Subset Sum: reduce special to general case

Problem Detail: 

Wikipedia states the subset sum problem as finding a subset of a given set of integers, whose sum is zero. Further it describes it as equivalent to finding a subset with sum $s$ for any given $s$.

So I believe as they are equivalent, there must be a reduction in either side. The one from $s$ to zero is trivial by setting $s = 0$. But I had no luck finding a reduction from zero to $s$, i.e. given a set of integers $A$, construct a set of integers $B$ containing a subset with sum $s$ (for any $s$), if and only if there is as subset of $A$ with sum zero.

Can you give me some pointers?

Asked By : ipsec

Answered By : Aryabhata

You actually already have a reduction from special to general. By setting $s=0$, you are basically using the general algorithm to solve the special problem.

For the other way round (i.e. a reduction from general to special):

Suppose you are given a set $S = \{x_1, \dots, x_n\}$ and a number $K$ and you have to determine if there is some subset of $S$ which sums to $K$.

Now you want to solve this problem, given an algorithm for the case where you can determine if some subset sums to $0$.

Now if $x_i \gt 0$, we have an easy reduction: $S' = \{x_1, x_2, \dots, x_n, -K\}$.

$S'$ has a subset of sum $0$ iff $S$ has a subset of sum $K$.

The problem occurs when we can have $x_i \le 0$ for some of the $i$.

We can assume that $K \gt 0$ (why?).

Suppose the sum of the positive $x_i$ is $P$ and the negative $x_i$ is $N$.

Now construct a new set $S' = \{y_1, y_2 \dots, y_n\}$ such that

$y_i = x_i + M$ where $M = P + |N| + K$.

Each $y_i \gt 0$.

Now run the zero-subset-sum algorithm on the sets

$ S' \cup \{-(K+M)\}$

$ S' \cup \{-(K+2M)\}$

$ S' \cup \{-(K+3M)\}$

$\dots$

$ S' \cup \{-(K+nM)\}$

It is easy to show that if $S$ has a subset of sum $K$, then at least one of the above sets has subset of sum zero.

I will leave the proof of the other direction to you.

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Question Source : http://cs.stackexchange.com/questions/10981