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  • Setting last N bits in an array

    - by Martin
    I'm sure this is fairly simple, however I have a major mental block on it, so I need a little help here! I have an array of 5 integers, the array is already filled with some data. I want to set the last N bits of the array to be random noise. [int][int][int][int][int] set last 40 bits [unchanged][unchanged][unchanged][24 bits of old data followed 8 bits of randomness][all random] This is largely language agnostic, but I'm working in C# so bonus points for answers in C#

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  • Effective way to calculate a similarity percentage between data sets

    - by UltraVi01
    I am currently working with User objects -- each of which have many Goal objects. The Goal objects are not User specific, that is, Users can share the same Goal. I am attempting to fashion a way to calculate a "similarity percentage" between two Users... (i.e., taking into account how many Goals they share as well as how many Goals they do not share) Does anyone have experience with this type of situation? I am using Grails with Mysql if that is helpful. Thanks

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  • 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

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  • Optimizing a Parking Lot Problem. What algorithims should I use to fit the most amount of cars in th

    - by Adam Gent
    What algorithms (brute force or not) would I use to put in as many cars (assume all cars are the same size) in a parking lot so that there is at least one exit (from the container) and a car cannot be blocked. Or can someone show me an example of this problem solved programmatically. The parking lot varies in shape would be nice but if you want to assume its some invariant shape that is fine. Another Edit: Assume that driving distance in the parking lot is not a factor (although it would be totally awesome if it was weighted factor to number of cars in lot). Another Edit: Assume 2 Dimensional (no cranes or driving over cars). Another Edit: You cannot move cars around once they are parked (its not a valet parking lot). I hope the question is specific enough now.

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  • Given a number N, find the number of ways to write it as a sum of two or more consecutive integers

    - by hilal
    Here is the problem (Given a number N, find the number of ways to write it as a sum of two or more consecutive integers) and example 15 = 7+8, 1+2+3+4+5, 4+5+6 I solved with math like that : a + (a + 1) + (a + 2) + (a + 3) + ... + (a + k) = N (k + 1)*a + (1 + 2 + 3 + ... + k) = N (k + 1)a + k(k+1)/2 = N (k + 1)*(2*a + k)/2 = N Then check that if N divisible by (k+1) and (2*a+k) then I can find answer in O(N) time Here is my question how can you solve this by dynamic-programming ? and what is the complexity (O) ? P.S : excuse me, if it is a duplicate question. I searched but I can find

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  • Elegant Method of Inserting Code Between Loops

    - by DeathMagus
    In web development, I often find I need to format and print various arrays of data, and separate these blocks of data in some manner. In other words, I need to be able to insert code between each loop, without said code being inserted before the first entry or after the last one. The most elegant way I've found to accomplish this is as follows: function echoWithBreaks($array){ for($i=0; $i<count($array); $i++){ //Echo an item if($i<count($array)-1){ //Echo "between code" } } } Unfortunately, there's no way that I can see to implement this solution with foreach instead of for. Does anyone know of a more elegant solution that will work with foreach?

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  • Find recipes that can be cooked from provided ingridients

    - by skaurus
    Sorry for bad English :( Suppose i can preliminary organize recipes and ingredients data in any way. How can i effectively conduct search of recipes by user-provided ingredients, preferably sorted by max match - so, first going recipes that use maximum of provided ingridients and do not contain any other ingrs, after them recipes that uses less of provided set and still not any other ingrs, after them recipes with minimum additional requirements and so on? All i can think about is represent recipe ingridients like bitmasks, and compare required bitmask with all recipes, but it is obviously a bad way to go. And related things like Levenstein distance i don't see how to use here. I believe it should be quite common task...

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  • Find a node in a Graph that minimizes the distance between two other nodes

    - by Andrés
    Here is the thing. I have a directed weighted graph G, with V vertices and E edges. Given two nodes in the graph, let's say A, and B, and given the weight of an edge A-B denoted as w(A, B), I need to find a node C so that max(w(A, C), w(B, C)) is minimal among all possibilities. By possibilities I mean all the values C can take. I don't know if it is completely clear, if it's not, I'll try to be more precise. Thanks in advance.

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  • 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?

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  • Finding if a string is an iterative substring?

    - by EsotericMe
    I have a string S. How can I find if the string follows S = nT. Examples: Function should return true if 1) S = "abab" 2) S = "abcdabcd" 3) S = "abcabcabc" 4) S = "zzxzzxzzx" But if S="abcb" returns false. I though maybe we can repeatedly call KMP on substrings of S and then decide. eg: for "abab": call on KMP on "a". it returns 2(two instances). now 2*len("a")!=len(s) call on KMP on "ab". it returns 2. now 2*len("ab")==len(s) so return true Can you suggest any better algorithms?

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  • 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?

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  • 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

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  • 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?

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  • 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

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  • 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?

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  • 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 :).

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  • 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?)

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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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  • 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).

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