Search Results

Search found 5298 results on 212 pages for 'marching cubes algorithm'.

Page 92/212 | < Previous Page | 88 89 90 91 92 93 94 95 96 97 98 99  | Next Page >

  • help implementing All Nearest Smaller Values algorithm

    - by davit-datuashvili
    http://en.wikipedia.org/wiki/All_nearest_smaller_values this is site of the problem and here is my code but i have some trouble to implement it import java.util.*; public class stack{ public static void main(String[]args){ int x[]=new int[]{ 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15 }; Stack<Integer> st=new Stack<Integer>(); for (int a:x){ while (!st.empty() && st.pop()>=a){ System.out.println( st.pop()); if (st.empty()){ break; } else{ st.push(a); } } } } } and here is pseudo code from site S = new empty stack data structure for x in the input sequence: while S is nonempty and the top element of S is greater than or equal to x: pop S if S is empty: x has no preceding smaller value else: the nearest smaller value to x is the top element of S push x onto S

    Read the article

  • LinkedList.contains execution speed

    - by Le_Coeur
    Why Methode LinkedList.contains() runs quickly than such implementation: for (String s : list) if (s.equals(element)) return true; return false; I don't see great difference between this to implementations(i consider that search objects aren't nulls), same iterator and equals operation

    Read the article

  • Number distribution

    - by Carra
    Problem: We have x checkboxes and we want to check y of them evenly. Example 1: select 50 checkboxes of 100 total. [-] [x] [-] [x] ... Example 2: select 33 checkboxes of 100 total. [-] [-] [x] [-] [-] [x] ... Example 3: select 66 checkboxes of 100 total: [-] [x] [x] [-] [x] [x] ... But we're having trouble to come up with a formula to check them in code, especially once you go 11/111 or something similar. Anyone has an idea?

    Read the article

  • Generate sample XML from composite hierarchical structure

    - by Jevgenij Nekrasov
    Let's say I have composite hierarchical structure. Each object in the structure has child collection of the same objects. Each object has XPath property, which stores the exact xpath to the element inside XML file. Right now I am trying to create an extension, which can generate XML string from that hierarchical structure, but first I want to make some sort of research what is the most efficient way to do that? ant what is the easiest way?

    Read the article

  • Using MySQL as a job queue

    - by user237815
    I'd like to use MySQL as a job queue. Multiple machines will be producing and consuming jobs. Jobs need to be scheduled; some may run every hour, some every day, etc. It seems fairly straightforward: for each job, have a "nextFireTime" column, and have worker machines search for the job with the nextFireTime, change the status of the record to "inProcess", and then update the nextFireTime when the job ends. The problem comes in when a worker dies silently. It won't be able to update the nextFireTime or set the status back to "idle". Unfortunately, jobs can be long-running, so a reaper thread that looks for jobs that have been inProcess too long isn't an option. There's no timeout value that would work. Can anyone suggest a design pattern that would properly handle unreliable worker machines?

    Read the article

  • Faster way to compare two sets of points in N-dimensional space?

    - by Amit
    List1 contains a high number (~7^10) of N-dimensional points (N <=10), List2 contains the same or fewer number of N-dimensional points (N <=10). My task is this: I want to check which point in List2 is closest (euclidean distance) to a point in List1 for every point in List1 and subsequently perform some operation on it. I have been doing it the simple- the nested loop way when I didn't have more than 50 points in List1, but with 7^10 points, this obviously takes up a lot of time. What is the fastest way to do this? Any concepts from Computational Geometry might help?

    Read the article

  • maximum of given function

    - by davit-datuashvili
    first of all i am doing programs in java language this code is merely taken from web site i have not question about divide and conqurer but about function and it's argument here is code of ternary search def ternarySearch(f, left, right, absolutePrecision): #left and right are the current bounds; the maximum is between them if (right - left) < absolutePrecision: return (left + right)/2 leftThird = (2*left + right)/3 rightThird = (left + 2*right)/3 if f(leftThird) < f(rightThird): return ternarySearch(f, leftThird, right, absolutePrecision) return ternarySearch(f, left, rightThird, absolutePrecision) i am not asking once again how implement it in java i am asking for example how define function?for example let y=x^+3 yes we can determine it as public static int y(int x){ return x*x+3; } but here return ternarySearch(f, leftThird, right, absolutePrecision) function f does not have argument and how do such?please help me

    Read the article

  • How to prevent overdrawing?

    - by afriza
    This is a difficult question to search in Google since it has other meaning in finance. Of course, what I mean here is "Drawing" as in .. computer graphics.. not money.. I am interested in preventing overdrawing for both 3D Drawing and 2D Drawing. (should I make them into two different questions?)

    Read the article

  • Efficient mapping of game entity positions in Java

    - by byte
    In Java (Swing), say I've got a 2D game where I have various types of entities on the screen, such as a player, bad guys, powerups, etc. When the player moves across the screen, in order to do efficient checking of what is in the immediate vicinity of the player, I would think I'd want indexed access to the things that are near the character based on their position. For example, if player 'P' steps onto element 'E' in the following example... | | | | | | | | | |P| | | | |E| | | | | | | | | ... would be to do something like: if(player.getPosition().x == entity.getPosition().x && entity.getPosition.y == thing.getPosition().y) { //do something } And thats fine, but that implies that the entities hold their positions, and therefor if I had MANY entities on the screen I would have to loop through all possible entities available and check each ones position against the player position. This seems really inefficient especially if you start getting tons of entities. So, I would suspect I'd want some sort of map like Map<Point, Entity> map = new HashMap<Point, Entity>(); And store my point information there, so that I could access these entities in constant time. The only problem with that approach is that, if I want to move an entity to a different point on the screen, I'd have to search through the values of the HashMap for the entity I want to move (inefficient since I dont know its Point position ahead of time), and then once I've found it remove it from the HashMap, and re-insert it with the new position information. Any suggestions or advice on what sort of data structure / storage format I ought to be using here in order to have efficient access to Entities based on their position, as well as Position's based on the Entity?

    Read the article

  • Creating objects makes the VM faster?

    - by Sudhir Jonathan
    Look at this piece of code: MessageParser parser = new MessageParser(); for (int i = 0; i < 10000; i++) { parser.parse(plainMessage, user); } For some reason, it runs SLOWER (by about 100ms) than for (int i = 0; i < 10000; i++) { MessageParser parser = new MessageParser(); parser.parse(plainMessage, user); } Any ideas why? The tests were repeated a lot of times, so it wasn't just random. How could creating an object 10000 times be faster than creating it once?

    Read the article

  • Looking for ideas on automatically arranging a set of objects (furniture) in a virtual room in AS3

    - by raf
    First of all, I don't want to visually arrange 3D models dragging them with the mouse, all I want is: Given a room of certain dimensions (L,W,H) and given a set of elements like beds, chairs, etc (with L,W,H dimensions, of course) I want to automatically arrange those elements to take advantage of the space as much as I can. So I want to be able to put as much furniture as I can in a given room. At the end I need to represent the arranged items visually, inside the room. My first thought was to use an array of items and sorting it with array.sortOn(["l","w","h"] Array.NUMERIC) and then define a gap between the objects and make the maths to put the objects one next to another, etc. but that isn't a good approach because some items may be placed on top of another ones (boxes of the same size, boxes on top of tables, etc). I really don't have experience on 3D programming, that's why I'm asking for help. Thanks in advance.

    Read the article

  • Formula for popularity? (based on "like it", "comments", "views")

    - by paullb
    I have some pages on a website and I have to create an ordering based on "popularity"/"activity" The parameters that I have to use are: views to the page comments made on the page (there is a form at the bottom where uses can make comments) clicks made to the "like it" icon Are there any standards for what a formula for popularity would be? (if not opinions are good too) (initially I thought of views + 10*comments + 10*likeit)

    Read the article

  • Credit Card checksums and validations that do not require connection to the financial institution

    - by cjavapro
    The validations I know of are: Checksum the whole card number should add up to zero. (range is 0-9) Check the first digit(s) against the card type Check the length against the card type Check the CCV length against the card type (I think all the major types are 3 anyway) Of course make sure it is accepted card type as well as non expired. Are there any other validations :) (I expect many folks did not know about all of these) The reason I ask is because I overheard there was one to checksum number against expiration or CCV.. I just wanted to check.

    Read the article

  • Counting bits set in a .Net BitArray Class

    - by Sam
    I am implementing a library where I am extensively using the .Net BitArray class and need an equivalent to the Java BitSet.Cardinality() method, i.e. a method which returns the number of bits set. I was thinking of implementing it as an extension method for the BitArray class. The trivial implementation is to iterate and count the bits set (like below), but I wanted a faster implementation as I would be performing thousands of set operations and counting the answer. Is there a faster way than the example below? count = 0; for (int i = 0; i < mybitarray.Length; i++) { if (mybitarray [i]) count++; }

    Read the article

  • How to find kth minimal element in the union of two sorted arrays?

    - by Michael
    This is a homework question. They say it takes O(logN + logM) where N and M are the arrays lengths. Let's name the arrays a and b. Obviously we can ignore all a[i] and b[i] where i k. First let's compare a[k/2] and b[k/2]. Let b[k/2] a[k/2]. Therefore we can discard also all b[i], where i k/2. Now we have all a[i], where i < k and all b[i], where i < k/2 to find the answer. What is the next step?

    Read the article

  • Recursion - Ship Battle

    - by rgorrosini
    I'm trying to write a little ship battle game in java. It is 100% academic, I made it to practice recursion, so... I want to use it instead of iteration, even if it's simpler and more efficient in most some cases. Let's get down to business. These are the rules: Ships are 1, 2 or 3 cells wide and are placed horizontally only. Water is represented with 0, non-hit ship cells are 1, hit ship cells are 2 and sunken ships have all it's cells in 3. With those rules set, I'm using the following array for testing: int[][] board = new int[][] { {0, 1, 2, 0, 1, 0}, {0, 0, 1, 1, 1, 0}, {0, 3, 0, 0, 0, 0}, {0, 0, 2, 1, 2, 0}, {0, 0, 0, 1, 1, 1}, }; It works pretty good so far, and to make it more user-friendly I would like to add a couple of reports. these are the methods I need for them: Given the matrix, return the amount of ships in it. Same as a), but separating them by state (amount of non-hit ships, hit and sunken ones). I will need a hand with those reports, and I would like to get some ideas. Remember it must be done using recursion, I want to understand this, and the only way to go is practice! Thanks a lot for your time and patience :).

    Read the article

  • Are fragments of hashes collision-resistent?

    - by Mark
    Let me see if someone would mind clearing up this elementary point about md5 and hashing. If you only use the first 4 bytes of an md5 hash, would that mean theoretically only 1 in 255^4 chance of collision. iow is that the intention with it (and other hash algorithms) - that you only have to use a small portion of the returned hash (say the hash is of a file of some size).

    Read the article

  • 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 }

    Read the article

< Previous Page | 88 89 90 91 92 93 94 95 96 97 98 99  | Next Page >