On an adjacency property of almost all tournaments

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1 Discrete Mathematics On an adjacency property of almost all tournaments Anthony Bonato, Kathie Cameron Department of Mathematics, Wilfrid Laurier University, Waterloo, Ont., Canada NL C Received March 00; received in revised form September 00; accepted September 00 Available online July 00 In loving memory of Claude Berge Abstract Let n be a positive integer. A tournament is called n-existentially closed or n-e.c. if for every subset S of n vertices and for every subset T of S, there is a vertex x/ S which is directed toward every vertex in T and directed away from every vertex in S\T.We prove that there is a -e.c. tournament with k vertices if and only if k and k = 8, and give explicit examples for all such orders k. We also give a replication operation which preserves the -e.c. property. 00 Elsevier B.V. All rights reserved. MSC: 0C; 0C80; 0C Keywords: Adjacency property; n-existentially closed; Tournament; Replication. Introduction A tournament is a directed graph with exactly one arc between each pair of distinct vertices. Consider the following adjacency property for tournaments. Definition. Let n be a positive integer. A tournament is called n-existentially closed or n-e.c. if for every n-element subset S of the vertices, and for every subset T of S, there is a vertex x/ S which is directed toward every vertex in T and directed away from every vertex in S\T. Note that T may be empty. Adjacency properties of tournaments were studied in [,8,,8,]. Much of the research on such properties is motivated by the fact that while almost all tournaments with arcs chosen independently and with probability p, where 0 <p< is a fixed real number are n-e.c. for any fixed positive integer n see [], few explicit examples of such tournaments are known. Adjacency properties of graphs were studied by numerous authors; see [9] for a survey. A graph is called n-existentially closed or n-e.c. if it satisfies the following adjacency property: for every n-element subset S of the vertices, and for every subset T of S, there is a vertex not in S which is joined to every vertex of T and to no vertex of S\T. The n-e.c. property is of addresses: abonato@rogers.com A. Bonato, kcameron@wlu.ca K. Cameron. 00-X/$ - see front matter 00 Elsevier B.V. All rights reserved. doi:0.0/j.disc

2 8 A. Bonato, K. Cameron / Discrete Mathematics Fig.. The tournament D. interest in part because the countable random graph is n-e.c. for all n ; in fact, the countable random graph is the unique up to isomorphism countable graph that is n-e.c. for all n. The countable random tournament is the analogue of the random graph for tournaments; see []. The countable random tournament is the unique up to isomorphism countable tournament that is n-e.c. for all n. The cases n =, for graphs were studied in [9,0,].Forn>, few explicit examples of n-e.c. graphs are known other than large Paley graphs see [,8]. A prolific construction of n-e.c. graphs for all n was recently given in []. In the present article, we concentrate on the -e.c. adjacency property. Note that a tournament is -e.c. if the following adjacencies hold: for every pair of vertices, u and v, there are four other vertices: one directed toward both u and v, one directed away from both u and v, one directed toward u and away from v, and one directed toward v and away from u. In Section, we prove that there is a -e.c. tournament with k vertices if and only if k and k = 8, and give explicit examples for all such orders k. We consider only finite and simple tournaments. For a tournament G, V G denotes its vertex-set and EG denotes its arc-set. The order of G is V G. We denote an arc directed from x to y by x, y. For a vertex x V G, we define N out x ={y : x, y EG}, and N in x ={y : y, x EG}. As usual, a vertex x with N in x = is called a source and a vertex x with N out x = is called a sink. If U V G, G U is the subgraph of G induced by U; for x V G, G x = G V G\{x}. For basic information on graphs and tournaments, see [,]. The Paley tournament of order q, written D q, where q is a prime power congruent to mod, is the tournament with vertices the elements of GF q, the finite field with q elements, and x, y ED q if and only if x y is a nonzero quadratic residue. For D, see Fig.. As discussed above for Paley graphs, for a fixed positive n, sufficiently large Paley tournaments are n-e.c. see [8]; however, no other explicit families of tournaments with these adjacency properties are known. The next lemma follows from the definitions. Lemma. Let G be an n-e.c. tournament for some n>. For a fixed v V G, the tournaments G v, G N in v, and G N out v are each n -e.c. Definition. A tournament G is n-e.c. minimal if G has the smallest number of vertices among all n-e.c. tournaments. An n-e.c. tournament is critical if deleting any vertex leaves a tournament which is not n-e.c. Clearly, an n-e.c. minimal tournament is n-e.c. critical. In Section, we show that there are exactly two -e.c. critical tournaments up to isomorphism. In Section, we give examples of -e.c. critical tournaments of all possible

3 A. Bonato, K. Cameron / Discrete Mathematics orders k and k = 8. Vertex-criticality for various properties has been studied by many authors, including Berge [,,,,,0,,].. The -e.c. critical tournaments We make the following trivial observations. Remark. A tournament is -e.c. if and only if it has no source or sink. Remark. A tournament with a directed hamilton cycle is -e.c. The tournament D is the directed circuit on three vertices. It is easy to see that D is the unique up to isomorphism -e.c. minimal tournament, and thus, it is -e.c. critical. Define T to be the tournament consisting of two copies of D, with arcs oriented from the first copy to the second. It is straightforward to check that T is -e.c. critical. Theorem. The only -e.c. critical tournaments up to isomorphism are D and T. Proof. Let G be a -e.c. critical tournament. We first observe that a strongly connected component S of G has exactly three vertices. To see this, suppose that S has at least k vertices. By a theorem of Moon [9], S has a directed circuit C of length k. Deleting the vertex that C misses in S leaves a -e.c. tournament, which is a contradiction. We claim that if G has exactly one or two strongly connected components, then G is isomorphic to D or T, respectively. Assume to the contrary that G has r strongly connected components. From G we construct an auxiliary tournament G, whose vertices are the strongly connected components of G with the induced adjacencies. Note that G is isomorphic to the r-element linear order. Let u be a vertex of G that is neither a least nor greatest element. If we delete a vertex x in the strongly connected component of G corresponding to u, then the remaining graph G x, is -e.c., which is a contradiction.. Examples of -e.c. tournaments In this section, our main theorem is the following. Theorem. There is a -e.c. tournament with k vertices if and only if k and k = 8. To prove Theorem, we first prove the following theorem. Theorem. There is a unique up to isomorphism-e.c. minimal tournament, the Paley tournament D. Proof. Let G be a -e.c. tournament. Then since the unique minimal -e.c. tournament has three vertices, V G by Lemma. Suppose now V G =, say V G={,,,,,, }. Say N in ={,, },,,,,, EG; N out ={,, };,,,,, EG. See Fig. a.vertex currently has outdegree two, but needs outdegree three, so without loss of generality, assume that, EG. Then by considering the degrees of and, we get,,, EG and,,, EG. See Fig. b. Since N in ={,, } and, EG, it follows that,,, EG. See Fig. c. Then, for degree of,, EG, and then for degree of,, EG. See Fig. d. Then f : V G VD is an isomorphism, where f = 0,f =,f =,f =, f =,f = and f =. Given a -e.c. tournament, another -e.c. tournament with two more vertices can be constructed using a tournament version of the replication operation which was instrumental in [9]. Definition. Let G be a tournament and let a, b EG. Add two new vertices a,b such that a has the same adjacencies to vertices of G other than b as a does, b has the same adjacencies to vertices of G other than a as b does,

4 0 A. Bonato, K. Cameron / Discrete Mathematics 0 00 a b c d D Fig.. The proof of Theorem. a,b,a,b,a is a directed circuit, a and a are joined either way and b and b are joined either way; that is, a replicate R = RG, e is a tournament with VR= V G {a,b } and ER = EG {a,v: v N out a\{b}} {v, a : v N in a} {b,v: v N out b} {v, b : v N in b\{a}} {b, a, a,b, b,a} {exactly one of a, a, a,a} {exactly one of b, b, b,b}. We observe that for each arc e, there are four nonidentical replicates RG, e that we may construct depending on how we orient the edges aa, bb. Definition. Let G be a tournament, and let n befixed. An n-e.c. tournament problem isa n matrix x... x n, i... i n where {x,...,x n } is an n-element subset of V G, and for j n, i j {, }. A solution to an n-e.c. tournament problem is a vertex z V G such that z N in x j if i j =and z N out x j if i j =. Note that a tournament G is n-e.c. if and only if each n-e.c. tournament problem in G has a solution. Theorem. IfGisa-e.c. tournament, then for every e EG, each replicate R = RG, e is -e.c.

5 A. Bonato, K. Cameron / Discrete Mathematics 0 00 Fig.. The unique -e.c. tournament of order. y Proof. Fix e = a, b EG. Fix distinct x,y VR. We show that each problem j, i, j {, } has a solution in R. Case : {a,b } {x,y} = 0. A solution to the problem in G is a solution to the problem in R. Case : {a,b } {x,y} =. Assume that x = a and y = b. First, suppose y = a.ifi, j =,, an in-neighbour of a in G solves the problem; if i, j =,, an out-neighbour of a in G other than b solves the problem. The vertex b solves a a and b solves a. a a i y j by say, c, ing. If c = b, then c also solves a i If y = a, first solve so b solves a y j. The case when x = b and y = a follows by a similar argument. Case : {a,b } {x,y} =. Where z is a solution of a b i j in G, z is a solution of a i b j in R. x i y j.ifc = b, then i= and y = b, Using tournament replication on D, we obtain -e.c. tournaments for any odd order k, k. Now we work on finding -e.c. tournaments of all possible even orders. Theorem. There is no -e.c. tournament of order 8. Proof. It is straightforward to see that there is a unique -e.c. tournament of order ; see Fig.. Let G be a -e.c. tournament of order 8. Then G has a vertex of degree. In fact, the outdegree sequence of G is completely determined. Claim. G has exactly four vertices of indegree and four vertices of indegree. Let v V G. Since both G N in v and G N out v are -e.c., it follows that N in v. Let x be the number of vertices of indegree, and let y be the number of vertices of indegree. Then since the sum of all indegrees is the number of arcs, x + y = 8, x + y = 8. Solving the system establishes the claim. Now suppose V G ={,...,8}. For each vertex v of G, one of the subgraphs induced by N in v and N out v is D and the other is the tournament of Fig.. Without loss of generality, suppose vertex has indegree and the subgraphs induced by N in and N out are as in Fig.. Case : Vertex 8 has indegree. Without loss of generality, by the symmetry of,, and in the directed graph in Fig.,, 8,, 8, 8, EG. N in 8 ={,, } and, EG, soforg N in 8 D, also,,, EG. N out 8 ={,,, } and,,, EG, so, EG. Now N out =, so all remaining arcs meeting must be directed toward, so, EG. Then N in ={,, } and, EG, so,,, EG. The vertices,, and 8 are in N in, but8,,, EG so N in =, so, EG.NowN in ={,,, 8} and,, 8, EG so, EG.

6 A. Bonato, K. Cameron / Discrete Mathematics Fig.. The in- and out-neighbours of. 8 Fig.. G missing one arc. Now we have all but one arc of G, either, or,. See Fig.. If that arc were,, then N out ={,,, 8} and,,,, 8, EG which is a contradiction. Otherwise, if that arc were,, then N out ={,,, } and,,,,, EG, which is a contradiction. Case : Vertex 8 has indegree. In this case, 8,, 8,, 8 EG. Then N out 8 ={,, }, but,,, EG, which is a contradiction. To find -e.c. tournaments of all possible even orders as described in Theorem, it is sufficient to give an example of a -e.c. tournament of order 0, and then use replication. For this, see the tournament R in Fig.. It is straightforward to verify that R is -e.c.: one need only check the vertices,, and 0 versus each of the other vertices. The details are tedious and are therefore omitted. In [] it was proved that whenever there is a -e.c. graph of order m, then there is an -e.c. graph of order m +, and the question of this type of monotonicity was raised in general for n-e.c. graphs. We remark that the gap for -e.c. tournaments supplies the first example of nonmonotonicity of a -e.c. property.

7 A. Bonato, K. Cameron / Discrete Mathematics ' ' 0 Fig.. The tournament R. Reverse the arc, in RD,, where, and, are arcs, and add a new vertex 0 so that N in 0 ={,,, }. Note that not all arcs are shown.. Examples of -e.c. critical tournaments Definition. An arc e = a, b of tournament G is good if every vertex v = a,b is the unique solution to some -e.c. tournament problem not involving a or b. Lemma. LetGbea-e.c. critical tournament and let arc e = a, b be good. Then each replicate R = RG, e is a -e.c. critical tournament. Proof. Note that: the unique solution of b b is a, of b b is a, of a a is b, and of a a is b. Now let x VR {a,a,b,b }. By hypothesis, x is the unique solution to some -e.c. tournament problem in G. Ifa were a solution to this problem in R then a would be a solution to it in G, and if b were a solution to this problem in R, then b would be a solution to it in G. Therefore, x is the unique solution to this problem in R. Definition. In the definition of replication of the arc e = a, b in tournament G, we insist that the arc between a and a be a, a, and the arc between b and b be b,b, then we call the replication a type- replication, and use a subscript to indicate the resulting tournament, R G, e. Lemma 8. Let G be a -e.c. critical tournament and let e = a, b EG be good. Repeatedly replicating e using type- replication gives a -e.c. critical tournament. Proof. Define G 0 = G.Fork 0, define G k+ = R G k,e, and call the replication arc e k+ = a k+,b k+. Then by Lemma, G k+ is a -e.c. tournament of order V G +k. We need to show that G k+ is -e.c. critical. We proceed by induction on k. Assume G k is -e.c. critical and that for j k, vertex a j uniquely solves bj b and vertex b j uniquely solves aj a. Consider G k+. Since a, b is good in G, each vertex v V G\{a,b} is the unique solution to some -e.c. tournament problem in G not involving a or b. InG k+, no vertex a j or b j, j k + can solve this problem, because otherwise, a or b would have solved it in G. Vertices a k+ and b k+ cannot solve the problems that a j and b j j kuniquely solve in the induction hypothesis: for the a j problem, b j,a k+ and b k+,bare arcs of G k+ ; for the b j problem, b k+,a j and a, a k+ are arcs of G k+. Vertex a k+ uniquely solves bk+ b since every vertex except a k+ and a and b is directed the same way with, since respect to b and b k+, and a is directed toward b. By a similar argument, vertex b k+ uniquely solves ak+ every vertex except b k+ and b and a is directed the same way with respect to a and a k+, and a is directed toward b. a

8 A. Bonato, K. Cameron / Discrete Mathematics 0 00 Finally, a uniquely solves bk+ b and b uniquely solves ak+ a. Using Lemmas and 8, we may construct -e.c. critical tournament for all the possible orders k, where k and k = 8 as follows. For the odd orders, we note that in D,, is a good arc, as demonstrated by the following table. Vertex Uniquely solves For the even orders, the following tables demonstrate that the tournament R introduced at the end of Section is -e.c. critical, and that 0, is a good arc. Vertex 0 Uniquely solves Vertex 0 0 Uniquely solves We close with the following problem: find examples of n-e.c. tournaments, where n, that are not Paley tournaments. Acknowledgements The authors gratefully acknowledge the support of the Natural Sciences and Engineering Research Council of Canada NSERC. References [] R. Aharoni, E. Berger, The number of edges in critical strongly connected graphs, Discrete Math [] W. Ananchuen, L. Caccetta, On the adjacency properties of Paley graphs, Networks 99. [] W. Ananchuen, L. Caccetta, On tournaments with a prescribed property, Ars Combin [] C. Berge, Graphs and Hypergraphs, North-Holland, Amsterdam, 9. [] C. Berge, Some Common properties for regularizable graphs, edge-critical graphs and B-graphs, Theory and Practice of Combinatorics, North-Holland Mathematical Studies, vol. 0, North-Holland, Amsterdam, 98, pp.. [] C. Berge, About vertex-critical nonbicolorable hypergraphs, Austral. J. Combin [] C. Berge, Some properties of non-bicolorable hypergraphs and the four-color problem, First International Colloquium on Graphs and Optimization GOI, 99 Grimentz, Discrete Appl. Math [8] B. Bollobás, Random Graphs, Academic Press, London, 98. [9] A. Bonato, K. Cameron, On an adjacency property of almost all graphs, Discrete Math [0] A. Bonato, K. Cameron, On -e.c. line-critical graphs, J. Combin. Math. Combin. Comput [] J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, North-Holland, New York, 9. [] L. Caccetta, P. Erdős, K. Vijayan, A property of random graphs, Ars Combin [] P.J. Cameron, The random graph, in: R.L. Graham, J. Nešetřil Eds., Algorithms and Combinatorics, vol., Springer, New York, 99, pp.. [] P.J. Cameron, D. Stark, A prolific construction of strongly regular graphs with the n-e.c. property, Electron. J. Combin. 9 00, Research Paper. [] P. Erdős, On a problem in graph theory, Math. Gaz [] T. Gallai, Kritische Graphen I, Magyar Tud. Akad. Mat. Kutató Int. Közl [8] R.L. Graham, J.H. Spencer, A constructive solution to a tournament problem, Canad. Math. Bull [9] J.W. Moon, On subtournaments of a tournament, Canad. Math. Bull [0] V. Neumann-Lara, Vertex critical -dichromatic circulant tournaments, Discrete Math

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