Is Measurable

measurable.js

// Actually, this is not going to work. Please ignore it.

function isMeasurable(target, array) {
    if (array.length < 1) { return false; }
    if (array.indexOf(target) >= 0) { return true; }
    
    var weights = array.concat(target);
    var length  = 1 << weights.length;
    
    for (var i = 0; i < length; i++) {
        var position = weights.length - 1;
        var bitmask  = i;
        var subset   = [];
        while(bitmask > 0) {
            if((bitmask & 1) == 1) { subset.push(weights[position]); }
            
            position--;
            bitmask >>= 1;
        }
        
        if (subset.length < 2) { continue; }
        if (subset.indexOf(target) < 0) { continue; }
        
        var total = sum(subset);
        if (total % 2 == 0 && total/2 >= target) { return true; }
    }
    
    return false;
}

function sum(subset) {
    var sum = 0;
    for (var i = 0; i < subset.length; i++) { sum += subset[i]; }
    
    return sum;
}

What are you trying to do?

Solve the following problem by iterating over the power-set (non-recursive): Given an array of weights, is it possible to measure the target on a traditional scale. You are allowed to put the weights on either side.

Ah, like finding out the following:
With [1, 2, 4, 8] you can measure targets between 0 and 15 by putting weights on one side and the target on the other side.
With [1, 3, 9] you can measure targets between 0 and 13 by putting weights on both sides and the target on the other side.

Sort of. To be precise:
isMeasurable(8, [5, 6, 9]) => true
isMeasurable(13, [5, 6, 9]) => false

Untested: https://gist.github.com/1021993

Cool, this works :-)

Can you come up with a recursive version?

It worked directly? There wasn't even a typo? Wow...
I'll think about a recursive version.

I've updated my gist with a recursive version (untested).