5
$\begingroup$

If $\sf{P} \ne \sf{NP}$ and $\sf{NP} \ne \sf{coNP}$ are both true then $\sf{P}$, $\sf{NP}$ and $\sf{coNP}$ are three separate complexity classes. In other words, verifying a solution, finding a solution and proving there is no solution to a formula are three separate tasks.

However, if $\sf{P} \ne \sf{NP}$ then both $\sf{NP}$ and $\sf{coNP}$ are in $\sf{EXPTIME}$ and not better. If $\sf{P} \ne \sf{NP}$ then it makes sense that $\sf{NP} = \sf{coNP}$, both finding a solution and proving there is no solution are exponentially hard.

From wikipedia:

  • It is suspected that $\sf{P} \ne \sf{NP}$
  • It is suspected that $\sf{NP} \ne \sf{coNP}$

I'm having trouble understanding how those two suspicions are compatible/consistent? Is it suspected that regardless the exponential difficulty for both, finding a solution is still easier than proving there is none in the general case?

$\endgroup$
11
$\begingroup$

However, if P != NP then both NP and coNP are in exptime and not better.

This is not neccesarily true. It just means that some NP (and coNP) problems cannot be solved in polynomial time. This is not necessarily the same as requiring exponential time (e.g. $n^{\log n}$ is not polynomial but also not exponential).

If P != NP then it makes sense that NP = coNP, both finding a solution and proving there is no solution are exponentially hard.

It does not matter that they are "equally hard" in the sense of both requiring a similar (whether that be exponential or some other) amount of time to solve. NP and coNP are classes of problems and even if the problems in NP and coNP have exactly "the same time complexity" the problems can still be different. NP and coNP are not defined in terms of time complexity.

Recall that NP is the set of problems for which

  • it can be proved (using an appropriate hint) that there is a solution in polynomial time

and coNP is the set of problems for which

  • it can be proved (using an appropriate hint) that there is no solution in polynomial time

To show that SAT is in NP, we can easily see that, if given a satisfying assignment as "hint", we can easily (in polynomial time) check that a given formula is a yes-instance of SAT. However, it is not clear how we can easily give a similar "hint" to allow one to easily check that a given formula does not have any satisfying assignment (which would be needed to show that SAT is in coNP). If one were to prove that P $\not =$ NP, that wouldn't change that. Why would it suddenly become possible to find such hints that allow you to easily verify the non-existence of a satisfying assignment?

In fact, the exact opposite of your intuition is true. If it turns out that P=NP then it immediately follows that NP=coNP.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you for the clear answer. It does seem that I didn't take the hint part into account in my definition of NP and coNP $\endgroup$
    – d3m4nz3
    Jan 1 at 17:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.