EDGE-DISJOINT HAMILTONIAN PATHS AND CYCLES IN TOURNAMENTS

Transcription

1 EDGE-DISJOINT HAMILTONIAN PATHS AND CYCLES IN TOURNAMENTS CARSTEN THOMASSEN [Received 21 May 1980 Revised 17 November 1980] ABSTRACT We describe sufficient conditions for the existence of Hamiltonian paths in oriented graphs and use these to provide a complete description of the tournaments with no two edge-disjoint Hamiltonian paths. We prove that tournaments with small irregularity have many edge-disjoint Hamiltonian cycles in support of Kelly's conjecture. 1. Introduction The existence of many edge-disjoint Hamiltonian cycles in undirected graphs of large minimum degree has been established by Nash-Williams [14] and (in the case of regular graphs) by Jackson [10]. It seems more difficult to obtain results of that type for directed graphs. Nash-Williams [15] conjectured that a digraph of order n in which each outdegree and indegree is at least jn has two edge-disjoint Hamiltonian cycles. For n odd this can be derived from the author's result in [18] but for n even the conjecture is open and Nincak 16] showed it to be false for n = 6. While it is easy to decompose the complete undirected graph of order 2/c + l into k edge-disjoint Hamiltonian cycles and Tillson [19] has solved the problem of decomposing the complete symmetric directed graph of order 2/c into Hamiltonian cycles, almost nothing is known about the following problem formulated about 20 years ago by P. J. Kelly (see also [1, 3, 13]). Kelly's conjecture. Every regular tournament of order 2/c+l can be decomposed into k edge-disjoint Hamiltonian cycles. The conjecture is easily verified for k ^ 3 (the three non-isomorphic regular tournaments of order 7 are described by Kotzig [12]) and the conjecture has been verified by B. Alspach (private communication) for k = 4 as well. The author [18] and Jackson [9,10] have formulated conjectures on Hamiltonian cycles in oriented graphs with restrictions on the degrees which, if true, imply the existence of many edgedisjoint Hamiltonian cycles in regular tournaments. It is easy to prove that Kelly's conjecture is equivalent to the statement that any tournament of order 2/c whose vertices all have outdegree k or k 1 can be decomposed into k edge-disjoint Hamiltonian paths. In view of this it seems worthwhile investigating edge-disjoint Hamiltonian paths in tournaments as well. A result of this type has been obtained by Jackson [11] who proved that a regular tournament has an edge-disjoint pair of a Hamiltonian path and a Hamiltonian cycle. A more general approach to Kelly's conjecture is suggested by Alspach et al. [1]. In the present paper we prove that if G is an undirected graph such that each component of the complement of G is bipartite or an odd cycle of length at least 5 (and at most one component of the complement is an odd cycle), then any strong Proc. London Math. Soc. (3), 45 (1982),

2 152 CARSTEN THOMASSEN orientation of G has a Hamiltonian path. In particular, if we delete the edges of a Hamiltonian path or cycle from a tournament and the resulting oriented graph is strong, then it has a Hamiltonian path. We prove that any 2-edge-connected tournament has a Hamiltonian path whose deletion results in a strong oriented graph and we use this result to give a complete characterization of the tournaments which have no two edge-disjoint Hamiltonian paths. The tournaments with no two edgedisjoint Hamiltonian cycles seem to be difficult to characterize. We describe an infinite family of such 2-connected tournaments with large minimum indegree and outdegree. We prove as a main result that, for some positive constant c, every oriented graph in which each indegree and outdegree is at least jn c^jn has a Hamiltonian cycle. The proof of this depends on the result in [17] that every 4-connected tournament is strongly Hamiltonian-connected combined with the Hajnal-Szemeredi theorem [8] on almost equipartite colourings. The result is of interest in connection with the conjecture of the author [18] asserting that the same conclusion holds under the weaker assumption that each outdegree and indegree is at least j/it and it implies that every regular or almost regular tournament of order n has [c^jri] edge-disjoint Hamiltonian cycles. We also provide short proofs of results of Grotschel and Harary [7] and Fink and Lesniak-Foster [6] on orientations of graphs and we conclude with a number of unsolved problems related to the results of this paper and prove in connection with one of these problems that the deletion of any set of k edges from a tournament always leaves a Hamiltonian oriented graph provided k is small compared to the connectivity of the tournament. This work was done while the author was visiting associate professor at the Department of Mathematics, Louisiana State University. The author thanks Professor K. B. Reid for his hospitality and the Danish Natural Science Research Council for partial financial support. Thanks are also due to Dr R. Haggkvist and Dr B. Jackson for valuable comments on the paper. 2. Terminology A digraph D is a pair V{D), E(D), where V(D) is a finite set of vertices and E(D) is a set of ordered pairs xy of vertices called edges. An oriented graph is a digraph with no cycle of length 2 and a tournament is an oriented graph with no non-adjacent vertices. A semi-complete digraph is a digraph with no two non-adjacent vertices. If xy e E(D) we say that x dominates y and the number of vertices y s A ^ V(D) dominated by x is denoted d%(x). We call dy (D) (x) the outdegree of x and also denote it by just d + (x). The indegree d~(y) and d^(y) are defined analogously. A digraph D is k-connected (respectively k-edge-connected) if the deletion of any set of fewer than k vertices (respectively edges) leaves a strong (i.e. strongly connected) digraph. An xy path is a path P: x x x 2... x k such that x = x x and y = x k. A path of the type x y x 2...x,zx x fc is an augmentation of P. A Hamiltonian path or cycle of a digraph is a path or cycle including all vertices of the digraph. A digraph is called Hamiltonian if it has a Hamiltonian cycle and strongly Hamiltonian-connected if for any distinct vertices x and y, it has a Hamiltonian path starting at x and terminating at y. It is well known that every strong semicomplete digraph has a Hamiltonian cycle. If tthis conjecture has recently been disproved by R. Haggkvist.

3 EDGE-DISJOINT HAMILTONIAN PATHS AND CYCLES 153 a semicomplete digraph D is not strong, we can label its (strong) components D 1,D 2,...,D k such that no vertex of D, dominates any vertex ofd,ifj < i. We say that D x (respectively D k ) is the initial (respectively terminal) component of D and that D, and D i+1 are consecutive components for each i, where 1 ^ i^k 1. If each component has order 1, D is a transitive tournament. An almost transitive tournament is obtained from a transitive tournament by reversing the edge from the vertex of indegree zero to the vertex of outdegree zero. If A <= V(D), D A is the digraph obtained from D by deleting all vertices of A and all edges incident with these vertices and D(A) = D (V(D)\A). The irregularity i(t) of a tournament Tis defined in [17] as max\d + (x) d~(x)\ where the maximum is taken over all vertices of T. Waldrop [20] defined the quasiregularity q{t) as msi\{d + {x)-d + (y)) where the maximum is taken over all pairs x,y of vertices of Tand observed that i(t) ^ q{t) ^ 2i(T) always. A tournament T with i(t) = 0 (respectively i(t) = 1) is called regular respectively almost regular). In 6 and 7 we shall use the following definition. We say that a semicomplete digraph D is obtained from T by an r-edge-addition of size m if there exists a sequence such that for each /, with 1 < i ^ m, D, is a semicomplete digraph obtained from D,_ j by adding an edge yx such that, in T, d + (x) ^ d + (y) + r and such that D,_i has a Hamiltonian path starting at x and terminating at y. 3. Hamiltonian paths in strong oriented graphs Grotschel and Harary [7] showed that only very few 2-edge-connected graphs have the property that any strong orientation has a Hamiltonian cycle. THEOREM 3.1 (Grotschel and Harary [7]). Let G be a 2-edge-connected undirected graph. JfG is neither a cycle nor a complete graph, then G has a strong orientation which has no Hamiltonian cycle. We shall present here a short alternative proof. If G has no Hamiltonian cycle there is nothing to prove. It is also easy to prove that the complete bipartite graph K k, k (k ^ 3) has a strong orientation with no Hamiltonian cycle. Now if G has a Hamiltonian cycle and G is not a complete graph, nor a cycle, nor a complete bipartite graph K k k, then by a result of Dirac and Thomassen [5], G has a Hamiltonian path x,x 2... x n which cannot be extended into a Hamiltonian cycle. By orienting the edges such that x, dominates Xj if and only if j = /+ 1 of j < i 1 we get a strong orientation with no Hamiltonian cycle. In this section we describe a large class of undirected graphs each strong orientation of which has a Hamiltonian path. THEOREM 3.2. Let G be an undirected graph such that each component of the complement of G is bipartite or an odd cycle of length at least 5 and assume that the complement ofg has at most one non-bipartite component. Then any strong orientation of G has a Hamiltonian path. Proof. Let H be any strong orientation of G. We write V(H) = Au B such that H{A)

4 1 54 CARSTEN THOMASSEN is a tournament and at most one edge is missing in H(B). Since this edge is part of an odd cycle of length at least 5 in the complement of G, we can choose the bipartition A\J B such that H contains a path xzy where x and y are the non-adjacent vertices of B. We first show that H has a cycle C including all vertices of A. This is clearly true if H(A) is strong. If H(A) is not strong, we consider a shortest path P: x l x 2...x k from the terminal component of H(A) to the initial component. Then H(A) {x 2,...,x k - 1 } is a tournament with the same initial (respectively terminal) component as H(A) and hence P can be extended into a cycle containing all vertices of A. We let C be a longest cycle containing all vertices of A. If H V(C) is a tournament, then some vertex of C dominates a vertex v of the initial component of H V(C) because H is strong. Now H V(C) has a Hamiltonian path starting at v and this can be extended into a Hamiltonian path of H. So we consider the case where H V(C) is not a tournament, that is, the complement of G has an odd cycle containing the edge xy. Let C: v l v 2...v m v l where the indices are expressed modulo m. Suppose first that z e V(C). Then C contains vertices v h v i+j such that y, dominates one of x,y and v i+j is dominated by the other, and 1 O ^ m. Choose v L and v i+j such that) is the smallest possible. We shall prove that; = 1. Suppose v t dominates x', and y' dominates v i+j where {x',y'} = {x,y}. Suppose j > 1. Because of the minimality of j, x' is not dominated by v i+s when 1 ^ s < j and because of the maximality of C, x' does not dominate v i+l. So x' is non-adjacent to v i+i and dominates each of v i + 2,v i + 3)...,v i+j _ i. Similarly we show that / is non-adjacent to v i+j. { and is dominated by each of v i+j. 2,..., v i+1. Since the complement of G has no 3-cycle,) ^ 3 and now the edges v i+l y' and x'v i + 2 contradict the minimality of). So we assume) = 1. We add to the oriented graph H-V{C) the edge y'x' and obtain thereby a tournament. As in a previous case we consider a Hamiltonian path of this tournament which starts at a vertex v dominated by some vertex of C and if this path does not include the edge y'x' we get a Hamiltonian path of H. On the other hand, if the path (starting at v) includes y'x', then we replace this edge by the path y'v i+i v i+2...vtx' and obtain a Hamiltonian path of//. We assume next that z is not in C. If H- V(C) is strong, we consider any edge joining x to C (such an edge exists since x has degree 2 in the complement of G). If this edge starts (respectively terminates) at x, we add to H V(C) the edge xy (respectively yx) and consider a Hamiltonian cycle in the resulting tournament. Using this together with C and the edge joining x to C, we can easily find a Hamiltonian path of H. So we assume H V(C) is not strong. Let // l5 // 2,...,// p be the components of H V(C) labelled in the usual way and assume without loss of generality that x and y are not both in // x. Then some vertex v of H x is dominated by some vertex of C. We can now find a Hamiltonian path of H as in the case where H V(C) is a tournament unless for some i, we have K(//,) = {x} and V(H i + l ) = {y} or ViH^^ = {y}. But this case does not occur because of the path xzy. (If the Hamiltonian path of H l terminates at x and V{H 2 ) = {y}, then we just consider H p instead of H x.) This completes the proof. In the above theorem it is important that the complement of G does not contain a 3- cycle. It is easy to describe a strong orientation of a complete graph of order at least 6 less a 3-cycle which has no Hamiltonian path. We just let the three vertices of the 3- cycle all have outdegree 1 such that they dominate the same vertex. However,

5 EDGE-DISJOINT HAMILTONIAN PATHS AND CYCLES 155 Theorem 3.2 is probably far from being the best possible. For example, by using the method of the proof of Theorem 3.2 it is not difficult to show that any strong orientation of the complete graph of order at least 10 less the edges of two disjoint 5- cycles has a Hamiltonian path. The proof of Theorem 3.2 is particularly short when the complement of H is bipartite, i.e. when H V{C) is a tournament. If none of A and B are empty, we can in this case describe a Hamiltonian path of// starting in A and terminating in B. If C is a Hamiltonian cycle of//, then this obvious. On the other hand, if C is not a Hamiltonian cycle, then the maximality of C implies that no vertex of H - V{C) is adjacent to each vertex of C. But then C has a vertex x such that x dominates a vertex y of the initial component of H V(C) and such that y is not dominated by the successsor z of x on C. Then y and z are non-adjacent and hence z is in A. Now the path C xz can be extended to a Hamiltonian path of H starting at z e A and terminating in V(H\\V(C) B. Note that if C is not a Hamiltonian cycle, then the Hamiltonian path above starts at a vertex x which is not adjacent to all vertices of B. Using this, we get a short proof of the following result: THEOREM 3.3 (Fink and Lesniak-Foster [6]). IfG is an undirected graph such that the complement of G has no vertex of degree greater than 2 and no path of length greater than 2, then every unilaterally connected orientation H of G has a Hamiltonian path. Proof Let H l,h 2,...,H m be the components of H such that no vertex of H } dominates a vertex of//, when i <j. Since H is unilaterally connected, some vertex of //, dominates some vertex of H i+l for each i = 1,2,...,m 1. We prove, by induction on m, that H has a Hamiltonian path. If m = 1, this follows from Theorem 3.2, so assume m ^ 2. By the induction hypothesis, H ViHi) has a Hamiltonian path P starting at, say, x, and H x has a Hamiltonian path P' terminating at y, say. Let z denote the first vertex of P in // 3 when m ^ 3. If V{H X ) \> 1, we can choose P' such that y dominates x, by the remark preceding Theorem 3.3, and we can extend PKJ F to a Hamiltonian path of H. So assume V(Hi) = {y} and that y does not dominate x. In particular, V(H 2 )\ ^ 3. Analogously, we can assume that V{H m )\ = 1 so m ^ 3. Using the assumption on G we can deduce that either V(H 2 ) has a bipartition A\j B such that y dominates each vertex of A, and z is dominated by each vertex of B, and H(A) and H(B) are tournaments, or else x is non-adjacent to y and z and hence adjacent to all vertices of H 2. By the remark preceding Theorem 3.3, H 2 has (in the former case) a Hamiltonian path starting in A and terminating in B and we thus get a Hamiltonian path of//. In the latter case we get a Hamiltonian path by using the last sentence preceding Theorem 3.3. Theorem 3.3 is the best possible in the sense that it becomes false if the complement of G is a K x 3 or a path of length 3 (see [6]). We shall also apply Theorem 3.2 to find pairs of edge-disjoint Hamiltonian paths and cycles in tournaments. 4. Sufficient conditions for the existence of a pair of edge-disjoint Hamiltonian paths and cycles in tournaments In this section we describe classes of tournaments with the property that the deletion of the edges of some Hamiltonian path or cycle leaves a strong oriented graph

6 156 CARSTEN THOMASSEN which, by Theorem 3.2, has a Hamiltonian path. For the first result we need the following lemma. LEMMA 4.1. Let T be a tournament of minimum outdegree at least 2. Then T contains a path P with five or six vertices such that T(V(P)) E(P) is strong. Proof. Let V be a subtournament of T such that 7" has minimum outdegree at least 2 and such that 7" is the smallest possible under this restriction. Clearly T has at least five vertices and also 7" is strong since the terminal component of 7" has minimum outdegree at least 2. On the other hand, T has at most six vertices. To see this we let A denote the set of vertices of 7" which have outdegree 2 (in T') and we let B be the set of vertices of V(T')\A which are dominated by some vertex of A. Then it is easy to see that A ^ 5 and A u B \ ^ 6 and if we delete any vertex of T (A vj B) from 7" we get a tournament of outdegree at least 2. So V is strong and has order 5 or 6 and minimum outdegree 2. Consider any Hamiltonian cycle C of T. If T' E(C) is strong we have finished, so assume 7" E(C) is not strong. Then T E(C) has two strong components each consisting of a 3-cycle (in the case where 7" E(C) has minimum indegree 1) or else 7" E(C) has a component of order 3 or 4 and has a path from a vertex of indegree zero (in T' E{C)) to this component including the remaining vertices of V. In each case it is easy to leave out an edge of C so that we obtain a path P such that T' E(P) is strong. THEOREM 4.2. Let The a 2-edge-connected tournament. Then Thas a Hamiltonian path such that T E(P) is strong. Proof. By Lemma 4.1, T contains a path P: x l x 2...x k such that T(V(P))-E(P) is strong. We shall choose P such that k is maximum and show that P is a Hamiltonian path. Suppose therefore (reductio ad absurdum) that this is not the case. Let v be any vertex of V(T)\ V(P). By augmentation or extension of P, we can obtain a path P' with vertex set V(P) u {v}. Since T(V(P')) -E(P') is not strong (by the maximality of P), we conclude that v either dominates or is dominated by k 1 vertices of P. Now let. A (respectively B) be the set of vertices not in P which are dominated by (respectively dominate) all vertices of P except possibly one. If a vertex v of A dominates a vertex of P, this vertex must be x k for otherwise the path P': x l x 2 --.x k v would contradict the maximality of P. Similarly, no vertex of B is dominated by a vertex of P other than x {. Without loss of generality we can assume that A ^ 0. We claim that Thas no A -> B edge. For suppose xy is such an edge. Then one of the paths below is contained in T: yx 1 x 2...x k x, x 1 yx 2...x k x, yx 1 x 2...x k - 1 xx k, x i yx 2...x k - l xx k. By considering this path instead of P, we obtain a contradiction to the maximality property of P. Now let A' be the set of vertices of A dominating x k. Then T(A') is transitive, for if xyzx is a cycle of T(A'), then we would get a contradiction to the maximality of P by considering the path x l x 2...x k - l xyzx k. So A' = {z 1> z 2,...,z m } where z,- dominates z i if

7 EDGE-DISJOINT HAMILTONIAN PATHS AND CYCLES 157 and only if i < j. If we delete all edges from A' to x k the resulting oriented graph is not strong, so we conclude that m ^ 2. No vertex v in A\A' dominates two vertices z { and z i (i <j) of A', for then we consider the path x l x 2...x k vz i z j and obtain a contradiction. Since T is 2-edge-connected, T contains a path from z m to x k not containing the edge z m x k. Any such path must contain an edge of the form vz ( where v e A\A' and i < m. Since v dominates at most one vertex of A', we conclude that z m dominates v and by using this, we can easily see that P': x 1 x 2.-.x k - l z i z m x k v has the same property as P. This contradiction proves the theorem. As we shall see in the last section, Theorem 4.2 is the best possible in the sense that for each natural number k there is a strong tournament of minimum indegree, and outdegree at least k, having no Hamiltonian path P such that the deletion of E(P) leaves a strong oriented graph. As part of the next proof it is shown that the deletion of the edge set of some Hamiltonian cycle of a tournament leaves a strong oriented graph provided the tournament has minimum outdegree at least 2 and every edge is contained in a 3- cycle. THEOREM 4.3. Let The a tournament of order at least 3 such that each edge oftis contained in a 3-cycle. Then T has an edge-disjoint pair C,P where C(respectively P) is a Hamiltonian cycle (respectively path) oftunless Tis a 3-cycle or the tournament of order 5 obtained from a 3-cycle by adding two vertices x,y and the edge xy and letting y (respectively x) dominate (respectively be dominated by) the vertices of the 3-cycle. Proof (by induction on n \ V(T)\). If n < 5, the theorem is easily verified so we proceed to the induction step and assume n ^ 5. We consider first the case where T has a vertex x of outdegree 1. Let y be the vertex dominated by x. Since every edge incident with x is in a 3-cycle, y has indegree 1. Also, every edge of T {x,y} is contained in a 3-cycle in T {x,y) so we can apply the induction hypothesis and verify the theorem in the case where T has a vertex of outdegree 1 or, by symmetry, a vertex of indegree 1. The case where T {x,y} is one of the exceptional tournaments described in the theorem needs separate treatment. We leave those details for the reader. So we assume that all vertices have outdegree and indegree at least 2 and we let C be any Hamiltonian cycle of T. We shall prove that C can be chosen such that T E(C) is strong so that the theorem follows from Theorem 3.2. Suppose therefore that T E(C) is not strong, i.e. we can partition V(T) = A u B such that in T E(C) no vertex of A dominates a vertex of B. We claim that either each vertex of B is dominated by a vertex of A or each vertex of A dominates some vertex of B (in T). For if there were a vertex x in A and a vertex y in B disproving this statement, then the edge yx would not be in a 3-cycle. So assume!? = {y\,y2,...,y k } and that each y { is dominated by a vertex x, in A. Then C contains all edges x ( y h where 1 ^ i ^ k. (In particular x,- ^ Xj whenever i ^j.) Now suppose x, dominates Xj in T Since y,x ; is in a 3-cycle, we conclude that y } dominates y (. Also, every vertex z of /txix^xj,...^,;} dominates each vertex of {x^x^.^x*} (because all edges y t z, with 1 < i ^ k, are in 3- cycles). Since T has minimum outdegree at least 2, we have k ^ 2. Assume without

8 158 CARSTEN THOMASSEN loss of generality that y x y 2.~y k is a Hamiltonian path of T(B). If n = 2k, then n ^ 6 and is a Hamiltonian cycle C such that T E(C) is strong and if n < 2k we extend the path x k x k - l...x l y l...y k into a Hamiltonian cycle C" and, for any such cycle, T E(C") is strong. This completes the proof. Jackson [7] proved a result implying that a regular tournament has an edgedisjoint pair C, P of a Hamiltonian cycle and path, respectively. This corollary is also contained in Theorem 4.3. But Theorem 4.3 goes much further since almost all tournaments satisfy the assumption of the theorem (see Moon [11]). From Theorems 4.2 and 3.2 it follows that every 2-edge-connected tournament contains two edge-disjoint Hamiltonian paths. We use this in the next section to prove a stronger result. 5. The tournaments with no two edge-disjoint Hamiltonian paths The almost transitive tournament of order n clearly has no two edge-disjoint Hamiltonian paths when n is odd. Also, if a tournament has two consecutive components of order 1, then any Hamiltonian path contains the edge between these components. Consequently, a tournament fails to have two edge-disjoint Hamiltonian paths if it has two consecutive components of order 1 or a component which is almost transitive and of odd order. We prove in this section that any other tournament has two edge-disjoint Hamiltonian paths. We shall use the following lemma. LEMMA 5.1. Let P x and P 2 be two edge-disjoint Hamiltonian paths of a tournament T terminating at the same vertex and suppose this vertex has outdegree at least 2. Then T has two edge-disjoint Hamiltonian paths P\, P' 2 with distinct terminal vertices. If the initial vertices of P y, P 2 are distinct, then P\ and P' 2 can be chosen such that they too have distinct initial vertices. Proof. Let P x : x l x 2...x n. l x n and P 2 : yiy 2...y n where x n y n. We must have n ^ 5. If x n dominates x x or y n dominates y l5 we get an edge-disjoint pair of a Hamiltonian path and cycle, respectively, and it is easy to complete the proof in this case. So assume x t dominates x n and y { dominates y n. Since x n has outdegree at least 2 there exists in T a path of length 2 of the form x i x n x i + 1 where 1 < i ^ n 3. By considering the path P\: x 1 x 2...x,x n x l x, ) _ 1 instead of P x we get the desired pair of Hamiltonian paths unless )> _! = x t. So assume this is the case and Thas only one path of the form x t x n x i+ i. This implies that x n is dominated by each Xj, where 1 ^y ^ i, and that there exists a number m ^ 1 such that x n dominates each vertex x jt with i+\ ^j^ i + m, and is dominated by each x J5 withy > i + m. Since x n has outdegree at least 2, m ^ 2. By considering P 2 instead of P^, we conclude that there exists a unique k, with 1 ^ k ^ n 3, such that Tcontains the path P' 2 \ yiy 2 ---y k y n y k +i---y n -i, and as above we deduce that y k = x n -i. Now P\ and P' 2 have the desired properties unless they have the edge x n x I + 1 in common, that is, y k +Y=x i+l. But now the paths P'I : x 1 x 2...x 1 x n x x n _ 1 x and P' 2 have the desired properties since P' 2 ter-

9 EDGE-DISJOINT HAMILTONIAN PATHS AND CYCLES 159 minates at y n -i = x,- (note that x,,-! dominates x l + 1 because y k = x n. 1 and.vfc+ 1 = x i+ l)- THEOREM 5.2. A tournament T fails to have two edge-disjoint Hamiltonian paths if and only if T has a component which is an almost transitive tournament of odd order or has two consecutive components of order 1. Proof It is easy to reduce Theorem 5.2 to the following statement: every tournament Twhich is strong and which is not an almost transitive tournament of odd order has two edge-disjoint Hamiltonian paths with distinct initial vertices and distinct terminal vertices. We shall prove this statement by induction on n = V(T)\. The statement is easily verified for n ^ 4 so we proceed to the induction step and assume that n ^ 5. By Theorems 3.2 and 4.2 and Lemma 5.1, we can assume that Tis not 2-edgeconnected. We consider first the case where Thas a vertex x of outdegree n 2 such that T x is strong, i.e. the vertex y which dominates x also dominates at least one vertex x' of the initial component of T {x,y). Since Tis strong, y is dominated by at least one vertex y' of the terminal component of T {x,y}. If T {x,y} is almost transitive and of odd order we let its vertices be labelled x lj x 2,...,x n _ 2 such that x ; dominates Xj whenever i < j except when i = 1 and j = n 2. Now T has a Hamiltonian path P of the form xx 1 x 2...x l -yx I+ x...x n _ 2 or xx n _ 2 yx 1 x 2...x n _3 and it is easy to find a Hamiltonian path P' of T E{P) starting with the edge yx and not terminating at the terminal vertex of P. If T {x,y} is strong and not both almost transitive and of odd order, it has, by the induction hypothesis, two edge-disjoint Hamiltonian paths P l :u l u 2..M n _ 2 and P 2 : v l v 2...v n - 2, where u x ^ u x and u n _ 2 ^ v n - 2. Now Thas the following two edge-disjoint Hamiltonian paths yxu i u 2...u n - 2 and xv i v 2...v i yv i+1...v n _ 2, where 1 ^ i ^ n 2. So we consider the case where T {x,y} is not strong and we let T' denote the tournament obtained from T {x,y] by reversing the edge x'y'. If T has two edgedisjoint Hamiltonian paths P X,P 2, we can assume that P x does not contain the edge y'x' and we now get two edge-disjoint Hamiltonian paths of Tas in the previous case replacing the edge y'x' by the path y'yx' if y'x' is in P 2. On the other hand, if T is almost transitive and of odd order, we label its vertices x li x 2,...,x n^2 such that x, dominates Xj when i < j except when i = 1 and; = n 2. Since the reversal of the edge y'x' results in a tournament which is not strong, we must have or or y' = x x and x' = x 2 y' = x n _ 3 and x' = x n _ 2 y' = x n _ 2 and x' = x,. If the third alternative holds, we can assume that y is dominated by some x,-, where 1 < i < n 2, for otherwise Tis almost transitive. Now if P denotes the path xx,yx 1 x 2...x,_ 1 x l x n _ 2

10 160 CARSTEN THOMASSEN it is easy to find a Hamiltonian path in T E(P) starting at x 2 and containing the path x n - 2 y xx \- If the third alternative does not hold we can assume that T {x,y] is not transitive and hence n ^ 7. In the case where the first alternative holds, T has the following edge-disjoint Hamiltonian paths: yxx 2 x i x 2 x^...x n. 2 and xx 2 x 5...x n _ 2 Xiyx 2 x 4...x n _ 3. When the second alternative holds, T has the following edge-disjoint Hamiltonian paths: xx l x 2...x n _ A x n. 2 x n. 3 y and yxx 2 x 4...x n _ 5 x /,_ 2 x 1 x 3...x, I _ 4 x n _3. So we can now assume that for any vertex x of outdegree n 2 and, by analogy, for any vertex x of indegree n 2, T x is not strong. If x and y both have outdegree n 2 and y dominates x it is easy to see that T x is strong. So we can assume that T has at most one vertex of outdegree (respectively indegree) n 2. We next consider the case where there is a partition V(T) = A u B such that A > 2, B\ ^ 2, and T has only one edge xy such that x e B, y e A. We then consider the tournaments T { and T 2, where 7i is obtained from T(A) by adding a vertex v { and letting t^ dominate y only, and T 2 is obtained from T(B) by adding a vertex y an 2 d letting v 2 dominate x only. Clearly both T x and T 2 are strong. Since we assume that T has at most one vertex of outdegree n 2, we conclude that T { has at most one vertex of indegree 1 and hence T x is not almost transitive. Similarly, T 2 is not almost transitive. By the induction hypothesis, 7] has two edge-disjoint Hamiltonian paths P lfl -, P 2J with distinct initial (respectively terminal) vertices for i = 1,2. We can assume that P lti terminates (respectively starts) at u, for / =1,2. We let P x be the unique Hamiltonian path of Tcontaining (P it i Vi)^j(Pi t2 v 2 ) and we let P 2 be the unique Hamiltonian path of Tcontaining (P 2 t uj u (P 2 2 v 2 )u {xy}. It is easy to see that P x and P 2 satisfy the conclusion of the theorem. We can therefore assume that V(T) has no partition A u B as described above. Since T is not 2-edge-connected this means that T has a vertex x of outdegree or indegree (say outdegree) n 2. Also we can assume that T x is not strong which means that the vertex y dominating x is dominated by each vertex of the initial component Tj of T-{x,y}. If we put A = K(T x )u {x} and B = V(T)\A, then yx is the only edge from B to A. Since \A\^2, we then conclude that B = {y}, that is T {x,y} = Tj is strong. By the induction hypothesis, either 7^ has two edge-disjoint Hamiltonian paths as described in the theorem or T y is almost transitive and of odd order. In each case it is easy to describe two edge-disjoint Hamiltonian paths of T satisfying the condition described in the theorem and the proof is complete. 6. Hamiltonian paths with prescribed initial and terminal vertex In this section we use the results of [17] to establish some auxiliary results to be used in the next section on edge-disjoint Hamiltonian cycles. We shall consider a Hamiltonian path from a vertex of large outdegree to a vertex of small outdegree and then add the edge from the terminating vertex to the initial vertex and by repeating this process we shall get a strongly Hamiltonian-connected semicomplete digraph. We begin with a result which is of independent interest (although only an easy part of this will be used in what follows). THEOREM 6.1. Let The a tournament with at least two vertices and let x (respectively y) be a vertex of T of maximum (respectively minimum) outdegree. Then T has a

11 EDGE-DISJOINT HAMILTONIAN PATHS AND CYCLES 161 Hamiltonian path from x to y unless one of the statements below holds. (i) The terminal component of T x consists of one vertex z dominating x, and the preceding component of T x consists of y only and y is dominated by x. (ii) The initial component oft y consists of one vertex z dominated by y, and the second component of T y consists of x only and x dominates y. (iii) Tis isomorphic to the regular tournament T 5 C of order 5 [17, Fig. 1] or to one of T 6 S,T 6 S of [\1, Fig. 21 Proof. If T is not strong, then x (respectively y) belongs to the initial (respectively terminal) component of Tand hence Tcontains a Hamiltonian path from x to y. If T is strong and T x is not strong, then either y belongs to the terminal component of T x (in which case it is easy to describe an xy Hamiltonian path) or else (i) holds. If T y fails to be strong we argue similarly so we assume that each of T, T x, T y is strong. If T {x,y} is not strong it is easy to find an xy Hamiltonian path (see, for example, the proof of Theorem 2.1 in [17]). So we can assume that T {x,y} is also strong and we let z l z 2...z n _ 2 z 1 be a Hamiltonian cycle of this tournament. Suppose that T has no xy Hamiltonian path. If x dominates z,-, then y cannot be dominated by z,_j. But this implies that d + (y) ^ d + (x) \ and hence Tis regular or almost regular. In [2,17] it is proved that every regular tournament of order at least 7 and every almost regular tournament of order at least 10 is strongly Hamiltonian-connected. With that result, it is not difficult to prove that (iii) holds (we omit the details). COROLLARY 6.2. Any tournament contains a Hamiltonian path from a vertex of maximum outdegree to a vertex of minimum outdegree. LEMMA 6.3. Let Tbe a tournament of order n > 8 such that Tdoes not contain two vertices x o,y o with the property that x 0 (respectively y 0 ) dominates (respectively is dominated by) every vertex of T {x o,y o }. Then we can obtain a 2-connected semicomplete digraph from Tby a y(n 5)-edge-addition of size at most 21. Proof. If Tis not strong, then we can obtain a strong semicomplete digraph D x by a jn-edge-addition of size 1. If Tis strong, put T= D t. Suppose D x has a vertex x of indegree 1. By the assumption of the lemma, the terminal component of D x x has more than one vertex and so D x has a Hamiltonian path starting at x and terminating at a vertex not dominating x and of outdegree at most j(n 1) in T. Since a semicomplete digraph has at most two vertices of indegree (respectively outdegree) 1, it follows from the reasoning above that we can obtain a semicomplete digraph D 2 of minimum indegree and outdegree at least 2 by a j(n 3)-edge-addition of size at most 5. We now consider eight vertices x lt x 2, x 3, x 4, y t, y 2, y 3, y* of T such that for each vertex z V(T)\{xi,x 2,x 3,x 4,y l,y 2,y 3,y 4 } we have (in T) d + ( Xl ) > d + (x 2 ) > d + (x 2 ) > d + (x 4 ) > d + (z) and we extend D 2 into a semicomplete digraph D 3 such that D 3 is a %n 5)-edgeaddition of Tand such that all edges in E(D 3 )\E(D 2 ) are of the form y.x, where i,j e {1,2, 3,4}. We assume D 3 is maximal under these conditions and the assertion of the lemma now follows by proving that D 3 is 2-connected.

12 162 CARSTEN THOMASSEN So we assume D 3 has a vertex JC such that D 3 x is not strong. Let D' (respectively D") be the initial (respectively terminal) component of D 3 x. Since no vertex of D 3 has indegree or outdegree 1, each of D', D" has order at least 2 and we let u l u 2...u l and v l v 2...v l be a Hamiltonian cycle of D' (respectively D"). Without loss of generality we can assume that x dominates u { and is dominated by i^. Also note that each edge u ( Vj of D 3 is also an edge of T. It is easy to see that for each vertex u e V(D')\{u l } and each vertex v e V(D")\{vj}, D 3 has a uv Hamiltonian path. Tf u (respectively v) is a vertex of K(D')\{MI} (respectively V{D")\{vi}) such that no other vertex of V(D')\{u 1 } (respectively V(D")\{v i }) has larger (respectively smaller) outdegree in T than u (respectively i>), then it is easy to see that, in T, Also, if u is any vertex of D'^^} and v',v" are any two vertices of D", then in T, d + (u)>m\n{d + (v'),d + (v")}. From this it follows that D' u x contains a vertex x { (1 ^ i < 4) and analogously D" v { contains a vertex y } (1 ^7^4). Since D 3 has an xjj Hamiltonian path and, in T, we have obtained a contradiction to the maximality property of D 3 and the proof is complete. LEMMA 6.4. Let k be an integer, k ^ 3, and let The a tournament of order n > 6k 4. // Tdoes not contain vertices x o,y o such that x 0 (respectively y 0 ) dominates {respectively is dominated by) all vertices oft {x o,y o }, then we can obtain a k-connected semicomplete digraph from Tby a j(n 3k + 3)-edge-addition of size at most (3k 3) Proof. By Lemma 6.3 let D Y be a 2-connected semicomplete digraph obtained from 7 by a j[n 5)-edge-addition of size at most 21. Let A, B be disjoint sets such that \A\ = \B\ = 3k 3 and such that for any vertices x e A, y e B, z e V(T)\(A u B) we have We then extend T into D 2 by a j{n 3fc + 3)-edge-addition such that D { <= D 2 and such that all edges in E(D 2 )\E(D l ) are of the form yx where y e B, x e A. We complete the proof by showing that D 2 is /c-connected. Suppose therefore that F is a set of vertices of D 2 such that D 2 F is not strong and F ^ k 1. Let D' and D" denote the initial and terminal component, respectively, of D 2 F. Then D' (respectively D") contains a vertex u (respectively v) which in T(V(D')) (respectively T(V(D"))) has outdegree at least {\\ V(D')\ 1) (respectively at most 1)). So in 7, We claim that A n V(D') and B n V(D") are both non-empty. For if A n V(D') = 0, say, then A n V(D") \ ^ 2k 2 and so A has a vertex which in T(V(D")) has indegree at least k 1 and now every vertex of V(D') has larger outdegree in T than that vertex of A, a contradiction. Now let x e An V(D') and y e Bn V(D"). If we select x and y such that they have

13 EDGE-DISJOINT HAMILTONIAN PATHS AND CYCLES 163 maximum (respectively minimum) outdegree in T under those restrictions we have, in T, Then x and y are connected by three paths of length 2 and, by [17, Corollary 3.5], D 2 has an xy Hamiltonian path. This contradicts the maximality property of Tand the proof is complete. LEMMA 6.5. Let T be a tournament with n ^ 24 vertices. Then we can obtain a semicomplete digraph D by a j(n \\)-edge-addition of size at most 103 such that T contains a set A ofo, 2, or 4 vertices and for each ordered pair x, y of vertices of V(T)\A, D has an xy Hamiltonian path. Proof. We use Lemma 6.4 with k = 4. If Tdoes not contain two vertices x o,y o as described in Lemma 6.4, we obtain by a^(/i 9)-edge-addition of size at most 102 a 4- connected semicomplete digraph which is strongly Hamiltonian-connected by [17, Corollary 5.7]. So assume that T contains such vertices x o,y o. If T {x o,y o } does not contain vertices x lt yx such that x x (respectively y { ) dominates (respectively is dominated by) every vertex of T {-K O 5.) ; o>- x i>.> ; i}> then by Lemma 6.4, we obtain from T {x o,y o }, by a j(n 1 l)-edge-addition of size at most 102, a 4-connected semicomplete digraph and we get the desired result (by also adding y o x o if necessary) with A = {x o,y o }. So we can assume that Tcontains vertices Xi,y x as described above. But now we get the desired result with A = {x Oi y Oi x l,y l }. For if x,y e V(T)\A, then any Hamiltonian path of T (A<u{x,y}) can be extended into an xy Hamiltonian path of Tu {yoxo^ixi} and the proof is complete. We conclude this section with an application of the technique used in the previous lemmas. THEOREM 6.6. Let D be a semicomplete digraph such that every Hamiltonian path ofd is contained in a Hamiltonian cycle. Then one of the following holds: (i) D is complete; or (ii) D is a 3-cycle; or (iii) D is obtained from a 3-cycle by adding an edge; or (iv) D is obtained from a complete digraph or a 3-cycle by adding two new vertices x o,y o and letting x 0 be dominated by y 0 and letting x 0 (respectively y 0 ) dominate (respectively be dominated by) all other vertices except possibly y 0 (respectively x 0 ). Proof (by induction on n = \ V(D)\). If w < 3, then (i), (ii), or (iii) holds, so assume n ^ 4. Using the same argument as in the proof of Lemma 6.3 we can assume that each vertex of D has indegree and outdegree at least 2 unless D contains vertices x o,y o as described in Lemma 6.3, in which case it is easy to see that (iv) holds (using the induction hypothesis on D {x o,y o }). So we can assume that D has minimum indegree and outdegree at least 2. By the same reasoning as in the proof of Lemma 6.3 we can even assume D to be 2-connected. Let x be any vertex of D. Now D x is strong and contains therefore, for any three vertices y, z, v, a Hamiltonian path connecting two of them, by [17, Corollary 2.2]. So

14 164 CARSTEN THOMASSEN x is dominated by at least one of y,z, v and by symmetry, x dominates at least one of y, z, v. So there are at most four vertices of D which are not joined to x by two edges. Suppose now that x is incident with at least two double edges. Then any Hamiltonian path of D-x can be augmented to a Hamiltonian path of D with the same initial and terminal vertex. So every Hamiltonian path of D x can be extended to a Hamiltonian cycle and it is easy to complete the proof using the induction hypothesis. It only remains to consider the case where every vertex of D is incident with at most one double edge. In particular, n < 6 in this case. We leave it to the reader to verify that D is strongly Hamiltonian-connected and hence complete. The above result is analogous to the aforementioned result of Dirac and Thomassen [5] and it extends the more special result for tournaments due to Griinbaum (see [3, 17]). 7. Hamiltonian cycles in oriented graphs with large indegrees and outdegrees THEOREM 7.1. IfG is an oriented graph of order n with each indegree and outdegree at least \n ^/(n/1000), then G is Hamiltonian. Proof. By the theorem by Hajnal and Szemeredi [8], any graph H of maximum degree A can be vertex coloured in A +1 colours such that each colour class has cardinality [ V{H) /(A + 1)] or [ V(H) /(A + 1)] +1. Hence the vertex set of G can be partitioned into sets A i,a 2,...,A 2k such that k ^ ^/(n/1000) and each 6(4,), for 1 ^ /' ^ n, is a tournament of order [n/2/c] or [n/2/c] +1. We claim that for each t = 1,2,..., 2k, there exists in G a cycle C, such that C, contains all vertices of A 1 u A 2 u... u A, and at most 104t vertices in A t+ 1 u A t + 2 u... u A 2k. We first show that C t exists. If G{A V ) is strong, this is obvious, so assume G{A V ) is not strong. Consider a Hamiltonian path of G(A l ) starting at a vertex x of maximum outdegree (in G{A l )) and terminating at a vertex y of minimum outdegree in G{A X ). Then in G{A X ), and in G, Since n/4/c > 2k, G has a path of length 2 from j; to x, and we get a cycle C x containing A x and one vertex not in A v Suppose we have already described C t, for 1 ^t^2k 1. Let T denote the tournament of G induced by A t + l \V(C { ). Then T has order n' where ri ^ [n/2fc]- 104r > n/2k-2q&k. By Lemma 6.5, we can obtain from 7" a semicomplete digraph D satisfying the assertion of that lemma by a-j(n' ll)-edge-addition of size at most 103. Let e l,e 2,... be the edges of E(D)\E{T) added in that order. Suppose first that, for each edge e, = y,x,,

15 EDGE-DISJOINT HAMILTONIAN PATHS AND CYCLES 165 Since in 7", or, equivalently, and in G, we conclude that G has at least paths of the form y ( zx,- where z K(C f ) v A t + l.(ln particular, t < 2k 1.) We replace e, by such a path in G and since G has more than 103 such paths for each e { we can assume that no two of them have an intermediate vertex in common. In this way we obtain a subgraph G' of G V(C t ) containing T and at most 103 vertices not in 7" such that for any vertices x, y of 7" A, G' has an xy path containing all vertices of T. We consider the case where some vertex xoft' Ais dominated by some vertex x' of C, in G and some vertex y of 7" A dominates some vertex y' of C f in G. Since each vertex of 7" is adjacent to some vertex of C f, we can assume that x ^ y. Also, each vertex of C, is in G adjacent to more than two vertices oft' A so it is easy to see that we can choose x,y, x',y' such that x' is the predecessor of y' on C, (the reason for these adjacencies is that V{C t ) and V{V) are large compared to k). But then C /+1 is obtained from C, by replacing the edge x'y' by an xy path of G', as described above. If all edges between T' A and C, have the same direction, say from C, towards 7" A, then we can find a path uzv where u is a vertex of smallest outdegree in V A, v is a vertex of largest outdegree in G(V(C,)), and z is a vertex in (A t + 2 u... u A 2k )\V(C t ). Now we can repeat the arguments above with G replaced by (G z)u {MI;} and we get the desired C r + 1 (we leave the details for the reader). Finally, we consider the case where for some edge e { = y,x,- we have 103 dvtcm + dy^xi) > \V(C t )\, and we can assume that this inequality is not satisfied for any e i with 1 ^ j < i. By the above inequality, it follows that y,- dominates some vertex y' of C, such that x t is dominated by its predecessor x'. But then C r + 1 is obtained from C t by replacing the edge x'y' by a Hamiltonian path from x t to y,- in Tu {e 1,e 2,...,e i - 1 } and then replacing the edges e 1,e 2,...,e i - l by appropriate paths of length 2, if necessary. This is possible because of the minimality of i. This completes the proof. COROLLARY 7.2. Every regular or almost regular tournament of order n has [^/(n/1000)] edge-disjoint Hamiltonian cycles. If we delete m 1 Hamiltonian cycles from the complete graph K n, then it is easy to see that the vertex set of the resulting graph G can be partitioned into 3.2 m ~ 1 sets such that each of these induces a complete subgraph of G and such that no two differ in cardinality by more than 1. Using this instead of the Hajnal-Szemeredi theorem [8] the above methods give the following result: THEOREM 7.3. Let k ^ 2 be a natural number. IfTis a tournament of order n > 2 2k +10 and with quasiregularity q(t) < n/2 k + l, then any oriented graph G, obtained from Tby

16 166 CARSTEN THOMASSEN deleting the edges of fewer than k edge-disjoint Hamiltonian cycles, is Hamiltonian. In particular, Thas k edge-disjoint Hamiltonian cycles. Since the proof of Theorem 7.3 is very similar to that of Theorem 7.1, we omit it. The possibility of applying the result of [8] was drawn to the author's attention by R. Haggkvist. 8. Further results and open problems J.-C. Bermond has informed the author that M. Las Vergnas has made the following conjecture: CONJECTURE 8.1. If a strong digraph D does not contain k + 2 independent vertices, then D contains a spanning arborescence with at most k sinks. If true, Conjecture 8.1 implies that Theorem 3.2 remains valid under the weaker condition that G has no three independent vertices. CONJECTURE 8.2. For each integer k ^ 2 there exists an integer tx(k) such that every a(/c)-connected tournament has k edge-disjoint Hamiltonian cycles. CONJECTURE 8.3. Let T be a /c-connected tournament. If A is any set of at most k 1 edges of T, then T A has a Hamiltonian cycle. It might be tempting to conjecture that a(k) = k, in which case Conjecture 8.3 would follow from Conjecture 8.2. However, that conjecture would be false. Consider two tournaments T l5 T 2. Take the disjoint union T x u T 2 and add a path x i x 2...x m, where m ^ 2. Now add all edges XjXj, with j < i 1, and all edges yz where y e V(T 2 ) or z e V(T X ) (or both). Suppose T x and T 2 are both /c-connected. Select vertices v l,v 2,...,v k in T x and u 1,u 2,...,u k in T 2 and reverse the direction of all edges x^, u 1 x m, u 2 v 2, u 3 v 3,...,u k v k. Then the resulting tournament is /c-connected and has no two edge-disjoint Hamiltonian cycles when k = 2. In fact a close inspection of the above tournaments shows that a is not bounded above by any linear function. We conjecture that a(2) = 3. Also note that the tournament above (with k = 1) has the property that the deletion of the edge set of any Hamiltonian path leaves an oriented graph that is not strong. This shows that Theorem 4.2 becomes false if we replace the condition on the edgeconnectivity by a condition on the minimum indegree and outdegree together with the condition that T is strong. It is easy to describe tournaments of connectivity 1 and high edge-connectivity containing an edge which is contained in all Hamiltonian cycles. Thus Conjectures 8.2 and 8.3 would be false if we only assume the tournaments in question to have high edge-connectivity. However, one might be tempted to conjecture that, for each integer k ^ 2, there exists an integer /?(/c) such that every /J(/c)-edge-connected tournament has k edge-disjoint Hamiltonian paths. By Theorems 3.2 and 4.2, /?(2) = 2. We do not know if /?(3) exists, but Dr B. Jackson has pointed out that P(k) does not exist for k ^ 4. To see this we consider a tournament Twith V(T) = A x u A 2 u A z u A x such that \A^ = A 3 1 ^ 1,1 ^41 = \A ^ 1, and such that each vertex of A x dominates each vertex of A 3, and each vertex of A { dominates each vertex of A i+1 for each

17 EDGE-DISJOINT HAMILTONIAN PATHS AND CYCLES 167 i = 1,2,3,4 (where A s = A x ). Any such tournament has large edge-connectivity if A x is large, but every Hamiltonian path of Tcontains an edge of T{A 2 ), so if /1 4 1 = 1, T has no four edge-disjoint Hamiltonian paths. The next conjecture is a weakening of Kelly's conjecture. CONJECTURE 8.4. If T is a regular tournament of order 2k + 1 and A is any set of at most k 1 edges of T, then T A is Hamiltonian. In the same spirit Alspach (private communication) suggested the following weakening of Conjectures 8.2 and 8.3. THEOREM 8.5. For each k ^ 2, there exists a number y(k) such that the deletion of any set of at most k 1 edges from any y(k)-connected tournament leaves a Hamiltonian oriented graph. This theorem shows that the theorem of Grotschel and Harary [7, Theorem 3.1] does not remain true if we impose connectivity conditions on the orientations. Theorem 8.5 is clearly a consequence of Theorem 8.6 below. THEOREM 8.6. For every k ^ 2 there exists an integer S(k) such that for any disconnected tournament and any set S of at most k vertices oft, Thas a Hamiltonian cycle not containing any edge joining two vertices of S. Proof. We prove that every 5/c-connected tournament T has the property described in the theorem. Let S = {x l,x 2,...,x k } such that "K(T)\K(S)( X l) ^ ^V(T)\V(S)( X 2) ^ ^ "K(D\K(S)( X k)- We claim that for each m, with 1 ^ m ^ k, T has a path P m from x x to x m of length at most 3(m 1) containing no edge joining two vertices of S and containing the vertices x 1,x 2,-.;X m and no other vertex of S. We suppose P m has been constructed and we construct P m + X as follows. Let A be the vertices of T (S u V(P m )) dominated by x m. If x m + 1 is dominated by some vertex of A, we obtain P m + 1 by adding a path of length 2 to P m. So assume that x m + 1 dominates all vertices of A and let the set of vertices of V(T)\(A u S u V(P m )) dominated by x m +1 be denoted by B. By the assumption on the ordering x l,x 2,... it follows that \B\ ^ 2(m 1). Hence K(PJuSu B\< 5k and so T-(F(/'JufiuS\{x m+1 }) contains an edge from A to a vertex not in A and we obtain P m + 1 by adding a path of length 3 to P m. We now take distinct vertices x,y in V(T)\ V(P k ) such that x dominates x x and y is dominated by x fc. Now T V(P k ) is 4-connected and contains by [17, Corollary 5.7] a Hamiltonian path from y to x, and hence we can extend P m into a Hamiltonian cycle with the desired property. By straightforward (but tedious) reasoning we can prove that y(2) = 3(2) = 2. It would also be easy to derive Theorem 8.6 from the following conjecture. CONJECTURE 8.7. For each k ^ 1, there exists an integer e(k) such that every e(k)- connected tournament has the property that any k independent edges are contained in a Hamiltonian cycle. In [17] it was shown that e(l) = 3.

18 168 EDGE-DISJOINT HAMILTONIAN PATHS AND CYCLES The fact that every strong tournament with at least four vertices has a vertex whose deletion results in a strong tournament is often used in inductive proofs. It would be convenient to have a similar result for tournaments of higher connectivity. However, for each k ^ 2, there are infinitely many minimally /e-connected tournaments. This can be seen by considering the /c-connected tournament described in connection with Conjecture 8.3. We delete successively vertices from that tournament until we get a minimally /c-connected tournament containing the path x 1 x 2...x m. We conclude with a problem of a probabilistic nature. For each > 0, almost all tournaments have minimum indegree and outdegree at least (j e)n and thus, by Theorem 7.3, almost all tournaments have k edge-disjoint Hamiltonian cycles, where k is any fixed natural number. Perhaps the following stronger assertion holds: CONJECTURE 8.8. For each e > 0 almost all tournaments of order n have [( e)ri] edge-disjoint Hamiltonian cycles. P. Erdos (private communication) suggested that perhaps almost all tournaments T have S(T) edge-disjoint Hamiltonian cycles where S(T) is the minimum of all indegrees and outdegrees of T. References 1. B. ALSPACH, D. W. MASON, and N. J. PULLMAN, 'Path numbers of tournaments', J. Combin. Theory Ser. B, 20 (1976), B. ALSPACH, K. B. REID, and D. P. ROSELLE, 'Bypasses in asymmetric digraphs', J. Combin. Theory Ser. B, 17(1974), L. W. BEINEKE and K. B. REID, 'Tournaments', Selected topics in graph theory (ed. L. W. Beineke and R. J. Wilson, Academic Press, New York, 1979), pp J.-C. BERMOND and C. THOMASSEN, 'Cycles in digraphs, a survey', J. Graph Theory, 5 (1981), G. A. DIRAC and C. THOMASSEN, 'Graphs in which every finitepath is contained in a circuit', Math. Ann., 203 (1973), J. F. FINK and L. LESNIAK-FOSTER, 'Graphs for which every unilateral orientation is traceable', Aequationes Math., to appear. 7. M. GROTSCHEL and F. HARARY, 'The graphs for which all strong orientations are Hamiltonian', J. Graph Theory, 3(1979), A. HAJNAL and E. SZEMEREDI, 'Proof of a conjecture of Erdos,' Combinatorial theory and it applications (ed. P. Erdos, A. Renyi, and V. T. Sos), Colloq. Math. Soc. J. Bolyai 4 (North-Holland, Amsterdam, 1970), pp B. JACKSON, 'Hamilton cycles in regular graphs', J. Graph Theory, 2 (1978), B. JACKSON, 'Edge-disjoint Hamiltonian cycles in regular graphs of large degree', J. London Math. Soc. (2), 19(1979), B. JACKSON, 'Paths and cycles in oriented graphs', Proc. Coll. Fr.-Can., Ann. Discrete Math., to appear. 12. A. KOTZIG, 'Des cycles dans les tournois', Theorie des graphes (ed. P. Rosenstiehl, Dunod, Paris, 1967), pp J. W. MOON, Topics in tournaments (Holt, Rinehart, and Winston, New York, 1968). 14. C. ST. J. A. NASH-WILLIAMS, 'Edge-disjoint Hamiltonian circuits in graphs with vertices of large valency', Studies in pure mathematics, papers presented to Richard Rado (Academic Press, London, 1971). 15. C. ST. J. A. NASH-WILLIAMS, 'Hamilton circuits in graphs and digraphs', The many facets of graph theory, Lecture Notes in Mathematics 110 (Springer, Berlin, 1969), pp J. NINCAK, 'On a conjecture by Nash-Williams', Comment. Math. Univ. Carolin., 14 (1973), C. THOMASSEN, 'Hamiltonian-connected tournaments', J. Combin. Theory Ser. B, 28 (1980), C. THOMASSEN, 'Long cycles in digraphs', Proc. London Math. Soc. (3), 42 (1981), T. TILLSON, 'A Hamiltonian decomposition of K$ m \ J. Combin. Theory Ser. B, 29 (1980), C. WALDROP, 'Arc reversals in tournaments', Ph.D. thesis, Louisiana State University, Present address: Matematisk Institut Mathematics Institute Aarhus Universitet Bygn Aarhus C The Technical University of Denmark Denmark 2800 Lyngby, Denmark