If true, would this be an actual 100% solution, or just an indicator that P != NP? This is all way over my head but the word "approximator" makes me wonder.
>> This implies P not equal NP.
I did some graduate level research on P =? NP, specifically in the SAT space " rel="nofollow">https://en.wikipedia.org/wiki/Satisfiability>.
In particular, I helped design MARMOSET (Marmoset Automated Reasoner Mostly Only Solves Easy Theorems), a competitive SAT problem solver. " rel="nofollow">http://www.cs.unb.ca/research-groups/argroup/marmoset/> . (It's a cool name... I didn't come up with it :))
The conclusion I drew was:
1. P != NP because you can convert in polynomial time every SAT problem down to Horn clauses, which are P to solve, plus non-Horn clauses that cannot be converted i.e. have intractable intrinsic NP complexity whose reduction to the polynomial space requires "clairvoyance" of the quantum computation variety.
2. Nobody's really interested in a proof that P != NP.
That said, I only spent a couple years at it, and my memory may be faulty and I might change my mind if I revisited the issue. Part of me has always felt that the Horn clause reduction is a first step to isolating problems for a next step, but again — it's been a long time.