(Theta, triangle)-free and (even hole, K 4 )-free graphs. Part 2 : bounds on treewidth
Marcin Pilipczuk, Ni Luh Dewi Sintiari, Stéphan Thomassé, Nicolas Trotignon
((Theta, triangle)-free and (even hole, K )-freegraphs. Part 2: Bounds on treewidth Marcin Pilipczuk ∗ , Ni Luh Dewi Sintiari † ,St´ephan Thomass´e † and Nicolas Trotignon † October 28, 2020
Abstract A theta is a graph made of three internally vertex-disjoint chordlesspaths P = a . . . b , P = a . . . b , P = a . . . b of length at least 2 and suchthat no edges exist between the paths except the three edges incident to a and the three edges incident to b . A pyramid is a graph made of threechordless paths P = a . . . b , P = a . . . b , P = a . . . b of length atleast 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 exist between the pathsexcept those of the triangle and the three edges incident to a . An evenhole is a chordless cycle of even length. For three non-negative integers i ≤ j ≤ k , let S i,j,k be the tree with a vertex v , from which start threepaths with i , j , and k edges respectively. We denote by K t the completegraph on t vertices.We prove that for all non-negative integers i, j, k , the class of graphsthat contain no theta, no K , and no S i,j,k as induced subgraphs havebounded treewidth. We prove that for all non-negative integers i, j, k, t ,the class of graphs that contain no even hole, no pyramid, no K t , andno S i,j,k as induced subgraphs have bounded treewidth. To bound thetreewidth, we prove that every graph of large treewidth must contain alarge clique or a minimal separator of large cardinality. In this article, all graphs are finite, simple, and undirected. A graph H is an induced subgraph of a graph G if some graph isomorphic to H can be obtained ∗ Institute of Informatics, University of Warsaw Banacha 2, 02-097 Warsaw, PolandThis research is a part of a project that has received funding from the European ResearchCouncil (ERC) under the European Union’s Horizon 2020 research and innovation programmeGrant Agreement no. 714704. † Univ Lyon, EnsL, UCBL, CNRS, LIP, F-69342, LYON Cedex 07, France.The last three authors are partially supported by the LABEX MILYON (ANR-10-LABX-0070)of Universit´e de Lyon, within the program “Investissements d’Avenir” (ANR-11-IDEX-0007)operated by the French National Research Agency (ANR) and by Agence Nationale de laRecherche (France) under research grant ANR DIGRAPHS ANR-19-CE48-0013-01. a r X i v : . [ c s . D M ] O c t igure 1: Pyramid, prism, theta, and wheel (dashed lines represent paths)from G by deleting vertices. A graph G contains H if H is an induced subgraphof G . A graph is H -free if it does not contain H . For a family of graphs H , G is H -free if for every H ∈ H , G is H -free.A hole in a graph is a chordless cycle of length at least 4. It is odd or even according to its length (that is its number of edges). We denote by K t thecomplete graph on t vertices.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 (see Fig. 1). Observe that a theta contains an even hole, becauseat least two paths in the theta have lengths of same parity and therefore inducean even hole.We are interested in understanding the structure of even-hole-free graphsand theta-free graphs. Our motivation for this is explained in the first paper ofthis series [24], where we give a construction that we call layered wheel, showingthat the cliquewith, the rankwidth, and the treewidth of (theta, triangle)-freeand (even hole, K )-free graphs are unbounded. We also indicate questions,suggested by this construction, about the induced subgraphs contained in graphswith large treewidth.In this second and last part, we prove that when excluding more inducedsubgraphs, there is an upper bound on the treewidth. Our results imply thatthe maximum independent set problem can be solved in polynomial time forsome classes of graphs that are possibly of interest because they are related toseveral well known open questions in the field. Results
We denote by P k the path on k vertices. For three non-negative integers i ≤ j ≤ k , let S i,j,k be the tree with a vertex v , from which start three paths with i , j , and k edges respectively. Note that S , ,k is a path of length k (so, isequivalent to P k +1 ) and that S ,i,j = S , ,i + j . The claw is the graph S , , .Note that { S i,j,k ; 1 ≤ i ≤ j ≤ k } is the set of all the subdivided claws and { S i,j,k ; 0 ≤ i ≤ j ≤ k } is the set of all subdivided claws and paths.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 exist2etween the paths except those of the triangle and the three edges incident to a (see Fig. 1).We do not not recall here the definition of treewidth and cliquewidth. Theyare parameters that measure how complex a graph is. See [19, 17] for surveysabout them.Our main result states that for every fixed non-negative integers i, j, k, t , thefollowing graph classes have bounded treewidth: • (theta, triangle, S i,j,k )-free graphs; • (even hole, pyramid, K t , S i,j,k )-free graphs.The exact bounds and the proofs are given in Section 4 (Theorems 4.6and 4.7). In fact, the class on which we actually work is larger. It is a commongeneralization C of the graphs that we have to handle in the proofs for the twobounds above. Also, we do not exclude S i,j,k , but some graphs that contain it,the so-called l -span-wheels for sufficiently large l . We postpone the definitionsof C and of span wheels to Section 4.To bound the treewidth, we prove that every graph of large treewidth mustcontain a large clique or a minimal separator of large cardinality. Let us definethem.For two vertices s, t ∈ V ( G ), a set X ⊆ V ( G ) is an st -separator if s, t / ∈ X and s and t lie in different connected components of G \ X . An st -separator X is a minimal st -separator if it is an inclusion-wise minimal st -separator. A set X ⊆ V ( G ) is a separator if there exist s, t ∈ V ( G ) such that X is an st -separatorin G . A set X ⊆ V ( G ) is a minimal separator if there exist s, t ∈ V ( G ) suchthat X is a minimal st -separator in G .Our graphs have no large cliques by definition, and by studying their struc-ture, we prove that they cannot contain large minimal separators, implying thattheir treewidth is bounded.Note that from the celebrated grid-minor theorem, it is easy to see that everygraph of large treewidth contains a subgraph with a large minimal separator (acolumn in the middle of the grid contains such a separator). But since we areinterested in the induced subgraph containment relation, we cannot delete edgesand we have to rely on our reinforcement. Treewidth and cliquewidth of some classes of graphs
We now survey results about the treewidth in classes of graphs related to thepresent work. Complete graphs provide trivial examples of even-hole-free graphsof arbitrarily large treewidth. In [11], it is proved that (even hole, triangle)-free graphs have bounded treewidth (this is based on a structural descriptionfrom [13]). In [10], it is proved that for every positive integer t , (even hole,pan, K t )-free graphs have bounded treewidth (where a pan is any graph thatconsists of a hole and a single vertex with precisely one neighbor on the hole).It is proved in [24] that the treewidth of (theta, triangle)-free graphs and (evenhole, pyramid, K )-free graphs are unbounded. Growing the treewidth in [24]3igure 2: A subdivision of a wall and its line graphrequires introducing in the graph a large clique minor and vertices of largedegree. It is therefore natural to ask whether these two conditions are reallyneeded, and the answer is yes for both of them, because in [2] it is proved thateven-hole-free graphs with no K t -minor have bounded treewidth, and in [1]it is proved that even-hole-free graphs with maximum degree t have boundedtreewidth.Since having bounded cliquewidth is a weaker property than having boundedtreewidth but still has nice algorithmic applications, we survey some resultsabout the cliquewidth in classes related to the present work.It is proved in [11] that (even hole, cap)-free graphs with no clique separatorhave bounded cliquewidth (where a cap is any graph that consists of a holeand a single vertex with precisely two adjacent neighbors on the hole, anda clique separator is a separator that is a clique). It is proved in [24], that(triangle, theta)-free and (even hole, pyramid, K )-free graphs have unboundedcliquewidth. It is proved in [3], that (even hole, diamond)-free graphs withno clique separator have unbounded cliquewidth (the diamond is the graphobtained from K by deleting an edge). The construction can be easily extendedto (even hole, pyramid, diamond)-free graphs as explained in [12]. It is easy toprovide (theta, K , S , , )-free graphs (or equivalently (claw, K )-free graphs)of unbounded cliquewidth. To do so, consider a wall W , subdivide all edges toobtain W (cid:48) , and take the line graph L ( W (cid:48) ) (see [24] for a definition and Fig. 2).The results mentioned in this paragraph are extracted from [15] (but someof them were first proved in other works). Let H U = { P , S , , , S , , } and H B = { P , S , , } . If H contains a graph from H U as an induced subgraph, thenthe class of (triangle, H )-free graph has unbounded cliquewidth (see Theorem7.ii.6 in [15]). If H is contained in a graph from H B , then the class of (triangle, H )-free graphs has bounded cliquewidth (see Theorem 7.i.3 in [15]).The cliquewidth of (triangle, S , , )-free graphs is bounded, see [7] or [16]. Algorithmic consequences
It is proved in [14] that in every class of graphs of bounded treewidth, manyproblems can be solved in polynomial time. Our result has therefore applicationsto several problems, but we here focus on one because the induced subgraphsthat are excluded in the most classical results and open questions about it seem4o be related to our classes.An independent set in a graph is a set of pairwise non-adjacent vertices.Our results imply that computing an independent set of maximum cardinalitycan be performed in polynomial time for (theta, triangle, S i,j,k )-free graphs and(even hole, pyramid, K t , S i,j,k )-free graphs.Finding an independent set of maximum cardinality is polynomial time solv-able for (even hole, triangle)-free graphs [11] and (even hole, pyramid)-freegraphs [12]. Its complexity is not known for (even hole, K )-free graphs and for(theta, triangle)-free graphs. Determining its complexity is also a well knownquestion for S i,j,k -free graphs. It is NP-hard for the class of H -free graphs when-ever H is not an induced subgraph of some S i,j,k [4]. It is solvable in polynomialtime for H -free graphs whenever H is contained in P k for k = 6 (see [20] for H = P and [18] for H = P ) or contained in S i,j,k with ( i, j, k ) ≤ (1 , , P , triangle)-free graphs [8] and for ( S , , , triangle)-free graphs [9]. The com-plexity is not known for H -free graphs whenever H is some S i,j,k that containseither P , S , , , or S , , . Bounding the number of minimal separators
One possible method to find maximum weight independent sets for a class ofgraphs is by proving that every graph in the class has polynomially many mini-mal separators (where the polynomial is in the number of vertices of the graph).This was for instance successfully applied to (even hole, pyramid)-free graphsin [12]. Therefore, our result on (even hole, pyramid, K t , S i,j,k )-free graphs doesnot settle a new complexity result for the Maximum Independent Set problem(but it still can be applied to other problems).Note that bounding the number of minimal separators cannot be applied to(even hole, K )-free graphs and to (theta, triangle)-free graphs since there existgraphs in both classes that contain exponentially many minimal separators.These graphs are called k -turtle and k -ladder , see Fig 3. It is straightforward tocheck that they have exponentially many minimal separators (the idea is thata separator can be built by making a choice in each horizontal edge, and thereare k of them). Moreover, k -turtles are (theta, triangle)-free (provided that theouter cycle is sufficiently subdivided) and k -ladders are (even hole, K )-free. Open questions
It is not known whether (even hole, K , diamond)-free graphs have boundedtreewidth (or cliquewidth). Also, for every fixed integer t ≥
4, it is notknown whether (theta, triangle)-free graphs of maximum degree t have boundedtreewidth (for t = 1 ,
2, the treewidth is trivially bounded and for t = 3 it followsfrom Corollary 4.3 in [2]). It is not known whether (triangle, S , , )-free graphshave bounded cliquewidth, see [16] for other open problems of the same flavor.5 .. ... Figure 3: k -turtle and k -ladder (dashed lines represent paths) Outline of the paper
In Section 2, we explain our method to bound the treewidth. In Section 3, wegive two technical lemmas that highlight structural similarities between (theta,triangle)-free and (even hole, pyramid)-free graphs. These will be used in Sec-tion 4 where we prove that graphs in our classes do not contain minimal sepa-rators of large cardinality, implying that their treewidth is bounded.
Notation By path we mean chordless (or induced) path. When a and b are vertices of apath P , we denote by aP b the subpath of P with ends a and b .When A, B ⊆ V ( G ), we denote by N B ( A ) the set of vertices from B \ A thathave at least one neighbor in A and N ( A ) means N V ( G ) ( A ). Note that N B ( A ) isdisjoint from A . We write N ( a ) instead of N ( { a } ) and N [ a ] for { a } ∪ N ( a ). Wedenote by G [ A ] the subgraph of G induced by A . To avoid too heavy notation,since there is no risk of confusion, when H is an induced subgraph of G , wewrite N H instead of N V ( H ) .A vertex x is complete (resp. anticomplete ) to A if x / ∈ A and x is adjacentto all vertices of A (resp. to no vertex of A ). We say that A is complete (resp. anticomplete ) to B if every vertex of A is complete (resp. anticomplete) to B (note that this means in particular that A and B are disjoint). If a graph has large treewidth, then it contains some sub-structure that is highlyconnected in some sense (grid minor, bramble, tangle, see [19]). Theorem 2.1seems to be a new statement of that kind. It says that graphs of large treewidthmust contain either a large clique or a minimal separator of large size. However,its converse is false, as shown by K ,t that has treewidth 2 (it is a series-parallelgraph) and contains a minimal separator of size t .6 variant of the following theorem can be obtained from the celebratedexcluded grid theorem of Robertson and Seymour. The idea is to use a largegrid to obtain a large minimal separator. But there are technicalities because weare not allowed to delete edges, so the grid might contain many crossing edges.To find two vertices that cannot be separated by a small separator, one needsto clean the grid. We do not include the details since the following provides abetter bound. Theorem 2.1.
Let G be a graph and let k ≥ and s ≥ be positive integers. If G does not contain a clique on k vertices nor a minimal separator of size largerthan s , then the treewidth of G is at most ( k − s − . Before proving Theorem 2.1, let us introduce some terminology and stateresults due to Bouchitt´e and Todinca [6]. For a graph G we denote by cc ( G )the set of all connected components of G (viewed as subsets of V ( G )). A set F ⊆ (cid:0) V ( G )2 (cid:1) \ E ( G ) is a fill-in or chordal completion if G + F = ( V ( G ) , E ( G ) ∪ F )is a chordal graph. A fill-in F is minimal if it is inclusion-wise minimal. If X ⊆ V ( G ), then every connected component D ∈ cc ( G \ X ) with N ( D ) = X is called a component full to X . Observe that a set X ⊆ V ( G ) is a minimalseparator if and only if there exist at least two connected components of G \ X that are full to X . An important property of minimal separators is that no newminimal separator appears when applying a minimal fill-in. Lemma 2.2 (see [6]) . For every graph G , minimal fill-in F , and minimal sep-arator X in G + F , X is a minimal separator in G as well. Furthermore, thefamilies of components cc (( G + F ) \ X ) and cc ( G \ X ) are equal (as familiesof subsets of V ( G ) ). A set Ω ⊆ V ( G ) is a potential maximal clique (PMC) if there exists a minimalfill-in F such that Ω is a maximal clique of G + F . A PMC is surrounded byminimal separators. Lemma 2.3 (see [6]) . For every PMC Ω in G and every component D ∈ cc ( G \ Ω) , the set N ( D ) is a minimal separator in G with D being a full component. The following characterizes PMCs.
Theorem 2.4 (see [6]) . A set Ω ⊆ V ( G ) is a PMC in G if and only if thefollowing two conditions hold:(i) for every D ∈ cc ( G \ Ω) we have N ( D ) (cid:40) Ω ;(ii) for every x, y ∈ Ω either x = y , xy ∈ E ( G ) , or there exists D ∈ cc ( G \ Ω) with x, y ∈ N ( D ) . In the second condition of Theorem 2.4, we say that a component D covers the nonedge xy . Lemma 2.5.
Let G be a graph, k ≥ and s ≥ be integers, and let Ω be aPMC in G with | Ω | > ( k − s . Then there exists in G either a clique of size k or a minimal separator of size larger than s . roof. By Lemma 2.3, we may assume that for every D ∈ cc ( G \ Ω) we have | N ( D ) | ≤ s .Assume first that for every x ∈ Ω the set of non-neighbors of x in Ω (i.e.,Ω \ N [ x ]) is of size less than s . Let A = Ω and consider the following iterativeprocess. Given A i for i ≥
0, pick x i ∈ A i , and set A i +1 = A i ∩ N ( x i ). Theprocess terminates when A i becomes empty. Clearly, the vertices x , x , . . . induce a clique. Furthermore, by our assumption, | A i \ A i +1 | ≤ s . Thereforethis process continues for at least k steps, giving a clique of size k in G .Thus we are left with the case when there exists x ∈ Ω with the set Ω \ N [ x ]of size at least s . Let Y = { x } ∪ (Ω \ N [ x ]); we have | Y | > s , Y ⊆ Ω, and G [ Y ] is disconnected.Consider the following iterative process. At step i , we will maintain a parti-tion A i of Y into at least two parts and for every A ∈ A i a set D i ( A ) ⊆ cc ( G \ Ω)with the following property: the sets { A ∪ (cid:83) D ∈D i ( A ) D | A ∈ A i } is the parti-tion of G [ Y ∪ (cid:83) A ∈A i (cid:83) D ∈D i ( A ) D ] into vertex sets of connected components. Inparticular, for every A ∈ A i and D ∈ D i ( A ) we have N ( D ) ∩ Y ⊆ A . We startwith A = cc ( G [ Y ]) and D ( A ) = ∅ for every A ∈ A .The process terminates when there exists A ∈ A i of size larger than s .Otherwise, we perform a step as follows. Pick two distinct A, B ∈ A i andvertices a ∈ A , b ∈ B . By the properties of A i , ab / ∈ E ( G ). By Theorem 2.4,there exists D ∈ cc ( G \ Ω) with a, b ∈ N ( D ). Let A = { C ∈ A i | N ( D ) ∩ C (cid:54) = ∅} .Note that A, B ∈ A . Furthermore, since | N ( D ) | ≤ s , we have 2 ≤ |A| ≤ s .We define A i +1 = ( A i \ A ) ∪ { (cid:83) C ∈A C } . For every C ∈ A i +1 ∩ A i we keep D i +1 ( C ) = D i ( C ). Furthermore, we set D i +1 ( (cid:83) C ∈A C ) = { D } ∪ (cid:83) C ∈A D i ( C ).It is straightforward to verify the invariant for A i +1 and D i +1 .Furthermore, since every set C ∈ A i is of size at most s while | Y | > s wehave that |A i | > s . Since 2 ≤ |A| ≤ s , we have 2 ≤ |A i +1 | < |A i | . Consequently,the process terminates after a finite number of steps with A i of size at least 2, D i , and some A ∈ A i of size greater than s .Let X = A ∪ (cid:83) D ∈D i ( A ) D and let y ∈ Y \ A . Note that G [ X ] is connectedby the invariant on A i and D i , y exists as |A i | ≥
2, and y is anticomplete to X .We use Theorem 2.4: for every a ∈ A fix a component D a ∈ cc ( G \ Ω) coveringthe nonedge ya . Since | N ( D a ) | ≤ s while | A | > s , the set D = { D a | a ∈ A } is of size greater than s . Since G [ X ] is connected and y is anticomplete to X ,there exists a minimal separator S with y in one full side and X in the otherfull side. However, then S ∩ D (cid:54) = ∅ for every D ∈ D . Hence, | S | ≥ |D| > s . Thisfinishes the proof of the lemma. Proof of Theorem 2.1.
Let G be a graph such that it does not contain a clique on k vertices and aminimal separator of size larger than s . Let F be a minimal chordal completionof G . By Lemma 2.5, every maximal clique of G + F is of size at most ( k − s .Therefore a clique tree of G + F is a tree decomposition of G of width at most( k − s −
1, as desired. 8
Nested -wheels Let k ≥ k -wheel is a graph formed by a hole H called the rim together with a set C of k vertices that are not in V ( H ) called the centers ,such that each center has at least three neighbors in the rim. We denote sucha k -wheel by ( H, C ). Observe that a 0-wheel is a hole. A 1-wheel is called a wheel (see Fig. 1). We often write (
H, u ) instead of ( H, { u } ).A 2-wheel ( H, { u, v } ) is nested if H contains two vertices a and b such thatall neighbors of u in H are in one path of H from a to b , while all the neighborsof v are in the other path of H from a to b . Observe that a and b may beadjacent to both u and v . As we will see in this section, the properties of 2-wheels highlight structural similarities between (theta, triangle)-free graphs and(even hole, pyramid)-free graphs, in the sense that in both classes, apart fromfew exceptions, every 2-wheel with non-adjacent centers is nested.For a center u of a k -wheel ( H, C ), a u -sector of H is a subpath of H oflength at least 1 whose ends are adjacent to u and whose internal vertices arenot. However, a u -sector may contain internal vertices that are adjacent to v for some center v (cid:54) = u . Observe that for every center u , the rim of a wheel isedgewise partitioned into its u -sectors. In (theta, triangle)-free graphs
The cube is the graph formed from a hole of length 6, say h h · · · h h togetherwith a vertex u adjacent to h , h , h and a vertex v non-adjacent to u andadjacent to h , h , h . Note that the cube is a non-nested 2-wheel with non-adjacent centers. Lemma 3.1.
Let G be a (theta, triangle)-free graph. If W = ( H, { u, v } ) is a2-wheel in G such that uv / ∈ E ( G ) , then W is either a nested wheel or the cube.Proof. Suppose that W is not a nested wheel. We will prove that W is the cube. Claim 1.
Every u -sector of H contains at most one neighbor of v and every v -sector of H contains at most one neighbor of u .Proof of Claim 1. For otherwise, without loss of generality, some u -sector P = x . . . y of H contains at least two neighbors of v . Let x (cid:48) , y (cid:48) be neighbors of v closest to x, y respectively along P . Note that x (cid:48) y (cid:48) / ∈ E ( G ) because G istriangle-free. Since W is not nested, H \ P contains some neighbors of v . Notealso that H \ P contains some neighbors of u .So, let Q = z . . . z (cid:48) be the path of H \ P that is minimal length and suchthat uz ∈ E ( G ) and vz (cid:48) ∈ E ( G ). Note that z (cid:48) is adjacent to either x or y , forotherwise uzQz (cid:48) v , uxP x (cid:48) v , and uyP y (cid:48) v form a theta from u to v . So supposeup to symmetry that z (cid:48) is adjacent to y . So, v is not adjacent to y since G is triangle-free. It then follows that the three paths vz (cid:48) y , vy (cid:48) P y , and vx (cid:48) P xuy form a theta, a contradiction. This proves Claim 1.
Claim 2. u and v have no common neighbors in H . roof of Claim 2. Otherwise, let x be such a common neighbor. Consider asubpath x . . . y of H of maximum length with the property of being a u -sector ora v -sector, and suppose up to symmetry that it is a u -sector. By its maximality,it contains a neighbor of v different from x . So in total it contains at least twoneighbors of v , a contradiction to Claim 1. This proves Claim 2.Claim 1 and 2 prove that | N H ( u ) | = | N H ( v ) | and the neighbors of u and v alternate along H . So, let x, y, z ∈ N H ( u ) and x (cid:48) , y (cid:48) , z (cid:48) ∈ N H ( v ) be dis-tinct vertices in H with x , x (cid:48) , y , y (cid:48) , z , z (cid:48) appearing in this order along H . If V ( H ) = { x, y, z, x (cid:48) , y (cid:48) , z (cid:48) } , then V ( H ) ∪ { u, v } induces the cube, so suppose { x, y, z, x (cid:48) , y (cid:48) , z (cid:48) } (cid:40) V ( H ). Hence, up to symmetry, we may assume that x , x (cid:48) , y , y (cid:48) , z and z (cid:48) are chosen such that: xz (cid:48) / ∈ E ( G ). But then the three paths vz (cid:48) ( H \ x ) z , vy (cid:48) ( H \ y ) z , and vx (cid:48) ( H \ y ) xuz form a theta, a contradiction.The following lemma of Radovanovi´c and Vuˇskovi´c shows that the presenceof the cube in a (theta, triangle)-free graph entails some structure. Lemma 3.2 (see [23]) . Let G be a (theta, triangle)-free graph. If G containsthe cube, then either it is the cube, or it has a clique separator of size at most 2. In even-hole-free graphs
Let us consider a classical generalization of even-hole-free graphs.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 are triangles and no edges exist between the paths except those ofthe two triangles (see Fig. 1). An even wheel is a wheel ( H, u ) such that u hasan even number of neighbors in H . A square is a hole of length 4.It is easy to see that all thetas, prisms, even wheels, and squares contain evenholes. The class of (theta, prism, even wheel, square)-free graphs is thereforea generalization of even-hole-free graphs that capture the structural propertiesthat we need here.A proof of the following lemma can be found in [12] (where it relies on manylemmas). We include here our self-contained proof for the sake of completeness.Call a wheel proper if it is not pyramid. A cousin wheel is a 2-wheel madeof a hole H = h h . . . h n h and two non-adjacent centers u and v , such that N H ( u ) = { h , h , h } and N H ( v ) = { h , h , h } . Lemma 3.3.
Let G be a (theta, prism, pyramid, even wheel, square)-free graph.If W = ( H, { u, v } ) is a 2-wheel in G such that uv / ∈ E ( G ) , then W is either anested or a cousin wheel. Moreover, if W is nested then | N H ( u ) ∩ N H ( v ) | ≤ .Proof. In the case where W = ( H, { u, v } ) is nested, it must be that | N H ( u ) ∩ N H ( v ) | ≤
1, for otherwise G would contain a square. Since G contains no evenwheel, it is sufficient to consider the following cases. Case 1: N H ( u ) = 3 or N H ( v ) = 3.Assume that W is not a nested wheel. We will prove that W is a cousinwheel. Without loss of generality, we may assume that | N H ( u ) | = 3, and let10 H ( u ) = { x, y, z } . We denote by P x = y . . . z , P y = x . . . z and P z = x . . . y thethree u -sectors of H .Suppose xyz is a path of H . Then v must be adjacent to y , for otherwise W is nested, a contradiction. Since V ( H ) ∪ { u } and V ( H \ y ) ∪ { u, v } do notinduce an even wheel, v has exactly two neighbors in P y . Moreover, the twoneighbors of v in P y are adjacent, for otherwise H \ y , u , and v form a theta.Since ( H, v ) is not a pyramid, this means that one of x or z is a neighbor of v .Therefore, W is a cousin wheel.Now suppose that { x, y, z } does not induce a path. So xy , yz , and zx arenon-edges. Note that v is adjacent to at most one of x , y , or z , because G contains no square. Up to symmetry, assume that vx / ∈ E ( G ). Let R be the v -sector of H which contains x (in its interior). Since ( H, { u, v } ) is not a nestedwheel, the ends of R are not both in P x , or both in P y , or both in P z . So assumethat R = y (cid:48) . . . z (cid:48) with z (cid:48) is in the interior of P z and y (cid:48) is not in P z . If y (cid:48) is in P x , then R , u , and v form a theta from x to z , a contradiction. Hence, y (cid:48) is notin P x , so y (cid:48) is in the interior of P y .Call x (cid:48) the neighbor of v in H different from y (cid:48) and z (cid:48) . If x (cid:48) is not in theinterior of P x , then P x is contained in the v -sector x (cid:48) Hz (cid:48) . Thus, there exists a v -sector S which contains P x . In particular, the hole made of S and v containstwo non adjacent neighbors of u , namely y and z . Hence, S , u , and v form atheta from y to z . So, x (cid:48) is in the interior of P x .This means x , y (cid:48) , z , x (cid:48) , y , z (cid:48) appear in this order along H . If x (cid:48) z / ∈ E ( G ),then the paths x (cid:48) ( H \ y ) z , x (cid:48) ( H \ z ) yuz , and x (cid:48) vy (cid:48) ( H \ x ) z form a theta from x (cid:48) to z , a contradiction. So, x (cid:48) z ∈ E ( G ). By symmetry, x (cid:48) y ∈ E ( G ). But then, { u, y, x (cid:48) z } induces a square, a contradiction. Case 2: N H ( u ) ≥ N H ( v ) ≥ H, { u, v } ) is not a nested wheel. First ofall, we have N H ( u ) (cid:54) = N H ( v ), for otherwise u , v , and two non-adjacent verticesof N H ( u ) would form a square. So in H , there exists a neighbor of v that is notadjacent to u . It is therefore well defined to consider the u -sector P = x . . . y of H whose interior contains k ≥ v , and to choose such a sector with k minimum. We denote by x (cid:48) the neighbor of x in H \ P , by y (cid:48) the neighbor of y in H \ P and by Q = x (cid:48) . . . y (cid:48) the path H \ P .Note that u has some neighbor in the interior of Q , because u has at least 5neighbors in H . We now show that v also has some neighbor in the interior of Q .Suppose that it is not the case. Then, the neighborhood of v in H is completelycontained in V ( P ) ∪{ x (cid:48) , y (cid:48) } . Since ( H, { u, v } ) is not a nested wheel, v is adjacentto x (cid:48) or y (cid:48) — and in fact to both of them, for otherwise the hole uxP yu wouldcontain an even number (at least 4) of neighbors of v , thus inducing an evenwheel, a contradiction. Now since { u, v, x, y } does not induce a square, up tosymmetry we may assume that vx / ∈ E ( G ). Since | N H ( v ) | ≥ v has at least2 neighbors in the interior of P , and so k ≥
2. Note that u is adjacent to x (cid:48) ,for otherwise, x (cid:48) would be the unique neighbor of v in the interior of a u -sector,contradicting the minimality of k . Since { u, v, x (cid:48) , y (cid:48) } does not induce a square,we know that u is not adjacent to y (cid:48) . But then, y (cid:48) is the unique neighbor of v
11n the interior of some u sector, a contradiction to the minimality of k . Thisproves that v has some neighbor in the interior of Q .By the fact that each of u and v has some neighbor in the interior of Q , apath S from u to v whose interior is in the interior of Q exists. Let x (cid:48)(cid:48) (resp. y (cid:48)(cid:48) ) be the neighbor of v in P closest to x (resp. y ) along P . If x (cid:48)(cid:48) = y (cid:48)(cid:48) , then x (cid:48)(cid:48) is an internal vertex of P , and so S and P form a theta from u to x (cid:48)(cid:48) . If x (cid:48)(cid:48) y (cid:48)(cid:48) ∈ E ( G ), then S and P form a pyramid. If x (cid:48)(cid:48) (cid:54) = y (cid:48)(cid:48) and x (cid:48)(cid:48) y (cid:48)(cid:48) / ∈ E ( G ),then S , uxP x (cid:48)(cid:48) v , and uyP y (cid:48)(cid:48) v form a theta from u to v . Each of the cases yieldsa contradiction; this completes the proof. In this section, we prove that the treewidth is bounded in (theta, triangle, S i,j,k )-free graphs and in (even hole, pyramid, K t , S i,j,k )-free graphs.For (theta, triangle)-free graphs, by Lemma 3.2, we may assume that thegraphs we work on are cube-free since the cube itself has small treewidth, andclique separators of size at most 2 in some sense preserve the treewidth (thiswill be formalized in the proofs). For (even hole, pyramid)-free graphs, recallthat we work from the start in a superclass, namely (theta, prism, pyramid,even wheel, square)-free graphs.Since our proof is the same for (theta, triangle, S i,j,k )-free graphs and (evenhole, pyramid, K t , S i,j,k )-free graphs, to avoid duplicating it, we introduce aclass C that contains all the graphs that we need to consider while entailing thestructural properties that we need.Call butterfly a wheel ( H, v ) such that N H ( v ) = { a, b, c, d } with ab ∈ E ( G ), bc / ∈ E ( G ), cd ∈ E ( G ) and da / ∈ E ( G ). Let C be the class of all (theta, prism,pyramid, butterfly)-free graphs such that every 2-wheel with non-adjacent cen-ters is either a nested or a cousin wheel. Lemma 4.1. If G is a (theta, triangle, cube)-free graph or a (theta, prism,pyramid, even wheel, square)-free graph, then G ∈ C .Proof. If G is a (theta, triangle, cube)-free graph, then G is theta-free and(prism, pyramid, butterfly)-free (because prisms, pyramids, and butterflies con-tain triangles). Furthermore, every 2-wheel with non-adjacent centers is a nestedwheel by Lemma 3.1.If G is a (theta, prism, pyramid, even wheel, square)-free graph, then G is(theta, prism, pyramid)-free and butterfly-free (because a butterfly is an evenwheel). Furthermore, every 2-wheel with non-adjacent centers is either a nestedor a cousin wheel by Lemma 3.3.Hence G ∈ C as claimed.For our proof, we need a special kind of k -wheel. A k -span-wheel is a k -wheel( H, C ) that satisfies the following properties.12
There exist two non-adjacent vertices x, y in H and we denote by P A = a . . . a α and P B = b . . . b β the two paths of H from x to y , with x = a = b and y = a α = b β . • C ∪ { x, y } is an independent set. • There exists an ordering of vertices in C , namely v , v , · · · , v k . • Every vertex of C has neighbors in the interiors of both P A and P B (andat least 3 neighbors in H since ( H, C ) is a k -wheel). • For every 1 ≤ i < j ≤ k and 1 ≤ i (cid:48) , j (cid:48) ≤ α , if v i a i (cid:48) ∈ E ( G ) and v j a j (cid:48) ∈ E ( G ) then i (cid:48) ≤ j (cid:48) . • For every 1 ≤ i < j ≤ k and 1 ≤ i (cid:48) , j (cid:48) ≤ β , if v i b i (cid:48) ∈ E ( G ) and v j b j (cid:48) ∈ E ( G ) then i (cid:48) ≤ j (cid:48) .Informally, a k -span-wheel is such that, walking from x to y along both P A and P B , one first meets all the neighbors of v , then all neighbors of v , and soon until v k . Observe that a 1-span-wheel is a wheel, 2-span-wheel is a nested2-wheel. Note that distinct v i and v j may share common neighbors on H (it iseven possible that N P A ( v ) = · · · = N P A ( v k ) = { a i } ).Observe that in the following theorem, thetas, pyramids, prisms, and but-terflies have to be excluded, since they do not satisfy the conclusion. Lemma 4.2.
Let G be a graph in C . Let C be a minimal separator in G of sizeat least 2 that is furthermore an independent set, and A and B be componentsof G \ C that are full to C . Then:1. There exist two vertices x and y in C , a path P A from x to y with interiorin A , and a path P B from x to y with interior in B such that all verticesin C \ { x, y } have neighbors in the interior of both P A and P B . Note that V ( P A ) ∪ V ( P B ) induces a hole that we denote by H .2. ( H, C \ { x, y } ) is a ( | C | − -span-wheel.Proof. We first prove 1, by induction on k = | C | .If k = 2, then x , y , P A , and P B exist from the connectivity of A and B ,and the conditions on C \ { x, y } vacuously hold. So suppose the result holdsfor some k ≥
2, and let us prove it for k + 1. Let z be any vertex from C , andapply the induction hypothesis to C \ z in G \ z . This provides two vertices x, y in C \ z and two paths P A and P B . We denote by H the hole formed by P A and P B . Claim 1.
Every vertex in C \ { x, y, z } has neighbors in the interior of both P A and P B .Proof of Claim 1. Follows directly from the induction hypothesis. This provesClaim 1. 13ince z has a neighbor in A and A is connected, there exists a path Q A = z . . . z A in A ∪ { z } , such that z A has a neighbor in the interior of P A . A similarpath Q B exists. We set Q = z A Q A zQ B z B . We suppose that x , y , P A , P B , Q A ,and Q B are chosen subject to the minimality of Q .Observe that Q is a chordless path by its minimality and the fact that A and B being anticomplete. The minimality of Q implies that the interior of Q is anticomplete to the interior of P A and to the interior of P B . Claim 2.
We may assume that Q has length at least 1.Proof of Claim 2. Otherwise, z = z A = z B , so z has neighbors in the interior ofboth P A and P B . Hence, by Claim 1, x , y , P A , and P B satisfy 1. This provesClaim 2.Let a (resp. a (cid:48) ) be the neighbor of z A in P A closest to x (resp. to y ) along P A .Let b (resp. b (cid:48) ) be the neighbor of z B in P B closest to x (resp. to y ) along P B . Claim 3. If a (cid:54) = a (cid:48) and aa (cid:48) / ∈ E ( G ) , then z = z A . If b (cid:54) = b (cid:48) and bb (cid:48) / ∈ E ( G ) ,then z = z B .Proof of Claim 3. We give a proof only for the statement of a , since the prooffor b is similar.For suppose a (cid:54) = a (cid:48) , aa (cid:48) / ∈ E ( G ), and z (cid:54) = z A , let z (cid:48) be the neighbor of z A in Q . Set P (cid:48) A = xP A az a a (cid:48) P A y and Q (cid:48) = z (cid:48) Qz B . Let us prove that x , y , P (cid:48) A , P B , and Q (cid:48) contradict the minimality of Q . Obviously, Q (cid:48) is shorter than Q , so we only have to prove that every vertex in C \ { z } has neighbors in theinterior of both P (cid:48) A and P B . For P B , it follows from Claim 1. So suppose for acontradiction that a vertex c ∈ C \ { z } has no neighbor in the interior of P (cid:48) A .Since by Claim 1 c has a neighbor c (cid:48) in the interior of P A , c (cid:48) is an internal vertexof aP A a (cid:48) . Since G is theta-free, ( H, z A ) is a wheel. Note that ( H, { c, z A } ) isnot nested because of c (cid:48) and some neighbor of c in the interior of P B (i.e. theneighborhood of c in H is not contained in a unique z A -sector). Since G ∈ C , byLemma 3.3, ( H, { c, z A } ) is a cousin wheel. Since c has neighbors in the interiorsof both P A and P B , this means that x or y is a common neighbor of c and z A , acontradiction to C being an independent set. The proof for the latter statement(with b ) is similar. This proves Claim 3. Claim 4.
We may assume that x has neighbors in the interior of Q and y hasno neighbor in the interior of Q .Proof of Claim 4. We show that if it is not the case, then there is a contradiction.For suppose both x and y have a neighbor in the interior of Q , then a path ofminimal length from x to y with interior in the interior of Q form a thetatogether with P A and P B , a contradiction.Now suppose that none of x and y has a neighbor in the interior of Q . Recallthat Claim 2 tells us that z A (cid:54) = z B . So either z (cid:54) = z A or z (cid:54) = z B . Up to symmetry,we may assume that z (cid:54) = z A . Hence by Claim 3, either a = a or aa (cid:48) ∈ E ( G ).14uppose a = a (cid:48) . This implies that a is in the interior of P A . If b = b (cid:48) , then b is in the interior of P B — so H and Q form a theta from a to b ; if bb (cid:48) ∈ E ( G ),then H and Q form a pyramid; and if b (cid:54) = b (cid:48) and bb (cid:48) / ∈ E ( G ), then aP A xP B bz B , aP A yP B b (cid:48) z B , az A Qz B form a theta from a to z B (note that az A Qz B has lengthat least 2 because z A (cid:54) = z B ), a contradiction. So, aa (cid:48) ∈ E ( G ).Suppose that bb (cid:48) ∈ E ( G ). Note that |{ a, a (cid:48) } ∩ { b, b (cid:48) } ∩ { x, y }| (cid:54) = 2, because x and y are not adjacent. Moreover, |{ a, a (cid:48) } ∩ { b, b (cid:48) } ∩ { x, y }| (cid:54) = ∅ , for otherwise H and Q form a prism. So, |{ a, a (cid:48) } ∩ { b, b (cid:48) } ∩ { x, y }| = 1. In this last case, wesuppose up to symmetry that x = a = b . So, z is in the interior of Q since itis non-adjacent to x — in particular Q has length at least 2. Hence, H and Q form a butterfly (with x = a = b being the center), a contradiction.So, bb (cid:48) / ∈ E ( G ). If b = b (cid:48) , then b is in the interior of P B ; thus P A , P B ,and Q form a pyramid (i.e. 3 P C ( az A a (cid:48) , b ), a contradiction. So, b (cid:54) = b (cid:48) , andhence by Claim 3, z B = z . This means that b (cid:54) = x and b (cid:48) (cid:54) = y (because C is an independent set). Therefore, aP A xP B bz , a (cid:48) P A yP B b (cid:48) z , and z A Qz form apyramid (i.e. 3 P C ( aa (cid:48) z A , z )), a contradiction.So, each case leads to a contradiction. Hence, exactly one of x or y hasneighbors in the interior of Q , and up to symmetry we may assume it is x . Thisproves Claim 4. Claim 5. a (cid:48) x ∈ E ( G ) and b (cid:48) x ∈ E ( G ) .Proof of Claim 5. First, suppose z A is adjacent to x , i.e. a = x . Then, z A (cid:54) = z since C is an independent set. Note that a = a (cid:48) is impossible since z A hasneighbors in the interior of P A . So, by Claim 3, a (cid:48) x ∈ E ( G ).Now suppose z A is not adjacent to x . By Claim 4, x has a neighbor in theinterior of Q , so we choose such a neighbor x (cid:48) closest to z A along Q . Note thatby the minimality of Q , no vertex in the interior of Q has neighbor in the interiorof P A and in the interior of P B . Since y is not adjacent to x (cid:48) (by Claim 4), x (cid:48) has no neighbors in ( P A ∪ P B ) \ { x } . We set R = xx (cid:48) Qz A and observe that R has length at least 2. If a (cid:54) = a (cid:48) and aa (cid:48) / ∈ E ( G ), then xP A az A , xP B yP A a (cid:48) z A ,and R form a theta from x to z A . If aa (cid:48) ∈ E ( G ), then P A , P B , and R form apyramid. Therefore a = a (cid:48) . Note that xa ∈ E ( G ), for otherwise, P A , P B , and R form a theta. Hence, a (cid:48) x ∈ E ( G ).The proof for b (cid:48) x ∈ E ( G ) is similar. This proves Claim 5.To conclude the proof of 1, set P (cid:48) A = zQz A a (cid:48) P A y and P (cid:48) B = zQz B b (cid:48) P B y . ByClaim 5, x has neighbors in the interior of both P (cid:48) A and P (cid:48) B (these neighborsare a (cid:48) and b (cid:48) ). Note that since a (cid:48) x, b (cid:48) x ∈ E ( G ), the interiors of P A and P B areincluded in the interiors of P (cid:48) A and P (cid:48) B respectively. Hence, by Claim 1, everyvertex of C \ z has neighbors in the interior of both P (cid:48) A and P (cid:48) B .Hence, the vertices z, y and the paths P (cid:48) A and P (cid:48) B show that 1 is satisfied.Let us now prove 2. Note that ( H, C \ { x, y } ) is ( | C | − G is theta-free, every vertex in C \ { x, y } has at least three neighbors in H ). It remains to prove that it is a ( | C | − | C | ≤
3. We set P A = a . . . a α and P B = b . . . b β with x = a = b and15 = a α = b β , as in the definition of a k -span-wheel. We just have to exhibit anordering of the vertices of C \ { x, y } that satisfies the rest of the definition.We first define v , the smallest vertex in the order we aim to construct. Notethat no vertex v ∈ C \ { x, y } is adjacent to x or y , because C is an independentset. We let v be a vertex of C that is adjacent to a i with i minimum. Let j bethe smallest integer such that v is adjacent to b j . We suppose that v is chosensubject to the minimality of j . Let i (cid:48) , j (cid:48) be the greatest integers such that v isadjacent to a i (cid:48) and b j (cid:48) . Note that 1 < i ≤ i (cid:48) < α and 1 < j ≤ j (cid:48) < β . Claim 6.
For every w ∈ C \ { x, y, v } , we have N H ( w ) ⊆ V ( a i (cid:48) P A yP B b j (cid:48) ) .Proof of Claim 6. We first note that the 2-wheel ( H, { v , w } ) is not a cousinwheel, because this may happen only when x ∈ N ( v ) or y ∈ N ( v ) (recall thatif it was a cousin wheel, N H ( v ) would induce a 3-vertex path in H ).Hence, ( H, { v , w } ) is a nested wheel. Suppose that N H ( w ) (cid:54)⊆ V ( a i (cid:48) P A a α ) ∪ V ( b j (cid:48) P B b β ). This means that w has a neighbor z in a i (cid:48) − P A xP B b j (cid:48) − . Since( H, { v , w } ) is a nested wheel, N H ( w ) is contained in a v -sector Q of ( H, v ).Moreover, since w has a neighbor in the interior of both P A and P B , we have Q = a i P A xP B b j . Since H and w form a wheel, w has neighbor in the interiorof Q . This contradicts the minimality of i or j . This proves Claim 6.The order of C \ { x, y } is now constructed as follows: we remove v from C ,define v as we defined v (minimizing i , and then minimizing j ), then remove v , define v , and so on. This iteratively constructs an ordering of C \ { x, y } showing that ( H, C \ { x, y } ) is a ( | C | − t, k ≥
1, the Ramsey number R ( t, k ) is the smallest integer n such that any graph on n vertices contains either a clique of size t , or anindependent set of size k . Theorem 4.3.
An ( l -span-wheel, K t )-free graph G ∈ C has treewidth at most ( t − R ( t, l + 2) − − .Proof. Suppose for a contradiction that the treewidth of G is at least( t − R ( t, l + 2) − . Since G is K t -free, by Theorem 2.1 G admits a minimalseparator D of size at least R ( t, l + 2). Let A and B be two connected compo-nents of G \ D that are full to D . By the definition of Ramsey number, G [ D ]contains an independent set C of size l + 2. We define G (cid:48) = G [ A ∪ C ∪ B ], andobserve that C is a minimal separator of G (cid:48) . Hence by Lemma 4.2 applied to G (cid:48) , the graph contains an l -span-wheel, a contradiction.The following shows that in C , an l -span-wheel with large l contains S i,j,k with large i, j, k . Lemma 4.4.
If a butterfly-free graph G contains a (4 k + 1) -span-wheel with k ≥ , then it contains S k +1 ,k +1 ,k +1 .Proof. Consider a (4 k + 1)-span-wheel in G , with x , y , P A , and P B be as inthe definition of span-wheel given in the beginning of the current section. Let16 , . . . , v k +1 be the centers of the span wheel. For each i = 1 , . . . , k + 1, let a i (resp. a (cid:48) i ) be the neighbor of v i in P A closest to x (resp. to y ) along P A . Let b i (resp. b (cid:48) i ) be the neighbor of v i in P B closest to x (resp. to y ) along P B . We set P i = a i P A xP B b i and Q i = a (cid:48) i P A yP B b (cid:48) i . Claim 1. P i has length at least i + 1 and Q i has length at least k + 3 − i .Proof of Claim 1. We prove this by induction on i for P i . It is clear that P has length at least 2 since x is not adjacent to v . Suppose the claim holds forsome fixed i ≥
1, and let us prove it for i + 1. From the induction hypothesis, P i has length at least i + 1, and since v i has a neighbor in the interior of P i +1 (because it has at least three neighbors in H ), the length of P i +1 is greater thanthe length of P i , so P i +1 has length at least i + 2.The proof for Q i is similar, except we start by proving that Q k +1 haslength at least 2, and that the induction goes backward down to Q . Thisproves Claim 1.We set l = 2 k +1. So, by Claim 1, P l and Q l both have length at least 2 k +2.We set v = v l , P = P l , Q = Q l , a = a l , a (cid:48) = a (cid:48) l , b = b l and b (cid:48) = b (cid:48) l . Since G isbutterfly-free, we do not have aa (cid:48) ∈ E ( G ) and bb (cid:48) ∈ E ( G ) simultaneously. So,up to symmetry we may assume that either a = a (cid:48) ; or a (cid:54) = a (cid:48) and aa (cid:48) / ∈ E ( G ).If a = a (cid:48) , let u , u (cid:48) , and u (cid:48)(cid:48) be three distinct vertices in P such that a , u , u (cid:48) ,and u (cid:48)(cid:48) appear in this order along P , aP u has length k + 1 and bP u (cid:48)(cid:48) has length k − P has length at least 2 k + 1). Let w be in Q and such that aQw has length k + 1 (which is possible because Q has length atleast 2 k + 1). The three paths aP u , avbP u (cid:48)(cid:48) , and aQw form an S k +1 ,k +1 ,k +1 .If a (cid:54) = a (cid:48) and aa (cid:48) / ∈ E ( G ), then let u , u (cid:48) , and u (cid:48)(cid:48) be three distinct verticesin P such that a , u , u (cid:48) , and u (cid:48)(cid:48) appear in this order along P , aP u has length k and bP u (cid:48)(cid:48) has length k . Let w be in Q and such that a (cid:48) Qw has length k . Thethree paths vaP u , vbP u (cid:48)(cid:48) and va (cid:48) Qw form an S k +1 ,k +1 ,k +1 .The following is a classical result on treewidth and we omit its proof. Lemma 4.5 ([21]) . The treewidth of a graph G is the maximum treewidth ofan induced subgraph of G that has no clique separator. Theorem 4.6.
For k ≥ , a (theta, triangle, S k,k,k )-free graph G has treewidthat most R (3 , k − − .Proof. By Lemma 4.5, it is enough to consider a graph G that does not have aclique separator. If G contains a cube, then Lemma 3.2 tells us that G itself is thecube. By classical results on treewidth, the treewidth of the cube is 3 (but thetrivial bound 8 would be enough for our purpose), which in particular achievesthe given bound. We may therefore assume that G is cube-free. Moreover, byLemma 4.1, G is in C . Since G is S k,k,k -free, by Lemma 4.4, G contains no(4 k − G contains no K by assumption. Hence, byTheorem 4.3, G has treewidth at most 2( R (3 , k − − heorem 4.7. For k ≥ , an (even hole, pyramid, K t , S k,k,k )-free graph G has treewidth at most ( t − R ( t, k − − .Proof. Since all thetas, prisms, even wheels, and squares contain even holes, G is (theta, prism, pyramid, even wheel, square)-free. So, by Lemma 4.1, G isin C . Since G is S k,k,k -free, by Lemma 4.4, G contains no (4 k − G contains no K t by assumption. Hence, by Theorem 4.3, G hastreewidth at most ( t − R ( t, k − − Acknowledgement
Thanks to ´Edouard Bonnet, Zdenˇek Dvoˇr´ak, Kristina Vˇuskovi´c, and R´emi Wa-trigant for useful discussions. We are also grateful to two anonymous refereesfor careful reading of the paper and valuable suggestions and comments, whichhave improved the presentation of the paper.
References [1] T. Abrishami, M. Chudnovsky and K. Vuˇskovi´c. Even-hole-free graphs withbounded degree have bounded treewidth.
CoRR , abs/2009.01297, 2020.[2] P. Aboulker, I. Adler, E. J. Kim, N. L. D. Sintiari and N. Trotignon Onthe tree-width of even-hole-free graphs.
CoRR , abs/2008.05504, 2020.[3] I. Adler, N.-K. Le, H. M¨uller, M. Radovanovi´c, N. Trotignon, andK. Vuˇskovi´c. On rank-width of even-hole-free graphs.
Discrete Mathe-matics & Theoretical Computer Science , 19(1), 2017.[4] V. E. Alekseev. On the local restrictions effect on the complexity of find-ing the graph independence number.
Combinatorial-algebraic methods inapplied mathematics , 132:3–13, 1983. Gorky University Press, Gorky, inRussian.[5] V. E. Alekseev. Polynomial algorithm for finding the largest independentsets in graphs without forks.
Discrete Applied Mathematics , 135(1-3):3–16,2004.[6] V. Bouchitt´e and I. Todinca. Treewidth and minimum fill-in: Groupingthe minimal separators.
SIAM J. Comput. , 31(1):212–232, 2001.[7] A. Brandst¨adt, S. Mahfud, and R. Mosca. Bounded clique-width of ( s , , ,triangle)-free graphs. CoRR , abs/1608.01820, 2016.[8] A. Brandst¨adt and R. Mosca. Maximum weight independent sets for ( P ,triangle)-free graphs in polynomial time. Discrete Applied Mathematics ,236:57–65, 2018. 189] A. Brandst¨adt and R. Mosca. Maximum weight independent sets for ( S , , ,triangle)-free graphs in polynomial time. CoRR , abs/1806.09472, 2018.[10] K. Cameron, S. Chaplick, and C. T. Ho`ang. On the structure of (pan, evenhole)-free graphs.
Journal of Graph Theory , 87(1):108–129, 2018.[11] K. Cameron, M. V. G. da Silva, S. Huang, and K. Vuˇskovi´c. Structureand algorithms for (cap, even hole)-free graphs.
Discrete Mathematics ,341(2):463–473, 2018.[12] M. Chudnovsky, S. Thomass´e, N. Trotignon, and K. Vuˇskovi´c. Comput-ing maximum stable sets in (pyramid, even hole)-free graphs.
CoRR ,abs/1912.11246, 2019.[13] M. Conforti, G. Cornu´ejols, A. Kapoor, and K. Vuˇskovi´c. Triangle-freegraphs that are signable without even holes.
Journal of Graph Theory ,34(3):204–220, 2000.[14] B. Courcelle. The monadic second-order logic of graphs. I. Recognizablesets of finite graphs.
Inf. Comput. , 85(1):12–75, 1990.[15] K. K. Dabrowski and D. Paulusma. Clique-width of graph classes definedby two forbidden induced subgraphs.
Comput. J. , 59(5):650–666, 2016.[16] K. K. Dabrowski, F. Dross, and D. Paulusma. Colouring diamond-freegraphs.
J. Comput. Syst. Sci. , 89:410–431, 2017.[17] K. K. Dabrowski, M. Johnson, and D. Paulusma. Clique-width for heredi-tary graph classes.
CoRR , abs/1901.00335, 2019.[18] A. Grzesik, T. Klimosova, M. Pilipczuk, and M. Pilipczuk. Polynomial-time algorithm for maximum weight independent set on P -free graphs. InTimothy M. Chan, editor, Proceedings of the Thirtieth Annual ACM-SIAMSymposium on Discrete Algorithms, SODA 2019, San Diego, California,USA, January 6-9, 2019 , pages 1257–1271. SIAM, 2019.[19] D. J. Harvey and D. R. Wood. Parameters tied to treewidth.
Journal ofGraph Theory , 84(4):364–385, 2017.[20] D. Lokshantov, M. Vatshelle, and Y. Villanger. Independent set in P Proceedings ofthe Twenty-Fifth Annual ACM-SIAM Symposium on Discrete Algorithms,SODA 2014, Portland, Oregon, USA, January 5-7, 2014 , pages 570–581.SIAM, 2014.[21] L. Lov´asz. Graph minor theory.
Bulletin of the American MathematicalSociety , 43(1):75–86, 2006.[22] V. V. Lozin and M. Milanic. A polynomial algorithm to find an indepen-dent set of maximum weight in a fork-free graph.
J. Discrete Algorithms ,6(4):595–604, 2008. 1923] M. Radovanovi´c and K. Vuˇskovi´c. A class of three-colorable triangle-freegraphs.
Journal of Graph Theory , 72(4):430–439, 2013.[24] N. L. D. Sintiari and N. Trotignon. (Theta, triangle)-free and (even hole,K4)-free graphs. Part 1: Layered wheels.