Proper Helly Circular-Arc Graphs Min Chih Lin 1, Francisco J. Soulignac 1 and Jayme L. Szwarcfiter 2 1 Universidad de Buenos Aires, Facultad de Ciencias Exactas y Naturales, Departamento de Computación, Buenos Aires, Argentina. 2 Universidade Federal do Rio de Janeiro, Inst. de Matemática, NCE and COPPE, Caixa Postal 2324, 20001-970 Rio de Janeiro, RJ, Brasil. {oscarlin, fsoulign}@dc.uba.ar, jayme@nce.ufrj.br Abstract. A circular-arc model M = (C, A) is a circle C together with a collection A of arcs of C. If no arc is contained in any other then M is a proper circular-arc model, if every arc has the same length then M is a unit circular-arc model and if A satisfies the Helly Property then M is a Helly circular-arc model. A (proper) (unit) (Helly) circular-arc graph is the intersection graph of the arcs of a (proper) (Helly) circular-arc model. Circular-arc graphs and their subclasses have been the object of a great deal of attention in the literature. Linear time recognition algorithms have been described both for the general class and for some of its subclasses. In this article we study the circular-arc graphs which admit a model which is simultaneously proper and Helly. We describe characterizations for this class, including one by forbidden induced subgraphs. These characterizations lead to linear time certifying algorithms for recognizing such graphs. Furthermore, we extend the results to graphs which admit a model which is simultaneously unit and Helly, also leading to characterizations and a linear time certifying algorithm. Key words: algorithms, forbidden subgraphs, Helly circular-arc graphs, proper circular-arc graphs, unit circular-arc graphs. 1 Introduction Circular-arc (CA) graphs and its subclasses are interesting families of graphs that have been receiving much of attention recently. The most common subclasses of circular-arc graphs [1, 3, 12] are proper circulararc (PCA) graphs, unit circular-arc (UCA) graphs and Helly circular-arc Partially supported by UBACyT Grants X184 and X212, CNPq under PROSUL project Proc. 490333/2004-4. Partially supported by UBACyT Grant X184 and CNPq under PROSUL project Proc. 490333/2004-4. Partially supported by CNPq and FAPERJ, Brasil.
(HCA) graphs. The classes of PCA and UCA graphs have been characterized by families of (infinite) forbidden subgraphs [14], and for HCA graphs there is a characterization by forbidden CA graphs [4]. Each of the graphs of the mentioned classes is the intersection graph of a collection of arcs of a circle. The circle and the collection of arcs form a model for the corresponding graph. The model is called CA,PCA,UCA or HCA, according to whether the corresponding graph is CA,PCA,UCA or HCA. For each of the four classes of graphs (CA, PCA, UCA, HCA) there is a linear time recognition algorithm (see [6,9, 11] for CA, [2,5] for PCA, [5, 8] for UCA and [4] for HCA). In this paper, we study the class of circular-arc graphs which admit a model which is simultaneously PCA and HCA. We call these graphs (models) as proper Helly circular-arc graphs (models). Clearly, such a class of graphs is contained in the intersection of the classes of PCA and HCA graphs. However, the containment is proper. Similarly, we define UHCA graphs as those admitting a model which is simultaneously UCA and HCA. We describe characterizations for the classes of PHCA and UHCA graphs, including a characterization by forbidden subgraphs. The characterizations lead to linear time algorithms for recognizing PHCA and UHCA graphs. Besides, we also describe how to obtain certificates, both positive and negative, and authenticate them in linear time. As a motivation for this work, besides its theoretical interest, we mention that PCA and HCA are two classes of graphs, which behave differently in various aspects. For instance, PCA graphs can be efficiently colored, while problems involving cliques tend to be easier for HCA graphs. For instance, interval graphs are exactly those admitting a clique matrix having the consecutive ones property on the columns. By extending the consecutive ones property to the circular consecutive ones property leads to a broader class, which is exactly the HCA graphs. In this context, arises the problem of characterizing the clique graphs of HCA graphs. The latter class corresponds precisely to the PHCA graphs [7]. Let G be a graph, V (G) and E(G) its sets of vertices and edges, respectively, V (G) = n and E(G) = m. For v V (G), denote by N(v) the set of vertices adjacent to v, and N[v] = N(v) {v}. A vertex v of G is universal if N[v] = V (G). Two vertices v and w are twins in G if N[v] = N[w]. A clique is a maximal subset of pairwise adjacent vertices. A circular-arc (CA) model M is a pair (C, A), where C is a circle and A is a collection of arcs of C. When traversing the circle C, we will always choose the clockwise direction, unless explicitly stated. If s, t are points of C, write (s, t) to mean the arc of C defined by traversing the
circle from s to t. Call s, t the extremes of (s, t), while s is the start point and t the end point of the arc. For A i A, write A i = (s i, t i ) and A i = (t i, s i ). The extremes of A are those of all arcs A i A. An s-sequence (t-sequence) is a maximal sequence of start points (end points) of A, in a traversal of C. Without loss of generality, all arcs of C are considered as open arcs, no two extremes of distinct arcs of A coincide and no single arc entirely covers C. We will say that ǫ > 0 is small enough if ǫ is smaller than the minimum arc distance between two consecutive extremes of A. When no arc of A contains any other, (C, A) is a proper circular-arc (PCA) model, while when every arc has the same length it is called unit circular-arc (UCA) model. When every set of pairwise intersecting arcs share a common point, (C, A) is called a Helly circular-arc (HCA) model. If no two arcs of A cover C then the model is called normal. A PHCA (UHCA) model is one which is both HCA and PCA (UCA). Finally, a (proper) interval model is a (proper) CA model where A A A C. A CA (PCA) (UCA) (HCA) (PHCA) (UHCA) graph is the intersection graph of a CA (PCA) (UCA) (HCA) (PHCA) (UHCA) model. We may use the same terminology used for vertices when talking about arcs. For example, we say an arc in a CA model is universal when its corresponding vertex in the intersection graph is universal. Similarly, a connected model is one whose intersection graph is connected. Say that two CA models are equivalent when they have the same intersection graph. If M is a CA model, denote by M 1 a model obtained from M by removing all its universal arcs, if existing, except precisely one. Also, M 0 represents the model where all universal arcs have been removed. Clearly, if M has no universal arcs then M = M 1 = M 0, while if M contains exactly one universal arc M = M 1. In a HCA graph, each clique Q V (G) can be represented by a clique-point q C, which is a point of the circle common to all those arcs of A, which correspond to the vertices of Q. 2 Preliminaries In this section we present some simple observations and basic propositions, that we employ throughout the article. Lemma 1. Let v, w be two twin vertices in a graph G. Then G is a PCA (UCA) (HCA) (proper interval) graph if and only if G \ {v} is a PCA (UCA) (HCA) (proper interval) graph. Moreover G has a normal PCA (UCA) (HCA) model if and only if G\{v} also has a normal PCA (UCA) (HCA) model.
Proof. =) Let M = (C, A) be a PCA (UCA) (HCA) (proper interval) model of H \ {v} and A i A be the arc corresponding to w. It is easy to see that M = (C, A {(s i + ǫ, t i + ǫ)}) is a PCA (UCA) (HCA) (proper interval) model of H, for every small enough ǫ. Moreover, if M is normal then we obtain that M is normal. The converse is clear. Lemma 2. If G is a PCA graph with at most one universal vertex then every PCA model of it is normal. Proof. If two arcs cover the circle of a PCA model then they must be both universal, thus the graph has more than one universal vertex. Theorem 1. [4] A circular-arc model (C, A) is HCA if and only if (i) if three arcs of A cover C then two of them also cover it, and (ii) the intersection graph of A is chordal. Corollary 1. If a normal PCA model (C, A) is not HCA then three arcs of A cover C. Proof. On the contrary, suppose that no three arcs of A cover C. Then, by Theorem 1, the intersection graph of A = {A : A A}) has a hole v 1,...,v k for some k 4. Thus, the arcs A 1, A 3 A corresponding to vertices v 1 and v 3 do not intersect and therefore A 1, A 3 are arcs of A that cover C, contradicting the fact that (C, A) is normal. 3 The Characterizations In this section, we describe characterizations for PHCA graphs including a characterization by forbidden subgraphs. The forbidden subgraphs for a PCA graph to be a PHCA graph are the 4-wheel, denoted by W 4 and depicted in Figure 1 and the 3-sun, which appears in Figure 2. Theorem 2. Let G be a PCA graph and M be a PCA normal model of it. The following affirmatives are equivalent: (i) M is equivalent to a PHCA model. (ii) M contains no 4-wheels nor 3-suns, as submodels. (iii) M 1 is HCA or M 0 is an interval model. Proof. (i) = (ii): By hypothesis, M is equivalent to a PHCA model of G. We know that a PCA model and a HCA model both exist for W 4 (Figure 1). However, we show that no PHCA model exists for W 4. Every
(a) HCA model (b) UCA model (c) Graph Fig. 1. HCA and PCA models of W 4. (a) HCA model (b) PCA model (c) Graph Fig. 2. HCA and PCA models of the 3-sun. HCA model for W 4 contains four clique points, such that the universal arc in the submodel corresponding to W 4 covers these four points, while each of the other arcs of the submodel covers exactly two of them. Consequently, no HCA model can be normal. By Theorem 1, the latter implies that whenever M is equivalent to a PHCA model, M does not contain a submodel of W 4. The proof for the 3-sun is similar. (ii) = (iii): Let M be a normal PCA model, containing no submodels of W 4 s nor 3-suns. Suppose M 1 is not a HCA model. Since M 1 is also a PCA normal model, Corollary 1 implies that M 1 contains three arcs A 1, A 2, A 3 covering C. By Lemma 2 no two arcs cover C, thus we may assume that in a traversal of C the order in which the extremes points of these arcs appear is s 1, t 3, s 2, t 1, s 3, t 2. First, we prove that one of the above three arcs must be universal. Suppose the contrary. Then there exist arcs B i, such that B i does intersect A i, for i {1, 2, 3}. However, since (C, A) is a proper model, it follows that B i intersects A j, A k for j, k {1, 2, 3} \ {i}. The latter leads to a contradiction because the intersection graph of {A i, B i } i {1,2,3} is a 3-sun, when B 1, B 2, B 3 are pairwise disjoint, or otherwise it contains a W 4. Consequently, one of A 1, A 2, A 3,
say A 1, is a universal arc. Without loss of generality, we may assume that A 1 is the universal arc of M 1. Next, we examine arc A 1 in M 1. Traverse (s 1, t 1 ) in the clockwise direction. We will prove that no start point s l A i can precede an end point t r A i. To obtain a contradiction for this fact, assume the contrary and discuss the following alternatives. Case 1 : A r = A 3 In this situation, A l, A 2, A 3 are three arcs covering C. Because A 1 is the unique universal arc of M 1, we know that A l, A 2, A 3 are not universal. Consequently, as above, M 1 contains a W 4 or a 3-sun, a contradiction (Figure 3(a)) Case 2 : A l = A 2 Similar to Case 1. Case 3 : A r A 3 and A l A 2 By Cases 1 and 2, above, it suffices to examine the situation where s l, t r (t 3, s 2 ). Suppose A l A 3 = A r A 2 =. In this case, the arcs A 1, A 2, A 3, A l, A r form a forbidden W 4, impossible (Figure 3(b)). Alternatively, let A l A 3. Then the arcs A 3, A l, A r cover the circle and none of them is the universal arc A 1, an impossibility (Figure 3(c)). The situation A r A 2 is similar. (a) (b) (c) Fig. 3. Theorem 2 By the above cases, we conclude that all end points must precede the start points in (s 1, t 1 ). Let t last and s first be the last end point and the first start point inside (s 1, t 1 ), respectively. Taking into account that A 1 is universal, we conclude that any point of the arc (t last, s first ) A 1 of C can not be contained in any arc of A except A 1. Hence M 0 = M 1 {A 1 } is an interval model.
(iii) = (i): Suppose M 1 is a HCA model. By Lemma 1, we can include in the model all the universal arcs that have been possibly removed from it, obtaining M as both a PCA and HCA model. Then M is a PHCA model and G a PHCA graph. Next, suppose M 0 is an interval model. Since M 0 is a PCA model, it is in fact a proper interval model. The extreme points of M 0 form a linear ordering. If M 0 = M there is nothing to prove. Otherwise, examine the alternatives. Case 1 : M 0 is connected Let t l be the first end point and s r the last start point in the ordering of the extreme points of M 0. Then insert the arc (t l ǫ, s r + ǫ), for small enough ǫ. It follows that such an arc must be universal and the model still proper, because M is so. In addition, include the possibly remaining universal arcs, as in Lemma 1. Hence M is a PHCA model. Case 2 : M 0 has two connected components Traversing C in the clockwise direction, let t l be the first end point in the first of the connected components, and s r the last start point in the second of the components. Then include the arc (s r + ǫ, t l ǫ), for small enough ǫ. The new model is equivalent to M 1. By Lemma 1, include the remaining universal arcs. Therefore M is equivalent to a PHCA model, i.e. G is a PHCA graph. Case 3 : M 0 has more than two connected components This case can not occur, as M would not be a PCA model. The proof is complete. 4 The Algorithms The characterizations described in the last section lead directly to an algorithm for recognizing PHCA graphs. Let G be a given graph. The algorithm answers YES, if G is a PHCA graph, and NO otherwise.the formulation is as follows.
1. Verify if G is a PCA graph, using algorithm [2] or [5]. If affirmative, let M be the PCA model given by the algorithm; otherwise answer NO. 2. Transform M into a normal model M, using algorithm [5] or [8]. 3. Compute M 1 and M 0, from M. 4. Verify if M 1 contains three arcs which cover the circle. If negative, answer YES. 5. Verify if M 0 is an interval model. If affirmative, answer YES; otherwise answer NO. The correctness of the algorithm follows from Theorems 1 and 2, Lemma 1 and Corollary 1. We employ the equivalence (i) (iii) of Theorem 2. However, the check of whether the model M 1 is HCA is replaced by the simpler check of whether M 1 contains three arcs covering the circle, according to Corollary 1. Next, we discuss the complexity of the algorithm. Step 1 requires O(n + m) time [2], [5]. All the remaining steps can be implemented in O(n) time, as follows. The construction of normal models can be done in O(n) time, using algorithms [5] or [8]. In order to implement Step 3, we need to find the set of universal arcs of M. Such a construction can be done easily, by observing that in a normal model, any arc A i is universal precisely when it contains exactly one of the extremes of each of the other arcs of A. Consequently, to check whether A i is universal, it suffices to check whether the n-th extreme point after s i, in the ordering, is the end point t i. This can be done in constant time, for each A i, hence in overall O(n) time. Step 4 can be implemented in time O(n), also in a simple manner, as follows. For any point p C, denote by PREV (p) the arc of A whose start point is closest to p, in the counterclockwise direction. Let t 1,...,t k be a t-sequence of a set of arcs A 1,...,A k. Clearly, PREV (t i ) = PREV (t 1 ) for every 1 i k. Therefore, by a simple examination of the t-sequences, we can find PREV (t i ) for every arc A i in O(n) time. If A 1 covers C with two other arcs, then A i, PREV (t i ), PREV (PREV (t i )) must cover C because the model is proper. Thus, it is enough to check if A i, PREV (t i ), PREV (PREV (t i )) cover the circle for every 1 i n, which can be done in O(n) time. Finally, for Step 5, we need to verify whether M 0 is an interval model, which can be easily checked in O(n) time. The above algorithm produces certificates, as a by-product. When the algorithm answers YES, we exhibit a PHCA model for the graph, while in the NO answer, we show either a forbidden subgraph or a corresponding argument for the graph not to be PHCA.
The YES answers appear in Steps 4 and 5. The certificate corresponding to Step 4 is just the model M, constructed in Step 2. For obtaining the YES certificate relative to Step 5, we refer to the proof of Theorem 2, in particular, the Case 2 of the implication (iii) = (i). In this situation, M 1 contains three arcs covering the circle, but M 0 is a disconnected interval model, formed by two connected components. To obtain the required model, we include in M 0 the arc (s r + ǫ, t l ǫ), where s r is the last start point in the second connected component of M 0, while t l is the first end point in the first component. Such a model contains exactly one universal arc. Finally, using Lemma 1, include the possibly remaining universal arcs that have been removed in Step 3. The model so obtained is a PHCA model for G. The complexity for producing such certificates is O(n). Next, we discuss the authentication of the YES certificates. Denote by M the certificate obtained by the algorithm, which we are required to authenticate, as a PHCA model of the input graph G. That is, we ought to verify that M is a model for G and that M is proper, normal and Helly. The first task is simple, just compare the adjacencies of the vertices of G with the intersections of the arcs of M. This requires O(n + m) time. The remaining authentications involve only operations on the arcs of M and can be done in O(n) time, as below described. We employ an observation that a model M is simultaneously proper and normal if and only if no arc A i of it contains both extremes of some other arc A j A i. This observation leads to an algorithm where we traverse each arc A i of M in the order i = 1,...,n, and if we detect both extremes of a same arc during the traversal of A i, the algorithm reports the certificate to be false and stops. After traversing the last arc A n it authenticates M as being proper and normal. In order to speed up the process, when checking arc A i, i > 1, we avoid traversing the extremes visited in the previous iteration, relative to A i 1. That is, for checking A i, we start at the extreme which immediately follows t i 1 in the circular ordering, whenever s i A i 1 ; otherwise, when s i A i 1, start at s i. This is correct, because we know that (s i, t i 1 ) does not contain the two extremes of a same arc, otherwise the iteration of A i 1 would have reported the invalidity of the certificate. In this way, the algorithm assures that each extreme is visited at most twice, during the entire process, meaning that the authentication for the model to be proper and normal requires O(n) time. Finally, for checking whether M is Helly, we apply Corollary 1 and just confirm that M does not contain three arcs covering
the circle. With this purpose, we run Step 4 of the Algorithm, terminating within O(n) time. The NO answers are in Steps 1 and 5. The negative answer in Step 1 occurs when G is not a PCA graph. A certificate of this fact is given by algorithm [5]. The task is divided into two cases. If G is co-bipartite then [5] employs the characterization proved in [13] and exhibits a forbidden subgraph for a co-bipartite graph to be a PCA graph. When G is not co-bipartite, the certificate obtained is that for a matrix not to have the consecutive ones property in its columns. Such a certificate is given by the algorithm [10]. In any of the cases, the certificates can be obtained in O(n + m) time. Finally, the NO answer in Step 5 corresponds to a certificate where we exhibit a forbidden subgraph for a PCA graph to be PHCA. According to Theorem 2, such a subgraph is either a W 4 or a 3-sun, and can be constructed as follows. Let A 1, A 2, A 3 be the three arcs which cover the circle, obtained in Step 4. If none of these arcs is universal then by Theorem 2, we know that there are three arcs B 1, B 2, B 3, such that B i intersects A j and not A i, for all 1 i j 3. Then the arcs A 1, A 2, A 3, B 1, B 2, B 3 either form a 3-sun or contain a W 4. Finally, if one among A 1, A 2, A 3, say A 1, is a universal arc then there are arcs A l, A r, such that s l precedes t r in A 1. In this situation, A 1, A 2, A 3, A l, A r form a forbidden W 4. There is no difficulty to obtain such certificates in O(n) time. Finally, we discuss the authentication of the NO certificates. The one obtained in Step 1 is given by algorithm [5] and can be authenticated in O(n) time [5]. The NO certificate constructed in Step 5 is one of the forbidden subgraphs W 4 or 3-sun. They can be easily authenticated in O(n) time, as being an induced subgraph of G. 5 Unit Helly Circular-Arc Graphs The characterization of PHCA graphs, described in Section 3, can be extended so as to characterize UHCA graphs. The formulation is below presented. The proof is similar to that of Theorem 2. Theorem 3. Let G be a UCA graph and M be a UCA normal model of it. The following affirmatives are equivalent: (i) M is equivalent to a UHCA model. (ii) M contains no 4-wheels, as submodels. (iii) M 1 is HCA or M 0 is an interval model.
The above characterization, together with Lemma 1 and Corollary 1, lead to a linear time algorithm for recognizing UHCA graphs and exhibiting certificates. The algorithm is similar to that for PHCA graphs, except that we employ algorithms [5] or [8] in Step 1. The positive certificate is a UHCA model and can be obtained by algorithm [8], whereas the negative certificate of Step 1 is obtained by [5]. 6 Conclusions We have described characterizations and recognition algorithms for PHCA and UHCA graphs. The characterizations imply a complete family of forbidden subgraphs for these classes. That is, adding W 4 and the 3-sun to Tucker s forbidden subgraphs for PCA graphs, we obtain the complete list of forbidden subgraphs for PHCA graphs. Similarly, adding W 4 to Tucker s forbidden family for UCA graphs, we obtain all the subgraphs forbidden for UHCA graphs. The algorithms have complexity O(n + m). Moreover, if the input consists of a circularly ordered PCA model, the complexity drops to O(n). References 1. Andreas Brandstädt, Van Bang Le, and Jeremy P. Spinrad, Graph classes: a survey, SIAM Monographs on Discrete Mathematics and Applications, Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA, 1999. 2. Xiaotie Deng, Pavol Hell, and Jing Huang, Linear-time representation algorithms for proper circular-arc graphs and proper interval graphs, SIAM J. Comput. 25(2) (1996), 390 403. 3. M. Golumbic, Algorithmic Graph Theory and Perfect Graphs, Academic Press, New York, 1980. 4. B. Joeris, M. C. Lin, R. McConnell, J. Spinrad, and J. L. Szwarcfiter, Linear time recognition and representation of helly circular-arc graphs, Manuscript. Presented at COCOON 2006, Taipei, and SIAM DM 2006, Victoria, Conferences, 2007. 5. Haim Kaplan and Yahav Nussbaum, Certifying algorithms for recognizing proper circular-arc graphs and unit circular-arc graphs, In: Graph-Theoretic Concepts in Computer Science, 32nd International Workshop, WG 2006, Bergen, Norway, June 22-24, 2006 (Fedor V. Fomin, ed.), Lecture Notes in Computer Science, vol. 4271, Springer, Berlin, 2006, pp. 289 300. 6. Haim Kaplan and Yahav Nussbaum, A simpler linear-time recognition of circulararc graphs, In: Algorithm Theory - SWAT 2006, 10th Scandinavian Workshop on Algorithm Theory, Riga, Latvia, July 6-8, 2006 (Lars Arge and Rusins Freivalds, eds.), Lecture Notes in Computer Science, vol. 4059, Springer, 2006, pp. 41 52. 7. Min Chih Lin, Francisco Juan Soulignac, and Jayme Luiz Szwarcfiter, Proper Helly circular-arc graphs, In: Graph-Theoretic Concepts in Computer Science, 33rd International Workshop, WG 2007, Dornburg, Germany, June 21-23, 2007 (Andreas Brandstädt, Dieter Kratsch, and Haiko Müller, eds.), Lecture Notes in Computer Science, vol. 4769, Springer, 2007, pp. 248 257.
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