This function, perm(xs), returns all the permutations of a given array: function perm(xs) { let ret = []; for (let i = 0; i < xs.length; i = i + 1) { let rest = perm(xs.slice(0, i).concat(xs.slice(i + 1))); if(!rest.length) { ret.push([xs[i]]) } else { for(let j = 0; j < rest.length; j = j … Read more
You should use the fact that when you want all permutations of N numbers there are N! possibilities. Therefore each number x from 1..N! encodes such a permutation. Here is a sample that iteratively prints out all permutations of a sting. private static void printPermutationsIterative(String string){ int [] factorials = new int[string.length()+1]; factorials[0] = 1; … Read more
In order to pick five characters from a string recursively, follow a simple algorithm: Your method should get a portion filled in so far, and the first position in the five-character permutation that needs a character If the first position that needs a character is above five, you are done; print the combination that you … Read more
Simplifying Amro’s answer, you could use this: %// Sample data x = ‘ABCD’; %// Set of possible letters K = 3; %// Length of each permutation %// Create all possible permutations (with repetition) of letters stored in x C = cell(K, 1); %// Preallocate a cell array [C{:}] = ndgrid(x); %// Create K grids of … Read more
you can also write something like this: let rec permutations list taken = seq { if Set.count taken = List.length list then yield [] else for l in list do if not (Set.contains l taken) then for perm in permutations list (Set.add l taken) do yield l::perm } The ‘list’ argument contains all the numbers … Read more
For a multiset, you can solve recursively by position (JavaScript code): function f(multiset,counters,result){ if (counters.every(x => x === 0)){ console.log(result); return; } for (var i=0; i<counters.length; i++){ if (counters[i] > 0){ _counters = counters.slice(); _counters[i]–; f(multiset,_counters,result + multiset[i]); } } } f([‘A’,’B’],[3,3],”);
Since I offered an ODBC approach I thought I should elaborate on it, as it is not immediately obvious how to do this. And, in honesty, I needed to relearn the process and document it for myself. This is a way to generate a Cartesian product of two or more one-dimensional data arrays using Excel … Read more
The algorithm basically works on this logic: All permutations of a string X is the same thing as all permutations of each possible character in X, combined with all permutations of the string X without that letter in it. That is to say, all permutations of “abcd” are “a” concatenated with all permutations of “bcd” … Read more
The total number of permutation of a given string of length n would be n! (if all characters are different), thus it would not be possible to explore all the combinations. This question is actually like the mathematics P & C question Find the rank of the word “stack” when arranged in dictionary order. Given … Read more
I can’t speak for the designer of itertools.permutations (Raymond Hettinger), but it seems to me that there are a couple of points in favour of the design: First, if you used a next_permutation-style approach, then you’d be restricted to passing in objects that support a linear ordering. Whereas itertools.permutations provides permutations of any kind of … Read more