Search Results

Search found 8262 results on 331 pages for 'optimization algorithm'.

Page 4/331 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • SQLAuthority News – Microsoft SQL Server 2005/2008 Query Optimization & Performance Tuning Training

    - by pinaldave
    Last 3 days to register for the courses. This is one time offer with big discount. The deadline for the course registration is 5th May, 2010. There are two different courses are offered by Solid Quality Mentors 1) Microsoft SQL Server 2005/2008 Query Optimization & Performance Tuning – Pinal Dave Date: May 12-14, 2010 Price: Rs. 14,000/person for 3 days Discount Code: ‘SQLAuthority.com’ Effective Price: Rs. 11,000/person for 3 days 2) SharePoint 2010 – Joy Rathnayake Date: May 10-11, 2010 Price: Rs. 11,000/person for 3 days Discount Code: ‘SQLAuthority.com’ Effective Price: Rs. 8,000/person for 2 days Download the complete PDF brochure. To register, either send an email to [email protected] or call +91 95940 43399. Feel free to drop me an email at pinal “at” SQLAuthority.com for any additional information and clarification. Training Venue: Abridge Solutions, #90/B/C/3/1, Ganesh GHR & MSY Plaza, Vittalrao Nagar, Near Image Hospital, Madhapur, Hyderabad – 500 081. Additionally there is special program of SolidQ India Insider. This is only available to first few registrants of the courses only. Read more details about the course here. Read my TechEd India 2010 experience here. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, SQL, SQL Authority, SQL Optimization, SQL Performance, SQL Query, SQL Server, SQL Tips and Tricks, SQL Training, SQLAuthority News, T SQL, Technology

    Read the article

  • Optimizing Jaro-Winkler algorithm

    - by Pentium10
    I have this code for Jaro-Winkler algorithm taken from this website. I need to run 150,000 times to get distance between differences. It takes a long time, as I run on an Android mobile device. Can it be optimized more? public class Jaro { /** * gets the similarity of the two strings using Jaro distance. * * @param string1 the first input string * @param string2 the second input string * @return a value between 0-1 of the similarity */ public float getSimilarity(final String string1, final String string2) { //get half the length of the string rounded up - (this is the distance used for acceptable transpositions) final int halflen = ((Math.min(string1.length(), string2.length())) / 2) + ((Math.min(string1.length(), string2.length())) % 2); //get common characters final StringBuffer common1 = getCommonCharacters(string1, string2, halflen); final StringBuffer common2 = getCommonCharacters(string2, string1, halflen); //check for zero in common if (common1.length() == 0 || common2.length() == 0) { return 0.0f; } //check for same length common strings returning 0.0f is not the same if (common1.length() != common2.length()) { return 0.0f; } //get the number of transpositions int transpositions = 0; int n=common1.length(); for (int i = 0; i < n; i++) { if (common1.charAt(i) != common2.charAt(i)) transpositions++; } transpositions /= 2.0f; //calculate jaro metric return (common1.length() / ((float) string1.length()) + common2.length() / ((float) string2.length()) + (common1.length() - transpositions) / ((float) common1.length())) / 3.0f; } /** * returns a string buffer of characters from string1 within string2 if they are of a given * distance seperation from the position in string1. * * @param string1 * @param string2 * @param distanceSep * @return a string buffer of characters from string1 within string2 if they are of a given * distance seperation from the position in string1 */ private static StringBuffer getCommonCharacters(final String string1, final String string2, final int distanceSep) { //create a return buffer of characters final StringBuffer returnCommons = new StringBuffer(); //create a copy of string2 for processing final StringBuffer copy = new StringBuffer(string2); //iterate over string1 int n=string1.length(); int m=string2.length(); for (int i = 0; i < n; i++) { final char ch = string1.charAt(i); //set boolean for quick loop exit if found boolean foundIt = false; //compare char with range of characters to either side for (int j = Math.max(0, i - distanceSep); !foundIt && j < Math.min(i + distanceSep, m - 1); j++) { //check if found if (copy.charAt(j) == ch) { foundIt = true; //append character found returnCommons.append(ch); //alter copied string2 for processing copy.setCharAt(j, (char)0); } } } return returnCommons; } } I mention that in the whole process I make just instance of the script, so only once jaro= new Jaro(); If you are going to test and need examples so not break the script, you will find it here, in another thread for python optimization.

    Read the article

  • How meaningful is the Big-O time complexity of an algorithm?

    - by james creasy
    Programmers often talk about the time complexity of an algorithm, e.g. O(log n) or O(n^2). Time complexity classifications are made as the input size goes to infinity, but ironically infinite input size in computation is not used. Put another way, the classification of an algorithm is based on a situation that algorithm will never be in: where n = infinity. Also, consider that a polynomial time algorithm where the exponent is huge is just as useless as an exponential time algorithm with tiny base (e.g., 1.00000001^n) is useful. Given this, how much can I rely on the Big-O time complexity to advise choice of an algorithm?

    Read the article

  • Checkers AI Algorithm

    - by John
    I am making an AI for my checkers game and I'm trying to make it as hard as possible. Here is the current criteria for a move on the hardest difficulty: 1: Look For A Block: This is when a piece is being threatened and another piece can be moved in behind it to protect it. Here is an example: Black Moves |W| |W| |W| |W| | | |W| |W| |W| |W| |W| | | |W| |W| | | | | |W| | | | | | | | | |B| | | | | |B| | | |B| |B| |B| |B| |B| |B| | | |B| |B| |B| |B| White Blocks |W| |W| |W| |W| | | |W| | | |W| |W| |W| |W| |W| |W| | | | | |W| | | | | | | | | |B| | | | | |B| | | |B| |B| |B| |B| |B| |B| | | |B| |B| |B| |B| 2: Move pieces out of danger: if any piece is being threatened, and a piece cannot block for that piece, then it will attempt to move out of the way. If the piece cannot move out of the way without still being in danger, the computer ignores the piece. 3: If the computer player owns any kings, it will attempt to 'hunt down' enemy pieces on the board, if no moves can be made that won't in danger the king or any other pieces, the computer ignores this rule. 4: Any piece that is owned by the computer that is in column 1 or 6 will attempt to go to a side. When a piece is in column 0 or 7, it is in a very strategic position because it cannot get captured while it is in either of these columns 5: It makes an educated random move, the move will not indanger the piece that is moving or any piece that is on the board. 6: If none of the above are possible it makes a random move. This question is not really specific to any language but if all examples could be in Java that would be great, considering this app is written in android. Does anyone see any room for improvement in this algorithm? Anything that would make it better at playing checkers?

    Read the article

  • C# XNA: Effecient mesh building algorithm for voxel based terrain ("top" outside layer only, non-destructible)

    - by Tim Hatch
    To put this bluntly, for non-destructible/non-constructible voxel style terrain, are generated meshes handled much better than instancing? Is there another method to achieve millions of visible quad faces per scene with ease? If generated meshes per chunk is the way to go, what kind of algorithm might I want to use based on only EVER needing the outer layer rendered? I'm using 3D Perlin Noise for terrain generation (for overhangs/caves/etc). The layout is fantastic, but even for around 20k visible faces, it's quite slow using instancing (whether it's one big draw call or multiple smaller chunks). I've simplified it to the point of removing non-visible cubes and only having the top faces of my cube-like terrain be rendered, but with 20k quad instances, it's still pretty sluggish (30fps on my machine). My goal is for the world to be made using quite small cubes. Where multiple games (IE: Minecraft) have the player 1x1 cube in width/length and 2 high, I'm shooting for 6x6 width/length and 9 high. With a lot of advantages as far as gameplay goes, it also means I could quite easily have a single scene with millions of truly visible quads. So, I have been trying to look into changing my method from instancing to mesh generation on a chunk by chunk basis. Do video cards handle this type of processing better than separate quads/cubes through instancing? What kind of existing algorithms should I be looking into? I've seen references to marching cubes a few times now, but I haven't spent much time investigating it since I don't know if it's the better route for my situation or not. I'm also starting to doubt my need of using 3D Perlin noise for terrain generation since I won't want the kind of depth it would seem best at. I just like the idea of overhangs and occasional cave-like structures, but could find no better 'surface only' algorithms to cover that. If anyone has any better suggestions there, feel free to throw them at me too. Thanks, Mythics

    Read the article

  • Which algorithm used in Advance Wars type turn based games

    - by Jan de Lange
    Has anyone tried to develop, or know of an algorithm such as used in a typical turn based game like Advance Wars, where the number of objects and the number of moves per object may be too large to search through up to a reasonable depth like one would do in a game with a smaller search base like chess? There is some path-finding needed to to engage into combat, harvest, or move to an object, so that in the next move such actions are possible. With this you can build a search tree for each item, resulting in a large tree for all items. With a cost function one can determine the best moves. Then the board flips over to the player role (min/max) and the computer searches the best player move, and flips back etc. upto a number of cycles deep. Finally it has found the best move and now it's the players turn. But he may be asleep by now... So how is this done in practice? I have found several good sources on A*, DFS, BFS, evaluation / cost functions etc. But as of yet I do not see how I can put it all together.

    Read the article

  • What is the best retort to "premature optimization is the root of all evil"

    - by waffles
    Often I hear the sentiment ... "Why worry about performance, write slow code, get your product to market ... don't worry about performance. You can sort that out later" The culmination of this sentiment is: "... premature optimization is the root of all evil ... #winning" I was wondering, does anybody have a good retort to this one liner. Ideally an equally strong one liner that encompasses the reverse of this sentiment?

    Read the article

  • Flood fill algorithm for Game of Go

    - by Jackson Borghi
    I'm having a hell of a time trying to figure out how to make captured stones disappear. I've read everywhere that I should use the flood fill algorithm, but I haven't had any luck with that so far. Any help would be amazing! Here is my code: package Go; import static java.lang.Math.*; import static stdlib.StdDraw.*; import java.awt.Color; public class Go2 { public static Color opposite(Color player) { if (player == WHITE) { return BLACK; } return WHITE; } public static void drawGame(Color[][] board) { Color[][][] unit = new Color[400][19][19]; for (int h = 0; h < 400; h++) { for (int x = 0; x < 19; x++) { for (int y = 0; y < 19; y++) { unit[h][x][y] = YELLOW; } } } setXscale(0, 19); setYscale(0, 19); clear(YELLOW); setPenColor(BLACK); line(0, 0, 0, 19); line(19, 19, 19, 0); line(0, 19, 19, 19); line(0, 0, 19, 0); for (double i = 0; i < 19; i++) { line(0.0, i, 19, i); line(i, 0.0, i, 19); } for (int x = 0; x < 19; x++) { for (int y = 0; y < 19; y++) { if (board[x][y] != YELLOW) { setPenColor(board[x][y]); filledCircle(x, y, 0.47); setPenColor(GRAY); circle(x, y, 0.47); } } } int h = 0; } public static void main(String[] args) { int px; int py; Color[][] temp = new Color[19][19]; Color[][] board = new Color[19][19]; Color player = WHITE; for (int i = 0; i < 19; i++) { for (int h = 0; h < 19; h++) { board[i][h] = YELLOW; temp[i][h] = YELLOW; } } while (true) { drawGame(board); while (!mousePressed()) { } px = (int) round(mouseX()); py = (int) round(mouseY()); board[px][py] = player; while (mousePressed()) { } floodFill(px, py, player, board, temp); System.out.print("XXXXX = "+ temp[px][py]); if (checkTemp(temp, board, px, py)) { for (int x = 0; x < 19; x++) { for (int y = 0; y < 19; y++) { if (temp[x][y] == GRAY) { board[x][y] = YELLOW; } } } } player = opposite(player); } } private static boolean checkTemp(Color[][] temp, Color[][] board, int x, int y) { if (x < 19 && x > -1 && y < 19 && y > -1) { if (temp[x + 1][y] == YELLOW || temp[x - 1][y] == YELLOW || temp[x][y - 1] == YELLOW || temp[x][y + 1] == YELLOW) { return false; } } if (x == 18) { if (temp[x - 1][y] == YELLOW || temp[x][y - 1] == YELLOW || temp[x][y + 1] == YELLOW) { return false; } } if (y == 18) { if (temp[x + 1][y] == YELLOW || temp[x - 1][y] == YELLOW || temp[x][y - 1] == YELLOW) { return false; } } if (y == 0) { if (temp[x + 1][y] == YELLOW || temp[x - 1][y] == YELLOW || temp[x][y + 1] == YELLOW) { return false; } } if (x == 0) { if (temp[x + 1][y] == YELLOW || temp[x][y - 1] == YELLOW || temp[x][y + 1] == YELLOW) { return false; } } else { if (x < 19) { if (temp[x + 1][y] == GRAY) { checkTemp(temp, board, x + 1, y); } } if (x >= 0) { if (temp[x - 1][y] == GRAY) { checkTemp(temp, board, x - 1, y); } } if (y < 19) { if (temp[x][y + 1] == GRAY) { checkTemp(temp, board, x, y + 1); } } if (y >= 0) { if (temp[x][y - 1] == GRAY) { checkTemp(temp, board, x, y - 1); } } } return true; } private static void floodFill(int x, int y, Color player, Color[][] board, Color[][] temp) { if (board[x][y] != player) { return; } else { temp[x][y] = GRAY; System.out.println("x = " + x + " y = " + y); if (x < 19) { floodFill(x + 1, y, player, board, temp); } if (x >= 0) { floodFill(x - 1, y, player, board, temp); } if (y < 19) { floodFill(x, y + 1, player, board, temp); } if (y >= 0) { floodFill(x, y - 1, player, board, temp); } } } }

    Read the article

  • FloodFill Algorithm for Game of Go

    - by Jackson Borghi
    I'm having a hell of a time trying to figure out how to make captured stones disappear. I've read everywhere that I should use the FloodFill algorithm, but I havent had any luck with that so far. Any help would be amazing! Here is my code: package Go; import static java.lang.Math.; import static stdlib.StdDraw.; import java.awt.Color; public class Go2 { public static Color opposite(Color player) { if (player == WHITE) { return BLACK; } return WHITE; } public static void drawGame(Color[][] board) { Color[][][] unit = new Color[400][19][19]; for (int h = 0; h < 400; h++) { for (int x = 0; x < 19; x++) { for (int y = 0; y < 19; y++) { unit[h][x][y] = YELLOW; } } } setXscale(0, 19); setYscale(0, 19); clear(YELLOW); setPenColor(BLACK); line(0, 0, 0, 19); line(19, 19, 19, 0); line(0, 19, 19, 19); line(0, 0, 19, 0); for (double i = 0; i < 19; i++) { line(0.0, i, 19, i); line(i, 0.0, i, 19); } for (int x = 0; x < 19; x++) { for (int y = 0; y < 19; y++) { if (board[x][y] != YELLOW) { setPenColor(board[x][y]); filledCircle(x, y, 0.47); setPenColor(GRAY); circle(x, y, 0.47); } } } int h = 0; } public static void main(String[] args) { int px; int py; Color[][] temp = new Color[19][19]; Color[][] board = new Color[19][19]; Color player = WHITE; for (int i = 0; i < 19; i++) { for (int h = 0; h < 19; h++) { board[i][h] = YELLOW; temp[i][h] = YELLOW; } } while (true) { drawGame(board); while (!mousePressed()) { } px = (int) round(mouseX()); py = (int) round(mouseY()); board[px][py] = player; while (mousePressed()) { } floodFill(px, py, player, board, temp); System.out.print("XXXXX = "+ temp[px][py]); if (checkTemp(temp, board, px, py)) { for (int x = 0; x < 19; x++) { for (int y = 0; y < 19; y++) { if (temp[x][y] == GRAY) { board[x][y] = YELLOW; } } } } player = opposite(player); } } private static boolean checkTemp(Color[][] temp, Color[][] board, int x, int y) { if (x < 19 && x > -1 && y < 19 && y > -1) { if (temp[x + 1][y] == YELLOW || temp[x - 1][y] == YELLOW || temp[x][y - 1] == YELLOW || temp[x][y + 1] == YELLOW) { return false; } } if (x == 18) { if (temp[x - 1][y] == YELLOW || temp[x][y - 1] == YELLOW || temp[x][y + 1] == YELLOW) { return false; } } if (y == 18) { if (temp[x + 1][y] == YELLOW || temp[x - 1][y] == YELLOW || temp[x][y - 1] == YELLOW) { return false; } } if (y == 0) { if (temp[x + 1][y] == YELLOW || temp[x - 1][y] == YELLOW || temp[x][y + 1] == YELLOW) { return false; } } if (x == 0) { if (temp[x + 1][y] == YELLOW || temp[x][y - 1] == YELLOW || temp[x][y + 1] == YELLOW) { return false; } } else { if (x < 19) { if (temp[x + 1][y] == GRAY) { checkTemp(temp, board, x + 1, y); } } if (x >= 0) { if (temp[x - 1][y] == GRAY) { checkTemp(temp, board, x - 1, y); } } if (y < 19) { if (temp[x][y + 1] == GRAY) { checkTemp(temp, board, x, y + 1); } } if (y >= 0) { if (temp[x][y - 1] == GRAY) { checkTemp(temp, board, x, y - 1); } } } return true; } private static void floodFill(int x, int y, Color player, Color[][] board, Color[][] temp) { if (board[x][y] != player) { return; } else { temp[x][y] = GRAY; System.out.println("x = " + x + " y = " + y); if (x < 19) { floodFill(x + 1, y, player, board, temp); } if (x >= 0) { floodFill(x - 1, y, player, board, temp); } if (y < 19) { floodFill(x, y + 1, player, board, temp); } if (y >= 0) { floodFill(x, y - 1, player, board, temp); } } } }

    Read the article

  • Web page database query optimization

    - by morpheous
    I am putting together a web page which is quite 'expensive' in terms of database hits. I don't want to start optimizing at this stage - though with me trying to hit a deadline, I may end up not optimizing at all. Currently the page requires 18 (that's right eighteen) hits to the db. I am already using joins, and some of the queries are UNIONed to minimize the trips to the db. My local dev machine can handle this (page is not slow) however, I feel if I release this into the wild, the number of queries will quickly overwhelm my database (MySQL). I could always use memcache or something similar, but I would much rather continue with my other dev work that needs to be completed before the deadline - at least retrieving the page works - its simply a matter of optimization now (if required). My question therefore is - is 18 db queries for a single page retrieval completely outrageous - (i.e. I should put everything on hold and optimize the hell of the retrieval logic), or shall I continue as normal, meet the deadline and release on schedule and see what happens? [Edit] Just to clarify, I have already done the 'obvious' things like using (single and composite) indexes for fields used in the queries. What I haven't yet done is to run a query analyzer to see if my indexes etc are optimal.

    Read the article

  • Why would this Lua optimization hack help?

    - by Ian Boyd
    i'm looking over a document that describes various techniques to improve performance of Lua script code, and i'm shocked that such tricks would be required. (Although i'm quoting Lua, i've seen similar hacks in Javascript). Why would this optimization be required: For instance, the code for i = 1, 1000000 do local x = math.sin(i) end runs 30% slower than this one: local sin = math.sin for i = 1, 1000000 do local x = sin(i) end They're re-declaring sin function locally. Why would this be helpful? It's the job of the compiler to do that anyway. Why is the programmer having to do the compiler's job? i've seen similar things in Javascript; and so obviously there must be a very good reason why the interpreting compiler isn't doing its job. What is it? i see it repeatedly in the Lua environment i'm fiddling in; people redeclaring variables as local: local strfind = strfind local strlen = strlen local gsub = gsub local pairs = pairs local ipairs = ipairs local type = type local tinsert = tinsert local tremove = tremove local unpack = unpack local max = max local min = min local floor = floor local ceil = ceil local loadstring = loadstring local tostring = tostring local setmetatable = setmetatable local getmetatable = getmetatable local format = format local sin = math.sin What is going on here that people have to do the work of the compiler? Is the compiler confused by how to find format? Why is this an issue that a programmer has to deal with? Why would this not have been taken care of in 1993? i also seem to have hit a logical paradox: Optimizatin should not be done without profiling Lua has no ability to be profiled Lua should not be optimized

    Read the article

  • What is an Efficient algorithm to find Area of Overlapping Rectangles

    - by namenlos
    My situation Input: a set of rectangles each rect is comprised of 4 doubles like this: (x0,y0,x1,y1) they are not "rotated" at any angle, all they are "normal" rectangles that go "up/down" and "left/right" with respect to the screen they are randomly placed - they may be touching at the edges, overlapping , or not have any contact I will have several hundred rectangles this is implemented in C# I need to find The area that is formed by their overlap - all the area in the canvas that more than one rectangle "covers" (for example with two rectangles, it would be the intersection) I don't need the geometry of the overlap - just the area (example: 4 sq inches) Overlaps shouldn't be counted multiple times - so for example imagine 3 rects that have the same size and position - they are right on top of each other - this area should be counted once (not three times) Example The image below contains thre rectangles: A,B,C A and B overlap (as indicated by dashes) B and C overlap (as indicated by dashes) What I am looking for is the area where the dashes are shown - AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA--------------BBB AAAAAAAAAAAAAAAA--------------BBB AAAAAAAAAAAAAAAA--------------BBB AAAAAAAAAAAAAAAA--------------BBB BBBBBBBBBBBBBBBBB BBBBBBBBBBBBBBBBB BBBBBBBBBBBBBBBBB BBBBBB-----------CCCCCCCC BBBBBB-----------CCCCCCCC BBBBBB-----------CCCCCCCC CCCCCCCCCCCCCCCCCCC CCCCCCCCCCCCCCCCCCC CCCCCCCCCCCCCCCCCCC CCCCCCCCCCCCCCCCCCC

    Read the article

  • Java code optimization on matrix windowing computes in more time

    - by rano
    I have a matrix which represents an image and I need to cycle over each pixel and for each one of those I have to compute the sum of all its neighbors, ie the pixels that belong to a window of radius rad centered on the pixel. I came up with three alternatives: The simplest way, the one that recomputes the window for each pixel The more optimized way that uses a queue to store the sums of the window columns and cycling through the columns of the matrix updates this queue by adding a new element and removing the oldes The even more optimized way that does not need to recompute the queue for each row but incrementally adjusts a previously saved one I implemented them in c++ using a queue for the second method and a combination of deques for the third (I need to iterate through their elements without destructing them) and scored their times to see if there was an actual improvement. it appears that the third method is indeed faster. Then I tried to port the code to Java (and I must admit that I'm not very comfortable with it). I used ArrayDeque for the second method and LinkedLists for the third resulting in the third being inefficient in time. Here is the simplest method in C++ (I'm not posting the java version since it is almost identical): void normalWindowing(int mat[][MAX], int cols, int rows, int rad){ int i, j; int h = 0; for (i = 0; i < rows; ++i) { for (j = 0; j < cols; j++) { h = 0; for (int ry =- rad; ry <= rad; ry++) { int y = i + ry; if (y >= 0 && y < rows) { for (int rx =- rad; rx <= rad; rx++) { int x = j + rx; if (x >= 0 && x < cols) { h += mat[y][x]; } } } } } } } Here is the second method (the one optimized through columns) in C++: void opt1Windowing(int mat[][MAX], int cols, int rows, int rad){ int i, j, h, y, col; queue<int>* q = NULL; for (i = 0; i < rows; ++i) { if (q != NULL) delete(q); q = new queue<int>(); h = 0; for (int rx = 0; rx <= rad; rx++) { if (rx < cols) { int mem = 0; for (int ry =- rad; ry <= rad; ry++) { y = i + ry; if (y >= 0 && y < rows) { mem += mat[y][rx]; } } q->push(mem); h += mem; } } for (j = 1; j < cols; j++) { col = j + rad; if (j - rad > 0) { h -= q->front(); q->pop(); } if (j + rad < cols) { int mem = 0; for (int ry =- rad; ry <= rad; ry++) { y = i + ry; if (y >= 0 && y < rows) { mem += mat[y][col]; } } q->push(mem); h += mem; } } } } And here is the Java version: public static void opt1Windowing(int [][] mat, int rad){ int i, j = 0, h, y, col; int cols = mat[0].length; int rows = mat.length; ArrayDeque<Integer> q = null; for (i = 0; i < rows; ++i) { q = new ArrayDeque<Integer>(); h = 0; for (int rx = 0; rx <= rad; rx++) { if (rx < cols) { int mem = 0; for (int ry =- rad; ry <= rad; ry++) { y = i + ry; if (y >= 0 && y < rows) { mem += mat[y][rx]; } } q.addLast(mem); h += mem; } } j = 0; for (j = 1; j < cols; j++) { col = j + rad; if (j - rad > 0) { h -= q.peekFirst(); q.pop(); } if (j + rad < cols) { int mem = 0; for (int ry =- rad; ry <= rad; ry++) { y = i + ry; if (y >= 0 && y < rows) { mem += mat[y][col]; } } q.addLast(mem); h += mem; } } } } I recognize this post will be a wall of text. Here is the third method in C++: void opt2Windowing(int mat[][MAX], int cols, int rows, int rad){ int i = 0; int j = 0; int h = 0; int hh = 0; deque< deque<int> *> * M = new deque< deque<int> *>(); for (int ry = 0; ry <= rad; ry++) { if (ry < rows) { deque<int> * q = new deque<int>(); M->push_back(q); for (int rx = 0; rx <= rad; rx++) { if (rx < cols) { int val = mat[ry][rx]; q->push_back(val); h += val; } } } } deque<int> * C = new deque<int>(M->front()->size()); deque<int> * Q = new deque<int>(M->front()->size()); deque<int> * R = new deque<int>(M->size()); deque< deque<int> *>::iterator mit; deque< deque<int> *>::iterator mstart = M->begin(); deque< deque<int> *>::iterator mend = M->end(); deque<int>::iterator rit; deque<int>::iterator rstart = R->begin(); deque<int>::iterator rend = R->end(); deque<int>::iterator cit; deque<int>::iterator cstart = C->begin(); deque<int>::iterator cend = C->end(); for (mit = mstart, rit = rstart; mit != mend, rit != rend; ++mit, ++rit) { deque<int>::iterator pit; deque<int>::iterator pstart = (* mit)->begin(); deque<int>::iterator pend = (* mit)->end(); for(cit = cstart, pit = pstart; cit != cend && pit != pend; ++cit, ++pit) { (* cit) += (* pit); (* rit) += (* pit); } } for (i = 0; i < rows; ++i) { j = 0; if (i - rad > 0) { deque<int>::iterator cit; deque<int>::iterator cstart = C->begin(); deque<int>::iterator cend = C->end(); deque<int>::iterator pit; deque<int>::iterator pstart = (M->front())->begin(); deque<int>::iterator pend = (M->front())->end(); for(cit = cstart, pit = pstart; cit != cend; ++cit, ++pit) { (* cit) -= (* pit); } deque<int> * k = M->front(); M->pop_front(); delete k; h -= R->front(); R->pop_front(); } int row = i + rad; if (row < rows && i > 0) { deque<int> * newQ = new deque<int>(); M->push_back(newQ); deque<int>::iterator cit; deque<int>::iterator cstart = C->begin(); deque<int>::iterator cend = C->end(); int rx; int tot = 0; for (rx = 0, cit = cstart; rx <= rad; rx++, ++cit) { if (rx < cols) { int val = mat[row][rx]; newQ->push_back(val); (* cit) += val; tot += val; } } R->push_back(tot); h += tot; } hh = h; copy(C->begin(), C->end(), Q->begin()); for (j = 1; j < cols; j++) { int col = j + rad; if (j - rad > 0) { hh -= Q->front(); Q->pop_front(); } if (j + rad < cols) { int val = 0; for (int ry =- rad; ry <= rad; ry++) { int y = i + ry; if (y >= 0 && y < rows) { val += mat[y][col]; } } hh += val; Q->push_back(val); } } } } And finally its Java version: public static void opt2Windowing(int [][] mat, int rad){ int cols = mat[0].length; int rows = mat.length; int i = 0; int j = 0; int h = 0; int hh = 0; LinkedList<LinkedList<Integer>> M = new LinkedList<LinkedList<Integer>>(); for (int ry = 0; ry <= rad; ry++) { if (ry < rows) { LinkedList<Integer> q = new LinkedList<Integer>(); M.addLast(q); for (int rx = 0; rx <= rad; rx++) { if (rx < cols) { int val = mat[ry][rx]; q.addLast(val); h += val; } } } } int firstSize = M.getFirst().size(); int mSize = M.size(); LinkedList<Integer> C = new LinkedList<Integer>(); LinkedList<Integer> Q = null; LinkedList<Integer> R = new LinkedList<Integer>(); for (int k = 0; k < firstSize; k++) { C.add(0); } for (int k = 0; k < mSize; k++) { R.add(0); } ListIterator<LinkedList<Integer>> mit; ListIterator<Integer> rit; ListIterator<Integer> cit; ListIterator<Integer> pit; for (mit = M.listIterator(), rit = R.listIterator(); mit.hasNext();) { Integer r = rit.next(); int rsum = 0; for (cit = C.listIterator(), pit = (mit.next()).listIterator(); cit.hasNext();) { Integer c = cit.next(); Integer p = pit.next(); rsum += p; cit.set(c + p); } rit.set(r + rsum); } for (i = 0; i < rows; ++i) { j = 0; if (i - rad > 0) { for(cit = C.listIterator(), pit = M.getFirst().listIterator(); cit.hasNext();) { Integer c = cit.next(); Integer p = pit.next(); cit.set(c - p); } M.removeFirst(); h -= R.getFirst(); R.removeFirst(); } int row = i + rad; if (row < rows && i > 0) { LinkedList<Integer> newQ = new LinkedList<Integer>(); M.addLast(newQ); int rx; int tot = 0; for (rx = 0, cit = C.listIterator(); rx <= rad; rx++) { if (rx < cols) { Integer c = cit.next(); int val = mat[row][rx]; newQ.addLast(val); cit.set(c + val); tot += val; } } R.addLast(tot); h += tot; } hh = h; Q = new LinkedList<Integer>(); Q.addAll(C); for (j = 1; j < cols; j++) { int col = j + rad; if (j - rad > 0) { hh -= Q.getFirst(); Q.pop(); } if (j + rad < cols) { int val = 0; for (int ry =- rad; ry <= rad; ry++) { int y = i + ry; if (y >= 0 && y < rows) { val += mat[y][col]; } } hh += val; Q.addLast(val); } } } } I guess that most is due to the poor choice of the LinkedList in Java and to the lack of an efficient (not shallow) copy method between two LinkedList. How can I improve the third Java method? Am I doing some conceptual error? As always, any criticisms is welcome. UPDATE Even if it does not solve the issue, using ArrayLists, as being suggested, instead of LinkedList improves the third method. The second one performs still better (but when the number of rows and columns of the matrix is lower than 300 and the window radius is small the first unoptimized method is the fastest in Java)

    Read the article

  • Algorithm for optimally choosing actions to perform a task

    - by Jules
    There are two data types: tasks and actions. An action costs a certain time to complete, and a set of tasks this actions consists of. A task has a set of actions, and our job is to choose one of them. So: class Task { Set<Action> choices; } class Action { float time; Set<Task> dependencies; } For example the primary task could be "Get a house". The possible actions for this task: "Buy a house" or "Build a house". The action "Build a house" costs 10 hours and has the dependencies "Get bricks" and "Get cement", etcetera. The total time is the sum of all the times of the actions required to perform. We want to choose actions such that the total time is minimal. Note that the dependencies can be diamond shaped. For example "Get bricks" could require "Get a car" (to transport the bricks) and "Get cement" would also require a car. Even if you do "Get bricks" and "Get cement" you only have to count the time it takes to get a car once. Note also that the dependencies can be circular. For example "Money" - "Job" - "Car" - "Money". This is no problem for us, we simply select all of "Money", "Job" and "Car". The total time is simply the sum of the time of these 3 things. Mathematical description: Let actions be the chosen actions. valid(task) = ?action ? task.choices. (action ? actions ? ?tasks ? action.dependencies. valid(task)) time = sum {action.time | action ? actions} minimize time subject to valid(primaryTask)

    Read the article

  • Algorithm for generating an array of non-equal costs for a transport problem optimization

    - by Carlos
    I have an optimizer that solves a transportation problem, using a cost matrix of all the possible paths. The optimiser works fine, but if two of the costs are equal, the solution contains one more path that the minimum number of paths. (Think of it as load balancing routers; if two routes are same cost, you'll use them both.) I would like the minimum number of routes, and to do that I need a cost matrix that doesn't have two costs that are equal within a certain tolerance. At the moment, I'm passing the cost matrix through a baking function which tests every entry for equality to each of the other entries, and moves it a fixed percentage if it matches. However, this approach seems to require N^2 comparisons, and if the starting values are all the same, the last cost will be r^N bigger. (r is the arbitrary fixed percentage). Also there is the problem that by multiplying by the percentage, you end up on top of another value. So the problem seems to have an element of recursion, or at least repeated checking, which bloats the code. The current implementation is basically not very good (I won't paste my GOTO-using code here for you all to mock), and I'd like to improve it. Is there a name for what I'm after, and is there a standard implementation? Example: {1,1,2,3,4,5} (tol = 0.05) becomes {1,1.05,2,3,4,5}

    Read the article

  • Optimize Binary Search Algorithm

    - by Ganesh M
    In a binary search, we have two comparisons one for greater than and other for less than, otherwise its the mid value. How would you optimize so that we need to check only once? bool binSearch(int array[], int key, int left, int right) { mid = left + (right-left)/2; if (key < array[mid]) return binSearch(array, key, left, mid-1); else if (key > array[mid]) return binSearch(array, key, mid+1, right); else if (key == array[mid]) return TRUE; // Found return FALSE; // Not Found }

    Read the article

  • Optimization of time-varying parameters

    - by brama
    I need to find an optimal set of "n" parameter values that minimize an objective function (a 2-hr simulation of a system). I have looked at genetic algorithm and simulated annealing methods, but was wondering if there are any better algorithms and guidance on their merits and limitations. With the above optimization methods I can find the optimal parameter values that hold true for the entire simulation duration. Incase, I want to find the optimal "time varying" parameter values (parameter values change with time during the 2-hr simulation), are there any methods/ideas other than making each time varying parameter value a variable to optimize? Any thoughts?

    Read the article

  • Optimization ended up in casting an object at each method call

    - by Aybe
    I've been doing some optimization for the following piece of code : public void DrawLine(int x1, int y1, int x2, int y2, int color) { _bitmap.DrawLineBresenham(x1, y1, x2, y2, color); } After profiling it about 70% of the time spent was in getting a context for drawing and disposing it. I ended up sketching the following overload : public void DrawLine(int x1, int y1, int x2, int y2, int color, BitmapContext bitmapContext) { _bitmap.DrawLineBresenham(x1, y1, x2, y2, color, bitmapContext); } Until here no problems, all the user has to do is to pass a context and performance is really great as a context is created/disposed one time only (previously it was a thousand times per second). The next step was to make it generic in the sense it doesn't depend on a particular framework for rendering (besides .NET obvisouly). So I wrote this method : public void DrawLine(int x1, int y1, int x2, int y2, int color, IDisposable bitmapContext) { _bitmap.DrawLineBresenham(x1, y1, x2, y2, color, (BitmapContext)bitmapContext); } Now every time a line is drawn the generic context is casted, this was unexpected for me. Are there any approaches for fixing this design issue ? Note : _bitmap is a WriteableBitmap from WPF BitmapContext is from WriteableBitmapEx library DrawLineBresenham is an extension method from WriteableBitmapEx

    Read the article

  • algorithm for Virtual Machine(VM) Consolidation in Cloud

    - by devansh dalal
    PROBLEM: We have N physical machines(PMs) each with ram Ri, cpu Ci and a set of currently scheduled VMs each with ram requirement ri and ci respectively Moving(Migrating) any VM from one PM to other has a cost associated which depends on its ram ri. A PM with no VMs is shut down to save power. Our target is to minimize the weighted sum of (N,migration cost) by migrating some VMs i.e. minimize the number of working PMs as well as not to degrade the service level due to excessive migrations. My Approach: Brute Force approach is choosing the minimum loaded PM and try to fit its VMs to other PMs by First Fit Decreasing algorithm or we can select the victim PMs and target PMs based on their loading level and shut down victims if possible by moving their VMs to targets. I tried this Greedy approach on the Data of Baadal(IIT-D cloud) but It isn't giving promising results. I have also tried to study the Ant colony optimization for dynamic VM consolidating but was unable to understand very much. I used the links. http://dumas.ccsd.cnrs.fr/docs/00/72/52/15/PDF/Esnault.pdf http://hal.archives-ouvertes.fr/docs/00/72/38/56/PDF/RR-8032.pdf Would anyone please explain the solution or suggest any new approach for better performance soon. Thanks in advance.

    Read the article

  • Efficient algorithm for Virtual Machine(VM) Consolidation in Cloud

    - by devansh dalal
    PROBLEM: We have N physical machines(PMs) each with ram Ri, cpu Ci and a set of currently scheduled VMs each with ram requirement ri and ci respectively Moving(Migrating) any VM from one PM to other has a cost associated which depends on its ram ri. A PM with no VMs is shut down to save power. Our target is to minimize the weighted sum of (N,migration cost) by migrating some VMs i.e. minimize the number of working PMs as well as not to degrade the service level due to excessive migrations. My Approach: Brute Force approach is choosing the minimum loaded PM and try to fit its VMs to other PMs by First Fit Decreasing algorithm or we can select the victim PMs and target PMs based on their loading level and shut down victims if possible by moving their VMs to targets. I tried this Greedy approach on the Data of Baadal(IIT-D cloud) but It isn't giving promising results. I have also tried to study the Ant colony optimization for dynamic VM consolidating but was unable to understand very much. I used the links. http://dumas.ccsd.cnrs.fr/docs/00/72/52/15/PDF/Esnault.pdf http://hal.archives-ouvertes.fr/docs/00/72/38/56/PDF/RR-8032.pdf Would anyone please clarify the solution or suggest any new approach/resources for better performance. I am basically searching for the algorithms not the physical optimizations and I also know that many commercial organizations have provided these solution but I just wanted to know more the underlying algorithms. Thanks in advance.

    Read the article

  • Algorithm for spreading labels in a visually appealing and intuitive way

    - by mac
    Short version Is there a design pattern for distributing vehicle labels in a non-overlapping fashion, placing them as close as possible to the vehicle they refer to? If not, is any of the method I suggest viable? How would you implement this yourself? Extended version In the game I'm writing I have a bird-eye vision of my airborne vehicles. I also have next to each of the vehicles a small label with key-data about the vehicle. This is an actual screenshot: Now, since the vehicles could be flying at different altitudes, their icons could overlap. However I would like to never have their labels overlapping (or a label from vehicle 'A' overlap the icon of vehicle 'B'). Currently, I can detect collisions between sprites and I simply push away the offending label in a direction opposite to the otherwise-overlapped sprite. This works in most situations, but when the airspace get crowded, the label can get pushed very far away from its vehicle, even if there was an alternate "smarter" alternative. For example I get: B - label A -----------label C - label where it would be better (= label closer to the vehicle) to get: B - label label - A C - label EDIT: It also has to be considered that beside the overlapping vehicles case, there might be other configurations in which vehicles'labels could overlap (the ASCII-art examples show for example three very close vehicles in which the label of A would overlap the icon of B and C). I have two ideas on how to improve the present situation, but before spending time implementing them, I thought to turn to the community for advice (after all it seems like a "common enough problem" that a design pattern for it could exist). For what it's worth, here's the two ideas I was thinking to: Slot-isation of label space In this scenario I would divide all the screen into "slots" for the labels. Then, each vehicle would always have its label placed in the closest empty one (empty = no other sprites at that location. Spiralling search From the location of the vehicle on the screen, I would try to place the label at increasing angles and then at increasing radiuses, until a non-overlapping location is found. Something down the line of: try 0°, 10px try 10°, 10px try 20°, 10px ... try 350°, 10px try 0°, 20px try 10°, 20px ...

    Read the article

  • AABB Sweeping, algorithm to solve "stacking box" problem

    - by Ivo Wetzel
    I'm currently working on a simple AABB collision system and after some fiddling the sweeping of a single box vs. another and the calculation of the response velocity needed to push them apart works flawlessly. Now on to the new problem, imagine I'm having a stack of boxes which are falling towards a ground box which isn't moving: Each of these boxes has a vertical velocity for the "gravity" value, let's say this velocity is 5. Now, the result is that they all fall into each other: The reason is obvious, since all the boxes have a downward velocity of 5, this results in no collisions when calculating the relative velocity between the boxes during sweeping. Note: The red ground box here is static (always 0 velocity, can utilize spatial partitioning ), and all dynamic static collisions are resolved first, thus the fact that the boxes stop correctly at this ground box. So, this seems to be simply an issue with the order the boxes are sweept against each other. I imagine that sorting the boxes based on their x and y velocities and then sweeping these groups correctly against each other may resolve this issues. So, I'm looking for algorithms / examples on how to implement such a system. The code can be found here: https://github.com/BonsaiDen/aabb The two files which are of interest are [box/Dynamic.lua][3] and [box/Manager.lua][4]. The project is using Love2D in case you want to run it.

    Read the article

  • Derive a algorithm to match best position

    - by Farooq Arshed
    I have pieces in my game which have stats and cost assigned to them and they can only be placed at a certain location. Lets say I have 50 pieces. e.g. Piece1 = 100 stats, 10 cost, Position A. Piece2 = 120 stats, 5 cost, Position B. Piece3 = 500 stats, 50 cost, Position C. Piece4 = 200 stats, 25 cost, Position A. and so on.. I have a board on which 12 pieces have to be allocated and have to remain inside the board cost. e.g. A board has A,B,C ... J,K,L positions and X Cost assigned to it. I have to figure out a way to place best possible piece in the correct position and should remain within the cost specified by the board. Any help would be appreciated.

    Read the article

  • Algorithm for approximating sihlouette image as polygon

    - by jack
    I want to be able to analyze a texture in real time and approximate a polygon to represent a silhouette. Imagine a person standing in front of a green screen and I want to approximately trace around their outline and get a 2D polygon as the result. Are there algorithms to do this and are they fast enough to work frame-to-frame in a game? (I have found algorithms to triangulate polygons, but I am having trouble knowing what to search for that describes my goal.)

    Read the article

  • Best algorithm for recursive adjacent tiles?

    - by OhMrBigshot
    In my game I have a set of tiles placed in a 2D array marked by their Xs and Zs ([1,1],[1,2], etc). Now, I want a sort of "Paint Bucket" mechanism: Selecting a tile will destroy all adjacent tiles until a condition stops it, let's say, if it hits an object with hasFlag. Here's what I have so far, I'm sure it's pretty bad, it also freezes everything sometimes: void destroyAdjacentTiles(int x, int z) { int GridSize = Cubes.GetLength(0); int minX = x == 0 ? x : x-1; int maxX = x == GridSize - 1 ? x : x+1; int minZ = z == 0 ? z : z-1; int maxZ = z == GridSize - 1 ? z : z+1; Debug.Log(string.Format("Cube: {0}, {1}; X {2}-{3}; Z {4}-{5}", x, z, minX, maxX, minZ, maxZ)); for (int curX = minX; curX <= maxX; curX++) { for (int curZ = minZ; curZ <= maxZ; curZ++) { if (Cubes[curX, curZ] != Cubes[x, z]) { Debug.Log(string.Format(" Checking: {0}, {1}", curX, curZ)); if (Cubes[curX,curZ] && Cubes[curX,curZ].GetComponent<CubeBehavior>().hasFlag) { Destroy(Cubes[curX,curZ]); destroyAdjacentTiles(curX, curZ); } } } } }

    Read the article

< Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >