Unions of Disjoint NP-Complete Sets [chapter]

Christian Glaßer, John M. Hitchcock, A. Pavan, Stephen Travers
2011 Lecture Notes in Computer Science  
We study the following question: if A and B are disjoint NP-complete sets, then is A ∪ B NP-complete? We provide necessary and sufficient conditions under which the union of disjoint NP-complete sets remain complete. Introduction A disjoint NP pair is a pair of languages (A, B) such that both A and B are in NP and are disjoint. Given a disjoint pair (A, B), we can view strings from A as "yes" instances and strings from B as "no" instances. We are interested in an algorithm that accepts all
more » ... nces from A and rejects all instances from B. Viewed this way, disjoint pairs are an equivalent formulation of promise problems, where A ∪ B is the promise. Promise problems were introduced by Even, Selman, and Yacobi [ESY84] . Promise problems and disjoint pairs arise naturally in many scenarios such as the study of complete problems for semantic classes and the study of hardness of approximation problems. In some instances promise problems more precisely capture the underlying computational problem rather than decision problems. Sometimes unresolved questions about complexity classes can be answered by considering promise versions of complexity classes. For example, we know that Promise-MA does not have fixed polynomial-size circuits whereas we do not have an analogous result for the class MA [San09] . For a recent survey on promise problems we refer the reader to [Gol06]. In addition to be able to capture several natural computational problems, disjoint pairs arise naturally in the study of public key cryptosystems and propositional proof systems. The computational problem capturing a public key cryptosystem can be formulated as a disjoint NP pair (A, B) [ESY84, GS88]. A separator of such a pair (A, B) is a set S with A ⊆ S and S ⊆ B. The class of pairs (A, B) whose separators do not belong to P are called P-inseparable pairs. The existence of P-inseparable disjoint NP pairs is closely related the existence of secure public key cryptosystems [ESY84, GS88] . Grollmann and Selman [GS88] showed that if P = UP, then there exist P-inseparable disjoint NP pairs. More recently Fortnow, Lutz, and Mayordomo [FLM10] showed that if NP does not have p-measure zero, then P-inseparable disjoint NP pairs exist. Works of Razborov [Raz94] and Pudlak [Pud01] show that disjoint NP pairs are also closely related to the study of propositional proof systems. Razborov identified a canonical disjoint NP pair 1 (SAT * , REF f ) for every propositional proof system f . Here SAT * is a padded version of SAT and REF f is the set of all formulas that have short proofs of unsatisfiability with respect to f . Glaßer, Selman, and Zhang [GSZ07] showed that for every disjoint NP pair (A, B) there is a propositional proof system f such that its canonical pair (SAT * , REF f ) is many-one equivalent to (A, B). Thus disjoint NP pairs and propositional systems have identical degree structure. There is a close relation between disjoint NP pairs and pairs whose both components are NPcomplete. For example, there is a P-inseparable disjoint NP pair if and only there is a P-inseparable pair whose both components are NP-complete [GS88] . We also know that if there is a complete pair for DisjNP, then there is such a pair where both components are NP-complete [GSS05] . In this article we focus on disjoint pairs whose both components are NP-complete. We investigate the following question: let (A, B) be a disjoint NP pair such that both A and B are NP-complete. Is the union A ∪ B NP-complete? This question was first explicitly raised by Selman [Sel88]. Apart from its connections to the study of public key cryptosystems and propositional proof systems, our question is also of independent interest. We are interested in a simple closure property of NP-complete sets-closure under disjoint unions. It is known that every NP-complete set can be split into two disjoint NP-complete sets [GPSZ08]. Here we are raising the converse question, is the combination of every two disjoint NP-complete sets NP-complete? Glaßer et al. [GSTW08] showed that if A and B are disjoint NP-complete sets, then A ∪ B is complete via strong nondeterministic Turing reductions. They also showed that if NP differs from co-NP at almost all lengths, then A ∪ B is many-one complete via P/poly-reductions. If we consider disjoint Turing complete sets, we know a little more. Glaßer et al. [GPSS06] showed that if UP ∩ co-UP contains bi-immune sets, then there exist disjoint Turing complete sets whose union is not Turing complete. The above mentioned results do not seem to shed light on the question of whether unions of disjoint NP-complete sets remain NP-complete (under polynomial-time many-one reductions). To date, we do not know of any reasonable hypothesis that either provides a positive answer or a negative answer to this question. In this paper we provide necessary and sufficient conditions under which the union of disjoint NP-complete sets remain NP-complete. We consider two statements and show that one of the statements yields a positive answer to our question, whereas the other statement yields a negative answer. Our statements relate to the complexity of SAT. Let us assume that NP differs from co-NP, thus there is no NP-algorithm for SAT. Could it still be the case that there is an NP-algorithm that solves SAT in some "average-case/approximate" sense? Let B be a set in NP that is disjoint from SAT. We can view B as an "approximate/average-case" NP-algorithm for SAT. Since B does not coincide with SAT, there must exist unsatisfiable formulas on which B errs. How easy/hard is it to produce such instances? Any machine that produces instances on which B differs from SAT is called a refuter. Given B, what is the complexity of the refuter? We can make two easy observations. If B can be decided in time 2 n k , then there exists a refuter that runs in time O(2 n k 2 n ). Using the fact that B is in NP we can also design a P Σ P 2 refuter. Can the complexity of these refuters be reduced further? We show that if the complexity of such refuters can be improved to polynomial-time, then unions of disjoint NP-complete sets remain NP-complete. On the other hand, we show that if the complexity of the refuters can not be reduced, then there exist disjoint NP-complete sets whose union is not NP-complete. More precisely, we show that if there exists a B ∈ NP that is disjoint
doi:10.1007/978-3-642-22685-4_22 fatcat:mwcc4za5hjdgdb5oag3xig23ae