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  • Performing text processing on flatpage content to include handling of custom tag

    - by Dzejkob
    Hi. I'm using flatpages app in my project to manage some html content. That content will include images, so I've made a ContentImage model allowing user to upload images using admin panel. The user should then be able to include those images in content of the flatpages. He can of course do that by manually typing image url into <img> tag, but that's not what I'm looking for. To make including images more convenient, I'm thinking about something like this: User edits an additional, let's say pre_content field of CustomFlatPage model (I'm using custom flatpage model already) instead of defining <img> tags directly, he uses a custom tag, something like [img=...] where ... is name of the ContentImage instance now the hardest part: before CustomFlatPage is saved, pre_content field is checked for all [img=...] occurences and they are processed like this: ContentImage model is searched if there's image instance with given name and if so, [img=...] is replaced with proper <img> tag. flatpage actual content is filled with processed pre_content and then flatpage is saved (pre_content is leaved unchanged, as edited by user) The part that I can't cope with is text processing. Should I use regular expressions? Apparently they can be slow for large strings. And how to organize logic? I assume it's rather algorithmic question, but I'm not familliar with text processing in Python enough, to do it myself. Can somebody give me any clues?

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  • Efficiently storing a list of prime numbers

    - by eSKay
    This article says: Every prime number can be expressed as 30k±1, 30k±7, 30k±11, or 30k±13 for some k. That means we can use eight bits per thirty numbers to store all the primes; a million primes can be compressed to 33,334 bytes "That means we can use eight bits per thirty numbers to store all the primes" This "eight bits per thirty numbers" would be for k, correct? But each k value will not necessarily take-up just one bit. Shouldn't it be eight k values instead? "a million primes can be compressed to 33,334 bytes" I am not sure how this is true. We need to indicate two things: VALUE of k (can be arbitrarily large) STATE from one of the eight states (-13,-11,-7,-1,1,7,11,13) I am not following how 33,334 bytes was arrived at, but I can say one thing: as the prime numbers become larger and larger in value, we will need more space to store the value of k. How, then can we fix it at 33,334 bytes?

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  • How to find the insertion point in an array using binary search?

    - by ????
    The basic idea of binary search in an array is simple, but it might return an "approximate" index if the search fails to find the exact item. (we might sometimes get back an index for which the value is larger or smaller than the searched value). For looking for the exact insertion point, it seems that after we got the approximate location, we might need to "scan" to left or right for the exact insertion location, so that, say, in Ruby, we can do arr.insert(exact_index, value) I have the following solution, but the handling for the part when begin_index >= end_index is a bit messy. I wonder if a more elegant solution can be used? (this solution doesn't care to scan for multiple matches if an exact match is found, so the index returned for an exact match may point to any index that correspond to the value... but I think if they are all integers, we can always search for a - 1 after we know an exact match is found, to find the left boundary, or search for a + 1 for the right boundary.) My solution: DEBUGGING = true def binary_search_helper(arr, a, begin_index, end_index) middle_index = (begin_index + end_index) / 2 puts "a = #{a}, arr[middle_index] = #{arr[middle_index]}, " + "begin_index = #{begin_index}, end_index = #{end_index}, " + "middle_index = #{middle_index}" if DEBUGGING if arr[middle_index] == a return middle_index elsif begin_index >= end_index index = [begin_index, end_index].min return index if a < arr[index] && index >= 0 #careful because -1 means end of array index = [begin_index, end_index].max return index if a < arr[index] && index >= 0 return index + 1 elsif a > arr[middle_index] return binary_search_helper(arr, a, middle_index + 1, end_index) else return binary_search_helper(arr, a, begin_index, middle_index - 1) end end # for [1,3,5,7,9], searching for 6 will return index for 7 for insertion # if exact match is found, then return that index def binary_search(arr, a) puts "\nSearching for #{a} in #{arr}" if DEBUGGING return 0 if arr.empty? result = binary_search_helper(arr, a, 0, arr.length - 1) puts "the result is #{result}, the index for value #{arr[result].inspect}" if DEBUGGING return result end arr = [1,3,5,7,9] b = 6 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = 6 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [] b = 60 arr.insert(binary_search(arr, b), b) p arr and result: Searching for 6 in [1, 3, 5, 7, 9] a = 6, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = 6, arr[middle_index] = 7, begin_index = 3, end_index = 4, middle_index = 3 a = 6, arr[middle_index] = 5, begin_index = 3, end_index = 2, middle_index = 2 the result is 3, the index for value 7 [1, 3, 5, 6, 7, 9] Searching for 6 in [1, 3, 5, 7, 9, 11] a = 6, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = 6, arr[middle_index] = 9, begin_index = 3, end_index = 5, middle_index = 4 a = 6, arr[middle_index] = 7, begin_index = 3, end_index = 3, middle_index = 3 the result is 3, the index for value 7 [1, 3, 5, 6, 7, 9, 11] Searching for 60 in [1, 3, 5, 7, 9] a = 60, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = 60, arr[middle_index] = 7, begin_index = 3, end_index = 4, middle_index = 3 a = 60, arr[middle_index] = 9, begin_index = 4, end_index = 4, middle_index = 4 the result is 5, the index for value nil [1, 3, 5, 7, 9, 60] Searching for 60 in [1, 3, 5, 7, 9, 11] a = 60, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = 60, arr[middle_index] = 9, begin_index = 3, end_index = 5, middle_index = 4 a = 60, arr[middle_index] = 11, begin_index = 5, end_index = 5, middle_index = 5 the result is 6, the index for value nil [1, 3, 5, 7, 9, 11, 60] Searching for -60 in [1, 3, 5, 7, 9] a = -60, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 1, middle_index = 0 a = -60, arr[middle_index] = 9, begin_index = 0, end_index = -1, middle_index = -1 the result is 0, the index for value 1 [-60, 1, 3, 5, 7, 9] Searching for -60 in [1, 3, 5, 7, 9, 11] a = -60, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 1, middle_index = 0 a = -60, arr[middle_index] = 11, begin_index = 0, end_index = -1, middle_index = -1 the result is 0, the index for value 1 [-60, 1, 3, 5, 7, 9, 11] Searching for -60 in [1] a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 0, middle_index = 0 the result is 0, the index for value 1 [-60, 1] Searching for 60 in [1] a = 60, arr[middle_index] = 1, begin_index = 0, end_index = 0, middle_index = 0 the result is 1, the index for value nil [1, 60] Searching for 60 in [] [60]

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  • big O notation algorithm

    - by niggersak
    Use big-O notation to classify the traditional grade school algorithms for addition and multiplication. That is, if asked to add two numbers each having N digits, how many individual additions must be performed? If asked to multiply two N-digit numbers, how many individual multiplications are required? . Suppose f is a function that returns the result of reversing the string of symbols given as its input, and g is a function that returns the concatenation of the two strings given as its input. If x is the string hrwa, what is returned by g(f(x),x)? Explain your answer - don't just provide the result!

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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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  • 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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  • 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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  • randomized quicksort: probability of two elements comparison?

    - by bantu
    I am reading "Probability and Computing" by M.Mitzenmacher, E.Upfal and I have problems understanding how the probability of comparison of two elements is calculated. Input: the list (y1,y2,...,YN) of numbers. We are looking for pivot element. Question: what is probability that two elements yi and yj (ji) will be compared? Answer (from book): yi and yj will be compared if either yi or yj will be selected as pivot in first draw from sequence (yi,yi+1,...,yj-1,yj). So the probablity is: 2/(y-i+1). The problem for me is initial claim: for example, picking up yi in the first draw from the whole list will cause the comparison with yj (and vice-versa) and the probability is 2/n. So, rather the "reverse" claim is true -- none of the (yi+1,...,yj-1) elements can be selected beforeyi or yj, but the "pool" size is not fixed (in first draw it is n for sure, but on the second it is smaller). Could someone please explain this, how the authors come up with such simplified conclusion? Thank you in advance

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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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  • 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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  • 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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  • 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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  • curious ill conditioned numerical problem

    - by aaa
    hello. somebody today showed me this curious ill conditioned problem (apparently pretty famous), which looks relatively simple ƒ = (333.75 - a^2)b^6 + a^2 (11a^2 b^2 - 121b^4 - 2) + 5.5b^8 + a/(2^b) where a = 77617 and b = 33096 can you determine correct answer?

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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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  • 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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  • 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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  • 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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