Permutation Javascript

Question

So I have this code now, and in input I have in ascending order my name\'s letters \"ahimrsu\". I need to show up the right number for \"mariush\" from all combinations whic

Answer 1

This is my solution from the following answer: https://stackoverflow.com/a/18879232/783743

var permute = (function () {
    return permute;

    function permute(list) {
        return list.length ?
            list.reduce(permutate, []) :
            [[]];
    }

    function permutate(permutations, item, index, list) {
        return permutations.concat(permute(
            list.slice(0, index).concat(
            list.slice(index + 1)))
            .map(concat, [item]));
    }

    function concat(list) {
        return this.concat(list);
    }
}());

You can use the permute function to find all the permutations of an array:

var array = "ahimrsu".split("");
var permutations = permute(array).map(join);
var index = permutations.indexOf("maruish");

function join(array) {
    return array.join("");
}

The algorithm is very simple to understand:

1. We want a function permute of the type [a] -> [[a]] (i.e. given a list of as it returns a list of permutations of the input).
2. Given the empty list ([]) an an input, the output is an empty list of permutations ([[]]).
3. Otherwise for every element:
   1. We remove the element from the list.
   2. We recursively find the permutations of the remaining elements.
   3. We add the element we removed to the beginning of every permutation.

For example, suppose we want to find the permutation of the array [1, 2, 3]:

1. permute([1, 2, 3]) === [1, 2, 3].reduce(permutate, [])
    1. permutate([], 1, 0, [1, 2, 3])
        1. permute([2, 3]) === [2, 3].reduce(permutate, [])
            1. permutate([], 2, 0, [2, 3])
                1. permute([3]) === [3].reduce(permutate, [])
                    1. permutate([], 3, 0, [3])
                        1. permute([]) === [[]]
                        2. [[]].map(concat, [3]) === [[3]]
                        3. [].concat([[3]]) === [[3]]
                2. [[3]].map(concat, [2]) === [[2, 3]]
                3. [].concat([[2, 3]]) === [[2, 3]]
            2. permutate([[2, 3]], 3, 1, [2, 3])
                1. permute([2]) === [2].reduce(permutate, [])
                    1. permutate([], 2, 0, [2])
                        1. permute([]) === [[]]
                        2. [[]].map(concat, [2]) === [[2]]
                        3. [].concat([[2]]) === [[2]]
                2. [[2]].map(concat, [3]) === [[3, 2]]
                3. [[2, 3]].concat([[3, 2]]) === [[2, 3], [3, 2]]
        2. [[2, 3], [3, 2]].map(concat, [1]) === [[1, 2, 3], [1, 3, 2]]
        3. [].concat([[1, 2, 3], [1, 3, 2]]) === [[1, 2, 3], [1, 3, 2]]
    2. permutate([[1, 2, 3], [1, 3, 2]], 2, 1, [1, 2, 3])
        1. permute([1, 3]) === [1, 3].reduce(permutate, [])
        2. [[1, 3], [3, 1]].map(concat, [2]) === [[2, 1, 3], [2, 3, 1]]
        3. [[1, 2, 3], [1, 3, 2]].concat([[2, 1, 3], [2, 3, 1]])
    3. permutate([[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1]], 3, 2, [1, 2, 3])
        1. permute([1, 2]) === [1, 2].reduce(permutate, [])
        2. [[1, 2], [2, 1]].map(concat, [3]) === [[3, 1, 2], [3, 2, 1]]
        3. [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1]].concat([[3, 1, 2], [3, 2, 1]])

Old explanation:

1. First we remove the first element of the list. Hence we have item 1 and list [2, 3].
   1. Next we find the permutations of [2, 3].
      1. We remove the first element. Hence we have item 2 and list [3].
         1. Next we find the permutations of [3].
            1. We remove the first element. Hence we have item 3 and list [].
               1. Next we find the permutations of [] which is [[]].
            2. We add 3 to the beginning of each permutation.
            3. The result is [[3]].
         2. We add 2 to the beginning of each permutation.
         3. The result is [[2, 3]].
      2. We remove the second element. Hence we have item 3 and list [[2]].
         1. Next we find the permutations of [2].
            1. We remove the first element. Hence we have item 2 and list [].
               1. Next we find the permutations of [] which is [[]].
            2. We add 2 to the beginning of each permutation.
            3. The result is [[2]].
         2. We add 3 to the beginning of each permutation.
         3. The result is [[3, 2]].
      3. We combine the two two lists.
      4. The result is [[2, 3], [3, 2]].
   2. We add 1 to the beginning of each permutation.
   3. The result is [[1, 2, 3], [1, 3, 2]].
2. Same for the second element: item 2 and list [1, 3].
3. Same for the third element: item 3 and list [1, 2].
4. We combine the three lists.
5. The result is [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]].

See the demo:

var permute = (function () {
    return permute;

    function permute(list) {
        return list.length ?
            list.reduce(permutate, []) :
            [[]];
    }

    function permutate(permutations, item, index, list) {
        return permutations.concat(permute(
            list.slice(0, index).concat(
            list.slice(index + 1)))
            .map(concat, [item]));
    }

    function concat(list) {
        return this.concat(list);
    }
}());

var array = "ahimrsu".split("");
var permutations = permute(array).map(join);
var index = permutations.indexOf("maruish");

alert("maruish is the " + (index + 1) + "th permutation of ahimrsu.");

function join(array) {
    return array.join("");
}

Hope that helps.

Answer 2

Well, 'mariush' is actually permutation 2220 if we are using your ordering scheme:

/*jslint white: true*/
var perm = function(s){
    'use strict';
    if(s.length === 1){
        return [s];
    }
    // For each character c in s, generate the permuations p of all
    // the other letters in s, prefixed with c.
    return [].reduce.call(s, function(p,c,i){ // permutations, char, index
        var other = s.slice(0,i) + s.slice(i+1);
        return p.concat(perm(other).map(function(oneperm){
            return c + oneperm;
        }));
    }, []);
};

alert(perm('ahimrsu').indexOf('mariush') + 1); // 2220

Answer 3

Algorithm for string permutation will be a little bit more complicated with recursive step (it's possible to code it without recursion though).

The next javascript implementation is based on the description of the algorithm from this answer:

1. Remove the first letter
2. Find all the permutations of the remaining letters (recursive step)
3. Reinsert the letter that was removed in every possible location.

Implementation then something like this:

function permutation(str) {

    if (str.length == 1) {
        return [str];
    }

    var first = str[0],  // Step #1
        perms = permutation(str.slice(1)), // Step #2
        result = [];

    // Step #3
    for (var i = 0; i < perms.length; i++) {
        for (var j = 0; j <= perms[i].length; j++) {
            result.push( perms[i].slice(0, j) + first + perms[i].slice(j) );
        }
    }

    return result;
}

console.log(permutation('ahimrsu'));

Above implementation gives 5040 combinations, which seems to be correct, since 7! == 5040 (number of permutations is a factorial of the number of chars).

Now when you have all possible permutations array you can easily find specific string occurrence:

var combinations = permutation('ahimrsu'); 
var index = combinations.indexOf('mariush'); // Index of the "mariush"
alert('"mariush" is the ' + (index + 1) + 'th permutation of "ahimrsu".');