SAT algorithm for determining if a graph is disjoint

Question

What are some good algorithms to have a SAT (CNF) solver determine if a given graph is fully-connected or disjoint?

The best one I can think of is this:

• Number the nodes 1..N, where N is the number of nodes in the graph.
• define N^2 variables with the ordered pair (P, Q), where P = 1..N and Q = 0..N-1.
• Set (1,0) to true.
• Set (A,P+1) to true iff there is an edge connecting node A and node B and (B,P) is true.
• If there exists a true (X,Y) variable for all possible nodes X, then the graph is connected.

Effectively, (X,Y) means "Node X is Y steps away from node X".

This seems inefficient at O(N^2) variables. Can this be improved?

A comment (from when I posted this on cstheory.stackexchange.com) asked why I would need a SAT-based algorithm when O(N) algorithms for connectivity are well-known. The reason is simple -- I have many other SAT-based constraints on the graph that also need to be satisfied at the same time.

Answer 1

Given a graph $G = (V,E)$, here is a SAT instance which is satisfiable iff the graph is not connected.

Pick an arbitrary vertex $v_0 \in V$, and add the following clauses, over the variables $x_v$ for $v \in V$:

• $x_{v_0}$.
• For every $(u,v) \in E$, $\lnot x_u \lor x_v$ and $\lnot x_v \lor x_u$.
• $\bigvee_{v \neq v_0} \lnot x_v$.

Here is a SAT instance which is satisfiable iff the graph is connected.

Pick an arbitrary vertex $v_0 \in V$, and add the following clauses, over the variables $x_{v,i}$ for $v \in V$ and $i \in \{0,\ldots,n-1\}$:

• $\lnot x_{v,0}$ for all $v \neq v_0$.
• For every vertex $v \in V$ and $i \in \{0,\ldots,n-2\}$, $\lnot x_{v,i+1} \lor x_{v,i} \lor \bigvee_{u\colon (u,v) \in E} x_{u,i}$.
• For every vertex $v \in V$, $x_{v,n-1}$.

Answer 2

Yuval describes a boolean CNF formula that is satisfiable iff the graph is not connected, using $|V|$ variables; and a boolean CNF formula that is satisfiable iff the graph is connected, using $|V|^2$ variables.

I describe here a boolean CNF formula that is satisfiable iff the graph is connected, using $O(|V| \lg |V| + |E|)$ variables. If $|V|$ is large enough and the graph is sparse, this is better than Yuval's solution.

Introduce variables $d_v$, one for each vertex $v \in V$, over the domain $d_v \in \{0,1,\dots,|V|-1\}$, and variables $y_{u,v}$, one for each edge $(u,v) \in E$. Add the following constraints:

• $d_{v_0} = 0$
• $d_v \le d_u + 1$ for each $(u,v) \in E$
• for each $v$, there exists $u$ such that $(u,v) \in E$ and $y_{u,v}$
• $y_{u,v} \implies d_v = d_u + 1$

Here the intended interpretation of $d_v$ is that it captures the distance (length of the shortest path) from $v_0$ to $v$; and the intended interpretation of $y_{u,v}$ is that it should be true if the edge $(u,v)$ is part of a shortest path $v_0 \leadsto v$. Note that this system of constraints has a satisfying assignment iff the graph is connected.

Now you can convert this to SAT using "bit-blasting". In particular, for each $d_v$, introduce variables $x_{v,i}$ for $i=0,1,\dots,\lceil \lg |V| \rceil -1$; the intended meaning is that $x_{v,i}$ is the $i$th bit of $d_v$. Now we can express the constraint $d_v \le d_u+1$ using $O(\lg |V|)$ additional variables and clauses (e.g., by building a circuit for this comparison and using Tseitin's transform). Also, we can express the constraint "for each $v$, there exists $u$ such that..." using $|V|$ clauses, one for each $v$, each of which is a disjunction over the appropriate set of $u$. In this way, we obtain a set of CNF clauses that uses $O(|V| \lg |V|)$ variables for the bit-blasting of the $d$'s, and $|E|$ variables for the $y$'s. This yields a CNF formula as claimed above.