## Monday, July 13, 2009

### Lesson 4: Stuck In the Middle

In lesson 2, I pointed out that, while two of Lekkerkerker and Boland's minimal asteroidal graphs also contain sATs, the two infinite families (except for the 3-sun) do not. This is because the lower path contains consecutive vertices adjacent to the neighbor(s) of the "top" vertex (and this is the only path between the "left" and "right" vertices not passing through these neighbors).

In their original paper, Monma, Reed, and Trotter showed that a graph is a co-TT graph if and only if each vertex x can be assigned two real numbers, ax and bx, such that vertices x and y are adjacent if and only if axby and aybx. If, for vertex x, axbx, then x is known as a bounded vertex; if axbx, then x is unbounded. A graph which contains only bounded vertices is equivalent to an interval graph (Jamison, "Cross Comparison Graphs," no citation currently exists), so the unbounded vertices are the key to the difference between interval graphs and co-TT graphs.

At left is a graph with the a and b values assigned to each vertex. The "top" vertex is the only unbounded vertex in this assignment, though any or all of the pendant vertices could be made unbounded.

If a graph does not contain an AT, it is an interval graph, so we are most interested in what happens with asteroidal triples. For a vertex v in an AT to be given an unbounded assignment, it must satisfy the requirement that every path between the other two vertices in the AT that does not contain a neighbor of v must be v-heavy. A vertex is v-heavy if it is adjacent to every neighbor of v; a path is v-heavy if it contains two consecutive heavy vertices. A vertex or path which is not v-heavy is v-light.

A vertex in an AT which satisfies these two properties is a middle vertex. In families III and IV, the "top" vertex is a middle vertex; to obtain sATs, we must add vertices to these graphs to create a light path between the other two asteroidal vertices. How do we do that? Stay tuned.