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[ "## Interface Function<F,T>\n\n• ### Method Summary\n\nMethods\nModifier and Type Method and Description\n`T` `apply(F input)`\nReturns the result of applying this function to `input`.\n`boolean` `equals(Object object)`\nIndicates whether another object is equal to this function.\n• ### Method Detail\n\n• #### apply\n\n```@Nullable\nT apply(@Nullable\nF input)```\nReturns the result of applying this function to `input`. This method is generally expected, but not absolutely required, to have the following properties:\n• Its execution does not cause any observable side effects.\n• The computation is consistent with equals; that is, `Objects.equal``(a, b)` implies that ```Objects.equal(function.apply(a), function.apply(b))```.\nThrows:\n`NullPointerException` - if `input` is null and this function does not accept null arguments\n• #### equals\n\n```boolean equals(@Nullable\nObject object)```\nIndicates whether another object is equal to this function.\n\nMost implementations will have no reason to override the behavior of `Object.equals(java.lang.Object)`. However, an implementation may also choose to return `true` whenever `object` is a `Function` that it considers interchangeable with this one. \"Interchangeable\" typically means that `Objects.equal(this.apply(f), that.apply(f))` is true for all `f` of type `F`. Note that a `false` result from this method does not imply that the functions are known not to be interchangeable.\n\nOverrides:\n`equals` in class `Object`\nParameters:\n`object` - the reference object with which to compare.\nReturns:\n`true` if this object is the same as the obj argument; `false` otherwise.\n`Object.hashCode()`, `HashMap`" ]

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[ "", null, "# Question & Answer: end tile1 = [0 2 2 0; % vertex points for the tile in homogeneous coordinates 0 0 1 1; 1 1 1 1]…..", null, "What I currently have for code:\n\nDon't use plagiarized sources. Get Your Custom Essay on\nQuestion & Answer: end tile1 = [0 2 2 0; % vertex points for the tile in homogeneous coordinates 0 0 1 1; 1 1 1 1]…..\nGET AN ESSAY WRITTEN FOR YOU FROM AS LOW AS \\$13/PAGE\n\nfunction [ T ] = translate( dx,dy )\nT = eye(3); % identity matrix\nT(1,3) = dx;\nT(2,3) = dy;\nend\n\ntile1 = [0 2 2 0; % vertex points for the tile in homogeneous coordinates 0 0 1 1; 1 1 1 1]\n\nx = tile1(1,:); % x coordinates for the tile\n\ny = tile1(2,:); % y coordinates for the tile\n\nfigure\n\nfill(x, y, ‘cyan’)\n\ngrid on\n\nhold on\n\naxis equal tight\n\naxis([-10 10 -10 10])\n\nset(gca, ‘FontSize’, 16)\n\nv1 = [2, 0]\n\nv2 = [1, 1]\n\ndx = v1(1);\n\ndy = v1(2);\n\nnext_tile = translate(dx, dy) * tile1\n\nx = next_tile(1,:);\n\ny = next_tile(2,:); % Get the x and y coordinates\n\nfill(x, y, ‘blue’)\n\nd. Now add a third tile in yellow. Translate tilel by v2, and plot it using fill with the color set to ‘yellow’. Just emulate the code in the above part. Rectangular Tessellation 8 6 4 2 -2 -6 -8 -10 10 -5 0 e. Generate the entire tessellation using two nested for loops. To create the entire rectangular tessellation, you will need to add some code inside this nested for loop. % Use nested for loops to tessellate the entire for m-10:10 for n–10:10 % add code here!! pause(0.05) % Animates the tessellation end end", null, "" ]

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[ "# #020af7 Color Information\n\nIn a RGB color space, hex #020af7 is composed of 0.8% red, 3.9% green and 96.9% blue. Whereas in a CMYK color space, it is composed of 99.2% cyan, 96% magenta, 0% yellow and 3.1% black. It has a hue angle of 238 degrees, a saturation of 98.4% and a lightness of 48.8%. #020af7 color hex could be obtained by blending #0414ff with #0000ef. Closest websafe color is: #0000ff.\n\n• R 1\n• G 4\n• B 97\nRGB color chart\n• C 99\n• M 96\n• Y 0\n• K 3\nCMYK color chart\n\n#020af7 color description : Vivid blue.\n\n# #020af7 Color Conversion\n\nThe hexadecimal color #020af7 has RGB values of R:2, G:10, B:247 and CMYK values of C:0.99, M:0.96, Y:0, K:0.03. Its decimal value is 133879.\n\nHex triplet RGB Decimal 020af7 `#020af7` 2, 10, 247 `rgb(2,10,247)` 0.8, 3.9, 96.9 `rgb(0.8%,3.9%,96.9%)` 99, 96, 0, 3 238°, 98.4, 48.8 `hsl(238,98.4%,48.8%)` 238°, 99.2, 96.9 0000ff `#0000ff`\nCIE-LAB 31.679, 75.75, -104.4 16.919, 6.944, 88.439 0.151, 0.062, 6.944 31.679, 128.986, 305.963 31.679, -9.347, -126.265 26.352, 68.489, -180.539 00000010, 00001010, 11110111\n\n# Color Schemes with #020af7\n\n• #020af7\n``#020af7` `rgb(2,10,247)``\n• #f7ef02\n``#f7ef02` `rgb(247,239,2)``\nComplementary Color\n• #0285f7\n``#0285f7` `rgb(2,133,247)``\n• #020af7\n``#020af7` `rgb(2,10,247)``\n• #7502f7\n``#7502f7` `rgb(117,2,247)``\nAnalogous Color\n• #85f702\n``#85f702` `rgb(133,247,2)``\n• #020af7\n``#020af7` `rgb(2,10,247)``\n• #f77502\n``#f77502` `rgb(247,117,2)``\nSplit Complementary Color\n• #0af702\n``#0af702` `rgb(10,247,2)``\n• #020af7\n``#020af7` `rgb(2,10,247)``\n• #f7020a\n``#f7020a` `rgb(247,2,10)``\n• #02f7ef\n``#02f7ef` `rgb(2,247,239)``\n• #020af7\n``#020af7` `rgb(2,10,247)``\n• #f7020a\n``#f7020a` `rgb(247,2,10)``\n• #f7ef02\n``#f7ef02` `rgb(247,239,2)``\n• #0107ab\n``#0107ab` `rgb(1,7,171)``\n• #0208c4\n``#0208c4` `rgb(2,8,196)``\n• #0209de\n``#0209de` `rgb(2,9,222)``\n• #020af7\n``#020af7` `rgb(2,10,247)``\n• #151dfd\n``#151dfd` `rgb(21,29,253)``\n• #2f35fd\n``#2f35fd` `rgb(47,53,253)``\n• #484efe\n``#484efe` `rgb(72,78,254)``\nMonochromatic Color\n\n# Alternatives to #020af7\n\nBelow, you can see some colors close to #020af7. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0247f7\n``#0247f7` `rgb(2,71,247)``\n• #0233f7\n``#0233f7` `rgb(2,51,247)``\n• #021ef7\n``#021ef7` `rgb(2,30,247)``\n• #020af7\n``#020af7` `rgb(2,10,247)``\n• #0e02f7\n``#0e02f7` `rgb(14,2,247)``\n• #2302f7\n``#2302f7` `rgb(35,2,247)``\n• #3702f7\n``#3702f7` `rgb(55,2,247)``\nSimilar Colors\n\n# #020af7 Preview\n\nThis text has a font color of #020af7.\n\n``<span style=\"color:#020af7;\">Text here</span>``\n#020af7 background color\n\nThis paragraph has a background color of #020af7.\n\n``<p style=\"background-color:#020af7;\">Content here</p>``\n#020af7 border color\n\nThis element has a border color of #020af7.\n\n``<div style=\"border:1px solid #020af7;\">Content here</div>``\nCSS codes\n``.text {color:#020af7;}``\n``.background {background-color:#020af7;}``\n``.border {border:1px solid #020af7;}``\n\n# Shades and Tints of #020af7\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, #00010e is the darkest color, while #f9f9ff is the lightest one.\n\n• #00010e\n``#00010e` `rgb(0,1,14)``\n• #000121\n``#000121` `rgb(0,1,33)``\n• #000234\n``#000234` `rgb(0,2,52)``\n• #010348\n``#010348` `rgb(1,3,72)``\n• #01045b\n``#01045b` `rgb(1,4,91)``\n• #01046f\n``#01046f` `rgb(1,4,111)``\n• #010582\n``#010582` `rgb(1,5,130)``\n• #010696\n``#010696` `rgb(1,6,150)``\n• #0107a9\n``#0107a9` `rgb(1,7,169)``\n• #0208bd\n``#0208bd` `rgb(2,8,189)``\n• #0208d0\n``#0208d0` `rgb(2,8,208)``\n• #0209e4\n``#0209e4` `rgb(2,9,228)``\n• #020af7\n``#020af7` `rgb(2,10,247)``\n• #1017fd\n``#1017fd` `rgb(16,23,253)``\n• #232afd\n``#232afd` `rgb(35,42,253)``\n• #363dfd\n``#363dfd` `rgb(54,61,253)``\n• #4a50fe\n``#4a50fe` `rgb(74,80,254)``\n• #5d63fe\n``#5d63fe` `rgb(93,99,254)``\n• #7175fe\n``#7175fe` `rgb(113,117,254)``\n• #8488fe\n``#8488fe` `rgb(132,136,254)``\n• #989bfe\n``#989bfe` `rgb(152,155,254)``\n• #abaefe\n``#abaefe` `rgb(171,174,254)``\n• #bfc1fe\n``#bfc1fe` `rgb(191,193,254)``\n• #d2d4ff\n``#d2d4ff` `rgb(210,212,255)``\n• #e6e6ff\n``#e6e6ff` `rgb(230,230,255)``\n• #f9f9ff\n``#f9f9ff` `rgb(249,249,255)``\nTint Color Variation\n\n# Tones of #020af7\n\nA tone is produced by adding gray to any pure hue. In this case, #757584 is the less saturated color, while #020af7 is the most saturated one.\n\n• #757584\n``#757584` `rgb(117,117,132)``\n• #6b6c8e\n``#6b6c8e` `rgb(107,108,142)``\n• #626497\n``#626497` `rgb(98,100,151)``\n• #585ba1\n``#585ba1` `rgb(88,91,161)``\n• #4f52aa\n``#4f52aa` `rgb(79,82,170)``\n• #4549b4\n``#4549b4` `rgb(69,73,180)``\n• #3b40be\n``#3b40be` `rgb(59,64,190)``\n• #3237c7\n``#3237c7` `rgb(50,55,199)``\n• #282ed1\n``#282ed1` `rgb(40,46,209)``\n• #1f25da\n``#1f25da` `rgb(31,37,218)``\n• #151ce4\n``#151ce4` `rgb(21,28,228)``\n• #0c13ed\n``#0c13ed` `rgb(12,19,237)``\n• #020af7\n``#020af7` `rgb(2,10,247)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #020af7 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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[ "# LeetCode75. 颜色分类\n\n输入: [2,0,2,1,1,0]\n\n• 一个直观的解决方案是使用计数排序的两趟扫描算法。\n首先，迭代计算出0、1 和 2 元素的个数，然后按照0、1、2的排序，重写当前数组。\n• 你能想出一个仅使用常数空间的一趟扫描算法吗？\n\nclass Solution {\npublic void sortColors(int[] nums) {\nfor(int i=0;i<nums.length-1;i++){\nfor(int j=0;j<nums.length-i-1;j++){\nif(nums[j]>nums[j+1]){\nint temp=nums[j+1];\nnums[j+1]=nums[j];\nnums[j]=temp;\n}\n}\n}\n}\n}\n\n方法二： 对数组元素中0,1,2进行计数，再重新对数组赋值。\n\nclass Solution {\npublic void sortColors(int[] nums) {\nint countZero=0;\nint countOne=0;\nint countTwo=0;\nfor(int i=0;i<nums.length;i++){\nif(nums[i]==0){\ncountZero++;\n}else if(nums[i]==1){\ncountOne++;\n}else if(nums[i]==2){\ncountTwo++;\n}\n}\n// System.out.println(countZero+\" \"+countOne+\" \"+countTwo);\nint i=0;\nfor(i=0;i<countZero;i++){\nnums[i]=0;\n}\nfor(i=countZero;i<countOne+countZero;i++){\nnums[i]=1;\n}\nfor(i=countOne+countZero;i<nums.length;i++){\nnums[i]=2;\n}\n}\n}\n\n06-24", null, "199", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "05-11", null, "682\n05-15", null, "645\n01-29", null, "8\n10-05", null, "195\n10-20", null, "603\n01-24", null, "58\n05-10", null, "78\n04-21", null, "122\n02-06", null, "17\n08-04", null, "541\n10-07", null, "19\n07-03", null, "107\n04-20", null, "63\n01-25", null, "128\n12-17\n03-10", null, "71万+\n©️2020 CSDN 皮肤主题: 编程工作室 设计师:CSDN官方博客" ]

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[ "# Choosing ArcMap Network Analyst Tool\n\nI have two polygon layers, and a roads layer that I've made into a network dataset. My goal is to find the closest polygon in one layer to each polygon in the other layer, but instead of just calculating the distance to find the closest I need to use the roads ND to see which of the polygons from the first layer is closest to the polygon from the second layer.\n\nWhat tool can I use to do this?\n\nI've assumed I can do it with something in the Network Analyst toolset, but I can't figure out what.\nFrom the picture, I want to find the closest green polygon to both of the red polygons using the yellow roads.\n\nI'm still relatively new to this, in my first course on the subject.", null, "Edit: I received this solution on GeoNet, the Esri forum:\n\n\" Convert the vertices of your polygon features to points (Feature Vertices To Points—Data Management toolbox | ArcGIS Desktop) Do this twice: once for the green polygons and once for the red polygons Calculate an OD cost matrix with the Network Analyst (OD cost matrix analysis layer—ArcGIS Pro | ArcGIS Desktop ) Load the vertices from the red polygons as origins and the vertices from the green polygons as destinations (or vice-versa) Use the OD cost matrix to find out the shortest path between any vertex of a green polygon and any vertex of a red polygon. Use these vertices geographic positions to determine to which green or red polygon the vertices belong.\"" ]

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[ "# TIL\n\n37 thoughts\nlast posted Jan. 23, 2018, 10:28 a.m.\n\n17 earlier thoughts\n\n0\n\nThe first log tables were laboriously hand calculated once, and then apparently cut'n'pasted from one application to another for the following three centuries:\n\nall logarithm tables for three hundred years were borrowed from Mr. Briggs’ tables by reducing the number of decimal places\n\nToday, computation is too cheap to meter and one can program the necessary calculation, even without import math, in a matter of minutes. Obtaining the patterns required to make the insights necessary to get to $$\\mathbb{C}$$ when given $$\\mathbb{N}$$ ... might take a bit longer.\n\n# after http://www.feynmanlectures.caltech.edu/I_22.html#footnote_source_1\n_=min\nsqrt = lambda n: _( cvg(nstep(n1))(n1)\nfor n1 in [float(n)]\nfor cvg in [lambda f: lambda x:\n_( x if y==x else cvg(f)(y)\nfor y in [f(x)] )]\nfor nstep in [lambda n: lambda g: (n/g+g)/2] )\n\nif __name__ == \"__main__\":\ndef D(x):print(x);return x\n# regenerate table 22-1 from feynman LoP\nn = 10\nfor i in range(16): n = sqrt(D(n))\n# check precision after inverting\nfor i in range(16): n = n*n\nprint(n,n - 10.)\n\n\n19 later thoughts" ]

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[ "# Decimal Worksheets Pdf Worksheets Library And Print On Comparing Decimals Worksheet Grade Decimal Place Value For Equivalent Dividing Decimals Worksheet Pdf Kuta", null, "decimal worksheets pdf worksheets library and print on comparing decimals worksheet grade decimal place value for equivalent dividing decimals worksheet pdf kuta.\n\ndecimal subtraction worksheets pdf dewey classification system worksheet operations review expanded form grade archives,decimal multiplication worksheets pdf word problems 6th grade worksheet addition subtraction,dividing decimals worksheet 7th grade pdf decimal word problems 6 rounding off numbers worksheets the best image collection,decimal operations worksheets pdf adding subtracting multiplying and dividing decimals addition common core,multiplying decimal worksheets fun free decimals horizontal operations review worksheet pdf adding grade 6 rounding numbers,multiplying decimals coloring worksheet pdf decimal multiplication and division worksheets practice kuta,dewey decimal worksheets pdf multiply divide decimals worksheet dividing 7th grade super teacher fractions and percents fresh,dividing decimals word problems worksheets pdf worksheet 7th grade multiplying decimal numbers converting fractions to,decimals worksheets grade 5 decimal addition and subtraction pdf dividing word problems multiply divide worksheet,dividing decimals worksheet pdf kuta naming worksheets for all download and share free decimal multiplication rounding." ]

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[ "Quiet Modem, to transmit data with sound\n\n# quiet.py", null, "Python ctypes bindings for libquiet to transmit data with sound.\n\n• numpy\n\n## Install\n\n• For ARM platform, binary package is available on pypi, just use `pip` to install it:\n\n``````sudo apt install python-numpy\npip install --no-deps quiet.py\n``````\n\nWe install `numpy` separately, as installing `numpy` via pip requires compiling numpy from source.\n\n• For x86/amd64\n\n``````sudo apt install cmake\ngit clone https://github.com/xiongyihui/quiet.py && cd quiet.py\n./scripts/libs.sh\npip install .\n``````\n\n## Usage\n\n1. Encode a message, and then decode it\n``````from quiet import Encode, Decoder\n\ndef test():\nencoder = Encoder()\ndecoder = Decoder()\n\nfor chunk in encoder.encode('hello, world'):\nmessage = decoder.decode(chunk)\nif message is not None:\nprint(message)\n\ntest()\n``````\n1. decode messages from recording in realtime\n``````import sys\nimport numpy\nimport pyaudio\nfrom quiet import Encode, Decoder\n\ndef decode():\nif sys.version_info < 3:\nimport Queue as queue\nelse:\nimport queue\n\nFORMAT = pyaudio.paFloat32\nCHANNELS = 1\nRATE = 44100\nCHUNK = 16384 # int(RATE / 100)\n\np = pyaudio.PyAudio()\nq = queue.Queue()\n\ndef callback(in_data, frame_count, time_info, status):\nq.put(in_data)\nreturn (None, pyaudio.paContinue)\n\nstream = p.open(format=FORMAT,\nchannels=CHANNELS,\nrate=RATE,\ninput=True,\nframes_per_buffer=CHUNK,\nstream_callback=callback)\n\ncount = 0\nwith Decoder(profile_name='ultrasonic-experimental') as decoder:\nwhile True:\ntry:\naudio = q.get()\naudio = numpy.fromstring(audio, dtype='float32')\n# audio = audio[::CHANNELS]\ncode = decoder.decode(audio)\nif code is not None:\ncount += 1\nprint(code.tostring().decode('utf-8', 'ignore'))\nexcept KeyboardInterrupt:\nbreak\n\ndecode()\n``````\n\n## Project details", null, "" ]

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[ "# 大厂技术实现 | 爱奇艺短视频推荐业务中的多目标优化实践 @推荐与计算广告系列", null, "# 一图读懂全文\n\n• 获取『多目标模型方法实现代码』，请前往GitHub项目 https://github.com/ShowMeAI-H...\n• 文章涉及到的部分论文以及『微信数据集』，请在公众号（AI算法研究所）后台回复关键字 『爱奇艺多任务』 获取。\n\n# 一、短视频推荐业务\n\n## 1.1 业务介绍", null, "## 1.2 用户反馈信息", null, "• 显示反馈：点击播放，点击up主头像、关注、点击/发布评论、收藏、点击圈子、分享等正向互动行为，点击不喜欢、举报等负向行为。\n• 隐式反馈：播放时长、完成率、用户快速划过等行为。\n\n## 1.3 业务优化目标", null, "# 二、[实现方式1] CTR预估模型融入权重\n\n### 2.1.1 方案介绍", null, "## 2.2 融合时长权重\n\n### 2.2.1 权重计算\n\n$$\\frac{1}{\\operatorname{count}\\left(D_{i}\\right)} * \\sum_{d \\in D_{i}} w(\\text { playtime, } \\text { duration })=C$$", null, "• 对一段时间窗内的播放样本按照duration（视频时长）排序分散到100个桶，确保同一桶中视频观看数相同。\n• 对每个duration桶按照playtime（播放时长）排序再次等频分散到100个桶，并将权重归一化到[0,99] 的区间整数。\n做完该处理之后，对任何给定的样本，可以依照（duration, playtime）确定分桶坐标进而确定权值。\n\n### 2.2.2 播放时长加权", null, "$$boost\\_sigmoid\\left ( playtime \\right ) =\\frac{Am}{1+e^-\\frac{playtime+offset}{slope} }+shift$$\n\n• $$Am$$ 是『上界值』\n• $$shift$$ 是『下界值』\n• $$offset$$ 是『时长偏移量』\n• $$slope$$ 是『斜率』\n\n### 2.2.3 视频年龄与用户习惯权重调整\n\n• 使用视频年龄（用户行为时间-视频发布时间）对样本降权（年龄越大，权重越低）。\n• 在保证任务产出效率的基础上，对不同平台用户生成特定权重配置，实现了周期性更新，及时拟合整体用户最近的消费习惯。\n\n### 2.2.4 方案优缺点", null, "# 三、[实现方式2] 多模型融合\n\n## 3.1 方案介绍", null, "## 3.2 方案优缺点", null, "• 单目标训练单模型，优化简单，容易调到『较优』的单模型。\n\n• 难以定量估算不同目标重要度，组合困难；\n• 线下训练多个模型的时间和计算资源消耗大，线上预估需请求多个模型，复杂度与时延可能增加；\n• 数据分布随时间有变化，需要做模型更新和组合参数更新，也需要确定更新时机；\n• 某个目标数据比较稀疏时，该目标训练时无法结合其他信息进行有效的训练和迭代。\n\n# 四、[实现方式3] 多任务学习：网络设计与调优\n\n• 保证用户观看时长、观看视频数、点击率等基础指标提升或稳定\n• 引导用户评论、点赞等互动\n\n（1）ESSM建模\n（2）MMoE建模\n\n## 4.1 ESSM建模\n\n### 4.1.1 方案介绍", null, "", null, "### 4.1.2 方案总结\n\n• 互动行为非常稀疏，训练效果较差；\n• 多个目标的loss直接相加，难以平衡对每个目标任务的影响，对模型训练造成扰动；\n• 不同目标可能差异较大，难以直接共享底层表示。\n\n## 4.2 MMoE+帕累托优化\n\n### 4.2.1 方案介绍", null, "MMOE模型底层通过采用Soft parameter sharing方式，能够有效解决两个任务相关性较差情况下的多任务学习。", null, "• 均匀设置『可更新目标权重值』和『权重边界值超参』，使用PE-LTR算法训练更新权重值;\n• 调整不同的『权重边界值超参』，进行多次任务训练，根据目标的重要性挑选效果最好的模型。\n\n# 五、[实现方式4] 多任务学习：融合方案\n\n## 5.1 多目标得分乘法融合", null, "$$\\text { score }=\\sum_{i=1}^{n} \\text { factor }\\left(\\alpha_{i}+score_{i}\\right)^{\\beta_{i}}$$\n\n• $$alpha_{i}$$：超参，灵敏度\n• $$score_{i}$$： 模型i的输出\n• $$beta_{i}$$：超参，提升比例，非线性处理；\n• $$factor$$：超参，组合权重；\n• $$n$$：模型数量。", null, "• ① 对于新增目标，加法融合受限于新目标的scale，需要进行调整；相比之下乘法融合具有一定的目标独立性。\n• ② 目标增多后，加法融合中，各子目标的重要性影响会减弱。但乘法融合并不会受其影响。\n\n$$\\text { score }=\\prod_{i=1}^{n} \\text { factor }\\left(\\alpha_{i}+score_{i}\\right)^{\\beta_{i}}$$\n\n## 5.2 更多关联业务目标建模", null, "• ① 通过限定『完播率』阈值构造完播『二分类目标』，以近似满足逻辑回归的假设条件；\n• ② 拟合平滑后的『播放时长』作为『回归目标』；\n• ③ 限定『播放时长』阈值，构建有效播放的『二分类目标』。\n\n## 5.3 PSO进化优化算法\n\n• 离线通过Grid Search得到离线较优的超参数组\n• 线上AB测试验证实际效果\n\nPSO算法通过初始化一群随机粒子，启发式地多次迭代求出最优解。每一次迭代，粒子通过个体极值（该粒子所经过的最优解）和群体极值（种群找到的最优解）来更新各自位置。最终所有粒子会兼顾个体的历史最优和群体共享的全局最优直至收敛。\n\n$$O b j=w_{1} * AUC(\\text {ctr})+w_{2} * AUC(\\text {comment})+w_{3} * PNR(\\text {playtime})+\\cdots$$", null, "PSO搜参过程能够使得模型和融合参数同步更新，大幅降低人工调参的成本。\n\n# 七、参考文献\n\n• Ma X, Zhao L, Huang G, et al. Entire space multi-task model: An effective approach for estimating post-click conversion rate[C]//The 41st International ACM SIGIR Conference on Research & Development in Information Retrieval. 2018: 1137-1140.\n• Ma J, Zhao Z, Yi X, et al. Modeling task relationships in multi-task learning with multi-gate mixture-of-experts[C]//Proceedings of the 24th ACM SIGKDD International Conference on Knowledge Discovery & Data Mining. 2018: 1930-1939.\n• Zhao Z, Hong L, Wei L, et al. Recommending what video to watch next: a multitask ranking system[C]//Proceedings of the 13th ACM Conference on Recommender Systems. 2019: 43-51.\n• Lin X, Chen H, Pei C, et al. A pareto-efficient algorithm for multiple objective optimization in e-commerce recommendation[C]//Proceedings of the 13th ACM Conference on Recommender Systems. 2019: 20-28." ]

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[ "# Jon Halapio RAS\n\n### Jon Halapio RAS\n\nJon Halapio was drafted by Patriots with pick 179 in round 6 in the 2014 NFL Draft out of Florida.\n\nHe recorded a Relative Athletic Score of 3.23, out of a possible 10.0. RAS is a composite metric on a 0 to 10 scale based on the average of all of the percentile for each of the metrics the player completed either at the Combine or pro day.\n\nHe had a recorded height of 6034 that season, recorded as XYYZ where X is feet, YY is inches, and Z is eighths of an inch. That correlates to 6 feet, 3 and 4/8 of an inch or 75.5 inches, or 191.77 centimeters. This correlates to a 4.35 score out of 10.0.\n\nHe recorded a weight of 323 in pounds, which is approximately 146 kilograms. This correlates to a 8.45 score out of 10.0.\n\nBased on his weight, he has a projected 40 yard dash time of 5.22. This is calculated by taking 0.00554 multiplied by his weight and then adding 3.433.\n\nAt the Combine, he recorded a 40 yard dash of 5.34 seconds. This was a difference of 0.12 seconds from his projected time. This forty time correlates to a 4.5 score out of 10.0.\n\nUsing Bill Barnwell’s calculation, this Combine 40 time gave him a Speed Score of 79.45.\n\nThe time traveled between the 20 and 40 yard lines is known as the Flying Twenty. As the distance is also known, we can calculate the player’s speed over that distance. The time he traveled the last twenty yards at the Combine was 2.28 seconds. Over 20 yards, we can calculate his speed in yards per second to 8.77. Taking into account the distance in feet (60 feet), we can calculate his speed in feet per second to 26.32. Breaking it down further, we can calculate his speed in inches per second to 315.79. Knowing the feet per second of 26.32, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 17.9 in the last 20 yards of his run.\n\nAt the Combine, he recorded a 20 yard split of 3.06 seconds. This correlates to a 5.34 score out of 10.0.\n\nWe can calculate the speed traveled over the second ten yards of the 40 yard dash easily, as the distance and time are both known. The time he traveled the second ten yards at the Combine was 1.23 seconds. Over 10 yards, we can calculate his speed in yards per second to 8.13. Taking into account the distance in feet (30 feet), we can calculate his speed in feet per second to 24.39. Breaking it down further, we can calculate his speed in inches per second to 292.68. Knowing the feet per second of 24.39, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 16.6 in the second ten yards of his run.\n\nAt the Combine, he recorded a 10 yard split of 1.83 seconds. This correlates to a 5.95 score out of 10.0.\n\nThe time he traveled the first ten yards at the Combine was 1.83 seconds. Over 10 yards, we can calculate his speed in yards per second to 5.0. Taking into account the distance in feet (30 feet), we can calculate his speed in feet per second to 16.0. Breaking it down further, we can calculate his speed in inches per second to 197.0. Knowing the feet per second of 16.0, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 10.9 in the first ten yards of his run.\n\nThis player did not have a recorded bench press for the Combine in the RAS database.\n\nAt the Combine, he recorded a vertical jump of 21.5 inches. This correlates to a 0.22 score out of 10.0.\n\nAt his pro day, he recorded a vertical jump of 27.0 seconds. Because he also recorded this metric at the Combine, his pro day did not count towards his RAS.\n\nAt the Combine, he recorded a broad jump of 804, which is recorded as FII or FFII . where F is feet and I is inches. This correlates to a 4.92 score out of 10.0.\n\nAt the Combine, he recorded a 5-10-5 or 20 yard short shuttle of 4.83 seconds. This correlates to a 3.6 score out of 10.0.\n\nAt the Combine, he recorded a 3 cone L drill of 8.26 seconds. This correlates to a 1.74 score out of 10.0.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]

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[ "Document (#17914)\n\nAuthor\nVos, N.C.C.\nTitle\n¬De toekomst van het titelbeschrijven\nSource\nInformatie professional. 1(1997) no.5, S.34-38\nYear\n1997\nAbstract\nAt a conference in the Netherlands in Dec 1995 cataloguing, which has traditionally been regarded as the heart of librarianship, was criticised for being slow and costly. However, the Durch Gemeenschappelijk Geautometiseerd Catalogiseersystem (Shared Online Catalogue Network) enables libraries to contribute and receive good quality catalogue records at substantial savings in cost and time. PICA (Project for Integrated Catalogue Automation), which is responsible for the GGC, is currently examining ways to speed up the process. Other sources for records include library suppliers and external agencies\nFootnote\nÜbers. d. Titels: The future of cataloging\nTheme\nFormalerschließung\nObject\nPICA\n\nSimilar documents (content)\n\n1. Oosterop, D.: PICA and the academic libraries in the Netherlands (1996) 0.18\n```0.17982924 = sum of:\n0.17982924 = product of:\n0.8991462 = sum of:\n0.024369847 = weight(abstract_txt:which in 911) [ClassicSimilarity], result of:\n0.024369847 = score(doc=911,freq=1.0), product of:\n0.066336475 = queryWeight, product of:\n2.938938 = idf(docFreq=6085, maxDocs=42306)\n0.022571582 = queryNorm\n0.36736724 = fieldWeight in 911, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n2.938938 = idf(docFreq=6085, maxDocs=42306)\n0.125 = fieldNorm(doc=911)\n0.120344445 = weight(abstract_txt:automation in 911) [ClassicSimilarity], result of:\n0.120344445 = score(doc=911,freq=1.0), product of:\n0.15268318 = queryWeight, product of:\n1.0727645 = boost\n6.305577 = idf(docFreq=209, maxDocs=42306)\n0.022571582 = queryNorm\n0.7881971 = fieldWeight in 911, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.305577 = idf(docFreq=209, maxDocs=42306)\n0.125 = fieldNorm(doc=911)\n0.1694685 = weight(abstract_txt:netherlands in 911) [ClassicSimilarity], result of:\n0.1694685 = score(doc=911,freq=1.0), product of:\n0.19182263 = queryWeight, product of:\n1.202427 = boost\n7.0677166 = idf(docFreq=97, maxDocs=42306)\n0.022571582 = queryNorm\n0.8834646 = fieldWeight in 911, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.0677166 = idf(docFreq=97, maxDocs=42306)\n0.125 = fieldNorm(doc=911)\n0.32245865 = weight(abstract_txt:pica in 911) [ClassicSimilarity], result of:\n0.32245865 = score(doc=911,freq=2.0), product of:\n0.23378317 = queryWeight, product of:\n1.3274416 = boost\n7.8025365 = idf(docFreq=46, maxDocs=42306)\n0.022571582 = queryNorm\n1.3793066 = fieldWeight in 911, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n7.8025365 = idf(docFreq=46, maxDocs=42306)\n0.125 = fieldNorm(doc=911)\n0.2625047 = weight(abstract_txt:catalogue in 911) [ClassicSimilarity], result of:\n0.2625047 = score(doc=911,freq=2.0), product of:\n0.29396504 = queryWeight, product of:\n2.578204 = boost\n5.051454 = idf(docFreq=735, maxDocs=42306)\n0.022571582 = queryNorm\n0.8929793 = fieldWeight in 911, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n5.051454 = idf(docFreq=735, maxDocs=42306)\n0.125 = fieldNorm(doc=911)\n0.2 = coord(5/25)\n```\n2. Plaister, J.: PICA : Centrum voor Bibliotheekautomatisering (1987) 0.17\n```0.16882668 = sum of:\n0.16882668 = product of:\n1.0551668 = sum of:\n0.18051668 = weight(abstract_txt:automation in 1589) [ClassicSimilarity], result of:\n0.18051668 = score(doc=1589,freq=1.0), product of:\n0.15268318 = queryWeight, product of:\n1.0727645 = boost\n6.305577 = idf(docFreq=209, maxDocs=42306)\n0.022571582 = queryNorm\n1.1822957 = fieldWeight in 1589, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.305577 = idf(docFreq=209, maxDocs=42306)\n0.1875 = fieldNorm(doc=1589)\n0.25420275 = weight(abstract_txt:netherlands in 1589) [ClassicSimilarity], result of:\n0.25420275 = score(doc=1589,freq=1.0), product of:\n0.19182263 = queryWeight, product of:\n1.202427 = boost\n7.0677166 = idf(docFreq=97, maxDocs=42306)\n0.022571582 = queryNorm\n1.3251969 = fieldWeight in 1589, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.0677166 = idf(docFreq=97, maxDocs=42306)\n0.1875 = fieldNorm(doc=1589)\n0.34201908 = weight(abstract_txt:pica in 1589) [ClassicSimilarity], result of:\n0.34201908 = score(doc=1589,freq=1.0), product of:\n0.23378317 = queryWeight, product of:\n1.3274416 = boost\n7.8025365 = idf(docFreq=46, maxDocs=42306)\n0.022571582 = queryNorm\n1.4629756 = fieldWeight in 1589, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.8025365 = idf(docFreq=46, maxDocs=42306)\n0.1875 = fieldNorm(doc=1589)\n0.2784283 = weight(abstract_txt:catalogue in 1589) [ClassicSimilarity], result of:\n0.2784283 = score(doc=1589,freq=1.0), product of:\n0.29396504 = queryWeight, product of:\n2.578204 = boost\n5.051454 = idf(docFreq=735, maxDocs=42306)\n0.022571582 = queryNorm\n0.9471476 = fieldWeight in 1589, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.051454 = idf(docFreq=735, maxDocs=42306)\n0.1875 = fieldNorm(doc=1589)\n0.16 = coord(4/25)\n```\n3. 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Huisman, F.W.: Gemeenschappelijke onderwerpsontsluiting (GOO) : een samenwerkingsproject in Pica-verband (1992) 0.15\n```0.14590794 = sum of:\n0.14590794 = product of:\n0.60794973 = sum of:\n0.015231155 = weight(abstract_txt:which in 2334) [ClassicSimilarity], result of:\n0.015231155 = score(doc=2334,freq=1.0), product of:\n0.066336475 = queryWeight, product of:\n2.938938 = idf(docFreq=6085, maxDocs=42306)\n0.022571582 = queryNorm\n0.22960453 = fieldWeight in 2334, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n2.938938 = idf(docFreq=6085, maxDocs=42306)\n0.078125 = fieldNorm(doc=2334)\n0.07521527 = weight(abstract_txt:automation in 2334) [ClassicSimilarity], result of:\n0.07521527 = score(doc=2334,freq=1.0), product of:\n0.15268318 = queryWeight, product of:\n1.0727645 = boost\n6.305577 = idf(docFreq=209, maxDocs=42306)\n0.022571582 = queryNorm\n0.49262318 = fieldWeight in 2334, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.305577 = idf(docFreq=209, maxDocs=42306)\n0.078125 = fieldNorm(doc=2334)\n0.10501214 = weight(abstract_txt:responsible in 2334) [ClassicSimilarity], result of:\n0.10501214 = score(doc=2334,freq=1.0), product of:\n0.19072759 = queryWeight, product of:\n1.19899 = boost\n7.047514 = idf(docFreq=99, maxDocs=42306)\n0.022571582 = queryNorm\n0.55058706 = fieldWeight in 2334, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.047514 = idf(docFreq=99, maxDocs=42306)\n0.078125 = fieldNorm(doc=2334)\n0.10591782 = weight(abstract_txt:netherlands in 2334) [ClassicSimilarity], result of:\n0.10591782 = score(doc=2334,freq=1.0), product of:\n0.19182263 = queryWeight, product of:\n1.202427 = boost\n7.0677166 = idf(docFreq=97, maxDocs=42306)\n0.022571582 = queryNorm\n0.5521654 = fieldWeight in 2334, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.0677166 = idf(docFreq=97, maxDocs=42306)\n0.078125 = fieldNorm(doc=2334)\n0.14250796 = weight(abstract_txt:pica in 2334) [ClassicSimilarity], result of:\n0.14250796 = score(doc=2334,freq=1.0), product of:\n0.23378317 = queryWeight, product of:\n1.3274416 = boost\n7.8025365 = idf(docFreq=46, maxDocs=42306)\n0.022571582 = queryNorm\n0.6095732 = fieldWeight in 2334, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.8025365 = idf(docFreq=46, maxDocs=42306)\n0.078125 = fieldNorm(doc=2334)\n0.16406544 = weight(abstract_txt:catalogue in 2334) [ClassicSimilarity], result of:\n0.16406544 = score(doc=2334,freq=2.0), product of:\n0.29396504 = queryWeight, product of:\n2.578204 = boost\n5.051454 = idf(docFreq=735, maxDocs=42306)\n0.022571582 = queryNorm\n0.5581121 = fieldWeight in 2334, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n5.051454 = idf(docFreq=735, maxDocs=42306)\n0.078125 = fieldNorm(doc=2334)\n0.24 = coord(6/25)\n```\n5. Bossers, A.: Cooperative library automation and the Pica experience (1989) 0.11\n```0.11100876 = sum of:\n0.11100876 = product of:\n0.6938048 = sum of:\n0.07039485 = weight(abstract_txt:enables in 2907) [ClassicSimilarity], result of:\n0.07039485 = score(doc=2907,freq=1.0), product of:\n0.14608797 = queryWeight, product of:\n1.0493395 = boost\n6.167887 = idf(docFreq=240, maxDocs=42306)\n0.022571582 = queryNorm\n0.48186618 = fieldWeight in 2907, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.167887 = idf(docFreq=240, maxDocs=42306)\n0.078125 = fieldNorm(doc=2907)\n0.10637046 = weight(abstract_txt:automation in 2907) [ClassicSimilarity], result of:\n0.10637046 = score(doc=2907,freq=2.0), product of:\n0.15268318 = queryWeight, product of:\n1.0727645 = boost\n6.305577 = idf(docFreq=209, maxDocs=42306)\n0.022571582 = queryNorm\n0.69667435 = fieldWeight in 2907, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n6.305577 = idf(docFreq=209, maxDocs=42306)\n0.078125 = fieldNorm(doc=2907)\n0.2850159 = weight(abstract_txt:pica in 2907) [ClassicSimilarity], result of:\n0.2850159 = score(doc=2907,freq=4.0), product of:\n0.23378317 = queryWeight, product of:\n1.3274416 = boost\n7.8025365 = idf(docFreq=46, maxDocs=42306)\n0.022571582 = queryNorm\n1.2191464 = fieldWeight in 2907, product of:\n2.0 = tf(freq=4.0), with freq of:\n4.0 = termFreq=4.0\n7.8025365 = idf(docFreq=46, maxDocs=42306)\n0.078125 = fieldNorm(doc=2907)\n0.23202358 = weight(abstract_txt:catalogue in 2907) [ClassicSimilarity], result of:\n0.23202358 = score(doc=2907,freq=4.0), product of:\n0.29396504 = queryWeight, product of:\n2.578204 = boost\n5.051454 = idf(docFreq=735, maxDocs=42306)\n0.022571582 = queryNorm\n0.7892897 = fieldWeight in 2907, product of:\n2.0 = tf(freq=4.0), with freq of:\n4.0 = termFreq=4.0\n5.051454 = idf(docFreq=735, maxDocs=42306)\n0.078125 = fieldNorm(doc=2907)\n0.16 = coord(4/25)\n```" ]

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[ "# Hands-on Machine Learning: What You Need to Know\n\nIf you’re interested in hands-on machine learning, then this blog post is for you! We’ll cover what you need to know in order to get started, including some key concepts and terminology.\n\nCheckout this video:\n\n## Introduction to Machine Learning\n\nMachine learning is a branch of artificial intelligence that deals with the design and development of algorithms that can learn from and make predictions on data. These algorithms are used to build models that can automatically improve given more data.\n\nMachine learning is a vast and complex field, but there are a few fundamental concepts that are essential to understanding how it works. In this article, we’ll introduce some of the key ideas behind machine learning, including supervised and unsupervised learning, feedback loops, training data, and test data.\n\n## What is Hands-on Machine Learning?\n\nHands-on machine learning is a process of learning by doing. It is a type of learning where students are actively engaged in the process of learning, rather than passively receiving information. This type of learning has been shown to be more effective than traditional methods of instruction, such as lectures and textbooks.\n\nIn hands-on machine learning, students are given opportunities to experiment with and apply the concepts they are learning. For example, rather than simply reading about how to build a machine learning model, students might be given data sets and be asked to build their own models. This type of learning allows students to better understand the material and to see its practical applications.\n\n## The Benefits of Hands-on Machine Learning\n\nHands-on machine learning is beneficial for a number of reasons. First, it allows you to gain a better understanding of the algorithms and models that you are working with. Second, it gives you the opportunity to try out different techniques and see what works best for your data. Third, it helps you build intuition about how the algorithms work and what their limitations are. Finally, hands-on machine learning is simply more fun than reading about machine learning!\n\n## The Different Types of Machine Learning\n\nThere are three different types of machine learning: supervised learning, unsupervised learning, and reinforcement learning. Supervised learning is when the computer is given a set of training data, and it is then able to learn and generalize from that data. Unsupervised learning is when the computer is given data but not told what to do with it, and it has to figure out patterns and relationships on its own. Reinforcement learning is when the computer is given a goal but not told how to achieve it, and it has to learn by trial and error.\n\n## The Different Types of Data\n\nIn machine learning, there are generally two types of data: structured data and unstructured data. Structured data is easy to work with because it is organized in a consistent format. This type of data can be found in databases and spreadsheets. Unstructured data is more difficult to work with because it is not organized in a consistent format. This type of data includes images, text, and audio files.\n\n## The Different Types of Algorithms\n\nAlgorithms are at the heart of machine learning. They are the engine that drives the training of models and the predictions that result from them. There are many different types of algorithms, each with its own strengths and weaknesses. In this section, we will take a brief look at some of the most common algorithms used in machine learning.\n\nLinear regression is one of the simplest and most widely used machine learning algorithms. It is used to predict a continuous value, such as a price or quantity. Linear regression works by finding the best fit line through a set of data points. The line is then used to make predictions for new data points.\n\nLogistic regression is another common algorithm, used for classification tasks (predicting whether an instance belongs to one class or another). It works by finding the best fit line through a set of data points, but unlike linear regression, the line is only used to make predictions for two classes (1 or 0).\n\nDecision trees are a popular type of algorithm used for both classification and regression tasks. They work by making predictions based on a series of yes/no questions. For each question, the model looks at a subset of the data and decides which answer is more likely to be correct. The model then makes its prediction based on the answer to the final question in the series. Decision trees are popular because they are easy to interpret and understand. They can also be very accurate, although they are sometimes prone to overfitting.\n\nRandom forest is an ensemble algorithm that combines multiple decision trees to make predictions. It works by training multiple decision trees on different subsets of the data and then averaging their predictions. This process often results in improved accuracy, as well as reduced overfitting.\n\n## The Different Types of Models\n\nIn machine learning, there are generally two types of models: supervised and unsupervised. Supervised models are those where we have a dataset that includes both the input data (X) as well as the corresponding expected output (y). Based on this data, the model “learns” a set of rules that can then be applied to new data in order to predict the expected output. In contrast, unsupervised models do not have an expected output associated with each input, but instead try to model the underlying structure or distribution of the data.\n\nSome common examples of supervised machine learning tasks are regression (predicting a continuous value), classification (predicting a class label), andsequence labeling (predicting a sequence of class labels). Unsupervised tasks include density estimation (modeling the probability density function of the input data) and clustering (partitioning the input data into groups).\n\nThere are also semi-supervised and reinforcement learning tasks, which fall somewhere in between supervised and unsupervised learning. In semi-supervised learning, we may have a dataset where only some of the labels are known; the goal is then to learn from both the labeled and unlabeled data. Reinforcement learning is similar in that there is no training dataset; instead, the model “learns” by interactively trying different actions and seeing what results they produce.\n\n## The Different Types of Learning\n\nIn machine learning, there are generally three different types of learning algorithms: supervised learning, unsupervised learning, and reinforcement learning. Each type of learning has its own advantages and disadvantages, and works best in different situations.\n\nSupervised Learning:\n\nSupervised learning is where you have a training set of data that you know the correct answers for. You use this training set to train your machine learning algorithm, and then you test it on a separate test set of data to see how well it performs. This type of learning is good for problems where you have a lot of known data, and you want to be able to generalize from that data. It can be either classification (predicting which category something belongs to) or regression (predicting a quantity).\n\nUnsupervised Learning:\n\nUnsupervised learning is where you have a dataset but no known correct answers. This can be used for things like clustering (grouping data points that are similar to each other) or dimensionality reduction (reducing the number of features in a dataset). This type of learning is good for exploratory data analysis, and can be used to find structure in data that might not be obvious at first glance.\n\nReinforcement Learning:\n\nReinforcement learning is where an algorithm interacts with an environment, trying to maximize some reward signal. This can be used for things like playing games or controlling robotic arms. Reinforcement learning is good for problems where there is a clear goal but no known path to get there.\n\n## The Different Types of Evaluation\n\nThere are many different types of evaluation, but we will focus on four of the most important and commonly used: cross-validation, bootstrapping, testing, and online learning.\n\nCross-validation is a technique that is used to assess the performance of a machine learning algorithm on a dataset. It is often used in conjunction with training the algorithm on a separate dataset. The idea behind cross-validation is to split the data into two parts: a training set and a test set. The machine learning algorithm is then trained on the training set and evaluated on the test set. This process is repeated multiple times, with each iteration using a different training/test split. The performance of the machine learning algorithm is then averaged over all iterations.\n\nBootstrapping is another technique that can be used to assess the performance of a machine learning algorithm. It works by randomly sampling data points from the dataset and training the algorithm on these sampled data points. The performance of the machine learning algorithm is then evaluated on the remaining data points. This process is repeated multiple times, with each iteration using a different sample of data points. The performance of the machine learning algorithm is then averaged over all iterations.\n\nTesting is another way of assessing the performance of a machine learning algorithm . It works by training the machine learning algorithm on one part of the data and testing it on another part. The performance of the machine learning algorithm is then measured by how well it performs on the test data .\n\nOnline learning is an approach to machine learning that allows algorithms to learn incrementally from data streams, without having to store all of the data in memory . Online learning algorithms can make predictions very quickly, which makes them well suited for applications where time is critical, such as credit card fraud detection or stock trading .\n\n## The Different Types of Deployment\n\nThere are four main types of deployment when it comes to machine learning: on-premises, cloud, hybrid, and edge.\n\nOn-premises refers to when an organization stores and manages its own data and infrastructure. This can be expensive and time-consuming, but some organizations prefer it for security or compliance reasons.\n\nCloud deployment means that an organization uses a cloud service provider to store and manage its data and infrastructure. This is often more cost-effective and efficient than on-premises deployment, but it can be less secure.\n\nHybrid deployment is a mix of on-premises and cloud deployment. This can be helpful for organizations that want the security of on-premises deployment but also want to take advantage of the cost savings of cloud deployment.\n\nEdge deployment is when machine learning is deployed at the edge of a network, closer to the devices that will be using it. This can be helpful for reducing latency and increasing security.\n\nScroll to Top" ]

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[ "Filter Content By\nVersion\n\naddMConstrs ( A, x, sense, b, names=\"\" )\n\nAdd a set of linear constraints to the model using matrix semantics. The added constraints are", null, "(except that the constraint sense is determined by the sense argument). The A argument must be a NumPy dense ndarray or a SciPy sparse matrix.\n\nNote that you will typically use overloaded operators to build and add constraints using matrix semantics. The overloaded @ operator can be used to build a linear matrix expression, which can then be used with an overloaded comparison operator to build a TempConstr object. This can then be passed to addConstr.\n\nArguments:\n\nA: The constraint matrix - a NumPy 2-D dense ndarray or a SciPy sparse matrix.\n\nx: Decision variables. Argument can be an MVar object, a list of Var objects, or None (None uses all variables in the model). The length of the argument must match the size of the second dimension of A.\n\nsense: Constraint senses, provided as a NumPy 1-D ndarray or as a single character. Valid values are", null, ",", null, ", or", null, ". The length of the array must be equal the size of the first dimension of A. A character will be promoted to an ndarray of the appropriate length.\n\nb: Right-hand side vector, stored as a NumPy 1-D ndarray. The length of the array must be equal the size of the first dimension of A.\n\nnames: Names for new constraints. The given name will be subscripted by the index of the constraint in the matrix.\n\nReturn value:\n\nList of Constr objects.\n\nExample usage:\n\n A = np.full((5, 10), 1)" ]

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[ "", null, "", null, "", null, "", null, "Next: The input flux wavelength Up: No Title Previous: Abbreviations and Acronyms\n\n# ETC Main Formula\n\nVery useful informations given by an ETC are the expected number of counts for an observed object and contribute of the sky, and the signal to noise ratio. The ETC gives also other informations which are usually instrument specific and so we prefer to describe them in other documents.\n\nThe main formulae used to evaluate the expected number of counts and the signal to noise ratio are given in the following table:", null, "where N is the number of electrons per bin, F is the incident flux in", null, ",", null, ", in spectroscopy, is the spectral bin in", null, ",", null, ", in imaging, is the filter band width", null, ", T the total exposure time in seconds, E the efficiency, S the telescope surface in m2,", null, "is the solid angle subtended by the integration element (in Spectroscopy),", null, "is the solid angle subtended by the integration element (in imaging), P the energy of one photon. The resulting dimension of N is", null, ".\n\nTo calculate the signal to noise ratio the ETC uses the formulae:", null, "(2.1)\n\nor in the infrared:", null, "(2.2)\n\nwhere", null, "(2.3)\n\nis the number of spatial pixels over which it is integrated the signal 2.1, where seeing is the seeing expressed in arcseconds;", null, "is the dimension of one pixel in arcseconds; NObj is the total number of counts per spectral bin coming from the observed object (point source or extended one); NSky is the similar term due to the sky; RON is the contribute to the detected counts due to the Read Out Noise; DARK is the similar term due to the dark counts.\n\nTable 2.2: Table with summary definitions of the terms present in the formula used to calculate the number of expected counts and the signal to noise ratio..\n simbol units meaning NObj", null, "object signal NSky", null, "sky signal DARK", null, "dark current RON", null, "read out noise npix number of integration pixels to evaluate the RON contribute to S/N nbin number of integration bins to evaluate the DARK contribute to S/N T s detector integration time", null, "number of detector integrations", null, "arcsec2 solid angle subtended by the integration element (in Spectroscopy)", null, "arcsec2 solid angle subtended by the integration element (in imaging)", null, "", null, "spectral bin (in Spectroscopy)", null, "", null, "filter band width (in imaging)", null, "The parameters present in Table 2.1 are defined or from user specifications, the ETC front page input parameters, or by other sources of information (instrument known characteristics, physical relations, the database).\n\nHow we obtain each result is described in its own following section.", null, "", null, "", null, "", null, "Next: The input flux wavelength Up: No Title Previous: Abbreviations and Acronyms\nPascal Ballester\n1999-07-29" ]

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[ "Showing 6 of 6 results\n\n#### Recoil Proton Polarization in $\\pi P$ Elastic Scattering at 547-{MeV}/c and 625-{MeV}/c\n\nSeftor, C.J. ; Adrian, S.D. ; Briscoe, W.J. ; et al.\nPhys.Rev.D 39 (1989) 2457-2463, 1989.\nInspire Record 282265\n\nThe polarization of the recoil proton in π+p and π−p elastic scattering using a liquid-hydrogen target has been measured for backward angles at 547 and 625 MeV/c. The scattered pion and recoil proton were detected in coincidence using the large-acceptance spectrometer to detect and analyze the momentum of the pions and the JANUS polarimeter to identify and measure the polarization of the protons. Results from this experiment agree with other measurements of the recoil polarization, with analyzing-power data previously taken by this group, and with predictions of partial-wave analyses.\n\n2 data tables match query\n\nNo description provided.\n\nNo description provided.\n\n#### Left-right Asymmetry in Inverse $\\pi^-$ Photoproduction From a Transversely Polarized Proton Target\n\nKim, G.J. ; Adrian, S.D. ; Arends, J. ; et al.\nPhys.Rev.Lett. 56 (1986) 1779-1782, 1986.\nInspire Record 232117\n\nAccurate measurements of the left-right asymmetry in π−p→γn at pπ=427−625 MeV/c with a transversely polarized target are reported. Results are compared with the predictions from the Arai and Fujii single-pion photoproduction partial-wave analysis and with data on the inverse process measured with a deuterium target. The agreement is poor, casting doubt on the correctness of the value for the radiative-decay amplitude of the neutral Roper resonance now in use.\n\n3 data tables match query\n\nNo description provided.\n\nNo description provided.\n\nNo description provided.\n\n#### rivet Analysis The $e^+e^- \\to 3(\\pi^+ \\pi^-), 2(\\pi^+ \\pi^- \\pi^0)$ and $K^+ K^- 2(\\pi^+ \\pi^-)$ cross sections at center-of-mass energies from production threshold to 4.5-GeV measured with initial-state radiation\n\nThe collaboration Aubert, Bernard ; Barate, R. ; Boutigny, D. ; et al.\nPhys.Rev.D 73 (2006) 052003, 2006.\nInspire Record 709730\n\nWe study the processes e+ e- --> 3(pi+pi-)gamma, 2(pi+pi-pi0)gamma and K+ K- 2(pi+pi-)gamma, with the photon radiated from the initial state. About 20,000, 33,000 and 4,000 fully reconstructed events, respectively, have been selected from 232 fb-1 of BaBar data. The invariant mass of the hadronic final state defines the effective e+e- center-of-mass energy, so that these data can be compared with the corresponding direct e+e- measurements. From the 3(pi+pi-), 2(pi+pi-pi0) and K+ K- 2(pi+pi-) mass spectra, the cross sections for the processes e+ e- --> 3(pi+pi-), e+ e- --> 2(pi+pi-pi0) and e+ e- --> K+ K- 2(pi+pi-) are measured for center-of-mass energies from production threshold to 4.5 GeV. The uncertainty in the cross section measurement is typically 6-15%. We observe the J/psi in all these final states and measure the corresponding branching fractions.\n\n2 data tables match query\n\nThe cross section for E+ E- --> 3PI+ 3PI- as measured with the ISR data. Errors are statistical only.\n\nThe cross section for E+ E- --> 2PI+ 2PI- 2PI0 as measured with the ISR data. Errors are statistical only.\n\n#### rivet Analysis The e+ e- ---> 2(pi+ pi-) pi0, 2(pi+ pi-) eta, K+ K- pi+ pi- pi0 and K+ K- pi+ pi- eta Cross Sections Measured with Initial-State Radiation\n\nThe collaboration Aubert, Bernard ; Bona, M. ; Boutigny, D. ; et al.\nPhys.Rev.D 76 (2007) 092005, 2007.\nInspire Record 758568\n\nWe study the processes $e^+ e^-\\to 2(\\pi^+\\pi^-)\\pi^0\\gamma$, $2(\\pi^+\\pi^-)\\eta\\gamma$, $K^+ K^-\\pi^+\\pi^-\\pi^0\\gamma$ and $K^+ K^-\\pi^+\\pi^-\\eta\\gamma$ with the hard photon radiated from the initial state. About 20000, 4300, 5500 and 375 fully reconstructed events, respectively, are selected from 232 fb$^{-1}$ of BaBar data. The invariant mass of the hadronic final state defines the effective $e^+ e^-$ center-of-mass energy, so that the obtained cross sections from the threshold to about 5 GeV can be compared with corresponding direct \\epem measurements, currently available only for the $\\eta\\pi^+\\pi^-$ and $\\omega\\pi^+\\pi^-$ submodes of the $e^+ e^-\\to 2(\\pi^+\\pi^-)\\pi^0$ channel. Studying the structure of these events, we find contributions from a number of intermediate states, and we extract their cross sections where possible. In particular, we isolate the contribution from $e^+ e^-\\to\\omega(782)\\pi^+\\pi^-$ and study the $\\omega(1420)$ and $\\omega(1650)$ resonances. In the charmonium region, we observe the $J/\\psi$ in all these final states and several intermediate states, as well as the $\\psi(2S)$ in some modes, and we measure the corresponding branching fractions.\n\n4 data tables match query\n\nMeasured cross section for E+ E- --> 2(PI+ PI-) PI0 with statistical errorsonly.\n\nMeasured cross section for E+ E- --> ETA PI+ PI- with statistical errors only.\n\nMeasured cross section for E+ E- --> OMEGA PI+ PI- with statistical errors only.\n\nMore…\n\n#### rivet Analysis Inclusive Lambda(c)+ Production in e+ e- Annihilations at s**(1/2) = 10.54-GeV and in Upsilon(4S) Decays\n\nThe collaboration Aubert, Bernard ; Bona, M. ; Boutigny, D. ; et al.\nPhys.Rev.D 75 (2007) 012003, 2007.\nInspire Record 725377\n\nWe present measurements of the total production rates and momentum distributions of the charmed baryon $\\Lambda_c^+$ in $e^+e^- \\to$ hadrons at a center-of-mass energy of 10.54 GeV and in $\\Upsilon(4S)$ decays. In hadronic events at 10.54 GeV, charmed hadrons are almost exclusively leading particles in $e^+e^- \\to c\\bar{c}$ events, allowing direct studies of $c$-quark fragmentation. We measure a momentum distribution for $\\Lambda_c^+$ baryons that differs significantly from those measured previously for charmed mesons. Comparing with a number of models, we find none that can describe the distribution completely. We measure an average scaled momentum of $\\left< x_p \\right> = 0.574\\pm$0.009 and a total rate of $N_{\\Lambda c}^{q\\bar{q}} = 0.057\\pm$0.002(exp.)$\\pm$0.015(BF) $\\Lambda_c^+$ per hadronic event, where the experimental error is much smaller than that due to the branching fraction into the reconstructed decay mode, $pK^-\\pi^+$. In $\\Upsilon (4S)$ decays we measure a total rate of $N_{\\Lambda c}^{\\Upsilon} = 0.091\\pm$0.006(exp.)$\\pm$0.024(BF) per $\\Upsilon(4S)$ decay, and find a much softer momentum distribution than expected from B decays into a $\\Lambda_c^+$ plus an antinucleon and one to three pions.\n\n1 data table match query\n\nThe integrated number of LAMBDA/C+'s per hadronic event for the continuum at cm energy 10.54 GeV.\n\n#### rivet Analysis The $e^+e^- \\to \\pi^+ \\pi^- \\pi^+ \\pi^-$, $K^+ K^- \\pi^+ \\pi^-$, and $K^+ K^- K^+ K^-$ cross sections at center-of-mass energies 0.5-GeV - 4.5-GeV measured with initial-state radiation\n\nThe collaboration Aubert, Bernard ; Barate, R. ; Boutigny, D. ; et al.\nPhys.Rev.D 71 (2005) 052001, 2005.\nInspire Record 676691\n\nWe study the process $e^+e^-\\to\\pi^+\\pi^-\\pi^+\\pi^-\\gamma$, with a hard photon radiated from the initial state. About 60,000 fully reconstructed events have been selected from 89 $fb^{-1}$ of BaBar data. The invariant mass of the hadronic final state defines the effective \\epem center-of-mass energy, so that these data can be compared with the corresponding direct $e^+e^-$ measurements. From the $4\\pi$-mass spectrum, the cross section for the process $e^+e^-\\to\\pi^+\\pi^-\\pi^+\\pi^-$ is measured for center-of-mass energies from 0.6 to 4.5 $GeV/c^2$. The uncertainty in the cross section measurement is typically 5%. We also measure the cross sections for the final states $K^+ K^- \\pi^+\\pi^-$ and $K^+ K^- K^+ K^-$. We observe the $J/\\psi$ in all three final states and measure the corresponding branching fractions. We search for X(3872) in $J/\\psi (\\to\\mu^+\\mu^-) \\pi^+\\pi^-$ and obtain an upper limit on the product of the $e^+e^-$ width of the X(3872) and the branching fraction for $X(3872) \\to J/\\psi\\pi^+\\pi^-$.\n\n1 data table match query\n\nMeasured PI+ PI- PI+ PI- cross sections. The errors are statistical only." ]

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[ "In our circuit, we use Boolean algebra simplification methods like the Quine-McCluskey algorithm to simplify the Boolean expression and display the output on the display. It works as a portable calculator to simplify the Boolean expression on the fly.\n\n• Portable\n• Fast\n• Low power\n• Low cost\n• Reliable\n\n#### Block Diagram Description:\n\nAbove figure shows the basic block diagram of the project. Now let us discuss all the blocks in detail:\n\n1. Power Supply: It can be defined as a device that supplies electrical energy to one or more electric loads. The term is most commonly applied to devices that convert one form of electrical energy to another, though it may also refer to devices that convert another form of energy (e.g., mechanical, chemical, solar) to electrical energy. In our project a supply mains that is 5volt d.C. is given to the microcontroller, LED’s , keypad, display.\n2. Microcontroller: Microcontroller ATMEGA 16L is used for the automation purpose and acts as brain of the project. It controls the output (Display) according to the input given to it. Read the post: Microcontroller Basics to get basic knowledge about microcontrollers.\n3. Display: The Display used here is 3 Bi-color LED’s. The Glowing Pattern of LED’s represent the desired minimized expression.\n4. Keypad: In this project series of switches have been used as keypad, is used to give the input (min-terms) expression. Each digit on the keypad corresponds to one min-term each.\n\n### Boolean Algebra Calculator Circuit Diagram:\n\n#### Circuit Schematic:\n\nThis circuit is a simple 3 variable Boolean expression minimizer. It uses the Quine McCluskey algorithm which was described in the chapter 1.In this the microcontroller plays a major role, it consists of code to implement the described algorithm as well as controlls other components in the circuit.\n\nInitially when the power is switched on an LED will glow which indicates that the microcontroller is ready to take the input. Here the input boolean expression is given in SOP form, i.e only min-terms are to be entered through the keypad provided.\n\nThe keypad consists of 9 switches of which 8 switches corresponds to one min-term each and the 9th one is used as next button.After entering the expression the input indicating LED will go OFF, now based on the algorithm microcontroller reduces the expression and the input representing LED glows which means that the expression has been minimized and is displayed.\n\nThe display consists of 3 Bi-color LED’s in which Green Light represents the variables in normal form and the rest Red Light represents the variables in the complemented form, the order of them is as shown in the circuit diagram.\n\nThe output is displayed as one min-term at a time, next min-term is displayed by pressing the next button and after reaching the last min-term of the reduced expression the input indicating LED is switched OFF which represents end of the output. After few seconds it is again switched ON automatically when microcontroller is ready to take the next input." ]

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[ "# #122b26 Color Information\n\nIn a RGB color space, hex #122b26 is composed of 7.1% red, 16.9% green and 14.9% blue. Whereas in a CMYK color space, it is composed of 58.1% cyan, 0% magenta, 11.6% yellow and 83.1% black. It has a hue angle of 168 degrees, a saturation of 41% and a lightness of 12%. #122b26 color hex could be obtained by blending #24564c with #000000. Closest websafe color is: #003333.\n\n• R 7\n• G 17\n• B 15\nRGB color chart\n• C 58\n• M 0\n• Y 12\n• K 83\nCMYK color chart\n\n#122b26 color description : Very dark (mostly black) cyan.\n\n# #122b26 Color Conversion\n\nThe hexadecimal color #122b26 has RGB values of R:18, G:43, B:38 and CMYK values of C:0.58, M:0, Y:0.12, K:0.83. Its decimal value is 1190694.\n\nHex triplet RGB Decimal 122b26 `#122b26` 18, 43, 38 `rgb(18,43,38)` 7.1, 16.9, 14.9 `rgb(7.1%,16.9%,14.9%)` 58, 0, 12, 83 168°, 41, 12 `hsl(168,41%,12%)` 168°, 58.1, 16.9 003333 `#003333`\nCIE-LAB 15.467, -11.256, 0.265 1.463, 1.996, 2.142 0.261, 0.356, 1.996 15.467, 11.259, 178.651 15.467, -8.676, 1.318 14.129, -6.241, 0.902 00010010, 00101011, 00100110\n\n# Color Schemes with #122b26\n\n• #122b26\n``#122b26` `rgb(18,43,38)``\n• #2b1217\n``#2b1217` `rgb(43,18,23)``\nComplementary Color\n• #122b1a\n``#122b1a` `rgb(18,43,26)``\n• #122b26\n``#122b26` `rgb(18,43,38)``\n• #12242b\n``#12242b` `rgb(18,36,43)``\nAnalogous Color\n• #2b1a12\n``#2b1a12` `rgb(43,26,18)``\n• #122b26\n``#122b26` `rgb(18,43,38)``\n• #2b1224\n``#2b1224` `rgb(43,18,36)``\nSplit Complementary Color\n• #2b2612\n``#2b2612` `rgb(43,38,18)``\n• #122b26\n``#122b26` `rgb(18,43,38)``\n• #26122b\n``#26122b` `rgb(38,18,43)``\n• #172b12\n``#172b12` `rgb(23,43,18)``\n• #122b26\n``#122b26` `rgb(18,43,38)``\n• #26122b\n``#26122b` `rgb(38,18,43)``\n• #2b1217\n``#2b1217` `rgb(43,18,23)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #030706\n``#030706` `rgb(3,7,6)``\n• #0a1916\n``#0a1916` `rgb(10,25,22)``\n• #122b26\n``#122b26` `rgb(18,43,38)``\n• #1a3d36\n``#1a3d36` `rgb(26,61,54)``\n• #214f46\n``#214f46` `rgb(33,79,70)``\n• #296156\n``#296156` `rgb(41,97,86)``\nMonochromatic Color\n\n# Alternatives to #122b26\n\nBelow, you can see some colors close to #122b26. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #122b20\n``#122b20` `rgb(18,43,32)``\n• #122b22\n``#122b22` `rgb(18,43,34)``\n• #122b24\n``#122b24` `rgb(18,43,36)``\n• #122b26\n``#122b26` `rgb(18,43,38)``\n• #122b28\n``#122b28` `rgb(18,43,40)``\n• #122b2a\n``#122b2a` `rgb(18,43,42)``\n• #122a2b\n``#122a2b` `rgb(18,42,43)``\nSimilar Colors\n\n# #122b26 Preview\n\nText with hexadecimal color #122b26\n\nThis text has a font color of #122b26.\n\n``<span style=\"color:#122b26;\">Text here</span>``\n#122b26 background color\n\nThis paragraph has a background color of #122b26.\n\n``<p style=\"background-color:#122b26;\">Content here</p>``\n#122b26 border color\n\nThis element has a border color of #122b26.\n\n``<div style=\"border:1px solid #122b26;\">Content here</div>``\nCSS codes\n``.text {color:#122b26;}``\n``.background {background-color:#122b26;}``\n``.border {border:1px solid #122b26;}``\n\n# Shades and Tints of #122b26\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, #010201 is the darkest color, while #f3faf8 is the lightest one.\n\n• #010201\n``#010201` `rgb(1,2,1)``\n• #060f0e\n``#060f0e` `rgb(6,15,14)``\n• #0c1d1a\n``#0c1d1a` `rgb(12,29,26)``\n• #122b26\n``#122b26` `rgb(18,43,38)``\n• #183932\n``#183932` `rgb(24,57,50)``\n• #1e473e\n``#1e473e` `rgb(30,71,62)``\n• #23544b\n``#23544b` `rgb(35,84,75)``\n• #296257\n``#296257` `rgb(41,98,87)``\n• #2f7063\n``#2f7063` `rgb(47,112,99)``\n• #357e6f\n``#357e6f` `rgb(53,126,111)``\n• #3b8c7c\n``#3b8c7c` `rgb(59,140,124)``\n• #409a88\n``#409a88` `rgb(64,154,136)``\n• #46a794\n``#46a794` `rgb(70,167,148)``\n• #4db4a0\n``#4db4a0` `rgb(77,180,160)``\n• #5bbaa7\n``#5bbaa7` `rgb(91,186,167)``\n• #68c0ae\n``#68c0ae` `rgb(104,192,174)``\n• #76c6b6\n``#76c6b6` `rgb(118,198,182)``\n• #84ccbd\n``#84ccbd` `rgb(132,204,189)``\n• #92d1c5\n``#92d1c5` `rgb(146,209,197)``\n• #a0d7cc\n``#a0d7cc` `rgb(160,215,204)``\n• #aeddd3\n``#aeddd3` `rgb(174,221,211)``\n• #bbe3db\n``#bbe3db` `rgb(187,227,219)``\n• #c9e8e2\n``#c9e8e2` `rgb(201,232,226)``\n• #d7eeea\n``#d7eeea` `rgb(215,238,234)``\n• #e5f4f1\n``#e5f4f1` `rgb(229,244,241)``\n• #f3faf8\n``#f3faf8` `rgb(243,250,248)``\nTint Color Variation\n\n# Tones of #122b26\n\nA tone is produced by adding gray to any pure hue. In this case, #1e1f1f is the less saturated color, while #023b30 is the most saturated one.\n\n• #1e1f1f\n``#1e1f1f` `rgb(30,31,31)``\n• #1b2220\n``#1b2220` `rgb(27,34,32)``\n• #192422\n``#192422` `rgb(25,36,34)``\n• #172623\n``#172623` `rgb(23,38,35)``\n• #142925\n``#142925` `rgb(20,41,37)``\n• #122b26\n``#122b26` `rgb(18,43,38)``\n• #102d27\n``#102d27` `rgb(16,45,39)``\n• #0d3029\n``#0d3029` `rgb(13,48,41)``\n• #0b322a\n``#0b322a` `rgb(11,50,42)``\n• #09342c\n``#09342c` `rgb(9,52,44)``\n• #06372d\n``#06372d` `rgb(6,55,45)``\n• #04392e\n``#04392e` `rgb(4,57,46)``\n• #023b30\n``#023b30` `rgb(2,59,48)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #122b26 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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[ "Cody\n\n# Problem 262. Swap the input arguments\n\nSolution 2792959\n\nSubmitted on 5 Aug 2020 by Sam Maltz\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\n[q,r] = swapInputs(5,10); assert(isequal(q,10)); assert(isequal(r,5));\n\n2   Pass\n[q,r] = swapInputs(magic(3), 'hello, world'); assert(isequal(q,'hello, world')); assert(isequal(r,magic(3)));\n\n3   Pass\n[q,r] = swapInputs({}, NaN); assert(isnan(q)); assert(iscell(r) && isempty(r));\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]

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[ "# Group\n\nزمرة (Ar). Groupe (Fr). Gruppe (Ge). Gruppo (It). (Ja). Группа (Ru). Grupo (Sp).\n\nA set G equipped with a binary operation *: G × GG, assigning to a pair (g,h) the product g*h is called a group if:\n\n1. The operation is associative, i.e. (a*b)*c = a*(b*c).\n2. G contains an identity element (neutral element) e: g*e = e*g = g for all g in G.\n3. Every g in G has an inverse element h for which g*h = h*g = e. The inverse element of g is written as g−1.\n\nOften, the symbol for the binary operation is omitted. The product of the elements g and h is then denoted by the concatenation gh.\n\nThe binary operation need not be commutative, i.e. in general one will have g*h ≠ h*g. In the case that g*h = h*g holds for all g,h in G, the group is an Abelian group.\n\nA group G may have a finite or infinite number of elements. In the first case, the number of elements of G is the order of G. In the latter case, G is called an infinite group. Examples of infinite groups are space groups and their translation subgroups, whereas crystallographic point groups are finite groups." ]

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[ "", null, "Theoretical Economics Letters, 2012, 2, 400-407 http://dx.doi.org/10.4236/tel.2012.24074 Published Online October 2012 (http://www.SciRP.org/journal/tel) Joint Characteristic Function of Stock Log-Price and Squared Volatility in the Bates Model and Its Asset Pricing Applications Oleksandr Zhylyevskyy Department of Economics, Iowa State University, Ames, USA Email: [email protected] Received July 17, 2012; revised August 17, 2012; accepted September 17, 2012 ABSTRACT The model of Bates specifies a rich, flexible structure of stock dynamics suitable for applications in finance and eco- nomics, including valuation of derivative securities. This paper analytically derives a closed-form expression for the joint conditional characteristic function of a stock’s log-price and squared volatility under the model dynamics. The use of the function, based on inverting it, is illustrated on examples of pricing European-, Bermudan-, and American-style options. The discussed approach for European-style derivatives improves on the option formula of Bates. The suggested approach for American-style derivatives, based on a compound-option technique, offers an alternative solution to exist- ing finite-difference methods Keywords: Bates Model; Stochastic Volatility; Jump-Diffusion; Characteristic Function; Option Pricing 1. Introduction Stochastic volatility and jump-diffusion are standard tools of modeling asset price dynamics in finance research (see Aït-Sahalia and Jacod, 2011 ). Popularity of stochastic volatility models, such as the continuous-time model of Heston (1993) , is partly due to their ability to account for several aspects of stock price data that are not cap- tured by analytically simpler geometric Brownian motion dynamics. For example, these models can help to account for an empirically relevant “leverage effect,” which re- fers to an increase in the volatility of a stock when its price declines, and a decrease in the volatility when the price rises. They also can help to partly correct for defi-ciencies of the famous Black and Scholes (1973) op- tion pricing formula (e.g., the implied volatility “smile”). The model of Bates (1996) extends the Heston model by incorporating jumps in stock dynamics. Allowing for jumps enables a more realistic representation of stock price time-series, which may feature discontinuities (for a discussion on jumps in asset data, see Aït-Sahalia and Jacod, 2009 ). In this paper, I analytically derive and provide exam- ples for the use of a closed-form expression for the joint conditional characteristic function of a stock’s log-price and squared volatility under the dynamics of the Bates model. The model offers a rich distributional structure of stock returns. For instance, a skewed distribution can arise due to a correlation between shocks to the stock price and shocks to the volatility or due to nonzero aver- age jumps. Excess kurtosis can arise from variable vola- tility or from a jump component. Also, the model can help to distinguish between two alternative explanations for skewness and excess kurtosis: stochastic volatility implies a positive relationship between the length of the holding period and the magnitude of skewness and kur-tosis, whereas jumps imply a negative relationship (Bates, 1996 , pp. 72-73). The flexibility of the model makes it particularly attractive for the task of valuation of de-rivative securities. As such, it is useful in applied research and practice. Under jump-diffusion dynamics with stochastic vola- tility, the values of derivative securities such as Euro- pean-style options are typically impossible to express in simple form. Instead, they may be computed numeri- cally by applying the transform methods of Duffie et al. (2000) and Bakshi and Madan (2000) , which require inverting a conditional characteristic function of an underlying state-price vector. Bates (1996) solved for the marginal conditional characteristic function of the logprice and derived a formula for the value of a European-style call option that involves two separate inversions. In contrast, the problem of finding the joint conditional characteristic function of the log-price and squared volatility was not posed, and to the best of my knowledge, a solution for this function is not available in Copyright © 2012 SciRes. TEL", null, "O. ZHYLYEVSKYY 401existing finance studies. This paper aims to fill in the gap by deriving a closed-form expression for the function, which is an analytically challenging task. In addition, I provide two practically relevant examples illustrating the use of the function. The first example revisits the prob-lem of the valuation of European-style options. I show that the marginal characteristic function is a special case of the joint characteristic function and then apply results from prior research to obtain formulas for European-style put and call options that require a single inversion; this approach is more efficient than the solution suggested by Bates involving two inversions. The second example addresses the problem of valuation of Bermudan- and American-style options by proposing an extension of the Geske-Johnson compound-option technique (Geske and Johnson, 1984 ). In this case, knowledge of the joint (rather than marginal) characteristic function is indis-pensable. The proposed approach provides an alternative to pricing American-style options using finite-difference methods (e.g., Chiarella et al., 2008 ), which can pose practical challenges when dealing with stochastic volatil-ity (for a review, see Zhylyevskyy, 2010 ). The em-pirical relevance of the example is due to a large num-ber of single name equity and commodity futures op-tions traded on organized exchanges being American- style. The remainder of the paper is organized as follows. Section 2 sets up the Bates model and outlines the as- sumptions and notation. Section 3 derives a stochastic differential equation for the stock’s log-price. Section 4 shows that the joint conditional characteristic function is a martingale and uses this result to derive a partial dif-ferential-integral equation for the function. Section 5 solves this equation analytically to obtain a closed-form expression for the function. Section 6 provides examples for the use of the function when pricing derivative secu-rities. Section 7 concludes. 2. The Bates Model I first outline the assumptions and introduce the notation. The financial market is assumed to admit no arbitrage opportunities. Thus, there is an equivalent martingale pro- bability measure (see Harrison and Kreps, 1979 ), denoted here as 1. Random variables and stochastic processes are defined on a probability space with as the probability measure. An expected value taken with respect to is denoted by . To rigorously analyze stochastic processes, I work with a filtered probability space 0t, where is the set of outcomes, indexes time,  is a filtration (i.e., a non-decreasing sequence of PtPP,,t[]Ett-f,P elds0i), and -fiel d . Stochastic processes are assumed to be adapted to tt0ttOne of the assets traded in the financial market is a riskless bond fund with a share worth . 0=rttMMe0r on date , where 0 is an initial value and is a risk-free interest rate, which is assumed to be constant over time. In contexts involving asset pricing (e.g., valuation of derivative securities), such riskless fund is often used as a numéraire asset, with prices of other assets being discounted by tt>0MM. Also, is often referred to as the “risk-neutral” probability measure. PI focus on a stock process 0tt, where t denotes the price of the stock on date . The stock is allowed to pay dividends continuously at a rate StS0=0, which is assumed to be constant over time ( in the case of no dividend). Since the stock process in the Bates model incorporates a jump component, which results in discon- tinuities in the stock price, it is helpful to introduce the notion of a “left limit” of a stochastic process. In particular, the left limit of 0ttS on date t is defined as 1mnt=liSStn, where is a positive integer. If there is a jump on date , then . ntttIn the Bates model, the dynamics of t under P are described by a system of two stochastic differential equa- tions: SSS1=,t ttdSdtv dWUdN ttS r (1) 2=.tttdv tvdtv dW  (2) Equation (1) shows that the instantaneous net return on the stock, ttdS S, is a sum of three distinct com- ponents: 1) a deterministic drift term ; 2) a stochastic diffusion term rd t1tt1WvdW , and (3) a stochastic jump term tUdN . A process 0tt underlying the stochastic diffusion term is a standard Brownian motion. A process N0tt underlying the stochastic jump term is a Poisson process with intensity 0, so that =tEN t. The processes 10tt and are independent of each other. A value of in- dicates that the stock price has undergone t jumps as of date . The magnitudes of such jumps are governed by independent and identically distributed (i.i.d.) ran- dom variables such that W0ttN>0tNNt12,,UU22ln1ln1/ 2,,UN  (3) where is a generic random variable having the same distribution as 12 , and U,,UU>1=1t and are the distribution parameters. In Equation (1), is the random percentage jump of the stock price given a jump occurring at (i.e., given ). The Bates model reduces to the Heston stochastic volatility model (Heston, 1993 ) if (1) 20UtdN=0, or (2) =0 and , since these cases effectively eliminate jumps from the stock dynamics. 2=0Equation (2) describes a mean-reverting square root 1P need not be unique, as the financial market may be incomplete. Copyright © 2012 SciRes. TEL", null, "O. ZHYLYEVSKYY 402 process for the squared volatility2. This equation is bor- rowed directly from the Heston model. A process 0tt is a standard Brownian motion possibly cor- related with 10{, so that 2W}ttW12,=tdWW dt, with 1. The process 2ttW12,...UU 0, process 0tt, and the random variables are mutually independent. Constants N,0, 0, and 0 are parameters. In order for to be almost surely (a.s.) positive so that t and t are real-valued a.s., tvvS and  are assumed to satisfy a restriction 22 (see Chernov and Ghysels, 2000 ). 3. Dynamics of Log-Price Let ts denote the stock’s log-price, t=lntsS. The dy- namics of ts under are derived using a generalized Itô formula for semimartingales, which allows me to properly account for possible discontinuities in the time path of the stock price. See Theorem 32 of Protter (1990 , p. 71) for details on the formula. Before applying the Itô formula, observe that Equation (1) implies that t and t are, in general, not equal to each other because of the presence of the jump term; more speci- fically, tt Thus, P=tSUSS SS.tdN =1tt tSS UdN, and therefore, lnln= ln1.tt tSS UdN Also, note that by the properties of the Poisson process, is effectively either 0 or 1. Therefore, tdNlnln =ln1=ln1.tt tSS UdN UdN t Hence, the generalized Itô formula applied to the function ln tS indicates that the dynamics of ts under are described by a stochastic differential equation P22111=2 lnln,tt tttttttttdsdSv SdtSSSSSSS  which is straightforward to simplify as: 1=2ln1.12,,UU-fi e l dtttttdsrvdtvdWUdN h(4) 4. Martingale Property and Dynamics of Joint Characteristic Function My main interest lies in deriving a closed-form ex- pression for the joint characteristic function of some future, date- log-price and squared volatility given their present, date-t values, where . Consider an arbitrary date such that , and note that since , . Equations (2) and (4), the properties of the Poisson process and standard Brownian motion, and the assumption of i.i.d. random variables imply that the information contained in the Thttnoted111,,ttsvTt. For example, 1st=max 0,Xe Copyright © 2012 SciRes. TEL", null, "O. ZHYLYEVSKYY 406 in the case of a put option and 1=max 0, steX in the case of a call option. Bermudan-style derivative se- curities obey a recursive relationship: 111 111 111 ,,= max,,,,,=maxnttrttttntttrttDsvTteEsvTtDsvTte where and 11 101,,,,, ,;,,,nttsvTtDsvTtfsvsvt tdvds00D1,; , ,ttfsvs v tt12 1,;,,tt can be com- putedrting by inve svt trelationship provny Bermudan-stylapproximate the price of atforwar2,,ttDsvT t3,,ttDsvTt, and t 2, as discussed es a way to comice of ae derivative se- curity, to n American- style onevalue of . In pr d to coute and , and ifm- hen 010 )is methodological approach is an alternative to pricing American-style derivative securities under the Bates mdynamics using a finite-difference-type scheme proposed by Chiarella et al. (2008) 4. 7. Conclusion This paper contributes to the literature by solving in closedform for the joint conditional characteristic function of nd d volatility undepates model. Theving a systeREFERENCES earlier. In theory, the pute the pras well as,,ttvT tputationally feasible, id by choosing a sufficiently large actice, it should be straighnmp couse these. Th1Dscomputed values to approximate the American-style option price ,,ttDsvTt by applying a Richardson extrapolation (e.g., see Zhylyevskyy,odel the log-price asquarer the jum-dif- fusion dynamics of the B model features a flexible distributional structure of asset returns. As such, it has a number of applications in finance and economics, including the problem of valuation of derivative se- curities. Obtaining a closed-form expression for the joint characteristic function is an analytically demanding task, which involves applying a generalized Itô formula for semimartingales and solm of differential equations, among other steps. The use of the derived function is illustrated on empirically relevant examples of pricing European-, Bermudan-, and American-style options. The proposed methodological approach is based on inverting the characteristic function, and may be em- ployed in practice as an alternative to pricing derivative securities using finite-difference techniques, particularly in the case of American-style options. Y. Aït-Sahalia and J. Jacod, “Analyzing the Spectrum of Asset Returns: Jump and Volatility Components in High Frequency Data,” Journal of Economic Literature, Vol. 50, No. 4, 2012, pp. 1007-1050. S. L. Heston, “A Closed-Form Solution for Options with Stochastic Volatility with Applications to Bond and Cur- rency Options,” Review of Financial Studies, Vol. 6, No. 2, 1993, pp. 327-343. doi:10.1093/rfs/6.2.327 F. Black and M. Scholes, “The Pricing of Options and Corporate Liabilities,” Journal of Political Economy, Vol. 81, No. 3, 1973, pp. 637-654. doi:10.1086/260062 D. S. Bates, “Jumps and Stochastic Volatility: Exchange Rate Processes Implicit in Deutsche Mark Options,” Re- view of Financial Studies, Vol. 9, No. 1, 1996, pp. 69-107. doi:10.1093/rfs/9.1.69 Y. Aït-Sahalia and J. Jacod, “Testing for Jumps in a Dis- cretely Observed Process,” Annals of Statistics, Vol. 37, No. 1, 2009, pp. 184-222. doi:10.1214/07-AOS568 D. Duffie, J. Pan and K. Singleton, “Transform Analysis and Asset Pricing for Affine Jump-Diffusions,” Econo- metrica, Vol. 68, No. 6, 2000, pp. 1343-1376. doi:10.1111/1468-0262.00164 G. Bakshi andand Derivative-Se- curity Valuationics, Vol. D. Madan, “Spanning ,” Journal of Financial Econom55, No. 2, 2000, pp. 205-238. doi:10.1016/S0304-405X(99)00050-1 R. Geske and H. E. Johnson, “The American Put Option Valued Analytically,” Journal of Finance, Vol. 39, No. 5, 1984, pp. 1511-1524. C. Chiarella, B. Kang, G. H. Meyer and A. Ziogas, “The Evaluation of American Option Prices under Stochastic Volatility and Jump-Diffusion Dynamics Using the Me- thod of Lines,” Quantitative Finance Research Centre, University of Technology, Sydney, 2008. O. Zhylyevskyy, “A Fast Fourier Transform Tec hnique for Pricing American Options under Stochastic Volatil- ity,” Review of Derivatives Research, Vol. 13, No. 1, 2010, pp. 1-24. doi:10.1007/s11147-009-9041-6 J. M. Harrison and D. M. Kreps, “Martingales and Arbi-trage in Multiperiod Securities Markets,” Journal of Eco- nomic Theory, Vol. 20, No. 3, 1979, pp. 381-408. doi:10.1016/0022-0531(79)90043-7 M. Chernov and E. Ghysels, “A Study towards a Unified Approach to the Joint Estimation of Objective and Risk Neutral Measures for the Purpose of Options VJournal of Financial Economicaluation,” s, Vol. 56, No. 3, 2000, pp. 407-458. doi:10.1016/S0304-405X(00)00046-5 P. Protter, “Stochastic Integration and Differential Equa-tions: A New Approach,” Springer-V1990. 4Chiarella et al. propose a method of lines, in which a partial differen-tial-integral equation is replaced with a system of simpler differential equations to be solved using a stabilized finite-difference scheme. The integral component of the equation is approximated using an Hermite-Gauss quadrature. erlag, New York, K. L. Chung, “A Course in Probability Theory,” 3rd Edi- Copyright © 2012 SciRes. TEL", null, "O. ZHYLYEVSKYY Copyright © 2012 SciRes. TEL 407ficient Pricing of European-Style tion, Academic Press, San Diego, 2001. O. Zhylyevskyy, “EfOptions under Heston’s Stochastic Volatility Model,” Theo- retical Economics Letters, Vol. 2, No. 1, 2012, pp. 16-20. doi:10.4236/tel.2012.21003 R. C. Merton, “Theory of Rational Option Pricing,” Bell Journal of Economics and Management Science, Vol. 4, No. 1, 1973, pp. 141-183. doi:10.2307/3003143 W. H. Press, S. A. Teukolsky, W. T. Vetterling and B. P. Flannery, “Numerical Recipes in Fortran 77: The Art of Scientific Computing,” 2nd Edition, Cambridge Univer-sity Press, Cambridge, 2001. for the Theory,” N. G. Shephard, “From Characteristic Function to Distri- bution Function: A Simple Framework Econometric Theory, Vol. 7, No. 4, 1991, pp. 519-529. doi:10.1017/S0266466600004746 T. W. Epps, “Pricing Derivative Securities,” World Sci- entific, River Edge, 2000." ]

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[ "DGtal::MeasureOfStraightLines Class Reference\n\n`#include <MeasureOfStraightLines.h>`\n\nPublic Member Functions\n\nMeasureOfStraightLines ()\n~MeasureOfStraightLines ()\nvoid selfDisplay (std::ostream &out) const\nbool isValid () const\ndouble computeMeasure (const std::vector< double > &a, const std::vector< double > &b)\ndouble computeCentroidA (const std::vector< double > &a, const std::vector< double > &b)\ndouble computeCentroidB (const std::vector< double > &a, const std::vector< double > &b)\nvoid setEpsilon (const double aValue)\n\nPrivate Member Functions\n\nMeasureOfStraightLines (const MeasureOfStraightLines &other)\nMeasureOfStraightLinesoperator= (const MeasureOfStraightLines &other)\ndouble computeMeasureEdge (double a0, double b0, double a1, double b1)\ndouble computeCentroidEdge_a (double a0, double b0, double a1, double b1)\ndouble computeCentroidEdge_b (double a0, double b0, double a1, double b1)\ndouble __computeCentroidTriApprox_b (double a0, double b0, double a1, double b1)\ndouble __computeCentroidEdgeApprox_b (double a0, double b0, double a1, double b1)\ndouble __computeCentroidSquare_b (double x1, double y1, double x2, double y2)\nint sign (double a)\n\ndouble myEpsilon\n\nDetailed Description\n\nDescription of class 'MeasureOfStraightLines'\n\nAim:\n\nThe aim of this class is to compute the measure in the Lebesgues sense of the set of straight lines associated to domains defined as polygons in the (a,b)-parameter space. This parameter space maps the line \\$ax-y+b=0\\$ to the point \\$(a,b)\\$.\n\n``` * @inproceedings{COEURJOLLY:2009:HAL-00432711:1,\n* title = { {M}easure of {S}traight {L}ines and its {A}pplications in {D}igital {G}eometry},\n* author = {{C}oeurjolly, {D}avid and {S}ivignon, {I}sabelle},\n* booktitle = {13th {I}nternational {W}orkshop on {C}ombinatorial {I}mage {A}nalysis 13th {I}nternational {W}orkshop on {C}ombinatorial {I}mage {A}nalysis },\n* publisher = {{R}esearch {P}ublishing {S}ervices },\n* pages = {1-12 },\n* address = {{C}ancun {M}exique },\n* audience = {internationale },\n* year = {2009},\n* URL = {http://hal.archives-ouvertes.fr/hal-00432711/PDF/mesure.pdf},\n* }\n* ```\ntestMeasure.cpp\nTodo:\nCreate a vector<Point> interface\n\nConstructor & Destructor Documentation\n\n DGtal::MeasureOfStraightLines::MeasureOfStraightLines ( )\n\nConstructor.\n\nDefault value\n\nReferences myEpsilon.\n\n DGtal::MeasureOfStraightLines::~MeasureOfStraightLines ( )\n\nDestructor.\n\n DGtal::MeasureOfStraightLines::MeasureOfStraightLines ( const MeasureOfStraightLines & other ) ` [private]`\n\nCopy constructor.\n\nParameters:\n other the object to clone. Forbidden by default.\n\nMember Function Documentation\n\n double DGtal::MeasureOfStraightLines::__computeCentroidEdgeApprox_b ( double a0, double b0, double a1, double b1 ) ` [private]`\n\nApproximate the centroid on 'b' on the trapezioid (a0,0)-(a0,b0)-(a1,b1)-(a1,0) (internal function)\n\n double DGtal::MeasureOfStraightLines::__computeCentroidSquare_b ( double x1, double y1, double x2, double y2 ) ` [private]`\n\nCompute the centroid on 'b' on the rectangular domain with vertices (x1,,y1) - (x2,y2) PRECONDITION: y1<y2\n\n double DGtal::MeasureOfStraightLines::__computeCentroidTriApprox_b ( double a0, double b0, double a1, double b1 ) ` [private]`\n\nApproximate the centroid on 'b' on the triangle (0,0)-(a0,b0)-(a1,b1) (internal function)\n\n double DGtal::MeasureOfStraightLines::computeCentroidA ( const std::vector< double > & a, const std::vector< double > & b )\n\nCompute the abscissa of the centroid of the polygon {(a_i,b_i)} in the (a,b)-parameter space with respect to the measure of lines.\n\nREQUIREMENTS:\n\n• The polygon is given counter-clockwise\n• a_i > 0\nParameters:\n a the a-value of polygon vertices b the b-value of polygon vertices\nReturns:\nthe measure value (positive value)\n\nReferenced by testUnitSquareCentroid().\n\n double DGtal::MeasureOfStraightLines::computeCentroidB ( const std::vector< double > & a, const std::vector< double > & b )\n\nCompute the ordinate of the centroid of the polygon {(a_i,b_i)} in the (a,b)-parameter space with respect to the measure of lines. Note that there is a numerical approximation is performed.\n\nREQUIREMENTS:\n\n• The polygon is given counter-clockwise\n• a_i > 0\nParameters:\n a the a-value of polygon vertices b the b-value of polygon vertices\nReturns:\nthe measure value (positive value)\n\nReferenced by testUnitSquareCentroid().\n\n double DGtal::MeasureOfStraightLines::computeCentroidEdge_a ( double a0, double b0, double a1, double b1 ) ` [private]`\n\nCompute the abscissa of the centroid associated to an edge (a0,b0)-(a1,b1) It returns the measure of the triangle defined by the origin and the edge.\n\nParameters:\n a0 b0 a1 b1\nReturns:\nthe measure\n double DGtal::MeasureOfStraightLines::computeCentroidEdge_b ( double a0, double b0, double a1, double b1 ) ` [private]`\n\nCompute the ordinate of the centroid associated to an edge (a0,b0)-(a1,b1) It returns the measure of the triangle defined by the origin and the edge.\n\nParameters:\n a0 b0 a1 b1\nReturns:\nthe measure\n double DGtal::MeasureOfStraightLines::computeMeasure ( const std::vector< double > & a, const std::vector< double > & b )\n\nCompute the measure of the polygon {(a_i,b_i)} in the (a,b)-parameter space\n\nREQUIREMENTS:\n\n• The polygon is given counter-clockwise\n• a_i > 0\nParameters:\n a the a-value of polygon vertices b the b-value of polygon vertices\nReturns:\nthe measure value (positive value)\n\nReferenced by testUnitSquare().\n\n double DGtal::MeasureOfStraightLines::computeMeasureEdge ( double a0, double b0, double a1, double b1 ) ` [private]`\n\nCompute the measure associated to an edge (a0,b0)-(a1,b1) It returns the measure of the triangle defined by the origin and the edge.\n\nParameters:\n a0 b0 a1 b1\nReturns:\nthe measure\n bool DGtal::MeasureOfStraightLines::isValid ( ) const\n\nChecks the validity/consistency of the object.\n\nReturns:\n'true' if the object is valid, 'false' otherwise.\n MeasureOfStraightLines& DGtal::MeasureOfStraightLines::operator= ( const MeasureOfStraightLines & other ) ` [private]`\n\nAssignment.\n\nParameters:\n other the object to copy.\nReturns:\na reference on 'this'. Forbidden by default.\n void DGtal::MeasureOfStraightLines::selfDisplay ( std::ostream & out ) const\n\nWrites/Displays the object on an output stream.\n\nParameters:\n out the output stream where the object is written.\n void DGtal::MeasureOfStraightLines::setEpsilon ( const double aValue )\n\nSet the internal Epsilon threshold for the numerical approximation.\n\nParameters:\n aValue the new epsilon value\n int DGtal::MeasureOfStraightLines::sign ( double a ) ` [private]`\nReturns:\nthe sign of a number (1 or -1)\n\nField Documentation\n\n double DGtal::MeasureOfStraightLines::myEpsilon` [private]`\n\nReferenced by MeasureOfStraightLines().\n\nThe documentation for this class was generated from the following files:" ]

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[ "# How do you normalize (2i -3j + 4k)?\n\nDec 26, 2015\n\nIf you have a vector $\\vec{a}$ with coordinates (x,y,z) then the normalized vector $\\vec{n}$ is\n\n$\\vec{n} = \\frac{\\vec{a}}{\\left\\mid a \\right\\mid} = \\frac{\\left\\langlex \\text{,\"y\",} z\\right\\rangle}{\\left\\mid a \\right\\mid}$\n\nwhere $\\left\\mid a \\right\\mid = \\sqrt{{x}^{2} + {y}^{2} + {z}^{2}}$\n\nHence for $x = 2 , y = - 3 , z = 4$ we have that\n\n$\\left\\mid a \\right\\mid = \\sqrt{{2}^{2} + {3}^{2} + {4}^{2}} = \\sqrt{4 + 9 + 16} = \\sqrt{29}$\n\nThus the normalized vector is\n\n$\\vec{n} = \\textcolor{b l u e}{\\left\\langle\\frac{2}{\\sqrt{29}} , - \\frac{3}{\\sqrt{29}} , \\frac{4}{\\sqrt{29}}\\right\\rangle}$" ]

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[ "django-db-constraints\npypi i django-db-constraints\n\n# django-db-constraints\n\n### by David Sanders\n\npypi i django-db-constraints\n\n# django-db-constraints\n\n## What is this?\n\nAdd database table-level constraints to your Django model's Meta class and have `makemigrations` add the appropriate migration.\n\n``````class Foo(models.Model):\nbar = models.IntegerField()\nbaz = models.IntegerField()\n\nclass Meta:\ndb_constraints = {\n'bar_equal_baz': 'check (bar = baz)',\n}\n``````\n\nThis should generate a migration like so:\n\n``````class Migration(migrations.Migration):\n\ninitial = True\n\ndependencies = [\n]\n\noperations = [\nmigrations.CreateModel(\nname='Foo',\nfields=[\n('id', models.AutoField(auto_created=True, primary_key=True, serialize=False, verbose_name='ID')),\n('bar', models.IntegerField()),\n('baz', models.IntegerField()),\n],\n),\ndjango_db_constraints.operations.AlterConstraints(\nname='Foo',\ndb_constraints={'bar_equal_baz': 'check (bar = baz)'},\n),\n]\n``````\n\nThe resulting SQL applied:\n\n``````CREATE TABLE \"sample_foo\" (\"id\" serial NOT NULL PRIMARY KEY, \"bar\" integer NOT NULL, \"baz\" integer NOT NULL)\nALTER TABLE \"sample_foo\" ADD CONSTRAINT \"bar_equal_baz\" check (bar = baz)\n``````\n\n## Composite foreign keys\n\nIt's possible to support composite foreign keys if you have a unique key on your reference model:\n\n``````class Bar(models.Model):\nbaz = models.IntegerField()\n\nclass Meta:\nunique_together = ('id', 'baz')\n\nclass Foo(models.Model):\nbar = models.ForeignKey(Bar)\nbaz = models.IntegerField()\n\nclass Meta:\ndb_constraints = {\n'composite_fk': 'foreign key (bar_id, baz) references sample_bar (id, baz)',\n}\n``````\n\nResults in:\n\n``````ALTER TABLE \"sample_foo\" ADD CONSTRAINT \"composite_fk\" foreign key (bar_id, baz) references sample_bar (id, baz)\n``````\n\n## Migration operation ordering\n\nGiven that nothing will depend on a constraint operation, they're simply added to the end of the list of operations for a migration. This includes operations that drop fields used in a constraint as the database drop will any related constraints as well (at least with PostgreSQL).\n\n## Caveats\n\nIt's possible to end up in a situation where the constraints are declared on the Meta class but do not exist in the database due to a database dropping a constraint implicitly when a field in the constraint is dropped.\n\n## Installation\n\n``````pip install django-db-constraints\n``````\n\n``````INSTALLED_APPS = [\n'django_db_constraints',\n…\n]\n``````\nVersionTagPublished\n0.3.0\n5yrs ago\n0.2.0\n5yrs ago\n\n## Rate & Review\n\n100", null, "", null, "No reviews found\nBe the first to rate" ]

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[ "", null, "Business\n31 August, 00:44\n\n# Straight-Line Depreciation A building acquired at the beginning of the year at a cost of \\$125,800 has an estimated residual value of \\$4,800 and an estimated useful life of 10 years. Determine the following: (a) The depreciable cost \\$ (b) The straight-line rate % (c) The annual straight-line depreciation \\$\n\n+2\n1.", null, "31 August, 01:20\n0\na. Depreciable cost = \\$121000\n\nb. The straight line rate = 10%\n\nc. Annual depreciation = \\$12100\n\nExplanation:\n\na.\n\nThe depreciable cost is the amount out of the total cost for an asset that is eligible for deprecaition. The deprecaible cost is the difference between the cost of an asset and its residual value.\n\nDepreciable cost = \\$125800 - 4800 = \\$121000\n\nb.\n\nThe straight line depreciation charges equal depreciation over an assets useful life. Let depreciable amount be considered as 100%\n\nThe straight line rate is = Depreciable amount in % / Useful life in years\n\nStright line rate = 100% / 10 = 10%\n\nc.\n\nAnnual straigh line depreciation = 121000 * 0.1 = 12100" ]

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[ "# A non Archimedean ordered field\n\nLet’s recall that an ordered field $$K$$ is said to be Archimedean if for any $$a,b \\in K$$ such that $$0 \\lt a \\lt b$$ it exists a natural number $$n$$ such that $$na > b$$.\n\nThe ordered fields $$\\mathbb Q$$ or $$\\mathbb R$$ are Archimedean. We introduce here the example of an ordered field which is not Archimedean. Let’s consider the field of rational functions\n$\\mathbb R(x) = \\left\\{\\frac{S(x)}{T(x)} \\ | \\ S, T \\in \\mathbb R[x] \\right\\}$ For $$f(x)=\\frac{S(x)}{T(x)} \\in \\mathbb R(x)$$ we can suppose that the polynomials have a constant polynomial greatest common divisor.\n\nNow we define $$P$$ as the set of elements $$f(x)=\\frac{S(x)}{T(x)} \\in \\mathbb R(x)$$ in which the leading coefficients of $$S$$ and $$T$$ have the same sign.\n\nOne can verify that the subset $$P \\subset \\mathbb R(x)$$ satisfies following two conditions:\n\nORD 1\nGiven $$f(x) \\in \\mathbb R(x)$$, we have either $$f(x) \\in P$$, or $$f(x)=0$$, or $$-f(x) \\in P$$, and these three possibilities are mutually exclusive. In other words, $$\\mathbb R(x)$$ is the disjoint union of $$P$$, $$\\{0\\}$$ and $$-P$$.\nORD 2\nFor $$f(x),g(x) \\in P$$, $$f(x)+g(x)$$ and $$f(x)g(x)$$ belong to $$P$$.\n\nThis means that $$P$$ is a positive cone of $$\\mathbb R(x)$$. Hence, $$\\mathbb R(x)$$ is ordered by the relation\n$f(x) > 0 \\Leftrightarrow f(x) \\in P.$\n\nNow let’s consider the rational fraction $$h(x)=\\frac{x}{1} \\in \\mathbb R(x)$$. $$h(x)$$ is a positive element, i.e. belongs to $$P$$ as $$h-1 = \\frac{x-1}{1}$$. For any $$n \\in \\mathbb N$$, we have\n$h – n 1=\\frac{x-n}{1} \\in P$ as the leading coefficients of $$x-n$$ and $$1$$ are both equal to $$1$$. Therefore, we have $$h \\gt n 1$$ for all $$n \\in \\mathbb N$$, proving that $$\\mathbb R(x)$$ is not Archimedean.\n\n# A field that can be ordered in two distinct ways\n\nFor a short reminder about ordered fields you can have a look to following post. We prove there that $$\\mathbb{Q}$$ can be ordered in only one way.\n\nThat is also the case of $$\\mathbb{R}$$ as $$\\mathbb{R}$$ is a real-closed field. And one can prove that the only possible positive cone of a real-closed field is the subset of squares.\n\nHowever $$\\mathbb{Q}(\\sqrt{2})$$ is a subfield of $$\\mathbb{R}$$ that can be ordered in two distinct ways. Continue reading A field that can be ordered in two distinct ways\n\n# An infinite field that cannot be ordered\n\n## Introduction to ordered fields\n\nLet $$K$$ be a field. An ordering of $$K$$ is a subset $$P$$ of $$K$$ having the following properties:\n\nORD 1\nGiven $$x \\in K$$, we have either $$x \\in P$$, or $$x=0$$, or $$-x \\in P$$, and these three possibilities are mutually exclusive. In other words, $$K$$ is the disjoint union of $$P$$, $$\\{0\\}$$, and $$-P$$.\nORD 2\nIf $$x, y \\in P$$, then $$x+y$$ and $$xy \\in P$$.\n\nWe shall also say that $$K$$ is ordered by $$P$$, and we call $$P$$ the set of positive elements. Continue reading An infinite field that cannot be ordered" ]

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[ "# #15281b Color Information\n\nIn a RGB color space, hex #15281b is composed of 8.2% red, 15.7% green and 10.6% blue. Whereas in a CMYK color space, it is composed of 47.5% cyan, 0% magenta, 32.5% yellow and 84.3% black. It has a hue angle of 138.9 degrees, a saturation of 31.1% and a lightness of 12%. #15281b color hex could be obtained by blending #2a5036 with #000000. Closest websafe color is: #003333.\n\n• R 8\n• G 16\n• B 11\nRGB color chart\n• C 48\n• M 0\n• Y 33\n• K 84\nCMYK color chart\n\n#15281b color description : Very dark (mostly black) cyan - lime green.\n\n# #15281b Color Conversion\n\nThe hexadecimal color #15281b has RGB values of R:21, G:40, B:27 and CMYK values of C:0.48, M:0, Y:0.33, K:0.84. Its decimal value is 1386523.\n\nHex triplet RGB Decimal 15281b `#15281b` 21, 40, 27 `rgb(21,40,27)` 8.2, 15.7, 10.6 `rgb(8.2%,15.7%,10.6%)` 48, 0, 33, 84 138.9°, 31.1, 12 `hsl(138.9,31.1%,12%)` 138.9°, 47.5, 15.7 003333 `#003333`\nCIE-LAB 14.151, -11.446, 6.167 1.266, 1.756, 1.309 0.292, 0.405, 1.756 14.151, 13.001, 151.683 14.151, -6.858, 6.044 13.252, -6.14, 3.419 00010101, 00101000, 00011011\n\n# Color Schemes with #15281b\n\n• #15281b\n``#15281b` `rgb(21,40,27)``\n• #281522\n``#281522` `rgb(40,21,34)``\nComplementary Color\n• #192815\n``#192815` `rgb(25,40,21)``\n• #15281b\n``#15281b` `rgb(21,40,27)``\n• #152825\n``#152825` `rgb(21,40,37)``\nAnalogous Color\n• #281519\n``#281519` `rgb(40,21,25)``\n• #15281b\n``#15281b` `rgb(21,40,27)``\n• #251528\n``#251528` `rgb(37,21,40)``\nSplit Complementary Color\n• #281b15\n``#281b15` `rgb(40,27,21)``\n• #15281b\n``#15281b` `rgb(21,40,27)``\n• #1b1528\n``#1b1528` `rgb(27,21,40)``\n• #222815\n``#222815` `rgb(34,40,21)``\n• #15281b\n``#15281b` `rgb(21,40,27)``\n• #1b1528\n``#1b1528` `rgb(27,21,40)``\n• #281522\n``#281522` `rgb(40,21,34)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #030704\n``#030704` `rgb(3,7,4)``\n• #0c1710\n``#0c1710` `rgb(12,23,16)``\n• #15281b\n``#15281b` `rgb(21,40,27)``\n• #1e3926\n``#1e3926` `rgb(30,57,38)``\n• #274932\n``#274932` `rgb(39,73,50)``\n• #2f5a3d\n``#2f5a3d` `rgb(47,90,61)``\nMonochromatic Color\n\n# Alternatives to #15281b\n\nBelow, you can see some colors close to #15281b. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #152816\n``#152816` `rgb(21,40,22)``\n• #152818\n``#152818` `rgb(21,40,24)``\n• #152819\n``#152819` `rgb(21,40,25)``\n• #15281b\n``#15281b` `rgb(21,40,27)``\n• #15281d\n``#15281d` `rgb(21,40,29)``\n• #15281e\n``#15281e` `rgb(21,40,30)``\n• #152820\n``#152820` `rgb(21,40,32)``\nSimilar Colors\n\n# #15281b Preview\n\nThis text has a font color of #15281b.\n\n``<span style=\"color:#15281b;\">Text here</span>``\n#15281b background color\n\nThis paragraph has a background color of #15281b.\n\n``<p style=\"background-color:#15281b;\">Content here</p>``\n#15281b border color\n\nThis element has a border color of #15281b.\n\n``<div style=\"border:1px solid #15281b;\">Content here</div>``\nCSS codes\n``.text {color:#15281b;}``\n``.background {background-color:#15281b;}``\n``.border {border:1px solid #15281b;}``\n\n# Shades and Tints of #15281b\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, #010101 is the darkest color, while #f4f9f5 is the lightest one.\n\n• #010101\n``#010101` `rgb(1,1,1)``\n• #070e0a\n``#070e0a` `rgb(7,14,10)``\n• #0e1b12\n``#0e1b12` `rgb(14,27,18)``\n• #15281b\n``#15281b` `rgb(21,40,27)``\n• #1c3524\n``#1c3524` `rgb(28,53,36)``\n• #23422c\n``#23422c` `rgb(35,66,44)``\n• #294f35\n``#294f35` `rgb(41,79,53)``\n• #305b3e\n``#305b3e` `rgb(48,91,62)``\n• #376846\n``#376846` `rgb(55,104,70)``\n• #3e754f\n``#3e754f` `rgb(62,117,79)``\n• #448258\n``#448258` `rgb(68,130,88)``\n• #4b8f60\n``#4b8f60` `rgb(75,143,96)``\n• #529c69\n``#529c69` `rgb(82,156,105)``\n• #59a872\n``#59a872` `rgb(89,168,114)``\n• #66af7d\n``#66af7d` `rgb(102,175,125)``\n• #73b588\n``#73b588` `rgb(115,181,136)``\n• #80bc93\n``#80bc93` `rgb(128,188,147)``\n• #8dc39e\n``#8dc39e` `rgb(141,195,158)``\n• #9acaa9\n``#9acaa9` `rgb(154,202,169)``\n• #a6d0b4\n``#a6d0b4` `rgb(166,208,180)``\n• #b3d7bf\n``#b3d7bf` `rgb(179,215,191)``\n• #c0deca\n``#c0deca` `rgb(192,222,202)``\n• #cde5d4\n``#cde5d4` `rgb(205,229,212)``\n• #daebdf\n``#daebdf` `rgb(218,235,223)``\n• #e7f2ea\n``#e7f2ea` `rgb(231,242,234)``\n• #f4f9f5\n``#f4f9f5` `rgb(244,249,245)``\nTint Color Variation\n\n# Tones of #15281b\n\nA tone is produced by adding gray to any pure hue. In this case, #1e1f1e is the less saturated color, while #023b14 is the most saturated one.\n\n• #1e1f1e\n``#1e1f1e` `rgb(30,31,30)``\n• #1c211e\n``#1c211e` `rgb(28,33,30)``\n• #1a231d\n``#1a231d` `rgb(26,35,29)``\n• #17261c\n``#17261c` `rgb(23,38,28)``\n• #15281b\n``#15281b` `rgb(21,40,27)``\n• #132a1a\n``#132a1a` `rgb(19,42,26)``\n• #102d19\n``#102d19` `rgb(16,45,25)``\n• #0e2f18\n``#0e2f18` `rgb(14,47,24)``\n• #0c3118\n``#0c3118` `rgb(12,49,24)``\n• #093417\n``#093417` `rgb(9,52,23)``\n• #073616\n``#073616` `rgb(7,54,22)``\n• #053815\n``#053815` `rgb(5,56,21)``\n• #023b14\n``#023b14` `rgb(2,59,20)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #15281b 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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[ "# 1, Principle of stack\n\nStack is a linear table (commonly known as stack) that is restricted to insert and delete operations at one end\n\nThe end at which operations are allowed is called the \"top of the stack\"\n\nThe other fixed end is called \"stack bottom\"\n\nWhen there are no elements in the stack, it is called \"empty stack\". Stack features: last in first out (LIFO).", null, "# 2, Implementation of sequential stack\n\nFor the principle and implementation of sequence table, see this article:\n[data structure and algorithm] program internal skill Part II - linear sequence table\n\n### 1. Sequential stack principle\n\nIt is a kind of sequential table, which has the same storage structure as the sequential table. It is defined by the array and completes various operations with the stack top pointer top (relative pointer) represented by the array subscript.", null, "Structure definition:\n\n```typedef int data_t ; /*Defines the data type of the data element in the stack*/\ntypedef struct\n{\ndata_t *data ; \t/*Point to the storage space of the stack with a pointer*/\nint maxlen;\t/*Maximum number of elements in the current stack*/\nint top ; \t/*A variable indicating the top position of the stack (array subscript)*/\n} sqstack; \t\t/*Sequential stack type definition*/\n```\n\n### 2. Stack creation\n\n```/**\n* @description: Creation of sequential stack\n* @param {int} len -User specified stack length\n* @return {sqstack-Stack top pointer}\n*/\nsqstack* stack_create(int len) {\nsqstack * s;\n\nif ((s =(sqstack *)malloc(sizeof(sqstack))) == NULL)\n{\nprintf(\"malloc sqstack failed\\n\");\nreturn NULL;\n}\n\nif ((s->data = (data_t *)malloc(len * sizeof(data_t)))==NULL)\n{\nprintf(\"malloc data failed\\n\");\nfree(s);\nreturn NULL;\n}\n\nmemset(s->data, 0, len*sizeof(data_t));\ns->maxlen = len;\ns->top = -1;\n\nreturn s;\n}\n```\n\n### 3. Sequential stack\n\n```/**\n* @description: Enter the stack\n* @param {sqstack* } s-Stack top pointer\n* @param {data_t} value-Stack value\n* @return {-1-Function failed, 0-function succeeded}\n*/\nint stack_push(sqstack *s, data_t value)\n{\nif (s == NULL) {\nprintf(\"s is NULL\\n\");\nreturn -1;\n}\n\nif (s->top == s->maxlen-1)\n{\nprintf(\"stack is full\\n\");\nreturn -1;\n}\n\ns->top++;\ns->data[s->top] = value;\n\nreturn 0;\n}\n```\n\n### 4. Sequential stack\n\n```/**\n* @description: Out of stack\n* @param {sqstack*} s-Stack top pointer\n* @return {Stack top value}\n*/\ndata_t stack_pop(sqstack *s) {\ns->top--;\nreturn (s->data[s->top+1]);\n}\n```\n\n### 5. Sequential stack deletion\n\n```/**\n* @description: Sequential stack deletion\n* @param {sqstack*} s-Stack top pointer\n* @return {-1-Function failed, 0-function succeeded}\n*/\nint stack_free(sqstack *s) {\nif (s == NULL) {\nprintf(\"s is NULL\\n\");\nreturn -1;\n}\n\nif (s->data != NULL)\nfree(s->data);\nfree(s);\n\nreturn 0;\n}\n```\n\n### 6. Empty stack and whether to empty stack\n\n```/**\n* @description: Stack emptying\n* @param {sqstack*} s-Stack top pointer\n* @return {-1-Function failed, 0-function succeeded}\n*/\nint stack_clear(sqstack *s) {\nif (s == NULL) {\nprintf(\"s is NULL\\n\");\nreturn -1;\n}\n\ns->top = -1;\nreturn 0;\n}\n\n/**\n* @description: Determine whether the stack is empty\n* @param {sqstack*} s-Stack top pointer\n* @return {-1-Function failed, 1-stack is empty, 0-stack is not empty}\n*/\nint stack_empty(sqstack *s)\n{\nif (s == NULL) {\nprintf(\"s is NULL\\n\");\nreturn -1;\n}\nreturn (s->top == -1 ? 1 : 0);\n}\n```\n\n# 3, Implementation of linked list stack\n\n[data structure and algorithm] program internal skill Part III - single linked list\n\n### 1. Single linked list implementation of stack\n\nThe insertion and deletion operations are carried out at the head of the linked list. The tail of the linked list is the bottom of the stack, and the pointer at the top of the stack is the head pointer.", null, "Node definition:\n\n```typedef int data_t ;\t /*Define the data types of data elements in the stack*/\ntypedef struct node {\ndata_t data ;\t\t /*Data domain*/\nstruct node *next ; /*Link pointer field*/\n}stacklist,*stacklink; \t\t /*Chain stack type definition*/\n```\n\n### 2. Create empty stack\n\n```/**\n* @description: Creation of single linked list stack\n* @param {*}\n* @return {Stack top pointer}\n*/\n{\nif((top = (stacklink)malloc(sizeof(stacklist))) == NULL)\n{\n#if DEBUG\nprintf(\"stacklist create error!\\n\");\n#endif\nreturn 0;\n}\n\ntop->data = 0;\ntop->next = NULL;\n\n}\n```\n\n### 3. Stack\n\n```/**\n* @description: Stack - header insertion\n* @param {stacklink} top-Stack top pointer\n* @param {data_t} value-Stack value\n* @return {0-Function succeeded, 1-function failed}\n*/\nint stacklist_top_insert(stacklink top, data_t value)\n{\nif(top == NULL)\n{\n#if DEBUG\nprintf(\"top is NULL!\\n\");\n#endif\nreturn 0;\n}\n\nif((sl = (stacklink)malloc(sizeof(stacklist))) == NULL)\n{\n#if DEBUG\nprintf(\"stacklist create error!\\n\");\n#endif\nreturn 0;\n}\n\nsl->data = value;\nsl->next = top->next;\ntop->next = sl;\n\nreturn 1;\n}\n```\n\n### 4. Out of stack\n\n```/**\n* @description: Out of stack\n* @param {stacklink} top-Stack top pointer\n* @return {0-Function succeeded, 1-function failed}\n*/\n{\nint value;\nif(top == NULL)\n{\n#if DEBUG\nprintf(\"top is NULL!\\n\");\n#endif\nreturn 0;\n}\nif(top->next == NULL)\n{\n#if DEBUG\nprintf(\"stacklist is empty!\\n\");\n#endif\nreturn 0;\n}\n\nsl = top->next;\ntop->next = sl->next;\nvalue = sl->data;\n\nfree(sl);\n\nreturn value;\n}\n```\n\n### 5. Delete linked list stack\n\n```/**\n* @description: Stack deletion\n* @param {stacklink} top-Stack top pointer\n* @return {0-Function succeeded, 1-function failed}\n*/\n{\nstacklink sl = top;\nif(top == NULL)\n{\n#if DEBUG\nprintf(\"top is NULL!\\n\");\n#endif\nreturn 0;\n}\n\nwhile(top)\n{\ntop = top->next;\nfree(sl);\nsl = top;\n}\n\nreturn 1;\n}\n```\n\n### 6. Judge whether it is an empty stack\n\n```/**\n* @description: Judge whether it is an empty table\n* @param {stacklink} top-Stack top pointer\n* @return {-1-Function failed, 1 - empty stack, 0 - not empty stack}\n*/\n{\nif(top == NULL)\n{\n#if DEBUG\nprintf(\"top is NULL!\\n\");\n#endif\nreturn -1;\n}\nreturn (top->next == NULL? 1:0);\n}\n```\n\n# 4, Application of stack\n\nCreate operand stack and operator stack. Operators have priority.\n\n① Scan the expression from left to right. When encountering an operand, it will enter the operand stack.\n② When an operator is encountered, if its priority is higher than that of the element at the top of the operator stack, it will be put on the stack. On the contrary, take out the two consecutive operands at the top of the stack, store the results in the operand stack, and then continue to compare the priority of the operator with the top of the stack operator.\n③ The left bracket will be put into the operator stack and the right bracket will not be put into the operator stack. Take out the operator at the top of the operator stack and the two operands at the top of the operand stack for operation, and press the result into the operand stack until the left bracket is taken out.\n\nFor example: calculation (4 + 8) × 2－3 ;\n\n```Operand stack: 4 8 \t| 12 2\t\t|24 3\t\t|21\nOperator stack:( ＋ \t|×\t\t |－\t\t |\n```", null, "To basically implement the program, click the following link for free:\nBasic C program of stack\n\nThat's it!", null, "Keywords: C Algorithm data structure linked list\n\nAdded by MNS on Thu, 10 Mar 2022 15:07:56 +0200" ]

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[ "# Absurd: An Introduction to C\n\n## Part XI, 20150610\n\nSome time has passed since I last did anything with my C tutorial. This is largely due to work and having several different projects going at any one time (learning Chinese, developing a hobby OS, a gaming community, etc...). However, I felt that I ought to cover pointers in a bit more detail. Quite a few people have asked me why one would ever use a pointer, so here's my attempt to explain that.\n\nDEFINITION\n\nA pointer is a variable that contains the address of some other variable.\n\nEXAMPLE\n\nLet's create a variable and print it:\n\n``````int foo = 0; printf(\"%d\",foo);\n``````\n\nNot too complicated at all really. Yet, we can refer to this variable using the address of the variable in memory. This will simply be another integer variable.\n\n``````int *bar;\n``````\n\nTo set this to the address of foo, we would do something like:\n\n``````bar = &foo;\n``````\n\nThe & infront of a variable means \"the address of\" and any time you see an asterisk in front of a variable is means \"the stuff at the address of\".\n\nSo, if want to copy foo using bar we can do:\n\n``````int foo2 = *bar;\n``````\n\nIf we want to copy the pointer we can do:\n\n``````int bar2 = bar;\n``````\n\nWe can also create a pointer to a pointer:\n\n``````int *barbar = &bar;\n``````\n\nSee, that's just it. Showing someone the above tells him/her none of what he/she really needs to know to make effective use of pointers.\n\nARRAYS\n\nC doesn't actually have array variables. C does have syntax for dealing with arrays, but that syntax is actually just a special pointer syntax.\n\n``````int randomArray;\nrandomArray = 25;\n``````\n\nSo, above we create a 10 unit array (0-9). We then assign the number 25 to the fourth array segment. This is really just a pointer assignment, as that line is identical in function to:\n\n``````*(randomArray+3) = 25;\n``````\n\nThat is, it's assigning 25 to the location three places right of randomarry which is the pointer.\n\nPerhaps more importantly for arrays, pointers allow us to create arrays of arbitrary lengths:\n\n``````int *randomArray;\nint raLength=20;\nrandomArray=(int*)malloc(raLength*sizeof(int));\n``````\n\nThis will create an integer array the length of \"raLength\". Note that because this memory was manually allocated with malloc(), we will need to release it with free(randomArray) when the program has completed its usage of this memory space.\n\nWe can use the array feature to create strings as well. These are just character arrays that are terminated with a null character.\n\n``````char *randomString;\nrandomString=\"Random Text\";\n``````\n\nOr, treating this like an array of characters:\n\n``````char randomString={'R','a','n','d','o','m',' ','T','e','x','t','\\0'};\n``````\n\nSo, Chris McClelland pointed out on G+ that it's kind of dishonest of me to imply that these are the same. They both do create the requested string, however:\n\n``````char *randomString = \"Random Text\";\n``````\n\nThis creates a pointer to the string literal.\n\n``````char randomString[] = \"Random Text\";\n``````\n\nAllocates space on stack (as opposed to heap which the malloc example below will do). So, as noted the first should be char *const randomString. The const means that randomString is not to be modified, yet what it points to can be; the pointer is constant- the data to which it points is not necessarily.\n\n``````const char *randomString;\n``````\n\nThis means that randomString is now a normal pointer, but the thing at which it points must not be modified.\n\nBack to arrays and such. We could do this manually:\n\n``````char *randomString;\nrandomString=(char*)malloc(12*sizeof(char));\nrandomString='R';\nrandomString='a';\nrandomString='n';\nrandomString='d';\nrandomString='o';\nrandomString='m';\nrandomString=' ';\nrandomString='T';\nrandomString='e';\nrandomString='x';\nrandomString='t';\nrandomString='\\0';\n``````\n\nThere are some down sides to working with strings in this manner. A string cannot contain a null value, as this would signify the end of the string. You're not going to have a good time with UTF8 or UTF16. The concatenation or copying of strings will require the writing of routines that work with each character individually. Remember the length! If you do not terminate the string you can get buffer overruns!\n\nWRITABLE FUNCTION PARAMETERS\n\nWhen a method/function/subroutine is called in C, a copy of its parameters are passed to the function. That is, the variable passed will be local in scope by default.\n\nAs one might guess, if you pass a pointer then the pointer gets copied. But, that's fine because you can still access the value at the address:\n\n``````int randomInt=1;\nrandomSubroutine(&randomInt);\nvoid randomSubroutine(int *randomParameter) {\n*randomParameter=*randomParameter+1;\n}\n``````\n\nThat will increase the value of randomInt by 1.\n\nSTRUCTS\n\nUp to this point, I haven't really introduced structures before. Structures (structs) are C's really crude implementation of objects. A struct is a collection of variables. So, you could create a struct foobar that consists of an array and an integer.\n\n``````struct randomDef {\nchar *randomString;\nint randomInt;\n}; // this defines the type\nstruct randomDef randomStruct; // this instantiates the definition as randomStruct\n``````\n\nThis will be familiar to people who use object oriented C-like languages. We access the array and integer as follows:\n\n``````randomStruct.randomString = \"Random Text\";\nrandomStruct.randomInt = 123;\n``````\n\nAs one might expect, there is a special syntax for this. It may be familiar to people who have used objects in other languages.\n\n``````struct randomDef *randomPointer;\nrandomPointer=&randomStruct;\n``````\n\nYou can now refer to the piece of the structure as follows:\n\n``````randomPointer->randomString=\"Random Text\";\nrandomPointer->randomInt=\"123\";\n``````\n\nThe above is functionally identical to:\n\n``````(*randomPointer).randomString=\"Random Text\";\n(*randomPointer).randomInt=\"123\";\n``````\n\nFor this example, the pointer to the struct isn't all that useful. However, if the structure were very large the use of the pointer would prevent consuming more memory just to pass the structure to a subroutine. The use of the pointer also cleans up the syntax a little bit should the structure itself contain pointers.\n\nCONCLUSION\n\nI realize that this was all very hasty. Obviously, this may require re-reading, but I hope that you now have a better idea about what pointers are, when they are used without your having to declare them, and when you should explicitly use them. In general, pointers are used to provide features in C that were not necessarily written into C. Why were the features not written in? C commands are directly translated to ASM for a given platform. We can assume that part of the reason was programmer time required for the creation of C. Another reason is to allow more flexibility to the programmer. This is evident when you look at languages implemented in C, and just how different those languages can be (Python, PHP, Ruby).\n\n⇠ back" ]

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[ "# Two way scatter plot stata\n\nTwo way scatter plot stata keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website\n\n## Keyword Suggestions\n\nTwo way tables\n\nTwo way street\n\nTwo way traffic\n\nTwo way mirror\n\nTwo way audio\n\nTwo way effect\n\nTwo way window\n\nTwo way audio cable\n\nTwo way tables worksheet\n\nTwo way anova matlab\n\nTwo ways home movie\n\nTwo way contract nba\n\nTwo way roadway with center lane\n\nTwo way anova in r\n\n( Please select at least 2 keywords )\n\n#### Most Searched Keywords\n\n## Websites Listing\n\nWe found at least 10 Websites Listing below when search with two way scatter plot stata on Search Engine\n\n### Graphics:Twoway Scatterplots Stata Learning Modules\n\nStats.idre.ucla.edu  DA: 19 PA: 37 MOZ Rank: 56\n\n• 28 rows · Graphics:Twoway Scatterplots | Stata Learning Modules This module shows some of the options when using the twoway command to produce scatterplots. This is illustrated by showing the command and the resulting graph\n• This includes hotlinks to the Stata Graphics Manual available over the web and from within Stata by typing help graph.\n\n### Graphics: Combining Twoway Scatterplots Stata Learning\n\nStats.idre.ucla.edu  DA: 19 PA: 45 MOZ Rank: 65\n\n• This module shows examples of combining twoway scatterplots\n• This is illustrated by showing the command and the resulting graph\n• This includes hotlinks to the Stata Graphics Manual available over the web and from within Stata by typing help graph\n• The data set used in these examples can be obtained using the following command:\n\n### Scatter Plots in Stata using the \"Twoway\" option\n\nYoutube.com  DA: 15 PA: 6 MOZ Rank: 23\n\nTutorial showing how to create scatter plots relating two variables across multiple sub-samples in Stata.Link to tutorial on Time-series plots in Statahttps:\n\n### Graph twoway lfit — Twoway linear prediction plots\n\nStata.com  DA: 13 PA: 33 MOZ Rank: 49\n\n• Graph twoway lfit— Twoway linear prediction plots 3 Results are visually the same as typing\n• scatter mpg weight || line fitted weight Cautions Do not use twoway lfit when specifying the axis scale options yscale(log) or xscale(log) to create log scales\n• scatter mpg weight, xscale(log) || lfit mpg\n\n### Stata for Students: Scatterplots\n\nSsc.wisc.edu  DA: 16 PA: 30 MOZ Rank: 50\n\n• A scatterplot is an excellent tool for examining the relationship between two quantitative variables\n• One variable is designated as the Y variable and one as the X variable, and a point is placed on the graph for each observation at the location corresponding to its values of those variables.\n\n### How to Create and Modify Scatterplots in Stata\n\nStatology.org  DA: 17 PA: 20 MOZ Rank: 42\n\n• You can create a scatterplot with more than two variables by simply typing more variables after the scatter command\n• Note that the last variable you type will be used for the x-axis\n• For example, the following command tells Stata to create a scatterplot using length as the x-axis variable and weight and displacement as the y-axis variables:\n\n### Stata Guide: Twoway (Bivariate) Charts\n\nWlm.userweb.mwn.de  DA: 18 PA: 19 MOZ Rank: 43\n\n• Organisation of data A line plot is a twoway graph, which implies that you need two variables\n• The one that (typically) represents \"time\" will be represented on the x axis, the other on the y axis\n• (In fact, a line plot is basically the same as a scatterplot, the only difference being the line that connects the data points.)\n\n### Stata twoway graph of means with confidence intervals\n\nStackoverflow.com  DA: 17 PA: 50 MOZ Rank: 74\n\n• I want to scatter plot mean score by group for each test (same graph) with confidence intervals (the real data has thousands of observations)\n• The resulting graph would have two sets of two dots\n• One set of dots for test==a (group==0 vs group==1) and one set of dots for test==b (group==0 vs group==1)\n• My current approach works but it is laborious.\n\n### Twoway scatter- different colors for\n\nStatalist.org  DA: 17 PA: 50 MOZ Rank: 75\n\n• I can easily make a scatter plot of this by using the, i.e\n• Twoway (scatter bloodresult age)\" However out half of the patients died and the other half survived, this is defined in a another variable called outcome\n• Now I would like to apply the a different colour to each dot on my scatterplot according to outcome, i.e\n\n### Graphs and Tables: Time to get visual (Part I)\n\nEconometricstutorial.com  DA: 24 PA: 47 MOZ Rank: 80\n\n• A two way scatter plot can be used to show the relationship between variables\n• Let’s open the example dataset census.dta and study the relationship among the urban population and the number of marriages accounting for every different regions.\n\n### Code to make a dot and 95% confidence interval figure in Stata\n\nBlog.uvm.edu  DA: 12 PA: 50 MOZ Rank: 72\n\n• In the same folder as the Excel file, copy/paste/save the code below as a .do file\n• Close Excel and close Stata then find the .do file from Windows Explorer and double click it\n• Doing this will force Stata to set the working directory …\n\n### Plotting regression coefficients and other estimates in Stata\n\nRepec.sowi.unibe.ch  DA: 19 PA: 33 MOZ Rank: 63\n\n• In Stata such plots can be produced by the marginsplot command ([R] marginsplot)\n• However, while marginsplot is very versatile and flexible, it has two major limitations: it can only process results left behind by margins ([R] margins bivariate distributions (e.g\n\n### Including Calculated Results In Stata Graphs\n\nSsc.wisc.edu  DA: 12 PA: 19 MOZ Rank: 43\n\n• Then create a scatter plot of mpg vs\n• Weight with a linear fit line overlaid on it, with confidence intervals: sysuse auto twoway (lfitci mpg weight) (scatter mpg weight) Be sure to put the linear fit first so the points from the scatter plot can overlay the shaded 95% confidence interval.\n\n### Q18a (6 Points). Write The Stata Command That Woul\n\nChegg.com  DA: 13 PA: 50 MOZ Rank: 76\n\nWrite the stata command that would let you create a two-way scatter plot with a fitted regression line as well as marker labels for countries (country) for the variables womenpar' (on the y-axis) and 'educatt_tertiary' (on the x-axis) but only including data for 2016 > [Answer here) 018b (2 points).\n\n### Learn to Test for Heteroscedasticity in Stata With Data\n\nMethods.sagepub.com  DA: 19 PA: 50 MOZ Rank: 83\n\n• Figure 2: Producing a Two-Way Scatterplot of Residuals and Predicted Values for a Regression Model in the Residual-Versus-Fitted Plot Dialog Box in Stata\n• To add a line at y = 0, select the “ Y axis” tab at the top of the dialog box and click on “Reference lines” as shown in Figure 3 .\n\n### UCD GEARY INSTITUTE FOR PUBLIC POLICY DISCUSSION …\n\nUcd.ie  DA: 10 PA: 50 MOZ Rank: 75\n\n• A basic knowledge of Stata is required\n• There are two ways to create graphs in Stata\n• You can either (a) use a written command which can be done interactively in the command line or written in a do-file or (b) you can use the dialogue boxes/pull-down menus at the top.\n\n### [STATA] Twoway scatter graph help Statistics Help @ Talk\n\nTalkstats.com  DA: 17 PA: 47 MOZ Rank: 80\n\n• I want to realize a twoway scatter plot in STATA, plotting id3 over the price, but I would like the y axis to be ordered from the lowest minimum price to the highest minimum price - now is ordered in alphabetic order of the ylabel\n• Seems an easy question, but I am really stuck For a better understanding, please see the attached graph.\n\n### Interpret all statistics and graphs for Two-way ANOVA\n\nSupport.minitab.com  DA: 19 PA: 50 MOZ Rank: 86\n\n• You can use a fitted line plot to graphically illustrate different R 2 values\n• The first plot illustrates a simple regression model that explains 85.5% of the variation in the response\n• The second plot illustrates a model that explains 22.6% of the variation in the response.\n\n### Week 1 : SCATTER AND LINE PLOTS\n\nYoutube.com  DA: 15 PA: 6 MOZ Rank: 39\n\n### Stata Graphics Data Science Workshops\n\nIqss.github.io  DA: 14 PA: 33 MOZ Rank: 66\n\ntwoway is basic Stata command for all two-way graphs; Use twoway anytime you want to make comparisons among variables; Can be used to combine graphs (i.e., overlay one graph with another) e.g., insert line of best fit over a scatter plot; Some basic examples:\n\n### How to Create and Modify Box Plots in Stata\n\nStatology.org  DA: 17 PA: 17 MOZ Rank: 54\n\n• A box plot is a type of plot that we can use to visualize the five number summary of a dataset, which includes:\n• The minimum; The first quartile; The median; The third quartile; The maximum This tutorial explains how to create and modify box plots in Stata\n\n### (PDF) Quantitative Data Management, Statistical Analysis\n\nAcademia.edu  DA: 16 PA: 50 MOZ Rank: 87\n\n• Stata is an interactive data management program and is used for the statistical analysis by statisticians, epidemiologists, economists, demographers and by researchers from most disciplines\n• two-way Regression Coefficients scatter, two-way line, two-way Hypothesis testing connected Problems with regression Using menu vs line commands\n\n### How to predict and forecast using ARIMA in STATA\n\nProjectguru.in  DA: 18 PA: 42 MOZ Rank: 82\n\n• two-way’ is for two-way graphs in STATAScatter’ means scatter plot\n• After running the command, the below graph will appear\n• It shows green dots as actual GDP values, the shaded region as a confidence interval, and straight-line as fitted values\n\n### Understanding 2-way Interactions University of Virginia\n\nData.library.virginia.edu  DA: 25 PA: 34 MOZ Rank: 82\n\n• Interactions can get yet more complicated\n• Two continuous variables can interact\n• You can have multiple two-way interactions\n• Even though software makes it easy to fit lots of interactions, Kutner, et al\n• (2005) suggest keeping two things in mind when fitting models with interactions: 1.\n\n### TWO-WAY TABLES & SCATTERPLOTS Graphs Quiz\n\nQuizizz.com  DA: 11 PA: 50 MOZ Rank: 85\n\n• The scatter plot shows the relationship between the number of chapters and the total number of pages for several books\n• Use the trend line to predict how many chapters would be in a …\n\n### How can i add label names per say states/countries names\n\nResearchgate.net  DA: 20 PA: 50 MOZ Rank: 95\n\n• How can i add label names per say states/countries names in scatter plot of two variables in Excel\n• I was using stata 9 which has xtbond2 command but now I am using" ]

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[ "# Category: Others\n\n## Find distinct prime factors from 1 to n in O((n+q)logn) (for q Queries)\n\nProblem: Find distinct prime factors of a number. You will be asked q times. Naive approach: Find out distinct prime factors for each query. It’s inefficient for large number of queries. As each query...\n\n## Conversion of BST(binary search tree) pre-order traversal to post-order traversal\n\nProblem Statement: Post-order traversal to be printed from Pre-order traversal of a Binary Search Tree(BST). Prerequisite: BST & traversal A pre-order traversal of a BST is given. In BST, value/data of all the children...\n\n## How you can add Mathematical formula in question – answer or in forum\n\nrendering mathematical formula using latex | syntax | latex code examples\n\n## QA : Question answer raaabnits\n\nask and answer | learn and let learn- you are also part of this online community. you can ask your queries here and the guy interested in your question or some experts from the..." ]

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[ "## Friday, June 19, 2020\n\n### Announcement: This Saturday at The Science Circle - \"The Equivalence Theorem\"\n\n\"The Equivalence Theorem\"\n\nSaturday June 20\n\n10AM to 11 AM\n\nThe EQUIVALENCE THEOREM (a.k.a. Bernstein-Schröder Theorem)\nModern set theory originated between 1874 – 1884 from the researches of Georg Cantor at Universität Halle, Germany. Around 1895, Cantor presented the EQUIVALENCE CONJECTURE:\n\nGiven sets A and B,\nif A is equivalent to a subset of B\nand B equivalent to a subset of A,\nthen A is equivalent to B.\n\nWhile this may seem obvious to many, it was not until 1898 that Cantor’s student, Felix Bernstein proved the theorem without flaw.\n\nIn this presentation we will review relevant concepts of set theory: types of mappings of sets and their behaviour when sets are combined by union, intersection and difference (relative complement); sizes of sets and equivalence; partitions of sets; special attention will be given to properties of chains of sets (nested subsets).\n\nThen we will demonstrate two approaches for proof of the Equivalence Theorem (the Cantor-Bernstein-Schröder Theorem): 1) à la Bernstein: 2) a stream-lined proof for contrast based on defining a bijective function (1-to-1 from A onto B).\n\nInteresting examples of applications of this theorem will be given that give insight to distinct “DEGREES OF INFINITY”.\n\nRelevant background for this is basic set theory, high school-level mathematics and a normal, human logic.\n\nThe goal of this is to help members of the audience understand the strategy of attack in the subtle proof of this important mathematical theorem whose statement is so deceptively simple. Secondarily, this instruction illuminates aspects of how one learns mathematics.\n\nRobert A Hendrix\n\nhttp://maps.secondlife.com/secondlife/The%20Science%20Circle/67/129/32" ]

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[ "Scientific calculators that have formula/equation memory\n12-14-2017, 04:23 PM\nPost: #27", null, "grsbanks", null, "Senior Member Posts: 904 Joined: Jan 2017\nRE: Scientific calculators that have formula/equation memory\n(12-14-2017 04:07 PM)pier4r Wrote:  interesting! Does the 35S, dm42 do the same? And others?\n\nThe 35S will lose its contents if you remove both batteries at the same time. They are connected in parallel (with a pair of diodes to ensure one battery doesn't feed into the other), so you remove and replace one, then the other, and your data stays intact.\n\nThe DM42 only has one battery. What you can do is save the machine state to flash memory then replace the battery. When you switch it on and if the battery was out for long enough for the machine to \"forget\" its contents, you can re-load it from the state file saved previously. So, the answer to your question for the DM42 is \"yes, kind of\". You have to save the machine state to flash, it won't do it automatically. The reason for the manual nature of this step is because you can save multiple state files that each include different \"virtual machines\" for different applications, eg. a state file with \"financial\" programs and data, a different \"engineering\" state file, a \"programming\" state file etc... You have to give the state file a name that reflects its purpose.\n « Next Oldest | Next Newest »\n\n Messages In This Thread Scientific calculators that have formula/equation memory - pier4r - 12-06-2017, 12:18 PM RE: Scientific calculators that have formula/equation memory - Thomas Okken - 12-06-2017, 12:24 PM RE: Scientific calculators that have formula/equation memory - pier4r - 12-06-2017, 12:29 PM RE: Scientific calculators that have formula/equation memory - grsbanks - 12-06-2017, 12:44 PM RE: Scientific calculators that have formula/equation memory - pier4r - 12-06-2017, 01:16 PM RE: Scientific calculators that have formula/equation memory - jebem - 12-06-2017, 01:00 PM RE: Scientific calculators that have formula/equation memory - Gene - 12-06-2017, 01:30 PM RE: Scientific calculators that have formula/equation memory - grsbanks - 12-06-2017, 01:33 PM RE: Scientific calculators that have formula/equation memory - jebem - 12-06-2017, 03:50 PM RE: Scientific calculators that have formula/equation memory - Dave Britten - 12-06-2017, 04:19 PM RE: Scientific calculators that have formula/equation memory - pier4r - 12-06-2017, 09:44 PM RE: Scientific calculators that have formula/equation memory - pier4r - 12-06-2017, 10:22 PM RE: Scientific calculators that have formula/equation memory - jebem - 12-14-2017, 04:04 PM RE: Scientific calculators that have formula/equation memory - Lonewolf - 06-06-2018, 11:53 PM RE: Scientific calculators that have formula/equation memory - ijabbott - 06-07-2018, 01:57 PM RE: Scientific calculators that have formula/equation memory - Dave Britten - 06-07-2018, 02:18 PM RE: Scientific calculators that have formula/equation memory - Lonewolf - 06-07-2018, 07:56 PM RE: Scientific calculators that have formula/equation memory - Dave Britten - 06-07-2018, 08:16 PM RE: Scientific calculators that have formula/equation memory - pier4r - 06-09-2018, 01:38 PM RE: Scientific calculators that have formula/equation memory - Lonewolf - 06-09-2018, 06:16 PM RE: Scientific calculators that have formula/equation memory - pier4r - 12-06-2017, 10:43 PM RE: Scientific calculators that have formula/equation memory - Massimo Gnerucci - 12-07-2017, 12:20 PM RE: Scientific calculators that have formula/equation memory - grsbanks - 12-07-2017, 12:29 PM RE: Scientific calculators that have formula/equation memory - pier4r - 12-07-2017, 12:59 PM RE: Scientific calculators that have formula/equation memory - grsbanks - 12-07-2017, 02:00 PM RE: Scientific calculators that have formula/equation memory - Massimo Gnerucci - 12-07-2017, 08:36 PM RE: Scientific calculators that have formula/equation memory - Chasfield - 12-07-2017, 02:01 PM RE: Scientific calculators that have formula/equation memory - jebem - 12-07-2017, 06:43 PM RE: Scientific calculators that have formula/equation memory - pier4r - 12-07-2017, 07:38 PM RE: Scientific calculators that have formula/equation memory - Csaba Tizedes - 12-09-2017, 11:47 AM RE: Scientific calculators that have formula/equation memory - pier4r - 12-09-2017, 12:06 PM RE: Scientific calculators that have formula/equation memory - Csaba Tizedes - 12-09-2017, 12:29 PM RE: Scientific calculators that have formula/equation memory - pier4r - 12-14-2017, 04:07 PM RE: Scientific calculators that have formula/equation memory - grsbanks - 12-14-2017 04:23 PM RE: Scientific calculators that have formula/equation memory - Massimo Gnerucci - 12-14-2017, 06:15 PM RE: Scientific calculators that have formula/equation memory - Guenter Schink - 12-14-2017, 10:10 PM RE: Scientific calculators that have formula/equation memory - Massimo Gnerucci - 12-14-2017, 11:01 PM RE: Scientific calculators that have formula/equation memory - klesl - 12-25-2017, 03:22 PM RE: Scientific calculators that have formula/equation memory - pier4r - 12-25-2017, 03:50 PM RE: Scientific calculators that have formula/equation memory - TheKaneB - 01-04-2018, 02:22 AM RE: Scientific calculators that have formula/equation memory - badaze - 01-04-2018, 09:49 PM RE: Scientific calculators that have formula/equation memory - grsbanks - 01-04-2018, 10:10 PM RE: Scientific calculators that have formula/equation memory - badaze - 01-04-2018, 10:24 PM RE: Scientific calculators that have formula/equation memory - pier4r - 01-05-2018, 10:11 AM RE: Scientific calculators that have formula/equation memory - martinot - 02-10-2018, 09:51 AM RE: Scientific calculators that have formula/equation memory - JeffJ - 02-16-2018, 06:13 PM RE: Scientific calculators that have formula/equation memory - martinot - 02-16-2018, 09:09 PM RE: Scientific calculators that have formula/equation memory - brickviking - 06-07-2018, 02:18 AM RE: Scientific calculators that have formula/equation memory - Nigel (UK) - 06-07-2018, 04:55 PM\n\nUser(s) browsing this thread: 1 Guest(s)" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: List Partition\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg53941] Re: [mg53895] List Partition\n• From: John Kiehl <john.kiehl at soundtrackny.com>\n• Date: Fri, 4 Feb 2005 04:12:26 -0500 (EST)\n• Reply-to: John Kiehl <john.kiehl at soundtrackny.com>\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```\nOn Wednesday, February 2, 2005 6:25 AM, zak <chocolatez at gmail.com> wrote:\n>what could i do to partition the list:\n>a={cat,E,in,the,hat,E,okay,fine,E}\n>into\n>a={{cat},{{in},{the},{hat}},{{okay},{fine}}}\n>\n>ie: every word in a sublist , and E determine the end of a\n>sentence\n>(a bigger list).\n>zak\n>\n>\n\nIn:= a={cat,E,in,the,hat,E,okay,fine,E}\nOut= {cat,E,in,the,hat,E,okay,fine,E}\n\nThis solution is an automation of this basic idea.\n\nIn:= Part[a,{1}]\nOut= {cat}\n\nIn:= Part[a,{3,4,5}]\nOut= {in,the,hat}\n\nIn:= Part[a,{7,8}]\nOut= {okay,fine}\n\nThe \"Position\" of E in the original list will be used to calculate the start and end of each sub-list.\n\n(*-------------------------------------------------*)\n(* so, first put brackets around each element as requested *)\nIn= aa=Map[List,a]\nOut= {{cat},{E},{in},{the},{hat},{E},{okay},{fine},{E}}\n\n(* in general, we want to find positions of {E} in list *)\nIn= pos = Position[aa,{E}]//Flatten\nOut= {2,6,9}\n\n(* one less than pos will be end of each sublist *)\nIn:= last=pos-1\nOut:= {1,5,8}\n\n(* one more than pos will be start of each sublist,\nalso drop last value and prepend a 1 *)\nIn:= first=Drop[Flatten[{1,pos}],-1]\nOut= {1,3,7}\n\n(* pair these up *)\nIn:= t=Transpose[{first,last}]\nOut= {{1,1},{3,5},{7,8}}\n\n(* map Part[...] over our list t *)\nIn:= Part[aa,Range[Sequence @@ #]]& /@ t\nOut= {{{cat}},{{in},{the},{hat}},{{okay},{fine}}}\n\n```\n\n• Prev by Date: Re: Problem with a sum\n• Next by Date: Re: Matrix equations\n• Previous by thread: Re: List Partition\n• Next by thread: Re: List Partition" ]

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[ "Pagina principale Mechanics of materials: an introduction to the mechanics of elastic and plastic deformation of solids..\n\n# Mechanics of materials: an introduction to the mechanics of elastic and plastic deformation of solids and structural materials\n\nOne of the most important subjects for any student of engineering to master is the behaviour of materials and structures under load. The way in which they react to applied forces, the deflections resulting and the stresses and strains set up in the bodies concerned are all vital considerations when designing a mechanical component such that it will not fail under predicted load during its service lifetime. All the essential elements of a treatment of these topics are contained within this course of study, starting with an introduction to the concepts of stress and strain, shear force and bending moments and moving on to the examination of bending, shear and torsion in elements such as beams, cylinders, shells and springs. A simple treatment of complex stress and complex strain leads to a study of the theories of elastic failure and an introduction to the experimental methods of stress and strain analysis. More advanced topics are dealt with in a companion volume - Mechanics of Materials 2. Each chapter contains a summary of the essential formulae which are developed in the chapter, and a large number of worked examples which progress in level of difficulty as the principles are enlarged upon. In addition, each chapter concludes with an extensive selection of problems for solution by the student, mostly examination questions from professional and academic bodies, which are graded according to difficulty and furnished with answers at the end. * Emphasis on practical learning and applications, rather than theory * Provides the essential formulae for each individual chapter * Contains numerous worked examples and problems\nCategories: Physics\nVolume: Volume 1\nAnno: 1997\nEdizione: 3rd ed\nEditore: Butterworth-Heinemann\nLingua: english\nPagine: 1038\nISBN 10: 0750632666\nISBN 13: 9780080523996\nFile: PDF, 38.89 MB\n\n## Most frequently terms\n\nPost a Review", null, "You can write a book review and share your experiences. Other readers will always be interested in your opinion of the books you've read. Whether you've loved the book or not, if you give your honest and detailed thoughts then people will find new books that are right for them.\n1\n\n### Mechanics of materials: an introduction to the mechanics of elastic and plastic deformation of solids and structural materials\n\nAnno: 1997\nLingua: english\nFile: PDF, 23.88 MB\n2\n\n### Handbook of Plastics, Elastomers and Composites\n\nAnno: 2002\nLingua: english\nFile: PDF, 10.03 MB\n```MECHANICS OF\nMATERIALS I\nAn Introduction to the Mechanics of Elastic and\nPlastic Deformation of Solids and Structural Materials\n\nTHIRD EDITION\n\nE. J. HEARN\nPh.D., B.Sc. (Eng.) Hons., C.Eng., F.I.Mech.E., F.I.Prod.E.,F.1.Diag.E.\n\nUniversity of Warwick\nUnited Kingdom\n\nE I N E M A N N\n\nOXFORD AUCKLAND BOSTON JOHANNESBURG MELBOURNE NEW DELHI\n\nButterworth-Heinemann\nLinacre House, Jordan Hill, Oxford OX2 8DP\n225 Wildwood Avenue, Woburn, MA 01801-2041\nA division of Reed Educational and Professional Publishing Ltd\n-@A\n\nmember of the Reed Elsevier plc group\n\nFirst published 1977\nReprinted with corrections 1980, 1981, 1982\nSecond edition 1985\nReprinted with corrections 1988\nReprinted 1989, 1991, 1993, 1995, 1996\nThird edition 1997\nReprinted 1998, 1999,2000\n\n0 E. J. Hearn 1977, 1985, 1997\nmay be reproduced in any material form (including\nphotocopying or storing in any medium by electronic\nmeans and whether or not transiently or incidentally to\nsome other use of this publication) without the written\npermission of the copyright holder except in accordance\nwith the provisions of the Copyright, Designs and\nPatents Act 1988 or under the terms of a licence issued\nby the Copyright Licensing Agency Ltd, 90 Tottenham\nCourt Road, London, England WIP 9HE.\nApplications for the copyright holder's written\npermission to reproduce any part of this publication\nto the publishers\nBritish Library Cataloguing in Publication Data\nA catalogue record for this book is available from the British Library\nISBN 0 7506 3265 8\nLibrary of Congress Cataloguing in Publication Data\nHearn, E. J. (Edwin John)\nMechanics of materials 1: an introduction to the mecahnics of elastic and plastic deformation of\nsolids and structural components/E. J. Hearn. - 3rd ed.\np. cm.\nIncludes bibliographical references and index.\nISBN 0 7506 3265 8\n1. Strength of materials. I. Title\nTA405.H3\n96-49967\n620.1'23-dc21\nCIP\n\nPrinted and bound in Great Britain by Scotprint, Musselburgh\n\nFOR EVERY TITLGTHAT WE PUBLISH.BU176RWORTHdlEINEMANN\nW U PAY FUR BTCV TO PLANT AND CARE FOR A TREE.\n\nAlso of interest\n\nASHBY\nMaterials Selection in Mechanical Design\nASHBY & JONES\nEngineering Materials 1\nEngineering Materials 2\nCAMPBELL\nCastings\nCHARLES, CRANE & FURNESS\nSelection and Use of Engineering Materials\nCRAWFORD\nPlastics Engineering\nHEARN\nMechanics of Materials 2\nHULL & BACON\nIntroduction to Dislocations, 3rd Edition\nJONES\nEngineering Materials 3\nLLEWELLYN\nSteels: Metallurgy & Applications\nSMALLMAN & BISHOP\nMetals and Materials\n\nINTRODUCTION\nThis text is the suitably revised and extended third edition of the highly successful text\ninitially published in 1977 and intended to cover the material normally contained in degree\nand honours degree courses in mechanics of materials and in courses leading to exemption\nfrom the academic requirements of the Engineering Council. It should also serve as a valuable\nreference medium for industry and for post-graduate courses. Published in two volumes, the\ntext should also prove valuable for students studying mechanical science, stress analysis, solid\nmechanics or similar modules on Higher Certificate and Higher Diploma courses in the UK\nor overseas and for appropriate NVQ* programmes.\nThe study of mechanics of materials is the study of the behaviour of solid bodies under\nload. The way in which they react to applied forces, the deflections resulting and the stresses\nand strains set up within the bodies, are all considered in an attempt to provide sufficient\nknowledge to enable any component to be designed such that it will not fail within its service\nlife. Typical components considered in detail in this volume include beams, shafts, cylinders,\nstruts, diaphragms and springs and, in most simple loading cases, theoretical expressions are\nderived to cover the mechanical behaviour of these components. Because of the reliance of\nsuch expressions on certain basic assumptions, the text also includes a chapter devoted to the\nimportant experimental stress and strain measurement techniques in use today with recommendations for further reading.\nEach chapter of the text contains a summary of essential formulae which are developed\nwithin the chapter and a large number of worked examples. The examples have been selected\nto provide progression in terms of complexity of problem and to illustrate the logical way in\nwhich the solution to a difficult problem can be developed. Graphical solutions have been\nintroduced where appropriate. In order to provide clarity of working in the worked examples\nthere is inevitably more detailed explanation of individual steps than would be expected in the\nmodel answer to an examination problem.\nAll chapters (with the exception of Chapter 16) conclude with an extensive list of problems\nfor solution of students together with answers. These have been collected from various\nsources and include questions from past examination papers in imperial units which have\nbeen converted to the equivalent SI values. Each problem is graded according to its degree of\ndifficulty as follows:\nA\nRelatively easy problem of an introductory nature.\nA/B Generally suitable for first-year studies.\nB\nGenerally suitable for second or third-year studies.\nC\nMore difficult problems generally suitable for third year studies.\n\n*National Vocational Qualifications\n\nxv\n\nxvi\n\nIntroduction\n\nGratitude is expressed to the following examination boards, universities and colleges who\nhave kindly given permission for questions to be reproduced:\nCity University\nEast Midland Educational Union\nEngineering Institutions Examination\nInstitution of Mechanical Engineers\nInstitution of Structural Engineers\nUnion of Educational Institutions\nUnion of Lancashire and Cheshire Institues\nUniversity of Birmingham\nUniversity of London\n\nC.U.\nE.M.E.U.\nE.I.E. and C.E.I.\n1.Mech.E.\n1.Struct.E.\nU.E.I.\nU.L.C.I.\nU.Birm.\nU.L.\n\nBoth volumes of the text together contain 150 worked examples and more than 500\nproblems for solution, and whilst it is hoped that no errors are present it is perhaps inevitable\nthat some errors will be detected. In this event any comment, criticism or correction will be\ngratefully acknowledged.\nThe symbols and abbreviations throughout the text are in accordance with the latest\nrecommendations of BS 1991 and PD 5686t.\nAs mentioned above, graphical methods of solution have been introduced where appropriate since it is the author’s experience that these are more readily accepted and understood\nby students than some of the more involved analytical procedures; substantial time saving\ncan also result. Extensive use has also been made of diagrams throughout the text since in the\nwords of the old adage “a single diagram is worth 1000 words”.\nFinally, the author is indebted to all those who have assisted in the production of this\nvolume; to Professor H. G. Hopkins, Mr R. Brettell, Mr R. J. Phelps for their work associated with the first edition and to Dr A. S. Tooth’, Dr N. Walke?, Mr R. Winters2for their\ncontributions to the second edition and to Dr M. Daniels for the extended treatment of the\nFinite Element Method which is the major change in this third edition. Thanks also go to the\npublishers for their advice and assistance, especially in the preparation of the diagrams and\nediting, to Dr. C. C. Perry (USA) for his most valuable critique of the first edition, and to Mrs\nJ. Beard and Miss S. Benzing for typing the manuscript.\nE. J. HEARN\n\nt Relevant Standards for use in Great Britain: BS 1991; PD 5686 Other useful SI Guides: The Infernational\nSystem of Units, N.P.L. Ministry of Technology, H.M.S.O. (Britain). Mechty, The International System of Units\n(Physical Constants and Conversion Factors), NASA, No SP-7012, 3rd edn. 1973 (U.S.A.) Metric Practice\nGuide, A.S.T.M. Standard E380-72 (U.S.A.).\nDr. A. S. Tooth, University of Strathclyde, Glasgow.\n1. \\$23.27.\nD. N. Walker and Mr. R. Winters, City of Birmingham Polytechnic.\n2. \\$26.\nDr M. M. Daniels, University of Central England.\n3. \\$24.4\n\nNOTATION\nQuantity\n\nS i Unit\n\nAngle\nLength\n\n0\n\nm (metre)\nmm (millimetre)\nmz\nm3\ns (second)\nm/s\nN (newton)\nkg (kilogram)\nkg/m3\nN\nNm\nPa (Pascal)\nN/m2\nbar ( = lo5N/m2)\nN/m2\n\nE\n\n-\n\nArea\nVolume\nTime\nAngular velocity\nVelocity\nWeight\nMass\nDensity\nForce\nMoment\nPressure\n\nStress\nStrain\nShear stress\nShear strain\nYoung's modulus\nShear modulus\nBulk modulus\nPoisson's ratio\nModular ratio\nPower\nCoefficient of linear expansion\nCoefficient of friction\nSecond moment of area\nPolar moment of area\nProduct moment of area\nTemperature\nDirection cosines\nPrincipal stresses\nPrincipal strains\nMaximum shear stress\nOctahedral stress\n\nA\nV\nt\n0\nV\n\nW\nm\nP\n\nF or P or W\nM\nP\n\nz\n\nN/m2\n\nY\n\n-\n\nE\nG\nK\n\nN/m2\nN/m2\nN/m2\n\nV\n\n-\n\nm\n\nW (watt)\nm/m \"C\n-\n\nm4\nm4\n\nm4\n\"C\n-\n\nN/m2\n\n-\n\nN/mz\nN/mZ\nxvii\n\nxviii\nQuantity\n\nNotation\nSymbol\n\nN/m2\n\nDeviatoric stress\nDeviatoric strain\nHydrostatic or mean stress\nVolumetric strain\nStress concentration factor\nStrain energy\nDisplacement\nDeflection\nPhotoelastic material fringe value\nNumber of fringes\nBody force stress\n\n-\n\nN/mz\n-\n\nJ\nm\nm\nm\nN/m2/fringe/m\n-\n\nN/m3\n\nSlenderness ratio\nGravitational acceleration\nCartesian coordinates\nCylindrical coordinates\nEccentricity\nNumber of coils or leaves of spring\nEquivalent J or effective polar moment\nof area\nAutofrettage pressure\nThick cylinder radius ratio R 2 / R 1\ninternal radius of thick cylinder R , / R 1\nResultant stress on oblique plane\nNormal stress on oblique plane\nShear stress on oblique plane\nDirection cosines of plane\nDirection cosines of line of action of\nresultant stress\nDirection cosines of line of action of\nshear stress\nComponents of resultant stress on\noblique plane\nShear stress in any direction 4 on\noblique plane\nInvariants of stress\nInvariants of reduced stresses\nAiry stress function\n\nSI Unit\n\nm4\nPA\n\nRP\nK\n\nN/m2 or bar\nm\n-\n\nm\n\nN/m2\nN/m2\nN/m2\n-\n\nN/m2\n\nxix\n\nNotation\nQuantity\n\nSI Unit\n\n‘Operator’ for Airy stress function\nbiharmonic equation\nStrain rate\nCoefficient of viscosity\nRetardation time (creep strain recovery)\nRelaxation time (creep stress relaxation)\nCreep contraction or lateral strain ratio\nMaximum contact pressure (Hertz)\nContact formulae constant\nContact area semi-axes\nMaximum contact stress\nSpur gear contact formula constant\nHelical gear profile contact ratio\n\nS\nS\n\n-\n\nN/mz\n(N/m2)m\nN/mZ\nN/mZ\n\nElastic stress concentration factor\nFatigue stress concentration factor\nPlastic flow stress concentration factor\nShear stress concentration factor\nEndurance limit for n cycles of load\nNotch sensitivity factor\nFatigue notch factor\nStrain concentration factor\nGriffith‘s critical strain energy release\nSurface energy of crack face\nPlate thickness\nStrain energy\nCompliance\nFracture stress\nStress Intensity Factor\nCompliance function\nPlastic zone dimension\nCritical stress intensity factor\n“J” Integral\nFatigue crack dimension\nCoefficients of Paris Erdogan law\nFatigue stress range\nFatigue mean stress\nFatigue stress amplitude\nFatigue stress ratio\nCycles to failure\nFatigue strength for N cycles\nTensile strength\nFactor of safety\n\nNm\nm\nNm\nmN-’\nN/m2\nN/m3I2\nm\nN/m3I2\nm\n-\n\nN/m2\nN/m2\nN/m2\n-\n\nN/m2\nN/m2\n-\n\nxx\nQuantity\n\nElastic strain range\nPlastic strain range\nTotal strain range\nDuctility\nSecondary creep rate\nActivation energy\nUniversal Gas Constant\nAbsolute temperature\nArrhenius equation constant\nLarson-Miller creep parameter\nSherby-Dorn creep parameter\nManson-Haford creep parameter\nInitial stress\nTime to rupture\nConstants of power law equation\n\nNo tut ion\nSymbol\n\nSI Unit\n-\n\nN/m2\nS\n\n-\n\nCONTENTS\n\nxv\n\nIntroduction\n\nXVii\n\nNotation\n\n1\n\n1 Simple Stress and Strain\n1.2 Direct or normal stress (a)\n1.3 Direct strain ( E )\n1.4 Sign conventionfor direct stress and strain\n1.5 Elastic materials - Hooke’s law\n1.6 Modulus of elasticity - Young’s modulus\n1.7 Tensile test\n1.8 Ductile materials\n1.9 Brittle materials\n1.10 Poisson’s ratio\n1.1 1 Application of Poisson’s ratio to a two-dimensional stress system\n1.12 Shear stress\n1.13 Shear strain\n1.14 Modulus of rigidity\n1,15 Double shear\n1.16 Allowable working stress -factor of safety\n1.18 Temperature stresses\n1.19 Stress concentrations -stress concentrationfactor\n1.20 Toughness\n1.21 Creep and fatigue\n\nExamples\nProblems\nBibliography\n\n1\n2\n2\n2\n3\n3\n4\n8\n8\n9\n10\n11\n11\n12\n12\n12\n13\n13\n14\n14\n15\n17\n25\n26\n\n27\n\n2 Compound Bars\nSummary\n2.1 Compound bars subjected to external load\nV\n\n27\n28\n\nvi\n\nContents\n2.2\n2.3\n2.4\n2.5\n2.6\n\nCompound bars - ‘<equivalent or “combined”modulus\nCompound bars subjected to temperature change\nCompound bar (tube and rod)\nCompound bars subjected to external load and temperature effects\nCompound thick cylinders subjected to temperature changes\nExamples\nProblems\n”\n\n3 Shearing Force and Bending Moment Diagrams\n\nSummary\n3.1 Shearing force and bending moment\n3.1.1 Shearing force (S.F.) sign convention\n3.1.2 Bending moment (B.M.) sign convention\n3.2 S.F. and B.M. diagrams for beams carrying concentrated loads only\n3.3 S.F. and B.M. diagrams for uniformly distributed loads\n3.4 S.F. and B.M. diagrams for combined concentrated and uniformly\n3.5 Points of contrafexure\n3.1 S.F. and B.M. diagrams for an applied couple or moment\n3.8 S.F. and B.M. diagrams for inclined loa&\n3.9 Graphical construction of S.F. and B.M. diagrams\n3.10 S.F. and B.M. diagrams for beams carrying distributed loads of\nincreasing value\n3.1 1 S.F. at points of application of concentrated loads\nExamples\nProblems\n4 Bending\n\nSummary\nIntroduction\n4.1 Simple bending theory\n4.2 Neutral axis\n4.3 Section modulus\n4.4 Second moment of area\n4.5 Bending of composite or fitched beams\n4.6 Reinforced concrete beams -simple tension reinforcement\n\n29\n30\n32\n34\n34\n34\n39\n41\n\n41\n41\n42\n42\n43\n46\n47\n48\n49\n50\n52\n54\n55\n55\n56\n59\n62\n\n62\n63\n64\n66\n68\n68\n70\n71\n\n73\n74\n\nContents\n4.9\n4.10\n4.1 1\n4.12\n\nvii\n76\n77\n78\n78\n79\n88\n\n“Middle-quarter and “middle-third rules\nShear stresses owing to bending\nStrain energy in bending\nLimitations of the simple bending theory\nExamples\nProblems\n”\n\n”\n\n92\n\n5 Slope and Deflection of Beams\nSummary\nIntroduction\n5.2 Direct integration method\n5.3 MacaulayS method\n5.4 Macaulay’s method for u.d.ls\n5.5 Macaulay’s method for beams with u.d.1. applied over part of the beam\n5.6 Macaulay’s method for couple applied at a point\n5.7 Mohr’s “area-moment” method\n5.8 Principle of superposition\n5.9 Energy method\n5.10 Maxwell’s theorem of reciprocal displacements\n5.1 1 Continuous beams - CIapeyron’s “three-moment equation\n5.12 Finite difference method\n5.13 Defections due to temperature effects\nExamples\nProblems\n”\n\n6 Built-in Beams\n\nSummary\nIntroduction\n6.1 Built-in beam carrying central concentrated load\n6.2 Built-in beam carrying uniformly distributed load across the span\n6.3 Built-in beam carrying concentrated load offset from the centre\n6.4 Built-in beam carrying a non-uniform distributed load\n6.6 Effect of movement of supports\nExamples\nProblems\n\n92\n94\n94\n97\n102\n105\n106\n106\n108\n112\n112\n112\n115\n118\n119\n123\n138\n\n140\n140\n141\n141\n142\n143\n145\n146\n146\n147\n152\n\n...\n\nContents\n\nVlll\n\n7 Shear Stress Distribution\n\nSummary\nIntroduction\n7.1 Distribution of shear stress due to bending\n7.2 Application to rectangular sections\n7.3 Application to I-section beams\n7.3.1 Vertical shear in the web\n7.3.2 Vertical shear in the flanges\n7.3.3 Horizontal she& in the flanges\n7.4 Application to circular sections\n7.5 Limitation of shear stress distribution theory\n7.6 Shear centre\nExamples\nProblems\n\n8 Torsion\nSummary\n8.1 Simple torsion theory\n8.2 Polar second moment of area\n8.3 Shear stress and shear strain in shafts\n8.4 Section modulus\n8.5 Torsional rigidity\n8.6 Torsion of hollow shafts\n8.7 Torsion of thin-walled tubes\n8.8 Composite shafts -series connection\n8.9 Composite shafts -parallel connection\n8.10 Principal stresses\n8.1 1 Strain energy in torsion\n8.12 Variation of data along shaft length -torsion of tapered shafts\n8.13 Power transmitted by shafts\n8,14 Combined stress systems -combined bending and torsion\n8.15 Combined bending and torsion - equivalent bending moment\n8.16 Combined bending and torsion -equivalent torque\n8.17 Combined bending, torsion and direct thrust\n8.18 Combined bending, torque and internal pressure\nExamples\nProblems\n\n154\n154\n155\n156\n157\n158\n159\n159\n160\n162\n\n164\n165\n166\n173\n\n176\n176\n177\n179\n180\n181\n182\n182\n182\n182\n183\n184\n184\n186\n186\n187\n187\n188\n189\n189\n190\n195\n\nContents\n\n9 Thin Cylinders and Shells\nSummary\n9.1 Thin cylinders under internal pressure\n9.1.1 Hoop or circumferential stress\n9.1.2 Longitudinal stress\n9.1.3 Changes in dimensions\n9.2 Thin rotating ring or cylinder\n9.3 Thin spherical shell under internal pressure\n9.3.1 Change in internal volume\n9.4 Vessels subjected to JIuid pressure\n9.5 Cylindrical vessel with hemispherical e n d\n9.6 Effects of end plates and joints\n9.7 Wire-wound thin cylinders\nExamples\nProblems\n10 Thick cylinders\nSummary\n10.1 Difference in treatment between thin and thick cylinders -basic\nassumptions\n10.2 Development of the Lame theory\n10.3 Thick cylinder - internal pressure only\n10.4 Longitudinal stress\n10.5 Maximum shear stress\n10.6 Change of cylinder dimensions\n10.7 Comparison with thin cylinder theory\n10.8 Graphical treatment - Lame line\n10.9 Compound cylinders\n10.10 Compound cylinders -graphical treatment\n10.11 Shrinkage or interference allowance\n10.12 Hub on solid shaji\n10.13 Force fits\n10.14 Compound cylinder -different materials\n10.15 Uniform heating of compound cylinders of different materials\n10.16 Failure theories -yield criteria\n10.17 Plastic yielding - “auto-frettage”\n10.18 Wire-wound thick cylinders\nExamples\nProblems\n\nix\n198\n198\n198\n199\n199\n200\n20 1\n202\n203\n203\n204\n205\n206\n208\n213\n\n215\n215\n216\n217\n219\n220\n22 1\n22 1\n222\n223\n224\n226\n226\n229\n229\n230\n23 1\n233\n233\n234\n236\n25 1\n\nContents\n\nX\n\n254\n\n11 Strain Energy\n\nSummary\nIntroduction\n1 1.1 Strain energy - tension or compression\n1 1.2 Strain energy -shear\n1 1.3 Strain energy -bending\n1 1.4 Strain energy - torsion\n1 1.5 Strain energy of a three-dimensionalprincipal stress system\n1 1.6 Volumetric or dilatational strain energy\n1 1.7 Shear or distortional strain energy\n1 1.10 Impact loads -bending applications\n1 1.11 Castigliano’sfirst theorem for deflection\n1 1.13 Application of Castigliano’s theorem to angular movements\n1 1.14 Shear deflection\nExamples\nProblems\n”\n\n12 Springs\nSummary\nIntroduction\n12.1 Close-coiled helical spring subjected to axial load W\n12.2 Close-coiled helical spring subjected to axial torque T\n12.3 Open-coiled helical spring subjected to axial load W\n12.4 Open-coiled helical spring subjected to axial torque T\n12.5 Springs in series\n12.6 Springs in parallel\n12.7 Limitations of the simple theory\n12.8 Extension springs - initial tension\n12.9 Allowable stresses\n12.10 Leaf or carriage spring: semi-elliptic\n12.11 Leaf or carriage spring: quarter-elliptic\n12.12 Spiral spring\nExamples\nProblems\n\n254\n256\n257\n259\n260\n26 1\n262\n262\n263\n263\n264\n265\n266\n268\n269\n269\n274\n292\n\n297\n\n297\n299\n299\n300\n30 1\n304\n305\n306\n306\n307\n308\n309\n312\n314\n316\n324\n\nContents\n\n13 Complex Stresses\nSummary\n13.1 Stresses on oblique planes\n13.2 Material subjected to pure shear\n13.3 Material subjected to two mutually perpendicular direct stresses\n13.4 Material subjected to combined direct and shear stresses\n13.5 Principal plane inclination in terms of the associated principal stress\n13.6 Graphical solution - Mohr 's stress circle\n13.7 Alternative representations of stress distributions at a point\n13.8 Three-dimensionalstresses -graphical representation\nExamples\nProblems\n14 Complex Strain and the Elastic Constants\nSummary\n14.1 Linear strain for tri-axial stress state\n14.2 Principal strains in terms of stresses\n14.3 Principal stresses in terms of strains -two-dimensional stress system\n14.4 Bulk modulus K\n14.5 Volumetric strain\n14.6 Volumetric strain for unequal stresses\n14.1 Change in volume of circular bar\n14.8 Effect of lateral restraint\n14.9 Relationship between the elastic constants E, G, K and v\n14.10 Strains on an oblique plane\n14.11 Principal strain - Mohr s strain circle\n14.12 Mohr 's strain circle -alternative derivation from the\ngeneral stress equations\n14.13 Relationship between Mohr 's stress and strain circles\n14.14 Construction of strain circle from three known strains\n(McClintock method) -rosette analysis\n14.15 Analytical determination of principal strains from rosette readings\n14.16 Alternative representations of strain distributions at a point\n14.1I Strain energy of three-dimensional stress system\nExamples\nProblems\n15 Theories of Elastic Failure\n\nSummary\nIntroduction\n\nxi\n326\n326\n326\n327\n329\n329\n331\n332\n334\n338\n342\n358\n\n361\n36 1\n36 1\n362\n363\n363\n363\n364\n365\n366\n361\n310\n312\n374\n375\n318\n38 1\n383\n385\n387\n397\n\n401\n401\n40 1\n\nContents\n\nxii\n15.1\n15.2\n15.3\n15.4\n15.5\n\n15.6\n15.7\n15.8\n15.9\n\n15.10\n15.1 1\n15.12\n15.13\n\nMaximum principal stress theory\nMaximum shear stress theory\nMaximum principal strain theory\nMaximum total strain energy per unit volume theory\nMaximum shear strain energy per unit volume (or distortion energy)\ntheory\nMohr 's modijied shear stress theory for brittle materials\nGraphical representation of failure theoriesfor two-dimensional\nstress systems (one principal stress zero)\nGraphical solution of two-dimensional theory of failure problems\nGraphical representation of the failure theoriesfor three-dimensional\nstress systems\n15.9.1 Ductile materials\n15.9.2 Brittle materials\nLimitations of the failure theories\nEflect of stress concentrations\nSafety factors\nModes of failure\nExamples\nProblems\n\n16 Experimental Stress Analysis\n\nIntroduction\n16.1 Brittle lacquers\n16.2 Strain gauges\n16.3 Unbalanced bridge circuit\n16.4 Null balance or balanced bridge circuit\n16.5 Gauge construction\n16.6 Gauge selection\n16.7 Temperature compensation\n16.8 Installation procedure\n16.9 Basic measurement systems\n16.10 D.C. and A.C. systems\n16.11 Other types of strain gauge\n16.12 Photoelasticity\n16.13 Plane-polarised light - basic polariscope arrangements\n16.14 Temporary birefringence\n16.15 Production of fringe patterns\n16.16 Interpretation of fringe patterns\n16.17 Calibration\n\n402\n403\n403\n403\n403\n404\n\n406\n410\n41 1\n41 1\n412\n413\n414\n414\n416\n417\n427\n\n430\n430\n43 1\n43 5\n437\n437\n437\n438\n439\n440\n441\n443\n\n444\n445\n446\n446\n448\n\n449\n450\n\nContents\n16.18\n16.19\n16.20\n16.21\n16.22\n16.23\n\nFractional fringe order determination - compensation techniques\nIsoclinics - circular polarisation\nStress separation procedures\nThree-dimensional photoelasticity\nReflective coating technique\nOther methods of strain measurement\nBibliography\n\nAppendix 1. Typical mechanical and physical pro'prties for engineering\nmaterials\nAppendix 2. Typical mechanical properties of non-metals\nAppendix 3. Other properties of non-metals\nIndex\n\nxiii\n45 1\n452\n454\n454\n454\n456\n456\n\nxxi\nxxii\nxxiii\nxxv\n\nCHAPTER 1\n\nSIMPLE STRESS AND STRAIN\n\nIn any engineering structure or mechanism the individual components will be subjected to\nexternal forces arising from the service conditions or environment in which the component\nworks. If the component or member is in equilibrium, the resultant of the external forces will\nbe zero but, nevertheless, they together place a load on the member which tends to deform\nthat member and which must be reacted by internal forces which are set up within the\nmaterial.\nIf a cylindrical bar is subjected to a direct pull or push along its axis as shown in Fig. 1.1,\nthen it is said to be subjected to tension or compression. Typical examples of tension are the\nforces present in towing ropes or lifting hoists, whilst compression occurs in the legs of your\nchair as you sit on it or in the support pillars of buildings.\n\n,Are0\n\nTension\n\nA\n\nCompression\n\nFig. 1.1. Types of direct stress.\n\nIn the SI system of units load is measured in newtons, although a single newton, in\nengineering terms, is a very small load. In most engineering applications, therefore, loads\nappear in SI multiples, i.e. kilonewtons (kN) or meganewtons (MN).\nThere are a number of different ways in which load can be applied to a member. Typical\n(b) Liue loads, as produced by, for example, lorries crossing a bridge.\n(c) Impact or shock loads caused by sudden blows.\n(d) Fatigue,fluctuating or alternating loads, the magnitude and sign of the load changing\nwith time.\n1\n\n2\n\nMechanics of Materials\n\n\\$1.2\n\n1.2. Direct or normal stress (a)\nIt has been noted above that external force applied to a body in equilibrium is reacted by\ninternal forces set up within the material. If, therefore, a bar is subjected to a uniform tension\nor compression, i.e. a direct force, which is uniformly or equally applied across the crosssection, then the internal forces set up are also distributed uniformly and the bar is said to be\nsubjected to a uniform direct or normal stress, the stress being defined as\nstress (a)= -= area A\nStress CT may thus be compressive or tensile depending on the nature of the load and will be\nmeasured in units of newtons per square metre (N/mZ)or multiples of this.\nIn some cases the loading situation is such that the stress will vary across any given section,\nand in such cases the stress at any point is given by the limiting value of 6 P / 6 A as 6 A tends to\nzero.\n1.3. Direct strain ( E )\nIf a bar is subjected to a direct load, and hence a stress, the bar will change in length. If the\nbar has an original length L and changes in length by an amount 6L, the strain produced is\ndefined as follows:\nstrain ( E ) = change in length =-6 L\noriginal length\nL\nStrain is thus a measure of the deformation of the material and is non-dimensional,Le. it has\nno units; it is simply a ratio of two quantities with the same unit (Fig. 1.2).\nStrain C = G L / L\n\nFig. 1.2.\n\nSince, in practice, the extensions of materials under load are very small, it is often\ni.e. microstrain, when the\nconvenient to measure the strains in the form of strain x\nsymbol used becomes /ALE.\nAlternatively, strain can be expressed as a percentage strain\ni.e.\n\n6L\nstrain ( E ) = - x 100%\nL\n1.4. Sign convention for direct stress and strain\n\nTensile stresses and strains are considered POSITIVE in sense producing an increase in\nlength. Compressive stresses and strains are considered NEGATIVE in sense producing a\ndecrease in length.\n\n\\$1.5\n\nSimple Stress and Strain\n\n3\n\n1.5. Elastic materials - Hooke’s law\n\nA material is said to be elastic if it returns to its original, unloaded dimensions when load is\nremoved. A particular form of elasticity which applies to a large range of engineering\nmaterials, at least over part of their load range, produces deformations which are\nproportional to the loads producing them. Since loads are proportional to the stresses they\nproduce and deformations are proportional to the strains, this also implies that, whilst\nmaterials are elastic, stress is proportional to strain. Hooke’s law, in its simplest form*,\ntherefore states that\nstress (a) a strain ( E )\ni.e.\n\nstress\n-- constant*\nstrain\n\nIt will be seen in later sections that this law is obeyed within certain limits by most ferrous\nalloys and it can even be assumed to apply to other engineering materials such as concrete,\ntimber and non-ferrous alloys with reasonable accuracy. Whilst a material is elastic the\ndeformation produced by any load will be completely recovered when the load is removed;\nthere is no permanent deformation.\nOther classifications of materials with which the reader should be acquainted are as\nfollows:\nA material which has a uniform structure throughout without any flaws or discontinuities\nis termed a homogeneous material. Non-homogeneous or inhomogeneous materials such as\nconcrete and poor-quality cast iron will thus have a structure which varies from point to point\ndepending on its constituents and the presence of casting flaws or impurities.\nIf a material exhibits uniform properties throughout in all directions it is said to be\nisotropic; conversely one which does not exhibit this uniform behaviour is said to be nonisotropic or anisotropic.\nAn orthotropic material is one which has different properties in different planes. A typical\nexample of such a material is wood, although some composites which contain systematically\norientated “inhomogeneities” may also be considered to fall into this category.\n1.6. Modulus of elasticity - Young’s modulus\n\nWithin the elastic limits of materials, i.e. within the limits in which Hooke’s law applies, it\nhas been shown that\nstress\n-- constant\nstrain\nThis constant is given the symbol E and termed the modulus of elasticity or Young’s modulus.\nThus\n\nE = - stress\n_ -0\nstrain E\nP 6 L --PL\nA ’ L\nASL\n\n=--I--\n\n* Readers should be warned that in more complex stress cases this simple form of Hooke’s law will not apply and\nmisapplication could prove dangerous; see 814.1, page 361.\n\nMechanics of Materials\n\n4\n\n\\$1.7\n\nYoung’s modulus E is generally assumed to be the same in tension or compression and for\nmost engineering materials has a high numerical value. Typically, E = 200 x lo9 N/m2 for\nsteel, so that it will be observed from (1.1) that strains are normally very small since\nE=-\n\n0\n\nE\n\nIn most common engineering applications strains do not often exceed 0.003 or 0.3 % so\nthat the assumption used later in the text that deformations are small in relation to original\ndimensions is generally well founded.\nThe actual value of Young’s modulus for any material is normally determined by carrying\nout a standard tensile test on a specimen of the material as described below.\n1.7. Tensile test\n\nIn order to compare the strengths of various materials it is necessary to carry out some\nstandard form of test to establish their relative properties. One such test is the standard tensile\ntest in which a circular bar of uniform cross-section is subjected to a gradually increasing\ntensile load until failure occurs. Measurements of the change in length of a selected gauge\nlength of the bar are recorded throughout the loading operation by means of extensometers\nand a graph of load against extension or stress against strain is produced as shown in Fig. 1.3;\nthis shows a typical result for a test on a mild (low carbon) steel bar; other materials will\nexhibit differentgraphs but of a similar general form see Figs 1.5 to 1.7.\nElastic\nP a r t i a l l y plastic\n\ntP\nExtension or strain\n\nFig. 1.3. Typical tensile test curve for mild steel.\n\nFor the first part of the test it will be observed that Hooke’s law is obeyed, Le. the material\nbehaves elastically and stress is proportional to strain, giving the straight-line graph\nindicated. Some point A is eventually reached, however, when the linear nature of the graph\nceases and this point is termed the limit of proportionality.\nFor a short period beyond this point the material may still be elastic in the sense that\ndeformations are completely recovered when load is removed (i.e. strain returns to zero) but\n\n51.7\n\n5\n\nSimple Stress and Strain\n\nHooke’s law does not apply. The limiting point B for this condition is termed the elastic limit.\nFor most practical purposes it can often be assumed that points A and B are coincident.\nBeyond the elastic limit plastic deformation occurs and strains are not totally recoverable.\nThere will thus be some permanent deformation or permanent set when load is removed.\nAfter the points C, termed the upper yield point, and D, the lower yield point, relatively rapid\nincreases in strain occur without correspondinglyhigh increases in load or stress. The graph\nthus becomes much more shallow and covers a much greater portion of the strain axis than\ndoes the elastic range of the material. The capacity of a material to allow these large plastic\ndeformations is a measure of the so-called ductility of the material, and this will be discussed\nin greater detail below.\nFor certain materials, for example, high carbon steels and non-ferrous metals, it is not\npossible to detect any difference between the upper and lower yield points and in some cases\nno yield point exists at all. In such cases a proof stress is used to indicate the onset of plastic\nstrain or as a comparison of the relative properties with another similar material. This\ninvolves a measure of the permanent deformation produced by a loading cycle; the 0.1 %\nproof stress, for example, is that stress which, when removed, produces a permanent strain or\n“set” of 0.1 % of the original gauge length-see Fig. 1.4(a).\n\nb\n”7\n\nr\n\nE\n\n-\n\n=\n\nF\n\ni\n5\nI\n\n’\n\nc\nI0 I % 1\n\nFig. 1.4. (a) Determination of 0.1 % proof stress.\n\n\\\n\nPermanent ‘ s e t ’\n\nFig. 1.4. (b) Permanent deformation or “set” after\nstraining beyond the yield point.\n\nThe 0.1 % proof stress value may be determined from the tensile test curve for the material\nin question as follows:\nMark the point P on the strain axis which is equivalent to 0.1 % strain. From P draw a line\nparallel with the initial straight line portion of the tensile test curve to cut the curve in N. The\nstress corresponding to Nis then the 0.1 %proof stress. A material is considered to satisfy its\nspecificationif the permanent set is no more than 0.1 %after the proof stress has been applied\nfor 15 seconds and removed.\nBeyond the yield point some increase in load is required to take the strain to point E on the\ngraph. Between D and E the material is said to be in the elastic-plastic state, some of the\nsection remaining elastic and hence contributing to recovery of the original dimensions if\nload is removed, the remainder being plastic. Beyond E the cross-sectional area of the bar\n\n6\n\n01.7\n\nMechanics of Materials\n\nbegins to reduce rapidly over a relatively small length of the bar and the bar is said to neck.\nThis necking takes place whilst the load reduces, and fracture of the bar finally occurs at\npoint F.\nThe nominal stress at failure, termed the maximum or ultimate tensile stress, is given by the\nload at E divided by the original cross-sectional area of the bar. (This is also known as the\ntensile strength of the material of the bar.) Owing to the large reduction in area produced by\nthe necking process the actual stress at fracture is often greater than the above value. Since,\nhowever, designers are interested in maximum loads which can be carried by the complete\ncross-section, the stress at fracture is seldom of any practical value.\nIf load is removed from the test specimen after the yield point C has been passed, e.g. to\nsome position S, Fig. 1.4(b), the unloading line STcan, for most practical purposes, be taken to\nbe linear. Thus, despite the fact that loading to S comprises both elastic (OC)and partially\ncommencing with the permanent elongation associated with the strain OT, would then follow\nthe line TS and continue along the previous curve to failure at F. It will be observed, however,\nthat the repeated load cycle has the effect of increasing the elastic range of the material, i.e.\nraising the effective yield point from C to S, while the tensile strength is unaltered. The\nprocedure could be repeated along the line PQ, etc., and the material is said to have been work\nhardened.\nIn fact, careful observation shows that the material will no longer exhibit true elasticity\nultimate stress value but the elongation or strain to failure will be much reduced.\nTypical stress-strain curves resulting from tensile tests on other engineering materials are\nshown in Figs 1.5 to 1.7.\n\n1500tr\n\nchrome steel\n\n/Nickel\n\nMedium carbon steel -\n\nu-\n\n?!\n&i\n600\n\n0\n\nY\n\nI\n\n0\n\nI\n20\n\nI\n40\n\nI\n\n30\nStrain.\n\n%\n\nFig. 1.5. Tensile test curves for various metals.\n\nI\n\n50\n\n7\n\nSimple Stress and Strain\n\n\\$1.7\n\nStrain,\n\n%\n\nFig. 1.6. Typical stressstrain curves for hard drawn wire material-note\nlarge reduction in strain values from those of Fig. 1.5.\n\nGlass remforced polycarbonate\n\neo\n\n’ II /\n?\n\n,Unreinforced\n\nc,\n\nY\n0\n\npdycnrbonote\n\nI\n\nI\n\nI\n\nI\n\nI\n\n10\n\n20\n\n30\n\n40\n\n50\n\nStrain,\n\n%\n\nFig. 1.7. Typical tension test results for various types of nylon and polycarbonate.\n\nAfter completing the standard tensile test it is usually necessary to refer to some “British\nStandard Specification” or “Code of Practice” to ensure that the material tested satisfies the\nrequirements, for example:\nBS 4360\nBS 970\nBS 153\nBS 449\n\nBritish Standard Specification for Weldable Structural Steels.\nBritish Standard Specification for Wrought Steels.\nBritish Standard Specification for Steel Girder Bridges.\nBritish Standard Specification for the use of Structural Steel in Building, etc.\n\n8\n\nMechanics of Materials\n\n51.8\n\n1.8. Ductile materials\n\nIt has been observed above that the partially plastic range of the graph of Fig. 1.3 covers a\nmuch wider part of the strain axis than does the elastic range. Thus the extension of the\nmaterial over this range is considerably in excess of that associated with elastic loading. The\ncapacity of a material to allow these large extensions, i.e. the ability to be drawn out\nplastically, is termed its ductility. Materials with high ductility are termed ductile materials,\nmembers with low ductility are termed brittle materials. A quantitative value of the ductility is\nobtained by measurements of the percentage elongation or percentage reduction in area, both\nbeing defined below.\nPercentage elongation =\nPercentage reduction in area =\n\nincrease in gauge length to fracture\nx loo\noriginal gauge length\nreduction in cross-sectionalarea of necked portion\nx 100\noriginal area\n\nThe latter value, being independent of any selected gauge length, is generally taken to be the\nmore useful measure of ductility for reference purposes.\nA property closely related to ductility is malleability, which defines a material's ability to be\nhammered out into thin sheets. A typical example of a malleable material is lead. This is used\nextensively in the plumbing trade where it is hammered or beaten into corners or joints to\nprovide a weatherproof seal. Malleability thus represents the ability of a material to allow\n1.9. Brittle materials\nA brittle material is one which exhibits relatively small extensions to fracture so that the\npartially plastic region of the tensile test graph is much reduced (Fig. 1.8). Whilst Fig. 1.3\nreferred to a low carbon steel, Fig. 1.8 could well refer to a much higher strength steel with a\nhigher carbon content. There is little or no necking at fracture for brittle materials.\n\nE\n\nFig. 1.8. Typical tensile test curve for a brittle material.\n\nTypical variations of mechanical properties of steel with carbon content are shown in\nFig. 1.9.\n\n\\$1.10\n\n9\n\nSimple Stress and Strain\n\ni90\n\nI\n\nI\n\n0 2\n\n04\n\nI\n08\n\n06\n\nYo\n\n10\n\nc\n\nFig. 1.9. Variation of mechanical properties of steel with carbon content.\n\n1.10. Poisson’s ratio\nConsider the rectangular bar of Fig. 1.10 subjected to a tensile load. Under the action of\nthis load the bar will increase in length by an amount 6 L giving a longitudinal strain in the bar\nof\n6L\nEL= L\n\nFig. 1.10.\n\nThe bar will also exhibit, however, a reduction in dimensions laterally, i.e. its breadth and\ndepth will both reduce. The associated lateral strains will both be equal, will be of opposite\nsense to the longitudinal strain, and will be given by\n&l,t\n\n=\n\n6b\n6d\n-- - - b\n\nd\n\nProvided the load on the material is retained within the elastic range the ratio of the lateral\nand longitudinal strains will always be constant. This ratio is termed Poisson’s ratio.\n1.e.\n\nPoisson’s ratio ( v ) =\n\nlateral strain - ( - 6 d / d )\nlongitudinal strain\n6LIL\n\n(1.4)\n\nThe negative sign of the lateral strain is normally ignored to leave Poisson’s ratio simply as\n\n10\n\nMechanics of Materials\n\n\\$1.11\n\na ratio of strain magnitudes. It must be remembered, however, that the longitudinal strain\ninduces a lateral strain of opposite sign, e.g. tensile longitudinal strain induces compressive\nlateral strain.\nFor most engineering materials the value of v lies between 0.25 and 0.33.\nSince\nlongitudinal stress - a\nlongitudinal strain =\n(1.4a)\nYoung’s modulus E\nHence\nd\n\nlateral strain = v E\n\n(1.4b)\n\n1.11. Application of Poisson’s ratio to a two-dimensional stress system\n\nA two-dimensional stress system is one in which all the stresses lie within one plane such as\nthe X-Y plane. From the work of \\$1.10 it will be seen that if a material is subjected to a tensile\nstress a on one axis producing a strain u / E and hence an extension on that axis, it will be\nsubjected simultaneously to a lateral strain of v times a/E on any axis at right angles. This\nlateral strain will be compressive and will result in a compression or reduction of length on\nthis axis.\nConsider, therefore, an element of material subjected to two stresses at right angles to each\nother and let both stresses, ux and c y ,be considered tensile, see Fig. 1.11.\n\nFig. 1.11. Simple twodimensional system of direct stresses.\n\nThe following strains will be produced\n(a)\n(b)\n(c)\n(d)\n\nin\nin\nin\nin\n\nthe X direction resulting from ax = a,/E,\nthe Y direction resulting from cy = a,/E.\nthe X direction resulting from 0, = - v(a,/E),\nthe Y direction resulting from ax = - v(a,/E).\n\nstrains (c) and (d) being the so-called Poisson’s ratio strain, opposite in sign to the applied\nstrains, i.e. compressive.\nThe total strain in the X direction will therefore be given by:\n&\n\n“\n\n6”\n\n1\nE\n\n= - - v o y = -(ax\n\nE\n\nE\n\n- va,)\n\n(1.5)\n\ng1.12\n\n11\n\nSimple Stress and Strain\n\nand the total strain in the Y direction will be:\n\nIf any stress is, in fact, compressive its value must be substituted in the above equations\ntogether with a negative sign following the normal sign convention.\n1.12. Shear stress\n\nConsider a block or portion of material as shown in Fig. 1.12a subjected to a set of equal and\nopposite forces Q.(Such a system could be realised in a bicycle brake block when contacted\nwith the wheel.) There is then a tendency for one layer of the material to slide over another to\nproduce the form of failure shown in Fig. 1.12b. If this failure is restricted, then a shear stress T\nis set up, defined as follows:\nshear stress (z)=\n\nQ\n-area resisting shear A\n\nThis shear stress will always be tangential to the area on which it acts; direct stresses, however,\nare always normal to the area on which they act.\n\nQ\n\n(a)\n\n(b)\n\nFig. 1.12. Shear force and resulting shear stress system showing typical form of failure by\nrelative sliding of planes.\n\n1.13. Shear strain\n\nIf one again considers the block of Fig. 1.12a to be a bicycle brake block it is clear that the\nrectangular shape of the block will not be retained as the brake is applied and the shear forces\nintroduced. The block will in fact change shape or “strain” into the form shown in Fig. 1.13.\nThe angle of deformation y is then termed the shear strain.\nShear strain is measured in radians and hence is non-dimensional, i.e. it has no units.\n\nT\n\nFig. 1.13. Deformation (shear strain) produced by shear stresses.\n\n12\n\nMechanics of Materials\n\n01.14\n\n1.14. Modulus of rigidity\n\nFor materials within the elastic range the shear strain is proportional to the shear stress\nproducing it,\ni.e.\nshear stress z\n= - = constant = G\nshear strain y\nThe constant G is termed the modulus of rigidity or shear modulus and is directly\ncomparable to the modulus of elasticity used in the direct stress application. The term\nmodulus thus implies a ratio of stress to strain in each case.\n\n1.15. Double shear\nConsider the simple riveted lap joint shown in Fig. 1.14a.When load is applied to the plates\nthe rivet is subjected to shear forces tending to shear it on one plane as indicated. In the butt\njoint with two cover plates of Fig. 1.14b, however, each rivet is subjected to possible shearing\non two faces, i.e. double shear. In such cases twice the area of metal is resisting the applied\nforces so that the shear stress set up is given by\nshear stress r (in double shear)\n\nSimple riveted lop p m t\n\n+--J&\n\nFailure of\nrivet ,n single\n\nP\n2A\n\n-\n\n+e+\n\nshear\n\nButt p m t wcrn t w o cover Plates\n\nla1\n\nlbl\n\nFig. 1.14. (a) Single shear. (b) Double shear.\n\n1.16. Allowable working stress-factor of safety\n\nThe most suitable strength or stiffness criterion for any structural element or component is\nnormally some maximum stress or deformation which must not be exceeded. In the case of\nstresses the value is generally known as the maximum allowable working stress.\netc., designers generally introduce a factor of safety into their designs, defined as follows:\nfactor of safety =\n\nmaximum stress\nallowable working stress\n\n(1.9)\n\nHowever, in view of the fact that plastic deformations are seldom accepted this definition is\nsometimes modified to\nfactor of safety =\n\nyield stress (or proof stress)\nallowable working stress\n\n51.17\n\n13\n\nSimple Stress and Strain\n\nIn the absence of any information as to which definition has been used for any quoted value\nof safety factor the former definition must be assumed. In this case a factor of safety of 3\nimplies that the design is capable of carrying three times the maximum stress to which it is\nexpected the structure will be subjected in any normal loading condition. There is seldom any\nrealistic basis for the selection of a particular safety factor and values vary significantly from\none branch of engineering to another. Values are normally selected on the basis of a\nconsideration of the social, human safety and economic consequences of failure. Typical\nvalues range from 2.5 (for relatively low consequence, static load cases) to 10 (for shock load\nand high safety risk applications)-see \\$15.12.\n\nIn some loading cases, e.g. buckling of struts, neither the yield stress nor the ultimate\nstrength is a realistic criterion for failure of components. In such cases it is convenient to\nreplace the safety factor, based on stresses, with a different factor based on loads. The load\nfactor is therefore defined as:\n\n(1.10)\n\nThis is particularly useful in applications of the so-called plastic limit design procedures.\n\n1.18. Temperature stresses\nWhen the temperature of a component is increased or decreased the material respectively\nexpands or contracts. If this expansion or contraction is not resisted in any way then the\nprocesses take place free of stress. If, however, the changes in dimensions are restricted then\nstresses termed temperature stresses will be set up within the material.\nConsider a bar of material with a linear coefficient of expansion a. Let the original length of\nthe bar be L and let the temperature increase be t. If the bar is free to expand the change in\nlength would be given by\nb L = Lat\n\n(1.11)\n\nand the new length\nL’ = L + Lat = L ( l + a t )\n\nIf this extension were totally prevented, then a compressive stress would be set up equal to\nthat produced when a bar of length L ( 1 + at) is compressed through a distance of Lat. In this\ncase the bar experiences a compressive strain\nAL\nL\n\nLat\nL(l +at)\n\nE=-=\n\nIn most cases at is very small compared with unity so that\nLat\nL\n\nE=---=\n\nat\n\n14\n\nMechanics of Materials\n\n81.19\n\nIS\n\n-=E\n\nBut\n\nE\n\n..\n\nstress\n\nt~\n\n= EE = Eat\n\n(1.12)\n\nThis is the stress set up owing to total restraint on expansions or contractions caused by a\ntemperature rise, or fall, t. In the former case the stress is compressive, in the latter case the\nstress is tensile.\nIf the expansion or contraction of the bar is partially prevented then the stress set up will be\nless than that given by eqn. (1.10).Its value will be found in a similar way to that described\nabove except that instead of being compressed through the total free expansion distance of\nLat it will be compressed through some proportion of this distance depending on the amount\nof restraint.\nAssuming some fraction n of Lat is allowed, then the extension which is prevented is\n( 1 - n)Lat. This will produce a compressive strain, as described previously, of magnitude\nE =\n\n(1-n)Lat\nL(l +at)\n\nor, approximately,\nE\n\n= ( 1 - n ) L a t / L = ( 1 -n)at.\n\nThe stress set up will then be E times E .\ni.e.\n\nIS =\n\n( 1 -n)Eat\n\n(1.13)\n\nThus, for example, if one-third of the free expansion is prevented the stress set up will be twothirds of that given by eqn. (1.12).\n1.19. Stress concentrations- stress concentration factor\n\nIf a bar of uniform cross-section is subjected to an axial tensile or compressive load the\nstress is assumed to be uniform across the section. However, in the presence of any sudden\nchange of section, hole, sharp corner, notch, keyway, material flaw, etc., the local stress will\nrise significantly. The ratio of this stress to the nominal stress at the section in the absence of\nany of these so-called stress concentrations is termed the stress concentration factor.\n1.20. Toughness\n\nToughness is defined as the ability of a material to withstand cracks, i.e. to prevent the\ntransfer or propagation of cracks across its section hence causing failure. Two distinct types\nof toughness mechanism exist and in each case it is appropriate to consider the crack as a very\nhigh local stress concentration.\nThe first type of mechanism relates particularly to ductile materials which are generally\nregarded as tough. This arises because the very high stresses at the end of the crack produce\nlocal yielding of the material and local plastic flow at the crack tip. This has the action of\nblunting the sharp tip of the crack and hence reduces its stressconcentration effect\nconsiderably (Fig. 1.15).\n\n§1.21\n\nSimple Stress and Strain\n\n15\n\nFig. 1.15. Toughness mechanism-type\n\nThe second mechanism refers to fibrous, reinforced or resin-based materials which have\nweak interfaces. Typical examples are glass-fibre reinforced materials and wood. It can be\nshown that a region of local tensile stress always exists at the front of a propagating crack and\nprovided that the adhesive strength of the fibre/resin interface is relatively low (one-fifth the\ncohesive strength of the complete material) this tensile stress opens up the interface and\nproduces a crack sink, i.e. it blunts the crack by effectively increasing the radius at the crack\ntip, thereby reducing the stress-concentration effect (Fig. 1.16).\nThis principle is used on occasions to stop, or at least delay, crack propagation in\nengineering components when a temporary \"repair\" is carried out by drilling a hole at the\nend of a crack, again reducing its stress-concentration effect.\n\nFig. 1.16. Toughness mechanism-type\n\n2.\n\n1.21. Creep and fatigue\nIn the preceding paragraphs it has been suggested that failure of materials occurs when the\nultimate strengths have been exceeded. Reference has also been made in §1.15 to caseswhere\nexcessive deformation, as caused by plastic deformation beyond the yield point, can be\nconsidered as a criterion for effective failure of components. This chapter would not be\ncan fail at stresses much less than the yield stress, namely creep and fatigue.\nCreep is the gradual increase of plastic strain in a material with time at constant load.\nParticularly at elevated temperatures some materials are susceptible to this phenomenon and\neven under the constant load mentioned strains can increase continually until fracture. This\nform of fracture is particularly relevant to turbine blades, nuclear reactors, furnaces, rocket\nmotors, etc.\n\n16\n\nMechanics of Materials\n\n51.21\n\nFracture\n\n/\nHigh stress\n\n/\n\n7\n\nFr oc t ure\n\nor temp/\n\n/\n\nc\n\n?\n\nm\nc\n\nIt-~tial creep\nstmin\n\nL\n\n/\n\n//\n\n//\nLOW\n\nstress\n\nTertiary\ncreep\n\ncreep\n\nI\n\nL\n\nTime\n\nFig. 1.17. Typical creep curve.\n\nThe general form of the strain versus time graph or creep curve is shown in Fig. 1.17 for two\ntypical operating conditions. In each case the curve can be considered to exhibit four\nprincipal features.\n(a) An initial strain, due to the initial application of load. In most cases this would be an\nelastic strain.\n(b) A primary creep region, during which the creep rate (slope of the graph) diminishes.\n(c) A secondary creep region, when the creep rate is sensibly constant.\n(d) A tertiary creep region, during which the creep rate accelerates to final fracture.\nIt is clearly imperative that a material which is susceptible to creep effects should only be\nsubjected to stresses which keep it in the secondary (straight line) region throughout its\nservice life. This enables the amount of creep extension to be estimated and allowed for in\ndesign.\nFatigue is the failure of a material under fluctuating stresses each of which is believed to\nproduce minute amounts of plastic strain. Fatigue is particularly important in components\nsubjected to repeated and often rapid load fluctuations, e.g. aircraft components, turbine\nblades, vehicle suspensions, etc. Fatigue behaviour of materials is usually described by a\nfatigue life or S-N curve in which the number of stress cycles N to produce failure with a\nstress peak of S is plotted against S. A typical S-N curve for mild steel is shown in Fig. 1.18.\nThe particularly relevant feature of this curve is the limiting stress S, since it is assumed that\nstresses below this value will not produce fatigue failure however many cycles are\napplied, i.e. there is injinite life. In the simplest design cases, therefore, there is an aim to keep\nall stresses below this limiting level. However, this often implies an over-design in terms of\nphysical size and material usage, particularly in cases where the stress may only occasionally\nexceed the limiting value noted above. This is, of course, particularly important in\napplications such as aerospace structures where component weight is a premium.\nAdditionally the situation is complicated by the many materials which do not show a defined\nlimit, and modern design procedures therefore rationalise the situation by aiming at a\nprescribed, long, but jinite life, and accept that service stresses will occasionally exceed the\nvalue S,. It is clear that the number of occasions on which the stress exceeds S , , and by how\n\n17\n\nSimple Stress and Strain\n\n\\$1.21\n\nt\n\nt\n\nof load or applied stress with time\n\n5\n\nFalse\nzero\n\nI os\n\nCycles\n\nI o6\n\nto\n\n10’\n\nIO’\n\nfoilure (N)\n\nFig. 1.18. Typical S-N fatigue curve for mild steel.\n\nmuch, will have an important bearing on the prescribed life and considerable specimen, and\noften full-scale,testing is required before sufficient statisticsare available to allow realistic life\nassessment.\nThe importance of the creep and fatigue phenomena cannot be overemphasised and the\ncomments above are only an introduction to the concepts and design philosophiesinvolved.\nFor detailed consideration of these topics and of the other materials testing topics introduced\nearlier the reader is referred to the texts listed at the end of this chapter.\n\nExamples\nExample 1.1\n\nDetermine the stress in each section of the bar shown in Fig. 1.19when subjected to anaxial\ntensile load of 20 kN. The central section is 30 mm square cross-section;the other portions are\nof circular section, their diameters being indicated. What will be the total extension of the\nbar? For the bar material E = 210GN/mZ.\n\n-\n\n20 k N\n\n* 20kN\nr\n\n20\n\nt\n\nI5\n\n30\nNot t o scale\nall dimensions rnrn\n\nFig. 1.19.\n\n18\n\nMechanics of Materials\n\nSolution\n\nforce P\nStress = __ = area\nA\nStress in section (1) =\n\n20 x 103 80 x 103\n~ ( 2 x0 1 0 - ~-) ~ 400 x\n\n= 63.66 MN/m2\n\n4\n\nStress in section (2) =\n\n20 x 103\n30 x 30 x\n\nStress in section (3) =\n\n20 x 103 80 x 103\nd 1 5 x 10-3)2- n x 225 x\n\n= 22.2 MN/m2\n\n= 113.2 MN/m2\n\n4\n\nNow the extension of a bar can always be written in terms of the stress in the bar since\n\nE=-\n\nB\nstress\n=strain 6 / L\n\nBL\n\n6=-\n\ni.e.\n\nE\n\nextension of section (1) = 63.66 x lo6 x\nextension of section (2) = 22.2 x lo6 x\n\nio0 x 10-3\n= 10.6 x 10-6m\n210 x 109\n\nextension of section (3) = 113.2 x lo6 x\n\n..\n\n250 x 10-3\n= 75.8 x 10-6m\n210 x 109\n\n400 x 10-3\n\n210 x 109\n\n= 215.6 x\n\nm\n\n+\n\ntotal extension = (75.8 10.6 + 215.6)10-6\n= 302 x\nm\n= 0.302mm\n\nExample 1.2\n(a) A 25 mm diameter bar is subjected to an axial tensile load of 100kN. Under the action\nof this load a 200mm gauge length is found to extend 0.19 x 10-3mm. Determine the\nmodulus of elasticity for the bar material.\n(b) If, in order to reduce weight whilst keeping the external diameter constant, the bar is\nbored axially to produce a cylinder of uniform thickness, what is the maximum diameter of\nbore possible given that the maximum allowable stress is 240MN/m2? The load can be\nassumed to remain constant at 100kN.\n(c) What will be the change in the outside diameter of the bar under the limiting stress\nquoted in (b)? (E = 210GN/m2and v = 0.3).\n\n19\n\nSimple Stress and Strain\nSolution\n\n(a) From eqn. (1.2),\nPL\nYoung’s modulus E = A6L\n100 x 103 x 200 x 10-3\n\n= 214 GN/mZ\n\n(b) Let the required bore diameter bed mm; the cross-sectional area of the bar will then be\nreduced to\n\n..\n\nP\n4x100~10~\nstress in bar = - =\nA 425’ -d2)10-6\n\nBut this stress is restricted to a maximum allowable value of 240 MN/m2.\n\n..\n\n240 x 106 =\n\n..\n\n252 -d2 =\n\n..\n\n4 x 100 x 103\n~ ( 2 5 ’- dZ)10-6\n4 x 100 x 103\n= 530.5\n240 x 106 x It x 10-6\n\nd = 94.48\n\nand d = 9.72 mm\n\nThe maximum bore possible is thus 9.72 mm.\n(c) The change in the outside diameter of the bar will be obtained from the lateral strain,\ni.e.\nBut\nand\n..\n\n..\n\n6d\nlateral strain = d\nPoisson’s ratio v\nlongitudinal strain\n\n=\n\nlateral strain\nlongitudinal strain\n0\n\n=- =\n\nE\n\n240x 106\n210 x 109\n\n6d _\n- - v - =0\nd\n\nE\n\nchange in outside diameter = =\n\n- 0.3 x 240 x lo6\n210 x 109\n\n0.3 x 240 x lo6\n210 x 109\n\n- 8.57 x\n\n25\n\nm (a reduction)\n\n20\n\nMechanics of Materials\n\nExample 1.3\n\nThe coupling shown in Fig. 1.20is constructed from steel of rectangular cross-section and\nis designed to transmit a tensile force of 50 kN. If the bolt is of 15 mm diameter calculate:\n(a) the shear stress in the bolt;\n(b) the direct stress in the plate;\n(c) the direct stress in the forked end of the coupling.\n\n50 kN\n\n50 rnrn\n\n50{:;\n\n6 mrn\n\n6 mrn\n\n6 mrn\n\nShear s y s t e m\non bolt\n\nFig. 1.20.\n\nSolution\n(a) The bolt is subjected to double shear, tending to shear it as shown in Fig. 1.14b.There is\nthus twice the area of the bolt resisting the shear and from eqn. (1.8)\nP\nshear stress in bolt = - -\n\n2A\n\n-\n\n50 x 103 x 4\n\n- 2 x a(15 x 10-3)2\n\nio0 x 103\n= 141.5MN/mZ\nn(i5 x 10-3)~\n\n(b) The plate will be subjected to a direct tensile stress given by\nP\na=-=\n\nA\n\n50 x 103\n= 166.7MN/mZ\n50x6~10-~\n\n(c) The force in the coupling is shared by the forked end pieces, each being subjected to a\ndirect stress\nP\n25 x 103\na=-=\n= 83.3MN/m2\nA 50x6~10-~\n\nExample 1.4\n\nDerive an expression for the total extension of the tapered bar of circular cross-section\nshown in Fig. 1.21 when it is subjected to an axial tensile load W.\n\n21\n\nSimple Stress and Strain\n\nFig. 1.21.\n\nSolution\n\nFrom the proportions of Fig. 1.21,\nd/2- (D-d)/2\n_\n\nL\n\nLO\n\nConsider an element of thickness d x and radius r, distance x from the point of taper A .\nW\nStress on the element = nr2\nr\nd\n- =But\nx\n2L0\nr =d\n\n..\n\n(z)\n\nx(D-d)\n2L\n\nx=\n\n~\n\n4WL2\n\nstress on the element =\n\nn ( D - d)’X2\na\nstrain on the element = E\nand extension of the element = E\n4WL2\ndx\nn ( D -d)’X2E\n\ns\n\nL,+L\n\n..\n\ntotal extension of bar =\n\n-\n\nn ( 4Dw-d)’E\nL 2 dx x2\n\nLO\n\n4 WL2\n\n-n(D-d)’E[-fI4\n- 4 WL2\nn ( D - d)’E\n\n[\n\nLo+L\n\n+\n\n(Lo L)\n\n22\n\nMechanics of Materials\n\nBut\n\n..\n\nL,+L=-\n\n+\n\nd\n\n(d D - d ) L=-----DL\n(D-d)L+L=\nD-d\n(D-d)\n\n... total extension\n-\n\n4WL\n(-d+D)\nn(D-d)E[\nDd\n\n]\n\n4WL\nnDdE\n\n=-\n\nExample 1.5\nThe following figures were obtained in a standard tensile test on a specimen of low carbon\nsteel:\ndiameter of specimen, 11.28mm;\ngauge length, 56mm;\nminimum diameter after fracture, 6.45 mm.\nUsing the above information and the table of results below, produce:\n(1) a load/extension graph over the complete test range;\n(2) a load/extension graph to an enlarged scale over the elastic range of the specimen.\n\nExtension\n(m x\n\n2.47\n5.6\n\n11.9\n\n18.2\n\nExtension\n(mx\n\n27.13\n\n29.6\n\n32.1\n\n73.5\n\n81.2\n\n89.6\n\nExtension\n(m~lO-~)\n\n35.8\n\n37\n\n38.7\n\n39.5\n\n1960\n\n2520\n\n3640\n\n5600\n\n4.91\n\n1.4\n\n9.86\n\n12.33\n\n14.8\n\n17.27\n\n19.14 22.2\n\n24.7\n\n24.5\n\n31.5\n\n38.5\n\n45.5\n\n52.5\n\n59.5\n\n66.5\n\n33.3\n\n31.2\n\n32\n\n31.5\n\n32\n\n32.2\n\n34.5\n\n840\n\n1120\n\n1680\n\n112\n\n224\n\n448\n\n672\n\n40\n\n39.6\n\n35.7\n\n7840\n\n11200\n\n13440\n\n28\n14560\n\nUsing the two graphs and other information supplied, determine the values of\n(a) Young's modulus of elasticity;\n(b) the ultimate tensile stress;\n(c) the stress at the upper and lower yield points;\n(d) the percentage reduction of area;\n(e) the percentage elongation;\n(f) the nominal and actual stress at fracture.\n\nSimple Stress and Strain\nSolution\n\nExtension\n\n(m x\n\nFig. 1.22. Load-extension graph for elastic range.\n\nYoung’s modulus E = - = -X\nE\narea\nextension\n\n-\n\nextension\n\nX\n\ngauge length\narea\n\nL\n56 x 10-3\nE = slope of graph x - = 3.636 x lo8 x\nA\n100 x 10-6\n\ni.e\n\n= 203.6 x\n\n..\n\nlo9N/mZ\n\nE = 203.6 GN/m2\n(b)\n\nUltimate tensile stress =\n\nmaximum load - 40.2 x lo3\n= 402 MN/m2\ncross-section area 100 x\n\n(see Fig. 1.23).\nUpper yield stress =\n\n33.3 x 103\n100 x 10-6\n\nLower yield stress =\n\n31.2 x 103\n= 312 MN/m2\n100 x 10-6\n\n= 333 MN/m2\n\n23\n\n24\n\nMechanics of Materials\n\n10000\n\n5000\n\nExtension\n\n15000\n\nm (lo6)\n\n- (11.282 -6.45’)\n11.28’\n\nPercentage elongation =\n\n= 67.3%\n\n(70.56 - 56)\n56\n\n= 26%\n\nNominal stress at fracture =\n\n28 x 103\n= 280 MN/mZ\nloo x 10-6\n\nActual stress at fracture = rr\n\n28\n\nlo3\n\n4(6.45)2x\n\n= 856.9MN/mZ\n\nSimple Stress and Strain\n\n25\n\nProblems\n1.1 (A). A 25mm squarecross-section bar of length 300mm carries an axial compressive load of 50kN.\nDetermine the stress set up ip the bar and its change of length when the load is applied. For the bar material\n[80 MN/m2; 0.12mm.l\nE = 200 GN/m2.\n\n1.2 (A). A steel tube, 25 mm outside diameter and 12mm inside diameter, cames an axial tensile load of 40 kN.\nWhat will be the stress in the bar? What further increase in load is possible if the stress in the bar is limited to\n[lo6 MN/m3; 45 kN.1\n225 MN/mZ?\n1.3 (A). Define the terms shear stress and shear strain, illustrating your answer by means of a simple sketch.\nTwo circular bars, one of brass and the other of steel, are to be loaded by a shear load of 30 kN. Determine the\nnecessary diameter of the bars (a)in single shear, (b) in double shear, if the shear stress in the two materials must not\nC27.6, 19.5, 19.5, 13.8mm.l\nexceed 50 MN/m2 and 100MN/mZ respectively.\n\n1.4 (A). Two forkend pieces are to be joined together by a single steel pin of 25mm diameter and they are\nrequired to transmit 50 kN. Determine the minimum cross-sectional area of material required in one branch of either\nfork if the stress in the fork material is not to exceed 180 MN/m2. What will be the maximum shear stress in the pin?\nC1.39 x 10e4mZ;50.9MN/mZ.]\n1.5 (A). A simple turnbuckle arrangement is constructed from a 40 mm outside diameter tube threaded internally\nat each end to take two rods of 25 mm outside diameter with threaded ends. What will be the nominal stresses set up\nin the tube and the rods, ignoring thread depth, when the turnbuckle cames an axial load of 30 kN? Assuming a\nsufficient strength of thread, what maximum load can be transmitted by the turnbuckle if the maximum stress is\nlimited to 180 MN/mz?\nC39.2, 61.1 MN/m2, 88.4 kN.1\n1.6 (A). An I-seetion girder is constructed from two 80mm x 12mm flanges joined by an 80mm x 12mm web.\nFour such girders are mounted vertically one at each corner of a horizontal platform which the girders support. The\nplatform is 4 m above ground level and weighs 10kN. Assuming that each girder supports an equal share of the load,\ndetermine the maximum compressive stress set up in the material of each girder when the platform supports an\nadditional load of 15kN. The weight of the girders may not be neglected. The density of the cast iron from which the\ngirders are constructed is 7470 kg/m3.\nC2.46 MN/mZ.]\n\n1.7 (A). A bar ABCD consists of three sections:AB is 25 mm square and 50 mm long, BC is of 20 mm diameter and\n40 mm long and CD is of 12mm diameter and 50 mm long.Determine the stress set up in each section of the bar when\nit is subjected to an axial tensile load of 20 kN. What will be the total extension of the bar under this load? For the bar\nmaterial, E = 210GN/m2.\n[32,63.7, 176.8 MN/mZ,0.062mrn.l\n1.8 (A). A steel bar ABCD consists of three sections: AB is of 20mm diameter and 200 mm long, BC is 25 mm\nsquare and 400mm long, and CD is of 12mm diameter and 200mm long. The bar is subjected to an axial compressive\nload which induces a stress of 30 MN/m2 on the largest cross-section. Determine the total decrease in the length of\nthe bar when the load is applied. For steel E = 210GN/m2.\nC0.272 mm.]\n1.9 (A). During a tensile test on a specimen the following results were obtained:\n\nExtension (mm)\n\n15\n0.05\n\n30\n0.094\n\n40\n0.127\n\n50\n0.157\n\n55\n\nExtension (mm)\n\n70\n5.08\n\n75\n7.62\n\n80\n12.7\n\n82\n16.0\n\n80\n19.05\n\n1.778\n\n60\n2.79\n\n65\n3.81\n70\n22.9\n\nDiameter of gauge length = 19mm\nGauge length = l00mm\nDiameter at fracture\nGauge length at fracture = 121 mm\n= 16.49mm\nPlot the complete load extension graph and the straight line portion to an enlarged scale. Hence determine:\n(a) the modulus of elasticity;\n(b) the percentage elongation;\n(c) the percentage reduction in area;\n\n(d) the nominal stress at fracture;\n(e) the actual stress at fracture;\n(f) the tensile strength.\n[116 GN/m2; 21 %; 24.7 %; 247 MN/m2; 328 MN/m2; 289 MN/mZ.]\n1.10 Figure 1.24 shows a special spanner used to tighten screwed components. A torque is applied at the tommybar and is transmitted to the pins which engage into holes located into the end of a screwed component.\n(a) Using the data given in Fig. 1.24 calculate:\n(i) the diameter D of the shank if the shear stress is not to exceed 50N/mm2,\n(u) the stress due to bending in the tommy-bar,\n(iii) the shear stress in the pins.\n(b) Why is the tommy-bar a preferred method of applying the torque?\n[C.G.] [9.14mm; 254.6 MN/m2; 39.8 MN/mZ.]\n\n26\n\nMechanics of Materials\n50N force opplied a t o distonce\nof 25mm from both ends of\ntomrny bor\n\nFig. 1.24.\n1.11 (a) A test piece is cut from a brass bar and subjected to a tensile test. With a load of 6.4 k N the test piece, of\ndiameter 11.28mm, extends by 0.04 mm over a gauge length of 50 mm. Determine:\n(i) the stress, (ii) the strain, (hi) the modulus of elasticity.\n(b) A spacer is turned from the same bar. The spacer has a diameter of 28 mm and a length of 250mm. both\nmeasurements being made at 20°C. The temperature of the spacer is then increased to 100°C,the natural expansion\nbeing entirely prevented. Taking the coefficientof linear expansion to be 18 x 10-6/”C determine:\n(i) the stress in the spacer, (ii) the compressive load on the spacer.\n[C.G.] [64MN/m2, 0.0008, 80GN/m2, 115.2 MN/m2, 71 kN.]\n\nBibliography\n1.\n2.\n3.\n4.\n5.\n\nJ. G. Tweedale, Mechanical Properties of Metal, George Allen & Unwin Ltd., 1964,\nE. N. Simons, The Testing of Metals, David & Charles, Newton Abbot, 1972.\nJ. Y. Mann, Fatigue of Materials- An Introductory Text, Melbourne University Press, 1967.\nP. G. Forrest, Fatigue of Metals, Pergamon, 1970.\nR. B. Heywood, Designing against Fatigue, Chapman & Hall, 1962.\n6. Fatigue- An Interdisciplinary Approach, 10th Sagamore Army Materials Research Conference Proceedings,\nSyracuse University Press, 1964.\nI. A. J. Kennedy, Processes of Creep and Fatigue in Metals, Oliver & Boyd, Edinburgh and London, 1962.\n8. R. K. Penny and D. L. Marriott, Design for Creep, McGraw-Hill (U.K.), 1971.\n9. A. I. Smith and A. M. Nicolson, Advances in Creep Design, Applied Science Publishers, London, 1971.\n10. J. F. Knott, Fundamentals of Fracture Mechanics, Butterworths, London, 1973.\n11. H. Liebowitz, Fracture-An Advanced Treatise, vols. 1 to 7, Academic Press, New York and London, 1972.\n12. W. D. Biggs, The Brittle Fracture of Steel, MacDonald & Evans Ltd., 1960.\n13. D. Broek, Elementary Engineering Fracture Mechanics, Noordhoff International Publishing, Holland, 1974.\n14. J. E. Gordon, The New Science of Strong Materials, Pelican 213, Penguin, 1970.\n15. R. J. Roark and W. C. Young, Formulas for Stress and Strain, 5th Edition, McGraw-Hill, 1975.\n\nCHAPTER 2\n\nCOMPOUND BARS\nSummary\nWhen a compound bar is constructed from members of different materials, lengths and\nareas and is subjected to an external tensile or compressive load W the load carried by any\nsingle member is given by\nElAl\nF1 =- L1\nc-EA\nL\n\nw\n\nEA\nwhere suffix 1refers to the single member and X - is the sum of all such quantities for all the\nL\nmembers.\nWhere the bars have a common length the compound bar can be reduced to a single\nequivalent bar with an equivalent Young’s modulus, termed a combined E.\n\nC EA\nCombined E = CA\nThe free expansion of a bar under a temperature change from Tl to T2 is\na(T2 -T1)L\n\nwhere a is the coefficient of linear expansion and L is the length of the bar.\nIf this expansion is prevented a stress will be induced in the bar given by\na(T2 -T#\n\nTo determine the stresses in a compound bar composed of two members of different free\nlengths two principles are used:\n(1) The tensile force applied to the short member by the long member is equal in magnitude\nto the compressive force applied to the long member by the short member.\n(2) The extension of the short member plus the contraction of the long member equals the\ndifference in free lengths.\n\nThis difference in free lengths may result from the tightening of a nut or from a temperature\nchange in two members of different material (Le. different coefficients of expansion) but of\nequal length initially.\nIf such a bar is then subjected to an additional external load the resultant stresses may be\nobtained by using the principle ofsuperposition. With this method the stresses in the members\n27\n\n28\n\nMechanics of Materials\n\n\\$2.1\n\narising from the separate effects are obtained and the results added, taking account of sign, to\ngive the resultant stresses.\nN.B.: Discussion in this chapter is concerned with compound bars which are symmetrically proportioned such that no bending results.\n\n2.1. Compound bars subjected to external load\nIn certain applications it is necessary to use a combination of elements or bars made from\ndifferent materials, each material performing a different function. In overhead electric cables,\nfor example, it is often convenient to carry the current in a set of copper wires surrounding\nsteel wires, the latter being designed to support the weight of the cable over large spans. Such\ncombinations of materials are generally termed compound burs. Discussion in this chapter is\nconcerned with compound bars which are symmetrically proportioned such that no bending\nresults.\nWhen an external load is applied to such a compound bar it is shared between the\nindividual component materials in proportions depending on their respective lengths, areas\nand Young’s moduli.\nConsider, therefore, a compound bar consisting of n members, each having a different\nlength and cross-sectional area and each being of a different material; this is shown\ndiagrammatically in Fig. 2.1. Let all members have a common extension x, i.e. the load is\npositioned to produce the same extension in each member.\n\nn’’mmernber\n\nFirst member\n\nModulus E ,\n\n7\n\nLength L,\nArea An\nModulus E,\n\nT’\n\nF r n m o n extension x\n\nW\n\nFig. 2.1. Diagrammatic representation of a compound bar formed of\ndifferent materials with different lengths, cross-sectional areas and\nYoung’s moduli.\n\nFor the nth member,\nstress -E,\nstrain\n\nLn\nAnxn\n\nf‘n\n\n=-\n\nwhere F, is the force in the nth member and A, and Lnare its cross-sectional area and length.\n\n52.2\n\n29\n\nCompound Bars\n\nThe total load carried will be the sum of all such loads for all the members\ni.e.\nNow from eqn. (2.1) the force in member 1 is given by\n\nBut, from eqn. (2.2),\n\n..\n\nF1=- L 1\n\nc-ELA\n\nw\n\n(2.3)\n\ni.e. each member carries a portion of the total load W proportional to its EAIL value.\nIf the wires are all of equal length the above equation reduces to\n\nThe stress in member 1 is then given by\n\n2.2. Compound bars - “equivalent” or “combined” modulus\n\nIn order to determine the common extension of a compound bar it is convenient to\nconsider it as a single bar of an imaginary material with an equivalent or combined modulus\nE,. Here it is necessary to assume that both the extension and the original lengths of the\nindividual members of the compound bar are the same; the strains in all members will then be\nequal.\nNow total load on compound bar = F 1 F z + F 3 + . . . + F , where F1, F2, etc., are the\nloads in members 1, 2, etc.\nforce = stress x area\nBut\n..\na ( A l + A z + . . . + A n ) = ~ l A l + ~ z A z .+. . + a , A ,\n\n+\n\nwhere 6 is the stress in the equivalent single bar.\nDividing through by the common strain E,\nd\n\n-(Al\nE\n\ni.e.\n\n+ Az + . . . + A n ) = - A l + :AZ + . . . + -o An ,\n6 1\n\nE\n\nE\n\nE c ( A i + A z + . . . + A n ) = E I A 1 + E z A z + . . . +E,An\n\nwhere E, is the equivalent or combined E of the single bar.\n\n30\n\nMechanics of Materials\n\n..\n\ncombined E = E i A i + E , A Z + . . . +E,An\nAl+A2+ . .. +An\n\n42.3\n\nXEA\nE, = EA\n\ni.e.\n\nWith an external load W applied,\nstress in the equivalent bar =\n\nW\nXA\n\n~\n\nand\n\nw\n\nx\n\nstrain in the equivalent bar = -= EJA\nL\n\n.’.\n\nsince\n\nstress\n--E\nstrain\n\nWL\nE,XA\n\ncommon extension x = = extension of single bar\n\n2.3. Compound bars subjected to temperature change\n\nWhen a material is subjected to a change in temperature its length will change by an\namount\naLt\nwhere a is the coefficient of linear expansion for the material, L is the original length and t the\ntemperature change. (An increase in temperature produces an increase in length and a\ndecrease in temperature a decrease in length except in very special cases of materials with zero\nor negative coefficients of expansion which need not be considered here.)\nIf, however, the free expansion of the material is prevented by some external force, then a\nstress is set up in the material. This stress is equal in magnitude to that which would be\nproduced in the bar by initially allowing the free change of length and then applying sufficient\nforce to return the bar to its original length.\nNow\nchange in length = aLt\n\n..\n\n.\naLt\nstrain = -= at\nL\n\nTherefore, the stress created in the material by the application of sufficient force to remove\nthis strain\n= strain x E\n= Eat\n\nConsider now a compound bar constructed from two different materials rigidly joined\ntogether as shown in Fig. 2.2 and Fig. 2.3(a). For simplicity of description consider that the\nmaterials in this case are steel and brass.\n\n§2.3\n\nCompound\n\nFig.\n\nBars\n\n31\n\n2.2.\n\nIn general, the coefficients of expansion of the two materials forming the compound bar\nwill be different so that as the temperature rises each material will attempt to expand by\ndifferent amounts. Figure 2.3b shows the positions to which the individual materials will\nextend if they are completely free to expand (i.e. not joined rigidly together as a compound\nbar). The extension of any length L is given by\ncxLt\n\nFig. 2.3. Thermal expansion of compound bar.\n\nThus the difference of \"free\" expansion lengths or so-called free lengths\n= IXBLt-IXsLt = (IXB-IXs)Lt\nsince in this casethe coefficient of expansion of the brass IXBisgreater than that for the steellXs'\nThe initial lengths L of the two materials are assumed equal.\nIf the two materials are now rigidly joined as a compound bar and subjected to the same\ntemperature rise, each material will attempt to expand to its free length position but each will\nbe affected by the movement of the other, The higher coefficient of expansion material (brass)\nwill therefore seek to pull the steel up to its free length position and conversely the lower\n\n32\n\nMechanics of Materials\n\n\\$2.4\n\ncoefficientof expansion material (steel)will try to hold the brass back to the steel “free length”\nposition. In practice a compromise is reached, the compound bar extending to the position\nshown in Fig. 2.3c, resulting in an effective compression of the brass from its free length\nposition and an effectiveextension of the steel from its free length position. From the diagram\nit will be seen that the following rule holds.\nRule 1.\nExtension of steel + compression of brass = dixerence in “free” lengths.\nReferring to the bars in their free expanded positions the rule may be written as\nExtension of “short” member compression of“1ong” member = dixerence infree lengths.\nApplying Newton’s law of equal action and reaction the following second rule also applies.\nRule 2.\nThe tensile force applied to the short member by the long member is equal in magnitude to\nthe compressive force applied to the long member by the short member.\nThus, in this case,\ntensile force in steel = compressive force in brass\n\n+\n\nNow, from the definition of Young’s modulus\n\nstress - o\nE=-strain 6 / L\nwhere 6 is the change in length.\n\n..\nAlso\n\nOL\nE\n\n6=-\n\nforce = stress x area = OA\n\nwhere A is the cross-sectional area.\nTherefore Rule 1 becomes\na,L tl L\n-++=(ag-as)Lt\nE,\nE,\nand Rule 2 becomes\n@,A, = c B A B\nWe thus have two equations with two unknowns osand oBand it is possible to evaluate the\nmagnitudes of these stresses (see Example 2.2).\n\n2.4. Compound bar (tube and rod)\n\nConsider now the case of a hollow tube with washers or endplates at each end and a central\nthreaded rod as shown in Fig. 2.4. At first sight there would seem to be no connection with the\nwork of the previous section, yet, in fact, the method of solution to determine the stresses set\nup in the tube and rod when one nut is tightened is identical to that described in \\$2.3.\nThe compound bar which is formed after assembly of the tube and rod, i.e. with the nuts\ntightened, is shown in Fig. 2.4c, the rod being in a state of tension and the tube in\ncompression. Once again Rule 2 applies, i.e.\ncompressive force in tube = tensile force in rod\n\n33\n\nCompound Bars\n\n\\$2.4\n\n7\n\nI\nI\n\nDifference in free lengths\n= distance moved by nut\n\nCompression of tube\n\n4\n\nI\n\n+€Extension\n\nof rod\n\nFig. 2.4. Equivalent “mechanical” system to that of Fig. 2.3.\n\nFigure 2.4a and b show, diagrammatically, the effectivepositions of the tube and rod before\nthe nut is tightened and the two components are combined. As the nut is turned there is a\nsimultaneous compression of the tube and tension of the rod leading to the final state shown\nin Fig. 2 . 4 ~As\n. before, however, the diagram shows that Rule 1 applies:\ncompression of tube +extension of rod\n\n= difference in\n\nfree lengths = axial advance of nut\n\ni.e. the axial movement of the nut ( = number of turns n x threads per metre) is taken up by\ncombined compression of the tube and extension of the rod.\nThus, with suffix t for tube and R for rod,\nOIL d L\n(2.10)\n- + R = n x threads/metre\nEl\n\nalso\nuRA, = a,A,\n(2.1 1)\nIf the tube and rod are now subjected to a change of temperature they may be treated as a\nnormal compound bar of \\$2.3 and Rules 1 and 2 again apply (Fig. 2.5),\n(2.12)\n\ni.e.\nDifference in\nfree lengths\n\n( a ) Free independent\nexpansion\nCompression of tube\nExtension of rod\n( b ) Cornpouna bar\nexpansoon\n\nt--GGz\n\nI\n\nd\n\nFig. 2.5.\n\n34\n\nMechanics of Materials\n\n§2.5\n\nwhere a;and 0; are the stresses in the tube and rod due to temperature change only and a, is\nassumed greater than aR. If the latter is not the case the two terms inside the final bracket\nshould be interchanged.\nAlso\na iA , = 0;A,\n2.5. Compound bars subjected to external load and temperature effects\nIn this case the principle ofsuperposition must be applied, i.e. provided that stresses remain\nwithin the elastic limit the effects of external load and temperature change may be assessed\nseparately as described in the previous sectionsand the results added, taking account of sign,\nto determine the resultant total effect;\ni.e.\n\ntotal strain = sum of strain due to external loads and temperature strain\n\n2.6. Compound thick cylinders subjected to temperature changes\nThe procedure described in \\$2.3 has been applied to compound cylinders constructed\nfrom tubes of different materials on page 230.\n\nExamples\nExample 2.1\n(a) A compound bar consists of four brass wires of 2.5 mm diameter and one steel wire of\n1.5 mm diameter. Determine the stresses in each of the wires when the bar supports a load of\n500 N. Assume all of the wires are of equal lengths.\n(b) Calculate the “equivalent” or “combined modulus for the compound bar and\ndetermine its total extension if it is initially 0.75 m long. Hence check the values of the stresses\nobtained in part (a).\nFor brass E = 100 GN/m’ and for steel E = 200 GN/m’.\nSolution\n\n(a) From eqn. (2.3) the force in the steel wire is given by\n\n200 x 109 x 2 x 1.52 x 10-6\n+4(1OO x lo9 x 2 x 2.5’ x\n= 200 x lo9 x 2 x 1.5’ x\n\n=\n\n[\n[\n\n2 x 1.5’\n(2 x 1.5’)+ (4 x 2.5’)\n\n1\n\n500 = 76.27 N\n\n35\n\nCompound Bars\n\n..\n\ntotal force in brass wires = 500 - 76.27 = 423.73 N\n\n..\n\n76.27\nstress in steel = -=\narea f x 1S2x\n\nand\n\n423.73\nstress in brass = -=\narea 4 x f x 2S2 x\n\n= 43.2 MN/m2\n\n= 21.6 MN/m2\n\n(b) From eqn. (2.6)\nC E A 200 x 109 x f x 1.52 x io-6+4(100 x 109 x 5 x 2S2 x 10-6)\ncombined E = -f(1S2 + 4 x 2.52)'0-6\nEA\n\nE = - stress\nstrain\n\nNow\n\nand the stress in the equivalent bar\n500\n=--\n\nZA\n\n..\n\n-\n\n500\n\n- 23.36 MN/m2\n\n\\$ ( 1 S 2 + 4 x 2.52)10-6 -\n\nstress\n23.36 x lo6\n= 0.216 x\nstrain in the equivalent bar = -=\nE\n108.26 x lo9\n\n..\n\ncommon extension = strain x original length\nx 0.75 = 0.162 x\n= 0.216 x\n= 0.162 mm\n\nThis is also the extension of any single bar, giving a strain in any bar\n\n-\n\n..\n\n0.162 x lo-'\n= 0.216 x\n0.75\n\nstress in steel = strain x E, = 0.216 x\n= 43.2 MN/m2\nstress in brass = strain x E, = 0.216 x\n= 21.6 MN/m2\n\nand\n\nas above\nx 200 x\n\nlo9\n\nx 100 x\n\nlo9\n\nThese are the same values as obtained in part (a).\nExample 2.2\n\n(a) A compound bar is constructed from three bars 50 mm wide by 12 mm thick fastened\ntogether to form a bar 50 mm wide by 36 mm thick. The middle bar is of aluminium alloy for\nwhich E = 70 GN/m2 and the outside bars are of brass with E = 100 GN/m2. If the bars are\ninitially fastened at 18°C and the temperature of the whole assembly is then raised to 5WC,\ndetermine the stresses set up in the brass and the aluminium.\nctg =\n\n18 x\n\nper \"C and u,., = 22 x\n\nper \"C\n\n36\n\nMechanics of Materials\n\n(b) What will be the changes in these stresses if an external compressive load of 15 kN is\napplied to the compound bar at the higher temperature?\nSolution\nWith any problem of this type it is convenient to let the stress in one of the component\nmembers or materials, e.g. the brass, be x.\nThen, since\nforce in brass = force in aluminium\nand\n\nforce = stress x area\nx x 2 x 50 x 12 x loF6= oAx 50 x 12 x\n\ni.e.\n\nstress in aluminium oA= 2x\n\nNow, from eqn. (2.Q\nextension of brass + compression of aluminium = difference in free lengths\n= (a” -ail)\nX L\n\n100 x 109\n\n2xL\n+ ___\n= (2270 x 109\n\n(Tz-TdL\n\n18)10-6(50- 18)L\n\n(7x + 2ox) = 4 x 10-6 x 32\n\n7 0 0 109\n~\n27x = 4 x l o v 6x 32 x 700 x lo9\nx = 3.32 MN/m2\n\nThe stress in the brass is thus 3.32 MN/m2 (tensile) and the stress in the aluminium is\n2 x 3.32 = 6.64 MN/mz (compressive).\n(b) With an external load of 15 kN applied each member will take a proportion of the total\nForce in aluminium = EAAA w\nCEA\n~\n\n=\n\n[\n\n70 x lo9 x 50 x 12 x\n\n+\n\n(70 x 50 x 12 2 x 100 x 50 x 12)109 x 10- 6\n\n= 3.89 kN\n\n..\n\nforce in brass = 15-3.89 = 11.11 kN\n\n..\n\n11.11 x 103\nstress in brass = -=\narea 2 x 50 x 12 x\n= 9.26 MN/m2 (compressive)\n\n115 x 103\n\n37\n\nCompound Bars\nStress in aluminium = area 50 x 12 x\n= 6.5 MN/m2 (compressive)\n\nThese stresses represent the changes in the stresses owing to the applied load. The total or\nresultant stresses owing to combined applied loading plus temperature effects are, therefore,\nstress in aluminium = - 6.64 - 6.5 = - 13.14 MN/m2\n= 13.14 MN/m2 (compressive)\n\nstress in brass =\n\n+ 3.32 - 9.26 = - 5.94 MN/m2\n\n= 5.94 MN/m2 (compressive)\n\nExample 2.3\nA 25 mm diameter steel rod passes concentrically through a bronze tube 400 mm long,\n50 mm external diameter and 40 mm internal diameter.The ends of the steel rod are threaded\nand provided with nuts and washers which are adjusted initially so that there is no end play at\n20°C.\n\n(a) Assuming that there is no change in the thickness of the washers, find the stress\nproduced in the steel and bronze when one of the nuts is tightened by giving it onetenth of a turn, the pitch of the thread being 2.5 mm.\n(b) If the temperature of the steel and bronze is then raised to 50°Cfind the changes that\nwill occur in the stresses in both materials.\nThe coefficient of linear expansion per \"C is 11 x\nfor steel and 18 x\nE for steel = 200 GN/m2. E for bronze = 100 GN/m2.\n\nfor bronze.\n\nSolution\n\n(a) Let x be the stress in the tube resulting from the tightening of the nut and o Rthe stress\nin the rod.\nThen, from eqn. (2.11),\nforce (stress x area) in tube = force (stress x area) in rod\nx x :(502 -402)10-6 = O R x\nOR\n\n=\n\n2 x 25'\n\n(502- 402)\n252\n\nx\n\nlow6\n\nx = 1.44x\n\n+ extension of rod = axial advance of nut, from eqn. (2.10),\nx x 400 x 10-3\nxx400 x 10-3 - x2.5 x\n+ i . ~200\n100 x 109\nx 109\n10\n\nAnd since compression of tube\n\n400\n\n..\n\n+\n\n(2x 1.44x)\n10-3 = 2.5 x 10-4\n200 x 109\n6 . 8 8 ~= 2.5 x 10'\nx = 36.3 MN/m2\n\nMechanics of Materials\n\n38\n\nThe stress in the tube is thus 36.3 MN/m2 (compressive) and the stress in the rod is\n1.44 x 36.3 = 52.3 MN/m2 (tensile).\n(b) Let p be the stress in the tube resulting from temperature change. The relationship\nbetween the stresses in the tube and the rod will remain as in part (a) so that the stress in the\nrod is then 1 . 4 4 ~In\n. this case, if free expansion were allowed in the independent members, the\nbronze tube would expand more than the steel rod and from eqn. (2.8)\ncompression of tube +extension of rod\n\n..\n\n100p L109\n\n= difference\n\n1.44pL\n+ 200\n=\nx 109\n\nin free length\n\n(aB-@s)(T2\n\n-TdL\n\n3.44p = 7 x 10-6 x 30 x 200 x 109\np = 12.21 MN/mZ\n= 17.6 MN/m2\n1.44~\n\nand\n\nThe changes in the stresses resulting from the temperature effects are thus 12.2 MN/m2\n(compressive) in the tube and 17.6 MN/m2 (tensile) in the rod.\nThe final, resultant, stresses are thus:\nstress in tube = - 36.3 - 12.2 = 48.5 MN/mZ (compressive)\nstress in rod = 52.3 17.6 = 69.9 MN/mZ (tensile)\n\n+\n\nExample 2.4\nA composite bar is constructed from a steel rod of 25 mm diameter surrounded by a copper\ntube of 50 mm outside diameter and 25 mm inside diameter. The rod and tube are joined by\ntwo 20 mm diameter pins as shown in Fig. 2.6. Find the shear stress set up in the pins if, after\npinning, the temperature is raised by 50°C.\n\nFor steel E = 210 GN/m2 and a = 11 x\nFor copper E\n\n=\n\n105 GN/m2 and a = 17 x\nCopper\n\nper \"C.\nper \"C.\n\nSteel\n\nFig. 2.6.\n\nSolution\n\nIn this case the copper attempts to expand more than the steel, thus tending to shear the\npins joining the two.\n\n39\n\nCompound Bars\nLet the stress set up in the steel be x, then, since\nforce in steel = force in copper\nx x \\$ x 25’ x\n\ni.e.\n\n= O, x\n\nstress in copper O, =\n\n4 (50’ - 25’)\n\nX\nx x 25’\n= 0 . 3 3 3 ~= (50’ - 25’)\n3\n\nNow the extension of the steel from its freely expanded length to its forced length in the\ncompound bar is given by\nXL\n\nOL\n_\n-\n\nE\n210x 109\nwhere L is the original length.\nSimilarly,the compressionof the copper from its freely expanded position to its position in\nthe compound bar is given by\nL\n\n-aL\n--x x\n\nE\n\n3\n\nio5 x 109\n\nNow the extension of steel +compression of copper\n= differencein\n= (Mz - a m ,\n\n..\n\nXL\n\nXL\n\n210 x 109 + 3 x 105 x 109\n\n-T,)L\n\n= (17 - 11)10-6 x 50 x L\n\n+\n\n3x 2x\n=6x\n6 x 105 x lo9\n\nx 50\n\n5x = 6 x\nx = 37.8 x\n\n..\n\n“free” lengths\n\nx 50 x 6 x 105 x\n\nlo9\n\nlo6 = 37.8 MN/m’\n\nload carried by the steel = stress x area\n= 37.8 x\n=\n\nlo6 x \\$ x 25’\n\nx\n\n18.56 kN\n\nThe pins will be in a state of double shear (see Ql.lS), the shear stress set up being given by\n7=-\n\n2 x area\n\n-\n\n18.56 x lo3\n2 x \\$ x 20’ x\n\n= 29.5 MN/m’\n\nProblems\n2.1 (A). A power transmission cable consists of ten copper wires each of 1.6 mm diameter surrounding three steel\nwires each of 3 mm diameter. Determine the combined E for the compound cable and hence determine the extension\nof a 30 m length of the cable when it is being laid with a tension of 2 kN.\nFor steel, E = 200 GN/mZ;for copper, E = 100 GN/mZ.\nC151.3 GN/mZ; 9.6 mm.]\n2.2 (A). If the maximum stress allowed in the copper of the cable of problem 2.1 is 60 MN/m2, determine the\nmaximum tension which the cable can support.\nC3.75 kN.1\n\n40\n\nMechanics of Materials\n\n2.3 (A). What will be the stress induced in a steel bar when it is heated from 15°C to W C , all expansion being\nprevented?\nFor mild steel, E = 210 GN/mZ and a = 11 x\nper \"C.\n[lo4 MN/m2\".]\n2.4 (A). A 75 mm diameter compound bar is constructed by shrinking a circular brass bush onto the outside of a\n50 mm diameter solid steel rod. If the compound bar is then subjected to an axial compressive load of 160 kN\n\ndetermine the load carried by the steel rod and the brass bush and the compressive stress set up in each material.\nFor steel, E = 210 GN/m2; for brass, E = 100 GN/m*.\n[I. Struct. E.] c100.3, 59.7 kN; 51.1, 24.3 MN/mZ.]\n2.5 (B). A steel rod of cross-sectional area 600mm2 and a coaxial copper tube of cross-sectional area loo0 mm2\nare firmly attached at their ends to form a compound bar. Determine the stress in the steel and in the copper when the\ntemperature of the bar is raised by 80°C and an axial tensile force of 60 kN is applied.\nFor steel, E = 200 GN/m2 with a = 11 x\nper \"C.\nFor copper, E = 100 GN/m2 with a = 16.5 x\nper \"C.\n[E.I.E.] C94.6, 3.3 MN/m2.]\n\n2.6 (B). A stanchion is formed by buttwelding together four plates of steel to form a square tube of outside crosssection 200 mm x 200 mm. The constant metal thickness is 10 mm. The inside is then filled with concrete.\n(a) Determine the cross-sectional area of the steel and concrete\n(b) If E for steel is 200 GN/m2 and this value is twenty times that for the concrete find, when the stanchion carries\n(i) The stress in the concrete\n(ii) The stress in the steel\n(iii) The amount the stanchion shortens over a length of 2m.\n[C.G.] [2, 40 MN/m2; 40 mm]\n\nCHAPTER 3\n\nSHEARING FORCE AND BENDING MOMENT\nDIAGRAMS\nSummary\nAt any section in a beam carrying transverse loads the shearing force is defined as the\nalgebraic sum of the forces taken on either side of the section.\nSimilarly, the bending moment at any section is the algebraic sum of the moments of the\nforces about the section, again taken on either side.\nIn order that the shearing-force and bending-moment values calculated on either side of\nthe section shall have the same magnitude and sign, a convenient sign convention has to be\nadopted. This is shown in Figs. 3.1 and 3.2 (see page 42).\nShearing-force (S.F.) and bending-moment (B.M.) diagrams show the variation of these\nquantities along the length of a beam for any fixed loading condition.\n\nPara bola\n\nBM\n\nI\n\nI\n\nw.\n\nSF\n\n-wL\n\n-w\nBM\n-WL\n\n2\n\n- W f\n\n3.1. Shearing force and bending moment\nAt every section in a beam carrying transverse loads there will be resultant forces on either\nside of the section which, for equilibrium, must be equal and opposite, and whose combined\n41\n\n42\n\nMechanics of Materials\n\n\\$3.1\n\naction tends to shear the section in one of the two ways shown in Fig. 3.la and b. The shearing\nforce (S.F.) at the section is defined therefore as the algebraic sum of theforces taken on one side\nof the section. Which side is chosen is purely a matter of convenience but in order that the\nvalue obtained on both sides shall have the same magnitude and sign a convenient sign\n3.1.1. Shearing force (S.F.) sign convention\n\nForces upwards to the left of a section or downwards to the right of the section are positive.\nThus Fig. 3.la shows a positive S.F. system at X-X and Fig. 3.lb shows a negative S.F.\nsystem.\n\ntX\n\nA!'?\n7\n2\n3\nIX\n\n( b ) Negative\nIx\n5.E\n\n( a ) Positive 5 F:\n\nFig. 3.1. S.F. sign convention.\n\nIn addition to the shear, every section of the beam will be subjected to bending, i.e. to a\nresultant B.M. which is the net effect of the moments of each of the individual loads. Again,\nfor equilibrium, the values on either side of the section must have equal values. The bending\nmoment (B.M.) is defined therefore as the algebraic sum of the moments of the forces about the\nsection, taken on either side of the section. As for S.F., a convenient sign convention must be\n\n3.1.2. Bending moment (B.M.) sign convention\nClockwise moments to the left and counterclockwise to the right are positive. Thus\nFig. 3 . h shows a positive bending moment system resulting in sagging of the beam at X-X\nand Fig. 3.2b illustrates a negative B.M. system with its associated hogging beam.\n\nIX\n\nWb\nIX\n\n( a ) Positive B M\n\nFig. 3.2.\n\nIX\n\ne\nIX\n\n( b ) Negative B.M\n\nB.M.sign convention.\n\nIt should be noted that whilst the above sign conventions for S.F. and B.M. are somewhat\narbitrary and could be completely reversed, the systems chosen here are the only ones which\nyield the mathematically correct signs for slopes and deflections of beams in subsequent work\nand therefore are highly recommended.\n\n\\$3.2\n\nShearing Force and Bending Moment Diagrams\n\n43\n\nDiagrams which illustrate the variation in the B.M. and S.F. values along the length of a\nbeam or structure for any fixed loading condition are termed B.M. and S.F. diagrams. They\nare therefore graphs of B.M. or S.F. values drawn on the beam as a base and they clearly\nillustrate in the early design stages the positions on the beam which are subjected to the\ngreatest shear or bending stresses and hence which may require further consideration or\nstrengthening.\nAt this point it is imperative to note that there are two general forms of loading to which\nstructures may be subjected, namely, concentrated and distributed loads. The former are\nassumed to act at a point and immediately introduce an oversimplification since all practical\nloading systems must be applied over a finite area. Nevertheless, for calculation purposes this\narea is assumed to be so small that the load can be justly assumed to act at a point. Distributed\nloads are assumed to act over part, or all, of the beam and in most cases are assumed to be\nequally or uniformly distributed; they are then termed uniformly distributed loads (u.d.1.).\nOccasionally, however, the distribution is not uniform but may vary linearly across the\nloaded portion or have some more complex distribution form.\n\n'X\n\nwx\n\nk\n\nFig. 3.3. S.F.-B.M. diagrams for standard cases.\n\nThus in the case of a cantilever carrying a concentrated load Wat the end (Fig. 3.3), the S.F.\nat any section X-X, distance x from the free end, is S.F. = - W. This will be true whatever the\nvalue of x, and so the S.F. diagram becomes a rectangle. The B.M. at the same section X-X is\n- Wx and this will increase linearly with x. The B.M. diagram is therefore a triangle.\nIf the cantilever now carries a uniformly distributed load, the S.F. at X-X is the net load to\none side of X-X, i.e. -wx. In this case, therefore, the S.F. diagram becomes triangular,\nincreasing to a maximum value of - W L at the support. The B.M. at X-X is obtained by\ntreating the load to the left of X-X as a concentrated load of the same value acting at the\ncentre of gravity,\ni.e.\n\nX\n\n- wx2\n\nB.M. at X - X = - w x - = - __\n2\n2\n\nPlotted against x this produces the parabolic B.M. diagram shown.\n3.2. S.F. and B.M. diagrams for beams carrying concentrated loads only\n\nIn order to illustrate the procedure to be adopted for the determination of S.F. and B.M.\nvalues for more complicated load conditions, consider the simply supported beam shown in\n\n44\n\nMechanics of Materials\n\n\\$3.2\n\nFig. 3.4.\n\nFig. 3.4 carrying concentrated loads only. (The term simply supported means that the beam\ncan be assumed to rest on knife-edges or roller supports and is free to bend at the supports\nwithout any restraint.)\nThe values of the reactions at the ends of the beam may be calculated by applying normal\nequilibrium conditions, i.e. by taking moments about F.\nThus\n\n+\n\n+\n\nRA x 12 = (10 x 10) (20 x 6) (30 x 2) - (20 x 8) = 120\nRA = 10kN\n\nFor vertical equilibrium\ntotal force up = total load down\nRA+RF = 10+20+30-20 = 40\nR F = 3OkN\nAt this stage it is advisable to check the value of RF by taking moments about A.\nSumming up the forces on either side of X-X we have the result shown in Fig. 3.5. Using\nthe sign convention listed above, the shear force at X-X is therefore +20kN, Le. the\nresultant force at X-X tending to shear the beam is 20 kN.\n\ni\n\nX\n\nI\n\nN\n\nI\n\nImkN\n\n1* 1 ~ 1 1 1\n\nI O kN\n\nx)kN/X\n\n, 20kN\n\n30kN\n\n20kN\n\n30kN\n\nI\nX\n\nFig. 3.5. Total S.F. at X-X.\n\nSimilarly,Fig. 3.6 shows the summation of the moments of the forces at X-X,the resultant\nB.M. being 40 kNm.\nIn practice only one side of the section is normally considered and the summations\ninvolved can often be completed by mental arithmetic. The complete S.F. and B.M. diagrams\nfor the beam are shown in Fig. 3.7, and the B.M. values used to construct the diagram are\nderived on page 45.\n\n\\$3.2\n\nShearing Force and Bending Moment Diagrams\n\nR,x5=50\n\n20x1,\n\n45\n\n30x5\n\nI\n\nIx\n\n'X\n\nFig. 3.6. Total B.M.at X-X.\n\nB.M. at A\n=\n=\nB.M. at B = + (10 x 2)\nB . M . a t C = +(lOx4)-(1Ox2)\n=\nB.M. at D = + ( l o x 6)+ (20 x 2)- (10 x 4) =\n=\nB.M. at E = + (30 x 2)\nB.M. at F\n=\n\no\n+20kNm\n+20kNm\n+60kNm\n+60kNm\n\no\n\nAll the above values have been calculated from the moments of the forces to the left of each\nsection considered except for E where forces to the right of the section are taken.\n\n10\n\nFig. 3.1.\n\nIt may be observed at this stage that the S.F. diagram can be obtained very quickly when\nworking from the left-hand side, since after plotting the S.F. value at the support all\nsubsequent steps are in the direction of and equal in magnitude to the applied loads, e.g.\n10kN up at A, down 10 kN at B, up 20 kN at C , etc., with horizontal lines joining the steps to\nshow that the S.F. remains constant between points of application of concentrated loads.\nThe S.F. and B.M. values at the left-hand support are determined by considering a section\nan infinitely small distance to the right of the support. The only load to the left (and hence the\n\nMechanics of Materials\n\n46\n\n53.3\n\nS.F.) is then the reaction of 10 kN upwards, Le. positive, and the bending moment = reaction\nx zero distance = zero.\nThe following characteristics of the two diagrams are now evident and will be explained\nlater in this chapter:\n(a) between B and C the S.F. is zero and the B.M. remains constant;\n(b) between A and B the S.F. is positive and the slope of the B.M. diagram is positive; vice\nversa between E and F;\n(c) the difference in B.M. between A and B = 20 kN m = area of S.F. diagram between A\nand B.\n3.3. S.F. and B.M. diagrams for uniformly distributed loads\nConsider now the simply supported beam shown in Fig. 3.8 carrying a u.d.1. w = 25 kN/m\nacross the complete span.\nA\n\nC\n\n25 kN/rn\nF\n\nE\n\nD\n\nG\n\n0\n\nRA\n\nI\n\n150\n\nR.2”\n\nI50\n0.M. dmgrorn (kN rn)\n450\n\nFig. 3.8.\n\nHere again it is necessary to evaluate the reactions, but in this case the problem is simplified\nby the symmetry of the beam. Each reaction will therefore take half the applied load,\ni.e.\n\nRA=Rs=\n\n25 x 12\n- 150 kN\n2\n\n~\n\n+\n\nThe S.F. at A, using the usual sign convention, is therefore 150kN.\nConsider now the beam divided into six equal parts 2 m long. The S.F. at any other point C\nis, therefore,\n150 - load downwards between A and C\n= 150 - (25 x 2) = 100kN\n\n+\n\nThe whole diagram may be constructed in this way, or much more quickly by noticing that\nthe S.F. at A is + 150kN and that between A and B the S.F. decreases uniformly, producing\nthe required sloping straight line, shown in Fig. 3.7. Alternatively, the S.F. at A is + 150kN\nand between A and B this decreases gradually by the amount of the applied load (Le. by\n25 x 12 = 300kN) to - 150kN at B.\n\nShearing Force and Bending Moment Diagrams\n\n\\$3.4\n\n47\n\nWhen evaluating B.M.’s it is assumed that a u.d.1. can be replaced by a concentrated load of\nequal value acting at the middle of its spread. When taking moments about C , therefore, the\nportion of the u.d.1. between A and C has an effect equivalent to that of a concentrated load of\n25 x 2 = 50 kN acting the centre of AC, i.e. 1m from C .\nB.M. at C = (RAx 2)- (50 x 1) = 300-50 = 250kNm\nSimilarly, for moments at D the u.d.1. on AD can be replaced by a concentrated load of\n25 x 4 = 100kN at the centre of AD, i.e. at C .\n\nB.M. at D = (R A x 4) - ( 100 x 2) = 600 - 200 = 400 kN m\nSimilarly,\nB.M. at E = (RAx 6)- (25 x 6)3 = 900-450 = 450kNm\nThe B.M. diagram will be symmetrical about the beam centre line; therefore the values of\nB.M. at F and G will be the same as those at D and C respectively. The final diagram is\ntherefore as shown in Fig. 3.8 and is parabolic.\nPoint (a) of the summary is clearly illustrated here, since the B.M. is a maximum when the\nS.F. is zero. Again, the reason for this will be shown later.\n3.4. S.F. and B.M. diagrams for combined concentrated and\nConsider the beam shown in Fig. 3.9 loaded with a combination of concentrated loads and\nu.d.1.s.\n(RA x 8) + (40 x 2) = (10 x 2 x 7) + (20 x 6) + (20 x 3) + (10 x 1)+ (20 x 3 x 1.5)\n8RA+ 80 = 420\nR A = 42.5 kN ( = S.F. at A )\nRA+RE= (10 x 2)+20+20 + 10+ (20 x 3)+40 = 170\nNow\nRE = 127.5 kN\nWorking from the left-hand support it is now possible to construct the S.F. diagram, as\nindicated previously, by following the direction arrows of the loads. In the case of the u.d.l.’s\nthe S.F. diagram will decrease gradually by the amount of the total load until the end of the\nu.d.1. or the next concentrated load is reached. Where there is no u.d.1. the S.F. diagram\nIn order to plot the B.M. diagram the following values must be determined:\nB.M. at\nB.M. at\nB.M. at\nB.M. at\n\nA\nB=\nC=\nD=\n\n=\n(42.5 x 2) - (10 x 2 x 1) = 85 - 20\n\n=\n\n(42.5 x 5 ) - (10 x 2 x 4) - (20 x 3) = 212.5 - 80 - 60\n(42.5 x 7) - (10 x 2 x 6 ) - (20 x 5 ) - (20 x 2)\n\n=\n\n- (20 x 2 x 1) = 297.5- 120- 1```" ]

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[ "Why is the time it takes for a vertically thrown ball to reach max height the same as the time it takes for the same ball to fall from max height to ground level?\n\nI agree with this logically but I can’t prove it mathematically...\n\nCan you please show me the mathematical proof for this fact without using calculus skills?\n\nThank you!! ^^\n\nRelated Introductory Physics Homework Help News on Phys.org\nharuspex\nHomework Helper\nGold Member\nWhy is the time it takes for a vertically thrown ball to reach max height the same as the time it takes for the same ball to fall from max height to ground level?\n\nI agree with this logically but I can’t prove it mathematically...\n\nCan you please show me the mathematical proof for this fact without using calculus skills?\n\nThank you!! ^^\nWhat formulae have you been taught for motion under constant acceleration?\n\nHmmm ... tough without using calculus. The usual proof is to start with Newton’s 2nd law of motion: F = ma and integrate twice to get\nx = x0 + v0 t + 1/2 a t^2\nPlug in the numbers and voila’. However without calculus I’m not sure what to say\n\nI guess one thing to say is that the laws of physics don’t change under time reversal, but that’s a statement, not a proof.\n\n•", null, "oofllama\nehild\nHomework Helper\nHmmm ... tough without using calculus. The usual proof is to start with Newton’s 2nd law of motion: F = ma and integrate twice to get\nx = x0 + v0 t + 1/2 a t^2\nPlug in the numbers and voila’. However without calculus I’m not sure what to say\n\nI guess one thing to say is that the laws of physics don’t change under time reversal, but that’s a statement, not a proof.\nWhat is the time dependence of the velocity v?\n\nDelta2\nHomework Helper\nGold Member\nWe have to use two equations that are offered to us by basic kinematic theory for the case of motion with constant acceleration a.\n\n$v=v_0+at$ (1)\n$s=v_0t+\\frac{1}{2}at^2$ (2)\n\nThe strict proof of these two equations requires calculus but can be done also with graphical methods, by the graph of acceleration versus time (for the proof of (1)) and by the graph of velocity versus time (for the proof of(2)).\n\nSo lets see how we apply these two equations in this problem. The constant acceleration is the gravitational acceleration g (will be changing sign depending how we apply the equations).\n\nWhen we throw the ball up with initial velocity $v_0$ its velocity $v$ at time $t$ is given by equation (1), where $a=-g$ because now the gravitation acceleration opposes the initial velocity $v_0$\nSo it will be $v=v_0-gt$ (3). If we put $v=0$ that means that the ball is at its highest point so the time $t_u$ to reach there can be found if we solve (3) for the time so it will be $0=v_0-gt_u \\Rightarrow t_u=\\frac{v_0}{g}$\nEquation (2) will become (because in this case acceleration is $a=-g$) $s=v_0t-\\frac{1}{2}gt^2$ . if in this last equation we plug $t=t_u=\\frac{v_0}{g}$ and after we do some algebra we ll find that the distance (height) $s_u$ that is travelled for the time duration of $t_u$ is equal to $s_u=\\frac{v_0^2}{2g}$.\n\nSo, the time to go up is $t_u=\\frac{v_0}{g}$ and the distance (maximum height) that it travels is $s_u=\\frac{v_0^2}{2g}$.\n\nNow we take the case where the ball starts falling from its highest point. We apply again equation (1) but this time acceleration is taken as positive $a=g$, and also the initial velocity $v_0=0$. So if we apply equation (1) for the velocity $v_d$ that it will have at the lowest point, and suppose that this will take time $t_d$ we ll have from (1)\n$v_d=0+gt_d$ (4)\nNow we know that the distance that it will travel going down $s_d$ is the same as $s_u$ so it will be $s_d=s_u=\\frac{v_0^2}{2g}$ (the $v_0$ here is the $v_0$ from the process of going up).\nThe time it takes to go down $t_d$ can be found if we apply equation (2) . It will be\n$s_d=0+\\frac{1}{2}gt_d^2 \\Rightarrow t_d=\\sqrt\\frac{2s_d}{g}$. If we plug $s_d=\\frac{v_0^2}{2g}$ and after some algebra we ll find that the time $t_d=\\frac{v_0}{g}$. So it will be $t_d=t_u$ which we wanted to proof.\n\nAnd also the velocity $v_d$ after it falls at the lowest point will be from (4) $v_d=gt_d=g\\frac{v_0}{g}=v_0$\n\nLast edited:\n•", null, "oofllama" ]

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[ "Error: no applicable method for RunTSNE applied to an object of class \"try-error\"\n1\n0\nEntering edit mode\nmoldach ▴ 20\n@moldach-8829\nLast seen 15 months ago\n\nI'm running into issues trying to catch errors thrown from Seurat::RunTSNE() inside of a function. When users input a small dataset we've noticed that this function will fail with the error:\n\nError in .checktsneparams(nrow(X), dims = dims, perplexity = perplexity, : perplexity is too large for the number of samples\n\nTo handle this condition I would like to run an alternative function and give the user a warning.\n\nTypically one does so using try() or tryCatch(), for example:\n\nexception_handling <- function() {\ntryCatch(\nexpr = {\nmessage(RunTSNE(object = dataSet, dims.use = 1:10, do.fast = TRUE))\nmessage(\"Successfully executed the RunTSNE call.\")\n},\nerror = function(e) {\nmessage(\"Caught an error!\")\nprint(e)\n},\nwarning = function(w) {\nmessage(\"Caught a warning!\")\nprint(w)\n},\nfinally = {\nmessage(\"Reduce perplexity\")\nwarning_msg <- \"Lowering perplexity to the lowest recommendation: 5.\"\nprint(warning_msg)\n\n# Works with some functions\nlmfit <- lm(mpg ~ wt, mtcars)\nreturn(lmfit)\n}\n)\n}\n\n\nThe output from exception_handling() is:\n\nCall:\nlm(formula = mpg ~ wt, data = mtcars)\n\nCoefficients:\n(Intercept) wt\n37.285 -5.344\n\n\nHowever, I run into an error when trying this with Seurat::RunTSEN():\n\nexception_handling2 <- function(x) {\ntryCatch(\nexpr = {\nmessage(RunTSNE(object = dataSet, dims.use = 1:10, do.fast = TRUE))\nmessage(\"Successfully executed the RunTSNE call.\")\n},\nerror = function(e) {\nmessage(\"Caught an error!\")\nprint(e)\n},\nwarning = function(w) {\nmessage(\"Caught a warning!\")\nprint(w)\n},\nfinally = {\nmessage(\"Reduce perplexity\")\nwarning_msg <- \"Lowering perplexity to the lowest recommendation: 5.\"\nprint(warning_msg)\n\n# Doesn't work for some function?\ndataSet_tSNE <- RunTSNE(object = dataSet, dims.use = 1:10, do.fast = TRUE, perplexity = 5)\nreturn(dataSet_tSNE)\n}\n)\n}\n\n\nError in UseMethod(generic = \"RunTSNE\", object = object) : no applicable method for RunTSNE applied to an object of class \"try-error\"\n\nUltimately, I think I want to run a while-loop that reduces the number of perplexity until the function runs successfully (but not sure if this is the optimal way to do it)....\n\nMaybe something like this:\n\ndataSet <- try(RunTSNE(object = dataSet, dims.use = 1:10, do.fast = TRUE))\nwhile (class(dataSet) == \"try-error\") {\ncat(\"Caught an error relating to 'perplexity', reducing the number from default = 50\")\nfor (i in 49:5) {\ndataSet <- RunTSNE(object = dataSet, dims.use = 1:10, do.fast = TRUE, perplexity = i)\n}\n}\n\nSeurat scRNA debugging exceptions unit-test • 513 views\n0\nEntering edit mode\n@martin-morgan-1513\nLast seen 9 weeks ago\nUnited States\n\nSeurat isn't a Bioconductor package and your example isn't reproducible. However, for your use of tryCatch()...\n\nAn 'error' stops a calculation, whereas a 'warning' should be handled and the calculation continued. One usually uses tryCatch() for errors, but withCallingHandlers() for warnings.\n\nSince the finally clause is always run, if one runs\n\ni <- 0\ntryCatch({\ni <- i + 1\nif (i == 1L)\nstop(\"oops\")\n}, finally = {\ni = i + 1\n})\n\n\nthen the result is i = 2. Changing the initial value to i = 1, so the error doesn't occur, then result is i = 3. So you don't want to 'recover' in the finally= clause; a typical use might be to close a file connection that you opened in expr and that needs to be closed whether an error occurs or not.\n\nIf one wanted to try a value and then provide a 'recovery' value in case of error, one might write something like\n\nf <- function(i) {\ntryCatch({\nif (i == 2)\nstop(\"oops\")\ni\n}, error = function(e) {\nwarning(e) # record error as warning\nNA # return NA\n})\n}\n\n\nf() returns its argument, unless an error occurs in which case it returns NA. It works as\n\n> sapply(1:5, f)\n\n 1 NA 3 4 5\nWarning message:\nIn doTryCatch(return(expr), name, parentenv, handler) : oops\n\n\nThis seems to be close to what you would like to do -- try a value of RunTSNE(), and if it fails recover in some way or another.\n\nCan you take these ideas and come up with a second iteration of your approach?\n\n0\nEntering edit mode\n\nHi Martin,\n\nI didn't realize Seurat wasn't under the Bioconductor umbrella and apologize for not including a reprex - I've done so now.\n\nlibrary(dplyr)\nlibrary(Seurat)\nlibrary(tibble)\n\n# remove genes that are not expressed in any cell\nraw_counts <- raw_counts[, colSums(raw_counts != 0) > 0]\n\n# convert to matrix\nrownames(raw_counts) <- raw_counts[,1]\nraw_counts <- raw_counts[,-1]\n\n# convert to Seurat object\ndataSet <- CreateSeuratObject(counts = raw_counts, min.cells = 3, min.features = 200)\n\ndataSet[[\"percent.mt\"]] <- PercentageFeatureSet(dataSet, pattern = \"^MT-\") # add percentage mitochondria\n\n# subset Seurat object\ndataSet <- Seurat:::subset.Seurat(dataSet, subset = nFeature_RNA > 200 & nFeature_RNA < 2500 & percent.mt < 5)\n\n# normalize the data\ndataSet <- Seurat::NormalizeData(dataSet, normalization.method = \"LogNormalize\", scale.factor = 10000)\n\n# find highly variable features\ndataSet <- FindVariableFeatures(dataSet, selection.method = \"vst\", nfeatures = 2000)\n\n# scale and cluster\nall.genes <- rownames(dataSet)\ndataSet <- ScaleData(dataSet, features = all.genes)\ndataSet <- RunPCA(dataSet, features = VariableFeatures(object = dataSet))\ndataSet <- FindNeighbors(dataSet, reduction = \"pca\", dims = 1:20)\ndataSet <- FindClusters(dataSet, resolution = 0.5, algorithm = 1)\n\nf <- function() {\ntryCatch({\nif(dataSet <- RunTSNE(object = dataSet, dims.use = 1:10, do.fast = TRUE, perplexity = 50))\nstop(\"oops\")\n}, error = function(e) {\nwarning(e) # record error as warning\ndataSet <- RunTSNE(object = dataSet, dims.use = 1:10, do.fast = TRUE, perplexity = 5)\ndataSet\n})\n}\n\ndataSet_tSNE <- f()\n\n\nThis suggestion to provide a 'recovery' value (or function) in case of an error is almost 100% of the way there. If a perplexity value of 50 results in the error perplexity is too large for the number of samples a back-up function runs.\n\nIdeally I would like a vectorized solution in the error = function(e){} that keeps trying lower values of perplexity until \"it works\", something like your sapply(49:5, f) example - although I cannot figure out how to implement this.\n\nOnly how to wrap multiple tryCatch()'s together but this is not optimal if your recursively trying perplexity parameters from 50 -> 5....\n\nf2 <- function() {\ntryCatch({\nif(dataSet <- RunTSNE(object = dataSet, dims.use = 1:10, do.fast = TRUE, perplexity = 50))\nstop(\"oops\")\n}, error = function(e) {\nwarning(e) # record error as warning\ntryCatch({\nif(dataSet <- RunTSNE(object = dataSet, dims.use = 1:10, do.fast = TRUE, perplexity = 49))\nstop(\"oops2\")\n},\n\n. . . # etc.\n\nerror = function(e) {\nwarning(e)\ndataSet <- RunTSNE(object = dataSet, dims.use = 1:10, do.fast = TRUE, perplexity = 5)\ndataSet\n})\n})\n}\n\n0\nEntering edit mode\n\nIf I had a function that only 'worked' for some small value of i, e.g.,\n\nf <- function(i) {\nif (i < 10) {\n\"ok\"\n} else {\nstop(\"yuck\")\n}\n}\n\n\nNaively I could write a loop that tried larger and then smaller values until it 'worked'\n\ni <- 50\nrepeat {\nresult <- tryCatch({\nf(i)\n}, error = function(e) {\nNA\n})\nif (!is.na(result)) # success!\nbreak # exit loop\ni <- i - 5 # try new value\n}\n\n\nand after the loop i would be the first value to succeed\n\n> i\n 9\n\n\nA better approach would be to write a function that calls the function you're interested in, but returns a numerical value indicating how close one is to the 'best' result (e.g., largest value that returns a computed result)\n\ng <- function(i) {\ntryCatch({\nf(i)\ni\n}, error = function(e) {\n0\n})\n}\n\n\nYou can see how g() evaluates for different values of i -- it's maximized at the largest value below 10.\n\n> sapply(1:20, g)\n 1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 0 0 0\n\n\nWe can find this maximum efficiently using optimize() or other base R functions (e.g., nlm(), uniroot(), perhaps re-defining g()), using\n\n> optimize(g, c(1, 20), maximum = TRUE)\n$maximum 9.999995$objective\n 9.999995\n\n\nI think though you should reflect on the overall strategy of choosing perplexity in this way." ]

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[ "# Snow Sprint – Multiplying Fractions\n\nSnow Sprint is an online game that helps students learn how to multiply fractions. The game is easy to play – students just need to enter the fractions they want to multiply and then click on the “multiply” button. The game will show them the result of the multiplication and then ask them to try again if they got it wrong.\n\nSnow Sprint – Multiplying Fractions is an online game that helps students practice multiplying fractions. The game is very easy to use – just select the level of difficulty, type in the fractions you want to multiply, and click on the “multiply” button. The game will then show you the answer.\n\n## Snow Sprint – Multiplying Fractions is an online game\n\nSnow Sprint is a fractions game that helps students learn how to multiply fractions. Players race against the clock to answer as many questions as they can in 60 seconds. The faster they answer, the more points they earn.", null, "Author: Nancy Smith" ]

[ null, "https://secure.gravatar.com/avatar/44534ecedd4a57470f37851b382a564a", null ]

[ "# If the pH is zero, what would be the normality of a sulphuric acid solution?\n\nIf pH of the solution is zero, what would be the normality of $250 \\, \\mathrm{ml}$ $\\ce{H2SO4}$ solution?\n\nMy work: $\\mathrm{pH} = 0$, hence the concentration of $\\ce{H+}$ ions is $\\pu{1 mol L^-1}$, which implies that the molarity of the solution is $\\pu{1M}$ hence the normality of the solution must be $2~\\mathrm{N}$. Is this correct?\n\n$\\ce{H2SO4}$ has 2 $\\mathrm pK_\\mathrm a$s, one that is strong whereby $\\ce{H2SO4}$ ionizes to $\\ce{HSO4-}$ and $\\ce{H+}$ and another that is weak whereby $\\ce{HSO4-}$ ionizies to $\\ce{SO4^{2-}}$ and $\\ce{H+}$\n\nAt $\\mathrm p[\\ce{H+}] = 0, [\\ce{H+}] = \\pu{1M}$\n\nAt $\\mathrm {pH} = 0$, sulfuric acid is substantially $\\ce{HSO4-}$ and $\\ce{H+}$, with little $\\ce{H2SO4}$ or $\\ce{SO4^{2-}}$.\n\nTherefore at $\\mathrm p[\\ce{H+}] = 0$, sulfuric acid concentration is $\\pu{1M}$, because each mole of sulfuric acid has released one mole of $\\ce{H+}$.\n\nNormality of an acid is the concentration of the acid times the number of acidic protons per molecule. Sulfuric acid has two acidic protons per molecule.\n\nNormality $= \\pu{1M} \\times 2 = \\pu{2N}$\n\nHere the $\\mathrm{pH}$ is $0$, it means it has concentration of $\\pu{1 M}$, i.e. $[\\ce{H+}]$ is $1$. $$[\\ce{H+}] = 10^{-\\mathrm{pH}} = 10^{-0} = \\pu{1 M}$$ Hence the molarity will be $1$. So its normality will be, $$\\text{normality} = \\text{molarity}\\times\\text{n-factor} = 1\\times 1 = \\pu{1 N}.$$ Here we have taken $\\text{n-factor}$ for $\\ce{H2SO4}$ will be $1$, because $\\mathrm{pH}$ is given as $0$, so it will be a strong acid and in strong condition the $\\text{n-factor}$ or valency factor becomes $1$ and so it forms to ions $\\ce{H+}$ and $\\ce{HSO4-}$, i.e. $\\text{n-factor} = \\text{no. of$\\ce{H+}$replaceable ions is 1}$, hence normality will be $\\pu{1N}$.\n\n• You might want to clean up this answer a bit. See the guidelines in the Help Center to assist you in formatting your answer properly. – Todd Minehardt Jun 9 '17 at 15:33\n• No of substitutable H in H2SO4 is 2 right? – Mockingbird Jun 9 '17 at 15:34" ]

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[ "# Convert character array to string in C++\n\nThis article shows how to convert a character array to a string in C++.\n\nThe std::string in c++ has a lot of inbuilt functions which makes implementation much easier than handling a character array. Hence, it would often be easier to work if we convert a character array to string.\n\nExamples:\n\n```Input: char s[] = { 'g', 'e', 'e', 'k', 's', 'f', 'o',\n'r', 'g', 'e', 'e', 'k', 's' } ;\nOutput: string s = \"geeksforgeeks\" ;\n\nInput: char s[] = { 'c', 'o', 'd', 'i', 'n', 'g' } ;\nOutput: string s = \"coding\" ;\n```\n• Method 1:\nApproach:\n1. Get the character array and its size.\n2. Create an empty string.\n3. Iterate through the character array.\n4. As you iterate keep on concatenating the characters we encounter in the character array to the string.\n5. Return the string.\n\nBelow is the implementation of the above approach.\n\n `// Demonstrates conversion ` `// from character array to string ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// converts character array ` `// to string and returns it ` `string convertToString(``char``* a, ``int` `size) ` `{ ` `    ``int` `i; ` `    ``string s = ``\"\"``; ` `    ``for` `(i = 0; i < size; i++) { ` `        ``s = s + a[i]; ` `    ``} ` `    ``return` `s; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``char` `a[] = { ``'C'``, ``'O'``, ``'D'``, ``'E'` `}; ` `    ``char` `b[] = ``\"geeksforgeeks\"``; ` ` `  `    ``int` `a_size = ``sizeof``(a) / ``sizeof``(``char``); ` `    ``int` `b_size = ``sizeof``(b) / ``sizeof``(``char``); ` ` `  `    ``string s_a = convertToString(a, a_size); ` `    ``string s_b = convertToString(b, b_size); ` ` `  `    ``cout << s_a << endl; ` `    ``cout << s_b << endl; ` ` `  `    ``return` `0; ` `} `\n\nOutput:\n```CODE\ngeeksforgeeks\n```\n• Method 2:\nThe std::string has an inbuilt constructor which does the work for us. This constructor takes in a null-terminated character sequence as it’s input. However, we can use this method only at the time of string declaration and it cannot be used again for the same string because it uses a constructor which is only called when we declare a string.\n\nApproach:\n\n1. Get the character array and its size.\n2. Declare a string (i.e, an object of the string class) and while doing so give the character array as its parameter for its constructor.\n3. Use the syntax: string string_name(character_array_name);\n4. Return the string.\n\nBelow is the implementation of the above approach.\n\n `// Demonstrates conversion ` `// from character array to string ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// uses the constructor in string class ` `// to convert character array to string ` `string convertToString(``char``* a, ``int` `size) ` `{ ` `    ``string s(a); ` ` `  `    ``// we cannot use this technique again ` `    ``// to store something in s ` `    ``// because we use constructors ` `    ``// which are only called ` `    ``// when the string is declared. ` ` `  `    ``// Remove commented portion ` `    ``// to see for yourself ` ` `  `    ``/* ` `    ``char demo[] = \"gfg\"; ` `    ``s(demo); // compilation error  ` `    ``*/` ` `  `    ``return` `s; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``char` `a[] = { ``'C'``, ``'O'``, ``'D'``, ``'E'` `}; ` `    ``char` `b[] = ``\"geeksforgeeks\"``; ` ` `  `    ``int` `a_size = ``sizeof``(a) / ``sizeof``(``char``); ` `    ``int` `b_size = ``sizeof``(b) / ``sizeof``(``char``); ` ` `  `    ``string s_a = convertToString(a, a_size); ` `    ``string s_b = convertToString(b, b_size); ` ` `  `    ``cout << s_a << endl; ` `    ``cout << s_b << endl; ` ` `  `    ``return` `0; ` `} `\n\nOutput:\n```CODE\ngeeksforgeeks\n```\n• Method 3:\nAnother way to do so would be to use an overloaded ‘=’ operator which is also available in the C++ std::string.\n\nApproach:\n\n1. Get the character array and its size.\n2. Declare a string.\n3. Use the overloaded ‘=’ operator to assign the characters in the character array to the string.\n4. Return the string.\n\nBelow is the implementation of the above approach.\n\n `// Demonstrates conversion ` `// from character array to string ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// uses overloaded '=' operator from string class ` `// to convert character array to string ` `string convertToString(``char``* a, ``int` `size) ` `{ ` `    ``string s = a; ` `    ``return` `s; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``char` `a[] = { ``'C'``, ``'O'``, ``'D'``, ``'E'` `}; ` `    ``char` `b[] = ``\"geeksforgeeks\"``; ` ` `  `    ``int` `a_size = ``sizeof``(a) / ``sizeof``(``char``); ` `    ``int` `b_size = ``sizeof``(b) / ``sizeof``(``char``); ` ` `  `    ``string s_a = convertToString(a, a_size); ` `    ``string s_b = convertToString(b, b_size); ` ` `  `    ``cout << s_a << endl; ` `    ``cout << s_b << endl; ` ` `  `    ``return` `0; ` `} `\n\nOutput:\n```CODE\ngeeksforgeeks\n```\n\n• A geek interested in Data Structures and Algorithm, passionate about competitive programming, who loves learning new things, solving interesting puzzles and sharing knowledge with other enthusiasts\n\nIf you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.\n\nPlease Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below.\n\nArticle Tags :\nPractice Tags :" ]

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[ "• - - - - - -\n• (must be a number from 0 to 1)\n\nWhat This Calculator Shows\n\nThis calculator gives you the estimated value of PageRank that should be passed on from a web page to each \"do follow\" link. This is value is calculated using the PageRank formula originally written by Google.\n\nDetermining the actual value of a link is important to know because a high PR web page does not mean that it's the best page to get a link from. Remember all PageRank is shared equally among all links on that page, therefore a web page with a high PR and many links could potentially pass only a small amount of PR in comparison to a lower PR page with little or almost no links on it.\n\nHow It Works\n\nGeneral Overview\nThe calculator retrieves the PageRank of the web page in question and counts out the number of links on that page, then takes the PageRank Formula and runs a small calculation to produce an estimated amount of PR that would get passed on to each link on a page based on the input you provide.\n\nDetails\nThe calculator works based off the Google PageRank formula from the original paper on PageRank by Google founders Sergey Brin and Lawrence Page. That formula can be expresed by the following equation 1.\n\nPR(A) = (1-d) + d (PR(T1)/C(T1) + PR(Tn)/C(Tn) + ...)\nUsing that formula and some algebra the formula was setup so that the it could be solved for the amount of PR passed on to links on that page. Basically it calculates how much PR or \"link juice\" is spilled over from that web page out to each link on the page. Only links that are \"do follow\" are counted as links that will take a piece of the PageRank\n\nWith the calculated link value you can really see how valuable a link from a specific web page is. A high PR page with many links might not be as valuable as a page with less links and a little lower PR because it all depends on the amount of link juice that gets passed on.\n\nDamping Factor\nThe damping factor is a number representing the probability that a Internet surfer randomly clicking on links will eventually stop clicking. And according to the original paper on PageRank by Google founders it's usually set to 0.85 1. We've left this as something you can change and fill in as the number might have changed since the original formulation of the PageRank formula. Changing that number can create radically different results and since we don't know exactly what it is we've left it open for experimentation.\n\nThis is value in terms of amount of PR that will be passed on to your each web page linked to from the web page your checking.\n\nReferences\n1 Sergey Brin and Lawrence Page (1998). \"The anatomy of a large-scale hypertextual Web search engine\". Proceedings of the seventh international conference on World Wide Web 7: 107-117 (Section 2.1.1 Description of PageRank Calculation)." ]

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[ "", null, "# Technical Analysis from A to Z\n\nby Steven B. Achelis\n\nSTOCHASTIC OSCILLATOR\n\nOverview\n\nSto.chas.tic (sto kas'tik) adj. 2. Math. designating a process having an infinite progression of jointly distributed random variables.\n- Webster's New World Dictionary\n\nThe Stochastic Oscillator compares where a security's price closed relative to its price range over a given time period.\n\nInterpretation\n\nThe Stochastic Oscillator is displayed as two lines. The main line is called \"%K.\" The second line, called \"%D,\" is a moving average of %K. The %K line is usually displayed as a solid line and the %D line is usually displayed as a dotted line.\n\nThere are several ways to interpret a Stochastic Oscillator. Three popular methods include:\n\n1. Buy when the Oscillator (either %K or %D) falls below a specific level (e.g., 20) and then rises above that level. Sell when the Oscillator rises above a specific level (e.g., 80) and then falls below that level.\n\n2. Buy when the %K line rises above the %D line and sell when the %K line falls below the %D line.\n\nLook for divergences. For example, where prices are making a series of new highs and the Stochastic Oscillator is failing to surpass its previous highs.\n\nExample\n\nThe following chart shows Avon Products and its 10-day Stochastic.", null, "I drew \"buy\" arrows when the %K line fell below, and then rose above, the level of 20. Similarly, I drew \"sell\" arrows when the %K line rose above, and then fell below, the level of 80.\n\nThis next chart also shows Avon Products.", null, "In this example I drew \"buy\" arrows each time the %K line rose above the %D (dotted). Similarly, \"sell\" arrows were drawn when the %K line fell below the %D line.\n\nThis final chart shows a divergence between the Stochastic Oscillator and prices.", null, "This is a classic divergence where prices are headed higher, but the underlying indicator (the Stochastic Oscillator) is moving lower. When a divergence occurs between an indicator and prices, the indicator typically provides the clue as to where prices will head.\n\nCalculation\n\nThe Stochastic Oscillator has four variables:\n\n1. %K Periods.\nThis is the number of time periods used in the stochastic calculation.\n\n2. %K Slowing Periods.\nThis value controls the internal smoothing of %K. A value of 1 is considered a fast stochastic;a value of 3 is considered a slow stochastic.\n\n3. %D Periods.\nThis is the number of time periods used when calculating a moving average of %K. The moving average is called \"%D\" and is usually displayed as a dotted line on top of %K.\n\n4. %D Method.\nThe method (i.e., Exponential, Simple, Time Series, Triangular, Variable, or Weighted) that is used to calculate %D.\n\nThe formula for %K is:", null, "For example, to calculate a 10-day %K, first find the security's highest-high and lowest-low over the last 10 days. As an example, let's assume that during the last 10 days the highest-high was 46 and the lowest-low was 38--a range of 8 points. If today's closing price was 41, %K would be calculated as:", null, "The 37.5% in this example shows that today's close was at the level of 37.5% relative to the security's trading range over the last 10 days. If today's close was 42, the Stochastic Oscillator would be 50%. This would mean that that the security closed today at 50%, or the mid-point, of its 10-day trading range.\n\nThe above example used a %K Slowing Period of 1-day (no slowing). If you use a value greater than one, you average the highest-high and the lowest-low over the number of %K Slowing Periods before performing the division.\n\nA moving average of %K is then calculated using the number of time periods specified in the %D Periods. This moving average is called %D.\n\nThe Stochastic Oscillator always ranges between 0% and 100%. A reading of 0% shows that the security's close was the lowest price that the security has traded during the preceding x-time periods. A reading of 100% shows that the security's close was the highest price that the security has traded during the preceding x-time periods.\n\nThis online edition of Technical Analysis from A to Z is reproduced here with permission from the author and publisher." ]

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[ "# #010024 Color Information\n\nIn a RGB color space, hex #010024 is composed of 0.4% red, 0% green and 14.1% blue. Whereas in a CMYK color space, it is composed of 97.2% cyan, 100% magenta, 0% yellow and 85.9% black. It has a hue angle of 241.7 degrees, a saturation of 100% and a lightness of 7.1%. #010024 color hex could be obtained by blending #020048 with #000000. Closest websafe color is: #000033.\n\n• R 0\n• G 0\n• B 14\nRGB color chart\n• C 97\n• M 100\n• Y 0\n• K 86\nCMYK color chart\n\n#010024 color description : Very dark (mostly black) blue.\n\n# #010024 Color Conversion\n\nThe hexadecimal color #010024 has RGB values of R:1, G:0, B:36 and CMYK values of C:0.97, M:1, Y:0, K:0.86. Its decimal value is 65572.\n\nHex triplet RGB Decimal 010024 `#010024` 1, 0, 36 `rgb(1,0,36)` 0.4, 0, 14.1 `rgb(0.4%,0%,14.1%)` 97, 100, 0, 86 241.7°, 100, 7.1 `hsl(241.7,100%,7.1%)` 241.7°, 100, 14.1 000033 `#000033`\nCIE-LAB 1.209, 8.345, -20.094 0.331, 0.134, 1.677 0.154, 0.062, 0.134 1.209, 21.758, 292.553 1.209, -0.287, -4.791 3.658, 9.746, -24.627 00000001, 00000000, 00100100\n\n# Color Schemes with #010024\n\n• #010024\n``#010024` `rgb(1,0,36)``\n• #232400\n``#232400` `rgb(35,36,0)``\nComplementary Color\n• #001124\n``#001124` `rgb(0,17,36)``\n• #010024\n``#010024` `rgb(1,0,36)``\n• #130024\n``#130024` `rgb(19,0,36)``\nAnalogous Color\n• #112400\n``#112400` `rgb(17,36,0)``\n• #010024\n``#010024` `rgb(1,0,36)``\n• #241300\n``#241300` `rgb(36,19,0)``\nSplit Complementary Color\n• #002401\n``#002401` `rgb(0,36,1)``\n• #010024\n``#010024` `rgb(1,0,36)``\n• #240100\n``#240100` `rgb(36,1,0)``\n• #002324\n``#002324` `rgb(0,35,36)``\n• #010024\n``#010024` `rgb(1,0,36)``\n• #240100\n``#240100` `rgb(36,1,0)``\n• #232400\n``#232400` `rgb(35,36,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #00000b\n``#00000b` `rgb(0,0,11)``\n• #010024\n``#010024` `rgb(1,0,36)``\n• #02003e\n``#02003e` `rgb(2,0,62)``\n• #020057\n``#020057` `rgb(2,0,87)``\n• #030071\n``#030071` `rgb(3,0,113)``\nMonochromatic Color\n\n# Alternatives to #010024\n\nBelow, you can see some colors close to #010024. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #000824\n``#000824` `rgb(0,8,36)``\n• #000524\n``#000524` `rgb(0,5,36)``\n• #000224\n``#000224` `rgb(0,2,36)``\n• #010024\n``#010024` `rgb(1,0,36)``\n• #040024\n``#040024` `rgb(4,0,36)``\n• #070024\n``#070024` `rgb(7,0,36)``\n• #0a0024\n``#0a0024` `rgb(10,0,36)``\nSimilar Colors\n\n# #010024 Preview\n\nThis text has a font color of #010024.\n\n``<span style=\"color:#010024;\">Text here</span>``\n#010024 background color\n\nThis paragraph has a background color of #010024.\n\n``<p style=\"background-color:#010024;\">Content here</p>``\n#010024 border color\n\nThis element has a border color of #010024.\n\n``<div style=\"border:1px solid #010024;\">Content here</div>``\nCSS codes\n``.text {color:#010024;}``\n``.background {background-color:#010024;}``\n``.border {border:1px solid #010024;}``\n\n# Shades and Tints of #010024\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, #000010 is the darkest color, while #fcfcff is the lightest one.\n\n• #000010\n``#000010` `rgb(0,0,16)``\n• #010024\n``#010024` `rgb(1,0,36)``\n• #020038\n``#020038` `rgb(2,0,56)``\n• #02004b\n``#02004b` `rgb(2,0,75)``\n• #03005f\n``#03005f` `rgb(3,0,95)``\n• #030072\n``#030072` `rgb(3,0,114)``\n• #040086\n``#040086` `rgb(4,0,134)``\n• #04009a\n``#04009a` `rgb(4,0,154)``\n``#0500ad` `rgb(5,0,173)``\n• #0500c1\n``#0500c1` `rgb(5,0,193)``\n• #0600d5\n``#0600d5` `rgb(6,0,213)``\n• #0600e8\n``#0600e8` `rgb(6,0,232)``\n• #0700fc\n``#0700fc` `rgb(7,0,252)``\n• #1710ff\n``#1710ff` `rgb(23,16,255)``\n• #2a24ff\n``#2a24ff` `rgb(42,36,255)``\n• #3d38ff\n``#3d38ff` `rgb(61,56,255)``\n• #504bff\n``#504bff` `rgb(80,75,255)``\n• #635fff\n``#635fff` `rgb(99,95,255)``\n• #7672ff\n``#7672ff` `rgb(118,114,255)``\n• #8986ff\n``#8986ff` `rgb(137,134,255)``\n• #9d9aff\n``#9d9aff` `rgb(157,154,255)``\n``#b0adff` `rgb(176,173,255)``\n• #c3c1ff\n``#c3c1ff` `rgb(195,193,255)``\n• #d6d5ff\n``#d6d5ff` `rgb(214,213,255)``\n• #e9e8ff\n``#e9e8ff` `rgb(233,232,255)``\n• #fcfcff\n``#fcfcff` `rgb(252,252,255)``\nTint Color Variation\n\n# Tones of #010024\n\nA tone is produced by adding gray to any pure hue. In this case, #111113 is the less saturated color, while #010024 is the most saturated one.\n\n• #111113\n``#111113` `rgb(17,17,19)``\n• #0f0f15\n``#0f0f15` `rgb(15,15,21)``\n• #0e0e16\n``#0e0e16` `rgb(14,14,22)``\n• #0d0c18\n``#0d0c18` `rgb(13,12,24)``\n• #0b0b19\n``#0b0b19` `rgb(11,11,25)``\n• #0a0a1a\n``#0a0a1a` `rgb(10,10,26)``\n• #09081c\n``#09081c` `rgb(9,8,28)``\n• #08071d\n``#08071d` `rgb(8,7,29)``\n• #06061e\n``#06061e` `rgb(6,6,30)``\n• #050420\n``#050420` `rgb(5,4,32)``\n• #040321\n``#040321` `rgb(4,3,33)``\n• #020123\n``#020123` `rgb(2,1,35)``\n• #010024\n``#010024` `rgb(1,0,36)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #010024 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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[ "## Arduino\n\nThe Arduino is an open-source electronic prototyping platform. It is easy to understand. Using this Arduino board here we can see how to connect the LCD to the Arduino and how we can program it.\n\nFigure 1: Arduino Board\n\nRequired Materials\n• Arduino UNO board\n• LCD display\n• Potentiometer\n• Hook up wires\n• USB cable.\nSTEP 1: LCD DISPLAY CONNECTION\n\nThe LCD display should be fixed or set as (16,2) i.e. the LCD consists of 16 rows and 2 columns\n1. VSS pin to the gnd\n2. VDD pin to the 5v\n3. LCD RS pin to digital pin 12\n4. LCD Enable pin to digital pin 11\n5. LCD D4 pin to digital pin 05\n6. LCD D5 pin to digital pin 04\n7. LCD D6 pin to digital pin 03\n8. LCD D4 pin to digital pin 02\n9. A pin to the 5v\n10. K pin to the gnd\n11. Then the potentiometer can be connected to the digital pin 3 and the gnd and vcc +5V are connected.\nSTEP 2:\n\nConnect the Arduino board to your computer with the USB cable and start with the following program,\n\nSTEP 3: PROGRAMMING\n1. #include <LiquidCrystal.h>               // Includes the library code\n2. LiquidCrystal lcd(12, 11, 5, 4, 3, 2);\n3. void setup()\n4. {\n5.       lcd.begin(16, 2);\n6.       lcd.print(\"C# Corner\");\n7. }\n8.\n9. void loop()\n10. {\n11.       lcd.print(millis() / 1000);\n12. }\nSTEP 4: Attach the following code in the Arduino\n\nAfter completion of the program, we want to upload our program in the Arduino to get the output.\n\n1) Compile it\n3) Output in the LCD\n\nExplanation\n\n<LiquidCrystal.h>\n\nIt is the library code before sketch we want to include this,\n\n<12, 11, 5, 4, 3, 2> are input pin to the Arduino. Initialize the library with the number of interface pins.\n\nHere we are using the two functions void setup() and void loop().\n• setup()\nIn the setup() function we have a code called lcd.begin(16, 2). It set the LCD number of columns and the rows.\n• loop()\nlcd.print(\"C# Corner\") - It prints the message to the LCD. In the void loop( ) function, we have a code called lcd.print (millis() / 1000) that prints the number of seconds since we want to reset the LCD." ]

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[ "# Quantum Theory Looks at Time Travel\n\nDaniel M. Greenberger Department of Physics, City College of the City University of New York\nNew York, NY 10031, USA\nKarl Svozil Institute of Theoretical Physics, Vienna University of Technology, Wiedner Hauptstraße 8-10/136, A-1040 Vienna, Austria\n###### Abstract\n\nWe introduce a quantum mechanical model of time travel which includes two figurative beam splitters in order to induce feedback to earlier times. This leads to a unique solution to the paradox where one could kill one’s grandfather in that once the future has unfolded, it cannot change the past, and so the past becomes deterministic. On the other hand, looking forwards towards the future is completely probabilistic. This resolves the classical paradox in a philosophically satisfying manner.\n\nTime travel, quantum information, foundations of quantum theory\n###### pacs:\n03.67.-a,03.30.+p\n\nClassically, time travel is inconsistent with free will. If one could visit the past, then one could change the past, and this would lead to an alternative present. So there is a paradox here, which is best illustrated by the famous scenario of a person going back in time to shoot his father before his father has met his mother, and thus negating the possibility of his having ever been born. It is for reasons like this that time travel has been considered impossible in principle .\n\nOf course, one can get around this problem if one considers the universe to be totally deterministic, and free will to be merely an illusion. Then the possibility of changing the past (or the future, for that matter) no longer exists. Since we prefer to think that the writing of this paper was not preordained at the time of the big bang, we shall reject this solution on psychological grounds, if not logical ones, and ask whether the paradoxes of classical physics can be gotten around, quantum mechanically.\n\nMost attempts to go beyond the confines of classical theory in order to study time travel have been in the framework of relativity theory, making use of freedom to warp the topological properties of spacetime. We shall not comment on these here, except to note that they are not incompatible with what we shall be saying, and might conceivably be combined with it.\n\nIt seems to us that time travel is very much in the spirit of quantum mechanics, and in fact, it seems quite arbitrary and outside the spirit of the subject to forbid it . For example, if one studies the propagation of a physical system from time to later time , one writes\n\n ψ(t2)=U(t2,t1)ψ(t1),t2>t1, (1)\n\nwhere is some unitary operator describing the dynamical unfolding of the system. To calculate , some sums over all possible paths leading from the initial state to the final state, but restricting these paths to the forward direction of time.\n\nFurthermore, it is well known that when one makes measurements in quantum theory, one’s simple sense of causality is violated, and so a classical sense of causality is a rather poor guide as to what should or should not be allowed quantum mechanically. And this restriction would seem to violate the spirit of the entire enterprise. Specifically, why should there not be some form of feedback into the past in determining what will happen in the future (see Fig. 1)?\n\nIn order to incorporate some form of feedback into the scheme, a simple feedback mechanism such as that used in electronic circuits would be impossible because in such a scheme, a simple feedback loop, such as that of Fig. 2 is used, and in such a loop, one has two circuit paths feeding into one, and quantum mechanically this would violate unitarity, because it could not be uniquely reversed. However, quantum mechanically, there is another way to introduce feedback, and that is through the introduction of beam splitters, which are unitary.\n\n## I Model of a Feedback System in Time\n\nThe model that we introduce is one which has two beam splitters, which allows us to generalize the classical scheme of Fig. 2, and at the same time to present a unitary scheme allowing the particle to sample earlier times. This should not be confused with the operation of time reversal, which is an anti-unitary operation. The scheme is shown in Fig. 3.\n\nIn this scheme, if there were no feedback, then the standard unitary time development would have evolving into\n\n ψ3(t2)=G1ψ(t1). (2)\n\nHere, the operator generates the effects of the feedback in time. These ’beam splitters’ are figurative, and their role is merely to couple the two incoming channels to two outgoing channels. The operator represents the ordinary time development in the absence of time feedback. The operator represents an alternate possible time evolution that can take place and compete with because there is feedback. We want to find in the presence of the feedback in time that is generated by the operator .\n\nAt the beam splitters, which are shown in more detail in Fig. 4, the forward amplitude is , while the reflected amplitude is i. One needs the factor of i because the two amplitudes must differ by 90 in order to preserve unitarity. Normally, we expect that , and in the limit , we should get the situation represented by (2).\n\nThe beam splitters perform the unitary transformation\n\n (3)\n\nHere we assume for simplicity that and are real. We can invert this to obtain\n\n |d⟩=α|a⟩−iβ|b⟩,|c⟩=α|b⟩−iβ|a⟩. (4)\n\nThe overall governing equations can be read directly from Fig. 3. At time the second beam splitter determines and . We have\n\n ψ3(t2)≡ψ′3=[αψ1(t2)−iβψ2(t2)]=αψ′1−iβψ′2, (5)\n\nwhere the prime indicates the time in the argument, and no prime indicates the time . The wave functions and are determined at time by\n\n ψ1(t2)=ψ′1=G1ψ1(t1)=G1ψ1, (6)\n ψ′2=G2ψ2. (7)\n\nSo that from (5),\n\n ψ′3=αG1ψ1−iβG2ψ2, (8)\n\nand equivalently\n\n ψ′4=αG2ψ2−iβG1ψ1. (9)\n\nThe propagator is what produces the feedback in time, propagating from back to , so that or\n\n ψ4=Mψ′4. (10)\n\nAt the beam splitter\n\n ψ1 = αψ−iβψ4, (11) ψ2 = αψ4−iβψ. (12)\n\n## Ii The Solution\n\nFirst, we want to eliminate the ub (11) and (12)., to get equations for and . Then from (8) we can obtain . From (9) and (10),\n\n ψ4=Mψ′4=αMG2ψ2−iβMG1ψ1. (13)\n\nWe plug this into (11) and (12),\n\n ψ1 = αψ−iβ(αMG2ψ2−iβMG1ψ1), (14) ψ2 = α(αMG2ψ2−iβMG1ψ1)−iβψ. (15)\n\nWe can rewrite these as\n\n ψ1 = (1+β2MG1)−1(−iαβMG2)ψ2+α(1+β2MG1)−1ψ, (16) ψ2 = (1−α2MG2)−1(−iαβMG1)ψ1−iβ(1−α2MG2)−1ψ. (17)\n\nThese are two simultaneous equations that we must solve to find and as functions of . To solve for substitute (17) into (16).\n\n ψ1 = (1+β2MG1)−1(−iαβMG2)[(1−α2MG2)−1(−αβMG1)ψ1 (18) −iβ(1−α2MG)−1ψ]+α(1+β2MG1)−1ψ.\n\nThis can be rewritten as\n\n [1+α2β2(1+β2MG1)−1(MG2)(1−α2MG2)−1(MG1)]ψ1 =(1+β2MG1)−1[−αβ2MG2(1−α2MG2)−1+α]ψ. (19)\n\nIf we write this as\n\n [X]ψ1=Y−1[Z]ψ, (20)\n\nthen we can simplify the equation as follows:\n\n YX = 1+β2MG1+α2β2MG2(1−α2MG2)−1MG1 (21) = 1+β2[1+(1−α2MG2)−1α2MG2]MG1 = 1+β2(1−α2MG2)−1MG1.\n\nand\n\n Z = α(1−α2MG2)−1(1−α2MG2−β2MG2) (22) = α(1−α2MG2)−1(1−MG2).\n\nThus,\n\n ψ1=α[1+β2(1−α2MG2)−1MG1]−1(1−α2MG2)−1(1−MG2)ψ. (23)\n\nThen, using the identity , we finally get\n\n ψ1=α(1−α2MG2+β2MG1)−1(1−MG2)ψ. (24)\n\nWe can solve for similarly, by substituting (16) into (17),\n\n ψ2=−iβ(1−α2MG2+β2MG1)−1(1+MG1)ψ. (25)\n\nNotice that in the denominator term in both (24) and (25), and have reversed the role of the operators they apply to. we can finally use (8) to solve for\n\n ψ3(t2)=[α2G1D(1−MG2)−β2G2D(1+MG1)]ψ(t1), (26)\n\nwhere\n\n## Iii Some Important Special Cases\n\nThe Case , This is the case where there is no feedback. Here\n\n ψ′3=G1(1−MG)−1(1−MG2)ψ=G1ψ (27)\n\nThe Case , This is the case where there is only feedback. Here\n\n ψ′3=−G2(1+MG1)−1(1+MG1)ψ=−G2ψ. (28)\n\nThe Case . This corresponds to the case where both paths lead to the same future. (the “” sign is due to the phasing effect of the beam splitters.)\n\n ψ′3 = [α2G1D(1+MG1)+β2G1D(1+MG1)]ψ, D = (1+β2MG1+α2MG1)−1=(1+MG1)−1. (29)\n\nThen\n\n ψ′3=G1ψ, (30)\n\nas we would expect.\n\nThe Case . This is expected to be the usual case. Then the answer only depends on Also, . Then to lowest order in , the denominator in (26) becomes\n\n D = [1+γMG1−(1−γ)MG2]−1 (31) = (1−MG2)−1−γ(1−MG2)−1(MG1+MG2)(1−MG2)−1,\n\nso that\n\n ψ′3 = {(1−γ)G1[1−γ(1−MG2)−1(MG1+MG2)] (32) = G1ψ−γ(G1+G2)(1−MG2)(1+MG1)ψ.\n\nThe most interesting case is the one that corresponds to the classical paradox where you shoot your father before he has met your mother, so that you can never be born. This case has a rather fascinating quantum-mechanical resolution. Actually, there are two possible realizations of this case. The first is the case , where there is a perfect absorber in the beam so that the system without any feedback would never get to evolve to time But quantum mechanically, we assume that there is another path along , the one where you do not shoot your father, that has a probability without feedback. In quantum theory we deal with probabilities, and as long as there is any chance that you may not meet your father, we must take this into account.\n\nThe solution in this case is\n\n ψ′3=−β2G2(1−α2MG2)−1ψ. (33)\n\nWe assume for simplicity that is just the standard time evolution operator\n\n G2=e−iE(t2−t1)/h, (34)\n\nand is just the simplest backwards in time evolution operator\n\n M=e−iE(t1−t2)/h+iϕ, (35)\n\nwhere we have also allowed for an extra phase shift. Then\n\n ψ′3=−β2e−iE(t2−t1)/h(1−α2eiϕ)−1ψ, (36)\n ∣∣ψ′3∣∣2=β4(1−α2eiϕ)(1−α2e−iϕ)|ψ|2=11+4(α2/β4)sin2(ϕ/2)|ψ|2. (37)\n\nNote that for for any value of . That means that no matter how small the probability of your ever having reached here in the first place, the fact that you are here, which can only happen because , guarantees that even though you are certain to have shot your father if you had met him nonetheless you will not have met him! You will have taken the other path, with 100% certainly. Obviously, this must be the case, if you are to be here at all.\n\nHow can we understand this result? In our model, with , we have and . Also, we will assume that , even though this is not necessary. The various amplitudes are\n\n |ψ1|=0,|ψ2/ψ|=1/β,|ψ4/ψ|=α/β,∣∣ψ′3/ψ∣∣=1. (38)\n\nSo we see that the two paths of the beam splitter at leading to the path cancel out. But of the original beam passes through to , while of the beam , only the fraction leaks through to . So the beam must have a very large amplitude, which it does, as we can see from (38), so that the two contributions can cancel at In fact has a much larger amplitude than the original beam! Similarly, in order to have , then must have a very large amplitude. Thus we see that there is a large current flowing around the system, between and . But does this not violate unitarity? The answer is that if they were both running forward in time, it would. But one of these currents is running forward in time, while the other runs backward in time, and so they do not in this case violate unitarity. This is how our solution is possible.\n\nThere is a second possible way to bring about this case, namely to allow any , but to make , so that the purpose of the time travel will be to undo . (Again the “” sign is due to the phasing of the beam splitters.) Then\n\n ψ′3 = [α2G1D(1+G−11G2)+0]ψ, D = (1−β2+α2G−11G2)−1=(1α2)(1+G−11G2)−1. (39)\n\nThus,\n\n ψ′3=G1ψ, (40)\n\nand instead of undoing , wipes out the alternative possible future, thus guaranteeing the future that has already happened.\n\n## V Conclusion\n\nAccording to our model, if you travel into the past quantum mechanically, you would only see those alternatives consistent with the world you left behind you. In other words, while you are aware of the past, you cannot change it. No matter how unlikely the events are that could have led to your present circumstances, once they have actually occurred, they cannot be changed. Your trip would set up resonances that are consistent with the future that has already unfolded.\n\nThis also has enormous consequences on the paradoxes of free will. It shows that it is perfectly logical to assume that one has many choices and that one is free to take any one of them. Until a choice is taken, the future is not determined. However, once a choice is taken, and it leads to a particular future, it was inevitable. It could not have been otherwise. The boundary conditions that the future events happen as they already have, guarantees that they must have been prepared for in the past. So, looking backwards, the world is deterministic. However, looking forwards, the future is probabilistic. This completely explains the classical paradox. In fact, it serves as a kind of indirect evidence that such feedback must actually take place in nature, in the sense that without it, a paradox exists, while with it, the paradox is resolved. (Of course, there is an equally likely explanation, namely that going backward in time is impossible. This also solves the paradox by avoiding it.)\n\nThe model also has consequences concerning the many-worlds interpretation of quantum theory. The world may appear to keep splitting so far as the future is concerned. However, once a measurement is made, only those histories consistent with that measurement are possible. In other words, with time travel, other alternative worlds do not exist, as once a measurement has been made confirming the world we live in, the other worlds would be impossible to reach from the original one. This explanation makes the von Neumann state reduction hypothesis much more reasonable, and in fact acts as a sort of justification of it.\n\nAnother interesting point comes from examining (37). For small angles , we see that\n\n ∣∣ψ′3∣∣2=11+4(α2/β4)sin2(ϕ/2)|ψ|2→11+α2ϕ2/β4|ψ|2, (41)\n\nso that the above result is strongly resonant, with a Lorentzian shape, and a width , since Thus less ’deterministic’ and fuzzier time traveling might be possible, a possibility we have not yet explored. Neither have we explored the possibility that feedback should be possible into the future as well as the past. Of course in this case, it ought to be called ’feedforward’ - rather than feedback.\n\n## References\n\n There are many books about the nature of time. The two main paradoxes are the question of reversing the direction of time so that time travel is possible, and the issue of the ’arrow of time’, namely, why time flows in one direction. A book of interesting essays on both questions is The Nature of Time, ed. by R. Flood and M. Lockwood, Basil Blackwell, Cambridge, Mass. (1986). Some other interesting references are: Time’s Arrow and Archimedes’ Point, by H. Price, Oxford University Press, New York (1996); The Physical Basis of the Direction of Time, by H.D. Zeh, Springer Verlag, Berlin (1999); and Time’s Arrows and Quantum Measurement, by L.S. Schulmann, Cambridge University Press, Cambridge (1997).\n\n This paper is an expanded version of an earlier paper on the subject, D.M. Greenberger and K. Svozil, in: Between Chance and Choice, ed. by H. Atmanspacher and R. Bishop, Imprint Academic, Thorverton England (2002), pp. 293-308.\n\n This paper contains minor changes to our paper published as Chapter 4 of Quo Vadis Quantum Mechanics?, ed. by A. Elitzur, S. Dolev and N. Kolenda, Springer Verlag, Berlin (2005), pp. 63-72.\n\n We have just become aware of an article by D. T. Pegg in Times’ arrows, quantum measurement and superluminal behavior, ed. by D. Mugnai, A. Ranfagni and L. S. Schulman, Consiglio Nazionale Delle Richerche, Roma, (2001) p. 113, eprint quant-ph/0506141, in which related ideas have been developed. Self-consistent causal loops arising from backward time travel have been discussed by K. Svozil, in Fundamental Problems in Quantum Theory: A Conference Held in Honor of Professor John A. Wheeler, ed. by D. M. Greenberger and A. Zeilinger, Annals of the New York Academy of Sciences 755 (1995), pp. 834-841, eprint quant-ph/9502008." ]

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[ "Home | New Examples | Popular | Top | Submit Examples | Request Examples\n\nFor example: Metaphors, hiatuses, adjectives, nouns,...\nExamples of...\nYou are here: Examples10.com > Science > Mathematics\n\nExamples of Mathematics\n\nFind the Median Value Posted on 2011-05-05 18:41:40\nThe Median is the \"middle number\" (in a sorted list of numbers). To find the Median, place the numbers you are given in value order and find...\nConvert Areas Posted on 2011-05-05 10:34:38\nTo convert area, simply remember that area is length by width: Area = Length × Width So, you need to convert once for the length and once a...\nConvert Volumes Posted on 2011-05-05 10:33:40\nConversion is simply a matter of multiplying by the right number. For example: Convert 30 cubic feet into cubic meters (30 ft3 to m3) The...\nConvert Lengths Posted on 2011-05-05 10:32:44\nConversion is simply a matter of multiplying by the right number. For example, the conversion for kilometers into miles is: 1 km = 0.6214 m...\nConvert Metric to imperial Posted on 2011-05-05 10:29:21\nConversion is simply a matter of multiplying by the right number. Some examples: What is 2,000 meters in Imperial? 1. In the chart it sho...\nInjective functions Posted on 2011-05-05 10:25:40\nA function f is injective if and only if whenever f(x) = f(y), x = y. Example: f(x) = x+5 from the set of real numbers naturals to natural...\nExpanding parentheses Posted on 2011-05-05 10:21:19\nExpanding parentheses or brackets means remove it. Whatever is inside the ( ) needs to be treated as a \"package\". When you multiply, you hav...\nDistributive Laws Posted on 2011-05-05 10:17:46\nThe Distributive Law says you get the same answer when you: 1. multiply a number by a group of numbers added together, or 2. do each multip...\nAssociative Laws Posted on 2011-05-05 10:14:56\nThe \"Associative Laws\" say that it doesn\"t matter how you group the numbers: 1) when you add: (a + b) + c = a + (b + c) Example: (6 + 3)...\nCommutative Laws Posted on 2011-05-05 00:27:15\nThe \"Commutative Laws\" say you can swap numbers over and still get the same answer: 1) when you add: a + b = b + a Example: 2 + 8 = 8 + ...\nQuadratic Equations Posted on 2011-05-04 18:30:50\nThe name Quadratic comes from \"quad\" meaning square, because the variable gets squared (like x2). Quadratic Equation looks like: ax2 + bx ...\nFibonacci Sequence Posted on 2011-05-03 12:39:04\nThis is the Fibonacci Sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ... The next number is found by adding the two numbers before it together:...\nCalculate terms of Sequences Posted on 2011-05-03 12:37:15\nA Sequence is a set of things (usually numbers) that are in order. Examples: Calculate the first 4 terms of this sequence: {an} = { (-1/n...\nInfinite or Finite sequences Posted on 2011-05-03 12:34:44\nA Sequence is a set of things (usually numbers) that are in order. If the sequence goes on forever it is called an infinite sequence, otherw...\nNatural Logarithms Posted on 2011-05-03 12:32:53\nA logarithm answers a simple question: How many of one number do we multiply to get another number? A natural logarithm is when base is e (...\nShowing 1 - 15 results of 81\n123\n\nTop Users\nnatttt (357 examples)\nkerhickerkris (38 examples)\nmeldkampphil (10 examples)\nMordecai (2 examples)\nthegirlnextfloor (1 examples)\ntsaliottathoma (1 examples)" ]

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[ "# #b293a6 Color Information\n\nIn a RGB color space, hex #b293a6 is composed of 69.8% red, 57.6% green and 65.1% blue. Whereas in a CMYK color space, it is composed of 0% cyan, 17.4% magenta, 6.7% yellow and 30.2% black. It has a hue angle of 323.2 degrees, a saturation of 16.8% and a lightness of 63.7%. #b293a6 color hex could be obtained by blending #ffffff with #65274d. Closest websafe color is: #999999.\n\n• R 70\n• G 58\n• B 65\nRGB color chart\n• C 0\n• M 17\n• Y 7\n• K 30\nCMYK color chart\n\n#b293a6 color description : Dark grayish pink.\n\n# #b293a6 Color Conversion\n\nThe hexadecimal color #b293a6 has RGB values of R:178, G:147, B:166 and CMYK values of C:0, M:0.17, Y:0.07, K:0.3. Its decimal value is 11703206.\n\nHex triplet RGB Decimal b293a6 `#b293a6` 178, 147, 166 `rgb(178,147,166)` 69.8, 57.6, 65.1 `rgb(69.8%,57.6%,65.1%)` 0, 17, 7, 30 323.2°, 16.8, 63.7 `hsl(323.2,16.8%,63.7%)` 323.2°, 17.4, 69.8 999999 `#999999`\nCIE-LAB 64.231, 14.852, -5.601 35.676, 33.086, 40.581 0.326, 0.303, 33.086 64.231, 15.873, 339.337 64.231, 17.081, -10.706 57.521, 10.049, -1.565 10110010, 10010011, 10100110\n\n# Color Schemes with #b293a6\n\n• #b293a6\n``#b293a6` `rgb(178,147,166)``\n• #93b29f\n``#93b29f` `rgb(147,178,159)``\nComplementary Color\n• #af93b2\n``#af93b2` `rgb(175,147,178)``\n• #b293a6\n``#b293a6` `rgb(178,147,166)``\n• #b29397\n``#b29397` `rgb(178,147,151)``\nAnalogous Color\n• #93b2af\n``#93b2af` `rgb(147,178,175)``\n• #b293a6\n``#b293a6` `rgb(178,147,166)``\n• #97b293\n``#97b293` `rgb(151,178,147)``\nSplit Complementary Color\n• #93a6b2\n``#93a6b2` `rgb(147,166,178)``\n• #b293a6\n``#b293a6` `rgb(178,147,166)``\n• #a6b293\n``#a6b293` `rgb(166,178,147)``\nTriadic Color\n• #9f93b2\n``#9f93b2` `rgb(159,147,178)``\n• #b293a6\n``#b293a6` `rgb(178,147,166)``\n• #a6b293\n``#a6b293` `rgb(166,178,147)``\n• #93b29f\n``#93b29f` `rgb(147,178,159)``\nTetradic Color\n• #916781\n``#916781` `rgb(145,103,129)``\n• #9d758d\n``#9d758d` `rgb(157,117,141)``\n• #a7849a\n``#a7849a` `rgb(167,132,154)``\n• #b293a6\n``#b293a6` `rgb(178,147,166)``\n• #bda2b2\n``#bda2b2` `rgb(189,162,178)``\n• #c7b1bf\n``#c7b1bf` `rgb(199,177,191)``\n• #d2c0cb\n``#d2c0cb` `rgb(210,192,203)``\nMonochromatic Color\n\n# Alternatives to #b293a6\n\nBelow, you can see some colors close to #b293a6. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #b293ae\n``#b293ae` `rgb(178,147,174)``\n• #b293ab\n``#b293ab` `rgb(178,147,171)``\n• #b293a9\n``#b293a9` `rgb(178,147,169)``\n• #b293a6\n``#b293a6` `rgb(178,147,166)``\n• #b293a3\n``#b293a3` `rgb(178,147,163)``\n• #b293a1\n``#b293a1` `rgb(178,147,161)``\n• #b2939e\n``#b2939e` `rgb(178,147,158)``\nSimilar Colors\n\n# #b293a6 Preview\n\nText with hexadecimal color #b293a6\n\nThis text has a font color of #b293a6.\n\n``<span style=\"color:#b293a6;\">Text here</span>``\n#b293a6 background color\n\nThis paragraph has a background color of #b293a6.\n\n``<p style=\"background-color:#b293a6;\">Content here</p>``\n#b293a6 border color\n\nThis element has a border color of #b293a6.\n\n``<div style=\"border:1px solid #b293a6;\">Content here</div>``\nCSS codes\n``.text {color:#b293a6;}``\n``.background {background-color:#b293a6;}``\n``.border {border:1px solid #b293a6;}``\n\n# Shades and Tints of #b293a6\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, #070506 is the darkest color, while #fbfafb is the lightest one.\n\n• #070506\n``#070506` `rgb(7,5,6)``\n• #120d10\n``#120d10` `rgb(18,13,16)``\n• #1d151a\n``#1d151a` `rgb(29,21,26)``\n• #291d24\n``#291d24` `rgb(41,29,36)``\n• #34252f\n``#34252f` `rgb(52,37,47)``\n• #402d39\n``#402d39` `rgb(64,45,57)``\n• #4b3643\n``#4b3643` `rgb(75,54,67)``\n• #573e4d\n``#573e4d` `rgb(87,62,77)``\n• #624657\n``#624657` `rgb(98,70,87)``\n• #6e4e61\n``#6e4e61` `rgb(110,78,97)``\n• #79566c\n``#79566c` `rgb(121,86,108)``\n• #845e76\n``#845e76` `rgb(132,94,118)``\n• #906780\n``#906780` `rgb(144,103,128)``\nShade Color Variation\n• #9a718a\n``#9a718a` `rgb(154,113,138)``\n• #a27c93\n``#a27c93` `rgb(162,124,147)``\n• #aa889d\n``#aa889d` `rgb(170,136,157)``\n• #b293a6\n``#b293a6` `rgb(178,147,166)``\n• #ba9eaf\n``#ba9eaf` `rgb(186,158,175)``\n• #c2aab9\n``#c2aab9` `rgb(194,170,185)``\n• #cab5c2\n``#cab5c2` `rgb(202,181,194)``\n• #d3c1cc\n``#d3c1cc` `rgb(211,193,204)``\n• #dbccd5\n``#dbccd5` `rgb(219,204,213)``\n• #e3d8df\n``#e3d8df` `rgb(227,216,223)``\n• #ebe3e8\n``#ebe3e8` `rgb(235,227,232)``\n• #f3eff1\n``#f3eff1` `rgb(243,239,241)``\n• #fbfafb\n``#fbfafb` `rgb(251,250,251)``\nTint Color Variation\n\n# Tones of #b293a6\n\nA tone is produced by adding gray to any pure hue. In this case, #a4a1a3 is the less saturated color, while #f94cb6 is the most saturated one.\n\n• #a4a1a3\n``#a4a1a3` `rgb(164,161,163)``\n• #ab9aa4\n``#ab9aa4` `rgb(171,154,164)``\n• #b293a6\n``#b293a6` `rgb(178,147,166)``\n• #b98ca8\n``#b98ca8` `rgb(185,140,168)``\n• #c085a9\n``#c085a9` `rgb(192,133,169)``\n• #c77eab\n``#c77eab` `rgb(199,126,171)``\n• #ce77ac\n``#ce77ac` `rgb(206,119,172)``\n• #d66fae\n``#d66fae` `rgb(214,111,174)``\n• #dd68b0\n``#dd68b0` `rgb(221,104,176)``\n• #e461b1\n``#e461b1` `rgb(228,97,177)``\n• #eb5ab3\n``#eb5ab3` `rgb(235,90,179)``\n• #f253b4\n``#f253b4` `rgb(242,83,180)``\n• #f94cb6\n``#f94cb6` `rgb(249,76,182)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #b293a6 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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[ "# 2019 AIME II Problems/Problem 6\n\n## Problem\n\nIn a Martian civilization, all logarithms whose bases are not specified are assumed to be base", null, "$b$, for some fixed", null, "$b\\ge2$. A Martian student writes down", null, "$$3\\log(\\sqrt{x}\\log x)=56$$", null, "$$\\log_{\\log x}(x)=54$$ and finds that this system of equations has a single real number solution", null, "$x>1$. Find", null, "$b$.\n\n## Solution 1\n\nUsing change of base on the second equation to base b,", null, "$$\\frac{\\log x}{\\log \\log_{b}{x}}=54$$", null, "$$\\log x = 54 \\cdot \\log\\log_{b}{x}$$", null, "$$x = (\\log_{b}{x})^{54}$$ Note by dolphin7 - you could also just rewrite the second equation in exponent form. Substituting this into the", null, "$\\sqrt x$ of the first equation,", null, "$$3\\log_{b}{((\\log_{b}{x})^{27}\\log_{b}{x})} = 56$$", null, "$$3\\log_{b}{(\\log_{b}{x})^{28}} = 56$$", null, "$$\\log_{b}{(\\log_{b}{x})^{84}} = 56$$\n\nWe can manipulate this equation to be able to substitute", null, "$x = (\\log_{b}{x})^{54}$ a couple more times:", null, "$$\\log_{b}{(\\log_{b}{x})^{54}} = 56 \\cdot \\frac{54}{84}$$", null, "$$\\log_{b}{x} = 36$$", null, "$$(\\log_{b}{x})^{54} = 36^{54}$$", null, "$$x = 6^{108}$$\n\nHowever, since we found that", null, "$\\log_{b}{x} = 36$,", null, "$x$ is also equal to", null, "$b^{36}$. Equating these,", null, "$$b^{36} = 6^{108}$$", null, "$$b = 6^3 = \\boxed{216}$$\n\n## Solution 2\n\nWe start by simplifying the first equation to", null, "$$3\\log_{b}{(\\sqrt{x}\\log x)}=\\log_{b}{(x^{\\frac{3}{2}}\\log^3x)}=56$$", null, "$$x^\\frac{3}{2}\\cdot \\log_b^3x=b^{56}$$ Next, we simplify the second equation to", null, "$$\\log_{\\log(x)}(x)=\\frac{\\log_b(x)}{\\log_b(\\log_b(x))}=54$$", null, "$$\\log_bx=54\\log_b(\\log_b(x))=\\log_b(\\log_b^{54}(x))$$", null, "$$x=\\log_b^{54}x$$ Substituting this into the first equation gives", null, "$$\\log_b^{54\\cdot \\frac{3}{2}}(x)\\cdot \\log_b^3x=\\log_b^{84}x=b^{56}$$", null, "$$x=b^{b^{\\frac{56}{84}}}=b^{b^{\\frac{2}{3}}}$$ Plugging this into", null, "$x=\\log_b^{54}x$ gives", null, "$$b^{b^{\\frac{2}{3}}}=\\log_b^{54}(b^{b^\\frac{2}{3}})=b^{\\frac{2}{3}\\cdot 54}=b^{36}$$", null, "$$b^{\\frac{2}{3}}=36$$", null, "$$b=36^{\\frac{3}{2}}=6^3=\\boxed{216}$$ -ktong\n\n## Solution 3\n\nApply change of base to", null, "$$\\log_{\\log x}(x)=54$$ to yield:", null, "$$\\frac{\\log_b(x)}{\\log_b(\\log_b(x))}=54$$ which can be rearranged as:", null, "$$\\frac{\\log_b(x)}{54}=\\log_b(\\log_b(x))$$ Apply log properties to", null, "$$3\\log(\\sqrt{x}\\log x)=56$$ to yield:", null, "$$3(\\frac{1}{2}\\log_b(x)+\\log_b(\\log_b(x)))=56\\Rightarrow\\frac{1}{2}\\log_b(x)+\\log_b(\\log_b(x))=\\frac{56}{3}$$ Substituting", null, "$$\\frac{\\log_b(x)}{54}=\\log_b(\\log_b(x))$$ into the equation", null, "$\\frac{1}{2}\\log_b(x)+\\log_b(\\log_b(x))=\\frac{56}{3}$ yields:", null, "$$\\frac{1}{2}\\log_b(x)+\\frac{\\log_b(x)}{54}=\\frac{28\\log_b(x)}{54}=\\frac{56}{3}$$ So", null, "$$\\log_b(x)=36.$$ Substituting this back in to", null, "$$\\frac{\\log_b(x)}{54}=\\log_b(\\log_b(x))$$ yields", null, "$$\\frac{36}{54}=\\log_b(36).$$ So,", null, "$$b^{\\frac{2}{3}}=36\\Rightarrow \\boxed{b=216}$$\n\n-Ghazt2002\n\n## Solution 4\n\n1st equation:", null, "$$\\log (\\sqrt{x}\\log x)=\\frac{56}{3}$$", null, "$$\\log(\\sqrt x)+\\log(\\log x)=\\frac{1}{2}\\log x+\\log(\\log x)=\\frac{56}{3}$$ 2nd equation:", null, "$$x=(\\log x)^{54}$$ So now substitute", null, "$\\log x=a$ and", null, "$x=b^a$:", null, "$$b^a=a^{54}$$", null, "$$b=a^{\\frac{54}{a}}$$ We also have that", null, "$$\\frac{1}{2}a+\\log_{a^\\frac{54}{a}} a=\\frac{56}{3}$$", null, "$$\\frac{1}{2}a+\\frac{1}{54}a=\\frac{56}{3}$$ This means that", null, "$\\frac{14}{27}a=\\frac{56}{3}$, so", null, "$$a=36$$", null, "$$b=36^{\\frac{54}{36}}=36^\\frac{3}{2}=\\boxed{216}$$.\n\n-Stormersyle\n\n## Solution 5 (Substitution)\n\nLet", null, "$y = \\log _{b} x$ Then we have", null, "$$3\\log _{b} (y\\sqrt{x}) = 56$$", null, "$$\\log _{y} x = 54$$ which gives", null, "$$y^{54} = x$$ Plugging this in gives", null, "$$3\\log _{b} (y \\cdot y^{27}) = 3\\log _{b} y^{28} = 56$$ which gives", null, "$$\\log _{b} y = \\dfrac{2}{3}$$ so", null, "$$b^{2/3} = y$$ By substitution we have", null, "$$b^{36} = x$$ which gives", null, "$$y = \\log _{b} x = 36$$ Plugging in again we get", null, "$$b = 36^{3/2} = \\fbox{216}$$\n\n--Hi3142\n\n## Solution 6 (Also Substitution)\n\nThis system of equations looks complicated to work with, so we let", null, "$a=\\log_bx$ to make it easier for us to read.\n\nNow, the first equation becomes", null, "$3\\log(\\sqrt x \\cdot a) = 56 \\implies \\log(\\sqrt{x}\\cdot a)=\\frac{56}3$.\n\nThe second equation,", null, "$\\log_{\\log(x)}(x)=54$ gives us", null, "$\\underline{a^{54} = x}$.\n\nLet's plug this back into the first equation to see what we get:", null, "$\\log_b(\\sqrt{a^{54}}\\cdot a)=\\frac{56}3$, and simplifying,", null, "$\\log_b(a^{27}\\cdot a^1)=\\log_b(a^{28})=\\frac{56}{3}$, so", null, "$b^{\\frac{56}3}=a^{28}\\implies \\underline{b^{\\frac 23}=a}$.\n\nCombining this new finding with what we had above", null, "$a^{54} = (b^{\\frac 23})^{54} = x\\implies \\mathbf{b^{36} =x}$.\n\nNow that we've expressed one variable in terms of the other, we can plug this into either equation, say equation 1. Then we get", null, "$\\log_b(\\sqrt{b^{36}}\\cdot\\log_b(b^{36})=\\frac{56}3\\implies$", null, "$\\log_b(b^{18}\\cdot 36)=\\frac{56}3\\implies b^{\\frac{56}3}=b^{18}\\cdot 36$.\n\nFinally, that gives us that", null, "$\\frac{b^{\\frac{56}3}}{b^{18}}=36\\implies b^{\\frac{56}{3}-18}=b^{\\frac{56}{3}-\\frac{54}{3}}=b^{\\frac 23}=36\\implies b=36^{\\frac 32}=6^3$. Thus,", null, "$b=\\boxed{216}$.\n\n~BakedPotato66\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.", null, "" ]

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[ "• 18\n\n## Learning Opportunities\n\nThis puzzle can be solved using the following concepts. Practice using these concepts and improve your skills.\n\n## Goal\n\nA (2-)neighbor-sum grid is a 5×5 matrix containing each number from 1 to 25 exactly once and where each value that is at least 3 can be obtained as the sum of two distinct values among its direct neighbors (horizontally, vertically and diagonally, so that an inner cell has 8 neighbors, a border cell has 5 neighbors and a corner cell has 3 neighbors).\n\nExample: (21 = 4+17, 14 = 4+10, 10 = 4+6, 16 = 6+10, 22 = 6+16, etc for every value >2 of the grid)\n21 14 10 16 2217 4 1 6 1912 3 5 11 1315 8 2 7 1823 24 9 20 25\n\nIt can be proven that there are 56816 such grids (or 7102 up to the 8 symmetries of the square).\n\nIn this problem, you are given a partially completed grid (in which unknown values are indicated by a 0 in the input) and you are asked to complete it. It is guaranteed for each given testcase that there exists a unique solution.\nInput\n5 lines of 5 space-separated numbers between 0 and 25, 0 indicating an unknown value.\nOutput\n5 lines of 5 space-separated numbers between 1 and 25 corresponding to the unique complete solution.\nConstraints\nExample\nInput\n21 14 10 0 0\n0 4 1 6 0\n0 3 5 11 13\n15 0 0 0 0\n23 24 0 20 25\n\nOutput\n21 14 10 16 22\n17 4 1 6 19\n12 3 5 11 13\n15 8 2 7 18\n23 24 9 20 25\n\n\nA higher resolution is required to access the IDE", null, "", null, "", null, "Join the CodinGame community on Discord to chat about puzzle contributions, challenges, streams, blog articles - all that good stuff!\nOnline Participants" ]

[ null, "data:image/svg+xml;base64,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", null, "data:image/svg+xml;base64,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", null, "data:image/svg+xml;base64,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", null ]

[ "# AEL-AI13 Electrotechnics Application (RLC Circuits, Electrostatics, Motors, Transformers, Lighting)", null, "INNOVATIVE SYSTEMS\n\nThe Electrotechnics Application (RLC Circuits, Electrostatics, Motors, Transformers, Lighting), \"AEL-AI13\", designed by EDIBON to study the basic knowledge about electric engineering.\n\nSee general description\n\nRELATED NEWS\n\n#### General Description\n\nThe Electrotechnics Application (RLC Circuits, Electrostatics, Motors, Transformers, Lighting), \"AEL-AI13\", designed by EDIBON to study the basic knowledge about electric engineering.\n\nThe application teaches the student several topics like electric circuits, static electricity, magnetism, electromagnetism, electromagnetic induction, electric capacity, dynamic electricity, lightning, motors, transformers, rectification and filtrate and electric circuits of application.\n\nOnce the user realizes all practical exercises, he must be able to do association of different loads, logic control circuits, rectification circuits, commissioning motors and transformers. Besides, the user gets wide knowledge in all this fields with the aid of the included manual with all necessary theory.\n\n### GUIDED PRACTICAL EXERCISES INCLUDED IN THE MANUAL\n\n1. Checking the operation of the Industrial Main Power Supply (N-ALI01).\n2. Checking the operation of the Auxiliary Power Supply (N-ALI10).\n3. Electrostatic demonstration on several materials.\n4. Verification of the electrotechnic effects with Acetate.\n5. Verification of the electrotechnic effects with Rabbit fur.\n6. Verification of the sign of the charge with the Electroscope.\n7. Resistance measurement.\n8. Resistors in series association.\n9. Resistors in parallel association.\n10. Coils in series association.\n11. Coils in parallel association.\n12. Star/delta transformation.\n13. Delta/star transformation.\n14. Capacity measurement of a capacitor.\n15. Capacitors series association.\n16. Capacitors parallel association.\n17. Analysis of the charge of a capacitor.\n18. Analysis of the discharge of a capacitor.\n19. Calculation of the time constant of discharge of the capacitor.\n20. Wiring and put into operation of a Single-Phase Motor.\n21. Wiring and put into operation of a Universal Motor.\n22. Wiring and put into operation of a Squirrel-Cage Three-Phase Motor.\n23. Electric energy into mechanic energy conversion.\n24. Mechanic energy into electric energy conversion.\n25. Electric energy into magnetic energy conversion.\n26. Magnetic induction: Lenz´s Law.\n27. Assembling the transformer.\n28. Analysis of a Back Transformer.\n29. Wiring of a Back Transformer.\n30. Verification of the transformer ratio of a Back Transformer.\n31. Analysis of a Boost Transformer.\n32. Wiring of a Boost Transformer.\n33. Verification of the transformer ratio of a Boost Transformer.\n34. Analysis of an Auto-Transformer.\n35. Wiring of an Auto-Transformer.\n36. Connection as single-phase transformer.\n37. Direct delta/delta three-phase connection.\n38. Star/delta three-phase connection.\n39. Delta /star three-phase connection.\n40. Three-phase/six-phase connection.\n41. Transformer with coils in series in phase.\n42. Calculation of the time constant in different configurations of RLC circuits.\n43. Analysis of a RL circuit in series.\n44. Analysis of a RL circuit in parallel.\n45. Analysis of a RC circuit in series.\n46. Analysis of a RC circuit in parallel.\n47. Analysis of a RCL circuit in series.\n48. Low-pass filter.\n49. High-pass filter.\n50. Lamp controlled by a switch and a push button.\n51. Lamp controlled from two points.\n52. Lamp controlled from three points.\n53. Lamp controlled by relays.\n54. Verification of the electrotechnic effects into an acoustic circuit.\n55. Verification of the electrotechnic effects into a fluorescent tube.\n\n#### Quality", null, "#### AFTER-SALES SERVICE", null, "" ]

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[ "Robotics 2\nJacobian Matrix\nQuiz", null, "", null, "", null, "", null, "", null, "", null, "", null, "Question 2:  The Jacobian matrix relates end-effector velocities to the joint velocities.  Which of these expresses the correct relationship between these three things?\nQuestion 3:  True or false: the positions of the joints have no effect on the velocity of the end-effector.\nQuestions 4-7:  Shown here is a kinematic diagram.  Find the Jacobian matrix, then use the drop-down boxes to fill in the missing values.\nQuestion 8:  Which of these equations does come from the Jacobian matrix in questions 4-7?\nQuestion 10:  Based on the Jacobian matrix in Questions 4-7, which of these end-effector velocities is controlled by only one joint?\nQuestion 1: I have a 4-degree-of-freedom manipulator.  What should be the dimensions of my Jacobian matrix?\nQuestion 9:  Based on the Jacobian matrix in Questions 4-7, which of these end-effector velocities will always be zero, no matter what the values of the joint variables are?", null, "", null, "6 rows and 4 columns\n4 rows and 6 columns\n2 rows and 4 columns\n4 rows and 2 columns\nJacobian Matrix = End-effector velocities * Joint velocities\nJacobian Matrix = Joint velocities * End-effector velocities\nJoint velocities = Jacobian Matrix * End-effector velocities\nEnd-effector velocities = Jacobian Matrix * Joint velocities\nTrue\nFalse", null, "The linear velocity in the Y direction\nThe linear velocity in the X direction\nThe rotational velocity around Z\nThe rotational velocity around X\nThe linear velocity in the Y direction\nThe linear velocity in the X direction\nThe rotational velocity around Z\nThe rotational velocity around X" ]

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[ "# Image Fusion\n\nThe principle of image fusion using wavelets is to merge the wavelet decompositions of the two original images using fusion methods applied to approximations coefficients and details coefficients. The two images must be of the same size and are supposed to be associated with indexed images on a common colormap (see wextend to resize images).\n\nTwo examples are examined: the first one merges two different images leading to a new image and the second restores an image from two fuzzy versions of an original image.\n\n### Fusion of Two Different Images\n\n```load mask; X1 = X; load bust; X2 = X;```\n\nMerge the two images from wavelet decompositions at level 1 using db2 by taking two different fusion methods: fusion by taking the mean for both approximations and details:\n\n`XFUSmean = wfusimg(X1,X2,'db2',1,'mean','mean');`\n\nand fusion by taking the maximum for approximations and the minimum for the details.\n\n`XFUSmaxmin = wfusimg(X1,X2,'db2',1,'max','min');`\n\nPlot original and synthesized images.\n\n```colormap(map); subplot(221), image(X1), axis square, title('Mask') subplot(222), image(X2), axis square, title('Bust') subplot(223), image(XFUSmean), axis square, title('Synthesized image, mean-mean') subplot(224), image(XFUSmaxmin), axis square, title('Synthesized image, max-min')```", null, "### Restoration by Fusion from Fuzzy Images\n\nLoad two fuzzy versions of an original image.\n\n```load cathe_1; X1 = X; load cathe_2; X2 = X;```\n\nMerge the two images from wavelet decompositions at level 5 using sym4 by taking the maximum of absolute value of the coefficients for both approximations and details.\n\n`XFUS = wfusimg(X1,X2,'sym4',5,'max','max');`\n\nPlot original and synthesized images.\n\n```figure('Color','white'),colormap(map); subplot(221), image(X1), axis square, title('Catherine 1') subplot(222), image(X2), axis square, title('Catherine 2') subplot(223), image(XFUS), axis square, title('Synthesized image')```", null, "## Support", null, "Get trial now" ]

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[ "# Canadian inflation rate in 1985: 3.96%\n\n\\$\n\n## Inflation in 1985 and its effect on dollar value\n\n\\$1 in 1984 is equivalent in purchasing power to about \\$1.04 in 1985. The dollar had an average inflation rate of 3.96% per year between 1984 and 1985, producing a cumulative price increase of 3.96%. Purchasing power decreased by 3.96% in 1985 compared to 1984. On average, you would have to spend 3.96% more money in 1985 than in 1984 for the same item.\n\nThis means that prices in 1985 are 1.04 times higher than average prices since 1984, according to Statistics Canada consumer price index.\n\nThe inflation rate in 1984 was 4.30%. The inflation rate in 1985 was 3.96%. The 1985 inflation rate is higher compared to the average inflation rate of 2.26% per year between 1985 and 2022.\n\nInflation rate is calculated by change in the consumer price index (CPI). The CPI in 1985 was 62.98. It was 60.58 in the previous year, 1984. The difference in CPI between the years is used by Statistics Canada to officially determine inflation.\n\n Average inflation rate 3.96% Converted amount (\\$1 base) \\$1.04 Price difference (\\$1 base) \\$0.04 CPI in 1984 60.575 CPI in 1985 62.975 Inflation in 1984 4.30% Inflation in 1985 3.96% \\$1 in 1984 \\$1.04 in 1985\n\n## How to calculate inflation rate for \\$1, 1984 to 1985\n\nOur calculations use the following inflation rate formula to calculate the change in value between 1984 and 1985:\n\nCPI in 1985 CPI in 1984\n×\n=\n\nThen plug in historical CPI values. The Canadian CPI was 60.575 in the year 1984 and 62.975 in 1985:\n\n62.97560.575\n×\n\\$1\n=\n\\$1.04\n\n\\$1 in 1984 has the same \"purchasing power\" or \"buying power\" as \\$1.04 in 1985.\n\nTo get the total inflation rate for the 1 years between 1984 and 1985, we use the following formula:\n\nCPI in 1985 - CPI in 1984CPI in 1984\n×\n100\n=\nCumulative inflation rate (1 years)\n\nPlugging in the values to this equation, we get:\n\n62.975 - 60.57560.575\n×\n100\n=\n4%\n\n## Data source & citation\n\nRaw data for these calculations comes from the government of Canada's annual Consumer Price Index (CPI), established in 1914 and computed by Statistics Canada (StatCan).\n\nYou may use the following MLA citation for this page: “Inflation Rate in 1985 | Canada Inflation Calculator.” Official Inflation Data, Alioth Finance, 28 Jun. 2022, https://www.officialdata.org/CAD-inflation-rate-in-1985.\n\nSpecial thanks to QuickChart for their chart image API, which is used for chart downloads.\n\nin2013dollars.com is a reference website maintained by the Official Data Foundation.", null, "" ]

[ null, "https://www.in2013dollars.com/assets/img/authors/ianwebster.jpg", null ]

[ "# Damped Oscillators and Binomial theorem step\n\n• I\n• Teachme\n\n#### Teachme\n\nI uploaded a picture of what I am stuck on. I understand the equation of motion 3.4.5a for a damped oscillator but I don't understand how to use binomial theorem to get the expanded equation 3.4.5b. I am no where near clever enough to figure this one out. I know how to use binomial theorem to expand something with two terms but the fact that this has three terms and is only to the first power throws me off and I don't know how to apply the binomial theorem. If anyone could give me any hints, I would very much appreciate it.\n\n#### Attachments\n\n•", null, "IMG_20160415_131247610.jpg\n39.9 KB · Views: 424\nI uploaded a picture of what I am stuck on. I understand the equation of motion 3.4.5a for a damped oscillator but I don't understand how to use binomial theorem to get the expanded equation 3.4.5b. I am no where near clever enough to figure this one out. I know how to use binomial theorem to expand something with two terms but the fact that this has three terms and is only to the first power throws me off and I don't know how to apply the binomial theorem. If anyone could give me any hints, I would very much appreciate it.\n\nwhat i can gather that you are writing the equivalent expressions for a 2nd order operator equation composed of two first order multiples and if you multiply they come out to be same...\nthe author refers this operation to a binomial expansion - may be the 'mathematicians' may be calling this by the name ' binomial.'\nthough its factorizing the quadratic equation in D. D being a differential operator-being a physics person we usually use maths as a tool.\nbut you can see Abramowitch and Stegun - mathematical functions and there is good info on binomials.\n\nthe two roots are say D1 and D2 and the two bracketed terms identically goes to zero.\n\nI uploaded a picture of what I am stuck on. I understand the equation of motion 3.4.5a for a damped oscillator but I don't understand how to use binomial theorem to get the expanded equation 3.4.5b. I am no where near clever enough to figure this one out. I know how to use binomial theorem to expand something with two terms but the fact that this has three terms and is only to the first power throws me off and I don't know how to apply the binomial theorem. If anyone could give me any hints, I would very much appreciate it.\n\nThe factoring has nothing to do with the binomial theorem. All that it is used is: $(a^2-b^2)=(a+b)(a-b)$.\n\nThe factoring has nothing to do with the binomial theorem. All that it is used is: $(a^2-b^2)=(a+b)(a-b)$.\nHey Mathman,\nthanks for your response. I think it is obvious that it is a2-b2=(a+b)(a-b) going from the second equation(3.4.5b) to the first equation(3.3.5a) but I don't see how it is ovbious going from equation(3.3.5a) to equation(3.4.5b).\n\nThanks for you help, much appreciated.\n\nwhat i can gather that you are writing the equivalent expressions for a 2nd order operator equation composed of two first order multiples and if you multiply they come out to be same...\nthe author refers this operation to a binomial expansion - may be the 'mathematicians' may be calling this by the name ' binomial.'\nthough its factorizing the quadratic equation in D. D being a differential operator-being a physics person we usually use maths as a tool.\nbut you can see Abramowitch and Stegun - mathematical functions and there is good info on binomials.\n\nthe two roots are say D1 and D2 and the two bracketed terms identically goes to zero.\n\nThank you drvrm,\nI will look into that book.\n\nThanks again for the help, I really appreciate it.\n\nHey Mathman,\nthanks for your response. I think it is obvious that it is a2-b2=(a+b)(a-b) going from the second equation(3.4.5b) to the first equation(3.3.5a) but I don't see how it is ovbious going from equation(3.3.5a) to equation(3.4.5b).\n\nThanks for you help, much appreciated.\nThe author had something in mind.\n$D^2+2\\gamma D+\\omega_0^2 = D^2+2\\gamma D+\\gamma^2+\\omega_0^2-\\gamma^2 = (D+\\gamma )^2-(\\gamma^2-\\omega_0^2)$\n\nThe author had something in mind.\n$D^2+2\\gamma D+\\omega_0^2 = D^2+2\\gamma D+\\gamma^2+\\omega_0^2-\\gamma^2 = (D+\\gamma )^2-(\\gamma^2-\\omega_0^2)$\nOh wow, ok, I see now. You are too smart. Thanks soo much! I Wouldn't have figured that one out!" ]

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[ "# QBasic/Basic Math\n\nThere are six numerical variables within QBasic:\n\nType Minimum Maximum\nInteger -32,768 32,767\nLong Integer -2,147,483,648 2,147,483,647\nFloat -3.37x10^38 3.37x10^38\nDouble -1.67x10^308 1.67x10^308\n64-bit Integer -9,223,372,036,854,775,808 9,223,372,036,854,775,807\n64-bit Float ±1.18E−4932 ±1.18E+4932\n\nPlease note that Integer and Float type variables for 64-bit are available only in QB64.\n\nA lot of programming is math. Don't let this scare you: a lot of the math is simple, but it's still math. In this section, we will look at doing some basic math (the same stuff you learned in the 3rd grade) and manipulating numbers.\n\n## Equation Setup\n\nIn QBasic an equation has a basic setup a right side and a left side. For instance X=5, as you can probably figure out, this sets the variable X to 5. But we can use variables on the right side too. Y=X*10 would set Y equal to 10 times X, in this situation, 50. In this next program I will show several equations to give you a feel for math.\n\n## 7MATH.BAS\n\n``` CLS\n\n'Set a-d to initial values\na = 10\nb = 6\nc = 3.1415\nd = 3.333333\n\ne = a + b\nPRINT a; \"+\"; b; \"=\"; e\n\nf = c * d\nPRINT c; \"*\"; d; \"=\"; f\n\ng = b - c\nPRINT b; \"-\"; c; \"=\"; g\n\nh = b / d\nPRINT b; \"/\"; d; \"=\"; h\n\ni = INT(d)\nPRINT \"Remove the decimal from \"; d; \"=\"; i\n```\n\n## Understanding 7MATH.BAS\n\nThe most important thing you can take away from this is the setup for math equations. I think you can figure out what all the symbols are and what they do, but QBasic is picky about equations. For 'e=a+b', if you try 'a+b=e' it will not work. The final thing I would like to address in 7MATH.BAS is the INT() function. As far as vocabulary, a function is something that takes in a piece of information and gives you another piece of information back. So PRINT, was a statement, and INT() is a function. The INT() function takes a number and truncates it's decimal, it does not round. So INT(5.1) is 5 and INT(5.999) is still 5. If you want to round a number use CINT().\n\n## 8MATH.BAS\n\n``` CLS\nINPUT \"Enter a number: \", x\nPRINT\n\nx = x + 5\nPRINT \"X is now: \"; x\n\nx = x * x\nPRINT \"X is now: \"; x\n\nx = x / 5\nPRINT \"X is now: \"; x\n\nx = x - 4\nPRINT \"X is now: \"; x\n\nx = x / x\nPRINT \"X should be 1: \"; x\n```\n\n## Understanding 8MATH.BAS\n\n8MATH.BAS shows one simple concept that is very important in programming, but impossible in math. The way that the computer calculates the equation is it does all the math on the right side of the equation and then sticks it in the variable on the left side. So the equation x=x+5 makes perfect sense, unlike math where it is a contradiction. Reassigning a value to a variable based on its current value is common and a good way to keep the number of variables down.\n\n## 9TIP.BAS\n\n``` CLS\nINPUT \"How much is your bill: \", bill\nINPUT \"What percent tip do you want to give: \", tip\n\ntip = tip / 100 'change percent to decimal\ntip = tip * bill 'change decimal to money\n\nPRINT\nPRINT \"The tip is\"; tip; \"\\$.\"\nPRINT \"Pay\"; tip + bill; \"\\$ total.\"\n```\n\n## Tip Calculator\n\n9TIP.BAS calculates your tip and total bill from the bill and percent tip you wish to give. The first three lines clear the screen and get the information from the user. The fifth line changes the tip from a percent to the correct decimal by dividing by 100 (ex. 20%=.2 because 20/100=.2) the next line takes that percent and turns it into a dollar value by multiplying the decimal value by the bill. So if your bill is \\$20.00 and you leave a 20% tip, it multiplies 20*.2 which is 4 or \\$4.00. The last three lines format the output.\nThis is a good example of a complete program. It collects information from the user, it processes the information and it gives the user feedback. Also, the middle section of the program is a good example of variable conservation. This is subject that will take some practice to get used to. In writing a program, if you use too many variables, it will become difficult to keep track of all of them. If you try and conserve too much, you code may become difficult to understand.\n\nYou may notice that the program may print more than two decimal places if you enter a bill that is not an exact dollar value. As an exercise, try modifying the program so that it only displays two decimal places - you can use the `CINT()` function or any other rounding method you intend to use.\n\n## 10OROP.BAS\n\n``` 'ORder of OPerations\nCLS\na = 15\nb = 10\nc = 12.2\nd = 1.618\n\nPRINT a * b + c 'these two are different\nPRINT a * (b + c)\n\nPRINT\n\nPRINT b - c / d 'these two are different\nPRINT (b - c) / d\n\nPRINT\n\nPRINT a * b - c * d / a + d 'these two are the same\nPRINT (a * b) - ((c * d) / a) + d\n```\n\n## Parentheses and Order of Operations\n\n10OROP.BAS is an example of order of operations and how parentheses can be used to manipulate it. I do not want to go into an indepth explanation of the order of operations here. The best advice I can give is unless you are sure of the order of operations, use parentheses to make sure the equation works how you want. All you need to know about parentheses is that the deepest nested parentheses calculate first. If you wish to know more, there are plenty of algebra resources available. On that note, you may wish to brush up on algebra. While it is not necessary for programming, it can help make programming easier and it can allow you to create more advanced programs.\n\n## Random Numbers\n\nThough we will not go into their use until the next section, I would like to discuss the generation of random numbers. QBasic has a random number statement, RND, that generates a random decimal between 0 and 1. You can think of it as a random percent. At first, this may seem like an odd way to generate random numbers. However, with a little math it is very easy to manipulate this to provide numbers in whatever range you want.\nThe first step is to multiply RND by a number (the range you want). For instance 'RND*10'. This will return random numbers (decimal numbers) between 0 and 10(both included). So, to pick a random number between zero and ten we would say '(RND*10)'\n\n## 11RND.BAS\n\n``` CLS\nRANDOMIZE TIMER\n\nPRINT \"Random number from 0-9:\"; RND * 10\nPRINT\n\nPRINT \"Random number from 1-10:\"; (RND * 10) + 1\nPRINT\n\nPRINT \"Random integer from 1-10:\"; INT(RND * 10) + 1\nPRINT\n\nPRINT \"Random even integer from 50-100:\"; INT(RND * 25) * 2 + 50\n```\n\n## More on RND\n\nA few notes on 11RND.BAS, the second line, RANDOMIZE TIMER, sets it so that the computer uses the current time to pick random number. If you don't do this, it picks the same random number every time (try it, write a one line program, PRINT RND, and run it over and over, your screen will fill up with the same number) this can prove useful for some applications, but not most. Stick RANDOMIZE TIMER in at the top of all your programs that use the RND statement and they will be far less predictable. This program just show some ways to choose what you want from your random number generator. The last line shows that you can be very specific in what you get. Make sure to run this program several times to see the different results." ]

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[ "author wenzelm Fri, 06 Oct 2000 01:04:56 +0200 changeset 10157 6d3987f3aad9 parent 10156 9d4d5852eb47 child 10158 00fdd5c330ea\n* HOL/Lattice: fundamental concepts of lattice theory and order structures;\n NEWS file | annotate | diff | comparison | revisions src/HOL/IsaMakefile file | annotate | diff | comparison | revisions src/HOL/Lattice/Bounds.thy file | annotate | diff | comparison | revisions src/HOL/Lattice/CompleteLattice.thy file | annotate | diff | comparison | revisions src/HOL/Lattice/Lattice.thy file | annotate | diff | comparison | revisions src/HOL/Lattice/Orders.thy file | annotate | diff | comparison | revisions src/HOL/Lattice/ROOT.ML file | annotate | diff | comparison | revisions src/HOL/Lattice/document/root.bib file | annotate | diff | comparison | revisions src/HOL/Lattice/document/root.tex file | annotate | diff | comparison | revisions\n--- a/NEWS\tThu Oct 05 14:04:56 2000 +0200\n+++ b/NEWS\tFri Oct 06 01:04:56 2000 +0200\n@@ -243,10 +243,17 @@\n* HOL/Algebra: new theory of rings and univariate polynomials, by\nClemens Ballarin;\n\n-* HOL/NumberTheory: Fundamental Theorem of Arithmetic, Chinese\n+* HOL/NumberTheory: fundamental Theorem of Arithmetic, Chinese\nRemainder Theorem, Fermat/Euler Theorem, Wilson's Theorem, by Thomas M\nRasmussen;\n\n+* HOL/Lattice: fundamental concepts of lattice theory and order\n+structures, including duals, properties of bounds versus algebraic\n+laws, lattice operations versus set-theoretic ones, the Knaster-Tarski\n+Theorem for complete lattices etc.; may also serve as a demonstration\n+for abstract algebraic reasoning using axiomatic type classes, and\n+mathematics-style proof in Isabelle/Isar; by Markus Wenzel;\n+\n* HOL/Prolog: a (bare-bones) implementation of Lambda-Prolog, by David\nvon Oheimb;\n\n--- a/src/HOL/IsaMakefile\tThu Oct 05 14:04:56 2000 +0200\n+++ b/src/HOL/IsaMakefile\tFri Oct 06 01:04:56 2000 +0200\n@@ -9,13 +9,38 @@\ndefault: HOL\nimages: HOL HOL-Real TLA\n\n-test: HOL-Isar_examples HOL-Induct HOL-Lambda HOL-AxClasses HOL-ex HOL-Subst HOL-IMP \\\n- HOL-IMPP HOL-NumberTheory HOL-Hoare HOL-Lex HOL-Algebra HOL-Auth \\\n- HOL-UNITY HOL-Modelcheck HOL-Prolog HOL-W0 HOL-MiniML HOL-BCV \\\n- HOL-MicroJava HOL-IOA HOL-Real-ex HOL-Real-HahnBanach \\\n- TLA-Inc TLA-Buffer TLA-Memory\n-\n-all: images test\n+#Note: keep targets sorted!\n+test: \\\n+ HOL-Algebra \\\n+ HOL-Auth \\\n+ HOL-AxClasses \\\n+ HOL-BCV \\\n+ HOL-Real-HahnBanach \\\n+ HOL-Real-ex \\\n+ HOL-Hoare \\\n+ HOL-IMP \\\n+ HOL-IMPP \\\n+ HOL-IOA \\\n+ HOL-Induct \\\n+ HOL-Isar_examples \\\n+ HOL-Lambda \\\n+ HOL-Lattice \\\n+ HOL-Lex \\\n+ HOL-MicroJava \\\n+ HOL-MiniML \\\n+ HOL-Modelcheck \\\n+ HOL-NumberTheory \\\n+ HOL-Prolog \\\n+ HOL-Subst \\\n+ TLA-Buffer \\\n+ TLA-Inc \\\n+ TLA-Memory \\\n+ HOL-UNITY \\\n+ HOL-W0 \\\n+ HOL-ex\n+ # ^ this is the sort position\n+\n+all: test images\n\n## global settings\n@@ -406,6 +431,16 @@\n@$(ISATOOL) usedir$(OUT)/HOL AxClasses\n\n+## HOL-Lattice\n+\n+HOL-Lattice: HOL $(LOG)/HOL-Lattice.gz + +$(LOG)/HOL-Lattice.gz: $(OUT)/HOL Lattice/Bounds.thy \\ + Lattice/CompleteLattice.thy Lattice/Lattice.thy Lattice/Orders.thy \\ + Lattice/ROOT.ML Lattice/document/root.tex + @$(ISATOOL) usedir $(OUT)/HOL Lattice + + ## HOL-ex HOL-ex: HOL$(LOG)/HOL-ex.gz\n@@ -502,6 +537,6 @@\n$(LOG)/HOL-W0.gz$(LOG)/HOL-MiniML.gz \\\n$(LOG)/HOL-BCV.gz$(LOG)/HOL-MicroJava.gz \\\n$(LOG)/HOL-IOA.gz$(LOG)/HOL-AxClasses \\\n-\t\t$(LOG)/HOL-Real-ex.gz \\ +$(LOG)/HOL-Lattice $(LOG)/HOL-Real-ex.gz \\$(LOG)/HOL-Real-HahnBanach.gz $(LOG)/TLA-Inc.gz \\$(LOG)/TLA-Buffer.gz $(LOG)/TLA-Memory.gz --- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/src/HOL/Lattice/Bounds.thy Fri Oct 06 01:04:56 2000 +0200 @@ -0,0 +1,325 @@ +(* Title: HOL/Lattice/Bounds.thy + ID:$Id$+ Author: Markus Wenzel, TU Muenchen +*) + +header {* Bounds *} + +theory Bounds = Orders: + +subsection {* Infimum and supremum *} + +text {* + Given a partial order, we define infimum (greatest lower bound) and + supremum (least upper bound) wrt.\\ @{text \\<sqsubseteq>} for two and for any + number of elements. +*} + +constdefs + is_inf :: \"'a::partial_order \\<Rightarrow> 'a \\<Rightarrow> 'a \\<Rightarrow> bool\" + \"is_inf x y inf \\<equiv> inf \\<sqsubseteq> x \\<and> inf \\<sqsubseteq> y \\<and> (\\<forall>z. z \\<sqsubseteq> x \\<and> z \\<sqsubseteq> y \\<longrightarrow> z \\<sqsubseteq> inf)\" + + is_sup :: \"'a::partial_order \\<Rightarrow> 'a \\<Rightarrow> 'a \\<Rightarrow> bool\" + \"is_sup x y sup \\<equiv> x \\<sqsubseteq> sup \\<and> y \\<sqsubseteq> sup \\<and> (\\<forall>z. x \\<sqsubseteq> z \\<and> y \\<sqsubseteq> z \\<longrightarrow> sup \\<sqsubseteq> z)\" + + is_Inf :: \"'a::partial_order set \\<Rightarrow> 'a \\<Rightarrow> bool\" + \"is_Inf A inf \\<equiv> (\\<forall>x \\<in> A. inf \\<sqsubseteq> x) \\<and> (\\<forall>z. (\\<forall>x \\<in> A. z \\<sqsubseteq> x) \\<longrightarrow> z \\<sqsubseteq> inf)\" + + is_Sup :: \"'a::partial_order set \\<Rightarrow> 'a \\<Rightarrow> bool\" + \"is_Sup A sup \\<equiv> (\\<forall>x \\<in> A. x \\<sqsubseteq> sup) \\<and> (\\<forall>z. (\\<forall>x \\<in> A. x \\<sqsubseteq> z) \\<longrightarrow> sup \\<sqsubseteq> z)\" + +text {* + These definitions entail the following basic properties of boundary + elements. +*} + +lemma is_infI [intro?]: \"inf \\<sqsubseteq> x \\<Longrightarrow> inf \\<sqsubseteq> y \\<Longrightarrow> + (\\<And>z. z \\<sqsubseteq> x \\<Longrightarrow> z \\<sqsubseteq> y \\<Longrightarrow> z \\<sqsubseteq> inf) \\<Longrightarrow> is_inf x y inf\" + by (unfold is_inf_def) blast + +lemma is_inf_greatest [elim?]: + \"is_inf x y inf \\<Longrightarrow> z \\<sqsubseteq> x \\<Longrightarrow> z \\<sqsubseteq> y \\<Longrightarrow> z \\<sqsubseteq> inf\" + by (unfold is_inf_def) blast + +lemma is_inf_lower [elim?]: + \"is_inf x y inf \\<Longrightarrow> (inf \\<sqsubseteq> x \\<Longrightarrow> inf \\<sqsubseteq> y \\<Longrightarrow> C) \\<Longrightarrow> C\" + by (unfold is_inf_def) blast + + +lemma is_supI [intro?]: \"x \\<sqsubseteq> sup \\<Longrightarrow> y \\<sqsubseteq> sup \\<Longrightarrow> + (\\<And>z. x \\<sqsubseteq> z \\<Longrightarrow> y \\<sqsubseteq> z \\<Longrightarrow> sup \\<sqsubseteq> z) \\<Longrightarrow> is_sup x y sup\" + by (unfold is_sup_def) blast + +lemma is_sup_least [elim?]: + \"is_sup x y sup \\<Longrightarrow> x \\<sqsubseteq> z \\<Longrightarrow> y \\<sqsubseteq> z \\<Longrightarrow> sup \\<sqsubseteq> z\" + by (unfold is_sup_def) blast + +lemma is_sup_upper [elim?]: + \"is_sup x y sup \\<Longrightarrow> (x \\<sqsubseteq> sup \\<Longrightarrow> y \\<sqsubseteq> sup \\<Longrightarrow> C) \\<Longrightarrow> C\" + by (unfold is_sup_def) blast + + +lemma is_InfI [intro?]: \"(\\<And>x. x \\<in> A \\<Longrightarrow> inf \\<sqsubseteq> x) \\<Longrightarrow> + (\\<And>z. (\\<forall>x \\<in> A. z \\<sqsubseteq> x) \\<Longrightarrow> z \\<sqsubseteq> inf) \\<Longrightarrow> is_Inf A inf\" + by (unfold is_Inf_def) blast + +lemma is_Inf_greatest [elim?]: + \"is_Inf A inf \\<Longrightarrow> (\\<And>x. x \\<in> A \\<Longrightarrow> z \\<sqsubseteq> x) \\<Longrightarrow> z \\<sqsubseteq> inf\" + by (unfold is_Inf_def) blast + +lemma is_Inf_lower [dest?]: + \"is_Inf A inf \\<Longrightarrow> x \\<in> A \\<Longrightarrow> inf \\<sqsubseteq> x\" + by (unfold is_Inf_def) blast + + +lemma is_SupI [intro?]: \"(\\<And>x. x \\<in> A \\<Longrightarrow> x \\<sqsubseteq> sup) \\<Longrightarrow> + (\\<And>z. (\\<forall>x \\<in> A. x \\<sqsubseteq> z) \\<Longrightarrow> sup \\<sqsubseteq> z) \\<Longrightarrow> is_Sup A sup\" + by (unfold is_Sup_def) blast + +lemma is_Sup_least [elim?]: + \"is_Sup A sup \\<Longrightarrow> (\\<And>x. x \\<in> A \\<Longrightarrow> x \\<sqsubseteq> z) \\<Longrightarrow> sup \\<sqsubseteq> z\" + by (unfold is_Sup_def) blast + +lemma is_Sup_upper [dest?]: + \"is_Sup A sup \\<Longrightarrow> x \\<in> A \\<Longrightarrow> x \\<sqsubseteq> sup\" + by (unfold is_Sup_def) blast + + +subsection {* Duality *} + +text {* + Infimum and supremum are dual to each other. +*} + +theorem dual_inf [iff?]: + \"is_inf (dual x) (dual y) (dual sup) = is_sup x y sup\" + by (simp add: is_inf_def is_sup_def dual_all [symmetric] dual_leq) + +theorem dual_sup [iff?]: + \"is_sup (dual x) (dual y) (dual inf) = is_inf x y inf\" + by (simp add: is_inf_def is_sup_def dual_all [symmetric] dual_leq) + +theorem dual_Inf [iff?]: + \"is_Inf (dual A) (dual sup) = is_Sup A sup\" + by (simp add: is_Inf_def is_Sup_def dual_all [symmetric] dual_leq) + +theorem dual_Sup [iff?]: + \"is_Sup (dual A) (dual inf) = is_Inf A inf\" + by (simp add: is_Inf_def is_Sup_def dual_all [symmetric] dual_leq) + + +subsection {* Uniqueness *} + +text {* + Infima and suprema on partial orders are unique; this is mainly due + to anti-symmetry of the underlying relation. +*} + +theorem is_inf_uniq: \"is_inf x y inf \\<Longrightarrow> is_inf x y inf' \\<Longrightarrow> inf = inf'\" +proof - + assume inf: \"is_inf x y inf\" + assume inf': \"is_inf x y inf'\" + show ?thesis + proof (rule leq_antisym) + from inf' show \"inf \\<sqsubseteq> inf'\" + proof (rule is_inf_greatest) + from inf show \"inf \\<sqsubseteq> x\" .. + from inf show \"inf \\<sqsubseteq> y\" .. + qed + from inf show \"inf' \\<sqsubseteq> inf\" + proof (rule is_inf_greatest) + from inf' show \"inf' \\<sqsubseteq> x\" .. + from inf' show \"inf' \\<sqsubseteq> y\" .. + qed + qed +qed + +theorem is_sup_uniq: \"is_sup x y sup \\<Longrightarrow> is_sup x y sup' \\<Longrightarrow> sup = sup'\" +proof - + assume sup: \"is_sup x y sup\" and sup': \"is_sup x y sup'\" + have \"dual sup = dual sup'\" + proof (rule is_inf_uniq) + from sup show \"is_inf (dual x) (dual y) (dual sup)\" .. + from sup' show \"is_inf (dual x) (dual y) (dual sup')\" .. + qed + thus \"sup = sup'\" .. +qed + +theorem is_Inf_uniq: \"is_Inf A inf \\<Longrightarrow> is_Inf A inf' \\<Longrightarrow> inf = inf'\" +proof - + assume inf: \"is_Inf A inf\" + assume inf': \"is_Inf A inf'\" + show ?thesis + proof (rule leq_antisym) + from inf' show \"inf \\<sqsubseteq> inf'\" + proof (rule is_Inf_greatest) + fix x assume \"x \\<in> A\" + from inf show \"inf \\<sqsubseteq> x\" .. + qed + from inf show \"inf' \\<sqsubseteq> inf\" + proof (rule is_Inf_greatest) + fix x assume \"x \\<in> A\" + from inf' show \"inf' \\<sqsubseteq> x\" .. + qed + qed +qed + +theorem is_Sup_uniq: \"is_Sup A sup \\<Longrightarrow> is_Sup A sup' \\<Longrightarrow> sup = sup'\" +proof - + assume sup: \"is_Sup A sup\" and sup': \"is_Sup A sup'\" + have \"dual sup = dual sup'\" + proof (rule is_Inf_uniq) + from sup show \"is_Inf (dual A) (dual sup)\" .. + from sup' show \"is_Inf (dual A) (dual sup')\" .. + qed + thus \"sup = sup'\" .. +qed + + +subsection {* Related elements *} + +text {* + The binary bound of related elements is either one of the argument. +*} + +theorem is_inf_related [elim?]: \"x \\<sqsubseteq> y \\<Longrightarrow> is_inf x y x\" +proof - + assume \"x \\<sqsubseteq> y\" + show ?thesis + proof + show \"x \\<sqsubseteq> x\" .. + show \"x \\<sqsubseteq> y\" . + fix z assume \"z \\<sqsubseteq> x\" and \"z \\<sqsubseteq> y\" show \"z \\<sqsubseteq> x\" . + qed +qed + +theorem is_sup_related [elim?]: \"x \\<sqsubseteq> y \\<Longrightarrow> is_sup x y y\" +proof - + assume \"x \\<sqsubseteq> y\" + show ?thesis + proof + show \"x \\<sqsubseteq> y\" . + show \"y \\<sqsubseteq> y\" .. + fix z assume \"x \\<sqsubseteq> z\" and \"y \\<sqsubseteq> z\" + show \"y \\<sqsubseteq> z\" . + qed +qed + + +subsection {* General versus binary bounds \\label{sec:gen-bin-bounds} *} + +text {* + General bounds of two-element sets coincide with binary bounds. +*} + +theorem is_Inf_binary: \"is_Inf {x, y} inf = is_inf x y inf\" +proof - + let ?A = \"{x, y}\" + show ?thesis + proof + assume is_Inf: \"is_Inf ?A inf\" + show \"is_inf x y inf\" + proof + have \"x \\<in> ?A\" by simp + with is_Inf show \"inf \\<sqsubseteq> x\" .. + have \"y \\<in> ?A\" by simp + with is_Inf show \"inf \\<sqsubseteq> y\" .. + fix z assume zx: \"z \\<sqsubseteq> x\" and zy: \"z \\<sqsubseteq> y\" + from is_Inf show \"z \\<sqsubseteq> inf\" + proof (rule is_Inf_greatest) + fix a assume \"a \\<in> ?A\" + hence \"a = x \\<or> a = y\" by blast + thus \"z \\<sqsubseteq> a\" + proof + assume \"a = x\" + with zx show ?thesis by simp + next + assume \"a = y\" + with zy show ?thesis by simp + qed + qed + qed + next + assume is_inf: \"is_inf x y inf\" + show \"is_Inf {x, y} inf\" + proof + fix a assume \"a \\<in> ?A\" + hence \"a = x \\<or> a = y\" by blast + thus \"inf \\<sqsubseteq> a\" + proof + assume \"a = x\" + also from is_inf have \"inf \\<sqsubseteq> x\" .. + finally show ?thesis . + next + assume \"a = y\" + also from is_inf have \"inf \\<sqsubseteq> y\" .. + finally show ?thesis . + qed + next + fix z assume z: \"\\<forall>a \\<in> ?A. z \\<sqsubseteq> a\" + from is_inf show \"z \\<sqsubseteq> inf\" + proof (rule is_inf_greatest) + from z show \"z \\<sqsubseteq> x\" by blast + from z show \"z \\<sqsubseteq> y\" by blast + qed + qed + qed +qed + +theorem is_Sup_binary: \"is_Sup {x, y} sup = is_sup x y sup\" +proof - + have \"is_Sup {x, y} sup = is_Inf (dual {x, y}) (dual sup)\" + by (simp only: dual_Inf) + also have \"dual {x, y} = {dual x, dual y}\" + by simp + also have \"is_Inf \\<dots> (dual sup) = is_inf (dual x) (dual y) (dual sup)\" + by (rule is_Inf_binary) + also have \"\\<dots> = is_sup x y sup\" + by (simp only: dual_inf) + finally show ?thesis . +qed + + +subsection {* Connecting general bounds \\label{sec:connect-bounds} *} + +text {* + Either kind of general bounds is sufficient to express the other. + The least upper bound (supremum) is the same as the the greatest + lower bound of the set of all upper bounds; the dual statements + holds as well; the dual statement holds as well. +*} + +theorem Inf_Sup: \"is_Inf {b. \\<forall>a \\<in> A. a \\<sqsubseteq> b} sup \\<Longrightarrow> is_Sup A sup\" +proof - + let ?B = \"{b. \\<forall>a \\<in> A. a \\<sqsubseteq> b}\" + assume is_Inf: \"is_Inf ?B sup\" + show \"is_Sup A sup\" + proof + fix x assume x: \"x \\<in> A\" + from is_Inf show \"x \\<sqsubseteq> sup\" + proof (rule is_Inf_greatest) + fix y assume \"y \\<in> ?B\" + hence \"\\<forall>a \\<in> A. a \\<sqsubseteq> y\" .. + from this x show \"x \\<sqsubseteq> y\" .. + qed + next + fix z assume \"\\<forall>x \\<in> A. x \\<sqsubseteq> z\" + hence \"z \\<in> ?B\" .. + with is_Inf show \"sup \\<sqsubseteq> z\" .. + qed +qed + +theorem Sup_Inf: \"is_Sup {b. \\<forall>a \\<in> A. b \\<sqsubseteq> a} inf \\<Longrightarrow> is_Inf A inf\" +proof - + assume \"is_Sup {b. \\<forall>a \\<in> A. b \\<sqsubseteq> a} inf\" + hence \"is_Inf (dual {b. \\<forall>a \\<in> A. dual a \\<sqsubseteq> dual b}) (dual inf)\" + by (simp only: dual_Inf dual_leq) + also have \"dual {b. \\<forall>a \\<in> A. dual a \\<sqsubseteq> dual b} = {b'. \\<forall>a' \\<in> dual A. a' \\<sqsubseteq> b'}\" + by (auto iff: dual_ball dual_Collect) (* FIXME !? *) + finally have \"is_Inf \\<dots> (dual inf)\" . + hence \"is_Sup (dual A) (dual inf)\" + by (rule Inf_Sup) + thus ?thesis .. +qed + +end --- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/src/HOL/Lattice/CompleteLattice.thy Fri Oct 06 01:04:56 2000 +0200 @@ -0,0 +1,409 @@ +(* Title: HOL/Lattice/CompleteLattice.thy + ID:$Id$+ Author: Markus Wenzel, TU Muenchen +*) + +header {* Complete lattices *} + +theory CompleteLattice = Lattice: + +subsection {* Complete lattice operations *} + +text {* + A \\emph{complete lattice} is a partial order with general + (infinitary) infimum of any set of elements. General supremum + exists as well, as a consequence of the connection of infinitary + bounds (see \\S\\ref{sec:connect-bounds}). +*} + +axclass complete_lattice < partial_order + ex_Inf: \"\\<exists>inf. is_Inf A inf\" + +theorem ex_Sup: \"\\<exists>sup::'a::complete_lattice. is_Sup A sup\" +proof - + from ex_Inf obtain sup where \"is_Inf {b. \\<forall>a\\<in>A. a \\<sqsubseteq> b} sup\" by blast + hence \"is_Sup A sup\" by (rule Inf_Sup) + thus ?thesis .. +qed + +text {* + The general @{text \\<Sqinter>} (meet) and @{text \\<Squnion>} (join) operations select + such infimum and supremum elements. +*} + +consts + Meet :: \"'a::complete_lattice set \\<Rightarrow> 'a\" + Join :: \"'a::complete_lattice set \\<Rightarrow> 'a\" +syntax (symbols) + Meet :: \"'a::complete_lattice set \\<Rightarrow> 'a\" (\"\\<Sqinter>_\" 90) + Join :: \"'a::complete_lattice set \\<Rightarrow> 'a\" (\"\\<Squnion>_\" 90) +defs + Meet_def: \"\\<Sqinter>A \\<equiv> SOME inf. is_Inf A inf\" + Join_def: \"\\<Squnion>A \\<equiv> SOME sup. is_Sup A sup\" + +text {* + Due to unique existence of bounds, the complete lattice operations + may be exhibited as follows. +*} + +lemma Meet_equality [elim?]: \"is_Inf A inf \\<Longrightarrow> \\<Sqinter>A = inf\" +proof (unfold Meet_def) + assume \"is_Inf A inf\" + thus \"(SOME inf. is_Inf A inf) = inf\" + by (rule some_equality) (rule is_Inf_uniq) +qed + +lemma MeetI [intro?]: + \"(\\<And>a. a \\<in> A \\<Longrightarrow> inf \\<sqsubseteq> a) \\<Longrightarrow> + (\\<And>b. \\<forall>a \\<in> A. b \\<sqsubseteq> a \\<Longrightarrow> b \\<sqsubseteq> inf) \\<Longrightarrow> + \\<Sqinter>A = inf\" + by (rule Meet_equality, rule is_InfI) blast+ + +lemma Join_equality [elim?]: \"is_Sup A sup \\<Longrightarrow> \\<Squnion>A = sup\" +proof (unfold Join_def) + assume \"is_Sup A sup\" + thus \"(SOME sup. is_Sup A sup) = sup\" + by (rule some_equality) (rule is_Sup_uniq) +qed + +lemma JoinI [intro?]: + \"(\\<And>a. a \\<in> A \\<Longrightarrow> a \\<sqsubseteq> sup) \\<Longrightarrow> + (\\<And>b. \\<forall>a \\<in> A. a \\<sqsubseteq> b \\<Longrightarrow> sup \\<sqsubseteq> b) \\<Longrightarrow> + \\<Squnion>A = sup\" + by (rule Join_equality, rule is_SupI) blast+ + + +text {* + \\medskip The @{text \\<Sqinter>} and @{text \\<Squnion>} operations indeed determine + bounds on a complete lattice structure. +*} + +lemma is_Inf_Meet [intro?]: \"is_Inf A (\\<Sqinter>A)\" +proof (unfold Meet_def) + from ex_Inf show \"is_Inf A (SOME inf. is_Inf A inf)\" + by (rule ex_someI) +qed + +lemma Meet_greatest [intro?]: \"(\\<And>a. a \\<in> A \\<Longrightarrow> x \\<sqsubseteq> a) \\<Longrightarrow> x \\<sqsubseteq> \\<Sqinter>A\" + by (rule is_Inf_greatest, rule is_Inf_Meet) blast + +lemma Meet_lower [intro?]: \"a \\<in> A \\<Longrightarrow> \\<Sqinter>A \\<sqsubseteq> a\" + by (rule is_Inf_lower) (rule is_Inf_Meet) + + +lemma is_Sup_Join [intro?]: \"is_Sup A (\\<Squnion>A)\" +proof (unfold Join_def) + from ex_Sup show \"is_Sup A (SOME sup. is_Sup A sup)\" + by (rule ex_someI) +qed + +lemma Join_least [intro?]: \"(\\<And>a. a \\<in> A \\<Longrightarrow> a \\<sqsubseteq> x) \\<Longrightarrow> \\<Squnion>A \\<sqsubseteq> x\" + by (rule is_Sup_least, rule is_Sup_Join) blast +lemma Join_lower [intro?]: \"a \\<in> A \\<Longrightarrow> a \\<sqsubseteq> \\<Squnion>A\" + by (rule is_Sup_upper) (rule is_Sup_Join) + + +subsection {* The Knaster-Tarski Theorem *} + +text {* + The Knaster-Tarski Theorem (in its simplest formulation) states that + any monotone function on a complete lattice has a least fixed-point + (see \\cite[pages 93--94]{Davey-Priestley:1990} for example). This + is a consequence of the basic boundary properties of the complete + lattice operations. +*} + +theorem Knaster_Tarski: + \"(\\<And>x y. x \\<sqsubseteq> y \\<Longrightarrow> f x \\<sqsubseteq> f y) \\<Longrightarrow> \\<exists>a::'a::complete_lattice. f a = a\" +proof + assume mono: \"\\<And>x y. x \\<sqsubseteq> y \\<Longrightarrow> f x \\<sqsubseteq> f y\" + let ?H = \"{u. f u \\<sqsubseteq> u}\" let ?a = \"\\<Sqinter>?H\" + have ge: \"f ?a \\<sqsubseteq> ?a\" + proof + fix x assume x: \"x \\<in> ?H\" + hence \"?a \\<sqsubseteq> x\" .. + hence \"f ?a \\<sqsubseteq> f x\" by (rule mono) + also from x have \"... \\<sqsubseteq> x\" .. + finally show \"f ?a \\<sqsubseteq> x\" . + qed + also have \"?a \\<sqsubseteq> f ?a\" + proof + from ge have \"f (f ?a) \\<sqsubseteq> f ?a\" by (rule mono) + thus \"f ?a : ?H\" .. + qed + finally show \"f ?a = ?a\" . +qed + + +subsection {* Bottom and top elements *} + +text {* + With general bounds available, complete lattices also have least and + greatest elements. +*} + +constdefs + bottom :: \"'a::complete_lattice\" (\"\\<bottom>\") + \"\\<bottom> \\<equiv> \\<Sqinter>UNIV\" + top :: \"'a::complete_lattice\" (\"\\<top>\") + \"\\<top> \\<equiv> \\<Squnion>UNIV\" + +lemma bottom_least [intro?]: \"\\<bottom> \\<sqsubseteq> x\" +proof (unfold bottom_def) + have \"x \\<in> UNIV\" .. + thus \"\\<Sqinter>UNIV \\<sqsubseteq> x\" .. +qed + +lemma bottomI [intro?]: \"(\\<And>a. x \\<sqsubseteq> a) \\<Longrightarrow> \\<bottom> = x\" +proof (unfold bottom_def) + assume \"\\<And>a. x \\<sqsubseteq> a\" + show \"\\<Sqinter>UNIV = x\" + proof + fix a show \"x \\<sqsubseteq> a\" . + next + fix b :: \"'a::complete_lattice\" + assume b: \"\\<forall>a \\<in> UNIV. b \\<sqsubseteq> a\" + have \"x \\<in> UNIV\" .. + with b show \"b \\<sqsubseteq> x\" .. + qed +qed + +lemma top_greatest [intro?]: \"x \\<sqsubseteq> \\<top>\" +proof (unfold top_def) + have \"x \\<in> UNIV\" .. + thus \"x \\<sqsubseteq> \\<Squnion>UNIV\" .. +qed + +lemma topI [intro?]: \"(\\<And>a. a \\<sqsubseteq> x) \\<Longrightarrow> \\<top> = x\" +proof (unfold top_def) + assume \"\\<And>a. a \\<sqsubseteq> x\" + show \"\\<Squnion>UNIV = x\" + proof + fix a show \"a \\<sqsubseteq> x\" . + next + fix b :: \"'a::complete_lattice\" + assume b: \"\\<forall>a \\<in> UNIV. a \\<sqsubseteq> b\" + have \"x \\<in> UNIV\" .. + with b show \"x \\<sqsubseteq> b\" .. + qed +qed + + +subsection {* Duality *} + +text {* + The class of complete lattices is closed under formation of dual + structures. +*} + +instance dual :: (complete_lattice) complete_lattice +proof intro_classes + fix A' :: \"'a::complete_lattice dual set\" + show \"\\<exists>inf'. is_Inf A' inf'\" + proof - + have \"\\<exists>sup. is_Sup (undual A') sup\" by (rule ex_Sup) + hence \"\\<exists>sup. is_Inf (dual undual A') (dual sup)\" by (simp only: dual_Inf) + thus ?thesis by (simp add: dual_ex [symmetric] image_compose [symmetric]) + qed +qed + +text {* + Apparently, the @{text \\<Sqinter>} and @{text \\<Squnion>} operations are dual to each + other. +*} + +theorem dual_Meet [intro?]: \"dual (\\<Sqinter>A) = \\<Squnion>(dual A)\" +proof - + from is_Inf_Meet have \"is_Sup (dual A) (dual (\\<Sqinter>A))\" .. + hence \"\\<Squnion>(dual A) = dual (\\<Sqinter>A)\" .. + thus ?thesis .. +qed + +theorem dual_Join [intro?]: \"dual (\\<Squnion>A) = \\<Sqinter>(dual A)\" +proof - + from is_Sup_Join have \"is_Inf (dual A) (dual (\\<Squnion>A))\" .. + hence \"\\<Sqinter>(dual A) = dual (\\<Squnion>A)\" .. + thus ?thesis .. +qed + +text {* + Likewise are @{text \\<bottom>} and @{text \\<top>} duals of each other. +*} + +theorem dual_bottom [intro?]: \"dual \\<bottom> = \\<top>\" +proof - + have \"\\<top> = dual \\<bottom>\" + proof + fix a' have \"\\<bottom> \\<sqsubseteq> undual a'\" .. + hence \"dual (undual a') \\<sqsubseteq> dual \\<bottom>\" .. + thus \"a' \\<sqsubseteq> dual \\<bottom>\" by simp + qed + thus ?thesis .. +qed + +theorem dual_top [intro?]: \"dual \\<top> = \\<bottom>\" +proof - + have \"\\<bottom> = dual \\<top>\" + proof + fix a' have \"undual a' \\<sqsubseteq> \\<top>\" .. + hence \"dual \\<top> \\<sqsubseteq> dual (undual a')\" .. + thus \"dual \\<top> \\<sqsubseteq> a'\" by simp + qed + thus ?thesis .. +qed + + +subsection {* Complete lattices are lattices *} + +text {* + Complete lattices (with general bounds) available are indeed plain + lattices as well. This holds due to the connection of general + versus binary bounds that has been formally established in + \\S\\ref{sec:gen-bin-bounds}. +*} + +lemma is_inf_binary: \"is_inf x y (\\<Sqinter>{x, y})\" +proof - + have \"is_Inf {x, y} (\\<Sqinter>{x, y})\" .. + thus ?thesis by (simp only: is_Inf_binary) +qed + +lemma is_sup_binary: \"is_sup x y (\\<Squnion>{x, y})\" +proof - + have \"is_Sup {x, y} (\\<Squnion>{x, y})\" .. + thus ?thesis by (simp only: is_Sup_binary) +qed + +instance complete_lattice < lattice +proof intro_classes + fix x y :: \"'a::complete_lattice\" + from is_inf_binary show \"\\<exists>inf. is_inf x y inf\" .. + from is_sup_binary show \"\\<exists>sup. is_sup x y sup\" .. +qed + +theorem meet_binary: \"x \\<sqinter> y = \\<Sqinter>{x, y}\" + by (rule meet_equality) (rule is_inf_binary) + +theorem join_binary: \"x \\<squnion> y = \\<Squnion>{x, y}\" + by (rule join_equality) (rule is_sup_binary) + + +subsection {* Complete lattices and set-theory operations *} + +text {* + The complete lattice operations are (anti) monotone wrt.\\ set + inclusion. +*} + +theorem Meet_subset_antimono: \"A \\<subseteq> B \\<Longrightarrow> \\<Sqinter>B \\<sqsubseteq> \\<Sqinter>A\" +proof (rule Meet_greatest) + fix a assume \"a \\<in> A\" + also assume \"A \\<subseteq> B\" + finally have \"a \\<in> B\" . + thus \"\\<Sqinter>B \\<sqsubseteq> a\" .. +qed + +theorem Join_subset_mono: \"A \\<subseteq> B \\<Longrightarrow> \\<Squnion>A \\<sqsubseteq> \\<Squnion>B\" +proof - + assume \"A \\<subseteq> B\" + hence \"dual A \\<subseteq> dual B\" by blast + hence \"\\<Sqinter>(dual B) \\<sqsubseteq> \\<Sqinter>(dual A)\" by (rule Meet_subset_antimono) + hence \"dual (\\<Squnion>B) \\<sqsubseteq> dual (\\<Squnion>A)\" by (simp only: dual_Join) + thus ?thesis by (simp only: dual_leq) +qed + +text {* + Bounds over unions of sets may be obtained separately. +*} + +theorem Meet_Un: \"\\<Sqinter>(A \\<union> B) = \\<Sqinter>A \\<sqinter> \\<Sqinter>B\" +proof + fix a assume \"a \\<in> A \\<union> B\" + thus \"\\<Sqinter>A \\<sqinter> \\<Sqinter>B \\<sqsubseteq> a\" + proof + assume a: \"a \\<in> A\" + have \"\\<Sqinter>A \\<sqinter> \\<Sqinter>B \\<sqsubseteq> \\<Sqinter>A\" .. + also from a have \"\\<dots> \\<sqsubseteq> a\" .. + finally show ?thesis . + next + assume a: \"a \\<in> B\" + have \"\\<Sqinter>A \\<sqinter> \\<Sqinter>B \\<sqsubseteq> \\<Sqinter>B\" .. + also from a have \"\\<dots> \\<sqsubseteq> a\" .. + finally show ?thesis . + qed +next + fix b assume b: \"\\<forall>a \\<in> A \\<union> B. b \\<sqsubseteq> a\" + show \"b \\<sqsubseteq> \\<Sqinter>A \\<sqinter> \\<Sqinter>B\" + proof + show \"b \\<sqsubseteq> \\<Sqinter>A\" + proof + fix a assume \"a \\<in> A\" + hence \"a \\<in> A \\<union> B\" .. + with b show \"b \\<sqsubseteq> a\" .. + qed + show \"b \\<sqsubseteq> \\<Sqinter>B\" + proof + fix a assume \"a \\<in> B\" + hence \"a \\<in> A \\<union> B\" .. + with b show \"b \\<sqsubseteq> a\" .. + qed + qed +qed + +theorem Join_Un: \"\\<Squnion>(A \\<union> B) = \\<Squnion>A \\<squnion> \\<Squnion>B\" +proof - + have \"dual (\\<Squnion>(A \\<union> B)) = \\<Sqinter>(dual A \\<union> dual B)\" + by (simp only: dual_Join image_Un) + also have \"\\<dots> = \\<Sqinter>(dual A) \\<sqinter> \\<Sqinter>(dual B)\" + by (rule Meet_Un) + also have \"\\<dots> = dual (\\<Squnion>A \\<squnion> \\<Squnion>B)\" + by (simp only: dual_join dual_Join) + finally show ?thesis .. +qed + +text {* + Bounds over singleton sets are trivial. +*} + +theorem Meet_singleton: \"\\<Sqinter>{x} = x\" +proof + fix a assume \"a \\<in> {x}\" + hence \"a = x\" by simp + thus \"x \\<sqsubseteq> a\" by (simp only: leq_refl) +next + fix b assume \"\\<forall>a \\<in> {x}. b \\<sqsubseteq> a\" + thus \"b \\<sqsubseteq> x\" by simp +qed + +theorem Join_singleton: \"\\<Squnion>{x} = x\" +proof - + have \"dual (\\<Squnion>{x}) = \\<Sqinter>{dual x}\" by (simp add: dual_Join) + also have \"\\<dots> = dual x\" by (rule Meet_singleton) + finally show ?thesis .. +qed + +text {* + Bounds over the empty and universal set correspond to each other. +*} + +theorem Meet_empty: \"\\<Sqinter>{} = \\<Squnion>UNIV\" +proof + fix a :: \"'a::complete_lattice\" + assume \"a \\<in> {}\" + hence False by simp + thus \"\\<Squnion>UNIV \\<sqsubseteq> a\" .. +next + fix b :: \"'a::complete_lattice\" + have \"b \\<in> UNIV\" .. + thus \"b \\<sqsubseteq> \\<Squnion>UNIV\" .. +qed + +theorem Join_empty: \"\\<Squnion>{} = \\<Sqinter>UNIV\" +proof - + have \"dual (\\<Squnion>{}) = \\<Sqinter>{}\" by (simp add: dual_Join) + also have \"\\<dots> = \\<Squnion>UNIV\" by (rule Meet_empty) + also have \"\\<dots> = dual (\\<Sqinter>UNIV)\" by (simp add: dual_Meet) + finally show ?thesis .. +qed + +end --- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/src/HOL/Lattice/Lattice.thy Fri Oct 06 01:04:56 2000 +0200 @@ -0,0 +1,609 @@ +(* Title: HOL/Lattice/Lattice.thy + ID:$Id$+ Author: Markus Wenzel, TU Muenchen +*) + +header {* Lattices *} + +theory Lattice = Bounds: + +subsection {* Lattice operations *} + +text {* + A \\emph{lattice} is a partial order with infimum and supremum of any + two elements (thus any \\emph{finite} number of elements have bounds + as well). +*} + +axclass lattice < partial_order + ex_inf: \"\\<exists>inf. is_inf x y inf\" + ex_sup: \"\\<exists>sup. is_sup x y sup\" + +text {* + The @{text \\<sqinter>} (meet) and @{text \\<squnion>} (join) operations select such + infimum and supremum elements. +*} + +consts + meet :: \"'a::lattice \\<Rightarrow> 'a \\<Rightarrow> 'a\" (infixl \"&&\" 70) + join :: \"'a::lattice \\<Rightarrow> 'a \\<Rightarrow> 'a\" (infixl \"||\" 65) +syntax (symbols) + meet :: \"'a::lattice \\<Rightarrow> 'a \\<Rightarrow> 'a\" (infixl \"\\<sqinter>\" 70) + join :: \"'a::lattice \\<Rightarrow> 'a \\<Rightarrow> 'a\" (infixl \"\\<squnion>\" 65) +defs + meet_def: \"x && y \\<equiv> SOME inf. is_inf x y inf\" + join_def: \"x || y \\<equiv> SOME sup. is_sup x y sup\" + +text {* + Due to unique existence of bounds, the lattice operations may be + exhibited as follows. +*} + +lemma meet_equality [elim?]: \"is_inf x y inf \\<Longrightarrow> x \\<sqinter> y = inf\" +proof (unfold meet_def) + assume \"is_inf x y inf\" + thus \"(SOME inf. is_inf x y inf) = inf\" + by (rule some_equality) (rule is_inf_uniq) +qed + +lemma meetI [intro?]: + \"inf \\<sqsubseteq> x \\<Longrightarrow> inf \\<sqsubseteq> y \\<Longrightarrow> (\\<And>z. z \\<sqsubseteq> x \\<Longrightarrow> z \\<sqsubseteq> y \\<Longrightarrow> z \\<sqsubseteq> inf) \\<Longrightarrow> x \\<sqinter> y = inf\" + by (rule meet_equality, rule is_infI) blast+ + +lemma join_equality [elim?]: \"is_sup x y sup \\<Longrightarrow> x \\<squnion> y = sup\" +proof (unfold join_def) + assume \"is_sup x y sup\" + thus \"(SOME sup. is_sup x y sup) = sup\" + by (rule some_equality) (rule is_sup_uniq) +qed + +lemma joinI [intro?]: \"x \\<sqsubseteq> sup \\<Longrightarrow> y \\<sqsubseteq> sup \\<Longrightarrow> + (\\<And>z. x \\<sqsubseteq> z \\<Longrightarrow> y \\<sqsubseteq> z \\<Longrightarrow> sup \\<sqsubseteq> z) \\<Longrightarrow> x \\<squnion> y = sup\" + by (rule join_equality, rule is_supI) blast+ + + +text {* + \\medskip The @{text \\<sqinter>} and @{text \\<squnion>} operations indeed determine + bounds on a lattice structure. +*} + +lemma is_inf_meet [intro?]: \"is_inf x y (x \\<sqinter> y)\" +proof (unfold meet_def) + from ex_inf show \"is_inf x y (SOME inf. is_inf x y inf)\" + by (rule ex_someI) +qed + +lemma meet_greatest [intro?]: \"z \\<sqsubseteq> x \\<Longrightarrow> z \\<sqsubseteq> y \\<Longrightarrow> z \\<sqsubseteq> x \\<sqinter> y\" + by (rule is_inf_greatest) (rule is_inf_meet) + +lemma meet_lower1 [intro?]: \"x \\<sqinter> y \\<sqsubseteq> x\" + by (rule is_inf_lower) (rule is_inf_meet) + +lemma meet_lower2 [intro?]: \"x \\<sqinter> y \\<sqsubseteq> y\" + by (rule is_inf_lower) (rule is_inf_meet) + + +lemma is_sup_join [intro?]: \"is_sup x y (x \\<squnion> y)\" +proof (unfold join_def) + from ex_sup show \"is_sup x y (SOME sup. is_sup x y sup)\" + by (rule ex_someI) +qed + +lemma join_least [intro?]: \"x \\<sqsubseteq> z \\<Longrightarrow> y \\<sqsubseteq> z \\<Longrightarrow> x \\<squnion> y \\<sqsubseteq> z\" + by (rule is_sup_least) (rule is_sup_join) + +lemma join_upper1 [intro?]: \"x \\<sqsubseteq> x \\<squnion> y\" + by (rule is_sup_upper) (rule is_sup_join) + +lemma join_upper2 [intro?]: \"y \\<sqsubseteq> x \\<squnion> y\" + by (rule is_sup_upper) (rule is_sup_join) + + +subsection {* Duality *} + +text {* + The class of lattices is closed under formation of dual structures. + This means that for any theorem of lattice theory, the dualized + statement holds as well; this important fact simplifies many proofs + of lattice theory. +*} + +instance dual :: (lattice) lattice +proof intro_classes + fix x' y' :: \"'a::lattice dual\" + show \"\\<exists>inf'. is_inf x' y' inf'\" + proof - + have \"\\<exists>sup. is_sup (undual x') (undual y') sup\" by (rule ex_sup) + hence \"\\<exists>sup. is_inf (dual (undual x')) (dual (undual y')) (dual sup)\" + by (simp only: dual_inf) + thus ?thesis by (simp add: dual_ex [symmetric]) + qed + show \"\\<exists>sup'. is_sup x' y' sup'\" + proof - + have \"\\<exists>inf. is_inf (undual x') (undual y') inf\" by (rule ex_inf) + hence \"\\<exists>inf. is_sup (dual (undual x')) (dual (undual y')) (dual inf)\" + by (simp only: dual_sup) + thus ?thesis by (simp add: dual_ex [symmetric]) + qed +qed + +text {* + Apparently, the @{text \\<sqinter>} and @{text \\<squnion>} operations are dual to each + other. +*} + +theorem dual_meet [intro?]: \"dual (x \\<sqinter> y) = dual x \\<squnion> dual y\" +proof - + from is_inf_meet have \"is_sup (dual x) (dual y) (dual (x \\<sqinter> y))\" .. + hence \"dual x \\<squnion> dual y = dual (x \\<sqinter> y)\" .. + thus ?thesis .. +qed + +theorem dual_join [intro?]: \"dual (x \\<squnion> y) = dual x \\<sqinter> dual y\" +proof - + from is_sup_join have \"is_inf (dual x) (dual y) (dual (x \\<squnion> y))\" .. + hence \"dual x \\<sqinter> dual y = dual (x \\<squnion> y)\" .. + thus ?thesis .. +qed + + +subsection {* Algebraic properties \\label{sec:lattice-algebra} *} + +text {* + The @{text \\<sqinter>} and @{text \\<squnion>} operations have to following + characteristic algebraic properties: associative (A), commutative + (C), and absorptive (AB). +*} + +theorem meet_assoc: \"(x \\<sqinter> y) \\<sqinter> z = x \\<sqinter> (y \\<sqinter> z)\" +proof + show \"x \\<sqinter> (y \\<sqinter> z) \\<sqsubseteq> x \\<sqinter> y\" + proof + show \"x \\<sqinter> (y \\<sqinter> z) \\<sqsubseteq> x\" .. + show \"x \\<sqinter> (y \\<sqinter> z) \\<sqsubseteq> y\" + proof - + have \"x \\<sqinter> (y \\<sqinter> z) \\<sqsubseteq> y \\<sqinter> z\" .. + also have \"\\<dots> \\<sqsubseteq> y\" .. + finally show ?thesis . + qed + qed + show \"x \\<sqinter> (y \\<sqinter> z) \\<sqsubseteq> z\" + proof - + have \"x \\<sqinter> (y \\<sqinter> z) \\<sqsubseteq> y \\<sqinter> z\" .. + also have \"\\<dots> \\<sqsubseteq> z\" .. + finally show ?thesis . + qed + fix w assume \"w \\<sqsubseteq> x \\<sqinter> y\" and \"w \\<sqsubseteq> z\" + show \"w \\<sqsubseteq> x \\<sqinter> (y \\<sqinter> z)\" + proof + show \"w \\<sqsubseteq> x\" + proof - + have \"w \\<sqsubseteq> x \\<sqinter> y\" . + also have \"\\<dots> \\<sqsubseteq> x\" .. + finally show ?thesis . + qed + show \"w \\<sqsubseteq> y \\<sqinter> z\" + proof + show \"w \\<sqsubseteq> y\" + proof - + have \"w \\<sqsubseteq> x \\<sqinter> y\" . + also have \"\\<dots> \\<sqsubseteq> y\" .. + finally show ?thesis . + qed + show \"w \\<sqsubseteq> z\" . + qed + qed +qed + +theorem join_assoc: \"(x \\<squnion> y) \\<squnion> z = x \\<squnion> (y \\<squnion> z)\" +proof - + have \"dual ((x \\<squnion> y) \\<squnion> z) = (dual x \\<sqinter> dual y) \\<sqinter> dual z\" + by (simp only: dual_join) + also have \"\\<dots> = dual x \\<sqinter> (dual y \\<sqinter> dual z)\" + by (rule meet_assoc) + also have \"\\<dots> = dual (x \\<squnion> (y \\<squnion> z))\" + by (simp only: dual_join) + finally show ?thesis .. +qed + +theorem meet_commute: \"x \\<sqinter> y = y \\<sqinter> x\" +proof + show \"y \\<sqinter> x \\<sqsubseteq> x\" .. + show \"y \\<sqinter> x \\<sqsubseteq> y\" .. + fix z assume \"z \\<sqsubseteq> x\" and \"z \\<sqsubseteq> y\" + show \"z \\<sqsubseteq> y \\<sqinter> x\" .. +qed + +theorem join_commute: \"x \\<squnion> y = y \\<squnion> x\" +proof - + have \"dual (x \\<squnion> y) = dual x \\<sqinter> dual y\" .. + also have \"\\<dots> = dual y \\<sqinter> dual x\" + by (rule meet_commute) + also have \"\\<dots> = dual (y \\<squnion> x)\" + by (simp only: dual_join) + finally show ?thesis .. +qed + +theorem meet_join_absorb: \"x \\<sqinter> (x \\<squnion> y) = x\" +proof + show \"x \\<sqsubseteq> x\" .. + show \"x \\<sqsubseteq> x \\<squnion> y\" .. + fix z assume \"z \\<sqsubseteq> x\" and \"z \\<sqsubseteq> x \\<squnion> y\" + show \"z \\<sqsubseteq> x\" . +qed + +theorem join_meet_absorb: \"x \\<squnion> (x \\<sqinter> y) = x\" +proof - + have \"dual x \\<sqinter> (dual x \\<squnion> dual y) = dual x\" + by (rule meet_join_absorb) + hence \"dual (x \\<squnion> (x \\<sqinter> y)) = dual x\" + by (simp only: dual_meet dual_join) + thus ?thesis .. +qed + +text {* + \\medskip Some further algebraic properties hold as well. The + property idempotent (I) is a basic algebraic consequence of (AB). +*} + +theorem meet_idem: \"x \\<sqinter> x = x\" +proof - + have \"x \\<sqinter> (x \\<squnion> (x \\<sqinter> x)) = x\" by (rule meet_join_absorb) + also have \"x \\<squnion> (x \\<sqinter> x) = x\" by (rule join_meet_absorb) + finally show ?thesis . +qed + +theorem join_idem: \"x \\<squnion> x = x\" +proof - + have \"dual x \\<sqinter> dual x = dual x\" + by (rule meet_idem) + hence \"dual (x \\<squnion> x) = dual x\" + by (simp only: dual_join) + thus ?thesis .. +qed + +text {* + Meet and join are trivial for related elements. +*} + +theorem meet_related [elim?]: \"x \\<sqsubseteq> y \\<Longrightarrow> x \\<sqinter> y = x\" +proof + assume \"x \\<sqsubseteq> y\" + show \"x \\<sqsubseteq> x\" .. + show \"x \\<sqsubseteq> y\" . + fix z assume \"z \\<sqsubseteq> x\" and \"z \\<sqsubseteq> y\" show \"z \\<sqsubseteq> x\" . +qed + +theorem join_related [elim?]: \"x \\<sqsubseteq> y \\<Longrightarrow> x \\<squnion> y = y\" +proof - + assume \"x \\<sqsubseteq> y\" hence \"dual y \\<sqsubseteq> dual x\" .. + hence \"dual y \\<sqinter> dual x = dual y\" by (rule meet_related) + also have \"dual y \\<sqinter> dual x = dual (y \\<squnion> x)\" by (simp only: dual_join) + also have \"y \\<squnion> x = x \\<squnion> y\" by (rule join_commute) + finally show ?thesis .. +qed + + +subsection {* Order versus algebraic structure *} + +text {* + The @{text \\<sqinter>} and @{text \\<squnion>} operations are connected with the + underlying @{text \\<sqsubseteq>} relation in a canonical manner. +*} + +theorem meet_connection: \"(x \\<sqsubseteq> y) = (x \\<sqinter> y = x)\" +proof + assume \"x \\<sqsubseteq> y\" + hence \"is_inf x y x\" .. + thus \"x \\<sqinter> y = x\" .. +next + have \"x \\<sqinter> y \\<sqsubseteq> y\" .. + also assume \"x \\<sqinter> y = x\" + finally show \"x \\<sqsubseteq> y\" . +qed + +theorem join_connection: \"(x \\<sqsubseteq> y) = (x \\<squnion> y = y)\" +proof + assume \"x \\<sqsubseteq> y\" + hence \"is_sup x y y\" .. + thus \"x \\<squnion> y = y\" .. +next + have \"x \\<sqsubseteq> x \\<squnion> y\" .. + also assume \"x \\<squnion> y = y\" + finally show \"x \\<sqsubseteq> y\" . +qed + +text {* + \\medskip The most fundamental result of the meta-theory of lattices + is as follows (we do not prove it here). + + Given a structure with binary operations @{text \\<sqinter>} and @{text \\<squnion>} + such that (A), (C), and (AB) hold (cf.\\ + \\S\\ref{sec:lattice-algebra}). This structure represents a lattice, + if the relation @{term \"x \\<sqsubseteq> y\"} is defined as @{term \"x \\<sqinter> y = x\"} + (alternatively as @{term \"x \\<squnion> y = y\"}). Furthermore, infimum and + supremum with respect to this ordering coincide with the original + @{text \\<sqinter>} and @{text \\<squnion>} operations. +*} + + +subsection {* Example instances *} + +subsubsection {* Linear orders *} + +text {* + Linear orders with @{term minimum} and @{term minimum} operations + are a (degenerate) example of lattice structures. +*} + +constdefs + minimum :: \"'a::linear_order \\<Rightarrow> 'a \\<Rightarrow> 'a\" + \"minimum x y \\<equiv> (if x \\<sqsubseteq> y then x else y)\" + maximum :: \"'a::linear_order \\<Rightarrow> 'a \\<Rightarrow> 'a\" + \"maximum x y \\<equiv> (if x \\<sqsubseteq> y then y else x)\" + +lemma is_inf_minimum: \"is_inf x y (minimum x y)\" +proof + let ?min = \"minimum x y\" + from leq_linear show \"?min \\<sqsubseteq> x\" by (auto simp add: minimum_def) + from leq_linear show \"?min \\<sqsubseteq> y\" by (auto simp add: minimum_def) + fix z assume \"z \\<sqsubseteq> x\" and \"z \\<sqsubseteq> y\" + with leq_linear show \"z \\<sqsubseteq> ?min\" by (auto simp add: minimum_def) +qed + +lemma is_sup_maximum: \"is_sup x y (maximum x y)\" (* FIXME dualize!? *) +proof + let ?max = \"maximum x y\" + from leq_linear show \"x \\<sqsubseteq> ?max\" by (auto simp add: maximum_def) + from leq_linear show \"y \\<sqsubseteq> ?max\" by (auto simp add: maximum_def) + fix z assume \"x \\<sqsubseteq> z\" and \"y \\<sqsubseteq> z\" + with leq_linear show \"?max \\<sqsubseteq> z\" by (auto simp add: maximum_def) +qed + +instance linear_order < lattice +proof intro_classes + fix x y :: \"'a::linear_order\" + from is_inf_minimum show \"\\<exists>inf. is_inf x y inf\" .. + from is_sup_maximum show \"\\<exists>sup. is_sup x y sup\" .. +qed + +text {* + The lattice operations on linear orders indeed coincide with @{term + minimum} and @{term maximum}. +*} + +theorem meet_mimimum: \"x \\<sqinter> y = minimum x y\" + by (rule meet_equality) (rule is_inf_minimum) + +theorem meet_maximum: \"x \\<squnion> y = maximum x y\" + by (rule join_equality) (rule is_sup_maximum) + + + +subsubsection {* Binary products *} + +text {* + The class of lattices is closed under direct binary products (cf.\\ + also \\S\\ref{sec:prod-order}). +*} + +lemma is_inf_prod: \"is_inf p q (fst p \\<sqinter> fst q, snd p \\<sqinter> snd q)\" +proof + show \"(fst p \\<sqinter> fst q, snd p \\<sqinter> snd q) \\<sqsubseteq> p\" + proof - + have \"fst p \\<sqinter> fst q \\<sqsubseteq> fst p\" .. + moreover have \"snd p \\<sqinter> snd q \\<sqsubseteq> snd p\" .. + ultimately show ?thesis by (simp add: leq_prod_def) + qed + show \"(fst p \\<sqinter> fst q, snd p \\<sqinter> snd q) \\<sqsubseteq> q\" + proof - + have \"fst p \\<sqinter> fst q \\<sqsubseteq> fst q\" .. + moreover have \"snd p \\<sqinter> snd q \\<sqsubseteq> snd q\" .. + ultimately show ?thesis by (simp add: leq_prod_def) + qed + fix r assume rp: \"r \\<sqsubseteq> p\" and rq: \"r \\<sqsubseteq> q\" + show \"r \\<sqsubseteq> (fst p \\<sqinter> fst q, snd p \\<sqinter> snd q)\" + proof - + have \"fst r \\<sqsubseteq> fst p \\<sqinter> fst q\" + proof + from rp show \"fst r \\<sqsubseteq> fst p\" by (simp add: leq_prod_def) + from rq show \"fst r \\<sqsubseteq> fst q\" by (simp add: leq_prod_def) + qed + moreover have \"snd r \\<sqsubseteq> snd p \\<sqinter> snd q\" + proof + from rp show \"snd r \\<sqsubseteq> snd p\" by (simp add: leq_prod_def) + from rq show \"snd r \\<sqsubseteq> snd q\" by (simp add: leq_prod_def) + qed + ultimately show ?thesis by (simp add: leq_prod_def) + qed +qed + +lemma is_sup_prod: \"is_sup p q (fst p \\<squnion> fst q, snd p \\<squnion> snd q)\" (* FIXME dualize!? *) +proof + show \"p \\<sqsubseteq> (fst p \\<squnion> fst q, snd p \\<squnion> snd q)\" + proof - + have \"fst p \\<sqsubseteq> fst p \\<squnion> fst q\" .. + moreover have \"snd p \\<sqsubseteq> snd p \\<squnion> snd q\" .. + ultimately show ?thesis by (simp add: leq_prod_def) + qed + show \"q \\<sqsubseteq> (fst p \\<squnion> fst q, snd p \\<squnion> snd q)\" + proof - + have \"fst q \\<sqsubseteq> fst p \\<squnion> fst q\" .. + moreover have \"snd q \\<sqsubseteq> snd p \\<squnion> snd q\" .. + ultimately show ?thesis by (simp add: leq_prod_def) + qed + fix r assume \"pr\": \"p \\<sqsubseteq> r\" and qr: \"q \\<sqsubseteq> r\" + show \"(fst p \\<squnion> fst q, snd p \\<squnion> snd q) \\<sqsubseteq> r\" + proof - + have \"fst p \\<squnion> fst q \\<sqsubseteq> fst r\" + proof + from \"pr\" show \"fst p \\<sqsubseteq> fst r\" by (simp add: leq_prod_def) + from qr show \"fst q \\<sqsubseteq> fst r\" by (simp add: leq_prod_def) + qed + moreover have \"snd p \\<squnion> snd q \\<sqsubseteq> snd r\" + proof + from \"pr\" show \"snd p \\<sqsubseteq> snd r\" by (simp add: leq_prod_def) + from qr show \"snd q \\<sqsubseteq> snd r\" by (simp add: leq_prod_def) + qed + ultimately show ?thesis by (simp add: leq_prod_def) + qed +qed + +instance * :: (lattice, lattice) lattice +proof intro_classes + fix p q :: \"'a::lattice \\<times> 'b::lattice\" + from is_inf_prod show \"\\<exists>inf. is_inf p q inf\" .. + from is_sup_prod show \"\\<exists>sup. is_sup p q sup\" .. +qed + +text {* + The lattice operations on a binary product structure indeed coincide + with the products of the original ones. +*} + +theorem meet_prod: \"p \\<sqinter> q = (fst p \\<sqinter> fst q, snd p \\<sqinter> snd q)\" + by (rule meet_equality) (rule is_inf_prod) + +theorem join_prod: \"p \\<squnion> q = (fst p \\<squnion> fst q, snd p \\<squnion> snd q)\" + by (rule join_equality) (rule is_sup_prod) + + +subsubsection {* General products *} + +text {* + The class of lattices is closed under general products (function + spaces) as well (cf.\\ also \\S\\ref{sec:fun-order}). +*} + +lemma is_inf_fun: \"is_inf f g (\\<lambda>x. f x \\<sqinter> g x)\" +proof + show \"(\\<lambda>x. f x \\<sqinter> g x) \\<sqsubseteq> f\" + proof + fix x show \"f x \\<sqinter> g x \\<sqsubseteq> f x\" .. + qed + show \"(\\<lambda>x. f x \\<sqinter> g x) \\<sqsubseteq> g\" + proof + fix x show \"f x \\<sqinter> g x \\<sqsubseteq> g x\" .. + qed + fix h assume hf: \"h \\<sqsubseteq> f\" and hg: \"h \\<sqsubseteq> g\" + show \"h \\<sqsubseteq> (\\<lambda>x. f x \\<sqinter> g x)\" + proof + fix x + show \"h x \\<sqsubseteq> f x \\<sqinter> g x\" + proof + from hf show \"h x \\<sqsubseteq> f x\" .. + from hg show \"h x \\<sqsubseteq> g x\" .. + qed + qed +qed + +lemma is_sup_fun: \"is_sup f g (\\<lambda>x. f x \\<squnion> g x)\" (* FIXME dualize!? *) +proof + show \"f \\<sqsubseteq> (\\<lambda>x. f x \\<squnion> g x)\" + proof + fix x show \"f x \\<sqsubseteq> f x \\<squnion> g x\" .. + qed + show \"g \\<sqsubseteq> (\\<lambda>x. f x \\<squnion> g x)\" + proof + fix x show \"g x \\<sqsubseteq> f x \\<squnion> g x\" .. + qed + fix h assume fh: \"f \\<sqsubseteq> h\" and gh: \"g \\<sqsubseteq> h\" + show \"(\\<lambda>x. f x \\<squnion> g x) \\<sqsubseteq> h\" + proof + fix x + show \"f x \\<squnion> g x \\<sqsubseteq> h x\" + proof + from fh show \"f x \\<sqsubseteq> h x\" .. + from gh show \"g x \\<sqsubseteq> h x\" .. + qed + qed +qed + +instance fun :: (\"term\", lattice) lattice +proof intro_classes + fix f g :: \"'a \\<Rightarrow> 'b::lattice\" + show \"\\<exists>inf. is_inf f g inf\" by rule (rule is_inf_fun) (* FIXME @{text \"from \\<dots> show \\<dots> ..\"} does not work!? unification incompleteness!? *) + show \"\\<exists>sup. is_sup f g sup\" by rule (rule is_sup_fun) +qed + +text {* + The lattice operations on a general product structure (function + space) indeed emerge by point-wise lifting of the original ones. +*} + +theorem meet_fun: \"f \\<sqinter> g = (\\<lambda>x. f x \\<sqinter> g x)\" + by (rule meet_equality) (rule is_inf_fun) + +theorem join_fun: \"f \\<squnion> g = (\\<lambda>x. f x \\<squnion> g x)\" + by (rule join_equality) (rule is_sup_fun) + + +subsection {* Monotonicity and semi-morphisms *} + +text {* + The lattice operations are monotone in both argument positions. In + fact, monotonicity of the second position is trivial due to + commutativity. +*} + +theorem meet_mono: \"x \\<sqsubseteq> z \\<Longrightarrow> y \\<sqsubseteq> w \\<Longrightarrow> x \\<sqinter> y \\<sqsubseteq> z \\<sqinter> w\" +proof - + { + fix a b c :: \"'a::lattice\" + assume \"a \\<sqsubseteq> c\" have \"a \\<sqinter> b \\<sqsubseteq> c \\<sqinter> b\" + proof + have \"a \\<sqinter> b \\<sqsubseteq> a\" .. + also have \"\\<dots> \\<sqsubseteq> c\" . + finally show \"a \\<sqinter> b \\<sqsubseteq> c\" . + show \"a \\<sqinter> b \\<sqsubseteq> b\" .. + qed + } note this [elim?] + assume \"x \\<sqsubseteq> z\" hence \"x \\<sqinter> y \\<sqsubseteq> z \\<sqinter> y\" .. + also have \"\\<dots> = y \\<sqinter> z\" by (rule meet_commute) + also assume \"y \\<sqsubseteq> w\" hence \"y \\<sqinter> z \\<sqsubseteq> w \\<sqinter> z\" .. + also have \"\\<dots> = z \\<sqinter> w\" by (rule meet_commute) + finally show ?thesis . +qed + +theorem join_mono: \"x \\<sqsubseteq> z \\<Longrightarrow> y \\<sqsubseteq> w \\<Longrightarrow> x \\<squnion> y \\<sqsubseteq> z \\<squnion> w\" +proof - + assume \"x \\<sqsubseteq> z\" hence \"dual z \\<sqsubseteq> dual x\" .. + moreover assume \"y \\<sqsubseteq> w\" hence \"dual w \\<sqsubseteq> dual y\" .. + ultimately have \"dual z \\<sqinter> dual w \\<sqsubseteq> dual x \\<sqinter> dual y\" + by (rule meet_mono) + hence \"dual (z \\<squnion> w) \\<sqsubseteq> dual (x \\<squnion> y)\" + by (simp only: dual_join) + thus ?thesis .. +qed + +text {* + \\medskip A semi-morphisms is a function$f$that preserves the + lattice operations in the following manner: @{term \"f (x \\<sqinter> y) \\<sqsubseteq> f x + \\<sqinter> f y\"} and @{term \"f x \\<squnion> f y \\<sqsubseteq> f (x \\<squnion> y)\"}, respectively. Any of + these properties is equivalent with monotonicity. +*} (* FIXME dual version !? *) + +theorem meet_semimorph: + \"(\\<And>x y. f (x \\<sqinter> y) \\<sqsubseteq> f x \\<sqinter> f y) \\<equiv> (\\<And>x y. x \\<sqsubseteq> y \\<Longrightarrow> f x \\<sqsubseteq> f y)\" +proof + assume morph: \"\\<And>x y. f (x \\<sqinter> y) \\<sqsubseteq> f x \\<sqinter> f y\" + fix x y :: \"'a::lattice\" + assume \"x \\<sqsubseteq> y\" hence \"x \\<sqinter> y = x\" .. + hence \"x = x \\<sqinter> y\" .. + also have \"f \\<dots> \\<sqsubseteq> f x \\<sqinter> f y\" by (rule morph) + also have \"\\<dots> \\<sqsubseteq> f y\" .. + finally show \"f x \\<sqsubseteq> f y\" . +next + assume mono: \"\\<And>x y. x \\<sqsubseteq> y \\<Longrightarrow> f x \\<sqsubseteq> f y\" + show \"\\<And>x y. f (x \\<sqinter> y) \\<sqsubseteq> f x \\<sqinter> f y\" + proof - + fix x y + show \"f (x \\<sqinter> y) \\<sqsubseteq> f x \\<sqinter> f y\" + proof + have \"x \\<sqinter> y \\<sqsubseteq> x\" .. thus \"f (x \\<sqinter> y) \\<sqsubseteq> f x\" by (rule mono) + have \"x \\<sqinter> y \\<sqsubseteq> y\" .. thus \"f (x \\<sqinter> y) \\<sqsubseteq> f y\" by (rule mono) + qed + qed +qed + +end --- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/src/HOL/Lattice/Orders.thy Fri Oct 06 01:04:56 2000 +0200 @@ -0,0 +1,294 @@ +(* Title: HOL/Lattice/Orders.thy + ID:$Id$+ Author: Markus Wenzel, TU Muenchen +*) + +header {* Orders *} + +theory Orders = Main: + +subsection {* Ordered structures *} + +text {* + We define several classes of ordered structures over some type @{typ + 'a} with relation @{text \"\\<sqsubseteq> \\<Colon> 'a \\<Rightarrow> 'a \\<Rightarrow> bool\"}. For a + \\emph{quasi-order} that relation is required to be reflexive and + transitive, for a \\emph{partial order} it also has to be + anti-symmetric, while for a \\emph{linear order} all elements are + required to be related (in either direction). +*} + +axclass leq < \"term\" +consts + leq :: \"'a::leq \\<Rightarrow> 'a \\<Rightarrow> bool\" (infixl \"[=\" 50) +syntax (symbols) + leq :: \"'a::leq \\<Rightarrow> 'a \\<Rightarrow> bool\" (infixl \"\\<sqsubseteq>\" 50) + +axclass quasi_order < leq + leq_refl [intro?]: \"x \\<sqsubseteq> x\" + leq_trans [trans]: \"x \\<sqsubseteq> y \\<Longrightarrow> y \\<sqsubseteq> z \\<Longrightarrow> x \\<sqsubseteq> z\" + +axclass partial_order < quasi_order + leq_antisym [trans]: \"x \\<sqsubseteq> y \\<Longrightarrow> y \\<sqsubseteq> x \\<Longrightarrow> x = y\" + +axclass linear_order < partial_order + leq_linear: \"x \\<sqsubseteq> y \\<or> y \\<sqsubseteq> x\" + +lemma linear_order_cases: + \"((x::'a::linear_order) \\<sqsubseteq> y \\<Longrightarrow> C) \\<Longrightarrow> (y \\<sqsubseteq> x \\<Longrightarrow> C) \\<Longrightarrow> C\" + by (insert leq_linear) blast + + +subsection {* Duality *} + +text {* + The \\emph{dual} of an ordered structure is an isomorphic copy of the + underlying type, with the @{text \\<sqsubseteq>} relation defined as the inverse + of the original one. +*} + +datatype 'a dual = dual 'a + +consts + undual :: \"'a dual \\<Rightarrow> 'a\" +primrec + undual_dual: \"undual (dual x) = x\" + +instance dual :: (leq) leq + by intro_classes + +defs (overloaded) + leq_dual_def: \"x' \\<sqsubseteq> y' \\<equiv> undual y' \\<sqsubseteq> undual x'\" + +lemma undual_leq [iff?]: \"(undual x' \\<sqsubseteq> undual y') = (y' \\<sqsubseteq> x')\" + by (simp add: leq_dual_def) + +lemma dual_leq [iff?]: \"(dual x \\<sqsubseteq> dual y) = (y \\<sqsubseteq> x)\" + by (simp add: leq_dual_def) + +text {* + \\medskip Functions @{term dual} and @{term undual} are inverse to + each other; this entails the following fundamental properties. +*} + +lemma dual_undual [simp]: \"dual (undual x') = x'\" + by (cases x') simp + +lemma undual_dual_id [simp]: \"undual o dual = id\" + by (rule ext) simp + +lemma dual_undual_id [simp]: \"dual o undual = id\" + by (rule ext) simp + +text {* + \\medskip Since @{term dual} (and @{term undual}) are both injective + and surjective, the basic logical connectives (equality, + quantification etc.) are transferred as follows. +*} + +lemma undual_equality [iff?]: \"(undual x' = undual y') = (x' = y')\" + by (cases x', cases y') simp + +lemma dual_equality [iff?]: \"(dual x = dual y) = (x = y)\" + by simp + +lemma dual_ball [iff?]: \"(\\<forall>x \\<in> A. P (dual x)) = (\\<forall>x' \\<in> dual A. P x')\" +proof + assume a: \"\\<forall>x \\<in> A. P (dual x)\" + show \"\\<forall>x' \\<in> dual A. P x'\" + proof + fix x' assume x': \"x' \\<in> dual A\" + have \"undual x' \\<in> A\" + proof - + from x' have \"undual x' \\<in> undual dual A\" by simp + thus \"undual x' \\<in> A\" by (simp add: image_compose [symmetric]) + qed + with a have \"P (dual (undual x'))\" .. + also have \"\\<dots> = x'\" by simp + finally show \"P x'\" . + qed +next + assume a: \"\\<forall>x' \\<in> dual A. P x'\" + show \"\\<forall>x \\<in> A. P (dual x)\" + proof + fix x assume \"x \\<in> A\" + hence \"dual x \\<in> dual A\" by simp + with a show \"P (dual x)\" .. + qed +qed + +lemma range_dual [simp]: \"dual UNIV = UNIV\" +proof (rule surj_range) + have \"\\<And>x'. dual (undual x') = x'\" by simp + thus \"surj dual\" by (rule surjI) +qed + +lemma dual_all [iff?]: \"(\\<forall>x. P (dual x)) = (\\<forall>x'. P x')\" +proof - + have \"(\\<forall>x \\<in> UNIV. P (dual x)) = (\\<forall>x' \\<in> dual UNIV. P x')\" + by (rule dual_ball) + thus ?thesis by simp +qed + +lemma dual_ex: \"(\\<exists>x. P (dual x)) = (\\<exists>x'. P x')\" +proof - + have \"(\\<forall>x. \\<not> P (dual x)) = (\\<forall>x'. \\<not> P x')\" + by (rule dual_all) + thus ?thesis by blast +qed + +lemma dual_Collect: \"{dual x| x. P (dual x)} = {x'. P x'}\" +proof - + have \"{dual x| x. P (dual x)} = {x'. \\<exists>x''. x' = x'' \\<and> P x''}\" + by (simp only: dual_ex [symmetric]) + thus ?thesis by blast +qed + + +subsection {* Transforming orders *} + +subsubsection {* Duals *} + +text {* + The classes of quasi, partial, and linear orders are all closed + under formation of dual structures. +*} + +instance dual :: (quasi_order) quasi_order +proof intro_classes + fix x' y' z' :: \"'a::quasi_order dual\" + have \"undual x' \\<sqsubseteq> undual x'\" .. thus \"x' \\<sqsubseteq> x'\" .. + assume \"y' \\<sqsubseteq> z'\" hence \"undual z' \\<sqsubseteq> undual y'\" .. + also assume \"x' \\<sqsubseteq> y'\" hence \"undual y' \\<sqsubseteq> undual x'\" .. + finally show \"x' \\<sqsubseteq> z'\" .. +qed + +instance dual :: (partial_order) partial_order +proof intro_classes + fix x' y' :: \"'a::partial_order dual\" + assume \"y' \\<sqsubseteq> x'\" hence \"undual x' \\<sqsubseteq> undual y'\" .. + also assume \"x' \\<sqsubseteq> y'\" hence \"undual y' \\<sqsubseteq> undual x'\" .. + finally show \"x' = y'\" .. +qed + +instance dual :: (linear_order) linear_order +proof intro_classes + fix x' y' :: \"'a::linear_order dual\" + show \"x' \\<sqsubseteq> y' \\<or> y' \\<sqsubseteq> x'\" + proof (rule linear_order_cases) + assume \"undual y' \\<sqsubseteq> undual x'\" + hence \"x' \\<sqsubseteq> y'\" .. thus ?thesis .. + next + assume \"undual x' \\<sqsubseteq> undual y'\" + hence \"y' \\<sqsubseteq> x'\" .. thus ?thesis .. + qed +qed + + +subsubsection {* Binary products \\label{sec:prod-order} *} + +text {* + The classes of quasi and partial orders are closed under binary + products. Note that the direct product of linear orders need + \\emph{not} be linear in general. +*} + +instance * :: (leq, leq) leq + by intro_classes + +defs (overloaded) + leq_prod_def: \"p \\<sqsubseteq> q \\<equiv> fst p \\<sqsubseteq> fst q \\<and> snd p \\<sqsubseteq> snd q\" + +lemma leq_prodI [intro?]: + \"fst p \\<sqsubseteq> fst q \\<Longrightarrow> snd p \\<sqsubseteq> snd q \\<Longrightarrow> p \\<sqsubseteq> q\" + by (unfold leq_prod_def) blast + +lemma leq_prodE [elim?]: + \"p \\<sqsubseteq> q \\<Longrightarrow> (fst p \\<sqsubseteq> fst q \\<Longrightarrow> snd p \\<sqsubseteq> snd q \\<Longrightarrow> C) \\<Longrightarrow> C\" + by (unfold leq_prod_def) blast + +instance * :: (quasi_order, quasi_order) quasi_order +proof intro_classes + fix p q r :: \"'a::quasi_order \\<times> 'b::quasi_order\" + show \"p \\<sqsubseteq> p\" + proof + show \"fst p \\<sqsubseteq> fst p\" .. + show \"snd p \\<sqsubseteq> snd p\" .. + qed + assume pq: \"p \\<sqsubseteq> q\" and qr: \"q \\<sqsubseteq> r\" + show \"p \\<sqsubseteq> r\" + proof + from pq have \"fst p \\<sqsubseteq> fst q\" .. + also from qr have \"\\<dots> \\<sqsubseteq> fst r\" .. + finally show \"fst p \\<sqsubseteq> fst r\" . + from pq have \"snd p \\<sqsubseteq> snd q\" .. + also from qr have \"\\<dots> \\<sqsubseteq> snd r\" .. + finally show \"snd p \\<sqsubseteq> snd r\" . + qed +qed + +instance * :: (partial_order, partial_order) partial_order +proof intro_classes + fix p q :: \"'a::partial_order \\<times> 'b::partial_order\" + assume pq: \"p \\<sqsubseteq> q\" and qp: \"q \\<sqsubseteq> p\" + show \"p = q\" + proof + from pq have \"fst p \\<sqsubseteq> fst q\" .. + also from qp have \"\\<dots> \\<sqsubseteq> fst p\" .. + finally show \"fst p = fst q\" . + from pq have \"snd p \\<sqsubseteq> snd q\" .. + also from qp have \"\\<dots> \\<sqsubseteq> snd p\" .. + finally show \"snd p = snd q\" . + qed +qed + + +subsubsection {* General products \\label{sec:fun-order} *} + +text {* + The classes of quasi and partial orders are closed under general + products (function spaces). Note that the direct product of linear + orders need \\emph{not} be linear in general. +*} + +instance fun :: (\"term\", leq) leq + by intro_classes + +defs (overloaded) + leq_fun_def: \"f \\<sqsubseteq> g \\<equiv> \\<forall>x. f x \\<sqsubseteq> g x\" + +lemma leq_funI [intro?]: \"(\\<And>x. f x \\<sqsubseteq> g x) \\<Longrightarrow> f \\<sqsubseteq> g\" + by (unfold leq_fun_def) blast + +lemma leq_funD [dest?]: \"f \\<sqsubseteq> g \\<Longrightarrow> f x \\<sqsubseteq> g x\" + by (unfold leq_fun_def) blast + +instance fun :: (\"term\", quasi_order) quasi_order +proof intro_classes + fix f g h :: \"'a \\<Rightarrow> 'b::quasi_order\" + show \"f \\<sqsubseteq> f\" + proof + fix x show \"f x \\<sqsubseteq> f x\" .. + qed + assume fg: \"f \\<sqsubseteq> g\" and gh: \"g \\<sqsubseteq> h\" + show \"f \\<sqsubseteq> h\" + proof + fix x from fg have \"f x \\<sqsubseteq> g x\" .. + also from gh have \"\\<dots> \\<sqsubseteq> h x\" .. + finally show \"f x \\<sqsubseteq> h x\" . + qed +qed + +instance fun :: (\"term\", partial_order) partial_order +proof intro_classes + fix f g :: \"'a \\<Rightarrow> 'b::partial_order\" + assume fg: \"f \\<sqsubseteq> g\" and gf: \"g \\<sqsubseteq> f\" + show \"f = g\" + proof + fix x from fg have \"f x \\<sqsubseteq> g x\" .. + also from gf have \"\\<dots> \\<sqsubseteq> f x\" .. + finally show \"f x = g x\" . + qed +qed + +end --- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/src/HOL/Lattice/ROOT.ML Fri Oct 06 01:04:56 2000 +0200 @@ -0,0 +1,8 @@ +(* Title: HOL/Lattice/ROOT.ML + ID:$Id$+ Author: Markus Wenzel, TU Muenchen + +Basic theory of lattices and orders. +*) + +time_use_thy \"CompleteLattice\"; --- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/src/HOL/Lattice/document/root.bib Fri Oct 06 01:04:56 2000 +0200 @@ -0,0 +1,48 @@ + +@InProceedings{Bauer-Wenzel:2000:HB, + author = {Gertrud Bauer and Markus Wenzel}, + title = {Computer-Assisted Mathematics at Work --- The {H}ahn-{B}anach Theorem in + {I}sabelle/{I}sar}, + booktitle = {Types for Proofs and Programs: TYPES'99}, + editor = {Thierry Coquand and Peter Dybjer and Bengt Nordstr{\\\"o}m + and Jan Smith}, + series = {LNCS}, + year = 2000 +} + +@Book{Davey-Priestley:1990, + author = {B. A. Davey and H. A. Priestley}, + title = {Introduction to Lattices and Order}, + publisher = {Cambridge University Press}, + year = 1990} + +@InProceedings{Wenzel:1999:TPHOL, + author = {Markus Wenzel}, + title = {{Isar} --- a Generic Interpretative Approach to Readable Formal Proof Documents}, + crossref = {tphols99}} + + +@Manual{Wenzel:2000:axclass, + author = {Markus Wenzel}, + title = {Using Axiomatic Type Classes in Isabelle}, + year = 2000, + institution = {TU Munich}, + note = {\\url{http://isabelle.in.tum.de/doc/axclass.pdf}} +} + +@Manual{Wenzel:2000:isar-ref, + author = {Markus Wenzel}, + title = {The {Isabelle/Isar} Reference Manual}, + year = 2000, + institution = {TU Munich}, + note = {\\url{http://isabelle.in.tum.de/doc/isar-ref.pdf}} +} + +@Proceedings{tphols99, + title = {Theorem Proving in Higher Order Logics: {TPHOLs} '99}, + booktitle = {Theorem Proving in Higher Order Logics: {TPHOLs} '99}, + editor = {Bertot, Y. and Dowek, G. and Hirschowitz, A. and + Paulin, C. and Thery, L.}, + series = {LNCS}, + volume = 1690, + year = 1999} --- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/src/HOL/Lattice/document/root.tex Fri Oct 06 01:04:56 2000 +0200 @@ -0,0 +1,56 @@ + +%$Id\\$\n+\n+\\documentclass[11pt,a4paper]{article}\n+\\usepackage{isabelle,isabellesym,pdfsetup}\n+\\usepackage[only,bigsqcap]{stmaryrd}\n+\n+\\urlstyle{rm}\\isabellestyle{it}\n+\n+\\hyphenation{Isabelle}\n+\\hyphenation{Isar}\n+\n+\n+\\begin{document}\n+\n+\\title{Lattices and Orders in Isabelle/HOL}\n+\\author{Markus Wenzel \\\\ TU M\\\"unchen}\n+\\maketitle\n+\n+\\begin{abstract}\n+ We consider abstract structures of orders and lattices. Many fundamental\n+ concepts of lattice theory are developed, including dual structures,\n+ properties of bounds versus algebraic laws, lattice operations versus\n+ set-theoretic ones etc. We also give example instantiations of lattices and\n+ orders, such as direct products and function spaces. Well-known properties\n+ are demonstrated, like the Knaster-Tarski Theorem for complete lattices.\n+\n+ This formal theory development may serve as an example of applying\n+ Isabelle/HOL to the domain of mathematical reasoning about axiomatic''\n+ structures. Apart from the simply-typed classical set-theory of HOL, we\n+ employ Isabelle's system of axiomatic type classes for expressing structures\n+ and functors in a light-weight manner. Proofs are expressed in the Isar\n+ language for readable formal proof, while aiming at its best-style'' of\n+ representing formal reasoning.\n+\\end{abstract}\n+\n+\\tableofcontents\n+\n+\\newpage\n+{\n+ \\parindent 0pt\\parskip 0.7ex\n+ \\renewcommand{\\isamarkupheader}{\\markright{THEORY~\\isabellecontext''}\\section{#1}}\n+\\end{document}" ]

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[ "Search for a tool\nNumber Partitions\n\nTool to generate partitions of a number (integer). A partition of an integer N is a decomposition of N into a set of numbers (inferior to N) which sum is N.\n\nResults\n\nNumber Partitions -\n\nTag(s) : Arithmetics\n\nShare", null, "dCode and more\n\ndCode is free and its tools are a valuable help in games, maths, geocaching, puzzles and problems to solve every day!\nA suggestion ? a feedback ? a bug ? an idea ? Write to dCode!\n\nPlease, check our dCode Discord community for help requests!\nNB: for encrypted messages, test our automatic cipher identifier!\n\nThanks to your feedback and relevant comments, dCode has developed the best 'Number Partitions' tool, so feel free to write! Thank you!\n\n# Number Partitions\n\n## Partitions Count/Enumeration\n\n### What is a partition of an integer number?\n\nDefinition: in mathematics, a partition $p(N)$ of a number $N$ is a set of numbers (less than or equal to $N$) whose sum is $N$.\n\nExample: The number $5$ can be decomposed into $7$ distinct partitions, the additions are: $5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1$\n\nPermutations are ignored: $4+1$ and $1+4$ are considered identical\n\nExample: The number $10$ has $42$ partitions decompositions, and the number $100$ has $190569292$.\n\nDue to servers' computation's cost with large lists, free generations are limited.\n\n### What is the Hardy-Ramanujan formula?\n\nIn 1918, Hardy and Ramanujan have found an approximation od $p(n)$ for big numbers $n$ :\n\n$$p(n) \\sim \\frac{1}{4n \\sqrt{3}} ~ e^{\\pi \\sqrt{\\frac{2n}{3}}}$$\n\n### How to list Coin Change-making problem solutions?\n\nPartitions of a number are used to solve the change-making problem and to list the ways of give back money.\n\nExample: There are 49 ways to make $100 with$5, $10,$20 or \\$50 notes\n\n### How remove the limits of the generator?\n\nThe generation is very costful in resources (which are expensive) as soon as the quantity of solution becomes large. dCode offers exhaustive lists, ask for prices !\n\n## Source code\n\ndCode retains ownership of the online \"Number Partitions\" source code. Except explicit open source licence (indicated CC / Creative Commons / free), the \"Number Partitions\" algorithm, the applet or snippet (converter, solver, encryption / decryption, encoding / decoding, ciphering / deciphering, translator), or the \"Number Partitions\" functions (calculate, convert, solve, decrypt / encrypt, decipher / cipher, decode / encode, translate) written in any informatic language (Python, Java, PHP, C#, Javascript, Matlab, etc.) and all data download, script, copy-paste, or API access for \"Number Partitions\" are not public, same for offline use on PC, tablet, iPhone or Android ! Remainder : dCode is free to use.\n\n## Need Help ?\n\nPlease, check our dCode Discord community for help requests!\nNB: for encrypted messages, test our automatic cipher identifier!" ]

[ null, "https://www.dcode.fr/images/share.png", null ]

[ "doclist 阅读(25) 评论(0)\n\n二、运行的过程中给对象绑定(添加)属性\n\n```class Person(object):\ndef __init__(self,name=None,age=None):\nself.name=name\nself.age=age\n\np = Person(\"小明\",\"24\")\nprint(p.name)\nprint(p.age)```\n\n```小明\n24```\n\n```class Person(object):\ndef __init__(self,name=None,age=None):\nself.name=name\nself.age=age\n\np = Person(\"小明\",\"24\")\nprint(p.name)\nprint(p.age)\n\n#动态添加属性\np.sex = \"男\"\nprint(p.sex)```\n\n```小明\n24\n\n三、运行的过程中给类绑定(添加)属性\n\n```class Person(object):\ndef __init__(self,name=None,age=None):\nself.name=name\nself.age=age\n\nP1 = Person(\"小明\",24)\nprint(P1.sex)```\n\n```Traceback (most recent call last):\nFile \"C:\\Users\\Se7eN_HOU\\Desktop\\test.py\", line 8, in <module>\nprint(P1.sex)\nAttributeError: 'Person' object has no attribute 'sex'```\n\n```class Person(object):\ndef __init__(self,name=None,age=None):\nself.name=name\nself.age=age\n\nP1 = Person(\"小明\",24)\n#动态给类添加属性\nPerson.sex = \"男\"\nprint(P1.sex)```\n\n四、运行的过程中给类绑定(添加)方法\n\n```class Person(object):\ndef __init__(self,name=None,age=None):\nself.name=name\nself.age=age\n\ndef eat(self):\nprint(\"正在吃东西\")\n\nP1 = Person(\"小明\",24)\nP1.eat()\nP1.run()```\n\n```正在吃东西\nTraceback (most recent call last):\nFile \"C:\\Users\\Se7eN_HOU\\Desktop\\test.py\", line 11, in <module>\n\nP1.run()\nAttributeError: 'Person' object has no attribute 'run'```\n\n```#动态添加方法需要导入types模块\nimport types\nclass Person(object):\ndef __init__(self,name=None,age=None):\nself.name=name\nself.age=age\n\ndef eat(self):\nprint(\"正在吃东西\")\n\n#定义好需要动态添加的方法\ndef run(self):\nprint(\"在跑步\")\n\nP1 = Person(\"小明\",24)\n#正常调用类里面的函数\nP1.eat()\n\n#给对象动态绑定方法\nP1.run = types.MethodType(run,P1)\n#对象调用动态绑定的方法\nP1.run()```\n\n```正在吃东西\n\n```#动态添加方法需要导入types模块\nimport types\nclass Person(object):\ndef __init__(self,name=None,age=None):\nself.name=name\nself.age=age\n\ndef eat(self):\nprint(\"正在吃东西\")\n\n#定义好需要动态添加的实例方法\ndef run(self):\nprint(\"在跑步\")\n\n#定义好需要动态添加的类方法\n@classmethod\ndef dynamicClassMethod(cls):\nprint(\"这是一个动态添加的类方法\")\n#定义好需要动态添加的静态方法\n@staticmethod\ndef dynamicStaticMethod():\nprint(\"这是一个动态添加的静态方法\")\n\nP1 = Person(\"小明\",24)\n#正常调用类里面的函数\nP1.eat()\n\n#给对象动态绑定方法\n#MethodType(参数1,参数2)\n#参数1：是动态绑定哪个方法，只写方法名即可\n#参数2：是把这个方法动态的绑定给谁\nP1.run = types.MethodType(run,P1)\nP1.run()\n\n#动态绑定类方法的使用\nPerson.dynamicClassMethod = dynamicClassMethod\nPerson.dynamicClassMethod()\n\n#动态绑定静态方法的使用\nPerson.dynamicStaticMethod = dynamicStaticMethod\nPerson.dynamicStaticMethod()```\n\n1. 给对象绑定属性直接在使用前进行赋值使用即可\n2. 给对象动态绑定方法需要import types模块\n3. 给对象动态绑定实例方法，需要使用type.MethodType（）方法\n4. 给类添加类方法和静态方法，也是直接在使用前赋值即可使用\n\n五、运行的过程中删除属性、方法\n\n1. del 对象.属性名\n2. delattr(对象, \"属性名\")\n```class Person(object):\ndef __init__(self,name=None,age=None):\nself.name=name\nself.age=age\n\nP1 = Person(\"小明\",24)\nprint(\"---------删除前---------\")\nprint(P1.name)\n\ndel P1.name\n\nprint(\"---------删除后---------\")\nprint(P1.name)```\n\n```---------删除前---------\n\n---------删除后---------\nprint(P1.name)AttributeError: 'Person' object has no attribute 'name'```\n\n六、__slots__\n\n```class Person(object):\n__slots__=(\"name\",\"age\")\n\np = Person()\np.name = \"老王\"\np.age = 40\nprint(p.name)\nprint(p.age)\n\n#slots之外的属性\np.sex = \"男\"\nprint(p.sex)```\n\n```老王\n40\np.sex = \"男\"\nAttributeError: 'Person' object has no attribute 'sex'```\n\n注意:\n\n• 使用__slots__要注意，__slots__定义的属性仅对当前类实例起作用，对继承的子类是不起作用的" ]

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[ "# Difference between revisions of \"G285 2021 MC10B\"\n\n## Problem 1\n\nFind", null, "$\\left \\lceil {\\frac{3!+4!+5!+6!}{2+3+4+5+6}} \\right \\rceil$", null, "$\\textbf{(A)}\\ 42\\qquad\\textbf{(B)}\\ 43\\qquad\\textbf{(C)}\\ 44\\qquad\\textbf{(D)}\\ 45\\qquad\\textbf{(E)}\\ 46$\n\n## Problem 2\n\nIf", null, "$deg(Q(x))=3$, and", null, "$deg(K(x))=2$, and", null, "$Q(x)=(x-2)K(x)$, what is", null, "$deg(Q(x)-2K(x))$?", null, "$\\textbf{(A)}\\ 0\\qquad\\textbf{(B)}\\ 1\\qquad\\textbf{(C)}\\ 2\\qquad\\textbf{(D)}\\ 3\\qquad\\textbf{(E)}\\ 4$\n\n## Problem 3\n\nA convex hexagon of length", null, "$s$ is inscribed in a circle of radius", null, "$r$, where", null, "$r \\neq s$. If", null, "$\\frac{s}{2r}=\\frac{21}{29}$, and", null, "$rs=58$, find the area of the hexagon.", null, "$\\textbf{(A)}\\ 42\\qquad\\textbf{(B)}\\ 60\\qquad\\textbf{(C)}\\ 84\\qquad\\textbf{(D)}\\ 90\\qquad\\textbf{(E)}\\ 120$\n\n## Problem 4\n\nFind the smallest", null, "$n$ such that:", null, "$$n \\equiv 3 \\pmod{9}$$", null, "$$2n \\equiv 7 \\pmod{13}$$", null, "$$5n \\equiv 14 \\pmod{17}$$", null, "$\\textbf{(A)}\\ 1560\\qquad\\textbf{(B)}\\ 1713\\qquad\\textbf{(C)}\\ 2211\\qquad\\textbf{(D)}\\ 3273\\qquad\\textbf{(E)}\\ 3702$\n\n## Problem 5\n\nA principal is pushing out an emergency COVID-19 alert to his school of", null, "$40$ teachers and", null, "$500$ students. Suppose the announcement is first approved by his", null, "$5$ aides. Then, each of the aides share the announcement to", null, "$n$ teachers and", null, "$t$ students, where", null, "$n,t \\in \\mathbb{Z}$ and for every aide", null, "$n \\neq t$. Moreover,", null, "$n+t = (u+1)^2$, where", null, "$u$ is the round number ( for the aides releasing info it is round 1, then round 2....) After every round", null, "$u$, some", null, "$k$ teachers in the previous round share the announcement to a new group of", null, "$n$ teachers and", null, "$t$ students, where", null, "$k=(u+1)^2$. How many rounds will it take until the entire school is informed? Assume that after all teachers are informed,", null, "$n=0$, but", null, "$t$ still grows as if", null, "$n \\neq 0$.", null, "$\\textbf{(A)}\\ 3\\qquad\\textbf{(B)}\\ 4\\qquad\\textbf{(C)}\\ 5\\qquad\\textbf{(D)}\\ 6\\qquad\\textbf{(E)}\\ 7$\n\n## Problem 6\n\nLet a plane parallel to the horizontal slice a sphere with radius", null, "$r$ at a random location to create a cross section, and two partial spheres. For what minimum radius", null, "$r$ will the area of the cross-section never be able to exceed the sum of the outer surface areas of the partial spheres?", null, "$\\textbf{(A)}\\ 0\\qquad\\textbf{(B)}\\ 1\\qquad\\textbf{(C)}\\ 2\\qquad\\textbf{(D)}\\ 4\\qquad\\textbf{(E)}\\ 5$\n\nInvalid username\nLogin to AoPS" ]

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[ "Importance of Complex Numbers\n\n###", null, "Author Topic: Importance of Complex Numbers  (Read 1780 times)\n\n####", null, "msu_math", null, "##### Importance of Complex Numbers\n« on: May 20, 2013, 08:44:25 AM »\nComplex number is a number that combines real and imaginary numbers. For example 2+3i, 1-5i, -2i etc are complex numbers where i is the imaginary unit defined as i2=-1.\n\nReal numbers are commonly used  where as complex numbers are mainly dealt by scientists, engineers and researchers. Every real life problem can be described by a special kind of mathematical model consisting group of simultaneous equations. The solution of a mathematical model provides a set of information (data consisting real and imaginary numbers) regarding the related problem. For these reasons development of complex numbers has a significant role in the areas of scientific research.\n\nHere are some examples of complex numbers.\n\nIn electronics, the state of a circuit element is described by two real numbers (the voltage V across it and the current I flowing through it). A circuit element also may possess a capacitance C and an inductance L that (in simplistic terms) describe its tendency to resist changes in voltage and current respectively.\n\nThese are much better described by complex numbers. Rather than the circuit element's state having to be described by two different real numbers V and I, it can be described by a single complex number z = V + i I. Similarly, inductance and capacitance can be thought of as the real and imaginary parts of another single complex number w = C + i L. The laws of electricity can be expressed using complex addition and multiplication.\n\nAnother example is electromagnetism. Rather than trying to describe an electromagnetic field by two real quantities (electric field strength and magnetic field strength), it is best described as a single complex number, of which the electric and magnetic components are simply the real and imaginary parts.\n\nLecturer in Mathematics\nDepartment of Natural Sciences\nFSIT, DIU\n\n####", null, "ashis3456\n\n• Newbie\n•", null, "• Posts: 20\n• Test", null, "##### Re: Importance of Complex Numbers\n« Reply #1 on: November 17, 2013, 10:06:28 PM »\nThank you sir.it's very useful for us.\nAshis Sarkar\nDept. of Textile Engineering\nDaffodil International University\nStudent ID-131-23-3456\nDiu [email protected]\[email protected]\n\n####", null, "Nargis Akter\n\n• Jr. Member\n•", null, "", null, "• Posts: 56\n• Test", null, "##### Re: Importance of Complex Numbers\n« Reply #2 on: April 08, 2017, 01:52:19 PM »\nThanks for sharing." ]

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[ "# cggsvd (l) - Linux Manuals\n\n## NAME\n\nCGGSVD - computes the generalized singular value decomposition (GSVD) of an M-by-N complex matrix A and P-by-N complex matrix B\n\n## SYNOPSIS\n\nSUBROUTINE CGGSVD(\nJOBU, JOBV, JOBQ, M, N, P, K, L, A, LDA, B, LDB, ALPHA, BETA, U, LDU, V, LDV, Q, LDQ, WORK, RWORK, IWORK, INFO )\n\nCHARACTER JOBQ, JOBU, JOBV\n\nINTEGER INFO, K, L, LDA, LDB, LDQ, LDU, LDV, M, N, P\n\nINTEGER IWORK( * )\n\nREAL ALPHA( * ), BETA( * ), RWORK( * )\n\nCOMPLEX A( LDA, * ), B( LDB, * ), Q( LDQ, * ), U( LDU, * ), V( LDV, * ), WORK( * )\n\n## PURPOSE\n\nCGGSVD computes the generalized singular value decomposition (GSVD) of an M-by-N complex matrix A and P-by-N complex matrix B:\nUaq*A*Q D1*( 0 R ),    Vaq*B*Q D2*( 0 R )\nwhere U, V and Q are unitary matrices, and Zaq means the conjugate transpose of Z. Let K+L = the effective numerical rank of the matrix (Aaq,Baq)aq, then R is a (K+L)-by-(K+L) nonsingular upper triangular matrix, D1 and D2 are M-by-(K+L) and P-by-(K+L) \"diagonal\" matrices and of the following structures, respectively:\nIf M-K-L >= 0,\n\nL\n\nD1      )\n\n)\n\nM-K-L  )\n\nL\n\nD2    )\n\nP-L  )\n\nN-K-L     L\n\n0 R    R11  R12 )\n\nR22 )\nwhere\n\ndiag( ALPHA(K+1), ... , ALPHA(K+L) ),\n\ndiag( BETA(K+1),  ... , BETA(K+L) ),\n\nC**2 S**2 I.\n\nR is stored in A(1:K+L,N-K-L+1:N) on exit.\nIf M-K-L < 0,\n\nK M-K K+L-M\n\nD1         )\n\nM-K       )\n\nK M-K K+L-M\n\nD2   M-K      )\n\nK+L-M      )\n\nP-L      )\n\nN-K-L    M-K  K+L-M\n\n0 R        R11  R12  R13  )\n\nM-K       R22  R23  )\n\nK+L-M          R33  )\nwhere\n\ndiag( ALPHA(K+1), ... , ALPHA(M) ),\n\ndiag( BETA(K+1),  ... , BETA(M) ),\n\nC**2 S**2 I.\n\n(R11 R12 R13 is stored in A(1:M, N-K-L+1:N), and R33 is stored\nR22 R23 )\n\nin B(M-K+1:L,N+M-K-L+1:N) on exit.\nThe routine computes C, S, R, and optionally the unitary\ntransformation matrices U, V and Q.\nIn particular, if B is an N-by-N nonsingular matrix, then the GSVD of A and B implicitly gives the SVD of A*inv(B):\n\nA*inv(B) U*(D1*inv(D2))*Vaq.\nIf ( Aaq,Baq)aq has orthnormal columns, then the GSVD of A and B is also equal to the CS decomposition of A and B. Furthermore, the GSVD can be used to derive the solution of the eigenvalue problem:\nAaq*A x lambda* Baq*B x.\nIn some literature, the GSVD of A and B is presented in the form\nUaq*A*X 0 D1 ),   Vaq*B*X 0 D2 )\nwhere U and V are orthogonal and X is nonsingular, and D1 and D2 are ``diagonalaqaq. The former GSVD form can be converted to the latter form by taking the nonsingular matrix X as\n\nQ*(       )\n\n0 inv(R) )\n\n## ARGUMENTS\n\nJOBU (input) CHARACTER*1\n= aqUaq: Unitary matrix U is computed;\n= aqNaq: U is not computed.\nJOBV (input) CHARACTER*1\n\n= aqVaq: Unitary matrix V is computed;\n= aqNaq: V is not computed.\nJOBQ (input) CHARACTER*1\n\n= aqQaq: Unitary matrix Q is computed;\n= aqNaq: Q is not computed.\nM (input) INTEGER\nThe number of rows of the matrix A. M >= 0.\nN (input) INTEGER\nThe number of columns of the matrices A and B. N >= 0.\nP (input) INTEGER\nThe number of rows of the matrix B. P >= 0.\nK (output) INTEGER\nL (output) INTEGER On exit, K and L specify the dimension of the subblocks described in Purpose. K + L = effective numerical rank of (Aaq,Baq)aq.\nA (input/output) COMPLEX array, dimension (LDA,N)\nOn entry, the M-by-N matrix A. On exit, A contains the triangular matrix R, or part of R. See Purpose for details.\nLDA (input) INTEGER\nThe leading dimension of the array A. LDA >= max(1,M).\nB (input/output) COMPLEX array, dimension (LDB,N)\nOn entry, the P-by-N matrix B. On exit, B contains part of the triangular matrix R if M-K-L < 0. See Purpose for details.\nLDB (input) INTEGER\nThe leading dimension of the array B. LDB >= max(1,P).\nALPHA (output) REAL array, dimension (N)\nBETA (output) REAL array, dimension (N) On exit, ALPHA and BETA contain the generalized singular value pairs of A and B; ALPHA(1:K) = 1,\nBETA(1:K) = 0, and if M-K-L >= 0, ALPHA(K+1:K+L) = C,\nBETA(K+1:K+L) = S, or if M-K-L < 0, ALPHA(K+1:M)= C, ALPHA(M+1:K+L)= 0\nBETA(K+1:M) = S, BETA(M+1:K+L) = 1 and ALPHA(K+L+1:N) = 0\nBETA(K+L+1:N) = 0\nU (output) COMPLEX array, dimension (LDU,M)\nIf JOBU = aqUaq, U contains the M-by-M unitary matrix U. If JOBU = aqNaq, U is not referenced.\nLDU (input) INTEGER\nThe leading dimension of the array U. LDU >= max(1,M) if JOBU = aqUaq; LDU >= 1 otherwise.\nV (output) COMPLEX array, dimension (LDV,P)\nIf JOBV = aqVaq, V contains the P-by-P unitary matrix V. If JOBV = aqNaq, V is not referenced.\nLDV (input) INTEGER\nThe leading dimension of the array V. LDV >= max(1,P) if JOBV = aqVaq; LDV >= 1 otherwise.\nQ (output) COMPLEX array, dimension (LDQ,N)\nIf JOBQ = aqQaq, Q contains the N-by-N unitary matrix Q. If JOBQ = aqNaq, Q is not referenced.\nLDQ (input) INTEGER\nThe leading dimension of the array Q. LDQ >= max(1,N) if JOBQ = aqQaq; LDQ >= 1 otherwise.\nWORK (workspace) COMPLEX array, dimension (max(3*N,M,P)+N)\nRWORK (workspace) REAL array, dimension (2*N)\nIWORK (workspace/output) INTEGER array, dimension (N)\nOn exit, IWORK stores the sorting information. More precisely, the following loop will sort ALPHA for I = K+1, min(M,K+L) swap ALPHA(I) and ALPHA(IWORK(I)) endfor such that ALPHA(1) >= ALPHA(2) >= ... >= ALPHA(N).\nINFO (output) INTEGER\n= 0: successful exit.\n< 0: if INFO = -i, the i-th argument had an illegal value.\n> 0: if INFO = 1, the Jacobi-type procedure failed to converge. For further details, see subroutine CTGSJA.\n\n## PARAMETERS\n\nTOLA REAL\nTOLB REAL TOLA and TOLB are the thresholds to determine the effective rank of (Aaq,Baq)aq. Generally, they are set to TOLA = MAX(M,N)*norm(A)*MACHEPS, TOLB = MAX(P,N)*norm(B)*MACHEPS. The size of TOLA and TOLB may affect the size of backward errors of the decomposition. Further Details =============== 2-96 Based on modifications by Ming Gu and Huan Ren, Computer Science Division, University of California at Berkeley, USA" ]

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[ "", null, "# G'MIC exercises\n\nGood example thanks, I’ve not yet used macros in the math parser! It’s obviously a commonly required algorithm type, therefore very useful to know", null, "Edit: should’ve added, yes it’s random single elements being removed so this fits the case perfectly.\n\nI’ll write one version that stores the data in an image stack instead, just to show it is as easy. As we can manage several images at the same time, this means having multiple dynamic lists is possible.\n\nHere is a version with an image rather than a vector :\n\n``````test :\n\ni[list] 1,1,1,3,-1 # List of RGB colors\n\neval \"\n\n# Macro that inserts a new element 'elt' into the image #ind, at index 'pos' ('pos' must be in [0,siz_list]).\ninsert(ind,siz_list,elt,pos) = (\n_pos = pos;\nif (_pos<=siz_list,\nsiz_list>=w(#ind)?resize(#ind,2*w(#ind)+1,1,1,s#ind,0);\nfor (k = 0, k<s#ind, ++k, copy(i(#ind,_pos+1,0,0,k),i(#ind,_pos,0,0,k),siz_list - _pos));\nI[#ind,_pos] = elt;\n++#siz_list;\n);\n);\n\n# Macro that removes an element from the image #ind, located at index 'pos' ('pos' must be in [0,siz_list]).\nremove(ind,siz_list,pos) = (\n_pos = pos;\nif (_pos<=siz_list,\n--#siz_list;\nfor (k = 0, k<s#ind, ++k, copy(I(#ind,_pos,0,0,k),I(#ind,_pos+1,0,0,k),siz_list - _pos));\nI[#ind,siz_list] = -1;\n)\n);\n\nsiz_list = 0;\n\necho('---- Initial state ------------');\nprint(#\"\\$list\"); print(siz_list); # Empty list.\n\ninsert(#\"\\$list\",siz_list,[ 0,128,0 ],0);\ninsert(#\"\\$list\",siz_list,[ 64,0,0 ],0);\ninsert(#\"\\$list\",siz_list,[ 0,0,255 ],2);\n\necho('---- After three insertions ---');\nprint(#\"\\$list\"); print(siz_list); # List with three elements.\n\nremove(#\"\\$list\",siz_list,0);\n\necho('--- After one removal ---------');\nprint(#\"\\$list\"); print(siz_list); # List with two elements.\n\"\n``````\n1 Like\n\nNo problem. Actually list and element manipulation were the first things I tried, with limited success, when I encountered G’MIC scripting (which was not that long ago; am still a relatively new user", null, "). If I stare at your discussions long enough, I might learn something. Still struggling to follow the math and code but it is fun nevertheless.", null, "PS All of you can make it up by looking into An Error Reduction Technique in Richardson-Lucy Deconvolution Method.", null, "Actually I already had a look, not certain whether it applies generally because it appears directed towards a particular form of noise. I suspect some of the maths will be outside what I know just now anyway, so much to learn!\n\nSome news about the management of dynamic arrays in the math parser.\nI’ve added new functions in command `math_lib` to ease handling this type of structure.\n\nBasically, you first define a 1-column image with the number of channels you want (filled with `0`). It will be used to represent your dynamic array. One example is worth a thousand words :\n\n``````test :\n\n1,1,1,3 # A dynamic array of RGB colors (initially empty).\n\nv -\neval \\${-math_lib}\"\n\nprint(dar_size(#0)); # Print number of elements -> '0'\n\n# 500 insertions at random positions.\nfor (k = 0, k<500, ++k,\ndar_insert(#0,u([255,255,255]),round(u(dar_size(#0))));\next('w 200,100%,0'); # Display array\n);\n\nprint(dar_size(#0)); # Print number of elements -> '500'\n\n# 200 removal at random positions.\nfor (k = 0, k<200, ++k,\ndar_remove(#0,round(u(dar_size(#0)-1)));\next('w 200,100%,0'); # Display array\n);\n\nprint(dar_size(#0)); # Print number of elements -> '300'\n\ndar_insert(#0,[1,2,3],125);\nprint(I[#0,125]);\n\"\nv +\n``````\n\nMaybe @garagecoder will be able to use those functions ?", null, "EDIT: How this internally work ?\nThe trick here is to store the current number of element in the image, at position `[w*h*d-1]`, and to have a image height that is always greater or equal than `dar_size() + 1`. The function `dar_insert()` resizes the image when needed, and the function `dar_remove()` also resizes to save memory usage.\nThe height of an image is not equal to the number of elements in your dynamic arrays, that’s why there is a function `dar_size()` to get this information.\n\nNice! I’ll certainly start with that, thanks. I have time today to begin writing it and that means I don’t have to worry about handling that part", null, "1 Like\n\nMight as well re-hijack the thread because I thought of an idea which is a compromise regarding what David and I were discussing earlier about linear versus nonlinear parameters attached to sliders: float and integer sliders which display and output linear values but move nonlinearly from left to right with specified modes and powers. It could have various modes such as n^x and x^n. Also, now that I think about it, step options for both integer and float sliders.\n\n1 Like\n\nWell this poisson disk noise turned out tougher than I expected", null, "Mainly because of dealing with 1D to 3D and not knowing enough about the math parser. Had to redesign the algorithm itself… this version is the first attempt so needs optimised and probably still has bugs (there’s even debug code still in). But it’s cool you can watch it drawing it point by point!\n\n``````gcd_poisson_disk : skip \\${1=2},\\${2=30}\n# input image to draw samples on\ndim={d>1?3:h>1?2:1} cw={0.999*\\$1/sqrt(\\$dim)}\n({[w,h,d,1]}) y. c # image dimensions vector\n{[ceil(I/\\$cw),-1]} # \"accelerator\" grid/cells\nr 1,1,1,\\$dim,-1 # keep only used dimensions in \n1,1,1,\\$dim 1,1,1,1 # samples list, active list\n{vector\\$dim(2*ceil(sqrt(\\$dim))+1)} # cell proximity kernel\nf. \"dot([x,y,z]-int([w/2,h/2,d/2]),[1,w#2,w#2*h#2])\" y. c\nnm dims nm grid nm samples nm active nm prox\n-v -\neval \\${-math_lib}\"\nconst N = \"\\$dim\";\nconst grid_cw = \"\\$cw\";\nconst max_sample_attempts = \\$2;\nprox = I#5;\nmag2(vec) = (dot(vec,vec));\n\ndar_insert(#3,I#1,0); # dummy sample to simplify bounds checks\ndar_insert(#3,u(I#1),1); # add initial sample to list\ndar_insert(#4,1,0); # add its index to active list\nI(#2,int(I[#3,1]/grid_cw)) = 1; # add its index to grid cell\ntr=0;\nwhiledo (dar_size(#4)>0,\nR = int(u(dar_size(#4)-1e-4)); # choose a random active list index\n\nP = i[#4,R]; # get the index of that sample\nT = I[#3,P]; # position vector of that sample\n\nfor (attempts=0, attempts < max_sample_attempts, ++attempts,\n\ndowhile (S=4*(u(vectorN(1))-0.5); M=mag2(S), M <= 1 || M > 4);\n\nX = T + radius * S; # potential sample from annulus around T\nif (min(X)<0 || min(I[#1,0]-X)<0, continue()); # check within bounds\n\n# check proximity of surrounding points\nG = int(X/grid_cw); # grid cell position vector\nGI = dot(G,[1,w#2,w#2*h#2]); # grid cell direct buffer index\n\nfor (K=0;rejected=0, K<size(prox), ++K,\nV = i[#2,GI+prox[K]]; # sample index from grid to check\n);\n\nif (!rejected,\nQ = dar_size(#3); # sample found, get new index\ndar_insert(#3,X,Q); # insert into samples list\ndar_insert(#4,Q,dar_size(#4)); # insert its index into active\nI(#2,G) = Q; # insert its index into grid\nI(#0,X)=1; # draw the point\nbreak();\n);\n);\n\nif (attempts == max_sample_attempts, dar_remove(#4,P));\n++tr; if(tr%2==0,ext('w 400,400,1'));\n);\n\" -v +\n``````\n\nusing:\n`gmic 128,128,1,1 gcd_poisson_disk 4`", null, "1 Like\n\nDecided to add other preview options. There’s a issue though, preview timeout.\n\n``````#@gui _\n#@gui <i>Reptorian</i>\n#@gui Modulus operations : fx_modulus, fx_modulus_preview(0)\n#@gui : note = note(\"This filters applies modulo operation after arithmetic operation. Note: Joan Rake's Sawtoothers are the expanded version of this filter.\")\n#@gui : sep = separator()\n#@gui : Multiply = float(1,0,32)\n#@gui : sep = separator(), Preview type = choice(\"Full\",\"Forward horizontal\",\"Forward vertical\",\"Backward horizontal\",\"Backward vertical\",\"Duplicate top\",\"Duplicate left\",\"Duplicate bottom\",\"Duplicate right\",\"Duplicate horizontal\",\"Duplicate vertical\",\"Checkered\",\"Checkered inverse\")\n#@gui : sep = separator(), note = note(\"<small>Author : <i>Reptorian</i>. Latest update : <i>2018/08/18</i>.</small>\")\nfx_modulo:\nrepeat \\$! l[\\$>] split_opacity l\nmul \\$1 add \\$2 mod 256\nendl a c endl done\nfx_modulus_preview :\n-gui_split_preview \"-fx_modulus \\${1--2}\", \\$-1``````\n\nThat really reminds me the kind of things you get with the farthest point sampling method.\nThe example below implements it:\n\n``````do_fps :\nv -\n\n# Initialisation\n400,400 noise. 0.01,2 !=. 0\n\n# Furthest point sampling.\nrepeat 500 +distance. 1 eval \"i(#0,xM,yM) = 1\" rm. done\n\nv +\n``````\n\nwhich generates this image:", null, "It’s a bit long to compute since a full distance function is computed before inserting each new point.\n\nAh yes that makes sense, it’s the exact same thing (random points a certain minimum distance apart) but a slightly different algorithm. In theory, once written properly, the one I’m working on should be faster due to using a proximity map. That depends a lot on whether I do a good job though", null, "I have the idea to re-update the distance function also only where it is necessary. I will be interesting to compare the computation times of the two approaches.\nGood luck !", null, "Hahah! If this is to become a competition it’s a very lopsided one in your favour. But we all win anyway because we hopefully get a fast poisson disk noise from it", null, "I’ve almost something ready", null, "Any suggestion for a command name ? `noise_farther` maybe ?\n\nHmm farther is perhaps a little vague (farther than what?)… some suggestions to give you more ideas (keeping the prefix):\n\nnoise_distance\nnoise_mindistance\nnoise_disk\nnoise_sample\nnoise_sampling\nnoise_sphere (if you’re dealing with the 3d usage)\n\nI’ll add more if I think of them\n\nI found a really dumb mistake by me (which explains why it was so slow), I was removing the wrong index from the active sample list! It should say R not P:\n\n`if (attempts == max_sample_attempts, dar_remove(#4,R));`\n\nNow it’s much faster!\n\nHere is my attempt:\n\n``````#@cli noise_maxdist : _nb_points[%]>=0\n#@cli : Add random noise points to selected image such that added points are set with the maximal possible distance.\n#@cli : \\$\\$ 300,300 noise_maxdist 1% dilate_circ 5\nnoise_maxdist : check \"\\$1>=0\"\ne[^-1] \"Add noise points to image\\$?, with maximal possible distance.\"\nv - repeat \\$! l[\\$>]\nN={round(\\${\"is_percent \\$1\"}?whd*\\$1:\\$1)}\nif \\$N\n100%,100%,100% +f. inf\nif {d==1} # 2d version\neval \"\n# Init with random point.\ni(#1,round(u([0,0],[w,h]-1))) = 1;\nx = w/2; y = h/2; r = max(w,h);\nconst N1 = \"\\$N\"-1;\nfor (n = 0, n<N1, ++n,\nx0 = max(0,x - r); y0 = max(0,y - r);\nx1 = min(w - 1,x + r); y1 = min(h - 1,y + r);\next('+z[^0] ',vtos([x0,y0,x1,y1]),' distance.. 1 min[-2,-1] j .,',vtos([x0,y0]),' rm.');\n\n# Find local maximum of distance function (faster than doing it globally, and improve randomization).\nxc = round(u(w-1)); yc = round(u(h-1)); r2 = r*2;\nx0 = max(0,xc - r2); y0 = max(0,yc - r2);\nx1 = min(w - 1,xc + r2); y1 = min(h - 1,yc + r2);\next('+z ',vtos([x0,y0,x1,y1]));\ns = stats(#-1); x = x0 + s; y = y0 + s;\nr = i(#2,x,y); ext('rm.');\n\n# Put new point on found location.\ni(#1,x,y) = 1;\n)\"\nelse # 3d version\neval \"\n# Init with random point.\ni(#1,round(u([0,0,0],[w,h,d]-1))) = 1;\nx = w/2; y = h/2; z = d/2; r = max(w,h,d);\nconst N1 = \"\\$N\"-1;\nfor (n = 0, n<N1, ++n,\nx0 = max(0,x - r); y0 = max(0,y - r); z0 = max(0,z - r);\nx1 = min(w - 1,x + r); y1 = min(h - 1,y + r); z1 = min(d - 1,z + r);\next('+z[^0] ',vtos([x0,y0,z0,x1,y1,z1]),' distance.. 1 min[-2,-1] j .,',vtos([x0,y0,z0]),' rm.');\n\n# Find local maximum of distance function (faster than doing it globally, and improve randomization).\nxc = round(u(w-1)); yc = round(u(h-1)); zc = round(u(d-1)); r2 = 2*r;\nx0 = max(0,xc - r2); y0 = max(0,yc - r2); z0 = max(0,zc - r2);\nx1 = min(w - 1,xc + r2); y1 = min(h - 1,yc + r2); z1 = min(d - 1,zc + r2);\next('+z ',vtos([x0,y0,z0,x1,y1,z1]));\ns = stats(#-1); x = x0 + s; y = y0 + s; z = z0 + s;\nr = i(#2,x,y,z); ext('rm.');\n\n# Put new point on found location.\ni(#1,x,y,z) = 1;\n)\"\nfi\nf \"init(val = vectors(iM==im?1:iM)); i#1?val:I\" k\nfi\nendl done v +\n``````\n\nworks in 2D and 3D (but slow in 3D", null, ").\n\nExample of use:\n\n``````\\$ gmic sp lena,128 noise_maxdist 3%\n``````", null, "But I think this is still a bit slow (and I’ve no ideas to speed it up right now).\n\n1 Like\n\nYes, just tested your code, @garagecoder, and it looks like you won the challenge", null, "Hah good jobs it’s not my algorithm", null, "I have several ideas to speed it up as well, so it could be very fast eventually…" ]

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[ "# Find time taken to execute the tasks in A based on the order of execution in B\n\nGiven two queues A and B, each of size N, the task is to find the minimum time taken to execute the tasks in A based on the order of execution in B where:\n\n1. If the task found at the front of queue B is at the front of queue A, then pop this task and execute it.\n2. If the task found at the front of queue B is not found at the front of queue A, then pop the current task from queue A and push it at the end.\n3. Push and Pop operation in a queue costs one unit of time and the execution of a task is done in constant time.\n\nExample\n\nInput: A = { 3, 2, 1 }, B = { 1, 3, 2 }\nOutput: 7\nExplanation:\nFor A = 3 and B = 1 => Since the front of queue A does not match with the front of queue B. Pop and Push 3 at the back of Queue. Then, A becomes { 2, 1, 3 } and time consumed = 2 ( 1 unit of time for each push and pop operation)\nFor A = 2 and B = 1 => Since the front of queue A does not match with the front of queue B. Pop and Push 2 at the back of Queue. Then, A becomes { 1, 3, 2 } and time = 2 + 2 = 4\nFor A = 1 and B = 1 => Since the front of queue, A equals to front of queue B. Pop 1 from both the queue and execute it, Then A becomes { 3, 2 } and B becomes { 3, 2 } and time = 4 + 1 = 5\nFor A = 3 and B = 3 => Since the front of queue, A equals to front of queue B. Pop 3 from both the queue and execute it, Then A becomes { 2 } and B becomes { 2 } and time = 5 + 1 = 6\nFor A = 2 and B = 2 => Since the front of the queue, A equals to front of queue B. Pop 2 from both the queue and execute it. All the tasks are executed. time = 6 + 1 = 7\nTherefore the total time is 7.\n\nInput: A = { 3, 2, 1, 4 }, B = { 4, 1, 3, 2 }\nOutput: 14\n\n## Recommended: Please try your approach on {IDE} first, before moving on to the solution.\n\nApproach:\nFor each task in queue A:\n\n1. If the front task of queue A is the same as the front task of queue B. Pop the task from both the queues and execute it. Increment the total time by one unit.\n2. If the front task of queue A is not the same as the front task of queue B. Pop the task from queue A and push it at the back of queue A and increment the total time by two units. (1 for pop operation and 1 for push operation)\n3. Repeat the above steps till all the task in queue A is executed.\n\nBelow is the implementation of the above approach:\n\n `// C++ program to find the total ` `// time taken to execute the task ` `// in given order ` ` `  `#include \"bits/stdc++.h\" ` `using` `namespace` `std; ` ` `  `// Function to calculate the ` `// total time taken to execute ` `// the given task in original order ` `int` `run_tasks(queue<``int``>& A, ` `              ``queue<``int``>& B) ` `{ ` ` `  `    ``// To find the total time ` `    ``// taken for executing ` `    ``// the task ` `    ``int` `total_time = 0; ` ` `  `    ``// While A is not empty ` `    ``while` `(!A.empty()) { ` ` `  `        ``// Store the front element of queue A and B ` `        ``int` `x = A.front(); ` `        ``int` `y = B.front(); ` ` `  `        ``// If the front element of the queue A ` `        ``// is equal to the front element of queue B ` `        ``// then pop the element from both ` `        ``// the queues and execute the task ` `        ``// Increment total_time by 1 ` `        ``if` `(x == y) { ` `            ``A.pop(); ` `            ``B.pop(); ` `            ``total_time++; ` `        ``} ` ` `  `        ``// If front element of queue A is not equal ` `        ``// to front element of queue B then ` `        ``// pop the element from queue A & ` `        ``// push it at the back of queue A ` `        ``// Increment the total_time by 2 ` `        ``//(1 for push operation and ` `        ``// 1 for pop operation) ` `        ``else` `{ ` `            ``A.pop(); ` `            ``A.push(x); ` `            ``total_time += 2; ` `        ``} ` `    ``} ` ` `  `    ``// Return the total time taken ` `    ``return` `total_time; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given task to be executed ` `    ``queue<``int``> A; ` `    ``A.push(3); ` `    ``A.push(2); ` `    ``A.push(1); ` `    ``A.push(4); ` ` `  `    ``// Order in which task need to be ` `    ``// executed ` `    ``queue<``int``> B; ` `    ``B.push(4); ` `    ``B.push(1); ` `    ``B.push(3); ` `    ``B.push(2); ` ` `  `    ``// Function the returns the total ` `    ``// time taken to execute all the task ` `    ``cout << run_tasks(A, B); ` ` `  `    ``return` `0; ` `} `\n\n `// Java program to find the total ` `// time taken to execute the task ` `// in given order ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to calculate the ` `// total time taken to execute ` `// the given task in original order ` `static` `int` `run_tasks(Queue A, ` `                    ``Queue B) ` `{ ` ` `  `    ``// To find the total time ` `    ``// taken for executing ` `    ``// the task ` `    ``int` `total_time = ``0``; ` ` `  `    ``// While A is not empty ` `    ``while` `(!A.isEmpty()) ` `    ``{ ` ` `  `        ``// Store the front element of queue A and B ` `        ``int` `x = A.peek(); ` `        ``int` `y = B.peek(); ` ` `  `        ``// If the front element of the queue A ` `        ``// is equal to the front element of queue B ` `        ``// then pop the element from both ` `        ``// the queues and execute the task ` `        ``// Increment total_time by 1 ` `        ``if` `(x == y)  ` `        ``{ ` `            ``A.remove(); ` `            ``B.remove(); ` `            ``total_time++; ` `        ``} ` ` `  `        ``// If front element of queue A is not equal ` `        ``// to front element of queue B then ` `        ``// pop the element from queue A & ` `        ``// push it at the back of queue A ` `        ``// Increment the total_time by 2 ` `        ``//(1 for push operation and ` `        ``// 1 for pop operation) ` `        ``else` `        ``{ ` `            ``A.remove(); ` `            ``A.add(x); ` `            ``total_time += ``2``; ` `        ``} ` `    ``} ` ` `  `    ``// Return the total time taken ` `    ``return` `total_time; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``// Given task to be executed ` `    ``Queue A = ``new` `LinkedList(); ` `    ``A.add(``3``); ` `    ``A.add(``2``); ` `    ``A.add(``1``); ` `    ``A.add(``4``); ` ` `  `    ``// Order in which task need to be ` `    ``// executed ` `    ``Queue B = ``new` `LinkedList(); ` `    ``B.add(``4``); ` `    ``B.add(``1``); ` `    ``B.add(``3``); ` `    ``B.add(``2``); ` ` `  `    ``// Function the returns the total ` `    ``// time taken to execute all the task ` `    ``System.out.print(run_tasks(A, B)); ` ` `  `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `\n\n `# Python3 program to find the total ` `# time taken to execute the task ` `# in given order ` `from` `collections ``import` `deque ` ` `  `# Function to calculate the ` `# total time taken to execute ` `# the given task in original order ` `def` `run_tasks(A, B): ` ` `  `    ``# To find the total time ` `    ``# taken for executing ` `    ``# the task ` `    ``total_time ``=` `0` ` `  `    ``# While A is not empty ` `    ``while` `(``len``(A) > ``0``): ` ` `  `        ``# Store the front element of queue A and B ` `        ``x ``=` `A.popleft() ` `        ``y ``=` `B.popleft() ` ` `  `        ``# If the front element of the queue A ` `        ``# is equal to the front element of queue B ` `        ``# then pop the element from both ` `        ``# the queues and execute the task ` `        ``# Increment total_time by 1 ` `        ``if` `(x ``=``=` `y): ` `            ``total_time ``+``=` `1` ` `  `        ``# If front element of queue A is not equal ` `        ``# to front element of queue B then ` `        ``# pop the element from queue A & ` `        ``# append it at the back of queue A ` `        ``# Increment the total_time by 2 ` `        ``#(1 for append operation and ` `        ``# 1 for pop operation) ` `        ``else``: ` `            ``B.appendleft(y) ` `            ``A.append(x) ` `            ``total_time ``+``=` `2` ` `  `    ``# Return the total time taken ` `    ``return` `total_time ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``# Given task to be executed ` `    ``A ``=` `deque() ` `    ``A.append(``3``) ` `    ``A.append(``2``) ` `    ``A.append(``1``) ` `    ``A.append(``4``) ` ` `  `    ``# Order in which task need to be ` `    ``# executed ` `    ``B ``=` `deque() ` `    ``B.append(``4``) ` `    ``B.append(``1``) ` `    ``B.append(``3``) ` `    ``B.append(``2``) ` ` `  `    ``# Function the returns the total ` `    ``# time taken to execute all the task ` `    ``print``(run_tasks(A, B)) ` ` `  `# This code is contributed by mohit kumar 29 `\n\n `// C# program to find the total ` `// time taken to execute the task ` `// in given order ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to calculate the ` `// total time taken to execute ` `// the given task in original order ` `static` `int` `run_tasks(Queue<``int``> A, ` `                    ``Queue<``int``> B) ` `{ ` ` `  `    ``// To find the total time ` `    ``// taken for executing ` `    ``// the task ` `    ``int` `total_time = 0; ` ` `  `    ``// While A is not empty ` `    ``while` `(A.Count != 0) ` `    ``{ ` ` `  `        ``// Store the front element of queue A and B ` `        ``int` `x = A.Peek(); ` `        ``int` `y = B.Peek(); ` ` `  `        ``// If the front element of the queue A ` `        ``// is equal to the front element of queue B ` `        ``// then pop the element from both ` `        ``// the queues and execute the task ` `        ``// Increment total_time by 1 ` `        ``if` `(x == y)  ` `        ``{ ` `            ``A.Dequeue(); ` `            ``B.Dequeue(); ` `            ``total_time++; ` `        ``} ` ` `  `        ``// If front element of queue A is not equal ` `        ``// to front element of queue B then ` `        ``// pop the element from queue A & ` `        ``// push it at the back of queue A ` `        ``// Increment the total_time by 2 ` `        ``//(1 for push operation and ` `        ``// 1 for pop operation) ` `        ``else` `        ``{ ` `            ``A.Dequeue(); ` `            ``A.Enqueue(x); ` `            ``total_time += 2; ` `        ``} ` `    ``} ` ` `  `    ``// Return the total time taken ` `    ``return` `total_time; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``// Given task to be executed ` `    ``Queue<``int``> A = ``new` `Queue<``int``>(); ` `    ``A.Enqueue(3); ` `    ``A.Enqueue(2); ` `    ``A.Enqueue(1); ` `    ``A.Enqueue(4); ` ` `  `    ``// Order in which task need to be ` `    ``// executed ` `    ``Queue<``int``> B = ``new` `Queue<``int``>(); ` `    ``B.Enqueue(4); ` `    ``B.Enqueue(1); ` `    ``B.Enqueue(3); ` `    ``B.Enqueue(2); ` ` `  `    ``// Function the returns the total ` `    ``// time taken to execute all the task ` `    ``Console.Write(run_tasks(A, B)); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `\n\nOutput:\n```14\n```\n\nTime Complexity: O(N2), where N is the number of tasks.\n\nAttention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.\n\nA computer Science ug student from lnmiit Interested in learning and sharing algorithmic knowledge Found near online Judges and food\n\nIf you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.\n\nPlease Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below.\n\nImproved By : princiraj1992, mohit kumar 29" ]

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[ "# Search Our Content Library\n\n37 filtered results\n37 filtered results\nSort by\nFood Math\nWorksheet\nFood Math\nThis playful worksheet is a fun way to work concepts including conversions, decimals and fractions, and nutrition guidelines.\nScience\nWorksheet\nFraction Review: Addition, Subtraction, and Inequalities\nWorksheet\nFraction Review: Addition, Subtraction, and Inequalities\nCovering everything from adding and subtracting fractions to fraction inequalities, this worksheet offers a great review.\nMath\nWorksheet\nScary Skeleton Word Problems\nWorksheet\nScary Skeleton Word Problems\nDon't let skeletons scare you from doing Halloween math! Practice addition, subtraction, multiplication, and even fractions with scary skeleton word problems.\nMath\nWorksheet\nPumpkin Patch Sequences\nWorksheet\nPumpkin Patch Sequences\nDo some pumpkin patch math with your child this fall. See if you can find the tricky pattern that weaves through this pumpkin patch math worksheet.\nMath\nWorksheet\nWorksheet\nAdding and subtracting mixed numbers can be daunting, but this worksheet helps by breaking the process down step by step.\nMath\nWorksheet\nFabulous Fraction Review\nWorksheet\nFabulous Fraction Review\nMath\nWorksheet\nWorksheet\nMath\nWorksheet\nFractions Review and Simplifying Fractions\nWorksheet\nFractions Review and Simplifying Fractions\nReview fraction math, including converting improper fractions and adding and subtracting fractions with like denominators.\nMath\nWorksheet\nFraction Concepts\nWorksheet\nFraction Concepts\nKids tackle various fraction concepts in this math worksheet.\nMath\nWorksheet\nWorksheet\nThis colorful math worksheet will jog your child's memory through the steps to complete mixed fraction addition!\nMath\nWorksheet\nTwo Methods for Finding Common Denominators\nWorksheet\nTwo Methods for Finding Common Denominators\nUse these cards to help students compare two strategies that involve finding common denominators.\nMath\nWorksheet\nButterfly Method for Fractions\nWorksheet\nButterfly Method for Fractions\nHaving trouble with fraction subtraction? Try the butterfly method for fractions!\nMath\nWorksheet\nFocus on Fractions\nWorksheet\nFocus on Fractions\nAssess fifth grade fraction concepts, including equivalent fractions, adding with unlike denominators, improper fractions, and mixed numbers.\nMath\nWorksheet\nAdding Mixed Fractions with Unlike Denominators\nWorksheet\nAdding Mixed Fractions with Unlike Denominators\nDoes your child need extra math practice? This worksheet can help her with the steps to adding mixed fractions with unlike denominators.\nMath\nWorksheet\nAdding with Unlike Denominators Word Problem\nWorksheet\nAdding with Unlike Denominators Word Problem\nYour child will learn how to master fraction word problems with this clear and simple worksheet.\nMath\nWorksheet\nSubtracting with Unlike Denominators Word Problem\nWorksheet\nSubtracting with Unlike Denominators Word Problem\nYour child can conquer fraction word problems with this simple, step-by-step worksheet.\nMath\nWorksheet\nRunning with Word Problems: Practicing Adding Mixed Number Fractions\nWorksheet\nRunning with Word Problems: Practicing Adding Mixed Number Fractions\nMath\nWorksheet\nVocabulary Cards: Explain Fraction Conversions\nWorksheet\nVocabulary Cards: Explain Fraction Conversions\nUse these vocabulary cards with the EL Support Lesson: Explain Fraction Conversions.\nWorksheet\nProblems, Fractions, and Tape Diagrams\nWorksheet\nProblems, Fractions, and Tape Diagrams\nStudents will follow the four steps explained in this worksheet, then practice solving mixed number addition problems using tape diagrams.\nMath\nWorksheet\nFraction Word Problems: Strawberry Stand\nWorksheet\nFraction Word Problems: Strawberry Stand\nAmy needs help calculating her strawberry sales! This math worksheet will be great for practical application of fractions in everyday life.\nMath\nWorksheet\nFraction Word Problems: Pie Time\nWorksheet\nFraction Word Problems: Pie Time\nWhat fraction of the pie do you want? This tasty math worksheet will help your student familiarize with fractions.\nMath\nWorksheet\nAdding Mixed Fractions with Different Denominators\nWorksheet\nAdding Mixed Fractions with Different Denominators\nDoes your child need extra math practice? This worksheet can help him with the steps to solving mixed fractions addition problems.\nMath\nWorksheet\nWorksheet\nUse these vocabulary cards with the EL Support Lesson: Explaining Fraction Addition.\nMath\nWorksheet" ]

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[ "", null, "# MACD Histogram Indicator for MT5\n\nTrend following and momentum-based technical indicators are very useful for technical traders as they make it easier for traders to spot trend-based trading opportunities.\n\nThe MACD histogram is one of the most popular types of trend-following technical indicators that traders use. Let's see how we can use this indicator.\n\n## What is MACD Histogram Indicator?\n\nThe MACD histogram is a version of the classic MACD indicator, or Moving Average Convergence and Divergence.\n\nThe MACD Histogram indicator is a momentum-based technical indicator that plots the trend and direction of momentum and reversal as an oscillator. It draws two lines, one moving faster than the other. These two lines oscillate freely around the center line, which is zero. It also draws a histogram bar based on two lines as well.\n\n## How does the MACD Histogram indicator work?\n\nThe MACD Histogram indicator is based on an underlying exponential moving average (EMA) line pair. It calculates the difference between the faster moving average line and the slower moving average line. The difference is then plotted in its indicator window as the MACD line, which is the faster line.\n\nThen the MACD line is averaged using the simple moving average (SMA) method. This is then displayed as a slower signal line.\n\nThe histogram bar is basically the difference between the MACD line and the signal line.\n\n## How to use MACD Histogram indicator for MT5\n\nThe direction of the trend can be based on whether the two lines are generally positive or negative. Short-term trends can also be identified based on the interaction of the two lines. A bullish trend reversal is shown when the faster line crosses the slower line, while a bearish trend reversal is shown when the faster line crosses below the slower line.\n\nThe same can be applied with histogram bars as well. An uptrend reversal is shown when the histogram bars cross above zero, and a negative trend reversal is shown when the bars cross below zero.\n\nBeing an oscillator indicator, the MACD Histogram can also be used effectively to identify divergences. This is usually based on the MACD line.\n\nThe MACD Histogram indicator has four modifiable variables.\n\n“Fast EMA Period” and “Slow EMA Period” refer to the number of periods on which the EMA line is based.\n\n“Signal SMA period” refers to the number of periods on which the signal line is based.\n\n“Applied Price” refers to the source of the price point for each candle.\n\nHow to use MACD Histogram indicator for MT5" ]

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

[ "Study of mathematics online.\nStudy math with us and make sure that \"Mathematics is easy!\"\n\n# Distance between two points\n\nDefinition. Distance between two points is the length of the line segment that connects this points.\n\n## Distance formulas:\n\n• The formula for calculating the distance between two points A(xaya) and B(xbyb) on plane:\nAB = √(xb - xa)2 + (yb - ya)2\n• The formula for calculating the distance between two points A(xayaza) and B(xbybzb) in space:\nAB = √(xb - xa)2 + (yb - ya)2 + (zb - za)2\n\n## Proof of the formula of distance between two points for the plane problem", null, "From the points A and B drop perpendiculars to the coordinate axes.\n\nConsider a right triangle ∆ABC. Legs of the triangle are equal to:\n\nAC = xb - xa;\nBC = yb - ya.\n\nUsing the Pythagorean theorem, calculate the length of the hypotenuse AB:\n\nAB = √AC2 + BC2.\n\nSubstituting to this expression the lengths of AC and BC which expressed in terms of the coordinates of points A and B, we obtain the formula for calculating the distance between points on the plane.\n\nProof of the formula for calculating the distance between two points in space is similar.\n\n## Examples of tasks with distance between two points\n\n### Examples of tasks with distance between two points on 2D\n\nExample 1.\nFind the distance between two points A(-1, 3) and B(6,2).\n\nSolution.\n\nAB = √(xb - xa)2 + (yb - ya)2 = √(6 - (-1))2 + (2 - 3)2 = √72 + 12 = √50 = 5√2\n\nExample 2.\nFind the distance between two points A(0, 1) and B(2,-2).\n\nSolution.\n\nAB = √(xb - xa)2 + (yb - ya)2 = (2 - 0)2 + (-2 - 1)2 = 22 + (-3)2 = √13\n\n### Examples of tasks with distance between two points on 3D\n\nExample 3.\nFind the distance between two points A(-1, 3, 3) and B(6, 2, -2).\n\nSolution.\n\nAB = √(xb - xa)2 + (yb - ya)2 + (zb - za)2 =\n\n= √(6 - (-1))2 + (2 - 3)2 + (-2 - 3)2 = √72 + 12 + 52 = √75 = 5√3\n\nExample 4.\nFind the distance between two points A(0, -3, 3) and B(3, 1, 3).\n\nSolution.\n\nAB = √(xb - xa)2 + (yb - ya)2 + (zb - za)2 =\n\n= √(3 - 0)2 + (1 - (-3))2 + (3 - 3)2 = 32 + 42 + 02 = √25 = 5" ]

[ null, "https://onlinemschool.com/pictures/coordinate/point_point_2d.png", null ]

[ "## ››Convert meganewton/square metre to gigabar\n\n meganewton/square meter gigabar\n\n## ››More information from the unit converter\n\nHow many meganewton/square meter in 1 gigabar? The answer is 100000000.\nWe assume you are converting between meganewton/square metre and gigabar.\nYou can view more details on each measurement unit:\nmeganewton/square meter or gigabar\nThe SI derived unit for pressure is the pascal.\n1 pascal is equal to 1.0E-6 meganewton/square meter, or 1.0E-14 gigabar.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between meganewtons/square meter and gigabars.\nType in your own numbers in the form to convert the units!\n\n## ››Want other units?\n\nYou can do the reverse unit conversion from gigabar to meganewton/square meter, or enter any two units below:\n\n## Enter two units to convert\n\n From: To:\n\n## ››Definition: Gigabar\n\nThe SI prefix \"giga\" represents a factor of 109, or in exponential notation, 1E9.\n\nSo 1 gigabar = 109 bars.\n\nThe definition of a bar is as follows:\n\nThe bar is a measurement unit of pressure, equal to 1,000,000 dynes per square centimetre (baryes), or 100,000 newtons per square metre (pascals). The word bar is of Greek origin, báros meaning weight. Its official symbol is \"bar\"; the earlier \"b\" is now deprecated, but still often seen especially as \"mb\" rather than the proper \"mbar\" for millibars.\n\n## ››Metric conversions and more\n\nConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!" ]

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[ "# fix gld command¶\n\n## Syntax¶\n\nfix ID group-ID gld Tstart Tstop N_k seed series c_1 tau_1 ... c_N_k tau_N_k keyword values ...\n\n• ID, group-ID are documented in fix command\n\n• gld = style name of this fix command\n\n• Tstart,Tstop = desired temperature at start/end of run (temperature units)\n\n• N_k = number of terms in the Prony series representation of the memory kernel\n\n• seed = random number seed to use for white noise (positive integer)\n\n• series = pprony is presently the only available option\n\n• c_k = the weight of the kth term in the Prony series (mass per time units)\n\n• tau_k = the time constant of the kth term in the Prony series (time units)\n\n• zero or more keyword/value pairs may be appended\n\nkeyword = frozen or zero\nfrozen value = no or yes\nno = initialize extended variables using values drawn from equilibrium distribution at Tstart\nyes = initialize extended variables to zero (i.e., from equilibrium distribution at zero temperature)\nzero value = no or yes\nno = do not set total random force to zero\nyes = set total random force to zero\n\n## Examples¶\n\nfix 1 all gld 1.0 1.0 2 82885 pprony 0.5 1.0 1.0 2.0 frozen yes zero yes\nfix 3 rouse gld 7.355 7.355 4 48823 pprony 107.1 0.02415 186.0 0.04294 428.6 0.09661 1714 0.38643\n\n\n## Description¶\n\nApplies Generalized Langevin Dynamics to a group of atoms, as described in (Baczewski). This is intended to model the effect of an implicit solvent with a temporally non-local dissipative force and a colored Gaussian random force, consistent with the Fluctuation-Dissipation Theorem. The functional form of the memory kernel associated with the temporally non-local force is constrained to be a Prony series.\n\nNote\n\nWhile this fix bears many similarities to fix langevin, it has one significant difference. Namely, fix gld performs time integration, whereas fix langevin does NOT. To this end, the specification of another fix to perform time integration, such as fix nve, is NOT necessary.\n\nWith this fix active, the force on the jth atom is given as\n\n$\\begin{split}{\\bf F}_{j}(t) = & {\\bf F}^C_j(t)-\\int \\limits_{0}^{t} \\Gamma_j(t-s) {\\bf v}_j(s)~\\text{d}s + {\\bf F}^R_j(t) \\\\ \\Gamma_j(t-s) = & \\sum \\limits_{k=1}^{N_k} \\frac{c_k}{\\tau_k} e^{-(t-s)/\\tau_k} \\\\ \\langle{\\bf F}^R_j(t),{\\bf F}^R_j(s)\\rangle = & \\text{k_\\text{B}T} ~\\Gamma_j(t-s)\\end{split}$\n\nHere, the first term is representative of all conservative (pairwise, bonded, etc) forces external to this fix, the second is the temporally non-local dissipative force given as a Prony series, and the third is the colored Gaussian random force.\n\nThe Prony series form of the memory kernel is chosen to enable an extended variable formalism, with a number of exemplary mathematical features discussed in (Baczewski). In particular, $$3N_k$$ extended variables are added to each atom, which effect the action of the memory kernel without having to explicitly evaluate the integral over time in the second term of the force. This also has the benefit of requiring the generation of uncorrelated random forces, rather than correlated random forces as specified in the third term of the force.\n\nPresently, the Prony series coefficients are limited to being greater than or equal to zero, and the time constants are limited to being greater than zero. To this end, the value of series MUST be set to pprony, for now. Future updates will allow for negative coefficients and other representations of the memory kernel. It is with these updates in mind that the series option was included.\n\nThe units of the Prony series coefficients are chosen to be mass per time to ensure that the numerical integration scheme stably approaches the Newtonian and Langevin limits. Details of these limits, and the associated numerical concerns are discussed in (Baczewski).\n\nThe desired temperature at each timestep is ramped from Tstart to Tstop over the course of the next run.\n\nThe random # seed must be a positive integer. A Marsaglia random number generator is used. Each processor uses the input seed to generate its own unique seed and its own stream of random numbers. Thus the dynamics of the system will not be identical on two runs on different numbers of processors.\n\nThe keyword/value option pairs are used in the following ways.\n\nThe keyword frozen can be used to specify how the extended variables associated with the GLD memory kernel are initialized. Specifying no (the default), the initial values are drawn at random from an equilibrium distribution at Tstart, consistent with the Fluctuation-Dissipation Theorem. Specifying yes, initializes the extended variables to zero.\n\nThe keyword zero can be used to eliminate drift due to the thermostat. Because the random forces on different atoms are independent, they do not sum exactly to zero. As a result, this fix applies a small random force to the entire system, and the center-of-mass of the system undergoes a slow random walk. If the keyword zero is set to yes, the total random force is set exactly to zero by subtracting off an equal part of it from each atom in the group. As a result, the center-of-mass of a system with zero initial momentum will not drift over time.\n\n## Restart, fix_modify, output, run start/stop, minimize info¶\n\nThe instantaneous values of the extended variables are written to binary restart files. Because the state of the random number generator is not saved in restart files, this means you cannot do “exact” restarts with this fix, where the simulation continues on the same as if no restart had taken place. However, in a statistical sense, a restarted simulation should produce the same behavior.\n\nNone of the fix_modify options are relevant to this fix. No global or per-atom quantities are stored by this fix for access by various output commands.\n\nThis fix can ramp its target temperature over multiple runs, using the start and stop keywords of the run command. See the run command for details of how to do this.\n\nThis fix is not invoked during energy minimization.\n\n## Restrictions¶\n\nThis fix is part of the EXTRA-FIX package. It is only enabled if LAMMPS was built with that package. See the Build package page for more info.\n\n## Default¶\n\nThe option defaults are frozen = no, zero = no.\n\n(Baczewski) A.D. Baczewski and S.D. Bond, J. Chem. Phys. 139, 044107 (2013)." ]

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[ "", null, "Sound: Decibel Levels", null, "The loudness of a sound is commonly expressed in decibels. Loudness is called the 'intensity level' of a sound. This is not to be confused with the intensity of a sound!", null, "The range of 'loudness', from the quietest sound that can be heard to a sound so loud that is produces pain rather than the sensation of hearing, is so large that an exponential scale is used. To measure 'loudness' we compare the sound's intensity to that of the least audible sound.\n\nThe quietest sound that can be heard is said to be at the threshold of hearing. The intensity of sound at the threshold of hearing is given the symbol Io Intensity of sound is a mesasure of the ound energy directed onto a metre square each second, so it is measured in watts per square metre (W m-2).\n\nI0 = 1.0 x 10-12 W m-2\n\nTo calculate the intensity level of a sound you have to find the ratio of the intensity of that sound to the threshold intensity: I/Io. But as the range is so large we use an exponential scale and you find the log of that ratio. You then measure that intensity level in Bels - the unit was named in recognition of Alexander Graham Bell - but they missed off the last 'l' for some reason!\n\nintensity level (B) = log10 (I/Io)\n\nThe bel is a large unit - we usually measure in decibels (dB)\n\nintensity level (dB) = 10 log10 (I/Io)\n\nExample 1\n\nThe intensity of sound in an exam room is about 1.0 x 10-7 W m-2. Calculate the decibel level of that sound.\n\nStep 1: Find the ratio of the sound intensity to the threshold intensity\n\nI/Io = (1.0 x 10-7)/(1.0 x 10-12)\n\n= 1.0 x 105 (no unit because it is a ratio)\n\nStep 2: Take the logarithm of the ratio\n\nlog10 (1.0 x 105) = 5\n\nStep 3: Multiply the ratio by 10\n\nIntensity level = 50 dB\n\nTo find the intensity of a given sound whose intensity level in decibels is known, perform the above steps in reverse order and perform the inverse operation.\n\nExample 2\n\nAn underground train produces an intensity level of 100 dB. What is the intensity of the sound which produces an intensity level of 100 dB?\n\nStep 1: Divide the decibel level by 10\n\n100/10 = 10\n\nStep 2: Use that value as the exponent of the ratio\n\nI/Io = 1010\n\nStep 3: Find the intensity in watts per square metre.\n\nI = 1010x Io\n\n= 1010x 1.0 x 10-12\n\n= 1.0 x 10-2 W m-2" ]

[ null, "https://www.cyberphysics.co.uk/graphics/animations/ani_spider214x700.gif", null, "https://www.cyberphysics.co.uk/ctainc/prof_2012.gif", null, "https://www.cyberphysics.co.uk/graphics/diagrams/waves/dB.png", null ]

[ "# problem with a plug-in (generate)\n\nHello. I try to write a plug - in from a script of P. Morales. It does not work. Do you understand the problem?\nBeforehand thank you for your help.\nRISSET TIBETAN.lsp (1.3 KB)\n\nThis is malformed:\n\n``````(defun sound\n(scale 0.1 (tibetan (freq offset dur rise dec)))\n``````\n\nA function definition must have its arguments listed in parentheses after the name of the function:\n\n``````(defun finction-name (arg1 arg2 ...)\nfunction body)\n``````\n\nIf there are no arguments, then it still needs the parentheses:\n\n``````(defun finction-name ()\nfunction body)\n``````\n\nYour function definition “sound” is missing a “)” at the end.\n\n``````(freq offset dur rise dec)\n``````\n\nThis is incorrect because:\na) “freq” is not a function\nb) If “freq” is supposed to be a variable (rather than a function), it needs to have a value, either scoped globally or passed to your “sound” function as an argument.\n\nAlso, best not to use “sound” as a function name as it is already a defined function in Nyquist.\n\nPerhaps you meant:\n\n``````(defun mysound ()\n(scale 0.1 (tibetan pitch offset dur rise dec)))\n\n(stretch 1 (mysound))\n``````\n\nThank you.\nI try to use the following code:\n\n``````(defun mytone ()\n(sine pitch dur))\n``````\n\nI added the code following in the plug-in “generate”\n\n``````;control pitch \"pitch\" real \"MIDI pitch\" 55 36 84\n``````\n\n.\nThe problem is in this code.\n\n``````(setf pitch (step–to–hz))\n``````\n\nThe nyquist console indicates the following error;\n\n``````error: illegal character - -30\nFunction: #<Subr-(null): #407d328>\n``````\n\nDo you understand my error?\n\n“step-to-hz” converts a MIDI note number (“step”) into Hz.\nFor example, A4 is MIDI note number 69 and the frequency of a sine wave at A4 is 440 Hz (standard tuning).\n\n``````(print (step-to-hz 69)) ; prints 440\n``````\n\n``````(print (hz-to-step 440)) ; prints 69\n``````\n\n(step-to-hz) is no good because you have not told the “step-to-hz” function what number to convert. “step-to-hz” takes one “argument” (the “step” number).\n\nHello.\nThank you.\nI would want to evoke the code of Pedro Jose Morales “b5.lsp”.\n\n``````(defun lfo-pitch-control ()\n(pwlv 0.25 0.6 0.25 1.4 0.5 2.0 0.95 2.2 0.25 3.4 0.5 3.8 0.78 dur -0.2))\n\n(defun starship (frq scl)\n(apply #' sim\n(mapcar #' (lambda (offset)\n(fmosc (hz-to-step (+ frq offset)) (scale scl (lfo-pitch-control))))\n'(0.0 4.3 9.5 23.0 39.0 84.0))))\n``````\n\nI would want to evoke the following part of the code:\n\n``````(lambda (offset)\n(fmosc (hz-to-step (+ frq offset)) (scale scl (lfo-pitch-control))))\n``````\n\nIn the user manual of R. B Dannenberg, the author indicates: (lambda args expr…) - make a function closure\nargs - formal argument list (lambda list) (quoted)\nMy question is thus the following one: apparently, the “lambda” function is followed by arguments.\nSo:\n\n``````(offset)\n``````\n\nIs it the first argument?\n\n``````(fmosc (hz-to-step (+ frq offset))\n``````\n\nIs it the second argument?\nand\n\n``````(scale scl (lfo-pitch-control)\n``````\n\nIs it the third argument of the “lambda” function?\n\n``````(lambda (offset)\n(fmosc (hz-to-step (+ frq offset)) (scale scl (lfo-pitch-control))))\n``````\n\nThe name of the function is: lambda\nThe argument list is enclosed in parentheses. There is just one argument: offset\nThe last part is the body: (fmosc (hz-to-step (+ frq offset)) (scale scl (lfo-pitch-control)))\n\nThe body contains one function: fmosc http://www.cs.cmu.edu/~rbd/doc/nyquist/part8.html#index378\n\n``````(fmosc (hz-to-step (+ frq offset))\n(scale scl(lfo-pitch-control)))\n``````\n\nfmosc requires at least two arguments. In this case it has exactly two arguments.\n\nThe first argument of fmosc defines the pitch:\n\n``````(hz-to-step (+ frq offset))\n``````\n\nThe second argument of fmosc defines the modulation:\n\n``````(scale scl(lfo-pitch-control))\n``````\n\nThank you.\nThe user manual of the XLISP function (mapcar indicate:\n(mapcar function list1 [list2 … ])\nIf the function (lambda with its list which has that an argument is the \"list1 “, I presume that the \" list 2” is:\n\n``````'(0.0 4.5 9.4 23.0 39.0 84.0)\n``````\n\nIt is right?\nThis is an unevaluated list.\nMy question is the following one. I tried to understand how the values of the “list 2” act on the generated sound wave. I did not understand the logic.\nHow do we choose these values?\n\nI presume that you are referring to this code:\n\n``````(mapcar #'(lambda (off)\n(osc (hz-to-step (+ frq (* off offset))) dur *tibetan-table*))\n'(0 1 2 3 4 -1 -2 -3 -4))\n``````\n\nThis is quite tricky to read, but we can break it down into sections by looking where the “(” match with “)”.\n\nNote that the function CAR returns the first element of a list, so when the XLISP manual says “successive CARs”, it means the first term, then the next term, then the next. It is simply taking each element of the list in turn.\n\nFrom the outer brackets we see:\n\n``````(mapcar #'(lambda .....)\n'(0 1 2 3 4 -1 -2 -3 -4))\n``````\n\nso '(0 1 2 3 4 -1 -2 -3 -4) is the list that the lambda expression is applied to.\n\nThen moving into the lambda expression:\n\n``````(lambda (off) (osc ....))\n``````\n\nwe have an unnamed function with a local variable “off” and the body of the lambda expression consists of the function OSC.\nThis function is applied to successive terms from the list.\n\nMoving in further to the OSC function we see that OSC has three arguments:\n(osc step duration table)\n\n``````(osc (hz-to-step (+ frq (* off offset)))\ndur\n*tibetan-table*)\n``````\n\nThe “step” argument is (hz-to-step (+ frq (* off offset)))\nThe “duration” argument is the variable “dur”\nThe “table” argument is the symbol “tibetan-table\n\nThe interesting thing here is that the “step” argument is a function that includes the variable “off” [which was defined as a local variable of the lambda expression]\n\nSo, moving back out again, the lambda expression has an argument “off” that is used within the function “hz-to-step”.\nMAPCAR applies the lambda expression to each term in the list. In other words, each of the numbers in the list becomes the “off” variable.\n\nTo see how this works, lets take a simple example that can be run in the Nyquist Prompt. Use the Debug button to see the output:\n\n``````(mapcar #'(lambda (x) (print x)) '(1 2 3 2 1))\n``````\n\nHere, the symbol “x” is a local variable of the unnamed function (the lambda expression) and '(1 2 3 2 1) is the list that the lambda expression is applied to.\nWithin the body of the lambda expression is the function (print x).\nSo what happens with MAPCAR is that the first element of the list [the number 1] is passed to the lambda expression and printed. Then the next element of the list [the number 2] and so on. The result is:\n\n``````1\n2\n3\n2\n1\n``````\n\nIn our more complex example, the numbers 0, 1, 2, 3, 4, -1, -2, -3, and -4 are passed to the lambda expression and are evaluated within the HZ-TO-STEP function, so that a series of “step” values is created which can be used by the OSC function. Thus the code creates a list of sounds." ]

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[ "Try the fastest way to create flashcards\nQuestion\n\n# Show that if available carpentry hours remain between 60 and 100, the current basis remains optimal. If between 60 and 100 carpentry hours are available, would Giapetto still produce 20 soldiers and 60 trains?\n\nSolution\n\nVerified\nStep 1\n1 of 3\n\nWe are to show that if available carpentry hours remain between $60$ and $100,$ the current basis remains optimal. Also we are to check if Giapetto would still produce $20$ soldiers and $60$ trains. Ofcourse, maximizing the objective function is $\\max z=3x_1+2x_2,$ where $x_1$ and $x_2$ are numbers (nonnegative) of soldiers and trains produced per week, respectively. It is subjected to previous constraints:\n\n$2x_1+x_2\\leq100,x_1+x_2\\leq80,x_1\\leq40.$\n\nLet us $b_2$ be the number of carpentry hours available. Obviously the change in $b_2$ shifts the carpentry constraint parallel to its current position. However, as long as both constraints intersect in the feasible region the current basis will remain optimal. From the figure preented below we see that if $b_2<60$ intersection of finishing and carepntry constraint is not in the feasible region, thus the current basis is not optimal. We see that for $b_2=60$ the point $(40,20)$ is intersection of all $3$ constraints (in feasible region ofcourse). If however $b_2<60$ the demand constraint would not be satisfied. If $b_2>100$ then finishing and carpentry constraint would never intersect, thus the current basis would not remain optimal. So the current basis will remain optimal if $60\\leq b_2\\leq100.$ However, the optimal solution will change. Let us $b_2=80+\\Delta.$ The current basis remains optimal if $-20\\leq\\Delta\\leq20.$ Thus it has to be:\n\n$2x_1+x_2=100,x_1+x_2=80+\\Delta,$\n\nsince we want to find a point where the constraints are binding. Subtracting the second from the first equation yields\n\n$x_1=20-\\Delta,x_2=60+2\\Delta.$\n\nSimply, increasing available carpentry hours decreases at the same rate the optimal number of soldiers produced. Similarly the number of trains increases two times the number of additional carpentry hours. Therefore, we are done.\n\n## Recommended textbook solutions", null, "#### Introduction to Operations Research\n\n10th EditionISBN: 9780073523453Frederick S. Hillier\nVerified solutions", null, "#### Operations Research: Applications and Algorithms\n\n4th EditionISBN: 9780534380588 (5 more)Wayne L Winston\n1,445 solutions", null, "#### STAT: Behavioral Sciences\n\n2nd EditionISBN: 9781305436923Gary Heiman", null, "#### A Mathematical Look at Politics\n\n1st EditionISBN: 9781439819838Daniel H. Ullman, E. Arthur Robinson Jr." ]

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[ "## Friday, 22 March 2013\n\n### Duration Calculator for Dates in VB.NET\n\nHello Guys,\nI was working on a project of a hotel management system. I had to calculate the day between two given dates by DateTimePicker in vb.NET.\nI named the DateTimePicker as \"DTPcheckin\" for the day when customer checks into the hotel and\n\"DTPcheckout\" for the date of checkout.\nTo calculate the date and to facilitate myself for the later calculation and SQL comparisons I converted the date to the NUMBER OF DAYS FROM THE BEGINNING OF THE CALENDER .\nLIKE \"20 MARCH 2013\" == 735328. that is 735328 days passed from date 01/01/0001.\nU can use this anywhere you want. here is the code------\n\nDim checkin_day As Integer\nDim checkout_day As Integer\nDim checkin_month As Integer\nDim checkout_month As Integer\nDim checkin_y As Integer\nDim checkout_y As Integer\nDim staying As Integer\nDim mdcin As Integer\nDim mdcout As Integer\nDim daycin As Integer\nDim daycout As Integer\n\nPrivate Sub calday()\ncheckin_day = CInt(DTPcheckin.Value.Day)\ncheckout_day = CInt(DTPcheckout.Value.Day)\ncheckin_month = CInt(DTPcheckin.Value.Month)\ncheckout_month = CInt(DTPcheckout.Value.Month)\ncheckin_y = CInt(DTPcheckin.Value.Year)\ncheckout_y = CInt(DTPcheckout.Value.Year)\n\nmdcin = mdcount(checkin_month)\nmdcout = mdcount(checkout_month)\ndaycin = checkin_y * 365 + mdcin + checkin_day\nDim temp2 As Integer = checkin_y - 1\ntemp2 = temp2 / 4\ndaycin = daycin + temp2\n\nIf (checkin_y Mod 400 = 0 Or (checkin_y Mod 100 <> 0 And checkin_y Mod 4 = 0)) And checkin_month > 2 Then\ndaycin = daycin + 1\nEnd If\n\ndaycout = checkout_y * 365 + mdcout + checkout_day\nDim temp As Integer = checkout_y - 1\ntemp = temp / 4\ndaycout = daycout + temp\nIf (checkout_y Mod 400 = 0 Or (checkout_y Mod 100 <> 0 And checkout_y Mod 4 = 0)) And checkout_month > 2 Then\ndaycout = daycout + 1\nEnd If\n//duration is calculate by subtracting the daycin from daycout\nduration = daycout - daycin\nEnd Sub\n\n//the mdcount function is\n//this  mdcount function calculate the number of days from 1st jan to the given month\nPublic Function mdcount(ByVal n As Integer) As Integer\nDim md As Integer\n\nIf n = 1 Then\nmd = 0\nElseIf n = 2 Then\nmd = 31\nElseIf n = 3 Then\nmd = 59\nElseIf n = 4 Then\nmd = 90\nElseIf n = 5 Then\nmd = 120\nElseIf n = 6 Then\nmd = 151\nElseIf n = 7 Then\nmd = 181\nElseIf n = 8 Then\nmd = 212\nElseIf n = 9 Then\nmd = 243\nElseIf n = 10 Then\nmd = 273\nElseIf n = 11 Then\nmd = 304\nElseIf n = 12 Then\nmd = 334\nEnd If\nReturn md\nEnd Function" ]

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[ "# Deriving boltzmann distribution\n\n• I\nI was reading the derivation of boltzmann distribution using the reservoir model.\nlets call the reservoir by index R and the tiny system by index A.\nIn the derivation they proposed that the probability for being at energy e (for A) is proportional to the number of states in reservoir. I didn't understand this completely and i would be happy to get some help!\nhere is my take on it, and please correct me if I'm wrong.\n- The temperature of the whole system is T and it's constant therefor the number of states for the whole systems g is also constant\n- both A and R are independent of each other therefor g = gA ⋅ gR\n- if gR goes up then gA has to go down meaning gA ∝ gR\n- P(e) ∝ 1/gA → P(e) ∝ gR\nI'm not really convinced by my explanation so if someone could explain it and perhaps give me an intuitive physical explanation, I'd be happy. Thank you" ]

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[ "Cody\n\n# Problem 5. Triangle Numbers\n\nSolution 1951372\n\nSubmitted on 28 Sep 2019 by Le Manh Long Le Manh Long\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\nn = 1; t = 1; assert(isequal(triangle(n),t))\n\nt = 1\n\n2   Pass\nn = 3; t = 6; assert(isequal(triangle(n),t))\n\nt = 6\n\n3   Pass\nn = 5; t = 15; assert(isequal(triangle(n),t))\n\nt = 15\n\n4   Pass\nn = 30; t = 465; assert(isequal(triangle(n),t))\n\nt = 465" ]

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[ "Memory leak in stack\n\nI'm implementing a stack by C++ and here I think it has memory leakage problem: in Stack::peek() and Stack::pop(), where I created heap space and returned pointer to function caller.\n\nUsers who call these two functions may perform deletion on returned pointer, which will recycle heap space once done. I am wondering what is a better approach to my current leakage prone approach.\n\n// this is the header file of stack data structure.\n\n#ifndef MY_STACK_H\n#define MY_STACK_H\n\nclass Node\n{\nprivate:\nint number;\nNode * next;\n\npublic:\nNode(){number = 0; next = NULL;}\n\nNode(int initialNumber, Node * initialNext = NULL)\n{\nnumber = initialNumber;\nnext = initialNext;\n}\n\n// copy constructor\nNode(Node & copyFromNode)\n{\nthis->number = copyFromNode.getNumber();\nthis->next = copyFromNode.getNext();\n}\n\n// setters & getters\nint getNumber() {return number;}\nNode * getNext() {return next;}\nvoid setNumber(int newNumber) {number = newNumber;}\nvoid setNext(Node * newNext) {next = newNext;}\n};\n\n{\nprivate:\n\npublic:\n\n// very similar to Stack::push()\n{\n}\n\n// somehow similar to Stack::pop()\nvoid deleteFirst()\n{\nif (head == NULL)\nreturn;\nelse\n{\nNode * temp = head;\ndelete temp;\n}\n}\n\nbool isEmpty() {return (head == NULL);}\n\n// there is no setter to head\n// since head should be maintained by addFirst() & deleteFirst() only.\n};\n\nclass Stack\n{\nprivate:\nNode * top;\npublic:\nStack(){top = NULL; }\n\n// push new node to the stack\n// new node will become the new top of stack\nvoid push(Node * newNode);\n\n// it checks if the stack is empty or not.\nbool isEmpty();\n\n// it copies the top node on stack,\n// without deleting it.\nNode * peek();\n\n// it copies the top node on stack,\n// and deletes the node from stack.\nNode * pop();\n};\n\n#endif\n\n.cpp file:\n\n// implementation for the stack\n\n#include <iostream>\n#include \"stack.h\"\n\nusing namespace std;\n\nvoid Stack::push(Node * newNode)\n{\n}\n\nbool Stack::isEmpty()\n{\nif(top == NULL)\nreturn true;\nelse\nreturn false;\n}\n\nNode * Stack::peek()\n{\n// make a copy, erase next link and return\n// copy is made by calling copy construtor\nNode * copy = new Node(*top); // ?? asks for reference but pointer is given\n\nreturn copy;\n}\n\nNode * Stack::pop()\n{\nif (isEmpty())\nreturn NULL;\nelse\n{\nNode * copy = new Node(*top);\nll.deleteFirst();\nreturn copy;\n}\n}\n\nint main()\n{\nStack stack;\n\nNode * nd1 = new Node(100);\nNode * nd2 = new Node(200);\n\n// test push()\nstack.push(nd1);\nstack.push(nd2);\n\n// test isEmpty()\ncout<<\"Test: isEmpty(): \"<< stack.isEmpty()<<endl;\n\n// test peek()\ncout<<\"Test: peek(): \"<<stack.peek()->getNumber()<<endl;\n\n// test pop()\ncout<<\"Test: pop()\"<<stack.pop()->getNumber()<<endl;\ncout<<\"Test: pop()\"<<stack.pop()->getNumber()<<endl;\n\ncout<<\"Test: isEmpty(): \"<<stack.isEmpty()<<endl;\n\nreturn 0;\n}\n\nDon't allocate nodes until you really need to.\n\nLinkedList should not expose that it has nodes. addFirst should take an int to be stored in the node and getHead should return the value in the head node.\n\nvoid addFirst(int number )\n{\nNode* newNode = new Node();\nnewNode->setNumber(number);\n}\n\nSame in stack don't expose that it deals in nodes and don't let calling code access them. That way you have more control over the lifetimes.\n\n• This is also what the C++ <stack>implementation does. You could return a reference from getHead/peek though. For complex types, copies should be avoided – which is also why <stack> doesn't return anything from pop, because it can't return a reference there. A copy constructor could throw and then the popped element would get lost. – Felix Dombek Oct 12 '16 at 11:37\n• @FelixDombek I'd been wondering why so many STL data structures are set up that way. Makes sense. – JAB Oct 12 '16 at 15:14\n\nDon't use NULL unless you have to\n\nSome IDEs will refuse to compile your code unless you include <cstdlib> so that NULL is available. If you compile using a C++ compiler using latest standards, you could use nullptr instead, since it is a language construct in those versions and does not require including anything.\n\nMinor improvements\n\n//bool Stack::isEmpty()\n//{\n// if(top == NULL)\n// return true;\n// else\n// return false;\n//}\n\nbool Stack::isEmpty()\n{\n}\n\nand\n\n//Node * Stack::pop()\n//{\n//\n// if (isEmpty())\n// return NULL;\n// else\n// {\n// Node * copy = new Node(*top);\n// ll.deleteFirst();\n// top = ll.getHead();\n// return copy;\n// }\n//}\n\nNode * Stack::pop()\n{\nif (isEmpty())\n{\nreturn nullptr;\n}\n\nNode * copy = new Node(*top);\nll.deleteFirst();\nreturn copy;\n}\n\nAlternative implementation\n\nIt's not difficult to devise a generic stack:\n\nmy_stack.h\n\n#ifndef MY_STACK_H\n#define MY_STACK_H\n\n#include <iostream>\n#include <stdexcept>\n#include <vector>\n\ntemplate<typename T>\nclass MyStack {\n\nstd::vector<T> storage;\nvoid check_not_empty()\n{\nif (storage.empty())\n{\nthrow std::runtime_error{\"Peeking from an empty stack.\"};\n}\n}\n\npublic:\nvoid push(const T& element)\n{\nstorage.push_back(element);\n}\n\nvoid pop()\n{\ncheck_not_empty();\nstorage.pop_back();\n}\n\nconst T& peek() const\n{\ncheck_not_empty();\nreturn storage[storage.size() - 1];\n}\n\nsize_t size()\n{\nreturn storage.size();\n}\n\nbool is_empty()\n{\nreturn storage.empty();\n}\n\nfriend std::ostream& operator<<(std::ostream& os, const MyStack<T>& stack)\n{\nos << \"[\";\nstd::string separator = \"\";\n\nfor (size_t i = 0; i != stack.storage.size(); ++i)\n{\nos << separator << stack.storage[i];\nseparator = \", \";\n}\n\nreturn os << \"]\";\n}\n};\n\n#endif\n\nmain.cpp\n\n#include \"my_stack.h\"\n#include <iostream>\n\nusing namespace std;\n\nint main()\n{\nMyStack<int> stack;\n\ncout << stack << endl;\n\nfor (int i = 1; i <= 5; ++i)\n{\nstack.push(i);\ncout << stack << endl;\n}\n\nwhile (!stack.is_empty())\n{\nstack.pop();\ncout << stack << endl;\n}\n}\n• nullptr is a C++11 language construct. If you are going to assume that the OP has access to C++11 then you should definitely point out smart pointers. – Sombrero Chicken Oct 12 '16 at 14:28\n• @GillBates Point. I made more explicit notice on modern C++. – coderodde Oct 12 '16 at 14:31\n\nThe other answers bring up good points, but, architecturally, the problem is that your linked list acquires control of Nodes and offers no way to relinquish it. Think about it: in deleteFirst, you delete your old list head , but how do you know that head was heapalloc'd in the first place?\n\nConsider the following:\n\nvoid example(void)\n{\nNode n(5, nullptr);\nll.deleteFirst();\n}\n\nNow you're going todelete a portion of the stack.\n\nInstead, deleteFirst should remove and return the head, instead of deleting it. Then the consumer can delete it as necessary (or not). Then a regular user can write something like\n\nvoid example(void)\n{\ndelete ll.deleteFirst();\n}\n\nand my example above compiles. So how do we implement this?\n\nNode *deleteFirst(void)\n{\n{\n}\nreturn temp;\n}\n\nAt this point, you may want to rename deleteFirst; for the sake of fidelity to the problem, I shan't. Now you can write your stack as\n\nNode &Stack::peek(void)\n{\nreturn *top;\n}\nNode *Stack::pop(void)\n{\nreturn isEmpty() ? nullptr : ll.deleteFirst();\n}\n\nYou say your problem is that you fear that the user may not delete the node once popped. There are many levels at which you may do something.\n\nGet your resource management straight\n\nHave a look at unique_ptr. A unique_ptr represents a unique owner of a resource, responsible for freeing the resource. It may not be copied, only moved. In your case, when popping, you want to give the ownership back to the user, have popreturn a unique_ptr<Node> by value. Reserve plain pointers as non-owners (a.k.a. observers) of the resource.\n\nAll your memory management problems will disappear unless the user explicitly does something dangerous. The dangerous will be made verbose.\n\nIf you want a quick look at resource management in C++11 and later, you may have a look at https://stackoverflow.com/a/28826952/2071258 (this is an answer of mine)\n\nFix your interface. Do not expose the nodes\n\nThe fact that there are Nodes is purely an implementation detail, worse, it fails to hide the fact that your Stack is implemented over a LinkedList (since Node belongs to LinkedList). Instead of returning a pointer to a node when popping, return a value. By value, since the node, containing the value, disappears." ]

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[ "# Creating new DF column based on average values from specific columns identified in second DF\n\nI apologize as I prefer to ask questions where I’ve made an attempt at the code needed to resolve the issue. Here, despite many attempts, I haven’t gotten any closer to a resolution (in part because I’m a hobbyist and self-taught). I’m attempting to use two dataframes together to calculate the average values in a specific column, then generate a new column to store that average.\n\nI have two dataframes. The first contains the players and their stats. The second contains a list of each player’s opponents during the season.\n\nWhat I’m attempting to do is use the two dataframes to calculate expected values when facing a specific opponent. Stated otherwise, I’d like to be able to see if a player is performing better or worse than the expected results based on the opponent but first need to calculate the average of their opponents.\n\nMy dataframes actually have thousands of players and hundreds of matchups, so I’ve shortened them here to have a representative dataframe that isn’t overwhelming.\n\nThe first dataframe (df) contains five columns. Name, STAT1, STAT2, STAT3, and STAT4.\n\nThe second dataframe (df_Schedule) has a Name column but then has a separate column for each opponent faced. The df_Schedule usually contains different numbers of columns depending on the week of the season. For example, after week 1 there may be four columns. After week 26 there might be 100 columns. For simplicity sake, I’ve included just five columns [‘Name’, ‘Opp1’, ‘Opp2’, ‘Opp3’, ‘Opp4’, ‘Opp5’].\n\nUsing these two dataframes I’m trying to create new columns in the first dataframe (df). EXP1 (for “Expected STAT1”), EXP2, EXP3, EXP4. The expected columns are simply an average of the STAT columns based on the opponents faced during the season. For example, Edgar faced Ralph three times, Marc once and David once. The formula to calculate Edgar’s EXP1 is simply:\n\n((Ralph.STAT1 * 3) + (Marc.STAT1 * 1) + (David.STAT1 * 1) / Number_of_Contests (which is five in this example) = 100.2\n\n```import pandas as pd\n\ndata = {'Name':['Edgar', 'Ralph', 'Marc', 'David'],\n'STAT1':[100, 96, 110, 103],\n'STAT2':[116, 93, 85, 100],\n'STAT3':[56, 59, 41, 83],\n'STAT4':[55, 96, 113, 40],}\n\ndata2 = {'Name':['Edgar', 'Ralph', 'Marc', 'David'],\n'Opp1':['Ralph', 'Edgar', 'David', 'Marc'],\n'Opp2':['Ralph', 'Edgar', 'David', 'Marc'],\n'Opp3':['Marc', 'David', 'Edgar', 'Ralph'],\n'Opp4':['David', 'Marc', 'Ralph', 'Edgar'],\n'Opp5':['Ralph', 'Edgar', 'David', 'Marc'],}\n\ndf = pd.DataFrame(data)\n\ndf_Schedule = pd.DataFrame(data2)\n\nprint(df)\n\nprint(df_Schedule)\n```\n\nI would like the result to be something like:\n\n```data_Final = {'Name':['Edgar', 'Ralph', 'Marc', 'David'],\n'STAT1':[100, 96, 110, 103],\n'STAT2':[116, 93, 85, 100],\n'STAT3':[56, 59, 41, 83],\n'STAT4':[55, 96, 113, 40],\n'EXP1':[100.2, 102.6, 101, 105.2],\n'EXP2':[92.8, 106.6, 101.8, 92.8],\n'EXP3':[60.2, 58.4, 72.8, 47.6],\n'EXP4':[88.2, 63.6, 54.2, 98],}\n\ndf_Final = pd.DataFrame(data_Final)\n\nprint(df_Final)\n```\n\nIs there a way to use the scheduling dataframe to lookup the values of opponents, average them, and then create a new column based on those averages?\n\nTry:\n\n```df = df.set_index(\"Name\")\ndf_Schedule = df_Schedule.set_index(\"Name\")\n\nfor i, c in enumerate(df.filter(like=\"STAT\"), 1):\ndf[f\"EXP{i}\"] = df_Schedule.replace(df[c]).mean(axis=1)\n\nprint(df.reset_index())\n```\n\nPrints:\n\n``` Name STAT1 STAT2 STAT3 STAT4 EXP1 EXP2 EXP3 EXP4\n0 Edgar 100 116 56 55 100.2 92.8 60.2 88.2\n1 Ralph 96 93 59 96 102.6 106.6 58.4 63.6\n2 Marc 110 85 41 113 101.0 101.8 72.8 54.2\n3 David 103 100 83 40 105.2 92.8 47.6 98.0\n```" ]

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[ "# onepole~\n\nSingle-pole lowpass filter\n\n## Description\n\nThe onepole~ implements the simple filter equation\n\noutput = previous input + cf * (input - previous input)\n\nwhere cf represents the cutoff frequency of the filter expressed in radians. The values for cf lie in the range -1.0-0. This produces a single-pole lowpass filter with a 6dB/octave attenuation, which can be useful to gently roll off harsh high end (e.g., the digital artifacts of downsampling). onepole~ is equivalent to a biquad~ object with the coefficients,\n\n[a0 = 1 + cf, a1 = 0, a2 = 0, b1 = cf, b2 = 0]\n\nIf you substitute these values into the biquad~ equation, you are left with the onepole~ object's algorithm. However, onepole~ will execute much faster, since biquad~ will still compute the unused portion of its equation.\n\n## Arguments\n\nName Type Opt Description\ncenter-frequency (Hz) float opt Sets the center frequency for the filter, as described above.\nHz/linear/radians symbol opt Using the symbols Hz, linear, or radians for an optional second argument sets the frequency input mode. The default mode is Hz (which is the same as providing no mode argument). Using the linear argument sets the frequency input mode to linear (0 - 1). Using the radians argument sets the frequency input mode to radians (0 - 1).\n\n## Messages\n\n int center-frequency (Hz) [int] Performs the same function as float. float center-frequency (Hz or 0 through 1 specified by object-argument) [float] In right inlet: Sets the frequency for the filter (if no signal is connected). By default, frequency is expressed in Hz, where the allowable range is from 0 to one fourth of the current sampling rate. For convenience, onepole~ has two additional input modes that use the more conventional input range, 0 - 1 (see the linear and radians messages). Hz In either inlet: Sets the frequency input mode to Hz (the default). clear In either inlet: Clears the internal state of onepole~. Since onepole~ does not have the inherent instability of other filter types, this should never be necessary. linear In either inlet: Sets the frequency input mode to linear (0 - 1). Linear mode is simply a scaled version of the standard Hz mode, except that values in the 0-1 range traverses the full frequency range. radians In either inlet: Sets the frequency input mode to radians (0 - 1). Radians mode lets you set the center frequency (cf) of the equation directly, while the input has the same range (0-1), the output has a curved frequency response that is closer to the exponential pitch scale of the human ear. signal In left inlet: Signal to be filtered. In right inlet: A signal can be used to set the frequency for the filter, with the same effect as a float. If a signal is connected to this inlet, its value is sampled once every signal vector.\n\n## Output\n\nsignal: The filtered signal.\n\n## Examples", null, "onepole~ provides efficient filtering for a simple sample player" ]

[ null, "https://docs.cycling74.com/max5/refpages/msp-ref/images/onepole~.png", null ]

[ "# DataFrame.strftime()\n\nPandas date & time Pandas\n\nGenerate Datetime format output .\n``````import pandas as pd\nmy_data=pd.date_range(start='10/20/2019', end='02/14/2020',freq='M')\ndf=my_data.strftime('%d-%b-%Y')\nprint(df)``````\nWe created one date range by using date_range() and then used strftime() to format the output. Here is the output\n``Index(['31-Oct-2019', '30-Nov-2019', '31-Dec-2019', '31-Jan-2020'], dtype='object')``\nHere is a list of directives can be used to create formatted output.\n Weekday %a Weekday Name ( local ) %A Weekday Name ( in Full ) %w Weekday as decimal number(0 Sun, 1 Mon,) %W Week number of the year (00,53) Mon as first day of week %U Week number of the year (00,53) Sun as first day of week Day %d Day of the month as number ( 01,02,31) %j Day of the year as number ( 001,284,336) Month %b Month name ( Oct, Nov) %B Month name ( October, November) %m Month number (01,12) Year %y Year short 2 digit ( 01,20) %Y Year Full 4 digit ( 2001,2020) %f Year Fiscal short 2 digit ( 01,20) %F Year Fiscal Full 4 digit ( 2001,2020) Hour %H Hour 24 hour clock ( 01,23) %I Hour 12 hour clock ( 01,11) Minute %M Minute as number ( 00,01,59) Second %S Second as decimal number ( 00,61) AM/PM %p AM / PM local time Time Zone %Z Time Zone name Local %x Local Date %X Local Time %c Local Date & Time Quater %q Quater as number ( 01,04)\n\nSubscribe to our YouTube Channel here\n\n## Subscribe\n\n* indicates required\nSubscribe to plus2net", null, "plus2net.com" ]

[ null, "https://www.plus2net.com/images/top2.jpg", null ]

[ "• Price\n\n• Type\n\n## Land & Lots\n\n• Beds\n\n• Baths\n\n• More\n• Days on Site: All 1 Day 2 Days 3 Days 4 Days 5 Days 6 Days 1 Week 2 Weeks 1 Month Price Change: All 1 Day 2 Days 3 Days 4 Days 5 Days 6 Days 1 Week 2 Weeks 1 Month Sale Type: All Foreclosures HUD Homes Short Sales No Short Sales Square Feet: No Preference 10,000+ 9,000+ 8,000+ 7,000+ 6,000+ 5,000+ 4,000+ 3,500+ 3,000+ 2,750+ 2,500+ 2,250+ 2,000+ 1,800+ 1,600+ 1,400+ 1,200+ 1,000+ 800+ 600+ 400+ 200+ Year Built: No Preference 2020+ 2019+ 2018+ 2017+ 2016+ 2015+ 2014+ 2013+ 2012+ 2011+ 2010+ 2009+ 2008+ 2007+ 2006+ 2005+ 2004+ 2003+ 2002+ 2001+ 2000+ 1999+ 1998+ 1997+ 1996+ 1995+ 1994+ 1993+ 1992+ 1991+ 1990+ 1989+ 1988+ 1987+ 1986+ 1985+ 1984+ 1983+ 1982+ 1981+ 1980+ 1979+ 1978+ 1977+ 1976+ 1975+ 1974+ 1973+ 1972+ 1971+ 1970+ 1969+ 1968+ 1967+ 1966+ 1965+ 1964+ 1963+ 1962+ 1961+ 1960+ 1959+ 1958+ 1957+ 1956+ 1955+ 1954+ 1953+ 1952+ 1951+ 1950+ 1949+ 1948+ 1947+ 1946+ 1945+ 1944+ 1943+ 1942+ 1941+ 1940+ 1939+ 1938+ 1937+ 1936+ 1935+ 1934+ 1933+ 1932+ 1931+ 1930+ 1929+ 1928+ 1927+ 1926+ 1925+ 1924+ 1923+ 1922+ 1921+ 1920+ 1919+ 1918+ 1917+ 1916+ 1915+ 1914+ 1913+ 1912+ 1911+ 1910+ 1909+ 1908+ 1907+ 1906+ 1905+ 1904+ 1903+ 1902+ 1901+ 1900+ 1899+ 1898+ 1897+ 1896+ 1895+ 1894+ 1893+ 1892+ 1891+ 1890+ 1889+ 1888+ 1887+ 1886+ 1885+ 1884+ 1883+ 1882+ 1881+ 1880+ 1879+ 1878+ 1877+ 1876+ 1875+ 1874+ 1873+ 1872+ 1871+ 1870+ 1869+ 1868+ 1867+ 1866+ 1865+ 1864+ 1863+ 1862+ 1861+ 1860+ 1859+ 1858+ 1857+ 1856+ 1855+ 1854+ 1853+ 1852+ 1851+ 1850+ 1849+ 1848+ 1847+ 1846+ 1845+ 1844+ 1843+ 1842+ 1841+ 1840+ 1839+ 1838+ 1837+ 1836+ 1835+ 1834+ 1833+ 1832+ 1831+ 1830+ 1829+ 1828+ 1827+ 1826+ 1825+ 1824+ 1823+ 1822+ 1821+ 1820+ 1819+ 1818+ 1817+ 1816+ 1815+ 1814+ 1813+ 1812+ 1811+ 1810+ 1809+ 1808+ 1807+ 1806+ 1805+ 1804+ 1803+ 1802+ 1801+ 1800+ Garage: No Preference 1+ 2+ 3+ 4+ 5+ 6+ 7+ 8+ 9+ 10+ Lot Size: No Preference .15+ acres .25+ acres .35+ acres .5+ acres .65+ acres .80+ acres 1+ acres 1.25+ acres 1.5+ acres 2+ acres 2.5+ acres 3+ acres 3.5+ acres 4+ acres 5+ acres 10+ acres 20+ acres 30+ acres 40+ acres 50+ acres 100+ acres 150+ acres 200+ acres 250+ acres 300+ acres 350+ acres 400+ acres 450+ acres 500+ acres Style: No Preference A-Frame Adobe Art Deco Bungalow Cabin Cape Cod Colonial Contemporary Cottage Craftsman Farm House French Hacienda Log Home Mediterranean Mid Century Modern Moorish Mooresque Other Post Modern Ranch Santa Fe Southwestern Spanish Spanish Colonial Traditional Tuscan Victorian High School: No Preference 29 Palms Big Bear Cathedral City Coachella Valley Desert Hot Springs Desert Mirage Hemet Hillcrest Indio La Quinta Lakeside Laquinta Menifee Other Palm Desert Palm Springs Paloma Rancho Mirage Rancho Verde Rim Of The World Shadow Hills Unknown Upland West Shores Mid. School: No Preference 1St Ave 29 Palms Amelia Arizona Big Bear Bobby Duke Cahuilla Desert Acad Colonel Mitchell Desert Ridge Desert Springs Glenn Hemet Indio James Workman Jefferson John Glenn La Contenta La Quinta Lakeside Lincoln Nellie N. Coffman Other Paige Painted Hills Palm Palm Desert Palm Desert Charter Pinacate Pioneer Raymond Cree Terra Cotta Thomas Jefferson Toro Canyon Wilson Ele. School: No Preference Abraham Lincoln Adams Agua Caliente Amelia Earhart Avalon Bear Valley Unified Benjamin Franklin Bubbling Wells Cabot Yerxa Cahuilla Carrillo Rancho Carter Cathedral City Cesar Chavez Charter Cielo Vista Dr. Carreon Foothill Ford Franklin George Washington Gerald Ford Hemet Hoover Jackson James Monroe John Adams Julius Corsini Katherine Finchy Kennedy Lake Hills Landau Las Palmitas Lincoln Machado Menifee Mountain Vista Oasis Oliphant Other Palm View Perris Rancho Mirage Reagan Richard R. Oliphant Ronald Reagan Sea View Sunny Sands Truman Two Bunch Palms Unknown Van Buren Vista Del Monte Washington Charter Westside Sort: No Preference Price - Low to High Price - High to Low Year Built - Low to High Year Built - High to Low Sq. Feet Lot Size \\$ Sq. Foot New Listings First\n\nSave Search\nMore Less\nMap List Gallery\n\nThis page contains ALL homes currently listed for sale in San Diego as of 12/15/2019.  You can change the search criteria at any time by utilizing the search bar above.\n\nProperty Saved\nProperty Removed", null, "", null, "", null, "Properties" ]

[ null, "https://www.ProAgentWebsites.com/idx/images/ia_msg_next_prev.png", null, "https://www.ProAgentWebsites.com/idx/images/ia_msg_more.png", null, "https://www.ProAgentWebsites.com/idx/images/ia_msg_area.png", null ]

[ "# The XY Spin-Glass with Slow Dynamic Couplings\n\nG. Jongen\\ftnote3E-mail:    D. Bollé\\ftnote1E-mail: and A.C.C. Coolen\\ftnote2E-mail: †§ Instituut voor Theoretische Fysica, K.U. Leuven, B-3001 Leuven, Belgium ‡ Department of Mathematics, King’s College London, London WC2R 2LS, UK\n###### Abstract\n\nWe investigate an XY spin-glass model in which both spins and couplings evolve in time: the spins change rapidly according to Glauber-type rules, whereas the couplings evolve slowly with a dynamics involving spin correlations and Gaussian disorder. For large times the model can be solved using replica theory. In contrast to the XY-model with static disordered couplings, solving the present model requires two levels of replicas, one for the spins and one for the couplings. Relevant order parameters are defined and a phase diagram is obtained upon making the replica-symmetric Ansatz. The system exhibits two different spin-glass phases, with distinct de Almeida-Thouless lines, marking continuous replica-symmetry breaking: one describing freezing of the spins only, and one describing freezing of both spins and couplings.\n\nRecently various models with a coupled dynamics of fast Ising spins and slow couplings have been studied (see e.g. [1, 5] and references therein). In addition to physical motivations, such as understanding the simultaneous learning and retrieval in recurrent neural networks or the influence of slow atomic diffusion processes in disordered magnetic systems, there is a more theoretical interest in such models in that they generate the replica formalism for a finite number of replicas . Moreover, the replica number is found to have a physical meaning as the ratio of two temperatures (characterizing the stochasticity in the spin dynamics and the coupling dynamics, respectively). In this letter we extend the methods and results obtained for Ising spin models to a classical XY spin-glass with dynamic couplings, whose spin variables are physically more realistic than Ising ones. In addition, the XY model is closely related to neural network models of coupled oscillators, which provide a phenomenological description of neuronal firing synchronization in brain tissue. We solve our model upon making the replica-symmetric Ansatz, and calculate the de Almeida-Thouless (AT) lines (of which here there are two types), where continuous transitions occur to phases with broken replica symmetry. In doing so we also improve the calculations of . As in the Ising case we find two qualitatively different types of spin-glass phases. In one spin-glass phase the spins do freeze in random directions, but on the time-scales of the coupling dynamics these ‘frozen directions’ change. In the second spin-glass phase the spins as well as the couplings freeze, such that even on the large time-scales the ‘frozen directions’ of the spins remain stationary.\n\nWe choose a system of classical two-component spin variables , , and symmetric exchange interactions (or couplings) , with a Glauber-type dynamics such that for stationary choices of the couplings the microscopic spin probability density would evolve towards a Boltzmann distribution, with the standard Hamiltonian and with inverse temperature . The couplings are taken to be of infinite range. They will now themselves be allowed to evolve in time in a stochastic manner, partially in response to the states of the spins and to externally imposed biases. However, we assume that the spin dynamics is very fast compared to that of the couplings, such that on the time-scales of the couplings, the spins are effectively in equilibrium (i.e. we take the adiabatic limit). For the dynamics of the couplings the following Langevin form is proposed (which is the natural adaptation to XY spins of the choices originally made in [1, 3] for Ising spins):\n\n ddtJij=⟨\\boldmathSi⋅\\boldmathSj⟩+KijN−μJij+ηij(t)N1/2            i\n\nThe term , representing local spin correlations associated with the coupling , is a thermodynamic average over the Boltzmann distribution of the spins, given the instantaneous couplings . External biases serve to steer the weights to some preferred values (note: this notation follows that of ). The are choosen to be quenched random variables, drawn independently from a Gaussian probability distribution with mean and variance . The decay term in (1) is added to limit the magnitude of the couplings. Finally, the terms represent Gaussian white noise contributions, of zero mean and covariance , with associated temperature . Factors of are introduced in order to ensure non-trivial behaviour in the thermodynamic limit .\n\nThe three independent global symmetries of our model, which can be expressed efficiently in terms of the Pauli spin matrices and , are the following:\n\n inversion of both spin axes:% \\boldmathSi→−\\boldmathSifor all iinversion of one spin axis:\\boldmathSi→σz\\boldmathSifor all ipermutation of spin axes:\\boldmathSi→σx\\boldmathSifor all i. (2)\n\nUpon using algebraic relations such as and we see that in the high (ergodic) regime these three global symmetries generate the following local identities, respectively:\n\n ⟨\\boldmathSi⟩=\\boldmath0,        ⟨\\boldmathSi⋅σx\\boldmathSj⟩=0,        ⟨\\boldmathSi⋅σz\\boldmathSj⟩=0. (3)\n\nThe equilibrium solution of the probability density associated with the stochastic equation (1) for the couplings follows from the fact that (1) is conservative, i.e. that it can be written as\n\n ddtJij=−1N∂∂Jij~H({Jij})+ηij(t)N1/2 (4)\n\nwith the following effective Hamiltionian for the couplings:\n\n ~H({Jij})=−1βlogZβ({Jij})+12μN∑k<ℓJ2kℓ−μN∑k<ℓBkℓJkℓ. (5)\n\nIn this expression is the partition function of the XY spins with instantaneous couplings . Thus the stationary probability density for the couplings is also of a Boltzmann form, with the Hamiltonian (5), and the thermodynamics of the slow system (the couplings) are generated by the partition function , leading to (modulo irrelevant prefactors):\n\n ~Z~β=∫∏k<ℓdJkℓ[Zβ({Jij})]nexp[μ~βN∑k<ℓBkℓJkℓ−12 μ~βN∑k<ℓJ2kℓ]. (6)\n\nFinally, we define the disorder-averaged free energy per site , in which is an average over the . In contrast to standard systems with frozen disorder, the replica number is here given by the ratio , and can take any real non-negative value. The limit corresponds to a situation in which the coupling dynamics is driven by the Gaussian white noise, rather than by spin correlations; in the limit the influence of spin correlations dominates.\n\nWe carry out the disorder average using the identity , evaluating the latter by analytic continuation from integer . Our system with partition function is thus replicated times; we label each replica by a Roman index. Each of the functions , in turn, is given by (6), and involves which is replaced by the product of further replicas, labeled by Greek indices. For non-integer , again analytic continuation is made from integer . Performing the disorder average in results in an expression involving coupled replicas of the original system: , with and . For this expression can be evaluated in the familiar fashion of replica mean-field theory , by saddle-point integration. This procedure induces the following order parameters:\n\n \\boldmathmαa=1N∑i⟨ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯⟨\\boldmathSαia⟩ ⟩B             qαβab=1N∑i⟨ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯⟨\\boldmathSαia⋅\\boldmathSβib⟩ ⟩B\n uαβab=1N∑i⟨ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯⟨% \\boldmathSαia⋅σx\\boldmathSβib⟩ ⟩B     vαβab=1N∑i⟨ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯⟨\\boldmathSαia⋅σz\\boldmathSβib⟩ ⟩B.\n\nThe horizontal bar denotes thermal averaging over the coupling dynamics with fixed biases . Comparison with (3) shows that the order parameters , and measure the breaking of the global symmetries (2). For simplicity we choose . We make the usual assumption that, in the absence of global symmetry-breaking forces, phase transitions can lead to at most local violation of the identities (3). Thus the latter will remain valid if averaged over all sites, at any temperature, which implies that and . The spin-glass order parameters , on the other hand, are not related to simple global symmetries, and serve to characterize the various phases.\n\nThe final stage of the calculation is to make the replica symmetry (RS) Ansatz. Since observables with identical Roman indices refer to system copies with identical couplings, whereas observables with identical Roman indices and identical Greek indices refer to system copies with identical couplings and identical spins, in the present model the RS Ansatz takes the form (note: ). The remaining two order parameters are determined as the solutions of the following coupled saddle-point equations:\n\n q0=∫dx P(x)⎧⎪ ⎪⎨⎪ ⎪⎩∫dz P(z)[I0(zΞ)]n−1I1(zΞ) I1(zxβΞ−1√12~Bq0)∫dz P(z)[I0(zΞ)]nI0(zxβΞ−1√12~Bq0)⎫⎪ ⎪⎬⎪ ⎪⎭2 (7)\n q1=∫dx P(x)⎧⎪ ⎪⎨⎪ ⎪⎩∫dz P(z)[I0(zΞ)]n−2[I1(zΞ)]2I0(zxβΞ−1√12~Bq0)∫dz P(z)[I0(zΞ)]nI0(zxβΞ−1√12~Bq0)⎫⎪ ⎪⎬⎪ ⎪⎭ (8)\n\nwith the two short-hands ,\n\n (9)\n\nIt is clear that .", null, "Figure 1: Phase diagram of the XY spin-glass with slow dynamic couplings, drawn in the n-T plane; for B0=0, ~B=1 and ~J=3. P: paramagnetic phase, q1=q0=0. SG1: first spin-glass phase, q1>0 and q0=0 (freezing on spin time-scales only). SG2: second spin-glass phase, q1>0 and q0>0 (freezing on all time-scales). AT lines: λR=0 (Roman replicon), λG=0 (Greek replicon).", null, "Figure 2: Phase diagram of the XY spin-glass with slow dynamic couplings, drawn in the ~T-T plane; for B0=0, ~B=1 and ~J=3. Further notation as in Figure 1.", null, "Figure 3: Phase diagram of the Ising spin-glass with slow dynamic couplings , drawn in the n-T plane; for B0=0, ~B=1 and ~J=3. Further notation as in Figure 1.\n\nWe have studied the fixed-point equations (7,8) after having first eliminated the parameter redundancy by putting and , which resulted in the phase diagram in the - plane as shown in figure 1. In addition to a paramagnetic phase (P), where , one finds two distinct spin-glass phases: SG1, where but , and SG2, where both and . The SG1 phase describes freezing of the spins on the fast time-scales only (where spin equilibration occurs); on the large time-scales, where coupling equilibration occurs, one finds that, due to the slow motion of the couplings, the frozen spin directions continually change. In the SG2 phase, on the other hand, both spins and couplings freeze, with the net result that even on the large time-scales the frozen spin directions are ‘pinned’. The SG1-SG2 transition is always second order. The transition SG1-P is second order for (in which case its location is given by ), but first order for . When further increases to , the SG1 phase disappears, and the system exhibits a first order transtion from P to SG2.\n\nAdditional transitions occur, corresponding to a continuous breaking of replica symmetry. The stability of the RS solutions is, as always, expressed in terms of the eigenvalues of the matrix of second derivatives of quadratic fluctuations at the saddle-point . We have calculated all eigenvalues and their multiplicity following [3, 6] (full details will be published elsewhere ). We find two replicon eigenvalues: , associated with the Greek replicas and , associated with the Roman replicas. We have drawn the corresponding AT-lines, where and are zero, respectively, in the phase diagram as dashed-dotted lines. At low spin temperature replica symmetry is broken with respect to the Greek replicas (), whereas at low coupling temperature replica symmetry is broken with respect to the Roman replicas (). This is illustrated more clearly by drawing the phase diagram in the - plane, as in figure 2. This second figure also shows that there is no re-entrance from SG1 to RSB, when is varied for fixed .\n\nQualitatively the phase diagram of the present system is very similar to that of the Ising spin-glass with dynamic couplings as studied in , see figure 3, including the behaviour of both the Greek AT-line and the Roman AT-line . Here our results differ from, and improve upon, those of (which is why we present figure 3, rather than just refer to ). The set of eigenvalues given in turn out to satisfy only part of the relevant orthogonality conditions used in their calculation. The replica symmetric solution in SG1 is always stable with respect to the Roman replicas. In fact, we can show analytically that the Roman AT-line coincides with the SG1-SG2 transition line. Our model, with dynamics on two different time-scales, is reminiscent of a simple XY model with one step replica-symmetry breaking (1RSB), and our eigenvalues formally resemble e.g. those describing the stability of the 1RSB solution in the perceptron model of .\n\nIn conclusion, we have solved a classical XY spin-glass model in which both the spins and their couplings evolve stochastically, according to coupled equations, but on different time-scales. The solution of our model in RS Ansatz is mathematically similar to that of the XY model with static couplings, but with one step RSB. Qualitatively, the phase diagram, which exhibits two different spin-glass types and both first and second order transitions, resembles closely that of the Ising spin-glass with dynamic couplings, provided appropriate adjustment of the calculation of the AT-line in is carried out. Our calculation shows how the methods used for solving the Ising case can be easily adapted to more complicated spin types, and illustrates the robustness of the structure and peculiarities of phase diagrams describing the behaviour of large spin systems with dynamic couplings.\n\n## Acknowledgments\n\nWe would like to thank R.W. Penney and D. Sherrington for helpful discussions. This work has been supported in part by the Research Fund of the K.U. Leuven (grant OT/94/9). DB and ACCC would like to thank the Fund for Scientific Research-Flanders (Belgium) and the British Council for financial support." ]

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[ "## 13086\n\n13,086 (thirteen thousand eighty-six) is an even five-digits composite number following 13085 and preceding 13087. In scientific notation, it is written as 1.3086 × 104. The sum of its digits is 18. It has a total of 4 prime factors and 12 positive divisors. There are 4,356 positive integers (up to 13086) that are relatively prime to 13086.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 5\n• Sum of Digits 18\n• Digital Root 9\n\n## Name\n\nShort name 13 thousand 86 thirteen thousand eighty-six\n\n## Notation\n\nScientific notation 1.3086 × 104 13.086 × 103\n\n## Prime Factorization of 13086\n\nPrime Factorization 2 × 32 × 727\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 4362 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 13,086 is 2 × 32 × 727. Since it has a total of 4 prime factors, 13,086 is a composite number.\n\n## Divisors of 13086\n\n1, 2, 3, 6, 9, 18, 727, 1454, 2181, 4362, 6543, 13086\n\n12 divisors\n\n Even divisors 6 6 3 3\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 28392 Sum of all the positive divisors of n s(n) 15306 Sum of the proper positive divisors of n A(n) 2366 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 114.394 Returns the nth root of the product of n divisors H(n) 5.53085 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 13,086 can be divided by 12 positive divisors (out of which 6 are even, and 6 are odd). The sum of these divisors (counting 13,086) is 28,392, the average is 2,366.\n\n## Other Arithmetic Functions (n = 13086)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 4356 Total number of positive integers not greater than n that are coprime to n λ(n) 726 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1558 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 4,356 positive integers (less than 13,086) that are coprime with 13,086. And there are approximately 1,558 prime numbers less than or equal to 13,086.\n\n## Divisibility of 13086\n\n m n mod m 2 3 4 5 6 7 8 9 0 0 2 1 0 3 6 0\n\nThe number 13,086 is divisible by 2, 3, 6 and 9.\n\n## Classification of 13086\n\n• Arithmetic\n• Abundant\n\n### Expressible via specific sums\n\n• Polite\n• Non-hypotenuse\n\n## Base conversion (13086)\n\nBase System Value\n2 Binary 11001100011110\n3 Ternary 122221200\n4 Quaternary 3030132\n5 Quinary 404321\n6 Senary 140330\n8 Octal 31436\n10 Decimal 13086\n12 Duodecimal 76a6\n20 Vigesimal 1ce6\n36 Base36 a3i\n\n## Basic calculations (n = 13086)\n\n### Multiplication\n\nn×i\n n×2 26172 39258 52344 65430\n\n### Division\n\nni\n n⁄2 6543 4362 3271.5 2617.2\n\n### Exponentiation\n\nni\n n2 171243396 2240891080056 29324300673612816 383737798614897310176\n\n### Nth Root\n\ni√n\n 2√n 114.394 23.5651 10.6955 6.65827\n\n## 13086 as geometric shapes\n\n### Circle\n\n Diameter 26172 82221.8 5.37977e+08\n\n### Sphere\n\n Volume 9.38662e+12 2.15191e+09 82221.8\n\n### Square\n\nLength = n\n Perimeter 52344 1.71243e+08 18506.4\n\n### Cube\n\nLength = n\n Surface area 1.02746e+09 2.24089e+12 22665.6\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 39258 7.41506e+07 11332.8\n\n### Triangular Pyramid\n\nLength = n\n Surface area 2.96602e+08 2.64092e+11 10684.7\n\n## Cryptographic Hash Functions\n\nmd5 5159f683253665f2122788b700686d8b f6933a1ba3fa8e545b8702d6429d85a4dff495d1 c229fa54c52290e09ad60d49e4e3f82335c0a679b1cda980492631ae8c67fef1 bcefaaeb875129bbbf61dba07c7ca5b42ab68b8f03f09edeabd73205a22b5532670780330ff94fa9423330f52a52e360088ed0c77476ee714c0c7b76a4bf3bd8 0f8e9a40a7a49682d8fea8e9075f60f77d1fcc39" ]

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[ "# Differentials\n\n## Prerequisites\n\nStudents should be able to:\n\n• Determine an ordinary derivative from a symbolic expression.\n• Hold variables constant when finding a derivative.\n• Sketch graphs and identify points and changes between points on those graphs.\n\n## In-class Content\n\n• QUIZ/Derivatives pretest (10 min)\n\n### Lecture: Differentials of 1D Functions (Lec - 20 min)\n\nMain ideas\n\n• Previewing the Math Bits theme of Variables and Representations\n• Teaching differentials, small changes, and zapping with d\n##### Lecture (30 minutes)\n\nAsk students to make a sketch of f = 7x2 on the big white board.\n\nWhat are the variables of interest? One idea here is to note that x is not the only variable here: f is also a variable from a physics perspective.\n\nWhat representations do we have for this relationship? The symbolic equation is one, and the graph is another.\n\nIntroduce the differential quantities df and dx as small changes in f and x, respectively. Ask students to add df and dx to their graphs.\n\nThen, ask students how df and dx are related to each other. Students should be able to articulate that this is a derivative.\n\nIs the relationship the same if we choose a different point on the graph (a different initial x)?\n\nThen relate df and dx using the symbolic representation: df = 14xdx.\n\nThe following two steps can be replaced by other activities on the relevant Hour page.\n• Give the students f = 5x2y3. What are the variables? What representations do we have (if time, hand out surfaces and have students make sketches similar to the above)? What does a differential relationship look like? For this last one, use the generic form df = A dx + B dy, and talk about the fact that this is another representation.\n• Do some specific examples of using the zapping with d strategy (see Zapping with d).\n\n### Activity: Total Differentials on a Surface\n\nActivity Highlights\n\n1. This small group activity is designed to help students understand differentials in multivariable functions.\n2. Students work in small groups to generalize symbolic and graphical representations of differentials for a single variable function to a multivariable case using a plastic surface as a test case.\n3. The whole class wrap-up discussion emphasizes that differentials are a way of linearizing the relationship between variables by expressing how small changes in those variables are related.\n\n### Activity: Evaluating Total Differentials\n\nActivity Highlights\n\n1. This small group activity is designed to help students understand differentials in multivariable functions.\n2. Students work in small groups to generalize symbolic and graphical representations of differentials for a single variable function to a multivariable case using a plastic surface as a test case.\n3. The whole class wrap-up discussion emphasizes that differentials are a way of linearizing the relationship between variables by expressing how small changes in those variables are related.\n\n### Activity: Covariation in Thermal Systems\n\nActivity Highlights\n\n1. This small group activity is designed to help students understand relationships between thermodynamic state variables.\n2. Students work in small groups to investigate states on a raising physics surface, and consider cycles.\n3. The whole class wrap-up discussion emphasizes the meaning of a “state” variable.\n\n## Homework for Energy and Entropy\n\n1. (mbZapWithD) Find the differential of special functions.\n\nFind the differential of each of the following expressions, ie. zap each of the following with $d$:\n\n1. $$f=3x-5z^2+2xy$$\n\n2. $$g=\\sin^2(\\omega t)$$\n\n3. $$h=\\frac{\\mu B}{k_B T}$$\n\n4. $$j=\\exp\\left(\\frac{\\mu B}{k_B T}\\right)$$\n\n5. $$k=\\ln\\left(e^{\\frac{\\mu B}{k_B T}} +e^{-\\frac{\\mu B}{k_B T}}\\right)$$\n\n2. (mbDifferentials) Finding the differential abstractly.\n\nFind the differential of $R$ where\\ldots\n\n1. $R(B,C) = B^2 + C^2$\n\n2. $R(B,C) = BC$\n\n3. $R(B,C) = e^{B^2+C^2}$\n\n4. $R(B,C) = e^{S(B,C)}$\n\n5. $R(B,C) = ST$, where $S = S(B,C)$ and $T = T(B,C)$\n\n3. (mbDerivVSDiff) Find the differential of special functions.\n\nConsider $V=-A\\cos(kx)$, where $A$ is a constant with dimensions of electrostatic potential.\n\n1. Find $dV$ and $\\frac{dV}{dx}$. Discuss the differences and similarities between the two.\n\n2. What are the dimensions of $k$, $V$, $dx$, $dV$, and $\\frac{dV}{dx}$?\n\n4. (mbMathematicaContours) Plot contours, Change of Variable.\n\nOn this problem, you will use Mathematica to graph and interpret some functions following the steps below. Mathematica can be found in the computers on campus, or you can download a student license for free, or even use online tools like sandbox.open.wolframcloud.com. Please include both your graphs and your source code to make grading easier!\n\n1. Create a contour plot of the function $G = A^4 + B^2$. Play around with the options until you have a graph that has all of the following features:\n\nThe axes are labeled. The contours are numbered. The plot is titled. There are 8 contours. The colored shading on the contours is different from the default.\n\n2. Describe how your contour plot changes if you modify the plot range (i.e., zooming in or out)?\n\n3. Suppose that $5A = 4U + 3V$ and $5B = 4V - 3U$.\n\nMake a contour plot of $G$ in terms of the new variables $U$ and $V$. Label and present your new graph as described above.\n\n4. Make a contour plot of $G$ in terms of the variables $A$ and $V$. Label and present your new graph as described above.\n\n5. Explain why all three of the contour plots you have made are valid representations of $G$. What is different about the three graphs?\n\n5. (CoffeeAndBagels) A thermo-like system. Requires Clairaut's theorem.\n\n\\newcommand\\juice{\\ensuremath{\\mathcal{O}}} \\newcommand\\cocoa{\\ensuremath{\\mathcal{C}}} \\newcommand\\priceorange{\\ensuremath{\\mathcal{P_O}}} \\newcommand\\pricecocoa{\\ensuremath{\\mathcal{P_C}}}\n\nIn economics, the term utility is roughly related to overall happiness. Many things affect your happiness, including the amount of money you have and the amount of coffee you drink. We cannot directly measure your happiness, but we can measure how much money you are willing to give up in order to obtain coffee or bagels. If we assume you choose wisely, we can thus determine that your happiness increases when you decrease your amount of money by that amount in exchange for increasing your coffee consumption. Thus money is a (poor) measure of happiness or utility.\n\nMoney is also a nice quantity because it is conserved—just like energy! You may gain or lose money, but you always do so by a transaction. (There are some exceptions to the conservation of money, but they involve either the Fed, counterfeiters, or destruction of cash money, and we will ignore those issues.)\n\nIn this problem, we will assume that you have bought all the coffee and bagels you want (and no more), so that your happiness has been maximized. Thus you are in equilibrium with the coffee shop. We will assume further that you remain in equilibrium with the coffee shop at all times, and that you can sell coffee and bagels back to the coffee shop at cost.\\footnote{Yes, this is ridiculous. It would be slightly less ridiculous if we were talking about nations and commodities, but also far less humorous.}\n\nThus your savings $S$ can be considered to be a function of your bagels $B$ and coffee $C$. In this problem we will also discuss the prices $P_B$ and $P_C$, which you may not assume are independent of $B$ and $C$. It may help to imagine that you have\n\n1. The prices of bagels and coffee $P_B$ and $P_C$ have derivative relationships between your savings and the quantity of coffee and bagels that you have. What are the units of these prices? What is the mathematical definition of $P_C$ and $P_B$?\n\n2. Write down the total differential of your savings, in terms of $B$, $C$, $P_B$ and $P_C$.\n\n3. Use the equality of mixed partial derivatives (Clairut's theorem) to find a relationship between $P_B$, $P_C$, $B$ and $C$. Write this relationship mathematically, and also describe in words what it means.\n\n4. Solve for the total differential of your net worth. Once again use Clairut's theorem considering second derivatives of $W$ to find a different partial derivative relationship between $P_B$, $P_C$, $B$ and $C$.\n\n##### Views\n\nNew Users\n\nCurriculum\n\nPedagogy\n\nInstitutional Change\n\nPublications\n\n##### Toolbox", null, "" ]

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[ "# Network Topology Matrices\n\nIn the previous chapter, we discussed how to convert an electric circuit into an equivalent graph. Now, let us discuss the Network Topology Matrices which are useful for solving any electric circuit or network problem by using their equivalent graphs.\n\n## Matrices Associated with Network Graphs\n\nFollowing are the three matrices that are used in Graph theory.\n\n• Incidence Matrix\n• Fundamental Loop Matrix\n• Fundamental Cut set Matrix\n\n## Incidence Matrix\n\nAn Incidence Matrix represents the graph of a given electric circuit or network. Hence, it is possible to draw the graph of that same electric circuit or network from the incidence matrix.\n\nWe know that graph consists of a set of nodes and those are connected by some branches. So, the connecting of branches to a node is called as incidence. Incidence matrix is represented with the letter A. It is also called as node to branch incidence matrix or node incidence matrix.\n\nIf there are ‘n’ nodes and ‘b’ branches are present in a directed graph, then the incidence matrix will have ‘n’ rows and ‘b’ columns. Here, rows and columns are corresponding to the nodes and branches of a directed graph. Hence, the order of incidence matrix will be n × b.\n\nThe elements of incidence matrix will be having one of these three values, +1, -1 and 0.\n\n• If the branch current is leaving from a selected node, then the value of the element will be +1.\n\n• If the branch current is entering towards a selected node, then the value of the element will be -1.\n\n• If the branch current neither enters at a selected node nor leaves from a selected node, then the value of element will be 0.\n\n### Procedure to find Incidence Matrix\n\nFollow these steps in order to find the incidence matrix of directed graph.\n\n• Select a node at a time of the given directed graph and fill the values of the elements of incidence matrix corresponding to that node in a row.\n\n• Repeat the above step for all the nodes of the given directed graph.\n\n### Example\n\nConsider the following directed graph.", null, "The incidence matrix corresponding to the above directed graph will be\n\n$$A = \\begin{bmatrix}-1 & 1 & 0 & -1 & 0 & 0\\\\0 & -1 & 1 & 0 & 1 & 0\\\\1 & 0 & -1 & 0 & 0 & 1 \\\\0 & 0 & 0 & 1 & -1 & -1 \\end{bmatrix}$$\n\nThe rows and columns of the above matrix represents the nodes and branches of given directed graph. The order of this incidence matrix is 4 × 6.\n\nBy observing the above incidence matrix, we can conclude that the summation of column elements of incidence matrix is equal to zero. That means, a branch current leaves from one node and enters at another single node only.\n\nNote − If the given graph is an un-directed type, then convert it into a directed graph by representing the arrows on each branch of it. We can consider the arbitrary direction of current flow in each branch.\n\n## Fundamental Loop Matrix\n\nFundamental loop or f-loop is a loop, which contains only one link and one or more twigs. So, the number of f-loops will be equal to the number of links. Fundamental loop matrix is represented with letter B. It is also called as fundamental circuit matrix and Tie-set matrix. This matrix gives the relation between branch currents and link currents.\n\nIf there are ‘n’ nodes and ‘b’ branches are present in a directed graph, then the number of links present in a co-tree, which is corresponding to the selected tree of given graph will be b-n+1.\n\nSo, the fundamental loop matrix will have ‘b-n+1’ rows and ‘b’ columns. Here, rows and columns are corresponding to the links of co-tree and branches of given graph. Hence, the order of fundamental loop matrix will be (b - n + 1) × b.\n\nThe elements of fundamental loop matrix will be having one of these three values, +1, -1 and 0.\n\n• The value of element will be +1 for the link of selected f-loop.\n\n• The value of elements will be 0 for the remaining links and twigs, which are not part of the selected f-loop.\n\n• If the direction of twig current of selected f-loop is same as that of f-loop link current, then the value of element will be +1.\n\n• If the direction of twig current of selected f-loop is opposite to that of f-loop link current, then the value of element will be -1.\n\n### Procedure to find Fundamental Loop Matrix\n\nFollow these steps in order to find the fundamental loop matrix of given directed graph.\n\n• Select a tree of given directed graph.\n\n• By including one link at a time, we will get one f-loop. Fill the values of elements corresponding to this f-loop in a row of fundamental loop matrix.\n\n• Repeat the above step for all links.\n\n### Example\n\nTake a look at the following Tree of directed graph, which is considered for incidence matrix.", null, "The above Tree contains three branches d, e & f. Hence, the branches a, b & c will be the links of the Co-Tree corresponding to the above Tree. By including one link at a time to the above Tree, we will get one f-loop. So, there will be three f-loops, since there are three links. These three f-loops are shown in the following figure.", null, "In the above figure, the branches, which are represented with colored lines form f-loops. We will get the row wise element values of Tie-set matrix from each f-loop. So, the Tieset matrix of the above considered Tree will be\n\n$$B = \\begin{bmatrix}1 & 0 & 0 & -1 & 0 & -1\\\\0 & 1 & 0 & 1 & 1 & 0\\\\0 & 0 & 1 & 0 & -1 & 1 \\end{bmatrix}$$\n\nThe rows and columns of the above matrix represents the links and branches of given directed graph. The order of this incidence matrix is 3 × 6.\n\nThe number of Fundamental loop matrices of a directed graph will be equal to the number of Trees of that directed graph. Because, every Tree will be having one Fundamental loop matrix.\n\n## Fundamental Cut-set Matrix\n\nFundamental cut set or f-cut set is the minimum number of branches that are removed from a graph in such a way that the original graph will become two isolated subgraphs. The f-cut set contains only one twig and one or more links. So, the number of f-cut sets will be equal to the number of twigs.\n\nFundamental cut set matrix is represented with letter C. This matrix gives the relation between branch voltages and twig voltages.\n\nIf there are ‘n’ nodes and ‘b’ branches are present in a directed graph, then the number of twigs present in a selected Tree of given graph will be n-1. So, the fundamental cut set matrix will have ‘n-1’ rows and ‘b’ columns. Here, rows and columns are corresponding to the twigs of selected tree and branches of given graph. Hence, the order of fundamental cut set matrix will be (n-1) × b.\n\nThe elements of fundamental cut set matrix will be having one of these three values, +1, -1 and 0.\n\n• The value of element will be +1 for the twig of selected f-cutset.\n\n• The value of elements will be 0 for the remaining twigs and links, which are not part of the selected f-cutset.\n\n• If the direction of link current of selected f-cut set is same as that of f-cutset twig current, then the value of element will be +1.\n\n• If the direction of link current of selected f-cut set is opposite to that of f-cutset twig current, then the value of element will be -1.\n\n### Procedure to find Fundamental Cut-set Matrix\n\nFollow these steps in order to find the fundamental cut set matrix of given directed graph.\n\n• Select a Tree of given directed graph and represent the links with the dotted lines.\n\n• By removing one twig and necessary links at a time, we will get one f-cut set. Fill the values of elements corresponding to this f-cut set in a row of fundamental cut set matrix.\n\n• Repeat the above step for all twigs.\n\n### Example\n\nConsider the same directed graph , which we discussed in the section of incidence matrix. Select the branches d, e & f of this directed graph as twigs. So, the remaining branches a, b & c of this directed graph will be the links.\n\nThe twigs d, e & f are represented with solid lines and links a, b & c are represented with dotted lines in the following figure.", null, "By removing one twig and necessary links at a time, we will get one f-cut set. So, there will be three f-cut sets, since there are three twigs. These three f-cut sets are shown in the following figure.", null, "We will be having three f-cut sets by removing a set of twig and links of C1, C2 and C3. We will get the row wise element values of fundamental cut set matrix from each f-cut set. So, the fundamental cut set matrix of the above considered Tree will be\n\n$$C = \\begin{bmatrix}1 & -1 & 0 & 1 & 0 & 0\\\\0 & -1 & 1 & 0 & 1 & 0\\\\1 & 0 & -1 & 0 & 0 & 1 \\end{bmatrix}$$\n\nThe rows and columns of the above matrix represents the twigs and branches of given directed graph. The order of this fundamental cut set matrix is 3 × 6.\n\nThe number of Fundamental cut set matrices of a directed graph will be equal to the number of Trees of that directed graph. Because, every Tree will be having one Fundamental cut set matrix.", null, "" ]

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[ "# Explicit and Recursive Sequences Practice Worksheet\n\nAn Explicit and Recursive Sequences worksheet is often one of the most important elements of any algebra class. The benefits to students of this type of worksheet are many, but it is also one of the things that most teachers find themselves looking for. The reason for this is simple – it makes learning the algebra basics more fun.", null, "Simplify and express in scientific notation from explicit and recursive sequences practice worksheet , source:pinterest.com\n\nThe explicit sequence worksheet gives students an organized way to practice addition and subtraction. In algebra, the order of operations is critical. To make the process of learning easier, the explicit sequence allows the student to complete their multiplication and division before doing addition or subtraction. This helps students build confidence in their ability to perform simple algebra lessons.\n\nWhen using the explicit sequence of a math worksheet, students learn that multiplication and division should be accomplished before addition. The sequence of operations should be in the correct order so that it is easy for students to remember and perform. Simply put, if you divide, you multiply; and if you multiply, you divide. The explicit sequence comes into play when a student is attempting to multiply two or more digits.", null, "Built in Types of Data from explicit and recursive sequences practice worksheet , source:introcs.cs.princeton.edu\n\nIt is important to note that using the explicit sequence on its own is not enough to practice multiplication and division. Students must then practice addition or subtraction as well. When this is done, students can begin to integrate the explicit sequence into their math lessons. This can be done by using the explicit sequence with other teaching tools such as math DVDs or online activities.\n\nAnother thing about the explicit sequence is that it can be used to teach different methods of multiplication and division. For example, a multiplication chart is often used in a math class. This worksheet allows students to practice addition and subtraction without having to be taught the math.", null, "Sequences And Recursive Formulas Best s About Formula Simages Org from explicit and recursive sequences practice worksheet , source:simages.org\n\nBy allowing students to use the explicit sequence as part of their math worksheet, students can also practice a different method of multiplying and dividing. Instead of the traditional four method, the explicit sequence can be used to teach one of the new methods of division and addition. In some ways, using this worksheet will allow students to create their own curriculum based on their specific needs.\n\nOf course, it should be noted that a lot of the benefits of using the explicit sequence come from how students are taught the math from the very beginning. One of the main advantages to using the explicit sequence is that students are taught all of the basics before they are able to apply these methods to more complex topics. This is not the case with traditional algebra classes. Because traditional math classes only teach a certain number of techniques, students must apply those techniques themselves.", null, "11 5 Recursive Rules for Sequences 681 687 from explicit and recursive sequences practice worksheet , source:yumpu.com\n\nAnother reason that the explicit sequence is considered valuable is because it provides a fun and interactive lesson. The learning process can be less stressful when students can just work with the explicit sequence to do addition and subtraction. Teachers are able to provide more practice and assignment material for students in an easier manner. All of these reasons, as well as the fact that using the explicit sequence allows students to add and subtract their own help make this worksheet a crucial part of math class.", null, "11 5 Recursive Rules for Sequences 681 687 from explicit and recursive sequences practice worksheet , source:yumpu.com", null, "AP Calculus practice questions from explicit and recursive sequences practice worksheet , source:khanacademy.org", null, "Sequences Practice Worksheet Answers Beautiful Autumn Fall Preschool from explicit and recursive sequences practice worksheet , source:kamada-maruyama.com", null, "Sequences And Recursive Formulas Best s About Formula Simages Org from explicit and recursive sequences practice worksheet , source:simages.org", null, "Geometric Sequence Brilliant Classroom Resources K 12 from explicit and recursive sequences practice worksheet , source:pinterest.com", null, "Fibonacci Rabbits from explicit and recursive sequences practice worksheet , source:jwilson.coe.uga.edu\n\nSHARE ON" ]

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[ "# Cmos Logic Diagram For Xor Gate\n\nCmos Logic Diagram For Xor Gate - circuit diagram munity circuit xor logic gate made using cmos the output of a xor gate is equal to 1 if the two inputs are both different and is 0 if they both have the same logic value the logic symbols jpq and can be used to denote xor in algebraic expressions c like languages use the caret symbol to denote bitwise xor note that the caret does not denote logical conjunction and in these languages despite the similarity of symbol pass gate logic wiring an xor gate can be constructed using mosfets here is a diagram of a pass transistor logic span class news dt jul 14 2015 span nbsp 0183 32 the ex or gate is defined as the hybrid logic gate with 2 or more inputs to perform the exclusive disjunction operation the xor circuit with 2 inputs is designed by using and or and not gates is shown above the output of 2 input xor gate is high only when one of its inputs are high if both the inputs are.\nsame then the output is low back what is logic xnor or exclusive nor gate xnor gate logic symbol boolean expression truth table xnor gate symbol boolean expression truth table xnor gate logic flow schematic diagram construction and working mechanism of xnor gate xnor gate using bjt and diodes xnor gate using mosfet and diodes xnor gate from other logic gates binational logic sum of product using not and or gate\n\ncircuit diagram xor gate wiring diagram table CMOS XOR Gate Schematic\nintegrated circuit simplifying cmos schematic to reduce number of CMOS XOR Gate Schematic\ncmos diagram for xor gate wiring diagram img CMOS XOR Gate Schematic\nxor gate wikipedia CMOS XOR Gate Schematic\nlogic gates (theory) digital vlsi design virtual lab CMOS XOR Gate Schematic\nwhy not switching extra inverters with opposite mosfets in cmos xor CMOS XOR Gate Schematic\ncmos xor gate circuit diagram download scientific diagram CMOS XOR Gate Schematic\nexclusive or gate(xor gate) CMOS XOR Gate Schematic\ncmos exor gate using gates as well as transistors CMOS XOR Gate Schematic\nactivity cmos logic circuits, transmission gate xor [analog devices CMOS XOR Gate Schematic" ]

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[ "Posted on Categories data science, Statistics\n\n## Estimating Rates using Probability Theory: Chalk Talk\n\nWe are sharing a chalk talk rehearsal on applied probability. We use basic notions of probability theory to work through the estimation of sample size needed to reliably estimate event rates. This expands basic calculations, and then moves to the ideas of: Sample size and power for rare events.\n\nPosted on Categories Pragmatic Machine Learning, Statistics, Tutorials1 Comment on Sample size and power for rare events\n\n## Sample size and power for rare events\n\nWe have written a bit on sample size for common events, we have written about rare events, and we have written about frequentist significance testing. We would like to specialize our sample size analysis to rare events (which allows us to derive a somewhat tighter estimate). Continue reading Sample size and power for rare events\n\nPosted on Categories data science, Statistics, Tutorials6 Comments on A bit more on sample size\n\n## A bit more on sample size\n\nIn our article What is a large enough random sample? we pointed out that if you wanted to measure a proportion to an accuracy “a” with chance of being wrong of “d” then a idea was to guarantee you had a sample size of at least:", null, "This is the central question in designing opinion polls or running A/B tests. This estimate comes from a quick application of Hoeffding’s inequality and because it has a simple form it is possible to see that accuracy is very expensive (to halve the size of difference we are trying to measure we have to multiply the sample size by four) and the cheapness of confidence (increases in the required confidence or significance of a result cost only moderately in sample size).\n\nHowever, for high-accuracy situations (when you are trying to measure two effects that are very close to each other) suggesting a sample size that is larger than is strictly necessary (as we are using an bound, not an exact formula for the required sample size). As a theorist or a statistician we like to error on the side of too large a sample (guaranteeing reliability), but somebody who is paying for each entry in a poll would want a smaller size.\n\nThis article shows a function that computes the exact size needed (using R). Continue reading A bit more on sample size" ]

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[ "# #136b64 Color Information\n\nIn a RGB color space, hex #136b64 is composed of 7.5% red, 42% green and 39.2% blue. Whereas in a CMYK color space, it is composed of 82.2% cyan, 0% magenta, 6.5% yellow and 58% black. It has a hue angle of 175.2 degrees, a saturation of 69.8% and a lightness of 24.7%. #136b64 color hex could be obtained by blending #26d6c8 with #000000. Closest websafe color is: #006666.\n\n• R 7\n• G 42\n• B 39\nRGB color chart\n• C 82\n• M 0\n• Y 7\n• K 58\nCMYK color chart\n\n#136b64 color description : Very dark cyan.\n\n# #136b64 Color Conversion\n\nThe hexadecimal color #136b64 has RGB values of R:19, G:107, B:100 and CMYK values of C:0.82, M:0, Y:0.07, K:0.58. Its decimal value is 1272676.\n\nHex triplet RGB Decimal 136b64 `#136b64` 19, 107, 100 `rgb(19,107,100)` 7.5, 42, 39.2 `rgb(7.5%,42%,39.2%)` 82, 0, 7, 58 175.2°, 69.8, 24.7 `hsl(175.2,69.8%,24.7%)` 175.2°, 82.2, 42 006666 `#006666`\nCIE-LAB 40.53, -26.142, -3.185 7.826, 11.573, 13.877 0.235, 0.348, 11.573 40.53, 26.335, 186.946 40.53, -30.297, -0.724 34.019, -18.472, -0.372 00010011, 01101011, 01100100\n\n# Color Schemes with #136b64\n\n• #136b64\n``#136b64` `rgb(19,107,100)``\n• #6b131a\n``#6b131a` `rgb(107,19,26)``\nComplementary Color\n• #136b38\n``#136b38` `rgb(19,107,56)``\n• #136b64\n``#136b64` `rgb(19,107,100)``\n• #13466b\n``#13466b` `rgb(19,70,107)``\nAnalogous Color\n• #6b3813\n``#6b3813` `rgb(107,56,19)``\n• #136b64\n``#136b64` `rgb(19,107,100)``\n• #6b1346\n``#6b1346` `rgb(107,19,70)``\nSplit Complementary Color\n• #6b6413\n``#6b6413` `rgb(107,100,19)``\n• #136b64\n``#136b64` `rgb(19,107,100)``\n• #64136b\n``#64136b` `rgb(100,19,107)``\nTriadic Color\n• #1a6b13\n``#1a6b13` `rgb(26,107,19)``\n• #136b64\n``#136b64` `rgb(19,107,100)``\n• #64136b\n``#64136b` `rgb(100,19,107)``\n• #6b131a\n``#6b131a` `rgb(107,19,26)``\nTetradic Color\n• #072a27\n``#072a27` `rgb(7,42,39)``\n• #0b403c\n``#0b403c` `rgb(11,64,60)``\n• #0f5550\n``#0f5550` `rgb(15,85,80)``\n• #136b64\n``#136b64` `rgb(19,107,100)``\n• #178178\n``#178178` `rgb(23,129,120)``\n• #1b968c\n``#1b968c` `rgb(27,150,140)``\n• #1faca1\n``#1faca1` `rgb(31,172,161)``\nMonochromatic Color\n\n# Alternatives to #136b64\n\nBelow, you can see some colors close to #136b64. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #136b4e\n``#136b4e` `rgb(19,107,78)``\n• #136b55\n``#136b55` `rgb(19,107,85)``\n• #136b5d\n``#136b5d` `rgb(19,107,93)``\n• #136b64\n``#136b64` `rgb(19,107,100)``\n• #136b6b\n``#136b6b` `rgb(19,107,107)``\n• #13636b\n``#13636b` `rgb(19,99,107)``\n• #135c6b\n``#135c6b` `rgb(19,92,107)``\nSimilar Colors\n\n# #136b64 Preview\n\nText with hexadecimal color #136b64\n\nThis text has a font color of #136b64.\n\n``<span style=\"color:#136b64;\">Text here</span>``\n#136b64 background color\n\nThis paragraph has a background color of #136b64.\n\n``<p style=\"background-color:#136b64;\">Content here</p>``\n#136b64 border color\n\nThis element has a border color of #136b64.\n\n``<div style=\"border:1px solid #136b64;\">Content here</div>``\nCSS codes\n``.text {color:#136b64;}``\n``.background {background-color:#136b64;}``\n``.border {border:1px solid #136b64;}``\n\n# Shades and Tints of #136b64\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, #010707 is the darkest color, while #f5fdfd is the lightest one.\n\n• #010707\n``#010707` `rgb(1,7,7)``\n• #041816\n``#041816` `rgb(4,24,22)``\n• #072826\n``#072826` `rgb(7,40,38)``\n• #0a3935\n``#0a3935` `rgb(10,57,53)``\n• #0d4a45\n``#0d4a45` `rgb(13,74,69)``\n• #105a54\n``#105a54` `rgb(16,90,84)``\n• #136b64\n``#136b64` `rgb(19,107,100)``\n• #167c74\n``#167c74` `rgb(22,124,116)``\n• #198c83\n``#198c83` `rgb(25,140,131)``\n• #1c9d93\n``#1c9d93` `rgb(28,157,147)``\n• #1faea2\n``#1faea2` `rgb(31,174,162)``\n• #22beb2\n``#22beb2` `rgb(34,190,178)``\n• #25cfc1\n``#25cfc1` `rgb(37,207,193)``\nShade Color Variation\n• #2edacc\n``#2edacc` `rgb(46,218,204)``\n• #3eddd0\n``#3eddd0` `rgb(62,221,208)``\n• #4fe0d4\n``#4fe0d4` `rgb(79,224,212)``\n• #5fe3d8\n``#5fe3d8` `rgb(95,227,216)``\n• #70e6dc\n``#70e6dc` `rgb(112,230,220)``\n• #81e9e0\n``#81e9e0` `rgb(129,233,224)``\n• #91ece4\n``#91ece4` `rgb(145,236,228)``\n• #a2efe8\n``#a2efe8` `rgb(162,239,232)``\n• #b3f1ec\n``#b3f1ec` `rgb(179,241,236)``\n• #c3f4f1\n``#c3f4f1` `rgb(195,244,241)``\n• #d4f7f5\n``#d4f7f5` `rgb(212,247,245)``\n• #e5faf9\n``#e5faf9` `rgb(229,250,249)``\n• #f5fdfd\n``#f5fdfd` `rgb(245,253,253)``\nTint Color Variation\n\n# Tones of #136b64\n\nA tone is produced by adding gray to any pure hue. In this case, #3f3f3f is the less saturated color, while #047a70 is the most saturated one.\n\n• #3f3f3f\n``#3f3f3f` `rgb(63,63,63)``\n• #3a4443\n``#3a4443` `rgb(58,68,67)``\n• #354947\n``#354947` `rgb(53,73,71)``\n• #304e4c\n``#304e4c` `rgb(48,78,76)``\n• #2b5350\n``#2b5350` `rgb(43,83,80)``\n• #265854\n``#265854` `rgb(38,88,84)``\n• #225c58\n``#225c58` `rgb(34,92,88)``\n• #1d615c\n``#1d615c` `rgb(29,97,92)``\n• #186660\n``#186660` `rgb(24,102,96)``\n• #136b64\n``#136b64` `rgb(19,107,100)``\n• #0e7068\n``#0e7068` `rgb(14,112,104)``\n• #09756c\n``#09756c` `rgb(9,117,108)``\n• #047a70\n``#047a70` `rgb(4,122,112)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #136b64 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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[ "Van Vu and I have just uploaded to the arXiv our paper “Random matrices: Localization of the eigenvalues and the necessity of four moments“, submitted to Probability Theory and Related Fields. This paper concerns the distribution of the eigenvalues", null, "$\\displaystyle \\lambda_1(M_n) \\leq \\ldots \\leq \\lambda_n(M_n)$\n\nof a Wigner random matrix", null, "${M_n}$. More specifically, we consider", null, "${n \\times n}$ Hermitian random matrices whose entries have mean zero and variance one, with the upper-triangular portion of the matrix independent, with the diagonal elements iid, and the real and imaginary parts of the strictly upper-triangular portion of the matrix iid. For technical reasons we also assume that the distribution of the coefficients decays exponentially or better. Examples of Wigner matrices include the Gaussian Unitary Ensemble (GUE) and random symmetric complex Bernoulli matrices (which equal", null, "${\\pm 1}$ on the diagonal, and", null, "${\\pm \\frac{1}{2} \\pm \\frac{i}{2}}$ off the diagonal). The Gaussian Orthogonal Ensemble (GOE) is also an example once one makes the minor change of setting the diagonal entries to have variance two instead of one.\n\nThe most fundamental theorem about the distribution of these eigenvalues is the Wigner semi-circular law, which asserts that (almost surely) one has", null, "$\\displaystyle \\frac{1}{n} \\sum_{i=1}^n \\delta_{\\lambda_i(M_n)/\\sqrt{n}} \\rightarrow \\rho_{sc}(x)\\ dx$\n\n(in the vague topology) where", null, "${\\rho_{sc}(x) := \\frac{1}{\\pi} (4-x^2)_+^{1/2}}$ is the semicircular distribution. (See these lecture notes on this blog for more discusssion of this law.)\n\nOne can phrase this law in a number of equivalent ways. For instance, in the bulk region", null, "${\\epsilon n \\leq i \\leq (1-\\epsilon) n}$, one almost surely has", null, "$\\displaystyle \\lambda_i(M_n) = \\gamma_i \\sqrt{n} + o(\\sqrt{n}) \\ \\ \\ \\ \\ (1)$\n\nuniformly for in", null, "${i}$, where the classical location", null, "${\\gamma_i \\in [-2,2]}$ of the (normalised)", null, "${i^{th}}$ eigenvalue", null, "${\\frac{1}{\\sqrt{n}} \\lambda_i}$ is defined by the formula", null, "$\\displaystyle \\int_{-2}^{\\gamma_i} \\rho_{sc}(x)\\ dx := \\frac{i}{n}.$\n\nThe bound (1) also holds in the edge case (by using the operator norm bound", null, "${\\| M_n\\|_{op} = (2+o(1)) \\sqrt{n}}$, due to Bai and Yin), but for sake of exposition we shall restriction attention here only to the bulk case.\n\nFrom (1) we see that the semicircular law controls the eigenvalues at the coarse scale of", null, "${\\sqrt{n}}$. There has been a significant amount of work in the literature in obtaining control at finer scales, and in particular at the scale of the average eigenvalue spacing, which is of the order of", null, "${\\sqrt{n}/n = n^{-1/2}}$. For instance, we now have a universal limit theorem for the normalised eigenvalue spacing", null, "${\\sqrt{n}(\\lambda_{i+1}(M_n) - \\lambda_i(M_n))}$ in the bulk for all Wigner matrices, a result of Erdos, Ramirez, Schlein, Vu, Yau, and myself. One tool for this is the four moment theorem of Van and myself, which roughly speaking shows that the behaviour of the eigenvalues at the scale", null, "${n^{-1/2}}$ (and even at the slightly finer scale of", null, "${n^{-1/2-c}}$ for some absolute constant", null, "${c>0}$) depends only on the first four moments of the matrix entries. There is also a slight variant, the three moment theorem, which asserts that the behaviour of the eigenvalues at the slightly coarser scale of", null, "${n^{-1/2+\\epsilon}}$ depends only on the first three moments of the matrix entries.\n\nIt is natural to ask whether these moment conditions are necessary. From the result of Erdos, Ramirez, Schlein, Vu, Yau, and myself, it is known that to control the eigenvalue spacing", null, "${\\lambda_{i+1}(M_n) - \\lambda_i(M_n)}$ at the critical scale", null, "${n^{-1/2}}$, no knowledge of any moments beyond the second (i.e. beyond the mean and variance) are needed. So it is natural to conjecture that the same is true for the eigenvalues themselves.\n\nThe main result of this paper is to show that this is not the case; that at the critical scale", null, "${n^{-1/2}}$, the distribution of eigenvalues", null, "${\\lambda_i(M_n)}$ is sensitive to the fourth moment, and so the hypothesis of the four moment theorem cannot be relaxed.\n\nHeuristically, the reason for this is easy to explain. One begins with an inspection of the expected fourth moment", null, "$\\displaystyle \\sum_{i=1}^n {\\bf E}(\\lambda_i(M_n)^4) = {\\bf E} \\hbox{tr} M_n^4.$\n\nA standard moment method computation shows that the right hand side is equal to", null, "$\\displaystyle 2n^3 + 2a n^2 + \\ldots$\n\nwhere", null, "${a}$ is the fourth moment of the real part of the off-diagonal coefficients of", null, "${M_n}$. In particular, a change in the fourth moment", null, "${a}$ by", null, "${O(1)}$ leads to a change in the expression", null, "${\\sum_{i=1}^n {\\bf E}(\\lambda_i(M_n)^4)}$ by", null, "${O(n^2)}$. Thus, for a typical", null, "${i}$, one expects", null, "${{\\bf E}(\\lambda_i(M_n)^4)}$ to shift by", null, "${O(n)}$; since", null, "${\\lambda_i(M_n) = O(\\sqrt{n})}$ on the average, we thus expect", null, "${\\lambda_i(M_n)}$ itself to shift by about", null, "${O(n^{-1/2})}$ by the mean-value theorem.\n\nTo make this rigorous, one needs a sufficiently strong concentration of measure result for", null, "${\\lambda_i(M_n)}$ that keeps it close to its mean value. There are already a number of such results in the literature. For instance, Guionnet and Zeitouni showed that", null, "${\\lambda_i(M_n)}$ was sharply concentrated around an interval of size", null, "${O(n^\\epsilon)}$ around", null, "${\\sqrt{n} \\gamma_i}$ for any", null, "${\\epsilon > 0}$ (in the sense that the probability that one was outside this interval was exponentially small). In one of my papers with Van, we showed that", null, "${\\lambda_i(M_n)}$ was also weakly concentrated around an interval of size", null, "${O(n^{-1/2+\\epsilon})}$ around", null, "${\\sqrt{n} \\gamma_i}$, in the sense that the probability that one was outside this interval was", null, "${O(n^{-c})}$ for some absolute constant", null, "${c>0}$. Finally, if one made an additional log-Sobolev hypothesis on the entries, it was shown by by Erdos, Yau, and Yin that the average variance of", null, "${\\lambda_i(M_n)}$ as", null, "${i}$ varied from", null, "${1}$ to", null, "${n}$ was of the size of", null, "${O(n^{-c})}$ for some absolute", null, "${c>0}$.\n\nAs it turns out, the first two concentration results are not sufficient to justify the previous heuristic argument. The Erdos-Yau-Yin argument suffices, but requires a log-Sobolev hypothesis. In our paper, we argue differently, using the three moment theorem (together with the theory of the eigenvalues of GUE, which is extremely well developed) to show that the variance of each individual", null, "${\\lambda_i(M_n)}$ is", null, "${O(n^{-c})}$ (without averaging in", null, "${i}$). No log-Sobolev hypothesis is required, but instead we need to assume that the third moment of the coefficients vanishes (because we want to use the three moment theorem to compare the Wigner matrix to GUE, and the coefficients of the latter have a vanishing third moment). From this we are able to make the previous arguments rigorous, and show that the mean", null, "${{\\bf E} \\lambda_i(M_n)}$ is indeed sensitive to the fourth moment of the entries at the critical scale", null, "${n^{-1/2}}$.\n\nOne curious feature of the analysis is how differently the median and the mean of the eigenvalue", null, "${\\lambda_i(M_n)}$ react to the available technology. To control the global behaviour of the eigenvalues (after averaging in", null, "${i}$), it is much more convenient to use the mean, and we have very precise control on global averages of these means thanks to the moment method. But to control local behaviour, it is the median which is much better controlled. For instance, we can localise the median of", null, "${\\lambda_i(M_n)}$ to an interval of size", null, "${O(n^{-1/2+\\epsilon})}$, but can only localise the mean to a much larger interval of size", null, "${O(n^{-c})}$. Ultimately, this is because with our current technology there is a possible exceptional event of probability as large as", null, "${O(n^{-c})}$ for which all eigenvalues could deviate as far as", null, "${O(n^\\epsilon)}$ from their expected location, instead of their typical deviation of", null, "${O(n^{-1/2})}$. The reason for this is technical, coming from the fact that the four moment theorem method breaks down when two eigenvalues are very close together (less than", null, "${n^{-c}}$ times the average eigenvalue spacing), and so one has to cut out this event, which occurs with a probability of the shape", null, "${O(n^{-c})}$. It may be possible to improve the four moment theorem proof to be less sensitive to eigenvalue near-collisions, in which case the above bounds are likely to improve." ]

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[ "Solutions by everydaycalculation.com\n\n## Compare 1/42 and 2/4\n\n1/42 is smaller than 2/4\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 42 and 4 is 84\n\nNext, find the equivalent fraction of both fractional numbers with denominator 84\n2. For the 1st fraction, since 42 × 2 = 84,\n1/42 = 1 × 2/42 × 2 = 2/84\n3. Likewise, for the 2nd fraction, since 4 × 21 = 84,\n2/4 = 2 × 21/4 × 21 = 42/84\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 2/84 < 42/84 or 1/42 < 2/4\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "Abstract\n\nThe present study proposed a time-space framework using discrete wavelet transform-based multiscale entropy (DWE) approach to analyze and spatially categorize the precipitation variation in Iran. To this end, historical monthly precipitation time series during 1960–2010 from 31 rain gauges were used in this study. First, wavelet-based de-noising approach was applied to diminish the effect of noise in precipitation time series which may affect the entropy values. Next, Daubechies (db) mother wavelets (db5–db10) were used to decompose the precipitation time series. Subsequently, entropy concept was applied to the sub-series to measure the uncertainty and disorderliness at multiple scales. According to the pattern of entropy across scales, each cluster was assigned an entropy signature that provided an estimation of the entropy pattern of precipitation in each cluster. Spatial categorization of rain gauges was performed using DWE values as input data to k-means and self-organizing map (SOM) clustering techniques. According to evaluation criteria, it was proved that k-means with clustering number equal to 5 with Silhouette coefficient=0.33, Davis–Bouldin=1.18 and Dunn index=1.52 performed better in determining homogenous areas. Finally, investigating spatial structure of precipitation variation revealed that the DWE had a decreasing and increasing relationship with longitude and latitude, respectively, in Iran.\n\nINTRODUCTION\n\nAssessment of precipitation variation over a large area (e.g., Iran) could provide valuable information for water resources management and engineering issues, particularly in a changing climate. The impact of global warming on different water cycle components is strongly variable across the globe and causes increases in average global precipitation, evaporation, and runoff (Clark et al. 1999; Pechlivanidis et al. 2017; Salvia et al. 2017; Sattari et al. 2017; Wei et al. 2017; Ba et al. 2018). Alteration of the hydrologic cycle will have significant impacts on the rate, timing, and distribution of rain, evaporation, temperature, snowfall, and runoff, the main causes of change in the accessibility of water resources (Mishra et al. 2009). The example of precipitation variation in Iran (during 1966–2005) could be referred to the rate of the significant decreasing trends in annual precipitation that varied from (−)1.999 mm/year in the northwest to (+)4.261 mm/year in the west of Iran. The significant negative trends mainly occurred in the northwest of Iran. These negative trends can affect agriculture and water supply of the regions. On the contrary, no significant trends were detected in the eastern, southern, and central parts of the country (Tabari & Hosseinzadeh Talaee 2011; Raziei 2017). By considering the high spatial and temporal variability of precipitation and frequent dry periods, the increasing water demands for growing population as well as for industry and economic development, including irrigation, aggravating water scarcity makes it difficult for a rationale water management. Hence, determination of sub-regions according to different precipitation regimes is important for water resources management and land use planning.\n\nIn recent decades, some studies have focused on studying precipitation across Iran (Domroes et al. 1998; Dinpashoh et al. 2004; Modarres 2006; Soltani et al. 2007; Raziei et al. 2008; Modarres & Sarhadi 2009; Tabari & Hosseinzadeh Talaee 2011). Domroes et al. (1998) applied principal component analysis (PCA) and cluster analysis (CA) on mean monthly precipitation of 71 stations and classified the precipitation regimes into five different sub-regions. On the other hand, applying the PCA and CA to 12 variables selected from 57 candidate variables for 77 stations distributed across the entire country, Dinpashoh et al. (2004) divided the country into seven climate sub-regions. Rainfall climates in Iran were also analyzed by Soltani et al. (2007) using monthly precipitation time series from 28 main sites. To determine regional climates, a hierarchical CA was applied to the autocorrelation coefficients at different lags, and three main climatic groups were found. Tabari & Hosseinzadeh Talaee (2011) analyzed trend over different sub-regions of Iran during 1966–2005. Raziei et al. (2008) analyzed the spatial distribution of the seasonal and annual precipitation in western Iran using data from 140 stations covering the period 1965–2000. Applying the precipitation concentration index (PCI), the intra-annual precipitation variability was also studied. The results suggest that five homogenous sub-regions can be identified based on different precipitation regimes. Modarres & Sarhadi (2009) performed spatial and temporal trend analysis of the annual and 24-hr maximum rainfall of a set of 145 precipitation gauging stations of Iran during the period of 1955–2000. The study showed that the annual rainfall is decreasing at 67% of the stations while the 24-hr maximum rainfall is increasing at 50% of the stations.\n\nWavelet analysis (WA), which has been widely applied in hydrology and hydrogeology, is capable of elucidating the localized characteristics of non-stationary time series both in temporal and frequency domains (Nourani et al. 2009, 2015; Kisi & Shiri 2012; Danandeh Mehr et al. 2015; Karimi et al. 2016; Danesh-Yazdi et al. 2017), and it is just suitable for hydrologic time series analyses. The wavelet entropy, combined by WA and information theory, is an important concept of describing the variability and complexity of hydrologic time series with non-stationary and multi-temporal characteristics (Zunino et al. 2007). It is used to first analyze a time series by WA, such as continuous wavelet transform (CWT) and multi-resolution analysis, and then calculate the entropy measures, mainly including Shannon entropy (Jaynes 1957), mutual information (Molini et al. 2006), and relative entropy (Abramov et al. 2005). Various studies have manifested the better performance of wavelet entropy in analyzing the variability and complexity of hydrologic variables compared with traditional methods (Simpson's index, McIntosh index, Berger–Parker index, Brillouin index, etc.) (Mishra et al. 2009; Brunsell 2010).\n\nThe proposed technique combines discrete wavelet transform (DWT) based multiscale entropy approach with k-means and self-organizing map (SOM) clustering techniques. The discrete wavelet multiscale entropy (DWE) which is a measure of the degree of order/disorder of the signal and carries information associated with multi-frequency signal, can provide useful information about the underlying dynamic processes associated with the signal and can help in precipitation-based studies (Cazelles et al. 2008). Therefore, this study tried to develop a precipitation-based regionalization-based DWE approach. In this study, the DWE method was applied to monthly precipitation data observed at 31 rain gauges in Iran. Higher entropy reflects more random and complicated systems and vice versa. Traditional entropy measures usually provide inaccurate or incomplete descriptions of climatic systems which generally operate over multi-resolution scales (Li & Zhang 2008). DWT was used to decompose each of the observed precipitation time series using the Daubechies (db) wavelet to capture the multiscale variability of the precipitation based on wavelet coefficients. Next, these wavelet coefficients for each scale are used to obtain the entropy for the respective scales (Sang 2012; Agarwal et al. 2016). The spatial organization of this multiscale variability in terms of DWE is identified using clustering methods.\n\nMATERIAL AND METHODS\n\nCase study and climatological dataset\n\nThis study used monthly climate data of 31 precipitation gauges all over Iran for studying precipitation regionalization (1960–2010) (Figure 1 and Table 1). Due to the variety of information involved in hydrologic processes and need to have accurate models, monthly precipitation time series was used which include various multivariate properties such as seasonality of process. Iran is a large country (approximately 1,600,000 km2), in which climate is mostly affected by the wide latitudinal extent. Iran is located in Southwest Asia (25° to 40°N and 44° to 63°E). There are three seas in Iran, in the north the Caspian Sea and in the south the Persian Gulf and Oman Sea (Araghi et al. 2014). The moisture coming from the Persian Gulf is usually trapped by the Zagros Mountains. The plateau is open to the cold (dry) continental currents flowing from the northeast and the mitigating influence of the Caspian Sea is limited to the northern regions of the Alborz Mountains. The Zagros chain, which stretches from northwest to southeast, is the source of several large rivers such as the Karkheh, Dez, and Karoon. Lowland areas receive surface water from these basins and are of great importance for agricultural applications (Raziei et al. 2008). Iran's climate is generally recognized as arid or semi-arid with an annual average precipitation of about 250 mm; however, its climate is very diverse, with annual precipitation and temperature variation over the country (Figure 1(a)). For instance, in different areas of the country annual precipitation changes from 0 to 2,000 mm (Domroes et al. 1998; Dinpashoh et al. 2004). The Caspian Sea coastal areas along with the northern and northwestern regions of the country are subjected to higher precipitation. On the other hand, the lowest values of annual precipitation are found in the southern, eastern, and the central desert regions (Ashraf et al. 2013). Generally, Iran is categorized as hyper-arid (35.5%), arid (29.2%), semi-arid (20.1%), Mediterranean (5%), and wet climate (10%). Also, temperature in Iran varies widely (−20 to +50 °C) (Saboohi et al. 2012). On the northern edge of the country (the Caspian coastal plain) temperatures rarely fall below freezing and the area remains humid for all of the year. Summer temperatures rarely exceed 29 °C (Nagarajan 2010; Weather & Climate Information 2015). To the west, settlements in the Zagros basin experience lower temperatures, severe winters with below zero average daily temperatures and heavy snowfall. The eastern and central basins are arid and have occasional deserts. Average summer temperatures rarely exceed 38 °C (Nagarajan 2010). The coastal plains of the Persian Gulf and Gulf of Oman in southern Iran have mild winters, and very humid and hot summers (Figure 1(a)). The dataset applied in this study was provided by the Iran Meteorological Organization (http://www.irimo.ir).\n\nTable 1\n\nSelection of optimum number of clusters based on Dunn, Davies–Bouldin, and Silhouette indices\n\nClustering technique Validity indices Cluster number\n\n10\nSOM Silhouette index 0.19 0.28 0.27 0.28 0.34 0.30 0.21 0.18 0.09\nDavies–Bouldin index 1.59 1.10 1.45 1.40 1.20 1.48 1.7 1.62 1.35\nDunn index 0.96 1.12 1.07 1.36 1.49 1.24 1.14 0.99 0.97\nK-means Silhouette index 0.22 0.23 0.26 0.33 0.21 0.25 0.25 0.22 0.22\nDavies–Bouldin index 1.34 1.33 1.33 1.18 1.18 1.51 1.14 1.42 1.34\nDunn index 1.37 1.19 1.32 1.52 1.2 0.83 0.90 0.78 1.07\nClustering technique Validity indices Cluster number\n\n10\nSOM Silhouette index 0.19 0.28 0.27 0.28 0.34 0.30 0.21 0.18 0.09\nDavies–Bouldin index 1.59 1.10 1.45 1.40 1.20 1.48 1.7 1.62 1.35\nDunn index 0.96 1.12 1.07 1.36 1.49 1.24 1.14 0.99 0.97\nK-means Silhouette index 0.22 0.23 0.26 0.33 0.21 0.25 0.25 0.22 0.22\nDavies–Bouldin index 1.34 1.33 1.33 1.18 1.18 1.51 1.14 1.42 1.34\nDunn index 1.37 1.19 1.32 1.52 1.2 0.83 0.90 0.78 1.07\n\nDiscrete wavelet transform (DWT)\n\nThe wavelet transform (WT) is a popular method and a very precise method for time series processing (Kisi & Shiri 2011; Nourani et al. 2014; Farajzadeh & Alizadeh 2017). While the general theory behind WT is quite analogous to that of the short-time Fourier transform (STFT), WT allows for a completely flexible window function (called the mother wavelet), which can be changed over time based on the shape and compactness of the signal. Given this property, WT can be used to analyze the time-frequency characteristics of any kind of time series. In recent years, WT has been widely used for the analysis of many hydro-meteorological time series (Adamowski et al. 2009; Partal 2010; Shiri & Kisi 2010; Nourani et al. 2013, 2015; Mehr et al. 2014). As the mother wavelet moves across the time series during the WT process, it generates several coefficients that represent the similarity between the time series and the mother wavelet (at any specific scale). There are two main types of WT: continuous and discrete. Use of the CWT can generate a large number of (often unnecessary) coefficients, making its use and interpretation more complicated. On the other hand, the DWT method simplifies the transformation process while still providing a very effective and precise analysis, since DWT is normally based on the dyadic calculation. DWT coefficients can be calculated by the following equation (Partal 2010):  where 2a represents the dyadic scale of the DWT. Applying DWT to a time series decomposes that time series into two ancillary time series shape components, called the approximation (A) and detail (D) components. Component A comprises the large-scale, low-frequency component of the time series, while component D represents the small-scale, high-frequency component.\n\nSignal de-noising with wavelets\n\nDe-noising a signal using WT is based on the observation that in many signals (e.g., rainfall signals) energy is mostly concentrated in a small number of wavelet dimensions. The coefficients of these dimensions are relatively large compared to the other dimensions or to noise, which has its energy spread over a large number of coefficients. Hence, by setting to zero, the coefficients smaller than a certain threshold, noise can nearly be optimally eliminated while preserving the important information of the original signal (Donoho 1995). Because amplitude de-noising is performed instead of frequency de-noising, the low frequency noise can also be suppressed (Nourani & Partoviyan 2017).\n\nTo de-noise a signal using WT, the detail coefficients are thresholded, since they represent mainly noise. One way to threshold the detail coefficients is to use ‘soft’ thresholding. In this case, the thresholded details are given by the following equation:\n\nIn Equation (2), λ and D(i) (j= 1, 2, … , M) indicate threshold value and absolute value of detailed sub-series at ith resolution level, respectively. The algorithm to de-noise a signal f(k) corrupted by a noise signal n(k) can be summarized by the following three steps:\n\n1. Apply DWT to a noisy signal to obtain approximations A(i) and details D(i).\n\n2. Apply a thresholding technique to detail coefficients D(i) to obtain the thresholded coefficients Dth(i).\n\n3. Transform the signal back based on A(i) and Dth(i) to obtain the de-noised signal (reconstruction).\n\nAccording to Donoho (1995), in the case of white Gaussian noise, the threshold (λ) can be estimated as follows (Donoho 1995):  where N is the length of the signal and σ is the noise level, which is calculated as σ = MAD/0.6745; and MAD is the median absolute value of the details coefficients estimated for the first level.\n\nTime series decomposition via the discrete wavelet transform\n\nThe conventional discrete WA of time series was performed on each rain gauge using the multilevel 1-D wavelet decomposition function in MATLAB (MATLAB Wavelet Toolbox). This produces the WT of the time series of the interest at all dyadic scales. The monthly precipitation input time series are all one-dimensional. Decomposing the time series using specified filters (wavelet and scaling functions) produces two types of coefficients: the approximation or residual, and detail vectors (Chou 2007). These coefficients resulted from the convolution of the original time series with a low-pass filter and a high-pass filter. The low-pass filter is the scaling function and the high-pass filter is the wavelet function. The convolutions of time series with the low-pass filter produced the approximation coefficients, which represent the large-scale or low-frequency components of the original time series. Convolutions with the high-pass filter produced the detail coefficients, which represent the low-scale or high-frequency components (Bruce et al. 2002). The process of time series decomposition was repeated multiple times, decomposing the original time series into several different lower-resolution components (Partal 2010). The detail and approximation coefficients produced from the time series decomposition were then reconstructed since they are merely intermediate coefficients. These have to be re-adjusted to the entire one-dimensional signal in order to enable the investigation of their contribution to the original time series (Dong et al. 2008). This contribution may be reflected in the different time scales such as intra-annual, inter-annual, decadal, and multi-decadal.\n\nSelection of an appropriate wavelet function poses significant challenges and is governed largely by the problem at hand and some of the distinctive properties of the wavelet function such as (i) its region of support and (ii) the number of vanishing moments (Maheswaran & Khosa 2012).\n\nThe region of support implies the length span of the given wavelet, which in turn affects its feature localization capabilities as it is understandable that a long and widely distributed wavelet function will calculate the instantaneous process amplitude while, at the same time, spanning a wider window of the underlying process resulting in a high degree of averaging of the process states. Vanishing moment, on the other hand, limits the wavelet's ability to suitably represent polynomial behavior or information in a time series. For example, the db2 wavelet encodes polynomials with two coefficients, i.e., a process having one constant and one linear time series component, and the db3 wavelet encodes a process having a constant, linear, and quadratic time series components. Within each family of wavelets are wavelet subclasses distinguished by their respective number of coefficients and the number of vanishing moments, as discussed below (Maheswaran & Khosa 2012).\n\nThe db wavelets were used in this study because they are commonly used mother wavelets for the DWT in hydro-meteorological wavelet-based studies (Mehr et al. 2013). The db wavelets provide compact support with extreme phase and highest number of vanishing moments for a given support width (Vonesch et al. 2007), indicating that the wavelets have non-zero basis functions over a finite interval, as well as full scaling and translational orthonormality properties (Popivanov & Miller 2002; de Artigas et al. 2006). These features are very important for localizing events in the time-dependent signals (Popivanov & Miller 2002). These properties are unique and cannot be found in other mother wavelets (i.e., Haar, Coife, Symlet, etc.). For the period of 612 months (51 years), in order to avoid unnecessary levels of time series decomposition in these larger datasets, the number of decomposition levels had to be determined first. This number is based upon the number of data points, as well as the mother wavelet used. The highest decomposition level should correspond to the data point at which the last subsampling becomes smaller than the filter length (de Artigas et al. 2006). There are several recommended methods to determine the most appropriate number of decomposition levels, of which one of the most commonly used is given by the following equation (de Artigas et al. 2006; Araghi et al. 2014):  where L is the number of decomposition levels, n is the number of data points in the time series and v is the number of vanishing moments of the db mother wavelet. In MATLAB, v is equal to the type number of the db. Smoother db wavelets (db5–db10) were then tried for each monthly time series. Smoother wavelets are preferred here because the trends are supposed to be gradual and represent slowly changing processes. Smoother wavelets should be better at detecting long-term time-varying behavior (good frequency-localization properties) (Adamowski et al. 2009). In addition to this, several trend studies used smoother db mother wavelets (e.g., Kallache et al. (2005) used least asymmetric LA(8); de Artigas et al. (2006) used db7). The border conditions were also taken into consideration when performing the DWT. This is because for time series with a limited length, convolution processes cannot proceed at both ends of the time series since there is no information available outside these boundaries (Su et al. 2011). This is referred to as the border effect (Su et al. 2011). As a result, an extension at both edges is needed. Border extensions that are commonly used are zero-padding, periodic extension, and symmetrization – all of which have their drawbacks, due to the discontinuities introduced at both ends of the time series (de Artigas et al. 2006; Su et al. 2011). The default extension method used in MATLAB is symmetrization, which assumes that time series outside the original support can be recovered by symmetric boundary replication (de Artigas et al. 2006). Zero-padding pads the time series with zeros beyond the original support of the wavelet; periodic padding assumes that time series can be recovered outside of the original support by periodic extension (de Artigas et al. 2006). The inverse discrete wavelet transform (IDWT) was then computed to ensure perfect signal reconstruction. There were three main parameters to determine for the DWT used in this study: (i) the appropriate type of db wavelet; (ii) the best method for time series border extension; and (iii) the most appropriate number of decomposition levels. In order to determine the smooth mother wavelet, optimal level of decomposition level, and the extension mode to be used in the time series analysis for each data type and dataset, two criteria were used. The first criterion used was proposed by de Artigas et al. (2006): all three extension modes for each db wavelet were employed in order to determine the extension method, and the db type, that would produce the lowest mean relative error (MRE). The MRE was calculated using Equation (5) (Popivanov & Miller 2002; de Artigas et al. 2006):  where xj is the original time series value of a time series whose number of records is n, and aj is the approximation value of xj. The second criterion used in this study is based on the relative error (er). Each of the extension modes for each of the smooth db wavelets was examined in order to determine the combination (of border condition and the mother wavelet) that would produce the lowest approximation Mann–Kendall Z-value relative error (er). The computation of the relative error was done using the following equation:  where Za is the MK Z-value of the last approximation for the decomposition level used, and Zo is the MK Z-value of the original time series. For the monthly time series, the MREs of the different border conditions did not show substantial differences. The differences in the relative errors were also more noticeable among the different border extensions and the different db wavelets. Since the monthly time series have 612 months of records, and according to the optimal MRE and er values, they could be decomposed up to six levels, which correspond to 64 months.\n\nDiscrete wavelet-based multiscale entropy approach (DWE)\n\nThis study proposed an approach based on hybrid DWT, entropy and k-means models to investigate the variation and regionalize the precipitation in Iran. Figure 2 shows the schematic of modeling in this study. The monthly time series of rain gauges used in this study were firstly pre-processed using DWT. For this end, Daubechies mother wavelet (db) and proper related parameters were selected for each precipitation time series.\n\nFurthermore, DWE was used to quantify the variability and complexity of monthly precipitation processes. In the information theories, the Shannon entropy (H) is calculated as (Brunsell 2010):  where p(xi) is the probability density function (PDF) used to describe the random characters of variable x with the length of n. H is a measure of information; more information results in lower entropy and vice versa. Therefore, bigger H value presents more disordered and complicated precipitation processes. When using the measure of DWE, the H value is calculated based on dyadic DWT results, and Equation (8) is used to compute the PDF, which is estimated according to the wavelet energy (i.e., variance) (Sang 2012):\n\nThe entropy of a random variable is a measure of the uncertainty of the random variable; it is a measure of the amount of information required on average to describe the random variable (Termini & Moramarco 2016; Werstuck & Coulibaly 2016). Next, entropy values of decomposed time series (detail (D1, D2Di) and approximation (Ai) components) were fed into k-means approaches in order to perform spatial clustering. Different statistical evaluation criteria were used to verify the validity of clustering, which is explained below.\n\nK-means clustering\n\nOne of the most popular clustering algorithms is the k-means method, in which the data is partitioned into k clusters, with each cluster represented by its centroid, which is the mean (weighted or otherwise) of feature vectors within the cluster (Agarwal et al. 2016). If Nk represents the number of feature vectors in cluster k, Ck is the mean of cluster k and Xp represents observed precipitation time series, then the centroid of each cluster is calculated using Equation (9):  The procedure follows a simple and easy way to classify a given dataset through a certain number of clusters (assume k clusters). The main idea is to define k centers, one for each cluster. The algorithm starts with the pre-defined initial number of clusters k chosen according to some criteria or some heuristic procedure. In each iteration, each cluster is assigned to its nearest cluster center according to the Euclidian distance measure between the two, and then the cluster centers (CC) are re-calculated (Rokach & Maimon 2005) until convergence of the algorithm occurs as per the defined criteria, e.g., when the algorithm exceeds the pre-defined number of iterations or when partitioning error is not going to reduce further on re-allocating cluster centroid, indicating that solution is locally optimal. The method is known for its low run time, its efficiency in clustering large datasets with numerical attributes (Rao & Srinivas 2008), and simple implementation and interpretation since no parameters (except the number of clusters) are involved. The linear complexity is also one of the reasons for the popularity of the k-means algorithm. In other words, since there are computational complexities in finding the optimal solution to the k-means clustering problem, a variety of heuristic algorithms such as Lloyd's algorithm (linear complexity) are generally used. More detailed information about the k-means clustering method can be obtained from Ball & Hall (1967) and MacQueen (1967), among others.\n\nSelf-organizing maps (SOM)\n\nThe self-organizing map is a powerful method used to explore and extract the inter-relationships of high-dimensional multivariate systems, and it is beneficial for clustering and forecasting in a widespread range of disciplines (Kohonen 1997). One of the main advantages of the SOM is its ability to extract implicit patterns from high-dimensional input dataset and classify the obtained patterns into a low-dimensional output layer, where similar inputs remain close together in the output neurons while preserving data structure (Hsu & Li 2010; Nourani et al. 2015). The neurons in the output layer are commonly arranged in two-dimensional grids so that the constructed topology can be visualized to give an insight into the system under investigation. The SOM has gained increasing interest and been successfully applied to hydrology and water resources management (Kalteh et al. 2008; Hsu & Li 2010; Nourani et al. 2015; Chang et al. 2016; Iwashita et al. 2018).\n\nEvaluation criteria\n\nIn the present study, three validation metrics, namely, Davies–Bouldin index (DBi), Dunn index, and Silhouette coefficient (SC) were utilized to validate the outcome of spatial clustering via k-means technique.\n\nIn hydrology, DBi is a widely applied internal evaluation criterion (Davies & Bouldin 1979; Kasturi et al. 2003), which is applied to distinguish the number of optimal clusters that are well-detached and well-set based on content and specification of dataset. A lower DBi value represents better clustering results. On the other hand, DBi has a disadvantage in that best information detection cannot be implied by a good reported DBi value.\n\nThe Dunn index's goal is to distinguish a category of clusters that are well-set, with a small variance among components of the cluster, and well detached, where the averages of the various clusters are adequately far apart when compared to the within cluster variance (Dunn 1973). A higher Dunn index's value shows better clustering outcome as it shows a well-compacted cluster (Agarwal et al. 2016). Computational cost increases when the number of clusters and dimensionality increases, which is a disadvantage for the Dunn index.\n\nThe SC index's goal is to show how analogous a member is to the related cluster (cohesion) in comparison to the other clusters (separation). The SC values vary from −1 to 1, where a high SC indicates that the member is well-adapted to the related cluster and insignificantly adapted to neighboring clusters.\n\nIf most members have a high SC value, then the formation of the clustering is suitable. On the other hand, if many members have a low or negative SC value, accordingly the clustering formation may have too many or too few clusters. Generally, studies have offered the applicability of SC (Hsu & Li 2010; Nourani et al. 2015). Nevertheless, the present study evaluated the outcome of spatial clustering based on all three indices to take advantage of them.\n\nRESULTS AND DISCUSSION\n\nPrecipitation time series pre-processing via DWT\n\nThe precipitation time series might include a degree of noise-contamination which could influence the calculation of wavelet-based entropy values. Hence, the noise in the time series was removed by WT de-noising approach; afterwards, the DWT was applied to the de-noised time series using the chosen db mother wavelet to decompose precipitation time series into approximation and detail components.\n\nAfter selecting the proper mother wavelet, boundary extensions, and decomposition level for each precipitation time series, an adequate threshold value should be chosen for the de-noising procedure. The range of threshold values within the local vicinity of the universal threshold value was acquired by Donoho's formula (Equation (3)) to determine ‘appropriate threshold value’ for precipitation decomposition via db mother wavelet. As an illustration, de-noised time series of rain gauge 4 (RG 4) is shown in Figure 3(a). Existence of noise in a time series can result in corruption and uncertainty by adding complexity to hydrologic time series, and the aforementioned issue becomes worse when the signal to noise ratio (SNR) value decreases. Therefore, the existence of noise in precipitation time series can significantly affect the results of the proposed model in both temporal pre-processing and spatial clustering stages. Besides, because the energy of noise mainly concentrates in small temporal scales, it has a more severe influence on the entropy values under small temporal scales than those under large temporal scales (Sang et al. 2011). Figure 3(b) and 3(c) show the power spectrum of original and de-noised time series, respectively. It is clearly observed that the power of de-noised time series in higher frequencies (lower temporal scales) has remarkably decreased in comparison to the original time series. Although WA analysis as a time series pre-processing method can also handle some degrees of noise included in the time series, as demonstrated in Figure 3, WD approach handle the noise in time series better, especially in higher frequencies.\n\nFigure 3\n\n(a) De-noised precipitation time series of rain gauge 4 (RG 4) via DWT, (b) power spectrum of original time series, and (c) power spectrum of de-noised time series.\n\nFigure 3\n\n(a) De-noised precipitation time series of rain gauge 4 (RG 4) via DWT, (b) power spectrum of original time series, and (c) power spectrum of de-noised time series.\n\nFigure 4 shows the results of decomposing RG 4 precipitation time series using db9 mother wavelet and zero-padding boundary extension. Each monthly precipitation dataset was decomposed into six lower resolution levels via the DWT approach. The detail components represent the 2-month periodicity (D1), 4-month periodicity (D2), 8-month periodicity (D3), 16-month periodicity (D4), 32-month periodicity (D5), and 64-month periodicity (D6). The A6 represents the approximation component (including the trend) at the sixth level of decomposition.\n\nIt was observed that as the transform progressed from low to high scale (short to long time scale), more boundary points became distorted due to the decimation process. In other words, as filter length increased, more points at the boundary become affected. For higher scales (trend), the distortion becomes visibly worse. Application of boundary extension can cause inconsistency in computations of sub-series captured from DWT since it can add some uncertainties into time series (Mun 2004). This inconsistency could affect the performance of the proposed model. Hence, it was attempted to minimize the effect of applied boundary extension by using the MRE and er criteria in order to select the efficient border extension (see the ‘Time series decomposition via the DWT’ section).\n\nRegionalization of rain gauges using the proposed model\n\nAt this stage, entropy-based values of the decomposed components of precipitation time series were calculated to be used as input layer of k-means. Spatial distribution of the seven entropy values of the sub-series (A6, D6, D5 … D1) in Iran are demonstrated in Figure 5. Based on Figure 5, highest DWE values generally were for A6 and D3 sub-series, whereas D1 sub-series had the least DWE values among all values calculated. It can be observed that there are compact counter lines on the north and northern west parts of Iran. It means that DWE values of various scales change rapidly on north and northern west parts of Iran and these zones are located in rainy and cold regions of Iran. Spatial changes of DWE becomes smoother for western and southern zones which are semi-arid and arid regions of Iran. Generally, it could be stated that rapid changes of entropy pattern are observed for the northern west parts of Iran, which are mostly cold areas.\n\nThese seven values as signature of decomposed time series were used as input data to SOM and k-means in order to perform precipitation regionalization. The number of clusters for the dynamic features of monthly precipitation time series was determined by three validity indices' values. Table 1 shows the validity values for various numbers of DWE-based clustering approaches.\n\nAs discussed, for spatial clustering of 31 rain gauges in Iran, the DWE value of each rain gauge was used as input data of k-means clustering technique. At first, k-means approach with a 1,000 trial was trained based on DWE values. The optimal number of clusters was determined using validation indices. The clustering number 5 with SC=0.33, DBi=1.18, and Dunn=1.52 showed a better performance in determining homogenous areas in comparison to other clustering numbers for the k-means approach. Therefore, clustering number equal to 5 was selected as the optimum value to categorize the rain gauges.\n\nOn the other hand, SOM models were used to cluster the 31 rain gauges into a visible 2-dimensional topology of regional RGL maps. For this end, map sizes of 2 × 2 to 10 × 10 were tried. The constructed topological maps coupled with related key features showed that clustering number 6 with SC=0.34, DBi=1.20, and Dunn=1.49 led to a better performance in determining homogenous areas of precipitation variation in comparison to other clustering numbers for the SOM approach. However, there was a failure in outcome of SOM with six clusters. It was observed that two clusters had only one rain gauge, and two clusters had more than ten rain gauges. Results of k-means in means of both evaluation criteria and classification of rain gauges in various clusters proved to be better than SOM, and therefore was applied for further analysis.\n\nSome studies took advantage of DWT-based clustering approaches as a modeling approach. For example, Hsu & Li (2010) used the WT and self-organizing map (WTSOM) framework to spatially cluster the precipitation time series. In the proposed approach, they combined the WT and a SOM neural network. WT was used to extract dynamic and multiscale features of the non-stationary precipitation time series, and SOM was employed to objectively identify spatially homogeneous clusters on the high-dimensional wavelet transformed feature space. Haar and Morlet wavelets were selected in the data pre-processing stage to preserve the desired characteristics of the precipitation data. In this study, decomposition was performed using smoother db mother wavelets (db5–db10) along with optimum parameters. The entropy-based dynamic features of the time series could improve the performance of the clustering approach. Sub-series (i.e., Ai, Di i= 1, 2 … 6) represent various monthly scales. Nevertheless, some of these components might not demonstrate enough correlation with rainfall original time series. For this end, DWE was calculated and used as input to k-means to perform spatial clustering.\n\nGeographic location of rain gauges based on clustering via DWE as input into k-means approaches is demonstrated in Figure 6. Also in Figure 6, the CC based on validity indices are presented. It was seen that some of the stations in a given cluster are spread across the study area, revealing that the basis of clustering is not geographic proximity. For example, the rain gauges located near the Caspian Sea (rain gauges 19, 18, 4, and 11) with highest precipitation values and geographical proximity, are assigned to various clusters due to the differences in entropies calculated for each rain gauge. The stations in each of these clusters are further examined for any common characteristics (in terms of multiscale entropy) they may have among themselves.\n\nRaziei et al. (2008) regionalized the precipitation of the western part of Iran. They found five zones based on the behavior of precipitation. As can be seen from Figure 7, the rain gauges located in the west of Iran were placed in five different clusters. Different from results of spatial classification of rain gauges, Raziei (2017) found eight sub-regions of precipitation in Iran, namely, mountainous regime (covering Zagrom and some part of Alborz mountains), central Alborz regime, monsoonal southeastern regime, Caspian regime, northwestern regime, central-eastern regime, south and southwestern regime, and costal southeastern regime (geographic neighborhood). Also, Domroes et al. (1998); Modarres (2006), and Raziei (2017) separated Iran's rainfall regions into eight groups. The outcome of these studies are very analogous to each other. Similar to the present study, precipitation in the west of Iran was subjected to various precipitation changes for these studies (Figure 5). However, the results of the clustering in this study are very different, since the aforementioned studies classified the precipitation regime in Iran based on neighborhood approximates. As can be seen, the clustering shows that there is hydrologic similarity (in terms of multiscale variation of precipitation) in the clusters apart from the geographic neighborhood. It was observed that some of the rain gauges in a given cluster are spread across the study area showing that the basis of clustering is not the geographic contiguity.\n\nAs can be seen in Figure 7, the multiscale entropy values are, to a great extent, similar within any given cluster and the basis of the clustering is the entropy signature of the precipitation observed at all the rain gauges for all clusters. For example, in Cluster 1 (Figure 7(a)), the entropy signatures for all the rain gauges are similar and the peaks in the plots indicate high values of entropy, which corresponds to high variability of the precipitation features at the specific scale across time. In addition, the pattern of the entropy in a given cluster across all scales for the rain gauges is unique for that cluster (homogeneity) but also different from every other cluster.\n\nIn order to ensure a more sensible and simpler analysis, the average entropies for all clusters were used instead of entropy values of single rain gauges, and this was considered as the representative value of entropy for all clusters at a specific scale. Figure 8 shows the DWE values obtained for detail and approximation components. It was observed that entropy values of D3 and approximation components had the highest values and lowest variation whereas D1 had lowest entropy values with highest variation among all components. The DWE values increased from D1 to D3, then decreased from D3 to D4. However, variation of DWE from D4 to D6 and A6 (Figures 7 and 8) was not constant. In order to prove the outcome of Figure 8, the wavelet power spectrum of central rain gauges of each cluster are presented in Figure 9. It can be observed, that for all the rain gauges, that there are very rapid changes for the period of 1 to 16 months. For the period of 1 month, mostly low powers were observed; however, for the periods up to 8 months, the change in power spectrum becomes rapid from low to high and vice versa was observed. These changes become smoother for the period of the 16-month band in comparison to the 8-month band. For the bands beyond 16 months, the changes become smoother in comparison to previous bands and also, power spectrum values are higher than the 1-month band. Therefore, it can be inferred that these entropies of D1 to D4 sub-series are the key variables in precipitation regionalization. Also, it can be stated that the precipitation variation is affected by different variables, such as timing, amount, and temporal distribution.\n\nBased on these observations, three distinct bands were determined for further analysis. Band 1 considered the features up to 8-month scales (D1, D2, and D3). Band 2 considered the features from 8 to 16 months. The features having a scale beyond 16 months were categorized as Band 3 (D5 and D6). Figure 10 shows the average normalized DWE for all the clusters at different bands (the DWE values were normalized for better comparisons). There is a clear distinction in the values of DWE for different clusters in the first two bands and, in view of this, information from the first two bands was further analyzed. The DWE of each cluster was further classified into ‘High’, ‘Medium’, and ‘Low’, by considering the condition of the individual DWE plot according to the mean level for that band. For instance, if the DWE of a cluster in a specific band fell below the mean of DWE of all clusters, then that particular cluster was assigned a signature of ‘Low’. Using this classification, an entropy signature was given to each cluster based on the entropy values in the three scale-based bands. For notational simplicity, the classifications ‘High’, ‘Medium’, and ‘Low’ were represented by ‘1’, ‘0’, and ‘ − 1’, respectively. This means, for example, that an entropy signature of (0, −1) would indicate that the cluster had a relatively moderate entropy up to 8 months and low entropy for 8–16 months. Based on these notations, the entropy signature for each of the 14 clusters is given in Table 2.\n\nTable 2\n\nEntropy signature of five clusters for 31 rain gauges in Iran\n\nCluster no. Comparative observation of DWE\n\nEntropy signature\nBand 1 Band 2\n(1,0)\n(− 1,0)\n(1, − 1)\n(0, − 1)\n(− 1, − 1)\nCluster no. Comparative observation of DWE\n\nEntropy signature\nBand 1 Band 2\n(1,0)\n(− 1,0)\n(1, − 1)\n(0, − 1)\n(− 1, − 1)\nFigure 10\n\nComparison of normalized DWE values for each scale for all clusters according to segregated bands: (a) Band 1: 2–8 months, (b) Band 2: 8–16 months, and (c) Band 3: 16–64 months.\n\nFigure 10\n\nComparison of normalized DWE values for each scale for all clusters according to segregated bands: (a) Band 1: 2–8 months, (b) Band 2: 8–16 months, and (c) Band 3: 16–64 months.\n\nAs a further step, it was attempted to connect the DWE values at different scale-based bands to their respective mean monthly precipitation of rain gauges. Boxplots of mean monthly precipitation (Figure 11) suggest that the clusters with ‘High’ entropy for the scale 9–13 months (i.e., Clusters 2 and 5) had smaller precipitation values. Clusters characterized by ‘Medium’ or ‘High’ entropy for the scale 8–16 months (i.e., Clusters 1, 3, and 4) had larger precipitation values. Hence, mean monthly precipitation values and relative entropy showed an inverse relationship.\n\nAs an important issue, the connection between the DWE with latitude and longitude was investigated to indicate the spatial structure of the precipitation variation, which is shown in Figure 12. For DWE latitude, R2 = 0.227 and P-value = 0.37 (not significant) were calculated and a downward relation was observed; however, for DWE longitude, R2 = 0.22 and P-value = 0.34 (not significant) were calculated and an upward relation was observed. For both of the relations no significant trend was detected. It can be inferred from Figure 12 that multiscale precipitation variation (DWE) possesses the latitude zonality, which implies that precipitation variability increases with the latitude from the west to the east. On the other hand, decrease of DWE values from north to south was observed.\n\nResults showed the capability of the present methodology for precipitation regionalization. When accessibility to recorded precipitation time series is limited at the region of interest, regionalization methods might lead to incorrect results. For the case of the discrete wavelet, although the wavelet power spectrum has successfully been used for capturing hydrological time series behavior, it becomes difficult to use the wavelet spectrum in cases of limited or incomplete time series. Nevertheless, entropy provides information about the uncertainty at a given scale, which can highlight the level of variation present at that scale. Further, entropy enables the determination of least-biased probability distributions with limited time series knowledge. Entropy theory can serve as a useful approach to study hydrologic and meteorological processes (Mishra et al. 2009; Agarwal et al. 2016). Sang (2012) also showed the usefulness of applying DWE in precipitation-based studies.\n\nThe obtained results are applicable in local scale, since various factors can affect the outcome of the proposed model (e.g., geographic location, precipitation variation, effect of climatic phenomena, precipitation gauges network, etc.). Due to the existence of uncertainties and various factors it is suggested to apply the proposed model for various case studies and to compare the outcome. Also, it is suggested to validate the capability of the proposed model on other hydrologic and climatic variables (i.e., evapotranspiration, temperature, runoff, etc.) with various time scales (e.g., daily, annual, etc.).\n\nCONCLUSION\n\nIn this study, the spatio-temporal variability of monthly precipitation in Iran during 1960–2010 was investigated using DWE, and the pattern of DWE changes along with regionalization of rain gauges were further analyzed. In order to meet the objectives of this study 31 rain gauges were selected.\n\nIn order to have a correct vision of decomposing precipitation time series, smoother db mother wavelets were applied (db5–db10). Also, optimal decomposition level and boundary extension treatment were applied. In order to classify the rain gauges, SOM and k-means clustering models were used. The methodology based on the DWE approach k-means clustering technique for precipitation regionalization proved to be robust for hydrologic regionalization.\n\nWavelet-based multiscale entropy values showed the distinct variation of precipitation dynamics at each rain gauge and allowed for the establishment of homogeneous areas (with no prior assumptions). Most of the previous studies in precipitation regionalization delineated the rain gauges based on geographic proximity; however, the present study categorized rain gauges according to the uncertainties (entropy) in a multiscale approach. The DWE was useful circumstantial evidence in capturing the precipitation characteristics. The 31 rain gauges studied were clustered into five groups, each one having a unique DWE pattern across different time scales. Based on the pattern of mean DWE for each cluster, a characteristic signature was assigned, which provided an estimation of DWE of a cluster across scales 2–8, 8–16, and 32–64 months relative to other stations. Fluctuations in DWE at different scales in this study were related to monthly precipitation.\n\nResults showed the capability of the present methodology for precipitation regionalization. When accessibility to recorded precipitation time series is limited at the region of interest, regionalization methods might lead to incorrect results. For the case of the discrete wavelet, although the wavelet power spectrum has successfully been used for capturing hydrological time series behavior, it becomes difficult to use the wavelet spectrum in cases of limited time series. Nevertheless, entropy provides information about the uncertainty at a given scale, which can highlight the level of variation present at that scale. Further, entropy enables the determination of least-biased probability distributions with limited time series knowledge. Entropy theory can serve as a useful approach to study hydrologic and meteorological processes (Mishra et al. 2009; Agarwal et al. 2016). 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B.\n&\nPollard\n,\nD.\n1999\nNorthern hemisphere ice-sheet influences on global climate change\n.\nScience\n286\n,\n1104\n1111\n.\nDanandeh Mehr\n,\nA.\n,\nKahya\n,\nE.\n,\nŞahin\n,\nA.\n&\n,\nM. J.\n2015\nSuccessive-station monthly streamflow prediction using different artificial neural network algorithms\n.\nInternational Journal of Environmental Science and Technology\n12\n(\n7\n),\n2191\n2200\n.\nDanesh-Yazdi\n,\nM.\n,\nTejedor\n,\nA.\n&\nFoufoula-Georgiou\n,\nE.\n2017\nSelf-dissimilar landscapes: probing into the causes and consequences via multi-scale analysis and synthesis\n.\nGeomorphology\n296\n,\n16\n27\n.\nDavies\n,\nD. L.\n&\nBouldin\n,\nD. W.\n1979\nA cluster separation measure\n.\nIEEE Transactions on Pattern Analysis and Machine Intelligence\n1\n(\n2\n),\n224\n227\n.\nde Artigas\n,\nM. Z.\n,\nElias\n,\nA. G.\n&\nde Campra\n,\nP. F.\n2006\nDiscrete wavelet analysis to assess long-term trends in geomagnetic activity\n.\nPhysics and Chemistry of the Earth\n31\n(\n1–3\n),\n77\n80\n.\nDinpashoh\n,\nY.\n,\nFakheri-Fard\n,\nA.\n,\n,\nM.\n,\nJahanbakhsh\n,\nS.\n&\nMirnia\n,\nM.\n2004\nSelection of variables for the purpose of regionalization of Iran's precipitation climate using multivariate methods\n.\nJournal of Hydrology\n29\n,\n109\n123\n.\nDomroes\n,\nM.\n,\nKaviani\n,\nM.\n&\nSchaefer\n,\nD.\n1998\nAn analysis of regional and intra-annual precipitation variability over Iran using multivariate statistical methods\n.\nTheoretical and Applied Climatology\n61\n,\n151\n159\n.\nDong\n,\nX.\n,\nNyren\n,\nP.\n,\nPatton\n,\nB.\n,\nNyren\n,\nA.\n,\nRichardson\n,\nJ.\n&\nMaresca\n,\nT.\n2008\nWavelets for agriculture and biology: a tutorial with applications and outlook\n.\nBioscience\n58\n(\n5\n),\n445\n453\n.\nDonoho\n,\nD. H.\n1995\nDe-noising by soft-thresholding\n.\nIEEE Transactions on Information Theory\n41\n(\n3\n),\n613\n617\n.\nIwashita\n,\nF.\n,\nFriede\n,\nM. J.\n,\nFrancisco\n,\nJ.\n&\nFerreira\n,\nJ. F.\n2018\nA self-organizing map approach to characterize hydrogeology of the fractured Serra-Geral transboundary aquifer\n.\nHydrology Research\n49\n(\n3\n),\n794\n814\n.\ndoi:10.2166/nh.2017.221\n.\nJaynes\n,\nE. T.\n1957\nInformation theory and statistical mechanics\n.\nPhysics Review\n106\n,\n620\n630\n.\nKallache\n,\nM.\n,\nRust\n,\nH. W.\n&\nKropp\n,\nJ.\n2005\nTrend assessment: applications for hydrology and climate research\n.\nNonlinear Processes in Geophysics\n12\n(\n2\n),\n201\n210\n.\nKalteh\n,\nA. M.\n,\nHjorth\n,\nP.\n&\nBerndtsson\n,\nR.\n2008\nReview of self-organizing map in water resources: analysis, modeling, and application\n.\nEnvironmental Modelling and Software\n23\n,\n835\n845\n.\nKarimi\n,\nS.\n,\nShiri\n,\nJ.\n,\nKisi\n,\nO.\n&\nShiri\n,\nA. A.\n2016\nShort-term and long-term streamflow prediction by using ‘wavelet-gene expression’ programming approach\n.\nISH Journal of Hydraulic Engineering\n22\n(\n2\n),\n148\n162\n.\nKasturi\n,\nJ.\n,\nAcharya\n,\nJ.\n&\nRamanathan\n,\nM.\n2003\nAn information theoretic approach for analyzing temporal patterns of gene expression\n.\nBioinformatics\n19\n(\n4\n),\n449\n458\n.\nKisi\n,\nO.\n&\nShiri\n,\nJ.\n2011\nPrecipitation forecasting using wavelet-genetic programming and wavelet-neuro-fuzzy conjunction models\n.\nWater Resources Management\n25\n(\n13\n),\n3135\n3152\n.\nKisi\n,\nO.\n&\nShiri\n,\nJ.\n2012\nWavelet and neuro-fuzzy conjunction model for predicting water table depth fluctuations\n.\nHydrology Research\n43\n(\n3\n),\n286\n300\n.\nKohonen\n,\nT.\n1997\nSelf-organizing Maps\n.\nSpringer-Verlag\n,\nBerlin\n.\nLi\n,\nZ. W.\n&\nZhang\n,\nY. K.\n2008\nMulti-scale entropy analysis of Mississippi River flow\n.\nStochastic Environmental Research and Risk Assessment\n22\n,\n507\n512\n.\nMacQueen\n,\nJ.\n1967\nSome methods for classification and analysis of multivariate observations\n.\nProceeding of Fifth Berkeley Symposium on Mathematical Statistics and Probability\n1\n,\n281\n297\n.\nMaheswaran\n,\nR.\n&\nKhosa\n,\nR.\n2012\nComparative study of different wavelets for hydrologic forecasting\n.\nComputers and Geosciences\n46\n,\n284\n295\n.\nMehr\n,\nA. D.\n,\nKahya\n,\nE.\n&\nOlyaei\n,\nE.\n2013\nStreamflow prediction using linear genetic programming in comparison with a neuro-wavelet technique\n.\nJournal of Hydrology\n505\n,\n240\n249\n.\nMehr\n,\nA. D.\n,\nKahya\n,\nE.\n&\nOzger\n,\nM.\n2014\nA gene–wavelet model for long lead time drought forecasting\n.\nJournal of Hydrology\n517\n,\n691\n699\n.\nMishra\n,\nA. K.\n,\nÖzger\n,\nM.\n&\nSingh\n,\nV. P.\n2009\nAn entropy-based investigation into the variability of precipitation\n.\nJournal of Hydrology\n370\n,\n139\n154\n.\nModarres\n,\nR.\n2006\nRegional precipitation climates of Iran\n.\nJournal of Hydrology: New Zealand\n45\n(\n1\n),\n13\n27\n.\nModarres\n,\nR.\n&\n,\nA.\n2009\nRainfall trends analysis of Iran in the last half of the twentieth century\n.\nJournal of Geophysical Research\n114\n,\nD03101\n.\nMolini\n,\nA.\n,\nBarbera\n,\nP. L.\n&\nLanza\n,\nL. G.\n2006\nCorrelation patterns and information flows in rainfall fields\n.\nJournal of Hydrology\n322\n,\n89\n104\n.\nMun\n,\nF. K.\n2004\nTime Series Forecasting Using Wavelet and Support Vector Machine\n.\nMS Thesis\n,\nNational University of Singapore\n,\nSingapore\n.\nNagarajan\n,\nR.\n2010\nDrought Assessment\n.\nSpringer Science & Business Media\n,\nNew York\n, p.\n383\n.\nNourani\n,\nV.\n&\nPartoviyan\n,\nA.\n2017\nHybrid denoising-jittering data pre-processing approach to enhance multi-step-ahead rainfall–runoff modeling\n.\nStochastic Environmental Research and Risk Assessment\n1\n18\n.\nNourani\n,\nV.\n,\nKomasi\n,\nM.\n&\nMano\n,\nA.\n2009\nA multivariate ANN-wavelet approach for rainfall–runoff modeling\n.\nWater Resources Management\n23\n(\n14\n),\n2877\n2894\n.\nNourani\n,\nV.\n,\nHosseini Baghanam\n,\nA.\n,\n,\nJ.\n&\nKisi\n,\nO.\n2014\nApplications of hybrid wavelet – Artificial intelligence models in hydrology: a review\n.\nJournal of Hydrology\n514\n,\n358\n377\n.\nNourani\n,\nV.\n,\nAlami\n,\nM. T.\n&\nVousoughi Daneshvar\n,\nF.\n2015\nWavelet-entropy data pre-processing approach for ANN-based groundwater level modeling\n.\nJournal of Hydrology\n524\n,\n255\n269\n.\nPartal\n,\nT.\n2010\nWavelet transform-based analysis of periodicities and trends of Sakarya basin (Turkey) streamflow data\n.\nRiver Research and Applications\n26\n(\n6\n),\n695\n711\n.\nPechlivanidis\n,\nI. G.\n,\nMcIntyre\n,\nN.\n&\nWheater\n,\nH. S.\n2017\nThe significance of spatial variability of rainfall on simulated runoff: an evaluation based on the Upper Lee catchment, UK\n.\nHydrology Research\n48\n(\n4\n),\n1118\n1130\n.\nPopivanov\n,\nI.\n&\nMiller\n,\nR. J.\n2002\nSimilarity search over time-series data using wavelets\n. In:\nProceedings 18th International Conference on Data Engineering\n,\nWashington, DC\n, pp.\n212\n221\n.\nRao\n,\nA. R.\n&\nSrinivas\n,\nV. V.\n2008\nRegionalization of Watersheds: an Approach Based on Cluster Analysis\n, Vol.\n58\n.\nSpringer Science & Business Media\n,\nNew York\n.\nRaziei\n,\nT.\n2017\nA precipitation regionalization and regime for Iran based on multivariate analysis\n.\nTheoretical and Applied Climatology\n131\n(\n3–4\n),\n1429\n1448\n.\nRaziei\n,\nT.\n,\nBordi\n,\nI.\n&\nPereira\n,\nL. S.\n2008\nA precipitation-based regionalization for Western Iran and regional drought variability\n.\nHydrology and Earth System Sciences\n12\n,\n1309\n1321\n.\nRokach\n,\nL.\n&\nMaimon\n,\nO.\n, (eds).\n2005\nClustering methods\n. In:\nData Mining and Knowledge Discovery Handbook\n.\nSpringer\n,\nNew York\n, pp.\n321\n352\n.\nSaboohi\n,\nR.\n,\nSoltani\n,\nS.\n&\nKhodagholi\n,\nM.\n2012\nTrend analysis of temperature parameters in Iran\n.\nTheoretical and Applied Climatology\n109\n,\n529\n547\n.\nSalvia\n,\nK.\n,\nVillarinia\n,\nG.\n&\nVecchib\n,\nG. A.\n2017\nHigh resolution decadal precipitation predictions over the continental United States for impacts assessment\n.\nJournal of Hydrology\n553\n,\n559\n573\n.\nSang\n,\nY. F.\n,\nWang\n,\nD.\n,\nWu\n,\nJ. C.\n,\nZhu\n,\nQ. P.\n&\nWang\n,\nL.\n2011\nWavelet-based analysis on the complexity of hydrologic series data under multi-temporal scales\n.\nEntropy\n13\n,\n195\n210\n.\nSattari\n,\nM.-T.\n,\n,\nA.\n&\nKusiak\n,\nA.\n2017\nAssessment of different methods for estimation of missing data in precipitation studies\n.\nHydrology Research\n48\n(\n4\n),\n1032\n1044\n.\nSoltani\n,\nS.\n,\nModarres\n,\nR.\n&\nEslamian\n,\nS. S.\n2007\nThe use of time series modelling for the determination of rainfall climates of Iran\n.\nInternational Journal of Climatology\n27\n,\n819\n829\n.\nSu\n,\nH.\n,\nLiu\n,\nQ.\n&\nLi\n,\nJ.\n2011\nAlleviating border effects in wavelet transforms for nonlinear time-varying signal analysis\n.\nAdvances in Electrical and Computer Engineering\n11\n(\n3\n),\n55\n60\n.\nTabari\n,\nH.\n&\n,\nP.\n2011\nAnalysis of trends in temperature data in arid and semi-arid regions of Iran\n.\nGlobal and Planetary Change\n79\n,\n1\n10\n.\nTermini\n,\nD.\n&\nMoramarco\n,\nT.\n2016\nApplication of entropic approach to estimate the mean flow velocity and Manning roughness coefficient in a high-curvature flume\n.\nHydrology Research\nnh2016106. doi: 10.2166/nh.2016.106\n.\nVonesch\n,\nC.\n,\nBlu\n,\nT.\n&\nUnser\n,\nM.\n2007\nGeneralized Daubechies wavelet families\n.\nIEEE Transactions on Signal Processing\n55\n(\n9\n),\n4415\n4429\n.\nWeather and Climate Information\n2015\nWeather and Climate: Iran, Average Monthly Rainfall, Sunshine, Temperature, Humidity and Wind Speed\n.\nWorld Weather and Climate Information\n,\nThe Netherlands\n.\nWei\n,\nQ.\n,\nSun\n,\nC.\n,\nWu\n,\nG.\n&\nPan\n,\nL.\n2017\nHaihe River discharge to Bohai Bay, North China: trends, climate, and human activities\n.\nHydrology Research\n48\n(\n4\n),\n1058\n1070\n.\nWerstuck\n,\nC.\n&\nCoulibaly\n,\nP.\n2016\nHydrometric network design using dual entropy multi-objective optimization in the Ottawa River Basin\n.\nHydrology Research\nnh2016344. doi: 10.2166/nh.2016.344\n.\nZunino\n,\nL.\n,\nPerez\n,\nD. G.\n,\nGaravaglia\n,\nM.\n&\nRosso\n,\nO. A.\n2007\nWavelet entropy of stochastic processes\n.\nPhysics A\n379\n,\n503\n512\n." ]

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[ ">\nSwitch to:\n\n# Netflix EBIT\n\n: \\$5,856 Mil (TTM As of Mar. 2021)\nView and export this data going back to 2002. Start your Free Trial\n\nNetflix's earnings before interest and taxes (EBIT) for the three months ended in Mar. 2021 was \\$2,229 Mil. Its earnings before interest and taxes (EBIT) for the trailing twelve months (TTM) ended in Mar. 2021 was \\$5,856 Mil.\n\nEBIT or Operating Income is linked to Return on Capital for both regular definition and Joel Greenblatt's definition. Netflix's annualized ROC % for the quarter that ended in Mar. 2021 was 18.34%. Netflix's annualized ROC (Joel Greenblatt) % for the quarter that ended in Mar. 2021 was 902.59%.\n\nEBIT is also linked to Joel Greenblatt's definition of earnings yield. Netflix's Earnings Yield (Joel Greenblatt) % for the quarter that ended in Mar. 2021 was 2.57%.\n\n## Netflix EBIT Historical Data\n\n* For Operating Data section: All numbers are indicated by the unit behind each term and all currency related amount are in USD.\n* For other sections: All numbers are in millions except for per share data, ratio, and percentage. All currency related amount are indicated in the company's associated stock exchange currency.\n\n* Premium members only.\n\n## Netflix EBIT Calculation\n\nEBIT, sometimes also called Earnings Before Interest and Taxes, is a measure of a firm's profit that includes all expenses except interest and income tax expenses. It is the difference between operating revenues and operating expenses. When a firm does not have non-operating income, then Operating Income is sometimes used as a synonym for EBIT and operating profit.\n\nEBIT for the trailing twelve months (TTM) ended in Mar. 2021 was 1357.928 (Jun. 2020 ) + 1314.863 (Sep. 2020 ) + 954.242 (Dec. 2020 ) + 2228.942 (Mar. 2021 ) = \\$5,856 Mil.\n\n* For Operating Data section: All numbers are indicated by the unit behind each term and all currency related amount are in USD.\n* For other sections: All numbers are in millions except for per share data, ratio, and percentage. All currency related amount are indicated in the company's associated stock exchange currency.\n\nNetflix  (NAS:NFLX) EBIT Explanation\n\n1. EBIT or Operating Income is linked to Return on Capital for both regular definition and Joel Greenblatt's definition.\n\nNetflix's annualized ROC % for the quarter that ended in Mar. 2021 is calculated as:\n\n ROC % (Q: Mar. 2021 ) = NOPAT / Average Invested Capital = Operating Income * ( 1 - Tax Rate % ) / ( (Invested Capital (Q: Dec. 2020 ) + Invested Capital (Q: Mar. 2021 )) / count ) = 7839.424 * ( 1 - 16.11% ) / ( (35566.185 + 36152.522) / 2 ) = 6576.4927936 / 35859.3535 = 18.34 %\n\nwhere\n\n Invested Capital (Q: Dec. 2020 ) = Total Assets - Accounts Payable & Accrued Expense - Excess Cash = Total Assets - Accounts Payable & Accrued Expense - ( Cash, Cash Equivalents, Marketable Securities - max(0, Total Current Liabilities - Total Current Assets + Cash, Cash Equivalents, Marketable Securities )) = 39280.359 - 1758.379 - ( 8205.55 - max(0, 7805.785 - 9761.58 + 8205.55 )) = 35566.185\n\n Invested Capital (Q: Mar. 2021 ) = Total Assets - Accounts Payable & Accrued Expense - Excess Cash = Total Assets - Accounts Payable & Accrued Expense - ( Cash, Cash Equivalents, Marketable Securities - max(0, Total Current Liabilities - Total Current Assets + Cash, Cash Equivalents, Marketable Securities )) = 40123.014 - 1824.754 - ( 8403.705 - max(0, 7961.77 - 10107.508 + 8403.705 )) = 36152.522\n\nNote: The Operating Income data used here is four times the quarterly (Mar. 2021) data.\n\n2. Joel Greenblatt's definition of Return on Capital:\n\nNetflix's annualized ROC (Joel Greenblatt) % for the quarter that ended in Mar. 2021 is calculated as:\n\n ROC (Joel Greenblatt) % (Q: Mar. 2021 ) = EBIT / Average of (Net fixed Assets + Net Working Capital) = EBIT / Average of (Property, Plant and Equipment + Net Working Capital) Q: Dec. 2020 Q: Mar. 2021 = EBIT / ( ( (Property, Plant and Equipment + Net Working Capital) + (Property, Plant and Equipment + Net Working Capital) ) / count ) = 8915.768 / ( ( (960.183 + max(-5749.877, 0)) + (1015.419 + max(-5559.179, 0)) ) / 2 ) = 8915.768 / ( ( 960.183 + 1015.419 ) / 2 ) = 8915.768 / 987.801 = 902.59 %\n\nwhere Working Capital is:\n\n Working Capital (Q: Dec. 2020 ) = (Accounts Receivable + Total Inventories + Other Current Assets) - (Accounts Payable & Accrued Expense + Defer. Rev. + Other Current Liabilities) = (610.819 + 0 + 945.211) - (1758.379 + 1117.992 + 4429.536) = -5749.877\n\n Working Capital (Q: Mar. 2021 ) = (Accounts Receivable + Total Inventories + Other Current Assets) - (Accounts Payable & Accrued Expense + Defer. Rev. + Other Current Liabilities) = (807.036 + 0 + 896.767) - (1824.754 + 1140.271 + 4297.957) = -5559.179\n\nWhen net working capital is negative, 0 is used.\n\nNote: The EBIT data used here is four times the quarterly (Mar. 2021) EBIT data.\n\n3. It is also linked to Joel Greenblatt's definition of Earnings Yield:\n\nNetflix's Earnings Yield (Joel Greenblatt) % for today is calculated as:\n\n Earnings Yield (Joel Greenblatt) % = EBIT (TTM) / Enterprise Value (Q: Mar. 2021 ) = 5855.975 / 228120.955 = 2.57 %\n\n* For Operating Data section: All numbers are indicated by the unit behind each term and all currency related amount are in USD.\n* For other sections: All numbers are in millions except for per share data, ratio, and percentage. All currency related amount are indicated in the company's associated stock exchange currency.\n\n## Netflix EBIT Headlines\n\nNo Headline\n\nGet WordPress Plugins for easy affiliate links on Stock Tickers and Guru Names | Earn affiliate commissions by embedding GuruFocus Charts\nGuruFocus Affiliate Program: Earn up to \\$400 per referral. ( Learn More)" ]

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[ "# mergeSort<E> function Null safety\n\nvoid mergeSort<E>(\n1. List<E> elements,\n2. {int start = 0,\n3. int? end,\n4. int compare(\n1. E,\n2. E\n)?}\n)\n\nSorts a list between `start` (inclusive) and `end` (exclusive) using the merge sort algorithm.\n\nIf `compare` is omitted, this defaults to calling Comparable.compareTo on the objects. If any object is not Comparable, that throws a TypeError.\n\nMerge-sorting works by splitting the job into two parts, sorting each recursively, and then merging the two sorted parts.\n\nThis takes on the order of `n * log(n)` comparisons and moves to sort `n` elements, but requires extra space of about the same size as the list being sorted.\n\nThis merge sort is stable: Equal elements end up in the same order as they started in.\n\n## Implementation\n\n``````void mergeSort<E>(List<E> elements,\n{int start = 0, int? end, int Function(E, E)? compare}) {\nend = RangeError.checkValidRange(start, end, elements.length);\ncompare ??= defaultCompare;\n\nvar length = end - start;\nif (length < 2) return;\nif (length < _mergeSortLimit) {\ninsertionSort(elements, compare: compare, start: start, end: end);\nreturn;\n}\n// Special case the first split instead of directly calling\n// _mergeSort, because the _mergeSort requires its target to\n// be different from its source, and it requires extra space\n// of the same size as the list to sort.\n// This split allows us to have only half as much extra space,\n// and allows the sorted elements to end up in the original list.\nvar firstLength = (end - start) >> 1;\nvar middle = start + firstLength;\nvar secondLength = end - middle;\n// secondLength is always the same as firstLength, or one greater.\nvar scratchSpace = List<E>.filled(secondLength, elements[start]);\n// TODO(linter/2097): Remove ignore when no longer required by linter.\n// See: https://github.com/dart-lang/linter/issues/2097\nE Function(E) id = identity; // ignore: omit_local_variable_types\n_mergeSort(elements, id, compare, middle, end, scratchSpace, 0);\nvar firstTarget = end - firstLength;\n_mergeSort(elements, id, compare, start, middle, elements, firstTarget);\n_merge(id, compare, elements, firstTarget, end, scratchSpace, 0, secondLength,\nelements, start);\n}``````" ]

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[ "# Boundary value problem, complex-variable methods\n\nMethods for studying boundary value problems for partial differential equations in which one uses representations of solutions in terms of analytic functions of a complex variable.\n\nGiven a second-order elliptic equation\n\n$$\\tag{1 } \\Delta u + a (x, y) \\frac{\\partial u }{\\partial x } + b (x, y) \\frac{\\partial u }{\\partial y } + c (x, y) u = 0,$$\n\nwhere $a, b, c$ are analytic functions of the real variables $x, y$ in some domain $D _ {0}$ of the $z$- plane, $z = x + iy$, consider the following boundary value problem: Find a solution of equation (1), regular in a simply-connected domain $S \\subset D _ {0}$, satisfying the boundary condition\n\n$$\\tag{2 } R (u) \\equiv \\ \\sum _ {0 \\leq j + k \\leq m } \\left [ a _ {jk} (t) \\frac{\\partial ^ {j + k } u }{\\partial x ^ {j} \\partial y ^ {k} } + T _ {jk} \\left ( \\frac{\\partial ^ {j + k } u }{\\partial x ^ {j} \\partial y ^ {k} } \\right ) \\right ] = f (t),$$\n\nwhere $a _ {jk} (t), f (t) \\in C _ \\alpha ( \\partial S), 0 < \\alpha < 1$, and the $T _ {jk}$ are linear operators mapping $C _ \\alpha ( \\partial S)$ into $C _ \\alpha ( \\partial S)$, with $T _ {m - k, k }$ completely continuous.\n\nThis problem includes as special cases the well-known classical boundary value problems of Dirichlet, Neumann, Poincaré, etc.\n\nUsing the formula for the general representation of solutions (see Differential equation, partial, complex-variable methods),\n\n$$u (x, y) = \\ \\mathop{\\rm Re} \\left \\{ G (z, \\overline{z}\\; _ {0} , z, \\overline{z}\\; ) \\Phi (z)\\right . -$$\n\n$$- \\left . \\int\\limits _ {z _ {0} } ^ { {z } } \\Phi (t) \\frac \\partial {\\partial t } G (t, \\overline{z}\\; _ {0} , z, \\overline{z}\\; ) dt \\right \\} ,$$\n\none reduces the problem to an equivalent boundary value problem for analytic functions:\n\n$$\\tag{3 } \\mathop{\\rm Re} \\sum _ {k = 0 } ^ { m } [a _ {k} (t) \\Phi ^ {(k)} (t) + T _ {k} ( \\Phi ^ {(k)} )] = f (t),$$\n\nwhere $a _ {k} (t)$ are given Hölder-continuous functions, $t \\in \\partial S$, $T _ {m}$ is a completely-continuous operator, and the $T _ {k}$ $(k = 0 \\dots m - 1 )$ are linear operators.\n\nSuppose that the finite simply-connected domain $S$ is bounded by a closed Lyapunov contour (see Lyapunov surfaces and curves) $\\partial S$ and that the $m$- th derivative, $m \\geq 0$, of the function $\\Phi (z)$( the latter is holomorphic in $S$), restricted to $\\partial S$, is a function of class $C _ \\alpha$, $0 < \\alpha < 1$. Then, assuming that the point $z = 0$ is in $S$, one can express $\\Phi (z)$ as follows:\n\n$$\\Phi (z) = \\ \\int\\limits _ {\\partial S } \\frac{t \\mu (t) dt }{t - z } + ic,\\ \\textrm{ if } m = 0;$$\n\n$$\\Phi (z) = \\int\\limits _ {\\partial S } \\mu (t) \\left ( 1 - { \\frac{z}{t} } \\right ) ^ {m - 1 } \\mathop{\\rm ln} \\left ( 1 - { \\frac{z}{t} } \\right ) dt +$$\n\n$$+ \\int\\limits _ {\\partial S } \\mu (t) dt + ic,\\ \\textrm{ if } m \\geq 1,$$\n\nwhere $\\mu (t)$ is a real function of class $C _ \\alpha ( \\partial S)$, $0 < \\alpha < 1$, and $c$ is a real constant; $\\mu (t)$ and $c$ are uniquely determined by $\\Phi (z)$.\n\nSubstituting these expressions into the boundary condition (3), one obtains a singular integral equation, equivalent to problem (2), for the unknown function $\\mu$:\n\n$$K ( \\mu ) = \\ A (t _ {0} ) \\mu ( t _ {0} ) + \\frac{B (t _ {0} ) }{\\pi i } \\int\\limits _ {\\partial S } \\frac{\\mu (t) dt }{t - t _ {0} } + T ( \\mu ) = f (t _ {0} ),$$\n\n$t _ {0} \\in \\partial S$, where $T$ is a completely-continuous operator.\n\nA necessary and sufficient condition for the boundary value problem (2) to be normally solvable is that\n\n$$\\tag{4 } a (t) = \\ \\sum _ {k = 0 } ^ { m } i ^ {k} a _ {m - k } (t) \\neq 0,\\ \\ t \\in \\partial S,\\ \\ m \\geq 0.$$\n\nThe Dirichlet problem ( $m = 0$) is always normally solvable. (Henceforth it is assumed throughout that condition (4) is satisfied.)\n\nThe index of the boundary value problem (2) is computed from the formula\n\n$$\\kappa = \\ 2 (m + p),\\ \\ m \\geq 1,$$\n\nwhere $p$ is the increment of the function $(1/2 \\pi ) \\mathop{\\rm arg} \\overline{ {a (t) }}\\;$ when the contour $\\partial S$ is described once in the positive sense. The index of the Dirichlet problem is zero.\n\nThe homogeneous boundary value problem $R (u) = 0$ has a finite number $k \\geq 0$ of linearly independent solutions, where $k \\geq \\kappa$; the inhomogeneous problem (2) has a solution if and only if\n\n$$\\int\\limits _ {\\partial S } f (t) \\nu _ {j} (t) dS = 0,\\ \\ j = 1 \\dots \\overline{k}\\; ,$$\n\nwhere $\\nu _ {j}$ is a complete system of linearly independent solutions of the associated homogeneous integral equation\n\n$$k ^ \\prime ( \\nu ) = \\ A (t _ {0} ) \\nu (t _ {0} ) - \\frac{1}{\\pi i } \\int\\limits _ {\\partial S } \\frac{B (t) \\nu (t) dt }{t - t _ {0} } + T ( \\nu ) = 0.$$\n\nThe boundary value problem (2) has a solution, whatever the free term on the right, if and only if there exist exactly $\\kappa$ linearly independent solutions of the associated homogeneous problem $R (u) = 0$. Consequently, if $\\kappa > 0$ the homogeneous boundary value problem $R (u) = 0$ always has at least $\\kappa$ linearly independent solutions; if $\\kappa < 0$ the inhomogeneous problem (2) is not solvable for an arbitrary free term on the right, but at least $| \\kappa |$ solvability conditions must be satisfied.\n\nNecessary and sufficient conditions for the solvability of an inhomogeneous boundary value problem may be formulated in terms of the completeness of a certain system of functions. The kernel and system of functions may be constructed explicitly using the Riemann function of equation (1) and the coefficients of the boundary conditions. For example, let $\\{ u _ {k} \\}$ be some complete system of solutions in the basic domain $D _ {0}$ of equation (1), and let $S \\subset D _ {0}$. Then a necessary and sufficient condition for problem (2) to be solvable for any free right-hand side is that the system of functions $\\{ R (u _ {k} ) \\}$ be complete on the boundary.\n\nVery complete results have been obtained for the following boundary value problem (the generalized Riemann–Hilbert problem): Find a solution of the equation\n\n$$\\tag{5 } \\partial _ {\\overline{z}\\; } w + A (z) w + B (z) \\overline{w}\\; = f (z),\\ \\ w = u + iv,$$\n\n$$2 \\partial _ {\\overline{z}\\; } = \\partial _ {x} - i \\partial _ {y} ,$$\n\nwhich is continuous in $S + \\partial S$ and satisfies the boundary condition\n\n$$\\tag{6 } \\mathop{\\rm Re} [ \\overline{ {\\lambda (z) }}\\; , w (z)] \\equiv \\ \\alpha u + \\beta v = \\gamma ,\\ \\ z \\in \\partial S,$$\n\nwhere $\\alpha , \\beta , \\gamma$ are real functions of class $C _ \\alpha ( \\partial S)$, $0 < \\alpha < 1$, with $\\alpha ^ {2} + \\beta ^ {2} = 1$. The domain $S$ may be multiply connected. A problem of this type may be reduced to an equivalent singular integral equation; this yields a full qualitative analysis of the boundary value problem (6).\n\nSuppose that the boundary $\\partial S$ of $S$ is the union of a finite number of simple closed curves $\\partial S _ {0} \\dots \\partial S _ {m}$ satisfying the Lyapunov conditions. Since the forms of the equation and the boundary condition are preserved under conformal mapping, it may be assumed without loss of generality that $\\partial S _ {0}$ is the unit circle with centre at $z = 0$, the latter being a point of $S$, while $\\partial S _ {1} \\dots \\partial S _ {m}$ are circles lying outside $\\partial S _ {0}$.\n\nThe index of problem (6) is defined as the integer $n$ equal to the increment of $(1/2 \\pi ) \\mathop{\\rm arg} [ \\alpha ( \\zeta ) + i \\beta ( \\zeta )]$ when the point $\\zeta$ describes $\\partial S$ once in the positive sense. The boundary condition can be reduced to the simpler form\n\n$$\\mathop{\\rm Re} [z ^ {-n} e ^ {ic (z) } w (z)] = \\gamma ,\\ \\ z \\in \\partial S,$$\n\nwhere $c (z) = c _ {j}$ on $\\partial S _ {j}$, with $c _ {0} = 0$, while $c _ {1} \\dots c _ {m}$ are real constants, uniquely expressible in terms of $\\alpha$ and $\\beta$. The index of the adjoint problem\n\n$$\\tag{7 } \\left . \\begin{array}{ll} \\partial _ {\\overline{z}\\; } w _ {*} - Aw _ {*} - B \\overline{w}\\; _ {*} = 0, & z \\in S, \\\\ \\mathop{\\rm Re} \\left \\{ ( \\alpha + i \\beta ) \\frac{d \\overline{z}\\; }{ds} w _ {*} (z) \\right \\} = 0, & z \\in \\partial S, \\\\ \\end{array} \\right \\}$$\n\nis calculated by the formula $n ^ \\prime = - n + m - 1$.\n\nProblem (6) has a solution if and only if\n\n$$\\int\\limits _ {\\partial S } ( \\alpha + i \\beta ) w _ {*} \\gamma ds = 0,$$\n\nwhere $w _ {*}$ is an arbitrary solution of the adjoint problem.\n\nLet $e$ and $e ^ \\prime$ be the numbers of linearly independent solutions of the homogeneous problems (6) and (7), respectively. Then\n\n$$e - e ^ \\prime = \\ n - n ^ \\prime = \\ 2n + 1 - m.$$\n\nIf $n < 0$, the homogeneous problem (6) has no non-trivial solutions. If $n > m - 1$, it has exactly $e = 2n + 1 - m$ linearly independent solutions, while the inhomogeneous problem (6) is always solvable. If $n < 0$, the inhomogeneous problem (6) is solvable if and only if\n\n$$\\int\\limits _ {\\partial S } ( \\alpha + i \\beta ) w _ {* j } \\gamma ds = 0,\\ \\ j = 1, 2 ,\\dots ; \\ \\ e ^ \\prime = m - 2n - 1,$$\n\nwhere $w _ {* j }$ is a complete system of solutions of the homogeneous problem (7). If $m = 0$ and $n = 0$, then $e = 1$ and all solutions of the homogeneous problem problem (6) are given by\n\n$$w (z) = \\ ice ^ {\\omega _ {0} (z) } ,$$\n\nwhere $c$ is a real constant and $\\omega _ {0}$ a continuous function on $S + \\partial S$.\n\nThe above results completely characterize the problem in the simply-connected ( $m = 0$) and multiply-connected ( $n < 0$, $n > m - 1$) cases. The cases $0 \\leq n \\leq m - 1$ require special examination; they have also been worked out in considerable detail.\n\nBoundary value problems of the type of the Poincaré problem have also been studied for equation (5).\n\nFor references see Differential equation, partial, complex-variable methods." ]

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[ "# Heat produced if earth stops rotating\n\nIn my textbook there is a question that is as follows:- If earth stops rotating about it's own axis,the increase in its temprature will be(Here R=radius of earth,ω=angular velocity of earth,J=mechanical equivalent of heat,C=average specific heat capacity of earth) Here i have doubt why heat is produced if earth stops rotating?and please explain how to solve it.\n\n• The question seems to assume, for some unexplained reason, that the rotational energy will become thermal energy. – G. Smith Sep 24 at 16:17\n• As it further assumes we know the Earth moment of inertia, what requires to know dependence of Earth density on radius., – Poutnik Sep 24 at 16:25\n• Since the density profile isn’t specified, the question must be assuming uniform density. – G. Smith Sep 24 at 16:58" ]

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[ "1-D Kinematics - Lesson 4 - Describing Motion with Velocity vs. Time Graphs\n\n# The Meaning of Slope for a v-t Graph\n\nAs discussed in the previous part of Lesson 4, the shape of a velocity versus time graph reveals pertinent information about an object's acceleration. For example, if the acceleration is zero, then the velocity-time graph is a horizontal line (i.e., the slope is zero). If the acceleration is positive, then the line is an upward sloping line (i.e., the slope is positive). If the acceleration is negative, then the velocity-time graph is a downward sloping line (i.e., the slope is negative). If the acceleration is great, then the line slopes up steeply (i.e., the slope is great). This principle can be extended to any motion conceivable. Thus the shape of the line on the graph (horizontal, sloped, steeply sloped, mildly sloped, etc.) is descriptive of the object's motion. In this part of the lesson, we will examine how the actual slope value of any straight line on a velocity-time graph is the acceleration of the object.\n\n### Analyzing a Constant Velocity Motion\n\nConsider a car moving with a constant velocity of +10 m/s. A car moving with a constant velocity has an acceleration of 0 m/s/s.", null, "", null, "The velocity-time data and graph would look like the graph below. Note that the line on the graph is horizontal. That is the slope of the line is 0 m/s/s. In this case, it is obvious that the slope of the line (0 m/s/s) is the same as the acceleration (0 m/s/s) of the car.\n\nTime\n(s)\nVelocity\n(m/s)\n0 10\n1 10\n2 10\n3 10\n4 10\n5 10", null, "So in this case, the slope of the line is equal to the acceleration of the velocity-time graph. Now we will examine a few other graphs to see if this is a principle that is true of all velocity versus time graphs.\n\n### Analyzing a Changing Velocity Motion\n\nNow consider a car moving with a changing velocity. A car with a changing velocity will have an acceleration.", null, "", null, "The velocity-time data for this motion show that the car has an acceleration value of 10 m/s/s. (In Lesson 6, we will learn how to relate position-time data such as that in the diagram above to an acceleration value.) The graph of this velocity-time data would look like the graph below. Note that the line on the graph is diagonal - that is, it has a slope. The slope of the line can be calculated as 10 m/s/s. It is obvious once again that the slope of the line (10 m/s/s) is the same as the acceleration (10 m/s/s) of the car.\n\nTime\n(s)\nVelocity\n(m/s)\n0 0\n1 10\n2 20\n3 30\n4 40\n5 50", null, "### Analyzing a Two-Stage Motion\n\nIn both instances above - the constant velocity motion and the changing velocity motion, the slope of the line was equal to the acceleration. As a last illustration, we will examine a more complex case - a two-stage motion. Consider the motion of a car that first travels with a constant velocity (a=0 m/s/s) of 2 m/s for four seconds and then accelerates at a rate of +2 m/s/s for four seconds. That is, in the first four seconds, the car is not changing its velocity (the velocity remains at 2 m/s) and then the car increases its velocity by 2 m/s per second over the next four seconds. The velocity-time data and graph are displayed below. Observe the relationship between the slope of the line during each four-second interval and the corresponding acceleration value.\n\nTime\n(s)\nVelocity\n(m/s)\n0 2\n1 2\n2 2\n3 2\n4 2\n5 4\n6 6\n7 8\n8 10", null, "From 0 s to 4 s: slope = 0 m/s/s\nFrom 4 s to 8 s: slope = 2 m/s/s\n\nA motion such as the one above further illustrates the important principle: the slope of the line on a velocity-time graph is equal to the acceleration of the object. This principle can be used for all velocity-time in order to determine the numerical value of the acceleration. A single example is given below in the Check Your Understanding section.\n\n## Investigate!\n\nThe widget below plots the velocity-time plot for an accelerating object. Simply enter the acceleration value, the intial velocity, and the time over which the motion occurs. The widget then plots the line with velocity on the vertical axis and time on the horizontal axis.\n\nTry experimenting with different signs for velocity and acceleration. For instance, try a positive initial velocity and a positive acceleration. Then, contrast that with a positive initial velocity and a negative acceleration.\n\n### We Would Like to Suggest ...", null, "Sometimes it isn't enough to just read about it. You have to interact with it! And that's exactly what you do when you use one of The Physics Classroom's Interactives. We would like to suggest that you combine the reading of this page with the use of our Graph That Motion or our Graphs and Ramps Interactives. Each is found in the Physics Interactives section of our website and allows a learner to apply concepts of kinematic graphs (both position-time and velocity-time) to describe the motion of objects.\n\nThe velocity-time graph for a two-stage rocket is shown below. Use the graph and your understanding of slope calculations to determine the acceleration of the rocket during the listed time intervals. When finished, use the buttons to see the answers. (Help with Slope Calculations)\n\n1. t = 0 - 1 second\n2. t = 1 - 4 second\n3. t = 4 - 12 second", null, "1.\n\n2.\nNext Section:" ]

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[ "# Questions tagged [gams-math]\n\nGAMS (General Algebraic Modeling System) is a high-level modeling system for mathematical optimization.\n\n315 questions\nFilter by\nSorted by\nTagged with\n13 views\n\n### How to call the same index in each of 2 dimensions in a GAMS variable?\n\nI am running a model in GAMS with a number of variables with 2 dimensions, calling the indexes i and j respectively. I have a few constraints relating to just the \"diagonal\" entries, so I ...\n38 views\n\n### Stationarity conditions in Pyomo\n\nI am trying to implement MCP (Mixed Complementarity Problem) problem in pyomo with PATH solver, and I found a way how to write complementarity conditions. I use pyomo.mpec and ComplementarityList ...\n26 views\n\n### Removing specific equation from GAMS\n\nI am working with the software, GAMS, and would appreciate some insight in a problem I am facing. Assume that we have the following structure (grossly simplified for exposition): Sets i ...\n61 views\n\n### Logical condition in set (GAMS) - how do this in pyomo?\n\nI'm passing code from Gams to pyomo and I don't know if this option to access an index has in python. In gams: k purifiers /PSA4241, PSA241, PSA241A, PSA3241, NEW/ oldk(k) = yes\\$(ord(k) le 4) ; ...\n16 views\n\n### Model for scheduling surgeries\n\nI am developing a model in GAMS to schedule surgeries. I think I am not getting the right results and think some equations may not be well formulated. Bellow you can find the entire model I developed. ...\n26 views\n\n### Solve multi-objective problem in Gurobi by using epsilon-constraint method\n\nI am trying to solve my multi-objective problem in Gurobi using the epsilon-constraint method. I didn't find any sample code in Gurobi(python). I am new to gurobi and have no experience in coding. ...\n56 views\n\n### How to limit the count of nonzero variables in GAMS mixed-integer nonlinear programming?\n\nThe background of the problem: There are 25 candidates on Neo blockchain that receive votes. Every voter except me has voted. The candidates ranked 1st to 7th each will give 200 dollars, ...\n41 views\n\n### Extract data to update model parameters and solve maximization problem in GAMS\n\nI have several farmers and I would like to conduct optimization (profit maximization) for each of them in GAMS. I can do this for a single farmer, so my problem is how to iteratively make it works for ...\n59 views\n\n### General problems\n\nBellow you can find the entire model I developed. I think my main issues lie in: eq08:it is supposed to only allow one surgery to be scheduled during the entire duration of said surgery. For example, ...\n42 views\n\n### I used Gams, it gives an error in the code\n\nI coded that model using gams, but when I pressed run, it an gives error on line 42.(line starting with cons2(i)) Before that it was giving an error in the last two lines. The last two lines are \"...\n19 views\n\n### Any automatic code formatting tool for GAMS (stand-alone or IDE-integrated)?\n\nI am mainly working in the python ecosystem and am used to automatic code formatters (e.g. Black). For a new project I have to work with GAMS and a lot of legacy GAMS code. The GAMS code is by ...\n54 views\n\n### How do set up something similar to GAMS' circular indexing for Pyomo\n\nI'm working on a pyomo constraint that has a rule in the first loop to do some calculation based on the last loop. So for T = 24, each loop is: def const1(model,t): return model.x[t] == model.x[t -...\n75 views\n\n### Need to write a set of linear equations for following condition:\n\nI have three variables : A, B, C. I want to write a set of linear equations such that X = 1 if atleast 2 of A,B,C are ones. X= 0 if only one of A,B,C is one. X = 0 if all of them are zero. A,B and C ...\n14 views\n\n### variable definition problem over sets in GAMS\n\nMy problem is with the definition of variable 'ds' used in equation 1. Eq 1 runs for each line. And each line has ns (ns is a parameter) number of steps. Variable 'ds' should be defined for each step ...\n25 views\n\n### How do I convert if statement to linear equation?\n\nI wanna convert this if statement to linear equation. for i,j -> 1 to n if D[i]>D[j] and f[i] > s[j] then w[i]+=c[j] The line below is what has come to my mind so far, but I do not know how ...\n19 views\n\n### Calculation with mapping in GAMS\n\nConsider the following data: set i / 1 * 4 / tid / 2019, 2020 / iagg / 1 * 2 /; parameter t(i,tid),tagg(iagg,tid); t(i,tid) = uniform(0,1); set itoiagg(i,iagg) / 1.1 2....\n24 views\n\n### Reading an excel file from gams by creating a .inc file\n\nI am trying to read an excel file from gams using the following code: set i /1*10/; Parameter x(i)/ \\$call xls2gms r=a1:j1 i=Data1.xlsx o=xi.inc \\$include xi.inc /; However .inc file is generated and ...\n82 views\n\n### gdx file is not created when trying to export data from excel to gams\n\nI'm trying to export data from Excel to GAMS and I'm using the next code for this aim: Set c row labels /c1*c10/ x column labels /x1*x2/; Parameter d(c,x); \\$call GDXXRW Data.xlsx trace=3 par=d ...\n16 views\n\n### Error in defining sets and indexes as alias\n\nI'm trying to calculate the euclidean distance between some points. The coordinates of the points are denoted by x(i) and y(i), and the indices i and j are alias. For this I use the following code: i &...\n21 views\n\n### Defining a dynamic set in GAMS, General Algebraic Modeling System\n\nI want to define a dynamic set in GAMS, General Algebraic Modeling System. For example, consider the next line: Set i \"Customers\" /1*100/; Sometimes it will be /1x50/ and sometimes /1x100/. ...\n29 views\n\n### create equation summing over variables with shifting index gaps in GAMS\n\nI want to add a constraint in gams that sums over a variable that takes indices with a shifting gap between them. For example, I have a set of hour (h) in the year. My variable is g(h). My desired ...\n36 views\n\n### GAMS I need help on a set covering problem\n\nI am trying to solve a set covering problem in GAMS. There are particular amount of nodes. I know the distance between the nodes. If the distance larger than 500m, I don't connect the nodes. I want to ...\n43 views\n\n59 views\n\n### Set up gams module for python on Linux\n\nI'm sure this problem is pretty basic but this is my first time trying to run gams via python on Linux so I'd appreciate some help. My problem seems to be that I cannot properly link the GAMS ...\n30 views\n\n### When I try to write a variable including different sets I get an error. How can I solve Gams Error 198?\n\nI get an error when I try to use a variable as v(i+1 , j+1). How can I write v(i+1 , j+1) in GAMS code if I defined only v(i , j) as a variable before? Related parts in the code are as following: ...\n16 views\n\n### reason that GAMS read 10 out of 11 variables from my model if they all have the same format?\n\nI'm running a Computable General Equilibrium Model supported by a Social Accounting Matrix. I have a subset of 11 variables, of which GAMS doesn't read one, I have tried to change the format of my ...\n23 views\n\n### GAMS- manipulating expression within a loop\n\nI have a matrix, of dimension, i rows and j columns, a specific element of which is called x(i,j), where say i are plants, and j are markets. In standard GAMS notation: Sets i canning plants /...\n13 views\n\n### GAMS - matrix row operations\n\nI have the following matrix in GAMS (call it a matrix x with rows i and columns j): I wish to transform this by summing elements in the same column from row 2 and row 1, and then setting row 2 to 0: ...\n24 views\n\n### GAMS translated nonlinear objective function looks different than defined objective\n\nI am studying the batchdes.lst file for MINLP model batchdes in GAMS library. The objective function is Defining objective function obj.. cost =g= sum(j, alpha(j)*(exp(n(j) + beta(j)*v(j)))); However, ...\n45 views\n\n### How to write piecewise function code in GAMS?\n\nI am fairly new to GAMS coding and would like to know how can I write code for this fucntion.\n16 views\n\n### How can i use function smax\n\ni have problem about use smax in GAMS model. how can i fixing. thankyou for ans. Positive Variable C(i,k) time in peocess Cmax(i) max time ; variable T(i) T(i)=max(0,Cmax(i) -...\n98 views\n\n### Multiply matrix with vector in GAMS\n\nConsider the following table in GAMS: Set i /i1*i4 /; Table a(i,j) 'original matrix' i1 i2 i3 i4 i1 2 0 0 0 i2 0.272727 2 0....\n18 views\n\n### Calculate specific matrix in GAMS\n\nConsider the following data in GAMS: Set i / 0*3 /; Parameters r(i) / 2 0.272727 3 0.8 / s(i) / 2 0.727273 3 0.2 /; I want to calculate the following matrix (...\n25 views\n\n### Insert parameter value into \\$evalGlobal in GAMS\n\nConsider the following \\$SetGlobal: \\$setglobal Path MyPath I use the following code to finding the numbers of characters in 'MyPath': set Length /%Path%/; Parameter report(Length,*); report(Length,'...\n26 views\n\n### How to create multidimensional sets in JuMP/Julia as in GAMS\n\nIn GAMS, we can create multidimensional sets as follows Set i \"mining regions\" / china, ghana, ee+ussr, s-leone / n \"ports\" / accra, freetown, ...\n25 views\n\n### Reading in data with already defined set in GAMS from excel\n\nI am learning how to use GAMS and am having trouble with reading in a single column using an already defined set. I managed to read in a table and it displays as needed. This is what I attempted: ...\n48 views\n\n### Problem with translating characters with tr (POSIX) in GAMS\n\nIn GAMS version 33.2.0, I am changing the semi-colons to commas in a csv-file 'file.csv' because GAMS demands commas as separators. Then I want to transfer the content of the csv-file into a table in ...\n31 views\n\n### Writing hyperbolic inequalities as second order conic inequalities in GAMS\n\nI coded a GAMS model that has hyperbolic inequalities and GAMS can solve it. However, when I transform the hyperbolic inequalities likesqr(th(j)) =l= u(j)*f(j); to second order conic inequalities as 2*...\n21 views\n\n### Logic programming in GAMS\n\nHow would I be able to perform in GAMS syntax the logical input p implies no q (p => ~q)? The context is that I have a product in a sequence where another product can not be right after the first ...\n99 views\n\n### GAMS compile to GDX drops column with all zero values\n\nNOTE: Update (but not solved) Below In my use case for GAMS I wish to define a collection of static variables in a .gdx file rather than in the main model.gms script. This is not an issue unless I ...\n31 views\n\n### Error while solving bilevel Optimization using GAMS with EMP\n\nI am solving bilevel nlp optimization with one follower using GAMS. I am getting this error while creating emp.info while using code given below: - Code: - \\$echo bilevel x(i,j) min of * eq1 eq2(j) eq3(...\n125 views\n\n### How to code particle swarm optimization with constraints in MATLAB?\n\nI have a GAMS code that represent a mathematical model which depends delay (i,p) for each job i and the period of day p. So, I have indices for represent delay and objective is minimizing of summing ...\n41 views\n\n### Writing equation for the last element in a set in GAMS\n\nI'm trying to model a job shop scheduling problem. In the model, L(i,j) represents the set of processes j for each job i. Each i has a different set of processes in different orders. In some of the ...\n242 views\n\n### GAMS - writing output to excel file using GDX\n\nI am in the midst of exporting data from GAMS to excel. As such, I am using the GDX utilities. Say that I have two variables, x and y, defined over the set i. The set i={1,2,3,4}. As such, x(i) and y(...\n33 views\n\n### Will GAMS optimize the default parameters of Solver?\n\nI have a MILP problem. I use CBC solver to solve it. If I call CBC through GAMS, the computation time will be short. But when I solve it directly using CBC.exe, it takes too much time. In both cases, ...\n44 views\n\n### Struggling in converting gams model to pyomo - objective function giving syntax error\n\nI would like to write the following function in Pyomo but I am struggling. I wrote it in GAMS as follows it works: objectivefunction.. NetP =e= sum(s\\$(Sp(s)), sum(n\\$(ord(n)=card(n)), Ps(s)*(ST(s,n) + ...\n91 views\n\n### How can I transfer my results from GAMS to Python?\n\nI built my first GAMS code, but I don't know how to transfer the results from the lst file to Python in order to process. If anyone could help with the issue I would be very grateful. For example, for ...\n18 views\n\n### Is there a GAMS function that returns related sets?\n\nguys! Does someone know how to create the following variable for a GAMS model? Sets p periods /p1*p4/ sp sub-periods /sp1*sp8/ subp(p,sp) ..." ]

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[ "54,108 questions with no upvoted or accepted answers\n1k views\n\n### Link Anomaly Detection in Temporal Network\n\nI came across this paper that uses link anomaly detection to predict trending topics, and I found it incredibly intriguing: The paper is \"Discovering Emerging Topics in Social Streams via Link Anomaly ...\n3k views\n\n### What problem does oversampling, undersampling, and SMOTE solve?\n\nIn a recent, well recieved, question, Tim asks when is unbalanced data really a problem in Machine Learning? The premise of the question is that there is a lot of machine learning literature ...\n652 views\n\n### Jaynes' $A_p$ distribution\n\nIn Jaynes' book \"Probability Theory: The Logic of Science\", Jaynes has a chapter (Ch 18) entitled \"The $A_p$ distribution and rule of succession\" in which he introduces the idea of $A_p$ distributions,...\n679 views\n\n### Bound for Arithmetic Harmonic mean inequality for matrices?\n\nNOTE: This question has originally been posted in MSE, but it did not generate any interest. It was first posted there, because the question itself is a pure matrix-algebra question. Nevertheless, ...\n953 views\n\n### How does a Relevance Vector Machine (RVM) work?\n\nRelevance Vector Machines (RVMs) are really interesting models when contrasted with the highly geometrical (and popular) SVMs. In the light of a question like How does a Support Vector Machine (SVM) ...\n662 views\n\n### Wavelet-domain gaussian processes: what is the covariance?\n\nI've been reading Maraun et al, \"Nonstationary Gaussian processes in wavelet domain: Synthesis, estimation, and significant testing\" (2007) which defines a class of non-stationary GPs that can be ...\n461 views\n\n### Blind source separation of convex mixture?\n\nSuppose I have $n$ independent sources, $X_1, X_2, ..., X_n$ and I observe $m$ convex mixtures: \\begin{align} Y_1 &= a_{11}X_1 + a_{12}X_2 + \\cdots + a_{1n}X_n\\\\ ...&\\\\ Y_m &= a_{m1}X_1 + ...\n877 views\n\n### Diagnostic plot for assessing homogeneity of variance-covariance matrices\n\nIs there a handy plot for comparing the variance-covariance matrices of two (or perhaps more) groups? An alternative to looking at lots of marginal plots, especially in the multivariate Normal case?\n1k views\n\n### $ARIMA(p,d,q)+X_t$, Simulation over Forecasting period\n\nI have time series data and I used an $ARIMA(p,d,q)+X_t$ as the model to fit the data. The $X_t$ is an indicator random variable that is either 0 (when I don’t see a rare event) or 1 (when I see the ...\n634 views\n\n### Fitting custom distributions by MLE\n\nMy question relates to fitting custom distributions in R but I feel it has enough of a probability element to remain on CV. I have an interesting set of data which has the following characteristics: ...\n3k views\n\n### Getting started with bayesian structural models using MCMC\n\nI'm trying to learn bayesian structural time series analysis. For a variety of reasons I need to use Python (mostly pymc3) not R so please do not suggest the ...\n4k views\n\n### Multivariate Beta distribution (no Dirichlet!)\n\nWhat is a multidimensional generalization of the Beta distribution, in compliance with the following specification? I am not looking for the Dirichlet distribution. I am looking for a generalization ...\n583 views\n\n### Distribution of inverse Wishart to a power?\n\nIn a related question, I had asked about the norm induced by an inverse Wishart matrix. I am interested in generalizing that result somewhat. Let $A\\sim\\mathcal{W}_p\\left(I,n\\right)$, a Wishart matrix ..." ]

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[ "# How to model two variables to NOT to belong to the same set partition using Constraint Programming\n\nAssume we have two variables $$x,y \\in S$$ where $$S=\\{1,2, \\dots, 1000\\}$$. Also, we are given a partition of set $$S$$ as:\n\n$$S_1 = \\{1,2, \\dots, 249\\}$$ $$,S_2 = \\{250, \\dots, 499\\}$$ $$,S_3 = \\{500, \\dots, 749\\}$$ $$,S_4 = \\{750, \\dots, 1000\\}$$\n\nHow to model a constraint that prevents variables $$x$$ and $$y$$ both belonging to the same partition. That said, $$x=1$$, $$y = 2$$ is an invalid assignment but $$x=1$$, $$y = 250$$ is allowed.\n\nI am using Google OR-Tools Constraint Programming.\n\n(Please note the intersection of any two of the subsets is empty, and their union is the whole set. Each partition is not necessarily a full range of integers, unlike the example. For instance, $$S_1=\\{1,\\dots,249,750,\\dots,1000\\}, \\quad S_2=\\{250,\\dots,749\\}$$ is too a valid partition.)\n\n• Yes, the partitions wouldn't necessarily be of the same size. – Mahmoud Oct 11 at 1:42\n\nHere's one way: $$x\\not=i \\lor y\\not=j \\text{ for } k\\in\\{1,2,3,4\\}, i\\in S_k, j\\in S_k$$\n\nHere's another way, using ELEMENT constraints, as suggested by @prubin. The following is SAS code, but maybe OR-Tools has something similar.\n\nproc optmodel;\nset S {k in 1..4} =\nif k = 1 then 1..249\nelse if k = 2 then 250..499\nelse if k = 3 then 500..749\nelse 750..1000;\nnum p {1..1000};\nfor {k in 1..4, i in S[k]} p[i] = k;\n\nvar X >= 1 <= 1000 integer;\nvar Y >= 1 <= 1000 integer;\n\nvar PX >= 1 <= 4 integer;\nvar PY >= 1 <= 4 integer;\n\n/* PX = p[X] */\ncon ElementConX:\nelement(X, p, PX);\n/* PY = p[Y] */\ncon ElementConY:\nelement(Y, p, PY);\n\ncon NotEqual:\nPX ne PY;\n\nsolve;\nprint X Y PX PY;\nquit;\n\n\nThe first solution found was\n\n(X, Y, PX, PY) = (1, 250, 1, 2)\n\n\nand specifying the FINDALLSOLNS option yields $$1000^2-249^2-250^2-250^2-251^2=749998$$ solutions, as expected.\n\nI don't use OR-Tools, so I cannot be specific about syntax, but I'm pretty sure it has a table lookup constraint. So you can create a table that associates each value from 1 to 1,000 with its partition index (1 to 4), and then just add a constraint that says the partition value of $$x$$ cannot equal the partition value of $$y$$.\n\nUsing intermediate booleans and AddLinearExpressionInDomain you get:\n\nfrom ortools.sat.python import cp_model\n\nmodel = cp_model.CpModel()\nsolver = cp_model.CpSolver()\n\nx = model.NewIntVar(1, 1000, \"x\")\ny = model.NewIntVar(1, 1000, \"y\")\n\nsx = {i: model.NewBoolVar(f\"x in S{i}\") for i in range(1, 5)}\nsy = {i: model.NewBoolVar(f\"y in S{i}\") for i in range(1, 5)}\nfor i in range(4):\nsi = cp_model.Domain.FromFlatIntervals([250 * i, 250 * (i + 1) - 1])\n\n• Thank you for the code snippet. One assumption I see you made is each domain $S_i$ is a full range of integers. However, the subsets could be from any partition of the full set. For instance, $S_1={1, 4, 7}, /quad S_2={2,3,5,12}$, then the sets are cannot be defined by range(start, end). – Mahmoud Oct 11 at 1:35" ]

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[ "All Collections\nAuthors\nEvaluation types\nEvaluation type: eval param rep & fb param rep\nEvaluation type: eval param rep & fb param rep\n\nEverything you need to know about the evaluation types: eval param rep and fb param rep\n\nUpdated over a week ago\n\nThe evaluation type eval param rep can be used to evaluate a parametric representation of the solution set of a list of linear equations. It does not have automated feedback itself, but instead, the evaluation type fb param rep can be used to give automated negative feedback on the student’s answer.\n\n### Evaluation type: eval param rep\n\nThe Definition field of this evaluation type should have two arguments, separated by a semicolon (;). The first argument should be a list of the linear equations, and the second should be a list of the coordinates.\n\nWhen this evaluation type is used in a Solution rule, the evaluation type fb param rep can be used in positive or negative feedback to give automated feedback on the student’s answer.\n\nWith the solution Definition above, the student is free to choose the parameters they use. If you don’t want the choice of parameters to be free, you can provide a third argument in the solution Definition which is a list of the parameters that may be used.\n\nExample:\n\nGreek symbols can be inputted by the student using the abc tab on the virtual keyboard:\n\nHowever, parameters don’t need to be Greek symbols.\n\n### Evaluation type: fb param rep\n\nThis evaluation type can only be used in negative feedback when a solution rule is defined that uses eval param rep. It gives feedback on one of the nine cases specified below.\n\nTo select a case, type the number (e.g. `2`) or the name (e.g. `no_vector`) in the Definition field. Every case supports automated feedback, and this automated feedback is shown in the examples below. The solution rule that is used for these feedback rules is:\n\n• 2 (`no_vector`)\n\nThe student’s answer is not a vector.\n\n• 3 (`wrong_length`)\n\nThe vector in the student’s answer has the wrong length.\n\n• 4 (`wrong_basevec`)\n\nThe base vector of the student’s answer is not a solution to the equations in the Solution.\n\n• 5 (`too_few_parms`)\n\nThe student’s answer has too few parameters.\n\n• 6 (`too_many_parms`)\n\nThe student’s answer has too many parameters.\n\n• 7 (`wrong_params`)\n\nWhen a third argument with allowed parameters is specified in the solution, this rule hits if the wrong parameters are used.\n\n• 8 (`non_linear`)\n\nThe student’s answer is non-linear in one of its parameters.\n\n• 9 (`wrong_vector`)\n\nThe students’s answer has at least one vector that is not a solution to the equations in the Solution.\n\n• 10 (`lin_dep`)\n\nAt least two direction vectors in the student’s answer are linearly dependent." ]

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[ "# Treynor Ratio\n\nThe Treynor Ratio was introduced by Jack L. Treynor. This ratio is used to make exact calculation of the amount of surplus return provided by a unit of risk.\n\nWe can also say that the Treynor ratio is used to calculate that extra amount of return that is excess to what would have been earned through safe investments or without taking any kind of risk. The Treynor ratio is developed upon the systematic risk.\n\nThe Treynor ratio is also termed as reward to volatility ratio as it measures the earnings made while facing the risks. Whenever the Treynor ratio is on the higher side, it denotes that the investor is provided with high yields by each unit of market risk.\n\nFormula of Treynor Ratio:\nT = (rp-rf )/ ß\n\nThe Treynor ratio is a kind of rating standard. There are several portfolios that are well diversified and consist of several sub portfolios. If these sub portfolios are ranked by Treynor ratio, then the sub portfolios would be benefited.\n\nIf there is no such portfolio, then those portfolios with same type of systematic risk but various types of total risk are going to be ranked as the same.\n\nAt the same time, there are a number of different portfolios that are not diversified and thus, these portfolios are subjected to high amount of risk. These portfolios are rarely preferred by the investors.", null, "" ]

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[ "", null, "", null, "", null, "Chapter 5.3, Problem 24E", null, "Elementary Geometry For College St...\n\n7th Edition\nAlexander + 2 others\nISBN: 9781337614085\n\nSolutions\n\nChapter\nSection", null, "Elementary Geometry For College St...\n\n7th Edition\nAlexander + 2 others\nISBN: 9781337614085\nTextbook Problem\n\nIn Exercises 17 to 24, complete each proof.Given: A B ¯ ∥ D C ¯ , A C ¯ ∥ D E ¯ Prove: A B D C = B C C E", null, "PROOF Statements Reasons 1. A B ¯ ∥ D C ¯ 2.3. ?4. ∠ A C B ≅ ∠ E 5. Δ A C B ∼ Δ D E C 6. ? 1. ?2. If 2 || lines are cut by a transversal, corresponding ∠ s are ≅ .3. Given4. ?5. ?6. ?\n\nTo determine\n\nTo prove:\n\nThe statement ABDC=BCCE if the sides AB and DC are parallel, AB¯DC¯ and the sides AC and DE are parallel, AC¯DE¯.\n\nExplanation\n\nDefinition:\n\nAA:\n\nIf two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.\n\nCSSTP:\n\nCorresponding sides of similar triangles are proportional.\n\nDescription:\n\nGiven that RS¯UV¯.\n\nThe given figure is shown below.\n\nFigure\n\nFrom the given figure, it is observed that the sides AB and DC are parallel, AB¯DC¯.\n\nIf two parallel lines are cut by a transversal, then the alternate interior angles are congruent.\n\nTherefore, BACD.\n\nFrom the given figure, it is also observed that the sides AC and DE are parallel, AC¯DE¯.\n\nIf two parallel lines are cut by a transversal, then the alternate interior angles are congruent.\n\nTherefore, ACBE.\n\nThe above mentioned AA definition, the two triangles ACB and DEC are similar since the two angles of one triangle are congruent to two angles of another triangle. Hence, ΔACBΔDEC.\n\nFrom the definition of CSSTP, corresponding sides of similar triangles are proportional\n\nStill sussing out bartleby?\n\nCheck out a sample textbook solution.\n\nSee a sample solution\n\nThe Solution to Your Study Problems\n\nBartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!\n\nGet Started\n\nEvaluate the expression sin Exercises 116. 22212424\n\nFinite Mathematics and Applied Calculus (MindTap Course List)\n\nEvaluate in problems 1-4. Write all the answers without using exponents. 1. (a) ...\n\nMathematical Applications for the Management, Life, and Social Sciences\n\nConvert from degrees to radians. 5. 900\n\nSingle Variable Calculus: Early Transcendentals, Volume I", null, "" ]

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