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Mathematicians Race To Debunk German Man Who Claimed To Solve The 'P Versus NP' Problem (vice.com)
A German man -- Norbert Blum -- who claimed that P is not equal to NP is seeing several challenges to his solution. From a report: Numerous mathematicians have begun to raise questions about whether the German mathematician solved it at all. Since Blum's paper was published, mathematicians and computer scientists worldwide have been racking their brains as to whether the Bonn-based researcher has, in fact, solved this Millennium Prize Problem. After an initially positive reaction, such as the one from Stanford mathematician Reza Zadeh, doubts are beginning to arise about whether Blum's reasoning is correct. In a forum for theoretical mathematics, a user named Mikhail reached out to Alexander Razborov -- the author of the paper on which Blum's proof is based -- to ask him about Blum's paper. Razborov purports to have discovered an error in Blum's paper: Blum's main argument contradicts one of Razborov's key assumptions. And mathematician Scott Aaronson, who is something of an authority in the math community when it comes to P vs. NP, said he would be willing to bet $200,000 that Blum's mathematical proof won't endure. "Please stop asking," Aaronson writes. If the proof hasn't been refuted, "you can come back and tell me I was a closed-minded fool." In the week since Aaronson's initial blog post, other mathematicians have begun trying to poke holes in Blum's proof. Dick Lipton, a computer science professor at Georgia Tech, wrote in a blog post that Blum's proof "passes many filters of seriousness," but suggested there may be some problems with it. A commenter on that blog post, known only as "vloodin," noted that there was a "single error on a subtle point" in the proof; other mathematicians have since chimed in and confirmed vloodin's initial analysis, and so the emerging consensus among many mathematicians is that a solve for P vs. NP remains elusive.
Proving that P!=NP only requires proof of one polynomial problem not being deterministic, it doesn't matter what it is, and it's proven.
Proving that P=NP, on the other hand, might be impossible without a new definition of polynomial.
I believe you are mistaken.
Finding one example of P = NP proves the classes are equal, because NP-complete problems can all be transformed to other NP problems in polynomial time.
So if you solve ANY NP-complete problem in polynomial time, you have a solution to ALL of them. If you solve 3-SAT in polynomial time you've solved TSP (travelling salesman) in polynomial time too, because TSP can be "mapped" to 3-SAT in polynomial time.
So basically, if you can prove OR disprove any NP-complete problem can be solved in polynomial time then you prove P=NP or P!=NP.
https://en.wikipedia.org/wiki/...
If he finds a solution in polynomial time of ANY NP-complete problem he proves P=NP.
By the definition of NP-complete:
"A problem p in NP is NP-complete if every other problem in NP can be transformed (or reduced) into p in polynomial time." (This makes sense, because if you can got from A to B in polytime, and A to C in polytime then you can go from A to C in polytime too, trviially by going from A to B and then B to C. (because polytime x polytime = polytime)
A P=NP proof would consist of finding a polynomial solution to *any* NP-complete problem (and therefore ALL NP-complete problems.)
There are a handful of problems in NP that are not known to be in P and not known to be NP-complete; but if P=NP they would also have polynomial time solutions.
NP-hard problems are NOT in NP. (e.g the halting problem) so they aren't at issue.
A P != NP proof is the harder proof since you can't use proof by example -- you have to formally prove that a polynomial solution does not exist for an NP-complete problem. (and therefore does not exist for all of them)