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.) $\endgroup$ Oct 8, 2015 at 4:11
  • $\begingroup$ @JoshuaGrochow, interesting. Could you please point me to a paper/resource on it? I would like to know more. $\endgroup$ Oct 8, 2015 at 4:45
  • $\begingroup$ I have asked the question here. math.stackexchange.com/questions/1470764/… $\endgroup$ Oct 8, 2015 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. $\endgroup$ Oct 8, 2015 at 20:50

1 Answer 1


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? $\endgroup$ Oct 7, 2015 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? $\endgroup$ Oct 7, 2015 at 21:11
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    $\begingroup$ It's clearly a collection of disjoint cycles, being a 2-regular graph. And Igor's algorithm obviously works. $\endgroup$ Oct 8, 2015 at 2:35

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