The ErdősHajnal conjecture for caterpillars and their complements
Abstract
The celebrated ErdősHajnal conjecture states that for every proper hereditary graph class there exists a constant such that every graph contains a clique or an independent set of size . Recently, there has been a growing interest in the symmetrized variant of this conjecture, where one additionally requires to be closed under complementation.
We show that any hereditary graph class that is closed under complementation and excludes a fixed caterpillar as an induced subgraph satisfies the ErdősHajnal conjecture. Here, a caterpillar is a tree whose vertices of degree at least three lie on a single path (i.e., our caterpillars may have arbitrarily long legs). In fact, we prove a stronger property of such graph classes, called in the literature the strong ErdősHajnal property: for every such graph class , there exists a constant such that every graph contains two disjoint sets of size at least each so that either all edges between and are present in , or none of them. This result significantly extends the family of graph classes for which we know that the strong ErdősHajnal property holds; for graph classes excluding a graph and its complement it was previously known only for paths [3] and hooks (i.e., paths with an additional pendant vertex at third vertex of the path) [4].
1 Introduction
Classical bounds on the Ramsey numbers due to Erdős and Szekeres [8] and Erdős [7] imply that every graph on vertices contains a clique or an independent set of size , and this bound is tight up to the constant factor. A longstanding conjecture by Erdős and Hajnal [9] states that this relation drastically changes if attention is restricted to a fixed hereditary graph class .
A graph class is called hereditary if it is closed under taking induced subgraphs. We say that a graph class has the ErdősHajnal property if there exists a constant such that every vertex graph contains a clique or an independent set of size at least . In 1989, Erdős and Hajnal [10] proved a weaker bound of and conjectured that every proper hereditary graph class has the ErdősHajnal property.
Up to today, this conjecture remains widely open, and the ErdősHajnal property remains proved only for a limited number of graph classes . We say a graph is free if does not contain the graph as an induced subgraph. It is known, for example, that the class of bullfree graphs has the ErdősHajnal property [6]. Yet, it is unknown whether the class of free graphs or the class of free graphs has the ErdősHajnal property. The substitution procedure by Alon, Pach, and Solymosi [1] provides more examples of graph classes with the ErdősHajnal property. For a full overview, we refer to a survey of Chudnovsky [5].
Substantial progress has been obtained by considering a weaker variant of the conjecture that asserts the ErdősHajnal property for every proper hereditary graph class that is additionally closed under complementation. In this line of direction, a breakthrough paper of Bousquet, Lagoutte, and Thomassé [3] proved the ErdősHajnal property for graph classes excluding a fixed path and its complement. Bonamy, Bousquet, and Thomassé extended this result to graph classes closed under complementation that exclude long holes [2], while Choromanski, Falik, Patel and the two authors extended the above result to hooks: paths with an additional pendant edge attached to the third vertex of the path [4]. Note that both the class of bullfree graphs and the class of free graphs are closed under complementation.
In fact, all the aforementioned results, except for the result for bullfree graphs, prove a much stronger property of a graph class in question, called the strong ErdősHajnal property. A graph class has the strong ErdősHajnal property if there exists a constant such that every vertex graph contains two disjoint sets of size at least each, such that and are fully adjacent (there are all possible edges between them) or antiadjacent (there are no edges between them). A simple inductive argument shows that a hereditary graph class with the strong ErdősHajnal property has also the (standard) ErdősHajnal property (cf. [12]). Further graph classes known to obey the strong ErdősHajnal property include line graphs [4] and certain subclasses of perfect graphs [4, 15].
Not all proper hereditary graph classes have the strong ErdősHajnal property: a certain poset construction shows that this property is false for comparability graphs (and hence also for perfect graphs) [11], and a simple randomized construction of [4] shows that if one considers the class of free graphs for a finite set (i.e., graphs that do not contain any graph in as an induced subgraph), then the strong ErdősHajnal property is false unless contains a forest and a complement of a forest. This, together with the positive results of [3, 4], motivates the following conjecture:
Conjecture 1.1.
For every tree , the class of graphs excluding and its complement as induced subgraphs has the strong ErdősHajnal property.
In this work we prove this conjecture for being a caterpillar which is a tree in which all vertices of degree at least three lie on a single path (i.e., our caterpillars may have long legs).
Theorem 1.2.
For every caterpillar , the class of graphs excluding and the complement of as induced subgraphs has the strong ErdősHajnal property.
2 Preliminaries
We use standard graph theoretic notation. All graphs in this paper are finite and simple. Let be a graph, let such that . By we denote the subgraph in induced by the subset . The set of vertices that have a neighbour in is denoted by , and denotes the closed neighbourhood of . If we abbreviate and write and , respectively. We also write for . The set of edges with one endpoint in and one endpoint in is denoted by . If , we say that and are antiadjacent, and if , then and are fully adjacent.
For a constant , an pair in a graph is a pair of two disjoint antiadjacent sets with . Recall that a caterpillar is a tree in which all vertices of degree at least three lie on a single path. For integers or we define a caterpillar as follows. Let be a path of length , say on vertex set , and let be a collection of pairwise vertexdisjoint paths and such that for all . The graph is formed by taking the disjoint union of and the paths , and by adding the edges , for all , , where is one of the two endpoints of . The graph , for example, is the graph consisting of a single vertex, while is a path on vertices. Note that every caterpillar is an induced subgraph of for sufficiently large parameters .
We need the following two auxiliary results. The first one by Fox and Sudakov [13] states that if a graph does not contain a graph as an induced subgraph then it contains either a very dense or a very sparse induced subgraph on a linear number of vertices.
Theorem 2.1 ([13]).
For every integer and every constant there exists a constant such that every vertex graph satisfies the following.

contains every graph on vertices as an induced subgraph;

there exists of size at least such that either or the complement of has maximum degree at most .
We need the following pathgrowing argument of Bousquet, Lagoutte, and Thomassé [3]. Since the paper [3] states weaker bounds on the constants, we provide the proof for completeness.
Lemma 2.2 ([3]).
Let be integers. Then, for every connected graph of maximum degree at most and more than vertices, and for every , either contains a pair of disjoint antiadjacent vertex sets of size at least each, or an induced path on vertices with one endpoint in .
Proof.
Assume that the graph does not contain the pair as in the lemma statement. This implies that for every set of size at least , the largest connected component of has more than vertices.
We construct an induced path on vertices inductively, starting with . Given an integer and vertices that induce a path in , we define and to be the vertex set of the largest connected component of . Note that , and thus . Consequently, . In the first step, we pick to be a neighbor of that is adjacent to a vertex of ; such a vertex exists by the connectivity of . In subsequent steps, we maintain the invariant that is adjacent to a vertex of . This allows us always to choose the next vertex within the neighborhood of , completing the construction.
Since , the construction process does not stop before defining the vertex , which gives the desired vertex path starting in . ∎
3 Proof of main theorem
Let be a caterpillar and let , , be such that is an induced subgraph of . Let be a large enough integer and let be small enough. (We will determine and later.) Let be a graph that does not contain nor the complement of as an induced subgraph. We show that contains an pair where is small enough. By Theorem 2.1, there exists a constant and a set of size at least such that either or the complement of has maximum degree at most . Let be equal to or the complement of , so that has maximum degree at most .
We partition arbitrarily the vertex set into parts such that for all . Now, either contains an pair, in which case we are done, or the following holds:

for all and all , and

there are no two sets , , for some , such that , and such that forms an antiadjacent pair.
By discarding at most vertices of we may in fact assume that is an equipartition, i.e. that for all and thus . Call a graph coloured if the vertex set of is partitioned into (not necessarily independent) sets ; and call an coloured graph clean if it satisfies (a) and (b).
We prove by induction on the maximum degree that an coloured clean graph contains the spine of a caterpillar which can be extended to an induced copy of , as long as is big enough and is small enough. To be precise, we need some notation.
For a set is called an bud if is connected and for some . A family of subsets of is called colourcompatible if for all there is an index such that and the indices for are pairwise distinct.
For integers , , and a constant , an junior caterpillar (in the underlying coloured graph ) is a pair , where is an induced path in on vertices and is a family of pairwise antiadjacent buds such that the family
is colourcompatible and for all , . Note that colourcompatibility implies in particular that the ’s are pairwise disjoint and that for all . See Figure 2 for an illustration.
We prove the following.
Lemma 3.1.
For all integers and there exist a constant and an integer such that every coloured clean graph contains an junior caterpillar.
Before proving the lemma let us show how this finishes the proof of ?THM? LABEL:thm:main. By choosing and , where and are the constants given by ?THM? LABEL:lem:main, we may assume that contains an junior caterpillar . (We may fix at this point, but we do need further restrictions on .)
Note that by definition, is an induced path and is a family of pairwise disjoint buds. Let for be an arbitrary enumeration of the buds and let be the unique vertex of in . For every , let be a minimal subset of with the following properties: is connected, , and
We claim that exists for all . Indeed, since is an bud we have that for some . Hence, we can pick vertex by vertex, starting with a neighbour of , keeping connected, until , which is possible if is chosen such that
(3.1) 
We now show that is not too big either. First note that by the minimality of and the bound on the maximum degree, we have that
Now is an antiadjacent pair. Hence, we can assume that for all . It follows that for every we have that
(3.2) 
Therefore, for all ,
where we used that by the bound on the maximum degree, and that Hence, the sets defined by
satisfy
(3.3) 
using (3.2) as well. By definition, and is antiadjacent to for all .
Now, for every in decreasing order, we define an induced path on vertices with and endpoint as follows. First, we define
Since and are antiadjacent for and since , we have that and, consequently, is connected. Note that
so that
by (3.3). Finally, since the maximum degree in is at most , we can apply ?THM? LABEL:lem:blt to to find an induced path on vertices rooted in .
From the construction, we infer that every path is induced with an endpoint in , is the unique vertex of , and and are antiadjacent for every . Consequently, induces the caterpillar in .
This finishes the proof of ?THM? LABEL:thm:main. It remains to prove ?THM? LABEL:lem:main.
We remark that the above proof does not use the assumptions on the parts of a junior caterpillar being colourcompatible; colourcompatibility will be used only in the inductive step in the proof of ?THM? LABEL:lem:main. In that proof we will iteratively find a more general structure which we call ferns. If such a fern is “deep” enough then it is easy to see that it contains the desired junior caterpillar. Otherwise, we construct the junior caterpillar inductively.
Let us recall some notation. A rooted tree is a tree with one vertex specified as the root. We refer to vertices of such trees as nodes as the rooted trees in this proof are auxiliary structures outside of the graph we are working with. For a node , the subtree of rooted at is denoted by . If is the neighbour of in on the path from the root to then is called the parent of whereas is called a child of . The height of a rooted tree is the maximum number of edges on a path from the root to a leaf.
For a constant and an integer , an fern (in an underlying coloured graph ) is a pair where is a rooted tree in which every internal node has exactly children, and where is a colourcompatible family of buds, one for each node in , that are pairwise antiadjacent unless is a parent of in in which case , i.e. every has a neighbour in . For a given fern , we refer to as , to as , and we set to be the vertex set on which the fern lives. The height of a fern is the height of the tree . See Figure 3 for an illustration.
Finally, we say that a fern grows on a set if is disjoint from every bud , is antiadjacent to every bud for nonroot nodes , and for the root of .
It is easy to observe that a high enough fern contains a junior caterpillar.
Lemma 3.2.
Let and be integers, and let be a constant. If an coloured graph contains an fern of height at least , then contains an junior caterpillar.
Proof.
Let be an fern of height at least . By the assumption on height, there exists a path in , say on nodes , such that is the root of and is a child of for all . Iteratively, for every in increasing order, pick such that for ; this is always possible since . Observe that the adjacency conditions on a fern imply that induce a path in with vertices in this order. To complete the construction of a junior caterpillar, for every , define buds , , to be the buds for children of except for . Finally, observe that the adjacency conditions on a fern imply that are pairwise antiadjacent with , while colourcompatibility of implies colourcompatibility of . ∎
We are now ready to prove our main technical lemma.
Proof of ?THM? LABEL:lem:main.
We prove the statement by induction on . Fix integers and and, if , assume that the statement is true for and with parameters and . Let be an coloured clean graph where
(3.4)  
(3.5) 
and where, for brevity, we set in the case . Assume that are the colour classes of .
We now describe an iterative process in which we find a family of ferns in , increasing the total number of buds by 1 in each step. For , we define a set of active colours, subsets for all , and a family of ferns, each one growing on for some . See Figure 4 for an illustration.
In each step, let be the set of all those ferns that grow on , and let be the collection of all buds in all ferns of .
The procedure.
We set , , and for every .
Now assume that for we have sets , ,
and a family of ferns as above.
We stop the process if for all we have that , or if .
Otherwise, we define , , and as follows. Pick an index as follows. If there exists a set of size , pick ; otherwise, pick so that . Now, let be the vertex set of the largest connected component of . Let be a minimal subset such that is connected and there exists an index with . Set
Finally, set for a fern defined as follows. For , suppose that are the ferns in (growing on ), and suppose that for , and let be the root of . We create a new node to be the root of , associate with it the set , and append all trees as subtrees rooted at the children of ; that is, every node becomes a child of . We then define .
Note that for and for the initial choices , , and , the following hold trivially.

.

Every is indeed an fern, growing on for some .

and is colourcompatible.

is antiadjacent to all buds of ferns .

and are antiadjacent for all distinct ferns .

, and for at most one family .
Assume now that for we have sets , , and a family of ferns that satisfy the conditions 1.6. We claim that, unless the process stops, the procedure is welldefined and that the properties 1.6. hold for , , and .
Note that since otherwise the process would stop (since ). Furthermore, we can assume that for some by the stopping condition, and hence the index is welldefined. Note also that there is at most one family with by 6., and that the index picks this unique family if it exists. By 1., , and hence, , as otherwise contains an antiadjacent pair , both and of size at least , contradicting being clean. This then implies that for all as otherwise would be an antiadjacent pair contradicting being clean. Consequently, since , the set as defined exists. It follows that , and are welldefined.
By definition, . Furthermore, we have that for all , by minimality of and the upper bound on for every (by part (a) of the definition of being clean). Hence, for all , , and thus 1. holds.
We now show that is an fern. By definition, is a rooted tree. If then consists of a single node (which is the root). Otherwise, and the trees are appended as subtrees rooted at the children of . Thus, the degree of is in . The family is colourcompatible since and is colourcompatible by 3. Furthermore, is an bud since . Note that is antiadjacent to for any bud unless is a bud corresponding to the root of for some . This is because and each grows on . For the same reason, for every corresponding to the root of for some .
We now claim that grows on . Indeed, is disjoint from since is colourcompatible by 3. By definition, . Finally, is antiadjacent to for any since and by 4. This proves 2. for . For 3. note that and that colourcompatibility of follows from for all , from , and from colourcompatibility of . For 4., note that is antiadjacent for all , , by 4. for . By definition, is antiadjacent to all for and 4. follows. Similarly, we only need to check 5. for . But this follows since and from 4. for . Finally, in step we add 1 fern to the family growing on (and to no other family), and the choice of guarantees that there is at most one family , , with exactly ferns.
Let us consider the case when first. Note that for every , every family has size 0 or 1, for . So the process only stops after step when and the process produces a fern in which the auxiliary tree is a path of length . That is, the fern is an fern of height at least , which contains an junior caterpillar by ?THM? LABEL:lem:fern2cat. This proves the statement for .
For , assume now that the process stops at step for some , that is, or for all . If there exists a fern of height at least then contains an junior caterpillar by ?THM? LABEL:lem:fern2cat, and we are done. So we may assume that every fern is of height less than , and thus it contains at most buds. Since every set is of size at most , and the total number of buds in is exactly , we have that Now since as well, we infer that by (3.4).
Pick any set of size exactly and consider an coloured graph with colour classes , . Since is clean and , we have that is clean by (3.5). By induction, there exists an junior caterpillar in . Let and let be the family of buds so that is colourcompatible in and so that is the set of private buds for vertex (in ). Note that an bud in is an bud in , since . Note also that colourcompatibility in implies colourcompatibility in .
For , let be such that , let be any fern in (which exists as ), and let be the bud associated with the root node of . Note that is antiadjacent to all for , by Property 4, and hence it is anti adjacent to all for , all . Furthermore, , and is antiadjacent to for any by Property 5. Hence, is an junior caterpillar in , where . This finishes the proof of the lemma. ∎
4 Conclusions
We would like to conclude with a number of future research directions.
Our result inscribes into a recent trend in studying the (strong) ErdősHajnal property for graph classes closed under complementation. Here, the progress is triggered by the result of Fox and Sudakov [13], see Theorem 2.1, which allows to concentrate on sparse graphs. For a fixed graph , let be the class of graphs that does not contain nor the complement of as an induced subgraph. As shown in [4], a simple randomized construction shows that does not have the strong ErdősHajnal property unless is a forest or the complement of a forest. In this work, we significantly extend the family of for which is known to have the strong ErdősHajnal property from paths [3] and hooks [4] to arbitrary caterpillars. We believe that Conjecture 1.1 is true in general. The natural next step would be to prove the conjecture for the tree depicted in Figure 1.
To prove the (strong) ErdősHajnal property for graph classes closed under complementation similar techniques are used as for proving boundedness. A graph class is called bounded if for every there exists a constant such that every graph with chromatic number at least contains a complete subgraph on vertices. The wellknown GyárfásSumner conjecture [14, 18] asserts that for every tree , the class of free graphs is bounded. The proof of Bousquet, Lagoutte, and Thomassé [3] can be considered as the ErdősHajnal analogue of the proof of Gyárfás that graphs excluding a fixed path are bounded.
In the world of boundedness, an old result of Scott [16] proves that if one excludes a fixed tree and all its subdivisions, then one obtains a bounded graph class. A weakening of Conjecture 1.1 would be the following.
Conjecture 4.1.
For every fixed tree , the set of graphs such that neither nor the complement of contains a subdivision of as an induced subgraph, has the strong ErdősHajnal property.
The proof in [16] considers two cases, depending on whether constantradius balls have small or large chromatic number. In the world of the ErdősHajnal property, this translates to whether the following property is true:
For some radius and a constant , every set of size at least