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See for example here: https://en.wikipedia.org/wiki/Graph_enumeration

I would have thought (naively) that the number of graphs on $n$ vertices would only grow as $\mathscr{O}\left( _nC_2\right)$, but it clearly grows much faster. Even the number of trees blows up faster than the factorial.

https://en.wikipedia.org/wiki/Cayley%27s_formula

Why does the number of graphs blow up so much faster than almost anything else?

I would have thought it would have just been a question of selecting which 2 vertices of $n$ to connect with an edge, which seemingly can be done with $\mathscr{O}\left( _nC_2\right)$ time. What am I missing?

Chill2Macht
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    $\binom{n}{2}$ is just the number of edges in a complete graph on $n$ vertices, but obviously there are way more graphs on $n$ vertices than such number of edges. An upper bound is given by $2^{\binom{n}{2}}$, i.e. the number of subsets in a set with $\binom{n}{2}$ elements. – Jack D'Aurizio May 19 '16 at 13:38
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    Why would you expect ${n\choose2}$? That is the total number of potential edges. Each of those can be present or not. So that gives $2^{n\choose2}$. But the number is reduced because of isomorphisms. – almagest May 19 '16 at 13:39
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    https://en.wikipedia.org/wiki/Combinatorial_explosion – mvw May 19 '16 at 13:39

2 Answers2

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$\binom{n}{2}$ is the number of potential edges in a graph with $n$ vertices. Every subset of the set of edges makes a graph, so there are $$ 2^{\binom{n}{2}} $$ graphs. That grows pretty quickly.

Ethan Bolker
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  • Oh this makes a lot of sense, thank you! I really like how this argument appeals to the intuition, so it's the same thing as why the cardinality of the reals is greater than the naturals (binary expansion), or why there are $2^n$ n-digit binary numbers and not $n$. Thank you so much! – Chill2Macht May 19 '16 at 13:46
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In addition to the first answer, I think it's worth mentioning that the more meaningful quantity here is not the number of all graphs on $n$ labeled vertices, which is $2^{{n\choose 2}}$ as already pointed out, but the number of non-isomorphic graphs on $n$ vertices, since for example there are $n!/2$ ways that the same graph $\cdot\to \cdot \to \ldots\to \cdot$ can be realized if you care about the labels. The number of nonisomorphic graphs on $n$ vertices is trivially lower bounded by $$ \frac{2^{{n\choose 2}}}{n!}\approx 2^{n^2/2-n\log n}$$ (see this question), so it's still a huge number

Community
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amakelov
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