Actually, SAT is an NP-complete problem (http://en.wikipedia.org/wiki/Boolean_satisfiability_problem#NP-completeness) so if it were calculatable in polynomial time, then P = NP.
On Jan 15, 2008 1:49 PM, Jim Bromer <[EMAIL PROTECTED]> wrote: > >> > In another message V. Nesov said: > Summarizing, you say that you might have proved P=NP, but don't give any > technical details, and there is God involved. It sounds really bad. > Vladimir Nesov mailto:[EMAIL PROTECTED] > ================ > > Ok, I would agree with you on that much. But I would also be wondering if > there was any possibility that it was true. > > At any rate, I should have a better idea if the idea will work or not by the > end of the year. > > But do you think that that SAT is not in p? Why not? (By the way, I did > not say that p=np. I think wikipedia gives examples of problems in np by > definition. So if SAT and the equivalent problems are in p that does not > mean that anything in np is in p.) > > Jim Bromer > > > > ________________________________ > Never miss a thing. Make Yahoo your homepage. > ________________________________ > This list is sponsored by AGIRI: http://www.agiri.org/email > To unsubscribe or change your options, please go to: > http://v2.listbox.com/member/?&; -- Robin Gane-McCalla YIM: Robin_Ganemccalla AIM: Robinganemccalla ----- This list is sponsored by AGIRI: http://www.agiri.org/email To unsubscribe or change your options, please go to: http://v2.listbox.com/member/?member_id=8660244&id_secret=86201040-9469e0
