Decidability of P = NP
- From: cplxphil@xxxxxxxxx
- Date: Fri, 15 Aug 2008 19:02:17 -0700 (PDT)
Here's an approach I thought of for to trying to show that the P vs NP
question might be undecidable. Any comments would be appreciated.
According to Baker-Gill-Solovay, there exist oracles A and B such that
P^A = NP^A and P^B != NP^B. As a result, it is concluded that P = NP
cannot be proven true or false by techniques that relativize.
Now consider a Turing machine that partially decides the language Q of
languages that are in P. By Rice's theorem, the language Q is
necessarily undecidable. However, that doesn't mean that it is
impossible for a TM to partially decide the language.
Now consider the language SAT. Assume, for the sake of contradiction,
that the TM that partially decides Q (call it R) is able to decide
whether or not SAT is in Q. If it can, then R^B, where B is the
oracle set s.t. P^B != NP^B, is also capable of deciding whether SAT
is in Q. The same holds for R^A, where A is defined as above also.
If this is true, however, then we have a proof technique that
relativizes. Running the same algorithm relative to different oracle
sets is a simple relativization of the proof technique; thus we have
found a proof that P = NP or that P != NP using a technique that
relativizes.
This contradiction shows that no TM can decide whether or not SAT is
in P, and thus no TM can decide if P = NP. Thus, either P = NP is
undecidable, or the computational model associated with the oracle
sets mentioned are inconsistent--which doesn't seem likely.
Any comments on this approach?
Thanks,
Phil
.
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