Search Results

Search found 7490 results on 300 pages for 'algorithm analysis'.

Page 99/300 | < Previous Page | 95 96 97 98 99 100 101 102 103 104 105 106  | Next Page >

  • What's the "Hello World!" of genetic algorithms good for?

    - by JohnIdol
    I found this very cool C++ sample , literally the "Hello World!" of genetic algorithms. I so decided to re-code the whole thing in C# and this is the result. Now I am asking myself: is there any practical application along the lines of generating a target string starting from a population of random strings? EDIT: my buddy on twitter just tweeted that "is useful for transcription type things such as translation. Does not have to be Monkey's". I wish I had a clue.

    Read the article

  • Given an even number of vertices, how to find an optimum set of pairs based on proximity?

    - by Alex Z
    The problem: We have a set of n vertices in 3D euclidean space, and there is an even number of these vertices. We want to pair them up based on their proximity. In other words, we'd like to be able to find a set of vertex pairs, where the vertices in each pair are as close as possible together. We want to minimise sacrificing the proximity between the vertices of any other pairs as much as possible in doing this. I am not looking for the most optimal solution (if it even strictly exists/can be done), just a reasonable one that can be computed relatively quickly. A relatively awful brute force approach involves choosing a vertex and looping through the rest to find its nearest neighbor and then repeating until there are none left. Of course as we near the end of the list the closest vertex could be very far away, but it is the only choice, therefore this can fail badly on the third point above.

    Read the article

  • A graph problem

    - by copperhead
    I am struggling to solve the following problem http://uva.onlinejudge.org/external/1/193.html However Im not able to get a fast solution. And as seen by the times of others, there should be a solution of maximum n^2 complexity http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&page=show_problem&problemid=129&page=problem_stats Can I get some help?

    Read the article

  • Convert arbitrary size of byte[] to BigInteger[] and then safely convert back to exactly the same by

    - by PatlaDJ
    I believe conversion exactly to BigInteger[] would be optimal in my case. Anyone had done or found this written in Java and willing to share? So imagine I have arbitrary size byte[] = {0xff,0x3e,0x12,0x45,0x1d,0x11,0x2a,0x80,0x81,0x45,0x1d,0x11,0x2a,0x80,0x81} How do I convert it to array of BigInteger's and then be able to recover it back the original byte array safely? ty in advance.

    Read the article

  • Clamping a vector to a minimum and maximum?

    - by user146780
    I came accross this: t = Clamp(t/d, 0, 1) but I'm not sure how to perform this operation on a vector. What are the steps to clamp a vector if one was writing their own vector implementation? Thanks clamp clamping a vector to a minimum and a maximum ex: pc = # the point you are coloring now p0 = # start point p1 = # end point v = p1 - p0 d = Length(v) v = Normalize(v) # or Scale(v, 1/d) v0 = pc - p0 t = Dot(v0, v) t = Clamp(t/d, 0, 1) color = (start_color * t) + (end_color * (1 - t))

    Read the article

  • How to select number of lines from large text files?

    - by MiNdFrEaK
    I was wondering how to select number of lines from a certain text file. As an example: I have a text file containing the following lines: branch 27 : rect id 23400 rect: -115.475609 -115.474907 31.393650 31.411301 branch 28 : rect id 23398 rect: -115.474907 -115.472282 31.411301 31.417351 branch 29 : rect id 23396 rect: -115.472282 -115.468033 31.417351 31.427151 branch 30 : rect id 23394 rect: -115.468033 -115.458733 31.427151 31.438181 Non-Leaf Node: level=1 count=31 address=53 branch 0 : rect id 42 rect: -115.768539 -106.251556 31.425039 31.717550 branch 1 : rect id 50 rect: -109.559479 -106.009361 31.296721 31.775299 branch 2 : rect id 51 rect: -110.937401 -106.226143 31.285870 31.771971 branch 3 : rect id 54 rect: -109.584412 -106.069092 31.285240 31.775230 branch 4 : rect id 56 rect: -109.570961 -106.000954 31.296721 31.780769 branch 5 : rect id 58 rect: -115.806213 -106.366188 31.400450 31.687519 branch 6 : rect id 59 rect: -113.173859 -106.244057 31.297440 31.627750 branch 7 : rect id 60 rect: -115.811478 -106.278252 31.400450 31.679470 branch 8 : rect id 61 rect: -109.953888 -106.020111 31.325319 31.775270 branch 9 : rect id 64 rect: -113.070969 -106.015968 31.331841 31.704750 branch 10 : rect id 68 rect: -113.065689 -107.034576 31.326300 31.770809 branch 11 : rect id 71 rect: -112.333344 -106.059860 31.284081 31.662920 branch 12 : rect id 73 rect: -115.071083 -106.309677 31.267879 31.466850 branch 13 : rect id 74 rect: -116.094414 -106.286308 31.236290 31.424770 branch 14 : rect id 75 rect: -115.423264 -106.286308 31.229691 31.415510 branch 15 : rect id 76 rect: -116.111656 -106.313110 31.259390 31.478300 branch 16 : rect id 77 rect: -116.247467 -106.309677 31.240231 31.451799 branch 17 : rect id 78 rect: -116.170792 -106.094543 31.156429 31.391781 branch 18 : rect id 79 rect: -116.225723 -106.292709 31.239960 31.442850 branch 19 : rect id 80 rect: -116.268013 -105.769913 31.157240 31.378111 branch 20 : rect id 82 rect: -116.215424 -105.827202 31.198441 31.383421 branch 21 : rect id 83 rect: -116.095734 -105.826439 31.197460 31.373819 branch 22 : rect id 84 rect: -115.423264 -105.815018 31.182640 31.368891 branch 23 : rect id 85 rect: -116.221527 -105.776512 31.160931 31.389830 branch 24 : rect id 86 rect: -116.203369 -106.473831 31.168350 31.367611 branch 25 : rect id 87 rect: -115.727631 -106.501587 31.189100 31.395941 branch 26 : rect id 88 rect: -116.237289 -105.790756 31.164780 31.358959 branch 27 : rect id 89 rect: -115.791344 -105.990044 31.072620 31.349529 branch 28 : rect id 90 rect: -115.736847 -106.495079 31.187969 31.376900 branch 29 : rect id 91 rect: -115.721710 -106.000130 31.160351 31.354601 branch 30 : rect id 92 rect: -115.792236 -106.000793 31.166620 31.378811 Leaf Node: level=0 count=21 address=42 branch 0 : rect id 18312 rect: -106.412270 -106.401367 31.704750 31.717550 branch 1 : rect id 18288 rect: -106.278252 -106.253387 31.520321 31.548361 I just want those lines which are in between Non-Leaf Node level=1 to Leaf Node Level=0 and also there are a lot of segments like this and I need them all.

    Read the article

  • Converting to a column oriented array in Java

    - by halfwarp
    Although I have Java in the title, this could be for any OO language. I'd like to know a few new ideas to improve the performance of something I'm trying to do. I have a method that is constantly receiving an Object[] array. I need to split the Objects in this array through multiple arrays (List or something), so that I have an independent list for each column of all arrays the method receives. Example: List<List<Object>> column-oriented = new ArrayList<ArrayList<Object>>(); public void newObject(Object[] obj) { for(int i = 0; i < obj.length; i++) { column-oriented.get(i).add(obj[i]); } } Note: For simplicity I've omitted the initialization of objects and stuff. The code I've shown above is slow of course. I've already tried a few other things, but would like to hear some new ideas. How would you do this knowing it's very performance sensitive?

    Read the article

  • question about api functions

    - by davit-datuashvili
    i have question we have API functions in java can user create it's own function and add to his java IDE? for example i am using netbeans can i create my own function add to netbean IDE?let say create binary function or something else thanks

    Read the article

  • Programming Contest Question: Counting Polyominos

    - by Martijn Courteaux
    Hi, An example question for a programming contest was to write a program that finds out how much polyominos are possible with a given number of stones. So for two stones (n = 2) there is only one polyominos: XX You might think this is a second solution: X X But it isn't. The polyominos are not unique if you can rotate them. So, for 4 stones (n = 4), there are 7 solutions: X X XX X X X X X X XX X XX XX XX X X X XX X X XX The application has to be able to find the solution for 1 <= n <=10 PS: Using the list of polyominos on Wikipedia isn't allowed ;) EDIT: Of course the question is: How to do this in Java, C/C++, C#

    Read the article

  • How to make disconnected closed curves connected by adding a shortest path using MATLAB?

    - by user198729
    bwlabel can be used to get disconnected objects in an image: [L Ne] = bwlabel(image); I want to make the objects(But my target is only the contours(closed curve) of these objects) connected by adding a shortest path where necessary. How do I approach this? UPDATE Or how to dilate the closed curves so that they get connected? How to calculate the shortest path between two disconnected closed curves?

    Read the article

  • Simple/Basic steganography algorithms and methods

    - by tomp
    What are the basic and simpliest steganography algorithms and methods? I mean the steganography applied to images. How does simple program that hides data to images work? How does the program recognize the encrypted message in image without the source image? What are the main techniques used?

    Read the article

  • How to find largest common sub-tree in the given two binary search trees?

    - by Bhushan
    Two BSTs (Binary Search Trees) are given. How to find largest common sub-tree in the given two binary trees? EDIT 1: Here is what I have thought: Let, r1 = current node of 1st tree r2 = current node of 2nd tree There are some of the cases I think we need to consider: Case 1 : r1.data < r2.data 2 subproblems to solve: first, check r1 and r2.left second, check r1.right and r2 Case 2 : r1.data > r2.data 2 subproblems to solve: - first, check r1.left and r2 - second, check r1 and r2.right Case 3 : r1.data == r2.data Again, 2 cases to consider here: (a) current node is part of largest common BST compute common subtree size rooted at r1 and r2 (b)current node is NOT part of largest common BST 2 subproblems to solve: first, solve r1.left and r2.left second, solve r1.right and r2.right I can think of the cases we need to check, but I am not able to code it, as of now. And it is NOT a homework problem. Does it look like?

    Read the article

  • Managed code (C#) vs Matlabs and C++ for speed

    - by dotnetdev
    Hi, I am about to start developing an edge detection system (once I've read through a couple of books, which I'm doing so at good speed), but one thing I am wondering is the speed of an app like Matlabs (which can compile code to C++) vs AFORGE.NET for edge detecton. Is unmanaged code generally faster? Thanks

    Read the article

  • how to elegantly duplicate a graph (neural network)

    - by macias
    I have a graph (network) which consists of layers, which contains nodes (neurons). I would like to write a procedure to duplicate entire graph in most elegant way possible -- i.e. with minimal or no overhead added to the structure of the node or layer. Or yet in other words -- the procedure could be complex, but the complexity should not "leak" to structures. They should be no complex just because they are copyable. I wrote the code in C#, so far it looks like this: neuron has additional field -- copy_of which is pointer the the neuron which base copied from, this is my additional overhead neuron has parameterless method Clone() neuron has method Reconnect() -- which exchanges connection from "source" neuron (parameter) to "target" neuron (parameter) layer has parameterless method Clone() -- it simply call Clone() for all neurons network has parameterless method Clone() -- it calls Clone() for every layer and then it iterates over all neurons and creates mappings neuron=copy_of and then calls Reconnect to exchange all the "wiring" I hope my approach is clear. The question is -- is there more elegant method, I particularly don't like keeping extra pointer in neuron class just in case of being copied! I would like to gather the data in one point (network's Clone) and then dispose it completely (Clone method cannot have an argument though).

    Read the article

  • Solving simultaneous equations

    - by Milo
    Here is my problem: Given x, y, z and ratio where z is known and ratio is known and is a float representing a relative value, I need to find x and y. I know that: x / y == ratio y - x == z What I'm trying to do is make my own scroll pane and I'm figuring out the scrollbar parameters. So for example, If the scrollbar must be able to scroll 100 values (z) and the thumb must consume 80% of the bar (ratio = 0.8) then x would be 400 and y would be 500. Thanks

    Read the article

  • How do I find all paths through a set of given nodes in a DAG?

    - by Hanno Fietz
    I have a list of items (blue nodes below) which are categorized by the users of my application. The categories themselves can be grouped and categorized themselves. The resulting structure can be represented as a Directed Acyclic Graph (DAG) where the items are sinks at the bottom of the graph's topology and the top categories are sources. Note that while some of the categories might be well defined, a lot is going to be user defined and might be very messy. Example: On that structure, I want to perform the following operations: find all items (sinks) below a particular node (all items in Europe) find all paths (if any) that pass through all of a set of n nodes (all items sent via SMTP from example.com) find all nodes that lie below all of a set of nodes (intersection: goyish brown foods) The first seems quite straightforward: start at the node, follow all possible paths to the bottom and collect the items there. However, is there a faster approach? Remembering the nodes I already passed through probably helps avoiding unnecessary repetition, but are there more optimizations? How do I go about the second one? It seems that the first step would be to determine the height of each node in the set, as to determine at which one(s) to start and then find all paths below that which include the rest of the set. But is this the best (or even a good) approach? The graph traversal algorithms listed at Wikipedia all seem to be concerned with either finding a particular node or the shortest or otherwise most effective route between two nodes. I think both is not what I want, or did I just fail to see how this applies to my problem? Where else should I read?

    Read the article

  • Time Complexities of recursive algorithms

    - by Peter
    Whenever I see a recursive solution, or I write recursive code for a problem, it is really difficult for me to figure out the time complexity, in most of the cases I just say its exponential? How is it exponential actually? How people say it is 2^n, when it is n!, when it is n^n or n^k. I have some questions in mind, let say find all permutations of a string (O(n!)) find all sequences which sum up to k in an array (exponential, how exactly do I calculate). Find all subsets of size k whose sum is 0 (will k come somewhere in complexity , it should come right?). Can any1 help me how to calculate the exact complexity of such questions, I am able to wrote code for them , but its hard understanding the exact time complexity.

    Read the article

  • algorithm q: Fuzzy matching of structured data

    - by user86432
    I have a fairly small corpus of structured records sitting in a database. Given a tiny fraction of the information contained in a single record, submitted via a web form (so structured in the same way as the table schema), (let us call it the test record) I need to quickly draw up a list of the records that are the most likely matches for the test record, as well as provide a confidence estimate of how closely the search terms match a record. The primary purpose of this search is to discover whether someone is attempting to input a record that is duplicate to one in the corpus. There is a reasonable chance that the test record will be a dupe, and a reasonable chance the test record will not be a dupe. The records are about 12000 bytes wide and the total count of records is about 150,000. There are 110 columns in the table schema and 95% of searches will be on the top 5% most commonly searched columns. The data is stuff like names, addresses, telephone numbers, and other industry specific numbers. In both the corpus and the test record it is entered by hand and is semistructured within an individual field. You might at first blush say "weight the columns by hand and match word tokens within them", but it's not so easy. I thought so too: if I get a telephone number I thought that would indicate a perfect match. The problem is that there isn't a single field in the form whose token frequency does not vary by orders of magnitude. A telephone number might appear 100 times in the corpus or 1 time in the corpus. The same goes for any other field. This makes weighting at the field level impractical. I need a more fine-grained approach to get decent matching. My initial plan was to create a hash of hashes, top level being the fieldname. Then I would select all of the information from the corpus for a given field, attempt to clean up the data contained in it, and tokenize the sanitized data, hashing the tokens at the second level, with the tokens as keys and frequency as value. I would use the frequency count as a weight: the higher the frequency of a token in the reference corpus, the less weight I attach to that token if it is found in the test record. My first question is for the statisticians in the room: how would I use the frequency as a weight? Is there a precise mathematical relationship between n, the number of records, f(t), the frequency with which a token t appeared in the corpus, the probability o that a record is an original and not a duplicate, and the probability p that the test record is really a record x given the test and x contain the same t in the same field? How about the relationship for multiple token matches across multiple fields? Since I sincerely doubt that there is, is there anything that gets me close but is better than a completely arbitrary hack full of magic factors? Barring that, has anyone got a way to do this? I'm especially keen on other suggestions that do not involve maintaining another table in the database, such as a token frequency lookup table :). This is my first post on StackOverflow, thanks in advance for any replies you may see fit to give.

    Read the article

  • BFS algorithm problem

    - by Gorkamorka
    The problem is as follows: A wanderer begins on the grid coordinates (x,y) and wants to reach the coordinates (0,0). From every gridpoint, the wanderer can go 8 steps north OR 3 steps south OR 5 steps east OR 6 steps west (8N/3S/5E/6W). How can I find the shortest route from (X,Y) to (0,0) using breadth-first search? Clarifications: Unlimited grid Negative coordinates are allowed A queue (linked list or array) must be used No obstacles present

    Read the article

  • Count Occurence of Needle String in Haystack String, most optimally?

    - by Taranfx
    The Problem is simple Find "ABC" in "ABCDSGDABCSAGAABCCCCAAABAABC" Here is the solution I propose, I'm looking for any solutions that might be better than this one. public static void main(String[] args) { String haystack = "ABCDSGDABCSAGAABCCCCAAABAABC"; String needle = "ABC"; char [] needl = needle.toCharArray(); int needleLen = needle.length(); int found=0; char hay[] = haystack.toCharArray(); int index =0; int chMatched =0; for (int i=0; i<hay.length; i++){ if (index >= needleLen || chMatched==0) index=0; System.out.print("\nchar-->"+hay[i] + ", with->"+needl[index]); if(hay[i] == needl[index]){ chMatched++; System.out.println(", matched"); }else { chMatched=0; index=0; if(hay[i] == needl[index]){ chMatched++; System.out.print("\nchar->"+hay[i] + ", with->"+needl[index]); System.out.print(", matched"); }else continue; } if(chMatched == needleLen){ found++; System.out.println("found. Total ->"+found); } index++; } System.out.println("Result Found-->"+found); } It took me a while creating this one. Can someone suggest a better solution (if any) P.S. Drop the sysouts if they look messy to you.

    Read the article

< Previous Page | 95 96 97 98 99 100 101 102 103 104 105 106  | Next Page >