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  • How to detect a image texture?

    - by Ole Jak
    So we have a photo like this How to detect that a red wall has a white figure painted on it and that that white figure is a texture and than how to cut that wall from the picture? I need an algorithm for performing such operation programaticly (not by hand)

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  • Estimating the size of a tree

    - by Full Decent
    I'd like to estimate the number of leaves in a large tree structure for which I can't visit every node exhaustively. Is this algorithm appropriate? Does it have a name? Also, please pedant if I am using any terms improperly. sum_trials = 0 num_trials = 0 WHILE time_is_not_up bits = 0 ptr = tree.root WHILE count(ptr.children) > 0 bits += log2(count(ptr.children)) ptr = ptr.children[rand()%count(ptr.children)] sum_trials += bits num_trials++ estimated_tree_size = 2^(sum_trials/num_trials)

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  • Else without if

    - by user2808951
    I'm trying to write a code for my computer programming class for a project due Monday, and I'm pretty new to Java, but I'm trying to write a program that will first determine if a number the user inputs is even or odd and then determine if the number is prime or not. I'm not sure if I did the algorithm right or not, so if anyone has any corrections on the program to my algorithm or anything else please say so, but my real issue is that the program is refusing to compile. Every time I try, it says it's having an else without if problem. Here's a link to my command box: http://s1341.photobucket.com/user/Emi_Nightshade/media/Capture_zps45f9a2ea.png.html Here's my code: import java.io.*; import java.util.*; public class Lesson9p1_ThuotteEmily { public static void main(String args[]) { Scanner kbReader0=new Scanner(System.in); System.out.print("\n\nPlease enter an integer. An integer is whole number, and it can be either negative or positive. Please enter your number: "); long num=kbReader0.nextLong(); if(num%2==0) //if and else with braces { System.out.println("Your integer " + num + " is even."); } else { System.out.println("Your integer " + num + " is odd."); } Scanner kbReader1=new Scanner(System.in); System.out.print("\n\nWould you like to know if your number is prime? Please enter yes or no: "); String yn=kbReader1.nextLine(); if(yn.equals.IgnoreCase("Yes")) { System.out.println("Okay. Give me a moment."); { if(num%2==0) { System.out.println("Your number isn't prime."); } else if(num==2) { System.out.println("Your number is 2, which is the only even prime number in existence. Cool, right?"); } for(int i=3;i*i<=n;i+=2) { if(n%1==0) { System.out.println("Your number isn't prime."); } } else { System.out.println("Your number is prime!"); } } } if(yn.equals.IgnoreCase("No")) { System.out.println("Okay."); } } } If anyone could help me out with this and also any problems I may have made elsewhere in the program, I'd be very grateful! Thanks.

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  • Calculate a set of concatenated sets of n sets

    - by Andras Zoltan
    Okay - I'm not even sure that the term is right - and I'm sure there is bound to be a term for this - but I'll do my best to explain. This is not quite a cross product here, and the order of the results are absolutely crucial. Given: IEnumerable<IEnumerable<string>> sets = new[] { /* a */ new[] { "a", "b", "c" }, /* b */ new[] { "1", "2", "3" }, /* c */ new[] { "x", "y", "z" } }; Where each inner enumerable represents an instruction to produce a set of concatenations as follows (the order here is important): set a* = new string[] { "abc", "ab", "a" }; set b* = new string[] { "123", "12", "1" }; set c* = new string[] { "xyz", "xy", "x" }; I want to produce set ordered concatenations as follows: set final = new string { a*[0] + b*[0] + c*[0], /* abc123xyz */ a*[0] + b*[0] + c*[1], /* abc123xy */ a*[0] + b*[0] + c*[2], /* abc123x */ a*[0] + b*[0], /* abc123 */ a*[0] + b*[1] + c*[0], /* abc12xyz */ a*[0] + b*[1] + c*[1], /* abc12xy */ a*[0] + b*[1] + c*[2], /* abc12x */ a*[0] + b*[1], /* abc12 */ a*[0] + b*[2] + c*[0], /* abc1xyz */ a*[0] + b*[2] + c*[1], /* abc1xy */ a*[0] + b*[2] + c*[2], /* abc1x */ a*[0] + b*[2], /* abc1 */ a*[0], /* abc */ a*[1] + b*[0] + c*[0], /* ab123xyz */ /* and so on for a*[1] */ /* ... */ a*[2] + b*[0] + c*[0], /* a123xyz */ /* and so on for a*[2] */ /* ... */ /* now lop off a[*] and start with b + c */ b*[0] + c*[0], /* 123xyz */ /* rest of the combinations of b + c with b on its own as well */ /* then finally */ c[0], c[1], c[2]}; So clearly, there are going to be a lot of combinations! I can see similarities with Numeric bases (since the order is important as well), and I'm sure there are permutations/combinations lurking in here too. The question is - how to write an algorithm like this that'll cope with any number of sets of strings? Linq, non-Linq; I'm not fussed. Why am I doing this? Indeed, why!? In Asp.Net MVC - I want to have partial views that can be redefined for a given combination of back-end/front-end culture and language. The most basic of these would be, for a given base view View, we could have View-en-GB, View-en, View-GB, and View, in that order of precedence (recognising of course that the language/culture codes could be the same, so some combinations might be the same - a Distinct() will solve that). But I also have other views that, in themselves, have other possible combinations before culture is even taken into account (too long to go into - but the fact is, this algo will enable a whole bunch of really cool that I want to offer my developers!). I want to produce a search list of all the acceptable view names, iterate through the whole lot until the most specific match is found (governed by the order that this algo will produce these concatenations in) then serve up the resolved Partial View. The result of the search can later be cached to avoid the expense of running the algorithm all the time. I already have a really basic version of this working that just has one enumerable of strings. But this is a whole different kettle of seafood! Any help greatly appreciated.

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  • Filling in gaps for outlines

    - by user146780
    I'm using an algorithm to generate quads. These become outlines. The algorithm is: void OGLENGINEFUNCTIONS::GenerateLinePoly(const std::vector<std::vector<GLdouble>> &input, std::vector<GLfloat> &output, int width) { output.clear(); if(input.size() < 2) { return; } int temp; float dirlen; float perplen; POINTFLOAT start; POINTFLOAT end; POINTFLOAT dir; POINTFLOAT ndir; POINTFLOAT perp; POINTFLOAT nperp; POINTFLOAT perpoffset; POINTFLOAT diroffset; POINTFLOAT p0, p1, p2, p3; for(unsigned int i = 0; i < input.size() - 1; ++i) { start.x = static_cast<float>(input[i][0]); start.y = static_cast<float>(input[i][1]); end.x = static_cast<float>(input[i + 1][0]); end.y = static_cast<float>(input[i + 1][1]); dir.x = end.x - start.x; dir.y = end.y - start.y; dirlen = sqrt((dir.x * dir.x) + (dir.y * dir.y)); ndir.x = static_cast<float>(dir.x * 1.0 / dirlen); ndir.y = static_cast<float>(dir.y * 1.0 / dirlen); perp.x = dir.y; perp.y = -dir.x; perplen = sqrt((perp.x * perp.x) + (perp.y * perp.y)); nperp.x = static_cast<float>(perp.x * 1.0 / perplen); nperp.y = static_cast<float>(perp.y * 1.0 / perplen); perpoffset.x = static_cast<float>(nperp.x * width * 0.5); perpoffset.y = static_cast<float>(nperp.y * width * 0.5); diroffset.x = static_cast<float>(ndir.x * 0 * 0.5); diroffset.y = static_cast<float>(ndir.y * 0 * 0.5); // p0 = start + perpoffset - diroffset //p1 = start - perpoffset - diroffset //p2 = end + perpoffset + diroffset // p3 = end - perpoffset + diroffset p0.x = start.x + perpoffset.x - diroffset.x; p0.y = start.y + perpoffset.y - diroffset.y; p1.x = start.x - perpoffset.x - diroffset.x; p1.y = start.y - perpoffset.y - diroffset.y; p2.x = end.x + perpoffset.x + diroffset.x; p2.y = end.y + perpoffset.y + diroffset.y; p3.x = end.x - perpoffset.x + diroffset.x; p3.y = end.y - perpoffset.y + diroffset.y; output.push_back(p2.x); output.push_back(p2.y); output.push_back(p0.x); output.push_back(p0.y); output.push_back(p1.x); output.push_back(p1.y); output.push_back(p3.x); output.push_back(p3.y); } } The problem is that there are then gaps as seen here: http://img816.imageshack.us/img816/2882/eeekkk.png There must be a way to fix this. I see a pattern but I just cant figure it out. There must be a way to fill the missing inbetweens. Thanks

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  • BFS Shortest Path: Edge weight either 1 or 2

    - by Hackster
    I am trying to implement a shortest path algorithm using BFS. That is I am trying to find the shortest path from a specified vertex to every other vertex. However, its a special case where all edge weights are either 1 or 2. I know it could be done with Dijkstra's algorithm but I must use Breadth First Search. So far I have a working version of BFS that searches first for a vertex connected with an edge of weight 1. If it cannot find it, then returns a vertex connected with an edge of weight 2. After thinking about it, this is not the correct way to find the shortest path. The problem is I cannot think of any reasoning why BFS would work with weights 1 or 2, as opposed to any weight. Here is the code: public void addEdge(int start, int end, int weight) { adjMat[start][end] = 1; adjMat[end][start] = 1; edge_weight[start][end] = weight; edge_weight[end][start] = weight; } // ------------------------------------------------------------- public void bfs() // breadth-first search { // begin at vertex 0 vertexList[0].wasVisited = true; // mark it displayVertex(0); // display it theQueue.insert(0); // insert at tail int v2; while( !theQueue.isEmpty() ) // until queue empty, { int v1 = theQueue.remove(); // remove vertex at head // until it has no unvisited neighbors while( (v2=getAdjUnvisitedVertex(v1)) != -1 ){// get one, vertexList[v2].wasVisited = true; // mark it displayVertex(v2); // display it theQueue.insert(v2); // insert it } } // end while(queue not empty) // queue is empty, so we're done for(int j=0; j<nVerts; j++) // reset flags vertexList[j].wasVisited = false; } // end bfs() // ------------------------------------------------------------- // returns an unvisited vertex adj to v -- ****WITH WEIGHT 1**** public int getAdjUnvisitedVertex(int v) { for (int j = 0; j < nVerts; j++) if (adjMat[v][j] == 1 && vertexList[j].wasVisited == false && edge_weight[v][j] == 1){ //System.out.println("Vertex found with 1:"+ vertexList[j].label); return j; } for (int k = 0; k < nVerts; k++) if (adjMat[v][k] == 1 && vertexList[k].wasVisited == false && edge_weight[v][k] == 2){ //System.out.println("Vertex found with 2:"+vertexList[k].label); return k; } return -1; } // end getAdjUnvisitedVertex() // ------------------------------------------------------------- } //////////////////////////////////////////////////////////////// public class BFS{ public static void main(String[] args) { Graph theGraph = new Graph(); theGraph.addVertex('A'); // 0 (start for bfs) theGraph.addVertex('B'); // 1 theGraph.addVertex('C'); // 2 theGraph.addEdge(0, 1,2); // AB theGraph.addEdge(1, 2,1); // BC theGraph.addEdge(2, 0,1); // AD System.out.print("Visits: "); theGraph.bfs(); // breadth-first search System.out.println(); } // end main() } The problem then is, that I don't know why BFS can work for the shortest path problem with edges of weight 1 or 2 as opposed to any edges of any weight. Any help is appreciated. Thanks!

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  • Can't get Jacobi algorithm to work in Objective-C

    - by Chris Long
    Hi, For some reason, I can't get this program to work. I've had other CS majors look at it and they can't figure it out either. This program performs the Jacobi algorithm (you can see step-by-step instructions and a MATLAB implementation here). BTW, it's different from the Wikipedia article of the same name. Since NSArray is one-dimensional, I added a method that makes it act like a two-dimensional C array. After running the Jacobi algorithm many times, the diagonal entries in the NSArray (i[0][0], i[1][1], etc.) are supposed to get bigger and the others approach 0. For some reason though, they all increase exponentially. For instance, i[2][4] should equal 0.0000009, not 9999999, while i[2][2] should be big. Thanks in advance, Chris NSArray+Matrix.m @implementation NSArray (Matrix) @dynamic offValue, transposed; - (double)offValue { double sum = 0.0; for ( MatrixItem *item in self ) if ( item.nonDiagonal ) sum += pow( item.value, 2.0 ); return sum; } - (NSMutableArray *)transposed { NSMutableArray *transpose = [[[NSMutableArray alloc] init] autorelease]; int i, j; for ( i = 0; i < 5; i++ ) { for ( j = 0; j < 5; j++ ) { [transpose addObject:[self objectAtRow:j andColumn:i]]; } } return transpose; } - (id)objectAtRow:(NSUInteger)row andColumn:(NSUInteger)column { NSUInteger index = 5 * row + column; return [self objectAtIndex:index]; } - (NSMutableArray *)multiplyWithMatrix:(NSArray *)array { NSMutableArray *result = [[NSMutableArray alloc] init]; int i = 0, j = 0, k = 0; double value; for ( i = 0; i < 5; i++ ) { value = 0.0; for ( j = 0; j < 5; j++ ) { for ( k = 0; k < 5; k++ ) { MatrixItem *firstItem = [self objectAtRow:i andColumn:k]; MatrixItem *secondItem = [array objectAtRow:k andColumn:j]; value += firstItem.value * secondItem.value; } MatrixItem *item = [[MatrixItem alloc] initWithValue:value]; item.row = i; item.column = j; [result addObject:item]; } } return result; } @end Jacobi_AlgorithmAppDelegate.m // ... - (void)jacobiAlgorithmWithEntry:(MatrixItem *)entry { MatrixItem *b11 = [matrix objectAtRow:entry.row andColumn:entry.row]; MatrixItem *b22 = [matrix objectAtRow:entry.column andColumn:entry.column]; double muPlus = ( b22.value + b11.value ) / 2.0; muPlus += sqrt( pow((b22.value - b11.value), 2.0) + 4.0 * pow(entry.value, 2.0) ); Vector *u1 = [[[Vector alloc] initWithX:(-1.0 * entry.value) andY:(b11.value - muPlus)] autorelease]; [u1 normalize]; Vector *u2 = [[[Vector alloc] initWithX:-u1.y andY:u1.x] autorelease]; NSMutableArray *g = [[[NSMutableArray alloc] init] autorelease]; for ( int i = 0; i <= 24; i++ ) { MatrixItem *item = [[[MatrixItem alloc] init] autorelease]; if ( i == 6*entry.row ) item.value = u1.x; else if ( i == 6*entry.column ) item.value = u2.y; else if ( i == ( 5*entry.row + entry.column ) || i == ( 5*entry.column + entry.row ) ) item.value = u1.y; else if ( i % 6 == 0 ) item.value = 1.0; else item.value = 0.0; [g addObject:item]; } NSMutableArray *firstResult = [[g.transposed multiplyWithMatrix:matrix] autorelease]; matrix = [firstResult multiplyWithMatrix:g]; } // ...

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  • Can I configure Wndows NDES server to use Triple DES (3DES) algorithm for PKCS#7 answer encryption?

    - by O.Shevchenko
    I am running SCEP client to enroll certificates on NDES server. If OpenSSL is not in FIPS mode - everything works fine. In FIPS mode i get the following error: pkcs7_unwrap():pkcs7.c:708] error decrypting inner PKCS#7 139968442623728:error:060A60A3:digital envelope routines:FIPS_CIPHERINIT:disabled for fips:fips_enc.c:142: 139968442623728:error:21072077:PKCS7 routines:PKCS7_decrypt:decrypt error:pk7_smime.c:557: That's because NDES server uses DES algorithm to encrypt returned PKCS#7 packet. I used the following debug code: /* Copy enveloped data from PKCS#7 */ bytes = BIO_read(pkcs7bio, buffer, sizeof(buffer)); BIO_write(outbio, buffer, bytes); p7enc = d2i_PKCS7_bio(outbio, NULL); /* Get encryption PKCS#7 algorithm */ enc_alg=p7enc->d.enveloped->enc_data->algorithm; evp_cipher=EVP_get_cipherbyobj(enc_alg->algorithm); printf("evp_cipher->nid = %d\n", evp_cipher->nid); The last string always prints: evp_cipher-nid = 31 defined in openssl-1.0.1c/include/openssl/objects.h #define SN_des_cbc "DES-CBC" #define LN_des_cbc "des-cbc" #define NID_des_cbc 31 I use 3DES algorithm for PKCS7 requests encryption in my code (pscep.enc_alg = (EVP_CIPHER *)EVP_des_ede3_cbc()) and NDES server accepts these requests, but it always returns answer encrypted with DES. Can I configure Wndows NDES server to use Triple DES (3DES) algorithm for PKCS#7 answer encryption?

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  • how to add bouncycastle algorithm to android

    - by Vamsi
    Hi friends, I am trying to write a small application using bouncycastle algorithm, from the http://tinyurl.com/ylclavn (BouncyCastleProvider.java) it says we have to import and add the provider during runtime by the following code import org.bouncycastle.jce.provider.BouncyCastleProvider; Security.addProvider(new BouncyCastleProvider()); error - The import org.bouncycastle cannot be resolved; during import error - BouncyCastleProvider cannot be resolved to a type; when calling addProvider I though bouncycastle is not provided with the Android 1.6 SDK, so thought of installing separately. how should i do this? If Bouncycastle is shipped along with SDK, what should i do to avoid these errors? I am using Android 1.6, eclipse-V3.4.0 on winXP . Thanks in advance

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  • reservoir sampling problem: correctness of proof

    - by eSKay
    This MSDN article proves the correctness of Reservoir Sampling algorithm as follows: Base case is trivial. For the k+1st case, the probability a given element i with position <= k is in R is s/k. The probability i is replaced is the probability k+1st element is chosen multiplied by i being chosen to be replaced, which is: s/(k+1) * 1/s = 1/(k+1), and prob that i is not replaced is k/k+1. So any given element's probability of lasting after k+1 rounds is: (chosen in k steps, and not removed in k steps) = s/k * k/(k+1), which is s/(k+1). So, when k+1 = n, any element is present with probability s/n. about step 3: What are the k+1 rounds mentioned? What is chosen in k steps, and not removed in k steps? Why are we only calculating this probability for elements that were already in R after the first s steps?

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  • Pathfinding Algorithm For Pacman

    - by user280454
    Hi, I wanted to implement the game Pacman. For the AI, I was thinking of using the A* algorithm, having seen it on numerous forums. However, I implemented the Breadth First Search for some simple pathfinding (going from point a to point b with certain obstacles in between) and found it gave the optimum path always. I guess it might be because in a game like pacman which uses simple pathfinding, there is no notion of costs in the graph. So, will it be OK if I use BFS instead of A* for pathfinding in Pacman?

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  • Formula for calculating Exotic wagers such as Trifecta and Superfecta

    - by JohnnyCantCode
    I am trying to create an application that will calculate the cost of exotic parimutuel wager costs. I have found several for certain types of bets but never one that solves all the scenarios for a single bet type. If I could find an algorithm that could calculate all the possible combinations I could use that formula to solve my other problems. Additional information: I need to calculate the permutations of groups of numbers. For instance; Group 1 = 1,2,3 Group 2 = 2,3,4 Group 3 = 3,4,5 What are all the possible permutation for these 3 groups of numbers taking 1 number from each group per permutation. No repeats per permutation, meaning a number can not appear in more that 1 position. So 2,4,3 is valid but 2,4,4 is not valid. Thanks for all the help.

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  • Brute force characters into a textbox in c#

    - by Fred Dunly
    Hey everyone, I am VERY new to programming and the only language I know is C# So I will have to stick with that... I want to make a program that "test passwords" to see how long they would take to break with a basic brute force attack. So what I did was make 2 text boxes. (textbox1 and textbox2) and wrote the program so if the text boxes had the input, a "correct password" label would appear, but i want to write the program so that textbox2 will run a brute force algorithm in it, and when it comes across the correct password, it will stop. I REALLY need help, and if you could just post my attached code with the correct additives in it that would be great. The program so far is extremely simple, but I am very new to this, so. Thanks in advance. private void textBox2_TextChanged(object sender, EventArgs e) { } private void button1_Click(object sender, EventArgs e) { if (textBox2.Text == textBox1.Text) { label1.Text = "Password Correct"; } else { label1.Text = "Password Wrong"; } } private void label1_Click(object sender, EventArgs e) { } } } `

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  • Text indexing algorithm

    - by Majd
    I am writing a C# winform application for an archiving system. The system has a huge database where some tables would have more than 1.5 million records. What i need is an algorithm that indexes the content of these records. Mainly, the files are Microsoft office, PDF and TXT documents. anyone can help? whether with ideas, links, books or codes, I appreciate it :) example: if i search for the word "international" in a certain folder in the database, i get all the files that contain that word ordered by a certain criteria such as relevance, modifying date...etc

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  • ClassNotFoundException error in implementing Bayesian algorithm in Apache Mahout on Hadoop

    - by Shweta
    Hi, I have a problem in executing the Bayesian algorithm in Mahout. I built it with Maven and the job file is in target directory. When run from terminal using hadoop, I'm getting the ClassNotFoundException error. What should be done? $HADOOP_HOME/bin/hadoop jar mahout-core-0.3-SNAPSHOT.job org.apache.mahout.classifier.bayes.mapreduce.bayes.bayesdriver -i test -o output Exception in thread "main" java.lang.ClassNotFoundException: org.apache.mahout.classifier.bayes.mapreduce.bayes.bayesdriver at java.net.URLClassLoader$1.run(URLClassLoader.java:200) at java.security.AccessController.doPrivileged(Native Method) at java.net.URLClassLoader.findClass(URLClassLoader.java:188) at java.lang.ClassLoader.loadClass(ClassLoader.java:307) at java.lang.ClassLoader.loadClass(ClassLoader.java:252) at java.lang.ClassLoader.loadClassInternal(ClassLoader.java:320) at java.lang.Class.forName0(Native Method) at java.lang.Class.forName(Class.java:247) at org.apache.hadoop.util.RunJar.main(RunJar.java:149)

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  • Finding all the shortest paths between two nodes in unweighted directed graphs using BFS algorithm

    - by andra-isan
    Hi All, I am working on a problem that I need to find all the shortest path between two nodes in a given directed unweighted graph. I have used BFS algorithm to do the job, but unfortunately I can only print one shortest path not all of them, for example if they are 4 paths having lenght 3, my algorithm only prints the first one but I would like it to print all the four shortest paths. I was wondering in the following code, how should I change it so that all the shortest paths between two nodes could be printed out? class graphNode{ public: int id; string name; bool status; double weight;}; map<int, map<int,graphNode>* > graph; int Graph::BFS(graphNode &v, graphNode &w){ queue <int> q; map <int, int> map1; // this is to check if the node has been visited or not. std::string str= ""; map<int,int> inQ; // just to check that we do not insert the same iterm twice in the queue map <int, map<int, graphNode>* >::iterator pos; pos = graph.find(v.id); if(pos == graph.end()) { cout << v.id << " does not exists in the graph " <<endl; return 1; } int parents[graph.size()+1]; // this vector keeps track of the parents for the node parents[v.id] = -1; // there is a direct path between these two words, simply print that path as the shortest path if (findDirectEdge(v.id,w.id) == 1 ){ cout << " Shortest Path: " << v.id << " -> " << w.id << endl; return 1; } //if else{ int gn; map <int, map<int, graphNode>* >::iterator pos; q.push(v.id); inQ.insert(make_pair(v.id, v.id)); while (!q.empty()){ gn = q.front(); q.pop(); map<int, int>::iterator it; cout << " Popping: " << gn <<endl; map1.insert(make_pair(gn,gn)); //backtracing to print all the nodes if gn is the same as our target node such as w.id if (gn == w.id){ int current = w.id; cout << current << " - > "; while (current!=v.id){ current = parents[current]; cout << current << " -> "; } cout <<endl; } if ((pos = graph.find(gn)) == graph.end()) { cout << " pos is empty " <<endl; continue; } map<int, graphNode>* pn = pos->second; map<int, graphNode>::iterator p = pn->begin(); while(p != pn->end()) { map<int, int>::iterator it; //map1 keeps track of the visited nodes it = map1.find(p->first); graphNode gn1= p->second; if (it== map1.end()) { map<int, int>::iterator it1; //if the node already exits in the inQ, we do not insert it twice it1 = inQ.find(p->first); if (it1== inQ.end()){ parents[p->first] = gn; cout << " inserting " << p->first << " into the queue " <<endl; q.push(p->first); // add it to the queue } //if } //if p++; } //while } //while } I do appreciate all your great help Thanks, Andra

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  • algorithm to combinatorics

    - by peiska
    I am trying to solve a combinatorics problem, it seems easy, but i am having some trouble with it. If i have at most X tables, and N persons to sit on the tables, Each table can have 1 to N seating places, and I can only sit persons in one side of a rectangular table( so the order how people sit matters). I want to make a code that can calculate all the distributions of seating places from 1 up to K tables. For example, if I have 12 persons and 1 table i have 479001600 ways of seating persons( thats easy to calculate I've used Factorial of 12). But if I have 12 persons and 3 tables i have 4390848000 ways of seating persons. I've tried different solutions but i was not able to find the correct one. I've tried to divided the 12 in 3, then o use factorial of the result (it didnt work), i've tried to use 12! * 3( it didn't work too). Can some one give me a tip in a algorithm that i can use?

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  • How to speed up calculation of length of longest common substring?

    - by eSKay
    I have two very large strings and I am trying to find out their Longest Common Substring. One way is using suffix trees (supposed to have a very good complexity, though a complex implementation), and the another is the dynamic programming method (both are mentioned on the Wikipedia page linked above). Using dynamic programming The problem is that the dynamic programming method has a huge running time (complexity is O(n*m), where n and m are lengths of the two strings). What I want to know (before jumping to implement suffix trees): Is it possible to speed up the algorithm if I only want to know the length of the common substring (and not the common substring itself)?

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  • linear interpolation on 8bit microcontroller

    - by JB
    I need to do a linear interpolation over time between two values on an 8 bit PIC microcontroller (Specifically 16F627A but that shouldn't matter) using PIC assembly language. Although I'm looking for an algorithm here as much as actual code. I need to take an 8 bit starting value, an 8 bit ending value and a position between the two (Currently represented as an 8 bit number 0-255 where 0 means the output should be the starting value and 255 means it should be the final value but that can change if there is a better way to represent this) and calculate the interpolated value. Now PIC doesn't have a divide instruction so I could code up a general purpose divide routine and effectivly calculate (B-A)/(x/255)+A at each step but I feel there is probably a much better way to do this on a microcontroller than the way I'd do it on a PC in c++ Has anyone got any suggestions for implementing this efficiently on this hardware?

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  • Ranking based string matching algorithm..for Midi Music

    - by Taha
    i am working on midi music project. What i am trying to do is:- matching the Instrument midi track with the similar instrument midi track... for example Flute track in a some midi music is matched against the Flute track in some other music midi file... After matching ,the results should come ranking wise according to their similarity.. Like 1) track1 2) track2 3) track3 I have this sort of string coming from my midi music .. F4/0.01282051282051282E4/0.01282051282051282Eb4/0.01282051282051282 D4/0.01282051282051282C#4/0.01282051282051282C4/0.01282051282051282 Which ranking algorithm with good metrics should i use for such data ? Thanking you in anticipation!

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  • reservoir sampling problem

    - by eSKay
    This MSDN article proves the correctness of Reservoir Sampling algorithm as follows: Base case is trivial. For the k+1st case, the probability a given element i with position <= k is in R is s/k. The probability i is replaced is the probability k+1st element is chosen multiplied by i being chosen to be replaced, which is: s/(k+1) * 1/s = 1/(k+1), and prob that i is not replaced is k/k+1. So any given element's probability of lasting after k+1 rounds is: (chosen in k steps, and not removed in k steps) = s/k * k/(k+1), which is s/(k+1). So, when k+1 = n, any element is present with probability s/n. about step 3: What are the k+1 rounds mentioned? What is chosen in k steps, and not removed in k steps? Why are we only calculating this probability for elements that were already in R after the first s steps?

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