I really don’t understand your argument at the end of the proof : “number of square-free monomials of degree p-1 in n variables which is .”

For example, , square-free monomials of degree 2 in 12 variables are , hence there are 13 square-free monomials in 12 variables. BUT your answer is . Could you please explain?

Many thanks, ]]>

Is Claim 1 correct? If I take p = 2, n = 4×2 = 8. Let x = (1,1,1,1,-1,-1,-1,-1), y = (1,-1,1,1,1,-1,-1,-1), then $ = 0$ (mod 2) but $=4$ (mod 8), a contradiction. ]]>

Is Claim 1 correct? If I take p = 2, n = 4×2 = 8. Let x = (1,1,1,1,-1,-1,-1,-1), y = (1,-1,1,1,1,-1,-1,-1), then = 0 mod 2 but =6 mod 8, a contradiction. ]]>

I really love your blog post. This is the best blog for combinatorics. Happy birthday to you ! ]]>

I have found a new complexity class that I called “equivalent-P” and I have showed it has a close relation with the P versus NP Problem. I have been making a paper that explain my ideas, and in the meantime, I decided to share them as a preprint in the web.

Here, I show you the abstract of the paper (so you can capture the idea):

“The P versus NP problem is one of the most important and unsolved problems in computer science. This consists in knowing the answer of the following question: Is P equal to NP? This incognita was first mentioned in a letter written by Kurt Gödel to John von Neumann in 1956. However, the precise statement of the P versus NP problem was introduced in 1971 by Stephen Cook in a seminal paper. We consider a new complexity class, called equivalent-P, which has a close relation with this problem. The class equivalent-P has those languages that contain ordered pairs of instances, where each one belongs to a specific problem in P, such that the two instances share a same solution, that is, the same certificate. We demonstrate that equivalent-P = NP and equivalent-P = P. In this way, we find the solution of P versus NP problem, that is, P = NP.”

In this preprint, I shall show that there is an NP-complete problem in equivalent-P and a P-complete problem in equivalent-P. Moreover, I shall show the complexity class equivalent-P is closed under reductions. Since P and NP are also closed under reductions, then we can conclude that P = NP.

You can read more on:

https://hal.archives-ouvertes.fr/hal-01161668/document

Kind regards,

Frank.