Question

I have a problem related to the subset sum problem and am wondering if the differences make it easier, i.e. solvable in a reasonable amount of time.

Given a value V, a set size L, and a sequence of numbers [1,N] S, how many size L subsets of S sum to less than V?

This is different than the subset sum problem in three ways:

1. I care how many subsets are less than a given value, not how many are equal.
2. The subset sizes are fixed.
3. I care how many sets sum to less than V, not just whether any exist.

Is there any reasonably efficient algorithm to solve this?

Edit: Obviously, this can be done in O(N choose L) using a combination generating algorithm. What I’m really interested in is clever hacks to significantly speed it up past that.

Answer 1

(The decision version of) your problem is still NP-complete. The idea is that if we could solve your problem, then (for each subset size, say) we could ask how many sets sum to less than V and how many sum to less than V-1, and the difference of those two numbers would tell us whether are subsets that sum to exactly V — thus we could solve the subset sum problem. [This is not a complete proof, because it’s a Turing reduction, not a many one reduction.]

However, there is a simple dynamic programming solution that runs in time O(nLV). [The reason this does not prove that P=NP is that V could be exponential in the input size: with n bits, you can represent values upto 2ⁿ. But assuming that your V is not exponential, this is not a problem.]

Let num[v][k][i] denote the number of size-k subsets of the first i elements of S that sum to v. You can calculate them as (for each i):

    num[0][0][i] = 1     for v = 1 to V:         for k = 1 to L:             num[v][k][i] = num[v][k][i-1] + num[v-S[i]][k-1][i-1] 

where S[i] is the ith element in your sequence. (Any set of size k that sums to v either doesn’t use S[i], so it’s counted in num[v][k][i-1], or it uses S[i] which means that the rest of the subset has k-1 elements, uses only the first i-1 numbers in the sequence, and sums to v-S[i].) Finally, count num[v][L][|S|] for each v less than V; that’s your answer.

Also, you can omit the third subscript if you do it carefully (run your loop downwards for each i, etc.); I only included it for clarity.