Derandomizing Valiant-Vazirani?

Question

The Valiant-Vazirani theorem says that if there is a polynomial time algorithm (deterministic or randomized) for distinguishing between a SAT formula that has exactly one satisfying assignment, and an unsatisfiable formula - then NP=RP. This theorem is proved by showing that UNIQUE-SAT is NP-hard under randomized reductions.

Subject to plausible derandomization conjectures, the Theorem can be strengthened to "an efficient solution to UNIQUE-SAT implies NP = P".

My first instinct was to think that implied there exists a deterministic reduction from 3SAT to UNIQUE-SAT, but it's not clear to me how this particular reduction can be derandomized.

My question is: what is believed or known about "derandomizing reductions"? Is it/should it be possible? What about in the case of V-V?

Since UNIQUE-SAT is complete for PromiseNP under randomized reductions, can we use a derandomization tool to show that "a deterministic polynomial time solution to UNIQUE-SAT implies that PromiseNP = PromiseP?

Answer 1

Under the right derandomization assumptions (see Klivans-van Melkebeek) you get the following: There is a polytime computable $f(\phi)=(\psi_1,\ldots,\psi_k)$ s.t. for all $\phi$,

• If $\phi$ is satisfiable then at least one of the $\psi_i$ has exactly one satisfying assignment.
• If $\phi$ is not satisfiable then all of the $\psi_i$ are unsatisfiable.

You need k polynomial in then length of $\phi$. Probably can't be done for $k=1$.

Answer 2

Just for reference, I stumbled across this really interesting paper today, which gives evidence that a deterministic reduction is unlikely:

Dell, H., Kabanets, V., Watanabe, O., & van Melkebeek, D. (2012). Is the Valiant-Vazirani Isolation Lemma Improvable? ECCC TR11-151

They argue that this is not possible unless NP is contained in P/poly.