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  • Stack overflow code golf

    - by Chris Jester-Young
    To commemorate the public launch of Stack Overflow, what's the shortest code to cause a stack overflow? Any language welcome. ETA: Just to be clear on this question, seeing as I'm an occasional Scheme user: tail-call "recursion" is really iteration, and any solution which can be converted to an iterative solution relatively trivially by a decent compiler won't be counted. :-P ETA2: I've now selected a “best answer”; see this post for rationale. Thanks to everyone who contributed! :-)

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  • c++ stl priority queue insert bad_alloc exception

    - by bsg
    Hi, I am working on a query processor that reads in long lists of document id's from memory and looks for matching id's. When it finds one, it creates a DOC struct containing the docid (an int) and the document's rank (a double) and pushes it on to a priority queue. My problem is that when the word(s) searched for has a long list, when I try to push the DOC on to the queue, I get the following exception: Unhandled exception at 0x7c812afb in QueryProcessor.exe: Microsoft C++ exception: std::bad_alloc at memory location 0x0012ee88.. When the word has a short list, it works fine. I tried pushing DOC's onto the queue in several places in my code, and they all work until a certain line; after that, I get the above error. I am completely at a loss as to what is wrong because the longest list read in is less than 1 MB and I free all memory that I allocate. Why should there suddenly be a bad_alloc exception when I try to push a DOC onto a queue that has a capacity to hold it (I used a vector with enough space reserved as the underlying data structure for the priority queue)? I know that questions like this are almost impossible to answer without seeing all the code, but it's too long to post here. I'm putting as much as I can and am anxiously hoping that someone can give me an answer, because I am at my wits' end. The NextGEQ function is too long to put here, but it reads a list of compressed blocks of docids block by block. That is, if it sees that the lastdocid in the block (in a separate list) is larger than the docid passed in, it decompresses the block and searches until it finds the right one. If it sees that it was already decompressed, it just searches. Below, when I call the function the first time, it decompresses a block and finds the docid; the push onto the queue after that works. The second time, it doesn't even need to decompress; that is, no new memory is allocated, but after that time, pushing on to the queue gives a bad_alloc error. struct DOC{ long int docid; long double rank; public: DOC() { docid = 0; rank = 0.0; } DOC(int num, double ranking) { docid = num; rank = ranking; } bool operator>( const DOC & d ) const { return rank > d.rank; } bool operator<( const DOC & d ) const { return rank < d.rank; } }; struct listnode{ int* metapointer; int* blockpointer; int docposition; int frequency; int numberdocs; int* iquery; listnode* nextnode; }; void QUERYMANAGER::SubmitQuery(char *query){ vector<DOC> docvec; docvec.reserve(20); DOC doct; //create a priority queue to use as a min-heap to store the documents and rankings; //although the priority queue uses the heap as its underlying data structure, //I found it easier to use the STL priority queue implementation priority_queue<DOC, vector<DOC>,std::greater<DOC>> q(docvec.begin(), docvec.end()); q.push(doct); //do some processing here; startlist is a pointer to a listnode struct that starts the //linked list cout << "Opening lists:" << endl; //point the linked list start pointer to the node returned by the OpenList method startlist = &OpenList(value); listnode* minpointer; q.push(doct); //more processing here; else{ //start by finding the first docid in the shortest list int i = 0; q.push(doct); num = NextGEQ(0, *startlist); q.push(doct); while(num != -1) cout << "finding nextGEQ from shortest list" << endl; q.push(doct); //the is where the problem starts - every previous q.push(doct) works; the one after //NextGEQ(num +1, *startlist) gives the bad_alloc error num = NextGEQ(num + 1, *startlist); q.push(doct); //if you didn't break out of the loop; i.e., all lists contain a matching docid, //calculate the document's rank; if it's one of the top 20, create a struct //containing the docid and the rank and add it to the priority queue if(!loop) { cout << "found match" << endl; if(num < 0) { cout << "reached end of list" << endl; //reached the end of the shortest list; close the list CloseList(startlist); break; } rank = calculateRanking(table, num); try{ //if the heap is not full, create a DOC struct with the docid and //rank and add it to the heap if(q.size() < 20) { doc.docid = num; doc.rank = rank; q.push(doct); q.push(doc); } } catch (exception& e) { cout << e.what() << endl; } } } Thank you very much, bsg.

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  • php: shortcut to writing this?

    - by YuriKolovsky
    what would be the shortest way to write this? if(strpos($haystack, $needle)!==false){ $len = strpos($haystack, $needle)+strlen($needle); }else{ $len = 0; } I remember that I saw some shortcut somewhere for this that checked and set a variable at the same time.

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  • Non greedy grep

    - by syker
    I want to grep the shortest match and the pattern should be something like: <car ... model=BMW ...> ... ... ... </car> ... means any character and the input is multiple lines.

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  • Java encryption : with method can get me a shorter message ?

    - by Frank
    I don't know too much about encryption, I just want to ask, which method can get me the shortest result message ? For instance, the message looks like this : "This is the secret input message", I wonder if the encrypted message can be shorter then the above 32 characters long ? Maybe something like "dfkfjkvf12". Frank

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  • Numeric equivalent of an Excel column name

    - by Vivin Paliath
    The challenge The shortest code by character count that will output the numeric equivalent of an Excel column string. For example, the A column is 1, B is 2, so on and so forth. Once you hit Z, the next column becomes AA, then AB and so on. Test cases: A: 1 B: 2 AD: 30 ABC: 731 WTF: 16074 ROFL: 326676 Code count includes input/output (i.e full program).

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  • Code Golf: Spider webs

    - by LiraNuna
    The challenge The shortest code by character count to output a spider web with rings equal to user's input. A spider web is started by reconstructing the center ring: \_|_/ _/ \_ \___/ / | \ Then adding rings equal to the amount entered by the user. A ring is another level of a "spider circles" made from \ / | and _, and wraps the center circle. Input is always guaranteed to be a single positive integer. Test cases Input 1 Output \__|__/ /\_|_/\ _/_/ \_\_ \ \___/ / \/_|_\/ / | \ Input 4 Output \_____|_____/ /\____|____/\ / /\___|___/\ \ / / /\__|__/\ \ \ / / / /\_|_/\ \ \ \ _/_/_/_/_/ \_\_\_\_\_ \ \ \ \ \___/ / / / / \ \ \ \/_|_\/ / / / \ \ \/__|__\/ / / \ \/___|___\/ / \/____|____\/ / | \ Input: 7 Output: \________|________/ /\_______|_______/\ / /\______|______/\ \ / / /\_____|_____/\ \ \ / / / /\____|____/\ \ \ \ / / / / /\___|___/\ \ \ \ \ / / / / / /\__|__/\ \ \ \ \ \ / / / / / / /\_|_/\ \ \ \ \ \ \ _/_/_/_/_/_/_/_/ \_\_\_\_\_\_\_\_ \ \ \ \ \ \ \ \___/ / / / / / / / \ \ \ \ \ \ \/_|_\/ / / / / / / \ \ \ \ \ \/__|__\/ / / / / / \ \ \ \ \/___|___\/ / / / / \ \ \ \/____|____\/ / / / \ \ \/_____|_____\/ / / \ \/______|______\/ / \/_______|_______\/ / | \ Code count includes input/output (i.e full program).

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  • Sliding Response after a Point-Square Collision

    - by mars
    In general terms and pseudo-code, what would be the best way to have a collision response of sliding along a wall if the wall is actually just a part of an entire square that a point is colliding into? The collision test method used is a test to see if the point lies in the square. Should I divide the square into four lines and just calculate the shortest distance to the line and then move the point back that distance?If so, then how can I determine which edge of the square the point is closest to after collision?

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  • Abstract attributes in Python

    - by deamon
    What is the shortest / most elegant way to implement the following Scala code with an abstract attribute in Python? abstract class Controller { val path: String } A subclass of Controller is enforced to define "path" by the Scala compiler. A subclass would look like this: class MyController extends Controller { override val path = "/home" }

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  • BFS algorithm problem

    - by Gorkamorka
    The problem is as follows: A wanderer begins on the grid coordinates (x,y) and wants to reach the coordinates (0,0). From every gridpoint, the wanderer can go 8 steps north OR 3 steps south OR 5 steps east OR 6 steps west (8N/3S/5E/6W). How can I find the shortest route from (X,Y) to (0,0) using breadth-first search? Clarifications: Unlimited grid Negative coordinates are allowed A queue (linked list or array) must be used No obstacles present

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  • jQuery cut link

    - by Happy
    Here is html: <a href="http://site.com/any/different/folders/picture_name.jpg">Go and win</a> <a href="http://site.com/not/similar/links/some_other_name.png">Go and win</a> How to cut all the data from a href, except picture_name.jpg? There can be any length of the link, we must take just the value from last / to last " And anybody does not the shortest way to compare, if alt and title of current link are equal? Thanks.

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  • Automatically Generate a FlowChart in Python

    - by fayce
    Dear All, I would like to automatically generate a flowchart similar to this one ( http://en.wikipedia.org/wiki/File:%281%29_2008-04-07_Information_Management-_Help_Desk.jpg ) with Python. Do you have any advice regarding the library I should use to draw boxes, arrows (with the shortest path), text and some colors. Many thanks in advance !

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  • declare or convert a string to array format

    - by Jamex
    Hi, How to convert a string format into an array format? I have a string, $string = 'abcde' I want to convert it to a 1 element array $string[0] = 'abcde' Is there a built in function for this task?? or the shortest way is to $string = 'abcde'; $array[0] = $string; $string = $array; TIA

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  • questions regarding the use of A* with the 15-square puzzle

    - by Cheeso
    I'm trying to build an A* solver for a 15-square puzzle. The goal is to re-arrange the tiles so that they appear in their natural positions. You can only slide one tile at a time. Each possible state of the puzzle is a node in the search graph. For the h(x) function, I am using an aggregate sum, across all tiles, of the tile's dislocation from the goal state. In the above image, the 5 is at location 0,0, and it belongs at location 1,0, therefore it contributes 1 to the h(x) function. The next tile is the 11, located at 0,1, and belongs at 2,2, therefore it contributes 3 to h(x). And so on. EDIT: I now understand this is what they call "Manhattan distance", or "taxicab distance". I have been using a step count for g(x). In my implementation, for any node in the state graph, g is just +1 from the prior node's g. To find successive nodes, I just examine where I can possibly move the "hole" in the puzzle. There are 3 neighbors for the puzzle state (aka node) that is displayed: the hole can move north, west, or east. My A* search sometimes converges to a solution in 20s, sometimes 180s, and sometimes doesn't converge at all (waited 10 mins or more). I think h is reasonable. I'm wondering if I've modeled g properly. In other words, is it possible that my A* function is reaching a node in the graph via a path that is not the shortest path? Maybe have I not waited long enough? Maybe 10 minutes is not long enough? For a fully random arrangement, (assuming no parity problems), What is the average number of permutations an A* solution will examine? (please show the math) I'm going to look for logic errors in my code, but in the meantime, Any tips? (ps: it's done in Javascript). Also, no, this isn't CompSci homework. It's just a personal exploration thing. I'm just trying to learn Javascript. EDIT: I've found that the run-time is highly depend upon the heuristic. I saw the 10x factor applied to the heuristic from the article someone mentioned, and it made me wonder - why 10x? Why linear? Because this is done in javascript, I could modify the code to dynamically update an html table with the node currently being considered. This allowd me to peek at the algorithm as it was progressing. With a regular taxicab distance heuristic, I watched as it failed to converge. There were 5's and 12's in the top row, and they kept hanging around. I'd see 1,2,3,4 creep into the top row, but then they'd drop out, and other numbers would move up there. What I was hoping to see was 1,2,3,4 sort of creeping up to the top, and then staying there. I thought to myself - this is not the way I solve this personally. Doing this manually, I solve the top row, then the 2ne row, then the 3rd and 4th rows sort of concurrently. So I tweaked the h(x) function to more heavily weight the higher rows and the "lefter" columns. The result was that the A* converged much more quickly. It now runs in 3 minutes instead of "indefinitely". With the "peek" I talked about, I can see the smaller numbers creep up to the higher rows and stay there. Not only does this seem like the right thing, it runs much faster. I'm in the process of trying a bunch of variations. It seems pretty clear that A* runtime is very sensitive to the heuristic. Currently the best heuristic I've found uses the summation of dislocation * ((4-i) + (4-j)) where i and j are the row and column, and dislocation is the taxicab distance. One interesting part of the result I got: with a particular heuristic I find a path very quickly, but it is obviously not the shortest path. I think this is because I am weighting the heuristic. In one case I got a path of 178 steps in 10s. My own manual effort produce a solution in 87 moves. (much more than 10s). More investigation warranted. So the result is I am seeing it converge must faster, and the path is definitely not the shortest. I have to think about this more. Code: var stop = false; function Astar(start, goal, callback) { // start and goal are nodes in the graph, represented by // an array of 16 ints. The goal is: [1,2,3,...14,15,0] // Zero represents the hole. // callback is a method to call when finished. This runs a long time, // therefore we need to use setTimeout() to break it up, to avoid // the browser warning like "Stop running this script?" // g is the actual distance traveled from initial node to current node. // h is the heuristic estimate of distance from current to goal. stop = false; start.g = start.dontgo = 0; // calcHeuristic inserts an .h member into the array calcHeuristicDistance(start); // start the stack with one element var closed = []; // set of nodes already evaluated. var open = [ start ]; // set of nodes to evaluate (start with initial node) var iteration = function() { if (open.length==0) { // no more nodes. Fail. callback(null); return; } var current = open.shift(); // get highest priority node // update the browser with a table representation of the // node being evaluated $("#solution").html(stateToString(current)); // check solution returns true if current == goal if (checkSolution(current,goal)) { // reconstructPath just records the position of the hole // through each node var path= reconstructPath(start,current); callback(path); return; } closed.push(current); // get the set of neighbors. This is 3 or fewer nodes. // (nextStates is optimized to NOT turn directly back on itself) var neighbors = nextStates(current, goal); for (var i=0; i<neighbors.length; i++) { var n = neighbors[i]; // skip this one if we've already visited it if (closed.containsNode(n)) continue; // .g, .h, and .previous get assigned implicitly when // calculating neighbors. n.g is nothing more than // current.g+1 ; // add to the open list if (!open.containsNode(n)) { // slot into the list, in priority order (minimum f first) open.priorityPush(n); n.previous = current; } } if (stop) { callback(null); return; } setTimeout(iteration, 1); }; // kick off the first iteration iteration(); return null; }

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  • version of ldap installed - one liner

    - by PJ
    HI, I am using LDAP which is installed in a solaris machine. To check the version of LDAP i go to /ldap and check the version installed as if it is version 5 then there is a directory of the name v5.0 and so on. After getting into the directory i check the directory structure. Can anybody tell me is there any shortest way or one liner to check the version of ldap installed in my machine. Thanks in advance.

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  • declare or convert a string to array format PHP

    - by Jamex
    How to convert a string format into an array format? I have a string, $string = 'abcde' I want to convert it to a 1 element array $string[0] = 'abcde' Is there a built in function for this task? Or the shortest way is to $string = 'abcde'; $array[0] = $string; $string = $array; TIA

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  • Simple pathfinding?

    - by pdx
    Does anyone know if there is a simple way to do pathfinding in PHP? I basically have a list of numbers (kinda like {origin:11485,outboundDirections:"11486,11487,11488"} and {origin:11487,outboundDirections:"11485,11676,94185"} and getting from 11485 to 94185 would result in 11485>11487>94185 with ways to "exixt", and I'm trying to figure out how to do this (it doesn't really have to be shortest path or anything AI-like, just a way to get from A to B) I have no idea where to start at all, unfortunately

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  • Routes on a sphere surface - Find geodesic?

    - by CaNNaDaRk
    I'm working with some friends on a browser based game where people can move on a 2D map. It's been almost 7 years and still people play this game so we are thinking of a way to give them something new. Since then the game map was a limited plane and people could move from (0, 0) to (MAX_X, MAX_Y) in quantized X and Y increments (just imagine it as a big chessboard). We believe it's time to give it another dimension so, just a couple of weeks ago, we began to wonder how the game could look with other mappings: Unlimited plane with continous movement: this could be a step forward but still i'm not convinced. Toroidal World (continous or quantized movement): sincerely I worked with torus before but this time I want something more... Spherical world with continous movement: this would be great! What we want Users browsers are given a list of coordinates like (latitude, longitude) for each object on the spherical surface map; browsers must then show this in user's screen rendering them inside a web element (canvas maybe? this is not a problem). When people click on the plane we convert the (mouseX, mouseY) to (lat, lng) and send it to the server which has to compute a route between current user's position to the clicked point. What we have We began writing a Java library with many useful maths to work with Rotation Matrices, Quaternions, Euler Angles, Translations, etc. We put it all together and created a program that generates sphere points, renders them and show them to the user inside a JPanel. We managed to catch clicks and translate them to spherical coords and to provide some other useful features like view rotation, scale, translation etc. What we have now is like a little (very little indeed) engine that simulates client and server interaction. Client side shows points on the screen and catches other interactions, server side renders the view and does other calculus like interpolating the route between current position and clicked point. Where is the problem? Obviously we want to have the shortest path to interpolate between the two route points. We use quaternions to interpolate between two points on the surface of the sphere and this seemed to work fine until i noticed that we weren't getting the shortest path on the sphere surface: We though the problem was that the route is calculated as the sum of two rotations about X and Y axis. So we changed the way we calculate the destination quaternion: We get the third angle (the first is latitude, the second is longitude, the third is the rotation about the vector which points toward our current position) which we called orientation. Now that we have the "orientation" angle we rotate Z axis and then use the result vector as the rotation axis for the destination quaternion (you can see the rotation axis in grey): What we got is the correct route (you can see it lays on a great circle), but we get to this ONLY if the starting route point is at latitude, longitude (0, 0) which means the starting vector is (sphereRadius, 0, 0). With the previous version (image 1) we don't get a good result even when startin point is 0, 0, so i think we're moving towards a solution, but the procedure we follow to get this route is a little "strange" maybe? In the following image you get a view of the problem we get when starting point is not (0, 0), as you can see starting point is not the (sphereRadius, 0, 0) vector, and as you can see the destination point (which is correctly drawn!) is not on the route. The magenta point (the one which lays on the route) is the route's ending point rotated about the center of the sphere of (-startLatitude, 0, -startLongitude). This means that if i calculate a rotation matrix and apply it to every point on the route maybe i'll get the real route, but I start to think that there's a better way to do this. Maybe I should try to get the plane through the center of the sphere and the route points, intersect it with the sphere and get the geodesic? But how? Sorry for being way too verbose and maybe for incorrect English but this thing is blowing my mind! EDIT: This code version is related to the first image: public void setRouteStart(double lat, double lng) { EulerAngles tmp = new EulerAngles ( Math.toRadians(lat), 0, -Math.toRadians(lng)); //set route start Quaternion qtStart.setInertialToObject(tmp); //do other stuff like drawing start point... } public void impostaDestinazione(double lat, double lng) { EulerAngles tmp = new AngoliEulero( Math.toRadians(lat), 0, -Math.toRadians(lng)); qtEnd.setInertialToObject(tmp); //do other stuff like drawing dest point... } public V3D interpolate(double totalTime, double t) { double _t = t/totalTime; Quaternion q = Quaternion.Slerp(qtStart, qtEnd, _t); RotationMatrix.inertialQuatToIObject(q); V3D p = matInt.inertialToObject(V3D.Xaxis.scale(sphereRadius)); //other stuff, like drawing point ... return p; } //mostly taken from a book! public static Quaternion Slerp(Quaternion q0, Quaternion q1, double t) { double cosO = q0.dot(q1); double q1w = q1.w; double q1x = q1.x; double q1y = q1.y; double q1z = q1.z; if (cosO < 0.0f) { q1w = -q1w; q1x = -q1x; q1y = -q1y; q1z = -q1z; cosO = -cosO; } double sinO = Math.sqrt(1.0f - cosO*cosO); double O = Math.atan2(sinO, cosO); double oneOverSinO = 1.0f / senoOmega; k0 = Math.sin((1.0f - t) * O) * oneOverSinO; k1 = Math.sin(t * O) * oneOverSinO; // Interpolate return new Quaternion( k0*q0.w + k1*q1w, k0*q0.x + k1*q1x, k0*q0.y + k1*q1y, k0*q0.z + k1*q1z ); } A little dump of what i get (again check image 1): Route info: Sphere radius and center: 200,000, (0.0, 0.0, 0.0) Route start: lat 0,000 °, lng 0,000 ° @v: (200,000, 0,000, 0,000), |v| = 200,000 Route end: lat 30,000 °, lng 30,000 ° @v: (150,000, 86,603, 100,000), |v| = 200,000 Qt dump: (w, x, y, z), rot. angle°, (x, y, z) rot. axis Qt start: (1,000, 0,000, -0,000, 0,000); 0,000 °; (1,000, 0,000, 0,000) Qt end: (0,933, 0,067, -0,250, 0,250); 42,181 °; (0,186, -0,695, 0,695) Route start: lat 30,000 °, lng 10,000 ° @v: (170,574, 30,077, 100,000), |v| = 200,000 Route end: lat 80,000 °, lng -50,000 ° @v: (22,324, -26,604, 196,962), |v| = 200,000 Qt dump: (w, x, y, z), rot. angle°, (x, y, z) rot. axis Qt start: (0,962, 0,023, -0,258, 0,084); 31,586 °; (0,083, -0,947, 0,309) Qt end: (0,694, -0,272, -0,583, -0,324); 92,062 °; (-0,377, -0,809, -0,450)

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