Search Results

Search found 5655 results on 227 pages for 'stl algorithm'.

Page 88/227 | < Previous Page | 84 85 86 87 88 89 90 91 92 93 94 95  | Next Page >

  • How to find two most distant points?

    - by depesz
    This is a question that I was asked on a job interview some time ago. And I still can't figure out sensible answer. Question is: you are given set of points (x,y). Find 2 most distant points. Distant from each other. For example, for points: (0,0), (1,1), (-8, 5) - the most distant are: (1,1) and (-8,5) because the distance between them is larger from both (0,0)-(1,1) and (0,0)-(-8,5). The obvious approach is to calculate all distances between all points, and find maximum. The problem is that it is O(n^2), which makes it prohibitively expensive for large datasets. There is approach with first tracking points that are on the boundary, and then calculating distances for them, on the premise that there will be less points on boundary than "inside", but it's still expensive, and will fail in worst case scenario. Tried to search the web, but didn't find any sensible answer - although this might be simply my lack of search skills.

    Read the article

  • Merging k sorted linked lists - analysis

    - by Kotti
    Hi! I am thinking about different solutions for one problem. Assume we have K sorted linked lists and we are merging them into one. All these lists together have N elements. The well known solution is to use priority queue and pop / push first elements from every lists and I can understand why it takes O(N log K) time. But let's take a look at another approach. Suppose we have some MERGE_LISTS(LIST1, LIST2) procedure, that merges two sorted lists and it would take O(T1 + T2) time, where T1 and T2 stand for LIST1 and LIST2 sizes. What we do now generally means pairing these lists and merging them pair-by-pair (if the number is odd, last list, for example, could be ignored at first steps). This generally means we have to make the following "tree" of merge operations: N1, N2, N3... stand for LIST1, LIST2, LIST3 sizes O(N1 + N2) + O(N3 + N4) + O(N5 + N6) + ... O(N1 + N2 + N3 + N4) + O(N5 + N6 + N7 + N8) + ... O(N1 + N2 + N3 + N4 + .... + NK) It looks obvious that there will be log(K) of these rows, each of them implementing O(N) operations, so time for MERGE(LIST1, LIST2, ... , LISTK) operation would actually equal O(N log K). My friend told me (two days ago) it would take O(K N) time. So, the question is - did I f%ck up somewhere or is he actually wrong about this? And if I am right, why doesn't this 'divide&conquer' approach can't be used instead of priority queue approach?

    Read the article

  • 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?

    Read the article

  • 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=)

    Read the article

  • 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

    Read the article

  • 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!

    Read the article

  • 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.

    Read the article

  • 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?)

    Read the article

  • 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 :)

    Read the article

  • 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)

    Read the article

  • 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

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

  • 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.

    Read the article

  • 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]

    Read the article

  • 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.

    Read the article

  • 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); } } }

    Read the article

  • 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

    Read the article

  • 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"

    Read the article

  • 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.

    Read the article

< Previous Page | 84 85 86 87 88 89 90 91 92 93 94 95  | Next Page >