Timeline for What do stronger circuit lower bounds give in terms of derandomization?

9 events

when toggle format	what		by	license	comment
Dec 12, 2017 at 18:08	comment	added	Or Meir		You are right of course. I fixed the answer. Thanks!
Dec 12, 2017 at 18:08	history	edited	Or Meir	CC BY-SA 3.0	typo
Dec 12, 2017 at 0:26	comment	added	Ryan O'Donnell		I believe the assumption only shows that BPP is in io-P. For BPP to equal P, E should fail to have 2^(eps n)-size circuits for almost all n.
Dec 8, 2017 at 2:01	vote	accept	Turbo
Dec 8, 2017 at 1:36	history	edited	Or Meir	CC BY-SA 3.0	Fixed typo.
Dec 8, 2017 at 1:35	comment	added	Or Meir		P/poly is indeed much smaller. Regarding the paper, see the last paragraph of Page 4 for the result, and Section 5.4 for the proof (but this proof really depends on more-or-less everything that comes before it).
Dec 7, 2017 at 23:13	comment	added	Turbo		which section are you talking in paper?
Dec 7, 2017 at 23:10	comment	added	Turbo		just for the record $P/poly$ is much smaller than $SIZE(2^{\epsilon\cdot n})$. right?
Dec 7, 2017 at 22:37	history	answered	Or Meir	CC BY-SA 3.0