# Circuit satisfiability problem

In theoretical computer science, the circuit satisfiability problem (also known as CIRCUIT-SAT, CircuitSAT, CSAT, etc.) is the decision problem of determining whether a given Boolean circuit has an assignment of its inputs that makes the output true.[1]

## Properties

CircuitSAT has been proven to be NP-complete.[2] In fact, it is a prototypical NP-complete problem; the Cook–Levin theorem is sometimes proved on CircuitSAT instead of on SAT for Boolean expressions and then reduced to the other satisfiability problems to prove their NP-completeness.[1][3]

The satisfiability of a circuit containing m arbitrary binary gates can be decided in time ${\displaystyle O(2^{0.4058m})}$.[4]

## Reduction

Arithmetization is to reduce a CircuitSAT problem into a set of arithmetic constrains. Currently, in the literature, there are two direction of reduction: reduce to a combination satisfiability problem or reduce to an algebraic satisfiability problem. An example of such reduction from boolean circuit to polynomial is presented here, and they are called Zhegalkin_polynomial.

### The Tseitin transformation

There is a straightforward reduction from CircuitSAT to SAT, known as the Tseitin transformation. The transformation is especially easy to describe if the circuit is wholly constructed from 2-input NAND gates (a functionally-complete set of Boolean operators): assign every net in the circuit a variable, then for each NAND gate, construct the conjunctive normal form clauses (v1v3) ∧ (v2v3) ∧ (¬v1 ∨ ¬v2 ∨ ¬v3) where v1 and v2 are the inputs to the NAND gate and v3 is the output. These clauses completely describe the relationship between the three variables. Conjoining the clauses from all the gates with an additional clause constraining the circuit's output variable to be true completes the reduction; an assignment of the variables satisfying all of the constraints exists if and only if the original circuit is satisfiable, and any solution is a solution to the original problem of finding inputs that make the circuit output 1.[1][5] (The converse, that SAT is reducible to CircuitSAT, is even easier—we simply rewrite the Boolean formula as a circuit and solve that.)