On tree decompositions whose trees are minors
Pablo Blanco, Linda Cook, Meike Hatzel, Claire Hilaire, Freddie, Illingworth, Rose McCarty

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 has a tree decomposition so that is a subgraph of and the width of is bounded by a function of the treewidth of . We prove that this is false, even when has treewidth and is allowed to be a minor of .
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Taxonomy
TopicsAdvanced Graph Theory Research · Limits and Structures in Graph Theory · Complexity and Algorithms in Graphs
On tree decompositions whose trees are minors
Pablo Blanco
Department of Mathematics, Princeton University, United States
Linda Cook Supported by the Institute for Basic Science (IBS-R029-C1). Discrete Math Group, Institute for Basic Science, Daejeon, Republic of Korea
Meike Hatzel Supported by the Federal Ministry of Education and Research (BMBF) and by a fellowship within the IFI programme of the German Academic Exchange Service (DAAD). National Institute of Informatics, Tokyo, Japan
Claire Hilaire
LaBRI, Université de Bordeaux, Bordeaux, France
Freddie Illingworth Supported by EPSRC grant EP/V007327/1. Mathematical Institute, University of Oxford, United Kingdom
Rose McCarty Supported by the National Science Foundation under Grant No. DMS-2202961. Department of Mathematics, Princeton University, United States
(February 23, 2023)
Abstract
In 2019, Dvořák asked whether every connected graph has a tree decomposition so that is a subgraph of and the width of is bounded by a function of the treewidth of . We prove that this is false, even when has treewidth and is allowed to be a minor of .
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20(-2.3, 6.5)
1 Introduction
Suppose that a graph has small treewidth, and consider all tree decompositions of whose width is not too much larger than the optimum. To what extent can we choose or manipulate the “shape” of ? For graphs with no long path, we can choose to also have no long path [NM12]111The reference gives us an elimination tree of of depth . Then we can obtain a tree decomposition of of width by letting the bag of each vertex be the set of all ancestors of in .; this gives rise to the parameter called treedepth. Similarly, for graphs of bounded degree, we can choose 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 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 to . In 2019, Dvořák suggested one way of accomplishing this goal. In the question below and throughout the paper, we write for the treewidth of .
Question 1** ([Dvo19]).**
Does there exist a polynomial such that every connected graph has a tree decomposition of width at most such that is a subgraph of ?
Unfortunately, we prove that the answer to Question 1 is “no” in the following strong sense.
Theorem 2**.**
For every positive integer , there is a connected graph of treewidth such that if is a tree decomposition of and is a minor of , then has width at least .
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 is replaced by , the pathwidth of . In other words, perhaps there exists a polynomial (or even just some function) so that every connected graph has a tree decomposition of width at most such that is a subgraph of . 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 -connected obstructions for treedepth [BJM*+*22] and pathwidth [DT17, HJMW20] in -connected graphs. It seems unlikely that requiring to be -connected would change the answer to Question 1, but the graphs we construct for Theorem 2 are not -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 is a subgraph of in Question 1 as it is to ensure is a minor of . Thus our main contribution is Lemma 3.3, which essentially shows that we can also force each vertex of 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 is an outerplanar graph [Hic19, Theorem 7.3.3]. Note that outerplanar graphs are the graphs with simple treewidth at most [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 is any subgraph of which is connected and acyclic.
Definition 2.1**.**
Let be a graph, let be a tree, and let be a family of subsets of indexed by the vertices of . For each vertex of , we define
[TABLE]
Then is a tree decomposition of if and only if the following conditions both hold.
- •
Each is a non-empty subtree of .
- •
If , then .
We use this notation throughout the paper. When there is no chance for confusion, we refer to and its vertex set interchangeably. The width of is then the maximum, over all , of . The treewidth of is the minimum width of a tree decomposition of . Note that, if we are given a tree and a collection of subtrees of which satisfy the conditions from Definition 2.1, then we can define a tree decomposition of by setting for each .
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 . 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 of , ” is equivalent to “for every vertex of , ”.
Conjecture 2.2**.**
There is a function such that every connected graph has a tree decomposition of width at most such that is a minor of .
Conjecture 2.3**.**
There is a function such that every connected graph has a tree decomposition of width at most such that is a spanning tree of .
Conjecture 2.4**.**
There is a function such that every connected graph has a tree decomposition of width at most such that is a spanning tree of and, for every vertex of , we have .
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 is a minor) implies Conjecture 2.3 (where is a spanning tree), for the sake of completeness.
Lemma 2.5**.**
If is a connected graph with a tree decomposition of width such that is a minor of , then there exists a tree decomposition of of width such that is a spanning tree of .
Proof.
Since is a minor of , there exists a collection of pairwise disjoint non-empty subtrees of such that, for each edge , there exists an edge with one end in and the other end in . Since is connected, we may assume that . Now, let be the spanning tree of which is obtained from by adding the edge for all .
For each , let be the subtree of which is induced by the union of all sets such that . This collection of subtrees of satisfies the conditions of Definition 2.1 and therefore yields a tree decomposition . Furthermore, this tree decomposition has the same width at , which completes the proof. ∎
3 Reduction to Conjecture 2.4
In this section we show that Conjecture 2.3 (where is a spanning tree) implies Conjecture 2.4 (where, additionally, for every vertex of , we have ).
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 and , if is a path with at least vertices and is a tree decomposition of of width at most , then contains a path of length .
Proof.
We consider a tree decomposition where is rooted at an arbitrary vertex . The height of is then the maximum length of a path which has as one of its ends. For fixed , we prove by induction on that, under the same hypothesis, we actually obtain the following stronger conclusion: that the height of is at least .
The base case of holds since has more than vertices and therefore has more than one bag. So we may assume that and the claim holds for . Observe that we can partition into sets, each of which induces in a path with at least vertices. Since has width at most , one of these sets is disjoint from the root bag . Thus, by the inductive hypothesis, one of the components of , when rooted at its neighbour of , has height at least . So has height at least , as desired. ∎
We use the following construction to show that Conjecture 2.3 implies Conjecture 2.4. Given a positive integer , a graph , and an arbitrary ordering of the vertices of , we define a new graph denoted . This graph is obtained from by attaching one rooted tree to each vertex of ; so has the same treewidth as (unless ). Moreover, in Lemma 3.3, we prove that if satisfies Conjecture 2.3 with a tree decomposition of width , then satisfies Conjecture 2.4 with a tree decomposition of width .
The attached trees are chosen such that no two have “comparable” tree decompositions. More formally, given two trees and , there is no tree decomposition of of width such that is a subgraph of , and likewise with the roles of and 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 and , making height of much larger than the height of , and the width of much larger than . See Figure 1 for a depiction.
With this intuition, we are ready to state the main definition.
Definition 3.2**.**
Fix a positive integer , a graph , and an arbitrary ordering of the vertices of . Then let be the graph which is constructed from as follows.
- •
First define integers as follows. Given , we define . Thus, by Lemma 3.1, if is a path on at least vertices and is a tree decomposition of of width at most , then contains a path of length .
- •
Next define integers and corresponding rooted trees as follows. Given , define to be the complete rooted -ary tree of height . Then, given and , define
[TABLE]
Finally, let be the graph which is obtained from the disjoint union of by, for each , identifying with the root of .
Note that the graph from Definition 3.2 can be obtained from by adding pendant vertices one at a time. It follows that . The next key lemma therefore completes the reduction from Conjecture 2.3 to Conjecture 2.4.
Lemma 3.3**.**
Let be a positive integer, let be a connected graph, let be an ordering of the vertices of , and let be the resulting graph constructed using Definition 3.2. Suppose that has a tree decomposition of width at most such that is a spanning tree of .
Then there exists a tree decomposition of of width at most such that is a spanning tree of and for every , we have .
Proof.
We use the notation introduced in Definition 3.2, except that we view each tree as an induced subgraph of which is rooted at the vertex . For the sake of convenience, we do not distinguish between and its vertex set.
In the first few claims we deduce roughly where each subtree (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 and every non-leaf vertex of , the set meets .
Proof.
Consider the union of the bags of that contain . Each bag has size at most , so this union has size at most . On the other hand, this union contains every neighbour of in and so, by the choice of ,
[TABLE]
In particular, is not a subgraph of . The claim follows. ∎
We say that a vertex of is free if meets or, equivalently, if . Otherwise, we call constrained. Note that if is constrained, then is a subgraph of some since is a subtree of . The number of free vertices is at most
[TABLE]
and so almost all vertices are constrained.
Claim 3.3.2**.**
For every , the vertex has a child in such that is a subgraph of .
Proof.
As is a complete -ary tree of height , there are vertex-disjoint paths that start at the children of and end at parents of leaves of . Since , at least one of these paths contains no free vertices – call this path . Let . So is a subtree of . Each is constrained, and so for each there is an such that is a subgraph of . But, as is connected and there is no edge between different , this must be the same for all . That is, there is some such that is a subgraph of . Let be the child of that is a vertex of (since , such a exists).
By Claim 3.3.1, we have that meets . Since is a subgraph of , we have . Next focus on the tree decomposition of where is restricted to the vertices of . This tree decomposition has width at most . The path contains vertices and so, by the choice of , the tree must contain a path of length at least . However, is a subgraph of whose longest paths have length less than . In particular, this implies that and so . Thus and is as required. ∎
We say that a vertex is grounded if contains .
Claim 3.3.3**.**
If a vertex is not grounded, then is a subgraph of and every neighbour of is both grounded and satisfies .
Proof.
Suppose that is not grounded. Then . Let be the child of given by Claim 3.3.2. As and are adjacent, meets . But is a subgraph of , and so meets . However, is connected and does not contain , so must be a subgraph of .
Let be a neighbour of . If is not grounded, then is a subgraph of . But then and do not meet, which is impossible as and are adjacent. Thus is grounded. Now and meet and so meets . So, since is connected, contains . ∎
We now define a tree decomposition of which satisfies Lemma 3.3. First, let be the subgraph of induced by ; notice that is a spanning tree of since is a spanning tree of . Next, delete all bags where and delete all vertices of that are not vertices of . Finally, if a vertex is not grounded, then add to the bag . Call the resulting collection of bags . We claim that is a tree decomposition of . This completes the proof of Lemma 3.3 since it is clear that has width at most , that is a spanning tree of , and that for every , we have .
Notice that if a vertex is grounded, then is just the induced subgraph of restricted to ; so is still connected. Likewise, if is not grounded, then by Claim 3.3.3, is connected. We are left to check that for every edge , the subtrees and meet. First suppose that is not grounded. Then, by Claim 3.3.3, and both contain the vertex . The case that is not grounded is symmetric, so we may assume that both and are grounded. As and are adjacent in , the trees and meet in . Let be such that they meet in . Now contains and is connected and contains and is connected, so and both contain . So both and contain , 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 , is the singleton graph with exactly one vertex and no edges. We call the vertex of its root vertex. Then, for any positive integer , the -th reflected-tree is constructed recursively as follows:
- •
Let and be two disjoint copies of , and let and be two new vertices, which we call the root vertices of . To construct , we start with and , then make adjacent to a root vertex of and a root vertex of . Finally, we make adjacent to the remaining root vertex of and the remaining root vertex of . See Figure 2 for a depiction.
Now we prove a lemma about the spanning trees of the reflected-tree. Whenever is a spanning tree of a graph , we denote the fundamental cycle of an edge with respect to by ; thus is the unique cycle in the graph obtained from by adding .
Lemma 4.2**.**
For any integer and any spanning tree of , there is a matching of size such that
[TABLE]
Proof.
Let and be the root vertices of , and denote the path between them in by . Under the same conditions, we prove the following stronger outcome holds by induction:
[TABLE]
for some matching of size . For the base case of , the graph is a cycle on four vertices; then any spanning tree of is a path on four vertices, and we can take to be the -edge matching .
Next, we may assume that and the claim holds for . By definition, has exactly two connected components both of which are isomorphic to . We denote these components by and . Exactly one of and is connected in ; without loss of generality, we assume that is connected in . We can apply the inductive hypothesis on , which is a spanning tree of , to find a matching of size with . The other subgraph is not connected. In fact, it contains exactly two components: one containing , and the other containing . Thus there exists an edge with one end in each of these two components of . Observe that lies in which is vertex-disjoint from . Thus is a matching since .
For convenience, let us define . is a matching of size , and we have . From here, it follows that
[TABLE]
Thus 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 , if is a tree decomposition of such that is a spanning tree of and, for every , we have , then the width of is at least .
Proof.
We begin by finding a matching of size satisfying the properties in Lemma 4.2. Let . By construction, is in the path between and in for every . From Definition 2.1, the trees and intersect; furthermore, since and are connected in , with and , we find that every vertex of is in . As a result, . That is, or for all . Since is a matching, we have that and the width of is at least . ∎
We are now ready to prove the main theorem, which is restated below for convenience.
See 2
Proof.
For convenience, we fix an integer . Now consider the -rd reflected-tree . Let be an arbitrary ordering of , and let be the graph obtained from the integer , the graph , and the ordering by applying Definition 3.2.
We now prove that satisfies the conditions of Theorem 2. First, recall that has treewidth equal to . Moreover, since is series parallel and not a tree. Thus is a connected graph of treewidth , as desired.
Next, suppose towards a contradiction that has a tree decomposition of width at most such that is a minor of . Since is connected, Lemma 2.5 says that has a tree decomposition of width at most such that is a spanning tree of . Thus, since is connected, Lemma 3.3 says that has a tree decomposition of width at most such that is a spanning tree of and for every , we have . However, Lemma 4.3 also says that has width at least . 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.
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