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(This is similar to computing n choose k (C(n,k)), as discussed in the notes for the previous class [/Notes 04 30.]?) |
The subset-sum problem is defined as follows:
You are given a set S of non-negative integers: {A[1], ..., A[n]} and a goal G (also a non-negative integer).
The question is whether, for any subset S' ⊆ S, the sum of the integers in S' is G.
For i=1,2,..,n and h=0,1,...,G, define
Claim: M satisfies the following recurrence:
This is because any subset S'' of {A[1],A[2],...,A[i]} summing to h either contains A[i] or doesn't. If the subset S'' doesn't contain A[i], then the subset is also a subset of {A[1],...,A[i-1]} summing to h. If the subset S'' does contain A[i], then removing A[i] yields a subset of {A[1],...,A[i-1]} that sums to h-A[i].
This gives the following recursive algorithm:
bool subset-sum(Array<int> A, int i, int h) {
if (i == 0) return (h == 0);
else return subset-sum(A, i-1, h) || subset-sum(A, i-1, h-A[i]);
}
As implemented, the running time is exponential (something like 2i). (What is the recursion tree?)
We can modify it to cache the answers:
bool subset-sum(Array<int> A, int i, int h) {
static Array<Array<bool> > M;
if (M[i].exists(h)) return M[i][h];
if (i == 0) return (h == 0);
else return (M[i][h] = (subset-sum(A, i-1, h) || subset-sum(A, i-1, h-A[i])));
}
Now what does the recursion diagram look like?
Since there are at most n*G distinct subproblems, and the time spent in each call (not counting the recursion) is O(1), the total time is O(n*G).