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  • RSA Factorization problem

    - by dada
    At class we found this programming problem, and currently, we have no idea how to solve it. The positive integer n is given. It is known that n = p * q, where p and q are primes, p<=q and |q-k*p|<10^5 for some given positive integer k. You must find p and q. Input: 35 1 121 1 1000730021 9 Output: 5 * 7 11 * 11 10007 * 100003 It's not a homework, we are just trying to solve some interesting problems. If you have some ideas, please post them here so we can try something, thanks.

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  • General Address Parser for Freeform Text

    - by Daemonic
    We have a program that displays map data (think Google Maps, but with much more interactivity and custom layers for our clients). We allow navigation via a set of combo boxes that prefill certain fields with a bunch of data (ie: Country: Canada, the Province field is filled in. Select Ontario, and a list of Counties/Regions is filled in. Select a county/region, and a city is filled in, etc...). While this guarantees accurate addresses, it's a pain for the users if they don't know where a street address or a city are located (ie, which county/region is kitchener in?). So we are looking at trying to do an address parser with a freeform text field. The user could enter something like this (similar to Google Maps, Bing Maps, etc...): 22 Main St, Kitchener, On And we could compartmentalize it into sections and do lookups on the data and get to the point they are looking for (or suggest alternatives). The problem with this is that how do we properly compartmentalize information? How do we break up the sections and find possible matches? I'm guessing we wouldn't be guaranteed that the user would enter data in a format we always expected (obviously). A follow up to this would be how to present the data if we don't find an exact match (or find multiple exact matches... two cities with the same street name in different counties, for example). We have a ton of data available in the mapping data (mapinfo tab format mostly). So we can do quick scans of street names, cities, states, etc. But I'm not sure about the best way to go about approaching this problem. Sure, using Google Maps would be nice, bue most of our clients are in closed in networks where outside access is not usually allowed and most aren't willing to rely on google maps (since it doesn't contain as much information as they need, such as custom map layers). They could, obviously, go to google and get the proper location then move to our software, but this would time consuming and speed of the process can be quite important.

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  • How to find nth element from the end of a singly linked list?

    - by Codenotguru
    The following function is trying to find the nth to last element of a singly linked list. For example: If the elements are 8->10->5->7->2->1->5->4->10->10 then the result is 7th to last node is 7. Can anybody help me on how this code is working or is there a better and simpler approach? LinkedListNode nthToLast(LinkedListNode head, int n) { if (head == null || n < 1) { return null; } LinkedListNode p1 = head; LinkedListNode p2 = head; for (int j = 0; j < n - 1; ++j) { // skip n-1 steps ahead if (p2 == null) { return null; // not found since list size < n } p2 = p2.next; } while (p2.next != null) { p1 = p1.next; p2 = p2.next; } return p1; }

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  • determine if intersection of a set with conjunction of two other sets is empty

    - by koen
    For any three given sets A, B and C: is there a way to determine (programmatically) whether there is an element of A that is part of the conjunction of B and C? example: A: all numbers greater than 3 B: all numbers lesser than 7 C: all numbers that equal 5 In this case there is an element in set A, being the number 5, that fits. I'm implementing this as specifications, so this numerical range is just an example. A, B, C could be anything.

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  • red black tree balancing?

    - by Anirudh Kaki
    i am working to generate tango tree, where i need to check whether every sub tree in tango is balanced or not. if its not balanced i need to make it balance? I trying so hard to make entire RB-tree balance but i not getting any proper logic so can any one help me out?? here i am adding code to check how to find my tree is balanced are not but when its not balanced how can i make it balance. static boolean verifyProperty5(rbnode n) { int left = 0, right = 0; if (n != null) { bh++; left = blackHeight(n.left, 0); right = blackHeight(n.right, 0); } if (left == right) { System.out.println("black height is :: " + bh); return true; } else { System.out.println("in balance"); return false; } } public static int blackHeight(rbnode root, int len) { bh = 0; blackHeight(root, path1, len); return bh; } private static void blackHeight(rbnode root, int path1[], int len) { if (root == null) return; if (root.color == "black"){ root.black_count = root.parent.black_count+1; } else{ root.black_count = root.parent.black_count; } if ((root.left == null) && (root.right == null)) { bh = root.black_count; } blackHeight(root.left, path1, len); blackHeight(root.right, path1, len); }

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  • maintaing a sorted list that is bigger than memory

    - by tcurdt
    I have a list of tuples. [ "Bob": 3, "Alice: 2, "Jane": 1, ] When incrementing the counts "Alice" += 2 the order should be maintained: [ "Alice: 4, "Bob": 3, "Jane": 1, ] When all is in memory there rather simple ways (some more or some less) to efficiently implement this. (using an index, insert-sort etc) The question though is: What's the most promising approach when the list does not fit into memory. Bonus question: What if not even the index fits into memory? How would you approach this?

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  • Find three numbers appeared only once

    - by shilk
    In a sequence of length n, where n=2k+3, that is there are k unique numbers appeared twice and three numbers appeared only once. The question is: how to find the three unique numbers that appeared only once? for example, in sequence 1 1 2 6 3 6 5 7 7 the three unique numbers are 2 3 5. Note: 3<=n<1e6 and the number will range from 1 to 2e9 Memory limits: 1000KB , this implies that we can't store the whole sequence. Method I have tried(Memory limit exceed): I initialize a tree, and when read in one number I try to remove it from the tree, if the remove returns false(not found), I add it to the tree. Finally, the tree has the three numbers. It works, but is Memory limit exceed. I know how to find one or two such number(s) using bit manipulation. So I wonder if we can find three using the same method(or some method similar)? Method to find one/two number(s) appeared only once: If there is one number appeared only once, we can apply XOR to the sequence to find it. If there are two, we can first apply XOR to the sequence, then separate the sequence into 2 parts by one bit of the result that is 1, and again apply XOR to the 2 parts, and we will find the answer.

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  • How Can I Improve This Algorithm (LCS)

    - by superguay
    (define (lcs lst1 lst2) (define (except-last-pair list) (if (pair? (cdr list)) (cons (car list) (except-last-pair (cdr list))) '())) (define (car-last-pair list) (if (pair? (cdr list)) (car-last-pair (cdr list)) (car list))) (if (or (null? lst1) (null? lst2)) null (if (= (car-last-pair lst1) (car-last-pair lst2)) (append (lcs (except-last-pair lst1) (except-last-pair lst2)) (cons (car-last-pair lst1) '())) **(if (> (length (lcs lst1 (except-last-pair lst2))) (length (lcs lst2 (except-last-pair lst1)))) (lcs lst1 (except-last-pair lst2)) (lcs lst2 (except-last-pair lst1)))))) I dont want it to run over and over.. Regards, Superguay

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  • Zoom image to pixel level

    - by zaf
    For an art project, one of the things I'll be doing is zooming in on an image to a particular pixel. I've been rubbing my chin and would love some advice on how to proceed. Here are the input parameters: Screen: sw - screen width sh - screen height Image: iw - image width ih - image height Pixel: px - x position of pixel in image py - y position of pixel in image Zoom: zf - zoom factor (0.0 to 1.0) Background colour: bc - background colour to use when screen and image aspect ratios are different Outputs: The zoomed image (no anti-aliasing) The screen position/dimensions of the pixel we are zooming to. When zf is 0 the image must fit the screen with correct aspect ratio. When zf is 1 the selected pixel fits the screen with correct aspect ratio. One idea I had was to use something like povray and move the camera towards a big image texture or some library (e.g. pygame) to do the zooming. Anyone think of something more clever with simple pseudo code? To keep it more simple you can make the image and screen have the same aspect ratio. I can live with that. I'll update with more info as its required.

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  • I am trying to build a list of limitations of all graph algorithms

    - by Jack
    Single Source shortest Path Dijkstra's - directed and undirected - works only for positive edge weights - cycles ?? Bellman Ford - directed - no cycles should exist All source shortest path Floyd Warshall - no info Minimum Spanning Tree ( no info about edge weights or nature of graph or cycles) Kruskal's Prim's - undirected Baruvka's

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  • Minimum cost strongly connected digraph

    - by Kazoom
    I have a digraph which is strongly connected (i.e. there is a path from i to j and j to i for each pair of nodes (i, j) in the graph G). I wish to find a strongly connected graph out of this graph such that the sum of all edges is the least. To put it differently, I need to get rid of edges in such a way that after removing them, the graph will still be strongly connected and of least cost for the sum of edges. I think it's an NP hard problem. I'm looking for an optimal solution, not approximation, for a small set of data like 20 nodes. Edit A more general description: Given a grap G(V,E) find a graph G'(V,E') such that if there exists a path from v1 to v2 in G than there also exists a path between v1 and v2 in G' and sum of each ei in E' is the least possible. so its similar to finding a minimum equivalent graph, only here we want to minimize the sum of edge weights rather than sum of edges. Edit: My approach so far: I thought of solving it using TSP with multiple visits, but it is not correct. My goal here is to cover each city but using a minimum cost path. So, it's more like the cover set problem, I guess, but I'm not exactly sure. I'm required to cover each and every city using paths whose total cost is minimum, so visiting already visited paths multiple times does not add to the cost.

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  • CODE1 Spoj - cannot solve it

    - by VaioIsBorn
    I am trying to solve the problem Secret Code and it's obviously math problem. The full problem For those who are lazy to go and read, it's like this: a0,a1,a2,...,an - sequence of N numbers B - some number known to us X = a0 + a1*B + a2*(B^2) + a3*(B^3) + ... + an*(B^n) So if you are given B and X, you should find a0,a1,..an. I don't know how or where to start, because not even N is known, just X and B. Can you help me ?

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  • Easiest way of checking if a string consists of unique characters?

    - by serg555
    I need to check in Java if a word consists of unique letters (case insensitive). As straight solution is boring, I came up with: For every char in a string check if indexOf(char) == lastIndexOf(char). Add all chars to HashSet and check if set size == string length. Convert a string to a char array, sort it alphabetically, loop through array elements and check if c[i] == c[i+1]. Currently I like #2 the most, seems like the easiest way. Any other interesting solutions?

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  • Brackets matching using BIT

    - by amit.codename13
    edit: I was trying to solve a spoj problem. Here is the link to the problem : http://spoj.pl/problems/BRCKTS I can think of two possible data structures for solving the problem one using segment tree and the other using a BIT. I have already implemented the solution using a segment tree. I have read about BIT but i can't figure out how to do a particular thing with it(which i have mentioned below) I am trying to check if brackets are balanced in a given string containing only ('s or )'s. I am using a BIT(Binary indexed tree) for solving the problem. The procedure i am following is as follows: I am taking an array of size equal to the number of characters in the string. I am assigning -1 for ) and 1 for ( to the corresponding array elements. Brackets are balanced in the string only if the following two conditions are true. The cumulative sum of the whole array is zero. Minimum cumulative sum is non negative. i.e the minimum of cumulative sums of all the prefixes of the array is non-negative. Checking condition 1 using a BIT is trivial. I am facing problem in checking condition 2.

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  • How to check if a integer is sum of given integers?

    - by p3trix
    Lets say I have a integer result and an array of integers, lets say [a,b,c] (not a fixed length). I need to detect if result=a*i +b*j + c*k, with i,j,k=0. I prefer a solution in C/C# if it is possible. PS The problem is from a reservation system, a trip can be sold if its durations is a combination of given durations. Thanks!

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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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  • what is the idea behind scaling an image using lanczos?

    - by banister
    Hi, I'm interested in image scaling algorithms and have implemented the bilinear and bicubic methods. However, I have heard of the lanczos and other more sophisticated methods for even higher quality image scaling and I am very curious how they work. Could someone here explain the basic idea behind scaling an image using lanczos (both upscaling and downscaling) and why it results in higher quality? I do have a background in fourier analysis and have done some signal processing stuff in the past, but not with relation to image processing, so don't be afraid to use terms like "frequency response" and such in your answer :) EDIT: I guess what i really want to know is the concept and theory behind using a convolution filter for interpolation. (Note: i have already read the wikipedia article on lanczos resampling but it didn't have nearly enough detail for me) thanks alot!

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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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  • Test if single linked list is circular by traversing it only once

    - by user1589754
    I am a fresher and I was asked this question in a recent interview I gave. The question was --- By traversing each element of linked list just once find if the single linked list is circular at any point. To this I answered that we will store reference of each node while traversing the list in another linked list and for every node in the list being tested we will find if the reference exists in the list I am storing the references. The interviewer said that he needs a more optimized way to solve this problem. Can anyone please tell me what would be a more optimized method to solve this problem.

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  • STL vector reserve() and copy()

    - by natersoz
    Greetings, I am trying to perform a copy from one vector (vec1) to another vector (vec2) using the following 2 abbreviated lines of code (full test app follows): vec2.reserve( vec1.size() ); copy(vec1.begin(), vec1.end(), vec2.begin()); While the call to vec2 sets the capacity of vector vec2, the copying of data to vec2 seems to not fill in the values from vec1 to vec2. Replacing the copy() function with calls to push_back() works as expected. What am I missing here? Thanks for your help. vectest.cpp test program followed by resulting output follows. Compiler: gcc 3.4.4 on cygwin. Nat /** * vectest.cpp */ #include <iostream> #include <vector> using namespace std; int main() { vector<int> vec1; vector<int> vec2; vec1.push_back(1); vec1.push_back(2); vec1.push_back(3); vec1.push_back(4); vec1.push_back(5); vec1.push_back(6); vec1.push_back(7); vec2.reserve( vec1.size() ); copy(vec1.begin(), vec1.end(), vec2.begin()); cout << "vec1.size() = " << vec1.size() << endl; cout << "vec1.capacity() = " << vec1.capacity() << endl; cout << "vec1: "; for( vector<int>::const_iterator iter = vec1.begin(); iter < vec1.end(); ++iter ) { cout << *iter << " "; } cout << endl; cout << "vec2.size() = " << vec2.size() << endl; cout << "vec2.capacity() = " << vec2.capacity() << endl; cout << "vec2: "; for( vector<int>::const_iterator iter = vec2.begin(); iter < vec2.end(); ++iter ) { cout << *iter << endl; } cout << endl; } output: vec1.size() = 7 vec1.capacity() = 8 vec1: 1 2 3 4 5 6 7 vec2.size() = 0 vec2.capacity() = 7 vec2:

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