# Weakest condition that connects $P=BPP$ and derandomizing $VV$

It is known $$P=BPP$$ is insufficient to derandomize $$VV$$ isolation lemma.

What does it mean to be 'insufficient' here? Is there some theorem which says $$P=BPP$$ $$+$$ 'condition $$A$$' gives derandomization of Valiant-Vazirani.

Of course $$P=NP$$ suffices as condition $$A$$. However I am looking for anything weaker that would be possible and if something weaker than $$P=NP$$ is not possible then why?

• I think it mostly means that no one knows how to derandomize Valiant-Vazirani if you assume only that P=BPP. If you look at Klivans-van Melkebeek, you'll see that because the Nisan-Wigderson generator relativizes, one can derandomize Valiant-Vazirani assuming existence of a sufficiently hard function (enough for the relativized NW generator). So from a relativizing statement (NW generator) we get a conditional derand of V-V, but from P=BPP alone (which doesn't relativize) we don't know how to. – Joshua Grochow Feb 16 at 3:10
• So, rather than "P=BPP + condition A" it's more like "condition A", which is known to imply P=BPP (but not conversely) also implies derand of V-V. – Joshua Grochow Feb 16 at 3:11
• Something weaker than $P=NP$ and stronger than $P=BPP$? – T.... Feb 16 at 3:51
• Existence of sufficiently hard functions for the (relativized) NW generator to work in this setting. I believe indeed that such functions imply P=BPP but don't imply P=NP. See Klivans-van Melkebeek for details. – Joshua Grochow Feb 16 at 4:30