Is 2-DNF is NP-complete?

Question

I want to know whether the 2-DNF problem is NP-complete or not? If it is NP-complete, can anyone provide a proof?

Answer 1

If you are referring to the problem of deciding whether a formula given in $2-DNF$ form is satisfiable, then it is in $P$, as well as general $DNF$ satisfiability. Indeed, such a formula is satisfiable iff there is a clause that does not contain an inner contradiction. That is, a clause that does not contain both $p$ and $\neg p$ for some atomic proposition $p$. This can be checked in linear time.

Perhaps you are referring to $2-CNF$? (which is also in $P$, but it's less trivial).

Answer 2

An additional note to Shaull's answer.

Checking if a DNF formula $\phi$ is valid (i.e. is a tautology) is co-NP-complete.

Indeed a formula $\phi$ is valid if and only if its negation is not satisfiable. But if you negate a DNF formula and apply De Morgan's laws you get a CNF formula. So the problem is equivalent to CNF unsatisfiability.

But if you restrict to 2-DNF, the validity can be checked in polynomial time because the corresponding negated formula is equivalent to the (un)satisfiability of a 2-CNF formula (that can be checked in polynomial time).