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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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  • Latent Dirichlet Allocation, pitfalls, tips and programs

    - by Gregg Lind
    I'm experimenting with Latent Dirichlet Allocation for topic disambiguation and assignment, and I'm looking for advice. Which program is the "best", where best is some combination of easiest to use, best prior estimation, fast How do I incorporate my intuitions about topicality. Let's say I think I know that some items in the corpus are really in the same category, like all articles by the same author. Can I add that into the analysis? Any unexpected pitfalls or tips I should know before embarking? I'd prefer is there are R or Python front ends for whatever program, but I expect (and accept) that I'll be dealing with C.

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  • what is order notation f(n)=O(g(n))?

    - by Lopa
    2 questions: question 1: under what circumstances would this[O(f(n))=O(k.f(n))] be the most appropriate form of time-complexity analysis? question 2: working from mathematical definition of O notation, show that O(f(n))=O(k.f(n)), for positive constant k.? My view: For the first one I think it is average case and worst case form of time-complexity. am i right? and what else do i write in that? for the second one I think we need to define the function mathematically, so is the answer something like because the multiplication by a constant just corresponds to a readjustment of value of the arbitrary constant 'k' in definition of O.

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  • How to calculate order (big O) for more complex algorithms (ie quicksort)

    - by bangoker
    I know there are quite a bunch of questions about big O notation, I have already checked Plain english explanation of Big O , Big O, how do you calculate/approximate it?, and Big O Notation Homework--Code Fragment Algorithm Analysis?, to name a few. I know by "intuition" how to calculate it for n, n^2, n! and so, however I am completely lost on how to calculate it for algorithms that are log n , n log n, n log log n and so. What I mean is, I know that Quick Sort is n log n (on average).. but, why? Same thing for merge/comb, etc. Could anybody explain me in a not to math-y way how do you calculate this? The main reason is that Im about to have a big interview and I'm pretty sure they'll ask for this kind of stuff. I have researched for a few days now, and everybody seem to have either an explanation of why bubble sort is n^2 or the (for me) unreadable explanation a la wikipedia Thanks!

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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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  • Using GA in GUI

    - by AlexT
    Sorry if this isn't clear as I'm writing this on a mobile device and I'm trying to make it quick. I've written a basic Genetic Algorithm with a binary encoding (genes) that builds a fitness value and evolves through several iterations using tournament selection, mutation and crossover. As a basic command-line example it seems to work. The problem I've got is with applying a genetic algorithm within a GUI as I am writing a maze-solving program that uses the GA to find a method through a maze. How do I turn my random binary encoded genes and fitness function (add all the binary values together) into a method to control a bot around a maze? I have built a basic GUI in Java consisting of a maze of labels (like a grid) with the available routes being in blue and the walls being in black. To reiterate my GA performs well and contains what any typical GA would (fitness method, get and set population, selection, crossover, etc) but now I need to plug it into a GUI to get my maze running. What needs to go where in order to get a bot that can move in different directions depending on what the GA says? Rough pseudocode would be great if possible As requested, an Individual is built using a separate class (Indiv), with all the main work being done in a Pop class. When a new individual is instantiated an array of ints represent the genes of said individual, with the genes being picked at random from a number between 0 and 1. The fitness function merely adds together the value of these genes and in the Pop class handles selection, mutation and crossover of two selected individuals. There's not much else to it, the command line program just shows evolution over n generations with the total fitness improving over each iteration. EDIT: It's starting to make a bit more sense now, although there are a few things that are bugging me... As Adamski has suggested I want to create an "Agent" with the options shown below. The problem I have is where the random bit string comes into play here. The agent knows where the walls are and has it laid out in a 4 bit string (i.e. 0111), but how does this affect the random 32 bit string? (i.e. 10001011011001001010011011010101) If I have the following maze (x is the start place, 2 is the goal, 1 is the wall): x 1 1 1 1 0 0 1 0 0 1 0 0 0 2 If I turn left I'm facing the wrong way and the agent will move completely off the maze if it moves forward. I assume that the first generation of the string will be completely random and it will evolve as the fitness grows but I don't get how the string will work within a maze. So, to get this straight... The fitness is the result of when the agent is able to move and is by a wall. The genes are a string of 32 bits, split into 16 sets of 2 bits to show the available actions and for the robot to move the two bits need to be passed with four bits from the agent showings its position near the walls. If the move is to go past a wall the move isn't made and it is deemed invalid and if the move is made and if a new wall is found then the fitness goes up. Is that right?

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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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  • Practical Python-based visual programming environment?

    - by Who8MyLunch
    I am looking for a practical visual programming environment based on Python. My primary application is algorithm development for processing remote-sensing imagery. I was initially inspired by LabVIEW from National Instruments, but that is more geared towards laboratory measurements and simulations. I write a lot of prototype code in Python and do a lot of interactive analysis with IPython. Does there exist a visual framework where a "program" is represented by connected nodes which each read data, do some work, and output data to the next node? I would like to use Python to write the code residing in each node. So far the best I've seen is Orange http://www.ailab.si/orange/, but it does not have the ability to start/stop individual nodes.

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  • Extracting rightmost N bits of an integer

    - by srandpersonia
    In the yester Code Jam Qualification round http://code.google.com/codejam/contest/dashboard?c=433101#s=a&a=0 , there was a problem called Snapper Chain. From the contest analysis I came to know the problem requires bit twiddling stuff like extracting the rightmost N bits of an integer and checking if they all are 1. I saw a contestant's(Eireksten) code which performed the said operation like below: (((K&(1<<N)-1))==(1<<N)-1) I couldn't understand how this works. What is the use of -1 there in the comparison?. If somebody can explain this, it would be very much useful for us rookies. Also, Any tips on identifying this sort of problems would be much appreciated. I used a naive algorithm to solve this problem and ended up solving only the smaller data set.(It took heck of a time to compile the larger data set which is required to be submitted within 8 minutes.). Thanks in advance.

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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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  • A detail question when applying genetic algorithm to traveling salesman

    - by burrough
    I read various stuff on this and understand the principle and concepts involved, however, none of paper mentions the details of how to calculate the fitness of a chromosome (which represents a route) involving adjacent cities (in the chromosome) that are not directly connected by an edge (in the graph). For example, given a chromosome 1|3|2|8|4|5|6|7, in which each gene represents the index of a city on the graph/map, how do we calculate its fitness (i.e. the total sum of distances traveled) if, say, there is no direct edge/link between city 2 and 8. Do we follow some sort of greedy algorithm to work out a route between 2 and 8, and add the distance of this route to the total? This problem seems pretty common when applying GA to TSP. Anyone who's done it before please share your experience. Thanks.

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