Derandomizing Valiant-Vazirani?
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 $k=1$.