Find all combinations of elements in an array

Question

The problem I am trying to solve is that I want to generate an array which represents all combinations of elements in a source array.

Given an input of 2 items, there are 4 combinations (represented here as binary)

Group 1 | 0 | 1 | 0 | 1
Group 2 | 0 | 0 | 1 | 1
--------------------------
&Result | 0 | 1 | 2 | 3

This can be generalised so the number of combinations is 2^{(number of groups)} (So 3 groups has 8 combinations, 4 has 16 etc).

So the question is; given a javascript array:

var groups = [
{
    name:"group1",
    bit: 1
},
{
    name:"group2",
    bit: 2
},
{
    name:"group3",
    bit: 4
}];

I need to generate an array where the index represents the and'ing of the bit property and the value of the array is arbitrary (further calculation - not relevant) so lets just make it an array of the group names (for the purpose of this question). This result is desirable:

var result = [
    {groups: []}, //0 
    {groups: ["group1"]}, //1
    {groups: ["group2"]},  //2
    {groups: ["group1","group2"]}, //3
    {groups: ["group3"]}, //4
    {groups: ["group1","group3"]}, //5
    {groups: ["group2","group3"]}, //6
    {groups: ["group1","group2","group3"]} //7
]

You can see in the comments there that each index in the array represents the act of and'ing the bit property from the original.

I have prepared a jsfiddle with the input and required output should it be useful in answering the question.

This is my current solution, based on mellamokb's answer but re-written in my prefered style. I was hoping there was a more elegant solution as this has a lot of iterations over the array which are unecessary. Any better solutions?

var resultCount = Math.pow(2,groups.length);
var result = [];
for(var i=0;i<resultCount;i++){
    result.push({
        groups: $.map(groups, function(e,idx){ 
           return  ((i & Math.pow(2,idx)) != 0)
               ? e.name
               : null
        })
    });
}

Does the solution need to worry about performance? What is the largest number of input groups you anticipate having? — mellamokb
– mellamokb, Commented Mar 31, 2014 at 20:12
Largest number of groups in practice is 4 (so 16 combinations). Realistically i'd err on the side of caution and expect there could be 6. — Jamiec
– Jamiec, Commented Mar 31, 2014 at 20:17
could you explain what iterations are unnecessary? i dont see how you could reduce the problem to less than (length^3) iterations if you want to enumerate all of the possibilities. — tau
– tau, Commented Mar 31, 2014 at 21:23
@tau - the only unnecessary combinations are reversed ones. All combinations are required. — Jamiec
– Jamiec, Commented Mar 31, 2014 at 22:58

mellamokb · Accepted Answer · 2014-03-31 21:35:51Z

4

Here is a relatively efficient solution that builds up the arrays by index:

var result = [];
var resultCount = Math.pow(2, groups.length);

for (var i = 0; i < resultCount; i++) {
    result[i] = { groups: [] };
    for (var g = 0; g < groups.length; g++) {
        if (i & groups[g].bit) result[i].groups.push(groups[g].name);
    }
}

Demo: http://jsfiddle.net/7ZsHL/9/

edited Mar 31, 2014 at 21:35

answered Mar 31, 2014 at 20:17

mellamokb

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2 Comments

Jamiec Over a year ago

This looks really good, but you can replace the iteration over the groups to calculate resultCount with var resultCount = Math.pow(2,groups.length)

mellamokb Over a year ago

@Jamiec: Good idea. I've always been a bit nervous about using Math.pow because it is a floating-point calculation.

tau · Accepted Answer · 2014-03-31 20:38:55Z

1

I think you can do it without having to store the bit and what not.

var groups = [
    "group1",
    "group2",
    "group3",
    "group4"
];

var output = [];
for (var i = 0; i < Math.pow(2, groups.length); i++) {
    var arr = [];
    output.push(arr);
    for (var j = 0; j < groups.length; j++) {
        if (i & (Math.pow(2, j))) {
            arr.push(groups[j]);
        } else {
            arr.push(0);
        }
    }
}

answered Mar 31, 2014 at 20:38

tau

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1 Comment

mellamokb Over a year ago

You might want to store Math.pow(2, groups.length) into a variable, otherwise it is getting recalculated with every iteration.

Levi Beckman · Accepted Answer · 2014-03-31 21:11:20Z

1

I like what others have done, but in a situation like this, I might choose to be a little mover verbose in my code, as the intent of the algorithm is hard to communicate.

An Update Of Your Fiddle

var groups = [
    {
        name:"group1",
        bit: 1
    },
    {
        name:"group2",
        bit: 2
    },
    {
        name:"group3",
        bit: 4
    }];

function parseIterations(groups){
    var aIterations = [];
    var aBits = [];
    var iGroups = groups.length;
    var iTotalIterations = Math.pow(2, iGroups);
    for (var i = 0; i < iTotalIterations; i++) {
        var oIteration = { groups: [] };
        for (var j = 0; j < iGroups; j++) {
            if (typeof aBits[j] == 'undefined')
                aBits[j] = true;
            // while you could infer the .bit value from j, 
            // i've chosen to use your .bit value here.
            aBits[j] = (i % groups[j].bit == 0) ? !aBits[j] : aBits[j];
            if (aBits[j])
                oIteration.groups.push(groups[j].name);
        }
        aIterations[i] = oIteration;
    }
    return aIterations;
}

var result = parseIterations(groups);

answered Mar 31, 2014 at 21:11

Levi Beckman

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1 Comment

Jamiec Over a year ago

I like the sentiment, but this still iterates multiple times so is essentially the same answer. Still +1 though. Thanks for your answer.

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