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We know that if there are $m$ edges, then the graph has at most $O(m^{3/2})$ triangles (proof). There is an easy construction that matches this upper bound asymptotically: create a clique of size $\sqrt{m}$.

Suppose the graph has arboricity $\alpha$ (for $\alpha \ll n$). Then my question is: what is an asymptotically tight upper bound on the number of triangles? Obviously, an upper bound would be $O((n\alpha)^{3/2})$ since there are at most $\alpha n$ edges. However, I am not sure how to match this upper bound. For example, if we create a clique of size $\sqrt{\alpha n}$, then the arboricity is at least $\alpha n/ \sqrt{\alpha n} = \sqrt{\alpha n} > \alpha$.

polar_bear_cheese
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1 Answers1

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Lemma. If $G$ has $n$ vertices, $m$ edges, and $t$ triangles, and $F$ is a forest on the same vertex set (edge-disjoint from $G$, without loss of generality), then $F \cup G$ has at most $t + 3m$ triangles.

Proof. There are, of course, $t$ triangles with all three edges in $G$.

There are at most $m$ triangles with one edge from $G$ and two edges from $F$. That's because for every edge $uv \in E(G)$, $F$ has at most one $u,v$-path of length $2$: it has at most one $u,v$-path of any length!

Finally, there are fewer than $2m$ triangles with two edges from $G$ and one edge from $F$. To see this, let $v$ be any vertex; there are fewer than $\deg_G(v)$ edges of $F$ in $N_G(v)$, so there are fewer than $\deg_G(v)$ triangles with two edges from $G$ incident to $v$, and one edge of $F$. Sum over all vertices $v$, and we get less than $2m$. $\square$

By induction with this lemma, we see that the union of $\alpha$ forests has at most $3 \binom{\alpha}{2}(n-1) = O(\alpha^2 n)$ triangles.

Asymptotically, this is matched by the union of $\alpha$ stars centered at $\alpha$ different vertices, which has $\binom \alpha 3 + \binom \alpha 2 (n-\alpha) = \Omega(\alpha^2 n)$ triangles.

Misha Lavrov
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