(Theta, triangle)-free and (even hole, K 4 )-free graphs. Part 1 : Layered wheels
((Theta, triangle)-free and (even hole, K )-freegraphs. Part 1 : Layered wheels Ni Luh Dewi Sintiari ∗ , Nicolas Trotignon ∗ July 23, 2019
Abstract
We present a construction called layered wheel. Layered wheels aregraphs of arbitrarily large treewidth and girth. They might be an out-come for a possible theorem characterizing graphs with large treewidthin term of their induced subgraphs (while such a characterization is wellunderstood in term of minors). They also provide examples of graphs oflarge treewidth and large rankwidth in well studied classes, such as (theta,triangle)-free graphs and even-hole-free graphs with no K (where a holeis a chordless cycle of length at least 4, a theta is a graph made of threeinternally vertex disjoint paths of length at least 2 linking two vertices,and K is the complete graph on 4 vertices). In this article, all graphs are finite, simple, and undirected. The vertex set ofa graph G is denoted by V ( G ) and the edge set by E ( G ). A graph H is an induced subgraph of a graph G if some graph isomorphic to H can be obtainedfrom G by deleting vertices. A graph H is a minor of a graph G if some graphisomorphic to H can be obtained from G by deleting vertices, deleting edges,and contracting edges.When we say that G contains H without specifying as a minor or as aninduced subgraph , we mean that H is an induced subgraph of G . A graph is H -free if it does not contain H (so, as an induced subgraph). For a family ofgraphs H , G is H -free if for every H ∈ H , G is H -free. A class of graph is hereditary if it is H -free for some H or, equivalently, if it is closed under takinginduced subgraphs. A hole in a graph is a chordless cycle of length at least 4. Itis odd or even according to its length (that is its number of edges). We denoteby K (cid:96) the complete graph on (cid:96) vertices. ∗ Univ Lyon, EnsL, UCBL, CNRS, LIP, F-69342, LYON Cedex 07, France.The authors are partially supported by the LABEX MILYON (ANR-10-LABX-0070) of Uni-versit´e de Lyon, within the program Investissements d’Avenir (ANR-11-IDEX-0007) operatedby the French National Research Agency (ANR) a r X i v : . [ c s . D M ] J u l igure 1: A grid and a wallThe present paper is originally motivated by a question asked by Cameron etal. in [3]: is the treewidth (or cliquewidth) of an even-hole-free graph boundedby a function of its clique number? We describe a construction called layeredwheel showing that the answer is no. It turns out that we found other resultsand questions, and we present them in a series of two articles. In this first part,there are three main motivations: • When considering the induced subgraph relation (instead of the minorrelation), is there a theorem similar to the celebrated grid-minor theoremof Robertson and Seymour? • A better understanding of the classes defined by excluding the so-calledTruemper configurations, that play an important role in hereditary classesof graphs. • The structure of even-hole-free graphs.We now give details on each of the three items.
The grid-minor theorem
The treewidth of a graph is an integer measuring how far is the graph from atree. We do not recall the classical definition, see [14] for a formal definitionand related notions. The ( k × k )- grid is the graph on { ( i, j ); 1 ≤ i, j ≤ k } wheretwo distinct ordered pairs ( i, j ) and ( i (cid:48) , j (cid:48) ) are adjacent whenever exactly oneof the following hold: | i − i (cid:48) | = 1 and j = j (cid:48) , or i = i (cid:48) and | j − j (cid:48) | = 1 (seefigure 1). Robertson and Seymour [17] proved that there exists a function f such that every graph with treewidth at least f ( k ) contains a ( k × k )-grid as aminor (see [7] for the best function known so far). This is called the grid-minortheorem . The ( k × k )- wall is the graph obtained from the ( k × k )-grid by deletingall edges with form (2 i + 1 , j ) − (2 i + 1 , j + 1) and (2 i, j + 1) − (2 i, j + 2). Subdividing k times an edge e = uv of a graph, where k ≥
1, means deleting e and adding a path uw . . . w k v . The k -subdivision of a graph G is the graphobtained from G by subdividing k -times all its edges (simultaneously). Notethat replacing “grid” by a more specific graph in the grid-minor theorem, suchas k -subdivision of a ( k × k )-grid, ( k × k )-wall, or k -subdivision of a ( k × k )-wall provides statements that are formally weaker (at the expense of a larger2igure 2: A subdivision of a grid, of a wall, and the line graphs of the formerfunction), because a large grid contains a large subdivision of a grid, a largewall, and a large subdivision of a wall. However, these trivial corollaries arein some sense stronger, because walls, subdivisions of walls, and subdivision ofgrids are graphs of large treewidth that are more sparse than grids. So theysomehow certify a large treewidth with less information. Since one can alwayssubdivide more, there is no “ultimate” theorem in this direction.It would be useful to have a similar theorem with “induced subgraph” in-stead of “minor”. Simply replacing “minor” with “induced subgraph” in thestatement is trivially false, and here is a list of known counter-examples: K k , K k,k , subdivisions of walls, line graphs of subdivisions of walls (see figure 2),where K k denotes the complete graph on k vertices, K k,k the complete bipar-tite graph with each side of size k , and where the line graph of a graph R isthe graph G on E ( R ) where two vertices in G are adjacent whenever they areadjacent edges of R .One of our results is that the simple list above is not complete. In section 3,we present a construction that we call layered wheel . Layered wheels have largetreewidth and large girth (the girth of a graph is the length of its shortestcycle). Large girth implies that they contain no K k , no K k,k and no line graphsof subdivisions of walls. Moreover, layered wheels contain no subdivisions of(3 , f every graph with treewidth at least f ( k ) contains either K k , K k,k , a subdivision of the ( k × k )-wall, the line graphof some subdivision of the ( k × k )-wall, or some variant of the layered wheel?We observe that to have a positive answer, we would need a definition of thelayered wheel with slightly more flexibility than what we propose now. However,in the next paragraphs, we give variants of Dvoˇr´ak’s question that are fully welldefined. Truemper configurations A prism is a graph made of three vertex-disjoint chordless paths P = a . . . b , P = a . . . b , P = a . . . b of length at least 1, such that a a a and b b b aretriangles and no edges exist between the paths except those of the two triangles.Such a prism is also referred to as a 3 P C ( a a a , b b b ) or a 3 P C (∆ , ∆) (3PC3igure 3: Pyramid, prism, theta, and wheel (dashed lines represent paths)stands for ).A pyramid is a graph made of three chordless paths P = a . . . b , P = a . . . b , P = a . . . b of length at least 1, two of which have length at least 2,vertex-disjoint except at a , and such that b b b is a triangle and no edges existbetween the paths except those of the triangle and the three edges incident to a . Such a pyramid is also referred to as a 3 P C ( b b b , a ) or a 3 P C (∆ , · ).A theta is a graph made of three internally vertex-disjoint chordless paths P = a . . . b , P = a . . . b , P = a . . . b of length at least 2 and such that no edgesexist between the paths except the three edges incident to a and the three edgesincident to b . Such a theta is also referred to as a 3 P C ( a, b ) or a 3 P C ( · , · ).Observe that the lengths of the paths in the three definitions above aredesigned so that the union of any two of the paths induces a hole. A wheel W = ( H, c ) is a graph formed by a hole H (called the rim ) together with avertex c (called the center ) that has at least three neighbors in the hole.A is a graph isomorphic to a prism, a pyramid, or atheta. A Truemper configuration is a graph isomorphic to a prism, a pyramid, atheta, or a wheel. They appear in a theorem of Truemper [19] that characterizesgraphs whose edges can be labeled so that all chordless cycles have prescribedparities (3-path-configurations seem to have first appeared in a paper Watkinsand Mesner [22]).Truemper configurations play an important role in the analysis of severalimportant hereditary graph classes, as explained in a survey of Vuˇskovi´c [21].Let us simply mention here that many decomposition theorems for classes ofgraphs are proved by studying how some Truemper configurations contained inthe graph attaches to the rest of the graph, and often, the study relies on the factthat some other Truemper configurations are excluded from the class. The mostfamous example is perhaps the class of perfect graphs . In these graphs, pyramidsare excluded, and how a prism contained in a perfect graphs attaches to the restof the graph is important in the decomposition theorem for perfect graphs, whosecorollary is the celebrated
Strong Perfect Graph Theorem due to Chudnovsky,Robertson, Seymour, and Thomas [5]. See also [18] for a survey on perfectgraphs, where a section is specifically devoted to Truemper configurations. Manyother examples exist, see [9] for a long list of them.Some researchers started to study systematically classes defined by the exclu-sion of some Truemper configurations [9]. We believe that among many classesthat can be defined in that way, the class of theta-free graphs is one of the4ost interesting. This is because it generalizes claw-free graphs (since a thetacontains a claw), and it seems that whether it might share its most interestingfeatures is maybe a good question: a structural description (see [6]), a polyno-mial time algorithm for the maximum stable set (see [10]), an approximationalgorithms for the chromatic number (see [15]), a polynomial time algorithm forthe induced linkage problem (see [11]), and a polynomial χ -bounding function(see [13]).In the attempt of finding a structural description of theta-free graphs, aseemingly easy case is when triangles are also excluded. Because then, everyvertex of degree at least 3 is the center of a claw, and therefore a possible startfor a theta, so that excluding theta and triangle should enforce some structure.Supporting this idea, Radovanovi´c and Vuˇskovi´c [16] proved that every (theta,triangle)-free is 3-colorable.Hence, we believed when starting this work that (theta, triangle)-free graphshave bounded treewidth. But this turned out to be false: layered wheels are(theta, triangle)-free graphs of arbitrarily large treewidth.However, on the positive side, we note that layered wheels needs many ver-tices to increase the treewidth. More specifically, a layered wheel G is madeof l + 1 layers, where l is an integer. Each layer is a path and | V ( G ) | ≥ l (see 3.2), l ≤ tw( G ) ≤ l (see Theorem 3.11 and 5.4). So, the treewidth ofa layered wheel is “small” in the sense that it is logarithmic in the number ofits vertices. We wonder whether such a behavior is general in the sense of thefollowing conjecture. Conjecture 1.1.
For some constant c , if G is a (theta, triangle)-free graph,then the treewidth of G is at most c log | V ( G ) | . This conjecture reflects our belief that constructions similar to the lay-ered wheel must have an exponential number of vertices (exponential in thetreewidth). It suggests the following variant of Dvoˇr´ak’s question: is it truethat for some constant c > f , every graph with treewidthat least f ( k ) contains either K k , K k,k , a subdivision of the ( k × k )-wall, the linegraph of some subdivision of the ( k × k )-wall, or contains at least c f ( k ) vertices?Kristina Vuˇskovi´c observed that K k,k is a (prism, pyramid, wheel)-free graph,or equivalently an only-theta graph (because the theta is the only Truemper con-figuration contained in it). Moreover, walls are only-theta graphs, line graphsof subdivisions of walls are only-prism graphs, and triangle-free layered wheelsare only-wheel graphs. Observe that complete graphs contain no Truemper con-figuration, so they are simultaneously only-prism, only-wheel, and only-theta.One may wonder whether a graph with large treewidth should contain an in-duced subgraphs of large treewidth with a restricted list of induced subgraphsisomorphic to Truemper configurations. Even-hole-free graphs
Our last motivation for this work is a better understanding of even-hole-freegraphs. These are related to Truemper configurations because thetas and prisms5bviously contain even holes (to see this, consider two paths of the same parityamong the three paths that form the configuration). Also, call even wheel awheel W = ( H, c ) where c has an even number of neighbors in H . It is easy tocheck that every even wheel contains an even hole.Even-hole-free graphs were originally studied to experiment techniques thatwould help to settle problems on perfect graphs. This has succeeded, in the sensethat the decomposition theorem for even-hole-free graphs (see [20]) is in somerespect similar to the one that was later on discovered for perfect graphs (see [5]).However, classical problems, like graph coloring or maximum stable set, arepolynomial time solvable for perfect graphs, while they are still open for even-hole-free graphs. This is a bit strange because the decomposition theorem foreven-hole-free graphs is in many respect simpler than the one for perfect graphs.Moreover, it is easy to provide perfect graphs of arbitrarily large treewidth(or even rankwidth), such as bipartite graphs, or their line graphs. While foreven-hole-free graphs, apart from complete graphs, it is not so easy. Someconstructions are known, see [1].But so far, every construction of even-hole-free graphs of arbitrarily largetreewidth (or rankwidth) contains large cliques. Moreover, it is proved in [4]that (even-hole, triangle)-free graphs have bounded treewidth. This is based ona structural description of the class from [8]. Hence, Cameron et al. [3] askedwhether (even hole, K )-free graphs have bounded treewidth. We prove in thisarticle that it is not the case, by a variant of the layered wheel construction.As for (theta, triangle)-free, we need a large number of vertices to grow thetreewidth, so that we propose the following conjecture. Conjecture 1.2.
There exists a constant c such that for any (even-hole, K )-free graph G , the treewidth of G is at most c log | V ( G ) | . Our construction of even-hole-free layered wheels contains diamonds andpyramids (this is explained at the end of Section 3). We therefore propose thefollowing conjectures.
Conjecture 1.3.
Even-hole-free graphs with no K and no pyramids havebounded treewidth. Conjecture 1.4.
Even-hole-free graphs with no K and no diamonds havebounded treewidth. Note that for the two classes where we prove unbounded treewidth, the cliquewidth (and therefore the rankwidth) is large also.
Outline of the article
In Section 2 we introduce the terminology used in our proofs.In Section 3, we describe the construction of layered wheels for two classes ofgraphs: (theta, triangle)-free graphs and (even-hole, K )-free graphs. We provethat the construction actually yields graphs in the classes (this is non-trivial,6ee Theorems 3.5 and 3.10). We then prove that layered wheels have unboundedtreewidth (see Theorem 3.11) and cliquewidth (see Theorem 3.14).In Section 4, we recall the definition of rankwidth. We exhibit (theta,triangle)-free graphs and (even-hole, K )-free graphs with large rankwidth. Thisis a trivial corollary of Theorem 3.14, but the computation is more accurate (seeTheorem 4.16).In Section 5, we give an upper bound on the treewidth of layered wheels.We prove a stronger result: the so-called pathwidth of layered wheels is boundedby some linear function of the number of its layers (see Theorem 5.4).The treewidth, cliquewidth, rankwidth, and pathwidth of a graph G aredenoted by tw( G ), cw( G ), rw( G ), and pw( G ) respectively. The following is wellknown. Lemma 1.5.
For every graph G , rw( G ) ≤ cw( G ) ≤ tw ( G ) ≤ pw ( G ) , and rw( G ) ≤ cw( G ) ≤ rw( G )+1 All the results presented in this article can be sum up in the next two theo-rems.
Theorem 1.6.
For every integers l ≥ and k ≥ , there exists a graph G l,k such that: • G l,k is theta-free and has girth at least k (in particular, G l,k is triangle-free). • l ≤ rw( G l,k ) ≤ cw( G l,k ) ≤ tw( G l,k ) ≤ pw( G l,k ) ≤ l ≤ l ≤ | V ( G l,k ) | . Theorem 1.7.
For every integers l ≥ and k ≥ , there exists a graph G l,k such that: • G l,k is (even hole, K )-free and every hole in G l,k has length at least k . • l ≤ rw( G l,k ) ≤ cw( G l,k ) ≤ tw( G l,k ) ≤ pw( G l,k ) ≤ l ≤ l ≤ | V ( G l,k ) | . A graph H is a subgraph of a graph G , denoted by H ⊆ G , if V ( H ) ⊆ V ( G )and E ( H ) ⊆ E ( G ). For a graph G and a subset X ⊆ V ( G ), we let G [ X ] denotethe subgraph of G induced by X , i.e. G [ X ] has vertex set X , and E ( G [ X ])consists of the edges of G that have both ends in X . In this case we say that G contains H . A graph H ⊆ G is an induced subgraph of G , if H = G [ X ] forsome X ⊆ V ( G ).For simplicity, sometimes we do not distinguish between a vertex set and thegraph induced by the vertex set. So we write G \ H instead of G [ V ( G ) \ V ( H )].7lso for a vertex v ∈ V ( G ), we write G \ v (instead of G [ V ( G ) \ { v } ]) andsimilarly, we write G \ S for some S ⊆ V ( G ). For v ∈ V ( G ), we denote by N H ( v ), the set of neighbors of v in H that is called the neighborhood of v , and N G ( v ) is also denoted by N ( v ).A path P is a graph formed by vertices p , . . . , p n , n ≥ p i p i +1 is an edge, for all 1 ≤ i < n . For two vertices p i , p j ∈ V ( P ) with j > i , thepath p i , p i +1 , . . . , p j is a subpath of P that is denoted by p i P p j . The subpath p · · · p n − is called the interior of P . The vertices p , p n are the ends of thepath, and the vertices in the interior of P are called the internal vertices of P . The length of a path P is the number of edges of P . Note that, | E ( P ) | = | V ( P ) − | .A cycle is defined similarly, with the additional properties that n ≥ p = p n . A path (or a cycle) in G is chordless if it is an induced subgraph of G . In this section, we describe the construction of layered wheels for two classesof graphs, namely the class of (theta, triangle)-free graphs and the class of(even-hole, K )-free graphs. We also give a lower bound on their treewidth. (Theta, triangle)-free layered-wheels We now present ttf-layered-wheels which are theta-free graphs of girth at least k ,containing K l as a minor, for all integers l ≥ , k ≥ Construction 3.1.
Let l ≥ and k ≥ be integers. An ( l, k ) -ttf-layered-wheel, denoted by G l,k , is a graph consisting of l + 1 layers, which are paths P , P , · · · , P l . The graph is constructed as follows.(A1) V ( G l,k ) is partitioned into l + 1 vertex-disjoint paths P , . . . , P l . So, V ( G l,k ) = V ( P ) ∪ · · · ∪ V ( P l ) . The paths are constructed in an inductiveway.(A2) The path P consists of a single vertex.(A3) For every ≤ i ≤ l and every vertex u in P i , we call ancestor of u anyneighbor of u in V ( P ) ∪ · · · ∪ V ( P i − ) . The type of u is the number of itsancestors (as we will see, the construction implies that every vertex hastype 0 or 1). Observe that the unique vertex of P has type 0. We will seethat the construction implies that for every ≤ i ≤ l , the ends of P i arevertices of type 1.(A4) Suppose inductively that l ≥ and layers P , . . . , P l − are constructed.The l th -layer P l is built as follows.For any u ∈ P l − we define a path Box u (that will be a subpath of P i ). G , • if u is of type 0, Box u contains three neighbors of u , namely u , u , u ,in such way that Box u = u . . . u . . . u . • if u is of type 1, let v be its unique ancestor. Box u contains sixneighbors of u , namely u , · · · , u , and three neighbors of v , namely v , v , v , in such a way that Box u = u . . . u . . . u . . . v . . . v . . . v . . . u . . . u . . . u . The neighbors of u in Box u are of type 1, the other vertices of Box u are of type 0. We now specify the lengths of the boxes and how they areconnected to form P l .(A5) The path P l goes through the boxes of P l in the same order as verticesin P l − . For instance, if uvw is a subpath of P l − , then P l goes through Box u , Box v , and Box w , in this order along P l . Note that the vertices of P l that are in none of the boxes are of type 0.(A6) Let w, w (cid:48) be vertices of type 1 in P l (so vertices from the boxes), andconsecutive in the sense that the interior of wP l w (cid:48) contains no vertex oftype 1. If w and w (cid:48) have the same ancestor, then wP l w (cid:48) is a path of lengthat least k − . If w and w (cid:48) have different ancestors, then wP l w (cid:48) is a pathof length at least k − .(A7) Observe that every vertex in P l has type 0 or 1.(A8) There are no other vertices or edges apart from the ones specified above. Observe that the construction is not fully deterministic because in (A6), wejust indicate a lower bound on the length of wP l w (cid:48) , so there may exists differentttf-layered-wheels G l,k . This flexibility will be convenient below to exhibit ttf-layered-wheels of arbitrarily large rankwidth. Lemma 3.2.
For ≤ i ≤ l − and i + 1 ≤ j ≤ l , every vertex u ∈ V ( P i ) hasat least j − i neighbors in P j .Proof. We prove the lemma by induction on j . If j = i +1, then (A4) implies thatfor every 1 ≤ i ≤ l − u in P i , u has 3 or 6 neighbors in P i +1 .9f j > i + 1, then by the induction hypothesis, every vertex u ∈ V ( P i ) has atleast 3 j − − i neighbors in P j − . Hence by (A4), it has at least 3 × j − − i = 3 j − i neighbors in P j .Lemma 3.2 implies in particular that every vertex of layer i has neighbors inall layers i + 1 , . . . , l . Construction 3.1 is in fact the description of an inductivealgorithm that constructs G l,k . So, the next lemma is clear. Lemma 3.3.
For every integers l ≥ and k ≥ , there exists an ( l, k ) -ttf-layered-wheel. We now prove that Construction 3.1 produces a theta-free graph with ar-bitrary large girth and treewidth. Observe that any subdivision of (3,5)-gridcontains a theta. Thus, Theorem 3.5 implies that a ttf-layered-wheel does notcontain any subdivision of (3,5)-grid as mentioned in the introduction.The next lemma is useful to prove Theorem 3.5. For a theta consisting ofthree paths P , P , P , the common ends of those paths are called the apexes ofthe theta. Let G be graph containing a path P . The path P is special if • there exists a vertex v ∈ V ( G \ P ) such that | N P ( v ) | ≥
3; and • in G \ v , every vertex of P has degree at most 2.Note that we make no assumption on G , that in particular may containtriangles. Lemma 3.4.
Let G be a graph containing a special path P . For any theta thatis contained in G (if any), no vertex of P can be an apex of the theta.Proof. For a contradiction, suppose that P contains some vertex u which is anapex of some theta Θ in G . Note that u must have degree 3, and is thereforea neighbor of v . Consider two subpaths of P , u P u and u P u such that u ∈ { u , u , u } ⊆ N ( v ) and both u P u , u P u have no neighbors of v in theirinterior. This exists since | N P ( v ) | ≥
3. Since u is an apex, either H = vu P u v or H = vu P u v is a hole of Θ. W.l.o.g. suppose that V ( H ) ⊆ V (Θ). Hencethe other apex of Θ must be also contained in H . Since u v, u v ∈ E ( G ) andall vertices of H \ { u , v, u } have degree 2, u , u must be the two apexes ofΘ. Since d ( u ) = 3, V ( u P u ) ⊆ Θ. But then v has degree 3 in Θ while notbeing an apex, a contradiction. This completes the proof. Theorem 3.5.
For every integer l ≥ and k ≥ , G l,k is a theta-free graphwith girth at least k .Proof. We first show by induction on l that G l,k has girth at least k . This isclear for l = 1, so suppose l ≥ H be a cycle in G l,k with length less than k . We may assume that layer P l contains some vertex of H , for otherwise H isa cycle in G l − ,k , so it has length at least k by the induction hypothesis. Let P = u . . . v be a path such that V ( P ) ⊆ V ( H ) ∩ V ( P l ) and with the maximumlength among such possible paths. Note that P contains at least two vertices.10ndeed, if P contains a single vertex, then such a vertex must have at least twoancestors, since it has degree 2 in H , which is impossible by the construction of G l,k . So u (cid:54) = v . Moreover, note that as P is contained in a cycle, both u and v must have an ancestor. Let u (cid:48) and v (cid:48) be the ancestor of u and v respectively.By (A6) of Construction 3.1, if u (cid:48) = v (cid:48) then P has length at least k − u (cid:48) (cid:54) = v (cid:48) then P has length at least k −
3. Hence u (cid:48) uP vv (cid:48) has length at least k ,so H has length at least k . This completes the proof.Now we show that G l,k is theta-free. For a contradiction, suppose that itcontains a theta. Let Θ be a theta with minimum number of vertices, andhaving u and v as apexes. As above, w.l.o.g., we may assume that P l containssome vertex of Θ. Note that every vertex of P l is contained in a special pathof G l,k . Hence, by Lemma 3.4, u, v / ∈ V ( P l ). In particular, every vertex of V ( P l ) ∩ V (Θ) has degree 2 in Θ.Let P = x . . . y for some x, y ∈ P l , be a path such that V ( P ) ⊆ V (Θ) ∩ V ( P l )and it is inclusion-wise maximal w.r.t. this property. Since every vertex of P l has at most one ancestor, x (cid:54) = y . Moreover, both x and y must have an ancestor,because every vertex of Θ has degree 2 or 3 in Θ. Let x (cid:48) and y (cid:48) be the ancestor of x and y respectively. By the maximality of P , both x (cid:48) and y (cid:48) are also in Θ. Notethat no vertex in the interior of P is adjacent to x (cid:48) or y (cid:48) , since otherwise sucha vertex would have degree 3 in Θ, meaning that it is an apex, a contradiction. Claim 1.
We have x (cid:48) (cid:54) = y (cid:48) , x (cid:48) y (cid:48) / ∈ E ( G l,k ) , and some internal vertex of P is oftype 1.Proof of Claim 1. Otherwise, x (cid:48) = y (cid:48) or x (cid:48) y (cid:48) ∈ E ( G l,k ), or every internalvertex of P is of type 0. In this last case, we also have x (cid:48) = y (cid:48) ∈ V ( P l − ) or x (cid:48) y (cid:48) ∈ E ( G l,k ) by the construction of G l,k . Hence, in all cases, V ( P ) ∪ { x (cid:48) , y (cid:48) } induces a hole in Θ, that must contain both u and v . Since u, v / ∈ V ( P l ), wehave u, v ∈ { x (cid:48) , y (cid:48) } . But this is not possible as x (cid:48) = y (cid:48) or x (cid:48) y (cid:48) ∈ E ( G l,k ). Thisproves Claim 1.We now set P (cid:48) = x (cid:48) xP l yy (cid:48) (that is a path by Claim 1). Claim 2.
There exists no vertex of type 0 in P l − that has a neighbor in theinterior of P .Proof of Claim 2. For a contradiction, let t ∈ V ( P l − ) be of type 0 that hasneighbors in the interior of P . Note that t / ∈ V (Θ) because internal vertices of P have degree 2 in Θ. Let Q be the shortest path from x (cid:48) to y (cid:48) in G l,k [ V ( P (cid:48) ) ∪{ t } ].Note that Q is shorter than P (cid:48) , because it does not go through one vertex of N P ( t ). So, P (cid:48) can be substituted for Q in Θ, which provides a theta from u to v with less vertices, a contradiction to the minimality of Θ. This proves Claim 2. Claim 3.
We may assume that: • x (cid:48) ∈ V ( P l − ) • y (cid:48) / ∈ V ( P l − ) 11 y (cid:48) has a neighbor w in P l − and x (cid:48) w ∈ E ( G l,k ) • Every vertex in P has type 0, except x , y , and three neighbors of w . Ob-serve that w has type 1 and has three more neighbors in P l that are notin P .Proof of Claim 3. Suppose first that x (cid:48) , y (cid:48) are both in P l − . Then by Claim 1, thepath x (cid:48) P l − y (cid:48) has length at least 2. Moreover, by Claim 2, all its internal verticesare of type 1, because they all have neighbors in the interior of P . It followsthat x (cid:48) P l − y (cid:48) has length exactly 2. We denote by z its unique internal vertex.Substituting x (cid:48) zy (cid:48) for P (cid:48) , we obtain a theta that contradicts the minimality ofΘ. Observe that the ancestor of z is not in V (Θ), because it has three neighborsin P . This proves that x (cid:48) , y (cid:48) are not both in P l − .So up to symmetry, we may assume y (cid:48) / ∈ V ( P l − ). Since y (cid:48) has neighbor in P l , it must be that y (cid:48) has a neighbor w ∈ V ( P l − P l , onevisits in order three neighbors of w , then y and two other neighbors o y (cid:48) , andthen three other neighbors of w .Let w (cid:48) be the neighbor of w in P l − , chosen so that w (cid:48) has neighbors in P . Since w (cid:48) has type 0, by Claim 2, we have w (cid:48) = x (cid:48) . Hence, as claimed, x (cid:48) ∈ V ( P l − ) and x (cid:48) w ∈ E ( G ). This proves Claim 3.Let a , b , c , a (cid:48) , b (cid:48) , c (cid:48) be the six neighbors of w in P l appearing in this orderalong P l , in such a way that a, b, c ∈ V ( P ) and a (cid:48) , b (cid:48) , c (cid:48) / ∈ V ( P ). We have { a (cid:48) , b (cid:48) , c (cid:48) } ∩ V (Θ) (cid:54) = ∅ , since otherwise we obtain a shorter theta from u to v byreplacing P (cid:48) with x (cid:48) wy (cid:48) , a contradiction to the minimality of Θ. Let y (cid:48)(cid:48) be theneighbor of y (cid:48) in P l closest to a (cid:48) along P l . Since w / ∈ V (Θ), V ( y (cid:48) y (cid:48)(cid:48) P l c (cid:48) ) ⊆ V (Θ).If y (cid:48) / ∈ { u, v } , then by replacing x (cid:48) P y (cid:48) y (cid:48)(cid:48) P l c (cid:48) with x (cid:48) wc (cid:48) , we obtain a theta,a contradiction to the minimality of Θ. So, y (cid:48) ∈ { u, v } . W.l.o.g. we may assumethat y (cid:48) = v .If u (cid:54) = x (cid:48) , then by replacing V ( x (cid:48) P (cid:48) y (cid:48) y (cid:48)(cid:48) P l c (cid:48) ) with { x (cid:48) , w, y (cid:48) , c (cid:48) } in Θ, weobtain a theta from w to u which contains less vertices than Θ, a contradictionto the minimality of Θ. So, u = x (cid:48) .Recall that x (cid:48) has type 0. Let z (cid:54) = w be the neighbor of x (cid:48) in P l − . Moreover,let z (cid:48) and z (cid:48)(cid:48) be the neighbor of z and x (cid:48) respectively, such that all vertices in theinterior of z (cid:48) P l z (cid:48)(cid:48) have degree 2. Since Θ goes through P , w / ∈ V (Θ). Therefore z, z (cid:48) , z (cid:48)(cid:48) ∈ V (Θ). This implies the hole zx (cid:48) z (cid:48)(cid:48) P l z (cid:48) z is a hole of Θ, a contradictionbecause the other apex v = y (cid:48) is not in the hole. This completes the proof that G l,k is theta-free. Even-hole-free layered-wheels
Recall that (even hole, triangle)-free graphs have treewidth at most 5 (see [4]),and as we will see, ttf-layered wheels of arbitrarily large treewidth exist. Hence,some ttf-layered wheels contain even holes (in fact, it can be checked that theycontain even wheels). We now provide a construction of layered wheel that is(even hole, K )-free, but that contains triangles (see Figure 5). Its structure issimilar to ttf-layered-wheel, but slightly more complicated. Here it is.12 onstruction 3.6. Let l ≥ and k ≥ be integers. An ( l, k ) -ehf-layered-wheel,denoted by G l,k , consists of l +1 layers, which are paths P , P , · · · , P l . We viewthese paths as oriented from left to right. The graph is constructed as follows.(B1) V ( G l,k ) is partitioned into l + 1 vertex-disjoint paths P , P , . . . , P l . So, V ( G l,k ) = V ( P ) ∪ · · · ∪ V ( P l ) . The paths are constructed in an inductiveway.(B2) The first layer P consists of a single vertex r . The nd layer P is a pathsuch that P = r P r P r , where { r , r , r } = N P ( r ) and for j = 1 , , r j P r j +1 is of odd length at least k − .(B3) For every ≤ i ≤ l and every vertex u in P i , we call ancestor of u anyneighbor of u in G l,k [ P ∪ · · · ∪ P i − ] . The type of u is the number of itsancestors (as we will see, the construction implies that every vertex hastype 0, 1, or 2). Observe that the unique vertex of P has type 0, and P consists only of vertices of type 0 or type 1. Moreover, we will see thatif u is of type 2, then its ancestors are adjacent. Also, the constructionimplies that for every ≤ i ≤ l , the ends of P i are vertices of type 1.(B4) Suppose inductively that l ≥ and P , . . . , P l − are constructed. The l th -layer P l is built as follows.For all ≤ i ≤ l − , any vertex u ∈ V ( P i ) has an odd number of neighborsin P l , that are into subpaths of P l that we call zones . These zones arelabeled by E u or O u according to their parity: a zone labeled E u contains4 neighbors of u , and a zone labeled O u contains 3 neighbors of u . Allthese 4 or 3 neighbors are of type 1, and all the other vertices of the zoneare of type 0.There are also zones that contain common neighbors of two vertices u, v .We label them E u,v (or O u,v ). A zone E u,v (resp. O u,v ) contains 4 (resp.3) common neighbors of u and v . All these 4 or 3 neighbors are of type 2,and all the other vertices of the zone are of type 0.(B5) For any u ∈ P l − , we define the box Box u , that is a subpath of P l , asfollows: • If u is of type 0 (so it is an internal vertex of P l − ), then let u (cid:48) and u (cid:48)(cid:48) be the neighbors of u in P l − , so that u (cid:48) uu (cid:48)(cid:48) is a subpath of P l − .In this case, Box u goes through three zones E u (cid:48) ,u , O u , E u,u (cid:48)(cid:48) thatappear in this order along P l , (see Figure 5). • If u is of type 1, then let v ∈ P i , i < l − be its ancestor .If u is an internal vertex of P l − , then let u (cid:48) and u (cid:48)(cid:48) be the neighborsof u in P l − , so that u (cid:48) uu (cid:48)(cid:48) is a subpath of P l − . In this case, Box u is made of five zones E u (cid:48) ,u , O u , O u,v , O u , E u,u (cid:48)(cid:48) (see Figure 5).If u is the left end of P l − , then let u (cid:48)(cid:48) be the neighbor of u in P l − .In this case, Box u is made of four zones O u , O u,v , O u , E u,u (cid:48)(cid:48) . f u is the right end of P l − , then let u (cid:48) be the neighbor of u in P l − .In this case, Box u is made of four zones E u (cid:48) ,u , O u , O u,v , O u . • If u is of type 2 (so it is an internal vertex of P l − ), then let v ∈ P i and w ∈ P j , j ≤ i be its ancestors. If i = j , we suppose that v and w appear in this order along P i (viewed from left to right). It turnsout that either w is an ancestor of v , or v, w are consecutive alongsome path P i (because as one can check, all vertices of type 2 that wecreate satisfy this statement). In this case, Box u is made of eightzones E u (cid:48) ,u , O u , O u,v , O w , O u,w , O v , O v,w , E u,u (cid:48)(cid:48) (see Figure 5).(B6) The path P l visits all the boxes Box − of P l in the same order as verticesin P l − . For instance, if uvw is a subpath of P l − , then Box u , Box v , and Box w appear in this order along P l .(B7) Let u and v be two vertices of P l , both of type 1 or 2, and consecutive inthe sense that every vertex in the interior of uP l v is of type 0. If u and v have a common ancestor, then uP l v has odd length, at least k − . If u and v have no common ancestor, then uP l v has even length, at least k − .(B8) Observe that every vertex in P l has type 0, 1, or 2. Moreover, as an-nounced, every vertex of type 2 has two adjacent ancestors.(B9) There are no other vertices or edges apart from the ones specified above. For the same reason as for ttf-layered-wheels, we allow flexibility in Con-struction 3.6, by just giving lower bounds for the lengths of paths describedin (B7). So there may exists different ehf-layered-wheels G l,k for the same valueof l and k . Lemma 3.7.
For ≤ i ≤ l − and i + 1 ≤ j ≤ l , every vertex u ∈ V ( P i ) hasat least j − i neighbors in P j .Proof. Similar to the proof of Lemma 3.2.Lemma 3.7 implies that every vertex of layer i has neighbors in all layers i + 1 , . . . , l . The next lemma is clear. Lemma 3.8.
For every integers l ≥ and k ≥ , there exists an ( l, k ) -ehf-layered-wheel. We need some properties of lengths of some paths in ehf-layered-wheel. Itis convenient to name specific subpaths of boxes first (see Figure 5). • Suppose that u is a vertex in P l − (of any type).A subpath of Box u is a shared part of Box u if it is either the zone E u (cid:48) ,u or the zone E u,u (cid:48)(cid:48) .The private part of Box u is the path from the rightmost vertex of E u (cid:48) ,u to the leftmost vertex of E u,u (cid:48)(cid:48) . 14 u (cid:48) E u (cid:48) ,u O u E u,u (cid:48)(cid:48) u (cid:48)(cid:48) P i P i +1 uu (cid:48) u (cid:48)(cid:48) P i P i +1 P j E u (cid:48) ,u E u,u (cid:48)(cid:48) O u O u O u,v v E u (cid:48) ,u E u,u (cid:48)(cid:48) O u O u,v O w O u,w O v O v,w u (cid:48) u u (cid:48)(cid:48) v w P i +1 P i P j P j (cid:48) Figure 5: The neighborhood of a type 0, type 1, or type 2 vertex u ∈ V ( P i ) in G l,k (dashed lines represent paths) 15bserve that Box u is edgewise partitioned into a private part, and someshared parts (namely zero if l = 1 and u is the unique vertex of layer P ,one if l > u is an end of P l − , two otherwise). • Suppose that u is of type 1 and v is its ancestor.If u is not the left end of P l − , then the left escape of v in Box u is thesubpath of Box u from the rightmost vertex of E u (cid:48) ,u to the leftmost vertexof O u,v .If u is not the right end of P l − , then the right escape of v in Box u is thesubpath of Box u from the rightmost vertex of O u,v to the leftmost vertexof E u,u (cid:48)(cid:48) . • Suppose that u is of type 2 and v, w are its ancestors as in Construction 3.6.If u is not the left end of P l − , then the left escape of v in Box u is thesubpath of Box u from the rightmost vertex of E u (cid:48) ,u to the leftmost vertexof O u,v and the left escape of w in Box u is the subpath of Box u from therightmost vertex of E u (cid:48) ,u to the leftmost vertex of O w .If u is not the right end of P l − , then the right escape of v in Box u is thesubpath of Box u from the rightmost vertex of O v,w to the leftmost vertexof E u,u (cid:48)(cid:48) and the right escape of w in Box u is the subpath of Box u fromthe rightmost vertex of O v,w to the leftmost vertex of E u,u (cid:48)(cid:48) . Lemma 3.9.
Suppose G l,k is an ehf-layered-wheel with l ≥ and u is a vertexin P l − . Then: • Shared parts of
Box u are paths of odd length. • The private part of
Box u is a path of even length. • If u has type 1 or 2, then all the left and right escapes of its ancestors in Box u are paths of even length.Proof. To check the lemma, it is convenient to follow the path
Box u on Figure 5from left to right.By (B7), shared parts of Box u have obviously odd length.If u has type 0, then along the private part of Box u , one meets 1 commonneighbor of u and u (cid:48) , then 3 private neighbors of u , and then 1 common neighborof u and u (cid:48)(cid:48) . In total, from the left most neighbor of u to its rightmost neighbor,one goes through 4 subpaths of Box u , each of odd length by (B7) (2 of the pathsare in zones, while 2 of them are between zones). The private part of Box u hastherefore even length.If u has type 1, then the proof is similar: one visits 10 subpaths (6 in zones,4 between zones), each of odd length by (B7).If u has type 2, there are more details to check. One visits 19 subpaths.Among them, 12 are in zones and have odd length by (B7). But 3 of thesubpaths between zones have even length by (B7), namely, the paths linking O u,v to O w (because { u, v } ∩ { w } = ∅ ), O u,w to O v and O v,w to E u,u (cid:48)(cid:48) . The 416emaining subpaths between zones have odd length by (B7). In total, the path Box u has even length as claimed.For the left and right escapes, the proof is similar. Note that if u is of type 2,then the rights escapes of v and w goes through the path linking O v,w to E u,u (cid:48)(cid:48) that has even length, and the left escape of w goes through the path linking O w to O u,v that has even length. Theorem 3.10.
For every integer l ≥ , k ≥ , every ( l, k ) -ehf-layered-wheel G l,k is (even-hole, K )-free and every hole in G l,k has length at least k .Proof. It is clear from the construction that G l,k does not contain K . Moreover,it follows from (B7) that apart from triangle, any cycle in G l,k is of length atleast k (we omit the formal proof that is similar to the proof that ttf-layered-wheels have girth at least k ).For a contradiction, consider an ehf-layered-wheel G l,k that contains an evenhole H . Suppose that l is minimal, and under this assumption that H hasminimum length. Hence, layer P l contains some vertex of H , for otherwise G l,k [ P ∪ · · · ∪ P l − ] would be a counterexample.Let P = s . . . t be a subpath of H in P l such that P is inclusion-wise maximal.So both s and t have an ancestor. If P contains a single vertex (i.e., s = t ),then s must have two ancestors, say, s and s , which are adjacent by (B3) ofConstruction 3.6. Thus { s, s , s } forms a triangle in H , which is not possible.So P contains at least two vertices and s (cid:54) = t . Let u and v be ancestors of s and t respectively, such that u, v ∈ V ( H ) (possibly u = v , or uv ∈ E ( G )).Recall that all layers are viewed as oriented from left to right. We supposethat s and t appear in this order, from left to right, along P l . Claim 1.
For every vertex p ∈ V ( P l − ) , N ( p ) ∩ V ( P l ) (cid:54)⊆ V ( P ) .Proof of Claim 1. Suppose p ∈ V ( P l − ) and N ( p ) ∩ V ( P l ) ⊆ V ( P ). By (B5),ancestors of p (if any) and the neighbors of p in P l − \ H must also have neighborsin P . Thus, all of such vertices do not belong to H because P is a subpath of H .By Lemma 3.9, the path Box p = p (cid:48) . . . p (cid:48)(cid:48) has an even length (regardless of thetype of p , it is made of two shared parts and one private part). It yields that Box p and p (cid:48) pp (cid:48)(cid:48) have the same parity, and hence replacing Box (cid:48)(cid:48) p in H with p (cid:48) pp (cid:48)(cid:48) yields an even hole with length strictly less than the length of H , a contradictionto the minimality of H . This proves Claim 1. Claim 2.
Exactly one of u and v is in P l − .Proof of Claim 2. Suppose that both u and v are not in P l − . Since u and v have neighbors in P then each of them has a neighbor in P l − , where suchneighbors also have some neighbor in P . Let u (cid:48) and v (cid:48) be such neighbors of u and v in P l − respectively. Note that by construction, the interior of u (cid:48) P l − v (cid:48) must contain a vertex w of type 0. It yields that N P l ( w ) is all contained in P ,a contradiction to Claim 1. 17uppose now that both u and v are in P l − . By Claim 1, no vertex of P l − has all its neighbors in P . So the interior of uP l − v contains at most twovertices.If u = v , then by (B7) P is of odd length, V ( H ) = { u } ∪ V ( H ) and H hasodd length, a contradiction.If uv ∈ E ( G ), then by (B7), P is of even length, V ( H ) = { u, v } ∪ V ( H ) and H has odd length, a contradiction.If the interior of uP l − v contains a single vertex, then let w be this vertex.Let w (resp. w ) be the neighbor of w (resp. w (cid:48) ) in P that is closest to s (resp. t ). Note that by (B5), s = w , t = w because both u and v are adjacent to w in P l − . So, sP t is the private part of Box w , and by Lemma 3.9, it has evenlength, as uwv . Moreover, if w has an ancestor, then such an ancestor musthave neighbors in P , and hence it does not belong to H . So, replacing uw P w u in P with uwv returns an even hole with length strictly less than the length of H , a contradiction to the minimality of H .So the interior of uP l − v contains two vertices. We let uP l − v = uww (cid:48) v , and w (resp. w (cid:48) ) be the neighbor of w (resp. w (cid:48) ) in P that is closest to s (resp. t ).By (B5), s = w , t = w (cid:48) . So, sP t is edgewise partioned into the private part of w , the part shared between w and w (cid:48) , and the private part of w (cid:48) . By Lemma 3.9, sP t has therefore odd length. In particular, the length of usP tv has the sameparity as the length of uww (cid:48) v . Moreover, if w or w (cid:48) has an ancestor, then suchan ancestor must have neighbors in P , and hence it does not belong to H . So,replacing uw P w (cid:48) u in P with uww (cid:48) v returns an even hole that is shorter than H , again a contradiction to the minimality of H . This proves Claim 2.If u / ∈ V ( P l − ), then by construction u has a neighbor u (cid:48) in P l − such that s ∈ Box u (cid:48) . Similarly, if v / ∈ V ( P l − ), then v has a neighbor v (cid:48) in P l − such that t ∈ Box v (cid:48) . By Claim 2, exactly one of u (cid:48) and v (cid:48) exists and we may break intothe following cases. Case 1: u / ∈ V ( P l − ) and v ∈ V ( P l − ).If the path u (cid:48) P l − v has length at least 3, then some vertex w ∈ V ( u (cid:48) P l − v )contradicts Claim 1.If u (cid:48) P l − v has length 2, so u (cid:48) P l − v = u (cid:48) wv for some vertex w ∈ V ( P l − ),then w is of type 0 because u (cid:48) is not of type 0. Hence, P is edgewise partitionedinto the right escape of u in Box u (cid:48) , the part of Box u (cid:48) shared between u (cid:48) and w , and the private part of w . In total, by Lemma 3.9, it has odd length. Hence,replacing usP tv with uu (cid:48) wv in H yields an even hole whose length is strictlyless than the length of H , a contradiction to the minimality of H .If u (cid:48) P l − v has length 1, so u (cid:48) P l − v = u (cid:48) v , then P is the right escape of v in Box u (cid:48) . Hence by Lemma 3.9, it has an even length. So replacing usP tv with uu (cid:48) v in H returns an even hole with length strictly less than the length of H , acontradiction to the minimality of H . Case 2: u ∈ V ( P l − ) and v / ∈ V ( P l − ).The proof is entirely similar, except that left escapes are used instead ofright escapes. 18 reewidth and cliquewidth For any l ≥
0, ttf-layered-wheels and ehf-layered-wheels on l + 1 layers contains K l +1 as a minor. To see this, note that each vertex in layer P i , i < l , hasneighbors in all layers i + 1 , . . . , l (see Lemma 3.2 and Lemma 3.7). Hence, bycontracting each layer into a single vertex, a complete graph on l + 1 verticesis obtained. Since when H be a minor of G we have tw( H ) ≤ tw( G ) and sincefor l ≥
1, a complete graph on l vertices has treewidth l −
1, we obtain thefollowing.
Theorem 3.11.
For any l ≥ , ttf-layered-wheels and ehf-layered-wheels on l + 1 layers have treewidth at least l . Gurski and Wanke [12] proved that the treewidth is in some sense equivalentto the cliquewidth when some complete bipartite graph is excluded as a sub-graph. Let us state and apply this formally (thanks to Sang-Il Oum for pointingthis out to us).
Theorem 3.12 (Gurski and Wanke [12]) . If a graph G contains no K , as asubgraph, then tw( G ) ≤ G ) − . Lemma 3.13.
A layered wheel (ttf or ehf ) contains no K , as a subgraph.Proof. Suppose a ttf-layered wheel G contains K , as a subgraph. Then, eitherit contains a theta (if K , is an induced subgraph of G ) or it contains a triangle(if K , is not an induced subgraph of G ). In both cases, there is contradiction.Suppose an ehf-layered wheel G contains K , as a subgraph. If one side ofthe K , is a clique, then G contains a K . Otherwise, each side of K , containsa non-edge, so G contains K , , that is isomorphic to a C . In both cases, thereis contradiction. Theorem 3.14.
For any integers l ≥ , k ≥ , the cliquewidth of a layeredwheel G l,k is at least l/ .Proof. Follows from Lemma 3.13 and Theorems 3.12 and 3.11.
Observations and open questions
It should be pointed out that by carefully subdividing, one may obtain bipartitettf-layered-wheels on any number l of layers. This is easy to prove by inductionon l . We just sketch the main step of the proof: when building the last layer,assuming that the previous layers induce a bipartite graph, only the verticeswith ancestors are assigned to one side of the bipartition (and only to one side,since a vertex has at most one ancestor in a ttf-layered-wheels). The parity ofthe paths linking vertices with ancestors can be adjusted to produce a bipartitegraph. 19 j P i P l − P l w p p (cid:48) v u u (cid:48)(cid:48) u ∗ u (cid:48) Figure 6: A pyramid in an ehf-layered-wheel G l,k It is easy to see that every prism, every theta, and every even wheel containsan even hole. Therefore, by Theorem 3.10, ehf-layered-wheels are (prism, theta,even wheel)-free, which is not obvious from their definitions.However, for any l ≥
3, an ehf-layered-wheel contains a pyramid. In Figure 6, u ∈ P l − is a type 2 vertex with ancestors v ∈ P i and w ∈ P j , j ≤ i ; u ∗ is acommon neighbor of v and w in P l − such that u and u ∗ are consecutive in azone labeled O v,w ; p is the rightmost vertex of a zone labelled O v,w ⊆ Box u in P l ; and p (cid:48) is the leftmost vertex of a zone labelled O u,u (cid:48)(cid:48) ⊆ Box u in P l with u (cid:48)(cid:48) is adjacent to u in P l − . The pyramid is made of triangle up (cid:48) u (cid:48)(cid:48) and the apex v . This motivates Conjecture 1.3. In this section, we prove that there exist ttf-layered-wheels and ehf-layered-wheels with arbitrarily large rankwidth. This follows directly from Theorem 3.14and Lemma 1.5, but by a direct computation, we provide a better bound. Letus first present some useful notion and definition about rankwidth.For a set X , let 2 X denote the set of all subsets of X . For sets R and C ,an ( R, C )-matrix is a matrix where the rows are indexed by elements in R andcolumns indexed by elements in C . For an ( R, C )-matrix M , if X (cid:40) R and Y (cid:40) C , we let M [ X, Y ] be the submatrix of M where the rows and the columnsare indexed by X and Y respectively. For a graph G = ( V, E ), let A G denotethe adjacency matrix of G over the binary field (i.e., A G is the ( V, V )-matrix,where an entry is 1 if the column-vertex is adjacent to the row-vertex, and 0otherwise). The cutrank function of G is the function cutrk G : 2 V → N , givenby cutrk G ( X ) = rank( A G [ X, V \ X ]) , where the rank is taken over the binary field.A tree is a connected, acyclic graph. A leaf of a tree is a node incident toexactly one edge. For a tree T , we let L ( T ) denote the set of all leaves of T . Atree node that is not a leaf is called internal . A tree is cubic , if it has at leasttwo vertices and every internal node has degree 3.20 rank decomposition of a graph G is a pair ( T, λ ), where T is a cubictree and λ : V ( G ) → L ( T ) is a bijection. If | V ( G ) | ≤
1, then G has norank decomposition. For every edge e ∈ E ( T ), the connected components of T \ e induce a partition ( A e , B e ) of L ( T ). The width of an edge e is defined ascutrk G ( λ − ( A e )). The width of ( T, λ ), denoted by width(
T, λ ), is the maximumwidth over all edges of T . The rankwidth of G , denoted by rw( G ), is the min-imum integer k , such that there is a rank decomposition of G of width k . (If | V ( G ) | ≤
1, we let rw( G ) = 0). The following lemma is well-known. Lemma 4.1.
Let G be a graph and H be an induced subgraph of G . Then rw( H ) ≤ rw( G ) . A class C of graphs has bounded rankwidth if there exists a constant k ∈ N ,such that every G ∈ C satisfies rw( G ) ≤ k . If such a constant does not exist,then C has unbounded rankwidth . In the following lemmas, we present somebasic properties related to rankwidth. Let T be a tree, we call an edge e ∈ E ( T ) balanced , if the partition ( A e , B e ) of L ( T ) satisfies | L ( T ) | ≤ | A e | and | L ( T ) | ≤ | B e | . The following is well known (we include a proof for the sake ofcompleteness). Lemma 4.2.
Every cubic tree has a balanced edge.Proof.
Let T be a cubic tree with n leaves. We may assume that n ≥
3, forotherwise, T is a path of length 1, and the unique edge of T is balanced.Let e = ab be an edge of T such that the component A e of T \ e thatcontains a satisfies | A e | ≥ | L ( T ) | /
3. Suppose that a and b are chosen subjectto the minimality of | A e | . If | A e | ≤ | L ( T ) | /
3, then e is balanced. Otherwise, | A e | > | L ( T ) | / ≥ a has two neighbors a (cid:48) , a (cid:48)(cid:48) different from b . Let A (cid:48) (resp. A (cid:48)(cid:48) ) be the set of leaves of T \ aa (cid:48) (resp. T \ aa (cid:48)(cid:48) ) that contains a (cid:48) (resp. a (cid:48)(cid:48) ). Since | A e | ≥ | L ( T ) | / A e = A (cid:48) ∪ A (cid:48)(cid:48) , either | A (cid:48) | > | L ( T ) | / | A (cid:48)(cid:48) | > | L ( T ) | /
3. Hence, one of A (cid:48) or A (cid:48)(cid:48) contradicts the minimality of | A e | .An n × n matrix M is fuzzy triangular if for every i ∈ { , · · · , n } , m i,i = 1and either m ,i = m ,i = · · · = m i − ,i = 0 or m i, = m i, = · · · = m i,i − = 0. Lemma 4.3.
Every n × n fuzzy triangular matrix has rank n .Proof. Let M be an n × n fuzzy triangular matrix. We prove by induction on n , that rank( M ) = n . For n = 1, this trivially holds. Suppose n ≥
2. If m ,n = m ,n = · · · = m n − ,n = 0, we show that rows r , · · · , r n of M arelinearly independent. Let λ , · · · , λ n ∈ { , } be such that Σ ni =1 λ i r i = 0. Since m n,n = 1, we have λ n = 0. This implies Σ n − i =1 λ i r (cid:48) i = 0, where r (cid:48) i is the rowobtained from r i by deleting its last entry. Since r (cid:48) , · · · , r (cid:48) n − are the rows ofan ( n − × ( n −
1) fuzzy triangular matrix, they are linearly independent bythe induction hypothesis, so λ = · · · = λ n − = 0.We can prove in the same way that, if m n, = m n, = · · · = m n,n − = 0,then the set of n columns of M are linearly independent. This shows thatrank( M ) = n . 21et G be a graph and ( X, Y ) be a partition of V ( G ). A path P in G is separated by ( X, Y ) if V ( P ) ∩ X and V ( P ) ∩ Y are both non-empty. Note thatin this case, there is an edge xy of P where x ∈ X and y ∈ Y . Lemma 4.4.
Let ( T, λ ) be a rank decomposition of width at most r of a layeredwheel with layers P , . . . , P l . Let e ∈ E ( T ) be an edge of T . Let ( X, Y ) be thepartition of V ( G ) induced by T \ e . There are at most r paths among { P , . . . , P l } that are separated by ( X, Y ) .Proof. Suppose for a contradiction that P i , . . . , P i r +1 are layers that are allseparated by ( X, Y ), where i < · · · < i r +1 . For each integer i j , consider anedge x i j y i j of P i j such that x i j ∈ X and y i j ∈ Y . Set S X = { x i , . . . , x i r +1 } and S Y = { y i , . . . , y i r +1 } .Consider M [ S X , S Y ], the adjacency matrix whose rows are indexed by S X and whose columns are indexed by S Y . The definition of layered wheels impliesthat when two vertices of some layer are adjacent, so at most one of themhas ancestors. It follows that M [ S X , S Y ] is fuzzy triangular. By Lemma 4.3, M [ S X , S Y ] has rank r + 1, a contradiction, becausewidth( T, λ ) ≥ cutrk G ( X ) = rank( M [ X, Y ]) ≥ r + 1 . Lemma 4.5 (See [1]) . Let G be a graph and ( T, λ ) be a rank decomposition of G of width at most r . Let P be an induced path of G and ( X, Y ) be the partitionof V ( G ) induced by T \ e where e ∈ E ( T ) . Then each of P [ X ] and P [ Y ] containsat most r + 1 connected components. Now we are ready to describe layered-wheels for which we can prove thatthe rankwidth is unbounded. Let us first define some useful terminology. RecallConstruction 3.1 of a ttf-layered-wheel G l,k . Let u and v be two vertices thatare adjacent in a layer P i for some i ∈ { , · · · , l − } , and that appear in thisorder (from left to right). Let a be the rightmost vertex of Box u and b be theleftmost vertex of Box v in P i +1 . Let a (cid:48) (resp. b (cid:48) ) be the neighbor of a (resp. b )in P i +1 \ Box u (resp. P i +1 \ Box v ). The path a (cid:48) P i +1 b (cid:48) is called the uv -bridge .An edge pq in a (cid:48) P i +1 b (cid:48) is called the middle edge of the bridge if the length ofthe paths a (cid:48) P i +1 p and qP i +1 b (cid:48) are equal.We have a similar definition for ehf-layered-wheel. For adjacent vertices u and v in P i +1 , the uv -bridge in P i +1 is the zone labelled E u,v ⊆ Box u ∩ Box v (that we called in the previous section a shared part). Note that in both layeredwheels, every internal vertex of some layer yields two bridges. Each end of alayers yields one bridge. We say a layered-wheel is special if every bridge inevery layer has odd length (and therefore admits a middle edge). The followinglemmas are a direct consequence of Construction 3.1 and Construction 3.6. Lemma 4.6.
For every integers l ≥ , k ≥ , there exists a special ttf-layered-wheel. roof. The result follows because by (A6) of Construction 3.1, the uv -bridgecan have any length (at least k − Lemma 4.7.
For every integers l ≥ , k ≥ , any ehf-layered-wheel is special.Proof. This follows from the fact that shared parts have odd length (see Lemma 3.9).Let G l,k be a special layered wheel. Let uv be an edge of some layer P i ,1 ≤ i < l , and suppose that u and v appear in this order from left to right.Then we denote by r u l v the middle edge of the uv -bridge (again, r u and l v appear in this order from left to right).For any vertex v ∈ P i , 1 ≤ i < l , the domain of v (or the v -domain ), denotedby Dom ( v ) is defined as follows: • if v ∈ V ( P ) then Dom v = V ( P ); • if v is an internal vertex of P i , then Dom ( v ) = V ( l v P i +1 r v ); • if v is the left end of P i then Dom ( v ) = V ( pP i +1 r v ), where p is the leftmostvertex of Box v ; and • if v is the right end of P i then Dom ( v ) = V ( l v P i +1 q ), where q is therightmost vertex of Box v .Note that for ttf-layered-wheels, Box v is completely contained in the v -domain, which is not the case for ehf-layered-wheels. We are now ready todescribe the layered-wheels that we need. Definition 4.8.
For some integer m , a special layered wheel G l,k is m -uniform ,if for every vertex v ∈ V ( P i ) , ≤ i ≤ l − , Dom ( v ) contains exactly m vertices. Observe that any m -uniform layered wheel is special. Lemma 4.9.
For every integers l ≥ , k ≥ and M , there exists an integer m ≥ M and a ttf-layered-wheel that is m -uniform.Proof. We construct an m -uniform ttf-layered-wheel G l,k by adjusting the lengthobtained in step (A6) of Construction 3.1. Lemma 4.10.
For every integers l ≥ , k ≥ and M , there exists an integer m ≥ M and an ehf-layered-wheel that is m -uniform.Proof. We construct an m -uniform ehf-layered-wheel G l,k by adjusting the lengthobtained in step (B7) of Construction 3.6.For an integer d ≥
1, and a vertex v ∈ P i , the v -domain of depth d , denotedby Dom d ( v ) is defined as follows. • Dom ( v ) = { v } ; • Dom d ( v ) = (cid:83) x ∈ Dom ( v ) Dom d − ( x ) for d ≥ bservation 4.11. For any v ∈ P i with ≤ i ≤ l − , and ≤ d ≤ l − i , wehave Dom d ( v ) ⊆ V ( P i + d ) , where the equality holds when i = 0 . Lemma 4.12.
For every ≤ d ≤ l , d ≤ i ≤ l , V ( P i ) = (cid:83) v ∈ P i − d +1 Dom d ( v ) .Moreover, for any distinct u, v ∈ V ( P i ) , Dom d ( u ) ∩ Dom d ( v ) = ∅ .Proof. By induction on d . Lemma 4.13.
Let for some integers l, k, m , G l,k be an m -uniform layered wheel.For every ≤ i ≤ l − , v ∈ P i , and ≤ d ≤ l − i , we have | Dom d ( v ) | = m d .Proof. Follows from Lemma 4.12 and m -uniformity: for any vertex v , | Dom ( v ) | = m and | Dom d ( v ) | = m · | Dom d − ( v ) | . Lemma 4.14.
Let for some integers l, k, m , G l,k be an m -uniform layered wheel.Denote by G i,k , the subgraph induced by the first i + 1 layers P , · · · , P i . Then | V ( G i,k ) | < m − | V ( P i +1 ) | for ≤ i ≤ l − .Proof. Recall that
Dom i ( r ) = V ( P i ) for every 1 ≤ i ≤ l , with r ∈ V ( P ). So byLemma 4.13, | V ( P i ) | = m i . Moreover, | V ( G i,k ) | = Σ id =0 | Dom d ( r ) | = m i +1 − m − . The result directly follows.
Lemma 4.15.
Let l ≥ , k ≥ , and m ≥ be integers, and ( T, λ ) be a rankdecomposition of an m -uniform layered wheel G l,k of width at most r . Let e bea balanced edge in T , and ( X, Y ) be the partition of V ( G l,k ) induced by e . Eachof X and Y contains an induced subpath of P l , namely P X and P Y where: | V ( P X ) | , | V ( P Y ) | ≥ (cid:22) | V ( P l ) | . r + 1) (cid:23) . Proof.
We will only prove the existence of P X (for P Y , the proof is similar).Since e is a balanced edge of T , we have | X | ≥ | V ( G l,k ) | . Clearly, | V ( P l ) ∩ X | ≥ | V ( G l,k ) |−| V ( G l − ,k ) | = ( | V ( P l ) − | V ( G l − ,k ) | ). By Lemma 4.5, X containsat most r + 1 connected components of P l . Hence: | V ( P X ) | ≥ | V ( P l ) ∩ X | r + 1 ≥ | V ( P l ) | − m − | V ( P l ) | r + 1) (1)= m − m − r + 1) × | V ( P l ) |≥ r + 1) × | V ( P l ) | (2)(1) is obtained from Lemma 4.14, and (2) follows because m ≥ Theorem 4.16.
For l ≥ , k ≥ , there exists an integer m such that therankwidth of an m -uniform layered wheel ( G l,k ) is at least l .Proof. Set M = 15 and consider an integer m as in Lemma 4.9 (or Lemma 4.10).Suppose for a contradiction, that rw( G l,k ) = r for some integer r ≤ l − T, λ ) be a rank decomposition of G l,k of width r , and e be a balanced edgeof T that partition V ( G l,k ) into ( X, Y ). Let P = { P , . . . , P l } , and S be the setof paths in P that are separated in ( X, Y ). By Lemma 4.4, |S| ≤ r .Let P j ∈ P \ S , i.e., the vertices of P j are completely contained either in X or Y . W.l.o.g., we let V ( P j ) ⊆ X . Moreover, note that j ≥ l + 1 − r (because |S| ≤ r ). By Lemma 4.15, there exists a subpath P Y of P l , such that V ( P Y ) ⊆ Y and | V ( P Y ) | ≥ (cid:106) | V ( P l ) | . r +1) (cid:107) .Let P (cid:48) be the subpath of P j containing vertices that have neighbors in P Y .By Lemma 4.12, for every v, v (cid:48) ∈ P j , Dom l − j +1 ( v ) ∩ Dom l − j +1 ( v (cid:48) ) = ∅ . So foreach vertex v ∈ P j , we can fix a vertex y v ∈ V ( Q Y ), such that y v (cid:54) = y v (cid:48) for any v (cid:54) = v (cid:48) . Let us denote S X = V ( P (cid:48) ) and S Y = { y v | v ∈ S X } . Observe thatthere is a bijection between X (cid:48) and Y (cid:48) , so M [ S X , S Y ] is an identity matrix ofsize | S X | .Moreover, we have | S X | ≥ (cid:106) | V ( P Y ) | Dom l − j ( v ) (cid:107) ≥ (cid:106) | V ( P Y ) | m l − j (cid:107) . By Lemma 4.13 and thefact that j ≥ l + 1 − r , r ≤ l −
1, the following holds. | S X | ≥ (cid:22) m l . r + 1) m l − j (cid:23) ≥ (cid:22) m . l − (cid:23) ≥ (cid:22) l . l − (cid:23) ≥ l which yields a contradiction, because r ≥ width( T, λ ) ≥ cutrk G l,k ( X ) = rank( M [ X, Y ]) ≥ rank( M [ S X , S Y ]) . Layered wheels have an exponential number of vertices in terms of the number oflayers l . In Section 3, we have seen that the treewidth of layered wheels is lower-bounded by l . In this section, we give an upper bound of the treewidth of layeredwheels. As mentioned in the introduction, we indeed prove a stronger result: theso-called pathwidth of layered wheels is upper-bounded by some linear functionof l . Since layered wheels G l,k contains an exponential number of vertices interms of the number of layers, this implies that tw( G l,k ) = Θ (log | V ( G l,k ) | ).Beforehand, let us state some useful notions. Pathwidth
The pathwidth of a graph G is denoted by pw( G ). We do not need the formaldefinition (recall that a path-decomposition of a graph G is defined similarly as25 tree-decomposition except that the underlying tree is required to be a path,the width of the path-decomposition is the size of a largest bag minus 1, andthe pathwidth is the minimum width of a path-decomposition of G ).To prove our results we only need two classical lemmas. Lemma 5.1.
For any graph G , tw( G ) ≤ pw( G ) . Let P be a path, and P , . . . , P k be subpaths of P . The interval graph assocciated to P , . . . , P k is the graph whose vertex set is { P , . . . , P k } with anedge between any pair of paths sharing at least one vertex. So, interval graphsare intersection graphs of subpaths of some path. Lemma 5.2.
Let G be a graph, and I be an interval graph that contains G asa subgraph (possibly not induced). Then pw( G ) ≤ ω ( I ) − , where ω ( I ) is thesize of the maximum clique of I . Now, for every layered wheel G l,k , we describe an interval graph I ( G l,k )such that G l,k is a subgraph of I ( G l,k ). We define the scope of a vertex. This issimilar to its domain, but slightly different (the main difference is that scopesmay overlap while domains do not). For v ∈ V ( P i ), for 0 ≤ i ≤ l −
1, the scopeof v , denoted by Scope ( v ), is defined as follows.For a ttf-layered-wheel: • if v ∈ P , Scope v = V ( P ); • if v is in the interior of P i , then Scope ( v ) = V ( L ) ∪ Box v ∪ V ( R ), where u and w are the left and the right neighbors of v in P i respectively, L isthe uv -bridge and R is the vw -bridge; • if v is the left end of P i then Scope ( v ) = Box v ∪ V ( R ) where w is theright neighbor of v in P i and R is the vw -bridge; • if v is the right end of P i then Scope ( v ) = V ( L ) ∪ Box v , where u is theleft neighbor of v in P i and L is the uv -bridge.For ehf-layered-wheel: • Scope ( v ) = Box v for every v ∈ P i , 0 ≤ i ≤ l − d scope of a vertex v ∈ P i , 0 ≤ i ≤ l −
1, denotedby
Scope d ( v ). We define Scope ( v ) = { v } , and Scope d ( v ) = (cid:91) x ∈ Scope ( v ) Scope d − ( x ) for 2 ≤ d ≤ l − i. For a layered wheel G l,k , we define the interval graph I ( G l,k ). For everyvertex v ∈ G l,k , define path P ( v ) associated to v as follows: • if v ∈ P l is not the right end of P l , then P ( v ) = vw where w is the rightneighbor of v ; 26 if v is the right end of P l then P ( v ) = { v }• if v ∈ P i with i < l , then P ( v ) = P l (cid:104) Scope l − i ( v ) (cid:105) .Note that P ( v ) is a subpath of P l . The graph I ( G l,k ) is the interval graphassociated to { P ( v ) | v ∈ V ( G l,k ) } . Lemma 5.3.
For any layered wheel G l,k and corresponding interval graph I ( G l,k ) , G l,k is a subgraph (possibly not induced) of I ( G l,k ) .Proof. It is clear by definition that there is a bijection between V ( I ( G l,k )) and V ( G l,k ). We show that E ( G l,k ) ⊆ E ( I ( G l,k )): for any two vertices u, v ∈ G l,k ,if uv ∈ E ( G l,k ) then the corresponding paths P ( u ) and P ( v ) share at least onevertex (i.e. V ( P ( u )) ∩ V ( P ( v )) (cid:54) = ∅ ).For u, v ∈ P l , this property trivially holds, because by definition, P ( u ) and P ( v ) both contain v . Moreover, if w.l.o.g., u ∈ V ( P i ) with i < l , and v ∈ P l ,then trivially V ( P ( v )) ⊆ V ( P ( u )) = Scope l − i ( u ).If u, v ∈ V ( P i ), i < l then by definition, Scope ( u ) ∩ Scope ( v ) (cid:54) = ∅ . Let x ∈ Scope ( u ) ∩ Scope ( v ). Note that for 2 ≤ d ≤ l − i , Scope d ( u ) and Scope d ( v )both contain Scope d − ( x ). Similarly if w.l.o.g., u ∈ P i and v ∈ P j where1 ≤ i < j < l then Scope ( v ) ⊆ Scope ( u ) and so Scope d ( v ) ⊆ Scope d ( u ) forevery 2 ≤ d ≤ l − i . Hence, V ( P ( u )) ∩ V ( P ( v )) (cid:54) = ∅ . Theorem 5.4.
For any layered wheel G l,k , tw( G l,k ) ≤ pw( G l,k ) ≤ l .Proof. By Lemmas 5.1, 5.2 and 5.3, it is enough to show that ω ( I ( G l,k )) ≤ l +1. Claim 1.
For every non adjacent vertices u and v in P i , ≤ i ≤ l − , we have Scope d ( u ) ∩ Scope d ( v ) = ∅ for every ≤ d ≤ l − i .Proof of Claim 1. Note that for every non adjacent vertices u and v in P i , 1 ≤ i ≤ l − Scope ( u ) ∩ Scope ( v ) = ∅ . Suppose by induction that Scope d ( u ) ∩ Scope d ( v ) = ∅ for any uv / ∈ E ( G l,k ) and some 1 ≤ d ≤ l − i −
1. We will showthat
Scope d +1 ( u ) ∩ Scope d +1 ( v ) = ∅ . Let u (cid:48) be the right end of Scope ( u )and v (cid:48) be the left end of Scope ( v ). Note that u (cid:48) P i +1 v (cid:48) is of length at least 2.By the induction hypothesis, Scope d ( u (cid:48) ) ∩ Scope d ( v (cid:48) ) = ∅ . So by construction, Scope d ( l ) ∩ Scope d ( r ) = ∅ for every vertex l ∈ V ( L ), and r ∈ V ( R ), where L = aP i +1 l and R = rP i +1 b with a and b are the left end and the right endof P i +1 respectively. Hence, Scope d +1 ( u ) ∩ Scope d +1 ( v ) = ∅ . This provesClaim 1.Let K be a maximum clique in I ( G l,k ). By Claim 1, if u, v ∈ V ( P i ) arenon-adjacent, then V ( P ( u )) ∩ V ( P ( v )) = ∅ . So K contains at most two verticesof layer P i , for every 1 ≤ i ≤ l . Since K may also contain the unique vertex in P , then ω ( I ( G l,k )) ≤ l + 1 as desired.27 Acknowledgement
Thanks to ´Edouard Bonnet, Zdenˇek Dvoˇr´ak, Serguei Norine, Marcin Pilipczuk,Sang-Il Oum, Natacha Portier, St´ephan Thomass´e, Kristina Vuˇskovi´c, and R´emiWatrigant for useful discussions.
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