If $\mathsf{NP}\subseteq \mathsf{DTIME}[n^{O(\log n)}]$ then what happens? Does it imply $\mathsf{NP}\neq \mathsf{EXP}$? Is there any other consequences such as $\mathsf{BPP}\neq \mathsf{EXP}$? Does it also give $\mathsf{PSPACE}\subseteq \mathsf{DTIME}[n^{O(\log n)}]$?

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    $\begingroup$ "Does it imply $\textbf{NP} \neq \textbf{EXP}$?" Yes. This follows from the time hierarchy theorem. $\endgroup$ – dkaeae Mar 1 '19 at 20:44

If $\textbf{NP} \subseteq \textbf{DTIME}(n^{O(\log n)})$, then we get $\textbf{P}^\textbf{NP} \subseteq \textbf{P}^{\textbf{DTIME}(n^{O(\log n)})} = \textbf{DTIME}(n^{O (\log n)})$. Continuing this reasoning, the entire polynomial hierarchy $\textbf{PH}$ is contained in $\textbf{DTIME}(n^{O(\log n)})$. By the time hierarchy theorem, $\textbf{DTIME}(n^{O(\log n)})$ is a proper subset of $\textbf{EXP}$ (and even of $\textbf{E} = \textbf{DTIME}(2^{O(n)})$), so $\textbf{PH} \neq \textbf{EXP}$. In particular, we also get $\textbf{NP} \neq \textbf{EXP}$ and $\textbf{BPP} \neq \textbf{EXP}$ because both $\textbf{NP}$ and $\textbf{BPP}$ are in the polynomial hierarchy. For $\textbf{BPP}$ this is not that surprising since we already suspect it is contained in subexponential time.

What happens for $\textbf{PSPACE}$? Not much, as long as I'm aware of. We have $\textbf{PH} \subseteq \textbf{PP}$ and $\textbf{PH} \subseteq \textbf{P}^\textbf{#P}$ (by Toda's theorem), but I don't see any direct consequence for $\textbf{PP}$, let alone $\textbf{PSPACE}$.

  • $\begingroup$ @dkaee I would have thought $PP$ would also be in quasipolynomial. $\endgroup$ – T.... Mar 2 '19 at 20:13
  • $\begingroup$ Why is it so? I am not aware of any reductions from $\mathbf{PP}$ to something below it. We can approximate $\#\mathbf{P}$ using $\mathbf{BPP}$ machines with $\mathbf{NP}$ oracle, but that's another story. $\endgroup$ – dkaeae Mar 4 '19 at 10:58
  • $\begingroup$ doesn't approximmation give msb? $\endgroup$ – T.... Mar 4 '19 at 11:46
  • $\begingroup$ Not necessarily. To distinguish $2^n$ from $2^n - 1$ you then need zero error. You could frame it as a promise problem, but then correspondence to the original classes is no longer guaranteed. $\endgroup$ – dkaeae Mar 4 '19 at 12:54
  • $\begingroup$ so $PH\subseteq \Sigma_{m}^P$ would be in $DTIME(n^{O(\log n)})$ if $m=O((\log n)^c)$ at any fixed $c>0$ correct? $\endgroup$ – T.... Mar 4 '19 at 13:04

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