How to prove polynomial time equivalence?

Question

Define the problem $W$:

Input: A multi-set of numbers $S$, and a number $t$.

Question: What is the smallest subset $s \subseteq S$ so that $\sum_{k \in s} k = t$, if there is one? (If not, return none.)

I am trying to find some polytime equivalent decision problem $D$ and provide a polytime algorithm for the non-decision problem $W$ assuming the existence of a polytime algorithm for $D$.

Here is my attempt at a related decision problem:

$\mathrm{MIN\text{-}W}$:

Input: A multi-set of numbers $S$, two numbers $t$ and $k$.

Question: Is there a subset $s \subseteq S$ so that $\sum_{k \in s} k = t$ and $|s| \leq k$?

Proof of polytime equivalence:

Assume $W \in \mathsf{P}$.

solveMIN-W(S, t, k):
1. S = sort(S)
2. Q = {}
3. for i=1 to k:
4.     Q.add(S_i)
5.     res = solveW(Q, t)
6.     if res != none and res = t: return Yes
7. return No

I'm not sure about this algorithm though. Can anyone help please?

Please check the definitions. In particular, you don't show equivalence but only one reduction. Note also that $D \in \mathsf{P}$ is trivial. You might also be interested in this post, and this related discussion. — Raphael, Mar 22 '13 at 13:46

score 0 · Accepted Answer · answered Mar 22 '13 at 06:19

The problems you mention here are variations of a problem called SUBSET-SUM, FYI if you want to read some literature on it.

What is $S_i$ in your algorithm? If it's the $i$-th element, then you assume that the minimal set will use the smaller elements, which is not necessarily true - there could be a large element that is equal to $t$, in which case there is a subset of size $1$. So the algorithm doesn't seem correct.

However, as with many optimization problems that correspond to NP-complete problems, you can solve the optimization problem given an oracle to the decision problem.

First, observe that by iterating over $k$ and calling the decision oracle, you can find what is the minimal size $k_0$ of a subset whose sum equals $k$ (but not the actual subset).

After finding the size, you can remove the first element in $S$, and check whether there is still a subset of size $k_0$ - if not, then $s_1$ is surely in a minimal subset, so compute $t-s_1$, and repeat the process with the rest of $S$.

If $s_1$ does not effect the size $k_0$, then repeat the process with $S\setminus \{s_1\}$.

It is not hard to verify that this process takes polynomial time.

How to prove polynomial time equivalence?

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