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  • Stack and queue operations on the same array.

    - by Passonate Learner
    Hi. I've been thinking about a program logic, but I cannot draw a conclusion to my problem. Here, I've implemented stack and queue operations to a fixed array. int A[1000]; int size=1000; int top; int front; int rear; bool StackIsEmpty() { return (top==0); } bool StackPush( int x ) { if ( top >= size ) return false; A[top++] = x; return true; } int StackTop( ) { return A[top-1]; } bool StackPop() { if ( top <= 0 ) return false; A[--top] = 0; return true; } bool QueueIsEmpty() { return (front==rear); } bool QueuePush( int x ) { if ( rear >= size ) return false; A[rear++] = x; return true; } int QueueFront( ) { return A[front]; } bool QueuePop() { if ( front >= rear ) return false; A[front++] = 0; return true; } It is presumed(or obvious) that the bottom of the stack and the front of the queue is pointing at the same location, and vice versa(top of the stack points the same location as rear of the queue). For example, integer 1 and 2 is inside an array in order of writing. And if I call StackPop(), the integer 2 will be popped out, and if I call QueuePop(), the integer 1 will be popped out. My problem is that I don't know what happens if I do both stack and queue operations on the same array. The example above is easy to work out, because there are only two values involved. But what if there are more than 2 values involved? For example, if I call StackPush(1); QueuePush(2); QueuePush(4); StackPop(); StackPush(5); QueuePop(); what values will be returned in the order of bottom(front) from the final array? I know that if I code a program, I would receive a quick answer. But the reason I'm asking this is because I want to hear a logical explanations from a human being, not a computer.

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  • How to use Haar wavelet to detect LINES on an image?

    - by Ole Jak
    So I have Image like this I want to get something like this (I hevent drawn all lines I want but I hope you can get my idea) I want to use SURF ( (Speeded Up Robust Features) is a robust image descriptor, first presented by Herbert Bay et al. in 2006 ) or something that is based on sums of 2D Haar wavelet responses and makes an efficient use of integral images for finding all straight lines on image. I want to get relative to picture pixel coords start and end points of lines. So on this picture to find all lines between tiles and thouse 2 black lines on top. Is there any such Code Example (with lines search capability) to start from? I love C and C++ but any other readable code will probably work for me=)

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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • Array sorting efficiency... Beginner need advice

    - by SoleSoft
    I'll start by saying I am very much a beginner programmer, this is essentially my first real project outside of using learning material. I've been making a 'Simon Says' style game (the game where you repeat the pattern generated by the computer) using C# and XNA, the actual game is complete and working fine but while creating it, I wanted to also create a 'top 10' scoreboard. The scoreboard would record player name, level (how many 'rounds' they've completed) and combo (how many buttons presses they got correct), the scoreboard would then be sorted by combo score. This led me to XML, the first time using it, and I eventually got to the point of having an XML file that recorded the top 10 scores. The XML file is managed within a scoreboard class, which is also responsible for adding new scores and sorting scores. Which gets me to the point... I'd like some feedback on the way I've gone about sorting the score list and how I could have done it better, I have no other way to gain feedback =(. I know .NET features Array.Sort() but I wasn't too sure of how to use it as it's not just a single array that needs to be sorted. When a new score needs to be entered into the scoreboard, the player name and level also have to be added. These are stored within an 'array of arrays' (10 = for 'top 10' scores) scoreboardComboData = new int[10]; // Combo scoreboardTextData = new string[2][]; scoreboardTextData[0] = new string[10]; // Name scoreboardTextData[1] = new string[10]; // Level as string The scoreboard class works as follows: - Checks to see if 'scoreboard.xml' exists, if not it creates it - Initialises above arrays and adds any player data from scoreboard.xml, from previous run - when AddScore(name, level, combo) is called the sort begins - Another method can also be called that populates the XML file with above array data The sort checks to see if the new score (combo) is less than or equal to any recorded scores within the scoreboardComboData array (if it's greater than a score, it moves onto the next element). If so, it moves all scores below the score it is less than or equal to down one element, essentially removing the last score and then places the new score within the element below the score it is less than or equal to. If the score is greater than all recorded scores, it moves all scores down one and inserts the new score within the first element. If it's the only score, it simply adds it to the first element. When a new score is added, the Name and Level data is also added to their relevant arrays, in the same way. What a tongue twister. Below is the AddScore method, I've added comments in the hope that it makes things clearer O_o. You can get the actual source file HERE. Below the method is an example of the quickest way to add a score to follow through with a debugger. public static void AddScore(string name, string level, int combo) { // If the scoreboard has not yet been filled, this adds another 'active' // array element each time a new score is added. The actual array size is // defined within PopulateScoreBoard() (set to 10 - for 'top 10' if (totalScores < scoreboardComboData.Length) totalScores++; // Does the scoreboard even need sorting? if (totalScores > 1) { for (int i = totalScores - 1; i > - 1; i--) { // Check to see if score (combo) is greater than score stored in // array if (combo > scoreboardComboData[i] && i != 0) { // If so continue to next element continue; } // Check to see if score (combo) is less or equal to element 'i' // score && that the element is not the last in the // array, if so the score does not need to be added to the scoreboard else if (combo <= scoreboardComboData[i] && i != scoreboardComboData.Length - 1) { // If the score is lower than element 'i' and greater than the last // element within the array, it needs to be added to the scoreboard. This is achieved // by moving each element under element 'i' down an element. The new score is then inserted // into the array under element 'i' for (int j = totalScores - 1; j > i; j--) { // Name and level data are moved down in their relevant arrays scoreboardTextData[0][j] = scoreboardTextData[0][j - 1]; scoreboardTextData[1][j] = scoreboardTextData[1][j - 1]; // Score (combo) data is moved down in relevant array scoreboardComboData[j] = scoreboardComboData[j - 1]; } // The new Name, level and score (combo) data is inserted into the relevant array under element 'i' scoreboardTextData[0][i + 1] = name; scoreboardTextData[1][i + 1] = level; scoreboardComboData[i + 1] = combo; break; } // If the method gets the this point, it means that the score is greater than all scores within // the array and therefore cannot be added in the above way. As it is not less than any score within // the array. else if (i == 0) { // All Names, levels and scores are moved down within their relevant arrays for (int j = totalScores - 1; j != 0; j--) { scoreboardTextData[0][j] = scoreboardTextData[0][j - 1]; scoreboardTextData[1][j] = scoreboardTextData[1][j - 1]; scoreboardComboData[j] = scoreboardComboData[j - 1]; } // The new number 1 top name, level and score, are added into the first element // within each of their relevant arrays. scoreboardTextData[0][0] = name; scoreboardTextData[1][0] = level; scoreboardComboData[0] = combo; break; } // If the methods get to this point, the combo score is not high enough // to be on the top10 score list and therefore needs to break break; } } // As totalScores < 1, the current score is the first to be added. Therefore no checks need to be made // and the Name, Level and combo data can be entered directly into the first element of their relevant // array. else { scoreboardTextData[0][0] = name; scoreboardTextData[1][0] = level; scoreboardComboData[0] = combo; } } } Example for adding score: private static void Initialize() { scoreboardDoc = new XmlDocument(); if (!File.Exists("Scoreboard.xml")) GenerateXML("Scoreboard.xml"); PopulateScoreBoard("Scoreboard.xml"); // ADD TEST SCORES HERE! AddScore("EXAMPLE", "10", 100); AddScore("EXAMPLE2", "24", 999); PopulateXML("Scoreboard.xml"); } In it's current state the source file is just used for testing, initialize is called within main and PopulateScoreBoard handles the majority of other initialising, so nothing else is needed, except to add a test score. I thank you for your time!

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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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  • 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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  • 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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  • 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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  • 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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