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3-satisfiability can be generalized to k-satisfiability (k-SAT, also k-CNF-SAT), when formulas in CNF are considered with each clause containing up to k literals. [ citation needed ] However, since for any k ≥ 3, this problem can neither be easier than 3-SAT nor harder than SAT, and the latter two are NP-complete, so must be k-SAT.
The partial Max-SAT problem is the problem where some clauses necessarily must be satisfied (hard clauses) and the sum total of weights of the rest of the clauses (soft clauses) are to be maximized or minimized, depending on the problem. Partial Max-SAT represents an intermediary between Max-SAT (all clauses are soft) and SAT (all clauses are ...
In computer science and formal methods, a SAT solver is a computer program which aims to solve the Boolean satisfiability problem.On input a formula over Boolean variables, such as "(x or y) and (x or not y)", a SAT solver outputs whether the formula is satisfiable, meaning that there are possible values of x and y which make the formula true, or unsatisfiable, meaning that there are no such ...
In computer science and mathematical logic, satisfiability modulo theories (SMT) is the problem of determining whether a mathematical formula is satisfiable.It generalizes the Boolean satisfiability problem (SAT) to more complex formulas involving real numbers, integers, and/or various data structures such as lists, arrays, bit vectors, and strings.
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#SAT is different from Boolean satisfiability problem (SAT), which asks if there exists a solution of Boolean formula. Instead, #SAT asks to enumerate all the solutions to a Boolean Formula. #SAT is harder than SAT in the sense that, once the total number of solutions to a Boolean formula is known, SAT can be decided in constant time.
A Horn formula is a propositional formula formed by conjunction of Horn clauses. Horn satisfiability is actually one of the "hardest" or "most expressive" problems which is known to be computable in polynomial time, in the sense that it is a P -complete problem. [ 2 ]
For every R, add clauses representing f R (x i1,...,x iq) using 2 q SAT clauses. Clauses of length q are converted to length 3 by adding new (auxiliary) variables e.g. x 2 ∨ x 10 ∨ x 11 ∨ x 12 = ( x 2 ∨ x 10 ∨ y R) ∧ ( y R ∨ x 11 ∨ x 12). This requires a maximum of q2 q 3-SAT clauses. If z ∈ L then there is a proof π such ...