COMS 482: unofficial class blog


Lecture 23: 3-SAT

Posted in Class Notes by Elliott Back on March 16th, 2005. [Del.icio.us]

Introductory Definitions:

A boolean variable is a variable that can either be assigned one of true or false. A term is either a boolean variable or its negation. A clause is built using terms and V (or operator).

SAT (Satisfiability):

Given a set of clauses c1 … ck over the variables x1 … xn, is there way to assign True and False to variables so that all clauses evaluate to true? SAT <=p 3-SAT <=p Independent Set.

3-SAT:

Given clauses c1 … ck each of length 3 over the variables x1 … xn, is there way to assign True and False to variables so that all clauses evaluate to true?

Claim: “3-SAT is NP-Complete”

Proof: Given a certificate (a set of assignments to variables) we can check in polynomial time that this satisfies all of the clauses. Next we want to show that 3-SAT reduces to independent set. Let’s map 3-SAT to a graph structure, obeying the following rules:

  • Make sure there is at least one true term in each clause so that all clauses are thus true.
  • Make sure x and !x are not both true

3-SAT Requirements

We can use these rules to construct a graph of k triangular clauses, connecting all x and !x between clauses. The independent set of this graph is the solution to 3-SAT:

3-SAT and Independent Set

Proof that 3-SAT reduces to Independent Set:

Claim: The original clauses can all be satisfied iff the graph has an independent set of size k.

Proof (=>): Suppose there exists a satisfying assignment. Then each clause has 1 or more true terms: Choose one of them. Put these into set S: we have k of them. For any 2 vertices in S there is no edge between them because 1) edges in the cause are not chosen and 2) edges between clauses are not chosen because a variable cannot be both true and false. So, S is an independent set.

Proof (<=): Assume we have an independent set S of size k. No 2 vertices can be in the same clause, so each clause has 1 vertex in S. For each vertex in S, we set the corresponding term to be true. X and !x are not both in S, so they cannot both be true. Make leftover variables either true or false. Each clause is satisfied by this assignment.

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