11. Satisfiability

The study of Boolean circuits is one way that we can hope to get more insights into the $$\mathbf{P}$$ versus $$\mathbf{NP}$$ problem. It also gives a great way to approach the study of $$\mathbf{NP}$$-complete problems.

Circuit Satisfiability

Consider the language

$\textsf{CircuitSAT} = \left\{ \left< C \right> : C \mbox{ is satisfiable}\right\}$

of all encodings of Boolean circuits that evaluate to True on at least one input. We can show directly that this language is $$\mathbf{NP}$$-complete.

Theorem. The language $$\textsf{CircuitSAT}$$ is $$\mathbf{NP}$$-complete.

Proof. We have already seen in the Non-Uniform Computation lecture that we can evaluate Boolean circuits efficiently with Turing machines. So there is polynomial-time verifier for $$\textsf{CircuitSAT}$$ that takes the (claimed) satisfying assignment as the certificate and simply simulates the encoded circuit $$C$$ on this input to verify that the circuit indeed evaluates to True. And so $$\textsf{CircuitSAT} \in \mathbf{NP}$$.

Consider now any language $$A \in \mathbf{NP}$$. Let $$V$$ be a polynomial-time verifier for $$A$$. As in the proof of Lemma 1 in the Non-Uniform Computation lecture, there is a polynomial-size circuit $$C$$ that can simulate $$V$$ on any input-certificate pair $$(x,c)$$ and evaluates to True if and only if $$V$$ accepts that pair. Furthermore, we can construct this circuit $$C$$ in polynomial time and hardcode the input $$x$$ in to obtain the circuit $$C'$$ that satisfies $$\left< C' \right> \in \textsf{CircuitSAT}$$ if and only if $$x \in A$$. Therefore, the construction gives a polynomial-time reduction from $$A$$ to $$\textsf{CircuitSAT}$$. Since this holds for all $$A \in \mathbf{NP}$$, by definition $$\textsf{CircuitSAT}$$ is $$\mathbf{NP}$$-hard.

As you may recall, we already established that another language, namely $$A_{TM}^{cert}$$ is $$\mathbf{NP}$$-complete. But the significance in this result is that the $$\textsf{CircuitSAT}$$ result is the first natural language that we have shown to be $$\mathbf{NP}$$-complete. From there, we can easily complete the proof of the Cook-Levin theorem that $$\textsf{SAT}$$ is $$\mathbf{NP}$$-complete.

Cook-Levin Theorem

The most natural satisfiability language is the language

$\textsf{SAT} = \{ \left< F \right> : F \mbox{ is a satisfiable formula}\}$

that considers the satisfiability of formulas instead of Boolean circuits. This language is also $$\mathbf{NP}$$-complete.

Cook-Levin Theorem. $$\textsf{SAT}$$ is $$\mathbf{NP}$$-complete.

Proof. The languages $$\textsf{SAT}$$ is in $$\mathbf{NP}$$ because we can easily verify in polynomial time whether a given assignment $$x$$ satisfies a formula $$F$$ given the encoding of $$F$$.

We now want to prove that $$\textsf{SAT}$$ is $$\mathbf{NP}$$-hard. We have already shown that $$\textsf{CircuitSAT}$$ is $$\mathbf{NP}$$-hard, so it suffices to show that $$\textsf{CircuitSAT} \le_{\mathbf{P}} \textsf{SAT}$$. For any Boolean circuit $$C$$, we want to construct in polynomial time a CNF formula that is satisfiable if and only if the circuit $$C$$ is satisfiable. To do this, the formula $$F$$ that we construct will have one variable for each of the $$n$$ inputs to $$C$$ as well as one variable $$x_g$$ for each gate $$g$$ of $$C$$. We want to construct constraints that are satisfied if and only if the variables for the gates are assigned the correct value given the circuit’s inputs. We can do this directly for the three types of gates.

A negation gate is simple. Let $$x_a$$ be the input to the negation gate associated to $$x_g$$. Then the two constraints

$(x_g \vee x_a) \wedge (\overline{x_g} \vee \overline{x_a})$

are both satisfied if and only if $$x_g = \neg x_a$$.

When $$x_g$$ is the variable associated to the $$\wedge$$ gate $$g$$ that takes as inputs $$x_a,x_b,\ldots,x_z$$ then the constraints

$(x_g \vee \overline{x_a} \vee \overline{x_b} \vee \cdots \vee \overline{x_z}) \wedge (\overline{x_g} \vee x_a) \wedge (\overline{x_g} \vee x_b) \wedge \cdots \wedge (\overline{x_g} \vee x_z)$

are all satisfied if and only if $$x_g = x_a \wedge x_b \wedge \cdots \wedge x_z$$.

Finally, when $$x_g$$ represents the $$\vee$$ gate with inputs $$x_a,\ldots,x_z$$ then

$(\overline{x_g} \vee x_a \vee x_b \vee \cdots \vee x_z) \wedge (x_g \vee \overline{x_a}) \wedge (x_g \vee \overline{x_b}) \wedge \cdots \wedge (x_g \vee \overline{x_z})$

are all satisfied if and only i f $$x_g = x_a \vee x_b \vee \cdots \vee x_z$$.

Combining the constraints for all the gates in $$C$$ and adding one additional constraint that consists simply of the variable for the ouput gate of $$C$$ gives us the final formula $$F$$. Its encoding can be constructed in polynomial time given the encoding of $$C$$. And $$F$$ is satisfiable if and only if $$C$$ is satisfiable, giving us the desired polynomial-time reduction from $$\textsf{CircuitSAT}$$ to $$\textsf{SAT}$$.

3SAT

We can show that an even more restricted variant of $$\textsf{SAT}$$ is also $$\mathbf{NP}$$-complete.

A CNF formula (shorthand for a Boolean formula in Conjunctive Normal Form) is a Boolean formula with unbounded fan-in of depth 2, where the bottom-most gate is a single $$\wedge$$ gate, the inputs to that output gate are $$\vee$$ gates, and the inputs to each $$\vee$$ gates are subsets of the input variables or their negations. Each $$\vee$$ gate in a CNF formula represents a clause. The number of inputs to a $$\vee$$ gate is the width of the corresponding clause. The size of a CNF formula is the number of clauses it contains, and the width of a CNF formula is the maximum width of all of its clauses.

A natural language that is most useful in the study of $$\mathbf{NP}$$-completeness is the satisfiability of CNF formulas of width at most 3, denoted by

$\textsf{3SAT} = \{ \left< F \right> : F \mbox{ is a satisfiable CNF formula of width } \le 3\}.$

Theorem. $$\textsf{3SAT}$$ is $$\mathbf{NP}$$-complete.

Proof. Looking back at the proof of the Cook-Levin theorem, we see that the polynomial-time reduction already converts general circuits into CNF formulas. The only missing step to modify the reduction so that it shows that $$\textsf{CircuitSAT} \le_{\mathbf{P}} \textsf{3SAT}$$ is to convert the resulting CNF formula into one that has width at most 3.

As it turns out, this final step is easy to accomplish. We can turn any clause $$x_1 \vee x_2 \vee x_3 \vee \cdots \vee x_k$$ of width $$k \ge 3$$ into a collection of clauses of width 3 by adding $$k-3$$ auxiliary variables $$y_1,\ldots,y_{k-3}$$ and taking the set of clauses

$(x_1 \vee x_2 \vee y_1) \wedge (\overline{y_1} \vee x_3 \vee y_2) \wedge (\overline{y_2} \vee x_4 \vee y_3) \wedge \cdots \wedge (\overline{y_{k-3}} \vee x_{k-1} \vee x_k).$

These clauses are all satisfied if and only if the original clause was satisfied, so we can convert the CNF formula $$\phi$$ produced by the reduction from $$\textsf{CircuitSAT}$$ to $$\textsf{SAT}$$ into one of width 3 that also satisfies the polynomial-time reduction conditions.

Eric Blais ©2024 — Last edited on Feb. 12, 2024