DFS on directed graph & Kosaraju's algorithm

Question

I'm having trouble to understand Kosaraju's algorithm for finding the strongly connected components of a directed graph. Here's what I have on my notebook (I'm a student :D):

1. Start from an arbitrary vertex (label it with #1) and perform a DFS. When you can't go any further, label the last visited vertex with #2, and start another DFS (skipping vertices already labeled), and so on.
2. Transpose the graph.
3. Do DFS starting from each vertex in reverse order, those vertices which end visited after each DFS belong to the same SCC.

I have this example:

And after the first step starting from E, the labels are:

1. E
2. G
3. K
4. J
5. I
6. H
7. F
8. C
9. D
10. B
11. A

So here comes the thing: Is there a difference for DFS in directed/undirected graphs? I did a mental test of the first step on my mind ignoring the arrows (just like it was undirected) and only got correct #1 for E (of course) and #2 for G, but #3 fell onto J, not K. So I thought maybe I should respect the arrows, and did a DFS considering that, but after the first pass starting from E, I can't go anywhere from G (which is #2), so I'm stuck there.

Is there anything about DFS on directed graphs that I'm not aware of? I've been taught DFS only on undirected graphs!

Answer 1

Your second step is incomplete. See Wikipedia:

Kosaraju's algorithm works as follows:

• Let G be a directed graph and S be an empty stack.
• While S does not contain all vertices:
  • Choose an arbitrary vertex v not in S. Perform a depth-first search starting at v. Each time that depth-first search finishes expanding a vertex u, push u onto S.
• Reverse the directions of all arcs to obtain the transpose graph.
• While S is nonempty:
  • Pop the top vertex v from S. Perform a depth-first search starting at v in the transpose graph. The set of visited vertices will give the strongly connected component containing v; record this and remove all these vertices from the graph G and the stack S. Equivalently, breadth-first search (BFS) can be used instead of depth-first search.

So you shouldn't only do something with the last vertex and first vertices, but with each vertex in the DFS.

Also note that you should be backtracking - when you can't go further, you go to the previous vertex and continue from there.

And no, you can't treat it as an undirected graph - the direction of the edges matter significantly.

So, starting from E, you'd, for example, go F, then G, then back to F, then H, then K, then I, then J, then back to I, K, H, F, and finally E, having pushed all visited vertices onto the stack.