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Unformatted text preview: CS 473 Homework 10 Solutions Spring 2010 1. Show that 2SAT is NP-hard, or describe a polynomial-time algorithm to solve it. Solution: We describe a polynomial-time algorithm to solve 2SAT. Given a 2CNF boolean formula φ , the implication graph G φ is defined as follows. For each variable x i in φ , the graph contains two vertices x i and x i . For each clause ( u ∨ v ) in φ , the graph contains two edges u → v and u → v . (Here, x i = x i .) The disjunction ( u ∨ v ) is logically equivalent to the implications u ⇒ v and v ⇒ u . Thus, the graph G φ is a direct encoding of φ with edges representing implications. w x y z w x y z The implication graph for the formula ( x ∨ y ) ∧ ( y ∨ z ) ∧ ( x ∨ z ) ∧ ( w ∨ y ) ∧ ( y ∨ z ) We say that two vertices are strongly connected if there is a directed cycle in G φ that passes through both of them. Strong connectivity is clearly an equivalence relation; the equivalence classes are called the strong components of G φ . y x z x z y w w 1 2 3 4 Strong components of the implication graph shown above. We claim that φ is satisfiable if and only if every variable x i is in a different strong component than its negation x i . = ⇒ Because implication is transitive, any path from a node x to another node z indicates the implication x ⇒ z . Thus, if G φ contains a path from a variable x i to its negation x i , then x i must be F ALSE in every satisfying assignment. Symmetrically, if there is a path from x i to x i in G φ , then x i must be T RUE in every satisfying assignment. Suppose G φ contains a cycle through both x i and x i . That cycle can be decomposed into a directed path from x i to x i and a path from x i to x i . The first path implies that x i must be T RUE in every satisfying assignment, and the second implies that x i must be F ALSE. We conclude that φ is not satisfiable....
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This note was uploaded on 04/18/2011 for the course CS 473 taught by Professor Chekuri,c during the Spring '08 term at University of Illinois, Urbana Champaign.
- Spring '08