Question

I need an algorithm for this problem:

Given a set of n natural numbers x1,x2,…,xn, a number S and k. Form the sum of k numbers picked from the set (a number can be pick many times) with sum S.

Stated differently: List every possible combination for S with Bounds: n<=256, x<=1000, k<=32

E.g.

  problem instance: {1,2,5,9,11,12,14,15}, S=30, k=3
  There are 4 possible combinations
  S=1+14+15, 2+14+14, 5+11+15, 9+9+12.   

With these bounds, it is unfeasible to use brute force but I think of dynamic programming is a good approach.

The scheme is: Table t, with t[m,v] = number of combinations of sum v formed by m numbers.

1. Initialize t[1,x(i)], for every i.   
2. Then use formula t[m,v]=Sum(t[m-1,v-x(i)], every i satisfied v-x(i)>0), 2<=m<=k.   
3. After obtaining t[k,S], I can trace back to find all the combinations.  

The dilemma is that t[m,v] can be increase by duplicate commutative combinations e.g., t[2,16]=2 due to 16=15+1 and 1+15. Furthermore, the final result f[3,30] is large, due to 1+14+15, 1+15+14, …,2+14+14,14+2+14,…

How to get rid of symmetric permutations? Thanks in advance.

Answer 1

You can get rid of permutations by imposing an ordering on the way you pick elements of x. Make your table a triple t[m, v, n] = number of combinations of sum v formed by m numbers from x1..xn. Now observe t[m, v, n] = t[m, v, n-1] + t[m-1, v-x_n, n]. This solves the permutation problem by only generating summands in reverse order from their appearance in x. So for instance it’ll generate 15+14+1 and 14+14+2 but never 14+15+1.

(You probably don’t need to fill out the whole table, so you should probably compute lazily; in fact, a memoized recursive function is probably what you want here.)