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What is the difference between NP and coNP? Is coNP a subset of NP ? or vice versa? What is the significance of NP $\cap$ coNP when it comes to mathematics?

user45118
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    Do you know what $NP$ means? If not, I think the book by Sipser on Computational Complexity is well written and suitable for self study. – Elle Najt Jun 24 '17 at 06:15
  • https://en.wikipedia.org/wiki/Co-NP – Elle Najt Jun 24 '17 at 06:17
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    You may get more in-depth answers posting to Computer Science Stack Exchange. – rb612 Jun 24 '17 at 06:21
  • The precise title is " Introduction to the Theory of Computation: Michael Sipser." – Elle Najt Jun 24 '17 at 19:53
  • NP means some decision branch returns true in polynomial time. co-NP means all decision branches return true in polynomial time. – DanielV Jun 24 '17 at 20:25
  • @ArtimisFowl No. What I said is correct. There are 2 kinds of non deterministic machines. One machine returns true if any branch returns true ($\exists$ type), the other kind only returns true if all branches return true ($\forall$ type) (Actually there is a kind of machine with both capabilities, but that's beyond the scope of the question). CoNP says "can this be solved in polytime on a $\forall$ type". NP asks "can this be solved in polytime on a $\exists$ type". – DanielV Jun 25 '17 at 12:14

2 Answers2

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NP is the class of decision problems for which there is a polynomial time algorithm that can verify "yes" instances given the appropriate certificate.

CoNP is the class of decision problems for which there is a polynomial time algorithm that can verify "no" instances given the appropriate certificate.

We don’t know whether coNP is different from NP.

There is a problem in NP for every problem in coNP, and vice versa. For example, the SAT problem asks "does there exist a boolean assignment which makes this formula evaluate to True?". The complement problem, which is in coNP, asks, "do all boolean assignments make this formula evaluate to False?"

Adam Rosenthal
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There are several definitions that all end up equivalent. However, my favorite has to do with verification. $NP$ is the class of problems where it's easy (read "takes polynomial time, if given help") to verify that there is a solution. For example, it's easy to convince someone that a SAT instance is solvable, just give them a satisfying assignment and they can check if all the clauses are satisfied.

Likewise, $CO-NP$ problems are easy to show that there isn't a solution. For example, checking if a Boolean function is unchanged when we switch the first two inputs. I only need to show you one case where $f(x,y,\cdots) \neq f(y,x,\ldots)$ to show that it's not symmetric, however it's not at all clear how to show if this function is symmetric.

Because we can't easily show a SAT instance is unsatisfiable, nor that a function is symmetric, neither $NP$ nor $CO-NP$ is known to be a subset of the other.

Being in both means that you can easily check either way, given a proof (like a satisfying assignment or a counterexample). Examples include every problem we can solve in polynomial time, which require no help to verify, the curious can quickly run a poly time algorithm themselves to verify an answer.

Artimis Fowl
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