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  • Simple Big O with lg(n) proof

    - by halohunter
    I'm attempting to guess and prove the Big O for: f(n) = n^3 - 7n^2 + nlg(n) + 10 I guess that big O is n^3 as it is the term with the largest order of growth However, I'm having trouble proving it. My unsuccesful attempt follows: f(n) <= cg(n) f(n) <= n^3 - 7n^2 + nlg(n) + 10 <= cn^3 f(n) <= n^3 + (n^3)*lg(n) + 10n^3 <= cn^3 f(n) <= N^3(11 + lg(n)) <= cn^3 so 11 + lg(n) = c But this can't be right because c must be constant. What am I doing wrong?

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  • User submitted content filtering

    - by Jim
    Hey all, Does anyone have any ideas on what could be used as a way to filter untrustworthy user submitted content? Take Yelp for instance, they would need to prevent competitors writing business reviews on their competitors. They would need to prevent business owners favourably reviewing their own business, or forcing friends/family to do so. They would need to prevent poor quality reviews from affecting a businesses rating and so on. I can't think what they might use to do this: Prevent multiple users from the same IP reviewing certain things Prevent business owners reviewing their own business (maybe even other businesses in the same categories as their own?) Somehow determine what a review is about and what the actual intentions behind it are Other than the first and second points, I can't think of any clever/easy way to filter potentially harmful reviews from being made available, other than a human doing it. Obviously for a site the size of Yelp this wouldn't be feasible, so what parameters could they take into consideration? Even with human intervention, how would anyone know it was the owners best buddy writing a fake review without knowing the people? I'm using this as an example in a larger study on the subject of filtering user content automatically. Does anyone have any ideas how these systems may work and what they take into consideration? Thanks!

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  • Comparing two speech sounds

    - by JessicaB
    I need to be able to determine if two sounds are very similar. The goal is to have a very limited vocabulary (10 or 15) of short one or two syllable words, then compare a captured sound to determine if it is one of those items with all the usual variability in environmental and capture conditions. The idea is that the user can issue a few simple commands by voice instead of keyboard or mouse. Does anyone know the best approach to this? I don't want to do full blown speech recognition, just something much more limited.

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  • Separating text and graphics in an image

    - by avd
    I dont know whether should I post this question here or not? But if someone knows it, please answer? What are the algorithms for determining which region in an image is text and which one is graphic? Means how to separate such regions? (figure or diagram)

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  • select i th smallest element from array

    - by davit-datuashvili
    i have divide and conqurer method to find i th smalles element from array here is code public class rand_select{ public static int Rand_partition( int a[],int p,int q,int i){ //smallest in a[p..q] if ( p==q) return a[p]; int r=partition (a,p,q); int k=r-p+1; if (i==k) return a[r]; if (i<k){ return Rand_partition(a,p,r-1,i); } return Rand_partition(a,r-1,q,i-k); } public static void main(String[]args){ int a[]=new int []{6,10,13,15,8,3,2,12}; System.out.println(Rand_partition(a,0,a.length-1,7)); } public static int partition(int a[],int p,int q){ int m=a[0]; while ( p<q){ while (p<q && a[p++] <m){ p++; } while (q>p && a[q--]>m){ q--; } int t=a[p]; a[p]=a[q]; a[q]=t; } int k=0; for (int i=0;i<a.length;i++){ if ( a[i]==m){ k=i; } } return k; } } but here is problem java.lang.ArrayIndexOutOfBoundsException please help me

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  • Adapting pseudocode to java implementation for finding the longest word in a trie

    - by user1766888
    Referring to this question I asked: How to find the longest word in a trie? I'm having trouble implementing the pseudocode given in the answer. findLongest(trie): //first do a BFS and find the "last node" queue <- [] queue.add(trie.root) last <- nil map <- empty map while (not queue.empty()): curr <- queue.pop() for each son of curr: queue.add(son) map.put(son,curr) //marking curr as the parent of son last <- curr //in here, last indicate the leaf of the longest word //Now, go up the trie and find the actual path/string curr <- last str = "" while (curr != nil): str = curr + str //we go from end to start curr = map.get(curr) return str This is what I have for my method public static String longestWord (DTN d) { Queue<DTN> holding = new ArrayQueue<DTN>(); holding.add(d); DTN last = null; Map<DTN,DTN> test = new ArrayMap<DTN,DTN>(); DTN curr; while (!holding.isEmpty()) { curr = holding.remove(); for (Map.Entry<String, DTN> e : curr.children.entries()) { holding.add(curr.children.get(e)); test.put(curr.children.get(e), curr); } last = curr; } curr = last; String str = ""; while (curr != null) { str = curr + str; curr = test.get(curr); } return str; } I'm getting a NullPointerException at: for (Map.Entry<String, DTN> e : curr.children.entries()) How can I find and fix the cause of the NullPointerException of the method so that it returns the longest word in a trie?

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  • GWT Calendrical Calculations

    - by Kyle Hayes
    We have a GWT application that needs to display various holidays. Is there a library available to do these calendrical calculations? If not, we'll have to do our own that we can ingest a set of rules to. Cheers

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  • question about permutation problem

    - by davit-datuashvili
    i have posted similar problem here http://stackoverflow.com/questions/2920315/permutation-of-array but i want following we know that with length n there is n! possible permutation from which one such that all element are in order they are in sorted variant so i want break permutation when array is in order and print result but something is wrong i think that problem is repeated of permutation here is my code import java.util.*; public class permut{ public static Random r=new Random(); public static void display(int a[],int n){ for (int i=0;i<n;i++){ System.out.println(a[i]); } } public static void Permut(int a[],int n){ int j=0; int k=0; while (j<fact(n)){ int s=r.nextInt(n); for (int i=0;i<n;i++){ k=a[i]; a[i]=a[s]; a[s]=k; } j++; if (sorted(a,n)) display(a,n); break; } } public static void main(String[]args){ int a[]=new int[]{3,4,1,2}; int n=a.length; Permut(a,n); } public static int fact(int n){ if (n==0 || (n==1) ) return 1; return n*fact(n-1); } public static boolean sorted(int a[],int n ){ boolean flag=false; for (int i=0;i<n-1;i++){ if (a[i]<a[i+1]){ flag=true; } else{ flag=false; } } return flag; } } can anybody help me? result is nothing

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  • Optimizing / simplifying a path

    - by user146780
    Say I have a path with 150 nodes / verticies. How could I simplify if so that for example a straight line with 3 verticies, would remove the middle one since it does nothing to add to the path. Also how could I avoid destroying sharp corners? And how could I remove tiny variations and have smooth curves remaining. Thanks

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  • Creating an adjacency List for DFS

    - by user200081
    I'm having trouble creating a Depth First Search for my program. So far I have a class of edges and a class of regions. I want to store all the connected edges inside one node of my region. I can tell if something is connected by the getKey() function I have already implemented. If two edges have the same key, then they are connected. For the next region, I want to store another set of connected edges inside that region, etc etc. However, I am not fully understanding DFS and I'm having some trouble implementing it. I'm not sure when/where to call DFS again. Any help would be appreciated! class edge { private: int source, destination, length; int key; edge *next; public: getKey(){ return key; } } class region { edge *data; edge *next; region() { data = new edge(); next = NULL; } }; void runDFS(int i, edge **edge, int a) { region *head = new region(); aa[i]->visited == true;//mark the first vertex as true for(int v = 0; v < a; v++) { if(tem->edge[i].getKey() == tem->edge[v].getKey()) //if the edges of the vertex have the same root { if(head->data == NULL) { head->data = aa[i]; head->data->next == NULL; } //create an edge if(head->data) { head->data->next = aa[i]; head->data->next->next == NULL; }//if there is already a node connected to ti } if(aa[v]->visited == false) runDFS(v, edge, a); //call the DFS again } //for loop }

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  • Finding if a string is an iterative substring?

    - by EsotericMe
    I have a string S. How can I find if the string follows S = nT. Examples: Function should return true if 1) S = "abab" 2) S = "abcdabcd" 3) S = "abcabcabc" 4) S = "zzxzzxzzx" But if S="abcb" returns false. I though maybe we can repeatedly call KMP on substrings of S and then decide. eg: for "abab": call on KMP on "a". it returns 2(two instances). now 2*len("a")!=len(s) call on KMP on "ab". it returns 2. now 2*len("ab")==len(s) so return true Can you suggest any better algorithms?

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  • public (static) swap() method vs. redundant (non-static) private ones...

    - by Helper Method
    I'm revisiting data structures and algorithms to refresh my knowledge and from time to time I stumble across this problem: Often, several data structures do need to swap some elements on the underlying array. So I implement the swap() method in ADT1, ADT2 as a private non-static method. The good thing is, being a private method I don't need to check on the parameters, the bad thing is redundancy. But if I put the swap() method in a helper class as a public static method, I need to check the indices every time for validity, making the swap call very unefficient when many swaps are done. So what should I do? Neglect the performance degragation, or write small but redundant code?

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  • Minimize the sequence by putting appropriate operations ' DP'

    - by Vikas
    Given a sequence,say, 222 We have to put a '+' or '* ' between each adjacent pair. '* ' has higher precedence over '+' We have to o/p the string whose evaluation leads to minimum value. O/p must be lexicographically smallest if there are more than one. inp:222 o/p: 2*2+2 Explaination: 2+2+2=6 2+2*2=6 2*2+2=6 of this 3rd is lexicographically smallest. I was wondering how to construct a DP solution for this.

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  • Find all numbers that appear in each of a set of lists

    - by Ankur
    I have several ArrayLists of Integer objects, stored in a HashMap. I want to get a list (ArrayList) of all the numbers (Integer objects) that appear in each list. My thinking so far is: Iterate through each ArrayList and put all the values into a HashSet This will give us a "listing" of all the values in the lists, but only once Iterate through the HashSet 2.1 With each iteration perform ArrayList.contains() 2.2 If none of the ArrayLists return false for the operation add the number to a "master list" which contains all the final values. If you can come up with something faster or more efficient, funny thing is as I wrote this I came up with a reasonably good solution. But I'll still post it just in case it is useful for someone else. But of course if you have a better way please do let me know.

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  • How does pattern matching work behind the scenes in F#?

    - by kryptic
    Hello Everyone, I am completely new to F# (and functional programming in general) but I see pattern matching used everywhere in sample code. I am wondering for example how pattern matching actually works? For example, I imagine it working the same as a for loop in other languages and checking for matches on each item in a collection. This is probably far from correct, how does it actually work behind the scenes?

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  • more efficient version of this?

    - by john connor
    i have this thingy here : function numOfPackets(bufferSize, packetSize) { if (bufferSize <= 0 || packetSize > bufferSize) return 0; if (packetSize < 0) throw Error(); var out = 0; for(;;){ out++; bufferSize = bufferSize - packetSize; if( packetSize > bufferSize ) break; } return out; } which i run at often , can u give me more efficent variant of it?

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  • question about quicksort

    - by davit-datuashvili
    i have write code of quicksort from programming pearls here is code public class Quick{ public static void quicksort(int x[], int l,int u) { if (l>=u) return ; int t=x[l]; int i=l; int j=u; do { i++; } while (i<=u && x[i]<t); do { j--; if (i>=j) break; } while ( x[j]>t); swap(x,i,j); swap(x, l,j); quicksort(x, l,j-1); quicksort(x, j+1,u); } public static void main(String[]args){ int x[]=new int[]{55,41,59,26,53,58,97,93}; quicksort(x,0,x.length-1); for (int i=0;i<x.length;i++){ System.out.println(x[i]); } } public static void swap(int x[], int i,int j){ int s=x[i]; x[i]=x[j]; x[j]=s; } } but it does not work here is output 59 41 55 26 53 97 58 93 any idea?

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  • Simple loop, which one I would get more performance and which one is recommended? defining a variable inside a loop or outside of it?

    - by Grego
    Variable outside of the loop int number = 0; for(int i = 0; i < 10000; i++){ number = 3 * i; printf("%d",number); } or Variable inside of the loop for(int i = 0; i < 10000; i++){ int number = 3 * i; printf("%d",number); } Which one is recommended and which one is better in performance? Edit: This is just an example to exhibit what I mean, All I wanna know is if defining a variable inside a loop and outside a loop means the same thing , or there's a difference.

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  • A question about matrix manipulation

    - by appi
    Given a 1*N matrix or an array, how do I find the first 4 elements which have the same value and then store the index for those elements? PS: I'm just curious. What if we want to find the first 4 elements whose value differences are within a certain range, say below 2? For example, M=[10,15,14.5,9,15.1,8.5,15.5,9.5], the elements I'm looking for will be 15,14.5,15.1,15.5 and the indices will be 2,3,5,7.

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  • Solving a recurrence T(n) = 2T(n/2) + n^4

    - by user563454
    I am studying using the MIT Courseware and the CLRS book Introduction to Algorithms. Solving recurrence T(n) = 2T(n/2) + n4 (page 107) If I make a recurrence tree I get: level 0 n^4 level 1 2(n/2)^4 level 2 4(n/4)^4 level 3 8(n/8)^4 The tree has lg(n) levels. Therefore the recurrence is T(n) = Theta(lg(n)n^4)) But, If I use the Master method I get. Apply case 3: T(n) = Theta(n^4) If I apply the substitution method both seem to hold. Which one is ri?

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  • How to find same-value rectangular areas of a given size in a matrix most efficiently?

    - by neo
    My problem is very simple but I haven't found an efficient implementation yet. Suppose there is a matrix A like this: 0 0 0 0 0 0 0 4 4 2 2 2 0 0 4 4 2 2 2 0 0 0 0 2 2 2 1 1 0 0 0 0 0 1 1 Now I want to find all starting positions of rectangular areas in this matrix which have a given size. An area is a subset of A where all numbers are the same. Let's say width=2 and height=3. There are 3 areas which have this size: 2 2 2 2 0 0 2 2 2 2 0 0 2 2 2 2 0 0 The result of the function call would be a list of starting positions (x,y starting with 0) of those areas. List((2,1),(3,1),(5,0)) The following is my current implementation. "Areas" are called "surfaces" here. case class Dimension2D(width: Int, height: Int) case class Position2D(x: Int, y: Int) def findFlatSurfaces(matrix: Array[Array[Int]], surfaceSize: Dimension2D): List[Position2D] = { val matrixWidth = matrix.length val matrixHeight = matrix(0).length var resultPositions: List[Position2D] = Nil for (y <- 0 to matrixHeight - surfaceSize.height) { var x = 0 while (x <= matrixWidth - surfaceSize.width) { val topLeft = matrix(x)(y) val topRight = matrix(x + surfaceSize.width - 1)(y) val bottomLeft = matrix(x)(y + surfaceSize.height - 1) val bottomRight = matrix(x + surfaceSize.width - 1)(y + surfaceSize.height - 1) // investigate further if corners are equal if (topLeft == bottomLeft && topLeft == topRight && topLeft == bottomRight) { breakable { for (sx <- x until x + surfaceSize.width; sy <- y until y + surfaceSize.height) { if (matrix(sx)(sy) != topLeft) { x = if (x == sx) sx + 1 else sx break } } // found one! resultPositions ::= Position2D(x, y) x += 1 } } else if (topRight != bottomRight) { // can skip x a bit as there won't be a valid match in current row in this area x += surfaceSize.width } else { x += 1 } } } return resultPositions } I already tried to include some optimizations in it but I am sure that there are far better solutions. Is there a matlab function existing for it which I could port? I'm also wondering whether this problem has its own name as I didn't exactly know what to google for. Thanks for thinking about it! I'm excited to see your proposals or solutions :)

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