RP is the class of problems decidable by a nondeterministic Turing machine that terminates in polynomial time, but that is also allowed one-sided error. P is the usual class of problems decidable by a deterministic Turing machine that terminates in polynomial time.

P = RP follows from a relationship in circuit complexity. Impagliazzo and Wigderson showed that P = BPP follows if some problem that can be decided in deterministic exponential time also requires exponential size circuits (note that P = BPP implies P = RP). Perhaps due to these results, there seems to be a feeling among some complexity theorists that probabilistic reductions can probably be derandomized.

What other specific evidence is there that P = RP?


4 Answers 4


The existence of problems in DTIME(2^O(n)) which require exponential-size circuits to compute (which is the assumption in IW) seems plausible since otherwise we would have non-uniformity giving a speedup on EVERY computational problem -- which goes completely against the current thinking that does not see a "too significant" gap between uniform and non-uniform complexity for "normal" problems. This thinking comes from the fact that there are very few examples where a "non-uniform" algorithm is known that is significantly better than the known uniform one (again except for derandomization).

Another piece of "evidence" is that relative to a random oracle we do have P=BPP.

  • $\begingroup$ I thought that was the precise paper I mentioned in the original question. What am I missing? $\endgroup$ Aug 17, 2010 at 17:50
  • $\begingroup$ oops, i guess i didn't read the question all the way... The reason that the assumption is plausible is that otherwise we would have non-uniformity giving a speedup on EVERY computational problem -- which goes completely against the current thinking that does not see a "too significant" gap between uniform and non-uniform complexity for "normal" problems. $\endgroup$
    – Noam
    Aug 17, 2010 at 18:37
  • 1
    $\begingroup$ edited the response now --- still getting to know the system... $\endgroup$
    – Noam
    Aug 17, 2010 at 18:59

Any concrete derandomization result gives evidence that P=BPP. As such PRIMES in P (Agrawal-Kayal-Saxena'02) is one good example. Generally, there are few natural problems in BPP that are not known to be in P (Polynomial Identity Testing is one notable exception.)

Similar in spirit to the result you mention, Hastad-Impagliazzo-Levin-Luby '99 showed that the existence of one-way functions implies the existence of pseudorandom generators. While this does not directly imply P=BPP based on the existence of one-way functions, it does show that pseudorandom generators follow from minimal cryptographic assumptions. This may be seen as another hint that BPP is not more powerful than P.


It's important to note that saying "probabilistic reductions can [probably] be derandomized" is much stronger than P=RP. In fact, one formalization of the notion of derandomizing all randomized reductions is that $P^X=RP^X$ relative to every oracle $X$, which we know is false (e.g. Heller. Relativized polynomial hierarchies extending two levels, Mathematical Systems Theory 17(2):71-84, 1984 gives an oracle where $ZPP=RP=EXP$ which is not equal to $P$ by the Time Hierarchy Theorem).

The above is, of course, talking about derandomizing randomized polynomial-time Turing reductions, rather than the usual polynomial-time many-one reductions. I wouldn't be surprised if Heller's oracle can be adapted to give a set X such that for all Y, Y is exponential-time many-one reducible to X iff Y is RP-reducible to X, but without going through the construction I couldn't swear to it.


Valiant and Vazirani showed in 1986 that there is a randomized reduction of SAT to $USAT_Q$, which is the decision problem based on SAT where only the difference between satisfiable and unsatisfiable instances matters. If $Q = \bot$ is the false predicate, then $USAT_{\bot}$ is the problem of deciding whether there is precisely one solution.

A solution to a Boolean formula $\phi$ is a (0,1)-vector assigning truth values to the free variables in $\phi$, so that $\phi$ is satisfied. A $k$-isolated solution $x$ to $\phi$ is a solution, with the additional property that any other solution differs in more than $k$ values. (Alternately, $x$ is a $k$-isolated solution if the Hamming distance of $x$ to any other solution exceeds $k$.)

Let $k$-ISOLATED SAT be the problem which requires deciding whether the input CNF formula has a solution that is $k$-isolated. If $n$ denotes the number of variables in an instance, then $USAT_{\bot}$ and $n$-ISOLATED SAT are precisely the same problem.

For any $\epsilon > 0$, by duplicating each variable polynomially many times, a deterministic reduction can be made from SAT to $n^{1-\epsilon}$-isolated SAT. (Details are available here.) This seems to provide further evidence that the gap between deterministic and randomized reductions is "small".

  • $\begingroup$ I would say this is evidence against P=RP. Analogously, the assumption $P\neq NP$ is supported by the fact that many good approximation algorithms are matched by NP-hardness results - if $P=NP$ it is difficult to explain why the techniques stop exactly at this small or non-existent gap. $\endgroup$ Nov 22, 2012 at 10:56
  • $\begingroup$ @Colin: No comment. :-) $\endgroup$ Nov 24, 2012 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.