Bounding the Clique-Width of H -Free Chordal Graphs

: A graph is H -free if it has no induced subgraph isomorphic to H . Brandst¨adt, Engelfriet, Le, and Lozin proved that the class of chordal graphs with independence number at most 3 has unbounded clique-width. Brandst¨adt, Le, and Mosca erroneously claimed that the gem and co-gem are the only two 1-vertex P 4 -extensions H for which the class of H -free chordal graphs has bounded clique-width. In fact we prove that bull-free chordal and co-chair-free chordal graphs have clique-width at most 3 and 4, respectively. In particular, we ﬁnd four new classes of H -free chordal graphs of bounded clique-width. Our main result, obtained by combining new and known results, provides a classiﬁcation of all but two stubborn cases, that is, with two potential exceptions we determine all graphs H for which the class of H -free chordal graphs has bounded clique-width. We illustrate the usefulness of this classiﬁcation for classifying other types of graph classes by proving that the class of ( 2 P 1 + P 3 , K 4 ) -free graphs has bounded clique-width via a reduction to K 4 -free chordal graphs. Finally, we give a complete classiﬁcation of the (un)boundedness of clique-width of H -free weakly chordal graphs. C (cid:2) 2017 Wiley Inc. 2017


Introduction
Clique-width is a well-studied graph parameter; see for example the surveys of Gurski [32] and Kamiński, Lozin and Milanič [35].In particular, there are numerous graph classes, such as those that can be characterized by one or more forbidden induced subgraphs, 1 for which it has been determined whether or not the class is of bounded clique-width (i.e.whether there is a constant c such that the clique-width of every graph in the class is at most c).Similar research has been done for variants of clique-width, such as linear clique-width [33] and powerbounded clique-width [3].Clique-width is also closely related to other graph width parameters.For instance, it is known that every graph class of bounded treewidth has bounded clique-width but the reverse is not true [14].Moreover, for any graph class, having bounded clique-width is equivalent to having bounded rank-width [47] and also equivalent to having bounded NLC-width [34].
Clique-width is one of the most difficult graph parameters to deal with and our understanding of it is still very limited.We do know that computing cliquewidth is NP-hard [27] but we do not know if there exist polynomial-time algorithms for computing the clique-width of even very restricted graph classes, such as unit interval graphs.Also the problem of deciding whether a graph has cliquewidth at most c for some fixed constant c is only known to be polynomial-time solvable if c ≤ 3 [13] and is a long-standing open problem for c ≥ 4. Identifying more graph classes of bounded clique-width and determining what kinds of structural properties ensure that a graph class has bounded clique-width increases our understanding of this parameter.Another important reason for studying these types of questions is that certain classes of NP-complete problems become polynomial-time solvable on any graph class G of bounded clique-width. 2Examples of such problems are those definable in Monadic Second Order Logic using quantifiers on vertices but not on edges.
In this paper we primarily focus on chordal graphs.The class of chordal graphs has unbounded clique-width, as it contains the class of proper interval graphs and the class of split graphs, both of which have unbounded clique-width as shown by Golumbic and Rotics [31] and Makowsky and Rotics [42], respectively.We study the clique-width of subclasses of chordal graphs, but before going into more detail we first give some necessary terminology and notation.

Notation
The disjoint union (V (G) ∪ V (H), E(G) ∪ E(H)) of two vertex-disjoint graphs G and H is denoted by G + H and the disjoint union of r copies of a graph G is denoted by rG.The complement of a graph G, denoted by G, has vertex set V (G) = V (G) and an edge between two distinct vertices if and only if these vertices are not adjacent in G.For two graphs G and H we write H ⊆ i G to indicate that H is an induced subgraph of G.The graphs C r , K r , K 1,r−1 and P r denote the cycle, complete graph, star and path on r vertices, respectively.The graph S h,i,j , for 1 ≤ h ≤ i ≤ j, denotes the subdivided claw, that is the tree that has only one vertex x of degree 3 and exactly three leaves, which are of distance h, i and j from x, respectively.For a set of graphs {H 1 , . . ., H p }, a graph G is (H 1 , . . ., H p )-free if it has no induced subgraph isomorphic to a graph in {H 1 , . . ., H p }.A graph is chordal if it is (C 4 , C 5 , . ..)-free.

Aim and Motivation
We want to determine all graphs H for which the class of H-free chordal graphs has bounded clique-width.Our motivation for this research is threefold.
Firstly, as discussed, such a classification might generate more graph classes for which a number of NP-complete problems can be solved in polynomial time.Although many of these problems, such as the Colouring problem [30], are polynomial-time solvable on chordal graphs, many others stay NP-complete for graphs in this class.To give an example, the well-known Hamilton Cycle problem is such a problem.It is NP-complete even for strongly chordal split graphs [43], but becomes polynomial-time solvable on any graph class of bounded clique-width [26,51].Of course, in order to find new "islands of tractability", one could consider superclasses of chordal graphs instead.However, we became interested in H-free chordal graphs in particular after realizing that a restriction from H-free graphs to H-free perfect graphs (perfect graphs form a superclass of chordal graphs) does not yield any new tractable graph classes.Indeed, the clique-width of the class of H-free graphs is bounded if and only if H is an induced subgraph of P 4 [21], and as we prove later, the induced subgraphs of P 4 are also the only graphs H for which the class of H-free perfect graphs has bounded clique-width.
Secondly, we have started to extend known results [2,5,6,7,8,9,11,17,19,42] on the clique-width of classes of (H 1 , H 2 )-free graphs in order to try to determine the boundedness or unboundedness of the clique-width of every such graph class [18,21].This led to a classification of all but 13 open cases (under some equivalence relation, see [21]).An important technique3 that we used for showing the boundedness of the clique-width of three new graph classes of (H 1 , H 2 )-free graphs [18] was to reduce these classes to some known subclass of perfect graphs of bounded clique-width.An example of such a subclass, which we used for one of the three cases, is the class of chordal diamond-free graphs (the diamond is the graph 2P 1 + P 2 , see also Fig. 1), which has bounded clique-width [31].We believe that a full classification of the boundedness of clique-width for H-free chordal graphs would be useful to attack some of the remaining open cases, just as the full classification for H-free bipartite graphs [20] has already proven to be [18,21].Examples of open cases included the class of (2P 1 + P 3 , K 4 )-free graphs and its superclass of (2P 1 + P 3 , 2P 1 + P 3 )-free graphs [21], the first of which turns out to have bounded clique-width, as we shall prove in this paper via a reduction to K 4 -free chordal graphs.The second case is still open.
Thirdly, a classification of those graphs H for which the clique-width of H-free chordal graphs is bounded would complete a line of research in the literature, which we feel is an interesting goal on its own.As a start, using a result of Corneil and Rotics [14] on the relationship between treewidth and clique-width it follows that the clique-width of the class of K r -free chordal graphs is bounded for all r ≥ 1. Brandstädt, Engelfriet, Le and Lozin [5] proved that the class of 4P 1 -free chordal graphs has unbounded clique-width.Brandstädt, Le and Mosca [9] considered forbidding the graphs P 1 + P 4 (gem) and P 1 + P 4 (cogem) as induced subgraphs (see also Fig. 2).They showed that (P 1 + P 4 )-free chordal graphs have clique-width at most 8 and also observed that P 1 + P 4 -free chordal graphs belong to the class of distance-hereditary graphs, which have clique-width at most 3 (as shown by Golumbic and Rotics [31]).Moreover, the same authors [9] erroneously claimed, that the gem and co-gem are the only two 1-vertex P 4 -extensions H for which the class of H-free chordal graphs has bounded clique-width.We prove that bull-free chordal graphs have clique-width at most 3, improving a known bound of 8, which was shown by Le [39].We also prove that S 1,1,2 -free chordal graphs have clique-width at most 4, which Le posed as an open problem.Results [31,37,42] for split graphs and proper interval graphs lead to other classes of H-free chordal graphs of unbounded clique-width, as we shall discuss in Section 2. However, in order to obtain our almost-full dichotomy for H-free chordal graphs new results also need to be proved.

Our Results
In Section 2, we collect all previously known results for H-free chordal graphs and use a result of Olariu [45] to prove that bull-free chordal graphs have cliquewidth at most 3.In Section 3 we present four new classes of H-free chordal graphs of bounded clique-width, 4 namely when H ∈ {K 1,3 + 2P 1 , P 1 + P 1 + P 3 , P 1 + 2P 1 + P 2 , S 1,1,2 } (see also Fig. 2).In particular, we show that S 1,1,2 -free graphs have clique-width at most 4. One of the algorithmic consequences of these results is that we have identified four new graph classes for which Hamilton Cycle is polynomial-time solvable.In Section 4 we prove that the three classes of H-free, (3P 1 + P 2 )-free and (K 3 + 2P 1 , K 4 + P 1 , P 1 + P 1 + P 4 )-free split graphs have unbounded clique-width (see also Fig. 3).In Section 5 we combine all The graphs H for which we prove that the class of H-free split graphs has unbounded clique-width.

. The class of H-free chordal graphs has bounded clique-width if and only if
In Section 5 we also show (using only previously known results) our aforementioned classification for H-free perfect graphs.

Theorem 2. Let H be a graph. The class of H-free perfect graphs has bounded clique-width if and only if H is an induced subgraph of P 4 .
In Section 6 we illustrate the usefulness of having a classification for H-free chordal graphs by proving that the class of (2P 1 + P 3 , K 4 )-free graphs has bounded clique-width via a reduction to K 4 -free chordal graphs.As such the number of (non-equivalent) pairs (H 1 , H 2 ) for which we do not know whether the clique-width of the class of (H 1 , H 2 )-free graphs is bounded is 13.These remaining cases are given in Section 7 (see also [21]).In Section 7, we mention a number of future research directions.

Preliminaries
All graphs considered in this paper are finite, undirected and have neither multiple edges nor self-loops.In this section we first define some more standard graph terminology, some additional notation and give some structural lemmas.We refer to the textbook of Diestel [24] for any undefined terminology.Afterwards, we give the definition of clique-width and present a number of known results on clique-width that we will use as lemmas for proving our results.
Let G = (V, E) be a graph.For S ⊆ V , we let G[S] denote the induced subgraph of G, which has vertex set S and edge set {uv | u, v ∈ S, uv ∈ E}.If S = {s 1 , . . ., s r } then, to simplify notation, we may also write Recall that for two graphs G and H we write We say that S dominates T if every vertex of T is adjacent to at least one vertex of S. We say that S is complete to T if every vertex in S is adjacent to every vertex in T , and we say that S is anti-complete to T if every vertex in S is non-adjacent to every vertex in T .Similarly, a vertex v ∈ V \ T is complete or anti-complete to T if it is adjacent or non-adjacent, respectively, to every vertex of T .A set of vertices M is a module if every vertex not in M is either complete or anti-complete to M .
A module in a graph is trivial if it contains zero, one or all vertices of the graph, otherwise it is non-trivial.A graph G is prime if every module in G is trivial.We say that a vertex v distinguishes two vertices x and y if v is adjacent to precisely one of x and y.Note that if a set M ⊆ V is not a module then there must be vertices x, y ∈ M and a vertex v ∈ V \ M such that v distinguishes x and y.
The following two structural lemmas, both of which we need for the proofs of our results, are about prime graphs containing some specific induced subgraph H.They show us that such prime graphs must also contain (as an induced subgraph) at least one of a list of possible extensions of H.The first lemma is due to Brandstädt, Le and de Ridder and the second one is due to Brandstädt.

Lemma 3 ([10]
).If a prime graph G contains an induced subgraph isomorphic to P 1 + P 4 then it contains one of the graphs in Fig. 5 as an induced subgraph.We also use the following structural lemma due to Olariu.

Lemma 5 ([45]
).Every prime (bull, house)-free graph is either K 3 -free or the complement of a 2P 2 -free bipartite graph (see also Figs. 2 and 6).Let G = (V, E) be a connected graph.An edge e ∈ E is a bridge if deleting it would make is disconnected.If G has at least three vertices, but no cut-vertices then it is 2-connected.It is well known that if G is a 2-connected graph then for any two vertices u and v in G, there are two paths from u to v in G that are internally vertex-disjoint (see e.g.[24]).A block of G is a maximal 2-connected subgraph, a bridge or a single vertex.Note that two blocks of G have at most one common vertex, which must be a cut-vertex of G.
The (r + 1)-vertex star K 1,r is the graph consisting of one vertex of degree r, called the central vertex, and r vertices of degree 1.A double-star is the graph formed from two stars K 1,s and K 1,r by joining the central vertices of each star with an edge. Let contains no edges.The independence number of G is the size of a largest independent set of G.If V can be partitioned into two (possibly empty) independent sets then G is bipartite.We say that G is complete multipartite if V can be partitioned into k independent sets V 1 , . . ., V k (called partition classes) for some integer k, such that two vertices are adjacent if and only if they belong to two different sets V i and V j .
The next result, which we will use later on, is due to Olariu [44] (note that the graph P 1 + P 3 is also called the paw).

Lemma 7. Every chordal graph has a simplicial vertex.
Let G = (V, E) be a graph.A set S ⊆ V is said to be a clique if G[S] is a complete graph.The clique number of G is the size of a largest clique of G.The chromatic number of G is the minimum number k for which G has a k-colouring, that is, for which there exists a mapping c : V → {1, . . ., k} such that c(u) = c(v) whenever u and v are adjacent.We say that G is perfect if, for every induced subgraph H ⊆ i G, the chromatic number of H equals its clique number.We say that G is a split graph if it has a split partition, that is, a partition of V into two (possibly empty) sets K and I, where K is a clique and I is an independent set; if K and I are complete to each other, then G is said to be a complete split graph.
It is well known that every split graph is chordal and that every chordal graph is perfect (see [30]).The first inclusion also follows from the next lemma, which is due to Földes and Hammer [28].

Lemma 8 ([28]). A graph is split if and only if it is
A graph G is a thin spider if its vertex set can be partitioned into a clique K, an independent set I and a set R such that |K| = |I| ≥ 2, the set R is complete to K and anti-complete to I and the edges between K and I form a induced matching (that is, every vertex of K has a unique neighbour in I and vice versa).Note that if a thin spider is prime then |R| ≤ 1.A thick spider is the complement of a thin spider.A graph is a spider if it is either a thin or a thick spider.
Spiders play an important role in our result for S 1,1,2 -free chordal graphs and we will need the following lemma (due to Brandstädt and Mosca).

Clique-width
The clique-width of a graph G, denoted cw(G), is the minimum number of labels needed to construct G by using the following four operations: 1. creating a new graph consisting of a single vertex v with label i (denoted by i(v)); 2. taking the disjoint union of two labelled graphs G 1 and G 2 (denoted by G 1 ⊕ G 2 ); 3. joining each vertex with label i to each vertex with label j (i = j, denoted by η i,j ); 4. renaming label i to j (denoted by ρ i→j ).
An algebraic term that represents such a construction of G and uses at most k labels is said to be a k-expression of G (i.e. the clique-width of G is the minimum k for which G has a k-expression).For instance, an induced path on four consecutive vertices a, b, c, d has clique-width equal to 3, and the following 3-expression can be used to construct it: A class of graphs G has bounded clique-width if there is a constant c such that the clique-width of every graph in G is at most c; otherwise the clique-width of G is unbounded.
Let G be a graph.We define the following operations.The subdivision of an edge uv replaces uv by a new vertex w with edges uw and vw.For an induced subgraph G ′ ⊆ i G, the subgraph complementation operation (acting on G with respect to G ′ ) replaces every edge present in G ′ by a non-edge, and vice versa.Similarly, for two disjoint vertex subsets S and T in G, the bipartite complementation operation with respect to S and T acts on G by replacing every edge with one end-vertex in S and the other one in T by a non-edge and vice versa.
We now state some useful facts about how the above operations (and some other ones) influence the clique-width of a graph.We will use these facts throughout the paper.Let k ≥ 0 be a constant and let γ be some graph operation.We say that a graph class G ′ is (k, γ)-obtained from a graph class G if the following two conditions hold: (i) every graph in G ′ is obtained from a graph in G by performing γ at most k times, and (ii) for every G ∈ G there exists at least one graph in G ′ obtained from G by performing γ at most k times.
If we do not impose a finite upper bound k on the number of applications of γ then we write that G ′ is (∞, γ)-obtained from G.
We say that γ preserves boundedness of clique-width if for any finite constant k and any graph class G, any graph class G ′ that is (k, γ)-obtained from G has bounded clique-width if and only if G has bounded clique-width.Fact 1. Vertex deletion preserves boundedness of clique-width [40].Fact 2. Subgraph complementation preserves boundedness of clique-width [35].Fact 3. Bipartite complementation preserves boundedness of clique-width [35] [35].
We also use a number of other elementary results on the clique-width of graphs.The first two are well known and straightforward to check.
Lemma 10.The clique-width of a forest is at most 3.
Lemma 11.The clique-width of a graph with maximum degree at most 2 is at most 4.
The following lemma tells us that if G is a graph class closed under vertex deletion then in order to determine whether G has bounded clique-width we may restrict ourselves to the graphs in G that are prime.

Lemma 12 ([16]
).Let G be a graph and let P be the set of all induced subgraphs of G that are prime.Then cw(G) = max H∈P cw(H).
We also need the well-known notion of a wall.We do not formally define this notion but instead refer to Fig. 7, in which three examples of walls of different height are depicted.The class of walls is well known to have unbounded cliquewidth; see for example [35].(Note that walls have maximum degree at most 3, hence the degree bound in Lemma 11 is tight.) A k-subdivided wall is the graph obtained from a wall after subdividing each edge exactly k times for some constant k ≥ 0.
The following lemma is well known and follows from combining Fact 5 with the aforementioned facts that walls have maximum degree at most 3 and unbounded clique-width.Lemma 13.For any constant k ≥ 0, the class of k-subdivided walls has unbounded clique-width.

Known Results on H-free Chordal Graphs
To prove our results, we need to use a number of known results.We present these results as lemmas in this subsection; a number of relevant graphs are displayed in Figs. 2 and 8.The first of these lemmas gives a classification for H-free graphs.

Lemma 14 ([21]). Let H be a graph. The class of H-free graphs has bounded clique-width if and only if H is an induced subgraph of P 4 .
We will use the following characterization of graphs H for which the class of H-free bipartite graphs has bounded clique-width (which is similar to a characterization of Lozin and Volz [41] for a different variant of the notion of H-freeness in bipartite graphs, see [20] for an explanation of the difference).

Lemma 15 ([20]). Let H be a graph. The class of H-free bipartite graphs has bounded clique-width if and only if one of the following cases holds:
For a graph G, let tw(G) denote the tree-width of G (see, for example, Diestel [24] for a definition of this notion).Corneil and Rotics [14] showed that cw(G) ≤ 3 × 2 tw(G)−1 for every graph G.Because the tree-width of a chordal graph is equal to the size of a maximum clique minus 1 (see e.g.[1]), this result leads to the following well-known lemma.

Lemma 16. The class of K r -free chordal graphs has bounded clique-width for all r ≥ 1.
The bull is the graph obtained from the cycle abca after adding two new vertices d and e with edges ad, be.In [9], Brandstädt, Le and Mosca erroneously mentioned that the clique-width of S 1,1,2 -free chordal graphs and of bull-free chordal graphs is unbounded.Using a general result of De Simone [23], Le [39] proved that every bull-free chordal graph has clique-width at most 8.In fact the following is true.

Lemma 17. Every bull-free chordal graph has clique-width at most 3.
Proof.Let G be a bull-free chordal graph.By Lemma 12, we may assume that G is prime.Note that the house contains an induced C 4 , so G is house-free.Then, by Lemma 5, G is either K 3 -free or the complement of a 2P 2 -free bipartite graph.Every K 3 -free chordal graph is a forest, so by Lemma 10 it has clique-width at most 3.We may therefore assume that G is a prime graph that is the complement of a 2P 2 -free bipartite graph.Such graphs are known as k-webs in [45], where k ≥ 2. A k-web consists of two cliques X = {x 1 , . . ., x k } and Y = {y 1 , . . ., y k } such that for i, j ∈ {1 . . ., k} the vertex x i is adjacent to y j if and only if i < j.We will show how to use the operations of clique-width 3-expressions to inductively construct a copy of a k-web in which every vertex in the set X is labelled 1 and every vertex in the set Y is labelled 2. Consider a k-web labelled as described above for some k ≥ 0 (if k = 0 this is the empty graph).Add a vertex labelled 3 to the graph, join it to every vertex of label 1 and to every vertex of label 2, then relabel it to have label 1. Next, add a vertex labelled 3 to the graph, join it to every vertex of label 2, then relabel it to have label 2. This is precisely the (k + 1)-web, also labelled as described above.We conclude that every k-web can be constructed using at most 3 labels, so G has clique-width at most 3.

⊓ ⊔
Since P 4 is a bull-free chordal graph and has clique-width 3, the bound in the above lemma is tight.
Next we recall the aforementioned results of Brandstädt et al.

Lemma 20 ([5]). The class of 4P 1 -free chordal graphs has unbounded cliquewidth.
Recall that Golumbic and Rotics [31] proved that the class of proper interval graphs has unbounded clique-width.Such graphs are well-known to be K 1,3 -free and chordal [49].

Lemma 21. The class of K 1,3 -free chordal graphs has unbounded clique-width.
The next lemma is obtained from combining Lemma 8 with the aforementioned result of Makowsky and Rotics [42], who showed that the class of split graphs has unbounded clique-width.

Note that Lemma 22 implies that the class of H-free chordal graphs has unbounded clique-width for
Korpelainen, Lozin and Mayhill [37] proved that the class of split permutation graphs has unbounded clique-width.A graph is a permutation graph if and only if both it and its complement are comparability graphs [25].Since comparability graphs are (net, net)-free [29], the following lemma follows immediately.
The graphs H for which the class of H-free chordal graphs was previously known to have unbounded clique-width.

New Classes of Bounded Clique-width
We present four new classes of H-free chordal graphs that have bounded cliquewidth, namely when H ∈ {K 1,3 + 2P 1 , P 1 + P 1 + P 3 , P 1 + 2P 1 + P 2 , S 1,1,2 }.We prove that these classes have bounded clique-width in the subsections below, making use of known results from Subsection 2.2.In particular we will often use Facts 1-5.Note that Facts 1 and 4 can be used safely, since every class of H-free chordal graphs is closed under vertex deletion (when applying the other three facts we need to be more careful).

The Case
Here is our first result.
Proof.Let G be a K 1,3 + 2P 1 -free chordal graph.By Fact 4 we may assume that G is 2-connected.Let K be a maximum clique in G on k vertices.We may assume that k ≥ 7, otherwise G is K 7 -free, in which case G has bounded cliquewidth by Lemma 16.We let S be the set of vertices outside K with at least two neighbours in K.Because K is maximum, k ≥ 7 and G is K 1,3 + 2P 1 -free, every vertex in S has either one non-neighbour or two non-neighbours in K.
We will prove that V (G) = K ∪ S. To this end, we first prove that G − S is connected.Suppose, for contradiction, that there exists a vertex x that is in a connected component D of G − S other than the component containing K. Let u ∈ K.Because G is 2-connected, it contains two paths P 1 and P 2 from x to u that are internally vertex-disjoint.Note that we may assume that each P i is induced.For i = 1, 2, let s i ∈ P i be the first vertex that is not in D and let x i be the predecessor of s i on P i .Note that s 1 , s 2 ∈ S. Since k ≥ 7, there must be a vertex u ′ ∈ K adjacent to both s 1 and s 2 .For i = 1, 2 let P ′ i be the path from x to u ′ formed by taking the part of the path P i from x to s i and adding u ′ .Note that P ′ 1 and P ′ 2 are both induced paths in G and each contains exactly one vertex from K and one from S. Since G is chordal, s 1 and s 2 must be adjacent and at least one of x 1 and x 2 must be adjacent to both s 1 and s 2 .Without loss of generality, we assume that x 1 is adjacent to both s 1 and s 2 .On the other hand, s 1 and s 2 have at least three common neighbours in K since k ≥ 7. Then x 1 , s 1 , s 2 , together with these three vertices in K form an induced Since G is 2-connected, there must exist an induced path P from y to u with v / ∈ V (P ).Then v is complete to V (P ) since G is chordal.Let y ′ be the last vertex (from y to u) on P that is not in K ∪ S (note that y ′ is not necessarily distinct from y).Let s be the successor of y ′ on P .Since y ′ / ∈ S and y ′ is adjacent to v, we find that y ′ is anti-complete to K \ {v}.Hence, s / ∈ K, so s ∈ S.Moreover, s and v have at least three common neighbours in K \ {v} since k ≥ 7. Then y ′ , s, v, together with these three vertices in K form an induced K 1,3 + 2P 1 , a contradiction.
For i = 1, 2, let S i consist of those vertices with exactly i non-neighbours in K.Because every vertex in S has either one non-neighbour or two non-neighbours, we find that S = S 1 ∪ S 2 .
We will now prove, via Claims 1-5, that G[S] is a forest.

Claim 1. Any two adjacent vertices in S 2 have the same pair of non-neighbours in K.
This follows directly from the fact that G is chordal.We will consider two cases depending on whether or not G[S] is 2P 2 -free.We need two more claims to deal with these.We are going to distinguish between two cases depending on whether or not G[S] is 2P 2 -free.For treating these two cases we need two more claims.Claim 6.If two vertices s 1 , s 2 ∈ S, together with a vertex w ∈ K form a triangle then w is complete to S \ (N (s 1 ) ∪ N (s 2 )).Indeed, suppose for contradiction that t ∈ S \ (N (s 1 ) ∪ N (s 2 )) is not adjacent to w.Since |K| ≥ 7, there must be vertices x, y ∈ K that are complete to {s 1 , s 2 , t}.Since t is non-adjacent to s 1 and s 2 , we find that {t, s 1 , s 2 , w, x, y} induces a K 1,3 + 2P 1 , a contradiction.

Claim 7. For any connected component D in G[S]
that contains at least one edge, there exist two vertices a and b in K such that K \ {a, b} is complete to S \ V (D).To see this, let D be a connected component with an edge st.Since S 1 is independent by Claim 4, we may assume that t ∈ S 2 .Let a and b in K be the two non-neighbours of t.It follows from Claims 1 and 3 that a and b are the only possible non-neighbours of s or t in K.In other words, K \ {a, b} is complete to {s, t}, and hence to S \ V (D) by Claim 6.
We are now ready to consider the two cases.

Case 1: G[S] contains an induced 2P 2 .
First suppose G[S] has only one connected component that contains an edge.Then, since G[S] is a forest and G[S] contains an induced 2P 2 , deleting one vertex from S, which we may do by Fact 1, yields two connected components D 1 and D 2 that contain edges s 1 t 1 and s 2 t 2 , respectively.It follows from Claim 7 that there exist vertices a and b in K such that S \ D 1 is complete to K \ {a, b}.In particular, s 2 and t 2 are complete to K \ {a, b}.Hence, K \ {a, b} is also complete to D 1 , by Claim 6.Thus, K \ {a, b} is complete to S. Deleting a and b (which we may do by Fact 1) and applying a bipartite complementation between K \ {a, b} and S (which we may do by Fact 3) splits the graph into two disjoint parts: a clique G[K \ {a, b}], which has clique-width 2, and a forest G[S], which has clique-width at most 3 by Lemma 10.We conclude that G has bounded clique-width.
In this case S contains at most one connected component with an edge.If such a connected component exists then it is a 2P 2 -free tree, and hence it must be a P 2 , K 1,r or a double-star.In all three cases deleting at most two vertices from S, which we may do by Fact 1, yields a split graph.If S 2 = ∅ then then let s be a vertex in S 2 and let k 1 and k 2 be its two (only) non-neighbours in K.By Claim 2, any other vertex of S 2 is non-adjacent to at least one of k 1 , k 2 .Hence, after removing k 1 and k 2 (which we may do by Fact 1), every vertex of S is adjacent to all but at most one vertex of K. (In the case where S 2 = ∅, we do not need to remove any vertices of K.) Next, we perform a bipartite complementation between K and S, which we may do by Fact 3.This results in a new split graph in which each vertex of S is adjacent to at most one vertex of K. Hence, this graph, and consequently G, has bounded clique-width by Fact 4.

3.2
The Case H = P 1 + P 1 + P 3 We first prove three useful lemmas.
Proof.Let G be an arbitrary (P This means that G is a complete split graph.

⊓ ⊔
Note that every induced P 1 + P 3 in a (P 1 + P 1 + P 3 )-free graph G is a dominating set of G.The proof of the next lemma, in which disconnected graphs are considered, heavily relies on this fact.We will also heavily exploit this property in the proof for the general case.
Proof.Let G be a disconnected (P 1 + P 1 + P 3 )-free chordal graph.Since G has at least two connected components and each connected component therefore contains a P 1 , every connected component of G must be P 1 + P 3 -free.By Lemma 26, every connected component of G must be a complete split graph or a tree.In the first case, the clique-width of the connected component is readily seen to be at most 2. In the second case, the clique-width of that connected component is at most 3 by Lemma 10.

⊓ ⊔
We are now ready to prove our second result.
Proof.Let G be a (P 1 + P 1 + P 3 )-free chordal graph.Let x be a simplicial vertex in G, which exists by Lemma 7. Let X = N (x) and Y = V (G) \ (X ∪ {x}).Note that no vertex of Y is adjacent to x, so G[Y ] must be P 1 + P 3 -free.By Lemma 26, every connected component of G[Y ] is either a tree or complete split graph.We say that a connected component of G[Y ] is trivial if it consists of a single vertex.
Otherwise it is non-trivial.
We will distinguish between two cases depending on whether or not G[Y ] is 2P 2 -free.In the first case we will need the following claim.

Claim 1. Suppose G[Y ] contains at least two non-trivial components and y
In order to prove this claim, suppose that y is not complete to X.We will show that z is complete to Y .Let D be the connected component of G[Y ] containing y. Since y is not complete to X, there must be a vertex z ′ ∈ X that is not adjacent to y.Now G[z, x, y, z ′ ] is a P 1 + P 3 .Since G is (P Let y 1 y 2 be an edge in some non-trivial component D ′ of Y other than D (recall that such a component exists by our assumption).If y 1 and y 2 are both adjacent to z ′ then G[y, z ′ , y 1 , x, y 2 ] would be a P 1 + P 1 + P 3 .Therefore we may assume without loss of generality that y 1 is not adjacent to z ′ .Since {z, z ′ } dominates y 1 , we find that y 1 must be adjacent to z.If y 2 is not adjacent to z then, since {z, z ′ } dominates y 2 , we find that y 2 must be adjacent to z ′ .In this case G[z, z ′ , y 2 , y 1 ] would be a C 4 , contradicting the fact that G is chordal.Hence both y 1 and y 2 are adjacent to z.Now G[z, y 1 , x, y 2 ] induces a P 1 + P 3 .Therefore z is complete to Y \ D ′ , since G is (P 1 + P 1 + P 3 )-free.Recall that y 1 is adjacent to z and non-adjacent to z ′ .By the same argument, with y 1 taking the role of y, since D is a non-trivial component of G[Y ], we find that z is complete to Y \ V (D).Hence, z is complete to Y .This completes the proof of Claim 1.
We are now ready to consider the two possible cases.First suppose that D is a tree.In this case G[D] must be a P 2 , K 1,r or a double-star.In all three cases, deleting at most two vertices in D (which we may do by Fact 1) makes Y an independent set, in which case we argue as before.By Lemma 26, we may therefore assume that G[Y ] is a complete split graph.We can partition V (D) into two sets D B and D W such that D B is a clique, D W is an independent set and D B is complete to D W in G.We may assume |D B | ≥ 3. Indeed, if |D B | ≤ 2 then by Fact 1 we may delete at most two vertices to obtain a graph in which G[Y ] has only trivial components, in which case we may argue as before.
Let X ′ be the set of vertices in X that have neighbours in D. We claim that As X is a clique and X ′ is complete to Y \ V (D), we find that (Y \ V (D)) ∪ (X \X ′ ) is complete to X ′ .By Fact 3, we may apply a bipartite complementation between (Y \ V (D)) ∪ (X \ X ′ ) and X ′ and another between X \ X ′ and {x}.
The first graph is a (P 1 + P 1 + P 3 )-free split graph, so it has bounded clique-width by Lemma 25.It remains to show that G[{x} ∪X ′ ∪V (D)] has bounded clique-width.
We partition the vertices of X ′ as follows: let Z be the set of vertices in X ′ that are complete to D B , let Z ′ be the set of vertices in X ′ \ Z that are complete to D W and let By definition, w must be non-adjacent to some vertex z ′ ∈ Z ′ ∪ Z ′′ and z must be non-adjacent to some vertex w ′ ∈ D W . Furthermore, z must be non-adjacent to some vertex b ∈ D B .Note that w is not adjacent to w ′ since D W is independent.Moreover, z and z ′ are adjacent because X ′ is a clique, and b is adjacent to both w and w ′ as D is a complete split graph.Then b and z ′ must be non-adjacent, otherwise G[b, w, z, z ′ ] would be a C 4 .Then w ′ must be adjacent to z ′ , otherwise G[w ′ , z, x, w, z ′ ] would be a P 1 + P 1 + P 3 .However, this means that G[z ′ , z, w, b, w ′ ] induces a C 5 , contradicting the fact that G is chordal.Therefore D ′′ W is indeed anti-complete to Z ′′ .By Fact 1, we may delete the vertex We say that a graph G = (V, E) is quasi-diamond-free if its vertex set V can be partitioned into a clique V 1 and some other (possibly empty) set We prove the following lemma, which will play an important role in our proof.

Lemma 29. The class of quasi-diamond-free graphs has bounded clique-width.
Proof.Let G be a quasi-diamond-free graph with corresponding clique V 1 .Let B be a block of G. Then B is either equal to V 1 or contains at most one vertex of V 1 with all its other vertices belonging to V 2 .In the first case, the clique-width of B is at most 2. In the second case, we may delete the vertex of B ∩ V 1 from B (if such a vertex exists) by Fact 1.This yields a 2P 1 + P 2 -free chordal graph G ′ .By Theorem 24, we find that G ′ has bounded clique-width.Therefore G has bounded clique-width by Fact 4.

⊓ ⊔
We are now ready to prove the following result.
Proof.Let G = (V, E) be a (P 1 + 2P 1 + P 2 )-free chordal graph.We may assume without loss of generality that G is connected.Let v be a simplicial vertex in G, which exists by Lemma 7. Let then we can apply Lemma 29 to G with V 1 = {v} ∪ L 1 .Thus, from now on we may assume that ∆ ≥ 2. This means that x and v have at least two common neighbours in L 1 .Hence, as G is (P 1 + 2P 1 + P 2 )-free, we find that Claim 2. Without loss of generality, every vertex in L 1 has a neighbour in L 2 .
In order to show this, let L ′ 1 ⊆ L 1 be the set of vertices with no neighbour in L 2 .We apply a bipartite complementation between (L 1 \ L ′ 1 ) ∪ {v} and L ′ 1 .We may do so due to Fact 3. As G[L ′  1 ] is a clique, it has clique-width at most 2, and we are left to consider We now consider two cases, depending on the difference between |L 1 | and ∆.
Let z be an arbitrary vertex in L 1 \ N L1 (x).Let A z be the set of neighbours of z in L 2 .By Claim 2, we find that A z = ∅.
Let u be an arbitrary vertex in A z .By our choice of x, we have that |N L1 (u)| ≤ |N L1 (x)|, and so u must have a non-neighbour y u ∈ N L1 (x).Then u is non-adjacent to x otherwise G[u, x, y u , z] would be a C 4 , contradicting the fact that G is chordal.Now by Claim 1, we find that

The above implies that A
We now show that y u = y u ′ for any two vertices u ∈ A z and u ′ ∈ A z ′ and for any two (not necessarily distinct) vertices z, z ′ ∈ L 1 \ N L1 (x).First, suppose z, z ′ ∈ L 1 \ N L1 (x) are distinct.Such vertices exist since ∆ ≤ |L 1 | − 3. Let u ∈ A z and u ′ ∈ A z ′ .We may assume such vertices exist since A z and A z ′ are not empty because of Claim 2. If y u = y u ′ then, since y u , y u ′ ∈ N L1 (x) and z, z ′ ∈ L 1 \ N L1 (x), we find that y u , y u ′ , z and z ′ are distinct vertices in L 1 .Since N L1 (u) = (N L1 (x) \ {y u }) ∪ {z} and N L1 (u ′ ) = (N L1 (x) \ {y u ′ }) ∪ {z ′ }, we find that u is adjacent to y u ′ and z, but u ′ is non-adjacent to both y u ′ and z.Therefore Claim 1 implies that u and u ′ must be adjacent; however then G[u, u ′ , y u , y u ′ ] is a C 4 , a contradiction.Hence, y u = y u ′ .Since the u-vertices in different sets A z and A z ′ share the same y vertex, and there are at least two such sets, this immediately implies that u-vertices from the same set A z also share the same y-vertex.Thus there exists a vertex y * ∈ N L1 (x) such that for every u ∈ A z and every z ∈ L 1 \ N L1 (u), we have Let A = N L1 (x)\{y * }.Let A y * be the set of vertices in L 2 whose neighbourhood in L 1 is N L1 (x) (so x ∈ A y * ).Now for each vertex z ∈ L 1 \ A (including the case where z = y * ) and every u ∈ A z , we have Let X be the set of vertices u ∈ L 2 ∪ L 3 whose neighbourhood in L 1 is properly contained in N L1 (x), that is, for which N L1 (u) N L1 (x) = A ∪ {y * }.Note that, as no vertex in L 3 has a neighbour in L 1 , we have L 3 ⊆ X.Also note that the sets X and A z , z ∈ L 1 \ A form a partition of L 2 ∪ L 3 .
Consider two distinct vertices w 1 , w 2 ∈ L 1 \ A. Note that w 1 and w 2 are not necessarily distinct from y * , but at least one of w 1 , w 2 is distinct from y * .Also note that if a vertex u ∈ X is adjacent to w i (i = 1, 2) then w i = y * .
Suppose there is a path P in G[L 2 ∪ L 3 ] from some vertex t 1 ∈ A w1 to some vertex t 2 ∈ A w2 .We shall choose P such that |V (P )| is minimum, where the minimum is taken over all choices of w 1 , w 2 , t 1 , t 2 and P .It follows from the minimality of P that V (P ) \ {t 1 , t 2 } ⊆ X.Moreover, since N L1 (t 1 ) = A ∪ {w 1 } and N L1 (t 2 ) = A∪{w 2 }, it follows that w 1 (respectively w 2 ) is non-adjacent to t 2 (respectively t 1 ).Thus, t 1 and t 2 must be non-adjacent, as otherwise G[t 1 , t 2 , w 2 , w 1 ] would be a C 4 .
Without loss of generality, we may assume w 1 = y * .Since V (P )\{t 1 , t 2 } ⊆ X, we find that w 1 must be anti-complete to V (P ) \ {t 1 , t 2 }.Let t 3 be the neighbour of w 2 on V (P ) that is nearest to t 1 .(If w 2 has no neighbours in V (P ) \ {t 1 , t 2 } then t 3 = t 2 .)Note that t 3 = t 1 , since w 2 is not adjacent to t 1 .Let P ′ be the part of the path P from t 1 to t 3 .The only neighbour of w 1 in V (P ′ ) is t 1 .The only neighbour of w 2 in V (P ′ ) is t 3 .Since w 1 and w 2 are adjacent and P ′ is an induced path on at least two vertices it follows that G[V (P ′ ) ∪ {w 1 , w 2 }] is a cycle on at least four vertices, contradicting the fact that G is chordal.We have so far shown that for any two distinct vertices w 1 , w 2 ∈ L 1 \ A, there is no path in G[L 2 ∪ L 3 ] from any vertex of A w1 to any vertex of A w2 .Now suppose that u ∈ X.As ∆ ≤ |L 1 | − 3, there exist three pairwise distinct vertices w 1 , w 2 , w 3 ∈ L 1 \ A. By Claim 2, there exist three vertices t 1 ∈ A w1 , t 2 ∈ A w2 and t 3 ∈ A w3 .Because the sets A wi are mutually disjoint, t 1 , t 2 and t 3 are also distinct.It follows from the conclusion above that u can be adjacent to at most one of t 1 , t 2 and t 3 .Without loss of generality, assume that u is nonadjacent to t 1 and t 2 and that w 1 = y * .Then u cannot be adjacent to w 1 .By Claim 1, u must be adjacent to every vertex of A. Since N L1 (u) N L1 (x), it follows that N L1 (u) = A. Since u was an arbitrary vertex in X, together with the observations made earlier, this shows that every vertex in L 2 ∪ L 3 is adjacent to every vertex of A and at most one other vertex in L 1 .Since ∆ ≥ 2, we have that |A| ≥ 1, and so L 3 must be empty.By Fact 3, we may apply a bipartite complementation between A and L 2 after which we may apply Lemma 29.This completes the proof of Case 1.

Case 2: ∆
there are at most two vertices in L 1 \ N L1 (x).By Fact 1, we may delete these vertices, if they exist.Note that this changes neither the value of ∆ nor the choice of x.Therefore we may assume that L 1 = N L1 (x).
Then N L1 (w) ⊆ N L1 (x) for all w ∈ L 2 .If ∆ = |L 1 | ≤ 3 then by deleting at most two vertices of L 1 (which we may do by Fact 1) we obtain a new graph for which we may apply Lemma 29.We may therefore assume without loss of generality that ∆ ≥ 4.
We distinguish three subcases depending on whether or not x dominates L 2 and whether or not G[L 2 ] is a clique.

Case 2a:
x does not dominate L 2 .Let y ∈ L 2 be a non-neighbour of x.Recall that N L1 (x) = L 1 .By Claim 1 we find that y must be adjacent to all but at most one vertex of L 1 .If y is not adjacent to some vertex of L 1 , we may delete this vertex by Fact 1.We may therefore assume that ∆ ≥ 3 and that y is complete to L 1 .
Suppose w ∈ L 2 has two non-neighbours a, b ∈ N L1 (x).As {x, y} is complete to L 1 , it follows that w is adjacent to both x and y by Claim 1.However, then G[x, w, y, a] is a C 4 , contradicting the fact that G is chordal.Therefore, every vertex in L 2 is adjacent to all but at most one vertex of L 1 .In particular, as ∆ ≥ 3, every vertex in L 2 has at least two neighbours in L 1 .This fact, together with the fact that no vertex in L 3 has neighbours in L 1 and Claim 1, implies that every vertex of L 2 is adjacent to every vertex of L 3 .By applying a bipartite complementation between L 2 and L 3 , we separate free chordal graph, so it has bounded clique-width by Theorem 24.By Fact 3, we may therefore assume that L 3 = ∅.
Let X be the set of vertices in L 2 that are complete to L 1 .For z ∈ L 1 , let U z be the set of vertices in L 2 that are complete to L 1 \ {z} and non-adjacent to z.As every vertex in L 2 is adjacent to all but at most one vertex of L 1 , we find that the sets X and U z , z ∈ L 1 , form a partition of vertices of L 2 .
Suppose there are at most six vertices z ∈ L 1 such that U z is not empty.By Facts 1 and 3, we may apply a bipartite complementation between L 1 and L 2 and then delete these vertices.In the resulting graph, no vertex of L 2 has a neighbour in L 1 and we can apply Lemma 29.We may therefore assume that there are at least seven vertices z ∈ L 1 such that U z is not empty.
Consider two distinct vertices z 1 , z 2 ∈ L 1 .We claim that U z1 must be anti-complete to U z2 .Indeed, if y 1 ∈ U z1 were adjacent to y 2 ∈ U z2 then G[y 1 , y 2 , z 1 , z 2 ] would be a C 4 , contradicting the fact that G is chordal.
We will now show that by deleting at most one vertex from L 2 (which we may do by Fact 1), we can make G[L 2 ] into a P 3 -free graph.Indeed, suppose that G[L 2 ] contains an induced Since there are at least seven non-empty sets U z , there must be at least four non-empty sets U z that do not contain a vertex in {v 1 , v 2 , v 3 }.Therefore there must be two sets U z1 and U z2 containing vertices y 1 and y 2 , respectively, such that y 1 and y 2 are adjacent to the same vertex in {v 1 , v 2 , v 3 }, say v i .Since U z1 and U z2 are anti-complete, y 1 and y 2 are non-adjacent.Hence, G[y 1 , v i , y 2 ] is a P 3 .Also note that v i ∈ X since v i has neighbours in both U z1 and U z2 .Now let z 3 ∈ L 1 \ {z 1 , z 2 } and suppose y 3 ∈ U z3 .By the same argument as above, y 3 must have a neighbour in {y 1 , v i , y 2 }.Moreover, as U z3 is anti-complete to both U z1 and U z2 , we find that y 3 is non-adjacent to both y 1 and y 2 .Hence, y 3 must be adjacent to v i .Now choose z 4 , z 5 ∈ L 1 \ {z 1 , z 2 } with y 4 ∈ U z4 and same arguments as at the start of the proof, we may assume that ∆ ≥ 2 and so V (G) = {v} ∪ L 1 ∪ L 2 ∪ L 3 .Again, by Claim 2, we may assume that every vertex of L 1 has a neighbour in L 2 in G. Then if ∆ ≤ |L 1 |− 3, we may apply Case 1.We may therefore assume that ∆ ≥ |L 1 |−2.By the same arguments as at the start of Case 2, we may assume that |L 1 | = ∆ and ∆ ≥ 4. To make this assumption, we may have to delete vertices from L 1 , which could cause vertices that were in L 2 previously to now be in L 3 for this modified graph.However, at no point above do we add vertices to L 2 , so it is still the case that every component of G[L 2 ] is a clique.Therefore Case 3 or Case 2a applies, depending on whether G[L 2 ] now contains one or more components, respectively.This completes the proof of Theorem 30.

The Case H = S 1,1,2
We now show that the clique-width of S . We partition M as follows: For i ∈ {1, . . ., 5}, let M i be the set of vertices in M with exactly i neighbours in V (N ).Let U be the set of vertices in M adjacent to every vertex of V (N ).Let Z be the set of vertices in M with no neighbours in V (N ).Note that Z is an independent set in G, since G is 2P 2 -free.
We now analyse the structure of G through a series of claims.
By symmetry, we may assume that x is adjacent to at least one vertex in Next, we prove that the vertices in M 3 and M 4 have a restricted type of neighbourhood in V (N ): Claim 2. If x ∈ M 3 then x is adjacent either to exactly one end-vertex a i and its two opposite mid-vertices b j and b k (j = i, k = i) or x is adjacent to all three every vertex in Z 0 is anti-complete to X (due to the definitions of Z, Z 0 and Z 1 together with the fact that Z is an independent set).Since G is prime, X must be a trivial module.Since X contains more than one vertex, it follows that By symmetry we may assume that i = 1 and j 2 then, by Claims 2 and 3, we find that G[x, a 1 , y, a 2 ] is a 2P 2 .This completes the proof of Claim 7.

By symmetry we may assume that
1 is non-adjacent to y ∈ M id 3 then, by Claims 2 and 3, we find that G[b 2 , a 2 , y, x, a 1 ] is an S 1,1,2 .This completes the proof of Claim 8.By Claims 2, 3, 5, 7 and 8 we find that, for every i ∈ {1, 2, 3}, Next, we show the following: 3 is adjacent to every midvertex of N and non-adjacent to every end-vertex of N (by definition).Hence there must be vertices x, y ∈ Q and z ∈ Z 1 such that z distinguishes x and y, say z is adjacent to x in G, but not to y.Because G[Q] is connected, we may assume that x and y are adjacent in G, in which case x and y are non-adjacent in G.However, then G[b 3 , a 3 , y, x, z] is an S 1,1,2 .This completes the proof of Claim 9.
By Claim 9 and the definition of M id 3 , we find that {b 1 , b 2 , b 3 }∪M id 3 is a clique.By the definition of Z and the fact that Z is independent, {a 1 , a 2 , a 3 } ∪ Z 1 is an independent set.Therefore G is a split graph.By Lemma 9, since G is prime and S 1,1,2 -free, it must be a spider.Since G contains an induced net, it must be a thin spider.
⊓ ⊔ Lemma 32.If G is a prime chordal S 1,1,2 -free graph then it is either a 2P 1 + P 2free graph or a thick spider.
Proof.Let G be a prime S 1,1,2 -free chordal graph.Note that since G is S 1,1,2free, it cannot contain d-A or d-domino as an induced subgraph (see also Figs. 5  and 6).If G is P 1 + P 4 -free then, by Lemma 4, it must therefore be 2P 1 + P 2 -free.Now suppose that G contains an induced copy of P 1 + P 4 .Since G is prime, G is also prime.Furthermore, G is (2P 2 , C 5 , S 1,1,2 )-free.By Lemma 3, G must contain one of the graphs in Fig. 5.The only graph in Fig. 5 which is (2P 2 , C 5 , S 1,1,2 )-free is the net, so G must contain a net.By Lemma 31, G is a thin spider, so G is a thick spider, completing the proof.

⊓ ⊔
As a corollary of the above lemma, we get the following: Theorem 33.Every S 1,1,2 -free chordal graph has clique-width at most 4.
Proof.Let G be an S 1,1,2 -free chordal graph.By Lemma 12, we may assume that G is prime.If G is 2P 1 + P 2 -free then it has clique-width at most 3 by Lemma 19.By Lemma 32, we may therefore assume that G is a thick spider.Note that since a thick spider is the complement of a thin spider (see also the definition of a thin spider), K is an independent set, I is a clique and R is complete to I and anti-complete to K. Every vertex in K has exactly one nonneighbour in K and vice versa.Since G is prime and R is a module, R contains at most one vertex.Let i 1 , . . ., i p be the vertices in I and let k 1 , . . ., k p be the vertices in K such that for each j ∈ {1, . . ., p}, the vertex i j is the unique non-neighbour of k j in I. Let G j be the labelled copy of G[i 1 , . . ., i j , k 1 , . . ., k j ] where every i h is labelled 1 and every k h is labelled 2. Now G 1 = 1(i 1 ) ⊕ 2(k 1 ) and for j ∈ {1, . . ., p − 1} we can construct G j+1 from G j as follows: If R = ∅ then using the above recursively we get a 4-expression for G p and therefore for G.If R = {x} then we obtain a 4-expression for G using η 1,4 (G p ⊕ 4(x)).Therefore G indeed has clique-width at most 4.This completes the proof.

New Classes of Unbounded Clique-width
We present three new subclasses of H-free split graphs that have unbounded clique-width.Recall that every split graph is chordal.Hence, every H-free split graph is also an H-free chordal graph.We first consider the class of H-free split graphs.Note that the graph H is the graph whose complement looks like a capital letter "H" (see also Fig. 3).

Theorem 34. The class of H-free split graphs has unbounded clique-width.
Proof.By Lemma 15, the class of 2P 3 -free bipartite graphs has unbounded cliquewidth.Consider the class of graphs obtained from 2P 3 -free bipartite graphs by complementing one of the bipartition classes.By Fact 2, this class of graphs also has unbounded clique-width.Since such graphs are H-free split graphs, this completes the proof.

⊓ ⊔
The proofs of the next two results are based on the class of 1-subdivided walls.For n ≥ 1, we let G n be the 1-subdivided wall of height n, where A = A n denotes the set of vertices in G n before the edge subdivisions, and B = B n denotes the set of vertices introduced by the edge subdivisions.
Proof.By Lemma 13, the class of graphs {G n } n≥1 has unbounded clique-width.For n ≥ 1, let H n be the graph obtained from G n by applying a subgraph complementation to G n [A].By Fact 2 the class of graphs {H n } n≥1 has unbounded clique-width.We claim that every H n is a 3P 1 + P 2 -free split graph.Since A is a clique in H n and B is an independent set, H n is a split graph.Every vertex of B has degree 2. Note that 3P 1 + P 2 has minimum degree 3. Hence, if H n contains an induced 3P 1 + P 2 , then every vertex of it is in A. This is not possible as A is a clique in H n .
Proof.By Lemma 13, the class of graphs {G n } n≥1 has unbounded clique-width.
For n ≥ 1, let H n be the graph obtained from G n by applying a subgraph complementation to G n [A] and a bipartite complementation between A and B. By Facts 2 and 3, the set of graphs {H n } n≥1 has unbounded clique-width.Because A is a clique in H n and B is an independent set, H n is a split graph.Below we show that every H n is (K 3 + 2P 1 , K 4 + P 1 , P 1 + P 1 + P 4 )-free.
For contradiction, suppose that, for some n ≥ 1, F is an induced subgraph of H n isomorphic to K 3 + 2P 1 , K 4 + P 1 or P 1 + P 1 + P 4 .Let x be an isolated vertex in F .If x ∈ A then every vertex of V (F ) \ {x} would be in B because A is a clique, contradicting the fact that B is an independent set.Therefore x ∈ B. Note that x (and every other vertex of B) has precisely two non-neighbours in A. Therefore, at most two vertices of V (F ) \ {x} may be in A. This has the following implications.If F = K 3 + 2P 1 then exactly two vertices of the K 3 must be in A, as B is an independent set.Therefore the two independent vertices (which must both be in B) of F have two common non-neighbours in A, which is not possible.If F = K 4 + P 1 then at least two vertices of the K 4 must be in B, but this contradicts the fact that B is an independent set.If F = P 1 + P 1 + P 4 then at least three vertices of the P 1 + P 4 must be in B, but this contradicts the fact that B is an independent set, since the independence number of P 1 + P 4 is 2.

The Classifications
In this section we first prove our main result, Theorem 1, which was presented in Section 1.In order to do so, we will use results from the previous sections and the following lemma.Recall that the graphs F 1 and F 2 are the graphs displayed in Figure 4.
Lemma 37.For any graph H / ∈ {F 1 , F 2 } but with F 1 ⊆ i H or F 2 ⊆ i H, the class of H-free chordal graphs has unbounded clique-width.
Proof.If H is not a split graph then we apply Lemma 22. Hence, we may assume that H is split with split partition (K, I).Note that F 1 and F 2 are split graphs that each have a split partition into a clique K ′ on four vertices and an independent set I ′ on two vertices.Let K ′ = {a 1 , a 2 , a 3 , a 4 } and I ′ = {b 1 , b 2 } be such subsets of V (H), with H[K ′ ∪ I ′ ] = F i for i = 1 or 2, where a 1 b 1 and a 2 b 2 are edges of the F i , whereas a 3 b 1 is also an edge is a P 4 , which has a unique split partition.Note that a 4 may be in K or I and if i = 1 then a 3 can be in K or I, but if i = 2 then a 3 ∈ K, since a 3 is adjacent to b 1 and I is independent.Since I is independent, at most one of a 3 and a 4 is in I. Without loss of generality, we may therefore assume that a 3 ∈ K.
First suppose that H contains a vertex x such that {a 1 , a We may now assume that there is no vertex x such that {a 1 , a 2 , a 3 , a 4 , x} is a clique on five vertices.Let y ∈ V (H) \ V (F i ) (this vertex exists as H = F i ).First suppose that y ∈ I, in which case y is non-adjacent to b 1 and b 2 .Suppose y is non-adjacent to a 1 and a 2 .If y is not adjacent to a 4 then H[a 4 , b 1 , b 2 , y] is a 4P 1 and we apply Lemma 20.If y is adjacent to a 4 then H[a 1 , a 2 , a 4 , b 1 , b 2 , y] is a net, in which case we apply Lemma 23.Therefore y must be adjacent to at least one of a 1 and a 2 .Choose the largest j ∈ {1, 2} such that y is adjacent to a j .Since {a 1 , a 2 , a 3 , a 4 , y} is not a clique, y must be non-adjacent to at least one vertex in {a 1 , a 2 , a 3 , a 4 }.Choose the smallest k ∈ {1, 2, 3, 4} such that y is not adjacent to a k .Now H[a j , a k , b j , y] is a K 1,3 , and we apply Lemma 21.We conclude that y ∈ K. Since a 1 , a 2 , a 3 ∈ K and K is a clique, y must be adjacent to a 1 , a 2 and a 3 .Since {a 1 , a 2 , a 3 , a 4 , y} is not a clique, y must be non-adjacent to a 4 .If y is not adjacent to b j for j = 1 or 2 then H[a j , a 4 , b j , y] is a K 1,3 , and we apply Lemma 21.We may therefore assume that y is adjacent to both b 1 and b 2 .Now H[a 1 , a 2 , y, a 4 , b 1 , b 2 ] is a net and we apply Lemma 23.This completes the proof.

⊓ ⊔
We are now ready to prove Theorem 1, which we restate below.
is adjacent to at least one of v 2 , v 3 , w. Suppose y is adjacent to v 2 or v 3 .Then, as H is K 1,3 -free, y is adjacent to v 2 and v 3 .Hence, as K is maximal, y is not adjacent to w.Then w must be adjacent to both v 1 and v 4 as otherwise H would contain an induced K 1,3 .Then H is isomorphic to net and we can use Lemma 23.Suppose y is non-adjacent to both v 2 and v 3 .Then y is adjacent to w, which implies, together with the K 1,3 -freeness of H, that w is not adjacent to v 1 or v 4 .Hence, H is isomorphic to the net and we can use Lemma 23 again.Now suppose that |K| ≥ 4. Let w ′ ∈ K \ {v 2 , v 3 , w}.Suppose w is nonadjacent to both v 1 and v 4 .If w ′ is adjacent to exactly zero or one of v 1 and v 4 then H contains an induced F 1 or F 2 , respectively, and we can apply Lemma 37.If w ′ is adjacent to both v 1 and v 4 then H[w ′ , v 1 , v 4 , w] is a K 1,3 , which is a contradiction.By symmetry, both w and w ′ must therefore each have at least one neighbour in {v 1 , v 4 }.If w or w ′ are both adjacent to v 1 or both adjacent to v 4 then H contains an induced 3P 1 + P 2 and we can use Theorem 35.Therefore w and w ′ must each be adjacent to exactly one of v 1 and v 4 , but not both to the same vertex.In this case H contains an induced H and we apply Theorem 34.
From now on suppose that H is P 4 -free.Let L 1 ⊆ K consist of those vertices of K that are adjacent to all vertices in I.
We claim that L 1 = ∅ and L 2 = ∅.As K is maximal and I = ∅, we have that L 2 = ∅.For contradiction, suppose L 1 = ∅.Let u ∈ K have maximum number of neighbours in I.Because L 1 = ∅, we find that u ∈ L 2 .So there exists a vertex v ∈ I that is not adjacent to u.Because H is connected, v has a neighbour u ′ in K \ {u}.By our choice of u there exists a vertex v ′ ∈ I adjacent to u but not to u ′ .Then H[v ′ , u, u ′ , v] is a P 4 , which is not possible.Hence, indeed, L 1 = ∅.
We claim that |J 1 | = 1.This can be seen as follows.As H is K 1,3 -free and both L 1 = ∅ and L 2 = ∅, we find that |J 1 | ≤ 1. Suppose |J 1 | = 0. Recall that L 2 = ∅.Let u ∈ L 2 have the maximum number of neighbours in I over all vertices in L 2 .Because u ∈ L 2 , there exists a vertex v ∈ I that is not adjacent to u.As |J 1 | = 0, we find that v ∈ J 2 .Then v has a neighbour u ′ in L 2 \ {u}.By our choice of u there exists a vertex v ′ ∈ I adjacent to u but not to u ′ .Then H[v ′ , u, u ′ , v] is a P 4 , which is not possible.Hence, indeed |J 1 | = 0, and thus We claim that |J 2 | = 0.For contradiction, suppose J 2 contains a vertex x.Because K is maximal, there exists a vertex y ∈ K not adjacent to x.Then y ∈ L 2 .As |J 1 | = 1, there exists a vertex w ∈ J 1 .By the definition of J 1 , we find that w is not adjacent to y.As L 1 = ∅, there exists a vertex z ∈ L 1 .By the definition of L 1 , we find that z is adjacent to both w and x.Then H[z, w, x, y] is a K 1,3 , a contradiction.Hence, we conclude that |J 2 | = 0.
If |L 1 | ≥ 3 then the facts that |J 1 | = 1 and L 2 = ∅ imply that 3P We claim that H has an induced 3P 1 .For contradiction, suppose H is 3P 1free.Then every connected component of H is a path.As H is 3P 1 -free, H has at most two connected components, each of which is a path on at most four vertices.Because H is not an induced subgraph of P 4 , this means that H has exactly two connected components.As H is 3P 1 -free, each of them is a path on at most two vertices.As H is 2P 2 -free, at most one of them has an edge.However, then H is an induced subgraph of P 4 , a contradiction.Now, as H has an induced 3P 1 , the complement of the class of H-free perfect graphs contains the class of bipartite graphs, and as such has unbounded clique-width.Applying Fact 2 completes the proof.

An Application
In this section we give an application of Theorem 1 by showing how to use it to prove that the class of (K 4 , 2P 1 + P 3 )-free graphs has bounded clique-width (see also Fig. 9), which means that only 13 (non-equivalent) cases in [21] remain open.Theorem 38.The class of (K 4 , 2P 1 + P 3 )-free graphs has bounded clique-width.
Proof.Suppose G is a (K 4 , 2P 1 + P 3 )-free graph.If G is chordal then it is a K 4 -free chordal graph, in which case it has bounded clique-width by Lemma 16.
We may therefore assume that G contains an induced cycle C with vertices v 1 , v 2 , . . ., v k in that order, such that k ≥ 4. We may also assume that this induced cycle is chosen such that k is minimal.Note that k ≤ 7, otherwise We partition the vertices not on the cycle C as follows.For S ∈ {1, . . ., k}, let V S contain those vertices x ∈ V (G) \ C such that N C (x) = {v i | i ∈ S}.We say that a set V S is large if it contains at least seven vertices, otherwise we say that it is small.We now prove some useful properties about these sets.

Claim 2. If v i and v j are consecutive vertices of the cycle and {i
Claim 4. Suppose S, T ⊆ {1, . . ., k} with S = T .If V S and V T are independent sets in G and V T is large then at most one vertex of V S has more than one non-neighbour in V T .Indeed, since |V T | ≥ 7 ≥ 4, by Claim 1 for any pair of vertices x, x ′ ∈ V S , at least one of these vertices must have at least two neighbours in V T .Therefore every vertex of V S except perhaps one has at least two neighbours in V T .Consider a vertex x ∈ V S that has two neighbours y, y ′ ∈ V T .The vertex x cannot have two non-neighbours z, z ′ ∈ V T , otherwise G[z, z ′ , y, x, y ′ ] would be a 2P 1 + P 3 .Therefore every vertex of V S except perhaps one has at most one non-neighbour in V T .Hence, at most one vertex of V S has more than one non-neighbour in V T .

Claim 5. Suppose S, T, U
is small then by Fact 1 we may assume it is empty.By Claim 4 and Fact 1, we may delete at most two vertices from each of V S , V T , V U after which every vertex in each of these sets will have at most one non-neighbour in each of the other two sets.In other words, every vertex in one of these sets will have at most two non-neighbours in total in the other two sets.Applying a bipartite complementation between each pair of sets (which we may do by Fact 3) yields a graph of maximum degree at most 2.This graph has bounded clique-width by Lemma 11.Claim 6. Suppose R, S, T, U ⊆ {1, . . ., k} are pairwise distinct.If V R , V S , V T , V U are all independent sets in G then at least one of V R , V S , V T , V U is small.Indeed, suppose for contradiction that all of V R , V S , V T , V U are large.Let V ′ R , V ′ S , V ′ T and V ′ U be the sets of those vertices in V R , V S , V T and V U , respectively, that do not have two non-neighbours in any of the three other sets.By Claim 4, each of there must be a vertex s ∈ V ′ S adjacent to r.Since |V ′ T | ≥ 3, there must be a vertex t ∈ V ′ T adjacent to r and s.Since |V ′ U | ≥ 4, there must be a vertex u ∈ V ′ U adjacent to r, s and t.Now G[r, s, t, u] is a K 4 , a contradiction.If any set V S is small then, by Fact 1, we may assume it is empty.We may therefore assume that every set V S is either large or empty.Furthermore, we may assume that some large set V S is not an independent set, otherwise we can apply Claim 6, to find that at most three sets V S are non-empty and then, after deleting the k ≤ 7 vertices of C (which we may do by Fact 1), we can apply Claim 5 to find that the clique-width of G is bounded.
We claim that k = 4.For contradiction, suppose that 5 ≤ k ≤ 7. Let S ⊆ {1, . . ., k} be a set such that G[V S ] is large and not independent.By Claim 3, it follows that |S| ≥ 2. By Claim 2, the vertices of V S cannot be adjacent to two consecutive vertices of C. Without loss of generality, assume that 1 ∈ S, which implies that 2, k / ∈ S. Then there must be a number j ∈ {3, . . ., k − 1} such that j ∈ S, and 2, . . .
contradicting the minimality of k.Hence, we conclude that indeed k = 4.
Again, let S ⊆ {1, . . ., k} be a set such that G[V S ] is large and not independent.By Claims 2 and 3, we find that S = {1, 3} or S = {2, 4}.If there exist vertices x, y, z ∈ V {1,3} that induce a P 3 then G[v 2 , v 4 , x, y, z] would be a 2P 1 + P 3 , which is not possible.Therefore G[V {1,3} ] must be P 3 -free, so it must be a disjoint union of cliques.If G[V {1,3} ] contained a K 3 on vertices x, y, z then G[v 1 , x, y, z] would be a K 4 , which is not possible.Thus every component of G[V {1,3} ] and (by symmetry) G[V {2,4} ] must be isomorphic to either P 1 or P 2 .
If G[V {1,3} ] and G[V {2,4} ] each contain at most one edge then, by deleting at most one vertex from each of V {1,3} and V {2,4} (which we may do by Fact 1), we obtain a graph in which every set V S is independent, in which case we find that G has bounded clique-width by proceeding as before: we first apply Claim 6, then delete the vertices of C by Fact 1 and finally apply Claim 5. Without loss of generality, we may therefore assume that G[V {1,3} ] contains two edges xx ′ and yy ′ (which together induce a 2P 2 ).
We claim that every set V T other than V {1,3} and V {2,4} is empty.Indeed, for contradiction, suppose such a set V T is non-empty.Then, as stated above, V T must be independent and large.By Claim 3, |T | ≥ 2. By symmetry we may therefore assume that {1, 2} ⊆ T .If z ∈ V T is adjacent to both x and x ′ then G[x, x ′ , v 1 , z] would be a K 4 , which is not possible.Therefore any vertex in V T can be adjacent to at most one vertex in each of {x, x ′ } and {y, y ′ }.Since |V T | ≥ 7 ≥ 5, we find that T contains two vertices z, z ′ , which are not adjacent to each other (as T is independent) and which are both non-adjacent to the same vertex in {x, x ′ } and to the same vertex in {y, y ′ }.By Claim 1, this is a contradiction, so V T must indeed be empty.
Recall that by Fact 1 we may delete the four vertices of C. We are therefore reduced to proving that G[V {1,3} ∪ V {2,4} ] has bounded clique-width.Note that if x ∈ V {1,3} is non-adjacent to two vertices y and y ′ in V {2,4} then y and y ′ must be adjacent, otherwise G[y, y ′ , v 1 , x, v 3 ] would be a 2P 1 + P 3 (which is not possible).This, together with the fact that G is K 4 -free, implies that any vertex in V {1,3} has at most two non-neighbours in V {2,4} , and vice versa.Let G ′ be the graph obtained from G[V {1,3} ∪ V {2,4} ] by applying a bipartite complementation between V {1,3} and V {2,4} .Then G ′ has maximum degree at most 3.By Fact 3, it remains to show that every connected component of G ′ has bounded cliquewidth.
Consider a connected component D of G ′ .We first prove that D contains at most four vertices of degree 3. Let x ∈ D be a vertex that has degree 3 in D. Without loss of generality assume that x ∈ V {1,3} .Then x has two neighbours y, y ′ ∈ V {2,4} and one neighbour x ′ ∈ V {1,3} .Recall that y is adjacent to y ′ due to the fact that G is 2P 1 +P 3 -free.For the same reason and because G[V {1,3} ] only has connected components isomorphic to P 1 or P 2 , we find that y and y ′ are adjacent to x ′ in D if they have degree 3 in D. Hence either V (D) = {x, x ′ , y, y ′ } or y, y ′ each have degree 2 in D and x ′ is a cut-vertex of D. In the first case, D has at most four vertices of degree 3.In the second case, we note that x ′ is adjacent to neither y nor y ′ in D (otherwise, for the same reason as before, x ′ would be adjacent to both of them if it had degree 3 in D, so V D would only contain the vertices x, x ′ , y, y ′ ).We then find that D is either obtained by identifying a vertex of a triangle and the end-vertex of a path, meaning that D has only one vertex of degree 3 (namely x), or else by connecting two vertex-disjoint triangles via a path between one vertex of one triangle and one of the other, meaning that D has exactly two vertices of degree 3.
Because D has at most four vertices of degree 3, we may remove these vertices by Fact 1 and then apply Lemma 11 to find that D has bounded clique-width.This completes the proof of Theorem 38.

Concluding Remarks
In our main result we characterized all but two graphs H for which the class of H-free chordal graphs has bounded clique-width.In particular we identified four new graph classes of bounded clique-width, namely the classes of H-free chordal graphs with H ∈ {K 1,3 + 2P 1 , P 1 + P 1 + P 3 , P 1 + 2P 1 + P 2 , S 1,1,2 }.We also showed that the restriction from H-free graphs to H-free perfect graphs does not yield any new classes of bounded clique-width.Moreover, we determined a new class of (H 1 , H 2 )-free graphs, namely the class of (K 4 , 2P 1 + P 3 )-free graphs, that has bounded clique-width via a reduction to chordal graphs.The latter means that only the following 13 cases, up to an equivalence relation, 6 are open in the classification for (H 1 , H 2 )-free graphs (see [21]).
1. We identify the following three main directions for future work.
Firstly, we still need, of course, to determine whether of not the classes of H-free chordal graphs have bounded clique-width when H ∈ {F 1 , F 2 }.For this purpose, we recently managed to show that the class of F 1 -free split graphs has bounded clique-width and we are currently exploring whether it is possible to generalize the proof of this result to the class of F 1 -free chordal graphs.This seems to be a challenging task.Determining whether the class of F 2 -free chordal graphs has bounded clique-width seems to be even more difficult.In particular, currently it is not even known whether or not the class of F 2 -free split graphs has bounded clique-width.
Secondly, we believe that the techniques developed in our paper may also be useful for attacking some of the other open cases in the classification for (H 1 , H 2 )-free graphs.In particular the case H 1 = 2P 1 +P 3 , H 2 = 2P 1 + P 3 seems a good candidate for a possible proof of bounded clique-width via a reduction to 2P 1 + P 3 -free chordal graphs (this subclass of chordal graphs has bounded cliquewidth by Theorem 1).For this direction we also note that it may be worthwhile to more closely examine the relationship between our study and the one on the computational complexity of the Graph Isomorphism problem (GI) for classes of (H 1 , H 2 )-free graphs, which was initiated by Kratsch and Schweitzer [38].Recently, Schweitzer [50] proved that for this study the number of open cases is finite and pointed out similarities between classifying boundedness of cliquewidth and solving GI for special graph classes.
Thirdly, the fact that the (un)boundedness of the clique-width of the class of H-free split graphs is known for so many graphs H raises the question whether we can obtain a full classification of all graphs H for which the class of H-free split graphs has bounded clique-width.We have now reduced this to four problematic cases (one of which is the case H = F 2 ).
Finally we pose the question of whether it is possible to extend the four newly found classes of H-free chordal graphs (when H ∈ {K 1,3 + 2P 1 , P 1 + P 1 + P 3 , P 1 + 2P 1 + P 2 , S 1,1,2 }) to larger classes of graphs for which Hamilton Cycle is polynomial-time solvable.

F1F2Fig. 4 .
Fig. 4. The graphs H for which the boundedness of clique-width of the class of H-free chordal graphs is open.

, c 3 ∈
S 2 .Then c 2 and c 3 must have the same pair of non-neighbours a, b ∈ K by Claim 1.If c 1 ∈ S 2 then by Claim 1, the non-neighbours of c 1 in K are also a and b.If c 1 ∈ S 1 then by Claim 3, the non-neighbour of c 1 in K is either a or b.Hence, in both these cases, K \ {a, b} ∪ {c 1 , c 2 , c 3 } is a clique of size more that |K|, contradicting the maximality of K.

Case 1 : 1 . 2 :
G[Y ] contains an induced 2P 2 .First suppose all vertices of this 2P 2 are in the same connected component D of G[Y ].Since split graphs are 2P 2 -free by Lemma 8, we find that D is a tree by Lemma 26.In this case, by Fact 1, we may delete one vertex in D so that the two edges of the 2P 2 are in two different connected components of G[Y ].We may therefore assume without loss of generality that G[Y ] contains two non-trivial components.Let Y ′ be the set of vertices in Y that are in non-trivial components of G[Y ].Let Y ′′ be the set of vertices in Y ′ that are complete to X. Let X ′ be the set of vertices in X that are complete to Y .It follows from Claim 1 that X \ X ′ is anticomplete to Y ′ \ Y ′′ .We can apply two bipartite complementation operations, one between X ′ and Y ′ ∪ {x} and the other between Y ′′ ∪ {x} and X \ X ′ .This will separate G[Y ′ ∪ {x}] from the rest of the graph.By Lemma 27, we find that G[Y ′ ∪ {x}] has bounded clique-width.Because G[V \ (Y ′ ∪ {x})] is a (P 1 + P 1 + P 3 )-free split graph, it has bounded clique-width by Lemma 25.By Fact 3, we find that G has bounded clique-width.This completes the proof of CaseCase G[Y ] is 2P 2 -free.If G[Y ] contains only trivial components then G is a (P 1 + P 1 + P 3 )-free split graph, so it has bounded clique-width by Lemma 25.Since G[Y ] is 2P 2 -free, it can contain at most one non-trivial component.We may therefore assume that G[Y ] contains exactly one non-trivial component D.
and anti-complete to D ′′ W .By Fact 3, we may apply two bipartite complementations: one between Z ′ ∪ D B and D ′ W ∪ D ′′ W ∪ Z and the other between Z ′′ and D ′ W ∪ Z.The resulting graph will be partitioned into two disjoint graphs: G[D W ∪ Z] and G[D B ∪ Z ′ ∪ Z ′′ ].The first of these is a (P 1 + P 1 + P 3 )-free split graph, so it has bounded clique-width by Lemma 25.Taking the complement of G[D B ∪ Z ′ ∪ Z ′′ ] (which we may do by Fact 2) yields the bipartite graph G[D B ∪ Z ′ ∪ Z ′′ ], which is 2P 2 -free since G is chordal and therefore has bounded clique-width by Lemma 15.We conclude that G has bounded clique-width.This completes the proof of Theorem 28.⊓ ⊔ 3.3 The Case H = P 1 + 2P 1 + P 2

, a 3
, b 1 , b 2 , b 3 such that a 1 , a 2 , a 3 is an independent set (the end-vertices of N ), b 1 , b 2 , b 3 is a clique (the mid-vertices of N ), and the only edges between a 1 , a 2 , a 3 and b 1 , b 2 , b 3 are
The graphs H for which the class of H-free chordal graphs has bounded cliquewidth; the four graphs at the top are new cases proved in this paper.
. For a class of graphs G of bounded maximum degree, let G ′ be a class of graphs that is (∞, es)-obtained from G, where es is the edge subdivision operation.Then G has bounded clique-width if and only if G ′ has bounded clique-width Suppose for contradiction that G[S] is not a forest.Then, since G is chordal, G[S] must contain a C 3 , on vertices c 1 , c 2 , c 3 , say.By Claim 4, we may assume without loss of generality that c 2 , c 3 / ∈ S 1 and thus c 2 S 2 have a common non-neighbour.Suppose that this is not the case.Then there exist two non-adjacent vertices t, t ′ ∈ S 2 and four distinct vertices a, b, c, d ∈ K with t non-adjacent to a and b and with t ′ non-adjacent to c and d.As t and t ′ belong to S 2 , it follows that t is adjacent to c and d, and that t ′ is adjacent to a and b.Since k ≥ 7, we find that t and t ′ have two common neighbours in K.These two common neighbours, together with c, d, t, t ′ form an induced K 1,3 + 2P 1 , a contradiction.Claim 3.If a vertex s ∈ S 1 is adjacent to a vertex t ∈ S 2 then s and t must have a common non-neighbour in K. Indeed, let v be the unique non-neighbour of s in K. Then v must be a nonneighbour of t otherwise a non-neighbour of t in K, together with s, t and v would induce a C 4 in G.This contradicts the fact that G is chordal.Claim 4. S 1 is an independent set.This holds as no two vertices in S 1 with a common non-neighbour in K are adjacent since K is maximum; while no two vertices in S 1 with different nonneighbours in K are adjacent since G is chordal.Claim 5. G[S] is a forest.
Lemma 31.If a prime (2P 2 , C 5 , S 1,1,2 )-free graph G contains an induced subgraph isomorphic to the net then G is a thin spider.Proof.Suppose that G is a prime (2P 2 , C 5 , S 1,1,2 )-free graph and suppose that G contains a net, say N with vertices a 1 , a 2 2 , a 3 , a 4 , x} is a clique on five vertices.Then x must be adjacent to b 2 , otherwise H[a 1 , a 3 , a 4 , x, b 2 ] would be a K 4 + P 1 and we could apply Theorem 36.This means that x must be non-adjacent to b 1 , otherwise H[x, a 4 , b 1 , b 2 ] would be a K 1,3 and we could apply Lemma 21.Now H[a 2 , a 3 , a 4 , x, b 1 ] is a K 4 + P 1 if i = 1 and H[a 1 , a 2 , a 3 , b 1 , b 2 , x] is an H if i = 2, and we apply Theorem 36 or 34, respectively.

Theorem 1 .
Let H be a graph with H / ∈ {F 1 , F 2 }.The class of H-free chordal graphs has bounded clique-width if and only if 1 + P 2 ⊆ i H, and we use Theorem 35.So we may assume that |L 1 | ≤ 2. We claim that |L 2 | ≥ 4. For contradiction, suppose that |L 2| ≤ 3.Then, as |J 1 | = 1, |J 2 | = 0 and |L 1 | ≤ 2, we find that H ⊆ i K 1,3 + 2P1 , a contradiction.Hence |L 2 | ≥ 4.Then, K 4 + P 1 ⊆ i H and we use Theorem 36.This completes the proof of Theorem 1. ⊓ ⊔ We now prove our dichotomy for H-free perfect graphs, which we recall below.Let H be a graph.The class of H-free perfect graphs has bounded clique-width if and only if H is an induced subgraph of P 4 .Proof.Let H be a graph.First suppose that H is an induced subgraph of P 4 .Then the class of H-free perfect graphs is contained in the class of P 4 -free graphs, which is known to have bounded clique-width (see also Lemma 14).Now suppose that H is not an induced subgraph of P 4 .Below we show that the class of Hfree perfect graphs has unbounded clique-width.Suppose that H is not a split graph.Then the class of H-free perfect graphs contains the class of split graphs, which has unbounded clique-width by Lemma 22. From now on assume that H is a split graph.Suppose that H contains a cycle C.As H is a split graph, it is (C 4 , C 5 , 2P 2 )-free by Lemma 8. Hence, C is isomorphic to C 3 .Then the class of H-free perfect graphs contains the class of bipartite graphs, which has unbounded clique-width (see also Lemma 15).From now on assume that H contains no cycle.