The Induced Separation Dimension of a Graph

Abstract

A linear ordering of the vertices of a graph G separates two edges of G if both the endpoints of one precede both the endpoints of the other in the order. We call two edges $\{a,b\}$ and $\{c,d\}$ of G strongly independent if the set of endpoints $\{a,b,c,d\}$ induces a $2K_2$ in G. The induced separation dimension of a graph G is the smallest cardinality of a family $\mathcal {L}$ of linear orders of V(G) such that every pair of strongly independent edges in G are separated in at least one of the linear orders in $\mathcal {L}$. For each $k \in \mathbb {N}$, the family of graphs with induced separation dimension at most k is denoted by ${\text {ISD}}(k)$. In this article, we initiate a study of this new dimensional parameter. The class ${\text {ISD}}(1)$ or, equivalently, the family of graphs which can be embedded on a line so that every pair of strongly independent edges are disjoint line segments, is already an interesting case. On the positive side, we give characterizations for chordal graphs in ${\text {ISD}}(1)$ which immediately lead to a polynomial time algorithm which determines the induced separation dimension of chordal graphs. On the negative side, we show that the recognition problem for ${\text {ISD}}(1)$ is NP-complete for general graphs. Nevertheless, we show that the maximum induced matching problem can be solved efficiently in ${\text {ISD}}(1)$. We then briefly study ${\text {ISD}}(2)$ and show that it contains many important graph classes like outerplanar graphs, chordal graphs, circular arc graphs and polygon-circle graphs. Finally, we describe two techniques to construct graphs with large induced separation dimension. The first one is used to show that the maximum induced separation dimension of a graph on n vertices is $\Theta (\lg n)$ and the second one is used to construct AT-free graphs with arbitrarily large induced separation dimension. The second construction is also used to show that, for every $k \ge 2$, the recognition problem for ${\text {ISD}}(k)$ is NP-complete even on AT-free graphs.