I want to know if checking whether a 3-SAT is satsifiable if and only if every clause has at least 2 or all three literal set to true NP-complete?

It seems similar to Exact-3SAT but it differs from it as the version of this problem which we'll call Exact2or3-3SAT ,where all literals are positive is trivial as its always true .That's why I want to know if its NP-complete or not ?

  • $\begingroup$ I can't understand your question. Can you edit it to break it down into smaller sentences? I can't understand what two propositions are being related by the "if and only if". I suggest you define the problem that you are interested in, by listing the inputs to the algorithm and the desired output. $\endgroup$
    – D.W.
    Jan 28, 2023 at 17:29

1 Answer 1


It is not currently known whether your problem is $\mathsf{NP}$-complete, but it can definitely be solved in polynomial time since it can be reduced to 2-SAT. If we assume $\mathsf{P} \neq \mathsf{NP}$, then your problem is not $\mathsf{NP}$-complete.

The reduction is as follows: a clause $(\ell_1, \ell_2, \ell_3)$ of your problem is satisfied iff there does not exist a pair of distinct literals $\ell_i, \ell_j$ that are both false. Equivalently, for all pairs $\ell_i, \ell_j$ of literals you must have that at least one of $\ell_i$ and $\ell_j$ is true.

This shows that you can replace each clause $(\ell_1, \ell_2, \ell_3)$ with: $$ (\ell_1 \vee \ell_2) \wedge (\ell_1 \vee \ell_3) \wedge (\ell_2 \vee \ell_3), $$ to obtain an equivalent 2-SAT instance.

As a note, there is also a straightforward reduction from 2-SAT to your problem: add a new variable $x$ and replace each 2-SAT clause $(\ell_1 \vee \ell_2)$ with $(\ell_1, \ell_2, x)$.


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