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  • Dijkstra shortest path algorithm with edge cost.

    - by Svisstack
    Hello, I have a directed, positive weighted graph. Each edge have a cost of use. I have only A money, i want to calculate shortest paths with dijkstra algorithm, but sum of edges costs on route must be less or equal to A. I want to do this with most smallest Dijstra modification (if I can do it with small modification of Dijkstra). Anyone can help me with this?

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  • Algorithm to calculate the number of divisors of a given number

    - by sker
    What would be the most optimal algorithm (performance-wise) to calculate the number of divisors of a given number? It'll be great if you could provide pseudocode or a link to some example. EDIT: All the answers have been very helpful, thank you. I'm implementing the Sieve of Atkin and then I'm going to use something similar to what Jonathan Leffler indicated. The link posted by Justin Bozonier has further information on what I wanted.

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  • Sort algorithm with fewest number of operations

    - by luvieere
    What is the sort algorithm with fewest number of operations? I need to implement it in HLSL as part of a pixel shader effect v2.0 for WPF, so it needs to have a really small number of operations, considering Pixel Shader's limitations. I need to sort 9 values, specifically the current pixel and its neighbors.

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  • Fastest sorting algorithm for a specific situation

    - by luvieere
    What is the fastest sorting algorithm for a large number (tens of thousands) of groups of 9 positive double precision values, where each group must be sorted individually? So it's got to sort fast a small number of possibly repeated double precision values, many times in a row. The values are in the [0..1] interval. I don't care about space complexity or stability, just about speed.

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  • finding shortest valid path in a colored-edge graphs

    - by user1067083
    Given a directed graph G, with edges colored either green or purple, and a vertex S in G, I must find an algorithm that finds the shortest path from s to each vertex in G so the path includes at most two purple edges (and green as much as needed). I thought of BFS on G after removing all the purple edges, and for every vertex that the shortest path is still infinity, do something to try to find it, but I'm kinda stuck, and it takes alot of the running time as well... Any other suggestions? Thanks in advance

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  • efficient algorithm for drawing circle arcs?

    - by banister
    I am using the mid-point circle algorithm (bresenham circle) to efficiently draw whole circles. Is there something similar to draw circle arcs? I would like to specify a start angle and end angle and have only that portion of the circle drawn. Thanks in advance! EDIT: I would like to draw filled circle arcs too, i.e pie-slices. :)

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  • Parallel curve like algorithm for graphs

    - by skrat
    Is there a well know algorithm for calculating "parallel graph"? where by parallel graph I mean the same as parallel curve, vaguely called "offset curve", but with a graph instead of a curve. Given this picture how can I calculate points of black outlined polygons?

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  • Good Java graph algorithm library?

    - by Nick Fortescue
    Has anyone had good experiences with any Java libraries for Graph algorithms. I've tried JGraph and found it ok, and there are a lot of different ones in google. Are there any that people are actually using successfully in production code or would recommend? To clarify, I'm not looking for a library that produces graphs/charts, I'm looking for one that helps with Graph algorithms, eg minimum spanning tree, Kruskal's algorithm Nodes, Edges, etc. Ideally one with some good algorithms/data structures in a nice Java OO API.

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  • Algorithm for a dice problem

    - by vivekeviv
    I was thinking what should be the best algorithm for finding all the solutions of this puzzle. http://1cup1coffee.com/puzzle/endice/ Could backtracking be the an approach for solving this or can you suggest any other approach? Thanks

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  • Algorithm for best positioning objects on a Visio model

    - by leandrosa81
    Hello, I am trying to map all network devices and create a visio file with the resulting network topology. I was wondering if there are any algorithm for best positioning the nodes on the graph, considering its connections. Connections are bidirectional, like this (may have many connections between the same nodes): --------- --------- | | | | | A |----------| B | |_______| |_______|

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  • An algorithm for pavement usage calculation

    - by student
    Given an area of specific size I need to find out how many pavement stones to use to completely pave the area. Suppose that I have an empty floor of 100 metre squares and stones with 20x10 cm and 30x10 cm sizes. I must pave the area with minimum usage of stones of both sizes. Anyone knows of an algorithm that calculates this? (Sorry if my English is bad) C# is preferred.

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  • What is the best algorithm for this problem?

    - by mark
    What is the most efficient algorithm to solve the following problem? Given 6 arrays, D1,D2,D3,D4,D5 and D6 each containing 6 numbers like: D1[0] = number D2[0] = number ...... D6[0] = number D1[1] = another number D2[1] = another number .... ..... .... ...... .... D1[5] = yet another number .... ...... .... Given a second array ST1, containing 1 number: ST1[0] = 6 Given a third array ans, containing 6 numbers: ans[0] = 3, ans[1] = 4, ans[2] = 5, ......ans[5] = 8 Using as index for the arrays D1,D2,D3,D4,D5 and D6, the number that goes from 0, to the number stored in ST1[0] minus one, in this example 6, so from 0 to 6-1, compare each res array against each D array My algorithm so far is: I tried to keep everything unlooped as much as possible. EML := ST1[0] //number contained in ST1[0] EML1 := 0 //start index for the arrays D While EML1 < EML if D1[ELM1] = ans[0] goto two if D2[ELM1] = ans[0] goto two if D3[ELM1] = ans[0] goto two if D4[ELM1] = ans[0] goto two if D5[ELM1] = ans[0] goto two if D6[ELM1] = ans[0] goto two ELM1 = ELM1 + 1 return 0 //bad row of numbers, if while ends two: EML1 := 0 start index for arrays Ds While EML1 < EML if D1[ELM1] = ans[1] goto two if D2[ELM1] = ans[1] goto two if D3[ELM1] = ans[1] goto two if D4[ELM1] = ans[1] goto two if D5[ELM1] = ans[1] goto two if D6[ELM1] = ans[1] goto two ELM1 = ELM1 + 1 return 0 three: EML1 := 0 start index for arrays Ds While EML1 < EML if D1[ELM1] = ans[2] goto two if D2[ELM1] = ans[2] goto two if D3[ELM1] = ans[2] goto two if D4[ELM1] = ans[2] goto two if D5[ELM1] = ans[2] goto two if D6[ELM1] = ans[2] goto two ELM1 = ELM1 + 1 return 0 four: EML1 := 0 start index for arrays Ds While EML1 < EML if D1[ELM1] = ans[3] goto two if D2[ELM1] = ans[3] goto two if D3[ELM1] = ans[3] goto two if D4[ELM1] = ans[3] goto two if D5[ELM1] = ans[3] goto two if D6[ELM1] = ans[3] goto two ELM1 = ELM1 + 1 return 0 five: EML1 := 0 start index for arrays Ds While EML1 < EML if D1[ELM1] = ans[4] goto two if D2[ELM1] = ans[4] goto two if D3[ELM1] = ans[4] goto two if D4[ELM1] = ans[4] goto two if D5[ELM1] = ans[4] goto two if D6[ELM1] = ans[4] goto two ELM1 = ELM1 + 1 return 0 six: EML1 := 0 start index for arrays Ds While EML1 < EML if D1[ELM1] = ans[0] return 1 //good row of numbers if D2[ELM1] = ans[0] return 1 if D3[ELM1] = ans[0] return 1 if D4[ELM1] = ans[0] return 1 if D5[ELM1] = ans[0] return 1 if D6[ELM1] = ans[0] return 1 ELM1 = ELM1 + 1 return 0 As language of choice, it would be pure c

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  • Algorithm for deciding price ranges.

    - by Paul Knopf
    I am looking for code that will take a huge list of numbers, and calculate price ranges correctly. There must be some algorithm that will choose the proper ranges, no? I am looking for this code in c#, but any language will do (I can convert). Thanks in advance!

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  • Latex: Listing all figures (tables, algorithm) once again at the end of the document

    - by Zlatko
    Hi all, I have been writhing a rather large document with latex. Now I would like to list all the figures / tables / algortihms once again at the end of the file so that I can check if they all look the same. For example, if every algorithm has the same notation. How can I do this? I know about \listofalgorithms and \listoffigures but they only list the names of the algorithms or figures and the pages where they are. Thanks.

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  • Reducing Integer Fractions Algorithm - Solution Explanation?

    - by Andrew Tomazos - Fathomling
    This is a followup to this problem: Reducing Integer Fractions Algorithm Following is a solution to the problem from a grandmaster: #include <cstdio> #include <algorithm> #include <functional> using namespace std; const int MAXN = 100100; const int MAXP = 10001000; int p[MAXP]; void init() { for (int i = 2; i < MAXP; ++i) { if (p[i] == 0) { for (int j = i; j < MAXP; j += i) { p[j] = i; } } } } void f(int n, vector<int>& a, vector<int>& x) { a.resize(n); vector<int>(MAXP, 0).swap(x); for (int i = 0; i < n; ++i) { scanf("%d", &a[i]); for (int j = a[i]; j > 1; j /= p[j]) { ++x[p[j]]; } } } void g(const vector<int>& v, vector<int> w) { for (int i: v) { for (int j = i; j > 1; j /= p[j]) { if (w[p[j]] > 0) { --w[p[j]]; i /= p[j]; } } printf("%d ", i); } puts(""); } int main() { int n, m; vector<int> a, b, x, y, z; init(); scanf("%d%d", &n, &m); f(n, a, x); f(m, b, y); printf("%d %d\n", n, m); transform(x.begin(), x.end(), y.begin(), insert_iterator<vector<int> >(z, z.end()), [](int a, int b) { return min(a, b); }); g(a, z); g(b, z); return 0; } It isn't clear to me how it works. Can anyone explain it? The equivilance is as follows: a is the numerator vector of length n b is the denominator vector of length m

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  • random, Graphics point ,searching- algorithm, via dual for loop set

    - by LoneXcoder
    hello and thanks for joining me in my journey to the custom made algorithm for "guess where the pixel is" this for Loop set (over Point.X, Point.Y), is formed in consecutive/linear form: //Original\initial Location Point initPoint = new Point(150, 100); // No' of pixels to search left off X , and above Y int preXsrchDepth, preYsrchDepth; // No' of pixels to search to the right of X, And Above Y int postXsrchDepth, postYsrchDepth; preXsrchDepth = 10; // will start search at 10 pixels to the left from original X preYsrchDepth = 10; // will start search at 10 pixels above the original Y postXsrchDepth = 10; // will stop search at 10 pixels to the right from X postYsrchDepth = 10; // will stop search at 10 pixels below Y int StopXsearch = initPoint.X + postXsrchDepth; //stops X Loop itarations at initial pointX + depth requested to serch right of it int StopYsearch = initPoint.Y + postYsrchDepth; //stops Y Loop itarations at initial pointY + depth requested below original location int CountDownX, CountDownY; // Optional not requierd for loop but will reports the count down how many iterations left (unless break; triggerd ..uppon success) Point SearchFromPoint = Point.Empty; //the point will be used for (int StartX = initPoint.X - preXsrchDepth; StartX < StopXsearch; StartX++) { SearchFromPoint.X = StartX; for (int StartY = initPoint.Y - preYsrchDepth; StartY < StpY; StartY++) { CountDownX = (initPoint.X - StartX); CountDownY=(initPoint.Y - StartY); SearchFromPoint.Y = StartY; if (SearchSuccess) { same = true; AAdToAppLog("Search Report For: " + imgName + "Search Completed Successfully On Try " + CountDownX + ":" + CountDownY); break; } } } <-10 ---- -5--- -1 X +1--- +5---- +10 what i would like to do is try a way of instead is have a little more clever approach <+8---+5-- -8 -5 -- +2 +10 X -2 - -10 -8-- -6 ---1- -3 | +8 | -10 Y +1 -6 | | +9 .... I do know there's a wheel already invented in this field (even a full-trailer truck amount of wheels (: ) but as a new programmer, I really wanted to start of with a simple way and also related to my field of interest in my project. can anybody show an idea of his, he learnt along the way to Professionalism in algorithm /programming having tests to do on few approaches (kind'a random cleverness...) will absolutely make the day and perhaps help some others viewing this page in the future to come it will be much easier for me to understand if you could use as much as possible similar naming to variables i used or implenet your code example ...it will be Greatly appreciated if used with my code sample, unless my metod is a realy flavorless. p.s i think that(atleast as human being) the tricky part is when throwing inconsecutive numbers you loose track of what you didn't yet use, how do u take care of this too . thanks allot in advance looking forward to your participation !

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  • Optimized OCR black/white pixel algorithm

    - by eagle
    I am writing a simple OCR solution for a finite set of characters. That is, I know the exact way all 26 letters in the alphabet will look like. I am using C# and am able to easily determine if a given pixel should be treated as black or white. I am generating a matrix of black/white pixels for every single character. So for example, the letter I (capital i), might look like the following: 01110 00100 00100 00100 01110 Note: all points, which I use later in this post, assume that the top left pixel is (0, 0), bottom right pixel is (4, 4). 1's represent black pixels, and 0's represent white pixels. I would create a corresponding matrix in C# like this: CreateLetter("I", new List<List<bool>>() { new List<bool>() { false, true, true, true, false }, new List<bool>() { false, false, true, false, false }, new List<bool>() { false, false, true, false, false }, new List<bool>() { false, false, true, false, false }, new List<bool>() { false, true, true, true, false } }); I know I could probably optimize this part by using a multi-dimensional array instead, but let's ignore that for now, this is for illustrative purposes. Every letter is exactly the same dimensions, 10px by 11px (10px by 11px is the actual dimensions of a character in my real program. I simplified this to 5px by 5px in this posting since it is much easier to "draw" the letters using 0's and 1's on a smaller image). Now when I give it a 10px by 11px part of an image to analyze with OCR, it would need to run on every single letter (26) on every single pixel (10 * 11 = 110) which would mean 2,860 (26 * 110) iterations (in the worst case) for every single character. I was thinking this could be optimized by defining the unique characteristics of every character. So, for example, let's assume that the set of characters only consists of 5 distinct letters: I, A, O, B, and L. These might look like the following: 01110 00100 00100 01100 01000 00100 01010 01010 01010 01000 00100 01110 01010 01100 01000 00100 01010 01010 01010 01000 01110 01010 00100 01100 01110 After analyzing the unique characteristics of every character, I can significantly reduce the number of tests that need to be performed to test for a character. For example, for the "I" character, I could define it's unique characteristics as having a black pixel in the coordinate (3, 0) since no other characters have that pixel as black. So instead of testing 110 pixels for a match on the "I" character, I reduced it to a 1 pixel test. This is what it might look like for all these characters: var LetterI = new OcrLetter() { Name = "I", BlackPixels = new List<Point>() { new Point (3, 0) } } var LetterA = new OcrLetter() { Name = "A", WhitePixels = new List<Point>() { new Point(2, 4) } } var LetterO = new OcrLetter() { Name = "O", BlackPixels = new List<Point>() { new Point(3, 2) }, WhitePixels = new List<Point>() { new Point(2, 2) } } var LetterB = new OcrLetter() { Name = "B", BlackPixels = new List<Point>() { new Point(3, 1) }, WhitePixels = new List<Point>() { new Point(3, 2) } } var LetterL = new OcrLetter() { Name = "L", BlackPixels = new List<Point>() { new Point(1, 1), new Point(3, 4) }, WhitePixels = new List<Point>() { new Point(2, 2) } } This is challenging to do manually for 5 characters and gets much harder the greater the amount of letters that are added. You also want to guarantee that you have the minimum set of unique characteristics of a letter since you want it to be optimized as much as possible. I want to create an algorithm that will identify the unique characteristics of all the letters and would generate similar code to that above. I would then use this optimized black/white matrix to identify characters. How do I take the 26 letters that have all their black/white pixels filled in (e.g. the CreateLetter code block) and convert them to an optimized set of unique characteristics that define a letter (e.g. the new OcrLetter() code block)? And how would I guarantee that it is the most efficient definition set of unique characteristics (e.g. instead of defining 6 points as the unique characteristics, there might be a way to do it with 1 or 2 points, as the letter "I" in my example was able to). An alternative solution I've come up with is using a hash table, which will reduce it from 2,860 iterations to 110 iterations, a 26 time reduction. This is how it might work: I would populate it with data similar to the following: Letters["01110 00100 00100 00100 01110"] = "I"; Letters["00100 01010 01110 01010 01010"] = "A"; Letters["00100 01010 01010 01010 00100"] = "O"; Letters["01100 01010 01100 01010 01100"] = "B"; Now when I reach a location in the image to process, I convert it to a string such as: "01110 00100 00100 00100 01110" and simply find it in the hash table. This solution seems very simple, however, this still requires 110 iterations to generate this string for each letter. In big O notation, the algorithm is the same since O(110N) = O(2860N) = O(N) for N letters to process on the page. However, it is still improved by a constant factor of 26, a significant improvement (e.g. instead of it taking 26 minutes, it would take 1 minute). Update: Most of the solutions provided so far have not addressed the issue of identifying the unique characteristics of a character and rather provide alternative solutions. I am still looking for this solution which, as far as I can tell, is the only way to achieve the fastest OCR processing. I just came up with a partial solution: For each pixel, in the grid, store the letters that have it as a black pixel. Using these letters: I A O B L 01110 00100 00100 01100 01000 00100 01010 01010 01010 01000 00100 01110 01010 01100 01000 00100 01010 01010 01010 01000 01110 01010 00100 01100 01110 You would have something like this: CreatePixel(new Point(0, 0), new List<Char>() { }); CreatePixel(new Point(1, 0), new List<Char>() { 'I', 'B', 'L' }); CreatePixel(new Point(2, 0), new List<Char>() { 'I', 'A', 'O', 'B' }); CreatePixel(new Point(3, 0), new List<Char>() { 'I' }); CreatePixel(new Point(4, 0), new List<Char>() { }); CreatePixel(new Point(0, 1), new List<Char>() { }); CreatePixel(new Point(1, 1), new List<Char>() { 'A', 'B', 'L' }); CreatePixel(new Point(2, 1), new List<Char>() { 'I' }); CreatePixel(new Point(3, 1), new List<Char>() { 'A', 'O', 'B' }); // ... CreatePixel(new Point(2, 2), new List<Char>() { 'I', 'A', 'B' }); CreatePixel(new Point(3, 2), new List<Char>() { 'A', 'O' }); // ... CreatePixel(new Point(2, 4), new List<Char>() { 'I', 'O', 'B', 'L' }); CreatePixel(new Point(3, 4), new List<Char>() { 'I', 'A', 'L' }); CreatePixel(new Point(4, 4), new List<Char>() { }); Now for every letter, in order to find the unique characteristics, you need to look at which buckets it belongs to, as well as the amount of other characters in the bucket. So let's take the example of "I". We go to all the buckets it belongs to (1,0; 2,0; 3,0; ...; 3,4) and see that the one with the least amount of other characters is (3,0). In fact, it only has 1 character, meaning it must be an "I" in this case, and we found our unique characteristic. You can also do the same for pixels that would be white. Notice that bucket (2,0) contains all the letters except for "L", this means that it could be used as a white pixel test. Similarly, (2,4) doesn't contain an 'A'. Buckets that either contain all the letters or none of the letters can be discarded immediately, since these pixels can't help define a unique characteristic (e.g. 1,1; 4,0; 0,1; 4,4). It gets trickier when you don't have a 1 pixel test for a letter, for example in the case of 'O' and 'B'. Let's walk through the test for 'O'... It's contained in the following buckets: // Bucket Count Letters // 2,0 4 I, A, O, B // 3,1 3 A, O, B // 3,2 2 A, O // 2,4 4 I, O, B, L Additionally, we also have a few white pixel tests that can help: (I only listed those that are missing at most 2). The Missing Count was calculated as (5 - Bucket.Count). // Bucket Missing Count Missing Letters // 1,0 2 A, O // 1,1 2 I, O // 2,2 2 O, L // 3,4 2 O, B So now we can take the shortest black pixel bucket (3,2) and see that when we test for (3,2) we know it is either an 'A' or an 'O'. So we need an easy way to tell the difference between an 'A' and an 'O'. We could either look for a black pixel bucket that contains 'O' but not 'A' (e.g. 2,4) or a white pixel bucket that contains an 'O' but not an 'A' (e.g. 1,1). Either of these could be used in combination with the (3,2) pixel to uniquely identify the letter 'O' with only 2 tests. This seems like a simple algorithm when there are 5 characters, but how would I do this when there are 26 letters and a lot more pixels overlapping? For example, let's say that after the (3,2) pixel test, it found 10 different characters that contain the pixel (and this was the least from all the buckets). Now I need to find differences from 9 other characters instead of only 1 other character. How would I achieve my goal of getting the least amount of checks as possible, and ensure that I am not running extraneous tests?

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  • Performance in backpropagation algorithm

    - by Taban
    I've written a matlab program for standard backpropagation algorithm, it is my homework and I should not use matlab toolbox, so I write the entire code by myself. This link helped me for backpropagation algorithm. I have a data set of 40 random number and initial weights randomly. As output, I want to see a diagram that shows the performance. I used mse and plot function to see performance for 20 epochs but the result is this: I heard that performance should go up through backpropagation, so I want to know is there any problem with my code or this result is normal because local minimums. This is my code: Hidden_node=inputdlg('Enter the number of Hidden nodes'); a=0.5;%initialize learning rate hiddenn=str2num(Hidden_node{1,1}); randn('seed',0); %creating data set s=2; N=10; m=[5 -5 5 5;-5 -5 5 -5]; S = s*eye(2); [l,c] = size(m); x = []; % Creating the training set for i = 1:c x = [x mvnrnd(m(:,i)',S,N)']; end % target value toutput=[ones(1,N) zeros(1,N) ones(1,N) zeros(1,N)]; for epoch=1:20; %number of epochs for kk=1:40; %number of patterns %initial weights of hidden layer for ii=1 : 2; for jj=1 :hiddenn; whidden{ii,jj}=rand(1); end end initial the wights of output layer for ii=1 : hiddenn; woutput{ii,1}=rand(1); end for ii=1:hiddenn; x1=x(1,kk); x2=x(2,kk); w1=whidden{1,ii}; w2=whidden{2,ii}; activation{1,ii}=(x1(1,1)*w1(1,1))+(x2(1,1)*w2(1,1)); end %calculate output of hidden nodes for ii=1:hiddenn; hidden_to_out{1,ii}=logsig(activation{1,ii}); end activation_O{1,1}=0; for jj=1:hiddenn; activation_O{1,1} = activation_O{1,1}+(hidden_to_out{1,jj}*woutput{jj,1}); end %calculate output out{1,1}=logsig(activation_O{1,1}); out_for_plot(1,kk)= out{1,ii}; %calculate error for output node delta_out{1,1}=(toutput(1,kk)-out{1,1}); %update weight of output node for ii=1:hiddenn; woutput{ii,jj}=woutput{ii,jj}+delta_out{1,jj}*hidden_to_out{1,ii}*dlogsig(activation_O{1,jj},logsig(activation_O{1,jj}))*a; end %calculate error of hidden nodes for ii=1:hiddenn; delta_hidden{1,ii}=woutput{ii,1}*delta_out{1,1}; end %update weight of hidden nodes for ii=1:hiddenn; for jj=1:2; whidden{jj,ii}= whidden{jj,ii}+(delta_hidden{1,ii}*dlogsig(activation{1,ii},logsig(activation{1,ii}))*x(jj,kk)*a); end end a=a/(1.1);%decrease learning rate end %calculate performance e=toutput(1,kk)-out_for_plot(1,1); perf(1,epoch)=mse(e); end plot(perf); Thanks a lot.

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