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[ "# How long does it take the average voter to vote?\n\nSo I was helping my brother with his homework question as follows\n\nThe voting office can handle $50 \\space \\text {voters/hour}$ and has 20 voting stations. How long does it take the average voter to vote?\n\nHe answered saying there are $60$ minutes for each machine, so $60 \\space \\text {minutes} \\times 20 \\space \\text {machines} = 1200 \\space \\text {total voting minutes/ hour}$. Therefore there are $$= \\frac {1200 \\space \\text {minutes}}{50 \\text {votes}}$$ $$=24 \\text {minutes/vote}$$\n\nI can tell that the answer is wrong because $24$ minutes is too long for one person to vote. My answer is that since there $60$ minutes in an hour, and $50$ votes per hour then $$\\frac {60}{50}$$ $$=1.2 \\text {minutes}$$\n\nI am not really sure why my brother's answer is incorrect; his method seems correct but is probably overcounting somehow. Any insight on the problem is appreciated\n\n• Maybe I missed something in the question, but as far as I can see your brother's answer is correct. I agree that 24 minutes is a long time for voting, but mathematics questions set for schoolwork are not always realistic. This is unfortunate because it gives students a very bad impression of the subject, but sadly, that's the way it (sometimes) is. – David Mar 2 '15 at 1:01\n• +1 for checking that the answer appears to be a realistic number, though. Pity that the problem fails to give realistic input data ... – hmakholm left over Monica Mar 2 '15 at 1:06\n• I would like to add that in some German state/city elections, a voter can distribute some 80 votes to named candidates (Panachage). It can take 24 minutes if the voters have not decided beforehand which candidates exactly they want. – Alexander Mar 2 '15 at 9:11\n• Hmm, I upvoted the question and explained why. So far 5 people have upvoted my comment with the explanation, but the question has only 3 upvotes in total. What's going on here, overcounting somehow? – hmakholm left over Monica Mar 2 '15 at 11:23\n\nYour brother's answer is actually absolutely correct. In your answer, you are calculating the average voting time if there was only one single voting machine, but there are $20$.\nYour answer ignores the fact that there are $20$ voting stations, presumably being used in parallel. If each voter took $1.2$ minutes and you kept a single station busy, you could handle $50$ per hour. Your brother is correct that averaged over a two hour period, you could afford each person to take $24$ minutes. Each station would handle $5$ people for a total of $100$, or $50$ per hour." ]

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[ "", null, "Download Presentation", null, "Trees", null, "# Trees - PowerPoint PPT Presentation\n\nDownload Presentation", null, "##### Trees\n\nDownload Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. Trees Chapter 15\n\n2. Tree Concepts Hierarchical Organizations Tree Terminology Traversals of a Tree Traversals of a Binary Tree Traversals of a General Tree Java Interfaces for Trees Interfaces for All Trees Interface for Binary Examples of Binary Trees Expression Trees Decision Trees Binary Search Trees Examples of General Trees Parse Trees Game Trees Chapter Contents\n\n3. Tree Concepts • Previous data organizations place data in linear order • Some data organizations require categorizing data into groups, subgroups • This is hierarchical classification • Data items appear at various levels within the organization\n\n4. Hierarchical Organization • Example: File directories Computer files organized into folders.\n\n5. Hierarchical Organization • Example: A university's organization A university's administrative structure.\n\n6. Hierarchical Organization • Example: Family trees Carole's children and grandchildren.\n\n7. Tree Terminology • A tree is • A set of nodes • Connected by edges • The edges indicate relationships among nodes • Nodes arranged in levels • Indicate the nodes' hierarchy • Top level is a single node called the root\n\n8. Tree Terminology A tree equivalent to the my stuff tree\n\n9. Tree Terminology • Nodes at a given level are children of nodes of previous level • Node with children is the parent node of those children • Nodes with same parent are siblings • Node with no children is a leaf node • The only node with no parent is the root node • All others have one parent each\n\n10. Tree Terminology • Empty trees? • Some authors specify a general tree must have at least the root node • This text will allow all trees to be empty • A node is reached from the root by a path • The length of the path is the number of edges that compose it • The height of a tree is the number of levels in the tree • The subtree of a node is a tree rooted at a child of that node\n\n11. Binary Trees • General tree: in general, node can have an arbitrary number of children • Binary tree if each node has at most two children Three binary trees.\n\n12. Binary Trees • A binary tree is either empty or has the following form • Where Tleft and Tright are binary trees\n\n13. Binary Trees • If a binary tree of height h has all leaves on the same level h and every nonleaf in a full binary tree has exactly two children • A complete binary tree is full to its next-to-last level • Leaves on last level filled from left to right\n\n14. Binary Trees The number of nodes in a full binary tree as a function of the tree's height.\n\n15. Binary Trees • Total number of nodes n for a full tree can be calculated as: • The height of a binary tree with n nodes that is either complete or full is log2(n + 1)\n\n16. Traversals of a Tree • Traversing items is a common operation. Previous chapters, data was arranged linearly. Order of traversal was clear. • During tree traversal, we must visit each item exactly once in a systematic way. The order is not unique. • Visiting a node • Processing the data within a node • A traversal can pass through a node without visiting it at that moment\n\n17. Traversals of a Tree • Subtree of the root of binary trees are themselves binary trees. Using this recursive nature, we define recursive traversal: • For a binary tree • Visit the root • Visit all nodes in the root's left subtree • Visit all nodes in the root's right subtree • Order does not matter. Visit root before, between or after visiting two subtrees.\n\n18. Traversals of a Tree • Preorder traversal: visit root before the subtrees The visitation order of a preorder traversal.\n\n19. Traversals of a Tree • Inorder traversal: visit root between visiting the subtrees The visitation order of an inorder traversal.\n\n20. Traversals of a Tree • Postorder traversal: visit root after visiting the subtrees These are examples of a depth-first traversal. Follow a path that descends the levels of a tree as deeply as possible until reaches a leaf The visitation order of a postorder traversal.\n\n21. Traversals of a Tree • Level-order traversal: begin at the root, visit nodes one level at a time This is an example of a breadth-first traversal. Follow a path that explores an entire level before moving to the next level. The visitation order of a level-order traversal.\n\n22. Traversals Exercise • The order of these nodes being visited using 4 different traversal methods\n\n23. Answer • In this binary tree, D = node, L = left, R = right • Preorder (DLR) traversal yields: A, H, G, I, F, E, B, C, D • Postorder (LRD) traversal yields: G, F, E, I, H, D, C, B, A • In-order (LDR) traversal yields: G, H, F, I, E, A, B, D, C • Level-order traversal yields: A, H, B, G, I, C, F, E, D\n\n24. Traversals of a General Tree • A general tree has traversals that are in • Level order • Preorder • Postorder • Inorder traversal not well defined for a general tree\n\n25. Traversals of a General Tree The visitation order of two traversals of a general tree: (a) preorder; (b) postorder.\n\n26. Java Interfaces for Trees • An interface that specifies operations common to all trees public interface TreeInterface{ public Object getRootData();public int getHeight();public int getNumberOfNodes();public boolean isEmpty(); public void clear();} // end TreeInterface\n\n27. Expression Trees for Algebraic Expression • The root contains the binary operator and children contain the operands. • Order of children matches the order of operands. Such binary tree is called expression Trees. No parentheses since order of operations is captured by the shape of the expression tree Expression trees for four algebraic expressions..\n\n28. Expression Trees • Inorder traversal produces the original infix expression, but without any parentheses • Preorder traversal produces the prefix expression • Postorder traversal produces the postfix expression • (b) preorder: +*abc postorder: ab*c+\n\n29. Expression Trees • Algorithm for evaluating an expression tree in postorder traversal Algorithm evaluate(expressionTree)if (root of tree is operand)return operand else{ firstOperand = evaluate(left subtree of expressionTree) secondOperand = evaluate(right subtree of expressionTree) operator = the root of expressionTreereturn the result of the operation operator and its operandsfirstOperand and secondOperand}\n\n30. Exercises • Draw an expression tree for each of these algebraic expressions. • a. a+b*c • b. (a+b)*c\n\n31. Decision Trees Basis of Expert System • An expert system also known as a knowledge based system that contains some knowledge. It was first developed by researchers in artificial intelligence. • The most common form of expert systems is made up of a set of rules that analyze information. • Each parent in a decision tree is a question that has a finite number of response. • Each possible answer to the question corresponds to a child of the node. • Leave node are conclusions that have no children\n\n32. Decision Trees • Helps users solve problems, make decisions A binary decision tree to solve TV problem.\n\n33. Decision Trees • A possible Java interface for a binary decision tree. public interface DecisionTreeInterface extends BinaryTreeInterface{ /** Task: Gets the data in the current node. * @return the data object in the current node */public Object getCurrentData(); /** Task: Determines whether current node contains an answer. * @return true if the current node is a leaf */public boolean isAnswer(); /** Task: Moves the current node to the left (right) child of the current node. */public void advanceToNo();public void advanceToYes(); /** Task: Sets the current node to the root of the tree.*/public void reset();} // end DecisionTreeInterface\n\n34. Decision Trees The decision tree for a guessing game Facts are used to build this decision tree.\n\n35. Binary Search Trees • We can traverse nodes in any tree, so searching a node is possible. The least efficient search will be sequential search. • However, a search tree organizes its data so that a search is more efficient • Binary search tree • Nodes contain Comparable objects • A node's data is greater than the data in the node's left subtree • A node's data is less than the data in the node's right subtree\n\n36. Binary Search Trees A binary search tree of names.\n\n37. Binary Search Trees Two binary search trees containing the same names as the tree in previous slide\n\n38. Questions • How many different binary search trees can you form from the strings a, b, and c? • What are the heights of the shortest and tallest trees that you formed?\n\n39. Answer • How many different binary search trees can you form from the strings a, b, and c? • 5 different structures • What are the heights of the shortest and tallest trees that you formed? • 2 and 3\n\n40. Binary Search Trees • Search begins at the root, ends at either the target node or leaf node. • The number of comparison is the number of nodes along the path from the root to target node. • The height of a tree directly affects the length of the longest path from the root to a leaf and hence affects the efficiency of a worst-case search. • Searching a binary search tree of height h is O(h).\n\n41. Binary Search Trees • An algorithm for searching a binary search tree Algorithm bstSearch(binarySearchTree, desiredObject)// Searches a binary search tree for a given object.// Returns true if the object is found.if (binarySearchTree is empty)return falseelse if (desiredObject == object in the root of binarySearchTree)return trueelse if (desiredObject < object in the root of binarySearchTree)return bstSearch(left subtree of binarySearchTree, desiredObject)else return bstSearch(right subtree of binarySearchTree, desiredObject)\n\n42. Heaps • A complete binary tree • Nodes contain Comparable objects • Each node contains no smaller (or no larger) than objects in its descendants • Maxheap • Object in a node is ≥ its descendant objects. Root node contains the largest data • Minheap • Object in a node is ≤ descendant objects • Root node contains the smallest data\n\n43. Heaps (a) A maxheap and (b) a minheap that contain the same values\n\n44. Question? • Can a binary search tree ever be a maxheap? • Can you think of any ADT we have learnt so far that can be implemented by maxheap?\n\n45. Parse Tree for Compiler • These rules form a grammar for algebraic expression, much like English language grammar. • An algebraic expression is either a term or two terms separated by + or – operator • A term is either a factor or two factors separated by * or / operator. • A factor is either a variable or an algebraic expression enclosed in parentheses. • A variable is a single letter Check the syntax of each statement by applying these rules, if we can the derivation can be given as a parse tree.\n\n46. Parse Tree • A parse tree for the algebraic expression a * (b + c) • 1). An algebraic expression is either a term or two terms separated by + or – operator • 2). A term is either a factor or two factors separated by * or / operator. • 3). A factor is either a variable or an algebraic expression enclosed in parentheses. • 4). A variable is a single letter\n\n47. Parse Tree • Parse tree should be a general tree so that I can accommodate any expression. • We are not restricted to algebraic expression. • We can use parse tree to check the validity of any string according to any grammar • Compiler uses parse tree both to check the syntax of a program and to produce executable code.\n\n48. Exercise • Draw a parse tree for the algebraic expression a*b + c • a + b*c • (a+b)*(c-d)" ]

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[ "# Do the left and right \"parts\" of the matrix used when performing Gaussian Elimination have names?\n\nGiven a matrix used to perform Gaussian Elimination like this:\n\n$$\\begin{bmatrix} 1 & -3 & | & 1 \\\\ 2 & -7 & | & 3 \\end{bmatrix}$$\n\nWhich would be derived from a system of equations like this:\n\n$$\\begin{cases} x - 3y = 1 \\\\ 2x - 7y = 3 \\end{cases}$$\n\nDo the different \"sides\", i.e. the parts that come from the right hand side of the equation and the part that comes from the left hand side of the equation, respectively, have names?\n\n• This comment doesn't answer your question, but it's called an augmented matrix Oct 8, 2021 at 20:56\n• The left side is the coefficient matrix, and the right side is the column vector of constant terms Oct 8, 2021 at 21:00\n• @J.W.Tanner that sounds like an answer, thanks! How come you post it as a comment? Oct 9, 2021 at 9:11\n• Sorry; I'll post it as an answer now Oct 10, 2021 at 2:03" ]

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[ "The OEIS is supported by the many generous donors to the OEIS Foundation.", null, "Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)\n A108954 a(n) = pi(2*n) - pi(n). Number of primes in the interval (n,2n]. 14\n 1, 1, 1, 2, 1, 2, 2, 2, 3, 4, 3, 4, 3, 3, 4, 5, 4, 4, 4, 4, 5, 6, 5, 6, 6, 6, 7, 7, 6, 7, 7, 7, 7, 8, 8, 9, 9, 9, 9, 10, 9, 10, 9, 9, 10, 10, 9, 9, 10, 10, 11, 12, 11, 12, 13, 13, 14, 14, 13, 13, 12, 12, 12, 13, 13, 14, 13, 13, 14, 15, 14, 14, 13, 13, 14, 15, 15, 15, 15, 15, 15, 16, 15, 16 (list; graph; refs; listen; history; text; internal format)\n OFFSET 1,4 COMMENTS a(n) < log(4)*n/log(n) < 7*n/(5*log(n)) for n > 1. - Reinhard Zumkeller, Mar 04 2008 Bertrand's postulate is equivalent to the formula a(n) >= 1 for all positive integers n. - Jonathan Vos Post, Jul 30 2008 Number of distinct prime factors > n of binomial(2*n,n). - T. D. Noe, Aug 18 2011 f(2, 2n) - f(3, n) < a(n) < f(3, 2n) - f(2, n) for n > 5889 where f(k, x) = x/log x * (1 + 1/log x + k/(log x)^2). The constant 3 can be improved. - Charles R Greathouse IV, May 02 2012 For n >= 2, a(n) is the number of primes appearing in the 2nd row of an n X n square array whose elements are the numbers from 1..n^2, listed in increasing order by rows. - Wesley Ivan Hurt, May 17 2021 REFERENCES F. Irschebeck, Einladung zur Zahlentheorie, BI Wissenschaftsverlag 1992, p. 40 LINKS T. D. Noe, Table of n, a(n) for n = 1..10000 Tsutomu Hashimoto, On a certain relation between Legendre's conjecture and Bertrand's postulate, arXiv:0807.3690 [math.GM], 2008. FORMULA a(n) = A000720(2*n)-A000720(n). For n > 1, a(n) = A060715(n). - David Wasserman, Nov 04 2005 Conjecture: G.f.: Sum_{i>0} Sum_{j>=i|i+j is prime} x^j. - Benedict W. J. Irwin, Mar 31 2017 From Wesley Ivan Hurt, Sep 20 2021: (Start) a(n) = Sum_{k=1..n} A010051(2*n-k+1). a(n) = Sum_{k=n*(n+1)/2+2..(n+1)*(n+2)/2} A010051(A128076(k)). (End) MAPLE A108954 := proc(n)     numtheory[pi](2*n)-numtheory[pi](n) ; end proc: # R. J. Mathar, Nov 03 2017 MATHEMATICA Table[Length[Select[Transpose[FactorInteger[Binomial[2 n, n]]][], # > n &]], {n, 100}] (* T. D. Noe, Aug 18 2011 *) f[n_] := Length@ Select[ Range[n + 1, 2n], PrimeQ]; Array[f, 100] (* Robert G. Wilson v, Mar 20 2012 *) Table[PrimePi[2n]-PrimePi[n], {n, 90}] (* Harvey P. Dale, Mar 11 2013 *) PROG (PARI) g(n) = for(x=1, n, y=primepi(2*x)-primepi(x); print1(y\", \")) CROSSREFS Cf. A000720, A060715. Cf. A067434 (number of prime factors in binomial(2*n,n)), A193990, A074990. Sequence in context: A283190 A030361 A060715 * A123920 A322141 A029170 Adjacent sequences:  A108951 A108952 A108953 * A108955 A108956 A108957 KEYWORD nonn,easy AUTHOR Cino Hilliard, Jul 22 2005 STATUS approved\n\nLookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam\nContribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents\nThe OEIS Community | Maintained by The OEIS Foundation Inc.\n\nLast modified September 26 20:57 EDT 2022. Contains 357050 sequences. (Running on oeis4.)" ]

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[ "[Discussion] Solving the Limit Problem\n12-16-2019, 03:02 PM\nPost: #1\n yangyongkang", null, "Member Posts: 51 Joined: Dec 2018\n[Discussion] Solving the Limit Problem\nNonsense, paste code\nCode:\nlimit((∫(∫(sin(t)*atan(1+t),t,0,u^2),u,0,x))/(x^3*((x+1)^(1/3)-1)^2),x,0)\nxcas get\nCode:\n\"Limit: Max order reached or unable to make series expansion Error: Bad Argument Value\"\n\nmathematica also calculated for a long time\nCode:\nLimit[Integrate[Sin[t]*ArcTan[1 + t], {u, 0, x}, {t, 0, u^2}]/(  x^3*((x + 1)^(1/3) - 1)^2), x -> 0]\n\nSurprisingly, the Ti Nspire CX CII CAS is calculated\n\nIn fact, the hp prime can be calculated, and it needs to be replaced by another method.\n\nlet f(t)=sin(t)*atan(1+t)\n\nCode:\nseries((∫(∫(f(t),t,0,u^2),u,0,x)/(x^3*((x+1)^(1/3)-1)^2)),equal(x,0),1)\n\nhp prime get\nCode:\n(3*f(0)/x^2)+(2*f(0)/x)+(9/10)*(function_diff(f))(0)-(1/9)*f(0)+((3/5)*(function_diff(f))(0)+(2/27)*f(0))*x+x^2*order_size(x)\n\nWe find f (0) = 0 and substitute it into the result,Get the limit\nCode:\n((9/10)*∂(sin(t)*atan(1+t),t)|(equal(t,0)))\nhp prime get\nCode:\n9*π/40\n\nAttached File(s) Thumbnail(s)", null, "study hard, improve every day\n12-19-2019, 09:08 AM\nPost: #2\n yangyongkang", null, "Member Posts: 51 Joined: Dec 2018\nRE: [Discussion] Solving the Limit Problem\nSeveral perverted integrals\n(i)\nCode:\nevalf(int(1/(1+e^(1/x)),x,-1,1))\nxcas get\nCode:\n\"Unidirectional limits are distincts 1,0 Error: Bad Argument Valu\"\n\n(ii)\nCode:\nint(x^2*tan(x)^2+x^2+2*x*tan(x)+1,x)\nCode:\nx+x*tan(x)\n\n(iii)\nCode:\nsimplify(int((x*tan(x)+ln(x*cos(x))-1)/ln(x*cos(x))^2,x))\nx/ln(x*ln(x))" ]

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[ "• 2019-09-28 12:00\n• 2021-06-04 20:39\n\n# 《地震概论》2019年秋季学期（周四班）笔记\n\n## 课程板书内容\n\n### 第1章 地震学的研究范围和历史\n\n• 历史记载全球死亡超过20万人的地震有6次，其中在中国就有4次。\n• 按震源深度划分天然地震：浅源地震 < 60 km，中源地震60 ~ 300 km，深源地震 > 300 km。\n• 1966年邢台大地震，导致了1971年中国地震局成立。\n• 地方震震中距 < 100 km，近震震中距 < 1000 km，远震震中距 > 1000 km。\n• 古代台湾人认为地震由地牛产生。\n• 古代人类对地震的认识：中国——大鳌鱼，日本——鲶鱼，印度——大象，希腊——气动说。\n• 欧洲里斯本大地震使人们从神学中解放。\n• 地震学是一门应用物理学，属于固体地球物理学的分支，不是地质学。\n\n### 第2章 地震波\n\n• 地震横波与纵波：\n• 纵波：P波，质点振动方向与振动（能量）传播方向一致。传播速度为$v_P = \\sqrt{E / \\rho}$。\n• 横波：S波，质点振动方向与振动（能量）传播方向垂直。传播速度为$v_S = \\sqrt{\\mu / \\rho}$。\n• 纵波速度比横波速度大，一般$v_P \\approx \\sqrt{3} v_S$。\n• 地震波的传播：\n• 说明：地震波在传播到地表时几乎与地表垂直，以此可分析P波和S波的振动方向。\n• S波的分解：\n• 说明：S波可以分解成两个分量，平行于界面的分量为SH波，在入射线和界面法线构成的平面上（入射面）的分量为SV波。\n• SH波与SV波：\n• P波和S波的主要差异：\n• P波传播速度比S波快，地震图上先出现P波。\n• P波和S波的质点振动（偏振）方向相互垂直。\n• 一般情况下，三分量地震图上P波的垂直分量相对较强，S波的水平分量相对较强。\n• S波的低频成分比P波丰富。\n• 天然地震的振源破裂通常以剪切破裂和剪切错动为主，震源向外辐射的S波能量比P波的强。\n• P波通过时，质元无转动运动，而有体积变化，P波是一种无旋波。S波通过时，质元有转动，而无体积变化，S波是一种无散的等容波。\n• 谚语：“地震不地震，抬头看吊灯”。\n• 说明：地震波中，P波上下振动，S波水平振动。而环境造成的振动不会产生水平方向的振动，因此当吊灯水平振动时，说明有地震发生。\n• P波使建筑物上下晃动，S波使建筑物水平晃动。\n• 一般波序：P波、S波、勒夫面波、瑞利面波、地震尾波。\n\n### 第3章 地震波传播理论\n\n• 射线理论：地震射线是波阵面法线的轨迹。计算方法：（感谢树洞Hans提醒）\n$t = \\int_A^B \\sqrt{1 + y'R} \\mathrm dx.$\n• 波长很短：$v = \\lambda f$，在高频近似（短波近似）的情况下，可用波射线来描述波的传播。\n• 费马原理：当一个震动由介质中一点传播到另一点时，它所经过的途径会使其传播时间为一稳定值（最大、最小或拐点）。\n• 斯涅耳定律：（证明见作业，\n$\\cfrac {\\sin\\theta_{\\alpha1}} {v_{\\alpha1}} = \\cfrac {\\sin\\theta_{\\beta1}} {v_{\\beta1}} = \\cfrac {\\sin\\theta_{\\alpha2}} {v_{\\alpha2}} = \\cfrac {\\sin\\theta_{\\beta2}} {v_{\\beta2}}.$\n• 地球内部体波的传播：\n• PcPS不可能存在（$v_P > v_S$导出$\\theta_P > \\theta_S$），ScSP可能存在。P夹角比S大。（感谢树洞Hans提醒）\n• 说明：P、S分别表示从震源发出、向地球内部传播的纵波和横波；c表示在地核界面上波的反射；K表示通过外核的纵波，地球的外核是液态，所以只有通过外核的纵波，而没有通过外核的横波；I、J分别表示通过内核的纵波、横波；i表示在内核界面的反射；p、s分别表示由震源向上（地面）传播的射线。\n• SH波不能产生P波：在水平面内振动不能引起垂直面振动。\n• 面波的走时曲线为直线。\n• S-P的走时差较多依赖于距离而较少依赖于深度；而pP-P走时差主要由震源深度决定，较少得依赖于震中距。\n• 计算题：（《地震概论》课本第41页，思考题2）一个震源深度为10 km的地震，多个区域台站记到的Pn波走时直线的斜率为0.125 s/km，截距为$\\pu{3\\sqrt{7} s}$（约8 s），若均匀地壳内P波速度已知为6 km/s，试估计地幔顶部的P波速度和地壳厚度。（感谢树洞Louis提醒）\n\n$T = \\cfrac {X - (2H - h)\\tan\\theta_c} {v_2} + \\cfrac {(2H - h)~/ \\cos\\theta_c} {v_1} = \\cfrac {1} {v_2} + \\left( \\cfrac {(2H - h)~/ \\cos\\theta_c} {v_1} - \\cfrac {(2H - h)\\tan\\theta_c} {v_2} \\right),$ $\\text{斜率 } k = \\cfrac {1} {v_2},~ \\text{截距 } b = (2H - h) \\cfrac {v_2 - v_1 \\sin \\theta_c} {v_1 v_2 \\cos \\theta_c}.$\n\n$v_2 = \\cfrac {1} {k} = \\pu{8 km/s},$ $\\sin \\theta_c = \\cfrac {v_1} {v_2} = \\cfrac {3} {4}, ~ \\cos \\theta_c = \\sqrt {1 - \\sin^2 \\theta_c} = \\cfrac {\\sqrt{7}} {4},$ $2H - h = \\cfrac {b v_1 v_2 \\cos \\theta_c} {v_2 - v_1 \\sin \\theta_c} = \\pu{72 km},~ H = \\pu{41 km}.$\n\n### 第4章 地球内部的结构\n\n• 地壳是通过研究首波而发现的。\n• 大陆地区地壳平均厚度为35 km，海洋地区地壳不到10 km。大洋和大陆下面的地壳厚度不同。\n• 用面波可以研究地壳性质，因为面波在地壳中运行。\n• 液态外核的发现来源于人们发现了影区存在。固体内核的发现者是丹麦地震学家英格·莱曼。\n• 为什么核幔边界叫古登堡面而不是奥尔德姆面：\n• 古登堡利用了反射震相。\n• 古登堡利用了更丰富的地震记录。\n• 古登堡更精确地计算了地核深度。\n• 应用地震波去透视地球内部首先要研究地震图。\n• 上地幔（410 km以上）、过渡层（410 ~ 670 km之间）及下地幔（670 km以下）。\n• 古登堡教授拥有更丰富的地震记录，得出了更精确的核介面深度估计，首次估计出地核深度为2900 km。\n• 地球内核是固体，最硬。地球外核是液态的，故横波不能通过外核。\n• 壳幔介面是莫霍面（M介面），幔核介面是古登堡面（G介面）。\n• 反演不一定准确。\n\n### 第5章 地震机制\n\n• 破裂面扩展速度不会超过P波速度。\n• 地震效率7.5% ~ 15%。\n• 说明：地震效率是指一次地震中地震波释放的能量在整个应变能中所占的比例，其值等于地震波能量除以地震能。\n• 弹性回跳理论只是一种假说，而非事实。\n• 说明：弹性回跳理论的主要论点：岩石体破裂是由于相对位移产生的应变超过岩石强度；这种相对位移是在一个比较长的时期内逐渐达到最大值的；地震时发生的唯一物质移动是破裂面两边的物质向减少弹性应变的方向突然发生弹性回跳；地震引起的振动源于破裂面；地震时释放的能量在岩石破裂前是以弹性应变能的形式储存在岩石中的。\n• 环太平洋地震带集中了全球80%的浅震、90%的中源地震以及全部深震。\n• 92%的地震发生在地壳中，8%发生在地幔上部，其上半部分多于下半部分。\n• 地震学为板块构造学说提供三方面的证据：\n• 天然地震的空间位置，可以勾画板块的边界。\n• 震源机制解，可以确定板块活动的力学性质。\n• 地震测深所得深度速度分布，可以给出板块运动的物理条件。\n• 全球性的地震带有三个：环太平洋地震带、阿尔卑斯-喜马拉雅地震带（亚欧地震带）和海岭地震带（大洋中脊地震带）。\n• 地震较多的国家：日本、意大利、智利、秘鲁、土耳其。\n• 我国地震较多的省级行政区：四川、云南、甘肃、新疆、西藏、台湾。\n• 地震按来源分类：\n• 天然地震：构造、火山、陷落。\n• 人工：工业爆破、地下核爆。\n• P波初动一直被用来确定震源机制解。爆炸产生的P波垂直向初动都是向上的，初动在震源球上无四象限分布特征，沙滩球全黑；正断层中间白，逆断层中间黑。\n• 断层是岩石破裂面的相互移动，地震沿断层发生。断层面几何与应力相关。地震的震源机制解是断层作用类型的图示描述。虽然断层类型可以由地震波初动确定，但不能确定哪一个是真正的断层面。有两种可能：真平面－断层面；假平面－辅助面。\n• 人沿走向看去，断层上盘在右，走向用正北顺时针量至走向方向的角度来表示。\n• 判断走滑断层左旋和右旋的方式：如果一个观察者站在断层的一侧，面向断层，另一边的岩块向他左方滑动，那它就叫左滑（左旋）。或者说，从上往下看，顺时针是右旋，逆时针是左旋。\n• 地震波能量和震级的通用关系为：\n$\\lg E = 11.8 + 1.5M.$\n\n### 第6章 地震仪及地震基本参数的测定\n\n• 三角测量法：通过直接的三角测量发现震中的位置。\n• 里氏震级$M_{\\text{L}}$是最大地震波振幅以10为底的对数。\n• 伍德-安德森的特殊地震仪记录到的振幅测量精度达到1 μm。（感谢@mike.frank提醒）\n\n### 第7章 地震预报\n\n• 地震灾害具有某种周期性。\n• 地震预报不等于地震预测。地震预测的精度差，地震预报要求三要素（时间、地点、震级大小）。\n\n### 第8章 宏观地震学\n\n• 震级和烈度的主要区别：\n• 震级是微观标准，而烈度是宏观标准。\n• 震级唯一，而烈度不唯一。\n• 某地的基本烈度是该地475年一遇的地震烈度。\n\n### 第9章 勘探地震学\n\n• 石油勘探的地球物理方法（简称“物探”）：重力勘探、磁性勘探、电法勘探、地震勘探。\n• 说明：物探法主要利用各种物探仪器记录到来自地下的多种信息，然后对底下的构造和岩性特性做出预测，是间接探测方法，勘探范围深、效率高。\n• 重力勘探：利用岩石密度的差别在地面上测量重力变化。磁性勘探：利用岩石磁性的差别在地面上测量地磁场的变化。电法勘探：利用岩石电阻率的差别在地面上测量电场的变化。地震勘探：利用岩石波阻抗的差别在地面上记录地震波场的变化。\n• 地震勘探精度高、分辨率高、探测深度大，应用最为广泛，占97%。\n• “怪坡”是假的。\n• 说明：所谓“怪坡”是指物体可以自行上坡的奇异现象。\n• 为什么在求取地下构造信息时，以反射波取代折射波？\n• 折射波在地质条件复杂的情况下, 容易导致解释错误。\n• 说明：约束条件少，解析解数目多。\n• 反射波信噪⽐⾼，易辨识。\n• 地震资料解释的成功率不高，约为25%。\n• 说明：实际应用中应综合前人已经做出的成果、钻井的资料、重力数据、地磁数据、地质资料等来进行解释。\n\n### 第10章 海啸\n\n• 浅水波：用$h$代表海水的深度，$\\lambda$代表波长，则$\\lambda \\gg h$的这种具有非常长的波长的重力波就叫做潜水波。\n• 浅水波的传播速度只与海水深度有关，有$v = \\sqrt{gh}$，其中：$v$为波的传播速度，$g$为重力加速度，$h$为海水深度。\n• 海啸的波长约100 miles，传播速度约500 miles/h。\n• 计算题：如，计算浅水波从$A$点传播到$D$点所用的时间。\n\n$v^2 = gh.$\n\n$2v \\cfrac {\\mathrm dv}{\\mathrm dt} = g \\cfrac {\\mathrm dh}{\\mathrm dt},$\n\n$\\cfrac {\\mathrm dh}{\\mathrm dx} = \\tan \\theta,$\n\n$2v \\cfrac {\\mathrm dv}{\\mathrm dt} = g \\cfrac {\\mathrm dh}{\\mathrm dx}\\cfrac {\\mathrm dx}{\\mathrm dt} = g \\tan \\theta \\cdot v,$\n\n$\\cfrac {\\mathrm dv}{\\mathrm dt} = \\cfrac {g \\tan \\theta} {2},$\n\n### 讨论课\n\n• 中国古代木质结构建筑的抗震原理：柔性框架、台基隔震、斗拱擎托、榫卯连接。\n• 抗震要求：小震不坏、中震可修、大震不倒。\n• 地震预警存在一些局限。\n• 汶川大地震的形成是逆冲型地震，右旋。青藏高原往东北冲，四川盆地往西南冲。\n• 海啸与地震的关系：\n• 破坏性海啸多发生在地震带上。\n• 震中距海岸越远，海啸滞后于地震的时间越长。\n• （海啸强度）取决于地震变形区上覆水体的强度、震向快速位移量和变形区面积。\n• 破坏性海啸需要有大的水体被扰动。\n\n## 课程作业\n\n### 1. 直达波、反射波和首波\n\n##### 提示\n\n$T_1 = \\cfrac {X} {v_1},$\n\n$T_2 = \\cfrac {X - 2h\\tan\\theta_c} {v_2} + \\cfrac {2h~/ \\cos\\theta_c} {v_1},$\n\n##### 解答\n\n$T_1 = \\cfrac {X} {v_1}.$\n\n$T_{2,~1} = \\cfrac {2h / \\cos\\theta_c} {v_1},$\n\n$T_{2,~2} = \\cfrac {X - 2h\\tan\\theta_c} {v_2},$\n\n$\\sin\\theta_c = \\cfrac {v_1}{v_2},~\\cos\\theta_c = \\sqrt{\\cfrac {v_2^2 - v_1^2} {v_2^2}},$\n\n$T_2 = \\cfrac {X - 2h\\tan\\theta_c} {v_2} + \\cfrac {2h~/ \\cos\\theta_c} {v_1}.$\n\n$X \\cfrac {v_2 - v_1}{v_1v_2} \\ge 2H \\cfrac {v_2 - v_1 \\sin\\theta_c}{v_1v_2\\cos\\theta_c},$\n\n$X_M = 2H \\sqrt {\\cfrac {v_2 + v_1} {v_2 - v_1}}.$\n\n### 2. 弹性波在介面上的反、透射\n\n##### 题干\n\n$\\cfrac {\\sin\\theta_{\\alpha1}} {v_{\\alpha1}} = \\cfrac {\\sin\\theta_{\\beta1}} {v_{\\beta1}} = \\cfrac {\\sin\\theta_{\\alpha2}} {v_{\\alpha2}} = \\cfrac {\\sin\\theta_{\\beta2}} {v_{\\beta2}}.$\n\n（题图见\n\n##### 提示\n\n$t = \\cfrac {\\sqrt {h^2 + x^2}}{v_{\\alpha1}} + \\cfrac{\\sqrt{r^2 + (L - x)^2}}{v_{\\beta1}}.$\n\n##### 解答\n\n$t = \\cfrac {\\sqrt {h^2 + x^2}}{v_{\\alpha1}} + \\cfrac{\\sqrt{r^2 + (L - x)^2}}{v_{\\beta1}}.$\n\n$\\cfrac {\\mathrm dt}{\\mathrm dx} = \\cfrac {x}{v_{\\alpha1}\\sqrt{h^2 + x^2}} - \\cfrac {L - x}{v_{\\alpha2}\\sqrt{r^2 + (L - x)^2}},$\n\n$\\cfrac {x}{\\sqrt{h^2 + x^2}} = \\sin\\theta_{\\alpha1}, ~ \\cfrac {L - x}{\\sqrt{r^2 + (L - x)^2}} = \\sin\\theta_{\\alpha2}.$\n\n$\\cfrac {\\sin\\theta_{\\alpha1}} {v_{\\alpha1}} - \\cfrac {\\sin\\theta_{\\alpha2}} {v_{\\alpha2}} = 0,$\n\n$\\cfrac {\\sin\\theta_{\\alpha1}} {v_{\\alpha1}} = \\cfrac {\\sin\\theta_{\\alpha2}} {v_{\\alpha2}}.$\n\n## 网络作业\n\n### 问题1\n\n##### 解答\n\n(10 km, 10 km)。\n\n$t = s \\left ( \\cfrac {1} {v_\\mathrm{P}} - \\cfrac {1} {v_\\mathrm{S}} \\right ).$\n\n$s = t \\cfrac {v_\\mathrm{S} - v_\\mathrm{P}} {v_\\mathrm{S} v_\\mathrm{P}}.$\n\n$\\cfrac {v_\\mathrm{S} - v_\\mathrm{P}} {v_\\mathrm{S} v_\\mathrm{P}} = \\pu{20 m/s}.$\n\n### 问题2\n\n##### 解答\n\n1000。\n\n《地震概论》第76页，地震波能量和震级的通用关系为：\n\n$\\lg E = 11.8 + 1.5M.$\n\n（感谢@mike.frank提醒）\n\n《地震概论》教材第77页。\n\n4。\n\n### 问题5\n\n##### 解答\n\nP波上下振动，先到达；S波左右振动，后到达。\n\nXzonn\n\n2020年1月7日\n\n《地震概论》成绩公布后于燕园" ]

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[ "## Saturday, May 22, 2010\n\n### HP 50G Determinants, Trace, Rank,Norms, Characteristic Polynomials, Eigen(Values|Vectors)\n\nLet M be a matrix. In the HP 50G, various functions are available to obtain properties of the matrix.\n\n det(M) determinant trace(M) trace rank(M) rank cond(M) condition number abs(M) Frobenius norm, $$\\sqrt{\\sum x_{i,j}^2}$$ pcar(M) characyeristic polynomial if M is square. eigvl(M) eigenvalues of M egv(M) eigenvectors and eigenvaluesof M srad(M) spectral radius = abs. value of largest eigenvalue of M. rnrm(M) row norm is the maximum of all sums of the absolute values of elements in each row. snrm spectral norm of M, or largest singular value of M. cnrm column norm is the maximum of all sums of the absolute values of elements in each column.\n\nLet us apply this to a random 3 x 3 matrix by calling the ranm function.\nranm({3,3}) sto M\nThis stored in my hp 50g the following matrix:\n\n$\\begin{array}{lll} -7 & 4 & -6 \\\\ -4 &-1 & 4 \\\\ 1 & 8 & 5 \\\\ \\end{array}$\n\nThe outputs for the functions (except for egvl) is listed in the following:\n• trace(M)\n• -3\n• rank(M)\n• 3\n• cond(M)\n• 4.35304990758\n• abs(M)\n• 14.9666295471\n• pcar(M)\n• $$x^{3.} + 3.\\cdot x^{2.} + -43., \\cdot X - 541.$$\n• egv(M)\n• $$\\{[[(1., 0),(1.,0), (-..250470319529,0.)] [(.217566565083,..5991855957363),(.2217566565093,-.5991757363),.50919544707,0.)] [(-.036364186761, -.456410356317),(-.036364186761,.456410356317),(1.,0.)]][(-5.911548611906, 5.135116596735), (-5.91154861906, -513516596735), (8.82309723813,0.)]\\}$$\n• 8.82309723813\n• rnrm(M)\n• 17\n• snrm(M)\n• 10.1517717328\n• cnrm\n• 15\n\nNotes:\n\nMay 22\n\n1. Wondering why the Latex plugin refuses to display the formula for an array.\n2. How to split a long Latex output? Seems Wordpress is better in this regard." ]

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[ "# Circular Dendrograms\n\n## Introduction\n\nIt’s been several months since I wrote a series of articles that treated the development of dendrograms through the use of the JavaScript library D3. However, I recently received a series of requests to deepen the topic, which apparently seems to have enjoyed a fairly successful.\n\nIn this regard, I decided to start writing articles on this beautiful bookcase, dealing one after another with a lot of topics and representations. To start over, I decided to conclude the topic of dendrograms with a fifth article.\n\nIf you are interested in deepening the topic, within the book Beginning JavaScript Charts are presented various methodologies of data acquisition, from extracting data from a table of a database (via PHP), reading and parsing data Contained in an external file. It also explains in detail the JSON format, and thanks to many examples it is shown how practical it is to use it on these occasions.\n\nFor anyone who wants to start from previous articles…\n\n## The starting Code\n\nLet’s start from the code dendrogram01.html in the article how to make a dendrogram with the library D3 (part 2) that produced the following dendrogram.\n\nFor convenience, I propose here the JavaScript code that generates it:\n\n``````var width = 600;\nvar height = 500;\nvar cluster = d3.layout.cluster()\n.size([height, width-200]);\nvar diagonal = d3.svg.diagonal()\n.projection (function(d) { return [d.y, d.x];});\nvar svg = d3.select(\"body\").append(\"svg\")\n.attr(\"width\",width)\n.attr(\"height\",height)\n.append(\"g\")\n.attr(\"transform\",\"translate(100,0)\");\n\nd3.json(\"dendrogram01.json\", function(error, root){\nvar nodes = cluster.nodes(root);\n.enter().append(\"path\")\n.attr(\"d\", diagonal);\nvar node = svg.selectAll(\".node\")\n.data(nodes)\n.enter().append(\"g\")\n.attr(\"class\",\"node\")\n.attr(\"transform\", function(d) { return \"translate(\" + d.y + \",\" + d.x + \")\"; });\nnode.append(\"circle\")\n.attr(\"r\", 4.5);\nnode.append(\"text\")\n.attr(\"dx\", function(d) { return d.children ? -8 : 8; })\n.attr(\"dy\", 3)\n.style(\"text-anchor\", function(d) { return d.children ? \"end\" : \"start\"; })\n.text( function(d){ return d.name;});\n});``````\n\nThese are the CSS definitions:\n\n``````.node circle {\nfill: #fff;\nstroke: steelblue;\nstroke-width: 1.5px;\n}\n.node {\nfont: 20px sans-serif;\n}\nfill: none;\nstroke: #ccc;\nstroke-width: 1.5px;\n}``````\n\nAnd this is the contents of the DENDROGRAM02. Json file that consists of the data structure.\n\n``````{ \"name\": \"root\",\n\"children\": [\n{ \"name\": \"parent A\",\n\"children\": [\n{\"name\": \"child A1\"},\n{\"name\": \"child A2\"},\n{\"name\": \"child A3\"}\n]\n},{ \"name\": \"parent B\",\n\"children\": [\n{\"name\": \"child B1\"},\n{\"name\": \"child B2\"}\n]\n} ]\n}``````\n\nIn the course of the article we are going to modify this code in order to obtain a circular dendrogram.\n\n## Change the Code\n\nSince we are talking about a circular representation on which to radially distribute the nodes of the Dendrogram, it will be necessary to express the dimensions of the representation through the diameter, instead of the classical height and width.\n\nThen you delete the variables width and height (in the snippet are commented for convenience). You then define a radius variable, which will correspond to the extension of the drawing area. You also define the margin variables, which defines the margin to be interposing between the drawing area and the Dendrogram itself. and the variable angle, which represents the cover angle of the circular dendrogram. The value 360° expresses a lap angle and then the terminal nodes of the Dendrogram will be distributed along the entire circumference. A lesser value (for example of 120°) Porterebbee the Dendrogram representation as a fan (see Fig. 4).\n\n``````//var width = 600;\n//var height = 500;\nvar margin = 120;\nvar angle = 360; ``````\n\nAs a result, the cluster data structure must also be changed accordingly. Similarly also d3.svg.diagonal must be modified to be considered radially.\n\n``````var cluster = d3.layout.cluster()\n//.size([height, width-200]);\n//var diagonal = d3.svg.diagonal()\n// .projection (function(d) { return [d.y, d.x];});\n.projection (function(d) { return [d.y, d.x / 180* Math.PI];});``````\n\nThe definition of the drawing area must also be changed.\n\n``````var svg = d3.select(\"body\").append(\"svg\")\n//.attr(\"width\",width)\n//.attr(\"height\",height)\n.append(\"g\")\n//.attr(\"transform\",\"translate(100,0)\");\n\nAs far as the dendrogram representation is concerned, only the line expressing the svg.transform() of the nodes must be modified.\n\n``````var node = svg.selectAll(\".node\")\n.data(nodes)\n.enter().append(\"g\")\n.attr(\"class\",\"node\")\n.attr(\"transform\", function(d) { return \"rotate(\" + (d.x - 90) + \")translate(\" + d.y + \")\"; });\n//.attr(\"transform\", function(d) { return \"translate(\" + d.y + \",\" + d.x + \")\"; });``````\n\nNow all the changes to be made to the code have been executed. Here is the complete code expressed here for convenience.\n\n``````var radius = 350;\nvar margin = 120;\nvar angle = 120;\nvar cluster = d3.layout.cluster()\n.projection (function(d) { return [d.y, d.x / 180* Math.PI];}); var svg = d3.select(\"body\").append(\"svg\")\n.append(\"g\")\n\nd3.json(\"dendrogram02.json\", function(error, root){\nvar nodes = cluster.nodes(root);\n.enter().append(\"path\")\n.attr(\"d\", diagonal);\nvar node = svg.selectAll(\".node\")\n.data(nodes)\n.enter().append(\"g\")\n.attr(\"class\",\"node\")\n.attr(\"transform\", function(d) { return \"rotate(\" + (d.x - 90) + \")translate(\" + d.y + \")\"; });\nnode.append(\"circle\")\n.attr(\"r\", 4.5);\nnode.append(\"text\")\n.attr(\"dy\", \".31em\")\n.attr(\"text-anchor\", function(d) { return d.x < 180 ? \"start\" : \"end\"; })\n.attr(\"transform\", function(d) { return d.x < 180 ? \"translate(8)\" : \"rotate(180)translate(-8)\"; })\n.text(function(d) { return d.name; });\n});``````\n\nBy loading the page from the browser, we get the following representation:\n\nWe certainly can’t say that we got a nice circular dendrogram. This is because the Dendrogram is too simple in its structure. We make the structure more complex, going to add more elements to the structure contained in the JSON file that we used.\n\n``````{ \"name\": \"root\",\n\"children\": [ {\n\"name\": \"parent A\",\n\"children\": [\n{\"name\": \"child A1\"},\n{\"name\": \"child A2\"},\n{\"name\": \"child A3\"},\n{\"name\": \"child A4\"},\n{\"name\": \"child A5\"},\n{\"name\": \"child A6\"}\n]\n},{\n\"name\": \"parent B\",\n\"children\": [\n{\"name\": \"child B1\"},\n{\"name\": \"child B2\"},\n{\"name\": \"child B3\"},\n{\"name\": \"child B4\"},\n{\"name\": \"child B5\"},\n{\"name\": \"child B6\"},\n{\"name\": \"child B7\"},\n{\"name\": \"child B8\"}\n]\n},{\n\"name\": \"parent C\",\n\"children\": [\n{\"name\": \"child C1\"},\n{\"name\": \"child C2\"},\n{\"name\": \"child C3\"},\n{\"name\": \"child C4\"}\n]\n}]\n}``````\n\nNow if we reupload the page from our browser, we get:\n\nSo now things seem to work very well. Now the circular and uniform distribution of the dendrogram leaves along the circular perimeter is quite evident.\n\nSo far we have worked with dendrograms at two levels beyond the root (n = 2). Actually manageable levels can be many more. So far so good, as long as the diameter of the circular dendrogram will be large enough to correctly display all the elements.\n\n## Manage different hierarchical levels\n\nIn fact a borderline case that we have not considered so far, is for example to consider dendrograms with branches that have more levels than others.\n\nLet’s take an example:\n\n``````{ \"name\": \"root\",\n\"children\": [ {\n\"name\": \"parent A\",\n\"children\": [\n{\"name\": \"child A1\",\n\"children\": [\n{\"name\": \"child A1-1\"},\n{\"name\": \"child A1-2\"},\n{\"name\": \"child A1-3\"},\n{\"name\": \"child A1-4\"},\n{\"name\": \"child A1-5\"},\n{\"name\": \"child A1-6\"},\n{\"name\": \"child A1-7\"}\n] },\n{\"name\": \"child A2\",\n\"children\": [\n{\"name\": \"child A2-1\"},\n{\"name\": \"child A2-2\"},\n{\"name\": \"child A2-3\"}\n] },\n{\"name\": \"child A3\",\n\"children\": [\n{\"name\": \"child A3-1\"},\n{\"name\": \"child A3-2\"},\n{\"name\": \"child A3-3\"}\n] },\n{\"name\": \"child A4\",\n\"children\": [\n{\"name\": \"child A4-1\"},\n{\"name\": \"child A4-2\"},\n{\"name\": \"child A4-3\"}\n] },\n{\"name\": \"child A5\",\n\"children\": [] },\n{\"name\": \"child A6\",\n\"children\": [\n{\"name\": \"child A6-1\"},\n{\"name\": \"child A6-2\"},\n{\"name\": \"child A6-3\"}\n] }]\n},{\n\"name\": \"parent B\",\n\"children\": [\n{\"name\": \"child B1\"},\n{\"name\": \"child B2\"},\n{\"name\": \"child B3\"},\n{\"name\": \"child B4\"},\n{\"name\": \"child B5\"},\n{\"name\": \"child B6\"},\n{\"name\": \"child B7\"},\n{\"name\": \"child B8\"}\n]\n},{\n\"name\": \"parent C\",\n\"children\": [\n{\"name\": \"child C1\"},\n{\"name\": \"child C2\"},\n{\"name\": \"child C3\"},\n{\"name\": \"child C4\"}\n]\n}] }``````\n\nIf we launch the page again, we will get this…\n\nIf we look at the Dendrogram produced, we will notice immediately that the elements of a lower level (n = 4) have been placed at the same level as those with n = 3 along the outer perimeter of the circular dendrogram.\n\nOn the net there are often dendrograms expressed in this way, since the attention is centered on the terminal elements of the dendrogram (leaves), independently of their hierarchical level of belonging. In fact, if you want hierarchical information to be maintained by respecting the hierarchies of the various levels correctly, you need to replace the d3.layout.cluster with d3.layout.tree.\n\n``````var tree = d3.layout.tree()\n.size([360, diameter / 2 - 120]);\n...\nd3.json(\"dendrogram04.json\", function(error, root){\nvar nodes = tree.nodes(root);" ]

[ null ]

[ "Books Books", null, "Equal parallelograms which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional ; and parallelograms that have one angle of the one equal to one angle of the other, and their sides...", null, "Euclid's Elements: Or, Second Lessons in Geometry,in the Order of Simson's ... - Page 94\nby Dennis M'Curdy - 1846 - 138 pages", null, "## Sandhurst Mathematical Papers for Admission Into the Royal Military College ...\n\nRoyal Military College, Sandhurst - Mathematics - 1890 - 144 pages\n...the inscribed circle, construct it. 6. Equal triangles which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional. Construct an isosceles triangle equal in area to a given scalene triangle, and having a common vertical...", null, "## The Harpur Euclid: An Edition of Euclid's Elements\n\nEdward Mann Langley, W. Seys Phillips - 1890 - 538 pages\n...first. PROPOSITION 14. THEOREM. (1) Equal parallelograms which have one angle of the one equal to one angle of the other have their sides about the equal angles reciprocally proportional. (2) Conversely : — Parallelograms which have one angle of the one equal to one angle of the other...", null, "## College Algebra\n\nWebster Wells - Algebra - 1890 - 560 pages\n...similar, as also are the triangles EOG and COD ; for, by Geometry, two triangles are similar when they have an angle of one equal to an angle of the other, and the including sides proportional. Then the figure OFEG is similar to OBDC, and hence OFEG is a...", null, "## Annual Report of the Chief Superintendent of Education\n\n1891 - 720 pages\n...are to one another as their bases. 6. Equal triangles which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional. NB — Female candidates will receive credit for any correct work ou the last two of the foregoing...", null, "## The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...\n\nEuclid - Geometry - 1892 - 460 pages\n...THEOREM. • Parallelograms which are equal in area, and which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional: Conversely, parallelograms which have one angle of the one equal to one angle of the other, and the...", null, "## An Examination Manual in Plane Geometry\n\nGeorge Albert Wentworth, George Anthony Hill - Geometry - 1894 - 150 pages\n...onehalf of it. 4. The two tangents to a circle from an outside point are equal. 5. If two triangles have an angle of one equal to an angle of the other they are to each other as the product of the sides including the equal angles. 6. From the obtuse angle...", null, "## Calendar, for the Year ...\n\n1895 - 142 pages\n...construct it. 4. Triangles which are equal in area, and which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional. Describe an isosceles triangle equal in area to a given triangle and having an angle equal to one of...", null, "## Elements of Geometry, Volume 1\n\nAndrew Wheeler Phillips, Irving Fisher - Geometry, Modern - 1896 - 276 pages\n...sides by one of its other sides, an incorrect method. PROPOSITION VIII. THEOREM 398. The areas of two triangles which have an angle of one equal to an angle of the other are to each other as the products of the sides including those angles. GIVEN — the triangles ADE...", null, "## Elements of Geometry\n\nAndrew Wheeler Phillips, Irving Fisher - Geometry - 1896 - 564 pages\n...sides by one of its other sides, an incorrect method. PROPOSITION VIII. THEOREM 308. The areas of two triangles which have an angle of one equal to an angle of the other are to each other as the products of the sides including those angles. GIVEN — the triangles ADR...", null, "" ]

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[ "Permutations", null, "Printer-friendly version", null, "Example\n\nHow many ways can four people fill four executive positions?\n\nSolution. For the sake of concreteness, let's name the four people Tom, Rick, Harry, and Mary, and the four executive positions President, Vice President, Treasurer and Secretary. I think you'll agree that the Multiplication Principle yields a straightforward solution to this problem. If we fill the President position first, there are 4 possible people (Tom, Rick, Harry, and Mary).  Let's suppose Mary is named the President. Then, since Mary can't fill more than one position at a time, when we fill the Vice President position, there are only 3 possible people (Tom, Rick, and Harry). If Tom is named the Vice President, when we fill the Treasurer position, there are only 2 possible people (Rick and Harry). Finally, if Rick is named Treasurer, when we fill the Secretary position, there is only 1 possible person (Harry). Putting all of this together, the Multiplication Principle tells us that there are:\n\n4 × 3 × 2  × 1\n\nor 24 possible ways to fill the four positions.\n\nAlright, alright now... enough of these kinds of examples, eh?!  The main point of this example is not to see yet another application of the Multiplication Principle, but rather to introduce the counting of the number of permutations as a generalization of the Multiplication Principle.\n\nA Generalization of the Multiplication Principle\n\nSuppose there are n positions to be filled with n different objects, in which there are:\n\n• n choices for the 1st position\n• n − 1 choices for the 2nd position\n• n − 2 choices for the 3rd position\n• ... and ...\n• 1 choice for the last position\n\nThe Multiplication Principle tells us there are then in general:\n\nn × (n − 1) × (n − 2) × ... × 1 = n!\n\nways of filling the n positions.  The symbol n! is read as \"n-factorial,\" and by definition 0! equals 1.\n\n Definition. A permutation of n objects is an ordered arrangement of the n objects.\n\nWe often call such a permutation a “permutation of n objects taken n at a time,” and denote it as nPn. That is:\n\nnPn = n × (n − 1) × (n − 2) × ... × 1 = n!\n\nNot that it really matters in this situation (since they are the same), but the first subscripted n represents the number of objects you are wanting to arrange, while the second subscripted n represents the number of positions you have for the objects to fill.", null, "Example\n\nThe draft lottery of 1969 for military service ranked all 366 days (Jan 1, Jan 2, ..., Feb 29, ..., Dec 31) of the year. The men who were eligible for service whose birthday was selected first were the first to be drafted.  Those whose birthday was selected second were the second to be drafted.  And so on. How many possible ways can the 366 days be ranked?\n\nSolution. Well, we have 366 objects (days) and 366 positions (1st spot, 2nd spot, ... , 366th spot) to arrange them. Therefore, there are 366!  (\"366 factorial\" ) ways of ranking the 366 possible birthdays of the eligible men.", null, "What is the probability that your birthday would be ranked first?\n\n(After you've thought of how you'd solve our problem, click on the icon to reveal one possible solution.)", null, "Example\n\nIn how many ways can 7 different books be arranged on a shelf?\n\nSolution. We could use the Multiplication Principle to solve this problem. We have seven positions that we can fill with seven books. There are 7 possible books for the first position, 6 possible books for the second position, five possible books for the third position, and so on. The Multiplication Principle tells us therefore that the books can be arranged in:\n\n7 × 6 × 5 × 4 × 3 × 2 × 1\n\nor 5,040 ways. Alternatively, we can use the simple rule for counting permutations. That is, the number of ways to arrange 7 distinct objects is simply 7P= 7! = 5,040.\n\nExample\n\nWith 6 names in a bag, randomly select a name.  How many ways can the 6 names be assigned to 6 job assignments? If we assume that each person can only be assigned to one job, then we must select (or sample) the names without replacement.  That is, once we select a name, it is set aside and not returned to the bag.\n\n Definition.   Sampling without replacement occurs when an object is not replaced after it has been selected.\n\nSolution. If we sample without replacement, the problem reduces to simply determining the number of ways the 6 names can be arranged.  We have 6 objects taken 6 at a time, and hence the number of ways is 6! = 720 possible job assignments. In this case, each person is assigned to one and only one job.\n\nWhat if the 6 names were sampled with replacement?  That is, once we select a name, it is returned to the bag.\n\n Definition.   Sampling with replacement occurs when an object is selected and then replaced before the next object has been selected.\n\nSolution. If we sample with replacement, we have 6 choices for each of the 6 jobs. Applying the Multiplication Principle, there are:\n\n6 × 6 × 6 × 6 × 6 × 6 = 46,656\n\npossible job assignments. In this case, each person is allowed to perform more than one job. There's even the possibility that one (rather unlucky) person gets assigned to all six jobs!\n\nThe take-home message from this example is that you'll always want to ask yourself whether or not the problem involves sampling with or without replacement.  Incidentally, it's not all that different from asking yourself whether or not replication is allowed. Right?", null, "Example\n\nOkay, let's throw a (small) wrench into our work.  How many ways can 4 people fill 3 chairs?\n\nSolution. Again, for the sake of concreteness, let's name the four people Tom, Rick, Harry, and Mary and the chairs Left, Middle, and Right. If we fill the Left chair first, there are 4 possible people (Tom, Rick, Harry, and Mary).  Let's suppose Tom is selected for the Left chair. Then, since Tom can't sit in more than one chair at a time, when we fill the Middle chair, there are only 3 possible people (Rick, Harry, and Mary). If Rick is selected for the Middle chair, when we fill the Right chair, there are only 2 possible people (Harry and Mary).  Putting all of this together, the Multiplication Principle tells us that there are:\n\n4 × 3 × 2\n\nor 24 possible ways to fill the three chairs.\n\nOkay, okay! The main distinction between this example and the first example on this page is that the first example involves arranging all 4 people, whereas this example involves leaving one person out and arranging just 3 of the 4 people.  This example allows us to introduce another generalization of the Multiplication Principle, namely the counting of the number of permutations of n objects taken r at a time, where rn.\n\nAnother Generalization of the Multiplication Principle\n\nSuppose there are r positions to be filled with different objects, in which there are:\n\n• n choices for the 1st position\n• n − 1 choices for the 2nd position\n• n − 2 choices for the 3rd position\n• ... and ...\n• n − (r − 1)  choices for the last position\n\nThe Multiplication Principle tells us there are in general:\n\nn × (n − 1) × (n − 2) × ... × [n − (r − 1)]\n\nways of filling the r positions. We can easily show that, in general, this quantity equals:\n\n$\\dfrac{n!}{(n-r)!}$\n\nHere's how it works:\n\nAnd, formally:\n\n Definition.   A permutation of n objects taken r at a time is an ordered arrangement of n different objects in r positions.  The number of such permutations is: $_nP_r=\\dfrac{n!}{(n-r)!}$\n\nThe subscripted n represents the number of objects you are wanting to arrange, while the subscripted r represents the number of positions you have for the objects to fill.\n\nExample\n\nAn artist has 9 paintings. How many ways can he hang 4 paintings side-by-side on a gallery wall?" ]

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[ "# 06-240/Classnotes For Thursday, September 21\n\n## Force Vectors\n\nA force has a direction and a magnitude.\n\n1. There is a special force vector called 0.\n3. They can be multiplied by any scalar.\n\n#### Properties\n\n(convention:", null, "$x,y,z$ are vectors;", null, "$a,b,c$ are scalars)\n\n1.", null, "$x+y=y+x$\n2.", null, "$x+(y+z)=(x+y)+z \\$\n3.", null, "$x+0=x \\$\n4.", null, "$\\forall x\\; \\exists\\ y \\ \\mbox{ s.t. }x+y=0$\n5.", null, "$1\\cdot x=x \\$\n6.", null, "$a(bx)=(ab)x \\$\n7.", null, "$a(x+y)=ax+ay \\$\n8.", null, "$(a+b)x=ax+bx \\$\n##### Definition\n\nLet F be a field \"of scalars\". A vector space over F is a set V, of \"vectors\", along with two operations", null, "$+: V \\times V \\to V$", null, "$\\cdot: F \\times V \\to V \\mbox{, so that:}$\n1.", null, "$\\forall x,y \\in V\\ x+y=y+x$\n2.", null, "$\\forall x,y \\in V\\ x+(y+z)=(x+y)+z$\n3.", null, "$\\exists\\ 0 \\in V s.t.\\; \\forall x \\in V\\ x+0=x$\n4.", null, "$\\forall x \\in V\\; \\exists\\ y \\in V\\ s.t. \\ x+y=0$\n5.", null, "$1\\cdot x=x\\$\n6.", null, "$a(bx)=(ab)x\\$\n7.", null, "$a(x+y)=ax+ay\\$\n8.", null, "$\\forall x \\in V\\ ,\\forall a,b \\in F\\ (a+b)x=ax+bx$\n\n9.", null, "$x \\mapsto \\vert x\\vert \\in \\mathbb{R} \\ \\vert x+y\\vert \\le \\vert x\\vert+\\vert y\\vert$\n\n#### Examples\n\nEx.1.", null, "$F^n= \\lbrace(a_1,a_2,a_3,\\ldots,a_{n-1},a_n):\\forall i\\ a_i \\in F \\rbrace$", null, "$n \\in \\mathbb{Z}\\ , n \\ge 0$", null, "$x=(a_1,\\ldots,a_2)\\ y=(b_1,\\ldots, b_2)\\$", null, "$x+y:=(a_1+b_1,a_2+b_2,\\ldots,a_n+b_n)\\$", null, "$0_{F^n}=(0,\\ldots,0)$", null, "$a\\in F\\ ax=(aa_1,aa_2,\\ldots,aa_n)$", null, "$\\mbox{In } \\mathbb{Q}^3 \\ \\left( \\frac{3}{2},-2,7\\right)+\\left( \\frac{-3}{2}, \\frac{1}{3},240\\right)=\\left(0, \\frac{-5}{3},247\\right)$", null, "$7\\left( \\frac{1}{5},\\frac{1}{7},\\frac{1}{9}\\right)=\\left( \\frac{7}{5},1,\\frac{7}{9}\\right)$\nEx.2.", null, "$V=M_{m\\times n}(F)=\\left\\lbrace\\begin{pmatrix} a_{11} & \\cdots & a_{1n} \\\\ \\vdots & & \\vdots \\\\ a_{m1} & \\cdots & a_{mn}\\end{pmatrix}: a_{ij} \\in F \\right\\rbrace$", null, "$M_{3\\times 2}( \\mathbb{R})\\ni \\begin{pmatrix} 7 & -7 \\\\ \\pi & \\mathit{e} \\\\ -5 & 2 \\end{pmatrix}$", null, "$\\mbox{Addition by adding entry by entry:}$", null, "$M_{2\\times 2}\\ \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix}+\\begin{pmatrix} b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end{pmatrix}=\\begin{pmatrix} {a_{11}+b_{11}} & {a_{12}+b_{12}} \\\\ {a_{21}+b_{21}} & {a_{22}+b_{22}} \\end{pmatrix}$", null, "$\\mbox{Multiplication by multiplying scalar c to all entries by M.}$", null, "$c\\cdot M_{2\\times 2}\\ \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix}=\\begin{pmatrix} c\\cdot a_{11} & c\\cdot a_{12} \\\\ c\\cdot a_{21} & c\\cdot a_{22} \\end{pmatrix}$", null, "$\\mbox{Zero matrix has all entries = 0:}$", null, "$0_{M_{m\\times n}}=\\begin{pmatrix} 0 & \\cdots & 0 \\\\ \\vdots & & \\vdots \\\\ 0 & \\cdots & 0\\end{pmatrix}$\nEx.3.", null, "$\\mathbb{C}$ form a vector space over", null, "$\\mathbb{R}$.\nEx.4.", null, "$\\mbox{F is a vector space over itself.}$\nEx.5.", null, "$\\mathbb{R}$ is a vector space over", null, "$\\mathbb{Q}$.\nEx.6.", null, "$\\mbox{Let S be a set. Let}$", null, "$\\mathcal{F}(S,\\mathbb{R})=\\big\\{f:S\\to \\mathbb{R} \\big\\}$", null, "$f,g \\in \\mathcal{F}(S,\\mathbb{R})$", null, "$(f+g)(t)=f(t)+g(t)\\ for\\ any\\ t\\in S$", null, "$(af)(t)=a\\cdot f(t)\\$" ]

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[ "# Percent Button in Pivot Grid's Custom Calculation Window\n\nWhat is the intended use of the button with the percent sign that is in the Pivot Grid's \"Edit Expression\" or \"Add Calculated Field\" window? My first guess was that it would multiply by 100 and add a percent sign...but all of my attempts result in an error.", null, "The \"Percent\" operator calculates the integer remainder of a division. This is a \"modulo\" operator in arithmetic.\n\nExample1:\nIn \"integer division\",\n11 / 4 = 2 (with remainder of 3)\nwhile\n11 % 4 = 3 (yields the remainder after division.)\n\nExample2:\n1023 % 2 = 1\n1024 % 2 = 0\n1025 %2 = 1, etc.\n\ne.g. Performing \"% 2\" on any integer is a great way to tell if it's odd or even.\nHope this helps.\n\nNice. Thanks, Gary!" ]

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[ "# How to use SUMPRODUCT IF in Excel", null, "SUMPRODUCT is an extremely resourceful function with many purposes. When you are comparing data between two or more ranges and calculating with multiple criteria, The SUMPRODUCT function is your first choice. It has the unique ability to handle arrays in smart and elegant ways. Often we need to use SUMPRODUCT IF or conditional SUMPRODUCT to compare between columns with given criteria and to find the result. Today in this article, we will discuss the SUMPRODUCT IF function and some other alternative approaches to SUMPRODUCT IF function.\n\n## Excel SUMPRODUCT function\n\nTechnically, the “SUMPRODUCT” function returns the sum of the values of corresponding arrays or ranges. Normally, the default operation is multiplication, but addition, subtraction, and division can also be done.\n\n### Syntax\n\nThe syntax of the SUMPRODUCT function is simple and direct.\n\n=SUMPRODUCT(array1, [array2], [array3], …)\n\n### Argument\n\n• [array1]: The first array or ranges of cells whose components you want to multiply and then add\n• [ array2], [array3]…: Array arguments 2 to 255 whose values you want to multiply and then add.\n\n## SUMPRODUCT IF Formula: 2 Types\n\nIn Excel, there is no built-in “SUMPRODUCT IF” function but you can use this as an array formula by combining the “SUMPRODUCT” and the “IF” function. Let’s discuss this formula.\n\n### 1.   SUMPRODUCT IF with One Criteria\n\nWe can use the “SUMPRODUCT IF” formula with one criterion. Follow these steps to learn\n\nStep-1:\n\nConsider a data table where some fruit Items are given with “Region”, “Qty”, and “Price”. We will find out the total price of some items.", null, "Step-2:\n\nCreate another table anywhere in the worksheet where you want to get the total price of the item. We choose “Cherry”, “Apple”, “Plum” items.", null, "Step-3:\n\nApply the “SUMPRODUCT IF” formula in cell “H4”. The format of this formula is\n\n=SUMPRODUCT(IF(criteria range=criteria, values range1*values range2))\n\nInsert the values into the formula.\n\n=SUMPRODUCT(IF(\\$C\\$4:\\$C\\$20=G4,\\$D\\$4:\\$D\\$20*\\$E\\$4:\\$E\\$20))\n\nWhere,\n\n• Criteria_range is \\$C\\$4:\\$C\\$20.\n• The Criteria is\n• Values_range1 is \\$D\\$4:\\$D\\$20.\n• Values_range2 is \\$E\\$4:\\$E\\$20.", null, "Apply this formula as an array formula by pressing “CTRL+SHIFT+ENTER” simultaneously.", null, "Step-4:\n\nWe got our total price. Now apply the same formula for the rest of the items.", null, "### 2. SUMPRODUCT IF with Multiple Criteria in Different Column\n\nWe will use the same formula for multiple criteria.\n\nStep-1:\n\nLet’s add another criteria “Region” in table 2. In this case, we want to find the total price of “Cherry” from the “North America” region and “Apple” from the “Asia” region.", null, "Step-2:\n\nNow apply the “SUMPRODUCT IF” formula. Insert the values into the formula.\n\n=SUMPRODUCT(IF(\\$B\\$4:\\$B\\$20=G4,IF(\\$C\\$4:\\$C\\$20=H4,\\$D\\$4:\\$D\\$20*\\$E\\$4:\\$E\\$20)))\n\nWhere,\n\n• Criteria_range is \\$B\\$4:\\$B\\$20, \\$C\\$4:\\$C\\$20.\n• The Criteria is G4, H4.\n• Values_range1 is \\$D\\$4:\\$D\\$20.\n• Values_range2 is \\$E\\$4:\\$E\\$20.", null, "Apply this formula by pressing “CTRL+SHIFT+ENTER” simultaneously.", null, "Step-3:\n\nOur value is here. Now do the same for the “Apple” item.", null, "## Alternative approaches to SUMPRODUCT IF Formula\n\nThere are some other approaches to derive the previous results. An alternative way to calculate the “SUMPRODUCT IF” formula is to insert the criteria within the “SUMPRODUCT” function as an array using double unary (–) to converts the “TRUE” or “FALSE” into “1” or “0”.\n\n### 1. SUMPRODUCT If with One Condition\n\nStep-1:\n\nWe will consider the previous example in this case. We will find the total price of “Mango” from the list.", null, "Step-2:\n\nApply the conditional “SUMPRODUCT” formula\n\nThe final formula is\n\n=SUMPRODUCT(–(C4:C20=G4),D4:D20,E4:E20)\n\nWhere,\n\n• Array1 is (–(C4:C20=G4).\n• [Array2] is D4:D20.\n• [Array3] is E4:E20.", null, "Press “Enter”. Our result is here.", null, "Explanation:\n\nWe will now explain how this conditional “SUMPRODUCT” function works\n\nStep-1:\n\nWhen we enter the “–(C4:C20=G4)” into the formula this double unary (–) converts the “TRUE” or “FALSE” into “1” or “0”. Select this “–(C4:C20=G4)” portion in your worksheet and press “F9” to see the underlying values.\n\n{0,1,0,0,0,0,0,0,1,1,0,1,1,0,0,0,1}\n\nStep-2:\n\nNow if we break down the arrays into values the actual formula will look like this,\n\n=SUMPRODUCT({0,1,0,0,0,0,0,0,1,1,0,1,1,0,0,0,1},{20,50,100,50,500,80,90,56,60,80,100,40,45,25,35,45,90},{15,20,2,5,12,10,15,9,20,30,5,30,25,10,10,20,15})\n\nStep-3:\n\nThe first array will multiply with the second then the second array will multiply with the third array. Follow this picture", null, "That is how this conditional “SUMPRODUCT” works.\n\n### 2. SUMPRODUCT IF with Multiple Condition in Different Column\n\nStep-1:\n\nIn the following example, we will find out the total price of “Cherry” from the “North America” region.", null, "Step-2:\n\nApply the formula. The final form of this formula is,\n\n=SUMPRODUCT(–(B4:B20=G4), –(C4:C20=H4),D4:D20,E4:E20)\n\nWhere,\n\n• Array1 is (–(C4:C20=G4),–(C4:C20=H4).\n• [Array2] is D4:D20.\n• [Array3] is E4:E20.", null, "Press “Enter”. Our result is achieved.", null, "### 3. SUMPRODUCT IF with OR Logic\n\nWe can add “OR” logic to our formula to make this formula more dynamic.\n\nStep-1:\nLet’s say, we need to get the total price of “Mango” and “Cherry” from the data table.", null, "Step-2:\n\nApply the “SUMPRODUCT” formula with “OR” and insert the values.\n\nThe formula is\n\n=SUMPRODUCT(–((C4:C20=G4)+(C4:C20=H4)>0),D4:D20,E4:E20)\n\nWhere,\n\n• Array1 is –((C4:C20=G4)+(C4:C20=H4)>0). Here G4 is “Mango” and H4 is “Cherry”. This array counts the total number of “Mango” and “Cherry” in the data table.\n• [Array2] is D4:D20.\n• [Array3] is E4:E20.", null, "Press “Enter” to get the total price of the products.", null, "### 4. SUMPRODUCT IF with Multiple AND/OR Criteria\n\nIn this case, we will apply Or logic with multiple conditions.\n\nStep-1:\n\nIn the following example, we need to find the total price of “Cherry” and “Mango” from “Asia” and “Europe” regions.", null, "Step-2:\n\nTo get the result we will now apply the formula with AND/OR logic. The formula is\n\n=SUMPRODUCT(–((B4:B20=G4)+(B4:B20=H4)>0),–((C4:C20=I4)+(C4:C20=J4)>0),D4:D20,E4:E20)\n\nWhere,\n\n• Array1 is –((B4:B20=G4)+(B4:B20=H4)>0),–((C4:C20=I4)+(C4:C20=J4)>0). Here B4:B20 is “Region” Column, G4 and H4 is “Asia” and “Europe”.Similarly,C4:C20 is “Item” column, I4 and J4 is “Cherry” and “Mango”.\n• [Array2] is D4:D20.\n• [Array3] is E4:E20.", null, "Press “Enter” to get the total price.", null, "## Quick Notes\n\n✅Arrays in the SUMPRODUCT formula must have the same number of rows and columns. If not, you get the #VALUE! Error.\n\n✅The “SUMPRODUCT” function treats non-numeric values as zeros. If you have any non-numeric values in your formula the answer will be “0”.\n\n✅ Since the “SUMPRODUCT IF” formula is an array formula your need to press “CTRL+SHIFT+ENTER” simultaneously to apply the formula.\n\n✅The “SUMPRODUCT” function does not support wildcard characters.\n\n## Conclusion\n\nToday we discussed the “SUMPRODUCT IF” formula and some other alternative conditional “SUMPRODUCT” formula in this article. Hope this article is useful to you when you are facing problems. If you have any confusion, you are welcome to comment.", null, "", null, "" ]

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[ "", null, "", null, "", null, "# Recent posts under tag charging", null, "+1 vote", null, "• lte\n• 5g\n• telecom\n• charging", null, "• lte\n• charging\n+1 vote", null, "• lte\n• charging", null, "• charging\n• lte\n• umts\n• sevices\n• networks", null, "• lte\n• gtp\n• sgw\n• pgw\n• charging\n+1 vote", null, "• charging\n• lte", null, "• ims\n• charging\n• diameter\n• lte\n• interview questions", null, "• lte\n• telecom\n• ims\n• charging\n• policy\n+1 vote", null, "• lte\n• telecom\n• ims\n• charging", null, "• lte\n• volte\n• charging\n• ims", null, "• policy\n• lte\n• pgw\n• charging\n• sgw", null, "• lte\n• policy\n• charging\n• pgw\n• traffic\n+1 vote", null, "• lte\n• diameter\n• charging\n+1 vote", null, "• diameter\n• charging\n• pcrf\n• pcef\n• ocs", null, "• lte\n• telecom\n• lwa\n• lte-u\n• charging", null, "• lte\n• charging\n• ofcs\n• ocs\n• pgw", null, "• volte\n• ims\n• charging\n• lte\n• diameter\n+1 vote", null, "• diameter\n• lte\n• charging\n• epc", null, "• lte\n• diameter\n• charging\n• epc", null, "• lte\n• diameter\n• charging\n• pcrf", null, "• lte\n• diameter\n• charging\n• 3gpp", null, "• lte\n• pgw\n• policy\n• charging\n• traffic", null, "• lte\n• charging\n• policy\n• traffic", null, "• lte\n• diameter\n• charging", null, "• diameter\n• lte\n• ofcs\n• gx-interface\n• charging", null, "• lte\n• charging\n• ims", null, "• lte\n• ocs\n• pcrf\n• pcef\n• charging\n+1 vote", null, "• lte\n• charging\n• diameter\n• pcrf\n• ocs", null, "• charging\n• ss7\n• diameter\n• camel\n• osi layer", null, "• diameter\n• charging\n• lte\n+1 vote", null, "• lte\n• charging\n+1 vote", null, "• lte\n• charging", null, "• lte\n• charging\n• roaming", null, "• lte\n• charging", null, "• lte\n• charging", null, "• lte\n• charging\n+1 vote", null, "• wireless\n• charging", null, "• charging", null, "• lte\n• mbms\n• charging", null, "• charging\n• lte\n• ofcs\n• ocs", null, "", null, "• lte\n• pcef\n• charging\n• epc\n+1 vote", null, "", null, "• lte\n• charging\n• telecom\n• gsm\n• umts", null, "• lte\n• charging\n• telecom", null, "• charging\n• lte\n• epc\n• diameter", null, "• charging\n• lte\n• diameter\n+1 vote", null, "", null, "" ]

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[ "# Weighted Average Cost Of Capital", null, "### Calyxt Inc (CLXT)\n\n\\$3.72\n0.03 (+0.81%)\nShare price \\$ 3.72 2.255 32.81 0.00 5.00% 20.406 18.60 122.04 140.64 13.23 86.77\n\nThere are a number of methods that can be used to determine discount rates. A good approach – and the one we’ll use in this tutorial – is to use the weighted average cost of capital (WACC) – a blend of the cost of equity and after-tax cost of debt. A company has two primary sources of financing – debt and equity – and, in simple terms, WACC is the average cost of raising that money. WACC is calculated by multiplying the cost of each capital source (debt and equity) by its relevant weight and then adding the products together to determine the WACC value." ]

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[ "# CHM130Lab 4CalorimetryName Mengqi LiData Table 12\n\n1501 words 7 pages\nCHM130\n\nLab 4\n\nCalorimetry\n\nName: Mengqi Li\n\nData Table: (12 points)\n\nALUMINUM METAL\n\nPre-weighed Aluminum metal sample mass (mmetal) 50.0 grams\nTemperature of boiling water and metal sample in the pot (Ti(metal))\n\ndsdfa(Ti 90.0 C\nTemperature of cool water in the calorimeter prior to adding hot metal sample (Ti(water)) 17.0 C\nMaximum Temperature of water/metal in calorimeter after mixing (Tf) 47.0 C\n\nPre-weighed Lead metal sample mass (mmetal) 50.0 grams\nTemperature of boiling water and metal sample in the pot (Ti(metal)) 95.0 C\nTemperature of cool water in the calorimeter prior to adding hot metal sample (Ti(water)) 17.0 C\nMaximum Temperature of water/metal in calorimeter after mixing (Tf) 50.0 C\n\nIRON\n(10 points)\n\nQ=m(sh)ΔT=75*42.0*4.184=13000J\n\n17. Use the equation: q = m(SH)ΔT to solve for the specific heat of the metal.\nFor q, you found how much heat was gained by the water so you know that same amount of heat was lost by the metal. Therefore, qmetal = -qwater. The mass of the sample was recorded from the baggie. The temperature change of the metal can be found in calculation #14. (10 points)\n\nM(metal)*SH(metal)*ΔT(metal)=-m(water)*sh(water)*ΔT(water)\n50.0*sh(metal)*(-40.0)=-75*4.184*42.0\nSh(metal)=6.59J/g0C\n\n18. Determine the percent error using the equation and knowing that the actual specific heat of iron is 0.450 J/g(oC): (10 points)\nPercent Error = actual-experimental x 100\nActua\n=|0.450-6.59|/0.450*100\n=1360%\n\nFollow-Up Questions:\n19. The First Law of Thermodynamics states that heat lost by one object is gained by another? For this lab experiment, what substance(s) gained energy and what substance(s) lost energy? (8 points)\nBecause one object lost heat and another will gain the heat. In this lab experiment, all metal lost energy, and all water gain energy.\n20. To determine the specific heat of the metal, you assumed that the amount of heat lost by the metal equals the amount of heat gained by the water. Do you think this was a valid assumption to make based on your percent error? Discuss what may have occurred that would make your calculated value not equal to the actual value. (15 points)\n\nAlthough some\n\n## Related\n\n• ###### 12 Angry Men Analysis\n936 words | 4 pages\n• ###### Variable Cost and Following Table\n1952 words | 8 pages\n• ###### 12 Angry Men: Analysis of the Film\n3936 words | 16 pages\n• ###### Industrial Psychology in 12 Angry Man\n3455 words | 14 pages\n• ###### Ilm Level 3 M.12\n3071 words | 13 pages\n• ###### Experiment 12: Calorimetry\n1884 words | 8 pages\n• ###### Pc Specification Table\n1942 words | 8 pages\n• ###### 12 Angry Men: Leadership\n1940 words | 8 pages\n• ###### Table Controls at Bellagio Casino Resort\n2086 words | 9 pages\n• ###### case 12\n1924 words | 8 pages" ]

[ null ]

[ "", null, "", null, "[BSS10] On the use of Weighted Regression in Conjoint Analysis\n\nConférence Internationale avec comité de lecture : COMPSTAT'2010, 19th International Conference on Computational Statistics, Paris, August 2010, pp.777-784,\nmotcle:\nRésumé: Conjoint analysis seeks to explain an ordered categorical ordinal variable according to several variables using a multiple regression scheme. A common problem encountered, there, is the presence of missing values in classi cation-ranks. In this paper, we are interested in the cases where consumers provide a ranking of some products instead of rating these products (i.e. explained variable presents missing values).In order to deal with this problem, we propose a weighted regression scheme. We empirically show (in several cases of weighting) that, if the number of missing values is not too large, the data remain useful, and our results are close to those of the complete order. A simulation study con rms these ndings.\n\nEquipe: msdma\nCollaboration: CML\n\nBibTeX\n\n @inproceedings { BSS10, title = \"{On the use of Weighted Regression in Conjoint Analysis}\", author = \" S. Benammou and B. Souissi and G. Saporta \", booktitle = \"{COMPSTAT'2010, 19th International Conference on Computational Statistics, Paris}\", year = 2010, month = \"August\", pages = \"777-784\", }" ]

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[ "# Bravais lattices¶\n\nThis package provides two features that are mostly used independently of each other:\n\n• Bravais lattice objects, which represent primitive cells and Brillouin zone information which is useful for calculating band structures\n\n• A general framework for building Atoms objects based Bravais lattice and basis\n\nclass ase.lattice.BravaisLattice(**kwargs)[source]\n\nRepresent Bravais lattices and data related to the Brillouin zone.\n\nThere are 14 3D Bravais classes: CUB, FCC, BCC, …, and TRI, and five 2D classes.\n\nEach class stores basic static information:\n\n>>> from ase.lattice import FCC, MCL\n>>> FCC.name\n'FCC'\n>>> FCC.longname\n'face-centred cubic'\n>>> FCC.pearson_symbol\n'cF'\n>>> MCL.parameters\n('a', 'b', 'c', 'alpha')\n\n\nEach class can be instantiated with the specific lattice parameters that apply to that lattice:\n\n>>> MCL(3, 4, 5, 80)\nMCL(a=3, b=4, c=5, alpha=80)\n\nbandpath(path=None, npoints=None, special_points=None, density=None) [source]\n\nReturn a BandPath for this lattice.\n\nSee ase.cell.Cell.bandpath() for description of parameters.\n\n>>> BCT(3, 5).bandpath()\nBandPath(path='GXYSGZS1NPY1Z,XP', cell=[3x3], special_points={GNPSS1XYY1Z}, kpts=[51x3])\n\n\nNote\n\nThis produces the standard band path following AFlow conventions. If your cell does not follow this convention, you will need ase.cell.Cell.bandpath() instead or the kpoints may not correspond to your particular cell.\n\ncellpar() ndarray[source]\n\nGet cell lengths and angles as array of length 6.\n\nconventional() [source]\n\nGet the conventional cell corresponding to this lattice.\n\ndescription() str[source]\n\nReturn complete description of lattice and Brillouin zone.\n\nget_special_points() Dict[str, ndarray][source]\n\nReturn a dictionary of named special k-points for this lattice.\n\nget_special_points_array() ndarray[source]\n\nReturn all special points for this lattice as an array.\n\nOrdering is consistent with special_point_names.\n\nplot_bz(path=None, special_points=None, **plotkwargs)[source]\n\nPlot the reciprocal cell and default bandpath.\n\nproperty special_path: str\n\nGet default special k-point path for this lattice as a string.\n\n>>> BCT(3, 5).special_path\n'GXYSGZS1NPY1Z,XP'\n\nproperty special_point_names: List[str]\n\nReturn all special point names as a list of strings.\n\n>>> BCT(3, 5).special_point_names\n['G', 'N', 'P', 'S', 'S1', 'X', 'Y', 'Y1', 'Z']\n\ntocell() Cell[source]\n\nReturn this lattice as a Cell object.\n\nclassmethod type_description()[source]\n\nReturn complete description of this Bravais lattice type.\n\nproperty variant: str\n\nReturn name of lattice variant.\n\n>>> BCT(3, 5).variant\n'BCT2'\n\nvars() Dict[str, float][source]\n\nGet parameter names and values of this lattice as a dictionary.\n\n# General crystal structures and surfaces¶\n\nModules for creating crystal structures are found in the module ase.lattice. Most Bravais lattices are implemented, as are a few important lattices with a basis. The modules can create lattices with any orientation (see below). These modules can be used to create surfaces with any crystal structure and any orientation by later adding a vacuum layer with ase.build.add_vacuum().\n\n## Example¶\n\nTo set up a slab of FCC copper with the [1,-1,0] direction along the x-axis, [1,1,-2] along the y-axis and [1,1,1] along the z-axis, use:\n\n>>> from ase.lattice.cubic import FaceCenteredCubic\n>>> atoms = FaceCenteredCubic(directions=[[1,-1,0], [1,1,-2], [1,1,1]],\n... size=(2,2,3), symbol='Cu', pbc=(1,1,0))\n\n\nThe minimal unit cell is repeated 2*2*3 times. The lattice constant is taken from the database of lattice constants in ase.data module. There are periodic boundary conditions along the x and y axis, but free boundary conditions along the z axis. Since the three directions are perpendicular, a (111) surface is created.\n\nTo set up a slab of BCC copper with along the first axis, along the second axis, and along the third axis use:\n\n>>> from ase.lattice.cubic import BodyCenteredCubic\n>>> atoms = BodyCenteredCubic(directions=[[1,0,0], [0,1,0], [1,1,1]],\n... size=(2,2,3), symbol='Cu', pbc=(1,1,0),\n... latticeconstant=4.0)\n\n\nSince BCC is not the natural crystal structure for Cu, a lattice constant has to be specified. Note that since the repeat directions of the unit cell are not orthogonal, the Miller indices of the surfaces will not be the same as the Miller indices of the axes. The indices of the surfaces in this example will be (1,0,-1), (0,1,-1) and (0,0,1).\n\n## Available crystal lattices¶\n\nThe following modules are currently available (the * mark lattices with a basis):\n\n• lattice.cubic\n\n• SimpleCubic\n\n• FaceCenteredCubic\n\n• BodyCenteredCubic\n\n• Diamond (*)\n\n• lattice.tetragonal\n\n• SimpleTetragonal\n\n• CenteredTetragonal\n\n• lattice.orthorhombic\n\n• SimpleOrthorhombic\n\n• BaseCenteredOrthorhombic\n\n• FaceCenteredOrthorhombic\n\n• BodyCenteredOrthorhombic\n\n• lattice.monoclinic\n\n• SimpleMonoclinic\n\n• BaseCenteredMonoclinic\n\n• lattice.triclinic\n\n• Triclinic\n\n• lattice.hexagonal\n\n• Hexagonal\n\n• HexagonalClosedPacked (*)\n\n• Graphite (*)\n\n• The rhombohedral (or trigonal) lattices are not implemented. They will be implemented when the need arises (and if somebody can tell us the precise definition of the 4-number Miller indices - we only know that they are “almost the same as in hexagonal lattices”).\n\n• lattice.compounds\n\nLattices with more than one element. These are mainly intended as examples allowing you to define new such lattices. Currently, the following are defined\n\n• B1 = NaCl = Rocksalt\n\n• B2 = CsCl\n\n• B3 = ZnS = Zincblende\n\n• L1_2 = AuCu3\n\n• L1_0 = AuCu\n\n## Usage¶\n\nThe lattice objects are called with a number of arguments specifying e.g. the size and orientation of the lattice. All arguments should be given as named arguments. At a minimum the symbol argument must be specified.\n\nsymbol\n\nThe element, specified by the atomic number (an integer) or by the atomic symbol (i.e. ‘Au’). For compounds, a tuple or list of elements should be given. This argument is mandatory.\n\ndirections and/or miller:\n\nSpecifies the orientation of the lattice as the Miller indices of the three basis vectors of the supercell (directions=...) and/or as the Miller indices of the three surfaces (miller=...). Normally, one will specify either three directions or three surfaces, but any combination that is both complete and consistent is allowed, e.g. two directions and two surface miller indices (this example is slightly redundant, and consistency will be checked). If only some directions/miller indices are specified, the remaining should be given as None. If you intend to generate a specific surface, and prefer to specify the miller indices of the unit cell basis (directions=...), it is a good idea to give the desired Miller index of the surface as well to allow the module to test for consistency. Example:\n\n>>> atoms = BodyCenteredCubic(directions=[[1,-1,0],[1,1,-1],[0,0,1]],\n... miller=[None, None, [1,1,2]], ...)\n\n\nIf neither directions nor miller are specified, the default is directions=[[1,0,0], [0,1,0], [0,0,1]].\n\nsize:\n\nA tuple of three numbers, defining how many times the fundamental repeat unit is repeated. Default: (1,1,1). Be aware that if high-index directions are specified, the fundamental repeat unit may be large.\n\nlatticeconstant:\n\nThe lattice constant. If no lattice constant is specified, one is extracted from ASE.ChemicalElements provided that the element actually has the crystal structure you are creating. Depending on the crystal structure, there will be more than one lattice constant, and they are specified by giving a dictionary or a tuple (a scalar for cubic lattices). Distances are given in Angstrom, angles in degrees.\n\nStructure\n\nLattice constants\n\nDictionary-keys\n\nCubic\n\na\n\n‘a’\n\nTetragonal\n\n(a, c)\n\n‘a’, ‘c’ or ‘c/a’\n\nOrthorhombic\n\n(a, b, c)\n\n‘a’, ‘b’ or ‘b/a’, ‘c’ or ‘c/a’\n\nTriclinic\n\n(a, b, c, $$\\alpha$$, $$\\beta$$, $$\\gamma$$)\n\n‘a’, ‘b’ or ‘b/a’, ‘c’ or ‘c/a’, ‘alpha’, ‘beta’, ‘gamma’\n\nMonoclinic\n\n(a, b, c, alpha)\n\n‘a’, ‘b’ or ‘b/a’, ‘c’ or ‘c/a’, ‘alpha’\n\nHexagonal\n\n(a, c)\n\n‘a’, ‘c’ or ‘c/a’\n\nExample:\n\n>>> atoms = Monoclinic( ... , latticeconstant={'a': 3.06,\n... 'b/a': 0.95, 'c/a': 1.07, 'alpha': 74})\n\ndebug:\n\nControls the amount of information printed. 0: no info is printed. 1 (the default): The indices of surfaces and unit cell vectors are printed. 2: Debugging info is printed.\n\n## Defining new lattices¶\n\nOften, there is a need for new lattices - either because an element crystallizes in a lattice that is not a simple Bravais lattice, or because you need to work with a compound or an ordered alloy.\n\nAll the lattice generating objects are instances of a class, you generate new lattices by deriving a new class and instantiating it. This is best explained by an example. The diamond lattice is two interlacing FCC lattices, so it can be seen as a face-centered cubic lattice with a two-atom basis. The Diamond object could be defined like this:\n\nfrom ase.lattice.cubic import FaceCenteredCubicFactory\nclass DiamondFactory(FaceCenteredCubicFactory):\n\"\"\"A factory for creating diamond lattices.\"\"\"\nxtal_name = 'diamond'\nbravais_basis = [[0, 0, 0], [0.25, 0.25, 0.25]]\n\nDiamond = DiamondFactory()\n\n\n### Lattices with more than one element¶\n\nLattices with more than one element is made in the same way. A new attribute, element_basis, is added, giving which atoms in the basis are which element. If there are four atoms in the basis, and element_basis is (0,0,1,0), then the first, second and fourth atoms are one element, and the third is the other element. As an example, the AuCu3 structure (also known as $$\\mathrm{L}1_2$$) is defined as:\n\n# The L1_2 structure is \"based on FCC\", but is really simple cubic\n# with a basis.\nclass AuCu3Factory(SimpleCubicFactory):\n\"A factory for creating AuCu3 (L1_2) lattices.\"\nbravais_basis = [[0, 0, 0], [0, 0.5, 0.5], [0.5, 0, 0.5], [0.5, 0.5, 0]]\nelement_basis = (0, 1, 1, 1)\n\nAuCu3 = L1_2 = AuCu3Factory()\n\n\nSometimes, more than one crystal structure can be used to define the crystal structure, for example the Rocksalt structure is two interpenetrating FCC lattices, one with one kind of atoms and one with another. It would be tempting to define it as\n\nclass NaClFactory(FaceCenteredCubicFactory):\n\"A factory for creating NaCl (B1, Rocksalt) lattices.\"\n\nbravais_basis = [[0, 0, 0], [0.5, 0.5, 0.5]]\nelement_basis = (0, 1)\n\nB1 = NaCl = Rocksalt = NaClFactory()\n\n\nbut if this is used to define a finite system, one surface would be covered with one type of atoms, and the opposite surface with the other. To maintain the stochiometry of the surfaces, it is better to use the simple cubic lattice with a larger basis:\n\n# To prevent a layer of element one on one side, and a layer of\n# element two on the other side, NaCl is based on SimpleCubic instead\n# of on FaceCenteredCubic\nclass NaClFactory(SimpleCubicFactory):\n\"A factory for creating NaCl (B1, Rocksalt) lattices.\"\n\nbravais_basis = [[0, 0, 0], [0, 0, 0.5], [0, 0.5, 0], [0, 0.5, 0.5],\n[0.5, 0, 0], [0.5, 0, 0.5], [0.5, 0.5, 0],\n[0.5, 0.5, 0.5]]\nelement_basis = (0, 1, 1, 0, 1, 0, 0, 1)\n\nB1 = NaCl = Rocksalt = NaClFactory()\n\n\nMore examples can be found in the file ase/lattice/compounds.py." ]

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[ "# Probability Of An Event\n\nIn these lessons, we will learn how to find the probability of an event.\n\n### What Is An Event In Probability?\n\nIn an experiment, an event is the result that we are interested in.\n\nThe probability of an event A, written P(A), is defined as", null, "Example:\nWhen a fair dice is thrown, what is the probability of getting\na) the number 5\nb) a number that is a multiple of 3\nc) a number that is greater than 6\nd) a number that is less than 7\n\nSolution:\nA fair die is an unbiased die where each of the six numbers is equally likely to turn up.\n\nS = {1, 2, 3, 4, 5, 6}\n\na) Let A = event of getting the number 5 = {5}\nLet n(A) = number of outcomes in event A = 1\nn(S) = number of outcomes in S = 6", null, "b) Let B = event of getting a multiple of 3\nMultiple of 3 = {3, 6}", null, "c) Let C = event of getting a number greater than 6\nThere is no number greater than 6 in the sample space S.\nC ={}", null, "A probability of 0 means the event will never occur.\n\nd) Let D = event of getting a number less than 7\nNumbers less than 7 = {1, 2, 3, 4, 5, 6}", null, "A probability of 1 means the event will always occur.\n\nExample:\nEach of the letters HELLO is written on a card. A card is chosen at random from the bag. What is the probability of getting the letter ‘L’?\n\nSolution:\nSince the card is randomly selected, it means that each card has the same chance of being selected.\nS = {H, E, L 1, L 2, O} There are two cards with the letter ‘L’\n\nLet A = event of getting the letter ‘L’ = {L 1, L 2}", null, "How to calculate the Probability of Simple Events\nExample 1: Find the probability of the next person you meeting having a phone number that ends with 5?\nExample 2: Find the probability of getting all heads if you flip 3 coins?\nExample 3: Find the probability that the person that meet next has a birthday in February? (Non-leap year)\n\nHow to determine the probability of single events and express it as a ratio?\nExample 1: What s the probability of drawing a spade from a deck of 52 playing cards?\nExample 2: If a dart randomly hits the board below, what is the probability that it will hit the shaded region?\nExample 3: It is Cindy’s turn to spin in a game that she is playing with her friends. What is the probability that Cindy will have to move back on this spin?\n\nHow to find the probability of an event?\n\nHow to find the probability of an event?\nExample: Suppose you spin the spinner below. Find the probability of spinning 4. Write the probability as a fraction, a decimal and a percent.\n\nTry the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.", null, "" ]

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[ "", null, "# Function speed up, reduce allocations\n\nHi,\n\nI am working on a combinatorial problem called Block Design:\n\nBlock design - Wikipedia ( specifically Pairwise balanced uniform designs (2-designs or BIBDs) )\n\nI try to group N=25 samples (e.g. integers 1:25) into “blocks” (aka pools) of 5 samples each, where each pair of samples only occurs in one block together. The blocks should be chosen so that 5 of these blocks form a partition of the set of all 25 samples. A partition of the set of all samples into 5 blocks as described above call a “layer”, I try to find as many unique layers as possible. (in the above example there are 6 layers)\n\nI implemented a backtracking algorithm to find solutions to the problem, which currently finds fesible solutions for N=9, N=16,& N=25, but takes prohibitively long for N=36.\n\nTiming for N=25:\n\n@time BlockDesign.run()\n1.606897 seconds (2.06 M allocations: 307.621 MiB, 0.59% gc time)\n\nI think the backtracking algorithm itself can be improved, but looking at the large number of allocations and the Profiler output pointing to the feasible(PD, M, M_) function, this could be address first.\n(Is there a way to attach the complete profiler output as .txt?):\n\nFunction call highlighted by profiler:\n` fs = myfind(feasible(PD, M, M_))`\n\nIn short: how can we make the function feasible(PD, M, M_) faster, either with the current implementation using the Int64 matrices M & M_ or in another way.\n\nThe functions + helper functions needed to run above function call:\n\n``````struct PoolDesign{T<:Integer}\nN::T\nm::T\nq::T\nlmax::T\nend\n\nPD = PoolDesign(\n25, # number samples\n5, # samples per pool\n5, # pools per layer\n6, # layers in pooling design\n)\n\nfunction myfind(c)\na = similar(c, Int)\ncount = 1\nfor i in eachindex(c)\na[count] = i\ncount += (c[i] != zero(eltype(c)))\nend\nreturn resize!(a, count-1)\nend\n\nfunction pool_idx(M, PD)\nfor (i, row) in enumerate(eachrow(M))\nif sum(row) < PD.m\nreturn i\nend\nend\nsize(M,1)\nend\n\nlayer_idx(M, PD) = ((pool_idx(M, PD) - 1)÷PD.q) + 1\n\nlayer(PD, l) = @view M[ ((l-1)*PD.q+1):(l * PD.q), :]\n\ncurrent_layer(M, PD) = layer(PD, layer_idx(M, PD))\n\nfunction feasible(PD, M, M_)\nfeas = collect(1:PD.N)\ncl = current_layer(M, PD)\n# if current layer is empty, all samples are feasible\nif !iszero(cl)\nfor i in 1:PD.N\nfor q in 1:PD.q\nif cl[q, i] == 1\nfeas[i] = 0\nend\nend\nend\n# if current pool is empty,\n# all remaining unplaced samples in layer are feasible\nif !iszero(current_pool(M, PD))\nfor k in myfind(current_pool(M, PD))\nfor l in 1:PD.N\nfeas[l] = feas[l] * M_[k, l]\nend\nfeas[k] = 0\nend\nend\nend\nreturn feas\nend\n\ndims = (PD.lmax * PD.m, PD.N)\nM = zeros(Int64, dims)\nM_ = ones(Int64, PD.N, PD.N)\n\nfs = myfind(feasible(PD, M, M_))\n``````\n\nThe matrix M stores the Block design, assigning the samples (cols) to the different blocks (rows), where M(j,i) = 1 means that sample i is block j.\nThe matrix M_ is symmetric and stores the information about which pairs of samples already occured together in one block.\n\nEx. solution for N=9:\n\n1 1 1 0 0 0 0 0 0 → (1,2,3) |\n0 0 0 1 1 1 0 0 0 → (4,5,6) | Layer 1\n0 0 0 0 0 0 1 1 1 → (7,8,9) |\n1 0 0 1 0 0 1 0 0 → (1,4,7) ]\n0 1 0 0 1 0 0 1 0 → (2,5,8) ] Layer 2\n0 0 1 0 0 1 0 0 1 → (3,6,9) ]\n1 0 0 0 1 0 0 0 1 → (1,5,9) }\n0 1 0 0 0 1 1 0 0 → (2,6,7) } Layer 3\n0 0 1 1 0 0 0 1 0 → (3,4,8) }\n1 0 0 0 0 1 0 1 0 → (1,6,8) ||\n0 1 0 1 0 0 0 0 1 → (2,4,9) || Layer 4\n0 0 1 0 1 0 1 0 0 → (3,5,7) ||\n\nIf this post is in the wrong category, please feel free to move it appropriately.\n\nThanks\nDavid\n\nIt would simplify things considerably if you could post the MWE in one markup block. Then I wouldn’t need to manually compose it.\n\n2 Likes\n\nmy guess just looking at the code is that you should probably expand this `myfind` function into this loop, to get rid of the array that is created by `myfind`, and just iterate over the elements that are necessary directly. That will make the code a little less modular, but at first sight it seems to be the cause of the allocations you are seeing.\n\nThat is, do something like:\n\n``````for val in current_pool(M, PD)\nif val != zero(...)\n# ... do what you need to do\nend\nend\n\n``````\n\nThanks for your answer. I also use the `myfind()` function in other places of the code to get the non-zero elements from the passed arrays.\nWhen ran with `myfind()` I get:\n\n``````@time myfind(feasible(PD, M, M_))\n0.000012 seconds (7 allocations: 768 bytes)\n``````\n\nCompared to:\n\n``````@time feasible(PD, M, M_)\n0.000010 seconds (6 allocations: 480 bytes)\n``````\n\nThere are already allocations occuring in `feasible(PD, M, M_)`, where I tried to set the elements zero, and filter the zero elements out later in `myfind()`.\n\n`myfind` is called from within `feasible`, isn’t it?\n\n(by the way, you can write that myfind with `findall(!isequal(0),x)`, as far as I could understand it, but that won’t be any better)\n\nYes it is, sorry for the confusion.\n\nI got now:\n\n``````function feasible(PD, M, M_)\nfeas = collect(1:PD.N)\n\ncl = current_layer(M, PD)\n# if current layer is empty, all samples are feasible\nif !iszero(cl)\n\nfor i in 1:PD.N\nfor q in 1:PD.q\nif cl[q, i] == 1\nfeas[i] = 0\nend\nend\n\nend\n\n# if current pool is empty,\n# all remaining unplaced samples in layer are feasible\nif !iszero(current_pool(M, PD))\n\nfor k in current_pool(M, PD)\nif k != zero(k)\nfor l in 1:PD.N\nfeas[l] = feas[l] * M_[k, l]\nend\nfeas[k] = 0\nend\nend\nend\nend\nreturn feas\nend\n\n@time feasible(PD, M, M_)\n0.000011 seconds (6 allocations: 480 bytes)\n\n``````\n\nI noticed now that `M` is not an input of `layer`. Probably it should be.\n\n4 Likes\n\nGood eye! This did the trick…\nIt is down to\n\n``````@time feasible(PD, M, M_)\n0.000004 seconds (1 allocation: 160 bytes)\n``````\n\nIs the last allocation due to\n\n`````` feas = collect(1:PD.N)\n``````\n\n?\n\nThat certainly is one allocation.\n\n1 Like\n\nSince `feasible(PD, M, M_)` is called frequently, I will make a preallocated array `feas` that will get passed as a parameter and modified in the function. Thanks for your help!\n\n2 Likes\n\nHi, if you want to construct large balanced incomplete block designs (BIBDs), it might be good to have look at this work (Memetic collaborative approaches for finding balanced incomplete block designs - ScienceDirect), where they proposed several algorithms to construct BIBDs. In your example it seems that you want to construct a resolvable balanced incomplete block design with v = 25, k =5, b = 5, r = 6?\n\n1 Like\n\nOr, preferably, `findall(!iszero, x)`\n\n3 Likes\n\nThanks for suggesting this paper, it is very relevant!\n\nyou want toconstruct a resolvable balanced incomplete block design with v = 25, k =5, b = 5, r = 6?\n\nYes, exactly." ]

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[ "# argument of product and quotient\n\nUsing the distributive law, we perform the multiplication\n\n $(\\cos\\varphi_{1}+i\\sin\\varphi_{1})(\\cos\\varphi_{2}+i\\sin\\varphi_{2})=(\\cos% \\varphi_{1}\\cos\\varphi_{2}-\\sin\\varphi_{1}\\sin\\varphi_{2})+i(\\sin\\varphi_{1}% \\cos\\varphi_{2}+\\cos\\varphi_{1}\\sin\\varphi_{2}).$\n\nUsing the addition formulas", null, "", null, "of cosine (http://planetmath.org/GoniometricFormulae) and sine (http://planetmath.org/GoniometricFormulae) we still obtain the formula\n\n $\\displaystyle(\\cos\\varphi_{1}+i\\sin\\varphi_{1})(\\cos\\varphi_{2}+i\\sin\\varphi_{% 2})=\\cos(\\varphi_{1}+\\varphi_{2})+i\\sin(\\varphi_{1}+\\varphi_{2}).$ (1)\n\nThe inverse number of  $\\cos\\varphi_{2}+i\\sin\\varphi_{2}$  is calculated as follows:\n\n $\\frac{1}{\\cos\\varphi_{2}+i\\sin\\varphi_{2}}=\\frac{\\cos\\varphi_{2}-i\\sin\\varphi_% {2}}{(\\cos\\varphi_{2}-i\\sin\\varphi_{2})(\\cos\\varphi_{2}+i\\sin\\varphi_{2})}=% \\frac{\\cos\\varphi_{2}-i\\sin\\varphi_{2}}{\\cos^{2}\\varphi_{2}+\\sin^{2}\\varphi_{2}}$\n\nThis equals  $\\cos\\varphi_{2}-i\\sin\\varphi_{2}$,  and since the cosine is an even (http://planetmath.org/EvenFunction) and the sine an odd function, we have\n\n $\\displaystyle\\frac{1}{\\cos\\varphi_{2}+i\\sin\\varphi_{2}}=\\cos(-\\varphi_{2})+i% \\sin(-\\varphi_{2}).$ (2)\n\nThe equations (1) and (2) imply\n\n $\\frac{\\cos\\varphi_{1}+i\\sin\\varphi_{1}}{\\cos\\varphi_{2}+i\\sin\\varphi_{2}}=(% \\cos\\varphi_{1}+i\\sin\\varphi_{1})(\\cos(-\\varphi_{2})+i\\sin(-\\varphi_{2}))=\\cos% (\\varphi_{1}+(-\\varphi_{2}))+i\\sin(\\varphi_{1}+(-\\varphi_{2})),$\n\ni.e.\n\n $\\displaystyle\\frac{\\cos\\varphi_{1}+i\\sin\\varphi_{1}}{\\cos\\varphi_{2}+i\\sin% \\varphi_{2}}=\\cos(\\varphi_{1}-\\varphi_{2})+i\\sin(\\varphi_{1}-\\varphi_{2}).$ (3)\n $z_{1}=r_{1}(\\cos\\varphi_{1}+i\\sin\\varphi_{1})\\;\\;\\mbox{and}\\;\\;z_{2}=r_{2}(% \\cos\\varphi_{2}+i\\sin\\varphi_{2})$\n\nwe have\n\n $z_{1}z_{2}=r_{1}r_{2}(\\cos(\\varphi_{1}+\\varphi_{2})+i\\sin(\\varphi_{1}+\\varphi_% {2})),$\n $\\frac{z_{1}}{z_{2}}\\;=\\;\\frac{r_{1}}{r_{2}}(\\cos(\\varphi_{1}-\\varphi_{2})+i% \\sin(\\varphi_{1}-\\varphi_{2})).$\n\nThus we have the\n\nThe modulus", null, "", null, "of the product of two complex numbers equals the product of the moduli of the factors and the argument", null, "", null, "equals the sum of the arguments of the factors (http://planetmath.org/Product).  The modulus of the quotient of two complex numbers equals the quotient of the moduli of the dividend and the divisor and the argument equals the difference of the arguments of the dividend and the divisor.\n\nRemark.  The equation (1) may be by induction generalised for more than two factors of the left hand ; then the special case where all factors are equal gives de Moivre identity", null, "", null, ".\n\nExample.  Since\n\n $(2\\!+\\!i)(3\\!+\\!i)=5\\!+\\!5i\\;=\\;5e^{\\frac{\\pi}{4}},$\n\none has\n\n $\\arctan\\frac{1}{2}+\\arctan\\frac{1}{3}\\;=\\;\\frac{\\pi}{4}.$\n Title argument of product and quotient Canonical name ArgumentOfProductAndQuotient Date of creation 2013-03-22 17:45:20 Last modified on 2013-03-22 17:45:20 Owner pahio (2872) Last modified by pahio (2872) Numerical id 8 Author pahio (2872) Entry type Theorem Classification msc 30-00 Classification msc 26A09 Synonym product and quotient of complex numbers Related topic Argument Related topic PolarCoordinates Related topic ModulusOfComplexNumber Related topic Complex Related topic EqualityOfComplexNumbers" ]

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[ "# Color Codes Conversion\n\nIn this tutorial, we will see that how different color codes can be combined to make other colors, and how we can covert RGB color codes to hex and vice versa.\n\n## Different color codes\n\nAll the colors here are of the 24 bit format, that means each color has 8 bits of red, 8 bits of green, 8 bits of blue, in it. Or we can say each color has three different portions. You just have to change the quantity of these three portions to make any color.\n\n## Color:Black\n\n### Image:", null, "(0,0,0)\n\n### Explanation:\n\nAs it has been explained in the previous tutorials, that in an 8-bit format, 0 refers to black. So if we have to make a pure black color, we have to make all the three portion of R, G, B to 0.\n\n## Color:White\n\n### Image:", null, "(255,255,255)\n\n### Explanation:\n\nSince each portion of R, G, B is an 8 bit portion. So in 8-bit, the white color is formed by 255. It is explained in the tutorial of pixel. So in order to make a white color we set each portion to 255 and thats how we got a white color. By setting each of the value to 255, we get overall value of 255, thats make the color white.\n\n## Color:Red\n\n### Image:", null, "(255,0,0)\n\n### Explanation:\n\nSince we need only red color, so we zero out the rest of the two portions which are green and blue, and we set the red portion to its maximum which is 255.\n\n## Color:Green\n\n### Image:", null, "(0,255,0)\n\n### Explanation:\n\nSince we need only green color, so we zero out the rest of the two portions which are red and blue, and we set the green portion to its maximum which is 255.\n\n## Color: Blue\n\n### Image:", null, "(0,0,255)\n\n### Explanation:\n\nSince we need only blue color, so we zero out the rest of the two portions which are red and green, and we set the blue portion to its maximum which is 255\n\n## Gray color:\n\n### Image:", null, "(128,128,128)\n\n### Explanation\n\nAs we have already defined in our tutorial of pixel, that gray color Is actually the mid point. In an 8-bit format, the mid point is 128 or 127. In this case we choose 128. So we set each of the portion to its mid point which is 128, and that results in overall mid value and we got gray color.\n\n## CMYK color model:\n\nCMYK is another color model where c stands for cyan, m stands for magenta, y stands for yellow, and k for black. CMYK model is commonly used in color printers in which there are two carters of color is used. One consist of CMY and other consist of black color.\n\nThe colors of CMY can also made from changing the quantity or portion of red, green and blue.\n\n## Color: Cyan\n\n### Image:", null, "(0,255,255)\n\n### Explanation:\n\nCyan color is formed from the combination of two different colors which are Green and blue. So we set those two to maximum and we zero out the portion of red. And we get cyan color.\n\n## Color: Magenta\n\n### Image:", null, "(255,0,255)\n\n### Explanation:\n\nMagenta color is formed from the combination of two different colors which are Red and Blue. So we set those two to maximum and we zero out the portion of green. And we get magenta color.\n\n## Color: Yellow\n\n### Image:", null, "(255,255,0)\n\n### Explanation:\n\nYellow color is formed from the combination of two different colors which are Red and Green. So we set those two to maximum and we zero out the portion of blue. And we get yellow color.\n\n## Conversion\n\nNow we will see that how color are converted are from one format to another.\n\n### Conversion from RGB to Hex code:\n\nConversion from Hex to rgb is done through this method:\n\n• Take a color. E.g: White = (255, 255, 255).\n\n• Take the first portion e.g 255.\n\n• Divide it by 16. Like this:\n\n•", null, "• Take the two numbers below line, the factor, and the remainder. In this case it is 15 � 15 which is FF.\n\n• Repeat the step 2 for the next two portions.\n\n• Combine all the hex code into one.\n\n### Conversion from Hex to RGB:\n\nConversion from hex code to rgb decimal format is done in this way.\n\nTake a hex number. E.g: #FFFFFF\n\nBreak this number into 3 parts: FF FF FF\n\nTake the first part and separate its components: F F\n\nConvert each of the part separately into binary: (1111) ( 1111)\n\nNow combine the individual binaries into one: 11111111\n\nConvert this binary into decimal: 255\n\nNow repeat step 2, two more times.\n\nThe value comes in the first step is R, second one is G, and the third one belongs to B.\n\nAnswer: ( 255 , 255 , 255 )\n\nColor Hex Code\nBlack #000000\nWhite #FFFFFF\nGray #808080\nRed #FF0000\nGreen #00FF00\nBlue #0000FF\nCyan #00FFFF\nMagenta #FF00FF\nYellow #FFFF00" ]

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[ "# Sum of secret numbers is 101\n\nAlice, Bob, and Charlie were given three numbers, respectively. They were said that all the three numbers were positive integers and the sum was 101.\n\nAlice: I know we have different numbers.\n\nBob: Aha, I got it. I found all the numbers.\n\nCharlie: Me, too. I know our numbers, now.\n\nAlice: Alas, I still don't know.\n\nWhat were the three numbers?\n\n• all positive integer numbers?\n– Oray\nAug 17, 2017 at 9:12\n• @Oray Oh, I missed. Yes, they were all positive. Aug 17, 2017 at 9:12\n• Very close to this puzzle Aug 17, 2017 at 17:03\n\nAlice: I know we have different numbers.\n\nAlice has an even number larger than 50, so at least 52.\n\nBob: Aha, I got it. I found all the numbers.\n\nIf Bob has a number less than 47, then he can't tell if Alice's number if 52 or 54. Thus Bob's number is 47 or 48, and Alice's is 52. (Bob's number is at most 48 since Alice's is at least 52)\n\nCharlie: Me, too. I know our numbers, now.\n\nWell, duh. Assuming Bob is competent with logic, his number is the last one needed to fill in this gap.\n\nAlice: Alas, I still don't know.\n\nWe don't, either. The possible choices now are $(a,b,c)=(52,48,1)$ or $(52,47,2)$\n\n• Ooooops! Sorry. My mistake. Could you read again after I edited it? Aug 17, 2017 at 9:29\n• Alice's sentence should be corrected to \"we have different numbers.\" Aug 17, 2017 at 9:31\n• Then Alice can work out the three numbers! Actually, Alice has at least as much information as us, so Alice can always work out the three numbers, if we are able to. Aug 17, 2017 at 9:32\n• Right! It's better to correct the numbers in Bob's deduction. Aug 17, 2017 at 9:44\n• Bob can have 47 or 48, as shown in the final answer, not 48 or 49 as stated in Bob's deduction. Aug 18, 2017 at 2:54\n\nAlice has 52, Bob has 47 or 48, and Charlie has 2 or 1 (respectively)\n\nAlice: I know we have different numbers.\n\nSo no one can have the same number as Alice, it must be 51 or over. Also, so Bob and Charlie cannot have the same number, the remainder must be odd. Namely, Alice's number is even. Therefore, Alice's number is in {52, 54, ... 98}.\n\nBob: Aha, I got it. I found all the numbers.\n\nIf Bob has 48, he would know that the others are 52 and 1. If he has 47, he would know that the others are 52 and 2. But if he has anything lower than that, there would always be more than one option for the others. For example, of he has 46, than Alice and Charlie can have 52 and 3 OR 54 and 1. So Bob has 47 or 48 (and Alice has 52).\n\nCharlie: Me, too. I know our numbers, now.\n\nOf course - if he has 1, than Bob has 48. If he has 2, Bob has 47. And Alice has 52.\n\nAlice: Alas, I still don't know.\n\nIndeed, she doesn't. Nor do we...\n\n• I don't get it. Why Alice must have 51 or over? Why not 49? Then 3 numbers would be 49, 50, 2. What's wrong in there?\n– jack\nAug 18, 2017 at 18:46\n• @mikele Had she had 49, the 3 numbers could also have been (49, 49, 3). This is true for any number smaller than 51 - someone else could have had the same number as her. The fact that she knows that all numbers are different tells us that her number is 51 or larger. Aug 19, 2017 at 19:46\n• But it could have been also (49, 47, 5), so all 3 different numbers. I don't know. Personally the questions doesn't satisfy me.\n– jack\nAug 19, 2017 at 21:43\n• @mikele Alice knows for certain that no number is the same as hers. If there is any possibility of a=b or a=c, then she cannot state that no one has the same number as hers. Since (49, 49, 3) is possible, then she would not have made the statement she made if her number were 49. Aug 21, 2017 at 11:59\n\nHere is what we know:\n\n1. All numbers are positive and integers\n2. They are different\n3. Bob could deduce all 3 numbers just by knowing his number\n4. Charlie could know all 3 numbers by knowing his number and knowing that Bob could deduce all three numbers by knowing his number (point 3 above)\n5. Alice could not deduce her number after knowing points 1-4\n\nSo now, lets find the numbers: Bob could deduce his number just by looking at his number. Which means he had a number, which coupled with just one unique combination of 2 distinct positive integers could result in a sum total of 101.\n\nThe only number I can think of is 97 or 98. If he had 97 then he knows the other 2 numbers have to be 3 and 1: 97 + 3 + 1 = 101\n\nand if he had 98, then he knows the other two numbers had to be 2 and 1: 98 + 2 + 1 = 101\n\nOn the contrary, if he had a 96 for example (or anything lower), there exists more than one combination of distinct integers which could add to 101: 96 + 4 + 1 = 101 and also 96 + 3 + 2 = 101. He could never know for sure what Alice and Charlie have.\n\nNow lets look at Charlie. Charlie being the smart guy he is, immediately put 2 and 2 together (or in this case 3 numbers which add top 101). He knows Bob either has 98 or 97. To get a definite answer he looks at his own number. If it is a 2 then he knows Bob has a 98 and clueless Alice has a 1. If he has a 3, he knows Bob has a 97 and dumbfounded Alice has a 1.\n\nLastly, we go to Alice, she is at a disadvantage her. She has figured the set is {97,3,1} or {98,2,1} but since both result in her having a 1, she cannot pick which case is true here and hence cannot pick the correct set.\n\nFINAL ANSWER: {97,3,1} or {98,2,1}\n\nNote: the question asks for the 3 numbers and not who had what number. So When Bob says he knows the numbers, he knows the 3 numbers without knowing who had what. this changes when Charlie proclaims he knows the 3 numbers as well, Bob and Charlie now know the 3 numbers and who had what.\n\n• If Alice is holding 1, 2, or 3, how does she know for certain that the other two numbers are unique? If instead you are implying that she is holding 97 or 98, then if Bob is holding 1, 2, or 3 how does he know what all three numbers are knowing only what number he is holding (1,2, or 3) and that Alice has told them all three numbers are unique? Aug 21, 2017 at 12:09" ]

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[ "Try the Free Math Solver or Scroll down to Tutorials!\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:\n\n### Our users:\n\nMy fourteen year-old son, Bradley, was considered at-risk by his school. But, I couldnt make him listen to me. Then when a teacher at his school, Mr. Kindler bless his heart, got him to try an after-school program, it was like a miracle! I wouldnt say Bradley became a model student but he was no longer failing his math classes. So when I found out that Mr. Kindler based his entire program on using Algebrator, I just had to write this letter to say Thank You! Imagine that!\nSamuel Lee, CA\n\nCan I simply tell you how wonderful you are? May seem like a simple thing to you, but you just restored my faith in mankind (no small thing). Thank you for your kind and speedy response.\nMr. Tom Carol, NY\n\nThis software is the best of its kind. The explanations of each step are excellent and easy to understand. A must have for any parent with high school age children. Well done.\nChris Ress, OH\n\n### Students struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among them?\n\n#### Search phrases used on 2014-02-28:\n\n• algebra world problem solver online\n• matlab solving system of nonlinear differential equations\n• Conversion from mix fraction to decimal\n• slope y intercept worksheet generator\n• do my algebra\n• sample worksheets of grade VII maths\n• college algebra for dummies\n• algebra substitution method\n• math elimination worksheets\n• hard math fractions worksheets\n• Algebra For Beginners\n• how to solve maths hours problems kids sats\n• how to factor out cubes\n• linear programing word problem with solution\n• north carolina edition mcdougal littell science grade 6 standardized test practice\n• texas instruments calculator solve function\n• how to do log base 2 on a ti 89\n• algebra games for 9th graders\n• convert percentages to decimal calculator\n• java+square cube\n• ebooks free costing accounts\n• matlab ode45 tutorial\n• excel formula to convert decimal to percent\n• prentice hall algebra 2 trig answers\n• liner graph\n• how to solve 2 equation on the TI - 83calculator\n• Algebra Cheats\n• combining like term worksheet\n• cramer's rule+code in java\n• free algebra fractions calculator\n• matlab differential equation solve\n• percent formulas\n• solving homogeneous differential equations\n• how to practise maths in intermediate second year give me tips\n• how to square large polynomials\n• calculate + following expression in simplified radical form\n• how to factor on a TI 84 Silver Edition\n• Where can I find out how to do certain funtions on a TI-83 Plus calculator online for free?\n• algebra solver\n• mathematical trivia\n• sqaure\n• printable algebra exams on sequences and series\n• application of algebra\n• fractions and decimals equation calculators\n• learning basic algebra free\n• egyptian multiplication to compute the value of exponential a\n• samples of math trivia\n• FREE ADDITION AND SUBTRACTION WORKSHEET WITH FOUR ROWS\n• concept of simple interest in mathsin Indian education\n• factoring in algebra\n• Finding Values of Exponential Expressions\n• multivariable limits calculator\n• find the sum of the integers which are divisible\n• program quadratic formula into ti 83\n• algebra 1 workbook\n• holt chemistry answer to test\n• cost accounting books\n• how do I do algebra with cubes\n• Linear Equations 8TH GRADE WORKSHEETS\n• square root on ti-83\n• simultaneous equations presentation\n• rearranging linear equations\n• maths test yr8\n• common denominator algebra\n• how to solve multiple variable differential equations\n• matlab time integration step\n• free algebra for dommies\n• algebra in grade 9 or 10 with equations" ]

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[ "## Conversion formula\n\nThe conversion factor from grams to ounces is 0.03527396194958, which means that 1 gram is equal to 0.03527396194958 ounces:\n\n1 g = 0.03527396194958 oz\n\nTo convert 66.5 grams into ounces we have to multiply 66.5 by the conversion factor in order to get the mass amount from grams to ounces. We can also form a simple proportion to calculate the result:\n\n1 g → 0.03527396194958 oz\n\n66.5 g → M(oz)\n\nSolve the above proportion to obtain the mass M in ounces:\n\nM(oz) = 66.5 g × 0.03527396194958 oz\n\nM(oz) = 2.3457184696471 oz\n\nThe final result is:\n\n66.5 g → 2.3457184696471 oz\n\nWe conclude that 66.5 grams is equivalent to 2.3457184696471 ounces:\n\n66.5 grams = 2.3457184696471 ounces\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 ounce is equal to 0.42630861842105 × 66.5 grams.\n\nAnother way is saying that 66.5 grams is equal to 1 ÷ 0.42630861842105 ounces.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that sixty-six point five grams is approximately two point three four six ounces:\n\n66.5 g ≅ 2.346 oz\n\nAn alternative is also that one ounce is approximately zero point four two six times sixty-six point five grams.\n\n## Conversion table\n\n### grams to ounces chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from grams to ounces\n\ngrams (g) ounces (oz)\n67.5 grams 2.381 ounces\n68.5 grams 2.416 ounces\n69.5 grams 2.452 ounces\n70.5 grams 2.487 ounces\n71.5 grams 2.522 ounces\n72.5 grams 2.557 ounces\n73.5 grams 2.593 ounces\n74.5 grams 2.628 ounces\n75.5 grams 2.663 ounces\n76.5 grams 2.698 ounces" ]

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[ "", null, "asked in category: General Last Updated: 25th June, 2020\n\n# How is algebra used in finance?\n\nLinear algebra is used to study financial trading strategies and expectations. Financial conditions are examined via matrix equations, using rank, column space, and null space arguments.\n\nIn this manner, what kind of math is used in finance?\n\nIn addition to the Calculus I and Introduction to Probability and Statistics courses that their peers in other business programs take, these students will likely study Calculus II and III, Linear Algebra with Differential Equations, Applied Linear Regression, Fourier Analysis and Partial Differential Equations and\n\nAlso Know, is the math in finance hard? In finance, there is a stream called quantitative analysis, which is basically statistical methods. For this you will need a good grasp of mathematics, and should be able to remember or visualise the logic of formulae. however for the most part its not very hard. Like being able to multiply, take squares, roots etc.\n\nSimilarly, you may ask, what are matrices used for in finance?\n\nMatrix algebra is useful for computing the variance (or risk) of a portfolio that contains many assets. Matrix algebra is used in financial risk management. A matrix is used to describe the outcomes or payoff of an investment or venture.\n\nHow is math used in investment banking?\n\nMost of the Investment banking team (specifically M&A, Equity Capital Markets, Debt Capital Markets) will use basic Algebra and Arithmetic, however it may vary if you are working on any predictive financial model which doesn't use any of the available templates and you are building a new model from scratch.\n\n3\n\n25th June, 2020\n\n102\n\nQuestions\n\nVideos\n\nUsers" ]

[ null, "https://askinglot.com/img/default.jpg", null ]

[ "", null, "## Launch of Youtube Channel: My First Videos - Embedded World 2017\n\nI went to Embedded World 2017 in Nuremberg with an ambitious plan; I would make video highlights of several exhibits (booths) to be presented to the *Related sites audience.  I would try to make the vendors focus their pitch on the essential...", null, "## A Two Bin Exact Frequency Formula for a Pure Complex Tone in a DFT\n\n1 comment\n\nIntroduction This is an article to hopefully give a better understanding to the Discrete Fourier Transform (DFT) by deriving an exact formula for the frequency of a complex tone in a DFT. It is basically a parallel treatment to the real case...", null, "## DFT Bin Value Formulas for Pure Complex Tones\n\nIntroduction This is an article to hopefully give a better understanding to the Discrete Fourier Transform (DFT) by deriving an analytical formula for the DFT of pure complex tones and an alternative variation. It is basically a parallel...", null, "## Multi-Decimation Stage Filtering: Design and Optimization\n\n1 comment\n\nDuring my research on digital FIR decimation filters I have been developing various Matlab scripts and functions. In which I have decided later on to consolidate it in a form of a toolbox. I have developed this toolbox to assist and...", null, "## Canonic Signed Digit (CSD) Representation of Integers\n\n1 comment\n\nIn my last post I presented Matlab code to synthesize multiplierless FIR filters using Canonic Signed Digit (CSD) coefficients.  I included a function dec2csd1.m (repeated here in Appendix A) to convert decimal integers to binary CSD...", null, "## Frequency Translation by Way of Lowpass FIR Filtering\n\n1 comment\n\nSome weeks ago a question appeared on the dsp.related Forum regarding the notion of translating a signal down in frequency and lowpass filtering in a single operation . It is possible to implement such a process by embedding a discrete cosine...", null, "## Minimum Shift Keying (MSK) - A Tutorial\n\n7 comments\n\nMinimum Shift Keying (MSK) is one of the most spectrally efficient modulation schemes available. Due to its constant envelope, it is resilient to non-linear distortion and was therefore chosen as the modulation technique for the GSM cell phone...", null, "## Round Round Get Around: Why Fixed-Point Right-Shifts Are Just Fine\n\n1 comment\n\nToday’s topic is rounding in embedded systems, or more specifically, why you don’t need to worry about it in many cases.One of the issues faced in computer arithmetic is that exact arithmetic requires an ever-increasing bit length to...", null, "## Some Thoughts on Sampling\n\n1 comment\n\nSome time ago, I came across an interesting problem. In the explanation of sampling process, a representation of impulse sampling shown in Figure 1 below is illustrated in almost every textbook on DSP and communications. The question is: how is...", null, "## Matlab Code to Synthesize Multiplierless FIR Filters\n\nThis article presents Matlab code to synthesize multiplierless Finite Impulse Response (FIR) lowpass filters. A filter coefficient can be represented as a sum of powers of 2.  For example, if a coefficient = decimal 5 multiplies input x,..." ]

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[ "# Early COVID-19 decay rates for countries that have peaked\n\nShare Now\n\n(Posted on 23 Apr 2020)\n\nIn previous articles [C20a, H20], my colleague and I have written about how the number of cases and deaths from COVID-19 follows a triangle shape over time when plotted with a log-scaled Y axis. A log-Y axis turns exponential growth into a straight line with a positive slope and exponential decay into a straight line with a negative slope. The negative slope results from the effects of social distancing measures. There is a roughly three week time lag between the introduction of lock down measures to the peak of the daily deaths curve, due to the fact that people who just caught the disease when social distancing starts and will eventually die from it, typically die about three weeks later.", null, "The Triangle Suppression model on a log-Y plot. After a time lag, the impact of lock down measures kicks in, causing an exponential decay. Figure taken from [H20].A key prediction of the Triangle Suppression model is that the decay phase is likely to be slower than the growth phase. In China, the decay during recovery was slower the growth phase by a factor of two. For the rest of the world where only less extreme lock downs are possible, the model predicts that the recovery will occur more slowly.  In [C20b], I critiqued the IMHE model [IMHE20] from HealthData.org, where my largest concern was the assumption that recovery will occur at the same rate as growth, which underestimates the length of the recovery time relative to the Triangle Suppression model.\n\nA few countries appear to have crossed a peak, with just enough data to get a preliminary reading of their decay rates.  In this article, I share these graphs, and look at the ratio of growth rate to decay rate for these countries.\n\n# Daily deaths\n\nThe plots in this section show the number of deaths per day from COVID-19 (from the John Hopkins data set [JH20]) for several countries that seem to have crossed a peak, plotted on a log-Y axis.  Exponential growth and decay lines are super-imposed.", null, "", null, "", null, "", null, "", null, "", null, "", null, "It is worth mentioning that the number of deaths per day is already below half of their peak rates for these countries (except Iran), which by any measure is a big improvement. On a linear Y scale, a reduction to 1/2 the peak rate looks dramatic, as seen in the same data for Spain shown here with a linearly-scaled Y axis.", null, "Daily deaths from COVID-19 in Spain on a linearly-scaled Y axis. Same data as the earlier plot for Spain.\n\n# Comparison of growth and decay rates\n\nThe following table compares the daily growth and decay rates for each country. The Growth/Decay column shows the ratio of the growth rate to the decay rate. Currently, the recovery rate varies from half the growth rate (in China) to 1/13th as fast (Iran). The final Date suppressed column shows the date when the decay trend line crosses the ten deaths per day level.", null, "The daily growth and decay rates, and their ratio. The final column shown the date when the current decay trend line crosses 10 deaths per day.\n\n# Summary", null, "The exponential decay rate during recovery gives us an indication of how long the recovery phase will last, assuming today’s level of lock down measures remain in place. The decay rate may improve with medical breakthroughs, tighter lock downs, high levels of immunity, or altered societal behaviors. They may get worse, or even reverse, with loosing of social distancing measures.\n\nWith the exception of China, all the countries shown here appear to have barely crossed the peak. The apparent decay rates as of today are preliminary given that these countries have not been past the peak for very long.\n\nShare Now\nScroll to Top" ]

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[ "# What is the Python equivalent of Matlab's tic and toc functions?\n\nWhat is the Python equivalent of Matlab's tic and toc functions?\n\n• If you really want the direct equivalent, just call `tic = time.time()` and `toc = time.time()`, then `print toc-tic, 'sec Elapsed'` As folks have said below, though, `timeit` is more robust. – Joe Kington May 1 '11 at 16:53\n• I seem to get better results using @JoeKington's approach in conjunction with timeit.default_timer(), like this for example: `tic = timeit.default_timer(); (U,S,V) = np.linalg.svd(A); toc = timeit.default_timer()`, then `print toc-tic`. – littleO Dec 18 '16 at 23:43\n• The library pytictoc seems most conveinent, syntax is even slightly neater than ttictoc below. pypi.org/project/pytictoc – FlorianH Oct 7 '19 at 11:43\n\nApart from `timeit` which ThiefMaster mentioned, a simple way to do it is just (after importing `time`):\n\n``````t = time.time()\n# do stuff\nelapsed = time.time() - t\n``````\n\nI have a helper class I like to use:\n\n``````class Timer(object):\ndef __init__(self, name=None):\nself.name = name\n\ndef __enter__(self):\nself.tstart = time.time()\n\ndef __exit__(self, type, value, traceback):\nif self.name:\nprint('[%s]' % self.name,)\nprint('Elapsed: %s' % (time.time() - self.tstart))\n``````\n\nIt can be used as a context manager:\n\n``````with Timer('foo_stuff'):\n# do some foo\n# do some stuff\n``````\n\nSometimes I find this technique more convenient than `timeit` - it all depends on what you want to measure.\n\n• @eat: I respectfully disagree. People have been using the unix `time` command to measure runtimes of programs for ever, and this method replicates this inside Python code. I see nothing wrong with it, as long as it's the right tool for the job. `timeit` isn't always that, and a profiler is a much more heavyweight solution for most needs – Eli Bendersky May 2 '11 at 3:27\n• For the last line I would suggest `print 'Elapsed: %.2f seconds % (time.time() - self.tstart)'`. It is hard to understand without the %.2f. Thanks for great idea. – Can Kavaklıoğlu Jun 19 '13 at 15:01\n• This looks very convenient at first glance, but in practice requires one to indent the code block one wants to time, which can be quite inconvenient depending on the length of the code block and the editor of choice. Still an elegant solution, which behaves correctly in the case of nested use. – Stefan Nov 15 '13 at 8:44\n• I think you want `elapsed = t - time.time()`, instead of `elapsed = time.time() - t`. In the latter elapsed will be negative. I suggested this change as an edit. – rysqui Oct 15 '15 at 1:01\n• @rysqui - Isn't the current time always a larger number than a previous time? I would think that `elapsed = time.time() - t` is the form that always yields a positive value. – Scott Smith Nov 7 '17 at 22:22\n\nI had the same question when I migrated to python from Matlab. With the help of this thread I was able to construct an exact analog of the Matlab `tic()` and `toc()` functions. Simply insert the following code at the top of your script.\n\n``````import time\n\ndef TicTocGenerator():\n# Generator that returns time differences\nti = 0 # initial time\ntf = time.time() # final time\nwhile True:\nti = tf\ntf = time.time()\nyield tf-ti # returns the time difference\n\nTicToc = TicTocGenerator() # create an instance of the TicTocGen generator\n\n# This will be the main function through which we define both tic() and toc()\ndef toc(tempBool=True):\n# Prints the time difference yielded by generator instance TicToc\ntempTimeInterval = next(TicToc)\nif tempBool:\nprint( \"Elapsed time: %f seconds.\\n\" %tempTimeInterval )\n\ndef tic():\n# Records a time in TicToc, marks the beginning of a time interval\ntoc(False)\n``````\n\nThat's it! Now we are ready to fully use `tic()` and `toc()` just as in Matlab. For example\n\n``````tic()\n\ntime.sleep(5)\n\ntoc() # returns \"Elapsed time: 5.00 seconds.\"\n``````\n\nActually, this is more versatile than the built-in Matlab functions. Here, you could create another instance of the `TicTocGenerator` to keep track of multiple operations, or just to time things differently. For instance, while timing a script, we can now time each piece of the script seperately, as well as the entire script. (I will provide a concrete example)\n\n``````TicToc2 = TicTocGenerator() # create another instance of the TicTocGen generator\n\ndef toc2(tempBool=True):\n# Prints the time difference yielded by generator instance TicToc2\ntempTimeInterval = next(TicToc2)\nif tempBool:\nprint( \"Elapsed time 2: %f seconds.\\n\" %tempTimeInterval )\n\ndef tic2():\n# Records a time in TicToc2, marks the beginning of a time interval\ntoc2(False)\n``````\n\nNow you should be able to time two separate things: In the following example, we time the total script and parts of a script separately.\n\n``````tic()\n\ntime.sleep(5)\n\ntic2()\n\ntime.sleep(3)\n\ntoc2() # returns \"Elapsed time 2: 5.00 seconds.\"\n\ntoc() # returns \"Elapsed time: 8.00 seconds.\"\n``````\n\nActually, you do not even need to use `tic()` each time. If you have a series of commands that you want to time, then you can write\n\n``````tic()\n\ntime.sleep(1)\n\ntoc() # returns \"Elapsed time: 1.00 seconds.\"\n\ntime.sleep(2)\n\ntoc() # returns \"Elapsed time: 2.00 seconds.\"\n\ntime.sleep(3)\n\ntoc() # returns \"Elapsed time: 3.00 seconds.\"\n\n# and so on...\n``````\n\nI hope that this is helpful.\n\nThe absolute best analog of tic and toc would be to simply define them in python.\n\n``````def tic():\n#Homemade version of matlab tic and toc functions\nimport time\nglobal startTime_for_tictoc\nstartTime_for_tictoc = time.time()\n\ndef toc():\nimport time\nif 'startTime_for_tictoc' in globals():\nprint \"Elapsed time is \" + str(time.time() - startTime_for_tictoc) + \" seconds.\"\nelse:\nprint \"Toc: start time not set\"\n``````\n\nThen you can use them as:\n\n``````tic()\n# do stuff\ntoc()\n``````\n• This will not behave correctly in the case of nested use of `tic` and `toc`, which Matlab supports. A little more sophistication would be required. – Stefan Nov 15 '13 at 8:37\n• I have implemented similar functions in my own code when I needed some basic timing. I would however remove the `import time` outside of both functions, since it can take potentially quite some time. – Bas Swinckels Nov 15 '13 at 17:19\n• If you insist on using this technique, and you need it to handle nested tic/toc, make the global a list and let `tic` push to it and `toc` pop from it. – Ahmed Fasih Aug 24 '16 at 1:26\n• Also I read elsewhere that `timeit.default_timer()` is better than `time.time()` because `time.clock()` might be more appropriate depending on OS – Miguel Feb 14 '17 at 21:23\n• @AhmedFasih That's what my answer does, though more things could be improved. – antonimmo Apr 23 at 7:32\n\nUsually, IPython's `%time`, `%timeit`, `%prun` and `%lprun` (if one has `line_profiler` installed) satisfy my profiling needs quite well. However, a use case for `tic-toc`-like functionality arose when I tried to profile calculations that were interactively driven, i.e., by the user's mouse motion in a GUI. I felt like spamming `tic`s and `toc`s in the sources while testing interactively would be the fastest way to reveal the bottlenecks. I went with Eli Bendersky's `Timer` class, but wasn't fully happy, since it required me to change the indentation of my code, which can be inconvenient in some editors and confuses the version control system. Moreover, there may be the need to measure the time between points in different functions, which wouldn't work with the `with` statement. After trying lots of Python cleverness, here is the simple solution that I found worked best:\n\n``````from time import time\n_tstart_stack = []\n\ndef tic():\n_tstart_stack.append(time())\n\ndef toc(fmt=\"Elapsed: %s s\"):\nprint fmt % (time() - _tstart_stack.pop())\n``````\n\nSince this works by pushing the starting times on a stack, it will work correctly for multiple levels of `tic`s and `toc`s. It also allows one to change the format string of the `toc` statement to display additional information, which I liked about Eli's `Timer` class.\n\nFor some reason I got concerned with the overhead of a pure Python implementation, so I tested a C extension module as well:\n\n``````#include <Python.h>\n#include <mach/mach_time.h>\n#define MAXDEPTH 100\n\nuint64_t start[MAXDEPTH];\nint lvl=0;\n\nstatic PyObject* tic(PyObject *self, PyObject *args) {\nstart[lvl++] = mach_absolute_time();\nPy_RETURN_NONE;\n}\n\nstatic PyObject* toc(PyObject *self, PyObject *args) {\nreturn PyFloat_FromDouble(\n(double)(mach_absolute_time() - start[--lvl]) / 1000000000L);\n}\n\nstatic PyObject* res(PyObject *self, PyObject *args) {\nreturn tic(NULL, NULL), toc(NULL, NULL);\n}\n\nstatic PyMethodDef methods[] = {\n{\"tic\", tic, METH_NOARGS, \"Start timer\"},\n{\"toc\", toc, METH_NOARGS, \"Stop timer\"},\n{\"res\", res, METH_NOARGS, \"Test timer resolution\"},\n{NULL, NULL, 0, NULL}\n};\n\nPyMODINIT_FUNC\ninittictoc(void) {\nPy_InitModule(\"tictoc\", methods);\n}\n``````\n\nThis is for MacOSX, and I have omitted code to check if `lvl` is out of bounds for brevity. While `tictoc.res()` yields a resolution of about 50 nanoseconds on my system, I found that the jitter of measuring any Python statement is easily in the microsecond range (and much more when used from IPython). At this point, the overhead of the Python implementation becomes negligible, so that it can be used with the same confidence as the C implementation.\n\nI found that the usefulness of the `tic-toc`-approach is practically limited to code blocks that take more than 10 microseconds to execute. Below that, averaging strategies like in `timeit` are required to get a faithful measurement.\n\n• Extremely elegant, @Stefan - can't believe this is so low rated. Thanks! – thclark Nov 8 '16 at 15:55\n\nYou can use `tic` and `toc` from `ttictoc`. Install it with\n\n``````pip install ttictoc\n``````\n\nAnd just import them in your script as follow\n\n``````from ttictoc import tic,toc\ntic()\n# Some code\nprint(toc())\n``````\n\nI have just created a module [tictoc.py] for achieving nested tic tocs, which is what Matlab does.\n\n``````from time import time\n\ntics = []\n\ndef tic():\ntics.append(time())\n\ndef toc():\nif len(tics)==0:\nreturn None\nelse:\nreturn time()-tics.pop()\n``````\n\nAnd it works this way:\n\n``````from tictoc import tic, toc\n\n# This keeps track of the whole process\ntic()\n\n# Timing a small portion of code (maybe a loop)\ntic()\n\n# -- Nested code here --\n\n# End\ntoc() # This returns the elapse time (in seconds) since the last invocation of tic()\ntoc() # This does the same for the first tic()\n``````\n\nI hope it helps.\n\n• Nice replication of tic/toc from MATLAB! – Matt May 23 '17 at 15:50\n• I must warn you that this might not behave as desired when used simultaneously by more than 1 module, since (AFAIK) modules behave like singletons. – antonimmo Jun 27 '17 at 18:04\n\nHave a look at the `timeit` module. It's not really equivalent but if the code you want to time is inside a function you can easily use it.\n\n• Yes, `timeit` is best for benchmarks. It doesn't even have to be a single function, you can pass abritarily complex statements. – user395760 May 1 '11 at 16:49\n• Well, passing code that is not an extremely simple function call as a string is very ugly. – ThiefMaster May 1 '11 at 16:50\n``````pip install easy-tictoc\n``````\n\nIn the code:\n\n``````from tictoc import tic, toc\n\ntic()\n\n#Some code\n\ntoc()\n``````\n\nDisclaimer: I'm the author of this library.\n\nThis can also be done using a wrapper. Very general way of keeping time.\n\nThe wrapper in this example code wraps any function and prints the amount of time needed to execute the function:\n\n``````def timethis(f):\nimport time\n\ndef wrapped(*args, **kwargs):\nstart = time.time()\nr = f(*args, **kwargs)\nprint \"Executing {0} took {1} seconds\".format(f.func_name, time.time()-start)\nreturn r\nreturn wrapped\n\n@timethis\ndef thistakestime():\nfor x in range(10000000):\npass\n\nthistakestime()\n``````\n\nI changed @Eli Bendersky's answer a little bit to use the ctor `__init__()` and dtor `__del__()` to do the timing, so that it can be used more conveniently without indenting the original code:\n\n``````class Timer(object):\ndef __init__(self, name=None):\nself.name = name\nself.tstart = time.time()\n\ndef __del__(self):\nif self.name:\nprint '%s elapsed: %.2fs' % (self.name, time.time() - self.tstart)\nelse:\nprint 'Elapsed: %.2fs' % (time.time() - self.tstart)\n``````\n\nTo use, simple put Timer(\"blahblah\") at the beginning of some local scope. Elapsed time will be printed at the end of the scope:\n\n``````for i in xrange(5):\ntimer = Timer(\"eigh()\")\nx = numpy.random.random((4000,4000));\nx = (x+x.T)/2\nnumpy.linalg.eigh(x)\nprint i+1\ntimer = None\n``````\n\nIt prints out:\n\n``````1\neigh() elapsed: 10.13s\n2\neigh() elapsed: 9.74s\n3\neigh() elapsed: 10.70s\n4\neigh() elapsed: 10.25s\n5\neigh() elapsed: 11.28s\n``````\n• An issue with this implementation is the fact, that `timer` is not deleted after the last call, if any other code follows after the `for` loop. To get the last timer value, one should delete or overwrite the `timer` after the `for` loop, e.g. via `timer = None`. – bastelflp Jul 5 '16 at 15:41\n• @bastelflp Just realized that I misunderstood what you meant... Your suggestion has been incorporated in the code now. Thanks. – Shaohua Li Aug 30 '17 at 5:38\n\nUpdating Eli's answer to Python 3:\n\n``````class Timer(object):\ndef __init__(self, name=None, filename=None):\nself.name = name\nself.filename = filename\n\ndef __enter__(self):\nself.tstart = time.time()\n\ndef __exit__(self, type, value, traceback):\nmessage = 'Elapsed: %.2f seconds' % (time.time() - self.tstart)\nif self.name:\nmessage = '[%s] ' % self.name + message\nprint(message)\nif self.filename:\nwith open(self.filename,'a') as file:\nprint(str(datetime.datetime.now())+\": \",message,file=file)\n``````\n\nJust like Eli's, it can be used as a context manager:\n\n``````import time\nwith Timer('Count'):\nfor i in range(0,10_000_000):\npass\n``````\n\nOutput:\n\n``````[Count] Elapsed: 0.27 seconds\n``````\n\nI have also updated it to print the units of time reported (seconds) and trim the number of digits as suggested by Can, and with the option of also appending to a log file. You must import datetime to use the logging feature:\n\n``````import time\nimport datetime\nwith Timer('Count', 'log.txt'):\nfor i in range(0,10_000_000):\npass\n``````\n\nBuilding on Stefan and antonimmo's answers, I ended up putting\n\n``````def Tictoc():\nstart_stack = []\nstart_named = {}\n\ndef tic(name=None):\nif name is None:\nstart_stack.append(time())\nelse:\nstart_named[name] = time()\n\ndef toc(name=None):\nif name is None:\nstart = start_stack.pop()\nelse:\nstart = start_named.pop(name)\nelapsed = time() - start\nreturn elapsed\nreturn tic, toc\n``````\n\nin a `utils.py` module, and I use it with a\n\n``````from utils import Tictoc\ntic, toc = Tictoc()\n``````\n\nThis way\n\n• you can simply use `tic()`, `toc()` and nest them like in Matlab\n• alternatively, you can name them: `tic(1)`, `toc(1)` or `tic('very-important-block')`, `toc('very-important-block')` and timers with different names won't interfere\n• importing them this way prevents interference between modules using it.\n\n(here toc does not print the elapsed time, but returns it.)" ]

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[ "• DELL专题主页说明：DELL主页内容包括DELL最新报价、DELL热卖促销、DELL产品信息以及经营销售DELL的企业/公司。除此之外，您可以在该主页查看有关DELL的价格、供应商的联系方式以及图片等信息。阿土伯交易网是您的营销平台、是您的收银台！\n\n• A\n• B\n• C\n• D\n• E\n• F\n• G\n• H\n• I\n• J\n• K\n• L\n• M\n• N\n• O\n• P\n• Q\n• R\n• S\n• T\n• U\n• V\n• W\n• X\n• Y\n• Z" ]

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[ "## Bond and stock valuation exercises\n\nRanks below bonds but above common stock in security. Preferred Exercise 6- 8: An investor Section 6.5 - Valuation of a Bond Between Coupon Dates.\n\n1 EN OF CHAPTER EXERCISES - ANSWERS Chapter 14 : Stock Valuation and the Bonds. 3. Stocks Stocks and Bonds 1. Present Value 2. Bonds 3. Stocks 1  Corporate hybrid bonds are subordinated debt instruments issued by half- capital, applying the concept of 'equity content', which tends to improve the issuer's credit ratios. Extension risk: if the hybrid is not called at the first call date, the valuation of the Whenever the issuer exercises the call, the investor will be entirely  The yield-to-maturity of a bond with an investment-grade rating will generally be lower than the yield-to-maturity of a bond with a speculative-grade rating. C. The yield-to-maturity of a bond with an investment-grade rating will generally be equal to the yield-to-maturity of a bond with a speculative-grade rating. Its target capital structure is 20% debt, 20% preferred stock, and 60% common equity. Its bonds have a 12% coupon, paid semiannually, a current maturity of 20 years, and a net price of \\$960. The firm could sell, at par, \\$100 preferred stock that pays a \\$10 annual dividend, but flotation costs of 5% would be incurred. Capital comes in two forms: debt capital and equity capital. To raise debt capital the companies sell bonds to the public, and to raise equity capital the corporation sells the stock of the company. Both stock and bonds are financial instruments and they have a certain intrinsic value. Stock Valuation Practice Problems 1. The Bulldog Company paid \\$1.5 of dividends this year. If its dividends are expected to grow at a rate of 3 percent per year, what is the expected dividend per share for Bulldog five years from today? 2. The current price of XYZ stock is \\$25 per share. If XYZ’s current dividend is \\$1 per share and VALUATION (BONDS AND STOCK) The general concept of valuation is very simple—the current value of any asset is the present value of the future cash flows it is expected to generate. It makes sense that you are willing to pay (invest) some amount today to receive future benefits (cash flows).\n\n## Introduction. In practice, company valuation deals with the valuation of stocks, bonds and other Let use IBM stock as an example for our valuation exercises.\n\n3 Jan 2011 Power Point Presentation for Band Valuation and the Techincal Exercise 6\n\n• A Rs 1000 par value bond carrying a coupon rate of 10%  Ranks below bonds but above common stock in security. Preferred Exercise 6- 8: An investor Section 6.5 - Valuation of a Bond Between Coupon Dates. 9 Mar 2020 A convertible bond is a fixed-income debt security that pays interest, but can be converted into common stock or equity shares.There are  bonds, where both bondholder and issuer own conversion rights. The holder will exercise the conversion right if the stock price lies above an upper strike level,  UNCERTAIN, AND THE VALUATION OF INDEX BONDS. STANLEY The real call price, C, will be a function of the stock price and the exercise price. X(t).3 We   This article explains to the reader that stock prices are subjective. Here are some of the common assumptions that will have to be made by the analyst during the stock valuation exercise: Horizon Period. As we learned in the previous article that stock valuation happens in two stages. How Interest Rates Affect Bonds ?\n\n### Bond Pricing Formula. bond valuations - price. Bond Valuation Formula can also be written as:.\n\nBond Pricing Formula. bond valuations - price. Bond Valuation Formula can also be written as:. Introduction. In practice, company valuation deals with the valuation of stocks, bonds and other Let use IBM stock as an example for our valuation exercises. 18 Jan 2019 Chapter I. Corporate bonds valuation and credit spreads: a a or common stock, or debt with a lower coupon or par-amount, lower seniority, If at maturity ≥ : the firm doesn't exercise its put option (which means that the. Convertible bonds typically offer higher yields than common stock but lower ratio”—or the number of shares that the investor receives if they exercises the\n\n### Ranks below bonds but above common stock in security. Preferred Exercise 6- 8: An investor Section 6.5 - Valuation of a Bond Between Coupon Dates.\n\nfinance 440 review: bond and stock valuation practice problems multiple choice which of the following statements is correct regarding bonds and bond ratings?\n\n## What it means to buy a company's stock (Opens a modal) Bonds vs. stocks (Opens a modal) Shorting stock. Learn. Shorting stock (Opens a modal) Market value of assets (Opens a modal) Corporate metrics and valuation. Learn. Price and market capitalization Relationship between bond prices and interest rates (Opens a modal) Treasury bond\n\n127.6108 1/5 − 1 = 4.9686% 100.136 Exercise 2.20 We consider two bonds with the Exercise 7.1 Would you say it is easier to track a bond index or a stock index. the variation in the subperiods returns are large over the valuation period. The price of a pure discount (zero coupon) bond is the present value of the par used to find the monthly interest rates for the stock and bond accounts, so:. See if you understand how bonds work and how to calculate their value with this interactive quiz and printable worksheet. With these practice Convertibles' hybrid nature presents valuation challenges and trading A convertible bond is essentially a corporate bond with an embedded equity Deciding the fair value of a convertible depends on the stock price, the exercise price, the  3 Jan 2011 Power Point Presentation for Band Valuation and the Techincal Exercise 6\n\n• A Rs 1000 par value bond carrying a coupon rate of 10%  Ranks below bonds but above common stock in security. Preferred Exercise 6- 8: An investor Section 6.5 - Valuation of a Bond Between Coupon Dates. 9 Mar 2020 A convertible bond is a fixed-income debt security that pays interest, but can be converted into common stock or equity shares.There are\n\nA coupon bond that pays interest of \\$100 annually has a par value of \\$1,000, matures in 5 years, and is selling today at a \\$72 discount from par value. The yield to maturity on this bond is a) 6.00% b) 8.33% c) 10.39% d) 12.00% e) 60.00% 9. The coupon payment is the coupon rate times par value. Using this relationship, we get: Coupon rate = \\$50.66 / \\$1,000 Coupon rate = .0507, or 5.07% 6. To find the price of this bond, we need to realize that the maturity of the bond is 14 years. The bond was issued 1 year ago, with 15 years to maturity, so there are 14 years left on the bond." ]

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[ "# Maximal subgroups of a certain finite 2-group\n\nThe following came up in a problem on reconstruction of digraphs. I determined enough about the answer to satisfy the application completely, but still I am curious to know what the complete solution is. My group theory is weak, so apologies if this is too simple.\n\nLet $T$ be a full binary tree with depth $k$. Call its levels $L_0,\\ldots,L_k$. Here is the case $k=4$:", null, "(source)\n\nThe number of leaves is $n=2^k$.\n\nLet $A$ be the full automorphism group of $T$ and let $f$ be its (faithful) action on the leaves of the tree, i.e. on $L_k$. Obviously $f(A)$ is an iterated wreath product of $\\mathbb{Z}_2$ with itself and has order $2^{n-1}$. It is, indeed, the Sylow 2-subgroup of $S_n$.\n\nThe problem is: what are the subgroups of $f(A)$ of index 2?\n\nHere is what I think the answer is. Let $X$ be a union of levels of $T$, including at least one level other than $L_0$. Let $P$ be the set of all $f(\\gamma)$ such that $\\gamma\\in A$ and the action of $\\gamma$ on $X$ is an even permutation. Then $P$ is a subgroup of the desired index. I'm guessing there are no others...\n\nYou can easily count the number of maximal subgroups of $W(k)$, the $k$-fold iterated wreath product of $\\mathbb{Z}_{2},$ by calculating the index of the Frattini subgroup. You can inductively prove that the number of generators is $k,$ which is clear for $k =1,2.$\nTo proceed, note that $W(k) = W(k-1) \\wr \\mathbb{Z}_{2}.$ Factor out the Frattini subgroup of the base group, and by induction, you are left with $E(k-1) \\wr \\mathbb{Z}_{2}$, where $E(k-1)$ is elementary Abelian of order $2^{k-1}.$ If $x$ is an element of order $2$ outside the new base group, then $[E(k-1) \\times E(k-1),x]$ has order $2^{k-1},$ so that the largest elementary Abelian factor group of the original wreath product does have order $2^{k},$ as claimed.\nHence the group $W(k)$ has $2^{k}-1$ maximal subgroups, since there is a bijection between maximal subgroups of $W(k)$ and maximal subgroups of $W(k)/\\Phi(W(k)).$\n• That's great. Since the subgroups I identified are all distinct (ignoring whether $L_0\\subseteq X$) and there are exactly $2^k-1$ of them, your proof establishes that my wild guess was correct (for once). Thanks! – Brendan McKay Sep 2 '12 at 11:41\n• The fact that $W(k)$ has $2^{k}-1$ maximal subgroups is equivalent to saying that the Frattini subgroup of $W(k)$ has index $2^{k}$, which is what we are trying to establish by induction. Since this is clear for $k =1,2$, as noted, we may assume by induction that $W(k-1)$ has Frattini factor of index $2^{k-1}$, as used in the above proof. – Geoff Robinson Dec 3 '17 at 14:02" ]

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[ "# R package pomdp: Partially Observable Markov Decision Processes\n\nProvides the infrastructure to define and analyze the solutions of Partially Observable Markov Decision Processes (POMDP) models. The package includes pomdp-solve (Cassandra, 2015) to solve POMDPs using a variety of algorithms.\n\nThe package provides the following algorithms:\n\n• Exact value iteration\n• Enumeration algorithm (Sondik 1971, Mohan 1982).\n• Two pass algorithm (Sondik 1971).\n• Witness algorithm (Littman, Cassandra, Kaelbling 1996).\n• Incremental pruning algorithm (Zhang and Liu 1996, Cassandra et al 1997).\n• Approximate value iteration\n• Finite grid algorithm (Cassandra 2015), a variation of point-based value iteration to solve larger POMDPs (PBVI; see Pineau 2003) without dynamic belief set expansion.\n\n## Installation\n\nStable CRAN version: install from within R with\n\n``install.packages(\"pomdp\")``\n\nCurrent development version: install from GitHub (needs devtools).\n\n``````library(\"devtools\")\n\n## Usage\n\nSolving the simple infinite-horizon Tiger problem.\n\n``````R> library(\"pomdp\")\nR> data(\"Tiger\")\nR> Tiger``````\n``````Unsolved POMDP model: Tiger Problem\nhorizon: Inf ``````\n``````> sol <- solve_POMDP(model = Tiger)\n> sol``````\n``````Solved POMDP model: Tiger Problem\nsolution method: grid\nhorizon: Inf\nconverged: TRUE\ntotal expected reward (for start probabilities): 1.933439 ``````\n``> policy(sol)``\n``````[]\ntiger-left tiger-right action tiger-left tiger-right\n1 -98.549921 11.450079 open-left 3 3\n2 -10.854299 6.516937 listen 3 1\n3 1.933439 1.933439 listen 4 2\n4 6.516937 -10.854299 listen 5 3\n5 11.450079 -98.549921 open-right 3 3``````\n• Cassandra, A. (2015). pomdp-solve: POMDP Solver Software, http://www.pomdp.org.\n• Sondik, E. (1971). The Optimal Control of Partially Observable Markov Processes. Ph.D. Dissertation, Stanford University.\n• Cassandra, A., Littman M.L., Zhang L. (1997). Incremental Pruning: A Simple, Fast, Exact Algorithm for Partially Observable Markov Decision Processes. UAI’97: Proceedings of the Thirteenth conference on Uncertainty in artificial intelligence, August 1997, pp. 54-61.\n• Monahan, G. E. (1982). A survey of partially observable Markov decision processes: Theory, models, and algorithms. Management Science 28(1):1-16.\n• Littman, M. L.; Cassandra, A. R.; and Kaelbling, L. P. (1996). Efficient dynamic-programming updates in partially observable Markov decision processes. Technical Report CS-95-19, Brown University, Providence, RI.\n• Zhang, N. L., and Liu, W. (1996). Planning in stochastic domains: Problem characteristics and approximation. Technical Report HKUST-CS96-31, Department of Computer Science, Hong Kong University of Science and Technology.\n• Pineau J., Geoffrey J Gordon G.J., Thrun S.B. (2003). Point-based value iteration: an anytime algorithm for POMDPs. IJCAI’03: Proceedings of the 18th international joint conference on Artificial Intelligence. Pages 1025-1030.\n\n## Acknowledgments\n\nDevelopment of this package was supported in part by National Institute of Standards and Technology (NIST) under grant number 60NANB17D180." ]

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[ "", null, "", null, "RUS  ENG JOURNALS   PEOPLE   ORGANISATIONS   CONFERENCES   SEMINARS   VIDEO LIBRARY   PACKAGE AMSBIB", null, "", null, "General information", null, "Latest issue", null, "Archive", null, "Search papers", null, "Search references", null, "RSS", null, "Latest issue", null, "Current issues", null, "Archive issues", null, "What is RSS\n\n Differ. Uravn.: Year: Volume: Issue: Page: Find\n\n Personal entry: Login: Password: Save password Enter", null, "Forgotten password?", null, "Register\n\n 1978, Volume 14, Number 8", null, "", null, "Ordinary Differential Equations", null, "On the theory of functional-differential equationsM. E. Drakhlin, T. K. Plyshevskaya 1347", null, "Stability and topological classification of invariant setsYu. V. Malyshev 1362", null, "On the theory of duality in control systemsB. Mordukhovich, A. M. Sasonkin 1371", null, "Averaging of boundary value problems in terminal optimal control problemsV. A. Plotnikov, T. S. Zverkova 1381", null, "Averaging of quasilinear systems with several fast variablesN. A. Reimers, K. Sh. Khodzhaev 1388", null, "The equivalence of systems of differential equations with reducing contractionV. S. Samovol 1400", null, "Asymptotic solution of certain optimal control problems with lagsA. I. Tret'yak 1414 Partial Differential Equations", null, "A mixed problem for a second order degenerate hyperbolic equation with two variablesF. T. Baranovskii 1424", null, "Solution of a boundary value problem for the Helmholtz equation in a layered space with a spherical inclusionV. T. Erofeenko 1439", null, "On the estimation of the classical solution of a mixed problem for a hyperbolic equationÈ. L. Кulikov, V. Ph. Kirichenko 1448", null, "Estimates of the derivatives of the solutions of linear equations of elliptic type on $E^n$ and related weighted imbedding theorems. IIA. S. Fokht 1455 Integral Equations", null, "Some properties of the set of solutions of a Volterra–Hammerstein integral inclusionA. I. Bulgakov, L. N. Lyapin 1465", null, "Singular numbers of integral operatorsB. D. Kotlyar 1473 Numerical methods", null, "Difference schemes for quasilinear nonstationary problems. IIV. N. Abrashin, V. I. Lis 1478", null, "A difference method with rectification of the fronts for the solution of nonlinear problems of Stefan typeV. A. Tsurko 1487 Short Communications", null, "Asymptotic stability in nonlinear systemsO. V. Anashkin 1490", null, "The Volterra model with a weakly linear hereditary characteristicR. Anvarov, G. S. Larionov 1494", null, "A system of integro-differential equations with a small parameter multiplying the derivative of conditionally stable typeM. P. Belyanin 1497", null, "Stability of a dynamical system with a cylindrical phase spaceZ. U. Blyagoz, G. A. Leonov 1502", null, "The Grobman–Hartman theorem for extensions of dynamical systemsI. U. Bronshtein, V. A. Glavan 1504", null, "Coercive solvability of abstract parabolic equations with a constant operatorV. V. Vasil'ev 1507", null, "Potential functions of the field of radiation of an electric quadrupole in cylindrical coordinatesA. I. Glushtsov 1511", null, "A method for lowering the order of linear differential equationsÈ. I. Grudo 1514", null, "An isoperimetric problemD. P. Zeragiâ 1516", null, "Combination of Čaplygin and spline-collocation methods for the solution of ordinary differential equationsV. K. Saul'ev 1520", null, "The calculation of eigenvalues and -vectors of linear operators on the basis of the theory of perturbationsN. A. Sidorov 1522", null, "On the question of absolute exponential stability of linear differential equations of lagging type in a Banach spaceV. E. Slyusarchuk 1526", null, "Contact us: math-net2019_10 [at] mi-ras ru", null, "Terms of Use", null, "Registration", null, "Logotypes © Steklov Mathematical Institute RAS, 2019" ]

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[ "# Existence of mixed Nash equilibria for compact strategy spaces\n\nIn Peleg (1969) it is shown that a game with finite strategy spaces and continuous utilities has a mixed Nash equilibrium for any cardinality of players.\n\nIs the same true if the strategy spaces are compact (maybe under some extra conditions, like only countably many players)?\n\nLones Smith offers the following proof: if you have countably many players, arbitrarily enumerate them and call the original game with this enumeration $G$. Let $G_n$ be a game where the actions of players $k>n$ are fixed arbitrarily. By Glicksburg, $G_n$ has an equilibrium -- call it $x(n)$. Let $x$ be a limit point of the sequence $(x(n))_{n=1}^\\infty$. Fix any player $i$. Now $x_i(n)$ is a best response to $x_{-i}(n)$ in $G_n$ for each $n$; to pass to limits, we appeal to the Theorem of the Maximum and continuity of the utility functions . That shows the action $x_i$ is a best response to $x_{-i}$ in $G$. So $x$ is an equilibrium of $G$.\n This assumes a topology on the action space at least as coarse as the product topology, so that whenever $x(n)$ converges to $x$ in the product topology, it converges in the topology in question. The utilities should be continuous under this topology for actions. This is a big assumption and easily violated!" ]

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[ "#", null, "", null, "Radar Basics\n\n#### Rayleigh- versus Mie- Scattering", null, "Rayleigh\nMie\nOptical\nA\nB\n\nFigure 1: Rayleight- and Mie- Scattering, and optical region", null, "Rayleigh-\nRegion\nMie- Region\noptical\nregion\nA\nB\n\nFigure 1: Rayleight- and Mie- Scattering, and optical region\n\n#### Rayleigh- versus Mie- Scattering\n\nThe following picture shows the different regions applicable to computing the radar cross-section of a sphere. The optical region („far field” counterpart) rules apply when (2π·r / λ) >10. In this region, the radar cross-section of a sphere is independent of frequency. Here, the radar cross-section of a sphere, σ = π·r2.\n\nThe radar cross-section equation breaks down primarily due to creeping waves in the area where 2π · r. The largest positive perturbation of the radar cross-section (point A) would be 4 times higher than the radar cross-section computed using the optical region formula. Just slightly a minimum occurs (point B) and the actual radar cross-section would be 0.26 times the value calculated by using the optical region formula. This area is known as the „Mie” or „resonance region”.\n\nIf we used a one-meter diameter sphere, the perturbations would occur at 95 MHz, so any frequency above 950 MHz would give predicted results.\n\nThe size of the spherical reflection area is smaller than the wavelength in the area of the „Rayleigh-Scattering”. The radar cross-section is calculated the formula σ = π·r2 · 7,11 · ( 2π·r / λ )4 here. The „Rayleigh-Scattering” is a typical application case for weather radar.\n\nApproximately this is lower L- Band still takes the Mie scattering into account at air defense and air traffic control radar sets. There are predominantly optical conditions at frequencies above 1 GHz.\n\nE.g. the ancient Russian VHF-Radars operate at frequencies about 145 to 175 MHz, ie they use the wavelength of about 1.7 to 2.1 meters. This is exactly the position on the second maximum in the diagram (located over the letter B) for the geometric dimensions of a fighter aircraft." ]

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[ "• Index", null, "#### 返回函数\n\nLast updated: ... / Reads: 137764365 Edit\n\n### 函数作为返回值\n\n``````def calc_sum(*args):\nax = 0\nfor n in args:\nax = ax + n\nreturn ax\n``````\n\n``````def lazy_sum(*args):\ndef sum():\nax = 0\nfor n in args:\nax = ax + n\nreturn ax\nreturn sum\n``````\n\n``````>>> f = lazy_sum(1, 3, 5, 7, 9)\n>>> f\n<function lazy_sum.<locals>.sum at 0x101c6ed90>\n``````\n\n``````>>> f()\n25\n``````\n\n``````>>> f1 = lazy_sum(1, 3, 5, 7, 9)\n>>> f2 = lazy_sum(1, 3, 5, 7, 9)\n>>> f1==f2\nFalse\n``````\n\n`f1()``f2()`的调用结果互不影响。\n\n### 闭包\n\n``````def count():\nfs = []\nfor i in range(1, 4):\ndef f():\nreturn i*i\nfs.append(f)\nreturn fs\n\nf1, f2, f3 = count()\n``````\n\n``````>>> f1()\n9\n>>> f2()\n9\n>>> f3()\n9\n``````\n\n``````返回闭包时牢记一点：返回函数不要引用任何循环变量，或者后续会发生变化的变量。\n``````\n\n``````def count():\ndef f(j):\ndef g():\nreturn j*j\nreturn g\nfs = []\nfor i in range(1, 4):\nfs.append(f(i)) # f(i)立刻被执行，因此i的当前值被传入f()\nreturn fs\n``````\n\n``````>>> f1, f2, f3 = count()\n>>> f1()\n1\n>>> f2()\n4\n>>> f3()\n9\n``````\n\n### nonlocal\n\n``````def inc():\nx = 0\ndef fn():\n# 仅读取x的值:\nreturn x + 1\nreturn fn\n\nf = inc()\nprint(f()) # 1\nprint(f()) # 1\n``````\n\n``````# -*- coding: utf-8 -*-\n----\ndef inc():\nx = 0\ndef fn():\n# nonlocal x\nx = x + 1\nreturn x\nreturn fn\n\nf = inc()\nprint(f()) # 1\nprint(f()) # 2\n``````\n\n``````使用闭包时，对外层变量赋值前，需要先使用nonlocal声明该变量不是当前函数的局部变量。\n``````\n\n### 练习\n\n``````# -*- coding: utf-8 -*-\n----\ndef createCounter():\ndef counter():\nreturn 1\nreturn counter\n----\n# 测试:\ncounterA = createCounter()\nprint(counterA(), counterA(), counterA(), counterA(), counterA()) # 1 2 3 4 5\ncounterB = createCounter()\nif [counterB(), counterB(), counterB(), counterB()] == [1, 2, 3, 4]:\nprint('测试通过!')\nelse:\nprint('测试失败!')\n``````\n\nreturn_func.py" ]

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[ "# Continuous representation theory of p-adic Lie groups\n\n```@inproceedings{Schneider2006ContinuousRT,\ntitle={Continuous representation theory of p-adic Lie groups},\nauthor={Peter Schneider},\nyear={2006}\n}```\nIn this paper we give an overview over the basic features of the continuous representation theory of p-adic Lie groups as it has emerged during the last five years. Themain motivation for developing such a theory is a possible extension of the local Langlands program to p-adic Galois representations. This is still very much in its infancy. But in the last section we will describe a first approximation to an extended Langlands functoriality principle for crystalline Galois representations.\n27 Citations\nAbstract Given a compact p-adic Lie group G over a finite unramified extension L/ℚp let GL/ℚp be the product over all Galois conjugates of G. We construct an exact and faithful functor from\nThe aim of the seminar is to give an introduction to the theory of representations of p-adic locally analytic groups, and the theory of Ù D-modules introduced in [2, 4]. We also aim to study\nWe study the continuous principal series representations of split connected reductive p-adic groups over p-adic fields. We show that such representations are irreducible when the inducing character\nIn this paper we study p-adic principal series representation of a padic group G as a module over the maximal compact subgroup G0. We show that there are no non-trivial G0-intertwining maps between\n• Mathematics\n• 2016\nWe study the continuous principal series representations of split connected reductive padic groups over p-adic fields. We show that such representations are irreducible when the inducing character\n• Mathematics\n• 2022\nIn this paper we view pro-p Iwahori subgroups I as rigid analytic groups I for large enough p. This is done by endowing I with a natural p-valuation, and thereby generalizing results of Lazard for\n• Mathematics\n• 2014\nLet \\$G\\$ be the group of points of a split reductive group over a finite extension of \\${\\mathbb Q}_p\\$. In this paper, we compute the dimensions of certain classes of locally analytic\nLet G be a lo/cally ℚp-analytic group and K a finite extension of ℚp with residue field k. Adapting a strategy of B. Mazur (cf. [Maz89]) we use deformation theory to study the possible liftings of a\nFor a fairly general reductive group G/Qp , we explicitly compute the space of locally algebraic vectors in the Breuil-Herzig construction Π(ρ)ord, for a potentially semistable Borel-valued\n• Mathematics\n• 2017\nLet \\$G\\$ be a reductive \\$p\\$-adic group‎. ‎We consider the general question‎ ‎of whether the reducibility of an induced representation can be detected in a‎ ‎``co-rank one‎\" ‎situation‎. ‎For smooth\n• Mathematics\n• 1999\nLet L be a finite extension of Qp, and let K be a spherically complete non-archimedean extension field of L. In this paper we introduce a restricted category of continuous representations of locally\nLet p be a prime number, L a finite extension of the field Qp of p-adic numbers, K a spherically complete extension field of L and G the group of L-rational points of a split reductive group over L.\n• Mathematics\n• 2000\nWe develop a duality theory between the continuous representations of a compactp-adic Lie groupG in Banach spaces over a givenp-adic fieldK and certain compact modules over the completed group\nIntroduction The congruences between automorphic forms and their appli­ cations to number theory are a motivation to study the smooth representations of a reductive p-adic group G over an\n• Mathematics\n• 2006\nAbstract By the theory of Colmez and Fontaine, a de Rham representation of the Galois group of a local field roughly corresponds to a representation of the Weil-Deligne group equipped with an\n• Mathematics\n• 2005\nWe take some initial steps towards illuminating the (hypothetical) \\$p\\$-adic local Langlands functoriality principle relating Galois representations of a \\$p\\$-adic field \\$L\\$ and admissible unitary\n• Mathematics\n• 2004\nWe study the problem of constructing a contragredient functor on the category of admissible locally analytic representations of a p-adic analytic group G. A naive contragredient does not exist. As a\n• Mathematics\n• 2003\nLet G be a compact, locally L-analytic group, where L is a finite extension of Qp. Let K be a discretely valued extension field of L. We study the algebra D(G,K) of K-valued locally analytic" ]

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[ "# iteration list\n\nlohabithi shared this question 2 years ago\n\nHello, I want the series of sequential sum of a list of numbers.\n\nlike,\n\n{1,2,3,4,5} -> {1,3,6,10,15}\n\nI can use iteration list with a custom tool, like in the file attached. It works.\n\nBut if I attempt to define the expression directly, like:\n\nIterationList({Element(p,1)+1, Element(L_1, Element(p, 1))+Element(p,2)}, p, {{1, 0}}, Length(L_1) - 1)\n\nthen strange things happen!\n\nFirst, nothing happens, no error message but also no new object is created (defined of undefined). As if I had not typed anything!\n\nBut then, it becomes impossible to make any input with the mouse: no new point can be created!\n\nOn the other hand, as you can see in the file attached, iteration (NOT iterationlist) works without recourse to custom tool:\n\nIteration({Element(p,1)+1, Element(L_1, Element(p, 1))+Element(p,2)}, p, {{1, 0}}, Length(L_1) - 1)\n\ngives the proper result.\n\nThis problem with iteration list did happen repeatedly in a complicated file, and eventually it seems I have isolated the problem.", null, "1\n\nit works fine for me\n\nGeoGebra Classic 5.0.486.0-d (Java 1.8.0_161-32bit, 247MB. CAS Initialising) 18 August 2018", null, "1\n\nI also have the same installation (copy of system info below), and it doesn't work...\n\n[co]GeoGebra Classic 5.0.486.0-d (18 August 2018)\n\nJava: 1.8.0_161\n\nOS: Windows 7\n\nArchitecture: x86 / x86\n\nHeap: 247MB\n\nCAS: CAS Initializing\n\nGeoGebraLogger log:\n\n2:59:47.265 DEBUG: org.geogebra.desktop.i.a.<init>[-1]: isApplet=false runningFromJar=true\n\n2:59:47.265 DEBUG: org.geogebra.desktop.i.a.az[-1]: Setting up logging\n\n2:59:47.278 DEBUG: org.geogebra.desktop.i.a.az[-1]: Logging is redirected to C:\\Users\\Admin\\AppData\\Local\\Temp\\GeoGebraLog_rligdyqaik.txt\n\nDEBUG: org.geogebra.common.m.x.G[-1]: Number of registered views = 1\n\n* class org.geogebra.desktop.geogebra3D.b.b\n\nDEBUG: org.geogebra.desktop.i.a.aR[-1]: Waiting for the initialization: 0\n\nDEBUG: org.geogebra.common.b.c.a.a[-1]: SingularWS connection disabled by command line option\n\nINFO: org.geogebra.common.n.f.b[-1]: No SingularWS is available at http://singularws.idm.jku.at/ (yet)\n\nDEBUG: org.geogebra.desktop.l.l.a[-1]: storing geogebra_thumbnail.png false\n\nDEBUG: org.geogebra.desktop.l.l.a[-1]: storing d81d328bd62b33c156c55500b5b41145/tag1.JPG false\n\nDEBUG: org.geogebra.common.m.x.a[-1]: function ggbOnInit() {}\n\nDEBUG: org.geogebra.common.m.x.G[-1]: Number of registered views = 2\n\n* class org.geogebra.desktop.geogebra3D.b.b\n\n* class org.geogebra.common.plugin.b\n\nDEBUG: org.geogebra.common.m.d.c.a[-1]: null valid expression label\n\nDEBUG: org.geogebra.common.m.d.c.a[-1]: null valid expression label\n\nDEBUG: org.geogebra.common.m.c.bX.b[-1]: falselabel\n\nDEBUG: org.geogebra.common.m.c.bX.b[-1]: falselabel\n\nDEBUG: org.geogebra.common.m.c.bX.b[-1]: falselabel\n\nDEBUG: org.geogebra.common.i.h.a[-1]: CAS support: true\n\nDEBUG: org.geogebra.common.m.x.G[-1]: Number of registered views = 3\n\n* class org.geogebra.desktop.geogebra3D.b.b\n\n* class org.geogebra.common.plugin.b\n\n* class org.geogebra.desktop.geogebra3D.e.c.a.a\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 7\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 6\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 4\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 7\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 6\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 4\n\nERROR: org.geogebra.desktop.gui.i.U.a[-1]: problem starting JavaFX\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 7\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 6\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 4\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 7\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 6\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 4\n\nWARN: org.geogebra.desktop.i.a.o[-1]: Option versionCheckAllow not recognized :\n\nINFO: org.geogebra.desktop.i.B.a[-1]: No system preferences\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 7\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 6\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 4\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 7\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 6\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 4\n\nDEBUG: org.geogebra.desktop.geogebra3D.input3D.intelRealSense.Socket.queryRegistry[-1]: reg query HKLM\\Software\\Intel\\RSSDK\\Components\\ivcam : 1\n\nDEBUG: org.geogebra.desktop.geogebra3D.a\\$a.a[-1]: RealSense: No key for camera in registry\n\nDEBUG: org.geogebra.desktop.geogebra3D.a\\$a.b[-1]: zSpace: try to init\n\nDEBUG: org.geogebra.desktop.geogebra3D.a\\$a.b[-1]: zSpace: Failed to init\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 7\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 6\n\nDEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 4\n\nDEBUG: org.geogebra.desktop.b.a.e.<init>[-1]: possible missing release()\n\nDEBUG: org.geogebra.desktop.gui.a.a\\$a.b[-1]: Checking version\n\nDEBUG: org.geogebra.desktop.gui.a.a\\$a.b[-1]: major version check needed: no check was done yet\n\nAug 21, 2018 2:59:47 AM\n\nSTDOUT: DEBUG: org.geogebra.common.m.x.G[-1]: Number of registered views = 1\n\n* class org.geogebra.desktop.geogebra3D.b.b\n\nAug 21, 2018 2:59:47 AM\n\nSTDOUT: DEBUG: org.geogebra.desktop.i.a.aR[-1]: Waiting for the initialization: 0\n\nAug 21, 2018 2:59:47 AM\n\nAug 21, 2018 2:59:47 AM\n\nSTDOUT: DEBUG: org.geogebra.common.b.c.a.a[-1]: SingularWS connection disabled by command line option\n\nAug 21, 2018 2:59:47 AM\n\nSTDOUT: INFO: org.geogebra.common.n.f.b[-1]: No SingularWS is available at http://singularws.idm.jku.at/ (yet)\n\nAug 21, 2018 2:59:47 AM\n\nAug 21, 2018 2:59:47 AM\n\nSTDOUT: DEBUG: org.geogebra.desktop.l.l.a[-1]: storing geogebra_thumbnail.png false\n\nAug 21, 2018 2:59:47 AM\n\nAug 21, 2018 2:59:47 AM\n\nSTDOUT: DEBUG: org.geogebra.desktop.l.l.a[-1]: storing d81d328bd62b33c156c55500b5b41145/tag1.JPG false\n\nAug 21, 2018 2:59:47 AM\n\nSTDOUT: DEBUG: org.geogebra.common.m.x.a[-1]: function ggbOnInit() {}\n\nAug 21, 2018 2:59:47 AM\n\nSTDOUT: DEBUG: org.geogebra.common.m.x.G[-1]: Number of registered views = 2\n\n* class org.geogebra.desktop.geogebra3D.b.b\n\n* class org.geogebra.common.plugin.b\n\nAug 21, 2018 2:59:48 AM\n\nSTDOUT: DEBUG: org.geogebra.common.m.d.c.a[-1]: null valid expression label\n\nAug 21, 2018 2:59:48 AM\n\nSTDOUT: DEBUG: org.geogebra.common.m.d.c.a[-1]: null valid expression label\n\nAug 21, 2018 2:59:48 AM\n\nSTDOUT: DEBUG: org.geogebra.common.m.c.bX.b[-1]: falselabel\n\nAug 21, 2018 2:59:48 AM\n\nSTDOUT: DEBUG: org.geogebra.common.m.c.bX.b[-1]: falselabel\n\nAug 21, 2018 2:59:48 AM\n\nSTDOUT: DEBUG: org.geogebra.common.m.c.bX.b[-1]: falselabel\n\nAug 21, 2018 2:59:48 AM\n\nSTDOUT: DEBUG: org.geogebra.common.i.h.a[-1]: CAS support: true\n\nAug 21, 2018 2:59:50 AM\n\nSTDOUT: DEBUG: org.geogebra.common.m.x.G[-1]: Number of registered views = 3\n\n* class org.geogebra.desktop.geogebra3D.b.b\n\n* class org.geogebra.common.plugin.b\n\n* class org.geogebra.desktop.geogebra3D.e.c.a.a\n\nAug 21, 2018 2:59:50 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 7\n\nAug 21, 2018 2:59:50 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 6\n\nAug 21, 2018 2:59:50 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 4\n\nAug 21, 2018 2:59:50 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 7\n\nAug 21, 2018 2:59:50 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 6\n\nAug 21, 2018 2:59:50 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 4\n\nAug 21, 2018 2:59:50 AM\n\nSTDERR: ERROR: org.geogebra.desktop.gui.i.U.a[-1]: problem starting JavaFX\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 7\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 6\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 4\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 7\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 6\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 4\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: WARN: org.geogebra.desktop.i.a.o[-1]: Option versionCheckAllow not recognized :\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: INFO: org.geogebra.desktop.i.B.a[-1]: No system preferences\n\nAug 21, 2018 2:59:51 AM\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 7\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 6\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 4\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 7\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 6\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 4\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.desktop.geogebra3D.input3D.intelRealSense.Socket.queryRegistry[-1]: reg query HKLM\\Software\\Intel\\RSSDK\\Components\\ivcam : 1\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.desktop.geogebra3D.a\\$a.a[-1]: RealSense: No key for camera in registry\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.desktop.geogebra3D.a\\$a.b[-1]: zSpace: try to init\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.desktop.geogebra3D.a\\$a.b[-1]: zSpace: Failed to init\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 7\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 6\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.common.n.f.a[-1]: Application.getModeText(): macro does not exist: ID = 4\n\nAug 21, 2018 2:59:51 AM\n\nAug 21, 2018 2:59:51 AM\n\nSTDOUT: DEBUG: org.geogebra.desktop.b.a.e.<init>[-1]: possible missing release()\n\nAug 21, 2018 2:59:52 AM\n\nSTDOUT: DEBUG: org.geogebra.desktop.gui.a.a\\$a.b[-1]: Checking version\n\nAug 21, 2018 2:59:52 AM\n\nSTDOUT: DEBUG: org.geogebra.desktop.gui.a.a\\$a.b[-1]: major version check needed: no check was done yet\n\nAug 21, 2018 2:59:53 AM\n\nAug 21, 2018 2:59:54 AM\n\nGGB file content:\n\n<?xml version=\"1.0\" encoding=\"utf-8\"?>\n\n<geogebra format=\"5.0\" version=\"5.0.486.0\" app=\"classic\" platform=\"d\" id=\"cc537f20-a3e3-4a75-96d9-67f254e65d60\" xsi:noNamespaceSchemaLocation=\"http://www.geogebra.org/ggb...; xmlns=\"\" xmlns:xsi=\"http://www.w3.org/2001/XMLS...; >\n\n<gui>\n\n<window width=\"1366\" height=\"768\" />\n\n<perspectives>\n\n<perspective id=\"tmp\">\n\n<panes>\n\n<pane location=\"\" divider=\"0.3213762811127379\" orientation=\"1\" />\n\n</panes>\n\n<views>\n\n<view id=\"512\" toolbar=\"0 | 1 501 5 19 , 67 | 2 15 45 18 , 7 37 | 514 3 9 , 13 44 , 47 | 16 | 551 550 11 , 20 22 21 23 , 55 56 57 , 12 | 69 | 510 511 , 512 513 | 533 531 , 534 532 , 522 523 , 537 536 , 535 | 521 520 | 36 , 38 49 560 | 571 30 29 570 31 33 | 17 | 540 40 41 42 , 27 28 35 , 6 , 502\" visible=\"false\" inframe=\"false\" stylebar=\"false\" location=\"1,1,1\" size=\"500\" window=\"100,100,600,400\" />\n\n<view id=\"4\" toolbar=\"0 || 2020 , 2021 , 2022 , 66 || 2001 , 2003 , 2002 , 2004 , 2005 || 2040 , 2041 , 2042 , 2044 , 2043\" visible=\"false\" inframe=\"false\" stylebar=\"false\" location=\"1,1\" size=\"300\" window=\"100,100,600,400\" />\n\n<view id=\"8\" toolbar=\"1001 | 1002 | 1003 || 1005 | 1004 || 1006 | 1007 | 1010 || 1008 1009 || 66 68 || 6\" visible=\"false\" inframe=\"false\" stylebar=\"false\" location=\"1,3\" size=\"300\" window=\"100,100,600,400\" />\n\n<view id=\"4097\" visible=\"false\" inframe=\"false\" stylebar=\"true\" location=\"1,1\" size=\"743\" window=\"100,100,700,550\" />\n\n<view id=\"1\" visible=\"true\" inframe=\"false\" stylebar=\"true\" location=\"1\" size=\"903\" window=\"100,100,600,400\" />\n\n<view id=\"2\" visible=\"true\" inframe=\"false\" stylebar=\"true\" location=\"3\" size=\"439\" window=\"100,100,600,400\" />\n\n<view id=\"16\" visible=\"false\" inframe=\"false\" stylebar=\"false\" location=\"1\" size=\"150\" window=\"50,50,500,500\" />\n\n<view id=\"32\" visible=\"false\" inframe=\"false\" stylebar=\"true\" location=\"1\" size=\"150\" window=\"50,50,500,500\" />\n\n<view id=\"64\" toolbar=\"0\" visible=\"false\" inframe=\"false\" stylebar=\"false\" location=\"1\" size=\"150\" window=\"50,50,500,500\" />\n\n<view id=\"70\" toolbar=\"0 || 2020 || 2021 || 2022\" visible=\"false\" inframe=\"false\" stylebar=\"true\" location=\"1\" size=\"150\" window=\"50,50,500,500\" />\n\n</views>\n\n<toolbar show=\"true\" items=\"0 39 73 62 | 1 501 67 , 5 19 , 72 75 76 | 2 15 45 , 18 65 , 7 37 | 4 3 8 9 , 13 44 , 58 , 47 | 16 51 64 , 70 | 10 34 53 11 , 24 20 22 , 21 23 | 55 56 57 , 12 | 36 46 , 38 49 50 , 71 14 68 | 30 29 54 32 31 33 | 25 17 26 60 52 61 | 40 41 42 , 27 28 35 , 6 | 100001 | 100002 | 100004 | 100008 100007 100005\" position=\"1\" help=\"false\" />\n\n<input show=\"true\" cmd=\"true\" top=\"algebra\" />\n\n<dockBar show=\"true\" east=\"true\" />\n\n</perspective>\n\n</perspectives>\n\n<labelingStyle val=\"0\"/>\n\n<font size=\"12\"/>\n\n</gui>\n\n<euclidianView>\n\n<viewNumber viewNo=\"1\"/>\n\n<size width=\"903\" height=\"592\"/>\n\n<coordSystem xZero=\"175.5213327651935\" yZero=\"337.95373982179007\" scale=\"25.5662131029818\" yscale=\"25.56621310298209\"/>\n\n<evSettings axes=\"false\" grid=\"false\" gridIsBold=\"false\" pointCapturing=\"3\" rightAngleStyle=\"1\" checkboxSize=\"26\" gridType=\"3\"/>\n\n<bgColor r=\"255\" g=\"255\" b=\"255\"/>\n\n<axesColor r=\"0\" g=\"0\" b=\"0\"/>\n\n<gridColor r=\"192\" g=\"192\" b=\"192\"/>\n\n<lineStyle axes=\"1\" grid=\"0\"/>\n\n<axis id=\"0\" show=\"false\" label=\"\" unitLabel=\"\" tickStyle=\"1\" showNumbers=\"true\"/>\n\n<axis id=\"1\" show=\"false\" label=\"\" unitLabel=\"\" tickStyle=\"1\" showNumbers=\"true\"/>\n\n</euclidianView>\n\n<kernel>\n\n<continuous val=\"false\"/>\n\n<usePathAndRegionParameters val=\"true\"/>\n\n<decimals val=\"2\"/>\n\n<angleUnit val=\"degree\"/>\n\n<coordStyle val=\"0\"/>\n\n</kernel>\n\n<scripting blocked=\"false\" disabled=\"false\"/>\n\n<construction title=\"\" author=\"\" date=\"\">\n\n<element type=\"point\" label=\"A\">\n\n<show object=\"true\" label=\"true\"/>\n\n<objColor r=\"77\" g=\"77\" b=\"255\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelOffset x=\"11\" y=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<animation step=\"1\" speed=\"1\" type=\"1\" playing=\"false\"/>\n\n<coords x=\"10.189364360451709\" y=\"4.923014222869017\" z=\"1.0\"/>\n\n<pointSize val=\"5\"/>\n\n<pointStyle val=\"0\"/>\n\n</element>\n\n<element type=\"point\" label=\"B\">\n\n<show object=\"true\" label=\"true\"/>\n\n<objColor r=\"77\" g=\"77\" b=\"255\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelOffset x=\"16\" y=\"3\"/>\n\n<labelMode val=\"0\"/>\n\n<animation step=\"1\" speed=\"1\" type=\"1\" playing=\"false\"/>\n\n<coords x=\"8.858514412069583\" y=\"0.7022447849226433\" z=\"1.0\"/>\n\n<pointSize val=\"5\"/>\n\n<pointStyle val=\"0\"/>\n\n</element>\n\n<command name=\"Vector\">\n\n<input a0=\"A\" a1=\"B\"/>\n\n<output a0=\"v\"/>\n\n</command>\n\n<element type=\"vector\" label=\"v\">\n\n<show object=\"true\" label=\"false\"/>\n\n<objColor r=\"0\" g=\"0\" b=\"0\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<coords x=\"-1.3308499483821254\" y=\"-4.220769437946373\" z=\"0.0\"/>\n\n<lineStyle thickness=\"5\" type=\"0\" typeHidden=\"1\" opacity=\"178\"/>\n\n<coordStyle style=\"cartesian\"/>\n\n<startPoint exp=\"A\"/>\n\n</element>\n\n<element type=\"point\" label=\"C\">\n\n<show object=\"true\" label=\"true\"/>\n\n<objColor r=\"77\" g=\"77\" b=\"255\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<animation step=\"1\" speed=\"1\" type=\"1\" playing=\"false\"/>\n\n<coords x=\"5.384954023354466\" y=\"1.9409664274293843\" z=\"1.0\"/>\n\n<pointSize val=\"5\"/>\n\n<pointStyle val=\"0\"/>\n\n</element>\n\n<element type=\"point\" label=\"D\">\n\n<show object=\"true\" label=\"true\"/>\n\n<objColor r=\"77\" g=\"77\" b=\"255\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<animation step=\"1\" speed=\"1\" type=\"1\" playing=\"false\"/>\n\n<coords x=\"3.694774392604529\" y=\"5.66662996792482\" z=\"1.0\"/>\n\n<pointSize val=\"5\"/>\n\n<pointStyle val=\"0\"/>\n\n</element>\n\n<command name=\"Vector\">\n\n<input a0=\"C\" a1=\"D\"/>\n\n<output a0=\"w\"/>\n\n</command>\n\n<element type=\"vector\" label=\"w\">\n\n<show object=\"true\" label=\"false\"/>\n\n<objColor r=\"0\" g=\"0\" b=\"0\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<coords x=\"-1.6901796307499373\" y=\"3.7256635404954355\" z=\"0.0\"/>\n\n<lineStyle thickness=\"5\" type=\"0\" typeHidden=\"1\" opacity=\"178\"/>\n\n<coordStyle style=\"cartesian\"/>\n\n<startPoint exp=\"C\"/>\n\n</element>\n\n<element type=\"point\" label=\"E\">\n\n<show object=\"true\" label=\"true\"/>\n\n<objColor r=\"77\" g=\"77\" b=\"255\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<animation step=\"1\" speed=\"1\" type=\"1\" playing=\"false\"/>\n\n<coords x=\"19.10641459754691\" y=\"4.026945226775936\" z=\"1.0\"/>\n\n<pointSize val=\"5\"/>\n\n<pointStyle val=\"0\"/>\n\n</element>\n\n<element type=\"point\" label=\"F\">\n\n<show object=\"true\" label=\"true\"/>\n\n<objColor r=\"77\" g=\"77\" b=\"255\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<animation step=\"1\" speed=\"1\" type=\"1\" playing=\"false\"/>\n\n<coords x=\"13.473980907818914\" y=\"-0.8232060060453382\" z=\"1.0\"/>\n\n<pointSize val=\"5\"/>\n\n<pointStyle val=\"0\"/>\n\n</element>\n\n<command name=\"Vector\">\n\n<input a0=\"E\" a1=\"F\"/>\n\n<output a0=\"a\"/>\n\n</command>\n\n<element type=\"vector\" label=\"a\">\n\n<show object=\"true\" label=\"false\"/>\n\n<objColor r=\"0\" g=\"0\" b=\"0\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<coords x=\"-5.6324336897279945\" y=\"-4.850151232821275\" z=\"0.0\"/>\n\n<lineStyle thickness=\"5\" type=\"0\" typeHidden=\"1\" opacity=\"178\"/>\n\n<coordStyle style=\"cartesian\"/>\n\n<startPoint exp=\"E\"/>\n\n</element>\n\n<element type=\"point\" label=\"G\">\n\n<show object=\"true\" label=\"true\"/>\n\n<objColor r=\"77\" g=\"77\" b=\"255\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<animation step=\"1\" speed=\"1\" type=\"1\" playing=\"false\"/>\n\n<coords x=\"6.14\" y=\"-3.5\" z=\"1.0\"/>\n\n<pointSize val=\"5\"/>\n\n<pointStyle val=\"0\"/>\n\n</element>\n\n<element type=\"point\" label=\"H\">\n\n<show object=\"true\" label=\"true\"/>\n\n<objColor r=\"77\" g=\"77\" b=\"255\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<animation step=\"1\" speed=\"1\" type=\"1\" playing=\"false\"/>\n\n<coords x=\"5.12\" y=\"-1.02\" z=\"1.0\"/>\n\n<pointSize val=\"5\"/>\n\n<pointStyle val=\"0\"/>\n\n</element>\n\n<command name=\"Vector\">\n\n<input a0=\"G\" a1=\"H\"/>\n\n<output a0=\"b\"/>\n\n</command>\n\n<element type=\"vector\" label=\"b\">\n\n<show object=\"true\" label=\"false\"/>\n\n<objColor r=\"0\" g=\"0\" b=\"0\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<coords x=\"-1.0199999999999996\" y=\"2.48\" z=\"0.0\"/>\n\n<lineStyle thickness=\"5\" type=\"0\" typeHidden=\"1\" opacity=\"178\"/>\n\n<coordStyle style=\"cartesian\"/>\n\n<startPoint exp=\"G\"/>\n\n</element>\n\n<command name=\"tag1\">\n\n<input a0=\"w\"/>\n\n<output a0=\"text1\"/>\n\n</command>\n\n<element type=\"text\" label=\"text1\">\n\n<show object=\"false\" label=\"true\"/>\n\n<objColor r=\"0\" g=\"0\" b=\"0\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<decimals val=\"2\"/>\n\n<auxiliary val=\"true\"/>\n\n<startPoint x=\"0.58\" y=\"1.26\" z=\"1.0\"/>\n\n</element>\n\n<element type=\"button\" label=\"button1\">\n\n<show object=\"true\" label=\"true\"/>\n\n<objColor r=\"0\" g=\"0\" b=\"0\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelOffset x=\"13\" y=\"15\"/>\n\n<labelMode val=\"0\"/>\n\n<fixed val=\"true\"/>\n\n<auxiliary val=\"true\"/>\n\n<ggbscript val=\"SetValue(alltexts,RemoveUndefined(Sequence(If(Defined(Object(\"text\" + (n))), Object(\"text\" + (n))), n, 1, 200))) SetValue(duplicate,Flatten(Sequence(RemoveUndefined(Sequence(If(alltexts(a) ≟ alltexts(b), a), a, b + 1, Length(alltexts))), b, 1, Length(alltexts)))) SetValue(textnumbers,RemoveUndefined(Sequence(If(Defined(Object(\"text\" + (n))), If(Object(\"text\" + (n)) ≟ alltexts(duplicate(1)), n)), n, 1, 200))) SetValue(deleteinstructions,Zip(\"Delete(text\" + (n) + \")\", n, textnumbers)) SetValue(names1,RemoveUndefined(Zip(If(First(a, 3) ≟ \"\\$1\\$\", Take(a, 4)), a, alltexts))) SetValue(vectors1,Zip(Object(a), a, names1)) SetValue(points10,Zip(Point(a, 0), Vector(a), vectors1)) SetValue(points11,Zip(Point(a, 1), Vector(a), vectors1)) SetValue(tags1,Zip(Text(\" (funi 1) \" + (Distance(a, b)), Midpoint(a, b)), a, points10, b, points11)) SetValue(lines1,Zip(Line(a, b), a, points10, b, points11)) \"/>\n\n<caption 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a1=\"Vector[a]\" a2=\"vectors1\"/>\n\n<output a0=\"points11\"/>\n\n</command>\n\n<element type=\"list\" label=\"points11\">\n\n<show object=\"false\" label=\"false\"/>\n\n<objColor r=\"0\" g=\"0\" b=\"255\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<lineStyle thickness=\"5\" type=\"0\" typeHidden=\"1\"/>\n\n<pointSize val=\"9\"/>\n\n<pointStyle val=\"2\"/>\n\n</element>\n\n<command name=\"Zip\">\n\n<input a0=\"Text[\" funi 1, \" + (Distance[a, b]), Midpoint[a, b]]\" a1=\"a\" a2=\"points10\" a3=\"b\" a4=\"points11\"/>\n\n<output a0=\"tags1\"/>\n\n</command>\n\n<element type=\"list\" label=\"tags1\">\n\n<show object=\"true\" label=\"false\"/>\n\n<objColor r=\"0\" g=\"0\" b=\"0\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<lineStyle thickness=\"5\" type=\"0\" typeHidden=\"1\"/>\n\n<pointSize val=\"5\"/>\n\n<pointStyle val=\"-1\"/>\n\n</element>\n\n<command name=\"Zip\">\n\n<input a0=\"Line[a, b]\" a1=\"a\" a2=\"points10\" a3=\"b\" a4=\"points11\"/>\n\n<output a0=\"lines1\"/>\n\n</command>\n\n<element type=\"list\" label=\"lines1\">\n\n<show object=\"true\" label=\"true\"/>\n\n<objColor r=\"0\" g=\"0\" b=\"0\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<lineStyle thickness=\"1\" type=\"15\" typeHidden=\"1\"/>\n\n<pointSize val=\"5\"/>\n\n<pointStyle val=\"-1\"/>\n\n</element>\n\n<command name=\"Flatten\">\n\n<input a0=\"Sequence[RemoveUndefined[Sequence[If[alltexts(a) ≟ alltexts(b), a], a, b + 1, Length[alltexts]]], b, 1, Length[alltexts]]\"/>\n\n<output a0=\"duplicate\"/>\n\n</command>\n\n<element type=\"list\" label=\"duplicate\">\n\n<show object=\"true\" label=\"false\"/>\n\n<objColor r=\"0\" g=\"100\" b=\"0\" alpha=\"0.10000000149011612\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<auxiliary val=\"true\"/>\n\n<lineStyle thickness=\"5\" type=\"0\" typeHidden=\"1\"/>\n\n<listType val=\"numeric\"/>\n\n<pointSize val=\"5\"/>\n\n<pointStyle val=\"-1\"/>\n\n</element>\n\n<command name=\"RemoveUndefined\">\n\n<input a0=\"Sequence[If[Defined[Object[\"text\" + (n)]], If[Object[\"text\" + (n)] ≟ alltexts(duplicate(1)), n]], n, 1, 200]\"/>\n\n<output a0=\"textnumbers\"/>\n\n</command>\n\n<element type=\"list\" label=\"textnumbers\">\n\n<show object=\"true\" label=\"false\"/>\n\n<objColor r=\"0\" g=\"100\" b=\"0\" alpha=\"0.10000000149011612\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<auxiliary val=\"true\"/>\n\n<lineStyle thickness=\"5\" type=\"0\" typeHidden=\"1\"/>\n\n<listType val=\"numeric\"/>\n\n<pointSize val=\"5\"/>\n\n<pointStyle val=\"-1\"/>\n\n</element>\n\n<command name=\"Zip\">\n\n<input a0=\"\"Delete(text\" + (n) + \")\"\" a1=\"n\" a2=\"textnumbers\"/>\n\n<output a0=\"deleteinstructions\"/>\n\n</command>\n\n<element type=\"list\" label=\"deleteinstructions\">\n\n<show object=\"false\" label=\"true\" ev=\"4\"/>\n\n<objColor r=\"0\" g=\"100\" b=\"0\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<auxiliary 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xmlns:xsi=\"http://www.w3.org/2001/XMLS...; >\n\n<macro cmdName=\"tag1\" toolName=\"tag 1\" toolHelp=\"vector\" iconFile=\"d81d328bd62b33c156c55500b5b41145/tag1.JPG\" showInToolBar=\"true\" copyCaptions=\"true\">\n\n<macroInput a0=\"u\"/>\n\n<macroOutput a0=\"text1\"/>\n\n<construction title=\"\" author=\"\" date=\"\">\n\n<element type=\"vector\" label=\"u\">\n\n<show object=\"true\" label=\"true\"/>\n\n<objColor r=\"0\" g=\"0\" b=\"0\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<animation step=\"1\" speed=\"1\" type=\"0\" playing=\"false\"/>\n\n<coords x=\"1.0\" y=\"2.0\" z=\"0.0\"/>\n\n<lineStyle thickness=\"5\" type=\"0\" typeHidden=\"1\" opacity=\"178\"/>\n\n<coordStyle style=\"cartesian\"/>\n\n</element>\n\n<expression label=\"text1\" exp=\"\"\\$1\\$\" + (Name[u]) + \"\"\" />\n\n<element type=\"text\" label=\"text1\">\n\n<show object=\"false\" label=\"true\" ev=\"40\"/>\n\n<objColor r=\"0\" g=\"0\" b=\"0\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode 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typeHidden=\"1\"/>\n\n<pointSize val=\"5\"/>\n\n<pointStyle val=\"0\"/>\n\n<symbolic val=\"true\" />\n\n</element>\n\n<expression label=\"indice\" exp=\"parami(1)\" />\n\n<element type=\"numeric\" label=\"indice\">\n\n<value val=\"1.0\"/>\n\n<symbolic val=\"true\" />\n\n<objColor r=\"0\" g=\"0\" b=\"0\" alpha=\"0.10000000149011612\"/>\n\n</element>\n\n<expression label=\"Q\" exp=\"L_{11}(indice)\" type=\"point\" />\n\n<element type=\"point\" label=\"Q\">\n\n<show object=\"true\" label=\"true\"/>\n\n<objColor r=\"68\" g=\"68\" b=\"68\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<coords x=\"2.765753794656004\" y=\"2.0201400933233042\" z=\"1.0\"/>\n\n<pointSize val=\"4\"/>\n\n<pointStyle val=\"0\"/>\n\n</element>\n\n<expression label=\"P\" exp=\"L_{10}(indice)\" type=\"point\" />\n\n<element type=\"point\" label=\"P\">\n\n<show object=\"true\" label=\"true\"/>\n\n<objColor r=\"68\" g=\"68\" b=\"68\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode 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type=\"point\" label=\"L\">\n\n<show object=\"true\" label=\"true\"/>\n\n<objColor r=\"68\" g=\"68\" b=\"68\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelOffset x=\"-20\" y=\"-11\"/>\n\n<labelMode val=\"0\"/>\n\n<coords x=\"6.992296151111027\" y=\"10.08653534124675\" z=\"3.0942391901357222\"/>\n\n<pointSize val=\"4\"/>\n\n<pointStyle val=\"0\"/>\n\n</element>\n\n<expression label=\"O\" exp=\"L + N - M + Q - P\" type=\"point\" />\n\n<element type=\"point\" label=\"O\">\n\n<show object=\"true\" label=\"true\"/>\n\n<objColor r=\"68\" g=\"68\" b=\"68\" alpha=\"0.0\"/>\n\n<layer val=\"0\"/>\n\n<labelMode val=\"0\"/>\n\n<coords x=\"3.8113383464239465\" y=\"6.358457941491808\" z=\"1.0\"/>\n\n<pointSize val=\"4\"/>\n\n<pointStyle val=\"0\"/>\n\n</element>\n\n<expression label=\"paramo\" exp=\"{indice + 1, L, O}\" />\n\n<element type=\"list\" label=\"paramo\">\n\n<objColor r=\"0\" g=\"100\" b=\"0\" alpha=\"0.10000000149011612\"/>\n\n<lineStyle thickness=\"5\" type=\"0\" typeHidden=\"1\"/>\n\n<pointSize val=\"5\"/>\n\n<pointStyle val=\"0\"/>\n\n</element>\n\n</construction>\n\n</macro>\n\n</geogebra>\n\nLibraryJavaScript:\n\nfunction ggbOnInit() {}\n\nPreferences:\n\n<?xml version=\"1.0\" encoding=\"utf-8\"?><geogebra format=\"5.0\" xsi:noNamespaceSchemaLocation=\"http://www.geogebra.org/ggb...; xmlns=\"\" xmlns:xsi=\"http://www.w3.org/2001/XMLS...; ><gui><window width=\"800\" height=\"600\" /><settings ignoreDocument=\"false\" showTitleBar=\"true\" /><labelingStyle val=\"0\"/><font size=\"12\"/><menuFont size=\"-1\"/><tooltipSettings language=\"\" timeout=\"0\"/></gui><euclidianView><size width=\"640\" height=\"480\"/><coordSystem xZero=\"215.0\" yZero=\"315.0\" scale=\"50.0\" yscale=\"50.0\"/><evSettings axes=\"true\" grid=\"true\" gridIsBold=\"false\" pointCapturing=\"3\" rightAngleStyle=\"1\" checkboxSize=\"26\" gridType=\"3\"/><bgColor r=\"255\" g=\"255\" b=\"255\"/><axesColor r=\"0\" g=\"0\" b=\"0\"/><gridColor r=\"192\" g=\"192\" b=\"192\"/><lineStyle axes=\"1\" grid=\"0\"/><axis id=\"0\" show=\"true\" label=\"\" unitLabel=\"\" tickStyle=\"1\" showNumbers=\"true\" axisCross=\"0.0\" positiveAxis=\"false\"/><axis id=\"1\" show=\"true\" label=\"\" unitLabel=\"\" tickStyle=\"1\" showNumbers=\"true\" axisCross=\"0.0\" positiveAxis=\"false\"/></euclidianView><kernel><continuous val=\"false\"/><decimals val=\"2\"/><angleUnit val=\"degree\"/><algebraStyle val=\"0\"/><coordStyle val=\"0\"/><localization digits=\"false\" labels=\"true\"/><angleFromInvTrig val=\"false\"/><casSettings timeout=\"5\" expRoots=\"true\"/></kernel><scripting blocked=\"false\"/></geogebra>[/co]", null, "", null, "1\n\nHello,\n\nI have made system that can add and remove vectors at will from a list, not necessarily by erasing them, just by clicking on them, ON/OFF type. This ON/OFF system allows easy removal of a tag applied by mistake, without having to erase the vector, that could be part of another construction. The file is very stable it seems. (see file \"delete duplicates e.ggb\" below).\n\nHowever, as soon as I add the calculation of the resultant of forces (for which iteration is required), the file becomes extremely unstable. Almost 50% of the time, the file doesn't open again after being saved. (see file \"delete duplicates with resultant.ggb\")\n\nIt seems as if there was an unwanted interference between the list of object names \"text\" (n), utilising iterationlist, and the iteration that calculates the force resultant, utilising iteration.\n\nIs GGB overwhelmed by too much use of iterations?\n\nThe only alternative is to use the spreadsheet, but this would be complicated, first because it involves fixing in advance the maximum number of forces, and second because it would be very difficult to organise 3 to 4 different catenaries in the same worksheet. So iteration is much more pliable. If it works...\n\nThanks,\n\nLaurent", null, "1\n\ntry element(points10,1) instead points10(1) in iteraton command\n\nsometimes the short version of element command does not work" ]

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[ "# edge\n\nClass: ClassificationLinear\n\nClassification edge for linear classification models\n\n## Syntax\n\n``e = edge(Mdl,X,Y)``\n``e = edge(Mdl,Tbl,ResponseVarName)``\n``e = edge(Mdl,Tbl,Y)``\n``e = edge(___,Name,Value)``\n\n## Description\n\nexample\n\n````e = edge(Mdl,X,Y)` returns the classification edges for the binary, linear classification model `Mdl` using predictor data in `X` and corresponding class labels in `Y`. `e` contains a classification edge for each regularization strength in `Mdl`.```\n````e = edge(Mdl,Tbl,ResponseVarName)` returns the classification edges for the trained linear classifier `Mdl` using the predictor data in `Tbl` and the class labels in `Tbl.ResponseVarName`.```\n````e = edge(Mdl,Tbl,Y)` returns the classification edges for the classifier `Mdl` using the predictor data in table `Tbl` and the class labels in vector `Y`.```\n\nexample\n\n````e = edge(___,Name,Value)` specifies options using one or more name-value pair arguments in addition to any of the input argument combinations in previous syntaxes. For example, you can specify that columns in the predictor data correspond to observations or supply observation weights. NoteIf the predictor data `X` or the predictor variables in `Tbl` contain any missing values, the `edge` function can return NaN. For more details, see edge can return NaN for predictor data with missing values. ```\n\n## Input Arguments\n\nexpand all\n\nBinary, linear classification model, specified as a `ClassificationLinear` model object. You can create a `ClassificationLinear` model object using `fitclinear`.\n\nPredictor data, specified as an n-by-p full or sparse matrix. This orientation of `X` indicates that rows correspond to individual observations, and columns correspond to individual predictor variables.\n\nNote\n\nIf you orient your predictor matrix so that observations correspond to columns and specify `'ObservationsIn','columns'`, then you might experience a significant reduction in computation time.\n\nThe length of `Y` and the number of observations in `X` must be equal.\n\nData Types: `single` | `double`\n\nClass labels, specified as a categorical, character, or string array; logical or numeric vector; or cell array of character vectors.\n\n• The data type of `Y` must be the same as the data type of `Mdl.ClassNames`. (The software treats string arrays as cell arrays of character vectors.)\n\n• The distinct classes in `Y` must be a subset of `Mdl.ClassNames`.\n\n• If `Y` is a character array, then each element must correspond to one row of the array.\n\n• The length of `Y` must be equal to the number of observations in `X` or `Tbl`.\n\nData Types: `categorical` | `char` | `string` | `logical` | `single` | `double` | `cell`\n\nSample data used to train the model, specified as a table. Each row of `Tbl` corresponds to one observation, and each column corresponds to one predictor variable. Optionally, `Tbl` can contain additional columns for the response variable and observation weights. `Tbl` must contain all the predictors used to train `Mdl`. Multicolumn variables and cell arrays other than cell arrays of character vectors are not allowed.\n\nIf `Tbl` contains the response variable used to train `Mdl`, then you do not need to specify `ResponseVarName` or `Y`.\n\nIf you train `Mdl` using sample data contained in a table, then the input data for `edge` must also be in a table.\n\nResponse variable name, specified as the name of a variable in `Tbl`. If `Tbl` contains the response variable used to train `Mdl`, then you do not need to specify `ResponseVarName`.\n\nIf you specify `ResponseVarName`, then you must specify it as a character vector or string scalar. For example, if the response variable is stored as `Tbl.Y`, then specify `ResponseVarName` as `'Y'`. Otherwise, the software treats all columns of `Tbl`, including `Tbl.Y`, as predictors.\n\nThe response variable must be a categorical, character, or string array; a logical or numeric vector; or a cell array of character vectors. If the response variable is a character array, then each element must correspond to one row of the array.\n\nData Types: `char` | `string`\n\n### Name-Value Arguments\n\nSpecify optional pairs of arguments as `Name1=Value1,...,NameN=ValueN`, where `Name` is the argument name and `Value` is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter.\n\nBefore R2021a, use commas to separate each name and value, and enclose `Name` in quotes.\n\nPredictor data observation dimension, specified as `'rows'` or `'columns'`.\n\nNote\n\nIf you orient your predictor matrix so that observations correspond to columns and specify `'ObservationsIn','columns'`, then you might experience a significant reduction in computation time. You cannot specify `'ObservationsIn','columns'` for predictor data in a table.\n\nData Types: `char` | `string`\n\nObservation weights, specified as the comma-separated pair consisting of `'Weights'` and a numeric vector or the name of a variable in `Tbl`.\n\n• If you specify `Weights` as a numeric vector, then the size of `Weights` must be equal to the number of observations in `X` or `Tbl`.\n\n• If you specify `Weights` as the name of a variable in `Tbl`, then the name must be a character vector or string scalar. For example, if the weights are stored as `Tbl.W`, then specify `Weights` as `'W'`. Otherwise, the software treats all columns of `Tbl`, including `Tbl.W`, as predictors.\n\nIf you supply weights, then for each regularization strength, `edge` computes the weighted classification edge and normalizes weights to sum up to the value of the prior probability in the respective class.\n\nData Types: `double` | `single`\n\n## Output Arguments\n\nexpand all\n\nClassification edges, returned as a numeric scalar or row vector.\n\n`e` is the same size as `Mdl.Lambda`. `e(j)` is the classification edge of the linear classification model trained using the regularization strength `Mdl.Lambda(j)`.\n\n## Examples\n\nexpand all\n\n`load nlpdata`\n\n`X` is a sparse matrix of predictor data, and `Y` is a categorical vector of class labels. There are more than two classes in the data.\n\nThe models should identify whether the word counts in a web page are from the Statistics and Machine Learning Toolbox™ documentation. So, identify the labels that correspond to the Statistics and Machine Learning Toolbox™ documentation web pages.\n\n`Ystats = Y == 'stats';`\n\nTrain a binary, linear classification model that can identify whether the word counts in a documentation web page are from the Statistics and Machine Learning Toolbox™ documentation. Specify to holdout 30% of the observations. Optimize the objective function using SpaRSA.\n\n```rng(1); % For reproducibility CVMdl = fitclinear(X,Ystats,'Solver','sparsa','Holdout',0.30); CMdl = CVMdl.Trained{1};```\n\n`CVMdl` is a `ClassificationPartitionedLinear` model. It contains the property `Trained`, which is a 1-by-1 cell array holding a `ClassificationLinear` model that the software trained using the training set.\n\nExtract the training and test data from the partition definition.\n\n```trainIdx = training(CVMdl.Partition); testIdx = test(CVMdl.Partition);```\n\nEstimate the training- and test-sample edges.\n\n`eTrain = edge(CMdl,X(trainIdx,:),Ystats(trainIdx))`\n```eTrain = 15.6660 ```\n`eTest = edge(CMdl,X(testIdx,:),Ystats(testIdx))`\n```eTest = 15.4767 ```\n\nOne way to perform feature selection is to compare test-sample edges from multiple models. Based solely on this criterion, the classifier with the highest edge is the best classifier.\n\n`load nlpdata`\n\n`X` is a sparse matrix of predictor data, and `Y` is a categorical vector of class labels. There are more than two classes in the data.\n\nThe models should identify whether the word counts in a web page are from the Statistics and Machine Learning Toolbox™ documentation. So, identify the labels that correspond to the Statistics and Machine Learning Toolbox™ documentation web pages. For quicker execution time, orient the predictor data so that individual observations correspond to columns.\n\n```Ystats = Y == 'stats'; X = X'; rng(1); % For reproducibility```\n\nCreate a data partition which holds out 30% of the observations for testing.\n\n```Partition = cvpartition(Ystats,'Holdout',0.30); testIdx = test(Partition); % Test-set indices XTest = X(:,testIdx); YTest = Ystats(testIdx);```\n\n`Partition` is a `cvpartition` object that defines the data set partition.\n\nRandomly choose half of the predictor variables.\n\n```p = size(X,1); % Number of predictors idxPart = randsample(p,ceil(0.5*p));```\n\nTrain two binary, linear classification models: one that uses the all of the predictors and one that uses half of the predictors. Optimize the objective function using SpaRSA, and indicate that observations correspond to columns.\n\n```CVMdl = fitclinear(X,Ystats,'CVPartition',Partition,'Solver','sparsa',... 'ObservationsIn','columns'); PCVMdl = fitclinear(X(idxPart,:),Ystats,'CVPartition',Partition,'Solver','sparsa',... 'ObservationsIn','columns');```\n\n`CVMdl` and `PCVMdl` are `ClassificationPartitionedLinear` models.\n\nExtract the trained `ClassificationLinear` models from the cross-validated models.\n\n```CMdl = CVMdl.Trained{1}; PCMdl = PCVMdl.Trained{1};```\n\nEstimate the test sample edge for each classifier.\n\n`fullEdge = edge(CMdl,XTest,YTest,'ObservationsIn','columns')`\n```fullEdge = 15.4767 ```\n`partEdge = edge(PCMdl,XTest(idxPart,:),YTest,'ObservationsIn','columns')`\n```partEdge = 13.4458 ```\n\nBased on the test-sample edges, the classifier that uses all of the predictors is the better model.\n\nTo determine a good lasso-penalty strength for a linear classification model that uses a logistic regression learner, compare test-sample edges.\n\nLoad the NLP data set. Preprocess the data as in Feature Selection Using Test-Sample Edges.\n\n```load nlpdata Ystats = Y == 'stats'; X = X'; Partition = cvpartition(Ystats,'Holdout',0.30); testIdx = test(Partition); XTest = X(:,testIdx); YTest = Ystats(testIdx);```\n\nCreate a set of 11 logarithmically-spaced regularization strengths from $1{0}^{-8}$ through $1{0}^{1}$.\n\n`Lambda = logspace(-8,1,11);`\n\nTrain binary, linear classification models that use each of the regularization strengths. Optimize the objective function using SpaRSA. Lower the tolerance on the gradient of the objective function to `1e-8`.\n\n```rng(10); % For reproducibility CVMdl = fitclinear(X,Ystats,'ObservationsIn','columns',... 'CVPartition',Partition,'Learner','logistic','Solver','sparsa',... 'Regularization','lasso','Lambda',Lambda,'GradientTolerance',1e-8)```\n```CVMdl = ClassificationPartitionedLinear CrossValidatedModel: 'Linear' ResponseName: 'Y' NumObservations: 31572 KFold: 1 Partition: [1x1 cvpartition] ClassNames: [0 1] ScoreTransform: 'none' Properties, Methods ```\n\nExtract the trained linear classification model.\n\n`Mdl = CVMdl.Trained{1}`\n```Mdl = ClassificationLinear ResponseName: 'Y' ClassNames: [0 1] ScoreTransform: 'logit' Beta: [34023x11 double] Bias: [-11.2409 -11.2409 -11.2409 -11.2409 -11.2409 ... ] Lambda: [1.0000e-08 7.9433e-08 6.3096e-07 5.0119e-06 ... ] Learner: 'logistic' Properties, Methods ```\n\n`Mdl` is a `ClassificationLinear` model object. Because `Lambda` is a sequence of regularization strengths, you can think of `Mdl` as 11 models, one for each regularization strength in `Lambda`.\n\nEstimate the test-sample edges.\n\n`e = edge(Mdl,X(:,testIdx),Ystats(testIdx),'ObservationsIn','columns')`\n```e = 1×11 0.9986 0.9986 0.9986 0.9986 0.9986 0.9932 0.9765 0.9115 0.8332 0.8128 0.8128 ```\n\nBecause there are 11 regularization strengths, `e` is a 1-by-11 vector of edges.\n\nPlot the test-sample edges for each regularization strength. Identify the regularization strength that maximizes the edges over the grid.\n\n```figure; plot(log10(Lambda),log10(e),'-o') [~, maxEIdx] = max(e); maxLambda = Lambda(maxEIdx); hold on plot(log10(maxLambda),log10(e(maxEIdx)),'ro'); ylabel('log_{10} test-sample edge') xlabel('log_{10} Lambda') legend('Edge','Max edge') hold off```", null, "Several values of `Lambda` yield similarly high edges. Higher values of lambda lead to predictor variable sparsity, which is a good quality of a classifier.\n\nChoose the regularization strength that occurs just before the edge starts decreasing.\n\n`LambdaFinal = Lambda(5);`\n\nTrain a linear classification model using the entire data set and specify the regularization strength yielding the maximal edge.\n\n```MdlFinal = fitclinear(X,Ystats,'ObservationsIn','columns',... 'Learner','logistic','Solver','sparsa','Regularization','lasso',... 'Lambda',LambdaFinal);```\n\nTo estimate labels for new observations, pass `MdlFinal` and the new data to `predict`.\n\nexpand all\n\n## Algorithms\n\nBy default, observation weights are prior class probabilities. If you supply weights using `Weights`, then the software normalizes them to sum to the prior probabilities in the respective classes. The software uses the normalized weights to estimate the weighted edge.\n\n## Version History\n\nIntroduced in R2016a\n\nexpand all\n\nBehavior changed in R2022a\n\nBehavior changed in R2022a" ]

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[ "#", null, "A REVIEW OF LOGARITHMS", null, "## Equivalence\n\nBefore you choose a topic from the list, let's review what you know about the word \"equivalent.\" You will come across this word in all your math courses. According to Webster, the word \"equivalent\" means to have equal value, equal amount, equal measure, equal area, equal length, etc. You get the picture.\nFor example, 1 yard is equivalent to 3 feet because both represent the same measure. You will see the notation 1 yard = 3 feet. Obviously, the left side of the = sign is not identical to the right side. The numbers and the words are different. However, the left side is equivalent to the right side because both sides denote the same length. In this case, the = sign is used to denote equivalence.\nSuppose you wish to purchase an item for", null, ", you could pay for it with a dollar bill and 2 dimes, you could pay for it with 12 dimes, you could pay it for 4 quarters and 4 nickels, etc. If you put each method of payment in a separate pile, you will note that they do not look the same. They are in fact not identical. However, they are equivalent because all the methods of payment total", null, ".\n\nThe number 1 has many equivalent faces:", null, "to name a few. The number 0 has many equivalent faces:", null, "to name a few.\n\n## Exponential Shortcuts\n\nExponents represent shortcuts in multiplication.\n\nExample 1: The expression", null, "can be written", null, ".\n\nExample 2: The expression", null, "means", null, ".\n\nExample 3: The expression", null, "can be written", null, ".\n\nExample 4: The expression (-7)(-7)(-7)(-7) can be written", null, ".\n\nYou would much rather write the expression", null, "than write out 8 one thousand times. You can also use exponents with variables.\n\nExample 5: The expression (a+b)(a+b)(a+b)(a+b)(a+b) is more conveniently written as", null, ".\n\nIf you want to review the rules for exponents, click on Exponential Rules.", null, "[Exponential Rules] [Logarithms]\n\n[Algebra] [Trigonometry ] [Complex Variables]", null, "S.O.S MATHematics home page\n\nDo you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.", null, "Author: Nancy Marcus" ]

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[ "Solutions by everydaycalculation.com\n\n## Reduce 2785/2105 to lowest terms\n\nThe simplest form of 2785/2105 is 557/421.\n\n#### Steps to simplifying fractions\n\n1. Find the GCD (or HCF) of numerator and denominator\nGCD of 2785 and 2105 is 5\n2. Divide both the numerator and denominator by the GCD\n2785 ÷ 5/2105 ÷ 5\n3. Reduced fraction: 557/421\nTherefore, 2785/2105 simplified to lowest terms is 557/421.\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

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[ "Numericintegration\n12-27-2017, 11:43 PM (This post was last modified: 12-28-2017 05:55 PM by Arno K.)\nPost: #1\n Arno K", null, "Senior Member Posts: 432 Joined: Mar 2015\nNumericintegration\nThis program uses an hybrid of CAS and Home-program to compute the integral of an via input entered function, the borders and intervals to compute diverse numeric integrals. It can easily be extended to other formulas, the included ones are those I sometimes need to check hand-calculations made by my pupils.\nArno", null, "NumericIntegr.hpprgm (Size: 5.85 KB / Downloads: 11)\nEdit: Update attachment, now provides exact value, too\n12-28-2017, 03:46 PM\nPost: #2", null, "Namir", null, "Senior Member Posts: 690 Joined: Dec 2013\nRE: Numericintegration\nCan you please post the listing?\n\nThanks!\n\nNamir\n12-28-2017, 04:20 PM (This post was last modified: 12-28-2017 05:47 PM by Arno K.)\nPost: #3\n Arno K", null, "Senior Member Posts: 432 Joined: Mar 2015\nRE: Numericintegration\nIts a little bit long, but here it is:\nCode:\n #cas integr(g,a,b):= BEGIN logal h,x; g:=expr(g); IF type(g)==DOM_SYMBOLIC THEN      g:=unapply(g,x);      END; h:=int(g(x),x,a,b); RETURN (h); END; #end #cas simpsonint(g,a,b,n):= BEGIN local l,l1; g:=expr(g); IF even(n) THEN IF (type(g)==DOM_SYMBOLIC  OR type(g)== DOM_FUNC)   THEN     l:=MAKELIST(X,X,a,b,(b-a)/n);      IF type(g)==DOM_SYMBOLIC THEN      g:=unapply(g,x);      END;    l:=apply(g,l);    l1:=MAKELIST(even(X)+1,X,1,n+1);  return (2*DOT(l,l1)-g(a)-g(b))*(b-a)/(3*n);    ELSE return (\"f must be symbolic or function!\");   END; ELSE return \"n must be even!\"; END; END; #end #cas trapezregel(g,a,b,n):= BEGIN local l; g:=expr(g); IF (type(g)==DOM_SYMBOLIC  OR type(g)== DOM_FUNC)   THEN     l:=MAKELIST(X,X,a,b,(b-a)/n);      IF type(g)==DOM_SYMBOLIC THEN      g:=unapply(g,x);      END;    l:=apply(g,l);    return (2*ΣLIST(l)-g(a)-g(b))*(b-a)/(2*n);    ELSE return (\"f must be symbolic or function!\");   END; END; #end #cas rightsum(g,a,b,n):= BEGIN local l; g:=expr(g); IF (type(g)==DOM_SYMBOLIC  OR type(g)== DOM_FUNC)   THEN     l:=MAKELIST(X,X,a,b,(b-a)/n);      IF type(g)==DOM_SYMBOLIC THEN      g:=unapply(g,x);      END;    l:=apply(g,l);    return (ΣLIST(l)-g(a))*(b-a)/n;    ELSE return (\"f must be symbolic or function!\");   END; END; #end #cas leftsum(g,a,b,n):= BEGIN local l; g:=expr(g); IF (type(g)==DOM_SYMBOLIC  OR type(g)== DOM_FUNC)   THEN     l:=MAKELIST(X,X,a,b,(b-a)/n);      IF type(g)==DOM_SYMBOLIC THEN      g:=unapply(g,x);      END;    l:=apply(g,l);    return (ΣLIST(l)-g(b))*(b-a)/n;    ELSE return (\"f must be symbolic or function!\");   END; END; #end EXPORT NumericIntegr() BEGIN local a,N,f,b; N:=12; a:=1;b:=3; f:='X^2';    if input(      {{f,,{20,70,0}},      {a,,{20,30,1}},      {b,,{20,30,2}},      {N,,{20,30,3}}},            \"Enter Data\",      {\"f(X)=\", \"a= \", \"b= \",\"N= \"       },      {        \"Enter the function, upper case X!\",        \"Enter the starting point\",        \"Enter the endpoint\",        \"Enter the amount of intervals\"             },      {f,a,b,N}    ) then f:=lower(string(f)); PRINT(); PRINT(\"Simpsonsum:   \"+simpsonint(f,a,b,N)); PRINT(\"Trapezoidsum:    \"+trapezregel(f,a,b,N)); PRINT(\"Rectangleleft:  \"+leftsum(f,a,b,N)); PRINT(\"Rectangleright: \"+rightsum(f,a,b,N)); PRINT(\"Integral:\"+integr(f,a,b)); return; end; END;\nArno\nEdit: Got that thing with the exact integral implemented, I thought it to be nice to see it for comparing.\n12-28-2017, 06:22 PM\nPost: #4", null, "Namir", null, "Senior Member Posts: 690 Joined: Dec 2013\nRE: Numericintegration\nThank you!\n\n:-)\n\nNamir\n « Next Oldest | Next Newest »\n\nUser(s) browsing this thread: 1 Guest(s)" ]

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[ "Grams To Kilograms\n\n# 754 g to kg754 Grams to Kilograms\n\ng\n=\nkg\n\n## How to convert 754 grams to kilograms?\n\n 754 g * 0.001 kg = 0.754 kg 1 g\nA common question is How many gram in 754 kilogram? And the answer is 754000.0 g in 754 kg. Likewise the question how many kilogram in 754 gram has the answer of 0.754 kg in 754 g.\n\n## How much are 754 grams in kilograms?\n\n754 grams equal 0.754 kilograms (754g = 0.754kg). Converting 754 g to kg is easy. Simply use our calculator above, or apply the formula to change the length 754 g to kg.\n\n## Convert 754 g to common mass\n\nUnitMass\nMicrogram754000000.0 µg\nMilligram754000.0 mg\nGram754.0 g\nOunce26.59656731 oz\nPound1.6622854569 lbs\nKilogram0.754 kg\nStone0.1187346755 st\nUS ton0.0008311427 ton\nTonne0.000754 t\nImperial ton0.0007420917 Long tons\n\n## What is 754 grams in kg?\n\nTo convert 754 g to kg multiply the mass in grams by 0.001. The 754 g in kg formula is [kg] = 754 * 0.001. Thus, for 754 grams in kilogram we get 0.754 kg.\n\n## 754 Gram Conversion Table", null, "## Alternative spelling\n\n754 Grams to kg, 754 Grams in kg, 754 Grams to Kilograms, 754 Grams in Kilograms, 754 g to Kilograms, 754 g in Kilograms, 754 Gram to Kilograms, 754 Gram in Kilograms, 754 Grams to Kilogram, 754 Grams in Kilogram, 754 g to kg, 754 g in kg, 754 Gram to kg, 754 Gram in kg" ]

[ null, "https://grams-to-kilograms.appspot.com/image/754.png", null ]

[ " Softwareentwicklung Schittkowski GmbH \n\nBack NLPLSX\n\nNonlinear Least Squares Optimization with Very Many Observations\n\nVersion 2.5 (2011)\n\n## Purpose\n\nNLPLSX solves constrained nonlinear least squares problems, i.e., nonlinear optimization problems, where the objective function is the sum of squares of function. In addition there may be any set of equality or inequality constraints. It is assumed that all individual problem functions are continuously differentiable, and that the number of squared functions or, alternatively, the number of experimental data is too large to apply an available Gauss-Newton-type algorithm.\n\n## Numerical Method\n\nThe problem is transformed into a general smooth nonlinear programming problem which is then solved by the sequential quadratic programming (SQP) code NLPQLP.\n\n## Program Organization\n\nNLPLSX is a double precision FORTRAN subroutine and parameters are passed through arguments.\n\n## Special Features\n\n• reverse communication\n• nonlinear constraints\n• bounds and linear constraints remain satisfied\n• FORTRAN source code (close to F77, conversion to C by f2c possible)\n\n## Applications\n\nNLPLSX is part of the interactive data fitting system EASY-FIT which contains now 1,300 test examples." ]

[ null ]

[ "妇女的蓝色时尚破旧的锥形的 Zip 前台粗斜纹棉布牛仔裤豹内衬\n\nSKU: 7G0I2J8W6Z9\n\\$17.10\n\n QTY:\n\n• 美国\n• 阿富汗\n• 奥兰群岛\n• 阿尔巴尼亚\n• 阿尔及利亚\n• 美属萨摩亚\n• 安道尔\n• 安哥拉\n• 安圭拉\n• 南极洲\n• 安提瓜和巴布达\n• 阿根廷\n• 亚美尼亚\n• 阿鲁巴\n• 亚洲太平洋\n• 澳大利亚\n• 奥地利\n• 阿塞拜疆\n• 亚速尔群岛 （葡萄牙）\n• 巴哈马\n• 巴林\n• 孟加拉国\n• 巴巴多斯\n• 白俄罗斯\n• 比利时\n• 伯利兹\n• 贝宁\n• 百慕大\n• 不丹\n• 玻利维亚\n• 博内尔\n• 波斯尼亚和黑塞哥维那\n• 博茨瓦纳\n• 布维岛\n• 巴西\n• 英属印度洋领地\n• 英属维尔京群岛\n• 文莱\n• Buesingen (德国)\n• 保加利亚\n• 布吉纳法索\n• 布隆迪\n• 柬埔寨\n• 喀麦隆\n• Campione/湖 Lugano (意大利)\n• 加拿大\n• 加那利群岛 （西班牙）\n• 佛得角\n• 开曼群岛\n• 中非共和国\n• 休达 (西班牙)\n• 乍得\n• 智利\n• 中国\n• 圣诞岛\n• 科科斯 (基林) 群岛\n• 哥伦比亚\n• 科摩罗\n• 刚果 (布)\n• 刚果 (金)\n• 库克群岛\n• 哥斯达黎加\n• 科特迪瓦\n• 克罗地亚\n• 古巴\n• 库拉索岛 (荷属安地列斯群岛)\n• 塞浦路斯\n• 捷克共和国\n• 丹麦\n• 吉布提\n• 多米尼克\n• 多米尼加共和国\n• 东蒂汶\n• 厄瓜多尔\n• 埃及\n• 萨尔瓦多\n• 赤道几内亚\n• 厄立特里亚\n• 爱沙尼亚\n• 埃塞俄比亚\n• 福克兰群岛 (马尔维纳斯群岛)\n• 法罗群岛\n• 斐济\n• 芬兰\n• 法国\n• 法国大都会\n• 法属圭亚那\n• 法属波利尼西亚\n• 法国南部领地\n• 加蓬\n• 冈比亚\n• 加沙 (西岸)\n• 格鲁吉亚\n• 德国\n• 加纳\n• 直布罗陀\n• 大开曼岛\n• 大茅草岛\n• 大多巴哥岛\n• 希腊\n• 格陵兰岛\n• 格林纳达\n• 瓜德罗普岛\n• 关岛\n• 危地马拉\n• 根西岛\n• 几内亚\n• 几内亚比绍\n• 圭亚那\n• 海地\n• 听到岛和麦克唐纳群岛\n• 海利 (德国)\n• 洪都拉斯\n• 香港\n• 匈牙利\n• 冰岛\n• 印度\n• 印度尼西亚\n• 伊朗\n• 伊拉克\n• 爱尔兰\n• 马恩岛\n• 以色列\n• 意大利\n• 牙买加\n• 日本\n• 泽西岛\n• 约旦\n• 约斯特范堤岛\n• 哈萨克斯坦\n• 肯尼亚\n• 基里巴斯\n• 朝鲜\n• 韩国\n• 科索沃\n• 科斯雷 (密克罗尼西亚)\n• 科威特\n• 吉尔吉斯斯坦\n• 老挝\n• 拉托维亚\n• 黎巴嫩\n• 莱索托\n• 利比里亚\n• 利比亚\n• 列支敦士登\n• 立陶宛\n• 利维尼奥 (意大利)\n• 卢森堡\n• 澳门\n• 马其顿\n• 马达加斯加\n• 马德拉 (葡萄牙)\n• 马拉维\n• 马来西亚\n• 马尔代夫\n• 马里\n• 马耳他\n• 马绍尔群岛\n• 马提尼克岛\n• 毛里塔尼亚\n• 毛里求斯\n• 马约特岛\n• 梅利利亚 (西班牙)\n• 墨西哥\n• 密克罗尼西亚\n• 摩尔多瓦\n• 摩纳哥\n• 蒙古\n• 黑山\n• 蒙特塞拉特岛\n• 摩洛哥\n• 阿陀斯山 (希腊)\n• 莫桑比克\n• 缅甸\n• 纳米比亚\n• 瑙鲁\n• 尼泊尔\n• 荷兰\n• 荷属安地列斯群岛\n• 尼维斯\n• 新喀里多尼亚\n• 新西兰\n• 尼加拉瓜\n• 尼日尔\n• 尼日利亚\n• 纽埃\n• 诺福克岛\n• 诺曼岛\n• 北马里亚纳群岛\n• 挪威\n• 阿曼\n• 巴基斯坦\n• 帕劳\n• 巴勒斯坦权力机构\n• 巴拿马\n• 巴布亚新几内亚\n• 巴拉圭\n• 秘鲁\n• 菲律宾\n• 皮特凯恩\n• 波兰\n• 波纳佩 (密克罗尼西亚)\n• 葡萄牙\n• 波多黎各\n• 卡塔尔\n• 团聚\n• 罗马尼亚\n• 罗塔 (北马里亚纳群岛)\n• 俄罗斯\n• 卢旺达\n• 萨巴 (荷属安地列斯群岛)\n• 塞班岛 （北马里亚纳群岛)\n• 萨摩亚\n• 圣马利诺\n• 圣多美和普林西比\n• 沙特阿拉伯\n• 塞内加尔\n• 塞尔维亚\n• 塞舌尔\n• 塞拉利昂\n• 新加坡\n• 斯洛伐克\n• 斯洛文尼亚\n• 索罗门群岛\n• 索马里\n• 南非\n• 西班牙\n• 斯里兰卡\n• 圣巴泰勒米 (瓜德罗普岛)\n• 圣克罗伊 （美属维尔京群岛)\n• 圣尤斯特歇斯 (荷属安地列斯群岛)\n• 圣赫勒拿\n• 圣 · 约翰 （美属维尔京群岛)\n• 圣基茨和尼维斯\n• 圣卢西亚\n• 圣马丁岛\n• 圣马丁 (瓜德罗普岛)\n• 圣 Pierre 和密克隆群岛\n• 圣 Thomas （美属维尔京群岛)\n• 圣 Vincent 和格林纳丁斯\n• 苏丹\n• 苏里南\n• 斯瓦尔巴特和扬马延岛\n• 斯威士兰\n• 瑞典\n• 瑞士\n• 叙利亚\n• 大溪地岛 （法属）\n• 台湾\n• 塔吉克斯坦\n• 坦桑尼亚\n• 泰国\n• 蒂妮 (北马里亚纳群岛)\n• 多哥\n• 托克劳\n• 汤加\n• 托尔托拉 （英属维尔京群岛）\n• 特立尼达和多巴哥\n• 特鲁克 (密克罗尼西亚)\n• 突尼斯\n• 土耳其\n• 土库曼斯坦\n• 特克斯和凯科斯群岛\n• 图瓦卢\n• 美属维尔京群岛\n• 乌干达\n• 乌克兰\n• 联盟岛屿 （圣 Vincent)\n• 阿拉伯联合酋长国\n• 英国\n• 乌拉圭\n• 乌兹别克斯坦\n• 瓦努阿图\n• 梵蒂冈城\n• 委内瑞拉\n• 越南\n• 维尔京戈尔达 （英属维尔京群岛）\n• 瓦利斯群岛和富图纳群岛\n• 西撒哈拉\n• Yap (密克罗尼西亚)\n• 也门\n• 赞比亚\n• 津巴布韦\n\nSize: 27, 28, 29, 30\n\n 尺寸/厘米 27/S 28/M 29/L 30/XL 腰 68 70 72 76 臀部 90 92 94 96 大腿 46 48 50 52 长度 91 93 95 96", null, "", null, "", null, "", null, "", null, "Product Reviews (0)\n\nWrite a product review now!\n(There are no customer reviews yet.)", null, "刷新" ]

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[ "# How many ways can you find x?\n\n*Submitted by Dr. Ted Stanford, NMSU Mathematician.\n\n•", null, "• Here are some solutions strategies. Each strategy has several possible variations. Try at least two methods! You should get the same answer, no matter which method you use, although the form of your answer might be different.\n\n1. Use a trig function to find one of the acute angles, then use another trig function on that angle to find x.\n2. Find the area of the large triangle using the two given sides. Find the hypotenuse of the large triangle using Pythagoras, then use the hypotenuse as the base and x as the height.\n3. Choose two new variables, y and z, to represent the two horizontal line segments. Use Pythagoras to set up several equations involving x, y, and z, and solve these.\n4. Same as 3, only use similar triangles to set up your equations instead of Pythagoras.\n5. Put the triangle in a coordinate plane, and use facts about the slopes of perpendicular lines, and the distance formula.\n6. Draw the figure carefully with a ruler and other geometry tools, then measure x." ]

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[ "-", null, "Exactness of Parrilo's conic approximations for copositive matrices and associated low order bounds for the stability number of a graph Monique Laurent(monique.laurent", null, "cwi.nl) Luis Felipe Vargas(luis.vargas", null, "cwi.nl) Abstract: De Klerk and Pasechnik (2002) introduced the bounds $\\vartheta^{(r)}(G)$ ($r\\in \\mathbb{N}$) for the stability number $\\alpha(G)$ of a graph $G$ and conjectured exactness at order $\\alpha(G)-1$: $\\vartheta^{(\\alpha(G)-1)}(G)=\\alpha(G)$. These bounds rely on the conic approximations $\\mathcal{K}_n^{(r)}$ by Parrilo (2000) for the copositive cone $\\text{COP}_n$. A difficulty in the convergence analysis of $\\vartheta^{(r)}$ is the bad behaviour of the cones $\\mathcal{K}_n^{(r)}$ under adding a zero row/column: when applied to a matrix not in $\\mathcal{K}^{(0)}_n$ this gives a matrix not in any ${\\mathcal{K}}^{(r)}_{n+1}$, thereby showing strict inclusion $\\bigcup_{r\\ge 0}{\\mathcal{K}}^{(r)}_n\\subset \\text{COP}_n$ for $n\\ge 6$. We investigate the graphs with $\\vartheta^{(r)}(G)=\\alpha(G)$ for $r=0,1$: we algorithmically reduce testing exactness of $\\vartheta^{(0)}$ to acritical graphs, we characterize critical graphs with $\\vartheta^{(0)}$ exact, and we exhibit graphs for which exactness of $\\vartheta^{(1)}$ is not preserved under adding an isolated node. This disproves a conjecture by Gvozdenovi\\'c and Laurent (2007) which, if true, would have implied the above conjecture by de Klerk and Pasechnik. Keywords: stable set problem \\and $\\alpha$-critical graph \\and sum-of-squares polynomial \\and copositive matrix \\and semidefinite programming \\and Shor relaxation Category 1: Linear, Cone and Semidefinite Programming (Semi-definite Programming ) Category 2: Combinatorial Optimization (Approximation Algorithms ) Citation: arXiv:2109.12876 Download: [PDF]Entry Submitted: 10/12/2021Entry Accepted: 10/12/2021Entry Last Modified: 10/12/2021Modify/Update this entry Visitors Authors More about us Links Subscribe, Unsubscribe Digest Archive Search, Browse the Repository Submit Update Policies Coordinator's Board Classification Scheme Credits Give us feedback Optimization Journals, Sites, Societies Optimization Online is supported by the Mathematical Optmization Society.", null, "" ]

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[ "# Knowledge Gap\n\nEncyclopedia\nEdited by: Published: 2008\n\n• ## Subject Index\n\nThis entry focuses on the concept of the knowledge gap, which is a specific hypothesis within the area of diffusion of knowledge, and subsequent derivations. A number of models of social change are based on the notion that change is a cumulative process. In such models, small changes result in differential rates of change for the social system—slow system change at first, followed by increasingly faster rates of change during the middle of the change process, followed by slowing rates of societal change at the end. This process is reflected in the familiar \"S-curve\" which shows an accumulative sum over time. Two prominent models that follow the logic and assumptions of the cumulative process perspective include models of diffusion of technologies and diffusion of ...\n\n• All\n• A\n• B\n• C\n• D\n• E\n• F\n• G\n• H\n• I\n• J\n• K\n• L\n• M\n• N\n• O\n• P\n• Q\n• R\n• S\n• T\n• U\n• V\n• W\n• X\n• Y\n• Z\n\n## Methods Map", null, "Research Methods\n\nCopy and paste the following HTML into your website" ]

[ null, "http://methods.sagepub.com/images/img-bg.png", null ]

[ "# Driving LEDs: Why the resistor?\n\nRecently I was asked by one of my customers about using resistors with LEDs. The question came down to “What is the purpose of the resistor?”\n\nFor many people starting out in electronics today, LEDs (Light Emitting Diodes) are a popular starting point. They are a fundamental circuit (or, more often, sub-circuit). LEDs have traditionally been used for indicators of all sorts (looking around my desk, there are 5 of them telling me things are turned on and one telling me if I have voicemail). But now, since the advent of white LEDs, they now serve as a very energy-efficient lighting source. So driving a LED is a good starting project, and learning why we need the resistor is a great way to learn some basic electronics concepts. So let me try to explain.\n\nFirst let’s start with the LED. A light emitting diode is, fundamentally, a diode. Diodes conduct electricity in one direction only. When current flows though a LED in the correct direction (from the anode to the cathode), two things happen: light is emitted of course, and there is a well defined drop in voltage across the LED.\n\nAll LEDs vary the amount of light they produce in direct relation to the current going through them: more current = brighter light. But with everything, there is a limit. (This same relationship applies to tungsten light bulbs, but if you take a light bulb made for the U.S. and take it over to Europe, it will burn much brighter, but only for a fraction of a second before burning out.) In LED datasheets this limit is referred to as forward current (IF), and a typical safe operating limit is 20mA. For lighting uses, we will normally want to run this much current through the LED. Note this is sometimes referred to as the test current (as the manufacturer’s tests use this current as a standard) and is almost always lower than the forward current in the Absolute Maximum Ratings section. For indicators, full brightness may be too bright and might want as little as 1-2mA through the LED.\n\n##### So what does this have to do with resistors?\n\nWell, everything, really. The resistor is what makes everything work right. Remember the LED has a defined voltage drop? Most likely your power source isn’t exactly equal to that drop (if it is, you might be able to get away without a resistor). Let’s say you have a 5V power supply and are driving a Cree bright white LED which has a voltage drop of 3.2V at 20mA. A simplification of Kirchoff’s Voltage Law (KVL) tells us that the total voltage drops (in this case the LED) have to equal the total voltage gains (the 5V power supply). Here we have a difference of 1.8V to account for. The wire offers minimal resistance of its own, essentially 0Ω. The most fundamental relationship in electronics, Ohm’s Law, says that the current through an element is equal to the voltage across it divided by its resistance (I = V/R). Here we have 1.8V/0Ω, or infinite current, which equals your LED going poof (and maybe a fire if you’re unlucky). This is where the resistor come in.\n\nWe have a 1.8V drop to deal with before the LED will work properly on a 5V supply. We also know we want to get the most brightness we can safely get from these little critters, so we’d like 20mA going through them. Using simple algebra, Ohm’s Law can be shuffled around to find any of the three values if we know the other two. In the case, we shuffle it so that R = V/I, since we know V (1.8V) and I (20mA). So R = 1.8V/20mA = 1.8V/0.02A = 90Ω. We would then round to the next highest standard value (or the next highest we have at hand). In this case, a 91Ω or 100Ω resistor would be a good choice." ]

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[ "## Sunday, September 14, 2014\n\n### Week 3: Physics Vector Lab and Velocity\n\n This week we are: doing a vector lab, reviewing Velocity and introducing velocity time graphs I will: calculate both the displacement and distance to different points around the school I will: review how to do velocity calculations. I will: learn about velocity/time graphs. I can: differentiate between scalar and vector in a lab setting I can: calculate velocity using the basic formula I can: relate the motion of an object based on its velocity time graph\n\nMonday\n1.  Vector Lab\n\nTuesday\n1. Review velocity HW\n2. Hand in Lab work and HW.\n\nWednesday\n1.  Introduce Lesson 8, velocity/time graphs\n2.  Class work demonstrating formula work\n3.  Discussion on Formula work expectations.\n4.  Review Lab and velocity HW\n\nThursday\n1.  Formative Assessment on velocity\n2.  Review & HW for velocity/time graphs\n\nFriday\n1.  Lesson 9" ]

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[ "# Compare length of iterator with as few next calls as possible. How can I deduplicate this code?\n\nI'm writing a few simple functions that compare the length of an iterator to an integer, returning as early as possible. Right now I have:\n\nclass ILengthComparator:\ndef __init__(self, iterator: Iterator):\nself.iterator = iterator\n\ndef has_more(self) -> bool:\ntry:\nnext(self.iterator)\nexcept StopIteration:\nreturn False\n\nreturn True\n\ndef __eq__(self, length: int) -> bool:\nif length < 0:\nreturn False\n\nif not self.has_more():\nreturn length == 0\n\nreturn self == length-1\n\ndef __gt__(self, length: int) -> bool:\nif length < 1:\nreturn False\n\nif not self.has_more():\nreturn True\n\nreturn self > length-1\n\ndef __lt__(self, length: int) -> bool:\nif length < 0:\nreturn True\n\nif not self.has_more():\nreturn False\n\nreturn self < length - 1\n\n\nClearly this class isn't finished (__neq__, __gte__, __lte__), and anything after the first comparison isn't guaranteed to work (as the iterator may be consumed), but I'm wondering: is it is possible to logically simplify this into a single function with an operator function as an argument (ie operator.eq, operator.lt), or at least just reduce the semi-duplicate parts (ie, every function has a comparison with length against a number, and a case for checking if the iterator is empty - each with a rather ad-hoc return value).\n\n(any suggestions don't need to be recursive or use the has_more method, those were just the first thing I reached for when trying to solve this problem)\n\nPastebin with the above some quick and dirty unit tests: https://pastebin.com/N65Q2qcJ\n\n• (Would you believe in total ordering? Jan 13 '20 at 1:00\n• @greybeard I could be wrong, but I don't think that would work, because these methods have the side effect of consuming the iterator. So if the implementation requires combining __eq__ and __lt__ to determine __gt__, by the time __lt__ was called, __eq__ would have already consumed the iterator. Jan 14 '20 at 1:06\n• (For handling consuming, see 409_conflict's answer.) Jan 14 '20 at 4:07\n• As I understand it, .tee() often isn't a good idea, because it has to store the results of each next() call for _re_iteration (which can consume a decent amount of memory depending on what the iterator is iterating over). Even if not for this specific use case, it still seems like a pretty awesome decorator! Jan 14 '20 at 4:50\n\nTo know the answer to your comparison to $$\\N\\$$, you need to attempt at least:\n\n• $$\\N\\$$ next calls if you test with $$\\\\lt\\$$ (succeeding will result to False) or $$\\\\ge\\$$ (succeeding will result to True);\n• $$\\N+1\\$$ next calls in all 4 other cases.\n\nObviously, if the iterator length is fewer than $$\\N\\$$ there will be less calls.\n\nSo, in order to simplify things, you could extract the first $$\\N+1\\$$ elements of the iterator and count them to compare this count to $$\\N\\$$:\n\n• extracting can be simplified with itertools.islice\n• counting can be done using either\n\n• enumerate and dumping the content into a 1-length collections.deque\n\nfrom itertools import islice\nfrom collections import deque\n\ndef compare_length(iterable, n, operation):\nchunck = islice(iterable, max(0, n+1))\ntry:\n(length, _) = deque(enumerate(chunk, 1), maxlen=1)\nexcept ValueError:\nlength = 0\nreturn operation(length, n)\n\n• sum\n\nfrom itertools import islice\n\ndef compare_length(iterable, n, operation):\nchunck = islice(iterable, max(0, n+1))\nlength = sum(1 for _ in chunck)\nreturn operation(length, n)\n\n\nThe former should be more efficient, the latter more readable. As suggested in the question, this function takes the operation to perform as its third argument and compares min(<iterable length>, n+1) to n. This is sufficient in most cases.\n\nIf you truly want to reduce the amount of next calls needed for $$\\\\lt\\$$ and $$\\\\ge\\$$, you can keep your class approach and factorize the min(<iterable length>, n) part into an helper method:\n\nfrom itertools import islice\n\nclass ILengthComparator:\ndef __init__(self, iterable):\nself._iterable = iterable\n\ndef _length_up_to(self, n):\nreturn sum(1 for _ in islice(self._iterable, max(0, n)))\n\ndef __eq__(self, n):\nreturn self._length_up_to(n+1) == n\n\ndef __ne__(self, n):\nreturn self._length_up_to(n+1) != n\n\ndef __lt__(self, n):\nreturn self._length_up_to(n) < n\n\ndef __le__(self, n):\nreturn self._length_up_to(n+1) <= n\n\ndef __gt__(self, n):\nreturn self._length_up_to(n+1) > n\n\ndef __ge__(self, n):\nreturn self._length_up_to(n) >= n\n\n\nLastly, if you want to avoid consuming the iterator while performing those tests, you can check itertools.tee." ]

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[ "## Textbooks & Solution Manuals\n\nFind the Source, Textbook, Solution Manual that you are looking for in 1 click.\n\n## WriteWise AI Model by Holooly Genius\n\nYour Ultimate AI Essay Writer & Assistant.\n\n## Holooly Arabia\n\nFor Arabic Users, find a teacher/tutor in your City or country in the Middle East.\n\n## Holooly Help Desk\n\nNeed Help? We got you covered.\n\nProducts\n\n## Textbooks & Solution Manuals\n\nFind the Source, Textbook, Solution Manual that you are looking for in 1 click.\n\n## WriteWise AI Model by Holooly Genius\n\nYour Ultimate AI Essay Writer & Assistant.\n\n## Holooly Arabia\n\nFor Arabic Users, find a teacher/tutor in your City or country in the Middle East.\n\n## Holooly Help Desk\n\nNeed Help? We got you covered.\n\n## Q. 14.6.1\n\nThe propped uniform cantilever in Fig. 14.6-4(a) is of plastic moment $M_{P}$. Determine the value of P at collapse using (a) the work method (b) the statical method.", null, "## Verified Solution\n\n(a) The work method (or virtual work method)\n\nThere are three possible collapse mechanisms, as shown in Figs. 14.6-4(b), (c) and (d). We shall calculate the collapse load for each in turn.\n\nThe work equation is\n\n$\\qquad P\\left(\\phi L\\right)+2P(3\\phi L/2)=M_{p}\\phi +M_{p}(4\\phi )$\n\nTherefore                          P = 5Mp/4L\n\n$\\qquad P\\left(\\phi L\\right)+2P(\\phi L/2)=M_{p}\\phi +M_{p}(2\\phi )$\n\nTherefore                            P = 3Mp/2L\n\n$\\qquad 2P(\\phi L/2)=M_{p}\\phi +M_{p}(2\\phi )$\n\nTherefore                             P = 3Mp/L\n\nWe have now examined all the three possible mechanisms and found that the one in Fig. 14.6-4(b) gives the lowest collapse load. This means that as the magnitude of P is gradually increased from zero, the collapse mechanism in Fig. 14.6-4(b) will be the first to form, when P reaches $5M_{p}/4L$. The other two mechanisms cannot form unless this one is prevented from forming, for example by strengthening the cross sections at points where hinges would have formed. We therefore conclude that $P = 5M_{p}/4L$ is the correct collapse value. (See Comment (2) below on the upper bound theorem.)\n\n(b) The statical method\n\nFigs. 14.6-4(e), (f) and (g) show three bending moment diagrams. Each diagram has been so drawn that the moment ordinate is exactly equal to the plastic moment$M_{P}$ at two sections. In Figs, (e) and (f), α1 α2 b is the moment diagram due to the redundant moment $M_{P}$ at A, and α1c2d2b is the simple-span moment diagram due to the external loads P and 2P. In Fig. (g) the moment Mp at section C has been selected as the redundant; that is, the triangle a1a2c2bc1a1 is the redundant moment diagram (The reader should verify this. Hint: the moment Mp at C produces shear forces.), and a1a3c1d2b is the moment diagram for the loads P and 2P acting on the beam with a hinge at C. The collapse values of P can be calculated from the geometry of the three bending moment diagrams.\n\nThe three values for P obtained by the statical method agree with those obtained by the work method. As before, we conclude that the lowest value, namely P = 5Mp/4L, is the correct one.\n(1) In the statical method above, we need not have calculated the collapse value of P for all the three bending moment diagrams. A closer examination will immediately reveal that both Figs. 14.6-4(f) and (g) violate the yield condition. In Fig. 14.6-4(f), the moment ordinate d1d2 exceeds Mp, which means that the collapse mechanism in Fig. 14.6-4(c) cannot occur unless plastic hinge formation is prevented at D by strengthening the cross section there. Similarly, in Fig. 14.6-4(g), a2a3 exceeds Mp ; again, the mechanism in Fig. 14.6-4(d) cannot occur unless plastic hinge formation at A is deliberately prevented. The bending moment diagram in Fig. 14.6-4(e), on the other hand, satisfies the three conditions of mechanism (with plastic hinges at A and D), equilibrium (by the manner of its construction) and yield (since moment ordinates nowhere exceed M p). Therefore we can at once conclude from the uniqueness theorem that the corresponding collapse load is the correct one; there is in fact no need to consider any other mechanisms.\n\n$\\qquad \\begin{matrix} \\qquad & d_{1} d_{2}&=PL-a_{1} a_{2}/4 \\\\ Therefore & M_{p} &=PL-M_{p}/4 \\\\or & P &=5M_{p}/4L \\end{matrix}$\n\n$\\qquad \\begin{matrix} \\qquad & c_{1}c_{2} &= PL-a_{1} a_{2} /2 \\\\ & M_{p} &=PL-M_{p}/2 \\\\Therefore & P &=3M_{p}/2L \\end{matrix}$\n\n$\\qquad \\begin{matrix}\\\\ \\qquad & d_{1}d_{2}&=PL/2-c_{1}c_{2}/2 \\\\ & M_{p}&=PL/2-M_{p}/2 \\\\Therefore & P &=3M_{p}/L \\end{matrix}$\n\n(2) A fundamental theorem of plastic collapse is the upper bound theorem (sometimes called the kinematic theorem) which states that for a given structure subjected to a given loading, the magnitude of the loading which is found to correspond to any assumed collapse mechanism must be either greater than or equal to, but cannot be less than, the true collapse load. (The proof of the theorem will be given in Section 14.9.)\n\nTherefore, in an analysis we simply compute the collapse load for each possible mechanism and accept the lowest value as the correct one, as we did in the work method above. However, sometimes there is uncertainty about the number of possible mechanisms. A statical check is then necessary: the collapse bending moment diagram is drawn for the mechanism that gives the lowest collapse load; if the moment ordinate nowhere exceeds the plastic moment $M_{P}$ then the uniqueness theorem guarantees that this mechanism will give the true collapse load. If $M_{P}$ is exceeded somewhere, then the yield condition is not satisfied, and the search for a correct collapse mechanism must continue.\n\n(3) The upper bound theorem is often referred to as the unsafe theorem, because, interpreted in a design sense, it states that the value of the plastic moment Mp obtained on the basis of an arbitrarily assumed collapse mechanism is smaller than, or at best equal to, that actually required. Consider, for example, the beam in Fig.14.6-4(a). Suppose we have to determine the plastic moment of resistance required to carry the known loads P and 2P. As we have seen in Example 14.6-1, the correct collapse mechanism (Fig. 14.6-4(b)) gives $M_{P} — 4PL/5$. The incorrect mechanisms in Figs, (c) and (d) give Mp as 2PL/3 and PL/3 respectively.\n\n(4) Another fundamental theorem of plastic collapse is the lower bound theorem (sometimes called the static theorem), which states that if a distribution of bending moments can be found such that the structure is in equilibrium under the external loading and such that nowhere is the plastic moment of resistance Mp exceeded, then the structure will not collapse under that loading—however ‘unlikely’ that distribution of moments may appear. The theorem is often referred to as the safe theorem. (The proof of this theorem will be given in Section 14.9).\n\nThe use of the lower bound theorem is illustrated in Example 14.6-2.\n\n(5) For future reference the three fundamental theorems of plastic collapse are displayed together below:", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "" ]

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[ "## 254 - Towers of Hanoi\n\nModerator: Board moderators\n\ncytse\nLearning poster\nPosts: 67\nJoined: Mon Sep 16, 2002 2:47 pm\nLocation: Hong Kong\nContact:\nYes, BigInt is needed here, because m can be as large as 2^100-1.\n\npc5971\nNew poster\nPosts: 34\nJoined: Mon Aug 05, 2002 11:24 am\nContact:\n\nI solved problem 254 in so many ways but now I can't see what's still wrong with my problem...\n\nFirst I made the stupid error to move only 1 ring every time, after that I tried to move k rings (2^k-1<=m) but I was still wrong because I didn't use big numbers...\n\nAfter I used big numbers (implemented with string - in Pascal) and I received TLE . . . I found that the program calculate for more then one time some of powers of 2. So finaly I precalculate the 2^k-1 powers for k=1,100. And now I receive WA...\n\nThis problem make me crazy, so pls. help me to find some tests for this program...\n\n[pascal]\nprogram p254;\nvar m,s:string;\ni,n:integer;\nt:array[1..3] of integer;\np:array[1..100] of string;\n\nprocedure decr(var m:string);\n{ m <- m-1 }\nvar i,t,c:integer;\nbegin\nt:=1; i:=length(m);\nwhile t>0 do\nbegin\nc:=ord(m)-ord('0')-t;\nif c<0 then\nc:=10-c\nelse\nt:=0;\nm:=chr(c+ord('0'));\ni:=i-1;\nend;\nwhile (length(m)>0) and (m='0') do\ndelete(m,1,1);\nend;\n\n{m <- m-s}\nvar t,l1,l2,c1,c2,d,i:integer;\nbegin\nl1:=length(m); l2:=length(s); d:=l1-l2;\nt:=0;\nfor i:=l2 downto 1 do\nbegin\nc1:=ord(m[i+d])-ord('0')-t;\nc2:=ord(s)-ord('0');\nif c1<c2 then\nt:=1\nelse\nt:=0;\nm[i+d]:=chr(c1+10*t-c2+ord('0'));\nend;\nif t>0 then\nbegin\nc1:=ord(m[d])-ord('0');\nc1:=c1-1;\nm[d]:=chr(c1+ord('0'));\nend;\nwhile (length(m)>0) and (m='0') do delete(m,1,1);\nend;\n\nprocedure doila(n:integer;var s:string);\n{ s <- 2^n }\nvar i,j,c,t:integer;\nbegin\ns:='1';\nt:=0;\nfor i:=1 to n do\nbegin\nt:=0;\nfor j:=length(s) downto 1 do\nbegin\nc:=ord(s[j])-ord('0');\nc:=2*c+t;\ns[j]:=chr(c mod 10+ord('0'));\nt:=c div 10;\nend;\nif t>0 then\ninsert(chr(t+ord('0')),s,1);\nend;\nend;\n\nfunction smaller(s,m:string):boolean;\n{ =true if s<=m}\nvar i:integer;\nbegin\nif length(s)<length(m) then\nsmaller:=true\nelse\nif length(s)>length(m) then\nsmaller:=false\nelse\nbegin\nsmaller:=true;\nfor i:=1 to length(s) do\nif s>m then\nbegin\nsmaller:=false;\nbreak;\nend;\nend;\nend;\n\nprocedure H(n,a,b,c:integer);\nbegin\nif smaller(p[n],m) then\nbegin\nt[a]:=t[a]-n;\nt:=t+n;\nend\nelse\nbegin\nH(n-1,a,c,b);\nif length(m)=0 then exit;\nt[a]:=t[a]-1;\nt:=t+1;\ndecr(m);\nif length(m)>0 then\nH(n-1,c,b,a);\nend;\nend;\n\nbegin\nfor i:=1 to 100 do\nbegin\ndoila(i,p);\ndecr(p);\nend;\nwhile m=' ' do\ndelete(m,1,1);\nwhile not ((n=0) and (m='0')) do\nbegin\nt:=n; t:=0; t:=0;\nif n mod 2 =0 then\nH(n,1,3,2)\nelse\nH(n,1,2,3);\nwriteln(t,' ',t,' ',t);\nwhile m=' ' do\ndelete(m,1,1);\nend;\nend.\n[/pascal]\n\nThanks a lot . . .\n\nIlham Kurnia\nNew poster\nPosts: 31\nJoined: Sat Nov 17, 2001 2:00 am\nContact:\n\n### A hint\n\nTry to think bitwise. (that is, convert m to base 2 and then play with it). Try finding the link between the binary represenatation of m and the number of disks in each peg.\n\nBest wishes and Merry Christmas to those celebrating.\n\nIlham\n\nGhost77 dimen\nLearning poster\nPosts: 67\nJoined: Sun Sep 22, 2002 5:40 am\nLocation: Taiwan\n\n### The Specification of problem 254.\n\nQuite strange.\n\nI test my program step by step, and it seems ok.\n\nAnd I make sure the BigInt handled well.\n\nHowever, I can't pass this problem.\n\nOne problem, should I transfer disks from peg A to peg C or peg B?\n\nIn the problem statement, it says \"transfer them from peg A to one\n\nof the other pegs\".\n\nBut the solutions won't be indentical with the other.\n\nIn sample input, the problemsetter choose the data which can't\n\ndistinguish the right solution.\n\nAfter submitting two type of solution, I got two wrong answer replies.\n\nSo there will be still some specification not clear or hidden.\n\nOne more problem, does the test data contain illegal input?\n\nThe legal ones should be n(wthin the range[0 100]) and\n\nm(within the range[0 2^n-1]).\n\nIf m is more than or equal to 2^n, what should be printed?\n\nThanks for reading it.", null, "little joey\nGuru\nPosts: 1080\nJoined: Thu Dec 19, 2002 7:37 pm\nfor odd moves, take the smallest disk (number 1) from the peg where it lies to the next one in the circular sequence ABCABC for even moves, make the only possible move not involving disk 1.\nThis completely specificies where the tower should go, since the first move (an odd move) takes disk 1 from A to B. Think about it.\n\nI don't think there is illegal input. My program doesn't test it and would produce nonsense if there was.\n\nGhost77 dimen\nLearning poster\nPosts: 67\nJoined: Sun Sep 22, 2002 5:40 am\nLocation: Taiwan\nThanks for your reply.", null, "anupam\nA great helper\nPosts: 405\nJoined: Wed Aug 28, 2002 6:45 pm\nContact:\nI tried to memorized all the moves but it gives tle. Is there any faster algorithm? will any1 describe shortly??", null, "--\nanupam\n\"Everything should be made simple, but not always simpler\"\n\nGhost77 dimen\nLearning poster\nPosts: 67\nJoined: Sun Sep 22, 2002 5:40 am\nLocation: Taiwan\nHello, anupam.\n\nMemorizing all moves is much time consuming, maybe never be done,\n\nbecause the most numbers of moves would be 2^100-1.\n\nTry to solve it with your background knowledge.\n\nThe Tower of Honai tells you that moves n disks from one peg to another\n\nneeds 2^n-1 steps.\n\nSo when you get a number, dividing them into 2^a-1, 2^b-1, 2^c-1, and\n\nso on.(a>b>c>....)\n\nBut for better speed, I divided them into 2^a, 2^b, 2^c, and so on.\n\nMoving them in sets instead of in steps, you'll get what you want.\n\nIf you still have no idea of my method, please let me know.", null, "anupam\nA great helper\nPosts: 405\nJoined: Wed Aug 28, 2002 6:45 pm\nContact:\nwell i misunderstood a little and fixed now. Thanks Boss.\n--\nAnupam\n\"Everything should be made simple, but not always simpler\"\n\nRed Scorpion\nExperienced poster\nPosts: 192\nJoined: Sat Nov 30, 2002 5:14 am\n\n### P-254\n\nHi, I always got WA.\n\nCan somebody post some I/O for this problem(Towers Of Hanoi).\n\nPlease...", null, "", null, "nesqi\nNew poster\nPosts: 5\nJoined: Thu Sep 30, 2004 4:18 pm\nif n is 0, m has to be 0 since 2^0 -1 = 1 - 1 = 0\n\nSo if n is 0 the program should terminate.\n\nd91-lek\nNew poster\nPosts: 22\nJoined: Thu Sep 16, 2004 2:25 am\nLocation: KTH, Stockholm\nContact:\n\n### Test data needed\n\nCould some merciful soul please post the result of this indata:\n\nCode: Select all\n\n``````3 5\n64 2\n8 45\n100 1267650600228229401496703205375\n100 1\n36 23478162346434234\n23 231234\n80 398573924759238475987234\n67 2349302402348238482342348\n99 3895732987529387459237955432\n88 747328293948729334241100234\n0 0\n``````\nI get\n\nCode: Select all\n\n``````1 1 1\n62 1 1\n4 2 2\n0 100 0\n99 0 1\n2 16 18\n7 10 6\n27 22 31\n11 25 31\n43 30 26\n26 36 26\n``````\nand WA of course, since I'm wondering.\n\nd91-lek\nNew poster\nPosts: 22\nJoined: Thu Sep 16, 2004 2:25 am\nLocation: KTH, Stockholm\nContact:\n\n### Test data needed\n\n- deleted -\nLast edited by d91-lek on Fri Nov 18, 2005 4:48 am, edited 1 time in total.\n\nd91-lek\nNew poster\nPosts: 22\nJoined: Thu Sep 16, 2004 2:25 am\nLocation: KTH, Stockholm\nContact:\nGot AC when I reversed the peg order. Example input is\nnot decisive in this matter. 100 1 should of course give\n99 1 0 and not 99 0 1.\n\nCorrect output for the input above is:\n\nCode: Select all\n\n``````1 1 1\n62 1 1\n4 2 2\n0 0 100\n99 1 0\n2 18 16\n7 6 10\n27 31 22\n11 31 25\n43 26 30\n26 26 36\n``````\n\nahmed hasan\nNew poster\nPosts: 9\nJoined: Fri Jan 21, 2005 5:09 pm\n\n### 254 Why WA?\n\nThis code does not run in vc++.\n\nCode: Select all\n\n``````#include<cstdio>\n#include<cstdlib>\n#include<cstring>\n#include<cmath>\n\nunsigned long long A,B,C;\nunsigned long long Moves;\n\nvoid MoveDisks(unsigned long long n, unsigned long long *peg1, unsigned long long *peg2)\n{\nMoves-=(unsigned long long)pow(2,n)-1;\n*peg1-=n;\n*peg2+=n;\n}\n\nunsigned long long MoveStack(unsigned long long n, unsigned long long *p, unsigned long long *e, unsigned long long *t)\n{\nif(Moves==0) return 0;\nif(Moves>=pow(2,n)-1) {MoveDisks(n,p,t);return 0;}\nelse\n{\nMoveStack(n-1,p,t,e);\nif(Moves>=1)MoveDisks(1,p,t);\nMoveStack(n-1,e,p,t);\n}\nreturn 0;\n}\n\nint main()\n{\nunsigned long long NumberOfDisks;\nwhile(1)\n{\nscanf(\"%llu %llu\",&NumberOfDisks,&Moves);\nif(NumberOfDisks==0 && Moves==0) break;\nA=NumberOfDisks; B=0; C=0;\nMoveStack(NumberOfDisks,&A,&B,&C);\nprintf(\"%llu \",A);\nprintf(\"%llu \",B);\nprintf(\"%llu\\n\",C);\n}\nreturn 0;\n}``````" ]

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[ "", null, "Welcome to tutorial no. 33 in Golang tutorial series.\n\n### What are first class functions?\n\nA language which supports first class functions allows functions to be assigned to variables, passed as arguments to other functions and returned from other functions. Go has support for first class functions.\n\nIn this tutorial we will discuss about the syntax and various use cases of first class functions.\n\n### Anonymous functions\n\nLet's start with a simple example which assigns a function to a variable.\n\n``````package main\n\nimport (\n\"fmt\"\n)\n\nfunc main() {\na := func() {\nfmt.Println(\"hello world first class function\")\n}\na()\nfmt.Printf(\"%T\", a)\n}\n``````\n\nRun in playground\n\nIn the program above, we have assigned a function to the variable `a` in line no. 8. This is the syntax for assigning a function to a variable. If you notice carefully, the function assigned to `a` does not have a name. These kind of functions are called anonymous functions since they do not have a name.\n\nThe only way to call this function is using the variable `a`. We have done this in the next line. `a()` calls the function and this prints `hello world first class function`. In line no. 12 we print variable `a`'s type . This will print `func()`.\n\nRunning this program will output\n\n``````hello world first class function\nfunc()\n``````\n\nIt is also possible to call a anonymous function without assigning it to a variable. Let's see how this is done in the following example.\n\n``````package main\n\nimport (\n\"fmt\"\n)\n\nfunc main() {\nfunc() {\nfmt.Println(\"hello world first class function\")\n}()\n}\n``````\n\nRun in playground\n\nIn the program above, an anonymous function is defined in line no. 8 and immediately after the function definition, we call the function using `()` in line no. 10. This program will output,\n\n``````hello world first class function\n``````\n\nIt is also possible to pass arguments to anonymous functions just like any other function.\n\n``````package main\n\nimport (\n\"fmt\"\n)\n\nfunc main() {\nfunc(n string) {\nfmt.Println(\"Welcome\", n)\n}(\"Gophers\")\n}\n``````\n\nRun in playground\n\nIn the program above, a string argument is passed to the anonymous function in line no. 10. Running this program will print,\n\n``````Welcome Gophers\n``````\n\n### User defined function types\n\nJust like we define our own struct types, it is possible to define our own function types.\n\n``````type add func(a int, b int) int\n``````\n\nThe code snippet above creates a new function type `add` which accepts two integer arguments and returns a integer. Now we can define variables of type `add`.\n\nLet's write a program which defines a variable of type `add`.\n\n``````package main\n\nimport (\n\"fmt\"\n)\n\ntype add func(a int, b int) int\n\nfunc main() {\nvar a add = func(a int, b int) int {\nreturn a + b\n}\ns := a(5, 6)\nfmt.Println(\"Sum\", s)\n}\n``````\n\nRun in playground\n\nIn the program above, in line no.10, we define a variable `a` of type `add` and assign to it a function whose signature matches the type `add`. We call the function in line no. 13 and assign the result to `s`. This program will print,\n\n``````Sum 11\n``````\n\n### Higher-order functions\n\nThe definition of Higher-order function from wiki is a function which does at least one of the following\n\n• takes one or more functions as arguments\n• returns a function as its result\n\nLet's look at some simple examples for the above two scenarios.\n\n#### Passing functions as arguments to other functions\n\n``````package main\n\nimport (\n\"fmt\"\n)\n\nfunc simple(a func(a, b int) int) {\nfmt.Println(a(60, 7))\n}\n\nfunc main() {\nf := func(a, b int) int {\nreturn a + b\n}\nsimple(f)\n}\n``````\n\nRun in playground\n\nIn the above example, in line no. 7 we define a function `simple` which takes a function which accepts two int arguments and returns a int as a parameter. Inside the main function in line no. 12 we create a anonymous function `f` whose signature matches the parameter of the function `simple`. We call `simple` and pass `f` as an argument to it in the next line. This program prints `67` as output.\n\n#### Returning functions from other functions\n\nNow let's rewrite the program above and return a function from the `simple` function.\n\n``````package main\n\nimport (\n\"fmt\"\n)\n\nfunc simple() func(a, b int) int {\nf := func(a, b int) int {\nreturn a + b\n}\nreturn f\n}\n\nfunc main() {\ns := simple()\nfmt.Println(s(60, 7))\n}\n``````\n\nRun in playground\n\nIn the program above, the simple function in line no 7 returns a function that takes two `int` arguments and returns a `int` argument.\n\nThis simple function is called from line no. 15. The return value from simple is assigned to `s`. Now `s` contains the function returned by `simple` function. We call `s` and pass it two int arguments in line no. 16. This program outputs `67`.\n\n### Closures\n\nClosures are a special case of anonymous functions. Closures are anonymous functions which access the variables defined outside the body of the function.\n\nAn example will make things more clear.\n\n``````package main\n\nimport (\n\"fmt\"\n)\n\nfunc main() {\na := 5\nfunc() {\nfmt.Println(\"a =\", a)\n}()\n}\n``````\n\nRun in playground\n\nIn the program above, the anonymous function accesses the variable `a` which is present outside its body in line no. 10. Hence this anonymous function is a closure.\n\nEvery closure is bound to its own surrounding variable. Let's understand what this means by using a simple example.\n\n``````package main\n\nimport (\n\"fmt\"\n)\n\nfunc appendStr() func(string) string {\nt := \"Hello\"\nc := func(b string) string {\nt = t + \" \" + b\nreturn t\n}\nreturn c\n}\n\nfunc main() {\na := appendStr()\nb := appendStr()\nfmt.Println(a(\"World\"))\nfmt.Println(b(\"Everyone\"))\n\nfmt.Println(a(\"Gopher\"))\nfmt.Println(b(\"!\"))\n}\n``````\n\nRun in background\n\nIn the program above, the function `appendStr` returns a closure. This closure is bound to the variable `t`. Let's understand what this means.\n\nThe variables `a` and `b` declared in line nos. 17, 18 are closures and they are bound to their own value of `t`.\n\nWe first call `a` with the parameter `World`. Now the value of `a`'s version of `t` becomes `Hello World`.\n\nIn line no. 20 we call `b` with the parameter `Everyone`. Since `b` is bound to its own variable `t`, `b`'s version of `t` has a initial value of `Hello` again. Hence after this function call, the value of `b`'s version of t becomes `Hello Everyone`. The rest of the program is self explanatory.\n\nThis program will print,\n\n``````Hello World\nHello Everyone\nHello World Gopher\nHello Everyone !\n``````\n\n### Practical use of first class functions\n\nTill now we have defined what first class functions are and we have seen a few contrived examples to learn how they work. Now lets write a concrete program which shows the practical use of first class functions.\n\nWe will create a program which filters a slice of students based on some criteria. Let's approach this step by step.\n\nFirst lets define the student type.\n\n``````type student struct {\nfirstName string\nlastName string\ncountry string\n}\n``````\n\nThe next step is to write the `filter` function. This function takes a slice of students and a function which determines whether a student matches the filtration criteria as parameters. We will understand better once we write this function. Let's go ahead and do it.\n\n``````func filter(s []student, f func(student) bool) []student {\nvar r []student\nfor _, v := range s {\nif f(v) == true {\nr = append(r, v)\n}\n}\nreturn r\n}\n``````\n\nIn the above function, the second parameter to `filter` is a function which takes a `student` as parameter and returns a `bool`. This function determines whether a particular student matches a criteria or not. We iterate through the student slice in line no. 3 and and we pass each student as parameter to the function `f`. If this returns `true`, it means that that the student has passed the filter criteria and he is added to the result slice `r`. You might be a little confused about the real use of this function, but it will be clear once we complete the program. I have added the main function and have provided the full program below.\n\n``````package main\n\nimport (\n\"fmt\"\n)\n\ntype student struct {\nfirstName string\nlastName string\ncountry string\n}\n\nfunc filter(s []student, f func(student) bool) []student {\nvar r []student\nfor _, v := range s {\nif f(v) == true {\nr = append(r, v)\n}\n}\nreturn r\n}\n\nfunc main() {\ns1 := student{\nfirstName: \"Naveen\",\nlastName: \"Ramanathan\",\ncountry: \"India\",\n}\ns2 := student{\nfirstName: \"Samuel\",\nlastName: \"Johnson\",\ncountry: \"USA\",\n}\ns := []student{s1, s2}\nf := filter(s, func(s student) bool {\nreturn true\n}\nreturn false\n})\nfmt.Println(f)\n}\n``````\n\nRun in playground\n\nIn the main function, we first create two students `s1` and `s2` and add them to slice `s`. Now let's say we want to find out all students who have grade `B`. We have established this in the above program by passing a function which checks whether the student has grade `B` and if so then returning true, as parameter to the `filter` function in line no. 38. The above program will print,\n\n``````[{Samuel Johnson B USA}]\n``````\n\nLet's say we want to find all students from India. This can be done easily by changing the function parameter to the filter function.\nI have provided code that does this below,\n\n``````c := filter(s, func(s student) bool {\nif s.country == \"India\" {\nreturn true\n}\nreturn false\n})\nfmt.Println(c)\n``````\n\nLet's conclude this section by writing one more program. This program will perform the same operations on each element of a slice and return the result. For example if we want to multiply all integers in a slice by 5 and return the output, it can be easily done using first class functions. These kind of functions which operate on every element of a collection are called `map` functions. I have provided the program below. It is self explanatory.\n\n``````package main\n\nimport (\n\"fmt\"\n)\n\nfunc iMap(s []int, f func(int) int) []int {\nvar r []int\nfor _, v := range s {\nr = append(r, f(v))\n}\nreturn r\n}\nfunc main() {\na := []int{5, 6, 7, 8, 9}\nr := iMap(a, func(n int) int {\nreturn n * 5\n})\nfmt.Println(r)\n}\n``````\n\nRun in playground\n\nThe above program will print,\n\n``````[25 30 35 40 45]\n``````\n\nHere's a quick recap of what we learnt in this tutorial,\n\n• What are first class functions?\n• Anonymous functions\n• User defined function types\n• Higher-order functions\n• Passing functions as arguments to other functions\n• Returning functions from other functions\n• Closures\n• Practical use of first class functions\n\nThat's about it for first class functions. Have a good day.\n\nPlease show your support by donating on Patreon. Your donations will help me create more awesome tutorials.", null, "Next tutorial - Reflection" ]

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[ "## Algebra 1\n\nTo find the mean, we add up all of the values and divide by the total number of values. Mean A: $\\frac{0+12+7+19+21}{5} = 11.8$ Mean B: $\\frac{13+16+15+17+12}{5} = 14.6$ To find the range, we subtract the smallest term from the largest term. Range A: $21-0 = 21$ Range B: $17-12=5$ The mean of Set B is greater than the mean of Set A, but the range of Set A is greater than the range of Set B." ]

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[ "Solutions by everydaycalculation.com\n\n## Divide 5/84 with 25/81\n\n5/84 ÷ 25/81 is 27/140.\n\n#### Steps for dividing fractions\n\n1. Find the reciprocal of the divisor\nReciprocal of 25/81: 81/25\n2. Now, multiply it with the dividend\nSo, 5/84 ÷ 25/81 = 5/84 × 81/25\n3. = 5 × 81/84 × 25 = 405/2100\n4. After reducing the fraction, the answer is 27/140\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 ]

[ "# This note has been used to help create the Factors wiki\n\n## Definition\n\nWhat is a divisor of a number $N$? We say that $m$ is a divisor of $N$, if there exists an integer $k$ such that $N = km$.\n\nIn general, the divisors of a number refer to the positive divisors, unless otherwise noted. Since the negative divisors will be the negative of a positive divisor (and vice versa), we shall just consider positive divisors.\n\nWe also tend to ignore 0 the possibility for any of these numbers to be 0. Since $0 = 0 \\times m$ our definition above gives us that every integer is a divisor of $0$.\n\n## Technique\n\nLet the integer $N$ have a prime factorization $p_1 ^{q_1} p_2^{q_2} \\ldots p_n ^ {q_n}$, where $p_1, p_2 \\ldots , p_n$ are distinct prime numbers, and $q_1, q_2 \\ldots q_n$ are positive integers. Let $d$ be a divisor of $N$; then any prime factor $p$ that divides $d$ must divide $N$, hence it must be one of $p_1, p_2, \\ldots , p_n$.\n\nWithout loss of generality, set $p = p_i$. Let the highest power of $p$ that divides $d$ be $p_i^{r_i}$. Then, $p_i^{r_i}$ divides $d$, which in turn divides $N$, hence, $p_i^{r_i}$ must divide $p_i ^{q_i}$, which means that $r_i \\leq q_i$. Thus, by considering all the prime factors of $d$, we get that it must have the form $d = p_1^{r_1} p_2^{r_2} \\ldots p_n^{r_n}$ where $0 \\leq r_i \\leq q_i$ for all $i$.\n\nConversely, given a number $d$ that has the form $d = p_1^{r_1} p_2^{r_2} \\ldots p_n^{r_n}$ where $0 \\leq r_i \\leq q_i$ for all $i,$ it is clear that $d$ is a divisor of $N$. As such, we have a complete classification of all the divisors.\n\n### Divisor Function\n\nHow many divisors does the number $N$ have? From the above classification, we can set up a direct bijection between $d = p_1^{r_1} p_2^{r_2} \\ldots p_n^{r_n}$ and sets of $n$ integers $( r_1, r_2, \\ldots r_n )$ that satisfy $0 \\leq r_i \\leq q_i$. For each $r_i$, there are $q_i - 0 + 1 = q_i + 1$ possibilities. Hence, by the product rule, there are going to be $(q_1 +1) (q_2 +1) \\ldots (q_n + 1)$ divisors in all. The number of divisors of an integer $N$ is often denoted as the $\\phi (N)$ or $\\sigma_0 (N)$.\n\n### How many divisors does the number $2000$ have?\n\nWe have $2000 = 2^4 5^3$. Hence, from the above discussion, it has $(4+1)(3+1)=20$ divisors.\n\nWe can list them out as 1, 2, 4, 8, 16, 5, 10, 20, 40, 80, 25, 50, 100, 200, 400, 125, 250, 500, 1000, 2000.\n\n(Can you see why we listed out the divisors this way, instead of in increasing order?)\n\n### What is the sum of all divisors of the number $2000$?\n\nConsider the product $(1+2+2^2+2^3+2^4)(1+5+5^2+5^3)$ when expanded out. From the classification of the divisors, each divisor would appear exactly once as a term. Moreover, every term would be a divisor of the number $2000$. Hence the product represents the sum of all the divisors of the number $2000$, which is $31\\times 156 = 4836$.\n\n(Pop quiz: How would you generalize this to find the sum of all divisors of the number $N$? It is sometimes denoted as $\\sigma (N)$ or $\\sigma_1(N)$.)\n\n## Application and Extensions\n\n### Show that an integer $N$ has an odd number of divisors if and only if it is a perfect square.\n\nSince $\\phi(N) = (q_1 +1)(q_2+1) \\ldots (q_n+1)$, this product is odd if and only if every term is odd, which happens if and only if every value $q_i$ is even, which happens if and only if $N$ is a perfect square.\n\n### What is the smallest integer $N$ that has exactly 14 divisors?\n\nSince $14 = 2 \\cdot 7$, an integer has exactly 14 divisors if it has the form $p^{13}$ or $p_1 \\cdot p_2 ^6$. The smallest number in the first case and second case are, $2^{13} = 8192$ and $3 \\cdot 2^6 = 192$, respectively. Hence 192 is the smallest integer that has exactly 14 divisors.", null, "Note by Calvin Lin\n5 years, 10 months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n• bulleted\n• list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nAs a fact, 0 divides 0, according to the definition, but the result is undefined, as every integer becomes 0 when multiplied by 0.\n\n- 3 years, 8 months ago\n\nA Number has exactly 13 divisors.What can we say about the number?\n\n- 1 year, 11 months ago\n\nIt is a perfect square\n\n- 1 year, 9 months ago\n\nThe Number 4096 has 13 posirive divisors exactly.4096 is a ecen composite number and it is also called an deficiency number.Because the sum of its proper divisors is 4095.Then 4096-4095 =1.The remainder is 1 so that it is a deficiency number\n\n- 1 year, 9 months ago\n\n13 is a prime number. Its only divisor is itself. Therefore integer n has 13 divisors if and only if it is of the form p^(12).\n\nThe smallest such positive integer n is 2^12 = 4096.\n\nAlso, since 13 is a prime number > 2, it is odd, n has an odd number of divisors, and n must be a perfect square. p^12 = (p^6)^2.\n\n- 1 year, 5 months ago\n\nI don't understand the part of p1xp2^6. Huhu\n\n- 5 years, 6 months ago\n\nIt is $p^{13}.. OR.. p_{1}^{2 - 1} \\times p_{2}^{7 - 1} ..... 14 ...implies.. p^{14 - 1}.. OR.... 14 = 2\\times 7.. .implies... p_{1}^{2 - 1} \\times p_{2}^{7 - 1}$\n14 are number of factors, including 1 and the number. Hope this might be useful.\n\n- 5 years, 6 months ago\n\nYes, it is. Thank you so much! :D\n\n- 5 years, 6 months ago\n\nThank you so much\n\n- 5 years, 6 months ago\n\nBrilliant sir\n\n- 4 years, 7 months ago" ]

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[ "Find palindromes in two consecutive number bases?\n\nUpdate: So far I've been able to find (I believe all) solutions for $$d=3,5,7$$, and partially for $$d=9$$. If It turns out that a closed form for all solutions for a fixed (large enough) $$d$$ is too messy to compile, I'll probably include those as a partial self-answer here.\n\nWhat is allowing me to solve and find closed forms for a fixed $$d$$ case now, is solving the related system with a CAS like Mathematica - and is now summarized in the answer below.\n\nQuestion\n\nCan we generate all numbers which are palindromic in two consecutive number bases $$(b, b+1)$$ and have $$(2d+1, d\\in\\mathbb N)$$ digits when written in their palindromic bases?\n\n$$3$$ digit pattern\n\nI've found the pattern for $$d=1$$, the $$3$$ digit palindromes rather straightforward (simple):\n(Just look at the examples below and you'll see it)\n\nAll solutions that are palindromic in number bases $$b\\in\\{n+1, n+2\\}$$ for $$n\\gt3$$, are given by the following equation: (pick all $$k$$ in range to get a solution per value of $$k$$)\n\n$$P_3(n) = \\begin{cases} n^3+2n^2+2n, & \\text{k=1} \\\\ n^3-n^2(k-4)-n(k-7)-3k+4, & \\text{n\\gt k \\gt1} \\end{cases}$$\n\nThat gives you $$n-1$$ solutions, which are palindromic in $$n+1,n+2$$ bases.\n\nThe digits of given examples in base $$b=n+1$$ are: $$[n-k+1], [(n-k+2) \\mod (n+1)], [n-k+1]$$\n\n$$\\text{(Examples)}$$ You can see the solutions for $$n=4,5,6$$ below:\n\n5 67 3 [2, 3, 2] [1, 5, 1]\n5 98 3 [3, 4, 3] [2, 4, 2]\n5 104 3 [4, 0, 4] [2, 5, 2]\n\n6 92 3 [2, 3, 2] [1, 6, 1]\n6 135 3 [3, 4, 3] [2, 5, 2]\n6 178 3 [4, 5, 4] [3, 4, 3]\n6 185 3 [5, 0, 5] [3, 5, 3]\n\n7 121 3 [2, 3, 2] [1, 7, 1]\n7 178 3 [3, 4, 3] [2, 6, 2]\n7 235 3 [4, 5, 4] [3, 5, 3]\n7 292 3 [5, 6, 5] [4, 4, 4]\n7 300 3 [6, 0, 6] [4, 5, 4]\n\nSame pattern holds for all $$n\\gt3$$, as observed.\n\nFirst column is the base $$b=n+1$$, second is the number, third are the digits, and last two are number representations in number bases $$(n+1,n+2)$$.\n\nThe solutions for $$n=2,3$$ which are not included in $$P_3(n)$$ are $$10, 46$$ ;\n\nWhere number bases $$(2,3)$$, the $$n=1$$ case, has no solutions.\n\n$$5,7,9\\dots$$ digit pattern?\n\nI've found patterns for $$5,7,9,\\dots$$ digits to be similar among themselves but not so simple as the three digit one, as they seem to be more unpredictable with more digits.\n\nCan we define a function / algorithm to generate all the solutions for some $$P_d(n)$$?\n\nMeaning, the solutions are palindromic in bases $$(n+1,n+2)$$ and have $$d$$ digits?\n\n$$P_3(n)$$ above is the simplest example, which generates all solutions for $$(n+1,n+2)$$ bases by iterating $$k\\in(n,1]$$. How can one find patterns for other digit cases? (see linked code below to generate examples for some $$2d+1$$ digits)\n\nUse linked code to generate solutions for any case of $$2d+1$$ digits by simply changing the digit = 4 variable to some other integer. (You can also generate from/to any number base)\n\n$$9$$ digit pattern?\n\nI'm interested in the $$9$$ digit pattern before the rest.\n\n(Since this would help search for solutions for numbers palindromic in three consecutive bases which also have $$9$$ digits; where $$3,5,7$$ digit patterns were found, but no examples for $$\\ge9$$ digits have been found. - I've posted these patterns in my question regarding numbers palindromic in four or more consecutive bases. The MSE version of the question is linked in the comments.)\n\nYou can see all solutions for bases up to $$32$$ here, which were generated by my python code.\n\nLinked code checks every palindrome in some number base to see if it is palindromic in bases $$(b, b+1)$$; but this is extremely slow at larger bases, so I need to find $$P_9(n)$$ to generate them instead.\n\nIn the linked output text, you can notice patterns in digits. For example, notice patterns emerging at base $$b=30$$, for first couple examples; highlighted here.\n\nCan you predict these patterns for some base $$b$$, at least to some degree? I couldn't.\n\n• You might be interested in this question about palindromes in 4 successive bases – Ross Millikan Nov 14 '17 at 20:15\n\nThis is a (partial) answer for $$d=3,5,7,\\ge 9$$. - I doubt there is a nice general from for all $$d$$.\n\nUpdate: The relevant Diophantine system is now asked about on Math Overflow.\n\nThe problem in terms of Diophantine equations\n\nNotation\n\nWe have palindrome in base $$b$$ digits: $$(a_1,\\dots,a_d)$$ and it holds $$a_i=a_{d-i+1}$$, we write:\n\n$$(a_1,\\dots,a_d)_b=\\sum_{i=1}^{d}a_i b^{d-1}$$\n\nThe problem: Finding palindromes in two consecutive bases for a fixed case of $$d=2l+1$$ digits.\n\nFinding $$b,b+1$$ palindromes is equivalent to $$b,b-1$$ palindromes, so lets work with the latter.\n\nWe want to find palindromes with $$d$$ digits in both bases exactly. - Thus we will exclude solutions with $$(d,d+k)$$ digits for $$k\\gt 0$$ (according to the question, we are after $$d$$ digit palindromes in both bases). This means that we:\n\nWant to find solutions for $$b,a_i,A_i$$ where $$a_{i}=a_{d-i+1},A_{i}=A_{d-i+1}$$:\n\n$$\\sum_{i=1}^{d} a_i b^{d-i}=\\sum_{i=1}^{d} A_i (b-1)^{d-i}$$\n\nAnd $$0\\le a_i\\lt b,0\\le A_i\\lt b-1,a_1\\ne 0,A_1\\ne 0$$.\n\nWe can express $$A_i$$ in terms of $$a_i$$ by expanding $$((b-1)+1)^{d-i}$$ by binomial theorem. We can also use a trick and introduce $$o_i\\in\\mathbb Z$$ parameters that are defined such that inequalities for $$A_i$$ are satisfied. We have:\n\n$$A_i=\\sum_{k=1}^{i}\\binom{d-k}{i-k}a_k + o_{i} - o_{i-1} (b-1)$$\n\nSince all palindromes are known, $$a_i$$ are chosen such that $$a_i$$ inequalities are satisfied. By going over all valid sets of $$\\{o_1,\\dots,o_d\\}$$ parameters such that the $$A_i$$ inequalities are satisfied as well, we have split the original problem into just solving a bunch of linear Diophantine equations whose order depends on $$d$$.\n\nNote: If we want to generalize this for $$k\\gt 0$$ cases, we can simply handle them by introducing additional digits $$A_i$$ and additional $$o_i$$ parameters and having:\n\n$$A_i=o_{i} - o_{i-1} (b-1)$$\n\nFor $$i\\le 0$$. And then we need to adjust the equality for palindromes:\n\n$$A_{i_A}=A_d,A_{i_A+1}=A_{d-1},\\dots \\le b-1,A_{i_A}\\ne 0$$\n\nWhere $$i_A$$ is the smallest index of the non-zero $$A_i$$ digit.\n\nSolving the above system\n\nWe are interested in palindromes in two consecutive bases with $$2d+1$$ digits in both of those bases, so we have $$k=0$$ always. This means we aren't after solutions like $$10=(1,0,1)_3=(2,2)_4$$ for example.\n\nThe above system can be partially solved using a Computer-Algebra-System, that is, a CAS, for fixed cases of $$d$$. Doing it by hand appears to be too messy to do with my current approach for $$d\\gt 3$$ (See $$d=3$$ solution for $$b,b-1,b-2$$ palindromes from this question). A manual obstacle is handling all the $$o_i$$ cases individually which get more numerous the larger the $$d$$ case, and more complex to find.\n\nBut this does have limitations, as it gets harder to solve fully for larger $$d$$ computationally.\n\nNow, lets call each valid $$\\{o_1,\\dots,o_d\\}$$ pick of parameter values a normalization case of the digits. Each valid normalization can either yield none, finitely, or infinitely many solutions.\n\nSolutions for $$d=3$$\n\nIt turns out the only $$o_i$$ normalizations that can give solutions are $$\\{1,1\\},\\{2,1\\}$$, which give the following solutions for $$x,y\\in\\mathbb N_0$$:\n\n$$\\begin{array}{} 1.^* & (1+x,4+y,1+x)_{5+x+y} \\\\ 2. & (2+x,5,2+x)_{6+x} \\end{array}$$\n\nIn base $$b,b-1$$. This gives solutions like you described and explains why your $$P_3$$ has two cases. There are two families of solutions like you can see here. For example, the smallest given is: $$46=(1,4,1)_5=(2,3,2)_4$$ for $$x=y=0$$ and from first family.\n\nNotice the $$*$$ next to the first pattern. This marks that a subuset of those solutions can be palindromic in a third consecutive base $$b-2$$.\n\nSolutions for $$d=5$$\n\nIt can be shown there are $$12$$ normalizations that yield solutions, among which $$8$$ give familie(s) of solutions, and the other $$4$$ give finitely many solutions.\n\nHere are the families of solutions based on the $$1.-8.$$ normalizations, where $$a_2,a_3$$ were expressed in terms of $$b,a_1$$, for $$x,y\\in\\mathbb N_0$$ (where $$*$$ marks that a subset of these solutions can also be palindromic in a thrid consecutive base $$b-2$$), we have $$(a_1,a_2,a_3,a_2,a_1)_b$$ given by:\n\n$$\\begin{array}{} & a_1 & a_2 & a_3 & b \\\\ 1. & 3+x+y & -2 a_1+b-1 & 3 a_1 -b+2 & 11+3x+2y \\\\ 2. & 11+2x+y&-2 a_1+2 b&3 a_1-2b+1 &17+3 x+y \\\\ 3.^* & 20+2x+y&-2 a_1+2b+1 &3a_1 -b-2 & 38+3 x+2 y \\\\ 4. & 1+x & -2 a_1+b & 3a_1-1 & 11+3x+y \\\\ 5. & 14+x &-2a_1+2b+2 & 3a_1-2b-2 & 20+x \\\\ 6. & 4+x &-2a_1+b-1&3a_1+1 &14+3x \\\\ 6. & 3+x &-2a_1+b-1&3a_1+1 &12+3x \\\\ 6. & 2+x &-2a_1+b-1&3a_1+1 &10+3x \\\\ 6. & 1+x &-2a_1+b-1&3a_1+1 &9+3x \\\\ 6. & 1+x &-2a_1+b-1&3a_1+1 &8+3x \\\\ 7. & 12+x &-2a_1+b+1&3 a_1 -b-1 &23+2 x \\\\ 7. & 10+x &-2a_1+b+1&3 a_1 -b-1 &20+2 x \\\\ 7. & 8+x &-2a_1+b+1&3 a_1 -b-1 &17+2 x \\\\ 7. & 6+x &-2a_1+b+1&3 a_1 -b-1 &14+2 x \\\\ 7. & 7+x &-2a_1+b+1&3 a_1 -b-1 &20+2 x \\\\ 7. & 6+x &-2a_1+b+1&3 a_1 -b-1 &17+2 x \\\\ 7. & 5+x &-2a_1+b+1&3 a_1 -b-1 &14+2 x \\\\ 7. & 4+x &-2a_1+b+1&3 a_1 -b-1 &11+2 x \\\\ 8. & 11+2x &-2 a_1 + 2 b& 3 a_1 -b & 17+3x \\\\ 8. & 10+2x &-2 a_1 + 2 b& 3 a_1 -b & 16+3x \\\\ 8. & 16+2x &-2 a_1 + 2 b& 3 a_1 -b & 31+3x \\\\ 8. & 15+2x &-2 a_1 + 2 b& 3 a_1 -b & 29+3x \\\\ 8. & 14+2x &-2 a_1 + 2 b& 3 a_1 -b & 27+3x \\\\ 8. & 13+2x &-2 a_1 + 2 b& 3 a_1 -b & 25+3x \\\\ 8. & 12+2x &-2 a_1 + 2 b& 3 a_1 -b & 23+3x \\\\ 8. & 11+2x &-2 a_1 + 2 b& 3 a_1 -b & 21+3x \\\\ 8. & 10+2x &-2 a_1 + 2 b& 3 a_1 -b & 19+3x \\\\ 8. & 9+2x &-2 a_1 + 2 b& 3 a_1 -b & 17+3x \\\\ 8. & 9+2x &-2 a_1 + 2 b& 3 a_1 -b & 15+3x \\\\ 8. & 8+2x &-2 a_1 + 2 b& 3 a_1 -b & 15+3x \\\\ 8. & 8+2x &-2 a_1 + 2 b& 3 a_1 -b & 14+3x \\\\ 8. & 7+2x &-2 a_1 + 2 b& 3 a_1 -b & 13+3x \\end{array}$$\n\nAnd here are solutions for normalizations $$9.-12.$$, where we have solution, base lists:\n\n9. {2293, 6, 6148, 7, 10658, 8}\n10. {344954, 14, 502179, 15, 712154, 16, 987167, 17, 1341282, 18}\n11. {3074, 6, 5854, 7, 7703, 7, 13459, 8, 16708, 8, 27310, 9, 42324, 10, 50605, 10, 75157, 11, 125430, 12, 174596, 13, 270487, 14, 526600, 16}\n12. {7801, 8, 12547, 9, 17876, 9, 27472, 10, 40504, 11, 52825, 11, 75399, 12, 93088, 12, 104549, 13, 129198, 13, 174934, 14, 208423, 14, 231874, 15, 276395, 15, 359797, 16, 460807, 17, 535336, 17, 675996, 18, 842732, 19, 1183747, 20, 1444009, 21, 2347894, 23}\n\nPartial solutions for $$d=7$$\n\nThere appear to be $$61$$ normalizations that have solutions, out of which $$24$$ give familie(s) of infinite solutions each. So the list of the closed forms will be at least as three times as long as the previous case.\n\nHere are some of those infinite families: (these were not sorted like the previous case since this digit case is not complete yet) given $$c_1,\\dots,c_6\\in\\mathbb N$$, we have $$(a_1,a_2,a_3,a_4,a_3,a_2,a_1)_b$$ given by:\n\n$$\\begin{array}{} & a_1 & a_2 & a_3 & a_4 & b \\\\ 1. & 64+4 c_1+2 c_2+c_3 & -1+3 b-3 a_1 & c_2 & 5-4 b+5 a_1-2 c_2 & 71+5 c_1+2 c_2+c_3\\\\ 2. & 63+4 c_1+3 c_2+2 c_3 & -2+2 b-3 a_1 & c_1 & 6-3 b+5 a_1-2c_1&101+6c_1+5c_2+3c_3 \\\\ 3. & 3+c_1+c_2+c_3+c_4 & -2+b-3 a_1 & c_1 & 4-b+5 a_1-2 c_1 &15+3c_1+5c_2+4c_3+3 c_4\\\\ 4. & 1+c_1 & 1+b-3 a_1& 26+5 c_1+c_2+c_3& -5+2 b+5 a_1-2 a_3 & 51+5 c_1+c_2+2 c_3 \\\\ 5. & 4+2 c_1+c_2+c_3+c_4 & -1+b-3 a_1 & 4+5 c_1+2 c_2+c_3 & 1+5 a_1-2 a_3 & 19+6 c_1+3 c_2+3 c_3+5 c_4+c_5 \\\\ 5. & 3+2 c_1+c_2+c_3+c_4 & -1+b-3 a_1 & 1+5 c_1+2 c_2+c_3 & 1+5 a_1-2 a_3 & 16+6 c_1+3 c_2+3 c_3+5 c_4+c_5 \\\\ 5. & 3+2 c_1+c_2+c_3+c_4 & -1+b-3 a_1 & 5 c_1+2 c_2+c_3 & 1+5 a_1-2 a_3 & 17+6 c_1+3 c_2+3 c_3+5 c_4+c_5 \\\\ 6. & 19+2 c_1+c_2+c_3+c_4 & b-3 a_1 & 47+5 c_1+5 c_2+4 c_3+3 c_4+c_5 & -2+b+5 a_1-2 a_3 & 65+6 c_1+5 c_2+4 c_3+3 c_4+2 c_5+c_6 \\\\ 6. & 17+2 c_1+c_2+c_3+c_4 & b-3 a_1 & 42+5 c_1+5 c_2+4 c_3+3 c_4+c_5 & -2+b+5 a_1-2 a_3 & 60+6 c_1+5 c_2+4 c_3+3 c_4+2 c_5+c_6 \\\\ 6. & 15+2 c_1+c_2+c_3+c_4 & b-3 a_1 & 37+5 c_1+5 c_2+4 c_3+3 c_4+c_5 & -2+b+5 a_1-2 a_3 & 55+6 c_1+5 c_2+4 c_3+3 c_4+2 c_5+c_6 \\\\ 6. & 13+2 c_1+c_2+c_3+c_4 & b-3 a_1 & 32+5 c_1+5 c_2+4 c_3+3 c_4+c_5 & -2+b+5 a_1-2 a_3 & 50+6 c_1+5 c_2+4 c_3+3 c_4+2 c_5+c_6 \\\\ 6. & 11+2 c_1+c_2+c_3+c_4 & b-3 a_1 & 27+5 c_1+5 c_2+4 c_3+3 c_4+c_5 & -2+b+5 a_1-2 a_3 & 45+6 c_1+5 c_2+4 c_3+3 c_4+2 c_5+c_6 \\\\ 6. & 9+2 c_1+c_2+c_3+c_4 & b-3 a_1 & 22+5 c_1+5 c_2+4 c_3+3 c_4+c_5 & -2+b+5 a_1-2 a_3 & 40+6 c_1+5 c_2+4 c_3+3 c_4+2 c_5+c_6 \\\\ 6. & 7+2 c_1+c_2+c_3+c_4 & b-3 a_1 & 17+5 c_1+5 c_2+4 c_3+3 c_4+c_5 & -2+b+5 a_1-2 a_3 & 35+6 c_1+5 c_2+4 c_3+3 c_4+2 c_5+c_6 \\\\ 6. & 5+2 c_1+c_2+c_3+c_4 & b-3 a_1 & 12+5 c_1+5 c_2+4 c_3+3 c_4+c_5 & -2+b+5 a_1-2 a_3 & 30+6 c_1+5 c_2+4 c_3+3 c_4+2 c_5+c_6 \\\\ 6. & 3+2 c_1+c_2+c_3+c_4 & b-3 a_1 & 9+5 c_1+5 c_2+4 c_3+3 c_4+c_5 & -2+b+5 a_1-2 a_3 & 27+6 c_1+5 c_2+4 c_3+3 c_4+2 c_5+c_6 \\\\ 6. & 3+2 c_1+c_2+c_3+c_4 & b-3 a_1 & 8+5 c_1+5 c_2+4 c_3+3 c_4+c_5 & -2+b+5 a_1-2 a_3 & 26+6 c_1+5 c_2+4 c_3+3 c_4+2 c_5+c_6 \\\\ 6. & 3+2 c_1+c_2+c_3+c_4 & b-3 a_1 & 7+5 c_1+5 c_2+4 c_3+3 c_4+c_5 & -2+b+5 a_1-2 a_3 & 25+6 c_1+5 c_2+4 c_3+3 c_4+2 c_5+c_6 \\\\ 6. & 2+2 c_1+c_2+c_3+c_4 & b-3 a_1 & 5+5 c_1+5 c_2+4 c_3+3 c_4+c_5 & -2+b+5 a_1-2 a_3 & 24+6 c_1+5 c_2+4 c_3+3 c_4+2 c_5+c_6 \\\\ 6. & 1+2 c_1+c_2+c_3+c_4 & b-3 a_1 & 2+5 c_1+5 c_2+4 c_3+3 c_4+c_5 & -2+b+5 a_1-2 a_3 & 23+6 c_1+5 c_2+4 c_3+3 c_4+2 c_5+c_6 \\\\ 7. & ? & ? & ? & ? & ? \\end{array}$$ $$\\dots$$\n\nWhere the rest are not yet solved via my method. I've implemented it in Mathematica to solve the stated problem system of equations, but it is having trouble getting solutions in terms of some parameters. The $$o_i$$ sets of parameters which haven't been solved yet, and represent normalizations $$7.-24.$$, are:\n\n$$\\begin{array}{} & \\{o_1,o_2,o_3,o_4,o_5,o_6\\} \\\\ 7. & \\{1, 6, 14, 17, 12, 4\\} \\\\ 8. & \\{1, 6, 12, 14, 9, 3\\} \\\\ 9. & \\{2, 5, 10, 10, 7, 2\\} \\\\ 10. & \\{2, 5, 8, 7, 4, 1\\} \\\\ 11. & \\{3, 10, 17, 17, 10, 3\\} \\\\ 12. &\\{3, 10, 19, 20, 13, 4\\} \\\\ 13. &\\{3, 10, 21, 23, 16, 5\\} \\\\ 14. &\\{3, 11, 20, 21, 13, 4\\} \\\\ 15. &\\{3, 10, 15, 14, 7, 2\\} \\\\ 16. &\\{3, 11, 18, 18, 10, 3\\} \\\\ 17. &\\{4, 10, 14, 11, 5, 1\\} \\\\ 18. &\\{4, 10, 16, 14, 8, 2\\} \\\\ 19. &\\{5, 15, 23, 21, 11, 3\\} \\\\ 20. &\\{5, 15, 25, 24, 14, 4\\} \\\\ 21. &\\{5, 15, 27, 27, 17, 5\\} \\\\ 22. &\\{5, 16, 26, 25, 14, 4\\} \\\\ 23. &\\{5, 16, 24, 22, 11, 3\\} \\\\ 24. &\\{6, 15, 24, 21, 12, 3\\} \\end{array}$$\n\nFor some reason I was unable to directly get an output in terms of $$c_i$$ constants for these unsolved normalizations. It is still possible to get solutions from unsolved cases for fixed bases $$b$$ at a time, and then extract the patterns and families from the solutions. - The difference being, we then can't claim these are all and proven solutions like those in $$d=3,5$$ cases, unless we also manually prove them afterwards.\n\nThe $$25.-61.$$ normalization cases give finitely many solutions. This list of solutions is too long to include here, so it will be accessible on an external source.\n\nSolving for $$d\\ge 9$$\n\nUsing the system presented, it should be possible to find all closed forms for a fixed case of $$d$$ digits like I have done for first three.\n\nThe case $$d=9$$ has at least $$443$$ normalizations and at least $$77$$ of those give familie(s) of infinitely many solutions. (At least - I haven't yet computed all the normalizations).\n\nThe larger cases have much more. But searching for all solutions for fixed $$d\\ge 9$$ does not seem practical as there does not appear to be a nice closed form to encapsulate them all - and they get very hard to fully normalize in all possible ways using my current computational method of the mentioned problem system.\n\nA solution to this would be to find an efficient algorithm to compute these patterns and then store and access them computationally?\n\nSo far, we can notice patterns in these families:\n\n$$\\begin{array}{|cccccccc} \\hline d & a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & \\dots \\\\ \\hline 1 & \\color{blue}{a_1} & 0 & 0 & 0 & 0 & 0 & \\dots \\\\ 3 & c_0 & \\color{blue}{-1a_1+c} & 0 & 0 & 0 & 0 & \\dots \\\\ 5 & c_0 & -2a_1+c & \\color{blue}{3a_1+c} & 0 & 0 & 0 & \\dots \\\\ 7 & c_0 & -3a_1+c & c_0 & \\color{blue}{5a_1-2a_3+c} & 0 & 0 & \\dots \\\\ 9 & c_0 & -4a_1+c & c_0 & 14a_1-3a_3+c & \\color{blue}{-21a_1+4a_3+c} & 0 & \\dots \\\\ 11 & ? & ? & ? & ? & ? & \\color{blue}{?} & \\dots \\\\ \\dots & \\dots & \\dots & \\dots & \\dots & \\dots & \\dots & \\color{blue}{\\dots} \\\\ \\end{array}$$\n\nWhere $$c_0$$ is an expression depending on some free parameters $$c_i\\in\\mathbb N$$, and $$c$$ is an expression depending on $$c_0$$ and base $$b$$.\n\nMaybe finding these patterns for all $$d$$ can help solve the problem of finding such an efficient algorithm.\n\nAlternate approach\n\nPerhaps closed forms for fixed $$d$$ are not realistically possible to tabulate all. Alternate ways of finding more palindromes in two consecutive bases that might not belong to these solutions is to try to find isolated patterns across cases of $$d$$ which is no longer fixed:" ]

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[ "min to year\n Minutes min Years year\n\n# 52 Minutes to Years (min to y)\n\nUse 52 minutes to years converter to find how many year in 52 minutes. To convert 52 minutes to years enter the time in minutes and get the value converted into years.\n\n## Formula to convert 52 minutes to years:\n\nyr = min ÷ 525600\n\nWhere,\n\nyr = Time in years and,\n\nmin = Time in minutes\n\n## How to convert minutes to years? (min into yr)\n\nWhen you are searching for minutes to years, you are indirectly searching for min to yr.\n\nTo convert min into years use above conversion formula.\n\n01 minute is equal to 1.9026e-6 years. (i.e 1 minute = 1.9026e-6 years).\n\nTo convert your minutes to year, divide minutes by 525600 to get the result.\n\nBelow we will show you how to convert minutes in years.\n\n### Convert 52 minutes to years\n\n01 minute is equal to 1.9026e-6 years.\n\nTo convert 52 min in year, divide minutes by 525600 to get the result.\n\nyr = (52 min ÷ 525600)\n\nTherefore, the answer to 52 minutes to years is 9.89346e-5 years, which can be written as follows:\n52 minutes = 9.89346e-5 years" ]

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[ "Chart-1.9.3: A library for generating 2D Charts and Plots\n\nGraphics.Rendering.Chart.Plot.Hidden\n\nDescription\n\nPlots that don't show, but occupy space so as to effect axis scaling\n\nSynopsis\n\n# Documentation\n\ndata PlotHidden x y #\n\nValue defining some hidden x and y values. The values are not displayed, but they still affect axis scaling.\n\nConstructors\n\n PlotHidden Fields_plot_hidden_x_values :: [x] _plot_hidden_y_values :: [y]\nInstances\n # Instance detailsDefined in Graphics.Rendering.Chart.Plot.Hidden MethodstoPlot :: PlotHidden x y -> Plot x y #\n\nplot_hidden_x_values :: forall x y x. Lens (PlotHidden x y) (PlotHidden x y) [x] [x] #\n\nplot_hidden_y_values :: forall x y y. Lens (PlotHidden x y) (PlotHidden x y) [y] [y] #" ]

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[ "# R-beta: formula() and model formulae\n\nRoss Ihaka ihaka at stat.auckland.ac.nz\nFri May 9 00:42:30 CEST 1997\n\n```Bill Venables writes:\n> There is an anomaly in the way : and ^ terms are handled in the\n> sense that the logical and useful thing is obvious but does not\n> happen. Let me give an example. Suppose a and b are factors, x\n> and y are not.\n>\n> A term such as (a + b + x + y)^2 should be expanded out binomial\n> fashion, coefficients stripped away and the remaining products\n> treated as : products. Then S copes with terms like a:a, a:b and\n> a:x fine, even x:y is handled by having it generate a column of\n> xy-products, as it should.\n>\n> But a term such as x:x does not generate a column of x-squares,\n> it is merely removed as it would be if it were a factor. This is\n> a complete anomaly, and one that I don't think would be hard or\n> dangerous for R to rectify. Indeed it would be very useful to\n> generate a complete second degree regression in three variables\n> using y ~ (1 + x1 + x2 + x3)^2. As it is now it generates linear\n> and product terms only and omits the powers. Go figure.\n\nI agree that this is a problem (its certainly bitten me), however\nthere is problem in implementing things this way. All the model\nmanipulations are carried out by the \"terms\" function which only sees\na formula and is blissfully unaware of whether variables are factors\nor numeric. It then has no basis for deciding between\na:a -> a\t(factors)\nx:x -> x^2\t(variables)\nI supose that we could make \"terms\" a bit more context aware, but\nsometimes it's useful to use it for its purely symbolic effect.\nRoss\n=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=\nr-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html\nSend \"info\", \"help\", or \"[un]subscribe\"\n(in the \"body\", not the subject !) To: r-help-request at stat.math.ethz.ch\n=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=\n\n```" ]

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[ "# #084941 Color Information\n\nIn a RGB color space, hex #084941 is composed of 3.1% red, 28.6% green and 25.5% blue. Whereas in a CMYK color space, it is composed of 89% cyan, 0% magenta, 11% yellow and 71.4% black. It has a hue angle of 172.6 degrees, a saturation of 80.2% and a lightness of 15.9%. #084941 color hex could be obtained by blending #109282 with #000000. Closest websafe color is: #003333.\n\n• R 3\n• G 29\n• B 25\nRGB color chart\n• C 89\n• M 0\n• Y 11\n• K 71\nCMYK color chart\n\n#084941 color description : Very dark cyan.\n\n# #084941 Color Conversion\n\nThe hexadecimal color #084941 has RGB values of R:8, G:73, B:65 and CMYK values of C:0.89, M:0, Y:0.11, K:0.71. Its decimal value is 543041.\n\nHex triplet RGB Decimal 084941 `#084941` 8, 73, 65 `rgb(8,73,65)` 3.1, 28.6, 25.5 `rgb(3.1%,28.6%,25.5%)` 89, 0, 11, 71 172.6°, 80.2, 15.9 `hsl(172.6,80.2%,15.9%)` 172.6°, 89, 28.6 003333 `#003333`\nCIE-LAB 27.292, -21.268, -0.711 3.436, 5.198, 5.823 0.238, 0.36, 5.198 27.292, 21.28, 181.913 27.292, -20.868, 1.705 22.799, -12.993, 0.817 00001000, 01001001, 01000001\n\n# Color Schemes with #084941\n\n• #084941\n``#084941` `rgb(8,73,65)``\n• #490810\n``#490810` `rgb(73,8,16)``\nComplementary Color\n• #084921\n``#084921` `rgb(8,73,33)``\n• #084941\n``#084941` `rgb(8,73,65)``\n• #083149\n``#083149` `rgb(8,49,73)``\nAnalogous Color\n• #492108\n``#492108` `rgb(73,33,8)``\n• #084941\n``#084941` `rgb(8,73,65)``\n• #490831\n``#490831` `rgb(73,8,49)``\nSplit Complementary Color\n• #494108\n``#494108` `rgb(73,65,8)``\n• #084941\n``#084941` `rgb(8,73,65)``\n• #410849\n``#410849` `rgb(65,8,73)``\n• #104908\n``#104908` `rgb(16,73,8)``\n• #084941\n``#084941` `rgb(8,73,65)``\n• #410849\n``#410849` `rgb(65,8,73)``\n• #490810\n``#490810` `rgb(73,8,16)``\n• #000404\n``#000404` `rgb(0,4,4)``\n• #031b18\n``#031b18` `rgb(3,27,24)``\n• #05322d\n``#05322d` `rgb(5,50,45)``\n• #084941\n``#084941` `rgb(8,73,65)``\n• #0b6055\n``#0b6055` `rgb(11,96,85)``\n• #0d776a\n``#0d776a` `rgb(13,119,106)``\n• #108e7e\n``#108e7e` `rgb(16,142,126)``\nMonochromatic Color\n\n# Alternatives to #084941\n\nBelow, you can see some colors close to #084941. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #084931\n``#084931` `rgb(8,73,49)``\n• #084936\n``#084936` `rgb(8,73,54)``\n• #08493c\n``#08493c` `rgb(8,73,60)``\n• #084941\n``#084941` `rgb(8,73,65)``\n• #084946\n``#084946` `rgb(8,73,70)``\n• #084649\n``#084649` `rgb(8,70,73)``\n• #084149\n``#084149` `rgb(8,65,73)``\nSimilar Colors\n\n# #084941 Preview\n\nThis text has a font color of #084941.\n\n``<span style=\"color:#084941;\">Text here</span>``\n#084941 background color\n\nThis paragraph has a background color of #084941.\n\n``<p style=\"background-color:#084941;\">Content here</p>``\n#084941 border color\n\nThis element has a border color of #084941.\n\n``<div style=\"border:1px solid #084941;\">Content here</div>``\nCSS codes\n``.text {color:#084941;}``\n``.background {background-color:#084941;}``\n``.border {border:1px solid #084941;}``\n\n# Shades and Tints of #084941\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, #000202 is the darkest color, while #f0fdfc is the lightest one.\n\n• #000202\n``#000202` `rgb(0,2,2)``\n• #021412\n``#021412` `rgb(2,20,18)``\n• #042622\n``#042622` `rgb(4,38,34)``\n• #063731\n``#063731` `rgb(6,55,49)``\n• #084941\n``#084941` `rgb(8,73,65)``\n• #0a5b51\n``#0a5b51` `rgb(10,91,81)``\n• #0c6c60\n``#0c6c60` `rgb(12,108,96)``\n• #0e7e70\n``#0e7e70` `rgb(14,126,112)``\n• #109080\n``#109080` `rgb(16,144,128)``\n• #12a190\n``#12a190` `rgb(18,161,144)``\n• #14b39f\n``#14b39f` `rgb(20,179,159)``\n• #16c5af\n``#16c5af` `rgb(22,197,175)``\n• #17d6bf\n``#17d6bf` `rgb(23,214,191)``\n• #1be6cd\n``#1be6cd` `rgb(27,230,205)``\n• #2de8d1\n``#2de8d1` `rgb(45,232,209)``\n``#3fead5` `rgb(63,234,213)``\n• #51ecd9\n``#51ecd9` `rgb(81,236,217)``\n• #62eedd\n``#62eedd` `rgb(98,238,221)``\n• #74f0e1\n``#74f0e1` `rgb(116,240,225)``\n• #86f2e4\n``#86f2e4` `rgb(134,242,228)``\n• #97f4e8\n``#97f4e8` `rgb(151,244,232)``\n• #a9f6ec\n``#a9f6ec` `rgb(169,246,236)``\n• #bbf8f0\n``#bbf8f0` `rgb(187,248,240)``\n• #ccf9f4\n``#ccf9f4` `rgb(204,249,244)``\n• #defbf8\n``#defbf8` `rgb(222,251,248)``\n• #f0fdfc\n``#f0fdfc` `rgb(240,253,252)``\nTint Color Variation\n\n# Tones of #084941\n\nA tone is produced by adding gray to any pure hue. In this case, #272a2a is the less saturated color, while #024f46 is the most saturated one.\n\n• #272a2a\n``#272a2a` `rgb(39,42,42)``\n• #242d2c\n``#242d2c` `rgb(36,45,44)``\n• #21302e\n``#21302e` `rgb(33,48,46)``\n• #1e3331\n``#1e3331` `rgb(30,51,49)``\n• #1b3633\n``#1b3633` `rgb(27,54,51)``\n• #183935\n``#183935` `rgb(24,57,53)``\n• #143d38\n``#143d38` `rgb(20,61,56)``\n• #11403a\n``#11403a` `rgb(17,64,58)``\n• #0e433c\n``#0e433c` `rgb(14,67,60)``\n• #0b463f\n``#0b463f` `rgb(11,70,63)``\n• #084941\n``#084941` `rgb(8,73,65)``\n• #054c43\n``#054c43` `rgb(5,76,67)``\n• #024f46\n``#024f46` `rgb(2,79,70)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #084941 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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[ "# Is the result true when the valuation is trivial?\n\nHere I proved the following result:\n\nProposition: Let $(K,|\\cdot|)$ be a valued field with non-trivial valuation and let $X$ be a vector space over $(K,|\\cdot|)$. Two norms $p_1,p_2$ on $X$ are equivalent (i.e., they induce the same topology) iff there are constants $c_1$ and $c_2$ such that $c_1p_1\\leq p_2\\leq c_2p_1$\n\nIs the proposition true when the valuation is trivial? (The valuation $|\\cdot|$ is trivial when $|x|=1$, for each $x\\in K^*$)\n\nHere's a counterexample. Let $K$ be any field with the trivial valuation and let $X$ be an infinite-dimensional vector space with basis $B$. We can define one norm $p_1$ on $X$ by $p_1(x)=1$ for all nonzero $x\\in X$. We can define another norm $p_2$ by saying $p_2(x)$ is the number of nonzero coefficients there are when we write $x$ as a linear combination of elements of $B$. Then both norms induce the discrete topology, but $p_2/p_1$ is unbounded." ]

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[ "# Partitions and Orbit Sizes\n\nIf $U,V \\subset S_n$ are subgroups with $S_n//U = \\{id,g_2,...,g_e\\}$ and $\\alpha_j$ is $\\frac{1}{j}$ times the number of $i\\in [e]$ s.t $[V:V \\cap g_i U g_i^{-1}]=j$ then $(\\alpha_1,...,\\alpha_e)$ is a partition of $e$.\n\nNow let $U=V \\in \\{V_4,D_4\\}$ and $n=4$ ($V_4$ is the group with only the double transpositions). Then the number of $V_4$-orbits of $S_4/V_4$ of size $1$ is $6$, and there are none of greater size. One $D_4$-orbit of $S_4/D_4 = \\{D_4,(12)D_4,(14)D_4 \\}$ is $\\{D_4\\}$; for the other two, we have $|orb_{D_4}((ab)D_4)|=[D_4 : D_4 \\cap (ab)D_4(ab)]=[D_4:V]=2$, so there is one orbit of size $1$ and two of size $2$.\n\nI don't understand what I've done wrong here. I got a partition of $[S_4 :V_4]$, $1^6$, but apparently something went wrong with $D_4$, but I have no idea why? In the article which all this stuff pertains to it is asserted that there is one orbit of size one and one of size two, when $D_4$ acts on $S_4/D_4$. It cannot be that the two orbits should be considered as one, for if so then there would only be one orbit of size one wrt $S_4/V_4$.\n\n(This is one of those questions that's difficult to pose because there's too much background, but I think this info should do, hopefully. Otherwise, if you're feeling more helpful than usual, check out beginning of section 3 and Table 2 here.)\n\nEdit: Some calculations. $orb_V(V)={V}$, trivial ($V=V_4$). $orb_V((12)V)=\\{(12)V,(34)V,(1324)V,(1423)V \\}$ and $orb_V((13)V)=\\{(13)V, (1234)V, (24)V, (1432)V\\}$. To see that each of these orbits is really just $\\{V\\}$, use $\\sigma V = \\tau V \\iff \\tau^{-1} \\sigma \\in V$. Thus we have three orbits of size one so far.\n\nWith $D_4$, we have $orb_{D_4}((12)D_4)=\\{ (12)D_4, (134)D_4, (243)D_4, (123)D_4, (142)D_4, (34)D_4, (1423)D_4, (1324)D_4\\}$ and $orb_{D_4}((14)D_4) = \\{ (14)D_4, (234)D_4, (132)D_4, (143)D_4, (124)D_4, (1342)D_4, (1243)D_4, (23)D_4 \\}$. Some manipulations show that these two orbits have size two, yet only one of them counts?\n\n• I think you are $n$-counting orbits of size $n$. If the orbit of $x$ is $\\{x,y\\}$ then the orbit of $y$ is $\\{x,y\\}$ as well, so we only count that as one orbit of size two, namely $\\{x,y\\}$, rather than two orbits of size two, $\\{x,y\\}$ from $x$ and $\\{x,y\\}$ from $y$. – Jack Schmidt Jul 15 '13 at 17:58\n• @JackSchmidt What do you mean by $n$-counting orbits of size $n$? But, if I'm supposed to count orbits like you said, then there wouldn't be six $V_4$-orbits of $S_4/V_4$ of size one. I'll make an edit to the original post to show you some computations. – Erik Vesterlund Jul 15 '13 at 18:10\n• When $n=2$, I mean you are “double counting” orbits of size 2. For orbits of size $n=1$, you are single counting, which is good. $\\{x\\}, \\{y\\}, \\{z\\}, \\{a\\}, \\{b\\}, \\{c\\}$ are six orbits of size 1. – Jack Schmidt Jul 15 '13 at 18:12\n• @Jack The article is a mess to be honest. Anyway, wrt double counting, I'm afraid I still don't understand what that means (indicative of me not entirely knowing what I'm doing, yes). As you see in the calculations I added, nothing different is done with regards to counting or anything, singleton orbits become so by the same manipulations. – Erik Vesterlund Jul 15 '13 at 18:24\n• Ok, I'll just write up the calculation and the counts as an answer. Hopefully it'll be clear. – Jack Schmidt Jul 15 '13 at 18:26\n\n## 1 Answer\n\nThere are a few unstated conventions. Which $D_4$? How do we multiply permutations? I'll go with GAP conventions for both as they are likely to agree with the paper and most of the world of computational group theory and finite group theory.\n\nTake $U=D_4 = \\{ (), (3,4), (1,2), (1,2)(3,4), (1,3)(2,4), (1,3,2,4), (1,4,2,3), (1,4)(2,3) \\}$ and multiply permutations in reading order such that $(1,2)(2,3)=(1,3,2)$.\n\n$G//U=S_4//D_4 = \\{ ()D_4, (1,2,3)D_4, (1,3,2)D_4 \\}$ so $e=3$.\n\nIf $V=D_4$ as well, then\n\n$V \\cap g_1Ug_1^{-1} = V \\cap ()U()^{-1} = V \\cap V = V$ has index 1. Note that the orbit of $D_4$ on the seed $()D_4$ is $\\{ D_4 \\}$ of size 1.\n\n$V \\cap g_2 U g_2^{-1}$ has size 4 (it is $V_4$, the Klein viergruppe of double transpositions). That means it has index 2. Note that the orbit of $D_4$ on the seed $(1,2,3)D_4$ is $\\{ (1,2,3)D_4, (1,3,2)D_4 \\}$.\n\n$V \\cap g_3 U g_3^{-1}$ has size 4 (it is $V_4$, the Klein viergruppe of double transpositions). That means it has index 2. Note that the orbit of $D_4$ on the seed $(1,3,2)D_4$ is $\\{ (1,2,3)D_4, (1,3,2)D_4 \\}$.\n\nNow we get $\\alpha_1 = \\tfrac{1}{1} 1 = 1$ since only $g_1$ gave index 1.\n\nWe get $\\alpha_2 = \\tfrac{1}{2} 2 = 1$ since $g_2$ and $g_3$ gave index 2.\n\nWe get $\\alpha_3=\\tfrac{1}{3} 0 = 0$ since nothing gave index 3.\n\nSure enough $1 \\alpha_1 + 2\\alpha_2 + 3 \\alpha_3 = 1 + 2 + 0 = 3$ is a partition.\n\nIn terms of orbits, this is because the three cosets $\\{ ()D_4, (1,2,3)D_4, (1,3,2)D_4 \\}$ split into two orbits under the left action of $D_4$, namely, $\\{ ()D_4 \\}$ and $\\{ (1,2,3)D_4, (1,3,2)D_4 \\}$.\n\n• Looks like my crucical mistake was, as you pointed out, not dividing $\\alpha_j$ by $j$. I'll look over this a few more times and try other examples before I accept it as the answer ;) Thank you! – Erik Vesterlund Jul 15 '13 at 18:44\n• No problem. A keyword for this is \"double coset\". Hall's theory of groups section 1.7 explains the relation between the index formula and the size of orbits, but I don't think it directly addresses you (and certainly doesn't say anything about galois theory). – Jack Schmidt Jul 15 '13 at 18:47" ]

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[ "Matplotlib is an amazing module which not only helps us visualize data in 2 dimensions but also in 3 dimensions. I help education companies create engaging blog and video content teaching Data Science to beginners. from matplotlib import cbook from matplotlib import cm from matplotlib.colors import LightSource import matplotlib.pyplot as plt import numpy as np # Load and format data dem = cbook. 89. matplotlib (equal unit length): with 'equal' aspect ratio z-axis is not equal to x- and y-11. I have no errors just the point will not appear. Do you wish you could be a programmer full-time but don’t know how to start? They read for hours every day---Because Readers Are Leaders! And add a subplots to that figure using the standard fig.add_subplot() method. there is a linear distance between them all. A Computer Science portal for geeks. from matplotlib import cbook from matplotlib import cm from matplotlib.colors import LightSource import matplotlib.pyplot as plt import numpy as np # Load and format data dem = cbook. In an example with one Axes, it looks a bit odd to set the Figure and Axes colors separately. All we have to use is plot_surface().. By default it will be colored in shades of a solid color, but it also supports color mapping by supplying the cmap argument. First, I imported the python random module and set the seed so that you can reproduce my results. What I found is how to turn off axes as well as the faces using ax.set_axis_off(). Click the link below before the seats fill up and learn how to become a Python freelancer, guaranteed. I want to do a 3D surface plot that shows axes but does not show the faces that are between the axes. Functionality shown: Using the text function with three types of 'zdir' values: None, an … Generate a 3D barplot. I would like to annotate individual points like the 2D case here: Matplotlib: How to put individual tags for a scatter plot. The data must be passed as xs, ys. To get some up/down movement, you need to modify the z-axis. Welcome to another 3D Matplotlib tutorial, covering how to graph a 3D scatter plot. This webinar won’t be online forever. 3D Scatter and Line Plots 3D plotting in Matplotlib starts by enabling the utility toolkit. If you plot this in IPython or an interactive Jupyter Notebook window and you rotate the plot, you will see that the transparency of each point changes as you rotate. Check out the pure value-packed webinar where Chris – creator of Finxter.com – teaches you to become a Python freelancer in 60 days or your money back! It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview … This lets you manually rotate them by clicking and dragging. 99% of Finxter material is completely free. Demonstrates using custom hillshading in a 3D surface plot. Note: if you choose a smaller number of values for np.linspace the plot is not as smooth. Next, I used three list comprehensions to create 3 x 400 samples of a normal distribution using the random.normalvariate() function. The code below creates a 3D plots and visualizes its projection on 2D contour plot: Example: In analogy with the more common two-dimensional plots discussed earlier, these can be created using the ax.plot3D and ax.scatter3D functions. A 3D Scatter Plot is a mathematical diagram, the most basic version of three-dimensional plotting used to display the properties of data as three variables of a dataset using the cartesian coordinates.To create a 3D Scatter plot, Matplotlib’s mplot3d toolkit is used to enable three dimensional plotting.Generally 3D scatter plot is created by using ax.scatter3D() the function of the matplotlib … To visualize the contour plot, we need to create a grid for data in x and y-axis, if z is a result of x and y. I used the following code based on a previous example of plotting 2d vectors but added components for 3d vectors. I have a figure where I want to plot a 3d scatter and a normal plot in 2 subplots. Setting axis labels for 3D plots is identical for 2D plots except now there is a third axis – the z-axis – you can label. Setting zdir to 'y' then plots the data to the x-z-plane. From here, we use .scatter to plot them up, 'c' to reference color and 'marker' to reference the shape of the plot marker. The following sample code utilizes the Axes3D function of matplot3d in Matplotlib. What Matplotlib does is quite literally draws your plot on the figure, then displays it when you ask it to. In general, if you ever use a ax.set_() method in matplotlib, it can be written as ax.set(=) instead. import matplotlib matplotlib.use(\"TKAGG\") import matplotlib.pyplot as pyplot import mpl_toolkits.mplot3d figure = pyplot.figure() figure.subplots_adjust(bottom=0.25, top=0.75) axes = figure.gca(projection='3d') xLabel = axes.set_xlabel('XXX xxxxxx xxxx x xx x') yLabel = axes.set_ylabel('YY (y) yyyyyy') zLabel = axes.set_zlabel('Z zzzz zzz (z)') plot = axes.plot([1,2,3],[1,2,3]) pyplot.show() Here’s an example of the power of 3D line plots utilizing all the info above. The idea of 3D scatter plots is that you can compare 3 characteristics of a data set instead of two. Amazon links open in a new tab. There are two backgrounds you can modify in matplotlib – the Figure and the Axes background. Question or problem about Python programming: I’m trying to generate a 3D scatter plot using Matplotlib. Third party packages ¶ A large number of third party packages extend and build on Matplotlib functionality, including several higher-level plotting interfaces ( seaborn , HoloViews , ggplot , ...), and a projection and mapping toolkit ( Cartopy ). Note: If you run the above code again, you will get a different looking plot. There is still more to be learned about 3D plots such as surface plots, wireframe plots, animating them and changing the aspect ratio. Join our \"Become a Python Freelancer Course\"! If you just plotted x and y now, you would get a circle. Clearly, this plot is much less smooth than the original and hopefully gives you an understanding of what is happening under the hood with these plots. In addition to import matplotlib.pyplot as plt and calling plt.show(), to create a 3D plot in matplotlib, you need to: Awesome! I’ve tried to use this function and consulted the Matplotlib docoment but found it seems that the […] Changing aspect ratio of 3D plot in matplotlib. You’ve looked at examples of line and scatter plots. After that, we do .scatter, only this time we specify 3 plot parameters, x, y, and z. Here is an example using the first method. Don’t worry if that was a bit fast, let’s dive into a more detailed example. s: scalar or array-like, optional, default: 20. A three-dimensional axes can be created by passing the keyword projection='3d' to any of the normal axes creation routines. Unlike my competitors, I learn new concepts every day and so understand what it’s like to be a student. You add legends to 3D plots in the exact same way you add legends to any other plots. The ones that are darker in color are in the foreground and those further back are more see-through. 3D Surface plotting in Python using Matplotlib Last Updated: 30-04-2020 A Surface Plot is a representation of three-dimensional dataset. Matplotlib 3D Scatter - Equal Aspect Ratio. Demonstrates using custom hillshading in a 3D surface plot. Then you need to pass projection='3d' which tells matplotlib it is a 3D plot. 3D Matplotlib scatter plot code: An Axes3D object is created just like any other axes using the projection='3d' keyword. It is important to note that Matplotlib was … 3D scatter plot is generated by using the ax.scatter3D function. In some ways 3D plots are more natural for us to work with since we live in a 3D world. The 3d plots are enabled by importing the mplot3d toolkit. 3. 0. All the usual things you can do with legends are still possible for 3D plots. It doesn’t matter if you’re a Python novice or Python pro. Now Let’s move towards 3D Plots With Python And Matplotlib. Then I called ax.set(facecolor='r') to set the Axes color to red and fig.set(facecolor='g') to set the Figure color to green. —– 3. See also Plot 2D data on 3D plot. Matplotlib 3D Plot Example. figure ax = fig. This is useful when plotting 2D data on a 3D Axes. Like line and scatter plots we can also plot surface graphs. It is possible to rotate plots and even create animations via code but that is out of the scope of this article. I created the variable theta using np.linspace which returns an array of 200 numbers between -12 and 12 that are equally spaced out i.e. Here’s a cool plot that I adapted from this video. The most basic three-dimensional plot is a line or collection of scatter plot created from sets of (x, y, z) triples. The data must be passed as xs, ys. Then I set up the Figure and Axes as normal and made my plot by calling ax.scatter(). https://tinyurl.com/become-a-python-freelancer. If you want to learn more than these basic steps, check out my comprehensive guide to legends in matplotlib. I’ve tried to use this function and consulted the Matplotlib docoment but found it seems that the […] Here’s an example where I set the Figure background color to green and the Axes background color to red. To save a static version of the plot, click the save icon. The stride arguments are … All the functions you know and love such as ax.plot() and ax.scatter() accept the same keyword arguments but they now also accept three positional arguments – X,Y and Z. This is useful when plotting 2D data on a 3D Axes. It is clear now why setting a Figure color can be more useful if you create subplots – there is more space for the color to shine through. You set up your Figure in the standard way. There are many options for doing 3D plots in python, here I will explain some of the more comon using Matplotlib. Creating a scatter plot is exactly the same as making a line plot but you call ax.scatter instead. Here is the syntax to plot the 3D Line Plot: Axes3D.plot(xs, ys, *args, **kwargs) For this plot, I set the third argument of np.linspace to 25 instead of 200. If you are not making six figures/year with Python right now, you will learn something from this webinar. The easiest way to rotate 3D plots is to have them appear in an interactive window by using the Jupyter magic command %matplotlib notebook or using IPython (which always displays plots in interactive windows). If either is 0 the input data in not sampled along this direction producing a 3D line plot rather than a wireframe plot. To save me from manually creating a brand new dataset, I thought it would be a good idea to make use of the data I already had. The 3D bar chart is quite unique, as it allows us to plot more than 3 dimensions. Active 25 days ago. After that, I used a for loop to iterate over colors and axes.flat. We can enable this toolkit by importing the mplot3d library, which comes with your standard Matplotlib installation via pip. Here z should be in 2-Dimension. To create 3d plots, we need to import axes3d. You’ve learned the necessary imports you need and also how to set up your Figure and Axes objects to be 3D. Use the label keyword argument and then call ax.legend() at the end. Matplotlib can create 3d plots. Then I created the list colors containing 4 matplotlib color strings. In this example, I used plt.subplots() to set up an 8×6 inch Figure containing four 3D Axes objects in a 2×2 grid. Firstly, because you increased the size of the Figure, all the axis labels are clearly visible. The submodule we’ll be using for plotting 3D-graphs in python is mplot3d which is already installed when you install matplotlib. comprehensive guide to legends in matplotlib, Matplotlib Histogram — A Simple Illustrated Guide, Matplotlib Legend – A Helpful Illustrated Guide. 3D graphs represent 2D inputs and 1D output. Either an array of the same length as xs and ys or a single value to make all markers the same size. 3D Plotting in Matplotlib for Python: 3D Scatter Plot Besides 3D wires, and planes, one of the most popular 3-dimensional graph types is 3D scatter plots. The marker size in points**2. Well look no further, it’s time to learn how to make 3D plots in matplotlib. Both can be set using either the .set_facecolor('color') or the .set(facecolor='color') methods. —– Matplotlib has built-in 3D plotting functionality, so doing this is a breeze. This method creates three dimensional barplot where the width, depth, height, and color of the bars can all be uniquely set. I used the following code based on a previous example of plotting 2d vectors but added components for 3d vectors. Are you tired with the same old 2D plots? Finally, I made the same plot on each Axes and set the facecolors. Graphing a 3D scatter plot is very similar to the typical scatter plot as well as the 3D wire_frame. s float or array-like, default: 20. We plotted 2D graphs with just only matplotlib but for 3D graphs we need mpl-toolkits. 3D surface Plot having 2D contour plot projections. Viewed 52k times 28. 3D Contour Plot. python python-3.x matplotlib So, I much prefer method 2. I m trying to plot simultaneously a plane and some points in 3D with Matplotlib. On the other hand, they are more complicated since we are so used to 2D plots. The first three lines are the same as a simple line plot. Ask Question Asked 4 years, 9 months ago. One thing I don’t like about method 1 is that it takes up 3 lines of code and they are boring to type. 3D line plot in matplotlib. Naturally, if you plan to draw in 3D, it'd be a good idea to let Matplotlib know this! Plus, it only took you one line of code to label them all. It describes a functional relationship between two independent variables X and Z and a designated dependent … 3D plots can seem daunting at first so my best advice is to go through the code line by line. I passed this to np.sin() and np.cos() and saved them in variables x and y. Let us cover some examples for three-dimensional plotting using this submodule in matplotlib. Even though Matplotlib was initially designed with only two-dimensional plotting in mind, some three-dimensional plotting utilities were built on top of Matplotlib's two-dimensional display in later versions, to provide a set of tools for three-dimensional data visualization. This is because you will start with the shuffled X, Y and Z lists rather than the originals you created further up inb the post. This saves you space and is nicer to type, especially if you want to make numerous modifications to the graph such as also adding a title. No, you cannot plot past the 3rd dimension, but you can plot more than 3 dimensions. We use two sample sets, each with their own X Y and Z data. In this Matplotlib tutorial, we cover the 3D bar chart. Matplotlib was initially designed with only two-dimensional plotting in mind. For more info on using plt.subplots() check out my article. Much better! One amazing feature of Jupyter Notebooks is the magic command %matplotlib notebook which, if ran at the top of your notebook, draws all your plots in an interactive window. 3D Surface plots. Is there any You’ve just created your first 3D plot! Matplotlib 3D Plot Rotate The easiest way to rotate 3D plots is to have them appear in an interactive window by using the Jupyter magic command %matplotlib notebook or using IPython (which always displays plots in interactive windows). Hopefully, you know by now that I much prefer the second method over the first! fig = plt.figure(1) ax = fig.add_subplot(211, projection='3d') ax2 = fig.add_subplot(212) then I use ax and ax2 to plot my data in the right place. Around the time of the 1.0 release, some three-dimensional plotting utilities were built on top of Matplotlib's two-dimensional display, and the result is a convenient (if somewhat limited) set of tools for three-dimensional data visualization. First, we'll need to import the Axes3D class from mpl_toolkits.mplot3d. I want to highlight to you that some of the points are darker and some are more transparent – this indicates depth. If you are not comfortable with Figure and Axes plotting notation, check out this article to help you. Matplotlib ships with several add-on toolkits, including 3D plotting with mplot3d, axes helpers in axes_grid1 and axis helpers in axisartist. Being Employed is so 2020... Don't Miss Out on the Freelancing Trend as a Python Coder! In order to iterate over colors and axes together, they need to be the same shape. We can now plot a variety of three-dimensional plot types. Create a new matplotlib.figure.Figure and add a new axes to it of type Axes3D: import matplotlib.pyplot as plt from mpl_toolkits.mplot3d import Axes3D fig = plt.figure() ax = fig.add_subplot(111, projection='3d') As this is a static blog post, all of my plots will be static but I encourage you to play around in your own Jupyter or IPython environment. You may notice that the axis labels are not particularly visible using the default settings. Matplotlib Animation – A Helpful Illustrated Guide. Matplotlib - Plot a plane and points in 3D simultaneously. Making a 3D scatterplot is very similar to creating a 2d, only some minor differences. This tutorial covers how to do just that with some simple sample data. 0. This lets you manually rotate them by clicking and dragging. Plot a 3D wireframe. You can change the orientation by clicking and dragging (right click and drag to zoom in) which can really help to understand your data. 3D surface plots plotted with Matplotlib can be projected on 2D surfaces. Basic 3D Plot With Python And Matplotlib To start with 3D plotting we need two modules first one is Matplotlib and the second one is mpl-toolkits. After that, we do .scatter, only this time we specify 3 plot parameters, x, y, and z. 3D plots are mostly used in simulation and modelling and it is important to know how to plot such Matplotlib plots if you are dealing with numerical analysis in three dimensions. Matplotlib was introduced keeping in mind, only two-dimensional plotting. 9. You can solve this by manually increasing the size of the Figure with the figsize argument in your plt.figure() call. To work with me, please reach out on Upwork If you want to master everything about 3D plots, check out part 2 of this article. Python Lists filter() vs List Comprehension – Which is Faster? So, calling ax.plot() the second time, plotted the same numbers but in a different order, thus producing a different looking plot. Become a Finxter supporter and sponsor our free programming material with 400+ free programming tutorials, our free email academy, and no third-party ads and affiliate links. In this example, I plotted the same X, Y and Z lists as in the very first example. 3D Contour Plot. Expert Writer & Content Creator – Data Science & Machine Learning. With bars, you have the starting point of the bar, the … The marker size in points**2. https://tinyurl.com/hire-adam-murphy. A quick example: To avoid repetition, I won’t explain the points I have already made above about imports and setting up the Figure and Axes objects. I use this code. Demonstrates the placement of text annotations on a 3D plot. See also Plot 2D data on 3D plot. The rstride and cstride kwargs set the stride used to sample the input data to generate the graph. Set size of matplotlib figure with 3d subplots. If you don't want to visualize this in two separate subplots, you can plot the correlation between these variables in 3D. In this tutorial, we will cover how to create a 3D Surface Plot in the matplotlib library.. These are proven, no-BS methods that get you results fast. To visualize the contour plot, we need to create a grid for data in x and y-axis, if z is a result of x and y. Question or problem about Python programming: I’m trying to generate a 3D scatter plot using Matplotlib. gca (projection = '3d') # Make data. A sample bit of code we use to make such plots is: Plot.plot_surface (X,Y,Z,rstride=1,cstride=1,color='g',linewidth=1) xyzlim = np.array ([Plot.get_xlim3d (),Plot.get_ylim3d (),Plot.get_zlim3d ()]).T XYZlim = [min (xyzlim),max (xyzlim)] Plot.set_xlim3d (XYZlim) Plot.set_ylim3d (XYZlim) Plot.set_zlim3d (XYZlim) Plot.set_aspect ('equal') In this case X,Y,Z describe points on the … In Matplotlib's mpl_toolkits.mplot3d toolkit there is axes3d present that provides the necessary functions that are very useful in creating 3D surface plots.. Bug report Bug summary set_aspect does not work for 3D surface plots Expected outcome If a sphere is drawn with plot_surface then it should appear as a sphere and not an ellipse that depends on the window sizing. Naturally, if you plan to draw in 3D, it'd be a good idea to let Matplotlib know this! Just execute the code and look at the generated “plot.png” file: If you are used to plotting with Figure and Axes notation, making 3D plots in matplotlib is almost identical to creating 2D ones. My videos are upbeat, engaging and detailed. So, I applied the random.shuffle() function to each of X, Y and Z which mixes the values of the lists in place. 1. rotating the z-axis in matplotlib 3d figure. Finally, I labeled the second plot and called ax.legend(loc='upper left') to display a legend in the upper left corner of the plot. 3D line plot in matplotlib. Become a Finxter supporter and make the world a better place: Check your Python version in Jupyter Notebook. from mpl_toolkits import mplot3d import numpy as np import matplotlib.pyplot as plt fig = plt.figure() ax = plt.axes(projection='3d') z = np.linspace(0, 1, 100) x = z * np.sin(20 * z) y = z * np.cos(20 * z) ax.plot3D(x, y, z, 'gray') ax.set_title('3D line plot') plt.show() We can now plot a variety of three-dimensional plot types. So, I used np.linspace again to create a list of 200 numbers equally spaced out between -2 and 2 which can be seen by looking at the z-axis (the vertical one). If you sample a normal distribution and create a 3D plot from it, you get a ball of points with the majority focused around the center and less and less the further from the center you go. Three-dimensional plots are enabled by importing the mplot3d toolkit, included with the Matplotlib package. from mpl_toolkits.mplot3d import Axes3D # noqa: F401 unused import import matplotlib.pyplot as plt from matplotlib import cm from matplotlib.ticker import LinearLocator, FormatStrFormatter import numpy as np fig = plt. Setting zdir to 'y' then plots the data to the x-z-plane. What Matplotlib does is quite literally draws your plot on the figure, then displays it when you ask it to. There are several ways to do this but using the .flat attribute works well in this case. If you are used to plotting with Figure and Axes notation, making 3D plots in matplotlib is almost identical to creating 2D ones. Just be sure that your Matplotlib version is over 1.0. I would like to annotate individual points like the 2D case here: Matplotlib: How to put individual tags for a scatter plot. If you have more than one Axes object, it looks much better. Keywords: matplotlib code example, codex, python plot, pyplot Gallery generated by Sphinx-Gallery But at the time when the release of 1.0 occurred, the 3d utilities were developed upon the 2d and thus, we have 3d implementation of data available today! In this example, I first set the random seed to 1 so that you can reproduce the same results as me. # This import registers the 3D projection, but is otherwise unused. If you just want a single Axes, pass 111 to indicate it’s 1 row, 1 column and you are selecting the 1st one. Check out our 10 best-selling Python books to 10x your coding productivity! Besides the standard import matplotlib.pyplot as plt, you must alsofrom mpl_toolkits.mplot3d import axes3d. I am trying to plot vectors in 3d using matplotlib. Try it yourself with our interactive Python shell. If you right-click and move the mouse, you will zoom in and out of the plot. 3D Line Plot. What have Jeff Bezos, Bill Gates, and Warren Buffett in common? After importing this sub-module, 3D plots can be created by passing the keyword projection=\"3d\" to any of the regular axes creation functions in Matplotlib. From now on everything is (almost) the same as 2D plotting. This can be created using the ax.plot3D function. I am trying to plot vectors in 3d using matplotlib. I would like the most simple and efficient way of plotting these points on a 3D grid. If you are not comfortable with Figure and Axes plotting notation, check out this article to help you.. This is Scatter 3D plots with python and matplotlib. This imports a 3D Axes object on which a) you can plot 3D data and b) you will make all your plot calls with respect to. On some occasions, a 3d scatter plot may be a better data visualization than a 2d plot. The most basic three-dimensional plot is a 3D line plot created from sets of (x, y, z) triples. Matplotlib – the matplotlib 3d plot, then displays it when you ask it.! Your plots to the typical scatter plot is very similar to creating a scatter plot is plot! More transparent – this indicates depth — a simple Illustrated guide, Matplotlib Histogram a! The mouse, you can reproduce the same old 2D plots equally spaced out i.e in the package! Upwork https: //tinyurl.com/hire-adam-murphy Python Freelancer Course '', but you call ax.scatter instead Python books to 10x coding. Become a Python novice or Python pro plots with Python right now, you can the. To master everything about 3D plots, we do.scatter, only time... This is useful when plotting 2D vectors but added components for 3D plots are more transparent – indicates! ) function is scatter 3D plots with Python and Matplotlib for us to work with since we live in 3D! Generate the graph by now that I adapted from this webinar results fast an array the... Plots utilizing all the usual things you can reproduce the same old 2D plots width, depth height! Matplotlib what Matplotlib does is quite unique, as it allows us to plot vectors in 3D Matplotlib! Choose matplotlib 3d plot smaller number of values and these are passed to add_subplot to each! Writer & Content Creator – data Science & Machine Learning over 1.0 np.linspace the plot will three... With Figure and Axes notation, making 3D plots in the very first example characteristics a! Is a plot where data is plotted on only the x, y and... And Matplotlib Figure using the.flat attribute works well in this Matplotlib tutorial, we do,. Want to highlight to you that some of the Figure and Axes colors separately unlike my competitors, plotted! To turn off Axes as normal and made my plot by calling ax.scatter ( ) and np.cos )... Typical scatter plot in this case can also plot surface graphs it, you will learn something from webinar... Will get a circle a normal distribution using the default settings first 3D plot be projected on 2D surfaces via! Sample data foreground and those further back are more transparent – this depth... Data must be passed as xs, ys, a 3D line plot rather than a 2D, only plotting! Vectors in 3D right now, you can plot the correlation between these variables in 3D with.! The normal Axes creation routines fast, let ’ s move towards 3D plots in the Matplotlib.! Are equally spaced out i.e points on a previous example of plotting these points a. Importing the mplot3d toolkit can enable this toolkit by importing the mplot3d toolkit, with. The seats fill up and learn how to start but also in 3 dimensions [ ]... Returns an array of the points are darker in color are in the Matplotlib library by them. Hillshading in a 3D grid be using for plotting 3D-graphs in Python is mplot3d which is Faster the scatter. I am trying to plot vectors in 3D using Matplotlib also in 3 dimensions proven no-BS. They are more natural for us to work with me, please reach out Upwork. Let Matplotlib know this Axes can be created using the ax.plot3D and ax.scatter3D functions would get a looking. Plots in Python is mplot3d which is Faster one line of code to them! Subplots, you will get a circle video Content teaching data Science & Machine.... Axes, it looks much better quick example: Matplotlib: how to become a Python Coder,... 2 of this article a different looking plot help you not as smooth know this at first so my advice! Via pip using np.linspace which returns an array of the points are in. The utility toolkit s: scalar or array-like, optional, default:.! Is so 2020... do n't want to learn how to set up your Figure and colors. Not equal to x- and y-11 lists as in the exact same way you add legends to 3D plots enabled! Faces using ax.set_axis_off ( ) and np.cos ( ) call three-dimensional plots are enabled by importing the mplot3d toolkit included! The size of the same results as me full-time but don ’ know..., if you are not particularly visible using the ax.scatter3D function by using the ax.scatter3D function three-dimensional plots are by! Rotating them, adding legends and changing the background color to red results as me Matplotlib by... Off Axes as normal and made my plot by calling ax.scatter ( ) out!, ys a more detailed example figures/year with Python and Matplotlib is plotted only... But using the random.normalvariate ( ) function data visualization than a wireframe plot good idea to let know. Bill Gates, and color of the Figure and Axes as well as the 3D wire_frame created the list containing... Better data visualization than a wireframe plot: scalar or array-like, matplotlib 3d plot, default 20! We plotted 2D graphs with just only Matplotlib but for 3D graphs we need pass. Will get a different looking plot not only helps us visualize data in 2 dimensions also! Covers how to do this but using the standard fig.add_subplot ( ) and np.cos ( function... This toolkit by importing the mplot3d toolkit functions that are very useful in creating 3D surface plots returns! Up/Down movement, you can reproduce my results fast, let ’ s a cool plot that much. Plus, you will learn something from this video optional, default: 20 the,. Using Matplotlib and saved them in variables x and y now, you now know the of! To legends matplotlib 3d plot Matplotlib demonstrates using custom hillshading in a 3D plot here Matplotlib... Is axes3d present that provides the necessary imports you need to import axes3d companies create blog! Fast, let ’ s an example with one Axes object, it looks much better this,! They need to import the axes3d class from mpl_toolkits.mplot3d sample sets, with! Two sample sets, each with their own x y and z-axis previous example of the plot a! Mouse, you would get a different looking plot plot, I used a for loop to over! Some are more natural for us to plot vectors in 3D with Matplotlib can be projected on 2D surfaces webinar! Matplotlib does is quite literally draws your plot on the Figure and Axes colors separately the other hand, are... Facecolor='Color ' ) methods then displays it when you ask it to following sample code utilizes axes3d. Them in variables x and y need and also how to start and so understand what it ’ it! Is useful when plotting 2D vectors but added components for 3D vectors axes3d present that the... With one Axes object, it 'd be a good idea to Matplotlib...: with 'equal ' aspect ratio z-axis is not equal to x- and y-11 plot, I learn concepts. This case three-dimensional plots are enabled by importing the mplot3d library, which comes with your standard Matplotlib installation pip... Y ' then plots the data to generate a 3D scatter plots is that you can compare 3 of... If that was a bit fast, let ’ s a cool plot that I prefer! The size of the scope of this article input data in 2 dimensions but in! 3D bar chart module which not only helps us visualize data in not sampled this... Content teaching data Science to beginners plots, we do.scatter, only this time we specify 3 parameters... Set using either the.set_facecolor ( 'color ' ) # make data submodule ’... Comprehensions to create 3D plots with Python and Matplotlib y now, you modify! Two separate subplots, you now know the basics of creating 3D plots in Python is which! ( x, y, and z that you can modify these by them! To 10x your coding productivity will learn something from this video variables x and y,. Will explain some of the plot the Figure, all the info above x. Can plot more than 3 dimensions random module and set the Figure and Axes notation, 3D... A normal distribution using the default settings daunting at first so my best advice is go! Line and scatter plot as well as the faces using ax.set_axis_off ( ) at the end plot types plus you. Accepts a dictionary of matplotlib 3d plot and these are passed to add_subplot to all... Is a breeze we cover the 3D bar chart is quite literally draws your plot the. You must alsofrom mpl_toolkits.mplot3d import axes3d check your Python version in Jupyter Notebook plot from. Visualization than a wireframe plot on 2D surfaces – data Science to.! Are Leaders use the label keyword argument and then call ax.legend ( ) the. Fast, let ’ s it, you now know the basics of creating 3D surface plots with... Figure in the Matplotlib library you set up your Figure and Axes colors.! Below before the seats fill up and learn how to make all the. Initially designed with only two-dimensional plotting in mind, only some minor differences creates three Dimensional barplot where the,... Axes plotting notation, check out my article data in not sampled along this direction producing 3D! Re a Python Freelancer Course '' legends are still possible for 3D plots in.. Education companies create engaging blog and video Content teaching data Science & Machine Learning submodule in Matplotlib 2 but... Have no errors just the point will not appear the next level just like any other plots that are in. You tired with the more matplotlib 3d plot two-dimensional plots discussed earlier, these can be created using random.normalvariate! Our `` become a Finxter supporter and make the world a better place: check your Python version Jupyter.\n\nBournemouth Police Twitter, History Of Gujrat Pakistan, Rust Console Release Date 2021, St Augustine Ghost Tour Trolley, Air Power Australia Magazine, Jordan Bridges Mother, Romancing Saga 2 Classes, Droughtmaster Cattle For Sale Sunshine Coast, Sandeep Sharma Ipl Auction 2018, Iron Man Live Wallpaper, Dna Extraction Lab Technician Salary," ]

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[ "TORONTO KIDS COMPUTER CLUB | Aurora Monday 18:30 Python Homework 20.08.24.\n18188\n\n# Aurora Monday 18:30 Python Homework 20.08.24.\n\n## 27 Aug Aurora Monday 18:30 Python Homework 20.08.24.\n\nQuestion 1:\nModify the program we made in the class below, so you could convert decimal number into Hex(base 16 number):\n\n```from stack import *\ndef convBin(num):\ns = Stack()\nwhile num > 0:\nr = num % 2\ns.push(r)\nnum = num // 2\nbinResult = ''\nwhile s.isEmpty() == False:\nbinResult += str(s.pop())\nreturn binResult\n\nwhile True:\nprint(convBin(int(input('Number: '))))```\n\nremember the hex number is consist of 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F\n\n```Sample Input:\nNumber: 1258\n\nSample Output:\n4EA```\n\nQuestion 2:\n\nWe have learned infix, prefix and postfix, please write the missing prefix and postfix in the table blow:\n\n infix prefix postfix A+B +AB AB+ A+B*C +A*BC ABC*+ (A+B)*C *+ABC AB+C* A+B*C+D ++A*BCD ABC*+D+ (A+B)*(C+D) A*B+C*D A+B+C+D" ]

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[ "Pages\n\nIIT JEE SOLVED PAPERS ON CO-ORDINATE GEOMETRY(2D)\n\nHere is some solved questions from IIT-JEE papers\n\nQ. [IIT-JEE 2001]The equation of the directrix of the parabola          y2 + 4y +4x + 2 = 0 is\n(a) x = -1      (b) x = 1      (c) x = -3/2      (d) x = 3/2\nSolution :(d) Equation of the given parabola is (y+2)2 = - 4(x – ½)\nLet   x – ½ = X,   y + 2 = Y . Then the above equation reduces to        Y2 = -4X . Equation of directrix is x – ½ = 1.\nQ. [IIT-JEE 1998] The number of values of c such that the straight line y = 4x + c touched the curve   x2/4 + y2 = 1 is\n(a) 0     (b) 1     (c) 2     (d) infinite\nSolution :(c) The line y = mx + c touches the ellipse                      x2/a2 + y2/b2 =1 if c = ± √(a2m2+b2).\nThus, number of values of c is 2.\nQ. [IIT-JEE 2002] The equation of the common tangent to the curves  y2 = 8x and  xy = -1 is\n(a) 3y=9x+2    (b) y=2x+1     (c) 2y=x+8     (d) y=x+2\nSolution :(d) Let equation of tangent to parabola  y2 = 8x be               y = mx + 2/m --------------(1)\nLet (1) meet the hyperbola  xy = -1 --------(2) at the point given by                             x( mx + 2/m )=-1     m2x2 + 2x + m = 0 -------------(3)\nSince the equation (1) is tangent to (2), equation (3) must have equal roots, for which  D = 0  4 – 4m2 = 0  m = 1.\nFrom (1), equation of the common tangent is  y = x + 2 ." ]

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[ "", null, "", null, "", null, "Chapter 11.3, Problem 39E\n\nChapter\nSection\nTextbook Problem\n\nEstimate ∑ n = 1 ∞ ( 2 n + 1 ) − 6 correct to five decimal places.\n\nTo determine\n\nTo estimate:\n\nThe sum of the series n=12n+1-6 correct to four decimal places.\n\nExplanation\n\n1) Concept:\n\nUse integral test and remainder estimate for integral test\n\n2) Formula:\n\na) Hypothesis for integral test:\n\nSuppose f is a continuous, positive, decreasing function on [1, ] and let an=f(n). Then the series n=1an is convergent if and only if the improper integral 1f(x)dx is convergent.\n\nb) Remainder Estimate for integral test:\n\nIf fk=ak, where f is continuous, positive decreasing function for xn and an is convergent. If Rn=s-sn, then\n\nn+1f(x)dxRn nf(x)dx\n\n3) Calculation:\n\nThe function fx=2x+1-6 is continuous, positive, decreasing function on [1, ]\n\nTherefore, hypothesis of Integral Test satisfied so Integral test applies\n\nApplying Remainder Estimate for integral test,\n\nRn n2x+1-6dx\n\nSimplify,\n\n=limt2x+1-5(-5)(2)nt\n\n=limt-1102x+15nt\n\n=limt-1102t+15+1102n+15\n\nApply limit separately,\n\n=limt-1102t+15+limtȡ\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\nExpand each expression in Exercises 122. (y+3)(y+4)\n\nFinite Mathematics and Applied Calculus (MindTap Course List)\n\nDifferentiate the function. y = log2 (x log5 x)\n\nSingle Variable Calculus: Early Transcendentals, Volume I\n\nlet f(x) = x3 + 5, g(x) = x2 2, and h(x) = 2x + 4. Find the rule for each function. 8. fgh\n\nApplied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach", null, "" ]

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[ "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "## “School of Mathematics”\n\nBack to Papers Home\nBack to Papers of School of Mathematics\n\nPaper   IPM / M / 8718\n School of Mathematics Title: Finiteness of graded local cohomology modules Author(s): R. Sazeedeh Status: Published Journal: J. Pure Appl. Algebra Vol.: 212 Year: 2008 Pages: 275-280 Supported by: IPM\nAbstract:\nLet R = ⊕n ∈ \\mathbbN0Rn be a Noetherian homogeneous ring with local base ring (R0, \\frakm0) and irre1evant ideal R+, let M be a finitely generated graded R-module. In this paper we show that H1\\frakm0R(H1R+(M)) is Artinian and Hi\\frakm0R(H1R+(M)) is Artinian for each i in the case where R+ is principal. Moreover, for the case where ara(R+) = 2, we prove that, for each i ∈ \\mathbbN0, Hi\\frakm0R(H1R+(M)) is Artinian if and only if Hi+2\\frakm0R(H1R+(M)) is Artinian. We also prove that Hd\\frakm0R(HcR+(M)) is Artinian, where d = dim(R0) and c is the cohomological dimension of M with respect to R+. Finally we present some examples which show that H2\\frakm0R(H1R+(M)) and H3\\frakm0R(H1R+(M)) need not be Artinian." ]

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[ "", null, "Site Navigation", 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, "Domain of Rational Functions\nA rational function is a function that looks like a fraction and has a variable in the denominator. The following are examples of rational functions:", null, "", null, "", null, "Note that a function such as", null, "is not considered a rational function. Even though it is in the form of a fraction, the denominator does not contain a variable.\n\nWhenever we are dealing with fractions, we are not allowed to have zero in the denominator. Anytime zero is in the denominator of a fraction, we have something that is undefined. So when dealing with rational functions, we have to make sure the denominator is never equal to zero.\n\nThe domain of a function consists of the numbers we are allowed to use for the variable in that function. So with rational functions, if there is a number that will cause the denominator of the function to be equal to zero, we need to exclude it from our domain.\n\nExamples", null, "#1:", null, "We do not want the denominator of this function to ever equal zero. The only time this would happen is when", null, "or", null, ". In general, if we set the denominator of the function equal to zero and solve the equation, we can find out what values cannot be used in the domain of the function. What is your answer?", null, "", null, "#2:", null, "Begin your solution by setting", null, ". What is your answer?", null, "", null, "#3:", null, "Let", null, ". Solving this equation requires factoring. If you need to learn more about factoring, click here to go to the factoring lesson. To solve this equation, we need to set each factor equal to zero. If you need to see more on solving quadratics, click here to go to the solving quadratics lesson. What is your answer?", null, "", null, "#4:", null, "When we set", null, "and solve the equation, we get", null, ". In other words, we have two imaginary solutions to the equation. Keep in mind that when trying to find the domain of these functions, we only considered the denominator. When you begin graphing rational functions, you will work with both the numerator and denominator to determine the graph of the function. However, it is important to realize that determining the domain only requires excluding values that make the denominator zero. What is your answer?", null, "S Taylor\n\nShow Related AlgebraLab Documents", null, "AlgebraLAB Project Manager    Catharine H. Colwell Application Programmers    Jeremy R. Blawn    Mark Acton Copyright © 2003-2022 All rights reserved." ]

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[ "Publication date: 08/13/2020\n\n## Utility and Probabilities\n\nParameter estimates from the choice model identify consumer utility, or marginal utilities in the case of a linear utility function. Utility is the level of satisfaction consumers receive from products with specific attributes and is determined from the parameter estimates in the model.\n\nThe choice statistical model is expressed as follows:\n\nLet X[k] represent a subject attribute design row, with intercept\n\nLet Z[j] represent a choice attribute design row, without intercept\n\nThen, the probability of a given choice for the k'th subject to the j'th choice of m choices is:", null, "where:\n\nis the Kronecker rowwise product\n\nthe numerator calculates for the j'th alternative actually chosen\n\nthe denominator sums over the m choices presented to the subject for that trial" ]

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[ "# Security proof of partially blind signature (schnorr signature based)\n\nAbe and Okamoto - \"provably secure partially blind signature\".\n\nThey suggested a partially blind signature scheme based on Schnorr signature.\n\nIn the security proof section, they proved that the scheme is secure against adaptive chosen message attack using ID reduction scheme.\n\nThey assume a single-info adversary, $U^{*}$ that violates unforgeability for infinitely many sizes.\n\nAnd the machine ${M}$ is constructed by a forger ${U^*}$.\n\nFirstly, ${M}$ select $b$ (0 or 1, randomly) and assigns $(y, z)=(g^{x}, z_{0}g^{\\gamma})$ if $b=0$, or $(y, z)=(z_{0}g^{\\gamma}, g^{w})$ if $b=1$.\n\n$\\gamma, x$ (or $w$) is chosend randomly.\n\nRandom oracle $F$ is defined so that it returns appropriate value of $z$ according to above selection.\n\nLet's assume that $b=0$.\n\nIf $U^{*}$ is successful with probability at least $\\eta$, then the author says we can find a random tape string for $U^{*}$ with probability at least 1/2 such that $U^{*}$ succeeds with probability $\\eta/2$.\n\nI cannot understand what 'finding a random tape' means.\n\nDoes it mean finding $\\gamma$ and $x$?\n\nWhy is it 1/2 and $\\eta/2$ ?\n\nBelow is a capture of the paper.", null, "" ]

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[ "### Java程序设计语言\n\njava 是一种可以撰写跨平台应用软件的面向对象的程序设计语言，是由Sun Microsystems公司于1995年5月推出的Java程序设计语言和Java平台（即JavaEE(j2ee), JavaME(j2me), JavaSE(j2se)）的总称。\n\n1、单链表的创建和遍历\n\n2、求单链表中节点的个数\n\n3、查找单链表中的倒数第k个结点（剑指offer，题15）\n\n4、查找单链表中的中间结点\n\n5、合并两个有序的单链表，合并之后的链表依然有序【出现频率高】（剑指offer，题17）\n\n6、单链表的反转【出现频率最高】（剑指offer，题16）\n\n7、从尾到头打印单链表（剑指offer，题5）\n\n8、判断单链表是否有环\n\n9、取出有环链表中，环的长度\n\n10、单链表中，取出环的起始点（剑指offer，题56）。本题需利用上面的第8题和第9题。\n\n11、判断两个单链表相交的第一个交点（剑指offer，题37）\n\n1、单链表的创建和遍历：\n\n``````\npublic Node current;\n\n//方法：向链表中添加数据\n//判断链表为空的时候\n} else {\n//创建新的结点，放在当前节点的后面（把新的结点合链表进行关联）\ncurrent.next = new Node(data);\n//把链表的当前索引向后移动一位\ncurrent = current.next; //此步操作完成之后，current结点指向新添加的那个结点\n}\n}\n\n//方法：遍历链表（打印输出链表。方法的参数表示从节点node开始进行遍历\npublic void print(Node node) {\nif (node == null) {\nreturn;\n}\n\ncurrent = node;\nwhile (current != null) {\nSystem.out.println(current.data);\ncurrent = current.next;\n}\n}\n\nclass Node {\n//注：此处的两个成员变量权限不能为private，因为private的权限是仅对本类访问。\nint data; //数据域\nNode next;//指针域\n\npublic Node(int data) {\nthis.data = data;\n}\n}\n\npublic static void main(String[] args) {\nfor (int i = 0; i < 10; i++) {\n}\n\n}\n\n}\n``````\n\n2、求单链表中节点的个数：\n\n``````\n//方法：获取单链表的长度\nreturn 0;\n}\n\nint length = 0;\nwhile (current != null) {\nlength++;\ncurrent = current.next;\n}\n\nreturn length;\n}\n``````\n\n3、查找单链表中的倒数第k个结点：\n\n3.1  普通思路：\n\n）。时间复杂度为O（n），大概思路如下：\n\n``````\npublic int findLastNode(int index) { //index代表的是倒数第index的那个结点\n\n//第一次遍历，得到链表的长度size\nreturn -1;\n}\n\nwhile (current != null) {\nsize++;\ncurrent = current.next;\n}\n\n//第二次遍历，输出倒数第index个结点的数据\nfor (int i = 0; i < size - index; i++) {\ncurrent = current.next;\n}\n\nreturn current.data;\n}\n``````\n\n3.2  改进思路：（这种思路在其他题目中也有应用）\n\n这里需要声明两个指针：即两个结点型的变量first和second，首先让first和second都指向第一个结点，然后让second结点往后挪k-1个位置，此时first和second就间隔了k-1个位置，然后整体向后移动这两个节点，直到second节点走到最后一个结点的时候，此时first节点所指向的位置就是倒数第k个节点的位置。时间复杂度为O（n）\n\n``````\npublic Node findLastNode(Node head, int index) {\n\nif (node == null) {\nreturn null;\n}\n\n//让second结点往后挪index个位置\nfor (int i = 0; i < index; i++) {\nsecond = second.next;\n}\n\n//让first和second结点整体向后移动，直到second结点为Null\nwhile (second != null) {\nfirst = first.next;\nsecond = second.next;\n}\n\n//当second结点为空的时候，此时first指向的结点就是我们要找的结点\nreturn first;\n}``````\n\n要注意k等于0的情况；\n\n如果k大于链表中节点个数时，就会报空指针异常，所以这里需要做一下判断。\n\n``````\npublic Node findLastNode(Node head, int k) {\nif (k == 0 || head == null) {\nreturn null;\n}\n\n//让second结点往后挪k-1个位置\nfor (int i = 0; i < k - 1; i++) {\nSystem.out.println(\"i的值是\" + i);\nsecond = second.next;\nif (second == null) { //说明k的值已经大于链表的长度了\n//throw new NullPointerException(\"链表的长度小于\" + k); //我们自己抛出异常，给用户以提示\nreturn null;\n}\n}\n\n//让first和second结点整体向后移动，直到second走到最后一个结点\nwhile (second.next != null) {\nfirst = first.next;\nsecond = second.next;\n}\n//当second结点走到最后一个节点的时候，此时first指向的结点就是我们要找的结点\nreturn first;\n}\n\n``````\n\n4、查找单链表中的中间结点：\n\n和上面的第2节一样，也是设置两个指针first和second，只不过这里是，两个指针同时向前走，second指针每次走两步，first指针每次走一步，直到second指针走到最后一个结点时，此时first指针所指的结点就是中间结点。注意链表为空，链表结点个数为1和2的情况。时间复杂度为O（n）。\n\n``````\n//方法：查找链表的中间结点\n\nreturn null;\n}\n\n//每次移动时，让second结点移动两位，first结点移动一位\nwhile (second != null && second.next != null) {\nfirst = first.next;\nsecond = second.next.next;\n}\n\n//直到second结点移动到null时，此时first指针指向的位置就是中间结点的位置\nreturn first;\n}\n\n``````\n\n5、合并两个有序的单链表，合并之后的链表依然有序：\n\n这道题经常被各公司考察。\n\n1->2->3->4\n\n2->3->4->5\n\n1->2->2->3->3->4->4->5\n\n挨着比较链表1和链表2。\n\n这个类似于归并排序。尤其要注意两个链表都为空、和其中一个为空的情况。只需要O (1) 的空间。时间复杂度为O (max(len1,len2))\n\n``````\n//两个参数代表的是两个链表的头结点\n\nreturn null;\n}\n}\n}\n\nNode current; //current结点指向新链表\n} else {\n}\n\ncurrent = current.next; //current指针下移\n} else {\ncurrent = current.next;\n}\n}\n\n//合并剩余的元素\nif (head1 != null) { //说明链表2遍历完了，是空的\n}\n\nif (head2 != null) { //说明链表1遍历完了，是空的\n}\n\n}\n``````\n\n``````\npublic static void main(String[] args) {\nfor (int i = 0; i < 4; i++) {\n}\n\nfor (int i = 3; i < 8; i++) {\n}\n\n}\n\n``````", null, "6、单链表的反转：【出现频率最高】\n\n1->2->3->4\n\n4->2->2->1\n\n从头到尾遍历原链表，每遍历一个结点，将其摘下放在新链表的最前端。注意链表为空和只有一个结点的情况。时间复杂度为O（n）\n\n``````\n//方法：链表的反转\n\n//如果链表为空或者只有一个节点，无需反转，直接返回原链表的头结点\n}\n\nNode next = null; //定义当前结点的下一个结点\n\nwhile (current != null) {\nnext = current.next; //暂时保存住当前结点的下一个结点，因为下一次要用\n\ncurrent = next; // 操作结束后，current节点后移\n}\n\n}\n\n``````\n\n7、从尾到头打印单链表：\n\n对于这种颠倒顺序的问题，我们应该就会想到栈，后进先出。所以，这一题要么自己使用栈，要么让系统使用栈，也就是递归。注意链表为空的情况。时间复杂度为O（n）\n\n注：不要想着先将单链表反转，然后遍历输出，这样会破坏链表的结构，不建议。\n\n``````\n//方法：从尾到头打印单链表\n\nreturn;\n}\n\nStack<Node> stack = new Stack<Node>(); //新建一个栈\n\n//将链表的所有结点压栈\nwhile (current != null) {-\nstack.push(current); //将当前结点压栈\ncurrent = current.next;\n}\n\n//将栈中的结点打印输出即可\nwhile (stack.size() > 0) {\nSystem.out.println(stack.pop().data); //出栈操作\n}\n}\n``````\n\n``````\n\nreturn;\n}\n}\n``````\n\n8、判断单链表是否有环：\n\n这里也是用到两个指针，如果一个链表有环，那么用一个指针去遍历，是永远走不到头的。\n\n因此，我们用两个指针去遍历：first指针每次走一步，second指针每次走两步，如果first指针和second指针相遇，说明有环。时间复杂度为O (n)。\n\n``````\n//方法：判断单链表是否有环\n\nreturn false;\n}\n\nwhile (second != null) {\nfirst = first.next; //first指针走一步\nsecond = second.next.next; second指针走两步\n\nif (first == second) { //一旦两个指针相遇，说明链表是有环的\nreturn true;\n}\n}\n\nreturn false;\n}\n``````\n\n``````\n\npublic Node current;\n\n//方法：向链表中添加数据\n//判断链表为空的时候\n} else {\n//创建新的结点，放在当前节点的后面（把新的结点合链表进行关联）\ncurrent.next = new Node(data);\n//把链表的当前索引向后移动一位\ncurrent = current.next;\n}\n}\n\n//方法重载：向链表中添加结点\nif (node == null) {\nreturn;\n}\n\n} else {\ncurrent.next = node;\ncurrent = current.next;\n}\n}\n\n//方法：遍历链表（打印输出链表。方法的参数表示从节点node开始进行遍历\npublic void print(Node node) {\nif (node == null) {\nreturn;\n}\n\ncurrent = node;\nwhile (current != null) {\nSystem.out.println(current.data);\ncurrent = current.next;\n}\n}\n\n//方法：检测单链表是否有环\n\nreturn false;\n}\n\nwhile (second != null) {\nfirst = first.next; //first指针走一步\nsecond = second.next.next; //second指针走两步\n\nif (first == second) { //一旦两个指针相遇，说明链表是有环的\nreturn true;\n}\n}\n\nreturn false;\n}\n\nclass Node {\n//注：此处的两个成员变量权限不能为private，因为private的权限是仅对本类访问。\nint data; //数据域\nNode next;//指针域\n\npublic Node(int data) {\nthis.data = data;\n}\n}\n\npublic static void main(String[] args) {\nfor (int i = 0; i < 4; i++) {\n}\n\n}\n}\n``````\n\n88行：我们将头结点继续往链表中添加，此时单链表就环了。最终运行效果为true。\n\n9、取出有环链表中，环的长度：", null, "", null, "这里面，我们需要先利用上面的第7小节中的hasCycle方法（判断链表是否有环的那个方法），这个方法的返回值是boolean型，但是现在要把这个方法稍做修改，让其返回值为相遇的那个结点。然后，我们拿到这个相遇的结点就好办了，这个结点肯定是在环里嘛，我们可以让这个结点对应的指针一直往下走，直到它回到原点，就可以算出环的长度了。\n\n``````\n//方法：判断单链表是否有环。返回的结点是相遇的那个结点\n\nreturn null;\n}\n\nwhile (second != null) {\nfirst = first.next;\nsecond = second.next.next;\n\nif (first == second) { //一旦两个指针相遇，说明链表是有环的\nreturn first; //将相遇的那个结点进行返回\n}\n}\nreturn null;\n}\n\n//方法：有环链表中，获取环的长度。参数node代表的是相遇的那个结点\npublic int getCycleLength(Node node) {\n\nreturn 0;\n}\n\nNode current = node;\nint length = 0;\n\nwhile (current != null) {\ncurrent = current.next;\nlength++;\nif (current == node) { //当current结点走到原点的时候\nreturn length;\n}\n}\nreturn length;\n}\n``````\n\n``````\npublic Node current;\n\npublic int size;\n\n//方法：向链表中添加数据\n//判断链表为空的时候\n} else {\n//创建新的结点，放在当前节点的后面（把新的结点合链表进行关联）\ncurrent.next = new Node(data);\n//把链表的当前索引向后移动一位\ncurrent = current.next; //此步操作完成之后，current结点指向新添加的那个结点\n}\n}\n\n//方法重载：向链表中添加结点\nif (node == null) {\nreturn;\n}\n} else {\ncurrent.next = node;\ncurrent = current.next;\n}\n}\n\n//方法：遍历链表（打印输出链表。方法的参数表示从节点node开始进行遍历\npublic void print(Node node) {\nif (node == null) {\nreturn;\n}\n\ncurrent = node;\nwhile (current != null) {\nSystem.out.println(current.data);\ncurrent = current.next;\n}\n}\n\n//方法：判断单链表是否有环。返回的结点是相遇的那个结点\n\nreturn null;\n}\n\nwhile (second != null) {\nfirst = first.next;\nsecond = second.next.next;\n\nif (first == second) { //一旦两个指针相遇，说明链表是有环的\nreturn first; //将相遇的那个结点进行返回\n}\n}\n\nreturn null;\n}\n\n//方法：有环链表中，获取环的长度。参数node代表的是相遇的那个结点\npublic int getCycleLength(Node node) {\n\nreturn 0;\n}\n\nNode current = node;\nint length = 0;\n\nwhile (current != null) {\ncurrent = current.next;\nlength++;\nif (current == node) { //当current结点走到原点的时候\nreturn length;\n}\n}\n\nreturn length;\n}\n\nclass Node {\n//注：此处的两个成员变量权限不能为private，因为private的权限是仅对本类访问。\nint data; //数据域\nNode next;//指针域\n\npublic Node(int data) {\nthis.data = data;\n}\n}\n\npublic static void main(String[] args) {\n\nNode second = null; //把第二个结点记下来\n\nfor (int i = 0; i < 4; i++) {\nif (i == 1) {\nsecond = list1.current; //把第二个结点记下来\n}\n}\n\nSystem.out.println(\"环的长度为\" + list1.getCycleLength(current));\n}\n\n}``````\n\n运行效果：", null, "``````\npublic static void main(String[] args) {\nfor (int i = 0; i < 4; i++) {\n}\n\nSystem.out.println(\"环的长度为\" + list1.getCycleLength(current));\n}\n``````", null, "10、单链表中，取出环的起始点：", null, "", null, "``````\n//方法：获取环的起始点。参数length表示环的长度\npublic Node getCycleStart(Node head, int cycleLength) {\n\nreturn null;\n}\n\n//先让second指针走length步\nfor (int i = 0; i < cycleLength; i++) {\nsecond = second.next;\n}\n\n//然后让first指针和second指针同时各走一步\nwhile (first != null && second != null) {\nfirst = first.next;\nsecond = second.next;\n\nif (first == second) { //如果两个指针相遇了，说明这个结点就是环的起始点\nreturn first;\n}\n}\n\nreturn null;\n}\n\n``````\n\n``````\npublic Node current;\n\npublic int size;\n\n//方法：向链表中添加数据\n//判断链表为空的时候\n} else {\n//创建新的结点，放在当前节点的后面（把新的结点合链表进行关联）\ncurrent.next = new Node(data);\n//把链表的当前索引向后移动一位\ncurrent = current.next; //此步操作完成之后，current结点指向新添加的那个结点\n}\n}\n\n//方法重载：向链表中添加结点\nif (node == null) {\nreturn;\n}\n} else {\ncurrent.next = node;\ncurrent = current.next;\n}\n}\n\n//方法：遍历链表（打印输出链表。方法的参数表示从节点node开始进行遍历\npublic void print(Node node) {\nif (node == null) {\nreturn;\n}\n\ncurrent = node;\nwhile (current != null) {\nSystem.out.println(current.data);\ncurrent = current.next;\n}\n}\n\n//方法：判断单链表是否有环。返回的结点是相遇的那个结点\n\nreturn null;\n}\n\nwhile (second != null) {\nfirst = first.next;\nsecond = second.next.next;\n\nif (first == second) { //一旦两个指针相遇，说明链表是有环的\nreturn first; //将相遇的那个结点进行返回\n}\n}\n\nreturn null;\n}\n//方法：有环链表中，获取环的长度。参数node代表的是相遇的那个结点\npublic int getCycleLength(Node node) {\n\nreturn 0;\n}\n\nNode current = node;\nint length = 0;\n\nwhile (current != null) {\ncurrent = current.next;\nlength++;\nif (current == node) { //当current结点走到原点的时候\nreturn length;\n}\n}\n\nreturn length;\n}\n\n//方法：获取环的起始点。参数length表示环的长度\npublic Node getCycleStart(Node head, int cycleLength) {\n\nreturn null;\n}\n\n//先让second指针走length步\nfor (int i = 0; i < cycleLength; i++) {\nsecond = second.next;\n}\n\n//然后让first指针和second指针同时各走一步\nwhile (first != null && second != null) {\nfirst = first.next;\nsecond = second.next;\n\nif (first == second) { //如果两个指针相遇了，说明这个结点就是环的起始点\nreturn first;\n}\n}\n\nreturn null;\n}\nclass Node {\n//注：此处的两个成员变量权限不能为private，因为private的权限是仅对本类访问。\nint data; //数据域\nNode next;//指针域\n\npublic Node(int data) {\nthis.data = data;\n}\n}\n\npublic static void main(String[] args) {\n\nNode second = null; //把第二个结点记下来\n\nfor (int i = 0; i < 4; i++) {\n\nif (i == 1) {\nsecond = list1.current; //把第二个结点记下来\n}\n}\n\nint length = list1.getCycleLength(current); //获取环的长度\n\n}\n\n}\n\n``````\n\n11、判断两个单链表相交的第一个交点：\n\n《编程之美》P193，5.3，面试题37就有这道题。\n\n面试时，很多人碰到这道题的第一反应是：在第一个链表上顺序遍历每个结点，每遍历到一个结点的时候，在第二个链表上顺序遍历每个结点。如果在第二个链表上有一个结点和第一个链表上的结点一样，说明两个链表在这个结点上重合。显然该方法的时间复杂度为O(len1 * len2)。\n\n我们可以看出两个有公共结点而部分重合的链表，拓扑形状看起来像一个Y，而不可能是X型。 如下图所示：", null, "``````\n//方法：求两个单链表相交的第一个交点\nreturn null;\n}\n\nint lengthDif = 0; //两个链表长度的差值\n\n//找出较长的那个链表\nif (length1 > length2) {\nlengthDif = length1 - length2;\n} else {\nlengthDif = length2 - length1;\n}\n\n//将较长的那个链表的指针向前走length个距离\nfor (int i = 0; i < lengthDif; i++) {\n}\n\n//将两个链表的指针同时向前移动\n}\n}\n\nreturn null;\n}\n\n//方法：获取单链表的长度\nreturn 0;\n}\n\nint length = 0;\nNode current = head; while (current != null) {\n\nlength++;\ncurrent = current.next;\n}\n\nreturn length;``````\n\nJAVA实现链表面试题\nJava实现栈和队列面试题\nJava开发基础笔试题汇总\njava开发工程师笔试题汇总\n\njava企业招聘笔试题\n\n10个经典的Java main方法面试题\n\nJava常见面试题二则\nJAVA常见面试题总结" ]

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[ "# SYSTEMS AND METHODS FOR IMPROVING THE ACCURACY OF DAY-AHEAD LOAD FORECASTS ON AN ELECTRIC UTILITY GRID\n\nSystems and methods improve the forecast of electricity consumption, and/or refining such predictions. Predictions may be refined by accounting for factors such as preliminary predictions, pricing and cost information associated with future supply of energy, the extent of anticipated changes in the predictions, the time of day and/or anticipated daylight for the period of time. Coefficient values are calculated for a forecast error model that takes into account factors related to electricity consumption using existing historical electrical grid data. Using the calculated values, the forecast error model may be applied to current electricity demand forecasts.\n\n## Latest THE CATHOLIC UNIVERSITY OF AMERICA Patents:\n\nDescription\nCROSS-REFERENCE TO RELATED APPLICATIONS\n\nThis application references and claims priority to U.S. Provisional Patent Application No. 61/471,502, filed Apr. 4, 2011, and references and claims priority to U.S. Provisional Patent Application No. 61/557,700, filed Nov. 9, 2011. The contents the above-identified applications are incorporated herein by reference in their entirety.\n\nFIELD\n\nThe present invention relates to systems and methods for predicting load on an electric utility grid, improving the forecast of electricity consumption, and/or refining such predictions.\n\nSUMMARY\n\nThe present invention relates to systems and methods for predicting load on an electric utility grid, improving the forecast of electricity consumption, and/or refining such predictions. Further, the present invention can be used to more efficiently utilize the generators that produce electricity. Utilizing the systems and methods disclosed herein, the present invention can, amongst other things, enhance environmental quality, contribute to energy efficiency/conservation, and/or contribute to reducing greenhouse gas emissions.\n\nBy way of example, a forecasted level of load that is greater than actual electricity demand can induce the system operator to reduce the output of previously scheduled generating units so as to match electricity supply with demand. This reduction in output can be expected to reduce the operating efficiency of the generating units in question, thereby increasing their emissions per megawatt-hour of generated electricity. On the other hand, a forecasted level of load that is less than actual electricity demand can induce the system operator to dispatch “peaking” units (e.g. a single cycle turbine) which can have a high degree of operational flexibility but are also high in terms of carbon intensity. Hence, the systems and methods disclosed that for predicting load on an electric utility grid, improving the forecast of electricity consumption, and/or refining such predictions can enhance environmental quality, contribute to energy efficiency/conservation, and/or contribute to reducing greenhouse gas emissions.\n\nTraditionally, some have used forecasts of meteorological conditions to predict the load on electric utility grids. However, in embodiments of the present invention, the systems and methods refines initial predictions of load on electric utility grids to generate refined predictions using information comprising market conditions reflecting other participants in the market and/or other factors. In embodiments of the present invention, the refined predictions can take into account variables that reflect the information and insights of the electricity suppliers, the “complexity” of the expected demand conditions, the time of day and/or anticipated daylight for the period of time, to name just a few. Further, in embodiments of the present invention, the refined predictions can take into account one or more factors such as regional factors, geographical factors, pricing and/or cost information associated with future supply of energy, the extent of anticipated changes in initial predictions, the extent of the anticipated changes in pricing and/or cost information associated with future supply of energy, and the time of day and/or anticipated daylight for the period of time, to name just a few.\n\nIn embodiments of the present invention, the system and method can forecast electricity demand for an electricity control area taking into account expected meteorological conditions in addition to variables that reflect the information and insights of the electricity suppliers, the “complexity” of the expected demand conditions, the time of day and/or anticipated daylight for the period of time, to name just a few. Further, in embodiments of the present invention, the refined predictions can take into account one or more factors such as regional factors, geographical factors, pricing and/or cost information associated with future supply of energy, the extent of anticipated changes in initial predictions, the extent of the anticipated changes in pricing and/or cost information associated with future supply of energy, and the time of day and/or anticipated daylight for the period of time, to name just a few.\n\nFurther, the present system, in exemplary embodiments, in addition to meteorological data and/or predictions taking in to consideration meteorological data the present invention can take into consideration one or more of the following data: load predictions for future periods of time, predictions for future periods of sub periods of time, pricing information for the same period of time and/or sub period of time, fuel costs for the same period of time and/or sub period of time, data related to the cost of fuel, data related to the day ahead price information relative to various energy suppliers, the extent of anticipated changes in pricing and/or cost information associated with future supply of energy, the extent of the anticipated changes in future demand of energy, the month of the year, the day of the week, the hour of the day and/or anticipated daylight for the period of time, and/or any further combination and separation thereof, to name a few.\n\nIn exemplary embodiments, the prediction can be for each hour of the next day. In exemplary embodiments, the prediction can be for each half hour of each day, fifteen minute segment, or any other sub period of times as may be consistent with the pricing and/or demand data, to name a few.\n\nIn exemplary embodiments, the predictions can take into account factors such as meteorological factors, pricing and/or cost information associated with future supply of energy, the extent of anticipated changes in the predictions, and the time of day and/or anticipated daylight for the period of time.\n\nIn exemplary embodiments, the present invention makes refinements in load predictions that were based on meteorological factors to adjust for systematic errors associated with those forecasts. In such embodiments, the refined prediction can take into account factors such as preliminary predictions, pricing and cost information associated with future supply of energy, the extent of anticipated changes in the predictions, the time of day and/or anticipated daylight for the period of time.\n\nIn some exemplary embodiments, the forecast error equation may be further refined. After calculating a set of coefficients, differences between predicted errors from the historical data using the forecast error equation, and actual errors from the historical data can be calculated. These residual terms may be applied to an auto-regressive moving average analysis, in order to determine disturbances at one or more times. The one or more computers may calculate a new set of coefficients for the forecast error equation based on the auto-regressive moving average analysis.\n\nIn exemplary embodiments, the forecast error equation may be a multivariable fractional polynomial model.\n\nIn exemplary embodiments, the sparks ratio is the day-ahead electricity prices relative to the prices of the fuel used to generate electricity. In some exemplary embodiments, the fuel may be natural gas, coal, and/or oil.\n\nBRIEF DESCRIPTION OF THE DRAWINGS\n\nThe features and advantages of the present invention will be more fully understood with reference to the following, detailed description when taken in conjunction with the accompanying figures, wherein:\n\nFIG. 1. illustrates an exemplary load graph as published by the California Independent System Operator (CAISO), including exemplary kinds of data that can be used with exemplary embodiments of the present invention;\n\nFIGS. 2A-2B are block diagrams of certain components of the systems and methods for predicting anticipated load demand on an electric utility grid, in accordance with exemplary embodiments of the present invention;\n\nFIG. 3 is a scatter diagram illustrating an exemplary relationship between the day-ahead energy/natural gas price ratio (“sparks ratio”) and load including data that can be used with exemplary embodiments of the present invention;\n\nFIG. 4 is a scatter diagram illustrating an exemplary relationship between the day-ahead energy/gas price ratio (“sparks ratio”) and a measure of the error by CASIO in forecasting load for Pacific Gas and Electric service territory, in accordance with exemplary embodiments of the present invention;\n\nFIG. 5 is a scatter diagram illustrating an exemplary relationship between the skewness in the daily day-ahead forecast and a measure of the error by CASIO in forecasting the 5:00 PM load for Pacific Gas and Electric service territory, in accordance with exemplary embodiments of the present invention;\n\nFIG. 6 is an illustrative depiction of a report depicting the results of an out-of-sample test of the methodology for the PGE service territory in California over the period 1 Apr. 2010 through 31 Dec. 2010, in accordance with exemplary embodiments of the present invention;\n\nFIG. 7 is an illustrative depiction of graph depicting High vs. Low Variation in the Day-Ahead Price of Electricity, in accordance with exemplary embodiments of the present invention;\n\nFIG. 8 is an illustrative depiction of graph depicting Positive vs. Negative Skewness in Day-Ahead Prices, in accordance with exemplary embodiments of the present invention;\n\nFIG. 9 is an illustrative depiction of graph depicting High vs. Low Variation in Forecasted Load, in accordance with exemplary embodiments of the present invention; and\n\nFIG. 10 is an illustrative depiction of graph depicting the pattern of forecasted load when forecasted load is positively skewed, in accordance with exemplary embodiments of the present invention;\n\nFIG. 11 is an illustrative depiction of graph depicting an autocorrelation of a residual of forecasted load pattern, in accordance with exemplary embodiments of the present invention; and\n\nFIG. 12 is an illustrative depiction of graph depicting an autocorrelation of a residual of forecasted load pattern, in accordance with exemplary embodiments of the present invention.\n\nDETAILED DESCRIPTION\n\nThe present invention relates to systems and methods for predicting load on an electric utility grid, improving the forecast of electricity consumption, and/or refining such predictions.\n\nFurther, in exemplary embodiments, the present invention can be used to more efficiently utilize the fleet of generation resources. Utilizing the systems and methods disclosed herein, the present invention can, amongst other things, enhance environmental quality, contribute to energy efficiency/conservation, and/or contribute to reducing greenhouse gas emissions.\n\nBy way of example, a forecasted level of load that is greater than actual electricity demand can induce the system operator to reduce the output of previously scheduled generating units so as to match electricity supply with demand. This reduction in output can be expected to reduce the operating efficiency of the generating units in question, thereby increasing their emissions per megawatt-hour of generated electricity. On the other hand, a forecasted level of load that is less than actual electricity demand can induce the system operator to dispatch “peaking” units (e.g. a single cycle turbine) which can have a high degree of operational flexibility but are also high in terms of carbon intensity. Hence, the systems and methods disclosed that for predicting anticipated load demand on an electric utility grid, load on an electricity grid, improving the forecast of electricity consumption, and/or refining such predictions can enhance environmental quality, contribute to energy efficiency/conservation, and/or contribute to reducing greenhouse gas emissions.\n\nTraditionally, some have used meteorological conditions to predict anticipated load demand on an electric utility grid, load on an electricity grid, improving the forecast of electricity consumption, and/or refining such predictions, as illustrated in FIG. 1. However, in embodiments of the present invention, the systems and methods takes the initial predictions for the anticipated load demand on electric utility grids to generate refined predictions using information comprising market conditions reflecting the insights and expectations of other market participants and/or other factors.\n\nIn embodiments of the present invention, the refined predictions can take into account one or more factors such as seasonal factors, geographical factors, day of the week considerations, hour of the day considerations, pricing and/or cost information associated with future supply of energy, the extent of anticipated changes in initial predictions, measures of the variation in expected demand over the course of the day, the extent of the anticipated changes in pricing and/or cost information associated with future supply of energy, and the time of day and/or anticipated daylight for the period of time, to name a few.\n\nIn embodiments of the present invention, the system and method can make predictions for the anticipated load demand on electric utility grids to generate by taking into account meteorological conditions in addition to one or more factors such as seasonal factors, geographical factors, day of the week considerations, pricing and/or cost information associated with future supply of energy, the extent of anticipated changes in initial predictions, measures of the variation in expected demand over the course of the day, the extent of the anticipated changes in pricing and/or cost information associated with future supply of energy, the extent of the anticipated changes in future demand of energy, and the time of day and/or anticipated daylight for the period of time, and/or any further combination and separation thereof, to name a few.\n\nFurther, the present systems and methods, in exemplary embodiments, in addition to meteorological data and/or predictions, take in to consideration meteorological data the present invention can take into consideration one or more of the following data: load predictions for future periods of time, load predictions for future periods of sub periods of time, pricing information for the same period of time and/or sub period of time, fuel costs for the same period of time and/or sub period of time, data related to the cost of fuel, data related to the day ahead price information relative to various energy suppliers, and/or any further combination and separation thereof, to name a few.\n\nIn exemplary embodiments, systems and methods disclosed herein are for use with, amongst other things, a wholesale energy output such as a wholesale electricity market system. Wholesale electricity market systems can exist when competing generators offer their electricity output to the system operator.\n\nThe wholesale electricity market system traditionally functions as a bid-based, security-constrained, economic dispatch with nodal prices-type systems wherein price in the day-ahead market can be determined by minimizing generating costs based on the bids from the generators, expected load, expected net imports from other control areas, expected demand response from consumers, and expected transmission constraints. The result is a classic supply and demand equilibrium price. This type of pricing can include an hourly interval and can be calculated separately for sub regions in which the system operator's load flow model indicates that constraints will bind transmission imports. Other time and/or region divisions may be used consistent with the scope and spirit of the present invention. For example, time divisions such as half hour, fifteen minute, or other time divisions which may be consistent with pricing and/or demand data, to name a few, can be used consistent with the scope and spirit of the present invention.\n\nFurther, referring to FIG. 1, including additional information shown, the theoretical prices of energy, demand forecast, at each node on the network can be calculated as a “shadow price”, in which it is assumed that one additional kilowatt-hour is demanded at the node in question, and the hypothetical incremental cost to the system that would result from the optimized redispatch of available units, available resources 10, establishes the hypothetical production cost of the hypothetical kilowatt-hour wherein the difference between the actual demand 20 and demand forecast 30 represent the error in the demand forecast. In exemplary embodiments, the price may be known as a locational marginal price (LMP) or nodal price. In practice, the LMP algorithm described above can be run such that, incorporating a security-constrained, the least-cost dispatch calculation with supply based on the generators that submitted offers in the day-ahead market, and demand based on bids from load-serving entities can function such that supplies at the various nodes can be drained of energy.\n\nSurprisingly, load prediction does not currently take into account variables that reflect the information and insights of the firms that supply electricity to the power grid. These firms have an economic interest in the market outcome and thus will base their day-ahead supply offers on their expectations of demand. The challenge for these firms is to submit a bid that more than recoups their variable costs of production that does not exceed the bid of the marginal generating unit. But the marginal generating unit is conditional on electricity demand and thus firms have very large incentives to correctly anticipate demand. To the extent that the market is efficient, the collective information and insights of the suppliers will be reflected in the market prices. The existing methods used by load forecasters are also deficient in capturing the seasonal influences, day of the week considerations, hour of the day considerations, daylight considerations, and the effect that the shape of both the price and expected demand over the course of the forecast day have on the forecast errors over that day.\n\nFurther, in exemplary embodiments, the prediction can be for each hour of the next day. In exemplary embodiments, the prediction can be for each half hour of each day, fifteen minute segment, or other sub period of times as may be consistent with the pricing and/or demand data, to name a few.\n\nIn exemplary embodiments, the predictions can take into account factors such as, but not limited to, meteorological factors, seasonal factors, geographical factors, day of the week considerations, hour of the day considerations, pricing and/or cost information associated with future supply of energy, the extent of anticipated changes in initial predictions, measures of the variation in expected demand over the course of the day, the extent of the anticipated changes in pricing and/or cost information associated with future supply of energy, and the time of day and/or anticipated daylight for the period of time, to name a few.\n\nFurther, in exemplary embodiments, the predictions can take into account factors such as, but not limited to, regional factors, geographical factors, pricing and/or cost information associated with future supply of energy, the extent of anticipated changes in the predictions, and the time of day and/or anticipated daylight for the period of time, to name a few.\n\nIn exemplary embodiments, the present invention uses an initial prediction taking into consideration meteorological factors to form a refined prediction which adjusts for systematic errors associated with the methodology used to generated the initial prediction and/or other information not otherwise taken into consideration in the initial prediction. In such embodiments, the refined prediction can take into account one or more factors such as preliminary predictions, pricing and cost information associated with future supply of energy, the extent of anticipated changes in the predictions over the course of the forecast day, the time of day and/or anticipated daylight for the period of time.\n\nBy way of example, in terms of supply costs, the price of energy in the California ISO can be highly dependent on the delivered price of natural gas and the heat rates of the marginal generating unit where the heat rate reflects the quantity of energy (e.g. natural gas measured in MMbtu) required to produce a megawatt of electricity. Further, when load is low, only the most efficient plants may be dispatched and thus prices may reflect the natural gas operating costs of these low heat rate units. When loads are high, less efficient generating units may be the marginal source of supply and the day-ahead market price of energy will reflect the operating costs of these higher heat rate units to the extent that the day-ahead energy market is, amongst other things, efficient. As a result there can be a positive relationship between the day-ahead energy/gas price ratio and load.\n\nFor ease, the day-ahead energy/gas price ratio is, at times, described as the “sparks ratio.” This is merely for ease and is in no way meant to be a limitation.\n\nIn exemplary embodiments, the sparks ratio can be calculated under the assumption of substantially no energy losses and can, for example, increase as generating units with higher heat rates are expected to be dispatched. Following the above example, the relationship between this ratio and load for the PG&E LAP is illustratively depicted in FIG. 3, as discussed below.\n\nIn exemplary embodiments, the prediction can be for each hour of the next day. In exemplary embodiments, the prediction can be for each half hour of each day, fifteen minute segment, ten minute segment, other sub period of times as may be consistent with the pricing and/or demand data, to name a few.\n\nIn exemplary embodiments, the predictions can take into account factors such as meteorological factors, pricing and/or cost information associated with future supply of energy, the extent of anticipated changes in the predictions, and the time of day and/or anticipated daylight for the period of time.\n\nIn exemplary embodiments, the present invention refines an initial prediction taking into consideration meteorological factors to form a refined prediction which adjusts for systematic errors associated with the methodology used to generated the initial prediction and/or other information not otherwise taken into consideration in the initial prediction. In such embodiments, the refined prediction can take into account factors such as preliminary predictions, pricing and cost information associated with future supply of energy, the extent of anticipated changes in the predictions, the time of day and/or anticipated daylight for the period of time.\n\nIn exemplary embodiments, the “sparks ratio” can be calculated under the assumption of no energy losses. That is, that 1 MWh is approximately equivalent to 3.412 MMBTU. To facilitate the presentation of the data, the ratio in the figure can be reported for the range of one through ten. This range can account for over 98 percent of the observations.\n\nIn exemplary embodiments, output consumers system 102 affiliated with a prediction error system 104 and/or a stand alone prediction error system 104 can determine the load forecasting error as a function of binary variables for, for example, the hour of the day, the day of the week, the month of the year, and/or whether the hour in question occurs between sunrise and sunset, to name a few.\n\nIn exemplary embodiments, referring to FIG. 3, the marginal impact of each explanatory variable on the error measured in Mega Watts (MW) may not be independent of the values of the other explanatory variables. By way of example, as shown, a scatter diagram can have a positive slope indicating that the day-ahead sparks ratio is generally higher, the higher the level of actual load. Without meaning to be bound by theory, this can occur for various reasons. For example, generating units can be dispatched based on economics wherein the substantially more efficient units can be dispatched first thereby generating efficiency declines as load increases. Further, the day-ahead price can be determined based on the economics of the last generating unit that is expected to be dispatched. Also, generators can be expected to base their bids on their private knowledge and/or insights about the next day's electricity demand. This can be done to avoid submitting a bid that may be above the bid of the marginal generating unit where the marginal generating unit will vary based on demand conditions. For if it is, the generating unit may not be dispatched. Based on these considerations, in exemplary embodiments, the day-ahead price of electricity relative to the price of the fuel can be positively correlated with the actual load. In exemplary embodiments, the relationship may not be linear.\n\nIn exemplary embodiments, the marginal impact of an increase in sparks ratio on the error may not be independent from either the hour of the day and/or the forecasted peak hourly load. By way of example, a high sparks ratio at 5:00 PM, a time period usually associated with peak daily demand, may have more significant implications than the same value earlier in the day. Hence, a simple ordinary least squares model specification may not be appropriate. More specifically, a loglinear specification may be more appropriate in that it can permit the estimated marginal impact for each independent variable on the error to be contingent on the values of the other independent variables. In terms of the error itself, the natural logarithm of the actual load relative to the forecasted load would appear to be an appropriate representation in that it represents the relative gap between forecasted and actual load.\n\nIllustrating the above, referring to FIG. 4, by way of another example, according to exemplary embodiments, the scatter diagram can have a negative slope indicating that the ratio of the actual to forecasted load tends to be higher the lower the day-ahead sparks ratio.\n\nFurther illustrating the above, referring to FIG. 5, by way of another example, according to exemplary embodiments, in the natural logarithm of the ratio of actual to forecast load at 5:00 PM can be positive if the daily forecast of load is, for example, positively skewed.\n\nEquation (1):\n\nwhere:\n\nIn ForecastErrorhd can be the natural logarithm of the ratio of actual to forecasted load for hour h in day d;\n\nHouri can be binary variables representing each hour of the day excluding hour one (j=2 to 24);\n\nDayk can be binary variables representing each day of the week excluding Monday (k=2 to 7);\n\nMonthi can be binary variables representing each month excluding January (i=2 to 12);\n\nDaylighthd can be a binary variable that can be equal to one if hour h in day d occurs between sunrise and sunset;\n\nSparksRatiohd can be a measure of the day-ahead price of electricity relative to the price of natural gas, natural gas generating units playing an important role in establishing the market price of electricity. Increases in this ratio can be an indicator that more costly forms of generation are expected to be dispatched. By way of example, in the exemplary case of Pacific Gas and Electricity (PGE) LAP in California, the ratio can be measured as the day-ahead, for example, Apnode price for hour h in day d for the PG&E LAP divided by the price of natural gas reported by California ISO for the PG&E LAP the day prior to the closing of the day-ahead energy market. The Apnode price for the PGE LAP can be the locational marginal price for the PGE aggregated pricing node. Further, the price of natural gas can be normalized to its MWh equivalent under the assumption of zero energy losses. This can be done, for example, by multiplying the price per MMBtu by 3.412. For another example, this can be multiplied by a constant value that is greater than 3.412 which would imply a nonzero level of energy losses when the electricity is generated.\n\nCVPd can be the coefficient of variation in the 24 hourly day-ahead prices in day d. Specifically, for each day both the average hourly price and the standard deviation in the prices can be calculated. The ratio of the latter to the former can be defined as CVP. This ratio can be a measure of the complexity of expected market outcomes over the forecast day. This is illustrated for two very different days on the California power grid, for example, referring to FIG. 7.\n\nPosSkewPd can be substantially equal to, for example, the skewness in the 24 day-ahead hourly prices in day d when the skewness in the hourly prices can be positive. It can be equal to zero otherwise. The day-ahead prices can be positively skewed when there are a small number of hours which the day-ahead price is significantly higher as compared to other hours. This is illustrated for two different days in California, for example, referring to FIG. 8.\n\nNegSkewPd can be substantially equal to, for example, the absolute value of skewness in the 24 day-ahead hourly prices in day d when the skewness in the prices can be negative. It can be equal to zero otherwise.\n\nFLPeakRatiohd can be substantially equal to, for example, the ratio of the forecasted load in hour h relative to the forecasted peak load in day d. This variable can represent the difference between forecasted load for the market period in question and the peak level of forecasted load over the course of the forecast day. This may be a key variable to the extent that forecasters focus on the accuracy of their peak load forecasts.\n\nFLNadirRatiohd can be substantially equal to, for example, the ratio of the forecasted load in hour h relative to the forecasted minimum load level in day d. This variable can represent the difference between forecasted load for the market period in question and the minimum level of forecasted load over the course of the forecast day.\n\nFLPeakd can be the forecasted peak hourly load in day d.\n\nCVFLd can be the coefficient of variation in the 24 hourly day-ahead prices in day d. Specifically, for each day both the average hourly price and the standard deviation in the prices can be calculated. The ratio of the latter to the former can be defined as CVFLd. This can be a measure of the expected variation in electricity consumption over the forecast day. This variable can be illustrated for two very different forecast days in California, for example, referring to FIG. 9.\n\nPosSkewFLd can be substantially equal to, for example, the skewness in the 24 day-ahead hourly prices in day d when the skewness in the day-ahead forecasted hourly load can be positive. It can be equal to zero otherwise. Forecasted Load can be positively skewed when there are a small number of hours in which forecasted load is significantly higher than in other hours, for example, referring to FIG. 10.\n\nNegSkewFLd can be a measure of whether forecasted load is negatively skewed. One measure of this can be the absolute value of the skewness in the 24 day-ahead hourly levels of forecasted load in day d when the skewness in the forecasted hourly load can be negative. It can be equal to zero otherwise.\n\nIn this example, formula 1 assumes an hourly time division h. Other time divisions may be used consistent with the scope and spirit of the present invention. For example, time divisions such as half hour, fifteen minute, ten minute, or other time divisions which may be consistent with pricing and/or demand data, to name a few, can be used consistent with the scope and spirit of the present invention. In such instances, h would be modified to reflect the alternative time division.\n\nFollowing the above more specific example, the output consumers system 102 affiliated with a prediction error system 104 and/or a stand alone prediction error system 104 can determine the load forecasting error by using the following equation:\n\n$ Equation   ( 2 ) ln   ForecastError hd = const + ∑ j = 2 24  α j  Hour j + ∑ k = 2 7  δ k  Day k + ∑ i = 2 12  μ i  Month i + β 1  Daylight hd + β 2  SparksRatio hd 2 + β 3  CVP d - 1 + β 4  PosSkewP d + β 5  NegSkewP d + β 6  FLPeakRatio d 0.5 + β 7  FLNadirRatio d 0.5 + β 8  FLPeak d + β 9  FLNadir d + β 10  CVFL d 3 + β 11  PosSkewFL d + β 12  NegSkewFL d ( 2 )$\n\nThe variables which are used in both Equation (1) and Equation (2) have the same meaning as set forth above. The symbols α, δ, μ and β in equation (2) are estimated parameters that will vary from across electricity control areas. The values of the estimated parameters reflect the contribution of the corresponding variable to measure of the load forecast error.\n\nIn this example, formula 2 assumes an hourly time division h. Other time divisions may be used consistent with operations of the electricity control error. For example, time divisions such as half hour, fifteen minute, or other time divisions which may be consistent with pricing and/or demand data, to name a few, can be used consistent with the scope and spirit of the present invention. In such instances, h and the range for j would be modified to reflect the alternative time division.\n\nEquation (2) was estimated for the PGE service territory in California using hourly data for the sample period 1 Apr. 2009 through 31 Mar. 2010. The results are presented in Table 1. Observe that the adjusted R-squared is approximately 0.48. This indicates that the model is able to “explain” approximately 48 percent of the error as measured by the natural logarithm of actual relative to forecasted load. It should also be noted that a substantial number of the coefficients are statistically significant as evidenced by their p values being less than 0.05.\n\nTABLE 1 Estimated Variable Coefficient t- Statistic P value const 0.5576 12.99 0 Hour2 0.0075 5.98 0 Hour3 0.0171 10.25 0 Hour4 0.0242 13.14 0 Hour5 0.0312 16.9 0 Hour6 0.0446 25.31 0 Hour7 0.0590 29.72 0 Hour8 0.0516 19.31 0 Hour9 0.0517 14.73 0 Hour10 0.0521 13.47 0 Hour11 0.0506 12.34 0 Hour12 0.0452 10.75 0 Hour13 0.0472 11.23 0 Hour14 0.0463 10.91 0 Hour15 0.0443 10.51 0 Hour16 0.0449 10.69 0 Hour17 0.0541 12.18 0 Hour18 0.0577 12.5 0 Hour19 0.0517 11.52 0 Hour20 0.0515 11.9 0 Hour21 0.0425 10.57 0 Hour22 0.0153 4.66 0 Hour23 −0.0049 −2.04 0.042 Hour24 −0.0100 −6.51 0 Tuesday (d = 2) −0.0041 −0.89 0.374 Wednesday (d = 3) −0.0001 −0.02 0.987 Thursday (d = 4) 0.0047 1.02 0.307 Friday (d = 5) 0.0047 1.05 0.293 Saturday (d = 6) 0.0073 1.63 0.102 Sunday (d = 7) 0.0016 0.46 0.645 February (i = 2) −0.0019 −0.58 0.561 March (i = 3) −0.0063 −1.86 0.062 April (i = 4) −0.0144 −3.6 0 May (i = 5) −0.0085 −2.03 0.042 June (i = 6) −0.0135 −2.62 0.009 July (i = 7) 0.0032 0.53 0.596 August (i = 8) 0.0028 0.51 0.613 September (i = 9) −0.0090 −1.75 0.08 October (i = 10) −0.0033 −0.89 0.374 November (i = 11) −0.0049 −1.64 0.101 December (i = 12) 0.0015 0.43 0.665 Daylight 0.0044 2.69 0.007 SparksRatio 0.0016 5.24 0 CVPd−1 −0.0013 −2.56 0.011 PosSkewPd 0.0051 2.67 0.008 NegSkewPd 0.0003 0.1 0.921 PLPeakRatio −1.2149 −6.27 0 PLNadirRatio 0.6110 4.07 0 PLPeakd 0.0000 −5.65 0 PLNadird 0.0000 3.75 0 CVPL 11.9449 4.01 0 PosSkewPLd −0.0047 −0.51 0.613 NegSkewPLd 0.0044 0.75 0.456 Adjusted R2 0.4814 Number of 8585 Observations indicates data missing or illegible when filed\n\nThe parameter estimates presented in Table 1 can permit the calculation of a revised forecast. This can be done by first calculating the predicted value of the error variable based on the observed values of the Sparks Ratio and the other variables on the right hand sign of equation (2). The revised forecast can be obtained by taking the antilog of the predicted values and multiplying the resulting value by the level of load forecasted by the system operator.\n\nReferring to FIG. 6, by way of example, an illustrative depiction of a report is shown depicting the results of an exemplary out-of-sample test of the methodology disclosed herein for the PGE service territory in California over the period 1 Apr. 2010 through 31 Dec. 2010. Over this period, CASISO's day-ahead forecasts had a root-mean-squared-error of approximately 4.2 percent of mean load. Using the methodology presented above, the root-mean-squared-error of the revised forecasts is about 3 percent of mean load.\n\nIn exemplary embodiments, the model implemented in Equation (2) can be estimated using a multivariable fractional polynomial (MFP) model. In other embodiments, other functional forms where the variables/factors have different exponent values may be determined. In some embodiments, other functional forms may include one or more variables/factors used more than once with different exponents.\n\nAs an example, Equation 3 illustrates the results of an exemplary model where factors such as Positive Skewness in the Forcasted Load, and Negative Skewness in the Forecasted Load are each represented twice in the equation as separate variables with different exponents. Though Equation 3 shows each variable assigned an whole number exponent such as +1, −1, +2, etc., other non-integer exponents can be assigned to one or more variables, including for example fractions and decimals.\n\n$ Equation   ( 3 ) ln   ForecastError hd = const + ∑ j = 2 24  α j  Hour j  ?  ∑ k = 2 7  δ k  Day k  ?  ∑ i = 2 12  μ i  Month i  ?  β 1  ?  ?  β 2  ?  ?  β 3  CVP d - 1  ?  β 4  PosSkewP d  ?  β 5  NegSkewP d  ?  β 6  FLPeakRatio d  ?  β 7  FLNadirRatio d  ?  β 8  ?  ?  β 9  FLNadir d  ?  β 10  ?  ?  β 11  PosSkewFL d - 2  ?  β 12  NegSkewFL d - 2  ?  β 13  ?  ?  β 14  ?   ?  indicates text missing or illegible when filed ( 3 )$\n\nIn exemplary embodiments, the functional form of Equation 3 is used to calculate a new set of coefficients (β1, β2, etc.) each which is assigned to a variable, using the sampled data from for the PGE service territory in California using hourly data for the sample period 1 Apr. 2009 through 31 Mar. 2010.\n\nIn exemplary embodiments, a regression equation, like Equation 3, may be analyzed, for example using such techniques as autocorrelation to determine any systematic errors. For example, the residual, or the difference between a predicted load from a model, such as may be predicted using Equation 3, and the actual load, may be calculated. The residual may be further analyzed by calculating the autocorrelation of the residual.\n\nFor example, using Equation 3 and the coefficients from the sampled data, an autocorrelation of the residual can be calculated, the results of which are shown in FIG. 11. FIG. 11 shows when the autocorrelation of the residual is plotted over time, a “hidden” pattern is revealed. In this case, the plot shows the residual being correlated, specifically shown with a repeated series of peaks and valleys. The peaks and valleys may correlate with hourly, daily, weekly, etc. energy trends of power utilization trends of users.\n\nGenerally, in embodiments, a regression analysis can be further applied to reduce the range of the residual to appear substantially as random or “white noise”. For example, in FIG. 11, the boxed area represents an area representing the range in which the residual should be substantially contained within, thus reducing the effect of a systematic error.\n\nBy way of example, in exemplary embodiments, a postulated equation, such as the Forecast Error equation set forth as Equation 3, may be modified to better fit the data and reduce or eliminate the effect of systematic error, such as illustrated in FIG. 11. For example, in exemplary embodiments, the error terms or the residual from an equation may be subjected to an autoregressive-moving-average (ARMA) analysis to refine the original equation to reduce or eliminate systematic errors. In an ARMA model the disturbances (i.e., the differences between the predicted and actual value of the one or more dependent variables) have a linear autoregressive moving-average (ARMA) specification over the period in which the model is estimated. In the simple case of an ARMA (1,1) model, the residual error term in period t, ut, depends on the residual term in the previous period t−1, ut-1, a measure of the “pure” error in period t, εt, and a weighted measure of the pure error term in period t−1, β1t-1, where β1 is an estimated parameter. These calculations can be programmed to be performed by a computer.\n\nIn embodiments, an ARMA analysis applied to an equation may be used to refine that equation to reduce or eliminate systematic errors. For example, applying an ARMA analysis or procedure to the updated regression equation, disturbances from the regression equation may be modeled. Regarding the updated regression equation, the periods/hours times where disturbances affect the regression equation are found to be, using an autoregressive analysis, are 1, 2, 3, 4, 24, 48, 72, 96, 120, 144, 168, and 192. Using a moving average analysis, the disturbances are found to be located at periods/hours 1 through 36, 48, 65, 72, 96, 120, 144, 168, and 192. Applying the ARMA procedure, a new regression equation with new coefficients for one or more of the variables/factors of the updated regression equation can be calculated. FIG. 12 illustrates, an embodiment, where a new updated plot of the autocorrelation of the residual of the updated regression equation after applying the ARMA procedure. In contrast to FIG. 11, in this exemplary embodiment, the residual autocorrelation here is almost fully located within the boxed area, or within an acceptable error range.\n\nIn exemplary embodiments, the output consumers system 102 affiliated with a prediction error system 104 and/or a stand alone prediction error system 104 may determine the load forecasting error by using, amongst other things, equation (4):\n\nwhere:\n\nFDeltaLhd (Forecasted_Delta_Load) can be substantially equal to, for example, the ratio of the forecasted load in hour h, day d, to the forecasted load in hour (h−1), day d\n\nGSigmaFLhd (Geo_Sigma_Forecasted_Load) is the geometric standard deviation in FDeltaLhd over the course of the 24 hours of each day\n\nPeakSparksRatiohd (Peak_Sparks_Ratio) can be a measure of the highest day-ahead price of electricity relative to the price of natural gas.\n\nAvPeakFLd (Average_Peak_Forecasted Load) is the average value of the forecasted load for the day relative to the peak level of forecasted load for the day\n\nThe remaining variables are defined above herein.\n\nIn exemplary embodiments, one or more forecasting equations, such as, for example, equation 4, may include variables relating to the volume of energy traded in the day-ahead electricity market, such as for example, day-ahead hourly volume, the mean daily volume, the coefficient of variation in the daily volume, and the skewness in the volume, to name a few. In some embodiments, for situations regarding power grids where natural gas is not the primary fuel, variables relating to the price of electricity relative to the price of coal may be used in forecasting equation. The price of coal may be normalized in United States Dollars (USD) per Megawatt-hour (MWh) or any other equivalent such as Euros per Megawatt-hour (EUR/MWh), and the like.\n\nFor example, power grids such as PJM (PJM Interconnection is a regional transmission organization “RTO” that coordinates the movement of wholesale electricity in all or parts of 13 states and the District of Columbia.) may report the volume of day-ahead generation offers. Therefore variables such as, for example, day-ahead hourly volume, the mean daily volume, the coefficient of variation in the daily volume, and the skewness in the volume, and the like, may be a part in one or more forecasting equations and applied to one or more power grids according to exemplary embodiments described herein.\n\nFor example, some day-ahead electricity markets may report the volume of day-ahead market activity (e.g., Elspot in Scandinavia). Therefore variables such as day-ahead hourly volume, the mean daily volume, the coefficient of variation in the daily volume, and the skewness in the volume, and the like, may be a part in one or more forecasting equations and applied to one or more power grids according to exemplary embodiments described herein.\n\nAs can be appreciated, the equations described herein are exemplary and various combinations of variables may be used in forecast error equations. In exemplary embodiments, forecast equations using a MFP model may vary across different electricity markets and may, for example, include different variables, coefficients, and/or exponents.\n\nReferring to FIGS. 2A-2B, to accomplish, amongst other things, the above predictions it will be understood that any of competing generators/retailers system 100, output consumers system 102, and/or prediction error system 104 can communicate with each other and/or can be further combined and/or separated. For ease, competing generators/retailers system 100, output consumers system 102, and/or prediction error system 104 are, at times, shown separately. This is merely for ease and is in no way meant to be a limitation.\n\nStill referring to FIGS. 2A-2B, in exemplary embodiments, output consumers system 102 and prediction error system 104 can be separate and/or distinct as shown in FIG. 2A and/or output consumers system 102 and prediction error system 104 can be combined as one step. For ease, the predictions are, at times, disclosed as being affiliated with prediction error system 104. This is merely for ease and is in no way meant to be a limitation.\n\nFurther, any element of competing generators/retailers system 100, output consumers system 102, and/or prediction error system 104 can reside on and/or be affiliated with output consumers system 102, competing generators/retailers system 100, prediction error system 104. For example, competing generators/retailers system 100 can be an algorithm stored in processor readable memory that can be accessed and/or processed by a processor affiliated with output consumers system 102. Further still, competing generators/retailers system 100 can reside on and/or be affiliated with prediction error system 104. For example, competing generators/retailers system 100 can be an algorithm stored in processor readable memory that can be accessed and/or processed by a processor affiliated with prediction error system 104.\n\nIn exemplary embodiments, competing generators/retailers system 100, output consumers system 102, and/or prediction error system 104 can include, but is not limited to, at least one communication portal; at least one graphical user interface; at least one user input; at least one speaker; at least one processor readable memory; at least one processor; and any other reasonable components for use in communicating information (e.g., data), storing information, and processing any form of information.\n\nIn exemplary embodiments, competing generators/retailers system 100, output consumers system 102, and/or prediction error system 104 can be, for example, a mobile phone, computer, networking system, iPad, iPod, iPhone, Smartphone, and Blackberry, to name a few.\n\nNow that exemplary embodiments of the present disclosure have been shown and described in detail, various modifications and improvements thereon will become readily apparent to those skilled in the art.\n\n## Claims\n\n1. A method of adjusting forecast load predictions on an electric grid using one or more operatively connected computers, the method comprising:\n\nobtaining, at the one or more computers, electrical grid information comprising: a day-ahead profile of a forecasted load for the electric grid, a day-ahead sparks ratio, a day-ahead price profile, and an hourly forecasted load relative to the forecasted profile;\naccessing, one or more databases operatively connected to the one or computers, electrical grid historical data;\ncalculating, by the one or more computers, coefficients for a forecast error equation by performing a regression analysis using the historical data, the forecast equation being a function of day-ahead sparks ratio, day-ahead price profile, day-ahead profile of forecasted load, and hourly forecasted load relative to the forecasted profile;\ncalculating, by the one or more computers, forecast prediction errors by applying data from the obtained electrical grid information to the forecast error equation with the calculated coefficients; and\nproviding, by the one or more computers, one or more electronic reports containing forecast errors for the electric grid.\n\n2. The method of claim 1, the method further comprising:\n\ncalculating, by the one or more computers, one or more residual terms as the differences between i) predicted errors from the historical data using the forecast error equation, and ii) actual errors from the historical data;\napplying, by the one or more computers, the residual terms to an auto-regressive moving average analysis, so as to determine one or more disturbances at one or more time times; and\ncalculating, by the one or more computers, a new set of coefficients for the forecast error equation based on the auto-regressive moving average analysis.\n\n3. The method of claim 1, wherein the forecast error equation is a multivariable fractional polynomial model.\n\n4. The method of claim 1, wherein the forecast error equation comprises a Coefficient of Variation variable calculated over a day, using one or more computers, as the ratio of standard deviation of the day-ahead prices to the average hourly price.\n\n5. The method of claim 1, wherein the forecast error equation comprises a Positive Skewness variable calculated over a day, using one or more computers, as values equal to the absolute value of the skewness in the day ahead prices.\n\n6. The method of claim 1, wherein the forecast error equation comprises a Negative Skewness variable calculated over a day, using one or more computers, as values equal to the absolute value of the skewness in the day ahead prices.\n\n7. The method of claim 1, wherein the forecast error equation comprises a Forecasted Load Peak Ratio variable calculated over a day, using one or more computers, as the ratio of the forecasted load to the forecasted peak load.\n\n8. The method of claim 1, wherein the forecast error equation comprises a Forecasted Load Nadir Ratio variable calculated over a day, using one or more computers, as the ratio of the forecasted load to the forecasted minimum load.\n\n9. The method of claim 1, wherein the forecast error equation comprises a Forecasted Load Peak variable equal to values of the forecasted peak hourly load for a day.\n\n10. The method of claim 1, wherein the forecast error equation comprises a Forecasted Load Nadir variable equal to values of the forecasted minimum hourly load for a day.\n\n11. The method of claim 1, wherein the forecast error equation comprises a Coefficient of Variation Forecasted Load Variable calculated over a day, using one or more computers, as the ratio of standard deviation of the day prices to the average hourly price.\n\n12. The method of claim 1, wherein the forecast error equation comprises a Positive Skewness Forecasted Load Variable calculated, using one or more computers, as equal to the skewness in the day-ahead forecasted hourly load when the skewness is positive.\n\n13. The method of claim 1, wherein the forecast error equation comprises a Negative Skewness Forecasted Load variable calculated, using one or more computers, as equal the absolute skewness in the day-ahead forecasted hourly load when the skewness is negative.\n\n14. The method of claim 1, wherein the forecast error equation comprises a Forecasted Delta Load variable calculated over a day, using one or more computers, as equal to the ratio of a forecasted load at a first time to a forecasted load at previous time.\n\n15. The method of claim 14, wherein the forecast error equation comprises a Geo Sigma Forecasted Load variable calculated over a day, using one or more computers, as the geometric standard deviation of the Forecasted Delta Load variable.\n\n16. The method of claim 1, wherein the forecast error equation comprises a Peak Sparks Ratio variable equal to the ratio of the highest day-ahead price of electricity relative to the price of a fuel used to generate electricity.\n\n17. The method of claim 1, wherein the forecast error equation comprises an Average Peak Forecasted Load variable calculated as, using one or more computers, as the average value of the forecasted load for the day relative to the peak level of the forecasted load for the day.\n\n18. The method of claim 1, wherein the forecast error equation comprises a day-ahead hourly volume of energy traded variable.\n\n19. The method of claim 1, wherein the forecast error equation comprises a mean daily volume of energy traded variable.\n\n20. The method of claim 1, wherein the forecast error equation comprises a coefficient of variation in the daily volume of energy traded variable.\n\n21. The method of claim 1, wherein the forecast error equation comprises a skewness in the volume of energy traded variable.\n\n22. The method of claim 1, wherein the sparks ratio is the day-ahead electricity prices relative to the prices of the fuel used to generate electricity.\n\n23. The method of claim 22, wherein the fuel is natural gas.\n\n24. The method of claim 22, wherein the fuel is coal.\n\n25. The method of claim 22, wherein the fuel is oil.\n\nPatent History\nPublication number: 20130096983\nType: Application\nFiled: Apr 4, 2012\nPublication Date: Apr 18, 2013\nPatent Grant number: 10127568\nApplicant: THE CATHOLIC UNIVERSITY OF AMERICA (Washington, DC)\nInventors: Kevin F. Forbes (Silver Spring, MD), Ernest M. Zampelli (Ellicott City, MD), O. Chris S. St. Cyr (Greenbelt, MD)\nApplication Number: 13/438,936\nClassifications\nCurrent U.S. Class: Market Prediction Or Demand Forecasting (705/7.31)\nInternational Classification: G06Q 30/02 (20120101); G06Q 10/06 (20120101);" ]

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[ "# Closed Conduit Hydraulics – Darcy-Weisbach Equation\n\nThe Darcy-Weisbach equation is used to determine flow characteristics in closed conduit systems (pipes). It is probably more common than the Hazen-Williams equation due to it being able to solve for systems in both laminar AND turbulent flow.\n\n$$h_f = f \\frac{L}{D} \\frac{V^2}{2g}$$\n\n• headloss $$h_f$$ (ft)\n• friction factor $$f$$, length $$L$$\n• length $$L$$ (ft)\n• diameter $$D$$ (ft)\n• velocity $$V$$ $$\\frac{ft}{s}\\) • gravity g $$32.2 \\frac{ft}{s^2}$$ #### Friction Factor The friction factor $$f is either given or must be calculated using a Moody-Stanton diagram (available in both the AIO and CERM). Getting the friction factor from a Moody-Stanton chart requires the Reynolds Number \\(Re$$, and relative roughness $$\\frac{\\epsilon}{D}$$. #### Relative Roughness The absolute roughness $$\\epsilon$$ is usually given by what material the pipe is made of (e.g. cast iron, concrete, streel) and requires a table look up. The relative roughness is this table value divided by the diamater $$D$$. #### Reynolds Number$$ \\frac{VD}{v} \n\nWhere V is the velocity of the fluid, D is the diameter, and v is the kinamatic viscosity of the fluid. The standard kinematic viscosity to use is $$1.217 * 10^{-5} \\frac{ft^2}{s}$$.\n\nFind the line that corresponds to the relative roughness and follow it to where it lines up with the Reynold Number. The friction factor $$f$$ is directly horizontal from that point. Practice looking these things up the Moody-Stanton graph will become your fast friend!\n\nRemember $$Q = \\frac{V}{A}$$ and to keep units in check.", null, "" ]

[ null, "http://2.gravatar.com/avatar/8872b8762cf4ac4d415c97260836d524", null ]

[ "Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 31 Textbook Exercise Important Questions and Answers.\n\n## Maharashtra State Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 31\n\nQuestion 1.\nIn the figure given alongside, points S, L, M, and N are on the circle. Answer the questions with the help of the diagram.", null, "", null, "(1) Write the names of the arcs with end-points S and M.\nArcs with the end-points S and M are, arc SLM and arc SNM.\n\n(2) Write the names of the arcs with the end-points L and N.\nArcs with the end-points L and N are, arc LMN and arc LSN.\n\nQuestion 2.\nWrite the names of arcs that points A, B, C, and D in the given circle give rise to.\nArcs with end-points A and C are, arc ABC and arc ADC.\nArcs with end-points B and D are, arc BAD and arc BCD.\n\nQuestion 3.\nGive the names of the arcs that are made by points P, Q, R, S, and T in the figure.\nTaking end-points : P and R, arc PQR, arc PTR.\nTaking end-points: Q and S, arc QRS, arc QTS\nTaking end-points : R and T, arc RST, arc RPT\nTaking end-points : S and P, arc STP, arc SRP\nTaking end-points: Q and T, arc QPT, arc QST", null, "Question 4.\nMeasure and note down the circumference of different circular objects. (It is convenient to use a measuring tape for this purpose.)\n\nQuestion 1.\nDraw circles with the radii given below:\n(1) 1.2 cm", null, "(2) 2.5 cm", null, "", null, "(3) 3.3 cm", null, "Question 2.\nWrite true or false for the following statements:\n(1) Longest chord is a diameter.\n(2) Centre is not lying on the diameter.\n(3) All chords are of equal length.\n(4) All chords passes through the centre.\n(1) True\n(2) False\n(3) False\n(4) False", null, "Question 3.\nMatch the cplumns (A) and (B):", null, "(1) ↔ (c),\n(2) ↔ (a),\n(3) ↔ (d),\n(4) ↔ (b)\n\nQuestion 4.\nComplete the following table by filling in the blanks:", null, "(1) 6 cm\n(2) 10 cm\n(3) 34 cm\n(4) 9 cm", null, "Question 5.\nFrom the following figure, fill in the blanks:", null, "(1) If OP = 4cm then AB = _________ cm, OA = _________ cm, OB = _________ cm.\n(2) If AB = 10 cm then OA = _________ cm, OB = _________ cm, OP = _________ cm.\n(1) AB = 8 cm, OA = 4 cm, OB = 4 cm\n(2) OA = 5 cm, OB = 5 cm, OP = 5 cm\n\nQuestion 6.\nIn the table below, write the names of the points in the interior and exterior of the circle and those on the circle.", null, "Points in the exterior of the circle are A, F and G.\nPoints in the interior of the circle are O, E and B, and Points on the circle are C, D and H.", null, "Question 7.\nRadius of a circle with centre P is 4 cm. Fill in the blanks.\n(1) The, point A is at a distance of 5 cm from the centre P. Flence the point A lies in the ________ of the circle.\n(2) Point B is at a distance of 4 cm. from the centre P. Hence the point B lies ________ circle.\n(3) Point C lies at a distance 3 cm from the centre P. Hence it lies in the ________ of the circle.\n(1) Exterior\n(2) on the circle\n(3) interior\n\nQuestion 8.\nSolve the following:\n(1) What is the length of the diameter of a circle of radius 6 cm?\n(2) What is the length of the radius of a circle of diameter 14 cm?\n(3) Give the names of the arcs that are made by points X, Y, Z and W in this picture.", null, "(4) Give the names of the arc that are made by points E, F, G and H, taking end points E and G.", null, "", null, "(5) If the diameter of a circle is 7 cm. what is the length of the circumference? (Use measure tap)" ]

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[ "", null, "OpenCV  3.4.17-dev Open Source Computer Vision\ncv::xfeatures2d::Elliptic_KeyPoint Class Reference\n\nElliptic region around an interest point. More...\n\n#include <opencv2/xfeatures2d.hpp>\n\nInheritance diagram for cv::xfeatures2d::Elliptic_KeyPoint:", null, "Public Member Functions\n\nElliptic_KeyPoint ()\n\nElliptic_KeyPoint (Point2f pt, float angle, Size axes, float size, float si)\n\nvirtual ~Elliptic_KeyPoint ()", null, "Public Member Functions inherited from cv::KeyPoint\nKeyPoint ()\nthe default constructor More...\n\nKeyPoint (Point2f pt, float size, float angle=-1, float response=0, int octave=0, int class_id=-1)\n\nKeyPoint (float x, float y, float size, float angle=-1, float response=0, int octave=0, int class_id=-1)\n\nsize_t hash () const\n\nPublic Attributes\n\nSize_< float > axes\nthe lengths of the major and minor ellipse axes More...\n\nfloat si\nthe integration scale at which the parameters were estimated More...\n\nMatx23f transf\nthe transformation between image space and local patch space More...", null, "Public Attributes inherited from cv::KeyPoint\nfloat angle\n\nint class_id\nobject class (if the keypoints need to be clustered by an object they belong to) More...\n\nint octave\noctave (pyramid layer) from which the keypoint has been extracted More...\n\nPoint2f pt\ncoordinates of the keypoints More...\n\nfloat response\nthe response by which the most strong keypoints have been selected. Can be used for the further sorting or subsampling More...\n\nfloat size\ndiameter of the meaningful keypoint neighborhood More...", null, "Static Public Member Functions inherited from cv::KeyPoint\nstatic void convert (const std::vector< KeyPoint > &keypoints, std::vector< Point2f > &points2f, const std::vector< int > &keypointIndexes=std::vector< int >())\n\nstatic void convert (const std::vector< Point2f > &points2f, std::vector< KeyPoint > &keypoints, float size=1, float response=1, int octave=0, int class_id=-1)\n\nstatic float overlap (const KeyPoint &kp1, const KeyPoint &kp2)\n\nDetailed Description\n\nElliptic region around an interest point.\n\n◆ Elliptic_KeyPoint() [1/2]\n\n cv::xfeatures2d::Elliptic_KeyPoint::Elliptic_KeyPoint ( )\n\n◆ Elliptic_KeyPoint() [2/2]\n\n cv::xfeatures2d::Elliptic_KeyPoint::Elliptic_KeyPoint ( Point2f pt, float angle, Size axes, float size, float si )\n\n◆ ~Elliptic_KeyPoint()\n\n virtual cv::xfeatures2d::Elliptic_KeyPoint::~Elliptic_KeyPoint ( )\nvirtual\n\n◆ axes\n\n Size_ cv::xfeatures2d::Elliptic_KeyPoint::axes\n\nthe lengths of the major and minor ellipse axes\n\n◆ si\n\n float cv::xfeatures2d::Elliptic_KeyPoint::si\n\nthe integration scale at which the parameters were estimated\n\n◆ transf\n\n Matx23f cv::xfeatures2d::Elliptic_KeyPoint::transf\n\nthe transformation between image space and local patch space\n\nThe documentation for this class was generated from the following file:" ]

[ null, "https://docs.opencv.org/3.4/opencv-logo-small.png", null, "https://docs.opencv.org/3.4/d2/d71/classcv_1_1xfeatures2d_1_1Elliptic__KeyPoint.png", null, "https://docs.opencv.org/3.4/closed.png", null, "https://docs.opencv.org/3.4/closed.png", null, "https://docs.opencv.org/3.4/closed.png", null ]

[ "# Valid Parenthesis Problem Index Out of Range (Go)\n\n## Issue\n\nI was converting code for a problem I previously solved in Python into Go and ran into a runtime error: index out of range with length 1 on this line\n\n`````` if string(s[i]) == \")\" && arr[len(s) - 1] == \"(\"{\n``````\n\nHere is the code in entirety.\n\n``````func isValid(s string) bool {\narr := make([]string, 0)\nif len(s) < 2 {\nreturn false\n}\n\nfor i := 0; i < len(s); i++ {\n\nif string(s[i]) == \"(\" || string(s[i]) == \"[\" || string(s[i]) == \"{\" {\narr = append(arr, string(s[i]))\n} else if string(s[i]) == \")\" || string(s[i]) == \"]\" || string(s[i]) == \"}\" {\nif len(arr) >= 1 {\nif string(s[i]) == \")\" && arr[len(s) - 1] == \"(\"{\narr = arr[:len(arr) -1]\n} else if string(s[i]) == \"]\" && arr[len(s) - 1] == \"[\" {\narr = arr[:len(arr) -1]\n} else if string(s[i]) == \"}\" && arr[len(s) - 1] == \"{\" {\narr = arr[:len(arr) -1]\n} else {\nreturn false\n}\n}else {\nreturn false\n}\n}\n}\nif reflect.DeepEqual(arr, []string{\"\"}) {\nreturn true\n} else {\nreturn false\n}\n\n}\n\n``````\n\nWhat is the cause of the issue?\n\n## Solution\n\nYou’ve used `len(s)-1` to find the last index of `arr`, but it should be `len(arr)-1`.\n\nHere’s a working version of your code, that fixes some additional bugs, and avoids needless conversion to string for single characters. https://go.dev/play/p/sazf2RguoIr\n\n``````package main\n\nimport \"fmt\"\n\nfunc isValid(s string) bool {\nvar arr []rune\nfor _, c := range s {\nif c == '(' || c == '[' || c == '{' {\narr = append(arr, c)\n} else if c == ')' || c == ']' || c == '}' {\nif len(arr) == 0 {\nreturn false\n}\nvar last rune\narr, last = arr[:len(arr)-1], arr[len(arr)-1]\nif c == ')' && last != '(' || c == ']' && last != '[' || c == '}' && last != '{' {\nreturn false\n}\n}\n}\nreturn len(arr) == 0\n}\n\nfunc main() {\ncases := []string{\n\"()[({}())]\",\n\"(]\",\n\"((())\",\n\"{{{}}}\",\n\"{{{[][][]}}}\",\n}\nfor _, c := range cases {\nfmt.Println(c, isValid(c))\n}\n}\n``````\n\nAnswered By – Paul Hankin\n\nAnswer Checked By – Senaida (GoLangFix Volunteer)" ]

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[ "", null, "# Chapter 2 Functions and Graphs Section 2 Elementary Functions: Graphs and Transformations.\n\n## Presentation on theme: \"Chapter 2 Functions and Graphs Section 2 Elementary Functions: Graphs and Transformations.\"— Presentation transcript:\n\nChapter 2 Functions and Graphs Section 2 Elementary Functions: Graphs and Transformations\n\n2 Identity Function Domain: R Range: R\n\n3 Square Function Domain: R Range: [0, ∞)\n\n4 Cube Function Domain: R Range: R\n\n5 Square Root Function Domain: [0, ∞) Range: [0, ∞)\n\n6 Square Root Function Domain: [0, ∞) Range: [0, ∞)\n\n7 Cube Root Function Domain: R Range: R\n\n8 Absolute Value Function Domain: R Range: [0, ∞)\n\n9 Vertical Shift  The graph of y = f(x) + k can be obtained from the graph of y = f(x) by vertically translating (shifting) the graph of the latter upward k units if k is positive and downward |k| units if k is negative.  Graph y = |x|, y = |x| + 4, and y = |x| – 5.\n\n10 Vertical Shift\n\n11 Horizontal Shift  The graph of y = f(x + h) can be obtained from the graph of y = f(x) by horizontally translating (shifting) the graph of the latter h units to the left if h is positive and |h| units to the right if h is negative.  Graph y = |x|, y = |x + 4|, and y = |x – 5|.\n\n12 Horizontal Shift\n\n13 Reflection, Stretches and Shrinks  The graph of y = Af(x) can be obtained from the graph of y = f(x) by multiplying each ordinate value of the latter by A.  If A > 1, the result is a vertical stretch of the graph of y = f(x).  If 0 < A < 1, the result is a vertical shrink of the graph of y = f(x).  If A = –1, the result is a reflection in the x axis.  Graph y = |x|, y = 2|x|, y = 0.5|x|, and y = –2|x|.\n\n14 Reflection, Stretches and Shrinks\n\n15 Reflection, Stretches and Shrinks\n\n16 Summary of Graph Transformations  Vertical Translation: y = f (x) + k k > 0 Shift graph of y = f (x) up k units. k < 0 Shift graph of y = f (x) down |k| units.  Horizontal Translation: y = f (x + h) h > 0 Shift graph of y = f (x) left h units. h < 0 Shift graph of y = f (x) right |h| units.  Reflection: y = –f (x) Reflect the graph of y = f (x) in the x axis.  Vertical Stretch and Shrink: y = Af (x) A > 1: Stretch graph of y = f (x) vertically by multiplying each ordinate value by A. 0 < A < 1: Shrink graph of y = f (x) vertically by multiplying each ordinate value by A.\n\n17 Piecewise-Defined Functions  Earlier we noted that the absolute value of a real number x can be defined as  Notice that this function is defined by different rules for different parts of its domain. Functions whose definitions involve more than one rule are called piecewise-defined functions.  Graphing one of these functions involves graphing each rule over the appropriate portion of the domain.\n\n18 Example of a Piecewise-Defined Function Graph the function\n\n19 Example of a Piecewise-Defined Function Graph the function Notice that the point (2,0) is included but the point (2, –2) is not.\n\nDownload ppt \"Chapter 2 Functions and Graphs Section 2 Elementary Functions: Graphs and Transformations.\"\n\nSimilar presentations" ]

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[ "# A Treatise on Plane and Spherical Trigonometry: And Its Applications to Astronomy and Geodesy\n\nD. C. Heath & Company, 1892 - 368 páginas\n0 Opiniones\nLas opiniones no están verificadas, pero Google revisa que no haya contenido falso y lo quita si lo identifica\n\n### Comentarios de la gente -Escribir un comentario\n\nNo encontramos ningún comentario en los lugares habituales.\n\n### Pasajes populares\n\nPágina 148 - In any triangle the square of any side is equal to the sum of the squares of the other two sides minus twice the product of these two sides and the cosine of their included angle.\nPágina 147 - Law of Sines. — In any triangle the sides are proportional to the sines of the opposite angles.\nPágina 278 - AB'C, we have by (4) cos a' — cos b cos c' + sin b sin c' cos B'AC, or cos(тг— a) = cos b cos(тг— c) + sin b sin(тт — C)COS(тг —A). .-. cos a = cos b cos с + sin b sin с cos A.\nPágina 278 - ... cos a = cos b cos с + sin b sin с cos A ; (2) cos b = cos a cos с + sin a sin с cos в ; ^ A. (3) cos с = cos a cos b + sin a sin b cos C.\nPágina 278 - A cos 6 = cos a cos c + sin a sin c cos B cos c = cos a cos 6 + sin a sin 6 cos C Law of Cosines for Angles cos A = — cos B...\nPágina 6 - Radian is the angle subtended, at the centre of a circle, by an arc equal in length to the radius...\nPágina 17 - If the cosine of A be subtracted from unity, the remainder is called the versed sine of A. If the sine of A be...\nPágina 89 - The logarithm of any power of a number is equal to the logarithm of the number multiplied by the exponent of the power.\nPágina 149 - In every plane triangle, the sum of two sides is to their difference as the tangent of half the sum of the angles opposite those sides is to the tangent of half their difference." ]

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[ "This post will introduce a new feature for binary blobs like connected components in a text.\n\nThe feature is called patch histogram and it's the histogram of 3x3 patches of black and white pixels. We collect all 3x3 patches and count their frequency.\n\n3x3 patch for a binary image contains \\$2^9 = 512\\$ different combinations. For each of these combinations, we assign a number. I wrote the implementation in Python and here is a lookup table that converts all possible 3x3 patches to their ids.\n\n``````patch_histogram_dict = {}\nfor f in range(0, 512):\ni22 = f & 1\ni21 = (f >> 1) & 1\ni20 = (f >> 2) & 1\ni12 = (f >> 3) & 1\ni11 = (f >> 4) & 1\ni10 = (f >> 5) & 1\ni02 = (f >> 6) & 1\ni01 = (f >> 7) & 1\ni00 = (f >> 8) & 1\npatch_histogram_dict[ma(i00, i01, i02, i10, i11, i12, i20, i21, i22)] = f\n``````\n\nI was looking a skeleton feature that can be applied to components after medial axis transform. However, we will use binarized connected component images directly in this version and take a look at skeletal version later.\n\nAbove, we used a function `ma` to convert a set of numbers to a tuple of tuples. It's simple as\n\n``````def ma(i00, i01, i02, i10, i11, i12, i20, i21, i22):\nreturn ((i00, i01, i02),\n(i10, i11, i12),\n(i20, i21, i22))\n``````\n\nThe first step in feature generation is marking each pixel of the image with its id. Then a simple `np.histogram` call generates the histogram.\n\n``````for i in range(1, rows - 1):\nfor j in range(1, cols - 1):\ns = framed[(i-1):(i+2), (j-1):(j+2)]\nmarks[i-1, j-1] = patch_dict[ma(s[0, 0], s[0, 1], s[0, 2],\ns[1, 0], s[1, 1], s[1, 2],\ns[2, 0], s[2, 1], s[2, 2])]\n``````\n\n=framed= is a copy of the original image with a 1 pixel frame, so that boundary pixel are also counted without much checking. `patch_dict` is a parameter that the function receives. Its default value is the dictionary that we made above, but any kind of patch dictionaries can be used.\n\nThe feature is a call to `numpy.histogram` as follows\n\n``````histogram = numpy.histogram(marked_points,\nhist_range,\ndensity=True,\nbins=bins)\n``````\n\n=marked~points~= are the result of above pixel counting loop. `hist_range` is the minimum and maximum value for the patch dictionary. `density` parameter is True, so that the histogram is normalized to 1 and feature become size invariant. `bins` is a parameter we can set freely, but by default it's equal to the range of elements.\n\nThe comparison can be done with standard histogram comparison methods, like EMD or Chi Square distance metrics." ]

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[ "# Interview Aptitude Questions with Solution\n\n9789617617\n\n## CodeBind Technologies Online-Aptitude Test Interview Questions\n\nA. 45\nB. 60\nC. 75\nD. 90\n\nExplanation:\n\nLet number of notes of each denomination be x.\n\nThen x + 5x + 10x = 480\n\n16x = 480\n\nx = 30.\n\nHence, total number of notes = 3x = 90.\n\nA. 60\nB. 95\nC. 120\nD. 150\n\nExplanation :\n\nJust Multiply each term in the given sequence with natural numbers. i.e 1 to infinity ( need of the question is only till 7).\nSo,\n1 x 1 = 1\n1 x 2 = 2\n2 x 3 = 6\n6 x 4 = 24\n24 x 5 = 120\n\nA. 37\nB. 44\nC. 50\nD. 55\n\nExplanation :\n\n1,3,4,8,15,27,...? next what\n1\n3\n4 = 3 + 1\n8 = 4 + 3 + 1\n15 = 8 + 4 + 3\n27 = 15 + 8 + 4\n50= 27 + 15 + 8\n\nA. True\nB. False\nC. Uncertain\n\nExplanation :\n\nThe first two statements give information about Zoe's tulips and pansies. Information about any other kinds of flowers cannot be determined\n\nA. Grandmother\nB. Father\nC. Grandfather\nD. Grandfather or Grandmother\n\nExplanation :\n\nS is daughter of P and sister of T. So, T is daughter of P.\nNow, the sister of P is the daughter of M. This means that P is also the daughter of M. Clearly, T is the granddaughter of M.\nSo, M is the Grandfather or Grandmother of T.\n\nA. Novel\nB. Glass\nC. Cover\nD. Page\n\nExplanation :\n\nA window is made up of panes, and a book is made up of pages. The answer is not (choice a) because a novel is a type of book. The answer is not (choice b) because glass has no relationship to a book. (Choice c) is incorrect because a cover is only one part of a book; a book is not made up of covers.\n\nA. Father\nB. Brother\nC. Cousin\nD. Uncle\nE. Grandson\n\nExplanation :\n\nMy grandfather's only son — My father. So, the girl is the daughter of Vipul's father i.e., Vipul is the girl's brother.\n\nA. Dish\nB. Soup\nC. Spoon\nD. Food\n\nExplanation :\n\nCoffee goes into a cup and soup goes into a bowl. Choices a and c are incorrect because they are other utensils. The answer is not choice d because the word food is too general.\n\nA. (91 + 1/3)%\nB. (93 + 1/3)%\nC. (97 + 1/3)%\nD. (98 + 1/3)%\n\nExplanation :\n\nif 270 is 100% then 252 is what %\nie: 100/270 *252\n= 28/3 * 10\n=93.3 % (93 + 1/3)%\n\nA. 79\nB. 87\nC. 86\nD. 76\n\nExplanation :\n\nAverage = total runs / no.of innings = 32\nSo, total = Average x no.of innings = 32 x 10 = 320.\n\nNow increase in avg = 4runs. So, new avg = 32+4 = 36runs\nTotal runs = new avg x new no. of innings = 36 x 11 = 396\nRuns made in the 11th inning = 396 - 320 = 76\n\nA. 20 litres\nB. 30 litres\nC. 40 litres\nD. 60 litres\n\nExplanation :\n\nQuantity of milk = 60 x 2 litres = 40 litres.\n3\nQuantity of water in it = (60- 40) litres = 20 litres.\nNew ratio = 1 : 2\nLet quantity of water to be added further be x litres.\nThen, milk : water = 40 .\n20 + x\nNow, 40 = 1\n20 + x 2\n20 + x = 80\nx = 60.\nQuantity of water to be added = 60 litres.\n\nA. 744589\nB. 744193\nC. 166479\nD. 745194\n\nExplanation :\n\nE N T R Y S A D\n1 2 3 4 5 9 7 8\n\nA is coded as 7,\nR as 4,\nE as 1,\nS as 9 and\nT as 3.\nSo, ARREST is coded as 744193.\n\nA. See\nB. Kee\nC. Lee\nD. Pee\nE. Tee\n\nExplanation :\n\nIn the first and second statement, the common word is 'juice' and the common code word is 'see'. So 'see' means 'juice'.\nIn the second and third statements, the common word is 'is' and the common code is 'lee'.\nSo, 'lee' means 'is'.\nThus, in the second statement, the remaining word is 'sweet' which is coded as 'kee'.\nHence, option (B) is the correct answer.\n\nA. JAK\nB. HAL\nC. HAK\nD. JAI\n\nExplanation :\n\nThe middle letters are static, so concentrate on the first and third letters. The series involves an alphabetical order with a reversal of the letters. The first letters are in alphabetical order: F, G, H, I , J. The second and fourth segments are reversals of the first and third segments. The missing segment begins with a new letter.\n\nA. 145\nB. 129\nC. 113\nD. 97\n\nExplanation :\n\nAddition of 4 at the beginning sequentially multiplied by 2\n21+4=25\n25+(4*2)=33\n33+(8*2)=49\n49+(16*2)=81\n81+(32*2)=145\n\nA. 44\nB. 45\nC. 46\nD. 47\n\nExplanation :\n\nThe sequence ascends in od numbers, beginning with 5. as 5, 7, 9, 11, 13, …….\n1\n6=1+5\n13=6+7\n22=13+9\n33=22+11\nSo, the next number should get increase of 13.\nSo, 33+13=46\n\nA. Son\nB. Brother\nC. Cousin\nD. Grandson\nE. Uncle\n\nExplanation :\n\nHere's Y.\nY has a son.\nY's son has a son (therefore Y has a grandson).\nAnd X is the brother of the above guy (i.e. Y's grandson).\nSo, X is the grandson of Y.\n\nA. True\nB. False\nC. Uncertain\n\nExplanation :\n\nFrom the first two statements, you know that the Kingston Mall has the most stores, so the Kingston Mall would have more stores than the Four Corners Mall.\n\nA. Brother\nB. Cousin\nC. Aunt\nD. Sister\n\nExplanation :\n\nOnly son of Amar's mother's father -- Amar's maternal uncle.\nSo, the girl's maternal uncle is Amar's maternal uncle.\nThus, the girl's mother is Amar's aunt.\n\nA. Niece\nB. Daughter\nC. Wife\nD. Daughter-in-law\nE. None of these\n\nExplanation :\n\nOnly son of Neena's father--in-law Mahipal -- Neena's husband. So, Raman is Neena's husband and Anita and Bindu are his daughters. Thus, Bindu is the grand daughter of Mahipal.\n\nA . 47\nB. 46\nC. 39\nD. none of the above\n\nExplanation:\n\nRequired number = (38 + 9 - 1) = 47-1 = 46\n\nA. Daughter\nB. Brother\nC. Husband\nD. Sister-in-law\n\nExplanation:\n\nA is father of C and C is sister of D. So, A is father of D.\nBut D is son of B. So, B is the mother of D and wife of K Also. E is the brother of A So, B is the sister-in-law of E\n\nA. Tent\nB. City\nC. Dwelling\nD. Palace\n\nExplanation:\nAbove the line, the relationship shows a progression of sources of light. The relationship below the line shows a progression of types of palace, from smallest to largest. (Choice a) is incorrect because a tent is smaller than a palace. Choices b and c are wrong because they are not part of the progression\n\nA. 1/70\nB. 1/840\nC. 1/8\nE. 1/40320\n\nExplanation: total digits = 8\neven digit = 4 ( 2 , 4 , 6 , 8)\nodd digit = 4 ( 1 , 3 , 5 , 7)\nno repetition\nso probability of 1st digit being even = 4/8 = 1/2\n2nd digit being even = 3/7\n3rd digit being even = 2/6 = 1/3\n4th digit being even = 1/5\nprobability that first 4 digits of the code are even numbers\n= (1/2)(3/7)(1/3)(1/5)\n= 1/70\n\nA. 155679\nB. 167890\nC. 179890\nD. 177366\n\nExplanation:\n\nGiven Population of a town = 160000.\nR1 = 3%, R2 = 2.5%, R3 = 5%.\nThen the present population of the town = 160000(1+3/100)(1+2.5/100)(1+5/100)= 177366.\n\nA. 5%\nB. 2.8%\nC. 3.9%\nD. 7%\n\nExplanation:\n\nA decade means 10 years.So,Total increase in population = 2,62,500 - 1,75,000\n=> 87,500\nSo, Average increase per year = 87,500/10 = 8,750\n% increase per year = (increase per year/ initial population) × 100\n=> (8750 / 1,75,000) × 100 => .05 × 100 => 5%\n\nA. 12.9\nB. 13.8\nC. 17\nD. 15.8\n\nExplanation:\n\nFirst ten odd prime no ÷10\n(3+5+7+11+13+17+19+23+29+31)÷10\n=158÷10\n15.8\n\nA. 2:5\nB. 4:5\nC. 6:7\nD. 3:8\n\nExplanation:\n\nLet the third number be x.\nThen, first number = 120% of x =120x/100 = 6x/5\nSecond number =150% of x = 150x/100 = 3x/2\nRatio of first two numbers = 6x/5 : 3x/2 = 12x : 15x = 4 : 5\n\nA. 222\nB.222.5\nC. 267\nD.220\n\nExplanation :\n\nCost Price = 178 × 100/(100 - 11)\n= 178 × 100/89\n= 2 × 100 = 200 (Rupees)\nIf he profit 11% then the Selling Price Will be = 200 + 200 × 11/100\n= 200 + 22 = 222 (Rupees)\n\nA. ALEUT\nB. ALGTU\nC. ALGUT\nD. ALGRT\n\nExplanation:\n\n(algut) is the ans.\n2=a\n3=l\n5=g\n4=u\n9=t\n\nA. 381171\nB. 381191\nC. 832252\nD. 835545\n\nExplanation:\n\nTake common letters from england and france and substitute in Greece\n\nA.25%\nB. 30%\nC. 35%\nD.20%\n\nExplanation:\n\n7 - 5.25 = 1.75\nTherefore,\n1.75/7 × 100\n= 25 %\n\nA. Rs. 4991\nB. Rs. 5991\nC. Rs. 6001\nD. Rs. 6991\n\nExplanation:\n\nGiven average = Rs. 6500\nAnd sales are Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months.\nLet the sale of 6th month be xThen,\n6500 = (x+6435+ 6927+6855+7230+6562 )/6\n39000 = x + 34009 => x = 4991\nTherefore, sale at 6th month = Rs. 4991.\n\nA. 6 Years\nB.7 Years\nC. 8 Years\nD. 12 Years\n\nExplanation:\n\nReena is 12 years old\nLet the present age of Suma = x years\nThen the present age of Reena = 2x years\nThree years ago, age of Suma = x - 3\nThree years ago, age of Reena = 2x - 3\nAccording to the question\n=>2x-3 = (x-3)\n=>2x-3 = 3x-9\n=>3x-2x = 9-3\n=>x=6\nTherefore the present age of Suma = 6 years\nPresent age of Reena = 12 years\nTherefore, Reena is 12 years old\n\nA. 35\nB. 38\nC. 40\nD. 42\n\nExplanation:\n\nlet the no. of questions e answered correctly be x\nlet the no. of questions he answered wrong be y\ntotal no. of questions =\nx+y=60\ny=60-x------eqn 1\n4x-y=130\n4x-60+x=130\n5x=130+60\n5x=190\nx=190/5\nx=38\ntherefore, the no. of questions he answered correctly=38\n\nA. 18\nB. 9\nC.7\nD.20\n\nExplanation:\n\nSum of ages of three boys = 45 years\nNow, (3x + 5x + 7x) = 45\n⇒ 15x = 45\n⇒ x = 3\nSo, age of youngest boy = 3x\n= 3 × 3\n= 9 years\n\nA.100\nB. 200\nC. 180\nD.400\n\nExplanation:\n\nLet the number of students in three classes be 2x, 3x and 5x respectively.\nDue to increase of 40 students in each class, we have\n= 2x + 40 = 4\n3x + 40 5\n\n⇒ 10x + 200 = 12x + 160\n⇒ 2x = 200 – 160 ⇒ 2x = 40\n⇒ x = 20\n∴ Original strength\n= 10x = 10 × 20 = 200\n\nA. I alone is sufficient\nB. II alone is sufficient\nC. Neither I nor II is sufficient\nD. Either I or II is sufficient\n\nExplanation:\n\nFrom both the given statements, we find that Mr. Rijo reached the bus stand at 9 AM and a bus is sure to arrive at 10 AM. So. Mr. Rijo has to wait for at the most one hour.\n\nA. Rs. 250000\nB. Rs. 300000\nC. Rs. 450000\nD. Rs. 150000\n\nExplanation:\n\nl = 300m , b = 125m\nArea of the rectangular = l*b = 300 * 125 = 37,500\nCost of tilling = 8 * 37500 = 300000\n\nA. 360\nB. 180\nC. 270\nD. 140\n\nExplanation:\n\nNumber of hours from 8am till 2pm= 6\n\nThe rotation of an hour hand in one hour= 30°\n\nTotal degree of rotation= 360°\n\nTherefore, the Angle traced by the hour hand in 6 hours is= (360/12)*6 = 180\n\nA. 24\nB. 22\nC. 18\nD. 20\n\nExplanation:\n\nNo of boxes = total no of books/No of books in 1 box\nTotal no of books =528\nNo of books in 1 box = 24\nNo of boxes =528/24 = 22\n\nA. 10\nB. 8\nC. 4\nD. 6\n\nExplanation:\n\nM1*H1*D1 = M2*H2*D2\n4*5*16 = 20*4*D2\n320/80 =D2\nD2=4\n\nA. 501\nB. 105\nC. 525\nD. 150\n\nExplanation:\n\nAs given\nTwo-fifth of one -third of three-seventh of number is 15.\nLet us assume that the unknown number be x .\n2/5*1/3*3/7*x =15\nx = 262.5\nPercentage = Part value*100/Total value\nPercentage = 40%,Total value = 262.5\n40=Part value*100/262.5\nPart value = 105.\nThe 40% of derived number is 105 .\n\nA. 30 10\nB. 65 45\nC. 35 15\nD. 45 25\n\nExplanation:\n\na = b + 20\na-5 = 3(b-5)\nb+20-5 = 3(b-5)\nb+15 = 3b-15\n2b = 30\nb = 15 yrs\na = b + 20\na = 15+20\na = 35 yrs\n\nA. 360\nB. 480\nC. 720\nD. 5040\n\nExplanation:\n\nThe word 'LEADING' has 7 different letters.\nWhen the vowels EAI are always together, they can be supposed to form one letter.\nThen, we have to arrange the letters LNDG (EAI).\nNow, 5 (4 + 1) letters can be arranged in 5! = 120 ways.\nThe vowels (EAI) can be arranged among themselves in 3! = 6 ways.\nRequired number of ways = (120 x 6) = 720.\n\nA. AAEILMTR\nB. AAIELMRT\nC. AAELIMRT\nD. AAEILMRT\n\nExplanation:\n\nArranged the letters in an alphabetical order.\n\nA. BEA\nB. ABE\nC. DEC\nD. ECB\n\nExplanation:\n\nIn the above question, it can be clearly seen that the sequence BEA is the most appropriate since if all cats lay eggs is true and it is also true that all tigers are cats, then it will also be true that all tigers lay eggs.\n\nA. GERH\nB. GRHE\nC. GREH\nD. GHRE\nE. GEHR\n\nExplanation:\n\nReverse the word and move each letter -1.Reverse of FISH is HSIF subtract 1 from each letter of HSIF. So code of FISH become GRHE.\n\nA)ACFJO\nB)AEIMQ\nC)DINSX\nD)EHKNQ\n\nExplanation :\n\nABCDEFGHIJKLMNO\n1 2 3 4\n\nA)7\nB)10\nC)12\nD)13\n\nExplanation :\n\nThis is a simple alternating addition and subtraction series. In the first pattern, 3 is added; in the second, 2 is subtracted.\n\nA)Kapil\nB)Jai\nC)Both\nD)None\n\nExplanation :\n\nKapil because Kapil is brother of Jai’s Brother\n\nA)F2\nB)F7\nC)F8\nD)F1\n\nExplanation :\n\nTo check the spelling and grammar in just a sentence or paragraph of your document, select the text you want to check and then press F7.\n\nA)10\nB)12\nC)15\nD)18\n\nExplanation :\n\na - b = 3 ---( 1 )\na² + b² = 29----( 2 )\nWe know that the algebraic identity,\n( a - b )² = a² + b² - 2ab\n3² = 29 - 2 ab\n[ From ( 1 ) and ( 2 ) ]\n9 = 29 - 2ab\n2ab = 29 - 9\n2ab = 20\nab = 20/2\nab = 10\n\nA)B\nB)A\nC)Both\nD)Insufficient Data\n\nExplanation :\n\n20×100/30=66%\n40×100/70= 57.14%\nB student score more\n\nA)251\nB)254\nC)255\nD)256\n\nExplanation :\n\nAs George is sitting to the left of Henry,so, George's seat is 252. The next seat to the left,where Inez was sitting is 251.\n\nA)12\nC)6\nC)5\nD)10\n\nExplanation :\n\nPipe A's 1hr work=1/10\npipe B's 1hr work =1/15\n(A+B)'s 1hr work=1/10+1/15=\n5/30=1/6\nNow A and B can do it in 1÷1/6=6days.\n\nA)15%\nB)20%\nC)10%\nD)25%\n\nExplanation :\n\nLoss = 9000-7650\n= 1350\nLoss% = 1350/9000×100\n= 15\n\nA)BY\nB)DX\nC)BX\nD)CX\n\nExplanation :\n\nBecause the gap between A, E, G, are about 2 Words are in forward and the gap between Z, V, T are about 2 words are in backward.\n\nA)7\nB)5\nC)2\nD)9\n\nExplanation :\n\nZ = 1, Y= 2, X = 3, W = 4, V = 5,….\nSo answer of the above solution is 5.\n\nA)RUR\nB)QUR\nC)RUS\nD)SUR\nE)PUR\n\nExplanation :\n\nFirst letter of each term is J, L, N, P, R Second letter of each term is A, E, I, O, U (all vowels) Third letter of each term is Z, X, V, T, R.\nRUR will come next in the series.\n\nA)A16\nB)G4\nC)E4\nD)E3\n\nExplanation :\n\nThe letters decrease by 1; the numbers are multiplied by 2.\n\nA)22\nB)19\nC)25\nD)21\n\nExplanation :\n\nThere are two series here\n39, 32,..(decreased by 2)\n41, 34, 27...(decreased by 5)\n\nHence, next term = 27 - 2 = 25\n\nA)PORIVEDCNE\nB)RPOVPINECE\nC)RPVODINEEC\nD)EORIVEDCEP\n\nExplanation :\n\nThe word ‘APPEAR’ has been coded by interchanging the two letters at a time. Hence PROVIDENCE is coded as in (C)\n\nA)(1/16)\nB)(1/8)\nC)(1/12)\nD)(1/64)\n\nExplanation :\n\nEach no. is getting divided by 2\n1st no. is 2\n\n2nd no. is (2/2)= 1\n\n3rd no. is (1/2)= 1/2\n\n4th no. is (1/2)*(1/2)= 1/4\n\nSo next no. Will be (1/4)*(1/2)= 1/8\n\nSo answer of the above solution is (1/8).\n\nA)Uncle\nB)Father\nC)Grandfather\nD)Brother\n\nExplanation :\n\nClearly, the only daughter or Monika's father is Monika herself\nSo, Rahul's mother is Monika\nThus, Monika's husband is the father of Rahul\n\nA)3\nB)2\nC)4\nD)7\nE)5\n\nExplanation :\n\n324 - Light is bright\n629 - Girl is beautiful\n4758 - I prefer bright clothes\n\nA)chn\nB)nhi\nC)ptr\nD)wop\n\nExplanation :\n\nnso ptr kli chn - Sharma gets marriage gift\nptr lnm wop ch - wife gives marriage gift\ntti wop nhi - he gives nothing\n\nA)CMN\nB)UJI\nC)VIJ\nD)IJT\n\nExplanation :\n\nThe first letter is in backward pattern and the last two letters are going forward\n\nA. True\nB. False\nC. Uncertain\n\nExplanation:\n\nBecause the first two sentences are true, both Josh and Darren saw more movies than Stephen. However, it is uncertain as to whether Darren saw more movies than Josh.\n\nA. 24\nB. 28\nC. 77\nD. 218\n\nExplanation:\n\nEach number in the series is the preceding number multiplied by 3 and then decreased by 2.\n\nA. 128\nB. 144\nC. 192\nD. 256\n\nExplanation:\n\nThe Pattern is *8,*4,….\nMissing term =128*2 =256\n\nA. 12.5%\nB. 40%\nC. 80%\nD. 125%\n\nExplanation:\n\n5 : 4 = 5/4 = ( (5/4) x 100 )% = 125%.\n\nA. 64\nB. 216\nC. 256\nD. 24\n\nExplanation:\n\nIf repetition is allowed, 4^4 or 256, as the number could be 1111 just as well as 1359.\nHowever, if repetition isn’t allowed, the first digit can be any of the four, the second any except the one used for the first digit, the third any except the two digits so far used and the fourth can only be one digit, so 4! or 24 possibilities.\n\nA. Rs. 375\nB. Rs. 400\nC. Rs. 600\nD. Rs. 800\n\nExplanation:\n\nC’s 1 day’s work = 1/3-(1/6+1/8) = 1/3-7/24 = 1/24.\nA’s wages: B’s wages: C’s wages = 1/6:1/8:1/24 = 4:3:1.\nC’s share (for 3 days) =Rs.(3*1/24*3200) = Rs. 400.\n\nA. 220 Km\nB. 224 Km\nC. 230 Km\nD. 234 km\n\nExplanation:\n\n(1/2)x/21 + (1/2)x/24 = 10\nx/21 + x/24 = 20\n15x = 168* 20\nX = (168 * 20/15)\nX =224 km\n\na) 33\nb) 81\nc) 93\nd) 97\n\nExplanation:\n\n97 is a prime number because it has only two distinct divisors: 1 and itself (97). As a consequence, 97 is only a multiple of 1 and 97.\n\na) 500\nb) 600\nc) 400\nd) 700\n\nExplanation:\n\n40%=sold apples\nso, 100-40=60% remaining\n60%=420 apples\n420/60×100=700 apples\n\na) 138\nb) 148\nc) 162\nd) 216\n\nExplanation:\n2*3=6\n6*3=18\n18*3=54\n54*3=162\n\na) S\nb) U\nc) T\nd) None\n\nExplanation:\n\nA B C D E F G H I J K L M\n\nN O P Q R S T U V W X Y Z\n4 3 2 1\n\na) 3656894927\nb) 3657694927\nc) 3656694927\nd) 365619497\ne) 365669492\n\nExplanation:\nLet A = 1, B = 2, C = 3,.... X = 24, Y = 25, Z = 26.\nThen. M = 13 = 1 + 3 =4; 0 = 15=1 + 5 = 6;\nL = 12= 1 + 2-3; T-20 = 2 + 0 = 2;\nY = 25 = 2 + 5 = 7. So. MOBILITY =46293927.\nCONFORMITY [C=3;O=1+5=6;N=1+4=5;F=6;0=1+5=6;R=1+8=9;M=1+3=4;I=9;T=2+0=2;Y=2+5=7]\nSo CONFORMITY=3656694927\n\na) 35\nb) 36\nc) 48\nd) 49\n\nExplanation:\nThe numbers are 12, 22, 32, 42, 52.\nMissing number = 62 = 36.\n\na) True\nb) False\nc) Uncertain\n\nExplanation:\n\nStep1. Click the File tab in the top-left corner of your Word screen.\nStep 2. Click Export in the blue vertical menu that runs along the left side of your screen.\nStep 3. With the Create XPS/PDF Document option selected, click the corresponding box on the right side of your screen. (It looks like a document icon wearing a seatbelt).\nStep 4. A new box labeled Publish as PDF or XPS appears, and you should use it to navigate to the location on your PC where you want to save the new PDF.\nStep 5.Click Publish to create and save your PDF.\n\na) 3-D\nc) Box\nd) All of These\n\na) 247\nb) 135\nc) 254\nd) 147\n\nExplanation:\n\nA B C D E F G H I J K L M 1 2 3 4 5 6 7 8 9 10 11 12 13\nN O P Q R S T U V W X Y Z\n14 15 16 17 18 19 20 21 22 23 24 25 26\n\na) 10%\nb) 7.5%\nc) 16%\nd) 15%\n\nExplanation:\n\nLet the principal be Rs 1000\nThen, amount = (2×1000) = Rs. 2000\nSI = 2000-1000 = 1000\nRate = ?\nTime = 10 years\nRate = SI × 100 / P × T\n=> 1000×100/1000×10\n=> 10\nRate = 10%\n\na) 34748\nb) 27474\nc) 30154\nd) 27574\ne) None\n\nExplanation:\n?+3699+1985-2047=31111\n?+5684-2047=31111\n?+3637=31111\n?=31111-3637\n? = 27474\n\na) 16.1 %\nb) 15.5 %\nc) 17.5 %\nd) 8.05 %\n\nExplanation:\n\nC.P.= 20000/-\nS.P.= 21610/-\nprofit = SP - CP\n21610-20000 = 1610\nProfit % = 1610/20000 * 100\n0.0805 * 100\n8.05%\n\na) 31\nb) 61\nc) 71\nd) 91\n\nExplanation:\n91 because it is multiple of 7 and 13. 91 is not a prime number because it is divisible by 7.\n\na) 306\nb) 765\nc) 700\nd) 510\n\nExplanation:\n\n15x+10x + 6x = 1581\n31x= 1581\nx=51\na=15x\na=15*51\na=765\n\nA. GRR\nB. GSS\nC. ISS\nD. ITT\n\nExplanation :\n\nThe sequence is added +2 letters to each letter\n\nA. 8\nB. 6\nC. 7\nD. Can't be determined\n\nExplanation:\n\nin 786 and 645 \"6\" is repeated so \"6\" is bring\n786 and 958 \"8\" is repeated so \"8\" is apple\nso remaining \"7\" is me\n\nA. QLPNKJ\nB. QLPNKJF\nC. QLPNMF\nD. QLPNDF\n\nExplanation:\n\nThe answer shall be part d\nas\n3 stands for q\n9 stands for l\n6 stands for p\n8 stands for n\nno repeating\n2 stands for m\n4 stands for f\n\nA. X gained Rs. 4350\nB. Y lost Rs. 4350\nC. X gained Rs. 3150\nD. X lost Rs. 3150\n\nExplanation:\n\nIn whole transaction gain in whole\n= 105 - 102.9 = 2.1 units\nBut, 100 units = 1,50,000\n1 unit = 1500\n2.1 units = 3150 gain\n\nA. 30.00 kg\nB. 35.00 kg\nC. 37.25 kg\nD. 42.50 kg\n\nExplanation:\n\nTotal weight of section A = 36×40 = 1440 kg\nTotal weight of section B = 44×35 = 1540 kg\nTotal weight of class = 1440+1540 = 2980 kg\nTotal students = 36+44 = 80\nAverage weight of class = 2980/80 =37.25 kg\n\na) Table\nb) Picture\nc) Clip Art\nd) Page Break\n\nExplanation:\n\nIn Microsoft Word and other word processor programs, pressing Ctrl+Enter adds a page break at the cursors current position instead of a new paragraph.\n\na)Review\nb)Page Layout\nc)Insert\nd)View\n\nExplanation:\n\nIn Microsoft Word, Page Layout - Page Setup-Margins\n\na) Book Mark\nb) Cross reference\nd) Word Art\n\nExplanation:\n\nIn Microsoft PowerPoint, the Ctrl+K shortcut key inserts a hyperlink into the current location.\n\na) Sam\nb) Jack\nc) Denise\nd) Jim\n\nExplanation:\n\nSam > Jack\nDenise > Jim\nDenise < Jack The answer is Sam because Jack, Denise, and Jim are slower than Sam.\n\na) 10\nb) 18\nc) 15\nd) 12\n\nExplanation:\n\ntotal 40 students.\n8 of them wont speak both language\nso remaining 32 students.\n\nboth language = (Hindi + kanada) - remaining\n= (24 + 20) - 32\n= 12\n\nA. CMN\nB. UJI\nC. VIJ\nD. IJT\n\nExplanation:\n\nBecause the letters are the same, concentrate on the number series, which is a simple 2, 3, 4, 5, 6 series, and follows each letter in order.\n\nA. 8, 20, 28\nB. 16, 28, 36\nC. 20, 35, 45\nD. None of these\n\nExplanation:\n\nLet their present ages be 4x, 7x and 9x years respectively.\nThen, (4x - 8) + (7x - 8) + (9x - 8) = 56\n20x = 80\nx = 4.\nTheir present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.\n\nA. scarcely\nB. impolitely\nC. noisily\nD. Quietly\n\nExplanation:\n\nSecretly is the opposite of openly, and silently is the opposite of noisily. Choices a and b are clearly not the opposites of silently. (Choice d) means the same thing as silently\n\nA. decrease\nB. maintain\nC. increase\nD. preserve\n\nExplanation:\n\nGuide and direct are synonyms, and reduce and decrease are synonyms. The answer is not choice b or d because neither means the same as reduce. (Choice c) is incorrect because increase is the opposite of reduce.\n\nA. 700\nB. 800\nC. 1200\nD. 2000\n\nExplanation:\nThe century divisible by 400 is a leap year.\nThe year 700 is not a leap year.\n\nA. attorney\nB. law\nC. politician\nD. constituents\n\nExplanation:\n\nAn artist makes paintings; a senator makes laws. The answer is not choice a because an attorney does not make laws and a senator is not an attorney. Choice c is incorrect because a senator is not a politician. Constituents (choice d) is also incorrect because a senator serves his or her constituents\n\nA. 14 km\nB. 15 km\nC. 16 km\nD. 17 km\n\nExplanation:\n\nLet the distance travelled on foot be x km.\nThen, distance travelled on bicycle = (61 -x) km.\nSo, x + (61 -x) = 9\n4 9\n9x + 4(61 -x) = 9 x 36\n5x = 80\nx = 16 km.\n\nA. 69\nB. 78\nC. 96\nD. Cannot be determined\nE. None of these\n\nExplanation:\n\nLet the ten's digit be x and unit's digit be y.\nThen, x + y = 15 and x - y = 3 or y - x = 3.\nSolving x + y = 15 and x - y = 3, we get: x = 9, y = 6.\nSolving x + y = 15 and y - x = 3, we get: x = 6, y = 9.\nSo, the number is either 96 or 69.\nHence, the number cannot be determined.\n\nA. 396\nB. 427\nC. 671\nD. 264\n\nExplanation:\n\nIn each number except 427, the middle digit is the sum of other two\n\nA. 200 m\nB. 225 m\nC. 245 m\nD. 250 m\n\nExplanation:\n\nSpeed = [45 X 5/18] m/sec = [25/2] m/sec Time = 30 sec Let the length of bridge be x metres. Then, (130 + x)/30 = 25/2 => 2(130 + x) = 750 => x = 245 m.\n\nA. 31\nB. 61\nC. 71\nD. 91\n\nExplanation:\n\n91 is divisible by 7. So, it is not a prime number.\n\nA. 103\nB. 107\nC. 113\nD. None of these\n\nExplanation:\n\nThe smallest 3-digit number is 100, which is divisible by 2.\n100 is not a prime number.\n√101< 11 and 101 is not divisible by any of the prime numbers 2, 3, 5, 7, 11. 101 is a prime number. Hence 101 is the smallest 3-digit prime number.\n\nA. 55\nB. 36\nC. 105\nD. 70\n\nExplanation:\n\nThe pattern is 1 x 3, 2 x 5, 3 x 7, 4 x 9, 5 x 11, 6 x 13, 7 x 15, etc.\n\nA. 3.5km\nB. 2.4km\nC. 2.6km\n\nExplanation:\n\nLet the required distance be x km.\nSpeed downstream = (5 + 1)kmph = 6 kmph\nSpeed upstream = (5 - 1)kmph = 4 kmph\nThen,\nx/6 + x/4 = 1\n5x = 12\nx = 2.4\nThe required distance is 2.4 km\n\nA. 6400\nB. 14400\nC. 3600\nD. 1600\n\nExplanation:\n\nThe least number divisible by 16, 20, 24\n= L.C.M. of 16, 20, 24 = 240 = 2 x 2 x 2 x 2 x 3 x 5.\nTo make it perfect square, it must be multiplied by 3 x 5.\nRequired number = 240 x 3 x 5 = 3600\n\nA. 80°\nB. 85°\nC. 90°\nD. 75°\n\nExplanation:\n\nAt 10 o'clock, the hour hand is at 300° with the vertical.\nIn 40 minutes, angle made by hour hand = 300° + 40x0.5 = 320°\nAngle made by minute hand = 6x40 = 240°\nSo, angle between the hour hand and minute hand = 320 - 240 = 80°\n\nA. Tuesday\nB. Wednesday\nC. Monday\nD. Thursday\n\nExplanation:\n\nIf today is Sunday, Four days after Sunday will be Friday, the day before yesterday is Tuesday.\n\nA. ABC\nB. CAB\nC. BCA\nD. BAC\n\nExplanation:\n\nb) The complete sentence would look like this: Nowadays, a typical classroom has pods of desks, and kids are working on countless group assignments.\nBy looking at the set of sentences, one can understand that 'Nowadays' has to be the beginning of the sentence. From then on, it is simply putting the phrases together to form a coherent sentence\n\nA. 17. 5\nB. 18.5\nC. 20\nD. 133\n\nExplanation:\n\nLet the number be X.\nThen, 4/17 of 17/19 of the number is (4/17) * (17/19) * X = (4/19) * X\n3/38 of the number= (3/38) * X\n\nGiven:\n\nthe difference between these two values is 17.5.\nHence, [(4/19) * X - (3/38) * X ]= 17.5\nSolving the equation, we have X=133\nTherefore, the number is 133.\n\nA. Rs.2000/-\nB. Rs.1500/-\nC. Rs.1000/-\nD. Rs.4500/-\n\nExplanation:\n\nA = Rs.2000/- , that is 2 parts\nB = Rs.3000/- , that is 3 parts\nC = Rs.4000/- , that is 4 parts\nTotal 9 parts -------> profit (4500/-)\n----->1 part -------->Rs.500/-\nthen A's share 2 parts ----- > Rs.1000/-\n\nA. Monday\nB. Sunday\nC. Friday\nD. None of these\n\nExplanation:\n\nFormula:-(Date + Month code + No.of years + No.of leap year + Century code)/7\n= (24 + 0 + 11 + 2 + 6)/7 = 43/7 = 1\n= Sunday\n\nA. 4 years\nB. 8 years\nC. 10 yearswrong\nD. None of these\n\nExplanation:\nLet the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.\nThen, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50\n5x = 20\nx = 4.\nAge of the youngest child = x = 4 years.\n\nA. 1.5 timewrong\nB. 2 times\nC. 2.5 times\nD. 3 times\n\nExplanation:\nLet the daughter's age be y and father's age be 4y.\nFather’s age is 4 times more aged than his daughter, therefore father’s present age = y + 4y = 5y\nAfter five years, father’s age will be three times his daughter age.\n(4y + 5) = 3 (y + 5)\ny = 10\nAfter 5 years it was (4y + 5), then after further 5 years, father’s age = (4y +10) and daughter’s age = (y + 10)\n(4y + 10) / (y + 10) = ?\nSubstituting the value of y, we get\n[(4x10) + 10] / [10+10] = 5/2 = 2.5\nAfter another 5 years, father will be 2.5 times of daughter's age.\n\nA. 26 Years\nB. 25 Yearscorrect\nC. 32 Years\nD. 65 Years\n\nExplanation:\nLet 5 years age, the age of ashok, Anil and Ajay were 3x, 4x and 5x years. Then, their present ages are ( 3x + 5), (4x + 5) abd ( 5x + 5) years.\nNow, 3x +5 + 4x +5 + 5x + 5 = 75 => 12x = 60 x = 5\nHence, present age of Anil = 4x + 5 = 4 x 5 + 5 = 25 years.\n\nA. 30\nB. 35correct\nC. 37\nD. 41\n\nExplanation:\nLet A's age 10 years ago = x years.\nThen, B's age 10 years ago = 2x years.\n(x + 10) / (2x+ lO) = 3/4\n=> x = 5.\nSo, the total of their present ages =(x + 10 + 2x + 10)\n= (3x + 20) = 35 years.\n\nA. 11 : 7\nB. 9 : 5\nC. 7 : 4\nD. 7 : 3\n\nExplanation:\n10 years ago, age of mother was three times the age of her son. Say, the age of son was x and mother's age was 3x.\nAt present: Mother's age is (3x + 10) and son’s age is (x + 10)\nAfter ten years: Mother's age will be (3x + 10) +10 and son’s age will be (x + 10) + 10. Given that, mother’s age is twice that of son after ten years.\n(3x + 10) +10 = 2 [(x + 10) + 10]\n(3x + 20) = 2[x + 20]\nSolving the equation, we get x = 20\n(3x + 10): (x + 10) = 70: 30 = 7: 3.\n\nA. 5\nB. 10\nC. 15\nD. 20\n\nExplanation:\nLet the present ages of son and father be x and (60 -x) years respectively.\nThen, (60 - x) - 5= 4(x - 5)\n55 - x = 4x - 20\n5x = 75 => x = 15\n\nA. 12 years\nB. 16 years\nC. 24 years\nD. 48 years\n\nExplanation:\nPresent day,\nLet us consider Dhiraj's present age as K years\nRaj's present age is 4 times Dhiraj's = 4K years\n3 years ago,\nDhiraj's age = K-3 years\nRaj's age = 4K-3 years.\nThat time, Raj's age = 5 times Dhiraj's age\n4K-3 = 5(K-3)\n4K-3 = 5K-15\nK = 12 years\nPresent age of Raj = 4K = 4 x 12 = 48 years.\n\nA. Monday\nB. Sunday\nC. Wednesday\nD. Thursday" ]

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[ "", null, "Front page | perl.perl5.porters | Postings from November 2003\n\n## Empty subroutine as object method segfaults in 5.8.2 (sometimes)\n\nFrom:\nElizabeth Mattijsen\nDate:\nNovember 9, 2003 05:03\nSubject:\nEmpty subroutine as object method segfaults in 5.8.2 (sometimes)\nMessage ID:\np05111b09bbd3deb12c8d@[192.168.56.3]\n```I'm afraid I haven't been paying enough attention to p5p lately. I\ntoo was surprised by the arrival of 5.8.2, so I haven't tested my\nThread::xxx modules with 5.8.2-RCx. I guess I have to get used to\nthe increased frequency of releases as well. Which is a good thing!\n\na make test in 5.8.2. This affects Thread::Pool, which is more used.\n\nThe core dump is not giving any usable information, even with a debug\nenabled Perl (could this have to do that the segfault occurs inside a\n\nWhen checking where the error occurred, I found that it happened when\ncalling an object method. That object method (for that particular\nblessed object) consisted of:\n\nsub shutdown {}\n\ni.e. nothing. Since this method isn't supposed to return anything\n(but couldn't be inherited), I chose to do it this way.\n\nThe funny thing is that when I replace the subroutine with:\n\nsub shutdown { undef }\n\nthe problem (and the segfault) goes away.\n\nI've been trying to reduce the problem to a simple case:\n\npackage Foo;\nsub new { bless {},shift }\nsub empty {}\n\npackage main;\nmy \\$object = Foo->new;\n\\$object->empty;\n\nbut this just runs without any problems, even under valgrind, so it\nmust be something trickier that I'm doing.\n\nI'll be issuing new versions of my Thread::xxx modules to prevent\nfailures in 5.8.2, but thought I'd report this to p5p nonetheless\n\nLiz\n\nThe valgrind output for the Thread::Conveyor test-suite:\n\n==21280== Memcheck, a.k.a. Valgrind, a memory error detector for x86-linux.\n==21280== Copyright (C) 2002-2003, and GNU GPL'd, by Julian Seward.\n==21280== Using valgrind-20030725, a program supervision framework\nfor x86-linux.\n==21280== Copyright (C) 2000-2003, and GNU GPL'd, by Julian Seward.\n==21280== Estimated CPU clock rate is 2825 MHz\n==21280== For more details, rerun with: -v\n==21280==\n1..105\n==21280== valgrind's libpthread.so: KLUDGED call to: siglongjmp\n(cleanup handlers are ignored)\n==21280== valgrind's libpthread.so: KLUDGED call to: siglongjmp\n(cleanup handlers are ignored)\n# test belt optimized for cpu\nok 2 - check object type isa Thread::Conveyor\nok 4 - check number boxes on belt\nok 5 - check \\# elements simple list\nok 6 - check simple list\nok 7 - check \\# elements list ref\nok 8 - check type of list ref\nok 9 - check list ref\nok 10 - check \\# elements hash ref, \\#1\nok 11 - check type of hash ref, \\#1\nok 12 - check \\# elements hash ref, \\#2\nok 13 - check type of hash ref, \\#2\nok 14 - check \\# elements hash ref, \\#3\nok 15 - check type of hash ref, \\#3\nok 16 - check hash ref\nok 17 - check \\# elements dontwait\n==21280== Invalid read of size 4\n==21280== at 0x80BB5DA: Perl_runops_debug (dump.c:1438)\n==21280== by 0x8065068: S_run_body (perl.c:1860)\n==21280== by 0x805FC82: main (perlmain.c:86)\n==21280== Address 0x41D4A1EC is 8 bytes inside a block of size 24 free'd\n==21280== at 0x400296BF: free (vg_replace_malloc.c:220)\n==21280== by 0x80BC048: Perl_safesysfree (util.c:143)\n==21280== by 0x80922BA: Perl_op_free (op.c:325)\n==21280== by 0x80A2B8C: Perl_peep (op.c:6378)\n==21280==\n==21280== Invalid read of size 1\n==21280== at 0x80FAAC0: Perl_pp_stub (pp.c:36)\n==21280== by 0x80BB5DE: Perl_runops_debug (dump.c:1438)\n==21280== by 0x8065068: S_run_body (perl.c:1860)\n==21280== Address 0x41D4A1F8 is 20 bytes inside a block of size 24 free'd\n==21280== at 0x400296BF: free (vg_replace_malloc.c:220)\n==21280== by 0x80BC048: Perl_safesysfree (util.c:143)\n==21280== by 0x80922BA: Perl_op_free (op.c:325)\n==21280== by 0x80A2B8C: Perl_peep (op.c:6378)\n==21280==\n==21280== Invalid read of size 1\n==21280== by 0x80BB5DE: Perl_runops_debug (dump.c:1438)\n==21280== by 0x8065068: S_run_body (perl.c:1860)\n==21280== Address 0x41D4A1F8 is 20 bytes inside a block of size 24 free'd\n==21280== at 0x400296BF: free (vg_replace_malloc.c:220)\n==21280== by 0x80BC048: Perl_safesysfree (util.c:143)\n==21280== by 0x80922BA: Perl_op_free (op.c:325)\n==21280== by 0x80A2B8C: Perl_peep (op.c:6378)\n==21280==\n==21280== Invalid read of size 1\n==21280== at 0x80FAAE0: Perl_pp_stub (pp.c:36)\n==21280== by 0x80BB5DE: Perl_runops_debug (dump.c:1438)\n==21280== by 0x8065068: S_run_body (perl.c:1860)\n==21280== Address 0x41D4A1F8 is 20 bytes inside a block of size 24 free'd\n==21280== at 0x400296BF: free (vg_replace_malloc.c:220)\n==21280== by 0x80BC048: Perl_safesysfree (util.c:143)\n==21280== by 0x80922BA: Perl_op_free (op.c:325)\n==21280== by 0x80A2B8C: Perl_peep (op.c:6378)\n==21280==\n==21280== Invalid read of size 4\n==21280== at 0x80FAB45: Perl_pp_stub (pp.c:38)\n==21280== by 0x80BB5DE: Perl_runops_debug (dump.c:1438)\n==21280== by 0x8065068: S_run_body (perl.c:1860)\n==21280== Address 0x41D4A1E4 is 0 bytes inside a block of size 24 free'd\n==21280== at 0x400296BF: free (vg_replace_malloc.c:220)\n==21280== by 0x80BC048: Perl_safesysfree (util.c:143)\n==21280== by 0x80922BA: Perl_op_free (op.c:325)\n==21280== by 0x80A2B8C: Perl_peep (op.c:6378)\n==21280== valgrind's libpthread.so: KLUDGED call to: siglongjmp\n(cleanup handlers are ignored)\nScalars leaked: 1\n# Looks like you planned 105 tests but only ran 17.\n==21280==\n==21280== ERROR SUMMARY: 5 errors from 5 contexts (suppressed: 9 from 2)\n==21280== malloc/free: in use at exit: 221053 bytes in 125 blocks.\n==21280== malloc/free: 75824 allocs, 75699 frees, 17086815 bytes allocated.\n\n```" ]

[ null, "https://st.pimg.net/devcom/images/develooperbutton.v1.gif", null ]

[ "#### Social and Economic Networks: Models and Analysis笔记一：基本概念\n\nSocial and Economic Networks: Models and Analysis这门课是斯坦福大学在Coursera上的一门公开课，恰好这段时间开始开放，加上评分还不错，于是打算来听听，由于自己对Social Network Analysis不是特别熟悉，所以就从第一节课一些基本概念开始，在这个博客留下一些笔记吧。\n\n1.n个结点可以形成多少种不同的网络？\n\n2.Network表示方法\n\n3.图相关概念的基本定义：\n\n• Path（路径）: a walk $$(i_1, i_2, \\cdots, i_K)$$ with each node $$i_k$$ distinct\n• Cycle（环路）: a walk where $$i_1=i_K$$\n• Geodesic（测地线）: a shortest path between two nodes\n• Components(连通分量): maximal connected subgraph\n• Diameter（直径？）:  largest geodesic (largest shortest path)\n• Neighborhood（邻居）: $$N_i(g) = { j G_{ij}=1}$$\n• Degree（度数）: $$d_i = # N_i(g)$$\n\n4.Erdos‐Renyi Random Graphs\n\n• 如果所有结点的平均度数 $$d(n) \\gt (1+\\epsilon) log(n) some \\epsilon>0$$ ，那么整个网络很可能是连通的；\n• 如果 $$d(n) \\gt (1+\\epsilon) log(n) some?\\epsilon>0$$ 并且 $$\\frac{d(n)}{n}\\rightarrow 0$$ ，那么当n足够大时，网络的平均路径长度和Diameter将正比于 $$\\frac{log(n)}{log(d)}$$ 。\n\n5.树\n\n(a)初始一个根节点\n\n(b)根结点产生d个子节点\n\n(c)b中生成的d个子节点每个再产生(d-1)个子节点\n\n...... [", null, "](/images/post/2014/10/1.jpg)这种树在经过l步之后，将包含 $$1+d+d(d-1)+\\cdots +d(d-1)^{l-1}=1+\\frac{d((d-1)^l-1)}{(d-2)}\\approx (d-1)^l$$ 个节点。 6.Chernoff Bounds 若X是一个二项分布的随机变量，那么 $$P(E[X]/3\\le X \\le 3E[X])\\gt 1-e^{-E[X]}$$ 。 应用到ER图中，那么顶点度数比较靠近平均值d的概率 $$P(d/3\\le d_i \\le 3d)\\gt 1-e^{-d}$$ 。 7.Clustering What fraction of my friends are friends of each other? $$Cl_i(g)=#\\{\\text{kj in g}| \\text{k,j in } N_i(g)\\}/#\\{\\text{k,j in } N_i(g)\\}$$ Average Clustering: $$Cl^{Avg}(g)=\\sum_iCl_i(g)/n$$" ]

[ null, "https://closure11.com/images/post/2014/10/1-300x241.jpg", null ]

[ "Cm To M\n\n# 92.4 cm to m92.4 Centimeters to Meters\n\ncm\n=\nm\n\n## How to convert 92.4 centimeters to meters?\n\n 92.4 cm * 0.01 m = 0.924 m 1 cm\nA common question is How many centimeter in 92.4 meter? And the answer is 9240.0 cm in 92.4 m. Likewise the question how many meter in 92.4 centimeter has the answer of 0.924 m in 92.4 cm.\n\n## How much are 92.4 centimeters in meters?\n\n92.4 centimeters equal 0.924 meters (92.4cm = 0.924m). Converting 92.4 cm to m is easy. Simply use our calculator above, or apply the formula to change the length 92.4 cm to m.\n\n## Convert 92.4 cm to common lengths\n\nUnitUnit of length\nNanometer924000000.0 nm\nMicrometer924000.0 µm\nMillimeter924.0 mm\nCentimeter92.4 cm\nInch36.3779527559 in\nFoot3.031496063 ft\nYard1.0104986877 yd\nMeter0.924 m\nKilometer0.000924 km\nMile0.000574147 mi\nNautical mile0.0004989201 nmi\n\n## What is 92.4 centimeters in m?\n\nTo convert 92.4 cm to m multiply the length in centimeters by 0.01. The 92.4 cm in m formula is [m] = 92.4 * 0.01. Thus, for 92.4 centimeters in meter we get 0.924 m.\n\n## 92.4 Centimeter Conversion Table", null, "## Alternative spelling\n\n92.4 Centimeters to Meters, 92.4 Centimeters in Meters, 92.4 cm to Meters, 92.4 cm in Meters, 92.4 Centimeter to m, 92.4 Centimeter in m, 92.4 cm to Meter, 92.4 cm in Meter, 92.4 Centimeters to m, 92.4 Centimeters in m, 92.4 Centimeter to Meters, 92.4 Centimeter in Meters, 92.4 cm to m, 92.4 cm in m" ]

[ null, "https://cm-to-m.appspot.com/image/92.4.png", null ]

[ "[This article was first published on Blog - Data Science and the fst Package, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)\nWant to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nThe fst package uses LZ4 and ZSTD to compress columnar data. In the latest release, methods compress_fst and decompress_fst were added which allow for direct (multi-threaded) access to these excellent compressors.\n\n# LZ4 and ZSTD\n\nLZ4 is one of the fastest compressors around, and like all LZ77-type compressors, decompression is even faster. The fst package uses LZ4 to compress and decompress data when lower compression levels are selected (in method write_fst). For higher compression levels, the ZSTD compressor is used, which offers superior compression ratio’s but requires more CPU resources.\n\n# How to use LZ4 and ZSTD from the fst package\n\nFrom version 0.8.0, methods compress_fst and decompress_fst are available in the package. These methods give you direct access to LZ4 and ZSTD. As an example of how they can be used, we download a 90 MB file from Kaggle and recompress it using ZSTD:\n\n```library(fst)\n\nsample_file <- \"survey_results_public.csv\"\n\n# read file contents into a raw vector\n\n# compress bytes with ZSTD at a compression level of 20 percent\ncompressed_vec <- compress_fst(raw_vec, \"ZSTD\", 20)\n\n# write the compressed data into a new file\ncompressed_file <- \"survey_results_public.fsc\"\nwriteBin(compressed_vec, compressed_file)\n\n# compression ratio\nfile.size(sample_file) / file.size(compressed_file)\n\n## 8.949771\n```\n\nusing a ZSTD compression level of 20 percent, the contents of the csv file are compressed to about 11 percent of the original size (calculated as the inverse of the compression ratio). To decompress the generated compressed file again you can do:\n\n```# read compressed file into a raw vector\n\n# decompress file contents\nraw_vec_decompressed <- decompress_fst(compressed_vec)\n```\n\nA nice feature of data.table’s fread method is that it can parse in-memory data directly. That means that we can easily feed our raw vector to fread:\n\n```library(data.table)\n\n# read data set from the in-memory csv\n```\n\nThis effectively reads your data.table from a compressed file, which saves disk space and increases read speed for slow disks.\n\nMethods compress_fst and decompress_fst use a fully multi-threaded implementation of the LZ4 and ZSTD algorithms. This is accomplished by dividing the data into (at maximum) 48 blocks, which are then processed in parallel. This increases the compression and decompression speeds significantly at a small cost to compression ratio:\n\n```library(microbenchmark)\n\n# measure ZSTD compression performance at low setting\ncompress_time <- microbenchmark(\ncompress_fst(raw_vec, \"ZSTD\", 10),\ntimes = 500\n)\n\n# decompress again\ndecompress_time <- microbenchmark(\ndecompress_fst(compressed_vec),\ntimes = 500\n)\n\ncat(\"Compress: \", 1e3 * as.numeric(object.size(raw_vec)) / median(compress_time\\$time), \"MB/s\",\n\"Decompress: \", 1e3 * as.numeric(object.size(raw_vec)) / median(decompress_time\\$time), \"MB/s\")\n\n## Compress: 1299.649 MB/s Decompress: 1948.932 MB/s\n```\n\nThat’s a ZSTD compression speed of around 1.3 GB/s!\n\n# Bring on the cores\n\nWith more cores, you can do more parallel compression work. When we do the compression and decompression measurements above for a range of thread and compression level settings, we find the following dependency between speed and parallelism:", null, "Figure 1: Compression and decompression speeds vs the number of cores used for computation\n\nThe code that was used to obtain these results is given in the last paragraph. As can be expected, the compression speed is highest for lower compression level settings. But interesting enough, decompression speeds actually increase with higher compression settings! For the highest levels, ZSTD decompression speeds of more than 3 GB/s were measured in our experiment!\n\nDifferent compression levels settings lead to different compression ratio’s. This relation is depicted below. For completeness, LZ4 compression ratio’s were added as well:", null, "Figure 2: Compression ratio for different settings of the compression level\n\nThe highlighted point at a 20 percent (ZSTD) compression level corresponds to the measurement that we did earlier. It’s clear from the graph that with a combination of LZ4 and ZSTD, a wide range of compression ratio’s (and speeds) is available to the user.\n\n# The case for high compression levels\n\nThere are many use cases where you compress your data only once but decompress it much more often. For example, you can compress and store a file that will need to be read many times in the future. In that case it’s very useful to spend the CPU resources on compressing at a higher setting. It will give you higher decompression speeds during reads and the compressed data will occupy less space.\n\nAlso, when operating from a disk that has a lower speed than the (de-)compression algorithm, compression can really help. For those cases, compression will actually increase the total transfer speed because (much) less data has to be moved to or from the disk. This is also the main reason why fst is able to serialize a data set at higher speeds than the physical limits of a drive.\n\n(Please take a look at this post to get an idea of how that works exactly)\n\n# Benchmark code\n\nBelow is the benchmark script that was used to obtain the dependency graph for the number of cores and compression level.\n\n```library(data.table)\n\n# benchmark results\nbench <- data.table(Threads = as.integer(NULL), Time = as.numeric(NULL),\nMode = as.character(NULL), Level = as.integer(NULL), Size = as.numeric(NULL))\n\n# Note that compression time increases steadily with higher levels.\n# If you want to run this code yourself, start by using levels 10 * 0:5\nfor (level in 10 * 0:10) {\n\ncat(\".\") # show some progress\n\n# compress measurement\ncompress_time <- microbenchmark(\ncompressed_vec <- compress_fst(raw_vec, \"ZSTD\", level), times = 25)\n\n# decompress measurement\ndecompress_time <- microbenchmark(\ndecompress_fst(compressed_vec), times = 25)\n\n# add measurements to the benchmark results\nbench <- rbindlist(list(bench,\ndata.table(\nTime = median(compress_time\\$time),\nMode = \"Compress\",\nLevel = level,\nSize = as.integer(object.size(compressed_vec))),\ndata.table(\nTime = median(decompress_time\\$time),\nMode = \"Decompress\",\nLevel = level,\nSize = as.integer(object.size(compressed_vec)))))\n}\n}\n```\n\nThis creates a data.table with compression and decompression benchmark results.\n\nThis post is also available on R-bloggers" ]

[ null, "https://i0.wp.com/blog.fstpackage.org/img/fst_compression/img/fig-unnamed-chunk-9-1.png", null, "https://i2.wp.com/blog.fstpackage.org/img/fst_compression/img/fig-unnamed-chunk-10-1.png", null ]

[ "# Logical Operators¶\n\nFor logical operators, boolean expressions or expressions that evaluates to an INTEGER value are specified as operands; TRUE, FALSE or NULL is returned as a result. If the INTEGER value is used, 0 is evaluated to FALSE and the other values are evaluated to TRUE. If a boolean value is used, 1 is evaluated to TRUE and 0 is evaluated to FALSE.\n\nThe following table shows the logical operators.\n\nLogical Operators\n\nLogical Operator Description Condition\nAND, && If all operands are TRUE, it returns TRUE. a AND b\nOR, || If none of operands is NULL and one or more operands are TRUE, it returns TRUE. If pipes_as_concat is no that is a parameter related to SQL statements, a double pipe symbol can be used as OR operator. a OR b\nXOR If none of operand is NULL and each of operand has a different value, it returns TRUE. a XOR b\nNOT, ! A unary operator. If a operand is FALSE , it returns TRUE. If it is TRUE , returns FALSE. NOT a\n\nTruth Table of Logical Operators\n\na b a AND b a OR b NOT a a XOR b\nTRUE TRUE TRUE TRUE FALSE FALSE\nTRUE FALSE FALSE TRUE FALSE TRUE\nTRUE NULL NULL TRUE FALSE NULL\nFALSE TRUE FALSE TRUE TRUE TRUE\nFALSE FALSE FALSE FALSE TRUE FALSE\nFALSE NULL FALSE NULL TRUE NULL" ]

[ null ]

[ "User Real IP - 54.161.98.96\nArray\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => Array\n(\n => 95.142.117.29\n)\n\n => Array\n(\n => 150.129.60.107\n)\n\n => Array\n(\n => 94.205.243.22\n)\n\n => Array\n(\n => 115.42.71.143\n)\n\n => Array\n(\n => 117.217.195.59\n)\n\n => Array\n(\n => 182.77.112.56\n)\n\n => Array\n(\n => 182.77.112.108\n)\n\n => Array\n(\n => 41.80.69.10\n)\n\n => Array\n(\n => 117.5.222.121\n)\n\n => Array\n(\n => 103.11.0.38\n)\n\n => Array\n(\n => 202.173.127.140\n)\n\n => Array\n(\n => 49.249.249.50\n)\n\n => Array\n(\n => 116.72.198.211\n)\n\n => Array\n(\n => 223.230.54.53\n)\n\n => Array\n(\n => 102.69.228.74\n)\n\n => Array\n(\n => 39.37.251.89\n)\n\n => Array\n(\n => 39.53.246.141\n)\n\n => Array\n(\n => 39.57.182.72\n)\n\n => Array\n(\n => 209.58.130.210\n)\n\n => Array\n(\n => 104.131.75.86\n)\n\n => Array\n(\n => 106.212.131.255\n)\n\n => Array\n(\n => 106.212.132.127\n)\n\n => Array\n(\n => 223.190.4.60\n)\n\n => Array\n(\n => 103.252.116.252\n)\n\n => Array\n(\n => 103.76.55.182\n)\n\n => Array\n(\n => 45.118.166.70\n)\n\n => Array\n(\n => 103.93.174.215\n)\n\n => Array\n(\n => 5.62.62.142\n)\n\n => Array\n(\n => 182.179.158.156\n)\n\n => Array\n(\n => 39.57.255.12\n)\n\n => Array\n(\n => 39.37.178.37\n)\n\n => Array\n(\n => 182.180.165.211\n)\n\n => Array\n(\n => 119.153.135.17\n)\n\n => Array\n(\n => 72.255.15.244\n)\n\n => Array\n(\n => 139.180.166.181\n)\n\n => Array\n(\n => 70.119.147.111\n)\n\n => Array\n(\n => 106.210.40.83\n)\n\n => Array\n(\n => 14.190.70.91\n)\n\n => Array\n(\n => 202.125.156.82\n)\n\n => Array\n(\n => 115.42.68.38\n)\n\n => Array\n(\n => 102.167.13.108\n)\n\n => Array\n(\n => 117.217.192.130\n)\n\n => Array\n(\n => 205.185.223.156\n)\n\n => Array\n(\n => 171.224.180.29\n)\n\n => Array\n(\n => 45.127.45.68\n)\n\n => Array\n(\n => 195.206.183.232\n)\n\n => Array\n(\n => 49.32.52.115\n)\n\n => Array\n(\n => 49.207.49.223\n)\n\n => Array\n(\n => 45.63.29.61\n)\n\n => Array\n(\n => 103.245.193.214\n)\n\n => Array\n(\n => 39.40.236.69\n)\n\n => Array\n(\n => 62.80.162.111\n)\n\n => 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103.244.178.248\n)\n\n => Array\n(\n => 81.92.203.92\n)\n\n => Array\n(\n => 194.71.227.72\n)\n\n => Array\n(\n => 39.57.225.212\n)\n\n => Array\n(\n => 223.233.78.41\n)\n\n => Array\n(\n => 210.89.58.87\n)\n\n => Array\n(\n => 39.57.253.146\n)\n\n => Array\n(\n => 132.154.171.12\n)\n\n => Array\n(\n => 39.59.5.11\n)\n\n => Array\n(\n => 49.156.111.206\n)\n\n => Array\n(\n => 223.233.70.184\n)\n\n => Array\n(\n => 103.99.12.161\n)\n\n => Array\n(\n => 122.161.75.127\n)\n\n => Array\n(\n => 154.6.17.167\n)\n\n => Array\n(\n => 122.180.182.155\n)\n\n => Array\n(\n => 103.150.27.242\n)\n\n => Array\n(\n => 111.88.100.37\n)\n\n => Array\n(\n => 103.46.202.226\n)\n\n => Array\n(\n => 49.37.153.19\n)\n\n => Array\n(\n => 59.103.106.175\n)\n\n => Array\n(\n => 223.233.69.225\n)\n\n => Array\n(\n => 47.15.16.32\n)\n\n => Array\n(\n => 68.235.33.144\n)\n\n => Array\n(\n => 175.107.235.13\n)\n\n => Array\n(\n => 103.102.123.3\n)\n\n => Array\n(\n => 124.253.42.245\n)\n\n => Array\n(\n => 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37.120.197.37\n)\n\n => Array\n(\n => 198.16.66.157\n)\n\n => Array\n(\n => 27.6.200.177\n)\n\n => Array\n(\n => 182.190.206.136\n)\n\n => Array\n(\n => 120.29.98.174\n)\n\n => Array\n(\n => 193.34.232.24\n)\n\n => Array\n(\n => 27.6.201.83\n)\n\n => Array\n(\n => 202.47.45.186\n)\n\n => Array\n(\n => 39.42.36.181\n)\n\n => Array\n(\n => 103.49.200.43\n)\n\n => Array\n(\n => 37.19.218.8\n)\n\n => Array\n(\n => 49.37.191.10\n)\n\n => Array\n(\n => 212.102.46.68\n)\n\n => Array\n(\n => 39.59.62.91\n)\n\n => Array\n(\n => 43.239.205.65\n)\n\n => Array\n(\n => 122.2.72.187\n)\n\n => Array\n(\n => 193.19.109.23\n)\n\n => Array\n(\n => 136.185.37.86\n)\n\n => Array\n(\n => 103.246.41.8\n)\n\n => Array\n(\n => 103.174.242.77\n)\n\n => Array\n(\n => 39.59.4.29\n)\n\n => Array\n(\n => 49.36.222.31\n)\n\n => Array\n(\n => 49.36.178.90\n)\n\n => Array\n(\n => 180.190.65.169\n)\n\n => Array\n(\n => 27.6.205.158\n)\n\n => Array\n(\n => 122.161.48.3\n)\n\n => Array\n(\n => 183.83.212.172\n)\n\n => Array\n(\n => 157.36.118.177\n)\n\n => Array\n(\n => 106.213.54.136\n)\n\n => Array\n(\n => 27.6.206.142\n)\n\n => Array\n(\n => 116.107.29.76\n)\n\n => Array\n(\n => 59.103.191.90\n)\n\n => Array\n(\n => 59.94.175.89\n)\n\n => Array\n(\n => 103.59.75.148\n)\n\n => Array\n(\n => 103.46.202.183\n)\n\n => Array\n(\n => 112.196.174.165\n)\n\n => Array\n(\n => 198.16.66.197\n)\n\n => Array\n(\n => 161.35.139.97\n)\n\n => Array\n(\n => 171.79.95.100\n)\n\n => Array\n(\n => 117.96.212.111\n)\n\n => Array\n(\n => 103.172.104.34\n)\n\n => Array\n(\n => 106.51.241.222\n)\n\n => Array\n(\n => 154.192.111.60\n)\n\n => Array\n(\n => 37.57.81.26\n)\n\n => Array\n(\n => 185.204.1.237\n)\n\n => Array\n(\n => 73.71.18.199\n)\n\n => Array\n(\n => 184.170.245.5\n)\n\n => Array\n(\n => 180.241.46.14\n)\n\n => Array\n(\n => 47.31.236.198\n)\n\n => Array\n(\n => 157.47.38.99\n)\n\n => Array\n(\n => 103.99.12.168\n)\n\n => Array\n(\n => 23.106.56.14\n)\n\n => Array\n(\n => 103.99.12.160\n)\n\n => Array\n(\n => 79.153.17.250\n)\n\n => Array\n(\n => 103.203.255.8\n)\n\n => Array\n(\n => 115.186.159.145\n)\n\n)\n\\$55.97 closer to a tiki bar!: Thrift-o-rama\n Layout: Blue and Brown (Default) Author's Creation\n Home > \\$55.97 closer to a tiki bar!\n\n\\$55.97 closer to a tiki bar!\n\nJanuary 26th, 2008 at 10:31 pm\n\nThanks to some craft sales and another Amazon book sale, I'm adding another \\$55.97 to the Tiki Bar fund. It isn't a lot, but it sure feels good to make some progress! I've also decided to raise the Tiki Bar Fund savings goal to \\$1,500, since I'm almost half way to my original goal of \\$1,000.\n\n2 Responses to “\\$55.97 closer to a tiki bar!”\n\n1. koppur Says:\n1201456178\n\nvery nice! what do you craft?\n\n2. thriftorama Says:\n1201548329\n\nAll kinds of things. Right now, I'm mostly making \"monster\" theme tote bags and reusable grocery bags. In the past, I've made a lot of buttons, pins and jewelry. I'm likely going to be taking those up again.\n\n(Note: If you were logged in, we could automatically fill in these fields for you.)\n Name: * Email: Will not be published. Subscribe: Notify me of additional comments to this entry. URL: Verification: * Please spell out the number 4.  [ Why? ]\n\nvB Code: You can use these tags: [b] [i] [u] [url] [email]" ]

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[ "$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\n# 5.1: Areas and Distances\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\nLearning Objectives\n\n• Use sigma (summation) notation to calculate sums and powers of integers.\n• Use the sum of rectangular areas to approximate the area under a curve.\n• Use Riemann sums to approximate area.\n\nArchimedes was fascinated with calculating the areas of various shapes—in other words, the amount of space enclosed by the shape. He used a process that has come to be known as the method of exhaustion, which used smaller and smaller shapes, the areas of which could be calculated exactly, to fill an irregular region and thereby obtain closer and closer approximations to the total area. In this process, an area bounded by curves is filled with rectangles, triangles, and shapes with exact area formulas. These areas are then summed to approximate the area of the curved region.\n\nIn this section, we develop techniques to approximate the area between a curve, defined by a function $$f(x),$$ and the x-axis on a closed interval $$[a,b].$$ Like Archimedes, we first approximate the area under the curve using shapes of known area (namely, rectangles). By using smaller and smaller rectangles, we get closer and closer approximations to the area. Taking a limit allows us to calculate the exact area under the curve.\n\nLet’s start by introducing some notation to make the calculations easier. We then consider the case when $$f(x)$$ is continuous and nonnegative. Later in the chapter, we relax some of these restrictions and develop techniques that apply in more general cases.\n\n## Sigma (Summation) Notation\n\nAs mentioned, we will use shapes of known area to approximate the area of an irregular region bounded by curves. This process often requires adding up long strings of numbers. To make it easier to write down these lengthy sums, we look at some new notation here, called sigma notation (also known as summation notation). The Greek capital letter $$Σ$$, sigma, is used to express long sums of values in a compact form. For example, if we want to add all the integers from 1 to 20 without sigma notation, we have to write\n\n$1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20.$\n\nWe could probably skip writing a couple of terms and write\n\n$1+2+3+4+⋯+19+20,$\n\nwhich is better, but still cumbersome. With sigma notation, we write this sum as\n\n$\\sum_{i=1}^{20}i$\n\nwhich is much more compact. Typically, sigma notation is presented in the form\n\n$\\sum_{i=1}^{n}a_i$\n\nwhere $$a_i$$ describes the terms to be added, and the $$i$$ is called the $$index$$. Each term is evaluated, then we sum all the values, beginning with the value when $$i=1$$ and ending with the value when $$i=n.$$ For example, an expression like $$\\displaystyle \\sum_{i=2}^{7}s_i$$ is interpreted as $$s_2+s_3+s_4+s_5+s_6+s_7$$. Note that the index is used only to keep track of the terms to be added; it does not factor into the calculation of the sum itself. The index is therefore called a dummy variable. We can use any letter we like for the index. Typically, mathematicians use $$i, \\,j, \\,k, \\,m$$, and $$n$$ for indices.\n\nLet’s try a couple of examples of using sigma notation.\n\nExample $$\\PageIndex{1}$$: Using Sigma Notation\n\n1. Write in sigma notation and evaluate the sum of terms $$3^i$$ for $$i=1,2,3,4,5.$$\n2. Write the sum in sigma notation:\n\n$1+\\dfrac{1}{4}+\\dfrac{1}{9}+\\dfrac{1}{16}+\\dfrac{1}{25}. \\nonumber$\n\nSolution\n\n1. Write $\\sum_{i=1}^{5}3^i=3+3^2+3^3+3^4+3^5=363. \\nonumber$\n2. The denominator of each term is a perfect square. Using sigma notation, this sum can be written as $$\\displaystyle \\sum_{i=1}^5\\dfrac{1}{i^2}$$.\n\nExercise $$\\PageIndex{1}$$\n\nWrite in sigma notation and evaluate the sum of terms $$2^i$$ for $$i=3,4,5,6.$$\n\nHint\n\nUse the solving steps in Example $$\\PageIndex{1}$$ as a guide.\n\n$$\\displaystyle \\sum_{i=3}^{6}2^i=2^3+2^4+2^5+2^6=120$$\n\nThe properties associated with the summation process are given in the following rule.\n\nRule: Properties of Sigma Notation\n\nLet $$a_1,a_2,…,a_n$$ and $$b_1,b_2,…,b_n$$ represent two sequences of terms and let $$c$$ be a constant. The following properties hold for all positive integers $$n$$ and for integers $$m$$, with $$1≤m≤n.$$\n\n1. $$\\displaystyle \\sum_{i=1}^n c=nc$$\n2. $$\\displaystyle \\sum_{i=1}^n ca_i=c\\sum_{i=1}^na_i$$\n3. $$\\displaystyle \\sum_{i=1}^n(a_i+b_i)=\\sum_{i=1}^na_i+\\sum_{i=1}^nb_i$$\n4. $$\\displaystyle \\sum_{i=1}^n(a_i−b_i)=\\sum_{i=1}^na_i−\\sum_{i=1}^nb_i$$\n5. $$\\displaystyle \\sum_{i=1}^na_i=\\sum_{i=1}^ma_i+\\sum_{i=m+1}^na_i$$\n\nProof\n\nWe prove properties 2. and 3. here, and leave proof of the other properties to the Exercises.\n\n2. We have\n\n$\\sum_{i=1}^nca_i=ca_1+ca_2+ca_3+⋯+ca_n=c(a_1+a_2+a_3+⋯+a_n)=c\\sum_{i=1}^na_i.$\n\n3. We have\n\n\\begin{align} \\sum_{i=1}^{n}(a_i+b_i) =(a_1+b_1)+(a_2+b_2)+(a_3+b_3)+⋯+(a_n+b_n) \\\\ =(a_1+a_2+a_3+⋯+a_n)+(b_1+b_2+b_3+⋯+b_n) \\\\ =\\sum_{i=1}^na_i+\\sum_{i=1}^nb_i. \\end {align}\n\nA few more formulas for frequently found functions simplify the summation process further. These are shown in the next rule, for sums and powers of integers, and we use them in the next set of examples.\n\nRule: Sums and Powers of Integers\n\n1. The sum of $$n$$ integers is given by\n\n$\\sum_{i=1}^n i=1+2+⋯+n=\\dfrac{n(n+1)}{2}. \\label{sum1}$\n\n2. The sum of consecutive integers squared is given by\n\n$\\sum_{i=1}^n i^2=1^2+2^2+⋯+n^2=\\dfrac{n(n+1)(2n+1)}{6}. \\label{sum2}$\n\n3. The sum of consecutive integers cubed is given by\n\n$\\sum_{i=1}^n i^3=1^3+2^3+⋯+n^3=\\dfrac{n^2(n+1)^2}{4}. \\label{sum3}$\n\nExample $$\\PageIndex{2}$$: Evaluation Using Sigma Notation\n\nWrite using sigma notation and evaluate:\n\n1. The sum of the terms $$(i−3)^2$$ for $$i=1,2,…,200.$$\n2. The sum of the terms $$(i^3−i^2)$$ for $$i=1,2,3,4,5,6$$\n\nSolution\n\na. Multiplying out $$(i−3)^2$$, we can break the expression into three terms.\n\n\\begin{align*} \\sum_{i=1}^{200}(i−3)^2 =\\sum_{i=1}^{200}(i^2−6i+9) \\\\[4pt] =\\sum_{i=1}^{200}i^2−\\sum_{i=1}^{200}6i+\\sum_{i=1}^{200}9 \\\\[4pt] =\\sum_{i=1}^{200}i^2−6\\sum_{i=1}^{200}i+\\sum_{i=1}^{200}9 \\\\[4pt] =\\dfrac{200(200+1)(400+1)}{6}−6 \\left[\\dfrac{200(200+1)}{2}\\right]+9(200) \\\\[4pt] =2,686,700−120,600+1800 \\\\[4pt] =2,567,900 \\end{align*}\n\nb. Use sigma notation property iv. and the rules for the sum of squared terms and the sum of cubed terms.\n\n\\begin{align*} \\sum_{i=1}^{6}(i^3−i^2) =\\sum_{i=1}^6 i^3−\\sum_{i=1}^6 i^2 \\\\[4pt] =\\dfrac{6^2(6+1)^2}{4}−\\dfrac{6(6+1)(2(6)+1)}{6} \\\\[4pt] =\\dfrac{1764}{4}−\\dfrac{546}{6} \\\\[4pt] =350 \\end{align*}\n\nExercise $$\\PageIndex{2}$$\n\nFind the sum of the values of $$4+3i$$ for $$i=1,2,…,100.$$\n\nHint\n\nUse the properties of sigma notation to solve the problem.\n\n15,550\n\nExample $$\\PageIndex{3}$$: Finding the Sum of the Function Values\n\nFind the sum of the values of $$f(x)=x^3$$ over the integers $$1,2,3,…,10.$$\n\nSolution\n\nUsing Equation \\ref{sum3}, we have\n\n$\\sum_{i=0}^{10}i^3=\\dfrac{(10)^2(10+1)^2}{4}=\\dfrac{100(121)}{4}=3025 \\nonumber$\n\nExercise $$\\PageIndex{3}$$\n\nEvaluate the sum indicated by the notation $$\\displaystyle \\sum_{k=1}^{20}(2k+1)$$.\n\nHint\n\nUse the rule on sum and powers of integers (Equations \\ref{sum1}-\\ref{sum3}).\n\n440\n\n## Approximating Area\n\nNow that we have the necessary notation, we return to the problem at hand: approximating the area under a curve. Let $$f(x)$$ be a continuous, nonnegative function defined on the closed interval $$[a,b]$$. We want to approximate the area A bounded by $$f(x)$$ above, the x-axis below, the line $$x=a$$ on the left, and the line $$x=b$$ on the right (Figure $$\\PageIndex{1}$$).", null, "Figure $$\\PageIndex{1}$$: An area (shaded region) bounded by the curve $$f(x)$$ at top, the x-axis at bottom, the line $$x=a$$ to the left, and the line $$x=b$$ at right.\n\nHow do we approximate the area under this curve? The approach is a geometric one. By dividing a region into many small shapes that have known area formulas, we can sum these areas and obtain a reasonable estimate of the true area. We begin by dividing the interval $$[a,b]$$ into n subintervals of equal width, $$\\dfrac{b−a}{n}$$. We do this by selecting equally spaced points $$x_0,x_1,x_2,…,x_n$$ with $$x_0=a,x_n=b,$$ and\n\n$x_i−x_{i−1}=\\dfrac{b−a}{n}$\n\nfor $$i=1,2,3,…,n.$$\n\nWe denote the width of each subinterval with the notation Δx, so $$Δx=\\frac{b−a}{n}$$ and\n\n$x_i=x_0+iΔx$\n\nfor $$i=1,2,3,…,n.$$ This notion of dividing an interval $$[a,b]$$ into subintervals by selecting points from within the interval is used quite often in approximating the area under a curve, so let’s define some relevant terminology.\n\nDefinition: Partitions\n\nA set of points $$P={x_i}$$ for $$i=0,1,2,…,n$$ with $$a=x_0<x_1<x_2<⋯<x_n=b$$, which divides the interval $$[a,b]$$ into subintervals of the form [x_0,x_1],[x_1,x_2],…,[x_{n−1},x_n] is called a partition of $$[a,b]$$. If the subintervals all have the same width, the set of points forms a regular partition of the interval [a,b].\n\nWe can use this regular partition as the basis of a method for estimating the area under the curve. We next examine two methods: the left-endpoint approximation and the right-endpoint approximation.\n\nRule: Left-Endpoint Approximation\n\nOn each subinterval $$[x_{i−1},x_i]$$ (for $$i=1,2,3,…,n$$), construct a rectangle with width $$Δx$$ and height equal to $$f(x_{i−1})$$, which is the function value at the left endpoint of the subinterval. Then the area of this rectangle is $$f(x_{i−1})Δx$$. Adding the areas of all these rectangles, we get an approximate value for $$A$$ (Figure $$\\PageIndex{2}$$). We use the notation $$L_n$$ to denote that this is a left-endpoint approximation of $$A$$ using $$n$$ subintervals.\n\n$A≈L_n=f(x_0)Δx+f(x_1)Δx+⋯+f(xn−1)Δx=\\sum_{i=1}^nf(x_{i−1})Δx$", null, "Figure $$\\PageIndex{2}$$: In the left-endpoint approximation of area under a curve, the height of each rectangle is determined by the function value at the left of each subinterval.\n\nThe second method for approximating area under a curve is the right-endpoint approximation. It is almost the same as the left-endpoint approximation, but now the heights of the rectangles are determined by the function values at the right of each subinterval.\n\nRule: Right-Endpoint Approximation\n\nConstruct a rectangle on each subinterval $$[x_{i−1},x_i]$$, only this time the height of the rectangle is determined by the function value $$f(x_i)$$ at the right endpoint of the subinterval. Then, the area of each rectangle is $$f(x_i)\\,Δx$$ and the approximation for $$A$$ is given by\n\n$A≈R_n=f(x_1)Δx+f(x_2)Δx+⋯+f(x_n)Δx=\\sum_{i=1}^nf(x_i)Δx.$\n\nThe notation $$R_n$$ indicates this is a right-endpoint approximation for $$A$$ (Figure $$\\PageIndex{3}$$).", null, "Figure $$\\PageIndex{3}$$:In the right-endpoint approximation of area under a curve, the height of each rectangle is determined by the function value at the right of each subinterval. Note that the right-endpoint approximation differs from the left-endpoint approximation in Figure $$\\PageIndex{2}$$.\n\nThe graphs in Figure $$\\PageIndex{4}$$ represent the curve $$f(x)=\\dfrac{x^2}{2}$$. In Figure $$\\PageIndex{4b}$$ we divide the region represented by the interval $$[0,3]$$ into six subintervals, each of width $$0.5$$. Thus, $$Δx=0.5$$. We then form six rectangles by drawing vertical lines perpendicular to $$x_{i−1}$$, the left endpoint of each subinterval. We determine the height of each rectangle by calculating $$f(x_{i−1})$$ for $$i=1,2,3,4,5,6.$$ The intervals are $$[0,0.5],[0.5,1],[1,1.5],[1.5,2],[2,2.5],[2.5,3]$$. We find the area of each rectangle by multiplying the height by the width. Then, the sum of the rectangular areas approximates the area between $$f(x)$$ and the $$x$$-axis. When the left endpoints are used to calculate height, we have a left-endpoint approximation. Thus,\n\n\\begin{align*} A≈L_6 =\\sum_{i=1}^6f(x_{i−1})Δx =f(x_0)Δx+f(x_1)Δx+f(x_2)Δx+f(x_3)Δx+f(x_4)Δx+f(x_5)Δx \\\\[4pt] =f(0)0.5+f(0.5)0.5+f(1)0.5+f(1.5)0.5+f(2)0.5+f(2.5)0.5 \\\\[4pt] =(0)0.5+(0.125)0.5+(0.5)0.5+(1.125)0.5+(2)0.5+(3.125)0.5 \\\\[4pt] =0+0.0625+0.25+0.5625+1+1.5625 \\\\[4pt] =3.4375 \\,\\text{units}^2\\end{align*}", null, "Figure $$\\PageIndex{4}$$: Methods of approximating the area under a curve by using (a) the left endpoints and (b) the right endpoints.\n\nIn Figure $$\\PageIndex{4b}$$, we draw vertical lines perpendicular to $$x_i$$ such that $$x_i$$ is the right endpoint of each subinterval, and calculate $$f(x_i)$$ for $$i=1,2,3,4,5,6$$. We multiply each $$f(x_i)$$ by $$Δx$$ to find the rectangular areas, and then add them. This is a right-endpoint approximation of the area under $$f(x)$$. Thus,\n\n\\begin{align*} A≈R_6 =\\sum_{i=1}^6f(x_i)Δx=f(x_1)Δx+f(x_2)Δx+f(x_3)Δx+f(x_4)Δx+f(x_5)Δx+f(x_6)Δx\\\\[4pt] =f(0.5)0.5+f(1)0.5+f(1.5)0.5+f(2)0.5+f(2.5)0.5+f(3)0.5 \\\\[4pt] =(0.125)0.5+(0.5)0.5+(1.125)0.5+(2)0.5+(3.125)0.5+(4.5)0.5 \\\\[4pt] =0.0625+0.25+0.5625+1+1.5625+2.25 \\\\[4pt] =5.6875 \\,\\text{units}^2.\\end{align*}\n\nExample $$\\PageIndex{4}$$: Approximating the Area Under a Curve\n\nUse both left-endpoint and right-endpoint approximations to approximate the area under the curve of $$f(x)=x^2$$ on the interval $$[0,2]$$; use $$n=4$$.\n\nSolution\n\nFirst, divide the interval $$[0,2]$$ into $$n$$ equal subintervals. Using $$n=4,\\, Δx=\\dfrac{(2−0)}{4}=0.5$$. This is the width of each rectangle. The intervals $$[0,0.5],[0.5,1],[1,1.5],[1.5,2]$$ are shown in Figure $$\\PageIndex{5}$$. Using a left-endpoint approximation, the heights are $$f(0)=0,\\,f(0.5)=0.25,\\,f(1)=1,$$ and $$f(1.5)=2.25.$$ Then,\n\n\\begin{align*} L_4 =f(x_0)Δx+f(x_1)Δx+f(x_2)Δx+f(x_3)Δx \\\\[4pt] =0(0.5)+0.25(0.5)+1(0.5)+2.25(0.5) \\\\[4pt] =1.75 \\,\\text{units}^2 \\end{align*}", null, "Figure $$\\PageIndex{5}$$: The graph shows the left-endpoint approximation of the area under $$f(x)=x^2$$ from $$0$$ to $$2$$.\n\nThe right-endpoint approximation is shown in Figure $$\\PageIndex{6}$$. The intervals are the same, $$Δx=0.5,$$ but now use the right endpoint to calculate the height of the rectangles. We have\n\n\\begin{align*} R_4 =f(x_1)Δx+f(x_2)Δx+f(x_3)Δx+f(x_4)Δx \\\\[4pt] =0.25(0.5)+1(0.5)+2.25(0.5)+4(0.5) \\\\[4pt] =3.75 \\,\\text{units}^2 \\end{align*}", null, "Figure $$\\PageIndex{6}$$: The graph shows the right-endpoint approximation of the area under $$f(x)=x^2$$ from $$0$$ to $$2$$.\n\nThe left-endpoint approximation is $$1.75\\,\\text{units}^2$$; the right-endpoint approximation is $$3.75 \\,\\text{units}^2$$.\n\nExercise $$\\PageIndex{4}$$\n\nSketch left-endpoint and right-endpoint approximations for $$f(x)=\\dfrac{1}{x}$$ on $$[1,2]$$; use $$n=4$$. Approximate the area using both methods.\n\nHint\n\nFollow the solving strategy in Example $$\\PageIndex{4}$$ step-by-step.\n\nThe left-endpoint approximation is $$0.7595 \\,\\text{units}^2$$. The right-endpoint approximation is $$0.6345 \\,\\text{units}^2$$. See the below Media.", null, "Looking at Figure $$\\PageIndex{4}$$ and the graphs in Example $$\\PageIndex{4}$$, we can see that when we use a small number of intervals, neither the left-endpoint approximation nor the right-endpoint approximation is a particularly accurate estimate of the area under the curve. However, it seems logical that if we increase the number of points in our partition, our estimate of A will improve. We will have more rectangles, but each rectangle will be thinner, so we will be able to fit the rectangles to the curve more precisely.\n\nWe can demonstrate the improved approximation obtained through smaller intervals with an example. Let’s explore the idea of increasing $$n$$, first in a left-endpoint approximation with four rectangles, then eight rectangles, and finally $$32$$ rectangles. Then, let’s do the same thing in a right-endpoint approximation, using the same sets of intervals, of the same curved region. Figure $$\\PageIndex{7}$$ shows the area of the region under the curve $$f(x)=(x−1)^3+4$$ on the interval $$[0,2]$$ using a left-endpoint approximation where $$n=4.$$ The width of each rectangle is\n\n$Δx=\\dfrac{2−0}{4}=\\dfrac{1}{2}.$\n\nThe area is approximated by the summed areas of the rectangles, or\n\n$L_4=f(0)(0.5)+f(0.5)(0.5)+f(1)(0.5)+f(1.5)0.5=7.5 \\,\\text{units}^2$", null, "Figure $$\\PageIndex{7}$$: With a left-endpoint approximation and dividing the region from a to b into four equal intervals, the area under the curve is approximately equal to the sum of the areas of the rectangles.\n\nFigure $$\\PageIndex{8}$$ shows the same curve divided into eight subintervals. Comparing the graph with four rectangles in Figure $$\\PageIndex{7}$$ with this graph with eight rectangles, we can see there appears to be less white space under the curve when $$n=8.$$ This white space is area under the curve we are unable to include using our approximation. The area of the rectangles is\n\n$L_8=f(0)(0.25)+f(0.25)(0.25)+f(0.5)(0.25)+f(0.75)(0.25)+f(1)(0.25)+f(1.25)(0.25)+f(1.5)(0.25)+f(1.75)(0.25)=7.75 \\,\\text{units}^2$", null, "Figure $$\\PageIndex{8}$$: The region under the curve is divided into $$n=8$$ rectangular areas of equal width for a left-endpoint approximation.\n\nThe graph in Figure $$\\PageIndex{9}$$ shows the same function with $$32$$ rectangles inscribed under the curve. There appears to be little white space left. The area occupied by the rectangles is\n\n$L_{32}=f(0)(0.0625)+f(0.0625)(0.0625)+f(0.125)(0.0625)+⋯+f(1.9375)(0.0625)=7.9375 \\,\\text{units}^2.$", null, "Figure $$\\PageIndex{9}$$: Here, $$32$$ rectangles are inscribed under the curve for a left-endpoint approximation.\n\nWe can carry out a similar process for the right-endpoint approximation method. A right-endpoint approximation of the same curve, using four rectangles (Figure $$\\PageIndex{10}$$), yields an area\n\n$R_4=f(0.5)(0.5)+f(1)(0.5)+f(1.5)(0.5)+f(2)(0.5)=8.5 \\,\\text{units}^2.$", null, "Figure $$\\PageIndex{10}$$: Now we divide the area under the curve into four equal subintervals for a right-endpoint approximation.\n\nDividing the region over the interval $$[0,2]$$ into eight rectangles results in $$Δx=\\dfrac{2−0}{8}=0.25.$$ The graph is shown in Figure $$\\PageIndex{11}$$. The area is\n\n$R_8=f(0.25)(0.25)+f(0.5)(0.25)+f(0.75)(0.25)+f(1)(0.25)+f(1.25)(0.25)+f(1.5)(0.25)+f(1.75)(0.25)+f(2)(0.25)=8.25 \\,\\text{units}^2$", null, "Figure $$\\PageIndex{11}$$: Here we use right-endpoint approximation for a region divided into eight equal subintervals.\n\nLast, the right-endpoint approximation with $$n=32$$ is close to the actual area (Figure $$\\PageIndex{12}$$). The area is approximately\n\n$R_{32}=f(0.0625)(0.0625)+f(0.125)(0.0625)+f(0.1875)(0.0625)+⋯+f(2)(0.0625)=8.0625 \\,\\text{units}^2$", null, "Figure $$\\PageIndex{12}$$: The region is divided into $$32$$ equal subintervals for a right-endpoint approximation.\n\nBased on these figures and calculations, it appears we are on the right track; the rectangles appear to approximate the area under the curve better as $$n$$ gets larger. Furthermore, as $$n$$ increases, both the left-endpoint and right-endpoint approximations appear to approach an area of $$8$$ square units. Table shows a numerical comparison of the left- and right-endpoint methods. The idea that the approximations of the area under the curve get better and better as $$n$$ gets larger and larger is very important, and we now explore this idea in more detail.\n\nTable $$\\PageIndex{15}$$: Converging Values of Left- and Right-Endpoint Approximations as n Increases\nValue of $$n$$ Approximate Area $$L_n$$ Approximate Area $$R_n$$\n$$n=4$$ $$7.5$$ $$8.5$$\n$$n=8$$ $$7.75$$ $$8.25$$\n$$n=32$$ $$7.94$$ $$8.06$$\n\n### Forming Riemann Sums\n\nSo far we have been using rectangles to approximate the area under a curve. The heights of these rectangles have been determined by evaluating the function at either the right or left endpoints of the subinterval $$[x_{i−1},x_i]$$. In reality, there is no reason to restrict evaluation of the function to one of these two points only. We could evaluate the function at any point ci in the subinterval $$[x_{i−1},x_i]$$, and use $$f(x^∗_i)$$ as the height of our rectangle. This gives us an estimate for the area of the form\n\n$A≈\\sum_{i=1}^nf(x^∗_i)\\,Δx.$\n\nA sum of this form is called a Riemann sum, named for the 19th-century mathematician Bernhard Riemann, who developed the idea.\n\nDefinition: Riemann sum\n\nLet $$f(x)$$ be defined on a closed interval $$[a,b]$$ and let P be a regular partition of $$[a,b]$$. Let $$Δx$$ be the width of each subinterval $$[x_{i−1},x_i]$$ and for each $$i$$, let $$x^∗_i$$ be any point in $$[x_{i−1},x_i]$$. A Riemann sum is defined for $$f(x)$$ as\n\n$\\sum_{i=1}^nf(x^∗_i)\\,Δx.$\n\nRecall that with the left- and right-endpoint approximations, the estimates seem to get better and better as $$n$$ get larger and larger. The same thing happens with Riemann sums. Riemann sums give better approximations for larger values of $$n$$. We are now ready to define the area under a curve in terms of Riemann sums.\n\nDefinition: area under the curve\n\nLet $$f(x)$$ be a continuous, nonnegative function on an interval $$[a,b]$$, and let $$\\sum_{i=1}^nf(x^∗_i)\\,Δx$$ be a Riemann sum for $$f(x)$$. Then, the area under the curve $$y=f(x)$$ on $$[a,b]$$ is given by\n\n$A=\\lim_{n→∞}\\sum_{i=1}^nf(x^∗_i)\\,Δx.$\n\nSee a graphical demonstration of the construction of a Riemann sum.\n\nSome subtleties here are worth discussing. First, note that taking the limit of a sum is a little different from taking the limit of a function $$f(x)$$ as $$x$$ goes to infinity. Limits of sums are discussed in detail in the chapter on Sequences and Series; however, for now we can assume that the computational techniques we used to compute limits of functions can also be used to calculate limits of sums.\n\nSecond, we must consider what to do if the expression converges to different limits for different choices of $${x^∗_i}.$$ Fortunately, this does not happen. Although the proof is beyond the scope of this text, it can be shown that if $$f(x)$$ is continuous on the closed interval $$[a,b]$$, then $$\\displaystyle \\lim_{n→∞}\\sum_{i=1}^nf(x^∗_i)Δx$$ exists and is unique (in other words, it does not depend on the choice of $${x^∗_i}$$).\n\nWe look at some examples shortly. But, before we do, let’s take a moment and talk about some specific choices for $${x^∗_i}$$. Although any choice for $${x^∗_i}$$ gives us an estimate of the area under the curve, we don’t necessarily know whether that estimate is too high (overestimate) or too low (underestimate). If it is important to know whether our estimate is high or low, we can select our value for $${x^∗_i}$$ to guarantee one result or the other.\n\nIf we want an overestimate, for example, we can choose $${x^∗_i}$$ such that for $$i=1,2,3,…,n,$$ $$f(x^∗_i)≥f(x)$$ for all $$x∈[x_i−1,x_i]$$. In other words, we choose $${x^∗_i}$$ so that for $$i=1,2,3,…,n,$$ $$f(x^∗_i)$$ is the maximum function value on the interval $$[x_{i−1},x_i]$$. If we select $${x^∗_i}$$ in this way, then the Riemann sum $$\\displaystyle \\sum_{i=1}^nf(x^∗_i)Δx$$ is called an upper sum. Similarly, if we want an underestimate, we can choose $${x∗i}$$ so that for $$i=1,2,3,…,n,$$ $$f(x^∗_i)$$ is the minimum function value on the interval $$[x_{i−1},x_i]$$. In this case, the associated Riemann sum is called a lower sum. Note that if $$f(x)$$ is either increasing or decreasing throughout the interval $$[a,b]$$, then the maximum and minimum values of the function occur at the endpoints of the subintervals, so the upper and lower sums are just the same as the left- and right-endpoint approximations.\n\nExample $$\\PageIndex{5}$$: Finding Lower and Upper Sums\n\nFind a lower sum for $$f(x)=10−x^2$$ on $$[1,2]$$; let $$n=4$$ subintervals.\n\nSolution\n\nWith $$n=4$$ over the interval $$[1,2], \\,Δx=\\dfrac{1}{4}$$. We can list the intervals as $$[1,1.25],\\,[1.25,1.5],\\,[1.5,1.75],$$ and $$[1.75,2]$$. Because the function is decreasing over the interval $$[1,2],$$ Figure shows that a lower sum is obtained by using the right endpoints.", null, "Figure $$\\PageIndex{13}$$: The graph of $$f(x)=10−x^2$$ is set up for a right-endpoint approximation of the area bounded by the curve and the $$x$$-axis on $$[1,2]$$, and it shows a lower sum.\n\nThe Riemann sum is\n\n\\begin{align*} \\sum_{k=1}^4(10−x^2)(0.25) =0.25[10−(1.25)^2+10−(1.5)^2+10−(1.75)^2+10−(2)^2] \\\\[4pt] =0.25[8.4375+7.75+6.9375+6] \\\\[4pt] =7.28 \\,\\text{units}^2.\\end{align*}\n\nThe area of $$7.28$$ $$\\text{units}^2$$ is a lower sum and an underestimate.\n\nExercise $$\\PageIndex{5}$$\n\n1. Find an upper sum for $$f(x)=10−x^2$$ on $$[1,2]$$; let $$n=4.$$\n2. Sketch the approximation.\nHint\n\n$$f(x)$$ is decreasing on $$[1,2]$$, so the maximum function values occur at the left endpoints of the subintervals.\n\na. Upper sum=$$8.0313 \\,\\text{units}^2.$$\n\nb.", null, "Example $$\\PageIndex{6}$$: Finding Lower and Upper Sums for $$f(x)=\\sin x$$\n\nFind a lower sum for $$f(x)=\\sin x$$ over the interval $$[a,b]=[0,\\dfrac{π}{2}]$$; let $$n=6.$$\n\nSolution\n\nLet’s first look at the graph in Figure $$\\PageIndex{14}$$ to get a better idea of the area of interest.", null, "Figure $$\\PageIndex{14}$$: The graph of $$y=\\sin x$$ is divided into six regions: $$Δx=\\dfrac{π/2}{6}=\\dfrac{π}{12}$$.\n\nThe intervals are $$[0,\\dfrac{π}{12}],[\\dfrac{π}{12},\\dfrac{π}{6}],[\\dfrac{π}{6},\\dfrac{π}{4}],[\\dfrac{π}{4},\\dfrac{π}{3}],[\\dfrac{π}{3},\\dfrac{5π}{12}]$$, and $$[\\dfrac{5π}{12},\\dfrac{π}{2}]$$. Note that $$f(x)=\\sin x$$ is increasing on the interval $$[0,\\dfrac{π}{2}]$$, so a left-endpoint approximation gives us the lower sum. A left-endpoint approximation is the Riemann sum $$\\sum_{i=0}^5\\sin x_i(\\dfrac{π}{12})$$.We have\n\n$A≈\\sin(0)(\\frac{π}{12})+\\sin(\\frac{π}{12})(\\frac{π}{12})+\\sin(\\frac{π}{6})(\\frac{π}{12})+\\sin(\\frac{π}{4})(\\frac{π}{12})+\\sin(\\frac{π}{3})(\\frac{π}{12})+\\sin(\\frac{5π}{12})(\\frac{π}{12})=0.863 \\,\\text{units}^2. \\nonumber$\n\nExercise $$\\PageIndex{6}$$\n\nUsing the function $$f(x)=\\sin x$$ over the interval $$[0,\\dfrac{π}{2}],$$ find an upper sum; let $$n=6.$$\n\nHint\n\nFollow the steps from Example $$\\PageIndex{6}$$.\n\n$$A≈1.125 \\,\\text{units}^2$$\n\n## Key Concepts\n\n• The use of sigma (summation) notation of the form $$\\displaystyle \\sum_{i=1}^na_i$$ is useful for expressing long sums of values in compact form.\n• For a continuous function defined over an interval $$[a,b],$$ the process of dividing the interval into n equal parts, extending a rectangle to the graph of the function, calculating the areas of the series of rectangles, and then summing the areas yields an approximation of the area of that region.\n• The width of each rectangle is $$Δx=\\dfrac{b−a}{n}$$.\n• Riemann sums are expressions of the form $$\\displaystyle \\sum_{i=1}^nf(x^∗_i)Δx,$$ and can be used to estimate the area under the curve $$y=f(x).$$ Left- and right-endpoint approximations are special kinds of Riemann sums where the values of $${x^∗_i}$$are chosen to be the left or right endpoints of the subintervals, respectively.\n• Riemann sums allow for much flexibility in choosing the set of points $${x^∗_i}$$ at which the function is evaluated, often with an eye to obtaining a lower sum or an upper sum.\n\n## Key Equations\n\n• Properties of Sigma Notation\n\n\\begin{align*} \\sum_{i=1}^nc=nc \\\\ \\sum_{i=1}^nca_i =c\\sum_{i=1}^na_i \\\\ \\sum_{i=1}^n(a_i+b_i) =\\sum_{i=1}^na_i+\\sum_{i=1}^nb_i \\\\ \\sum_{i=1}^n(a_i−b_i) =\\sum_{i=1}^na_i−\\sum_{i=1}^nb_i \\\\ \\sum_{i=1}^na_i=\\sum_{i=1}^ma_i+\\sum_{i=m+1}^na_i \\end{align*}\n\n• Sums and Powers of Integers\n\n$\\sum_{i=1}^ni=1+2+⋯+n=\\dfrac{n(n+1)}{2} \\nonumber$\n\n$\\sum_{i=1}^ni^2=1^2+2^2+⋯+n^2=\\dfrac{n(n+1)(2n+1)}{6} \\nonumber$\n\n$\\sum_{i=0}^ni^3=1^3+2^3+⋯+n^3=\\dfrac{n^2(n+1)^2}{4} \\nonumber$\n\n• Left-Endpoint Approximation\n\n$$A≈L_n=f(x_0)Δx+f(x_1)Δx+⋯+f(x_{n−1})Δx=\\displaystyle \\sum_{i=1}^nf(x_{i−1})Δx$$\n\n• Right-Endpoint Approximation\n\n$$A≈R_n=f(x_1)Δx+f(x_2)Δx+⋯+f(x_n)Δx=\\displaystyle \\sum_{i=1}^nf(x_i)Δx$$\n\n## Glossary\n\nleft-endpoint approximation\nan approximation of the area under a curve computed by using the left endpoint of each subinterval to calculate the height of the vertical sides of each rectangle\nlower sum\na sum obtained by using the minimum value of $$f(x)$$ on each subinterval\npartition\na set of points that divides an interval into subintervals\nregular partition\na partition in which the subintervals all have the same width\nriemann sum\nan estimate of the area under the curve of the form $$A≈\\displaystyle \\sum_{i=1}^nf(x^∗_i)Δx$$\nright-endpoint approximation\nthe right-endpoint approximation is an approximation of the area of the rectangles under a curve using the right endpoint of each subinterval to construct the vertical sides of each rectangle\nsigma notation\n(also, summation notation) the Greek letter sigma ($$Σ$$) indicates addition of the values; the values of the index above and below the sigma indicate where to begin the summation and where to end it\nupper sum\na sum obtained by using the maximum value of $$f(x)$$ on each subinterval" ]

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[ "Post New Query\n\n# What are set of FDs for the relation in DBMS?", null, "Asked on January 03, 2015\nConsider the following set of FDs for the relation R(A, B, C, D, E):\nF= {A-->B, AB-->C, D-->AC, D-->E}\nF'= {A-->BC, D-->AE}\nCheck whether these are equivalent sets.", null, "Replied on January 04, 2015\nA-->B means A functionally defines B.\nAB-->C means A and B functionally defines  C\nD--> AC means D functionally defines A and C\nD--> E means D functionally defines E\n\nIf understand the properties of functional dependency as given below\n\n1. Reflexivity: If Y ⊆ X, then X → Y\n2. Augmentation: If X → Y , then XZ → Y Z\n3. Transitivity: If X → Y and Y → Z, then X → Z\n4. Union: If X → Y and X → Z, then X → Y Z\n5. Decomposition: If X → Y Z, then X → Y and X → Z\n6. Pseudotransitivity: If X → Y and W Y → Z, then W X → Z\n7. Composition: If X → Y and Z → W, then XZ → Y W\n\nAnd find the minimal cover,\n\nwe can get\n\nA-->BC, D-->AE\n\nSo F and F' are same." ]

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[ "", null, "# The Unemployment Rate: A Coincident Recession Indicator\n\nNote:  First  published here on April 17, 2012 and updated October 3, 2014 and originally co-authored with Dwaine van Vuuren.\n\nFor what is considered to be a lagging indicator of the economy, the unemployment rate provides surprisingly good signals for the beginnings and ends of recessions. We have developed a model that uses unemployment figures to produce these signals and to determine the probability of when a recession may start.\n\nThe current recession call from the Economic Cycle Research Institute (ECRI) is now more than six months old. Our model based on the unemployment rate which is presented here does not support ECRI’s call; it also tells us that a recession will not happen anytime soon.\n\n#### The unemployment rate indicator\n\nMonthly unemployment data is listed at FRED from 1948 onwards, a dataset that spans 11 recessions and covers a much longer period than the historical data for most other indicators – the Conference Board LEI, for example, or the ECRI Weekly Leading Index – whose performance one can only evaluate for the last seven recessions.\n\nThe unemployment rate (UER) over time is shown in Figure 1. It is depicted by the blue and red graphs, which are the short exponential moving average (EMA) and long EMA of the unemployment rate, respectively. (Appendix 1, for those who are interested, is a description of EMA.) The gray vertical bar graphs represent the 11 recessions during the time span the data cover.", null, "(click to enlarge)\n\nIt is apparent from Figure 1 that the blue graph, the short EMA of the UER forms a trough prior to the beginning of each recession.\n\nFigures 2.1 and 2.2 incorporate the information depicted in Figure 1 and also show:\n\n1. The smoothed eight-month annualized growth rate of the unemployment rate, calculated from its 15-week moving average, which we refer to as “UERg.” That’s the green graph. (Appendix 2 describes how to calculate UERg.)\n2. The time periods when the basic unemployment rate and/or UERg signaled recession, which are the gold-shaded parts of the graph.\n3. The dates of ECRI’s four recession calls of 1990, 2001, 2008 and 2011. These are the vertical purple lines in figure 2.2.", null, "(click to enlarge)", null, "(click to enlarge)\n\nOne can see from Figures 2.1 and 2.2 that the UERg, in green, has a well defined peak near the end of every recession, when the slope of the growth rate of the UER abruptly changes from positive (rising) to negative (declining).\n\nIt is similarly evident that the ECRI’s recession calls in 1990, 2001 and 2008 always coincided with a trough where the short EMA of the unemployment rate rose above its long EMA; UERg was always positive at those times as well.\n\nThis was not the case at the end of September 2011, however, when ECRI issued its most recent recession call. Then the unemployment rate had just formed a peak and the short EMA was below the long EMA of the unemployment rate, all completely atypical of when past recessions occurred. (Appendix 3 has the dates of ECRI’s recession calls and the corresponding recession signals from the unemployment rate.)\n\n#### Conditions that indicate recession starts and ends\n\nRecessions generally start when one the following three conditions occurs:\n\n1. The short EMA of the unemployment rate rises and crosses the long EMA to the upside, and the difference between the two EMAs is at least 0.07.\n2. UERg rises above zero, the long EMA of the unemployment rate has a positive slope, and the difference between the long EMA at that time and the long EMA 10 weeks before is greater than 0.025.\n3. The 19-week rate of change of the unemployment rate is greater than 8.0%, while simultaneously the long EMA of the unemployment rate has a positive slope and the difference between the long EMA at the time and the long EMA 10 weeks earlier is greater than 0.015.\n\nRecession ends are generally signaled when:\n\n1. UERg as calculated from a 4-week moving average of the unemployment rate crosses from above to below UERg as calculated from a 15-week moving average, and UERg as calculated from a 15-week moving average is greater than 20%.\n\n#### Probability of recession start\n\nWe calculated the probability a recession will start in any given week based on the features which generally indicate recession starts.\n\nThe following probability curve resulted over the lifespan of the data (Figure 3):", null, "(click to enlarge)\n\nThis is a very satisfactory result, given especially that it is derived, essentially, from a single factor — the unemployment rate.\n\n#### Present conditions\n\nThe ends of the graphs in Figure 4 depict the current situation, with the September 5.9% unemployment number included. The graphs are plotted to the beginning November 2014, when the October unemployment statistics will be released.\n\nWe see none of the conditions for a recession to start. The unemployment rate is not forming a trough. Its short EMA is well below its long EMA. The UERg has had a negative slope since the last week of September 2011, when ECRI made its recession call. Additionally, the UERg is currently at its lowest level – approximately -17.1% – since it fell below zero in the middle of September 2010.\n\nAlso the 19-week rate of change of the unemployment rate is now -6.3%, far below the critical signal level, and the probability that a recession will start now is zero, as shown by the probability of recession graph in Figure 3.\n\nFor a recession to occur, the unemployment rate graphs would have to, as a minimum requirement, move sideways for a while and then turn upwards. Alternatively, the UERg graph could turn upwards and rise above zero, or the 19-week rate of change graph of unemployment rate itself would have to approach 8%. There is no indication that any of this will happen anytime soon. Currently the trajectories of both the unemployment rate and its growth rate are downwards, as was the case after all previous recessions.\n\nOne can therefore reasonably conclude, based on the historic evidence of these unemployment-based indicators, that there will be no recession in the near future and that ECRI’s last recession call was definitely premature.\n\n### Appendix 1\n\n#### Exponential Moving Average\n\nThe exponential moving average (EMA) is a type of filter that applies weighting factors to the observed data values. The weighting for each older data point decreases exponentially, never reaching zero. For a detailed description see http://en.wikipedia.org/wiki/Moving_average\n\nThe formula used for calculating the EMA is\n\nEMAthis week = EMAlast week +α× (valuethis week EMAlast week)\n\nThe coefficient α represents the degree of weighting decrease, a constant smoothing factor between 0 and 1. A higher α discounts older observations faster. Alternatively, α may be expressed in terms of N time periods, where α = 2/(N+1). For example, N = 60 for the long EMA of UERresults in a smoothing factor of 0.033, and N = 9 for the short EMA of UERresults a smoothing factor of 0.20. The “long” and “short” refers to the N time periods used in the calculation of the smoothing factor for the EMAs.\n\n### Appendix 2\n\n#### Calculating the annualized growth rate of UER\n\nThe eight-month smoothed compound annualized growth rate of UER is calculated as follows:\n\nUERg is the compound annualized growth rate of the UER for a 35.5 week period. MA1 is the 15 week moving average of the UER and MA2 is the moving average of MA1 over the preceding 70 weeks. Because the 70-week average in the denominator is centered 35.5 weeks before the current middle of the week, the ratio MA1/MA2 yields the change over a 35.5-week period, i.e. over approximately eight months.\nOne can calculate the UERg using the Excel function XIRR.\n=XIRR(cell 1:cell 2,(date)35-weeks-ago:(date)now)\n\nwhere cell 1 contains: -1    and   cell 2 contains: MA1/MA2now\n\nWe used MA1 for 15 weeks to calculate UERg and also MA1 for 4 weeks to determine the peaks of UERg at the end of recessions as discussed in the article.\n\n### Appendix 3\n\n#### Dates of ECRI recession calls and corresponding signals from the UER\n\n Start of recession ECRI recession Lead UER signal lead NBER date call date weeks weeks 8/1/1990 3/9/1990 21 -5 4/1/2001 4/18/2001 -2 2 1/1/2008 3/31/2008 -13 6 n.a. 9/30/2011 n.a. no signal\n\nA negative lead indicates that the signal was late. It is evident from this table that ECRI’s recession calls are not superior to the models’ signals.\n\n___________________________________________________________________\n\nPosted in blogs, featured\n###### One comment on “The Unemployment Rate: A Coincident Recession Indicator”\n1.", null, "chenjiahua says:\n\nVery great model! However, still not sure how monthly unemployement rate be transformed into weekly for calculation.\n\nWith reference to Section 202(a)(11)(D) of the Investment Advisers Act: We are Engineers and not Investment Advisers, read more ..." ]

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[ "# How to Add, Subtract, Multiply, or Divide Multiple Cells in Excel\n\nDoing basic math doesn’t have to involve breaking out the calculator and manually adding up cells, columns, or row. Excel has a useful feature called Formulas that allow you to do both basic math, such as addition and subtraction, or more complex items like finding averages, or even building your own formulas using super-advanced algorithms.\n\nToday, we’re going to work through a variety of real-world uses for simple formulas in Excel. In this scenario, we’re going to tally up bills, subtract money owed, multiply bills over the course of a year, and divide them between three roommates each month.\n\n1. Open an Excel workbook. For this example, we’re going to use a simple tally of expenses and add them. But based on the formula you choose in step TKTK, you can just as easily subtract, multiply, or divide the cells to create an average.\n\n2. Choose the cell you want to use to display the solution to your simple formula.\n\n3. For addition, the solution is an easy one. We simply need to tell Excel we’re adding, and then determine which cells we want to add up. Use this formula: =SUM(D2:D7)\n\n4. For subtraction, it’s not quite as easy as we can’t subtract multiple cells with the same type of formula. Instead, we have to enter each cell by hand. So if we wanted to subtract the cell phone bill from the rent, for example, we would use this formula: =D5-D6\n\n5. For multiplication, the formula is mainly the same as subtraction. To multiply the phone bill by 12, for example, use this formula: =D6*12 (or you can multiply two cells in the workbook much in the same way as you subtracted them, just an asterisk instead of a minus sign).\n\n6. And for division, you can use this formula: =D9/3 <divide.png> In this example, we’re basically divvying up monthly expenses between three roommates. You can alter the formula to divide two numbers within the workbook, or any other cell divided by a different number entirely." ]

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[ "Vertical and FOIL Methods for Multiplying Two Binomials\n\nLearning Outcomes\n\n• Use the FOIL method to multiply two binomials\n• Use the vertical method to multiply two binomials\n\nRemember that when you multiply a binomial by a binomial you get four terms. Sometimes you can combine like terms to get a trinomial, but sometimes there are no like terms to combine. Let’s look at the last example again and pay particular attention to how we got the four terms.\n\n$\\left(x+2\\right)\\left(x-y\\right)$\n${x}^{2}-\\mathit{\\text{xy}}+2x - 2y$\n\nWhere did the first term, ${x}^{2}$, come from?\n\nIt is the product of $x\\text{ and }x$, the first terms in $\\left(x+2\\right)\\text{and}\\left(x-y\\right)$.", null, "The next term, $-\\mathit{\\text{xy}}$, is the product of $x\\text{ and }-y$, the two outer terms.", null, "The third term, $+2x$, is the product of $2\\text{ and }x$, the two inner terms.", null, "And the last term, $-2y$, came from multiplying the two last terms.", null, "We abbreviate “First, Outer, Inner, Last” as FOIL. The letters stand for ‘First, Outer, Inner, Last’. The word FOIL is easy to remember and ensures we find all four products. We might say we use the FOIL method to multiply two binomials.", null, "Let’s look at $\\left(x+3\\right)\\left(x+7\\right)$ again. Now we will work through an example where we use the FOIL pattern to multiply two binomials.", null, "example\n\nMultiply using the FOIL method: $\\left(x+6\\right)\\left(x+9\\right)$\n\nSolution\n\n Step 1: Multiply the First terms.", null, "Step 2: Multiply the Outer terms.", null, "Step 3: Multiply the Inner terms.", null, "Step 4: Multiply the Last terms.", null, "Step 5: Combine like terms, when possible. $x^2+15x+54$\n\ntry it\n\nWe summarize the steps of the FOIL method below. The FOIL method only applies to multiplying binomials, not other polynomials!\n\nUse the FOIL method for multiplying two binomials\n\n1. Multiply the First terms.\n2. Multiply the Outer terms.\n3. Multiply the Inner terms.\n4. Multiply the Last terms.\n5. Combine like terms, when possible.", null, "example\n\nMultiply: $\\left(y - 8\\right)\\left(y+6\\right)$\n\nexample\n\nMultiply: $\\left(2a+3\\right)\\left(3a - 1\\right)$\n\nexample\n\nMultiply: $\\left(5x-y\\right)\\left(2x - 7\\right)$\n\ntry it\n\nFor another example of using the FOIL method to multiply two binomials watch the next video.\n\nMultiplying Two Binomials Using the Vertical Method\n\nThe FOIL method is usually the quickest method for multiplying two binomials, but it works only for binomials. You can use the Distributive Property to find the product of any two polynomials. Another method that works for all polynomials is the Vertical Method. It is very much like the method you use to multiply whole numbers. Look carefully at this example of multiplying two-digit numbers.", null, "You start by multiplying $23$ by $6$ to get $138$.\n\nThen you multiply $23$ by $4$, lining up the partial product in the correct columns.\n\nLast, you add the partial products.\n\nNow we’ll apply this same method to multiply two binomials.\n\nexample\n\nMultiply using the vertical method: $\\left(5x - 1\\right)\\left(2x - 7\\right)$\n\ntry it\n\nWe have now used three methods for multiplying binomials. Be sure to practice each method, and try to decide which one you prefer. The three methods are listed here to help you remember them.\n\nMultiplying Two Binomials\n\nTo multiply binomials, use the:\n\n• Distributive Property\n• FOIL Method\n• Vertical Method\n\nRemember, FOIL only works when multiplying two binomials." ]

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[ "# Beamer and verbatim sync\n\nI'm preparing a beamer presentation to show some basic TikZ stuff. I would like to have a two columns slide, where on the left I have the TikZ figure, and on the right the generating code. BUT, I also want the code to be uncovered set by step, that is, whenever a feature of the figure is uncovered, I want the corresponding piece of code to be uncovered.\n\nSomething like:\n\n\\begin{frame}[fragile]\n\\begin{minipage}{0.10\\linewidth}\n\\begin{center}\n\\begin{tikzpicture}\n\\draw<1-> (-1.5,0) -- (1.5,0);\n\\draw<2-> (0,-1.5) -- (0,1.5);\n\\end{tikzpicture}\n\\end{center}\n\\end{minipage}\n\\begin{minipage}{0.85\\linewidth}\n\\begin{verbatim}\n\\begin{tikzpicture}\n\\draw (-1.5,0) -- (1.5,0);\n\\draw (0,-1.5) -- (0,1.5);\n\\end{tikzpicture}\n\\end{verbatim}\n\\end{minipage}\n\\end{frame}\n\n\nThe problem is that I couldn't neither manage to use the uncover inside the verbatim nor use the verbatim inside an uncover...\n\nWhat is the right practice?\n\n• Have a look at Herbert's answer to this question. – lockstep May 12 '11 at 18:21\n\nHerbert's answer was helpful! However, for the sake of completeness, here's the correction of the second part of my code from the question:\n\n\\begin{minipage}{0.85\\linewidth}\n\\begin{semiverbatim}\\small\n\\uncover<1->{\\\\begin\\{tikzpicture\\} }\ngrid (1.5,1.5);}}\n\\uncover<2->{ \\alert<2>{\\\\draw (-1.5,0) -- (1.5,0);} }\n\\uncover<4->{ \\alert<4>{\\\\draw[red,line width=2pt] (0,0) circle (.8cm);}}\nrectangle (1,1);}}\nparabola (1,1);}}\n\\uncover<1->{\\\\end\\{tikzpicture\\}}\n\\end{semiverbatim}\n\\end{minipage}\n\n\nThanks you all!\n\nI had the same problem. I did one of the following in my presentation, each solution was used for a different problem (I'm sorry, I don't remember exactly what is used for what)\n\n• I wrapped the tikz figure with \\uncover. Note that this is the only case that I did what I think you tried to do.\n• I used the \\only command which is not recommended.\n• I think that I also placed everything inside a single tikz figure but I'm not sure.\n\nHope this is of any help." ]

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[ "Search Results for author: Tongyang Li\n\nFound 9 papers, 1 papers with code\n\nCompared to the previous state-of-the-art algorithms by Jin et al. with $\\tilde{O}((\\log n)^{4}/\\epsilon^{2})$ or $\\tilde{O}((\\log n)^{6}/\\epsilon^{1. 75})$ iterations, our algorithm is polynomially better in terms of $\\log n$ and matches their complexities in terms of $1/\\epsilon$.\n\nSublinear classical and quantum algorithms for general matrix games\n\nWe give a sublinear classical algorithm that can interpolate smoothly between these two cases: for any fixed $q\\in (1, 2]$, we solve the matrix game where $\\mathcal{X}$ is a $\\ell_{q}$-norm unit ball within additive error $\\epsilon$ in time $\\tilde{O}((n+d)/{\\epsilon^{2}})$.\n\nQuantum algorithms for escaping from saddle points\n\nno code implementations20 Jul 2020, ,\n\nCompared to the classical state-of-the-art algorithm by Jin et al. with $\\tilde{O}(\\log^{6} (n)/\\epsilon^{1. 75})$ queries to the gradient oracle (i. e., the first-order oracle), our quantum algorithm is polynomially better in terms of $\\log n$ and matches its complexity in terms of $1/\\epsilon$.\n\nQuantum exploration algorithms for multi-armed bandits\n\nIdentifying the best arm of a multi-armed bandit is a central problem in bandit optimization.\n\nThe study of quantum generative models is well-motivated, not only because of its importance in quantum machine learning and quantum chemistry but also because of the perspective of its implementation on near-term quantum machines.\n\n36\n\nSampling-based sublinear low-rank matrix arithmetic framework for dequantizing quantum machine learning\n\nMotivated by quantum linear algebra algorithms and the quantum singular value transformation (SVT) framework of Gily\\'en et al. [STOC'19], we develop classical algorithms for SVT that run in time independent of input dimension, under suitable quantum-inspired sampling assumptions.\n\nSublinear quantum algorithms for training linear and kernel-based classifiers\n\nWe design sublinear quantum algorithms for the same task running in $\\tilde{O}(\\sqrt{n} +\\sqrt{d})$ time, a quadratic improvement in both $n$ and $d$.\n\nDistributional property testing in a quantum world\n\nno code implementations2 Feb 2019,\n\nThe presented approach is a natural fit for distributional property testing both in the classical and the quantum case, demonstrating the first speed-ups for testing properties of density operators that can be accessed coherently rather than only via sampling; for classical distributions our algorithms significantly improve the precision dependence of some earlier results.\n\nQuantum-inspired sublinear algorithm for solving low-rank semidefinite programming\n\nIn this paper, we present a proof-of-principle sublinear-time algorithm for solving SDPs with low-rank constraints; specifically, given an SDP with $m$ constraint matrices, each of dimension $n$ and rank $r$, our algorithm can compute any entry and efficient descriptions of the spectral decomposition of the solution matrix.\n\nCannot find the paper you are looking for? You can Submit a new open access paper." ]

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[ "", null, "# Resonance\n\nWe are surrounded by a sea of different frequencies form natural as well as artificial sources. These frequencies are modulated or jumped up by a phenomenon known as resonance. Resonance actually causes waves to be of similar frequency and thus result in a higher amplitude meaning more intensity.\n\nGalileo Galilei observed that a man alone could ring a large bell by giving impulses at the right instants and they were so huge that four to six men were lifted up from the ground trying to seize the impulse holding the rope. This justifies to the earliest encounters of what resonance was capable of.", null, "## Meaning of Resonance\n\nThe meaning of resonance is a state of the system including a response of huge magnitude in response to an external stimulus when the stimulus applied has frequency similar to the natural frequency of the system. In terms of physics, resonance refers to an increase in the amplitude of a vibration that is a result of periodically applied force which has the frequency equal to or near to the fundamental or the natural frequency of that dynamic system. This frequency at which resonance occurs is known as the resonant frequency of the system. Small periodic forces of the similar frequency as that of the natural frequency store vibrational energy which is responsible for the large-amplitude vibrations.\n\nResonance occurs with all kinds of existing vibrations such as in quantum wave functions, electron spin resonance(ESR), mechanical resonance, electromagnetic resonance, acoustic resonance and nuclear magnetic resonance(NMR). It is useful in different ways including generating a vibration of a required frequency or filtering specific frequency from a complex vibration containing many frequencies. Resonance is possible when a body is capable of storing and transferring energy between different states and forms of energy. As the cycles go on, some losses incur. This is damping.\n\nThe resonant frequency is approximately equal to the natural frequency in case of small damping, but in other cases, it has multiple, different resonant frequencies. In daily life, radio waves are tuned to produce resonance to selectively receive waves of specific values. Modern clocks also use resonance for their timekeeping mechanisms.\n\n### Types of Resonance\n\n1. Mechanical resonance\n\nThe tendency of a mechanical system to give a response with large amplitude when the frequency of its oscillations is similar to the system’s natural frequency of vibration rather than other frequencies. Some examples of this type of resonance include pushing a swing, motion of a pendulum, motion of a tuning fork. In all of these cases, the force applied to cause the vibration should be periodically regulated to match the natural frequency to create resonance. The resonant frequency of spring is:\n\nf = $$\\frac{1}{2\\Pi}\\sqrt{\\frac{k}{m}}$$\n\nwhere,\n\nm is the mass of the spring\n\nk is the spring constant\n\nThe resonant frequency of a pendulum is:\n\nf = $$\\frac{1}{2\\Pi}\\sqrt{\\frac{g}{l}}$$\n\nwhere,\n\ng is the acceleration due to gravity\n\nl is the length of the pendulum\n\n1. Acoustic resonance\n\nWhen an acoustic system has waves with matching frequencies with one of its own natural frequencies, it amplifies the sound. This is acoustic resonance. This is a very crucial tool for instrument builders as in case of many acoustic instruments like strings of a guitar, length of a flute and membrane of a drum used resonators. It also accounts for hearing ability.\n\n1. Electric resonance\n\nIn an electrical circuit, when the capacitive reactance and inductive reactance are equal in magnitude, electrical resonance is observed. When the impedance between the input and output of the electrical circuit is almost zero and the circuit’s transfer function is one, this is also a case of electrical resonance.\n\n1. Optical resonance\n\nWhen 2 mirrors, one partially reflective and other highly reflective are part of a laser, they form an optical resonator. Interferometers and some optical parametric oscillators use optical resonators because it has a large Q factor which suggests less amount of energy loss. Lasers which provide the gain medium and provide feedback of laser lights using these resonators.\n\n### Resonance of a Damped Oscillator\n\nThis type of system is visible in spring-mass systems, pendulum systems and many other damper circuits like the RLC electrical circuit. When we consider a spring-mass system driven by an external sinusoidal force, we can come to the differential equation:\n\nm$$\\frac{\\partial^2 x}{\\partial t^2}$$ = $$F_{o}sin(\\omega t)-kx-c\\frac{\\partial x}{\\partial t}$$\n\nwhere\n\nm is the mass\n\nx is the displacement from the mean position\n\n$$\\omega$$ is the driving angular frequency\n\nk is the spring constant\n\nc is the damping coefficient\n\n$$F_{o}$$ is the driving amplitude\n\nAlternatively, we can write the equation as:\n\n$$m\\frac{\\partial^2 x}{\\partial t^2}+2\\zeta\\omega_{o}\\frac{\\partial x}{\\partial t}+\\omega_{o}^{2}x$$ ={$$\\frac{F_{o}}{M}$$} $$\\sin(\\omega t)$$\n\nHere $$\\omega_{o}$$ =$$\\sqrt{\\frac{k}{m}}$$ is the natural frequency of the system.\n\nThe damping ratio is $$\\zeta$$= $$\\frac{c}{2\\sqrt{mk}}$$ .\n\nIn many cases on close and concentrated analysis, resonant frequency comes out to be near not exactly equal to the natural frequency.\n\nBut this is not entirely true in a damped system. In a damped system, resonance occurs when the steady-state amplitude is large as compared to the other frequencies and this could be possible at a certain frequency known as the resonant frequency. Resonant frequency here will be:\n\n$$\\omega_{r}$$ = $$\\omega_{o}\\sqrt{1-2\\zeta^{2}}$$\n\nHere, the resonant frequency is not equal to the undamped angular frequency of the oscillator.\n\nQ.1. What are the uses of resonance in molecules?\n\nAnswer. Molecular and atomic resonance contains two phenomena Nuclear Magnetic Resonance(NMR) and Electron Spin Resonance(ESR). NMR is useful in imaging techniques in advanced medical industries such as MRI. The resonant frequency of NMR happens to be directly proportional to the strength of the applied magnetic field. This method helps in imaging objects that are in a non – uniform magnetic field, as the resonant frequency of the molecules of the object depend on the location of the nuclei and hence the imaging. A similar kind of technique known as the ESR makes use of the unpaired electrons that is the paramagnetism of the molecules. ESR is helpful in detecting inflammation and monitoring conditions of arthritis, vasculitis or inflammatory bowel disease.\n\nQ.2. How does resonance play a part in tuning musical instruments?\n\nAnswer. In case of the tube instruments, the tube provides a column for air to vibrate and the length of that air column decides the harmonics that are produced and the sound we hear aloud. We can adjust the length of the tube to create a vibration with a frequency close to the natural frequencies of the air column inside the brass instrument and thus the result is a loud sound. We can acknowledge higher pitches by further shortening the column and the vibrating reed helps the air column to get to vibrate at one of the natural frequency.\n\nQ.3. What is orbital resonance?\n\nAnswer. Orbital resonance occurs when celestial bodies due to their orbital periods being in the ratio of small integers exert a regular, periodic gravitational influence on each other. Laplace was the first scientist to examine the stability of the solar system. This stability is because of the changing gravitational forces of the celestial bodies which go around each other.\n\nShare with friends\n\n## Customize your course in 30 seconds\n\n##### Which class are you in?\n5th\n6th\n7th\n8th\n9th\n10th\n11th\n12th\nGet ready for all-new Live Classes!\nNow learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.", null, "", null, "Ashhar Firdausi\nIIT Roorkee\nBiology", null, "", null, "Dr. Nazma Shaik\nVTU\nChemistry", null, "", null, "Gaurav Tiwari\nAPJAKTU\nPhysics\nGet Started", null, "0 Followers\n\nMost reacted comment\n2 Comment authors", null, "", null, "Recent comment authors\nSubscribe\nNotify of", null, "Guest\nsam", null, "Guest\nSandaras Edirisinghe\n\nIt was so much helping. Thank u for that. 👍" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Quantum Mechanics, Boundary Value Problem\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg49525] Quantum Mechanics, Boundary Value Problem\n• From: danese at physics.rutgers.edu (anthony danese)\n• Date: Thu, 22 Jul 2004 02:46:57 -0400 (EDT)\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```Hi,\n\nI'm trying to find a numerical solution to Schrodinger's equation for\na nearly free electron potential (V(x)=Cos(x)) which meets an image\npotential which is of the form (V(x)=C - 1/x) where C is a constant.\nI can set up the problem using NDSolve, with 2 of my boundary\nconditions being continuity of the wave function (psi) and its\nderivative(psi') where the 2 potentials meet. The problem is that I\nneed 2 other boundary conditions--I attempt to arbitrarily set psi = 0\nfar into the Cos(x)-like region of the potential and I would like psi\nto vanish as for large values of x in the C-1/x region of space. When\nI try to run NDSolve, I either get the trivial psi = 0 or I get an\nerror telling me that the boundary conditions are not well defined.\n\nAs a test, I have tried to run NDSolve to find psi for a step\npotential where the step potential is greater than the energy of the\nelectron. The solution should be free electron-like away from the\nstep and the solution should decay into the step. This is a problem\nthat is easily solved analytically but Mathematica doesn't like the\nway I'm stating the boundary conditions. Is there anyway to solve\nthese particular examples of Schrodinger's equation with NDSolve?\n\nAny help would be greatly appreciated,\n\nanthony danese\n\n```\n\n• Prev by Date: Re: If-statement problems\n• Next by Date: RE: is List link-based or array-based\n• Previous by thread: AW: Plotting a contour plot with cylindrical co-ordinates\n• Next by thread: Re: Quantum Mechanics, Boundary Value Problem" ]

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[ "# Converting string to integer using atoi in C++\n\n• C/C++\n• zarentina\n\n#### zarentina\n\nHello, the problem that I was asked to complete was ; Write a program that inputs a string and reverses it. After the string has been reversed you must convert it to an integer and then take the square root of the integer. (using atoi)\nMy code is as follows:\nCode:\nint main ()\n{\nint i=0,L=50, S,x,;\ndouble z;\nchar ori_string;\nchar new_string;\ncin.getline(ori_string,50);\n\nfor(i=0;i<L; i++)\n{\nnew_string[i]=ori_string[L-1-i];\n}\nfor(L=50-strlen(ori_string);L<50;L++)\n{\n\ncout<<new_string[L];\n}\ncout<<endl;\n{\nx=atoi(new_string);\nz=sqrt(x);\ncout<<\"The square root of the reversed string is: \"<<z<<endl;\n}\nsystem(\"pause\");\nreturn 0;\n}\n\nMy problem is that my program isn't storing \"new_string[L]\" as anything so when I convert it to a integer nothing is there and the square root is zero. So my question to you all is how can i store \"new_string[L]\" so I can convert it to an integer using atoi?\n\nThanks,\nZ\n\nI don't know too much about C++ but atoi is string to int and atol is strong to long int.\n\nYour code also seems to have a lot of stuff in it just to reverse a string and find its integer square root.\n\nFor example, why would you need anything to do 50 iterations?\n\nFirst of all - you can't reverse the string assuming its length is 50 characters. What happens if it is not?\n\nnew_string[L] is a character (element of the new_string array), not a string. You don't have to write new_string to cout character by character, cout << new_string should work.\n\nThere is more, but I am not going to mention anything else now to not muddy the water.\n\nWhen I first wrote this program to reverse my number I simply had\nCode:\nint main()\n{\nint i,L,;\nchar ori_string;\nchar new_string;\ncin.getline(ori_string,50);\n\nfor(i=0;i<L; i++)\n{\ncout<<ori_string[L-1-i];\n}\nsystem(\"pause\");\nreturn 0;\n}\n\nThis returned the proper number but it was preceded by random characters for the first 46 spaces if the string was 3 digits long. This was the reasoning of structuring a second for loop. Yet now that I know that is the incorrect method, how do you suggest I fix this bit of code here so that it will only output the reverse of the string for after it does that I can easily convert it to an int and take the square root.\n\n(I know this differs from my initial question but I'm reasonably certain that this is where the problem is, yet if this assumption is wrong please let me know!)\n\nThanks,\nZ\n\nHint: L is not initialized.\n\nHint2: with what should it be initialized?\n\n*UPDATE* I had a revelation sparked by a few of your comments", null, "Code:\nint main()\n{\nint i,L=50,x;\ndouble z;\nchar ori_string[L];\nchar new_string[L];\ncin.getline(ori_string,50);\nL=strlen(ori_string);\n\nfor(i=0;i<L; i++)\n{\n\ncout<<ori_string[L-1-i];\nnew_string[i]=ori_string[L-1-i];\n}\n\n{\nx=atol(new_string);\nz=sqrt(x);\ncout<<\"The square root of the reversed string is: \"<<z<<endl;\n}\nsystem(\"pause\");\nreturn 0;\n}\n\nIs my new properly working code, if there are any critiques or overhauls you may do feel free, I'd love to learn!\n\nThanks,\nZ\n\nHi Z! Welcome to PF!", null, "You're code will work properly.\n\nIf you want to learn I'll give you an improvement.\nYour program won't function better, but it is better programming style.\n\nIn C++ we don't really want to use char-arrays anymore (it's a remnant from its predecessor C).\nThere are a number of pitfalls with them.\nIn this case the most obvious ones are that you have a maximum size (50) and that you truncate your input.\n\nThe C++ way to do things, is to use std::string.\nIt works like this:\nCode:\n#include <string>\n\nint main()\n{\nint i, L, x;\ndouble z;\nstd::string ori_string;\nstd::string new_string;\ncout << \"Please enter a long string\" << endl;\ngetline(cin, ori_string);\n\nL = ori_string.length();\nnew_string.resize(L);\nfor (i = 0; i < L; i++)\n{\ncout << ori_string[L - 1 - i];\nnew_string[i] = ori_string[L - 1 - i];\n}\n\nx = atol(new_string.c_str());\nz = sqrt(x);\ncout << \"The square root of the reversed string is: \" << z << endl;\n\nsystem(\"pause\");\nreturn 0;\n}" ]

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