JavaScript functions to calculate combinations of elements in Array.

combinations.js

/**
 * Copyright 2012 Akseli Palén.
 * Created 2012-07-15.
 * Licensed under the MIT license.
 * 
 * <license>
 * Permission is hereby granted, free of charge, to any person obtaining
 * a copy of this software and associated documentation files
 * (the "Software"), to deal in the Software without restriction,
 * including without limitation the rights to use, copy, modify, merge,
 * publish, distribute, sublicense, and/or sell copies of the Software,
 * and to permit persons to whom the Software is furnished to do so,
 * subject to the following conditions:
 * 
 * The above copyright notice and this permission notice shall be
 * included in all copies or substantial portions of the Software.
 * 
 * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
 * EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
 * MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND
 * NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS
 * BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN
 * ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN
 * CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
 * SOFTWARE.
 * </lisence>
 * 
 * Implements functions to calculate combinations of elements in JS Arrays.
 * 
 * Functions:
 *   k_combinations(set, k) -- Return all k-sized combinations in a set
 *   combinations(set) -- Return all combinations of the set
 */


/**
 * K-combinations
 * 
 * Get k-sized combinations of elements in a set.
 * 
 * Usage:
 *   k_combinations(set, k)
 * 
 * Parameters:
 *   set: Array of objects of any type. They are treated as unique.
 *   k: size of combinations to search for.
 * 
 * Return:
 *   Array of found combinations, size of a combination is k.
 * 
 * Examples:
 * 
 *   k_combinations([1, 2, 3], 1)
 *   -> [[1], [2], [3]]
 * 
 *   k_combinations([1, 2, 3], 2)
 *   -> [[1,2], [1,3], [2, 3]
 * 
 *   k_combinations([1, 2, 3], 3)
 *   -> [[1, 2, 3]]
 * 
 *   k_combinations([1, 2, 3], 4)
 *   -> []
 * 
 *   k_combinations([1, 2, 3], 0)
 *   -> []
 * 
 *   k_combinations([1, 2, 3], -1)
 *   -> []
 * 
 *   k_combinations([], 0)
 *   -> []
 */
function k_combinations(set, k) {
	var i, j, combs, head, tailcombs;
	
	// There is no way to take e.g. sets of 5 elements from
	// a set of 4.
	if (k > set.length || k <= 0) {
		return [];
	}
	
	// K-sized set has only one K-sized subset.
	if (k == set.length) {
		return [set];
	}
	
	// There is N 1-sized subsets in a N-sized set.
	if (k == 1) {
		combs = [];
		for (i = 0; i < set.length; i++) {
			combs.push([set[i]]);
		}
		return combs;
	}
	
	// Assert {1 < k < set.length}
	
	// Algorithm description:
	// To get k-combinations of a set, we want to join each element
	// with all (k-1)-combinations of the other elements. The set of
	// these k-sized sets would be the desired result. However, as we
	// represent sets with lists, we need to take duplicates into
	// account. To avoid producing duplicates and also unnecessary
	// computing, we use the following approach: each element i
	// divides the list into three: the preceding elements, the
	// current element i, and the subsequent elements. For the first
	// element, the list of preceding elements is empty. For element i,
	// we compute the (k-1)-computations of the subsequent elements,
	// join each with the element i, and store the joined to the set of
	// computed k-combinations. We do not need to take the preceding
	// elements into account, because they have already been the i:th
	// element so they are already computed and stored. When the length
	// of the subsequent list drops below (k-1), we cannot find any
	// (k-1)-combs, hence the upper limit for the iteration:
	combs = [];
	for (i = 0; i < set.length - k + 1; i++) {
		// head is a list that includes only our current element.
		head = set.slice(i, i + 1);
		// We take smaller combinations from the subsequent elements
		tailcombs = k_combinations(set.slice(i + 1), k - 1);
		// For each (k-1)-combination we join it with the current
		// and store it to the set of k-combinations.
		for (j = 0; j < tailcombs.length; j++) {
			combs.push(head.concat(tailcombs[j]));
		}
	}
	return combs;
}


/**
 * Combinations
 * 
 * Get all possible combinations of elements in a set.
 * 
 * Usage:
 *   combinations(set)
 * 
 * Examples:
 * 
 *   combinations([1, 2, 3])
 *   -> [[1],[2],[3],[1,2],[1,3],[2,3],[1,2,3]]
 * 
 *   combinations([1])
 *   -> [[1]]
 */
function combinations(set) {
	var k, i, combs, k_combs;
	combs = [];
	
	// Calculate all non-empty k-combinations
	for (k = 1; k <= set.length; k++) {
		k_combs = k_combinations(set, k);
		for (i = 0; i < k_combs.length; i++) {
			combs.push(k_combs[i]);
		}
	}
	return combs;
}

I think it doesnt work for 4 elements

It works great. Thanks. I tested it for 4 elements as well and it works great. Thanks.

So it works but only generates combinations in the same original order - how could you get all combinations in any order?

thanks a lot

@chrisgoddard you're looking for a permutations function, not a combinations function.

@chrisgoddard you can run each combination through a permutation algorithm with something like this: https://github.com/eriksank/permutation-engine
Or, use permutation/combination engine, like this one:
https://github.com/dankogai/js-combinatorics

Is the above gist available on npm somewhere? I want to use it in my library, right now I've just copy & pasted it.

@danbars Thanks for the links. +1

Does Not work for this set "ababb" it repeats elements and also "abbba", "babab" are missing

@foluis all member elements must be considered unique in order to form a set mathematically

possible combination of 1, 2, 3 are following,may be i am wrong

[1], [2], [3], [1, 2], [1, 3], [2, 3], [2, 1], [3, 1], [3, 2], [1, 2, 3], [3, 2, 1]

Thanks a lot!

Thanks!

@imVinayPandya Looks like he is using N choose K which should only return unique combinations. Such as 1,2 and 2,1 are the same. Also, you are setting the number of members for the k combination. Thus this will only return single member combinations, or double member combinations or triple member combinations etc etc etc.

thanks!

Here is a short hand version using underscore and some es6 syntax for readability

"use strict";
//https://gist.github.com/axelpale/3118596

const _ = require('underscore');

const elemTransform = elem => [elem];
const tailcombPush = (combs, elem) => tailcomb => combs.push([elem, ...tailcomb]);
const k_combPush = combs => k_komb => combs.push(k_komb);

function k_combinations(set, k) {
  const setLen = set.length;
  if (k > setLen || k <= 0) {
    return [];
  }
  if (k === setLen) {
    return [set];
  }
  if (k === 1) {
    return _.map(set, elemTransform);
  }
  const combs = [];
  for (let i = 0; i < setLen - k + 1; i++) {
    _.each(k_combinations(set.slice(i + 1), k - 1), tailcombPush(combs, set[i]));
  }
  return combs;
}

function combinations(set) {
  const combs = [];
  for (let k = 1; k <= set.length; k++) {
    _.each(k_combinations(set, k), k_combPush(combs));
  }
  return combs;
}

module.exports = {
  k_combinations,
  combinations
};

Here's an es6 version

const k_combinations = (set, k) => {                                            
  if (k > set.length || k <= 0) {                                               
    return []                                                                   
  }                                                                             
  if (k === set.length) {                                                       
    return [set]                                                    
  }                                                                             
  const combs = []                                                              
  if (k === 1) {                                                                
    for (let i = 0; i < set.length; i++) {                                      
      combs.push([set[i]])                                                      
    }                                                                           
    return combs                                                                
  }                                                                             
  for (let i = 0; i < set.length - k + 1; i++) {                                
    const head = set.slice(i, i + 1)                                            
    const tailcombs = k_combinations(set.slice(i + 1), k - 1)              
    for (let j = 0; j < tailcombs.length; j++) {                                
      combs.push(head.concat(tailcombs[j]))                                     
    }                                                                           
  }                                                                             
  return combs                                                                  
}                                                                               
                                                                                
const combinations = (set) => {                                                 
  const combs = [];                                                             
  for (let k = 1; k <= set.length; k++) {                                       
    const k_combs = k_combinations(set, k)                                      
    for (let i = 0; i < k_combs.length; i++) {                                  
      combs.push(k_combs[i])                                                    
    }                                                                           
  }                                                                             
  return combs                                                                  
}

ES6 version:

const k_combinations = (set, k) => {
  if (k > set.length || k <= 0) {
    return []
  }
  
  if (k == set.length) {
    return [set]
  }
  
  if (k == 1) {
    return set.reduce((acc, cur) => [...acc, [cur]], [])
  }
  
  let combs = [], tail_combs = []
  
  for (let i = 0; i <= set.length - k + 1; i++) {
    tail_combs = k_combinations(set.slice(i + 1), k - 1)
    for (let j = 0; j < tail_combs.length; j++) {
      combs.push([set[i], ...tail_combs[j]])
    }
  }
  
  return combs
}

const combinations = set => {
  return set.reduce((acc, cur, idx) => [...acc, ...k_combinations(set, idx + 1)], [])
}

And here is a (generic) typescript version, based on the ES6 version by @luispaulorsl:

export const k_combinations = <T>(set: T[], k: number) => {
    if (k > set.length || k <= 0) {
        return [];
    }

    if (k === set.length) {
        return [set];
    }

    if (k === 1) {
        return set.reduce((acc, cur) => [...acc, [cur]], [] as T[][]);
    }

    const combs = [] as T[][];
    let tail_combs = [];

    for (let i = 0; i <= set.length - k + 1; i++) {
        tail_combs = k_combinations(set.slice(i + 1), k - 1);
        for (let j = 0; j < tail_combs.length; j++) {
            combs.push([set[i], ...tail_combs[j]]);
        }
    }

    return combs;
};

export const combinations = <T>(set: T[]) => {
    return set.reduce((acc, _cur, idx) => [...acc, ...k_combinations(set, idx + 1)], [] as T[][]);
};