Strong Tournaments with the Fewest Hamiltonian Paths

Strong Tournaments with the Fewest Hamiltonian Paths J. W. Moon and Laura L. M. Yang Department of Mathematical Sciences University of Alberta, Edmonton, Alberta, Canada T6G 2G1 Abstract Busch recently determined the minimum number of Hamiltonian paths a strong tournament can have. We characterize the strong tournaments that realize this minimum. 1. Introduction A tournament, or an oriented complete graph, T = T n consists of a set of n nodes 1, 2,..., n such that each pair of distinct nodes i and j is joined by exactly one of the (oriented) arcs ij or ji. If the arc ij is in T we say that i dominates j and write i j. It is known (cf. [1; p. 14], [4; p. 21], [6; p. 173], or [7]) that every tournament T has a Hamiltonian path, i.e., a path that passes through each node T exactly once. Let h(t n ) denote the number of Hamiltonian paths in the tournament T n. Moon [5] gave upper and lower bounds for the minimum value of h(t n ) over all strong tournaments T n, that is, tournaments such that for every ordered pair of nodes i and j, there exists a path from i to j in T n. Busch [2] recently showed that the upper bound given in [5] was the actual value of the minimum. More precisely, he showed that if T n is a strong tournament with n 3 nodes, then h(t n ) f(n), where f(n) equals 3 5 (n 3)/3, 5 (n 1)/3, or 9 5 (n 5)/3, according as n is congruent to 0, 1, or 2 modulo 3. The bound is best possible in the sense that for any n 3 certain strong tournaments T n for which h(t n ) = f(n) are known (see [5]). We define these special tournaments in Section 2 and give a preliminary lemma in Section 3. In Section 4 we refine Busch s argument and show that these known special tournaments are the only tournaments T n for which h(t n ) = f(n). (For definitions not given here or for additional material on tournaments, see, e.g., [1], [4], or [6].) 1

2. Special Chains of Nearly-Transitive Tournaments We recall [2] that a tournament T = T n with n ( 3) nodes is nearlytransitive if its nodes can be labelled z 1,..., z n in such a way that z i z j when i < j except that z n z 1. We refer to z 1 and z n as the top and bottom nodes and to the remaining nodes as intermediate nodes. Note that if n = 3 and T is a 3-cycle, then any one of the nodes could be regarded as the top node, but once the choice has been made the next nodes of the cycle would be the intermediate node and the bottom node. If n 4, then the top node, bottom node, and intermediate nodes in a nearly-transitive tournament are uniquely determined. Suppose W and X are two disjoint nearly-transitive tournaments. Let us identify the top node of W with the bottom node of X and add arcs of the form xw for all nodes x and w in X and W (other than the node X and W now have in common). We call the resulting tournament a chain H = H(W, X) of the two nearly-transitive tournaments W and X; the top and bottom nodes of H are the top and bottom nodes of X and W, respectively; the intermediate nodes of H are the intermediate nodes of W and X; and the intersection node of H is the node W and X now have in common. These definitions can be extended to chains of m nearly-transitive tournaments in the obvious way for all m 1 (except that a chain consisting of a single nearly-transitive tournament does not have an intersection node). We sometimes refer to the nearly-transitive subtournaments in such a chain as its links. It is not difficult to see that the number of Hamiltonian paths in a nearly-transitive tournament T n is 2 n 2 + 1 and that the number of Hamiltonian paths in a chain is the product of the number of Hamiltonian paths in its links. It is a straightforward exercise (see [5; p. 102]) to verify that if H n is a chain with n ( 3) nodes, then h(h n ) f(n) with equality holding if and only if all the links of H n are of order 4 except for certain exceptional links: when n 0 (mod 3) there is one exceptional link of order 3 and when n 2 (mod 3) either there are two exceptional links of order 3 or there is one of order 5; we call such a chain H n a special chain. 3. A Preliminary Result Lemma. Let T = T n be a strong tournament with n 5 nodes and let R = R r+1 and S = S s+1 be two strong subtournaments of T that have a single node v in common, where r, s > 2 and r + s + 1 = n. If h(t ) = h(r)h(s) (3.1) 2

and R and S are both chains of nearly-transitive tournaments then T is also a chain of nearly-transitive tournaments. Before giving the details of the argument, we make some observations and introduce some notation. Busch [2; Lemma 2] showed that for any disjoint paths P and Q in a tournament T, there exists a path [P : Q] containing the nodes of P and Q and no other nodes such that the nodes of P and Q occur in the same order in [P :Q] as they do in P and Q. From this it follows, as he pointed out, that for each pair of Hamiltonian paths of R and S of the form BvA and B va, there exists at least one Hamiltonian path of T of the form [B :B ]v[a:a ]. This implies that h(t ) h(r)h(s). (3.2) This inequality was a key step in Busch s proof of the inequality h(t n ) f(n). Every strong tournament contains a Hamiltonian cycle (cf. [1; p. 16], [3], [4; p. 6], or [6; p. 173]). So we may assume the nodes of T are labelled so that vp 1 p 2... p r v and vq 1 q 2... q s v are Hamiltonian cycles of R and S. Let B i := p i+1... p r and A i := p 1... p i, for 0 i r and where B r and A 0 are the empty paths; let B j and A j be similarly defined with respect to the analogous Hamiltonian path of S. Then B i va i and B j va j are Hamiltonian paths of R and S, for 0 i r and 0 j s; so T has at least one Hamiltonian path of the form [B i :B j ]v[a i :A j ]. We now establish some conclusions that follow from our assumption that strict equality holds in (3.2). Our general approach will be to show that if these conditions did not hold, then either there would be at least two Hamiltonian paths of T of the form [B : B ]v[a : A ], for some pair of Hamiltonian paths BvA and B va of R and S; or there would be at least one Hamiltonian path of T that was not of this form for any pair of Hamiltonian paths of R and S. This would imply that strict equality would not hold in (3.2), bearing in mind its derivation. We formulate these conclusions in the following assertions. Assertion 1. Suppose that T, R, and S are as described above, that h(t ) = h(r)h(s), and that p 1 q 1. Then (a) p i q j, for 1 i r and 1 j s. Furthermore, (b) q 1 q s, and (c) v q s 1. 3

Proof of (a). Suppose that q j p i for some i and j, where i, j 1. We may assume that if i > 1, then p k q m for 1 k < i and all m; and that if j > 1, then p i q m for 1 m < j. We will show that this gives rise to a contradiction. There are two cases to consider. Case (i). i 1 and j > 1. We may suppose that p i q 1 and, if i > 1, that p i 1 q 1. So if q j p i, then p 1... p i q 1... q j and p 1... p i 1 q 1... q j p i would be two different [A i : A j ] paths (where it is understood that the second path starts with node q 1 if i = 1). Then there would be at least two Hamiltonian paths of T of the form [B i : B j ]v[a i : A j ]. This would contradict assumption (3.1), in view of the earlier observations. Case (ii). i > 1 and j = 1. We may suppose that p i 1 q 2. If q 1 p i, then q 1 p i... p r vp 1... p i 1 q 2... q s would be a Hamiltonian path of T that was not of the form [B : B ]v[a : A ] for any Hamiltonian paths BvA and B va of R and S; for, B would consist of the single node q 1, but there is no Hamiltonian path of S of the form q 1 va since v q 1. This also leads to a contradiction of assumption (3.1), so conclusion (a) must hold. Proof of (b). We may assume that s 3. We know, from part (a), that p r q s. If q s q 1, then vp 1... p r q s q 1... q s 1 would be a Hamiltonian path of T that was not of the form [B : B ]v[a : A ] for any Hamiltonian paths BvA and B va of R and S; for, A would necessarily be the path q s q 1... q s 1 and B would be empty, but va is not a Hamiltonian path of S since q s v. Thus the assumption that q s q 1 would imply a contradiction of (3.1). This implies conclusion (b). Proof of (c). If q s 1 v, then q 1... q s 1 vp 1... p r q s would be a Hamiltonian path of T that was not of the form [B :B ]v[a:a ] for any Hamiltonian paths BvA and B va of R and S; for B would necessarily be the path q 1... q s 1 and A would consist of the node q s, but B va is not a Hamiltonian path of S since q s v. This contradicts assuumption (3.1), so conclusion (c) must hold. Assertion 2. Suppose the assumptions of Assertion 1 hold. If, in addition, the subtournament S = S s+1 is a chain of nearly-transitive tournaments, then v is the top node of S. Proof of Assertion 2. We may assume that s 3. Since S = S s+1 is a chain of nearly-transitive tournaments, S has a unique Hamiltonian cycle C that proceeds from the bottom node through the intersection nodes if there are any to the top node and then through the intermediate nodes back to the bottom node. We know, from parts (b) and (c) of Assertion 1, that q 1 q s and v q s 1. It is not difficult to see, bearing in mind the definitions of C and S and the assumption that s 3, that these facts preclude v from being an intermediate node or the bottom node. Now suppose the chain S has at least two links and that v is one of its intersection nodes. Let q j denote the top node of S (so q 1,..., q j 1 are the intersection nodes between v and q j ). It follows from the definitions of C 4

and S and our labelling conventions that q j q j+2 and q j+1 v. Then q 1... q j q j+2... q s vp 1... p r q j+1 would be a Hamiltonian path of T. (Here we use the fact that p r q j+1, by Assertion 1(a).) But this Hamiltonian path is not of the form [B :B ]v[a:a ] for any Hamiltonian paths BvA and B va of R and S; for A would consist of the single node q j+1, but there is no Hamiltonian path of S of the form B vq j+1 since q j+1 v. Thus the assumption that v is an intersection node is incompatible with assumption (3.1). Therefore, v must be the top node, as asserted. The required Lemma now follows immediately from Assertions 1(a) and 2 and their duals. 4. Main Result Theorem. Let T = T n be a strong tournament with n 3 nodes. Then h(t n ) f(n), with equality holding if and only if T is a special chain of nearly-transitive tournaments. Proof. The only strong tournaments with 3 and 4 nodes are nearlytransitive tournaments; and h(t n ) = f(n) for these tournaments. So we may assume the theorem holds for all strong tournaments with fewer than n nodes, for some n 5. Now consider any strong tournament T n with n nodes. If T n is nearly-transitive then h(t n ) = 2 n 2 + 1 f(n), with strict inequality holding when n 6. So we may assume that T n is a strong tournament with n 5 nodes that is not nearly-transitive. Busch [2; Lemma 1] showed that such a tournament contains two non-trivial strong subtournaments R = R r+1 and S = S s+1 with the properties described in the first sentence of the Lemma. Following Busch, we may apply inequality (3.2), (part of) the induction hypothesis, and the definition of f(m) to conclude that h(t n ) h(r)h(s) f(r + 1)f(s + 1) f(n). It remains to determine when equality holds throughout here. It follows from the induction hypothesis that the equality h(r)h(s) = f(r + 1)f(s + 1) holds only if both R and S are special chains of nearlytransitive tournaments. And if h(t n ) = h(r)h(s), then it follows from the Lemma that the tournament T is a chain of nearly-transitive tournaments. The required conclusion now follows from the observation at the end of Section 2. The authors are indebted to a referee for some helpful suggestions. 5

References [1] J. Bang-Jensen and G. Gutin, Digraphs: Theory, Algorithms and Applications, Springer-Verlag London Ltd. 2001. [2] A. H. Busch, A note on the number of hamiltonian paths in strong tournaments, Electron. J. Comb. 13 (2006), #N3. [3] P. Camion, Chemins et circuits hamiltoniens des graphes complets, C. R. Acad. Sci. Paris 249 (1959), 2151-2152. [4] J. W. Moon, Topics on Tournaments, Holt, Rinehart and Winston, New York, 1968. [5] J. W. Moon, The minimum number of spanning paths in a strong tournament, Publ. Math. Debrecen 19 (1972), 101-104. [6] K. B. Reid and L. W. Beineke, Tournaments, Selected Topics in Graph Theory, L. W. Beineke and R. J. Wilson, (Editors), Academic Press, London, 1978, pp. 169-204. [7] L. Rédei, Ein Kombinatorischer Satz, Acta Litt. Szeged 7 (1934), 39-43. 6