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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
Updated on: 29-Jan-2020
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