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Is it true that the number of distinct (i.e., non-isomorphic) $n$-vertex $O(n)$-edge graphs is $2^{O(n)}$? The bound holds for sparse graphs like trees and planar graphs. Does it hold for general sparse graphs?

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  • $\begingroup$ How could we tell whether we were in possession of such a graph? I mean, consider for instance some graph $G=(V,E)$ where $|V| = 6,178,372,847$ and $|E| = 20,724,0161,002,050$. Should it be included in your count or no? Exactly what test should we use to decide that question? $\endgroup$ Commented Feb 1 at 18:25
  • $\begingroup$ @PaulTanenbaum, Nice question! One can rephrase my question as: ‘Is P(c) true for each positive constant c?’ Here, P(c) stands for: ‘There exists a positive constant d (depending on c) such that for each sufficiently large n, the number of distinct n-vertex graphs having at most cn edges is at most 2^{dn}.’ So, whether or not your graph should be included depends on the value of c. $\endgroup$ Commented Feb 2 at 2:39
  • $\begingroup$ My point, @plm, was merely that, as it is expressed, having $O(n)$ edges is an ill-defined property. After all for any graph $G$, both $|V(G)|$ and $|E(G)|$ are constants. @sisylana’s rephrasing (in his comment) resolves his meaning. And if his original abuse of terminology is common practice in exploring the asymptotic behavior of graphs, I certainly wouldn’t presume to dictate or legislate against it. But I am entitled to an opinion about its advisedness. $\endgroup$ Commented Feb 2 at 16:00
  • $\begingroup$ Cross-posted: mathoverflow.net/questions/486992/… $\endgroup$ Commented Feb 2 at 17:09

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Perhaps the second answer here answers your question. If you can verify it then the answer is negative https://mathoverflow.net/questions/298421/an-upper-bound-for-the-number-of-non-isomorphic-graphs-having-exactly-m-edges

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  • $\begingroup$ It does - in fact it shows that there are (for some absolute constant $c>0$ and all large $n$) at least $(n/c)^n$ many distinct graphs with $n$ vertices and at most $6n$ edges. $\endgroup$ Commented Feb 2 at 10:00

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