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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • How to perform a Depth First Search iteratively using async/parallel processing?

    - by Prabhu
    Here is a method that does a DFS search and returns a list of all items given a top level item id. How could I modify this to take advantage of parallel processing? Currently, the call to get the sub items is made one by one for each item in the stack. It would be nice if I could get the sub items for multiple items in the stack at the same time, and populate my return list faster. How could I do this (either using async/await or TPL, or anything else) in a thread safe manner? private async Task<IList<Item>> GetItemsAsync(string topItemId) { var items = new List<Item>(); var topItem = await GetItemAsync(topItemId); Stack<Item> stack = new Stack<Item>(); stack.Push(topItem); while (stack.Count > 0) { var item = stack.Pop(); items.Add(item); var subItems = await GetSubItemsAsync(item.SubId); foreach (var subItem in subItems) { stack.Push(subItem); } } return items; } EDIT: I was thinking of something along these lines, but it's not coming together: var tasks = stack.Select(async item => { items.Add(item); var subItems = await GetSubItemsAsync(item.SubId); foreach (var subItem in subItems) { stack.Push(subItem); } }).ToList(); if (tasks.Any()) await Task.WhenAll(tasks); UPDATE: If I wanted to chunk the tasks, would something like this work? foreach (var batch in items.BatchesOf(100)) { var tasks = batch.Select(async item => { await DoSomething(item); }).ToList(); if (tasks.Any()) { await Task.WhenAll(tasks); } } The language I'm using is C#.

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  • Simple loop, which one I would get more performance and which one is recommended? defining a variable inside a loop or outside of it?

    - by Grego
    Variable outside of the loop int number = 0; for(int i = 0; i < 10000; i++){ number = 3 * i; printf("%d",number); } or Variable inside of the loop for(int i = 0; i < 10000; i++){ int number = 3 * i; printf("%d",number); } Which one is recommended and which one is better in performance? Edit: This is just an example to exhibit what I mean, All I wanna know is if defining a variable inside a loop and outside a loop means the same thing , or there's a difference.

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  • question about function derivatives

    - by davit-datuashvili
    hi i have question for example i have some function int sumefunction(//parameters here let say int x){ return something let say x*x+2*x+3 or does not matter } i am interesting how find derivative of this function ?if i have int f(int x){ return sin(x); } after derivative it must return cos(x) thanks

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  • more efficient version of this?

    - by john connor
    i have this thingy here : function numOfPackets(bufferSize, packetSize) { if (bufferSize <= 0 || packetSize > bufferSize) return 0; if (packetSize < 0) throw Error(); var out = 0; for(;;){ out++; bufferSize = bufferSize - packetSize; if( packetSize > bufferSize ) break; } return out; } which i run at often , can u give me more efficent variant of it?

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  • How does pattern matching work behind the scenes in F#?

    - by kryptic
    Hello Everyone, I am completely new to F# (and functional programming in general) but I see pattern matching used everywhere in sample code. I am wondering for example how pattern matching actually works? For example, I imagine it working the same as a for loop in other languages and checking for matches on each item in a collection. This is probably far from correct, how does it actually work behind the scenes?

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  • how to fast compute distance between high dimension vectors

    - by chyojn
    assume there are three group of high dimension vectors: {a_1, a_2, ..., a_N}, {b_1, b_2, ... , b_N}, {c_1, c_2, ..., c_N}. each of my vector can be represented as: x = a_i + b_j + c_k, where 1 <=i, j, k <= N. then the vector is encoded as (i, j, k) wich is then can be decoded as x = a_i + b_j + c_k. my question is, if there are two vector: x = (i_1, j_1, k_1), y = (i_2, j_2, k_2), is there a method to compute the euclidian distance of these two vector without decode x and y.

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  • public (static) swap() method vs. redundant (non-static) private ones...

    - by Helper Method
    I'm revisiting data structures and algorithms to refresh my knowledge and from time to time I stumble across this problem: Often, several data structures do need to swap some elements on the underlying array. So I implement the swap() method in ADT1, ADT2 as a private non-static method. The good thing is, being a private method I don't need to check on the parameters, the bad thing is redundancy. But if I put the swap() method in a helper class as a public static method, I need to check the indices every time for validity, making the swap call very unefficient when many swaps are done. So what should I do? Neglect the performance degragation, or write small but redundant code?

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  • Distributing players to tables

    - by IVlad
    Consider N = 4k players, k tables and a number of clans such that each member can belong to one clan. A clan can contain at most k players. We want to organize 3 rounds of a game such that, for each table that seats exactly 4 players, no 2 players sitting there are part of the same clan, and, for the later rounds, no 2 players sitting there have sat at the same table before. All players play all rounds. How can we do this efficiently if N can be about ~80 large? I thought of this: for each table T: repeat until 4 players have been seated at T: pick a random player X that is not currently seated anywhere if X has not sat at the same table as anyone currently at T AND X is not from the same clan as anyone currently at T seat X at T break I am not sure if this will always finish or if it can get stuck even if there is a valid assignment. Even if this works, is there a better way to do it?

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  • Adapting pseudocode to java implementation for finding the longest word in a trie

    - by user1766888
    Referring to this question I asked: How to find the longest word in a trie? I'm having trouble implementing the pseudocode given in the answer. findLongest(trie): //first do a BFS and find the "last node" queue <- [] queue.add(trie.root) last <- nil map <- empty map while (not queue.empty()): curr <- queue.pop() for each son of curr: queue.add(son) map.put(son,curr) //marking curr as the parent of son last <- curr //in here, last indicate the leaf of the longest word //Now, go up the trie and find the actual path/string curr <- last str = "" while (curr != nil): str = curr + str //we go from end to start curr = map.get(curr) return str This is what I have for my method public static String longestWord (DTN d) { Queue<DTN> holding = new ArrayQueue<DTN>(); holding.add(d); DTN last = null; Map<DTN,DTN> test = new ArrayMap<DTN,DTN>(); DTN curr; while (!holding.isEmpty()) { curr = holding.remove(); for (Map.Entry<String, DTN> e : curr.children.entries()) { holding.add(curr.children.get(e)); test.put(curr.children.get(e), curr); } last = curr; } curr = last; String str = ""; while (curr != null) { str = curr + str; curr = test.get(curr); } return str; } I'm getting a NullPointerException at: for (Map.Entry<String, DTN> e : curr.children.entries()) How can I find and fix the cause of the NullPointerException of the method so that it returns the longest word in a trie?

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  • Find a common element within N arrays

    - by kunjaan
    If I have N arrays, what is the best(Time complexity. Space is not important) way to find the common elements. You could just find 1 element and stop. Edit: The elements are all Numbers. Edit: These are unsorted. Please do not sort and scan. This is not a homework problem. Somebody asked me this question a long time ago. He was using a hash to solve the problem. I was thinking if SO has solved similar problems.

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  • Algorithms for finding the intersections of intervals

    - by tomwu
    I am wondering how I can find the number of intervals that intersect with the ones before it. for the intervals [2, 4], [1, 6], [5, 6], [0, 4], the output should be 2. from [2,4] [5,6] and [5,6] [0,4]. So now we have 1 set of intervals with size n all containing a point a, then we add another set of intervals size n as well, and all of the intervals are to the right of a. Can you do this in O(nlgn) and O(nlg^2n)?

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  • Help to learn Image Search algorithm

    - by R Manu
    I am a beginner in image processing. I want to write an application in C++ or in C# for Searching an image in a list of images Searching for a particular feature (for e.g. face) in a list of images. Can anybody suggest where should I start from? What all should I learn before doing this? Where can I find the correct information regarding this?

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  • question about in -place sort

    - by davit-datuashvili
    for example we have following array char data[]=new char[]{'A','S','O','R','T','I','N','G','E','X','A','M','P','L','E'}; and index array int a[]=new int[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14}: void insitu(char data[],int a[],N){ for (int i=0;i<N;i++) { char v=data[i]; int j,int k; for (k=i;a[k]!=i;k=a[j];a[j]=j) { j=k;data[k]=data[a[k]; } data[k]=v; a[k]=k; } i have question what is initialize value of j? when i run this code it asks me to initialize j and what should do? }

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  • F# insert/remove item from list

    - by Timothy
    How should I go about removing a given element from a list? As an example, say I have list ['A'; 'B'; 'C'; 'D'; 'E'] and want to remove the element at index 2 to produce the list ['A'; 'B'; 'D'; 'E']? I've already written the following code which accomplishes the task, but it seems rather inefficient to traverse the start of the list when I already know the index. let remove lst i = let rec remove lst lst' = match lst with | [] -> lst' | h::t -> if List.length lst = i then lst' @ t else remove t (lst' @ [h]) remove lst [] let myList = ['A'; 'B'; 'C'; 'D'; 'E'] let newList = remove myList 2 Alternatively, how should I insert an element at a given position? My code is similar to the above approach and most likely inefficient as well. let insert lst i x = let rec insert lst lst' = match lst with | [] -> lst' | h::t -> if List.length lst = i then lst' @ [x] @ lst else insert t (lst' @ [h]) insert lst [] let myList = ['A'; 'B'; 'D'; 'E'] let newList = insert myList 2 'C'

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  • question about siftdown operation on heap

    - by davit-datuashvili
    i have following pseudo code which execute siftdown operation on heap array suppose is x void siftdown(int n) pre heap(2,n) && n>=0 post heap(1,n) i=1; loop /*invariant heap(1,n) except perhaps between i and it's (0,1,or 2) children*/ c=2*i; if (c>n) break; // c is left child of i if (c+1)<=n /* c+1 is rigth child of i if (x[c+1]<x[c]) c++ /* c is lesser child of i if (x[i]<=x[c]) break; swap(c,i) i=c; i have wrote following code is it correct? public class siftdown{ public static void main(String[]args){ int c; int n=9; int a[]=new int[]{19,100,17,2,7,3,36,1,25}; int i=1; while (i<n){ c=2*i; if (c>n) break; //c is the left child of i if (c+1<=n) //c+1 ir rigth child of i if (a[c+1]<a[c]) c++; if (a[i]<=a[c]) break; int t=a[c]; a[c]=a[i]; a[i]=t; i=c; } for (int j=0;j<a.length;j++){ System.out.println(a[j]); } } } // result is 19 2 17 1 7 3 36 100 25

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