I define a long CNF to contain at least $2^\frac{n}{2}$ clauses, where $n$ is the number of its variables. Let $\text{Long-SAT}=\{\phi: \phi$ is a satisfiable long CNF formula$\}$.
I'd like to know why $\text{Long-SAT} \in P$. First I thought it is $\text{NPC}$ since I can do a polynomial-time reduction from $\text{SAT}$ to $\text{Long-SAT}$, no?
But maybe I can reduce $\text{2-SAT}$ to $\text{Long-SAT}$? How do I do that?
Asked By : Numerator
Answered By : Luke Mathieson
Unless I'm missing something, it's trivially in P as the length of the formula is exponential in the number of variables. Hence all $2^{n}$ truth assignments can be generated and checked in polynomial time in the length of the formula.
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Question Source : http://cs.stackexchange.com/questions/2704
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