# A class of graphs with large rankwidth

We describe several graphs of arbitrarily large rankwidth (or equivalently of arbitrarily large cliquewidth). Korpelainen, Lozin, and Mayhill [Split permutation graphs, Graphs and Combinatorics, 30(3):633–646, 2014] proved that there exist split graphs with Dilworth number 2 of arbitrarily large rankwidth, but without explicitly constructing them. Our construction provides an explicit construction. Maffray, Penev, and Vušković [Coloring rings, arXiv:1907.11905, 2019] proved that graphs that they call rings on n sets can be colored in polynomial time. Our construction shows that for some fixed integer n≥ 3, there exist rings on n sets of arbitrarily large rankwidth. When n≥ 5 and n is odd, this provides a new construction of even-hole-free graphs of arbitrarily large rankwidth.

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## 1 Introduction

The cliquewidth of a graph is an integer intended to measure how complex is the graph. It was defined by Courcelle, Engelfriet and Rozenberg in [6] and is successful in the sense that many hard problems on graphs become tractable on graph classes of bounded cliquewidth [7]. This includes for instance finding the largest clique or independent set, and deciding if a colouring with at most colors exists (for fixed ). This makes cliquewidth particularly interesting in the study of algorithmic properties of hereditary graph classes.

The notion of rankwidth was defined by Oum and Seymour in [16], where they use it for an approximation algorithm for cliquewidth. They also show that rankwidth and cliquewidth are equivalent, in the sense that a graph class has bounded rankwidth if, and only if, it has bounded cliquewidth. In the rest of this article, we only use rankwidth, and we therefore refer to results in the literature with this notion, even if in the original papers, the notion of cliquewidth is used (recall the two notions are equivalent as long as we care only for a class being bounded or not by the parameter).

Determining whether a given class of graphs has bounded rankwidth has been well studied lately, and let us survey the main results in this direction. For every classe of graphs defined by excluding one or two graphs as induced subgraphs, it is known whether the class has bounded or unbounded rankwidth, apart for a very small number of open cases, see [8] for the most recent results.

Similar classifications were obtained for chordal graphs [3] and split graphs [2]. Recall that a chordal graph is a graph such that every cycle of length at least 4 has a chord, and a split graph is a graph whose vertex set can be partitioned into a clique and a stable set.

A graph is even-hole-free if every cycle of even length has a chord. Determining whether several subclasses of even-hole-free graphs have bounded rankwidth also attracted some attention. See [17, 13, 5, 4] for subclasses of bounded rankwidth and [1, 18] for subclasses of unbounded rankwidth.

When is a graph and a vertex of , we denote by the set of all neighbors of . We set . The Dilworth number of a graph is the maximum number of vertices in a set such that for all distinct , the two sets and are non-empty. In [11], it is proved that graphs with Dilworth number 2 and arbitrarily large rankwidth exist. It should be pointed out that these graphs are split, and that only their existence is proved, no explicit construction is given.

For an integer , a ring on sets is a graph whose vertex set can be partitioned into cliques , with three additional properties:

• For all and all , either or .

• For all and all , (where the addition of subscripts is modulo ).

• For all , there exists a vertex that is adjacent to all vertices of .

Rings were studied in [15], where a polynomial algorithm to color them is given. The notion of ring in [15] is slightly more restricted than ours, but it makes no essential difference. Observe that the Dilworth number of a ring on sets is at most . As explained in [13], a construction from [10] shows that there exist rings of arbitrarily large rankwidth. Also in [12], it is proved that the so-called twisted chain graphs, that are similar in some respect to rings on 3 sets, have unbounded rankwidth. However, for any fixed integer , it is not known whether there exist rings on sets of arbitrarily large rankwidth.

Our main result is a new way to build graphs of arbitrarily large rankwidth. The construction has some flexibility, so it allows us to reach several goals. First, we give split graphs with Dilworth number 2 of arbitrarily large rankwidth, and we describe them explicitly. By tuning the construction differently, we show that for each integer , there exist rings on sets of arbitrarily large rankwidth. For odd integers , this provides new even-hole-free graphs or arbitrarily large rankwidth. It should be pointed out that our construction does not rely on modifying a grid (a classical method to obtain graphs or arbitrarily large rankwidth).

In Section 2, we recall the definition of rankwidth together with lemmas related to it. In Section 3, we give the main pieces needed to construct our graphs, called carousels, to be defined in Section 4. In Section 5, we give an overview of the proof that carousels have unbounded rankwidth. In Section 6, we give several technical lemmas about the rank of matrices that arise form partitions of the vertices in carousels. In Sections 7 and 8, we prove that carousels have unbounded rankwidth (we need two sections because there are two kinds of carousels, the even ones and the odd ones). In Section 9, we show how to tune carousels in order to obtain graphs split graphs with Dilworth number 2 or rings. We conclude the paper by open questions in Section 10.

## 2 Rankwidth

When is a graph and a partition of some subset of , we denote by the matrix whose rows are indexed by , whose columns are indexed by and such when and when . We define , where the rank is computed on the binary field. When clear from the context we may refer to as the rank of .

A tree is a connected acyclic graph. A leaf of a tree is a node incident to exactly one edge. For a tree , we let denote the set of all leaves of . A tree node that is not a leaf is called internal. A tree is cubic, if it has at least two nodes and every internal node has degree .

A tree decomposition of a graph is a cubic tree , such that . Note that if , then has no tree decomposition. For every edge , has two connected components, and (that we view as trees). The width of an edge is defined as . The width of is the maximum width over all edges of . The rankwidth of , denoted by , is the minimum integer , such that there is a tree decomposition of of width . If , we let .

Let be a graph. A partition of is balanced if

 |V(G)|/3≤|Y|,|Z|≤2|V(G)|/3

and unbalanced otherwise. An edge of a tree decomposition is (un)-balanced if the partition of as defined above is (un)-balanced.

###### Lemma 2.1

Every tree decomposition of a graph has a balanced edge.

###### Proof.

For every edge , removing from yields two components and . We orient from to if . If there is a non-oriented edge , then is balanced. So, assume that all edges are oriented. Since is a tree, some node must be a sink. Note that cannot be a leaf. But is cubic, so each of the three subtrees obtained from by deleting contains less than of the vertices of , a contradiction. ∎

Interestingly, we do not need the full definition of the rankwidth of a graph, the following property is enough for our purpose.

###### Lemma 2.2

Let be a graph and an integer. If every partition of with rank less than is unbalanced, then .

###### Proof.

Suppose for a contradiction that . Then there exists a tree decomposition of with width less that . Consider a balanced edge of , and let and be the two connected components of . Then, is a partition of that is balanced and has rank less than , a contradiction to our assumptions. ∎

We do not need many definitions from linear algebra. In fact the following basic lemma and the fact that the rank does not increase when taking submatrices are enough for our purpose. A (0-1) matrix is diagonal if whenever and whenever . It is antidiagonal if whenever and whenever . It is triangular if whenever and whenever .

###### Lemma 2.3

For every integer , the rank of a diagonal, antidiagonal or triangular matrix is at least (in fact it is when is odd and the matrix is antidiagonal, and it is otherwise).

###### Proof.

Clear. ∎

A (0-1) matrix is near-triangular if whenever and whenever . The values on the diagonal are not restricted.

###### Lemma 2.4

For every integer , the rank of a near-triangular matrix is at least .

###### Proof.

The submatrix of formed by the rows of even indexes and the columns of odd indexes is a triangular matrix (formally for every , ). By Lemma 2.3, . ∎

## 3 Matchings, antimatchings and crossings

We here describe the basic adjacencies that we need to build our graphs. Suppose that a graph contains two disjoint ordered sets of vertices of same cardinality , say and .

• We say that is a regular matching when:

for all , if and only if .

• We say that is a regular antimatching when:

for all , if and only if .

• We say that is a regular crossing when:

for all , if and only if .

• We say that is a expanding matching when:

for all , if and only if or .

• We say that is a expanding antimatching when:

for all , if and only if and .

• We say that is a expanding crossing when:

for all , if and only if .

• We say that is a skew expanding matching when equal 2 modulo 4 and:

For all and , if and only if or ; and

For all and , ;

For all and , if and only if or .

• We say that is a skew expanding antimatching when equal 2 modulo 4 and:

for all and , if and only if or ;

for all and , ; and

for all and , if and only if or .

• We say that is a skew expanding crossing when equal 2 modulo 4 and:

for all and , if and only if ;

for all and , if and only if ;

for all and , if and only if ; and

for all and , if and only if .

A triple is regular if it is a regular matching, a regular antimatching or a regular crossing, see Figure 1. On the figures, vertices are partitioned into boxes and several sets receive names. These will be explained in the next section.

A triple is expanding if it is an expanding matching, an expanding antimatching or an expanding crossing, see Figure 2.

A triple is skew expanding if it is a skew expanding matching, a skew expanding antimatching or a skew expanding crossing, see Figure 3.

A triple is a parallel triple if it is a matching or an antimatching (regular, expanding or skew expanding). A triple is a cross triple if it is a crossing (regular, expanding or skew expanding).

## 4 Carousels

The graphs that we construct are called carousels and are built from sets of vertices of equal cardinality : , …, . So, let be a graph such that . Throughout the rest of the paper, the subscripts for sets ’s are considered modulo . The graph is a carousel on sets of cardinality if:

1. is a regular crossing;

2. for every , is a regular triple and

3. is an expanding triple or a skew expanding triple.

Whether should be an expanding triple or a skew expanding triple depends on how many crossing triples there are among the triples for . Let us explain this.

Let be an integer. An even carousel of order on sets is a carousel on sets of cardinality such that:

1. ;

2. the total number of crossing triples among the triples for is even and

3. is an expanding triple.

Let be an integer. An odd carousel of order on sets is a carousel on sets of cardinality such that:

1. (so is equal to 2 modulo 4);

2. the total number of crossing triples among the triples for is odd and

3. is a skew expanding triple.

In carousels, the edges inside the sets or between sets and such that are not specified, they can be anything. Our main result is the following.

###### Theorem 4.1

For every integers and , there exists an integer such that every even carousel and every odd carousel of order on sets has rankwidth at least .

## 5 Sketch of the proof

For all , we set . There is a symmetry in every set and we need some notation for it. Let be the function defined for each integer by . We will use an horizontal bar to denote it as follows. When is a vertex in some set , we denote by the vertex and by the vertex . We use a similar notation for sets of vertices: if then

 ¯¯¯¯Si={¯¯¯xji|xji∈Si}

and

 ¯¯¯¯Si+1={¯¯¯xji+1|xji∈Si}.

Note that for any object such that means something.

Each set is partitioned into parts and this differs for the even and the odd case.

When is an even carousel, we designate by induction each set as a top set or a botom set. The set is by definition a top set, and the status of the next ones change at every cross triple. More formally, for every :

• If is top set and is a parallel triple, then is a top set.

• If is top set and is a cross triple, then is a botom set.

• If is botom set and is a parallel triple, then is a botom set.

• If is botom set and is a cross triple, then is a top set.

Then, for all and , we set . Observe that for some fixed , the ’s (resp. the ’s) form a partition of . We view every top set as partitioned by the ’s and every botom set as partitioned by the ’s.

In an odd carousel of order , there are also top sets and botom sets (defined as above), but they are all partitioned in the same way. For all and , we set . Observe that for some fixed , the ’s and ’s form a partition of .

To prove Theorem 4.1, we fix the integers and . We then compute a large integer (depending on and ) and we consider a carousel of order on sets. We then study how behaves a partition of of rank less than , our goal being to prove that it is unbalanced (this proves the rankwidth of is at least by Lemma 2.2).

To check whether a partition is balanced, we need a notation to measure how many elements of some set contains. For an integer , a set receive label if

 (m−1)r<|S∩Y|≤mr.

Note that has label if and only if , and having label for a larger means be more in being more far away from being included in . Having label means that . Observe that for every subset of , there exists a unique integer such that has label .

The first step of the proof is to note that when going from to (so in ), there are not too many changes from to or from to . Because a big number of changes would imply that the rank between and is high, this is formally stated and proved in Lemma 6.1. For this to be true, we need that is a crossing, this is why there is no flexibility in the definition for the adjacency between and in the definition of carousels.

Since (the common cardinality of the sets ’s) is large enough by our choice of , we then know that in there must be large sequences of vertices in or in , say in up to symmetry. So, one of the sets that partition is fully in , say for some large integer , and has therefore label . The rest of the proof shows that this label propagates in the rest of the graph. By a propagation lemma (stated in the next section), we prove that (if is a parallel triple) or (if is a cross triple) has a label very close to , namely or . This is because another label, so with , would allow to find a matrix of rank at least between and . And we continue like that until we reach (or ). By and by, the label may be , , …, , but not worse.

After we apply other propagation lemmas, that handle the triple (that is expanding or skew expanding by definition of carousels). Here the adjacency is designed so that some part that is twice larger, namely if is an even carousel, or if is an odd carousel, has a label under control.

For even carousels, by repeating the procedure above, we prove that for each , and have a label with not too big. And since the size of the sets is exponential in , these two sets and represent a proportion more than of all the set . And since and have label with small, they contains mostly vertices from , so that the partition is unbalanced.

For odd carousels, it is similar, except that we prove that , , and have label . These sets represent a proportion more than of all the set .

## 6 Blocks and propagation

Throughout the rest of this section, , and is a an even or an odd carousel of order on sets. Also, is an integer and is a partition of of rank less than .

Suppose is an ordered set and is a partition of . We call interval of any subset of of the form . A block of w.r.t.  is any non-empty interval of that is fully contained in or in and that is maximal w.r.t. this property. Clearly, is partitioned into its blocks w.r.t. .

###### Lemma 6.1

has less than blocks w.r.t. .

###### Proof.

Suppose that has at least blocks, and let , …, be the first ones. Up to symmetry, we may assume that for every , and . For each , we choose some element and some element . We denote by and the symmetric in of and respectively. Formally, for some integer , and the definiton of is similar. We set .

Suppose , and . From the definition of regular crossings, we have the following key observation:

• If , then .

• If , then .

• Note that if , the adjacency between and is not specified, it depends on whether is or and on whether is or .

Suppose first that . Then, there exist at least sets among , … that have a non-empty intersection with . This means that there exist distinct integers , …, such that for every , some vertex from is in . We denote by such a vertex, and let . We then set .

By definition, and . Also, by the key observation above, the matrix is a near-triangular matrix. By Lemma 2.4, it has rank at least . This proves that , a contradiction.

When , the proof is similar. ∎

The following lemma tells how labels propagate in regular triples (represented in Figure 1).

###### Lemma 6.2

Let , and be integers.

1. Suppose that is a regular parallel triple. If has label , then has label , or .

If has label , then has label , or .

2. Suppose that is a regular cross triple. If has label , then has label , or .

If has label , then has label , or .

###### Proof.

We first deal with the case when is a cross triple and has label . Suppose that the conclusion does not hold. This means that has label with or .

If , then . Since , we have . So there exists a subset of such that , and . The matrix is triangular and has rank by Lemma 2.3, a contradiction to being a partition of of rank less than .

If , then . Since , we have . So, there exists a subset of such that , and . The matrix is triangular and has rank by Lemma 2.3, a contradiction to being a partition of of rank less than .

All the other cases are similar. When has label , the proof is symmetric. When is a regular matching, we obtain a diagonal matrix, and when is a regular antimatching, we obtain a antidiagonal matrix. ∎

The following lemma tells how labels propagate in expanding triples (see Figure 2).

###### Lemma 6.3

Let and be integers such that .

1. Suppose that is a parallel expanding triple. If has label , then has label , , , or .

2. Suppose that is a cross expanding triple. If has label , then has label , , , or .

###### Proof.

We first deal with the case when is a cross expanding triple and has label . Suppose that the conclusion does not hold. This means that has label with or .

If , then . Since , we have . So there exists a subset of such that , and . So, contains a subset such that , , and the matrix is triangular and has rank by Lemma 2.3, a contradiction to being a partition of of rank less than .

If , then . Since , we have . So, there exists a subset of such that , and . So, contains a subset such that , , and the matrix is triangular and has rank by Lemma 2.3, a contradiction to being a partition of of rank less than . ∎

The following lemma tells how labels propagate in skew expanding triples (see Figure 3). We omit the proof that is similar to the previous one.

###### Lemma 6.4

Let and be integers such that .

1. Suppose that is a parallel skew expanding triple.

If has label , then has label , , , or .

If has label , then has label , , , or .

2. Suppose that is a cross triple.

If has label , then has label , , , or .

If has label , then has label , , , or .

## 7 Even carousels

In this section, and are fixed integers. We prove Theorem 4.1 for even carousels. We therefore look for an integer such that every even carousel of order on sets has rankwidth at least . We define as follows. Let be an integer such that:

 2q+8r−1 ≥10(n+1)(q+8r+1)r (1)

Clearly exists. We set . We now consider an even carousel of order on sets. To prove that it has rankwidth at least , it is enough by Lemma 2.2 to prove that every partition of with rank less than is unbalanced, so let us consider a partition of or rank less than .

###### Lemma 7.1

There exists such that has label or .

###### Proof.

Otherwise, for every the set is an interval of that must contain elements of and elements of . Hence, the sets