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Valiant & Vazirani proved SAT is reducible to UNIQUE SAT under randomized probabilistic reductions in polynomial time. Calabro et al. showed that UNIQUE k-SAT is as hard as k-SAT. Now the question is, if someone shows that UNIQUE k-SAT is in P, does it imply P=NP?

References

  1. L. G. Valiant and V. V. Vazirani, "NP is as easy as detecting unique solutions." Theoretical Computer Science 47:85–93, 1986. (PDF on ScienceDirect.)

  2. C. Calabro, R. Impagliazzo, V. Kabanets and R. Paturi, "The Complexity of Unique k-SAT: An Isolation Lemma for k-CNFs". Journal of Computer and System Sciences 74(3):386–393, 2008. (PDF at ACM Digital Library; free PDF.)

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  • $\begingroup$ Valiant & Vazirani showed that SAT is reducible to UNAMBIGUOUS SAT under RP reductions. In second case you are speaking about UNIQUE SAT which is $\mathsf{NP}$-hard. $\endgroup$ – rus9384 Sep 13 '17 at 20:57
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This is still an open question; UP is not known to be equivalent to NP. In the paper "NP Might Not Be As Easy As Detecting Unique Solutions," Beigel, Burhman and Fortnow construct an oracle under which P contains UP but P is still not equivalent to NP.

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