Can someone give me the resolution procedure for 2 SAT, which is O (nˆ2)

Question

According to Wikipedia, Even, S .; Itai, A .; Shamir, A. cited it in "On the complexity of time table and multi-commodity flow problems".

The paper can be found here: http://www.cs.technion.ac.il/users/wwwb/cgi-bin/tr-get.cgi/1975/CS/CS0059.pdf

EDIT: Reference [5] of the above paper can be found here: https://www.jstor.org/stable/2310460

Answer 1

Actually, Wikipedia cites the Strongly connected components algorithm which is linear:

1. Consider a 2SAT problem:

$$(¬a ∨ b) ∧ (¬b ∨ c) ∧ (¬c ∨ b) ∧ (y ∨ b) ∧ (¬z ∨ y)$$

2. For each clause, convert it into two material conditionals where the negative of the first literal implies the second literal and vice versa (i.e. the contraposition):

$$(¬a ∨ b)$$ $$=>$$ $$a {\displaystyle \rightarrow } b,$$ $$¬b {\displaystyle \rightarrow } ¬a$$

3. You get a graph where the nodes are the literals; you can note that this graph is skew symmetric, due to the nature of the contraposition:

    (a) → (b) → (¬y) → (z)
      ↖  ↙
       (c)
(¬a) ← (¬b) ← (y) ← (¬z)
  ↘   ↗
   (¬c)

4. For the Kosaraju's algorithm, let's create an ordered list L of graph vertices.

5. For each vertex u of the graph, mark u as unvisited. Let L be empty.

6. Start to visit the graph from one node (like ¬b) following the arrow and avoiding the visited node.

7. Each time you're blocked and you have to backtrack to continue, you add the last node in the list:

    (a) → (b) → (¬y) → (z)
      ↖  ↙
       (c)
(¬a✓) ← (¬b✓) ← (y) ← (¬z)
  ↘    ↗
   (¬c✓)


L = [¬c, ¬a, ¬b]

8. When you are completely blocked, start to explore from an unvisited node and again and again:

    (a✓) → (b✓) → (¬y✓) → (z✓)
      ↖   ↙
       (c✓)
(¬a✓) ← (¬b✓) ← (y✓) ← (¬z✓)
  ↘    ↗
   (¬c✓)


L = [¬c, ¬a, ¬b, y, ¬z, c, a, b, ¬y, z]

9. Return the list:

  L = [z, ¬y, b, a, c, ¬z, y, ¬b, ¬a, ¬c]

10. Create the transpose graph; it's just the same with reversed arrows:

    (a) ← (b) ← (¬y) ← (z)
      ↘  ↗
       (c)
(¬a) → (¬b) → (y) → (¬z)
  ↖   ↙
   (¬c)

11. Visit again the transpose graph starting from the first node of the list:

    (a) ← (b) ← (¬y) ← (z)
      ↘  ↗
       (c)
(¬a) → (¬b) → (y) → (¬z✓)
  ↖   ↙
   (¬c)


L = [(z), ¬y, b, a, c, ¬z, y, ¬b, ¬a, ¬c]

12. Each time you are blocked, you have found a strongly connected components; you note it and you remove all its nodes from the list; repeat until the list is empty:

  L = []

13. The Kosaraju's algorithm tells you that you have a first strongly connected component {a, b, c}, a second one {¬y}, a third one {z} and their opposites:

    (a) → (b)   (¬y)   (z)
      ↖  ↙
       (c)
(¬a) ← (¬b)   (y)   (¬z)
  ↘   ↗
   (¬c)

14. If a strongly connected component contains a literal and its opposite, the problem is unsolveable and we halt.

15. Otherwise, replace the nodes by their strongly connected components:

    (a, b, c) ← (¬y) ← (z)
(¬z) → (y) → (¬a, ¬b, ¬c)

16. Use the topological order defined by the Kosaraju's algorithm:

    (4:a, b, c) ← (6:¬y) ← (5:z)
(2:¬z) → (3:y) → (1:¬a, ¬b, ¬c)

17. Reverse the order:

    (3:a, b, c) ← (1:¬y) ← (2:z)
(5:¬z) → (4:y) → (6:¬a, ¬b, ¬c)

18. Starting from the order, if the variables are not assigned, set the variables of the component to their literal value:

    (3:a=false, b=false, c=false) ← (1:¬y) ← (2:z)
(5:¬z) → (4:y) → (6:a=false, b=false, c=false)

19. Repeat until all the variables are assigned:

    (3:a=true, b=true, c=true) ← (1:y=false) ← (2:z=true)
(5:z=true) → (4:y=false) → (6:a=true, b=true, c=true)

One solution is a=true, b=true, c=true, y=true, z=false