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  • about Master theorem

    - by matin1234
    Hi this is the link http://www.cs.mcgill.ca/~cs251/OldCourses/1997/topic5/ is written that for T(n)<=2n+T(n/3)+T(n/3) the T(n) is not O(n) but with master theorem we can use case 3 and we can say that its T(n) is theta(n) please help me! thanks how can we prove that T(n) is not O(n)

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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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 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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  • 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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  • 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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  • Detecting Singularities in a Graph

    - by nasufara
    I am creating a graphing calculator in Java as a project for my programming class. There are two main components to this calculator: the graph itself, which draws the line(s), and the equation evaluator, which takes in an equation as a String and... well, evaluates it. To create the line, I create a Path2D.Double instance, and loop through the points on the line. To do this, I calculate as many points as the graph is wide (e.g. if the graph itself is 500px wide, I calculate 500 points), and then scale it to the window of the graph. Now, this works perfectly for most any line. However, it does not when dealing with singularities. If, when calculating points, the graph encounters a domain error (such as 1/0), the graph closes the shape in the Path2D.Double instance and starts a new line, so that the line looks mathematically correct. Example: However, because of the way it scales, sometimes it is rendered correctly, sometimes it isn't. When it isn't, the actual asymptotic line is shown, because within those 500 points, it skipped over x = 2.0 in the equation 1 / (x-2), and only did x = 1.98 and x = 2.04, which are perfectly valid in that equation. Example: In that case, I increased the window on the left and right one unit each. My question is: Is there a way to deal with singularities using this method of scaling so that the resulting line looks mathematically correct? I myself have thought of implementing a binary search-esque method, where, if it finds that it calculates one point, and then the next point is wildly far away from the last point, it searches in between those points for a domain error. I had trouble figuring out how to make it work in practice, however. Thank you for any help you may give!

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  • Return a number between 0 and 4

    - by munchine
    How do I return a number between 0 and 4, depending the input number? For example if I pass it number 23 it will return 3. The number set should look like 0 5 10 15 20 .. 1 6 11 16 21 .. 2 7 12 17 22 .. 3 8 13 18 23 .. 4 9 14 19 24 What's the math for this?

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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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  • 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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  • Reverse factorial

    - by dada
    Well, we all know that if N is given it's easy to calculate N!. But what about reversing? N! is given and you are about to find N - Is that possible ? I'm curious.

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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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