Comment Re:Answering the actual question (Score 1) 525
I'm not quite sure I follow all of your logic here. If you have shown that you have a polynomial algorithm that solves an NP-complete problem you have proven that P=NP. Your algorithm is the proof. P=NP iff there exists a polynomial time algorithm to solve an np-complete problem in polynomial time. There may be a way to prove it that doesn't imply an algorithm, but I'd be impressed if I saw it...
A problem is NP-complete iff it is in NP, and all problems in NP are polynomial reducable to it. Thus is you solve one NP-complete problem, you have polynomial solved any problem in NP. Now, the coefficients may be very large, especially when you start multiplying things together as you do reductions, and you're also probably increasing the size considerably (think of the reduction from 3-SAT, to 3 coloring), so it's not necessarily very convenient to do this.
P=NP, or P!=NP is an important theoretical result, as much as it is a practical one. In fact, it's probably very unlikely that we'd be able to find efficient algorithms to solve any of the NP-complete problems any time soon even if P=NP were proven. However it would mean that we not only have a polynomial solution to these problems, but also to the optomization versions -- the NP-Hard problems, which is more what we're worried about when we talk about things such as factoring numbers, or finding set-partitions. NP-complete problems are just decision (yes, no) problems. .
As far as proving it -- if you do it, go ahead and publish! It's an important theoretical result, probably far more important in theory than it is in practice for now. If people just hoarded their discoveries we'd never have seen the advancement in the area. Heck, if Cook hadn't released his work, we'd probably not be arguing about this now. . .
A problem is NP-complete iff it is in NP, and all problems in NP are polynomial reducable to it. Thus is you solve one NP-complete problem, you have polynomial solved any problem in NP. Now, the coefficients may be very large, especially when you start multiplying things together as you do reductions, and you're also probably increasing the size considerably (think of the reduction from 3-SAT, to 3 coloring), so it's not necessarily very convenient to do this.
P=NP, or P!=NP is an important theoretical result, as much as it is a practical one. In fact, it's probably very unlikely that we'd be able to find efficient algorithms to solve any of the NP-complete problems any time soon even if P=NP were proven. However it would mean that we not only have a polynomial solution to these problems, but also to the optomization versions -- the NP-Hard problems, which is more what we're worried about when we talk about things such as factoring numbers, or finding set-partitions. NP-complete problems are just decision (yes, no) problems. .
As far as proving it -- if you do it, go ahead and publish! It's an important theoretical result, probably far more important in theory than it is in practice for now. If people just hoarded their discoveries we'd never have seen the advancement in the area. Heck, if Cook hadn't released his work, we'd probably not be arguing about this now. . .
