Isn't NP problems always also co-NP problems?

Question

I think I have a hard time understanding the definition for NP. It says: "All decision problem where every yes-instance can be verified in polynomial time". But doesn't this just mean that every possible solution can be verified to be yes or no in polynomial time? If co-NP then is verifying no-instances isn't NP already doing that?

So simply put, if you can verifiy yes-instances can't you implicitly verifiy no-instances?

score 3 · Accepted Answer · answered Dec 12 '17 at 15:47

The definition you give in the quesition is very imprecise so it's understandable that you've misunderstood it. The formal definitions are given at our reference question on NP.

A decision problem is in NP if there's some polynomial $p$ and an ordinary, deterministic polynomial-time Turing machine $M$ such that:

For every "yes" instance $x$, there is some $y$ with $|y|\leq p(|x|)$ such that $M$ accepts the string $x,y$;
For every "no" instance $x$, $M$ rejects all strings $x,y$.

So, for example, 3-COLOURABILITY is in NP: the string $y$ is just a description of a $3$-colouring and $M$ accepts $x,y$ if $y$ is a valid $3$-colouring of the graph $x$ and rejects otherwise. The point is not that you can "verify 'yes' instances" but, rather, that every "yes" instance has a certificate $y$ that you can use to easily prove that it really is a "yes" instance.

But there's a significant asymmetry in the definition. If you give $M$ a "yes" input and a certificate $y$, then $M$ will say "I agree – that's a 'yes' instance." However, there's no input you can give it that will cause it to say "I agree – that's a 'no' instance." For any certificate you try, it will just say "That doesn't convince me it's a 'yes'." You could, of course, try all possible certificates. That would certainly prove that you had a "no" instance but there's a problem: there are exponentially many possible certificates to try, so trying all of them doesn't give you a polynomial-time algorithm.

score -1 · Answer 2 · answered Dec 12 '17 at 15:32

-1

No. As an intuitive argument, consider verifying primality. It is easy to show you that $8633$ is composite by $8633 = 89 \times 97$, whereas it is not evident that how this helps to develop a method to persuade you that some number is indeed a prime.

answered Dec 12 '17 at 15:32

Lwins

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Bad example: primality is in P and, even before that was known, it was known to be in co-NP. – David Richerby Dec 12 '17 at 15:34
@DavidRicherby Surely I know that. – Lwins Dec 12 '17 at 15:35
@Ferus I am sorry that I could not get your point. Would you please explain more specifically? – Lwins Dec 12 '17 at 15:45
@Lwins Never mind, realized the co-NP would have different looking solutions. – Ferus Dec 12 '17 at 15:50

Isn't NP problems always also co-NP problems?

2 Answers2