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The NP-completeness of NAE3SAT can be proven by a reduction from 3-satisfiability (3SAT). [2] First the nonsymmetric 3SAT is reduced to the symmetric NAE4SAT by adding a common dummy literal to every clause, then NAE4SAT is reduced to NAE3SAT by splitting clauses as in the reduction of general -satisfiability to 3SAT.
A variant of the 3-satisfiability problem is the one-in-three 3-SAT (also known variously as 1-in-3-SAT and exactly-1 3-SAT). Given a conjunctive normal form with three literals per clause, the problem is to determine whether there exists a truth assignment to the variables so that each clause has exactly one TRUE literal (and thus exactly two ...
Notice that the 3SAT formula is equivalent to the circuit designed above, hence their output is same for same input. Hence, If the 3SAT formula has a satisfying assignment, then the corresponding circuit will output 1, and vice versa. So, this is a valid reduction, and Circuit SAT is NP-hard. This completes the proof that Circuit SAT is NP ...
There is often only a small difference between a problem in P and an NP-complete problem. For example, the 3-satisfiability problem, a restriction of the Boolean satisfiability problem, remains NP-complete, whereas the slightly more restricted 2-satisfiability problem is in P (specifically, it is NL-complete), but the slightly more general max ...
He constructs a PCP Verifier for 3-SAT that reads only 3 bits from the Proof. For every ε > 0, there is a PCP-verifier M for 3-SAT that reads a random string r of length ( ()) and computes query positions i r, j r, k r in the proof π and a bit b r. It accepts if and only if 'π(i r) ⊕ π(j r) ⊕ π(k r) = b r.
The proof is by reduction to planar maximum cut. [8] Planar circuit SAT: This is a variant of circuit SAT in which the circuit, computing the SAT formula, is a planar directed acyclic graph. Note that this is a different graph than the adjacency graph of the formula. This problem is NP-complete. [9]