Team Diagonalization

TL;DR

This paper explores the implications of NP-completeness, demonstrating that the union of two non-NP-complete NP sets can be NP-complete if P ≠ NP, using a novel team diagonalization technique.

Contribution

It provides a detailed proof of Ladner's result on splitting NP sets, introducing a new team diagonalization method for such proofs.

Findings

Union of two non-NP-complete NP sets can be NP-complete if P ≠ NP

Introduces a team diagonalization technique for complexity proofs

Existence of OptP functions whose composition is NP-hard but neither is NP-hard

Abstract

Ten years ago, Gla{\ss}er, Pavan, Selman, and Zhang [GPSZ08] proved that if P $\neq$ NP, then all NP-complete sets can be simply split into two NP-complete sets. That advance might naturally make one wonder about a quite different potential consequence of NP-completeness: Can the union of easy NP sets ever be hard? In particular, can the union of two non-NP-complete NP sets ever be NP-complete? Amazingly, Ladner [Lad75] resolved this more than forty years ago: If P $\neq$ NP, then all NP-complete sets can be simply split into two non-NP-complete NP sets. Indeed, this holds even when one requires the two non-NP-complete NP sets to be disjoint. We present this result as a mini-tutorial. We give a relatively detailed proof of this result, using the same technique and idea Ladner [Lad75] invented and used in proving a rich collection of results that include many that are more general than this result: delayed diagonalization. In particular, the proof presented is based on what one can call team diagonalization (or if one is being playful, perhaps even tag-team diagonalization): Multiple sets are formed separately by delayed diagonalization, yet those diagonalizations are mutually aware and delay some of their actions until their partner(s) have also succeeded in some coordinated action. We relatedly note that, as a consequence of Ladner's result, if P $\neq$ NP, there exist OptP functions f and g whose composition is NP-hard yet neither f nor g is NP-hard.