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  • Seaching for an element in a circular sorted array

    - by guirgis
    I wanted to share this with you, i had this problem in a google interview. we want to search for a given element in a circular sorted array in complexity not greater than O(Log n). ex: search for 13 in {5,9,13,1,3}. My idea was to convert the circular array into a regular sorted array then do a binary search on the resulting array, but my problem was the algorithm i came up was stupid that it takes O(n) in the worst case: for(i = 1; i < a.length; i++){ if (a[i] < a[i-1]){ minIndex = i; break; } } then the corresponding index of ith element will be determined from the following relation: (i + minInex - 1) % a.length it is clear that my conversion (from circular to regular) algorithm may take O(n), so we need a better one, any suggestions?

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  • All minimum spanning trees implementation

    - by russtbarnacle
    I've been looking for an implementation (I'm using networkx library.) that will find all the minimum spanning trees (MST) of an undirected weighted graph. I can only find implementations for Kruskal's Algorithm and Prim's Algorithm both of which will only return a single MST. I've seen papers that address this problem (such as http://fano.ics.uci.edu/cites/Publication/Epp-TR-95-50.html) but my head tends to explode someway through trying to think how to translate it to code. In fact i've not been able to find an implementation in any language!

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  • Evenly distribute items on the screen

    - by abolotnov
    I am trying to solve this little puzzle (the algorithm): I have N image icons and I want to distribute them evenly on users screen. Say, I put them in a table. If there is one image, there will be one cell in a table. If two - one row with two columns, if three - one row and three columns, if four - two rows, two columns... and so on until row space is gone and since then the table should only grow in columns without adding extra rows. I'm trying to figure an algorithm for this and perhaps this is something that has a solution already somewhere? My attempt is so far something like this: obtain_max_rows() obtain_visible_columns() if (number_of_pictures > max_rows*max_columns) { columns = roundup(number_of_pictures/max_rows) for(max_rows){generate row;for columns{generate column}} } else { **here comes to trouble...** } This logic is bit silly though - it somehow needs to think cases where there are 12 pictures on first screen and 2 on the other trying to balance it say 8/6 or somehow like that.

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  • Is it a solvable problem to generate a regular expression that matches some input set?

    - by Roman
    I provide some input set which contains known separated number of text blocks. I want to make a program that automatically generate 1 or more regular expressions each of which matches every text block in the input set. I see some relatively easy ways to implement a brute-force search. But I'm not an expert in compilers theory. That's why I'm curious: 1) is this problem solvable? or there are some principle impossibility to make such algorithm? 2) is it possible to achieve polynomial complexity for this algorithm and avoid brute forcing?

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  • A simple algorithm for polygon intersection

    - by Elazar Leibovich
    I'm looking for a very simple algorithm for computing the polygon intersection/clipping. That is, given polygons P, Q, I wish to find polygon T which is contained in P and in Q, and I wish T to be maximal among all possible polygons. I don't mind the run time (I have a few very small polygons), I can also afford getting an approximation of the polygons' intersection (that is, a polygon with less points, but which is still contained in the polygons' intersection). But it is really important for me that the algorithm will be simple (cheaper testing) and preferably short (less code). edit: please note, I wish to obtain a polygon which represent the intersection. I don't need only a boolean answer to the question of whether the two polygons intersect.

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  • Fastest way of converting a quad to a triangle strip?

    - by Tina Brooks
    What is the fastest way of converting a quadrilateral (made up of foyr x,y points) to a triangle strip? I'm well aware of the general triangulation algorithms that exist, but I need a short, well optimized algorithm that deals with quadrilaterals only. My current algorithm does this, which works for most quads but still gets the points mixed up for some: #define fp(f) bounds.p##f /* Sort four points in ascending order by their Y values */ point_sort4_y(&fp(1), &fp(2), &fp(3), &fp(4)); /* Bottom two */ if (fminf(-fp(1).x, -fp(2).x) == -fp(2).x) { out_quad.p1 = fp(2); out_quad.p2 = fp(1); } else { out_quad.p1 = fp(1); out_quad.p2 = fp(2); } /* Top two */ if (fminf(-fp(3).x, -fp(4).x) == -fp(3).x) { out_quad.p3 = fp(3); out_quad.p4 = fp(4); } else { out_quad.p3 = fp(4); out_quad.p4 = fp(3); }

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  • Given a number of rectangles that can be rotated, find an enclosing rectangle of minimum area

    - by efficiencyIsBliss
    So, I'm trying to implement an algorithm that takes in a number of rectangles as input and tries to pack them into a rectangle of minimum area. The rectangles can all be rotated by 90 degrees. I realize that this is similar to the bin packing problem, but I am unable to find a good algorithm that accounts for the rotation. I found a paper that discusses this at length here and while I understand the article itself, I was hoping to find something simpler. Any suggestions? -Edit- I think I misstated the problem earlier. We are given a number of rectangles, such that each can be rotated by 90 degrees. We need to find a rectangle that fits all the given rectangles such that no two rectangles overlap, while minimizing the area of the enclosing rectangle. The problem I face here is that we are asked to find the minimum, as opposed to being given an enclosing rectangle and checking if the given rectangles fit or something of that sort.

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  • Support-function in the GJK-algorithm.

    - by Marcus Johansson
    I am trying to implement the GJK-algorithm but I got stuck instantly. The problem is to implement the Support-function that isn't O(n^2). As it is now I'm computing the complete Minkowski difference, and then there is really no point in doing the GJK-algorithm. (or is it?) What I mean by Support-function is the function that returns the point in the Minkowski difference that is furthest away in a specified direction. I assume this shouldn't be O(n^2) as it is in my current implementation.

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  • Why increase pointer by two while finding loop in linked list, why not 3,4,5?

    - by GG
    I had a look at question already which talk about algorithm to find loop in a linked list. I have read Floyd's cycle-finding algorithm solution, mentioned at lot of places that we have to take two pointers. One pointer( slower/tortoise ) is increased by one and other pointer( faster/hare ) is increased by 2. When they are equal we find the loop and if faster pointer reaches null there is no loop in the linked list. Now my question is why we increase faster pointer by 2. Why not something else? Increasing by 2 is necessary or we can increase it by X to get the result. Is it necessary that we will find a loop if we increment faster pointer by 2 or there can be the case where we need to increment by 3 or 5 or x.

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  • Measuring the performance of classification algorithm

    - by Silver Dragon
    I've got a classification problem in my hand, which I'd like to address with a machine learning algorithm ( Bayes, or Markovian probably, the question is independent on the classifier to be used). Given a number of training instances, I'm looking for a way to measure the performance of an implemented classificator, with taking data overfitting problem into account. That is: given N[1..100] training samples, if I run the training algorithm on every one of the samples, and use this very same samples to measure fitness, it might stuck into a data overfitting problem -the classifier will know the exact answers for the training instances, without having much predictive power, rendering the fitness results useless. An obvious solution would be seperating the hand-tagged samples into training, and test samples; and I'd like to learn about methods selecting the statistically significant samples for training. White papers, book pointers, and PDFs much appreciated!

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  • Pure functional bottom up tree algorithm

    - by Axel Gneiting
    Say I wanted to write an algorithm working on an immutable tree data structure that has a list of leaves as its input. It needs to return a new tree with changes made to the old tree going upwards from those leaves. My problem is that there seems to be no way to do this purely functional without reconstructing the entire tree checking at leaves if they are in the list, because you always need to return a complete new tree as the result of an operation and you can't mutate the existing tree. Is this a basic problem in functional programming that only can be avoided by using a better suited algorithm or am I missing something?

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  • C optimization breaks algorithm

    - by Halpo
    I am programming an algorithm that contains 4 nested for loops. The problem is at at each level a pointer is updated. The innermost loop only uses 1 of the pointers. The algorithm does a complicated count. When I include a debugging statement that logs the combination of the indexes and the results of the count I get the correct answer. When the debugging statement is omitted, the count is incorrect. The program is compiled with the -O3 option on gcc. Why would this happen?

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  • optimized grid for rectangular items

    - by peterchen
    I have N rectangular items with an aspect ratio Aitem (X:Y). I have a rectangular display area with an aspect ratio Aview The items should be arranged in a table-like layout (i.e. r rows, c columns). what is the ideal grid rows x columns, so that individual items are largest? (rows * colums = N, of course - i.e. there may be "unused" grid places). A simple algorithm could iterate over rows = 1..N, calculate the required number of columns, and keep the row/column pair with the largest items. I wonder if there's a non-iterative algorithm, though (e.g. for Aitem = Aview = 1, rows / cols can be approximated by sqrt(N)).

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  • Divide and Conquer Algo to find maximum difference between two ordered elements

    - by instance
    Given an array arr[] of integers, find out the difference between any two elements such that larger element appears after the smaller number in arr[]. Examples: If array is [2, 3, 10, 6, 4, 8, 1, 7] then returned value should be 8 (Diff between 10 and 2). If array is [ 7, 9, 5, 6, 3, 2 ] then returned value should be 2 (Diff between 7 and 9) My Algorithm: I thought of using D&C algorithm. Explanation 2, 3, 10, 6, 4, 8, 1, 7 then 2,3,10,6 and 4,8,1,7 then 2,3 and 10,6 and 4,8 and 1,7 then 2 and 3 10 and 6 4 and 8 1 and 7 Here as these elements will remain in same order, i will get the maximum difference, here it's 6. Now i will move back to merege these arrays and again find the difference between minimum of first block and maximum of second block and keep doing this till end. I am not able to implement this in my code. can anyone please provide a pseudo code for this?

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  • Getting the most frequent items without counting every item

    - by DeadMonkeyWalkin
    Hi. I was wondering if there was an algorithm for counting "most frequent items" without having to keep a count of each item? For example, let's say I was a search engine and wanted to keep track of the 10 most popular searches. What I don't want to do is keep a counter of every query since there could be too many queries for me to count (and most them will be singletons). Is there a simple algorithm for this? Maybe something that is probabilistic? Thanks!

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  • Balanced Search Tree Query, Asymtotic Analysis..

    - by AGeek
    Hi, The situation is as follows:- We have n number and we have print them in sorted order. We have access to balanced dictionary data structure, which supports the operations serach, insert, delete, minimum, maximum each in O(log n) time. We want to retrieve the numbers in sorted order in O(n log n) time using only the insert and in-order traversal. The answer to this is:- Sort() initialize(t) while(not EOF) read(x) insert(x,t); Traverse(t); Now the query is if we read the elements in time "n" and then traverse the elements in "log n"(in-order traversal) time,, then the total time for this algorithm (n+logn)time, according to me.. Please explain the follow up of this algorithm for the time calculation.. How it will sort the list in O(nlogn) time?? Thanks.

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  • (Python) algorithm to randomly select a key based on proportionality/weight

    - by LaundroMat
    Hi - I'm a bit at a loss as to how to find a clean algorithm for doing the following: Suppose I have a dict k: >>> k = {'A': 68, 'B': 62, 'C': 47, 'D': 16, 'E': 81} I now want to randomly select one of these keys, based on the 'weight' they have in the total (i.e. sum) amount of keys. >>> sum(k.values()) >>> 274 So that there's a >>> 68.0/274.0 >>> 0.24817518248175183 24.81% percent change that A is selected. How would you write an algorithm that takes care of this? In other words, that makes sure that on 10.000 random picks, A will be selected 2.481 times?

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  • Construct A Polygon Out of Union of Many Polygons

    - by Ngu Soon Hui
    Supposed that I have many polygons, what is the best algorithm to construct a polygon--maybe with holes- out of the union of all those polygons? For my purpose, you can imagine each piece of a polygon as a jigsaw puzzle piece, when you complete them you will get a nice picture. But the catch is that a small portion <5% of the jigsaw is missing, and you are still require to form a picture as complete as possible; that's the polygon-- maybe with holes-- that I want to form. My naive approach is to take two polygons, union them, and take another polygon, union it with the union of the two polygons, and repeat this process until every single piece is union. Then I will run through the union polygon list and check whether there are still some polygons can be combined, and I will repeat this process until a satisfactory result is achieved. But this seems to be like an extremely naive approach. I just wonder is there any other better algorithm?

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  • Will Algorithm written in OCaml compiled from C be Faster than Algorithm written in Pure C code?

    - by Ole Jak
    So I have some cool Image Processing algorithm. I have written it in OCaml. It performs well. I now I can compile it as C code with such command ocamlc -output-obj -o foo.c foo.ml (I have a situation where I am not alowed to use OCaml compiler to bild my programm for my arcetecture, I can use only specialy modified gcc. so I will compile that programm with sometyhing like gcc -L/usr/lib/ocaml foo.c -lcamlrun -lm -lncurses and Itll run on my archetecture.) I want to know in general case will my OCaml code compiled into C run faster than algorithm implemented in pure C?

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  • How can I compute the average cost for this solution of the element uniqueness problem?

    - by Alceu Costa
    In the book Introduction to the Design & Analysis of Algorithms, the following solution is proposed to the element uniqueness problem: ALGORITHM UniqueElements(A[0 .. n-1]) // Determines whether all the elements in a given array are distinct // Input: An array A[0 .. n-1] // Output: Returns "true" if all the elements in A are distinct // and false otherwise. for i := 0 to n - 2 do for j := i + 1 to n - 1 do if A[i] = A[j] return false return true How can I compute the average cost (i.e. number of comparisons for a given n) for this algorithm? What is a reasonable assumption about the input?

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  • Autoclick security for a like button

    - by Ali Davut
    Hi everyone I want to develop a button like 'facebook like button'. I am going to use it on my website and thinking it to share as iframe like facebook but I cannot think its securty because someone can develop a script that can click on it automatically. I thought a solution using sessions but I couldn't make an algorithm completely. How can I disallow autoclicks and which solution is the best? It can be any language I just want algorithm. Thanks, have a nice day.

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