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  • bidirectional buble sort

    - by davit-datuashvili
    Here is the code I'm using for shacker sort (or bidirectional buble sort). Something is wrong; the error is java.lang.ArrayIndexOutOfBoundsException. Can anybody help me? public class bidirectional{ public static void main(String[]args){ int x[]=new int[]{12,9,4,99,120,1,3,10}; int j; int n=x.length; int st=-1; while (st<n){ st++; n--; for (j=st;j<n;j++) { if (x[j]>x[j+1]) { int t=x[j]; x[j]=x[j+1]; x[j+1]=t; } } for (j=n;--j>=st;) { if (x[j]>x[j+1]) { int t=x[j]; x[j]=x[j+1]; x[j+1]=t; } } } for (int k=0;k<x.length;k++) { System.out.println(x[k]); } } }

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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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  • 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 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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  • Word Counter Implementation

    - by kenny
    Is there a better way than the following brute foce implementation of a c# word counting class? UPDATED CODE: Sorry! /// <summary> /// A word counting class. /// </summary> public class WordCounter { Dictionary<string, int> dictTest = new Dictionary<string, int> (); /// <summary> /// Enters a word and returns the current number of times that word was found. /// </summary> /// <param name="word">The word or string found.</param> /// <returns>Count of times Found() was called with provided word.</returns> public int Found ( string word ) { int count = 1; return dictTest.TryGetValue ( word, out count ) ? ++dictTest[word] : dictTest[word] = 1; } }

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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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  • 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?

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  • Best way to reverse a string in C# 2.0

    - by Guy
    I've just had to write a string reverse function in C# 2.0 (i.e. LINQ not available) and came up with this: public string Reverse(string text) { char[] cArray = text.ToCharArray(); string reverse = String.Empty; for (int i = cArray.Length - 1; i > -1; i--) { reverse += cArray[i]; } return reverse; } Personally I'm not crazy about the function and am convinced that there's a better way to do it. Is there?

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  • question about quicksort 3 way partition

    - by davit-datuashvili
    i want implement quicksort 3 way partition here is code public class quick3{ public static void quicksort3(int a[],int l,int r){ int k; int v=a[r]; if (r<=l) return; int i=l; int j=r; int p=l-1; int q=r; for (;;) { while (a[++i]<v); while (v<a[--j]) if (j==i) break; if (i>=j) break; swap( a,i, j); if (a[i]==v){ p++; swap(a,p,i);} if (v==a[j]){ q--; swap( a,q,j); } } swap(a,i,r); j=i-1; i=i+1; for (k=1;k<=p;k++,j--) swap(a,k,j); for (k=r-1;k>=q;k--,i++) swap(a,k,i); quicksort3(a,l,j); quicksort3(a,i,r); } public static void main(String[]args){ int a[]=new int[]{4,6,5,9,7,8,3}; quicksort3(a,0,a.length-1); for (int i=0;i<a.length;i++){ System.out.println(a[i]); } } public static void swap(int a[],int i,int j){ int t=a[i]; a[i]=a[j]; a[j]=t; } } after change result is 4 8 7 6 3 5 9 any suggestion?please help

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  • 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?

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  • 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?

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  • 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?

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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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  • 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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  • please help me to choose good books on algorithms [closed]

    - by davit-datuashvili
    Possible Duplicate: What is the best book for learning about Algorithms? i want to help me to choose good books on algorithms many people from this site say me that show me your code and now i ask u to help me to choose good books on algorithms please i have not books on algorithms and in case i decide to buy it of course must buy book which has high quality yes? so please any ideas ?links everything

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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 prove worst-case number of inversions in a heap is O(nlogn)?

    - by Jacques
    I am busy preparing for exams, just doing some old exam papers. The question below is the only one I can't seem to do (I don't really know where to start). Any help would be appreciated greatly. Use the O(nlogn) comparison sort bound, the theta(n) bound for bottom-up heap construction, and the order complexity if insertion sort to show that the worst-case number of inversions in a heap is O(nlogn).

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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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  • extra storage merge sort

    - by davit-datuashvili
    I need make a merge sort using an additional array. Here is my code: public class extra_storage{ public static void main(String[]args) { int x[]=new int[]{12,9,4,99,120,1,3,10}; int a[]=new int[x.length]; mergesort(x,0,x.length-1,a); for (int i=0;i<x.length;i++){ System.out.println(x[i]); } } public static void mergesort(int x[],int low,int high, int a[]){ if (low>=high) { return; } int middle=(low+high)/2; mergesort(x,low,middle,a); mergesort(x,middle+1,high,a); int k; int lo=low; int h=high; for (k=low;k<high;k++) if ((lo<=middle ) && ((h>high)||(x[lo]<x[h]))){ a[k]=x[lo++]; } else { a[k]=x[h++]; } for (k=low;k<=high;k++){ x[k]=a[k]; } } } But something is wrong. When I run it the output is this: 1 0 3 0 4 0 9 0 What is the problem?

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  • 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.

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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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  • 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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