The Domination and Competition Graphs of a Tournament. University of Colorado at Denver, Denver, CO K. Brooks Reid 3

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1 The omination and Competition Graphs of a Tornament avid C. Fisher, J. Richard Lndgren 1, Sarah K. Merz University of Colorado at enver, enver, CO 817 K. rooks Reid California State University, San Marcos, C 99 bstract. Vertices x and y dominate a tornament T if for all vertices z = x; y; either x beats z or y beats z. Let dom(t ) be the graph on the vertices of T with edges between pairs of vertices that dominate T. We show dom(t ) is either an odd cycle with possible pendant vertices or a forest of caterpillars. Since dom(t ) is the complement of the competition graph of the tornament formed by reversing the arcs of T, complementary reslts are obtained for the competition graph of a tornament. 1. Introdction. Sppose n tennis players compete in a \rond robin" tornament where each players plays every other players exactly once. If players are roghly even in ability and n is fairly large, it is nlikely that one player will beat every other player. It is more likely there are two players so that every other player is beaten by at least one of the two (sch a pair might form a good \dobles" team). We show that regardless of the reslts, there are at most n sch pairs. The reslts can be modeled by a \tornament". digraph is a set V () of vertices and a set () of ordered pair of vertices called arcs. We will denote an arc from x to y by (x; y) () and say x beats y. For all vertices x, let O (x) or O(x) (the ot-set of x) be the set of the vertices that x beats. Similarly, let I (x) or I(x) (the in-set) be the set of vertices that beat x. Let d + (x) = jo(x)j be the ot-degree of x. Let + () be the maximm of d + (x) over all vertices x in V (). tornament T is a digraph withot loops (i.e., arcs of the form (x; x)) in which for all x and y (distinct) in V (T ), either (x; y) (T ), or (y; x) (T ), bt not both. n n-tornament is a tornament with n vertices. reglar tornament is one in which d + (x) is constant for all vertices x. See Moon [8] and Reid and eineke [1] for more abot tornaments. Given a digraph, vertices x and y dominate if O(x) [ O(y) [ fx; yg = V (). Let the domination graph of, denoted dom(), be the graph with vertices V () and edges between the pairs of vertices that dominate (see Figre 1). The domination graph is closely related to the \competition graph". Given a digraph, the competition graph of, denoted C(), is the graph with the same vertices as and an edge between vertices x and y if and only if O(x) \ O(y) = ;. The domination graph of a tornament T is the complement of the competition graph of the tornament formed by reversing the arcs of T (this is not necessarily tre for all digraphs). So for tornaments, reslts on domination graphs correspond to reslts on competition graphs. owever, since the domination graph of a tornament generally 1 This research was partially spported by Research Contract N1-91-J-11 of the Oce of Naval Research. This research was partially spported by Research Contract N of the Oce of Naval Research. This research was partially spported by Research Contract N1-9-J-1 of the Oce of Naval Research. 1

2 T = g h e f 1 P I I PPPPPPPq - Y I N P - I - ) N U PPPPPPPq a b d c dom(t ) = g h f e a b d c Fig. 1. tornament and its domination graph. The edges of the domination graph dom(t ) are all pairs of vertices that dominate the tornament T. For example, vertices c and e are adjacent in dom(t ) becase in T vertex c beats a, g, f and d, and vertex e beats b, c, f and h. has fewer edges than the competition graph, it is more convenient to state and prove reslts on domination graphs. Competition graphs were introdced by Cohen [, ] in the stdy of food webs. food web can be modeled by a digraph whose vertices represent varios species, with an arc from vertex x to vertex y if the species represented by x preys pon the species represented by y. If two vertices have arcs to a common vertex, this represents two species competing for common prey, hence the name \competition graph". Competition graphs and their generalizations have been extensively stdied, for example, by righam and tton [1], Kim, McKee, McMorris, and Roberts [], and Roberts and Raychadhri [9]. Competition graphs of tornaments were rst considered by Lndgren, Merz, and Rasmssen [7]. Comprehensive srveys on competition graphs are provided by Kim [] and Lndgren [].. omination Graphs of Tornaments. Which graphs can be the domination graph of a tornament We partially answer this by nding graphs that are not sbgraphs of dom(t ) for all tornaments T. sbset of the vertices of a graph G is independent if there are no edges between the vertices in the sbset. Proposition.1. Let T be a tornament with z V (T ). Then O(z) is an independent set of dom(t ). Proof. Let x; y O(z). Then z O(x)[O(y)[fx; yg. So x and y do not dominate T and hence are not adjacent in dom(t ). Let (G) (the independence nmber of G) denote the maximm cardinality of an independent set of G. Corollary.. For any tornament T, we have + (T ) (dom(t )). Let C k denote the ndirected cycle on k vertices. Lemma.. Let k be an even nmber and T be any k-tornament. Then C k is not a sbgraph of dom(t ).

3 (a) I j * a 1 aaaaaaaaq!!!!!!!!1 C So CC S! S U C S I CC CW W S * S / S j (b) j * - O ai aaaaaaaa Ṟ U S Y C SS I C Sw C C 1 S!!!!!!! 7 SS CO C C!)! (c) Y O I!!!!!!!!1 a 1 S CO 7 C SS K C! aaaaaaaaq - Y C S C SS C RC Sw a 1 a a! aaaaaa!!!!!!!! Fig.. Reglar tornaments on 7 vertices and their domination graphs. Proof. Sppose C k is a sbgraph of dom(t ). Let 1; ; : : : ; k be the consectively labeled vertices of C k. s k is even, + (T ) k. Corollary. shows + (T ) (dom(t )) (C k ) = k. So + (T ) = k. The only two possible independent sets of order k in dom(t ) are the two maximal independent sets of C k given by = f1; ; : : : ; k 1g and = f; ; : : : ; kg. Say d + (1) = k. Then by Proposition.1, O(1) =. Then for any i, we have O(i) = becase 1 beats i, and O(i) = becase i. For any i f1g, we have O(i) = becase i, and O(i) = becase i beats 1. Ths d + (i) < k for all i = 1. So by symmetry, at most one vertex of T can have ot-degree k. Ths the sm of the ot-degrees of the vertices of T is at most k + (k 1) k 1. For k, this is less than the k(k1) arcs in T, a contradiction. Let k be an odd integer. Let S be a k1 -set contained in Zk (the integers mod k) where S and s 1 + s for all s 1 ; s S. For sch sets, we can form a reglar tornament T (S) called the rotational tornament with symbol S whose vertices are labeled by the elements of Z k and with arcs (i; j) if j i s where s S. Let U k = T (f1; ; : : : ; k g). Figre (a) shows U 7. Figre (b) shows T (f1; ; g). The reglar 7-tornament in Figre (c) is not rotational as O() and O() indce distinct -tornaments. Lemma.. Let k be an odd nmber and T a k-tornament. Then C k is a sbgraph of dom(t ) if and only if T is isomorphic to U k. Proof. (() The only dominating pairs in U k are i and j where j i 1 or k 1. So dom(u k ) is the k-cycle with vertices labelled consectively by ; 1; ; : : : ; k 1. ()) ssme C k is a sbgraph of dom(t ). Consectively label the vertices of C k

4 g f a b c d e Fig.. This graph is not the indced sbgraph of the domination graph of a tornament. by ; 1; ; : : : ; k 1. Corollary. shows + (T ) (dom(t )) (C k ) = k1. Since the average ot-degree in a k-tornament is k1, we have that d+ (i) = k1 for all i. Withot loss of generality, assme beats 1. The only independent set of C k of order k1 withot either or 1 is f; ; : : : ; k 1g, so O(1) is this set. In particlar, 1 beats. The only independent set of C k of order k1 withot either 1 or is f; ; : : : ; kg, so O() is this set. Contining in this way along the vertices of C k shows that T is isomorphic to U k. tree is a connected acyclic graph. caterpillar is a tree sch that the removal of all pendant (degree one) vertices yields a path. Figre shows the smallest tree that is not a caterpillar. This tree will be called NC7 (non-caterpillar on 7 vertices). It is well known that a tree which is not a caterpillar mst contain a copy of NC7. Lemma.. Let T be a 7-tornament. Then NC7 is not a sbgraph of dom(t ). Proof. Sppose NC7 is a sbgraph of dom(t ) with the vertices of T labeled as in Figre. Corollary. shows + (T ) (dom(t )) (NC7) =. Since the average ot-degree in a tornament with 7 vertices is, either T is reglar or T has a vertex with ot-degree. Figre shows the domination graphs of the only three nonisomorphic reglar 7-tornaments. None of these graphs has NC7 as a sbgraph. Ths, T has a vertex with ot-degree. Since fa; c; e; gg is the only vertex independent set in NC7, by Proposition.1 only b, d, or f can have ot-degree in T. y symmetry, assme withot loss of generality that b has ot-degree. Then b beats a, c, e and g, and loses to d and f. Since b and c dominate T, we have c beats d and f. Now frther assme withot loss of generality that d beats f. Since f and g dominate T and c and d both beat f, we see that g beats c and d. t then c and d do not dominate T becase neither beats g, a contradiction. Ths NC7 is not a sbgraph of the domination graph of a 7-tornament. Lemma.. Let S be an indced sbdigraph of a digraph. Then the indced sbgraph of dom() on the vertices of S is a sbgraph of dom(s). Proof. Let x; y S where fx; yg is an edge in dom(). Then O (x) [ O (y) [ fx; yg = V (). s V (S) V (), we have O S (x)[o S (y)[fx; yg = V (S). Ths fx; yg is an edge in dom(s). spiked cycle is a connected graph sch that the removal of all pendant vertices yields a cycle.

5 Theorem.7. Let T be an n-tornament. Then dom(t ) is either a spiked odd cycle, with or withot isolated vertices, or a forest of caterpillars. Proof. Lemmas. and. show that dom(t ) has no even cycles. First assme dom(t ) has a k-cycle C where k is odd. Lemmas. and. show that the sbtornament of T on the vertices of C is U k. y Lemma., the indced sbgraph of dom(t ) on C is a sbgraph of dom(u k ) = C k. So C = C k is an indced sbgraph of dom(t ). y Proposition.1, if x is not on C, then O(x) \ V (C) is an independent set. Since the independent sets in a k-cycle have at most k1 vertices, two vertices not in C cannot beat all k vertices in C. So the sbgraph indced on the vertices that are not on C has no edges. If some vertex x that is not on C is adjacent in dom(t ) to at least two vertices on C, say y and z, then edges fx; yg and fx; zg together with one of the two path connecting y and z form an even cycle in dom(t ), a contradiction. Ths dom(t ) is a spiked odd cycle possibly with isolated vertices. Otherwise, assme dom(t ) is cycle-free. Then by Lemmas. and., each component mst be a caterpillar. So, dom(t ) is a forest of caterpillars. Proposition.8. ll graphs G consisting of a spiked odd cycle C with possible isolated vertices are the domination graph of some tornament. Proof. Let G be sch a graph on n vertices with a cycle, C, of length k. Consectively label the vertices of C as f; 1; : : : ; k 1g. For i f; 1; : : : ; k 1g, let N i be the set of vertices pendant to i. Let J be the set of isolated vertices. We then constrct a tornament T with dom(t ) = G as follows. Create arcs between vertices in V (C) so that the indced sbgraph on V (C) is U k. Let i beat all vertices in N i. Let all vertices in N i beat all vertices in V (C) which dominate i and let all vertices in V (C) that are dominated by i beat all vertices in N i. For all i; j f; 1; : : : ; k 1g with i = j, if i beats j, then let each vertex in N i beat all vertices in N j. Let each vertex not in J beat all vertices in J. Pairs of vertices in N i and pairs of vertices in J are joined by arcs in an arbitrary manner. Then dom(t ) = G. Not all forests of caterpillars are the domination graph of some tornament (e.g., a path on vertices is not the domination graph of any tornament). In a sbseqent paper we will address the problem of which forests of caterpillars occr as domination graphs of tornaments.. Conseqences of the Characterization. Since Theorem.7 gives sch a sharp characterization, it is straightforward to dedce reslts abot varios graph parameters for domination and competition graphs of tornaments. elow are some examples. The bonds in Corollaries.1 to. are achieved for all allowed vales of n. Corollary.1. For n, then the maximm possible nmber of edges in the domination graph of an n-tornament is n. Corollary.. For n, the minimm possible nmber of edges in the competition graph of an n-tornament is n n.

6 sbset of the vertices of G form a cliqe if there are edges between every pair of vertices in the sbset. Let!(G) (the cliqe nmber of G) be the maximm cardinality of a cliqe of G. Clearly, (G) =!(G) where G is the complement of G. coloring of G is a labeling of its vertices so that adjacent vertices do not have the same label. Let (G) (the chromatic nmber of G) be the minimm possible nmber of labels in a coloring of G. Corollary.. For n, let T be an n-tornament. The cliqe nmber and chromatic nmber of the domination graph of T are at most. cliqe cover of G is a labeling of its vertices so that nonadjacent vertices do not have the same label. Let cc(g) (the cliqe cover nmber of G) be the minimm possible nmber of labels in a cliqe cover of G. Clearly, cc(g) = (G). Corollary.. For n, let T be an n-tornament. The independence nmber and the cliqe cover nmber of the competition graph of T are at most. Corollary.. For n, let T be an n-tornament. The independence nmber and the cliqe cover nmber of the domination graph of T are at least bn=c. Corollary.. For n, let T be an n-tornament. The cliqe nmber and the chromatic nmber of the competition graph of T are at least bn=c. digraph or graph is vertex transitive if for every pair of vertices i and j, there is an atomorphism that maps i to j. Rotational tornaments are vertex transitive. Corollary.7 shows that the tornaments U k are the only vertex transitive tornaments that have a dominating pair of vertices. The three reglar 7-tornaments illstrate this. Figre (a) is U 7 (which is vertex-transitive) and its domination graph is a 7- cycle. Figre (b) is also vertex transitive and its domination graph is edgeless. The domination graph of Figre (c) is neither a cycle nor is it edgeless, bt this reglar tornament is not vertex transitive. Corollary.7. Let T be a vertex-transitive k-tornament. Then either dom(t ) is C k and T is U k or dom(t ) is edgeless. Proof. Since a vertex transitive tornament is reglar and reglar tornaments have an odd nmber of vertices, jt j is odd. Frther, since T is vertex transitive, dom(t ) is also vertex transitive. So dom(t ) is a reglar graph on an odd nmber of vertices. Ths, the common degree of the vertices of dom(t ) is even. Corollary.1 shows that dom(t ) has at most n edges. Therefore this degree is either or. If it is, then dom(t ) is the disjoint nion of cycles. owever, Theorem.7 states that dom(t ) can have only one cycle. Ths dom(t ) is C k. y Lemma., T = U k. Otherwise, the degree is and dom(t ) is edgeless.

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