3
$\begingroup$

I am very interested in the maximum number of triangles could a connected graph with $n$ vertices and $m$ edges have. For example, if $m\leq n−1$, this number is $0$, if $m=n$, this number is $1$, if $m=n+1$, this number is $2$, and if $m=n+2$, this number is $4$.

$\endgroup$
2
  • 1
    $\begingroup$ If we omit the restriction of connectedness, see this question: math.stackexchange.com/questions/823481/… Also note that we can achieve connectedness with a few number of edges, so the order of magnitude of the number of triangles is the same without your restriction. $\endgroup$ Commented Jun 22, 2015 at 10:55
  • $\begingroup$ @Soltész I want to get a good bound in terms of $m-n+1$. I believe this number could be bounded by $(m-n+1)^2$ roughly. $\endgroup$ Commented Jun 22, 2015 at 11:10

2 Answers 2

7
$\begingroup$

It is a bound and since it is very long, I wrote it an answer, may be it can be helpful.

Let $G$ be a connected graph with $n$ vertices and $m$ edges. Suppose the eigenvalues of this graph are $\lambda_1\geq \lambda_2\geq\ldots\geq\lambda_n$. We know that $\sum{\lambda_i^3}=6\Delta_G$, where $\Delta_G$ counts the total number of triangles of the graph $G$.

Also,we have:

$$\lambda_1\leq\sqrt{2m-\delta(n-1)+\Delta(\delta-1)}.$$

Since your graph is connected, we can set $\delta=1$ and obtain: $$\lambda_1\leq\sqrt{2m-n+1}.$$

So we have:

$$\Delta_G\leq\frac{n}{6}(2m-n+1)^{\frac{3}{2}},$$

as you wanted in your comments.

Actually, you can get more information from this method since we exactly know when the upper inequalities which I used are equality for which graphs. You can search for "SHARP UPPER BOUNDS OF SPECTRAL RADIUS OF GRAPHS" or similar keywords.

$\endgroup$
2
  • $\begingroup$ Thanks, this method is very helpful. This bound can be easily improved as $\Delta_G\leq\frac{n-1}{6}(2m-n+1)^{\frac{3}{2}}$. $\endgroup$ Commented Jun 27, 2015 at 8:07
  • $\begingroup$ This can be further improved with $\Delta_G \leq m^{3/2}$. $\endgroup$
    – orezvani
    Commented Feb 16, 2017 at 2:58
3
$\begingroup$

This question (together with massive generalizations) is answered in I. Rivin's 2001 paper.

$\endgroup$
1
  • $\begingroup$ The paper seems interesting. It has some trivial results, but one up vote for the very nice style of writing. $\endgroup$
    – Shahrooz
    Commented Apr 16, 2016 at 9:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .