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• Javascript algorithm to find elements in array that are not in another array

  - by Tauren

  I'm looking for a good algorithm to get all the elements in one array that are not elements in another array. So given these arrays: var x = ["a","b","c","t"]; var ?????????y = [???????"d","a","t","e","g"]; I want to end up with this array: var z = ["d","e","g"]; I'm using jquery, so I can take advantage of $.each() and $.inArray(). Here's the solution I've come up with, but it seems like there should be a better way. // goal is to get rid of values in y if they exist in x var x = ["a","b","c","t"]; var y = ["d","a","t","e","g"]; var z = []; $.each(y, function(idx, value){ if ($.inArray(value,x) == -1) { z.push(value); } }); ?alert(z); // should be ["d","e","g"] Here is the code in action. Any ideas?

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• PHP: simplify with algorithm?

  - by tshabala

  Hello. Here's a piece of PHP code I think is not very "pretty", I'm sure it's possible to simplify it with for or something. I'm trying to find and algorithm that would work for this, but I can't figure it out, please help me. Here's the code: if(isset($four)) { if(isset($navi[$one][$two][$three][$four])) echo "/content/" . $one . "/" . $two . "/" . $three . "/" .$four . ".php"; else echo "error"; } else if(isset($three)) { if(isset($navi[$one][$two][$three])) echo "/content/" . $one . "/" . $two . "/" . $three . ".php"; else echo "error"; } else if(isset($two)) { if(isset($navi[$one][$two])) echo "/content/" . $one . "/" . $two . ".php"; else echo "error"; } else if(isset($one)) { if(isset($navi[$one]))echo "/content/" . $one . ".php"; else echo "error"; } else { echo "error"; } Thanks!

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• Algorithm Question

  - by Ravi

  Hi, I am trying to find a O (n) algorithm for this problem but unable to do so even after spending 3 - 4 hours. The brute force method times out (O (n^2)). I am confused as to how to do it ? Does the solution requires dynamic programming solution ? http://acm.timus.ru/problem.aspx?space=1&num=1794 In short the problem is this: There are some students sitting in circle and each one of them has its own choice as to when he wants to be asked a question from a teacher. The teacher will ask the questions in clockwise order only. For example: 5 3 3 1 5 5 This means that there are 5 students and : 1st student wants to go third 2nd student wants to go third 3rd student wants to go first 4th student wants to go fifth 5th student wants to go fifth. The question is as to where should teacher start asking questions so that maximum number of students will get the turn as they want. For this particular example, the answer is 5 because 3 3 1 5 5 2 3 4 5 1 You can see that by starting at fifth student as 1st, 2 students (3 and 5) are getting the choices as they wanted. For this example the answer is 12th student : 12 5 1 2 3 6 3 8 4 10 3 12 7 because 5 1 2 3 6 3 8 4 10 3 12 7 2 3 4 5 6 7 8 9 10 11 12 1 four students get their choices fulfilled. Thanks Ravi

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• barebones sort algorithm

  - by user309322

  i have been asked to make a simple sort aglorithm to sort a random series of 6 numbers into numerical order. However i have been asked to do this using "Barebones" a theoretical language put forward in the Book Computer Science an overview. Some information on the language can be found here http://www.brouhaha.com/~eric/software/barebones/ Just to clarify i am a student teacher and have been doing anaysis on "mini-programing languages" and their uses in a teaching environment, i suggested to my tutor that i look at barebones and asked what sort of exmaple program i should write . He suggested a simple sort algorithm. Now since looking at the language i cant understand how i can do this without using arrays and if statements. The code to swap the value of variables would be while a not 0 do; incr Aux1; decr a; end; while b not 0 do; incr Aux2 decr b end; while Aux1 not 0 do; incr a; decr Aux1; end; while Aux2 not 0 do; incr b; decr Aux2; end; however the language does not provide < or operators

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• C# algorithm for figuring out different possible combinations...

  - by ttomsen

  I have 10 boxes, each box can hold one item from a group/type of items, each 'group' type only fits in one of the 10 box types. The pool can have n number of items. The groups have completely distinct items. Each item has a price, i want an algorithm that will generate all the different possibilities, so i can figure out different price points. so a smaller picture of the problem BOX A - can have item 1,2,3,4 in it BOX B - can have item 6,7,8,9,10,11,12 BOX C - can have item 13,15,16,20,21 The items are stored in a db, they have a column which denotes which box they can go in. All box types are stored in an array, and I can put the items in a generic list. Anyone see a straightforward way to do this. I have tried doing 10 nested foreach's to see if i could find a simpler way. The nested loops will take MANY hours to run. the nested for each's basically pull all combinations, then calculate a rank for each combination, and store the top 10 ranked combination of items for output

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• search algorithm using sentinel

  - by davit-datuashvili

  i am trying to do search algorithm using sentinel which reduce time to 3.87n nanoseconds for example compare to this code int search (int t ){ for (int i=0;i<n;i++) if (x[i]==t) return i; return -1; } it takes 4.06 nanoseconds so i am trying to optimize it here is code public class Search{ public static int search(int a[],int t){ int i; int p=0; int n=a.length; int hold; hold=a[n-1]; a[n-1]=t; for ( i=0;;i++) if (a[i]==t) break; a[n-1]=t; if (i==n){ p= -1; } else{ p= i; } return p; } public static void main(String[]args){ int t=-1; int a[]=new int[]{4,5,2,6,8,7,9}; System.out.println(search(a,t)); } } but is show me that 9 is at position 6 which is correct but if t =1 or something else which is not array it show me position 6 too please help

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• Questions about the Backpropogation Algorithm

  - by Colemangrill

  I have a few questions concerning backpropogation. I'm trying to learn the fundamentals behind neural network theory and wanted to start small, building a simple XOR classifier. I've read a lot of articles and skimmed multiple textbooks - but I can't seem to teach this thing the pattern for XOR. Firstly, I am unclear about the learning model for backpropogation. Here is some pseudo-code to represent how I am trying to train the network. [Lets assume my network is setup properly (ie: multiple inputs connect to a hidden layer connect to an output layer and all wired up properly)]. SET guess = getNetworkOutput() // Note this is using a sigmoid activation function. SET error = desiredOutput - guess SET delta = learningConstant * error * sigmoidDerivative(guess) For Each Node in inputNodes For Each Weight in inputNodes[n] inputNodes[n].weight[j] += delta; // At this point, I am assuming the first layer has been trained. // Then I recurse a similar function over the hidden layer and output layer. // The prime difference being that it further divi's up the adjustment delta. I realize this is probably not enough to go off of, and I will gladly expound on any part of my implementation. Using the above algorithm, my neural network does get trained, kind of. But not properly. The output is always XOR 1 1 [smallest number] XOR 0 0 [largest number] XOR 1 0 [medium number] XOR 0 1 [medium number] I can never train the [1,1] [0,0] to be the same value. If you have any suggestions, additional resources, articles, blogs, etc for me to look at I am very interested in learning more about this topic. Thank you for your assistance, I appreciate it greatly!

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• Search and Matching algorithm

  - by Tony

  Hello everyone. I am trying to come up with an algorithm to do the following: I have total 12 cells that I need to fill until program stops. I have 3 rows and each row has 4 columns. As an example, let me illustrate this as in airplane. So you have 3 rows and each row has 4 columns and you have window/aisle seats. Each row will have a window seat, aisle seat, aisle seat and window seat (|WA AW| Just like seat arrangement in airplane). At each iteration (different set of passengers), there would be some number of passengers (between 1 and 12) and I need to seat them closest together possible (Seat together). And I do this for next group (each iteration) until program stops (It will stop when I am done with every group). For example, I have 3 passengers (A,B,and C) and A wants to seat in Window, B wants to seat in Aisle and C wants to seat in Window. Assuming that all the seats (all 12) are available, I could place them like |A# BC| or |CB #A| and mark the seats dirty (so I don’t pick same seats again for next passengers). And I do this for next group (iteration). I am not sure if this right forum, but if somebody can advise me how I should accomplish, I would really appreciate it. Thanks.

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• Algorithm: Find smallest subset containing K 0's

  - by Vishal

  I have array of 1's and 0's only. Now I want to find contiguous subset/subarray which contains at least K 0's. Example Array is 1 1 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 1 1 0 0 1 0 0 0 and K(6) should be 0 0 1 0 1 1 0 0 0 or 0 0 0 0 1 0 1 1 0.... My Solution Array: 1 1 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 1 1 0 0 Index: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 Sum: 1 2 2 3 4 4 5 6 6 6 6 6 7 7 8 9 9 9 9 10 11 11 11 Diff(I-S): 0 0 1 1 1 2 2 2 3 4 5 6 6 7 7 7 8 9 10 10 10 11 12 For K(6) Start with 9-15 = Store difference in diff. Next increase difference 8-15(Difference in index) 8-14(Compare Difference in index) So on keep moving to find element with least elements... I am looking for better algorithm for this solution.

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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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• 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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• 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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• 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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• 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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• 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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