Show that $pw(T_{d}) \geq d- 1$. Feel free to suggest a completely different approach, if you know of a better one.
A complete ternary tree $T_{d}$ is a rooted tree in which each vertex has a 3 descendants and has depth $d$. Depth $d$ indicates the distance from the root vertex to a vertex at the deepest level of a tree. For example the complete ternary tree $T_{1}$ is a root node with 3 descendants (a 3 claw tree).
A path decomposition of a tree is a tree decomposition, in which the resulting tree is a path. Path width is defined as the bag with the largest amount of vertices minus 1.
My attempt:
I have attempted the problem through induction on $d$. The claim holds trivially for $d=1$ and $d=2$, as any connected graph must have $pw \geq 1$. Any edge in a connected graph must be in some bag, meaning there exists a bag with 2 vertices. Suppose the claim holds for depth $<d$. The tree $T_{d}$ is formed of a root vertex $v_{0}$ and three connected components, all of which are complete ternary trees of depth $d-1$. In other words the tree $T_{d}$ is formed of 3 tress $T_{d-1}$ and a root vertex $v_{0}$. By the induction hypothesis the path width of $T_{d-1}$ must be $\geq d-2$.
How can I calculate the path width of $T_{d}$ given that I know that I have 3 subtrees of path width $d-2$?
Best Answer
Here's a sketch: The key idea is that a path decomposition which covers all three principal subtrees of $T_d$ must pass through the subtrees in some order, and it has to "thicken" the middle one a bit.
Call the three subtrees of $T_d$ adjacent to the root $a$, $b$, and $c$, and let $P=(\mathcal{B},E)$ be a tree decomposition of $T_d$ which is a path, having bags $B$ and edges $E$. The induction step follows from these observations:
We have a dichotomy. Either:
Or: