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Dec 31, 2015 at 10:02 history tweeted twitter.com/StackCSTheory/status/682502026048212992
Dec 23, 2015 at 9:00 vote accept Turbo
Dec 22, 2015 at 23:01 history edited Turbo CC BY-SA 3.0
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Dec 22, 2015 at 16:46 history edited Turbo CC BY-SA 3.0
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Dec 22, 2015 at 15:29 vote accept Turbo
Dec 23, 2015 at 8:51
Dec 22, 2015 at 15:28 history edited Turbo CC BY-SA 3.0
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Dec 22, 2015 at 13:10 history edited Turbo CC BY-SA 3.0
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Dec 22, 2015 at 7:53 vote accept Turbo
Dec 22, 2015 at 15:23
Dec 22, 2015 at 7:48 answer added user6973 timeline score: 7
Dec 22, 2015 at 7:40 comment added Thomas Steinke If $\mathsf{BPP} \setminus (\mathsf{NP} \cup \mathsf{coNP}) \ne \emptyset$, then $\mathsf{BPP} \ne \mathsf{P}$. So the belief that $\mathsf{BPP}=\mathsf{P}$ rules out the possibility of finding $L \in \mathsf{BPP} \setminus (\mathsf{NP} \cup \mathsf{coNP})$.
Dec 22, 2015 at 5:38 comment added Joshua Grochow While it's true that it's possible that $\mathsf{NP} \cup \mathsf{coNP}$ could be contained in $\mathsf{BPP}$, if this is true then $\mathsf{NP}=\mathsf{RP}$ (exercise), and $\mathsf{PH}$ then collapses down to $\mathsf{BPP}$...
Dec 22, 2015 at 5:09 history edited Turbo CC BY-SA 3.0
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Dec 22, 2015 at 4:51 history asked Turbo CC BY-SA 3.0