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  • Suggested GA operators for a TSP problem?

    - by Mark
    I'm building a genetic algorithm to tackle the traveling salesman problem. Unfortunately, I hit peaks that can sustain for over a thousand generations before mutating out of them and getting better results. What crossover and mutation operators generally do well in this case?

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  • Finding backedges in a graph, with special conditions.

    - by Morteza M.
    There is a vertex v, such that from the subtree rooted at v, there are at least two backedges to proper ancestors of v. The problem is finding whether such backedges exist or not ( finding v is not important at all). I run DFS algorithm, I can find backedges and save them in an array. I know which backedges in this array belong to a common tree. but I have problem on matching this condition in O(E) time. can anyone help me with that?

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  • Drawing Directed Acyclic Graphs: Minimizing edge crossing?

    - by Robert Fraser
    Laying out the verticies in a DAG in a tree form (i.e. verticies with no in-edges on top, verticies dependent only on those on the next level, etc.) is rather simple without graph drawing algorithms such as Efficient Sugimiya. However, is there a simple algorithm to do this that minimizes edge crossing? (For some graphs, it may be impossible to completely eliminate edge crossing.) A picture says a thousand words, so is there an algorithm that would suggest: instead of: EDIT: As the picture suggests, a vertex's inputs are always on top and outputs are always below, which is another barrier to just pasting in an existing layout algorithm.

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  • graph and all pairs shortest path in java

    - by Sandra
    I am writing a java program using Flyod-Warshall algorithm “All pairs shortest path”. I have written the following : a0 is the adjacency matrix of my graph, but has infinity instead of 0. vList is the list of vertexes and the cost for each edge is 1. Path[i][j] = k+1 means for going from I to j you first go to k then j int[][] path = new int[size][size]; for(int i = 0; i<path.length;i++) { for(int j = 0; j<path.length; j++) { if(adjM[i][j]==1) path[i][j]=j+1; } } //*************** for (int k = 0; k < vList.size(); k++) for (int i = 0; i < vList.size(); i++) for (int j = 0; j < vList.size(); j++) { if (a0[i][j]>a0[i][k]+ a0[k][j]) path[i][j] = k + 1; a0[i][j] = Math.min(a0[i][j], a0[i][k] + a0[k][j]); } After running this code, in the result a0 is correct, but path is not correct and I don’t know why!. Would you please help me?

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  • Infinite loop during A* algorithm

    - by Tashu
    The A* algorithm is used by enemies to have a path to the goal. It's working but when sometimes I placed a tower in a grid (randomly) it produces a stack overflow error. The A* algorithm would iterate the enemy and find its path and pass the list to the enemy's path. I added debug logs and the list that I'm getting it looks like it would arrive from start cell to goal cell. Here's the log - 06-19 19:26:41.982: DEBUG/findEnemyPath, enemy X:Y(4281): X2.8256836:Y3.5 06-19 19:26:41.990: DEBUG/findEnemyPath, grid X:Y(4281): X3:Y2 06-19 19:26:41.990: DEBUG/START CELL ID:(4281): 38 06-19 19:26:41.990: DEBUG/GOAL CELL ID:(4281): 47 06-19 19:26:41.990: DEBUG/Best : 38(4281): passThrough:0.0 06-19 19:26:41.990: DEBUG/Neighbor's Parent:(4281): 38 06-19 19:26:41.990: DEBUG/Neighbor's Parent:(4281): 38 06-19 19:26:41.990: DEBUG/Neighbor's Parent:(4281): 38 06-19 19:26:41.990: DEBUG/Neighbor's Parent:(4281): 38 06-19 19:26:41.990: DEBUG/Best : 39(4281): passThrough:8.875 06-19 19:26:41.990: DEBUG/Neighbor's Parent:(4281): 39 06-19 19:26:41.990: DEBUG/Neighbor's Parent:(4281): 39 06-19 19:26:41.990: DEBUG/Neighbor's Parent:(4281): 39 06-19 19:26:41.990: DEBUG/Best : 40(4281): passThrough:7.9375 06-19 19:26:41.990: DEBUG/Neighbor's Parent:(4281): 40 06-19 19:26:41.990: DEBUG/Neighbor's Parent:(4281): 40 06-19 19:26:41.990: DEBUG/Best : 52(4281): passThrough:8.9375 06-19 19:26:41.990: DEBUG/Neighbor's Parent:(4281): 52 06-19 19:26:41.990: DEBUG/Neighbor's Parent:(4281): 52 06-19 19:26:41.990: DEBUG/Best : 53(4281): passThrough:7.96875 06-19 19:26:41.990: DEBUG/Neighbor's Parent:(4281): 53 06-19 19:26:41.990: DEBUG/Best : 28(4281): passThrough:8.9375 06-19 19:26:41.990: DEBUG/Neighbor's Parent:(4281): 28 06-19 19:26:41.990: DEBUG/Best : 65(4281): passThrough:8.984375 06-19 19:26:41.990: DEBUG/Neighbor's Parent:(4281): 65 06-19 19:26:41.990: DEBUG/Neighbor's Parent:(4281): 65 06-19 19:26:41.990: DEBUG/Best : 66(4281): passThrough:7.9921875 06-19 19:26:41.990: DEBUG/Neighbor's Parent:(4281): 66 06-19 19:26:42.000: DEBUG/Best : 78(4281): passThrough:8.99609375 06-19 19:26:42.000: DEBUG/Neighbor's Parent:(4281): 78 06-19 19:26:42.000: DEBUG/Best : 79(4281): passThrough:7.998046875 06-19 19:26:42.000: DEBUG/Neighbor's Parent:(4281): 79 06-19 19:26:42.000: DEBUG/Best : 80(4281): passThrough:6.9990234375 06-19 19:26:42.000: DEBUG/Neighbor's Parent:(4281): 80 06-19 19:26:42.000: DEBUG/Neighbor's Parent:(4281): 80 06-19 19:26:42.000: DEBUG/Best : 81(4281): passThrough:5.99951171875 06-19 19:26:42.000: DEBUG/Neighbor's Parent:(4281): 81 06-19 19:26:42.000: DEBUG/Neighbor's Parent:(4281): 81 06-19 19:26:42.000: DEBUG/Best : 82(4281): passThrough:4.999755859375 06-19 19:26:42.000: DEBUG/Neighbor's Parent:(4281): 82 06-19 19:26:42.000: DEBUG/Neighbor's Parent:(4281): 82 06-19 19:26:42.000: DEBUG/Best : 83(4281): passThrough:3.9998779296875 06-19 19:26:42.000: DEBUG/Neighbor's Parent:(4281): 83 06-19 19:26:42.000: DEBUG/Best : 71(4281): passThrough:2.99993896484375 06-19 19:26:42.000: DEBUG/Neighbor's Parent:(4281): 71 06-19 19:26:42.000: DEBUG/Best : 59(4281): passThrough:1.99951171875 06-19 19:26:42.000: DEBUG/Neighbor's Parent:(4281): 59 06-19 19:26:42.000: DEBUG/Neighbor's Parent:(4281): 59 06-19 19:26:42.000: DEBUG/Neighbor's Parent:(4281): 59 06-19 19:26:42.000: DEBUG/Best : 47(4281): passThrough:0.99951171875 Then, the goal cell would be iterating its parent till start cell to break off the loop. private void populateBestList(Cell cell, List<Cell> bestList) { bestList.add(cell); if (cell.parent.start == false) { Log.d("ID:", ""+cell.id); Log.d("ParentID:", ""+cell.parent.id); populateBestList(cell.parent, bestList); } return; } The log with error above would show like this - 06-19 19:26:42.010: DEBUG/ID:(4281): 47 06-19 19:26:42.010: DEBUG/ParentID:(4281): 59 06-19 19:26:42.010: DEBUG/ID:(4281): 59 06-19 19:26:42.010: DEBUG/ParentID:(4281): 71 06-19 19:26:42.010: DEBUG/ID:(4281): 71 06-19 19:26:42.010: DEBUG/ParentID:(4281): 59 06-19 19:26:42.010: DEBUG/ID:(4281): 59 06-19 19:26:42.010: DEBUG/ParentID:(4281): 71 06-19 19:26:42.010: DEBUG/ID:(4281): 71 71 and 59 would switch over and goes on. I thought the grid is the issue due to the fact that enemies are using the single grid so I make the parent, start, and goal clear before starting the A* algorithm for an enemy. for(int i = 0; i < GRID_HEIGHT; i++) { for(int j = 0; j < GRID_WIDTH; j++) { grid[i][j].parent = null; grid[i][j].start = false; grid[i][j].goal = false; } } That didn't work. I thought it might be something related to this code, but not sure if I'm on right track - neighbor.parent = best; openList.remove(neighbor); closedList.remove(neighbor); openList.add(0, neighbor); Here's the code of the A* algorithm - private List<Cell> findEnemyPath(Enemy enemy) { for(int i = 0; i < GRID_HEIGHT; i++) { for(int j = 0; j < GRID_WIDTH; j++) { grid[i][j].parent = null; grid[i][j].start = false; grid[i][j].goal = false; } } List<Cell> openList = new ArrayList<Cell>(); List<Cell> closedList = new ArrayList<Cell>(); List<Cell> bestList = new ArrayList<Cell>(); int width = (int)Math.floor(enemy.position.x); int height = (int)Math.floor(enemy.position.y); width = (width < 0) ? 0 : width; height = (height < 0) ? 0 : height; Log.d("findEnemyPath, enemy X:Y", "X"+enemy.position.x+":"+"Y"+enemy.position.y); Log.d("findEnemyPath, grid X:Y", "X"+height+":"+"Y"+width); Cell start = grid[height][width]; Cell goal = grid[ENEMY_GOAL_HEIGHT][ENEMY_GOAL_WIDTH]; if(start.id != goal.id) { Log.d("START CELL ID: ", ""+start.id); Log.d("GOAL CELL ID: ", ""+goal.id); //Log.d("findEnemyPath, grid X:Y", "X"+start.position.x+":"+"Y"+start.position.y); start.start = true; goal.goal = true; openList.add(start); while(openList.size() > 0) { Cell best = findBestPassThrough(openList, goal); //Log.d("ID:", ""+best.id); openList.remove(best); closedList.add(best); if (best.goal) { System.out.println("Found Goal"); System.out.println(bestList.size()); populateBestList(goal, bestList); /* for(Cell cell : bestList) { Log.d("ID:", ""+cell.id); Log.d("ParentID:", ""+cell.parent.id); } */ Collections.reverse(bestList); Cell exit = new Cell(13.5f, 3.5f, 1, 1); exit.isExit = true; bestList.add(exit); //Log.d("PathList", "Enemy ID : " + enemy.id); return bestList; } else { List<Cell> neighbors = getNeighbors(best); for (Cell neighbor : neighbors) { if(neighbor.isTower) { continue; } if (openList.contains(neighbor)) { Cell tmpCell = new Cell(neighbor.position.x, neighbor.position.y, 1, 1); tmpCell.parent = best; if (tmpCell.getPassThrough(goal) >= neighbor.getPassThrough(goal)) { continue; } } if (closedList.contains(neighbor)) { Cell tmpCell = new Cell(neighbor.position.x, neighbor.position.y, 1, 1); tmpCell.parent = best; if (tmpCell.getPassThrough(goal) >= neighbor.getPassThrough(goal)) { continue; } } Log.d("Neighbor's Parent: ", ""+best.id); neighbor.parent = best; openList.remove(neighbor); closedList.remove(neighbor); openList.add(0, neighbor); } } } } Log.d("Cannot find a path", ""); return null; }

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  • Disk Search / Sort Algorithm

    - by AlgoMan
    Given a Range of numbers say 1 to 10,000, Input is in random order. Constraint: At any point only 1000 numbers can be loaded to memory. Assumption: Assuming unique numbers. I propose the following efficient , "When-Required-sort Algorithm". We write the numbers into files which are designated to hold particular range of numbers. For example, File1 will have 0 - 999 , File2 will have 1000 - 1999 and so on in random order. If a particular number which is say "2535" is being searched for then we know that the number is in the file3 (Binary search over range to find the file). Then file3 is loaded to memory and sorted using say Quick sort (which is optimized to add insertion sort when the array size is small ) and then we search the number in this sorted array using Binary search. And when search is done we write back the sorted file. So in long run all the numbers will be sorted. Please comment on this proposal.

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  • Help with Algorithm chinese auction

    - by sam munkes
    Hi, i am designing a Chinese auction website. Tickets ($5, $10 & $20) are sold either individually, or via packages to receive discounts. There are various Ticket packages for example: 5-$5 tickets = receive 10% off 5-$10 tickets = receive 10% off 5-$20 tickets = receive 10% off 5-$5 tickets + 5-$10 tickets + 5-$20 tickets = receive 15% off When users add tickets to their cart, i need to figure out the cheapest package(s) to give them. the trick is that if a user adds 4-$5 tickets + 5-$10 tickets + 5-$20 tickets, it should still give him package #3 since that would be the cheapest for him. Any help in figuring out a algorithm to solve this, or any tips would be greatly appreciate it. thanks

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  • Algorithm to price bulk discounts

    - by sam munkes
    Hi, i am designing a Chinese auction website. Tickets ($5, $10 & $20) are sold either individually, or via packages to receive discounts. There are various Ticket packages for example: 5-$5 tickets = receive 10% off 5-$10 tickets = receive 10% off 5-$20 tickets = receive 10% off 5-$5 tickets + 5-$10 tickets + 5-$20 tickets = receive 15% off When users add tickets to their cart, i need to figure out the cheapest package(s) to give them. the trick is that if a user adds 4-$5 tickets + 5-$10 tickets + 5-$20 tickets, it should still give him package #4 since that would be the cheapest for him. Any help in figuring out a algorithm to solve this, or any tips would be greatly appreciate it. thanks

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  • Creating a "crossover" function for a genetic algorithm to improve network paths

    - by Dave
    Hi, I'm trying to develop a genetic algorithm that will find the most efficient way to connect a given number of nodes at specified locations. All the nodes on the network must be able to connect to the server node and there must be no cycles within the network. It's basically a tree. I have a function that can measure the "fitness" of any given network layout. What's stopping me is that I can't think of a crossover function that would take 2 network structures (parents) and somehow mix them to create offspring that would meet the above conditions. Any ideas? Clarification: The nodes each have a fixed x,y coordiante position. Only the routes between them can be altered.

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  • Suggestion on algorithm to distribute objects of different value

    - by Unknown
    Hello, I have the following problem: Given N objects of different values (N < 30, and the values are multiple of a "k" constant, i.e. k, 2k, 3k, 4k, 6k, 8k, 12k, 16k, 24k and 32k), I need an algorithm that will distribute all items to M players (M <= 6) in such a way that the total value of the objects each player gets is as even as possible (in other words, I want to distribute all objects to all players in the fairest way possible). I don't need (pseudo)code to solve this (also, this is not a homework :) ), but I'll appreciate any ideas or links to algorithms that could solve this. Thanks!

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  • help with number calculation algorithm [hw]

    - by sa125
    Hi - I'm working on a hw problem that asks me this: given a finite set of numbers, and a target number, find if the set can be used to calculate the target number using basic math operations (add, sub, mult, div) and using each number in the set exactly once (so I need to exhaust the set). This has to be done with recursion. So, for example, if I have the set {1, 2, 3, 4} and target 10, then I could get to it by using ((3 * 4) - 2)/1 = 10. I'm trying to phrase the algorithm in pseudo-code, but so far haven't gotten too far. I'm thinking graphs are the way to go, but would definitely appreciate help on this. thanks.

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  • Graph layouting with Perl

    - by jonny
    Ok, I have a flowchart definition (basically, array of nodes and edges for each node). Now I want to calculate coordinates for every task in the flow, preferably hierarchycal style. I need something like Graph::Easy::Layout but I have no idea how to get nodes coordinates: I render nodes myself and I only want to retrieve box coordinates/size. Any suggestions? What I need is a cpan module avialable even in Debian repository.

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  • Open Source Graph Layout Library

    - by James Westgate
    I'm looking for an open source (GPL, LGPL etc) graph layout library for .net framework, preferably fully managed code. Im not worried about the visualisation aspect of things. I can find lots of them for Java, but none for .net... Thanks!

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  • Drawing Directed Acyclic Graphs: Using DAG property to improve layout/edge routing?

    - by Robert Fraser
    Hi, Laying out the verticies in a DAG in a tree form (i.e. verticies with no in-edges on top, verticies dependent only on those on the next level, etc.) is rather simple. However, is there a simple algorithm to do this that minimizes edge crossing? (For some graphs, it may be impossible to completely eliminate edge crossing.) A picture says a thousand words, so is there an algorithm that would suggest: instead of:

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  • Algorithm for dynamic combinations

    - by sOltan
    My code has a list called INPUTS, that contains a dynamic number of lists, let's call them A, B, C, .. N. These lists contain a dynamic number of Events I would like to call a function with each combination of Events. To illustrate with an example: INPUTS: A(0,1,2), B(0,1), C(0,1,2,3) I need to call my function this many times for each combination (the input count is dynamic, in this example it is three parameter, but it can be more or less) function(A[0],B[0],C[0]) function(A[0],B[1],C[0]) function(A[0],B[0],C[1]) function(A[0],B[1],C[1]) function(A[0],B[0],C[2]) function(A[0],B[1],C[2]) function(A[0],B[0],C[3]) function(A[0],B[1],C[3]) function(A[1],B[0],C[0]) function(A[1],B[1],C[0]) function(A[1],B[0],C[1]) function(A[1],B[1],C[1]) function(A[1],B[0],C[2]) function(A[1],B[1],C[2]) function(A[1],B[0],C[3]) function(A[1],B[1],C[3]) function(A[2],B[0],C[0]) function(A[2],B[1],C[0]) function(A[2],B[0],C[1]) function(A[2],B[1],C[1]) function(A[2],B[0],C[2]) function(A[2],B[1],C[2]) function(A[2],B[0],C[3]) function(A[2],B[1],C[3]) This is what I have thought of so far: My approach so far is to build a list of combinations. The element combination is itself a list of "index" to the input arrays A, B and C. For our example: my list iCOMBINATIONS contains the following iCOMBO lists (0,0,0) (0,1,0) (0,0,1) (0,1,1) (0,0,2) (0,1,2) (0,0,3) (0,1,3) (1,0,0) (1,1,0) (1,0,1) (1,1,1) (1,0,2) (1,1,2) (1,0,3) (1,1,3) (2,0,0) (2,1,0) (2,0,1) (2,1,1) (2,0,2) (2,1,2) (2,0,3) (2,1,3) Then I would do this: foreach( iCOMBO in iCOMBINATIONS) { foreach ( P in INPUTS ) { COMBO.Clear() foreach ( i in iCOMBO ) { COMBO.Add( P[ iCOMBO[i] ] ) } function( COMBO ) --- (instead of passing the events separately) } } But I need to find a way to build the list iCOMBINATIONS for any given number of INPUTS and their events. Any ideas? Is there actually a better algorithm than this? any pseudo code to help me with will be great. C# (or VB) Thank You

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  • how to tackle this combinatorial algorithm problem

    - by Andrew Bullock
    I have N people who must each take T exams. Each exam takes "some" time, e.g. 30 min (no such thing as finishing early). Exams must be performed in front of an examiner. I need to schedule each person to take each exam in front of an examiner within an overall time period, using the minimum number of examiners for the minimum amount of time (i.e. no examiners idle) There are the following restrictions: No person can be in 2 places at once each person must take each exam once noone should be examined by the same examiner twice I realise that an optimal solution is probably NP-Complete, and that I'm probably best off using a genetic algorithm to obtain a best estimate (similar to this? http://stackoverflow.com/questions/184195/seating-plan-software-recommendations-does-such-a-beast-even-exist). I'm comfortable with how genetic algorithms work, what i'm struggling with is how to model the problem programatically such that i CAN manipulate the parameters genetically.. If each exam took the same amount of time, then i'd divide the time period up into these lengths, and simply create a matrix of time slots vs examiners and drop the candidates in. However because the times of each test are not necessarily the same, i'm a bit lost on how to approach this. currently im doing this: make a list of all "tests" which need to take place, between every candidate and exam start with as many examiners as there are tests repeatedly loop over all examiners, for each one: find an unscheduled test which is eligible for the examiner (based on the restrictions) continue until all tests that can be scheduled, are if there are any unscheduled tests, increment the number of examiners and start again. i'm looking for better suggestions on how to approach this, as it feels rather crude currently.

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  • Merge method in MergeSort Algorithm .

    - by Tony
    I've seen many mergeSort implementations .Here is the version in Data Structures and Algorithms in Java (2nd Edition) by Robert Lafore : private void recMergeSort(long[] workSpace, int lowerBound,int upperBound) { if(lowerBound == upperBound) // if range is 1, return; // no use sorting else { // find midpoint int mid = (lowerBound+upperBound) / 2; // sort low half recMergeSort(workSpace, lowerBound, mid); // sort high half recMergeSort(workSpace, mid+1, upperBound); // merge them merge(workSpace, lowerBound, mid+1, upperBound); } // end else } // end recMergeSort() private void merge(long[] workSpace, int lowPtr, int highPtr, int upperBound) { int j = 0; // workspace index int lowerBound = lowPtr; int mid = highPtr-1; int n = upperBound-lowerBound+1; // # of items while(lowPtr <= mid && highPtr <= upperBound) if( theArray[lowPtr] < theArray[highPtr] ) workSpace[j++] = theArray[lowPtr++]; else workSpace[j++] = theArray[highPtr++]; while(lowPtr <= mid) workSpace[j++] = theArray[lowPtr++]; while(highPtr <= upperBound) workSpace[j++] = theArray[highPtr++]; for(j=0; j<n; j++) theArray[lowerBound+j] = workSpace[j]; } // end merge() One interesting thing about merge method is that , almost all the implementations didn't pass the lowerBound parameter to merge method . lowerBound is calculated in the merge . This is strange , since lowerPtr = mid + 1 ; lowerBound = lowerPtr -1 ; that means lowerBound = mid ; Why the author didn't pass mid to merge like merge(workSpace, lowerBound,mid, mid+1, upperBound); ? I think there must be a reason , otherwise I can't understand why an algorithm older than half a center ,and have all coincident in the such little detail.

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  • Sparse parameter selection using Genetic Algorithm

    - by bgbg
    Hello, I'm facing a parameter selection problem, which I would like to solve using Genetic Algorithm (GA). I'm supposed to select not more than 4 parameters out of 3000 possible ones. Using the binary chromosome representation seems like a natural choice. The evaluation function punishes too many "selected" attributes and if the number of attributes is acceptable, it then evaluates the selection. The problem is that in these sparse conditions the GA can hardly improve the population. Neither the average fitness cost, nor the fitness of the "worst" individual improves over the generations. All I see is slight (even tiny) improvement in the score of the best individual, which, I suppose, is a result of random sampling. Encoding the problem using indices of the parameters doesn't work either. This is most probably, due to the fact that the chromosomes are directional, while the selection problem isn't (i.e. chromosomes [1, 2, 3, 4]; [4, 3, 2, 1]; [3, 2, 4, 1] etc. are identical) What problem representation would you suggest? P.S If this matters, I use PyEvolve.

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  • Algorithm to match list of regular expressions

    - by DSII
    I have two algorithmic questions for a project I am working on. I have thought about these, and have some suspicions, but I would love to hear the community's input as well. Suppose I have a string, and a list of N regular expressions (actually they are wildcard patterns representing a subset of full regex functionality). I want to know whether the string matches at least one of the regular expressions in the list. Is there a data structure that can allow me to match the string against the list of regular expressions in sublinear (presumably logarithmic) time? This is an extension of the previous problem. Suppose I have the same situation: a string and a list of N regular expressions, only now each of the regular expressions is paired with an offset within the string at which the match must begin (or, if you prefer, each of the regular expressions must match a substring of the given string beginning at the given offset). To give an example, suppose I had the string: This is a test string and the regex patterns and offsets: (a) his.* at offset 0 (b) his.* at offset 1 The algorithm should return true. Although regex (a) does not match the string beginning at offset 0, regex (b) does match the substring beginning at offset 1 ("his is a test string"). Is there a data structure that can allow me to solve this problem in sublinear time? One possibly useful piece of information is that often, many of the offsets in the list of regular expressions are the same (i.e. often we are matching the substring at offset X many times). This may be useful to leverage the solution to problem #1 above. Thank you very much in advance for any suggestions you may have!

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  • How to gain accurate results with Painter's algorithm?

    - by pimvdb
    A while ago I asked how to determine when a face is overlapping another. The advice was to use a Z-buffer. However, I cannot use a Z-buffer in my current project and hence I would like to use the Painter's algorithm. I have no good clue as to when a surface is behind or in front of another, though. I've tried numerous methods but they all fail in edge cases, or they fail even in general cases. This is a list of sorting methods I've tried so far: Distance to midpoint of each face Average distance to each vertex of each face Average z value of each vertex Higest z value of vertices of each face and draw those first Lowest z value of vertices of each face and draw those last The problem is that a face might have a closer distance but is still further away. All these methods seem unreliable. Edit: For example, in the following image the surface with the blue point as midpoint is painted over the surface with the red point as midpoint, because the blue point is closer. However, this is because the surface of the red point is larger and the midpoint is further away. The surface with the red point should be painted over the blue one, because it is closer, whilst the midpoint distance says the opposite. What exactly is used in the Painter's algorithm to determine the order in which objects should be drawn?

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  • Triangulation A* (TA*) pathfinding algorithm

    - by hyn
    I need help understanding the Triangle A* (TA*) algorithm that is described by Demyen in his paper Efficient Triangulation-Based Pathfinding, on pages 76-81. He describes how to adapt the regular A* algorithm for triangulation, to search for other possibly more optimal paths, even after the final node is reached/expanded. Regular A* stops when the final node is expanded, but this is not always the best path when used in a triangulated graph. This is exactly the problem I'm having. The problem is illustrated on page 78, Figure 5.4: I understand how to calculate the g and h values presented in the paper (page 80). And I think the search stop condition is: if (currentNode.fCost > shortestDistanceFound) { // stop break; } where currentNode is the search node popped from the open list (priority queue), which has the lowest f-score. shortestDistanceFound is the actual distance of the shortest path found so far. But how do I exclude the previously found paths from future searches? Because if I do the search again, it will obviously find the same path. Do I reset the closed list? I need to modify something, but I don't know what it is I need to change. The paper lacks pseudocode, so that would be helpful.

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  • need explanation on amortization in algorithm

    - by Pradeep
    I am a learning algorithm analysis and came across a analysis tool for understanding the running time of an algorithm with widely varying performance which is called as amortization. The autor quotes " An array with upper bound of n elements, with a fixed bound N, on it size. Operation clear takes O(n) time, since we should dereference all the elements in the array in order to really empty it. " The above statement is clear and valid. Now consider the next content: "Now consider a series of n operations on an initially empty array. if we take the worst case viewpoint, the running time is O(n^2), since the worst case of a sigle clear operation in the series is O(n) and there may be as many as O(n) clear operations in the series." From the above statement how is the time complexity O(n^2)? I did not understand the logic behind it. if 'n' operations are performed how is it O(n ^2)? Please explain what the autor is trying to convey..

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  • How does flocking algorithm work?

    - by Chan
    I read and understand the basic of flocking algorithm. Basically, we need to have 3 behaviors: 1. Cohesion 2. Separation 3. Alignment From my understanding, it's like a state machine. Every time we do an update (then draw), we check all the constraints on both three behaviors. And each behavior returns a Vector3 which is the "correct" orientation that an object should transform to. So my initial idea was /// <summary> /// Objects stick together /// </summary> /// <returns></returns> private Vector3 Cohesion() { Vector3 result = new Vector3(0.0f, 0.0f, 0.0f); return result; } /// <summary> /// Object align /// </summary> /// <returns></returns> private Vector3 Align() { Vector3 result = new Vector3(0.0f, 0.0f, 0.0f); return result; } /// <summary> /// Object separates from each others /// </summary> /// <returns></returns> private Vector3 Separate() { Vector3 result = new Vector3(0.0f, 0.0f, 0.0f); return result; } Then I search online for pseudocode but many of them involve velocity and acceleration plus other stuffs. This part confused me. In my game, all objects move at constant speed, and they have one leader. So can anyone share me an idea how to start on implement this flocking algorithm? Also, did I understand it correctly? (I'm using XNA 4.0)

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  • finding shortest valid path in a colored-edge graphs

    - by user1067083
    Given a directed graph G, with edges colored either green or purple, and a vertex S in G, I must find an algorithm that finds the shortest path from s to each vertex in G so the path includes at most two purple edges (and green as much as needed). I thought of BFS on G after removing all the purple edges, and for every vertex that the shortest path is still infinity, do something to try to find it, but I'm kinda stuck, and it takes alot of the running time as well... Any other suggestions? Thanks in advance

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  • Visualizing Undirected Graph That's Too Large for GraphViz?

    - by Gabe
    Hi Everyone, I was wondering if anyone has any advice for rendering an undirected graph with 178,000 nodes and 500,000 edges. I've tried Neato, Tulip, and Cytoscape. Neato doesn't even come remotely close, and Tulip and Cytoscape claim they can handle it but don't seem to be able to. (Tulip does nothing and Cytoscape claims to be working, and then just stops.) Does anyone have any ideas? I'd just like a vector format file (ps or pdf) with a remotely reasonable layout of the nodes. Thanks!

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