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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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  • 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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  • Java array of arry [matrix] of an integer partition with fixed term

    - by user335209
    Hello, for my study purpose I need to build an array of array filled with the partitions of an integer with fixed term. That is given an integer, suppose 10 and given the fixed number of terms, suppose 5 I need to populate an array like this 10 0 0 0 0 9 0 0 0 1 8 0 0 0 2 7 0 0 0 3 ............ 9 0 0 1 0 8 0 0 1 1 ............. 7 0 1 1 0 6 0 1 1 1 ............ ........... 0 6 1 1 1 ............. 0 0 0 0 10 am pretty new to Java and am getting confused with all the for loops. Right now my code can do the partition of the integer but unfortunately it is not with fixed term public class Partition { private static int[] riga; private static void printPartition(int[] p, int n) { for (int i= 0; i < n; i++) System.out.print(p[i]+" "); System.out.println(); } private static void partition(int[] p, int n, int m, int i) { if (n == 0) printPartition(p, i); else for (int k= m; k > 0; k--) { p[i]= k; partition(p, n-k, n-k, i+1); } } public static void main(String[] args) { riga = new int[6]; for(int i = 0; i<riga.length; i++){ riga[i] = 0; } partition(riga, 6, 1, 0); } } the output I get it from is like this: 1 5 1 4 1 1 3 2 1 3 1 1 1 2 3 1 2 2 1 1 2 1 2 1 2 1 1 1 what i'm actually trying to understand how to proceed is to have it with a fixed terms which would be the columns of my array. So, am stuck with trying to get a way to make it less dynamic. Any help?

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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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  • How to find kth minimal element in the union of two sorted arrays?

    - by Michael
    This is a homework question. They say it takes O(logN + logM) where N and M are the arrays lengths. Let's name the arrays a and b. Obviously we can ignore all a[i] and b[i] where i k. First let's compare a[k/2] and b[k/2]. Let b[k/2] a[k/2]. Therefore we can discard also all b[i], where i k/2. Now we have all a[i], where i < k and all b[i], where i < k/2 to find the answer. What is the next step?

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  • Collision free hash function for a specific data structure

    - by Max
    Is it possible to create collision free hash function for a data structure with specific properties. The datastructure is int[][][] It contains no duplicates The range of integers that are contained in it is defined. Let's say it's 0..1000, the maximal integer is definitely not greater than 10000. Big problem is that this hash function should also be very fast. Is there a way to create such a hash function? Maybe at run time depending on the integer range?

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  • finding the maximum in series

    - by peril brain
    I need to know a code that will automatically:- search a specific word in excel notes it row or column number (depends on data arrangement) searches numerical type values in the respective row or column with that numeric value(suppose a[7][0]or a[0][7]) it compares all other values of respective row or column(ie. a[i][0] or a[0][i]) sets that value to the highest value only if IT HAS GOT NO FORMULA FOR DERIVATION i know most of coding but at a few places i got myself stuck... i'm writing a part of my program upto which i know: using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.IO; using System.Threading; using Microsoft.Office.Interop; using Excel = Microsoft.Office.Interop.Excel; Excel.Application oExcelApp; namespace a{ class b{ static void main(){ try { oExcelApp = (Excel.Application)System.Runtime.InteropServices.Marshal.GetActiveObject("Excel.Application"); ; if(oExcelApp.ActiveWorkbook != null) {Excel.Workbook xlwkbook = (Excel.Workbook)oExcelApp.ActiveWorkbook; Excel.Worksheet ws = (Excel.Worksheet)xlwkbook.ActiveSheet; Excel.Range rn; rn = ws.Cells.Find("maximum", Type.Missing, Excel.XlFindLookIn.xlValues, Excel.XlLookAt.xlPart,Excel.XlSearchOrder.xlByRows, Excel.XlSearchDirection.xlNext, false, Type.Missing, Type.Missing); }}} now ahead of this i only know tat i have to use cell.value2 ,cell.hasformula methods..... & no more idea can any one help me with this..

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  • Length of Encrypted String

    - by Agnel Kurian
    I need to create a database column which will store a string encrypted using Triple DES. How do I determine the length of the encrypted string column? (Answers for algorithms other than Triple DES are also welcome.)

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  • [C++] std::tring manipulation: whitespace, "newline escapes '\'" and comments #

    - by rubenvb
    Kind of looking for affirmation here. I have some hand-written code, which I'm not shy to say I'm proud of, which reads a file, removes leading whitespace, processes newline escapes '\' and removes comments starting with #. It also removes all empty lines (also whitespace-only ones). Any thoughts/recommendations? I could probably replace some std::cout's with std::runtime_errors... but that's not a priority here :) const int RecipeReader::readRecipe() { ifstream is_recipe(s_buffer.c_str()); if (!is_recipe) cout << "unable to open file" << endl; while (getline(is_recipe, s_buffer)) { // whitespace+comment removeLeadingWhitespace(s_buffer); processComment(s_buffer); // newline escapes + append all subsequent lines with '\' processNewlineEscapes(s_buffer, is_recipe); // store the real text line if (!s_buffer.empty()) v_s_recipe.push_back(s_buffer); s_buffer.clear(); } is_recipe.close(); return 0; } void RecipeReader::processNewlineEscapes(string &s_string, ifstream &is_stream) { string s_temp; size_t sz_index = s_string.find_first_of("\\"); while (sz_index <= s_string.length()) { if (getline(is_stream,s_temp)) { removeLeadingWhitespace(s_temp); processComment(s_temp); s_string = s_string.substr(0,sz_index-1) + " " + s_temp; } else cout << "Error: newline escape '\' found at EOF" << endl; sz_index = s_string.find_first_of("\\"); } } void RecipeReader::processComment(string &s_string) { size_t sz_index = s_string.find_first_of("#"); s_string = s_string.substr(0,sz_index); } void RecipeReader::removeLeadingWhitespace(string &s_string) { const size_t sz_length = s_string.size(); size_t sz_index = s_string.find_first_not_of(" \t"); if (sz_index <= sz_length) s_string = s_string.substr(sz_index); else if ((sz_index > sz_length) && (sz_length != 0)) // "empty" lines with only whitespace s_string.clear(); } Some extra info: std::string s_buffer is a class data member, so is std::vector v_s_recipe. Any comment is welcome :)

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  • Recursion - Ship Battle

    - by rgorrosini
    I'm trying to write a little ship battle game in java. It is 100% academic, I made it to practice recursion, so... I want to use it instead of iteration, even if it's simpler and more efficient in most some cases. Let's get down to business. These are the rules: Ships are 1, 2 or 3 cells wide and are placed horizontally only. Water is represented with 0, non-hit ship cells are 1, hit ship cells are 2 and sunken ships have all it's cells in 3. With those rules set, I'm using the following array for testing: int[][] board = new int[][] { {0, 1, 2, 0, 1, 0}, {0, 0, 1, 1, 1, 0}, {0, 3, 0, 0, 0, 0}, {0, 0, 2, 1, 2, 0}, {0, 0, 0, 1, 1, 1}, }; It works pretty good so far, and to make it more user-friendly I would like to add a couple of reports. these are the methods I need for them: Given the matrix, return the amount of ships in it. Same as a), but separating them by state (amount of non-hit ships, hit and sunken ones). I will need a hand with those reports, and I would like to get some ideas. Remember it must be done using recursion, I want to understand this, and the only way to go is practice! Thanks a lot for your time and patience :).

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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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  • 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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  • 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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  • 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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  • 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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  • Interview question : What is the fastest way to generate prime number recursively ?

    - by hilal
    Generation of prime number is simple but what is the fastest way to find it and generate( prime numbers) it recursively ? Here is my solution. However, it is not the best way. I think it is O(N*sqrt(N)). Please correct me, if I am wrong. public static boolean isPrime(int n) { if (n < 2) { return false; } else if (n % 2 == 0 & n != 2) { return false; } else { return isPrime(n, (int) Math.sqrt(n)); } } private static boolean isPrime(int n, int i) { if (i < 2) { return true; } else if (n % i == 0) { return false; } else { return isPrime(n, --i); } } public static void generatePrimes(int n){ if(n < 2) { return ; } else if(isPrime(n)) { System.out.println(n); } generatePrimes(--n); } public static void main(String[] args) { generatePrimes(200); }

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  • question about Batcher odd-even sort

    - by davit-datuashvili
    hi i ave question about Batcher's odd-even sort i have following code public class Batcher{ public static void batchsort(int a[],int l,int r){ int n=r-l+1; for (int p=1;p<n;p+=p) for (int k=p;k>0;k/=2) for (int j=k%p;j+k<n;j+=(k+k)) for (int i=0;i<n-j-k;i++) if ((j+i)/(p+p)==(j+i+k)/(p+p)) exch(a,l+j+i,l+j+i+k); } public static void main(String[]args){ int a[]=new int[]{2,4,3,4,6,5,3}; batchsort(a,0,a.length-1); for (int i=0;i<a.length;i++){ System.out.println(a[i]); } } public static void exch(int a[],int i,int j){ int t=a[i]; a[i]=a[j]; a[j]=t; } } //result is 3 3 4 4 5 2 6 what i missed ? hat is wrong?

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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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  • 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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  • algorithms that destruct and copy_construct

    - by FredOverflow
    I am currently building my own toy vector for fun, and I was wondering if there is something like the following in the current or next standard or in Boost? template<class T> void destruct(T* begin, T* end) { while (begin != end) { begin -> ~T(); ++begin; } } template<class T> T* copy_construct(T* begin, T* end, T* dst) { while (begin != end) { new(dst) T(*begin); ++begin; ++dst; } return dst; }

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  • phrase split algorithm in PHP

    - by Eric Sim
    Not sure how to explain. Let's use an example. Say I want to split the sentence "Today is a great day." into today today is today is a today is a great today is a great day is is a is a great is a great day a a great a great day great great day day The idea is to get all the sequential combination in a sentence. I have been thinking what's the best way to do it in PHP. Any idea is welcome.

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