Using human intuition I think the graphs are not isomorphic because the loops are on opposite sides but I do not have a formal proof for it. I ran the basic checks and I still cant disprove it.

  1. Check if vertex set cardinalities differ - 5 for both
  2. Check if edge set cardinalities differ - 8 for both
  3. Compare degree sequences - (4,3,3,3,2) for both
  4. Compare number of connected components - 1 for both
  5. Compare cycle lengths - 3 and 4 for both

What other conditions can I use to disprove graph isomorphism?

EDIT: Can I use the sequence of degrees of vertices in cycles?

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    $\begingroup$ There is a point that has an edge from itself to itself. For the graph on the right the shortest path connecting it a point that has a cycle of length 2 is two edges, on the left this length is one edge. $\endgroup$ – s.harp Feb 4 '17 at 14:10
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    $\begingroup$ There is no general conclusive list of criteria you can use in every situation, although your list is a great place to start. If the list fails, you have to look at the graphs and see what property make them differ. For instance, in this case, you could use the distance from the vertex connected to itself to the nearest vertex that's part of the $2$-cycle. Or, as you've suggested, in one graph there are two neighbouring vertices of degree 4. $\endgroup$ – Arthur Feb 4 '17 at 14:11

I would say that in both there are exactly two vertices with valency $4$. The two are adjacent in one graph, but not in the other.


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