Search Results

Search found 5070 results on 203 pages for 'algorithm'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 19/203 | < Previous Page | 15 16 17 18 19 20 21 22 23 24 25 26  | Next Page >

  • Algorithm for Determining Variations of Differing Lengths

    - by joseph.ferris
    I have four objects - for the sake of arguments, let say that they are the following letters: A B C D I need to calculate the number of variations that can be made for these under the following two conditions: No repetition Objects are position agnostic Taking the above, this means that with a four object sequence, I can have only one sequence that matches the criteria (since order is not considered for being unique): ABCD There are four variations for a three object combination from the four object pool: ABC, ABD, ACD, and BCD There are six variations for a two object combination from the four object pool: AB, AC, AD, BC, BD, and CD And the most simple one, if taken on at a time: A, B, C, and D I swear that this was something covered in school, many, many years ago - and probably forgotten since I didn't think I would use it. :-) I am anticipating that factorials will come into play, but just trying to force an equation is not working. Any advice would be appreciated.

    Read the article

  • sql server replication algorithm.

    - by reggie
    Anyone know how the underlying replication model in sql server works? Do they essentially depend on UTC datetime values to determine if something is new or do they keep a table of all the changes (like a table of tableID+rowid that have changed). I am building my own "replication" system and was planning on using the dates to know what to replicate. Then I started wondering what would happen if the date got off in the computer for some reason. The obvious choice is to keep a log of the changes as you go and once you replicate those changes, you remove from the log of changes. But thats a lot of extra work, instead of just checking dates. I figure if sql server replication works by just checking the dates, then that should be good enough for me. Any wisdom here? thanks

    Read the article

  • Algorithm for generating an array of non-equal costs for a transport problem optimization

    - by Carlos
    I have an optimizer that solves a transportation problem, using a cost matrix of all the possible paths. The optimiser works fine, but if two of the costs are equal, the solution contains one more path that the minimum number of paths. (Think of it as load balancing routers; if two routes are same cost, you'll use them both.) I would like the minimum number of routes, and to do that I need a cost matrix that doesn't have two costs that are equal within a certain tolerance. At the moment, I'm passing the cost matrix through a baking function which tests every entry for equality to each of the other entries, and moves it a fixed percentage if it matches. However, this approach seems to require N^2 comparisons, and if the starting values are all the same, the last cost will be r^N bigger. (r is the arbitrary fixed percentage). Also there is the problem that by multiplying by the percentage, you end up on top of another value. So the problem seems to have an element of recursion, or at least repeated checking, which bloats the code. The current implementation is basically not very good (I won't paste my GOTO-using code here for you all to mock), and I'd like to improve it. Is there a name for what I'm after, and is there a standard implementation? Example: {1,1,2,3,4,5} (tol = 0.05) becomes {1,1.05,2,3,4,5}

    Read the article

  • K-nearest neighbour with closed-loop dimensions

    - by Tomas
    Hi, I've got a K-nearest neighbour problem where some of the dimensions are closed loops. For example one is 'time of day' and I'm matching for similarity so 'very early morning' is close to 'late evening', you can't just make it a linear scale from 'very early morning' at one end to 'late evening' at the other. How can I represent this in the data model? Is there an established way to handle this or a way to work around it?

    Read the article

  • Any better algorithm possible here?

    - by Cupidvogel
    I am trying to solve this problem in Python. Noting that only the first kiss requires the alternation, any kiss that is not a part of the chain due to the first kiss can very well have a hug on the 2nd next person, this is the code I have come up with. This is just a simple mathematical calculation, no looping, no iteration, nothing. But still I am getting a timed-out message. Any means to optimize it? import psyco psyco.full() testcase = int(raw_input()) for i in xrange(0,testcase): n = int(raw_input()) if n%2: m = n/2; ans = 2 + 4*(2**m-1); ans = ans%1000000007; print ans else: m = n/2 - 1 ans = 2 + 2**(n/2) + 4*(2**m-1); ans = ans%1000000007 print ans

    Read the article

  • Algorithm complexity question

    - by Itsik
    During a recent job interview, I was asked to give a solution to the following problem: Given a string s (without spaces) and a dictionary, return the words in the dictionary that compose the string. For example, s= peachpie, dic= {peach, pie}, result={peach, pie}. I will ask the the decision variation of this problem: if s can be composed of words in the dictionary return yes, otherwise return no. My solution to this was in backtracking (written in Java) public static boolean words(String s, Set<String> dictionary) { if ("".equals(s)) return true; for (int i=0; i <= s.length(); i++) { String pre = prefix(s,i); // returns s[0..i-1] String suf = suffix(s,i); // returns s[i..s.len] if (dictionary.contains(pre) && words(suf, dictionary)) return true; } return false; } public static void main(String[] args) { Set<String> dic = new HashSet<String>(); dic.add("peach"); dic.add("pie"); dic.add("1"); System.out.println(words("peachpie1", dic)); // true System.out.println(words("peachpie2", dic)); // false } What is the time complexity of this solution? I'm calling recursively in the for loop, but only for the prefix's that are in the dictionary. Any idea's?

    Read the article

  • Algorithms behind FarmVille Game

    - by Vadi
    What are all the algorithms involved in Farmville game, specifically I am interested in drawing trees that has fruits based on user's activities. I am into a project which has a specific need to draw a tree-type image in SVG. I am not sure how to go about the algorithms to define the tree and based on certain business rules the leafs in the tree grows etc., I think you get the idea. Farmville is just an example I took to explain. Any help is greatly appreciated..

    Read the article

  • Blit Queue Optimization Algorithm

    - by martona
    I'm looking to implement a module that manages a blit queue. There's a single surface, and portions of this surface (bounded by rectangles) are copied to elsewhere within the surface: add_blt(rect src, point dst); There can be any number of operations posted, in order, to the queue. Eventually the user of the queue will stop posting blits, and ask for an optimal set of operations to actually perform on the surface. The task of the module is to ensure that no pixel is copied unnecessarily. This gets tricky because of overlaps of course. A blit could re-blit a previously copied pixel. Ideally blit operations would be subdivided in the optimization phase in such a way that every block goes to its final place with a single operation. It's tricky but not impossible to put this together. I'm just trying to not reinvent the wheel. I looked around on the 'net, and the only thing I found was the SDL_BlitPool Library which assumes that the source surface differs from the destination. It also does a lot of grunt work, seemingly unnecessarily: regions and similar building blocks are a given. I'm looking for something higher-level. Of course, I'm not going to look a gift horse in the mouth, and I also don't mind doing actual work... If someone can come forward with a basic idea that makes this problem seem less complex than it does right now, that'd be awesome too. EDIT: Thinking about aaronasterling's answer... could this work? Implement customized region handler code that can maintain metadata for every rectangle it contains. When the region handler splits up a rectangle, it will automatically associate the metadata of this rectangle with the resulting sub-rectangles. When the optimization run starts, create an empty region handled by the above customized code, call this the master region Iterate through the blt queue, and for every entry: Let srcrect be the source rectangle for the blt beng examined Get the intersection of srcrect and master region into temp region Remove temp region from master region, so master region no longer covers temp region Promote srcrect to a region (srcrgn) and subtract temp region from it Offset temp region and srcrgn with the vector of the current blt: their union will cover the destination area of the current blt Add to master region all rects in temp region, retaining the original source metadata (step one of adding the current blt to the master region) Add to master region all rects in srcrgn, adding the source information for the current blt (step two of adding the current blt to the master region) Optimize master region by checking if adjacent sub-rectangles that are merge candidates have the same metadata. Two sub-rectangles are merge candidates if (r1.x1 == r2.x1 && r1.x2 == r2.x2) | (r1.y1 == r2.y1 && r1.y2 == r2.y2). If yes, combine them. Enumerate master region's sub-rectangles. Every rectangle returned is an optimized blt operation destination. The associated metadata is the blt operation`s source.

    Read the article

  • Algorithm to suggest a list of tags to users

    - by Itay Moav
    Given a free text, I need to analyse this this text and suggest a list of tags from a pre existing list. What algorithms are out there in the market? Can they handle a case where, for example, the text have a word like high cholesterol and I would like it so suggest heart disease although "high cholesterol" might not exists (initially) in the pre defined list.

    Read the article

  • Optimize Binary Search Algorithm

    - by Ganesh M
    In a binary search, we have two comparisons one for greater than and other for less than, otherwise its the mid value. How would you optimize so that we need to check only once? bool binSearch(int array[], int key, int left, int right) { mid = left + (right-left)/2; if (key < array[mid]) return binSearch(array, key, left, mid-1); else if (key > array[mid]) return binSearch(array, key, mid+1, right); else if (key == array[mid]) return TRUE; // Found return FALSE; // Not Found }

    Read the article

  • Algorithm to find added/removed elements in an array

    - by jj
    Hi all, I am looking for the most efficent way of solving the following Problem: given an array Before = { 8, 7, 2, 1} and an array After ={1, 3, 8, 8} find the added and the removed elements the solution is: added = 3, 8 removed = 7, 2, 1 My idea so far is: for i = 0 .. B.Lenghtt-1 { for j= 0 .. A.Lenght-1 { if A[j] == B[i] A[j] = 0; B[i] = 0; break; } } // B elemnts different from 0 are the Removed elements // A elemnts different from 0 are the Added elemnts Does anyone know a better solution perhaps more efficent and that doesn't overwrite the original arrays

    Read the article

  • Question about my sorting algorithm in C++

    - by davit-datuashvili
    i have following code in c++ #include <iostream> using namespace std; void qsort5(int a[],int n){ int i; int j; if (n<=1) return; for (i=1;i<n;i++) j=0; if (a[i]<a[0]) swap(++j,i,a); swap(0,j,a); qsort5(a,j); qsort(a+j+1,n-j-1); } int main() { return 0; } void swap(int i,int j,int a[]) { int t=a[i]; a[i]=a[j]; a[j]=t; } i have problem 1>c:\users\dato\documents\visual studio 2008\projects\qsort5\qsort5\qsort5.cpp(13) : error C2780: 'void std::swap(std::basic_string<_Elem,_Traits,_Alloc> &,std::basic_string<_Elem,_Traits,_Alloc> &)' : expects 2 arguments - 3 provided 1> c:\program files\microsoft visual studio 9.0\vc\include\xstring(2203) : see declaration of 'std::swap' 1>c:\users\dato\documents\visual studio 2008\projects\qsort5\qsort5\qsort5.cpp(13) : error C2780: 'void std::swap(std::pair<_Ty1,_Ty2> &,std::pair<_Ty1,_Ty2> &)' : expects 2 arguments - 3 provided 1> c:\program files\microsoft visual studio 9.0\vc\include\utility(76) : see declaration of 'std::swap' 1>c:\users\dato\documents\visual studio 2008\projects\qsort5\qsort5\qsort5.cpp(13) : error C2780: 'void std::swap(_Ty &,_Ty &)' : expects 2 arguments - 3 provided 1> c:\program files\microsoft visual studio 9.0\vc\include\utility(16) : see declaration of 'std::swap' 1>c:\users\dato\documents\visual studio 2008\projects\qsort5\qsort5\qsort5.cpp(14) : error C2780: 'void std::swap(std::basic_string<_Elem,_Traits,_Alloc> &,std::basic_string<_Elem,_Traits,_Alloc> &)' : expects 2 arguments - 3 provided 1> c:\program files\microsoft visual studio 9.0\vc\include\xstring(2203) : see declaration of 'std::swap' 1>c:\users\dato\documents\visual studio 2008\projects\qsort5\qsort5\qsort5.cpp(14) : error C2780: 'void std::swap(std::pair<_Ty1,_Ty2> &,std::pair<_Ty1,_Ty2> &)' : expects 2 arguments - 3 provided 1> c:\program files\microsoft visual studio 9.0\vc\include\utility(76) : see declaration of 'std::swap' 1>c:\users\dato\documents\visual studio 2008\projects\qsort5\qsort5\qsort5.cpp(14) : error C2780: 'void std::swap(_Ty &,_Ty &)' : expects 2 arguments - 3 provided 1> c:\program files\microsoft visual studio 9.0\vc\include\utility(16) : see declaration of 'std::swap' 1>c:\users\dato\documents\visual studio 2008\projects\qsort5\qsort5\qsort5.cpp(16) : error C2661: 'qsort' : no overloaded function takes 2 arguments 1>Build log was saved at "file://c:\Users\dato\Documents\Visual Studio 2008\Projects\qsort5\qsort5\Debug\BuildLog.htm" please help

    Read the article

  • Recursion problem in algorithm

    - by Marthin
    I'm not sure if this is the right place to post this, but the problem actually belongs to a programming assignment. Solve the recursion: T(0) = 2; T(n) = T(n-1) + 2; Solution: T(n) = 2(n+1) Could someone please show me how they got to that solution?

    Read the article

  • algorithm to find longest non-overlapping sequences

    - by msalvadores
    I am trying to find the best way to solve the following problem. By best way I mean less complex. As an input a list of tuples (start,length) such: [(0,5),(0,1),(1,9),(5,5),(5,7),(10,1)] Each element represets a sequence by its start and length, for example (5,7) is equivalent to the sequence (5,6,7,8,9,10,11) - a list of 7 elements starting with 5. One can assume that the tuples are sorted by the start element. The output should return a non-overlapping combination of tuples that represent the longest continuos sequences(s). This means that, a solution is a subset of ranges with no overlaps and no gaps and is the longest possible - there could be more than one though. For example for the given input the solution is: [(0,5),(5,7)] equivalent to (0,1,2,3,4,5,6,7,8,9,10,11) is it backtracking the best approach to solve this problem ? I'm interested in any different approaches that people could suggest. Also if anyone knows a formal reference of this problem or another one that is similar I'd like to get references. BTW - this is not homework. Edit Just to avoid some mistakes this is another example of expected behaviour for an input like [(0,1),(1,7),(3,20),(8,5)] the right answer is [(3,20)] equivalent to (3,4,5,..,22) with length 20. Some of the answers received would give [(0,1),(1,7),(8,5)] equivalent to (0,1,2,...,11,12) as right answer. But this last answer is not correct because is shorter than [(3,20)].

    Read the article

  • What is a good measure of strength of a link and influence of a node?

    - by Legend
    In the context of social networks, what is a good measure of strength of a link between two nodes? I am currently thinking that the following should give me what I want: For two nodes A and B: Strength(A,B) = (neighbors(A) intersection neighbors(B))/neighbors(A) where neighbors(X) gives the total number of nodes directly connected to X and the intersection operation above gives the number of nodes that are connected to both A and B. Of course, Strength(A,B) != Strength(B,A). Now knowing this, is there a good way to determine the influence of a node? I was initially using the Degree Centrality of a node to determine its "influence" but I somehow think its not a good idea because just because a node has a lot of outgoing links does not mean anything. Those links should be powerful as well. In that case, maybe using an aggregate of the strengths of each node connected to this node is a good idea to estimate its influence? I'm a little confused. Does anyone have any suggestions?

    Read the article

  • Finding good heuristic for A* search

    - by Martin
    I'm trying to find the optimal solution for a little puzzle game called Twiddle (an applet with the game can be found here). The game has a 3x3 matrix with the number from 1 to 9. The goal is to bring the numbers in the correct order using the minimum amount of moves. In each move you can rotate a 2x2 square either clockwise or counterclockwise. I.e. if you have this state 6 3 9 8 7 5 1 2 4 and you rotate the upper left 2x2 square clockwise you get 8 6 9 7 3 5 1 2 4 I'm using a A* search to find the optimal solution. My f() is simply the number of rotations need. My heuristic function already leads to the optimal solution but I don't think it's the best one you can find. My current heuristic takes each corner, looks at the number at the corner and calculates the manhatten distance to the position this number will have in the solved state (which gives me the number of rotation needed to bring the number to this postion) and sums all these values. I.e. You take the above example: 6 3 9 8 7 5 1 2 4 and this end state 1 2 3 4 5 6 7 8 9 then the heuristic does the following 6 is currently at index 0 and should by at index 5: 3 rotations needed 9 is currently at index 2 and should by at index 8: 2 rotations needed 1 is currently at index 6 and should by at index 0: 2 rotations needed 4 is currently at index 8 and should by at index 3: 3 rotations needed h = 3 + 2 + 2 + 3 = 10 But there is the problem, that you rotate 4 elements at once. So there a rare cases where you can do two (ore more) of theses estimated rotations in one move. This means theses heuristic overestimates the distance to the solution. My current workaround is, to simply excluded one of the corners from the calculation which solves this problem at least for my test-cases. I've done no research if really solves the problem or if this heuristic still overestimates in same edge-cases. So my question is: What is the best heuristic you can come up with? (Disclaimer: This is for a university project, so this is a bit of homework. But I'm free to use any resource if can come up with, so it's okay to ask you guys. Also I will credit Stackoverflow for helping me ;) )

    Read the article

  • what is the best algorithm to use for this problem

    - by slim
    Equilibrium index of a sequence is an index such that the sum of elements at lower indexes is equal to the sum of elements at higher indexes. For example, in a sequence A: A[0]=-7 A[1]=1 A[2]=5 A[3]=2 A[4]=-4 A[5]=3 A[6]=0 3 is an equilibrium index, because: A[0]+A[1]+A[2]=A[4]+A[5]+A[6] 6 is also an equilibrium index, because: A[0]+A[1]+A[2]+A[3]+A[4]+A[5]=0 (sum of zero elements is zero) 7 is not an equilibrium index, because it is not a valid index of sequence A. If you still have doubts, this is a precise definition: the integer k is an equilibrium index of a sequence if and only if and . Assume the sum of zero elements is equal zero. Write a function int equi(int[] A); that given a sequence, returns its equilibrium index (any) or -1 if no equilibrium indexes exist. Assume that the sequence may be very long.

    Read the article

  • Algorithm to Group All the Cycles Together

    - by Ngu Soon Hui
    I have a lot of cycles ( indicated by numeric values, for example, 1-2-3-4 corresponds to a cycle, with 4 edges, edge 1 is {1:2}, edge 2 is {2:3}, edge 3 is {3,4}, edge 4 is {4,1}, and so on). A cycle is said to be connected to another cycle if they share one and only one edge. For example, let's say I have two cycles 1-2-3-4 and 5-6-7-8, then there are two cycle groups because these two cycles are not connecting to each other. If I have two cycles 1-2-3-4 and 3-4-5-6, then I have only one cycle group because these two cycles share the same edge. What is the most efficient way to find all the cycle groups?

    Read the article

  • An algorithm to find common edits

    - by Tass
    I've got two word lists, an example: list 1 list 2 foot fuut barj kijo foio fuau fuim fuami kwim kwami lnun lnun kizm kazm I'd like to find o ? u # 1 and 3 i ? a # 3 and 7 im ? ami # 4 and 5 This should be ordered by amount of occurrences, so I can filter the ones that don't appear often. The lists currently consist of 35k words, the calculation should take about 6h on an average server.

    Read the article

  • Help with a algorithm in linq to resolve a query

    - by Deumber
    I have been some time thinking how to resolve this problem, but out of ideas i prefer expose it for help. I have 2 tables (linq to sql) A and B, A have a many relation with B, so A have a property EntitySet of B A have the following properties: CreateDate (Datetime) ModificateDate (Datetime) Bs (EntitySet<B>) B have the following properties: CreateDate (Datetime) ModificateDate (Datetime) All that i want is return a ordered collection of A by the Max date between : A.CreateDate, A.ModificateDate, The Max B.CreateDate of all B in A The Max B.ModificateDate of all B in A if i someone need a little example, just ask for it.

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 15 16 17 18 19 20 21 22 23 24 25 26  | Next Page >