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] In other words, it asks whether the inputs to a given Boolean Circuit can be consistently set to 1 or 0 such that the circuit outputs 1. If that is the case, the circuit is called satisfiable. Otherwise, the circuit is called unsatisfiable. For instance, in the figure above, the circuit on the left can be satisfied by setting both inputs to be 1, but the circuit on the right is unsatisfiable.
CircuitSAT is closely related to Boolean satisfiability problem (SAT), and likewise, 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 the SAT and then reduced to the other satisfiability problems to prove their NP-completeness.[1][3] The satisfiability of a circuit containing arbitrary binary gates can be decided in time .[4]
Proof of NP-Completeness
Given a circuit and a satisfying set of inputs, one can compute the output of each gate in constant time. Hence, the output of the circuit is verifiable in polynomial time. Thus Circuit SAT belongs to complexity class NP. To show NP-hardness, we'll reduce 3SAT to Circuit SAT. The reduction goes as follows:
Suppose the original 3SAT formula has variables , and operators (AND, OR, NOT) . Design a circuit such that it has an input corresponding to every variable and a gate corresponding to every operator. Connect the gates according to the 3SAT formula. For instance, if the 3SAT formula is the circuit will have 3 inputs, one AND, one OR, and one NOT gate. The input corresponding to will be inverted before sending to an AND gate with and the output of the AND gate will be sent to an OR gate with
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-Complete.
Restricted Variants and Related Problems
Planar Circuit SAT
Assume that we are given a planar boolean circuit (i.e. a boolean circuit whose underlying graph is planar) containing only NAND gates with exactly two inputs. Planar Circuit SAT is the decision problem of determining whether this circuit has an assignment of its inputs that makes the output true. This problem is NP-complete.[5] In fact, if the restrictions are changed so that any gate in the circuit is a NOR gate, the resulting problem remains NP-complete.[5]
Circuit UNSAT
Circuit UNSAT is the decision problem of determining whether a given boolean circuit outputs false for all possible assignments of its inputs. This is the complement of the Circuit SAT problem, and is therefore Co-NP-complete.
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 Tseytin transformation
There is a straightforward reduction from CircuitSAT to SAT, known as the Tseytin 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 (v1 ∨ v3) ∧ (v2 ∨ v3) ∧ (¬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][6] (The converse, that SAT is reducible to CircuitSAT, is even easier—we simply rewrite the Boolean formula as a circuit and solve that.)
Minesweeper Inference Problem
This problem asks whether it is possible to locate all the bombs given a Minesweeper board. It has been proven to be CoNP-Complete via a reduction from Circuit UNSAT problem.[7]
See Also
- Circuit Value Problem
- Structured Circuit Satisfiability
- Satisfiability problem
References
- ^ a b c David Mix Barrington and Alexis Maciel (July 5, 2000). "Lecture 7: NP-Complete Problems" (PDF).
- ^ Luca Trevisan (November 29, 2001). "Notes for Lecture 23: NP-completeness of Circuit-SAT" (PDF).
- ^ See also, for example, the informal proof given in Scott Aaronson's lecture notes from his course Quantum Computing Since Democritus.
- ^ Sergey Nurk (December 1, 2009). "An O(2^{0.4058m}) upper bound for Circuit SAT".
- ^ a b "Algorithmic Lower Bounds: Fun With Hardness Proofs at MIT" (PDF).
{{cite web}}: Cite has empty unknown parameter:|dead-url=(help) - ^ Marques-Silva, João P. and Luís Guerra e Silva (1999). "Algorithms for Satisfiability in Combinational Circuits Based on Backtrack Search and Recursive Learning" (PDF).
- ^ Scott, Allan; Stege, Ulrike; van Rooij, Iris (2011-12-01). "Minesweeper May Not Be NP-Complete but Is Hard Nonetheless". The Mathematical Intelligencer. 33 (4): 5–17. doi:10.1007/s00283-011-9256-x. ISSN 1866-7414.