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We are given a set of $n$ numbers and want to know whether it can be partitioned to two sets with an equal sum.

To prove that an equal partition exists, it is sufficient to show a partition.

What is the shortest way to prove that an equal partition does NOT exist? Apparently the only way to prove it is to show all the partitions and show that they are not equal. Is there a shorter proof?

Erel Segal-Halevi
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    If a universally applicable, polynomially verifiable certificate were available to certify negative instances of PARTITION (or, any other NP-complete problem), then we would have NP=co-NP which would be a massive breakthrough. Given that, it may be worth exploring methods that are either 1) super-polynomial in time complexity, or 2) not universally applicable (i.e.: only useful in some subset of cases). – mhum Jul 09 '18 at 20:10
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    The paper Semantic versus syntactic cutting planes gives an unsolvable instance of subset sum which is provably hard to disprove in a particular proof system, cutting planes. – Yuval Filmus Jul 09 '18 at 23:08

1 Answers1

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The current expectation is that most likely there is no efficient (polynomial-size, polynomially-verifiable) way to do that.

The partition problem is NP-complete. For a NP-complete problem, we have an efficient way to certify that the answer to the question is YES, but not necessarily any efficient way to certify that the answer is NO. If you find an efficient way to verify that the answer is NO (i.e., a polynomial-length proof that can be verified in polynomial time), then you have proven that the problem is in co-NP. If you prove that any NP-complete problem is in co-NP, then you have proven NP = co-NP. Currently, it is widely conjectured that NP $\ne$ co-NP. Assuming that conjecture is correct, there is no polynomial-length polynomially-verifiable proof that an equal partition does not exist.

D.W.
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  • It might be worth noting that NP = co-NP if P = NP. (That's not an iff, so not quite as exciting, but still relevant.) – Draconis Jul 09 '18 at 22:03
  • Thanks! So, assuming NP $\neq$ co-NP, is there way to prove non-existence that is more efficient than just enumerating all partitions? (e.g, super-polynomial but sub-exponential, or even exponential with a small exponent) – Erel Segal-Halevi Jul 10 '18 at 05:03
  • @ErelSegal-Halevi, depending on the size of the integers, the standard pseudo-polynomial time algorithm using dynamic programming may be more efficient than enumerating all partitions. I don't know what is known about the exact complexity of this problem or whether subexponential-time algorithms are known. Related: https://en.wikipedia.org/wiki/Exponential_time_hypothesis, https://cs.stackexchange.com/q/84458/755, https://cstheory.stackexchange.com/q/4644/5038, https://cs.stackexchange.com/q/9813/755. – D.W. Jul 10 '18 at 05:35