1

On one hand, Horn-SAT is known to be tractable in linear time - where Horn-SAT is the problem of deciding whether a given set of propositional Horn clauses (with at most one positive literal) is satisfiable or not. On the other hand, Double-SAT is NP-complete (see this post : Proving Double-SAT is NP-complete) - where Double-SAT is the problem of deciding whether a given set of propositionnal clauses has at least two models.

Let Double-Horn-SAT be the problem of deciding whether a given set of propositional Horn clauses has at least two models.

What is the complexity of Double-Horn-SAT ?

Community
  • 1
Xavier Labouze
  • 551
  • 2
  • 12

1 Answers1

4

Double-Horn-SAT is polynomial-time solvable.

Given a set of Horn-clauses $X = \{C_1,C_2,...,C_n\}$ and a solution $x_1=b_1,x_2=b_2,...,x_n=b_n$, $b_i \in \{true,false\}$ we can observe that a second solution must differ in at least one variable $x_i$. So we can run the Horn-SAT algorithm $n$ times on the augmented set $X_i' = X \cup \{ \lnot x_i \}$ if in the first solution $x_i = true$ or on the augmented set $X_i' = X \cup \{ x_i \}$ if in the first solution $x_i = false$. If we find a solution for one of the $X_i'$ then it is a valid solution for the original set $X$, too and by construction it differs from the first one; otherwise the original set doesn't have a second solution, as can be easily proved by contradiction.

See also the question "Complexity of finding a second solution given a correct solution to an NP-complete problem" on cstheory and Laurent Juban, Dichotomy theorem for the generalized unique satisfiability problem (1999)

Community
  • 1
Vor
  • 12,743
  • 1
  • 31
  • 62