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  • How to find the longest contiguous subsequence whose reverse is also a subsequence

    - by iecut
    Suppose I have a sequence x1,x2,x3.....xn, and I want to find the longest contiguous subsequence xi,xi+1,xi+2......xi+k, whose reverse is also a subsequence of the given sequence. And if there are multiple such subsequences, then I also have to find the first. ex:- consider the sequences: abcdefgedcg here i=3 and k=2 aabcdddd here i=5, k=3 I tried looking at the original longest common subsequence problem, but that is used to compare the two sequences to find the longest common subsequence.... but here is only one sequence from which we have to find the subsequences. Please let me know what is the best way to approach this problem, to find the optimal solution.

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  • Fastest way to calculate an X-bit bitmask?

    - by Virtlink
    I have been trying to solve this problem for a while, but couldn't with just integer arithmetic and bitwise operators. However, I think its possible and it should be fairly easy. What am I missing? The problem: to get an integer value of arbitrary length (this is not relevant to the problem) with it's X least significant bits sets to 1 and the rest to 0. For example, given the number 31, I need to get an integer value which equals 0x7FFFFFFF (31 least significant bits are 1 and the rest zeros). Of course, using a loop OR-ing a shifted 1 to an integer X times will do the job. But that's not the solution I'm looking for. It should be more in the direction of (X << Y - 1), thus using no loops.

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  • Chess algorithm

    - by Ockonal
    Hi guys, I want to create chess application without AI. I just need in checking available ways for chosen chess-object and checkmate for the king. What is the best way to implement this?

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  • Permutation algorithm without recursion? Java

    - by Andreas Hornig
    Hi, I would like to get all combination of a number without any repetation. Like 0.1.2, 0.2.1, 1.2.0, 1.0.2, 2.0.1, 2.1.0. I tried to find an easy scheme but couldn't find so I drawed a graph/tree for it and this screams to use recursion. But I would like to do it without, if this is possible. So could anyone please help me how to do that? Thank you in advance, Andreas

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  • How to use Boost 1.41.0 graph layout algorithmes

    - by daniil-k
    Hi I have problem using boost graph layout algorithmes. boost verision 1_41_0 mingw g++ 4.4.0. So there are issues I have encountered Can you suggest me with them? The function fruchterman_reingold_force_directed_layout isn't compiled. The kamada_kawai_spring_layout compiled but program crashed. Boost documentation to layout algorithms is wrong, sample to fruchterman_reingold_force_directed_layout isn't compiled. This is my example. To use function just uncomment one. String 60, 61, 63. #include <boost/config.hpp> #include <boost/graph/adjacency_list.hpp> #include <boost/graph/graph_utility.hpp> #include <boost/graph/simple_point.hpp> #include <boost/property_map/property_map.hpp> #include <boost/graph/circle_layout.hpp> #include <boost/graph/fruchterman_reingold.hpp> #include <boost/graph/kamada_kawai_spring_layout.hpp> #include <iostream> //typedef boost::square_topology<>::point_difference_type Point; typedef boost::square_topology<>::point_type Point; struct VertexProperties { std::size_t index; Point point; }; struct EdgeProperty { EdgeProperty(const std::size_t &w):weight(w) {} double weight; }; typedef boost::adjacency_list<boost::listS, boost::listS, boost::undirectedS, VertexProperties, EdgeProperty > Graph; typedef boost::property_map<Graph, std::size_t VertexProperties::*>::type VertexIndexPropertyMap; typedef boost::property_map<Graph, Point VertexProperties::*>::type PositionMap; typedef boost::property_map<Graph, double EdgeProperty::*>::type WeightPropertyMap; typedef boost::graph_traits<Graph>::vertex_descriptor VirtexDescriptor; int main() { Graph graph; VertexIndexPropertyMap vertexIdPropertyMap = boost::get(&VertexProperties::index, graph); for (int i = 0; i < 3; ++i) { VirtexDescriptor vd = boost::add_vertex(graph); vertexIdPropertyMap[vd] = i + 2; } boost::add_edge(boost::vertex(1, graph), boost::vertex(0, graph), EdgeProperty(5), graph); boost::add_edge(boost::vertex(2, graph), boost::vertex(0, graph), EdgeProperty(5), graph); std::cout << "Vertices\n"; boost::print_vertices(graph, vertexIdPropertyMap); std::cout << "Edges\n"; boost::print_edges(graph, vertexIdPropertyMap); PositionMap positionMap = boost::get(&VertexProperties::point, graph); WeightPropertyMap weightPropertyMap = boost::get(&EdgeProperty::weight, graph); boost::circle_graph_layout(graph, positionMap, 100); // boost::fruchterman_reingold_force_directed_layout(graph, positionMap, boost::square_topology<>()); boost::kamada_kawai_spring_layout(graph, positionMap, weightPropertyMap, boost::square_topology<>(), boost::side_length<double>(10), boost::layout_tolerance<>(), 1, vertexIdPropertyMap); std::cout << "Coordinates\n"; boost::graph_traits<Graph>::vertex_iterator i, end; for (boost::tie(i, end) = boost::vertices(graph); i != end; ++i) { std::cout << "ID: (" << vertexIdPropertyMap[*i] << ") x: " << positionMap[*i][0] << " y: " << positionMap[*i][1] << "\n"; } return 0; }

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  • Sparse O(1) array with indices being consecutive products

    - by Kos
    Hello, I'd like to pre-calculate an array of values of some unary function f. I know that I'll only need the values for f(x) where x is of the form of a*b, where both a and b are integers in range 0..N. The obvious time-optimized choice is just to make an array of size N*N and just pre-calculate just the elements which I'm going to read later. For f(a*b), I'd just check and set tab[a*b]. This is the fastest method possible - however, this is going to take a lot of space as there are lots of indices in this array (starting with N+1) which will never by touched. Another solution is to make a simple tree map... but this slows down the lookup itself very heavily by introducing lots of branches. No. I wonder - is there any solution to make such an array less sparse and smaller, but still quick branchless O(1) in lookup?

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  • fast sphere-grid intersection

    - by Mat
    hi! given a 3D grid, a 3d point as sphere center and a radius, i'd like to quickly calculate all cells contained or intersected by the sphere. Currently i take the the (gridaligned) boundingbox of the sphere and calculate the two cells for the min anx max point of this boundingbox. then, for each cell between those two cells, i do a box-sphere intersection test. would be great if there was something more efficient thanks!

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  • Permutatation algorithm without recursion? Java

    - by Andreas Hornig
    Hi, I would like to get all combination of a number without any repetation. Like 0.1.2, 0.2.1, 1.2.0, 1.0.2, 2.0.1, 2.1.0. I tried to find an easy scheme but couldn't find so I drawed a graph/tree for it and this screams to use recursion. But I would like to do it without, if this is possible. So could anyone please help mw how to do that? Thank you in advance, Andreas

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  • Grouping consecutive identical items: IEnumerable<T> to IEnumerable<IEnumerable<T>>

    - by Romain Verdier
    I've got an interresting problem: Given an IEnumerable<string>, is it possible to yield a sequence of IEnumerable<string> that groups identical adjacent strings in one pass? Let me explain. Considering the following IEnumerable<string> (pseudo representation): {"a","b","b","b","c","c","d"} How to get an IEnumerable<IEnumerable<string>> that would yield something of the form: { // IEnumerable<IEnumerable<string>> {"a"}, // IEnumerable<string> {"a","b","b"}, // IEnumerable<string> {"c","c"}, // IEnumerable<string> {"d"} // IEnumerable<string> } The method prototype would be: public IEnumerable<IEnumerable<string>> Group(IEnumerable<string> items) { // todo } Important notes : Only one iteration over the original sequence No intermediary collections allocations (we can assume millions of strings in the original sequence, and millions consecutives identicals strings in each group) Keeping enumerators and defered execution behavior Is it possible, and how would you write it?

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  • calculating the potential effect of inaccurate triangle vertex positions on the triangle edge lenght

    - by stingrey
    i'm not sure how to solve the following problem: i have a triangle with each of the three known vertex positions A,B,C being inaccurate, meaning they can each deviate up to certain known radii rA, rB, rC into arbitrary directions. given such a triangle, i want to calculate how much the difference of two specific edge lengths (for instance the difference between lengths of edge a and edge b) of the triangle may change in the worst case. is there any elegant mathematical solution to this problem? the naive way i thought of is calculating all 360^3 angle combinations and measuring the edge differences for each case, which is a rather high overhead.

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  • What is validating a binary search tree?

    - by dotnetdev
    I read on here of an exercise in interviews known as validating a binary search tree. How exactly does this work? What would one be looking for in validating a binary search tree? I have written a basic search tree, but never heard of this concept. Thanks

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  • Latent Dirichlet Allocation, pitfalls, tips and programs

    - by Gregg Lind
    I'm experimenting with Latent Dirichlet Allocation for topic disambiguation and assignment, and I'm looking for advice. Which program is the "best", where best is some combination of easiest to use, best prior estimation, fast How do I incorporate my intuitions about topicality. Let's say I think I know that some items in the corpus are really in the same category, like all articles by the same author. Can I add that into the analysis? Any unexpected pitfalls or tips I should know before embarking? I'd prefer is there are R or Python front ends for whatever program, but I expect (and accept) that I'll be dealing with C.

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  • Maximum bipartite graph (1,n) "matching"

    - by Imre Kelényi
    I have a bipartite graph. I am looking for a maximum (1,n) "matching", which means that each vertex from partitation A has n associated vertices from partition B. The following figure shows a maximum (1,3) matching in a graph. Edges selected for the matching are red and unselected edges are black. This differs from the standard bipartite matching problem where each vertex is associate with only one other vertex, which could be called (1,1) matching with this notation. If the matching cardinality (n) is not enforced but is an upper bound (vertices from A can have 0 < x <= n associated vertices from B), then the maximum matching can be found easily by transforming the graph to a flow network and finding the max flow. However, this does not guarantee that the maximum number of vertices from A will have n associated pairs from B.

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  • Advice on String Similarity Metrics (Java). Distance, sounds like or combo?

    - by andreas
    Hello, A part of a process requires to apply String Similarity Algorithms. The results of this process will be stored and produce lets say SS_Dataset. Based on this Dataset, further decisions will have to be made. My questions are: Should i apply one or more string similarity algorithms to produce SS_Dataset ? Any comparisons between algorithms that calculate the 'distance' and the 'Sounds Like' similarity ? Does one family of algorithms produces more accurate results over the other? Does a combination give more accurate results on similarity? Can you recommend implementations that you have worked with? My implementation will include packages from the following libraries http://www.dcs.shef.ac.uk/~sam/simmetrics.html http://jtmt.sourceforge.net/ Regards,

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  • Drawing the call stack in a recursive method

    - by Shaza
    Hey, I want to draw the call stack for any recursive method, so I've created a schema like this, recursiveMethod(){ //Break recursion condition if(){ // Add value here to the return values' list- No drawing return } else{ //Draw stack with the value which will be pushed to the stack here variable <- recursiveMethod() //Clear the drawing which represents the poped value from the stack here return variable }} Notes: This schema can draw recursive methods with n recursive call by making the recursive calls in a separate return statements. returnValues list, is a list which save all the return values, just for viewing issues. What do you think of this? any suggestions are extremely welcomed.

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  • Artificial Inteligence library in python

    - by João Portela
    I was wondering if there are any python AI libraries similar to aima-python but for a more recent version of python... and how they are in comparison to aima-python. I was particularly interested in search algorithms such as hill-climbing, simulated annealing, tabu search and genetic algorithms. edit: made the question more clear.

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  • Statistical approach to chess?

    - by Chinmay Kanchi
    Reading about how Google solves the translation problem got me thinking. Would it be possible to build a strong chess engine by analysing several million games and determining the best possible move based largely (completely?) on statistics? There are several such chess databases (this is one that has 4.5 million games), and one could potentially weight moves in identical (or mirrored or reflected) positions using factors such as the ratings of the players involved, how old the game is (to factor in improvements in chess theory) etc. Any reasons why this wouldn't be a feasible approach to building a chess engine?

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  • Red-Black trees - Erasing a node with two non-leaf children

    - by SalamiArmi
    Hi all, I've been implementing my own version of a red-black tree, mostly basing my algorithms from Wikipedia (http://en.wikipedia.org/wiki/Red-black_tree). Its fairly concise for the most part, but there's one part that I would like clarification on. When erasing a node from the tree that has 2 non-leaf (non-NULL) children, it says to move either side's children into the deletable node, and remove that child. I'm a little confused as to which side to remove from, based on that. Do I pick the side randomly, do I alternate betweek sides, or do I stick to the same side for every future deletion?

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  • recurrence maths

    - by Tony
    Hi all! I have the following: T(n) <= c floor(n/2) + c ceiling(n/2) + 1 = cn + 1 T(n) = O(n) I don't understand how it gets from the first equation to the second equation? What part of the maths am I missing to understand how this comes to be? Is it done using 'Simplifying Equations' or some other rules? Can someone help me?

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  • Optimizing multiple dispatch notification algorithm in C#?

    - by Robert Fraser
    Sorry about the title, I couldn't think of a better way to describe the problem. Basically, I'm trying to implement a collision system in a game. I want to be able to register a "collision handler" that handles any collision of two objects (given in either order) that can be cast to particular types. So if Player : Ship : Entity and Laser : Particle : Entity, and handlers for (Ship, Particle) and (Laser, Entity) are registered than for a collision of (Laser, Player), both handlers should be notified, with the arguments in the correct order, and a collision of (Laser, Laser) should notify only the second handler. A code snippet says a thousand words, so here's what I'm doing right now (naieve method): public IObservable<Collision<T1, T2>> onCollisionsOf<T1, T2>() where T1 : Entity where T2 : Entity { Type t1 = typeof(T1); Type t2 = typeof(T2); Subject<Collision<T1, T2>> obs = new Subject<Collision<T1, T2>>(); _onCollisionInternal += delegate(Entity obj1, Entity obj2) { if (t1.IsAssignableFrom(obj1.GetType()) && t2.IsAssignableFrom(obj2.GetType())) obs.OnNext(new Collision<T1, T2>((T1) obj1, (T2) obj2)); else if (t1.IsAssignableFrom(obj2.GetType()) && t2.IsAssignableFrom(obj1.GetType())) obs.OnNext(new Collision<T1, T2>((T1) obj2, (T2) obj1)); }; return obs; } However, this method is quite slow (measurable; I lost ~2 FPS after implementing this), so I'm looking for a way to shave a couple cycles/allocation off this. I thought about (as in, spent an hour implementing then slammed my head against a wall for being such an idiot) a method that put the types in an order based on their hash code, then put them into a dictionary, with each entry being a linked list of handlers for pairs of that type with a boolean indication whether the handler wanted the order of arguments reversed. Unfortunately, this doesn't work for derived types, since if a derived type is passed in, it won't notify a subscriber for the base type. Can anyone think of a way better than checking every type pair (twice) to see if it matches? Thanks, Robert

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  • Use of for_each on map elements

    - by Antonio
    I have a map where I'd like to perform a call on every data type object member function. I yet know how to do this on any sequence but, is it possible to do it on an associative container? The closest answer I could find was this: Boost.Bind to access std::map elements in std::for_each. But I cannot use boost in my project so, is there an STL alternative that I'm missing to boost::bind? If not possible, I thought on creating a temporary sequence for pointers to the data objects and then, call for_each on it, something like this: class MyClass { public: void Method() const; } std::map<int, MyClass> Map; //... std::vector<MyClass*> Vector; std::transform(Map.begin(), Map.end(), std::back_inserter(Vector), std::mem_fun_ref(&std::map<int, MyClass>::value_type::second)); std::for_each(Vector.begin(), Vector.end(), std::mem_fun(&MyClass::Method)); It looks too obfuscated and I don't really like it. Any suggestions?

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  • Algoritms for "fuzzy matching" strings

    - by Alexey Romanov
    By fuzzy matching I don't mean similar strings by Levenshtein distance or something similar, but the way it's used in TextMate/Ido/Icicles: given a list of strings, find those which include all characters in the search string, but possibly with other characters between, preferring the best fit.

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  • given two bits in a set of four, fine position of two other bits

    - by aaa
    hello I am working on a simple combinatorics part, and found that I need to recover position of two bits given position of other two bits in 4-bits srring. for example, (0,1) maps to (2,3), (0,2) to (1,3), etc. for a total of six combinations. My solution is to test bits using four nested ternary operators: ab is a four bit string, with two bits set. c = ((((ab & 1) ? (((ab & 2) ? ... ))) : 0) abc = ab | c recover the last bit in the same fashion from abc. can you think of a better way/more clever way? thanks

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