Suppose that there are $n$ vertices, we want to construct a regular graph with degree $p$, which, of course, is less than $n$. My question is how many possible such graphs can we get?
$\begingroup$
$\endgroup$
3
asked Oct 10, 2011 at 19:30
-
1$\begingroup$ Why the "of course"? $\endgroup$– Autumn KentOct 10, 2011 at 19:36
-
$\begingroup$ Because he defines "graph" as "simple graph", I am guessing. $\endgroup$– Igor RivinOct 10, 2011 at 19:38
-
1$\begingroup$ This question appeared before on MO: mathoverflow.net/questions/22089/enumeration-of-regular-graphs/… $\endgroup$– Andy BOct 11, 2011 at 16:03
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
5
McKay and Wormald conjectured that the number of simple $d$-regular graphs of order $n$ is asymptotically $$\sqrt 2 e^{1/4} (\lambda^\lambda(1-\lambda)^{1-\lambda})^{\binom n2}\binom{n-1}{d}^n,$$ where $\lambda=d/(n-1)$ and $d=d(n)$ is any integer function of $n$ with $1\le d\le n-2$ and $dn$ even.
Bender and Canfield, and independently Wormald, proved this for bounded $d$ in 1978, and Bollobás extended this to $d=O(\sqrt{\log n})$ in 1980.
McKay and Wormald proved the conjecture in 1990-1991 for $\min\{d,n-d\}=o(n^{1/2})$ [1], and $\min\{d,n-d\}>cn/\log n$ for constant $c>2/3$ [2]. These remain the best results.
The gap between these ranges remains unproved, though the computer says the conjecture is surely true there too. The formula apart from the $\sqrt2e^{1/4}$ has a simple combinatorial interpretation, and the universality of the constant $\sqrt2e^{1/4}$ is an enigma crying out for an explanation.
Incidentally this conjecture is for labelled regular graphs. For isomorphism classes, divide by $n!$ for $3\le d\le n-4$, since in that range almost all regular graphs have trivial automorphism groups (references on request). For $d=0,1,2,n-3,n-2,n-1$, this isn't true.
[1] Combinatorica, 11 (1991) 369-382. http://cs.anu.edu.au/~bdm/papers/nickcount.pdf
[2] European J. Combin., 11 (1990) 565-580. http://cs.anu.edu.au/~bdm/papers/highdeg.pdf
ADDED in 2018: The "gap between those ranges" mentioned above was filled by Anita Liebenau and Nick Wormald [3]. Another proof, by Mikhail Isaev and myself, is not ready for distribution yet.
answered Oct 10, 2011 at 23:21
-
$\begingroup$ Is there an asymptotic value for all d-regular graphs on n vertices (not necessarily simple)? $\endgroup$– user24576Aug 23, 2014 at 19:36
-
$\begingroup$ @Amudhan: The sparse case is here: arxiv.org/abs/1303.4218 . $\endgroup$ Aug 24, 2014 at 2:33
-
$\begingroup$ I am sorry, I am a bit confused about, what is the meaning of $d=d(n)$, why is it a function? is $d$ not fixed before hand ? $\endgroup$– SagarMApr 20, 2021 at 9:49
-
1$\begingroup$ @SagarM $d$ could be fixed, but suppose you want the number of regular graphs with $n$ vertices and degree $d=n/2$? Then $d$ is not fixed. The most general case is to take $d$ to be a function of $n$. $\endgroup$ Apr 20, 2021 at 12:20
-
$\begingroup$
$\endgroup$
There is no closed formula (that anyone knows of), but there are asymptotic results, due to Bollobas, see A probabilistic proof of an asymptotic formula for the number of labelled regular graphs (1980) by B Bollobás (European Journal of Combinatorics) or Random Graphs (by the selfsame Bollobas).
answered Oct 10, 2011 at 19:37
$\begingroup$
$\endgroup$
And 'of course', if you really want those graphs you might have a look at genreg by Markus Meringer: http://www.mathe2.uni-bayreuth.de/markus/reggraphs.html
