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  • integer division properties

    - by aaa
    hi. does the following integer arithmetic property hold? (m/n)/l == m/(n*l) At first I thought I knew answer (does not hold), but now am not sure. Does it hold for all numbers or only for certain conditions, i.e. n > l?

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  • Next month, same day in PHP

    - by Guillermo
    I got a last time (and urgent) requeriment of an "event" that needs to be scheduled the same day of every month. Say of you set the start date on the 1st May you should get the next events on the 1st of Jun, 1 Jul etc. The problem comes with a start date on the 31st (the next ones could be 30 or 28 depending on the month) Considering that there are months with different numbers of days (28,30,31) depending on the month itself and the year... what would ba an easy way to setup this? Consider the following (and flawed) nextmonth function: $events = array() function nextmonth($date) { return $date+(60*60*24*30); } $curr = $start; while($curr < $end) { $events[ = $curr; $curr = nextmonth($curr); } Edited to add: The problem for me is, simply enought, is to solve what the number of days of any given month is and thus get the next corresponding date..

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  • Calculate shortest path through a grocery store

    - by Bart
    Hi, I'm trying to find a way to find the shortest path through a grocery store, visiting a list of locations (shopping list). The path should start at a specified startposition and can end at multiple endpositions (there are multiple checkout counters). Also, I have some predefined constraints on the path, such as "item x on the shopping list needs to be the last, second last, or third last item on the path". There is a function that will return true or false for a given path. Finally, this needs to be calculated with limited cpu power (on a smartphone) and within a second or so. If this isn't possible, then an approximation to the optimal path is also ok. Is this possible? So far I think I need to start by calculating the distance between every item on the list using something like A* or Dijkstra's. After that, should I treat it like the travelling salesman problem? Because in my problem there is a specified startnode, specified endnodes, and some constraints, which are not in the travelling salesman problem. Any help would be appreciated :)

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  • Formula needed: Sort array to array-"snaked"

    - by aw
    After the you guys helped me out so gracefully last time, here is another tricky array sorter for you. I have the following array: a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16] I use it for some visual stuff and render it like this: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Now I want to sort the array to have a "snake" later: // rearrange the array according to this schema 1 2 3 4 12 13 14 5 11 16 15 6 10 9 8 7 // the original array should look like this a = [1,2,3,4,12,13,14,5,11,16,15,6,10,9,8,7] Now I'm looking for a smart formula / smart loop to do that ticker = 0; rows = 4; // can be n cols = 4; // can be n originalArray = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]; newArray = []; while(ticker < originalArray.length) { //do the magic here ticker++; } Thanks again for the help.

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  • Help me understand Inorder Traversal without using recursion

    - by vito
    I am able to understand preorder traversal without using recursion, but I'm having a hard time with inorder traversal. I just don't seem to get it, perhaps, because I haven't understood the inner working of recursion. This is what I've tried so far: def traverseInorder(node): lifo = Lifo() lifo.push(node) while True: if node is None: break if node.left is not None: lifo.push(node.left) node = node.left continue prev = node while True: if node is None: break print node.value prev = node node = lifo.pop() node = prev if node.right is not None: lifo.push(node.right) node = node.right else: break The inner while-loop just doesn't feel right. Also, some of the elements are getting printed twice; may be I can solve this by checking if that node has been printed before, but that requires another variable, which, again, doesn't feel right. Where am I going wrong? I haven't tried postorder traversal, but I guess it's similar and I will face the same conceptual blockage there, too. Thanks for your time! P.S.: Definitions of Lifo and Node: class Node: def __init__(self, value, left=None, right=None): self.value = value self.left = left self.right = right class Lifo: def __init__(self): self.lifo = () def push(self, data): self.lifo = (data, self.lifo) def pop(self): if len(self.lifo) == 0: return None ret, self.lifo = self.lifo return ret

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  • An extended Bezier Library or Algorithms of bezier operations

    - by Sorush Rabiee
    Hi, Is there a library of data structures and operations for quadratic bezier curves? I need to implement: bezier to bitmap converting with arbitrary quality optimizing bezier curves common operations like subtraction, extraction, rendering etc. languages: c,c++,.net,python Algorithms without implementation (pseudocode or etc) could be useful too. (especially optimization)

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  • bug/error in basis set path algorithm i can't figure out

    - by Roy McAvoy
    The following looks through a 2d array to find basis set paths. It is supposed to print out the individual paths but not repeat any and end when all paths are found. It however doesn't stop at the last path and has a bug in it somewhere in which the following happens: It goes halfway through the path and then goes to zero and ends the path for some reason. For example the table is filled with the following: all 0s, except for [1][2], [1][3], [2][4], [2][5], [3][5], [4][6], [5][6], [6][0] which all have a 1 in them. The desired paths are P1: 1 2 4 6 0 P2: 1 3 5 6 0 P3: 1 2 5 6 0. The output I get when i run the program is 12460 13560 1250 124 Any and all help on this is much appreciated, this is just the function that scans through the array looking for paths, I can add the entire program if that would be helpful. Thanks.. void find_path(int map[][MAX], int x){ int path =0; int m=1; int blah=0; bool path_found = false; do { for(int n=0;n<(x+1);n++){ if(map[m][n]==-1){ blah=(n+1); if(blah<(x+1)){ for(blah;blah<(x+1);blah++){ if(map[m][blah]==1){ map[m][blah]=-1; path=m; path_found = true; cout<<path; m=blah; n=0; } } } else{ path=m; path_found=false; cout<<path; m=n; if(m==0){ path=0; cout<<path<<endl; m=1; path_found=false; } } } else if(map[m][n]==1){ map[m][n]=-1; path=m; path_found = true; cout<<path; m=n; if(m==0){ path=0; cout<<path<<endl; m=1; path_found=false; } } } } while(m<(x+1) && path_found); }

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  • Number Random Shuffle

    - by stjowa
    Hi, I need a Javascript random number shuffler for my website. Seems simple, but I can not figure out how to do it. Can anyone help me out? I have the following array of numbers: 1 2 3 4 5 6 7 8 9 I would like to be able to have these numbers shuffled randomly. Like the following: 3 6 4 2 9 5 1 8 7 or 4 1 7 3 5 9 2 6 8 So, specifically, I would like a function that takes in an array of numbers (1 - n) and then returns that same array of numbers - shuffled randomly with different calls to the function. Maybe a noob function, but can't seem to figure it out. Thanks! NOTE: Thanks for the clarification on "Shuffle". Have found a lot more online about this with that term.

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  • Substring and its reverse in a string

    - by christa
    My professor was talking about this in a Dynamic programming class and asked us to think over it. She gave us some examples as well. Given a string, we were to find the longest continuous subsequence whose reverse is also a subsequence present in the given string. Example: INPUT: pqrstuvtsrv OUTPUT: i=3, k=2 rst -> tsr (rst found first at i=3 and for 2 more positions) INPUT: mpqrsrqp OUTPUT: i=2, k=6 pqrsrqp in reverse INPUT: mmpqssss OUTPUT: i=5, k=3 I thought of putting the string and its reverse into 2 different arrays and comparing character by character. But I'm sure this is not the best way to do it. Any suggestions as to what could be the most efficient ?

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  • Pong: How does the paddle know where the ball will hit?

    - by Roflcoptr
    After implementing Pacman and Snake I'm implementing the next very very classic game: Pong. The implementation is really simple, but I just have one little problem remaining. When one of the paddle (I'm not sure if it is called paddle) is controlled by the computer, I have trouble to position it at the correct position. The ball has a current position, a speed (which for now is constant) and a direction angle. So I could calculate the position where it will hit the side of the computer controlled paddle. And so Icould position the paddle right there. But however in the real game, there is a probability that the computer's paddle will miss the ball. How can I implement this probability? If I only use a probability of lets say 0.5 that the computer's paddle will hit the ball, the problem is solved, but I think it isn't that simple. From the original game I think the probability depends on the distance between the current paddle position and the position the ball will hit the border. Does anybody have any hints how exactly this is calculated?

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  • 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?

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  • 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?

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  • Permutations of Varying Size

    - by waiwai933
    I'm trying to write a function in PHP that gets all permutations of all possible sizes. I think an example would be the best way to start off: $my_array = array(1,1,2,3); Possible permutations of varying size: 1 1 // * See Note 2 3 1,1 1,2 1,3 // And so forth, for all the sets of size 2 1,1,2 1,1,3 1,2,1 // And so forth, for all the sets of size 3 1,1,2,3 1,1,3,2 // And so forth, for all the sets of size 4 Note: I don't care if there's a duplicate or not. For the purposes of this example, all future duplicates have been omitted. What I have so far in PHP: function getPermutations($my_array){ $permutation_length = 1; $keep_going = true; while($keep_going){ while($there_are_still_permutations_with_this_length){ // Generate the next permutation and return it into an array // Of course, the actual important part of the code is what I'm having trouble with. } $permutation_length++; if($permutation_length>count($my_array)){ $keep_going = false; } else{ $keep_going = true; } } return $return_array; } The closest thing I can think of is shuffling the array, picking the first n elements, seeing if it's already in the results array, and if it's not, add it in, and then stop when there are mathematically no more possible permutations for that length. But it's ugly and resource-inefficient. Any pseudocode algorithms would be greatly appreciated. Also, for super-duper (worthless) bonus points, is there a way to get just 1 permutation with the function but make it so that it doesn't have to recalculate all previous permutations to get the next? For example, I pass it a parameter 3, which means it's already done 3 permutations, and it just generates number 4 without redoing the previous 3? (Passing it the parameter is not necessary, it could keep track in a global or static). The reason I ask this is because as the array grows, so does the number of possible combinations. Suffice it to say that one small data set with only a dozen elements grows quickly into the trillions of possible combinations and I don't want to task PHP with holding trillions of permutations in its memory at once.

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  • LZMA for Delphi

    - by SaCi
    I got a LZMA library on 7-zip site, but that didn't worked. I'm not using files, just stream. And for some why the library on 7-zip site just write the header on the stream but not compress the stream. Some one faced the some problem ? Have some example ? Know other LZMA library for Delphi ? Tks

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  • Time complexity of a certain program

    - by HokageSama
    In a discussion with my friend i am not able to predict correct and tight time complexity of a program. Program is as below. /* This Function reads input array "input" and update array "output" in such a way that B[i] = index value of nearest greater value from A[i], A[i+1] ... A[n], for all i belongs to [1, n] Time Complexity: ?? **/ void createNearestRightSidedLargestArr(int* input, int size, int* output){ if(!input || size < 1) return; //last element of output will always be zero, since no element is present on its right. output[size-1] = -1; int curr = size - 2; int trav; while(curr >= 0){ if(input[curr] < input[curr + 1]){ output[curr] = curr + 1; curr--; continue; } trav = curr + 1; while( input[ output [trav] ] < input[curr] && output [trav] != -1) trav = output[trav]; output[curr--] = output[trav]; } } I said time complexity is O(n^2) but my friend insists that this is not correct. What is the actual time complexity?

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  • 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.

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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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  • How to determine visibility in 2D

    - by Jack
    Hello, I'm developing an AI sandbox and I would like to calculate what every living entity can see. The rule is to simply hide what's behind the edges of the shapes from the point of view of the entity. The image clarifies everything: I need it either as an input to the artificial intelligence either graphically, to show it for a specific entity while it moves.. Any cool ideas?

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  • how to manage a "resource" array efficiently

    - by Haiyuan Zhang
    The senario of my question is that one need to use a fixed size of array to keep track of certain number of "objects" . The object here can be as simply as a integer or as complex as very fancy data structure. And "keep track" here means to allocate one object when other part of the app need one instance of object and recyle it for future allocation when one instance of the object is returned .Finally ,let me use c++ to put my problme in a more descriptive way . #define MAX 65535 /* 65535 just indicate that many items should be handled . performance demanding! */ typedef struct { int item ; }Item_t; Item_t items[MAX] ; class itemManager { private : /* up to you.... */ public : int get() ; /* get one index to a free Item_t in items */ bool put(int index) ; /* recyle one Item_t indicate by one index in items */ } how will you implement the two public functions of itemManager ? it's up to you to add any private member .

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  • How to get predecessor and successors from an adjacency matrix

    - by NickTFried
    Hi I am am trying to complete an assignment, where it is ok to consult the online community. I have to create a graph class that ultimately can do Breadth First Search and Depth First Search. I have been able to implement those algorithms successfully however another requirement is to be able to get the successors and predecessors and detect if two vertices are either predecessors or successors for each other. I'm having trouble thinking of a way to do this. I will post my code below, if anyone has any suggestions it would be greatly appreciated. import java.util.ArrayList; import java.util.Iterator; import java.util.LinkedList; import java.util.Queue; import java.util.Stack; public class Graph<T> { public Vertex<T> root; public ArrayList<Vertex<T>> vertices=new ArrayList<Vertex<T>>(); public int[][] adjMatrix; int size; private ArrayList<Vertex<T>> dfsArrList; private ArrayList<Vertex<T>> bfsArrList; public void setRootVertex(Vertex<T> n) { this.root=n; } public Vertex<T> getRootVertex() { return this.root; } public void addVertex(Vertex<T> n) { vertices.add(n); } public void removeVertex(int loc){ vertices.remove(loc); } public void addEdge(Vertex<T> start,Vertex<T> end) { if(adjMatrix==null) { size=vertices.size(); adjMatrix=new int[size][size]; } int startIndex=vertices.indexOf(start); int endIndex=vertices.indexOf(end); adjMatrix[startIndex][endIndex]=1; adjMatrix[endIndex][startIndex]=1; } public void removeEdge(Vertex<T> v1, Vertex<T> v2){ int startIndex=vertices.indexOf(v1); int endIndex=vertices.indexOf(v2); adjMatrix[startIndex][endIndex]=1; adjMatrix[endIndex][startIndex]=1; } public int countVertices(){ int ver = vertices.size(); return ver; } /* public boolean isPredecessor( Vertex<T> a, Vertex<T> b){ for() return true; }*/ /* public boolean isSuccessor( Vertex<T> a, Vertex<T> b){ for() return true; }*/ public void getSuccessors(Vertex<T> v1){ } public void getPredessors(Vertex<T> v1){ } private Vertex<T> getUnvisitedChildNode(Vertex<T> n) { int index=vertices.indexOf(n); int j=0; while(j<size) { if(adjMatrix[index][j]==1 && vertices.get(j).visited==false) { return vertices.get(j); } j++; } return null; } public Iterator<Vertex<T>> bfs() { Queue<Vertex<T>> q=new LinkedList<Vertex<T>>(); q.add(this.root); printVertex(this.root); root.visited=true; while(!q.isEmpty()) { Vertex<T> n=q.remove(); Vertex<T> child=null; while((child=getUnvisitedChildNode(n))!=null) { child.visited=true; bfsArrList.add(child); q.add(child); } } clearVertices(); return bfsArrList.iterator(); } public Iterator<Vertex<T>> dfs() { Stack<Vertex<T>> s=new Stack<Vertex<T>>(); s.push(this.root); root.visited=true; printVertex(root); while(!s.isEmpty()) { Vertex<T> n=s.peek(); Vertex<T> child=getUnvisitedChildNode(n); if(child!=null) { child.visited=true; dfsArrList.add(child); s.push(child); } else { s.pop(); } } clearVertices(); return dfsArrList.iterator(); } private void clearVertices() { int i=0; while(i<size) { Vertex<T> n=vertices.get(i); n.visited=false; i++; } } private void printVertex(Vertex<T> n) { System.out.print(n.label+" "); } }

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