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8 Batteries Discrete Mathematics Level 2 You have 8 batteries, numbered 1 through 8. Four of them are good, and four are bad. But you don't know which are which. You have a flashlight that requires 4 batteries to work. What is the minimum number of times you need to put in the batteries to guarantee to get the flashlight working? Assume that the flashlight either works (when all four are correct) or not (for any other combination). Nothing in between. And the total number of times includes the final time you put them in, even if at that point you know you have a workable set. Image credit: http://www.evilcontrollers.com/ Problem Loading... Note Loading... Set Loading...
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Zentralblatt MATH Publications of (and about) Paul Erdös Zbl.No: 774.05050 Autor: Erdös, Paul; Hajnal, András; Simonovits, M.; Sós, V.T.; Szemerédi, E. Title: Turán-Ramsey theorems and simple asymptotically extremal structures. (In English) Source: Combinatorica 13, No.1, 31-56 (1993). Review: Let L1, L2,...,Lr be given graphs (``forbidden'' graphs), and let n be a positive integer and f a given function; \alpha(G) denotes the maximum number of independent vertices in a graph G; RT(n,L1,L2,...,Lr,f(n)) denotes the maximum number of edges in a graph Gn on n vertices, having \alpha(Gn) \leq f(n), whose edges may be coloured in r colours so that the subgraph of the ith colour contains no Li (i = 1,2,...,r); the results of this paper generally apply to the case f(n) = o(n), and the maximum is usually denoted by RT(n,L1,L2,...,Lr,o(n)). A sequence (Sn) of graphs for which \alpha(Sn) \leq f(n) and Sn has RT(n,L1,L2,...,Lr,f(n))+o(n2) edges, is asymptotically extremal for RT(n,L1,L2,...,Lr,f(n)) if the edges of Sn may be r-coloured so that the subgraph of the ith colour contains no Li (i = 1,...,r). In Theorem 2 a construction of B. Bollobás and P. Erdös [On a Ramsey-Turán type problem, J. Comb. Theory, Series B 21, 166-168 (1976; Zbl 337.05134)] used to prove that RT(n,K4,o(n)) \geq 1/8 n2-o(n2) is generalized to prove the existence of a sequence of graphs that is asymptotically extremal for RT(n,Kk1,Kk2,...,Kkr,o(n2)), where k1, k2,...,kr are integers each exceeding 2. Let \theta(L1,L2,...,Lr) denote the minimum real number such that RT(n,L1,L2,...,Lr,f(n)) \leq \theta(L1,L2,...,Lr)n2+o(n2); in Theorem 3 the values of \theta(K3,K3), \theta(K3,K4), \theta(K3,K5), \theta(K4,K4) are determined, as well as an asymptotically extremal sequence for each case; it is shown that the distance between two such sequences - - i.e. the minimum number of edge additions/deletions needed to transform one such sequence into another -- is o(n2) in each case. Theorem 4: If p and q are odd integers, then RT(n,Cp,Cq,o(n)) = 1/4 n2+o(n2). The paper concludes with a list of open problems. Reviewer: W.G.Brown (Montreal) Classif.: * 05C35 Extremal problems (graph theory) 05C55 Generalized Ramsey theory 05C38 Paths and cycles Keywords: Turán-Ramsey theorems; asymptotically extremal structures; colour; asymptotically extremal sequence Citations: Zbl 337.05134 © European Mathematical Society & FIZ Karlsruhe & Springer-Verlag Books Problems Set Theory Combinatorics Extremal Probl/Ramsey Th. Graph Theory Add.Number Theory Mult.Number Theory Analysis Geometry Probabability Personalia About Paul Erdös Publication Year Home Page
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Science / Algebraic morsels Random Science Quiz Can you name the answers to these problems from first-year algebra/pre-algebra? Quiz not verified by Sporcle Score 0/13 Timer 15:00 Of 2^2, -2^2, (-2)^2, 2^(-2), -2^(-2), and (-2)^(-2), find the sum of the numbers of positive numbers, negative numbers, integers, and real numbers. Jason has 10 coins in total. He has only quarters and dimes. The total value of the coins is $2.20. How many dimes does Jason have? Find the discriminant of 4x^2 - 5x + 3 = 0. Find (1+2)(3+4)÷5+6 as an improper fraction or decimal. Let x be the greatest real solution of |x^2-16|=9. Find x^x / x. When the repeating decimal 0.242424... is expressed as a fraction, find the sum of the numerator and denominator if they are coprime. The sum of the squares of two positive numbers is 58 and their product is 21. Find their sum. Find the coefficient of the x^2 term in the expansion of 6(x+1)(x+2)(x+3) If 32*sqrt(1 + sqrt(1 + sqrt(1 + ... ))) is expressed in the form a + b sqrt( c ), find a + b + c. Compute the sum of the first 6 terms of the Fibonacci sequence. Give the word describing the proper subset of the real numbers for which the identity sqrt(a^2) = a is false. Name the property that states that order of addition does not matter. The sum of the distances from any point P in the set L to (-3, 0) and to (3, 0) is equal to 10. Find the greatest possible ordinate (y-coordinate) of a point in L. You're not logged in! Compare scores with friends on all Sporcle quizzes. Sign Up with Email Log In You Might Also Like... Show Comments Your Account Isn't Verified!
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What are perfect square trinomials? Quick Answer A perfect square trinomial is a polynomial with three terms that is expanded or FOILed form of a binomial squared. Taking the square root of a perfect square trinomial will result in a rational expression, unlike the roots of other trinomials. Continue Reading Full Answer A polynomial is a mathematical expression with multiple terms that usually cannot be simplified or combined anymore than they are. Trinomials have three terms. These terms frequently include one or more variables that are either different themselves or raised to different powers. x^2 + 6x + 9 is a trinomial, because it has three terms, none of which can be combined. It is also a perfect square. Perfect square trinomials follow the rule that a^2 + 2ab + b^2 = (a + b)^2. Therefore, x^2 + 6x + 9 must equal (x + 3)^2. (x + 3)*(x + 3) can be expanded using a method called FOIL, which stands for first, outside, inside, last. The square of the first term is x^2 because x*x = x^2. The product of the inside terms and that of outside terms can be added together: 3x + 3x = 6x. The square of the last term is 9 because 3*3 = 9. To factor perfect-square trinomials, students determine the square roots of the outside and inside terms. If twice the product of their square roots is the middle term, then the original trinomial is a perfect square. Learn more about Exponents Related Questions
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How to Convert String to Long in PHP By Kevin Walker Strings can be converted and parsed into their numerical values in PHP simply by using them within a mathematical context. PHP will determine the appropriate numerical data type upon conversion. If the number is small enough, it will be converted into an integer data type. If it is larger or contains decimal or scientific notation, then it will be converted into the float data type, which combines the roles for the long, double and float data types in C and Java. Step 1 Open a text editor and immediately save with the name "longParse.php." Step 2 Paste the following code to assign a string to a variable named "$str":$astring = "100"; Step 3 Paste the following to convert that string to an integer and add 1 to it:$anumber = $astring + 1;Because the string is being used in a mathematical context and contains information that can be understood as a number, PHP will convert it automatically to the appropriate numerical type (in this case an integer because the number is so small) and continue forward. This can feel uncomfortable for programmers who want more precise control over the resulting data, so it is possible to explicitly perform the conversion using a cast command. Step 4 Paste the following to explicitly cast the string into either an integer or a float:$anumber = (int) $astring;$anumber = (float) $astring;If you are certain that the numbers are relatively small and will never contain decimals, you should use "int," as it saves memory. However, if you think the numbers may be quite large or contain decimals, you should use "float."
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Čeština Slovenčina Français English Deutsch Description: Blosics 2 Level Pack Shoot the green blocks out of the screen using spheres of various dimensions. Earn enough points to advance to the next level. This game is really great and calls far physics skills. If you like it play also the other instalments such as Blosics, Blosics 2 or Blosics 3. Add a comment Sign in to comment. More items from the category: Puzzle Smart Chess Rollercoaster Creator 2 The second instalment of a popular Rollercoaster Creator game features even more options than the first... Your task is to collect all stars with a car. In each level you have a set of items you can use to help... Lofty Tower 2 Lofty Tower 2 is a block building game (well a kind of). Your task is to build a tower, which must be... Monkey Go Happy This Monkey is pretty sad. Can you cheer him up? Solve puzzles and make the monkey smile. In this game you have to click on the cubes, which are not paired. If you click on a paired cube you...
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Abakus-Func is a computer algebra program which performs symbolic calculations for mathematical functions. The current Android version 1.0 contains the following features: - Calculate function values - Determine exact zeros, singularities, extremal points, and turn points - Calculate symbolic limits - Graphical representation (2D and 3D). As an example, Abakus-Func yields the following exact results for the function f(x) = x/(x+1)^2+2. Zeros: x = -2, x = -1/2; Singularities: x = -1; Extremal points: Maximum (1,9/4); Turn points: (2,20/9). You can easily check the symbolic results by viewing the plot of the function. Several functions can be displayed simultaneously within the graphical window. You can select different colors for the individual functions. For functions with two variables, Abakus-Func generates a 3D-like graphical representation. Moreover Abakus-Func contains a special editor for mathematical formulas. So you can enter even complex expressions very fast.
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Artemis, heiress to the moon Picture ok yes artemis is the greek goddess of the moon and she is not an elf but this isnt the artemis of greek mythology. i was thinking and came up with an idea that there is a diferent world with two races, elves and humans. There are 10 councel members and then a king and queen that govern the elf race, humans do didily squat. Each is guardian of something important to their survival and each have a certian animal species that they can control and turn into, but the animal is white. Artemis' mother, Diana, is curently the councel member trusted with the moon and her father, apollo, is the sun. Once her mother dies she will take her place, right now she is a heiress. Now of course there are others and each has personalites and govern the same things as the gods but that is about it. Like the greek goddess she is excelent with the bow and arrow and is a great hunteress. But unlike the goddess she summons ice, controls the forest and turns into a white wolf. Rock on, Prometheus Artemis, heiress to the moon You were my brother Apollo...
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Game Objects Game Object Guide Game Objects in Game Programming A game object is any object in a game that the player can see and/or interact with. The player object, power ups, enemies, platforms, walls, weapons (if collision detection is considered) and projectiles are all game objects. Depending on the capabilities of the programming language that you are using, a game object may or may not be constructed in code as an actual object, as in the case with object oriented programming. However, no matter what programming language you are using, all game objects are constructed from a number of variables, subroutines, and functions that give the game object it's qualities and capabilities. Creating Game Objects How you create your game objects will depend on the programming language you are using. However, the following description of game object programming assumes only that your programming language has variables, types (or structs), and some kind of user-defined subroutines or functions. These features have been standard in programming languages for almost two decades and if yours doesn't support these features, you may want to consider finding a new programming language. If you are using an object oriented programming language, such as Brutus2D, you will be able to easily transfer these concepts to OOP. Game Object Qualities Game objects have qualities, that is, information about the game object that your game engine needs to interact with that object. For example, in a 2D game, the most basic qualities (properties, in OO lingo) is an object's horizontal and vertical location. Qualities such as these are contained in variables. Almost always, the variables x and y are used to hold an object's horizontal and vertical location. Without them, your game would have no way of knowing where the object was. Other examples of qualities that a game object may have is damage capacity, such as hit points, and damage rating, which is how much damage a game object is able to inflict. Types are used to put all of these qualities in an easy to manage package. Here is a FreeBASIC example : type playerObject x as single y as single hitPoints as integer damageRating as integer end type
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Channels ▼ Padding and Rearranging Structure Members Padding and sizeof For any type T, an object of type "pointer to T" can point to any T object, whether that object is an individual object, a structure member, or an array element. For example, given: widget w; widget x[10]; widget *p = &w; widget *q = &x[2]; the assignment: *q = *p; copies the value of a standalone widget object into a widget that's an array element. For such assignments to work, all widget objects must have the same size and alignment. The size of each structure must be a multiple of its alignment. Compilers add trailing padding to each structure if necessary to make this so. For example, since widget must be word aligned, the compiler adds trailing padding to bring the size of a widget up to the next multiple of the word size, which in my examples, is four. The sizeof operator applied to a structure such as widget yields the total number of bytes in the object, including any padding. In the case of widget compiled for a target processor where sizeof(int) equals four, the cumulative size of widget's three members would be six: sizeof(char) + sizeof(int) + sizeof(char) = 6 However, the padding adds six more bytes, so for our hypothetical target processor, sizeof(widget) would be 12. Type widget has an alignment of four, and 12 is a multiple of four. Why must the size of a structure be a multiple of its alignment? Well, suppose it weren't. Suppose that widget had no trailing padding so that sizeof(widget) were nine. However, each element in an array of widgets would still have to be word aligned, so each array element would have to be padded with three bytes. That padding just wouldn't be counted in the size of each element. What's wrong with that? According to both the C and C++ Standards, "When applied to an array, the result [of the sizeof operator] is the total number of bytes in the array." The C++ Standard adds that "This implies that the size of an array of n elements is n times the size of an element." If the padding after each array weren't counted in the size of each element, then the size of an array of n elements would in fact be greater than n times the size of each element. Various common programming techniques would cease to work. For example, the common C idiom for allocating storage for an array of n widgets is to call malloc in an expression of the form: pw = malloc(n * sizeof(widget)); If sizeof(widget) didn't account for the trailing padding after each widget, then this expressions could easily allocate too little storage for the entire array. As another example, suppose that gadget is a structure defined as: struct gadget char m1; double m2; char m3; and that objects of type double must be double-word aligned. The compiler will add 7 bytes of padding after member m1 and also after member m3, as if the structure had been defined as: struct gadget char m1; char padding_after_m1[7]; double m2; char m3; char padding_after_m3[7]; The most strictly aligned member of gadget is m2, which is double-word aligned. Thus, each gadget must also be double-word aligned. The padding after m1 ensures that member m2 is double-word aligned relative to the beginning of each gadget. The padding after m3 ensures that the next gadget in an array of gadgets is also double-word aligned. Rearranging Members to Reduce Padding You can reduce the size of each widget by rearranging the members to reduce the number of padding bytes. Specifically, you can rearrange the member declarations so that the char members m1 and m3 are adjacent to each other, as in: struct widget char m1; char m3; int m2; In this case, the compiler will add only two padding bytes, as if you had declared widget as: struct widget char m1; char m3; char padding[2]; int m2; Consequently, sizeof(widget) would be only eight, rather than 12. (Again, I'm calculating these sizes assuming an int is a 4-byte object aligned to an address that's a multiple of four. The sizes could be different on a machine with different object sizes and alignment requirements.) You can rearrange widget's members in other ways to reduce the number of padding bytes. For example, defining widget as: struct widget int m2; char m1; char m3; also reduces the number of padding bytes to two and sizeof(widget) to eight. A Standard C compiler won't rearrange the members of a structure automatically to reduce the structure's size. According to the Standard: "Within a structure object, the non-bit-field members … have addresses that increase in the order in which they are declared." This effectively prohibits compilers from rearranging structure members from the order in which they're declared. Class Padding in C++ A C++ class is a generalization of a C structure. For the most part, C code that defines and uses structures behaves the same when compiled and executed as C++. A C++ class can have elements that a C structure cannot, such as access specifiers (public, protected, and private), member functions, static data members, and base classes. Some of these elements alter the physical layout of class objects and complicate the rules for padding. Those rules will be the subject of a future article. Thanks to Joel Saks for all his help. Related Reading More Insights Currently we allow the following HTML tags in comments: Single tags <br> Defines a single line break <hr> Defines a horizontal line Matching tags <a> Defines an anchor <b> Defines bold text <big> Defines big text <blockquote> Defines a long quotation <caption> Defines a table caption <cite> Defines a citation <code> Defines computer code text <em> Defines emphasized text <fieldset> Defines a border around elements in a form <h1> This is heading 1 <h2> This is heading 2 <h3> This is heading 3 <h4> This is heading 4 <h5> This is heading 5 <h6> This is heading 6 <i> Defines italic text <p> Defines a paragraph <pre> Defines preformatted text <q> Defines a short quotation <samp> Defines sample computer code text <small> Defines small text <span> Defines a section in a document <s> Defines strikethrough text <strike> Defines strikethrough text <strong> Defines strong text <sub> Defines subscripted text <sup> Defines superscripted text <u> Defines underlined text
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The Algebraic and Transcendental Numbers A real or complex number z is called algebraic if it is the root of a polynomial equation z n + a n - 1 z n - 1 + … + a 1 z + a 0 = 0, where the coefficients a 0, a 1, … a n - 1 are all rational; if z cannot be a root of such an equation, it is said to be transcendental. The number 2art/square-root-of-2.gifthe square root of 2 is algebraic because it is a root of the equation z 2 + 2 = 0; similarly, i, a root of z 2 + 1 = 0, is also algebraic. However, F. Lindemann showed (1882) that π is transcendental, and using this fact he proved the impossibility of "squaring the circle" by straight edge and compass alone (see geometric problems of antiquity). The number e has also been found to be transcendental, although it still remains unknown whether e + π is transcendental. See more Encyclopedia articles on: Mathematics Play Hangman Play Poptropica Play Same Game Try Our Math Flashcards
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counter tube • An electron tube which produces an output pulse after counting a specified number of pulses. Also called counting tube (1). When one output occurs for every ten input pulses it is called decade counter tube (1). • An electron tube with one input electrode and multiple output electrodes, and in which with each successive input pulse the conduction is transferred in sequence to the next output electrode. Also called counting tube (2). When there are ten output electrodes it is called decade counter tube (2). • An electron tube which registers ionizing radiation, such as alpha rays, and produces an output electric pulse which can be counted. For example, a Geiger-Müller tube. Also called radiation counter tube. • synonym counting tube Not what you were looking for?
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(Skin inflammation) If one sees his skin thinned or inflamed between his legs in a dream, it means that he may marry a rich woman who will give him her wealth. Ifone sees himself shedding his skin or skinning himselfin a dream, it represents money he will part with, and if he is sick, it means his death.
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The Partition platform recursively partitions data according to a relationship between the X and Y values, creating a tree of partitions. It finds a set of cuts or groupings of X values that best predict a Y value. It does this by exhaustively searching all possible cuts or groupings. These splits (or partitions) of the data are done recursively forming a tree of decision rules until the desired fit is reached. This is a powerful platform, because it chooses the optimum splits from a large number of possible splits. Example of a Partition Plot
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Singularity spectrum From Wikipedia, the free encyclopedia Jump to: navigation, search The singularity spectrum is a function used in Multifractal analysis to describe the fractal dimension of a subset of points of a function belonging to a group of points that have the same Holder exponent. Intuitively, the singularity spectrum gives a value for how "fractal" a set of points are in a function. More formally, the singularity spectrum of a function, , is defined as: Where is the function describing the Holder exponent, of at the point . is the Hausdorff dimension of a point set. See also[edit]
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Ocena brak Discuss pros and cons of living in the country Autor /Zachariasz Dodano /15.11.2011 Living in the country has both pros and cons. One advantage is that there are not many cars. There’s no traffic, no noise, air is not polluted. What is more, if you want to relax, you can go to forest, sit under a tree and breath with fresh air. You can go to the forest for a mushrooms or berries. You know everybody, you say hello to everyone, it makes friendly tone around you. However, there are also disadvantages. One of them is that you can’t find job, because there’s not many places to work. It is a very big problem for the country. There is a big problem with public transport. It is important to have a car if you really want to go somewhere, do shopping in the nearest city etc. And I don’t like old people, which are very superstitious. Podobne prace Do góry
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Homesick, Paperback It is 1995 and Noa and Amir have decided to move in together. Noa is studying photography in Jerusalem and Amir is a psychology student in Tel Aviv, so they choose a tiny flat in a village called El-Kastel. Noa and Amir's flat is separated from that of their landlords, by a thin wall, but on each side we find a completely different world. Free P&Pwhen you spend £20 or over
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New User Special Price Expires in Let's log you in. Sign in with Facebook Don't have a StudySoup account? Create one here! Create a StudySoup account Be part of our community, it's free to join! Sign up with Facebook Create your account Already have a StudySoup account? Login here Applied Graph Theory and Algorithms by: Buck Ankunding Applied Graph Theory and Algorithms CMPE 177 Buck Ankunding GPA 3.56 Almost Ready Purchase these notes here, or revisit this page. Either way, we'll remind you when they're ready :) Preview These Notes for FREE Unlock Preview Unlock Preview Preview these materials now for free View Preview About this Document Class Notes 25 ? Popular in Course Popular in Computer Engineering This 30 page Class Notes was uploaded by Buck Ankunding on Monday September 7, 2015. The Class Notes belongs to CMPE 177 at University of California - Santa Cruz taught by Staff in Fall. Since its upload, it has received 67 views. For similar materials see /class/182222/cmpe-177-university-of-california-santa-cruz in Computer Engineering at University of California - Santa Cruz. Reviews for Applied Graph Theory and Algorithms Report this Material What is Karma? Karma is the currency of StudySoup. Date Created: 09/07/15 Applied Graph Theory Class 2 Sept 29 2005 Complete Graph A complete graph is one where every vertex is connected to every other vertex Vu7vEV7u7 v HeEElpeu7v u a v ensures that we only consider discrete vertices The opposite of a complete graph is an empty graph with E g5 Bipartite Graph Bipartite graphs originate in situations where we need to use binary edges to describe relations with arity more than two7 such as the is a member of the family relation Here we are forced to create two kinds of nodes7 one representing family and the other representing family member Edges only connect nodes of different kinds VV1UV27V10V2 VeEE7peEV1 XVZUVZ XVl Note They have no self loops For the matching problem7 which is to match vertices in one set with compatible vertices in the other7 if we have a complete bipartite graph7 then the solution time is linear in the number of vertices Degree of a vertex Degree of a vertex v E V is the number of edges which have v either as their source or destination Note that self loops are counted twice dltvgt we l we W we l we W This above applies to the undirected case For directed graphs7 we make the distinction between out degree edges for which the given vertex is the source and in degree edges for which the given vertex is the destination dumb le l We Vlv dmV le l Me Vl First theorem of Graph Theory The sum of the degrees of all vertices equals twice the number of edges This applies to both directed as well as undirected graphs 26V dltvgt 2m Corollary There is an even number of odd vertices vertices with an odd degree This follows from the fact that the sum of an odd number of odd numbers is odd 2n 1 X oddNumber 2m oddNumber which is odd The usefullness of the the corollary If you know of a vertex in a graph with odd degree then there must be another vertex in that graph with odd degree For a walk on an arbitrary graph You will never stop at a node of even degree if you can take an unvisited edge to reach that node So if you start at an odd vertex you will eventually stop at an odd vertex This is very useful for proofs and is good to docket Connected graph For the undirected case there always exists a path from a node to another node in the graph For directed graphs there are two notions A weakly connected directed graph is one where the underlying undirected graph is a connected graph A strongly connected directed graph is one where there always exists a path from a node to another node in the graph Subgraph If G V E p then a subgraph of G G1 V1 E1 m l V1 Q V E1 Q E p1 p on E1 The last condition says that edges in the subgraph lead exactly where they used to in the original graph Further G1 is also a graph Note that when you pick an edge from G you pick both end points Graphs don7t have dangling edges A graph is a subgraph of itself With a slight abuse of notation if G1 Q G and G2 Q G denoting sub graphs of G then G1 U G2 and G1 0 G2 are Q G Note that G1 G2 which is a graph formed by removing all the vertices and edges that are common to G1 and G2 may not produce a graph as it may have dangling edges Strongly connected component SCC A strongly connected component in a graph is a subgraph which is a con nected graph This means that one can reach any vertex from any given vertex in that subgraph A maximal SCC is one that cannot be made larger by including anymore vertices G1 G2 Q G G1 0 G2 a Z5 and G1 G2 are strongly connected implies G1 U G2 is also strongly connected For a proof start from the result that the union is an SCC This implies there always exists a path from a node to any other node The interesting case is the existence of a path from a node in G1 to a node in G2 Since G1 is strongly connected there always exists a path from a node in G1 to a node in G1 0 G2 Similarly since G2 is strongly connected there always exists a path from a node in G1 0 G2 to G2 hence there always exists a path from any node in G1 to one in G2 and vice versa lnsight For such proofs always start from the result in this case that in the union there always exists a path from one node to another and for all interesting cases show the result using the pre conditions in this case that a path exists from any node in one SCC to any node in the other SCC Complement of a graph Given a graph G G v is a graph that results when we remove v and all the edges leading to it or leaving it Similarly G e is the graph that results when we remove edge e If U Q V the G U is the graph that results when we iteratively remove all vertices in U from G G G1 is the graph formed by removing all vertices and edges in G1 0 G from G The complement of a graph G is a graph that has edges between all vertices that are not connected in G Graphs are considered disjoint if there are no vertices in common Graphs are considered edge disjoint if there are no edges in common Walk in a graph A walk in a graph is a sequence of alternating vertices and edges of the form v0 e1 v1 e2 v2vn Walks for now are nite There are two kinds of walks Open walks where v0 a vn and closed walks where v0 V The length of a walk is the number of edges in the walk Therefore there can be walks of length zero A trail is a walk with no repeated vertices Theorem If there is a v u walk implying a walk from vertex v to vertex u then there is a v u trail If you have a path with loops then chop off the loops to get the corre sponding trail For an algorithm that does this build a stack of vertices visited as you walk the graph At each new vertex reached check if it has been encountered before already on stack If it has then clear the stack all the way to the earlier occurence of this vertex and continue walk This guarantees that loops are chopped off as soon as they are encountered The stack at the end of the walk will be the trail Each time we get rid of a loop we generate a trail pre x lVl n i a v u trail has length at most n 1 Theorem An undirected graph G is bipartite i it has no odd cycles The theorem as stated is incorrect It states that a graph is bipartite implies it has no odd cycles and a graph has no odd cycles implies that it is bipartite The latter implication should have been a graph has no odd cycles implies it is 2 colorable Proof If a graph is bipartite then in any walk the vertices are of alter nating kinds Therefore we can never close a loop using an even number of edges as we will then have an edge connecting vertices of the same kind which will make the graph not bipartite For proving the graph has no odd cycles i it is 2 colorable consider the following intuitive proof no odd cycles 2 colorable77 is the same as proving the statement not 2 colorable i there are odd cycles Consider a node v1 in conflict that has two incoming edges Along one edge we want to color the node White whereas along the other we want to color the node Black This can only happen if the immediate predecessor vertex say v2 along the latter edge is White and the immediate predecessor vertex along the former edge say V3 is Black Now if the two paths originate from the same Black node say v then the length of v V3 has to be even and the length of v v2 has to be odd This follows from the fact that even paths connect nodes of the same kind Since there is one edge from v2 to v1 and one from V3 to v1 we have a cycle of length even odd 1 1 which is odd Hence the proof Eccentricity Radius and Diameter of a vertex du v distance of u v the length of the shortest walk same as trail since shortest from u to v Eccentricity of a vertex v ev maxdu v l u E V RadiusG minev l v E V DiameterG maxev l v E V which is the same as maxdu v l u v e V Eggma C 1 41M Ionso mzm News Rama 1 Mk Algonkfhm I Slwvd td MvFA VV V Huh 0 uivp Reps l i Pl ck U chq I BFU gigW p04 UA H I m chmitz em De t k mls war by QrFRnAW Wchi 0quot9 30W I 7WW WWO Accgwal 0106 Ne bhrs 0 WE39er LMLW Hu had twat 295 en n39aly fwnqu B e quot w cv s39f V g ll bar 9 quot39 Why DU ks hA Works do 96 th 9mm 10 1 v0 j t 0 i f 1 r5 4 RH Unfit 7yl lo 10 r vm m we a fme H1 0 xivMR Pow 306 Guquot39 0 u 5 to TMquot WM aim L7 o Inc FMM vb 0 to 1 C934 ijiuj IUOgt lt 695 Vejro 0 LQquot LV 1ij lt Voirojdo TIM W Nahquot t Luau bigger 14quot uo WkJok 3 M 1 uz 61 40 r4 wM W7 RheeLed Wi U 0 um W39I blew91 neng 46 U0 Z39r ugoiuAm39m 90m V9 Vab9190 Alfemne pthlq Lo L19 w39L V0 pun A to 0 H in U sz bus 139 3H 0 d c If pen I no MH V Al I a wcxj rum V0 19 0 5 Fa 9155 bgt cw 2 o 139 1n I39 C0 VDfolda gt 605 Va odu Hvt U0 NodJ1 how1 lawn Ma sFaltemy Calch LAQ Iquot We Cir l 4 m N 9wer U w Vu AC WCUDu3 RQPM Pi39Lk X e I V H no n U V94 I WW w cl u UIZUUiuS HunWAofoJw j I For A 6 dCr Min our 1 DICu3 w qu 3 UA H UIV SIMJL la wwnf n9 M39H o n 39r M R 39 fr 695 04 D53ksml per WW OOP lVi plUlng GA 469 00 Cu I Cor U1 Uuiuzs v 97 AVmln Ed r Acugtwcw 11gt OCV JECVHH W39W gt 00 mlVl 3931th 2 C6M1tbiefAIK9a quot fatall CSKYJSK 39 l C v h litHal Hi a 539 c 4 Z cc r S 2 P0P l Z 3 QQH39AMJ m W llnlmuy4 O 09 n gt O CMmlog n 1 3 WWW queue ovHvk ME ruleaHlu MMWI 5 399 MMHY Linyuawm wke Hwr m 9 anng kun n S u as l adnly 009479Wnce 9A NAL H VCF arrkka 395 SDoge passzC 0 Ginrng axlge 39H m knewr mg graph J MHS tux g imam V V0 Va Um 3 WW lleHc be I 0 J r LP mfmeIInIIA H39Qn 0C oplls l n beau 2 Me 5 576919544 5 Chem I I wm ksanva AHst m MW Mas AKJ3gt 19 539 0 Shark3 prov Ir0 J Wk I C Van l39t 1513 9 n Mares A VI Vlj HVK LJ 9quot 1 GM 393 JK1 PW k EMRL 01quot 2 m5 ig wgAKOJ k13A 4l3 g cm rm I AW km foam MRA 1 The Algorf an VVDJUjrgzVn V0233 4039 UN 1 WC For kilo 0 14quot V551 AK39C5JJM Aquot JJAk5krdkkn53 end n mm A as i C490quot 5 0 Cm axPeMmVa J bu CAMs 0x 511911923139 Pa s M ax J Pailrs Itquot A 1W Re wlow EXPFIL9 VS Par I UI Ven39 r c gt L M SC H 33 can W 1 5403quot A 743 7 r bc 4 He39t H ff f 0r bCi CI PMstI Q xrrem 2 0 3 noJe WCI39JJ 31 19 0 3 alp mbi From I 39Io C 90 39 H 1 s LU Am So AHChan 3a 1753 tvHvk A U n CMPE 177 Fall 2002 Induction Proofs Let Pn be a propositional function de ned on the set of integers In other words P is a function P Z gt True False where Z stands for the set of integers or possibly a subset of the integers Informally this means that Pn is some statement whose truth or falsity depends on the value of an integer n Mathematical Induction is a proof technique which can be used to prove statements of the form Vn 21 Pn for all positive n Pn is true More generally we seek to prove Vn 2 n0 Pn for some integer n0 Such a proof consists of two steps 1 Base Step Prove Pno In other words prove directly that the propositionPn0 is true IIa Induction Step Prove Vn 2 n0 Pn gt Pn 1 Pick an arbitrary n 2 n0 and assume that for this n Pn is true Then show as a consequence that Pn1 is true The statement Pn is often called the induction hypothesis since it is what is assumed in the induction step When I and II are complete we conclude that Pn is true for all integers n 2 n0 Induction is sometimes explained in terms of a domino analogy Consider an infinite set of dominos which are lined up and ready to fall Let Pn be the statement the nth domino falls First prove Pl which is the first domino falls then prove Vn 21Pn gtPn1 which says if any particular domino falls then the next domino must also fall When this is done we may conclude Vn 2 1 Pn all dominos fall There are a number of variations on the induction step The first is just a reparametrization of Ila IIb Induction Step Prove Vn gt n0 Pn 1 gt Pn Let n gt n0 assume Pn 1 is true then prove Pn is true Forms Ila and 11b are said to be based on the first principle of mathematical induction Another important variation is the second principle of mathematical induction also called strong induction IIc Induction Step Prove Vn 2 n0 Vk S n Pk gt Pn 1 Let n 2 n0 assume that for all k in the range n0 S k S n Pk is true Then prove as a consequence that Pn1 is true In this case the term induction hypothesis refers to the stronger assumption Vk S n Pk The strong induction form can and often is parameterized as in IIb IId Induction Step Prove Vn gt n0 Vk lt n Pk gt Pn Let n gt n0 assume that for all k in the range n0 S k lt n Pk is true then prove as a consequence that Pn is true In this case the induction hypothesis is Vk lt n Pk Example 1 Prove that for all n 2 1 quot 2 nn12n1 6 Proof Let Pn be the boxed equation above We begin the induction at no 1 1 Base step 1 Clearly 2139 2 1 11 showing that P1 is true IIa Induction Step Let n 2 1 and assume Pn is true That is for this particular value of n the boxed equation holds Then quot1 21 1392 n12 11 11 W 71 12 by the induction hypothesis nn 12n 1 6n 12 6 bysome algebra showing that Pn 1 is true It follows from the principle of mathematical induction that Pn is true for all n 2 1 One should always state explicitly what is being assumed in the induction step ie what is the induction hypothesis One should also make note of the point in the proof at which the induction hypothesis is used Example 2 Let xe R and x 1Showthatforalln20 n1 1 n I x 2x 11 x 1 Proof Again let Pn be the boxed equation We begin the induction at no 0 1 Base step 0 Z x x0 1 x j showing that P1 is true x 10 IIb Induction Step Let n gt 0 and assume that Pn l is true ie assume for this particular 71 that I xquot l x x l39 H 3 Then l o by induction hypothesis showing that Pn is true Steps I and II prove that Pn holds for all n 2 0 Often the propositional function Pn is not a formula but some assertion concerning other types of mathematical structures Recall that a graph G is a collection of vertices and edges where each edge joins two not necessarily distinct vertices A path in G is a sequence of distinct vertices joined by edges and a cycle in G is a closed path G is called acyclic if it contains no cycles G is called connected if any two vertices are joined by a path in G A graph T is called a tree if it is both connected and acyclic The next example is a case in which strong induction is necessary Example 3 Let n 2 l and suppose T be atree on n vertices Prove that T necessarily has n l edges Proof Let Pn be the statement if T is a tree on n vertices then T contains 71 l edges We begin at n l I Base step If T has just one vertex then being acyclic it can have no edges whence Pl holds IId Induction Step Strong Induction Let 71 gt1 and assume that for all k in the range 1 S k lt n Pk is true That is for any such k all trees on k vertices contain k l edges Now let T be a tree on n vertices Pick any edge e in T and remove it The removal of e splits T into two subtrees each having fewer than n vertices Say the two subtrees have ml and n2 vertices respectively Then by our inductive hypothesis the two subtrees have n1 1 and n2 1 edges respectively Now replacing the edge e we see that T contains a total of I11 ln2 ll n1 n2 ln l edges Observe that no vertices were removed so that n1 r12 n The result now follows for all trees by induction There are many other variations on the induction technique Occasionally we must use double induction which involves a modi cation of both the base and induction steps I Base Step Prove Pno and Pn0 l II Induction Step Prove Vn 2 n0 2 Pn 2 APn l gt Pn When this is accomplished we may conclude Vn 2 n0 Pn In terms of our domino analogy we prove that I the rst two dominos fall and II if any two consecutive dominos fall then the very next domino falls From this we deduce that all dominos fall The Fibonacci numbers Fquot are de ned by the recurrence 0 if n 0 F 1 ifn1 Fquot1 FH if n 2 2 ie each term is the sum of the previous two Using this recurrence the rst few terms of the sequence are readily computed as F0 0F1 lF2 lF3 2F4 3 etc Example 4 Prove that for all n 2 0 F 15 aquot bquotl 1 43 2 SI 3 where 6112 and b Proof Let Pn denote the boxed equation above I Base Observe that PO and Pl are true since i61 b 0 F and LL b1 1 F 5 0 J3 1 II Double Induction Let n 2 2 and assume that both Pn 2 and Pn l are true ie we assume FH 2a bH and FM 2a bH Thus 1 F F71F72 aquot 2a1 bquot 2b1 n n n 5 One checks that a and b are roots of the quadratic equation x2 x l 0 Therefore a2 61 1 and b2 bl whence l 1 F anilra2bnilrb2 anbn I 1 I 1 showing that Pn is true Together steps I and II imply that Fquot i 61quot bquot for all n 2 0 J3 Often the proposition to be proved is an inequality as in the next example Example 5 De ne Tn for n e Z l by the recurrence T0 0 if n 1 TIn211 1f n22 Prove that for all n 2 5 showing that Tn Olg n Proof Let Pn be the boxed inequality above 1 Base One checks easily that T5 3 S 2 r lg5 using the recurrence and the fact that 8 S 25 so P5 holds IId Strong Induction Let n gt 5 and assume for all k in the range 5 S k lt n that Pk is true ie Tk S 21gk In particular TIn2DS 2lgn2D for k rt2 One shows easily that lgxDS l2 lgx for all x21J2 l Since 71 gt 5 2 2J2 l we have n2 21J2 l whence TIn2DS 212 lgn2 Now by the de nition of Tn Tn Tfn21 S 21 2 lgn 2 l by the above inequality 21gn showing that Pn is true Therefore Tn S 2lgn for all n 2 5 as claimed Induction Fallacies The next few examples illustrate some pitfalls in constructing induction proofs The result in Example A below was proved correctly in Example 3 We give an invalid proof of the same fact which illustrates an argument some authors have called the induction trap Example A Show that for all n 21 if T is a tree on n vertices then T has n l edges Proof Invalid Base If n 1 then Thas no edges being acyclic Induction Let n 2 l and let T be a tree on n vertices Assume that T has n l edges Add a new vertex and join it to T with a new edge The resulting graph has n1 vertices and n edges and is clearly a tree since connectedness is maintained and no cycles were created By the principle of mathematical induction all trees on n vertices have n l edges B Let us first observe that this argument does not follow the induction paradigm In this example Pn is of the form An gtBn where An is the statement T is a tree on n vertices and Bn is T has n l edges The induction step should therefore be to prove for all n 2 l Pn gt Pn 1 That is An gt Bn gt An 1 gt Bn 1 To prove this we should assume An gtBn then assume Anl then show as a consequence that Bn 1 is true In other words we should 0 Assume all trees on n vertices have n l edges 0 Assume T has 71 1 vertices 0 Show as a consequence that T has n edges The above argument did not follow this format however Instead the arguer does the following 0 Assume T has n vertices 0 Assume T has n l edges 0 Construct a new tree from T having 71 1 vertices and n edges Therefore the above argument was not a proof by induction Some students would nevertheless hold that the above argument is still valid even though it is not a true induction proof The next example shows convincingly that it cannot be valid First we introduce a few more definitions related to graphs A graph G is called simple if it contains no loops edges whose end vertices are identical and no multiple edges pairs of edges with the same end vertices Example B For all n 2 l ifG is a simple graph on n vertices then G has n l edges We notice right away that the above statement is false since the graph below provides an elementary counter example But consider the following proof in light of Example A Proof Invalid Base If n 1 then G has no edges being simple Induction Let n 2 1 and let G be a simple graph on n vertices Assume that G has n 1 edges Add a new vertex and join it to G with a new edge The resulting graph has n1 vertices and n edges and is clearly simple since no loops or multiple edges were created By the principle of mathematical induction all simple graphs on n vertices have n 1 edges Observe that Example B follows the format of Example A word for word Thus if A is valid so must B be valid But the assertion proved in B is false Therefore B cannot be a valid argument and so neither is A Another fallacy comes about by not proving the induction step for all n 2 no Example C Prove that all horses are of the same color Proof Invalid We prove that for all n 2 1 ifS is a set ofn horses then all horses in S have the same color The result follows on letting S be the set of all horses Let Pn be the boxed statement and proceed by induction on n Base Let n 1 Obviously if S is a set consisting of just one horse then all horses in S must have the same color Thus P1 is true Induction Let n gt 1 and assume that in any set of 71 horses all horses are of the same color Let S be a set of 71 1 horses say S h1 h2 h3 hn1 Then the sets Sh h35quotquothn1Sh1 and S39hlh3hn1S hz each contain exactly 71 horses and so by the induction hypothesis all horses in S are of one color and likewise for S 39 Observe that h3 e S m S 39 and that h3 can have only one color Therefore the color of the horses in S is identical to that of the horses in S Note 71 gt 1 gt 71 2 2 gt 71 1 2 3 so there is in fact a third horse and he can have only one color Since S S U S 39 it follows that all horses in S are of the same color Therefore Pn gt Pn 1 for all n gt 1 The result now follows by induction D Obviously the proposition being proved is false so there is something wrong with the proof but what The base step is certainly correct and the induction step as stated is also correct The problem is that the induction step was not quantified properly We should have proved Vn 21Pn gtPn1 Instead we proved correctly that Vn gt1 Pn gt Pn1 Indeed it is true that P2 gtP3 for instance but we never proved and it is false that P1 gt P2 Gm RQEW DQ a HMs T mm P MQ HRX 39 Ni V TaVes 1 6che 66gt wow Mad Rx when lama L 0 d 5V mew DST rpm MA mam m h1Ta4ec BOA Vva Waco Nof bard QW Bad rmd ina em E1 951 when watm space Cam be emei DVKCAA as Mask m U AT Wait DDAM WPX W JYWUT e m0 A MW 39VLV 6V 39 maxCA VDNW lusf 4g x v gt awe3 MARK 3quotva 3V CLV 3 vie erCL we came warng Memm mxi A 1 was walk Li 5 1 In CK e L quot cobv vj wWi e color CV1gtQCK u awagbh n u m e XLaLw M Qm ow claim aWM WWmce T am 3 1 30 L L QMObSQg Jars t Uw ma ox X a k KCLC 39 PxC K L Iii PMS Cw LB 39mw 50 L I PTCK QW05 9H i demWWMxL UVwL P LK wok Mwe Jf od vo Elmaa m Meai ewWCl w dew WL D F5 I 3me Maciaskquot TX L s CK Mr X a n WW 06v 9 m m r w W Mv F3 P4103 Cole FOO39339 wW i wokAC O 4613 ex f7 AQMX CL TY DIEgtvi 3 Vk 7 MCLCVB 4 abrCVo39li bT COLOCCJAB VOW V c l Ow LOWPDQ U CV01quot V aC Q 7 4 Cv gt I C Ty CM z VCL CW u I Lab r Q1mckl 3 I OCVH39WQ mcfma Msoxcenca Md I Me c1 39 6mg It epmsevx ygi a ahHow V QQVlequ I 6 66V Ver c anti R M 61 6X S MWQ A G katk at ESQ gt PAM 866MW 0L n E TWs 6an x JOK TM E wank ltVNM wc Q hoo BHAOK fexajrrm ex 3on Cwq lvk KIA1Wqu tic0x 1 Zquot L jonx Mxcq 3 NLACOK ASL Cows uo avg er were Mair sWd meec b vt A B 51gt B 75 kc a 5 13 UXNSUA A C CVCAJV Doou A D h D WWWth A6 4e 539T 9 rec QM vo mesecl COMtwat WV QL X 5W9 res mlvefs Wt FQMOVR L mlvx xmwwk mmmber 0 1665 ST W Wk xs NbKQW wt ru 3mm or mam ou vmwe 3 gtWquot 5 my ew 5 bwkg 2 an KAKWQWQC 33 ma a worM As at v a TM 0L1 HAWIS WK ecum V quot M Jags MC e VO O o j w nd Me ex ondr mlLorwome MAR R lac m ed 0 cm Sammie a gfmmms 4w rech 535 M Twp q 12lm6 beemms Pl 6 GFQPM GCE gt QvLEgt IZM V Ni quot 955 Cum3 erV Z 4 ak 2 1 h 3 39 E 2xDQampEgt gtgt 41 6V7 Equot Qquot a 2 ADCQ smch W IQ V 3 77 quot9b a 0A 6V 79 CC 3 3 fb We citch uwou th Caricaeg M r i gtka WCMTV W w LMOQT P900K 5 Commx meb W e Eam gGbC zK Q 496 L09 39 65 a veW K4 gt e 30 a x e A No PM a M No sew Imps W 6 CVIJEngt39 oxWA 622CVZJE3bz Ggt QW GZ 0J8 NSoww Alg K Q G quotWere u39 N6IXLL Wxafavt b xge ms AA 6E1 XE Z WI 130950 hVLGVlequot V7 VL OMA g quot1 1 WWW0493 quotAceA amk 5 AlAD LQZ39B 44w AMIL M CRWFerL DKO edggg Applied Graph Theory Glass 9 Oct 27 2005 Articulation Points If G is an undirected graph a vertex V E V is an articulation point if and only if G V is a disconnected graph A graph G is biconnected i it has no articulation points A graph is n connected i removing any n l or lesser number of vertices does not disconnect it lntuitive idea behind an algorithm to nd articulation points If we do a DFS since the graph we are considering is undirected we do Visit all vertices in one pass Do DFS The Root V is an articulation point i it has more than one child children are always connected by tree edges Remember that DFS builds a tree Therefore if the root has two or more Tree edges it is an articulation point For all other nodes we need to check if removing the node will separate the ancestors of the node from its children This can only happen if there are no Backedges during DFS from the children of the node to its ancestors De ne a function low V a N as follows l0wv minuevdulv Li u lowv computes the smallest arrival time over the set of all ancestors of the node V such that they are connected to the node v through paths originating at node V that have zero of more tree edges followed by a single back edge For all such ancestor vertices u E V lowv returns the smallest discovery time Therefore if we nd a child t of v having a lowt less than the discovery time of V then we have found a path originating from v leading to one of its ancestors This implies v cannot be an articulation point since it does not disconnect its ancestors from its children Therefore v is not an articulation point i v has a child t with lowt S dv the discovery time of the vertex v Using recursive DFS Assume we have computed lowwi for i 123 where w are children of v implies they are connected by tree edges We 1 Applied Graph Theory Class 7 Oct 18 2005 Reducibility Reduction is used to reduce a given problem to a known problem Say we have an e icient algorithm7 say A7 that has complexity 0n2 If we know of a reduction for a new problem in space and time Oklogk where k is the size of the new problem7 then we can solve it in time Oklogk2 Oklogk7 which has the asymptotic complexity of Oklogk2 Point to be taken Strive to cast a new problem in terms of one you know and can solve e iciently Communication network Consider a communication network with loss probabilities along edges The rst thing we do is convert all loss probablities to success probabilities This is because a packet being lost along an edge i the packet is lost along any path that starts from its destination vertex Probability of successful transmission HeeE 1 P8 The problem Given u7 v nd a sequence u uO7 uh uk v that maximizes Hf11 Pu 1L7 ui Remember that V u7 v 6 V7 Pu7 v E 07 1 Convert the product into a sum using log Transformed problem7 k mazimize logH1 7 Pu 17 2391 k mazimize Z log1 7 Pu 17 21 Since the Shortest Path SP algorithm computes the shortest path by minimizing the sum of the weights7 invert the sign of log1 Pu L7 The problem then becomes7 k minimize Z 7 log1 7 Pu 1 21 The terms above will always be positive since the domain of the log function is in the range 0 I So set wr s where r s E the set of edges to log1 Pu 1 and solve using SP Note In the above discussion for paths that don7t exist set the loss probability to 1 Interesting note on logs In general sums are better than products Even from the way we have been taught multiplication and addition in school multiplication is quadratic whereas addition is linear So when you have products try to convert them into sums logs are very useful in this regard FloydWarshall Proof that the algorithm computes the shortest path between any two vertices i and j Note that the re nement step in the algorithm where we check for a lower cost path between any two vertices i and j is given by dk1l7 mmfdevJ39LdeC 1 dkk 171 Proof We know that d0i j wi j Assume that dki j computes minimum paths At step k1 there are two cases to consider Use the node k1 since there is a path from i to k1 and from k1 to j which has a total cost less than dki j In this case since all nodes along the paths i k1 and k1 j are below k1 by induction hypothesis paths i k1 and k1 j are each minimal paths Hence their concatenation is a minimum path If we don7t use the node at k1 then dk1i j dki j which again by induction hypothesis is minimal BellmanFord Let d be a function that maps each vertex with a real number So d V a IR Now BFv d mindv Winner101 u We consider two versions of the algorithm The Strati ed version and the in place version optimized version Strati ed BellmanFord o d0u 0 if u source 00 otherwise 0 For k 0 1 2 n 1 For all u E V dk1u BFu dk 0 Output V u dnu We need to prove that the algorithm computes the shortest path between any vertex and the source We use induction to prove that at step k1 the algorithm computes the shortest path using only edges 3 k1 When we are done since we would have considered all n edges dnu will then give us the shortest path from u to the source Proof Assume that dku is the minimum weight of a path to the source from u that has at most k edges This is our induction hypothesis There are two cases to consider The rst case is when we need 3 k edges for the shortest path This means that BFu dk does not return a smaller value by using an additional edge k1 By our induction hypothesis since dk u is optimal in this case dk1u is optimal as well The second case is when there is a better path with k1 edges Then if V7 is the set of vertices that are immediate neighbours of u the cost of such a path will be minvEVdkv wv u This will be a shortest path since dk is the shortest path to v by our induction hypothesis and to this shortest path we are adding the edge with the smallest weight that leads to u so dk1u should be shortest as well In place Bellman Ford 0 du 0 if u source 00 otherwise 0 For k 0 1 2 n 1 For all u E V du BFu d 0 Output V u du We have simply dropped the subscript We need to prove that this version of the algorithm solves exactly the same problem as the Strati ed Bellman Ford described above lntuition for a proof If the vertices are traversed in the same order as their occurence in the optimal path then we need only one pass to compute the values du for all vertices along that path So the in place algorithm may work faster based on the order in which vertices are updated Euler Tours A tourof G is a closed walk which includes every edge at least once An Euler tourof G is a tour which includes every edge exactly once An Euler trail of G is a trail which includes every edge exactly once swag CMPE177 Fall 2mm A graph is Eulerian if it has an Eulertour Eulerian 2 Does it have an Eulertrail 2 swag CMPE177 Fal 2mm Thm For a connected graph G the following are equivalent 1 G is Eulerian 2All vertices of G have even degree 3The edges of G can be partitioned into cycles Thm 32 1 if and only if 2 Thm 33 1 if and only if 3 Show 1L 2L 3L 1 swag CMPE177 Fall 2mm Eulerian 2 Does it have an Eulertrail 2 swag CMPE177 Fal 2mm Buy Material Are you sure you want to buy this material for 25 Karma Buy Material BOOM! Enjoy Your Free Notes! You're already Subscribed! Why people love StudySoup Steve Martinelli UC Los Angeles Kyle Maynard Purdue Bentley McCaw University of Florida Parker Thompson 500 Startups Become an Elite Notetaker and start selling your notes online! Refund Policy
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Play free online games! On you'll find the best collection of Zuma games! You'll find no less than 13 different Zuma games, such as Zuma's Revenge & Maya. Shoot the balls at the moving row and make sure that at least 3 balls of the same color are next to each other in order to remove them from the game! Can you remove the entire row of balls before they reach the end? Send feedback
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Busy. Please wait. Forgot Password? Don't have an account? Sign up show password Already a StudyStack user? Log In Reset Password Remove ads Don't know (0) Know (0) remaining cards (0) Pass complete! "Know" box contains: Time elapsed: restart all cards Normal Size Small Size show me how Algebra I EOC Vocab Vocabulary STAAR Practice 2 of 3 Dependent Variable the output variable (y) whose value in a function is determined by the input variable (x) Independent Variable the input variable (x) whose value in a function affects the output variable (y) Negative Slope if the line goes down and to the right on a graph Zero Slope (y = #) is a horizontal line Slope A measure of steepness of a straight line defined as rise over run Undefined Slope (x = #) is a vertical line y-intercept where the line or parabola crosses the y-axis Δx indicates change in the domain Δy indicates change in the vertical axis Line of Best Fit a line drawn through or near data points that most accurately represents a set of data Linear Function Data sets that have a constant rate of change and form a straight line, have x as the input variable where x is raised only to the first power Point-Slope Form equations written as y-y1=m(x-x1) Slope-Intercept Form y = mx + b, where m= slope b = y intercept and (x , y) shows the coordinate of a point on the line Standard Form ax + by = c where a, b and c are integers Infinitely Many Solutions linear systems of equations that have an infinite number of solutions are the same line No solution when a system of equations do not have an intersection System equations that you solve at the same time Created by: SuzyMahoney
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String Quartet No.2, Op.30 String Quartet No.2, Op.30 is a piece composed by Carl Reinecke. It was first published in 1852. The piece has four movements. This sheet contains the following parts: Sheet Title Download Complete Parts (Leipzig: Friedrich Hofmeister, n.d.) Carl Heinrich Carsten Reinecke
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Howdy, Stranger! Programming algorithm tanteltantel Member Posts: 2 You are going on a long horse trip. You start on the road at mile post 0. Along the way there are n horse stations at mile posts m1 < m2 < ..< mn where each mi is measured from the starting point. The only places you are allowed to stop and change horses are at these stations, but you can chose which of the stations you stop at. You must stop at the final station(at distance mn) which is your destination. Assume that at a station it takes 2 hours to change horses and a horse takes x^2/400 hours to travel x miles. Design a dynamic programming algorithm to determine a sequence of stations at which to stop so as to minimize the total hours. please help me with solving this algorithm can be just an algo or a C program Sign In or Register to comment.
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From Ultronomicon Jump to: navigation, search Starbox is an attempt at presenting each star system and its planets in a standardized format. Deemed excessive, it was decided it would only be used for the Sol system, for now. {{StarBox|Star Name|Star Color|Star Size|Number of Planets in system| PlanentLine parameters}} Color: Yellow Size: Medium Name Type Temp. (°c) Weather Tectonic Gravity (g.) Moons Mercury Metal (Metallic) 165 0 3 0.37 0 Venus Primordial 457 8 2 0.90 0 Earth Water 22 2 2 1.00 1 Mars Dust -53 2 2 0.38 0 Jupiter Gas Giant -143 8 0 2.68 4 Saturn Gas Giant -197 8 0 1.22 1 Uranus Gas Giant -117 8 0 0.86 0 Neptune Gas Giant -229 8 0 1.10 1 Pluto Pellucid -235 0 1 0.05 0 Possibilities for Expansion[edit] • It is possible to add columns for total bio points and RUs on each planet. • It is possible to include the moons of each planet within the table, since they contain the same data. • It is possible to include a column for Special item/encounter ... like a checkbox (Y/N). • Unless someone knows of a way to dynamically link N number of template entries, as in PlanetDataForN, this template is severely limited by requiring a number of statically created "helper" templates for systems with {0, 1, 2, ..., N} number of planets. I do not know enough wiki to know if the equivalent of a for-loop or something similar exists. • A solution to the preceding limitation is likely pointless, aside from personal curiosity, since creating a page for every star system within the game was seen as filling the Ultronomicon with potentially useless data. See Also[edit]
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A loan of $950 helped a member to buy recycled items for resale. Sophal's Group's story Sophal, a father of four schoolchildren, makes a living through buying recycled products for resale. To earn extra income, his wife raises chickens. Having been with this business for five years, he is able to make 42,800 KHR per day and he is able to save about 6,000 KHR per day. Sophal's group consists of three members. Two are fourth-time borrowers and one is the first-time borrower. Sophal has managed to overcome his family's difficulty from the past loan that he has paid back completely. Now, he and one of his co-members are applying for 1,400,000 KHR each while the remaining member is seeking 1,000,000 KHR. As the leader, he is going to use his portion to buy recycled items for resale and also to buy chicken feed. He hopes to make more income from his current business so he can maintain his living. In this group: Sophal, Savuth, Lin *not pictured Loan details Lenders and lending teams Loan details
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Infinite Descent is a proof technique of use in solving problems from number theory, ranging from elementary problems to vital theorems of modern algebraic number theory. Infinite Descent was first described by name by the 17th century mathematician Fermat in the paper La méthode de la 'descente infinie ou indéfinie. Proof by Infinite Descent is a special case of proof by contradiction (reductio ad absurdum), although from a mathematical logic point of view it can be seen as equivalent to the axiom of induction, or the well-order property of the natural numbers. The technique is usually employed to demonstrate non-existence of collections of numbers with a particular property. To do so, it is shown that if one set exists, then another exists which is in some sense 'smaller'. In this way, infinitely many successively smaller collections can be found, yet the problem is posed in such a way that there cannot be infinitely many such collections- usually because of the existence of a bound. This is a contradiction, so the assumption of existence of a solution isn't tenable and we thus conclude no solutions exist. This is probably best understood by way of an example. So consider the following problem: Show that there is no quadruple (x,y,z,u) of positive integers satisfying x2 + y2 = 3(z2 + u2) The important thing here is that we are looking for positive integers -the counting numbers 1,2,3,...- and thus an infinite descent cannot exist: a sequence of such values cannot keep strictly decreasing forever. By contrast, infinite strictly decreasing sequences of rational numbers or real numbers are not inherently paradoxical- the sequence obtained by successive halving (1, 1/2, 1/4, 1/8,...) being an example. Descent Proof We proceed therefore to give a proof by infinite descent. Suppose we have four positive integer values x,y,z,u satisfying the required condition. Then, since 3(z2 +u2) is divisible by three, and x2 + y2 is equal to that expression, it must be that x2+y2 is divisible by 3. Formally, we say that x2 + y2 is congruent to 0 mod 3, which is a fancy way of saying we are left with remainder 0 after division by 3. An aside on working mod 3. Here a number is essentially either 0, 1 or 2: 3 is 0 mod 3 (as are 6,9,12 etc.) , 4 is 1 mod 3 (as are 7,10,13,etc) and so on. Addition works as usual except you discard multiples of 3, so, for example, 2+2 = 4 = 3+1 so 2+2 =1. Knowing that x2+y2=0 mod 3, we conclude that one of three cases holds: • Both x2 and y2 are 0 mod 3 • x2 = 1 mod 3 and y2 = 2 mod 3 • x2 = 2 mod 3 and y2 = 1 mod 3 However, you can easily verify (test all cases!) that the square of any number mod 3 is 0 or 1. That is, there are no numbers which square to 2 mod 3, so the second two cases above are impossible. We know therefore that x2 and y2 are both 0 mod 3, meaning they are divisible by 3. But if the square of an integer can be divided by 3, then the integer itself can be divided by 3. So we can write x=3x' and y=3y' for some x' and y'. Since x and y were strictly positive, x' and y' are definitely smaller (x' being a third of x, y' a third of y). We can plug these new versions into the equation to get (3x')2 + (3y')2 = 3(z2 + u2). That simplifies to 3.3(x'2+y'2) = 3(z2+u2), and cancelling a 3 from each side we get So we can deduce from the presence of a 3 on the left hand side that z2+u2 =0 mod 3, and by the same reasoning as for x and y earlier can find ourselves z' and u' such that z=3z' and u=3u'. Plugging into the equation again, we find 3(x'2 +y'2)= (3z')2 + (3u')2 So 3(x'2 +y'2)= 9(z'2 + u'2) i.e x'2+y'2=3(z'2 + u'2) Hence we're right back where we started, except all the numbers are a third of those we started with. There's nothing to stop us applying the same thought process to find a set x'',y'',z'',u''. In fact, we can do this infinitely many times, with the terms strictly decreasing in size each time. Thus admitting one solution gives rise to an infinite descent, so there can be no solutions. Direct proof by contradiction To confirm proof by descent is a contradiction argument in disguise (even though we needn't explicitly appeal to that contradiction to complete a proof; as in the example above the observation of an infinite descent of integers is sufficient), here's a proof of the same result by slightly different means. Suppose for contradiction there exist positive integer solutions to the problem. Let x,y,z,u be a solution such that V=x2 + y2 = 3(z2 + u2) is minimal. Then, by the earlier reasoning, we can find x',y,z',u' satisfying the conditions. But they give rise to V'=x'2 + y'2 = 3(z'2 + u'2) = V/9. So V'<V, contradicting the minimality of V. So positive integer solutions do not exist. Constructive proof One last proof can be given, with a more constructivist feel- it demonstrates that any non-negative integer solution must be the solution 0,0,0,0. Thus, there cannot be a positive integer solution, since x,y,z,u a solution implies x=y=z=u=0 and x,y,z,u arbitrary positive integers implies x,y,z,u non-zero; the method of proof by contraposition then ensures x,y,z,u aren't a solution and their arbitrary nature proves there are no such solutions. To reason in this way, let X,Y,Z,U be a non-negative integer solution (so zero is allowed). Then we can construct a new solution X',Y',Z',U' with X=3X'. Repeating, we get X'' st X=3X' = 3.3X'' and so on; concluding that 3n divides X for every n. But the only number which is infinitely divisible by 3 is zero, so X=0. Further, the same reasoning gives each of Y,Z and U =0. We are done. Some notable results by Infinite Descent Fermat's theorem on sums of two squares (the Christmas Theorem) An odd prime p can be written as p=x2 + y2 iff p is congruent to 1 mod 4 Typically for Fermat, he didn't prove this. Euler gave the first proof, by means of infinite descent. No solution to x4+y4=z2 This was known to Fermat, and allows an elementary proof of his last theorem in the case n=4 (clearly, if the sum cannot be a square it cannot be a fourth power). Irrationality of the square root of 2 That the square root of 2 cannot be rational was known to the Greeks. Euclid gave a proof by infinite descent, so the technique actually predates Fermat by nearly 2000 years! The result generalises to the square root of any prime.
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시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율 1 초 128 MB 1 1 1 100.000% Many of you have heard the story of Turing’s bicycle: Seems the sprocket on the crank of the bicycle had a broken prong. Also his chain had one link that was bent. When the bent link on the chain came to hook up with the broken prong, the chain would fall off and Turing would stop and put the chain back on. But Turing, being who he was, could predict just when this was going to happen — meaning he would know how many pedal strokes it would be — and so would hop off his bike just before it happened and gently move the pedals by hand as the undesired coupling passed. Then he’d be merrily (we imagine) on his way. (A picture of the sprocket-chain set up is shown below.) Your job here is to calculate the number of revolutions required in such a situation as Turing’s: You’ll be given the number of prongs on the front sprocket, the number of links on the chain, the location of the broken prong and the location of the bent link in the chain. The top prong is at location 0, then the next one forward on the sprocket is location 1 and so on until prong numbered s − 1. (See the diagram. Notice that prong s − 1 is the next prong that moves to the top of the sprocket as Turing pedals.) Location of the links is similar: The link at the top of the sprocket is location 0 and so on forward until c − 1. The chain falls off when broken prong and bent link are both at location 0. Input for each test case will be one line of the form s c p l, where s is the number of prongs on the front sprocket (1 < s < 100) , c is the number of links in the chain (200 > c > s), p is the initial position of the broken prong, and l is the initial position of the bent link. The line 0 0 0 0 will follow the last line of input. Broken prong and bent link will never both start at position 0. For each test case output a single one line as follows: Case n: r m/s if it requires r m/s revolutions to first fail, or Case n: Never if this can never happen. Note that the denominator of the fraction will always be the number of prongs on the sprocket; the fraction will not necessarily be in lowest terms. Always print the values of r and m, even if 0. 예제 입력 40 71 32 23 20 40 4 24 40 71 8 33 20 40 3 17 0 0 0 0 예제 출력 Case 1: 1 8/40 Case 2: 0 16/20 Case 3: 25 32/40 Case 4: Never
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Euclidean algorithm (another Prime Pages' Glossary entries) The Prime Glossary Glossary: Prime Pages: Top 5000: The Euclidean Algorithm to compute the greatest common divisor for two integers a and b (not zero) is based on the following fact: If r is the remainder when a is divided by b (see the division algorithm), then gcd (a,b)=gcd(b,r). For example, gcd(356,96) = gcd(96,68) (because 68 = 356 - 3.96). Continuing this process of dividing, then continuing with the remainder and divisor, we have gcd(356,96) = gcd(96,68) = gcd(68,28) = gcd(28,12) = gcd(12,4) = gcd(4,0) = 4. This ancient algorithm was stated by Euclid in his Elements over 2000 years ago, and is still one of the most efficient ways to find the greatest common divisor of two integers. The algorithm can be written in pseudo-code as follows Euclid(a,b) { while (b not 0) { b := b mod a On modern computers the binary gcd algorithm is usually faster even though it takes more (but simpler) steps. One of the uses of the Euclidean algorithm is to solve the diophantine equation ax+by = c. This is solvable (for x and y) whenever gcd(a,b) divides c. If we keep track of the quotients in the Euclidean algorithm while finding gcd(a,b), we can reverse the steps to find x and y. See Also: LamesTheorem
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시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율 1 초 256 MB 6 6 6 100.000% En route to Rigel 7, Chief Engineer Geordi Laforge and Data were discussing favorite numbers. Geordi exclaimed he preferred Narcissistic Numbers: those numbers whose value is the same as the sum of the digits of that number, where each digit is raised to the power of the number of digits in the number. Data agreed that Narcissistic Numbers were interesting, but not as good as his favorite: Perfect Numbers. Geordi had never heard of a Perfect Number, so Data elaborated, “A positive integer is said to be Perfect if it is equal to the sum of its positive divisors less than itself. For example, 6 is Perfect because 6 = 1 + 2 + 3.” Geordi began thinking about an algorithm to determine if a number was Perfect, but did not have the raw computing ability of Data. He needs a program to determine if a given number is Perfect. Help Geordi write that program. Input consists of a single entry per line. Each line contains a single positive integer n, where 2 < n < 100000 for each case. A line containing -1 denotes the end of input and should not be processed. For each case, determine whether or not the number is Perfect. If the number is Perfect, display the sum of its positive divisors less than itself. The ordering of the terms of the sum must be in ascending order. If a number is not Perfect, print “<NUM> is NOT perfect.” where <NUM> is the number in question. There must be a single space between any words, symbols, or numbers in all output, with the exception of the period at the end of the sentence when a number is not perfect. 예제 입력 예제 출력 6 = 1 + 2 + 3 12 is NOT perfect. 28 = 1 + 2 + 4 + 7 + 14
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Pure per stirpes Pure per stirpes is a system of determining the descendants of individuals who have died intestate. Any part of the intestate estate not passing to the decedent’s surviving spouse is passed to the descendants of the decedent. In the pure per stirpes system, the estate is divided into primary shares at the generation nearest to the decedent (the decedent’s children). If a person from the “primary share” generation has predeceased the decedent, that one primary share is further divided by the number of people at the next generation (the decedent’s grandchildren). Any deceased children who have no living descendants are disregarded in determining the number of primary shares. However, if the decedent has no living children, the number of primary shares is still determined at the children generation instead of skipping to the grandchildren generation. Taxonomy upgrade extras:
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Prove that a graph G with at least four vertices is 2-connected if and only if for every pair X, Y of disjoint vertex subsets with |X|, |Y| is greater than or equal to 2, there exist two completely disjoint paths P1, P2 in G such that each has an endpoint in X and an endpoint in Y and no internal vertex in X or Y.
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A number theory problem by Maricel Montanez Number Theory Level 3 What is the largest number that cannot be expressed as: A. A multiple of 3, or B. A multiple of 10, or C. A sum of a multiple of 3 and a multiple of 10? Problem Loading... Note Loading... Set Loading...
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Real Numbers: Normalisation From Wikibooks, open books for an open world Jump to: navigation, search UNIT 3 - ⇑ Fundamentals of data representation ⇑ ← Floating point numbers floating point normalisation Errors → When storing numbers we need to use the space we are given in the most efficient way. For instance if we take a denary floating point number such as (Planck's constant) If we were to rewrite it as: (Planck's constant) Then you can see the representation takes up an extra 2 characters, the two extra 0's, even though it represents exactly the same number. This may be acceptable when you are not worried about how many characters a number makes up, but in binary and with limited computer memories, the space that numbers take up is very important. We need the most efficient representation we can. With a fixed number of bits, a normalised representation of a number will display the number to the greatest accuracy possible. In summary normalised numbers: • Give only one representation of a number. • Save space. • Give the most accurate representation of a number in a given number of bits. As a rule of thumb: when dealing with Floating point numbers in binary you must make sure that the first two bits are different. That is: And most definitely NOT Let's look at an example. Taking a binary floating point number: To make sure you have normalised it correctly, check that Lets try a more complicated example: Summary: Normalising numbers 1. Normalise the left hand side (mantissa). 2. Record the number of ‘bounces’ it has taken to normalise. 3. Work out the exponent of the normalised number by using: original exponent – ‘bounce’. • Normalised numbers start with 2 bits that are different. • Make sure that your normalisation does not change the sign of the mantissa. • Normalisation provides the maximum precision for a given number of bits. • Normalisation makes sure there is only one representation for each number Exercise: Normalisation Questions Are the Following numbers normalised? 0.010000000 111111 Answer : No, as it starts with 0.0 0.111111000 111111 Answer : Yes, as it starts with 0.1 1.100000010 111111 Answer : No, as it starts with 1.1 Normalise the following numbers: 0 010000000 111111 Answer : 1. 0.010000000 111111 -> 00.10000000 111111 2. One place to the right 3. 111111 - 1 = -1 -1 = -2 = 000010 (+2) = 111110 (-2) 00.10000000 111110 = 0.100000000 111110 0 001101000 000110 Answer : 1. 0.001101000 000110 -> 000.1101000 000110 2. Two places to the right 3. 000110 - 2 = 6 - 2 = 4 = 000100 (+4) 000.1101000 000100 = 0.110100000 000100 1 111111010 000011 Answer : 1. 1.111111010 000011 -> 1111111.010 000011 2. Six places to the right 3. 000011 - 6 = 3 - 6 = -3 = 111101 (-3) 111111.010 111101 = 1.01000000 111101
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1. Top Ten 2. Popular Pet Names 3. Popular Animal Breeds dog Names: muggy Muggy is the number 100+ most popular dog name on Is your dog named Muggy? Sign up today and make some new friends! Back to Dog Names This is Muggy. He is a Boxer. He is very cool and calm. Gets hyper spells and runs around the house. He is a big man and weigh's 90 lbs! He is 3 yrs old now. And loves other animals.
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Your observation might be valid empirically without too many exceptions. (Thought there are corrupts democracies, too.) But it's rubbish as a definition. See it as (a very simplified model in) game theory: What is the interest of some game players, given their positions on the board? Here is an interesting take:
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Sentence Examples with the word BIPARTITE He held that the Son was a torch lighted at the torch of the Father, that Father and Son are a bipartite light. The weight of the function is bipartite and consists of the two numbers Ep and Eq; the symbolic expression of the symmetric function is a partition into biparts (multiparts) of the bipartite (multipartite) number Ep, Eq. View more Neither in the plan of Smith's university course nor in the wellknown passage at the end of his Moral Sentiments is there any indication of his having conceived such a bipartite scheme. Jonah (Abulwalid), which was cast in a similar bipartite form; and it was chiefly due to I imhi's grammar and lexicon that, while the contents of Abulwalid's works were common knowledge, they themselves remained in oblivion for centuries. In geometry, a bipartite curve consists of two distinct branches (see Parabola, figs.
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(mathematics) (Or "whole number") One of the numbers in the set There are an infinite number of integers, though each one is finite. An inductive definition of an integer is a number that is either zero or an integer plus or minus one. An integer has no fractional part. If written as a real number, e.g. 42.0, the part after the decimal point will be zero. A natural number is a non-negative integer. Computers usually store integers in binary. Natural numbers can be stored as unsigned integers and integers that may be negative require a sign bit and typically use twos complement representation. Other representations have been used, such as binary-coded decimal. Last updated: 2002-04-07
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Waste less time on Facebook — follow Brilliant. Absolute Value Absolute Value is the magnitude of a number without regard to its sign. In other words to take the absolute value of a number is to measure its distance from zero. Because distance is a positive measurement of course the values will be positive but this is in regards to distance and not direction. (In this sense absolute value is to relative value as speed is to velocity.) If we have a number line: The values of -10 and +10 are both 10 units from 0, although -10 is to the left and +10 is to the right. Using a number line gives a good visual for absolute value but can be time consuming. A more practical method to calculate absolute values would be squaring the number and then taking the square root. This works because squaring a negative will result in a positive value. Thus when you try to revert back to the root of the square, you cannot. You can only determine its absolute value. Yes, simply changing the sign to positive would also work. Find the absolute value of -11 \[(-11)^{2} = 121\] \[\sqrt{121} = 11\] This is my personal method for calculating absolute value. Although I must say it is a slippery slope and is too be used wisely because it doesn't work for many absolute value problems. Although it can be helpful at times such as inputting a formula for absolute value into a graphing calculator. Note by Brody Acquilano 1 year, 10 months ago No vote yet 1 vote There are no comments in this discussion. Problem Loading... Note Loading... Set Loading...
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시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율 4 초 128 MB 78 20 20 27.027% You are building advanced chips for machines. Making the chips is easy, but the power supply turns out to be an issue since the available batteries have varied power outputs. Consider the problem of n machines, each with two chips, where each chip is powered by k batteries. Surprisingly, it does not matter how much power each chip gets, but a machine works best when its two chips have power outputs as close as possible. The power output of a chip is simply the smallest power output of its k batteries. You have a stockpile of 2nk batteries that you want to assign to the chips. It might not be possible to allocate the batteries so that in every machine both chips have equal power outputs, but you want to allocate them so that the differences are as small as possible. To be precise, you want to tell your customers that in all machines the difference of power outputs of the two chips is at most d, and you want to make d as small as possible. To do this you must determine an optimal allocation of the batteries to the machines. Consider Sample Input 1. There are 2 machines, each requiring 3 batteries per chip, and a supply of batteries with power outputs 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. You can, for instance, assign the batteries with power outputs 1, 3, 5 to one chip, those with power 2, 4, 12 to the other chip of the same machine, those with power 6, 8, 9 to the third chip, and those with power 7, 10, 11 to the fourth. The power outputs of the chips are 1, 2, 6, and 7, respectively, and the difference between power outputs is 1 in both machines. Note that there are many other ways to achieve this result. The input consists of a single test case. A test case consists of two lines. The first line contains two positive integers: the number of machines n and the number of batteries per chip k (2nk ≤ 106). The second line contains 2nk integers pi specifying the power outputs of the batteries (1 ≤ pi ≤ 109). Display the smallest number d such that you can allocate the batteries so that the difference of power outputs of the two chips in each machine is at most d. 예제 입력 2 3 1 2 3 4 5 6 7 8 9 10 11 12 예제 출력
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Adds two numbers. This addition arithmetic operator also can add a number, in days, to a date. expression + expression Any valid expression in Microsoft SQL Server Compact of any of the data types in the numeric category, except the bit data type. Returns the data type of the argument with the higher precedence. In the following example, you want to create a target total, which is the current number of customers plus 50. The following example adds 50 to the total number of current customers. This result of the query in this example assumes that there are 90 current customers listed in the Customers table. SELECT COUNT(CustomerID) + 50 AS "Target Total" FROM Customers This is the result set: Target Total
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A loan of $375 helped to buy fertilizers for his crops. Roy's story Roy is a hardworking man. He is 40 years old and lives with his family in San Martin. Roy has five children who are his greatest source of pride. For this reason, he strives each day to provide them a better future. Roy has worked in agriculture for many years. He plants mainly corn during favorable seasons. Some time ago, he began dabbling in cacao crops and is doing very well. As a boy, Roy learned how to farm by helping his parents in the fields. He gained a great deal of knowledge from those experiences. In addition, he is passionate about cultivating different crops and is concerned about providing a quality product to his customers. Roy is grateful to the Kiva lenders for the support they have provided. He will be able to buy supplies so that he can fertilize his crops, which include the cacao plants that he cultivating in order to generate more income. As a result, his family’s quality of life will improve. Thanks to this support, Roy can continue to move forward so that he can become the most important producer in the area. Translator profile picture Translated from Spanish by Kiva volunteer Ronan Reodica Loan details Lenders and lending teams Loan details
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Notable Properties of Specific Numbers First page . . . Back to page 3 . . . Forward to page 5 . . . Last page (page 25) Magic Squares 15 is the magic constant for a 3×3 magic square, the smallest possible nontrivial magic square. An N×N magic square consists of the N2 integers from 1 to N arranged in an N×N square grid, such that the sum of any row, or any column, or of one of the two diagonals, is equal to any other. There is only one solution for a 3×3 square, not counting its rotations and reflections: 4 9 2 3 5 7 8 1 6 This magic square was known to the Chinese at least 3000 years ago and is called the Lo Shu. It appears in legends and artwork; for an example see here. As the magic constant of a magic square, 15 is a member of the sequence A6003 (my MCS26973): 1, 5, 15, 34, 65, 111, 175, 260, 369, .... For more about magic squares see the entries for 34, 38, 59, 65, 177, 199, 1514, 1665, 1035369, 1172421, and 276951438. Triangular Numbers 15 is the sum of the first 5 numbers: 1+2+3+4+5=15. Such numbers are called triangular because 15 things can be arranged in a triangular shape, by putting 1 object in a top "row", 2 in the 2nd row, and so on down to 5 in the 5th row. In the OEIS, such numbers are sequence A0217: 1, 3, 6, 10, 15, 21, 28, 35, 45, ... The general formula for the Nth triangular number is N(N+1)/2. Because either N or N+1 is even, we know that either N/2 or (N+1)/2 is an integer. It follows that every triangular number is the product of two integers — either N × (N+1)/2 or N/2 × (N+1) — and therefore, every triangular number above 3 is composite. The sum 1+2+3+4+5=15 is a little more significant because the first and last numbers, 1 and 5, are also the digits of the total. See 27. 15.154262241479... = ee On the x-y plane, the point where the line x=y crosses the curve connecting the nontrivial solutions of xy=yx is at x=y=e. At that point xy equals ee, and the slope of the xy=yx curve is -1. See also 3814279.10476024 and 1010101.0126×101656520. 15.438887358552... = (27/8)9/4 = (9/4)27/8 This is the value of xy for a non-trivial solution of xy=yx where x and y are both rational. First, the general solution: Define L to be the log base x of y: L = logxy then xy=yx becomes: xxL = (xL)x = xL x which (for x>1) reduces to: xL = L x xL-1 = L Solving for x, and substituting back for y, we get formulas for x and y in terms of L: x = L(1/(L-1)) y = xL = (L(1/(L-1)))L = L(L/(L-1)) This is the general solution for xy=yx with x not equal to y — use any value of L except L=1 and the formulas will give you values for x and y. We can make x and y both rational by making sure 1/(L-1) is an integer. That is true whenever L is of the form L=(n+1)/n where n is any nonzero integer. Then x = ((n+1)/n)n y = x(n+1)/n = ((n+1)/n)(n+1) Here are the first few rational (x,y) along with the value of xy: n L x y xy=yx 1 2 2 4 16 2 3/2 (3/2)2 = 9/4 (3/2)3 = 27/8 (27/8)9/4 =(3/2)(3×9/4) =(3/2)27/4 =(3/2)(2×27/8) =(9/4)27/8 =15.438887358552... 3 4/3 (4/3)3 = 64/27 (4/3)4 = 256/81 (256/81)64/27 =15.296931343617... . . . infinite 1 e e ee As n gets bigger, L gets closer to 1 and x and y get closer and closer to each other. For example, when n is 100, x=2.704813... and y=2.731861.... Both converge on e, because e = lim (1 + 1/n)n See also the solution of xy=x y, xy=xy=xy, 1.632526919438... and xx=10. 16 = 24 = 222 = 23 = (22)2 = 23, where and are the higher and lower hyper4 operators. The only non-trivial solution of xy=yx for integer x and y. (This was proved by Euler, and this entry describes a proof.) Base 16, called hexadecimal, is the most popular base among computer programmers for representing raw data in computer memory. The "digits" A, B, C, D, E and F are used to represent values of 10 through 15. So, for example, the hexadecimal number 6A16 is 6×16+10 = 106. Until the mid-1970's, base 8 was the most common base in computer programming applications. The primary reason base 16 overtook base 8 is that it uses 4 bits per digit and 4 itself is a power of 2. All the popular microprocessors, from the very early ones in the 1970's, have been based on a power of 2 bits per machine word. 17 is the only prime number that is the sum of four consecutive primes (2 + 3 + 5 + 7). 17 is a Fermat prime (a prime of the form 22N+1) and it is also the exponent of a Mersenne prime (a prime p for which 2p-1 is prime). The corresponding perfect number is 8589869056. There are 17 planar crystallographic groups. Peter, a film critic, enjoys going to eclectic cocktail parties to meet the directors and producers. He has noticed that for any two people in a party, they have always either A) met each other at the studio, B) met each other at the Academy Awards, or C) met each other here because they haven't met before the party. Peter got to wondering, how many people would need to be at a party to guarantee that there is at least one group of three people who have met each other in the same place (A B or C)? The answer is 17. 17 is the smallest number that can be written as A2 + B3 in two different ways: 17 = 32 + 23 = 42 + 13. By the way, the pair (8, 9) is the only pair of consecutive numbers where one is a square and the other is a cube (Euler proved this.) Catalan's conjecture (proven in 2002) states that 23 and 32 is the only pair of consecutive powers, aside from trivial cases where one of the numbers is a "1st power". Any convex polyhedron has at least one face that it can rest on without falling over (proof: if it didn't, it would be a perpetual-motion machine!). Most have more than one stable face. The minimum number of faces on a polyhedron that has only one stable face is 17. (The assumption is that the polyhedron is rigid, solid (not hollow) and of uniform density.) The reciprocal of 17, 1/17=0.05882352941176470588235..., has a 16-digit repeating decimal, which is the longest possible. This is a property that 17 shares with 7 and with many higher prime numbers. It results from the fact that, when you perform long division to compute 1/17, every remainder except 0 comes up exactly once. Also because of this, the multiples 2/17, 3/17, and so on have decimal fractions that use the same set of 16 digits, but starting in a different place (just as seen with 1/7, 2/7, 3/7 etc.). A reader named Jeremy pointed out to me that these 16 digits reduce to 9 using the casting out 9's technique, and thus 0588235294117647 is a multiple of 9. This holds for all reciprocals of primes (starting with 7) and is a consequence of Fermat's little theorem. cult numbers and psychologically random numbers These are two closely-related types of numbers. 17 belongs to both classes. A "cult" number has a "following", a group of "fans", many of whom have set up web pages for the number, a sort of virtual shrine to the number. Typically, a cult-number fan is someone who has one favourite number, and who delights in noticing that number, whenever it occurs in a place that seems to be more than just coincidence. I seem to have chosen 27, because I notice that number a lot more than I "should". Other cult numbers include 23, 37, 42, 47, 69, and 666; closely related to these are numbers that attract an abnormal amount of quasi-scientific interest, such as the fine structure constant. A "psychologically random" number, is one that "sounds random", or is chosen more often when someone is asked to pick a random number. 17 is the most often picked number in response to the request "Pick a random number from 1 to 20." Psychologically random numbers are used by writers when some large number is needed but no particular value is better than any other. For example, see 37 for some examples from movies. Psychologically random numbers are usually odd and don't end in 5, because there is a natural psychological bias to thinking even numbers and numbers that end in 5 are "less random". This means that most psychologically random numbers are prime. An extended argument along the same lines (attributed to Hilary Putnam by Mark Kalderon), adds that 7, 11 and 13 are nonrandom because of their lucky and unlucky associations, 9 is a square and 3 is "for the Trinity", leaving 17 as the only number less than 20 that isn't special. When a number is not consciously chosen but just happens by accident, it is more likely to be noticed and perceived as "more than just a coincidence" if the number is psychologically random. When such numbers are noticed repeatedly, they can then become cult numbers. This is why many cult numbers are also psychologically-random numbers. Here are some links to "cult number" websites: 17: A, B, C 23: A 27: A (formerly here), B, C 37: A Seventeen (XVII) is considered unlucky in Italy, perhaps because of the resemblance of the Roman numeral to Latin vixi, a euphemism for death. In this way the phobia resembles east-Asian tetraphobia. (the saros cycle) 18 years is the amount of time it takes for the tilt of the Moon's orbit to rotate a full-circle. Another way of saying the same thing is that the eclipse season slips back an entire year — so every 18 years (more precisely, 18 years 11 days), there have been 38=2×19 eclipse seasons. Viewed simply as a close match between a multiple of years (18) and eclipse seasons (38), it isn't such a big deal. It's 11 days off, and if you wait a year, you actually get a closer match (40 eclipse seasons is 8 days shorter than 19 tropical years). What makes the 18-year period so special is that the number of synodic months involved (223) is only 0.04 days short of 242 draconic months and only 0.2 days short of 239 anomalistic months. This means that the eclipses, in addition to being at the same time of year, also have the same positioning of Moon and Earth in the other two dimensions (distance and north-south positioning). On each eclipse, the Moon is the same distance away from Earth, and in the same position north-to-south, as it way 18 years 11 days previously. The distance governs whether solar eclipses are annular or total, and the north-south positioning determines where the shadow crosses for both types of eclipses (lunar and solar). This is the saros, an incredibly rare coincidence that makes eclipses easy to predict. It was discovered by the Babylonians and the knowledge was passed on to the Greeks and thence to later civilizations. The coincidence is not exact; each saros the Moon is a little further north (on a descending node) or south (on an ascending node) than the previous time, by a distance about equal to 1/70 of the Earth's diameter. So, each repeated eclipse repeats about 70 times, or about 70×18=1260 years. Also, the saros period is not an exact number of days; the Earth has turned about 1/3 of the way around, so the eclipses (particularly solar) are not seen by the same people. See also 161178. To test a number (example 660364) for divisibility by 19: The Metonic Cycle 19 is the number of years in the repeating pattern of a lunisolar calendar (such as the Hebrew calendar) designed to use months that stay in phase with the moon while also having the year stay in phase with the seasons. 19 tropical years is 6939.60160373 mean solar days, only 2 hours 5 minutes shorter than 235 synodic months (which is 6939.68838046 mean solar days). Due to this rather happy coincidence, the phases of the moon fall on the same dates every 19 years. It takes over 200 years for the error to get to be more than a day (however, in order to accomplish this accuracy, the lengths of at least one of the 235 months must differ from one 19-year cycle to another, see 6940). This cycle was known to several cultures at least as far back as the 4th century BCE. The period of 19 years also figures in the calculation of the date of Easter (but this is complicated by an additional multiple of 7 due to the fact that Easter must fall on a Sunday, see 133). Due to another amazing coincidence, 19 tropical years is also (within less than a day) equal to 255 draconic months, which means that eclipses also repeat every 19 years. (However, the pattern only repeats 4 or 5 times before an eclipse stops happening on a given date; the saros is a much better match) 19 is the numerical value of the Arabic word wahid ("one"), one of the names of God. 19 is considered sacred by the Baha'is, an Islamic sect, and they divided the year up into 19 "months" of 19 days each (this makes 192=361 days; to round out the year an intercalary period of 4 or 5 days is added)9. They also group years into 19-year and 361-year cycles (see Bahai calendar). Someone wrote to me pointing out that 19 has a "special" property: It is the sum of 9 and 10, and also the difference of their squares: 100 - 81 = 19. Of course, this property is true for any number that can be expressed in the form X + X + 1, which means pretty nearly any number (to prove, expand (X+1)2 and then subtract X2.). This is an example of the type of property that is often reported for cult numbers like 23 and psychologically random numbers like 37, because it is often desirable to find as many properties of such numbers as possible. eπ-π. See 23.140692... and xkcd 217 for details. Joerg Arndt came up with this pretty expression relating this number to 20: 20 = eπ - π + 1/(1111 + 1/(11 + 1/√2)) - ε where the "error" ε is about 1.21×10-12 (and noting that in handwriting a pair of 1's can be made to look a bit like π). See also 262537412640768743.999999.... aboutcontact mrob mrob27 @mrob_27 mrob27
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Where is Plutonium Found? Plutonium is a radioactive metal, named after the planet Pluto, and is essentially man made. Plutonium was discovered in 1940 after the bombardment of uranium-238 with neutrons in a device called a cyclotron. Plutonium-239 is one of very few materials whose atoms can be split causing an instant release of massive energy.
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Wheel Rim Codes TAKASAGO 185 X19 4K 303 First stamp is the manufacturer Takasago, then the rim size, then a number, then a letter, then 3 numbers. The number after the rim size is the year the rim was made. 2, 3, 4, 5 etc....197(2) 7(3) etc. The letter denotes the month of manufacture.....A-L. Jan- Dec. The last 3 numbers tell you if the rim is H1 or H2 or other model. The H1's are 305 rear, H2 is 303, S3 front is 308, S3 rear 307. S1/2=303 fnt S1/2=307 rr S3=308 fnt S3=307 rr H1=305 rr H1/H2=303 fnt H2=303 rr
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Enumerable.Average<TSource> Method (IEnumerable<TSource>, Func<TSource, Nullable<Int64>>) Computes the average of a sequence of nullable Int64 values that are obtained by invoking a transform function on each element of the input sequence. Namespace: System.Linq Assembly: System.Core (in System.Core.dll) public static Nullable<double> Average<TSource>( this IEnumerable<TSource> source, Func<TSource, Nullable<long>> selector Type Parameters The type of the elements of source. Type: System.Collections.Generic.IEnumerable<TSource> A sequence of values to calculate the average of. Type: System.Func<TSource, Nullable<Int64>> A transform function to apply to each element. Return Value Type: System.Nullable<Double> The average of the sequence of values, or null if the source sequence is empty or contains only values that are null. Usage Note The following code example demonstrates how to use Average<TSource>(IEnumerable<TSource>, Func<TSource, Int64>) to calculate an average. string[] numbers = { "10007", "37", "299846234235" }; double average = numbers.Average(num => Convert.ToInt64(num)); outputBlock.Text += String.Format("The average is {0}.", average) + "\n"; // This code produces the following output: // The average is 99948748093. Supported in: 5, 4, 3 Silverlight for Windows Phone Supported in: Windows Phone OS 7.1, Windows Phone OS 7.0 XNA Framework Supported in: Xbox 360, Windows Phone OS 7.0 Community Additions
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How does one play the dice game Bones? Quick Answer The object of Bones is to accumulate 10,000 points by throwing six dice, whose combinations earn a certain score. A straight (the same number on each of six dice) is worth 2,500 points, rolling five of a kind is worth 2,000 and rolling four of a kind is worth 1,500. Three of a kind is worth 1,000 points, while ones are worth 100 and fives count as 50. Continue Reading Full Answer The order of play is determined by a pre-game roll of five dice. Players have the option of re-rolling any dice that don't carry a score; only ones, threes, and fives count towards the point totals throughout the game. For example, if a player rolls three fives and three fours, only the fives count as 1,000 points earned for a three-of-a-kind roll. The player with the highest number of points is the one who goes first when starting the game. If a player ends up making a roll with no counting dice, he loses all points for that turn. If a player rolls five of a kind on his first roll, he scores 10,000 points (which is called a "bones"). Unless the other players can do the same, the first player wins the game automatically. To get on the scoreboard, a player must choose a minimum number of points and then reach that number on his first turn. Beginners are advised to choose 50 so that any score admits them into the game, but 500 is the common minimum. Learn more about Contests & Gambling Related Questions
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Find Out Who Called You From 615-671-XXXX 615-671-XXXX is in Davidson County, TN in or around Nashville (37201) Reverse Phone Look up for Tennessee: (xxx)xxx-xxxx or xxx-xxx-xxxx With a massive list of numbers nationwide... the number of sequences is limitless So, if you have been searching for a specific phone number that starts with a 615-671 area code exchange, you you now can. With the help of, all you have to do to find information on a person with a 615-671 is enter the entire nine digit phone numberr into the provided search area. That is all you need to start your search. The days of trying to find background information from multiple sources are over.
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How To Set up a Purchase Order Number System Shipping company There are basically two methods of setting up a purchase order number system: Use pre-printed purchase order forms with sequential numbers. Or, use your basic bookkeeping software to print out purchase orders with your company purchase order number system. Purchase order numbers are extremely important and should be tied to original sales order numbers to maintain good order track records. A purchase order number system serves another purpose as a reference to your clients and vendors of any order shipment and it acts as a secondary order track to bookkeeping systems that receive your purchase orders. Here's how to come up with a numbering system for your purchase orders. For practical purposes in a business where numbered sales order forms are regularly used, it's easy to incorporate the sales order number as part of the purchase order track number. Simply add a digit number that indicates it's now a purchase order. As an example: If the sales order number is 1044, use a mathematics number that consistently indicates it's a purchase order, such as the number 2. The "1" would indicate it began as a sales order, the first step in the sales process. The number "2" indicates it's now Purchase Order No. 21044, the second step in the tracking process. This will only work if purchase orders are not preprinted with a number system and can be added when printing from bookkeeping software. The best and most accurate systems for all business transactions are those that utilize consistent number tracking. In some companies, each customer is given an account number or special customer ID number. If purchasing in-house supplies from vendors, each vendor can also have an account number that helps create a purchase order number system. If purchases are made from a secondary vendor for a client, tie the client's purchase order number to all future bookkeeping transactions. For example, you, the vendor, purchase a widget for a client from another vendor. Your purchase order to your outside vendor has your company purchase order track number that will be used when the widget is shipped to you or to your client, whichever happens to be the case. All shipping documentation relies on your company purchase order number for identification. This method of ordering tracking is crucial when shipping internationally. Start your company purchase order number system with a relevant and easily identifiable number code. Many companies identify their clients with a customer code of numbers. This code may include a salesman's number, date of order and sequential number of the purchases. Share this article! Follow us! Find more helpful articles:
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Connect Dots (1 - 9) In this connect the dots worksheet, students use their counting skills to connect the dots numbered 1-9 in order to complete a picture. Pre-K - 1st Math 3 Views 0 Downloads Resource Details 1 more... Resource Types 2 more... What Members Say Kathy S. Kathy S. Norwood, NJ
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Free online games for girls Anna Secret Kiss Anna Secret Kiss Title: Anna Secret Kiss Description: Anna and Kristoff love each other very much, but Elsa wouldn't like to see them together. They can't do anything but hiding from Elsa, so will you help them? When they are kissing, warn Anna and Kristoff whenever Elsa or Olaff are getting close, so they can live a beautiful, secret love story. Enjoy! Categories: Love Games,
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This is machine translation Translated by Microsoft Class: opc.hda.Data Package: opc.hda Convert OPC HDA data object array to uint8 matrix V = uint8(DObj) V = uint8(DObj) converts the OPC HDA data object array DObj into an uint8 matrix. V is constructed as an M-by-N array of uint8 values, where M is the number of items in DObj and N is the number of time stamps in the array. DObj must have the same time stamps for each of the item IDs (elements of DObj), othewise an error is generated. Use tsunion, tsintersect, or resample to generate an OPC HDA data object containing the same time stamp for all items in the object. Load the OPC HDA example data file, convert the hdaDataSmall object to have the same time stamps, and create an uint8 matrix from the result: load opcdemoHDAData; dUnion = tsunion(hdaDataSmall); vUInt8 = uint8(dUnion); Was this topic helpful?
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We gratefully acknowledge support from the Simons Foundation and member institutions Full-text links: Current browse context: Change to browse by: References & Citations (what is this?) Computer Science > Data Structures and Algorithms Title: There is no 16-Clue Sudoku: Solving the Sudoku Minimum Number of Clues Problem Abstract: We apply our new hitting set enumeration algorithm to solve the sudoku minimum number of clues problem, which is the following question: What is the smallest number of clues (givens) that a sudoku puzzle may have? It was conjectured that the answer is 17. We have performed an exhaustive search for a 16-clue sudoku puzzle, and we did not find one, thereby proving that the answer is indeed 17. This article describes our method and the actual search. The hitting set problem is computationally hard; it is one of Karp's twenty-one classic NP-complete problems. We have designed a new algorithm that allows us to efficiently enumerate hitting sets of a suitable size. Hitting set problems have applications in many areas of science, such as bioinformatics and software testing. Comments: 36 pages Subjects: Data Structures and Algorithms (cs.DS) Cite as: arXiv:1201.0749 [cs.DS] (or arXiv:1201.0749v1 [cs.DS] for this version) Submission history From: Gary McGuire [view email] [v1] Sun, 1 Jan 2012 19:04:10 GMT (28kb) [v2] Sun, 1 Sep 2013 16:43:56 GMT (321kb,D)
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Pieter Breed Pieter Breed is a person. He laments the fact that he needs to code in C++, VB and .NET technologies to be able to make a living in South Africa. He sometimes pops into #lisp and asks stupid questions under the nickname pwab. He has a webpage/blog too and can be reached at [email protected].
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The miniseries was five hours long including commercials, each hour dedicated to relating the events of an entire day. It ran for four days, the first episode detailing the events of the first two days. Overview from themoviedb.org
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Every student is certain to face the necessity to prepare a great number of the academic assignments of different types. In order to help you to gain an understanding of the different types of the academic written assignments, we offer you to get acquainted with a list of written assignments. In this list, you can find the definitions of the academic written assignments of the different types. Types of Written Assignments Hinterlasse eine Antwort
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Discrete Time Fourier Transform (DTFT) $\displaystyle X(\tilde{\omega}) \isdef \sum_{n=-\infty}^\infty x(n) e^{-j\tilde{\omega}n} where $ \tilde{\omega}\isdef \omega T\in[-\pi,\pi)$ denotes the continuous normalized radian frequency variable,B.1 and $ x(n)$ is the signal amplitude at sample number $ n$. The inverse DTFT is $\displaystyle x(n) = \frac{1}{2\pi}\int_{-\pi}^\pi X(\tilde{\omega}) e^{j\tilde{\omega}n} d\tilde{\omega} which can be derived in a manner analogous to the derivation of the inverse DFT (see Chapter 6). Instead of operating on sampled signals of length $ N$ (like the DFT), the DTFT operates on sampled signals $ x(n)$ defined over all integers $ n\in{\bf Z}$. As a result, the DTFT frequencies form a continuum. That is, the DTFT is a function of continuous frequency $ \tilde{\omega}\in[-\pi,\pi)$, while the DFT is a function of discrete frequency $ \omega_k$, $ k\in[0,N-1]$. The DFT frequencies $ \omega_k=2\pi k/N$, $ k=0,1,2,\ldots,N-1$, are given by the angles of $ N$ points uniformly distributed along the unit circle in the complex plane (see Fig.6.1). Thus, as $ N\to\infty$, a continuous frequency axis must result in the limit along the unit circle in the $ z$ plane. The axis is still finite in length, however, because the time domain remains sampled. Next Section: Fourier Transform (FT) and Inverse Previous Section: FFT Software
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Warner D. Miller was an amateur ethnographer, and artist. He also was a teacher on the Standing Rock, Cheyenne, and Rosebud Reservations in the first part of the twentieth century. Almost immediately, he began to record the stories he heard and events he witnessed. He was also an amateur artist, and often illustrated the legends he recounted.
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1911 Encyclopædia Britannica/Magic Square From Wikisource Jump to: navigation, search MAGIC SQUARE, a square divided into equal squares, like a chess-board, in each of which is placed one of a series of consecutive numbers from 1 up to the square of the number of cells in a side, in such a manner that the sum of the numbers in each row or column and in each diagonal is constant. From a very early period these squares engaged the attention of mathematicians, especially such as possessed a love of the marvellous, or sought to win for themselves a superstitious regard. They were then supposed to possess magical properties, and were worn, as in India at the present day, engraven in metal or stone, as amulets or talismans. According to the old astrologers, relations subsisted between these squares and the planets. In later times such squares ranked only as mathematical curiosities; till at last their mode of construction was systematically investigated. The earliest known writer on the subject was Emanuel Moscopulus, a Greek (4th or 5th century). Bernard Frenicle de Bessy constructed magic squares such that if one or more of the encircling bands of numbers be taken away the remaining central squares are still magical. Subsequently Poignard constructed squares with numbers in arithmetical progression, having the magical summations. FIG. 1. The later researches of Phillipe de la Hire, recorded in eral methods of construction. He has there collected the results of the labours of earlier pioneers; but the subject has now been fully systematized, and extended to cubes. Two interesting magical arrangements are said to have been given by Benjamin Franklin; these have been termed the "magic square of squares" and the "magic circle of circles." FIG. 2. The first (fig. 1) is a square divided into 256 squares, i.e. 16 squares along a side, in which are placed the numbers from 1 to 256. The chief properties of this square are (1) the sum of the 16 numbers in any row or column is 2056; (2) the sum of the 8 numbers in half of any row or column is 1028, i.e. one half of 2056; (3) the sum of the numbers in two half-diagonals equals 2056; (4) the sum of the four corner numbers of the great square and the four central numbers equals 1028; (5) the sum of the numbers in any 16 cells of the large square which themselves are disposed in a square is 2056. This square has other curious properties. The "magic circle of circles" (fig. 2) consists of eight annular rings and a central circle, each ring being divided into eight cells by radii drawn from the centre; there are therefore 65 cells. The number 12 is placed in the centre, and the consecutive numbers 13 to 75 are placed in the other cells. The properties of this figure include the following: (1) the sum of the eight numbers in any ring together with the central number 12 is 360, the number of degrees in a circle; (2) the sum of the eight numbers in any set of radial cells together with the central number is 360; (3) the sum of the numbers in any four adjoining cells, either annular, radial, or both radial and two annular, together with half the central number, is 180. Construction of Magic Squares.[edit] FIG. 3. A square of 5 (fig. 3) has adjoining it one of the eight equal squares by which any square may be conceived to be surrounded, each of which has two sides resting on adjoining squares, while four have sides resting on the surrounded square, and four meet it only at its four angles. 1, 2, 3 are placed along the path of a knight in chess; 4, along the same path, would fall in a cell of the outer square, and is placed instead in the corresponding cell of the original square; 5 then falls within the square. a, b, c, d are placed diagonally in the square; but e enters the outer square, and is removed thence to the same cell of the square it had left. α, β, γ, δ, ε pursue another regular course; and the diagram shows how that course is recorded in the square they have twice left. Whichever of the eight surrounding squares may be entered, the corresponding cell of the central square is taken instead. The 1, 2, 3, . . . ., a, b, c, . . . ., α, β, γ, . . . . are said to lie in "paths." Squares whose Roots are Odd.[edit] Figs 4, 5, and 6 exhibit one of the earliest methods of constructing magic squares. FIG. 4. FIG. 5. FIG. 6. Here the 3's in fig. 4 and 2's in fig. 5 are placed in opposite diagonals to secure the two diagonal summations; then each number in fig. 5 is multiplied by and added to that in the corresponding square in fig. 4, which gives the square of fig. 6. Figs. 7, 8 and 9 give De la Hire's method; the squares of figs. 7 and 8, being combined, give the magic square of fig. 9. C. G. Bachet arranged the numbers as in fig. 10, where there are three numbers in each of four surrounding squares; these being placed in the corresponding cells of the central square, the square of fig. 11 is formed. FIG. 7. FIG. 8. FIG. 9. He also constructed squares such that if one or more outer bands of numbers are removed the remaining central squares are magical. His method of forming them may be understood from a square of 5. Here each summation is 5×13; if therefore 13 is subtracted from each number, the summations will be zero, and the twenty-five cells will contain the series ±1, ±2, ±3, . . . . ±12, the odd cell having 0. The central square of 3 is formed with four of the twelve numbers with + and - signs and zero in the middle; the band is filled up with the rest, as in fig. 12; then, 13 being added in each cell, the magic square of fig. 23 is obtained. Squares whose Roots are Even.[edit] These were constructed in various ways, similar to that of 4 in figs. 24, 25 and 16. The numbers in fig. 15 being multiplied by 4, and the squares of figs. 14 and 15 being superimposed, give fig. 16. FIG. 10. FIG. 11. The application of this method to squares the half of whose roots are odd requires a complicated adjustment. Squares whose half root is a multiple of 4, and in which there are summations along all the diagonal paths, FIG. 10. FIG. 12. may be formed, by observing, as when the root is 4, that the series 1 to 16 may be changed into the series 15, 13, . . . . 3, 1, -1, -3, . . . . -13, -15, by multiplying each number by 2 and subtracting 17; FIG. 14. FIG. 15. FIG. 16. and, vice versa, by adding 17 to each of the latter, and dividing by 2. The diagonal summations of a square, filled as in fig. 27, make zero; FIG. 17. FIG. 18. FIG. 19. and, to obtain the same in the rows and columns, we must assign such values to the p's and q's as satisfy the equations p1+p2+a1+a2=0, p3+p4+a3+a4=0, p1+p3-a1-a3=0, and p2+p4-a2-a4=0, — a solution of which is readily obtained by inspection, as in fig. 18; this leads to the square, fig. 19. FIG. 20. When the root is 8, the upper four subsidiary rows may at once be written, as in fig, 20; then, if 6 be added to each, and the sums halved, the square is completed. In such squares as these, the two opposite squares about the same diagonal (except that of 4) may be turned through any number of right angles, in the same direction, without altering the summations. Nasik Squares.[edit] Squares that have many more summations than in rows, columns and diagonals were investigated by A. H. Frost (Cambridge Math. Jour., 1857), and called Nasik squares, from the town in India where he resided; and he extended the method to cubes, various sections of which have the same singular properties. In order to understand their construction it will be necessary to consider carefully fig. 21, which shows that, when the root is a prime, and not composite, number, as 7, eight letters a, b, . . . h may proceed from any, the same, cell, suppose that marked 0, each letter being repeated in the cells along different paths. These eight paths arc called "normal paths," their number being one more than the root. FIG. 21. Observe here that, excepting the cells from which any two letters start, they do not occupy again the same cell, and that two letters, starting from any two different cells along different paths, will appear together in one and only one cell. Hence, if p1 be placed in the cells of one of the n+1 normal paths, each of the remaining n normal paths will contain one, and only one, of these p1's. If now we fill each row with p2, p3, . . . pn in the same order, commencing from the p1 in that row, the p2's, p3's and pn's will lie each in a path similar to that of p1 and each of the n normal paths will contain one, and only one, of the letters p1, p2, p3, . . . pn, whose sum will be Σp. Similarly, if q1 be placed along any of the normal paths, different from that of the p's, and each row filled as above with the letters q2, q3, . . . qn, the sum of the q's along any normal path different from that of the q1 will be Σq. The n2e cells of the square will now be found to contain all the combinations of the p's and q's; and if the q's be multiplied by n, the p's made equal to 1, 2, . . . n, and the q's to 0, 1, 2,. . . (n—1) in any order, the Nasik square of n will be obtained, and the summations along, all the normal paths, except those traversed by the p's and q's, will be the constant Σnq + Σp. When the root is an odd composite number, as 9, 15, &c., it will be found that in some paths, different from the two along which the p1 and q1 were placed, instead of having each of the p's and q's, some will be wanting, while some are repeated. Thus, in the case of 9, the triplets, p1p4p7, p2p5p8, p3p6p9, and q1q4q7, q2q5q8, q3q6q9 occur, each triplet thrice, along paths whose summation should be —Σp 45 and Σr 36. But if we make p1, p2, . . . p9 = 1, 3, 6, 5, 4, 7, 9, 8, 2, and the r1, r2, . . . r9= 0, 2, 5, 4, 3, 6, 8, 7, 1, thrice each of the above sets of triplets will equal Σp and Σq respectively. FIG. 22. If now the q's are multiplied by 9, and added to the p's in their several cells, we shall have a Nasik square, with a constant summation along eight of its ten normal paths. In fig. 22 the numbers are in the nonary scale; that in the centre is the middle one of 1 to 92, and the sum of pair of numbers equidistant from and opposite to the central 45 is twice 45; and the sum of any number and the 8 numbers 3 from it, diagonally, and in its row and column, is the constant Nasical summation, e.g. 72 and 32, 22, 76, 77, 26, 37, 36, 27. The numbers in fig. 22 being kept in the nonary scale, it is not necessary to add any nine of them together in order to test the Nasical summation; for, taking the first column, the figures in the place of units are seen at once to form the series, 1, 2, 3, . . . .9, and those in the other place three triplets of 6, 1, 5. For the squares of 15 the p's and q's may be respectively 1, 2, 10, 8, 6, 14, 15, 11, 4, 13, 9, 7, 3, 12, 5, and 0, 1, 9, 7, 5, 13, 14, 10, 3, 12, 8, 6, 2, 11, 4, where five times the sum of every third number and three times the sum of every fifth number makes Σp and Σq; then, if the q's are multiplied by 15, and added to the p's, the Nasik square of 15 is obtained. FIG. 23. FIG. 24. When the root is the multiple of 4, the same process gives us, for the square of 4, fig. 23. Here the columns give Σp, but alternately 2q1, 2q3 and 2q2, 2q4; and the rows give Σq, but alternately 2p1, 2p3 and 2p2, 2p4; the diagonals giving Σp and Σq. If p1, p2, p3, p4 and q1, q2, q3, q4 be 1, 2, 4, 3, and 0, 1, 3, 2, we have the Nasik square of fig. 24. A square like this is engraved in the Sanskrit character on the gate of the fort of Gwalior, in India. The squares of higher multiples of 4 are readily obtained by a similar adjustment. Nasik Cubes[edit] A Nasik cube is composed of n3 small equal cubes, here called cubelets, in the centres of which the natural numbers from 1 to n3 are so placed that every section of the cube by planes to an edge has the properties of a Nasik square; also sections by planes perpendicular to a face, and passing through the cubelet centres of any path of Nasical summation in that face. Fig. 25 shows by dots the way in which these cubes are constructed. A dot is here placed on three faces of a cubelet at the corner, showing that this cubelet belongs to each of the faces AOB, BOC, COA, of the cube. Dots are placed on the cubelets of some path of AOB (here the knight's path), beginning from O, also on the cubelets of a knight's path in BOC. Dots are now placed in the cubelets of similar paths to that on BOC in the other six sections parallel to BOC, starting from their dots in AOB. Forty-nine of the three hundred and forty-three cubelets will now contain a dot; and it will be observed that the dots in sections perpendicular to BO have arranged themselves in similar paths. FIG. 25. - Nasik Cube In this manner, p1, q1, r1 being placed in the corner cubelet O, these letters are severally placed in the cubelets of three different paths of AOB, and again along any similar paths in the seven sections perpendicular to AO, starting from the letters' position in AOB. Next, p2q2r2, p3q3r3, . . . p7q7r7 are placed in the other cubelets of the edge AO, and dispersed in the same manner as p1q1r1. Every cubelet will then be found to contain a different combination of the p's, q's and r's. If therefore the p's are made equal to 1, 2, 7, and the q's and r's to 0, 1, 2, . . . 6, in any order, and the q's multiplied by 7, and the r's by 72, then, as in the case of the squares, the 73 cubelets will contain the numbers from 1 to 73, and the Nasical summations will be Σ72r+Σ7q+p. If 2, 4, 5 be values of r, p, q, the number for that cubelet is written 245 in the septenary scale, and if all the cubelet numbers are kept thus, the paths along which summations are found can be seen without adding, as the seven numbers would contain 1, 2, 3, . . . 7 in the unit place, and 0, 1, 2, . . . 6 in each of the other places. In all Nasik cubes, if such values are given to the letters on the central cubelet that the number is the middle one of the series 1 to n3, the sum of all the pairs of numbers opposite to and equidistant from the middle number is the double of it. FIG. 26. Also, if around a Nasik cube the twenty-six surrounding equal cubes be placed with their cells filled with the same numbers, and their corresponding faces looking the same way, — and if the surrounding space be conceived thus filled with similar cubes, and a straight line of unlimited length be drawn through any two cubelet centres, one in each of any two cubes, — the numbers along that line will be found to recur in groups of seven, which (except in the three cases where the same p, q or r recur in the group) together make the Nasical summation of the cube. Further, if we take n similarly filled Nasik cubes of n, n new letters, s1, s2, . . . sn can be so placed, one in each of the n4 cubelets of this group of n cubes, that each shall contain a different combination of the p's, q's, r's and s's. This is done by placing s1 on each of the n2 cubelets of the first cube that contain p1 and on the n2 cubelets of the 2d, 3d, . . . and nth cube that contain s2, s3, . . . sn, respectively. FIG. 27. FIG. 28. This process is repeated with s2, beginning with the cube at which we ended, and so on with the other s's; the n4 cubelets, after multiplying the q's, r's and s's by n, n2, and n3 respectively, will now be filled with the numbers from 1 to n4, and the constant summation will be Σn3s + Σn2r + Σnq + Σp. This process may be carried on without limit; for, if the n cubes are placed in a row with their faces resting on each other, and the corresponding faces looking the same way, n such parallelepipeds might be put side by side, and the n5 cubelets of this solid square be Nasically filled by the introduction of a new letter t; while, by introducing another letter, the n6 cubelets of the compound cube of n3 Nasik cubes might be filled by the numbers from 1 to n6, and so ad infinitum. When the root is an odd composite number the values of the three groups of letters have to be adjusted as in squares, also in cubes of an even root. A similar process enables us to place successive numbers in the cells of several equal squares in which the Nasical summations are the same in each, as in fig. 26. Among the many ingenious squares given by various writers, this article may justly close with two by L. Euler, in the Histoire de l’académie royale des sciences (Berlin, 1759). In fig. 27 the natural numbers show the path of a knight that moves within an odd square in such a manner that the sum of pairs of numbers opposite to and equidistant from the middle figure is its double. In fig. 28 the knight returns to its starting cell in a square of 6, and the difference between the pairs of numbers opposite to and equidistant from the middle point is 18. A model consisting of seven Nasik cubes, constructed by A. H. Frost, is in the South Kensington Museum. The centres of the cubes are placed at equal distances in a straight line, the similar faces looking the same way in a plane parallel to that line. Each of the cubes has seven parallel glass plates, to which, on one side, the seven numbers in the septenary scale are fixed, and behind each, on the other side, its value in the common scale. 1201, the middle number from 1 to 74, occupies the central cubelet of the middle cube. Besides each cube having separately the same Nasical summation, this is also obtained by adding the numbers in any seven similarly situated cubelets, one in each cube. Also, the sum of all pairs of numbers, in a straight line, through the central cube of the system, equidistant from it, in whatever cubes they are, is twice 1201. (A. H. F.) Fennell's Magic Ring.[edit] It has been noticed that the numbers of magic squares, of which the extension by repeating the rows and columns of n numbers so as to form a square of 2n—1 sides yields n2 magic squares of n sides, are arranged as if they were all inscribed round a cylinder and also all inscribed on another cylinder at right angles to the first. C. A. M. Fennell explains this apparent anomaly by describing such magic squares as Mercator's projections, so to say, of "magic rings." The surface of these magic rings is symmetrically divided into n2 quadrangular compartments or cells by n equidistant zonal circles parallel to the circular axis of the ring and by n transverse circles which divide each of the n zones between any two neighbouring zonal circles into n equal quadrangular cells, while the zonal circles divide the sections between two neighbouring transverse circles into n unequal quadrangular cells. The diagonals of cells which follow each other passing once only through each zone and section, form similar and equal closed curves passing once quite round the circular axis of the ring and once quite round the centre of the ring. The position of each number is regarded as the intersection of two diagonals of its cell. The numbers are most easily seen if the smallest circle on the surface of the ring, which circle is concentric with the axis, be one of the zonal circles. In a perfect magic ring the sum of the numbers of the cells whose diagonals form any one of the 2n diagonal curves aforesaid is ½n(n2+1) with or without increment, i.e. is the same sum as that of the numbers in each zone and each transverse section. But if n be 3 or a multiple of 3, only from 2 to n of the diagonal curves carry the sum in question, so that the magic rings are imperfect; and any set of numbers which can be arranged to make a perfect magic ring or magic square can also make an imperfect magic ring, e.g. the set 1 to 16 if the numbers 1, 6, 11, 16 lie thus on a diagonal curve instead of in the order 1, 6, 16, 11. From a perfect magic ring of n2 cells containing one number each, n2 distinct magic squares can be read off; as the four numbers round each intersection of a zonal circle and a transverse circle constitute corner numbers of a magic square. The shape of a magic ring gives it the function of an indefinite extension in all directions of each of the aforesaid n2 magic squares. (C. A. M. F.) • F. E. A. Lucas, Récréations mathématiques (1891—1894) • W. W. R. Ball, Mathematical Recreations (1892) • W. E. M. G. Ahrens, Mathematische Unterhaltungen und Spiele (1901) • H. C. H. Schubert, Mathematische Mussestunden (1900) • A very detailed work is B. Violle, Traité complet des carrés magiques (3 vols., 1837—1838). • The theory of "path nasiks" is dealt with in a pamphlet by C. Planck (1906)
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5.7.2 All pairs Minimum Routes problem In this problem we want the minimum routes (m.r.) between all the pairs of the peaks of G. The Floyd algorithm solves this problem.This algorithm is an expansion of another algorithm,the Warshall algorithm,who was first defined for the solution of another problem: In a digraph G (whether there are costs or not,is of no importance) find whether there is a route from V(i) to V(j),for all pairs of (i,j) , i<>j. To solve this problem we find an array A.The elements of this array are A(i,j)=1 if there is a route from i to j,otherwise A(i,j)=0.Because the cost is not important we define the Adjoining Array as if all the costs were 1 that means C(i,j)=1 if there is eij belonging to E and otherwise C(i,j)=0.The requested array A is called transitive closure of the Adjoining Array. We notice that the elements of the A array are Boolean variables(0 or 1),which means that the operations AND and OR are valid.The Warshall algorithm initializes the A array at the value of C: A(i,j)=C(i,j), i,j=1,...,n At this point the A array shows only the direct connections as existing routes.Then the algorithm goes through the A array n times,one time for every node k=1,....,n.For every node V(k) the main thinking is : Is there a route from V(i) to V(j) ,if it has already been found {that is A(i,j)=1] or if a route is found through V(k),that is if the routes from V(i) to V(k) and from V(k) to V(j)[that is if A(i,k)=1 and A(k,j)=1]. If the BOOLEAN characteristics of the elements of A are taken under consideration,then the rule in the k pass is: A(i,j)=A(i,j) OR {A(i,k) AND A(k,j) }. Click WARSHALL for the Warshall Algorithm. We now come back to the m.r. problem for all pairs.This time we are talking about a graph,and the Adjoining Array is defined by the costs C(i,j)=c(eij).The A array will finally consist of all the costs of the minimum routes. During the k pass the following formula is valid: A(i,j)=min{A(i,j),A(i,k) + A(k,j) } which means that if the route through V(k) is cheaper will be the winner.That gives us the Floyd algorithm. Click FLOYD for the Floyd Algorithm. The complexity of the Floyd Algorithm is (in the worst case):O(n3). For the same problem the Dijkstra Algorithm could be used n times(every time another starting point).Because every use of the Adjoining Array demands complexity time of O(n2),the algorithm will solve the problem in time : O(n3).By use of other data structures the Dijkstra Algorithm will require O(mnlogn) and for great n with few sides,it is a much better algorithm. The replacement of "A(i,j):=s",with the: begin A(i,j):=s;P(i,j):=k end creates a new array with the values k(equivalent to n(i) in the Prim and Dijkstra Algorithm),which is the pointer of the peak through which the connection for the minimum route is made.The initialization is P(i,j)=0. HTML PAGE DIRECTOR :Papaioannou Panagiotis
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What do Prisoners do All Day? It will depend on the prison they are in but prisoners are given jobs they have to work at for a certain number of hours each day. They also will be given an hour or so outdoors in the yard.Many will do some studying They spend time eating meals and sleeping, but it still leaves them lots of time to think about what they did.
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Published 2002 in Proceedings, SCI (World Multiconference on Systemics, Cybernetics and Informatics), V. II, 318-325. Clustering algorithms have been widely applied to gene expression data. For both hierarchical and partitioning clustering algorithms, selecting the number of significant clusters is an important problem and many methods have been proposed. Existing methods for selecting the number of clusters tend to find only the global patterns in the data (e.g.: the over and under expressed genes). We have noted the need for a better method in the gene expression context, where small, biologically meaningful clusters can be difficult to identify. In this paper, we define a new criteria, Mean Split Silhouette (MSS), which is a measure of cluster heterogeneity. We propose to choose the number of clusters as the minimizer of MSS. In this way, the number of significant clusters is defined as that which produces the most homogeneous clusters. The power of this method compared to existing methods is demonstrated on simulated microarray data. The minimum MSS method is an example of a general approach that can be applied to any clustering routine with any global criteria. Microarrays | Multivariate Analysis | Statistical Methodology | Statistical Theory
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【poj1704】Georgia and Bob Georgia and Bob decide to play a self-invented game. They draw a row of grids on paper, number the grids from left to right by 1, 2, 3, …, and place N chessmen on different grids, as shown in the following figure for example: Georgia and Bob move the chessmen in turn. Every time a player will choose a chessman, and move it to the left without going over any other chessmen or across the left edge. The player can freely choose number of steps the chessman moves, with the constraint that the chessman must be moved at least ONE step and one grid can at most contains ONE single chessman. The player who cannot make a move loses the game. Georgia always plays first since “Lady first”. Suppose that Georgia and Bob both do their best in the game, i.e., if one of them knows a way to win the game, he or she will be able to carry it out. Given the initial positions of the n chessmen, can you predict who will finally win the game? The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case contains two lines. The first line consists of one integer N (1 <= N <= 1000), indicating the number of chessmen. The second line contains N different integers P1, P2 … Pn (1 <= Pi <= 10000), which are the initial positions of the n chessmen. For each test case, prints a single line, “Georgia will win”, if Georgia will win the game; “Bob will win”, if Bob will win the game; otherwise ‘Not sure’. Sample Input Sample Output
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Hyperdimension Neptunia Victory Hyperdimension Neptunia Victory Tips Chapter 4+: Quick Money From chapter 4 on-wards, you have a very quick and easy way to make a lot of credits in little to no time. During chapter 4 a new area called "Hello Continent" appears, which only has one dungeon called "Suaho Mountain Range". In here you can find an enemy called Pinky which drops Invisible Cloth. Collect as many of these items as you can. Once you've had your fill, go back to town, and purchase a dress from the shop called "Monochrome" which should cost you 5000C. Buy as many of these as you have Invisible Cloths You can now go into Item Creation menu and make the Monochrome Style dress. (Make as many as you can). The Monochrome Style dress has a purchase price of 2,500,000C from the store once it's made, and therefore the sale price of each one of these is 1,250,000C. Therefore by simply collecting the Invisible Cloths, you are in fact making 1,245,000C, which is a very quick and easy way to make money FAST!
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ABC Conjecture September 18, 2012 The ABC Conjecture has recently been in the news on math blogs because of the claim that it has been proved by Shinichi Mochizuki. Though the proof is being taken seriously, due to Mochizuki’s reputation, it is five hundred pages long, and confirmation will take several months. The conjecture states that given two positive integers a and b and their sum c with no common factors, the product of the distinct prime factors of abc is rarely much smaller than c. The radical of a number n is the product of the distinct prime factors of n; for instance, rad(18) = 6 because 18 = 2 × 3 × 3 and, eliminating the duplicate occurrence of 3, 2 × 3 = 6. The quality of an (a,b,c) triple is given by q(a,b,c) = log(c) / log(rad(abc)). The precise statement of the ABC conjecture is that for every ε > 0 there are only finitely many triples (a,b,c) with a and b coprime positive integers and a + b = c such that rad(abc)1+ε < c, or equivalently, such that q(a,b,c) > 1. Your task is to write the functions rad and q and find all the triples with c < 1000 for which q(a,b,c) > 1. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below. Pages: 1 2
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Classify Triangles: Practice For this classifying triangles worksheet, students name given triangles, then use given triangles to solve 5 additional problems. Houghton Mifflin text is referenced. 3rd - 5th Math 14 Views 81 Downloads Resource Details Arithmetic & Pre-Algebra 1 more... Resource Types Problem Solving 1 more... What Members Say Kathy S. Kathy S. Norwood, NJ
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Dismiss Notice Dismiss Notice Join Physics Forums Today! Proving a polynomial is constant 1. Nov 16, 2007 #1 Can someone give me a hint on problem 5: 2. Relevant equations 3. The attempt at a solution 2. jcsd 3. Nov 16, 2007 #2 what have you tried? Last edited: Nov 16, 2007 4. Nov 16, 2007 #3 It is a nice problem. First, you need to know that is a polynomial of degree n takes the same value (n+1) times then it must be be a constant polynomial. The problem says P(x) is a polynomial with integer coefficients so it means P(x) is an integer whenever x is an integer. We know that |P(x)|<n^2 whenever |x|<n so it means for x = -(n-1),...,-1,0,1,...,(n-1) we get P(x) = -(n^2-1),...,-1,0,1,...,(n^2-1). Let P(x) be the "pigeons" and x be the "x". By strong pigeonhole at least n+1 of the pigeons end up in the same hole. So it must be constant. 5. Nov 17, 2007 #4 User Avatar Science Advisor Homework Helper P(x) = x n=1, and |P(x)|<1 whenever |x|<1^2. But P(x) is not constant. 6. Nov 17, 2007 #5 First of all it should be |P(x)|<n whenever |x|<n^2 and secondly I get that there are 2n^2-1 pigeons flying into n-1 pigeon-holes? I have never heard of the strong pigeonhole principle but I do not see why one of the pigeonholes needs to get n+1 pigeons? 7. Nov 17, 2007 #6 User Avatar Staff Emeritus Science Advisor Gold Member If each pigeonhole has at most n pigeons, then at most how many pigeons are there? 8. Nov 17, 2007 #7 Less than or equal to n times the number of pigeon-holes. In our case, there must be less than or equal to 2n^2 - n total pigeons, which is a contradiction when n is greater than 1. So that explains morphism's counterexample. But what exactly is the strong pigeon-hole principle? How is it different than "if kn+1 pigeon fly into n pigeonholes, than one of the pigeonholes gets at least k+1 pigeons"? Can you instead show that ceiling( (2n^2-1)/(2n-1)) is greater than or equal to n+1 somehow? Last edited: Nov 17, 2007 9. Nov 17, 2007 #8 I did the problem backward, I am sorry. But the way it should be as Ehrenfest posted is correct if you apply my argument. Okay, it is very simple. If you have 6 pigeonholes and 32 pigeons then there is a pigeonhole that has at least 6 pigeons. In general given h pigeonholes and p pigeons then the number is [n/p] where [ ] here is the ceiling function. 10. Nov 17, 2007 #9 That is what I would just call the normal pigeonhole principle. Is there a reason you called is "strong"? 11. Nov 17, 2007 #10 I call the "basic" pigeonhole to be the one that says that there exists at least one hole having two pigeons. The "strong" one is the generalized argument. I am not sure if that is how it is officially called but that is how I refer to it. Here is a problem to try for you to solve: "Let S be the set {2,3,....,100} what is the largest subset that can be chosen of non-prime numbers so that all are pairwise co-prime?" (Here, 'largest' means the largest number of elements in a set). 12. Nov 17, 2007 #11 That's hard. So the pigeon-holes would have to be the number of primes less than 100. And the pigeons would have to be the elements of S. But the problem is pigeon's can fly into multiple holes! 13. Nov 18, 2007 #12 User Avatar Science Advisor Homework Helper Am I missing something, or isn't P(x)=x a counterexample to the original problem? 14. Nov 18, 2007 #13 It is a counterexample. See post #7. 15. Nov 21, 2007 #14 @ehrenfest. It really is not so hard. If n is a number in {2,3,...,100} it not a prime number then we can write n = p*m where p is a prime number. So for any n there exists a smallest possible prime divisor. Given any n the smallest prime divisor is 7 because it cannot be 11 because if it were its other prime divisor (which it must have because the number is not prime) must be at least 11 again but then 11*11=121 > 100 which is too large. So 7 is the smallest prime divisor of n. So the smallest prime divisors of n can be: 2,3,5,7. That means if you have 5 (non-prime) numbers by pigeonhole it means two of them share a prime factor so they are not co-prime. That means if you have at least 5 non-prime numbers then they all cannot be pairwise coprime. That means 4 is the largest possible subset, i.e. {4,9,25,49} Have something to add?
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运行式和表达式 - 语言基础 - MQL4参考 MQL4参考 语言基础 运行式和表达式 Operations and Expressions Some characters and character sequences are of a special importance. These are so-called operation symbols, for example: + - * / % Symbols of arithmetic operations && || Symbols of logical operations = += *= Characters assignment operators Operation symbols are used in expressions and have sense when appropriate operands are given to them. Punctuation marks are emphasized, as well. These are parentheses, braces, comma, colon, and semicolon. Operation symbols, punctuation marks, and spaces are used to separate language elements from each other. This section contains the description of the following topics:
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English version stringed instrument in Music topic From Longman Dictionary of Contemporary Englishldoce_322_fstringed instrumentˌstringed ˈinstrument noun [countable] APMa musical instrument such as a violin, that produces sound from a set of strings Examples from the Corpus stringed instrumentWriting music, the bottom line is that it's a stringed instrument but rhythmically you are playing a keyboard.Flittern Rattletrap hammered the strings of a low-throated stringed instrument, his feet stamping time.
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Quick Reference The symbols used to enclose a group of symbols or numbers that are to be taken together. Brackets can be nested so that the whole contents of the inner bracket is treated as a single term in the larger bracket. Brackets can be used to change the order in which operations are to be done. For example, 7×2+3 will be 14+3=17. If the sum intended is 7×5=35 then brackets can be put round 2+3 so that the sum is taken before the product, i.e. 7×(2+3). Subjects: Mathematics. Reference entries
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You may also like problem icon Equation Matcher Can you match these equations to these graphs? problem icon Curve Fitter Can you fit a cubic equation to this graph? problem icon Guess the Function Real-life Equations Stage: 5 Challenge Level: Challenge Level:1 This is a list of many of the most important equations in science. In each case, we have labelled the two variable quantities $x$ and $y$. The letters $a, b$ stand for constants in each case Constant motion $\quad\quad\quad\quad\quad a = \frac{x}{y}$ Constant acceleration $\quad\quad\quad x = uy + \frac{1}{2} ay^2$ Beer Lambert Law $\quad\quad\quad\quad a=bxy$ Exponential decay $\quad\quad\quad\quad x=a e^{by}$ Michaelis-Menton $\quad\quad\quad\quad x = \frac{ay}{b+y}$ pH $\quad\quad\quad\quad\quad\quad\quad\quad\quad x = -\log_{10}(y)$ Can you identify the possible meanings of the variables $x$ and $y$ and the constants in each case? Four graphs are shown above, where the two axes intersect at the origin $(0, 0)$. The red crosses show four measurements. Although we do not know the numerical values (because there are no scales on the graphs), we can see whether the values are positive or negative in each variable. For example, the first measurement is positive in $x$ and positive in $y$; the second measurement is positive in $y$, negative in $x$. For processes evolving according to each of the equations above, which measurements are possible?
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Challenge Discussion Sheet 6 Students are challenged to determine values that are guaranteed by the mean value theorem. They sketch the graph of a function and show that it satisfies the conditions of the mean value theorem over a given interval. Students determine the domain of each of five functions and identify the relative and absolute extrema, inflection points, intercepts, and asymptotes. They find the volume of a hemisphere as a function of a cylinder that it's sitting on. 12th - Higher Ed Math 3 Views 3 Downloads Resource Details 2 more... Resource Types 2 more... What Members Say Linda D. Linda D. Floresville, TX
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Quick Reference A finger or toe. In the basic limb structure of terrestrial vertebrates there are five digits (see pentadactyl limb). This number is retained in humans and other primates, but in some other species the number of digits is reduced. Frogs, for example, have four fingers and five toes, and in ungulate (hooved) mammals, the digits are reduced and their tips are enclosed in horn, forming hooves. Subjects: Biological Sciences — Medicine and Health. Reference entries
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Back to Questions There are three prizes to be distributed among five students. If no students gets more than one prize, then this can be done in: $10$ ways $30$ ways $60$ ways $80$ ways Hide Ans Option(A) is correct $3$ prizes among $5$ students can be distributed in $^5C_3$ ways $= 10$ ways. Edit: Thank you Rakesh for explaining why ans is not $5 \times 4 \times 3 = 60$ and the assumption behind this solution. (5) Comment(s) Why the answer is not 60 As prize1 can be given in 5 ways prize2 in 4 ways prize3 in 3 ways so overall 60 ways While solving it is assumed that prizes are identical. Thus the problem reduces to choosing 3 students among 5 students (and there is only 1 way of distributing, since prizes are identical). Think about it and you will understand why it is true. Had the prizes been different, there would have been $3!$ ways of distributing it among the three students. Making total ways to be: $^5C_3 \times 3! = 10 \times 3 \times 2$ $=\textbf{ 60 ways}$ As all the prizes are same . So if any 3 students get these prizes no need to find the way he can get other prize. So if we consider 5 students as A,B,C,D,E and F. Then 3 prizes would be given to--> A B C ---1 way A B D ----2 ways A B E ----3 ways A B F ----4 ways B C D ----5 ways B C E ----6 ways B C F ----7 ways C D E ----8 ways C D F ----9 ways D E F ----10 ways NOTE:- We should not consider A B C as one way and B A C as other way, because prizes are same. As here no. of boys to select from 5 i.e 3 is equal to no. of prizes i.e 3 and we have to select not arrange after getting the prize so Combination will be used. By formula Total ways, $^5C_3=10$ ways. who said that the prizes are equal.........they may be 1st,2nd and 3rd prizes of running competetion.........i can assume like this also........so 60 should be correct from my perspective.........is there any notice that if nothing mentioned assume all are identical........???? One prize can be given 5 ways. Next, 2nd prize is given in 4 ways to any of the 4 students. 3rd prize can be given in 3 ways. In all, $5+4+3= 12$ ways,Again, 4 prizes can be given amongst the 3 students in 3! ways. Hence, total number of ways $=12*6=72$? Please clarify.
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Explore BrainMass Prove that R is reflexive. 8. Let R be a relation on a set S such that R is symmetric and transitive and for each x ε S there is an element y ε S such that x R y. Prove that R is an equivalence relation (i.e. prove that R is reflexive) Solution Summary Reflexiveness is proven. The response received a rating of "5" from the student who posted the question.
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Pets awaiting Adoption Juneau Gastineau Humane Society Shelter (907) 789-0260 Chevy is a 2-year-old, yellow lab mix, a neutered male. A housebroken housedog, he loves children of all ages. Chevy came from Washington with his family, but they couldn't find a place to live that would allow a dog. So now this friendly-family pet is looking for a new family. Neveah is mild-mannered, trusting and loves to sleep on her person's bed. A people-oriented, spayed female, Neveah would be an affectionate companion for an older person. She has short hair, black-and-brown tiger stripes, and a cute, one-of-a-kind corkscrew tail. Haines Animal Rescue Kennel (907) 766-3334 R2 is a chocolate lab Pitt-bull mix. He is two months old and is ready to find a permanent home. R2 is not fully potty trained, but he is very smart and in need of a good training family. He will be neutered soon. R2 came from a family of seven so he loves to play with other animals. Sid is a handsome neutered 10-month-old puppy who still looking for a new home. He is a very energetic and playful little guy. He loves the outdoors and is housebroken. He likes to sleep inside at night, but prefers being outside during the day. He is a very excitable dog that loves to walk, run, and play with other dogs and humans.
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Retention and Retrieval of Unidirectional, Paired-associate Verbal Input Mehmet OZCAN This study is designed to investigate whether the sequence of input is influential in the recollection of the retained information. Eighty first year students, 40 girls and 40 boys, studying in ELT department participated in the study. Three tasks were administered to the participants. In the first test three different lists of words and their counterparts are given. The participants were asked to learn the words in a column and their counterparts in the next one. They were informed that they were going to be given a test two days later. In the second test, administered two days later following the first one, they were given the lists as the sequence of the words was reversed. In the third test, the participants were given the words tea, window, spill and break as a free association test. While the failure in the recollection of paired-associate words in Forward Recall tests was low, it was significantly high in Backward Recollection tests. The findings show that the sequence of the input at the time of retention is crucial in the recollection of the information stored in memory. Full Text: Copyright © Canadian Center of Science and Education
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he Olsen Gang has finally made it. They are in Mallorca, having fulfilled their dream. Almost, that is. They don't have any money, so Egon has to open a safe at a restaurant to get some. As usual, however, Egon ends up in prision. When they return to Denmark, he has a new plan. Overview from themoviedb.org
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Quadratic equations, which are expressed in the form of `ax^2 +bx+c= 0` , where a does not equal 0, may have how many solutions? 1 Answer | Add Yours lemjay's profile pic lemjay | High School Teacher | (Level 2) Senior Educator Posted on The number of solutions an equation has depends on the degree of the polynomial. Note that degree refers to the highest exponent of the variable. If the degree is one, the maximum number of solutions is 1. If it is two, then the maximum number of solutions is 2. So in `ax^2 + bx + c = 0` , the equation can have two solutions. Or, it may only be one. Or none at all. We’ve answered 317,587 questions. We can answer yours, too. Ask a question
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Explore BrainMass If three parts are selected at random from the bin, what is the probability that exactly two are defective? I am having trouble understanding the correct equation to use and then how to use it? Please explain step by step. Solution Preview The probability equation is P(event happening)=The number of ways even can occur/The total number of possible outcomes In this case, the event is a defective part. So here we have: P (defective part)= number of ...
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Cruise Itinerary Details We're sorry. There are no cruises available for the requested dates. Viking Skadi Departing from: Amsterdam, Netherlands Ports of Call: View more 3.2 / 5 (5 reviews) Cruise Itinerary Day Ports of Call Arrival Departure 1 Amsterdam, Netherlands --- --- 2 Kinderdijk, Netherlands --- --- 3 Cologne, Germany --- --- 4 Koblenz, Germany --- --- 4 Cruise Rhine River --- --- 5 Miltenberg, Germany --- --- 6 Wurzburg, Germany --- --- 7 Bamberg, Germany --- --- 8 Nuremberg, Germany --- --- 9 Regensburg, Germany --- --- 10 Passau, Germany --- --- 11 Melk, Austria --- --- 11 Krems, Austria --- --- 12 Vienna, Austria --- --- 13 Bratislava, Slovakia --- --- 14 Budapest, Hungary --- --- 15 Budapest, Hungary --- --- • With only 95 cabins, intimate, elegant accommodations create a comfortable casual atmosphere similar to that of a boutique hotel. • Enjoy ever-changing views and a delicious five-course gourmet meal featuring regional specialties at the Restaurant. • Past guests loved the ease of stepping aboard and unpacking once, while enjoyhing impeccable service and expert guides who reveal hidden gems in each destination. • This innovative ship's "green" features include energy efficient hybrid engines, solar panels and an organic herb garden. Your Package Includes: View more Accessibility: Viking Skadi Cruise Deals by Email
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Calculate the first N emirp (prime, spelled backwards) numbers, where N is a positive number that the user provides as input. An Emirp is a prime number whose reversal is also a prime. For example, 17 is a prime and 71 is a prime, so 17 and 71 are emirps. Write a program that prints out the first N emirps, five on each line. For example: For this assignment, you are required to make use of 2 functions (which you must write). bool isPrime(int value); // Returns true if value is a prime number. int reverse (int value); // Returns the reverse of the value (i.e. if value is 35, returns 53). You should follow a top-down design, using these functions in conjunction with logic in main, to perform the computation of N emirps and print them out according to the screenshot above. The general outline for main would be as follows: Step 1: Ask user for positive number (input validation) Step 2: Initialize a variable Test to 2 Step 3: While # emirps found is less than the input: Call isPrime with Test, and call it again with reverse(Test). If both are prime, print and increment number of emirps found.
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ELA, Volume 8, pp. 140-157, December 2001, abstract. Minimal CP rank Naomi Shaked-Monderer For every completely positive matrix A, cp-rankA >= rankA. Let cp-rankG be the maximal cp-rank of a CP matrix realization of G. Then for every graph G on n vertices, cp-rankG >= n. In this paper the graphs G on n vertices for which equality holds in the last inequality, and graphs G such that cp-rankA = rankA for every CP matrix realization A of G, are characterized.
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LinkedIn Google + Facebook YouTube Twitter Build infinite skills with project-based training videos FileMaker - Practical Techniques Free Tutorial: Planning Your Solution Course: FileMaker - Practical Techniques Other Free Videos From The FileMaker - Practical Techniques Training Course FileMaker - Practical Techniques Video Title: About The Course Duration: 00:04:59 Video Title: About The Author Duration: 00:03:00 Video Title: How To Use The Tutorials Duration: 00:03:02 Video Title: Why A Contact Manager? Duration: 00:02:25 Video Title: Goal For This Tutorial Duration: 00:02:31 Video Title: Using The Included Working Files Duration: 00:01:13 Video Title: Data Modelling Duration: 00:05:48 Video Title: Entity Relationship Diagram 101 Duration: 00:04:41 Video Title: Example ERDs Duration: 00:04:54 Video Title: Contact Manager ERD Duration: 00:05:48 Video Title: Source Tables Duration: 00:01:52 Video Title: Housekeeping Fields Duration: 00:04:08 Video Title: Key Fields Duration: 00:04:59 Video Title: Data Entry Fields Duration: 00:02:49 Video Title: Relationships Duration: 00:03:50 Video Title: Themes Duration: 00:02:01 Video Title: Form Layout Duration: 00:05:14 Video Title: List Layout Duration: 00:04:10 Video Title: Harmonious Layouts Duration: 00:02:41
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note Marshall <blockquote><i> how I can convert the time format such as "2012-10-02T12:34:08.94Z" to a UNIX format?</blockquote></i><p> You cannot do that because the Unix time format is an integer seconds from the "epoch time". ".94" hundredth's seconds is meaningless.<p> A conversion from "2012-10-02 12:34:08Z" to a Unix time_t integer is possible.<p> It is possible to track higher resolutions that one second. But you need a different data structure than a time_t value.<p> I would look at Date::Time. 997304 997304
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handra Shekhar aka Chandu (Nitin) is a dare devil youth and is a fearless boy. Once he comes across a beautiful girl Nandu (Hansika Motwani), daughter of a dreaded factionalist Peddi Reddy (Suman). Soon he loses his heart. He starts teasing her and wants to prove that he is sincerely loving Nandu. He tells Nandu that he could do anything for her. Then the latter tells him that he should not follow her until she calls him. In order to grab her attention, he sends his father (Chandramohan) and mother (Pragati) to express his love and let her know that they too encourage him instead of trying to divert his attention from love. Later Nandu realises that she too is truly loving him. At this juncture, Jayaprakash Reddy (Jayaprakash Reddy) goes to Peddi Reddy for an alliance for his brother's son Veera Pratap Reddy (Saleem Panda). Peddi Reddy refuses to marry Nandu with Veera Pratap Reddy. Overview from themoviedb.org
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시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율 1 초 128 MB 0 0 0 0.000% For several years now, the Nordic Conference on Partitions and Combinatorics, NCPC, has had a growing number of participants. This year the organizing team is expecting an all time high record in the hundreds. Due to the politics of arranging this prestigious event, the conference site was decided a long time ago to be the Grand Hôtel in Stockholm. The hotel has two large dining halls, but unfortunately, each of these halls alone can only fit up to two thirds of the NCPC participants, so the participants are going to have to be divided in two groups. This constraint calls for some thinking on behalf of the organizing team for the conference dinner: could they come up with some division of the participants in two parts, none of which is larger than 2/3 of the entire group, meeting some witty division rule suitable for the occasion, which they could tell the participants for their amusement? After all, as long as there is some grand logic rule to which of the two dining halls you are being seated in, you (as a mathematician) would be happy! They thought for a while and came up with the following idea for the division: Is there a year Y and a division of the participants in two parts such that every pair in the first part met for the first time some time before year Y , and every pair in the second part met for the first time some time in or after year Y ? Now this clearly qualified as an appropriate rule to all of them, but the question was whether it would be possible? The first line of input contains an integer 4 ≤ n ≤ 400, the number of participants, and c, the number of known first encounters. The next c lines are each in the format a b y, meaning participant a and b (1 ≤ a < b ≤ n) met for the first time in year y (1948 ≤ y < 2008). No pair of participants will appear more than once on the list, and every pair of participants not in the list is assumed to have meet only now (in the year 2008). For each test case, either the smallest year Y such that it is possible to divide the participants in two parts, neither of which contains more than 2n/3 people, such that all people in the first part first met before year Y , and all people in the second part first met in or after year Y . If there is no such year, output the string ’Impossible’. 예제 입력 6 3 1 2 1970 3 4 1980 5 6 1990 예제 출력
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A loan of $550 helped to purchase bulls. Ozbekov's story Ozbekov A. is the resident of Balykchi, a nice city in Isykul region. A. is 45 years old and he is married. He has two sons, both of them are at school. In order to earn some money A. purchased cement in the factory and then sold it at a higher price near the home. In this way he earned 10000 soms per month. That profit made it possible for him to save money and purchase cattle. He breeds cattle and then resells them after fattening at the higher price. With money received from the loan he wants to purchase two bulls for breeding. In the future he wants to rent a place and start selling building materials. Loan details Lenders and lending teams Loan details
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Dismiss Notice Join Physics Forums Today! Some of simple machines 1. Jul 18, 2009 #1 The simple pulley :- A simple pulley consists essentially of a wheel with a rope round it which it contained in a suspended framework . suppose W is the load or weight in lb.wt. attached to one end of the rope , and P is the effort in lb.wt. applied downwards at the other end which can just raise W . if the rope is very light and there is no friction round the wheel , the tension in the rope is W lb.wt. , and P must hence equal W . however , there is considerably less strain in pulling the rope of a pulley downward to lift heavy loads than in raising them to the same height by a direct upward force , and the hauler can also use his own weight in this case . The mechanical advantage (M.A.) of a machine is defined as the ratio (load / effort) = W/P . we shall see later that the load raised in a practical system of pulleys is much greater than the effort , so that the mechanical advantage is much greater than one . in the case one of the simple pulley , however , W = P , neglecting friction ; hence the mechanical advantage , W/P , is one . The velocity ration(V.R.) of a machine is defined as the ratio r/s , where r,s are the distances moved by the effort and load respectively in the same time . in general , the effort (applied force) moves a much greater distance than the load when the latter is raised , and the velocity ratio is thus much greater than one . in the simple pulley , the distance (r) moved by the effort is equal to the distance (s) moved by the load , since the rope which raises the weight is also used to apply the effort . thus the velocity ratio , r/s , is one in this case . it should be carefully noted that the magnitude of the velocity ratio , unlike the mechanical advantage , is not affected by friction in the pulley system . The block and tackle :- In practice , pulleys are designed to provide a large mechanical advantage , so that a large load or weight can be raised with a small effort . this figure :- illustrates a useful system of pulleys , known as a block and tackle system , which contains tow sets of pulleys with one continuous rope round them . the load of weight W is attached to the lower set of pulleys , which is movable , while the upper set is supported from a beam and is fixed in position . assuming that the pulleys and rope are light , and that no friction is present , the tension in every part of the rope is equal to P , the applied effort . hence , since there are four portions of rope round the lower set of pulleys , which support W , it follows that 4P = W . thus the mechanical advantage , W/P , is 4 in this case . in practice the mechanical advantage is less than this figure , since there is friction between the rope and the pulleys , and , moreover , the pulleys have weight . for example , if the two lower pulleys in the figure above have a total weight of 50 lb.wt. , then 4P = W+50 , considering the equilibrium of the two lower pulleys . consequently 4P is greater in magnitude than W , so that W/P is less than 4 . to find the velocity ratio (V.R.) of the system of pulleys , suppose that the weight W is raised a distance x when the effort P is applied . if we imagine the lowest pulley A raised this distance from QR to PS (in the figure below :-) , the length of rope made available = PQ + SR = x + x = 2x . the upper pulley , also rises a distance x , so that the movement of this pulley also makes a length 2x of rope available . thus a total length , 4x , of rope slips round the pulleys when the load is raised a distance x , and hence 4x is the distance moved by the effort . then V.R. = distance moved by effort / distance moved by load in same time = 4x/x = 4 . this is the magnitude of the (V.R.) obtained even when friction and the weights of the pulleys are taken into account , since the distance moved by the effort must always be 4x when the load is raised a distance x . it should again be noted that the magnitude of the mechanical advantage is affected by friction and the weights of the pulleys . in general , the V.R. of this system of pulleys is n , where n is the total number of pulleys in the system . the mechanical advantage is also equal to n when friction and the weights of the pulleys are neglected . in the case of an odd number of pulleys , the upper fixed block has one more pulley in it than the lower movable . thus , if there were 5 pulleys , the upper fixed block in the first figure would have 3 pulleys , and the string would be connected to the lower limb at the lower block . this system of pulleys is used at railway stations and engineering works for hauling heavy loads . archimedean system of pulleys :- the basic form of another pulley system , sometimes known as the Archimedean or first system of pulleys , is shown in the figure(a) :- [IMG] the load W is attached to a pulley a A1 , which has a rope passing round it . one end of the rope is attached to a fixed point on a beam , while the downward effort P is applied to the other end of the rope with the aid of a small fixed pulley F . suppose that the pulleys are light and smooth , and that the rope is light . the tension in the rope is then equal to P lb.wt. the upward tensions in the two portions of the rope on either side of the pulley A1 support the load W , and hence W = 2P . the mechanical advantage is thus ideally 2 , although in practice it is less . the V.R. can be deduced by imagining W raised a distance y , when a length 2y of rope moves round the pulley A . the effort , P , thus moves a distance 2y , and hence the V.R. is 2y/y , or 2 . it will be noted that the fixed pulley F offers a convenient means of applying the force P in a downward direction ; as it is fixed , F is not a part of the pulley system , which comprises only A1 in this case . a more practical form of the same pulley system , often used by builders , is shown in figure (b) . in this case there are three movable pulleys , A1 , A2 , A3 , with a load of weight W attached to the lowest pulley . we shall assume the pulleys are very light and that friction is absent . the relation between the effort P and W can be found by noting that the tension in the rope round A1 is W/2 , since the tension in the two parts of the rope round A1 support W . the two parts of the rope round A2 must each supply a tension of ½ of W/2 , or W/4 , by considering the pulley A2 ; and , similarly , the tension in the rope supporting the pulley A3 must be ½ of W/4 , or W/8 . but the tension in the latter rope is P , the effort applied . thus P = W/8 . consequently a theoretical mechanical advantage of 8 is obtained with this pulley system . by considering the lowest pulley A1 and the attached weight W to be raised a vertical distance x , and then following the lengths of rope moving round A1 , A2 , A3 , it can be shown that the effort P moves a distance 8x . in general , the V.R. is 2^n , where n is the number of pulleys ; thus if n = 3 , as in the figure above , the V.R. is 2^3 , or 8 . the mechanical advantage is 2^n only if friction and the weights of the pulleys are neglected . sources :- 1- principles of physics . 2. jcsd 3. Jul 18, 2009 #2 User Avatar Staff Emeritus Science Advisor There are a lot of words here. Is there a question? 4. Jul 19, 2009 #3 no there is not a question. Have something to add? Similar Discussions: Some of simple machines 1. Running machines (Replies: 16) 2. Atwood's Machine (Replies: 10) 3. Carnot Machine (Replies: 1)
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Updated 2014-06-07 17:54:50 by pooryorick someone on the chat was asking for a topological sort algorithm. Here's one I have lying around, for what it's worth. # [TopologicalSort] takes an alternative (which is represented by a directed acyclic graph, DAG) # and returns a topological sort of it. A topological sort of a DAG is # a list of all the vertices of the DAG (here a list of package names), # such that if A -> B is in the DAG, then the index of A in the list is # less than the index of B in the list. In addition, when constructing # the list, this proc includes the corresponding requirement along with # each package name. Thus, the result is a requirement list, ordered so # that each requirement is listed after its prerequisites. # proc ::package::TopologicalSort alt { proc TopologicalSort alt { # Unpack the alternative data structure foreach {DAG MAP} $alt {break} array set dag [lindex $DAG 0] set max [lindex $DAG 1] array set map $MAP set answer {} foreach vertex $max { unset dag($vertex) while {[llength $max]} { set pkg [lindex $max 0] set max [lrange $max 1 end] set answer [linsert $answer 0 [ linsert [lindex $map($pkg) 1] 0 $pkg]] foreach vertex [array names dag] { set idx [lsearch -exact $dag($vertex) $pkg] set dag($vertex) [lreplace $dag($vertex) $idx $idx] if {[llength $dag($vertex)] == 0} { lappend max $vertex unset dag($vertex) set answer # Test: catch {console show} set Result {} lappend Topo foo 1 bar 2 puts "Topo: $Topo" set Result [ TopologicalSort $Topo ] puts $Result The above code may make little sense out of its context. For that, see [1]. Look in mkdepend.tcl. HJG: An example with some data would be nice. The above test gives an error "list must have an even number of elements..."
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\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage[latin1]{inputenc} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \begin{center} \vskip 1cm{\LARGE\bf The Number of Labelled $k$-Arch Graphs } \vskip 1cm \large C\'edric Lamathe\\ LORIA - INRIA Lorraine\\ Campus Scientifique, B.P. 239\\ 54506 Vandoeuvre-l\`es-Nancy Cedex, France\\ \href{mailto:[email protected]}{\tt [email protected]} \\ \end{center} \vskip .2 in \begin{abstract} In this note, we deal with $k$-arch graphs, a generalization of trees, which contain $k$-trees as a subclass. We show that the number of vertex-labelled $k$-arch graphs with $n$ vertices, for a fixed integer $k\geq 1$, is ${n\choose k}^{n-k-1}$. As far as we know, this is a new integer sequence. We establish this result with a one-to-one correspondence relating $k$-arch graphs and words whose letters are $k$-subsets of the vertex set. This bijection generalises the well-known Pr\"{u}fer code for trees. We also recover Cayley's formula $n^{n-2}$ that counts the number of labelled trees. \end{abstract} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}{Proposition}[section] \newtheorem{corollary}{Corollary}[section] \newtheorem{lemma}{Lemma}[section] %\usepackage[dvips]{epsfig} %\usepackage{color} %\usepackage{float} %%%%%%%%%%%%%%%%%%PERSONAL DEFINITIONS%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\R}{\mathbb R} \newcommand{\N}{\mathbb N} \newcommand{\Z}{\mathbb Z} \newcommand{\Q}{\mathbb Q} \newcommand{\K}{\mathbb K} \newcommand{\A}{\mathbb A} \newcommand{\Sym}{\mathbb S} %\theoremstyle{plain} %\theoremstyle{remark} \newtheorem{theo}{\bf Theorem} \newtheorem{de}{\bf Definition} \newtheorem{lem}{\bf Lemma} \newtheorem{cor}{\bf Corolary} \newtheorem{prop}{\bf Proposition} \newtheorem{rem}{\bf Remark} \newtheorem{ex}{\bf Example} \newenvironment{preuve}{\begin{trivlist}\item{\bf{Proof.}}} {\hfill\rule{2mm}{2mm}\end{trivlist}} \newcommand{\ark}{\mathcal{G}} \section{Introduction} We recursively define the class of $k$-{\em arch graphs}, for $k\geq 1$, as the smallest class of simple graphs such that: \begin{itemize} \item[1.] a $(k-1)$-simplex (i.e., a complete graph on $k$ vertices) is a $k$-arch graph; \item[2.] if a simple graph $G$ has a vertex $v$ of degree $k$ such that the graph $G-\lbrace v \rbrace $ obtained from $G$ by removing $v$ and its incident edges is a $k$-arch graph, then $G$ is a $k$-arch graph. \end{itemize} Figure~\ref{fig:arch} shows an unlabelled $2$-arch graph (we simply say {\em arch graph} in this case) and a vertex-labelled $2$-arch graph, each one built over 12 vertices. Note that, when $k=1$, $1$-arch graphs coincide with (Cayley) trees. In a constructive way, to build a $k$-arch graph of $n+1$ vertices from one on $n$ vertices, we have to choose $k$ vertices and join the new vertex to these selected vertices. The term {\it arch} evokes attaching the new vertex over the $k$ chosen vertices. % \begin{figure}[ht] \centerline{\includegraphics[width=.99\textwidth]{arch-unl-lab}} \caption{Arch graphs on 12 vertices a) unlabelled, b) vertex-labelled.} \label{fig:arch} \end{figure} % There are few papers about $k$-arch graphs, apart from \cite{T}, where these graphs seem to appear, and \cite{L}. However, there is an abundant literature about a subclass of $k$-arch graphs, called $k$-trees, studied since the later 1960's. The essential difference between $k$-trees and $k$-arch graphs lie in the fact that for $k$-trees, we assume that the vertex $v$ is attached to $k$ mutually adjacent vertices (that is, it forms a complete graph on $k+1$ vertices). For instance, see \cite{BP,M-69} for the labelled enumeration of $k$-trees, and \cite{BM,P} for the particular case $k=2$. Harary and Palmer \cite{HP} treat the unlabelled enumeration as well as Fowler et al. \cite{TF}, who provide in addition asymptotic formulas. We also mention \cite{LLL-JCT,BL} for the enumeration of generalizations of 2-trees and \cite{these}, where various specializations of 2-trees are enumerated. Finally, Labelle et al. \cite{LLL-TCS} propose a classification of outerplanar 2-trees according to their symmetries. Although $k$-trees have been extensively studied, unlabelled enumeration of these structures is still an open problem. Apart from enumerative aspects of 2-trees, Leclerc and Makarenkov \cite{LM} give a correspondence between tree functions and 2-trees in the framework of tree metrics and tree dissimilarities (see \cite{BG1,BG2}). In \cite{T,L}, $k$-arch graphs are defined as maximal $kd$-acyclic graphs, where a graph is said $kd$-acyclic, if it contains no $kd$-cycle (see \cite{L}, definition 3.1). Todd shows that this definition is equivalent to the recursive one given above. Leclerc uses valuated arch graphs (that is, arch graph with valuated edges) to encode tree distances or tree functions (see \cite{BG1,BG2}).\\ We recall that the number of labelled $k$-trees over $n$ vertices is given by (\cite{BP,M-69}) \begin{equation}\label{ktrees} a_n^k = {n\choose k}\left(k(n-k)+1\right)^{n-k-2}. \end{equation} When $k=1$, we recover the well-known Cayley formula $a_n=n^{n-2}$ counting labelled (Cayley) trees.\\ The aim of this note is to obtain the number of labelled $k$-arch graphs. We show this number is \[ {n\choose k}^{n-k-1}. \] To achieve our goal, we propose in Section 2 a one-to-one correspondence between $k$-arch graphs on $n$ vertices and words of length $n-k-1$ whose letters are $k$-subsets of the set of vertex labels. This correspondence generalizes the Pr\"{u}fer code for trees (\cite{Prufer}). %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{The number of labelled $k$-arch graphs} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% In this section, we establish a formula giving the number of labelled $k$-arch graphs, for $k\geq 1$. We prove this formula using a bijective argument based on a generalization of the Pr\"{u}fer code for vertex-labelled Cayley trees (\cite{Prufer}) which has been generalized for $k$-trees (\cite{RR}). Note that formula (\ref{number}) first appear (without proof) in the conclusion of \cite{these}. We call {\em leaf} of a $k$-arch graph, a vertex of degree $k$. For instance, in Figure~\ref{fig:arch} b), there are four leaves, respectively labelled 2,3,4,9. This definition of leaf is legitimate since, for the special case $k=1$, a vertex of degree one in a Cayley tree is a leaf, in the common sense of graph theory. % \medskip % \begin{prop}\label{prop} Let $k\geq 1$ be a fixed integer. Then, the number $\ark^k_n$ of $k$-arch graphs over $n$ labelled vertices, for $n>k$, is given by \begin{equation}\label{number} \ark_n^k = {n\choose k}^{n-k-1}\quad {\rm and}\qquad \ark_k^k=1. \end{equation} \end{prop} % \begin{preuve} We construct a one-to-one correspondence between $k$-arch graphs on $n$ labelled vertices and words $w=w_1w_2\ldots w_{n-k-1}$ of length $n-k-1$ where each $w_i$ is a (valid) $k$-letter of the following form: \begin{equation} \begin{pmatrix} v_{i_1}\\ v_{i_2}\\ \vdots \\ v_{i_k} \end{pmatrix} \end{equation} such that \begin{enumerate} \item for all $1\leq j\leq k$, $1\leq i_j\leq n$; \item for all $1\leq j\leq n$, $v_j$ is a vertex of the $k$-arch graph; \item $i_1
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Atomic Weight The atomic weight of an element (as it appears in the periodic table) is the weighted average of the atomic mass numbers of all of the isotopes for that element. For example, the atomic weight of carbon (as shown on the periodic table) is 12.011 a.m.u. (atomic mass units). However, there is no one carbon atom that has a mass of 12.011 a.m.u. Carbon exists as four different isotopes: carbon-11, carbon-12, carbon-13, and carbon-14, which have approximate atomic mass numbers of 11, 12, 13, and 14 a.m.u., respectively. If you know the percent abundance of each of those isotopes, you can calculate the atomic weight of carbon by determining the weighted average of the atomic mass numbers of the four isotopes. That weighted average comes to 12.011 a.m.u. The reason that the atomic weight is closer to 12 than it is to the other atomic mass numbers is that carbon-12 is the most common isotope of carbon. For most elements, the atomic mass number of the most common isotope for that element can be determined by rounding the atomic weight for that element to the nearest whole number. For example, the atomic weight of aluminum is 26.98154 a.m.u. Therefore, the most common isotope for aluminum can be assumed to be aluminum-27. © Copyright 2015 Kurt M. Wicks All Rights Reserved
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Sign in Chat Now 24/7 Examples and Samples Critical Essay on Shakespeare’s Macbeth Macbeth is considered to be one of the most unusual and mysterious tragedies of Shakespeare. Being one of the very few plays, which was never performed while Shakespeare was alive, and the shortest of his plays, it is notorious among actors as the most unlucky theatrical works. Another peculiarity of Macbeth is that it focuses on the theme of the evil, but with a peculiar difference. The main character, Macbeth, is wicked, no question, but he is not an ordinary villain. He knows too well the difference between the good and the evil. He realizes that having killed Duncan he violates moral laws, which he deeply respects. Not having committed the crime yet, he knows what horrible pains of remorse he is doomed to experience. But still he decides to commit a crime. His longing for power appears to be stronger than conscience and fear of moral scruples. Having committed the crime, Macbeth loses his inner peace completely. He stops believing in friendship, his soul is tortured by suspicions. As he feels that everyone around him is his enemy, he becomes cruel and meticulous. Macbeth achieves the long desired power, but he cannot feel satisfied with it. The people and the aristocracy hate him and he has to fight them. Although he knows that his life is his worst nightmare and a meaningless massacre, he does not surrender, deep down being a hero. Macbeth is an unusual story of a good, moral person committing a… We know how difficult it is to write an essay. Paul, CO 1 2 3 4 Steve, NJ 1 2 3 4 5 View All Testimonials Want to place an order via the phone? It's free
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Loading presentation... Present Remotely Send the link below via email or IM Present to your audience Start remote presentation • Invited audience members will follow you as you navigate and present • People invited to a presentation do not need a Prezi account • This link expires 10 minutes after you close the presentation • A maximum of 30 users can follow your presentation • Learn more about this feature in our knowledge base article Do you really want to delete this prezi? Make your likes visible on Facebook? You can change this under Settings & Account at any time. No, thanks Whats the secret code? No description Francis Lim on 3 October 2014 Comments (0) Please log in to add your comment. Report abuse Transcript of Whats the secret code? What's the secret code? Clue number 1 The number is between 8,500 and 8,800 so there is three hundred numbers. Therefore the number in the thousands place starts with an eight. Clue number 3 Question number 3 There are no zeros in the decimal places.This means there are also decimal places to worry about. Question number 4 The tens place is equivalent to the ones place and the hundreds place is three times the tens place. The answer for the new clue is 8,622.125. By : Carter, Hannah, Francis, and Mark. When multiplied by 8, the result is a whole number. In order for a decimal to be a whole number, there has to be a zero in the decimal. We need to create a zero by multiplying 8x5 to create 40. We can start with 8 x 0.5, 8x 0.25, or 8 x0.125. Clue number 4 Clue number 2 The digit in the hundreds place is 3/4 the digit in the thousands place, The digit in the thousands place is 8, and 3/4 of 8 is 6. So, the first two digits are 8 and 6. Clue number 5 The digit in the hundredths place is 200% of the digit in the tenths place. This means that the digit in the hundredths place is 2 times of the tenths place. Clue number 6 The sum of all the digits in the number is 26. Question number 2 What code numbers fit these clues? We knew that 300 numbers were in between 8,500 and 8,800. We had to find numbers or decimal numbers that when multiplied, the product was a whole number. Then we took 8 and found that 3/4 of it was 6. Since the total number had to equal 26 and the digit in the hundredths place is 2 times the digit in the tenths place, so the decimal place is 0.125. After we found most of the numbers and added them. The sum was 22. Then we subtracted 26 from 22 to get 4. So then we knew that the tens and ones place had to equal 4 when added. We found 5 answers: 8,640.125; 8,604.125; 8,613.125; 8,631.125; 8,622.125. Explain how you used all of these clues to find these possibilities Full transcript
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bridges Balibridges Bali However, if you are very enthusiastic to work for a retreat company, please send your latest resume to We will contact you once there is a post that suits your profile. bridges Bali team
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