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JavaScript functions to calculate combinations of elements in Array.
/**
* 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;
}
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agrimnigam21 Nov 6, 2013

I think it doesnt work for 4 elements

I think it doesnt work for 4 elements

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hardcoremore Nov 8, 2013

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

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

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chrisgoddard Aug 9, 2014

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

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

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@ghost

ghost Aug 18, 2014

thanks a lot

ghost commented Aug 18, 2014

thanks a lot

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skibulk Sep 10, 2015

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

skibulk commented Sep 10, 2015

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

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danbars Sep 28, 2015

@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

danbars commented Sep 28, 2015

@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

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rhalff Feb 23, 2016

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

rhalff commented Feb 23, 2016

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

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sallar Feb 24, 2016

@danbars Thanks for the links. +1

sallar commented Feb 24, 2016

@danbars Thanks for the links. +1

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foluis Aug 5, 2016

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

foluis commented Aug 5, 2016

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

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ComethTheNerd Oct 25, 2016

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

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

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imVinayPandya Mar 31, 2017

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]

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]

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douglasFranco Apr 2, 2017

Thanks a lot!

Thanks a lot!

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ragamufin May 5, 2017

Thanks!

Thanks!

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t1mduma5 Jul 22, 2017

@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.

@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.

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regevbr Feb 20, 2018

thanks!

regevbr commented Feb 20, 2018

thanks!

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regevbr Feb 20, 2018

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
};

regevbr commented Feb 20, 2018

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
};
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morenoh149 Feb 28, 2018

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                                                                  
}

morenoh149 commented Feb 28, 2018

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                                                                  
}
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