2-Satisfiability (2-SAT) Problem in C/C++

Let f = (x1 ∨ y1) ∧ (x2 ∨ y2) ∧ ... ∧ (xn ∨ yn).

Problem: Is f satisfiable?

xi ∨ yi and and 

  are all equivalent. So we are converting each of (xi ∨ yi) s into those two statements.

Now assume a graph with 2n vertices. In case of each of (xi∨yi) s two directed edges are added

• From ¬xi to yi

• From ¬yi to xi

f is not treated as satisfiable if both ¬xi and xi are in the same SCC (Strongly Connected Component)

Assume that f is treated as satisfiable. Now we want to provide values to each variable in order to satisfy f. It can be performed with a topological sort of vertices of the graph we construct. If ¬xi is after xi in topological sort, xi should be treated as FALSE. It should be treated as TRUE otherwise.

Pseudo Code

func dfsFirst1(vertex v1):
   marked1[v1] = true
   for each vertex u1 adjacent to v1 do:
      if not marked1[u1]:
            dfsFirst1(u1)
      stack.push(v1)
   func dfsSecond1(vertex v1):
      marked1[v1] = true
      for each vertex u1 adjacent to v1 do:
         if not marked1[u1]:
            dfsSecond1(u1)
   component1[v1] = counter
for i = 1 to n1 do:
      addEdge1(not x[i], y[i])
      addEdge1(not y[i], x[i])
for i = 1 to n1 do:
   if not marked1[x[i]]:
      dfsFirst1(x[i])
   if not marked1[y[i]]:
      dfsFirst1(y[i])
   if not marked1[not x[i]]:
      dfsFirst1(not x[i])
   if not marked1[not y[i]]:
      dfsFirst1(not y[i])
set all marked values false
counter = 0
flip directions of edges // change v1 -> u1 to u1 -> v1
while stack is not empty do:
   v1 = stack.pop
   if not marked1[v1]
      counter = counter + 1
      dfsSecond1(v1)
for i = 1 to n1 do:
   if component1[x[i]] == component1[not x[i]]:
      it is unsatisfiable
      exit
   if component1[y[i]] == component1[not y[i]]:
      it is unsatisfiable
      exit
it is satisfiable
exit