Balancedness of subclasses of circular-arc graphs

A graph is balanced if its clique-vertex incidence matrix contains no square submatrix of odd order with exactly two ones per row and per column. There is a characterization of balanced graphs by forbidden induced subgraphs, but no characterization by mininal forbidden induced subgraphs is known, not even for the case of circular-arc graphs. A circular-arc graph is the intersection graph of a family of arcs on a circle. In this work, we characterize when a given graph G is balanced in terms of minimal forbidden induced subgraphs, by restricting the analysis to the case where G belongs to certain classes of circular-arc graphs, including Helly circular-arc graphs, claw-free circular-arc graphs, and gem-free circular-arc graphs. In the case of gem-free circular-arc graphs, analogous characterizations are derived for two superclasses of balanced graphs: clique-perfect graphs and coordinated graphs.


Introduction
Two fundamental combinatorial optimization problems are set packing and set covering, which can be expressed by and respectively, where A is some {0, 1}-matrix.The matrix A is perfect (resp.ideal) if no integrality requirements are needed in (1) (resp.( 2)) as the polytope P (A) = {x ∈ R n + : Ax ≤ 1} (resp.the polyhedron ) has integral extreme points only.The matrix A is balanced if all its submatrices are both perfect and ideal or, equivalently, if it contains no submatrix of odd order with exactly two ones per row and per column [2,18].Well-known examples of balanced matrices are totally unimodular matrices where even no integrality requirements are needed in (1) and (2) for varying right hand side vectors.
A graph G is balanced if its clique-matrix is balanced.Here, a clique Q in a graph G = (V, E) is an inclusion-wise maximal subset of pairwise adjacent vertices and given an enumeration Q 1 , . . ., Q k of all cliques of G and an order v 1 , . . ., v n of all vertices of G, a clique-matrix of G is the k × n {0, 1}-matrix A = (a ij ) such that a ij = 1 if and only if v j ∈ Q i .The clique-matrix of a graph is unique up to permutations of rows and/or columns.The name 'balanced graphs' appeared explicitly in [3], but these graphs were already considered in [2, see Theorem 5 therein].
The class of balanced graphs is closed under taking induced subgraphs.Examples of balanced graphs are interval graphs (the intersection graphs of intervals of a line) and bipartite graphs (having a partition of their nodes into two stable sets) as their clique-matrices are totally unimodular [17,22] and, thus, balanced.
Well-known superclasses of balanced graphs are perfect graphs and hereditary clique-Helly graphs.A graph is perfect if its clique-matrix is perfect [15].Some years ago, the minimal forbidden induced subgraphs of perfect graphs were characterized [14], settling affirmatively a conjecture posed more than 40 years before by Berge [1].The minimal forbidden induced subgraphs of perfect graphs are the chordless cycles of odd length having at least 5 vertices, called odd holes C 2k+1 , and their complements, the odd antiholes C 2k+1 .
A graph is hereditary clique-Helly if, in any of its induced subgraphs, every nonempty subfamily of pairwise intersecting cliques has a common vertex.It follows from [2] that balanced graphs are hereditary clique-Helly.Prisner [28] characterized hereditary clique-Helly graphs as those graphs containing no induced 0-, 1-, 2-, or 3-pyramid (see Figure 1).Hence, no balanced graph contains an odd hole, odd antihole, or any pyramid as induced subgraph.In addition, balanced graphs were characterized by means of forbidden induced subgraphs as follows.For a graph G = (V, E) and W ⊆ V , let N (W ) = w∈W N (w) and use N (e) as shorthand for N ({u, v}) for an edge e = uv, whereas N (∅) = V (G).An unbalanced cycle of G is an odd cycle C such that, for each edge e ∈ E(C) (i.e., joining two consecutive vertices of C), there exists a (possibly empty) complete subgraph W e of G such that W e ⊆ N (e) \ V (C) and N (W e ) ∩ N (e) ∩ V (C) = ∅.Notice that it is possible for the sets W e and W e for different edges e and e to overlap.It is not hard to see that unbalanced cycles are obstructions for balanced graphs.Indeed, if, for each e ∈ E(C), we choose any clique Q e of G containing W e as well as the endpoints of e, then the submatrix M of any clique-matrix of G formed by the rows corresponding to the cliques Q e for every e ∈ E(C) and the columns corresponding to the vertices of C has exactly two ones per row and per column.More precisely, for each e ∈ E(C), the ones in the row of M corresponding to the clique Q e are exactly in the columns corresponding to the endpoints of e, while the ones in the column of M corresponding to each vertex v of C are exactly in the rows corresponding to the cliques Q e and Q e where e and e are the two edges of C incident to v.An extended odd sun is a graph G with an unbalanced cycle C such that V (G) = V (C) ∪ e∈E(C) W e and |W e | ≤ |N (e) ∩ V (C)| for each edge e ∈ E(C).The extended odd suns with the smallest number of vertices are C 5 and the pyramids.Moreover, every odd hole is an extended odd sun (by letting W e = ∅ for each e).Notice that C 3 is not an extended odd sun, since otherwise we would be forced to choose W e = ∅ for each edge e, but then N (W e ) ∩ N (e) ∩ V (C) = {v} where v is the only vertex non-incident to e (because N (W e ) = N (∅) = V (C) and N (e) = {v}).The characterization of balancedness by forbidden induced subgraphs is the following.
Theorem 1 ([2, 9]) A graph is balanced if and only if it has no unbalanced cycle, or, equivalently, if and only if it contains no induced extended odd sun.However, the above characterization is not by minimal forbidden induced subgraphs because some extended odd suns contain some other extended odd suns as induced subgraphs, as Figure 2 shows.Thus, excluding extended odd suns suffices to guarantee balancedness, but it is not necessary to exclude all of them.We address the problem of finding the minimal forbidden induced subgraphs, i.e., those graphs that are not balanced but all their proper induced subgraphs are balanced.This problem is still open.Partial answers are obtained in [10] where minimal forbidden induced subgraph characterizations of balanced graphs restricted to the following graph classes are found: P 4 -tidy graphs, paw-free graphs, line graphs, and complements of line graphs.In this paper, we study balanced graphs restricted to some subclasses of circular-arc graphs.An extended abstract containing the main results of this work appeared in [11].
Let F be a family of sets.The intersection graph of F is a graph whose vertices represent the members of F, where two members of F are adjacent if and only if they intersect.A circular-arc (CA) graph is the intersection graph of a family of open arcs on a circle [24].Since it is always possible to perturb a set of arcs on a circle so that they do not share any endpoints while preserving their intersection graph, the assumption that the arcs are open is not restrictive.Clearly, CA graphs can be seen as an extension of interval graphs.But while interval graphs form a subclass of balanced graphs, this is not the case for CA graphs.Note that CA graphs are neither perfect nor hereditary clique-Helly in general as odd holes, odd antiholes, and pyramids can be easily seen to be CA graphs.Perfectness of CA graphs was addressed in [32], but the study of balancedness of CA graphs is still in order.This paper is organized as follows.In Section 2, we present minimal forbidden induced subgraph characterizations of balanced graphs within a superclass of the class of Helly circular-arc graphs and the classes of claw-free circular-arc graphs and gem-free circular-arc graphs.In Section 3, we additionally characterize, within gem-free circular-arc graphs, two further superclasses of balanced graphs: cliqueperfect and coordinated graphs (see Section 3 for the definitions).

Basic definitions
We close this section by providing some basic definitions.All graphs in this paper are undirected, without loops and without multiple edges.Let G be a graph.We denote by V (G) its vertex set, by E(G) its edge set, and by A universal vertex of G is adjacent to all the other vertices of G and an isolated vertex of G is adjacent to no vertex of G.The neighborhood of a vertex v in G consists of all vertices that are adjacent to v and is denoted by A complete is a set of mutually adjacent vertices.The complete on n vertices will be denoted by K n .A complete on 3 vertices is said a triangle.An inclusion-wise maximal complete is a clique.A stable set of a graph is a set of pairwise nonadjacent vertices.A set Paths and cycles are assumed to be simple; i.e., with no repeated vertices aside from the starting and ending vertices in the case of cycles.By the edges of a cycle C we mean those edges joining two consecutive vertices of C. The set of edges of a cycle C will be denoted by E(C).A chord of a cycle (resp.path) is an edge joining two nonconsecutive vertices.We denote the chordless cycle on n vertices by C n and the chordless path on n vertices by P n .A chord of a cycle is short if its endpoints are at distance two within the cycle, and is long otherwise.Two chords ab and cd of a cycle C such that their endpoints are four different vertices of C that appear in the order a, c, b, d in C are called crossing.A cycle is odd if it has an odd number of vertices, and is called even otherwise.A hole is a chordless cycle of length at least 4.An odd antihole is the complement of an odd hole.
Let G and H be two graphs with The disjoint union is clearly an associative operation, and for each nonnegative integer t we will denote by tG the disjoint union of t copies of G.A class G of graphs is called hereditary if, for every graph G of G, each induced subgraph of G belongs to G. The class of balanced graphs is hereditary (see [9]).Let G and H be two graphs.We say that G is H-free to mean that G contains no induced H.If H is a collection of graphs we say that G is H-free to mean that G contains no induced H for any H ∈ H.A graph H is a forbidden induced subgraph for graph class C if no graph of C contains an induced H.Moreover, if C is a hereditary class, H is said a minimal forbidden induced subgraph for C or a minimally non-C graph if H does not belong to C but each proper induced subgraph of H belongs to C. Some small graphs to be referred to in this Fig.3: Some small graphs context are depicted in Figure 3.
For each t ≥ 3, the complete t-sun S t is a graph whose vertex set can be partitioned into a clique {q 1 , q 2 , . . ., q t } and a stable set {s 1 , s 2 , . . ., s t } such that N (s i ) = {q i , q i+1 } for each i ∈ {1, 2, . . ., t} (where q t+1 stands for q 1 ).The graph S 3 is simply called 3-sun.The 3-sun is also called tent in the literature.

Balancedness of some subclasses of circular-arc graphs
Recall that a circular-arc (CA) graph is the intersection graph of a family of open arcs on a circle.Such a family of arcs is called a CA model of the graph.CA graphs were first studied by Tucker [30,31,32,33] and can be recognized in linear time [27].Some minimal forbidden induced subgraphs for the class of CA graphs are Figure 4).Since C n ∪ K 1 is not a CA graph for any n ≥ 4, if G is a CA graph and H is a hole of G, then V (H) is dominating in G.We state the following slightly more general result for future reference (cf.[7]).
Lemma 2 Let G be a CA graph and H be a hole of G.

Balancedness of a superclass of Helly circular-arc graphs
A family A of nonempty sets has the Helly property if every nonempty subfamily of A consisting of pairwise intersecting sets has a nonempty intersection.A Helly circular-arc (HCA) graph [19] is a circulararc graph admitting a circular-arc model A that satisfies the Helly property.We call A a HCA model of the graph.The class of HCA graphs contains all interval graphs because every set of intervals of a line has the Helly property [21].Let G be a HCA graph and let A be a HCA model of G. Let us denote by A v the arc of A that corresponds to the vertex v ∈ V (G).For a clique Q of G, we call any point p ∈ v∈Q A v an anchor of Q.Since Q is inclusion-wise maximal, for each anchor p of Q and each v ∈ V (G), it holds that p ∈ A v if and only if v ∈ Q.In [23], a linear-time recognition algorithm for HCA graphs was proposed, as well as a characterization by forbidden induced subgraphs of those CA graphs that are HCA graphs (see Theorem 5 on page 11).
In this subsection, we give a minimal forbidden induced subgraph characterization of balancedness for a superclass of HCA graphs.In order to do so, we introduce the graph families below, which are schematically represented in Figure 5.
• For each t ≥ 2, let D 2t+1 be the graph with {v 1 , v 2 , . . ., v 2t+1 , u 1 , u 2 , u 3 } as vertex set such that v 1 v 2 . . .v 2t+1 v 1 is a cycle whose only chords are v 2t+1 v 2 and • For each t ≥ 2 and each even p with 4 ≤ p ≤ 2t, let X 2t+1 be the graph with vertex set In the three families of graphs above, C = v 1 v 2 . . .v 2t+1 v 1 is an unbalanced cycle and consequently all their members are not balanced.In fact, we will see later that all these graphs are minimally non-balanced (see Corollary 4).
Our first result below is the minimal forbidden induced subgraph characterization of balanced graphs restricted to HCA graphs.
Theorem 3 Let G be a HCA graph.Then, G is balanced if and only if G has no odd holes and contains no induced 3-sun, 1-pyramid, 2-pyramid, C 7 , V 2t+1 p , D 2t+1 , or X 2t+1 p for any t ≥ 2 and any valid p.
Proof: The 'only if' part is clear because the class of balanced graphs is hereditary.Conversely, suppose that G is not balanced.Then, G contains some induced subgraph H that is minimally non-balanced; i.e., H is not balanced but each proper induced subgraph of H is balanced.Since G is a HCA graph, H also is so.The proof will be complete as soon as we prove that H is an odd hole, 3-sun, 1-pyramid, 2-pyramid, C 7 , V 2t+1 p , D 2t+1 , or X 2t+1 p for some t ≥ 2 and some valid p.Since H is not balanced, the clique-matrix of H contains some square submatrix of odd order with two ones per row and per column.Clearly, this implies that the clique-matrix of H contains some square submatrix that is the clique-matrix of an odd chordless cycle.Therefore, there are some cliques for each i ∈ {1, 2, . . ., 2t + 1} (all along the proof, subindices are to be understood modulo 2t + 1) Suppose first that t = 1.Since C is an unbalanced cycle, there is some vertex there must be at least one vertex of W v1v2 that is not adjacent to v 3 ; clearly, it suffices to let u 1 be any such vertex.Similarly, there is some vertex , then H equals a 3-sun, 1-pyramid or 2-pyramid (because the 3-pyramid is not a HCA graph).
From now on, we will assume, without loss of generality, that t ≥ 2. Let A be a HCA model of H in a circle C. Denote by A i the arc of A corresponding to the vertex v i for each i ∈ {1, 2, . . ., 2t + 1}.Fix an anchor p j of the clique Q j for each j ∈ {1, . . ., 2t + 1}.By construction, p j ∈ A i if and only if . ., A 2t+1 are arcs of C, there are only two possible orders for the anchors when traversing C in clockwise direction, either p 1 , p 2 , . . ., p 2t+1 or p 2t+1 , . . ., p 2 , p 1 .So, we can assume, without loss of generality, that the anchors p 1 , p 2 , . . ., p 2t+1 appear exactly in that order when traversing C in clockwise direction.Hence, A i ∩ {p 1 , p 2 , . . ., p 2t+1 } = {p i−1 , p i } implies that A i is contained in the clockwise open arc of C that starts in p i−2 and ends in p i+1 for each i ∈ {1, . . ., 2t+1}.We now prove the following three claims about C.
Claim 1 All chords of C are short.
If t = 2, all possible chords of C are short.So, suppose that t ≥ 3. Since A i is contained in the clockwise open arc of C that starts in p i−2 and ends in p i+1 for each i ∈ {1, . . ., 2t + 1}, it follows that if the arc A i intersects A j for some j ∈ {1, . . ., 2t + 1} then i = j − 2, j − 1, j, j + 1, or j + 2 (modulo 2t + 1).We conclude that each chord of C is short, as claimed.
Claim 2 Any set of three vertices of C that induces a triangle in H consists of three consecutive vertices of C.
Suppose, for the purpose of contradiction, that there is some set S of three vertices of C that induces a triangle T in H but, nevertheless, S does not consist of three consecutive vertices of C. Notice that if each vertex of S were consecutive in C to some other vertex of S, then S would consist of three consecutive vertices of C. So, necessarily, there must be some vertex s 1 of S such that s 1 is not consecutive in C to any vertex of S \ {s 1 }.By symmetry, we can assume that s 1 = v 1 and, since all chords of C are short, S = {v 1 , v 3 , v 2t }.As C is odd and each of its chords is short, necessarily t = 2. Consequently, S = {v 1 , v 3 , v 4 } is contained in some clique of H, that should have some anchor q.Nevertheless, since A 1 is contained in the clockwise open arc of C that starts in p 4 and ends in p 2 , A 3 is contained in the clockwise open arc of C that starts in p 1 and ends in p 4 , and A 4 is contained in the clockwise open arc of C that starts in p 2 and ends in p 5 , there is no suitable position in C for q.This contradiction proves that indeed any set of three vertices of C that induces a triangle in H consists of three consecutive vertices of C, as claimed.
Claim 3 Every two chords of C are crossing.
Suppose, for the purpose of contradiction, that C has two different chords e i = v i−1 v i+1 and e j = v j−1 v j+1 that are not crossing (recall that all chords are short by Claim 1).Notice that it is possible that e i and e j share one endpoint.We will show that H − {v i , v j } is not balanced.Indeed, consider the cycle With the help of the three previous claims, we complete the proof of Theorem 3. Notice that if C has no chords, then, by the minimality of H, H = C 2t+1 , as required.Therefore, we will assume that C contains at least one chord.Since all chords of C are short and crossing by Claims 1 and 3, either C has exactly one chord that is short or C has exactly two chords that are short and are crossing.We divide the remaining proof into two cases corresponding to the former and the latter cases.
Case 1 C has exactly one chord that is short.
Without loss of generality, let v 1 v 3 be the only chord of C. Since C is an unbalanced cycle, there exists Suppose, for the purpose of contradiction, that u 1 is adjacent to u 2 .If u 2 were adjacent to v 2t+1 , then either 3 in H, depending on whether u 1 is adjacent to v 2t+1 or not, respectively.Since H is a minimally non-balanced CA graph and K 2,3 is not a CA graph, we conclude that u 2 is not adjacent to v 2t+1 .If u 1 were adjacent to v 2t+1 , then {v 2t+1 , v 1 , v 2 , v 3 , u 1 , u 2 } would induce a proper 1-pyramid in H.This contradiction shows that u 1 is not adjacent to v 2t+1 , and this means that q = 2. Symmetrically, p = 2.But then, either t = 2 and {v H, which are not CA graphs, a contradiction.This contradiction arose from assuming that u 1 and u 2 were adjacent, so we conclude that u 1 is not adjacent to u 2 .
If p were odd, then u 2 v p+1 v p+2 . . .v 2t+1 v 1 v 2 u 2 would be an odd hole in H, contradicting the minimality of H. Thus, p is even and, by symmetry, q is also even.If t = 2, then, up to symmetry, either p = q = 4 and H = C 7 , or q = 2 and H = V 5 p for some p ∈ {2, 4}, as desired.So, without loss of generality, assume that t ≥ 3.If N H (u 1 ) ∩ N H (u 2 ) = {v 2 }, then, since p and q are even, there would exist some k would be an odd hole in H, contradicting the minimality of H. Therefore, we can assume that q = 2, and finally H = V 2t+1 p for some p even such that 2 ≤ p ≤ 2t.
Case 2 C has exactly two chords that are short and are crossing.
Since the two chords are crossing, we assume, without loss of generality, that the chords of C are v 1 v 3 and v 2t+1 v 2 .Since C is an unbalanced cycle, there is some By construction, 2 ≤ r ≤ 2t and, by Lemma 2 applied to the hole induced by would induce a proper 1-, 2-or 3-pyramid in H (depending on the existence or not of the edges u 1 u 2 and u 1 v 3 ), a contradiction with the minimality of H.If r is even and 2 < r < 2t, then the cycle u 2 v r+1 v r+2 . . .v 2t v 2t+1 v 2 u 2 would be a proper odd hole in H, a contradiction.If r were odd and r = 3, then the cycle u 2 v r+1 v r+2 . . .v 2t v 2t+1 v 1 v 3 u 2 would be a proper odd hole in H, a contradiction.So, r = 2 or 3. Symmetrically, if s = |N H (u 1 ) ∩ V (C)|, then s = 2 or 3 and, by Lemma 2 applied to the hole induced by Suppose, for the purpose of contradiction, that u 1 and u 2 are adjacent.Then, the set {u 1 , v 1 , v 2 , u 2 } induces a C 4 in H, which must be dominating because H is a CA graph.If t = 2, then at least one of u 1 and u 2 should be adjacent to v 4 and either So, in all cases we reach a contradiction with the minimality of H.These contradictions prove that u 1 and u 2 are nonadjacent.
We claim that r = s = 2. Indeed, if r = s = 3, then v 1 v 2 u 2 v 4 v 5 . . .v 2t u 1 v 1 would be an odd hole in H, a contradiction.Alternatively, if r = 3 and s from the discussion in the paragraph above.)The case r = 2 and s = 3 is symmetric.We conclude that our claim, r = s = 2, is true; in other words, Suppose that there is some Then, by minimality, If u 1 were adjacent to u 3 , then either t = 2 and {v 2 , v 3 , v 4 , v 5 , u 1 , u 3 } would induce a domino in H, or t ≥ 3 and {u 1 , v 2t+1 , v 2 , u 3 , v 5 } would induce C 4 ∪ K 1 in H, which are not CA graphs, a contradiction.So, u 1 is nonadjacent to u 3 and, symmetrically, u 2 is nonadjacent to u 3 .We conclude that, if (*) holds, H = D 2t+1 , as desired.
It only remains to consider the case when (*) does not hold.Since C is an unbalanced cycle, this means that there are two adjacent vertices u 3 and u 4 such that u 3 , u 4 ∈ N H (v 1 v 2 ) \ V (C), u 3 is adjacent to v 2t+1 but not to v 3 , and u 4 is adjacent to v 3 but not to v 2t+1 .
If p were odd and p = 3, then v 1 u 4 v p v p+1 . . .v 2t+1 v 1 would be a proper odd hole in H, a contradiction.So, p = 3 or p is even.Symmetrically, q = 3 or q is even.If p and q had the same parity, then u 3 u 4 v p v p+1 . . .v 2t−q+4 u 3 would be a proper odd hole of H (recall that ), a contradiction.By symmetry, we will assume, without loss of generality, that p is even, p ≥ 4, and q = 3.In particular, u 4 is adjacent to v 4 .
Notice that u 2 is not adjacent to u 3 , since otherwise u 2 v 3 v 4 . . .v 2t+1 u 3 u 2 would be a proper odd hole of H.Moreover, u 2 is adjacent to u 4 , since otherwise {v 2 , v 3 , v 4 , u 3 , u 4 , u 2 } would induce a proper 3-sun in H. So, N H (u 2 ) = {v 2 , v 3 , u 4 }.(Recall that we already proved that u 1 and u 2 are nonadjacent.) If u 1 were adjacent to u 4 , then {v 2t+1 , v 1 , v 2 , u 1 , u 2 , u 4 } would induce a proper 1-pyramid in H, contradicting the minimality of H. So, u 1 is nonadjacent to u 4 .Finally, if u 1 were adjacent to u 3 , then and, therefore, since u 1 and u 4 are nonadjacent, V (C )∪{u 1 , u 4 } would induce a proper V 2t+1 p in H, a contradiction.This contradiction shows that u 1 is nonadjacent to u 3 and we conclude that N H (u 1 ) = {v 2t+1 , v 1 }.We proved that H = X 2t+1 p where p is even and 4 ≤ p ≤ 2t, as required. 2 It is easy to see that among the forbidden induced subgraphs that characterize balancedness in Theorem 3 there are no two of them such that one is a proper induced subgraph of the other.Therefore, Theorem 3 is indeed a characterization by minimal forbidden induced subgraphs.In particular, we obtain the following result.

Corollary 4
The graphs V 2t+1 p , D 2t+1 , and X 2t+1 p are minimally non-balanced for any t ≥ 2 and any valid p.
We will extend Theorem 3 to a superclass of HCA graphs; namely, the class of {net,U 4 ,S 4 }-free CA graphs (cf. Figure 3).This extension will also serve as a basis for the characterizations in the following two subsections.
For that, let us firstly present the forbidden induced subgraph characterization of those CA graphs that are HCA graphs given in [23].Let an obstacle be a graph H containing a clique Q = {v 1 , v 2 , . . ., v t } where t ≥ 3 and such that for each i ∈ {1, . . ., t}, at least one of the following assertions holds (where in both assertions, v t+1 means v 1 ): With this definition, the characterization of those CA graphs that are HCA graphs runs as follows.
Theorem 5 ([23]) Let G be a CA graph.Then, G is a HCA graph if and only if G contains no induced obstacle.
Notice that obstacles are not necessarily minimal; i.e., there are obstacles that contain proper induced obstacles.For instance, 2C 5 is an obstacle and contains a proper induced 2P 4 , which is also an obstacle.In addition, there are minimal obstacles that are not CA graphs; e.g., antenna and C 6 are minimal obstacles that are not CA graphs.Our next result determines all the {1-pyramid,2-pyramid}-free minimal obstacles that are CA graphs.Recall that for each t ≥ 3, S t denotes the complete t-sun.
Theorem 6 Let H be a {1-pyramid,2-pyramid}-free minimal obstacle which is a CA graph.Then, H is the 3-pyramid, U 4 , or S t for some t ≥ 3.
Proof: Let Q = {v 1 , . . ., v t }, the w i 's, the u i 's, and the z i 's as in the definition of an obstacle.All along the proof, subindices should be understood modulo t.
Let us consider first the case where t = 3. Suppose that (Y 2 ) holds for at least two values of i, say i = 1 and i = 2.Then, {u 1 , z 1 , z 2 } is a complete and {v 1 , v 2 , v 3 , u 1 , z 1 , z 2 } induces a 3-pyramid, since otherwise {v 1 , v 2 , v 3 , u 1 , z 1 , z 2 } would induce a 1-pyramid or a 2-pyramid.Hence, by minimality, H = 3-pyramid.Consider now the case where (Y 2 ) holds for exactly one value of i, say i = 1, and, consequently, (Y 1 ) holds for i = 2 and i = 3.We claim that {u 1 , z 1 } is anticomplete to w 2 .Indeed, if w 2 were adjacent to z 1 , then {v 1 , v 2 , v 3 , w 2 , z 1 , u 1 } would induce a 1-pyramid or a 2-pyramid in H, a contradiction.In addition, if w 2 were adjacent to u 1 , then {v 1 , v 2 , u 1 , z 1 , w 2 } would induce a K 2,3 in G, which is not a CA graph.We proved that {u 1 , z 1 } is anticomplete to w 2 and, symmetrically, to w 3 .Also notice that w 2 and w 3 are nonadjacent, since otherwise {v 1 , v 2 , w 2 , w 3 , u 1 , z 1 } would induce a domino, which is not a CA graph.Then, by minimality, H = U 4 , as desired.Finally, assume that (Y 1 ) holds for each i ∈ {1, 2, 3}.Necessarily {w 1 , w 2 , w 3 } is a stable set, since otherwise G would contain an induced C 4 ∪ K 1 , G 3 (see Figure 4), or C 6 which are not CA graphs.By minimality, H = net = S 3 , as desired.
From now on, we assume that t ≥ 4. Suppose, by way of contradiction, that (Y 2 ) holds for some i, say i = 1.On the one hand, if (Y 1 ) held for i = 3, then {v 1 , v 2 , v 3 , u 1 , z 1 , w 3 } would induce a 1-pyramid, 2-pyramid, or a proper 3-pyramid in H, a contradiction.On the other hand, if (Y 2 ) held for i = 3, then {v 1 , v 2 , v 3 , u 1 , z 1 , u 3 } would induce a 1-pyramid, 2-pyramid or a proper 3-pyramid in H, a contradiction.These contradictions arose from assuming that (Y 2 ) held for some i.We conclude that, if t ≥ 4, then (Y 2 ) does not hold for any i ∈ {1, . . ., t} and, by definition of obstacle, (Y 1 ) holds for each i ∈ {1, . . ., t}.By minimality, the vertices of H are Q ∪ W where W = {w 1 , w 2 , . . ., w t }.We claim that W is a stable set and, consequently, H = S t .We divide the proof of the claim into two cases: t = 4 and t ≥ 5.
Assume that t = 4. Suppose, for the purpose of contradiction, that W is not a stable set.Suppose first that w i is adjacent to w i+1 for some i, say w 3 is adjacent to w 4 .Necessarily w 1 is nonadjacent to w 4 , since otherwise {v 1 , v 2 , v 3 , w 1 , w 3 , w 4 } would induce a 1-pyramid or a 2-pyramid in H (depending on the adjacency between w 1 and w 3 ), a contradiction.In addition, w 1 is nonadjacent to w 3 , since otherwise {w 1 , v 1 , w 4 , v 3 , w 3 } would induce a K 2,3 , which is not a CA graph.Symmetrically, w 2 is nonadjacent to w 3 and w 4 .On the one hand, if w 1 and w 2 are adjacent, then {w 2 , v 1 , w 3 , w 4 , v 3 , w 1 } induces a domino in G, which is not a CA graph.On the other hand, if w 1 and w 2 are nonadjacent, then {v 1 , v 2 , v 3 , w 1 , w 2 , w 3 , w 4 } induces a proper U 4 in H, a contradiction with the minimality of H.These contradictions prove that w i is not adjacent to w i+1 for any i.Notice that also w i and w i+2 are nonadjacent, since otherwise {v i , w i , w i+2 , v i+3 , w i+3 } would induce C 4 ∪ K 1 in G, which is not a CA graph.We conclude that W is a stable set and H = S 4 , as claimed.
It only remains to consider the case where t ≥ 5. Let S be any unordered pair of vertices from W . Since t ≥ 5, S can be extended to a set S = {w i , w j , w j+1 } of three vertices where i and j are not consecutive modulo t and neither are i and j + 1.Notice that S is a stable set in H, since otherwise {v i , v j , v j+2 , w i , w j , w j+1 } would induce a 1-pyramid, a 2-pyramid, or a proper 3-pyramid in H, a contradiction.Since S is a stable set, so is S. Since S is any pair of vertices from W , W is a stable set and H = S t , as claimed.
Finally, notice that 3-pyramid, U 4 and S t for t ≥ 3 are obstacles, are CA graphs, and none of them is a proper induced subgraph of any of the others. 2 As a corollary of Theorems 5 and 6, we obtain a minimal forbidden induced subgraph characterization of HCA graphs within the class of {1-pyramid,2-pyramid}-free CA graphs.
Corollary 7 Let G be a {1-pyramid,2-pyramid}-free CA graph.Then, G is a HCA graph if and only if it contains no induced 3-pyramid, U 4 , or S t for any t ≥ 3.
Since net, U 4 , and S 4 are obstacles, the class of {net,U 4 ,S 4 }-free CA graphs is indeed a superclass of HCA graphs.We now prove the main result of this subsection, which is an extension of the characterization of Theorem 3 to the class of {net,U 4 ,S 4 }-free CA graphs.Proof: If G is a HCA graph, the result reduces to Theorem 3. So, assume that G is not a HCA graph.Then, by Corollary 7 and since G is {net,U 4 ,S 4 }-free, G contains an induced 1-pyramid, 2-pyramid, or 3-pyramid or an induced S t for some t ≥ 5 (notice that S 3 = net and S 4 = S 4 ).Since S t contains an induced 3-sun for every t ≥ 5, we conclude that G is not balanced and contains an induced pyramid. 2 Corollary 8 is crucial in the proof of the main results in the next two subsections.

Balancedness of claw-free circular-arc graphs
In this subsection we will characterize, by minimal forbidden induced subgraphs, those claw-free CA graphs that are balanced.A proper circular-arc graph (PCA) is a CA graph admitting a CA model in which no arc properly contains another.The class of claw-free CA graphs is a superclass of the class of PCA graphs, as follows from the forbidden induced subgraph characterization of PCA graphs in [31].
By Corollary 8, in order to characterize balanced graphs within claw-free CA graphs, it will be enough to study the balancedness of those claw-free CA graphs containing an induced net (because claw-free graphs contain neither induced U 4 's nor induced S 4 's).The following lemma will be of help in analyzing the structure of claw-free CA graphs containing an induced net.

Lemma 9 ([7]
) Let G be a claw-free CA graph containing a net induced by H = {t 1 , t 2 , t 3 , s 1 , s 2 , s 3 }, where {t 1 , t 2 , t 3 } induces a triangle and s i is adjacent to (where subindices should be understood modulo 3).
A graph G arises from a graph H by vertex replication if G can be obtained from H by replacing each vertex x of H by a nonempty complete graph M x and adding all possible edges between M x and M y if and only if x and y are adjacent in H.In [7], a slightly stronger variant of the above lemma is used to study the structure of chordal claw-free CA graphs containing an induced net.The proof in [7] can be easily adapted to prove the following related result in which chordality is not required.For the sake of completeness, we give the adapted proof.
Theorem 10 ( [7]) If G is a claw-free CA graph containing an induced net and containing no induced 3-sun, then G arises from the net by vertex replication.
Proof: The proof will be by induction on the number of vertices of G.If |V (G)| = 6, then G equals a net, which trivially arises from the net by vertex replication.So, assume that |V (G)| > 6.Then, there is some vertex v of G such that G − {v} contains an induced net.Since G − {v} is also a claw-free graph containing an induced net and containing no induced 3-sun, by induction hypothesis, G − {v} arises from a net by vertex replication; i.e., the vertices of V (G − {v}) can be partitioned into nonempty completes S 1 , S 2 , S 3 , T 1 , T 2 , T 3 such that T 1 , T 2 , T 3 are mutually complete and T i is complete to S i and anticomplete to S i+1 and S i+2 , for each i ∈ {1, 2, 3} (where subindices along the proof should be understood modulo 3).Let If s j ∈ S j , then, applying Lemma 9 to the net induced by {t 1 , t 2 , t 3 , s j , s j+1 , s j+2 }, it follows that v is adjacent to s j if and only if j = i.Thus, v is complete to S i and anticomplete to S i+1 and S i+2 .Using the same strategy, we can prove that v is complete to T i and anticomplete to T i+1 and T i+2 .Thus, we can obtain a partition of the vertices of G showing that G is arises from a net by vertex replication (replacing S i by S i ∪ {v}).
Finally, consider that Reasoning as in the above paragraph, it follows that v is complete to T 1 , T 2 , T 3 , and S i , and v is anticomplete to S i+1 and S i+2 .Thus, we obtain a partition of the vertices of G showing that G arises from a net by vertex replication (replacing T i by T i ∪ {v}). 2 Recall that a graph is minimally non-balanced if it is not balanced but each of its proper induced subgraphs is balanced.It is clear that minimally non-balanced graphs have no true twins, since the removal of one of the true twins only eliminates a repeated column in the clique-matrix.Now we state and prove the main result of this subsection.
Theorem 11 Let G be a claw-free CA graph.Then, G is balanced if and only if G has no odd holes and contains no induced pyramids and no induced C 7 .
Proof: The 'only if' part is clear.In order to prove the 'if' part, suppose that G is not balanced.Then, G contains some induced subgraph H that is minimally non-balanced.Since G is a claw-free CA graph, H also is so.The proof will be complete if we prove that H is an odd hole, a pyramid, or C 7 .Suppose, for the purpose of contradiction, that H is not net-free.By Theorem 10, H is a net, has true twins, or contains an induced 3-sun.Since the net is balanced and since minimally non-balanced graphs have no true twins, G contains an induced 3-sun.By minimality, H is a 3-sun, a contradiction with the fact that H is not net-free.This contradiction proves that H is net-free.Since U 4 and S 4 are not claw-free, H is {net,U 4 ,S 4 }-free and Corollary 8 implies that H has an odd hole or contains an induced pyramid or C 7 (because each of V 2t+1 p , D 2t+1 , and X 2t+1 p contains an induced claw for each t ≥ 2 and each valid p).By the minimality of H, we conclude that H is an odd hole, a pyramid, or C 7 , as required. 2 As PCA graphs are claw-free, and the odd holes, the pyramids, and C 7 are all PCA graphs, the minimal forbidden induced subgraphs for balancedness within PCA graphs are the same as those within claw-free CA graphs.

Balancedness of gem-free circular-arc graphs
In this subsection we will give minimal forbidden induced subgraph characterizations of balanced gemfree CA graphs.
Lemma 12 Let G be a gem-free CA graph that contains an induced net or an induced U 4 .Then, G either has true twins or has a cutpoint.
Proof: Assume that G has no true twins.We will prove that G has a cutpoint.
Consider first the case where G contains an induced U 4 .That is, there is some chordless cycle C = u 1 u 2 u 3 u 4 u 1 in G, some vertex z that is complete to V (C), and a pair of nonadjacent vertices We will analyze the possibilities for the nonempty set N G (v) ∩ V (C).
Suppose, for the purpose of contradiction, that v has two neighbors on C.Then, they are consecutive vertices of C by Lemma 2. So, N G (v) ∩ V (C) = {u i , u i+1 } for some i ∈ {1, 2, 3, 4} (from now on, subindices should be understood modulo 4).If v were not adjacent to z, then {v, u i , z, u i+2 , u i+1 } would induce a gem in G.If v were adjacent to z, then {v, u i+1 , u i+2 , u i+3 , z} would induce a gem in G. Since G is gem-free, we conclude that Now, for each i ∈ {1, 2, 3, 4}, let V i be the set of vertices not in V (C) whose only neighbor in C is u i .In particular, p i ∈ V i for each i ∈ {1, 2}.Let Z be the set of vertices not in V (C) that are complete to V (C), so z ∈ Z. Finally, for each i ∈ {1, 2, 3, 4}, let Vi be the set of vertices not in V (C) whose only non-neighbor in C is u i .Notice that every v ∈ V (G) \ V (C) belongs to one of the sets in {V i , Vi } i∈{1,2,3,4} or Z.
Claim 1 V i is anticomplete to V j for every i = j.Indeed, if v i ∈ V i and v j ∈ V j were adjacent, then V (C) ∪ {v i , v j } would induce either a domino or the graph G 2 in Figure 4, which are not CA graphs, a contradiction.
Indeed, if v i ∈ V i and w ∈ Z were adjacent, then {v i , u i , u i+1 , u i+2 , w} would induce a gem in G, a contradiction.
Claim 3 Z is a complete.Indeed, if w, w in Z were nonadjacent, then, by the previous claim, both of them would be nonadjacent to p 2 and {u 1 , w, u 3 , w , p 2 } would induce C 4 ∪ K 1 in G, which is not a CA graph, a contradiction.Claim 4 Vi is a complete and is complete to Z for every i ∈ {1, 2, 3, 4}.Indeed, if vi , v i in Vi were nonadjacent, then {v i , v i , u i , u i−1 , u i+1 } would induce K 2,3 in G, which is not a CA graph, a contradiction.And, if vi ∈ Vi and w ∈ Z were nonadjacent, then {v i , u i+2 , w, u i , u i+1 } would induce a gem in G, also a contradiction.
By the previous claims, all the vertices in Z are true twins.So, since G has no true twins, we conclude that Z = {z}.

Claim 6
Vi is anticomplete to V j for every j = i + 2.
Let vi ∈ Vi and v j ∈ V j and suppose, by way of contradiction, they are adjacent.If j = i, then {u i , vi , u i+1 , u i+3 , v j } induces a K 2,3 in G, that is not a CA graph, a contradiction.If j = i ± 1, then {v j , u i+1 , u i+2 , u i+3 , vi } induces a gem in G, also a contradiction.These contradictions prove that vi and v j are nonadjacent unless j = i + 2.
Suppose, by way of contradiction, that Vi is nonempty for some i ∈ {1, 2, 3, 4} and let vi ∈ Vi .Since vi is not a true twin of v i+2 , by the previous claims, there must be a vertex v i+2 in V i+2 that is nonadjacent to vi .But then, {v i , u i+3 , u i , u i+1 , v i+2 } induces a C 4 ∪ K 1 in G, that is not a CA graph, a contradiction.By all the previous claims, u 1 and u 2 are cutpoints of G, as required.This completes the proof when G contains an induced U 4 .
It only remains to consider the case where G contains no induced U 4 but a net induced by

Suppose now that |N
Finally, if |N G (v)∩H| = 1, then the neighbor of v in H belongs to T ; since otherwise G would contain the graph G 5 in Figure 4 as induced subgraph, and it is not a CA graph.
This shows that we can classify the vertices in G − H as follows: Let S i be the set of vertices in G − H whose only neighbor in T is t i (i.e., the set of neighbors in H is either {t i } or {t i , s i }), T i be the set of vertices in G − H whose neighbors in H are {t 1 , t 2 , t 3 , s i }, and Z i be the set of vertices in G − H whose neighbors in H are T − {t i }.Since G is gem-free, at most one of the Z i 's is nonempty.So, without loss of generality, assume that Z 2 and Z 3 are empty., for each t ≥ 2 and each valid p, contains an induced gem).The minimality of H ensures that H is an odd hole or a 3-pyramid, which concludes the proof. 2 We conclude this section with Table 1 that summarizes the characterizations of balanced graphs by minimal forbidden induced subgraphs within each studied subclass of CA graphs.

Considering further superclasses of balanced graphs
Recall that well-known superclasses of balanced graphs are perfect graphs and hereditary clique-Helly graphs.As perfect graphs are exactly the odd hole-and odd antihole-free graphs [14] and hereditary clique-Helly graphs exactly the pyramid-free graphs [28], the list of minimal forbidden induced subgraphs from Theorems 11 and 14 implies the following for claw-free and gem-free CA graphs: Corollary 15 Let G be a claw-free CA graph.Then, G is balanced if and only if G is perfect and hereditary clique-Helly.

Corollary 16
Let G be a gem-free CA graph.Then, G is balanced if and only if G is perfect and hereditary clique-Helly.
This motivates us to also consider the relationship between CA graphs and further superclasses of balanced graphs.Note that perfect graphs are characterized in several different ways, among them in terms of a min-max relation of two graph parameters: The size of a largest clique in a graph is a trivial lower bound for the minimum number of colors required to assign different colors to adjacent vertices; perfect graphs are exactly the graphs where both parameters coincide for all induced subgraphs.Some other graph classes were defined in a similar way and turned out to be superclasses of balanced graphs.
A clique-independent set of a graph G is a set of pairwise disjoint cliques of G.A clique-transversal of G is a subset of vertices meeting all the cliques of G.The maximum cardinality α c (G) of a cliqueindependent set is a trivial lower bound for the minimum cardinality τ c (G) of a clique-transversal of G.A graph G is clique-perfect if and only if α c (H) = τ c (H) holds for each induced subgraph H of G.The term 'clique-perfect' was coined in [20], but the equality of these parameters has been studied long before in the context of hypergraphs [4].It is important to mention that clique-perfect graphs do not need to be perfect; e.g., odd antiholes of length 6n + 3 are clique-perfect for each n ≥ 1 and are not perfect (Reed, 2001, cf.[16]).
Claim 2 For each i ∈ {1, 2, 3, 4}, every vertex of V i is a true twin of w i in H.
In fact, notice that the following statements hold for each i ∈ {1, 2, 3, 4} (subindices should be understood modulo 4): • V i is complete to U : If there were vertices v i ∈ V i and u ∈ U which were nonadjacent, then {u, v i , w i , w i+1 , w i+3 } would induce a K 2,3 in H, a contradiction.
• V i is a complete set: If there were two different nonadjacent vertices v i and v i in V i , then the set {v i , v i , w i , w i+1 , w i+3 } would induce a K 2,3 in H, a contradiction.
• V i is complete to V i+1 : If there were vertices v i in V i and v i+1 in V i+1 which were nonadjacent, then {v i , w i+2 , v i+1 , w i , w i+3 } would induce a gem in H, a contradiction.
• V i is anticomplete to V i+2 : If there were vertices v i in V i and v i+2 in V i+2 which were adjacent, then {w i , v i+2 , v i , w i+2 , w i+1 } would induce a gem in H, a contradiction.
By Claim 1, the definition of V i , and the four above statements, it follows that every vertex of V i is a true twin of w i in H, as claimed.
Claim 3 The sets V 1 , V 2 , V 3 , and V 4 are all empty.
Indeed, since α c , τ c , γ c , and ∆ c are invariant under the addition of a true twin, no graph in C 1 ∪ C 2 has true twins.In particular, H has no true twins and, by Claim 2, V i is empty for each i ∈ {1, 2, 3, 4}, as claimed.
Claim 4 H is the disjoint union of three or more complete bipartite graphs.(We regard an isolated vertex as a trivial complete bipartite graph.) In fact, since H is gem-free and K 2,3 -free, H[U ] is P 4 -free and 3K 1 -free.Moreover, in [7], it is proved that if a graph is {3K 1 , P 4 }-free, then its complement is the disjoint union of complete bipartite graphs.Therefore, H[U ] is the disjoint union of complete bipartite graphs.By Claims 1 and 3, V (H) = U ∪ W .
Since H[W ] = C 4 = 2K 2 and U is anticomplete to W in H, H is the disjoint union of at least three complete bipartite graphs, as claimed.
With the help of Claim 4 we now complete the proof of the theorem.Notice that Claim 4 implies that H / ∈ C 1 because no minimally non-clique-perfect graph is the complement of a disconnected graph (cf.[25,Lemma 1]).So, necessarily, H ∈ C 2 ; i.e., H is minimally non-coordinated.Then, γ c (H) = ∆ c (H) and, in particular, H has no universal vertices; i.e., each connected component of H has at least two vertices.Let H 1 , H 2 , . . ., H t be the connected components of H and, for each i ∈ {1, 2, . . ., t}, let {A 1 i , A 2 i } be the partition into two stable sets of the vertices of the complete bipartite graph H i .Then, the cliques of H are of the form A j1 1 ∪ A j2 2 ∪ • • • ∪ A jt t where j 1 , . . ., j t ∈ {1, 2}.Notice that γ c (H) = 2 t−1 and clearly ∆ c (H) = 2 t−1 (indeed, each vertex of H belongs to 2 t−1 cliques of H), which contradicts the fact that H ∈ C 2 .This contradiction arose from assuming that there was some graph in C 1 ∪ C 2 which was not an odd hole, and thus completes the proof of the theorem. 2

Further remarks
In this work, we considered the problem of characterizing balanced graphs by minimal forbidden subgraphs within different subclasses of CA graphs, including a superclass of each of two of the most studied subclasses of CA graphs: the class of HCA graphs and the class of PCA graphs.The complete characterization of balanced graphs by minimal forbidden induced subgraphs within CA graphs in general, remains unknown.The sun S 5 is the only example of a minimally non-balanced CA graph not belonging to any of the here studied classes of CA graphs that we know.(The graphs S t with t odd and t ≥ 7 are not circular-arc graphs.)A careful reading of the proof of Theorem 3 reveals that the hypothesis that the graph is HCA (and not merely a CA graph) is only used in the proofs of Claim 1 and 2, and in the latter case only for t = 2. So, along the proof we indeed identified all CA graphs that are minimally non-balanced and whose unbalanced cycles have length at least 7 and have only short chords.Therefore, a possible road towards extending the proof of Theorem 3 to the entire class of CA graphs could be that of finding some property of the chords of the unbalanced cycles within CA graphs in general that could serve as a substitute for Claim 1.A different approach would be to take Theorem 6 as a starting point and to study the balancedness of CA graphs containing net, U 4 , or S 4 as induced subgraph.We managed to do so when restricting ourselves to claw-free and gem-free graphs.A better understanding of the structure of CA graphs would be of help to overcome these restrictions.
We also considered the problem of characterizing clique-perfect and coordinated graphs by forbidden induced subgraphs within CA graphs.In [6], clique-perfect graphs were characterized by minimal forbidden induced subgraphs within the class of HCA graphs.It is easy to see that the approach used here to extend Theorem 3 to Corollary 8, works also for extending the characterization of clique-perfect graphs within HCA graphs in [6] to a characterization of clique-perfect graphs within the class of {net,U 4 ,S 4 }free ∩ {1-pyramid,2-pyramid,3-pyramid}-free CA graphs.Nevertheless, characterizing clique-perfect graphs by forbidden induced subgraphs within all {net,U 4 ,S 4 }-free CA graphs seems much harder.The characterization of clique-perfect graphs by forbidden induced subgraphs is open even within PCA graphs.For coordinated graphs, the characterization remains unresolved even within both HCA graphs and PCA graphs.

Fig. 2 :
Fig.2: On the left, an extended odd sun that is not minimal.Bold lines correspond to the edges of a proper induced extended odd sun, depicted on the right.

Fig. 4 :
Fig. 4: Some forbidden induced subgraphs for the class of circular-arc graphs

Fig. 5 :
Fig. 5: Families of minimally non-balanced HCA graphs: (a) Family V 2t+1 p : The dotted paths joining v3 and vp+1 resp.vp+2 and v1 represent chordless even paths, not simultaneously empty.All vertices of the dotted path joining v3 to vp+1 are adjacent to u2.(b) Family D 2t+1 : The dotted path joining v3 and v2t+1 represents a nonempty even path of length 2t − 2. (c) Family X 2t+1 p : The dotted paths joining v4 and vp resp.vp+1 and v2t+1 represent any chordless even paths, both of them possibly empty, even simultaneously.The vertices of the dotted path joining v4 to vp are all adjacent to u4.
For each edge e of C , define W e = ∅ if either e = e i or e = e j , and W e = W e otherwise.Since all the triangles of C are induced by three consecutive vertices of C by Claim 2, C and the W e 's satisfy the definition of an unbalanced cycle.Indeed, for each edge e of C , either W e = W e and N (W e ) ∩ N (e) ∩ V (C ) ⊆ N (W e ) ∩ N (e) ∩ V (C) = ∅, or e = e k for k ∈ {i, j} and N (W e ) ∩ N (e) ∩ V (C ) ⊆ N (e) ∩ (V (C) \ {v k }) = ∅ because, by Claim 2, the only vertex of C with which vertices v k−1 and v k+1 can form a triangle in H is v k .Therefore, H − {v i , v j } is not balanced, a contradiction with the minimality of H.This contradiction shows that indeed every two chords of C are crossing, as claimed.

Corollary 8
Let G be a {net,U 4 ,S 4 }-free CA graph.Then, G is balanced if and only if G has no odd holes and contains no induced pyramid, C 7 , V 2t+1 p , D 2t+1 , or X 2t+1 p for any t ≥ 2 and any valid p.
then G would contain an induced gem, so |N G (v) ∩ H| ≤ 4. Suppose that |N G (v) ∩ H| = 4.If |N G (v) ∩ S| = 3 then G would contain the graph G 3 in Figure 4 as induced subgraph, which is not a CA graph.If |N G (v) ∩ S| = 2, then G would contain an induced gem.So, if |N G (v) ∩ H| = 4, then |N G (v) ∩ S| = 1.Suppose, by way of contradiction, that |N G (v) ∩ H| = 3.If |N G (v) ∩ S| = 3, then G would contain the graph G 9 in Figure 4 as induced subgraph, which is not a CA graph.If |N G (v) ∩ S| = 2, then G would contain either C 5 ∪ K 1 or C 4 ∪ K 1 as induced subgraph, and neither of them is a CA graph.If |N G (v)∩S| = 1, then G would contain either a gem or C 4 ∪K 1 as induced subgraph.If |N G (v)∩S| = 0, then G would contain the graph G 6 in Figure 4 as induced subgraph, which is not a CA graph.We conclude that |N G (v) ∩ H| = 3.
Tab. 1: Minimal forbidden induced subgraphs for balancedness in each studied subclass of CA graphs