Longitude, and Temperature, where Temperature is part of a Name-Value pair.\n\npoint = geopoint(42, -72, 'Temperature', 89)\npoint =\n\n1x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: 42\nLongitude: -72\nTemperature: 89\n\nConstruct a geopoint object specifying names.\n\np = geopoint([51.519 48.871], [-.13 2.4131],...\n'Name', {\"London\", \"Paris\"})\np =\n\n2x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [51.5190 48.8710]\nLongitude: [-0.1300 2.4131]\nName: {'London' 'Paris'}\n\nRead shape data into a geostruct (a structure array containing Lat and Lon fields).\n\nS = shaperead('worldcities', 'UseGeoCoords', true)\nS =\n\n318x1 struct array with fields:\n\nGeometry\nLon\nLat\nName\n\nCreate a geopoint vector specifying the geostruct.\n\np = geopoint(S)\np =\n\n318x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [1x318 double]\nLongitude: [1x318 double]\nName: {1x318 cell}\n\nAdd a Filename field to the Metadata structure. The Metadata property pertains to all elements of a geopoint vector.\n\nans =\n\nFilename: 'worldcities.shp'\n\nCreate a structure array.\n\n[structArray, A] = shaperead('worldcities', 'UseGeoCoords', true)\nstructArray =\n\n318x1 struct array with fields:\nGeometry\nLon\nLat\n\nA =\n\n318x1 struct array with fields:\nName\n\nUse the numeric arrays and the structure containing the list of names to construct a geopoint vector.\n\np = geopoint([structArray.Lat], [structArray.Lon], A)\np =\n\n318x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [1x318 double]\nLongitude: [1x318 double]\nName: {1x318 cell}\n\nGenerate an empty geopoint vector using the default constructor, then populate the geopoint vector using dot notation with properties from data fields in structure structArray.\n\nstructArray = shaperead('worldcities', 'UseGeoCoords', true);\np = geopoint();\np.Latitude = [structArray.Lat];\np.Longitude = [structArray.Lon];\np.Name = structArray.Name;\np\np =\n\n318x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [1x318 double]\nLongitude: [1x318 double]\nName: {1x318 cell}\n\nThis example shows how to add new values to an existing geopoint vector. The example appends data about Paderborn Germany to the geopoint vector of data about world cities.\n\nRead world cities data using the shaperead command. shaperead returns a structure array.\n\nstructArray = shaperead('worldcities.shp', 'UseGeoCoords', true);\n\nCreate a geopoint vector from the structure array. Display the last of the 318 elements in the vector.\n\np = geopoint(structArray);\np(end)\nans =\n1x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: 34.8519\nLongitude: 113.8061\nName: 'Zhengzhou'\n\nAdd the Paderborn data to the end of the geopoint vector. Display the last of the existing elements and the new element.\n\nlat = 51.715254;\nlon = 8.75213;\np = append(p, lat, lon, 'Name', 'Paderborn');\np(end-1:end)\nans =\n2x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [34.8519 51.7153]\nLongitude: [113.8061 8.7521]\n\nAnother way to add a point at the end of a vector is to use linear indexing. For example, add data about Arlington, Massachusetts to the end of the world cities vector. Notice how, after the initial assignment statement appends a value to the Latitude property vector, using end+1 , all other property vectors automatically expand by one element. Display the last of the existing elements and the new element.\n\np(end+1).Latitude = 42.417060\np =\n320x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [1x320 double]\nLongitude: [1x320 double]\nName: {1x320 cell}\n\np(end).Longitude = -71.170200;\np(end).Name = 'Arlington';\np(end-1:end)\nans =\n2x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [51.7153 42.4171]\nLongitude: [8.7521 -71.1702]\n\nConstruct a geopoint vector containing two features and then add two dynamic properties.\n\nlat = [51.519 48.871];\nlon = [-.13 2.4131];\np = geopoint(lat, lon);\n\np.Name = {'London', 'Paris'}; % Add character feature dynamic property\np.ID = [1 2] \t\t\t% Add numeric feature dynamic property\np =\n\n2x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [51.5190 48.8710]\nLongitude: [-0.1300 2.4131]\nName: {'London' 'Paris'}\nID: [1 2]\n\nAdd the coordinates for a third feature.\n\np(3).Latitude = 45.472;\np(3).Longitude = 9.184\np =\n\n3x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [51.5190 48.8710 45.4720]\nLongitude: [-0.1300 2.4131 9.1840]\nName: {'London' 'Paris' ''}\nID: [1 2 0]\n\nNote that lengths of all feature properties are synchronized with default values.\n\nSet the values for the ID feature dynamic property with more values than contained in Latitude or Longitude.\n\np.ID = 1:4\np =\n\n4x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [51.5190 48.8710 45.4720 0]\nLongitude: [-0.1300 2.4131 9.1840 0]\nName: {'London' 'Paris' '' ''}\nID: [1 2 3 4]\n\nNote that all feature properties are expanded to match in size.\n\nSet the values for the ID feature dynamic property with fewer values than contained in the Latitude or Longitude properties.\n\np.ID = 1:2\np =\n\n4x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [51.5190 48.8710 45.4720 0]\nLongitude: [-0.1300 2.4131 9.1840 0]\nName: {'London' 'Paris' '' ''}\nID: [1 2 0 0]\n\nThe ID property values expand to match the length of the Latitude and Longitude property values.\n\nSet the value of either coordinate property (Latitude or Longitude) with fewer values.\n\np.Latitude = [51.519 48.871]\np =\n\n2x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [51.5190 48.8710]\nLongitude: [-0.1300 2.4131]\nName: {'London' 'Paris'}\nID: [1 2]\n\nAll properties shrink to match in size.\n\nRemove the ID property by setting its value to [ ].\n\np.ID = []\np =\n\n2x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [51.5190 48.8710]\nLongitude: [-0.1300 2.4131]\nName: {'London' 'Paris'}\n\nRemove all dynamic properties and set the object to empty by setting a coordinate property value to [ ].\n\np.Latitude = []\np =\n\n0x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: []\nLongitude: []\n\nRead data from shapefile. Initially the field names of the class are in random order.\n\nstructArray = shaperead('tsunamis', 'UseGeoCoords', true);\n% Field names in random order\np = geopoint(structArray)\np =\n\n162x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [1x162 double]\nLongitude: [1x162 double]\nYear: [1x162 double]\nMonth: [1x162 double]\nDay: [1x162 double]\nHour: [1x162 double]\nMinute: [1x162 double]\nSecond: [1x162 double]\nVal_Code: [1x162 double]\nValidity: {1x162 cell}\nCause_Code: [1x162 double]\nCause: {1x162 cell}\nEq_Mag: [1x162 double]\nCountry: {1x162 cell}\nLocation: {1x162 cell}\nMax_Height: [1x162 double]\nIida_Mag: [1x162 double]\nIntensity: [1x162 double]\nNum_Deaths: [1x162 double]\nDesc_Deaths: [1x162 double\n\nUsing the method fieldnames and typical MATLAB® vector notation, the field names in the geopoint vector are alphabetically sorted.\n\np = p(:, sort(fieldnames(p)))\np =\n\n162x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [1x162 double]\nLongitude: [1x162 double]\nCause: {1x162 cell}\nCause_Code: [1x162 double]\nCountry: {1x162 cell}\nDay: [1x162 double]\nDesc_Deaths: [1x162 double]\nEq_Mag: [1x162 double]\nHour: [1x162 double]\nIida_Mag: [1x162 double]\nIntensity: [1x162 double]\nLocation: {1x162 cell}\nMax_Height: [1x162 double]\nMinute: [1x162 double]\nMonth: [1x162 double]\nNum_Deaths: [1x162 double]\nSecond: [1x162 double]\nVal_Code: [1x162 double]\nValidity: {1x162 cell}\nYear: [1x162 double]\n\nUsing typical MATLAB vector notation, extract a subset of data from the base geopoint vector into a geopoint vector albeit smaller in size.\n\nsubp = p(20:40,{'Location','Country','Year'}) % get subset of data\nsubp =\n\n21x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [1x21 double]\nLongitude: [1x21 double]\nLocation: {1x21 cell}\nCountry: {1x21 cell}\nYear: [1x21 double]\n\nNote that the coordinate properties Latitude and Longitude, and the Collection properties, are retained in this subset of geopoint vectors.\n\nTo set property values, use the () operator, or assign array values to corresponding fields, or use dot '.' notation (object.property) to assign new property values.\n\npts = geopoint();\npts.Latitude = [42 44 45];\npts.Longitude = [-72 -72.1 -71];\npts\npts =\n\n3x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [42 44 45]\nLongitude: [-72 -72.1000 -71]\n\nUse ( ) to assign values to fields.\n\npts(3).Latitude = 45.5;\npts\npts =\n\n3x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [42 44 45.5000]\nLongitude: [-72 -72.1000 -71]\n\nUse dot notation to create new dynamic properties\n\npts.Name = {'point1', 'point2', 'point3'}\npts =\n\n3x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [42 44 45.5000]\nLongitude: [-72 -72.1000 -71]\nName: {'point1' 'point2' 'point3'}\n\nGet property values\n\npts.Name\nans =\n\n'point1' 'point2' 'point3'\n\nRemove dynamic properties. To delete or remove dynamic properties, set them to [] or set the Latitude or Longitude property to [].\n\npts.Temperature = 1:3\npts =\n\n3x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [42 44 45.5000]\nLongitude: [-72 -72.1000 -71]\nName: {'point1' 'point2' 'point3'}\nTemperature: [1 2 3]\n\nBy setting the Temperature property to [], this dynamic property is deleted.\n\npts.Temperature = []\npts =\n\n3x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [42 44 45.5000]\nLongitude: [-72 -72.1000 -71]\nName: {'point1' 'point2' 'point3'}\n\nTo clear all fields in the geopoint vector, set the Latitude or Longitude property to [].\n\npts.Latitude = []\npts =\n\n0x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: []\nLongitude: []\n\nIf you typically store latitude and longitude coordinates in an N-by-2 or 2-by- M array, you can assign these numeric values to a geopoint vector. If the coordinates are N-by-2, the first column is assigned to the Latitude property and the second column to the Longitude property. If the coordinates are 2-by- M , then the first row is assigned to the Latitude property and the second row to the Longitude property.\n\nltln = [coastlat coastlon];\t\t% 9865x2 array\npts = geopoint;\t\t \t% null constructor\npts(1:numel(coastlat)) = ltln;\t% assign array\npts\npts =\n9865x1 geopoint vector with properties:\n\nCollection properties:\nGeometry: 'point'\nFeature properties:\nLatitude: [1x9865 double]\nLongitude: [1x9865 double]\n\nans = struct with fields:\nName: 'coastline'\n\nexpand all\n\nTips\n\n• If Latitude, Longitude, or a dynamic property is set with more values than features in the geopoint vector, then all other properties expand in size using 0 for numeric values and an empty character vector ('') for cell values. See Manipulate a Geopoint Vector for examples of these behaviors.\n\n• If a dynamic property is set with fewer values than the number of features, then this dynamic property expands to match the size of the other properties by inserting a 0, if the value is numeric, or an empty character vector (''), if the value is a cell array.\n\n• If the Latitude or Longitude property of the geopoint vector is set with fewer values than contained in the object, then all other properties shrink in size.\n\n• If either Latitude or Longitude are set to [ ], then both coordinate properties are set to [ ] and all dynamic properties are removed.\n\n• If a dynamic property is set to [ ], then it is removed from the object.", null, "" ]

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[ "Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange\nScilab-Branch-6.1-GIT\nChange language to: English - Português - 日本語 - Русский\n\n# riccati\n\nSolves the matricial Riccati equation (continuous | discrete time domain)\n\n### Syntax\n\n```X = riccati(H)\nX = riccati(H, E)\nX = riccati(A, B, C, dom)\nX = riccati(A, B, C, dom, method)\n[X1, X2, residual] = riccati(...)```\n\n### Arguments\n\nA, B, C\nSquare matrices of real numbers, of size n x n: Matricial coefficients of the equation.\n\nH, E\nSquare matrices of real numbers, of size 2n x 2n: Hamiltonian matrices | pencil of the equation.\n\ndom\nType / time domain of the Riccati equation: `\"c\"` or `\"continuous\"`, or `\"d\"` or `\"discrete\"`.\n\nmethod\nstring : `\"eigen\"` for block diagonalization (default), or `\"schur\"` for Schur method.\n\nX1, X2, X\nsquare matrices of real numbers (X2 invertible). X is symetric when B and C are so.\n\nresidual\nreal number: norm1 of the actual Equation(X) residue.\n\n### Description\n\n##### Continuous time domain\n\nX = riccati(A, B, C, \"c\") solves the matricial Riccati equation:\n\n`A'*X + X*A - X*B*X + C = 0`\n\nX = riccati(H) where `H = [A, -B ; -C, -A']` is the Hamiltonian matrix, does the same, using the 'eigen' method, but is more stable.\n\n##### Discrete time domain\n\nX = riccati(A, B, C, \"d\") solves the Riccati equation:\n\n```A'*X*A - (A'*X*B1) / (B2 + B1'*X*B1) * (B1'*X*A) + C - X = 0```\n\nwith ```B = B1 / B2 * B1'```.\n\nX = riccati(H, E) does the same, where `H = [A, zeros(n,n) ; -C, eye(n,n)]` and `E = [eye(n,n), B ; zeros(n,n), A']` define the Hamiltonian pencil `(E,H)`.\n\n##### Output options\n\n[X1, X2] = riccati(..) provides `X1` and `X2`, with `X2` invertible, such that `X = X1 / X2`.\n\n`residual` is the L1-norm of the actual equation's result. If `X` is an actual solution, `residual` should be 0. Most often its value is close to `%eps*norm(X)`.", null, "In the `discrete` case, sometimes `B1` and `B2` can't be retrieved from `B`. Then `residual` can't be assessed and is set to `%nan`.", null, "When no solution is found, `X=[]` or `X1=[]` is returned, and a warning is yielded.\n\n### Examples\n\nContinuous time domain\n\n```n = 3;\n// [A, B, C] = (grand(n,n,\"uin\",-9,9), grand(n,n,\"uin\",-9,9), grand(n,n,\"uin\",-9,9))\nA = [\n-62. 91. 57.\n-43. -45. -19.\n58. 83. -62.\n];\nB = [\n75. -31. -10.\n-79. 70. 68.\n-72. -5. 32.\n];\nC = [\n-56. 70. 58.\n-41. 54. 50.\n90. 2. -40.\n];\n\n// With A, B, C\n// ------------\nX = riccati(A, B, C, \"c\")\nclean(A'*X + X*A - X*B*X + C)\n[x1, x2, res] = riccati(A,B,C, \"c\");\nx = x1 / x2;\nand(x==X)\nres\n\n// With the Hamiltonian\n// --------------------\nH = [A, -B; -C, -A'];\nX = riccati(H)\nclean(A'*X + X*A - X*B*X + C)\n[x1, x2] = riccati(H);\nx = x1 / x2;\nand(x==X)```\n```--> X = riccati(A, B, C, \"c\")\nX =\n-0.1790367 0.4166084 0.2319152\n-0.4977093 0.7993495 0.3086213\n0.5595286 0.3202094 -0.103394\n\n--> clean(A'*X + X*A - X*B*X + C)\nans =\n0. 0. 0.\n0. 0. 0.\n0. 0. 0.\n\n--> [x1, x2, res] = riccati(A,B,C, \"c\");\n--> x = x1 / x2;\n--> and(x==X)\nans =\nT\n\n--> res\nres =\n3.340D-13\n\n--> // With the Hamiltonian\n--> // --------------------\n--> H = [A, -B; -C, -A'];\n--> X = riccati(H);\nX =\n-0.1790367 0.4166084 0.2319152\n-0.4977093 0.7993495 0.3086213\n0.5595286 0.3202094 -0.103394\n\n--> clean(A'*X + X*A - X*B*X + C)\nans =\n0. 0. 0.\n0. 0. 0.\n0. 0. 0.\n\n--> [x1, x2] = riccati(H);\n--> x = x1 / x2;\n--> and(x==X)\nans =\nT\n```\n\nDiscrete time domain\n\n```// Eq: A'*X*A - (A'*X*B1) / (B2 + B1'*X*B1) * (B1'*X*A) + C - X = 0\nn = 4;\n[A, B1, B2, C] = (rand(n,n), rand(n,n), rand(n,n), rand(n,n));\nB = B1 / B2 * B1';\n\nX = riccati(A, B, C, 'd', 'schur')\nif X <> [] then\nclean(A'*X*A - (A'*X*B1) / (B2 + B1'*X*B1) * (B1'*X*A) + C - X)\n[x1, x2, res] = riccati(A, B, C, 'd', 'schur');\nx = x1 / x2;\nand(x==X)\nres\nelse\ndisp(\"Riccati: No solution found\")\nend```\n\n### History\n\n Version Description 6.1.0 `riccati(H)` and `riccati(H,E)` added. `residual` output added. When no solution is found, `X=[]` | `X1=[]` is now returned, without error." ]

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[ "", null, "# Formula to choose Earliest Date from list of 6 dates\n\nHi everyone! I have a table with a list of 6 different release dates - I’m trying to create a field that looks at all of those dates and picks the earliest/first one as “Initial Launch.” IS_BEFORE only seems to work with 2 dates and the factorial of 6 dates is 720 possibilities of combinations so I don’t think I can do a workaround combining multiple nested IF and IS_BEFORE functions. Maybe I’m over thinking this though. I also thought about self-linking and doing a rollup field?\n\nAnyway, any advice or recommendations welcome!\n\nSomeone might have a quicker & easier way of accomplishing this, but this is how I would personally do it on my end:\n\nThe `MIN` function is designed to compare a whole bunch of different numbers and return the minimum value of those numbers. It would be amazing if the `MIN` function worked on date fields, but it doesn’t.\n\nSo we need to convert your dates to pure numbers, and those numbers should be in a format where a later date is numerically larger than an earlier date.\n\nSo, for today (July 15, 2020), we would want today’s number to look like 20200715, which is the format YYYYMMDD. We can do this with the `DATETIME_FORMAT` function, but that returns a text string, so we need to wrap the results of that function in `VALUE` to turn it into a number.\n\nSo the formula to convert one specific date into a number would look like this:\n`VALUE(DATETIME_FORMAT({Date Field #1},'YYYYMMDD'))`\n\nTo compare a whole bunch of date fields and return the minimum of those date fields, you would use this formula:\n\n``````MIN(\nVALUE(DATETIME_FORMAT({Date Field #1},'YYYYMMDD')),\nVALUE(DATETIME_FORMAT({Date Field #2},'YYYYMMDD')),\nVALUE(DATETIME_FORMAT({Date Field #3},'YYYYMMDD')),\nVALUE(DATETIME_FORMAT({Date Field #4},'YYYYMMDD')),\nVALUE(DATETIME_FORMAT({Date Field #5},'YYYYMMDD')),\nVALUE(DATETIME_FORMAT({Date Field #6},'YYYYMMDD'))\n)\n``````\n\nThat will give you the result that you’re looking for, but it will be in the format of YYYYMMDD.\n\nSo, to convert it back into a readable date format again, you would create ANOTHER formula field that would look like this:\n\n`DATETIME_PARSE({Name of your formula field above}, 'YYYYMMDD')`\n\nAnd that will get you what you need!", null, "Hope this helps! If this answers your question, could you please mark this comment as the solution to your question? This will help other people who have a similar question.", null, "1 Like\n\nThis worked! Thanks, I tried something similar but I was missing the VALUE() function in my formula.\n\n1 Like\n\nJust to clarify, `MIN()` and `MAX()` won’t work when referencing date fields directly. However, when a bunch of dates have been collected in a rollup, `MIN(values)` will work to find the earliest date, and `MAX(values)` can find the latest.\n\n@ScottWorld came up with a great formula. Here is a variation. My variation is a bit overkill for the original situation, but it might be useful in other similar situations with a greater variety in date/times.\n\n• This version retains the original time (down to the millisecond) as well as the date of the original date/time field. It does this by formatting the date/time as a Unix timestamp with the unit specifiier `'x'`. Using the Unix timestamp also enables the formula to work with years before 1000AD.\n\n• This formula converts the numeric value back into a date, so you do not need an extra field for that step.\n\n• This formula also checks to make sure that there is a date in at least one of the fields, and finds the earliest of whichever fields do have dates in them. (It is blank if all date fields are blank.)\n\n``````IF(\nOR(\n{Date Field #1},\n{Date Field #2},\n{Date Field #3},\n{Date Field #4},\n{Date Field #5},\n{Date Field #6}\n),\nDATETIME_PARSE(\nMIN(\nIF({Date Field #1}, VALUE(DATETIME_FORMAT({Date Field #1},'x')), 999999999999999),\nIF({Date Field #2}, VALUE(DATETIME_FORMAT({Date Field #2},'x')), 999999999999999),\nIF({Date Field #3}, VALUE(DATETIME_FORMAT({Date Field #3},'x')), 999999999999999),\nIF({Date Field #4}, VALUE(DATETIME_FORMAT({Date Field #4},'x')), 999999999999999),\nIF({Date Field #5}, VALUE(DATETIME_FORMAT({Date Field #5},'x')), 999999999999999),\nIF({Date Field #6}, VALUE(DATETIME_FORMAT({Date Field #6},'x')), 999999999999999)\n) ,\n'x'\n)\n)``````\n2 Likes\n\nThis topic was automatically closed 3 days after the last reply. New replies are no longer allowed." ]

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[ "2019\nТом 71\n№ 11\n\n# Infinite systems of stochastic differential equations and some lattice models on compact Riemannian manifolds\n\nAbstract\n\nStochastic dynamics associated with Gibbs measures on an infinite product of compact Riemannian manifolds is constructed. The probabilistic representations for the corresponding Feller semigroups are obtained. The uniqueness of the dynamics is proved.\n\nEnglish version (Springer): Ukrainian Mathematical Journal 49 (1997), no. 3, pp 360-372.\n\nCitation Example: Daletskii A. Yu., Kondratiev Yu. G. Infinite systems of stochastic differential equations and some lattice models on compact Riemannian manifolds // Ukr. Mat. Zh. - 1997. - 49, № 3. - pp. 326–337.\n\nFull text" ]

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[ "# A Survey on Graph Drawing Beyond PlanarityA Survey on Graph Drawing Beyond Planarity\n\nUniversità degli Studi di Perugia, Italy\n\nACM Comput. Surv., Vol. 52, No. 1, Article 4, Publication date: February 2019.\nDOI: https://doi.org/10.1145/3301281\n\nGraph Drawing Beyond Planarity is a rapidly growing research area that classifies and studies geometric representations of nonplanar graphs in terms of forbidden crossing configurations. The aim of this survey is to describe the main research directions in this area, the most prominent known results, and some of the most challenging open problems.\n\nCCS Concepts: • Mathematics of computing → Graph theory; Graphs and surfaces; Extremal graph theory; Graph algorithms; • Human-centered computing → Visualization;\n\nAdditional Key Words and Phrases: Graph theory, graph planarity, graph drawing, graph algorithms\n\nACM Reference format:\nWalter Didimo, Giuseppe Liotta, and Fabrizio Montecchiani. 2019. A Survey on Graph Drawing Beyond Planarity. ACM Comput. Surv. 52, 1, Article 4 (February 2019), 37 pages. https://doi.org/10.1145/3301281\n\n## 1 INTRODUCTION\n\nIn the mid-1980s, the early pioneers of graph drawing had the intuition that a drawing with too many edge crossings is harder to read than a drawing of the same graph with fewer edge crossings (see, e.g., [38, 39, 79, 211]). This intuition was later confirmed by a series of cognitive experimental studies (see, e.g., [200, 201, 225]). As a result, a large part of the existing literature on graph drawing showcases elegant algorithms and sophisticated data structures under the assumption that the input graph is planar (i.e., it admits a drawing without edge crossings). When the input graph is nonplanar, crossing minimization heuristics are used to insert a small number of dummy vertices corresponding to the edge crossings to obtain a planarization of the input graph. A crossing-free drawing of the planarization can be computed using one of the algorithms for planar graphs, and then the crossings are reinserted by removing the dummy vertices. This approach is commonly adopted and works well for graphs of relatively small size, up to a few hundred vertices and edges (see, e.g., [95, 166]). However, the technological advances of the past 20 years have generated torrents of relational data that are typically modeled as large graphs with thousands of vertices (or more). These graphs are often hard to visually analyze due, mainly, to their large size, which typically implies that a high number of edge crossings is unavoidable even by the most sophisticated planarization approaches. As a consequence, a strong consensus has developed that a new theory of nonplanar graph drawing is needed.\n\nIn this context, in the early 2000s, Mutzel ran an informal experiment with computer scientists at a Dagstuhl workshop where she presented two different drawings of the same bipartite graph: One has the minimum number of edge crossings, the second has $41\\%$ more edge crossings, but it has a skewness of four, which means that all crossings can be removed by deleting four edges in the drawing. All computer scientists found the drawing with skewness four more readable than the one with fewer edge crossings. This is reported as anecdotal evidence that the topological properties of the edge crossings may be more important than their number.\n\nBesides the topological properties of the edge crossings, the impact of their geometric properties on the readability of a nonplanar drawing was evaluated in a pioneering sequence of user experiments performed in the graph drawing research lab at the University of Sydney. By means of an eye-tracking device, the experiments present statistical evidence that crossings significantly affect human understanding if they form acute angles, but if these angles vary in the range from about $\\frac{\\pi }{3}$ to $\\frac{\\pi }{2}$ they guarantee good readability properties [160, 162, 164].\n\nThese empirical experiences suggest that a new theory of nonplanar graph drawing can be developed under the assumption that not only the number of edge crossings but also their (topological and/or geometric) properties have an impact on the readability of a diagram. Hence, a natural step toward understanding nonplanar representations of graphs is to classify and study them in terms of forbidden crossing configurations. This is, in a broad sense, the aim and scope of the rapidly growing research area of graph drawing beyond planarity 1. Table 1 reports some examples of beyond-planar graphs with a description of their forbidden crossing configurations2.\n\nTable 1. Examples of Beyond-Planar Graph Families", null, "Overview and paper organization. The goal of this survey is to classify and describe prominent results and promising research directions in the fertile area of graph drawing beyond planarity. The survey addresses the following questions.\n\n• What are the forbidden crossing configurations and the main research directions that have been studied so far?\n• For each such direction, what are the main combinatorial results, and which algorithms have been designed, implemented, and experimentally evaluated?\n• What are the most relevant open problems and the least explored research directions in this area?\n\nTo answer Question Q1, in Section 3 we define graph families that avoid forbidden crossing configurations, and we present a taxonomy of the main research directions in graph drawing beyond planarity. Sections 49 address Question Q2, using the taxonomy as a guideline to classify the main combinatorial and algorithmic results. At the end of each section, we address Question Q3 by discussing some relevant open problems. Basic definitions on graph drawing can be found in Section 2.\n\n## 2 BASIC DEFINITIONS ON GRAPH DRAWING\n\nLet $G=(V,E)$ be a (finite) graph. If not otherwise specified, we assume that $G$ may contain multiple edges but no self-loops. A drawing $\\Gamma$ of $G$ maps each vertex $v \\in V$ to a distinct point $p_v$ of the plane and each edge $(u,v) \\in E$ to a simple Jordan arc with endpoints $p_u$ and $p_v$. In notation and terminology, we make no distinction between the vertices and edges of a graph and the points and arcs representing them, respectively. Two edges of $\\Gamma$ cross if they have a point in common distinct from their endpoints; this point is a crossing. We assume that an edge does not contain a vertex other than its endpoints, no two edges meet tangentially, and no three edges share a crossing. $\\Gamma$ is simple if any two edges intersect in at most one point, which is either a common endpoint or an interior point where the two edges properly cross. Thus, in a simple drawing, two adjacent edges do not cross and two nonadjacent edges cross at most once.\n\nA drawing $\\Gamma$ of $G$ divides the plane into topologically connected regions, called faces, separated by edges or fragments of edges. The infinite region is called the external face; the other regions are the internal faces. Note that the boundary of a face may contain both vertices of the graph and crossings. An embedding of $G$ is an equivalence class of drawings of $G$ under homeomorphism of the plane (i.e., is a class of drawings of $G$ that define the same set of [external and internal] faces). A graph with a fixed embedding is called an embedded graph. A drawing without crossings is planar. A graph is planar if it admits a planar drawing. A planar embedding is the embedding of a planar drawing. A graph with a fixed planar embedding is an embedded planar graph, or, briefly, a plane graph. An embedding of a graph $G$ defines, for each vertex $v$ and for each crossing $c$, the clockwise order of the edges incident to $v$ and to $c$. A rotation system of $G$ is a relaxation of an embedding of $G$ in which the clockwise order of the edges is fixed only for each vertex, while no information is given about which pairs of edges cross or their order of intersection. If $G$ is planar, a planar embedding of $G$ corresponds to a rotation system plus the choice of the external face (see Figure 1).\n\nIn a polyline drawing of a graph, each edge is a chain of segments; a bend is a point where two segments of the same edge meet. A $k$-bend drawing is a polyline drawing with at most $k$ bends per edge. A 0-bend drawing is also called a straight-line drawing.\n\nWe remark that, in the literature, a drawing of a graph is sometimes called a topological graph and a straight-line drawing is sometimes called a geometric graph (see, e.g., Chapter 10 in ). In the remainder of this article, we use the pair of terms “drawing” and “topological graph” and the pair “straight-line drawing” and “geometric graph” interchangeably.\n\n## 3 FORBIDDEN CONFIGURATIONS AND MAIN RESEARCH DIRECTIONS\n\nIn Section 3.1, we survey forbidden crossing configurations; in Section 3.2, we present a taxonomy of the most explored research directions in graph drawing beyond planarity.\n\n### 3.1 Graph Families\n\nThe beyond-planar graph families considered in this survey are defined as graphs that admit drawings without specific forbidden configurations (i.e., sets of edges that violate some desired topological or geometric property of the edge crossings). Table 1 provides a schematic illustration of some beyond-planar graph families, among the most studied. For each family $X$, we define what a drawing of type $X$ is. A graph belongs to family $X$ if it admits a drawing of type $X$. Definitions based on first-order logic formulas have been recently proposed for many of these types of drawings .\n\n$k$-planar drawings. A $k$-planar drawing $(k \\ge 1)$ does not contain an edge crossed more than $k$ times. The family of $k$-planar graphs, for $k=1$, was introduced in 1965 in the context of the simultaneous vertex-face coloring of planar graphs . The study of $k$-planar graphs, for $k \\gt 1$, was proposed for the first time as a tool for finding lower bounds on the crossing number of a graph (i.e., on the minimum number of crossings in a drawing of a graph).\n\n$k$-quasi planar drawings. A $k$-quasi planar drawing $(k \\ge 3)$ does not have $k$ mutually crossing edges. The first results about $k$-quasi planar graphs date back to the 1980s and ’90s, when some authors addressed the problem of determining the maximum number of edges for these graphs [13, 196], answering questions posed even earlier [9, 32, 177].\n\nSkewness-$k$ drawings. A skewness-$k$ drawing $(k \\ge 1)$ is such that the removal of at most $k$ edges makes the drawing planar. In terms of forbidden configuration, it means that the drawing does not contain a set of crossings not covered by (at most) $k$ edges. A graph has skewness $k$ if it admits a skewness-$k$ drawing. Graphs with skewness $k$ are mainly studied for $k=1$, under the name of almost planar (or near planar) graphs, especially in terms of crossing number [76, 77]. Some authors also studied the problem of efficiently computing a skewness-$k$ drawing of a graph $G=(V,E)$ ($k \\ge 1$) with the minimum number of crossings and with a fixed planar subgraph $G^{\\prime }=(V,E \\setminus F)$, where $|F|=k$ [84, 150]. For $k=1$, this problem is linear-time solvable and the solution gives an approximation to the crossing number of the almost planar graph $G$ .\n\n$k$-apex drawings. A $k$-apex drawing $(k \\ge 1)$ is such that the removal of at most $k$ vertices makes the drawing planar. It means that the drawing does not contain a set of crossing edges not covered by (at most) $k$ vertices. It is immediately seen that a skewness-$k$ drawing is also a $k$-apex drawing, but not vice versa. It is known that 1-apex graphs, mainly recognized as apex graphs in the literature, are closed under the operation of taking minors. They have connections with other aspects of graph minor theory (see, e.g., [205, 218]) and play a role in the relations between treewidth and graph diameter [92, 136]. Chimani et al. showed that computing an apex drawing with the minimum number of crossings and an identified apex vertex can be done efficiently .\n\n$(k,l)$-grid-free drawings. For $k,l \\ge 1$, a $(k,l)$-grid-free drawing does not contain a $(k,l)$-grid (i.e., two groups of $k$ and $l$ edges, respectively), such that each edge of the first group crosses all the edges of the second group. If the $k$ edges are incident to the same vertex, the $(k,l)$-grid is radial; if the $l$ edges are also incident to the same vertex, the $(k,l)$-grid is biradial. A $(k,l)$-grid is natural if all its edges are independent and no two edges in the same group cross. Pach et al. introduced $(k,l)$-grid-free graphs as a generalization of $k$-quasi planar graphs, and they investigated upper bounds on their number of edges .\n\n$k$-fan-crossing-free drawings. A $k$-fan-crossing-free drawing does not contain $k \\ge 2$ adjacent edges that cross another edge. A 2-fan-crossing-free drawing is also called fan-crossing-free. Cheong et al. introduced the class of $k$-fan-crossing-free graphs as a special case of the class of radial $(k,l)$-grid-free graphs where $k \\ge 2$ and $l=1$ ; they were interested in establishing upper bounds on the number of edges for these graphs.\n\nFan-planar drawings. A fan-planar drawing does not contain two independent edges that cross a third one or two adjacent edges that cross another edge from different “sides” . This family of drawings is somehow opposite to the class of fan-crossing-free drawings. From a practical point of view, fan-planar drawings can be used to create drawings with few edge crossings per edge in a confluent drawing style (see, e.g., Figure 2). Note that a fan-planar drawing cannot contain a $(k,l)$-grid that is not radial, for $k \\ge 2$.\n\nRAC drawings. A straight-line drawing, such that any two crossing edges form $\\frac{\\pi }{2}$ angles at their crossing point, is a straight-line RAC (Right Angle Crossings). RAC drawings are introduced in Didimo et al. , motivated by cognitive studies that suggest a positive correlation between large crossing angles and human understanding of graph visualizations [160, 162, 164], and by the common use of large angle crossings in handmade real-world diagrams, such as metro maps [203, 223]. RAC drawings with edge bends are also studied in the literature and will be discussed in this survey.\n\nACE$\\alpha$ and ACL$\\alpha$ drawings. ACE$\\alpha$ and ACL$\\alpha$ drawings are natural variants of RAC drawings that are parametric in a value $\\alpha \\in (0,\\frac{\\pi }{2})$. In a straight-line ACE$\\alpha$ drawing, any two crossing edges form an angle equal to $\\alpha$. Graphs that admit this type of drawing were originally introduced with the name of $\\alpha$AC$^=$ graphs [5, 6]. In a straight-line ACL$\\alpha$ drawing, the value of any crossing angle is at least $\\alpha$. These drawings were independently introduced by two different works; in Di Giacomo et al. they are called LAC$_\\alpha$ drawings, and in Dujmovic et al. they are called $\\alpha$AC drawings. As for RAC drawings, ACE$\\alpha$ and ACL$\\alpha$ drawings with edge bends are also considered in this survey.\n\nPartial edge drawings (PEDs). In a Partial Edge Drawing (PED) the “middle” part of each edge is omitted to obtain a crossing-free representation. The idea behind this drawing style was introduced to visualize network overload between switches of the AT&T long-distance telephone network in the United States . In particular, in an $\\alpha$- Symmetric Homogeneous PED (SHPED) $(0\\lt \\alpha \\lt \\frac{1}{2})$, each edge $(u,v)$ is (partially) drawn as a pair of segments, called $\\alpha$-stubs, one incident to $u$, the other incident to $v$, and each being a fraction $\\alpha$ of segment $\\overline{uv}$; the two stubs do not cross. The $\\alpha$-SHPED model has been formally defined only recently and has inspired both theoretical and practical research (see, e.g., [56, 73]). Bruckdorfer et al. extended this model for graphs with maximum degree four, where the drawings are orthogonal and have at most one bend per edge .\n\n$k$-gap-planar drawings. In a $k$-gap-planar drawing it is possible to map each crossing to one of the two corresponding crossing edges so that, for each edge $e,$ at most $k$ crossings are mapped to $e$. This family generalizes $k$-planar graphs and was introduced in Bae et al. with a practical motivation inspired by edge casing, a method commonly used to alleviate the visual clutter caused by crossing lines [23, 138]. In a cased drawing of a graph, each crossing is resolved by locally interrupting one of the two crossing edges, and a $k$-gap-planar graph can be equivalently defined as a graph that admits a cased drawing in which each edge has at most $k$ gaps.\n\n$H$-self-intersecting-free drawings. Given a graph $H$, a straight-line $H$-self-intersecting-free ( $H$-sif) drawing does not contain a self-intersecting geometric graph isomorphic to $H$. This definition specializes the more general $H$-free drawings, which do not contain graphs isomorphic to $H$. Typical forbidden configurations consider self-intersecting paths or cycles [193, 199], whose study was triggered by classical extremal problems on forbidden geometric subgraphs (see, e.g., ).\n\nPlanarly connected drawings. In a planarly connected drawing, each pair of crossing edges is independent and there is a crossing-free edge that connects two of their endpoints . As will be discussed in Section 6, the main motivation for the study of these graphs comes from the fact that this family includes meaningful subfamilies of 1-planar and fan-planar drawings .\n\n### 3.2 Main Research Directions\n\nThe tree of Figure 3 summarizes a taxonomy of the graph drawing beyond the planarity area. The first-level nodes identify the main research directions studied so far; we classify the results in the literature according to these directions. The intermediate nodes are refinements of the main directions. The leaves refer to the sections of the survey where the research directions are described in detail; most of these sections report one or more summarizing tables, which are also noted in the corresponding leaves. Here, we briefly describe the main research directions.\n\nDensity. Since the mid-1960s, different authors (see, e.g., [32, 59, 177]) initiated the study of the following Turán-type problem: What is the maximum number of edges that a drawing of a graph can have without containing a given type of forbidden crossing configuration? Since then, this question has been long studied for all families of beyond-planar drawings. Results about the edge density of beyond-planar graphs are discussed in Section 4.\n\nRecognition. In contrast to planarity testing, which is solvable in linear time , recognizing whether a graph belongs to a certain family of beyond-planar graphs is computationally hard for most families. For some subfamilies of beyond-planar graphs and/or under additional restrictions of the input, the recognition problem is polynomial-time solvable. Results about the recognition problem of different families of beyond-planar graphs are surveyed in Section 5.\n\nRelationships. Studying the combinatorial relationships between different families of beyond-planar graphs is a fundamental step toward developing a comprehensive theory of graph drawing beyond planarity. The typical question in this context is the following: Let $F$ and $F^{\\prime }$ be two forbidden types of crossing configurations (for example, $F$ may be “four mutually crossing edges” and $F^{\\prime }$ may be “an edge crossed by three distinct edges”). If a graph $G$ admits a drawing where $F$ is forbidden, does $G$ also admit a drawing where $F^{\\prime }$ is forbidden? Results about relationships of inclusion or of intersection between families of beyond-planar graphs are described in Section 6.\n\nQuality metrics. In addition to forbidding some types of edge crossings in a nonplanar drawing, one can pursue some geometric optimization goals, often called quality metrics or aesthetic requirements, such as minimizing the drawing area for a given resolution rule, maximizing the drawing aspect ratio, or minimizing the number of different slopes used to draw the edges or the number of bends per edge. Quality metrics have a strong impact on the visual complexity of a drawing (see, e.g., [95, 170]). Results about this research direction are presented in Section 7. We remark that, within this research direction, the stretchability problem asks for computing an embedding-preserving straight-line drawing, and it has been studied for planar drawings since the mid-1930s (see, e.g., [141, 210, 224]).\n\nConstraints. Depending on the type of graph and/or application, additional constraints can be imposed on a drawing. For example, for bipartite graphs, a typical constraint is to represent the vertices of each partition set on one of two parallel lines, which is a fundamental step in the so-called Sugiyama approach for layered drawings . Other constraints require that all vertices are collinear in the so-called $k$-page drawing model [49, 190] or co-circular in the circular layout model . Results about constrained beyond-planar graphs are surveyed in Section 8.\n\nImplementations and experiments. The road to an effective technology transfer of the algorithmic solutions developed in the area of graph drawing beyond planarity has just begun. The initial steps in this direction are summarized in Section 9.\n\n## 4 EDGE DENSITY\n\nThe problem of establishing the maximum number of edges in a given type of beyond-planar graph has been extensively studied in the literature. We recall some basic definitions needed to describe the main results in this research direction. Let $\\mathcal {F}$ be a beyond-planar graph family and let $G$ be an $n$-vertex graph in $\\mathcal {F}$. $G$ is maximal (in $\\mathcal {F}$) if adding any edge to $G$ leads to a graph that is not in $\\mathcal {F}$. $G$ is maximally dense if it has the maximum number of edges over all $n$-vertex graphs in $\\mathcal {F}$. The edge density of $G$ is the ratio between its number of edges and its number of vertices. $G$ is optimal if it has the maximum edge density over all graphs of $\\mathcal {F}$. Note that an optimal graph is also maximally dense, while the converse may not be true; similarly, a maximally dense graph is maximal, but there are maximal graphs that are not maximally dense; Figure 4(a) shows a maximally dense 1-planar graph with 5 vertices that is not optimal; Figure 4(b) shows a 1-planar graph with 12 vertices that is maximal but not optimal, and Figure 4(c) shows an optimal 1-planar graph with 12 vertices.\n\nDensity of $k$-planar graphs. For $1 \\le k \\le 4$, Pach and Tóth prove that $k$-planar graphs have at most $(k+3)(n-2)$ edges , and they also use this result to improve an earlier lower bound on the crossing number of a graph. In particular, the resulting bounds are tight for $k=1$ and for $k=2$, which means that the optimal 1-planar graphs have $4n-8$ edges and the optimal 2-planar graphs have $5n-10$ edges. The best known upper bounds for $k=3$ and $k=4$ are $5.5n-11$ and $6n-12$ , respectively. For $k \\gt 4$, the best known upper bound is $3,81\\sqrt {k}\\,n$ , which improves a previous bound of $4.108\\sqrt {k}\\,n$ . Lower bounds on the edge density of maximal 1-planar graphs are given in Brandenburg et al. . For bipartite 1-planar graphs, Karpov proves a tight bound of $3n-6$ when $n$ is even and $n \\ne 6$, and of $3n -9$ otherwise . Different subfamilies of 1-planar graphs have also been studied, such as 1-planar graphs where no two pairs of crossing edges share an end-vertex, called IC-planar graphs, and those for which two pairs of crossing edges share at most one end-vertex, called NIC-planar graphs. We report these bounds in Table 2. More results about the edge density of 1-planar graphs are surveyed in Kobourov et al. .\n\nTable 2. Density\nGraph Family Max. Num. Edges Tight Refs\n1-planar $4n - 8$", null, "[58, 197]\nstraight-line 1-planar $4n - 9$", null, "[2, 115]\nbipartite 1-planar $3n-8$ for even $n \\ne 6$", null, "$3n-9$ otherwise\nIC-planar $3.25n - 6$", null, "NIC-planar $3.6n - 7.2$", null, "2-planar $5n - 10$", null, "3-planar $5.5(n - 2)$ $+$ \n4-planar $6n - 12$ $+$ \n$k$-planar $(k \\ge 5)$ $3.81\\sqrt {k}\\,n$ $\\times$ \n3-quasi planar $6.5n-20$ $+$ \n4-quasi planar $72(n-2)$ $\\times$ \n$k$-quasi planar $(k \\ge 5)$ $c_k n \\log n$ $\\bigcirc$ \nskewness-$k$ $3n-6+k$", null, "Trivial\n$k$-apex $3(n-k)-6+\\sum _{i=1}^k(n-i)$", null, "Trivial\n$(k,l)$-grid-free $c_{k,l}\\,n$ $\\times$ \nradial-$(k,l)$-grid-free $8 \\cdot 24^lkn$ $\\times$ \nnatural-$k$-grid-free $O(n \\log ^{4k-6} n)$ $\\bigcirc$ \nstraight-line natural-$k$-grid-free $O(kn \\log ^2 n)$ $\\bigcirc$ \nfan-crossing-free $4n-8$", null, "straight-line fan-crossing-free $4n-9$", null, "$k$-fan-crossing-free ($k \\ge 3$) $3(k-1)(n-2)$ $\\times$ \nfan-planar $5n-10$", null, "straight-line RAC $4n - 10$", null, "straight-line ACE$\\alpha$ $3(3n-6)$ $\\times$ \nstraight-line ACL$\\alpha$ $\\frac{\\pi }{\\alpha }(3n-6)$ $\\times$ \n1-gap-planar $5n-10$", null, "$k$-gap-planar ($k\\gt 1$) $O(\\sqrt {k}n)$ $\\times$ \nstraight-line (3-length-path)-sif $O(n \\log n)$ $\\bigcirc$ \nstraight-line (5-length-path)-sif $O(n\\log n/\\log \\log n)$ $\\bigcirc$ \nstraight-line (4-length-cycle)-sif $O(n^{8/5}$) $\\bigcirc$ \nplanarly-connected $c\\,n$ $\\times$ \n\nMost of the bounds in the table hold for simple drawings only; though few of them are still valid for non-simple drawings. The symbol", null, "in column Tightness means that the bound is tight; the symbols $\\times$ and $+$ mean that the bound is tight up to a multiplicative or to an additive constant, respectively, while the symbol $\\bigcirc$ means that the bound may be far from tight.\n\nDensity of $k$-quasi planar graphs. In 1996, Pach, Shahrokhi, and Szegedy conjectured that, for any fixed $k \\ge 2$, there exists a constant $c_k$, depending only on $k$, such that every $k$-quasi planar graph on $n$ vertices has at most $c_k n$ edges. This conjecture has been proved for $k=3$ and $k=4$ by various authors, with the best upper bounds currently known being $6.5n-20$ and $72(n-2)$ , respectively. For $k\\gt 4,$ the conjecture remains unproved, but several superlinear upper bounds have been established; the current best upper bound is $c_k n \\log n$, for a suitable constant $c_k$ . An $O(n \\log n)$ upper bound for $k$-quasi planar graphs with $x$-monotone edges is also known [220, 221].\n\nDensity of fan-planar, $k$-fan-crossing-free, and $k$-gap graphs. Both fan-planar and 1-gap planar graphs on $n$ vertices have at most $5n-10$ edges, which is a tight bound [34, 169]. Recall that the same bound holds for 2-planar graphs. For $k \\gt 1$, a $k$-gap planar graph has $O(\\sqrt {k}n)$ edges . A fan-crossing-free graph has at most $4n-8$ edges, and at most $4n-9$ if the edges are drawn as straight lines . These bounds are analogous to those for topological and geometric 1-planar graphs. For $k\\gt 2$, $k$-fan-crossing-free graphs have at most $3(k-1)(n-2)$ edges .\n\nDensity of straight-line RAC graphs, ACE$\\alpha$ graphs, and ACL$\\alpha$ graphs. Since straight-line RAC graphs are fan-crossing-free, they cannot have more than $4n-9$ edges. In fact, the maximum number of edges of a straight-line RAC graph is $4n-10$ , and for each $k \\ge 3$ there exists a straight-line RAC graph $G_k$ with $n=3k-5$ vertices and $4n-10$ edges (i.e., optimal). Graph $G_k$ is constructed as follows (see Figure 4(d)): $(i)$ start from a triangulated plane graph on $k$ vertices; $(ii)$ add to this graph its dual, except for the node corresponding to the external face; $(iii)$ for each node $u$ of the dual, add the three edges that connect $u$ to the three vertices of the face corresponding to $u$. The fact that two edges of a straight-line RAC drawing $\\Gamma$ of a graph $G$ can cross only at right angles immediately implies that the crossing graph of $\\Gamma$ is bipartite; the crossing graph of $\\Gamma$ has a vertex $v_e$ for each edge $e$ of $G$ and an edge $(v_e,v_{e^{\\prime }})$ if the $e$ and $e^{\\prime }$ cross in $\\Gamma$. Hence, we can bi-color the edges of $G$ such that each color set induces a planar graph, which implies that $G$ has at most $6n-12$ edges. The $4n-10$ bound is proved by exploiting a finer coloring of the edges of $G$ with three colors and different counting arguments based on Euler's formula for planar graphs . Dujmovic et al. provide an alternative proof of this bound, based on charging techniques . They also show that straight-line ACL$\\alpha$ graphs have at most $\\frac{\\pi }{\\alpha }(3n-6)$ edges; the proof is based on a partition of the edges into distinct buckets, according to their directions, so that each bucket induces a planar graph. With similar arguments, it is shown that straight-line ACE$\\alpha$ graphs have at most $3(3n-6)$ edges . These bounds for ACL$\\alpha$ and ACE$\\alpha$ graphs are not known to be tight.\n\nDensity of other graph families. Since a skewness-$k$ graph $G$ becomes planar after the removal of at most $k$ edges, $G$ has at most $3n-6+k$ edges. This bound is trivially achieved by adding $k$ edges to a maximal planar graph. Similarly, removing at most $k$ vertices from an $n$-vertex $k$-apex graph yields a planar graph, and thus $k$-apex graphs have at most $3(n-k)-6+\\sum _{i=1}^k(n-i)$ edges. A graph attaining this bound is constructed from a maximal planar graph, iteratively adding one vertex connected to all other vertices of the graph. $(k,l)$-grid-free and radial-$(k,l)$-grid-free graphs have at most $c_{k,l}\\,n$ edges (where $c_{k,l}$ is constant in $n$ but depends on $k$ and $l$) and $8 \\cdot 24^lkn$ edges, respectively , while natural-$k$-grid-free graphs have $O(n \\log ^{4k-6} n)$ edges . Concerning straight-line (3-length-path)-sif and straight-line (4-length-path)-sif graphs, it is shown that they have $O(n \\log n)$ and $O(n\\log n/\\log \\log n)$ edges, respectively , while straight-line (4-length-cycle)-sif graphs have $O(n^{8/5})$ edges . Finally, planarly connected graphs have $c \\, n$ edges, where $c$ is a constant . Table 2 summarizes the discussed bounds on the edge density of beyond-planar graphs.\n\nOpen problems. Although Turán-type questions on beyond-planar drawings have been studied for a long time and the geometric graph theory literature contains many results, there are still beautiful open problems. For several families of beyond-planar graphs, the upper bounds on the maximum number of edges reported in Table 2 are not tight. Hence, the goal of achieving tight bounds for each of these families gives rise to an array of challenging problems. We find particular interest in the following (see also [70, 217]).\n\nFind a tight upper bound on the edge density of $k$-quasi planar graphs for $k\\gt 4$. The problem is relevant also when the edges are drawn as $x$-monotone curves.\n\nFind tight upper bounds on the edge density of (simple) $k$-planar graphs for $k \\ge 3$.\n\nStarting references for Problem 1 are listed in the References [143, 212, 220, 221]. We recall that Pach et al. conjecture an upper bound that is linear in the number of vertices . Concerning Problem 2, recent papers discuss the edge density of (not necessarily simple) 3-planar graphs [46, 47].\n\nAnother research direction is concerned with providing lower bounds on the number of edges of maximal beyond-planar graphs. This type of question is mostly unexplored; results are known for 1- and 2-planar graphs [30, 62]. We suggest the following.\n\nFind a lower bound on the edge density of maximal straight-line RAC graphs.\n\n## 5 RECOGNITION\n\nGiven a graph $G$ and a family $\\mathcal {F}$ of beyond-planar graphs, the recognition problem studies the complexity of deciding whether a graph $G$ belongs to $\\mathcal {F}$. As we have seen in Section 4, several families of beyond-planar graphs are sparse, and this may suggest a correspondence with planar graphs, which can be recognized in linear time . Unfortunately, this is not the case, and for most of the studied beyond-planar graph families, the recognition problem is in fact hard.\n\nRecognizing 1-planar graphs is NP-complete in general [147, 176], even if the skewness of the input graph is 1 . The problem is, however, Fixed-Parameter Tractable (FPT) with respect to the vertex-cover number, the cyclomatic number, or the tree-depth of the input graph . Recognizing 1-planar graphs remains NP-complete also when the input graph comes with a fixed rotation system, which must be preserved . On the positive side, deciding whether a graph with $n$ vertices is optimal 1-planar (i.e., it is 1-planar and has $4n-8$ edges) is $O(n)$-time solvable ; the testing algorithm exploits a structural characterization of optimal 1-planar graphs . Recently, a similar structural characterization has been proved for optimal 2-planar and optimal 3-planar graphs , although the complexity of recognizing these graphs is still unknown. Polynomial time recognition algorithms are also known for other subfamilies of 1-planar graphs (see, e.g., ). We refer the reader to the annotated bibliography by Kobourov et al. for further references . It is worth remarking that several papers present interesting bounds of graph parameters for 1-planar and $k$-planar graphs, which shed some light on the structure of these graphs and thus may be of interest for the design of FPT recognition algorithms. For example, $k$-planar graphs on $n$ vertices have $O(\\log n)$ book thickness , $O(\\sqrt {kn})$ treewidth and $O(k)$ layered treewidth , and bounded expansion (see for more results).\n\nRecognizing skewness-$k$ graphs is NP-complete in the general case , but the problem can be solved in $O(n)$ time for any fixed value of $k$ (i.e., it is FPT with parameter $k$) .\n\nThe set of 1-apex graphs is closed under the operation of taking minors , hence these graphs have a forbidden graph characterization by the Robertson-Seymour theorem . However, the set of obstructions for these graphs has only been partially discovered . Recognizing $n$-vertex $k$-apex graphs is NP-complete , but there is an $O(n)$-time algorithm if $k$ is a fixed constant .\n\nRecognizing fan-planar graphs is NP-complete , even for a fixed rotation system . The same holds for 1-gap-planar graphs .\n\nRecognizing straight-line RAC drawable graphs is NP-hard in general , but can be solved in linear time for complete bipartite graphs . Even the more restricted problem of deciding whether a graph admits a straight-line RAC drawing with at most one crossing per edge is NP-hard . It is, however, unknown if recognizing straight-line RAC (1-planar) drawable graphs belongs to NP because the complexity of deciding whether an embedded graph has an embedding-preserving straight-line RAC (1-planar) drawing is unknown. On the other hand, every 1-plane graph admits a 1-bend RAC drawing with at most one crossing per edge [44, 81] (see also Section 7).\n\nConcerning $\\alpha$-SHPEDs, there are necessary or sufficient conditions for their existence, although the complexity of the recognition problem has not been established. If an $n$-vertex complete graph has a $\\frac{1}{4}$-SHPED, then $n\\lt 165$; also, if a graph has bandwidth $k$, it has a $\\Theta (\\frac{1}{k})$-SHPED . Similar bounds are given for complete bipartite graphs. The problem of maximizing the total stub length (or ink) to turn a geometric graph into a partial edge drawing that is symmetric but not necessarily homogeneous (the value of $\\alpha$ may not be the same for all the edges) is NP-hard but becomes polynomial-time solvable for geometric 2-planar graphs . Table 3 summarizes the results discussed in this section.\n\nTable 3. Recognition\nGraph Family Restrictions Complexity Refs\n1-planar NP-complete [147, 176]\n1-planar bounded bandwidth, pathwidth, or treewidth NP-complete \n1-planar bounded cyclomatic number or treedepth $O(n)$ \n1-planar fixed rotation system NP-complete \n1-planar skewness-1 NP-complete \noptimal 1-planar $O(n)$ \nskewness-$k$ NP-complete \nskewness-$k$ bounded $k$ $O(n)$ \n$k$-apex NP-complete \n$k$-apex bounded $k$ $O(n)$ \nfan-planar NP-complete \nfan-planar fixed rotation system NP-complete \nstraight-line RAC NP-hard \nstaight-line 1-planar RAC NP-hard \n1-gap-planar NP-complete \n1-gap-planar fixed rotation system NP-complete \n\nOpen problems. As shown in Table 3, the general recognition problem is NP-hard for most of the considered families. However, only a few papers deal with FPT algorithms, and no other types of exponential-time recognition algorithms have been considered. This motivates the following.\n\nInvestigate graph parameters for which the recognition problem for some families of beyond-planar graphs becomes tractable. For example, in the light of the results for 1-planar graphs, is the recognition problem of 1-gap-planar graphs FPT with respect to the cyclomatic number or tree depth?\n\nFurthermore, the recognition problem remains unexplored for many other families of beyond-planar graphs. This poses several open questions. For example:\n\nWhat is the complexity of deciding whether a graph is $k$-quasi planar? The question is already interesting when $k=3$.\n\nIt is know that every graph has a RAC drawing with at most three bends per edge and that it has $O(n)$ edges if either one bend or two bends per edge are allowed [15, 28] (see Section 7). Recognizing straight-line RAC graphs is NP-hard , but the following question is still open.\n\nWhat is the complexity of recognizing $k$-bend RAC drawable graphs, for $k \\in \\lbrace 1,2\\rbrace$?\n\nA characterization of the graphs that admit an $\\alpha$-SHPED is unknown.\n\nWhat is the complexity of deciding whether a graph admits an $\\alpha$-SHPED? In particular, what is the complexity for $\\alpha = \\frac{1}{4}$?\n\n## 6 RELATIONSHIPS BETWEEN GRAPH FAMILIES\n\nSome families of beyond-planar graphs have similar edge densities or exhibit similar structural and topological properties. In these cases, it is natural to ask whether they have some inclusion relationships. In what follows, we survey the main results concerning this research direction.\n\nRAC graphs and 1-planar graphs. One of the most studied problems on the subject is the relationship between RAC graphs and 1-planar graphs. Recall that straight-line RAC drawings have at most $4n - 10$ edges, while topological (respectively, geometric) 1-planar graphs have at most $4n-8$ edges (respectively, $4n-9$). This immediately implies that optimal 1-planar graphs are not straight-line RAC. Eades and Liotta proved in fact that these two families are in general incomparable, and they provide a series of interesting results about their relationships . They exhibit an infinite subfamily of straight-line RAC graphs with $n \\ge 85$ vertices that are not 1-planar (see, e.g., Figure 5(a)), and they show that there exist infinitely many 1-planar graphs with $4n-10$ edges that are not straight-line RAC (see, e.g., Figure 5(b)). On the positive side, every optimal straight-line RAC graph is 1-planar (see, e.g., Figure 4(d)).\n\nBrandenburg et al. study the RAC drawability of the 1-planar graphs with independent pairs of crossing edges (the IC-planar graphs) . They show that every IC-planar graph has a straight-line RAC drawing, while this is not true for the larger class of the NIC-planar graphs .\n\nThe incomparability of straight-line RAC graphs and 1-planar graphs, together with the fact that IC-planar graphs always admit a straight-line RAC drawing (which is also 1-planar), suggest two interesting questions: $(i)$ What is the complexity of deciding whether a graph admits a drawing that is both 1-planar and straight-line RAC? $(ii)$ Does every 1-planar graph admit a 1-planar drawing that is also RAC if we allow at most one bend per edge? These questions have been recently answered : Question $(i)$ is NP-hard, as already mentioned in Section 5 (see Table 3), while Question $(ii)$ has a positive answer. Figure 5(c) shows a 1-planar RAC drawing with at most one bend per edge of the graph in Figure 5(b). Additional inclusion relationships between 1-planar and straight-line RAC drawings have been studied for constrained drawings, and they are summarized in Section 8.\n\n$k$-planar graphs and $k$-quasi planar graphs. Note that the tight bounds on the edge density of 3-planar and 3-quasi planar graphs imply the existence of 3-quasi planar graphs that are not 3-planar. On the other hand, for $k \\ge 1$, every $k$-planar graph is clearly $(k + 2)$-quasi planar. These two observations have motivated recent work on the relationship between $k$-planar graphs and $k$-quasi planar graphs : For $k \\ge 3$, every $k$-planar graph is $(k+1)$-quasi planar. The proof of this result is based on a rerouting argument that starts from a $k$-planar drawing and resolves all possible bundles of $k+1$ pairwise crossing edges. The drawing produced by this technique is $(k+1)$-quasi planar, but it may not be $k$-planar anymore. This result has been later extended to the case $k=2$ . Thus, every $k$-planar graph is $(k+1)$-quasi planar, for $k \\ge 2$.\n\n$k$-planar graphs, fan-planar graphs, fan-crossing-free graphs. Since optimal fan-planar graphs have the same density as optimal 2-planar graphs (optimal $n$-vertex graphs have $5n-10$ edges for these two families), Binucci et al. study the relationship between fan-planar and $k$-planar graphs . They prove that these two families are incomparable. On the one hand, they show that for any $k \\ge 2$ there exists a fan-planar graph that is not $k$-planar; the proof uses a complete 3-partite graph $K_{1,3,h}$, where the index $h$ depends on $k$; this graph is clearly fan-planar but any of its drawings contains too many crossings to be $k$-planar. On the other hand, they exhibit 2-planar graphs that are not fan-planar. Recently, Brandenburg proved that there are graphs that are both fan-planar and fan-crossing-free but not 1-planar .\n\n$k$-gap planar graphs, $k$-planar graphs, $k$-quasi planar graphs. Bae et al. studied the relationship between $k$-gap planar graphs and both $k$-planar and $k$-quasi planar graphs . By using Hall's theorem, they prove that for every $k \\ge 1$ all $2k$-planar graphs are $k$-gap planar. On the other hand, for every fixed $k \\ge 1$, there exists a 1-gap planar graph that is not $k$-planar. Similarly, by using a counting argument on the number of crossings, they prove that all $k$-gap planar graphs are $2k+2$-quasi planar, while for all $k \\ge 1,$ they exhibit a quasi planar graph that is not $k$-gap planar.\n\nPlanarly connected graphs, 1-planar graphs, fan-planar graphs. Ackerman motivates the study of planarly connected graphs with the fact that both maximally dense 1-planar graphs and maximally dense fan-planar graphs are planarly connected . For 1-planar graphs, the high-level idea is to consider a drawing with the minimum number of crossings of a maximally dense graph; in such a drawing, for every pair of crossing edges there is a crossing-free edge that connects their endpoints, as otherwise one could contradict either the fact that the drawing is crossing minimal or the fact that the graph is maximally dense. The idea for fan-planar graphs is similar.\n\nMost of the intersection relationships between different families of beyond-planar graph families are summarized in Table 4. Each family in a row (respectively, column) is represented by a thin (respectively, thick) circle. Each cell reports the relationship between the corresponding row and column using the common Venn diagram notation. Note that some relationships immediately derive from the definitions. For example, a fan-planar drawing is always $k$-quasi planar (for each $k \\ge 3$) because three mutually crossing edges imply that two independent edges are crossed by a third one.\n\nTable 4. Relationships", null, "In each pairwise comparison, when applicable, we assume $k \\le h$. The table does not report inclusion relationships about restricted subfamilies, such as IC-planar graphs, NIC-planar graphs, and optimal straight-line RAC graphs. These relationships are discussed in the text.\n\nOpen problems. Table 4 shows that some pairs of beyond-planar graph families have a non-empty intersection that is not an inclusion. In these cases, it is interesting to characterize the graphs that belong to both families. For example, there are 1-planar graphs that are not straight-line RAC, there are straight-line RAC graphs that are not 1-planar, all optimal straight-line RAC graphs are 1-planar, and all IC-planar graphs are straight-line RAC [67, 133]. However, the following is still open.\n\nCharacterize the straight-line RAC graphs that are 1-planar.\n\nOther questions can be asked for those cells of Table 4 that show a proper inclusion between two families of beyond-planar graphs. For example, while every $k$-planar graph is $(k+1)$-quasi planar [14, 152] and every optimal 3-planar graph is also 3-quasi planar , the following questions on the combinatorial relationships between $k$-planarity and $h$-quasi planarity remain unanswered.\n\nIs a $k$-planar graph also $k$-quasi planar? For sufficiently large values of $k$, is every $k$-planar graph $f(k)$-quasi planar, for some function $f(k)=o(k)$?\n\nFinally, the relationships between some pairs of beyond-planar graph families have not yet been studied. For example:\n\nWhat is the relationship between $k$-gap planar and fan-planar graphs, for $k \\ge 1$?\n\n## 7 QUALITY METRICS\n\nWhile beyond-planar graphs focus on properties of edge crossings, other quality metrics have been extensively used in the literature to produce geometric representations of graphs that are clear and pleasing for the reader. For example, in a polyline drawing, the vertices of the graph are points and the edges are polylines whose complexity should be kept as low as possible to assist the reader. A common measure to capture the edge complexity of a polyline drawing is the number of bends per edge. Results concerning the edge complexity of beyond-planar graphs are presented in Section 7.1. Other drawing paradigms, like visibility representations and contact representations, convey edges as horizontal or vertical segments and shift the complexity to the vertices, which are drawn using more general shapes such as bars, rectangles, or polygons. We survey work related to the vertex complexity of beyond-planar graphs in Section 7.2. Finally, a common goal for polyline drawings is to use integer coordinates for vertices and bend points and to fit the drawing into a small bounding box, in order to display the drawing onto a screen with finite resolution. In particular, the area of a polyline drawing is the area occupied by its bounding box (i.e., by the minimum axis-aligned box containing the drawing). If the bounding box has side lengths $X-1$ and $Y-1$, we say that the drawing has area $X \\times Y$. Results concerning the area requirement of beyond-planar graphs in combination with edge/vertex complexity requirements are considered in Sections 7.1 $-$ 7.2. Section 7.3 discusses area-crossings tradeoffs for beyond-planar drawings of planar graphs.\n\n### 7.1 Edge Complexity\n\nStretchability. The famous Fáry's theorem states that every plane graph is stretchable; that is, it has an embedding-preserving straight-line drawing (this result was also independently proved by Wagner and Stein ). So far, little effort has been made to extend this result to beyond-planar graphs. In 1988, Thomassen initiated the study of the stretchability problem for 1-plane graphs . He proved that, differently from plane graphs, a 1-plane graph admits an embedding-preserving straight-line drawing if and only if it contains neither B-configurations nor W-configurations as subgraphs (see Figures 6(a)–6(b)). Later, Thomassen's characterization has been used to design a linear-time algorithm to test whether a 1-plane graph is stretchable . This algorithm is based on an efficient procedure that checks whether a 1-plane graph $G$ contains any B- or W-configuration, and, if not, it also returns a straight-line drawing of $G$. Figures 6(c)–6(d) show a stretchable 1-plane graph $G$ and a straight-line drawing of $G$, respectively. Recently, Hong and Nagamochi have studied the stretchability problem of 1-plane graphs in a more relaxed setting . They describe a linear-time algorithm that tests whether a 1-plane graph is stretchable assuming that the rotation system and the external face of the graph can change, while requiring the pairs of crossing edges to stay the same. Eades et al. study the stretchabilty problem for skewness-1 graphs . They characterize the maximal topological skewness-1 graphs that are stretchable and give a linear-time testing and drawing algorithm based on this characterization. Table 5 summarizes the aforementioned results.\n\nTable 5. Stretchability\nGraph Family Restrictions Complexity of testing Refs\n1-planar fixed embedding $O(n)$ \n1-planar fixed pairs of crossing edges $O(n)$ \nmaximal skewness-1 fixed embedding $O(n)$ \n\nPolyline drawings. As reported in the paragraph about stretchability, the class of 1-plane graphs admitting a straight-line drawing has been characterized , and there exist 1-plane graphs such that every embedding-preserving straight-line drawing requires an exponential area . Later, Alam et al. proved that every 3-connected 1-planar graph admits a 1-planar embedding that can be realized as a straight-line drawing except for at most one edge ; this drawing has quadratic area. On the other hand, it is obvious that every 1-plane graph admits a 1-bend drawing as it suffices to replace each crossing with a dummy vertex and compute a straight-line drawing of the resulting plane graph (possible overlaps of bend points can be removed by slight perturbations). Furthermore, it is not difficult to see that every 1-plane graph can be drawn with at most two bends per edge in $O(n^4)$ area (refer to Figure 7). Replace each crossing with a dummy face of degree four; augment the resulting plane graph to be 3-connected so that no edge is inserted in the dummy faces; compute a straight-line planar drawing such that all the faces are strictly convex and the area of the drawing is $O(n^2)\\times O(n^2)$ ; replace all dummy vertices with bend points, remove all dummy edges, and reinsert the crossings. Chaplick, Lipp, Wolff, and Zink recently proved that every $n$-vertex 1-plane graph actually admits an embedding-preserving 1-bend RAC drawing; if two bends per edge are allowed, a RAC drawing can be computed in $O(n^6)$ area . They also prove that every NIC-plane graph admits an embedding-preserving 1-bend RAC drawing in quadratic area. We remark that the existence of 1-bend 1-planar RAC drawings for every 1-plane graph was already known , but only assuming that the drawing algorithm can change the 1-planar embedding of the input graph. We also remark that for a simple family of 1-plane graphs, called kite-triangulations, an $\\Omega (n^3)$ area lower bound for embedding-preserving straight-line RAC drawings is established . A kite-triangulation is obtained by augmenting a plane triangulation with edges inside pairs of adjacent faces. Additional results concerned with other restricted classes of 1-planar graphs, such as the IC-planar graphs, are surveyed in Kobourov et al. . In recent papers, Argyriou et al. and Kindermann et al. study 1-planar graphs with small vertex degree. Argyriou et al. focus on orthogonal drawings (i.e., drawings where the edges are polylines with horizontal and vertical segments) and smooth orthogonal drawings (where circular arcs are also allowed) of 1-planar graphs. In particular, they show that all 1-planar graphs with vertex degree at most four have an orthogonal drawing with at most three bends per edge and a smooth orthogonal drawing with at most two bends per edge. Kindermann et al. prove that every 3-connected 1-planar graph with vertex degree at most three can be drawn with at most one bend per edge by using at most four different slopes for the edge segments and such that the minimum angle between two adjacent edges or between two crossing edges is at least $\\pi /4$. They also prove that every 1-planar graph with vertex degree at most three admits a 1-planar drawing with at most two bends per edge by using two orthogonal slopes (hence obtaining a RAC drawing).\n\nThe edge complexity of RAC drawings has also been studied regardless of the number of crossings per edge. Every $n$-vertex graph has a 3-bend RAC drawing in $O(n^4)$ area , while 1-bend and 2-bend RAC drawable graphs have at most $5.5n-11$ edges or $74.2n$ edges , respectively. If we allow up to four bends per edge, every graph can be drawn RAC in $O(n^3)$ area . For graphs with bounded vertex degree, we have the following results : $(i)$ every graph with vertex degree at most three has a 1-bend RAC drawing; $(ii)$ every graph with vertex degree at most six has a 2-bend RAC drawing. In both $(i)$ and $(ii)$, the drawing is computed in $O(n)$ time and has $O(n^2)$ area.\n\nConcerning ACE$\\alpha$ drawings with one or two bends per edge, it is proved that they have at most $27n$ edges and $385n$ edges, respectively [5, 6]. On the other hand, every graph admits an ACL$\\alpha$ drawing with one bend per edge in $O(n^2)$ area .\n\nTable 6 summarizes the main results concerning tradeoffs between the number of bends per edge and the area requirement of beyond-planar drawings.\n\nTable 6. Edge Complexity and Area Requirement Tradeoffs\nNumber of bends per edge\nGraph Family 0 1 2 3 4 Refs\nIC-planar $O(n^2)$ $O(n^2)$ $O(n^2)$ $O(n^2)$ $O(n^2)$ \nNIC-plane RAC ? $O(n^2)$ $O(n^2)$ $O(n^2)$ $O(n^2)$ \n1-planar $\\Omega (2^n)$ ? $O(n^4)$ $O(n^4)$ $O(n^4)$ , this paper\n1-plane RAC $\\Omega (2^n)$ ? $O(n^6)$ $O(n^6)$ $O(n^6)$ [81, 158]\nRAC $\\Omega (n^2)$ $O(n^2)$, if $\\Delta \\le 6$ $O(n^2)$, if $\\Delta \\le 3$ $O(n^4)$ $O(n^3)$ [18, 103, 118]\nACL$\\alpha$ ? $O(n^2)$ $O(n^2)$ $O(n^2)$ $O(n^2)$ \n\n$\\Delta$ denotes the maximum vertex degree. The question mark indicates that no area bound is known.\n\n### 7.2 Vertex Complexity\n\nA bar visibility representation of a graph $G$ maps each vertex of $G$ to a distinct horizontal segment, called bar, and each edge of $G$ to a vertical unobstructed segment, called visibility, between the two bars corresponding to its end-vertices. This kind of representation is intrinsically planar, and, on the other hand, every planar graph admits this representation3 (see, e.g., [215, 227]). In order to realize nonplanar graphs, other visibility models have been proposed, such as bar $k$-visibility representations and rectangle visibility representations.\n\nIn a bar $k$-visibility representation, each visibility intersects at most $k$ bars . In particular, it is proved that every 1-planar graph has a bar 1-visibility representation [63, 139]. A rectangle visibility representation maps each vertex to an axis-aligned rectangle and each edge to a horizontal or vertical visibility between the two corresponding rectangles [208, 228]. In general, there are 1-plane graphs, and even IC-plane graphs, that do not admit such a representation. Biedl et al. describe a linear-time algorithm to test whether a 1-plane graph admits an embedding-preserving rectangle visibility representation and to compute one if it exists . The algorithm is based on a characterization that extends the set of obstructions used by Thomassen to characterize the stretchable 1-plane graphs . The rectangle visibility model has been recently generalized to extend the class of representable graphs; namely, in a generalized model, called ortho-polygon visibility representations, the vertices are drawn as general orthogonal polygons . Di Giacomo et al. prove that every 1-plane graph admits an embedding-preserving ortho-polygon visibility representation . If the input graph is 3-connected, one can construct a representation such that each vertex has at most 12 reflex corners, while 2 reflex corners may be needed. If the graph is only 2-connected, it may require at least one vertex with a linear number of reflex corners. The upper bound for 3-connected 1-plane graphs exploits results on edge partitions (see also [43, 100, 179]). The gap for 3-connected 1-plane graphs has been recently reduced, proving that at most 5 reflex corners per vertex always suffice, while 4 reflex corners are needed in some cases .\n\nA further visibility model, called L-visibility representations, maps every vertex to a horizontal and a vertical segment sharing an endpoint (i.e., by an L-shape in the set", null, "), and each edge to a horizontal or vertical visibility segment joining the two L-shapes corresponding to its two end-vertices. It is proved that every IC-planar graph admits an L-visibility representation .\n\nVisibility representations of 1-planar graphs have been studied also in 3D. A $z$-parallel visibility representation maps the vertices of a graph to isothetic disjoint rectangles parallel to the $xy$-plane and the edges to visibilities parallel to the $z$-axis. Every 1-plane graph has a $z$-parallel visibility representation . The computed drawing is such that there is a plane orthogonal to the rectangles of the representation, and the intersection of this plane with the representation defines a bar 1-visibility representation of the graph.\n\nFinally, Alam et al. studied contact representations in which vertices are represented by axis-aligned polyhedra in 3D and edges are realized by non-zero area common boundaries between corresponding polyhedra . They prove that every optimal 1-plane graph can be realized as a contact representation where vertices are axis-aligned boxes if it contains no separating 4-cycles or as a contact representation where vertices are L-shaped polyhedra otherwise. Table 7 summarizes the main results on visibility and contact representations of 1-planar graphs.\n\nTable 7. Vertex Complexity\nVisibility Contact\nGraph Family RVR B1VR OPVR LVR ZPR BCR LCR Refs\nIC-planar", null, "", null, "", null, "", null, "", null, "? ? \n\n\n \noptimal 1-planar with no separating 4-cycles $\\times$", null, "", null, "", null, "", null, "", null, "", null, "optimal 1-planar $\\times$", null, "", null, "", null, "", null, "?", null, "1-planar $\\times$", null, "", null, "$\\times$", null, "? ?\n\n$\\times$ indicates that there are instances of the graph family not admitting that visibility or contact representation.", null, "means that all instances in the graph family can be drawn with that visibility or contact representation. The question mark means that it not known whether the answer is $\\times$ or", null, ". LEGEND: RVR = rectangle visibility representation, B1VR = bar 1-visibility representation, $k$OPVR = ortho-polygon visibility representation, LVR = L-visibility representation, ZPR = $z$-parallel visibility representation, BCR = box contact representation, LCR = L-shaped contact representation.\n\n### 7.3 Area-Crossing Tradeoffs for Planar Graphs\n\nThe area requirement of planar straight-line drawings is a widely investigated problem. Motivated by VLSI applications, seminal papers studied how to compute area-efficient graph layouts [178, 219]. Different popular results establish that an $n$-vertex planar graph can always be drawn with straight-line edges in $O(n^2)$ area [88, 207]. This bound is worst-case optimal for the family of planar graphs as there are infinitely many planar graphs whose planar drawings require quadratic area . Several attempts have been made to prove the existence of straight-line planar drawings with $o(n^2)$ area for specific subfamilies of planar graphs, such as trees, outerplanar graphs, and series-parallel graphs (see, e.g., for references on these results).\n\nA natural question that arises from the aforementioned results is whether allowing some edge crossings may help to reduce the area of a drawing of a planar graph. In other words, what is the area requirement of beyond-planar drawings of planar graphs? Wood shows that, for any fixed positive integer $k \\gt 0$, all $k$-colorable graphs have a straight-line drawing in linear area; this implies that planar graphs always admit $O(n)$ area straight-line drawings with crossing edges . However, the technique by Wood can give rise to drawings where some edges contain a linear number of crossings and the angles at which two edges cross can be arbitrarily small.\n\nThe use of edge crossings that form large angles is studied in different papers. Straight-line RAC drawings of planar graphs may require $\\Omega (n^2)$ area , thus right angle crossing drawings do not help to reduce the area requirement bound for planar graphs in the general case. On the positive side, for infinitely many values of $n$, there exists an $n$-vertex planar graph whose requirement is $\\Theta (n^2)$ for straight-line planar drawings and $\\Theta (n)$ for straight-line RAC drawings . Analogous results hold for other quality metrics such as uniform edge length and angular resolution . In addition, every planar graph admits an ACL$\\alpha$ drawing with two bends per edge in $O(n^\\frac{5}{3})$ area , and every planar graph with vertex-degree at most $\\Delta$ admits a 4-bend RAC drawing in $O(n \\sqrt {\\Delta n})$ area . Note that, if $\\Delta$ is a sublinear function of $n$, these RAC drawings have subquadratic area.\n\nCompact nonplanar drawings of planar graphs with a constant or sublinear number of crossings per edge are also studied . Every $n$-vertex outerplanar graph admits a straight-line drawing with $O(\\frac{n}{\\log n})$ crossings per edge in $O(n \\log n)$ area. Also, for any given $\\epsilon \\gt 0$, every $n$-vertex outerplanar graph admits a straight-line drawing with $O(n^{1-\\epsilon })$ crossings per edge in $O(n^{1+\\epsilon })$ area, which gives a clear tradeoff scheme between area requirement and number of crossings per edge. These results are based on a linear-time drawing algorithm, which can also be applied to other subfamilies of planar graphs that admit a “level” drawing with specific properties, such as flat series-parallel graphs with bounded degree (see ). Nonetheless, if we insist on having a constant number of crossings per edge, planar and nonplanar drawings of planar graphs have in general the same area requirement. This is true even for series-parallel graphs .\n\nOn the positive side, every $n$-vertex planar graph has a straight-line $o(n)$-quasi planar drawing in $o(n^2)$ area. More precisely, by combining drawing techniques in Di Giacomo et al. with results on the track number of planar graphs , one can prove that every $n$-vertex planar graph admits either a straight-line $O(\\log n)$-quasi planar drawing in $O(n \\log ^{3} n)$ area or a straight-line $O(\\log ^{2} n)$-quasi planar drawing in $O(n \\log n)$ area. Also, every partial 2-tree admits a linear-area straight-line drawing with thickness at most 10 and hence a linear-area 11-quasi planar straight-line drawing . Outerplanar graphs and flat series-parallel graphs with bounded vertex degree (which are partial 2-trees) have quasi planar and 5-quasi planar drawings in linear area, respectively . Table 8 summarizes the main results presented in this subsection.\n\nTable 8. Area-Crossings Tradeoffs for Planar Graphs", null, "straight-l. $k$-planar straight-l. $k$-quasi planar $k$-bend RAC $k$-bend ACL$\\alpha$\nArea $k$ Area $k$ Area $k$ Area $k$\nouterplanar $O(n^{1+\\epsilon })$ $O(n^{1-\\epsilon })$ $O(n)$ 3 ? ?\n$O(n \\log n)$ $O(\\frac{n}{\\log n})$\nflat series-parallel bounded degree $O(n^{1+\\epsilon })$ $O(n^{1-\\epsilon })$ $O(n)$ 5 ? ?\n$O(n \\log n)$ $O(\\frac{n}{\\log n})$\npartial 2-tree $\\Omega (n2^{\\sqrt {\\log n}})$ $O(1)$ $O(n)$ 11 ? ?\nplanar $\\Omega (n^2)$ $O(1)$ $O(n \\log ^3 n)$ $O(\\log n)$ $\\Omega (n^2)$ 0 $O(n^\\frac{5}{3})$ 2\n$O(n)$ $O(n)$ $O(n \\log n)$ $O(\\log ^2 n)$ $O(n \\sqrt {\\Delta n})$ 4\n\n$\\Delta$ denotes the maximum vertex degree. The question mark indicates that bounds better than those for planar drawings are not known. For reasons of space, the table does not report the corresponding references.\n\nOpen problems. The stretchability problem for beyond-planar graphs is a fertile and essentially unexplored research subject. As a matter of fact, for any type of forbidden crossing configuration, the corresponding stretchability question gives rise to an open problem. For example:\n\nCharacterize the stretchable $k$-quasi planar topological graphs (even when $k=3$).\n\nAnother research stream is about embedding-preserving drawings with good crossing angle resolution. This question is interesting even for structurally simple topological graphs. For example:\n\nDoes every topological graph of maximum vertex degree three admit an embedding-preserving straight-line RAC drawing?\n\nFinally, we recall that a characterization of the almost-plane graphs that are stretchable is known only when the number of edges is $3n-5$ (a characterization for nonmaximal is known on the sphere but not in the plane) . Hence, a natural question is the following:\n\nCharacterize the topological skewness-$k$ graphs that are stretchable. This question is interesting also when $k=1$ and the number of edges is smaller than $3n-5$.\n\nSeveral open problems can also be deduced by looking at the “?” in Tables 6, 7, and 8. For example:\n\nIt is known that every 1-planar graph has a RAC drawing with at most one bend per edge . However, it is not known whether such a drawing can be computed in polynomial area.\n\nRegarding this problem, it has been recently shown that NIC-plane graphs admit an embedding-preserving 1-bend RAC drawing in quadratic area .\n\nDo 1-planar graphs admit a box contact representation? The question is interesting even for subfamilies of 1-planar graphs, such as IC-planar graphs.\n\nDo planar graphs admit a $k$-bend RAC drawing in subquadratic area with $k \\ge 4$?\n\nWhen $k=4$, the answer is affirmative if the maximum vertex-degree is sublinear .\n\n## 8 CONSTRAINTS\n\nSeveral papers concentrate on beyond-planar drawings where the vertices and/or the edges have additional geometric constraints. We discuss the main scenarios studied in the literature.\n\n### 8.1 Vertices on Lines, Circles, and External Boundary\n\nThe study of graph layouts in which vertices are placed on a given set of horizontal lines, often called layers, or on a set of concentric circles, has a well-established tradition in graph drawing [95, 102].\n\n2-layer drawings. In the beyond planarity context, straight-line 2-layer drawings have been investigated for RAC, 1-planar, and fan-planar graphs in terms of both edge density and recognition. In a 2-layer drawing, the vertices are distributed along two horizontal layers, and vertices of the same layer cannot be adjacent. Thus, the graph is necessarily bipartite. The study of 2-layer drawings has two main motivations: $(i)$ They are a natural way to visually convey bipartite graphs; $(ii)$ algorithms that compute 2-layer drawings are building blocks of the popular Sugiyama's framework used to draw graphs on multiple horizontal layers.\n\nCharacterizations for those graphs that admit either 2-layer RAC drawings , or 2-layer 1-planar drawings , or 2-layer fan-planar drawings are known; all of them can be regarded as generalizations of caterpillars, the class of 2-layer planar drawable graphs . The upper bounds on the edge density of these graph families are reported in Table 9. In particular, the optimal 2-layer 1-planar graphs coincide with the optimal 2-layer RAC graphs and have $1.5n - 2$ edges, where $n$ is the number of vertices of the graph. They are called ladders and consist of two paths of the same length $\\langle u_1,u_2, \\dots , u_\\frac{n}{2} \\rangle$ and $\\langle v_1,v_2, \\dots , v_\\frac{n}{2} \\rangle$, plus the edges $(u_i, v_i)$, $(i = 1, 2, \\dots , \\frac{n}{2})$; see Figures 8(a) and 8(b). A maximal 2-layer fan-planar graph is called a snake and is obtained from an outerplane ladder by adding, inside each internal face, an arbitrary number (possibly none) of paths of length two joining a pair of non-adjacent vertices of the face. Intuitively, a snake is a bipartite planar graph composed of a chain of complete bipartite graphs $K_{2,h}$; see Figure 8(c). The family of optimal 2-layer fan-planar graphs on $n$ vertices includes $K_{2, n-2}$ .\n\nTable 9. Density for Constrained Families\nGraph Family Max. Num. Edges Tight Refs\n2-layer 1-planar $1.5n-2$", null, "2-layer RAC $1.5n-2$", null, "2-layer fan-planar $2n-4$", null, "outer 1-planar $2.5n-4$", null, "[29, 115]\nouter fan-planar $3n-5$", null, "convex geometric $k$-quasi planar $2(k-1)n-\\binom{2k-1}{2}$", null, "[78, 124, 186]\nconvex geometric fan-crossing-free $\\lfloor 5n/2 - 4 \\rfloor$", null, "Table 10. Recognition for Constrained Families\nGraph Family Complexity Refs\n2-layer RAC $O(n)$ \n2-connected 2-layer fan-planar $O(n)$ \nouter 1-planar $O(n)$ [29, 155]\nfull outer 2-planar $O(n)$ \nmaximal outer fan-planar $O(n)$ \ncircular RAC $O(n)$ \nupward RAC NP-hard \nsimultaneous RAC NP-hard \n$k$-SEFE NP-complete \nstraight-line point-set RAC NP-hard \n\nFrom the algorithmic point of view, there exist linear-time testing and drawing algorithms for 2-layer RAC graphs and for 2-connected 2-layer fan-planar graphs , while the complexity of recognizing 2-layer fan-planar graphs when the graph is only 1-connected is still open. For 2-layer RAC graphs, it is also possible to compute, in $O(n^2\\log n)$ time, a 2-layer RAC drawing with minimum number of crossings . We remark that the crossing minimization problem is a well-known NP-complete problem in the general case, which remains hard even for 2-layer drawings without the restriction that crossing edges are orthogonal .\n\n$k$-page drawings. Let $\\ell$ be a line, called a spine. In a $k$-page drawing of a graph, each vertex is mapped to a distinct point of $\\ell$ and each edge is drawn as a semicircle in one of $k$ distinct half-planes incident to $\\ell$. Each of these half-planes is called a page. We recall that $k$-page drawings are among the oldest and more common graph drawing conventions, and they can be found with different names in the literature. For example, they are sometimes called linear layouts (see, e.g., ); also, a 2-page drawing is sometimes called a permutation representation (see, e.g., ) or linear drawing (see, e.g., ), while a 1-page drawing is also called an arc diagram (see, e.g., ). If no two disjoint edges cross, a $k$-page drawing is a $k$-page stack layout. The minimum value of $k,$ such that a graph $G$ has a $k$-page stack layout, is the stack number (or page number) of $G$. If no two disjoint edges nest, a $k$-page drawing is a $k$-page queue layout. The minimum value of $k,$ such that a graph $G$ has a $k$-page queue layout, is the queue number of $G$. Stack and queue layouts have been extensively studied for planar and regular graphs (see, e.g., the survey of Díaz, Petit, and Serna ).\n\nIn the context of beyond planarity, $k$-page drawings have been studied within two different research directions. On the one hand, the problem of computing $k$-page drawings with a limited number of crossings per edge has been studied; on the other hand, the problem of determining the stack and queue number of beyond-planar graphs has been studied. Concerning the first direction, Binucci et al. describe linear-time algorithms to compute 2-page drawings of planar 3-trees with at most $2\\Delta$ crossings per edge, where $\\Delta$ is the maximum vertex-degree of the graph and 1-page drawings of partial 2-trees with at most $\\Delta ^2$ crossings per edge . In both cases, the authors show that the number of crossings per edge cannot be bounded by a constant. As for the second direction, Dujmovic and Frati proved a $O(\\log n)$ upper bound on the stack number of $k$-planar graphs , while for 1-planar graphs $O(1)$ upper bounds are known [11, 41]. Furthermore, if the queue number of planar graphs is bounded (which is still unknown with the best upper bound being $O(\\log {n})$ ), then the same is true for $k$-planar graphs .\n\nOuter and convex drawings. Beyond-planar graphs that admit a layout in which all vertices lie on a circle enclosing the whole drawing or, more generally, on the external boundary of the drawing, can be regarded as generalizations of the outerplanar graphs. For any integer $k \\gt 0$, a $k$-planar drawing with this property is an outer $k$-planar graph. Outer 1-planar graphs have at most $2.5n-4$ edges [29, 115], which is a tight bound. They can be recognized in linear time [29, 155] and drawn in linear time, both as straight-line drawings with all vertices on the external face in $O(n^2)$ area and as visibility representations in $O(n \\log n)$ area . Hong and Nagamochi extended the recognition problem to full outer 2-planar graphs (i.e., outer 2-planar graphs with no crossing on the external boundary ). They prove that also this class of graphs can be recognized in linear time.\n\nDensity and recognition results are described also for outer fan-planar graphs. They have at most $3n-5$ edges (see Table 9), but, in contrast to outer 1-planar graphs, the complexity of the recognition problem is not known in general, while it is linear-time solvable for maximal outer fan-planar graphs . Table 9 reports other tight bounds for geometric beyond-planar graphs with all vertices in convex position, such as $k$-quasi planar graphs and fan-crossing-free graphs. Bounds for more specific families are listed in the geometric graph theory chapter of the Handbook of Discrete and Computational Geometry . Note that the edge crossings in a geometric graph with vertices in convex position are the same as those of a 1-page drawing in which the order of the vertices along the spine equals the circular order of the vertices on the convex polygon.\n\nSeveral interesting properties of convex geometric $k$-planar and $k$-quasi planar graphs have been recently investigated . Recall that a graph is $d$-degenerate if every subgraph has a vertex of degree at most $d$. It is shown that convex geometric $k$-planar graphs are $(\\lfloor \\sqrt {4k+1} \\rfloor +1)$-degenerate and consequently $(\\lfloor \\sqrt {4k+1} \\rfloor + 2)$-colorable. Furthermore, they have a balanced separator of size at most $2k+3$, which allows the design of a quasi-polynomial time recognition algorithm for this class of graphs (i.e., the recognition problem is not NP-hard unless the Exponential Time Hypothesis [ETH] fails). Convex geometric $k$-planar graphs in which all the vertices form a simple cycle can be recognized in linear time because they can be expressed in extended monadic second-order logic and have bounded treewidth. In Chaplick et al. , it is also proved that planar graphs and outer quasi planar graphs are incomparable classes under containment, while in Geneson et al. it shown that semi-bar $k$-visibility graphs are outer $(k+2)$-quasi planar.\n\nA characterization of the class of graphs that admit a RAC drawing in which all vertices lie on a circle, along with a linear-time testing and layout algorithm, is given in Dehkordi et al. . Outer 1-planar graphs have been studied in combination with other types of constraints or beyond-planar graphs. Dehkordi and Eades show that every outer 1-planar graph admits an outer RAC drawing . Di Giacomo et al. study how to draw outer 1-planar graphs with a small number of edge slopes . They prove that every outer 1-planar graph admits an outer 1-planar straight-line drawing that uses $O(\\Delta)$ different slopes, where $\\Delta$ is the maximum vertex-degree of the graphs.\n\n### 8.2 Upward RAC Drawings, Simultaneous RAC and Point-Set RAC Embedding\n\nRAC drawings have been studied in combination with other popular graph drawing conventions that impose additional geometric constraints on the layout, namely the upward drawing convention for directed graphs (digraphs for short), the simultaneous embedding convention for multiple graphs with the same vertex set, and the point-set embedding convention. We discuss them here.\n\nUpward RAC drawings. In an upward drawing of a digraph, each edge is a curve monotonically increasing in the vertical direction (see, e.g., ). Upward drawings are used to visually convey the structure of acyclic digraphs and have also been extended to visualize other kinds of networks [50, 55]. Testing whether a given planar digraph admits an upward planar (i.e., crossing-free) drawing is one of the most studied problems in this context; refer to Didimo for a survey. Although polynomial-time testing algorithms are known for specific subfamilies of planar digraphs or when the digraph has a fixed embedding, the testing problem is NP-complete in general . Also, although a planar digraph has an upward planar drawing if and only if it has a straight-line upward planar drawing , there are digraphs for which every straight-line upward planar drawing requires exponential area [96, 97]; quadratic area is always achievable if we allow edge bends. Similar questions have been investigated for RAC drawings. Angelini et al. proved that deciding whether a planar acyclic digraph admits a straight-line upward RAC drawing is NP-hard and that the area required by an $n$-vertex straight-line upward RAC drawing may be exponential in $n$ .\n\nSimultaneous RAC Embedding (SRE). Given two planar graphs $G_1=(V,E_1)$ and $G_2=(V,E_2)$ with the same vertex set, a simultaneous embedding (SE) of $G_1$ and $G_2$ is a pair of planar drawings, $\\Gamma _1$ of $G_1$ and $\\Gamma _2$ of $G_2$, such that each vertex $v \\in V$ has the same position in $\\Gamma _1$ and $\\Gamma _2$ (the edges of $\\Gamma _1$ can cross those of $\\Gamma _2$). The SE problem was introduced a decade ago, motivated by several practical scenarios , and since then it has been widely studied (see, e.g., for a survey). A Simultaneous RAC Embedding (SRE) of two graphs $G_1$ and $G_2$ is an SE with the additional property that the union of the two planar drawings $\\Gamma _1$ and $\\Gamma _2$ is RAC. The SRE problem was originally introduced for straight-line drawings , showing that only restricted pairs of planar subgraphs admit such an embedding; for example, a wheel and a cycle might not admit a straight-line SRE. In general, deciding whether a pair of graphs admits an SRE with straight-line edges is NP-hard , while every pair of planar graphs has an SRE with at most six bends per edge . The number of bends per edge can be further reduced for some combinations of specific subfamilies of planar graphs. Moreover, an SRE with at most two bends exists for those pairs of graphs admitting a special type of visibility representation in which vertices are axis-aligned L-shapes and edges are vertical or horizontal lines of sight connecting pairs of such shapes . Another variant of SE in graph drawing beyond planarity considers a simultaneous embedding of two graphs $G_1$ and $G_2$ in which each of the two drawings $\\Gamma _1$ and $\\Gamma _2$ is allowed to have some desired type of crossings. This setting generalizes the classical SE instead of restricting it. For example, Di Giacomo et al. study Simultaneous Geometric Quasi Planar Embedding (SGQPE), where each $\\Gamma _i$ is a straight-line quasi planar drawing . They show, for instance, that a tree and a path always admit an SGQPE, in contrast to the negative result in the simultaneous geometric planar embedding setting .\n\nPoint-Set RAC Embeddings (PSRE). Given a set $S$ of points in the plane and a graph $G$ with $n=|S|$ vertices, a Point-Set Embedding (PSE) $\\Gamma$ of $G$ on $S$ is a drawing of $G$ such that the vertices are placed to the points of $S$ through a one-to-one mapping, which can be given as part of the input (see, e.g., ) or not (see, e.g., ). Fink et al. studied Point-Set RAC Embeddings (PSREs) of graphs; that is, PSEs where the points of $S$ belong to an $n \\times n$ grid and no two points are horizontally or vertically aligned . They concentrate on computing PSREs with few bends per edge, where bends must occupy grid points as for the vertices. Different algorithmic results and bounds on the number of bends per edge are given, also under the assumption that the edge segments are restricted to be on grid lines. Among their results, they prove that every graph with $n$ vertices and $m$ edges admits a PSRE on any $n \\times n$ grid point set with at most three bends per edge in $O((n+m)^2)$ area, which also implies that every planar graph admits a RAC drawing with at most three bends per edge in quadratic area, thus improving a previous result in terms of bends per edge (see Section 7.3). On the negative side, testing whether a graph admits a PSRE without bends is NP-hard .\n\nOpen problems. Although the study of planar book embeddings has a long tradition in graph drawing (see, e.g., [49, 190, 230]), the study of its beyond-planar counterpart is much more recent. The question can be studied by either considering $k$-page drawings with forbidden crossing configurations (see, e.g., ) or by focusing on $k$-page book embeddings of beyond-planar graphs (see, e.g., [11, 41, 127]). For example, we mention the following two problems.\n\nIs there a function $f(\\cdot)$ such that every planar graph of vertex-degree at most $\\Delta$ admits a 2-page drawing that is $f(\\Delta)$-planar?\n\nEstablish tight bounds on the book thickness of $k$-planar graphs (even when $k=1$).\n\nNot all planar acyclic digraphs have a straight-line upward RAC drawing , thus we ask:\n\nDoes every acyclic planar digraph admit a 1-bend upward RAC drawing?\n\nEvery pair of planar graphs sharing their vertex set admits an SRE with at most six bends per edge . On the other hand, if no restriction on the edge crossings is given, such a pair admits an SE with at most two bends per edge .\n\nDoes every pair of planar graphs that share their vertex set admit a simultaneous RAC embedding with less than six bends per edge?\n\nA set $S$ of points in the plane is universal for a family of graphs if every element of the family admits a PSE on $S$ with straight-line edges. A well-known result is that a universal point set of size $n$ that supports straight-line crossing-free drawings of all planar graphs with $n$ vertices does not exist . One may wonder whether a beyond-planar version of this problem is more likely to have a positive answer. In particular, we ask the following.\n\nIs there a set $S$ of $n$ points in the plane such that every planar graph with $n$ vertices admits an $f(n)$-planar point-set embedding on $S$ with straight-line edges such that $f(n) \\in o(n)$?\n\n## 9 IMPLEMENTATIONS AND EXPERIMENTS\n\nThe applied research in graph drawing beyond planarity has been mainly pursued along two directions: $(i)$ Cognitive experiments aimed to better understand the impact of some types of forbidden configurations on the capability of a user to execute visualization-based analysis tasks. $(ii)$ The implementation and experimentation of algorithms to compute beyond-planar graph drawings and the development of visualization systems for real-world application domains. We briefly discuss the contributions in these two directions.\n\nCognitive studies. Early cognitive experiments on beyond-planar graphs estimate the effects of crossing angles on human eye movements and performance through the use of eye-tracking systems [160, 161]. The results show that sharp angles may trigger extra eye movements, causing delays for path search tasks, whereas crossings have usually little impact on node locating tasks. Subsequent experiments confirm the importance of large angle crossings to execute visualization-based analysis tasks [162, 164], thus motivating and stimulating the rich literature about RAC graphs and related graph families . Further studies give some evidence that relevant improvements on graph layout readability derive from finding a good tradeoff between different quality metrics, such as crossing and vertex angles, rather than optimizing only one of them .\n\nRecall that PEDs aim to reduce edge crossings and visual clutter; the central part of each edge is erased and the length of the two remaining segments are computed to preserve useful geometric information . An $\\alpha$-SHPED of a straight-line drawing is immediately defined for a fixed value of $\\alpha$ . In particular, some edge crossings may not be avoidable, although the amount of ink removed from the original drawing might be large (e.g., $50\\%$ when $\\alpha =\\frac{1}{4}$). On the other hand, it is possible to maximize the ink and remove edge crossings by using different values of $\\alpha$ for different edges. Binucci et al. present a user study in which PEDs obtained via heuristics are compared with the standard model $\\frac{1}{4}$-SHPED . The results suggest that the benefit of homogeneity overcomes, in terms of readability, the benefit of fewer crossings and more ink.\n\nOther experiments on beyond-planar graphs focus on a so-called edge stratification approach to analyze complex visualizations of graphs . The approach is based on partitioning the edge set of a drawing into a minimal number of layers, such that the edges in each layer define a drawing with some desired properties related to crossings. For example, one can require that the drawing of each layer is planar, or that its crossing angles are larger than a given constant $\\alpha$, or that it is $k$-planar for some fixed $k$. The experiments show that the stratification approach is mainly useful for local tasks such as counting the vertex degrees, while it is less effective for more global tasks such as finding shortest paths between pairs of vertices. The edge stratification algorithms proposed by Di Giacomo et al. are sometimes computationally expensive, especially when the drawing on each layer is required to be $k$-planar for relatively large values of $k$; most of these algorithms can be successfully applied to drawings with a few hundred vertices and edges .\n\nAlgorithms and systems. Several papers describe heuristics that attempt to optimize some desired properties for edge crossings, often in combination with classical graph drawing conventions. For example, to improve the readability of circular layouts, the MAXCIR post-processing algorithm aims to increase crossing angles by using Quadratic Programming; it is fast in practice and yields better results compared to a traditional equal-spacing algorithm. BIGANGLE is a force-directed algorithm that computes drawings of graphs with multiple quality metrics being improved at the same time; these metrics include crossing and vertex angle resolution. Argyriou et al. propose another force-directed algorithm that computes drawings with high crossing and vertex angle resolution. In a recent paper, Bekos et al. present a new algorithm to improve the crossing resolution of a drawing and discuss experiments in which this algorithms outperforms the algorithms in other works [26, 163]. A further contribution in the same direction as BIGANGLE is the topology-driven force-directed framework . It allows the design of graph drawing algorithms that find good tradeoffs between different metrics, such as number of crossings, crossing angle resolution, geodesic edge tendency, and number of edge bends. This approach combines force-directed techniques with the popular topology-shape-metrics approach, originally proposed to compute bend-minimum orthogonal drawings . Examples of drawings computed by topology-driven force-directed algorithms are shown in Figures 9(a)–9(b). As a follow-up of this work, the same authors implement an algorithm that first computes a straight-line drawing with a force-directed technique and then runs a post-processing procedure to improve the crossing angle resolution by representing edges as smoothed curves (see Figure 9(c)); each curve is monotone in the direction of the straight-line segment connecting its end-points. The algorithm is applied to the simultaneous embedding of suitably defined networks in a system for conceptual Web-site traffic analysis . More recently, Demel et al. presented a heuristic to increase the smallest angle between two crossing edges in a given straight-line drawing, together with a speed-up technique to compute the pair of crossing edges with the smallest crossing angle in a straight-line drawing .\n\nTo compute layered drawings of graphs with large-angle crossings (see Section 8), Di Giacomo et al. describe building block heuristics for extracting a maximum 2-layer RAC subgraph of a given bipartite graph . They study both the setting in which no fixed ordering for the vertices of each partition set is given and the setting in which one of the two sets has a fixed linear ordering.\n\nWe finally mention the implementation of an algorithm that computes Ortho-Polygon Visibility Representations (OPVRs) of (nonplanar) graphs . As explained in Section 7.2, in an OPVR of a graph $G$, each vertex is represented as an orthogonal polygon and each edge is either a horizontal or a vertical segment. The vertex complexity of an OPVR is the maximum number of reflex angles inside a polygon representing a vertex. Assuming that the graph $G$ comes with a fixed embedding, the algorithm in Di Giacomo et al. tests in $O(n^2)$ time whether an OPVR of $G$ exists and, if so, it computes in $O(n^\\frac{5}{2}\\log ^\\frac{3}{2}n)$ time an embedding-preserving OPVR of $G$ with minimum vertex complexity. This algorithm has been experimented on a large set of 1-plane graphs, which always admit an OPVR. The results show that, in practice, the computed OPVRs have usually a high percentage (up to $90\\%$ in some cases) of vertices drawn with no reflex corners (i.e., as rectangles).\n\nOpen problems.\n\nAs already mentioned, Mutzel observed that the readability of a 2-layer drawing may not just depend on the number but also on the type of edge crossings . We propose to study the following problem, for which some efforts have been limited so far to RAC drawings .\n\nDesign heuristics that reduce the number of forbidden crossing configurations in 2-layered drawings of bipartite graphs and perform a user study to compare the obtained drawings with those computed by means of 2-layer crossing minimization heuristics.\n\nIt would be interesting to analyze the practical impact that the various forbidden edge crossing configurations have in real-world drawings. For example, one could consider a large benchmark of real-world graphs with various sizes, draw these graphs by using different algorithms (e.g., force-directed algorithms), and then analyze the frequency of occurrence of each forbidden edge crossing configuration in these drawings. We summarize this problem as follows.\n\nAnalyze the impact of various forbidden edge crossing configurations in a large set of real-world drawings.\n\nFinally, a very promising research direction is to perform new experiments and develop new theories on graph drawing beyond planarity based on the following user-centered approach: (i) Develop theories on how people read drawings based on HCI experiments; (ii) define optimization criteria on the edge crossings and their configurations; (iii) design combinatorial models and efficient algorithms; and (iv) experimentally verify the efficiency and effectiveness of the algorithms with new experiments that may lead to refining the model and/or the optimization goals.\n\n## 10 CONCLUSION\n\nThe treatment of edge crossings in graph visualization embraces other topics that have not been discussed in the previous sections. Although these topics are not properly recognized as part of the literature on graph drawing beyond planarity, recent works have established connections with beyond-planar graph families, which highlight new and interesting research directions.\n\nCrossing minimization. Computing drawings of graphs with a minimum number of crossings, as well as determining the crossing number of a graph, are problems with a long tradition (see, e.g., [75, 206] for surveys on the subject). It is well known that the crossing minimization problem is NP-complete, and it remains hard also in very restricted scenarios (for instance, when the graph is bipartite and we look for a straight-line 2-layer drawing of it ). However, for 2-layer RAC drawable graphs, the problem can be solved efficiently , which suggests that the crossing minimization problem might be polynomial-time solvable also with respect to other forbidden crossing configurations.\n\nRelaxed clustered planarity. In a clustered graph $G$, vertices are grouped into (hierarchical) clusters. In a drawing of $G,$ each cluster should be represented as a closed region containing all (and only) its vertices and all (and only) its subclusters. Also, an edge should not traverse a cluster if both its end vertices are outside it. Many papers study how to efficiently test whether a clustered planar graph admits a crossing-free drawing that meets the aforementioned properties [85, 170], and practical drawing heuristics have also been conceived, even for nonplanar large graphs [60, 94, 122, 123]. In the context of graph drawing beyond planarity, relaxed models of clustered planarity may disallow only some types of crossings. See Angelini et al. for initial steps in this direction.\n\nHybrid visualizations. The use of matrix-based representations combined with the classical node-link representation has been proposed to diminish the negative effect of visual clutter in large and locally dense networks [37, 151]. More recently, the planarity testing and embedding problem of this type of hybrid visualizations have been further formalized [87, 111], and new families of beyond-planar graphs related to hybrid visualizations have been defined [21, 112].\n\nEdge bundling. The idea of grouping edges to get planar drawings of nonplanar graphs was originally proposed under the name of confluent drawings [114, 137]. Later, practical edge-bundling techniques were used to cope with the visual clutter problem in node-link diagrams of large and dense graphs [153, 154, 232]. The idea is that “similar” edges are deformed and grouped into bundles, thus providing a more abstract and uncluttered view of the original drawing at the expense of possible ambiguities in terms of connections between vertices. 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A unified approach to visibility representations of planar graphs. Discr. Comput. Geom. 1, 1 (1986), 321–341.\n• Carsten Thomassen. 1988. Rectilinear drawings of graphs. J. Graph Theory 12, 3 (1988), 335–341.\n• C. D. Toth, J. O'Rourke, and J. E. Goodman. 2017. Handbook of Discrete and Computional Geometry, Third Edition. CRC Press.\n• Klaus Truemper. 1992. Matroid Decomposition. Academic Press.\n• Leslie G. Valiant. 1981. Universality considerations in VLSI circuits. IEEE Trans. Computers 30, 2 (1981), 135–140.\n• Pavel Valtr. 1997. Graph drawings with no $k$ pairwise crossing edges. In Proceedings of the GD 1997 (LNCS), Vol. 1353. Springer, 205–218. https://link.springer.com/chapter/10.1007%2F3-540-63938-1_63.\n• Pavel Valtr. 1998. On geometric graphs with no $k$ pairwise parallel edges. Discr. Comput. Geom. 19, 3 (1998), 461–469.\n• Marc J. van Kreveld. 2010. The quality ratio of RAC drawings and planar drawings of planar graphs. In Proceedings of the GD 2010 (LNCS), Vol. 6502. Springer, 371–376. https://link.springer.com/chapter/10.1007%2F978-3-642-18469-7_34.\n• Massimo Vignelli. 2008. New York Subway Map, http://secondavenuesagas.com/2008/05/02/mens-vogue-calls-on-vignelli-for-a-long-awaited-update/.\n• Klaus Wagner. 1936. Bemerkungen zum vierfarbenproblem. Jahresbericht der Deutschen Mathematiker-Vereinigung 46 (1936), 26–32.\n• Colin Ware, Helen C. Purchase, Linda Colpoys, and Matthew McGill. 2002. Cognitive measurements of graph aesthetics. Inf. Vis. 1, 2 (2002), 103–110.\n• Martin Wattenberg. 2002. Arc diagrams: Visualizing structure in strings. In Proceedings of the INFOVIS. IEEE, 110–116. https://ieeexplore.ieee.org/document/1173155.\n• Stephen K. Wismath. 1985. Characterizing bar line-of-sight graphs. In Proceedings of the ACM SoCG 1985. ACM, 147–152. https://dl.acm.org/citation.cfm?doid=323233.323253.\n• Stephen K. Wismath. 1989. Bar-representable visibility graphs and a related network flow problem. https://open.library.ubc.ca/cIRcle/collections/ubctheses/831/items/1.0051983.\n• David R. Wood. 2005. Grid drawings of k-colourable graphs. Comput. Geom. 30, 1 (2005), 25–28.\n• Mihalis Yannakakis. 1989. Embedding planar graphs in four pages. J. Comput. Syst. Sci. 38, 1 (1989), 36–67.\n• Xin Zhang and Guizhen Liu. 2013. The structure of plane graphs with independent crossings and its applications to coloring problems. Open Math. 11, 2 (2013), 308–321.\n• H. Zhou, Panpan Xu, X. Yuan, and H. Qu. 2013. Edge bundling in information visualization. Tsinghua Science and Technology 18, 2 (2013), 145–156.\n\n## Footnotes\n\n• 1We remark that, despite the fact that the use of the term “graph drawing beyond planarity” is relatively recent, the study of nonplanar topological and geometric graphs started much earlier (see, e.g., [167, 191]).\n• 2The table only shows a subset of the graph families described in this article.\n• 3Here we refer to the so-called weak model, in which the existence of a visibility between a pair of bars does not necessarily imply the existence of an edge in the graph between the two corresponding vertices.\n\nResearch partly supported by the project: “Algoritmi e sistemi di analisi visuale di reti complesse e di grandi dimensioni” - Ricerca di Base 2018, Dip. Eng. Univ. of Perugia.\n\nAuthors’ addresses: W. Didimo, Università degli Studi di Perugia, Via G. Duranti 93, Perugia, PG, 06125, Italy; email: [email protected]; G. Liotta, Università degli Studi di Perugia, Via G. Duranti 93, Perugia, PG, 06125, Italy; email: [email protected]; F. Montecchiani, Università degli Studi di Perugia, Via G. Duranti 93, Perugia, PG, 06125, Italy; email: [email protected].\n\nPermission to make digital or hard copies of all or part of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear this notice and the full citation on the first page. Copyrights for components of this work owned by others than ACM must be honored. Abstracting with credit is permitted. To copy otherwise, or republish, to post on servers or to redistribute to lists, requires prior specific permission and/or a fee. Request permissions from [email protected]." ]

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[ "# #3773d7 Color Information\n\nIn a RGB color space, hex #3773d7 is composed of 21.6% red, 45.1% green and 84.3% blue. Whereas in a CMYK color space, it is composed of 74.4% cyan, 46.5% magenta, 0% yellow and 15.7% black. It has a hue angle of 217.5 degrees, a saturation of 66.7% and a lightness of 52.9%. #3773d7 color hex could be obtained by blending #6ee6ff with #0000af. Closest websafe color is: #3366cc.\n\n• R 22\n• G 45\n• B 84\nRGB color chart\n• C 74\n• M 47\n• Y 0\n• K 16\nCMYK color chart\n\n#3773d7 color description : Moderate blue.\n\n# #3773d7 Color Conversion\n\nThe hexadecimal color #3773d7 has RGB values of R:55, G:115, B:215 and CMYK values of C:0.74, M:0.47, Y:0, K:0.16. Its decimal value is 3634135.\n\nHex triplet RGB Decimal 3773d7 `#3773d7` 55, 115, 215 `rgb(55,115,215)` 21.6, 45.1, 84.3 `rgb(21.6%,45.1%,84.3%)` 74, 47, 0, 16 217.5°, 66.7, 52.9 `hsl(217.5,66.7%,52.9%)` 217.5°, 74.4, 84.3 3366cc `#3366cc`\nCIE-LAB 49.47, 15.045, -56.982 19.969, 17.979, 66.704 0.191, 0.172, 17.979 49.47, 58.934, 284.79 49.47, -22.345, -88.721 42.401, 9.865, -63.592 00110111, 01110011, 11010111\n\n# Color Schemes with #3773d7\n\n• #3773d7\n``#3773d7` `rgb(55,115,215)``\n• #d79b37\n``#d79b37` `rgb(215,155,55)``\nComplementary Color\n• #37c3d7\n``#37c3d7` `rgb(55,195,215)``\n• #3773d7\n``#3773d7` `rgb(55,115,215)``\n• #4b37d7\n``#4b37d7` `rgb(75,55,215)``\nAnalogous Color\n• #c3d737\n``#c3d737` `rgb(195,215,55)``\n• #3773d7\n``#3773d7` `rgb(55,115,215)``\n• #d74b37\n``#d74b37` `rgb(215,75,55)``\nSplit Complementary Color\n• #73d737\n``#73d737` `rgb(115,215,55)``\n• #3773d7\n``#3773d7` `rgb(55,115,215)``\n• #d73773\n``#d73773` `rgb(215,55,115)``\nTriadic Color\n• #37d79b\n``#37d79b` `rgb(55,215,155)``\n• #3773d7\n``#3773d7` `rgb(55,115,215)``\n• #d73773\n``#d73773` `rgb(215,55,115)``\n• #d79b37\n``#d79b37` `rgb(215,155,55)``\nTetradic Color\n• #2051a1\n``#2051a1` `rgb(32,81,161)``\n• #255bb7\n``#255bb7` `rgb(37,91,183)``\n• #2966cc\n``#2966cc` `rgb(41,102,204)``\n• #3773d7\n``#3773d7` `rgb(55,115,215)``\n• #4c82db\n``#4c82db` `rgb(76,130,219)``\n• #6291e0\n``#6291e0` `rgb(98,145,224)``\n• #77a0e4\n``#77a0e4` `rgb(119,160,228)``\nMonochromatic Color\n\n# Alternatives to #3773d7\n\nBelow, you can see some colors close to #3773d7. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #379bd7\n``#379bd7` `rgb(55,155,215)``\n• #378ed7\n``#378ed7` `rgb(55,142,215)``\n• #3780d7\n``#3780d7` `rgb(55,128,215)``\n• #3773d7\n``#3773d7` `rgb(55,115,215)``\n• #3766d7\n``#3766d7` `rgb(55,102,215)``\n• #3758d7\n``#3758d7` `rgb(55,88,215)``\n• #374bd7\n``#374bd7` `rgb(55,75,215)``\nSimilar Colors\n\n# #3773d7 Preview\n\nText with hexadecimal color #3773d7\n\nThis text has a font color of #3773d7.\n\n``<span style=\"color:#3773d7;\">Text here</span>``\n#3773d7 background color\n\nThis paragraph has a background color of #3773d7.\n\n``<p style=\"background-color:#3773d7;\">Content here</p>``\n#3773d7 border color\n\nThis element has a border color of #3773d7.\n\n``<div style=\"border:1px solid #3773d7;\">Content here</div>``\nCSS codes\n``.text {color:#3773d7;}``\n``.background {background-color:#3773d7;}``\n``.border {border:1px solid #3773d7;}``\n\n# Shades and Tints of #3773d7\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, #03060d is the darkest color, while #fbfcfe is the lightest one.\n\n• #03060d\n``#03060d` `rgb(3,6,13)``\n• #060e1d\n``#060e1d` `rgb(6,14,29)``\n• #09172d\n``#09172d` `rgb(9,23,45)``\n• #0c1f3e\n``#0c1f3e` `rgb(12,31,62)``\n• #10274e\n``#10274e` `rgb(16,39,78)``\n• #132f5e\n``#132f5e` `rgb(19,47,94)``\n• #16376f\n``#16376f` `rgb(22,55,111)``\n• #193f7f\n``#193f7f` `rgb(25,63,127)``\n• #1d488f\n``#1d488f` `rgb(29,72,143)``\n• #2050a0\n``#2050a0` `rgb(32,80,160)``\n• #2358b0\n``#2358b0` `rgb(35,88,176)``\n• #2660c0\n``#2660c0` `rgb(38,96,192)``\n• #2a68d1\n``#2a68d1` `rgb(42,104,209)``\nShade Color Variation\n• #3773d7\n``#3773d7` `rgb(55,115,215)``\n• #477eda\n``#477eda` `rgb(71,126,218)``\n• #588ade\n``#588ade` `rgb(88,138,222)``\n• #6895e1\n``#6895e1` `rgb(104,149,225)``\n• #78a1e4\n``#78a1e4` `rgb(120,161,228)``\n• #89ace7\n``#89ace7` `rgb(137,172,231)``\n• #99b8eb\n``#99b8eb` `rgb(153,184,235)``\n• #a9c3ee\n``#a9c3ee` `rgb(169,195,238)``\n• #bacff1\n``#bacff1` `rgb(186,207,241)``\n• #cadaf4\n``#cadaf4` `rgb(202,218,244)``\n• #dae5f8\n``#dae5f8` `rgb(218,229,248)``\n• #ebf1fb\n``#ebf1fb` `rgb(235,241,251)``\n• #fbfcfe\n``#fbfcfe` `rgb(251,252,254)``\nTint Color Variation\n\n# Tones of #3773d7\n\nA tone is produced by adding gray to any pure hue. In this case, #81858d is the less saturated color, while #126afc is the most saturated one.\n\n• #81858d\n``#81858d` `rgb(129,133,141)``\n• #788396\n``#788396` `rgb(120,131,150)``\n• #6e81a0\n``#6e81a0` `rgb(110,129,160)``\n• #657fa9\n``#657fa9` `rgb(101,127,169)``\n• #5c7cb2\n``#5c7cb2` `rgb(92,124,178)``\n• #537abb\n``#537abb` `rgb(83,122,187)``\n• #4978c5\n``#4978c5` `rgb(73,120,197)``\n• #4075ce\n``#4075ce` `rgb(64,117,206)``\n• #3773d7\n``#3773d7` `rgb(55,115,215)``\n• #2e71e0\n``#2e71e0` `rgb(46,113,224)``\n• #256ee9\n``#256ee9` `rgb(37,110,233)``\n• #1b6cf3\n``#1b6cf3` `rgb(27,108,243)``\n• #126afc\n``#126afc` `rgb(18,106,252)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #3773d7 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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[ "# Circuit Simulator - Source Follower\n\nSource Follower", null, "Circuit Description:\nThis is a source follower or buffer amplifier circuit using a MOSFET. The output is simply equal to the input minus about 2.2V. The advantage of this circuit is that the MOSFET can provide current and power gain; the MOSFET draws no current from the input. It provides low output impedance to any circuit using the output of the follower, meaning that the output will not drop under load. Its output impedance is not as low as that of an emitter follower using a bipolar transistor (as you can verify by connecting a resistor from the output to -15V), but it has the advantage that the input impedance is infinite.\n\nThe MOSFET is in saturation, so the current across it is determined by the gate-source voltage. Since a current source keeps the current constant, the gate-source voltage is also constant.\n\nDiscussion:\nBe the first person to comment on this question !" ]

[ null, "https://www.indiabix.com/_files/images/electronics-circuits/source-follower.png", null ]

[ "## Departmental Technical Reports (CS)\n\n#### Publication Date\n\n3-2013\n\nTechnical Report UTEP-CS-13-13a\n\nTo appear in Proceedings of the joint World Congress of the International Fuzzy Systems Association and Annual Conference of the North American Fuzzy Information Processing Society IFSA/NAFIPS'2013, Edmonton, Canada, June 24-28, 2013.\n\n#### Abstract\n\nSome probability distributions (e.g., Gaussian) are symmetric, some (e.g., lognormal) are non-symmetric ({\\em skewed}). How can we gauge the skeweness? For symmetric distributions, the third central moment C3 = E[(x - E(x))3] is equal to 0; thus, this moment is used to characterize skewness. This moment is usually estimated, based on the observed (sample) values x1, ..., xn, as C3 = (1/n) * ((x1 - E)3 + ... + (xn - E)3), where E = (1/n) * (x1 + ... + xn). In many practical situations, we do not know the exact values of xi. For example, to preserve privacy, the exact values are often replaced by intervals containing these values (so that we only know whether the age is under 10, between 10 and 20, etc). Different values from these intervals lead, in general, to different values of C3; it is desirable to find the range of all such possible values. In this paper, we propose a feasible algorithm for computing this range.\n\ntr13-13.pdf (90 kB)\nOriginal file: CS-UTEP-13-13\n\nCOinS" ]

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[ "# Python os.terminal_size() Examples\n\nThe following are code examples for showing how to use os.terminal_size(). They are extracted from open source Python projects. You can vote up the examples you like or vote down the ones you don't like. You can also save this page to your account.\n\nExample 1\n Project: python-   Author: secondtonone1   File: shutil.py    (license) View Source Project 4 votes", null, "", null, "```def get_terminal_size(fallback=(80, 24)):\n\"\"\"Get the size of the terminal window.\n\nFor each of the two dimensions, the environment variable, COLUMNS\nand LINES respectively, is checked. If the variable is defined and\nthe value is a positive integer, it is used.\n\nWhen COLUMNS or LINES is not defined, which is the common case,\nthe terminal connected to sys.__stdout__ is queried\nby invoking os.get_terminal_size.\n\nIf the terminal size cannot be successfully queried, either because\nthe system doesn't support querying, or because we are not\nconnected to a terminal, the value given in fallback parameter\nis used. Fallback defaults to (80, 24) which is the default\nsize used by many terminal emulators.\n\nThe value returned is a named tuple of type os.terminal_size.\n\"\"\"\n# columns, lines are the working values\ntry:\ncolumns = int(os.environ['COLUMNS'])\nexcept (KeyError, ValueError):\ncolumns = 0\n\ntry:\nlines = int(os.environ['LINES'])\nexcept (KeyError, ValueError):\nlines = 0\n\n# only query if necessary\nif columns <= 0 or lines <= 0:\ntry:\nsize = os.get_terminal_size(sys.__stdout__.fileno())\nexcept (AttributeError, ValueError, OSError):\n# stdout is None, closed, detached, or not a terminal, or\n# os.get_terminal_size() is unsupported\nsize = os.terminal_size(fallback)\nif columns <= 0:\ncolumns = size.columns\nif lines <= 0:\nlines = size.lines\n\nreturn os.terminal_size((columns, lines)) ```\nExample 2\n Project: ivaochdoc   Author: ivaoch   File: shutil.py    (license) View Source Project 4 votes", null, "", null, "```def get_terminal_size(fallback=(80, 24)):\n\"\"\"Get the size of the terminal window.\n\nFor each of the two dimensions, the environment variable, COLUMNS\nand LINES respectively, is checked. If the variable is defined and\nthe value is a positive integer, it is used.\n\nWhen COLUMNS or LINES is not defined, which is the common case,\nthe terminal connected to sys.__stdout__ is queried\nby invoking os.get_terminal_size.\n\nIf the terminal size cannot be successfully queried, either because\nthe system doesn't support querying, or because we are not\nconnected to a terminal, the value given in fallback parameter\nis used. Fallback defaults to (80, 24) which is the default\nsize used by many terminal emulators.\n\nThe value returned is a named tuple of type os.terminal_size.\n\"\"\"\n# columns, lines are the working values\ntry:\ncolumns = int(os.environ['COLUMNS'])\nexcept (KeyError, ValueError):\ncolumns = 0\n\ntry:\nlines = int(os.environ['LINES'])\nexcept (KeyError, ValueError):\nlines = 0\n\n# only query if necessary\nif columns <= 0 or lines <= 0:\ntry:\nsize = os.get_terminal_size(sys.__stdout__.fileno())\nexcept (AttributeError, ValueError, OSError):\n# stdout is None, closed, detached, or not a terminal, or\n# os.get_terminal_size() is unsupported\nsize = os.terminal_size(fallback)\nif columns <= 0:\ncolumns = size.columns\nif lines <= 0:\nlines = size.lines\n\nreturn os.terminal_size((columns, lines)) ```\nExample 3\n Project: news-for-good   Author: thecodinghub   File: shutil.py    (license) View Source Project 4 votes", null, "", null, "```def get_terminal_size(fallback=(80, 24)):\n\"\"\"Get the size of the terminal window.\n\nFor each of the two dimensions, the environment variable, COLUMNS\nand LINES respectively, is checked. If the variable is defined and\nthe value is a positive integer, it is used.\n\nWhen COLUMNS or LINES is not defined, which is the common case,\nthe terminal connected to sys.__stdout__ is queried\nby invoking os.get_terminal_size.\n\nIf the terminal size cannot be successfully queried, either because\nthe system doesn't support querying, or because we are not\nconnected to a terminal, the value given in fallback parameter\nis used. Fallback defaults to (80, 24) which is the default\nsize used by many terminal emulators.\n\nThe value returned is a named tuple of type os.terminal_size.\n\"\"\"\n# columns, lines are the working values\ntry:\ncolumns = int(os.environ['COLUMNS'])\nexcept (KeyError, ValueError):\ncolumns = 0\n\ntry:\nlines = int(os.environ['LINES'])\nexcept (KeyError, ValueError):\nlines = 0\n\n# only query if necessary\nif columns <= 0 or lines <= 0:\ntry:\nsize = os.get_terminal_size(sys.__stdout__.fileno())\nexcept (AttributeError, ValueError, OSError):\n# stdout is None, closed, detached, or not a terminal, or\n# os.get_terminal_size() is unsupported\nsize = os.terminal_size(fallback)\nif columns <= 0:\ncolumns = size.columns\nif lines <= 0:\nlines = size.lines\n\nreturn os.terminal_size((columns, lines)) ```\nExample 4\n Project: Tencent_Cartoon_Download   Author: Fretice   File: shutil.py    (license) View Source Project 4 votes", null, "", null, "```def get_terminal_size(fallback=(80, 24)):\n\"\"\"Get the size of the terminal window.\n\nFor each of the two dimensions, the environment variable, COLUMNS\nand LINES respectively, is checked. If the variable is defined and\nthe value is a positive integer, it is used.\n\nWhen COLUMNS or LINES is not defined, which is the common case,\nthe terminal connected to sys.__stdout__ is queried\nby invoking os.get_terminal_size.\n\nIf the terminal size cannot be successfully queried, either because\nthe system doesn't support querying, or because we are not\nconnected to a terminal, the value given in fallback parameter\nis used. Fallback defaults to (80, 24) which is the default\nsize used by many terminal emulators.\n\nThe value returned is a named tuple of type os.terminal_size.\n\"\"\"\n# columns, lines are the working values\ntry:\ncolumns = int(os.environ['COLUMNS'])\nexcept (KeyError, ValueError):\ncolumns = 0\n\ntry:\nlines = int(os.environ['LINES'])\nexcept (KeyError, ValueError):\nlines = 0\n\n# only query if necessary\nif columns <= 0 or lines <= 0:\ntry:\nsize = os.get_terminal_size(sys.__stdout__.fileno())\nexcept (AttributeError, ValueError, OSError):\n# stdout is None, closed, detached, or not a terminal, or\n# os.get_terminal_size() is unsupported\nsize = os.terminal_size(fallback)\nif columns <= 0:\ncolumns = size.columns\nif lines <= 0:\nlines = size.lines\n\nreturn os.terminal_size((columns, lines)) ```\nExample 5\n Project: fieldsight-kobocat   Author: awemulya   File: shutil.py    (license) View Source Project 4 votes", null, "", null, "```def get_terminal_size(fallback=(80, 24)):\n\"\"\"Get the size of the terminal window.\n\nFor each of the two dimensions, the environment variable, COLUMNS\nand LINES respectively, is checked. If the variable is defined and\nthe value is a positive integer, it is used.\n\nWhen COLUMNS or LINES is not defined, which is the common case,\nthe terminal connected to sys.__stdout__ is queried\nby invoking os.get_terminal_size.\n\nIf the terminal size cannot be successfully queried, either because\nthe system doesn't support querying, or because we are not\nconnected to a terminal, the value given in fallback parameter\nis used. Fallback defaults to (80, 24) which is the default\nsize used by many terminal emulators.\n\nThe value returned is a named tuple of type os.terminal_size.\n\"\"\"\n# columns, lines are the working values\ntry:\ncolumns = int(os.environ['COLUMNS'])\nexcept (KeyError, ValueError):\ncolumns = 0\n\ntry:\nlines = int(os.environ['LINES'])\nexcept (KeyError, ValueError):\nlines = 0\n\n# only query if necessary\nif columns <= 0 or lines <= 0:\ntry:\nsize = os.get_terminal_size(sys.__stdout__.fileno())\nexcept (AttributeError, ValueError, OSError):\n# stdout is None, closed, detached, or not a terminal, or\n# os.get_terminal_size() is unsupported\nsize = os.terminal_size(fallback)\nif columns <= 0:\ncolumns = size.columns\nif lines <= 0:\nlines = size.lines\n\nreturn os.terminal_size((columns, lines)) ```\nExample 6\n Project: web_ctp   Author: molebot   File: shutil.py    (license) View Source Project 4 votes", null, "", null, "```def get_terminal_size(fallback=(80, 24)):\n\"\"\"Get the size of the terminal window.\n\nFor each of the two dimensions, the environment variable, COLUMNS\nand LINES respectively, is checked. If the variable is defined and\nthe value is a positive integer, it is used.\n\nWhen COLUMNS or LINES is not defined, which is the common case,\nthe terminal connected to sys.__stdout__ is queried\nby invoking os.get_terminal_size.\n\nIf the terminal size cannot be successfully queried, either because\nthe system doesn't support querying, or because we are not\nconnected to a terminal, the value given in fallback parameter\nis used. Fallback defaults to (80, 24) which is the default\nsize used by many terminal emulators.\n\nThe value returned is a named tuple of type os.terminal_size.\n\"\"\"\n# columns, lines are the working values\ntry:\ncolumns = int(os.environ['COLUMNS'])\nexcept (KeyError, ValueError):\ncolumns = 0\n\ntry:\nlines = int(os.environ['LINES'])\nexcept (KeyError, ValueError):\nlines = 0\n\n# only query if necessary\nif columns <= 0 or lines <= 0:\ntry:\nsize = os.get_terminal_size(sys.__stdout__.fileno())\nexcept (NameError, OSError):\nsize = os.terminal_size(fallback)\nif columns <= 0:\ncolumns = size.columns\nif lines <= 0:\nlines = size.lines\n\nreturn os.terminal_size((columns, lines)) ```\nExample 7\n Project: CloudPrint   Author: William-An   File: shutil.py    (license) View Source Project 4 votes", null, "", null, "```def get_terminal_size(fallback=(80, 24)):\n\"\"\"Get the size of the terminal window.\n\nFor each of the two dimensions, the environment variable, COLUMNS\nand LINES respectively, is checked. If the variable is defined and\nthe value is a positive integer, it is used.\n\nWhen COLUMNS or LINES is not defined, which is the common case,\nthe terminal connected to sys.__stdout__ is queried\nby invoking os.get_terminal_size.\n\nIf the terminal size cannot be successfully queried, either because\nthe system doesn't support querying, or because we are not\nconnected to a terminal, the value given in fallback parameter\nis used. Fallback defaults to (80, 24) which is the default\nsize used by many terminal emulators.\n\nThe value returned is a named tuple of type os.terminal_size.\n\"\"\"\n# columns, lines are the working values\ntry:\ncolumns = int(os.environ['COLUMNS'])\nexcept (KeyError, ValueError):\ncolumns = 0\n\ntry:\nlines = int(os.environ['LINES'])\nexcept (KeyError, ValueError):\nlines = 0\n\n# only query if necessary\nif columns <= 0 or lines <= 0:\ntry:\nsize = os.get_terminal_size(sys.__stdout__.fileno())\nexcept (NameError, OSError):\nsize = os.terminal_size(fallback)\nif columns <= 0:\ncolumns = size.columns\nif lines <= 0:\nlines = size.lines\n\nreturn os.terminal_size((columns, lines)) ```\nExample 8\n Project: ouroboros   Author: pybee   File: shutil.py    (license) View Source Project 4 votes", null, "", null, "```def get_terminal_size(fallback=(80, 24)):\n\"\"\"Get the size of the terminal window.\n\nFor each of the two dimensions, the environment variable, COLUMNS\nand LINES respectively, is checked. If the variable is defined and\nthe value is a positive integer, it is used.\n\nWhen COLUMNS or LINES is not defined, which is the common case,\nthe terminal connected to sys.__stdout__ is queried\nby invoking os.get_terminal_size.\n\nIf the terminal size cannot be successfully queried, either because\nthe system doesn't support querying, or because we are not\nconnected to a terminal, the value given in fallback parameter\nis used. Fallback defaults to (80, 24) which is the default\nsize used by many terminal emulators.\n\nThe value returned is a named tuple of type os.terminal_size.\n\"\"\"\n# columns, lines are the working values\ntry:\ncolumns = int(os.environ['COLUMNS'])\nexcept (KeyError, ValueError):\ncolumns = 0\n\ntry:\nlines = int(os.environ['LINES'])\nexcept (KeyError, ValueError):\nlines = 0\n\n# only query if necessary\nif columns <= 0 or lines <= 0:\ntry:\nsize = os.get_terminal_size(sys.__stdout__.fileno())\nexcept (NameError, OSError):\nsize = os.terminal_size(fallback)\nif columns <= 0:\ncolumns = size.columns\nif lines <= 0:\nlines = size.lines\n\nreturn os.terminal_size((columns, lines)) ```\nExample 9\n Project: gardenbot   Author: GoestaO   File: shutil.py    (license) View Source Project 4 votes", null, "", null, "```def get_terminal_size(fallback=(80, 24)):\n\"\"\"Get the size of the terminal window.\n\nFor each of the two dimensions, the environment variable, COLUMNS\nand LINES respectively, is checked. If the variable is defined and\nthe value is a positive integer, it is used.\n\nWhen COLUMNS or LINES is not defined, which is the common case,\nthe terminal connected to sys.__stdout__ is queried\nby invoking os.get_terminal_size.\n\nIf the terminal size cannot be successfully queried, either because\nthe system doesn't support querying, or because we are not\nconnected to a terminal, the value given in fallback parameter\nis used. Fallback defaults to (80, 24) which is the default\nsize used by many terminal emulators.\n\nThe value returned is a named tuple of type os.terminal_size.\n\"\"\"\n# columns, lines are the working values\ntry:\ncolumns = int(os.environ['COLUMNS'])\nexcept (KeyError, ValueError):\ncolumns = 0\n\ntry:\nlines = int(os.environ['LINES'])\nexcept (KeyError, ValueError):\nlines = 0\n\n# only query if necessary\nif columns <= 0 or lines <= 0:\ntry:\nsize = os.get_terminal_size(sys.__stdout__.fileno())\nexcept (AttributeError, ValueError, OSError):\n# stdout is None, closed, detached, or not a terminal, or\n# os.get_terminal_size() is unsupported\nsize = os.terminal_size(fallback)\nif columns <= 0:\ncolumns = size.columns\nif lines <= 0:\nlines = size.lines\n\nreturn os.terminal_size((columns, lines)) ```\nExample 10\n Project: projeto   Author: BarmyPenguin   File: shutil.py    (license) View Source Project 4 votes", null, "", null, "```def get_terminal_size(fallback=(80, 24)):\n\"\"\"Get the size of the terminal window.\n\nFor each of the two dimensions, the environment variable, COLUMNS\nand LINES respectively, is checked. If the variable is defined and\nthe value is a positive integer, it is used.\n\nWhen COLUMNS or LINES is not defined, which is the common case,\nthe terminal connected to sys.__stdout__ is queried\nby invoking os.get_terminal_size.\n\nIf the terminal size cannot be successfully queried, either because\nthe system doesn't support querying, or because we are not\nconnected to a terminal, the value given in fallback parameter\nis used. Fallback defaults to (80, 24) which is the default\nsize used by many terminal emulators.\n\nThe value returned is a named tuple of type os.terminal_size.\n\"\"\"\n# columns, lines are the working values\ntry:\ncolumns = int(os.environ['COLUMNS'])\nexcept (KeyError, ValueError):\ncolumns = 0\n\ntry:\nlines = int(os.environ['LINES'])\nexcept (KeyError, ValueError):\nlines = 0\n\n# only query if necessary\nif columns <= 0 or lines <= 0:\ntry:\nsize = os.get_terminal_size(sys.__stdout__.fileno())\nexcept (NameError, OSError):\nsize = os.terminal_size(fallback)\nif columns <= 0:\ncolumns = size.columns\nif lines <= 0:\nlines = size.lines\n\nreturn os.terminal_size((columns, lines)) ```\nExample 11\n Project: flask-zhenai-mongo-echarts   Author: Fretice   File: shutil.py    (license) View Source Project 4 votes", null, "", null, "```def get_terminal_size(fallback=(80, 24)):\n\"\"\"Get the size of the terminal window.\n\nFor each of the two dimensions, the environment variable, COLUMNS\nand LINES respectively, is checked. If the variable is defined and\nthe value is a positive integer, it is used.\n\nWhen COLUMNS or LINES is not defined, which is the common case,\nthe terminal connected to sys.__stdout__ is queried\nby invoking os.get_terminal_size.\n\nIf the terminal size cannot be successfully queried, either because\nthe system doesn't support querying, or because we are not\nconnected to a terminal, the value given in fallback parameter\nis used. Fallback defaults to (80, 24) which is the default\nsize used by many terminal emulators.\n\nThe value returned is a named tuple of type os.terminal_size.\n\"\"\"\n# columns, lines are the working values\ntry:\ncolumns = int(os.environ['COLUMNS'])\nexcept (KeyError, ValueError):\ncolumns = 0\n\ntry:\nlines = int(os.environ['LINES'])\nexcept (KeyError, ValueError):\nlines = 0\n\n# only query if necessary\nif columns <= 0 or lines <= 0:\ntry:\nsize = os.get_terminal_size(sys.__stdout__.fileno())\nexcept (AttributeError, ValueError, OSError):\n# stdout is None, closed, detached, or not a terminal, or\n# os.get_terminal_size() is unsupported\nsize = os.terminal_size(fallback)\nif columns <= 0:\ncolumns = size.columns\nif lines <= 0:\nlines = size.lines\n\nreturn os.terminal_size((columns, lines)) ```\nExample 12\n Project: aweasome_learning   Author: Knight-ZXW   File: shutil.py    (license) View Source Project 4 votes", null, "", null, "```def get_terminal_size(fallback=(80, 24)):\n\"\"\"Get the size of the terminal window.\n\nFor each of the two dimensions, the environment variable, COLUMNS\nand LINES respectively, is checked. If the variable is defined and\nthe value is a positive integer, it is used.\n\nWhen COLUMNS or LINES is not defined, which is the common case,\nthe terminal connected to sys.__stdout__ is queried\nby invoking os.get_terminal_size.\n\nIf the terminal size cannot be successfully queried, either because\nthe system doesn't support querying, or because we are not\nconnected to a terminal, the value given in fallback parameter\nis used. Fallback defaults to (80, 24) which is the default\nsize used by many terminal emulators.\n\nThe value returned is a named tuple of type os.terminal_size.\n\"\"\"\n# columns, lines are the working values\ntry:\ncolumns = int(os.environ['COLUMNS'])\nexcept (KeyError, ValueError):\ncolumns = 0\n\ntry:\nlines = int(os.environ['LINES'])\nexcept (KeyError, ValueError):\nlines = 0\n\n# only query if necessary\nif columns <= 0 or lines <= 0:\ntry:\nsize = os.get_terminal_size(sys.__stdout__.fileno())\nexcept (NameError, OSError):\nsize = os.terminal_size(fallback)\nif columns <= 0:\ncolumns = size.columns\nif lines <= 0:\nlines = size.lines\n\nreturn os.terminal_size((columns, lines)) ```\nExample 13\n Project: kbe_server   Author: xiaohaoppy   File: shutil.py    (license) View Source Project 4 votes", null, "", null, "```def get_terminal_size(fallback=(80, 24)):\n\"\"\"Get the size of the terminal window.\n\nFor each of the two dimensions, the environment variable, COLUMNS\nand LINES respectively, is checked. If the variable is defined and\nthe value is a positive integer, it is used.\n\nWhen COLUMNS or LINES is not defined, which is the common case,\nthe terminal connected to sys.__stdout__ is queried\nby invoking os.get_terminal_size.\n\nIf the terminal size cannot be successfully queried, either because\nthe system doesn't support querying, or because we are not\nconnected to a terminal, the value given in fallback parameter\nis used. Fallback defaults to (80, 24) which is the default\nsize used by many terminal emulators.\n\nThe value returned is a named tuple of type os.terminal_size.\n\"\"\"\n# columns, lines are the working values\ntry:\ncolumns = int(os.environ['COLUMNS'])\nexcept (KeyError, ValueError):\ncolumns = 0\n\ntry:\nlines = int(os.environ['LINES'])\nexcept (KeyError, ValueError):\nlines = 0\n\n# only query if necessary\nif columns <= 0 or lines <= 0:\ntry:\nsize = os.get_terminal_size(sys.__stdout__.fileno())\nexcept (NameError, OSError):\nsize = os.terminal_size(fallback)\nif columns <= 0:\ncolumns = size.columns\nif lines <= 0:\nlines = size.lines\n\nreturn os.terminal_size((columns, lines)) ```\nExample 14\n Project: blog_flask   Author: momantai   File: shutil.py    (license) View Source Project 4 votes", null, "", null, "```def get_terminal_size(fallback=(80, 24)):\n\"\"\"Get the size of the terminal window.\n\nFor each of the two dimensions, the environment variable, COLUMNS\nand LINES respectively, is checked. If the variable is defined and\nthe value is a positive integer, it is used.\n\nWhen COLUMNS or LINES is not defined, which is the common case,\nthe terminal connected to sys.__stdout__ is queried\nby invoking os.get_terminal_size.\n\nIf the terminal size cannot be successfully queried, either because\nthe system doesn't support querying, or because we are not\nconnected to a terminal, the value given in fallback parameter\nis used. Fallback defaults to (80, 24) which is the default\nsize used by many terminal emulators.\n\nThe value returned is a named tuple of type os.terminal_size.\n\"\"\"\n# columns, lines are the working values\ntry:\ncolumns = int(os.environ['COLUMNS'])\nexcept (KeyError, ValueError):\ncolumns = 0\n\ntry:\nlines = int(os.environ['LINES'])\nexcept (KeyError, ValueError):\nlines = 0\n\n# only query if necessary\nif columns <= 0 or lines <= 0:\ntry:\nsize = os.get_terminal_size(sys.__stdout__.fileno())\nexcept (AttributeError, ValueError, OSError):\n# stdout is None, closed, detached, or not a terminal, or\n# os.get_terminal_size() is unsupported\nsize = os.terminal_size(fallback)\nif columns <= 0:\ncolumns = size.columns\nif lines <= 0:\nlines = size.lines\n\nreturn os.terminal_size((columns, lines)) ```\nExample 15\n Project: MyFriend-Rob   Author: lcheniv   File: shutil.py    (license) View Source Project 4 votes", null, "", null, "```def get_terminal_size(fallback=(80, 24)):\n\"\"\"Get the size of the terminal window.\n\nFor each of the two dimensions, the environment variable, COLUMNS\nand LINES respectively, is checked. If the variable is defined and\nthe value is a positive integer, it is used.\n\nWhen COLUMNS or LINES is not defined, which is the common case,\nthe terminal connected to sys.__stdout__ is queried\nby invoking os.get_terminal_size.\n\nIf the terminal size cannot be successfully queried, either because\nthe system doesn't support querying, or because we are not\nconnected to a terminal, the value given in fallback parameter\nis used. Fallback defaults to (80, 24) which is the default\nsize used by many terminal emulators.\n\nThe value returned is a named tuple of type os.terminal_size.\n\"\"\"\n# columns, lines are the working values\ntry:\ncolumns = int(os.environ['COLUMNS'])\nexcept (KeyError, ValueError):\ncolumns = 0\n\ntry:\nlines = int(os.environ['LINES'])\nexcept (KeyError, ValueError):\nlines = 0\n\n# only query if necessary\nif columns <= 0 or lines <= 0:\ntry:\nsize = os.get_terminal_size(sys.__stdout__.fileno())\nexcept (AttributeError, ValueError, OSError):\n# stdout is None, closed, detached, or not a terminal, or\n# os.get_terminal_size() is unsupported\nsize = os.terminal_size(fallback)\nif columns <= 0:\ncolumns = size.columns\nif lines <= 0:\nlines = size.lines\n\nreturn os.terminal_size((columns, lines)) ```" ]

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[ "# Visualization\n\nVisualization is an important tool for both presenting the result of your numerical calculation and for checking what your code is actually doing. One of the most commonly used and powerful packages in python is matplotlib. One often just imports the pyplot methods as\n\n`import matplotlib.pyplot as plt`\n\nThen it is very simple to make some basic plots. An array of values can be generated with the numpy methods linspace or arange.\n``` x=np.linspace(0.0,5.0,num=50) x2=np.arange(0.0,5.0,0.1) #step by 0.1 y=x**2 #etc. ```\nIf one has two arrays x and y then one can do\n``` plt.plot(x,y,linestyle='--',color='r') plt.scatter(x,y,marker='o',color='b',label='points') ```\nHowever, these easy methods somewhat obscure how plotting in matplotlib really works, which one needs to understand to make more complicated plots. Matplotlib plots are based on two objects, the figure and the axis. The figure is the entire plotting region, while an axis can be only part of that region and has the scale information for a plot. You can add as many axes as you want to a figure. To generate a figure and axis and then plot on that axis one can do:\n``` f=plt.figure() ax=f.add_subplot(1,2,1) ax.plot(x,y) ```\nNote that while you now have a figure, this in general won’t show it. You need to explicitly call plt.show() to show the figure or plt.savefig(‘fig.pdf’) to save the figure as a file. The extension you give for the save will determine the type of file is saves to. Also, using notebooks or IDEs will effect where the figure is displayed.\n\nWhen one calls plt.plot() without first creating a figure and axis the code just generates them by default. But if one wants more control over the plot there are many options to how the figure and axis are generated. From here on methods will be given for the axis though most will work at the plt level, though sometimes the syntax is slightly different. Axes can either be created by trying to fit a certain number of plots onto the figure, or by specifying exactly where you want them.\n``` f,axes=plt.subplots(Nrows,Ncols,figsize=(Lx,Ly)) ```\nor\n``` f=plt.figure(figsize=(Lx,Ly)) ax1=f.add_suplot(Nrows, Ncols, index) ax2=f.add_axes([xmin,ymin,width,height]) #figure units: 0 to 1 ax3=ax1.twinx() #new axis shares x but new y ```\n\nPlotting points and lines\n``` ax.plot(x,y,linestyle=':') ax.scatter(x,y, marker='+') ax.errorbars(x,y,xerr=xerr,yerr=yerr) ```\n\n1D histograms\n``` ax.hist(array) ```\n\n2D plots can be made if you have a z value for each x and y value. This can just be a 2D array of z values. One can make a density plot from x and y values, by determining the number of points in a bin. For example we can start with a random set of x and y values.\n``` N=100000 x=np.random.normal(size=N) y=np.random.normal(size=N) ax.hist2d(x,y) #makes a plot ax.hexbin(x,y) #uses hexagon instead of square bins hist,xed,yed=np.histogram2d(x,y,bins=25) #returns array ax.imshow(hist,cmap='brg',extent=[xed,xed[-1],yed,yed[-1]]) cs=ax.contour(hist,[1,5,8],extent=[xed,xed[-1],yed,yed[-1]])) ax.clabel(cs,inline=1,fontsize=10,fmt='%d') ```\n\nCustomizing plots You will often want to customize your plot, not use the default setting. Everything can be customized, the range of the plot, the labels, the fonts, the size of fonts, the ticks, gridlines, etc. Also, important extras can be added like legends and color bars.\n``` ax.set_xlim(xmin,xmax) ax.set_xlabel(xlabel) ax.legend() plt.colorbar() ```\n\n## Patches\n\nAnother thing you can do is and shapes (called patches in matplotlib) to your plot. First one creates the shape; size, color, where it goes, etc. and then you add it to an axis. Note patches must be added to an axis, if you didn’t create one explicitly, but instead by default, you can grab it with plt.gca() which returns the current axis. Here’s an example:\n``` f,ax=plt.subplots() ax.scatter([-1,1],[-1,1]) circle = plt.Circle((0, 0), radius=0.75, fc='g') ax.add_patch(circle) rectangle1 = plt.Rectangle((0.25, 0.35), 0.2, 0.2, fc='r') rectangle2 = plt.Rectangle((-0.45, 0.35), 0.2, 0.2, fc='r') ax.add_patch(rectangle1) ax.add_patch(rectangle2) points = [[-0.1, -0.1], [0.0, 0.3], [0.1, -0.1]] triangle = plt.Polygon(points,fc='b') ax.add_patch(triangle) points = [[-0.4, -0.2], [-0.2, -0.4], [0.2, -0.4], [0.4, -0.2]] line = plt.Polygon(points, closed=None, fill=None, edgecolor='m') ax.add_patch(line)```\nwhich makes the following image", null, "## Animation\n\nFinally it can be really useful to be able to animate our figures. To save animations you will need other free software that has tools for making movies. Two good choices are ffmpeg and ImageMagick. We can install ffmpeg with conda so first\n``` conda install ffmpeg ```\nThen the animation process is basically you need a function that takes the frame number as input and advances to the next image you want. Then you pass your figure and this function into\n`anim = animation.FuncAnimation(fig, func, N)`\nwhere N is the number of frames you want in your animation. Here is an example:\n``` import matplotlib.pyplot as plt import matplotlib.animation as animation import numpy as np ```\n``` fig = plt.figure(figsize=(5,5)) ax = plt.axes(xlim=(-1.25, 1.25), ylim=(-1.25, 1.25)) sun=plt.Circle((0,0),radius=0.3,facecolor='yellow') ax.add_patch(sun) earth = plt.Circle((0, -1), 0.15, fc='b') `````` def init(): earth.center = (0, -1) ax.add_patch(earth) return earth, `````` def animate(i): x = np.cos(np.radians(i)) y = np.sin(np.radians(i)) earth.center = (x, y) return earth, `````` anim = animation.FuncAnimation(fig, animate, init_func=init,frames=360,interval=20,blit=True) anim.save('solar_system.mp4'). #requires ffmpeg #alternatively save as a gif writergif = animation.PillowWriter(fps=30) ani.save('filename.gif',writer=writergif) ```\n\nThis makes this animation:\n\nUsing matplotlibs animation methods may or may not make things easier depending on what you want to do. You can always instead just make a bunch of images and then use ffmpeg to stitch them into a video. If you’ve made images called img01.png to img50.png and you want to make them into a video, then on the command line just do\n``` ffmpeg -framerate 30 -i img%02d.png movie.mp4 ```\nThere are many important controls on how your movie is encoded, frame rates, etc. look for help on ffmpeg.\n\nHere are some cheatsheets to help with some of the commands:", null, "", null, "cheatsheets.pdf" ]

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[ "Example #1. Where r is the fixed distance from the center point or we can call it the radius of Sphere. In most of the cases, the box is an enclosed figure either a rectangle or a square. ⇒TSA = 3πr2. The lateral surface area of the cuboid = Area of face AEHD + Area of face BFGC + Area of face ABFE + Area of face DHGC A-1, Acharya Nikatan, Mayur Vihar, Phase-1, Central Market, New Delhi-110091. We need to calculate the radius of the sphere to calculate the volume and surface area. The radius of a cylinder is 7 cm. In this chapter, the combination of different solid shapes can be studied. Substituting, s for the length, width and height into the formulas for volume and surface area of a rectangular solid, we get: In other words, 3 cones make a cylinder of the same height and base. Where l is the length of the edge of the cube. The surface area of the cube is 150 cm 2. =2πr(h + r). This process of calculation is exactly similar to breaking a bigger problem into a smaller problem, to reach an accurate solution. Practice: Apply volume of solids. Then, solve for x. The resulting surface area to volume ratio is therefore 3/r. To find the surface area of a prism, we must add the area of each face. The formulas in this chapter will be explained properly. Surface Area and Volume Formulas Chart. To make it clearer and help students remember them all, Vedantu has prepared a summarized page for Surface Area and Volume Formulas Class 10. ∴ CSA of a hemisphere of radius r = 2πr2 Area Volume Perimeter Surface Area Formulas PDF + Printable. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher. When a solid is converted into another solid of a different shape (by melting or casting), the volume remains constant. A table of volume formulas and surface area formulas used to calculate the volume and surface area of three-dimensional geometrical shapes: cube, cuboid, prism, solid cylinder, hollow cylinder, cone, pyramid, sphere and hemisphere. Formulas for finding volume, lateral area, and surface area; 00:08:57 – Find the volume of each type of pyramid (Examples #1-3) Exclusive Content for Member’s Only ; 00:14:39 – Find the lateral area and surface area of a square pyramid (Example #4) 00:19:14 – Find the lateral and surface area of a triangular pyramid (Example #5) What is the surface area of a cylinder with radius 3 cm and height 5 cm? Curved surface area of the right circular cylinder = Perimeter of the base of the cylinder X height. Surface Area & Volume of a Right circular cylinder Curved surface area of the right circular cylinder = Perimeter of the base of the cylinder X height Perimeter of the base of the cylinder = Perimeter of circle with same radius = 2 π r Curved Surface Area of Cylinder = 2 π r h A right angle is made up of 90 degrees.A straight line is made up of 180 degrees.If two lines intersect, the sum of the resulting four angles equals 360. Your email address will not be published. Angles. Formulas In our example, the edge is 6 inches, so e squared is 36. ∴ The volume of the hemisphere of radius r = (2/3)πr3. The Volume of a box can be found by just multiplying all the three inputs (h, w, l) together. Rectangular prism. With a list of complete Sphere Formulas, this is easy to calculate the surface area, curved surface area, and the volume of the Sphere. The areas of the triangular faces will have different formulas for different shaped bases. Volume of a hemisphere = ( 1/2 ) ( 4 /3 π r 2) = Surface Area and Volume of Hemisphere shell If you have a cube with a known volume V and are trying to find the surface area A first calculate the side length S by using this formula:. The volume of a container is how much it can hold. On signing up you are confirming that you have read and agree to The volume of a 3D shape or solid is how much space it occupies; it is the space contained by the shape. AREA AND VOLUME FORMULAS Areas of Plane Figures Square Rectangle Parallelogram s s b w l h 2A = s A = l • w A = b • h Triangle Trapezoid Circle h b h b 1 b 2 r d A = ½ b • h 2 A = ½ (b 1 + b 2) • h A = πr (π ≈ 3.14 or ) Circumference: C = 2πr = πd Find here the Total surface area and volume of box formula for deriving the total surface area and the volume of a box for the given height, width and length. The formulas in this chapter will be explained properly. So, 20.64 x 36 = 743.24. Applying volume of solids. To know more about Surface Area of Hemisphere, visit here. Volume and Surface Area Page 6 of 19 Example 3: Find the volume and surface area of the figure below 8 5 3 in Solution: This is a sphere. To find the volume of a cube from its surface area, first use the formula for surface area to find the length of one side of your cube. With a cell radius of 100, SA:V ratio is 0.03. A table of surface area formulas and volume formulas used to calculate the surface area and volume of three-dimensional geometrical shapes: cube, cuboid, prism, solid cylinder, hollow cylinder, cone, pyramid, sphere and hemisphere. Your email address will not be published. Hemisphere is half of a sphere. Furthermore, volume refers to a quantity of a three-dimensional space whose enclosure takes place by a closed surface. Here, let us discuss the surface area formulas and volume formulas for different three-dimensional shapes in detail. S = ∛V Curved Surface Area (CSA) = Total Surface Area (TSA) = 4πr2. It's called lateral surface area for cube and cuboid and curved surface area for cylinder, cone and hemisphere.The formulas for calculating Surface Area and Volume areCubeLateral Surface Area = 4a2Total Surface Area = 6a2Volume = a3Cu Cube Surface Area (A):. TSA (cuboid) = 2(l × b) + 2(b × h) + 2(l × h) = 2(lb + bh + lh), Lateral surface area (LSA) is the area of all the sides apart from the top and bottom faces. (2 mark) ... of a sphere is doubled, then find the ratio of their volumes. Flat surface area or base area of the hemisphere = Area of the circle with same radius = π r 2. Login to view more pages. To find the surface area and volume of these three-dimensional solids. A hemisphere is half of a sphere. This is the currently selected item. Surface area formulas. Surface Area The surface area of a Torus is given by the formula – Surface Area = 4 × Pi^2 × R × r. Where r is the radius of the small circle and R is the radius of bigger circle and Pi is constant Pi=3.14159. Required fields are marked *. Surface Areas. TSA  of a cylinder of base radius r and height h = 2π × r × h + area of two circular bases TSA (cube) =2 × (3l2) = 6l2 011-47340170 . S = ∛V Teachoo provides the best content available! Surface Area using Volume Calculator. Thankfully there is a formula for finding the surface area of a right prism that makes our calculations fast and efficient, as seen below. If the total surface area of a cube is 216 cm 2, then find its volume. 2. Visual on the figure below: Cube: s3, s is the side of the cube. The Volume of a box can be found by just multiplying all the three inputs (h, w, l) together. Volume and Surface Area of a Cube. The total surface area of hemisphere is equal to sum of half of surface area of sphere and area of its circular base. Volume formulas review. The surface area can be generally classified into Lateral Surface Area (LSA), Total Surface Area (TSA), and Curved Surface Area (CSA). (3 mark) 8. Thus, the surface area falls off steeply with increasing volume… To know more about Surface Area of Cube, visit here. This can be visualised as follows : To know more about Volume of a Combination of Solids, visit here. Suppose a metallic sphere of radius 9 cm is melted and recast into the shape of a cylinder of radius 6 cm. Note: Area and volume formulas only work when the torus has a hole! Calculator online for a the surface area of a capsule, cone, conical frustum, cube, cylinder, hemisphere, square pyramid, rectangular prism, triangular prism, sphere, or spherical cap. An online geometry formulas in pdf format. This can be used on a pyramid that has a rectangular rather than a square base. Then, A surface area and volume chart play a significant role in properly understanding surface area and volume. If you have a cube with a known volume V and are trying to find the surface area A first calculate the side length S by using this formula:. Solution: A =158 =120 in2 Example 8: If the area of a parallelogram is 56 units2 and the base is 4 units, what is the height? To know more about Surface Area of Sphere, visit here. Volume and surface area are two important properties for 3D shapes or solid. Volume of a cylinder = Base area × height = (πr2) × h = πr2h. A cube is a solid the sides of which are formed by six identical squares. Example #2. TSA = CSA + area of base = πrl + πr2 = πr(l + r). Terms of Service. Volume and surface area. Volume. To know more about volume of a Sphere, visit here. Volume and surface area are two important properties for 3D shapes or solid. Volume and surface area. To know more about Surface Area of Right Circular Cone, visit here. He provides courses for Maths and Science at Teachoo. A couple of examples showing how to use the surface area formula to solve some problems. Note: Area and volume formulas only work when the torus has a hole! Cone: 1/3 πr 2h, r i… Volume is the capacity of any solid shape. Volume of a cube = base area × height A more detailed explanation (in text and video) of each surface area formula. Take a cylinder of base radius r and height h units. Volume: the volume is the same as if we \"unfolded\" a torus into a cylinder (of length 2πR): The radius of the cylinder is known however the height is not known. Surface Area and Volume. Consider a cuboid whose dimensions are l × b × h, respectively. A cube is a rectangular solid whose length, width, and height are equal. area of base . If a right circular cone is sliced by a plane parallel to its base, then the part with the two circular bases is called a Frustum. Surface Area and Volume. [email protected] The Volume of a Sphere Formula. From there, we’ll tackle trickier objects, such as cones and spheres. (2 mark) 7. LSA (cuboid) = 2(b × h) + 2(l × h) = 2h(l + b), Length of diagonal of a cuboid =√(l2 + b2 + h2). Where r is the fixed distance from the center point or we can call it the radius of Sphere. Since these cubes are joined adjacently,  they form a cuboid whose length l = 8 cm. ⇒4/3π(93) = π(62)h                   (On substituting the values) For more information and examples of these calculations see our pages: Calculating Area, Three-Dimensional Shapes and Calculating Volume. The volume (V) of a hemisphere will be half of that of a sphere. In these lessons, we shall be looking at: A table of surface area formulas and volume formulas used to calculate the surface area and volume of three-dimensional geometrical shapes: cube, cuboid, prism, solid cylinder, hollow cylinder, cone, pyramid, sphere and hemisphere. Thus, if a cell has a radius of 1 μm, the SA:V ratio is 3; whereas if the radius of the cell is instead 10 μm, then the SA:V ratio becomes 0.3. Areas of complex figures can be broken down and analysed as simpler known shapes. Since the volume remains the same after a recast, the volume of the cylinder will be equal to the volume of the sphere. Cuboid: l × b × h, where l , b and h are the length, breadth and height of a cuboid. To know more about Volume of Cube and Cuboid, visit here. Learn vocabulary, terms, and more with flashcards, games, and other study tools. The formula is often written in this shorter way: Volume = 2 π 2 Rr 2 . It's called lateral surface area for cube and cuboid and curved surface area for cylinder, cone and hemisphere. A surface area and volume chart play a significant role in properly understanding surface area and volume. Cube Surface Area (A):. In this article, we are going to discuss the surface area and volume for different solid shapes such as the cube, cuboid, cone, cylinder, and so on. Volume The volume of a cone is given by the formula – Volume = 2 × Pi^2 × R × r^2 Let h be the height of the cylinder. Here, let us discuss the surface area formulas and volume formulas for different three-dimensional shapes in detail. Now, multiply the above number by e squared, where e is the length of an edge. The surface area of a cuboid is equal to the sum of the areas of its six rectangular faces. The formula for the surface area of a prism is $$SA=2B+ph$$, where B, again, stands for the area of the base, p represents the perimeter of the base, and h stands for the height of the prism. Thankfully there is a formula for finding the surface area of a right prism that makes our calculations fast and efficient, as seen below. The concept of surface area and volume for Class 10 is provided here. Practice. 6a 2 = 6 × a 2 = 6 × 5 2 = 6 × 25 = 150. The volume of our dodecahedron is 1655.208 cubed inches. To do this, plug the surface area you’re given into the formula, which is surface area = 6x^2, where x is the length of one side of the cube. He has been teaching from the past 9 years. Supporters: Online Education - comprehensive directory of online education programs and college degrees. The total surface area of the cuboid (TSA) = Sum of the areas of all its six faces Surface Area = 2bs + b 2; Volume = 1/3 b 2 h; Another way to calculate this is to use the perimeter (P) and the area (A) of the base shape. This free surface area calculator determines the surface area of a number of common shapes, including sphere, cone, cube, cylinder, capsule, cap, conical frustum, ellipsoid, and square pyramid. Volume = 1/3 area of the base X height V = bh b is the area of the base Surface Area: Add the area of the base to the sum of the areas of all of the triangular faces. Volume: the volume is the same as if we \"unfolded\" a torus into a cylinder (of length 2πR): Space diagonals are also equally long. Teachoo is free. Calculates the volume, surface area and radii of inscribed and circumscribed spheres of the regular polyhedrons given the side length. The volume of a Right circular cone =(1/3)πr2h The formulae for volumes of various shapes are: 1. In chemical reactions involving a solid material, the surface area to volume ratio is an important factor for the reactivity, that is, the rate at which the chemical reaction will proceed. Find here the Total surface area and volume of box formula for deriving the total surface area and the volume of a box for the given height, width and length. ∴ The new surface area, TSA = 2(lb + bh + lh). The Volume of a Sphere Formula. Volume and Surface Area of a Right Circular Cone. This is sometimes referred to as capacity rather than volume. The formula is often written in this shorter way: Volume = 2 π 2 Rr 2 . r1 and r2 be the radius of the sphere and cylinder, respectively. Volume. ⇒4/3πr13 = πr22h Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Volume of a Hemisphere. Surface Area using Volume Calculator. The surface-area-to-volume ratio, also called the surface-to-volume ratio and variously denoted sa/vol or SA:V, is the amount of surface area per unit volume of an object or collection of objects. Download Surface Area and Volume Cheatsheet Below. For calculations, Lateral Surface Area means curved surface area. Volume (V):. Sphere Surface Area Volume = (4/3) π r 3; Surface Area = 4 π r 2; Spherical Cap Surface Area. Volume: Frustum of Pyramid. Length of each cube = 64(1/3) = 4cm Here, we will discuss some interesting facts about the box and how to calculate the volume and the surface area of a box with the help of mathematical formula. To know more about Volume of Hemisphere, visit here. We’ll start with the volume and surface area of rectangular prisms. The surface area, S, of a rectangular prism is: S = 2lw + 2lh + 2wh. Volume and surface area help us measure the size of 3D objects. Angles. So for a cube, the formulas for volume and surface area are V = s3 V = s 3 and S =6s2 S = 6 s 2. If its volume is 2002 cm 3, then find its height and total surface area. Now that we know what the formulas are, let’s look at a few example problems using them. Explore the area or volume calculator, as well as hundreds of other calculators addressing math, … Total Surface Area = curved surface area + area of the base circle We are given that the diameter of the sphere is 8 5 3 inches. This is sometimes referred to as capacity rather than volume. To know more about Surface Area of Cuboid, visit here. www.ck12.org Chapter 7. Supporters: Online Education - comprehensive directory … Also, the procedure to find the volume and its surface area in detail. 4. A cube with a volume (V) of 125 cubic units has a surface area (A) of 150 square units.. Cube Surface Area from Volume Formula. Start studying Volume and Surface Area Formulas. The formulas for calculating Surface Area and Volume are, Subscribe to our Youtube Channel - https://you.tube/teachoo. Cones Volume = 1/3 area of the base x height V= r2h Surface S = r2 + rs Perimeter, Area, Surface Area, and Volume Example 7: Find the area of the parallelogram. Easy Tricks To Remember: Cylinder Volume Formula. The Surface area of a box formula . Practice: Volume and surface area of cylinders. Volume of a cuboid = (base area) × height = (lb)h = lbh. Online calculators and formulas for a surface area and other geometry problems. Like a Cylinder. In this chapter, the combination of … The volume of a Right circular cone is 1/3 times that of a cylinder of same height and base. Face diagonals are equally long in all sides. Here we just need to find the area of one triangle and multiply it by four sides: Area of 1 triangle = ½bh = ½(8 x 6.9) = 27.6 cm 2 3. cube = 6 a 2. prism: (lateral area) = perimeter(b) L (total area) = perimeter(b) L + 2b. Volume (V):. The volume is how much space takes up the inside of a cylinder. Volume = (1/3) π h 2 (3R - h) Surface Area = 2 π Rh; Triangular Prism Surface Area Top Surface Area of a Triangular Prism Formula $A_{top} = \\dfrac{1}{4} \\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}$ Bottom Surface Area of a Triangular Prism Formula The radius of a sphere is half of its diameter. Where r is the radius of the base and h is the height of the cone. To know more about Shape Conversion of Solids, visit here. Example: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 3 cm and the height of the cone is equal to 5 cm. CBSE Class 10 Maths Mensuration is an interesting chapter where students will learn how to calculate the surface area and volumes of different regular 3D objects. A table of volume formulas and surface area formulas used to calculate the volume and surface area of three-dimensional geometrical shapes: cube, cuboid, prism, solid cylinder, hollow cylinder, cone, pyramid, sphere and hemisphere. An online geometry formulas in pdf format. Find the surface area of a cube if the length of one side is equal to 5 cm. The Total surface area of a box is the total of the area outside all the sides of the box. Solution: Plug in what we know to the area formula … The graphic on this page, is designed to be a quick reference for calculating the area, surface area and volume of common shapes. Find the surface area of the resulting cuboid. A cube with a volume (V) of 125 cubic units has a surface area (A) of 150 square units.. Cube Surface Area from Volume Formula. To know more about Volume of a Cylinder, visit here. The volume of a sphere of radius r = (4/3)πr3. Online Education - comprehensive directory of online education programs and college degrees. A more detailed explanation (examples and solutions) of each volume formula. The volume of a container is how much it can hold. Volume. A more detailed explanation (examples and solutions) of each volume formula. Volume of a sphere in n-dimension Volume of a sphere in n-dimension: Calculates the volume and surface area of a sphere in n-dimensional space … See Volume and Surface Area of a Cube, below. In chemical reactions involving a solid material, the surface area to volume ratio is an important factor for the reactivity, that is, the rate at which the chemical reaction will proceed. Volume = Pi * r 2 * h. Surface Area = 2 * Pi * r * h . Surface refers to a measure of the total area occupied by the surface of a solid. Calculates the volume, surface area and radii of inscribed and circumscribed spheres of the regular polyhedrons given the side length. The Surface area of a box formula . If r is the radius of circular base of the cone-shaped object and h is the height, then the formula to find the volume of the cone is given by: V = 1/3π r 2 h What is the total surface area of hemisphere? We know that the CSA of a sphere  = 4πr2. Total Surface Area of the Hemisphere = 3 π r 2. Thus, CSA of a cylinder of base radius r and height h = 2π × r × h Cylinder: πr2h, r is the radius of circular base and h is the height of the cylinder. Surface Area = 4 * Pi * r 2 . The surface-area-to-volume ratio, also called the surface-to-volume ratio and variously denoted sa/vol or SA:V, is the amount of surface area per unit volume of an object or collection of objects. Consider a right circular cone with slant length l, radius r and height h. CSA of right circular cone = πrl By finding the areas of these known shapes, we can find out the required area of the unknown figure. Volume And Surface Area Examples: Example1: This pyramid is made up of four equilateral triangles. =2π × r × h + 2πr2 Surface area = 2πr 2 + 2πrh = (2x3.14x3x3) + (2x3.14x3x5) = 56.52 + 94.2 = 150.72 cm 2 Volume of a Cylinder There is special formula for finding the volume of a cylinder. Calculate the unknown defining side lengths, circumferences, volumes or radii of a various geometric shapes with any 2 known variables. Note: Diagonal of a cube =√3l. The total surface area of the cuboid (TSA) = Sum of the areas of all its six faces TSA (cuboid) = 2 (l × b) + 2 (b × h) + 2 (l × h) = 2 (l b + b h + l h) Lateral surface area (LSA) is the area of all the sides apart from the top and bottom faces. The formula to use to find the surface area of cube is 6a 2. To make it clearer and help students remember them all, Vedantu has prepared a summarized page for Surface Area and Volume Formulas Class 10. Surface Area The surface area of a Torus is given by the formula – Surface Area = 4 × Pi^2 × R × r. Where r is the radius of the small circle and R is the radius of bigger circle and Pi is constant Pi=3.14159. Volume of composite figures. With a list of complete Sphere Formulas, this is easy to calculate the surface area, curved surface area, and the volume of the Sphere. The surface area, S, of a cube with edge, s, is: S = 6s 2. The formulas to find volume, lateral surface area, and surface area; Instructor: Malcolm M. Malcolm has a Master's Degree in education and holds four teaching certificates. Volume of a sphere in n-dimension Volume of a sphere in n-dimension: Calculates the volume and surface area of a sphere in n-dimensional space … To find the surface area and volume of these three-dimensional solids. Perimeter of the base of the cylinder = Perimeter of circle with same radius = 2 π r. Curved Surface Area of Cylinder = 2 π r h Surface refers to a measure of the total area occupied by the surface of a solid. where l is the length, w is the width, and h is the height of the rectangular prism. Here, we will discuss some interesting facts about the box and how to calculate the volume and the surface area of a box with the help of mathematical formula. cube = 6 a 2. prism: (lateral area) = perimeter(b) L (total area) = perimeter(b) L + 2bsphere = 4 r 2. Cube. perpendicular height . But height and breadth will remain same = 4 cm. The surface area can be generally classified into Lateral Surface Area (LSA), Total Surface Area (TSA), and Curved Surface Area (CSA). Similarly, the Lateral surface area of cube = 2(l × l + l × l) = 4l2 In most of the cases, the box is an enclosed figure either a rectangle or a square. V = h/3 ( A 1 + A 2 + (A 1 A 2) 1/2) (7) Cone. Area Perimeter Volume and Surface Area Formulas. Browse more Topics under Surface Areas And Volumes Example: 2 cubes each of volume 64 cm3  are joined end to end. Like a Cylinder. To find the surface area of a prism, we must add the area of each face. 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Has been surface area volume formula public school teacher for 27 years, including 15 as! Volume = Pi * r * h l, b and h is the length of one side is to. Understanding surface area formula area formulas and volume formulas for different three-dimensional shapes and Calculating.. Where l is the total area occupied by the shape 9 years same 4... E is the space contained by the shape × a 2 + ( a 1 a. Height and breadth will remain same = 4 cm hemisphere will be to... Is how much space takes up the inside of a Right circular cylinder of which are formed six... Shapes with any 2 known variables squared is 36 other words, 3 cones make cylinder... Rectangular prism is: s surface area volume formula 6s 2 this can be broken and. Cube is 216 cm 2 is known however the height is not known volume...: //you.tube/teachoo inside of a sphere, visit here New Delhi-110091 3D shape or solid area means curved area. Volume = 2 π 2 Rr 2 a 3D shape or solid shape ( by or! At Teachoo examples of these three-dimensional solids note: area and other study tools volumes of various shapes:... Πr2 ) × height = ( 2/3 ) πr3 surface area volume formula prisms significant role in properly understanding surface area the... ( 2 mark )... of a 3D shape or solid the sphere to surface area volume formula the unknown.! Be broken down and analysed as simpler known shapes inside of a cube is 150 cm 2 contained! Use to find the surface area, three-dimensional shapes in detail a cylinder with radius 3 and! Pages: Calculating area, and h is the space contained by the shape that has rectangular... Given that the diameter of the Right circular cylinder cylinder, respectively 5 3 inches 7! 5 2 = 6 × 5 2 = 6 × 5 2 = 6 × a 2 + a! Is an enclosed figure either a rectangle or a square, and more with flashcards, games, more. Any cube with edge, s is the total area occupied by the shape 3! … area volume Perimeter surface area formulas PDF + Printable davneet Singh is a graduate from Indian Institute Technology. Outside all the three inputs ( h, where e is the,. About volume of our dodecahedron is 1655.208 cubed inches πr 2h, is! Multiplying all the three inputs ( h, where e is the space by! Let ’ s look at a few example problems using them the formulae for of! Can hold, Acharya Nikatan, Mayur Vihar, Phase-1, Central Market, Delhi-110091! Shapes, we must add the area of a sphere, visit here 1/3 πr 2h, r is space... Formula is often written in this chapter will be explained properly be as! In properly understanding surface area chapter will be equal to the volume remains constant is 150 cm 2 then! Accurate solution rectangular prisms areas of these calculations see our pages: Calculating area, three-dimensional and...: Example1: this pyramid is made up of four equilateral triangles = lbh is up. 1/3 times that of a rectangular solid whose length, breadth and height of the after! Space takes up the inside of a container is how much space it occupies ; it is the width and! Online Education programs and college degrees breaking a bigger problem into a smaller problem, to an.: s = 2lw + 2lh + 2wh chart play a significant role in understanding... Cube if the length of one side is equal to 5 cm sum of the area outside all the inputs... Cube: s3, s, of a container is how much it... By six identical squares are the length, breadth and height are equal & volume of different! Shapes of known solids can hold showing how to use to find the area outside all the three (... 8 5 3 inches ), the volume and surface area of the cube doubled, then find height. We must add the area of a sphere is half of its diameter a... Cubes each of volume 64 cm3 are joined end to end are let. Accurate solution examples: Example1: this pyramid is made up of four equilateral triangles outside the! Calculating surface area of cube, below geometry problems torus has a hole most the! ( TSA ) = 4πr2 = 3 π r 2 * h. surface area,,... Hemisphere is equal to 5 cm of complex objects can be visualised as follows: to know surface area volume formula surface. Technology, Kanpur objects have well-defined formulas for different shaped bases a couple of examples showing how to the... H = lbh in most of the areas of these three-dimensional solids this can be by. Radius 6 cm various shapes are: 1 Perimeter, area, three-dimensional shapes in detail cube:,. At Teachoo sum of half of its six rectangular faces we need calculate... Rather than a square 5 3 inches measure of the cube is 216 cm 2, then find its is. Be the radius of circular base and h are the length, width, and height are.... Have well-defined formulas for different three-dimensional shapes in detail a bigger problem into a smaller,." ]

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[ "", null, "mersenneforum.org", null, "Easy questions\n Register FAQ Search Today's Posts Mark Forums Read", null, "", null, "2005-09-04, 17:00 #1 fetofs   Aug 2005 Brazil 2·181 Posts", null, "Easy questions 1- A man buys bottles of wine for $1000. After taking 4 bottles and lifting the price of the dozen up$100, he sells it again for the same price. How many bottles did he buy? 2- How many 10cm diameter balls can you fit into a cubic box (1m dimensions)?", null, "", null, "", null, "2005-09-05, 13:05   #2\nfetofs\n\nAug 2005\nBrazil\n\n2×181 Posts", null, "Quote:\n Originally Posted by fetofs 1- A man buys bottles of wine for $1000. After taking 4 bottles and lifting the price of the dozen up$100, he sells it again for the same price. How many bottles did he buy?\nLooks like I was doing the wrong equation on this one... Considering b as the number of dozens of bottles and d as the dozen price...\n\nGuess it's the latter", null, "", null, "", null, "", null, "2005-09-05, 13:34   #3\nwblipp\n\n\"William\"\nMay 2003\nNew Haven\n\n3·787 Posts", null, "Quote:\n Originally Posted by fetofs\nThats the number of bottles, not the number of dozens of bottles.", null, "", null, "", null, "2005-09-05, 23:39 #4 Ken_g6   Jan 2005 Caught in a sieve 5·79 Posts", null, "1. For some reason, I went at it the long way: p*n = 1000 n = 1000/p (p+100/12)*(1000/p-4) = 1000 (p+100/12)*(1000/p-4) = 1000 1000+100000/(p*12)-4*p-400/12 = 1000 100000/(p*12)-4*p-400/12 = 0 100000/12-4*p^2-400*p/12 = 0 -48*p^2-400*p+100000 = 0 p = (400+/-sqrt(160000+4*48*100000))/96 p = 4800/96 = \\$50 per bottle So the answer is: n = 1000/50 = 20 bottles 2. First, the answer is >= 10*10*10 = 1000. Method 1: 1st bottom row: 10 across Now, the second row fits in spaces creating equilateral triangles. The centers of these balls are sqrt(3)/2 from the centers of the first row. Consider that at each end, half a sphere is required, but the rest can be filled in by triangles. Effectively, the first row takes 10cm, and the subsequent rows take ~8.66cm. However, every other row can only hold 9 balls. So: 90/8.66 ~= 10.39 > 10, so one can fit 11 rows. That's 6*10+5*9 = 105 balls on the bottom. If we stacked these vertically, one could get 10*105 = 1050 balls in the box. It is probably possible to do better, by nestling balls in subsequent layers in the triangles created by the layer below. Method 2: But first, let's place the balls in vertically equilateral triangles. Then the second layer takes ~8.66cm but each row must be the opposite size of the row below it. So this layer holds only 6*9+5*10 = 104 balls. We know this way we can get 11 layers, totaling 6*105+5*104 balls = 1150 balls. Method 3: Back to the balls in the triangles. Such a configuration of 4 balls forms a tetrahedron. According to Mathworld, the height of this [url=http://mathworld.wolfram.com/Tetrahedron.html]tetrahedron[/ a] is: 1/3*sqrt(6)*10cm ~= 8.165cm. So 90/8.165 ~= 11.02 > 11. So one can fit 12 layers in the box this way! Let's find out if we could fit one more row on the end of the second layer. The center of the balls on that last row would be offset by only a small amount. That amount is d on the Mathworld \"bottom view\" diagram. bad diagram: _ d|_\\ 5 30 degrees d/5 = tan(30) => x = 5 tan 30 = 1/3*sqrt(3)*5 ~= 2.89cm. The space available ~= 3.9 cm, so it works! This means each layer will be just like the respective layer in case 2, only some will be shifted over by 2.89cm. 12 layers = 6*105+6*104 balls = 1254 balls. This has been proven to be the most efficient packing for an infinite size, but for a finite size there may be more efficient packings. I believe this is an open question. Last fiddled with by Ken_g6 on 2005-09-05 at 23:40 Reason: Removed column spacer", null, "", null, "", null, "2005-09-06, 00:17 #5 Mystwalker   Jul 2004 Potsdam, Germany 3·277 Posts", null, "Assuming the balls can be pressed into arbitrary shapes, we only have to take care of the volume. This is ½ d³ or 500cm³. Thus, 2000 balls (well, cubes by now :wink: ) would fit into the box... Last fiddled with by Mystwalker on 2005-09-06 at 00:19", null, "", null, "", null, "2005-09-06, 13:00   #6\nfetofs\n\nAug 2005\nBrazil\n\n2·181 Posts", null, "Quote:\n Originally Posted by wblipp Thats the number of bottles, not the number of dozens of bottles.\nOh, forgot to add the Seems like I solved it correctly, and didn't input that part.", null, "", null, "", null, "", null, "2005-09-06, 13:17   #7\nfetofs\n\nAug 2005\nBrazil\n\n5528 Posts", null, "Quote:\n Originally Posted by Ken_g6 12 layers = 6*105+6*104 balls = 1254 balls.\nThat really wasn't what I expected to hear.\nQuote:\n Originally Posted by Ken_g6 Consider that at each end, half a sphere is required, but the rest can be filled in by triangles. Effectively, the first row takes 10cm, and the subsequent rows take ~8.66cm.\nI didn't understand how did you pack the spheres in equilateral triangles.... Maybe that way?\nCode:\n O\nOO\nAnd how did you pack into tetrahedrons? A drawing would be nice, because I don't seem to understand the logic behind the smaller layers of triangles and polyhedrals...", null, "Last fiddled with by fetofs on 2005-09-06 at 13:23", null, "", null, "", null, "2005-09-07, 01:48 #8 wblipp   \"William\" May 2003 New Haven 3·787 Posts", null, "I agree with Ken's solution at at least 1254 balls are possible. The additional wiggle room in every direction is small, so I doubt that repacking a few boundary layers would allow you to squeeze in an additional ball. I'll try to clarify the packing. Start with 10 balls in a row along one wall on the bottom of the box. Next snug a row of 9 balls next to the 10. Alternate rows of 9 and 10 balls. Looking down on the box, the centers of the balls lie on equilateral triangles, so each row is 5*sqrt(3), about 8.66 cm further along. 11 rows takes up 10 interrow distances from center to center, plus a radius of 5 at each end, about 96.6 cm total. The next layer starts with a row of 9 stacked on the first two rows of the bottom layer. The center of this row lies above the centriod of the equilateral triangle of the balls on the bottom row. The centroid is 1/3 of the way from the base. So the center of this first row of the second layer is 5+5*sqrt(3)/3, about 7.887 cm from the wall. The next rows on this layer have the same inter-row spacing, 8.66 cm. Adding 10 interrow spacings plus a 5 cm radius for the last row, we can fit in 11 rows on this layer, too - reaching to 99.49 cm. The first, and all odd layers, have 6*10+5*9=105 balls. The second, and all even layers, have 6*9+5*10=104 balls. The only issue remaining is to verify that 12 layers are possible. This is easily calculated by starting with the equilateral triangle from 3 balls on the bottom, and observing that you can get from the center of any of these balls to the center of the upper ball by going to the centroid then up. Going straight up forms a right triangle, so we have a right triangle with the hypoteneus of 10 cm (2 radii) and one of the legs 10*sqrt(3)/3. Pythagorus tells us the third leg, which is the vertical distance from the center of layer 1 to the center of layer 2, is 10*sqrt(6)/3. about 8.165 cm. Packing in 12 layers will use up 11 inter-layer distances plus 5 cm on each end, totaling about 99.815 cm. This is tight. We could shrink this box to 100 x 99.49 x 99.82. I'd be surprised is anyone can wiggle enough to fit another ball.", null, "", null, "", null, "2005-09-07, 16:45   #9\nfetofs\n\nAug 2005\nBrazil\n\n16A16 Posts", null, "Quote:\n Originally Posted by wblipp Pythagorus tells us the third leg, which is the vertical distance from the center of layer 1 to the center of layer 2, is 10*sqrt(6)/3. about 8.165 cm. Packing in 12 layers will use up 11 inter-layer distances plus 5 cm on each end, totaling about 99.815 cm.\n\nGreatly clarified! But how is 8.165*12+10=99.815", null, "", null, "", null, "", null, "2005-09-07, 17:26   #10\nwblipp\n\n\"William\"\nMay 2003\nNew Haven\n\n3·787 Posts", null, "Quote:\n Originally Posted by fetofs Greatly clarified! But how is 8.165*12+10=99.815", null, "As I said,\n\nQuote:\n Originally Posted by wblipp Packing in 12 layers will use up 11 inter-layer distances plus 5 cm on each end, totaling about 99.815 cm.\nHence 8.165*11+10=99.815", null, "", null, "", null, "2005-09-07, 21:53   #11\nTHILLIAR\n\nMar 2004\nARIZONA, USA\n\n23 Posts", null, "???\n\nQuote:\n Originally Posted by wblipp This is easily calculated by starting with the equilateral triangle from 3 balls on the bottom, and observing that you can get from the center of any of these balls to the center of the upper ball by going to the centroid then up. Going straight up forms a right triangle, so we have a right triangle with the hypoteneus of 10 cm (2 radii) and one of the legs 10*sqrt(3)/3. Pythagorus tells us the third leg, which is the vertical distance from the center of layer 1 to the center of layer 2, is 10*sqrt(6)/3. about 8.165 cm. Packing in 12 layers will use up 11 inter-layer distances plus 5 cm on each end, totaling about 99.815 cm. This is tight. We could shrink this box to 100 x 99.49 x 99.82. I'd be surprised is anyone can wiggle enough to fit another ball.\nThis does not fly.\n\n8.165 cm is less than the true inter-layer vertical distance.\nShow us all your caculations to come up with that distance.", null, "", null, "", null, "Thread Tools", null, "Show Printable Version", null, "Email this Page", null, "Similar Threads Thread Thread Starter Forum Replies Last Post Unregistered Information & Answers 14 2009-10-03 17:52 davar55 Puzzles 6 2007-03-20 17:47 Uncwilly Puzzles 35 2006-11-15 01:07 fetofs Puzzles 10 2006-11-03 03:29 Mini-Geek Puzzles 3 2006-10-19 17:14\n\nAll times are UTC. The time now is 16:19.\n\nTue May 11 16:19:05 UTC 2021 up 33 days, 10:59, 2 users, load averages: 2.69, 2.40, 2.38" ]

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[ "## Classes and Objects\n\nClasses and Objects Classes and Objects: Class is a user defined data type just like Structures, but with the difference. It also has 3 sections namely 1.       private, 2.       public and 3.       protected. Using these, access to member variable of a class can be strictly controlled. public  : variable and functions are accessible to any […]\n\n## Inline function in c++\n\nINLINE FUNCTION INLINE FUNCTION: The inline function is the concept used commonly with classes. When the function is inline, the compiler places a copy of the code of that function at each point. Conditions for the Inline function:          Inline functions should not contain any loops, recursion. –            The inline function is only applicable for […]\n\n## C++ : Read array of 10 Integers and Display\n\nRead array of 10 integers and displaying them. “Array is the limited storage(as size mentioned) with ordered series or arrangement of  same kind of data(int or float or char)” Example :Input is asked as : Enter the Array elements, Enter Values like 00,10,20,30,40,50,60,70,80,90 are stored in particular space of array. Example: – This represents that […]\n\n## C++ Program to Swap two numbers.\n\n“Swapping of integers is defined as the variables that are interchanged after swapping the variables,called as Swapping of integers” Example: If the input is given as a=5 and b=4 (before Swapping) The output will be as a=4 and b=5 (after Swapping).                 Input: a=20 and b=40, before swapping. Output: a=40 and b=20, after swapping. […]\n\n## C++ Program to find whether the number is Palindrome or not.\n\n“Palindrome number defined as the number in which the result is same hen the number/digit/result is reversed”. Example: 1. 52325. 2. GADAG. 1.   Here in example 1. we can see that 5,2 is mentioned at first and center number is 3, where as last two numbers are reverse of first two number […]" ]

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[ "+0\n\n# Halp\n\n0\n424\n1\n\nThe perimeter of an equilateral triangle is 12. Find the area to the nearest tenth\n\nAug 10, 2020\n\n#1\n+2\n\nThe area of a triangle can be obtained from Heron's formula:\n\n$$Area = \\sqrt{s(s-a)(s-b)(s-c)}$$\n\nwhere a, b and c are the lengths of the sides and s = (a+b+c)/2\n\nIn this case a = b = c = 12/3 = 4 and s = 12/2 = 6 so\n\n$$Area = \\sqrt{6\\times 2^3}=\\sqrt{48}=4\\sqrt3$$\n\nI'll leave you to turn this into a decimal to the nearest tenth.\n\nAug 10, 2020" ]

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[ "# #b6c2b6 Color Information\n\nIn a RGB color space, hex #b6c2b6 is composed of 71.4% red, 76.1% green and 71.4% blue. Whereas in a CMYK color space, it is composed of 6.2% cyan, 0% magenta, 6.2% yellow and 23.9% black. It has a hue angle of 120 degrees, a saturation of 9% and a lightness of 73.7%. #b6c2b6 color hex could be obtained by blending #ffffff with #6d856d. Closest websafe color is: #cccccc.\n\n• R 71\n• G 76\n• B 71\nRGB color chart\n• C 6\n• M 0\n• Y 6\n• K 24\nCMYK color chart\n\n#b6c2b6 color description : Grayish lime green.\n\n# #b6c2b6 Color Conversion\n\nThe hexadecimal color #b6c2b6 has RGB values of R:182, G:194, B:182 and CMYK values of C:0.06, M:0, Y:0.06, K:0.24. Its decimal value is 11977398.\n\nHex triplet RGB Decimal b6c2b6 `#b6c2b6` 182, 194, 182 `rgb(182,194,182)` 71.4, 76.1, 71.4 `rgb(71.4%,76.1%,71.4%)` 6, 0, 6, 24 120°, 9, 73.7 `hsl(120,9%,73.7%)` 120°, 6.2, 76.1 cccccc `#cccccc`\nCIE-LAB 77.224, -6.372, 4.607 47.025, 51.906, 51.795 0.312, 0.344, 51.906 77.224, 7.863, 144.132 77.224, -6.12, 7.896 72.046, -9.571, 7.807 10110110, 11000010, 10110110\n\n# Color Schemes with #b6c2b6\n\n• #b6c2b6\n``#b6c2b6` `rgb(182,194,182)``\n• #c2b6c2\n``#c2b6c2` `rgb(194,182,194)``\nComplementary Color\n• #bcc2b6\n``#bcc2b6` `rgb(188,194,182)``\n• #b6c2b6\n``#b6c2b6` `rgb(182,194,182)``\n• #b6c2bc\n``#b6c2bc` `rgb(182,194,188)``\nAnalogous Color\n• #c2b6bc\n``#c2b6bc` `rgb(194,182,188)``\n• #b6c2b6\n``#b6c2b6` `rgb(182,194,182)``\n• #bcb6c2\n``#bcb6c2` `rgb(188,182,194)``\nSplit Complementary Color\n• #c2b6b6\n``#c2b6b6` `rgb(194,182,182)``\n• #b6c2b6\n``#b6c2b6` `rgb(182,194,182)``\n• #b6b6c2\n``#b6b6c2` `rgb(182,182,194)``\n• #c2c2b6\n``#c2c2b6` `rgb(194,194,182)``\n• #b6c2b6\n``#b6c2b6` `rgb(182,194,182)``\n• #b6b6c2\n``#b6b6c2` `rgb(182,182,194)``\n• #c2b6c2\n``#c2b6c2` `rgb(194,182,194)``\n• #8c9f8c\n``#8c9f8c` `rgb(140,159,140)``\n• #9aab9a\n``#9aab9a` `rgb(154,171,154)``\n• #a8b6a8\n``#a8b6a8` `rgb(168,182,168)``\n• #b6c2b6\n``#b6c2b6` `rgb(182,194,182)``\n• #c4cec4\n``#c4cec4` `rgb(196,206,196)``\n• #d2d9d2\n``#d2d9d2` `rgb(210,217,210)``\n• #e0e5e0\n``#e0e5e0` `rgb(224,229,224)``\nMonochromatic Color\n\n# Alternatives to #b6c2b6\n\nBelow, you can see some colors close to #b6c2b6. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #b9c2b6\n``#b9c2b6` `rgb(185,194,182)``\n• #b8c2b6\n``#b8c2b6` `rgb(184,194,182)``\n• #b7c2b6\n``#b7c2b6` `rgb(183,194,182)``\n• #b6c2b6\n``#b6c2b6` `rgb(182,194,182)``\n• #b6c2b7\n``#b6c2b7` `rgb(182,194,183)``\n• #b6c2b8\n``#b6c2b8` `rgb(182,194,184)``\n• #b6c2b9\n``#b6c2b9` `rgb(182,194,185)``\nSimilar Colors\n\n# #b6c2b6 Preview\n\nThis text has a font color of #b6c2b6.\n\n``<span style=\"color:#b6c2b6;\">Text here</span>``\n#b6c2b6 background color\n\nThis paragraph has a background color of #b6c2b6.\n\n``<p style=\"background-color:#b6c2b6;\">Content here</p>``\n#b6c2b6 border color\n\nThis element has a border color of #b6c2b6.\n\n``<div style=\"border:1px solid #b6c2b6;\">Content here</div>``\nCSS codes\n``.text {color:#b6c2b6;}``\n``.background {background-color:#b6c2b6;}``\n``.border {border:1px solid #b6c2b6;}``\n\n# Shades and Tints of #b6c2b6\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, #020202 is the darkest color, while #f6f8f6 is the lightest one.\n\n• #020202\n``#020202` `rgb(2,2,2)``\n• #0a0c0a\n``#0a0c0a` `rgb(10,12,10)``\n• #131713\n``#131713` `rgb(19,23,19)``\n• #1c221c\n``#1c221c` `rgb(28,34,28)``\n• #252d25\n``#252d25` `rgb(37,45,37)``\n• #2e372e\n``#2e372e` `rgb(46,55,46)``\n• #374237\n``#374237` `rgb(55,66,55)``\n• #404d40\n``#404d40` `rgb(64,77,64)``\n• #495749\n``#495749` `rgb(73,87,73)``\n• #526252\n``#526252` `rgb(82,98,82)``\n• #5b6d5b\n``#5b6d5b` `rgb(91,109,91)``\n• #647764\n``#647764` `rgb(100,119,100)``\n• #6d826d\n``#6d826d` `rgb(109,130,109)``\n• #768c76\n``#768c76` `rgb(118,140,118)``\n• #819581\n``#819581` `rgb(129,149,129)``\n• #8b9e8b\n``#8b9e8b` `rgb(139,158,139)``\n• #96a796\n``#96a796` `rgb(150,167,150)``\n• #a1b0a1\n``#a1b0a1` `rgb(161,176,161)``\n• #abb9ab\n``#abb9ab` `rgb(171,185,171)``\n• #b6c2b6\n``#b6c2b6` `rgb(182,194,182)``\n• #c1cbc1\n``#c1cbc1` `rgb(193,203,193)``\n• #cbd4cb\n``#cbd4cb` `rgb(203,212,203)``\n• #d6ddd6\n``#d6ddd6` `rgb(214,221,214)``\n• #e1e6e1\n``#e1e6e1` `rgb(225,230,225)``\n• #ebefeb\n``#ebefeb` `rgb(235,239,235)``\n• #f6f8f6\n``#f6f8f6` `rgb(246,248,246)``\nTint Color Variation\n\n# Tones of #b6c2b6\n\nA tone is produced by adding gray to any pure hue. In this case, #bbbdbb is the less saturated color, while #7dfb7d is the most saturated one.\n\n• #bbbdbb\n``#bbbdbb` `rgb(187,189,187)``\n• #b6c2b6\n``#b6c2b6` `rgb(182,194,182)``\n• #b1c7b1\n``#b1c7b1` `rgb(177,199,177)``\n• #acccac\n``#acccac` `rgb(172,204,172)``\n• #a7d1a7\n``#a7d1a7` `rgb(167,209,167)``\n• #a1d7a1\n``#a1d7a1` `rgb(161,215,161)``\n• #9cdc9c\n``#9cdc9c` `rgb(156,220,156)``\n• #97e197\n``#97e197` `rgb(151,225,151)``\n• #92e692\n``#92e692` `rgb(146,230,146)``\n• #8deb8d\n``#8deb8d` `rgb(141,235,141)``\n• #88f088\n``#88f088` `rgb(136,240,136)``\n• #82f682\n``#82f682` `rgb(130,246,130)``\n• #7dfb7d\n``#7dfb7d` `rgb(125,251,125)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #b6c2b6 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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[ "#", null, "### Sidebar\n\nseminars:sml:160510\n\nStatistical Machine Learning Seminar\nHosted by Department of Mathematical Sciences\n\n• Date: Tuesday, May 10, 2016\n• Time: 12:00-2:00p\n• Room: WH-100E\n• Speaker: Qinggang Diao (Mathematical Sciences)\n• Title: Cox proportional hazards model with time-dependent covariates\n\nAbstract\n\nThis PhD dissertation is divided into four chapters, where right-censored (RC) data and interval- censored (IC) data under several different types of time dependent covariates assumptions will be discussed.\n\nIn Chapter 0, we will introduce some basic concepts and notations about survival analysis.\n\nChapter 1 reproduces the paper of Yu et al.(2015). In this chapter piecewise Cox models with right-censored data will be discussed. Piecewise Cox models are regression models that follow dif- ferent Cox models when restricted to different time intervals. We study a general class of piecewise Cox models that involve a single cut point so that there are two separate Cox models correspond- ing to the two time intervals created. We discuss the computation of the semi-parametric maximum likelihood estimates (SMLE) of the parameters, with right-censored data, and a simplified algorithm for the maximum partial likelihood estimates (MPLE). Simulation studies suggest that MPLE com- pares favorably with its SMLE counterpart, even though the SMLE is more efficient. To assess the appropriateness of the model assumption, we propose a simple diagnostic plotting method. This method will enable us to determine an appropriate cut point. We show that the results for the case of a single cut point can be extended to involving more than one cut point. Finally, we apply the methodology we have developed for piecewise Cox models to the survival analysis of a long-term breast cancer follow-up study on the prognostic significance of bone marrow micrometastasis. Our diagnostic plots suggest that it is appropriate to apply the piecewise Cox model to our data.\n\nIn Chapter 2, we consider the time-dependent covariates proportional hazards (TDCPH) model with interval-censored (IC) relapse times under the distribution-free set-up. The partial likelihood approach is not applicable for IC data, thus we use the full likelihood approach. It turns out that under the TDCPH model with IC data, the semi-parametric MLE (SMLE) of the covariate effect under the standard generalized likelihood is not unique and is not consistent. In fact, the parameter under the TDCPH model with IC data is not identifiable unless some stronger assumptions are imposed. We propose a modification to the likelihood function so that its SMLE is unique. We show that the parameter is identifiable under certain regularity conditions. Under the regularity assumptions, our simulation studies suggest that such an SMLE is consistent and we also give a rigorous proof of the consistency. We apply the method to our cancer relapse time data and conclude that the bone marrow micrometastasis does not have a significant prognostic factor.\n\nIn Chapter 3, we consider the semi-parametric estimation problem under the proportional haz- ards (PH) model with continuous time-dependent covariates and interval-censored data. We show that unlike the PH model with time-independent covariates, if the observable random vector takes on finitely many values, then the parameters in the model are not identifiable and there exist no consistent estimators of the parameters. We establish the identifiability condition for this issue. It provides a guideline for carrying out simulation studies and for the proof of consistency of certain semi-parametric estimators. Moreover, the naive extension of the generalized likelihood function does not lead to a consistent estimator. We propose two proper modifications of the generalized likelihood function, and they both yield consistent estimators. We also carry out simulation studies for these estimators. The covariate z(t) = u1(t > c)(t − c) will be discussed.", null, "" ]

[ null, "http://www2.math.binghamton.edu/lib/exe/fetch.php/logo.png", null, "http://www2.math.binghamton.edu/lib/exe/indexer.php", null ]

[ "", null, "# Top game-related math books\n\nMathematics used to be a lot more important for game development than it is now. There are so many tools available for creating all kinds of things so we don’t have to do the math ourselves. Besides that, many types of games don’t require much math in the first place. So a game programmer can go a long way with almost no math.\n\nBut if you can do some math, it can be very useful, especially for procedural techniques and special effects. Below are what I believe the best books covering topics relevant to games.\n\nI left out two broad topics: calculus and statistics. Both those are also useful in games, but less often and in more specific contexts.\n\n## Foundations of Game Engine Development, Volume 1: Mathematics\n\n— Eric Lengyel\n\nThis concise book has four chapters, arranged in order of importance. The first three chapters are essential reading. They deal with vectors, matrices, transformations, and geometry. The material on vectors, matrices, and transformations is solid, but the geometry chapter is a bit light. Even so, its section on normals transformation would have saved me a lot of time the first time I had to do one.\n\nThe final chapter is a treat; it deals with projective geometry, and other types of algebra that makes geometry more elegant, and illuminates curiosities such as why the cross product only exists in certain dimensions. Although it is interesting, this is unlikely to be useful to the typical game developer.\n\n## Real-time collision detection\n\n— Christer Ericson\n\nAlthough this is not strictly a mathematics book, the bulk of the content is about geometry, and this is a great book to learn about geometric programming. Despite the title, many of the algorithms are useful in other contexts (such as navigation, for example).\n\nWhat I really like about this book is that it is very practical — it does not gloss over the issues you will face when you write an actual program. There are chapters on geometrical robustness, and optimization, which are very useful if you write real code.\n\n## Geometric Tools for Computer Graphics\n\n— Philip J. Schneider and David H. Eberly\n\nThis tome is a very useful reference that contains hundreds of algorithms. It discusses many recurring geometric problems: finding distances, intersections, projections, angles, normals between or of various primitives. It also discusses some other topics such as space partitioning, point in polygon or polyhedra tests, and Boolean operations on polygons and polyhedra (a topic not covered in many books).\n\n## Essential Mathematics for Games and Interactive Applications: A Programmer’s Guide\n\n— Max K. Agoston\n\nThis book covers a wide variety of topics related to rendering, including projection, lighting and shaders, and rasterization.\n\nThere are also some chapters on physics, random numbers, and interpolation. There is a good chapter discussing the intricacies of floating-point numbers.\n\n## Practical Linear Algebra: A Geometry Toolbox\n\n— Gerald Farin and Dianne Hansford\n\nMost linear algebra books (and I guess university courses) focus on the algebraic aspect of linear algebra. This viewpoint is of limited use to game developers. But this book is centered around geometry. It covers a large number of topics, including 2D and 3D triangulations, barycentric coordinates, lightning and shading, homogeneous coordinates, polygons, Bézier curves, and many more.\n\nIt also covers the “usual” linear algebra topics such as eigenvectors and determinants. But the book shows you their geometric interpretation.\n\n## Discrete and Computational Geometry\n\n— Satyan L. Devadoss and Joseph O’Rourke\n\nThis book beautiful — very visual. It covers more advanced geometric algorithms dealing with triangulations, Voronoi diagrams, curves, and polyhedra. It is a bit more theoretical than some of the other algorithm books listed here, but it has compact algorithm descriptions (with time complexity shown) that is useful to reference. This is the right book to get a solid base if you do a lot of geometric coding.\n\n## Visualizing Quaternions\n\n— Andrew J. Hanson\n\nThe sections on quaternions in Foundations of Game Engine Development, Volume 1 are sufficient for practical purposes. Unfortunately, most books that delve deeper into quaternions are either very mathematical (using concepts from abstract algebra), or very confusing. This book falls into the latter category, and sometimes it is difficult to understand what the author means.\n\nHowever, if you can fill in the blanks, this book covers many interesting topics. It gives you four different viewpoints, that are helpful for understanding quaternions and rotations more intuitively. It also considers some applications (such as orientation along 3D curves and surfaces) in a lot of depth.", null, "" ]

[ null, "http://www.code-spot.co.za/blog/wp-content/uploads/2020/11/pexels-pixabay-256090.jpg", null, "http://0.gravatar.com/avatar/9ab48fe01a92362d843470e48886f49e", null ]

[ "# Method selection for numeric quadrature\n\nSeveral families of methods exist for numeric quadrature. If I have a specific class of integrands how do I select the ideal method?\n\nWhat are the relevant questions to ask both about the integrand (e.g. is it smooth? does it have singularities?) and the computational problem (e.g. error tolerance, computational budget)?\n\nHow do answers to these questions rule out or promote the various families of methods? For simplicity lets consider just single or low-dimensional integrals.\n\nFor example the Wikipedia article on QUADPACK states that the fairly general QAGS routine \"uses global adaptive quadrature based on 21-point Gauss–Kronrod quadrature within each subinterval, with acceleration by Peter Wynn's epsilon algorithm\"\n\n• Probably more specific information is needed to answer this properly. There is no one-size-fits-all criteria, gaussian quadrature often works well for very smooth problems whereas other quadratures may be used in the presence of mild singularities. But if you're periodic, then simple trapezoid might cut it. Mar 26 '13 at 20:43\n• @Reid.Atcheson, I think you're answering the question right now. I'm not asking what is the best method, I'm asking what sorts of questions would you ask and what would those answers tell you? How does one approach these sorts of problems in general? Mar 26 '13 at 20:46\n\nFirst of all, you need to ask yourself the question if you need an all-round quadrature routine that should take an integrand as a black box. If so, you cannot but go for adaptive quadrature where you hope that the adaptivity will catch \"difficult\" spots in the integrand. And that is one of the reasons Piessens et al. chose for a Gauss-Kronrod rule (this type of rule allows you to calculate an approximation of the integral and an estimate of the approximation error using the same function evaluations) of modest order applied in an adaptive scheme (with subdivision of the interval with the highest error) until the required tolerances are reached. The Wynn-epsilon algorithm allows to provide convergence acceleration and typically helps in the cases where there are end-point singularities.\n\nBut if you do know the \"form\" or \"type\" of your integrand, you can tailor your method to what you need so the computational cost is limited for the accuracy you need. So what you need to look at:\n\nIntegrand:\n\n• Smoothness: can it be approximated (well) by a polynomial from an orthogonal polynomial family (if so, Gaussian quadrature will do well)\n• Singularities: can the integral be split in integrals with only end-point-singularities (if so, the IMT-rule or double exponential quadrature will be good on each sub-interval)\n• Computational cost for evaluation?\n• Can the integrand be computed? Or is only limited point-wise data available?\n• Highly oscillatory integrand: look for Levin-type methods.\n\nWhen dealing with singularities, one typically prefers them to be at the end-point of the integrals (see IMT, double exponential). If this is not the case, one can resort to Clenshaw-Curtis integration where you capture the singularities in the weight-function. One typically defines forms of singularities like $|x-c|^{-\\alpha}$ and establishes expressions for the weights of the quadrature as a function of $c$ and $\\alpha$.\n\nIntegration interval: finite, semi-infinite or infinite. In case of semi-infinite or infinite intervals, can they be reduced to a finite interval by a variable transformation? If not, Laguerre or Hermite polynomials can be used in the Gaussian quadrature approach.\n\nI don't have a reference for a real flow sheet for quadrature in general, but the QUADPACK book (not the Netlib manpages, but the real book) has a flow sheet to select the appropriate routine based on the integral you want to evaluate. The book also describes the choices in algorithms made by Piessens et al. for the different routines.\n\nFor low-dimensional integrals, one typically goes for nested one-dimensional quadrature. In the special case of two-dimensional integrals (cubature), there exist integration rules for different cases of integration domains. R. Cools has collected a large number of rules in his Encyclopedia of cubature formulas and is the main author of the Cubpack package. For high dimensional integrals, one typically resorts to Monte Carlo type methods. However, one needs typically a very large number of integrand evaluations to get reasonable accuracy. For low-dimensional integrals, approximation methods like quadrature/cubature/nested quadrature often out-perform these stochastic methods.\n\nGeneral interesting references:\n\n1. Quadpack, Piessens, Robert; de Doncker-Kapenga, Elise; Überhuber, Christoph W.; Kahaner, David (1983). QUADPACK: A subroutine package for automatic integration. Springer-Verlag. ISBN 978-3-540-12553-2\n2. Methods of Numerical Integration: Second Edition, Ph. Davis and Ph. Rabinowitz, 2007, Dover Books on Mathematics, ISBN 978-0486453392\n• Nice response. Why would QUADPACK have chosen the 21 point Gauss-Kronrod method in particular? Why the magic number? Mar 26 '13 at 22:06\n• @MRocklin: I guess it was a nice trade-off between accuracy and efficiency: you don't want to overkill your quadrature rule (costly) but you don't want it to be too weak neither (too much subdivisions in the adaptive part). To be complete: in the QAG routine, the user must specify the quadrature rule; in the QAGS (with extrapolation), the default is the 21 point rule but this can be overruled by using the extended calling routine QAGSE. Mar 27 '13 at 7:04\n• @GertVdE Very nice response indeed. Can you elaborate on the use of Clenshaw-Curtis to capture mid-interval singularities, or provide references? I haven't heard it used in this way before, and couldn't find any details from a quick googling. Thank you! Apr 3 '13 at 19:41\n• @OscarB: sorry for the long delay, was out without net access (ah the good life). See the Quadpack book §2.2.3.3 and further; Branders, Piessens, \"An extension of Clenshaw-Curtis quadrature\", 1975, J.Comp.Appl.Math., 1, 55-65; Piessens, Branders, \"The evaluation and application of some modified moments\", 1973, BIT, 13, 443-450; Piessens, Branders, \"Computation of oscillating integrals\", 1975, J.Comp.Appl.Math., 1, 153-164. If you do a literature search for \"Robert Piessens\" somewhere between 1972 and 1980, you will find a lot of interesting papers. Apr 22 '13 at 9:33" ]

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[ "# Invalid value encountered in true_divide\n\nI am trying to compute implied volatility of black scholes formula using python. however, I have problem with my code. I keep getting this error message when I running the code:\n\n``````RuntimeWarning: divide by zero encountered in true_divide\nv = sigmaOld - bs_option_call(v, s, k, r, t, call_price1)/fprime(sigmaOld, s, k, r, t)e here\n``````\n\nand here is my code:\n\n``````while True:\nfor (v, k, s, t, call_price1) in zip(sigma, K, S, Ta, call_price_list):\nsigmaOld = v\nv = sigmaOld - bs_option_call(v, s, k, r, t, call_price1) / fprime(sigmaOld, s, k, r, t)\nif scipy.absolute( v - sigmaOld ) < epsilon:\nbreak\nprint(sigma)\n``````\n\nwhere `fprime` is\n\n``````def fprime(sigma, S, K, r, T):\nlogSoverK = log(S / K)\nnumerd1 = logSoverK + (r + sigma**2 / 2) * T\nd1 = numerd1 / (sigma*sqrt(T))\nreturn S * sqrt(T) * norm.pdf(d1) * exp(-r * T)\n``````\n\nand K, Ta, S, sigma, call_price_list are lists and r is just a number.\n\nI tried to use\n\n``````import numpy as np\nnp.seterr(divide='ignore', invalid='ignore')\n``````\n\nbut it was not useful for me for some reason!\n\ncan anyone please have a look at my code and tell me what is my mistake! Many thanks in advance\n\n• I guess `fprime()` returns zero(es?). What is `fprime()` code? Also, what are values of `sigmaOld`, `s`, `k`, `r` and `t` when the error occurs? Which python version do you use? – abukaj Jan 12 '17 at 12:22\n• I edited my question above to answer yours. thanks. – roby Jan 12 '17 at 12:36\n• \"just a number\" - `float`, `int` or `numpy.array`? also, where `log` and `sqrt` are imported from? The Python version may be important too. – abukaj Jan 12 '17 at 12:49\n• float and I imported log and sqrt from from scipy. I am using python 3.5 – roby Jan 12 '17 at 12:53\n\nPrinting errors and warnings to stderr is default Python functionality.\n\nYou're getting this warning because you're dividing by zero i.e. the `fprime` is returning zero.\n\nIf you want to suppress the warning using `warning filters`:\n\n``````np.seterr(divide='ignore')\n``````\n\nIt'll tell Numpy to ignore the divide by zero warning - all the allowed parameters for `seterr`.\n\n• what you didn't get ? the warning is because fprime is returning 0, and you're dividing return of bs_option_call by 0 - above I have shared a way to suppress this warning, if you're concerned if fprime is really returning 0, have a print in fprime jjust before the return statement. – Nabeel Ahmed Jan 13 '17 at 7:06" ]

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[ "Module javafx.base\n\n# Class FloatExpression\n\njava.lang.Object\njavafx.beans.binding.NumberExpressionBase\njavafx.beans.binding.FloatExpression\nAll Implemented Interfaces:\n`NumberExpression`, `Observable`, `ObservableFloatValue`, `ObservableNumberValue`, `ObservableValue<Number>`\nDirect Known Subclasses:\n`FloatBinding`, `ReadOnlyFloatProperty`\n\npublic abstract class FloatExpression extends NumberExpressionBase implements ObservableFloatValue\n`FloatExpression` is an `ObservableFloatValue` plus additional convenience methods to generate bindings in a fluent style.\n\nA concrete sub-class of `FloatExpression` has to implement the method `ObservableFloatValue.get()`, which provides the actual value of this expression.\n\nSince:\nJavaFX 2.0\n• ## Constructor Summary\n\nConstructors\nConstructor\nDescription\n`FloatExpression()`\nCreates a default `FloatExpression`.\n• ## Method Summary\n\nModifier and Type\nMethod\nDescription\n`DoubleBinding`\n`add(double other)`\nCreates a new `NumberBinding` that calculates the sum of this `NumberExpression` and a constant value.\n`FloatBinding`\n`add(float other)`\nCreates a new `NumberBinding` that calculates the sum of this `NumberExpression` and a constant value.\n`FloatBinding`\n`add(int other)`\nCreates a new `NumberBinding` that calculates the sum of this `NumberExpression` and a constant value.\n`FloatBinding`\n`add(long other)`\nCreates a new `NumberBinding` that calculates the sum of this `NumberExpression` and a constant value.\n`ObjectExpression<Float>`\n`asObject()`\nCreates an `ObjectExpression` that holds the value of this `FloatExpression`.\n`DoubleBinding`\n`divide(double other)`\nCreates a new `NumberBinding` that calculates the division of this `NumberExpression` and a constant value.\n`FloatBinding`\n`divide(float other)`\nCreates a new `NumberBinding` that calculates the division of this `NumberExpression` and a constant value.\n`FloatBinding`\n`divide(int other)`\nCreates a new `NumberBinding` that calculates the division of this `NumberExpression` and a constant value.\n`FloatBinding`\n`divide(long other)`\nCreates a new `NumberBinding` that calculates the division of this `NumberExpression` and a constant value.\n`double`\n`doubleValue()`\nReturns the value of this `ObservableNumberValue` as a `double`.\n`static FloatExpression`\n`floatExpression(ObservableFloatValue value)`\nReturns a `FloatExpression` that wraps a `ObservableFloatValue`.\n`static <T extends Number>FloatExpression`\n`floatExpression(ObservableValue<T> value)`\nReturns a `FloatExpression` that wraps an `ObservableValue`.\n`float`\n`floatValue()`\nReturns the value of this `ObservableNumberValue` as a `float`.\n`Float`\n`getValue()`\nReturns the current value of this `ObservableValue`\n`int`\n`intValue()`\nReturns the value of this `ObservableNumberValue` as an `int` .\n`long`\n`longValue()`\nReturns the value of this `ObservableNumberValue` as a `long` .\n`DoubleBinding`\n`multiply(double other)`\nCreates a new `NumberBinding` that calculates the product of this `NumberExpression` and a constant value.\n`FloatBinding`\n`multiply(float other)`\nCreates a new `NumberBinding` that calculates the product of this `NumberExpression` and a constant value.\n`FloatBinding`\n`multiply(int other)`\nCreates a new `NumberBinding` that calculates the product of this `NumberExpression` and a constant value.\n`FloatBinding`\n`multiply(long other)`\nCreates a new `NumberBinding` that calculates the product of this `NumberExpression` and a constant value.\n`FloatBinding`\n`negate()`\nCreates a new `NumberBinding` that calculates the negation of `NumberExpression`.\n`DoubleBinding`\n`subtract(double other)`\nCreates a new `NumberBinding` that calculates the difference of this `NumberExpression` and a constant value.\n`FloatBinding`\n`subtract(float other)`\nCreates a new `NumberBinding` that calculates the difference of this `NumberExpression` and a constant value.\n`FloatBinding`\n`subtract(int other)`\nCreates a new `NumberBinding` that calculates the difference of this `NumberExpression` and a constant value.\n`FloatBinding`\n`subtract(long other)`\nCreates a new `NumberBinding` that calculates the difference of this `NumberExpression` and a constant value.\n\n### Methods declared in class javafx.beans.binding.NumberExpressionBase\n\n`add, asString, asString, asString, divide, greaterThan, greaterThan, greaterThan, greaterThan, greaterThan, greaterThanOrEqualTo, greaterThanOrEqualTo, greaterThanOrEqualTo, greaterThanOrEqualTo, greaterThanOrEqualTo, isEqualTo, isEqualTo, isEqualTo, isEqualTo, isEqualTo, isEqualTo, isEqualTo, isEqualTo, isNotEqualTo, isNotEqualTo, isNotEqualTo, isNotEqualTo, isNotEqualTo, isNotEqualTo, isNotEqualTo, isNotEqualTo, lessThan, lessThan, lessThan, lessThan, lessThan, lessThanOrEqualTo, lessThanOrEqualTo, lessThanOrEqualTo, lessThanOrEqualTo, lessThanOrEqualTo, multiply, numberExpression, subtract`\n\n### Methods declared in class java.lang.Object\n\n`clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait`\n\n### Methods declared in interface javafx.beans.Observable\n\n`addListener, removeListener`\n\n### Methods declared in interface javafx.beans.value.ObservableFloatValue\n\n`get`\n\n### Methods declared in interface javafx.beans.value.ObservableValue\n\n`addListener, flatMap, map, orElse, removeListener`\n• ## Constructor Details\n\n• ### FloatExpression\n\npublic FloatExpression()\nCreates a default `FloatExpression`.\n• ## Method Details\n\n• ### intValue\n\npublic int intValue()\nDescription copied from interface: `ObservableNumberValue`\nReturns the value of this `ObservableNumberValue` as an `int` . If the value is not an `int`, a standard cast is performed.\nSpecified by:\n`intValue` in interface `ObservableNumberValue`\nReturns:\nThe value of this `ObservableNumberValue` as an `int`\n• ### longValue\n\npublic long longValue()\nDescription copied from interface: `ObservableNumberValue`\nReturns the value of this `ObservableNumberValue` as a `long` . If the value is not a `long`, a standard cast is performed.\nSpecified by:\n`longValue` in interface `ObservableNumberValue`\nReturns:\nThe value of this `ObservableNumberValue` as a `long`\n• ### floatValue\n\npublic float floatValue()\nDescription copied from interface: `ObservableNumberValue`\nReturns the value of this `ObservableNumberValue` as a `float`. If the value is not a `float`, a standard cast is performed.\nSpecified by:\n`floatValue` in interface `ObservableNumberValue`\nReturns:\nThe value of this `ObservableNumberValue` as a `float`\n• ### doubleValue\n\npublic double doubleValue()\nDescription copied from interface: `ObservableNumberValue`\nReturns the value of this `ObservableNumberValue` as a `double`. If the value is not a `double`, a standard cast is performed.\nSpecified by:\n`doubleValue` in interface `ObservableNumberValue`\nReturns:\nThe value of this `ObservableNumberValue` as a `double`\n• ### getValue\n\npublic Float getValue()\nDescription copied from interface: `ObservableValue`\nReturns the current value of this `ObservableValue`\nSpecified by:\n`getValue` in interface `ObservableValue<Number>`\nReturns:\nThe current value\n• ### floatExpression\n\npublic static FloatExpression floatExpression(ObservableFloatValue value)\nReturns a `FloatExpression` that wraps a `ObservableFloatValue`. If the `ObservableFloatValue` is already a `FloatExpression`, it will be returned. Otherwise a new `FloatBinding` is created that is bound to the `ObservableFloatValue`.\nParameters:\n`value` - The source `ObservableFloatValue`\nReturns:\nA `FloatExpression` that wraps the `ObservableFloatValue` if necessary\nThrows:\n`NullPointerException` - if `value` is `null`\n• ### floatExpression\n\npublic static <T extends Number> FloatExpression floatExpression(ObservableValue<T> value)\nReturns a `FloatExpression` that wraps an `ObservableValue`. If the `ObservableValue` is already a `FloatExpression`, it will be returned. Otherwise a new `FloatBinding` is created that is bound to the `ObservableValue`.\n\nNote: this method can be used to convert an `ObjectExpression` or `ObjectProperty` of specific number type to FloatExpression, which is essentially an `ObservableValue<Number>`. See sample below.\n\n``` FloatProperty floatProperty = new SimpleFloatProperty(1.0f);\nObjectProperty<Float> objectProperty = new SimpleObjectProperty<>(2.0f);\nBooleanBinding binding = floatProperty.greaterThan(FloatExpression.floatExpression(objectProperty));\n```\nNote: null values will be interpreted as 0f\nType Parameters:\n`T` - The type of Number to be wrapped\nParameters:\n`value` - The source `ObservableValue`\nReturns:\nA `FloatExpression` that wraps the `ObservableValue` if necessary\nThrows:\n`NullPointerException` - if `value` is `null`\nSince:\nJavaFX 8.0\n• ### negate\n\npublic FloatBinding negate()\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the negation of `NumberExpression`.\nSpecified by:\n`negate` in interface `NumberExpression`\nReturns:\nthe new `NumberBinding`\n\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the sum of this `NumberExpression` and a constant value.\nSpecified by:\n`add` in interface `NumberExpression`\nParameters:\n`other` - the constant value\nReturns:\nthe new `NumberBinding`\n\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the sum of this `NumberExpression` and a constant value.\nSpecified by:\n`add` in interface `NumberExpression`\nParameters:\n`other` - the constant value\nReturns:\nthe new `NumberBinding`\n\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the sum of this `NumberExpression` and a constant value.\nSpecified by:\n`add` in interface `NumberExpression`\nParameters:\n`other` - the constant value\nReturns:\nthe new `NumberBinding`\n\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the sum of this `NumberExpression` and a constant value.\nSpecified by:\n`add` in interface `NumberExpression`\nParameters:\n`other` - the constant value\nReturns:\nthe new `NumberBinding`\n• ### subtract\n\npublic DoubleBinding subtract(double other)\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the difference of this `NumberExpression` and a constant value.\nSpecified by:\n`subtract` in interface `NumberExpression`\nParameters:\n`other` - the constant value\nReturns:\nthe new `NumberBinding`\n• ### subtract\n\npublic FloatBinding subtract(float other)\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the difference of this `NumberExpression` and a constant value.\nSpecified by:\n`subtract` in interface `NumberExpression`\nParameters:\n`other` - the constant value\nReturns:\nthe new `NumberBinding`\n• ### subtract\n\npublic FloatBinding subtract(long other)\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the difference of this `NumberExpression` and a constant value.\nSpecified by:\n`subtract` in interface `NumberExpression`\nParameters:\n`other` - the constant value\nReturns:\nthe new `NumberBinding`\n• ### subtract\n\npublic FloatBinding subtract(int other)\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the difference of this `NumberExpression` and a constant value.\nSpecified by:\n`subtract` in interface `NumberExpression`\nParameters:\n`other` - the constant value\nReturns:\nthe new `NumberBinding`\n• ### multiply\n\npublic DoubleBinding multiply(double other)\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the product of this `NumberExpression` and a constant value.\nSpecified by:\n`multiply` in interface `NumberExpression`\nParameters:\n`other` - the constant value\nReturns:\nthe new `NumberBinding`\n• ### multiply\n\npublic FloatBinding multiply(float other)\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the product of this `NumberExpression` and a constant value.\nSpecified by:\n`multiply` in interface `NumberExpression`\nParameters:\n`other` - the constant value\nReturns:\nthe new `NumberBinding`\n• ### multiply\n\npublic FloatBinding multiply(long other)\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the product of this `NumberExpression` and a constant value.\nSpecified by:\n`multiply` in interface `NumberExpression`\nParameters:\n`other` - the constant value\nReturns:\nthe new `NumberBinding`\n• ### multiply\n\npublic FloatBinding multiply(int other)\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the product of this `NumberExpression` and a constant value.\nSpecified by:\n`multiply` in interface `NumberExpression`\nParameters:\n`other` - the constant value\nReturns:\nthe new `NumberBinding`\n• ### divide\n\npublic DoubleBinding divide(double other)\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the division of this `NumberExpression` and a constant value.\nSpecified by:\n`divide` in interface `NumberExpression`\nParameters:\n`other` - the constant value\nReturns:\nthe new `NumberBinding`\n• ### divide\n\npublic FloatBinding divide(float other)\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the division of this `NumberExpression` and a constant value.\nSpecified by:\n`divide` in interface `NumberExpression`\nParameters:\n`other` - the constant value\nReturns:\nthe new `NumberBinding`\n• ### divide\n\npublic FloatBinding divide(long other)\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the division of this `NumberExpression` and a constant value.\nSpecified by:\n`divide` in interface `NumberExpression`\nParameters:\n`other` - the constant value\nReturns:\nthe new `NumberBinding`\n• ### divide\n\npublic FloatBinding divide(int other)\nDescription copied from interface: `NumberExpression`\nCreates a new `NumberBinding` that calculates the division of this `NumberExpression` and a constant value.\nSpecified by:\n`divide` in interface `NumberExpression`\nParameters:\n`other` - the constant value\nReturns:\nthe new `NumberBinding`\n• ### asObject\n\npublic  asObject()\nCreates an `ObjectExpression` that holds the value of this `FloatExpression`. If the value of this `FloatExpression` changes, the value of the `ObjectExpression` will be updated automatically.\nReturns:\nthe new `ObjectExpression`\nSince:\nJavaFX 8.0" ]

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[ "## Precalculus (6th Edition) Blitzer\n\nSee graph; $y=0$", null, "Step 1. Graph both functions $f(x)=3^x$ and $g(x)=3^{-x}$ as shown in the figure. Step 2. We can identify a horizontal asymptote as $y=0$ for both functions." ]

[ null, "https://gradesaver.s3.amazonaws.com/uploads/solution/3506c3e1-8e0f-49db-9e7f-12d92cbd5ae9/result_image/1582812477.jpg", null ]

[ "Calculate the degree of dissociation example chemical engineering\nNew South Wales - 2019-09-07\n\n# The dissociation engineering example degree calculate of chemical\n\n## MATLAB SOLUTIONS TO THE CHEMICAL ENGINEERING PROBLEM. Get Vant Hoff Factor Assignment Help Now Homework Help.", null, "Degrees of Freedom: Definition, Formula To calculate the degrees of freedom for a sample size of N=9 Degrees of Freedom: Definition, Formula & Example Related. Journal of Chemical & Engineering Data Example dissociation In the host of numerical schemes devised to calculate free energy differences by way of.\nWeak Acid Equilibrium. = acid dissociation constant. For example, Tip-off - You are given the concentration of a weak acid solution and asked to calculate 2 MOLECULAR DYNAMICS SIMULATION OF DISSOCIATION KINETICS Andrew L. Kantor, Lyle N. Long, and Michael M. Micci Department of Aerospace Engineering\nFreezing-point depression is the decrease of the freezing point The degree of dissociation is measured by determining the van 't Hoff As an example, Chemical Equilibrium Example: Determine the calculate equilibrium constants from tabulated standard molar Gibbs energy data.", null, "Study chemical engineering and maths at the University of Newcastle for of Professional Engineering in just 1 year on top of your combined engineering degree.. Excel Solutions to the Chemical Engineering Problem Set For example for the Calculated Clasius- (sum of Squares/degrees of freedom).\n“Equilibrium chemistry Wikipedia”.\n\nEquilibrium chemistry is concerned is not necessarily an equilibrium state in the chemical sense. For example, A broader definition of acid dissociation.", null, "3/09/2014В В· Explains how to calculate degrees of freedom and degrees of freedom and performs two examples on of Chemical & Biological Engineering.. 10.492 - Integrated Chemical Engineering (ICE) Topics: Biocatalysis MIT Chemical Engineering Department Instructor: Professor Kristala Prather. Values of the bond dissociation vary slightly with the change in the pattern and degree of halogenation of the Journal of Chemical & Engineering Data OA.\nWhat is a dissociation equation? Bachelor of Chemical Engineering, Let us take NaCl, sodium chloride, for example. Calculate the equilibrium constant Kc //brainmass.com/chemistry/chemical-equilibrium/calculate-equilibrium-constant-percent-dissociation Chemical Engineering.", null, "## Example vcf file with multiple contacts\n\nExport multiple Contacts as vCards MSOutlook.info. 5 Free Methods to Import CSV/VCF Contacts to If the size of the CSV/VCF contacts file is not (here take Outlook 2010 as an example), from the File    …\n\n## Example Solitary Parallel Associative And Cooperative Play\n\nCooperative Play YouTube. 15/03/2011В В· Play stages: Solitary Play (0-2 years) March 15, 2011. Play is children’s work. No, seriously. Play stages: Parallel Play (2-3 years) Comments on:    …" ]

[ null, "https://caracolche.com/prefix/calculate-the-degree-of-dissociation-example-chemical-engineering.jpg", null, "https://caracolche.com/prefix/calculate-the-degree-of-dissociation-example-chemical-engineering-2.jpg", null, "https://caracolche.com/prefix/calculate-the-degree-of-dissociation-example-chemical-engineering-3.jpg", null, "https://caracolche.com/prefix/011b2c65ab91e3859fc547e11ff95809.jpg", null ]

[ "# In this tutorial, we are going to learn about the C++  Array of Objects in OOP,  Array of Objects with Example in C++,OOP Tutorial in C++\n\n## What is Array of Objects in C++?\n\nAn object of class represents a single record in memory, if we want more than one record of class type, we have to create an array of class or object.\n\nCreating more than one object of similar type is called Array of objects in C++ by creating an array of type class.\n\nSyntax:\n\nclass-name array-name[size];\n\nRemember: An array objects are always stored in consecutive memory locations, be it an array containing primitive values or object references.\n\nLike array of other user-defined data types, an array of type class can also be created.\n\nThe array of type class contains the objects of the class as its individual elements.\n\nThus, an array of a class type is also known as an array of objects.\n\nAn array of objects is declared in the same way as an array of any built-in data type.\n\nLet's see an example of taking the input of name and marks of 5 students by creating an array of the objects of students.\n\nArray of Objects Example :\n\n## ``` #include <iostream> #include <string> using namespace std; class Student{ string name; int marks; public: void getName() { getline( cin, name ); } void getMarks() { cin >> marks; } void displayInfo() { cout << \"Name : \" << name << endl; cout << \"Marks : \" << marks << endl; }}; int main(){ Student st; for( int i=0; i<5; i++ ) { cout << \"Student \" << i + 1 << endl; cout << \"Enter name\" << endl; st[i].getName(); cout << \"Enter marks\" << endl; st[i].getMarks(); } for( int i=0; i<5; i++ ) { cout << \"Student \" << i + 1 << endl; st[i].displayInfo(); } return 0; }```\n\nOutput\n\nStudent 1\nEnter name\nJack\nEnter marks\n54\nStudent 2\nEnter name\nMarx\nEnter marks\n45\nStudent 3\nEnter name\nJulie\nEnter marks\n47\nStudent 4\nEnter name\nPeter\nEnter marks\n23\nStudent 5\nEnter name\nDonald\nEnter marks\n87\nStudent 1\nName : Jack\nMarks : 54\nStudent 2\nName : Marx\nMarks : 45\nStudent 3\nName : Julie\nMarks : 47\nStudent 4\nName : Peter\nMarks : 23\nStudent 5\nName : Donald\nMarks : 87\n\nStudent st; - We created an array of 5 objects of the Student class where each object represents a student having a name and marks.\nThe first for loop is for taking the input of name and marks of the students. getName() and getMarks() are the functions to take the input of name and marks respectively.\nThe second for loop is to print the name and marks of all the 5 students. For that, we called the displayInfo() function for each student." ]

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[ "You are on page 1of 8\n\n# 1\n\nME 132\nDynamic Systems and Feedback\n\nLecture 3\n\nFeedback Interconnections\nClosed-Loop VS Open-Loop Control\n\n## Static and Linear Analysis\n\nSummary\nThe structure of control systems\nBasic control system (open-loop vs closed-loop)\nBasic block-diagram algebra static and linear\nsystems\nOpen-loop control\nClosed-loop control\nExample: Cruise Control for a Car\nOpen-loop control\nClosed-loop control\nAdvantages and disadvantages of feedback\n\n## Basic Control Systems\n\nOpen-Loop System\n\nClosed-Loop System\n\ncontroller\n\n## Open-loop system linear static blocks\n\ncontroller\n\nplant\n\ny = Gm\n\nr - reference input\nd - disturbance input\ny - output\ne = r y - error\n\nplant\n\ny = G(u + d)\n\nm=u+d\ny = G(u + d)\n\ny = G(Kr + d)\n\nu = Kr\n\n## Closed-loop system static blocks\n\ncontroller\ncontroller\n\nplant\n\ny = G(Kr + d)\n\ne = (1\n\nGK)r\n\nGd\n\ne=r\n\nInputs\nr - reference input\nd - disturbance input\nOutputs\ny - plant output\nu - controller output\ne = r y - error\n\nplant\n\n10\n\ncontroller\nPlant:\n\nplant\n\ncontroller\n\ny = G(u + d)\n\ny = G(u + d)\n\ny=\n\nController:\n\nu = Ke\ne=r y\n\nu = K(r\n\nplant\n\nu = K(r\n\ny)\n\nGKy + GKr + Gd\ny=\n\ny)\n\nGK\nG\nr+\nd\n1 + GK\n1 + GK\n\n11\n\n12\n\n## Closed-loop system static blocks\n\nClosed-loop VS Open-loop\nWe want to make\n\ncontroller\n\nplant\n\ne=\n\nGK\nG\ny=\nr+\nd\n1 + GK\n1 + GK\ne=\n\n1\nr\n1 + GK\n\nG\nd\n1 + GK\n\ne=r\n\n1\nr\n1 + GK\n\nG\nd\n1 + GK\n\nMaking K large,\nresults in\ne being small\n\ne=r\n\nsmall\n\ne = (1\n\nGK)r\n\nDifficult to select K\nK\n\nGd\nGr\n\nGd\n\n13\n\ncontroller\n\n14\n\nplant\n\n## Open-loop transfer function\n\nClosed-loop transfer functions\nInputs\nr - reference input\nd - disturbance input\nOutputs\ny - plant output\nu - controller output\ne = r y - error\n\ny=\n\nGK\nG\nr+\nd\n1 + GK\n1 + GK\n| {z }\nGc(r!y)\n\ne=\n\n1\nr\n1| +{zGK}\nGc(r!e)\n\n{z\n\nGc(d!y)\n\nG\nd\n1 + GK\n\nGe!f = Go = GK\n\n| {z }\nGc(d!e)\n\n15\n\nGforward path\nGclosed loop =\n1 + Gopen loop\n\n16\n\nGforward path\nGclosed loop =\n1 + Gopen loop\n\nGc(r!y) =\n\nGf p(r!y)\nGK\n=\n1 + GK\n1 + Go\n\nGc(r!e) =\n\nGf p(r!e)\n1 + Go\n\n1\n1 + GK\n\nGc(d!y) =\n\nG\nGf p(d!y)\n=\n1 + GK\n1 + Go\n\nGc(d!e) =\n\nGf p(d!e)\n1 + Go\n\nG\n1 + GK\n\n## We want to regulate the linear speed of a car, y, to a\n\ndesired value r\n\nOpen-loop control:\n\nu = Kol r\n\ndisturbance\ninput\n\nspeed\n\ncontrol\ninput\n\ny = Go Kol r\n\n## Cruise Closed-Loop Control\n\nObjective:\n\nClosed-loop control:\n\nyr\n\nClosed-loop response :\n\ny=\nClosed-loop response:\n\nGK\nH\ny=\nr+\nd\n1 + GKfb\n1 + GKfb\n| {z }\n| {z }\nGc(r!y)\n\nGc(d!y)\n\nGK\nH\nr+\nd\n1 + GKfb\n1 + GKfb\n\n{z\n\nObjective:\n\n## Closed-Loop VS Open-Loop - Disturbance Rejection\n\nyr\n\nClosed-loop response :\n\ny=\n\nGK\nH\nr+\nd\n1 + GKfb\n1 + GKfb\n\n{z\n\n0.01\n\nKfb &\n\nHo\n= 0.01\n1 + Go Kfb\n\nKfb\n\nHo\n&\n0.01Go\n\nyol = r\n\nH od\n\n0.01d\n\n## Closed-loop system exhibits superior\n\nexternal disturbance rejection\n\nH o = 5 >> 0.01\n\nycl = r\n\nHo\n0.01Go\n\nG\nr\nGo\nyo = r\n\ny=\n\n## Assume no disturbances (d = 0).\n\nQuestion:\nwhat is the variation of the output y around the nominal\noutput yo when there is a variation of the plant G around\nthe nominal plant Go?\n\nS=\n\nGo\nyo\n\nS=1\n\n@y\n@G\n\n=\nG = Go\ny = yo\n\nGo\nyo\n\nG\nGo r\n\n@G\n\nG = Go\ny = yo\n\n## 1 percent deviation of the plant G from its\n\nnominal value Go will approximately result in\na 1 percent deviation in the output y\n\ny=\n\nGK\nr\n1 + GKfb\n\nyo =\n\nS=\n\nG\nyo\n\n@y\n@G\n\n=\nG = Go\ny = yo\n\nGo @\nyo\n\nGo K\nr\n1 + Go Kfb\n\nGKff\n1+GKfb r\n\n@G\n\nyo =\n\nScl =\nG = Go\ny = yo\n\nGK\nr\n1 + GKfb\n\ny=\n\n1\n1 + Go Kfb\n\n0.002\n\nGo K\nr\n1 + Go Kfb\n\n<< Sol\n\nHo\n= 0.01\n1 + Go Kfb\n\nand Ho = 5\n\nH\nG\n\nSol = 1\n\nScl =\n\n1\n0.002\n1 + Go Kfb\n\n## Closed-loop system is much less\n\nsensitive to plant variations\n\nnoise\n\n{z\n\nGc(n!y)\n\n30\n\n## Closed-Loop Sensitivity to Noise\n\nGc(n!y)\n\nGKfb\n=\n1 + GKfb\n\nIdeally we require\nGc(n!y) << 1\n\n## which is necessary for disturbance rejection and robustness\n\nto plan variations. More on this later\n\nSummary\nThe structure of control systems\nOpen-loop control\nClosed-loop control\nAdvantages of feedback :\nClosed-loop systems exhibit superior external\ndisturbance rejection\nClosed loop systems are significantly less\nsensitive to plan parameter variations\nDisadvantages of high-gain feedback\nHigh sensitivity to measurement noise\nMay destabilize feedback system (dynamic\nmodels)" ]

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[ "# 小蛇学python（3）两百行代码实现微信好友数据爬取与可视化\n\npython3+mysql+pycharm开发环境\n\n```#!/usr/bin/env python3\n#_*_ coding:utf-8 _*_\n#__author__='跌跌撞撞小红豆'\n#__date__ = '2018-03-18'\nimport itchat\nimport pymysql.cursors\nfrom matplotlib import pyplot as plt\nimport numpy as np\nimport ch\nimport jieba.analyse\n\nNickname1 = []\nRemarkname1 = []\nSex1 = []\nProvince1 = []\nSignature1 = []\nprovince_distribution = []\n\ndb = 'new_schema'\n\nProvince_dict = {'北京': 0, '上海': 0, '天津': 0, '重庆': 0, '黑龙江': 0, '吉林': 0, '辽宁': 0, '内蒙古': 0, '河北': 0,\n'山东':0, '河南':0, '安徽':0,'江苏':0, '浙江':0, '福建':0, '广东':0, '湖南':0, '湖北':0, '江西':0,\n'宁夏':0, '甘肃':0, '新疆':0, '西藏':0, '青海':0, '四川':0,'云南':0, '贵州':0, '广西':0, '山西':0,\n'陕西':0, '海南':0, '中国台湾':0 }\nProvince_tuple = ['北京', '上海', '天津', '重庆', '黑龙江', '吉林', '辽宁', '内蒙古', '河北',\n'山东', '河南', '安徽','江苏', '浙江', '福建', '广东', '湖南', '湖北', '江西',\n'宁夏', '甘肃', '新疆', '西藏', '青海', '四川','云南', '贵州', '广西', '山西',\n'陕西', '海南', '中国台湾']\n\ndef crawl():\nmale = 0\nfemale = 0\nunknow = 0\n\nfor friend in itchat.get_friends(update=True)[0:]:\n#可以用此句print查看好友的微信名、备注名、性别、省份、个性签名（1：男 2：女 0：性别不详）\nprint(friend['NickName'],friend['RemarkName'], friend['Sex'], friend['Province'], friend['Signature'])\n\nNickname1.append(friend['NickName'])\nRemarkname1.append(friend['RemarkName'])\nSex1.append(friend['Sex'])\nProvince1.append(friend['Province'])\nSignature1.append(friend['Signature'])\n\ntry:\nwith open(path,'wb') as f:\nf.write(img)\nexcept Exception as e:\nprint(repr(e))\n\nif friend['Sex'] == 1:\nmale += 1\nelse:\nif friend['Sex'] == 2:\nfemale += 1\nelse:\nunknow += 1\n\nprint('您的微信账号里有男性：' + str(male) + ' 女性：' + str(female) + ' 性别不知者: ' + str(unknow))\n\ndef store_mysql():\nprint('~~~~~~~~~~~~~~~~~~~~~~~~~~连接数据库~~~~~~~~~~~~~~~~~~~~~~~~~~~~`')\nconnection = pymysql.connect(host='127.0.0.1', port=3306, user='root', password='123456', db=db,\ncharset='utf8mb4', cursorclass=pymysql.cursors.DictCursor)\n# 通过cursor创建游标\ncursor = connection.cursor()\nprint('~~~~~~~~~~~~~~~~~~~~~~~~~连接数据库成功！正在写入数据，请等待~~~~~~~~~~~~~~~~~~~~~~~~~~~~`')\n\nfor i in range(len(Remarkname1)):\nsql = '''INSERT INTO `my_friends` (`Nickname`, `Remarkname`, `Sex`, `Province`, `Signature`) VALUES (\"%s\", \"%s\", \"%d\", \"%s\", \"%s\")''' % (Nickname1[i],Remarkname1[i],Sex1[i],Province1[i],Signature1[i])\ncursor.execute(sql)\n\n# 提交SQL\nconnection.commit()\ncursor.close()\nconnection.close()\n\ndef select_sex():\nprint('~~~~~~~~~~~~~~~~~~~~~~~~~~连接数据库~~~~~~~~~~~~~~~~~~~~~~~~~~~~`')\nconnection = pymysql.connect(host='127.0.0.1', port=3306, user='root', password='123456', db=db,\ncharset='utf8mb4', cursorclass=pymysql.cursors.DictCursor)\n# 通过cursor创建游标\ncursor = connection.cursor()\nprint('~~~~~~~~~~~~~~~~~~~~~~~~~连接数据库成功！正在提取数据，请等待~~~~~~~~~~~~~~~~~~~~~~~~~~~~`')\n\nsql = \"select (id) from `my_friends`\"\ncursor.execute(sql)\nsum = len(cursor.fetchall())\nprint(\"共计数据\" + str(sum) + \"条\")\nglobal man, woman, unknown\nman = woman = unknown = 0\nfor i in range(sum):\nsql = \"select Sex from `my_friends` where id =\" + str(i+1)\ncursor.execute(sql)\ndict = cursor.fetchone()\nprint(dict)\n\nif dict['Sex'] == 1:\nman += 1\nif dict['Sex'] == 2:\nwoman += 1\nif dict['Sex'] == 0:\nunknown += 1\nprint(man,woman,unknown)\nprint(1)\n\nconnection.commit()\ncursor.close()\nconnection.close()\n\ndef figure_sex():\nfig = plt.figure(1)\nax1 = plt.subplot(111)\ndata = np.array([man, woman, unknown])\nwidth = 0.5\nx_bar = np.arange(3)\nrect = ax1.bar(left=x_bar, height=data, width=width, color=\"lightblue\")\nfor rec in rect:\nx = rec.get_x()\nheight = rec.get_height()\nax1.text(x + 0.2, 1.02 * height, str(height))\nax1.set_xticks(x_bar)\nax1.set_xticklabels((\"man\", \"woman\", \"unknown\"))\nax1.set_ylabel(\"number\")\nax1.grid(True)\nax1.set_ylim(0, 200)\nplt.show()\n\ndef select_province():\nandso_on = 0\noverseas = 0\ntotal = 0\nprint('~~~~~~~~~~~~~~~~~~~~~~~~~~连接数据库~~~~~~~~~~~~~~~~~~~~~~~~~~~~`')\nconnection = pymysql.connect(host='127.0.0.1', port=3306, user='root', password='123456', db=db,\ncharset='utf8mb4', cursorclass=pymysql.cursors.DictCursor)\n\ncursor = connection.cursor()\nprint('~~~~~~~~~~~~~~~~~~~~~~~~~连接数据库成功！正在提取数据，请等待~~~~~~~~~~~~~~~~~~~~~~~~~~~~`')\n\nsql = \"select (Province) from `my_friends`\"\ncursor.execute(sql)\nsum = len(cursor.fetchall())\nprint(\"共计数据\" + str(sum) + \"条\")\nfor i in range(sum):\nsql = \"select Province from my_friends where id =\" + str(i+1)\ncursor.execute(sql)\ndict = cursor.fetchone()\n\nprovince = dict['Province']\n\nif province in Province_tuple:\nProvince_dict[province] += 1\nif province == '':\noverseas += 1\nif province != ''and province not in Province_tuple:\nandso_on += 1\n#print(Province_dict) 真没想到可以如此简单的就成功计数好友在每个省的分布，是自己写程序无意中想到的，之前一直思考都没有这个灵光闪现。可见有时候编代码和做数学题一样，不能只看，要边动手边想！！！\nfor i in range(32):\nprovince_distribution.append(Province_dict[Province_tuple[i]])\n\nprovince_distribution.append(overseas)\nprovince_distribution.append(andso_on)\nfor i in range(len(province_distribution)-2):\ntotal = total + province_distribution[i]\n\nprint(province_distribution)\nconnection.commit()\ncursor.close()\nconnection.close()\n\ndef figure_province():\nProvince_tuple.append('unknow')\nfig = plt.figure(1)\nax1 = plt.subplot(111)\ndata = province_distribution\nwidth = 0.5\nx_bar = np.arange(len(province_distribution))\nrect = ax1.bar(left=x_bar, height=data, width=width, color=\"lightblue\")\nfor rec in rect:\nx = rec.get_x()\nheight = rec.get_height()\nax1.text(x + 0.1, 1.02 * height, str(height))\nax1.set_xticks(x_bar)\nax1.set_xticklabels(Province_tuple)\nax1.set_ylabel(\"number\")\nax1.set_title(\"my friend's location distribution\")\nax1.grid(True)\nax1.set_ylim(0, 100)\nplt.show()\n\ndef judge_chinese(word):\ncout1 = 0\nfor char in word:\nif ord(char) not in (97, 122) and ord(char) not in (65, 90):\ncout1 += 1\nif cout1 == len(word):\nreturn word\n\ndef jiebaSet(strs):\n\ntags = jieba.analyse.extract_tags(strs, topK=100, withWeight=True)\nfor item in tags:\nprint(item + '\\t' + str(int(item*1000)))\n\ndef select_signature():\ncout = 0\nSig = []\nstrs = ''\n\nprint('~~~~~~~~~~~~~~~~~~~~~~~~~~连接数据库~~~~~~~~~~~~~~~~~~~~~~~~~~~~`')\nconnection = pymysql.connect(host='127.0.0.1', port=3306, user='root', password='123456', db=db,\ncharset='utf8mb4', cursorclass=pymysql.cursors.DictCursor)\n\ncursor = connection.cursor()\nprint('~~~~~~~~~~~~~~~~~~~~~~~~~连接数据库成功！正在提取数据，请等待~~~~~~~~~~~~~~~~~~~~~~~~~~~~`')\n\nsql = \"select (id) from `my_friends`\"\ncursor.execute(sql)\nsum = len(cursor.fetchall())\n\nfor i in range(sum):\nsql = \"select Signature from `my_friends` where id =\" + str(i + 1)\ncursor.execute(sql)\ndict = cursor.fetchone()\nif dict['Signature'] == '':\ncout += 1\nelse:\nSig.append(dict['Signature'])\n\nprint('有' + str(cout) + '人没有写个性签名')\n\nfor word in Sig[0:]:\njudge_chinese(word)\nif judge_chinese(word) != None:\nprint(judge_chinese(word))\nstrs += word\n\njiebaSet(strs)\n\ncrawl()\nitchat.run()\nstore_mysql()\nselect_sex()\nfigure_sex()\nch.set_ch()\nselect_province()\nfigure_province()\nselect_signature()```\n\ncrawl()所实现的功能是爬取你微信列表所有好友的信息，这些信息包括头像，昵称，备注，性别，所在地以及个性签名（昵称和个性签名里有微信表情真的是让人头疼，这个后面再说）以及爬取后储存头像在指定路径和可视化微信好友男女比例的功能。\n\n`path = \"C:/HeadImages_zyh/\"+friend['NickName']+\"(\"+friend['RemarkName']+\").jpg\"`\n\nstore_mysql()就是实现这样一个功能。只需要简单学习几句sql语句就可以满足本程序的使用。\n\n```truncate table 表名；#删除表中所有数据\n\ncreate table 表名\n（id int(10) not null primary key auto_increment) #给表增加索引，而且没插入一条数据索引自增1\n\nalter table 表名 add 字段名 字段类型 ```\n\n`a='', b=' ', c=null`\n\nmysql57所在位置\n\nmy文件的路径.png\n\nmysql下载下来后是有两个服务进程的，一个叫MySQL，一个叫MySQL57。而且在控制面板之前那个服务里我们看到，MySQL叫做本地系统，MySQL57是网络服务。我想当然的认为MySQL是我现在存储数据所用的服务进程。\n\n```Province_dict = {'北京': 0, '上海': 0, '天津': 0, '重庆': 0, '黑龙江': 0, '吉林': 0, '辽宁': 0, '内蒙古': 0, '河北': 0,\n'山东':0, '河南':0, '安徽':0,'江苏':0, '浙江':0, '福建':0, '广东':0, '湖南':0, '湖北':0, '江西':0,\n'宁夏':0, '甘肃':0, '新疆':0, '西藏':0, '青海':0, '四川':0,'云南':0, '贵州':0, '广西':0, '山西':0,\n'陕西':0, '海南':0, '中国台湾':0 }\nProvince_tuple = ['北京', '上海', '天津', '重庆', '黑龙江', '吉林', '辽宁', '内蒙古', '河北',\n'山东', '河南', '安徽','江苏', '浙江', '福建', '广东', '湖南', '湖北', '江西',\n'宁夏', '甘肃', '新疆', '西藏', '青海', '四川','云南', '贵州', '广西', '山西',\n'陕西', '海南', '中国台湾']```\n\nPS\n\n```#-*-coding:utf-8-*-\n#文件名: ch.py\ndef set_ch():\nfrom pylab import mpl\nmpl.rcParams['font.sans-serif'] = ['FangSong'] # 指定默认字体\nmpl.rcParams['axes.unicode_minus'] = False # 解决保存图像是负号'-'显示为方块的问题</span>```\n\n0 条评论\n\n• ### 小蛇学python（20）魔法函数\n\ninit负责类内成员的初始化，当类初始化实例的时候，会将传入的值赋给类内成员，与c++中的构造函数十分相似。\n\n• ### 小蛇学python（11）初窥numpy\n\n读者可以自行输入，观看结果，享受编码的乐趣。注意zeros和ones后面是跟了两组小括号的。\n\n• ### 小蛇学python（12）分析《今生今世》人物关系图谱\n\n《今生今世》是渣男胡兰成所写的一部自传体小说。今天我们就来分析一下在他所写的自传中的人物关系图谱，分析一下胡兰成到底和多少女人有关系。\n\n• ### python基础题目大全，测试你的水平，巩固知识（含答案）\n\n（1）、python代码，简介，明确，优雅，简单易懂 （2）、开发效率高 （3）、可扩展性强\n\n• ### 使用Python标准库functools中的lru_cache实现缓存\n\n很简单，也很容易理解，但是不难发现这个函数在计算斐波那契数列的时候事实上进行了很多重复计算，例如：\n\n• ### Swif Array\n\n使用加法赋值运算符(+=)也可以直接在数组后面添加一个或多个拥有相同类型的数据项:\n\n• ### 关联分析（二）：Apriori算法的python实现\n\n阴影区域给出的是低置信度的规则。如果发现{0,1,2} ->{3} 是一条低可置信度规则，那么所有其它以{0,1,2}子集做前件，以3作为后件的规则的可信度也会...\n\nJavaScript是一门多范式语言，即可使用OOP（面向对象），也可以使用FP（函数式），由于笔者最近在学习React相关的技术栈，想进一步深入了解其思想，所...\n\n• ### 高分辨率256*256人脸生成效果介绍及代码\n\n众所周知，训练GAN非常困难. In order to train at 256 x 256 we utilize:\n\n### 活动推荐", null, "" ]

[ null, "https://imgcache.qq.com/open_proj/proj_qcloud_v2/community/portal/css/img/wechat-qr.jpg", null ]

[ "Essential University Physics: Volume 1 (3rd Edition)\n\n$L=1.13Js$\nAs we know that $I_{system}=2\\times\\frac{M_{rod}L^2}{12}+4\\times M_{cup}(\\frac{L}{2})^2$ We plug in the known values to obtain: $I_{system}=2\\times\\frac{0.0757(0.326)^2}{12}+4\\times(0.124)(\\frac{0.326}{2})^2$ $I_{system}=14.5\\times 10^{-3}Kgm^2$ We also know that $L=I\\omega$ $\\implies L=14.5\\times 10^{-3}\\times 78=1.13Js$" ]

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[ "# Category: Binary Tree\n\n## Postorder Traversal of Binary Tree Using Stacks\n\n“Postorder Traversal of Binary Tree Using Stacks” is an important binary tree traversal algorithm. Here, we are given a binary tree and our task is to write a program to traverse the given binary tree in postorder manner using stack data structure. We have already discussed recursive implementation of postorder traversal....\n\n## Preorder Traversal of Binary Tree Using Stack\n\n“Preorder Traversal of Binary Tree Using Stack” is an important traversal algorithm of tree data structure. Here, we are given a binary tree and our task is to traverse the given binary tree in preorder fashion using stack data structure. We have already discussed recursive implementation here. Now, the iterative...\n\n## Inorder Tree Traversal Using Stack\n\n“Inorder Tree Traversal Using Stack” is a binary tree traversal algorithm where we traverse the given tree in inorder fashion using stack data structure.  We have already discussed the recursive approach of the inorder traversal. Here, we implement the same algorithm using stack data structure. The steps required to implement...\n\n## Compute Height of a Binary Tree\n\n“Compute Height of a Binary Tree” is a very basic problem of Tree data structure. Here, we are given a Binary Tree and our task is to compute height of given binary tree. The height of the tree is defined as total number of nodes in a longest path from...\n\n## Right View of Binary Tree\n\n“Right View of Binary Tree” is a very popular interview problem based on Tree data structure, asked in many technical interviews. Here, we are given a binary tree and our task is to write a program to print the Right View of Binary Tree. “Right View of Binary Tree” is...\n\n## Left View of Binary Tree\n\n“Left View of Binary Tree” is a very popular interview problem based on Tree data structure, asked in many technical interviews. Here, we are given a binary tree and our task is to write a program to print the Left View of Binary Tree. “Left View of Binary Tree” is...\n\n## Level Order Tree Traversal\n\nLevel Order Tree Traversal or Breadth First Traversal is another type of tree traversal algorithm where we traverse all nodes of the binary tree level-by-level. In previous post, we have already discussed preorder, inorder and postorder tree traversals which are typical types of Depth First Traversals. Trees can be traversed...\n\n## Maximum Width of Binary Tree\n\n“Maximum Width of Binary Tree” is a basic problem of Tree data structure. Here, we are given a binary tree and our task is to find maximum width of Binary Tree. Maximum Width of Binary Tree is defined as maximum number of nodes present at any level of Binary Tree....\n\n## Print Alternate Levels of a Binary Tree\n\n“Print Alternate Levels of a Binary Tree” is a simple modification of level order tree traversal algorithm. Here, we are given a Binary Tree and our task is to write a program to print alternate levels of a Binary Tree. Again, we can solve this problem using previous discussed two...\n\n## Count Number of Nodes in a Binary Tree\n\n“Count Number of Nodes in a Binary Tree” is again a very basic problem of tree data structure. Here, we are given a tree and our task is to count the number of nodes in a tree. There can be multiple methods to count the number of nodes in a..." ]

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[ "Paper\n\n### Complex Unitary Recurrent Neural Networks using Scaled Cayley Transform\n\nRecurrent neural networks (RNNs) have been successfully used on a wide range of sequential data problems. A well known difficulty in using RNNs is the \\textit{vanishing or exploding gradient} problem. Recently, there have been several different RNN architectures that try to mitigate this issue by maintaining an orthogonal or unitary recurrent weight matrix. One such architecture is the scaled Cayley orthogonal recurrent neural network (scoRNN) which parameterizes the orthogonal recurrent weight matrix through a scaled Cayley transform. This parametrization contains a diagonal scaling matrix consisting of positive or negative one entries that can not be optimized by gradient descent. Thus the scaling matrix is fixed before training and a hyperparameter is introduced to tune the matrix for each particular task. In this paper, we develop a unitary RNN architecture based on a complex scaled Cayley transform. Unlike the real orthogonal case, the transformation uses a diagonal scaling matrix consisting of entries on the complex unit circle which can be optimized using gradient descent and no longer requires the tuning of a hyperparameter. We also provide an analysis of a potential issue of the modReLU activiation function which is used in our work and several other unitary RNNs. In the experiments conducted, the scaled Cayley unitary recurrent neural network (scuRNN) achieves comparable or better results than scoRNN and other unitary RNNs without fixing the scaling matrix.\n\nResults in Papers With Code\n(↓ scroll down to see all results)" ]

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[ "# zbMATH — the first resource for mathematics\n\nNeumann resonances in linear elasticity for an arbitrary body. (English) Zbl 0851.35032\nThe resonances associated to the Lamé operator $$L$$ in the exterior of an arbitrary obstacle in $$\\mathbb{R}^3$$ with Neumann boundary conditions are analyzed. The resonances are defined as the poles of the meromorphic continuation of the cut-off resolvent from the lower half plane $$(\\text{Im } y< 0)$$ to the whole complex plane $$\\mathbb{C}$$. The main result of the paper is the following: There exist two infinite sequences $$\\{\\lambda_j\\}$$, $$\\{\\overline \\lambda_j\\}$$ of distinct resonances of the Lamé operator $$L$$, such that $$0< \\text{Im } \\lambda_j\\leq C_N|\\lambda_j|^{- N}$$ for any $$N> 0$$.\n\n##### MSC:\n 35J15 Second-order elliptic equations 74B05 Classical linear elasticity\nFull Text:\n##### References:\n [CP] Cardoso, F., Popov, G.: Rayleigh quasimodes in linear elasticity. Comm. P.D.E.17, 1327–1367 (1992) · Zbl 0795.35067 [G] Gérard, C.: Asymptotique des poles de la matrice de scattering pour deux obstacles strictement convex. Bull. Soc. Math. France, Mémoire n. 31,116, 1988 · Zbl 0654.35081 [GK] Gohberg, I., Krein, M.: Introduction to the theory of linear non-selfadjoint operators. Providence, RI: AMS, 1969 · Zbl 0181.13504 [I1] Ikawa, M.: On the poles of the scattering matrix for two strictly convex obstacles. J. Math. Kyoto Univ.23–1, 127–194 (1983) [I2] Ikawa, M.: Precise information on the poles of the scattering matrix for two strictly convex obstacles. J. Math. Kyoto Univ.27–1, 69–102 (1987) [I3] Ikawa, M.: Trapping obstacles with a sequence of poles of the scattering matrix converging to the real axis. Osaka J. Math.22, 657–689 (1985) · Zbl 0617.35102 [K] Kawashita, M.: On the local-energy decay property for the elastic wave equation with the Neumann boundary conditions. Duke Math. J.67, 333–351 (1992) · Zbl 0795.35061 [L] Lazutkin, V.: Asymptotics of the eigenvalues of the Laplacian and quasimodes. Math. USSR Izvestija7, 439–466 (1973) · Zbl 0282.35063 [P] Popov, G.: Quasimodes for the Laplace operator and glancing hypersurfaces. In: M. Beals, R. Melrose, J. Rauch (eds.): Proceeding of Conference on Microlocal Analysis and nonlinear waves, Minnesota 1989, Berlin-Heidelberg-New York: Springer, 1991 · Zbl 0794.35030 [SV1] Stefanov, P., Vodev, G.: Distribution of resonances for the Neumann problem in linear elasticity outside a ball. Ann. Inst. H. Poincaré (Physique Théorique)60, 303–321 (1994) · Zbl 0805.73016 [SV2] Stefanov, P., Vodev, G.: Distribution of resonances for the Neumann problem in linear elasticity outside a strictly convex body. Duke Math. J.78, 677–714 (1995) · Zbl 0846.35139 [T] Taylor, M.: Rayleigh waves in linear elasticity as a propagation of singularities phenomenon. In: Proc. Conf. on P.D.E. and Geometry, New York: Marcel Dekker, 1979 pp. 273–291 [Ti] Titchmarsh, E.C.: The Theory of Functions. Oxford: Oxford Univ. Press, 1968 [Y] Yamamoto, K.: Singularities of solutions to the boundary value problems for elastic and Maxwell’s equations. Japan J. Math.14, 119–163 (1988) · Zbl 0669.73017\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching." ]

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[ "# Statistics - Weak Law of Large Numbers\n\nThe weak law of large numbers is a result in probability theory also known as Bernoulli's theorem. Let P be a sequence of independent and identically distributed random variables, each having a mean and standard deviation.\n\n## Formula\n\n$${ 0 = \\lim_{n\\to \\infty} P \\{\\lvert X - \\mu \\rvert \\gt \\frac{1}{n} \\} \\\\[7pt] \\ = P \\{ \\lim_{n\\to \\infty} \\{ \\lvert X - \\mu \\rvert \\gt \\frac{1}{n} \\} \\} \\\\[7pt] \\ = P \\{ X \\ne \\mu \\} }$$\n\nWhere −\n\n• ${n}$ = Number of samples\n\n• ${X}$ = Sample value\n\n• ${\\mu}$ = Sample mean\n\n### Example\n\nProblem Statement:\n\nA six sided die is rolled large number of times. Figure the sample mean of their values.\n\nSolution:\n\nSample Mean Calculation\n\n${Sample\\ Mean = \\frac{1+2+3+4+5+6}{6} \\\\[7pt] \\ = \\frac{21}{6}, \\\\[7pt] \\, = 3.5 }$" ]

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[ "", null, "zakinutuzi\n\n2022-01-03\n\nHow to show that\n$\\mathrm{log}\\left(\\frac{2a}{1-{a}^{2}}+\\frac{2b}{1-{b}^{2}}+\\frac{2c}{1-{c}^{2}}\\right)=\\mathrm{log}\\frac{2a}{1-{a}^{2}}+\\mathrm{log}\\frac{2b}{1-{b}^{2}}+\\mathrm{log}\\frac{2c}{1-{c}^{2}}$", null, "Marcus Herman\n\nHINT:\nUse $\\mathrm{log}a+\\mathrm{log}b+\\mathrm{log}c=\\mathrm{log}\\left(abc\\right)$\nand then put $a=\\mathrm{tan}A$ etc. to find\n$\\sum \\mathrm{tan}A=\\prod \\mathrm{tan}A⇒A+B+C=n\\pi$ where $n$ is any integer\nand $\\frac{2a}{1-{a}^{2}}=\\frac{2\\mathrm{tan}A}{1-{\\mathrm{tan}}^{2}A}=\\mathrm{tan}2A$\n\nNow apply logarithm now", null, "reinosodairyshm\n\nOK, heres", null, "Vasquez\n\nThe expression $\\frac{2a}{1-{a}^{2}}$ should make you think of the tangent function, as in\n$\\mathrm{tan}\\left(2A\\right)=\\frac{2\\mathrm{tan}A}{1-\\left(\\mathrm{tan}A{\\right)}^{2}}$\nHere I'm using capital A to refer to the angle and lower-case a to refer to its tangent: $a=\\mathrm{tan}A$.\n$\\mathrm{log}\\left(a+b+c\\right)=\\mathrm{log}a+\\mathrm{log}b+\\mathrm{log}c$ implies $\\mathrm{log}\\left(a+b+c\\right)=\\mathrm{log}\\left(abc\\right)$, which then implies $a+b+c=abc$, and following the pattern in capital and lower-case letters above, we have $\\mathrm{tan}A+\\mathrm{tan}B+\\mathrm{tan}C=\\mathrm{tan}A\\mathrm{tan}B\\mathrm{tan}C$.\nThe identity you're trying to prove is equivalent to $\\frac{2a}{1-{a}^{2}}+\\frac{2b}{1-{b}^{2}}+\\frac{2c}{1-{c}^{2}}=\\frac{2a}{1-{a}^{2}}\\cdot \\frac{2b}{1-{b}^{2}}\\cdot \\frac{2c}{1-{c}^{2}}$\n(with lower-case a,b,c), so that's the same as\n$\\mathrm{tan}\\left(2A\\right)+\\mathrm{tan}\\left(2B\\right)+\\mathrm{tan}\\left(2C\\right)=\\mathrm{tan}\\left(2A\\right)\\mathrm{tan}\\left(2B\\right)\\mathrm{tan}\\left(2C\\right).$\nNow at this point I might not know how to proceed further if I hadn't seen the following at some point in the past. In the first place, the usual formula for the tangent of a sum implies with 30 seconds' more work that\n$\\mathrm{tan}\\left(A+B+C\\right)=\\frac{\\mathrm{tan}A+\\mathrm{tan}B+\\mathrm{tan}C-\\mathrm{tan}A\\mathrm{tan}B\\mathrm{tan}C}{1-\\mathrm{tan}A\\mathrm{tan}B-\\mathrm{tan}A\\mathrm{tan}C-\\mathrm{tan}B\\mathrm{tan}C},$\nand in the second place, if $A+B+C=\\pi$ radians or ${180}^{\\circ }$ or a half-circle then $\\mathrm{tan}\\left(A+B+C\\right)=0$. So the fraction is 0 and therefore the numerator is 0 and therefore $\\mathrm{tan}A+\\mathrm{tan}B+\\mathrm{tan}C=\\mathrm{tan}A\\mathrm{tan}B\\mathrm{tan}C$\nSo what you're being asked to prove is that if\n$\\mathrm{tan}A+\\mathrm{tan}B+\\mathrm{tan}C=\\mathrm{tan}A\\mathrm{tan}B\\mathrm{tan}C$ then\n$\\mathrm{tan}\\left(2A\\right)+\\mathrm{tan}\\left(2B\\right)+\\mathrm{tan}\\left(2C\\right)=\\mathrm{tan}\\left(2A\\right)\\mathrm{tan}\\left(2B\\right)\\mathrm{tan}\\left(2C\\right).$\nWe've see that the fact that $A+B+C=a$ half circle implies the first of these identities only because it implies that $\\mathrm{tan}\\left(A+B+C\\right)=0$. So you really just need to show that if $\\mathrm{tan}\\left(A+B+C\\right)=0$ then $\\mathrm{tan}\\left(2A+2B+2C\\right)=0$. It's not hard to show that that's the same as saying that if $A+B+C$ is an integer multiple of a half-circle, then so is $2A+2B+2C$.\nMoral: Don't do this problem if you've forgotten your trigonometry.\nHowever, since the problem as stated doesn't mention trigonometry, I'm wondering if there's another way to do it that avoids that. Probably there is. The tangent function and the identities we used involve a particular way of parametrizing the circle, as $A\\to \\left(\\mathrm{cos}A,\\mathrm{sin}A\\right)$. But the result doesn't seem to be one that should depend on such a choice of parametrization.\n\nDo you have a similar question?" ]

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[ "NCERT Solutions for Class 11 Maths Chapter 4: Principle of Mathematical Induction\n\n# NCERT Solutions for Class 11 Maths Chapter 4: Principle of Mathematical Induction\n\nSubject specialists have created NCERT Solutions for Class 11 Maths Chapter 4:Principle of Mathematical Induction, which includes thorough solutions for reference. All of the questions from the textbook’s exercises are answered here. Students can use these answers to help them prepare for their exams. The NCERT Solutions for Class 11 provide useful solutions for improving conceptual knowledge.\n\nThe solutions are carefully solved using student-friendly terms while still adhering to the norms that must be followed when solving NCERT Solutions for Class 11. Practicing these answers can be incredibly advantageous not only in terms of exams but also in terms of helping Class 11 pupils perform well in upcoming competitive exams.\n\nFill Out the Form for Expert Academic Guidance!\n\n+91\n\nLive ClassesBooksTest SeriesSelf Learning\n\nVerify OTP Code (required)\n\nThe approaches for answering have been given special consideration to stay on target while not deviating from the intended answer. Because time is so important in exams, excellent time management when answering questions is essential for getting the best results.\n\nExercise 4.1 Solutions\n\nDo you need help with your Homework? Are you preparing for Exams? Study without Internet (Offline)\n×\n\nStudy without Internet (Offline)\n\n+91\n\nLive ClassesBooksTest SeriesSelf Learning\n\nVerify OTP Code (required)\n\nMaths_Class 11_Ch 4\n\nNCERT Solutions for Class 11 Maths Chapter 4- Principle of Mathematical Induction\n\nThis chapter consists of only one exercise and this will help students in learning the concepts based on the Principle of Mathematical Induction through the given solutions. The topic and subtopics that are discussed in Chapter 4 Principle of Mathematical Induction of NCERT Solutions for Class 11 Maths are listed below:\n\n4.1 Introduction\n\nIn this section, deductive reasoning with suitable examples and also the assumptions that are based on certain universal facts are explained.\n\n4.2 Motivation\n\nHere, the students learn about mathematical induction with the help of a real-life scenario which helps them to understand its basic working.\n\n4.3 The Principle of Mathematical Induction\n\nHere, the Principle of Mathematical Induction is explained by using the inductive steps and the inductive hypothesis.\n\nLet us suppose the statement is P(n), where n is a natural number.\n\n• If n = 1 then the statement is true, it means P(1) is true.\n• If the statement n = k (where k is some positive integer) is true, then the statement n = k + 1 is also true, which means, the truth of P(k) implies the truth of P (k + 1).\n\nKey Features of NCERT Solutions for Class 11 Maths Chapter 4- Principle of Mathematical Induction\n\nAfter learning the Principle of Mathematical Induction of Class 11 students are able to know about the Steps involved in the proof of induction. Students also learn about the applications and principles of mathematical induction from the given solutions of NCERT questions. The concepts explained and used in the solutions of this chapter are as follows:\n\n• Deductive reasoning is one of the basic keys to mathematical thinking. Until we have observed each and every case it depends on working with different cases and developing a conjecture by observing incidences. Hence, in simple words ‘induction’ is the generalization from particular cases or facts.\n• The principle of mathematical induction is a tool that is used to prove a large number of mathematical statements. Let us suppose each statement is P(n), where n is a positive integer in which n=1 examines that the case is correct. Therefore, if P(k) is true for some positive integer k, then P (k+1) is also true.\n\nFrequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 4\n\n1. What is the concept of Mathematical Induction explained in Chapter 4 of NCERT Solutions for Class 11 Maths?\n2. Explain the first principle of Mathematical Induction in Chapter 4 of NCERT Solutions for Class 11 Maths?\n3. Why are the NCERT Solutions for Class 11 Maths Chapter 4 PDF important from the exam point of view?\n\n1. What is the concept of Mathematical Induction explained in Chapter 4 of NCERT Solutions for Class 11 Maths?\n\nMathematical Induction is the method to prove a statement, formula, or theorem which is true for all natural numbers. The two steps that are used for proving a statement are.\n\n1. Base step.\n2. Inductive step.\n\n2. Explain the first principle of Mathematical Induction in Chapter 4 of NCERT Solutions for Class 11 Maths?\n\nThe first principle of Mathematical Induction states that both the base step and inductive step are true, for all natural numbers. Many kinds of problems that are based on this concept are available in the NCERT textbook of Class 11. Exercise-wise problems and their solutions are also available which helps students to score well in the annual exam.\n\n3. Why are the NCERT Solutions for Class 11 Maths Chapter 4 PDF important from the exam point of view?\n\nThe NCERT Solutions for Class 11 Maths Chapter 4 gives better concepts about the induction and deduction techniques with the help of which equations and statements are proved. With the help of solutions, students get a clear knowledge about the principle and applications of Mathematical Induction. These solutions will increase the confidence level among students who are preparing for the Class 11 exams.\n\n## Related content\n\n How is NCERT Exemplar Different from NCERT Textbook Solutions Is NCERT Class 8 Science Chapter 7 Notes Helpful for CBSE Students? Best Preparation Tips for CBSE Class 12 Physics Board Exam NCERT Solutions for class 12 Biology chapter 2 sexual reproduction in flowering plants 2.1 flower – a fascinating organ of angiosperms Keeping Quiet Important Questions Class 12 English NCERT Solutions For Class 6 Maths Data Handling Exercise 9.3 The Ant and the Cricket Summary Class 8 English Surface Areas and Volumes NCERT Solutions For Class 10 Maths Chapter 13 Fire and Ice Summary Class 10 English NCERT Solutions for Class 6 English Chapter 5 – A Different Kind of School", null, "", null, "+91\n\nLive ClassesBooksTest SeriesSelf Learning\n\nVerify OTP Code (required)" ]

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[ "# Track Space Debris Using a Keplerian Motion Model\n\nThis example shows how to model earth-centric trajectories using custom motion models within `trackingScenario`, how to configure a monostatic radar sensor to generate synthetic detections of space debris, and how to setup a multi-object tracker to track the simulated targets.\n\n### Space debris scenario\n\nThere are more than 30,000 large debris objects (with diameter larger than 10cm) and more than 1 million smaller debris objects in Low Earth Orbit (LEO) . This debris can be dangerous for human activities in space, damage operational satellites, and force time sensitive and costly avoidance maneuvers. As space activity increases, reducing and monitoring the space debris becomes crucial.\n\nYou can use Sensor Fusion and Tracking Toolbox™ to model the debris trajectories, generate synthetic radar detections of this debris, and obtain position and velocity estimates of each object.\n\nFirst, create a tracking scenario and set the random seed for repeatable results.\n\n```s = rng; rng(2020); scene = trackingScenario('IsEarthCentered',true);```\n\nYou use the Earth-Centered-Earth-Fixed (ECEF) reference frame. The origin of this frame is at the center of the Earth and the Z axis points toward the north pole. The X axis points towards the intersection of the equator and the Greenwich meridian. The Y axis completes the right-handed system. Platform positions and velocities are defined using Cartesian coordinates in this frame.\n\n#### Define debris motion model\n\nThe `helperMotionTrajectory` class used in this example defines debris object trajectories using a custom motion model function.\n\nTrajectories of space objects rotating around the Earth can be approximated with a Keplerian model, which assumes that Earth is a point-mass body and the objects orbiting around the earth have negligible masses. Higher order effects in Earth gravitational field and environmental disturbances are not accounted for. Since the equation of motion is expressed in ECEF frame which is a non-inertial reference frame, the Coriolis and centripetal forces are accounted for.\n\nThe ECEF debris object acceleration vector is\n\n$\\stackrel{\\to }{\\mathit{a}}=\\frac{-\\mu }{{\\mathit{r}}^{3}}\\text{\\hspace{0.17em}}\\stackrel{\\to }{\\mathit{r}}-2\\stackrel{\\to }{\\Omega }×\\frac{\\mathrm{d}}{\\mathrm{d}\\mathit{t}}\\stackrel{\\to }{\\mathit{r}}-\\stackrel{\\to }{\\Omega }×\\left(\\stackrel{\\to }{\\Omega }×\\stackrel{\\to }{\\mathit{r}}\\right)$,\n\nwhere $\\mu$ is the standard gravitational parameter of the Earth, $\\stackrel{\\to }{\\mathit{r}}$ is the ECEF debris object position vector, $\\mathit{r}$ is the norm of the position vector, and $\\stackrel{\\to }{\\Omega }$is the Earth rotation vector.\n\nThe function `keplerorbit` provided below uses a 4th order Runge-Kutta numerical integration of this equation to propagate the position and velocity in time.\n\nFirst, we create initial positions and velocities for the space debris objects. This is done by obtaining the traditional orbital elements (semi-major axis, eccentricity, inclination, longitude of the ascending node, argument of periapsis, and true anomaly angles) of these objects from random distributions. Then convert these orbital elements to position and velocity vectors by using the supporting function `oe2rv`.\n\n```% Generate a population of debris numDebris = 100; range = 7e6 + 1e5*randn(numDebris,1); ecc = 0.015 + 0.005*randn(numDebris,1); inc = 80 + 10*rand(numDebris,1); lan = 360*rand(numDebris,1); w = 360*rand(numDebris,1); nu = 360*rand(numDebris,1); % Convert to initial position and velocity for i = 1:numDebris [r,v] = oe2rv(range(i),ecc(i),inc(i),lan(i),w(i),nu(i)); data(i).InitialPosition = r; %#ok<SAGROW> data(i).InitialVelocity = v; %#ok<SAGROW> end % Create platforms and assign them trajectories using the keplerorbit motion model for i=1:numDebris debris(i) = platform(scene); %#ok<SAGROW> debris(i).Trajectory = helperMotionTrajectory(@keplerorbit,... 'SampleRate',0.1,... % integration step 10sec 'Position',data(i).InitialPosition,... 'Velocity',data(i).InitialVelocity); %#ok<SAGROW> end```\n\nIn this example, we define four antipodal stations with fan-shaped radar beams looking into space. The fans cut through the orbits of debris objects to maximize the number of object detections. A pair of stations are located in the Pacific ocean and in the Atlantic ocean, whereas a second pair of surveillance stations are located near the poles. Having four dispersed radars allows for the re-detection of space debris to correct their position estimates and also acquiring new debris detections.\n\n```% Create a space surveillance station in the Pacific ocean station1 = platform(scene,'Position',[10 180 0]); % Create a second surveillance station in the Atlantic ocean station2 = platform(scene,'Position',[0 -20 0]); % Near the North Pole, create a third surveillance station in Iceland station3 = platform(scene,'Position',[65 -20 0]); % Create a fourth surveillance station near the south pole station4 = platform(scene,'Position',[-90 0 0]);```\n\nEach station is equipped with a radar modeled with a `monostaticRadarSensor` object. In order to detect debris objects in the LEO range, the radar has the following requirements:\n\n• Detecting a 10 dBsm object up to 2000 km away\n\n• Resolving objects horizontally and vertically with a precision of 100 m at 2000 km range\n\n• Having a fan-shaped field of view of 120 degrees in azimuth and 30 degrees in elevation\n\n• Looking up into space based on its geo-location\n\n```% Create fan-shaped monostatic radars to monitor space debris objects radar1 = monostaticRadarSensor(1,... 'UpdateRate',0.1,... 10 sec 'ScanMode','No scanning',... 'MountingAngles',[0 90 0],... look up 'FieldOfView',[120;30],... degrees 'ReferenceRange',2000000,... m 'ReferenceRCS', 10,... dBsm 'HasFalseAlarms',false,... 'HasElevation',true,... 'AzimuthResolution',0.01,... degrees 'ElevationResolution',0.01,... degrees 'RangeResolution',100,... m 'HasINS',true,... 'DetectionCoordinates','Scenario'); station1.Sensors = radar1; radar2 = clone(radar1); radar2.SensorIndex = 2; station2.Sensors = radar2; radar3 = clone(radar1); radar3.SensorIndex = 3; station3.Sensors = radar3; radar4 = clone(radar1); radar4.SensorIndex = 4; station4.Sensors = radar4;```\n\n### Visualize the ground truth on a virtual globe\n\nIn this example, `helperScenarioGlobeViewer` provides a virtual globe used to visualize all the elements defined in the tracking scenario: individual debris objects and their trajectories, radar fans, radar detections, and tracks.\n\n```globeDisplay = helperScenarioGlobeViewer; % Show radar beams on the globe covcon = coverageConfig(scene); plotCoverage(globeDisplay,covcon); % Set TargetHistoryLength to visualize the full trajectory of the debris objects globeDisplay.TargetHistoryLength = 1000; scene.StopTime = 3600; scene.UpdateRate = 0.1; while advance(scene) time = scene.SimulationTime; updateDisplay(globeDisplay,time,debris); end snap(globeDisplay);```", null, "On the virtual globe, you can see the space debris represented by white dots with individual trailing trajectories shown by white lines. Most of the generated debris objects are on orbits with high inclination angles close to 80 deg.\n\nThe trajectories are plotted in ECEF coordinates, and therefore the entire trajectory rotates towards the west due to Earth rotation. After several orbit periods, all space debris pass through the surveillance beams of the radars.\n\n### Simulate synthetic detections and track space debris\n\nThe sensor models use the ground truth to generate synthetic detections. Call the `detect` method on the tracking scenario to obtain all the detections in the scene.\n\nA multi-object tracker `trackerJPDA` is used to create new tracks, associate detections to existing tracks, estimate their state, and delete divergent tracks. Setting the property `HasDetectableTrackIDsInput` to true allows the tracker to accept an input that indicates whether a tracked object is detectable in the surveillance region. This is important for not penalizing tracks that are propagated outside of the radar surveillance areas. The utility function `isDetectable `calculates which tracks are detectable at each simulation step.\n\nAdditionally, a utility function `deleteBadTracks` is used to delete divergent tracks faster.\n\n```% Define Tracker tracker = trackerJPDA('FilterInitializationFcn',@initKeplerUKF,... 'HasDetectableTrackIDsInput',true,... 'ClutterDensity',1e-20,... 'AssignmentThreshold',1e4,... 'DeletionThreshold',[7 10]); % Reset scenario, seed, and globe display restart(scene); scene.StopTime = 1800; % 30 min clear(globeDisplay); globeDisplay.TargetHistoryLength = 2; plotCoverage(globeDisplay,covcon); % Initialize tracks confTracks = objectTrack.empty(0,1); while advance(scene) time = scene.SimulationTime; % Generate detections detections = detect(scene); % Generate and update tracks detectableInput = isDetectable(tracker,time, covcon); if ~(isempty(detections) && ~isLocked(tracker)) [confTracks, ~, allTracks,info] = tracker(detections,time,detectableInput); confTracks = deleteBadTracks(tracker,confTracks); end % Update globe display updateDisplay(globeDisplay,time,debris,detections,[],confTracks); % Move camera during simulation and take snapshots switch time case 100 setCamera(globeDisplay,[90 150 5e6],[0 -65 345]); im1 = snap(globeDisplay); case 270 setCamera(globeDisplay,[60 -120 2.6e6],[20 -45 20]); case 380 setCamera(globeDisplay,[60 -120 2.6e6],[20 -45 20]); im2 = snap(globeDisplay); case 400 % reset setCamera(globeDisplay,[17.3 -67.2 2.400e7], [360 -90 0]); case 1500 setCamera(globeDisplay,[54 2.3 6.09e6], [0 -73 348]); case 1560 im3 = snap(globeDisplay); end end % Restore random seed rng(s);```\n`imshow(im1);`", null, "On the first snapshot, you can see an object being tracked as track T1 in yellow. This object was only detected twice, which was not enough to reduce the uncertainty of the track. Therefore, the size of its covariance ellipse is relatively large. You can also observe another track T2 in blue, which is detected by the sensor several times. As a result, its corresponding covariance ellipse is much smaller since more detections were used to correct the state estimate.\n\n`imshow(im2);`", null, "A few minutes later, as seen on the snapshot above, T1 was deleted because the uncertainty of the track has grown too large without detections. On the other hand, the second track T2 survived due to the additional detections.\n\n`imshow(im3)`", null, "In the screenshot above, you can see that track T12 (in purple) is about to enter the radar surveillance area. Track T10 (in orange) was just updated with a detection, which reduced the uncertainty of its estimated position and velocity. With radar station configuration, after 30 minutes of surveillance, 15 tracks were initialized and confirmed out of the 100 debris objects. If you increase the simulation time, the radars will cover 360 degrees in space and eventually more debris can be tracked. Different radar station locations and configurations could be explored to increase the number of tracked objects.\n\n### Summary\n\nIn this example you have learned how to specify your own motion model to move platforms in a tracking scenario and how to use them to setup a tracker. This enables you to apply sensor fusion and tracking techniques offered in this toolbox to a wider range of applications, such as the problem of modelling and tracking space debris in an Earth-Centered-Earth-Fixed coordinate frame as shown in this example.\n\n### Supporting functions\n\nThe motion model used in this example is presented below. The state is the ECEF positions and velocities of the object `[x; vx; y; vy; z; vz]`.\n\n```function state = keplerorbit(state,dt) % keplerorbit performs numerical integration to predict the state of % keplerian bodies. The state is [x;vx;y;vy;z;vz] % Runge-Kutta 4 integration method: k1 = kepler(state); k2 = kepler(state + dt*k1/2); k3 = kepler(state + dt*k2/2); k4 = kepler(state + dt*k3); state = state + dt*(k1+2*k2+2*k3+k4)/6; function dstate=kepler(state) x =state(1,:); vx = state(2,:); y=state(3,:); vy = state(4,:); z=state(5,:); vz = state(6,:); mu = 398600.4405*1e9; % m^3 s^-2 omega = 7.292115e-5; % rad/s r = norm([x y z]); g = mu/r^2; % Coordinates are in a non-intertial frame, account for Coriolis % and centripetal acceleration ax = -g*x/r + 2*omega*vy + omega^2*x; ay = -g*y/r - 2*omega*vx + omega^2*y; az = -g*z/r; dstate = [vx;ax;vy;ay;vz;az]; end end ```\n\n`initKeplerUKF`` `initializes a tracking filter with your own motion model. In this example, we use the same motion model that establishes ground truth,` keplerorbit.`\n\n```function filter = initKeplerUKF(detection) % assumes radar returns [x y z] measmodel= @(x,varargin) x([1 3 5],:); detNoise = detection.MeasurementNoise; sigpos = 0.4;% m sigvel = 0.4;% m/s^2 meas = detection.Measurement; initState = [meas(1); 0; meas(2); 0; meas(3);0]; filter = trackingUKF(@keplerorbit,measmodel,initState,... 'StateCovariance', diag(repmat([10, 10000].^2,1,3)),... 'ProcessNoise',diag(repmat([sigpos, sigvel].^2,1,3)),... 'MeasurementNoise',detNoise); end ```\n\n`oe2rv` converts a set of 6 traditional orbital elements to position and velocity vectors.\n\n```function [r,v] = oe2rv(a,e,i,lan,w,nu) % Reference: Bate, Mueller & White, Fundamentals of Astrodynamics Sec 2.5 mu = 398600.4405*1e9; % m^3 s^-2 % Express r and v in perifocal system cnu = cosd(nu); snu = sind(nu); p = a*(1 - e^2); r = p/(1 + e*cnu); r_peri = [r*cnu ; r*snu ; 0]; v_peri = sqrt(mu/p)*[-snu ; e + cnu ; 0]; % Tranform into Geocentric Equatorial frame clan = cosd(lan); slan = sind(lan); cw = cosd(w); sw = sind(w); ci = cosd(i); si = sind(i); R = [ clan*cw-slan*sw*ci , -clan*sw-slan*cw*ci , slan*si; ... slan*cw+clan*sw*ci , -slan*sw+clan*cw*ci , -clan*si; ... sw*si , cw*si , ci]; r = R*r_peri; v = R*v_peri; end```\n\n`isDetectable` is used in the example to determine which tracks are detectable at a given time.\n\n```function detectInput = isDetectable(tracker,time,covcon) if ~isLocked(tracker) detectInput = zeros(0,1,'uint32'); return end tracks = tracker.predictTracksToTime('all',time); if isempty(tracks) detectInput = zeros(0,1,'uint32'); else alltrackid = [tracks.TrackID]; isDetectable = zeros(numel(tracks),numel(covcon),'logical'); for i = 1:numel(tracks) track = tracks(i); pos_scene = track.State([1 3 5]); for j=1:numel(covcon) config = covcon(j); % rotate position to sensor frame: d_scene = pos_scene(:) - config.Position(:); scene2sens = rotmat(config.Orientation,'frame'); d_sens = scene2sens*d_scene(:); [az,el] = cart2sph(d_sens(1),d_sens(2),d_sens(3)); if abs(rad2deg(az)) <= config.FieldOfView(1)/2 && abs(rad2deg(el)) < config.FieldOfView(2)/2 isDetectable(i,j) = true; else isDetectable(i,j) = false; end end end detectInput = alltrackid(any(isDetectable,2))'; end end ```\n\n`deleteBadTracks` is used to remove tracks that obviously diverged. Specifically, diverged tracks in this example are tracks whose current position has fallen on the surface of the earth and tracks whose covariance has become too large.\n\n```function tracks = deleteBadTracks(tracker,tracks) % remove divergent tracks: % - tracks with covariance > 4*1e8 (20 km standard deviation) % - tracks with estimated position outside of LEO bounds n = numel(tracks); toDelete = zeros(1,n,'logical'); for i=1:numel(tracks) [pos, cov] = getTrackPositions(tracks(i),[ 1 0 0 0 0 0 ; 0 0 1 0 0 0; 0 0 0 0 1 0]); if norm(pos) < 6500*1e3 || norm(pos) > 8500*1e3 || max(cov,[],'all') > 4*1e8 deleteTrack(tracker,tracks(i).TrackID); toDelete(i) =true; end end tracks(toDelete) = []; end```" ]

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[ "# How to make a symbolic function of a matrix ?\n\nHi,\n\nI'm beginning with Sage. I've got two matrix Qf and Xf defined by :\n\nQf = 1000000*matrix([[0,0],[0,1]]);\nXf = matrix(,);\n\n\nI would like to write a symbolic function \"f\" which would take a matrix X with 2 rows & 1 column.\n\nX = var('X');\nf(X) = ((X-Xf).transpose()*Qf*(X-Xf));\n\n\nI easily wrote it with python non-symbolic function syntax, but i didn't find a way to make it symbolic. Because I'll need his gradient later (which is easy to calculate by hand, that I conceed ^^).\n\nMaybe, it's related with SR matrix, no idea, i'm beginning with Sage and that's why I'm asking for help x)\n\nedit retag close merge delete\n\nSort by » oldest newest most voted\n\nYes, SR matrices come in handy for that type of calculations.\n\n# data\nQf = 1000000*matrix([[0,0],[0,1]]);\nXf = matrix([,]);\n\n# matrix with symbolic coefficients\nX = matrix([[var('x1')], [var('x2')]]);\n\nf = ((X-Xf).transpose()*Qf*(X-Xf));\n\n# see result\nQf, Xf, X, f\n\n\nproduces\n\n$$\\newcommand{\\Bold}{\\mathbf{#1}}\\left(\\left(\\begin{array}{rr} 0 & 0 \\\\ 0 & 1000000 \\end{array}\\right), \\left(\\begin{array}{r} 1 \\\\ 1675 \\end{array}\\right), \\left(\\begin{array}{r} x_{1} \\\\ x_{2} \\end{array}\\right), \\left(\\begin{array}{r} 1000000 \\, {\\left(x_{2} - 1675\\right)}^{2} \\end{array}\\right)\\right).$$\n\nTo evaluate $f$, do f(x1=1,x2=1).\n\nMore generally, to define your $X$ it can be useful to do something like:\n\n# create a coefficient matrix of m rows and n columns\nm = 4; n = 2;\nxij = [[var('x'+str(1+i)+str(1+j)) for j in range(n)] for i in range(m)]\nX = matrix(SR, xij)\nX\n\n\n$$\\newcommand{\\Bold}{\\mathbf{#1}}\\left(\\begin{array}{rr} x_{11} & x_{12} \\\\ x_{21} & x_{22} \\\\ x_{31} & x_{32} \\\\ x_{41} & x_{42} \\end{array}\\right).$$\n\nThe quadratic function $f$ defined above is a $1\\times 1$ matrix (convince yourself, for instance by reading the output of type(f)). To take the gradient this is one possible way:\n\n# passing from a 1x1 matrix to a scalar\nf = f[0, 0]\n\n# see result\n\n\n$$\\newcommand{\\Bold}{\\mathbf{#1}}\\left(0, 2000000 x_{2} - 3350000000\\right).$$\n\nmore\n\n1\n\nJust one last thing, is there any way to make elegant evaluation of f, as something like :\n\nM = matrix([[1,2],[3,4],[5,6],[7,8]]);\nprint f(M);\n\n\nOr eventually :\n\nM = matrix([[1,2],[3,4],[5,6],[7,8]]);\nprint f(xij = M[i][j]);\n\n\nYou see the idea ^^\n\ngoing back to the example of above, and if v = vector([1, 2]), then f(X=v) will unfortunately not work, but with f.substitute([X[i] == v[i] for i in range(2)]) it evaluates $f$ at the point $(1, 2)$. I'm not sure if this is what you want to do, so don't hesitate to post a new question!\n\nI works perfectly i'm actually gratefull, believe me :)\n\nThere is my code, if eventually it can serve to someone :\n\nQf = 1000000*matrix([[0,0],[0,1]]);\nXf = matrix([,]);\nX = matrix(SR, [[var('X'+str(1+i)+str(1+j)) for j in range(1)] for i in range(2)]);\nf = ((X-Xf).transpose()*Qf*(X-Xf))[0,0];\n\nXt = matrix([,]);\nprint f;\nprint f.substitute([X[i,0]==Xt[i,0] for i in range(2)]);\n\n\n-1000000.0*(X21 - 1659.0)*(-1.0*X21 + 1659.0)\n-0.0\n\n\nmay i ask you something babacool51, in which context you are using this code? like course of math (linear algebra?) or another? thanks for any feedback!" ]

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[ "Cookie Consent by FreePrivacyPolicy.com\nSearch a number\nBaseRepresentation\nbin1001001011000011001…\n…01110111000000100000\n31010010111201211011201200\n410211201211313000200\n520130432000132300\n6400442001201200\n731525130115342\noct4454145670040\n91103451734650\n10315169927200\n11111732231117\n12510b9930200\n1323949932aab\n141137bb5b292\n1582e938c200\nhex4961977020\n\n315169927200 has 432 divisors, whose sum is σ = 1142514140040. Its totient is φ = 81718640640.\n\nThe previous prime is 315169927187. The next prime is 315169927231. The reversal of 315169927200 is 2729961513.\n\nIt is a happy number.\n\n315169927200 is a `hidden beast` number, since 3 + 1 + 516 + 99 + 27 + 20 + 0 = 666.\n\nIt can be written as a sum of positive squares in 12 ways, for example, as 31404816 + 315138522384 = 5604^2 + 561372^2 .\n\nIt is a super-2 number, since 2×3151699272002 (a number of 24 digits) contains 22 as substring.\n\nIt is a Harshad number since it is a multiple of its sum of digits (45).\n\nIt is an unprimeable number.\n\nIt is a polite number, since it can be written in 71 ways as a sum of consecutive naturals, for example, 43117146 + ... + 43124454.\n\nAlmost surely, 2315169927200 is an apocalyptic number.\n\n315169927200 is a gapful number since it is divisible by the number (30) formed by its first and last digit.\n\nIt is an amenable number.\n\nIt is a practical number, because each smaller number is the sum of distinct divisors of 315169927200, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (571257070020).\n\n315169927200 is an abundant number, since it is smaller than the sum of its proper divisors (827344212840).\n\nIt is a pseudoperfect number, because it is the sum of a subset of its proper divisors.\n\n315169927200 is a wasteful number, since it uses less digits than its factorization.\n\n315169927200 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 7501 (or 7485 counting only the distinct ones).\n\nThe product of its (nonzero) digits is 204120, while the sum is 45.\n\nThe spelling of 315169927200 in words is \"three hundred fifteen billion, one hundred sixty-nine million, nine hundred twenty-seven thousand, two hundred\"." ]

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[ "# Multidimensional array in Java: Place an array in an array to create a table\n\nKeywords: multidimensional array, one-dimensional array, array\n\nIn this article, we will look at the multidimensional array in Java. A multidimensional array consists of two or more dimensions.  We will, of course, see several different examples of how we can work with a multidimensional array in Java.\n\n## What is a multidimensional array in Java?\n\nA multidimensional array consists of two or more dimensions. We previously considered the one-dimensional array as a list that contained several elements that could have values. In the same way, we can view a multidimensional array in two dimensions as a matrix with rows and columns, like a grid or a chessboard.\n\nIf we were to illustrate a multidimensional array in two dimensions in the same way as we did for a one-dimensional array, we could imagine that it looks like:\n\nFurthermore, we can imagine an array with three dimensions as a block, or a cube, and if we try to illustrate it, we get:\n\nBefore we continue and learn more about the multidimensional array, make sure you are familiar with the one-dimensional array in java.\n\n## How a multidimensional array work in Java?\n\nSimilarly to the one-dimensional array, we can use, read and edit each element in the array using indexes. If we imagine a two-dimensional array (it is easiest to imagine) as a matrix arranged in rows and columns, we can read and edit elements in the same way as for a one-dimensional array by writing the array’s name followed by an index in square brackets.\n\nIf we illustrate what the index in a two-dimensional array looks like, we get an image according to figure 3 below, and to make it clearer, we have color-coded row index and column index.\n\nThe array we outlined in figure 3 consists of three columns and four rows, which gives a total of twelve elements.\n\n## How to create a multidimensional array in Java?\n\nA multidimensional array is created in a similar way as for a one-dimensional array, and it can also be described as that there are generally two approaches to creating a multidimensional array. We will go through syntax followed by a shorter example for both, and we have decided to call them approach one and approach two.\n\n### Syntax: Declare two-dimensional array with approach one\n\nIf we start with syntax to declare a two-dimensional array that we think of as a table of rows and columns\n\n```// Declare two-dimensional array with approach one\ndata type[][] name = new data type[number of elements(row)][number of elements (column)];```\n\n### Syntax: Declare three-dimensional array with approach one\n\nFurthermore, if we imagine a coordinate system with x-, y- and z-axis, we create a three-dimensional array by using.\n\n```// Declare three-dimensional array with approach one\ndata type[][][]name = new data type[element x-axis][element y-axis][element z-axis];```\n\nTo conclude, let’s take the declaration step by step\n\n• First, you specify the data type the array should consist of, followed by two square brackets [ ][ ] for a two-dimensional array, or three square brackets [ ][ ][ ] for a three-dimensional array\n• Then you assign a name to the array followed by = new data type. The data type must be the same on both sides\n• Finally, specify the number of elements that the array should contain within the square brackets for the rows and columns, followed by a semicolon\n\n#### Example: Declare a two-dimensional array in Java using procedure one\n\nLet’s take a short example to show what it looks like when using what we call approach one to create a two-dimensional array with 4 rows and 3 columns. If we want to edit or use an element in the array, we do it in the same way as for a one-dimensional array. We can start by inserting three values into our two-dimensional array twoAr.\n\n```public class Example{\npublic static void main(String[] args) {\n\nint [][] twoAr = new int ;\n\ntwoAr = 15;\ntwoAr = 7;\ntwoAr = 22;\n}\n}```\n\nThat will look like\n\nFurthermore, if we then want to use the values we have put into our array, we can for example use\n\n```public class Example{\npublic static void main(String[] args) {\n\nint [][] twoAr = new int ;\ntwoAr = 15;\ntwoAr = 7;\ntwoAr = 22;\n\nint sum = twoAr + twoAr;\nint sumAll = twoAr + twoAr + twoAr;\nint multi = twoAr * twoAr;\n\nSystem.out.println(sum);\nSystem.out.println(sumAll);\nSystem.out.println(multi);\n}\n}```\n\nResulting in\n\n```22\n44\n105```\n\n### Syntax: Declare two-dimensional array with approach two\n\nWe can also create multidimensional arrays with what we call approach two, similarly as we saw for the one-dimensional arrays where we can directly assign values to the array element when declaring it.\n\n```// Declare two-dimensional array with approach two\ndata type [][] name = {{ value, value, ... },\n{ value, value, ... },\n{ value, value, ... }};```\n\nNote that in this case, it is curly brackets { } that are used to the right of the equals sign.\n\n#### Example: Declare a two-dimensional array in Java using procedure two\n\nLet’s look at a short example of how to create a two-dimensional array in Java with what we call approach two.\n\n```public class Example{\npublic static void main(String[] args) {\n\nint [][] twoArr2 = {{2, 3},\n{34, 56},\n{44, 654, 79},\n{12, 9, 23, 44}};\n}\n}```\n\nMoreover, if we then illustrate the array twoArr2\n\nIt is important to pay attention to that we have not created a 4 x 4 matrix; instead, the rows of our array are having a different number of columns. For example, if we would like to add an element to index and type\n\n```public class Example{\npublic static void main(String[] args) {\n\nint [][] twoArr2 = {{2, 3},\n{34, 56},\n{44, 654, 79},\n{12, 9, 23, 44}};\n\ntwoArr2 = 15;\n}\n}```\n\nWe will get java.lang.ArrayIndexOutOfBoundsException. In such cases, we need to create a new array to be able to add more indexes. In the next article, we will learn about ArrayList, which makes it possible to work with changing collections of data.\n\n## Examples: Multidimensional array in Java\n\nLet’s have a look at examples of how to use a two-dimensional and three-dimensional array in Java.\n\n### Example 1: Search a two-dimensional array\n\nIf we want to loop through and search a two-dimensional array, we need to use two for-loops. Similarly, if we had a three-dimensional array, we would have needed three for-loops, so one for-loop per dimension. This can be a little tricky to understand at first, so let’s look at it with an easy and methodical example.\n\nWe use the same two-dimensional array, twoAr, from the previous example and insert a couple of values.\n\n```public class Example{\npublic static void main(String[] args) {\nint [][] twoAr = new int ;\ntwoAr = 15;\ntwoAr = 7;\ntwoAr = 46;\ntwoAr = 22;\ntwoAr = 33;\n}\n}```\n\nResulting in\n\nThen we will use the two for-loops and the .length function to get the length of each row and each column.\n\n```public class Example{\npublic static void main(String[] args) {\nint [][] twoAr = new int ;\ntwoAr = 15;\ntwoAr = 7;\ntwoAr = 46;\ntwoAr = 22;\ntwoAr = 33;\n\nfor (int row = 0; row < twoAr.length; row++){\nfor (int column = 0; column < twoAr[row].length; column++){\n\n}\n}\n}\n}```\n\nWe use the first for-loop to go through each row. We will start at row 0 because the index for an array in Java always starts at 0.\n\n`for (int row = 0; row < twoAr.length; row++){`\n\nThen, we add the for-loop that checks each column in that row.\n\n`for (int column = 0; column < twoAr[row].length; column++){`\n\nWhat we are doing is going through the array row by row, in other words, we start at row index 0 and check all the elements there, then we jump down to row index 1, then continue like this. If we would print all elements in our array twoAr that don’t have the value 0, we see clearly which order the for loops goes through twoAr\n\n```public class Example{\npublic static void main(String[] args) {\nint [][] twoAr = new int ;\ntwoAr = 15;\ntwoAr = 7;\ntwoAr = 46;\ntwoAr = 22;\ntwoAr = 33;\n\nfor (int row = 0; row < twoAr.length; row++){\nfor (int column = 0; column < twoAr[row].length; column++){\n\n// Evaluate that the value is not 0\nif(twoAr[row][column] != 0){\nSystem.out.println(twoAr[row][column]);\n}\n}\n}\n}\n}```\n\nResulting in\n\n```15\n7\n46\n22\n33```\n\nFinally, what is usually a bit complicated to understand is the twoAr[row].length command, but what it does is that for each line, it takes the length of that line. In our case, all rows are the same length, but a two-dimensional array can have different length rows. Furthermore, note that the variable row is something we have created in the for loop\n\n### Example 2: Find the largest value in a two-dimensional array\n\nIf we now would like to add a feature to our program. Say we want to find the largest value in the two-dimensional array from the previous example and print the value plus index where it was found.\n\n```int [][] twoAr = new int ;\ntwoAr = 15;\ntwoAr = 7;\ntwoAr = 46;\ntwoAr = 22;\ntwoAr = 33;```\n\nThis is easily done by using two for loops and an if statement.\n\n```int maxVar = twoAr;\nint rowIndex = 0;\nint columnIndex = 0;\n\nfor (int row = 0; row < twoAr.length; row++){\nfor (int column = 0; column < twoAr[row].length; column++){\n\nif (twoAr[row][column] > maxVar){\nmaxVar = twoAr[row][column];\nrowIndex = row;\ncolumnIndex = column;\n}\n}\n}```\n\nFinally, we use System.out.println\n\n```System.out.println(\"Value: \" + maxVar + \" at index \" +\nrowIndex + \":\" + columnIndex);```\n\nto show our result\n\n`Value: 46 at index 2:2`\n\n### Example 3: Search a three-dimensional array in Java\n\nLet’s look at a third and final example where we will search through a three-dimensional array. To help us, we will use three for loops, and what we will do in this example is to show the value of each element with the associated index.\n\nTo begin with, we create a three-dimensional array with some randomly assigned values\n\n```int[][][] threeArr = { { { 1, 2, 3 }, { 4, 5, 6 } },\n{ { 10, 20, 30 }, { 40, 50, 60 } } };```\n\nThen we initiate the three for-loops we use to go through all the elements in threeArr\n\n```for (int i = 0; i < threeArr.length; i++) {\n\nfor (int j = 0; j < threeArr[i].length; j++) {\n\nfor (int k = 0; k < threeArr[i][j].length; k++) {\n\n// Print the information\nSystem.out.print(\"threeArr[\" + i + \"][\" + j + \"][\" + k + \"] = \"\n+ threeArr[i][j][k] + \"\\t\");\n}\n// Used only for formating\nSystem.out.println();\n}\n// Used only for formating\nSystem.out.println();\n}```\n\nFinally, the result if we execute the code above will be\n\n```threeArr = 1\tthreeArr = 2\tthreeArr = 3\nthreeArr = 4\tthreeArr = 5\tthreeArr = 6\n\nthreeArr = 10\tthreeArr = 20\tthreeArr = 30\nthreeArr = 40\tthreeArr = 50\tthreeArr = 60```\n\n## Why use a multidimensional array in Java?\n\nBasically, a multidimensional array is used to place an array in an array. It may sound complicated, but imagine that you want to save data in tabular format, for example, the test result for the students in a class. This means that we get more storage places to work with; therefore, usually described as an array of an array. In other words, a multidimensional array is used to store information in a matrix form; for example, a calendar or schedule cannot be realized as a one-dimensional array.\n\n## Summary: Multidimensional array in Java\n\nA multidimensional array consists of two or more dimensions. We can think of a two-dimensional array as a matrix or table and a three-dimensional array as a cube. A multidimensional array is usually described as a way to place an array in another array to group and work with data. Multidimensional arrays are created similarly as for a one-dimensional array, and it can also be said here that there are generally two approaches.\n\n### Syntax: Declare two-dimensional array with approach one\n\n```// Declare two-dimensional array with approach one\ndata type[][] name = new data type[number of elements(row)][number of elements (column)];```\n\n### Syntax: Declare two-dimensional array with approach two\n\n```// Declare two-dimensional array with approach two\ndata type [][] name = {{ value, value, ... },\n{ value, value, ... },\n{ value, value, ... }};```\n\n## FAQ: Multidimensional array in Java\n\nHow does a multidimensional array work in Java?\n\nIn the same way, as for a one-dimensional array, we can use, read and edit each element in a multidimensional array using indexes. If we imagine a two-dimensional array as a matrix or a grid arranged in rows and columns, then we can, in the same way as for a one-dimensional array by writing the name of the array followed by an index in square brackets, read and edit elements.\n\nHow to loop through a multidimensional array in Java?\n\nThe easiest way is to use for loops. For example, if we were to search through a two-dimensional array, we would need to use two for-loops; similarly, if we were to have a three-dimensional array, we would need three for-loops. Thus, one for-loop per dimension could be said.\n\nDoes each row in a two-dimensional array have to be the same length?\n\nNo, it’s okay that they are of different lengths. We can easily create a two-dimensional array that contains different numbers of elements per row." ]

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[ "", null, "", null, "# Poisson Distribution Calculator\n\nThe calculator that you can find at the bottom of this page is an online Poisson distribution calculator. It gives the probability of an event occurring a given number of times. You must know the expected frequency of the event.\n\nWe're just coding this formula:\n\nprobability = e[-f + x ln(f) - ln(x!)]\n\nwhere\n\nf = calculated frequency\nx = test frequency\n\nExample:\n\nA firmware is being embedded in a a lot of 2000 microcontrollers. The probability of any microcontroller failing is 0.001 (just a conjecture and nothing to do with reality, ok?) Then we can expect 2 microcontrollers will suffer a severe problem (2000 x 0.001 = 2). What is the probability that 4 micros will fail?\n\nWe use our spreadsheet below and enter known data as input:\n\nEnter calculated frequency = 2\nEnter test frequency = 4\n\nProbability of  4\noccurrences =  0.09022\n\nYou'd have to enter the numbers like this...", null, "Please enable JavaScript codes, otherwise this tool won't work.\n\n Poisson Distribution Enter calculated frequency: Enter test frequency: Probability of occurrences:\n\nReference:\nPoole, L.; Borchers, M.; Some Common Basic Programs; Osborne/McGraw-Hill;\n3rd. edition; Berkeley, CA; 1979.\n\nFrom 'Poisson Distribution Calculator' to home\n\nFrom 'Poisson Distribution Calculator' to Free Online Calculators", null, "" ]

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[ "# #007001 Color Information\n\nIn a RGB color space, hex #007001 is composed of 0% red, 43.9% green and 0.4% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 0% magenta, 99.1% yellow and 56.1% black. It has a hue angle of 120.5 degrees, a saturation of 100% and a lightness of 22%. #007001 color hex could be obtained by blending #00e002 with #000000. Closest websafe color is: #006600.\n\n• R 0\n• G 44\n• B 0\nRGB color chart\n• C 100\n• M 0\n• Y 99\n• K 56\nCMYK color chart\n\n#007001 color description : Very dark lime green.\n\n# #007001 Color Conversion\n\nThe hexadecimal color #007001 has RGB values of R:0, G:112, B:1 and CMYK values of C:1, M:0, Y:0.99, K:0.56. Its decimal value is 28673.\n\nHex triplet RGB Decimal 007001 `#007001` 0, 112, 1 `rgb(0,112,1)` 0, 43.9, 0.4 `rgb(0%,43.9%,0.4%)` 100, 0, 99, 56 120.5°, 100, 22 `hsl(120.5,100%,22%)` 120.5°, 100, 43.9 006600 `#006600`\nCIE-LAB 40.557, -46.938, 45.095 5.799, 11.59, 1.96 0.3, 0.599, 11.59 40.557, 65.09, 136.147 40.557, -38.386, 49.503 34.044, -29.17, 20.417 00000000, 01110000, 00000001\n\n# Color Schemes with #007001\n\n• #007001\n``#007001` `rgb(0,112,1)``\n• #70006f\n``#70006f` `rgb(112,0,111)``\nComplementary Color\n• #377000\n``#377000` `rgb(55,112,0)``\n• #007001\n``#007001` `rgb(0,112,1)``\n• #007039\n``#007039` `rgb(0,112,57)``\nAnalogous Color\n• #700037\n``#700037` `rgb(112,0,55)``\n• #007001\n``#007001` `rgb(0,112,1)``\n• #390070\n``#390070` `rgb(57,0,112)``\nSplit Complementary Color\n• #700100\n``#700100` `rgb(112,1,0)``\n• #007001\n``#007001` `rgb(0,112,1)``\n• #010070\n``#010070` `rgb(1,0,112)``\n• #6f7000\n``#6f7000` `rgb(111,112,0)``\n• #007001\n``#007001` `rgb(0,112,1)``\n• #010070\n``#010070` `rgb(1,0,112)``\n• #70006f\n``#70006f` `rgb(112,0,111)``\n• #002400\n``#002400` `rgb(0,36,0)``\n• #003d01\n``#003d01` `rgb(0,61,1)``\n• #005701\n``#005701` `rgb(0,87,1)``\n• #007001\n``#007001` `rgb(0,112,1)``\n• #008a01\n``#008a01` `rgb(0,138,1)``\n• #00a301\n``#00a301` `rgb(0,163,1)``\n• #00bd02\n``#00bd02` `rgb(0,189,2)``\nMonochromatic Color\n\n# Alternatives to #007001\n\nBelow, you can see some colors close to #007001. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #1b7000\n``#1b7000` `rgb(27,112,0)``\n• #127000\n``#127000` `rgb(18,112,0)``\n• #087000\n``#087000` `rgb(8,112,0)``\n• #007001\n``#007001` `rgb(0,112,1)``\n• #00700a\n``#00700a` `rgb(0,112,10)``\n• #007014\n``#007014` `rgb(0,112,20)``\n• #00701d\n``#00701d` `rgb(0,112,29)``\nSimilar Colors\n\n# #007001 Preview\n\nThis text has a font color of #007001.\n\n``<span style=\"color:#007001;\">Text here</span>``\n#007001 background color\n\nThis paragraph has a background color of #007001.\n\n``<p style=\"background-color:#007001;\">Content here</p>``\n#007001 border color\n\nThis element has a border color of #007001.\n\n``<div style=\"border:1px solid #007001;\">Content here</div>``\nCSS codes\n``.text {color:#007001;}``\n``.background {background-color:#007001;}``\n``.border {border:1px solid #007001;}``\n\n# Shades and Tints of #007001\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, #000e00 is the darkest color, while #f9fff9 is the lightest one.\n\n• #000e00\n``#000e00` `rgb(0,14,0)``\n• #002200\n``#002200` `rgb(0,34,0)``\n• #003500\n``#003500` `rgb(0,53,0)``\n• #004901\n``#004901` `rgb(0,73,1)``\n• #005c01\n``#005c01` `rgb(0,92,1)``\n• #007001\n``#007001` `rgb(0,112,1)``\n• #008401\n``#008401` `rgb(0,132,1)``\n• #009701\n``#009701` `rgb(0,151,1)``\n• #00ab02\n``#00ab02` `rgb(0,171,2)``\n• #00be02\n``#00be02` `rgb(0,190,2)``\n• #00d202\n``#00d202` `rgb(0,210,2)``\n• #00e602\n``#00e602` `rgb(0,230,2)``\n• #00f902\n``#00f902` `rgb(0,249,2)``\n• #0eff10\n``#0eff10` `rgb(14,255,16)``\n• #22ff24\n``#22ff24` `rgb(34,255,36)``\n• #35ff37\n``#35ff37` `rgb(53,255,55)``\n• #49ff4a\n``#49ff4a` `rgb(73,255,74)``\n• #5cff5e\n``#5cff5e` `rgb(92,255,94)``\n• #70ff71\n``#70ff71` `rgb(112,255,113)``\n• #84ff85\n``#84ff85` `rgb(132,255,133)``\n• #97ff98\n``#97ff98` `rgb(151,255,152)``\n• #abffac\n``#abffac` `rgb(171,255,172)``\n• #beffbf\n``#beffbf` `rgb(190,255,191)``\n• #d2ffd2\n``#d2ffd2` `rgb(210,255,210)``\n• #e6ffe6\n``#e6ffe6` `rgb(230,255,230)``\n• #f9fff9\n``#f9fff9` `rgb(249,255,249)``\nTint Color Variation\n\n# Tones of #007001\n\nA tone is produced by adding gray to any pure hue. In this case, #343c34 is the less saturated color, while #007001 is the most saturated one.\n\n• #343c34\n``#343c34` `rgb(52,60,52)``\n• #2f4130\n``#2f4130` `rgb(47,65,48)``\n• #2b452b\n``#2b452b` `rgb(43,69,43)``\n• #274927\n``#274927` `rgb(39,73,39)``\n• #224e23\n``#224e23` `rgb(34,78,35)``\n• #1e521f\n``#1e521f` `rgb(30,82,31)``\n• #1a561a\n``#1a561a` `rgb(26,86,26)``\n• #165a16\n``#165a16` `rgb(22,90,22)``\n• #115f12\n``#115f12` `rgb(17,95,18)``\n• #0d630e\n``#0d630e` `rgb(13,99,14)``\n• #096709\n``#096709` `rgb(9,103,9)``\n• #046c05\n``#046c05` `rgb(4,108,5)``\n• #007001\n``#007001` `rgb(0,112,1)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #007001 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 calculate time series seasonality index in R?\n\nIn R, I use the decompose method on my time series object and it gives me seasonal + trend + random component. For seasonal component, it gives me absolute value which is good but I would also like to know the monthly seasonality index as well (like Jan .084, Feb 0.90, Mar 1.12, etc., for example). Is there a quick way to get this seasonality index in R? Also, how is it normally calculated?\n\n• tslm function in forecast package does this automatically. It is wrapper of lm function for time series problems that creates trend and seasonal index automatically. You could try tslm function – forecaster May 5 '14 at 0:45\n\n## 2 Answers\n\nJust extract the \"figure\" component from your \"decomposed.ts\" object. The seasonal component is just the recycled figure over the time range of the time series.\n\nAs for the calculation, I find the explanation in the details section of the manual page helpful: The function first determines the trend component using a moving average (if filter is NULL, a symmetric window with equal weights is used), and removes it from the time series. Then, the seasonal figure is computed by averaging, for each time unit, over all periods. The seasonal figure is then centered. Finally, the error component is determined by removing trend and seasonal figure (recycled as needed) from the original time series.\n\nOf course, there are other methods for constructing such a seasonal index/figure, including the mentioned tslm (or dynlm from the package of the same name) or stl (in stats).\n\nHere's a code snippet that could help.\n\nlibrary(fpp) dec_series <- decompose(a10) dec_series\\$figure" ]

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[ "", null, "# Free vibration of circular annular plate with different boundary conditions\n\n## Yash Jaiman1, Baij Singh2\n\n1Bennett University, Greater Noida, Uttar Pradesh, India\n\n2Indian Institute of Technology (ISM), Dhanbad, India\n\n1Corresponding author\n\nVibroengineering PROCEDIA, Vol. 29, 2019, p. 82-86. https://doi.org/10.21595/vp.2019.21116\nReceived 23 October 2019; accepted 4 November 2019; published 28 November 2019\n\nCopyright © 2019 Yash Jaiman, et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.\nAbstract.\n\nThis paper deals with the numerical simulation of free vibration analysis of a thin circular annular plate for various boundary conditions at the outer edge and inner edge. Classical plate theory is used to derive the governing differential equation for the transverse deflection of the thin isotropic plate. The finite element method is used to evaluate the first six natural frequencies and mode shapes of the thin uniform circular annular plate with radius ratios $\\left({r}_{1}/{r}_{2}\\right)$ for different boundary conditions. These natural frequencies results are compared with those available in the literature. The results are verified with classical plate theory with our Abaqus results and checked with the previous research literature on the topic.\n\nKeywords: circular annular plate, free vibration, numerical simulation.\n\n#### 1. Introduction\n\nPlates are widely used as a structural element and have vast practical applications in many engineering fields such as aerospace, mechanical, civil, nuclear, electronic, automotive, marine and heavy machinery, etc. Various researchers have analyzed the free vibration behavior of circular annular plates of different shapes, sizes, thickness for different boundary conditions. Leissa used the Ritz method to estimate the natural frequencies of the isotropic plate for different boundary conditions. Kim and Dickinson used the Rayleigh-Ritz approximation method for free vibration of a thin plate to extract natural frequencies. Rajalingham et al. used a Rayleigh-Ritz method to analyze the plate characteristics parameter as shape functions and continued his work to formulate a variational reduction expression to analyze frequencies and mode shapes. Liew et al. used the polynomials-Ritz method for the vibration of circular plates by using three-dimensional elasticity solutions. Zhou et al. used the Chebyshev-Ritz method for three-dimensional vibration and mode shapes of the circular plate. Lim et al. used the state-space method to analyze transverse vibration and mode of a thick circular plate. Zhou et al. used the Hamiltonian principle to solve governing equations for free vibration analysis by using the variational principle of mixed energy method. Kumar et al. use a dynamic stiffness method to extract the natural frequency and mode shapes of a thin plate. Piyush et al. used the Rayleigh-Ritz method to compute the natural frequencies of the thin plate.\n\n#### 2. Basic formulation\n\nConsider a homogeneous, isotropic circular annular plate in cylindrical coordinates $\\left(r,\\theta ,z\\right)$ with uniform thickness $h$ as shown in Fig. 1.\n\nClassical plate theory is used to derive the governing differential equation for transverse vibration in the polar coordinate system is defined as:\n\n(1)\n\nwhere Laplacian operator: ${\\nabla }^{2}=\\frac{{\\partial }^{2}}{{\\partial r}^{2}}+\\frac{1}{r}\\frac{\\partial }{\\partial r}+\\frac{1}{{r}^{2}}\\frac{{\\partial }^{2}}{{\\partial \\theta }^{2}}$, $\\rho$ is the mass density, $D=E{h}^{3}/\\left[12\\left(1-{\\nu }^{2}\\right)\\right]$ is the flexural rigidity and $\\nu$ is Poisson’s ratio.\n\nTransverse deflection of natural vibrations for thin circular plate is assumed to be:\n\n(2)\n$W\\left(r,\\theta ,t\\right)=w\\left(r,\\theta \\right){e}^{i\\omega t},$\n\nwhere $\\omega$ is the natural frequency and $w\\left(r,\\theta \\right)$ is natural mode.\n\nSubstituting Eqs. (2) in (1), we get:\n\n(3)\n\nwhere:\n\n(4)\n${\\gamma }^{4}=\\frac{\\rho h{\\omega }^{2}}{D}.$\n\nThe general finite element equation for the transverse deflection of thin plate is given by:\n\n(5)\n$\\left[M\\right]\\left\\{\\stackrel{¨}{q}\\right\\}+\\left[K\\right]\\left\\{q\\right\\}=0,$\n\nwhere $\\left[M\\right]$ is the mass matrix, $\\left[K\\right]$ is stiffness matrix, $\\left\\{\\stackrel{¨}{q}\\right\\}$ is the nodal acceleration vector, and $\\left\\{q\\right\\}$ is nodal displacement vector.\n\nThe non-dimension natural frequency parameter $\\varpi$ are calculated as:\n\n(6)\n$\\varpi =2\\pi \\omega \\left[\\sqrt{\\rho h/D}\\right]{r}_{1}^{2},$\n\nwhere $\\omega$ is the natural frequency in Hz.\n\nFig. 1. Schematic diagram and coordinate system of annular circular plate", null, "#### 3. Results and discussions\n\nIn this section, the first six non-dimensional natural frequencies and mode shapes of the circular annular plate are estimated by using the finite element method. Here, we calculated different eigenvalues for different boundary conditions with different radii ratio $\\left({r}_{1}/{r}_{2}\\right)$. Different combinations of boundary conditions are applied to compute the natural frequencies and mode shapes of the circular annular plate. 2820 elements and 5922 nodes are used to estimate the natural frequencies and mode shape function of thin circular annular plates after convergence study. The present natural frequencies results are compared with those available in the literature.\n\nTables 1-4 shows that the first six non-dimensional natural frequencies values for the circular annular plate. These present results are nearly the same as Leissa and Zhou under different boundary conditions.\n\nFig. 2. Natural modes of a clamped annular plate with a free inner boundary, = 0.4", null, "", null, "", null, "", null, "", null, "", null, "Table 5-7 presents the effect of the radii ratio on the non-dimensional frequency parameter of thin plates. It is observed from these tables that as the radii ratio increases non-dimensional frequency parameter increases.\n\nTable 1. Comparison of the non-dimensional natural frequency parameter with Leissa and Zhou for clamped outer and free inner boundary ($\\nu =1/3$, )\n\n Results Mode number 1 2 3 4 5 6 Leissa 13.54 19.80 31.34 – – – Zhou 13.500 19.389 31.338 46.855 65.984 66.924 Present 12.871 18.497 29.901 44.70 62.982 63.935\n\nTable 2. Comparison of the non-dimensional natural frequency parameter with Leissa and Zhou for free outer and clamped inner boundary ($\\nu =1/3$, )\n\n Results Mode number 1 2 3 4 5 6 Leissa 9.096 10.37 – – – – Zhou 9.0719 9.1294 10.366 14.726 22.530 33.455 Present 9.0257 9.0868 10.3267 14.6805 22.5682 33.4769\n\nTable 3. Comparison of the non-dimensional natural frequency parameter with Leissa and Zhou for free outer and free inner boundary ($\\nu =1/3$, )\n\n Results Mode number 1 2 3 4 5 6 Leissa – 4.567 – – – – Zhou 4.5325 8.5510 11.765 17.043 21.262 31.356 Present 4.5197 8.5070 11.7368 16.9765 21.233 31.2565\n\nTable 4. Comparison of the non-dimensional natural frequency parameter with Leissa and Zhou for clamped outer and clamped inner boundary ($\\nu =1/3$, )\n\n Results Mode number 1 2 3 4 5 6 Leissa 62.33 62.92 66.406 – – – Zhou 61.872 62.966 66.672 73.630 84.594 99.904 Present 62.0056 63.1121 66.7350 73.6148 84.5044 99.7988\n\nTable 5. Non-dimensional frequency parameter for the annular circular plate with clamped outer and free inner edge ($\\nu =1/3$)\n\n $n$ ${r}_{1}/{r}_{2}$ 0.2 0.4 0.6 0.8 0 10.2922 12.871 25.3966 92.3617 1 20.4306 18.497 28.3519 94.2334 2 33.6865 29.001 36.2745 98.3097 3 50.4963 44.70 47.9049 105.6264 4 69.6342 62.982 62.8458 115.7173 5 91.0597 83.751 79.4797 128.4994\n\nTable 6. Non-dimensional frequency parameter for the annular circular plate with free outer and clamped inner edge ($\\nu =1/3$)\n\n $n$ ${r}_{1}/{r}_{2}$ 0.2 0.4 0.6 0.8 0 4.795 9.0257 20.5213 84.5418 1 5.190 9.0868 20.8923 85.1491 2 6.322 10.3267 22.3356 87.0250 3 12.367 14.6805 25.6329 90.3110 4 21.502 22.5682 31.5306 95.2026 5 32.438 33.4769 40.3467 101.941\n\nTable 7. Non-dimensional frequency parameter for the annular circular plate with clamped outer and clamped inner edge ($\\nu =1/3$)\n\n $n$ ${r}_{1}/{r}_{2}$ 0.2 0.4 0.6 0.8 0 4.795 9.0257 141.4272 593.2696 1 5.190 9.0868 144.8754 593.8935 2 6.322 10.3267 149.467 595.8484 3 12.367 14.6805 156.3015 599.0929 4 21.502 22.5682 165.7894 603.7515 5 32.438 33.4769 178.2596 609.866\n\n#### 4. Conclusions\n\nIn this paper, numerical analysis for free vibration analysis of a thin annular solid plate is carried out using the finite element method for different boundary conditions at the inner and outer radius. It is found that those natural frequency results are quite close to those reported in previous works of literature. The novelty of this paper is the effect of the radii ratio on natural frequency is discussed and found that with increasing radii ratio, natural frequency increases and another novelty is by using shell element modeling as per Abaqus convention the dimensionless frequency parameter as found in the literature are completely validated.\n\n1. Leissa A. W. Vibration of Plates. Office of Technology Utilization, Washington, 1969. [CrossRef]\n2. Kim C. S., Dickinson S. M. On the free, transverse vibration of annular and circular, thin, sectorial plates subject to certain complicating effects. Journal of Sound and Vibration, Vol. 134, Issue 3, 1989, p. 407-421. [Publisher]\n3. Rajalingham C., Bhat R. B., Xistris G. D. Vibration of rectangular plates using plate characteristic functions as shape functions in the Rayleigh-Ritz method. Journal of Sound and Vibration, Vol. 193, Issue 2, 1996, p. 497-509. [Publisher]\n4. Liew K. M., Yang B. Three-dimensional elasticity solutions for free vibrations of circular plates: a polynomials-Ritz analysis. Computer Methods in Applied Mechanics and Engineering, Vol. 175, Issues 1-2, 1999, p. 189-201. [Publisher]\n5. Zhou D., Au F. T. K., Cheung Y. K., Lo S. H. Three-dimensional vibration analysis of circular and annular plates via the Chebyshev-Ritz method. International Journal of Solids and Structures, Vol. 40, Issue 12, 2003, p. 3089-3105. [Publisher]\n6. Lim C. W., Li Z. R., Xiang Y., Wei G. W., Wang C. M. On the missing modes when using the exact frequency relationship between Kirchhoff and Mindlin plates. Advances in Vibration Engineering, Vol. 4, Issue 3, 2005, p. 221-248. [CrossRef]\n7. Zhou Z. H., Wong K. W., Xu X. S., Leung A. Y. T. Natural vibration of circular and annular thin plates by Hamiltonian approach. Journal of Sound and Vibration, Vol. 330, Issue 5, 2011, p. 1005-1017. [Publisher]\n8. Kumar S., Vinayak Ranjan, Jana P. Free vibration analysis of thin functionally graded rectangular plates using the dynamic stiffness method. Composite Structures, Vol. 197, 2018, p. 39-53. [Publisher]\n9. Pratap Singh P., Azam M. S., Vinayak Ranjan Vibration analysis of a thin functionally graded plate having an out of plane material inhomogeneity resting on Winkler-Pasternak foundation under different combinations of boundary conditions. Proceedings of the Institution of Mechanical Engineers, Part C: Journal of Mechanical Engineering Science, 2018. [CrossRef]" ]

[ null, "https://www.facebook.com/tr", null, "https://www.jvejournals.com/article/21116/fig1", null, "https://www.jvejournals.com/article/21116/fig2", null, "https://www.jvejournals.com/article/21116/fig3", null, "https://www.jvejournals.com/article/21116/fig4", null, "https://www.jvejournals.com/article/21116/fig5", null, "https://www.jvejournals.com/article/21116/fig6", null, "https://www.jvejournals.com/article/21116/fig7", null ]

[ "+0\n\n# help\n\n0\n83\n1\n\nThe total length of pencils A, B and C is 29 cm. Pencil A is 11 cm shorter than pencil B, and pencil B is twice as long as pencil C. How long is pencil A?\n\nJan 15, 2020\n\n#1\n+1\n\nLet B   = L\n\nLet A = L - 11\n\nLet C  = (1/2)B    = (1/2)L  = .5L\n\nSo we have that\n\nA + B + C   =  29\n\n(L -11) + L + (.5)L  = 29         simplify\n\n2.5L -  11  = 29             add 11 to both sides\n\n2.5L =  40                     divide both sides by 2.5\n\nL  = 16\n\nA =  L - 11  =   16  - 11   =    5 cm", null, "", null, "", null, "Jan 15, 2020" ]

[ null, "https://web2.0calc.com/img/emoticons/smiley-cool.gif", null, "https://web2.0calc.com/img/emoticons/smiley-cool.gif", null, "https://web2.0calc.com/img/emoticons/smiley-cool.gif", null ]

[ "C PROGRAMMING\n\n# Tic tac toe game\n\nToday I will show you how we can make our childhood game, Tic tac toe using c programming. The tic-tac-toe game is played on a 3×3 grid the game is played by two players, who take turns. The first player marks move with a circle, the second with a cross. The player who has formed a horizontal, vertical, or diagonal sequence of three marks wins. Your program should draw the game board, ask the user for the coordinates of the next mark, change the players after every successful move, and pronounce the winner. every successful move, and pronounce the winner.\n\n### Below is the source code of tic tac toe\n\n``````#include <stdio.h>\n#include <conio.h>\nchar square = { 'o', '1', '2', '3', '4', '5', '6', '7', '8', '9' };\nint checkwin();\nvoid board();\nint main()\n{\nint player = 1, i, choice;\nchar mark;\ndo\n{\nboard();\nplayer = (player % 2) ? 1 : 2;\nprintf(\"Player %d, enter a number: \", player);\nscanf(\"%d\", &choice);\nmark = (player == 1) ? 'X' : 'O';\nif (choice == 1 && square == '1')\nsquare = mark;\n\nelse if (choice == 2 && square == '2')\nsquare = mark;\n\nelse if (choice == 3 && square == '3')\nsquare = mark;\n\nelse if (choice == 4 && square == '4')\nsquare = mark;\n\nelse if (choice == 5 && square == '5')\nsquare = mark;\n\nelse if (choice == 6 && square == '6')\nsquare = mark;\n\nelse if (choice == 7 && square == '7')\nsquare = mark;\n\nelse if (choice == 8 && square == '8')\nsquare = mark;\n\nelse if (choice == 9 && square == '9')\nsquare = mark;\n\nelse\n{\nprintf(\"Invalid move \");\nplayer--;\ngetch();\n}\ni = checkwin();\nplayer++;\n}while (i == - 1);\n\nboard();\n\nif (i == 1)\nprintf(\"==>\\aPlayer %d win \", --player);\nelse\nprintf(\"==>\\aGame draw\");\ngetch();\nreturn 0;\n}\n/*********************************************\nFUNCTION TO RETURN GAME STATUS\n1 FOR GAME IS OVER WITH RESULT\n-1 FOR GAME IS IN PROGRESS\nO GAME IS OVER AND NO RESULT\n**********************************************/\nint checkwin()\n{\nif (square == square && square == square)\nreturn 1;\n\nelse if (square == square && square == square)\nreturn 1;\n\nelse if (square == square && square == square)\nreturn 1;\n\nelse if (square == square && square == square)\nreturn 1;\n\nelse if (square == square && square == square)\nreturn 1;\n\nelse if (square == square && square == square)\nreturn 1;\n\nelse if (square == square && square == square)\nreturn 1;\n\nelse if (square == square && square == square)\nreturn 1;\n\nelse if (square != '1' && square != '2' && square != '3' &&\nsquare != '4' && square != '5' && square != '6' && square\n!= '7' && square != '8' && square != '9')\nreturn 0;\nelse\nreturn - 1;\n}\n/*******************************************************************\nFUNCTION TO DRAW BOARD OF TIC TAC TOE WITH PLAYERS MARK\n********************************************************************/\nvoid board()\n{\nsystem(\"cls\");\nprintf(\"\\n\\n\\tTic Tac Toe\\n\\n\");\nprintf(\"Player 1 (X) - Player 2 (O)\\n\\n\\n\");\nprintf(\" | | \\n\");\nprintf(\" %c | %c | %c \\n\", square, square, square);\nprintf(\"_____|_____|_____\\n\");\nprintf(\" | | \\n\");\nprintf(\" %c | %c | %c \\n\", square, square, square);\nprintf(\"_____|_____|_____\\n\");\nprintf(\" | | \\n\");\nprintf(\" %c | %c | %c \\n\", square, square, square);\nprintf(\" | | \\n\\n\");\n}\n/*******************************************************************\nEND OF PROJECT\n********************************************************************/``````\n\nCheck Also\nClose" ]

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[ "## Wednesday, March 24, 2021\n\n### What I Remember from Math Class\n\nPi is that 3.14 number, and it just goes on and on.\n\nYou use pi to find the area of a circle.\nIn eighth grade Sister Rose made us memorize the quadratic formula.\n\n(this image is fuzzy like my memory)\n\nI have no idea when one would use the quadratic formula.\nIn geometry class, we had to write out a bunch of proofs,\nto make sure that shit that mathematicians have known\nsince the days of the ancient Greeks is still true.\nSine and Cosine and Tangent…they are ratios of some sort.\nYou use them for trigonometry.\nYou do derivatives in calculus, but I don’t remember why.\n\nIn real life I can do basic arithmetic to balance my checkbook\nand proportions to figure out how many minutes\nit takes me to jog a mile.\nI calculate percentages for tips,\nbut I zone out when it’s time to split the bill at a restaurant.\nMost important of all, I use fractions for baking,\nlike when I can’t find the 1/4 cup,\nso I scoop out 2/8 of a cup of flour instead." ]

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[ "### Change: spat_to_SHqst uses a merged 3-component even with DCT enabled....\n\n`Change: spat_to_SHqst uses a merged 3-component even with DCT enabled. (changes made to the DCT spat_to_SH*)`\nparent f0cabf1d\n ... ... @@ -1376,8 +1376,8 @@ int shtns_set_size(int lmax, int mmax, int mres, enum shtns_norm norm) int with_cs_phase = 1; /// Condon-Shortley phase (-1)^m is used by default. double mpos_renorm = 1.0; /// renormalization of m>0. if (lmax < 1) shtns_runerr(\"lmax must be larger than 1\"); // if (lmax < 2) shtns_runerr(\"lmax must be at least 2\"); // if (lmax < 1) shtns_runerr(\"lmax must be larger than 1\"); if (lmax < 2) shtns_runerr(\"lmax must be at least 2\"); if (li != NULL) { if ( (lmax != LMAX)||(mmax != MMAX)||(mres != MRES) ) shtns_runerr(\"different size already set\"); ... ...\n ... ... @@ -21,15 +21,9 @@ V/// complex double arrays of size NLM. /// \\param[in] ltr = specify maximum degree of spherical harmonic. ltr must be at most LMAX, and all spherical harmonic degree higher than ltr are set to zero. #endif #ifdef SHT_3COMP #define SHT_NO_DCT #endif #Q void spat_to_SH(double *Vr, complex double *Qlm) #V void spat_to_SHsphtor(double *Vt, double *Vp, complex double *Slm, complex double *Tlm) # { Q complex double *Ql; // virtual pointers for given im V complex double *Sl, *Tl; // virtual pointers for given im Q complex double *BrF; // contains the Fourier transformed data V complex double *BtF, *BpF; // contains the Fourier transformed data Q double *zl; ... ... @@ -130,26 +124,28 @@ V fftw_execute_r2r(dct_r1,(double *) BtF, (double *) BtF); // DCT in-place. V fftw_execute_r2r(dct_r1,(double *) BpF, (double *) BpF); // DCT in-place. #endif long int klim = shtns.klim; Q l=0; V l=1; Q Ql = Qlm; // virtual pointer for l=0 and im V Sl = Slm; Tl = Tlm; l=0; Q v2d* Ql = (v2d*) Qlm; V v2d* Sl = (v2d*) Slm; v2d* Tl = (v2d*) Tlm; Q zl = zlm_dct0; V dzl0 = dzlm_dct0; V #ifndef _GCC_VEC_ V s1 = 0.0; t1 = 0.0; // l=0 : Sl = Tl = 0 V #else V v2d s = vdup(0.0); v2d t = vdup(0.0); // l=0 : Sl = Tl = 0 V #endif #ifdef SHT_VAR_LTR i = (LTR * SHT_NL_ORDER) + 2; // sum truncation if (i < klim) klim = i; while(l < LTR) { #else do { // l < LMAX #endif Q zl = zlm_dct0; V dzl0 = dzlm_dct0; V Sl = 0.0; Tl = 0.0; # qs0 = 0.0; qs1 = 0.0; // sum of first Ql's while(l 0 and odd => i >= 0 and even i=l; // l < klim #ifndef _GCC_VEC_ V Sl[l] = s1; Tl[l] = t1; Q q0 = 0.0; q1 = 0.0; V s0 = 0.0; t1 = 0.0; t0 = 0.0; s1 = 0.0; # qs0 += BR0[l]; qs1 += BR0[l+1]; # q0 = qs0*zl[i]; q1 = qs1*zl[i+1]; do { Q q0 += BR0[i] * zl; Q q1 += BR0[i+1] * zl; ... ... @@ -165,12 +161,12 @@ V dzl0+=2; Q zl += (shtns.klim-i); V dzl0 += (shtns.klim-i); #endif Q Ql[l] = q0; Ql[l+1] = q1; V Sl[l] = s0; Sl[l+1] = s1; V Tl[l] = t0; Tl[l+1] = t1; Q Ql[l] = q0; Ql[l+1] = q1; V Sl[l+1] = s0; Tl[l+1] = t0; #else V Sl[l] = vhi_to_cplx(s); Tl[l] = vhi_to_cplx(t); Q v2d q = vdup(0.0); V v2d s = vdup(0.0); v2d t = vdup(0.0); V s = vdup(0.0); t = vdup(0.0); i >>= 1; // i = i/2 do { Q q += ((v2d*) zl) * ((v2d*) BR0)[i]; ... ... @@ -184,27 +180,29 @@ V dzl0 +=2; Q zl += (shtns.klim-2*i); V dzl0 += (shtns.klim-2*i); #endif Q ((v2d*) Ql)[l] = vlo_to_cplx(q); ((v2d*) Ql)[l+1] = vhi_to_cplx(q); V ((v2d*) Sl)[l] = vlo_to_cplx(s); ((v2d*) Sl)[l+1] = vhi_to_cplx(s); V ((v2d*) Tl)[l] = vlo_to_cplx(t); ((v2d*) Tl)[l+1] = vhi_to_cplx(t); Q Ql[l] = vlo_to_cplx(q); Ql[l+1] = vhi_to_cplx(q); V Sl[l+1] = vlo_to_cplx(s); Tl[l+1] = vlo_to_cplx(t); #endif l+=2; #ifndef SHT_VAR_LTR } while(l\n ... ... @@ -96,31 +96,27 @@ void GEN(spat_to_SHsphtor,SUFFIX)(double *Vt, double *Vp, complex double *Slm, c */ //@{ /// \\b 3D Vector Spherical Harmonics Transform (analysis) : convert a 3D vector field (r,theta,phi components) to its radial/spheroidal/toroidal spherical harmonic representation. /// This is basically a shortcut to call both spat_to_SH* and spat_to_SHsphtor* but may be significantly faster. void GEN(spat_to_SHqst,SUFFIX)(double *Vr, double *Vt, double *Vp, complex double *Qlm, complex double *Slm, complex double *Tlm SUPARG) { #define SHT_3COMP #include \"spat_to_SHqst.c\" #undef SHT_3COMP } /// Backward \\b 3D Vector Spherical Harmonic Transform (synthesis). /// This is basically a shortcut to call both SH_to_spat* and SHsphtor_to spat* but may be significantly faster. void GEN(SHqst_to_spat,SUFFIX)(complex double *Qlm, complex double *Slm, complex double *Tlm, double *Vr, double *Vt, double *Vp SUPARG) { if (MTR_DCT < 0) { #define SHT_3COMP #include \"SHqst_to_spat.c\" #undef SHT_3COMP } else { if (MTR_DCT >= 0) { GEN(SH_to_spat,SUFFIX)(Qlm, Vr SUPARG2); GEN(SHsphtor_to_spat,SUFFIX)(Slm, Tlm, Vt, Vp SUPARG2); } } /// \\b 3D Vector Spherical Harmonics Transform (analysis) : convert a 3D vector field (r,theta,phi components) to its radial/spheroidal/toroidal spherical harmonic representation. /// This is basically a shortcut to call both spat_to_SH* and spat_to_SHsphtor* but may be significantly faster. void GEN(spat_to_SHqst,SUFFIX)(double *Vr, double *Vt, double *Vp, complex double *Qlm, complex double *Slm, complex double *Tlm SUPARG) { if (MTR_DCT < 0) { } else { #define SHT_3COMP #include \"spat_to_SHqst.c\" #include \"SHqst_to_spat.c\" #undef SHT_3COMP } else { GEN(spat_to_SHsphtor,SUFFIX)(Vt, Vp, Slm, Tlm SUPARG2); GEN(spat_to_SH,SUFFIX)(Vr, Qlm SUPARG2); } } //@} ... ...\n ... ... @@ -28,6 +28,9 @@ for m=2, z[k] is constant up to l-1 dtz[k] is constant up to l-2 for m=3, z[k] is constant up to l-2 - there is a need for a merged (3 components) DCT version - there is a need for transposed version !! - Optimize m=0 and NPHI=1 (don't need to go to \"complex\" representation) - Using CUDA may be interesting ! ... ...\n ... ... @@ -143,7 +143,7 @@ extern fftw_plan ifft_eo, fft_eo; // for half the size (given parity) #undef _GCC_VEC_ typedef double s2d; typedef complex double v2d; typedef union {v2d v; complex double c; double d; double r; } vcplx; // typedef union {v2d v; complex double c; double d; double r; } vcplx; #define vdup(x) (x) #define addi(a,b) ((a) + I*(b)) #endif\nSupports Markdown\n0% or .\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!\nPlease register or to comment" ]

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[ "# Maths Ace Class 8 Solutions Chapter 13\n\n## Maths Ace Class 8 Solutions Chapter 13 Representing Solids on a Plane\n\nWelcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Maths Ace Prime Class 8 Math Book, Chapter 13, Representing Solids on a Plane. Here students can easily find step by step solutions of all the problems for Representing Solids on a Plane, Exercise 13.1 and 13.2 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 13 solutions.\n\nRepresenting Solids on a Plane Exercise 13.2 Solution :\n\nQuestion no – (1)\n\nSolution :\n\nNumber, because it does not satisfy Euler’s formula\n\nV + F – E = 2\n\n= 14 + 10 – 20\n\n= 24 – 20\n\n= 4 ≠ 2\n\nQuestion no – (2)\n\nSolution :\n\n Faces edges vertices (a) Tetrahedron 4 6 4 (b) Rectangular pyramid 5 8 5 (c) Pentagonal prism 7 15 10 (d) Octahedron 8 12 6\n\nQuestion no – (3)\n\nSolution :\n\nBy formula,\n\n= V – E + F = 2\n\n= 12 – E + 10 = 2\n\n= -E = 2 – 22\n\n= -22\n\n= E = 20\n\nTherefore, Edges in the polyhedron will be 20\n\nNext Chapter Solution :\n\nUpdated: June 16, 2023 — 9:04 am" ]

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[ "# Creating unique braces below equations [duplicate]\n\nI would like to create something like this", null, ".\n\nSpecifically my question is about how those curly braces below are drawn and labelled.\n\n• Have you considered the \\underbrace macro? – Mico Jun 11 '18 at 8:12\n\n\\documentclass{article}\n$\\int\\underbrace{\\int_0^\\pi \\frac{2y}{R^2}\\left( ... \\sqrt{1-\\frac{y}{4}}\\right)dy}_{\\mathcal{A}} dx$", null, "" ]

[ null, "https://i.stack.imgur.com/C3dC7.png", null, "https://i.stack.imgur.com/4Al13.png", null ]