Question

Problem 2. In the Subset-Sum problem the input consists of a set of positive integers X = {x1, . . . , xn}, and some integer k. The answer is YES if and only if there exists some subset of X that sums to k.

In the Bipartition problem the input consists of a set of positive integers Y = {y1, . . . , yn}.

The answer is YES if and only if there exists some subset of X that sums to (Pn i=1 xi)/2.

We saw in class that the Subset-Sum problem is NP-complete. Show that the Bipartition problem is also NP-complete.

Hint: Devise a polynomial reduction from Subset-Sum to Bipartition. In your reduction you start with an input X, k of Subset-Sum and you construct an input Y of Bipartition. The goal is that there exists some subset of Y that sums to (P y∈Y y)/2 if and only if there exists some subset of X that sums to k. To that end, it might be a good idea to set Y to be equal to X, together with some additional integers that somehow encode k.

Answer 1

It is easy to see that SET-PARTITION can be verified in polynomial time; given a partition P1,P2P1,P2just sum the two and verify that their sums equal each other, which is obviously a polynomial time verification (because summation is a polynomial operation and we are only performing at most |X||X| many summations).

The core of the proof is in reducing SUBSETSUM to PARTITION; to that end given set XX and a value tt (the subset sum query) we form a new set X′=X∪{s−2t}X′=X∪{s−2t} where s=∑x∈Xxs=∑x∈Xx. To see that this is a reduction:

• (⟹⟹ ) assume there exists some S⊂XS⊂X such that t=∑x∈Sxt=∑x∈Sx then we would have that

  s−t=∑x∈S∪{s−2t}x,s−t=∑x∈S∪{s−2t}x,

  s−t=∑x∈X′∖(S∪{s−2t})xs−t=∑x∈X′∖(S∪{s−2t})x

  and we would have that S∪{s−2t}S∪{s−2t} and X′∖(S∪{s−2t})X′∖(S∪{s−2t}) form a partition of X′X′

• (⟸⟸) Suppose that there is a partition P′1,P′2P1′,P2′ of X′X′ such that ∑x∈P′1x=∑x∈P′2x∑x∈P1′x=∑x∈P2′x. Notice that this induces a natural partition P1P1 and P2P2 of XX such that WLOG we have that

  s−2t+∑x∈P1x=∑x∈P2xs−2t+∑x∈P1x=∑x∈P2x

  ⟹s−2t+∑x∈P1x+∑x∈P1x=∑x∈P2x+∑x∈P1x=s⟹s−2t+∑x∈P1x+∑x∈P1x=∑x∈P2x+∑x∈P1x=s

  ⟹s−2t+2∑x∈P1x=s⟹s−2t+2∑x∈P1x=s

  ⟹∑x∈P1x=t⟹∑x∈P1x=t

Hence from a solution t=∑x∈Sxt=∑x∈Sx we can form a parition P1=S∪{s−2t}P1=S∪{s−2t}, P2=X′∖(S∪{s−2t})P2=X′∖(S∪{s−2t}) and conversely from a partition P′1,P′2P1′,P2′ we can form a soltuion t=∑x∈P′1∖{s−2t}xt=∑x∈P1′∖{s−2t}x and therefore the mapping f:(X,t)→X′f:(X,t)→X′ is a reduction (because (X,t)(X,t) is in the language/set SUBSETSUM ⇔X′=f(X,t)⇔X′=f(X,t) is in the language/set PARTITION) and it is clear to see that the transformation was done in polynomial time.