G G[S] G[D]

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1 Edge colouring reduced indierence graphs Celina M. H. de Figueiredo y Celia Picinin de Mello z Jo~ao Meidanis z Carmen Ortiz x Abstract The chromatic index problem { nding the minimum number of colours required for colouring the edges of a graph { is still unsolved for indierence graphs, whose vertices can be linearly ordered so that the vertices contained in the same maximal clique are consecutive in this order. Two adjacent vertices are twins if they belong to the same maximal cliques. A graph is reduced if it contains no pair of twin vertices. We prove that every indierence graph having no twin maximum degree vertices is Class 1. We present two decomposition methods for edge colouring reduced indierence graphs. Key words. edge colourings, interval graphs, proper interval graphs, minimum proper interval graphs, reduced proper interval graphs. AMS subject classication. 05C85, 05C15, 68R10, 90C27 Submitted to WG'99 on 27 February This work was partially supported by CNPq, CAPES, FAEP, FAPESP and FAPERJ, Brazilian research agencies. y Instituto de Matematica and COPPE, Universidade Federal do Rio de Janeiro, Brazil. celina@cos.ufrj.br. z Instituto de Computac~ao, Universidade Estadual de Campinas, Brazil. fmeidanis,celiag@dcc.unicamp.br. x Universid de Chile, Chile. cortiz@ibanez.uai.cl. 1

2 1 Introduction In this note, G denotes a simple, undirected, nite, connected graph. The sets V (G) and E(G) are the vertex and edge sets of G. Denote jv (G)j by n and je(g)j by m. A graph with one vertex is called trivial. A clique is a set of vertices pairwise adjacent in G. A maximal clique of G is a clique not properly contained in any other clique. A subgraph of G is a graph H with V (H) V (G) and E(H) E(G). For X V (G), denote by G[X] the subgraph induced by X, that is, V (G[X]) = X and E(G[X]) consists of those edges of E(G) having both ends in X. For Y E(G), the subgraph induced by Y is the subgraph of G whose vertex set is the set of endpoints of edges in Y and whose edge set is Y ; this subgraph is denoted by G[Y ]. The notation G ny denotes the subgraph of G induced by E(G) n Y. A matching M of G is a set of pairwise non adjacent edges of G. A matching M of G covers a vertex set X of G when each vertex of X is incident with some edge of M. The graph G[M] is also called a matching. For each vertex v of a graph, Adj(v) denotes the set of vertices which are adjacent to v. The degree of a vertex v is deg(v) = jadj(v)j. The maximum degree of a graph G is then (G) = max v2v (G) deg(v). We call a vertex with maximum degree -vertex. N[v] denotes the neighbourhood of v, that is, N[v] = Adj(v) [ fvg. A subgraph induced by the neighbourhood of a vertex is simply called a neighbourhood. Two vertices v e w are twins when N[v] = N[w]. Equivalently, two vertices are twins when they belong to the same set of maximal cliques. A graph is reduced if it contains no pair of twin vertices. The chromatic index 0 (G) of a graph G is the minimum number of colours needed to colour the edges of G such that no adjacent edges get the same colour. A celebrated theorem by Vizing [8, 10] states that 0 (G) is always or +1, where is the maximum vertex-degree of G. Graphs with 0 (G) = are said to be in Class 1; graphs with 0 (G) = + 1 are said to be in Class 2. A graph G satisfying the inequality m > (G) bn=2c, is said to be an overfull graph [6]. A graph G is subgraph-overfull [6] when it has an overfull subgraph H with (H) = (G). When the overfull subgraph H can be chosen to be a neighbourhood, we say that G is neighbourhood-overfull [3]. Overfull, subgraph-overfull and neighbourhoodoverfull graphs are in Class 2. It is well known that the set of graphs in Class 1 is NP-complete [7]. The problem remains NP-complete for several classes, including comparability graphs [1]. On the other hand, the problem remains unsolved for indierence graphs: graphs whose vertices can be linearly ordered so that 2

3 the vertices contained in the same maximal clique are consecutive in this order [9]. We call such an order an indierence order. Indierence graphs form an important subclass of interval graphs: they are also called unitary interval graphs or proper interval graphs. Our previous work has shown that every odd maximum degree indierence graph is Class 1 [3] and that every subgraph-overfull indierence graph is in fact neighbourhood-overfull [2]. We conjectured that every Class 2 indierence graph is neighbourhood-overfull [2, 3]. The purpose of this note is to show that every reduced indierence graph is Class 1. First we prove that an indierence graph with no twin maximum degree vertices cannot be neighbourhood-overfull. Then we exhibit an edge colouring for a graph in this subclass with maximum degree colours. Thus our result shows that reduced indierence graphs are in Class 1 and gives positive evidence for the above mentioned conjecture on edge colouring general indierence graphs. Since it is known that every odd maximum degree indierence graph is Class 1, we focus on even maximum degree indierence graphs. In Section 2 we decompose an indierence graph into a bipartite graph and a smaller indierence graph. In Section 3 we decompose an indierence graph with no twin maximum degree vertices into a matching covering all maximum degree vertices and an odd maximum degree indierence graph. Our conclusions are in Section 4. 2 Decomposing indierence graphs into a bipartite and a smaller indierence graph Let E 1 ; ; E k be a partition of the edge set of a graph G. If the subgraphs G i = G[E i ] satisfy (G) = P i G i and all the subgraphs G i belong to Class 1, then G is also in Class 1. Let G be an indierence graph. Consider v 1 ; v 2 ; : : : ; v n an indierence order for G. Label v i by i, i.e., for each i, set label(v i ) = i. Let S be the set of edges of G that join vertices with labels with equal parity, and let D be the set of edges of G that join vertices with labels with distinct parity. First we establish in Lemma 1 two properties satised by a general indierence graph G decomposed into graphs G[S] and G[D]. Lemma 1 Let G be a non trivial indierence graph. Then G[S] is an indierence graph and G[D] is a bipartite graph. 3

4 Proof: Let G be a non trivial indierence graph. Consider v 1 ; v 2 ; : : : ; v n an indierence order for G and a labeling for the vertices of G dened as above. First dene the following indierence order on G[S]: order the vertices of G[S] by placing rst and consecutively the vertices with even labels with respect to the indierence order for G, and then placing consecutively the vertices with odd labels with respect to the indierence order for G. Let v i v j 2 S. Then i 6= j, and i; j have equal parity. Since G is an indierence graph, for each k with the same parity of i, with i < k < j, we have that v i v k and v k v j are edges of G. On the other hand, if G[D] contained an odd cycle, then D would contain an edge joining vertices with equal parity. The equality (G) = (G[S]) + (G[D]) does not hold for the indierence graph G depicted in Figure 1: in this example, (G) = 6, (G[S]) = 3, and (G[D]) = 4. We dene next a subclass of reduced indierence graphs for which the desired equality (G) = (G[S]) + (G[D]) holds. G G[S] G[D] Figure 1: An indierence graph G such that (G) 6= (G[S]) + (G[D]). An indierence graph is a red line when, for a given indierence order v 1 ; v 2 ; : : : ; v n, there is a constant x 0 such that n 2x + 1 and, for each v i, 1 i n? x, its rightmost neighbour is v i+x. The constant x is called the size of the red line. The maximal cliques of a red line are in one-to-one correspondence with the edges v i v i+x, 1 i n? x. An edge v i v i+x is called a maximum edge. A red line is a reduced graph since each vertex belongs to a dierent set of maximal cliques. In addition, every vertex of a red line is the rst or the last vertex of a maximal clique with respect to its 4

5 unique indierence order. Corollary 1 Let G be a red line with size x = 2i or x = 2i + 1, with i 1. Then G[S] is a red line with size i. Proof: Let G be a red line. By Lemma 1, G[S] is an indierence graph. If x = 2i, then v j v j+x 2 S, 1 j n? x. Hence the maximum edges of the red line G have endpoints of equal parity and are in fact maximum edges of G[S]. If v j v j+x is an edge joining two vertices of odd (even) labels, then there are i + 1 vertices v k of odd (even) labels such that j k j + x. Therefore, the size of the red line G[S] is i. If x = 2i + 1, then the maximum edges of G have distinct parity. Hence, for each v j 2 G, we have that v j v j+x?1 2 S, for 1 j n? x, and G[S] is a red line of size i. Lemma 2 Let G be a red line decomposed into G[S] and G[D]. (G) = (G[S]) + (G[D]). Then Proof: The maximum degree of a red line G with size x is (G) = 2x. Corollary 1 says the size of the red line G[S] is bx=2c. Hence, (G[S]) = x, if x is even; and (G[S]) = x? 1, if x is odd. On the other hand, by the denition of G[D], the maximum degree (G[D]) = x, if x is even; and (G[D]) = x + 1, if x is odd. Therefore, if G is a red line, then (G) = (G[S]) + (G[D]). Theorem 1 If G is a red line, then G is Class 1. Proof: We argue by induction on the size x of a red line G. If x 1, then G is isomorphic to G[D] and G is Class 1, being a bipartite graph. If x 2, consider the decomposition of G into graphs G[S] and G[D]. Since by Lemma 2, we have the equality (G) = (G[S]) + (G[D]), it is enough to prove that G[S] and G[D] are both Class 1. By Corollary 1, G[S] is a red line of size bx=2c. Hence, by induction, G[S] is Class 1. In addition, by Lemma 1, G[D] is a bipartite graph. 5

6 3 Indierence graphs having no twin maximum degree vertices In this section we prove that every indierence graph with no twin -vertices is in Class 1. This result implies that all reduced indierence graphs are in Class 1. First we establish in Lemma 3 a necessary condition for an indierence graph with no twin -vertices to be Class 1. Lemma 3 An indierence graph with no twin -vertices cannot be neighbourhood-overfull. Proof: An odd maximum degree indierence graph is not neighbourhoodoverfull [3]. Hence, it is sucient to show that an indierence graph with no twin -vertices and even maximum degree is not neighbourhood-overfull. Let G be an indierence graph free of twin -vertices and with even maximum degree and v a -vertex of G. Recall that G is connected. If G is a trivial graph containing only the vertex v, then G is not neighbourhoodoverfull. Assume that G is not trivial. Consider any indierence order on the vertex set of G and let v 1 ; v 2 ; : : : ; v +1 be the indierence order induced on N[v]. Note that the N[v] is a subgraph of K +1, the complete graph on + 1 vertices. Because v is a -vertex of G and because G does not contain twin -vertices, we have v = v j, with 1 < j < + 1. Vertex v 1 is not adjacent to vertex v j+1, as otherwise, since G does not contain twin -vertices, vertex v j+1 would have to see vertex v +2 and vertex v j+1 would have degree greater than, a contradiction. An analogous argument shows that vertex v +1 is not adjacent to vertex v j?1. Thus v 1 does not see + 1? j vertices and v +1 does not see j? 1 vertices. This gives (j? 1) + ( + 1? j) = > =2 edges missing between vertices of N[v]. Therefore the number of edges of N[v] is lower than (G) 2 =2 and N[v] is not overfull. We partition the edge set of an indierence graph with even maximum degree and free of twin -vertices into two sets E 1 ; E 2, such that G 1 = G[E 1 ] is an odd maximum degree indierence graph and G 2 = G[E 2 ] is a matching. Let G be an indierence graph and v 1 ; v 2 ; : : : ; v n an indierence order for G. An edge v i v j is maximal if does not exist another edge v k v l with k i and j l. Note that an edge v i v j is maximal if and only if the edges v i?1 v j and v i v j+1 do not exist. In addition, every maximal edge v i v j denes a maximal clique having v i as its rst vertex and v j as its last vertex. 6

7 Thus every vertex is incident to zero, one or two maximal edges. Moreover, given an indierence graph with an indierence order and an edge that is maximal with respect to this order, the removal of this edge gives a smaller indierence graph: the original indierence order is an indierence order for the smaller indierence graph. Based on Lemma 4 below, we formulate an algorithm for choosing a matching of an indierence graph free of twin -vertices that covers every -vertex of G. Lemma 4 Let G be a non trivial graph. If G is an indierence graph with no twin -vertices, then every -vertex of G is incident to precisely two maximal edges. Proof: Let G be a non trivial indierence graph without twin -vertices and let v be a -vertex of G. Consider v 1 ; v 2 ; : : : ; v n an indierence order for G. Because v is a -vertex of G and G is not a clique, we have v = v j, for 1 < j < n. Let v i and v k be the leftmost and the rightmost vertices with respect to the indierence order that are adjacent to v j, respectively. Suppose that v i v j is not a maximal edge. Then v i?1 v j or v i v j+1 is an edge in G. The existence of v i?1 v j contradicts v i being the leftmost neighbour of v j. The existence of v i v j+1 implies deg(v j+1 ) + 1 since v j and v j+1 are not twins. Analogously, we have that v j v k is also a maximal edge. Now we describe an algorithm for choosing a set of maximal edges that covers all -vertices of G. Let G be an indierence graph free of twin -vertices and let v 1 ; : : : ; v n be an indierence order of G. Construct an edge set E as follows. For each -vertex of G, say v j in the indierence order, choose the edge v i v j, where v i is its leftmost neighbour with respect to the indierence order. Each component of the graph G[E] is a path. (Each component H of G[E] has (H) 2 and none of them is a cycle, by the maximality of the chosen edges.) For each path component of G[E], number each edge with consecutive integers starting from 1. If a path component contains an odd number of edges, then form a matching M of G[E] choosing the edges numbered by odd integers. If a path component contains an even number of edges, then form a matching M choosing the edges numbered by even integers. We claim that the above dened matching M covers all -vertices of G. For, if a path component of G[E] contains an odd number of edges, then M covers all of its vertices. If a path component of G[E] contains an even 7

8 number of edges, then the only vertex not covered by M is the rst vertex of this path component. However, by denition of G[E], this vertex is not a -vertex of G. Theorem 2 If G is an indierence graph free of twin -vertices, then G is Class 1. Proof: Let G be an indierence graph free of twin -vertices. If G has odd maximum degree, then G is Class 1 [3]. Suppose that G is an even maximum degree graph. Let v 1 ; : : : ; v n be an indierence order of G. Use the algorithm described above to nd a matching M for G that covers all -vertices of G. The graph G n M is an indierence graph with odd maximum degree because the vertex sets of G and GnM are the same and the indierence order of G is also an indierence order for G n M. Moreover, since M is a matching that covers all -vertices of G, we have that (G n M) =? 1 is odd. Hence, the edges of G n M can be coloured with? 1 colours and one additional colour is needed to colour the edges in the matching M. This implies that G is Class 1. Corollary 2 All reduced indierence graphs are Class 1. 4 Conclusions We have dened two decompositions for edge colouring indierence graphs. One method decomposes an indierence graph into a bipartite graph and a smaller indierence graph. The other method decomposes an indierence graph with no twin maximum degree vertices into a matching covering all maximum degree vertices and an odd maximum degree indierence graph. A graph G is a minimum indierence graph if G is a reduced indierence graph and, for some indierence order of G, every vertex of G is the rst or the last element of a maximal clique of G [4]. Minimum indierence graphs were dened in the context of clique graphs. Given two distinct minimum indierence graphs, their clique graphs are also distinct [4]. This property is not true for general indierence graphs [5]. The class of graphs edge coloured in Section 2 is a subclass of minimum indierence graphs. More generally, Corollary 2 of Section 3 says that every minimum indierence graph is Class 1. 8

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