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 /…
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 ||…
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’])…
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…
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…
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 solved How to get summation of i ∈∑ nCr…
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,…
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…
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 solved nCr mod 10^9 + 7 for n
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…
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…
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)); solved Combination formula…
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…
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…
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,…