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Let $\mathcal{G}$ be the family of simple connected graphs on $n$ vertices, where each graph has more than $m$ edges, and the maximal degree is less than $\frac{n-1}{2}$. We can assume that m is large, i.e., $m \geq c\frac{n(n-1)}{2}$ for some $0 < c <0.5$.

I don't need a specific formula, but rather a bound on the number of graphs. Can it be that $|\mathcal{G}| \in O(2^{C\binom{n}{2}})$ for some $0 < C < 1$? Or is this claim false?

I know that almost all of the possible $2^{\binom{n}{2}}$ are connected (see the answer here). And that for $c = 0.5$ we have a bound a bit less than $\frac{1}{2}2^{\binom{n}{2}}$. So there is not much hope in the multiplicative edge constant. But, maybe something can be done with the fact that for each $i \in [n]:=\{1,\ldots,n\}$ we have $\deg(i)\leq \frac{n-1}{2}$.

Maybe there is some literature you can refer me to.

RobPratt
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Kuzja
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1 Answers1

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I posted this question on mathoverflow, and there I got a comment by Fedor Petrov that there is no way to find the needed constant $0 < C < 1$ even if all degrees are equal to $(n-1)/2$.

Here is a detailed explanation on why is that.

If a graph has $2k-1$ vertices, and is $k$-regular, it has to be connected. In our case the degree $(n-1)/2$ is a bit smalle than half of the vertice, relative part of diconnected graphs on $n$ vertices with degree $(n-1)/2$ goes to zero as $n$ goes to infinity, as each vertex gets closer to be connected to half of the vertices of the graph.

Now, let's compute the bound on the number of $(n−1)/2$-regular graphs. In this paper the number of $d$-regular graphs on $n$ vertices is asymptotically equal to $$\frac{\binom{n-1}{d}^n \binom{\binom{n}{2}}{m}}{\binom{n(n-1)}{2m}}\cdot e^{\frac{1}{4}},$$ where $m = dn/2$.

Applying $d = (n-1)/2$ we get $$\frac{\binom{n-1}{(n-1)/2}^n \binom{\binom{n}{2}}{\frac{1}{2}\binom{n}{2}}}{\binom{2\binom{n}{2}}{\binom{n}{2}}}e^{\frac{1}{4}}.$$

Applying the Stirling's approximation to central binomial coefficient we get $$\binom{2N}{N} \approx \frac{2^{2N}}{\sqrt{\pi N}}.$$

Thus, $$\frac{\binom{n-1}{(n-1)/2}^n \binom{\binom{n}{2}}{\frac{1}{2}\binom{n}{2}}}{\binom{2\binom{n}{2}}{\binom{n}{2}}} = \frac{ \frac{2^{n(n-1)}}{\pi^{n/2} (n-1)^{n/2}} \frac{2^{\frac{n(n-1)}{2}}}{\sqrt{\pi \frac{n(n-1)}{4}}}}{ \frac{2^{2\frac{n(n-1)}{2}}}{\sqrt{\pi \frac{n(n-1)}{2}}}}e^{\frac{1}{4}} = e^{\frac{1}{4}} \cdot 2^{\frac{1}{2}} 2^{\frac{n(n-1)}{2} - \frac{n\log(n-1)}{2} - \frac{n\log(\pi)}{2}},$$ and it is eacy to check that for any $C < 1$ for any sufficiently large $n$ it holds that $$ \frac{n(n-1)}{2} - \frac{n\log(n-1)}{2} - \frac{n\log(\pi)}{2} = \frac{n}{2}(n - \log\pi(n-1))> C \frac{n(n-1)}{2}.$$

Kuzja
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