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  • recursion tree and binary tree cost calculation

    - by Tony
    Hi all, I've got the following recursion: T(n) = T(n/3) + T(2n/3) + O(n) The height of the tree would be log3/2 of 2. Now the recursion tree for this recurrence is not a complete binary tree. It has missing nodes lower down. This makes sense to me, however I don't understand how the following small omega notation relates to the cost of all leaves in the tree. "... the total cost of all leaves would then be Theta (n^log3/2 of 2) which, since log3/2 of 2 is a constant strictly greater then 1, is small omega(n lg n)." Can someone please help me understand how the Theta(n^log3/2 of 2) becomes small omega(n lg n)?

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  • Need help iteratating over an array, retrieve two possibilites, no repeats, for Poker AI

    - by elguapo-85
    Can't really think of a good way to word this question, nor a good title, and maybe the answer is so ridiculously simple that I am missing it. I am working on a poker AI, and I want to calculate the number of hands that exist which are better then mine, I understand how to that, but want I can't figure out is the best way to iterate over a group of cards. So I am at the flop, I know what my two cards are and there are 3 cards on the board. So there are 47 unknown cards and I want to iterate over all possible combination of those 47 cards assuming that two are passed out, so you can't have two cards of the same rank and suit, and you if you have previously calculated a set you don't want to do it over again, because I will being wasting time, and this will be called many times. If you don't understand want I am asking please tell me and I will clarify more. So I can set something up like this, if that element equals one, it means it is not in my hand and not on the board, 4 for each suit, and 13 for each rank. setOfCards[4][13] If I do a simple set of for loops like this: (pseudocode) //remove cards I know are in play from setOfCards by setting values to zero for(int i = 0; i < 4; i++) for(int j = 0; j < 13; j++) for(int k = 0; k < 4; k++) for(int l = 0; l < 4; l++) //skip if values equal zero card1 = setOfCards[i][j] card2 = setOfCards[k][l] //now compare card1, card2 and set of board cards So this is actually going to repeat many values, for example: card1 = AceOfHearts, card2 = KingOfHearts is the same as card1 = KingOfHearts, card2 = AceOfHearts. It will also alter my calculations. How should I go about avoiding this? Also is there a name for this technique? Thank you.

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  • return first non repeating character in a string

    - by Amm Sokun
    I had to solve this question in which given a string, i had to return the first non repeating character present in the string. I solved it using hashtable and wrote a method which takes a constant reference to the string and returns the first non repeating character. However, when there is no non repeating character present in the string, i return -1 and in the main program i check as follows char c = firstNonRepeating( word ); if (static_cast<int> (c) == -1) cout<<"no non repeating character present\n"; else cout<<c<<endl; is that the correct way to return -1 when required character is not present?

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  • C++ find largest BST in a binary tree

    - by fonjibe
    what is your approach to have the largest BST in a binary tree? I refer to this post where a very good implementation for finding if a tree is BST or not is bool isBinarySearchTree(BinaryTree * n, int min=std::numeric_limits<int>::min(), int max=std::numeric_limits<int>::max()) { return !n || (min < n->value && n->value < max && isBinarySearchTree(n->l, min, n->value) && isBinarySearchTree(n->r, n->value, max)); } It is quite easy to implement a solution to find whether a tree contains a binary search tree. i think that the following method makes it: bool includeSomeBST(BinaryTree* n) { if(!isBinarySearchTree(n)) { if(!isBinarySearchTree(n->left)) return isBinarySearchTree(n->right); } else return true; else return true; } but what if i want the largest BST? this is my first idea, BinaryTree largestBST(BinaryTree* n) { if(isBinarySearchTree(n)) return n; if(!isBinarySearchTree(n->left)) { if(!isBinarySearchTree(n->right)) if(includeSomeBST(n->right)) return largestBST(n->right); else if(includeSomeBST(n->left)) return largestBST(n->left); else return NULL; else return n->right; } else return n->left; } but its not telling the largest actually. i struggle to make the comparison. how should it take place? thanks

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  • Generating a URL pattern when provided a set of 5 or so URLs

    - by ryan
    Provided with a set of URLs, I need to generate a pattern, For example: http://www.buy.com/prod/disney-s-star-struck/q/loc/109/213724402.html http://www.buy.com/prod/samsung-f2380-23-widescreen-1080p-lcd-monitor-150-000-1-dc-8ms-1920-x/q/loc/101/211249863.html http://www.buy.com/prod/panasonic-nnh765wf-microwave-oven-countertop-1-6-ft-1250w-panasonic/q/loc/66357/202045865.html http://www.buy.com/prod/escape-by-calvin-klein-for-women-3-4-oz-edp-spray/q/loc/66740/211210860.html http://www.buy.com/prod/v-touch-8gb-mp3-mp4-2-8-touch-screen-2mp-camera-expandable-minisd-w/q/loc/111/211402014.html Pattern is http://www.buy.com/prod/[^~]/q/loc/[^~].html

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  • Texture coordintes for a polygon and a square texture

    - by user146780
    basically I have a texture. I also have a lets say octagon (or any polygon). I find that octagon's bounding box. Let's say my texture is the size of the octagon's bounding box. How could I figure out the texture coordinates so that the texture maps to it. To clarify, lets say you had a square of tin foil and cut the octagon out you'd be left with a tin foil textured polygon.I'm just not sure how to figure it out for an arbitrary polygon. Thanks

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  • ListAdapters and WrapperListAdapter algorithm

    - by Matty F
    This logic is written in a function with signature private void showDialog(final AdapterView<? extends Adapter> parent, String title, String message, final Tag subject) Is there a better way of doing this? // refresh adapter SimpleCursorAdapter adapter; if (parent.getAdapter() instanceof WrapperListAdapter) { adapter = (SimpleCursorAdapter) ((WrapperListAdapter) parent.getAdapter()).getWrappedAdapter(); } else { adapter = (SimpleCursorAdapter) parent.getAdapter(); } adapter.getCursor().requery(); adapter.notifyDataSetChanged(); Also, is there any point in having AdapterView<? extends Adapter> in the signature and not just AdapterView<?>?

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  • What is the most efficient/elegant way to parse a flat table into a tree?

    - by Tomalak
    Assume you have a flat table that stores an ordered tree hierarchy: Id Name ParentId Order 1 'Node 1' 0 10 2 'Node 1.1' 1 10 3 'Node 2' 0 20 4 'Node 1.1.1' 2 10 5 'Node 2.1' 3 10 6 'Node 1.2' 1 20 What minimalistic approach would you use to output that to HTML (or text, for that matter) as a correctly ordered, correctly intended tree? Assume further you only have basic data structures (arrays and hashmaps), no fancy objects with parent/children references, no ORM, no framework, just your two hands. The table is represented as a result set, which can be accessed randomly. Pseudo code or plain English is okay, this is purely a conceptional question. Bonus question: Is there a fundamentally better way to store a tree structure like this in a RDBMS? EDITS AND ADDITIONS To answer one commenter's (Mark Bessey's) question: A root node is not necessary, because it is never going to be displayed anyway. ParentId = 0 is the convention to express "these are top level". The Order column defines how nodes with the same parent are going to be sorted. The "result set" I spoke of can be pictured as an array of hashmaps (to stay in that terminology). For my example was meant to be already there. Some answers go the extra mile and construct it first, but thats okay. The tree can be arbitrarily deep. Each node can have N children. I did not exactly have a "millions of entries" tree in mind, though. Don't mistake my choice of node naming ('Node 1.1.1') for something to rely on. The nodes could equally well be called 'Frank' or 'Bob', no naming structure is implied, this was merely to make it readable. I have posted my own solution so you guys can pull it to pieces.

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  • Re-adjusting a binary heap after removing the minimum element

    - by BeeBand
    After removing the minimum element in a binary heap, i.e. after removing the root, why is the last leaf then assigned to the root and sifted down? Why not take the lesser child of what used to be the root and just keep sifting up all the children? Isn't this the same amount of operations, so why is the "sift down" method preferred?

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  • Python script to calculate aded combinations from a dictionary

    - by dayde
    I am trying to write a script that will take a dictionary of items, each containing properties of values from 0 - 10, and add the various elements to select which combination of items achieve the desired totals. I also need the script to do this, using only items that have the same "slot" in common. For example: item_list = { 'item_1': {'slot': 'top', 'prop_a': 2, 'prop_b': 0, 'prop_c': 2, 'prop_d': 1 }, 'item_2': {'slot': 'top', 'prop_a': 5, 'prop_b': 0, 'prop_c': 1, 'prop_d':-1 }, 'item_3': {'slot': 'top', 'prop_a': 2, 'prop_b': 5, 'prop_c': 2, 'prop_d':-2 }, 'item_4': {'slot': 'mid', 'prop_a': 5, 'prop_b': 5, 'prop_c':-5, 'prop_d': 0 }, 'item_5': {'slot': 'mid', 'prop_a':10, 'prop_b': 0, 'prop_c':-5, 'prop_d': 0 }, 'item_6': {'slot': 'mid', 'prop_a':-5, 'prop_b': 2, 'prop_c': 3, 'prop_d': 5 }, 'item_7': {'slot': 'bot', 'prop_a': 1, 'prop_b': 3, 'prop_c':-4, 'prop_d': 4 }, 'item_8': {'slot': 'bot', 'prop_a': 2, 'prop_b': 2, 'prop_c': 0, 'prop_d': 0 }, 'item_9': {'slot': 'bot', 'prop_a': 3, 'prop_b': 1, 'prop_c': 4, 'prop_d':-4 }, } The script would then need to select which combinations from the "item_list" dict that using 1 item per "slot" that would achieve a desired result when added. For example, if the desired result was: 'prop_a': 3, 'prop_b': 3, 'prop_c': 8, 'prop_d': 0, the script would select 'item_2', 'item_6', and 'item_9', along with any other combination that worked. 'item_2': {'slot': 'top', 'prop_a': 5, 'prop_b': 0, 'prop_c': 1, 'prop_d':-1 } 'item_6': {'slot': 'mid', 'prop_a':-5, 'prop_b': 2, 'prop_c': 3, 'prop_d': 5 } 'item_9': {'slot': 'bot', 'prop_a': 3, 'prop_b': 1, 'prop_c': 4, 'prop_d':-4 } 'total': 'prop_a': 3, 'prop_b': 3, 'prop_c': 8, 'prop_d': 0 Any ideas how to accomplish this? It does not need to be in python, or even a thorough script, but just an explanation on how to do this in theory would be enough for me. I have tried working out looping through every combination, but that seems to very quickly get our of hand and unmanageable. The actual script will need to do this for about 1,000 items using 20 different "slots", each with 8 properties. Thanks for the help!

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  • How do I modify an attribute across all rows in a table?

    - by prgmatic
    Hi folks, My apologies for asking such a novice question but, I need help building a script using either PHP or directly in MySQL that can do the following: Take the values of a column in a table (text) Change them into capitalized words (from "this is a title" to "This Is A Title") Replace the old values (uncapitalized) with the new values (capitalized). Thanks for the help and support.

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  • Given 4 objects, how to figure out whether exactly 2 have a certain property

    - by Cocorico
    Hi guys! I have another question on how to make most elegant solution to this problem, since I cannot afford to go to computer school right so my actual "pure programming" CS knowledge is not perfect or great. This is basically an algorhythm problem (someone please correct me if I am using that wrong, since I don't want to keep saying them and embarass myself) I have 4 objects. Each of them has an species property that can either be a dog, cat, pig or monkey. So a sample situation could be: object1.species=pig object2.species=cat object3.species=pig object4.species=dog Now, if I want to figure out if all 4 are the same species, I know I could just say: if ( (object1.species==object2.species) && (object2.species==object3.species) && (object3.species==object4.species) ) { // They are all the same animal (don't care WHICH animal they are) } But that isn't so elegant right? And if I suddenly want to know if EXACTLY 3 or 2 of them are the same species (don't care WHICH species it is though), suddenly I'm in spaghetti code. I am using Objective C although I don't know if that matters really, since the most elegant solution to this is I assume the same in all languages conceptually? Anyone got good idea? Thanks!!

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  • Fastest primality test

    - by Grigory Javadyan
    Hi. Could you suggest a fast, deterministic method that is usable in practice, for testing if a large number is prime or not? Also, I would like to know how to use non-deterministic primality tests correctly. For example, if I'm using such a method, I can be sure that a number is not prime if the output is "no", but what about the other case, when the output is "probably"? Do I have to test for primality manually in this case? Thanks in advance.

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  • Learning Algorithms and Data Structures Fundamentals

    - by valya
    Can you recommend me a book or (better!) a site with many hard problems and exercises about data structures? I'm already answering project Euler questions, but these questions are about interesting, but uncommon algorithms. I hardly used even a simple tree. Maybe there is a site with exercises like: hey, you need to calculate this: ... . Do it using a tree. Now do it using a zipper. Upload your C (Haskell, Lisp, even Pascal or Fortress go) solution. Oh, your solution is so slow! Self-education is very hard then you trying to learn very common, fundamental things. How can I help myself with them without attending to courses or whatever?

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  • Find subset with K elements that are closest to eachother

    - by Nima
    Given an array of integers size N, how can you efficiently find a subset of size K with elements that are closest to each other? Let the closeness for a subset (x1,x2,x3,..xk) be defined as: 2 <= N <= 10^5 2 <= K <= N constraints: Array may contain duplicates and is not guaranteed to be sorted. My brute force solution is very slow for large N, and it doesn't check if there's more than 1 solution: N = input() K = input() assert 2 <= N <= 10**5 assert 2 <= K <= N a = [] for i in xrange(0, N): a.append(input()) a.sort() minimum = sys.maxint startindex = 0 for i in xrange(0,N-K+1): last = i + K tmp = 0 for j in xrange(i, last): for l in xrange(j+1, last): tmp += abs(a[j]-a[l]) if(tmp > minimum): break if(tmp < minimum): minimum = tmp startindex = i #end index = startindex + K? Examples: N = 7 K = 3 array = [10,100,300,200,1000,20,30] result = [10,20,30] N = 10 K = 4 array = [1,2,3,4,10,20,30,40,100,200] result = [1,2,3,4]

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  • substitution cypher with different alphabet length

    - by seanizer
    I would like to implement a simple substitution cypher to mask private ids in URLs I know how my IDs will look like (combination of upperchase ascii, digits and underscore), and they will be rather long, as they are composed keys. I would like to use a longer alphabet to shorten the resulting codes (I'd like to use upper and lower case ascii letters, digits and nothing else). So my incoming alphabet would be [A-Z0-9_] (37 chars) and my outgoing alphabet would be [A-Za-z0-9] (62 chars) so a compression of almost 50% would be available. let's say my URLs look like this: /my/page/GFZHFFFZFZTFZTF_24_F34 and I want them to look like this instead: /my/page/Ft32zfegZFV5 Obviously both arrays would be shuffled to bring some random order in. This does not have to be secure. if someone figures it out: fine, but I don't want the scheme to be obvious. My desired solution would be to convert the string to an integer representation of radix 37, convert the radix to 62 and use the second alphabet to write out that number. is there any sample code available that does something similar? Integer.parseInt ( http://java.sun.com/javase/6/docs/api/java/lang/Integer.html#parseInt%28java.lang.String,%20int%29 ) has some similar logic, but it is hard-coded to use standard digit behavior Any hints? I am using java to implement this but code or pseudo-code in any other language is of course also helpful

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  • problem with evolutionary algorithms degrading into simulated annealing: mutation too small?

    - by Schnalle
    i have a problem understanding evolutionary algorithms. i tried using this technique several times, but i always ran into the same problem: degeneration into simulated annealing. lets say my initial population, with fitness in brackets, is: A (7), B (9), C (14), D (19) after mating and mutation i have following children: AB (8.3), AC (12.2), AD (14.1), BC(11), BD (14.7), CD (17) after elimination of the weakest, we get A, AB, B, AC next turn, AB will mate again with a result around 8, pushing AC out. next turn, AB again, pushing B out (assuming mutation changes fitness mostly in the 1 range). now, after only a few turns the pool is populated with the originally fittest candidates (A, B) and mutations of those two (AB). this happens regardless of the size of the initial pool, it just takes a bit longer. say, with an initial population of 50 it takes 50 turns, then all others are eliminated, turning the whole setup in a more complicated simulated annealing. in the beginning i also mated canditates with themselves, worsening the problem. so, what do i miss? are my mutation rates simply too small and will it go away if i increase them? here's the project i'm using it for: http://stefan.schallerl.com/simuan-grid-grad/ yeah, the code is buggy and the interface sucks, but i'm too lazy to fix it right now - and be careful, it may lock up your browser. better use chrome, even thought firefox is not slower than chrome for once (probably the tracing for the image comparison pays off, yay!). if anyone is interested, the code can be found here. here i just dropped the ev-alg idea and went for simulated annealing. ps: i'm not even sure about simulated annealing - it is like evolutionary algorithms, just with a population size of one, right?

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  • Best data-structure to use for two ended sorted list

    - by fmark
    I need a collection data-structure that can do the following: Be sorted Allow me to quickly pop values off the front and back of the list Remain sorted after I insert a new value Allow a user-specified comparison function, as I will be storing tuples and want to sort on a particular value Thread-safety is not required Optionally allow efficient haskey() lookups (I'm happy to maintain a separate hash-table for this though) My thoughts at this stage are that I need a priority queue and a hash table, although I don't know if I can quickly pop values off both ends of a priority queue. I'm interested in performance for a moderate number of items (I would estimate less than 200,000). Another possibility is simply maintaining an OrderedDictionary and doing an insertion sort it every-time I add more data to it. Furthermore, are there any particular implementations in Python. I would really like to avoid writing this code myself.

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  • Number of different elements in an array.

    - by AB
    Is it possible to compute the number of different elements in an array in linear time and constant space? Let us say it's an array of long integers, and you can not allocate an array of length sizeof(long). P.S. Not homework, just curious. I've got a book that sort of implies that it is possible.

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  • compare function for upper_bound / lower_bound

    - by Martin Beckett
    I want to find the first item in a sorted vector that has a field less than some value x. I need to supply a compare function that compares 'x' with the internal value in MyClass but I can't work out the function declaration. Can't I simply overload '<' but how do I do this when the args are '&MyClass' and 'float' ? float x; std::vector< MyClass >::iterator last = std::upper_bound(myClass.begin(),myClass.end(),x);

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  • Chain call in clojure?

    - by Konrad Garus
    I'm trying to implement sieve of Eratosthenes in Clojure. One approach I would like to test is this: Get range (2 3 4 5 6 ... N) For 2 <= i <= N Pass my range through filter that removes multiplies of i For i+1th iteration, use result of the previous filtering I know I could do it with loop/recur, but this is causing stack overflow errors (for some reason tail call optimization is not applied). How can I do it iteratively? I mean invoking N calls to the same routine, passing result of ith iteration to i+1th.

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  • Grouping php array items based on user and created time

    - by Jim
    This is an array of objects showing a user uploading photos: Array ( [12] => stdClass Object ( [type] => photo [created] => 2010-05-14 23:36:41 [user] => stdClass Object ( [id] => 760 [username] => mrsmith ) [photo] => stdClass Object ( [id] => 4181 ) ) [44] => stdClass Object ( [type] => photo [created] => 2010-05-14 23:37:15 [user] => stdClass Object ( [id] => 760 [username] => mrsmith ) [photo] => stdClass Object ( [id] => 4180 ) ) ) However instead of showing: mr smith uploaded one photo mr smith uploaded one photo I'd like to display: mr smith uploaded two photos by grouping similar items, grouping by user ID and them having added them within, let's say 15 minutes of each other. So I'd like to get the array in this sort of shape: Array ( [12] => stdClass Object ( [type] => photo [created] => 2010-05-14 23:36:41 [user] => stdClass Object ( [id] => 760 [username] => mrsmith ) [photos] => Array ( [0] => stdClass Object ( [id] => 4181 ) [1] => stdClass Object ( [id] => 4180 ) ) ) ) preserving the first item of the group and it's created time, and supplementing it with any other groupable photos and then unsetting any items that were grouped (so the final array doesn't have key 44 anymore as it was grouped in with 12). The array contains other actions than just photos, hence the original keys of 12 and 44. I just can't figure out a way to do this efficiently. I used to use MySQL and PHP to do this but am trying to just use pure PHP for caching reasons. Can anyone shed any insights? I thought about going through each item and seeing if I can group it with the previous one in the array but the previous one might not necessarily be relevant or even a photo. I've got total brain freeze :(

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