Question

I’m trying to generate all possible combinations for pair of 1’s within given bit width.

Let’s say the bit width is 6, i.e. number 32. This is what I would like to generate:

000000
000011
000110
001100
001111
011000
011011
011110
110000
110011
110110
111100
111111

If I have variables:

var a = 1,
    b = 2;
    num = a | b;

and create a loop that I’ll loop over width - 1 times, and where I shift both a << 1 and b << 1, I’ll get all combinations for one pair. After that, I’m pretty much stuck.

Could someone , please, provide some help.

Update: working example
Based on Barmar’s mathematical approach, this is what I managed to implement

var arr = [],
    arrBits = [];

function getCombs(pairs, startIdx) {
    var i, j, val = 0, tmpVal, idx;

    if (startIdx + 2 < pairs) {
        startIdx = arr.length - 1;
        pairs -= 1;
    }

    if (pairs < 2) {
        return;
    }

    for (i = 0; i < pairs-1; i++) {
        idx = startIdx - (i * 2);
        val += arr[idx];

    }

    for (j = 0; j < idx - 1; j++) {
        arrBits.push((val + arr[j]).toString(2));
    }

    getCombs(pairs, startIdx-1);
}

(function initArr(bits) {
    var i, val, pairs, startIdx;

    for (i = 1; i < bits; i++) {
        val = i == 1 ? 3 : val * 2;
        arr.push(val);
        arrBits.push(val.toString(2));
    }

    pairs = Math.floor(bits / 2);
    startIdx = arr.length - 1;

    getCombs(pairs, startIdx);
    console.log(arrBits);

}(9));

Working example on JSFiddle
http://jsfiddle.net/zywc5/

Answer 1

The numbers with exactly one pair of 1’s are the sequence 3, 6, 12, 24, 48, …; they start with 3 and just double each time.

The numbers with two pairs of 1’s are 12+3, 24+3, 24+6, 48+3, 48+6, 48+12, …; these are the above sequence starting at 12 + the original sequence up to n/4.

The numbers with three pairs of 1’s are 48+12+3, 96+12+3, 96+24+3, 96+24+6, …

The relationship between each of these suggests a recursive algorithm making use of the original doubling sequence. I don’t have time right now to write it, but I think this should get you going.