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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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This feels like the sort of thing that would be in Harary and Palmer's Graphical Enumeration. –  Michael Lugo Jan 27 '10 at 3:02
    
Also, I changed the title; I almost didn't read this question because the original title was too vague. –  Michael Lugo Jan 27 '10 at 3:23
    
It looks like the vast majority of them have one disconnected vertex. –  Douglas Zare Jan 27 '10 at 3:58
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1 Answer 1

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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Thanks, this is helpful. It is related to the topological properties of random simplicial complexes. –  Richard Stanley Jan 28 '10 at 16:32
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