Hi Hai,

Using the classic graph connectivity DFS algorithm:

int FindConnectedComponents()
{
	Dictionary<DiagramNode, int> nodeConnMap = new Dictionary<DiagramNode, int>();
	int component = 0;

	foreach (DiagramNode node in diagram.Nodes)
	{
		if (!nodeConnMap.ContainsKey(node))
		{
			TraverseComponent(nodeConnMap, node, component);
			component++;
		}
	}

	return component;
}

void TraverseComponent(Dictionary<DiagramNode, int> nodeConnMap, DiagramNode node, int component)
{
	nodeConnMap[node] = component;

	foreach (DiagramLink l in node.OutgoingLinks)
	{
		DiagramNode child = l.Destination;
		if (!nodeConnMap.ContainsKey(child))
			TraverseComponent(nodeConnMap, child, component);
	}

	foreach (DiagramLink l in node.IncomingLinks)
	{
		DiagramNode child = l.Origin;
		if (!nodeConnMap.ContainsKey(child))
			TraverseComponent(nodeConnMap, child, component);
	}
}
 

If FindComponents returns a value greater than 1, then the workflow graph is not entirely connected. You might make nodeConnMap an out argument to get more information on connectivity from the method. You will know that there is a path between a pair of nodes if they have the same value in the dictionary. You could remove the loop over IncomingLinks if you want to treat the graph as directed.

I hope that helps,
Stoyan