# $P=BPP$ without good PRGs?

We know that the existence of good pseudorandom generators (PRGs) does not only imply $$P=BPP$$, but also $$PromiseP=PromiseBPP$$.

Let us assume $$PromiseP\ne PromiseBPP$$. Then good PRGs do not exist. However it is still possible that $$P=BPP$$ holds. If good PRGs do not exist then does $$P=BPP$$ imply $$P=NP$$ or is there a possibility $$P\neq NP$$ still might be true?

Is it possible to have $$P=BPP$$ without good PRGs and with $$P\neq NP$$ and with $$PromiseP=PromiseBPP$$?

• If $PromiseBPP \neq PromiseP$ then $P \neq NP$. Proof: If $P = NP$ then $PH$ collapses to $P$, so (due to approximate counting in $PH$) we can solve the problem "given a circuit, approximate its acceptance probability" in polynomial time. This in turn implies $PromiseBPP$ is in $PromiseP$ (because this circuit approximation problem is "complete" for $PromiseBPP$). – Ryan Williams Oct 26 '18 at 17:54