On tree decompositions whose trees are minors

TL;DR

This paper disproves a 2019 conjecture by showing that not all connected graphs, even with small treewidth, admit a tree decomposition with a tree that is a minor of the graph and has bounded width.

Contribution

The paper provides a counterexample to the conjecture, demonstrating that such tree decompositions do not always exist for graphs of treewidth 2.

Findings

Counterexample for graphs with treewidth 2

Disproof of the conjecture for minor-based tree decompositions

Shows limitations of tree decompositions with minor trees

Abstract

In 2019, Dvo\v{r}\'{a}k asked whether every connected graph $G$ has a tree decomposition $(T, \mathcal{B})$ so that $T$ is a subgraph of $G$ and the width of $(T, \mathcal{B})$ is bounded by a function of the treewidth of $G$. We prove that this is false, even when $G$ has treewidth $2$ and $T$ is allowed to be a minor of $G$.

Equations

$$T_v:=T[\{x:v\in B_x\}].$$

$$w_{j}\coloneqq(k+1)\biggl{(}n+\sum_{i=j+1}^{n}\lvert V(S_{i})\rvert\biggr{)}+1.$$

$$(k+1)\lvert V(T_{v})\rvert\geqslant\deg_{\widetilde{G}}(v)\geqslant\deg_{S_{j}}(v)\geqslant w_{j}>(k+1)\biggl{(}\lvert V(G)\rvert+\sum_{i=j+1}^{n}\lvert V(S_{i})\rvert\biggr{)}.$$

$$|B_{a_1}\cup \cdots \cup B_{a_n}|⩽(k+1)n,$$

$$\bigcap _{e\in M}V\left(C_T^e\right)\ne \varnothing .$$

$$\bigcap _{e\in M}E\left(C_T^e\right)\cap E(P_{uv})\ne \varnothing ,$$

$$\bigcap _{e\in M}E\left(C_T^e\right)\cap E(P_{uv})\ne \varnothing .$$

Full text

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20(-2.3, 6.5)

1 Introduction

Suppose that a graph $G$ has small treewidth, and consider all tree decompositions $(T,\mathcal{B})$ of $G$ whose width is not too much larger than the optimum. To what extent can we choose or manipulate the “shape” of $T$? For graphs with no long path, we can choose $T$ to also have no long path [NM12]111The reference gives us an elimination tree $F$ of $G$ of depth $k$. Then we can obtain a tree decomposition $(F,\mathcal{B})$ of $G$ of width $k$ by letting the bag of each vertex $v\in V(F)$ be the set of all ancestors of $v$ in $F$.; this gives rise to the parameter called treedepth. Similarly, for graphs of bounded degree, we can choose $T$ to also have bounded degree [DO95]; this relates to the parameters of congestion and dilation. Moreover, for graphs excluding any tree as a minor, we can choose $T$ to just be a path [BRST91]; this results in the parameter called pathwidth.

It would be wonderful if we could unify all such results into a single theorem which relates the shape of $T$ to $G$. In 2019, Dvořák suggested one way of accomplishing this goal. In the question below and throughout the paper, we write $\operatorname{tw}(G)$ for the treewidth of $G$.

Question 1** ([Dvo19]).**

Does there exist a polynomial $P$ such that every connected graph $G$ has a tree decomposition $(T,\mathcal{B})$ of width at most $P(\operatorname{tw}(G))$ such that $T$ is a subgraph of $G$?

Unfortunately, we prove that the answer to Question 1 is “no” in the following strong sense.

Theorem 2**.**

For every positive integer $k$, there is a connected graph $G$ of treewidth $2$ such that if $(T,\mathcal{B})$ is a tree decomposition of $G$ and $T$ is a minor of $G$, then $(T,\mathcal{B})$ has width at least $k$.

Intriguingly, in our proof of Theorem 2, it seems crucial that the constructed graphs contain all trees as minors; perhaps Question 1 could be true when $\operatorname{tw}(G)$ is replaced by $\operatorname{pw}(G)$, the pathwidth of $G$. In other words, perhaps there exists a polynomial (or even just some function) $P$ so that every connected graph $G$ has a tree decomposition $(T,\mathcal{B})$ of width at most $P(\operatorname{pw}(G))$ such that $T$ is a subgraph of $G$. We leave this as an open problem.

There has been strong interest in obtaining good bounds for treedepth [CNP21, HJM*+*23, KR22], pathwidth [GJNW21], and treewidth [CC16, CT21] as a function of the natural obstructions (which are paths, trees, and grids, respectively222Formally, a class of graphs has bounded treedepth/pathwidth/treewidth if and only if it does not contain all paths/trees/grids as minors, respectively. See [NM12], [BRST91], and [RS90] for the respective proofs. Note that sometimes the obstructions are considered as subgraphs or subdivisions rather than minors. This occurs when the two definitions are equivalent, for instance when considering paths as minors (or equivalently as subgraphs).). These problems were in large part motivated by the desire to obtain better approximation algorithms and better win-win algorithms based on the obstructions. An affirmative answer to Question 1 would have unified these approaches, but unfortunately Theorem 2 shows that this is not possible.

There has also been recent interest in finding the $2$-connected obstructions for treedepth [BJM*+*22] and pathwidth [DT17, HJMW20] in $2$-connected graphs. It seems unlikely that requiring $G$ to be $2$-connected would change the answer to Question 1, but the graphs we construct for Theorem 2 are not $2$-connected, thus leaving this as an open possibility.

We present a self-contained proof of Theorem 2, however some steps were discovered independently by Hickingbotham [Hic19]. In particular, Hickingbotham [Hic19, Lemma 7.2.1] noticed that it is just as hard to ensure $T$ is a subgraph of $G$ in Question 1 as it is to ensure $T$ is a minor of $G$. Thus our main contribution is Lemma 3.3, which essentially shows that we can also force each vertex of $T$ to be in its own bag. Hickingbotham [Hic19, Theorem 7.5.1] already proved that this stronger condition can blow up the width. Moreover, Hickingbotham proved some positive results, including that the answer to Question 1 is “yes” if $G$ is an outerplanar graph [Hic19, Theorem 7.3.3]. Note that outerplanar graphs are the graphs with simple treewidth at most $2$ [KU12], and so in this sense Theorem 2 is optimal.

We outline our approach to proving Theorem 2 in more detail in the next section.

2 Preliminaries

We use the following “subtree view” of tree decompositions. Recall that a subtree of a graph $G$ is any subgraph of $G$ which is connected and acyclic.

Definition 2.1**.**

Let $G$ be a graph, let $T$ be a tree, and let $\mathcal{B}=\{B_{x}\colon x\in V(T)\}$ be a family of subsets of $V(G)$ indexed by the vertices of $T$. For each vertex $v$ of $G$, we define

[TABLE]

Then $(T,\mathcal{B})$ is a tree decomposition of $G$ if and only if the following conditions both hold.

• •

Each $T_{v}$ is a non-empty subtree of $T$.

• •

If $uv\in E(G)$, then $V(T_{u})\cap V(T_{v})\neq\varnothing$.

We use this notation $T_{v}$ throughout the paper. When there is no chance for confusion, we refer to $T_{v}$ and its vertex set $V(T_{v})$ interchangeably. The width of $(T,\mathcal{B})$ is then the maximum, over all $x\in V(T)$, of $\lvert\{v\in V(G)\colon x\in T_{v}\}\rvert-1$. The treewidth of $G$ is the minimum width of a tree decomposition of $G$. Note that, if we are given a tree $T$ and a collection $(T_{v}\colon v\in V(G))$ of subtrees of $T$ which satisfy the conditions from Definition 2.1, then we can define a tree decomposition $(T,\mathcal{B})$ of $G$ by setting $B_{x}\coloneqq\{v\in V(G)\colon x\in T_{v}\}$ for each $x\in V(T)$.

We now outline our overall strategy for proving Theorem 2. This theorem equivalently says that Conjecture 2.2 below is false, even for connected graphs of treewidth $2$. We disprove Conjecture 2.2 by reducing each of the three conjectures below to the next one, and then disproving the final conjecture. Afterwards, we evaluate the treewidth of the constructed counterexamples more carefully. Note that in Conjecture 2.4, the condition “for every vertex $v$ of $G$, $v\in T_{v}$” is equivalent to “for every vertex $x$ of $T$, $x\in B_{x}$”.

Conjecture 2.2**.**

There is a function $f$ such that every connected graph $G$ has a tree decomposition $(T,\mathcal{B})$ of width at most $f(\operatorname{tw}(G))$ such that $T$ is a minor of $G$.

Conjecture 2.3**.**

There is a function $f$ such that every connected graph $G$ has a tree decomposition $(T,\mathcal{B})$ of width at most $f(\operatorname{tw}(G))$ such that $T$ is a spanning tree of $G$.

Conjecture 2.4**.**

There is a function $f$ such that every connected graph $G$ has a tree decomposition $(T,\mathcal{B})$ of width at most $f(\operatorname{tw}(G))$ such that $T$ is a spanning tree of $G$ and, for every vertex $v$ of $G$, we have $v\in T_{v}$.

Hickingbotham proved that Conjecture 2.2 implies Conjecture 2.3 in [Hic19, Lemma 7.2.1]. In Section 3, we show that Conjecture 2.3 implies Conjecture 2.4; this crucial new step is our main contribution. Finally, in Section 4, we construct a graph that does not satisfy Conjecture 2.4. Hickingbotham [Hic19, Theorem 7.5.1] independently discovered a counterexample to Conjecture 2.4 which actually contains our counterexample. However, we include ours since it is slightly simpler and makes the paper self-contained. We now conclude this section by providing a short proof that Conjecture 2.2 (where $T$ is a minor) implies Conjecture 2.3 (where $T$ is a spanning tree), for the sake of completeness.

Lemma 2.5**.**

If $G$ is a connected graph with a tree decomposition $(T,\mathcal{B})$ of width $k$ such that $T$ is a minor of $G$, then there exists a tree decomposition $(T^{\prime},\mathcal{B}^{\prime})$ of $G$ of width $k$ such that $T^{\prime}$ is a spanning tree of $G$.

Proof.

Since $T$ is a minor of $G$, there exists a collection $(Q_{x}\colon x\in V(T))$ of pairwise disjoint non-empty subtrees of $G$ such that, for each edge $xy\in E(T)$, there exists an edge $e_{xy}\in E(G)$ with one end in $V(Q_{x})$ and the other end in $V(Q_{y})$. Since $G$ is connected, we may assume that $V(G)=\cup_{x\in V(T)}V(Q_{x})$. Now, let $T^{\prime}$ be the spanning tree of $G$ which is obtained from $\cup_{x\in V(T)}Q_{x}$ by adding the edge $e_{xy}$ for all $xy\in E(T)$.

For each $v\in V(G)$, let $T^{\prime}_{v}$ be the subtree of $T^{\prime}$ which is induced by the union of all sets $V(Q_{x})$ such that $x\in T_{v}$. This collection of subtrees of $T^{\prime}$ satisfies the conditions of Definition 2.1 and therefore yields a tree decomposition $(T^{\prime},\mathcal{B}^{\prime})$. Furthermore, this tree decomposition has the same width at $(T,\mathcal{B})$, which completes the proof. ∎

3 Reduction to Conjecture 2.4

In this section we show that Conjecture 2.3 (where $T$ is a spanning tree) implies Conjecture 2.4 (where, additionally, for every vertex $v$ of $G$, we have $v\in T_{v}$).

We use the following well-known fact about tree decompositions of paths. We include a proof for the sake of completeness. The bounds are not optimal; we aim for simplicity instead.

Lemma 3.1**.**

For any positive integers $h$ and $k$, if $P$ is a path with at least $(k+2)^{h}$ vertices and $(T,\mathcal{B})$ is a tree decomposition of $P$ of width at most $k$, then $T$ contains a path of length $h$.

Proof.

We consider a tree decomposition $(T,\mathcal{B})$ where $T$ is rooted at an arbitrary vertex ${r\in V(T)}$. The height of $T$ is then the maximum length of a path which has $r$ as one of its ends. For fixed $k$, we prove by induction on $h$ that, under the same hypothesis, we actually obtain the following stronger conclusion: that the height of $T$ is at least $h$.

The base case of $h=1$ holds since $P$ has more than $k+1$ vertices and therefore $(T,\mathcal{B})$ has more than one bag. So we may assume that $h>1$ and the claim holds for $h-1$. Observe that we can partition $V(P)$ into $k+2$ sets, each of which induces in $P$ a path with at least $(k+2)^{h-1}$ vertices. Since $(T,\mathcal{B})$ has width at most $k$, one of these sets is disjoint from the root bag $B_{r}$. Thus, by the inductive hypothesis, one of the components of $T-\{r\}$, when rooted at its neighbour of $r$, has height at least $h-1$. So $T$ has height at least $h$, as desired. ∎

We use the following construction to show that Conjecture 2.3 implies Conjecture 2.4. Given a positive integer $k$, a graph $G$, and an arbitrary ordering of the vertices of $G$, we define a new graph denoted $\widetilde{G}$. This graph $\widetilde{G}$ is obtained from $G$ by attaching one rooted tree to each vertex of $G$; so $\widetilde{G}$ has the same treewidth as $G$ (unless $E(G)=\varnothing$). Moreover, in Lemma 3.3, we prove that if $\widetilde{G}$ satisfies Conjecture 2.3 with a tree decomposition of width $k$, then $G$ satisfies Conjecture 2.4 with a tree decomposition of width $k+1$.

The attached trees are chosen such that no two have “comparable” tree decompositions. More formally, given two trees $T_{1}$ and $T_{2}$, there is no tree decomposition $(T^{\prime}_{2},\mathcal{B}^{\prime}_{2})$ of $T_{2}$ of width $k$ such that $\mathcal{T}^{\prime}_{2}$ is a subgraph of $T_{1}$, and likewise with the roles of $T_{1}$ and $T_{2}$ reversed. We do not frame the argument in this way, but it is the underlying reason our proof works. We accomplish this condition by, up to symmetry between $T_{1}$ and $T_{2}$, making height of $T_{2}$ much larger than the height of $T_{1}$, and the width of $T_{1}$ much larger than $\lvert V(T_{2})\rvert$. See Figure 1 for a depiction.

With this intuition, we are ready to state the main definition.

Definition 3.2**.**

Fix a positive integer $k$, a graph $G$, and an arbitrary ordering $a_{1},\dotsc,a_{n}$ of the vertices of $G$. Then let $\widetilde{G}$ be the graph which is constructed from $G$ as follows.

• •

First define integers $2=h_{1}\ll h_{2}\ll\dotsb\ll h_{n}$ as follows. Given $h_{j-1}$, we define $h_{j}\coloneqq(k+2)^{2h_{j-1}}+1$. Thus, by Lemma 3.1, if $P$ is a path on at least $h_{j}-1$ vertices and $(T,\mathcal{B})$ is a tree decomposition of $P$ of width at most $k$, then $T$ contains a path of length $2h_{j-1}$.

• •

Next define integers $(k+1)n+1=w_{n}\ll w_{n-1}\ll\dotsb\ll w_{1}$ and corresponding rooted trees $S_{n},S_{n-1},\dotsc,S_{1}$ as follows. Given $w_{j}$, define $S_{j}$ to be the complete rooted $w_{j}$-ary tree of height $h_{j}$. Then, given $w_{n},w_{n-1},\dotsc,w_{j+1}$ and $S_{n},S_{n-1},\dotsc,S_{j+1}$, define

[TABLE]

Finally, let $\widetilde{G}$ be the graph which is obtained from the disjoint union of $G,S_{1},S_{2},\dotsc,S_{n}$ by, for each $j\in\{1,2,\dots,n\}$, identifying $a_{j}$ with the root of $S_{j}$.

Note that the graph $\widetilde{G}$ from Definition 3.2 can be obtained from $G$ by adding pendant vertices one at a time. It follows that $\operatorname{tw}(\widetilde{G})=\max\left(\operatorname{tw}(G),1\right)$. The next key lemma therefore completes the reduction from Conjecture 2.3 to Conjecture 2.4.

Lemma 3.3**.**

Let $k$ be a positive integer, let $G$ be a connected graph, let $a_{1},\dotsc,a_{n}$ be an ordering of the vertices of $G$, and let $\widetilde{G}$ be the resulting graph constructed using Definition 3.2. Suppose that $\widetilde{G}$ has a tree decomposition $(T,\mathcal{B})$ of width at most $k$ such that $T$ is a spanning tree of $\widetilde{G}$.

Then there exists a tree decomposition $(T^{\prime},\mathcal{B}^{\prime})$ of $G$ of width at most $k+1$ such that $T^{\prime}$ is a spanning tree of $G$ and for every $v\in V(G)$, we have $v\in T^{\prime}_{v}$.

Proof.

We use the notation introduced in Definition 3.2, except that we view each tree $S_{j}$ as an induced subgraph of $\widetilde{G}$ which is rooted at the vertex $a_{j}\in V(G)$. For the sake of convenience, we do not distinguish between $S_{j}$ and its vertex set.

In the first few claims we deduce roughly where each subtree $T_{v}$ (as defined in Definition 2.1) lies. We say that two sets meet if their intersection is non-empty.

Claim 3.3.1**.**

For every $j\in\{1,2,\dotsc,n\}$ and every non-leaf vertex $v$ of $S_{j}$, the set $T_{v}$ meets $(S_{1}\cup\dotsb\cup S_{j})\setminus V(G)$.

Proof.

Consider the union of the bags of $(T,\mathcal{B})$ that contain $v$. Each bag has size at most $k+1$, so this union has size at most $(k+1)\lvert V(T_{v})\rvert$. On the other hand, this union contains every neighbour of $v$ in $\widetilde{G}$ and so, by the choice of $w_{j}$,

[TABLE]

In particular, $T_{v}$ is not a subgraph of $S_{j+1}\cup\dotsb\cup S_{n}\cup G$. The claim follows. ∎

We say that a vertex $v$ of $\widetilde{G}$ is free if $T_{v}$ meets $V(G)$ or, equivalently, if $v\in B_{a_{1}}\cup\dotsb\cup B_{a_{n}}$. Otherwise, we call $v$ constrained. Note that if $v$ is constrained, then $T_{v}$ is a subgraph of some $S_{j}-a_{j}$ since $T_{v}$ is a subtree of $\widetilde{G}$. The number of free vertices is at most

[TABLE]

and so almost all vertices are constrained.

Claim 3.3.2**.**

For every $j\in\{1,2,\dotsc,n\}$, the vertex $a_{j}$ has a child $b_{j}$ in $S_{j}$ such that $T_{b_{j}}$ is a subgraph of $S_{j}-a_{j}$.

Proof.

As $S_{j}$ is a complete $w_{j}$-ary tree of height $h_{j}$, there are $w_{j}$ vertex-disjoint paths that start at the children of $a_{j}$ and end at parents of leaves of $S_{j}$. Since $w_{j}>(k+1)n$, at least one of these paths contains no free vertices – call this path $P$. Let $T_{P}=\cup_{v\in V(P)}T_{v}$. So $T_{P}$ is a subtree of $\widetilde{G}$. Each $v\in V(P)$ is constrained, and so for each $v$ there is an $i$ such that $T_{v}$ is a subgraph of $S_{i}-a_{i}$. But, as $T_{P}$ is connected and there is no edge between different $S_{i}-a_{i}$, this $i$ must be the same for all $v\in V(P)$. That is, there is some $i$ such that $T_{P}$ is a subgraph of $S_{i}-a_{i}$. Let $b_{j}$ be the child of $a_{j}$ that is a vertex of $P$ (since $h_{1}\geqslant 2$, such a $b_{j}$ exists).

By Claim 3.3.1, we have that $T_{b_{j}}$ meets $(S_{1}-a_{1})\cup\dotsb\cup(S_{j}-a_{j})$. Since $T_{b_{j}}$ is a subgraph of $T_{P}$, we have $i\leqslant j$. Next focus on the tree decomposition $(T_{P},\mathcal{B}_{P})$ of $P$ where $\mathcal{B}_{P}$ is $\mathcal{B}$ restricted to the vertices of $P$. This tree decomposition has width at most $k$. The path $P$ contains $h_{j}-1$ vertices and so, by the choice of $h_{j}$, the tree $T_{P}$ must contain a path of length at least $2h_{j-1}$. However, $T_{P}$ is a subgraph of $S_{i}-a_{i}$ whose longest paths have length less than $2h_{i}$. In particular, this implies that $h_{i}>h_{j-1}$ and so $i\geqslant j$. Thus $i=j$ and $b_{j}$ is as required. ∎

We say that a vertex $a_{j}\in V(G)$ is grounded if $T_{a_{j}}$ contains $a_{j}$.

Claim 3.3.3**.**

If a vertex $a_{j}\in V(G)$ is not grounded, then $T_{a_{j}}$ is a subgraph of $S_{j}-a_{j}$ and every neighbour $a_{i}\in V(G)$ of $a_{j}$ is both grounded and satisfies ${a_{j}}\in T_{a_{i}}$.

Proof.

Suppose that $a_{j}\in V(G)$ is not grounded. Then $a_{j}\notin T_{a_{j}}$. Let $b_{j}$ be the child of $a_{j}$ given by Claim 3.3.2. As $a_{j}$ and $b_{j}$ are adjacent, $T_{a_{j}}$ meets $T_{b_{j}}$. But $T_{b_{j}}$ is a subgraph of $S_{j}-a_{j}$, and so $T_{a_{j}}$ meets $S_{j}-a_{j}$. However, $T_{a_{j}}$ is connected and does not contain $a_{j}$, so $T_{a_{j}}$ must be a subgraph of $S_{j}-a_{j}$.

Let $a_{i}\in V(G)$ be a neighbour of $a_{j}$. If $a_{i}$ is not grounded, then $T_{a_{i}}$ is a subgraph of $S_{i}-a_{i}$. But then $T_{a_{i}}$ and $T_{a_{j}}$ do not meet, which is impossible as $a_{i}$ and $a_{j}$ are adjacent. Thus $a_{i}$ is grounded. Now $T_{a_{i}}$ and $T_{a_{j}}$ meet and so $T_{a_{i}}$ meets $S_{j}-a_{j}$. So, since $T_{a_{i}}$ is connected, $T_{a_{i}}$ contains $a_{j}$. ∎

We now define a tree decomposition of $G$ which satisfies Lemma 3.3. First, let $T^{\prime}$ be the subgraph of $T$ induced by $V(G)$; notice that $T^{\prime}$ is a spanning tree of $G$ since $T$ is a spanning tree of $\widetilde{G}$. Next, delete all bags $B_{x}$ where $x\notin V(T^{\prime})$ and delete all vertices of $\widetilde{G}$ that are not vertices of $G$. Finally, if a vertex $a_{j}$ is not grounded, then add $a_{j}$ to the bag $B_{a_{j}}$. Call the resulting collection of bags $\mathcal{B}^{\prime}=(B^{\prime}_{a_{j}})_{1\leqslant j\leqslant n}$. We claim that $(T^{\prime},\mathcal{B}^{\prime})$ is a tree decomposition of $G$. This completes the proof of Lemma 3.3 since it is clear that $(T^{\prime},\mathcal{B}^{\prime})$ has width at most $k+1$, that $T^{\prime}$ is a spanning tree of $G$, and that for every $a_{j}\in V(G)$, we have $a_{j}\in T^{\prime}_{a_{j}}$.

Notice that if a vertex $a_{j}\in V(G)$ is grounded, then $T^{\prime}_{a_{j}}$ is just the induced subgraph of $T_{a_{j}}$ restricted to $V(G)$; so $T^{\prime}_{a_{j}}$ is still connected. Likewise, if $a_{j}$ is not grounded, then by Claim 3.3.3, $T^{\prime}_{a_{j}}=\{a_{j}\}$ is connected. We are left to check that for every edge $a_{i}a_{j}\in E(G)$, the subtrees $T^{\prime}_{a_{i}}$ and $T^{\prime}_{a_{j}}$ meet. First suppose that $a_{j}$ is not grounded. Then, by Claim 3.3.3, $T^{\prime}_{a_{i}}$ and $T^{\prime}_{a_{j}}$ both contain the vertex $a_{j}$. The case that $a_{i}$ is not grounded is symmetric, so we may assume that both $a_{i}$ and $a_{j}$ are grounded. As $a_{i}$ and $a_{j}$ are adjacent in $\widetilde{G}$, the trees $T_{a_{i}}$ and $T_{a_{j}}$ meet in $\widetilde{G}$. Let $\ell\in\{1,2,\dotsc,n\}$ be such that they meet in $S_{\ell}$. Now $T_{a_{i}}$ contains $a_{i}$ and is connected and $T_{a_{j}}$ contains $a_{j}$ and is connected, so $T_{a_{i}}$ and $T_{a_{j}}$ both contain $a_{\ell}$. So both $T^{\prime}_{a_{i}}$ and $T^{\prime}_{a_{j}}$ contain $a_{\ell}$, as required. This completes the proof of Lemma 3.3. ∎

4 Construction

In this section we disprove Conjecture 2.4 and then combine the previous reductions in order to prove Theorem 2.

We begin by defining the relevant graphs. Then we prove that they are counterexamples in Lemmas 4.2 and 4.3.

Definition 4.1**.**

The first reflected-tree, which we denote by $G_{1}$, is the singleton graph with exactly one vertex and no edges. We call the vertex of $G_{1}$ its root vertex. Then, for any positive integer $r\geqslant 2$, the $r$-th reflected-tree $G_{r}$ is constructed recursively as follows:

• •

Let $H$ and $H^{\prime}$ be two disjoint copies of $G_{r-1}$, and let $u$ and $v$ be two new vertices, which we call the root vertices of $G_{r}$. To construct $G_{r}$, we start with $H$ and $H^{\prime}$, then make $u$ adjacent to a root vertex of $H$ and a root vertex of $H^{\prime}$. Finally, we make $v$ adjacent to the remaining root vertex of $H$ and the remaining root vertex of $H^{\prime}$. See Figure 2 for a depiction.

Now we prove a lemma about the spanning trees of the reflected-tree. Whenever $T$ is a spanning tree of a graph $G$, we denote the fundamental cycle of an edge $e\in E(G)\setminus E(T)$ with respect to $T$ by $C^{e}_{T}$; thus $C^{e}_{T}$ is the unique cycle in the graph obtained from $T$ by adding $e$.

Lemma 4.2**.**

For any integer $r\geqslant 2$ and any spanning tree $T$ of $G_{r}$, there is a matching $M\subseteq E(G_{r})\setminus E(T)$ of size $r-1$ such that

[TABLE]

Proof.

Let $u$ and $v$ be the root vertices of $G_{r}$, and denote the path between them in $T$ by $P_{uv}$. Under the same conditions, we prove the following stronger outcome holds by induction:

[TABLE]

for some matching $M\subseteq E(G_{r})\setminus E(T)$ of size $r-1$. For the base case of $r=2$, the graph $G_{r}$ is a cycle on four vertices; then any spanning tree $T$ of $G_{2}$ is a path on four vertices, and we can take $M$ to be the $1$-edge matching $E(G_{2})\setminus E(T)$.

Next, we may assume that $r>2$ and the claim holds for $r-1$. By definition, $G_{r}-\{u,v\}$ has exactly two connected components both of which are isomorphic to $G_{r-1}$. We denote these components by $H$ and $H^{\prime}$. Exactly one of $T_{H}\coloneqq T[V(H)\cup\{u,v\}]$ and $T_{H^{\prime}}\coloneqq T[V(H^{\prime})\cup\{u,v\}]$ is connected in $T$; without loss of generality, we assume that $T_{H}$ is connected in $T$. We can apply the inductive hypothesis on $T[V(H)]$, which is a spanning tree of $H$, to find a matching $M_{H}\subseteq E(H)\setminus E(T[V(H)])$ of size $r-2$ with $\bigcap_{e\in M_{H}}E\left(C^{e}_{T}\right)\cap E(P_{uv}-\{u,v\})\neq\varnothing$. The other subgraph $T_{H^{\prime}}$ is not connected. In fact, it contains exactly two components: one containing $u$, and the other containing $v$. Thus there exists an edge $e^{\prime}\in E(G_{r})\setminus E(T)$ with one end in each of these two components of $T_{H^{\prime}}$. Observe that $e^{\prime}$ lies in $G[V(H^{\prime})\cup\{u,v\}]$ which is vertex-disjoint from $H$. Thus $M_{H}\cup\{e^{\prime}\}$ is a matching since $M_{H}\subseteq H$.

For convenience, let us define $M\coloneqq M_{H}\cup\{e^{\prime}\}$. $M$ is a matching of size $r-1$, and we have $E(P_{uv})\subseteq E(C^{e^{\prime}}_{T})$. From here, it follows that

[TABLE]

Thus $M$ is our desired matching. ∎

We are now ready to prove the following lemma, which shows that reflected-trees are a counterexample to Conjecture 2.4.

Lemma 4.3**.**

For every $k\in\mathbb{N}$, if $(T,\mathcal{B})$ is a tree decomposition of $G_{k+2}$ such that $T$ is a spanning tree of $G_{k+2}$ and, for every $v\in V(G_{k+2})$, we have $v\in T_{v}$, then the width of $(T,\mathcal{B})$ is at least $k$.

Proof.

We begin by finding a matching $M\coloneqq\{u_{1}v_{1},\ldots,u_{k+1}v_{k+1}\}\subseteq E(G_{k+2})\setminus E(T)$ of size $k+1$ satisfying the properties in Lemma 4.2. Let $x\in\bigcap_{e\in M}V\left(C^{e}_{T}\right)$. By construction, $x$ is in the path $P_{u_{i}v_{i}}$ between $u_{i}$ and $v_{i}$ in $T$ for every $i\in\{1,\ldots,k+1\}$. From Definition 2.1, the trees $T_{u_{i}}$ and $T_{v_{i}}$ intersect; furthermore, since $T_{u_{i}}$ and $T_{v_{i}}$ are connected in $T$, with $u_{i}\in V(T_{u_{i}})$ and $v_{i}\in V(T_{v_{i}})$, we find that every vertex of $P_{u_{i}v_{i}}$ is in $V(T_{u_{i}})\cup V(T_{v_{i}})$. As a result, $x\in V(T_{u_{i}})\cup V(T_{v_{i}})$. That is, $u_{i}\in B_{x}$ or $v_{i}\in B_{x}$ for all $i\in\{1,\ldots,k+1\}$. Since $M$ is a matching, we have that $|B_{x}|\geqslant k+1$ and the width of $(T,\mathcal{B})$ is at least $k$. ∎

We are now ready to prove the main theorem, which is restated below for convenience.

See 2

Proof.

For convenience, we fix an integer $k\geqslant 2$. Now consider the $(k+3)$-rd reflected-tree $G_{k+3}$. Let $a_{1},\dotsc,a_{n}$ be an arbitrary ordering of $V(G_{k+3})$, and let $\widetilde{G}_{k+3}$ be the graph obtained from the integer $k-1$, the graph $G_{k+3}$, and the ordering $a_{1},\dotsc,a_{n}$ by applying Definition 3.2.

We now prove that $\widetilde{G}_{k+3}$ satisfies the conditions of Theorem 2. First, recall that $\widetilde{G}_{k+3}$ has treewidth equal to $\max(\operatorname{tw}(G_{k+3}),1)$. Moreover, $\operatorname{tw}(G_{k+3})=2$ since $G_{k+3}$ is series parallel and not a tree. Thus $\widetilde{G}_{k+3}$ is a connected graph of treewidth $2$, as desired.

Next, suppose towards a contradiction that $\widetilde{G}_{k+3}$ has a tree decomposition $(T,\mathcal{B})$ of width at most $k-1$ such that $T$ is a minor of $\widetilde{G}_{k+3}$. Since $\widetilde{G}_{k+3}$ is connected, Lemma 2.5 says that $\widetilde{G}_{k+3}$ has a tree decomposition $(T^{\prime},\mathcal{B}^{\prime})$ of width at most $k-1$ such that $T^{\prime}$ is a spanning tree of $\widetilde{G}_{k+3}$. Thus, since $G_{k+3}$ is connected, Lemma 3.3 says that $G_{k+3}$ has a tree decomposition $(T^{\prime\prime},\mathcal{B}^{\prime\prime})$ of width at most $k$ such that $T^{\prime\prime}$ is a spanning tree of $G_{k+3}$ and for every $v\in V(G_{k+3})$, we have $v\in T^{\prime\prime}_{v}$. However, Lemma 4.3 also says that $(T^{\prime\prime},\mathcal{B}^{\prime\prime})$ has width at least $k+1$. This contradiction completes the proof of Theorem 2. ∎

Acknowledgements

The authors would like to thank Sang-il Oum for his suggestions about where to look for a counterexample to Conjecture 2.4; Zdeněk Dvořák for sharing the original problem and helpful comments which improved the paper; David Wood for pointing out all of the progress in [Hic19], and Sophie Spirkl for her helpful feedback on our proof that Conjecture 2.4 is false. Aristotelis Chaniotis, Linda Cook, Sepehr Hajebi, and Sophie Spirkl asked about a counterexample to a strengthening of Conjecture 2.4 at the Barbados Graph Theory Workshop in December 2022, which started this whole work. As such, the authors would like to thank Aristotelis Chaniotis, Sepehr Hajebi, Sophie Spirkl, and the organizers of the Barbados Graph Theory Workshop – Sergey Norin, Paul Seymour, and David Wood.