Interés, importancia y conocimiento ambiental

Recall that M is the hereditary closure of the set {Mn(2) : n ≥ 1}. By analogy with Theorem 3.2.1, one can prove the following result.

Theorem 3.2.6. The clique-width of Mn(2) is at least n/4.

Theorem 3.2.6 shows that the clique-width of graphs in the class M is un- bounded. However, unlike the class F studied in the previous section, M is not a minimal hereditary class of unbounded clique-width. To show this, let us provide an alternative definition of the graph Mn(2). Let Kn,n be the complete bipartite graph with verticesa1, . . . an in one part and vertices b1, . . . bn in the other part. Denote byMn,n the graph obtained from the Kn,n by subdividing each edgeaibj by a new vertex cij (i.e., by introducing vertex cij on the edge aibj). It is not difficult to see that Mn,n coincides with Mn(2). Therefore, every graph in M is obtained from a bipartite graph by subdividing each of its edges exactly once (or is an induced subgraph of such a graph). We will call the vertices of type ai orbi in Mn,n black and the vertices of typeci,j white.

We intend to show that M is not a minimal hereditary class of unbounded clique-width. To this end denote bySk be the class of (C3, . . . , Ck, H1, . . . , Hk)-free bipartite graphs of vertex degree at most 3 and by M_k the intersection S_k∩ M. (Hi is defined as in Figure 2.6.)

Lemma 3.2.7. For any natural k, the clique-width of graphs in M_k is unbounded. Proof. It is known that both the clique-width and tree-width are unbounded in the class Sk for any value ofk [Lozin and Rautenbach, 2006]. Since subdivision of an edge does not change the tree-width of a graph (see e.g. [Lozin and Rautenbach, 2006]), by subdividing each edge of graphs in S_k exactly once we obtain a class of graphs X of unbounded tree-width. Moreover, it is known that for graphs of bounded vertex degree, the tree-width is bounded if and only if the clique-width is bounded [Courcelle and Olariu, 2000]. Therefore, the clique-width of graphs in X

Lemma 3.2.7 shows that M is not a minimal hereditary class of unbounded clique-width. Indeed, for anyk, the classMk is a subclass of Msimply because in

M_k the vertex degree is bounded by 3, while in M it is not. Moreover, it is not difficult to see that M2,2 =M

(2)

2 is a C8, i.e., M8 is a subclass ofM2,2-free graphs inM.

Let us denote the limit class of the sequence S₁ ⊃ S₂ ⊃ S₃. . . by S, i.e.,

S = T

k≥1

S_k. It is not difficult to see that S is the class of graphs every con- nected component of which is of the form Si,j,k represented in Figure 2.6. Obvi- ously,S is a subclass ofM_k for eachk. Therefore,S is a limit class of the sequence

M1 ⊃ M2 ⊃ M3. . . as well. In the rest of the section, we show that S is a min- imal limit subclass of M, i.e., for any graph H ∈ S, the clique width of graphs inM ∩F ree(H) is bounded by a constant. This will be done through a sequence of auxiliary lemmas. The first lemma in this sequence was proved in [Lozin and Rautenbach, 2004b].

Lemma 3.2.8 (Lozin and Rautenbach [2004b]). For a class of graphs X and an integerρ, let[X]ρbe the class of graphsGsuch thatG−U belongs toXfor some

subsetU ⊆V(G) of cardinality at most ρ, and let[X]B be the class of graphs every

block of which belongs toX. If the clique-width of graphs inX is bounded by p, then the clique-width of graphs in [X]ρ is bounded by 2ρ(p+ 1), and the clique-width in [X]B is bounded by p+ 2.

In the proofs of the next lemmas we will frequently use the following obvious observation.

Observation. Any cycle in any graph G∈ M is chordless.

Lemma 3.2.9. Fork≥3, the clique-width of graphs inLk:=M∩F ree(Ck, Ck+1, . . .)

is bounded by a function of k.

Proof. For k= 3, the proposition follows from the fact that every graph in L3 is a forest. Fork >3, we use induction onk.

Let G be a graph in Lk+1. By Lemma 3.2.8, we can assume without loss of generality thatG is 2-connected. If Gcontains no cycles of length k, then G∈Lk in which case the lemma follows by induction. Now letC be a cycle of lengthk in

G. We will show that any other cycle C0 of length k in G (if any) has a common vertex with C. Assume the contrary: C and C0 are vertex disjoint. Consider two edges e∈ C and e0 ∈C0. Since G is 2-connected, there is a cycle containing both

which contains the endpoints in C and C0, and the remaining vertices outside the cycles. The endpoints of the paths P and Q partition each of the cycles C and C0

into two parts. The larger parts in both cycles together with paths P and Qform a cycle of length at leastk+ 2, contradicting the assumption thatG∈Lk+1. This contradiction shows that any two cycles of lengthkinG have a vertex in common. Therefore, removing the vertices of any cycle of lengthkfrom H results in a graph inLk, as required.

Lemma 3.2.10. For each k≥1, the clique-width of graphs in M ∩F ree(Sk,k,k) is

bounded by a function ofk.

Proof. LetG∈ M ∩F ree(Sk,k,k). Consider a chordless pathP of length 2k−2 and a chordless cycle C of length at least 2k+ 2 in G. If G does not contain such P

orC, the the clique-width of G is bounded according to Lemma 3.2.9. Assume P

and C are vertex disjoint. SinceG is connected, there must exist a chordless path

P0 connectingC toP. Since only black vertices ofC can have neighbors outsideC, the vertex of P0 that has a neighbor on C is white, and therefore this vertex has exactly one neighbor on C. Similarly, it is not difficult to see that the vertex ofP0

that has a neighbor onP is adjacent to exactly one vertex ofP. But now the reader can easily find an inducedSk,k,k. This contradiction shows that P and C contain a vertex in common. Therefore, the graph obtained fromGby deletion of the vertices ofP belongs to L2k+2, and the proposition follows from Lemmas 3.2.8 and 3.2.9.

Theorem 3.2.11. For any graphH∈ S, the clique-width of graphs inM∩F ree(H)

is bounded by a constant.

Proof. Without loss of generality we will assume that every connected component of

His of the formSk,k,kfor someevenk≥2 (obviously every graph inSis an induced subgraph of a graph of this form). Let p be the number of connected components of H, i.e., H = pSk,k,k. We will show that the clique-width of any graph G in

M ∩F ree(H) is bounded by a function of k and p. The proof will be given by induction on the minimum number m ≤ p such that G is mSk,k,k-free. If m = 1, then the clique-width of G is bounded according to Lemma 3.2.10. If G contains an induced copy of Sk,k,k, then by deleting this copy we obtain a graph G0 which is (m−1)Sk,k,k-free. Indeed, ifG0 contains an induced copy of (m−1)Sk,k,k, then there are no edges between this copy and the deleted copy of Sk,k,k in G, because

kis even, which means white vertices in both copies have no neighbors outside the copies. By the induction hypothesis, the clique-width ofG0 is bounded by a function of k and p. Therefore, by Lemma 3.2.8, the clique-width of G is bounded as well, since the number of deleted vertices is 3k+ 1.

3.3

Bipartite Double Bichain Graphs and Split Permu-