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I guess the answer is that this unknown. Maybe this implies some "lowness" result on NP relative to BPP?

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up vote 8 down vote accepted

Note that it would suffice to prove $EXP \subseteq P/poly$ implies $NP = BPP$ (since the latter implies $NP = RP$).

I am pretty sure this is not known. $NP \neq RP$ does not seem to imply any circuit lower bounds for $EXP$.

Note that $EXP \subseteq P/poly$ does imply $P \neq NP$. If $EXP \subseteq P/poly$ then there is some fixed $k$ for which $TIME[2^n]$ has $n^k$-size circuits. Therefore $P \subseteq TIME[2^n]$ has $n^k$-size circuits. But $P = NP$ implies that $\Sigma_2 P = P$, and we know by Kannan's theorem that $\Sigma_2 P$ does not have $n^k$-size circuits for any fixed $k$ -- this is a contradiction. (This result is due to Meyer, cited in the famous Karp-Lipton paper.)

I would suspect that one can prove $EXP \subseteq P/poly$ implies $RP \neq NP$ or maybe $ZPP \neq NP$, but the above proof doesn't do the job. (Technically, since I believe $EXP \not\subseteq P/poly$, the negation should imply everything...)

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Thanks! Regarding $EXP \subseteq P/poly$ imply $P\neq NP$, from the assumption it follows that $EXP=\sigma_2$(Karp-Lipton due to Meyer) and $P=NP$ collapse PH to P, which imply that $P=EXP$, contradiction (Hartmanis and Stern using diagnolization). I think it's maybe simpler. Proving what you suggest could be interesting, I think $BPP\neq EXP$ is a consequence. – Sebastian Ben Daniel Aug 10 '10 at 23:14
More is known: if $EXP\subseteq P/\text{poly}$, then $EXP=PH=MA$, due to Babai, Fortnow, Nisan, and Wigderson. – Emil Jeřábek Mar 3 '11 at 12:25
@Emil, of course (in fact I used this in some recent papers)... but that doesn't strengthen the above arguments, does it? – Ryan Williams Mar 4 '11 at 3:21
No, it doesn't, but I thought it was relevant. Your post starts by contemplating whether $EXP\subseteq P/\mathrm{poly}$ implies the collapse of PH to BPP; we don't known how to do that, but we can at least collapse the hierarchy to the somewhat bigger class MA (though that's admittedly not a big deal, already $NP\subseteq P/\mathrm{poly}$ collapses the hierarchy to $S^P_2$ by the Karp–Lipton argument). – Emil Jeřábek Mar 4 '11 at 11:24

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