Search Results

Search found 816 results on 33 pages for 'bingo star'.

Page 6/33 | < Previous Page | 2 3 4 5 6 7 8 9 10 11 12 13  | Next Page >

  • How to find minimum cut-sets for several subgraphs of a graph of degrees 2 to 4

    - by Tore
    I have a problem, Im trying to make A* searches through a grid based game like pacman or sokoban, but i need to find "enclosures". What do i mean by enclosures? subgraphs with as few cut edges as possible given a maximum size and minimum size for number of vertices that act as soft constraints. Alternatively you could say i am looking to find bridges between subgraphs, but its generally the same problem. Given a game that looks like this, what i want to do is find enclosures so that i can properly find entrances to them and thus get a good heuristic for reaching vertices inside these enclosures. So what i want is to find these colored regions on any given map. The reason for me bothering to do this and not just staying content with the performance of a simple manhattan distance heuristic is that an enclosure heuristic can give more optimal results and i would not have to actually do the A* to get some proper distance calculations and also for later adding competitive blocking of opponents within these enclosures when playing sokoban type games. Also the enclosure heuristic can be used for a minimax approach to finding goal vertices more properly. Do you know of a good algorithm for solving this problem or have any suggestions in things i should explore?

    Read the article

  • What does the * symbol do near a function argument and how to use that in others scenarios?

    - by user502052
    I am using Ruby on Rails 3 and I would like to know what means the presence of a *simbol near a function argument and to understand its usages in others scenarios. Example scenario (this method was from the Ruby on Rails 3 framework: def find(*args) return to_a.find { |*block_args| yield(*block_args) } if block_given? options = args.extract_options! if options.present? apply_finder_options(options).find(*args) else case args.first when :first, :last, :all send(args.first) else find_with_ids(*args) end end end

    Read the article

  • Why does A* path finding sometimes go in straight lines and sometimes diagonals? (Java)

    - by Relequestual
    I'm in the process of developing a simple 2d grid based sim game, and have fully functional path finding. I used the answer found in my previous question as my basis for implementing A* path finding. (http://stackoverflow.com/questions/735523/pathfinding-2d-java-game). To show you really what I'm asking, I need to show you this video screen capture that I made. I was just testing to see how the person would move to a location and back again, and this was the result... http://www.screenjelly.com/watch/Bd7d7pObyFo Different choice of path depending on the direction, an unexpected result. Any ideas?

    Read the article

  • A* pathfinder obstacle collision problem

    - by Cheesebaron
    I am working on a project with a robot that has to find its way to an object and avoid some obstacles when going to that object it has to pick up. The problem lies in that the robot and the object the robot needs to pick up are both one pixel wide in the pathfinder. In reality they are a lot bigger. Often the A* pathfinder chooses to place the route along the edges of the obstacles, sometimes making it collide with them, which we do not wish to have to do. I have tried to add some more non-walkable fields to the obstacles, but it does not always work out very well. It still collides with the obstacles, also adding too many points where it is not allowed to walk, results in that there is no path it can run on. Do you have any suggestions on what to do about this problem?

    Read the article

  • How do I create a graph from this datastructure?

    - by Shawn Mclean
    I took this data structure from this A* tutorial: public interface IHasNeighbours<N> { IEnumerable<N> Neighbours { get; } } public class Path<TNode> : IEnumerable<TNode> { public TNode LastStep { get; private set; } public Path<TNode> PreviousSteps { get; private set; } public double TotalCost { get; private set; } private Path(TNode lastStep, Path<TNode> previousSteps, double totalCost) { LastStep = lastStep; PreviousSteps = previousSteps; TotalCost = totalCost; } public Path(TNode start) : this(start, null, 0) { } public Path<TNode> AddStep(TNode step, double stepCost) { return new Path<TNode>(step, this, TotalCost + stepCost); } public IEnumerator<TNode> GetEnumerator() { for (Path<TNode> p = this; p != null; p = p.PreviousSteps) yield return p.LastStep; } IEnumerator IEnumerable.GetEnumerator() { return this.GetEnumerator(); } } I have no idea how to create a simple graph with. How do I add something like the following undirected graph using C#: Basically I'd like to know how to connect nodes. I have my own datastructures that I can already determine the neighbors and the distance. I'd now like to convert that into this posted datastructure so I can run it through the AStar algorithm. I was seeking something more like: Path<EdgeNode> startGraphNode = new Path<EdgeNode>(tempStartNode); startGraphNode.AddNeighbor(someOtherNode, distance);

    Read the article

  • A* algorithm works OK, but not perfectly. What's wrong?

    - by Bart van Heukelom
    This is my grid of nodes: I'm moving an object around on it using the A* pathfinding algorithm. It generally works OK, but it sometimes acts wrongly: When moving from 3 to 1, it correctly goes via 2. When going from 1 to 3 however, it goes via 4. When moving between 3 and 5, it goes via 4 in either direction instead of the shorter way via 6 What can be wrong? Here's my code (AS3): public static function getPath(from:Point, to:Point, grid:NodeGrid):PointLine { // get target node var target:NodeGridNode = grid.getClosestNodeObj(to.x, to.y); var backtrace:Map = new Map(); var openList:LinkedSet = new LinkedSet(); var closedList:LinkedSet = new LinkedSet(); // begin with first node openList.add(grid.getClosestNodeObj(from.x, from.y)); // start A* var curNode:NodeGridNode; while (openList.size != 0) { // pick a new current node if (openList.size == 1) { curNode = NodeGridNode(openList.first); } else { // find cheapest node in open list var minScore:Number = Number.MAX_VALUE; var minNext:NodeGridNode; openList.iterate(function(next:NodeGridNode, i:int):int { var score:Number = curNode.distanceTo(next) + next.distanceTo(target); if (score < minScore) { minScore = score; minNext = next; return LinkedSet.BREAK; } return 0; }); curNode = minNext; } // have not reached if (curNode == target) break; else { // move to closed openList.remove(curNode); closedList.add(curNode); // put connected nodes on open list for each (var adjNode:NodeGridNode in curNode.connects) { if (!openList.contains(adjNode) && !closedList.contains(adjNode)) { openList.add(adjNode); backtrace.put(adjNode, curNode); } } } } // make path var pathPoints:Vector.<Point> = new Vector.<Point>(); pathPoints.push(to); while(curNode != null) { pathPoints.unshift(curNode.location); curNode = backtrace.read(curNode); } pathPoints.unshift(from); return new PointLine(pathPoints); } NodeGridNode::distanceTo() public function distanceTo(o:NodeGridNode):Number { var dx:Number = location.x - o.location.x; var dy:Number = location.y - o.location.y; return Math.sqrt(dx*dx + dy*dy); }

    Read the article

  • Correct formulation of the A* algorithm

    - by Eli Bendersky
    Hello, I'm looking at definitions of the A* path-finding algorithm, and it seems to be defined somewhat differently in different places. The difference is in the action performed when going through the successors of a node, and finding that a successor is on the closed list. One approach (suggested by Wikipedia, and this article) says: if the successor is on the closed list, just ignore it Another approach (suggested here and here, for example) says: if the successor is on the closed list, examine its cost. If it's higher than the currently computed score, remove the item from the closed list for future examination. I'm confused - which method is correct ? Intuitively, the first makes more sense to me, but I wonder about the difference in definition. Is one of the definitions wrong, or are they somehow isomorphic ?

    Read the article

  • Searching in graphs trees with Depth/Breadth first/A* algorithms

    - by devoured elysium
    I have a couple of questions about searching in graphs/trees: Let's assume I have an empty chess board and I want to move a pawn around from point A to B. A. When using depth first search or breadth first search must we use open and closed lists ? This is, a list that has all the elements to check, and other with all other elements that were already checked? Is it even possible to do it without having those lists? What about A*, does it need it? B. When using lists, after having found a solution, how can you get the sequence of states from A to B? I assume when you have items in the open and closed list, instead of just having the (x, y) states, you have an "extended state" formed with (x, y, parent_of_this_node) ? C. State A has 4 possible moves (right, left, up, down). If I do as first move left, should I let it in the next state come back to the original state? This, is, do the "right" move? If not, must I transverse the search tree every time to check which states I've been to? D. When I see a state in the tree where I've already been, should I just ignore it, as I know it's a dead end? I guess to do this I'd have to always keep the list of visited states, right? E. Is there any difference between search trees and graphs? Are they just different ways to look at the same thing?

    Read the article

  • Problems with with A* algorithm

    - by V_Programmer
    I'm trying to implement the A* algorithm in Java. I followed this tutorial,in particular, this pseudocode: http://theory.stanford.edu/~amitp/GameProgramming/ImplementationNotes.html The problem is my code doesn't work. It goes into an infinite loop. I really don't know why this happens... I suspect that the problem are in F = G + H function implemented in Graph constructors. I suspect I am not calculate the neighbor F correclty. Here's my code: List<Graph> open; List<Graph> close; private void createRouteAStar(Unit u) { open = new ArrayList<Graph>(); close = new ArrayList<Graph>(); u.ai_route_endX = 11; u.ai_route_endY = 5; List<Graph> neigh; int index; int i; boolean finish = false; Graph current; int cost; Graph start = new Graph(u.xMap, u.yMap, 0, ManhattanDistance(u.xMap, u.yMap, u.ai_route_endX, u.ai_route_endY)); open.add(start); current = start; while(!finish) { index = findLowerF(); current = new Graph(open, index); System.out.println(current.x); System.out.println(current.y); if (current.x == u.ai_route_endX && current.y == u.ai_route_endY) { finish = true; } else { close.add(current); neigh = current.getNeighbors(); for (i = 0; i < neigh.size(); i++) { cost = current.g + ManhattanDistance(current.x, current.y, neigh.get(i).x, neigh.get(i).y); if (open.contains(neigh.get(i)) && cost < neigh.get(i).g) { open.remove(open.indexOf(neigh)); } else if (close.contains(neigh.get(i)) && cost < neigh.get(i).g) { close.remove(close.indexOf(neigh)); } else if (!open.contains(neigh.get(i)) && !close.contains(neigh.get(i))) { neigh.get(i).g = cost; neigh.get(i).f = cost + ManhattanDistance(neigh.get(i).x, neigh.get(i).y, u.ai_route_endX, u.ai_route_endY); neigh.get(i).setParent(current); open.add(neigh.get(i)); } } } } System.out.println("step"); for (i=0; i < close.size(); i++) { if (close.get(i).parent != null) { System.out.println(i); System.out.println(close.get(i).parent.x); System.out.println(close.get(i).parent.y); } } } private int findLowerF() { int i; int min = 10000; int minIndex = -1; for (i=0; i < open.size(); i++) { if (open.get(i).f < min) { min = open.get(i).f; minIndex = i; System.out.println("min"); System.out.println(min); } } return minIndex; } private int ManhattanDistance(int ax, int ay, int bx, int by) { return Math.abs(ax-bx) + Math.abs(ay-by); } And, as I've said. I suspect that the Graph class has the main problem. However I've not been able to detect and fix it. public class Graph { int x, y; int f,g,h; Graph parent; public Graph(int x, int y, int g, int h) { this.x = x; this.y = y; this.g = g; this.h = h; this.f = g + h; } public Graph(List<Graph> list, int index) { this.x = list.get(index).x; this.y = list.get(index).y; this.g = list.get(index).g; this.h = list.get(index).h; this.f = list.get(index).f; this.parent = list.get(index).parent; } public Graph(Graph gp) { this.x = gp.x; this.y = gp.y; this.g = gp.g; this.h = gp.h; this.f = gp.f; } public Graph(Graph gp, Graph parent) { this.x = gp.x; this.y = gp.y; this.g = gp.g; this.h = gp.h; this.f = g + h; this.parent = parent; } public List<Graph> getNeighbors() { List<Graph> aux = new ArrayList<Graph>(); aux.add(new Graph(x+1, y, g,h)); aux.add(new Graph(x-1, y, g,h)); aux.add(new Graph(x, y+1, g,h)); aux.add(new Graph(x, y-1, g,h)); return aux; } public void setParent(Graph g) { parent = g; } } Little Edit: Using the System.out and the Debugger I discovered that the program ALWAYS is check the same "current" graph, (15,8) which is the (u.xMap, u.yMap) position. Looks like it keeps forever in the first step.

    Read the article

  • Finding good heuristic for A* search

    - by Martin
    I'm trying to find the optimal solution for a little puzzle game called Twiddle (an applet with the game can be found here). The game has a 3x3 matrix with the number from 1 to 9. The goal is to bring the numbers in the correct order using the minimum amount of moves. In each move you can rotate a 2x2 square either clockwise or counterclockwise. I.e. if you have this state 6 3 9 8 7 5 1 2 4 and you rotate the upper left 2x2 square clockwise you get 8 6 9 7 3 5 1 2 4 I'm using a A* search to find the optimal solution. My f() is simply the number of rotations need. My heuristic function already leads to the optimal solution but I don't think it's the best one you can find. My current heuristic takes each corner, looks at the number at the corner and calculates the manhatten distance to the position this number will have in the solved state (which gives me the number of rotation needed to bring the number to this postion) and sums all these values. I.e. You take the above example: 6 3 9 8 7 5 1 2 4 and this end state 1 2 3 4 5 6 7 8 9 then the heuristic does the following 6 is currently at index 0 and should by at index 5: 3 rotations needed 9 is currently at index 2 and should by at index 8: 2 rotations needed 1 is currently at index 6 and should by at index 0: 2 rotations needed 4 is currently at index 8 and should by at index 3: 3 rotations needed h = 3 + 2 + 2 + 3 = 10 But there is the problem, that you rotate 4 elements at once. So there a rare cases where you can do two (ore more) of theses estimated rotations in one move. This means theses heuristic overestimates the distance to the solution. My current workaround is, to simply excluded one of the corners from the calculation which solves this problem at least for my test-cases. I've done no research if really solves the problem or if this heuristic still overestimates in same edge-cases. So my question is: What is the best heuristic you can come up with? (Disclaimer: This is for a university project, so this is a bit of homework. But I'm free to use any resource if can come up with, so it's okay to ask you guys. Also I will credit Stackoverflow for helping me ;) )

    Read the article

  • How to avoid that the robot gets trapped in local minimum?

    - by nesmoht
    Hi, I have some time occupying myself with motion planning for robots, and have for some time wanted to explore the possibility of improving the opportunities as "potential field" method offers. My challenge is to avoid that the robot gets trapped in "local minimum" when using the "potential field" method. Instead of using a "random walk" approach to avoid that the robot gets trapped I have thought about whether it is possible to implement a variation of A* which could act as a sort of guide for precisely to avoid that the robot gets trapped in "local minimum". Is there some of the experiences of this kind, or can refer to literature, which avoids local minimum in a more effective way than the one used in the "random walk" approach.

    Read the article

  • Structure of Astar (A*) graph search data in C#

    - by Shawn Mclean
    How do you structure you graphs/nodes in a graph search class? I'm basically creating a NavMesh and need to generate the nodes from 1 polygon to the other. The edge that joins both polygons will be the node. I'll then run A* on these Nodes to calculate the shortest path. I just need to know how to structure my classes and their properties? I know for sure I wont need to create a fully blown undirected graph with nodes and edges.

    Read the article

  • What is a mantainable way of saving "star rating" in a database?

    - by Montecristo
    I'll use the jQuery plugin for presenting the user with a nice interface The request is to display 5 stars, up to a total score of 10 (2 points per star). By now I thought about using 7/10 as a format for that value, but what if at some point in the future I'll receive a request like We would like to give users more choice, let's increase the total score to 20 (so that each star contributes with a maximum of 4 points) I'll end up with a table with mixed values for the "star rating" column: some will be like 7/10 while others will be like 14/20. Is it ok for you to have this difference in the database and deal with it in the logic layer to have it consistent? Or is preferred another way so that querying the table will not result in inconsistent results outside the application? Maybe floating point values could help me, is it better to store that value as a number less than or equal to one? So in each of the two examples the resulting value stored in the database would be 0,7, as a number, not a varchar, which can be queried also outside the application. What do you think?

    Read the article

  • Mediamonkey ratings imported into Rhythmbox/Banshee/Quod Libet

    - by JoshK
    I recently made a complete shift from Windows 7 to Ubuntu 12.04. Everything went smoothly until I scanned my music files in some of the Ubuntu players... All of my ratings added in Windows, using Mediamonkey, were out-of-five-stars (some with a half-star precision) - but when imported into RhythmBox, Banshee and Quod Libet, they all changed to a out-of-four-stars rating, with no five-star songs and no indication of how the old ratings were mapped into the new system. Does anyone know how to fix this? Even getting to know which way the ratings were mapped from 5-star system (with a half-star increment) to a 4-star basis will be very helpful. Thank you!

    Read the article

  • How to install wgt files (widgets) on Samsung Star without an internet connection?

    - by Koning Baard XIV
    Hi, I just created a new widget by following a tutorial. I created a zip containing all files and renamed it to HelloWorld.wgt instead of HelloWorld.zip. I sent it to my phone via Bluetooth, but when I try to open the wgt file on my phone it says it can't open it, because it doesn't know the filetype. Is there a way to install homebrew widgets on a Samsung Star without using a webserver? Thanks,

    Read the article

  • How do I read old star office 5.x spreadcheet documents today? (*.sdc)

    - by Johan
    Hi I found a couple of old spreadsheet documents that I created a couple of years ago that I would like to read again. I think I used the old Star Office 5.2 to create those Spreadsheets. They all have names like *.sdc I have tried to use Open Office to open them, but he can't recognise what it is. Does anybody have any idea how I can open those documents? Thanks Johan

    Read the article

  • Flash error 1084: "Syntax error"

    - by Elliot Broomhall
    Hi I'm getting two error messages in Flash when using actionscropt 3.0 "Topbar,Layer 'Action Layer',Frame 1,line 12 1084: syntax error: expection semicolon before add. "Topbar,Layer 'Action Layer',Frame 1,line 12 1084: syntax error: expection rightbrace before semicolon Here is my code could anyone give some insight to what is actually happening thanks and help on rectifying the issue thanks. clip = Number(random(7)) + 1; while (Number(clip) <= 7) { clip = Number(clip) + 1; Scale = Number(random(80)) + 1; setProperty("/star", _x, Number(random(800)) + 10); setProperty("/star", _rotation, Number(random(330)) + 50); setProperty("/star", _xscale, Scale); setProperty("/star", _yscale, Scale); setProperty("/star", _y, Number(random(800)) + 50); n = Number(n) + 1; bn = "star" add n; duplicateMovieClip("star", bn, n); set(bn add ":n", n); } // end while clip = "0";

    Read the article

  • Desktop Customization: Sci-Fi Icon Packs

    - by Asian Angel
    Are you a sci-fi fan who has been looking for some great custom icons for your desktop or favorite app launcher? Then you will want to have a look through our sci-fi icon packs collection. Over the past few months we’ve been showing you collections of cool desktop wallpapers you can use to liven up your computer. Today we extend the customization collections with a series of cool icon packs for you to use for folders and shortcut icons. Star Wars 1.0 Download Star Wars the Icons Download Star Wars Vehicles Download Star Wars Icons Download Star Trek Download Trek Tech Note: Contains “.png files” for use in Linux. Download Refresh Trek Download Star Trek Folders Download Battlestar Galactica Vol. 1 Download Battlestar Galactica Vol. 2 Download Battlestar Galactica Vol. 3 Download Battlestar Galactica Vol. 4 Download Baby Spaceships Download Space: 1999 Download War of the Worlds   Download Conclusion Now that you have some of these cool icons downloaded, be sure to check out our tutorial on how to customize your icons in Vista and Windows 7. If you’re still using XP check out our article on customizing icons in Windows XP. Also, be you might want to visit our new Desktop Fun section for more customization goodness! Similar Articles Productive Geek Tips Restore Missing Desktop Icons in Windows 7 or VistaWindows 7 Welcome Screen Taking Forever? Here’s the Fix (Maybe)Add Home Directory Icon to the Desktop in Windows 7 or VistaQuick Help: Downloadable Show Desktop Icon for XPDisplay My Computer Icon on the Desktop in Windows 7 or Vista TouchFreeze Alternative in AutoHotkey The Icy Undertow Desktop Windows Home Server – Backup to LAN The Clear & Clean Desktop Use This Bookmarklet to Easily Get Albums Use AutoHotkey to Assign a Hotkey to a Specific Window Latest Software Reviews Tinyhacker Random Tips Revo Uninstaller Pro Registry Mechanic 9 for Windows PC Tools Internet Security Suite 2010 PCmover Professional Windows 7 Easter Theme YoWindoW, a real time weather screensaver Optimize your computer the Microsoft way Stormpulse provides slick, real time weather data Geek Parents – Did you try Parental Controls in Windows 7? Change DNS servers on the fly with DNS Jumper

    Read the article

  • New P6 Reporting Database R2

    - by mark.kromer
    Along with our announced GA release of P6 Analytics R1 recently, you may have noticed that when you purchase P6 Analytics, we provide a restricted use license for P6 Reporting Database R2. This represent an updated version of the previous P6 Reporting Database 6.2 and can be purchased individually on a per-CPU basis. Typically, you will want just the reporting database if you would like the P6 data warehouse components such as the ETL, data models, ODS and star schemas in order to report on that data with another reporting tool other than Oracle. The P6 Analytics solution will only work on Oracle BI (OBI). But I pasted below some examples of a simplistic matrix report that I built from the P6 Reporting Database using Microsoft SQL Server Reporting Services. This is the Report Builder tool which is very similar to other similar tools to build reports on the market today such as Crystal Reports or Oracle BI Publisher. This is an example of what you can do (in a very simple format) by using the P6 Reporting Database without P6 Analytics: Here is a quick run-down of some of the key new features in P6 Reporting Database R2 that were added as enhancements to the 6.2 version: • 4 new star schemas (improved projects star, project history, resource utilization and resource allocation) • Improved ETL performance and reliability • P6 security is inherited at the star schema level • Custom P6 project, activity & resource codes are now available as customizable dimensions in the star schemas • Time-phase data down to the data is now available from the star schemas • An updated Operational Data Store (ODS) for operational reporting that includes the WBS hierarchy • The ODS now includes daily spreads for activity and resource assignments

    Read the article

  • How can I better implement A star algorithm with a very large set of nodes?

    - by Stephen
    I'm making a game with nodejs in which many enemies must converge on the player as the player moves around a relatively open space (right now it is an open field with few obstacles, but eventually there may be some small buildings in the field with 1 or 2 rooms). It's a multiplayer game using websockets, so the server needs to keep track of enemies and players. I found this javascript A* library which I've modified to be used on the server as a nodejs module. The library utilizes a Binary Heap to track the nodes for the algorithm, so it should be pretty fast (and indeed, with a small grid, say 100x100 it is lightning fast). The problem is that my game is not really tile-based. As the player moves around the map, he is moving on a more or less 1-to-1 per-pixel coordinate system (the player can move in 8 directions, 1 or 2 pixels at a time). In preliminary tests, on an 800x600 field, the path-finding can take anywhere from 400 to 1000 ms. Multiply that by 10 enemies and the game starts to get pretty choppy. I have already set it up so that each enemy will only do a path-finding call once per second or even as slow as once every 2 seconds (they have to keep updating their path because the players can move freely). But even with this long interval, there are noticeable lag spikes or chops every couple of seconds as the enemies update their paths. I'm willing to approach the problem of path-finding differently, if there's another option. I'm assuming that the real problem is the enormous grid (800x600). It also occurs to me that maybe the large arrays are to blame, as I've read that V8 has trouble with large arrays.

    Read the article

  • What is the significance of '*' (star, asterisk) in the file listing results?

    - by vfclists
    I have noticed that some of my files have an asterisk at end. Does the asterisk at the end have any particular significance? I think they are mostly executable and displayed in green by the ls command. You will see that ./bkmp* and ./bkmp0* have an asterisk at the end. They are executable bash scripts. Here's my output: drwxr-xr-x 7 username username 4096 Oct 2 18:28 ./ drwxr-xr-x 8 root root 4096 Oct 2 09:25 ../ -rw-r--r-- 1 username username 3724 Sep 22 03:06 .bashrc -rwxr--r-- 1 username username 319 Sep 22 03:42 .bkmp* -rwxr--r-- 1 username username 324 Sep 29 23:30 .bkmp0* drwx------ 2 username username 4096 Sep 17 13:52 .cache/ -rw-r--r-- 1 username username 675 Sep 17 13:37 .profile drwx------ 2 username username 4096 Sep 22 10:10 .ssh/ drwx------ 2 username username 4096 Sep 24 19:49 .ssh.local/ drwxr-xr-x 2 username username 4096 Sep 22 04:10 archives/ drwxr-xr-x 3 username username 4096 Sep 24 19:51 home/ -rw-r--r-- 1 username username 27511 Sep 24 19:51 username_backup.20120924_1908.tar.gz

    Read the article

  • javascript disabling a div or a link? (for a 5-star rating system)

    - by Cyprus106
    Basically, I've created a 5-star rating system. Pretty typical. It shows how many stars other people have given the item, and then when a user hovers over the stars, it lights up x number of stars based on how many they're over. It's all run by AJAX. They click 5 stars it automatically adds their 5-star rating to the group. The problem is that after they rate it I want to turn the system off, but I can't seem to be able to do that. I've tried everything I can think of. I've tried using element.disable for the a hrefs and for the div, but it still lets them vote away, over and over again, at least in firefox.... Can anyone help me out with a method to simply "freeze" the stars on what the user voted?? If I need to add code that's cool! I figured it probably wasn't necessary in this situation!

    Read the article

  • In a star schema, are foreign key constraints between facts and dimensions neccessary?

    - by Garett
    I'm getting my first exposure to data warehousing, and I’m wondering is it necessary to have foreign key constraints between facts and dimensions. Are there any major downsides for not having them? I’m currently working with a relational star schema. In traditional applications I’m used to having them, but I started to wonder if they were needed in this case. I’m currently working in a SQL Server 2005 environment.

    Read the article

< Previous Page | 2 3 4 5 6 7 8 9 10 11 12 13  | Next Page >