I would like to know the asymptotic number of labelled disconnected (simple) graphs with n vertices and $\lfloor \frac 12{n\choose 2}\rfloor$ edges.
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$\begingroup$ This feels like the sort of thing that would be in Harary and Palmer's Graphical Enumeration. $\endgroup$– Michael LugoJan 27, 2010 at 3:02
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$\begingroup$ Also, I changed the title; I almost didn't read this question because the original title was too vague. $\endgroup$– Michael LugoJan 27, 2010 at 3:23
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$\begingroup$ It looks like the vast majority of them have one disconnected vertex. $\endgroup$– Douglas ZareJan 27, 2010 at 3:58
1 Answer
The vast majority of disconnected graphs have a single isolated vertex.
Let $A$ be a nonempty proper subset of $\{1,...,n\}$ of size $a$. Let $s(a)$ be the number of graphs with $e=\lfloor \frac12 {n \choose 2}\rfloor$ edges which have no edges from $A$ to $A^c$.
We want to count the union of all of these. Inclusion-exclusion works, with the dominant terms coming from when $a=1$.
An upper bound is the sum of $s(a)$ over all $A$ of size at most $n/2$, which is at most $n ~s(1)$ + ${n\choose 2}s(1)$ + $2^ns(3)$.
To get a lower bound, subtract the number of graphs with no edges connecting $A$ to $A^c$ or edges connecting $B$ to $B^c$ for all disjoint $\{A,B\}$. Denote this by $s(\#A,\#B)$. So, subtract
${n\choose2}s(1,1) + 3^ns(1,2)$ from $n~s(1)$.
The rest should be routine estimates on $s(1)$, $s(2)$, $s(3)$, $s(1,1)$, and $s(1,2)$.
$s(a,b) \le s(a+b)$.
$s(a) = ({n\choose 2} -a(n-a))$ choose $e$.
Let the total number of graphs with $e$ edges be $\#G = s(0)$.
$$s(a)/\#G = \prod_{i=0}^{a(n-a)-1} \frac{\lceil{n\choose2}/2\rceil-i}{{n\choose2}-i}.$$
$s(2)/s(1) \le 2^{-n+3}$.
$s(3)/s(1) \le 2^{-2n+8}$.
The dominant term in both the upper bound and the lower bound is $n~s(1)$.
If I calculated correctly, that's asymptotic to $\frac 2 e n 2^{-n} ~\#G$.
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$\begingroup$ Thanks, this is helpful. It is related to the topological properties of random simplicial complexes. $\endgroup$ Jan 28, 2010 at 16:32