Permutations without recursive function call

Question

Requirement: Algorithm to generate all possible combinations of a set , without duplicates , or recursively calling function to return results.

The majority , if not

Answer 1

Here's a snippet for an approach that I came up with on my own, but naturally was also able to find it described elsewhere:

generatePermutations = function(arr) {
  if (arr.length < 2) {
    return arr.slice();
  }
  var factorial = [1];
  for (var i = 1; i <= arr.length; i++) {
    factorial.push(factorial[factorial.length - 1] * i);
  }

  var allPerms = [];
  for (var permNumber = 0; permNumber < factorial[factorial.length - 1]; permNumber++) {
    var unused = arr.slice();
    var nextPerm = [];
    while (unused.length) {
      var nextIndex = Math.floor((permNumber % factorial[unused.length]) / factorial[unused.length - 1]);
      nextPerm.push(unused[nextIndex]);
      unused.splice(nextIndex, 1);
    }
    allPerms.push(nextPerm);
  }
  return allPerms;
};

Enter comma-separated string (e.g. a,b,c):






  Generate permutations

Explanation

Since there are factorial(arr.length) permutations for a given array arr, each number between 0 and factorial(arr.length)-1 encodes a particular permutation. To unencode a permutation number, repeatedly remove elements from arr until there are no elements left. The exact index of which element to remove is given by the formula (permNumber % factorial(arr.length)) / factorial(arr.length-1). Other formulas could be used to determine the index to remove, as long as it preserves the one-to-one mapping between number and permutation.

Example

The following is how all permutations would be generated for the array (a,b,c,d):

#    Perm      1st El        2nd El      3rd El    4th El
0    abcd   (a,b,c,d)[0]   (b,c,d)[0]   (c,d)[0]   (d)[0]
1    abdc   (a,b,c,d)[0]   (b,c,d)[0]   (c,d)[1]   (c)[0]
2    acbd   (a,b,c,d)[0]   (b,c,d)[1]   (b,d)[0]   (d)[0]
3    acdb   (a,b,c,d)[0]   (b,c,d)[1]   (b,d)[1]   (b)[0]
4    adbc   (a,b,c,d)[0]   (b,c,d)[2]   (b,c)[0]   (c)[0]
5    adcb   (a,b,c,d)[0]   (b,c,d)[2]   (b,c)[1]   (b)[0]
6    bacd   (a,b,c,d)[1]   (a,c,d)[0]   (c,d)[0]   (d)[0]
7    badc   (a,b,c,d)[1]   (a,c,d)[0]   (c,d)[1]   (c)[0]
8    bcad   (a,b,c,d)[1]   (a,c,d)[1]   (a,d)[0]   (d)[0]
9    bcda   (a,b,c,d)[1]   (a,c,d)[1]   (a,d)[1]   (a)[0]
10   bdac   (a,b,c,d)[1]   (a,c,d)[2]   (a,c)[0]   (c)[0]
11   bdca   (a,b,c,d)[1]   (a,c,d)[2]   (a,c)[1]   (a)[0]
12   cabd   (a,b,c,d)[2]   (a,b,d)[0]   (b,d)[0]   (d)[0]
13   cadb   (a,b,c,d)[2]   (a,b,d)[0]   (b,d)[1]   (b)[0]
14   cbad   (a,b,c,d)[2]   (a,b,d)[1]   (a,d)[0]   (d)[0]
15   cbda   (a,b,c,d)[2]   (a,b,d)[1]   (a,d)[1]   (a)[0]
16   cdab   (a,b,c,d)[2]   (a,b,d)[2]   (a,b)[0]   (b)[0]
17   cdba   (a,b,c,d)[2]   (a,b,d)[2]   (a,b)[1]   (a)[0]
18   dabc   (a,b,c,d)[3]   (a,b,c)[0]   (b,c)[0]   (c)[0]
19   dacb   (a,b,c,d)[3]   (a,b,c)[0]   (b,c)[1]   (b)[0]
20   dbac   (a,b,c,d)[3]   (a,b,c)[1]   (a,c)[0]   (c)[0]
21   dbca   (a,b,c,d)[3]   (a,b,c)[1]   (a,c)[1]   (a)[0]
22   dcab   (a,b,c,d)[3]   (a,b,c)[2]   (a,b)[0]   (b)[0]
23   dcba   (a,b,c,d)[3]   (a,b,c)[2]   (a,b)[1]   (a)[0]

Note that each permutation # is of the form:

(firstElIndex * 3!) + (secondElIndex * 2!) + (thirdElIndex * 1!) + (fourthElIndex * 0!)

which is basically the reverse process of the formula given in the explanation.