Does a path exist through all vertices of a directed graph?

Question

Given G, a directed graph, is there a path (not necessarily a simple path) that goes through all the vertices in G?

I first need to examine what happens in an acyclic graph and in a strongly-connected graph, and then to find a solution for a general graph, using the graph of the strongly connected components.

So far I have figured out that for a strongly connected graph there is always a path. For an acyclic graph, if there is more than one source a path will never exist. Also, if there is a vertex for which D out is greater than 1, a path will never exist.

The problem is, I'm not sure that last one is right, and if it's wrong, my algorithm is wrong.

Answer 1

The last assumption is not true, for example have a look at the graph G = (V,E), where E = {(v_i,v_j) | i < j } The graph is obviously a DAG. so finding the maximal strongly connected component won't change the graph. Also - the graph has a hamiltonian path, however d_out(v_1) > 1 [assuming |V| > 3].

However - you are on the right track.

Algorithm in high level pseudo code:

1. find Maximal Strongly Connected Components in the graph - the resulting graph is a DAG.
2. Use topological sort on the resulting graph¹.
3. Check if the ordered sorting creates a hamiltonian path
4. if it is - return true, else return false

Claim:

A path with all vertices exists if and only if the DAG representing the MSCC of the graph has hamiltonian path

Proof of claim:
<-
if there is a hamiltonian path - then such a path is trivial, for each MSCC - the path goes through all vertices, and then to the out edge that is representing the our edge that was chosen in the hamiltonian path in the MSCC graph.
->
If such a path exists, let it be v0->v1->...vm.
Let's denote V_i the maximal strongly connected component in which v_i lays.
Now, for the path in the original graph v0->v1->...->vm, there is also a path in the MSCC graph²: V_0->V_1->...->V_m.
Note that if V_i appears twice [or more] in the above path - the two occurances are adjacent to each other, otherwise the MSCC found is not maximal because if V_i->V_k->...->V_i is feasible path - then V_i and V_k are not maximal, since you can unite them into one bigger SCC.
Now, for each V_i collapse all occurances of it into one, and you get yourself a path - where each V_i appears at most once [we just showed why], and exactly one [since every v_i was on the original path and we did not remove MSCC's completely - just collapsed them].
Thus - the generated path is a hamiltonian path in the MSCC graph.

Proof of correctness of the suggested algrotithm:
Since we showed hamiltonian path in MSCC graph exists if and only if there exists a requested path in the original graph - then:
->
The algorithm returned true -> The algorithm found a Hamiltonian path in the DAG -> There is a Hamiltonian Path in the Dag [foot-note 1] -> There is a path as requested in the original graph.
<-
The algorithm returned false -> The algorithm did not find a Hamiltonian pat in the DAG -> There is no hamiltonian path in the DAG [foot-note 1] -> There is no path as requested in the original path.

Q.E.D.

---

1: in a DAG, there is a unique topological sort if there exists a hamiltonian path, and if there is a hamiltonian path - it is the order of the vertices in the topological sort. Thus, in a DAG - finding a hamiltonian path is easy.
2: Actually, it is a bit modification, V_i->V_i is not really an edge on the MSCC, graph but for now let's denote it as one.