Clique-Width for Four-Vertex Forbidden Subgraphs

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1 Clique-Width for Four-Vertex Forbidden Subgraphs Andreas Brandstädt 1 Joost Engelfriet 2 Hoàng-Oanh Le 3 Vadim V. Lozin 4 March 15, Institut für Informatik, Universität Rostock, D Rostock, Germany. ab@informatik.uni-rostock.de 2 LIACS, Leiden University, P.O. Box 9512, 2300 RA Leiden, The Netherlands. engelfri@liacs.nl 3 Fachbereich Informatik, Technische Fachhochschule Berlin, D Berlin, Germany. oanhle@tfh-berlin.de 4 RUTCOR, Rutgers University, 640 Bartholomew Rd., Piscataway, NJ , U.S.A. lozin@rutcor.rutgers.edu Abstract Clique-width of graphs is a major new concept with respect to efficiency of graph algorithms. The notion of clique-width extends the one of treewidth, since bounded treewidth implies bounded clique-width. We give a complete classification of all graph classes defined by forbidden induced subgraphs of at most four vertices with respect to bounded or unbounded clique-width. Keywords: Clique-width of graphs; efficient graph algorithms. 1 Introduction Recently, in connection with graph grammars, in [17] the notion of clique-width of a graph was introduced which, by now, has attracted much attention since, in [18], Courcelle, Makowsky and Rotics have shown that every graph problem expressible in LinEMSOL(τ 1,L ) (a variant of Monadic Second Order Logic) is linear-time solvable on graphs with bounded clique-width if the input graph is given together with a k- expression defining it. (The time complexity may be sublinear with respect to the size of the input graph G if the input is just a k-expression of G). Research of the third author partially supported by German Research Community DFG Br /1, /2 1

2 Various NP-complete problems such as Vertex Cover, Maximum Weight Stable Set (MWS), Maximum Weight Clique, Steiner Tree, Domination, k-colorability for fixed k 3 and Maximum Induced Matching are LinEMSOL(τ 1,L ) expressible. Restricting the input to some graph classes defined by forbidding small graphs leads to polynomial time algorithms for some problems; for example, Minty [29] gave a polynomial time algorithm for the MWS problem on claw-free graphs (see also [34]), Randerath [32] and Randerath et al. [33] discussed k-colorability of graph classes defined by small forbidden subgraphs, and Corneil et al. [15] described how MWS and related problems can be solved bottom-up along the cotree of a P 4 -free graph (also called cograph). It is known that a graph is P 4 -free if and only if its clique-width is at most 2, and a 2-expression can be found in linear time along its cotree. Thus, it is a natural question to ask which other forbidden 4-vertex graphs (and which of their combinations) will lead to bounded clique-width. Figure 1 contains all 4-vertex graphs. Figure 1: All four-vertex graphs In [11], the clique-width (and some structure results) of (H,co-H)-free graphs was described for any 4-vertex graph H. Thus e.g., the (diamond,co-diamond)-free graphs and the (claw,co-claw)-free graphs have bounded clique-width. Independently, it was shown in [1] in a different way that (claw,co-claw)-free graphs as well as (claw,paw)-free graphs have bounded clique-width (and simple structure). We extend these results by giving a complete classification of all graph classes defined in terms of some forbidden graphs with at most four vertices with respect to bounded or unbounded clique-width. This is done by identifying 14 inclusion-maximal classes of bounded clique-width and four inclusion-minimal classes of unbounded clique-width (see Figure 2). In particular, it will turn out that for a 4-vertex graph H, the class 2

3 Figure 2: Essential classes for all combinations of forbidden 4-vertex graphs; + ( ) denotes bounded (unbounded) clique-width of H-free graphs has bounded clique-width if and only if H is the P 4, and every class defined by six forbidden 4-vertex graphs has bounded clique-width. This also continues research done in [2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13] and is partially based on some of the results of these papers. This paper is organized as follows: In Section 2, some basic notions are given. In Section 3, the notion of clique-width is described, and facts on clique-width are presented. Section 4 contains the results on bounded clique-width for classes defined by two forbidden 4-vertex graphs. Section 5 contains the results on bounded clique-width for classes defined by three forbidden 4-vertex graphs. Section 6 contains the results on unbounded clique-width. Section 7 summarizes the results and discusses some consequences. 3

4 2 Basic notions Throughout this paper, let G = (V, E) be a finite undirected graph without self-loops and multiple edges and let V = n, E = m. Let V (G) = V denote the vertex set of graph G. For a vertex v V, let N(v) = {u uv E} denote the (open) neighborhood of v in G, let N[v] = {v} {u uv E} denote the closed neighborhood of v in G, and for a subset U V and a vertex v / U, let N U (v) = {u u U, uv E} denote the neighborhood of v in U. Let u v if uv E and u v otherwise. Disjoint vertex sets X, Y form a join, denoted by X 1 Y (co-join, denoted by X 0 Y ) if for all pairs x X, y Y, xy E (xy / E) holds. We will also say that X has a join to Y, that there is a join between X and Y, or that X and Y are connected by join (and similarly for co-join). Subsequently, we will consider join and co-join also as operations, i.e., the co-join operation for disjoint vertex sets X and Y is the disjoint union of the subgraphs induced by X and Y, and the join operation for X and Y consists of the co-join operation for X and Y followed by adding all edges xy, x X, y Y. A vertex z V distinguishes vertices x, y V if zx E and zy / E. We also say that vertex z distinguishes a vertex set U V, z / U, if z has a neighbor and a nonneighbor in U. A vertex set M V is a module if no vertex from V \ M distinguishes M, i.e., every vertex v V \ M has either a join or a co-join to M. A module is trivial if it is either the empty set, a one-vertex set or the entire vertex set V. Nontrivial modules are called homogeneous sets. A graph is prime if it contains only trivial modules. The notion of module plays a crucial role in the modular (or substitution) decomposition of graphs (and other discrete structures) which is of basic importance for the design of efficient algorithms - see e.g. [30] for modular decomposition of discrete structures and its algorithmic use and [28] for a linear-time algorithm constructing the modular decomposition tree of a given graph. For U V, let G[U] denote the subgraph of G induced by U. Throughout this paper, all subgraphs are understood to be induced subgraphs. Let F denote a set of graphs. A graph G is F-free if none of its induced subgraphs is in F. A vertex set U V is stable (or independent) in G if the vertices in U are pairwise nonadjacent. Let co-g = G = (V, E) denote the complement graph of G. A vertex set U V is a clique in G if U is a stable set in G. Let K l denote the clique with l vertices, and let lk 1 denote the stable set with l vertices. For k 1, let P k denote a chordless path with k vertices and k 1 edges, and for k 3, let C k denote a chordless cycle with k vertices and k edges. The 2K 2 is the complement of C 4 (see Figure 1). A graph is chordal if it contains no induced C k, k 4. Note that the P 4 is the smallest nontrivial prime graph with at least three vertices and the complement of a P 4 is a P 4 itself. For a subgraph H of G, a vertex not in H is a k-vertex for H if it has exactly k neighbors in H. We say that H has no k-vertex if there is no k-vertex for H. For a 4

5 set S V (H) with S = k let N S (H) (N S for short if H is understood) denote the set of k-vertices for H adjacent to vertices in S. We also write N ab respectively N x for S = {a, b} respectively S = {x}, etc. The subgraph H dominates the graph G if H has no 0-vertex in G. See Figure 3 for the definition of chair, P, bull, gem and their complements. In particular, the house is the co-p 5. See Figure 1 for the definition of diamond, paw, claw and their complements. Figure 3: All one-vertex extensions of a P 4 A graph is a split graph if its vertex set is partitionable into a clique and a stable set. Lemma 1 ([21]) G is a split graph if and only if G is (2K 2, C 4, C 5 )-free. A graph is a k-sun for k 3 if it consists of 2k vertices, say u 1,...,u k, v 1,...,v k, such that {u 1,...,u k } is a clique, and v i is exactly adjacent to u i and u i+1, i {1,...,k} (index arithmetic modulo k). A graph is sun-free if it is k-sun-free for all k 3. A graph is strongly chordal [20] if it is chordal and sun-free. A vertex w is simple in graph G if its open neighborhood N(w) = {w 1,...,w l } is a clique and the closed neighborhoods of w 1,..., w l can be linearly ordered by set inclusion. In [20], it is shown that every strongly chordal graph has a simple vertex. Figure 4: 3-sun and its complement graph, the net We will also need the following classes of graphs: 5

6 G is a thin spider if its vertex set is partitionable into a clique C and a stable set S with C = S or C = S + 1 such that the edges between C and S are a matching and at most one vertex is not covered by the matching. The thin spider with 6 vertices is also called net (see Figure 4). A graph is a thick spider if it is the complement of a thin spider. The complement of the net is the 3-sun (see Figure 4). G is matched co-bipartite if its vertex set is partitionable into two cliques C 1, C 2 with C 1 = C 2 or C 1 = C 2 1 such that the edges between C 1 and C 2 are a matching and at most one vertex is not covered by the matching. G is co-matched bipartite if G is the complement of a matched co-bipartite graph. A bipartite graph B = (X, Y, E) is a bipartite chain graph [35] if there is an ordering x 1, x 2,...,x k of all vertices in X such that N(x i ) N(x j ) for all 1 i < j k. (Note that then also the neighborhoods of the vertices from Y are linearly ordered by set inclusion.) If, moreover, X = Y = k and N(x i ) = {y 1,...,y i } for all 1 i k, then B is prime. G is a co-bipartite chain graph if it is the complement of a bipartite chain graph. G is an enhanced co-bipartite chain graph if it is partitionable into a co-bipartite chain graph with cliques C 1, C 2 and three additional vertices a, b, c (a and c optional) such that N(a) = C 1 C 2, N(b) = C 1, and N(c) = C 2, and there are no other edges in G. G is an enhanced bipartite chain graph if it is the complement of an enhanced co-bipartite chain graph. G is a tractable graph if G is (4K 1,C 4,claw)-free and its vertex set can be partitioned into four (possibly empty) pairwise disjoint vertex sets Q 1, Q 2, Q 3 and Q 4 which induce cliques in G such that there are no edges between Q i and Q i+2 for both i = 1 and i = 2. Note that G[Q i Q i+1 ], i {1,..., 4} (index arithmetic modulo 4) are co-bipartite chain graphs since G is C 4 -free. The P 6 and the C 7 are examples of (prime) tractable graphs. See Figure 5 for C 7 and a larger example. 3 Cographs, clique-width and logical expressibility of problems The P 4 -free graphs (also called cographs) play a fundamental role in graph decomposition; see [16] for linear time recognition of cographs, [14, 15, 16] for more information on P 4 -free graphs and [9] for a survey on this graph class and related ones. 6

7 Figure 5: Examples of tractable graphs For a cograph G, either G or its complement is disconnected, and the cotree of G expresses how the graph is recursively generated from single vertices by repeatedly applying join and co-join operations. Note that the cographs are those graphs whose modular decomposition tree contains only join and co-join nodes as internal nodes. Based on the following operations on vertex-labeled graphs, namely (i) create a vertex u labeled by integer l, denoted by l(u), (ii) disjoint union (i.e., co-join), denoted by, (iii) join between all vertices with label i and all vertices with label j for i j, denoted by η i,j, and (iv) relabeling all vertices of label i by label j, denoted by ρ i j, the notion of clique-width cwd(g) of a graph G is defined in [17] as the minimum number of labels which are necessary to generate G by using the operations (i) (iv). It is easy to see that cographs are exactly the graphs whose clique-width is at most two. A k-expression for a graph G of clique-width k describes how G is recursively generated by repeatedly applying the operations (i) (iv) using at most k different labels. Observe that, trivially, the clique-width of a graph with n vertices is at most n. The following result by Johansson gives a slightly sharper bound. Lemma 2 ([25]) If G has n vertices then cwd(g) n k as long as 2 k + 2k n. Thus, for instance, the clique-width of a graph with nine vertices is at most seven. Lemma 3 ([1]) Let G = (V, E) be a graph and V = F 1 F 2 be a partition of V with F 2 s. If there is a t-expression for G[F 1 ] then there is a (2 s (t + 1))-expression for G. 7

8 This lemma which is contained in Theorem 2 of [1] means that adding a constant number s of vertices to a graph H from a class of bounded clique-width maintains bounded clique-width. In some of our proofs this allows us to disregard certain specific vertices and thus to reduce graph G to its essential part G. Proposition 1 ([18, 19]) (i) The clique-width cwd(g) of a graph G is the maximum of the clique-width of its prime induced subgraphs. (ii) cwd(g) 2 cwd(g). In [18], it is shown that every problem expressible in a certain kind of Monadic Second Order Logic, called LinEMSOL(τ 1,L ), is linear-time solvable on any graph class with bounded clique-width for which a k-expression can be constructed in linear time. Roughly speaking, MSOL(τ 1 ) is Monadic Second Order Logic with quantification over subsets of vertices but not of edges; MSOL(τ 1,L ) is MSOL(τ 1 ) with additional vertex labels, and LinEMSOL(τ 1,L ) is the variant of MSOL(τ 1,L ) which allows to search for sets of vertices which are optimal with respect to some linear evaluation functions. Theorem 1 ([18]) Let C be a class of graphs of clique-width at most k such that there is an O(f( E, V )) algorithm, which for each graph G in C, constructs a k-expression defining it. Then for every LinEMSOL(τ 1,L ) problem on C, there is an algorithm solving this problem in time O(f( E, V )). The next, straightforward, proposition was already stated in [3]. Proposition 2 The clique-width is at most (i) 3 for chordless paths as well as for their complements, (ii) 4 for chordless cycles as well as for their complements, (iii) 3 for thin spiders, 4 for thick spiders, (iv) 3 for bipartite chain graphs, 3 for co-bipartite chain graphs, (v) 4 for matched co-bipartite as well as for co-matched bipartite graphs, (vi) 4 for enhanced bipartite chain graphs as well as for enhanced co-bipartite chain graphs, and corresponding k-expressions, k {3, 4}, can be obtained in linear time. Theorem 2 ([2]) If G is a prime (P 5,diamond)-free graph then G is matched cobipartite or G is a thin spider or G is an enhanced bipartite chain graph or G has at most nine vertices. 8

9 Thus, by Propositions 1 and 2 and Lemma 2, the clique-width of (P 5,diamond)-free graphs as well as their complements is at most 7. For a similar purpose we will use the next result. Theorem 3 ([11]) Let G be a prime graph. (i) If G is (diamond,co-diamond)-free then G or G is a matched co-bipartite graph or G has at most nine vertices. (ii) If G is (claw,co-claw)-free then G or G is an induced path or cycle or G has at most nine vertices. (iii) If G is (paw,co-paw)-free then G is a P 4 or C 5. Theorem 4 ([6]) The clique-width of (P 5, gem)-free graphs is at most 5. Theorem 5 ([8]) The clique-width of (chair, co-p, gem)-free graphs is at most 7. Actually, the clique-width bound in [8] is nine instead of seven but the better bound follows immediately from Theorem 2 in [8] and Lemma 2. Finally in this section, we show that tractable graphs have bounded clique-width which is used later for the proof that (4K 1,C 4,claw)-free graphs have bounded clique-width. As a preparing step, we define typical graphs T n (see Figure 6 for an example): Figure 6: The typical graph T 4 Let T n be the graph with vertex set {1, 2,..., n} {1, 2, 3, 4} and edge set {(s, t)(x, y) (y = t + 1 and x s) or (x s and y = t)} (index arithmetic modulo 4). We call {1, 2,..., n} {i} the ith column of T n, i {1, 2, 3, 4}. Note that the four columns of T n are cliques, T n is (4K 1,claw)-free and there are no edges between non-consecutive columns but T n is not tractable since it contains C 4. 9

10 Lemma 4 The clique-width of typical graphs is at most 8. Proof. We give an 8-expression τ n for the typical graph T n : For 1 i n, let τ i be the expression defined inductively as follows: τ 1 := η 4,1 (η 3,4 (η 2,3 (η 1,2 (1(1, 1) 2(1, 2) 3(1, 3) 4(1, 4))))); for i := 2 to n do begin α := τ i 1 5(i, 1) 6(i, 2) 7(i, 3) 8(i, 4); β := η 1,5 (η 2,6 (η 3,7 (η 4,8 (α)))); γ := η 5,2 (η 5,6 (η 6,3 (η 6,7 (η 7,4 (η 7,8 (η 8,1 (η 8,5 (β)))))))); τ i := ρ 5 1 (ρ 6 2 (ρ 7 3 (ρ 8 4 (γ)))) end Obviously, τ n constructs T n. Lemma 5 Every tractable graph with n vertices is an induced subgraph of T n. Proof. (by Sang-il Oum, [31]) Let G be a tractable graph having a partition of V (G) into four cliques Q 1, Q 2, Q 3, and Q 4 as in the definition of tractable graphs (index arithmetic in Q i+1, Q i 1 modulo 4). Thus, n = V (G). This proof consists of two parts. First, we show that there is an ordering of each Q i having a certain property. In the second part, we prove Lemma 5 by induction using the ordering. For x, y Q i, i {1, 2, 3, 4}, we write x i y if N Qi 1 (x) N Qi 1 (y) and N Qi+1 (x) N Qi+1 (y). Obviously i is reflexive and transitive. Claim 3.1 For all x, y Q i, either x i y or y i x. Proof of Claim 3.1. Suppose that x i y. We have to consider two cases. In the first case, let us assume that there is a vertex w Q i 1 adjacent to y but not adjacent to x. Then every vertex u Q i 1 adjacent to x must be adjacent to y, because the subgraph induced by {x, y, u, w} is not isomorphic to C 4. Furthermore every vertex v Q i+1 adjacent to y must be adjacent to x, because the subgraph induced by {x, y, v, w} is not isomorphic to the claw. Therefore y i x in this case. In the second case, we assume that there is a vertex w Q i+1 adjacent to x but not adjacent to y. By a similar argument as in the previous case, we also obtain y i x. This shows Claim 3.1. Claim 3.2 G is an induced subgraph of T n such that Q i is a subset of the ith column for all i {1, 2, 3, 4}. Proof of Claim 3.2. By induction on n = V (G). The induction basis is trivial. For the induction step we show: 10

11 Claim 3.3 If there exists a vertex x Q j such that every vertex in Q j+1 is adjacent to x and every vertex in Q j 1 is nonadjacent to x, then Claim 3.2 is fulfilled. Proof of Claim 3.3. By induction, G \ {x} is contained in T n 1 such that Q i \ {x} is a subset of the ith column for all i {1, 2, 3, 4}. We identify x with vertex (n, j) of T n. Since every vertex of the jth and (j + 1)th column of T n 1 is adjacent to (n, j), and every vertex of the (j 1)th column of T n 1 is nonadjacent to (n, j), T n contains G as an induced subgraph. This shows Claim 3.3. If Q i for all i {1, 2, 3, 4} then we may pick x i Q i such that every y Q i satisfies y i x i. Since the subgraph of G induced by {x 1, x 2, x 3, x 4 } is not isomorphic to 4K 1 or C 4, there is an index j such that x j is adjacent to x j+1 but not adjacent to x j 1. We observe that N[x j ] = Q j Q j+1 by definition of i. Therefore T n contains G by Claim 3.3. If more than two of Q 1, Q 2, Q 3, and Q 4 are empty, then Claim 3.2 is true because G is a complete graph. Therefore we may assume that at most two of Q 1, Q 2, Q 3, and Q 4 are empty. If Q i = Q i+2 = for some i {1, 2}, then we are done by Claim 3.3. Thus we may assume that either Q i or Q i+2 is nonempty for all i {1, 2}. Suppose that Q i, Q i+1 and Q i 1, Q i+2 = for some i {1, 2, 3, 4}. We pick x i Q i and x i+1 Q i+1 as before. If x i and x i+1 are adjacent, then every vertex of Q i+1 is adjacent to x i. By using Claim 3.3 with j = i, we obtain a typical graph. If x i and x i+1 are not adjacent, then every vertex of Q i is not adjacent to x i+1, and therefore by using Claim 3.3 with j = i + 1, we obtain a typical graph. We may now assume that Q j+2 = and Q j 1, Q j, Q j+1. By Claim 3.3, we may assume that there is no vertex in Q j 1 that is adjacent to every vertex of Q j and there is no vertex in Q j+1 that is adjacent to no vertex of Q j. As before, we pick x i Q i such that every y Q i satisfies y i x for all i {j 1, j, j + 1}. Since there exists y 1 Q j which is nonadjacent to x j 1, we deduce that x j is nonadjacent to x j 1, and therefore x j is adjacent to no vertex in Q j 1. Similarly, since there exists y 2 Q j adjacent to x j+1, we deduce that x j is adjacent to x j+1, and therefore x j is adjacent to every vertex of Q j+1. By Claim 3.3, we obtain a typical graph containing G. Corollary 1 The clique-width of tractable graphs is at most 8. Proof. We have shown that every tractable graph is an induced subgraph of a typical graph T n for some n. By Lemma 4 and Proposition 1 (i), this implies Corollary 1. 11

12 4 Bounded clique-width of classes defined by two forbidden 4-vertex subgraphs The following theorem describes the inclusion-maximal classes of bounded clique-width defined by two forbidden 4-vertex subgraphs. Theorem 6 The following classes have bounded clique-width: (i) (diamond,co-diamond)-free graphs; (ii) (claw,co-claw)-free graphs; (iii) (paw,co-paw)-free graphs; (iv) (diamond,co-paw)-free graphs; (v) (diamond,2k 2 )-free graphs; (vi) (2K 2,paw)-free graphs; (vii) (claw,paw)-free graphs; (viii) (K 4,K 4 )-free graphs; (ix) (K 4,co-paw)-free graphs; (x) (K 4,co-diamond)-free graphs. Proof. (i), (ii) and (iii) follow from Theorem 3, Propositions 1 and 2 and Lemma 2. (iv), (v) and (vi) follow from Theorem 4. (vii) follows from Theorem 5. (viii) follows from Ramsey theory: (K 4,K 4 )-free graphs have at most 17 vertices [23]. (ix): We claim that prime (K 4, co-paw)-free graphs have at most eight vertices. First, assume that G is (K 4, K 3 )-free. Then, by Ramsey theory, G has at most eight vertices. Now let G be a prime (K 4, co-paw)-free graph containing a 3K 1 H with vertices a 1, a 2, a 3. Moreover, let Z be the set of 0-vertices for H and let U be the set of 3- vertices for H. Then for all i {1, 2, 3}, the following facts are easy to see, since G is co-paw-free (index arithmetic modulo 3): (ix.1) N ai a i+1 =, (ix.2) N ai has a co-join to N ai+1 N ai 1, (ix.3) U has a join to N ai and Z has a co-join to N ai, (ix.4) Z has a join to U. 12

13 It follows from (ix.1) (ix.3) that for each i, {a i } N ai is a module of G and thus N ai =. By (ix.4), both Z and U are modules of G, i.e., G has at most five vertices. (x): We will prove that (4K 1, diamond)-free graphs have bounded clique-width. Then by Proposition 1 (ii), (x) follows. First, assume that G is 3K 1 -free; then G is (P 5, diamond)-free and thus by Theorem 2, G has bounded clique-width. Now let G be a (4K 1, diamond)-free graph containing a 3K 1 H with vertices a 1, a 2, a 3 and let U be the set of 3-vertices for H. Since G is diamond-free, U and N ai a i+1 for all i {1, 2, 3} each induces a stable set (index arithmetic modulo 3). Moreover, as G is 4K 1 -free, H has no 0-vertex, and for all i {1, 2, 3}, N ai induces a clique, N ai a i+1 2 and U 3. By Lemma 3, it suffices to show that the essential part G := G[N a1 N a2 N a3 ] of G has bounded clique-width. Since G is diamond-free, no two vertices in N ai have a common neighbor in N ai+1 (in N ai 1, respectively). Thus, for each i {1, 2, 3}, the edges between N ai and N ai+1 form a matching. This implies that the edge set between N a1, N a2, and N a3 is a vertexdisjoint union of induced cycles of length 3k, k 1, and induced paths of arbitrary length. The principle of our construction of G with an 8-expression is as follows: All the cycles (paths, respectively) can be constructed successively by an 8-expression such that after constructing any of the cycles (paths), all cycle (path) vertices in N ai get label i, i {1, 2, 3}. This will be done in such a way that during the construction also all the clique edges between a newly constructed vertex in N ai and all already existing vertices in N ai, i {1, 2, 3} are added. Obviously, the induced paths are induced subgraphs of corresponding cycles. Thus, for bounded clique-width, it would suffice to discuss the case of cycles instead of paths, but for convenience, we construct paths as an intermediate step. Let A = N a1 = {a 1,...,a k1 }, B = N a2 = {b 1,...,b k2 }, and C = N a3 = {c 1,...,c k3 }. As a first step, we construct 8-expressions τ k for paths (a 1, b 1, c 1,..., a k, b k, c k ), k 1, between A, B and C (0 is temporarily the label for start vertex a 1 and 7 is temporarily the label for end vertex c i, i k). τ 1 := η 2,7 (η 0,2 (0(a 1 ) 2(b 1 ) 7(c 1 ))); for j := 2 to k do τ j := ρ 6 7 (ρ 7 3 (ρ 5 2 (ρ 4 1 (η 5,6 (η 4,5 (η 7,4 (η 7,6 (η 3,6 (η 2,5 (η 1,4 (η 0,4 ( (4(a j ) 5(b j ) 6(c j ) τ j 1 ))))))))))))). If we finally add an edge between the start and end vertex (and correspondingly relabel them), we obtain a cycle (a 1, b 1, c 1,...,a k, b k, c k, a 1 ) of length 3k: τ k := ρ 7 3(ρ 0 1 (η 0,7 (τ k ))). 13

14 If more than one cycle between A, B and C has to be created, we proceed inductively. Assume that τ is an 8-expression constructing G with i 1 cycles between the cliques A, B, C such that the final label of already constructed vertices in A (B, C, respectively) is 1 (2, 3, respectively). Then an (i + 1)-th cycle (a 1, b 1, c 1,...,a k, b k, c k, a 1 ), a i A, b i B, c i C, of length 3k is added by an 8-expression ν k (which is very similar to τ k above) as follows: ν 1 := ρ 5 2 (η 5,7 (η 0,5 (η 7,3 (η 5,2 (η 0,1 (0(a 1) 5(b 1) 7(c 1) τ)))))); for j := 2 to k do ν j := ρ 6 7 (ρ 7 3 (ρ 5 2 (ρ 4 1 (η 5,6 (η 4,5 (η 7,4 (η 7,6 (η 3,6 (η 2,5 (η 1,4 (η 0,4 ( (4(a j) 5(b j) 6(c j) ν j 1 ))))))))))))). and ν k := ρ 7 3(ρ 0 1 (η 0,7 (ν k ))). This proves (x). 5 Bounded clique-width of classes defined by three forbidden 4-vertex subgraphs The inclusion-maximal classes of bounded clique-width defined by three forbidden 4- vertex graphs are described in Theorems 7 and 8. Theorem 7 (i) Prime (K 4,C 4,2K 2 )-free graphs have at most nine vertices; (ii) Prime (C 4,claw,2K 2 )-free graphs are thin spiders or have at most six vertices; (iii) Prime (K 4,claw,2K 2 )-free graphs have at most nine vertices. Proof. (i): Let G be a prime (K 4,C 4,2K 2 )-free graph. If G is C 5 -free then, by Lemma 1, it is a split graph. Prime K 4 -free split graphs, however, have at most 9 vertices: Let V = C S be a partition of V (G) into a clique C and a stable set S. Then C 3, and if C = 3, no vertex of S has a join to C. Since G is prime, every vertex of S has a neighbor in C, and the neighborhoods of distinct vertices x, y S in C are distinct. Thus, S has at most six vertices. Now assume that the prime graph G contains a C 5 C with vertices v 1,..., v 5 and edges v i v i+1 (index arithmetic modulo 5). Since G is (C 4,2K 2 )-free, C has no k-vertex for k {1, 2, 3, 4}. Let U denote the set of 5-vertices for C. Since G is (K 4,C 4 )-free, 14

15 U 1 holds: If u 1, u 2 U with u 1 u 2 then if u 1 u 2, v 1 v 2 u 1 u 2 is a K 4, and if u 1 u 2 then v 1 u 1 v 3 u 2 is a C 4, a contradiction. Since G is prime, every 0-vertex for C must be adjacent to U; thus a vertex in U would be universal in G which is impossible in a prime graph and implies that U =, and there is no 0-vertex for C, i.e., G itself is a C 5. (ii): Let G be a prime (C 4,claw,2K 2 )-free graph. If G is C 5 -free then, by Lemma 1, it is a split graph. Prime claw-free split graphs, however, are either the 3-sun or a thin spider (see e.g. [12]). Now assume that G contains a C 5 C as in the proof of (i); then C has no k-vertex for k {1, 2, 3, 4}. There is no edge between any 0- and any 5-vertex: If x is a 0-vertex and u is a 5-vertex such that x u then v 1 v 3 xu is a claw. Since G is prime, C has no 0- and also no 5-vertex, i.e., G is the C 5 itself. (iii): Let G be a prime (K 4,claw,2K 2 )-free graph. Case 1. G is K 3 -free. If G is a (prime) cograph then it has one vertex. Now assume that G contains a P 4 Q with vertices a, b, c, d and edges ab, bc and cd. Then Q has no 1-vertex since G is (claw,2k 2 )-free, and Q has no 3- and no 4-vertex since G is K 3 -free. Moreover, the only 2-vertices for Q are adjacent to a and d since G is (K 3,claw)-free. A 0-vertex, however, cannot be adjacent to such a 2-vertex. Thus, Q has also no 0-vertex, and the set N ad of 2-vertices has at most one element, i.e., G is either a C 5 or the P 4 Q itself. Case 2. G contains K 3 with vertices a, b, c. Since G is (K 4,claw)-free, N xy 1 for x y, x, y {a, b, c}. Since G is (claw,2k 2 )-free, N x 1 for x {a, b, c}. Since G is 2K 2 -free, N 0 N x for x {a, b, c}. Claim 5.1 N 1 N xy for x y, x, y {a, b, c}. Proof. Since G, as a prime graph, is connected, every vertex in N must have a neighbor in some N xy for x y, x, y {a, b, c}. Let r v for r N, {v} = N ab. If r u for u N ac then u v, since rv, uc is not a 2K 2 but then rubv is a claw - contradiction. Thus, N 1 N ac and similarly for N bc. This proves Claim 5.1. Claim 5.1 implies that N is a module and thus has at most one vertex. Claim 5.2 If N then N z = for z {a, b, c}. Proof. Since G is claw-free, N x 1 N xy for x y, x, y {a, b, c}. Let r N and {v} = N ab with r v. If x N a then v x and so rxbv is a claw, and similarly for x N b. If x N c then, since rv, xc is not a 2K 2, v x but then rxbv is a claw, a contradiction. This proves Claim 5.2. Thus, G has at most nine vertices. Theorem 8 The clique-width of (4K 1,C 4,claw)-free graphs is bounded. 15

16 The remainder of this section consists of the proof of Theorem 8. Let G be a prime (4K 1,C 4,claw)-free graph. Then G is C j -free for j 8. We will consider the following cases: (i) G is chordal; (ii) G contains C 5 ; (iii) G contains C 6 but is C 5 -free; (iv) G contains C 7 but is C 5 - and C 6 -free. Case (i): In Lemma 8, we show that G is a tractable graph or has at most seven vertices. Cases (ii) and (iii): We first reduce graph G to its essential part G by using Lemma 3. Then, for Case (ii), in Lemma 9, we give a partition of G into five tractable graphs G i = (Q 1 i Q2 i Q3 i Q4 i, E i), i {1,..., 5}, such that for every i, j {1,..., 5}, and every k, k {1, 2, 3, 4}, if i j then Q k i and Q k j form a join or a co-join. By the proof of Lemma 4, Lemma 5 and Corollary 1, for each i, G i can be constructed with eight labels such that finally, for every k {1, 2, 3, 4}, all vertices of one Q k i get the same label. Moreover, for 1 i, j 5, k, k {1, 2, 3, 4}, we can construct G i such that the labels l, l of every two distinct Q k i, Qk j are different. Then the remaining edges of G between G i, G j for i j can be easily inserted with η l,l. For Case (iii), we describe in Lemma 10 a similar partition of G into six tractable graphs, more precisely into twelve tractable co-bipartite chain graphs. Case (iv): In Lemma 11, we show that G itself is a C 7. That is, in each case (i) (iv), the clique-width of G is bounded, proving Theorem Clique-width of (4K 1, claw)-free chordal graphs The main result of this subsection is the fact that prime (4K 1, claw)-free chordal graphs are tractable or have at most seven vertices. For its proof we need the subsequent Lemmas 6 and 7. Remark. By Proposition 1, 4K 1 -free chordal graphs have unbounded clique-width, since K 4 -free co-chordal graphs have unbounded clique-width as will be shown in Theorem 11; also, claw-free chordal graphs have unbounded clique-width since unit interval graphs are claw-free chordal and have unbounded clique-width as shown in [24]. Lemma 6 Prime (4K 1, claw)-free chordal graphs containing a net have at most seven vertices. Proof. Let G be a prime (4K 1,claw)-free chordal graph containing a net Q with vertices v 1,..., v 6 and edges v 1 v 2, v 3 v 4, v 5 v 6, v 2 v 4, v 4 v 6, v 6 v 2. The vertices v 1, v 3, v 5 16

17 of degree 1 in Q are called end vertices and the vertices v 2, v 4, v 6 of degree 3 in Q are called mid vertices of Q. Since G is (4K 1,claw)-free, every vertex v / V (Q) is adjacent to exactly one or two of the end vertices. If v / V (Q) is adjacent to exactly one end vertex v j, j {1, 3, 5} then, as G is 4K 1 -free, v must be adjacent to the neighbor v j+1 of v j. Since G is claw-free, v is either a 2-vertex for Q (with neighbors v j and v j+1 ) or v is a 4-vertex for Q (with neighbors v j, v j+1, v j+3, v j+5 - index arithmetic modulo 6). If v / V (Q) is adjacent to exactly two end vertices v j, v j+2 then, as G is claw-free, v v j 1. Moreover, since G is chordal, v must be adjacent to the two other mid vertices, v j+1, v j+3, i.e., v is a 4-vertex for Q. It follows that Q has no k-vertex for k {0, 1, 3, 5, 6}. The next claim shows that actually, Q has no 2-vertex. Let N ij denote the set of 2-vertices for Q being adjacent to v i and v j, and similarly for other N S. Claim 5.3 The sets N 12 {v 1 }, N 34 {v 3 }, N 56 {v 5 } are modules and thus, N 12 = N 34 = N 56 =. Proof. Assume that x, y N 12 {v 1 } and z / N 12 {v 1 } with z x and z y. Let Q x (Q y ) denote the net induced by x, v 2, v 3,..., v 6 (y, v 2, v 3,...,v 6, respectively). Then z is a k-vertex for Q x if and only if z is a (k 1)-vertex for Q y, a contradiction, since z is a 2- or 4-vertex for Q x, Q y. This shows Claim 5.3. In a similar way one can show that for every j {1, 3, 5}, there is no 4-vertex v for Q with neighbors v j, v j+1, v j+3, v j+5. In this case, Q v is obtained from Q by changing v j+1 into v. Claim 5.4 The sets N 1246 {v 2 }, N 2346 {v 4 }, N 2456 {v 6 } are modules, and thus, N 1246 = N 2346 = N 2456 =. Since G is prime, the next claim implies that there is at most one 4-vertex for Q: Claim 5.5 At most one set of 4-vertices with neighbors v j, v j+1, v j+2, v j+3, j {1, 3, 5}, is nonempty. Proof. Assume that x N 1234 and y N Since xv 2 v 6 yv 3 is not a C 5, x y but now xyv 6 v 2 is a C 4, a contradiction. This shows Claim 5.5. Now, since a nonempty 4-vertex set is a module, G has at most seven vertices. Lemma 7 A prime (4K 1, claw, net)-free chordal graph containing a 3-sun is a 3-sun itself. Proof. Let G be a prime (4K 1, claw, net)-free chordal graph containing a 3-sun S with vertices v 1,...,v 6 and edges v i v i+1 (index arithmetic modulo 6), 1 i 6, and v 2 v 4, v 4 v 6, v 6 v 2. The vertices v 1, v 3, v 5 of degree 2 are called end vertices of S. 17

18 Since G is (4K 1,claw)-free, every vertex v / V (S) is adjacent to exactly one or two of the end vertices of S. If v / V (S) is adjacent to exactly one end vertex, say v 1 then, since G is claw-free, v v 4, and since G is (claw,net)-free, v v 2 and v v 6. In this case, v is a 3-vertex for S. If v / V (S) is adjacent to exactly two end vertices, say v 1, v 3 then also v v 2 since v 1 v 2 v 3 v is no C 4, and v v 4, v v 6 since v 1 v 6 v 4 v 3 v is no C 5 and G is C 4 -free. In this case, v is a 5-vertex for S. This implies that S has no k-vertex for k {0, 1, 2, 4, 6}. Claim 5.6 The sets N 612 {v 1 }, N 234 {v 3 }, N 456 {v 5 } are modules and thus, N 612 = N 234 = N 456 =. Proof. Assume that x, y N 612 {v 1 } and z / N 612 {v 1 } with z x and z y. Let S x (S y ) denote the 3-sun induced by x, v 2, v 3,...,v 6 (y, v 2, v 3,...,v 6, respectively). Then z is a k-vertex for S x if and only if z is a (k 1)-vertex for S y, a contradiction, since z is a 3- or 5-vertex for S x, S y. This shows Claim 5.6. In a similar way one can show: Claim 5.7 The sets N {v 2 }, N {v 4 }, N {v 6 } are modules and thus, N = N = N =. Thus, G is a 3-sun. Lemma 8 Prime (4K 1, claw)-free chordal graphs are tractable or have at most seven vertices. Proof. Let G be a prime (4K 1, claw)-free chordal graph. If G contains a net or is net-free but contains a 3-sun then, by Lemmas 6 and 7, G has at most seven vertices. Now let G be a prime (4K 1, 3-sun, net, claw)-free chordal graph. Then G is sun-free chordal, i.e., G is strongly chordal since it is 3-sun-free and claw-free (note that larger suns contain claw). Thus, G has a simple vertex w, i.e., the open neighborhood of w is a clique U = {u 1,...,u k } and there is a linear ordering of the closed neighborhoods of u i, say N[u i ] N[u j ] for i j. Since G, as a chordal graph, is perfect, and the stability number of the subgraph H induced by the set of nonneighbors V (G)\N[w] of w is at most two since G is 4K 1 -free, H is partitionable into two cliques X, Y. Let X 1 (Y 1 ) denote the X (Y ) vertices which have neighbors in U and let X 2 (Y 2 ) denote the X (Y ) vertices which have no neighbors in U. Then u k, as a maximum neighbor, has a join to X 1 Y 1. Since G is claw-free, X 1 Y 1 is a clique. We claim that also X 2 Y 2 is a clique. Assume that x 2 y 2 for x 2 X 2 and y 2 Y 2. Then x 2 and y 2 have no common neighbor in X 1 Y 1 since G is claw-free. If X 1 = or Y 1 = then G is tractable (with the partition into the three cliques N[w], X and Y ). Now assume that X 1 and Y 1 ; let x 1 X 1 and y 1 Y 1. Then u k is adjacent to x 1 and y 1 but now wu k x 1 y 1 x 2 y 2 is a net - contradiction. Thus, V (G) is partitionable into cliques Q 1 = N[w], Q 2 = X 1 Y 1 and Q 3 = X 2 Y 2 such that there are no edges between Q 1 and Q 3. By definition, G is tractable. 18

19 5.2 Clique-width of (4K 1,C 4,claw)-free graphs containing C 5 The main result of this subsection is the following Lemma 9 The clique-width of prime (4K 1,C 4,claw)-free graphs containing C 5 is bounded. Proof. Let G = (V, E) be a prime (4K 1,C 4,claw)-free graph containing a C 5 C with vertices v 1,...,v 5 and edges v i v i+1 (index arithmetic modulo 5). Since G is claw-free, C has no 1-vertex, and since G is C 4 -free, C has no 4-vertex. Moreover, the neighbors of 2- and of 3-vertices for C are consecutive in C. Let U denote the set of 5-vertices for C and let Z denote the set of 0-vertices for C. For i {1,..., 5}, let X i denote the set of 2-vertices for C adjacent to v i+2 and v i+3, and let Y i denote the set of 3-vertices for C adjacent to v i 1, v i and v i+1. Since G is claw-free, X i as well as Y i are cliques for every i {1,...,5}. For the remaining part of this subsection, let x i X i, y i Y i, i {1,...,5}, and u U, z Z (if the corresponding set is nonempty, respectively). In the proof of the following claims, when symmetry is obvious, we only show the first parts. Claim 5.8 For all i {1,..., 5} : (i) U 0 X i, U 1 Y i, U 0 Z, Z 1 X i, and Z 0 Y i ; (ii) X i 0 X i 2 X i+2, and Y i 0 Y i 2 Y i+2 ; (iii) X i 0 Y i, and X i 1 Y i 2 Y i+2 ; (iv) U and Z are modules in G; (v) X i is a module in G[V \ (X i 1 X i+1 Y i 1 Y i+1 )]; (vi) Y i is a module in G[V \ (X i 1 X i+1 Y i 1 Y i+1 )]. Proof. (i): Since G is claw-free, U has no edges to 0- and 2-vertices, and since G is C 4 -free, U has a join to all 3-vertices. Since G is 4K 1 -free, Z has a join to all 2-vertices, and since G is claw-free, Z has no edges to 3-vertices. (ii): Obviously fulfilled since G is C 4 -free. (iii): If x i y i then x i v i+1 v i+2 y i is a C 4. If x i y i 2 then x i y i 2 v i+2 v i+1 is a claw. (iv), (v), (vi): Immediate from (i), (ii), and (iii). This shows Claim 5.8. By Claim 5.8 (iv), U 1 and Z 1. Let A i := {x i X i x i 1 X i 1 X i+1 and x i 0 Y i 1 Y i+1 }. By Claim 5.8 (v), for every i {1,..., 5}, A i is a module in G, and thus, A i 1. 19

20 By Lemma 3, for proving Lemma 9, it suffices to show that the essential part G := G[V \({v 1,...,v 5 } U Z 5 i=1 A i)] of G has bounded clique-width. Let X i := X i \A i. Thus: For i {1,..., 5}, every vertex in X i has a nonneighbor in X i 1 X i+1 or has a neighbor in Y i 1 Y i+1. (1) Let Yi := {y i y i 0 X i 1 X i+1 and y i 1 Y i 1 Y i+1 }. By Claim 5.8 (vi), Yi {v i } is a module in G for every i {1,...,5}, and thus, Yi =, i.e., for i {1,..., 5}, every vertex in Y i has a neighbor in X i 1 X i+1 or has a nonneighbor in Y i 1 Y i+1. Before presenting a partition of G = G[ 5 i=1 X i Y i] into five tractable subgraphs, we collect some facts which justify the intended partition: Claim 5.9 For all i {1,..., 5} : (i) If x i has a nonneighbor in X i 1 (in X i+1, respectively) then x i 1 X i+1 (x i 1 X i 1, respectively) and x i 0 Y i+1 (x i 0 Y i 1, respectively). (ii) If y i has a nonneighbor in Y i 1 (Y i+1 ) then y i 1 Y i+1 (y i 1 Y i 1 ) and y i 0 X i+1 (y i 0 X i 1 ). (iii) If x i has a neighbor in Y i+1 (Y i 1 ) then x i 0 Y i 1 (x i 0 Y i+1 ) and x i 1 X i 1 (x i 1 X i+1 ). (iv) If y i has a neighbor in X i+1 (X i 1 ) then y i 0 X i 1 (y i 0 X i+1 ) and y i 1 Y i 1 (y i 1 Y i+1 ). Proof. (i): Let x i have a nonneighbor x i 1. If x i has a nonneighbor x i+1 then x i x i 1 x i+1 v i is a 4K 1. If x i has a neighbor y i+1 then x i x i 1 v i y i+1 is a claw. (ii): Let y i have a nonneighbor y i 1. If y i has a nonneighbor y i+1 then y i y i 1 y i+1 v i is a claw. If y i has a neighbor x i+1 then y i y i 1 x i+1 v i is a C 4. (iii): Let x i have a neighbor y i+1. If x i has a neighbor y i 1 then x i y i 1 y i+1 v i is a C 4. The second part follows from (i). (iv): Let y i have a neighbor x i+1. If y i has a neighbor x i 1 then y i x i 1 x i+1 v i is a claw. The second part follows from (ii). This shows Claim 5.9. We subdivide the 2-vertex sets X i and 3-vertex sets Y i in G as follows. For each i {1,..., 5}, let X i := {v X i v has a nonneighbor in X i 1 or a neighbor in Y i 1 }; (2) 20

21 X + i := {v X i v has a nonneighbor in X i+1 or a neighbor in Y i+1 }; Y i := {v Y i v has a neighbor in X i 1 or a nonneighbor in Y i 1 }; Y + i := {v Y i v has a neighbor in X i+1 or a nonneighbor in Y i+1 }. Moreover, let G i := G [X i X + i 1 Y i Y + i 1 ]. Claim 5.10 For all i {1,..., 5} : (i) X i = X i X + i (ii) X i X + i and Y i = Y i Y + i ; = and Y i Y + i = ; (iii) G i is a tractable graph; (iv) If M {X i, X+ i 1, Y i, Y + i 1 } and M M 1 M or M 0 M. {X j, X+ j 1, Y j, Y + j 1 } for i j then Proof. (i): Immediate from (1) and (2). (ii): Let x X i. If x has a nonneighbor in X i 1 then, by Claim 5.9 (i), x 1 X i+1 and x 0 Y i+1, and thus, x X + i. If x has a neighbor in Y i 1 then, by Claim 5.9 (iii), x 0 Y i+1 and x 1 X i+1 and thus again, x X + i. Let y Y i. If y has a neighbor in X i 1 then, by Claim 5.9 (iv), y 0 X i+1 and y 1 Y i+1, i.e., y Y + i. If y has a nonneighbor in Y i 1 then, by Claim 5.9 (ii), y 1 Y i+1 and y 0 X i+1, i.e. again, y Y + i. (iii): Obviously, G i is tractable since, by Claim 5.8 (iii), X i 0 Y i for i {1,..., 5}. (iv): It follows from Claim 5.9 (i) (iii) that Xi 1 X i+1, Xi 0 Y i+1, X i + 1 X i 1, and X + i 0 Y i 1, and from Claim 5.9 (ii), (iv) the same statements follow with X and Y interchanged. Thus, for V, W {X, Y }, V i has a join or co-join with W i+1, and V + i has a join or co-join with W i 1. Together with Claim 5.8 (ii) (iii) this shows that M and M form a join or co-join. This shows Claim Now, as in the proof of Lemma 4, Lemma 5 and Corollary 1, for each i {1,..., 5}, G i can be constructed with 8 labels, say 1 i, 2 i,...,8 i, such that finally, all vertices in the nonempty set M i {X i, X+ i 1, Y i, Y + i 1 } get the same label l M i and for distinct sets M i, M i {X i, X+ i 1, Y i, Y + i 1 }, l M i l M i. By Claim 5.10, the remaining edges of G can be inserted by the corresponding join operations, proving that G has bounded clique-width. This completes the proof of Lemma 9. 21

22 5.3 Clique-width of (4K 1, C 4, C 5, claw)-free graphs containing C 6 The main result of this subsection is: Lemma 10 The clique-width of prime (4K 1,C 4,C 5,claw)-free graphs containing C 6 is bounded. Proof. Let G = (V, E) be a prime (4K 1,C 4,C 5,claw)-free graph containing a C 6 C with vertices v 1,..., v 6 and edges v i v i+1 (index arithmetic modulo 6). Obviously, C has no k-vertex for k {0, 1, 4, 5, 6}. Moreover, 2- and 3-vertices for C have consecutive neighbors in C. For i {1,...,6}, let X i,i+1 denote the set of 2-vertices adjacent to v i and v i+1, and let Y i denote the set of 3-vertices adjacent to v i 1, v i and v i+1. Since G is claw-free, X i,i+1 and Y i are cliques. Claim 5.11 For all i {1,..., 6} : (i) X i,i+1 1 X i+1,i+2 X i 1,i ; (ii) X i,i+1 0 X i+2,i+3 X i 2,i 1 ; (iii) X i,i+1 = or X i+3,i+4 = ; (iv) Y i 0 Y i+2 Y i+3 Y i 2 ; (v) X i,i+1 1 Y i Y i+1 ; (vi) X i,i+1 0 Y i+3 Y i 2 ; (vii) X i,i+1 is a module in G[V \ (Y i+2 Y i 1 )]; (viii) Y i is a module in G[V \ (X i+1,i+2 X i 2,i 1 Y i+1 Y i 1 )]. Proof. The negation of each of the points (i) (vi) leads to, respectively, a 4K 1, a C 4, a C 5 or a 4K 1, a C 5 (for i + 2 and i 2) or a C 4 (for i + 3), a claw, and a C 4. The points (vii) and (viii) are a simple consequence of (i) (vi). This shows Claim By (iii), at most three of the 2-vertex sets are nonempty. Claim 5.12 For all i {1,..., 6} : (i) If x X i,i+1 has a neighbor in Y i+2 (in Y i 1, respectively) then x 0 Y i 1 (x 0 Y i+2, respectively). (ii) If y Y i has a neighbor in X i 2,i 1 (in X i+1,i+2 ) then y 1 Y i 1 (y 1 Y i+1 ). 22

23 (iii) If y Y i has a nonneighbor in X i 2,i 1 (in X i+1,i+2 ) then y 1 Y i+1 (y 1 Y i 1 ). (iv) If y Y i has a nonneighbor in Y i+1 (in Y i 1 ) then y 1 Y i 1 (y 1 Y i+1 ). Proof. The negation of each of the points (i) (iv) leads to, respectively, a C 5 (using Claim 5.11 (iv)), a C 4 (using Claim 5.11 (v)), a 4K 1 (using Claim 5.11 (vi)), and a 4K 1 (using Claim 5.11 (iv)). This shows Claim Let A i,i+1 := {x X i,i+1 x 0 Y i+2 Y i 1 }. By Claim 5.11 (vii), for every i {1,..., 6}, A i,i+1 is a module in G and thus, A i,i+1 1. By Lemma 3, for proving Lemma 10, it suffices to show that the essential part G := G[V \({v 1,..., v 6 } 6 i=1 A i,i+1)] of G has bounded clique-width. Let X i,i+1 := X i,i+1 \ A i,i+1. Then: For i {1,...,6}, every vertex in X i,i+1 has a neighbor in Y i 1 Y i+2. (3) On the other hand, by Claim 5.12 (i), no vertex x X i,i+1 has neighbors in both Y i+2 and Y i 1. Thus, we obtain the following partition of X i,i+1, i {1,...,6}: Let X i,i+1 := {x X i,i+1 x has a neighbor in Y i 1 }, and X + i,i+1 := {x X i,i+1 x has a neighbor in Y i+2}. Now we define partitions of the 3-vertex sets. Let Y i := {y Y i y 0 X i+1,i+2 X i 2,i 1 and y 1 Y i+1 Y i 1 }. By Claim 5.11 (iv) (viii), Yi {v i } is a module in G. Thus, Yi {1,..., 6}, i.e., = for all i every vertex in Y i has a neighbor in X i+1,i+2 X i 2,i 1 or has a nonneighbor in Y i+1 Y i 1. (4) For i {1,..., 6}, let: Y + i := {y Y i y has a nonneighbor in Y i+1 }; Y i := {y Y i y has a nonneighbor in Y i 1 }; Y R+ i := {y Y i y 1 Y i 1 Y i+1 and y has a neighbor in X i+1,i+2 }. Y R i := {y Y i y 1 Y i 1 Y i+1 and y has a neighbor in X i 2,i 1 }. 23

24 By (4), Y i = Y + i Y i Y R+ i Y R i. On the other hand, by definition, Claim 5.12 (iv), and Claim 5.11 (iii), Y + i, Y i, Y R+ i and Y R i are mutually disjoint. Now we define co-bipartite chain graphs G j = (Q 1 j, Q 2 j, E j ), j {1,..., 12}, covering G as follows (recall that by Claim 5.11 (iii), only one of X i 2,i 1, X i+1,i+2 might be nonempty, and thus, at most 12 of them are nontrivial): For i {1,..., 6}, let G i := G [Y + i Y i+1 ]; G i,i+1 = G [X i,i+1 Y R+ i 1 ], and G + i,i+1 = G [X + i,i+1 Y R i+2 ]. Each G j is, in fact, a tractable graph with a partition into only two cliques Q 1 j and Q2 j. By all the properties shown before, it is easy to see that if Q k j, Q k j are two of the 24 cliques which do not belong to the same of the 12 tractable graphs then the edge set between them is a join or co-join. Thus, in a similar way as in the proof of Lemma 9, each of the 12 tractable graphs can be constructed with a bounded number of labels such that finally, each of the 24 cliques Q 1 j, Q 2 j, j {1,..., 12}, gets its personal label. Then the remaining join edges can be added. Therefore, the clique-width of a prime (4K 1, C 4, C 5, claw)-free graph containing C 6 is bounded, proving Lemma Clique-width of (4K 1, C 4, C 5, C 6, claw)-free graphs containing C 7 The last case for (4K 1, C 4, claw)-free graphs is: Lemma 11 A prime (4K 1, C 4, C 5, C 6, claw)-free graph containing C 7 is the C 7 itself. Proof. Let G be a prime (4K 1, C 4, C 5, C 6, claw)-free graph containing a C 7 C with vertices v 1,...,v 7 and edges v i v i+1, i {1,..., 7} (index arithmetic modulo 7). Since G is (4K 1, C 4, C 5, C 6, claw)-free, C has no k-vertex for k 3, and 3-vertices have consecutive neighbors in C. Let N i, i {1,..., 7}, denote the sets of 3-vertices adjacent to v i 1, v i, and v i+1. Then, since G is 4K 1 -free, N i 1 N i+1 holds, and since G is (C 4, C 5, C 6 )-free, N i 0 N i+j for i j 2. Thus, N i {v i } are modules, i.e., N i = which implies that G is a C 7. Now, Lemmas 8, 9, 10, and 11 show that (4K 1,C 4,claw)-free graphs have bounded clique-width proving Theorem 8. 24

25 6 Unbounded clique-width In this section, we identify four inclusion-minimal classes of unbounded clique-width which are defined by forbidden 4-vertex graphs. Two of them are an immediate consequence of the following result by Makowsky and Rotics: Theorem 9 ([27]) The following graph classes have unbounded clique-width: (i) split graphs; (ii) H n,q grids. The H n,q grid is constructed from an n n square grid G n as follows (see Figure 7) [27]: Let n 4 and q 2. Replace every edge of G n by a simple path with three edges, introducing two new vertices which are the internal vertices of the path. Let G n denote the resulting graph. For all vertices v of degree 4 in G n, do the following: For the four clockwise neighbors u 1, u 2, u 3, u 4 of v, omit v from G n and add the edges u iu i+1 (index arithmetic modulo 4) such that u 1 u 2 u 3 u 4 induce a C 4 in G n. Let G n denote the resulting graph. Replace every edge of G n by a simple path with q edges, introducing q 1 new vertices which are the internal vertices of the path. Then H n,q is the resulting graph. Theorem 9, Lemma 1 and the fact that H n,q grids contain no C 4 and no K 3, and thus no K 4, diamond, paw, and co-claw, imply: Corollary 2 The following classes have unbounded clique-width: (i) (C 4,2K 2 )-free graphs; (ii) (K 4, diamond, C 4, paw, co-claw)-free graphs. For the other two cases, we will show the following theorem: Theorem 10 The following classes have unbounded clique-width: (i) (K 4,2K 2 )-free graphs; (ii) (K 4,diamond,C 4,claw)-free graphs. 25