Does P = NP?

[Penis Drager]

I kinda doubt it...

 

[King Fructose]

According to my CS theory homework: no. Proof?
The N variable is only on one side of the equation.

 

[Penis Drager]

According to my CS theory homework: no. Proof?
The N variable is only on one side of the equation.

N could = 1...

 

[Clout $ Monei]

Yes.
Source : dude trust me.

Now give me my million dollars.

 

[King Fructose]

N could = 1...

Well shit.

 

[Screed]

N-I+G×G/E < R

 

[Haunted Gambler]

The answer is probably Pineapple.

 

[Anstiv]

Probably not. If it did that'd be pretty based though

 

[Solid Snek]

I like socks.

These are the socks I wear when I do computer stuff. They make me look like a super cute hacker (or, as my friends say on the internet; Le Kawaii H4xx0r!)

But enough about my credentials.

N could = 1...

Yes, if it was a simple algebraic equation, but I don't think NP means N*P.

I believe P means "polynomial time" - as in, a problem that can be solved in a reasonably efficient number of steps - and NP means "nondeterministic polynomial time" - as in, a problem whose solution is not currently know, but whose answer (once attained by, for example, a hypothetical, nondeterministic computer) can be verified in polynomial time. Both of these are sets of problems, not a specific problem (so proving one instance of NP is P would not, in and of itself, be sufficient to show N=NP) , and the N in NP is not a separate variable, but just part of the way we choose to identify that particular set (so N will not equal 1, such that P=1P)

But this stuff isn't covered on my sock, so how the fuck should I know?!

Go ask that Indian guy, the one who keeps staring at me; he likes all that brainy math stuff. I've got a speedrun tournament to practice for, and I can't help you solve fundamental issues in computer science.

 

[King Fructose]

I'll ask my buddy once he gets back from the designated shitting street.

 

[deso2y]

No, and I have discovered a truly marvellous demonstration of this proposition that this margin is too narrow to contain.

 

[byuu]

Both of these are sets of problems, not a specific problem (so proving one instance of NP is P would not, in and of itself, be sufficient to show N=NP)

That is not quite right. If you can show that one NP-hard problem in NP is also in P, you have proven P=NP.

So you have to find some really clever algorithm that solves such a problem in polynomial time.
Or prove that such an algorithm can not possibly exist to proof P /= NP.