This question is supplementary to the question asked here.

One of the answers give a class of graphs for which the adjacency matrices are invertible which is defined as follows.

Given a permutation $\pi$ of a finite set $V$, form its cycle graph $G$ as follows: the vertex set is $V$ and the edges are pairs $(v,w)$ for which $\pi(v)=w$. (This is a simple directed graph.) The adjacency matrix will in fact be the permutation matrix corresponding to $\pi$, which is invertible.

Let's consider a class of graph isomorphism problems where it is promised that both of the input graphs will be from the class defined above.

Can the graph isomorphism problems from this class be solved efficiently?

  • 2
    $\begingroup$ Note that if you work with the larger class of all graphs whose adjacency matrices are invertible, I believe the problem becomes GI-complete. (Namely, given any graph, add a new neighbor of degree 1 to each vertex. The resulting graph has invertible adjacency matrix.) $\endgroup$ – Joshua Grochow Oct 8 '15 at 4:11
  • $\begingroup$ @JoshuaGrochow, interesting. Could you please point me to a paper/resource on it? I would like to know more. $\endgroup$ – Omar Shehab Oct 8 '15 at 4:45
  • $\begingroup$ I have asked the question here. math.stackexchange.com/questions/1470764/… $\endgroup$ – Omar Shehab Oct 8 '15 at 18:17
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    $\begingroup$ I do not know a paper on this particular question. Is there something basic you didn't understand about my comment? If you understand the basics (e.g. what GI-complete means), then I'd treat the rest of my comment as a good exercise. $\endgroup$ – Joshua Grochow Oct 8 '15 at 20:50

Isn't this graph just a collection of cycles? So we only need to compare that all the lengths in the two graphs match (which can be done by sorting the lengths).

  • $\begingroup$ You mean lengths of all cycles? $\endgroup$ – Omar Shehab Oct 7 '15 at 20:28
  • $\begingroup$ I am not sure what definition of 'cycle graph' he has used here. According the the definition in Wikipedia (en.wikipedia.org/wiki/Cycle_graph), a cycle graph can have only one cycle. In that case, there cannot be any cycle graph for the permutation $(1 2)(3 4)$, right? $\endgroup$ – Omar Shehab Oct 7 '15 at 21:11
  • 4
    $\begingroup$ It's clearly a collection of disjoint cycles, being a 2-regular graph. And Igor's algorithm obviously works. $\endgroup$ – Sasho Nikolov Oct 8 '15 at 2:35

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