1
$\begingroup$

Let $G=(V,A)$ be a directed acyclic graph, for which the underlying undirected graph is chordal (so that every induced cycle in the underlying undirected graph is a triangle).

It is known that in a chordal graph the number of maximum cliques is linearly bounded in the number $|V|$ of vertices.

Let $G'=(V,A')$ be the transitive closure of $G$, so that for every directed path $p= (v_i,\ldots,v_j)$ from $v_i$ to $v_j$ in $G$, there exists a directed edge $(v_i,v_j)$ in $A'$.

Question: Is there any polynomial/linear bound on the number of maximum cliques in the undirected graph that underlies $G'$?

$\endgroup$
1
  • $\begingroup$ You are right, but that is a description of the structure of $G$, independent of the edge orientation. $\endgroup$
    – LukasB97
    Mar 22, 2021 at 16:18

2 Answers 2

4
$\begingroup$

I think you can extend Vinicius dos Santos' idea to show that no polynomial bound is possible.

Consider a graph on $n$ vertices divided into $d\geq 1$ groups of size about $n/d$ as follows:

enter image description here

Its transitive closure has about $(\frac{n}{d})^d$ maximal (undirected) cliques.

$\endgroup$
3
  • 1
    $\begingroup$ @Laakeri Then I must have my definitions mixed up somehow, it seems chordal to me :/ $\endgroup$
    – Tassle
    May 20, 2021 at 8:58
  • 1
    $\begingroup$ I'm sorry, of course it's a chordal graph. Somehow I didn't see the edge in the middle. $\endgroup$
    – Laakeri
    May 20, 2021 at 9:15
  • $\begingroup$ @Laakeri All good! $\endgroup$
    – Tassle
    May 20, 2021 at 18:08
1
$\begingroup$

A linear bound is impossible. Suppose your graph is a star, with half of the edges oriented towards the universal vertex $u$ and the rest oriented outward. In the transitive closure, the maximal cliques will be triangles, each of them containing $u$ and exactly one of its inneighbors and one of its outneighbors. This gives a quadratic number of maximal (and maximum) cliques.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.