2-Sat

Created: 2019-12-05 Thu 16:31

Objectives

An “and” expression is called a conjunction.

E.g. \( x_1 \wedge x_2 \)
An “or” is called a disjunction.

E.g. \( x_1 \vee x_2 \)
An expression of the form \( (x_1 \vee x_2 \vee \cdots \vee x_n) \wedge (y_1 \vee y_2 \vee \cdots \vee y_n) \cdots \)

is said to be in conjunctive normal form (CNF).

If each of the disjucts have only 2 elements, then we have (2-CNF).
Solving CNF is NP-Complete.
Solving 2-CNF is much easier!
We will convert the 2-CNF into a graph.
- Each variable \(x\) become two nodes: one for \(x\) and one for \(\neg x\).
- Implications become directed edges.
- Find strongly connected components, and check that \(x\) and \(\neg x\) are never in the same component.
Where to get implications? Basic idea:
- Remember that \( a \rightarrow b \equiv a \vee \neg b \)
- So.. we convert every \(x \vee y\) into \( \neg x \rightarrow y \)

To keep track of nodes, you need \(2n\) entries.
Let \(x\) and \(\neg x\) be adjacent:
E.g. \(x_2\) in position 4 and \(\neg x_2\) in position 5.
Then you can get the index of the negation by xor with 1.
To find strongly connected components, you can use the DFS method presented.
Or use Kosaraju’s Algorithm
- DFS the original graph, push nodes visited to a stack \(s\).
- DFS the inverse graph, starting with the top of stack.
- Each time you have to get a new element from the stack, increment the SCC counter.