[ad_1] So you want to generate all permutations from size = 1 to size = N? Stack Overflow user Andy Carson has created a JavaScript library with a permutation generator function that lets you specify size. See: GitHub / acarl005 / Generatorics // Adapted from: https://stackoverflow.com/a/41232141/1762224 // Uses: https://github.com/acarl005/generatorics const allPermutations = (arr) => Array.from({ … Read more
[ad_1] static void Main() { var cnk = comb(new [] {1,2,3},2); foreach ( var c in cnk) { } } public static IEnumerable<int[]> comb(int[] a, int k) { if (a == null || a.Length == 0 || k < 1 || k > a.Length) yield break; int n = a.Length; // 1 if ( k … Read more
[ad_1] Alright I’ve done the hard part hoping you can finish it up: tmp = [] def recal(_list): n = [] if ‘-‘ in _list: for i in 0,1: t = _list[:] t[t.index(‘-‘)] = i n.append(recall(t)) else: tmp.append(l) return l recall([‘-‘,’-‘,’0′,’1’]) for l in tmp: print int(”.join(l),2) 0 [ad_2] solved combinations program help in python … Read more
[ad_1] I being a number, there is 10 possibilities. S being a character, there is 26 possibilities. The total number of combinations is 10 power (TheNumberOfI) * 26 power (TheNumberOfS) This can be dynamically solved using a simple function that counts the number of S and the number of I and uses the results in … Read more
[ad_1] You haven’t specified in what order the keys of the dictionary should be processed in the output. If one assumes reverse sorting order, you can do this trivially with itertools.product(): from itertools import product combinations = product(*([‘{0}{1}’.format(v, i) for i in range(dictElement[v])] for v in sorted(dictElement, reverse=True)) Demo: >>> from itertools import product >>> … Read more
[ad_1] Use the hockey stick rule bCr + (b-1)Cr + (b-2)Cr + …. + aCr = bCr + (b-1)Cr + (b-2)Cr + …. + rCr – ((a-1)Cr+(a-2)Cr+…+rCr) = (b+1)C(r+1)-aC(r+1) For example, 6+10+15=4C2+5C2+6C2=7C3-4C3=35-4 [ad_2] solved How to get summation of i ∈∑ nCr where ( a
[ad_1] Here’s what I’ve got, took me a while but this will do what I was looking for: it takes each combination of the arrays (t1, t2, t3, t4) and adds their elements and returns whatever combinatorial you choose for n, (in this example i left it as 3). If there’s any more optimizations you … Read more
[ad_1] First thing you need to explode the string by _ $str=”one_two_three_four_five_six”; $array = explode(‘_’, $str); Add an empty array to store result there $result = []; Define a recursive function that takes an array, implode array values, remove last element and recall the same array until length is 0 function visitArray($array, &$result) { if(count($array) … Read more
[ad_1] n is large while r is small, you are better off compute nCr by n(n-1)…(n-r+1)/(1*2*…*r) You may need to find multiplicate inverse of 1, 2, … r mod 10^9+7 [ad_2] solved nCr mod 10^9 + 7 for n
[ad_1] Visual Studio was taking too long to load, so I did it in JavaScript since I could test in my console. This will print out all the choices. (Also it sounds more like a “what’s the algorithm for this?” not “what’s the algorithm for this in C#?”) function makeGroup(id, count) { var result = … Read more
[ad_1] If you have F unique first names and L unique last names, then overall number of combinations is N = F * L So you can generate needed number of non-repeating random integer values in range 0..N-1 (for example, with Fisher-Yates sampling), sort them, and get corresponding name combinations: for i = 0..M-1 Generate … Read more
[ad_1] This line: result = n! / ((n-k)!*k!); …is not valid C code. ! in C means “not”. You will need to provide a factorial function so that you can call: result = factorial(n) / (factorial(n-k) * factorial(k)); [ad_2] solved Combination formula “n! / ((n-k)!*k!)” doesn’t work [closed]
[ad_1] You’re looking for a Cartesian product, not a combination or permutation of [0, 1]. For that, you can use itertools.product. from itertools import product items = [0, 1] for item in product(items, repeat=3): print(item) This produces the output you’re looking for (albeit in a slightly different order): (0, 0, 0) (0, 0, 1) (0, … Read more
[ad_1] Patrick, I’m not sure this is the best way to do this, it does what you want but again I’m not really sure if using 3 consecutive loops is good practice. In real world environment, your txts files may contain hundreds or thousands of entries wich will make this painful for your server to … Read more
[ad_1] You caught me on a day where I don’t want to do what I’m supposed to be doing1. The code below only generates token output; it doesn’t attempt to store tokens anywhere. You can redirect the output to a file, but as others have pointed out, you’re going to need a bigger boat hard … Read more