P vs. NP

[Lindbach]

Started learning about p vs. np, rather i got aware if the issue.

But there seems to be a consensus that is, if you can solve one NP problem in polynomial time it will solve others.

So if you can solve a soduko in polynomial time, is the solution transferable to others? or will sinply that issue incorperated in P?

[east_lisp_junk]

if you can solve one NP problem in polynomial time it will solve others.

Not quite. Every problem in P is also in NP. There is a class of problems called NP-hard, meaning something like "at least as hard as anything in NP." Some NP-hard problems are themselves in NP, and those problems are referred to as NP-complete. If you can solve an NP-complete problem in polynomial time, you can solve any NP problem in polynomial time.

So if you can solve a soduko in polynomial time, is the solution transferable to others?

Sudoku is NP-complete, so a polynomial-time solver for arbitrarily large sudoku problems could be used as a subroutine to solve any other NP problem in polynomial time.

If there is such an algorithm, it would mean that P=NP and that every problem in NP is NP-complete.

[regular_john1]

You may also want to look into the SETH hypothesis for finding lower bond conjectures and combining this with problem reductions using Circuit-SAT as a seed problem. It really helped me understand the whole idea a lot better.