Get the closest value for combinations of an array (JS)

Question

I'm looking for an algorithm that I can use for combining values in array, to get as close as possible to "another value".

For instance, the number I want to find out what combination that gives the closes result to is 2.5. And my array is [0.5, 1.0, 1.5, 2.0, 3.0]. The combination in this case would be 2.0+0.5.

2.7 would yield the same combo (2.5 is the closest), while 3.7 would yield 3.0+0.5 and 7.0 would be 3.0+3.0+1.0.

I've been reading up on different algorithms to create available combinations and such – for instance this one: https://codereview.stackexchange.com/questions/7001/better-way-to-generate-all-combinations However, I'm having difficulties to write a function that allows for the same value to be used multiple times (like my example with 7.0). This makes the number of combinations quite large.

Anyone having a good example tucked away? Or have any pointers to give?

EDIT @zkar told me about the "knapsack problem". I may add that for my example, the sought after value are in a specified range (1.0 and 10.0) – which limits the the combinations somewhat.

Answer 1

You can try this simple algorithm (JSFiddle demo):

/**
 * @param src {Array} List of available values
 * @param val {Number} Target value
 * @returns {Array}
 */
function get_combinations(src, val)
{
    var result = [];
    var source = src.slice();
    source.sort();

    while (val > 0)
    {
        for (var i = source.length - 1; i >= 0; i--)
        {
            if (source[i] <= val || i == 0)
            {
                val = val - source[i];
                result.push(source[i]);
                break;
            }
        }
    }

    return result;
}

Answer 2

Your problem is a mixture of Coin Problem and Knapsack Problem

If Coins are used only Once:

Given a set of values S, n = |S|, m: value to approximate

DEFINE BEST = { }
DEFINE SUM = 0
DEFINE K = 0

WHILE S IS NOT EMPTY DO
    K = K + 1
    FIND MIN { Si : |(SUM+Si) - m| is minimal }
    ADD TUPLE < Si, |(SUM+Si) - m|, K > to BEST
    SUM = SUM + Si
    REMOVE Si from S
END-FOR

RETURN BEST

This algorithm runs in Time: O(|S|²) ~ O(n²)

The Set BEST will have n solutions, for each K: 1..n

for K: you have the optimal choice at that stage

to find the complete solution:

GIVEN BEST = { < COIN:X, DISTANCE:Y, DEGREE:K > }
DEFINE SOLUTION = { }
Y" = MINIMUM Y IN BESTi.Y for i: 1..n
KEEP ADDING BESTj.X to SOLUTION UNTILL BESTj.Y = Y" FOR j: 1..n

If Coins can be re-used:

DEFINE SOLUTION = { }
DEFINE SUM = 0
LESS = { Si : Si < m }
SORT LESS IN DESCENDING ORDER
FOR Li in LESS DO
    WHILE (SUM+Li) <= m DO
        SUM = SUM + Li
        ADD Li TO SOLUTION
    END-WHILE

    IF SUM = m THEN BREAK-FOR
END-FOR
RETURN SOLUTION

In JavaScript:

function coinProblem (var coins, var value)
{
    var solution = new Array();
    var sum = 0;
    var less = new Array();

    for (var i in coins)
        if (i <= value)
            less.push(i);

    // sort in descending order
    less.sort();
    less.reverse();

    for (var i in less)
    {
        while ((sum+i) <= value)
        {
            solution.push(i);
            sum = sum + i;
        }

        if (sum == value) break;
    }

    return solution;
}