What is complexity theoretic implication of following possibilities - $NP=coNP=BPP=RP$ or $coNP\neq NP=BPP=RP$ (consensus is these seem impossible)?
Consequences of $NP=coNP=BPP=RP$
complexity theory
Unless I'm missing something obvious (it is getting late here), the second isn't possible - $BPP = coBPP$, so $NP=BPP \rightarrow NP = coBPP \rightarrow NP = coNP$.In the first case, some of the consequences would be:Feasible algorithms for $NP$-complete problems, for at least the randomised version of feasible.$PH \subseteq NP$Unless $P = NP$, adding randomness adds computational power.Strong pseudorandom number generators do not exist (nor do one way functions). (we would need $BPP = P$ for them to exist, but that would imply $P=NP$, in which case they don't).