On Hamiltonian Bypasses in one Class of Hamiltonian Digraphs
aa r X i v : . [ m a t h . C O ] A p r On Hamiltonian Bypasses in one Class ofHamiltonian Digraphs
Samvel Kh. Darbinyan and Iskandar A. Karapetyan
Institute for Informatics and Automation Problems, Armenian National Academy of SciencesE-mails: [email protected], [email protected]
Abstract
Let D be a strongly connected directed graph of order n ≥ x, y with a common in-neighbour d ( x ) + d ( y ) ≥ n − min { d ( x ) , d ( y ) } ≥ n −
1. In [2] (J. of Graph Theory 22 (2) (1996) 181-187)) J. Bang-Jensen, G. Gutinand H. Li proved that D is Hamiltonian. In [9] it was shown that if D satisfies the condition (*) andthe minimum semi-degree of D at least two, then either D contains a pre-Hamiltonian cycle (i.e., a cycleof length n −
1) or n is even and D is isomorphic to the complete bipartite digraph (or to the completebipartite digraph minus one arc) with partite sets of cardinalities of n/ n/
2. In this paper we showthat if the minimum out-degree of D at least two and the minimum in-degree of D at least three, then D contains also a Hamiltonian bypass, (i.e., a subdigraph is obtained from a Hamiltonian cycle by reversingexactly one arc). Keywords:
Digraphs, cycles, Hamiltonian cycles, Hamiltonian bypasses.
The directed graph (digraph) D is Hamiltonian if it contains a Hamiltonian cycle, i.e., a cycle thatincludes every vertex of D . A Hamiltonian bypass in D is a subdigraph obtained from a Hamiltoniancycle by reversing exactly one arc. We recall the following well-known degree conditions (Theorems 1-5)that guarantee that a digraph is Hamiltonian. Theorem 1 (Nash-Williams [14]). Let D be a digraph of order n such that for every vertex x , d + ( x ) ≥ n/ d − ( x ) ≥ n/
2, then D is Hamiltonian. Theorem 2 (Ghouila-Houri [12]). Let D be a strong digraph of order n . If d ( x ) ≥ n for all vertices x ∈ V ( D ), then D is Hamiltonian. Theorem 3 (Woodall [16]). Let D be a digraph of order n ≥
2. If d + ( x ) + d − ( y ) ≥ n for all pairs ofvertices x and y such that there is no arc from x to y , then D is Hamiltonian. Theorem 4 (Meyniel [13]). Let D be a strong digraph of order n ≥
2. If d ( x ) + d ( y ) ≥ n − D , then D is Hamiltonian.It is easy to see that Meyniel’s theorem is a common generalization of Ghouila-Houri’s and Woodall’stheorems. For a short proof of Theorem 1.3, see [5].C. Thomassen [15] (for n = 2 k + 1) and S. Darbinyan [6] (for n = 2 k ) proved the following: Theorem 5 [15, 6]. If D is a digraph of order n ≥ n − n/ −
1, then D is Hamiltonian (unless some extremal cases which arecharacterized).In view of the next theorems we need the following definitions. Definition 1 . Let D denote any digraph of order n ≥ n odd, such that V ( D ) = A ∪ B , where A ∩ B = ∅ , A is an independent set with ( n + 1) / B is a set of ( n − / D has ( n + 1)( n − / A and B . Note that D has noHamiltonian bypass. Definition 2 . For any k ∈ [1 , n −
2] let D denote a digraph of order n ≥
4, obtained from K ∗ n − k and K ∗ k +1 by identifying a vertex of the first with a vertex of the second. Note that D has no Hamiltonianbypass. Definition 3 . By T (5) we denote a tournament of order 5 with vertex set V ( T (5)) = { x , x , x , x , y } and arc set A ( T (5)) = { x i x i +1 /i ∈ [1 , } ∪ { x x , x y, x y, yx , yx , x x , x x } . T (5) has no Hamilto-nian bypass.In [4] it was proved that if a digraph D satisfies the condition of Nash-Williams’ or Ghouila-Houri’s orWoodall’s theorem, then D contains a Hamiltonian bypass. In [4] the following theorem was also proved: Theorem 6 (Benhocine [4]). Every strongly 2-connected digraph of order n and minimum degree atleast n − D is isomorphic to a digraph of type D .In [7] the first author proved the following theorem: Theorem 7 (Darbinyan [7]). Let D be a strong digraph of order n ≥
3. If d ( x ) + d ( y ) ≥ n − D , then D contains a Hamiltonian bypass unless it is isomorphic toa digraph of the set D ∪ { D , T , C } , where C is a directed cycle of length 3.For n ≥ k ∈ [2 , n ], D ( n, k ) denotes the digraph of order n obtained from a directed cycle C oflength n by reversing exactly k − D ( n,
3) in digraphs with the condition of Meyniel’s theorem and in oriented graphs withlarge in-degrees and out-degrees.
Theorem 8 (Darbinyan [7]). Let D be a strong digraph of order n ≥
4. If d ( x ) + d ( y ) ≥ n − D , then D contains a D ( n, Theorem 9 (Darbinyan [8]). Let D be an oriented graph of order n ≥
10. If the minimum in-degreeand out-degree of D at least ( n − /
2, then D contains a D ( n, Theorem 10 [2] (Bang-Jensen, Gutin, H.Li [2]). Let D be a strong digraph of order n ≥
2. Supposethat min { d ( x ) , d ( y ) } ≥ n − d ( x ) + d ( y ) ≥ n − ∗ )for every pair of non-adjacent vertices x, y with a common in-neighbour, then D is Hamiltonian.In [9] the following results were obtained: Theorem 11 [9]. Let D be a strong digraph of order n ≥ D atleast two. Suppose that D satisfies the condition (*). Then either D contains a pre-Hamiltonian cycleor n is even and D is isomorphic to the complete bipartite digraph or to the complete bipartite digraphminus one arc with partite sets of cardinalities n/ n/ Theorem 12 (Main Result). Let D be a strong digraph of order n ≥ min { d ( x ) , d ( y ) } ≥ n − d ( x ) + d ( y ) ≥ n − ∗ )2or every pair of non-adjacent vertices x, y with a common in-neighbour. Then D contains a Hamiltonianbypass. We shall assume that the reader is familiar with the standard terminology on the directed graphs (digraph)and refer the reader to the monograph of Bang-Jensen and Gutin [1] for terminology not discussed here.In this paper we consider finite digraphs without loops and multiple arcs. For a digraph D , we denoteby V ( D ) the vertex set of D and by A ( D ) the set of arcs in D . The order of D is the number of itsvertices. Often we will write D instead of A ( D ) and V ( D ). The arc of a digraph D directed from x to y is denoted by xy or x → y . If x, y, z are distinct vertices in D , then x → y → z denotes that xy and yz ∈ D . Two distinct vertices x and y are adjacent if xy ∈ A ( D ) or yx ∈ A ( D ) (or both). By a ( x, y ) we denote the number of arcs with end vertices x and y , in particular, a ( x, y ) means that thevertices x and y are non-adjacent. For disjoint subsets A and B of V ( D ) we define A ( A → B ) as theset { 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 write x instead of { x } . If A and B are two distinct subsets of V ( D ) such that every vertex of A dominates every vertex of B , then we say that A dominates B , denoted by A → B . The out-neighborhoodof a vertex x is the set N + ( x ) = { y ∈ V ( D ) /xy ∈ A ( D ) } and N − ( x ) = { y ∈ V ( D ) /yx ∈ A ( D ) } is the in-neighborhood of x . Similarly, if A ⊆ V ( D ), then N + ( x, A ) = { y ∈ A/xy ∈ A ( D ) } and N − ( x, A ) = { y ∈ A/yx ∈ A ( D ) } . The out-degree of x is d + ( x ) = | N + ( x ) | and d − ( x ) = | N − ( x ) | is thein-degree of x . Similarly, d + ( x, A ) = | N + ( x, A ) | and d − ( x, A ) = | N − ( x, A ) | . The degree of the vertex x in D is defined as d ( x ) = d + ( x )+ d − ( x ) (similarly, d ( x, A ) = d + ( x, A )+ d − ( x, A )). The path (respectively,the cycle) consisting of the distinct vertices x , x , . . . , x m ( m ≥
2) and the arcs x i x i +1 , i ∈ [1 , m − x i x i +1 , i ∈ [1 , m − x m x ), is denoted by x x · · · x m (respectively, x x · · · x m x ).We say that x x · · · x m is a path from x to x m or is an ( x , x m )-path. For a cycle C k := x x · · · x k x of length k , the subscripts considered modulo k , i.e., x i = x s for every s and i such that i ≡ s (mod k ).If P is a path containing a subpath from x to y we let P [ x, y ] denote that subpath. Similarly, if C is acycle containing vertices x and y , C [ x, y ] denotes the subpath of C from x to y . A digraph D is stronglyconnected (or, just, strong) if there exists a path from x to y and a path from y to x for every pair ofdistinct vertices x, y . For an undirected graph G , we denote by G ∗ the symmetric digraph obtained from G by replacing every edge xy with the pair xy , yx of arcs. K p,q denotes the complete bipartite graphwith partite sets of cardinalities p and q . For integers a and b , a ≤ b , let [ a, b ] denote the set of all integerswhich are not less than a and are not greater than b . By D ( n ; 2) = [ x x n ; x x . . . x n ] is denoted theHamiltonian bypass obtained from a Hamiltonian cycle x x . . . x n x by reversing the arc x n x . The following well-known simple Lemmas 1 and 2 are the basis of our results and other theorems ondirected cycles and paths in digraphs. They will be used extensively in the proof of our result.
Lemma 1 [11]. Let D be a digraph of order n ≥ C m , m ∈ [2 , n − 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 is a slight modification of a lemma by Bondy and Thomassen [5]. Lemma 2 . Let D be a digraph of order n ≥ P := x x . . . x m , m ∈ [2 , n −
1] andlet x be a vertex not contained in this path. If one of the following conditions holds:3i) 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 (the arc x i x i +1 is a partner of x ), i.e., D contains a path x x . . . x i xx i +1 . . . x m of length 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 ). Definition 4 ([1], [2]). Let Q = y y . . . y s be a path in a digraph D (possibly, s = 1) and let P = x x . . . x t , t ≥
2, be a path in D − V ( Q ). Q has a partner on P if there is an arc (the partnerof Q ) x i x i +1 such that x i y , y s x i +1 ∈ D . In this case the path Q can be inserted into P to give a new( x , x t )-path with vertex set V ( P ) ∪ V ( Q ). The path Q has a collection of partners on P if there areintegers i = 1 < i < · · · < i m = s + 1 such that, for every k = 2 , , . . . , m the subpath Q [ y i k − , y i k − ]has a partner on P . Lemma 3 ([1], [2], Multi-Insertion Lemma). Let Q = y y . . . y s be a path in a digraph D (possibly, s = 1) and let P = x x . . . x t , t ≥
2, be a path in D − V ( Q ). If Q has a collection of partners on P ,then there is an ( x , x t )-path with vertex set V ( P ) ∪ V ( Q ).The following lemma is obvious. Lemma 4 . Let D be a digraph of order n ≥ C := x x . . . x n − x be an arbitrary cycle oflength n − D and let y be the vertex not on C . If D contains no Hamiltonian bypass, then(i) d + ( y, { x i , x i +1 } ) ≤ d − ( y, { x i , x i +1 } ) ≤ i ∈ [1 , n − d + ( y ) ≤ ( n − / d − ( y ) ≤ ( n − / d ( y ) ≤ n − x k y, yx k +1 ∈ D , then x i +1 x i / ∈ D for all x i = x k .Let D be a digraph of order n ≥ C n − be a cycle of length n − D . If for the vertex y / ∈ C n − , d ( y ) ≥ n , then we say that C n − is a good cycle. Notice that, by Lemma 4(ii), if a digraph D contains a good cycle, then D also contains a Hamiltonian bypass.We now need to state and prove some general lemmas. Lemma 5 . Let D be a digraph of order n ≥ C := x x . . . x n − x be an arbitrary cycle of length n − D and let y be the vertexnot on C . Then for any i ∈ [1 , n −
1] the following holds:(i) If yx i / ∈ D and x i − x i / ∈ D , then x i has a partner on C [ x i +1 , x i − ] or d ( x i ) ≥ n − yx i / ∈ D and d ( x i ) ≤ n −
2, then x i has a partner on C [ x i +1 , x i − ] or there is a vertex x k ∈ C [ x i +1 , x i − ] such that { x k , x k +1 , . . . , x i − } → x i .(iii) If yx i ∈ D , x i − x i / ∈ D and d − ( x i ) ≥
3, then x i has a partner on C [ x i +1 , x i − ] or d ( x i ) ≥ n − Proof . (i) The proof is by contradiction. Assume that x i has no partner on C [ x i +1 , x i − ] and d ( x i ) ≤ n −
2. Since d − ( x i , { y, x i − } ) = 0 and d − ( x i ) ≥
2, there is an x k ∈ C [ x i +1 , x i − ] such that x k x i ∈ D . From x k → { x k +1 , x i } , d ( x i ) ≤ n − x i x k +1 / ∈ D and the condition (*) it follows that x k +1 x i ∈ D . By a similar argument we conclude that x i − x i ∈ D , which is a contradiction.For the proofs of (ii) and (iii) we can use precisely the same arguments as in the proof of (i). Lemma 6 . Let D be a digraph of order n ≥ C := x x . . . x n − x be an arbitrary cycle of length n − D and let y be the vertexnot on C . Then 4i) If for some i ∈ [1 , n − x i y ∈ D and x i +1 , y are non-adjacent or (ii) a ( x i , y ) = 2 or(iii) d ( y ) ≥ n −
1, then D contains a Hamiltonian bypass. Proof . (i) Assume that (i) is not true. Without loss of generality, we assume that x n − y ∈ D , d ( y, { x , x , . . . , x a } ) = 0 and x a +1 , y are non-adjacent, where a ≥
1. Then x and y is a dominated pairof non-adjacent vertices with a common in-neighbour x n − . Therefore, by condition (*), d ( y ) ≥ n − d ( x ) ≥ n −
1. On the other hand, using Lemma 4(i) we obtain that d ( y ) ≤ n − a and hence, a = 1 and d ( y ) = n −
1. This together with the condition (*) implies that d ( x ) ≥ n . If yx ∈ D , then x n − yx x . . . x n − x n − is a good cycle in D and therefore, D contains a Hamiltonian bypass. Assumetherefore that x y ∈ D . Since d ( x ) ≥ n , by Lemma 2, x has a partner on the path C [ x , x n − ], i.e,there is an ( x , y )-Hamiltonian path which together with the arc x y forms a Hamiltonian bypass, whichis a contradiction and completes the proof of (i).(ii) It follows immediately from Lemmas 6(i) and 4(i).(iii) Suppose, on the contrary, that d ( y ) ≥ n − D contains no Hamiltonian bypass as well asno good cycle. By Lemma 4(ii), d ( y ) = n −
1. From Lemma 6(ii) it follows that a ( y, x i ) = 1 for all i ∈ [1 , n − N + ( y ) = { x , x , . . . , x n − } and N − ( y ) = { x , x , . . . , x n − } . (1)Notice that(2) for every vertex x i , x i x i − / ∈ D and x i has no partner on the path C [ x i +1 , x i − ] (for otherwise, D contains a Hamiltonian bypass).Assume first that x x ∈ D . Then it is not difficult to show that x , x n − are non-adjacent and x x / ∈ D . Indeed, by (1) if x n − x ∈ D , then D ( n,
2) = [ x x ; x x x yx . . . x n − x ]; if x x ∈ D , then D ( n,
2) = [ x x ; x x x . . . x n − yx x ]; and if x x n − ∈ D , then D ( n,
2) = [ x x n − ; x yx x x . . . x n − ],in each case we have a contradiction. Now, since x n − x / ∈ D , yx / ∈ D and x has no partner on C [ x , x ],Lemma 5(i) implies that d ( x ) ≥ n −
1. On the other hand, using Lemma 2(ii), a ( x , x n − ) = 0, x x / ∈ D and (2), we obtain n − ≤ d ( x ) = d ( x , { y, x , x } ) + d ( x , C [ x , x n − ]) ≤ n − , a contradiction.Assume second that x x / ∈ D . By the symmetry of the vertices x n − and x (by (1)), we also mayassume that x n − x / ∈ D . Since x has no partner on C [ x , x n − ], again using Lemma 2(iii) and (2) weobtain that d ( x ) = d ( x , { x n − , x , y } ) + d ( x , C [ x , x n − ]) ≤ n − . Therefore, by condition (*), we have that x is adjacent with x and x n − , i.e., x x , x x n − ∈ D ,since y → { x n − , x , x } . Now it is easy to see that { x , x , . . . , x n − } → x , which contradicts that x n − x / ∈ D . In each case we obtain a contradiction, and hence the proof of Lemma 6(iii) is completed.The following simple observation is of importance in the rest of the paper. Remark . Let D be a digraph of order n ≥ C := x x . . . x n − x be an arbitrary cycle of length n − D and let y be the vertexnot on C . If D contains no Hamiltonian bypass, then(i) There are two distinct vertices x k and x l such that { x k , x k +1 } ∩ { x l , x l +1 } = ∅ , x k → y → x k +1 and x l → y → x l +1 (by Lemmas 4(i) and 5(i)).(ii) x i +1 x i / ∈ D for all i ∈ [1 , n − y → { x i − , x i +1 } or { x i − , x i +1 } → y , then x i has no partner on the path C [ x i +1 , x i − ].5iv) If x i +1 x i − ∈ D , then d ( x i ) ≤ n − Lemma 7 . Let D be a digraph of order n ≥ C := x x . . . x n − x be an arbitrary cycle of length n − D and let y be thevertex not on C . Assume that y → { x , x n − } , x → y and d ( y, { x , x n − } ) = 0. Then D contains aHamiltonian bypass. Proof . The proof is by contradiction. Assume that D contains no Hamiltonian bypass. From Remark(i) and Lemmas 6(i), 4(i) it follows that for some j ∈ [4 , n − x j → y → x j +1 .Now we show that x n − x / ∈ D . Assume that this is not the case. Then x n − x ∈ D and d ( x n − ) ≤ n − y → { x , x n − } , the condition (*) implies that x and x n − are adjacent,i.e., x x n − ∈ D or x n − x ∈ D . If x x n − ∈ D , then D ( n,
2) = [ x x n − ; x x . . . x n − x yx n − ], andif x n − x ∈ D , then D ( n,
2) = [ x n − x ; x n − x x . . . x j yx j +1 . . . x n − x ]. In both cases we have aHamiltonian bypass, a contradiction. Therefore x n − x / ∈ D .Now, since x has no partner on C [ x , x n − ], by Lemma 5(i), d ( x ) ≥ n −
1. On the other hand, from d ( y, { x , x n − } ) = 0 , d ( y ) ≤ n − x x / ∈ D and x x n − / ∈ D (inparticular, a ( x , x n − ) = 0). Now using Lemma 2(ii) and Remark (ii) we obtain n − ≤ d ( x ) = d ( x , { y, x , x n − } ) + d ( x , C [ x , x n − ]) ≤ n − , which is a contradiction. Lemma 7 is proved. Lemma 8 . Let D be a digraph of order n ≥ C := x x . . . x n − x be an arbitrary cycle of length n − D and let y be the vertexnot on C . If d − ( y ) ≥ y is adjacent with four consecutive vertices of the cycle C , then D containsa Hamiltonian bypass. Proof . Suppose, on the contrary, that D contains no Hamiltonian bypass and no good cycle. UsingLemmas 6(i) and 4(i), without loss of generality, we can assume that { x n − , x } → y and y → { x , x } .By Remarks (ii) and (iii) we have(3) x i x i − / ∈ D for each i ∈ [1 , n −
1] and x (respectively, x ) has no partner on the path C [ x , x n − ](respectively, C [ x , x ]).If x n − x / ∈ D and x x / ∈ D , then using Lemma 2(iii) and (3) we obtain d ( x ) = d ( x , { y, x , x n − } ) + d ( x , C [ x , x n − ]) ≤ n − . Therefore, by condition (*), the vertices x , x are adjacent, since y → { x , x } ( x and x has a commonin-neighbour y ). This means that x x ∈ D . Since x has no partner on C [ x , x n − ], it follows from x → { x , x } and d ( x ) ≤ n − x x ∈ D . Similarly, we conclude that x n − x ∈ D whichcontradicts the assumption that x n − x / ∈ D . Assume therefore that x n − x ∈ D or x x ∈ D. (4)Now we prove that d ( x ) ≥ n −
1. Assume that this is not the case, that is d ( x ) ≤ n −
2. Thenagain by condition (*) x , x are adjacent because of y → { x , x } . Therefore x x ∈ D or x x ∈ D .If x x ∈ D , then it is not difficult to show that { x , x , . . . , x n − } → x , i.e., d ( x ) ≥ n −
1, acontradiction. Assume therefore that x x / ∈ D and x x ∈ D . Then C ′ := x x x . . . x n − yx is acycle of length n − x . Then d ( x ) ≤ n − x has no partner on C [ x , x ] (by (3)) and d − ( x , { x i , x i +1 } ) ≤ i ∈ [3 , n −
2] (by Lemma 4(i)),it follows that d − ( x , { y, x , x , . . . , x n − } ) = 0. Then x n − x ∈ D because of d − ( x ) ≥
2. From d − ( y ) ≥ x j ∈ C [ x , x n − ] such that x j → y → x j +1 .6herefore D ( n,
2) = [ x x ; x x x . . . x j yx j +1 . . . x n − x ] is a Hamiltonian bypass, a contradiction. Thiscontradiction proves that d ( x ) ≥ n − x n − x / ∈ D , by Remark (iv). From (4) it follows that the following two cases are possible: x x ∈ D (Case 1) or x x / ∈ D and x n − x ∈ D (Case 2). Case 1. x x ∈ D . Then d ( x ) ≤ n − x x / ∈ D and x n − x / ∈ D (if x n − x ∈ D , then D has a cycle of length n − x , and hence d ( x ) ≤ n − d ( x ) ≥ n − d ( x ) ≤ n − x n − x / ∈ D and x has no partner on C [ x , x n − ] (by (3)). Case 2. x x / ∈ D and x n − x ∈ D . It is easy to see that x n − x / ∈ D and x n − x n − / ∈ D .If yx n − ∈ D , then x n − has no partner on C [ x , x n − ]. This together with d ( x n − ) ≤ n −
2, and x n − x n − / ∈ D contradicts Lemma 5(i). Assume therefore that y and x n − are non-adjacent. Then, since d ( y ) ≤ n − x y ∈ D , we have that x x n − / ∈ D .Assume first that x x n − ∈ D . Then x n − x / ∈ D (for otherwise, the arc x n − x n − ∈ C [ x , x n − ] isa partner of x on C [ x , x n − ], a contradiction against (3)). Therefore x and x n − are non-adjacent.Now we have x i x ∈ D , where i ∈ [4 , n −
3] since d − ( x ) ≥ d − ( x , { y, x , x n − , x n − } ) = 0. It isnot difficult to see that d ( x ) ≥ n − n − ≤ d ( x ) = d ( x , { y, x , x , x n − } ) + d ( x , C [ x , x n − ]) ≤ n − , i.e., d ( x ) = n − d ( x , C [ x , x n − ]) = n −
5. By Lemma 2, x x and x n − x ∈ D . From d ( x ) = n − d ( x n − ) ≥ n , since x and x n − are non-adjacent and have a commonin-neighbour x n − . If x x n − ∈ D , then D ( n,
2) = [ x x ; x x n − x n − yx x . . . x n − x ], a contradiction.Assume therefore that x x n − / ∈ D . Now we consider the cycle C ′ := x n − x x n − yx x . . . x n − oflength n − x n − and x . Since d ( x n − ) ≥ n and x x n − / ∈ D (i.e., a ( x , x n − ) = 1), then d ( x n − , C ′ ) ≥ n −
1. Therefore, by Lemma 1, there is a cycle, say C ′′ , of length n − x . Then, since d ( x , C ′′ ) ≥ n −
1, by Lemma 4(ii) D contains a Hamiltonianbypass.Assume second that x and x n − are non-adjacent. Then, since d ( x n − ) ≤ n −
2, the condition (*)implies that x n − x / ∈ D . Then by Remark (ii) and Lemma 2(ii), we have d ( x ) = d ( x , { y, x , x } ) + d ( x , C [ x , x n − ]) ≤ n − . This contradicts Lemma 5(i) (because of (3)) and completes the proof of Lemma 8.From Lemmas 6, 7 and 8 immediately the following lemma follows:
Lemma 9 . Let D be a digraph of order n ≥ C := x x . . . x n − x be an arbitrarycycle of length n − D and let y be the vertex not on C . If the vertex y is adjacent with threeconsecutive vertices of the cycle C , then D contains a Hamiltonian bypass. Lemma 10 . Let D be a digraph of order n ≥ C := x x . . . x n − x be an arbitrarycycle of length n − D and let y be the vertex not on C . If D contains no Hamiltonian bypass and x i − x i +1 ∈ D for some i ∈ [1 , n − d ( x i , { x i − , x i +2 } ) = 0. Proof . The proof is by contradiction. Without loss of generality, we may assume that D has noHamiltonian bypass, x n − x ∈ D and a ( x , x ) ≥ a ( x , x n − ) ≥
1. If a ( x , x ) ≥ a ( x , x n − ) ≥ y is not adjacent with three consecutive vertices of C , by Remark (i) thereexists a vertex x k ∈ C [ x , x n − ] (respectively, x k ∈ C [ x , x n − ]) such that x k → y → x k +1 . It is not dif-ficult to see that C ′ := C [ x , x k ] yC [ x k +1 , x n − ] x is a cycle of length n − x , and x
7s adjacent with three consecutive vertices of C ′ , namely with x n − , x , x (respectively, x n − , x n − , x ),which is a contradiction against Lemma 9. Lemma 10 is proved. Lemma 11 . Let D be a digraph of order n ≥ C := x x . . . x n − x be an arbitrarycycle of length n − D and let y be the vertex not on C . D contains no Hamiltonian bypass, then x i +1 x i − / ∈ D for all i ∈ [1 , n − Proof . The proof is by contradiction. Without loss of generality, we may assume that x x ∈ D .Assume first that the vertex x has a partner on C [ x , x n − ], i.e., there is an x j ∈ C [ x , x n − ]such that x j → x → x j +1 . From d − ( y ) ≥ x k ∈ C [ x , x n − ] distinct from x j such that x k → y → x k +1 . Therefore, if k ≥ j + 1, then D ( n,
2) =[ x x ; x x . . . x j x x j +1 . . . x k yx k +1 . . . x n − x ], and if k ≤ j −
1, then D ( n,
2) = [ x x ; x x . . . x k yx k +1 . . . x j x x j +1 . . . x n − x ], a contradiction.Assume second that x has no partner on C [ x , x n − ]. Since x x ∈ D , Lemma 10 implies that x x / ∈ D and x n − x / ∈ D . Now using Lemma 2(iii) and Remark (ii) we obtain d ( x ) = d ( x , { y, x , x } ) + d ( x , C [ x , x n − ]) ≤ n − . This together with the condition (*) implies that d − ( x , C [ x , x n − ]) = 0. Therefore d − ( x ) ≤
2, whichcontradicts that d − ( x ) ≥
3. Lemma 11 is proved.
Proof of Theorem 12.
By Theorem 11 the digraph D contains a cycle of length n − n is even and D is isomorphic to the complete bipartite digraph (or to the complete bipartite digraph minus one arc)with partite sets of cardinalities n/ n/
2. If n ≤ D contains no cycle of length n −
1, then it isnot difficult to check that D contains a Hamiltonian bypass. Assume therefore that n ≥ D contains acycle of length n − C is an arbitrarycycle of length n − D and the vertex y is not on C , then there are not three consecutive vertices of C which are adjacent with y . Let C := x x . . . x n − x be an arbitrary cycle of length n − D and let y be the vertex not on C . Then, by Lemma 6(i), the following two cases are possible: There is a vertex x i and an integer a ≥ d ( y, { x i +1 , x i +2 , . . . , x i + a } ) = 0, x i − → y → x i and the vertices y , x i + a +1 are adjacent (Case I) or d ( y, { x i +1 , x i +2 , . . . , x i + a } ) = 0, y → { x i , x i + a +2 } , x i + a +1 y ∈ D and thevertices y , x i − are non-adjacent, where a ∈ [1 , n − a . We will first show that the theorem is true for a = 1. Case I. a = 1. Without loss of generality, we may assume that x n − → y → x n − , x , y are adjacentand y , x are non-adjacent. Since the vertex y is not adjacent with three consecutive vertices of C (Lemma 9), it follows that y, x n − also are non-adjacent. The condition (*) implies that x n − x / ∈ D ,since d ( y ) ≤ n − x n − y ∈ D .We show that x has a partner on C [ x , x n − ]. Assume that this is not the case. Then by Lemma5(i) we have d ( x ) ≥ n −
1, since x n − x / ∈ D and yx / ∈ D . On the other hand, using Lemma 2(ii) andRemark (ii), we obtain n − ≤ d ( x ) = d ( x , { x , x n − } ) + d ( x , C [ x , x n − ]) ≤ n − , which is a contradiction.So, indeed x has a partner on C [ x , x n − ]. Let the arc x k x k +1 ∈ C [ x , x n − ] be partner of x ,i.e., x k → x → x k +1 . Notice that k ∈ [4 , n −
4] (by Lemma 11). If yx ∈ D , then D ( n,
2) =8 yx n − ; yx x . . . x k x x k +1 . . . x n − x n − ], a contradiction. Assume therefore that yx / ∈ D . Then x y , yx ∈ D and y , x are non-adjacent, by Lemmas 9 and 6(i). This together with the condition (*) impliesthat x x / ∈ D , since x y ∈ D and d ( y ) ≤ n −
2. If x n − x ∈ D , then C ′ := x n − x yx . . . x k x x k +1 . . .x n − is a cycle of length n − x n − for which { x n − , y } → x n − . Then x n − x ∈ D ,by Lemmas 6(i) and 4, i.e., x n − is adjacent with three consecutive vertices of C ′ , which is contrary toLemma 9. Assume therefore that x n − x / ∈ D . Now we show that x also has a partner on C [ x , x n − ].Assume that this is not the case. Then, since x x n − / ∈ D (by Lemma 11) and x x / ∈ D , using Lemma2(iii) and Remark (ii) we obtain d ( x ) = d ( x , { y, x , x , x n − } ) + d ( x , C [ x , x n − ]) ≤ n − . This together with x → { x , x k +1 } and the condition (*) implies that x , x k +1 are adjacent. It is easyto see that x k +1 x ∈ D . By a similar argument, we conclude that x n − x ∈ D , which contradicts thefact that x n − x / ∈ D . Thus, x also has a partner on C [ x , x n − ]. Therefore by Multi-Insertion Lemmathere is a ( x , x n − )-path with vertex set V ( C ), which together with the arcs yx n − and yx forms aHamiltonian bypass. This completes the discussion of induction first step for ( a = 1) Case I.Now we consider the induction first step for Case II. Case II. a = 1. Without loss of generality, we may assume that y → { x , x n − } , x y ∈ D and d ( y, { x , x , x n − } ) = 0. By induction first step of Case I, we may assume that y, x also are non-adjacent. This together with d ( y ) ≤ n − x y ∈ D and the condition (*) implies that d + ( x , { x , x , x n − } ) = 0 , (5)and hence, by Lemma 11, in particular, the vertices x , x are non-adjacent. If x n − x ∈ D , then thecycle C ′ := x n − x x yx . . . x n − has length n − x n − and { x n − , y } → x n − → x ,i.e., for the cycle C ′ and vertex x n − the considered induction first step of Case I holds. Assume thereforethat x n − x / ∈ D . Then x , x n − are non-adjacent (Lemma 11). It is not difficult to see that x has apartner on C [ x , x n − ]. Indeed, for otherwise from Lemma 5(i) it follows that d ( x n − ) ≥ n − n − ≤ d ( x ) = d ( x , { x , x n − } ) + d ( x , C [ x , x n − ]) ≤ n − , which is a contradiction. Thus, indeed x has a partner on C [ x , x n − ]. Let the arc x k x k +1 ∈ C [ x , x n − ]be a partner of x . Note that k ∈ [4 , n −
4] (by Lemma 11). Therefore neither the vertex x nor the arc x x has a partner on C [ x , x n − ] (for otherwise, by Multi-Insertion Lemma, there is an ( x , x n − )-pathwith vertex set V ( C ), which together with the arcs yx and yx n − forms a Hamiltonian bypass). Recallthat a ( x , x ) = 0 and x x / ∈ D (by (5). Now using Lemma 2(ii) and Remark (ii) we obtain d ( x ) = d ( x , { y, x , x } ) + d ( x , C [ x , x n − ]) ≤ n − . This together with x → { x , x k +1 } and the condition (*) implies that x and x k +1 are adjacent. Then x k +1 x ∈ D (if x x k +1 ∈ D , then the arc x x has a partner on C [ x , x n − ]). By a similar argument,we conclude that { x n − , x n − } → x . Then C ′ := x n − x yx x . . . x k x x k +1 . . . x n − is a cycle of length n −
1, which does not contain the vertex x n − and d ( x n − , { x n − , x , y } ) = 3, a contradiction againstLemma 9 and hence, the discussion of case a = 1 is completed. The induction hypothesis . Now we suppose that the theorem is true if D contains a cycle C := x x . . . x n − x of length n − y for which there is a vertex x i such that d ( y, { x i +2 , x i +3 , . . . , x i + j } ) = 0 and (i) x i → y → x i +1 and the vertices y, x i + j +1 are adjacent or (ii)9 → { x i +1 , x i + j +2 } and x i + j +1 y ∈ D , where 2 ≤ j ≤ a ≤ n − Claim . Let C := x x . . . x n − x be an arbitrary cycle of length n − D and let y be the vertexnot on C and let d ( y, { x , x , . . . , x a } ) = 0, where a ≥ x n − y, yx n − ∈ D and the vertices y and x a +1 are non-adjacent or (ii) x a +1 y, yx a +2 and yx n − ∈ D , then x k − x k +1 / ∈ D for all k ∈ [1 , a ]. Proof of the claim . Suppose, on the contrary, that x k − x k +1 ∈ D for some k ∈ [1 , a ], then C ′ := x n − yx n − x . . . x k − x k +1 . . . x n − or C ′′ := x n − x . . . x k − x k +1 . . . x a +1 yx a +2 . . . x n − x n − is a cycle oflength n − x k for (i) and (ii), respectively. Therefore, d ( x k ) ≤ n − x k is not adjacent with vertices x k +2 , x k +3 , . . . , x k + a , x k − , x k − , . . . x k − a .In particular, d − ( x k , { x k +1 , x k +2 , . . . , x a +1 } ) = 0 and d − ( x k , C [ x n − , x k − ]) = 0 (it is easy to show thatin both cases x n − x k / ∈ D ). Since d ( x k ) ≤ n − x k − x k / ∈ D , by Lemma 5(i), the vertex x k has apartner on C [ x a +2 , x n − ], say the arc x j x j +1 ∈ C [ x a +2 , x n − ] is a partner of x k , i.e., x j x k , x k x j +1 ∈ D .Therefore x n − x . . . x k − x k +1 . . . x a x a +1 . . . x j x k x j +1 . . . x n − x n − is a cycle of length n − y , for which d ( y, C [ x , x a ] − { x k } ) = 0 and x n − y, yx n − ∈ D , x a +1 , y are non-adjacent or yx n − , x a +1 y, yx a +2 ∈ D for (i) and (ii), respectively. Therefore, by the induction hypothesis D containsa Hamiltonian bypass, a contradiction to our assumption. The claim is proved. Case I . Without loss of generality, we may assume that d ( y, { x , x , . . . , x a +1 } ) = 0, where a ≥ x n − y, yx ∈ D and the vertices y, x a +2 are non-adjacent.Notice, the condition (*) implies that for all i ∈ [2 , a + 1], x n − x i / ∈ D , since x n − y ∈ D , d ( y ) ≤ n − x i , y are non-adjacent. Subcase I.1 . There are integers k and l with 1 ≤ l < k ≤ a + 2 such that x k x l ∈ D . Without loss ofgenerality, we assume that k − l is as small as possible. From Remark (ii) and Lemma 11 it follows that k − l ≥
3. If every vertex x i ∈ C [ x l +1 , x k − ] has a partner on the path P := x k x k +1 . . . x n − yx . . . x l ,then by Multi-Insertion Lemma there exists an ( x k , x l )-Hamiltonian path, which together with the arc x k x l forms a Hamiltonian bypass. Assume therefore that some vertex x i ∈ C [ x l +1 , x k − ] has no partneron P . From the minimality of k − l ≥ x i − ∈ C [ x l , x k ] and a ( x i , x i − ) = 0or x i +2 ∈ C [ x l , x k ] and a ( x i , x i +2 ) = 0. Therefore by the minimality of k − l we have d ( x i , C [ x l , x k ]) ≤ k − l − . (6)Since x i has no partner on the path C [ x k +1 , x n − ], and if l ≥ C [ x , x l − ], using Lemma 2 withthe fact that x n − x i / ∈ D we obtain d ( x i , C [ x k +1 , x n − ] ≤ n − k − l ≥ , then d ( x i , C [ x , x l − ]) ≤ l. The last two inequalities together with (6) give: if l ≥
2, then d ( x i ) ≤ n −
2, and if l = 1, then d ( x i ) ≤ n −
3. Thus, d ( x i ) ≤ n −
2. In addition, Claim 1 and x n − x i / ∈ D imply that x i − x i / ∈ D .Therefore, by Lemma 5(i), x i has a partner on P , which is contrary to our assumption. Subcase I.2 . For any pair of integers k and l with 1 ≤ l < k ≤ a + 2, x k x l / ∈ D . Then it is easy tosee that for each x i ∈ C [ x , x a +1 ]), d ( x i , C [ x , x a +2 ]) ≤ a, (7)since x i − ∈ C [ x , x a +2 ] and a ( x i , x i − ) = 0 or x i +2 ∈ C [ x , x a +2 ] and a ( x i , x i +2 ) = 0.We first show that every vertex x i ∈ C [ x , x a +1 ] has a partner on C [ x a +3 , x n − ]. Assume that this isnot the case, i.e., some vertex x i ∈ C [ x , x a +1 ] has no partner on C [ x a +3 , x n − ]. Then, since x n − x i / ∈ D ,by Lemma 2(ii) we have that d ( x i , C [ x a +3 , x n − ]) ≤ n − a −
3. This inequality together with (7) gives d ( x i ) ≤ n −
3, a contradiction against Lemma 5(i), since x i − x i / ∈ D . Thus each vertex x i ∈ C [ x , x a +1 ]has a partner on C [ x a +3 , x n − ]. Therefore, by Multi-Insertion Lemma there is an ( x a +3 , x n − )-path, say10 , with vertex set V ( C ) − { x , x a +2 } . If yx a +2 ∈ D , then [ yx ; yx a +2 Rx ] is a Hamiltonian bypass.Assume therefore that yx a +2 / ∈ D . Then x a +2 y ∈ D . By Lemma 6(i) and by the induction hypothesis,we have yx a +3 ∈ D and d ( y, { x a +4 , x a +5 } ) = 0. This together with x a +2 y ∈ D , d ( y ) ≤ n − d + ( x a +2 , , { x a +4 , x a +5 } ) = 0 , (8)in particular, by Lemma 11, a ( x a +2 , a a +4 ) = 0. Since yx a +3 ∈ D and each vertex x i ∈ C [ x , x a +1 ] hasa partner on C [ x a +3 , x n − ], to show that D contains a Hamiltonian bypass, by Multi-Insertion Lemmait suffices to prove that x a +2 also has a partner on C [ x a +3 , x n − ]. Assume that x a +2 has no partner on C [ x a +3 , x n − ]. Then, since the vertices x a and x a +2 are non-adjacent (Claim 1 and Lemma 11), fromLemma 5(i) it follows that d ( x a +2 ) ≥ n −
1. On the other hand, using (7), (8), d ( x a +2 , { x a , x a +4 } ) = 0and Lemma 2, we obtain n − ≤ d ( x a +2 ) = d ( x a +2 , C [ x , x a +1 ]) + d ( x a +2 , { y, x a +3 ) + d ( x a +2 , C [ x a +5 , x n − ]) ≤ n − , a contradiction. So, x a +2 also has a partner on C [ x a +3 , x n − ] and the discussion of Case I is completed. Case II . Without loss of generality, we assume that d ( y, { x , x , . . . , x a } ) = 0, where a ≥ x a +1 y ∈ D and y → { x n − x a +2 } .By the considered Case I, without loss of generality, we may assume that d ( y, { x n − , x a +3 , x a +4 } ) = 0.Since x a +1 y ∈ D , d ( y ) ≤ n − d ( y, { x , x , . . . , x a , x a +3 , x a +4 , x n − } ) = 0, the condition (*) impliesthat d + ( x a +1 , { x , x , . . . , x a , x a +3 , x a +4 , x n − } ) = 0 . (9) Subcase II.1 . There are integers k and l with 1 ≤ l < k ≤ a + 1 such that x k x l ∈ D . By (9), k = a + 1. Without loss of generality, we assume that k − l is as small as possible. By Remark (ii) andLemma 11 we have k − l ≥ C [ x l +1 , x k − ] has a partner on the path P := x k x k +1 . . . x a +1 yx a +2 . . . x n − x . . . x l . Assume that this is not the case and let x i ∈ C [ x l +1 , x k − ] have no partner on P . Then, since x i − x i / ∈ D (Claim 1), from Lemma 5(i) and the minimality of k − l it follows that d ( x i ) ≥ n −
1. On the other hand, using the minimality of k − l and the fact that x i − ∈ C [ x l , x k ]) and a ( x i , x i − ) = 0 or x i +2 ∈ C [ x l , x k ] and a ( x i , x i +2 ) = 0 we obtain d ( x i , C [ x l , x k ]) ≤ k − l − . In addition, by Lemma 2 and x a +1 x i / ∈ D we also have d ( x i , C [ x k +1 , x a +1 ]) ≤ a − k + 1 and d ( x i , C [ x a +2 , x l − ]) ≤ n − a + l − . Summing the last three inequalities gives d ( x i ) ≤ n −
2, which contradicts that d ( x i ) ≥ n −
1. Thus,indeed each vertex x i ∈ C [ x l +1 , x k − ] has a partner on P . Then, by Multi-Insertion Lemma, there is an( x k , x l )-Hamiltonian path , which together with the arc x k x l forms a Hamiltonian bypass. Subcase II.2 . There are no i and j such that 1 ≤ i < j ≤ a + 1 and x j x i / ∈ D . If every vertex x i ∈ C [ x , x a +1 ]) has a partner on C [ x a +2 , x n − ], then by Multi-Insertion Lemma there is an ( x a +2 , x n − )-path, say R , with vertex set V ( C ). Therefore [ yx n − ; yR ] is a Hamiltonian bypass. Assume thereforethat there is a vertex x i ∈ C [ x , x a +1 ] which has no partner on C [ x a +2 , x n − ].Let x i − x i / ∈ D , then from Lemma 5(i) it follows that d ( x i ) ≥ n − d ( x i , C [ x , x a +1 ]) = a −
1. Using Lemma 2 we obtain that if x i = x a +1 , then n − ≤ d ( x i ) = d ( x i , C [ x , x a +1 ]) + d ( x i , C [ x a +2 , x n − ]) ≤ n − , x a +1 x a +3 / ∈ D , if x i = x a +1 , then n − ≤ d ( x a +1 ) = d ( x a +1 , C [ x , x a +1 ]) + d ( x a +1 , { y, x a +2 ) + d ( x a +1 , C [ x a +3 , x n − ]) ≤ n − , a contradiction.Assume second that d ( x i , C [ x , x a +1 ]) = a . Then from Claim 1 and Lemma 11 it follows that a = 2, x i = x , d ( x , { x , x } ) = 2 and d ( x , { x n − } ) = 0. Then n − ≤ d ( x ) = d ( x , { x , x } ) + d ( x , C [ x , x n − ]) ≤ n − , a contradiction.Let now x i − x i ∈ D . Then, by Claim 1, x i = x and x n − x ∈ D . We consider the cycle C ′ := x n − x n − x x . . . x a x a +1 yx a +2 . . . x n − of length n − x n − . Then { x n − , y } → x n − and x n − x ∈ D , i.e., for the cycle C ′ and the vertex x n − Case I holds since |{ x , x , . . . , x a +1 }| = a .The discussion of Case II is completed and with it the proof of the theorem is also completed. The following two examples of digraphs show that if the minimal semi-degree of a digraph is equal toone, then the theorem is not true:(i) Let D (7) be a digraph with vertex set { x , x , . . . , x , y } and let x x . . . x x be a cycle of length6 in D (7). Moreover, N + ( y ) = { x , x , x } , N − ( y ) = { x , x , x } , x x , x x , x x ∈ D (7) and D (7)has no other arcs. Note that d − ( x ) = d − ( x ) = d − ( x ) = 1 and D (7) contains no dominated pair ofnon-adjacent vertices. It is not difficult to check that D (7) contains no Hamiltonian bypass.(ii) Let D ( n ) be a digraph with vertex set { x , x , . . . , x n } and let x x . . . x n x be a Hamiltonian cy-cle in D ( n ). Moreover, D ( n ) also contains the arcs x x , x x , . . . , x n − x n (or x x , x x , . . . , x n − x n − ,x n − x and D ( n ) has no other arcs. Note that D ( n ) contains no dominated pair of non-adjacent vertices, d − ( x ) = d + ( x ) = 1. It is not difficult to check that D ( n ) contains no Hamiltonian bypass.We believe that Theorem 12 also is true if we require that the minimum in-degree at least two, insteadof three .In [2] and [3] Theorem 13 and Theorem 14 were proved, respectively. Theorem 13 (Bang-Jensen, Gutin, H. Li [2]). Let D be a strong digraph of order n ≥
3. Supposethat min { d + ( x ) + d − ( y ) , d − ( x ) + d + ( y ) } ≥ n for any pair of non-adjacent vertices x, y with a commonout-neighbour or a common in-neighbour, then D is Hamiltonian. Theorem 14 (Bang-Jensen, Guo, Yeo [3]). Let D be a strong digraph of order n ≥
3. Suppose that d ( x ) + d ( y ) ≥ n − min { d + ( x ) + d − ( y ) , d − ( x ) + d + ( y ) } ≥ n − x, y with a common out-neighbour or a common in-neighbour, then D is Hamiltonian.In [9] and [10] the following results were proved: Theorem 15 ( [9]). Let D be a strong digraph of order n ≥ min { d + ( x ) + d − ( y ) , d − ( x ) + d + ( y ) } ≥ n for any pair of non-adjacent vertices x, y with a commonout-neighbour or a common in-neighbour. Then either D contains a pre-Hamiltonian cycle or n is evenand D = K ∗ n/ ,n/ . Theorem 16 ( [10]). Let D be a strong digraph of order n ≥ d ( x )+ d ( y ) ≥ n − min { d + ( x )+ d − ( y ) , d − ( x )+ d + ( y ) } ≥ n − x, y with a common out-neighbour or a common in-neighbour. Then D contains a pre-Hamiltonian12ycle or a cycle of length n − Problem . Characterize those digraphs which satisfy the condition of Theorem 13 or 14 but have noHamiltonian bypass.
Acknowledgment . The autors will be grateful to the colleagues who will make any mathematicaland grammatical comments.
References [1] J. Bang-Jensen, G. Gutin, Digraphs: Theory, Algorithms and Applications, Springer, 2000.[2] J. Bang-Jensen, G. Gutin, H. Li, ”Sufficient conditions for a digraph to be Hamiltonian”,
J. GraphTheory , vol. 22 no. 2, pp. 181-187, 1996.[3] J. Bang-Jensen, Y. Guo, A.Yeo, ”A new sufficient condition for a digraph to be Hamiltonian”,
Discrete Applied Math. , vol. 95, pp. 77-87, 1999.[4] A.Benhocine, ”On the existence of a specified cycles in digraphs with constraints on degrees”,
Journalof Graph Theory , vol. 8, pp.101-107, 1984.[5] J.A. Bondy, C. Thomassen, ”A short proof of Meyniel’s theorem”,
Discrete Mathematics , vol. 19,no. 1, pp. 85-92, 1977.[6] S.Kh. Darbinyan, ”A sufficient condition for the Hamiltonian property of digraphs with large semide-grees”,
Akad. Nauk Armyan. SSR Dokl. , vol. 82, no. 1, pp. 6-8, 1986 (see also arXiv: 1111.1843v1[math.CO] 8 Nov 2011).[7] S.Kh. Darbinyan, ”On Hamiltonian bypasses in digraphs satisfying Meyniel-like conditions”,
Math.Problems in Computer Science , vol. 20, pp.7-19, 1998 (in Russian)(see also Proceedings of 5-thscience-technical conference for yang researchers, p. 23, Tsaghkadzor, Armenia, 1986).[8] S.Kh. Darbinyan, ”On the specified cycles in oriented graphs”,
Akad. Nauk Armyan. SSR Dokl. , vol.84, no. 1, pp. 51-55, 1987 (in Russian).[9] S.Kh. Darbinyan, I.A. Karapetyan, ”On longest non-Hamiltonian cycles in digraphs with the condi-tions of Bang-Jensen, Gutin and Li”,
Preprint available at htte: arXiv 1207.5643v2 [math.CO] , 20Sep 2012.[10] S.Kh. Darbinyan, I.A. Karapetyan, ”A note on long non-Hamiltonian cycles in one class of digraphs”,
Preprint available at htte: arXiv 1209.4456v1 [math.CO] , 20 Sep 2012.[11] R. H¨aggkvist, C. Thomassen, ”On pancyclic digraphs”,
J. Combin. Theory Ser. B , vol. 20, pp. 20-40,1976.[12] A. Ghouila-Houri, ”Une condition suffisante d’existence d’un circuit hamiltonien”,
C. R. Acad. Sci.Paris Ser. A-B , no. 251, pp. 495-497, 1960.[13] M. Meyniel, ”Une condition suffisante d’existence d’un circuit hamiltonien dans un graphe oriente”,
J. Combin. Theory Ser. B , vol. 14, pp. 137-147, 1973.[14] C.St.J.A. Nash-Williams, ”Hamilton circuits in graphs and digraphsit. The many facts of graphtheory”,
Springer Lecture Notes . 110, pp.237-243, 1969.1315] C. Thomassen, ”Long cycles in digraphs”,
Proc. London Math. Soc. , vol. 3, no. 42, pp. 231-251,1981.[16] D.R. Woodall, ”Sufficient conditions for circuits in graphs”,