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  • sorting using recursion

    - by user310587
    I have the following function to sort an array with even numbers in the front and odd numbers in the back. Is there a way to get it done without using any loops? //front is 0, back =array.length-1; arrangeArray (front, back); public static void arrangeArray (int front, int back) { if (front != back || front<back) { while (numbers [front]%2 == 0) front++; while (numbers[back]%2!=0) back--; if (front < back) { int oddnum = numbers [front]; numbers[front]= numbers[back]; numbers[back]=oddnum; arrangeArray (front+1, back-1); } } }

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  • Media recommendation engine - Single user system - How to start

    - by Microkernel
    Hi guys, I want to implement a media recommendation engine. I saw a similar posts on this, but I think my requirements are bit different from those, so posting here. Here is the deal. I want to implement a recommendation engine for media players like VLC, which would be an engine that has to care for only single user. Like, it would be embedded in a media player on a PC which is typically used by single user. And it will start learning the likes and dislikes of the user and gradually learns what a user likes. Here it will not be able to find similar users for using their data for recommendation as its a single user system. So how to go about this? Or you can consider it as a recommendation engine that has to be put in say iPods, which has to learn about a single user and recommend music/Movies from the collections it has. I thought of start collecting the genre of music/movies (maybe even artist name) that user watches and recommend movies from the most watched Genre, but it look very crude, isn't it? So is there any algorithms I can use or any resources I can refer up to? Regards, MicroKernel :)

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  • C++: building iterator from bits

    - by gruszczy
    I have a bitmap and would like to return an iterator of positions of set bits. Right now I just walk the whole bitmap and if bit is set, then I provide next position. I believe this could be done more effectively: for example build statically array for each combination of bits in single byte and return vector of positions. This can't be done for a whole int, because array would be too big. But maybe there are some better solutions? Do you know any smart algorithms for this?

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  • Detecting Singularities in a Graph

    - by nasufara
    I am creating a graphing calculator in Java as a project for my programming class. There are two main components to this calculator: the graph itself, which draws the line(s), and the equation evaluator, which takes in an equation as a String and... well, evaluates it. To create the line, I create a Path2D.Double instance, and loop through the points on the line. To do this, I calculate as many points as the graph is wide (e.g. if the graph itself is 500px wide, I calculate 500 points), and then scale it to the window of the graph. Now, this works perfectly for most any line. However, it does not when dealing with singularities. If, when calculating points, the graph encounters a domain error (such as 1/0), the graph closes the shape in the Path2D.Double instance and starts a new line, so that the line looks mathematically correct. Example: However, because of the way it scales, sometimes it is rendered correctly, sometimes it isn't. When it isn't, the actual asymptotic line is shown, because within those 500 points, it skipped over x = 2.0 in the equation 1 / (x-2), and only did x = 1.98 and x = 2.04, which are perfectly valid in that equation. Example: In that case, I increased the window on the left and right one unit each. My question is: Is there a way to deal with singularities using this method of scaling so that the resulting line looks mathematically correct? I myself have thought of implementing a binary search-esque method, where, if it finds that it calculates one point, and then the next point is wildly far away from the last point, it searches in between those points for a domain error. I had trouble figuring out how to make it work in practice, however. Thank you for any help you may give!

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  • How to get predecessor and successors from an adjacency matrix

    - by NickTFried
    Hi I am am trying to complete an assignment, where it is ok to consult the online community. I have to create a graph class that ultimately can do Breadth First Search and Depth First Search. I have been able to implement those algorithms successfully however another requirement is to be able to get the successors and predecessors and detect if two vertices are either predecessors or successors for each other. I'm having trouble thinking of a way to do this. I will post my code below, if anyone has any suggestions it would be greatly appreciated. import java.util.ArrayList; import java.util.Iterator; import java.util.LinkedList; import java.util.Queue; import java.util.Stack; public class Graph<T> { public Vertex<T> root; public ArrayList<Vertex<T>> vertices=new ArrayList<Vertex<T>>(); public int[][] adjMatrix; int size; private ArrayList<Vertex<T>> dfsArrList; private ArrayList<Vertex<T>> bfsArrList; public void setRootVertex(Vertex<T> n) { this.root=n; } public Vertex<T> getRootVertex() { return this.root; } public void addVertex(Vertex<T> n) { vertices.add(n); } public void removeVertex(int loc){ vertices.remove(loc); } public void addEdge(Vertex<T> start,Vertex<T> end) { if(adjMatrix==null) { size=vertices.size(); adjMatrix=new int[size][size]; } int startIndex=vertices.indexOf(start); int endIndex=vertices.indexOf(end); adjMatrix[startIndex][endIndex]=1; adjMatrix[endIndex][startIndex]=1; } public void removeEdge(Vertex<T> v1, Vertex<T> v2){ int startIndex=vertices.indexOf(v1); int endIndex=vertices.indexOf(v2); adjMatrix[startIndex][endIndex]=1; adjMatrix[endIndex][startIndex]=1; } public int countVertices(){ int ver = vertices.size(); return ver; } /* public boolean isPredecessor( Vertex<T> a, Vertex<T> b){ for() return true; }*/ /* public boolean isSuccessor( Vertex<T> a, Vertex<T> b){ for() return true; }*/ public void getSuccessors(Vertex<T> v1){ } public void getPredessors(Vertex<T> v1){ } private Vertex<T> getUnvisitedChildNode(Vertex<T> n) { int index=vertices.indexOf(n); int j=0; while(j<size) { if(adjMatrix[index][j]==1 && vertices.get(j).visited==false) { return vertices.get(j); } j++; } return null; } public Iterator<Vertex<T>> bfs() { Queue<Vertex<T>> q=new LinkedList<Vertex<T>>(); q.add(this.root); printVertex(this.root); root.visited=true; while(!q.isEmpty()) { Vertex<T> n=q.remove(); Vertex<T> child=null; while((child=getUnvisitedChildNode(n))!=null) { child.visited=true; bfsArrList.add(child); q.add(child); } } clearVertices(); return bfsArrList.iterator(); } public Iterator<Vertex<T>> dfs() { Stack<Vertex<T>> s=new Stack<Vertex<T>>(); s.push(this.root); root.visited=true; printVertex(root); while(!s.isEmpty()) { Vertex<T> n=s.peek(); Vertex<T> child=getUnvisitedChildNode(n); if(child!=null) { child.visited=true; dfsArrList.add(child); s.push(child); } else { s.pop(); } } clearVertices(); return dfsArrList.iterator(); } private void clearVertices() { int i=0; while(i<size) { Vertex<T> n=vertices.get(i); n.visited=false; i++; } } private void printVertex(Vertex<T> n) { System.out.print(n.label+" "); } }

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  • What kind of data processing problems would CUDA help with?

    - by Chris McCauley
    Hi, I've worked on many data matching problems and very often they boil down to quickly and in parallel running many implementations of CPU intensive algorithms such as Hamming / Edit distance. Is this the kind of thing that CUDA would be useful for? What kinds of data processing problems have you solved with it? Is there really an uplift over the standard quad-core intel desktop? Chris

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  • Drawing N-width lines?

    - by user146780
    Given a series of points, how could I calculate the vector for that line 5 pixels away? Ex: Given: \ \ \ How could I find the vector for \ \ \ \ \ \ The ones on the right. But bear in mind that it may not always be a nice straight line. I'm trying to figure out how programs like Flash can make thick outlines. Thanks

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  • How does Batcher Merge work at a high level?

    - by Mike
    I'm trying to grasp the concept of a Batcher Sort. However, most resources I've found online focus on proof entirely or on low-level pseudocode. Before I look at proofs, I'd like to understand how Batcher Sort works. Can someone give a high level overview of how Batcher Sort works(particularly the merge) without overly verbose pseudocode(I want to get the idea behind the Batcher Sort, not implement it)? Thanks!

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  • Performing text processing on flatpage content to include handling of custom tag

    - by Dzejkob
    Hi. I'm using flatpages app in my project to manage some html content. That content will include images, so I've made a ContentImage model allowing user to upload images using admin panel. The user should then be able to include those images in content of the flatpages. He can of course do that by manually typing image url into <img> tag, but that's not what I'm looking for. To make including images more convenient, I'm thinking about something like this: User edits an additional, let's say pre_content field of CustomFlatPage model (I'm using custom flatpage model already) instead of defining <img> tags directly, he uses a custom tag, something like [img=...] where ... is name of the ContentImage instance now the hardest part: before CustomFlatPage is saved, pre_content field is checked for all [img=...] occurences and they are processed like this: ContentImage model is searched if there's image instance with given name and if so, [img=...] is replaced with proper <img> tag. flatpage actual content is filled with processed pre_content and then flatpage is saved (pre_content is leaved unchanged, as edited by user) The part that I can't cope with is text processing. Should I use regular expressions? Apparently they can be slow for large strings. And how to organize logic? I assume it's rather algorithmic question, but I'm not familliar with text processing in Python enough, to do it myself. Can somebody give me any clues?

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  • Number of ways to place kings on chess board

    - by Rakesh
    You have an N x N chessboard and you wish to place N kings on it. Each row and column should contain exactly one king, and no two kings should attack each other (two kings attack each other if they are present in squares which share a corner). The kings in the first K rows of the board have already been placed. You are given the positions of these kings as an array pos[ ]. pos[i] is the column in which the king in the ith row has already been placed. All indices are 0-indexed. In how many ways can the remaining kings be placed? Input: The first line contains the number of test cases T. T test cases follow. Each test case contains N and K on the first line, followed by a line having K integers, denoting the array pos[ ] as described above. Output: Output the number of ways to place kings in the remaining rows satisfying the above conditions. Output all numbers modulo 1000000007. Constraints: 1 <= T <= 20 1 <= N <= 16 0 <= K <= N 0 <= pos_i < N The kings specified in the input will be in different columns and not attack each other. Sample Input: 5 4 1 2 3 0 5 2 1 3 4 4 1 3 0 2 6 1 2 Sample Output: 1 0 2 1 18 Explanation: For the first example, there is a king already placed at row 0 and column 2. The king in the second row must belong to column 0. The king in the third row must belong to column 3, and the last king must beong to column 1. Thus there is only 1 valid placement. For the second example, there is no valid placement. How should i approach this problem

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  • Reverse factorial

    - by dada
    Well, we all know that if N is given it's easy to calculate N!. But what about reversing? N! is given and you are about to find N - Is that possible ? I'm curious.

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  • randomized quicksort: probability of two elements comparison?

    - by bantu
    I am reading "Probability and Computing" by M.Mitzenmacher, E.Upfal and I have problems understanding how the probability of comparison of two elements is calculated. Input: the list (y1,y2,...,YN) of numbers. We are looking for pivot element. Question: what is probability that two elements yi and yj (ji) will be compared? Answer (from book): yi and yj will be compared if either yi or yj will be selected as pivot in first draw from sequence (yi,yi+1,...,yj-1,yj). So the probablity is: 2/(y-i+1). The problem for me is initial claim: for example, picking up yi in the first draw from the whole list will cause the comparison with yj (and vice-versa) and the probability is 2/n. So, rather the "reverse" claim is true -- none of the (yi+1,...,yj-1) elements can be selected beforeyi or yj, but the "pool" size is not fixed (in first draw it is n for sure, but on the second it is smaller). Could someone please explain this, how the authors come up with such simplified conclusion? Thank you in advance

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  • C# - split String into smaller Strings by length variable

    - by tyndall
    I'd like to break apart a String by a certain length variable. It needs to bounds check so as not explode when the last section of string is not as long as or longer than the length. Looking for the most succinct (yet understandable) version. Example: string x = "AAABBBCC"; string[] arr = x.SplitByLength(3); // arr[0] -> "AAA"; // arr[1] -> "BBB"; // arr[2] -> "CC"

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  • C++ string array binary search

    - by Jose Vega
    string Haystack[] = { "Alabama", "Alaska", "American Samoa", "Arizona", "Arkansas", "California", "Colorado", "Connecticut", "Delaware", "District of Columbia", "Florida", "Georgia", "Guam", "Hawaii", "Idaho", "Illinois", "Indiana", "Iowa", "Kansas", "Kentucky", "Louisiana", "Maine", "Maryland", "Massachusetts", "Michigan", "Minnesota", "Mississippi", "Missouri", "Montana", "Nebraska", "Nevada", "New Hampshire", "New Jersey", "New Mexico", "New York", "North Carolina", "North Dakota", "Northern Mariana Islands", "Ohio", "Oklahoma", "Oregon", "Pennsylvania", "Puerto Rico", "Rhode Island", "South Carolina", "South Dakota", "Tennessee", "Texas", "US Virgin Islands", "Utah", "Vermont", "Virginia", "Washington", "West Virginia", "Wisconsin", "Wyoming"}; string Needle = "Virginia"; if(std::binary_search(Haystack, Haystack+56, Needle)) cout<<"Found"; If I also wanted to find the location of the needle in the string array, is there an "easy" way to find out?

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  • What's the best general programming book to review basic development concepts?

    - by Charles S.
    I'm looking for for a programming book that reviews basic concepts like implementing linked lists, stacks, queues, hash tables, tree traversals, search algorithms, etc. etc. Basically, I'm looking for a review of everything I learned in college but have forgotten. I prefer something written in the last few years that includes at least a decent amount of code in object-oriented languages. This is to study for job interview questions but I already have the "solving interview questions" books. I'm looking for something with a little more depth and explanation. Any good recommendations?

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  • KD-Trees and missing values (vector comparison)

    - by labratmatt
    I have a system that stores vectors and allows a user to find the n most similar vectors to the user's query vector. That is, a user submits a vector (I call it a query vector) and my system spits out "here are the n most similar vectors." I generate the similar vectors using a KD-Tree and everything works well, but I want to do more. I want to present a list of the n most similar vectors even if the user doesn't submit a complete vector (a vector with missing values). That is, if a user submits a vector with three dimensions, I still want to find the n nearest vectors (stored vectors are of 11 dimensions) I have stored. I have a couple of obvious solutions, but I'm not sure either one seem very good: Create multiple KD-Trees each built using the most popular subset of dimensions a user will search for. That is, if a user submits a query vector of thee dimensions, x, y, z, I match that query to my already built KD-Tree which only contains vectors of three dimensions, x, y, z. Ignore KD-Trees when a user submits a query vector with missing values and compare the query vector to the vectors (stored in a table in a DB) one by one using something like a dot product. This has to be a common problem, any suggestions? Thanks for the help.

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  • How To Generate Parameter Set for the Diffie-Hellman Key Agreement Algorithm in Android

    - by sebby_zml
    Hello everyone, I am working on mobile/server security related project. I am now stuck in generating a Diffie-Hellman key agreement part. It works fine in server side program but it is not working in mobile side. Thus, I assume that it is not compactible with Android. I used the following class to get the parameters. It returns a comma-separated string of 3 values. The first number is the prime modulus P. The second number is the base generator G. The third number is bit size of the random exponent L. My question is is there anything wrong with the code or it is not compactible for android?What kind of changes should I do? Your suggestion and guidance would be very much help for me. Thanks a lot in advance. public static String genDhParams() { try { // Create the parameter generator for a 1024-bit DH key pair AlgorithmParameterGenerator paramGen = AlgorithmParameterGenerator.getInstance("DH"); paramGen.init(1024); // Generate the parameters AlgorithmParameters params = paramGen.generateParameters(); DHParameterSpec dhSpec = (DHParameterSpec)params.getParameterSpec(DHParameterSpec.class); // Return the three values in a string return ""+dhSpec.getP()+","+dhSpec.getG()+","+dhSpec.getL(); } catch (NoSuchAlgorithmException e) { } catch (InvalidParameterSpecException e) { } return null; } Regards, Sebby

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  • open source smooth particle hydrodynamics

    - by user325181
    Anyone know of any open source libraries for particle based large scale smooth particle hydrodynamics. I am looking for a easier way of simulating large scale planetary body impacts with rotation. I was also wondering if you had any ideas on how to visualize the output from said simulation. I have tried using IBM graphviz, but it is very difficult to work with. Any pointers would be appreciated. Thanks!

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  • Deletion procedure for a Binary Search Tree

    - by Metz
    Consider the deletion procedure on a BST, when the node to delete has two children. Let's say i always replace it with the node holding the minimum key in its right subtree. The question is: is this procedure commutative? That is, deleting x and then y has the same result than deleting first y and then x? I think the answer is no, but i can't find a counterexample, nor figure out any valid reasoning. EDIT: Maybe i've got to be clearer. Consider the transplant(node x, node y) procedure: it replace x with y (and its subtree). So, if i want to delete a node (say x) which has two children i replace it with the node holding the minimum key in its right subtree: y = minimum(x.right) transplant(y, y.right) // extracts the minimum (it doesn't have left child) y.right = x.right y.left = x.left transplant(x,y) The question was how to prove the procedure above is not commutative.

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