Search Results

Search found 5638 results on 226 pages for 'scheduling algorithm'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 105/226 | < Previous Page | 101 102 103 104 105 106 107 108 109 110 111 112  | Next Page >

  • java random percentages

    - by erw
    I need to generate n percentages (integers between 0 and 100) such that the sum of all n numbers adds up to 100. If I just do nextInt() n times, each time ensuring that the parameter is 100 minus the previously accumulated sum, then my percentages are biased (i.e. the first generated number will usually be largest etc.). How do I do this in an unbiased way?

    Read the article

  • question on HyperOperation

    - by davit-datuashvili
    i am trying to solve following recurence program http://en.wikipedia.org/wiki/Hyper_operator here is my code i know it has mistakes but i have done what i could public class hyper{ public static int Hyper(int a,int b,int n){ int t=0; if ( n==0) return b+1; if ((n==1) &&(b==0)) return a; if ((n==2) && (b==0)) return 0; if ((n>=3) && (b==0)) return 1; t=Hyper(a,b-1,n); return Hyper (a,t,n-1); } public static void main(String[]args){ int n=3; int a=5; int b=7; System.out.println(Hyper(a,b,n)); } } please help

    Read the article

  • All possibilities in 2d array

    - by valli-R
    I have this array: $array = array ( array('1', '2', '3'), array('!', '@'), array('a', 'b', 'c', 'd'), ); And I want to know all character combination of sub arrays.. for example : 1!a 1!b 1!c 1!d 1@a 1@b 1@c 1@d 2!a 2!b 2!c 2!d 2@a 2@b ... Currently I am having this code : for($i = 0; $i < count($array[0]); $i++) { for($j = 0; $j < count($array[1]); $j++) { for($k = 0; $k < count($array[2]); $k++) { echo $array[0][$i].$array[1][$j].$array[2][$k].'<br/>'; } } } It works, but I think it is ugly, and when I am adding more arrays, I have to add more for. I am pretty sure there is a way to do this recursively, but I don't know how to start/how to do it. A little help could be nice! Thanks you!

    Read the article

  • Shortest distance between points on a toroidally wrapped (x- and y- wrapping) map?

    - by mstksg
    I have a toroidal-ish Euclidean-ish map. That is the surface is a flat, Euclidean rectangle, but when a point moves to the right boundary, it will appear at the left boundary (at the same y value), given by x_new = x_old % width Basically, points are plotted based on: (x_new, y_new) = ( x_old % width, y_old % height) Think Pac Man -- walking off one edge of the screen will make you appear on the opposite edge. What's the best way to calculate the shortest distance between two points? The typical implementation suggests a large distance for points on opposite corners of the map, when in reality, the real wrapped distance is very close. The best way I can think of is calculating Classical Delta X and Wrapped Delta X, and Classical Delta Y and Wrapped Delta Y, and using the lower of each pair in the Sqrt(x^2+y^2) distance formula. But that would involve many checks, calculations, operations -- some that I feel might be unnecessary. Is there a better way?

    Read the article

  • How to implement Pentago AI algorithm

    - by itsho
    Hi, i'm trying to develop Pentago-game in c#. right now i'm having 2 players mode which working just fine. the problem is, that i want One player mode (against computer), but unfortunately, all implements of minimax / negamax are for one step calculated. butin Pentago, every player need to do two things (place marble, and rotate one of the inner-boards) I didn't figure out how to implement both rotate part & placing the marble, and i would love someone to guide me with this. if you're not familiar with the game, here's a link to the game. if anyone want's, i can upload my code somewhere if that's relevant. thank you very much in advance

    Read the article

  • What is the complexity of this c function

    - by Bunny Rabbit
    what is the complexity of the following c Function ? double foo (int n) { int i; double sum; if (n==0) return 1.0; else { sum = 0.0; for (i =0; i<n; i++) sum +=foo(i); return sum; } } Please don't just post the complexity can you help me in understanding how to go about it . EDIT: It was an objective question asked in an exam and the Options provided were 1.O(1) 2.O(n) 3.O(n!) 4.O(n^n)

    Read the article

  • Fastest method to define whether a number is a triangular number

    - by psihodelia
    A triangular number is the sum of the n natural numbers from 1 to n. What is the fastest method to find whether a given positive integer number is a triangular one? I suppose, there must be a hidden pattern in a binary representation of such numbers (like if you need to find whether a number is even/odd you check its least significant bit). Here is a cut of the first 1200th up to 1300th triangular numbers, you can easily see a bit-pattern here (if not, try to zoom out): (720600, '10101111111011011000') (721801, '10110000001110001001') (723003, '10110000100000111011') (724206, '10110000110011101110') (725410, '10110001000110100010') (726615, '10110001011001010111') (727821, '10110001101100001101') (729028, '10110001111111000100') (730236, '10110010010001111100') (731445, '10110010100100110101') (732655, '10110010110111101111') (733866, '10110011001010101010') (735078, '10110011011101100110') (736291, '10110011110000100011') (737505, '10110100000011100001') (738720, '10110100010110100000') (739936, '10110100101001100000') (741153, '10110100111100100001') (742371, '10110101001111100011') (743590, '10110101100010100110') (744810, '10110101110101101010') (746031, '10110110001000101111') (747253, '10110110011011110101') (748476, '10110110101110111100') (749700, '10110111000010000100') (750925, '10110111010101001101') (752151, '10110111101000010111') (753378, '10110111111011100010') (754606, '10111000001110101110') (755835, '10111000100001111011') (757065, '10111000110101001001') (758296, '10111001001000011000') (759528, '10111001011011101000') (760761, '10111001101110111001') (761995, '10111010000010001011') (763230, '10111010010101011110') (764466, '10111010101000110010') (765703, '10111010111100000111') (766941, '10111011001111011101') (768180, '10111011100010110100') (769420, '10111011110110001100') (770661, '10111100001001100101') (771903, '10111100011100111111') (773146, '10111100110000011010') (774390, '10111101000011110110') (775635, '10111101010111010011') (776881, '10111101101010110001') (778128, '10111101111110010000') (779376, '10111110010001110000') (780625, '10111110100101010001') (781875, '10111110111000110011') (783126, '10111111001100010110') (784378, '10111111011111111010') (785631, '10111111110011011111') (786885, '11000000000111000101') (788140, '11000000011010101100') (789396, '11000000101110010100') (790653, '11000001000001111101') (791911, '11000001010101100111') (793170, '11000001101001010010') (794430, '11000001111100111110') (795691, '11000010010000101011') (796953, '11000010100100011001') (798216, '11000010111000001000') (799480, '11000011001011111000') (800745, '11000011011111101001') (802011, '11000011110011011011') (803278, '11000100000111001110') (804546, '11000100011011000010') (805815, '11000100101110110111') (807085, '11000101000010101101') (808356, '11000101010110100100') (809628, '11000101101010011100') (810901, '11000101111110010101') (812175, '11000110010010001111') (813450, '11000110100110001010') (814726, '11000110111010000110') (816003, '11000111001110000011') (817281, '11000111100010000001') (818560, '11000111110110000000') (819840, '11001000001010000000') (821121, '11001000011110000001') (822403, '11001000110010000011') (823686, '11001001000110000110') (824970, '11001001011010001010') (826255, '11001001101110001111') (827541, '11001010000010010101') (828828, '11001010010110011100') (830116, '11001010101010100100') (831405, '11001010111110101101') (832695, '11001011010010110111') (833986, '11001011100111000010') (835278, '11001011111011001110') (836571, '11001100001111011011') (837865, '11001100100011101001') (839160, '11001100110111111000') (840456, '11001101001100001000') (841753, '11001101100000011001') (843051, '11001101110100101011') (844350, '11001110001000111110') For example, can you also see a rotated normal distribution curve, represented by zeros between 807085 and 831405?

    Read the article

  • question about partition

    - by davit-datuashvili
    i have question about hoare partition method here is code and also pseudo code please if something is wrong correct pseudo code HOARE-PARTITION ( A, p, r) 1 x ? A[ p] 2 i ? p-1 3 j ? r +1 4 while TRUE 5 do repeat j ? j - 1 6 until A[ j ] = x 7 do repeat i ? i + 1 8 until A[i] = x 9 if i < j 10 then exchange A[i] ? A[ j ] 11 else return j and my code public class Hoare { public static int partition(int a[],int p,int r) { int x=a[p]; int i=p-1; int j=r+1; while (true) { do { j=j-1; } while(a[j]>=x); do { i=i+1; } while(a[i]<=x); if (i<j) { int t=a[i]; a[i]=a[j]; a[j]=t; } else { return j; } } } public static void main(String[]args){ int a[]=new int[]{13,19,9,5,12,8,7,4,11,2,6,21}; partition(a,0,a.length-1); } } and mistake is this error: Class names, 'Hoare', are only accepted if annotation processing is explicitly requested 1 error any ideas

    Read the article

  • Java: Efficient Equivalent to Removing while Iterating a Collection

    - by Claudiu
    Hello everyone. We all know you can't do this: for (Object i : l) if (condition(i)) l.remove(i); ConcurrentModificationException etc... this apparently works sometimes, but not always. Here's some specific code: public static void main(String[] args) { Collection<Integer> l = new ArrayList<Integer>(); for (int i=0; i < 10; ++i) { l.add(new Integer(4)); l.add(new Integer(5)); l.add(new Integer(6)); } for (Integer i : l) { if (i.intValue() == 5) l.remove(i); } System.out.println(l); } This, of course, results in: Exception in thread "main" java.util.ConcurrentModificationException ...even though multiple threads aren't doing it... Anyway. What's the best solution to this problem? "Best" here means most time and space efficient (I realize you can't always have both!) I'm also using an arbitrary Collection here, not necessarily an ArrayList, so you can't rely on get.

    Read the article

  • question about heapsearch order

    - by davit-datuashvili
    ia have meet following problem suppose we have sorted array of size 2^k-1 where k is given number we should copy this array into heapsearch array b the elements in odd positions of a go in order into last half of the positions of b positions congruent to 2 modul0 4 go into b's secodn quarter and so on this is not homework and please nobody tag it as homework it is from programming pearls please any ideas

    Read the article

  • question about interface

    - by davit-datuashvili
    i have posted this question http://stackoverflow.com/questions/2874487/how-can-i-implement-this-python-snippet-in-java i have compiled it now i need to use in main project public static void main(String[]args){ } ? can anybody show me example?

    Read the article

  • question about interface

    - by davit-datuashvili
    i have posted this question http://stackoverflow.com/questions/2874487/how-can-i-implement-this-python-snippet-in-java i have compiled it now i need to use in main project public static void main(String[]args){ } ? can anybody show me example?

    Read the article

  • question about interface

    - by davit-datuashvili
    i have posted this question http://stackoverflow.com/questions/2874487/how-can-i-implement-this-python-snippet-in-java i have compiled it now i need to use in main project public static void main(String[]args){ } ? can anybody show me example?

    Read the article

  • Minimizing distance to a weighted grid

    - by Andrew Tomazos - Fathomling
    Lets suppose you have a 1000x1000 grid of positive integer weights W. We want to find the cell that minimizes the average weighted distance.to each cell. The brute force way to do this would be to loop over each candidate cell and calculate the distance: int best_x, best_y, best_dist; for x0 = 1:1000, for y0 = 1:1000, int total_dist = 0; for x1 = 1:1000, for y1 = 1:1000, total_dist += W[x1,y1] * sqrt((x0-x1)^2 + (y0-y1)^2); if (total_dist < best_dist) best_x = x0; best_y = y0; best_dist = total_dist; This takes ~10^12 operations, which is too long. Is there a way to do this in or near ~10^8 or so operations?

    Read the article

  • Algorithms after load-balancer?

    - by Vimvq1987
    I need to study about load-balancers, such as Network Load Balancing, Linux Virtual Server, HAProxy,...There're somethings under-the-hood I need to know: What algorithms/technologies are used in these load-balancers? Which is the most popular? most effective? I expect that these algorithms/technologies will not be too complicated. Are there some resources written about them? Thank you very much for your help.

    Read the article

  • Doubling a number - shift left vs. multiplication

    - by ToxicAvenger
    What are the differences between int size = (int)((length * 200L) / 100L); // (1) and int size = length << 1; // (2) (length is int in both cases) I assume both code snippets want to double the length parameter. I'd be tempted to use (2) ... so are there any advantages for using (1)? I looked at the edge cases when overflow occurs, and both versions seem to have the same behavior. Please tell me what am I missing.

    Read the article

  • Given a 2d array sorted in increasing order from left to right and top to bottom, what is the best w

    - by Phukab
    I was recently given this interview question and I'm curious what a good solution to it would be. Say I'm given a 2d array where all the numbers in the array are in increasing order from left to right and top to bottom. What is the best way to search and determine if a target number is in the array? Now, my first inclination is to utilize a binary search since my data is sorted. I can determine if a number is in a single row in O(log N) time. However, it is the 2 directions that throw me off. Another solution I could use, if I could be sure the matrix is n x n, is to start at the middle. If the middle value is less than my target, then I can be sure it is in the left square portion of the matrix from the middle. I then move diagnally and check again, reducing the size of the square that the target could potentially be in until I have honed in on the target number. Does anyone have any good ideas on solving this problem? Example array: Sorted left to right, top to bottom. 1 2 4 5 6 2 3 5 7 8 4 6 8 9 10 5 8 9 10 11

    Read the article

  • generate k distinct number less then n

    - by davit-datuashvili
    hi i have following question task is this generate k distinct positive numbers less then n without duplication my method is following first create array size of k where we should write these numbers int a[]=new int[k]; //now i am going to cretae another array where i check if (at given number position is 1 then generate number again else put this number in a array and continue cycle i put here a piece of code and explanations int a[]=new int[k]; int t[]=new int[n+1]; Random r=new Random(); for (int i==0;i<t.length;i++){ t[i]=0;//initialize it to zero } int m=0;//initialize it also for (int i=0;i<a.length;i++){ m=r.nextInt(n);//random element between 0 and n if (t[m]==1){ //i have problem with this i want in case of duplication element occurs repeats this steps afain until there will be different number else{ t[m]=1; x[i]=m; } } so i fill concret my problem if t[m]==1 it means that this element occurs already so i want to generate new number but problem is that number of generated numbers will not be k beacuse if i==0 and occurs duplicate element and we write continue then it will switch at i==1 i need like goto for repeat step or for (int i=0;i<x.length;i++){ loop: m=r.nextInt(n); if ( x[m]==1){ continue loop; } else{ x[m]=1; a[i]=m; continue;//continue next step at i=1 and so on } } i need this code in java please help

    Read the article

  • yet another logic.

    - by Sunil
    I'm working on a research problem out of curiosity and I don't know how to program the logic that I've in mind. Let me explain it to you : I've 4 vectors say for example, v1 = 1 1 1 1 v2 = 2 2 2 2 v3 = 3 3 3 3 v4 = 4 4 4 4 Now what I want to do is to add them combination-wise. i.e v12 = v1+v2 v13 = v1+v3 v14 = v1+v4 v23 = v2+v3 v24 = v2+v4 v34 = v3+v4 Till this step it is just fine. The problem/trick is now, at the end of each iteration I give the obtained vectors into a black box function and it returns only few of the vectors say v12, v13 and v34. Now, I want to add each of these vectors one vector from v1,v2,v3,v4 which it hasn't added before. For example v3 and v4 hasn't been added to v12 so I want to create v123 and v124. similarly for all the vectors like, v12 should become : v123 = v12+v3 v124 = v12+v4 v13 should become : v132 // this should not occur because I already have v123 v134 = v13+v4; v14,v23 and v24 cannot be considered because it was deleted in the black box function so all we have in our hands to work with is v12,v13 and v34. v34 should become : v341 // cannot occur because we have 134 v342 = v34+v2 It is important that I do not do all at one step at the start like for example I can do (4 choose 3) 4C3 and finish it off but I want to do it step by step at each iteration. I've asked a modified version of this question before (without including the black box function) and got answers here. Can anybody tell me how to do it when the black box function is included ? A modification of the previous answer would also be great. Thanks in advance.

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 101 102 103 104 105 106 107 108 109 110 111 112  | Next Page >