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HPL-2010-95 P ≠ NP - Deolalikar, Vinay
Keyword(s):P, NP, complexity theory
Abstract:We demonstrate the separation of the complexity class NP from its subclass P.Throughout our proof, we observe that the ability to compute a property on structures in polynomial time is intimately related to an atypical property of the space of solutions ...
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I can prove P = NP.
Though i am wondering at first, because there are many ways to arrive at numbers.

-But now, if i see any number with this same formula i can solve for its source without apparently using that same number itself.
In fact, i have solved P is not NP and now P = NP (This is the origin of numbers).

Hints: P can be any number and if you give me any number to look for its source, i will tell you.

PLEASE WHERE CAN I SUMMIT MY PROVE?
Thanks.

P and NP, as discussed in this proof, are not numbers, but collections of problems. P is the collection of problems that can be solved in polynomial time (in reference to it's complexity), while NP is the collection that can't be solved in that time. NP problems would be problems like The Traveling Salesman or the Knapsack Problem.

Serious doubts have been raised about the correctness of the proof as initially published. See Lipton's summary at Issues In The Proof That P?NP

Has the biggest question in computer science been solved? On 6 August, Vinay Deolalikar, a mathematician at Hewlett-Packard Labs in Palo Alto, California, sent out draft copies of a paper titled simply "P ≠ NP".

I can prove P = NP:
Given that P is not N ( a + b )
Where P is not ( a + b ) because P is not NP.

If P = ( a + b ) then P = NP and N = 1.

Proof that P = ( a + b )?

If we can prove this we have answer the question.

This algerbra doesn't really relate to the news announcement. P and NP aren't variables!