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  • 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?

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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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  • Firefox: how to block cookies by name, not by site?

    - by deepc
    Firefox allows to block all cookies on a site-by-site level. This is ok for the most part. However, it does not help with blocking only Google Analytics cookies. The GA cookie names start with __ut. Is there a Firefox add-on which can block all __ut* cookies? I know there are many cookie related add-ons for Firefox - but apparently all of them simply fine tune cookie site-by-site blocking, according to their descriptions. Hopefully I missed the one who can do this... I also know about Google's plugin to opt out of analytics. Installing a specific plug-in for that purpose (as opposed to an add-on) seems a bit overdone. Plus, I would have to trust Google with that and that is exactly what I don't.

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  • How to calculate the y-pixels of someones weight on a graph? (math+programming question)

    - by RexOnRoids
    I'm not that smart like some of you geniuses. I need some help from a math whiz. My app draws a graph of the users weight over time. I need a surefire way to always get the right pixel position to draw the weight point at for a given weight. For example, say I want to plot the weight 80.0(kg) on the graph when the range of weights is 80.0 to 40.0kg. I want to be able to plug in the weight (given I know the highest and lowest weights in the range also) and get the pixel result 400(y) (for the top of the graph). The graph is 300 pixels high (starts at 100 and ends at 400). The highest weight 80kg would be plot at 400 while the lowest weight 40kg would be plot at 100. And the intermediate weights should be plotted appropriately. I tried this but it does not work: -(float)weightToPixel:(float)theWeight { float graphMaxY = 400; //The TOP of the graph float graphMinY = 100; //The BOTTOM of the graph float yOffset = 100; //Graph itself is offset 100 pixels in the Y direction float coordDiff = graphMaxY-graphMinY; //The size in pixels of the graph float weightDiff = self.highestWeight-self.lowestWeight; //The weight gap float pixelIncrement = coordDiff/weightDiff; float weightY = (theWeight*pixelIncrement)-(coordDiff-yOffset); //The return value return weightYpixel; }

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  • 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

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  • Trie vs B+ tree

    - by Fakrudeen
    How does Trie and B+ tree compare for indexing lexicographically sorted strings [on the order some billions]? It should support range queries as well. From perf. as well as implementation complexity point of view.

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  • question about tree

    - by davit-datuashvili
    i have question for example i want to implement binary tree with array i want understand what will indexes for left child and rigth child ?my array is 0 based i want implement searching in tree using array can anybody help me?

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  • 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)

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  • Python: Determine whether list of lists contains a defined sequence

    - by duhaime
    I have a list of sublists, and I want to see if any of the integer values from the first sublist plus one are contained in the second sublist. For all such values, I want to see if that value plus one is contained in the third sublist, and so on, proceeding in this fashion across all sublists. If there is a way of proceeding in this fashion from the first sublist to the last sublist, I wish to return True; otherwise I wish to return False. In other words, for each value in sublist one, for each "step" in a "walk" across all sublists read left to right, if that value + n (where n = number of steps taken) is contained in the current sublist, the function should return True; otherwise it should return False. (Sorry for the clumsy phrasing--I'm not sure how to clean up my language without using many more words.) Here's what I wrote. a = [ [1,3],[2,4],[3,5],[6],[7] ] def find_list_traversing_walk(l): for i in l[0]: index_position = 0 first_pass = 1 walking_current_path = 1 while walking_current_path == 1: if first_pass == 1: first_pass = 0 walking_value = i if walking_value+1 in l[index_position + 1]: index_position += 1 walking_value += 1 if index_position+1 == len(l): print "There is a walk across the sublists for initial value ", walking_value - index_position return True else: walking_current_path = 0 return False print find_list_traversing_walk(a) My question is: Have I overlooked something simple here, or will this function return True for all true positives and False for all true negatives? Are there easier ways to accomplish the intended task? I would be grateful for any feedback others can offer!

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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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  • Calculating color shades

    - by matejv
    I have the next problem. I have a base color with couple of different shades of that color. Example: Base color: #4085c5 Shade: #005cb1 Now, I have a different color (let's say #d60620), but no shades of it. From the color I would like to calculate shades, that have similar difference as colors mentioned in first paragraph. First I tried calculating difference of RGB elements and applying them to second color, but the result was not like I expected to be. Than I tried with converting color to HSV, reading saturation value and applying the difference to second color, but again the resulting color was still weird. The formula was something like: (HSV(BaseColor)[S] - HSV(Shade)[S]) + HSV(SecondColor)[H] Does anyone know how this problem could be solved? I know I am doing something wrong, but I don't know what. :)

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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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  • 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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  • Why is i-- faster than i++ in loops? [closed]

    - by Afshin Mehrabani
    Possible Duplicate: JavaScript - Are loops really faster in reverse…? I don't know if this question is valid in other languages or not, but I'm asking this specifically for JavaScript. I see in some articles and questions that the fastest loop in JavaScript is something like: for(var i = array.length; i--; ) Also in Sublime Text 2, when you try to write a loop, it suggests: for (var i = Things.length - 1; i >= 0; i--) { Things[i] }; I want to know, why is i-- faster than i++ in loops?

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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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  • 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

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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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