Split permutation graphs of large clique-width

In this section, we prove that the clique-width of split permutation graphs can be arbitrarily large. We quote the following theorem by Courcelle [Courcelle, 2004] which deals with infinite countable graphs (i.e. graph whose vertex set is countable):

Theorem 3.3.13. If a countable graph Ghas clique-width greater than22k+1_{, then}

some finite induced subgraph of Ghas clique-width greater than k.

We consider a countable grid of vertices{v(i, j) :i, j∈_N∪ {0}}. We say that vertex

v(i, j) belongs to rowiand columnj. We define the following sets of vertices for all

i, j∈_N∪ {0}and all n∈_N.

• Xi,n:={v(i, in+ 1), v(i, in+ 2), . . . , v(i,(i+ 1)n)}(A horizontal block)

• Yj,n :={v(jn+ 1, j), v(jn+ 2, j), . . . , v((j+ 1)n, j)}(A vertical block)

• X0 :={x0, x1, x2, . . .}

• Y0 :={y0, y1, y2, . . .}

It is not difficult to see that the blocks are pairwise disjoint for fixedn. Now let us define the graphGn by

V(Gn) =X0∪Y0∪(∪∞i=0Xi,n)∪(∪ ∞ j=0Yj,n)

andE(Gn) =E0∪Ex∪Ey, where

E0 ={(xi, xj) :i, j∈N∪{0}}∪{(xi, yj) :i, j∈N∪{0}}∪{(yi, yj) :i, j∈N∪{0}}

Ex ={(xi, v(r, s)),:i, r, s,∈N∪ {0}, r≥iand v(r, s)∈V(Gn)}

Ey ={(yj, v(t, u)) :j, t, u∈N∪ {0}, u≥j andv(t, u)∈V(Gn)}

Lemma 3.3.14. Gn is a split permutation graph.

Proof. Since X0∪Y0 is a clique (the edge setE0) and the remaining vertices form an independent set, the graph Gn is a split graph. Also, if i < j, then N(xj) ⊂

N(xi)∪ {xi} and N(yj) ⊂ N(yi)∪ {yi}. Therefore, the set X0 ∪Y0 can be split into two vicinal chains. Finally, it is not difficult to see that the set of vertices in horizontal blocks and the set of vertices in vertical blocks each forms a vicinal chain. Therefore, the Dilworth number ofGnis at most 2 and hence by Theorem 3.3.8,Gn is a split permutation graph.

Lemma 3.3.15. cwd(Gn)≥n/4.

Proof. LetT be a parse tree definingGn. For a nodeainT, we denote byT(a) the subtree ofT rooted at a. The label of a vertexv of the graph Gn at the nodea is defined as the label thatv has immediately prior to applying the operation a.

Leta be a lowest L

-node in T such that T(a) contains a full block (i.e. a full set of form Xi,n orYj,n) of Gn, and denote by b and c the two sons of ain T. We colour the vertices ofT(b) and T(c) by red and blue, respectively, and all the other vertices by white. If u and v are red vertices and there exists a non-red vertex w

which is adjacent tou but not to v, we say that w distinguishes between u and v. The following fact is easy to deduce:

• Ifw distinguishes betweenuandv, thenu andv have different labels at node

a.

In order to justify this fact, we simply note that the operation of type η for introducing the edge (w, u) is located outside T(a), so u and v must have different labels to avoid creating an edge (w, v) under the same operation. Of course, the respective fact holds for blue verticesu and v, wherew is non-blue.

Due to the choice ofa, the graphGndoes not contain an entirely red block or an entirely blue block, but it contains a block with no white vertex. We may assume without loss of generality that there exists such a block which is horizontal and

contains at least n/2 red vertices. (Otherwise we could swap the roles of columns and rows and/or the colors blue and red.) LetXk,n be such a horizontal block.

To prove the lemma, we will show that Gn contains a subset of red vertices of size at leastn/4 whose members have pairwise different labels at nodea. Any subset of this type will be called good.

Consider the setY₀0 :={yj ∈Y0 : there exists a red vertex ofXk,n in columnj}. We split our proof into two cases.

Case 1. At least half of the vertices in Y₀0 are non-red. Let us denote the set of non-red vertices of Y₀0 by Y₀∗ and let X∗ := {v(k, j) ∈ Xk,n :yj ∈Y0∗}. Note that

|X∗|=|Y₀∗| ≥n/4. It suffices to show thatX∗ is a good set of red vertices. Choose any pair of vertices v(k, j1), v(k, j2) ∈ X∗ such that j1 < j2. Then, by definition, the non-red vertex yj2 distinguishes between v(k, j1) and v(k, j2). By our earlier observation, this implies thatv(k, j1) andv(k, j2) have different labels at nodeaof

T. Thus X∗ is a good set of red vertices.

Case 2. At least half of the vertices inY₀0 are red. Let us denote the set of red vertices ofY₀0 by Y₀∗∗. For eachyj ∈Y0∗∗, choose a non-red vertex v(i(j), j)∈Yj,n. This is possible, since by minimality of nodea, there exist no entirely red blocks in

Gn.

We denote X∗∗ := {v(i(j), j) : yj ∈ Y0∗∗}. Note that |Y0∗∗| = |X∗∗| ≥ n/4. It suffices to show thatY₀∗∗ is a good set of red vertices. Choose any pair of vertices

yj1, yj2 ∈Y ∗∗

0 such that j1 < j2. Then, by definition, the non-red vertexv(i(j1), j) distinguishes betweenyj1 and yj2. By our earlier observation, this implies that yj1 andyj2 have different labels at nodeaof T. Thus Y

∗∗

0 is a good set of red vertices. Since we attained a good set of red vertices in both cases, we are done.

Combining Theorem 3.3.13 and Lemmas 3.3.14 and 3.3.15, we obtain the fol- lowing conclusion:

Theorem 3.3.16. The class of (finite) split permutation graphs has unbounded clique-width.

With each split graph G= (K, I, E) we can associate a bipartite graph β(G), as was done in Definition 3.3.5.

From Propositions 3.3.9 and 3.3.10, we can derive the following conclusion:

Corollary 3.3.17. The class of(P7, C6,3K2)-free bipartite graphs (i.e. the class of