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I know a few graph invariants and it seemed that these two graphs do not have the same amount of circuit length $3$.
(our definition of a circuit is a closed path, a path being a walk with no repeated edges)

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In particular, it seems that $G_1$ has $9$ circuits of length $3$ in the anti-clockwise direction (so there are $18$ circuits of length $3$ in total).

$G_2$ has only $6$ circuits of length $3$ in the anti-clockwise direction so it has $12$ circuits in total of length $3$.

So $G_1$ is not isomorphic to $G_2$, however, my solutions say that they are.

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    $\begingroup$ Where do you find $6$ or $9$ circuits of length $3$ in either? I found only two in $G_1$ ($a,c,d$ and $c,d,e$) and two in $G_2$ ($x,y,z$ and $x,y,t$). $\endgroup$ – Misha Lavrov Oct 23 '17 at 1:02
  • $\begingroup$ The one I'm looking at for $G_1$. $adca,dcad,cadc$ (so $3$ for that triangle) $cdec,decd,ecde$ (so $3$ for that triangle) Actually, I jsut found my mistake in the counting of $G_1$ after writing that, thanks! $\endgroup$ – Natash1 Oct 23 '17 at 1:06
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Actually these graphs are isomorphic. If you take f(a)=z f(b)=u f(e)=t f(d)=x f(c)=y f is an isomorphism because if ab is an edge in G1, f(a)f(b) is an edge in G2

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