/// <summary>
/// Check if it is possible to assign electrons to the subgraph (specified by
/// the set bits in of <paramref name="bs"/>). Each connected subset is counted up and
/// checked for odd cardinality.
/// </summary>
/// <param name="g"> graph</param>
/// <param name="bs">binary set indicated vertices for the subgraph</param>
/// <returns>there is an odd cardinality subgraph</returns>
private static bool ContainsOddCardinalitySubgraph(Graph g, BitArray bs)
{
// mark visited those which are not in any subgraph
bool[] visited = new bool[g.Order];
for (int i = BitArrays.NextClearBit(bs, 0); i < g.Order; i = BitArrays.NextClearBit(bs, i + 1))
{
visited[i] = true;
}
// from each unvisited vertices visit the connected vertices and count
// how many there are in this component. if there is an odd number there
// is no assignment of double bonds possible
for (int i = BitArrays.NextSetBit(bs, 0); i >= 0; i = BitArrays.NextSetBit(bs, i + 1))
{
if (!visited[i] && IsOdd(Visit(g, i, 0, visited)))
{
return(true);
}
}
return(false);
}