2
$\begingroup$

Consider a $k$ regular graph of $n$ vertices, where $3 \leq k \leq (n-1)$. Is there any upper or lower bound, in the worst case, known for either the tree-width or the clique width of each $k$ regular family?

$\endgroup$

1 Answer 1

5
$\begingroup$

For each $k$-regular family, the treewidth and cliquewidth can be both $\Theta(n)$, due to the existence of expanders.

By On Balanced Separators, Treewidth, and Cycle Rank Thm. 2.1, $tw(G) \geq \tilde s(G) -1$, and by the definition of the strict balanced separator number, $\tilde s(G)$ is at least the size of a balanced separator of a graph. A random regular graph is an expander, which in turn makes the size of balanced separators $\Theta(n)$. Thus $tw(G)=\Theta(n)$.

By The Tree-Width of Clique-Width Bounded Graphs without $K_{n,n}$, every graph of clique-width $k$ which does not contain the complete bipartite graph $K_{n,n}$ for some $n > 1$ as a subgraph has tree-width at most $3k(n − 1) − 1$. It's possible to take $n=3$ for random regular graphs, thus the clique-width is also $\Theta(n)$.

$\endgroup$
3
  • $\begingroup$ Is there a reference to the fact that no random regular graph has $K_{3 \times 3}$ as a subgraph? Also, for what regularity does this fact break down? It is obviously false for a complete ($n-1$ regular graph), because the tree width of a complete graph is linear, but the clique width is bounded. It is also false for $n-2$ and $n-3$ regular graphs for a similar reason. Is it true from $n-4$ regular graphs? $\endgroup$ Commented Sep 15, 2022 at 5:16
  • 1
    $\begingroup$ @RandomMatrices It's true for n-4 regular graphs. As the complement of an (n-4)-regular graph is a cubic graph, and the complement graph of a graph of clique-width k has clique-width at most 2k, the (n-4)-regular graph has clique-width Θ(n) if the cubic graph is taken to be a random cubic graph. $\endgroup$ Commented Sep 15, 2022 at 5:24
  • 1
    $\begingroup$ The reference for $K_{3,3}$ is N. C. Wormald's survey, Models of random regular graphs, Lemma 2.7. The lemma states that for fixed d and fixed graph F with more edges than vertices, almost all random d-regular graphs do not contain F as a subgraph. $\endgroup$ Commented Sep 15, 2022 at 5:27

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.