Since P is a subset of NP, clearly there are problems in NP that can
be solved in polynomial time.  NP problems must be solvable by a
non-deterministic Turing machine (in simple terms a turing machine
that executes multiple actions for each symbol but only chooses the
symbols which lead to an accept state) in polynomial time and have
their solutions verified by a deterministic turing machine in
polynomial time.  NP-complete problems are just problems for which no
polynomial time deterministic turing machine algorithm is known.  All
that is required to prove P = NP is to prove that NP is a subset of P,
which would require finding a polynomial time algorithm for solving
any problem which is NP-complete.

As for your problem involving SAT, it's not applicable to P-NP because
they are classes of decisions problems
(http://en.wikipedia.org/wiki/Decision_problem), which means problems
that can be answered yes or no.

And what you quoted should have been

By the definition of NP-complete, any problem in NP must be reducible
to any problem in NP-COMPLETE through a polynomial transformation.

On Jan 18, 2008 1:47 PM, Jim Bromer <[EMAIL PROTECTED]> wrote:
> Robin Gane-McCalla <[EMAIL PROTECTED]> wrote: Can you give me an
> example of any such problems (or a class of such problems)? By the
> definition of NP-complete, any problem in NP must be reducible to any
> problem in NP through a polynomial transformation.
> ---------------------------------------------------------
> It took me a while to figure this one out.  I was using the term np as a
> measure of the time (or time-data space) for computational processes and it
> was not very precise as such.  However,  the definition of np is more
> precise which is why your comment forced me to think about it.
>
> I was just trying to point out that even if a np-complete program was proven
> to be solvable in p-time, there still would be problems that would take much
> longer to solve.  And some of them might look like np-complete problems to
> the casual onlooker.
>
> For instance the SAT problem asks only if a certain logical formula has a
> solution (that is, a True solution).  At this time, in the worse case the
> solution to this problem for n variables could take more than 2^n 'steps' to
> complete, which is the same time it would take to try every possible
> combination of  logical (True or False) values in the formula.
>
> Let's say that a method (might as well say my method) produces a solution to
> the SAT problem where even the worse cases can be determined in polynomial
> time.
>
> Now let's say that the problem is not to determine whether or not there is a
> solution or not, but to output every possible combination of values for a
> logical formula and give the logical result as well.  Since there are 2^n
> combinations for a formula of n distinct logical variables, the output would
> have to take more than 2^n 'steps'.  So even supposing the np-complete
> problem could be solved in polynomial time, there are still problems that
> cannot be completed in polynomial time.
>
> Interestingly, if a polynomial time solution to the SAT problem was found,
> it might be possible to output a couple of formulas that could define all of
> the true and all of the false combinations without distinctly outputting
> every one separately.  However, the problem I defined above specifically
> called for each possible combination to be output and therefore it cannot be
> solved in polynomial time -by definition-.  This is a trivial example, but
> it shows that there are problems that cannot be solved in polynomial time,
> and because this problem has something in common with the np-complete
> problem of Satisfiability, some people might be confused by the problem.
> Jim Bromer
>
>
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-- 
Robin Gane-McCalla
YIM: Robin_Ganemccalla
AIM: Robinganemccalla

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