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• Issue with angular gradient algorithm

  - by user146780

  I have an angular gradient algorithm: GLuint OGLENGINEFUNCTIONS::CreateAngularGradient( std::vector<ARGBCOLORF> &input,POINTFLOAT start, POINTFLOAT end, int width, int height ) { std::vector<GLubyte> pdata(width * height * 4); float pi = 3.1415; float cx; float cy; float r1; float r2; float ang; float px; float py; float t; cx = start.x; cy = start.y; r1 = 0; r2 = 500; ang = end.x / 100; ARGBCOLORF color; for (unsigned int i = 0; i < height; i++) { for (unsigned int j = 0; j < width; j++) { px = j; py = i; if( px * px + py * py <= r2 * r2 && px * px + py * py >= r1 * r1 ) { t= atan2(py-cy,px-cx) + ang; t= t+ pi; if (t > 2* pi) { t=t-2*pi; t=t/(2*pi); } } //end + start color.r = (0 * t) + (120 * (1 - t)); color.g = (50 * t) + (255 * (1 - t)); color.b = (0 * t) + (50 * (1 - t)); color.a = (255 * t) + (200 * (1 - t)); pdata[i * 4 * width + j * 4 + 0] = (GLubyte) color.r; pdata[i * 4 * width + j * 4 + 1] = (GLubyte) color.g; pdata[i * 4 * width + j * 4 + 2] = (GLubyte) 12; pdata[i * 4 * width + j * 4 + 3] = (GLubyte) 255; } } It works fine except the ang variable only controls the end sweep, not the start sweep. The start sweep is always facing the middle left as seen here: http://img810.imageshack.us/img810/9623/uhoh.png Basically I have no control over the end one going this <- way. How could I control this one? Thanks

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• Algorithm to Render a Horizontal Binary-ish Tree in Text/ASCII form

  - by Justin L.

  It's a pretty normal binary tree, except for the fact that one of the nodes may be empty. I'd like to find a way to output it in a horizontal way (that is, the root node is on the left and expands to the right). I've had some experience expanding trees vertically (root node at the top, expanding downwards), but I'm not sure where to start, in this case. Preferably, it would follow these couple of rules: If a node has only one child, it can be skipped as redundant (an "end node", with no children, is always displayed) All nodes of the same depth must be aligned vertically; all nodes must be to the right of all less-deep nodes and to the left of all deeper nodes. Nodes have a string representation which includes their depth. Each "end node" has its own unique line; that is, the number of lines is the number of end nodes in the tree, and when an end node is on a line, there may be nothing else on that line after that end node. As a consequence of the last rule, the root node should be in either the top left or the bottom left corner; top left is preferred. For example, this is a valid tree, with six end nodes (node is represented by a name, and its depth): [a0]------------[b3]------[c5]------[d8] \ \ \----------[e9] \ \----[f5] \--[g1]--------[h4]------[i6] \ \--------------------[j10] \-[k3] Which represents the horizontal, explicit binary tree: 0 a / \ 1 g * / \ \ 2 * * * / \ \ 3 k * b / / \ 4 h * * / \ \ \ 5 * * f c / \ / \ 6 * i * * / / \ 7 * * * / / \ 8 * * d / / 9 * e / 10 j (branches folded for compactness; * representing redundant, one-child nodes; note that *'s are actual nodes, storing one child each, just with names omitted here for presentation sake) (also, to clarify, I'd like to generate the first, horizontal tree; not this vertical tree) I say language-agnostic because I'm just looking for an algorithm; I say ruby because I'm eventually going to have to implement it in ruby anyway. Assume that each Node data structure stores only its id, a left node, and a right node. A master Tree class keeps tracks of all nodes and has adequate algorithms to find: A node's nth ancestor A node's nth descendant The generation of a node The lowest common ancestor of two given nodes Anyone have any ideas of where I could start? Should I go for the recursive approach? Iterative?

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• Algorithm to retrieve every possible combination of sublists of a two lists

  - by sgmoore

  Suppose I have two lists, how do I iterate through every possible combination of every sublist, such that each item appears once and only once. I guess an example could be if you have employees and jobs and you want split them into teams, where each employee can only be in one team and each job can only be in one team. Eg List<string> employees = new List<string>() { "Adam", "Bob"} ; List<string> jobs = new List<string>() { "1", "2", "3"}; I want Adam : 1 Bob : 2 , 3 Adam : 1 , 2 Bob : 3 Adam : 1 , 3 Bob : 2 Adam : 2 Bob : 1 , 3 Adam : 2 , 3 Bob : 1 Adam : 3 Bob : 1 , 2 Adam, Bob : 1, 2, 3 I tried using the answer to this stackoverflow question to generate a list of every possible combination of employees and every possible combination of jobs and then select one item from each from each list, but that's about as far as I got. I don't know the maximum size of the lists, but it would be certainly be less than 100 and there may be other limiting factors (such as each team can have no more than 5 employees) Update Not sure whether this can be tidied up more and/or simplified, but this is what I have ended up with so far. It uses the Group algorithm supplied by Yorye (see his answer below), but I removed the orderby which I don't need and caused problems if the keys are not comparable. var employees = new List<string>() { "Adam", "Bob" } ; var jobs = new List<string>() { "1", "2", "3" }; int c= 0; foreach (int noOfTeams in Enumerable.Range(1, employees.Count)) { var hs = new HashSet<string>(); foreach( var grouping in Group(Enumerable.Range(1, noOfTeams).ToList(), employees)) { // Generate a unique key for each group to detect duplicates. var key = string.Join(":" , grouping.Select(sub => string.Join(",", sub))); if (!hs.Add(key)) continue; List<List<string>> teams = (from r in grouping select r.ToList()).ToList(); foreach (var group in Group(teams, jobs)) { foreach (var sub in group) { Console.WriteLine(String.Join(", " , sub.Key ) + " : " + string.Join(", ", sub)); } Console.WriteLine(); c++; } } } Console.WriteLine(String.Format("{0:n0} combinations for {1} employees and {2} jobs" , c , employees.Count, jobs.Count)); Since I'm not worried about the order of the results, this seems to give me what I need.

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• Php algorithm - How to achieve that without eval

  - by Marcelo

  I have a class that keeps data stores/access data by using words.separated.by.dots keys and it behaves like the following: $object = new MyArray() $object->setParam('user.name','marcelo'); $object->setParam('user.email','[email protected]'); $object->getParams(); /* array( 'user' => array( 'name' => 'marcelo', 'email' => '[email protected]' ) ); */ It is working, but the method unsetParam() was horribly implemented. That happened because i didn't know how to achieve that without eval() function. Although it is working, I found that it was a really challenging algorithm and that you might find fun trying to achieve that without eval(). class MyArray { /** * @param string $key * @return Mura_Session_Abstract */ public function unsetParam($key) { $params = $this->getParams(); $tmp = $params; $keys = explode('.', $key); foreach ($keys as $key) { if (!isset($tmp[$key])) { return $this; } $tmp = $tmp[$key]; } // bad code! $eval = "unset(\$params['" . implode("']['", $keys) . "']);"; eval($eval); $this->setParams($params); return $this; } } The test method: public function testCanUnsetNestedParam() { $params = array( '1' => array( '1' => array( '1' => array( '1' => 'one', '2' => 'two', '3' => 'three', ), '2' => array( '1' => 'one', '2' => 'two', '3' => 'three', ), ) ), '2' => 'something' ); $session = $this->newSession(); $session->setParams($params); unset($params['1']['1']['1']); $session->unsetParam('1.1.1'); $this->assertEquals($params, $session->getParams()); $this->assertEquals($params['1']['1']['2'], $session->getParam('1.1.2')); }

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• Algorithm for finding symmetries of a tree

  - by Paxinum

  I have n sectors, enumerated 0 to n-1 counterclockwise. The boundaries between these sectors are infinite branches (n of them). The sectors live in the complex plane, and for n even, sector 0 and n/2 are bisected by the real axis, and the sectors are evenly spaced. These branches meet at certain points, called junctions. Each junction is adjacent to a subset of the sectors (at least 3 of them). Specifying the junctions, (in pre-fix order, lets say, starting from junction adjacent to sector 0 and 1), and the distance between the junctions, uniquely describes the tree. Now, given such a representation, how can I see if it is symmetric wrt the real axis? For example, n=6, the tree (0,1,5)(1,2,4,5)(2,3,4) have three junctions on the real line, so it is symmetric wrt the real axis. If the distances between (015) and (1245) is equal to distance from (1245) to (234), this is also symmetric wrt the imaginary axis. The tree (0,1,5)(1,2,5)(2,4,5)(2,3,4) have 4 junctions, and this is never symmetric wrt either imaginary or real axis, but it has 180 degrees rotation symmetry if the distance between the first two and the last two junctions in the representation are equal. Edit: This is actually for my research. I have posted the question at mathoverflow as well, but my days in competition programming tells me that this is more like an IOI task. Code in mathematica would be excellent, but java, python, or any other language readable by a human suffices. Here are some examples (pretend the double edges are single and we have a tree) http://www2.math.su.se/~per/files.php?file=contr_ex_1.pdf http://www2.math.su.se/~per/files.php?file=contr_ex_2.pdf http://www2.math.su.se/~per/files.php?file=contr_ex_5.pdf Example 1 is described as (0,1,4)(1,2,4)(2,3,4)(0,4,5) with distances (2,1,3). Example 2 is described as (0,1,4)(1,2,4)(2,3,4)(0,4,5) with distances (2,1,1). Example 5 is described as (0,1,4,5)(1,2,3,4) with distances (2). So, given the description/representation, I want to find some algorithm to decide if it is symmetric wrt real, imaginary, and rotation 180 degrees. The last example have 180 degree symmetry. (These symmetries corresponds to special kinds of potential in the Schroedinger equation, which has nice properties in quantum mechanics.)

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• Special scheduling Algorithm (pattern expansion)

  - by tovare

  Question Do you think genetic algorithms worth trying out for the problem below, or will I hit local-minima issues? I think maybe aspects of the problem is great for a generator / fitness-function style setup. (If you've botched a similar project I would love hear from you, and not do something similar) Thank you for any tips on how to structure things and nail this right. The problem I'm searching a good scheduling algorithm to use for the following real-world problem. I have a sequence with 15 slots like this (The digits may vary from 0 to 20) : 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 (And there are in total 10 different sequences of this type) Each sequence needs to expand into an array, where each slot can take 1 position. 1 1 0 0 1 1 1 0 0 0 1 1 1 0 0 1 1 0 0 1 1 1 0 0 0 1 1 1 0 0 0 0 1 1 0 0 0 1 1 1 0 0 0 1 1 0 0 1 1 0 0 0 1 1 1 0 0 0 1 1 The constraints on the matrix is that: [row-wise, i.e. horizontally] The number of ones placed, must either be 11 or 111 [row-wise] The distance between two sequences of 1 needs to be a minimum of 00 The sum of each column should match the original array. The number of rows in the matrix should be optimized. The array then needs to allocate one of 4 different matrixes, which may have different number of rows: A, B, C, D A, B, C and D are real-world departments. The load needs to be placed reasonably fair during the course of a 10-day period, not to interfere with other department goals. Each of the matrix is compared with expansion of 10 different original sequences so you have: A1, A2, A3, A4, A5, A6, A7, A8, A9, A10 B1, B2, B3, B4, B5, B6, B7, B8, B9, B10 C1, C2, C3, C4, C5, C6, C7, C8, C9, C10 D1, D2, D3, D4, D5, D6, D7, D8, D9, D10 Certain spots on these may be reserved (Not sure if I should make it just reserved/not reserved or function-based). The reserved spots might be meetings and other events The sum of each row (for instance all the A's) should be approximately the same within 2%. i.e. sum(A1 through A10) should be approximately the same as (B1 through B10) etc. The number of rows can vary, so you have for instance: A1: 5 rows A2: 5 rows A3: 1 row, where that single row could for instance be: 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 etc.. Sub problem* I'de be very happy to solve only part of the problem. For instance being able to input: 1 1 2 3 4 2 2 3 4 2 2 3 3 2 3 And get an appropriate array of sequences with 1's and 0's minimized on the number of rows following th constraints above.

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• Making an efficient algorithm

  - by James P.

  Here's my recent submission for the FB programming contest (qualifying round only requires to upload program output so source code doesn't matter). The objective is to find two squares that add up to a given value. I've left it as it is as an example. It does the job but is too slow for my liking. Here's the points that are obviously eating up time: List of squares is being recalculated for each call of getNumOfDoubleSquares(). This could be precalculated or extended when needed. Both squares are being checked for when it is only necessary to check for one (complements). There might be a more efficient way than a double-nested loop to find pairs. Other suggestions? Besides this particular problem, what do you look for when optimizing an algorithm? public static int getNumOfDoubleSquares( Integer target ){ int num = 0; ArrayList<Integer> squares = new ArrayList<Integer>(); ArrayList<Integer> found = new ArrayList<Integer>(); int squareValue = 0; for( int j=0; squareValue<=target; j++ ){ squares.add(j, squareValue); squareValue = (int)Math.pow(j+1,2); } int squareSum = 0; System.out.println( "Target=" + target ); for( int i = 0; i < squares.size(); i++ ){ int square1 = squares.get(i); for( int j = 0; j < squares.size(); j++ ){ int square2 = squares.get(j); squareSum = square1 + square2; if( squareSum == target && !found.contains( square1 ) && !found.contains( square2 ) ){ found.add(square1); found.add(square2); System.out.println( "Found !" + square1 +"+"+ square2 +"="+ squareSum); num++; } } } return num; }

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• 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.

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• Algorithm to select groups of similar items in 2d array

  - by mafutrct

  There is a 2d array of items (in my case they are called Intersections). A certain item is given as a start. The task is to find all items directly or indirectly connected to this item that satisfy a certain function. So the basic algorithm is like this: Add the start to the result list. Repeat until no modification: Add each item in the array that satisfies the function and touches any item in the result list to the result list. My current implementation looks like this: private IList<Intersection> SelectGroup ( Intersection start, Func<Intersection, Intersection, bool> select) { List<Intersection> result = new List<Intersection> (); Queue<Intersection> source = new Queue<Intersection> (); source.Enqueue (start); while (source.Any ()) { var s = source.Dequeue (); result.Add (s); foreach (var neighbour in Neighbours (s)) { if (select (start, neighbour) && !result.Contains (neighbour) && !source.Contains (neighbour)) { source.Enqueue (neighbour); } } } Debug.Assert (result.Distinct ().Count () == result.Count ()); Debug.Assert (result.All (x => select (x, result.First ()))); return result; } private List<Intersection> Neighbours (IIntersection intersection) { int x = intersection.X; int y = intersection.Y; List<Intersection> list = new List<Intersection> (); if (x > 1) { list.Add (GetIntersection (x - 1, y)); } if (y > 1) { list.Add (GetIntersection (x, y - 1)); } if (x < Size) { list.Add (GetIntersection (x + 1, y)); } if (y < Size) { list.Add (GetIntersection (x, y + 1)); } return list; } (The select function takes a start item and returns true iff the second item satisfies.) This does its job and turned out to be reasonable fast for the usual array sizes (about 20*20). However, I'm interested in further improvements. Any ideas? Example (X satisfies in relation to other Xs, . does never satisfy): .... XX.. .XX. X... In this case, there are 2 groups: a central group of 4 items and a group of a single item in the lower left. Selecting the group (for instance by starting item [2, 2]) returns the former, while the latter can be selected using the starting item and sole return value [0, 3]. Example 2: .A.. ..BB A.AA This time there are 4 groups. The 3 A groups are not connected, so they are returned as separate groups. The bigger A and B groups are connected, but A does not related to B so they are returned as separate groups.

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• Add a Graphical User Interface (GUI) to the Microsoft Robocopy Command Line Tool

  - by Lori Kaufman

  Robocopy, or “Robust File Copy,” is a command line directory replication tool from Microsoft. It is available as part of Windows 7 and Vista as a standard feature, and was available as part of the Windows Server 2003 Resource Kit. NOTE: For Windows XP, you can obtain Robocopy by downloading the resource kit. Robocopy allows you to setup simple or advanced backup strategies. It provides such features as multi-threaded copying, mirroring or synchronization mode, automatic retry, and the ability to resume the copying process. If you are comfortable with using command line tools, you can run Robocopy directly on the command line using the command syntax and options. You can also download the command line reference and usage notes for Robocopy as a PDF file. If you are more comfortable using a graphical user interface, or GUI, rather than the command line, there are a couple of options for adding a GUI to the Robocopy command line tool, making it easier to use. Both tools, RoboMirror and RichCopy, are discussed below and links to download each tool are provided. How to Factory Reset Your Android Phone or Tablet When It Won’t Boot Our Geek Trivia App for Windows 8 is Now Available Everywhere How To Boot Your Android Phone or Tablet Into Safe Mode

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• Restart problem after installing graphical card driver

  - by Tim

  My laptop is Lenovo T400, running Ubuntu 10.10. My problem: I just run jockey-gtk and installed ATI/AMD proprietary FGLRX graphics driver. But after reboot, there is a short period of graphical "Ubuntu" and then instead of starting X-window it completely changed to command line to ask me to login. Even after login and then issuing xinit, it still failed to start X. To solve this problem: I followed this post, where one person suggested to you can simply write sudo apt-get remove fglrx This worked for me. If it doesn't work, then try sudo apt-get remove xorg-driver-fglrx and restart. I actually don't need the driver anyway, so I issued the first command after login under command line. But after reboot, the situation is even worse, and there is now even no command-line interface to ask me login, instead the screen is completely blank with just some ambient light in the background and Ubuntu is hanging there probably forever. So I have no chance to try the second command the person suggested. I was wondering what I can do now to solve my problem? Thanks and regards!

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• High resolution graphical representation of the Earth's surface

  - by Simon

  I've got a library, which I inherited, which presents a zoomable representation of the Earth. It's a Mercator projection and is constructed from triangles, the properties of which are stored in binary files. The surface is built up, for any given view port, by drawing these triangles in an overlapping fashion to produce the image. The definition of each triangle is the lat/long of the vertices. It looks OK at low values of zoom but looks progressively more ragged as the user zooms in. The view ports are primarily referenced though a rectangle of lat/long co-ordinates. I'd like to replace it with a better quality approach. The problem is, I don't know where to begin researching the options as I am not familiar either with the projections needed nor the graphics techniques used to render them. For example, I imagine that I could acquire high resolution images, say Mercator projections although I'm open to anything, break them into tiles and somehow wrap them onto a graphical representation of a sphere. I'm not asking for "how do I", more where should I begin to understand what might be involved and the techniques I will need to learn. I am most grateful for any "Earth rendering 101" pointers folks might have.

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• Java game design question (graphical objects)

  - by vemalsar

  Hello Guys, I'm beginner in game development, in Java and here on this site too and I have a game design question. Please comment my idea: I have a main loop which call update and draw method. I want to use an ArrayList which store graphical objects, they have coordinate and image or text to draw and my game objects extends this class. In update, I can choose which objects should be put in the array and in draw method I'll display the elements of array on the screen. I'm using a buffer and draw first there, but it is not important now I guess...Here is a simple (not full) code, only the logic: public class GamePanel extends JPanel implements KeyListener { ArrayList<graphicalObjects> graphArray = new ArrayList<graphicalObjects>(); public void update() { //change the game scene, update the graphArray, process input etc. } public void draw() { //draws every element of graphArray to a JPanel } public static main(String[] args) { while(true) { update(); draw(); } } } My questions: Should have I use interface or abstract class for graphicalObjects? graphicalObjects class and the ArrayList really needs or there is some better solution? How to draw objects? They draw themself with their own method or in the draw method I have to draw manually based on graphicalObjects variables (x,y coordinates, image etc.)? If this conception is wrong, please suggest another one! All comments are welcome and sorry if this is dumb question, thanks!

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• Problem Solving: Algorithm Required Urgently, Plz Help

  - by user616417

  Problem Solving: I've been working on something since last week. I am stuck at a point, where I want to find the minimum number of airplanes required to carry out a flight schedule given below. Plz, try out the brainstorming, i need the algorithm really badly, i'm also short of time. Thank u in advance. The Schedule---- Flight #,From,To,Departure,Arrival,Days,Via 6E 204,Agartala,Delhi,10:15:00,13:55:00,Daily,Kolkata 6E 360,Agartala,Imphal,13:50:00,14:35:00,Mo Th Sa, 6E 204,Agartala,Kolkata,10:15:00,11:00:00,Daily, 6E 360,Agartala,Kolkata,13:50:00,16:15:00,Mo Th Sa,Imphal 6E 362,Agartala,Kolkata,15:25:00,16:15:00,Tu We Fr Su, 6E 153,Ahmedabad,Bangalore,17:00:00,19:00:00,Daily, 6E 212,Ahmedabad,Chennai,9:00:00,12:55:00,Daily,Mumbai 6E 154,Ahmedabad,Delhi,12:30:00,14:00:00,Daily, 6E 211,Ahmedabad,Jaipur,19:10:00,20:20:00,Daily, 6E 410,Ahmedabad,Kolkata,15:00:00,17:30:00,Daily, 6E 212,Ahmedabad,Mumbai,9:00:00,10:10:00,Daily, 6E 409,Ahmedabad,Pune,14:25:00,15:40:00,Ex Sat, 6E 154,Bangalore,Ahmedabad,10:00:00,12:00:00,Daily, 6E 277,Bangalore,Chennai,15:35:00,16:25:00,Daily, 6E 132,Bangalore,Delhi,6:00:00,8:25:00,Daily, 6E 102,Bangalore,Delhi,9:50:00,13:45:00,Ex Sat,Pune 6E 154,Bangalore,Delhi,10:00:00,14:00:00,Daily,Ahmedabad 6E 104,Bangalore,Delhi,11:30:00,14:10:00,Sat, 6E 122,Bangalore,Delhi,17:20:00,20:00:00,Daily, 6E 108,Bangalore,Delhi,19:20:00,23:10:00,Sat,Pune 6E 106,Bangalore,Delhi,19:30:00,22:00:00,Ex Sat, 6E 275,Bangalore,Goa,12:15:00,13:15:00,Daily, 6E 351,Bangalore,Hyderabad,8:25:00,9:25:00,Daily, 6E 152,Bangalore,Hyderabad,19:10:00,20:10:00,Ex Sat, 6E 152,Bangalore,Hyderabad,19:30:00,20:35:00,Sat, 6E 152,Bangalore,Jaipur,19:10:00,22:30:00,Ex Sat,Hyderabad 6E 152,Bangalore,Jaipur,19:30:00,22:30:00,Sat,Hyderabad 6E 351,Bangalore,Kolkata,8:25:00,11:55:00,Daily,Hyderabad 6E 277,Bangalore,Kolkata,15:35:00,19:15:00,Daily,Chennai 6E 402,Bangalore,Mumbai,6:05:00,7:45:00,Daily, 6E 275,Bangalore,Mumbai,12:15:00,14:45:00,Daily,Goa 6E 414,Bangalore,Mumbai,12:45:00,14:20:00,Daily, 6E 412,Bangalore,Mumbai,21:20:00,23:20:00,Daily, 6E 102,Bangalore,Pune,9:50:00,11:10:00,Ex Sat, 6E 108,Bangalore,Pune,19:20:00,20:40:00,Sat, 6E 258,Bhubaneshwar,Delhi,18:55:00,20:55:00,Daily, 6E 257,Bhubaneshwar,Hyderabad,10:40:00,12:05:00,Daily, 6E 257,Bhubaneshwar,Mumbai,10:40:00,13:50:00,Daily,Hyderabad 6E 211,Chennai,Ahmedabad,15:10:00,18:40:00,Daily,Mumbai 6E 275,Chennai,Bangalore,10:50:00,11:40:00,Daily, 6E 302,Chennai,Delhi,11:35:00,15:20:00,Daily,Hyderabad 6E 282,Chennai,Delhi,19:45:00,22:30:00,Daily, 6E 275,Chennai,Goa,10:50:00,13:15:00,Daily,Bangalore 6E 302,Chennai,Hyderabad,11:35:00,12:40:00,Daily, 6E 211,Chennai,Jaipur,15:10:00,20:20:00,Daily,Mumbai/Ahmedabad 6E 523,Chennai,Kolkata,8:20:00,10:30:00,Daily, 6E 277,Chennai,Kolkata,16:55:00,19:15:00,Daily, 6E 211,Chennai,Mumbai,15:10:00,16:50:00,Daily, 6E 524,Chennai,Pune,21:15:00,23:00:00,Daily, 6E 273,Delhi,Agartala,6:15:00,9:45:00,Daily,Kolkata 6E 153,Delhi,Ahmedabad,14:45:00,16:30:00,Daily, 6E 101,Delhi,Bangalore,6:30:00,9:10:00,Ex Sat, 6E 103,Delhi,Bangalore,6:45:00,10:40:00,Sat,Pune 6E 121,Delhi,Bangalore,9:30:00,12:10:00,Daily, 6E 105,Delhi,Bangalore,14:20:00,18:30:00,Ex Sat,Pune 6E 153,Delhi,Bangalore,14:45:00,19:00:00,Daily,Ahmedabad 6E 107,Delhi,Bangalore,15:55:00,18:40:00,Sat, 6E 131,Delhi,Bangalore,20:45:00,23:15:00,Daily, 6E 257,Delhi,Bhubaneshwar,8:10:00,10:10:00,Daily, 6E 301,Delhi,Chennai,7:00:00,11:05:00,Daily,Hyderabad 6E 283,Delhi,Chennai,16:30:00,19:05:00,Daily, 6E 181,Delhi,Goa,9:15:00,13:35:00,Daily,Mumbai 6E 333,Delhi,Goa,11:45:00,14:15:00,Daily, 6E 201,Delhi,Guwahati,5:30:00,7:50:00,Daily, 6E 301,Delhi,Hyderabad,7:00:00,9:00:00,Daily, 6E 257,Delhi,Hyderabad,8:10:00,12:05:00,Daily,Bhubaneshwar 6E 305,Delhi,Hyderabad,14:00:00,15:55:00,Daily, 6E 307,Delhi,Hyderabad,21:00:00,22:55:00,Daily, 6E 201,Delhi,Imphal,5:30:00,9:10:00,Daily,Guwahati 6E 305,Delhi,Kochi,14:00:00,18:25:00,Daily,Hyderabad 6E 273,Delhi,Kolkata,6:15:00,8:20:00,Daily, 6E 203,Delhi,Kolkata,15:00:00,17:05:00,Daily, 6E 209,Delhi,Kolkata,18:30:00,20:45:00,Daily, 6E 183,Delhi,Mumbai,6:45:00,8:35:00,Daily, 6E 181,Delhi,Mumbai,9:15:00,11:35:00,Daily, 6E 481,Delhi,Mumbai,10:50:00,13:50:00,Daily,Vadodara 6E 189,Delhi,Mumbai,14:45:00,16:50:00,Daily, 6E 187,Delhi,Mumbai,17:50:00,19:50:00,Daily, 6E 185,Delhi,Mumbai,20:15:00,22:20:00,Daily, 6E 135,Delhi,Nagpur,8:55:00,10:40:00,Ex Sat, 6E 103,Delhi,Pune,6:45:00,8:45:00,Sat, 6E 135,Delhi,Pune,8:55:00,12:30:00,Ex Sat,Nagpur 6E 105,Delhi,Pune,14:20:00,16:30:00,Ex Sat, 6E 481,Delhi,Vadodara,10:50:00,12:20:00,Daily, 6E 277,Goa,Bangalore,14:05:00,15:00:00,Daily, 6E 277,Goa,Chennai,14:05:00,16:25:00,Daily,Bangalore 6E 334,Goa,Delhi,14:45:00,17:10:00,Daily, 6E 277,Goa,Kolkata,14:05:00,19:15:00,Daily,Bangalore/Chennai 6E 275,Goa,Mumbai,13:45:00,14:45:00,Daily, 6E 202,Guwahati,Delhi,11:00:00,13:25:00,Daily, 6E 201,Guwahati,Imphal,8:25:00,9:10:00,Daily, 6E 208,Guwahati,Jaipur,12:40:00,16:55:00,Daily,Kolkata 6E 208,Guwahati,Kolkata,12:40:00,14:00:00,Daily, 6E 322,Guwahati,Kolkata,15:30:00,16:50:00,Daily, 6E 322,Guwahati,Mumbai,15:30:00,20:20:00,Daily,Kolkata 6E 151,Hyderabad,Bangalore,8:20:00,9:20:00,Daily, 6E 352,Hyderabad,Bangalore,19:40:00,20:40:00,Daily, 6E 258,Hyderabad,Bhubaneshwar,16:40:00,18:20:00,Daily, 6E 301,Hyderabad,Chennai,9:50:00,11:05:00,Daily, 6E 308,Hyderabad,Delhi,6:10:00,8:00:00,Daily, 6E 302,Hyderabad,Delhi,13:10:00,15:20:00,Daily, 6E 258,Hyderabad,Delhi,16:40:00,20:55:00,Daily,Bhubaneshwar 6E 306,Hyderabad,Delhi,21:00:00,23:05:00,Daily, 6E 152,Hyderabad,Jaipur,20:50:00,22:30:00,Ex Sat, 6E 152,Hyderabad,Jaipur,21:10:00,22:30:00,Sat, 6E 305,Hyderabad,Kochi,16:45:00,18:25:00,Daily, 6E 351,Hyderabad,Kolkata,9:55:00,11:55:00,Daily, 6E 257,Hyderabad,Mumbai,12:35:00,13:50:00,Daily, 6E 362,Imphal,Agartala,14:15:00,14:55:00,Tu We Fr Su, 6E 202,Imphal,Delhi,9:40:00,13:25:00,Daily,Guwahati 6E 202,Imphal,Guwahati,9:40:00,10:25:00,Daily, 6E 362,Imphal,Kolkata,14:15:00,16:15:00,Tu We Fr Su,Agartala 6E 360,Imphal,Kolkata,15:05:00,16:15:00,Mo Th Sa, 6E 212,Jaipur,Ahmedabad,7:30:00,8:35:00,Daily, 6E 151,Jaipur,Bangalore,6:00:00,9:20:00,Daily,Hyderabad 6E 212,Jaipur,Chennai,7:30:00,12:55:00,Daily,Mumbai/Ahmedabad 6E 207,Jaipur,Guwahati,8:20:00,12:10:00,Daily,Kolkata 6E 151,Jaipur,Hyderabad,6:00:00,7:40:00,Daily, 6E 207,Jaipur,Kolkata,8:20:00,10:10:00,Daily, 6E 323,Jaipur,Kolkata,17:35:00,23:00:00,Daily,Mumbai 6E 212,Jaipur,Mumbai,7:30:00,10:10:00,Daily,Ahmedabad 6E 323,Jaipur,Mumbai,17:35:00,19:15:00,Daily, 6E 306,Kochi,Delhi,19:00:00,23:05:00,Daily,Hyderabad 6E 306,Kochi,Hyderabad,19:00:00,20:30:00,Daily, 6E 273,Kolkata,Agartala,8:50:00,9:45:00,Daily, 6E 360,Kolkata,Agartala,12:30:00,13:20:00,Mo Th Sa, 6E 362,Kolkata,Agartala,12:30:00,14:55:00,TuWeFrSu,Imphal 6E 409,Kolkata,Ahmedabad,11:10:00,13:50:00,Daily, 6E 275,Kolkata,Bangalore,7:30:00,11:40:00,Daily,Chennai 6E 352,Kolkata,Bangalore,16:50:00,20:40:00,Daily,Hyderabad 6E 275,Kolkata,Chennai,7:30:00,9:50:00,Daily, 6E 524,Kolkata,Chennai,18:15:00,20:25:00,Daily, 6E 210,Kolkata,Delhi,7:45:00,10:05:00,Daily, 6E 204,Kolkata,Delhi,11:40:00,13:55:00,Daily, 6E 274,Kolkata,Delhi,19:45:00,22:10:00,Daily, 6E 275,Kolkata,Goa,7:30:00,13:15:00,Daily,Chennai/Bangalore 6E 207,Kolkata,Guwahati,10:50:00,12:10:00,Daily, 6E 321,Kolkata,Guwahati,13:00:00,14:20:00,Daily, 6E 352,Kolkata,Hyderabad,16:50:00,19:00:00,Daily, 6E 362,Kolkata,Imphal,12:30:00,13:45:00,Tu We Fr Su, 6E 360,Kolkata,Imphal,12:30:00,14:35:00,MoThSa,Agartala 6E 208,Kolkata,Jaipur,14:35:00,16:55:00,Daily, 6E 320,Kolkata,Mumbai,6:00:00,8:30:00,Daily, 6E 322,Kolkata,Mumbai,17:35:00,20:20:00,Daily, 6E 404,Kolkata,Mumbai,18:35:00,21:55:00,Daily,Nagpur 6E 404,Kolkata,Nagpur,18:35:00,20:05:00,Daily, 6E 409,Kolkata,Pune,11:10:00,15:40:00,Ex Sat,Ahmedabad 6E 524,Kolkata,Pune,18:15:00,23:00:00,Daily,Chennai 6E 211,Mumbai,Ahmedabad,17:40:00,18:40:00,Daily, 6E 411,Mumbai,Bangalore,6:20:00,7:50:00,Daily, 6E 413,Mumbai,Bangalore,15:00:00,16:40:00,Daily, 6E 415,Mumbai,Bangalore,21:05:00,22:40:00,Daily, 6E 258,Mumbai,Bhubaneshwar,14:30:00,18:20:00,Daily,Hyderabad 6E 212,Mumbai,Chennai,11:00:00,12:55:00,Daily, 6E 184,Mumbai,Delhi,6:15:00,8:15:00,Daily, 6E 180,Mumbai,Delhi,8:25:00,10:35:00,Daily, 6E 482,Mumbai,Delhi,9:25:00,12:35:00,Daily,Vadodara 6E 188,Mumbai,Delhi,14:25:00,16:35:00,Daily, 6E 186,Mumbai,Delhi,17:50:00,19:55:00,Daily, 6E 182,Mumbai,Delhi,21:15:00,23:20:00,Daily, 6E 181,Mumbai,Goa,12:35:00,13:35:00,Daily, 6E 321,Mumbai,Guwahati,9:20:00,14:20:00,Daily,Kolkata 6E 258,Mumbai,Hyderabad,14:30:00,16:00:00,Daily, 6E 207,Mumbai,Jaipur,5:55:00,7:40:00,Daily, 6E 211,Mumbai,Jaipur,17:40:00,20:20:00,Daily,Ahmedabad 6E 207,Mumbai,Kolkata,5:55:00,10:10:00,Daily,Jaipur 6E 321,Mumbai,Kolkata,9:20:00,12:00:00,Daily, 6E 403,Mumbai,Kolkata,15:35:00,18:50:00,Daily,Nagpur 6E 323,Mumbai,Kolkata,20:05:00,23:00:00,Daily, 6E 403,Mumbai,Nagpur,15:35:00,16:50:00,Daily, 6E 482,Mumbai,Vadodara,9:25:00,10:25:00,Daily, 6E 136,Nagpur,Delhi,18:10:00,19:40:00,Ex Sat, 6E 403,Nagpur,Kolkata,17:20:00,18:50:00,Daily, 6E 404,Nagpur,Mumbai,20:35:00,21:55:00,Daily, 6E 135,Nagpur,Pune,11:20:00,12:30:00,Ex Sat, 6E 410,Pune,Ahmedabad,13:10:00,14:30:00,Ex Sat, 6E 103,Pune,Bangalore,9:15:00,10:40:00,Sat, 6E 105,Pune,Bangalore,17:00:00,18:30:00,Ex Sat, 6E 523,Pune,Chennai,5:55:00,7:40:00,Daily, 6E 102,Pune,Delhi,11:45:00,13:45:00,Ex Sat, 6E 136,Pune,Delhi,16:15:00,19:40:00,Ex Sat,Nagpur 6E 108,Pune,Delhi,21:10:00,23:10:00,Sat, 6E 523,Pune,Kolkata,5:55:00,10:30:00,Daily,Chennai 6E 410,Pune,Kolkata,13:10:00,17:30:00,Ex Sat,Ahmedabad 6E 136,Pune,Nagpur,16:15:00,17:40:00,Ex Sat, 6E 482,Vadodara,Delhi,10:55:00,12:35:00,Daily, 6E 481,Vadodara,Mumbai,12:50:00,13:50:00,Daily,

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• 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.

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• 12.04 boots fine, with graphical splash screen, but then Monitor "out of range"

  - by Jim Bednar

  I see dozens of posts from people whose monitors are saying "out of range" under Ubuntu; seems like there are some serious problems in Ubuntu with autodetection of monitor capabilities. :-( But none of the many, many suggestions I found have solved my problem, and right now I can't use anything graphical on this machine! History: I installed Ubuntu 12.04 on my HP Proliant Microserver N40L, which worked reasonably at the default resolution across several reboots. At some point I noticed that the proprietary video driver was not in use, and tried to install one to get better window-drawing speeds, but it failed with some sort of error, and I gave up on that. A few weeks later when I next rebooted, it showed the usual BIOS screen and various boot loading screens (including GRUB), and then the usual purple Ubuntu splash screen with the dots showing that things were loading, but when it finished booting the monitor went black and eventually showed "Out of range" (with no other information). Given that there were several weeks between reboots (it's a server, after all), I've no idea if it was some system update, trying to install the proprietary drivers, or something else that caused the problem. Anyway, the system has booted fine, as I can do Ctrl-Alt-F1 to get a text prompt and can log in there. But Ctrl-Alt-F7 goes back to the out of range error. Some posters said to try Ctrl-Alt-- (minus) to cycle through resolutions until one works, but that didn't have any visible effect. Many, many others said it was a grub problem, which seems unlikely given that grub's screen looks fine, but I tried editing /etc/default/grub to set a particular resolution (trying many of them) and running update-grub, with no apparent effect. Rebooting into failsafe mode works the same as regular mode. Replacing xorg.conf with xorg.conf.failsafe works the same too. I'm at my wits' end! Isn't there anything I can do to convince Ubuntu to choose a mode that the monitor supports? E.g. the one that it is using for the splash screen? I don't need great resolution on this machine, just anything that works!!!!! Help!!!!!! Please!!!!

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• Tetris Piece Rotation Algorithm

  - by coppercoder

  What are the best algorithms (and explanations) for representing and rotating the pieces of a tetris game? I always find the piece rotation and representation schemes confusing. Most tetris games seem to use a naive "remake the array of blocks" at each rotation: http://www.codeplex.com/Project/ProjectDirectory.aspx?ProjectSearchText=tetris However, some use pre-built encoded numbers and bit shifting to represent each piece: http://www.codeplex.com/wintris Is there a method to do this using mathematics (not sure that would work on a cell based board)?

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• 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

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• Probability algorithm: Finding probable correct item in a list (e.g John, John, Jon)

  - by Andrew White

  Hi, Take for example the list (L): John, John, John, John, Jon We are to presume one item is to be correct (e.g. John in this case), and give a probability it is correct. First (and good!) attempt: MostFrequentItem(L).Count / L.Count (e.g. 4/5 or 80% likelihood) But consider the cases: John, John, Jon, Jonny John, John, Jon, Jon I want to consider the likelihood of the correct item being John to be higher in the first list! I know I have to count the SecondMostFrequent Item and compare them. Any ideas? This is really busting my brain! Thx, Andrew

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• Mysql Algorithm for Determining Closest Colour Match

  - by buggedcom

  I'm attempting to create a true mosaic application. At the moment I have one mosaic image, ie the one the mosaic is based on and about 4000 images from my iPhoto library that act as the image library. I have already done my research and analysed the mosaic image. I've converted it into 64x64 slices each of 8 pixels. I've calculated the average colour for each slice and assertain the r, g, b and brightness (Luminance (perceived option 1) = (0.299*R + 0.587*G + 0.114*B)) value. I have done the same for each of the image library photos. The mosaic slices table looks like so. slice_id, slice_image_id, slice_slice_id, slice_image_column, slice_image_row, slice_colour_hex, slice_rgb_red, slice_rgb_blue, slice_rgb_green, slice_rgb_brightness The image library table looks like so. upload_id, upload_file, upload_colour_hex, upload_rgb_red, upload_rgb_green, upload_rgb_blue, upload_rgb_brightness So basically I'm reading the image slices from the slices table into PHP and then pulling out the appropriate images from the library table based on the colour hexs. My trouble is that I've been on this too long and probably had too many energy drinks so am not concentrating properly, I can't figure out the way to pick out the nearest colour neighbor if the appropriate hex code doesn't exist. Any ideas on the perfect query? NB: I know pulling out the slices one by one is not ideal however the mosaic is only rebuilt periodically so a sudden burst in the mysql load doesn't really bother me, however if there us a way to pull the images out all at once that would also be a massive bonus.

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• Al Zimmermann's Son of Darts

  - by polygenelubricants

  There's about 2 months left in Al Zimmermann's Son of Darts programming contest, and I'd like to improve my standing (currently in the 60s) to something more respectable. I'd like to get some ideas from the great community of stackoverflow on how best to approach this problem. The contest problem is known as the Global Postage Stamp Problem in literatures. I don't have much experience with optimization algorithms (I know of hillclimbing and simulated annealing in concept only from college), and in fact the program that I have right now is basically sheer brute force, which of course isn't feasible for the larger search spaces. Here are some papers on the subject: A Postage Stamp Problem (Alter & Barnett, 1980) Algorithms for Computing the h-Range of the Postage Stamp Problem (Mossige, 1981) A Postage Stamp Problem (Lunnon, 1986) Two New Techniques for Computing Extremal h-bases Ak (Challis, 1992) Any hints and suggestions are welcome. Also, feel free to direct me to the proper site if stackoverflow isn't it.

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• Working out global tab order algorithmically?

  - by Mrgreen

  We have a proprietry system where we can configure fields on indiviual forms. However these fields have a global tab order (we cannot specify for a specific form). We have a bunch of forms (35 in total) which share a lot of different fields. Each form has a specific tab/edit order that needs to be configured. Example: Form 1 has fields A,B,C,D in that order. Form 2 has fields E,F,G,A in that order. Form 3 has fields E,B,H,I in that order. The global tab orders would be E,F,G,A,B,C,D,H,I Notice how A needs to come before B yet after G. Is there any easy way to work this out using the tab order lists for each form? I need to merge this tab order information into a single global tab order list. I have over 200 fields in total and it is near impossible to do by hand.

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• Hashing Algorithm, its uses?

  - by ritch

  I don't fully understand hashing algorithms. Anybody care to explain it to me in a very simply understandable way. Thanks EDIT: Using it with Usernames from a text file.

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• Algorithm shortest path between all points

  - by Jeroen

  Hi, suppose I have 10 points. I know the distance between each point. I need to find the shortest possible route passing trough all points. I have tried a couple of algorithms (Dijkstra, Floyd Warshall,...) and the all give me the shortest path between start and end, but they don't make a route with all points on it. Permutations work fine, but they are to resource expensive. What algorithms can you advise me to look into for this problem? Or is there a documented way to do this with the above mentioned algorithms? Tnx Jeroen

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• algorithm analysis - orders of growth question

  - by cchampion

  I'm studing orders of growth "big oh", "big omega", and "big theta". Since I can't type the little symbols for these I will denote them as follows: ORDER = big oh OMEGA = big omega THETA = big theta For example I'll say n = ORDER(n^2) to mean that the function n is in the order of n^2 (n grows at most as fast n^2). Ok for the most part I understand these: n = ORDER(n^2) //n grows at most as fast as n^2 n^2 = OMEGA(n) //n^2 grows atleast as fast as n 8n^2 + 1000 = THETA(n^2) //same order of growth Ok here comes the example that confuses me: what is n(n+1) vs n^2 I realize that n(n+1) = n^2 + n; I would say it has the same order of growth as n^2; therefore I would say n(n+1) = THETA(n^2) but my question is, would it also be correct to say: n(n+1) = ORDER(n^2) please help because this is confusing to me. thanks.

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