// Counts the number of independent sets in a graph, such that:
// - All vertices in R are included in the independent set, initially empty
// - Some of the vertices in P are included in the independent set, initially all vertices of the graph
// - None of the vertices in X are included in the independent set, initially empty
private static int Compute(Graph graph, BitSet R, BitSet P, BitSet X)
{
// Base case, when P and X are both empty we cannot expand the IS
if (P.IsEmpty && X.IsEmpty)
return 1;
int count = 0;
BitSet copy = P.Copy();
// Foreach vertex v in P we include it in the IS and compute how many maximal IS will include v by going into recursion.
foreach (int v in copy)
{
count += Compute(graph, R + v, P - graph.ClosedNeighborhood(v), X - graph.OpenNeighborhood(v));
P.Remove(v);
X.Add(v);
}
return count;
}
// Counts the number of independent sets in a graph, such that:
// - All vertices in R are included in the independent set, initially empty
// - Some of the vertices in P are included in the independent set, initially all vertices of the graph
// - None of the vertices in X are included in the independent set, initially empty
private static long Compute(Graph graph, BitSet R, BitSet P, BitSet X)
{
// Base case, when P and X are both empty we cannot expand the IS
if (P.IsEmpty && X.IsEmpty)
return 1;
long count = 0;
int pivot = -1;
int min = int.MaxValue;
// Procedure to find a pivot
// The idea is that in any maximal IS, either vertex u or a neighbor of u is included (else we could expand by adding u to the IS)
// We find the u with the smallest neighborhood, so that we will keep the number of branches low
foreach (int u in (P + X))
{
int size = (P * graph.OpenNeighborhood(u)).Count;
if (size < min)
{
min = size;
pivot = u;
}
}
// There should always be a pivot after the selection procedure, else P and X should both have been empty
if (pivot == -1)
throw new Exception("Pivot has not been selected");
// Foreach vertex v in the set containing the legal choices of the the closed neighborhood of the pivot,
// we include each choice in the IS and compute how many maximal IS will include v by going into recursion
foreach (int v in (P * graph.ClosedNeighborhood(pivot)))
{
count += Compute(graph, R + v, P - graph.ClosedNeighborhood(v), X - graph.OpenNeighborhood(v));
P.Remove(v);
X.Add(v);
}
return count;
}
