# SUBSET SUM reduction to PARTITION

This is the PARTITION problem:

Given a multiset S of positive integers, decide if it can be partitioned into two equal-sum subsets.

This is the SUBSET SUM problem:

Given a multiset S of integers and an integer T, decide if any subset of S sums to T.

The positive variant of SUBSET SUM is NP-complete.

SUBSET SUM can be reduced to PARTITION as follows:

Define S' = S + {c}, where c = 2T - sum(S). S' can be partitioned into two equal-sum subsets iff there is a subset in S summing to T.

Proof:

Partition S into A and B:

S = A + B

If there exists a subset in S summing to T, then S can be partitioned so either sum(A) = T or sum(B) = T. Suppose sum(A) = T. Let S' = A + B' where B' = B + {c}.

sum(B') = sum(S) - sum(A) + c
sum(B') = sum(S) - sum(A) + 2T - sum(S)
sum(B') = -sum(A) + 2T


By assumption, T = sum(A) therefore sum(B') = sum(A). This means S' can be partitioned into two equal-sum subsets.

If there does not exist a subset in S summing to T, then there cannot exist a c such that S' can be partitioned into two equal-sum subsets.

Suppose for contradiction c can exist.

Adding c to B creates S' = A + B'.

What value of T produces c such that sum(A) = sum(B') = sum(B) + c?

sum(A) = sum(B) + c
sum(A) = sum(B) + 2T - sum(S)
sum(A) = sum(B) + 2T - (sum(A) + sum(B))
sum(A) = sum(B) + 2T - sum(A) - sum(B)
2sum(A) = 2T
sum(A) = T


But A already sums to sum(A), contradicting the assumption that there is no subset of S summing to T.

If 2T < sum(S), then c < 0, but PARTITION only applies for positive integer multisets. Does this reduction still hold?

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– D.W.
Jul 19, 2021 at 5:15

You should make $$c=|2T-sum(S)|$$. Show that the reduction then holds no matter what $$T$$ is.