The question is in the title, I suppose. I am studying complexity classes, and I understand that NP-Hard is the set of problems that are at least as hard as the hardest problems in NP. Therefore, it will naturally contain PSPACE problems.

However, I was specifically wondering if there were any PSPACE problems that were not in NP-Hard? (from my understanding, implying that they are easier than the hardest problems in NP).


1 Answer 1


Assuming we use polytime reductions in both, all PSPACE-hard problems are NP-hard, which follows directly from the definition and from the easy fact NP$\subseteq$PSPACE: a languages $L$ is PSPACE-hard if all languages in PSPACE polytime-reduce to $L$. If $L$ is PSPACE-hard then in particular, all languages in NP polytime-reduce to $L$, and so $L$ is also NP-hard.

On the other hand, there are problems in PSPACE which are not NP-hard, for example $\emptyset$ and $\Sigma^*$. If P=NP then all other problems in PSPACE are NP-hard, since every non-trivial language (a language different from $\emptyset,\Sigma^*$) is NP-hard in this case. If P$\neq$NP then every problem in P also belongs to PSPACE and is not NP-hard.

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    $\begingroup$ The core point is that NP-hardness is a lower bound while PSPACE-ness is an upper bound. $\endgroup$
    – Raphael
    May 3, 2014 at 9:36
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    $\begingroup$ I've gone through most of the related questions/answers here on stackexchange and I'm not sure so asking here: are there non-trivial languages outside of NP that are not NP-hard? $\endgroup$
    – starflyer
    Apr 2, 2015 at 18:53
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    $\begingroup$ @starflyer This can only happen if P$\neq$NP. If you make the stronger assumption that NP$\neq$coNP, then any coNP-complete program is an example. I imagine that you can construct an example under the hypothesis P$\neq$NP using diagonalization, as in Ladner's theorem. $\endgroup$ Apr 2, 2015 at 19:05

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