It is known that $\mathsf{NL}=\mathsf{Co{-}NL}$ and unknown if $\mathsf{NP}=\mathsf{Co{-}NP}$. But what about $$\mathsf{NEXP}=\mathsf{Co{-}NEXP}?$$ Is it known whether these two classes are equal?


1 Answer 1


It is known that $\mathsf{NP} = \mathsf{coNP}$ implies $\mathsf{NEXP} = \mathsf{coNEXP}$, using a padding argument. However, both are considered unlikely.

The difference between classes like $\mathsf{NP}$ and $\mathsf{NEXP}$ and the class $\mathsf{NL}$ is that the former are defined by time constraints while the latter is defined by space constraints. The Immerman–Szelepcsényi argument used to prove $\mathsf{NL}=\mathsf{coNL}$ only works for space-constrained classes.

  • $\begingroup$ That makes perfect sense. Thanks for putting it so clearly! $\endgroup$ May 5, 2016 at 7:49
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    $\begingroup$ Actually, an Immerman–Szelepcsényi-like argument shows the fun fact $\mathrm{NEXP/poly=coNEXP/poly}$. $\endgroup$ May 5, 2016 at 17:58
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    $\begingroup$ @EmilJeřábek : ​ ​ ​ In fact, that can easily be strengthened to ​ NE/poly = coNE/poly . ​ ​ ​ ​ ​ ​ ​ ​ $\endgroup$
    – user12859
    May 5, 2016 at 19:30
  • $\begingroup$ Of course. I stated it for NEXP as that’s what the question is about. $\endgroup$ May 5, 2016 at 20:26
  • $\begingroup$ @RickyDemer Could $\mathsf{NP/poly}=\mathsf{coNP/poly}$ be true? Would this have any implications? $\endgroup$
    – Turbo
    Oct 23, 2016 at 21:54

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