Induced subgraphs and tree decompositions
XIII. Basic obstructions in -free graphs for finite
Abstract
Unlike minors, the induced subgraph obstructions to bounded treewidth come in a large variety, including, for every , the -basic obstructions: the graphs and , along with the subdivisions of the -by- wall and their line graphs. But this list is far from complete. The simplest example of a “non-basic” obstruction is due to Pohoata and Davies (independently). For every , they construct certain graphs of treewidth and with no -basic obstruction as an induced subgraph, which we call -arrays.
Let us say a graph class is clean if the only obstructions to bounded treewidth in are in fact the basic ones. It follows that a full description of the induced subgraph obstructions to bounded treewidth is equivalent to a characterization of all families of graphs for which the class of all -free graphs is clean (a graph is -free if no induced subgraph of is isomorphic to any graph in ).
This remains elusive, but there is an immediate necessary condition: if -free graphs are clean, then there are only finitely many such that there is an -array which is -free. The above necessary condition is not sufficient in general. However, the situation turns out to be different if is finite: we prove that for every finite set of graphs, the class of all -free graphs is clean if and only if there is no -free -array except possibly for finitely many values of .
title = Induced subgraphs and tree decompositions
XIII. Basic obstructions in -free graphs for finite , author = Bogdan Alecu, Maria Chudnovsky, Sepehr Hajebi, and Sophie Spirkl,
plaintextauthor = Bogdan Alecu, Maria Chudnovsky, Sepehr Hajebi, Sophie Spirkl,
plaintexttitle = Induced subgraphs and tree decompositions XIII. Basic obstructions in H-free graphs for finite H, runningtitle = Induced subgraphs and tree decompositions XIII.,
copyrightauthor = B. Alecu, M. Chudnovsky, S. Hajebi and S. Spirkl,
keywords = Tree decompositions, Treewidth, Induced subgraphs, graph minors.,
\aicEDITORdetailsyear=2024,
number=6,
received=26 November 2023, published=29 November 2024, doi=10.19086/aic.2024.6,
\DeclareMathOperator\twtw
\DeclareMathOperator\ta\texttree-α
[classification=text]
1 Introduction
The set of all positive integers is denoted by , and for every integer , we write for the set of all positive integers less than or equal to (so if ). Graphs in this paper have finite vertex sets, no loops and no parallel edges. Let be a graph. An induced subgraph of is the graph for some , that is, the graph obtained from by removing the vertices in . For , we use both and to denote the subgraph of induced on , which is the same as . We say contains a graph if is isomorphic to an induced subgraph of ; otherwise, we say is -free. We also say is -free for a family of graphs if is -free for all .
The treewidth of a graph (denoted by ) is the smallest for which can be represented as (a subgraph of) an intersection graph of subtrees of a tree, such that each vertex of the underlying tree appears in at most subtrees. For instance, the “Helly property of subtrees” [8] implies that for every , the complete graph and the complete bipartite graph both have treewidth .
There are also sparse graphs with arbitrarily large treewidth, the most well-known example of which is the hexagonal grid (see Figure 1). For every , the -by- hexagonal grid, also called the -by- wall, denoted , has treewidth [14], and the same remains true for all subdivisions of , as well. Indeed, Robertson and Seymour proved in 1986 [14] that containing a hexagonal grid as a minor (or a subdivided one as a subgraph) is qualitatively the only reason why a graph may have large treewidth:

Theorem 1.1 (Robertson and Seymour [14])
For every , every graph of sufficiently large treewidth contains a subdivision of as a subgraph.
For induced subgraphs, however, excluding just the walls is not enough to guarantee bounded treewidth: note that complete graphs, complete bipartite graphs and subdivided walls are three pairwise “independent” types of graphs with arbitrarily large treewidth, in the sense that no graph from one type contains an induced subgraph of another type with large treewidth. There is also a fourth example, namely the line-graphs of subdivided walls, where the line-graph of a graph is the graph with vertex set , such that two vertices of are adjacent if the corresponding edges of share an end.
It is useful to group all these graphs together: given , we say a graph is a -basic obstruction if is isomorphic to one the following: the complete graph , the complete bipartite graph , a subdivision of , or the line-graph of a subdivision of (see Figure 2). For every , every -basic obstruction has treewidth , and each induced subgraph of large treewidth in a basic obstruction of a given type contains a basic obstruction of the same type and of (relatively) large treewidth. Let us say a graph is -clean if contains no -basic obstruction. It follows that every graph of treewidth less than is -clean.

One may hope for the basic obstructions to be the only induced subgraph obstructions to bounded treewidth. In other words, the neatest possible analog of Theorem 1.1 for induced subgraphs would be the following: for every , there is a constant such that every -clean graph has treewidth less than . However, there are now several counterexamples to this statement [3, 6, 12, 17]. The simplest construction is due to Pohoata and Davies (independently) [6, 12], consisting of -clean graphs of arbitrarily large treewidth. We call these graphs arrays.
Specifically, for , an -array is a graph consisting of pairwise disjoint paths with no edges between them as well as pairwise non-adjacent vertices , such that for every :
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each of has at least one neighbor in ; and
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for every , all neighbors of in appear before all neighbors of in (in particular, each vertex of is adjacent to at most one of ).

See Figure 3. As mentioned earlier, arrays are -clean graphs with unbounded treewidth:
Theorem 1.2 (Pohoata [12], Davies [6]; see also Theorem 3.1 in [2])
For every , every -array is a -clean graph of treewidth at least .
This motivates the following definition: A graph class is said to be clean if the only induced subgraph obstructions to bounded treewidth in are in fact the basic ones, that is, for every , there is a constant such that every -clean graph in has treewidth less than . An exact analog of Theorem 1.1 for induced subgraphs is therefore equivalent to a characterization of all families of graphs for which the class of all -free graphs is clean.
This appears to be out of reach of the current techniques, and even formulating a conjecture seems rather difficult. Nevertheless, the known “non-basic” obstructions may provide some insight into the structural properties of a family with the above property. For example, from Theorem 1.2 combined with the definition of a clean class, we deduce that:
Observation 1.3
Let be a family of graphs such that the class of -free graphs is clean. Then there exists such that for every and every -array , there is a graph which is isomorphic to an induced subgraph of .
The converse to Observation 1.3 is not true in general; indeed, all other non-basic obstructions discovered so far [3, 17] are counterexamples to this converse:
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Let be the family of all graphs which are the disjoint unions of two cycles. Then every -array with contains a graph in , while it is proved in [3] that there are -clean -free graphs of arbitrarily large treewidth.
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Let be the family of all subdivisions of , also known as the thetas. Then it is readily observed that every -array with contains a theta, whereas it is proved in [17] that the class of theta-free graphs is not clean.
It turns out that what allows these counterexamples to occur is the fact that the corresponding family of graphs that we are forbidding is infinite. Our main result in this paper shows for a finite set of graphs, the necessary condition from Observation 1.3 is in fact sufficient:
Theorem 1.4
Let be a finite set of graphs. Then the class of all -free graphs is clean if and only if there are only finitely many for which there is an -free -array.
This may also be regarded as a natural strengthening of the main result from [1], which handles the special case where is a singleton:
Theorem 1.5 (Abrishami, Alecu, Chudnovsky, Hajebi and Spirkl [1])
Let be a graph. Then the class of all -free graphs is clean if and only if every component of is a subdivided star, that is, a tree with at most one vertex of degree more than two.
1.1 Tassels and tasselled families
Note that by Observation 1.3, we only need to prove the “if” implication of Theorem 1.4. To that end, we find it technically most convenient to reformulate the “if” implication in terms of “the columns of the arrays,” which we refer to as “tassels.”
Let us make this precise. A strand is a graph obtained from a path by adding a new vertex with at least one neighbor in , and we say is a -strand, for , if is not adjacent to the first and last vertices of . We call and the neck of and the path of , respectively. For , by a -tassel we mean a graph obtained from at least pairwise disjoint copies of a -strand by identifying their necks into a single vertex, called the neck of . We also refer to each copy of in as a strand of , and to the path of each copy of as a path of (see Figure 4).

We say that a family of graphs is tasselled if there is a constant with the property that for every -tassel there is such that contains each component of . Also, if is finite, we define . It follows that:
Theorem 1.6
Let be a finite set of graphs such that there is no -free -array except possibly for finitely many . Then is tasselled.
Proof 1.7.
By the assumption, there exists such that there is no -free -array for any . Let . In order to prove that is tasselled, we show that for every -tassel , there is a graph such that contains each component of .
Let be the number of paths of . Construct an -array as follows. Start with pairwise disjoint copies of . For each , let be the neck of and fix an enumeration of the paths of . For every , fix a labelling of the ends of . Then, for every and every , add an edge between and (see Figure 5).

Since is an -array with , it follows that is not -free, and so we may choose a graph which is contained in .
Let be a component of . Our goal is to show that is contained in . This is immediate if is a path, because and is a -tassel. Thus, we may assume that is not a path. Since is contained in , it follows that some induced subgraph of is isomorphic to . In particular, is a connected graph on at most vertices which is not a path. Also, from the construction of , it is readily observed that the necks of are pairwise at distance at least in . Consequently, there exists exactly one for which belongs to .
Now, since is connected, it follows that for every component of , we have and has a neighbor in . This, along with the fact that is connected and , implies that for some . In conclusion, we have shown that , and so . Hence, is contained in , as desired.
Theorem 1.8.
Let be a finite set of graphs which is tasselled. Then the class of -free graphs is clean.
The rest of the paper is devoted to the proof of Theorem 1.8, which is completed in Section 7. Note also that Theorems 1.6 and 1.8 combined with Observation 1.3 imply the following (see also Figure 6).
Corollary 1.9.
The following are equivalent for every finite set of graphs.
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There is no -free -array except possibly for finitely many .
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is tasselled.
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The class of all -free graphs is clean.

1.2 Outline
To prove the Theorem 1.8, we proceed as follows. First, in Section 2, we show that a finite family is tasselled if and only if it is “hassled,” which is described by containment not in a -tassel, but in a less restricted -hassle: while similar to a tassel, the paths are replaced by walks such that any short stretch is an induced path, the adjacency from the neck to the walks need not be the same, and we may have edges between walks.
Then, our goal is to show that -clean graphs of large treewidth contain -hassles. To do so, we start with some preliminary Ramsey-type results in Section 3. In Section 4, we show that in “blocks,” which are sets of vertices with pairwise many paths between them, we can arrange that those paths are long. This is useful for getting many long induced paths, which will form part of our -hassles.
In Section 5, we get a structure of large treewidth consisting of one of the following:
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A large set of paths, and a large set of vertices, each with a neighbor in every path; or
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Two large sets of paths such that each path in one set has an edge to each path in the other, but no vertex has neighbors in many paths.
In Section 6, we show how each of these configurations leads to a -hassle. In the former case, this is almost immediate, whereas the latter case is more involved: we find a block whose vertex set is contained in the union of the paths, and leverage the interaction between the two structures, along with the fact that blocks are long, to find our -hassle.
In Section 7, we put everything together and prove our main result.
2 Hassled families
In this section, we introduce the notion of a “hassled family” as a variant of tasselled families, and show that for finite families of graphs, these two properties are in fact equivalent. This in turn reduces Theorem 1.8 to: for every finite and hassled family of graphs, the class of all -free graphs is clean. The remainder of this paper is then occupied with a proof of the latter statement, which will appear as Theorem 7.1 in Section 7. In order to define hassled families, first we need another notion, that of a “-hassle,” which is similar to a -tassel except its paths are replaced by walks that are “locally” isomorphic to a path, and, up to sparsity, there may be additional edges between these walks.
Let us define all this formally. Let and let be an -vertex path (as a graph). Then we write to mean that and is adjacent to if and only if . We call the vertices and the ends of , and refer to as the interior of , denoted . For vertices , we denote by the subpath of from to . Recall that the length of a path is the number of edges in it.
Let be a graph. A path in is an induced subgraph of which is a path. For . We say is complete to in if every vertex of is adjacent to every vertex in , and we say and are anticomplete in if there is no edge in with an end in and an end in . If , we say is complete (anticomplete) to if and are complete (anticomplete) in .
For , by an -segment we mean a set of at most consecutive integers; in particular, is an -segment. A graph is a walk if there is and a surjective map such that for every , we have (one may in fact observe that a graph is a walk if and only if it is connected). Given a walk along with choices of and , for , we refer to and as the first vertices of and the last vertices of , respectively. We also say is -stretched if is a path in for every -segment (see Figure 7 – intuitively, this means a snake of length traversing through can never see/hit itself).
We now define a -hassle, where , to be a graph obtained from at least pairwise disjoint -stretched walks, called the walks of , by adding edges arbitrarily between the walks, and then adding a vertex , called the neck of , which has a neighbor in each walk and which is anticomplete to the first and last vertices of each walk (see Figure 8).

For a family of graphs, we say is hassled if for every , there is with the property that for every -free -hassle there exists such that contains every component of . In particular, observe that every -tassel is a -free -hassle, and so every hassled family is tasselled. More importantly, for finite families, the converse is also true:
Theorem 2.1.
Let be a finite set of graphs. Then is tasselled if and only if is hassled.

The goal of this section is to prove Theorem 2.1, beginning with two lemmas:
Lemma 2.2.
Let , let be a family of -tassels, each with exactly paths, such that . For each , let be the neck of and fix an enumeration of the paths of . Let be a connected graph on at most vertices which is not a path, and assume that for every , there is an isomorphism from to an induced subgraph of ; in particular, contains . Then there exist and with for which the following hold.
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(a)
For every , we have .
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(b)
For every component of , there exists such that for every , we have .
Proof 2.3.
For each , since is not a path and is an isomorphism, it follows that there is a unique vertex such that . Also, since has at most vertices, it follows that:
(1) There exist and with such that for every , we have .
By \eqrefst:sameneck, for each and every component of , we have , which in turn implies that there exists for which we have . Now, since has at most components and since , it follows that there exists with , as well as for every component of , such that for each , we have . Hence, and satisfy 2.2b. Moreover, from \eqrefst:sameneck, it follows that and satisfy 2.2a, as desired.
Lemma 2.4.
For every finite and tasselled set of graphs, there is a constant with the following property. For every -hassle with neck , there is a graph such that for every component of , one of the following holds.
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(a)
is a path.
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There exists and a map with , such that for every component of , the restriction of to is an isomorphism from to .
Proof 2.5.
Since is tasselled, it follows that there is a constant with the property that for every -tassel there is such that contains every component of . We claim that
satisfies the lemma. To see this, let be a -hassle with neck , and let be the collection of all walks of . Recall that each walk in is -stretched.
For each , let and be as in the definition of a walk, and construct a -tassel as follows. Let be pairwise disjoint and anticomplete paths, each on vertices, and for every , choose a bijection such that are adjacent in if and only if (note that there are only two such bijections). Let be the graph obtained from by adding a vertex such that for every and every , the vertex is adjacent to in if and only if is adjacent to in (see Figure 9).

Note that for every , in the graph , the vertex has a neighbor in and no neighbor among the first and last vertices of . In particular, from the construction and the fact that , it follows that for every , has a neighbor in and no neighbor among the first and last vertices of . Thus, for every , the graph is a -tassel with neck and paths .
Also, since is tasselled, it follows from the choice of that for every there exists such that the -tassel contains every component of . Consequently, since and , it follows that there exists and with such that for every , the -tassel contains every component of .
We now prove that satisfies 2.4. Let be a component of which is not a path. We wish to show that satisfies 2.4b. Note that for every , the -tassel contains , and so there is an isomorphism from to an induced subgraph of . This allows for an application of Lemma 2.2 to and . Since , we deduce that there exists as well as , such that and satisfy Lemma 2.2a and b.
Henceforth, for each , we write
Also, for , we write
The outcomes of Lemma 2.2 can now be rewritten as:
(2) The following hold.
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For every , we have .
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For every component of , there exists such that for every , we have .
On the other hand, since , it follows that every component of is a path on less than vertices. This, combined with the second bullet of \eqrefst:lemtasselagain, yields the following:
(3) For every and every component of , there exists a -segment for which we have .
Let us now finish the proof. Define a map as follows. Let , and for every component of and every , let

See Figure 10. We prove that satisfies Lemma 2.4b. Let be a component of . By \eqrefst:getsegment, we have , and so we have . This, along with the assumption that is a -stretched walk, implies that is a path in . In particular, the restriction of to is an isomorphism from to .
It remains to show that for every , the vertices are adjacent in if and only if are adjacent in . To that end, note that since is an isomorphism from to an induced subgraph of , it follows from the first bullet of \eqrefst:lemtasselagain that is adjacent to in if and only if is adjacent to in . In addition, from the definition of , it follows that is adjacent to in if and only if is adjacent to in . This completes the proof of Lemma 2.4.
We also need the following well-known result.
Lemma 2.6 (See Lemma 2 in [11]).
For all there is a constant with the following property. Let be a -free graph. Let be a collection of pairwise disjoint subsets of , each of cardinality at most , with . Then there are distinct sets which are pairwise anticomplete in .
We can now prove Theorem 2.1, which we restate:
Theorem 2.1.
Let be a finite set of graphs. Then is tasselled if and only if is hassled.
Proof 2.2.
The “if” implication is clear as discussed at the beginning of this section. To prove the “only if” implication, assume that is a finite set of graphs which is tasselled. Let be as in Lemma 2.4. For every , let be as in Lemma 2.6, and let
We prove that for every -free -hassle , there exists a graph such that contains each component of . This will show that is hassled.
Let be the neck of . By the choice of , we may choose a set of pairwise disjoint families of walks of , each of cardinality . It follows that for every , the graph is a -hassle with neck , and with as its set of walks.
Since is tasselled, and by the choice of , for each , we can apply Lemma 2.4 to and , and obtain a graph satisfying Lemma 2.4a and b. Moreover, since and , it follows that there exist and with such that for every , we have . More explicitly, for every component of , one of the following holds.
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is a path.
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For every , there exists and an injective map with , such that for every component of , the restriction of to is an isomorphism from to .
To conclude the proof, it suffices to show that contains every component of . Assume that is a path. Since is a -hassle, it follows that contains a path on vertices. But now we are done because . Consequently, we may assume that is not a path, and so the second bullet above holds for and every . In addition, from and , it follows that there exist and with such that for every , we have .
On the other hand, is a collection of pairwise disjoint subsets of , each of cardinality less than . This, along with the choice of and the assumption that is -free, allows for an application of Lemma 2.6. We deduce that:
(4) There exist for which the sets are pairwise anticomplete in .
Now, let be an enumeration of the components of ; then we have . Let be as in \eqrefst:anticomphassle. By the second bullet above, for each , the restriction of to is an isomorphism from and with . Moreover, by \eqrefst:anticomphassle, the sets are pairwise disjoint and anticomplete in . Hence, is isomorphic to
This completes the proof of Theorem 2.1.
3 A lemma about pairs of sets of vertices
We now begin to make our way to the proof of Theorem 7.1. In particular, here we prove a Ramsey-type result that we will use several times later. We need the following product version of Ramsey’s Theorem.
Theorem 3.1 (Graham, Rothschild and Spencer [9]).
For all , there is a constant with the following property. Let be sets, each of cardinality at least and let be a non-empty set of cardinality at most . Let be a map from the Cartesian product to . Then there exist and with for each , such that for every , we have .
For an induced subgraph of a graph and a vertex , we denote by the set of all neighbors of in , and write . Let be a set and let . An -pair over is a pair of subsets of with . Two -pairs are said to be disjoint if .
Lemma 3.2.
For all and , there is a constant with the following property. Let be a graph. Let be collections of pairwise disjoint -pairs over , each of cardinality at least . Then for every , there exists with such that for all distinct , the following hold.
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We have for all and .
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Either is anticomplete to in for all and , or for every , there exists such that has a neighbor in for every .
Proof 3.3.
We prove that
satisfies the theorem, where is as in Theorem 3.1. Let . For every -pair , fix an enumeration of the elements of ; recall that . For every two -pairs , let
Let be the set of all -by- matrices whose entries are subsets of ; so we have . Consider the product . For every , define such that for all , we have
It follows that for every , are unique, and so the map with is well-defined. This, along with the choice of and Theorem 3.1, implies that there exists with for each , as well as , such that for every , we have and . Moreover, we deduce:
(5) Let be distinct. Then we have for all and
Suppose for a contradiction that there are distinct such that for some and , we have . Then we have . Also, since , we may choose . It follows that is non-empty. But then , a contradiction with the assumption that are disjoint. This proves \eqrefst:disjointramsey.
(6) Let be distinct. Then either is anticomplete to in for all and , or for every , there exists such that has a neighbor in for every .
Note that for all and , we have . If , then is anticomplete to in for all and . Otherwise, one may choose , and so for each , the vertex has a neighbor in for every . This proves \eqrefst:antiornot.
Now the result follows from \eqrefst:disjointramsey and \eqrefst:antiornot. This completes the proof of Lemma 3.2.
4 Blocks
This section collects several results from the literature about “blocks” in -clean graphs of large treewidth. We begin with a couple of definitions. Given a set and , we denote by the power set of and by the set of all -subsets of . Let be a graph. For a collection of paths in , we adopt the notation and . Let . A -block in is a set of at least vertices in such that for every -subset of , there exists a collection of pairwise internally disjoint paths in from to . In addition, we say is strong if the collections can be chosen such that for all distinct -subsets of , we have . In [1], with Abrishami we proved the following:
Theorem 4.1 (Abrishami, Alecu, Chudnovsky, Hajebi and Spirkl [1]).
For all , there is a constant such that every -clean graph of treewidth more than contains a strong -block.
Given a graph and , we say a (strong) -block in is -short if for every -subset of , every path is of length at most . The following shows that large enough short blocks contain strong (and short) “sub-blocks.”
Theorem 4.2.
For all , there is a constant with the following property. Let be a graph and let be a -short -block in . Then every -subset of is a -short strong -block in .
Proof 4.3.
Let , where is as in Lemma 3.2. Let be a -short -block in , and let with . It follows that for every -subset of , there is a collection of pairwise internally disjoint paths in , each of length at most . Let . Then is a collection of pairwise disjoint -pairs over . Thus, the choice of allows for an application of Lemma 3.2 to the collections . We deduce that for every , there exists with , such that for all distinct , the collections and satisfy the outcomes of Lemma 3.2. In particular, for all distinct , it follows from Lemma 3.2a that for every and every , we have . Equivalently, we have . Hence, is a -short strong -block in with respect to . This completes the proof of Theorem 4.2.
Recall that a subdivision of a graph is a graph obtained from by replacing the edges of with pairwise internally disjoint paths of non-zero lengths between the corresponding ends. Let . An -subdivision of is a subdivision of in which the path replacing each edge has length at most . Also, a proper subdivision of is a subdivision of in which the path replacing each edge has length at least two. We need the following immediate corollary of a result of [7]:
Theorem 4.4 (Dvořák [7]).
For every graph and all and , there is a constant with the following property. Let be a graph with no induced subgraph isomorphic to a subdivision of . Assume that contains a -subdivision of as a subgraph. Then contains either or .
Theorem 4.5.
For all , there is a constant with the following property. Let be a -clean graph. Then there is no -short -block in .
Proof 4.6.
Let be as in Theorem 4.4. We show that satisfies the theorem, where is as in Theorem 4.2. Let be a -clean graph; that is, has no induced subgraph isomorphic to a -basic obstruction, and in particular a subdivision of . Suppose for a contradiction that there is a -short -block in . Since , we may choose with . It follows from Theorem 4.2 that is a -short strong -block in . For every -subset of , let be a collection of pairwise internally disjoint paths in from to as in the definition of a strong -block, and fix a path . Now the union of the paths for all forms a subgraph of isomorphic to a -subdivision of . This, together with the choice of and the assumption that is -clean, violates Theorem 4.4. This contradiction completes the proof of Theorem 4.5.
5 Obtaining complete bipartite minor models
The main result of this section, Theorem 5.1, shows that -clean graphs of sufficiently large treewidth contain large complete bipartite minor models in which every “branch set” is a path, such that for every branch set on at least two vertices, each vertex in has neighbors in only a small number of branch sets in the “opposite side” of the bipartition. The exact statement of 5.1 is somewhat technical and involves a number of definitions which we give below.
Let be a graph. For , a -polypath in is a set of pairwise disjoint paths in . Let be a -polypath in . For , we say is -loose if for every , each vertex has neighbors (in ) in fewer than paths in (this is particular implies that a vertex of “large degree” has “most” of its neighbors on just one path; we will use this property extensively in Section 6). Also, for , we say is -fancy if there exists with such that for every and every , is not anticomplete to in . It follows that if is a -polypath in which is -fancy, then has a -minor (see Figure 11).
For , an -cluster in is a pair where with and is an -polypath in , such that every vertex has at least one neighbor in every path . If , say , we also denote the -cluster by the pair . For , we say is -meager if every vertex in has fewer than neighbors in . Again, it is easily seen that if is an -cluster in , then has a -minor (see Figure 11).

Our goal is to prove:
Theorem 5.1.
For all , there is a constant with the following property. Let be a -clean graph of treewidth more than . Then one of the following holds.
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(a)
There exists a -meager -cluster in .
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(b)
There exists a -polypath in which is both -fancy and -loose.
The proof of Theorem 5.1 is split into a number of lemmas. To begin with, we need the following multicolor version of Ramsey’s Theorem for (uniform) hypergraphs.
Theorem 5.2 (Ramsey [13]).
For all , there is a constant with the following property. Let be a set of cardinality at least and let be a non-empty set of cardinality at most . Let be a map. Then there exist and with such that for every , we have .
For graphs (and two colors), it is also convenient to use a quantified version of Ramsey’s classical result (recall that a stable set is a set of pairwise non-adjacent vertices).
Theorem 5.3 (Ramsey [13]).
For all , every graph on at least vertices contains either a clique of cardinality or a stable set of cardinality .
From Theorem 5.2, we deduce:
Lemma 5.4.
For all , there is a constant with the following property. Let be a graph and let be a -polypath in . Then one of the following holds.
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(a)
There exists an -cluster in such that and .
-
(b)
There exists an -loose -polypath in with .
Proof 5.5.
Assume that 5.4a does not hold. Fix an enumeration of the elements of . For every -subset of with , let be the set of all for which there exists a vertex that has a neighbor in for every . It follows that the map is well-defined. Therefore, by the choice of , we can apply Theorem 5.2 and obtain with as well as such that for every , we have . Now we claim that:
(7) is empty.
Suppose not. Then we may choose . Since , it follows that there exist with , and , such that and . For every , define ; it follows that and is the smallest element of . In particular, for every , we have and so , which in turn implies that there is a vertex that has a neighbor in for every . Let and let . Then , is an -polypath in , and every vertex in has a neighbor in every path in . But now is an -cluster in satisfying 5.4a, a contradiction. This proves \eqrefst:Fisempty.
The following two lemmas have similar proofs:
Lemma 5.6.
Let , let be a graph and let be an -cluster in . Then one of the following holds.
-
(a)
contains or .
-
(b)
There exists with such that is a -meager -cluster in .
Proof 5.7.
Suppose that 5.6a does not hold. For every , let be the largest integer in for which there exists a vertex with at least neighbors in . It follows that:
(8) We have .
Suppose not. Let with . Then for every , we may choose and such that and is complete to in . Since , it follows that there exist and such that , and for every , we have . Let . Then is disjoint from and complete to in . Also, since is -free and , it follows from Theorem 5.3 that there are two stable sets and in with . But then is isomorphic to , a contradiction. This proves \eqrefst:fewtL.
Now the result is immediate from \eqrefst:fewtL and the fact that . This completes the proof of Lemma 5.6.
Lemma 5.8.
Let , let be a graph and let be -polypaths in with . Then one of the following holds.
-
(a)
There exists an -cluster in such that either and , or and .
-
(b)
There exist and with such that every vertex in has neighbors in fewer than paths in , and every vertex in has neighbors in fewer than paths in .
Proof 5.9.
Suppose 5.8a does not hold. Let . Then we have
In particular, one may choose with . For every , let be the largest integer in for which there exists a vertex which has neighbors in at least paths in . It follows that:
(9) We have .
Suppose not. Let with . Then for every , we may choose and such that and has a neighbor in every path in . Since , it follows that there exist and such that , , and for every , we have . Let ; so we have . But now is an -cluster in with and , contrary to the assumption that 5.8a does not hold. This proves \eqrefst:polybipround1.
By \eqrefst:polybipround1 and since , we may choose with such that every vertex in has neighbors in fewer than paths in .
Next, for every , let be the largest integer in for which there exists a vertex which has neighbors in at least paths in .
(10) We have .
Suppose not. Let with . Then for every , we may choose and such that and has a neighbor in every path in . Since , it follows that there exist and such that , , and for every , we have . Let ; so we have . But then is an -cluster in with and , contrary to the assumption that 5.8a does not hold. This proves \eqrefst:polybipround2.
The last tool we need is a recent result from [10], the statement of which requires another definition. Let be a graph and let . By a -web in we mean a pair where
-
(W1)
is a -subset of ;
-
(W2)
is a map such that for every -subset of , is a path in with ends ; and
-
(W3)
for all distinct -subsets of , we have .
It follows that has a subgraph isomorphic to a subdivision of if and only if there is a -web in . It is proved in [10] that:
Theorem 5.10 (Hajebi [10]).
For all , there is a constant with the following property. Let be a graph and let be a -web in . Then one of the following holds.
-
(a)
There exists with and a collection of pairwise disjoint -subsets of with such that for every and every , has a neighbor in .
-
(b)
There are disjoint subsets and of with such that for every and every , is not anticomplete to in .
-
(c)
There exists with such that:
-
-
is a stable set in ;
-
-
for any three vertices , is anticomplete to ; and
-
-
for all distinct -subsets of , is anticomplete to .
-
-
We are now in a position to prove our main result in this section, which we restate:
Theorem 5.1.
For all , there is a constant with the following property. Let be a -clean graph of treewidth more than . Then one of the following holds.
-
(a)
There exists a -meager -cluster in .
-
(b)
There exists a -polypath in which is both -fancy and -loose.
Proof 5.2.
Let , let be as in Lemma 5.4 and let be as in Theorem 5.10. We claim that
satisfies the theorem, where is as in Theorem 4.1.
Let be a -clean graph of treewidth more than . Suppose 5.1a does not hold. This, along with the choice of , the assumption that is -clean, and Lemma 5.6, implies that:
(11) There is no -cluster in .
From the choice of and Theorem 4.1, we obtain a strong -block in . In particular, for every -subset of , we may choose a path in from to , such that for all distinct -subsets of , we have . It follows that is a -web in , and so we can apply Theorem 5.10 to . Note that Theorem 5.10a yields an -cluster in , which violates \eqrefst:nobigcluster. Also, Theorem 5.10c implies that contains a proper subdivision of (as an induced subgraph), which in turn contains a proper subdivision of every graph on vertices. But this violates the assumption that is -clean because .
We conclude that Theorem 5.10b holds. In particular, there are two -polypaths in such that for every and every , is not anticomplete to in . Now, from \eqrefst:nobigcluster, the choice of and Lemma 5.4, it follows that there are two -loose -polypaths and in . Furthermore, from \eqrefst:nobigcluster and Lemma 5.8, it follows that there exist and with such that every vertex in has neighbors in fewer than paths in , and every vertex in has neighbors in fewer than paths in . But now is a -polypath in which is both -fancy and -loose, and so 5.1b holds. This completes the proof of Theorem 5.1.
6 Dealing with complete bipartite minor models
Here we take the penultimate step of our proof by showing that:
Theorem 6.1.
For all , there is a constant such that every -clean graph of treewidth more than contains a -hassle.
From Theorem 5.1, we know that every -clean graph of sufficiently large treewidth contains, omitting the corresponding parameters, either a meager cluster or a polypath which is both loose and fancy. So it suffices to prove Theorem 6.1 separately in each of these two cases. First, we show:
Theorem 6.2.
Let and let be a graph. Assume that there is a -meager -cluster in . Then contains a -hassle.
Proof 6.3.
Let be a -meager -cluster in . For every path in , let be the set of all vertices in with a neighbor in . For every and each end of , let be the longest path in containing such that ; then is an end of . It follows that:
(12) For every and every end of , we have .
Suppose for a contradiction that . Then since is -meager, it follows that . Let be the end of other than . Since and every vertex in has a neighbor in , it follows that . Let be the unique neighbor of in , and let . Then , and since is -meager, it follows that . This violates the choice of , and proves \eqrefst:Tsmall.
From \eqrefst:Tsmall, it follows that for every , there are fewer than vertices in with a neighbor among the first or last vertices of . This, along with and the fact that every vertex in has a neighbor in every path in , implies that for every , there is a vertex with a neighbor in and no neighbor among the first and last vertices of . On the other hand, we have . Thus, there exist and with such that for every , the vertex has a neighbor in and no neighbor among the first and last vertices of . But now is a -hassle in with neck and with as its set of walks (each of which is a path in ). This completes the proof of Theorem 6.2.
Handling the second outcome of Theorem 5.1 is more demanding. We prove:
Theorem 6.4.
For all , there is a constant with the following property. Let be a -clean graph. Assume that there exists a -polypath in which is both -fancy and -loose. Then contains a -hassle.
We need to prepare for the proof of Theorem 6.3. Let be a graph, let be distinct and non-adjacent, and let be a collection of pairwise internally disjoint paths in from to . An -slash for in is a path in such that for every , the unique neighbor of in belongs to . Our first lemma says that:
Lemma 6.5.
Let . Let be a graph, let be distinct and non-adjacent, and let be a collection of pairwise internally disjoint paths in from to . Let be an -slash for in . Then one of the following holds.
-
(a)
There exists with and a path , such that:
-
•
is an -slash for in ; and
-
•
there is no path of length at most in from to .
-
•
-
(b)
There exists a collection of pairwise internally disjoint paths in from to , each of length at most .
Proof 6.6.
Let be a maximal set of pairwise internally disjoint paths in from to , each of length at most . In particular, for every , is a path in on at most vertices.
Note that if , then 6.5b holds, as required. Thus, we may assume that , and so . In particular, has at most components, and since the paths in are pairwise internally disjoint, it follows that there are at most paths in for which . This, along with the assumption that and is an -slash for in , implies that there exist with and , as well as a component of , such that for every , the unique neighbor of in belongs to . In other words, is an -slash for . Moreover, we have , and so by the maximality of , there is no path of length at most in from to . Now and satisfy 6.5a, as desired.
The next lemma is the main step in the proof of Theorem 6.3.
Lemma 6.7.
For all , there is a constant with the following property. Let be a graph, let be distinct and non-adjacent, and let be a collection of pairwise internally disjoint paths in from to . Assume that there is an -slash for in . Then one of the following holds.
-
(a)
contains or .
-
(b)
contains a -hassle.
-
(c)
There is a collection of pairwise internally disjoint paths in from to , each of length at most .
Proof 6.8.
Let . Let be as in Lemma 3.2, and let
be as in Lemma 3.2. We will show that
satisfies the lemma. Let be a graph, let be distinct and non-adjacent, let be a collection of pairwise internally disjoint paths in from to , and let be an -slash for in .
Assume that none of the three outcomes of 6.7 hold. We apply Lemma 6.5 to and . In particular, since 6.7c is equivalent to Lemma 6.5b, it follows from the choice of that:
(13) There exists with and a path such that:
-
•
is an -slash for in ; and
-
•
there is no path of length at most in from to .
From now on, let and be as in \eqrefst:getP&W. For every vertex , we denote by the unique path in for which is the unique neighbor of in . Since is an -slash for , it follows that . Let and be the ends of . Since , it follows that one may choose pairwise disjoint -subsets of , such that the following hold.
-
•
For each , there are pairwise disjoint paths in , each on vertices, such that for every :
-
–
contains ; and
-
–
Traversing from to , every vertex in appears before every vertex in , and every vertex in appears before every vertex in .
In particular, traversing from to , every vertex in appears before every vertex in , and every vertex in appears before every vertex in (see Figure 12).
-
–
-
•
For every , traversing from to , every vertex in appears before every vertex in (see Figure 13).


We deduce that:
(14) For every , there exist , and , for which is disjoint from and anticomplete to .
To see this, let
Then are three collections of pairwise disjoint -pairs over , each of cardinality . By the choice of , we can apply Lemma 3.2 to . It follows that there exist , and with such that the collections , and satisfy Lemma 3.2a and 3.2b. In fact, Lemma 3.2a along with the assumption that , implies that for every , every and every , we have . It remains to show that there exist , and , for which is anticomplete to . Suppose not. Note that for every , since is a neighbor of , it follows from the second bullet of \eqrefst:getP&W that is anticomplete to . Consequently, by Lemma 3.2b, there exists such that for every , there exists a vertex which has a neighbor in for every . On the other hand, since , it follows that there exists with . Now, let and let . Then is a -cluster in . This, combined with the choice of , Lemma 5.6, and the assumption that is -free, implies that there exists a -meager -cluster in . But then by Theorem 6.2, contains a -hassle. This violates the assumption that 6.7b does not hold, and so proves \eqrefst:disjointantislash.
Henceforth, for each , let , and be as in \eqrefst:disjointantislash. We write , and . Also, we denote by the ends of , such that traverses the vertices in this order. It follows that traverses the vertices in this order, and are the only vertices among which can be the same (only if ).
Let be fixed. For each , traversing starting at , let be the first vertex in with a neighbor in (note that exists because has a neighbor in ). From \eqrefst:disjointantislash, we know that is disjoint from and anticomplete to ; in particular, for every , has a neighbor in the interior of . It follows that for every , the vertex belongs to the interior of , and there is a path in from to whose interior is contained in .
For every and each , let be the longest path of length at most in containing . It follows that:
(15) For every and each , we have . Consequently, the sets are pairwise disjoint in .
Observe that by the choice of , either , in which case \eqrefst:pathininterior follows immediately from the fact that , or is a path on vertices with . This proves \eqrefst:pathininterior.
Now, for each , let , and let be the walk such that traversing from to , we first traverse the path starting at the end distinct from and stopping at , then we traverse the path from to , and then we traverse starting at (so we have ). In particular, we have . We claim that:
(16) The following hold.
-
•
For all , the walk is -stretched and has a neighbor in .
-
•
For all , the vertex has no neighbors among the first and last vertices of .
-
•
The paths are pairwise disjoint.
-
•
The sets are pairwise disjoint.
The first bullet is immediate from and the observation that both and are paths in containing . Also, the third bullet is trivial, and the fourth is immediate from \eqrefst:pathininterior. It remains to prove the second bullet. Note that by the definition of and , either has a neighbor among the first or the last vertices of , or the first vertices of are contained in , and the last vertices of are contained in . In the former case, the result follows directly from the second bullet of \eqrefst:getP&W, and in the latter case, the result follows from the fact that has exactly one neighbor in and exactly one neighbor in . This proves \eqrefst:Wi’s.
We can now finish the proof. For every , let
From the choice of and the third bullet of \eqrefst:Wi’s, it follows that are collections of pairwise disjoint -pairs over , each of cardinality . The choice of in turn allows for an application of Lemma 3.2 to . In particular, Lemma 3.2 a implies that for every , there exists such that and are disjoint for all distinct . This, combined with the third and the fourth bullet of \eqrefst:Wi’s, implies that are pairwise disjoint. But now by the first two bullets of \eqrefst:Wi’s, the subgraph of induced on is a -hassle with neck and walks , contradicting the assumption that 6.7b does not hold. This completes the proof of Lemma 6.7.
From Lemma 6.7, we deduce the following:
Lemma 6.9.
For all , there is a constant with the following property. Let be a graph and let be a -loose polypath in . Let be distinct and non-adjacent, and let be a collection of pairwise internally disjoint paths in from to with . Then one of the following holds.
-
(a)
contains or .
-
(b)
contains a -hassle.
-
(c)
There exists a collection of pairwise internally disjoint paths in from to , each of length at most .
Proof 6.10.
Let be as in Lemma 6.7. We show that
satisfies the lemma. Let such that . Then has at most two neighbors in . Also, since is -loose, it follows that has neighbors in at most paths in . This, along with the fact that , implies that there exists with and a path , such that for every , the unique neighbor of in belongs to . In particular, we have because are distinct (hence disjoint) and . Since has one or two components (depending on whether or not), it follows that there exist with as well as a component of , such that for every , the unique neighbor of in belongs to . Therefore, is a collection of pairwise internally disjoint paths in from to , and is an -slash for in . Now the result follows from Lemma 6.7 applied to .
We can now restate and prove Theorem 6.3:
Theorem 6.3.
For all , there is a constant with the following property. Let be a -clean graph. Assume that there exists a -polypath in which is both -fancy and -loose. Then there is a -hassle in .
Proof 6.4.
Let be as in Theorem 4.5. Let be as in Lemma 6.9. We define
where is as in Theorem 4.1. Let be a -clean graph and let be a -polypath in which is both -fancy and -loose. Suppose for a contradiction that does not contain a -hassle.
Let . Then is a -clean graph which has a -minor. It follows that has treewidth at least , which along with Theorem 4.1 implies that there is a strong -block in . In particular, we may choose a -subset of such that for every -subset of , there exists a collection of pairwise internally disjoint paths in from to . Since and since is -free, it follows from Theorem 5.3 that there exists a stable set in of cardinality .
Now, fix a -subset of . Then are non-adjacent in , and so by the choice of , we can apply Lemma 6.9 to and . Note that Lemma 6.9a violates the assumption that is -clean, and Lemma 6.9b violates the assumption that contains no -hassle. Thus, Lemma 6.9c holds, that is, there exists a collection of pairwise internally disjoint paths in from to , each of length at most . But now is a -short (not necessarily strong) -block in . Combined with the choice of and Theorem 4.5, this violates the assumption that is -clean, hence completing the proof of Theorem 6.3.
Theorem 6.1.
For all , there is a constant such that every -clean graph of treewidth more than contains a -hassle.
Proof 6.2.
Let and let . Let be as in Theorem 6.3. Let
where is as in Theorem 5.1. Let be a -clean graph of treewidth more than . It follows from Theorem 5.1 that either there exists a -meager -cluster in , or there is -polypath in which is both -fancy and -loose. In the former case, by Theorem 6.2, contains a -hassle. Also, in the latter case, the choice of along with Theorem 6.3 implies that contains a -hassle. This completes the proof of Theorem 6.1.
7 Part assembly
Finally, let us bring everything together and prove Theorem 1.8. First, from Theorem 6.1, we deduce that:
Theorem 7.1.
Let be a finite set of graphs which is hassled. Then the class of all -free graphs is clean.
Proof 7.2.
Let be a finite set of graphs which is hassled. In order to show that the class of all -free graphs is clean, it is enough to prove, for every , that there is a constant such that every -clean graph of treewidth more that contains some graph .
We begin with setting the value of . Since is hassled, it follows that for every , there is a constant such that every -free -hassle contains all components of some graph in . Let be as in Theorem 6.1, and let be as in Lemma 2.6. Define
Let be a -clean graph of treewidth more than . Since , by Theorem 6.1, there is an induced subgraph of which is a -hassle. This, combined with the assumption that is hassled and is -free, implies that there exists with such that contains all components of some graph in .
Let be maximum such that there are pairwise disjoint subsets of , each of cardinality at most , such that for every there exists such that contains every component of . We claim that:
(17) .
Suppose not. Let . Then we have . Thus, by Theorem 6.1, there is an induced subgraph of which is a -hassle. Since is hassled and is -free, it follows that there exists with and such that contains every component of . But this violates the maximality of , and so proves \eqrefst:manycopies.
By \eqrefst:manycopies, we can choose pairwise disjoint subsets of , each of cardinality at most , such that for every , there exists such that contains every component of . Since , it follows that there exists a graph , as well as with , such that for every , we have . Also, since is -free, from the choice of , it follows that there exist such that are pairwise anticomplete in .
Let be the components of . Then , and for every , there exists such that is isomorphic to . Now is isomorphic to , as desired.
8 Connection to unavoidable binary strings
We conclude the paper with a brief discussion around explicit constructions of tasselled families. There is one particularly nice example of a tasselled family, described in the following corollary of Theorem 1.6 (we omit the proof as it is straightforward).
Corollary 8.1.
Given , let be the set of all strands such that the path of has vertices in this order, and the neck of is non-adjacent to and adjacent to . (Thus, contains distinct graphs, corresponding to all possible ways for to be adjacent to .) Then is tasselled.
While a full description of all tasselled families remains unknown, a promising approach is to attempt to model the “tasselled” property word-combinatorially, and then use results from the literature on formal languages in order to gain insight into the graph problem.
One way to view a strand is as a binary string denoting neighbors of the neck in the path. More explicitly, for a strand with neck and path , if are the vertices of in order, then we can define the binary string where if , and otherwise (note that the string is uniquely defined up to reversal). Moreover, every strand of a -tassel with neck corresponds to a binary string (again, unique up to reversal) which starts and ends with at least zeroes, but which is not an all-zero string; let us say such a string is -padded.
For a graph , we say a vertex is a neck for if every component of is a path, and we denote by the set of all strings where is a component of . It is easy to see that:
Theorem 8.2.
Let be a set of graphs. Then is tasselled if and only if there exists such that, for every -padded string , there is a graph with the following property. For every component of , one may choose a neck of such that for every string , the string contains either or its reverse as a consecutive substring.
There is an interesting special case of this where every graph in is a -tassel for some . Note that in this case, every is connected and the set has cardinality one for each neck of . So the question of whether is tasselled translates into:
Question 8.3.
For which sets of strings is it true that there is a constant such that every -padded string contains either a string in or the reverse of a string in as a consecutive substring?
If the above holds for some fixed , let us call such a set a -unavoidable language. In view of our aim, it is of particular interest to identify the minimal -unavoidable languages (with respect to the inclusion). In particular, the only sets of cardinality one which are -unavoidable for some are of the form (up to reversal) and , and one may observe that this matches the outcome of Theorem 1.5. However, in general, larger -unavoidable languages seem harder to describe; for example, the following is -unavoidable:
On the other hand, for a finite set of strings, it is finitely testable if is -unavoidable for some . To see this, note that if there is a -padded string which does not contain any string in nor the reversal of a string in , then, assuming to be the length of the longest string in , one may choose such that and has length at most (which follows by noticing that a repetition of a consecutive substring of length in not containing the first or last bits can be used to shorten ).
The following, closely related question, has been studied before:
Question 8.4.
For which sets of strings is it true that every sufficiently long string contains a string in as a consecutive substring?
These sets are called unavoidable languages (see, for example [15, 16]). In particular, in [4, 5] it is shown that every minimal unavoidable language is finite. This is not the case for -unavoidable languages; for example, is -unavoidable. However, if we omit the string for some , then avoids . It would be interesting to consider which results on unavoidable languages generalize to -unavoidable languages.
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[balec]
Bogdan Alecu
School of Computing, University of Leeds
Leeds, UK
{authorinfo}[mchud]
Maria Chudnovsky
Princeton University
Princeton, New Jersey, USA
{authorinfo}[laci]
Sepehr Hajebi
Department of Combinatorics and Optimization, University of Waterloo
Waterloo, Ontario, Canada
{authorinfo}[andy]
Sophie Spirkl
Department of Combinatorics and Optimization, University of Waterloo
Waterloo, Ontario, Canada