Colouring Square-Free Graphs without Long Induced Paths

The complexity of Colouring is fully understood for H-free graphs, but there are still major complexity gaps if two induced subgraphs H_1 and H_2 are forbidden. Let H_1 be the s-vertex cycle C_s and H_2 be the t-vertex path P_t. We show that Colouring is polynomial-time solvable for s=4 and t≤ 6, strengthening several known results. Our main approach is to initiate a study into the boundedness of the clique-width of atoms (graphs with no clique cutset) of a hereditary graph class. We first show that the classifications of boundedness of clique-width of H-free graphs and H-free atoms coincide. We then show that this is not the case if two graphs are forbidden: we prove that (C_4,P_6)-free atoms have clique-width at most 18. Our key proof ingredients are a divide-and-conquer approach for bounding the clique-width of a subclass of C_4-free graphs and the construction of a new bound on the clique-width for (general) graphs in terms of the clique-width of recursively defined subgraphs induced by homogeneous pairs and triples of sets. As a complementary result we prove that Colouring is -complete for s=4 and t≥ 9, which is the first hardness result on Colouring for (C_4,P_t)-free graphs. Combining our new results with known results leads to an almost complete dichotomy for restricted to (C_s,P_t)-free graphs.

There are no comments yet.

Authors

• 20 publications
• 9 publications
• 43 publications
• Graph Isomorphism for (H_1,H_2)-free Graphs: An Almost Complete Dichotomy

We consider the Graph Isomorphism problem for classes of graphs characte...
11/29/2018 ∙ by Marthe Bonamy, et al. ∙ 0

• Clique-Width: Harnessing the Power of Atoms

Many NP-complete graph problems are polynomially solvable on graph class...
06/05/2020 ∙ by Konrad K. Dabrowski, et al. ∙ 0

• Hereditary Graph Classes: When the Complexities of Colouring and Clique Cover Coincide

A graph is (H_1,H_2)-free for a pair of graphs H_1,H_2 if it contains no...
07/22/2016 ∙ by Alexandre Blanché, et al. ∙ 0

• Clique-width and Well-Quasi-Ordering of Triangle-Free Graph Classes

Daligault, Rao and Thomassé asked whether every hereditary graph class t...
11/23/2017 ∙ by Konrad K. Dabrowski, et al. ∙ 0

• Coloring even-hole-free graphs with no star cutset

A hole is a chordless cycle of length at least 4. A graph is even-hole-f...
05/04/2018 ∙ by Ngoc Khang Le, et al. ∙ 0

• Wheel-free graphs with no induced five-vertex path

A 4-wheel is the graph consisting of a chordless cycle on four vertices ...
04/03/2020 ∙ by Arnab Char, et al. ∙ 0

• List k-Colouring P_t-Free Graphs with No Induced 1-Subdivision of K_1,s: a Mim-width Perspective

A colouring of a graph G=(V,E) is a mapping c V→{1,2,…} such that c(u)≠ ...
08/03/2020 ∙ by Nick Brettell, et al. ∙ 0

This week in AI

Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.

1 Introduction

Graph colouring has been a popular and extensively studied concept in computer science and mathematics since its introduction as a map colouring problem more than 150 years ago due to its many application areas crossing disciplinary boundaries and to its use as a benchmark problem in research into computational hardness. The corresponding decision problem, Colouring, is to decide, for a given graph and integer , if admits a -colouring, that is, a mapping such that whenever . Unless , it is not possible to solve Colouring in polynomial time for general graphs, not even if the number of colours is limited to 3 [45]. To get a better understanding of the borderline between tractable and intractable instances of Colouring, it is natural to restrict the input to some special graph class. Hereditary graph classes, which are classes of graphs closed under vertex deletion, provide a unified framework for a large collection of well-known graph classes. It is readily seen that a graph class is hereditary if and only if it can be characterized by a unique set of minimal forbidden induced subgraphs. Graphs with no induced subgraph isomorphic to a graph in a set are called -free.

Over the years, the study of Colouring for hereditary graph classes has evolved into a deep area of research in theoretical computer science and discrete mathematics (see, for example, [8, 29, 38, 53]). One of the best-known results is the classical result of Grötschel, Lovász, and Schrijver [31], who showed that Colouring is polynomial-time solvable for perfect graphs. Faster, even linear-time, algorithms are known for subclasses of perfect graphs, such as chordal graphs, bipartite graphs, interval graphs, and comparability graphs; see for example [29]. All these classes are characterized by infinitely many minimal forbidden induced subgraphs.

Král’, Kratochvíl, Tuza, and Woeginger [43] initiated a systematic study into the computational complexity of Colouring restricted to hereditary graph classes characterized by a finite number of minimal forbidden induced subgraphs. In particular they gave a complete classification of the complexity of Colouring for the case where consists of a single graph .

Theorem 1 ([43]).

If is an induced subgraph of or of , then colouring restricted to -free graphs is polynomial-time solvable, otherwise it is NP-complete.

Theorem 1 led to two natural directions for further research:

1. Is it possible to obtain a dichotomy for Colouring on -free graphs if the number of colours is fixed (that is, no longer belongs to the input)?

2. Is it possible to obtain a dichotomy for Colouring on -free graphs if has size 2?

We briefly discuss known results for both directions below and refer to [26] for a detailed survey. Let and denote the cycle on vertices and path on vertices, respectively. We start with the first question. If is fixed, then we denote the problem by -Colouring. It is known that for every , the -Colouring problem on -free graphs is NP-complete whenever contains a cycle [22] or an induced claw [35, 44]. Therefore, only the case when is a disjoint union of paths remains. In particular, the situation where has been thoroughly studied. On the positive side, -Colouring on -free graphs [5], -Colouring on -free graphs [11, 12] and -Colouring on -free graphs for any  [33] are polynomial-time solvable. On the negative side, Huang [36] proved NP-completeness for (, ) and for (, ). The case (, ) remains open, although some partial results are known [13].

In this paper we focus on the second question, that is, we restrict the input of Colouring to -free graphs for . For two graphs and , we write for the disjoint union of two vertex-disjoint graphs and , and for the disjoint union of copies of . As a starting point, Král’, Kratochvíl, Tuza, and Woeginger [43] identified the following three main sources of NP-completeness:

• both and contain a claw;

• both and contain a cycle; and

• both and contain an induced subgraph from the set .

They also showed additional NP-completeness results by mixing the three types. Since then numerous papers [3, 9, 10, 17, 18, 20, 32, 34, 36, 40, 43, 46, 50, 51, 52, 56] have been devoted to this problem, but despite all these efforts the complexity classification for Colouring on -free graphs is still far from complete, and even dealing with specific pairs may require substantial work.

One of the “mixed” results obtained in [43] is that Colouring is NP-complete for -free graphs when and . This, together with the well-known result that Colouring can be solved in linear time for -free graphs (see also Theorem 1) implies the following dichotomy.

Theorem 2 ([43]).

Let be a fixed integer. Then Colouring for -free graphs is polynomial-time solvable when and NP-complete when .

Theorem 2 raises the natural question: what is the complexity of Colouring on -free graphs when ?

For , Huang, Johnson and Paulusma [37] proved that 4-Colouring, and thus Colouring, is NP-complete for -free graphs. A result of Brandstädt, Klembt and Mahfud [7] implies that Colouring is polynomial-time solvable for -free graphs.

For , it is only known that Colouring is polynomial-time solvable for -free graphs [50]. This is unless we fix the number of colours: for every and , it is known that -Colouring is polynomial-time solvable for -free graphs for every , and thus for -free graphs [28] (take ). The underlying reason for this is a result of Atminas, Lozin and Razgon [2], who proved that every -free graph either has small treewidth or contains a large biclique as a subgraph. Then Ramsey arguments can be used but only if the number of colours  is fixed. The result for and fixed is in contrast to the result of [37] that for all and , there exists a constant  such that -Colouring is NP-complete even for -free graphs.

Our Main Results

We show, in Section 4, that Colouring is polynomial-time solvable for -free graphs. The class of -free graphs generalizes the classes of split graphs (or equivalently, -free graphs) and pseudosplit graphs (or equivalently, -free graphs). The case of -free graphs was explicitly mentioned as a natural case to consider in [26]. Our result unifies several previous results on colouring -free graphs, namely: the polynomial-time solvability of Colouring for -free graphs [50]; the polynomial-time solvability of -Colouring for -free graphs for every  [28]; and the recent -approximation algorithm for Colouring for -free graphs [24]. It also complements a recent result of Karthick and Maffray [41] who gave tight linear upper bounds of the chromatic number of a -free graph in terms of its clique number and maximum degree that strengthen a similar bound given in [24].

It was not previously known if there exists an integer such that Colouring is NP-complete for -free graphs. In Section 5 we complement our positive result of Section 4 by giving an affirmative answer to this question: already the value makes the problem NP-complete.

Our Methodology

The general research goal of our paper is to increase, in a systematic way, our insights in the computational hardness of Colouring by developing new techniques. In particular we aim to narrow the complexity gaps between the hard and easy cases. Clique-width is a well-known width parameter and having bounded clique-width is often the underlying reason for a large collection of NP-complete problems, including Colouring, to become polynomial-time solvable on a special graph class; this follows from results of [14, 23, 42, 54, 55]. For this reason we want to use clique-width to solve Colouring for -free graphs, However, the class of -free graphs has unbounded clique-width, as it contains the class of split graphs, or equivalently, -free graphs, which may have arbitrarily large clique-width [49].

To overcome this obstacle we first preprocess the -free input graph. An atom is a graph with no clique cutset. Clique cutsets were introduced by Dirac [21], who proved that every chordal graph is either complete or has a clique cutset. Later, decomposition into atoms became a very general tool for solving combinatorial problems on chordal graphs and other hereditary graph classes, such as those that forbid some Truemper configuration [4]. For instance, Colouring and also other problems, such as Independent Set and Clique, are polynomial-time solvable on a hereditary graph class  if they are so on the atoms of  [58]. Hence, we may restrict ourselves to the subclass of -free atoms in order to solve Colouring for -free graphs.

Adler et al. [1] proved that (diamond, even-hole)-free atoms have unbounded clique-with. However, so far, (un)boundedness of the cliquewidth of atoms in special graph classes has not been well studied. It is known that a class of -free graphs has bounded clique-width if and only if is an induced subgraph of (see [19]). As a start of a more systematic study, we show in Section 3 that the same result holds fo atoms: a class of -free atoms has bounded clique-width if and only if is an induced subgraph of . In contrast, we observe that, although split graphs have unbounded clique-width [49], split atoms are cliques [21] and thus have clique-width at most 2. Recall that split graphs are characterized by three forbidden induced subgraphs. This yields the natural question whether one can prove the same result for a graph class characterized by two forbidden induces subgraphs. In this paper we give an affirmative answer to this question by showing that the class of -free atoms has bounded clique-width. As mentioned, this immediately yields a polynomial-time algorithm for Colouring on -free graphs,

In order to prove that -free atoms have bounded clique-width, we further develop the approach of [24] used to bound the chromatic number of -free graphs as a linear function of their maximum clique size and to obtain a -approximation algorithm for Colouring for -free graphs. The approach of [24] is based on a decomposition theorem for -free atoms. We derive a new variant of this decomposition theorem for so-called strong atoms, which are atoms that contain no universal vertices and no pairs of twin vertices. We use this decomposition to prove that -free strong atoms have bounded clique-width. To obtain this result we also apply a divide-and-conquer approach for bounding the clique-width of a subclass of -free graphs. As another novel element of our proof, we show a new bound on the clique-width for (general) graphs in terms of the clique-width of recursively defined subgraphs induced by homogeneous triples and pairs of sets. Our techniques may be of independent interest and can possibly be used to prove polynomial-time solvability of Colouring on other graph classes.

Remark. The Independent Set problem is to decide if a given graph has an independent set of at least  vertices for some given integer . Brandstädt and Hoàng [6] proved that Independent Set is polynomial-time solvable for -free graphs. As mentioned, just as for Colouring, it suffices to consider only the atoms of a hereditary graph class in order to solve Independent Set [58]. Brandstädt and Hoàng followed this approach. Although we will use one of their structural results as lemmas, their method does not yield a polynomial-time algorithm for Colouring on -free graphs.

2 Preliminaries

Let be a graph. For , the subgraph induced by , is denoted by . The complement of is the graph with vertex set and edge set . A clique is a clique cutset if has more connected components than . If has no clique cutsets, then is called an atom. The edge subdivision of an edge removes from and replaces it by a new vertex  and two new edges and .

The neighbourhood of a vertex is denoted by and its degree by . For a set , we write . For and , we let be the set of neighbours of that are in , that is, . A subset is a dominating set of if every vertex not in has a neighbour in . A vertex  is universal in if it is adjacent to all other vertices, that is, is a dominating set of .

For , we say that is complete (resp. anti-complete) to if every vertex in is adjacent (resp. non-adjacent) to every vertex in . Let be two distinct vertices. We say that a vertex distinguishes and if is adjacent to exactly one of and . A set is a homogeneous set if no vertex in can distinguish two vertices in . A homogeneous set is proper if . A graph is prime if it contains no proper homogeneous set.

We say that and are (true) twins if and are adjacent and have the same set of neighbours in . Note that the binary relation of being twins is an equivalence relation on , and so can be partitioned into equivalence classes of twins. The skeleton of is the subgraph induced by a set of vertices, one from each of . A blow-up of is a graph obtained by replacing each vertex by a clique of size at least , such that two distinct cliques and are complete in if and are adjacent in , and anti-complete otherwise. Since each equivalence class of twins is a clique and any two equivalence classes are either complete or anti-complete, every graph is a blow-up of its skeleton.

Let be a set of graphs for some integer . We say that is -free if contains no induced subgraph isomorphic to for some . If , we may write that is -free instead. If can be partitioned into a clique and an independent set , then is a split graph.

The clique-width of a graph , denoted by , is the minimum number of labels required to construct  using the following four operations:

• : create a new graph consisting of a single vertex with label ;

• : take the disjoint union of two labelled graphs and ;

• : join each vertex with label to each vertex with label (for );

• : rename label to .

A clique-width expression of is an algebraic expression that describes how can be recursively constructed using these operations. An -expression of is a clique-width expression using at most distinct labels. For instance, is a -expression for the path on vertices in that order. A class of graphs  has bounded clique-width if there is a constant  such that the clique-width of every graph in  is at most ,and unbounded otherwise.

Clique-width is of fundamental importance in computer science, since every problem expressible in monadic second-order logic using quantifiers over vertex subsets but not over edge subsets becomes polynomial-time solvable for graphs of bounded clique-width [14]. Although this meta-theorem does not directly apply to Colouring, a result of Espelage, Gurski and Wanke [23] (see also [42, 55]) combined with the polynomial-time approximation algorithm of Oum and Seymour [54] for finding an -expression of a graph, showed that Colouring can be added to the list of such problems.

Theorem 3 ([23, 54]).

Colouring can be solved in polynomial time for graphs of bounded clique-width.

3 Atoms and Clique-Width

Recall that a graph is an atom if it contains no clique cutset. It is a natural question whether the clique-width of a graph class of unbounded clique-width becomes bounded after restricting to the atoms of the class. For classes of -free graphs we note that this is not the case though. In order to explain this we need the notion of a wall; see Figure 1 for three examples (for a formal definition we refer to, for example, [16]). A -subdivided wall is a graph obtained from a wall after subdividing each edge exactly  times for some constant . The following lemma is well known.

Lemma 1 ([48]).

For every constant , the class of -subdivided walls has unbounded clique-width.

We also need the following lemma.

Lemma 2.

For every constant , every -subdivided wall and every complement of a -subdivided wall is an atom.

Proof.

Let . Let be a -subdivided wall. As is -free, a largest clique has size 2. It is readily seen that contains no set of at most two vertices that disconnect .

Now consider the complement of . For contradiction, assume that is not an atom. Then has a clique cutset . Let and be two connected components of . If and both have at least two vertices and , respectively, then contains a , which is not possible. Hence, one of , say , only contains one vertex . As the neighbourhood of in is a clique, the non-neighbourhood of in is an independent set. However, no vertex in has this property. ∎

Recall that a class of -free graphs has bounded clique-width if and only if is an induced subgraph of (see [19]). We show that the same classification holds for -free atoms.

Proposition 1.

Let  be a graph. The class of -free atoms has bounded clique-width if and only if  is an induced subgraph of .

Proof.

If  is an induced subgraph of , then the class of -free graphs, which contains all -free atoms, has clique-width at most  [15].

Now suppose that is not an induced subgraph of . For every , every -subdivided wall is an atom by Lemma 2. First suppose that contains a cycle. Then the class of -subdivided walls is contained in the class of -free atoms for some appropriate value of . Hence, the class of -free atoms has unbounded clique-width due to Lemma 1.

Now suppose that does not contain a cycle. Hence is a forest. As is not an induced subgraph of , we find that must contain an induced or an induced . Let be the class of -free atoms, and let be the class that consists of the complements of -free atoms. As every wall is -free, the complement of every wall is -free. By Lemma 2, the complement of every wall is an atom as well. Hence, contains all complements of walls. It is well known that complementing all graphs in a class of unbounded clique-width results in another class of unbounded clique-width [39]. Hence, complements of walls have unbounded clique-width due to Lemma 1. This means that , and thus , has unbounded clique-width. ∎

In contrast to Proposition 1, we recall that there exist classes of -free graphs of unbounded clique-width whose atoms have bounded clique-width. Namely, the class of split graphs, or equivalently, the class of -free graphs, has unbounded clique-width [49], whereas split atoms are cliques and thus have clique-width at most 2. In the next section, we will prove that there exist even classes of -free graphs with this property by showing that the property holds even for the class of -free graphs.

4 The Polynomial-Time Result

In this section, we will prove our main result.

Theorem 4.

Colouring is polynomial-time solvable for -free graphs.

The main ingredient for proving Theorem 4 is a new structural property of -free atoms, which asserts that -free atoms have bounded clique-width. The following result is due to Tarjan.

Theorem 5 ([58]).

If Colouring is polynomial-time solvable on atoms in an hereditary class , then it is polynomial-time solvable on all graphs in .

As the class of -free graphs is hereditary, we can apply Theorem 5 and may restrict ourselves to -free atoms. Then, due to Theorem 3, it suffices to show the following result in order to prove Theorem 4.

Theorem 6.

The class of -free atoms has bounded clique-width. More precisely, every -free atom has clique-width at most .

Note that -free atoms are an example of a class of -free graphs of unbounded clique-width, whose atoms have bounded clique-width.

The remainder of the section is organised as follows. In Section 4.1, we present the key tools on clique-width that play an important role in the proof of Theorem 6. In Section 4.2, we list structural properties around a 5-cycle in a -free graph that are frequently used in later proofs. We then present the proof of Theorem 6 in Section 4.3.

4.1 Key Tools for Clique-Width

Let be a graph and be a proper homogeneous set in . Then is partitioned into two subsets and where is complete to and is anti-complete to . Let be an arbitrary vertex and . We say that and are factors of with respect to . Suppose that is an -expression for using labels and is an -expression for using labels . Then substituting in with results in an -expression for where . Moreover, all vertices in have the same label in this -expression for .

Lemma 3 ([15]).

The clique-width of any graph is the maximum clique-width of any prime induced subgraph of .

A bipartite graph is a chain graph if it is -free. A co-bipartite chain graph is the complement of a bipartite chain graph. Let be a (not necessarily bipartite) graph such that is partitioned into two subsets and . We say that an -expression for is nice if all vertices in end up with the same label and all vertices in end up with the same label with . It is well-known that any co-bipartite chain graph whose vertex set is partitioned into two cliques has a nice -expression (see Appendix A for a proof).

Lemma 4 (Folklore).

There is a nice -expression for any co-bipartite chain graph.

We now use a divide-and-conquer approach to show that a special graph class has a nice 4-expression. This plays a crucial role in our proof of the main theorem (Theorem 6).

Lemma 5.

A -free graph has a nice -expression if can be partitioned into two (possibly empty) subsets and that satisfy the following conditions:

1. is a clique;

2. is -free;

3. no vertex in has two non-adjacent neighbours in ;

4. there is no induced in that starts with a vertex in followed by three vertices in .

Proof.

We use induction on . If contains at most one vertex, then is a co-bipartite chain graph and the lemma follows from Lemma 4. Assume that contains at least two vertices. Since is -free, either or is disconnected [57]. Suppose first that is disconnected. Then can be partitioned into two nonempty subsets and that are anti-complete to each other. Let and . Then , with partition , satisfies conditions iiv for . Note also that, by iii, is anti-complete to and is anti-complete to . By the inductive hypothesis there is a nice -expression for in which all vertices in and have labels and , respectively. Now is a nice -expression for .

Suppose now that is disconnected. This means that can be partitioned into two subsets and that are complete to each other. Since is -free, either or is a clique. Without loss generality, we may assume that is a clique. Moreover, we choose the partition such that is maximal, so . Then every vertex in is not adjacent to some vertex in , for otherwise we could have moved such a vertex to . If then is a co-bipartite chain graph and so the lemma follows from Lemma 4. Therefore, we assume that . Let and . Note that is anti-complete to .

We claim that is complete to . Suppose, by contradiction, that and are not adjacent. By definition, has a neighbour . Recall that is not adjacent to some vertex . Now induces either a or a , depending on whether and are adjacent. This contradicts iv or the -freeness of . This proves the claim. Since is complete to , we find that is anti-complete to and (see Figure 2).

Note that , with the partition satisfies conditions iiv. By the inductive hypothesis there is a nice -expression for in which all vertices in and have labels and , respectively. As and are cliques and is -free, we find that is a co-bipartite chain graph. It then follows from Lemma 4 that there is a nice -expression for it in which all vertices in and have labels and , respectively.

We now are going to use the adjacency between the different sets as displayed in Figure 2. We first deduce that

 σ=ρ3→4(ρ1→2(η3,4(η2,3(η1,2(ϵ⊕τ)))))

is a nice -expression for . Let be a -expression for in which all vertices in have label . Then is a nice -expression for . This completes the proof. ∎

Let be a graph and , , and are three pairwise disjoint subsets of . We say that is a homogeneous triple if no vertex in can distinguish any two vertices in , or . A pair of sets is a homogeneous pair if is a homogeneous triple. If both and are cliques, then is a homogeneous pair of cliques. Note that homogeneous sets are special cases of homogeneous pairs and triples. An -expression for a homogeneous triple is nice if two vertices of have the same label if and only if they belong to the same set , or . We establish a new bound on the clique-width of a graph  in terms of the number of pairwise disjoint homogenous pairs and triples of .

Lemma 6.

Let be a graph. If can be partitioned into a subset , with , and homogeneous pairs and homogeneous triples such that there is a nice -expression for each homogeneous pair and a nice -expression for each homogeneous triple, then .

Proof.

We first construct the homogenous pairs and triples and the edges inside these pairs and triples one by one using nice -expressions and nice -expressions, respectively. So we need at most four different labels for each homogenous pair and at most six different labels for each homogenous triple. As soon as we have constructed a homogenous pair (triple) with its internal edges using a nice -expression (-expression), we introduce a new label for all vertices of each of its two (three) sets before considering the next homogenous pair or triple. We can do so, because all vertices of each set in a homogeneous pair received the same label by the definition of a nice -expression for homogenous sets and triples. Consequently, we may use the previous labels over and over again as auxiliary labels. Afterwards, we can view each set in a homogeneous pair or triple as a single vertex, each with its own unique label.

So far we used at most different labels. By using the auxiliary labels as unique labels for the sets of the last pair or triple and by considering pairs before triples, we need in fact at most distinct labels if and at most labels if . We now assign a unique label to each vertex in  after first using all the remaining auxiliary labels.

So far we only constructed edges of that are within a homogenous pair or triple. From our labelling procedure and the definitions of homogenous pairs and triples it follows that we can put in all the remaining edges of using only join and disjoint union operations. Hence, as , the total number of distinct labels is at most . ∎

4.2 Structure around a 5-Cycle

Let be a graph and be an induced subgraph of . We partition into subsets with respect to as follows: for any , we denote by the set of vertices in that have as their neighbourhood among , i.e.,

 S(X)={v∈V∖V(H):NV(H)(v)=X}.

For , we denote by the set of vertices in that have exactly neighbours among . Note that . We say that a vertex in is a -vertex. Let be a -free graph and be an induced in . We partition with respect to as above. All indices below are modulo . Since is -free, there is no vertex in that is adjacent to vertices and but not to vertex . In particular, , , etc. are empty. The following properties a-i of were proved in [32] using the fact that is -free.

1. is a clique.

2. is complete to and anti-complete to . Moreover, if neither nor are empty then both sets are cliques.

3. is complete to and anti-complete to . Moreover, if neither nor are empty then both sets are cliques.

4. is anti-complete to .

5. is anti-complete to if . Moreover, if a vertex in is not anti-complete to then it is universal in .

6. is anti-complete to .

7. is anti-complete to .

8. Either or is empty. By symmetry, either or is empty.

9. At least one of , and is empty.

We now prove some further properties that are used in Lemma 10.

1. For each connected component of , each vertex in is either complete or anti-complete to .

Proof. It suffices to prove the property for . Suppose that some vertex is neither complete nor anti-complete to a connected component of . By symmetry, we may assume that . By the connectivity of , there exists an edge in such that is adjacent to but not to . Then induces a , a contradiction.

2. No vertex in can distinguish an edge between and .

Proof. It suffices to prove the property for . Let and be adjacent. If a vertex is adjacent to exactly one of and , then either or induces a .

3. If a vertex has a neighbour in , then is complete to .

Proof. It suffices to prove the property for . Suppose that is not adjacent to some . Since has a neighbour , say , it follows that induces a .

4. Each vertex in is anti-complete to either or .

Proof. It suffices to prove the property for . Suppose that has a neighbour and . By d, and are not adjacent. Then induces a .

5. Each vertex in and is either complete or anti-complete to each connected component of .

Proof. It suffices to prove the property for . Suppose that distinguishes an edge in , say is adjacent to but not to . By symmetry, we may assume that . Then induces a .

6. If both and are not empty, then each vertex in is either complete or anti-complete to .

Proof. It suffices to prove the property for . Let and be two arbitrary vertices. If distinguishes and , say is adjacent to but not to , then induces a , since is not adjacent to by c.

4.3 Proof of Theorem 6

In this section, we give a proof of Theorem 6, which states that -free graphs have clique-width at most 18.

A graph is chordal if it does not contain any induced cycle of length at least 4. The following structure of -free graphs discovered by Brandstädt and Hoàng [6] is of particular importance in our proofs below.

Theorem 7 ([6]).

Let be a -free atom. Then the following statements hold: (i) every induced is dominating; (ii) if contains an induced which is not dominating, then is the join of a blow-up of the Petersen graph (Figure 3) and a (possibly empty) clique.

We say that an atom is strong if it has no pair of twin vertices or universal vertices. Note that a pair of twin vertices and a universal vertex in a graph give rise to two special kinds of proper homogeneous sets such that one of the factors decomposed by these homogeneous sets is a clique. Therefore, removing twin vertices and universal vertices does not change the clique-width of the graph by Lemma 3. So, to prove Theorem 6 it suffices to prove the theorem for strong atoms.

We follow the approach of [24]. In [24], the first and second author showed how to derive a useful decomposition theorem for -free atoms by eliminating a sequence , , and (see Figure 4 for the graphs and ) of induced subgraphs and then employing Dirac’s classical theorem [21] on chordal graphs. Here we adopt the same strategy and show in Lemma 7Lemma 10 below that if a -free strong atom contains an induced or , then it has clique-width at most 18. The remaining case is therefore that is a chordal atom, and so is a clique by Dirac’s theorem [21]. Since cliques have clique-width , Theorem 6 follows. It turns out that we can easily prove Lemma 7 and Lemma 8 via the framework formulated in Lemma 6 using the structure of the graphs discovered in [24]. The difficulty is, however, that we have to extend the structural analysis in [24] extensively for Lemma 9 and Lemma 10 and provide new insights on bounding the clique-width of certain special graphs using divide-and-conquer (see Lemma 5).

Lemma 7.

If a -free strong atom contains an induced , then has clique-width at most .

Proof.

Let be a -free strong atom that contains an induced subgraph that is isomorphic to with where induces the underlying 5-cycle of and is adjacent to and , is adjacent to and , is adjacent to and , and is adjacent to and , see Figure 4. We partition with respect to . We choose such that maximized. Note that , and . All indices below are modulo . Since is an atom, it follows from Theorem 7 that . Moreover, it follows immediately from the -freeness of that . If , then is a blow-up of the graph (see Figure 5) [24]. Since contains no twin vertices, is isomorphic to and so has clique-width at most . If then has the structure prescribed in Figure 6 [24]. Note that is a homogeneous clique in and so . We partition into two subsets and . Note that is anti-complete to . In addition, is anti-complete to since is -free. It is routine to check that each of , , and is a homogeneous pair of cliques in . Now is partitioned into a subset of size and four homogeneous pairs of cliques. Since each pair of homogeneous cliques induces a co-bipartite chain graph and so has a nice 4-expression by Lemma 4. So, by Lemma 6. ∎

Lemma 8.

If a -free strong atom contains an induced , then has clique-width at most .

Proof.

Let be an induced six-cycle of . We partition with respect to . If is not dominating, then it follows from Theorem 7 that is the join of a blow-up of the Petersen graph and a (possibly empty) clique. Since has no twin vertices or universal vertices, is isomorphic to the Petersen graph and so has clique-width at most . In the following, we assume that is dominating, i.e., . It was shown in [24] that is a blow-up of a graph of order at most . Therefore, has clique-width at most . ∎

Lemma 9.

If a -free strong atom contains an induced , then has clique-width at most .

Proof.

Let be a -free strong atom that contains an induced subgraph that is isomorphic to with such that induces the underlying 5-cycle , and is adjacent to , and , is adjacent to and is adjacent to and . Moreover, is adjacent to both and , see Figure 4. We partition with respect to . We choose such that has maximized. Note that , and .

The overall strategy is to first decompose into a subset of constant size and constant number of homogeneous pairs of sets, and then finish off the proof via Lemma 6 by showing that each homogeneous pair of sets has a nice 4-expression using Lemma 5.

We start with the decomposition. Since and are not empty, it follows from h that . If both and are not empty, say and , then induces either a or a , depending on whether and are adjacent. This shows that for some . Now we argue that . If contains a vertex , then is adjacent to and by c but not adjacent to by g. This implies that induces a , So, . If contains a vertex , then is adjacent to by c and so induces a , a contradiction. This shows that . By symmetry, . Therefore, . The following properties among subsets of were proved in [24].

1. Each vertex in is either complete or anti-complete to .

2. and are cliques.

3. Each vertex in