$(2+\varepsilon)$-Sat Is NP-hard

We prove the following hardness result for a natural promise variant of the classical CNF-satisfiability problem: Given a CNF-formula where each clause has width $w$ and the guarantee that there exists an assignment satisfying at least $g = \lceil \frac{w}{2}\rceil -1$ literals in each clause, it is NP-hard to find a satisfying assignment to the formula (that sets at least one literal to true in each clause). On the other hand, when $g = \lceil \frac{w}{2}\rceil$, it is easy to find a satisfying assignment via simple generalizations of the algorithms for 2-Sat. Viewing 2-Sat $\in \mathrm{P}$ as tractability of Sat when 1 in 2 literals are true in every clause, and NP-hardness of 3-Sat as intractability of Sat when 1 in 3 literals are true, our result shows, for any fixed $\varepsilon > 0$, the difficulty of finding a satisfying assignment to instances of “$(2+\varepsilon)$-Sat” where the density of satisfied literals in each clause is guaranteed to exceed $\frac{1}{2+\varepsilon}$. We also strengthen the ...