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We know $\mathsf{P=BPP} \implies \mathsf{NEXP\not\subseteq P/Poly}$ or permanent does not have polynomial sized circuits. However permanent needs superpoly circuits imply $\mathsf{NEXP\not\subseteq P/Poly}$ since $\mathsf{P^{\#P}\subseteq PSPACE\subseteq NEXP}$. So why dont we just say $\mathsf{P=BPP} \implies \mathsf{NEXP\not\subseteq P/Poly}$?

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    $\begingroup$ Where have you seen "... or permanent does not have polynomial sized circuits"? (I've only seen "... or permanent does not have polynomial sized arithmetic circuits".) $\endgroup$ – user12859 Dec 30 '15 at 3:38
  • $\begingroup$ @RickyDemer is it not true if permanent does not have poly size arith ckts then nexp is not in p/poly? $\endgroup$ – T.... Dec 30 '15 at 3:47
  • $\begingroup$ That implication is probably true, since its conclusion is probably true. $\endgroup$ – user12859 Dec 30 '15 at 4:33
  • $\begingroup$ @RickyDemer ' probably true' is different from already known. is it not already known? $\endgroup$ – T.... Dec 30 '15 at 4:37
  • $\begingroup$ (Bear in mind that your earlier comment only asked about truth.) Yes, that's not already known. $\endgroup$ – user12859 Dec 30 '15 at 4:51
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The conclusion is that the permanent does not have polynomial-size (constant-free) arithmetic circuits or $\mathsf{NEXP} \not\subseteq \mathsf{P/poly}$. Permanent not having polynomial-size constant-free arithmetic circuits is not known to imply $\mathsf{NEXP} \not\subseteq \mathsf{P/poly}$.

However, there are a couple more recent similar results whose conclusion is not a disjunction, but simply a lower bound on some polynomial family in $\mathsf{NEXP}$:

  • Jansen and Santhanam showed that if PIT is in $\mathsf{NSUBEXP}$ then there is a family of polynomial whose integer evaluation problem is in $\mathsf{NEXP}$ but which do not have poly-size constant-free arithmetic circuits.

  • Carmosino, Impagliazzo, Kabanets, and Kolokolova show that if PIT over a fixed finite field is in $\mathsf{NSUBEXP}$ then there is a multilinear polynomial whose evaluation problem over that same finite field is in $\mathsf{NE}$ that does not have poly-size arithmetic circuits. (This paper also has many other results of interest, including true hardness-to-randomness equivalences, avoiding many of the usual caveats.)

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  • $\begingroup$ @joshuagrchow does $\mathsf{NEXP}$ not having poly-size constant-free arithmetic circuits with some additional condition known to imply $\mathsf{NEXP}\not\subseteq\mathsf{P/poly}$? If so what is that additional condition? $\endgroup$ – T.... Dec 31 '15 at 1:10
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    $\begingroup$ @Turbo: Not as far as I know. Most of the big implications show that a Boolean circuit lower bound implies an arithmetic circuit lower bound, not the other way around (which makes sense intuitively, since arithmetic circuits are a restricted model). $\endgroup$ – Joshua Grochow Dec 31 '15 at 2:04
  • $\begingroup$ so moral is the original P=BPP implications is the real deal or else (or anything lower than that) will not have NEXP not in P/Poly as an implication. $\endgroup$ – T.... Dec 31 '15 at 2:20

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