Re: Proof of P and NP

Suresh Devanathan <idatarm@NOJUNKgmx.com> writes:

Suppose you are trying to solve a Boolean satisfiability problem C[x]. Now
convert the problem into probabilistic domain.
>
For probability vector x
C[x] = p
     = p + 0*K*(B - p)
     = lim K->infinity p + 1/K*K(B- p)
     = B
>
>
For friend
  in case a satisfiability, B = 1
  in case untestable, assume x_n=  0.5, B = 0
>
For enemy,
 in case a satisfiability, B = 0
  in case untestable, assume x_n = 0.5, B = 1
>
>
since by cook's thesis, even in 1 problem in NP proven to be P , every
problem proven P.

Not right.  If a problem proven to be NP-complete is shown to be in P,
then P = NP.

If
reverse result is always true of enemy, the number of permutation is at
least as great as 2^N
where N is number of inputs, showing problem is be NP.

The problem must be NP-complete, not just in NP.

complete proof

Not yet.

--
Ben.