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  • submatrix from a matrix

    - by Grv
    A matrix is of size n*n and it consists only 0 and 1 find the largest submatrix that consists of 1's only eg 10010 11100 11001 11110 largest sub matrix will be of 3*2 from row 2 to row 4 please answer with best space and time complexity

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  • Finding unreachable sections of a 2D map

    - by dada
    I don't want you to solve this problem for me, i just want to ask for some ideas. This is the input below, and it represents a map. The 'x' represents land, and the dots - water. So with the 'x' you can represent 'islands' on the map. xxx.x...xxxxx xxxx....x...x ........x.x.x ..xxxxx.x...x ..x...x.xxx.x ..x.x.x...x.. ..x...x...xxx ...xxxxxx.... x............ As you can see, there are some islands which are closed, i.e. if some boat is inside its territory, it won't be able to get out, for ex: ..xxxxx. ..x...x. ..x.x.x. ..x...x. ..xxxxx. And there are some open islands which is possible to get out of them, ex: .xxxxx .x...x .x.x.x .xxx.x The problem is this: For a given NxM map like those above, calculate howm any of the islands are open, and how many are closed. I repeat: I don't want you to solve it, just need some sugestions, ideas for solving. thanks

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  • Media recommendation engine - Single user system - How to start

    - by Microkernel
    Hi guys, I want to implement a media recommendation engine. I saw a similar posts on this, but I think my requirements are bit different from those, so posting here. Here is the deal. I want to implement a recommendation engine for media players like VLC, which would be an engine that has to care for only single user. Like, it would be embedded in a media player on a PC which is typically used by single user. And it will start learning the likes and dislikes of the user and gradually learns what a user likes. Here it will not be able to find similar users for using their data for recommendation as its a single user system. So how to go about this? Or you can consider it as a recommendation engine that has to be put in say iPods, which has to learn about a single user and recommend music/Movies from the collections it has. I thought of start collecting the genre of music/movies (maybe even artist name) that user watches and recommend movies from the most watched Genre, but it look very crude, isn't it? So is there any algorithms I can use or any resources I can refer up to? Regards, MicroKernel :)

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  • C# - split String into smaller Strings by length variable

    - by tyndall
    I'd like to break apart a String by a certain length variable. It needs to bounds check so as not explode when the last section of string is not as long as or longer than the length. Looking for the most succinct (yet understandable) version. Example: string x = "AAABBBCC"; string[] arr = x.SplitByLength(3); // arr[0] -> "AAA"; // arr[1] -> "BBB"; // arr[2] -> "CC"

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  • What's the best general programming book to review basic development concepts?

    - by Charles S.
    I'm looking for for a programming book that reviews basic concepts like implementing linked lists, stacks, queues, hash tables, tree traversals, search algorithms, etc. etc. Basically, I'm looking for a review of everything I learned in college but have forgotten. I prefer something written in the last few years that includes at least a decent amount of code in object-oriented languages. This is to study for job interview questions but I already have the "solving interview questions" books. I'm looking for something with a little more depth and explanation. Any good recommendations?

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  • Is there any simple way to test two PNGs for equality?

    - by Mason Wheeler
    I've got a bunch of PNG images, and I'm looking for a way to identify duplicates. By duplicates I mean, specifically, two PNG files whose uncompressed image data are identical, not necessarily whose files are identical. This means I can't do something simple like compare CRC hash values. I figure this can actually be done reliably since PNGs use lossless compression, but I'm worried about speed. I know I can winnow things down a little by testing for equal dimensions first, but when it comes time to actually compare the images against each other, is there any way to do it reasonably efficiently? (ie. faster than the "double-for-loop checking pixel values against each other" brute-force method?)

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  • sorting using recursion

    - by user310587
    I have the following function to sort an array with even numbers in the front and odd numbers in the back. Is there a way to get it done without using any loops? //front is 0, back =array.length-1; arrangeArray (front, back); public static void arrangeArray (int front, int back) { if (front != back || front<back) { while (numbers [front]%2 == 0) front++; while (numbers[back]%2!=0) back--; if (front < back) { int oddnum = numbers [front]; numbers[front]= numbers[back]; numbers[back]=oddnum; arrangeArray (front+1, back-1); } } }

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  • Word Jumble Algorithm

    - by MasterMax1313
    Given a word jumble (i.e. ofbaor), what would be an approach to unscramble the letters to create a real word (i.e. foobar)? I could see this having a couple of approaches, and I think I know how I'd do it in .NET, but I curious to see what some other solutions look like (always happy to see if my solution is optimal or not). This isn't homework or anything like that, I just saw a word jumble in the local comics section of the paper (yes, good ol' fashioned newsprint), and the engineer in me started thinking.

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  • Capturing time intervals when somebody was online? How would you impement this feature?

    - by Kirzilla
    Hello, Our aim is to build timelines saying about periods of time when user was online. (It really doesn't matter what user we are talking about and where he was online) To get information about onliners we can call API method, someservice.com/api/?call=whoIsOnline whoIsOnline method will give us a list of users currently online. But there is no API method to get information about who IS NOT online. So, we should build our timelines using information we got from whoIsOnline. Of course there will be a measurement error (we can't track information in realtime). Let's suppose that we will call whoIsOnline method every 2 minutes (yes, we will run our script by cron every 2 minutes). For example, calling whoIsOnline at 08:00 will return Peter_id Michal_id Andy_id calling whoIsOnline at 08:02 will return Michael_id Andy_id George_id As you can see, Peter has gone offline, but we have new onliner - George. Available instruments are Db(MySQL) / text files / key-value storage (Redis/memcache); feel free to choose any of them (or even all of them). So, we have to get information like this George_id was online... 12 May: 08:02-08:30, 12:40-12:46, 20:14-22:36 11 May: 09:10-12:30, 21:45-23:00 10 May: was not online And now question... How would you store information to implement such timelines? How would you query/calculate information about periods of time when user was online? Additional information.. You cannot update information about offline users, only users who are "currently" online. Solution should be flexible: timeline information could be represented relating to any timezone. We should keep information only for last 7 days. Every user seen online is automatically getting his own identifier in our database. Uff.. it was really hard for me to write it because my English is pretty bad, but I hope my question will be clear for you. Thank you.

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  • what is order notation f(n)=O(g(n))?

    - by Lopa
    2 questions: question 1: under what circumstances would this[O(f(n))=O(k.f(n))] be the most appropriate form of time-complexity analysis? question 2: working from mathematical definition of O notation, show that O(f(n))=O(k.f(n)), for positive constant k.? My view: For the first one I think it is average case and worst case form of time-complexity. am i right? and what else do i write in that? for the second one I think we need to define the function mathematically, so is the answer something like because the multiplication by a constant just corresponds to a readjustment of value of the arbitrary constant 'k' in definition of O.

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  • What kind of data processing problems would CUDA help with?

    - by Chris McCauley
    Hi, I've worked on many data matching problems and very often they boil down to quickly and in parallel running many implementations of CPU intensive algorithms such as Hamming / Edit distance. Is this the kind of thing that CUDA would be useful for? What kinds of data processing problems have you solved with it? Is there really an uplift over the standard quad-core intel desktop? Chris

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  • Refining data stored in SQLite - how to join several contacts?

    - by Krab
    Problem background Imagine this problem. You have a water molecule which is in contact with other molecules (if the contact is a hydrogen bond, there can be 4 other molecules around my water). Like in the following picture (A, B, C, D are some other atoms and dots mean the contact). A B . . O / \ H H . . C D I have the information about all the dots and I need to eliminate the water in the center and create records describing contacts of A-C, A-D, A-B, B-C, B-D, and C-D. Database structure Currently, I have the following structure in the database: Table atoms: "id" integer PRIMARY KEY, "amino" char(3) NOT NULL, (HOH for water or other value) other columns identifying the atom Table contacts: "acceptor_id" integer NOT NULL, (the atom near to my hydrogen, here C or D) "donor_id" integer NOT NULL, (here A or B) "directness" char(1) NOT NULL, (this should be D for direct and W for water-mediated) other columns about the contact, such as the distance Current solution (insufficient) Now, I'm going through all the contacts which have donor.amino = "HOH". In this sample case, this would select contacts from C and D. For each of these selected contacts, I look up contacts having the same acceptor_id as is the donor_id in the currently selected contact. From this information, I create the new contact. At the end, I delete all contacts to or from HOH. This way, I am obviously unable to create C-D and A-B contacts (the other 4 are OK). If I try a similar approach - trying to find two contacts having the same donor_id, I end up with duplicate contacts (C-D and D-C). Is there a simple way to retrieve all six contacts without duplicates? I'm dreaming about some one page long SQL query which retrievs just these six wanted rows. :-) It is preferable to conserve information about who is donor where possible, but not strictly necessary. Big thanks to all of you who read this question to this point.

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  • Url shortening algorithm

    - by Bozho
    Now, this is not strictly about URL shortening, but my purpose is such anyway, so let's view it like that. Of course the steps to URL shortening are: Take the full URL Generate a unique short string to be the key for the URL Store the URL and the key in a database (a key-value store would be a perfect match here) Now, about the 2nd point. Here's what I've come up with: ByteArrayOutputStream baos = new ByteArrayOutputStream(); DataOutputStream dos = new DataOutputStream(baos); UUID uuid = UUID.randomUUID(); dos.writeLong(uuid.getMostSignificantBits()); String encoded = new String(Base64.encodeBase64(baos.toByteArray()), "ISO-8859-1"); String shortUrl = StringUtils.left(6); // returns the leftmost 6 characters // check if exists in database, repeat until it does not I wonder if this is good enough. Is it?

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  • red black tree balancing?

    - by Anirudh Kaki
    i am working to generate tango tree, where i need to check whether every sub tree in tango is balanced or not. if its not balanced i need to make it balance? I trying so hard to make entire RB-tree balance but i not getting any proper logic so can any one help me out?? here i am adding code to check how to find my tree is balanced are not but when its not balanced how can i make it balance. static boolean verifyProperty5(rbnode n) { int left = 0, right = 0; if (n != null) { bh++; left = blackHeight(n.left, 0); right = blackHeight(n.right, 0); } if (left == right) { System.out.println("black height is :: " + bh); return true; } else { System.out.println("in balance"); return false; } } public static int blackHeight(rbnode root, int len) { bh = 0; blackHeight(root, path1, len); return bh; } private static void blackHeight(rbnode root, int path1[], int len) { if (root == null) return; if (root.color == "black"){ root.black_count = root.parent.black_count+1; } else{ root.black_count = root.parent.black_count; } if ((root.left == null) && (root.right == null)) { bh = root.black_count; } blackHeight(root.left, path1, len); blackHeight(root.right, path1, len); }

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  • How can I further optimize this color difference function?

    - by aLfa
    I have made this function to calculate color differences in the CIE Lab colorspace, but it lacks speed. Since I'm not a Java expert, I wonder if any Java guru around has some tips that can improve the speed here. The code is based on the matlab function mentioned in the comment block. /** * Compute the CIEDE2000 color-difference between the sample color with * CIELab coordinates 'sample' and a standard color with CIELab coordinates * 'std' * * Based on the article: * "The CIEDE2000 Color-Difference Formula: Implementation Notes, * Supplementary Test Data, and Mathematical Observations,", G. Sharma, * W. Wu, E. N. Dalal, submitted to Color Research and Application, * January 2004. * available at http://www.ece.rochester.edu/~gsharma/ciede2000/ */ public static double deltaE2000(double[] lab1, double[] lab2) { double L1 = lab1[0]; double a1 = lab1[1]; double b1 = lab1[2]; double L2 = lab2[0]; double a2 = lab2[1]; double b2 = lab2[2]; // Cab = sqrt(a^2 + b^2) double Cab1 = Math.sqrt(a1 * a1 + b1 * b1); double Cab2 = Math.sqrt(a2 * a2 + b2 * b2); // CabAvg = (Cab1 + Cab2) / 2 double CabAvg = (Cab1 + Cab2) / 2; // G = 1 + (1 - sqrt((CabAvg^7) / (CabAvg^7 + 25^7))) / 2 double CabAvg7 = Math.pow(CabAvg, 7); double G = 1 + (1 - Math.sqrt(CabAvg7 / (CabAvg7 + 6103515625.0))) / 2; // ap = G * a double ap1 = G * a1; double ap2 = G * a2; // Cp = sqrt(ap^2 + b^2) double Cp1 = Math.sqrt(ap1 * ap1 + b1 * b1); double Cp2 = Math.sqrt(ap2 * ap2 + b2 * b2); // CpProd = (Cp1 * Cp2) double CpProd = Cp1 * Cp2; // hp1 = atan2(b1, ap1) double hp1 = Math.atan2(b1, ap1); // ensure hue is between 0 and 2pi if (hp1 < 0) { // hp1 = hp1 + 2pi hp1 += 6.283185307179586476925286766559; } // hp2 = atan2(b2, ap2) double hp2 = Math.atan2(b2, ap2); // ensure hue is between 0 and 2pi if (hp2 < 0) { // hp2 = hp2 + 2pi hp2 += 6.283185307179586476925286766559; } // dL = L2 - L1 double dL = L2 - L1; // dC = Cp2 - Cp1 double dC = Cp2 - Cp1; // computation of hue difference double dhp = 0.0; // set hue difference to zero if the product of chromas is zero if (CpProd != 0) { // dhp = hp2 - hp1 dhp = hp2 - hp1; if (dhp > Math.PI) { // dhp = dhp - 2pi dhp -= 6.283185307179586476925286766559; } else if (dhp < -Math.PI) { // dhp = dhp + 2pi dhp += 6.283185307179586476925286766559; } } // dH = 2 * sqrt(CpProd) * sin(dhp / 2) double dH = 2 * Math.sqrt(CpProd) * Math.sin(dhp / 2); // weighting functions // Lp = (L1 + L2) / 2 - 50 double Lp = (L1 + L2) / 2 - 50; // Cp = (Cp1 + Cp2) / 2 double Cp = (Cp1 + Cp2) / 2; // average hue computation // hp = (hp1 + hp2) / 2 double hp = (hp1 + hp2) / 2; // identify positions for which abs hue diff exceeds 180 degrees if (Math.abs(hp1 - hp2) > Math.PI) { // hp = hp - pi hp -= Math.PI; } // ensure hue is between 0 and 2pi if (hp < 0) { // hp = hp + 2pi hp += 6.283185307179586476925286766559; } // LpSqr = Lp^2 double LpSqr = Lp * Lp; // Sl = 1 + 0.015 * LpSqr / sqrt(20 + LpSqr) double Sl = 1 + 0.015 * LpSqr / Math.sqrt(20 + LpSqr); // Sc = 1 + 0.045 * Cp double Sc = 1 + 0.045 * Cp; // T = 1 - 0.17 * cos(hp - pi / 6) + // + 0.24 * cos(2 * hp) + // + 0.32 * cos(3 * hp + pi / 30) - // - 0.20 * cos(4 * hp - 63 * pi / 180) double hphp = hp + hp; double T = 1 - 0.17 * Math.cos(hp - 0.52359877559829887307710723054658) + 0.24 * Math.cos(hphp) + 0.32 * Math.cos(hphp + hp + 0.10471975511965977461542144610932) - 0.20 * Math.cos(hphp + hphp - 1.0995574287564276334619251841478); // Sh = 1 + 0.015 * Cp * T double Sh = 1 + 0.015 * Cp * T; // deltaThetaRad = (pi / 3) * e^-(36 / (5 * pi) * hp - 11)^2 double powerBase = hp - 4.799655442984406; double deltaThetaRad = 1.0471975511965977461542144610932 * Math.exp(-5.25249016001879 * powerBase * powerBase); // Rc = 2 * sqrt((Cp^7) / (Cp^7 + 25^7)) double Cp7 = Math.pow(Cp, 7); double Rc = 2 * Math.sqrt(Cp7 / (Cp7 + 6103515625.0)); // RT = -sin(delthetarad) * Rc double RT = -Math.sin(deltaThetaRad) * Rc; // de00 = sqrt((dL / Sl)^2 + (dC / Sc)^2 + (dH / Sh)^2 + RT * (dC / Sc) * (dH / Sh)) double dLSl = dL / Sl; double dCSc = dC / Sc; double dHSh = dH / Sh; return Math.sqrt(dLSl * dLSl + dCSc * dCSc + dHSh * dHSh + RT * dCSc * dHSh); }

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  • C++: building iterator from bits

    - by gruszczy
    I have a bitmap and would like to return an iterator of positions of set bits. Right now I just walk the whole bitmap and if bit is set, then I provide next position. I believe this could be done more effectively: for example build statically array for each combination of bits in single byte and return vector of positions. This can't be done for a whole int, because array would be too big. But maybe there are some better solutions? Do you know any smart algorithms for this?

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  • Clojure - tail recursive sieve of Eratosthenes

    - by Konrad Garus
    I have this implementation of the sieve of Eratosthenes in Clojure: (defn sieve [n] (loop [last-tried 2 sift (range 2 (inc n))] (if (or (nil? last-tried) (> last-tried n)) sift (let [filtered (filter #(or (= % last-tried) (< 0 (rem % last-tried))) sift)] (let [next-to-try (first (filter #(> % last-tried) filtered))] (recur next-to-try filtered)))))) For larger n (like 20000) it ends with stack overflow. Why doesn't tail call elimination work here? How to fix it?

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  • Algorithm for Negating Sentences

    - by Kevin Dolan
    I was wondering if anyone was familiar with any attempts at algorithmic sentence negation. For example, given a sentence like "This book is good" provide any number of alternative sentences meaning the opposite like "This book is not good" or even "This book is bad". Obviously, accomplishing this with a high degree of accuracy would probably be beyond the scope of current NLP, but I'm sure there has been some work on the subject. If anybody knows of any work, care to point me to some papers?

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  • Grouping rectangles (getting the bounding boxes of rects)

    - by hyn
    What is a good, fast way to get the "final" bounding boxes of a set of random (up to about 40, not many) rectangles? By final I mean that all bounding boxes don't intersect with any other. Brute force way: in a double for loop, for each rect, test for intersection against every other rect. The intersecting rects become a new rect (replaced), indicating the bounding box. Start over and repeat until no intersection is detected. Because the rects are random every time, and the rect count is relatively small, collision detection using spatial hashing seems like overkill. Is there a way to do this more effectively?

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  • How does Batcher Merge work at a high level?

    - by Mike
    I'm trying to grasp the concept of a Batcher Sort. However, most resources I've found online focus on proof entirely or on low-level pseudocode. Before I look at proofs, I'd like to understand how Batcher Sort works. Can someone give a high level overview of how Batcher Sort works(particularly the merge) without overly verbose pseudocode(I want to get the idea behind the Batcher Sort, not implement it)? Thanks!

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