# Isomorphic regular graphs

How many non-isomorphic classes of regular graphs on $(2n+1)^{m}$ vertices with $m(2n+1)^{m}$ edges with vertex degree $2m$, where $n,m \in \mathbb{N}$ are there? Is there a classification known? Can there can be more than one such class (that is are they all isomorphic)?

Is there an example of such non-isomorphic graphs if there are any?

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Since for regular graphs, number of vertices times degree is twice the number of edges, your condition implies $m=1$? Either there is a typo or this looks like homework. –  Chris Godsil Nov 29 '11 at 17:44
@Chris this is not a hw. Corrected! –  user16007 Nov 29 '11 at 17:52
If you don't assume connected, then there are many non-isomorphic examples. For example, for $m=1$ and $n=3$ the 7-cycle and the disjoint union of a 4-cycle and a 3-cycle are not isomorphic. –  Tony Huynh Nov 29 '11 at 18:41

The asymptotic number of $m$-regular graphs on $N$ vertices is well understood and can be found, for example, in Bollobas' Random Graphs (the argument uses Bollobas' "configuration model"). With probability $1$ a graph has no automorphisms, so this is also the number of isomorphism classes as long as $N$ is large. In your case $N=(2n+1)^m.$ So, for a reasonably sized $n$ (since yours is a natural number, $n>0$ should be fine), if you pick two random graphs, they will be non-isomorphic.

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Degree at least 3 is needed. Also, you might like to divide by $N!$ to convert the asymptotic number of labelled graphs into the asymptotic number of isomorphism classes. This is ok since the number of graphs grows faster than $N!$. In my answer at mathoverflow.net/questions/77730 I surveyed what is known about this enumeration problem. –  Brendan McKay Nov 29 '11 at 23:53
$5=6{}{}{}{}{}$? –  Gerry Myerson Nov 29 '11 at 22:07
It is the limit as $5\rightarrow 6.$ –  Igor Rivin Nov 30 '11 at 4:16