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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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  • 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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  • How To Generate Parameter Set for the Diffie-Hellman Key Agreement Algorithm in Android

    - by sebby_zml
    Hello everyone, I am working on mobile/server security related project. I am now stuck in generating a Diffie-Hellman key agreement part. It works fine in server side program but it is not working in mobile side. Thus, I assume that it is not compactible with Android. I used the following class to get the parameters. It returns a comma-separated string of 3 values. The first number is the prime modulus P. The second number is the base generator G. The third number is bit size of the random exponent L. My question is is there anything wrong with the code or it is not compactible for android?What kind of changes should I do? Your suggestion and guidance would be very much help for me. Thanks a lot in advance. public static String genDhParams() { try { // Create the parameter generator for a 1024-bit DH key pair AlgorithmParameterGenerator paramGen = AlgorithmParameterGenerator.getInstance("DH"); paramGen.init(1024); // Generate the parameters AlgorithmParameters params = paramGen.generateParameters(); DHParameterSpec dhSpec = (DHParameterSpec)params.getParameterSpec(DHParameterSpec.class); // Return the three values in a string return ""+dhSpec.getP()+","+dhSpec.getG()+","+dhSpec.getL(); } catch (NoSuchAlgorithmException e) { } catch (InvalidParameterSpecException e) { } return null; } Regards, Sebby

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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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  • 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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  • 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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  • 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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  • 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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  • Code bacteria: evolving mathematical behavior

    - by Stefano Borini
    It would not be my intention to put a link on my blog, but I don't have any other method to clarify what I really mean. The article is quite long, and it's in three parts (1,2,3), but if you are curious, it's worth the reading. A long time ago (5 years, at least) I programmed a python program which generated "mathematical bacteria". These bacteria are python objects with a simple opcode-based genetic code. You can feed them with a number and they return a number, according to the execution of their code. I generate their genetic codes at random, and apply an environmental selection to those objects producing a result similar to a predefined expected value. Then I let them duplicate, introduce mutations, and evolve them. The result is quite interesting, as their genetic code basically learns how to solve simple equations, even for values different for the training dataset. Now, this thing is just a toy. I had time to waste and I wanted to satisfy my curiosity. however, I assume that something, in terms of research, has been made... I am reinventing the wheel here, I hope. Are you aware of more serious attempts at creating in-silico bacteria like the one I programmed? Please note that this is not really "genetic algorithms". Genetic algorithms is when you use evolution/selection to improve a vector of parameters against a given scoring function. This is kind of different. I optimize the code, not the parameters, against a given scoring function.

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  • A simple explanation of Naive Bayes Classification

    - by Jaggerjack
    I am finding it hard to understand the process of Naive Bayes, and I was wondering if someone could explained it with a simple step by step process in English. I understand it takes comparisons by times occurred as a probability, but I have no idea how the training data is related to the actual dataset. Please give me an explanation of what role the training set plays. I am giving a very simple example for fruits here, like banana for example training set--- round-red round-orange oblong-yellow round-red dataset---- round-red round-orange round-red round-orange oblong-yellow round-red round-orange oblong-yellow oblong-yellow round-red

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  • B-Tree Revision

    - by stan
    Hi, If we are looking for line intersections (horizontal and vertical lines only) and we have n lines with half of them vertical and no intersections then Sorting the list of line end points on y value will take N log N using mergesort Each insert delete and search of our data structue (assuming its a b-tree) will be < log n so the total search time will be N log N What am i missing here, if the time to sort using mergesort takes a time of N log N and insert and delete takes a time of < log n are we dropping the constant factor to give an overal time of N log N. If not then how comes < log n goes missing in total ONotation run time? Thanks

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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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