Re: Hamiltonian cycle problem in 2-connected bipartite cubic graph is Polynomial

On Dec 27 2007, 10:06 pm, "Fra" <fra.cristi...@xxxxxxxxx> wrote:
Thank you for review my homepage about the Hamiltonian cycle problem
on bicubic graph.
In fact, I had submitted several paper to some journal, but they
always refuse to review and rejected as some reason such as "not not
fall within the scope of the journal or meet all the criteria required
by the Journal... ".

So I think that the editor of most journals regards the P vs NP as
not important problem or others idea on my paper, I had to build up a
website and show my work to public.

The second reason to build up a website is that I hope every one could
challenge the P vs NP problem fair. Although many researchers had been
review my work, most of them are honest, only very very few of them
wish to jump new idea base my paper, but they don't admit my work.

Many researcher had cost many time on this kind of problem, as I known
from China, although they are very poor and without any funding. So
it is a good path to successful by building up a website.

I also wish some researcher funding could open the gate to this kind
of researcher (such as scholarship in university or others),
otherwise, this talent researcher maybe disappear by the poor life, or
maybe be employee by some security government or some computer virus
company. The latter will be a nightmare to world ( can image that
crack RSA in polynomial by a unsecurity government) .

-----------------------------------------------------
Guohun Zhu

Why you don't try to write a program that takes in input istance of SAT
(that is the paradigmatic np-complete problem), reduces this istance in
Hamiltonian cycle problem in 2-connected bipartite graph istances, and then
resolve it? (giving in output a solution for the SAT problem). It will be a
good heuristics evidence of your work.

Thank you for your comments about SAT problem.
I have study it from 2006 and I will show some results after two
weeks.
-----------------------------------------------------
Guohun Zhu
.

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