Forbidden induced subgraph characterization of circle graphs within split graphs
Flavia Bonomo-Braberman, Guillermo A. Durán, Nina Pardal, Martín D. Safe
FForbidden induced subgraph characterization of circlegraphs within split graphs
Flavia Bonomo-Braberman a,b , Guillermo Dur´an c,d,e , Nina Pardal b,c ,Mart´ın D. Safe f,g a Universidad de Buenos Aires. Facultad de Ciencias Exactas y Naturales. Departamento deComputaci´on. Buenos Aires, Argentina. b CONICET-Universidad de Buenos Aires. Instituto de Investigaci´on en Ciencias de laComputaci´on (ICC). Buenos Aires, Argentina. c CONICET-Universidad de Buenos Aires. Instituto de C´alculo (IC). Buenos Aires,Argentina. d Universidad de Buenos Aires. Facultad de Ciencias Exactas y Naturales. Departamento deMatem´atica. Buenos Aires, Argentina. e Departamento de Ingenier´ıa Industrial, Facultad de Ciencias F´ısicas y Matem´aticas,Universidad de Chile, Santiago, Chile. f Departamento de Matem´atica, Universidad Nacional del Sur (UNS), Bah´ıa Blanca,Argentina g INMABB, Universidad Nacional del Sur (UNS)-CONICET, Bah´ıa Blanca, Argentina
Abstract
A graph is circle if its vertices are in correspondence with a family of chordsin a circle in such a way that every two distinct vertices are adjacent if and onlyif the corresponding chords have nonempty intersection. Even though thereare diverse characterizations of circle graphs, a structural characterization byminimal forbidden induced subgraphs for the entire class of circle graphs is notknown, not even restricted to split graphs (which are the graphs whose vertexset can be partitioned into a clique and a stable set). In this work, we givea characterization by minimal forbidden induced subgraphs of circle graphs,restricted to split graphs.
Keywords: circle graphs, forbidden induced subgraphs, split graphs,structural characterization
1. Introduction
A graph is circle if its vertices are in correspondence with a family of chordsin a circle in such a way that two vertices are adjacent if and only if the cor-responding chords have nonempty intersection. These graphs were defined byEven and Itai [7] in 1971 to solve an ordering problem stated by Knuth, using
Email addresses: [email protected] (Flavia Bonomo-Braberman), [email protected] (Guillermo Dur´an), [email protected] (Nina Pardal), [email protected] (Mart´ın D. Safe)
Preprint submitted to Elsevier a r X i v : . [ c s . D M ] J u l he minimum number of parallel intermediate stacks without the restriction oncompletion of loading before unloading. They proved that this problem can betranslated into the problem of finding the chromatic number of a circle graph.In 1985, Naji [13] characterized circle graphs in terms of the solvability of asystem of linear equations, yielding a O ( n )-time recognition algorithm for thisclass. More recently, it was shown in [11] that circle graphs can be recognizedin almost linear time. For a survey on circle graphs, see [6].All graphs in this work are simple, undirected, with no loops or multipleedges. Let G = ( V, E ) be a graph. Given V (cid:48) ⊆ V , the subgraph of G inducedby V (cid:48) , denoted G [ V (cid:48) ], is the graph whose vertex set is V (cid:48) and whose edge setconsists of all the edges in E that have both endpoints in V (cid:48) . The neighborhood of a vertex u ∈ V is the subset N G ( u ) (or simply N ( u )) consisting of the verticesof G that are adjacent to u . The complement of G is the graph G having V as vertex set and such that every two distinct vertices of G are adjacentif and only if they are non-adjacent in G . The local complement of G withrespect to a vertex u ∈ V is the graph G ∗ u that arises from G by replacingthe induced subgraph G [ N ( u )] by its complement. Two graphs G and H are locally equivalent if and only if G arises from H by a finite sequence of localcomplementations. Circle graphs were characterized by Bouchet [3] in 1994 interms of forbidden induced subgraphs of some locally equivalent graph. Inspiredby this result, Geelen and Oum [10] gave a new characterization of circle graphsin terms of pivoting . The result of pivoting a graph G with respect to an edge uv is the graph G × uv = G ∗ u ∗ v ∗ u . Let G and G be two graphs suchthat | V ( G i ) | ≥
3, for each i = 1 ,
2, and assume that V ( G ) ∩ V ( G ) = ∅ . Let v i be a distinguished vertex of G i , for each i = 1 ,
2. The split composition of G and G with respect to v and v is the graph G ∗ G whose vertex set is V ( G ∗ G ) = ( V ( G ) ∪ V ( G )) \ { v , v } and whose edge set is E ( G ∗ G ) = E ( G − { v } ) ∪ E ( G − { v } ) ∪ { uv : u ∈ N G ( v ) and v ∈ N G ( v ) } . Thevertices v and v are called the marker vertices . We say that G has a splitdecomposition if there exist two graphs G and G with | V ( G i ) | ≥
3, for each i = 1 ,
2, such that G = G ∗ G with respect to some pair of marker vertices; ifso, G and G are called the factors of the split decomposition. Those graphsthat do not admit a split decomposition are called prime graphs . Notice that,if any of the factors of a split decomposition admits a split decomposition, wecan continue the process until every factor is prime, a star or a complete graph.Bouchet proved that circle graphs are closed under split composition [2].In spite of all these results, no characterizations for the entire class of circlegraphs by forbidden induced subgraphs is known. Some partial results in thisdirection were found by restricting the problem to the classes of P -tidy graphs,tree-cographs and linear-domino graphs [1].In this work, we consider the problem of characterizing circle graphs byminimal forbidden induced subgraphs restricted to split graphs. The motivationto study circle graphs restricted to this particular graph class comes from chordalgraphs , which are those graphs that contain no induced cycle of length greaterthan 3, and constitute a widely studied graph class with very nice structuralproperties. Something similar happens with the class of split graphs , which is2n interesting subclass of chordal graphs. Split graphs are those graphs whosevertex set can be partitioned into a clique and a stable set. Equivalently, splitgraphs are those chordal graphs whose complement is also a chordal graph.Hence, studying those split graphs that are also circle is a good first step towardsa characterization of those chordal graphs that are also circle. If G is a splitgraph, the pair ( K, S ) is a split partition of G if { K, S } is a partition of thevertex set of G and the vertices of K (resp. S ) are pairwise adjacent (resp.nonadjacent); i.e., K is a clique and S is a stable set. We denote it G = ( K, S ).Let us consider a split graph G , and suppose that G is minimally non-circle ; i.e., G is not circle but every proper induced subgraph of G is circle.Since G is not circle, in particular G is not a permutation graph. Permutationgraphs are exactly those comparability graphs whose complement graph is also acomparability graph [8]. Comparability graphs were characterized by forbiddeninduced subgraphs in [9]. This characterization of comparability graphs leadsto a forbidden induced subgraph characterization for the class of permutationgraphs. Hence, given that permutation graphs are a subclass of circle graphs(see, e.g. [12, p. 252]), in particular G is not a permutation graph. Using the listof minimal forbidden induced subgraphs for permutation graphs and the factthat G is a split graph, we conclude that G contains either a tent , a 4 -tent , a co- -tent or a net as an induced subgraph (see Figure 1). It is not difficult tosee that these four graphs are also circle graphs. Figure 1: The forbidden induced subgraphs for permutation graphs within split graphs.
In Figure 2 we define some graph families that will be central throughoutthe sequel. The odd k -suns with center are defined for each odd k ≥
3. The even k -sun is defined for each even k ≥
4. We denote by F sc the graph classconsisting of the graphs belonging to any of the families depicted in Figure 2.None of the graphs in F sc is a circle graph (see Lemma 3.43).The theorem below, which is the main result of this work, gives the charac-terization of circle graphs by minimal forbidden induced subgraphs, restrictedto split graphs. Theorem 1.1.
Let G be a split graph. Then, G is a circle graph if and only if G contains none of the graphs in F sc (depicted in Figure 2) as induced subgraph. This work is organized as follows. In Section 2, given a split graph G withsplit partition ( K, S ) and an induced subgraph H of G isomorphic to tent, 4-tentor co-4-tent, we introduce partitions of K and S according to the adjacenciesand prove that these partitions are well defined. In Section 3, we address the3 igure 2: The family F sc of graphs. The 3-sun with center is also known as tent-with-center. problem of characterizing the forbidden induced subgraphs of a circle graph thatcontains an induced tent, 4-tent, co-4-tent or net, and we finish by giving theguidelines to draw a circle model for each case. In each subsection we address acase for proof of Theorem 1.1, as explained in Subsection 3.4. In Section 4, we setout some final remarks and future challenges about structural characterizationsof circle graphs.
2. Preliminaries
Throughout this section, we define some subsets in both K and S dependingon whether G contains an induced tent, 4-tent or co-4-tent H as an inducedsubgraph. We prove that these subsets induce a partition of both K and S .In each case, we first partition the vertices in the complete set K into subsets,according to the adjacencies with the vertices of V ( T ) ∩ S , and then we partitionthe vertices of the independent set S into subsets, according to the adjacencieswith the partition defined on K . We give the full proof when G contains aninduced tent, and state which parts of K and S are nonempty when G containsa 4-tent or a co-4-tent, since the proof is very similar in these cases. For moredetails on this, see [14]. These partitions will be useful in Section 3, when wegive the proof of the characterization by forbidden induced subgraphs for splitcircle graphs. Notice that we do not consider the case in which G contains aninduced net. We explain in detail in Section 3.4 how this case can be reducedto one of the other cases.Let A = ( a ij ) be a n × m (0 , a i. and a .j the i th rowand the j th column of matrix A . From now on, we associate each row a i. withthe set of columns in which a i. has a 1. For example, the intersection of tworows a i. and a j. is the subset of columns in which both rows have a 1. Two rows4 i. and a k. are disjoint if there is no j such that a ij = a kj = 1. We say that a i. is contained in a k. if for each j such that a ij = 1 also a kj = 1. We say that a i. and a k. are nested if a i. is contained in a k. or a k. is contained in a i. . We saythat a row a i. is empty if every entry of a i. is 0, and we say that a i. is nonempty if there is at least one entry of a i. equal to 1. We say that two nonempty rows overlap if they are non-disjoint and non-nested. For every nonempty row a i. , let l i = min { j : a ij = 1 } and r i = max { j : a ij = 1 } for each i ∈ { , . . . , n } . Finally,we say that a i. and a k. start (resp. end ) in the same column if l i = l k (resp. r i = r k ), and we say a i. and a k. start (end) in different columns , otherwise.Let G be a split graph with split partition ( K, S ), n = | S | , and m = | K | .Let s , . . . , s n and v , . . . , v m be linear orderings of S and K , respectively. Let A = A ( S, K ) be the n × m matrix defined by A ( i, j ) = 1 if s i is adjacent to v j and A ( i, j ) = 0, otherwise. From now on, we associate the row (resp. column) ofthe matrix A ( S, K ) with the corresponding vertex in the independent set (resp.vertex in the complete set) of the partition. S and K for a graph containing an induced tent Let G = ( K, S ) be a split graph where K is a clique and S is an independentset. Let H be an induced subgraph of G isomorphic to a tent. Let V ( T ) = { k ,k , k , s , s , s } where k , k , k ∈ K , s , s , s ∈ S , and the neighborsof s ij in H are precisely k i and k j .We introduce sets K , K , . . . , K as follows. • For each i ∈ { , , } , let K i be the set of vertices of K whose neighbors in V ( T ) ∩ S are precisely s ( i − i and s i ( i +2) (where subindexes are modulo 6). • For each i ∈ { , , } , let K i be the set of vertices of K whose only neighborin V ( T ) ∩ S is s ( i − i +1) (where subindexes are modulo 6).See Figure 3 for a graphic idea of this. Notice that K , K and K are alwaysnonempty sets.We say a vertex v is complete to the set of vertices X if v is adjacent toevery vertex in X , and we say v is anticomplete to X if v has no neighbor in X .We say that v is adjacent to X if v has at least one neighbor in X . Notice thatcomplete to X implies adjacent to X if and only if X is nonempty. For v in S ,let N i ( v ) = N ( v ) ∩ K i . Given two vertices v and v in S , we say that v and v are nested if either N ( v ) ⊆ N ( v ) or N ( v ) ⊆ N ( v ). In particular, given i ∈ { , . . . , } , if either N i ( v ) ⊆ N i ( v ) or N i ( v ) ⊆ N i ( v ), then we say that v and v are nested in K i . Additionally, if N ( v ) ⊆ N ( v ), then we say that v is contained in v .Given a graph H , the graph G is H -free if it does not contain H as inducedsubgraph. For a family of graphs H , a graph G is H -free if G is H -free for every H ∈ H . Lemma 2.1. If G is F sc -free, then { K , K , . . . , K } is a partition of K .Proof. Every vertex of K is adjacent to precisely one or two vertices of V ( T ) ∩ S ,for if not we find either a tent ∨ K or a 3-sun with center as induced subgraphof G , a contradiction. 5 igure 3: Tent H and the split graph G according to the given extensions. Let i, j ∈ { , . . . , } and let S ij be the set of vertices of S that are adjacent tosome vertex in K i and some vertex in K j , are complete to K i +1 , K i +2 , . . . , K j − ,and are anticomplete to K j +1 , K j +2 , . . . , K i − (where subindexes are modulo 6).The following claims are necessary to prove Lemma 2.8, that states, on the onehand, which sets of { S ij } i,j ∈{ ,..., } may be nonempty, and, on the other hand,and that the sets { S ij } i,j ∈{ ,..., } indeed induce a partition of S . This showsthat the adjacencies of a vertex of S have a circular structure with respect tothe defined partition of K . Claim 2.2. If G is F sc -free, then there is no vertex v in S such that v issimultaneously adjacent to K , K and K . Moreover, there is no vertex v in S adjacent to K , K and K such that v is anticomplete to any two of K j , for j ∈ { , , } . Let v in S and let w i in K i for each i ∈ { , , } , such that v is adjacent toeach w i . Hence, { w , w , w , s , s , s , v } induce in G a 3-sun with center,a contradiction.To prove the second statement, let w i in K i such that v is adjacent to w i for every i ∈ { , , } . Suppose that v is anticomplete to K and K . Thus, wefind a 4-sun induced by the set { w , k , k , w , s , s , s , v } . If instead v isanticomplete to K and K , then we find a 4-sun induced by { k , k , w , w , s , s , s , v } , and if v is anticomplete to K and K , then a 4-sun is inducedby { k , w , w , k , s , s , s , v } . (cid:3) Claim 2.3. If G is F sc -free and v in S is adjacent to K i and K i +3 , then v iscomplete to K j , either for j ∈ { i + 1 , i + 2 } or for j ∈ { i − , i − } . Let w i in K i , w i +3 in K i +3 such that v is adjacent to w i and w i +3 . Noticethat the statement is exactly the same for i = j and for i = j + 3, so let usassume that i is even. 6f w j in K j is a non-neighbor of v for each j ∈ { i − , i + 1 } , then we find aninduced M III (3). Hence, v is complete to K j for at least one of j ∈ { i − , i +1 } .Suppose that v is complete to K i +1 . If K i +2 = ∅ , then the claim holds. Hence,suppose that K i +2 (cid:54) = ∅ and suppose w i +2 in K i +2 is a non-neighbor of v . Inparticular, since v is adjacent to w i +3 and k i +1 , then v is anticomplete to K i − by Claim 2.2. However, in this case we find M III (3) induced by { s ( i − i +3) , s ( i +3)( i − , v , k i − , k i +1 , w i +2 , w i +3 } . It follows analogously if instead v iscomplete to K i − and is not complete to K i − , for we find the same inducedsubgraphs. Notice that the proof is independent on whether K j = ∅ or not, forevery even j . (cid:3) Claim 2.4. If G is F sc -free and v in S is adjacent to K i and K i +2 , with i odd,then v is complete to K i +1 . Given the symmetry of the odd-indexed and even-indexed sets K j , we ana-lyze the case in which v is adjacent to K and K . Let w , w be the respectiveneighbors. By Claim 2.2, v is anticomplete to K . If v is nonadjacent to somevertex w in K , then the set { s , v , s , w , w , k , w } induces a 3-sun withcenter. Hence, v is complete to K . (cid:3) Claim 2.5. If G is F sc -free and v in S is adjacent to K i and K i +2 , with i even,then either v is complete to K i +1 and one of { K i − , K i +3 } , or v is complete to K j for j ∈ { i − , i − , i − } . Given the symmetry of the odd-indexed and even-indexed sets K j , we ana-lyze the case in which v is adjacent to K and K . Let w , w be the respectiveneighbors. First, notice that v is complete to either K or K , for if not wefind a 4-sun induced by { s , s , s , v , w , w , w , w } , where w and w arenon-neighbors of v in K and K , respectively. Suppose that v is complete to K . If v is not complete to K , then v is complete to K and K , for if not thereis M III (3) induced by { s , s , v , k , w , w , w j } for both j = 5 ,
6, where w (cid:96) is a non-neighbor of v in K (cid:96) , for (cid:96) ∈ { , , } . (cid:3) Remark . As a consequence of the previous claims we also proved that, if G is F sc -free, then: • For each i ∈ { , , . . . , } , the sets S i ( i − are empty, for if not, there isa vertex v in S such that v is adjacent to K , K and K (contradictingClaim 2.2). Moreover, the same holds for S i ( i − , for each i ∈ { , , } . • For each i ∈ { , , } , the sets S i ( i +2) are empty since every vertex v in S such that v is adjacent to K i and K i +2 is necessarily complete to either K i − or K i +3 (Claim 2.5). Claim 2.7. If G is F sc -free, then for each i ∈ { , , } , every vertex in S i ( i +3) ∪ S ( i +3) i is complete to K i . We will prove this claim without loss of generality for i = 1.Let v in S . By definition, v is adjacent to k and nonadjacent to k .Towards a contradiction, let w and w in K such that v is nonadjacent to7 and v is adjacent to w , and let w in K such that v is adjacent to w . Inthis case, we find F induced by the set { s , s , v , w , w , k , w , k } .Analogously, if v is in S , then F is induced by { s , s , v , w , w , k , w , k } . (cid:3) The following lemma is a straightforward consequence of Claims 2.2 to 2.7.
Lemma 2.8.
Let G = ( K, S ) be a split graph that contains an induced tent. If G is F sc -free, then all the following assertions hold: • { S ij } i,j ∈{ , ,..., } is a partition of S . • For each i ∈ { , , } , S i ( i − and S i ( i − are empty. • For each i ∈ { , , } , S i ( i − and S i ( i +2) are empty. • For each i ∈ { , , } , S i ( i +3) and S ( i +3) i are complete to K i . i \ j (cid:88) (cid:88) (cid:88) (cid:88) ∅ ∅ ∅ (cid:88) (cid:88) ∅ (cid:88) (cid:88) ∅ ∅ (cid:88) (cid:88) (cid:88) (cid:88) (cid:88) (cid:88) ∅ (cid:88) (cid:88) ∅ (cid:88) (cid:88) ∅ ∅ (cid:88) (cid:88) (cid:88) ∅ (cid:88) (cid:88) ∅ (cid:88) Table 1: The (possibly) nonempty parts of S in the tent case. The orange checkmarks denotethose S ij for which every vertex is complete to K i or K j . S and K for a graph containing an induced 4-tent Let G = ( K, S ) be a split graph where K is a clique and S is an independentset. Let H be an induced subgraph of G isomorphic to a 4-tent. Let V ( T ) = { k ,k , k , k , s , s , s } where k , k , k , k ∈ K , s , s , s ∈ S , and theneighbors of s ij in H are precisely k i and k j .We introduce sets K , K , . . . , K as follows. • Let K be the set of vertices of K whose only neighbor in V ( T ) ∩ S is s . Analogously, let K be the set of vertices of K whose only neighbor in V ( T ) ∩ S is s , and let K be the set of vertices of K whose only neighborin V ( T ) ∩ S is s . • For each i ∈ { , } , let K i be the set of vertices of K whose neighbors in V ( T ) ∩ S are precisely s ji and s ik , for i = 2, j = 1 and k = 2 or i = 4, j = 2 and k = 5. • Let K be the set of vertices of K that are anticomplete to V ( T ) ∩ S .Let i, j ∈ { , . . . , } and let S ij defined as in the previous section. We denoteby S [ ij (resp. S ij ] ) the set of vertices in S that are adjacent to K j and complete to K i , K i +1 , . . . , K j − (resp. adjacent to K i and complete to K i +1 , . . . , K j − , K j ).We denote by S [ ij ] the set of vertices in S that are complete to K i , . . . , K j .Consider those vertices in S that are complete to K , . . . , K and are adjacentto al least one vertex in both K and K . We consider these vertices divided into8wo distinct subsets: we denote by S [16 the subset that contains those verticesthat are complete to K , K , . . . , K , and S to the subset of those that areadjacent but not complete to K . Furthermore, we denote by S the verticesin S that are adjacent but not complete to K .In an analogous way as in the tent case, we obtain the following lemma fora split graph that contains an induced 4-tent. The details of the proof can befound in [14]. Lemma 2.9.
Let G = ( K, S ) be a split graph that contains an induced -tentand contains no induced tent. If G is F sc -free, then all of the following assertionshold: • { K , K , . . . , K } is a partition of K . • { S ij } i,j ∈{ , ,..., } is a partition of S . • For each i ∈ { , , , } , S i is empty. • For each i ∈ { , , } , S i is empty. • The subsets S , S and S are empty. • The following subsets coincide: S = S [13 , S = S , S = S [25 , S = S [26 , S = S , S = S [46 , S = S and S = S . i \ j (cid:88) (cid:88) (cid:88) (cid:88) (cid:88) (cid:88) ∅ (cid:88) (cid:88) (cid:88) (cid:88) (cid:88) ∅ ∅ (cid:88) (cid:88) (cid:88) (cid:88) ∅ ∅ ∅ (cid:88) (cid:88) (cid:88) ∅ ∅ ∅ ∅ (cid:88) (cid:88) (cid:88) (cid:88) (cid:88) (cid:88) (cid:88) (cid:88) Table 2: The (possibly) nonempty parts of S in the 4-tent case. The orange checkmarksdenote those sets S ij complete to either K i or K j . S and K for a graph containing an induced co-4-tent Let G = ( K, S ) be a split graph where K is a clique and S is an independentset, and suppose that G contains no induced tent or 4-tent. Let H be an inducedsubgraph of G isomorphic to a co-4-tent. Let V ( T ) = { k , k , k , s , s , s ,s } where k , k , k ∈ K , s , s , s , s in S such that the neighbors of s ij in H are precisely k i and k j and the neighbor of s i in H is precisely k i .We introduce sets K , K , . . . , K as follows. • Let K be the set of vertices of K whose only neighbors in V ( T ) ∩ S are s and s . Analogously, let K be the set of vertices of K whose onlyneighbors in V ( T ) ∩ S are s and s , and let K be the set of vertices of K whose only neighbors in V ( T ) ∩ S are s and s . Let K be the setof vertices of K whose only neighbors in V ( T ) ∩ S are s and s , K bethe set of vertices of K whose only neighbors in V ( T ) ∩ S are s and s and K be the set of vertices of K whose only neighbors in V ( T ) ∩ S are s and s . 9 Let K be the set of vertices of K whose neighbors in V ( T ) ∩ S are precisely s , s and s , and let K be the set of vertices of K whose neighbors in V ( T ) ∩ S are precisely s , s and s . Let K be the set of vertices of K whose neighbors in V ( T ) ∩ S are precisely s , s and s , and let K bethe set of vertices of K whose neighbors in V ( T ) ∩ S are precisely s , s and s . • Let K be the set of vertices of K whose only neighbor in V ( T ) ∩ S isprecisely s , and let K be the set of vertices of K whose only neighborin V ( T ) ∩ S is precisely s . Let K be the set of vertices of K whoseonly neighbor in V ( T ) ∩ S is precisely s , and let K be the set of verticesof K whose only neighbor in V ( T ) ∩ S is precisely s . • Let K be the set of vertices of K that are anticomplete to V ( T ) ∩ S . Remark . If K = ∅ , then there is a split decomposition of G . Let usconsider the subset K on the one hand, and on the other hand a vertex u (cid:54)∈ G such that u is complete to K and is anticomplete to V ( G ) \ K . Let G and G be the subgraphs induced by the vertex subsets V = V ( G ) \ S and V = { u } ∪ K ∪ S , respectively. Hence, G is the result of the split compositionof G and G with respect to K and u . The same holds if K = ∅ considering thesubgraphs induced by the vertex subsets V = V ( G ) \ S and V = { u }∪ K ∪ S ,where in this case u is complete to K and is anticomplete to V ( G ) \ K .If we consider H a minimally non-circle graph, then H is a prime graph, forif not one of the factors should be non-circle and thus H would not be minimallynon-circle [2]. Hence, in order to characterize those circle graphs that containan induced co-4-tent, we will assume without loss of generality that G is a primegraph, and therefore K (cid:54) = ∅ and K (cid:54) = ∅ .In an analogous way as in the tent case, we obtain the following lemma fora split graph that contains an induced co-4-tent. The details of the proof canbe found in [14]. Lemma 2.11.
Let G = ( K, S ) be a split graph that contains an induced co- -tent and contains no induced tent or -tent. If G is F sc -free, then all thefollowing assertions hold: • K , . . . , K are empty sets and { K , K , . . . , K } is a partition of K . • { S ij } i,j ∈{ , ,..., } is a partition of S . • For each i ∈ { , , , , , , } , S i is empty. • For each i ∈ { , , , , } , S i is empty. • For each i ∈ { , , , } , S i is empty, and S is also empty. • For each i ∈ { , , , } , S i is empty. • For each i ∈ { , , , , , } , S i is empty. • The subsets S , S and S are empty. • The following subsets coincide: S i = S [1 i for i = 3 , , ; S = S , S = S , S = S [27 , S = S , S = S [46 , S = S and S = S [85 (as the case may be, according to whether K i (cid:54) = ∅ or not, for i = 6 , , ). Since S = S [18 , we will consider these vertices as those in S that arecomplete to K and S = ∅ . Moreover, those vertices that are complete to10 , . . . , K , K and are adjacent to K will be considered as in S , thus S is the set of vertices of S that are complete to K , . . . , K and are adjacent butnot complete to K . These results are summarized in Table 3. i \ j (cid:88) (cid:88) (cid:88) (cid:88) ∅ (cid:88) (cid:88) ∅ ∅ (cid:88) (cid:88) ∅ (cid:88) (cid:88) (cid:88) ∅ ∅ ∅ (cid:88) (cid:88) (cid:88) (cid:88) ∅ ∅ ∅ ∅ ∅ (cid:88) (cid:88) (cid:88) ∅ ∅ ∅ ∅ ∅ ∅ (cid:88) ∅ ∅ ∅ ∅ ∅ ∅ ∅ ∅ (cid:88) ∅ ∅ ∅ ∅ ∅ (cid:88) (cid:88) (cid:88) (cid:88) ∅ ∅ (cid:88) (cid:88) (cid:88) (cid:88) (cid:88) (cid:88) (cid:88) Table 3: The (possibly) nonempty parts of S in the co-4-tent case. The orange checkmarksdenote those subsets S ij that are either complete to K i or K j .
3. Characterization by forbidden induced subgraphs of circle graphswithin split graphs
In this section, we give the proof of Theorem 1.1. The proof strongly relieson a characterization of 2 -nested matrices given in [14]. The 2-nested matricesare those matrices having an ordering of its columns such that the ones in eachrow appear in at most two blocks and for which there is certain color assignmentfor every row using 2 colors (for the details, see Definition 3.10).This section is organized as follows. First, we define and give the charac-terization of nested and 2-nested matrices by forbidden subconfigurations (forthe complete proof, see [14], [15]). Afterwards, in Sections 3.1 to 3.4 we provethe characterization given in Theorem 1.1, which gives the complete list of for-bidden induced subgraphs for those split graphs that are also circle. This proofis divided into four cases, depending on whether the split graph contains aninduced tent, 4-tent, co-4-tent or net.A (0 , consecutive-ones property (C1P) for the rows if thereis a permutation of its columns such that the ones in each row appear consecu-tively. We say that a matrix B is a subconfiguration of a matrix A if B equalssome submatrix of A up to permutation of rows and/or columns. Given a setof rows R of a matrix A , we say that R induces a matrix B if B is a sub-configuration of the submatrix of A given by selecting only those rows in R .Tucker characterized all the minimal forbidden subconfigurations for the C1P,later known as Tucker matrices (a graphic representation of which can be foundin [16]).
Definition 3.1.
Let A be a (0 , -matrix. We say A is nested if there is aconsecutive-ones ordering for the rows and every two rows are disjoint or nested. Definition 3.2.
A split graph G = ( K, S ) is nested if and only if A ( S, K ) is anested matrix. efinition 3.3. Let A be a (0 , -matrix. We say A is an enriched matrix ifall of the following conditions hold:1. Each row of A is either unlabeled or labeled with one of the following labels:L or R or LR. We say that a row is an LR-row (resp. L-row, R-row) if itis labeled with LR (resp. L, R). An unlabeled row is called U-row.2. Each row of A is either uncolored or colored with either blue or red.3. The only colored rows may be those labeled with L or R, and those LR-rowshaving a in every column.4. The LR-rows having a in every column are all colored with the samecolor. The 0-gem, 1-gem and 2-gem are the following enriched matrices: (cid:16) (cid:17) , (cid:18) (cid:19) , (cid:18) LR LR (cid:19) respectively. Definition 3.4.
Let A be an enriched matrix. We say that A contains a gem (resp. doubly-weak gem ) if it contains a -gem (resp. a -gem) as a subconfig-uration. We say that A contains a weak gem if it contains a -gem such that,either the first is an L-row (resp. R-row) and the second is a U-row, or the firstis an LR-row and the second is a non-LR-row. We say that a -gem is badly-colored if the entries in the column in which both rows have a are in blockscolored with the same color. Theorem 3.5 ([14, 15]) . A (0 , -matrix is nested if and only if it contains no -gem as a subconfiguration. Definition 3.6.
Let A be an enriched matrix. We say A is LR-orderable ifthere is a linear ordering Π for the columns of A such that each of the followingassertions holds: • Π is a consecutive-ones ordering for every non-LR row of A . • The ordering Π is such that the ones in every nonempty row labeled withL (resp. R) start in the first column (resp. end in the last column). • Π is a consecutive-ones ordering for the complements of the LR-rows of A .Such an ordering is called an LR-ordering . Definition 3.7.
For each row of A labeled with L or LR and having a in thefirst column of Π , we define its L-block (with respect to Π) as the maximal setof consecutive columns of Π starting from the first one on which the row has a1. R-blocks are defined on an entirely analogous way. For each unlabeled rowof A , we say its U-block (with respect to Π) is the set of columns having a in the row. The blocks of A with respect to Π are its L-blocks, its R-blocks andits U-blocks. We say an L-block (resp. R-block, U-block) is colored if there isa -color assignment for every entry of the block.An LR-ordering Π is suitable if the L-blocks of those LR-rows with exactlytwo blocks are disjoint with every R-block, the R-blocks of those LR-rows with xactly two blocks are disjoint with the L-blocks and for each LR-row the inter-section with any U-block is empty with either its L-block or its R-block. Definition 3.8.
Let A be an enriched matrix and let Π be a LR-ordering. Wedefine A ∗ as the enriched matrix that arises from A by: • Replacing each LR-row by its complement. • Adding two distinguished rows: both rows have a in every column, oneis labeled with L and the other is labeled with R. Definition 3.9. A tagged matrix is a (0 , -matrix, each of whose rows areeither uncolored or colored with blue or red, together with a set of at most twodistinguished columns. The distinguished columns will be referred to as tagcolumns . Let A be an enriched matrix. We define the tagged matrix of A asa tagged matrix, denoted by A tag , whose underlying matrix is obtained from A by adding two columns, c L and c R , such that: (1) the column c L has a if f is labeled L or LR and otherwise, (2) the column c R has a if f is labeledR or LR and otherwise, and (3) the set of distinguished columns of A tag is { c L , c R } . We denote A ∗ tag to the tagged matrix of A ∗ . By simplicity we willconsider column c L as the first and column c R as the last column of A tag and A ∗ tag . Notice that for every enriched matrix, the only colored rows are those labeledwith L or R and those empty LR-rows. Moreover, for every LR-orderable matrix,there is an ordering of the columns such that every row labeled with L (resp.R) starts in the first column (resp. ends in the last column), and thus all its 1’sappear consecutively. Thus, if an enriched matrix is also LR-orderable, then thegiven coloring induces a partial block bi-coloring, in which every empty LR-rowremains the same, whereas for every nonempty colored labeled row, we color allits 1’s with the color given in the definition of the matrix.
Definition 3.10.
Let A be an enriched matrix. We say A is if thereexists an LR-ordering Π of the columns and an assignment of colors red or blueto the blocks of A such that all of the following conditions hold:1. If an LR-row has an L-block and an R-block, then they are colored withdistinct colors.2. For each colored row r in A , any of its blocks is colored with the samecolor as r in A .3. If an L-block of an LR-row is properly contained in the L-block of an L-row,then both blocks are colored with different colors.4. Every L-block of an LR-row and any R-block are disjoint. The same holdsfor an R-block of an LR-row and any L-block.5. If an L-block and an R-block are not disjoint, then they are colored withdistinct colors.6. Each two U-blocks colored with the same color are either disjoint or nested.7. If an L-block and a U-block are colored with the same color, then eitherthey are disjoint or the U-block is contained in the L-block. The sameholds replacing L-block for R-block. . If two distinct L-blocks of non-LR-rows are colored with distinct colors,then every LR-row has an L-block. The same holds replacing L-block forR-block.9. If two LR-rows overlap, then the L-block of one and the R-block of theother are colored with the same color.An assignment of colors red and blue to the blocks of A that satisfies all theseproperties is called a (total) block bi-coloring . These matrices admit the following characterization by forbidden subconfig-urations.
Theorem 3.11 ([14, 15]) . Let A be an enriched matrix. Then, A is -nestedif and only if A contains none of the following listed matrices or their dualmatrices as subconfigurations: • M , M II (4) , M V or S ( k ) for every even k ≥ (See Figure 4) • Every enriched matrix in the family D (See Figure 5) • Every enriched matrix in the family F (See Figure 6) • Every enriched matrix in the family S (See Figure 7) • Every enriched matrix in the family P (See Figure 8) • Monochromatic gems, monochromatic weak gems, badly-colored doubly-weak gemsand A ∗ contains no Tucker matrices and none of the enriched matrices in M or their dual matrices as subconfigurations. (See Figure 9). M = M II (4) = M V = S ( k ) = ... ... ... ............... ... ... Figure 4: The matrices M , M II (4), M V and S ( k ) ∈ { , } (( k +1) × k for any even k ≥ D . = (cid:32) (cid:33) F ( k ) = ... ... ... ... ............... ... F ( k ) = ... ... ... ............... ... F (cid:48) = L (LR) F (cid:48)(cid:48) = L R F (cid:48) ( k ) = . . . L (LR) . . . . . . . . . ... L (LR) . . . F (cid:48) ( k ) = . . . L (LR) . . . . . . ... . . . Figure 6: The enriched matrices of the family F . The matrices F represented in Figure 6 are defined as follows: F ( k ) ∈{ , } k × ( k − , F ( k ) ∈ { , } k × k , F (cid:48) ( k ) ∈ { , } k × ( k − and F (cid:48) ( k ) ∈ { , } k × ( k − ,for every odd k ≥
5. In the case of F (cid:48) , F (cid:48) ( k ) and F (cid:48) ( k ), the labeled rows maybe either L or LR indistinctly, and in the case of their dual matrices, the labeledrows may be either R or LR indistinctly.The matrices S in Figure 7 are defined as follows. If k is odd, then S ( k ) ∈{ , } ( k +1) × k for k ≥
3, and if k is even, then S ( k ) ∈ { , } k × ( k − for k ≥
4. The remaining matrices have the same size whether k is even or odd: S ( k ) ∈ { , } k × ( k − for k ≥ S ( k ) ∈ { , } k × ( k − for k ≥ S ( k ) ∈{ , } k × ( k − for k ≥ S ( k ) ∈ { , } k × ( k − , S ( k ) ∈ { , } k × k for k ≥ S ( k ) ∈ { , } k × ( k +1) for every k ≥ S (2 j ) ∈ { , } j × (2 j ) for j ≥
2. If k is even, then the first and last row of S ( k ) and S ( k ) are colored with the samecolor, and in S ( k ) and S ( k ) are colored with distinct colors.15 (2 j ) = L . . . . . . ... . . . LR . . . L . . . S (2 j + 1) = L . . . . . . ... . . . LR . . . S ( k ) = L . . . . . . ... . . . L . . . •• S ( k ) = L . . . . . . ... . . . R . . . •• S ( k ) = LR . . . L . . . . . . ... . . . R . . . •• S ( k ) = L . . . . . . ... . . . LR . . . L . . . •• S (3) = LR R S (cid:48) (3) = LR R S ( k ) = LR . . . R . . . . . . ... . . . • S (3) = LR LR S (2 j ) = LR . . . LR . . . . . . ... . . . S (2 j ) = LR . . . . . . ... . . . Figure 7: The family of matrices S for every j ≥ k ≥ In the matrices P , the integer l represents the number of unlabeled rowsbetween the first row and the first LR-row. The matrices P described in Fig-ure 8 are defined as follow: P ( k, ∈ { , } k × k for every k ≥ P ( k, l ) ∈{ , } k × ( k − for every k ≥ l > P ( k, ∈ { , } k × ( k − for every k ≥ P ( k, l ) ∈ { , } k × ( k − for every k ≥ l > P ( k, ∈ { , } k × ( k − for every k ≥ P ( k, l ) ∈ { , } k × ( k − for every k ≥ l >
0. If k is even,then the first and last row of every matrix in P are colored with distinct colors.16 ( k,
0) = L . . . LR . . . . . . ... . . . R . . . •• P ( k, l ) = L . . . . . . . . . . . . ... . . . . . . LR . . . . . . . . . . . . ... . . . . . . R . . . . . . •• P ( k,
0) = L . . . LR . . . LR . . . . . . ... . . . R . . . •• P ( k, l ) = L . . . . . . . . . . . . ... . . . . . . LR . . . . . . LR . . . . . . . . . . . . ... . . . . . . R . . . . . . •• P ( k,
0) = L . . . LR . . . LR . . . LR . . . LR . . . . . . ... . . . R . . . •• P ( k, l ) = L . . . . . . . . . . . . ... . . . . . . LR . . . . . . LR . . . . . . LR . . . . . . LR . . . . . . . . . . . . ... . . . . . . R . . . . . . •• Figure 8: The family of enriched matrices P for every odd k . The following lemmas, theorem and definition will be useful in the sequel.
Lemma 3.12 ([14, 15]) . Let A be an enriched matrix. Then, A is if A with its partial block bi-coloring contains none of the matrices listed inTheorem 3.11 and the given partial block bi-coloring of A can be extended to atotal block bi-coloring of A . Lemma 3.13 ([14, 15]) . Let A be an enriched matrix. If A with its partialblock bi-coloring contains none of the matrices listed in Theorem 3.11 and B isobtained from A by extending its partial coloring to a total block bi-coloring, then B is -nested if and only if for each LR-row its L-block and R-block are coloredwith distinct colors and B contains no monochromatic gems, monochromaticweak gems or badly-colored doubly-weak gems as subconfigurations. (cid:48) ( k ) = . . . L . . . . . . ... . . . L . . . M (cid:48)(cid:48) ( k ) = R . . . L . . . . . . ... . . . R . . . L . . . M (cid:48) ( k ) = L . . . . . . ... . . . . . . M (cid:48)(cid:48) ( k ) = . . . . . . ... . . . R . . . M (cid:48) = L M (cid:48)(cid:48) = L R M (cid:48) = R M (cid:48)(cid:48) = L L Figure 9: The enriched matrices in family M : M (cid:48) ( k ), M (cid:48) ( k ), M (cid:48)(cid:48) ( k ), M (cid:48)(cid:48)(cid:48) ( k ) for k ≥
4, and M (cid:48)(cid:48) ( k ) for k ≥ Theorem 3.14 ([14, 15]) . If A is admissible, LR-orderable and contains no M , M II (4) , M V or S ( k ) for every even k ≥ , then there is at least one suitableLR-ordering. Definition 3.15.
An enriched matrix A is admissible if A is {D , S , P} -free. The proof of Theorem 1.1 is organized as follows. In each of the followingsections we consider a split graph G that contains an induced subgraph H ,where H is either a tent, a 4-tent, a co-4-tent or a net (Sections 3.1, 3.2, 3.3and 3.4, respectively), and each of these is a case of the proof of Theorem 1.1.Using the partitions of K and S described in the previous section, we defineone enriched (0 , K i of K and four auxiliary non-enriched (0 , G . At the end ofeach section, we prove that G is circle if and only if these enriched matrices are2-nested and the four non-enriched matrices are nested, giving the guidelinesfor a circle model in each case. From now on, we will refer indistinctly to arow r (resp. a column c ) of a matrix and the vertex in the independent (resp.complete) set of the split partition of G whose adjacency is represented by therow (resp. column). In this section we address the first case of the proof of Theorem 1.1, whichis the case where G contains an induced tent. This section is subdivided asfollows. In Section 3.1.1, we use the partitions of K and S given in Section 2.118o define the matrices A i for each i = 1 , , . . . , A i for G to be a F sc -free graph and the guidelines toobtain a circle model for a F sc -free split graph G containing an induced tentgiven in Theorem 3.21. A , A , . . . , A Let G = ( K, S ) and H as in Section 2.1. For each i ∈ { , , . . . , } , let A i be an enriched (0 , s ∈ S such that s belongs to S ij or S ji for some j ∈ { , , . . . , } , and one column for each vertex k ∈ K i and such that the entry corresponding to the row s and the column k is1 if and only if s is adjacent to k in G . For each j ∈ { , , . . . , } − { i } , we markthose rows corresponding to vertices of S ji with L and those corresponding tovertices of S ij with R.Moreover, we color some of the rows of A i as follows. • If i ∈ { , , } , then we color each row corresponding to a vertex s ∈ S ij for some j ∈ { , , . . . , } − { i } with color red and each row correspondingto a vertex s ∈ S ji for some j ∈ { , , . . . , } − { i } with color blue. • If i ∈ { , , } , then we color each row corresponding to a vertex s ∈ S ij ∪ S ji for some j ∈ { , , . . . , } with color red if j = i + 1 or j = i − A = K S R · · · S R · · · S · · · S L · · · S L · · · •••• A = K S L · · · S R · · · S · · · S L · · · S L · · · S R · · · S R · · · •••••• The following results are useful in the sequel.
Claim 3.16.
Let v in S ij and v in S ik , for i, j, k ∈ { , , . . . , } such that i (cid:54) = j, k . If A i contains no D for each i ∈ { , , . . . , } , then the followingassertions hold: • If j (cid:54) = k , then v and v are nested in K i . Moreover, if j = k , then v and v are nested in both K i and K j . • For each i ∈ { , , . . . , } , there is a vertex v ∗ i in K i such that for every j ∈ { , , . . . , } − { i } and every s in S ij , the vertex s is adjacent to v ∗ i . Let v , v in S ij , for some i, j ∈ { , . . . , } . Towards a contradiction, supposewithout loss of generality that v and v are not nested in K i . Since v and v are both adjacent to at least one vertex in K i , then there are vertices w , w in K i such that w is adjacent to v and nonadjacent to v , and w is adjacentto v and nonadjacent to v . Moreover, since v and v lie in S ij and i (cid:54) = j , itfollows from the definition of A i that the corresponding rows are labeled withthe same letter and colored with the same color. Therefore, we find D inducedby the rows corresponding to v and v , and the columns w and w , which19 a) A (b) A Figure 10: Sketch model of G with some of the chords associated to rows in A and A . results in a contradiction. The proof is analogous by symmetry for K j and if j (cid:54) = k . Moreover, the second statement of the claim follows from the previousargument and the fact that there is a C1P for the columns of A i . (cid:3) In this section, we will use the matrix theory developed in [14, 15] to charac-terize the forbidden induced subgraphs that arise in a split graph that containsan induced tent when this graph is not a circle graph. We will start by provingthat, given a split graph G that contains an induced tent, if G is F sc -free, then A i is 2-nested for each i = 1 , , . . . , Lemma 3.17. If A i is not -nested, for some i ∈ { , . . . , } , then G containsan induced subgraph of the families depicted in Figure 2.Proof. Based on the symmetry of the subsets K i of K , it suffices to see whathappens when i = 3 or i = 4, since the proof depends solely on the parity of i .The proof is organized as follows. First, we assume that A i is not admis-sible, thus A contains one of the forbidden subconfigurations in D , S or P .Once we reach a contradiction, we will assume that A ∗ i contains either a Tuckermatrix or one of the forbidden subconfigurations in M , once again reaching acontradiction. The next step is to assume that A i contains no monochroma-tic gems, monochromatic weak-gems nor badly colored doubly-weak-gems, andfinally that A i is not 2-nested.Notice that, if G is F sc -free, then each A i ( i = 1 , . . . ,
6) contains no M (3-sun with center), M II (4), M V or S ( k ) for every even k ≥ k -sun), sincethese matrices are matrices A ( S, K ) of graphs in F sc . Hence, we assume that A i contains none of the matrices depicted in Figure 4.20 ase (1 ) Suppose first that A i is not admissible. Since A i contains no LR-rows,then A i contains either D , D , D or S ( k ), S ( k ) for some k ≥ Case (1.1)
Suppose A i contains D . Let v and v in S be the vertices whoseadjacency is represented by the first and second row of D , respectively, andlet k i and k i in K i be the vertices whose adjacency is represented by the firstand second column of D , respectively. Both rows of D are labeled with thesame letter and the coloring given to each row is indistinct. We assume withoutloss of generality that both rows are labeled with L, due to the symmetry of theproblem. Case (1.1.1)
Suppose first that i = 3 . In this case, v and v lie in S or S . By Claim 3.16 there is a vertex k in K (resp. k in K ) adjacent to everyvertex in S (resp. S ). Thus, if both v and v lie in S , then we find a M III (3) induced by { k , k , k , v , v , s , k } . If instead both v and v liein S , then we find a 3-sun with center induced by { k , k , k , v , v , s , s } .Suppose that v in S and v in S . Let k in K adjacent to v , let k in K adjacent to v and let k be any vertex in K . Thus, v and v arenonadjacent to k and v is nonadjacent to k . Hence, we find a 4-sun inducedby the set { s , s , v , v , k , k , k , k } . Case (1.1.2)
Suppose now that i = 4 . Thus, the vertices v and v belongto either S , S or S . Suppose v in S and v in S , and let k in K and k in K such that v is adjacent to k . Since v is complete to K , then v is adjacent to k , and both v and v are nonadjacent to k . Hence, we find M II (4) induced by { s , s , v , v , k , k , k , k } . The same holds if v liesin S .If instead v and v lie in S , then we find a M III (3) induced by the set { k , k , k , v , v , s , k } .Finally, if v and v lie in S ∪ S , then we find a tent ∨ K induced by { k , k , k , k , v , v , s } , where k in K is adjacent to v and v . Case (1.2)
Suppose A i contains D . Both rows of D are labeled withdistinct letters and are colored with the same color. Let v and v in S bethe vertices whose adjacency is represented by the first and second row of D ,respectively, and let k i in K i be the vertex whose adjacency is represented bythe column of D . We assume without loss of generality that v is labeled withL and v is labeled with R. It follows from the definition of A i that, if i is odd,then there are no two rows labeled with distinct letters and colored with thesame color, thus we assume that i is even and hence i = 4.In this case, either v in S and v in S , or v in S ∪ S and v in S ∪ S .If v in S and v in S , then we find a 4-sun induced by { v , v , s , s , k , k , k , k } , where k in K is adjacent to v and nonadjacent to v , k in K is adjacent to both v and v , k in K is adjacent to v and nonadjacent to v , and k in K is nonadjacent to both v and v .Suppose that v lies in S and v lies in S . We find a tent ∨ K inducedby { v , v , s , k , k , k , k } , where k , k , k and k are vertices analogousas those described in the previous paragraph. Analogously, we find the same21orbidden induced subgraph in G if v in S or v in S . Case (1.3)
Suppose D in A i . Let v and v in S be the vertices whoseadjacency is represented by the first and second row of D , respectively, and let k i and k i in K i be the vertices whose adjacency is represented by the first andsecond column of D , respectively. Both rows of D are labeled with distinctletters and colored with distinct colors, for the “same color” case is covered sincewe proved that there is no D as a submatrix of A i . We assume without loss ofgenerality that v is labeled with L and v is labeled with R. Case (1.3.1)
Suppose that i = 4 . Thus, v in S and v in S ∪ S . Wefind a 3-sun with center induced by { v , v , s , k , k , k , k } , where k in K is adjacent to v and nonadjacent to v and k in K is adjacent to v andnonadjacent to v . We find the same forbidden induced subgraph if v in S or S . Case (1.3.2)
Suppose that i = 3 . In this case, v in S ∪ S , and v in S ∪ S .Suppose first that K (cid:54) = ∅ . If v in S and v in S , then we find M II (4)induced by { v , v , s , s , k , k , k , k } . If instead v in S , then wefind M II (4) induced by the same subset of vertices with the exception of k ,considering an analogous vertex k in K . Moreover, the same forbidden inducedsubgraph can be found if v in S , as long as K (cid:54) = ∅ .If instead K = ∅ , then necessarily v in S . If v in S , then we find a3-sun with center induced by the subset { v , v , s , k , k , k , k } . If v in S , then we find M III (4) induced by { v , v , s , s , k , k , k , k , k } . Case (1.4)
Suppose S ( j ) is a subconfiguration of A i for some j ≥ . Let v , v , . . . , v j be the vertices in S represented by the rows of S ( j ) and k i , k i , . . . , k i ( j − be the vertices in K i that represent columns 1 to j − S ( j ).Notice that v and v j are labeled with the same letter, and depending on whether j is odd or even, then v and v j are colored with distinct colors or with the samecolor, respectively. We assume without loss of generality that v and v j are bothlabeled with L. Case (1.4.1)
Suppose j is odd. If i = 3, then there are no vertices v and v j labeled with the same letter and colored with distinct colors as in S ( j ). Thus,let i = 4. In this case, v in S and v j in S ∪ S . Let k in K be a vertexadjacent to both v and v j , and let k in K adjacent to v j . Thus, we find F ( j + 2) induced by { s , s , v , . . . , v j , k , k , k i , . . . , k i ( j − } . Case (1.4.2)
Suppose j is even. We split this in two cases, depending on theparity of i . If i = 3, then v and v j lie in S ∪ S . Suppose that v in S and v j in S . Let k in K adjacent to v and v j . Hence, we find F ( j + 2) inducedby the subset { v , . . . , v j , k , k i , . . . , k i ( j − , s } . The same holds if both v and v j lie in S . If instead v and v j both lie in S , then we find F ( j + 2)induced by the same subset but replacing k for a vertex k in K adjacent toboth v and v j .Suppose now that i = 4. In this case, v and v j lie in S ∪ S . In eithercase, there is a vertex k in K that is adjacent to both v and v j . We find F ( j + 1) induced by { v , . . . , v j , k , k i , . . . , k i ( j − , s } . Case (1.5)
Suppose S ( j ) is a subconfiguration of A i for some j ≥ . Let22 , v , . . . , v j be the vertices represented by the rows of S ( j ) and k i , . . . , k i ( j − be the vertices represented by columns 1 to j − S ( j ). Notice that v and v j are labeled with distinct letters, and depending on whether j is odd or even, v and v j are either colored with distinct colors or with the same color, respectively.We assume without loss of generality that v is labeled with L and v j is labeledwith R. Case (1.5.1)
Suppose j is odd. If i = 3, then v lies in S ∪ S , and v j liesin S ∪ S . If v lies in S and v j lies in S , then we find F ( j + 2) inducedby { v , . . . , v j , k , k , k i , . . . , k i ( j − , s , s } . If v lies in S and v j lies in S , then we find F ( j + 2) induced by { v , . . . , v j , k , k , k i , . . . , k i ( j − , s , s } . If instead v lies in S and v j lies in S , then we find F ( j + 2) inducedby { v , . . . , v j , k , k , k i , . . . , k i ( j − , s , s } .If instead i = 4, then v in S and v j in S ∪ S . In either case, we find a( j + 1)-sun induced by { v , . . . , v j , k i , . . . , k i ( j − , k , k , s } . Case (1.5.2)
Suppose j is even. If i = 3, then there no two rows in A labeledwith distinct letters and colored with the same color. Hence, assume i = 4 andthus either v in S and v j in S , or v in S ∪ S and v j in S ∪ S .If v in S and v j in S , then we find a ( j + 2)-sun induced by { v , . . . , v j , k , k , k , k i , . . . , k i ( j − , s , s } , where k in K is nonadjacent to both v and v j .If instead v in S ∪ S and v j in S ∪ S , then we find a j -sun inducedby { v , . . . , v j , k , k i , . . . , k i ( j − } .Therefore, A i is admissible for every j ∈ { , , . . . , } . Case (2 ) A i is admissible but not LR-orderable. Then A i contains a Tuckermatrix, or one of the following submatrices: M (cid:48) , M (cid:48)(cid:48) , M (cid:48) , M (cid:48)(cid:48) , M (cid:48) ( k ), M (cid:48)(cid:48) ( k ), M (cid:48) ( k ), M (cid:48)(cid:48) ( k ) or their corresponding dual matrices, for some k ≥
4. We assumethroughout the rest of the proof that, for each pair of vertices x and y in the S ij of S , there are vertices k i in K i and k j in K j such that both x and y areadjacent to k i and k j . This follows from Claim 3.16 and the fact that A i isadmissible.Suppose there is M I ( j ) as a submatrix of A i . Let v , . . . , v j be the verticesof S represented by rows 1 to j of M I ( k ), and let k i , . . . , k ij be the vertices in K represented by columns 1 to j . Thus, if j is even, then we find either a j -suninduced by { v , . . . , v j , k i , . . . , k ij } , and if j is odd, then we find a j -sun withcenter induced by the subset { v , . . . , v j , k i , . . . , k ij , s i ( i +2) } .For any other Tucker matrix, we find the homonym forbidden induced sub-graph induced by the subset { v , . . . , v j , k i , . . . , k ij } .Suppose that A i contains either M (cid:48) , M (cid:48)(cid:48) , M (cid:48) , M (cid:48)(cid:48) , M (cid:48) ( k ), M (cid:48)(cid:48) ( k ), M (cid:48) ( k ), M (cid:48)(cid:48) ( k ) or one of their corresponding dual matrices, for some k ≥
4. Let M besuch a submatrix. Notice that, for any tag column c of M that denotes whichvertices are labeled with L, there is a vertex k (cid:48) in either K i − or K i − such thatthe vertices represented by a labeled row in c are adjacent to k (cid:48) in G . If insteadthe tag column c denotes which vertices are labeled with R, then we find ananalogous vertex k (cid:48)(cid:48) in either K i +1 or K i +2 .Depending on whether there is one or two tag columns in M , we find the23omonym forbidden induced subgraph induced by the vertices in S and K rep-resented by the rows and the non-tagged columns of M plus one or two vertices k (cid:48) and k (cid:48)(cid:48) as those described in the previous paragraph. Case (3 ) A i is LR-orderable but contains some gem. Moreover, since A i contains no LR-rows, then A i contains either a monochromatic gem or a mono-chromatic weak gem.Let v and v in S be the vertices represented by the rows of the monochro-matic gem. Notice that both rows are labeled rows, since every unlabeled rowin A i is uncolored. It follows that a monochromatic gem or a monochromaticweak gem may be induced only by two rows labeled with L or R, and henceboth are the same case. Case (3.1) If i = 3, since both vertices are colored with the same color, then v in S and v in S . In that case, we find D in A i since both rows arelabeled with the same letter, which results in a contradiction for we assumedthat A i is admissible. The same holds if both vertices belong to either S or S . Case (3.2)
If instead i = 4, then we have three possibilities. Either v in S and v in S , or v in S and v in S , or v in S and v in S . The firstcase is analogous to the i = 3 case stated above. For the second and third case,since both rows are labeled with distinct letters, then we find D as a submatrixof A i . This results once more in a contradiction, for A i is admissible.Therefore, A i contains no monochromatic gems, monochromatic weak gemsor badly-colored doubly-weak gems. Case (4 ) A i contains none of the matrices listed in Theorem 3.11, but A i is not -nested. It follows from Lemma 3.12 that, for every suitable LR-ordering and2-color assignment of all the blocks of A i that extends the given pre-coloring,we find either a monochromatic gem or a monochromatic weak gem. Moreover,at least one of the rows of such a gem is unlabeled, for there are no LR-rowsin A i . Consider the columns of the matrix A i ordered according to a suitableLR-ordering. Suppose there is a monochromatic gem given by the rows v j and v j +1 , and suppose that both rows are colored with red. If is not possible to colorthese two rows with distinct colors, then there is at least one more row v j − colored with blue and forces v j to be colored with red. If v j − is unlabeled, then v j − and v j overlap. If v j − is labeled with L or R, then v j and v j − induce aweak gem.If v j − forces the coloring only on v j , let v j +2 be a distinct row that forces v j +1 to be colored with red. Suppose first that v j +2 forces the coloring only tothe row v j +1 . Hence, there is a submatrix as the following in A i : v j − v j v j +1 v j +2 •••• If there are no other rows forcing the coloring of v j − and v j +2 , then thissubmatrix can be colored blue-red-blue-red. Since this is not possible, there isa sequence of overlapping rows v l , . . . , v j − and v j +3 , . . . , v k such that each row24orces the coloring of the next one, and this sequence includes v j − , v j , v j +1 and v j +2 . Moreover, suppose that this is the longest sequence of vertices with thisproperty. Hence, v l and v k are labeled rows, for if not we could color againthe rows and thus extending the pre-coloring, which results in a contradiction.However, we find either S ( k − l + 1) or S ( k − l + 1) in A i , and this also resultsin a contradiction, for A i is admissible.Suppose now that v j − forces the red color on both v j and v j +1 . If v j − is unlabeled, then v j − overlaps with both v j and v j +1 . Since v j and v j +1 overlap, either v j [ r j ] = v j +1 [ r j ] = 1 or v j [ l j ] = v j +1 [ l j ] = 1. Suppose withoutloss of generality that v j [ r j ] = v j +1 [ r j ] = 1. Since v j − overlaps with v j , theneither v j − [ l j ] = 1 or v j − [ r j ] = 1, and the same holds for v j − [ l j +1 ] = 1 or v j − [ r j +1 ] = 1. If v j − [ l j ] = 1, then v j − [ l j +1 ] = 1 and v j [ l j +1 ] = 1, and thuswe find F induced by { v j − , v j , v j +1 , l j − , l j +1 − l j +1 , r j , r j + 1 } , which is acontradiction. Analogously, if v j − [ r j ] = 1, then v j − [ l j +1 ] = 1 and v j − [ l j ] = 1,and thus we find F induced by { v j − , v j , v j +1 , l j , l j +1 , r j , r j + 1, r j − } .It follows analogously if v j − is labeled with L or R, except that we find F (cid:48) instead of F as a subconfiguration in A i . Moreover, the proof is analogous if v j and v j +1 induce a weak-gem instead of a gem.Therefore, we reached a contradiction in every case and thus A i is 2-nested.Let G = ( K, S ) and H as in Section 2.1, and the matrices A i for each i = 1 , , . . . , A i is 2-nested for each i = 1 , , . . . ,
6. Hence, there is a suitable LR-ordering Π i for each i = 1 , , . . . , χ i of the given block bi-coloring. Since A i containsno LR-rows, then every row in each matrix A i is colored with either red or blue.Let Π be the ordering of the vertices of K given by concatenating the LR-orderings Π , Π , . . . , Π . Let A = A ( S, K ) and consider the columns of A ordered according to Π. Given s in S ij , we denote throughout the following by s i the row corresponding to s in A i . For each vertex s in S ij , if i ≤ j , then s i in A i and s j in A j are colored with the same color. Thus, we consider therow corresponding to s in A colored with that color. Notice that, if i < l < j ,then s is complete to each K l . If instead i > j , then s i in A i and s j in A j arecolored with distinct colors. Moreover, the row corresponding to s in A has anL-block and an R-block. Thus, we consider its L-block colored with the samecolor assigned to s i and the R-block colored with the same color assigned to s j .Notice that the assignment of distinct colors in A i and A j makes sense, sincewe are describing vertices whose chords must have one of its endpoints drawnin the K + i portion of the circle and the other endpoint in the K − j portion of thecircle.Let s ∈ S . Hence, s lies in S ij for some i, j ∈ { , , . . . , } . Notice that, arow representing a vertex s in S ii is entirely colored with the same color. Definition 3.18.
We define the (0 , -matrix A r as the matrix obtained byconsidering only those rows representing vertices in S \ (cid:83) i =1 S ii and adding twodistinct columns c L and c R such that the entry A r ( s, k ) is defined as follows: If i < j and s i is colored with red, then the entry A r ( s, k ) has a if s is adjacent to k and a otherwise, for every k in K , and A r ( s, c R ) = A r ( s, c L ) = 0 . • If i > j and s i is colored with red, then the entry A r ( s, k ) has a if s is adjacent to k and a otherwise, for every k in K i ∪ . . . K , and A r ( s, c R ) = 1 , A r ( s, c L ) = 0 . If instead s j is colored with red, then theentry A r ( s, k ) has a if s is adjacent to k and a otherwise, for every k in K ∪ . . . K j , and A r ( s, c R ) = 0 , A r ( s, c L ) = 1 .The matrix A b is defined in an entirely analogous way, changing red for bluein the definition.We define the (0 , -matrix A r − b as the submatrix of A obtained by consid-ering only those rows corresponding to vertices s in S ij with i > j for which s i is colored with red. The matrix A b − r is defined as the submatrix of A obtainedby considering those rows corresponding to vertices s in S ij with i > j for which s i is colored with blue. Lemma 3.19.
Suppose that A i is -nested for every = 1 , , . . . , . If A r , A b , A r − b or A b − r are not nested, then G contains either tent ∨ K or F as inducedsubgraphs.Proof. Suppose first that A r is not nested. Then, there are two rows f and f in A r that induce a 0-gem. Let v in S ij and v in S lm be the verticescorresponding to such rows in G . Since A i is 2-nested for every = 1 , , . . . , A i . First we need thefollowing claim. Claim 3.20. If v and v lie in the same S ij , then v and v are nested. Suppose by simplicity that i < j . Towards a contradiction, suppose that v and v are not nested. Thus, there are vertices k i and k i in K i , and k j and k j in K j such that v is adjacent to k i , k i and k j and nonadjacent to k j ,and v is adjacent to k i , k j and k j and nonadjacent to k i . Let l ∈ { , , } such that v i is nonadjacent to K l . The existence of such index follows fromClaim 2.2, since there is no vertex of S simultaneously adjacent to K , K and K . We find F induced by { v , v , s ∗ , k i , k i , k j , k j , k l } , where s ∗ = s jl or s ∗ = s ( j +1) l depending on the parity of j . (cid:3) We assume from now on that v and v lie in distinct subsets of S . The rowsin A r represent vertices in the following subsets of S : S , S , S , S , S , S , S , S , S , S , S or S . Notice that S = S [36 , S = S . Case (1 ) v in S ∪ S . Suppose that v in S , thus v in S since A isadmissible. We find F induced by { v , v , s , k , k , k , k , k } . It followsanalogously if v in S , for the only possibility is v in S since S is completeto K . Case (2 ) v in S ∪ S . Since S is complete to K , S is complete to K and A is admissible, then necessarily v in S . In that case, either v in S ∪ S or v in S ∪ S . We find F induced by { v , v , s , k , k , k ,26 , k } if v in S , or { v , v , s , k , k , k , k , k } if v in S . We findtent ∨ K induced by { v , v , s , k , k , k , k } if v in S ∪ S . Case (3 ) v in S ∪ S . Since A and A are admissible, then the onlypossibility is v in S and v in S . We find F induced by { v , v , s , k , k , k , k , k } and therefore A r is nested.Let us suppose that A b is not nested. The rows in A b represent vertices inthe following subsets of S : S , S , S , S , S , S , S , S , S , S , S or S . Notice that S = S [14 , S = S [52 and S = S . Case (1 ) v in S ∪ S . Thus, v in S since A is admissible. If v in S , then we find F induced by { v , v , s , k , k , k , k , k } . It followsanalogously by symmetry if v in S . Case (2 ) v in S . Since S is complete to K , the only possibility is v in S . We find F induced by { v , v , s , k , k , k , k , k } . Case (3 ) v in S . Since A is admissible, then v in S ∪ S . We find F induced by { v , v , s , k , k , k , k , k } if v in S and by { v , v , s , k , k , k , k , k } if v in S .Suppose now that A b − r is not nested. The rows in A b − r represent verticesin the following subsets of S : S , S , S , S or S . Notice that S = S and S = S [52 . If v in S and v in S , then we find F induced by { v , v , s , k , k , k , k , k } . The proof is analogous if the vertices lie in S ∪ S .If instead v in S , then v in S . We find F induced by { v , v , s , k , k , k , k , k } and therefore A b − r is nested.Suppose that A r − b is not nested. The rows in A r − b represent vertices in S or S . If v in S and v in S , then we find F induced by { v , v , s , k , k , k , k , k } . This finishes the proof and therefore A r , A b , A b − r and A r − b are nested. Theorem 3.21.
Let G = ( K, S ) be a split graph containing an induced tent.Then, the following are equivalent:1. G is circle;2. G is F sc -free;3. A , A , . . . , A are -nested and A r , A b , A b − r and A r − b are nested.Proof. It is not hard to see that (1) ⇒ (2), and that (2) ⇒ (3) is a consequenceof the previous lemmas. We will show (3) ⇒ (1). Suppose that each of thematrices A , A , . . . , A is 2-nested, and that the matrices A r , A b , A b − r and A r − b are nested. Let Π i be a suitable LR-ordering for the columns of A i foreach i = 1 , , . . . ,
6, and let Π be the ordering obtained by concatenation of Π i for all the vertices in K . Consider the circle divided into twelve pieces as inFigure 10a. For each vertex in K i , we place a chord having one endpoint in K + i and the other endpoint in K − i , considering the endpoints of the chords in both K + i and K − i ordered according to Π i . We denote by a − i and a + i the positionon the circle of the endpoints of the chords corresponding to the first and lastvertex of K i –ordered according to Π i – respectively. We denote by s + i,i +2 the27lacement of the chord corresponding to the vertex s i,i +2 of the tent H , whichlies between a + i − and a − i , and s + i,i +2 to the placement of the chord of the vertex s i,i +2 that lies between a + i +1 and a − i +2 .Let us see how to place the chords for the vertices in every subset S ij of S . Claim 3.22.
The following assertions hold:1. If i (cid:54) = j , then all the vertices in each S ij are nested.2. If k ≤ i and j ≤ l , then every vertex in S ij is contained in every vertex of S kl .3. For each i ∈ { , , } , the vertex set N K i ( S ( i − i ) ∩ N K i ( S i ( i +1) ) is empty.Moreover, N K i ( S ij ) ∩ N K i ( S ( i +3) i ) = ∅ and N K i ( S ij ) ∩ N K i ( S ( i +2) i ) = ∅ ,for j = i + 3 , i + 4 .4. For each i ∈ { , , } , if S i ( i +3) (cid:54) = ∅ , then S ( i − i +2) (cid:54) = ∅ , and viceversa. Every statement follows directly from Claim 3.16 and the proof of Lemma 3.19. (cid:3)
Let us consider the subsets S ij such that i (cid:54) = j and i = 1 or j = 1. Thesesubsets are S , S , S , S , S and S j . It follows from Claim 3.22 that everyvertex in S is contained in every vertex of S , which is in turn contained in S , and the same holds for S , S and S . Furthermore, it follows that S (cid:54) = ∅ only if S = ∅ and that every vertex in S is contained in every vertexin S . Now consider those subsets S ij for which i (cid:54) = j and either i = 2 or j = 2,namely S , S , S , S , S and S . It follows from the previous claim thata vertex in S is disjoint with any vertex in S . Moreover, every vertex in S is nested in every vertex in S and every vertex in S is nested in every vertexin S . The same analysis follows by symmetry for those subsets such that i (cid:54) = j and either i ∈ { , , , } or j ∈ { , , , } . Furthermore, since A i is nested forevery i ∈ { , . . . , } , then every vertex in S ii is either contained or disjoint withevery vertex in S jk , for every j, k ∈ { , . . . , } such that j ≤ i ≤ k . Therefore,Claim 3.22 ensures that we can place the chords corresponding to vertices of S in such a way that they do not intersect. We consider the vertices in each S ij ordered by inclusion. The guidelines for each subset S ij are as follows. • For those vertices in S i ( i +1) : if i = 1 , ,
5, then we place one endpointin K − i and the other endpoint in K − i +1 . If i = 3 ,
4, then we place oneendpoint in K + i and the other endpoint in K + i +1 . If i = 6, then we placeone endpoint in K − and the other endpoint in K +1 . • For the vertices in S ( i − i +1) : if i = 2, then we place one endpoint in K − and the other endpoint in K − . If i = 4, then we place one endpoint in K +3 and the other endpoint in K +5 . If i = 6, then we place one endpointin K − and the other endpoint in K +1 . • For S ( i − i +2) : if i = 1, we place one endpoint in K +6 , and the otherendpoint between s − and the chord corresponding to a − in K − . If i = 2,we place one endpoint between the chord corresponding to a +6 in K +6 and s +13 , and the other endpoint in K − . If i = 3, we place one endpoint in K +2 ,and the other endpoint between s − and the chord corresponding to a − in K +6 . If i = 4, we place one endpoint between the chord corresponding28o a +2 in K +2 and s +35 , and the other endpoint in K +6 . If i = 5, we placeone endpoint in K − , and the other endpoint between s − and the chordcorresponding to a − in K +2 . If i = 6, we place one endpoint between thechord corresponding to a +4 in K − and s +51 , and the other endpoint in K +2 . • For S i − ,i +2 : if i = 2, we place one endpoint in K +6 and the other endpointin K − . If i = 4, we place one endpoint in K +2 and the other endpoint in K +6 . If i = 6, we place one endpoint in K − and the other endpoint in K +2 .This gives a circle model for the given split graph G . In this section we address the second case of the proof of Theorem 1.1, whichis when G contains an induced 4-tent and contains no induced tent. The maindifference with the previous case is that one of the enriched matrices that wewill define may contain LR-rows. This section is subdivided as follows. InSection 3.2.1, we define the matrices B i for each i = 1 , , . . . ,
6. In Section 3.2.2,we prove the necessity of the 2-nestedness of each B i for G to be a F sc -graph andgive the guidelines to draw a circle model for a F sc -free split graph G containingan induced 4-tent in Theorem 3.35. B , B , . . . , B Let G = ( K, S ) and H as in Section 2.2. For each i ∈ { , , . . . , } , let B i be an enriched (0 , s ∈ S such that s belongs to S ij or S ji for some j ∈ { , , . . . , } and one column for each vertex k ∈ K i and such that such that the entry corresponding to row s and column k is 1 if and only if s is adjacent to k in G . For each j ∈ { , , . . . , } − { i } , welabel those rows corresponding to vertices of S ji with L and those correspondingto vertices of S ij with R, with the exception of S [15] and S [16 , whose verticesare labeled with LR. As in the previous section, some of the rows of B i arecolored. Given the symmetry of the partitions of K , it suffices to study B i for i = 1 , , , B = K S R · · · S · · · S R · · · S R · · · S R · · · S L · · · ••••• B = K S L · · · S · · · S R · · · S R · · · ••• B = K S R · · · S R · · · S L · · · S · · · S R · · · S L · · · ••••• B = K S R · · · S R · · · S R · · · S L · · · S L · · · S · · · S R · · · S R · · · S L · · · S L · · · S L · · · S [15] LR · · · S [16 LR · · · •••••••••• Figure 11: The matrices B , B , B and B . Since S , S , S and S are complete to K , then they are not consideredfor the definition of the matrix B . The same holds for S with regard to B and S with regard to B .Notice that we consider S and S [16 as two distinct29ubsets of S . Moreover, every vertex in S [16 is labeled with LR and every vertexin S is labeled with L. We consider S as the subset of vertices in S that arecomplete to K , . . . , K , are adjacent to K and K but are not complete to K . Furthermore, any vertex in S [15] is represented by an empty LR-row in B .It follows from the definition of enriched matrix that every row correspondingto a vertex in S [15] in B must be colored with the same color. We give moredetails on this further on in Section 3.2.2. Remark . Claim 3.16 holds for any two vertices v in S ij and v in S ik ,for every B i , i ∈ { , . . . , } . and the proof is analogous as in the tent case. In this section, we state a result analogous to Lemma 3.17. The matrices B i contain no LR-rows, for each i ∈ { , . . . , } , hence the proof is very similar tothe one given in Section 3.1.2 for the tent case. We leave out the details of theproof, which can be consulted in [14]. Afterwards, we state and prove a lemmaanalogous to Lemma 3.24 but for the matrix B . The main difference betweenthis matrix and the matrices B i for each i = 1 , , . . . , B containsLR-rows Lemma 3.24 ([14]) . If B i is not -nested, for some i ∈ { , . . . , } , then G contains one of the forbidden induced subgraphs in Figure 2. Now we will focus on the matrix B . First, we define how to color those rowsthat correspond to vertices in S [15] , since we defined B as an enriched matrixand these are the only empty LR-rows in B . Remember that all the emptyLR-rows must be colored with the same color. Hence, if there is at least one redrow labeled with L or one blue row labeled with R (resp. blue row labeled withL or red row labeled with R), then we color every LR-row in S [15] with blue(resp. with red). This gives a 1-color assignment to each empty LR-row only if G is F sc -free. Lemma 3.25.
Let G be a split graph that contains an induced -tent and suchthat G contains no induced tent, and let B as defined in the previous section.If S [15] (cid:54) = ∅ and one of the following holds: • There is at least one red row f and one blue row f , both labeled with L(resp. R) • There is at least one row f labeled with L and one row f labeled with R,both colored with red (resp. blue).Then, we find either F (5) , -sun with center or -sun as an induced subgraphof G .Proof. We assume that B contains no D , for we will prove this in Lemma 3.26.Let v , v and w be three vertices corresponding to rows of B , v is coloredwith red and labeled with L, v is colored with blue and labeled with L and w in S [15] . Thus, v in S ∪ S and v in S ∪ S ∪ S . In either case, we find F (5) induced by { k , k , k , k , v , v , w , s , s } or { k , k , k , k , v , v , w , s , s } , depending on whether v in S or in S ∪ S , respectively. Suppose30ow that v corresponds to a red row labeled with R. Thus, v in S ∪ S and v in S ∪ S ∪ S . If v in S , then there is a 4-sun induced by { k , k , k , k , v , v , s , s } . If instead v in S ∪ S , then we find a 3-sun with centerinduced by { k , k , k , k , v , v , w } . This finishes the proof since the othercases are analogous by symmetry.To prove the following lemma, we assume without loss of generality that S [15] = ∅ . Lemma 3.26.
Let G = ( K, S ) be a split graph containing an induced -tentsuch that G contains no induced tent and let B = B . If B is not -nested, then G contains one of the graphs listed in Figure 2 as an induced subgraph.Proof. We assume proven Lemma 3.24 by simplicity. This is, we assume thatthe matrices B , . . . , B are 2-nested. In particular, any pair of vertices v in S ij and v in S ik such that i (cid:54) = 6 and j (cid:54) = k are nested in K i . Moreover, thereis a vertex v ∗ i in K i adjacent to both v and v . Also notice that B containsno M , M II (4), M V , S ( k ) for even k ≥ F for wefind the homonymous forbidden induced subgraphs in each case.The structure of the proof is analogous as in Lemmas 3.17 and 3.24. Themain difference is that B may have some LR-rows and thus we also have toconsider what happens if B contains every subconfiguration with at least oneLR-row in each case. Case (1 ) Suppose that B is not admissible. Hence, B contains at least one ofthe matrices D , D , . . . , D , S ( j ) , S ( j ) , . . . , S ( j ) for some j ≥ P ( j, l ), P ( j, l ) for some l ≥ , j ≥ P ( j, l ), for some l ≥ , j ≥ Case (1.1) B contains D . Let v and v be the vertices represented by thefirst and second row of D respectively, and k , k in K represented by thefirst and second column of D , respectively. Case (1.1.1)
Suppose first that v and v are colored with the same color.Since the case is symmetric with regard of the coloring, we may assume thatboth rows are colored with red. Hence, either v and v lie in S ∪ S ∪ S ,or v and v lie in S ∪ S . If v and v lie in S and k in K is adjacentto both v and v , then we find a M III (3) induced by { k , k , k , k , v , v , s } . We find M III (3) if either v and v lie in S ∪ S changing k for some k in K adjacent to both v and v , k for some k in K nonadjacent to both v and v and s for s . We also find the same subgraph if v and v lie in S ∪ S , changing k for some k in K adjacent to both v and v and s for s . If instead v in S and v in S ∪ S , since we defined S as thosevertices adjacent but not complete to K , then there are vertices k in K and k in K such that v is nonadjacent to both, and v is adjacent to k and isnonadjacent to k . Thus, we find F (5) induced by { k , k , k , k , k , v , v , s , s , s } . Case (1.1.2)
Suppose now that v is colored with red and v is colored withblue. Hence, v lies in S ∪ S , and v lies in S ∪ S ∪ S . If v in S , thenthere is a vertex k in K nonadjacent to v and v . Hence, we find M III (4)31nduced by { k , k , k , k , k , v , v , s s } . If instead v in S ∪ S , thenwe find M III (4) induced by { k , k , k , k , k , v , v , s , s } . Case (1.2) B contains D . Let v and v be the vertices that represent therows of D and let k in K be the vertex that represents the column of D .Suppose without loss of generality that both rows are colored with red, hence v in S ∪ S and v in S ∪ S ∪ S . Notice that v is complete to K since S = S [46 . If v in S , then we find a 4-sun induced by { k , k , k , k , v , v , s , s } . If v in S is not complete to K , then we find a tent induced by { k , k , k , v , v , s } . If instead v in S ∪ S is complete to K , then wefind a M II (4) induced by { k , k , k , k , v , v , s , s } . Case (1.3) B contains D . Let v and v be the first and second row of D ,and let k and k be the vertices corresponding to first and second column of D , respectively. Suppose that v is colored with blue and v is colored withred, thus v lies in S ∪ S ∪ S and v lies in S ∪ S ∪ S . If v in S and v in S , then we find a 5-sun with center induced by { k , k , k , k , k , k , v , v , s , s , s } . If instead v in S ∪ S , since v is not complete to K and we assume that B is admissible and thus contains no D , then there isa vertex k in K adjacent to v and nonadjacent to v We find a tent inducedby { k , k , k , v , v , s } . The same holds if v in S and v in S , for B is admissible and v is adjacent but not complete to K . Moreover, if v in S and v in S , then we find a tent induced by { k , k , k , v , v , s } . Finally,if v in S ∪ S and v in S ∪ S , then there are vertices k in K and k in K such that k is nonadjacent to v and adjacent to v , and k is nonadjacentto v and adjacent to v . Hence, we find F (5) induced by { k , k , k , k , v , v , s , s , s } . Remark . If G is circle and S ∪ S (cid:54) = ∅ , then S ∪ S = ∅ , and viceversa. Case (1.4) B contains D . Let v and v be the vertices corresponding to therows of D labeled with L and R, respectively, w be the vertex correspondingto the LR-row, and k , k and k in K be the vertices corresponding to thecolumns of D . An uncolored LR-row in B represents a vertex in S [16 . Noticethat there is no vertex k i in K i for some i ∈ { , . . . , } adjacent to both v and v , since w is complete to K i , thus we find M III (3) induced by { k , k , k , k i , v , v , w } .If v is colored with red and v is colored with blue, then v in S ∪ S and v in S ∪ S . We find M III (4) induced by { k , k , k , k , k , v , v , w , s } .Conversely, if v is colored with blue and v is colored with red, then v in S ∪ S ∪ S and v in S ∪ S ∪ S . It follows by symmetry that it sufficesto see what happens if v in S and v in either S or S . If v in S , thenwe find M III (6) induced by { k , k , k , k , k , k , k , v , v , s , s , s , w } . If instead v in S , then we find M III (4) induced by { k , k , k , k , k , v , v , s , w } . Case (1.5) B contains D . Let v and v be the vertices represented by therows labeled with L, w be the vertex represented by the LR-row and k in K corresponding to the column of D . Since B contains no D , suppose that v
32s colored with red and v is colored with blue. Thus, v lies in S ∪ S ∪ S and v lies in S ∪ S . In either case, we find F (5): if v in S , then it isinduced by { k , k , k , k , v , v , w , s , s } , and if v in S ∪ S , then it isinduced by { k , k , k , k , v , v , w , s , s } . Case (1.6) B contains D . Let v and v be the vertices representing therows labeled with L and R, respectively, w be the vertex corresponding to theLR-row, and k in K corresponding to the column of D . Suppose v is coloredwith blue and v is colored with red. Notice that for any two vertices x in S ij and x in S jk , we may assume that there are vertices k j and k j in K j suchthat x is adjacent to k j and is nonadjacent to k j and x is adjacent to k j and is nonadjacent to k j , since we assume B i admissible for each i ∈ { , . . . , } .It follows that, if v in S ∪ S and v in S , then there is a tent induced by { k , k , k , v , v , s } , where k is a vertex nonadjacent to v . The same holdsif v in S and v in S , where the tent is induced by { k , k , k , v , v , s } ,with k in K adjacent to v and nonadjacent to v . Finally, if v in S and v in S , then we find a 5-sun with center induced by { k , k , k , k , k , v , v , w , s , s , s } . Remark . If G is circle and contains no induced tent, then any two vertices v in S and v in S are disjoint in K . The same holds for any two vertices v in S and v in S in K . Case (1.7) B contains D . Let v and v be the vertices represented by therows labeled with L and R, respectively, w be the vertex corresponding to theLR-row, and k and k in K corresponding to the first and second columnof D , respectively. Suppose without loss of generality that v and v are bothcolored with red, thus v lies in S ∪ S and v lies in S ∪ S ∪ S . Since B i is admissible for every i ∈ { , . . . , } , then there is no vertex in K i adjacentto both v and v . It follows that, since v is complete to K , then v in S .However, we find F (5) induced by { k , k , k , k , k , v , v , w , s , s } .The following remark is a consequence of the previous statement. Remark . If G is circle, v in S ∪ S and v in S ∪ S ∪ S , then forevery vertex w in S [16 either N K ( v ) ⊆ N K ( w ) or N K ( v ) ⊆ N K ( w ). Thesame holds for v in S ∪ S ∪ S and v in S ∪ S . Case (1.8) B contains D or D . Then, there is a vertex k i in some K i with i (cid:54) = 6 such that k i is adjacent to the three vertices corresponding to every rowof D . We find a net ∨ K . Case (1.9) B contains D or D . There is a tent induced by all three rowsand columns of D or D Case (1.10) B contains D or D . It is straightforward that in this case wefind F . Case (1.11) B contains D . Let v and v be the vertices represented by therows labeled with L and R, respectively, w and w be the vertices representedby the LR-rows and k , . . . , k in K be the vertices corresponding to thecolumns of D . Suppose that v is colored with red and v is colored with blue,hence v lies in S ∪ S and v lies in S ∪ S . Let k in K adjacent to v v and let k in K adjacent to v and nonadjacent to v .We find F (5) induced by { v , v , w , w , s , k , k , k , k } . Case (1.12) B contains S ( j ) . Case (1.12.1) If j ≥ v , v , . . . , v j be the vertices represented bythe rows of S ( j ), where v and v j are labeled both with L or both with R, v j − is a vertex corresponding to the LR-row, and k , . . . , k j − in K the verticescorresponding to the columns. Suppose without loss of generality that v and v j are labeled with L. It follows that either v and v j lie in S ∪ S , or v and v j lie in S ∪ S or v lies in S ∪ S ∪ S and v j lies in S ∪ S . In eithercase, there is k in K adjacent to both v and v j and k is also adjacent to v j − since it lies in S [16 . Thus, this vertex set induces a ( j − Case (1.12.2) If j is odd, since S ( j ) has j − v , . . . , v j − vertices), then the subset of vertices given by { v , . . . , v j − , k , . . . , k j − , k } induces an even ( j − Case (1.13) B contains S ( j ). Let v and v j be the vertices correspondingto the labeled rows, k , . . . , k j − ) in K be the vertices corresponding to thecolumns of S ( j ), and suppose without loss of generality that v and v j arelabeled with R. Case (1.13.1) j is odd. Suppose first that v is colored with red and v j iscolored with blue. Thus, v in S ∪ S ∪ S and v j in S ∪ S . If v in S , then let k i in K i for i = 1 , , k is adjacent to v and v j , k is adjacent to v j and nonadjacent to v , and k is nonadjacent to both v and v j . We find F ( j + 2) induced by { k , k , k , k , . . . , k j − ), v , . . . , v j , s , s } . If v in S ∪ S , then we find F ( j ) induced by { k , k , . . . , k j − , v , . . . , v j } , with k in K adjacent to v and nonadjacent to v j . Conversely,suppose v is colored with blue and v is colored with red, thus v in S ∪ S and v j in S ∪ S ∪ S . If v j lies in S ∪ S , then we find F ( j + 2) inducedby { k , k , k , k , . . . , k j − , v , . . . , v j , s , s } , with k i in K i for i = 2 , , k is adjacent to v and v k , k is adjacent to v j and nonadjacent to v ,and k is nonadjacent to both v and v j . If instead v j in S , then it is inducedby { k , k , k , k , . . . , k j − , v , . . . , v j , s , s } . Case (1.13.2) j is even. Hence, v and v j are colored with the same color.Suppose without loss of generality that are both colored with red, and thus v and v j lie in S ∪ S ∪ S . If v and v j in S , then we find F ( j + 1) inducedby { k , k , k , . . . , k j − , v , . . . , v j , s } . We find F ( j + 1) if v and v j liein S or S , changing s for s , and k and k for k and k , where k isnonadjacent to both v and v j and k is adjacent to both. If only v lies in S ,then we find F ( j + 3) induced by { k , k , k , k , k , . . . , k j − , v , . . . , v j , s , s , s } , with k i in K i for i = 1 , , ,
5. If only v j lies in S , then we find F (5) induced by { k , k , k , k , k , v , v j , s , s , s } , with k i in K i for i = 1 , , , Case (1.14)
Suppose that B contains S ( j ). Let v and v j be the verticescorresponding to the labeled rows, k , . . . , k j − ) in K be the vertices corre-sponding to the columns of S ( j ). Case (1.14.1) j is odd. Suppose that v is labeled with L and coloredwith blue and v j is labeled with R and colored with red. In this case, v in34 ∪ S ∪ S and v j in S ∪ S ∪ S . If v in S , then we find a ( j + 3)-sunif v j in S , induced by { k , k , k , k , k , . . . , k j − , v , . . . , v j , s , s , s } .If v j in S ∪ S , then we find a ( j + 1)-sun induced by { k , k , k , . . . , k j − , v , . . . , v j , s } . Moreover, if v j in S and v in S ∪ S , then we find a ( j +1)-sun induced by { k , k , . . . , k j − , k , v , . . . , v j , s } . Finally, it follows fromRemark 3.27 that it is not possible that v j in S ∪ S and v in S ∪ S . Case (1.14.2) j is even. Suppose without loss of generality that v and v j are both colored with red. Thus, v in S ∪ S ∪ S and v j in S ∪ S .If v in S , then we find ( j + 2)-sun induced by { k , k , k , k , . . . , k j − , v , . . . , v j , s , s } . If instead v in S ∪ S , then we find j -sun induced by { k , k , . . . , k j − , v , . . . , v j } . Case (1.15) B contains S ( j ) . Let v be the vertex corresponding to the rowlabeled with LR, v corresponding to the row labeled with L, v j labeled with Rand k , . . . , k j − ) in K the vertices corresponding to the columns of S ( j ). Case (1.15.1) j is even. Hence, v and v j are colored with the same color.Suppose without loss of generality that they are both colored with red, thus v in S ∪ S and v j in S ∪ S ∪ S . If v j lies in S ∪ S , then we find a( j − { k , k , . . . , k j − , v , 2 , . . . , v j } . If instead v j in S , then we find a ( j + 1)-sun with center induced by { k , . . . , k j − , k , k , k , v , 2 , . . . , v j , s , s } . Case (1.15.2) j is odd. Suppose without loss of generality that v is coloredwith red and v j is colored with blue. Hence, v in S ∪ S and v j in S ∪ S .We find a j -sun with center induced by { k , k , . . . , k j − , k , v , 2 , . . . , v j , s } . Case (1.16) B contains S ( j ) . Let v and v j be the vertices representingthe rows labeled with L, v j − the vertex corresponding to the row labeled withLR and k , . . . , k j − ) in K be the vertices corresponding to the columns of S ( j ). Case (1.16.1) j is even. Hence v and v j lie in S ∪ S . We find F ( j + 1)induced by { k , k , k , . . . , k j − , v , . . . , v j − , v j , s } . Case (1.16.2) j is odd. Suppose v is colored with red and v j is colored withblue, thus v in S ∪ S and v j in S ∪ S ∪ S . If v j in S , then we find F ( j ) induced by { k , k , k , . . . , k j − , v , . . . , v j − , v j , s } . If instead v j lies in S ∪ S , then we find F ( j + 2) induced by { k , , k , k , k , . . . , k j − , v , . . . , v j − , v j , s , s } . Case (1.17) B contains S ( j ) . Case (1.17.1)
Suppose first that B contains S (3) or S (cid:48) (3). Let v , v and v be the vertices that represent the LR-row, the R-row and the unlabeled row,respectively. Independently on where lies v , there is vertex v in K \ K suchthat v is adjacent to v and v and is nonadjacent to v and thus we find aninduced 3-sun with center. Case (1.17.2) If B contains S ( j ) for some even j ≥
4, then we find F ( j )induced by every row and column of S ( j ). If instead j is odd, then we find M II ( j ) induced by every row and column of S ( j ) and a vertex k i in some K i with i (cid:54) = 6. We choose such a vertex k i adjacent to v , and thus since v in S [16 , v is also adjacent to k i and v , . . . , v j are nonadjacent to k i for they represent35ertices in S . Case (1.18) B contains S ( j ) . Suppose B contains S (3). It is straightfor-ward that the rows and columns induce M II (4). Furthermore, if j >
3, then j is even. The rows and columns of S ( j ) induce a j -sun. Case (1.19) B contains S (2 j ) . If j = 2, then we find a tent induced by thelast three columns and the last three rows. If instead j >
2, then we find a(2 j − K –which will be the center–, since K (cid:54) = ∅ . Case (1.20) B contains P ( j, l ) . Let v , . . . , v j in S and k , . . . , k j in K bethe vertices represented by the rows and the columns of P ( j, l ), respectively.The rows v and v j are labeled with L and R, respectively, and v l +2 is anLR-row.Suppose l = 0. If j is even, then v and v j are colored with distinct colors.Suppose without loss of generality that v is colored with red, thus v lies in S ∪ S and v j lies in S ∪ S . In that case, there are vertices k i in K i for i = 2 , k is adjacent to v j and nonadjacent to v and k is adjacentto v and nonadjacent to v j . We find F ( j + 1) induced by { k , , k , k , . . . , k j , v , . . . , v j , s } If instead j is odd, then v and v j are colored with the same color. Supposewithout loss of generality that they are both colored with red. Hence, v liesin S ∪ S and v j lies in S ∪ S ∪ S . We find F ( j + 2) induced by { k , , k , k , k , . . . , k j , v , . . . , v j , s , s } if v j lies in S , and by { k , , k , k , k , . . . , k j , v , . . . , v j , s , s } if v j lies in S ∪ S .The proof is analogous if l > Case (1.21) B contains P ( j, l ) . Let v , . . . , v j and k , . . . , k j − in K bethe vertices represented by the rows and the columns of P ( j, l ), respectively,where v and v j are labeled with L and R, respectively, and v l +2 and v l +3 areLR-rows. Case (1.21.1)
Suppose first that l = 0. If j is odd, then v and v j are coloredwith the same color. Suppose without loss of generality that they are coloredwith red, thus, v lies in S ∪ S and v j lies in S ∪ S ∪ S . In eithercase, v is anticomplete to K . Hence, we find F ( j ) induced by every row andcolumn of P ( j,
0) and an extra column that represents a vertex in K adjacentto v j , v and v and nonadjacent to v i , for 1 ≤ i ≤ j − i (cid:54) = j, ,
3. If instead j is even, then suppose v and v j are colored with red and blue, respectively.Thus, v lies in S ∪ S and v j lies in S ∪ S . We find F ( j + 1) inducedby the vertices that represent every row and column of P ( j, s and twovertices k in K and k in K such that k is adjacent to v j , v and v and isnonadjacent to v i , and k is adjacent to v , v and v and is nonadjacent to v i ,for each 1 ≤ i ≤ j − i (cid:54) = j, , Case (1.21.2)
Suppose l >
0. The proof is analogous to the previous case if j is even. If instead j is odd, then v lies in S ∪ S and v j lies in S ∪ S ∪ S .If v j in S , then we find F ( j + 2) induced by { k , k , . . . , k j − , k , k , v , . . . , v j , s , s } . If instead v j (cid:54)∈ S , then we find F ( j ) induced by thevertices corresponding to every row and column of P ( j, l ) and a vertex in K adjacent to every vertex represented by a labeled row.36 ase (1.22) B contains P ( j, l ) . Let v , . . . , v j and k , . . . , k j − in K bethe vertices represented by the rows and the columns of P ( j, l ), respectively,where v and v j are labeled with L and R, respectively, and v l +2 , v l +3 , v l +4 and v l +5 are LR-rows.Suppose l = 0. If j is even, then we find F ( j −
1) induced by { k , k , . . . , k j − , v , v , v , . . . , v j , s } . The same subgraph arises if l > j is odd, then v and v j are colored with the same color, thus we assumethat v lies in S ∪ S and v j lies in S ∪ S ∪ S . If v j (cid:54)∈ S , then we find F ( j −
2) induced by { k , k , k , . . . , k j − , v , v , v , . . . , v j , k } , where k in K is adjacent to v , v , v and v j . The same subgraph arises if l >
0. If v j in S , then there are vertices k i in K i , for i = 1 , , k is adjacentto v j and is nonadjacent to v , k is nonadjacent to both and k is adjacentto v and nonadjacent to v j . If l = 0, we find M II ( j ) induced by { k , k , k , . . . , k j − , v , v , v , . . . , v j , k , k , k , s , s } . If instead l >
0, then wefind F ( j ) induced by { k , k , k , . . . , k j − , k , k , k , v , v , v , v , . . . , v j , s , s } . Case (2 ) Suppose now that B is admissible but B ∗ contains either a Tuckermatrix, or one of the matrices in Figure 9. It suffices to see that B ∗ tag contains noTucker matrix, for in the case of the matrices listed in Figure 9, each labeled rowadmits a vertex belonging to the same subsets of K considered in the analysisfor a Tucker matrix having at least one LR-row. Towards a contradiction, let M be a Tucker matrix contained in B ∗ tag . Throughout the proof, when we referto an LR-row in M , we refer to the row in B , this is, the complement of therow that appears in M . Case (2.1)
Suppose first that M = M I ( j ), for some j ≥
3. Let v , . . . , v j and k , . . . , k j in K be the vertices corresponding to the rows and the columns of M , respectively. Remark . If two non-LR-rows in M = M I ( j ) are labeled with the sameletter, then they induce D . Moreover, any pair of consecutive non-LR-rowslabeled with distinct letters induce D or D . Since we assume B admissible,then there are at most two labeled non-LR-rows in M I ( j ) and such rows arenon-consecutive and labeled with distinct letters. Furthermore, it is easy to seethat there are at most two LR-rows in M I ( j ), for if not such rows induce D , D or D . Case (2.1.1)
Suppose M = M I (3). Suppose first that v is the only LR-rowin M .If v and v are unlabeled, then we find M III (3) induced by { v , v , v , k , k , k , k l } , where k l is any vertex in K l (cid:54) = K . The same holds if either v or v are labeled rows, by accordingly replacing k l for some l such that k l is nonadjacent to both v and v (there are no labeled rows complete to eachpartition K i (cid:54) = K of K ). By Remark 3.30, if both v and v are labeled rows,then they are labeled with distinct letters. Thus, we find F induced by { v , v , v , k , k , k , k , k } , where k in K is adjacent to v and nonadjacentto v and k in K is adjacent to v and nonadjacent to v , or viceversa. Such37ertices exist since B i is admissible for every i ∈ { , . . . , } . If instead v and v are LR-rows, then we find a tent induced by { v , v , v , k , k , k l } , considering k l in K l for some l ∈ { , . . . , } such that v is nonadjacent to k l . Every othercase is analogous by symmetry. Moreover, if v , v and v are LR-rows, thenthere is a vertex k l in K l with l (cid:54) = 6 such that v , v and v are adjacent to k l ,hence we find a M III (3) induced by { v , v , v , k , k , k , k l } . Case (2.1.2)
Suppose now that M = M I ( j ) for some j ≥
4, and let ussuppose first that there is exactly one LR-row in M and that v is the suchLR-row. Notice first that, if j is odd, then we find F ( j ) in B induced by thevertices represented by every row and column of M . Hence, suppose j is even.By Remark 3.30, there are at most two labeled rows in M and they are labeledwith distinct letters. If either there are no labeled rows or there is exactly onelabeled row, then we find M III ( j ) induced by { v , . . . , v j , k , . . . , k j , k l } , where k l is any vertex in some K l (cid:54) = K , nonadjacent to the labeled row. Suppose thereare two labeled rows v i and v l in M . It suffices to see what happens if v i belongsto S ∪ S and v l belongs to either S , S ∪ S or S ∪ S . If v l in S ,then there is a vertex k in K nonadjacent to both v i and v l , and thus we find M III ( j ) induced by the same vertex set from the previous paragraph. If instead v l in S ∪ S , then there are vertices k in K and k in K such that k isadjacent to both v l and v i . Hence, if | l − i | is even, then we find an ( l − i )-sun.If instead | l − i | is odd, then we find a ( l − i )-sun with center, where the centeris given by the LR-vertex v . Using a similar argument, if v l lies in S ∪ S ,then we find an even sun or an odd sun with center considering the same vertexset as before plus s .Suppose now that v and v are LR-rows. If j ≥ v i with i > M II ( j )induced by { v , . . . , v j , k , k , . . . , k j , k l } , where k l is any vertex in some K l (cid:54) = K such that each v i is nonadjacent to k l for every i ≥
3. Moreover, if j ≥ F ( j ) induced by { v , . . . , v j , k , k , . . . , k j } . Thesame holds if there is exactly one labeled row since we can always find a vertexin some K l with l (cid:54) = 6 that is nonadjacent to such labeled vertex, if necessary.Let us suppose there are exactly two labeled rows v i and v l . By Remark 3.30,these rows are non-consecutive and are labeled with distinct letters. As in theprevious case, v i belongs to S ∪ S and v l belongs to either S or S ∪ S .If v l belongs to S , then there is a vertex k in K nonadjacent to both v i and v l , and thus we find { v , . . . , v j , k , k , . . . , k j , k } . If instead v l lies in S ∪ S , then we find k in K adjacent to both v i and v l and hence we findeither an even sun or an odd sun with center as in the previous case. Usinga similar argument, if v l lies in S ∪ S , then we find an even sun if | l − i | is even or an odd sun with center if | l − i | is odd. Finally, suppose v and v i are LR-rows, where i >
2. If j = 4, then we find a 4-sun as an inducedsubgraph, hence, suppose that j >
5. In that case, we find a tent contained inthe subgraph induced by { v , v , v } if i = 3 and { v , v j − , v j } if i = j −
1. Let3 < i < j −
1. However, in that case we find M II ( i ) induced by { v , v , . . . , v i , k , . . . , k j − , k j } . Therefore, there is no M I ( j ) in B ∗ tag . Case (2.2)
Suppose that B ∗ tag contains M = M II ( j ). Let v , . . . , v j and38 , . . . , k j in K be the vertices corresponding to the rows and the columns of M . If j is odd and there are no labeled rows, then we find F ( j ) by considering { v , . . . , v j , k . . . , k j − } . Moreover, if there are no LR-rows and j is odd,then we find M II ( j ) as an induced subgraph. Hence, we assume from now onthat there is at least one LR-row. Remark . There are at most two rows labeled with L or R in M = M II ( k ),for any three LR-rows induce an enriched submatrix that contains either D , D or D . Moreover, since B is admissible, then there are at most three LR-rows.If v i and v l with 1 < i < l < j are two rows labeled with either L or R, thenthey are labeled with distinct letters for if not we find D . Moreover, they arenon-consecutive since we there is no D or D in B . Thus, since v i belongsto S ∪ S and v l belongs to either S or S ∪ S or S ∪ S , one of thefollowing holds: • If v l in S , then we find a ( l − i + 2)-sun if l − i is even or a ( l − i + 2)-sunwith center if | l − i | is odd (the center is k j ) induced by { v i , . . . , v l , s , s , k i +1) . . . , k l , k , k , k , k j } . • If v l in S ∪ S (resp. S ∪ S ), then we find a ( l − i )-sun if | l − i | iseven or a ( l − i )-sun with center if | l − i | is odd (the center is k j ) inducedby { v i , . . . , v l , k i +1) . . . , k l , k , k j } (resp. k , k ).Furthermore, let v and v i labeled with either L or R, where 1 < i ≤ j . If i = 2 , j ,then they are labeled with distinct letters for if not we find D . Moreover, theyare colored with distinct colors for if not we find D . If instead 2 < i < j , thenthey are labeled with the same letter for if not we find D or D .By Remark 3.31, we may assume without loss of generality that, if thereare rows labeled with either L or R, then these rows are either v j and v j − , v and v j or v j − and v j for every other case is analogous or already covered.Moreover, if v j and v j − (resp. v ) are labeled rows, then we may assume theyare colored with distinct colors. Case (2.2.1)
There is exactly one LR-row. Suppose first that v is the onlyLR-row. If every non-LR row is unlabeled or v j − and v j are labeled rows, thenthey are labeled with the same letter for if not we find D or D considering v , v j − and v j . Then, we find M III ( j ) induced by { k l , v , . . . , v j , k , . . . , k j } ,where k l is any vertex in K l (cid:54) = K . Moreover, if v j − is a labeled row, thenwe find either a ( j − j − j is even or odd, induced by { v , . . . , v j − , k l , k , . . . , k j − , k j } . If v isan LR-row, then we find M II ( j −
1) or F ( j −
1) –depending on whether j isodd or even– induced by every column of B and the rows v to v j . This holdsdisregarding on whether there are or not rows labeled with L or R.Suppose v i is an LR-row for some 2 < i < j −
1. Let r i be the first columnin which v i has a 0 and c i be column in which v j has a 0, then we find a tentinduced by { k , k r i ) , k c i ) , v , v i , v j } . If v j − is an LR-row, then we find M II ( j −
1) induced by { v , . . . , v j − , k , . . . , k j − , k j } .If v j is an LR-row and either every other row is unlabeled or there is exactlyone labeled row, then we find M III ( j ) induced by { k l , v , . . . , v j , k , . . . , k j } ,where k l is any vertex in K l (cid:54) = K such that the vertex representing the only39abeled row is nonadjacent to k l . If instead there are two labeled rows, then itfollows from Remark 3.31 that such rows are either v and v or v and v i forsome 2 < i < j . However, if v i is a labeled row for some 1 < i < j −
1, thenwe find either an even sun or an odd sun with center analogously as we havein Remark 3.31. If instead v j − and v are labeled rows, then they are labeledwith the same letter and thus we are in the same situation as if there were nolabeled rows in B , since we can find a vertex that is nonadjacent to both v and v j − . Case (2.2.2)
There are two LR-rows. If v and v are LR-rows, then wefind M II ( j −
1) as in the case where only v is an LR-row. Suppose v and v are LR-rows. If j = 4, then we find M II (4) induced by { v , . . . , v , k , k , k , k l } where k l in K l (cid:54) = K . Such a vertex exists, since v and v are eitherunlabeled rows or are rows labeled with the same letter, for if they were labeledwith distinct letters we would find D or D . Thus, there is a vertex that isnonadjacent to both v and v and is adjacent to v and v . If j >
4, then wefind a tent induced by { v , v j − , v j , k j − , k j − , k j } . Moreover, if v i is anLR-row for 1 < < j − v j − and v j are non-LR-rows, then we find a tentinduced by { v i , v j − , v j , k j − , k j − , k j } .It remains to see what happens if v and v j − and v and v j are LR-rows. If v and v j − are LR-rows, then we find M II ( j ) induced by every row of M andevery column except for column j −
1, which is replaced by some vertex k l in K l (cid:54) = K . This follows since, if there are two labeled rows, then they must be v i for some 1 < i < j − v j , hence they are labeled with the same letterand therefore there is a vertex k l nonadjacent to both. Finally, if v and v j areLR-rows, then we find a j -sun or a j -sun with center, depending on whether j is even or odd, contained in the subgraph induced by { v , . . . , v j , k , . . . , k j , k l } , where k l in K l (cid:54) = K is nonadjacent to every non-LR row. Therefore, thereis no M = M II ( j ) in B ∗ tag . Case (2.3)
Suppose that B contains M = M III ( j ), let v , . . . v j and k , . . . , k j +1) be the rows and the columns of M . If there are no LR-rows, then we find M III ( j ), hence we assume there is at least one LR-row. It follows analogouslyas in the previous cases that there are at most two LR-rows in M , since B is ad-missible. Notice that every pair of labeled rows v i and v l with 1 ≤ < i, l < j − D . Once more, ifsuch rows are labeled with distinct letters, then they are non-consecutive, forif not we find D or D . Furthermore, if such v i and v l are labeled rows, thenwe find either an even sun or an odd sun with center. It follows using the samearguments that, if i = 1 , j − l = j , then v i and v l are not both labeledrows. Hence, if there are two labeled rows, then such rows must be v j and v i for some i such that 2 < i < j − Case (2.3.1)
There is exactly one LR-row. If v i is an LR-row for some 1 ≤ i 1, then we find M II ( j − i + 1) induced by { v i , . . . , v j , k i +1) , . . . , k j +1) } . If v j − is an LR-row, then we find M II ( j ), induced by { v , . . . , v j , k , . . . , k j − , k j +1) } . If instead v j is an LR-row, then we find an even j -sun or an odd j -sunwith center k j +1) . Case (2.3.2) There are two LR-rows v i and v l . If 1 ≤ i < l < j − v i v l are non-consecutive rows, then we find a tent induced by v i , v l , v j , k s in K s (cid:54) = K adjacent to both v i and v l and nonadjacent to v j , k i (or k i +1) if i = 1) and k l (or k l +1) if l = j − l = i + 1 and i > 1. Ifinstead i = 1, or i = j − l = i + 1, then we find F (or M III (3) if j = 3)induced by { v i , v i +1 , k i , k i +1) , k i +2) , k ( j + 1), k s } with k s in K s (cid:54) = K adjacent to both v i and v i +1 . Finally, if v and v j are LR-rows, then we find M III ( j ) induced by { v , . . . , v j , k , . . . , k ( j + 1) } . If instead v i and v j areLR-rows for some i > 1, then we find M V induced by { v i , v j , v , v j − , k , k , k i , k i +1) , k j } , therefore there is no M III ( j ) in B ∗ tag . Case (2.4) Suppose B ∗ tag contains M = M IV , let v , . . . , v and k , . . . , k be the rows and the columns of M . If there are no labeled rows, then we find M IV as an induced subgraph, and since B is admissible and any three rows arenot pairwise nested, then there are at most two LR-rows. If v i is an LR-row for i = 1 , , 3, then we find M V induced by { v , v , v , k , . . . , k } . Moreover, ifonly v is an LR-row, then we find M IV induced by all the rows and columns of M . Thus, we assume there are exactly two LR-rows. If v and v are LR-rows,then we find M V induced by { v , v , v , v , k , k , . . . , k } . The same holdsif v i and v are LR-rows, with i = 2 , 3. Finally, if v and v are LR-rows, thenwe find a tent induced by { v , v , v , k , k , k } . It follows analogously bysymmetry if v and v or v and v are LR-rows, therefore there is no M IV in B ∗ tag . Case (2.5) Suppose B ∗ tag contains M = M V , let v , . . . , v and k , . . . , k bethe rows and the columns of M . Once more, if there are no LR-rows, then wefind M V as an induced subgraph, thus we assume there is at least one LR-row.Moreover, since any three rows are not pairwise nested, there are at most twoLR-rows. Case (2.5.1) There is exactly one LR-row. If v is the only LR-row, then wefind a tent induced by { v , v , v , k , k , k } . The same holds if v is theonly LR-row. If v is the only LR-row and every other row is unlabeled (or areall labeled with the same letter), then we find M IV induced by { v , v , v , v , k , . . . , k , k l } where k l in K l (cid:54) = K adjacent only to v . Suppose there are atleast two rows labeled with either L or R. Notice that, if v and v are labeled,then they are labeled with distinct letters for if we find D in B . Moreover, v (resp. v ) and v are not both labeled, for in that case we find either D , D or D in B . Hence, there are at most two rows labeled with either L or R,and they are necessarily v and v . In that case, there is a vertex k l in some K l (cid:54) = K such that v and v are adjacent to k l and v is nonadjacent to k l .We find a tent induced by v , v , v , k l , k and k . If v is the only LR-rowand every other row is unlabeled or are (one, two or) all labeled with the sameletter, then we find M V induced by { v , v , v , v , k , . . . , k , k l } where k l in K l (cid:54) = K adjacent only to v . Case (2.5.2) There are exactly two LR-rows. If v and v are LR-rows, thenwe find a tent induced by { v , v , v , k , k , k } . If instead v and v areLR-rows and every other row is unlabeled or (one or) all are labeled with thesame letter, then we find M V induced by every row and column plus a vertex k l in some K l (cid:54) = K such that both v and v are nonadjacent to k l . Moreover,41ince v and v overlap and there is a column in which both rows have a 0, thenthey are not labeled with distinct letters –disregarding of the coloring– for inthat case we find D or D in B . If v and v are LR-rows and every otherrow is unlabeled or are (one or) all labeled with the same letter, then we find atent induced by every row and column plus a vertex k l in some K l (cid:54) = K suchthat both v and v are nonadjacent to k l . Notice that v and v are labeledwith the same letter for if not we find either D or D in B . If v and v areLR-rows and every other row is unlabeled or v (resp. v ) is labeled with L orR, then we find M IV induced by every row and column plus a vertex k l in some K l (cid:54) = K such that both v and v are nonadjacent to k l . Notice that v and v are labeled with distinct letters for if not they induce D . However, they cannotbe labeled with distinct letters since in that case we find either D or D . Case (2.5.3) There are exactly three LR-rows. If v , v and v are LR-rows,since there is a vertex k l ∈ K l with l (cid:54) = 6 such that v is nonadjacent to k l , thenwe find a tent induced by { v , v , v , k , k , k l } . Analogously, if v , v and v are LR-rows and v is not, then the tent is induced by { v , v , v , k , k , k } . The same holds if all 4 rows are LR-rows, where the tent is induced by { v , v , v , k , k , k } . Finally, if v , v and v are LR-rows, since there is avertex k l ∈ K l with l (cid:54) = 6 such that v is nonadjacent to k l , then we find M V induced by { v , v , v , v , k , k , k , k , k l } . Case (3 ) B is admissible and B ∗ tag is M -free, but B contains a monochromaticgem, or a monochromatic weak-gem or a badly-colored doubly weak-gem. Sincethere are no uncolored labeled rows and those colored rows are labeled witheither L or R and do not induce any of the matrices in D , then in particularno two pre-colored rows of B induce a monochromatic gem or a monochromaticweak gem, and there are no badly-colored gems since every LR-row is uncolored,therefore this case is not possible. Case (4 ) Finally, let us suppose that B and B ∗ tag contain none of the matriceslisted in Theorem 3.11, but B is not 2-nested. We consider B ordered accordingto a suitable LR-ordering. Let B (cid:48) be a matrix obtained from B by extendingits partial pre-coloring to a total 2-coloring. It follows from Lemma 3.13 that,if B (cid:48) is not 2-nested, then either there is an LR-row for which its L-block andR-block are colored with the same color, or B (cid:48) contains a monochromatic gem ora monochromatic weak gem or a badly-colored doubly weak gem. If B (cid:48) containsa monochromatic gem where the rows that induce such a gem are not LR-rows,then the proof is analogous as in the tent case. Thus, we may assume that atleast one of the rows of the monochromatic gem is an LR-row. Case (4.1) Let us suppose first that there is an LR-row w for which its L-block w L and R-block w R are colored with the same color. If these two blocksare colored with the same color, then there is either one odd sequence of rows v , . . . , v j that force the same color on each block, or two distinct sequences,one that forces the same color on each block. Case (4.1.1) Suppose first that there is one odd sequence v , . . . , v j thatforces the color on both blocks. If k = 1, then notice this is not possible sincewe are coloring B (cid:48) using a suitable LR-ordering. If there is not a suitable LR-ordering, then it follows from Lemma 3.14 that B is either not admissible or not42R-orderable, which is a contradiction. Thus, let j > v intersects w L and v j intersects w R . Moreover, we assumethat each of the rows in the sequence v , . . . , v j is colored with a distinct colorand forces the coloring on the previous and the next row of the sequence. If v , . . . , v j are all unlabeled rows, then we find an even ( j + 1)-sun. If instead v is an L-row, then w L is properly contained in v . Thus, v , . . . , v j − are notcontained in v , since at least v j intersects w R . If v j is unlabeled or labeledwith R, then we find an even ( j + 1)-sun. If instead v j is labeled with L, since j is odd, then we find S ( j + 1) in B which is not possible since we are assuming B admissible. Case (4.1.2) Suppose there are two independent sequences v , . . . , v j and x , . . . , x l that force the same color on w L and w R , respectively. Suppose with-out loss of generality that w L and w R are colored with red. If j = 1 and l = 1,then we find D , which is not possible. Hence, we assume that either j > l > 1. Suppose that j > l > 1. In this case, there is a labeled row ineach sequence, for if not we can change the coloring for each row in one of thesequences and thus each block of w can be colored with distinct colors. We mayassume that v j is labeled with L and x l is labeled with R (for the LR-orderingused to color B (cid:48) is suitable and thus there is no R-row intersecting w L , andthe same holds for each L-block and w R ). As in the previous paragraphs, weassume that each row in each sequence forces the coloring on both the previousand the next row in its sequence. In that case, v , . . . , v j is contained in w L and x , . . . , x l is contained in w R . Moreover, w represents a vertex in S [16 , v j lies in S ∪ S or S ∪ S ∪ S and x l lies in S ∪ S ∪ S or S ∪ S (depending on whether they are colored with red or blue, respectively). Supposefirst that they are both colored with red, thus v j lies in S ∪ S and x l liesin S ∪ S ∪ S . In this case j and l are both even. If x l lies in S ∪ S ,since there is a k i in some K i (cid:54) = K adjacent to both v j and x l , then we find F ( j + l + 1) contained in the subgraph induced by k i and each row and columnon which the rows in w and both sequences are not null. If instead x l lies in S ,we find F ( k + l + 3) contained in the same submatrix but adding three vertices k i in K i for i = 1 , , 4. The same holds if v j and x l are both blue. Suppose nowthat v j is colored with red and v l is colored with blue. Thus, j is even and l isodd. In this case, we find F ( j + l + 2) contained in the submatrix induced bythe row that represents s , two columns representing any two vertices in K and K and each row and column on which the rows in w and both sequencesare not null. The proof is analogous if either j = 1 or l = 1.Hence, we may assume there is either a monochromatic weak gem in whichone of the rows is an LR-row or a badly-colored doubly-weak gem in B (cid:48) , for thecase of a monochromatic gem or a monochromatic weak gem where one of therows is an L-row (resp. R-row) and the other is unlabeled is analogous to thetent case. Case (4.2) Let us suppose there is a monochromatic weak gem in B (cid:48) , and let v and v be the rows that induce such gem, where v is an LR-row. Case (4.2.1) Suppose first that v is a pre-colored row. Suppose withoutloss of generality that the monochromatic weak gem is induced by v and the43-block of v and that v and v are both colored with red. We denote by v L the L-block of v . If v is labeled with R, then v is the L-block of some LR-row r in B and v is the R-block of itself. However, since the LR-ordering we areconsidering to color B (cid:48) is suitable, then the L-block of an LR-row has emptyintersection with the R-block of a non-LR row and thus this case is not possible.If v is labeled with L, since both rows induce a weak gem, then v L is properlycontained in v . Since v is a row labeled with L in B , then v is a pre-coloredrow. Moreover, since v L is colored with the same color as v , then there is eithera blue pre-colored row, or a sequence of rows v , . . . , v j where v j forces the redcoloring of v L . In either case, there is a pre-colored row in that sequence thatforces the color on v L , and such row is either labeled with L or with R. Supposefirst that such row is labeled with L. If v is a the blue pre-colored row thatforces the red coloring on v L , then v L is properly contained in v . However,in that case we find D which is not possible since B is admissible. Hence, weassume v , . . . , v j − is a sequence of unlabeled rows and that v j is a labeled rowsuch that this sequence forces v L to be colored with red, and each row of thesequence forces the color on both its predecessor and its successor. If j − v j is colored with blue, and if j − v j is colored withred. In either case, we find S ( j ) contained in the submatrix induced by rows v , v , v , . . . , v j . If instead the row v that forces the coloring on v L is labeledwith R, since the LR-ordering used to color B is suitable, then the intersectionbetween v L and v is empty. Hence, v (cid:54) = v , thus we assume that v , . . . , v j − are unlabeled rows and v j = v . If j − v j is colored with red, andif j − v j is colored with blue. In either case we also find S ( j ),which is not possible since B is admissible. Case (4.2.2) Suppose now that v is an unlabeled row. Notice that, since v and v induce a weak gem, then v is not nested in v . Hence, either thecoloring of both rows is forced by the same sequence of rows or the coloring of v and v is forced for each by a distinct sequence of rows. As in the previouscases, we assume that the last row of each sequence represents a pre-coloredlabeled row. Suppose first that both rows are forced to be colored with red bythe same row v . Thus, v is a labeled row pre-colored with blue. Moreover,since v forces v to be colored with red, then v is not contained in v andthus there is a column k in which v has a 1 and v has a 0. We may alsoassume that v has a 0 in such a column since v is also not contained in v .Moreover, since v forces v to be colored with red, then v is labeled with thesame letter than v and v is not contained in v , thus we can find a column k in which v has a 0 and v and v both have a 1. Furthermore, since v and v are both labeled with the same letter and the three rows have pairwisenonempty intersection, then there is a column k in which all three rows havea 1. Since v is a row labeled with either L or R in B , then there are vertices k l ∈ K l , k m ∈ K m with l (cid:54) = m , l, m (cid:54) = 6 such that v is adjacent to k l andnonadjacent to k m . Moreover, since v is an LR-row, then v is adjacent toboth k l and k m and v j is nonadjacent to k l and k m . Hence, we find F inducedby { v , v , v , k l , k , k , k , k m } . Suppose instead there is a sequence ofrows v , . . . , v j that force the coloring of both v and v , where v , . . . , v j − are44nlabeled rows and v j is labeled with either L or R and is pre-colored. We havetwo possibilities: either v j is labeled with L or with R. If v j is labeled withL and v j forces the coloring of v , then we have the same situation as in theprevious case. Thus we assume v j is nested in v . In this case, since v j and v are labeled with L, the vertices v , . . . , v j − are nested in v and thus they arechained from right to left. Moreover, since v and v are colored with the samecolor, then there is an odd index 1 ≤ l ≤ j − v contains v , . . . , v l and does not contain v l +1 , . . . , v j . Hence, we find F ( l + 1) considering the rows v , v , . . . , v l +1 . Suppose now that v j is labeled with R. Since B (cid:48) is colored usinga suitable LR-ordering, then v j and v have empty intersection, thus there is asequence of unlabeled rows v , . . . , v j − , chained from left to right. Notice thatit is possible that v = v . Suppose first that j is even. If v = v , then thereis an odd number of unlabeled rows between v and v j . In this case we find a( j − v , v = v , v , . . . , v j . Ifinstead v (cid:54) = v , then v and v and v and v both induce a 0-gem, and thuswe find a ( j − j is odd, then there is an evennumber of unlabeled rows between v and v j . Once more, we find a ( j − v , v , . . . , v j .Notice that these are all the possible cases for a weak gem. This followsfrom the fact that, if there is a pre-colored labeled row that forces the coloringupon v then it forces the coloring upon v and viceversa. Moreover, if there isa sequence of rows that force the coloring upon v , then one of these rows of thesequence also forces the coloring upon v , and viceversa. Furthermore, since thelabel of the pre-colored row of the sequence determines a unique direction inwhich the rows overlap in chain, then there is only one possibility in each case,as we have seen in the previous paragraphs. It follows that the case in whichthere is a sequence forcing the coloring upon each v and v can be reduced tothe previous case. Case (4.3) Suppose there is a badly-colored doubly-weak gem in B (cid:48) . Let v and v be the LR-rows that induce the doubly-weak gem. Since the suitable LR-ordering determines the blocks of each LR-row, then the L-block of v properlycontains the L-block of v and the R-block of v is properly contained in theR-block of v , or viceversa. Moreover, the R-block of v may be empty. Let usdenote by v L and v L (resp. v R and v R ) the L-blocks (resp. R-blocks) of v and v . There is a sequence of rows that forces the coloring on both LR-rowssimultaneously or there are two sequences of rows and each forces the coloringupon the blocks of v and v , respectively. Whenever we consider a sequenceof rows that forces the coloring upon the blocks of v and v , we will considera sequence in which every row forces the coloring upon its predecessor and itssuccessor, a pre-colored row is either the first or the last row of the sequence,the first row of the sequence forces the coloring upon the corresponding blockof v and the last row forces the coloring upon the corresponding block of v .It follows that, in such a sequence, every pair of consecutive unlabeled rowsoverlap. We can also assume that there are no blocks corresponding to LR-rows in such a sequence, for we can reduce this to one of the cases. Supposefirst there is a sequence of rows v , . . . , v j that forces the coloring upon both45R-rows simultaneously. We assume that v intersects v and v j intersects v .If v , . . . , v j forces the coloring on both L-blocks, then we have four cases: (1)either v , . . . , v j are all unlabeled rows, (2) v is the only pre-colored row, (3) v j is the only pre-colored row or (4) v and v j are the only pre-colored rows. Ineither case, if v , . . . , v j is a minimal sequence that forces the same color uponboth v L and v L , then j is odd. Case (4.3.1) Suppose v , . . . , v j are unlabeled. If j = 3, then we find S (3)contained in the submatrix induced by v , v and v . Suppose j > 3, thus wehave two possibilities. If v ∩ v (cid:54) = ∅ , since j is odd, then we find a ( j − v , v , v , . . . , v j .If instead v ∩ v = ∅ , then we find F ( j ) contained in the same submatrix. Case (4.3.2) Suppose v is the only pre-colored row. Since v is a pre-coloredrow and forces the color red upon the L-block of v , then v contains v L and v is colored with blue. If v ∩ v L (cid:54) = ∅ , then we find F in the submatrixgiven by considering the rows v , v , v , having in mind that there is a columnrepresenting some k i in K i (cid:54) = K in which the row corresponding to v has a1 and the rows corresponding to v and v both have 0. This follows since v is unlabeled and thus represents a vertex that lies in S , and v is pre-coloredand labeled with L or R and, thus it represents a vertex that is not adjacentto every partition K i of K . If instead v ∩ v L = ∅ , then we find F ( j − v , v , . . . , v j − if v ∩ v R = ∅ ,and induced by the rows v , v , v , . . . , v j if v ∩ v R (cid:54) = ∅ . Case (4.3.3) Suppose v j is the only pre-colored row. In this case, v j properlycontains v L and we can assume that the rows v , . . . , v j − are contained in v L .If v ∩ v (cid:54) = ∅ , then we find an even ( j − v , v , . . . , v j . If instead v ∩ v = ∅ , then we find F ( j ) in the submatrixgiven by rows v , . . . , v j . Case (4.3.4) Suppose that v and v j are the only pre-colored rows. Thus,we can assume that v j properly contains v L and v properly contains v L , thus v properly contains v L . Hence, we find D induced by the rows v , v and v which is not possible since B is admissible. The only case we have left iswhen v , . . . , v j forces the coloring upon v L and v R . This follows from thefact that, if v , . . . , v j forces the color upon v L and v R (cid:54) = ∅ , then this case canbe reduced to case (4.3.3). Hence, either (1) v , . . . , v j are unlabeled rows, (2) v is the only pre-colored row, or (3) v and v j are the only pre-colored rows.Notice that in either case, j is even and thus for (1) we find S ( j ), which resultsin a contradiction since B is admissible. Moreover, in the remaining cases, v properly contains v L and v L . Since v and v overlap, we find D which is notpossible for B is admissible.This finishes the proof.Let G = ( K, S ), H as in Section 2.2 and the matrices B i for i = { . . . , } as defined in the previous subsection. Suppose B i is 2-nested for each i ∈{ , , . . . , } , let χ i be a total block bi-coloring and Π i a suitable LR-orderingfor B i , for each i ∈ { , , . . . , } . 46et Π be the ordering of the vertices of K given by concatenating the order-ings Π , Π , . . . , Π , as defined in Section 3.1.2. Definition 3.32. We define the (0 , -matrices B r , B b , B r − b and B b − r as inthe tent case, but considering only those vertices of S that are not in S [16 . Notice that the only nonempty subsets S ij with i > j that we are consideringare those where i = 6. Hence, the rows of B r − b are those representing vertices in S ∪ S ∪ S and the rows of B b − r are those representing vertices in S ∪ S . Lemma 3.33. Suppose B i is -nested for each i = 1 , . . . , . If B r , B b , B r − b or B b − r are not nested, then G contains F , F (5) or F (5) as induced subgraphs.Proof. Suppose first that B r is not nested, thus there is a 0-gem. Every row in B r represents a vertex that belongs to one of the following subsets of S : S , S , S , S , S , S , S , S , S or S . Recall that S = S [13 , S = S , S = S and S = S Let f and f be two rows that induce a gem in B r and v in S ij and v in S lm be the corresponding vertices in G . Analogouslyas in the tent case we obtain the following claim, strongly using that no row in B r , B b , B r − b and B b − r represents a vertex of S complete to K , for we do notconsider S [16] to define them. Claim 3.34. Let v and v be two vertices that induce a -gem in B r , B b , B r − b or B b − r . If v and v lie in the same S ij , then v and v are nested. Moreover, since B i is 2-nested for every i ∈ { , , . . . , , } , in particularthere are no monochromatic gems in each B i . Moreover, if j = l , then we find D in K i or K j , respectively. It follows from these remarks that we have onlytwo possibilities for the gem: (1) v in S and v in S ∪ S [46 and (2) v in S ∪ S and v in S [13 . Case (1 ) v in S and v in S ∪ S [46 . Suppose that v in S . Let k in K nonadjacent to both. There are vertices k , k in K such that k is adjacentonly to v and k is adjacent to both. Moreover, there are vertices k in K and k in K such that k is adjacent to both and k is adjacent only to v . Wefind F induced by { v , v , s , k , k , k , k , k } . If instead v in S , thenwe also find F changing k for some vertex k in K in the same subset. Case (2 ) v in S ∪ S and v in S [13 . Since v is also complete to K , thenone of the columns of the 0-gem is induced by the column c L of B r . Thus, thereis a vertex k in K adjacent to v and nonadjacent to v . Moreover, the gemis induced by a column corresponding to a vertex k in K nonadjacent to v and adjacent to v . We find F induced by { v , v , s , k , k , k , k , k } .Hence B r is nested. Suppose now that B b is not nested, and let v in S ij and v in S lm two vertices for which its rows in B b induce a 0-gem. Every rowin B b represents a vertex belonging to either S , S , S , S , S , S , S , S , S , S , S , S or S . Recall that S = S , S = S [25 , S = S [26 , S = S and S = S . It follows from this and Claim 3.34 that the casesare: (1) v in S ∪ S and v in S , (2) v in S ∪ S and v in S ∪ S ,or v in S ∪ S , or v in S ∪ S , (3) v in S and v in S ∪ S and (4) v in S and v in S ∪ S . 47 ase (1 ) v in S ∪ S and v in S . Suppose v in S . We find F (5) inducedby { v , v , s , s , s , k , k , k , k , k } , where v and v are nonadjacent to k and k . It follows analogously by symmetry if v in S . Case (2 ) v in S ∪ S . Case (2.1) v in S ∪ S If v in S and v in S , then we find F (5)induced by { v , v , s , s , s , k , k , k , k } . The same holds if instead v in S . Moreover, we find F (5) induced by the same subset if v in S and v in S , since there is a vertex in K that is nonadjacent to v . Case (2.2) v in S ∪ S . We find F (5) induced by { v , v , s , s , s , k , k , k , k } . Case (2.3) v in S ∪ S . Since S is complete to K and K , the 0-gemcannot be induced by v and a vertex v in S since v is also complete to K .If v in S , then we find F induced by { v , v , s , k , k , k , k , k } . Case (3 ) v in S and v in S ∪ S . Since v is not complete to K bydefinition, we find F (5) induced by { v , v , s , s , s , k , k , k , k , k } ,where k represents the third column of the gem and thus, k is adjacent to v and nonadjacent to v . Case (4 ) v in S and v in S ∪ S . We find F (5) induced by { v , v , s , s , s , k , k , k , k } since v is not complete to K and v is not completeto K by definition.Hence B b is nested. Suppose that B b − r is not nested. Every row in B b − r represents a vertex in S or S . It follows from Claim 3.34 that it sufficesto consider a 0-gem induced by two vertices v in S and v in S . Let k , k and k be the vertices represented by the columns of the gem. We find F induced by { v , v , s , k , k , k , k , k } .Finally, suppose that B r − b is not nested. Every row in B r − b represents avertex in S , S or S . Let v and v in S ∪ S ∪ S be two vertices whoserows induce a 0-gem. If v in S and v in S ∪ S , then we find F (5) inducedby { v , v , s , s , s , k , k , k , k , k } . Similarly, we find F induced by { v , v , s , k , k , k , k , k } if v in S and v in S , and this finishesthe proof. Theorem 3.35. Let G = ( K, S ) be a split graph containing an induced -tent.Then, the following are equivalent:1. G is circle;2. G is F sc -free;3. B , B , . . . , B are -nested and B r , B b , B r − b and B b − r are nested.Proof. It is not hard to see that (1) ⇒ (2), and that (2) ⇒ (3) is a consequenceof the previous lemmas. We will show (3) ⇒ (1). Suppose that each of thematrices B , B , . . . , B is 2-nested and the matrices B r , B b , B r − b or B b − r arenested. Let Π be the ordering for all the vertices in K obtained by concatenatingeach suitable LR-ordering Π i for i ∈ { , , . . . , } .Consider the circle divided into twelve pieces as in Figure 12. For each i ∈ { , , . . . , } and for each vertex k i ∈ K i we place a chord having one48 igure 12: Sketch model of G with some of the chords associated to the rows in B . endpoint in K + i and the other endpoint in K − i , in such a way that the orderingof the endpoints of the chords in K + i and K − i is Π i .Let us see how to place the chords for each subset S ij of S . First, someuseful remarks. Remark . The following assertions hold: • By Lemma 3.33, all the vertices in S ij are nested, for every pair i, j = { , , . . . , } , i (cid:54) = j . This follows since any two vertices in S ij are nondis-joint. Moreover, in each S ij , all the vertices are colored with either onecolor (the same), or they are colored red-blue or blue-red. Hence, thesevertices are represented by rows in the matrices B r − b and B b − r and there-fore they must be nested since each of these matrices is a nested matrix. • It follows from the previous and Claim 3.16 that, if i ≥ k and j ≤ l , thenevery vertex in S ij is nested in every vertex of S kl . • Also as a consequence of the previous and Lemma 3.33, if we consideronly those vertices labeled with the same letter in some B i , then there isa total ordering of these vertices. This follows from the fact that, if twovertices v and v are labeled with the same letter in some B i , since B i is –in particular– admissible, then they are nested in K i . Moreover, if v and v are labeled with L in B i , then they are either complete to K i − orlabeled with R in B i − . Thus, there is an index j l such that v i is labeledwith R in B j l , for l = 1 , 2. Therefore, we can find in such a way a totalordering of all these vertices. • If v and v are labeled with distinct letters in some B i , then they areeither disjoint in K i (if they are colored with the same color) or N K i ( v ) ∪ N K i ( v ) = K i (if they are nondisjoint and colored with distinct colors),for there are no D or D in B i for every i ∈ { , , . . . , } .Notice that, when we define the matrix B , we pre-color every vertex in S [15] with the same color. Since we are assuming B is 2-nested and thus in particular49s admissible, the subset S [15] (cid:54) = ∅ if and only if the vertices represented in B areeither all vertices in S ∪ S [16 and vertices that are represented by labeled rows r , all of them colored with the same color and labeled with the same letter L orR. Moreover, since B is admissible, the sets N K ( S i ) ∩ N K ( S j ) are empty,for i = 1 , , j = 3 , 4. The same holds for the sets N K ( S i ) ∩ N K ( S j ),for i = 2 , j = 2 , , S [16 = ∅ , then the placing of the chords that represent vertices with one orboth endpoints in K is very similar as in the tent case. Suppose that S [16 (cid:54) = ∅ .Before proceeding with the guidelines to draw the circle model, let us see a fewremarks on the relationship between the vertices in S ij with either i = 6 or j = 6 and those vertices in S [16 , which follow from the claims stated throughoutthe proof of Lemma 3.26: Remark . Let G be a circle graph that contains no induced tent but con-tains an induced 4-tent, and such that each matrix B i is 2-nested for every i = 1 , , . . . , 6. Then, all of the following statements hold: • If S ∪ S (cid:54) = ∅ , then S ∪ S = ∅ , and viceversa. • If v in S ∪ S and v in S ∪ S ∪ S , then for every vertex w in S [16 either N K ( v ) ⊆ N K ( w ) or N K ( v ) ⊆ N K ( w ). Moreover, v and v are disjoint in K . The same holds for v in S ∪ S ∪ S and v in S ∪ S . • If v in S and w in S , then N K ( v ) ∩ N K ( w ) = ∅ . The same holds for v in S and w in S .Let v in S ij (cid:54) = S [16 and w in S [16 , with either i = 6 or j = 6. Suppose firstthat i = j = 6. Since B is 2-nested, the submatrix induced by the rows thatrepresent v and w in B contains no monochromatic gems or monochromaticweak gems. If instead i < j , since B is admissible, then the submatrix inducedby the rows that represent v and w in B contains no monochromatic weak gem,and thus we can place the endpoint of w corresponding to K in the arc portion K +6 and the K endpoint of v in K − , or viceversa.Remember that, since we are considering a suitable LR-ordering, there is anL-row m L such that any L-row and every L-block of an LR-row are contained in m L and every R-row and R-block of an LR-row are contained in the complementof m L . Moreover, since we have a block bi-coloring for B , then for each LR-rowone of its blocks is colored with red and the other is colored with blue. Hence,for any LR-row, we can place one endpoint in the arc portion K +6 using theordering given for the block that colored with red, and the other endpoint inthe arc portion K − using the ordering given for the block that is colored withblue. Notice that, if B is 2-nested, then all the rows labeled with L (resp. R)and colored with the same color and those L-blocks (resp. R-blocks) of LR-rowsare nested. In particular, the L-block (resp. R-block) of every LR-row containsall the L-blocks of those rows labeled with L (resp. R) that are colored with thesame color. Equivalently, let r be an LR-row in B with its L-block r L coloredwith red and its R-block r R colored with blue, r (cid:48) be a row labeled with L and r (cid:48)(cid:48) be a row labeled with R. Hence, if r L , r (cid:48) and r (cid:48)(cid:48) are colored with the same color,then r contains r (cid:48) and r ∩ r (cid:48)(cid:48) = ∅ . This holds since we are considering a suitable50R-ordering and a total block bi-coloring of the matrix B , thus it contains no D , D , D , D or D . Since every matrix B r , B b , B r − b and B b − r is nested, thereis a total ordering for the nondisjoint rows in each of these matrices. In otherwords, there is a total ordering for all the rows that intersect that are coloredwith the same color, or with red-blue or with blue-red, respectively. Moreover,if v and w are two vertices in S such that they both have rows representingthem in one of these matrices –hence, they are colored with the same color orsequence of colors–, then either v and w are disjoint or nested.Notice that there are no other conditions besides being disjoint or nestedoutside each of the following subsets: S , S , S , S , S , S . For thesubset S , we only need to consider if every vertex in S ∪ S ∪ S aredisjoint or nested. The same holds for the subsets S and S , considering S ∪ S and S ∪ S , respectively. Since each matrix B i is 2-nested for every i = 1 , , . . . , 6, if there are vertices in both S and S , then they are disjointin K . The same holds for vertices in S and S , and S and S ∪ S ∪ S .This is in addition to every property seen in Remark 3.37.With this in mind, we give guidelines to build a circle model for G . Weplace first the chords corresponding to every vertex in K , using the orderingΠ. For each subset S ij , we order its vertices with the inclusion ordering of theneighborhoods in K and the ordering Π. When placing the chords correspondingto the vertices of each subset, we do it from lowest to highest according to thepreviously stated ordering given for each subset.First, we place those vertices in S ii for each i = 1 , , . . . , 6, considering theordering given by inclusion. If v in S ii and the row that represents v is coloredwith red, then both endpoints of the chord corresponding to v are placed in K + i .If instead the row is colored with blue, then both endpoints are placed in K − i .For each v in S ij (cid:54) = S [16 , if the row that represents v in B i is colored withred (resp. blue) , then we place the endpoint corresponding to K i in the portion K + i (resp. K − i ) . We apply the same rule for the endpoint corresponding to K j .Let us consider now the vertices in S [15] . If G is circle, then all the rows in B are colored with the same color. Moreover, if S [15] (cid:54) = ∅ , then either everyrow labeled with L or R in B is labeled with L and colored with red or labeledwith R and colored with blue, or viceversa. Suppose first that every row labeledwith L or R in B is labeled with L and colored with red or labeled with R andcolored with blue. In that case, every row representing a vertex v in S [15] iscolored with blue, hence we place one endpoint of the chord corresponding to v in K +6 and the other endpoint in K − . In both cases, the endpoint of the chordcorresponding to v is the last chord of a vertex of S that appears in the portionof K +6 and is the first chord of a vertex of S that appears in the portion of K − .We place all the vertices in S [15] in such a manner. If instead every row labeledwith L or R in B is labeled with L and colored with blue or labeled with R andcolored with red, then every row representing a vertex in S [15] is colored withred. We place the endpoints of the chord in K − and K +6 , as the last and firstchord that appears in that portion, respectively.Finally, let us consider now a vertex v in S [16 . Here we have two possibilities:511) the row that represents v has only one block, (2) the row that representsthe row that represents v has two blocks of 1’s. Let us consider the first case. Ifthe row that represents v has only one block, then it is either an L-block or anR-block. Suppose that it is an L-block. If the row in B is colored with red, thenwe place one endpoint of the chord as the last of K − and the other endpointin K +6 , considering in this case the partial ordering given for every row thathas an L-block colored with red in B . If instead the row in B is colored withblue, then we place one endpoint of the chord as the first of K +6 and the otherendpoint in K − , considering in this case the partial ordering given for every rowthat has an L-block colored with blue in B . The placement is analogous forthose LR-rows that are an R-block. Suppose now that the row that represents v has an L-block v L and an R-block v R . If v L is colored with red, then v R iscolored with blue. We place one endpoint of the chord in K +6 , considering thepartial ordering given by every row that has an L-block colored with red in B ,and the other endpoint of the chord in K − , considering the partial orderinggiven by every row that has an R-block colored with blue in B . The placementis analogous if v L is colored with blue.This gives a circle model for the given split graph G . In this section we will address the last case of the proof of Theorem 1.1, whichis the case where G contains an induced co-4-tent. This case is mostly similar tothe 4-tent case, with two particular difference. First of all, the co-4-tent is not aprime graph. This implies that there is more than one possible circle model forthis graph. Moreover, if a non-circle graph G is also not a prime graph, then forany split decomposition one of its factors is a non-circle graph. This follows fromthe fact that circle graphs are closed under split decomposition [2]. Hence, aminimally non-circle graph is necessarily a prime graph. However, this problemwas solved in Section 2, more precisely in Remark 2.10. We assume K (cid:54) = ∅ and K (cid:54) = ∅ in order to work with a minimally non-circle graph throughout theproof. Second of all, even when considering these extra hypothesis to work witha prime graph, we have to divide the proof in 5 cases. However, we will see thatone of these 5 cases is indeed more general than the other four.Let G = ( K, S ) and H as in Section 2.3. For each i ∈ { , , . . . , } , let C i be a (0 , s ∈ S such that s belongs to S ij or S ji for some j ∈ { , , . . . , } and one column for each vertex k ∈ K i andsuch that such that the entry corresponding to row s and column k is 1 if andonly if s is adjacent to k in G . For each j ∈ { , , . . . , } − { i } , we label thoserows corresponding to vertices of S ji with L and those corresponding to verticesof S ij with R, with the exception of those rows in C that represent vertices in S and S [86] which are labeled with LR. Notice that we have considered thosevertices that are complete to K , . . . , K and K and are also adjacent to K and K divided into two distinct subsets. Thus, S are those vertices that arenot complete to K and therefore the corresponding rows are labeled with R in C and with L in C . As in the 4-tent case, there are LR-rows in C . Moreover,there may be some empty LR-rows, which represent those vertices of S that52re complete to K , . . . , K and K and are anticomplete to K . These verticesare all pre-colored with the same color, and that color is assigned depending onwhether S ∪ S ∪ S (cid:54) = ∅ or S ∪ S (cid:54) = ∅ . We color some of the remainingrows of C i as we did in the previous sections, to denote in which portion ofthe circle model the chords have to be drawn. In order to characterize theforbidden induced subgraphs of G and using an argument of symmetry, we willonly analyse the properties of the matrices C , C , C , C and C , since thematrices C i i = 4 , , C , C and C , respectively.We consider 5 distinct cases, according to whether the subsets K , K and K are empty or not, for the matrices we need to define may be different ineach case. Using the symmetry of the subclasses K and K , the cases we needto study are the following: (1) K , K , K (cid:54) = ∅ , (2) K , K (cid:54) = ∅ , K = ∅ , (3) K , K (cid:54) = ∅ , K = ∅ , (4) K (cid:54) = ∅ , K , K = ∅ , (5) K (cid:54) = ∅ , K , K = ∅ . We havethe following lemma. For more details on its proof, see [14]. Lemma 3.38. Let C j , . . . , C j be the matrices defined for each of the previouslystated cases j = 1 , . . . , . Then, for each i ∈ { , . . . , } , the matrix C ji is asubmatrix of C i for every j = 2 , , , . In (1), the subsets are given as described in Table 3, and thus the matrices C i as are follows: C = K S L · · · S · · · S L · · · S L · · · ••• C = K S R · · · S · · · S L · · · S L · · · S L · · · •••• C = K S R · · · S L · · · S · · · S L · · · S L · · · S R · · · ••••• C = K S · · · S R · · · S R · · · S [46 R · · · S R · · · S [86 R · · · ••••• C = K S R · · · S [27 R · · · S · · · S L · · · S L · · · S L · · · S R · · · S [86] LR · · · S LR · · · •••••• Let us suppose that K , K , K (cid:54) = ∅ . The Claims in Section 2 and thefollowing prime circle model allow us to assume that some subsets of S are empty.We denote by S the set of vertices in S that are complete to K , . . . , K , areadjacent to K and K but are not complete to K , and analogously S is theset of vertices in S that are complete to K , . . . , K , K , are adjacent to K and K but are not complete to K . Hence, S denotes the vertices of S that arecomplete to K , . . . , K , K and are adjacent to K .We state some results analogous as the ones seen in the previous sections fora graph that contains an induced co-4-tent and contains no 4-tent or tent. Theseresults and its proofs are analogous to Lemmas 3.24, 3.26 and Theorem 3.35,thus we will just state the results (for the complete proof of each of these results,see [14]).— 53 igure 13: A circle model for the co-4-tent graph. Lemma 3.39 ([14]) . If C , C , . . . , C are not -nested, then G contains oneof the graphs listed in Figure 2 as induced subgraphs. We define the matrices C r , C b , C r − b and C b − r as in Section 3.2.2. Similarly,we have the following lemma for these matrices. Lemma 3.40 ([14]) . Suppose that C i is -nested for each i = 1 , . . . , . If C r , C b , C r − b or C b − r are not nested, then G contains F as a minimal forbiddeninduced subgraph for the class of circle graphs. The main result of this section is the following theorem. Theorem 3.41 ([14]) . Let G = ( K, S ) be a split graph containing an inducedco- -tent. Then, the following are equivalent:1. G is circle;2. G is F sc -free;3. C , C , . . . , C are -nested and C r , C b , C r − b and C b − r are nested.3.4. Split circle graphs containing an induced net and proof of Theorem 1.1 Let G = ( K, S ) be a split graph. If G is a minimally non-circle graph,then it contains either a tent, or a 4-tent, or a co-4-tent, or a net as inducedsubgraphs. In the previous sections, we addressed the problem of having a splitminimally-non-circle graph that contains an induced tent, 4-tent and co-4-tent,respectively. Let us consider a split graph G that contains no induced tent,4-tent or co-4-tent, and suppose there is a net subgraph in G .We define K i as the subset of vertices in K that are adjacent only to s i if i = 1 , , 5, and if i = 2 , , K that are adjacent to s i − and s i +1 . We define K as the subset of vertices in K that are nonadjacent to s , s and s . Let s in S . We denote by T ( s ) the vertices that are false twins of s .54 emark . The net is not a prime graph. Moreover, if K i = ∅ , K j = ∅ for any pair i, j ∈ { , , } , then G is not prime. For example, if K = ∅ and K = ∅ , then a split decomposition can be found considering the inducedsubgraphs H = { u } ∪ K ∪ T ( s ) and H = V ( G ) \ T ( s ), where u (cid:54)∈ V ( G ) iscomplete to K and anticomplete to V ( G ) \ T ( s ).Since in the proof we consider a minimally non-circle graph G , it follows fromthe previous remark that at least two of K , K and K must be nonempty sothat G results prime. However, in either case we find a 4-tent as an inducedsubgraph. Therefore, as a consequence of this and the previous sections, we havenow proven that any graph containing no graph in F sc as an induced subgraphis a circle graph. Hence, since the class of circle graphs is hereditary, in orderto complete the proof of Theorem 1.1, it suffices to verify that no graph in F sc is a circle graph. We do so in the lemma below. Lemma 3.43. None of the graphs in F sc is a circle graph.Proof. We show this for the odd k -suns with center and the even k -suns; theproofs for the remaining graphs are similar (for more details, see [14]). Weuse throughout the proof that BW , W and W (see Figure 14) are non-circlegraphs [3], for every k ≥ Figure 14: The graphs W , W and BW . Let us first consider an odd k -sun with center, where v , . . . , v k are thevertices of the clique of size k , w , . . . , w k are the petals (the vertices of degree2) and x is the center (the vertex adjacent to every vertex of the clique of size k ). If k = 3, we consider the local complement with respect to the center and weobtain BW . If k = 5 or k = 7, we apply local complementation with respect to x , w , . . . and w k , obtaining W and W , respectively. If instead k = 9, we firstapply local complementation with respect to x , w , . . . and w k and we obtain W . Once we have this wheel, we apply local complement with respect to v , v and v and obtain W induced by { v , . . . , v } . If we now consider the localcomplement with respect to v , v and x (in that order), we find C , which is nota circle graph because it is locally equivalent to W . More in general, for every k ≥ 11 we can obtain a W k considering the sequence described at the beginningof the paragraph. Once we have a wheel, if we apply local complementationby v , v and v k , then we obtain W k − . We can repeat this until k − < k = 5, k = 6 or k = 7 and we reduce this to one of theprevious cases.Let us consider now an even k -sun, where v , . . . , v k are the vertices of theclique of size k and w , . . . , w k are the petals (the vertices of degree 2), where w i is adjacent to v i and v i +1 . If k = 4, then we apply local complementation withrespect to the sequence w , w , w , w , v , w , w , w , v , w and v , and we55btain C , which is locally equivalent to W . If k = 6 and we apply local comple-mentation to the sequence w , w , . . . , w k , then we obtain C . Let us consideran even k ≥ 8, thus k = 2 j for some j ≥ l = k − . We apply localcomplementation with respect to the sequence v , v j +1 , v , v k , . . . , v l , v k − l and we find W or W induced by { v , v j +1 , v j +1 − , v j +1 − , v j +1+1 , v j +1+2 } or by { v , v j +1 , v j +1 − , v j +1 − , v j +1 − , v j +1+1 , v j +1+2 , v j +1+3 } , depending onwhether k ≡ k ≡ Theorem 3.44. All the graphs in F sc are minimally non-circle.Proof. Since all the graphs depicted in Figure 2 are non-circle graphs, it sufficesto see that none of these graphs is an induced subgraph of any of the others.Moreover, if we consider the matrix A ( S, K ) for each of these graphs, then itsuffices to see that none of these matrices is a subconfiguration of any othercorresponding to a graph on the list. Let U be the set containing all thesematrices. We call the matrix A ( S, K ) for each graph in U with the name asthe corresponding graph, and we denote with M I ( k ) the matrix A ( S, K ) forthe k -sun, for each k ≥ 3. Notice that the matrices corresponding to k -suns, M II (2 j ), M III (3), M III (2 j ), M IV and M V are all Tucker matrices, for every j ≥ 2. In particular, the matrix A ( S, K ) that represents an even k -sun is M I ( k ). For its part, the matrices corresponding to the graphs F , F (2 j + 1)and F (2 j + 1) for every j ≥ A ( S, K )matrix corresponding to an odd k -sun with center is S ( k ), which is the Tuckermatrix M I ( k ) plus one row having a 1 in each column, and the A ( S, K ) matrixof tent ∨ K is M , which is the Tucker matrix M I (3) plus one column havinga 1 in every row (see Figure 4).If M is a Tucker matrix, then M is not a subconfiguration of any other Tuckermatrix since these matrices are minimal forbidden subconfigurations for theC1P. It follows that M is not a subconfiguration of F , F (2 j +1) and F (2 j +1)for every j ≥ M and S (2 j )contain M I (3) and M I (2 j − 1) as subconfigurations, respectively, it follows usingthe same arguments that neither M nor S (2 j ) is a subconfiguration of any ofthe matrices in U , for every j ≥ 2. Finally, S (2 j ) is not a subconfiguration of M since S (2 j ) always has more rows that M , and conversely, M is not asubconfiguration of S (2 j ) for in that case M I (3) would be a subconfigurationof M I ( k ) for some odd k ≥ F , F (2 j + 1) and F (2 j + 1) are not subconfigurations of any other matrix in U , for every j ≥ U has three or more rows that haveat least three 1’s, hence it is not possible for F to be a subconfiguration ofany of these matrices. Toward a contradiction, suppose there is M in U suchthat F (2 j + 1) is a subconfiguration of M , for some j ≥ 2. Since j ≥ M has at least five rows in total, and in particular, at least two distinct rows that56ave at least three 1’s. It follows that M is either M IV , M V or M II (2 j ) forsome j ≥ 2. Moreover, F (2 j + 1) is not a subconfiguration of M IV or M V since these matrices have only four rows and F (2 j + 1) has at least five rows.Let us suppose that M = M II ( k ) for some even k ≥ 4. Notice that k ≥ k ≥ j + 1 and k is even. Thus, the first and second rows of F (2 j + 1)correspond to the first and last rows of M since these are the only rows in M having more than two 1’s. Furthermore, since the first and last row of M haveonly one 0, then columns 1 and k − M correspond to columns 1 and 2 j of F (2 j + 1). Hence, rows 2 to 2 j − F (2 j + 1) correspond either to rows2 to 2 j − M or to rows k − − j to k − 1. However, k ≥ j − k − − j of M has a 0 in the first or last column, andtherefore F (2 j + 1) cannot be a subconfiguration of M . It follows analogouslyfor F (2 j + 1) and M III ( k ), and since none of the matrices has at least one rowwith at least three 1’s and at least two 0’s, with the exception of M III ( k ), thisfinishes the proof. 4. Final remarks and future challenges To conclude, we leave some possible future challenges about structural char-acterizations on circle graphs. • Recall that split graphs are those chordal graphs whose complement is alsoa chordal graph. Moreover, the graph depicted in Figure 15 is a chordalgraph that is neither circle nor a split graph. It follows that the list ofgraphs given in Theorem 1.1 is not enough to characterize those chordalgraphs that are also circle. However, Theorem 1.1 is indeed a good firststep to characterize circle graphs by forbidden induced subgraphs withinthe class of chordal graphs, which remains as an open problem. Figure 15: Example of a chordal graph that is neither circle nor split. • Bouchet [4] showed that if a circle graph is the complement of a bipartitegraph, then its complement is also a circle graph. Another possible conti-nuation of this work would be studying the characterization of those circlegraphs whose complement is a circle graph as well. • Characterize Helly circle graphs (graphs that have a model of chords withthe Helly property) by forbidden induced subgraphs. The class of Hellycircle graphs was characterized by forbidden induced subgraphs withincircle graphs [5], but the problem of finding such a characterization forthe whole class of Helly circle graphs is still open. Moreover, it would be57nteresting to find a decomposition analogous as the split decompositionis for circle graphs, this is, such that Helly circle graphs are closed underthis decomposition. Acknowledgements This work was partially supported by ANPCyT PICT-2015-2218, UBACyTGrants 20020170100495BA and 20020160100095BA, PIO CONICET UNGS-144-20140100011-CO, Universidad Nacional del Sur Grants PGI 24/L103 andPGI 24/L115, and ANPCyT PICT-2017-1315 (Argentina), and ISCI CONICYTPIA FB0816; ICM P-05-004-F (Chile). Nina Pardal is partially supported by aCONICET doctoral fellowship. References [1] F. Bonomo, G. Dur´an, L.N. Grippo, and M.D. Safe. Partial characteriza-tions of circle graphs. Discrete Applied Mathematics , 159(16):1699–1706,2011.[2] A. Bouchet. Reducing prime graphs and recognizing circle graphs. Combi-natorica , 7(3):243–254, 1987.[3] A. Bouchet. Circle graphs obstructions. Journal of Combinatorial Theory.Series B , 60(1):107–144, 1994.[4] A. Bouchet. Bipartite graphs that are not circle graphs. Ann. Inst. Fourier(Grenoble) , 49(3):809–814, 1999.[5] J. Daligault, D. Gon¸calves, and M. Rao. Diamond-free circle graphs areHelly circle. 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