A class of graphs with large rankwidth
AA class of graphs with large rankwidth
Chính T. Hoàng ∗ and Nicolas Trotignon † September 1, 2020
Abstract
We describe several graphs with arbitrarily large rankwidth (or equivalently witharbitrarily large cliquewidth). Korpelainen, Lozin, and Mayhill [Split permutationgraphs,
Graphs and Combinatorics , 30(3):633–646, 2014] proved that there exist splitgraphs with Dilworth number 2 with arbitrarily large rankwidth, but without explicitlyconstructing them. Our construction provides an explicit construction. Maffray, Penev,and Vušković [Coloring rings,
CoRR , abs/1907.11905, 2019] proved that graphs thatthey call rings on n sets can be colored in polynomial time. Our construction showsthat for some fixed integer n ≥ , there exist rings on n sets with arbitrarily largerankwidth. When n ≥ and n is odd, this provides a new construction of even-hole-free graphs with arbitrarily large rankwidth. 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 thesense that many hard problems on graphs become tractable on graph classes of boundedcliquewidth [7]. This includes for instance finding the largest clique or independent set, anddeciding if a colouring with at most k colors exists (for fixed k ∈ N ). This makes cliquewidthparticularly 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 useit for an approximation algorithm for cliquewidth. They also show that rankwidth andcliquewidth are equivalent, in the sense that a graph class has bounded rankwidth if, andonly if, it has bounded cliquewidth. In the rest of this article, we only use rankwidth, andwe 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 careonly for a class being bounded or not by the parameter). ∗ Department of Physics and Computer Science, Wilfrid Laurier University, Waterloo, Ontario, Canada. † Univ Lyon, EnsL, UCBL, CNRS, LIP, F-69342, LYON Cedex 07, France. Partially supported bythe LABEX MILYON (ANR-10-LABX-0070) of Université de Lyon, within the program “Investissementsd’Avenir” (ANR-11-IDEX-0007) operated by the French National Research Agency (ANR). a r X i v : . [ c s . D M ] A ug etermining whether a given class of graphs has bounded rankwidth has been wellstudied lately, and let us survey the main results in this direction. For every classe ofgraphs defined by forbidding one or two graphs as induced subgraphs, it is known whetherthe class has bounded or unbounded rankwidth, apart for a very small number of opencases, see [8] for the most recent results.Similar classifications were obtained for chordal graphs [3] and split graphs [2]. Recallthat a chordal graph is a graph such that every cycle of length at least 4 has a chord, anda 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 whetherseveral subclasses of even-hole-free graphs have bounded rankwidth also attracted someattention. See [17, 13, 5, 4] for subclasses of bounded rankwidth and [1, 18] for subclasseswith unbounded rankwidth.When G is a graph and x a vertex of G , we denote by N ( x ) the set of all neighborsof x . We set N [ x ] = N ( x ) ∪ { x } . The Dilworth number of a graph G is the maximumnumber of vertices in a set D such that for all distinct x, y ∈ D , the two sets N ( x ) \ N [ y ] and N ( y ) \ N [ x ] are non-empty. In [11], it is proved that graphs with Dilworth number 2and 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 n ≥ , a ring on n sets is a graph G whose vertex set can be partitionedinto n cliques X , . . . , X n , with three additional properties: • For all i ∈ { , . . . , n } and all x, x (cid:48) ∈ X i , either N ( x ) ⊆ N ( x (cid:48) ) or N ( x (cid:48) ) ⊆ N ( x ) . • For all i ∈ { , . . . , n } and all x ∈ X i , N ( x ) ⊆ X i − ∪ X i ∪ X i +1 (where the additionof subscripts is modulo n ). • For all i ∈ { , . . . , n } , there exists a vertex x ∈ X i that is adjacent to all vertices of X i − ∪ X i +1 .Rings were studied in [15], where a polynomial time algorithm to color them is given.The notion of ring in [15] is slightly more restricted than ours (at least 4 sets are required),but it makes no essential difference here. Observe that the Dilworth number of a ring on n sets is at most n . As explained in [13], a construction from [10] shows that there exist ringswith arbitrarily large rankwidth. Also in [12], it is proved that the so-called twisted chaingraphs , that are similar in some respect to rings on 3 sets, have unbounded rankwidth.However, for any fixed integer n , it is not known whether there exist rings on n sets witharbitrarily large rankwidth.Our main result is a new way to build graphs with arbitrarily large rankwidth. Theconstruction has some flexibility, so it allows us to reach several goals. First, we givesplit graphs with Dilworth number 2 with arbitrarily large rankwidth, and we describethem explicitly. By “tuning” the construction differently, we will show that for each integer n ≥ , there exist rings on n sets with arbitrarily large rankwidth. For odd integers n , this2rovides new even-hole-free graphs with arbitrarily large rankwidth. It should be pointedout that our construction does not rely on modifying a grid (a classical method to obtaingraphs with 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 ingredients needed to construct our graphs, called carousels ,to be defined in Section 4. In Section 5, we give an overview of the proof that carouselshave unbounded rankwidth. In Section 6, we give several technical lemmas about the rankof matrices that arise form partitions of the vertices in carousels. In Sections 7 and 8, weprove that carousels have unbounded rankwidth (we need two sections because there aretwo kinds of carousels, the even ones and the odd ones). In Section 9, we show how totune carousels in order to obtain graphs split graphs with Dilworth number 2 or rings. Weconclude the paper by open questions in Section 10. When G is a graph and ( Y, Z ) a partition of some subset of V ( G ) , we denote by M G,Y,Z thematrix M whose rows are indexed by Y , whose columns are indexed by Z and such M y,z = 1 when yz ∈ E ( G ) and M y,z = 0 when yz / ∈ E ( G ) . We define rank G ( Y, Z ) = rank( M G,Y,Z ) ,where the rank is computed on the binary field. When the context is clear, we may referto rank( G Y,Z ) as the rank of ( Y, Z ) .A tree is a connected acyclic graph. A leaf of a tree is a node incident to exactly oneedge. For a tree T , we let L ( T ) denote the set of all leaves of T . A tree node that is not aleaf is called internal . A tree is cubic , if it has at least two nodes and every internal nodehas degree .A tree decomposition of a graph G is a cubic tree T , such that L ( T ) = V ( G ) . Note thatif | V ( G ) | ≤ , then G has no tree decomposition. For every edge e ∈ E ( T ) , T \ e has twoconnected components, Y e and Z e (that we view as trees). The width of an edge e ∈ E ( T ) is defined as rank( M G,L ( Y e ) ,L ( Z e ) ) . The width of T is the maximum width over all edgesof T . The rankwidth of G , denoted by rw( G ) , is the minimum integer k , such that there isa tree decomposition of G of width k . If | V ( G ) | ≤ , we let rw( G ) = 0 .Let G be a graph. A partition ( Y, Z ) of V ( G ) is balanced if | V ( G ) | / ≤ | Y | , | Z | ≤ | V ( G ) | / and unbalanced otherwise. An edge of a tree decomposition T is (un)-balanced if thepartition ( L ( Y e ) , L ( Z e )) of G as defined above is (un)-balanced. Lemma 2.1
Every tree decomposition T of a graph G has a balanced edge.Proof. For every edge e ∈ E ( T ) , removing e from T yields two components Y e and Z e . Weorient e from Y e to Z e if | L ( Y e ) | < | L ( Z e ) | . If there is a non-oriented edge e , then e isbalanced. So, assume that all edges are oriented. Since T is a tree, some node s ∈ V ( T ) s cannot be a leaf. But T is cubic, so each of the three subtreesobtained from T by deleting s contains less than of the vertices of L ( T ) , a contradiction. (cid:50) Interestingly, we do not need the full definition of the rankwidth of a graph, the followingproperty is enough for our purpose.
Lemma 2.2
Let G be a graph and r ≥ an integer. If every partition ( Y, Z ) of V ( G ) withrank less than r is unbalanced, then rw( G ) ≥ r .Proof. Suppose for a contradiction that rw( G ) < r . Then there exists a tree decomposition T of G with width less that r . Consider a balanced edge e of T , and let Y e and Z e be thetwo connected components of T \ e . Then, ( L ( Y e ) , L ( Z e )) is a partition of V ( G ) that isbalanced and has rank less than r , a contradiction to our assumptions. (cid:50) We do not need many definitions from linear algebra. In fact the following basic lemmaand the fact that the rank does not increase when taking submatrices are enough for ourpurpose. A (0-1) n × n matrix M is diagonal if M i,j = 1 whenever i = j and M i,j = 0 whenever i (cid:54) = j . It is antidiagonal if M i,j = 0 whenever i = j and M i,j = 1 whenever i (cid:54) = j .It is triangular if M i,j = 1 whenever i ≥ j and M i,j = 0 whenever i < j . Lemma 2.3
For every integer r ≥ , the rank of a diagonal, antidiagonal or triangular r × r matrix is at least r − (in fact it is r − when r is odd and the matrix is antidiagonal,and it is r otherwise).Proof. Clear. (cid:50)
A (0-1) n × n matrix M is near-triangular if M i,j = 1 whenever i > j and M i,j = 0 whenever i < j . The values on the diagonal are not restricted. Lemma 2.4
For every integer r ≥ , the rank of a near-triangular r × r matrix M is atleast r .Proof. The submatrix N of M formed by the rows of even indexes and the columns of oddindexes is a triangular r × r matrix (formally for every i, j ∈ { , . . . r } , N i,j = M i, j − ).By Lemma 2.3, rank( M ) ≥ rank( N ) = r . (cid:50) In this section, we describe the particular adjacencies that we need to build our graphs.Suppose that a graph G contains two disjoint ordered sets of vertices of same cardinality k , say X = { u , . . . , u k } and X (cid:48) = { v , . . . , v k } .4 We say that ( G, X, X (cid:48) ) is a regular matching when:for all j, j (cid:48) ∈ { , . . . , k } , u j v j (cid:48) ∈ E ( G ) if and only if j = j (cid:48) . • We say that ( G, X, X (cid:48) ) is a regular antimatching when:for all j, j (cid:48) ∈ { , . . . , k } , u j v j (cid:48) ∈ E ( G ) if and only if j (cid:54) = j (cid:48) . • We say that ( G, X, X (cid:48) ) is a regular crossing when:for all j, j (cid:48) ∈ { , . . . , k } , u j v j (cid:48) ∈ E ( G ) if and only if j + j (cid:48) ≥ k + 1 . • We say that ( G, X, X (cid:48) ) is a expanding matching when:for all j, j (cid:48) ∈ { , . . . , k } , u j v j (cid:48) ∈ E ( G ) if and only if j (cid:48) = 2 j or j (cid:48) = 2 j + 1 . • We say that ( G, X, X (cid:48) ) is a expanding antimatching when:for all j, j (cid:48) ∈ { , . . . , k } , u j v j (cid:48) ∈ E ( G ) if and only if j (cid:48) (cid:54) = 2 j and j (cid:48) (cid:54) = 2 j + 1 . • We say that ( G, X, X (cid:48) ) is a expanding crossing when:for all j, j (cid:48) ∈ { , . . . , k } , u j v j (cid:48) ∈ E ( G ) if and only if j + j (cid:48) ≥ k + 2 . • We say that ( G, X, X (cid:48) ) is a skew expanding matching when k equal 2 modulo 4 and:For all j ∈ { , . . . , ( k − / } and j (cid:48) ∈ { , . . . , k } , u j v j (cid:48) ∈ E ( G ) if and only if j (cid:48) = 2 j or j (cid:48) = 2 j + 1 ; andFor all j ∈ { ( k − / , . . . , (3 k + 2) / } and j (cid:48) ∈ { , . . . , k } , u j v j (cid:48) / ∈ E ( G ) ;For all j ∈ { (3 k + 2) / , . . . , k } and j (cid:48) ∈ { , . . . , k } , u j v j (cid:48) ∈ E ( G ) if and only if j (cid:48) = 2 j − k − or j (cid:48) = 2 j − k − . • We say that ( G, X, X (cid:48) ) is a skew expanding antimatching when k equal 2 modulo 4and:for all j ∈ { , . . . , ( k − / } and j (cid:48) ∈ { , . . . , k } , u j v j (cid:48) ∈ E ( G ) if and only if j (cid:48) = 2 j or j (cid:48) = 2 j + 1 ;for all j ∈ { ( k − / , . . . , (3 k + 2) / } and j (cid:48) ∈ { , . . . , k } , u j v j (cid:48) ∈ E ( G ) ; andfor all j ∈ { (3 k + 2) / , . . . , k } and j (cid:48) ∈ { , . . . , k } , u j v j (cid:48) ∈ E ( G ) if and only if j (cid:48) = 2 j − k − or j (cid:48) = 2 j − k − . • We say that ( G, X, X (cid:48) ) is a skew expanding crossing when k equal 2 modulo 4 and:for all j ∈ { , . . . , ( k − / } and j (cid:48) ∈ { , . . . , k } , u j v j (cid:48) ∈ E ( G ) if and only if j + j (cid:48) ≥ k ;for all j ∈ { ( k − / , . . . , k/ } and j (cid:48) ∈ { , . . . , k } , u j v j (cid:48) ∈ E ( G ) if and only if j (cid:48) ≥ k/ ;for all j ∈ { k/ , . . . , (3 k + 2) / } and j (cid:48) ∈ { , . . . , k } , u j v j (cid:48) ∈ E ( G ) if and only if j (cid:48) ≥ k/ − ; and 5 i x i x i x i x i x i x i X i, X i, X i, X i x i +1 x i +1 x i +1 x i +1 x i +1 x i +1 x i +1 X i +1 , X i +1 , X i +1 , X i +1 x i x i x i x i x i x i x i X i, X i, X i, X i x i +1 x i +1 x i +1 x i +1 x i +1 x i +1 x i +1 X i +1 , X i +1 , X i +1 , X i +1 x i x i x i x i x i x i x i X i, X i, X i, X i x i +1 x i +1 x i +1 x i +1 x i +1 x i +1 x i +1 X i +1 , X i +1 , X i +1 , X i +1 Figure 1: A regular matching, a regular antimatching and a regular crossing (on the regularantimatching, only non-edges are represented)for all j ∈ { (3 k + 2) / , . . . , k } and j (cid:48) ∈ { , . . . , k } , u j v j (cid:48) ∈ E ( G ) if and only if j + j (cid:48) − ≥ k .A triple ( G, X, X (cid:48) ) is regular if it is a regular matching, a regular antimatching or aregular crossing, see Figure 1. On the figures, vertices are partitioned into boxes and severalsets receive names. These will be explained in the next section.A triple ( G, X, X (cid:48) ) is expanding if it is an expanding matching, an expanding antimatch-ing or an expanding crossing, see Figure 2.A triple ( G, X, X (cid:48) ) is skew expanding if it is a skew expanding matching, a skew ex-panding antimatching or a skew expanding crossing, see Figure 3.A triple ( G, X, X (cid:48) ) is a parallel triple if it is a matching or an antimatching (regular,expanding or skew expanding). A triple ( G, X, X (cid:48) ) is a cross triple if it is a crossing (regular,expanding or skew expanding). The graphs that we construct are called carousels and are built from n ≥ sets of vertices ofequal cardinality k ≥ : X , . . . , X n . So, let G be a graph such that V ( G ) = X ∪ · · · ∪ X n .Throughout the rest of the paper, the subscripts for sets X i ’s are considered modulo n .The graph G is a carousel on n sets of cardinality k if:(i) ( G, X , X ) is a regular crossing;(ii) for every i ∈ { , . . . , n − } , ( G, X i , X i +1 ) is a regular triple and(iii) ( G, X n , X ) is an expanding triple or a skew expanding triple.6 n x n x n x n x n x n x n X n, X n, X n, X n x x x x x x x X , X , X , X x n x n x n x n x n x n x n X n, X n, X n, X n x x x x x x x X , X , X , X x n x n x n x n x n x n x n X n, X n, X n, X n x x x x x x x X , X , X , X Figure 2: An expanding matching, an expanding antimatching and an expanding crossing x n x n x n x n x n x n x n x n x n x n x n x n x n x n X n, X n, X n, X n, X n, X n, X n x x x x x x x x x x x x x x X , X , X , X , X , X , X x n x n x n x n x n x n x n x n x n x n x n x n x n x n X n, X n, X n, X n, X n, X n, X n x x x x x x x x x x x x x x X , X , X , X , X , X , X x n x n x n x n x n x n x n x n x n x n x n x n x n x n X n, X n, X n, X n, X n, X n, X n x x x x x x x x x x x x x x X , X , X , X , X , X , X Figure 3: A skew expanding matching, a skew expanding antimatching and a skew expand-ing crossing (on the crossing, for each i ∈ { , . . . , k } , only the edge x in x j such that j isminimum is represented) 7hether ( G, X n , X ) should be an expanding triple or a skew expanding triple dependson how many crossing triples there are among the triples ( G, X i , X i +1 ) for i ∈ { , . . . , n } .Let us explain this.Let s ≥ be an integer. An even carousel of order s on n sets is a carousel on n setsof cardinality k such that:(i) k = 2 s − ;(ii) the total number of crossing triples among the triples ( G, X i , X i +1 ) for i ∈ { , . . . , n } is even and(iii) ( G, X n , X ) is an expanding triple.Let s ≥ be an integer. An odd carousel of order s on n sets is a carousel on n sets ofcardinality k such that:(i) k = 2 × (2 s − (so k is equal to 2 modulo 4);(ii) the total number of crossing triples among the triples ( G, X i , X i +1 ) for i ∈ { , . . . , n } is odd and(iii) ( G, X n , X ) is a skew expanding triple.In carousels, the edges inside the sets X i or between sets X i and X j such that j / ∈{ i − , i, i + 1 } are not specified, they can be anything. Our main result is the following. Theorem 4.1
For every integers n ≥ and r ≥ , there exists an integer s such that everyeven carousel and every odd carousel of order s on n sets has rankwidth at least r . In this section, we provide an outline of the proof of the main theorem. The detail will begiven in later sections.For all i ∈ { , ..., n } , we let X i = { x i , ..., x ki } . There is a symmetry in every set X i and we need some notation for it. Let f be the function defined for each integer j by f ( j ) = k − j + 1 . We will use an horizontal bar to denote it as follows. When x ji is a vertexin some set X i , we denote by x ji the vertex x f ( j ) i and by x ji +1 the vertex x f ( j ) i +1 . We use asimilar notation for sets of vertices: if S i ⊆ X i then S i = { x ji | x ji ∈ S i } and S i +1 = { x ji +1 | x ji ∈ S i } . a = a for any object a such that a is defined.Each set X i is partitioned into parts and this differs for the even and the odd case.When G is an even carousel, we designate by induction each set X i as a top set or a botom set . The set X is by definition a top set, and the status of the next ones change atevery cross triple. More formally, for every i ∈ { , . . . , n − } : • If X i is top set and ( G, X i , X i +1 ) is a parallel triple, then X i +1 is a top set. • If X i is top set and ( G, X i , X i +1 ) is a cross triple, then X i +1 is a botom set. • If X i is botom set and ( G, X i , X i +1 ) is a parallel triple, then X i +1 is a botom set. • If X i is botom set and ( G, X i , X i +1 ) is a cross triple, then X i +1 is a top set.Then, for all i ∈ { , . . . , n } and j ∈ { , . . . , s } , we set X i,j = { x j − i , . . . , x j − i } . Observethat for some fixed i , the X i,j ’s (resp. the X i,j ’s) form a partition of X i . We view every topset X i as partitioned by the X i,j ’s and every botom set X i as partitioned by the X i,j ’s.In an odd carousel of order s , there are also top sets and botom sets (defined as above),but they are all partitioned in the same way. For all i ∈ { , . . . , n } and j ∈ { , . . . , s } , weset X i,j = { x j − i , . . . , x j − i } . Observe that for some fixed i , the X i,j ’s and X i,j ’s form apartition of X i .To prove Theorem 4.1, we fix the integers n and r . We then compute a large integer s (depending on n and r ) and we consider a carousel G of order s on n sets. We then studythe behavior of a partition ( Y, Z ) of V ( G ) of rank less than r , our goal being to prove thatit is unbalanced (this proves the rankwidth of G is at least r by Lemma 2.2).To check whether a partition is balanced, we need a notation to measure how manyelements of Y some set contains. For an integer m ≥ , a set S ⊆ V ( G ) receive label Y m if ( m − r < | S ∩ Y | ≤ mr. Note that S has label Y if and only if S ⊆ Z . Having label Y m means that (cid:100)| S ∩ Y | /r (cid:101) = m .Observe that for every subset S of V ( G ) , there exists a unique integer m ≥ such that S has label Y m .The first step of the proof is to note that when going from x to x k (so in X ), thereare not too many changes from Y to Z or from Z to Y . Because a big number of changeswould imply that the rank between X and X is high, this is formally stated and provedin Lemma 6.1. For this to be true, we need that ( G, X , X ) is a crossing, this is why thereis no flexibility in the definition for the adjacency between X and X in the definition ofcarousels.Since k (the common cardinality of the sets X i ’s) is large enough by our choice of s , wethen know that in X there must be large intervals of consecutive vertices in Y or in Z ,say in Z up to symmetry. So, one of the sets that partition X is fully in Z , say X ,t forsome large integer t , and has therefore label Y . The rest of the proof shows that this label9ropagates in the rest of the graph. By a propagation lemma (stated in the next section),we prove that X ,t (if ( G, X , X ) is a parallel triple) or X ,t (if ( G, X , X ) is a cross triple)has a label very close to Y , namely Y or Y . This is because a larger label, say, Y m with m > , would give rise to a matrix of rank at least r between Y and Z . And we continueto apply the propagation mechanism until we reach X n, (or X n, ). The label may be Y , Y , . . . , Y n − , but not larger.Now, we apply other propagation lemmas to handle the triple ( G, X n , X ) (that isexpanding or skew expanding by definition of carousels). Here the adjacency is designedso that some part that is twice larger than X i,t , namely X ,t +1 if G is an even carousel, or X ,t +1 if G is an odd carousel, has a label being not too large.For even carousels, by repeating the procedure above, we prove that for each i ∈{ , . . . , n } , X i,s − and X i,s have a label Y m with m not too large. And since the sizeof the sets X i,j is exponential in j , these two sets X i,s − and X i,s represent a proportionmore than of all the set X i . And since X i,s − and X i,s have label Y m with m small, theycontains mostly vertices from Z , so that the partition ( Y, Z ) is unbalanced.For odd carousels, the proof is similar, except that we prove that X i,s − , X i,s , X i,s − and X i,s have label Y m with m being not too large. These n sets represent a proportionmore than of all the set X i .In the next section, we supply the detail of this proof. Throughout the rest of this section, n ≥ , s ≥ and G is a an even or an odd carousel oforder s on n sets. Also, we assume r ≥ is an integer and ( Y, Z ) is a partition of V ( G ) ofrank less than r .Suppose X = { x , . . . , x k } is an ordered set and ( Y, Z ) is a partition of X . We call interval of X any subset of X of the form { x i , x i +1 , . . . , x j } . A block of X w.r.t. ( Y, Z ) isany non-empty interval of X that is fully contained in Y or in Z and that is maximal w.r.t.this property. Clearly, X is partitioned into its blocks w.r.t. ( Y, Z ) . Lemma 6.1 X has less than r blocks w.r.t. ( Y, Z ) .Proof. Suppose that X has at least r blocks, and let B , . . . , B r be the r first ones. Upto symmetry, we may assume that for every i ∈ { , . . . , r } , B i − ⊆ Y and B i ⊆ Z . Foreach i ∈ { , . . . , r } , we choose some element y i ∈ B i − and some element z i ∈ B i . Wedenote by y i and z i the corresponding vertices in X of y i and z i respectively. Formally, y i = x j for some integer j ∈ { , . . . , k } , y i = x j = x k − j +12 and the definiton of z i is similar.We set X (cid:48) = { y , z , . . . , y r , z r } .Consider i, j ∈ { , . . . , r } , u ∈ { y i , z i } and v ∈ { y j , z j } . From the definition of regularcrossings, we have the following key observation:10 If i + j < r + 1 , then uv / ∈ E ( G ) . • If i + j > r + 1 , then uv ∈ E ( G ) . • Note that if i + j = 4 r + 1 , the adjacency between u and v is not specified, it dependson whether u is y i or z i and on whether v is y j or z j .Suppose first that | Y ∩ X (cid:48) | ≥ | Z ∩ X (cid:48) | . Then, there exist at least r sets among { y , z } , . . . { y r , z r } that have a non-empty intersection with Y . This means that thereexist r distinct integers i , . . . , i r such that for every j ∈ { , . . . , r } , some vertex from { y i j , z i j } is in Y . We denote by v j such a vertex, and let Y (cid:48) = { v , . . . , v r } . We then set Z (cid:48) = { z i , . . . , z i r } .By definition, Z (cid:48) ⊆ Z and Y (cid:48) ⊆ Y . Also, by the key observation above, the matrix M G [ Y (cid:48) ∪ Z (cid:48) ] ,Y (cid:48) ,Z (cid:48) is a near-triangular r × r matrix. By Lemma 2.4, it has rank at least r .This proves that rank G ( Y, Z ) ≥ r , a contradiction.When | Z ∩ X | ≥ | Y ∩ X | , the proof is similar. (cid:50) The following lemma explains how labels propagate in regular triples (represented inFigure 1).
Lemma 6.2
Let m ≥ , ≤ i < n and ≤ j ≤ s be integers.(i) Suppose that ( G, X i , X i +1 ) is a regular parallel triple. If X i,j has label Y m , then X i +1 ,j has label Y max( m − , , Y m or Y m +1 .If X i,j has label Y m , then X i +1 ,j has label Y max( m − , , Y m or Y m +1 .(ii) Suppose that ( G, X i , X i +1 ) is a regular cross triple. If X i,j has label Y m , then X i +1 ,j has label Y max( m − , , Y m or Y m +1 .If X i,j has label Y m , then X i +1 ,j has label Y max( m − , , Y m or Y m +1 .Proof. We first deal with the case when ( G, X i , X i +1 ) is a cross triple and X i,j has label Y m . Suppose that the conclusion does not hold. This means that X i +1 ,j has label Y j with j > m + 1 or ≤ j < m − .If j > m + 1 , then | X i +1 ,j ∩ Y | > ( m + 1) r . Since | X i,j ∩ Y | ≤ mr , we have | X i +1 ,j ∩ Y | − | X i,j ∩ Y | ≥ r + 1 . So there exists a subset S i of X i,j such that | S i | = r + 1 , S i ⊆ Z and S i +1 ⊆ Y . The matrix M G [ S i ∪ S i +1 ] ,S i ,S i +1 is triangular and has rank at least r byLemma 2.3, a contradiction to ( Y, Z ) being a partition of V ( G ) of rank less than r .If ≤ j < m − , then | X i +1 ,j ∩ Y | ≤ ( m − r . Since | X i,j ∩ Y | > ( m − r , we have | X i,j ∩ Y | − | X i +1 ,j ∩ Y | ≥ r + 1 . So, there exists a subset S i of X i,j such that | S i | = r + 1 , S i ⊆ Y and S i +1 ⊆ Z . The matrix M G [ S i ∪ S i +1 ] ,S i ,S i +1 is triangular and has rank at least r by Lemma 2.3, a contradiction to ( Y, Z ) being a partition of V ( G ) of rank less than r .11ll the other cases are similar. When X i,j has label Y m , the proof is symmetric. When ( G, X i , X i +1 ) is a regular matching, we obtain a diagonal matrix, and when ( G, X i , X i +1 ) is a regular antimatching, we obtain a antidiagonal matrix. (cid:50) The following lemma describes how labels propagate in expanding triples (see Figure 2).
Lemma 6.3
Let m ≥ and j be integers such that ≤ j ≤ s − .(i) Suppose that ( G, X n , X ) is a parallel expanding triple. If X n,j has label Y m , then X ,j +1 has label Y max( m − , , Y max( m − , , Y m , Y m +1 or Y m +2 .(ii) Suppose that ( G, X n , X ) is a cross expanding triple. If X n,j has label Y m , then X ,j +1 has label Y max( m − , , Y max( m − , , Y m , Y m +1 or Y m +2 .Proof. We first deal with the case when ( G, X n , X ) is a cross expanding triple and X n,j has label Y m . Suppose that the conclusion does not hold. This means that X ,j +1 has label Y j with j > m + 2 or ≤ j < m − .If j > m + 2 , then | X ,j +1 ∩ Y | > ( m + 2) r . Since | X n,j ∩ Y | ≤ mr , we have | X ,j +1 ∩ Y | − | X n,j +1 ∩ Y | ≥ r + 1 . So there exists a subset S n of X n,j such that | S n | = 2 r + 1 , S n ⊆ Z and S ⊆ Y . So, S n contains a subset S (cid:48) n such that | S (cid:48) n | = r + 1 , S (cid:48) n ⊆ Z , S (cid:48) ⊆ Y and the matrix M G [ S (cid:48) n ∪ S (cid:48) ,j +1 ] ,S (cid:48) n ,S (cid:48) ,j +1 is triangular and has rank at least r by Lemma 2.3,a contradiction to ( Y, Z ) being a partition of V ( G ) of rank less than r .If ≤ j < m − , then | X ,j +1 ∩ Y | ≤ ( m − r . Since | X n,j ∩ Y | > ( m − r , wehave | X n,j ∩ Y | − | X ,j +1 ∩ Y | ≥ r + 1 . So, there exists a subset S n of X n,j such that | S n | = 2 r + 1 , S i ⊆ Y and S ⊆ Z . So, S n contains a subset S (cid:48) n such that | S (cid:48) n | = r + 1 , S (cid:48) n ⊆ Y , S (cid:48) ⊆ Z and the matrix M G [ S (cid:48) n ∪ S (cid:48) ,j +1 ] ,S (cid:48) n ,S (cid:48) ,j +1 is triangular and has rank at least r by Lemma 2.3, a contradiction to ( Y, Z ) being a partition of V ( G ) of rank less than r .When ( G, X n , X ) is a parallel expanding triple, the proof is similar. We obtain adiagonal, or anti-diagonal matrix of rank at least r , a contradiction. (cid:50) The following lemma describes how labels propagate in skew expanding triples (seeFigure 3). We omit the proof since it is similar to the previous one.
Lemma 6.4
Let m ≥ and j be integers such that ≤ j ≤ s − .(i) Suppose that ( G, X n , X ) is a parallel skew expanding triple.If X n,j has label Y m , then X ,j +1 has label Y max( m − , , Y max( m − , , Y m , Y m +1 or Y m +2 .If X n,j has label Y m , then X ,j +1 has label Y max( m − , , Y max( m − , , Y m , Y m +1 or Y m +2 . ii) Suppose that ( G, X i , X i +1 ) is a cross triple.If X n,j has label Y m , then X ,j +1 has label Y max( m − , , Y max( m − , , Y m , Y m +1 or Y m +2 .If X n,j has label Y m , then X ,j +1 has label Y max( m − , , Y max( m − , , Y m , Y m +1 or Y m +2 . In this section, n ≥ and r ≥ are fixed integers. We prove Theorem 4.1 for even carousels.We therefore look for an integer s such that every even carousel of order s on n sets hasrankwidth at least r . We define s as follows. Let q be an integer such that: q +8 r − ≥ n + 1)( q + 8 r + 1) r (1)Clearly q exists. We set s = q + 8 r + 1 . We now consider an even carousel G of order s on n sets. To prove that it has rankwidth at least r , it is enough by Lemma 2.2 to provethat every partition ( Y, Z ) of V ( G ) with rank less than r is unbalanced, so let us consider ( Y, Z ) a partition of V ( G ) or rank less than r . Lemma 7.1
There exists t ∈ { q, . . . , q + 8 r − } such that X ,t has label Y or Z .Proof. Otherwise, for every t ∈ { q, . . . , q + 8 r − } the set X ,i is an interval of X that mustcontain elements of Y and elements of Z . Hence, the r sets X ,q , . . . , X ,q +8 r − show thatthe interval S = ∪ q +8 r − j = q X ,j has at least r blocks, a contradiction to Lemma 6.1. (cid:50) Up to symmetry, we may assume that there exists t ∈ { q, . . . , q + 8 r − } such that X ,t has label Y . We denote by (cid:101) X i,j the set that is equal to X i,j if X i is a top set, and that isequal to X i,j if X i is a botom set.The idea is now to apply Lemmas 6.2 and 6.3 repeatedly to show that for all i ∈{ , . . . , n } and for all j ∈ { t, . . . , q + 8 r + 1 } , the set (cid:101) X i,j has a label Y m with m not too big.We obtain that m ≤ ( n +1)(8 r +1) , because there are at most ( n − r +1) applications ofLemmas 6.2 that each augments the index of the label by at most , and r + 1 applicationsof Lemma 6.3 that each augment the index of the label by at most 2.Let us now prove that ( Y, Z ) is unbalanced. To do so, we focus on the sets (cid:101) X i,j for i ∈ { , . . . , n } and j ∈ { s − , s } . We denote by X their union. By elementary propertiesof powers of 2, we have | X | > | V ( G ) | . Also, by the discussion above, each of the set X i,j has label Y m with m ≤ ( n + 1)(8 r + 1) . Hence, for i ∈ { , . . . , n } and j ∈ { r, r + 1 } , | X i,j ∩ Y | ≤ ( n + 1)( q + 8 r + 1) r . So, by inequality (1), the proportion of vertices from Z in X is at least . Hence, | Z | ≥ × | V ( G ) | = 2740 | V ( G ) | > | V ( G ) | . ( Y, Z ) is therefore unbalanced as claimed. In this section, n ≥ and r ≥ are fixed integers. We prove Theorem 4.1 for odd carousels.We therefore look for an integer s such that every odd carousel of order s on n sets hasrankwidth at least r . We define s as follows. Let q be an integer such that: q +16 r − ≥ n + 1)( q + 16 r + 1) r (2)Clearly q exists. We set s = q + 16 r + 1 . We now consider a skew carousel G of order s on n sets. To prove that it has rankwidth at least r , it is enough by Lemma 2.2 to provethat every partition ( Y, Z ) of V ( G ) with rank less than r is unbalanced, so let us consider ( Y, Z ) a partition of V ( G ) of rank less than r . Lemma 8.1
There exists t ∈ { q, . . . , q + 16 r − } such that X ,t ∪ X ,t +1 has label Y or Z .Proof. Otherwise, for every i ∈ { q, . . . , q + 16 r − } the set X ,i ∪ X ,i +1 is an interval of X that must contain elements of Y and elements of Z . Hence, the r sets X ,q ∪ X ,q +1 , X ,q +2 ∪ X ,q +3 , . . . , X ,q +16 r − ∪ X ,q +16 r − show that the interval of X S = ∪ q +16 r − j = q X ,j has at least r blocks a contradiction to Lemma 6.1. (cid:50) Up to symmetry, we may assume that there exists t ∈ { q, . . . , q + 16 r − } such that X ,t ∪ X ,t +1 has label Y . We denote (cid:101) X i,j the set that is equal to X i,j if X i is a top set,and that is equal to X i,j if X i is a botom set.The idea is now to apply Lemmas 6.2 and 6.4 repeatedly to show that all i ∈ { , . . . , n } and all j ∈ { t, . . . , q + 16 r + 1 } , the sets X i,j and X i,j both have a label Y m with m nottoo big. We obtain that m ≤ ( n + 1)(16 r + 1) , because there are at most ( n − r + 1) applications of Lemmas 6.2 that each augment the index of the label by at most , and r + 1 applications of Lemma 6.3 that each augment the index of the label by at most 2.Let us now prove that ( Y, Z ) is unbalanced. To do so, we focus on the n sets X i,j , X i,j for i ∈ { , . . . , n } and j ∈ { s − , s } . We denote by X their union. By elementaryproperties of powers of 2, we have | X | > | V ( G ) | . Also, each of the set X i,j or X i,j haslabel Y m with m ≤ ( n +1)(16 r +1) . Hence, for i ∈ { , . . . , n } and j ∈ { s − , s } , | X i,j ∩ Y | ≤ ( n + 1)( q + 16 r + 1) r (and a similar inequality holds for X i,j ). So, by inequality (2), theproportion of vertices from Z in X is at least . Hence, | Z | ≥ × | V ( G ) | = 2740 | V ( G ) | > | V ( G ) | . The partition ( Y, Z ) is therefore unbalanced as claimed.14 Applications
In this section, we explain how to tune our carousels to obtain the results announced in theintroduction.
Split permutation graphs with Dilworth number 2
In [11], it is proved that there exist split permutation graphs with Dilworth number 2 andarbitrarily large rankwidth. Theorem 3 in [11] states that the class of split permutationgraphs is precisely the class of split graphs of Dilworth number at most 2 . Therefore, it isenough for us to study split graphs with Dilworth number 2.We now provide an explicit construction. Consider an even carousel on 4 sets X , X , X , X and the following additional properties: • X ∪ X is a clique. • X ∪ X is a stable set. • The triples ( G, X , X ) , ( G, X , X ) , ( G, X , X ) ; are regular crossings. • The triple ( G, X , X ) is an expanding crossing.It is clear that graphs constructed in that way are split graphs because X ∪ X is aclique and X ∪ X is a stable set. To check that they have Dilworth number 2, it is enoughto notice that for all vertices x, y ∈ X ∪ X , either N ( x ) \ N [ y ] = ∅ or N ( y ) \ N [ x ] = ∅ ,and for all vertices x, y ∈ X ∪ X , either N ( x ) \ N [ y ] = ∅ or N ( y ) \ N [ x ] = ∅ . Rings
We consider a carousel on n ≥ sets, all are cliques cliques, with all triples being crossings.If n even, this is an even carousel; and if n is odd, this is an odd carousel. For a fixed n ,this gives rings with arbitrarily large rankwidth. Flexibility of carousels
Observe that in this section, we never use matchings and antimatchings, that were includedin the definition for possible later use of carousels. More generality could be allowed (forinstance by not forcing ( G, X , X ) to be crossing, and instead forcing some ( G, X i , X i +1 ) to be crossing). This might be useful, but we felt that it would make the explanation toocomplicated. 15igure 4: Graphs that are not induced subgraphs of rings on 3 sets, and that are minimalwith respect to this property
10 Open questions
It would be nice to characterize the class of induced subgraphs of rings on n sets by forbiddeninduced subgraphs. For n = 3 this seems to be non-trivial. It might be easier for larger n .On Figure 4, a list of obstructions for n = 3 is given but we do not know whether it iscomplete.Let S , , be the graph represented on Figure 5, and let C the class of graphs thatcontain no triangle and no S , , as an induced subgraph. A famous open question [8] is todetermine whether graphs in C have bounded rankwidth. We tried many ways to tune ourconstruction, but each time, there was either a triangle or an S , , in it. So, to the best ofour knowledge, our construction does not help to prove that the rankwidth is unboundedfor this class. Similarly, whatever tuning we try, it seems that our construction does nothelp to solve the open problems from [2, 3, 8]. This might be because we did not try hard16igure 5: S , , enough to tune our construction, or because it is not the right approach. And of course,the rankwidth might be bounded for all these classes.In [9], a conjecture is given, stating that in each class of graphs that is closed undertaking induced subgraphs and where the rankwidth is unbounded, there should be aninfinite sequence of graphs G , G , . . . such that for each pair of distinct integers i, j , G i isnot an induced subgraph of G j . This was later disproved, see [14], but we wonder whethercarousels may provide new counter-examples. This is difficult to check, because it not clearwhether “big” carousels contain smaller ones as induced subgraphs.
11 Acknowledgement
We are grateful to O-joung Kwon and Sang-il Oum for useful discussions.
References [1] Isolde Adler, Ngoc-Khang Le, Haiko Müller, Marko Radovanović, Nicolas Trotignon,and Kristina Vušković. On rank-width of even-hole-free graphs.
Discrete Mathematics& Theoretical Computer Science , 19(1), 2017.[2] Andreas Brandstädt, Konrad Dabrowski, Shenwei Huang, and Daniël Paulusma.Bounding the clique-width of H -free split graphs. Discret. Appl. Math. , 211:30–39,2016.[3] Andreas Brandstädt, Konrad Kazimierz Dabrowski, Shenwei Huang, and DaniëlPaulusma. Bounding the clique-width of H -free chordal graphs. Journal of GraphTheory , 86(1):42–77, 2017.[4] Kathie Cameron, Steven Chaplick, and Chính T. Hoàng. On the structure of (pan,even hole)-free graphs.
Journal of Graph Theory , 87(1):108–129, 2018.175] Kathie Cameron, Murilo V. G. da Silva, Shenwei Huang, and Kristina Vušković. Struc-ture and algorithms for (cap, even hole)-free graphs.
Discrete Mathematics , 341(2):463–473, 2018.[6] Bruno Courcelle, Joost Engelfriet, and Grzegorz Rozenberg. Handle-rewriting hyper-graph grammars.
J. Comput. Syst. Sci. , 46(2):218–270, 1993.[7] Bruno Courcelle, Johann A. Makowsky, and Udi Rotics. Linear time solvable optimiza-tion problems on graphs of bounded clique-width.
Theory Comput. Syst. , 33(2):125–150, 2000.[8] Konrad K. Dabrowski, François Dross, and Daniël Paulusma. Colouring diamond-freegraphs.
J. Comput. Syst. Sci. , 89:410–431, 2017.[9] Jean Daligault, Michaël Rao, and Stéphan Thomassé. Well-quasi-order of relabel func-tions.
Order , 27(3):301–315, 2010.[10] Martin Charles Golumbic and Udi Rotics. On the clique-width of some perfect graphclasses.
Int. J. Found. Comput. Sci. , 11(3):423–443, 2000.[11] Nicholas Korpelainen, Vadim V. Lozin, and Colin Mayhill. Split permutation graphs.
Graphs and Combinatorics , 30(3):633–646, 2014.[12] O-joung Kwon, Michal Pilipczuk, and Sebastian Siebertz. On low rank-width colorings.
Eur. J. Comb. , 83, 2020.[13] Ngoc-Khang Le.
Detecting and coloring some graph classes . PhD thesis, École NormaleSupérieure de Lyon, 2018.[14] Vadim V. Lozin, Igor Razgon, and Viktor Zamaraev. Well-quasi-ordering versus clique-width.
Journal of Combinatorial Theory, Series B , 130:1–18, 2018.[15] Frédéric Maffray, Irena Penev, and Kristina Vušković. Coloring rings.
CoRR ,abs/1907.11905, 2019.[16] Sang-il Oum and Paul D. Seymour. Approximating clique-width and branch-width.
J.Comb. Theory, Ser. B , 96(4):514–528, 2006.[17] Ana Silva, Aline Alves da Silva, and Cláudia Linhares Sales. A bound on the treewidthof planar even-hole-free graphs.
Discret. Appl. Math. , 158(12):1229–1239, 2010.[18] Ni Luh Dewi Sintiari and Nicolas Trotignon. (Theta, triangle)-free and (even hole,K4)-free graphs. Part 1 : Layered wheels.