Question

This is in regards to a problem for finding a number of combinations of an ordered set, where the elements have constraints.

As an example:

a+b+c+d+e=635, which may be…

[0-90] + [1-120] + [50-150] + [20-200] + [30-250] = 635

One solution uses multiple summations, as it was answered in the mathematics stack exchange.

https://math.stackexchange.com/questions/159197/combinatorics-using-constraints-and-ordered-set

Can someone please give a general idea of the procedure or pseudo-code to solving this type of problem?

Thank you very much!

Answer 1

A bunch of nested for-loops is the simplest way to do it.

Pseudocode:

let combinations = 0;
for a = 0 to 90
    for b = max(a+1, 1) to 120
        for c = max(b+1, 50) to 150
            for d = max(c+1, 20) to 200
                let e = 635 - a - b - c - d;
                if max(d+1, 50) <= e <= 250
                    let combinations = combinations + 1

Update

The above can be optimised a bit, but you end up with a specific, rather than general, solution.

You can observe that (a+1) >= 1 is always true, so we can get rid of the max call in the assignment to b. Likewise, (c+1) >= 20 is always true, so the assignment to d can be simplified.

You can also see that the maximum possible value of a + b + c + d is 540, which gives a minimum possible value of 95 for e. This is greater than the stated lower bound for e, so we just have to check that e >= (d+1).

We end up with:

let combinations = 0;
for a = 0 to 90
    for b = a+1 to 120
        for c = max(b+1, 50) to 150
            for d = c+1 to 200
                let e = 635 - a - b - c - d;
                if d+1 <= e <= 250
                    let combinations = combinations + 1