A graph is quasi-triangulated if each of its induced subgraphs has a vertex which is either simplicial (its neighbors form a clique) or cosimplicial (its nonneighbors form an independent set). We prove that a graph G is quasi-triangulated if and only if each induced subgraph H of G contains a vertex that does not lie in a hole, or an antihole, where a hole is a chordless cycle with at least four vertices, and an antihole is the complement of a hole. We also present an algorithm that recognizes a quasi-triangulated graph in O(nm) time

Gorgos, Ion

Hoàng, Chính T.

Voloshin, Vitaly

English

Wilfrid Laurier University

Wilfrid Laurier University Scholars Commons @ Laurier Physics and Computer Science Faculty Publications Physics and Computer Science 2006 A Note on Quasi-Triangulated Graphs Ion Gorgos Academy of Economic Studies of Moldova Chính T. Hoàng Wilfrid Laurier University, choang@wlu.ca Vitaly Voloshin Troy University Follow this and additional works at: https://scholars.wlu.ca/phys_faculty Recommended Citation Gorgos, Ion; Hoàng, Chính T.; and Voloshin, Vitaly, "A Note on Quasi-Triangulated Graphs" (2006). Physics and Computer Science Faculty Publications. 72. https://scholars.wlu.ca/phys_faculty/72 This Article is brought to you for free and open access by the Physics and Computer Science at Scholars Commons @ Laurier. It has been accepted for inclusion in Physics and Computer Science Faculty Publications by an authorized administrator of Scholars Commons @ Laurier. For more information, please contact scholarscommons@wlu.ca. SIAM J. DISCRETE MATH. c© 2006 Society for Industrial and Applied MathematicsVol. 20, No. 3, pp. 597–602A NOTE ON QUASI-TRIANGULATED GRAPHS∗ION GORGOS† , CHI´NH T. HOA`NG‡ , AND VITALY VOLOSHIN§Abstract. A graph is quasi-triangulated if each of its induced subgraphs has a vertex which iseither simplicial (its neighbors form a clique) or cosimplicial (its nonneighbors form an independentset). We prove that a graph G is quasi-triangulated if and only if each induced subgraph H of Gcontains a vertex that does not lie in a hole, or an antihole, where a hole is a chordless cycle with atleast four vertices, and an antihole is the complement of a hole. We also present an algorithm thatrecognizes a quasi-triangulated graph in O(nm) time.Key words. triangulated graphs, chordal graphs, quasi-triangulated graphsAMS subject classiﬁcations. 05C75, 05C85DOI. 10.1137/S08954801044443991. Introduction. In a graph G, a vertex x is simplicial if its neighborhoodN(x) induces a complete subgraph of G. A graph is triangulated (chordal) if itdoes not contain a chordless cycle of length at least four (a hole) as an inducedsubgraph. A famous theorem of Dirac [3] states that every triangulated graph hasa simplicial vertex. Actually, Dirac proved more: every triangulated graph diﬀerentfrom a clique contains two nonadjacent simplicial vertices. Let us say that a vertexis cosimplicial if its nonneighbors form an independent subset of vertices and thata graph is cotriangulated if it does not contain the complement of a chordless cycleon at least four vertices (an antihole). Dirac’s theorem says equivalently that everycotriangulated graph has a cosimplicial vertex. Our purpose is to investigate the largerclass of graphs which are called quasi-triangulated graphs (QT for short), deﬁned asfollows: a graph G is in class QT if and only if every induced subgraph H of G has avertex which is either simplicial or cosimplicial in H. Quasi-triangulated graphs havebeen introduced by the third author in [9, 11] as a generalization of chordal graphs.The problem of characterizing the class QT was raised in [9] and, independently, in[7] (where they are called good). The reader is referred to [1] for more information onthe class QT .Following [7], we say that an order v1 < v2 < · · · < vn on a graph G is good if, forany induced subgraph H of G, either the largest vertex of (H,<) is simplicial or thesmallest vertex of (H,<) is cosimplicial. Good orders are perfect in the sense of [2].Simplicial vertices cannot lie in a hole; and cosimplicial vertices cannot lie in anantihole. A graph with each vertex belonging to some hole and some antihole is calledlatticed.The third author conjectured (unpublished) and the ﬁrst author proved in [4, 5]the following.∗Received by the editors June 10, 2004; accepted for publication (in revised form) November 9,2005; published electronically August 25, 2006.http://www.siam.org/journals/sidma/20-3/44439.html†Academy of Economic Studies of Moldova, 61 Banulescu-Bodoni str. MD-2005, Chisinau,Moldova.‡Department of Physics and Computer Science, Wilfrid Laurier University, 75 University Ave. W.,Waterloo, ON N2L 3C5, Canada (choang@wlu.ca). This author’s research was supported by theNSERC.§Department of Mathematics and Physics, Troy University, Troy, AL 36082 (vvoloshin@troy.edu).This author’s research was partially supported by a Troy University research grant.597598 ION GORGOS, CHI´NH T. HOA`NG, AND VITALY VOLOSHINTheorem 1. For a graph G, the following three conditions are equivalent:(i) G is quasi-triangulated.(ii) G does not contain a latticed subgraph as an induced subgraph.(iii) G admits a good order.As usual, n (respectively, m) denote the number of vertices (respectively, edges) ofthe input graph. For the quasi-triangulated graph recognition problem, the third au-thor [10] proposed an O(n4) algorithm, Spinrad [12] proposed an O(n2.77) algorithm,and the second author [6] independently proposed an O(nm) algorithm.Theorem 2. There is an O(nm)-time O(n2)-space algorithm to recognize a quasi-triangulated graph.Theorems 1 and 2 are known by researchers in the ﬁeld and have been referredto in the literature, but their proofs have never been published. The purpose of thispaper is to provide the proofs of these two theorems.2. Proof of Theorem 1. To prove Theorem 1, we will need the following lemma,which was included in the original proof in [4] and was rediscovered independently in[8].Lemma 1. Let G be a graph and x be a vertex of G that does not lie in a hole.Then any minimal cutset C of G which is contained in the neighborhood N(x) of x isa clique.Proof of Lemma 1. Deﬁne G, x,C as in the lemma. Let Y be a component ofG− C that does not contain x. We may assume that there are nonadjacent verticesu, v in C, for otherwise we are done. Since C is a minimal cutset, each of u and vhas a neighbor in Y . It follows that there is a chordless path of length at least twojoining u to v whose interior vertices lie in Y . This path together with x forms a hole,a contradiction to our assumption on x.Proof of Theorem 1. It is easy to see that (i) and (iii) are equivalent, and (i)implies (ii). So, we need only to prove that (ii) implies (i). We shall prove this byinduction on the number of vertices. Let G be a graph satisfying (ii). We may assumeG contains no simplicial vertex and no cosimplicial vertex, for otherwise we are doneby the induction hypothesis. If G is disconnected, then each component of G containsa hole (for otherwise, it is triangulated and contains a simplicial vertex that remainssimplicial in G); thus, G contains the union of two disjoint holes, a contradiction to(ii). So, G must be connected.By replacing G by its complement G if necessary, we may assume that G containsa vertex that does not lie in a hole.Deﬁne X = {x | x does not lie in a hole of G}.Our assumption on G implies that X = ∅. Let G′ = G−X. G′ is nonempty, forotherwise G is triangulated and thus contains a simplicial vertex by Dirac’s theorem.By the induction hypothesis, G′ contains a simplicial or cosimplicial vertex y. Sinceevery vertex of G′ lies in a hole, y is cosimplicial. We shall prove that y is adjacentto all vertices of X (this will imply y is cosimplicial in G, a contradiction).Let x be a vertex in X. Since G is connected and x is not cosimplicial (byassumption), there is a nonempty set C of vertices in N(x) that is a minimal cutsetof G. By Lemma 1, C is a clique. Let G1, G2 be induced subgraphs of G such thatG = G1 ∪G2, G1 ∩G2 = C, and there is no edge between G1 − C and G2 − C.Suppose G1 is triangulated. We claim that there is a simplicial vertex s in G1−C.If G1 is a clique, then the claim obviously holds; otherwise, by Dirac’s theorem, G1contains two nonadjacent simplicial vertices, one of which must lie in G1 −C since CA NOTE ON QUASI-TRIANGULATED GRAPHS 599is a clique. But s remains a simplicial vertex of G, a contradiction to our assumptionon G. Thus G1, and similarly G2, cannot be triangulated.Therefore, G1 contains a hole. Since C is a clique, one edge, say e1, of this holelies completely in G1 − C. Similarly, there is an edge, say e2, that lies completely inG2 −C and belongs to a hole. Since y is cosimplicial in G′ and all endpoints of e1, e2are in G′, y must be in C, and therefore adjacent to x, as desired.3. A recognition algorithm for quasi-triangulated graphs. In this section,we prove Theorem 2 by describing an algorithm that recognizes a quasi-triangulatedgraph in O(nm) time using O(n2) space.For a vertex x, an S-obstruction is a triple (a, b, x) that induces a P3 with x beingthe interior vertex of the path; a C-obstruction is a triple (a, b, x) that induces anS-obstruction (a, b, x) in the complement.A straightforward algorithm to recognize quasi-triangulated graphs proceeds asfollows. First, for all vertices x, list all S- and C-obstructions. Then ﬁnd a vertex ywith no S- or C-obstructions; if no such vertex exists, then the graph is not quasi-triangulated. Remove y and update the lists of obstructions for the remaining vertices.Repeat this process to eliminate all vertices. If all vertices can be eliminated in thisway, then the graph is quasi-triangulated; otherwise, it is not.Since a vertex has O(n2) obstructions, we will need a data structure to storeO(n3) obstructions of the graph. Thus the algorithm runs in O(n3) time using O(n3)space. We are going to show that the algorithm can be reﬁned to run in time O(nm)using O(n2) space.Proof of Theorem 2. We may suppose there is a total order < on the vertices of agiven graphG. We say that (a, b, x) is less than (c, d, x), denoted by (a, b, x) < (c, d, x),if a < c, or a = c and b < d. To achieve the O(nm) time bound, we will list only thesmallest S-obstruction and C-obstruction for each vertex. When removing a vertex y,if a vertex x loses an obstruction, then we will ﬁnd a smallest obstruction for x in theremaining graph. We shall show that, over the life of the algorithm, the time neededto list the (currently) smallest obstructions for a vertex x is O(m). The outline of ouralgorithm is as follows.Outline of algorithm. We begin with the input graph G and proceed to elimi-nate vertices one by one using the following steps.1. For each vertex x of graph G, list a smallest S-obstruction (a, b, x) and asmallest C-obstruction (g, d, x).2. If every remaining vertex has an S-obstruction and a C-obstruction, then G isnot quasi-triangulated.3. If a vertex z has no S-obstruction or no C-obstruction, eliminate z from G, andfor each remaining vertex x that loses an S-obstruction (respectively, C-obstruction),generate a new smallest S-obstruction (respectively, C-obstruction). Replace G byG− z, and repeat step 2.The graph G is quasi-triangulated if and only if recursive applications of step 3eliminate all vertices. To anticipate, our algorithm lists the S-obstructions in O(nm)time using O(n+m) space and the C-obstructions in O(nm) time using O(n2) space.Let N(x) be the adjacency list of vertex x. Without loss of generality, we mayassume for all x that the lists N(x) are sorted in increasing order.Listing the smallest S-obstruction for a vertex x. For each vertex x,we use two pointers, α(x) and β(x). Initially α(x) points to the ﬁrst vertex α inN(x) and β(x) points to the immediate successor β of α in N(x) (for simplicity, welet α (respectively, β) denote the name of the vertex pointed to by the pointer α(x)600 ION GORGOS, CHI´NH T. HOA`NG, AND VITALY VOLOSHIN(respectively, β(x)). If α(x) or β(x) cannot be initialized, then x has no S-obstruction.We simply advance β(x) on N(x) until we ﬁnd that α and β are nonadjacent. Whenβ(x) reaches the end of N(x) (i.e., it has value null), we advance α(x) in N(x) andinitialize β(x) (making β(x) point to the immediate successor of α(x) in N(x)). Ifα(x) = null, then x has no S-obstruction, and a message “No S-obstruction” isproduced. We can summarize this process as follows. (In the following procedure, thefunction IsEdge(a, b) returns true if and only if ab is an edge.)Procedure ListSmallest-S-Obstruction(x).while true{ if α(x) = nullthen return “no S-obstruction for x”if IsEdge(α, β) = truethen advance β(x) in N(x)elsereturn (α, β, x)while ((α(x) = null) and β(x) = null)){ advance α(x) in N(x)initialize β(x)}}Suppose we eliminate a vertex z and x loses its S-obstruction (a, b, x) (because a = z orb = z). If b (respectively, a) is eliminated, then we advance β(x) (respectively, α(x))and call Procedure ListSmallest-S-Obstruction(x) to get the smallest S-obstructionfor x. The number of movements of the pointers α(x), β(x) in N(x) is proportional toO(n+m) since we advance β(x) only in the presence of an edge, and α(x) is reset atmost the degree of x times. If we have the incidence matrix of G at our disposal, theneach call to Procedure IsEdge takes only constant time, but this method needs O(n2)space. We are going to show that for each vertex x we can implement ProcedureIsEdge in O(n+m) time using only the adjacency lists of G (O(n+m) space).Now we describe Procedure IsEdge(α, β), which returns true if and only if αβ isan edge. For a vertex x, there is a pointer p(x) which initially points to the ﬁrst vertexin N(α) (recall that α(x) is the pointer associated with vertex x). If p(x) cannot beinitialized, then αβ is not an edge. Pointer p(x) is advanced in N(α) until it pointsto either (i) β (αβ is an edge) or (ii) the smallest vertex in N(α) that is greater thanβ (αβ is not an edge). The vertex pointed to by p(x) is denoted by p.Procedure IsEdge(α, β).while true{ if p(x) = nullreturn falseif p < βadvance p(x) in N(α)else if p = βreturn trueelse if p > βreturn false}For each vertex x and each α(x), p(x) scans N(α) only once. Thus, for each x, thecost of testing for all edges αβ is O(n+m).A NOTE ON QUASI-TRIANGULATED GRAPHS 601Listing the smallest C-obstruction for a vertex x. Assume that the verticesare numbered 1, 2, . . . , n. For each vertex x, we maintain an integer variable counterγ(x) that refers to the smallest nonneighbor of x. We need to generate the smallestC-obstruction of the form (γ, δ, x) for some vertex δ (that must be adjacent to γ andnonadjacent to x). This can be done as follows.For each vertex x, we maintain a 0-1 characteristic vector I(x) of size n to rep-resent the neighborhood of x (the jth entry of I(x) is 1 if and only if vertex j is aneighbor of x; in other words, I(x) is the xth row of the incidence matrix of G). Thisis necessary so that testing of an edge of the form xy can be done in constant time.Given γ(x), we ﬁnd the smallest neighbor δ of γ (the vertex referred to by γ(x)) suchthat γ < δ and δ is nonadjacent to x by scanning the list N(γ) and, for each vertexy in this list, testing whether yx is an edge. We use a pointer δ(x) to point to thelocation of δ in N(γ). If δ(x) cannot be initialized, then there is no C-obstruction ofthe form (γ, δ, x); in this case, we increase γ(x) by one and repeat the process (theinitial value of γ(x) is one). This is summarized in the following procedure (we leavethe pointer initialization problem to the reader).Procedure ListSmallest-C-Obstruction(x).while true{ if δ(x) = nullrepeatincrease γ(x) by oneif (γ > n)return “No C-obstruction for x”let δ(x) point to the ﬁrst vertex in N(γ)until (xγ is not an edge) and δ(x) is not nullif (xδ is not an edge) and (γ < δ)return the C-obstruction (γ, δ, x)advance δ(x) in N(γ)}Suppose we eliminate a vertex z and x loses its C-obstruction (g, d, x) (because g = zor d = z). If d is eliminated, then we advance δ(x) in N(γ(x)) and call ProcedureListSmallest-C-Obstruction. If g is eliminated, then we repeatedly increase γ(x) byone until we get the next smallest nonneighbor of x and call Procedure ListSmallest-C-Obstruction.For each vertex x and each γ(x), the list N(γ(x)) is scanned at most once. Thus,for each x, we can list the smallest C-obstruction in O(n + m) time over the life ofthe algorithm. This method requires O(n2) space.In the case of listing the C-obstructions, we do not know how to implement ouralgorithm in O(nm) time using linear space. We leave this as an open problem. Wenote Spinrad’s algorithm [12] uses O(n2) space since it relies on matrix multiplications.REFERENCES[1] A. Brandsta¨dt, V. B. Le, J. P. Spinrad, Graph Classes: A Survey, SIAM Monogr. DiscreteMath. Appl. 3, SIAM, Philadelphia, 1999.[2] V. Chva´tal, Perfectly ordered graphs, in Topics on Perfect Graphs, C. Berge and V. Chva´tal,eds., Ann. Discrete Math. 21, North–Holland, Amsterdam, 1984, pp. 63–65.[3] G. A. Dirac, On rigid circuit graphs, Abh. Math. Sem. Univ. Hamburg, 25 (1961), pp. 71–76.[4] I. M. Gorgos, A Characterization of Quasi-triangulated Graphs, Preprint 11B494, KishinevState University, Kishinev, Moldova, 1984 (in Russian).602 ION GORGOS, CHI´NH T. HOA`NG, AND VITALY VOLOSHIN[5] I. M. Gorgos, Method of Alternating Chains and Its Applications, Ph.D. Thesis, KishinevState University, Kishinev, Moldova, 1985 (in Russian).[6] C. T. Hoa`ng, Recognizing Quasi-triangulated Graphs in O(nm) Time, manuscript.[7] C. T. Hoa`ng and N. V. R. Mahadev, A note on perfect orders, Discrete Math., 74 (1989),pp. 77–84.[8] C. T. Hoa`ng, S. Hougardy, F. Maffray, and N. V. R. Mahadev, On simplicial and co-simplicial vertices in graphs, Discrete Appl. Math., 138 (2004), pp. 117–132.[9] V. I. Voloshin, Quasi-triangulated Graphs, Preprint 5569-81, Kishinev State University,Kishinev, Moldova, 1981 (in Russian).[10] V. I. Voloshin, Quasi-triangulated Graphs Recognition Program, Algorithms and ProgramsP006124, Moscow, Russia, 1983 (in Russian).[11] V. I. Voloshin, Triangulated Graphs and Their Generalizations, Ph.D. Thesis, Kishinev StateUniversity, Kishinev, Moldova, 1983 (in Russian).[12] J. Spinrad, Recognizing quasi-triangulated graphs, Discrete Appl. Math., 138 (2004), pp. 203–213.

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