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I am thinking of the equality "PCP(O(log(n)),0) = P"

Say I have a deterministic polynomial time algorithm $A$ whose correctness I can't prove immediately. But say I create a probabilistic version of this algorithm say $A_P$ such that it uses log(n) random bits. Now if I prove somekind of (exactly what ?) expectation correctness of $A_P$ then does it imply that $A$ was correct?

If the above is not right then I would like if someone can point me to what is the closest correct thing to this!

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The statement you state is not the PCP theorem. The PCP theorem is a characterization of NP.

The statement you mention is much easier to prove. It says in effect that a polytime algorithm that uses $O(\log n)$ random bits can be made deterministic while keeping it polytime (given reasonable completeness and soundness properties). The proof is very simple – we can run the algorithm over the $2^{O(\log n)} = n^{O(1)}$ many choices of random bits, check how many times the algorithm accepts, and so determine whether the input belongs to the language.

At most you can think of this as a derandomization procedure – if your algorithm uses only logarithmically many random bits, then you can make it deterministic by "exhaustive search".

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  • $\begingroup$ Yes. Thanks! Are there other polytime derandomizations possible? Like if there were linear number of coin tosses? $\endgroup$ – user6818 May 3 '15 at 22:41
  • $\begingroup$ It is suspected that P=BPP, that is, that we can deal with any polynomial number of coin tosses. But we don't know how to prove it right now. $\endgroup$ – Yuval Filmus May 3 '15 at 22:43
  • $\begingroup$ I understand that. But I hear of of this implicit idea which I am not able to trace an exact reference to learn from - that even within our current knowledge there are ways to convert a probabilistic algorithm proof into a proof for a deterministic one - under what conditions and how? (any pedagogic reference to read would be helpful!) [...I actually currently have at hand a deterministic polynomial algorithm which is doing something amazing but I don't know how to prove it directly and hence I am looking for this approach to a possible proof...] $\endgroup$ – user6818 May 4 '15 at 20:32
  • $\begingroup$ There is an entire area dealing with this problem, known as derandomization. $\endgroup$ – Yuval Filmus May 4 '15 at 21:04
  • $\begingroup$ :D Anything simple/pedagogic I can read off immediately to learn enough to try to use something in a special case? $\endgroup$ – user6818 May 4 '15 at 23:01

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