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  • C# compare algorithms

    - by public static
    Hi, Are there any open source algorithms in c# that solve the problem of creating a difference between two text files? It would be super cool if it had some way of highlighting what exact areas where changed in the text document also.

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  • 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?

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  • Weakly connected balanced digraph

    - by user1074557
    How can I prove that if a balanced digraph is weakly connected, then it is also strongly connected? (balanced digraph means that for every node, it's indegree and outdegree is the same and weakly connected means the non-directed version of this graph is connected). What I can think of so far is: if the graph is balanced, it means it is a union of directed cycles. So if I remove any cycle, it will stay balanced. Also each vertex in the cycle has one edge coming into it and one edge leading out of it.. Then I guess I need to use some contradiction or induction to prove that the graph is strongly connected.. That's where I confused.

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  • 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

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  • determine if line segment is inside polygon

    - by dato
    suppose we have simple polygon(without holes) with vertices (v0,v1,....vn) my aim is to determine if for given point p(x,y) any line segment connecting this point and any vertices of polygon is inside polygon or even for given two point p(x0,y0) `p(x1,y1)` line segment connecting these two point is inside polygon? i have searched many sites about this ,but i am still confused,generally i think we have to compare coordinates of vertices and by determing coordinates of which point is less or greater to another point's coordinates,we could determine location of any line segment,but i am not sure how correct is this,please help me

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  • How to find the largest power of 2 less than the given number

    - by nazar_art
    I need to find the largest power of 2 less than the given number. And I stuck and can't find any solution. Code: public class MathPow { public int largestPowerOf2 (int n) { int res = 2; while (res < n) { res =(int)Math.pow(res, 2); } return res; } } This doesn't work correctly. Testing output: Arguments Actual Expected ------------------------- 9 16 8 100 256 64 1000 65536 512 64 256 32 How to solve this issue?

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  • 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

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  • 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

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  • Special simple random number generator

    - by psihodelia
    How to create a function, which on every call generates a random integer number? This number must be most random as possible (according to uniform distribution). It is only allowed to use one static variable and at most 3 elementary steps, where each step consists of only one basic arithmetic operation of arity 1 or 2. Example: int myrandom(void){ static int x; x = some_step1; x = some_step2; x = some_step3; return x; } Basic arithmetic operations are +,-,%,and, not, xor, or, left shift, right shift, multiplication and division. Of course, no rand(), random() or similar staff is allowed.

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  • 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.

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  • What guarantees are there on the run-time complexity (Big-O) of LINQ methods?

    - by tzaman
    I've recently started using LINQ quite a bit, and I haven't really seen any mention of run-time complexity for any of the LINQ methods. Obviously, there are many factors at play here, so let's restrict the discussion to the plain IEnumerable LINQ-to-Objects provider. Further, let's assume that any Func passed in as a selector / mutator / etc. is a cheap O(1) operation. It seems obvious that all the single-pass operations (Select, Where, Count, Take/Skip, Any/All, etc.) will be O(n), since they only need to walk the sequence once; although even this is subject to laziness. Things are murkier for the more complex operations; the set-like operators (Union, Distinct, Except, etc.) work using GetHashCode by default (afaik), so it seems reasonable to assume they're using a hash-table internally, making these operations O(n) as well, in general. What about the versions that use an IEqualityComparer? OrderBy would need a sort, so most likely we're looking at O(n log n). What if it's already sorted? How about if I say OrderBy().ThenBy() and provide the same key to both? I could see GroupBy (and Join) using either sorting, or hashing. Which is it? Contains would be O(n) on a List, but O(1) on a HashSet - does LINQ check the underlying container to see if it can speed things up? And the real question - so far, I've been taking it on faith that the operations are performant. However, can I bank on that? STL containers, for example, clearly specify the complexity of every operation. Are there any similar guarantees on LINQ performance in the .NET library specification?

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  • 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!

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  • 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.

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  • longest increasing subsequent

    - by davit-datuashvili
    i have write this code is it correct? public class subsequent{ public static void main(String[] args){ int a[]=new int[]{0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15}; int a_b[]=new int[a.length]; a_b[0]=1; int int_max=0; int int_index=0; for (int i=0;i<a.length;i++){ for (int j=0;j<i;j++){ if (a[i]>a[j] && a_b[i]<(a_b[j]+1)){ a_b[i]=a_b[j]+1; } } if (a_b[i]>int_max){ int_max=a_b[i]; int_index=i; } } int k=int_max+1; int list[]=new int[k]; for (int i=int_index;i>0;i--){ if (a_b[i]==k-1){ list[k-1]=a[i]; k=a_b[i]; } } for (int g=0;g<list.length;g++){ System.out.println(list[g]); } } }

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  • i have question about job

    - by davit-datuashvili
    hi i will explain my situation i am study algorithms and data structures and i know many thing practical and others most thoereticaly and trying to implement in practise i want to get job where i will study more quickly i am interested if i have chance to start job when i have not expierence? thanks please advise what to do

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  • question about Littles Law

    - by davit-datuashvili
    I know that Little's Law states (paraphrased): the average number of things in a system is the product of the average rate at which things leave the system and the average time each one spends in the system, or: n=x*(r+z); x-throughput r-response time z-think time r+z - average response time now i have question about a problem from programming pearls: Suppose that system makes 100 disk accesses to process a transaction (although some systems require fewer, some systems will require several hundred disk access per transaction). How many transactions per hour per disk can the system handle? please help

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  • Programmatically determine the relative "popularities" of a list of items (books, songs, movies, etc

    - by Horace Loeb
    Given a list of (say) songs, what's the best way to determine their relative "popularity"? My first thought is to use Google Trends. This list of songs: Subterranean Homesick Blues Empire State of Mind California Gurls produces the following Google Trends report: (to find out what's popular now, I restricted the report to the last 30 days) Empire State of Mind is marginally more popular than California Gurls, and Subterranean Homesick Blues is far less popular than either. So this works pretty well, but what happens when your list is 100 or 1000 songs long? Google Trends only allows you to compare 5 terms at once, so absent a huge round-robin, what's the right approach? Another option is to just do a Google Search for each song and see which has the most results, but this doesn't really measure the same thing

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  • 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?

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