A sufficient condition for the hamiltonian property of digraphs with large semi-degrees
aa r X i v : . [ m a t h . C O ] N ov A sufficient condition for the hamiltonian property ofdigraphs with large semi-degrees
S. Kh. Darbinyan
Institute for Informatics and Automation ProblemsArmenian National Academy of Sciences, P. Sevak 1, Yerevan 0014, Armeniae-mail: [email protected]
Abstract
Let D be a digraph on p ≥ p − p/ −
1. For D (unless some extremal cases) we present a detailed proof of the followingresults [12]: (i) D contains cycles of length 3, 4 and p −
1; (ii) if p = 2 n , then D is hamiltonian.Keywords: Digraphs; semi-degrees; cycles; Hamiltonian cycles
1. Introduction and Terminology
Ghouila-Houri [18] proved that every strong digraph on p vertices and with minimum degree at least p is hamiltonian. There are many extentions of this theorem for digraphs and orgraphs. In particular,in many papers, various degree conditions have been obtained for digraphs (orgraphs) to be hamiltonianor pancyclic or vertex pancyclic (see e.g. [2]-[33]). C. Thomassen [31] proved that any digraph on p = 2 n + 1 vertices with minimum semi-degree at least n is hamiltonian unless some extremal cases,which are characterized. In [9], we proved that if a digraph D on 2 n + 1 vertices satisfies the conditions ofthis Tomassen’s theorem, then D also is pancyclic (the extremal cases are characterized). For additionalinformation on hamiltonian and pancyclic digraphs, see the book [1] by J. Bang-Jenssen and G. Gutin.In this paper we present a detailed proof of the following results.Every digraph D (unless some extremal cases) on p ≥ p − p/ −
1: (i) D has cycles of length 3, 4 and p −
1; (ii) if p = 2 n ,then D is hamiltonian (in [12], we gave only a short outline of the proofs of this results).In this paper we will consider finite digraphs without loops and multiple arcs. We denote the vertexset of digraph D by V ( D ) and its arc set by A ( D ). We will often use D instead of A ( D ) and V ( D ). Thearc from a vertex x to a vertex y will be denoted by xy . If xy is an arc, then we say that x dominates y (or y is dominated by x ). For A , B ⊂ V ( D ), we define A ( A → B ) as the set { xy ∈ A ( D ) /x ∈ A, y ∈ B } and A ( A, B ) = A ( A → B ) ∪ A ( B → A ). If x ∈ V ( D ) and A = { x } , we often write x instead of { x } . For disjoint subsets A and B of V ( D ), A → B means that every vertex of A dominates everyvertex of B . If C ⊂ V ( D ), A → B and B → C , then we write A → B → C . The outset of vertex x is the set O ( x ) = { y ∈ V ( D ) /xy ∈ A ( D ) } and I ( x ) = { y ∈ V ( D ) /yx ∈ A ( D ) } is the inset of x .Similarly, if A ⊆ V ( D ) then O ( x, A ) = { y ∈ A/xy ∈ A ( D ) } and I ( x, A ) = { y ∈ A/yx ∈ A ( D ) } . Theout-degree of x is od ( x ) = | O ( x ) | and id ( x ) = | I ( x ) | is the in-degree of x . Similarly, od ( x, A ) = | O ( x, A ) | and id ( x, A ) = | I ( x, A ) | . The degree of the vertex x in D is defined as d ( x ) = id ( x ) + od ( x ). Thesubdigraph of D induced by a subset A of V ( D ) is denoted by h A i . All paths and cycles we considerin this paper are directed and simple. The path ( respectively, the cycle ) consisting of distinct vertices x , x , . . . , x n ( n ≥
2) and arcs x i x i +1 , i ∈ [1 , n −
1] ( respectively, x i x i +1 , i ∈ [1 , n − x n x ), isdenoted by x x . . . x n (respectively, x x . . . x n x ). The cycle on k vertices is denoted C k . For a cycle C k = x x . . . x k x , we take the indices modulo k , i.e., x s = x i for every s and i such that i ≡ s mod k .Two distinct vertices x and y are adjacent if xy ∈ A ( D ) or yx ∈ A ( D ) (or both) (i.e, x is adjacent1ith y and y is adjacent with x ). The notation A ( x, y ) = ∅ (respectively, A ( x, y ) = ∅ ) means that thevertices x and y are adjacent (respectively, are not adjacent).The converse digraph ←− D of a digraph D is the digraph obtained from D by reversing all arcs of D .For an undirected graph G , we denote by G ∗ symmetric digraph obtained from G by replacing everyedge xy with the pair xy , yx of arcs. Further, C ∗ (5) is a symmetric digraph obtained from undirected cycleof length 5. K n (respectively, K n,m ) denotes the complete undirected graph on n vertices (respectively,undirected complete bipartite graph, with partite sets of cardinalities n and m ), and K n denotes thecomplement of K n . If G and G are undirected graphs, then G ∪ G is the disjoint union of G and G . The join of G and G , denoted by G + G , is the union of G ∪ G and of all the edges between G and G .For integers a and b , let [ a, b ] denote the set of all integers which are not less than a and are not greaterthan b . We refer the reader to J.Bang-Jensens and G.Gutin’s book [1] for notations and terminology notdefined here.
2. Preliminaries and Additional notations
Let us recall some well-known lemmas used in this paper.
Lemma 1 ([21]). Let D be a digraph on p ≥ C m , m ∈ [2 , p − x be avertex not contained in this cycle. If d ( x, C m ) ≥ m +1, then D contains a cycle C k for all k ∈ [2 , m +1].The following Lemma will be used extensively in the proofs our results. Lemma 2 ([6]). Let D be a digraph on p ≥ P := x x . . . x m , m ∈ [2 , p − x be a vertex not contained in this path. If one of the following holds:(i) d ( x, P ) ≥ m + 2;(ii) d ( x, P ) ≥ m + 1 and xx / ∈ D or x m x / ∈ D ;(iii) d ( x, P ) ≥ m , xx / ∈ D and x m x / ∈ D ;then there is an i ∈ [1 , m −
1] such that x i x, xx i +1 ∈ D , i.e., D contains a path x x . . . x i xx i +1 . . . x m oflength m (we say that x can be inserted into P or the path x x . . . x i xx i +1 . . . x m is extended from P with x ).As an immediate consequence of Lemma 2, we get the following: Lemma 3 . Let D be a digraph on p ≥ P := x x . . . x m , m ∈ [2 , p − x to x m in D . If the induced subdigraph h V ( D ) \ V ( P ) i is strong and d ( x, V ( P )) = m + 1 for every vertex x ∈ V ( D ) \ V ( P ), then there is an integer l ∈ [1 , m ] such that O ( x, V ( P )) = { x , x , . . . , x l } and I ( x, V ( P )) = { x l , x l +1 , . . . , x m } .Now we introduce the following notations. Notation . For any positive integer n , let H ( n, n ) denote the set of digraphs D on 2 n vertices such that V ( D ) = A ∪ B , h A i ≡ h B i ≡ K ∗ n , A ( B → A ) = ∅ and for every vertex x ∈ A (respectively, y ∈ B ) A ( x → B ) = ∅ (respectively, A ( A → y ) = ∅ ). Notation . For any integer n ≥
2, let H ( n, n − ,
1) denote the set of digraphs D on 2 n vertices suchthat V ( D ) = A ∪ B ∪ { a } , | A | = | B | + 1 = n , A ( h A i ) = ∅ , h B ∪ { a }i ⊆ K ∗ n , yz, zy ∈ D for each pair of2ertices y ∈ A , z ∈ B and either I ( a ) = B and a → A or O ( a ) = B and A → a . Notation . For any integer n ≥ H (2 n ) as follows: V ( H (2 n )) = A ∪ B ∪ { x, y } , h A i ≡ h B i ≡ K ∗ n − , A ( A, B ) = ∅ , O ( x ) = { y } ∪ A , I ( x ) = O ( y ) = A ∪ B and I ( y ) = { x } ∪ B . H ′ (2 n ) is a digraph obtained from H (2 n ) by adding the arc yx . Notation.
Let D be a digraph with vertex set { x , x , . . . , x , x } and arc set { x i x i +1 / ≤ i ≤ } ∪ { xx i / ≤ i ≤ } ∪ { x x , x x , x x , x x , x x , x x, x x , x x } . By D ′ we denote a digraph obtained from D by adding the arc x x .Note that the digraphs D and D ′ both are not hamiltonian and each of D and D ′ contains a cycleof length 5. Lemma 4 . Let D be a digraph on p ≥ p − p/ −
1. Then(i) either D is strong or D ∈ H ( n, n );(ii) if B ⊂ V ( D ), | B | ≥ ( p + 1) / x ∈ V ( D ) \ B , then A ( x → B ) = ∅ and A ( B → x ) = ∅ .
3. A sufficient condition for the existence of cycles of length | V ( D ) | − in digraph D Theorem 1 . Let D be a digraph on p ≥ p − p/ −
1. Then D has a cycle of length p − D ∈ H ( n, n ) ∪ { [( K n ∪ K n ) + K ] ∗ , H (2 n ) , H ′ (2 n ) , C ∗ (5) } or else p = 2 n and D ⊆ K ∗ n,n . Proof . By Lemma 4(i), the result is easily verified if D is not strong. Assume that D is strong. Suppose,on the contrary, that the theorem is not true. In particular, D contains no cycle of length p −
1. Let C := C m := x x . . . x m x be an arbitrary non-hamiltonian cycle of maximum length in D . It is easy tosee that m ∈ [3 , p − m it follows that for each vertex y ∈ B := V ( D ) \ V ( C ) andfor each i ∈ [1 , m ], d ( y, C ) ≤ m, d ( y, B ) ≥ p − m − x i y ∈ D, then yx i +1 / ∈ D. (1)Using d ( y, B ) ≥ p − m − Claim 1 . For any two distinct vertices x, y ∈ B if in subdigraph h B i there is no path from x to y , thenin h B i there is a path from y to x of length at most 2.We first prove the following Claims 2 and 3. Claim 2 . The induced subdigraph h B i is strongly connected. Proof . Suppose, on the contrary, that h B i is not strong. Let D , D , . . . , D s ( s ≥
2) be the strongcomponents of h B i labeled in such a way that no vertex of D i dominates a vertex of D j whenever i > j .From Claim 1 it follows that for each pair of vertices y ∈ V ( D ) and z ∈ V ( D s ) in h B i there is a pathfrom y to z of length 1 or 2. We choose the vertices y ∈ V ( D ) and z ∈ V ( D s ) such that the path y y . . . y k , where y := y and y k := z , will have minimum length among all paths in h B i with origin3ertex in D and terminus vertex in D s . By Claim 1, k = 2 or k = 3. We consider the following tree cases. Case 1 . k < | B | = p − m. It follows from the maximality of C that if x i y ∈ D , where i ∈ [1 , m ], then A ( y k → { x i +1 , x i +2 ,. . . , x i + k } ) = ∅ . Since D is strong, we see that C I ( y ). Therefore the vertex y k dose not dom-inate at least id ( y , C ) + 1 vertices of C . On the other hand, we have A ( y k → V ( D )) = ∅ and I ( y ) ⊂ C ∪ V ( D ). Hence the vertex y k dose not dominate at least id ( y ) + 3 vertices. From thiswe obtain od ( y k ) ≤ p − id ( y ) − ≤ p/ −
2, which is a contradiction.
Case 2 . k = | B | = 2.It is easy to see that s = 2, m = p − V ( D ) = { y } , V ( D ) = { y } , I ( y ) ⊂ C and | A ( x i → y ) | + | A ( y → x i +2 ) | ≤ i ∈ [1 , m ]. Hence the vertex y dose not dominate at least id ( y ) + 2 vertices. Therefore od ( y ) ≤ p − id ( y ) − ≤ p/ −
1. It follows that p = 2 n , id ( y ) = od ( y ) = n − y x i ∈ D if and only if x i − y / ∈ D. (2)By Lemma 1, it is easy to see that d ( y ) = d ( y ) = 2 n − od ( y ) = id ( y ) = n , m ≥
4. Now wedivide this case into two subcases.
Subcase 2.1 . y → { x i , x i +1 } for some i ∈ [1 , m ].Note that, by Lemma 2, without loss of generality, we may assume that x m y ∈ D, y → { x , x } and A ( x , y ) = ∅ . From this, (1) and (2) it follows that x y / ∈ D , y x , y x ∈ D and A ( x , y ) = ∅ . There-fore, by Lemma 2 we have x y ∈ D since d ( y , C ) = 2 n − y cannot be inserted into thepath x x . . . x m x . If x x ∈ D , then C n − = x m y x x y x . . . x m . This contradicts our suppositionthat D contains no cycle of length p −
1. Hence, x x / ∈ D . From this and A ( x , y ) = A ( x → y ) = ∅ it follows that d ( x , { x , x , . . . , x m } ) ≥ n −
3. Therefore by Lemma 2, x m x ∈ D since the vertex x cannot be inserted into the path x x . . . x m . Now it is easy to see that | A ( x i → y ) | + | A ( y → x i +1 ) | ≤ i ∈ [2 , m − x y / ∈ D since y x ∈ D . From this and (2) it follows that y x ∈ D and x y / ∈ D . Continuing in this manner, we obtain that A ( { x , x , . . . , x m − } → y ) = ∅ . Therefore A ( { x , x , . . . , x m − } → y ) = ∅ , which is a contradiction. Subcase 2.2 . | A ( y → { x i , x i +1 ) | ≤ i ∈ [1 , m ].Since od ( y ) = n , we can assume that O ( y ) = { x , x , . . . , x n − , y } . Using this and od ( y ) = id ( y ) = n −
1, we obtain I ( y ) = { x , x , . . . , x n − } . Therefore by (2), O ( y ) = { x , x , . . . , x n − } and I ( y ) = { y , x , x , . . . , x n − } . If x i x j ∈ D for distinct vertices x i , x j ∈ { x , x , . . . , x n − } , then C n − = y x i x j x j +1 . . . x i − y x i +1 . . .x j − y , when |{ x i +1 , x i +2 , . . . , x j − }| ≥ C n − = x i x j y y x j +1 x j +2 . . . x i − x i , when |{ x i +1 , x i +2 ,. . . , x j − }| = 1. This contradicts that C p − D . Thus we have A ( h{ x , x , . . . , x n − , y }i ) = ∅ . Considering the digraph ←− D , by the same arguments we obtain A ( h{ x , x , . . . , x n − , y }i ) = ∅ . Therefore D ⊆ K ∗ n,n , which contradicts our supposition that the theorem is not true.4 ase 3 . k = | B | = 3.From the minimality of k it follows that y y / ∈ D , s = 3 , A ( { y , y } → y ) = ∅ and V ( D ) = { y } .Hence I ( y ) ⊂ C . On the other hand, from the maximality of the cycle C it follows that for each i ∈ [1 , m ]if x i y ∈ D, then A ( y → { x i +1 , x i +2 } ) = ∅ . Therefore y dose not dominate at least id ( y ) + 3 vertices, a contradiction. Claim 2 is proved. Claim 3 . At least two distinct vertices of C are adjacent with some vertices of B . Proof . Assume that Claim 3 is not true. Then exactly one vertex, say x , of C is adjacent with somevertices of B . Hence for each vertex x i ∈ C \ { x } and for each vertex y ∈ B we have d ( x i ) = d ( x i , C ) ≤ m − d ( y ) = d ( y, B ) + d ( y, x ) ≤ p − m. Since d ( x i ) + d ( y ) ≥ p −
2, we conclude that the inequalities above are equalities. This implies that thesubdigraphs h C i and h B ∪ { x }i are complete. From d ( x i ) = 2 m − ≥ p − d ( y ) = 2 p − m ≥ p − p = 2 m −
1. Therefore G ≡ [( K m − ∪ K m − ) + K ] ∗ , which contradicts our supposition.This proves Claim 3.Since D is strong, then A ( C → B ) = ∅ and A ( B → C ) = ∅ . Together with Claim 3 this implies thatthere are vertices x a = x b , x a , x b ∈ C and x, y ∈ B such that x a x , yx b ∈ D and A ( { x a +1 , x a +2 , . . . , x b − } , B ) = ∅ , if x b = x a +1 . (3)To be definite, assume that x b := x and x a := x m − h (0 ≤ h ≤ m − Case 1 . x m − h +1 = x (i.e., h ≥ P , P , . . . , P k (0 ≤ k ≤ h and k is as maximum as possible), where P := P := x x . . . x m − h and the path P i , i ∈ [1 , k ], is extended from the path P i − with a vertex z i ∈{ x m − h +1 , x m − h +2 , . . . , x m } \ { z , z , . . . , z i − } . Note that the path P i , i ∈ [0 , k ], contains m − h + i vertices. It follows that some vertices y , y , . . . , y d ∈ { x m − h +1 , x m − h +2 , . . . , x m } , where 1 ≤ d ≤ h , dosenot containing the extended path P k . Therefore, using (3) and Lemma 2, for each z ∈ B and for each y i we obtain d ( z ) = d ( z, B ) + d ( z, C ) ≤ p − m − m − h + 1 = 2 p − m − h − d ( y i ) = d ( y i , C ) ≤ m + d − . Hence it is clear that 2 p − ≤ d ( z ) + d ( y i ) ≤ p + d − h − . It is not difficult to see that h = d , d ( z, C ) = m − h + 1, d ( y i , C ) = m + h − h B i and h{ x m − h +1 , x m − h +2 , . . . , x m }i are complete. By Lemma 2(ii), we also have x m − h → B ∪ { x m − h +1 , x m − h +2 , . . . , x m } → x . It is easy to see that h = | B | = p − m ≥ P = x x . . . x m − h has maximum length among all paths from x to x m − h in subdigraph h C i and insubdigraph h B ∪ { x , x , . . . , x m − h }i . Therefore by Lemma 3, there are integers l ∈ [1 , m − h ] and r ∈ [1 , m − h ] such that O ( u, P ) = { x , x , . . . , x l } , I ( u, P ) = { x l , x l +1 , . . . , x m − h } ,O ( z, P ) = { x , x , . . . , x r } , I ( z, P ) = { x r , x r +1 , . . . , x m − h } . (4)for all u ∈ B and for all z ∈ { x m − h +1 , x m − h +2 , . . . , x m } .5ithout loss of generality, we may assume that l ≤ r (otherwise we will consider the digraph ←− D ).Let l = 1. Then from od ( u ) ≥ p/ − h ≥ p/ − p ≥ p/ −
1) + m − h = p − m − h . Since m − h ≥
2, we see that p = 2 n , m − h = 2, h = n − r = 2. Therefore G ∈ { H (2 n ) , H ′ (2 n ) } , which contradicts the our supposition.Let now l ≥
2. We can assume that r ≤ m − h − ←− D we will have theconsidered case l = 1). Since h{ x m − h +1 , x m − h +2 , . . . , x m }i are complete and (4), for each vertex z ∈{ x m − h +1 , x m − h +2 , . . . , x m } we have I ( z ) = { x r , x r +1 , . . . , x m } \ { z } . This implies that m − r ≥ p/ −
1. If i ∈ [ r + 1 , m − h ] and x x i ∈ D then by (4) and 2 ≤ l ≤ r ≤ m − h − C m +1 = x x i x i +1 . . . x m x . . . x i − xx , where x ∈ B , a contradiction. Because of this and 2 ≤ l ≤ r , we mayassume that A ( x → B ∪ { x r +1 , x r +2 , . . . , x m } ) = ∅ . Therefore, since m − r ≥ p/ − | B | = h ≥
2, we obtain od ( x ) ≤ p − − h − ( m − r ) ≤ p/ − h ,which contradicts the condition that od ( x ) ≥ p/ − Case 2 . x m − h +1 = x (i.e., h = 0).Then any path from x to y in h B i is a hamiltonian path. Let u u . . . u p − m be a hamiltonian path in h B i , where u := x , u p − m := y . From this, if 1 ≤ i < j ≤ p − m , then u i u j ∈ D if and only if j = i + 1.For this case ( h = 0) we first prove Claims 4-9. Claim 4 . p − m = 2 (i.e., m = p − Proof . Suppose, to the contrary, that p − m ≥
3. It follows from observations above that u u p − m / ∈ D and od ( u , B ) = id ( u p − m , B ) = 1. From this and (1), we obtain p − ≤ d ( u ) ≤ m + 1 + id ( u , B ) and p − ≤ d ( u p − m ) ≤ m + 1 + od ( u p − m , B ) . This implies that id ( u , B ) and od ( u p − m , B ) ≥ p − m −
2. Therefore in h B i there is a path from u p − m to u of length k = 1 or k = 2 since p − m ≥
3. For any integer l ≥ I + l := { x j / x j − l u p − m ∈ D } . Since id ( u p − m , C ) = id ( u p − m ) − C I ( u p − m ), we see that for each l ∈ [1 , | I + l ∪ I + l +1 | ≥ id ( u p − m ) . From the maximality of the cycle C it follows that A ( u → I + k ∪ I + k +1 ) = ∅ . Together with A ( u →{ u , u , . . . , u p − m } ) = ∅ this implies that p/ − ≤ od ( u ) ≤ p − − | I + k ∪ I + k +1 | − ( p − m − ≤ m + 1 − id ( u p − m ) ≤ m + 1 − p/ . Therefore, since m ≤ p −
3, we obtain that p − m = 3, p = 2 n and od ( u ) = id ( u ) = n −
1. Hence, id ( u ) and od ( u ) ≥ n . We now claim that u u and u u ∈ D . Indeed, otherwise id ( u , C ) ≥ n − x i u ∈ D , then A ( u → { x i +2 , x i +3 } ) = ∅ . From this it is not difficult to see that od ( u ) ≤ n − od ( u ) ≥ n .Similarly, we can see that u u ∈ D . So we have u u , u u , u u ∈ D . Then, since id ( u , C ) = n − m ≥ n , m ≥ id ( u , C )+2 and C is a non-hamiltonian cycle of maximal length, it follows that |∪ i =1 I + i | ≥ n and A ( u → ∪ i =1 I + i ) = ∅ . Together with u u / ∈ D this implies that od ( u ) ≤ n −
2, a contradiction.This completes the proof of Claim 4.Note that, by Claims 4 and 2 we have m = p − B := { u, v } and uv , vu ∈ D .6 emark . By symmetry of the vertices u and v , Claims 5-9 are also true for the vertex v . Claim 5 . If x i u , ux i +2 ∈ D , i ∈ [1 , m ], then | A ( x i +1 , v ) | = 2 (i.e., x i +1 v and vx i +1 ∈ D ). Proof . Since the cycle x i ux i +2 x i +3 . . . x i has length m and the vertices v and x i +1 are not on this cycle,the subdigraph h{ v, x i +1 }i is strong by Claim 2. Therefore vx i +1 and x i +1 v ∈ D .From Claim 5, uv , vu ∈ D and the maximality of the cycle C we have the following: Claim 6 . If i ∈ [1 , m ], then | A ( { x i , x i +1 } → u ) | + | A ( u → x i +3 ) | ≤ | A ( x i − → u ) | + | A ( u → { x i , x i +1 } ) | ≤ . Claim 7 . If k ∈ [1 , m ], then | A ( { x k − , x k } → u ) | ≤ Proof . Suppose, to the contrary, that is k ∈ [1 , m ] and { x k − , x k } → u . Without loss of generality,we may assume that A ( u, x k +1 ) = ∅ . To be definite, assume that x k +2 := x and x m := x k +1 . Then ux / ∈ D by Claim 6.First suppose that x u ∈ D . It is easy to see that p ≥ A ( u → { x m − , x m , x , x } ) = ∅ . Using thistogether with od ( u ) ≥ p/ − A ( u → { x , x , . . . , x m − } ) = ∅ and d ( u, { x , x , . . . , x m − } ) ≥ p −
3. Note that m ≥ j ∈ [3 , m − | A ( u → { x j , x j +1 }| ≤ . (5)Assume that (5) is not true. Then u → { x j , x j +1 } for some j ∈ [3 , m − j isas small as possible. Then A ( u, x j − ) = ∅ and x j − u / ∈ D by Claim 6. Hence j ≥
4. Since the vertex u cannot be inserted into the cycle C , ux / ∈ D and x j − u / ∈ D , by Lemma 2 we have d ( u, { x , x , . . . , x j − } ) ≤ j − d ( u, { x j , x j +1 , . . . , x m − } ) ≤ m − j + 1 . Hence d ( u ) ≤ p −
2, a contradiction, which proves (5).From A ( u → { x m − , x m , x , x } ) = ∅ and (5) it follows that od ( u ) ≤ p/ −
2, a contradiction.So suppose next that x u / ∈ D . Then A ( u, x ) = ∅ by Claim 6, m ≥ d ( u, { x , x , . . . , x m − } ) ≥ p −
3. Hence, ux ∈ D by Lemma 2(ii). Note that A ( v, x m ) = ∅ and vx / ∈ D . By Lemma 2(iii), it iseasy to see that x m − v ∈ D and d ( v, { x , x , . . . , x m − } ) = p −
3. If x v / ∈ D , then A ( v, x ) = ∅ , and byLemma 2, vx ∈ D . Now we have x m − u , vx ∈ D and A ( { u, v } , { x m , x } ) = ∅ , and the considered Case1 ( h ≥
1) holds. So we may assume that this is not the case. Then x v ∈ D . We also can assume that x m − v / ∈ D (otherwise { x m − , x m − , x } → v and for the vertex v the considered case x u ∈ D holds).From x m − v / ∈ D and vx / ∈ D , by Lemma 2(iii), it follows that d ( v, { x , x , . . . , x m − } ) ≤ p −
5. Hence vx m − ∈ D . From A ( x m , { u, v } ) = ∅ and d ( x m ) ≥ p − d ( x m , { x , x , . . . , x m − } ) ≥ p − x m cannot be inserted into the path x x . . . x m − (otherwise we obtain a cycle of length m , which does not contain the vertices v and x and therefore, by Claim 2, the subdigraph h{ v, x }i is strong, which contradicts the fact that vx / ∈ D ). It follows that x m x ∈ D by Lemma 2, and C m +1 = x m − uvx m − x m x . . . x m − , a contradiction. Claim 7 is proved.Similarly to Claim 7, we can show the following: Claim 8 . If i ∈ [1 , m ], then | A ( u → { x i , x i +1 } ) | ≤ Claim 9 . If k ∈ [1 , m ], then | A ( x k → u ) | + | A ( u → x k − ) | ≤ Proof . Suppose, to the contrary, that is k ∈ [1 , m ] and x k u , ux k − ∈ D . To be definite, assume that x k := x . From Claims 7, 8 and (1) it follows that A ( u, { x m , x } ) = A ( x → u ) = A ( u → x ) = ∅ . m ≥ d ( u, { x , x , . . . , x m − ) ≥ p −
5. Since the vertex u cannotbe inserted into the path x x . . . x m − , by Lemma 2 we have ux and x m − u ∈ D . Hence, | A ( v, x ) | = | A ( v, x m ) | = 2 by Claim 5. Therefore A ( v, { x , x } ) = ∅ by Claims 7 and 8. Since C m +1 D it is notdifficult to see that x m x , x x , x x and x x / ∈ D (if x x ∈ D , then C m +1 = x m vx x ux . . . x m ).So we have d ( x , { u, v, x , x } ) ≤ d ( x , { x , x , . . . , x m } ≥ p −
5. Therefore, since x m x / ∈ D and x x / ∈ D , applying Lemma 2(iii), we can insert x into the path x x . . . x m (i.e., x i x , x x i +1 ∈ D for some i ∈ [4 , m − x m vux . . . x i x x i +1 . . . x m of length m , which does notcontain the vertices x and x . By Claim 2, the subdigraph h{ x , x }i is strong. Hence x x ∈ D , whichcontradicts the fact that x x / ∈ D . This completes the proof of Claim 9.We now divide Case 2 ( h = 0) into two subcases. Subcase 2.1 . p = 2 n + 1.From (1) and Claims 7- 9 it follows that the vertex u (respectively, v ) is adjacent with at most onevertex of two consecutive vertices of the cycle C and O ( u, C ) = I ( u, C ) and O ( v, C ) = I ( v, C ). Hence,without loss of generality, we may assume that A ( u, { x , x } ) = ∅ and O ( u ) = I ( u ) = { x , x , x , . . . , x p − , v } . (6)If m = 3, then Claim 3 implies that | A ( v, x ) | = 2 or | A ( v, x ) | = 2 and C ⊂ D , a contradiction.Assume that m = 2 n − ≥
5. Since x m − u ∈ D , by Claim 5 we have | A ( x m , v ) | = 2. Therefore, by anargument similar to (6), we get that ether | A ( v, x ) | = 2 or | A ( v, x ) | = 2. From this and (6) it is easy tosee that if vx ∈ D , then C m +1 = x uvx x . . . x m x and if x v ∈ D , then C m +1 = x vux x . . . x m x x ,a contradiction. Subcase 2.2. p = 2 n .From d ( u ) ≥ n − od ( u ) ≥ n or id ( u ) ≥ n . Without loss of generality, we mayassume that od ( u ) ≥ n (otherwise we will consider the digraph ←− D ). Now from Claims 7-9 it follows that u → { x , x , . . . , x n − } , A ( u, { x , x , . . . , x n − } ) = ∅ , (7) I ( u ) ⊆ { v, x , x , . . . , x n − } . (8)Since id ( u ) ≥ n −
1, without loss of generality, we may assume that { x , x , . . . , x n − } → u . Hence , by(7) and Claim 5, it follows that for each i ∈ [1 , n − | A ( u, x i − ) | = | A ( v, x i ) | = 2 . (9)Then by Claims 7-9 we have A ( v, { x , x , . . . , x n − } ) = ∅ . Therefore A ( v, x n − ) = ∅ since d ( v ) ≥ n − vx n − ∈ D or v n − v ∈ D . If vx n − ∈ D , then x n − u ∈ D and x n − v ∈ D by Claim 5, (8) and(9). So, in any case we have that x n − v ∈ D . Then id ( v ) ≥ n .We will now show that A ( h{ x , x , . . . , x n − }i ) = ∅ . (10) Proof of (10) . Assume that (10) is not true. Then x i x j ∈ D for some distinct vertices x i , x j ∈{ x , x , . . . , x n − } . Assume that |{ x i +1 , x i +2 , . . . , x j − }| = 1. Then from (9) and x n − v ∈ D we have:a) if j = 2 n −
3, then i = 2 n − C m +1 = x n − x n − x n − vux x . . . x n − ; b) if j = 2 n −
3, then C m +1 = x i x j uvx j +1 . . . x i − x i . Now assume that |{ x i +1 , x i +2 , . . . , x j − }| ≥
2. Then n ≥
6. Using (9) wecan see that: c) if i = 2 n − j = 1, then C m +1 = x i x j x j +1 . . . x i − vx i +1 . . . x j − ux i ; d) if i = 2 n − j = 1, then C m +1 = x i x j x j +1 . . . x i − ux i +2 . . . x j − vx i − x i . Hence in each case we have a C m +1 ⊂ D ,8hich is a contradiction and (10) is proved.Using an analogous argument for ←− D , similarly to (10), we can show that A ( h{ x , x , . . . , x n − }i ) = ∅ . Therefore A ( h{ v, x , x , . . . , x n − }i ) = A ( h{ u, x , x , . . . , x n − }i ) = ∅ and D ⊆ K ∗ n,n . This contradicts the our supposition. The discussion of Case 2 is completed and Theorem 1 is proved.
3. A sufficient condition for a digraph to be hamiltonianTheorem 2 . Let D be a digraph on 2 n ≥ n − n −
1. Then D is hamiltonian unless D ∈ H ( n, n ) ∪ H ( n, n − , ∪ { H (2 n ) , H ′ (2 n ) , D , D ′ , ←− D , ←− D ′ } . Proof . By Lemma 4(i), the result is easily verified if D is not strong. Now assume that D is strong.The proof is by contradiction. Suppose that Theorem 2 is false, in particular, D is not hamiltonian.Then it is not difficult to see that D K ∗ n,n . By Theorem 1, D has a cycle of length 2 n −
1. Let C := C n − := x x . . . x n − x be an arbitrary cycle of length 2 n − D and let the vertex x is notcontaining this cycle C . Since C is a longest cycle, using Lemmas 1 and 2, we obtain the following claim: Claim 1. (i) d ( x ) = 2 n − x l , l ∈ [1 , n −
1] such that A ( x, x l ) = ∅ .(ii) If x i x / ∈ D , then xx i +1 ∈ D and if xx i / ∈ D , then x i − x ∈ D , where i ∈ [1 , n − A ( x, x i ) = ∅ , then x i − x, xx i +1 ∈ D and d ( x i ) = 2 n − A ( x, x n − ) = ∅ . For convenience, let p := 2 n − y := x n − . We have yx , x p y ∈ D and x p x , xx ∈ D by Claim 1(iii), and d ( y ) = 2 n − { u, v } := { x, y } and for each z ∈ { x, y } let O − ( z ) := { x i /zx i +1 ∈ D, i ∈ [1 , p − } , I + ( z ) := { x i /x i − z ∈ D, i ∈ [2 , p − } . We first prove the following Claims 2-11.
Claim 2 . If x p − u ∈ D , then A ( O − ( v ) → x p ) = ∅ . Proof . Assume, to the contrary, that x p − u ∈ D and x i x p ∈ D , where x i ∈ O − ( v ). Then by the defini-tion of O − ( v ), vx i +1 ∈ D and x x . . . x i x p vx i +1 . . . x p − ux is a hamiltonian cycle, a contradiction. Claim 3 . If x p − u, vx p ∈ D , then A ( x p → I + ( v )) = ∅ . Proof . Assume, to the contrary, that x p − u , vx p and x p x i ∈ D , where x i ∈ I + ( v ). Then by the definitionof I + ( v ), x i − v ∈ D and x x . . . x i − vx p x i . . . x p − ux is a hamiltonian cycle, a contradiction. Claim 4 . If x p − → { x, y } , then od ( x ) = od ( y ) = n − id ( x ) = id ( y ) = n and O ( x ) = O ( y ). Proof . Let x p − → { x, y } . Since C is longest cycle of D , we have A ( { x, y } → x p ) = ∅ . By Claim2, A ( O − ( x ) ∪ O − ( y ) → x p ) = ∅ . Hence, | O − ( x ) ∪ I − ( y ) ∪ { x, y }| ≤ n by Lemma 4(ii). Therefore,since | O − ( u ) | = od ( u ) −
1, we deduce that od ( x ) = od ( y ) = n − O ( x ) = O ( y ). Together with d ( x ) = d ( y ) = 2 n − id ( x ) = id ( y ) = n . Claim 4 is proved.9imilarly to Claim 4, we can show the following claim: Claim 5 . If { x, y } → x , then id ( x ) = id ( y ) = n − od ( x ) = od ( y ) = n and I ( x ) = I ( y ). Claim 6 . | A ( x p − → { x, y } ) | ≤ Proof . Assume, on the contrary, that x p − → { x, y } . Then id ( x ) = id ( y ) = n , od ( x ) = od ( y ) = n − O ( x ) = O ( y ) by Claim 4. Hence, xx ∈ D if and only if yx ∈ D . Therefore, A ( { x, y } → x ) = ∅ by Claim 5. Hence, x → { x, y } by Claim 1(ii). Together with od ( x ) ≥ k ∈ [2 , p −
2] such that x k − x , xx k +1 ∈ D and A ( x, x k ) = ∅ . Applying Claim 1(iii) we find that d ( x k ) = 2 n −
1. From O ( x ) = O ( y ) it is not difficult to see that A ( x k , y ) = ∅ , yx k +1 , x k − y ∈ D . Thenby Lemma 2, since x k cannot be inserted into the path x x . . . x k − and into the path x k +1 x k +2 . . . x p ,we have d ( x k , { x , x , . . . , x k − } ) ≤ k and d ( x k , { x k +1 , x k +2 , . . . , x p } ) ≤ p − k + 1 . Using this, d ( x k ) = p + 1, A ( x k , { x, y } ) = ∅ and Lemma 2(ii), we obtain x k x , x p x k ∈ D and d ( x k , { x , x , . . . , x k − } ) = k, d ( x k , { x k +1 , x k +2 , . . . , x p } ) = p − k + 1 . (11)We now show that A ( { x, y } → x k +2 ) = ∅ . (12) Proof of (12) . Suppose that (12) is false. From O ( x ) = O ( y ) it is clear that { x, y } → x k +2 . Since x k x ∈ D we see that x k +1 x k / ∈ D (otherwise x k +1 x k ∈ D and x x . . . x k − xx k +2 . . . x p yx k +1 x k x is ahamiltonian cycle, a contradiction). Note that x x . . . x k − xx k +1 . . . x p yx is a cycle of length 2 n − x k cannot be inserted into this cycle. Then, since x k +1 x k / ∈ D and d ( x k ) = 2 n −
1, usingClaim 1(ii) we get x k x k +2 ∈ D . From this it follows that { x, x k } → { x k +1 , x k +2 } and x k − → { x, x k } for the path x k +1 x k +2 . . . x p yx x . . . x k − . Therefore id ( x ) = n − id ( x ) = n . This proves (12).From x → { x, y } and (12) it follows that | A ( x → { x i , x i +1 } ) | ≤ i ∈ [1 , p − od ( y ) = od ( x ) = n − O ( x ) = O ( y ), it is not difficult to see that { x, y } → { x , x , . . . , x p − } → { x, y } , (13) A ( { x, y } , { x , x , . . . , x p − } ) = ∅ . (14)Together with Claim 2 this implies that A ( { x , x , . . . , x p − } → x p ) = ∅ . (15)It is not difficult to show that A ( h{ x , x , . . . , x p − }i ) = ∅ . (16)Indeed, if (16) is false, then x i x j ∈ D for some distinct vertices x i , x j ∈ { x , x , . . . , x p − } . It is easyto see that if i < j , then C n = x x . . . x i x j . . . x p yx i +1 . . . x j − xx and if i > j , then x j x ∈ D by (11),and C n = x x . . . x j − xx i +1 . . . x p yx j +1 . . . x i x j x , a contradiction.From (14) and (16) it follows that A ( h{ x, y, x , x , . . . , x n − }i ) = ∅ . By (15), now it is not diffi-cult to see that D ∈ H ( n, n − , a := x p , A := { x, y, x , x , . . . , x n − } and B := { x , x , . . . ,x n − } . This contradicts to our supposition that D / ∈ H ( n, n − , laim 7 . | A ( { x, y } → x ) | ≤ Claim 8 . There is a vertex x i , i ∈ [2 , p − A ( x, x i ) = ∅ (i.e., if C n − , p = 2 n −
2, is anarbitrary cycle of D and the vertex x / ∈ C n − , then x is not adjacent with at least two vertices). Proof . Suppose, on the contrary, that the vertex x is adjacent with each vertex x i , i ∈ [2 , p − n ≥ d ( x ) = p + 1 and D is not hamiltonian, there is an l ∈ [2 , p −
1] such that O ( x ) = { x , x , . . . , x l } and I ( x ) = { x l , x l +1 , . . . , x p } . (17)Since od ( x ) and id ( x ) ≥ n −
1, we see that l = n − l = n . Hence x p − y / ∈ D and yx / ∈ D by Claims6 and 7. Now x y and yx p ∈ D by Claim 1(ii). Therefore, since y cannot be inserted into the path x x . . . x p , there is a vertex x k , k ∈ [2 , p − A ( y, x k ) = ∅ . Using Claim 1(iii), we get x k − y, yx k +1 ∈ D and d ( x k ) = p + 1 . (18)Choose k is as large as possible. It follows that y → { x k +1 , x k +2 , . . . , x p } and k ≥ l −
1. We can assumethat k ≥ l (if k = l −
1, then in digraph ←− D we will have the case k ≥ l + 1).Suppose first that k ≥ l + 1. If x i x k ∈ D , where i ∈ [1 , l − x x . . . x i x k . . . x p xx i +1 . . . x k − yx is a hamiltonian cycle, a contradiction. So we may assume that A ( { x , x , . . . , x l − } → x k ) = ∅ . Using this together with A ( { x, y } → x k ) = ∅ , l ≥ n − id ( x k ) ≥ n −
1, we obtain x p x k ∈ D .Therefore x x . . . x k − yx k +1 . . . x p x k xx is a hamiltonian cycle, a contradiction.Now suppose that k = l . Assume, without loss of generality, that A ( x i , y ) = ∅ for each i ∈ [2 , l − ←− D we will have the considered case k ≥ l + 1). Then from x y ∈ D it follows that { x , x , . . . , x l − } → y. (19)We also can assume that l = n (if l = n −
1, then in ←− D we will have the case l = n ). So, we have k = l = n . It is not difficult to see that A ( x , x n ) = A ( { x , x , . . . , x n − , x p } → x n ) = A ( x n → { x n +2 , x n +3 , . . . , x p − } ) = ∅ . (20)Indeed, if it is not true, then by (17) and (19) we haveif x i x n ∈ D and i ∈ [1 , n − C n = x x . . . x i x n . . . x p xx i +1 . . . x n − yx ;if x n x i ∈ D and i ∈ [ n + 2 , p − C n = x x . . . x n x i x i +1 . . . x p yx n +1 . . . x i − xx ;if x p x n ∈ D , then C n = x x . . . x n − yx n +1 . . . x p x n xx ;if x n x ∈ D , then C n = x x . . . x n − yx n +1 . . . x p xx n x . In each case we have a hamiltonian cycle,a contradiction, and (20) holds.Therefore from d ( x n ) = 2 n − x n cannot be inserted into the paths x x . . . x n − and x n +1 x n +2 . . . x p , it follows that (by Lemma 2) { x n +1 , x n +2 , . . . , x p − } → x n → { x , x , . . . , x n − } . (21)If x i x ∈ D for some i ∈ [2 , p − \{ n } , then by (17), (18), (19) and (21) we have if i ∈ [2 , n − C n = x x . . . x i − yx n +1 . . . x p xx i +1 . . . x n x i x and if i ∈ [ n + 1 , p − C n = x x . . . x n − yx i +1 . . . x p xx n . . . x i x , a contradiction. So, we may assume that A ( { x , x , . . . , x p − } → x ) = ∅ . (22)Hence, by Lemma 4(ii), 2 n − ≤ n , i.e. n ≤
4. Let n = 4. Then by (22,) id ( x ) ≤
3. On theother hand, from A ( x → { x, x , x , x } ) = ∅ it follows (if x x ∈ D , then x x ∈ D by (21), and C = x x x xx x x yx ) that od ( x ) ≤
3. So d ( x ) ≤
6, a contradiction. Let now n = 3. From (21) wesee that x x ∈ D . Hence it is easy to see that x x / ∈ D by (20), x x / ∈ D and A ( x → { x, x , x } ) = A ( { x , x } → x ) = ∅ . x x ∈ D . Now, it is not difficult to check that D is isomorphic to one of the digraphs D , D ′ ,a contradiction. This completes the proof of Claim 8.Similarly to Claim 8, we can show the following claim: Claim 9 . There is a vertex x i , i ∈ [2 , p − A ( y, x i ) = ∅ . Claim 10 . x p − y / ∈ D . Proof of claim 10 . Suppose, on the contrary, that x p − y ∈ D . By Claim 6 we have x p − x / ∈ D .Therefore xx p ∈ D by Claim 1(ii) . By Claim 8 there is a vertex x l , l ∈ [2 , p − A ( x, x l ) = ∅ .Using Claim 1(iii), we obtain x l − x, xx l +1 ∈ D and d ( x l ) = 2 n − p + 1 . (23)For the vertex x l we first will prove the following statements (a)-(i) . (a). x p x l / ∈ D . Proof . Indeed, if (a) is not true, then x p x l ∈ D and C n = x x . . . x l − xx p x l . . . x p − yx by (23), acontradiction. (b) . If l ≤ p −
2, then A ( x l , x p ) = ∅ . Proof . From l ≤ p − x l ∈ O − ( x ). Hence x l x p / ∈ D by Claim 2. Therefore by statement (a) , A ( x l , x p ) = ∅ . (c) . If l ≤ p −
2, then x p − x l and x l y ∈ D . Proof . Note that A ( x l , x p ) = ∅ by statement (b) , and the cycle x x . . . x l − xx l +1 . . . x p yx has length2 n −
1. Therefore x p − x l and x l y ∈ D by Claim 1(iii). (d) . If l ≤ p −
2, then A ( y → { x l , x l +1 , . . . , x p } ) = ∅ . Proof . Suppose, on the contrary, that A ( y → { x l , x l +1 , . . . , x p } ) = ∅ . It follows that O − ( y ) ⊆{ x , x , . . . , x l − } . If x i ∈ O − ( y ) and x i x l ∈ D , then C n = x x . . . x i x l . . . x p yx i +1 . . . x l − xx is ahamiltonian cycle in D , a contradiction. So we can assume that A ( O − ( y ) → x l ) = ∅ . Together with A ( { x, y } → x l ) = ∅ and | O − ( y ) | ≥ n − x p x l ∈ D . But this contradicts (a) , and hence (d) is proved. (e) . If l ≤ p −
2, then x p x l +1 / ∈ D and x l − x p / ∈ D . Proof . Recall that xx p , x p x ∈ D , and x p − x l , x l y ∈ D by (c) . Then by (23) we have, if x p x l +1 ∈ D ,then C n = x x . . . x l − xx p x l +1 . . . x p − x l yx and if x l − x p ∈ D , then C n = x x . . . x l − x p xx l +1 . . .x p − x l yx . Therefore D is hamiltonian, a contradiction. (f ) . If l ≤ p − xx l +2 ∈ D , then x l x l +2 / ∈ D and x l +1 x l ∈ D . Proof . Indeed, if x l x l +2 ∈ D , then for the path x l +1 x l +2 . . . x p yx x . . . x l − we have { x, x l } →{ x l +1 , x l +2 } and x l − → { x, x l } , which contradicts Claim 7. So x l x l +2 / ∈ D . Now from Claim 1(ii)it follows that x l +1 x l ∈ D . (g) . If l ≥ x l − x ∈ D , then x l − x l / ∈ D and x l x l − ∈ D . Proof . Indeed, if x l − x l ∈ D , then for the path x l +1 x l +2 . . . x p yx . . . x l − x l − we have { x l − , x l − }→ { x, x l } and { x, x l } → x l +1 , which contradicts Claim 6. So x l − x l / ∈ D . From this and Claim 1(ii) it12ollows that x l x l − ∈ D . Statement (g) is proved. (h) . If l ≤ p −
2, then x i x p ∈ D if and only if x i / ∈ { x l − } ∪ O − ( x ); and x p x i ∈ D if and only if x i / ∈ { x l +1 } ∪ I + ( x ). Proof . By Claims 2, 3 and statement (e) we have A ( O − ( x ) ∪ { x l − , y } → x p ) = A ( x p → { x l +1 } ∪ I + ( x )) = ∅ . (24)From x p − x / ∈ D and xx p ∈ D , we get that | I + ( x ) | = id ( x ) − | O − ( x ) | = od ( x ) − . Therefore id ( x p ) ≤ n − − od ( x ) and od ( x p ) ≤ n − − id ( x ) by (24). Hence id ( x p ) = 2 n − − od ( x )and od ( x p ) = 2 n − − id ( x ) (otherwise d ( x ) + d ( x p ) < n −
2, which is a contradiction). Now from thisit is not difficult to see that statement (h) is true.Recall that the proof of statement (h) implies the following statement: (i) . The vertex x is not adjacent with at most one vertex of the path x x . . . x p − , in particular, thevertex x is not adjacent with at most 3 vertices (i.e., if C n − is an arbitrary cycle of D and the vertex x / ∈ C n − , then x is not adjacent with at most tree vertices).By Claim 8 there is a vertex x k , k ∈ [2 , p − A ( x, x k ) = ∅ . Without loss of generality,assume that k is as large as possible. From the maximality of k and Claim 1(iii) it is easy to see that x k − x ∈ D, d ( x k ) = p + 1 , x → { x k +1 , x k +2 , . . . , x p } , A ( { x k +1 , x k +2 , . . . , x p − } → x ) = ∅ . (25)We now consider two cases. Case 1 . k ≤ p − (c) we have x p − x k ∈ D and x k y ∈ D. (26)From statement (i) and (25) it follows that if i ∈ [1 , p ] and i = k , then A ( x, x i ) = ∅ . (27)It is easy to see that n ≥
4. Indeed, if n = 3, then k = 2 and by (26) the vertex y is not adjacent onlywith one vertex of the cycle x x . . . x p xx , which contradicts Claim 9.Suppose first that k ≤ p −
3. Then x / ∈ I + ( x ), and x k +2 / ∈ I + ( x ) by (25). Together with statement (h) this implies that x p → { x , x k +2 } . (28)If x k +1 y ∈ D , then using (25), (26) and (28), we obtain C n = x x . . . x k − xx p x k +2 . . . x p − x k x k +1 yx , a contradiction. So, we may assume that x k +1 y / ∈ D . Since x k y ∈ D by (26), we see that A ( y, x k +1 ) = ∅ . Therefore yx k +2 ∈ D by Claim 1(iii). Recall that x k +1 x k ∈ D by statement (f ) , and hence by (25)and (28) we have a hamiltonian cycle x x . . . x k − xx k +1 x k yx k +2 . . . x p x , a contradiction.Suppose next that k = p −
2. Then by x p − y ∈ D and statements (d) , (c) , yx p − ∈ D . If xx ∈ D ,then x x / ∈ D , x / ∈ I + ( x ) and x p x ∈ D by statement (h) . Since xx ∈ D , by Claim 7 we have yx / ∈ D .Therefore x y ∈ D by Claim 1(ii), and we get a hamiltonian cycle x yx p − x p − x p x . . . x p − xx , acontradiction. So we may assume that xx / ∈ D . From this and (27) it follows that13 x , x , . . . , x p − } → x and A ( x → { x , x , . . . , x p − } ) = ∅ . Therefore n = 4 (i.e., p = 6). Then x x ∈ D by statement (g) . We can assume that yx / ∈ D (otherwise yx ∈ D and for ←− D we will have the considered case k ≤ p − od ( y ) ≥ x y, yx ∈ D and A ( y, x ) = ∅ . Since x / ∈ I + ( x ), we have x x ∈ D and x x / ∈ D by statement (h) . Then x x / ∈ D (otherwise x x ∈ D and C n = x yx x x x xx x ).Now we have A ( { x, y, x , x } → x ) = ∅ . Hence x x ∈ D and C n = x yx x x xx x x , a contradiction. Case 2 . k = p − yx p − / ∈ D . Then it is not difficult to see that A ( O − ( y ) → x p − ) = ∅ (otherwiseif x i ∈ O − ( y ) and x i x p − ∈ D , then C n = x x . . . x i x p − x p yx i +1 . . . x p − xx ). This together with | O − ( y ) | = od ( y ) − A ( { x, y, x p } → x p − ) = ∅ implies that id ( x p − ) ≤ n −
2, a contradiction.Suppose next that yx p − ∈ D . We assume that n ≥ n = 3 and 4. We leave its proof to the reader). Subcase 2.1 . xx ∈ D .Then x x / ∈ D . Using Claims 7 and 1(ii), we obtain yx / ∈ D and x y ∈ D . We may assume that A ( x , y ) = ∅ (otherwise for the vertex y in digraph ←− D we have the considered Case 1 ( k ≤ p − yx ∈ D . Similarly to yx p − ∈ D , we also may assume that x x ∈ D . From n ≥ A ( x, x s ) = ∅ for some s ∈ [3 , p −
3] (otherwise O ( x ) = { x , x , x p } , i.e., od ( x ) ≤
3, a contra-diction). Since x / ∈ { x s +1 } ∪ I + ( x ), using statement (h) , we see that x p x ∈ D and x yx . . . x p x xx is a hamiltonian cycle, a contradiction. Subcase 2.2 . xx / ∈ D .Then x x ∈ D by Claim 1(ii). By statement (i) , the vertex x is not adjacent with at most one vertexof { x , x , . . . , x p − } . From this and n ≥ A ( x, x s ) = ∅ exactly for one s ∈ [2 , p − A ( x, x i ) = ∅ for each i ∈ [2 , p −
4] and by xx ∈ D , A ( x → { x , x , . . . , x p − } ) = ∅ , i.e. O ( x ) ⊆ { x , x p − , x p } and od ( x ) ≤
3, which contradicts that n ≥ s = 2 (i.e., A ( x, x ) = ∅ ). Note that x x , xx ∈ D by Claim 1(iii). From statement (c) it followsthat x p − x and x y ∈ D . Since x / ∈ I + ( x ), by statement (h) we have x p x ∈ D . If yx / ∈ D , then x y ∈ D by Claim 1(ii), and x yx p − x . . . x p − xx p x is a hamiltonian cycle, a contradiction. So we mayassume that yx ∈ D . Also we may assume that id ( x ) = n (for otherwise we will consider the digraph ←− D ).Then, since n ≥
4, we see that { x p − , x p − } → x and by (g) , x p − x p − ∈ D . Then x p − / ∈ O − ( x ) ∪ { x } and by (h) ( l = 2) we see that x p − x p ∈ D . Thus x x . . . x p − x p yx p − x p − xx is a hamiltonian cycle,a contradiction.Let now s ∈ [3 , p − x x ∈ D it follows that { x , x , . . . , x s − } → x . Together withstatement (i) this implies that { x s − , x s − } → x → { x s +1 , x s +2 } . By statements (f ) and (g) we have x s +1 x s , x s x s − ∈ D , x s x s +2 / ∈ D and x s − x s / ∈ D .It is not difficult to show that A ( x s , { x s − , x s +2 } ) = ∅ . (29)Indeed, if x s x s − ∈ D , then, since { x s − , x s − } → x , for the path x s +1 x s +2 . . . x p yx x . . . x s − andfor the vertex x s we will have the considered Case 1, and if x s +2 x s ∈ D , then in digraph ←− D for thepath x s − x s − . . . x yx p x p − . . . x s +2 x s +1 and for the vertex x s again we will have the considered Case 1( k ≤ p −
2) and (29) holds. 14ecall that A ( x s , x p ) = ∅ by statement (b) . Together with (29) and A ( x s , x ) = ∅ this implies that A ( x s , { x, x p , x s − , x s +2 } = ∅ , i.e., the vertex x s is not adjacent with at least 4 vertices of cycle x x . . . x s − xx s +1 . . . x p yx , this iscontrary to statement (i) . The proof of Claim 10 is completed.Similarly to Claim 10 ( x p − y / ∈ D ), we can show the following claim: Claim 11 . x p − x / ∈ ( G ), xx / ∈ D and yx / ∈ D .Now let us complete the proof of Theorem 2. Without loss of generality, we may assume that od ( x ) = n . It follows that x → { x i , x i +1 } for some i ∈ [1 , p − i ≥ x → { x, y } . Therefore A ( x, x l ) = ∅ , x → { x l +1 , x l +2 } and x l − x ∈ D for some l ∈ [2 , i − x l +1 x l +2 . . . x p yx . . . x l − we have A ( x, x l ) = ∅ , x → { x l +1 , x l +2 } and x l − x l , x l x l +1 , x l − x ∈ D . This is a contradiction to Claim 11 ( xx / ∈ D ) that xx l +2 / ∈ D . The proof ofTheorem 2 is completed.
4. Cycles of length 3 and 4 in digraph D . The next two results will be used in the proof of Theorem 3.
Theorem A (R. H¨aggkvist, R. J. Faudree, R.H. Schelp [20]). Let G be an undirected graph on 2 n +1 ≥ n . Then precisely one of the following hold: (i) G is pancyclic;(ii) G ≡ ( K n ∪ K n ) + K ; or (iii) K n,n +1 ⊆ G ⊆ K n + K n +1 . Theorem B (C. Tomassen [30]). Let D be a strongly connected digraph on p ≥ x, y of nonadjacent distinct vertices d ( x ) + d ( y ) ≥ p , then D is pancyclic or p is even and D ≡ K ∗ p/ ,p/ .Now we difine the digraphs C ∗ (1), H ′ and H ′′ as folllows:(i) V ( C ∗ (1)) = { x , x , . . . , x } and A ( C ∗ (1)) = { x i x i +1 , x i +1 x i /i ∈ [1 , } ∪ { x x , x x , x x , x x ,x x , x x } ;(ii) V ( H ′ ) = V ( H ′′ ) = { x, y, z, u, v, w } , A ( H ′ ) = { ux, xu, xv, vx, yz, zy, zw, wz, xw, xy, uz, vz, wu,wv, yu, yv } and A ( H ′′ ) = { ux, xu, xw, xy, vx, vz, vw, wv, wu, zw, zy, yz, uz, yu, yv } . Theorem 3 . Let D be a digraph on p ≥ p − p/ −
1. Then the following hold:(i) D contains a cycle of length 3 or p = 2 n and D ⊆ K ∗ n,n or else D ∈ { C ∗ , K ∗ n,n +1 } ;(ii) D contains a cycle of length 4 or D ∈ { C ∗ , H ′ , H ′′ , C ∗ (1) , H (3 , , [( K ∪ K ) + K ] ∗ } . Proof . Using Theorems A and B, we see that Theorem 3 is true if D is a symmetric digraph. Supposethat D is not symmetric digraph. If D contains no cycle of length 3, then it is not difficult to show that p = 2 n and D ⊆ K ∗ n,n (we leave the details to the reader).Assume that D contains no cycle of length 4. For each arc xy ∈ D put S ( x, y ) := I ( x ) ∩ O ( y ) and E ( x, y ) := V ( D ) \ ( O ( y ) ∪ I ( x ) ∪ { x, y } ) . Since D has no cycle of length 4, we see that A ( O ( y ) \ { x } → I ( x ) \ { y } ) = ∅ . (30)15et us consider the following cases. Case 1 . There is an arc xy ∈ D such that yx / ∈ D and od ( y ) ≥ n or id ( x ) ≥ n , where n := ⌊ p/ ⌋ .Without loss of generality, we can assume that od ( y ) ≥ n (if id ( x ) ≥ n , then we will consider thedigraph ←− D ). Then from (30) and Lemma 4(ii) it follows that I ( x ) ⊆ O ( y ) , I ( x ) = S ( x, y ) and A ( h S ( x, y ) i ) = ∅ . (31)We now shall prove that I ( x ) = O ( y ) . (32) Proof of (32) . Assume that (32) is not true. Then O ( y ) \ I ( x ) = ∅ by (31), and let z ∈ O ( y ) \ I ( x ). By(30), A ( z → { x } ∪ I ( x )) = ∅ . From this and Lemma 4(ii) it follows that |{ x } ∪ I ( x ) | = p/ p = 2 n ≥ id ( x ) = n − z → V ( D ) \ ( { x, z } ∪ I ( x )) , (33)in particular, zy ∈ D , od ( z ) = n − id ( z ) ≥ n . If xz ∈ D , then C = xzyux , where u ∈ S ( x, y ),contradicting the our assumption. Therefore xz / ∈ D . From id ( x ) = n − id ( z ) ≥ n and Lemma 4(ii)it follows that uz ∈ D for some vertex u ∈ I ( x ). Now it is not difficult to see that O ( y ) \ I ( x ) = { z } , E ( x, y ) = ∅ and A ( E ( x, y ) → y ) = ∅ .Suppose first that for each vertex v ∈ I ( x ) there is a vertex v ∈ E ( x, y ) such that v v ∈ D . Hence A ( I ( x ) → y ) = ∅ by (33) and C D . Therefore A ( I ( x ) ∪ E ( x, y ) → y ) = ∅ , | E ( x, y ) | = 1 and n = 3. Let E ( x, y ) := { w } and I ( x ) := { u, v } . Note that w → { u, v } . Now it is not difficult to see that if wz ∈ D ,then D ≡ H ′ and if wz / ∈ D , then D ≡ H ′′ .Suppose next that A ( E ( x, y ) → v ) = ∅ for some v ∈ I ( x ). Then | E ( x, y ) | = 1 by (31) and n = 3, xv ∈ D , vy / ∈ D and v → { z, w } , where w ∈ E ( x, y ). Now it is easy to see that O ( w ) = { z } . Therefore od ( w ) ≤
1, a contradiction. This proves (32), i.e., I ( x ) = O ( y ) = S ( x, y ). Subcase 1.1 . A ( x → S ( x, y )) = ∅ .Let xu ∈ A ( x → S ( x, y )). If uy ∈ D , then C = xuyu x , where u ∈ S ( x, y ) \ { u } , a contradiction.So we may assume that uy / ∈ D . From od ( y ) ≥ n , (32) and (31) we get that u → E ( x, y ) and od ( y ) = n .It is not difficult to see that A ( E ( x, y ) → ( S ( x, y ) \ { u } )) = ∅ (otherwise C ⊂ D ). Then E ( x, y ) := { w } , wu, wy ∈ D since wx / ∈ D , and xv ∈ D , where v ∈ S ( x, y ) \ { u } . Then vy ∈ D or vw ∈ D . In both casewe obtain a cycle of length 4, which is a contradiction. Subcase 1.2 . A ( x → S ( x, y )) = ∅ .We can assume that A ( S ( x, y ) → y ) = ∅ (otherwise in ←− D we will have Subcase 1.1). From od ( y ) ≥ n ,by (32) and Lemma 4(ii) we have E ( x, y ) = ∅ and x → E ( x, y ) → y . Therefore C = xzyux , where z ∈ E ( x, y ) and u ∈ S ( x, y ), which is a contradiction and completes the discussion of Case 1. Case 2 . For each arc xy ∈ D if yx / ∈ D , then od ( y ) < n and id ( x ) < n .From conditions of theorem it follows easily that od ( y ) = id ( x ) = n − p = 2 n ≥
6. If S ( x, y ) = ∅ ,then using (30) and Lemma 4 (ii) it is easy to see that C ⊂ D or D ∈ H (3 , S ( x, y ) = ∅ .Since id ( y ) and od ( x ) ≥ n , we can assume that O ( y ) → y and x → I ( x ) (otherwise for some arc ux or yv we have the considered Case 1). Hence it is easy to see that | S ( x, y ) | = 1, I ( x ) = O ( y ) and | E ( x, y ) | = 1.Let E ( x, y ) := { w } and S ( x, y ) := { z } . From (30) it follows that O ( y ) \ { z } → w → I ( x ) \ { z } . Fromthis, we obtain A ( z, w ) = ∅ since C D . Now it is not difficult to see that for some v ∈ I ( x ) − { z } (or u ∈ O ( y ) − { z } ) vz ∈ D (or zu ∈ D ). Without loss of generality we may assume that zu ∈ D . From thiswe have A ( I ( x ) \ { z } → z ) = ∅ , wy / ∈ D and n = 3. Hence wu, vw, xw, vu ∈ D and D ≡ C ∗ (1). This16ompletes the proof of Theorem 3.In [13], we proved the following: Theorem.
Let D be a digraph on p ≥