Question

I’m having trouble solving this problem. I have to find all simple paths starting from a source vertex s containing a simple cycle in a directed graph. i.e. No repeats allowed, except of course for the single repeated vertex where the cycle joins back on the path.

I know how to use a DFS visit to find if the graph has cycles, but I can’t find a way to use it to find all such paths starting from s.

For example, in this graph

        +->B-+
        |    v
s-->T-->A<---C
        |    ^
        +->D-+

Starting from s, the path S-T-A-B-C-A will correctly be found. But the path S-T-A-D-C-A will not be found, because the vertex C is marked as Visited by DFS.

Can someone hint me how to solve this problem?
Thanks

Answer 1

This is actually quite an easy algorithm, simpler than DFS. You simply enumerate all paths in a naive recursive search, remembering not to recurse any further any time the path loops back on itself:

(This is just a Python-inspired pseudocode. I hope it’s clear enough.)

 def find_paths_with_cycles(path_so_far):
       node_just_added = path_so_far.back()
       for neigh in out_neighbours(node_just_added):
           if neigh in path_so_far:
               # this is a cycle, just print it
               print path_so_far + [neigh]
           else:
               find_paths_with_cycles(path_so_far + [neigh])


 initial_path = list()
 initial_path.append(s)
 find_paths_with_cycles(initial_path)