Equivalence between two definitions of Tree width

Question

Treewidth :

1) By chordal graphs : size of the largest clique $(\omega (G))$ - 1 in a chordal completion of the graph $G$.

2) By tree decomposition :

A tree decomposition of $G = (V , E)$ consists of a tree $T$ (on a different node set from $G$), and a subset $V_t ⊆ V$ associated with each node $t$ of $T$. (We will call these subsets $V_t$ the “pieces” of the tree decomposition.) We will sometimes write this as the ordered pair $(T , {V_t : t ∈ T })$. The tree T and the collection of pieces $\{V_t : t ∈ T \}$ must satisfy the following three properties

(Node Coverage) Every node of $G$ belongs to at least one piece $V_t$.

(Edge Coverage) For every edge $e$ of $G$, there is some piece $V_t$ containing both ends of $e$.

(Coherence) Let $t_1, t_2,$ and $t_3$ be three nodes of $T$ such that $t_2$ lies on the path from $t_1$ to $t_3$. Then, if a node $v$ of $G$ belongs to both $V_{t_{1}}$ and $V_{t_{3}}$, it also belongs to $V_{t_2}$

So we define the width of a tree decomposition $(T , {V_t })$ to be one less than the maximum size of any piece $V_t$ (over all $t$): $$width (T , {V_t}) = max |V_t| − 1$$

Claim 1: If size of largest clique in a chordal decomposition of graph is say $k$ then by tree decomposition we will get tree width $k$

Proof : let us assume that largest clique size in chordal completion graph is $k$, so there exist a bag (subset of vertices of graph) that contain the clique ( due to edge coverage ). so we are done.

claim 2 : If tree width is $k$ by tree decomposition method then by chordal completion method also is $k$

Question : How to prove the claim 2 ?High level proof will be welcomed.

Answer 1

First, let me note that the claim, as stated in the question, is false: Consider the following graph:

$\begin{matrix} & & 2& - & 3\\ & \diagup\\ 1 & & | & & |\\ & \diagdown \\ & &4 &-& 5 \end{matrix} $

The complete graph on $5$ vertices is a chordal completion of this graph. However, this graph has treewidth $2$. (A decomposition is $\{1,2,4\}$, $\{2,3,4\}$, $\{3,4,5\}$ )

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The correct claim is

The treewidth of $G$ of is the size of the largest clique $\omega(G) - 1$ in a chordal completion with the minimum largest clique of the graph $G$.

Now, a proof is as follows: if the treewidth of $G$ by tree decomposition is $k$, then every tree decomposition has a bag of size at least $k$. We use the concept of a separator graph from a paper by Parra and Scheffler, in particular the fact that every maximal clique in the separator graph of $G$ is a tree decomposition of $G$ (This can be seen by comparing the definition of tree decompositions with those in the paper)

Then, by Theorem 4.7$^*$ from the same paper, every minimal chordal completion of $G$ has all the bags of some tree decomposition as a clique. This means every minimal chordal completion of $G$ has a clique of size $k$, so the chordal completion method also gives a tree-width of $k$. $\square$

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*: Paraphrased in our notation, Theorem 4.7 states:

A graph $H$ is a minimal chordal completion of $G$ if and only if $H$ is the graph $G$ with exactly the edges added such that all vertex sets in $\mathcal{S}$ are cliques. Here $\mathcal{S}$ is a maximal clique in the separator graph, which consists of minimal separators of $G$.

I attempted to find a proof using only elementary techniques, but I don't think an easy proof without any deeper theory would be easy to find.