The structure of strong k -quasi-transitive digraphs with large diameters
aa r X i v : . [ m a t h . C O ] J un The structure of strong k -quasi-transitive digraphs withlarge diameters ✩ Ruixia Wang ∗ , Hui Zhang School of Mathematical Sciences, Shanxi University, Taiyuan, Shanxi, PR China
Abstract
Let k be an integer with k ≥
2. A digraph D is k -quasi-transitive, if for anypath x x . . . x k of length k , x and x k are adjacent. Suppose that there existsa path of length at least k + 2 in D . Let P be a shortest path of length k + 2in D . Wang and Zhang [Hamiltonian paths in k -quasi-transitive digraphs,Discrete Mathematics, 339(8) (2016) 2094–2099] proved that if k is even and k ≥
4, then D [ V ( P )] and D [ V ( D ) \ V ( P )] are both semicomplete digraphs.In this paper, we shall prove that if k is odd and k ≥
5, then D [ V ( P )]is either a semicomplete digraph or a semicomplete bipartite digraph and D [ V ( D ) \ V ( P )] is either a semicomplete digraph, a semicomplete bipartitedigraph or an empty digraph. Keywords: k -quasi-transitive digraph; semicomplete digraph; semicompletebipartite digraph
1. Terminology and introduction
We shall assume that the reader is familiar with the standard terminologyon digraphs and refer the reader to [2] for terminology not defined here. Weonly consider finite digraphs without loops or multiple arcs. Let D be adigraph with vertex set V ( D ) and arc set A ( D ). For any x, y ∈ V ( D ), wewill write x → y if xy ∈ A ( D ), and also, we will write xy if x → y or y → x .For disjoint subsets X and Y of V ( D ), X → Y means that every vertex of ✩ This work is supported by the National Natural Science Foundation for Young Scien-tists of China (11401354)(11501490)(11501341). ∗ Corresponding author.
Email address: [email protected] (Ruixia Wang)
Preprint submitted to ** June 12, 2020 dominates every vertex of Y , X ⇒ Y means that there is no arc from Y to X and X Y means that both of X → Y and X ⇒ Y hold. For subsets X, Y of V ( D ), we define ( X, Y ) = { xy ∈ A ( D ) : x ∈ X, y ∈ Y } . Let H be asubset of V ( D ) and x ∈ V ( D ) \ H . We say that x and H are adjacent if x and some vertex of H are adjacent. For S ⊆ V ( D ), we denote by D [ S ] thesubdigraph of D induced by the vertex set S . The converse of D , ←− D , is thedigraph which is obtained from D by reversing all arcs.Let x and y be two vertices of V ( D ). The distance from x to y in D ,denoted d ( x, y ), is the minimum length of an ( x, y )-path, if y is reachablefrom x , and otherwise d ( x, y ) = ∞ . The distance from a set X to a set Y ofvertices in D is d ( X, Y ) = max { d ( x, y ) : x ∈ X, y ∈ Y } . The diameter of D is diam( D ) = d ( V ( D ) , V ( D )). Clearly, D has finite diameter if and only if itis strong.Let P = y y . . . y m be a path or a cycle of D . For i < j , y i , y j ∈ V ( P )we denote the subpath y i y i +1 . . . y j of P by P [ y i , y j ]. Let Q = q q . . . q n bea vertex-disjoint path or cycle with P in D . If there exist y i ∈ V ( P ) and q j ∈ V ( Q ) such that y i q j ∈ A ( D ), then we will use P [ y , y i ] Q [ q j , q n ] to denotethe path y y . . . y i q j q j +1 . . . q n .A digraph is quasi-transitive, if for any path x x x of length 2, x and x are adjacent. The concept of k -quasi-transitive digraphs was introducedin [6] as a generalization of quasi-transitive digraphs. A digraph is k -quasi-transitive , if for any path x x . . . x k of length k , x and x k are adjacent. The k -quasi-transitive digraph has been studied in [4, 6, 7, 8, 9].The following theorem completely characterizes strong quasi-transitivedigraphs in recursive sense. Theorem 1.1. [1] Let D be a strong quasi-transitive digraph. Then there ex-ists a strong semicomplete digraph S with vertices { v , v , . . . , v s } and quasi-transitive digraphs { Q , Q , . . . , Q s } such that Q i is either a vertex or isnon-strong and D = S [ Q , Q , . . . , Q s ] , where Q i is substituted for v i , i =1 , , . . . , s . In [3], Galeana-S´anchez et al. characterized strong 3-quasi-transitive di-graphs. Let F n be a digraph with vertex set { x , x , . . . , x n } and arc set { x x , x x , x x } ∪ { x i x , x x i , i = 3 , , . . . , n } , where n ≥ Theorem 1.2. [3] Let D be a strong 3-quasi-transitive digraph of order n .Then D is either a semicomplete digraph, a semicomplete bipartite digraph,or isomorphic to F n . D be a strong k -quasi-transitive digraph with diam( D ) ≥ k +2. Thenthere exist two vertices u, v ∈ V ( D ) such that d ( u, v ) = k + 2. Let P be ashortest ( u, v )-path in D . In [9], Wang and Zhang proved that if k ( ≥
4) iseven, then D [ V ( P )] and D [ V ( D ) \ V ( P )] are both semicomplete digraphs.In Section 2, we prove the following theorem. Theorem 1.3.
Let k be an odd integer with k ≥ , D be a strong k -quasi-transitive digraph with diam ( D ) ≥ k + 2 and u, v ∈ V ( D ) such that d ( u, v ) = k + 2 . If P is a shortest ( u, v ) -path, then D [ V ( P )] is either a semicompletedigraph or a semicomplete bipartite digraph and D [ V ( D ) \ V ( P )] is either asemicomplete digraph, a semicomplete bipartite digraph or an empty digraph.
2. Main results
We begin with a rather trivial observation.
Remark 2.1.
A digraph D is k -quasi-transitive if and only if ←− D is k -quasi-transitive. Proposition 2.2. [4] Let k be an integer with k ≥ , D be a k -quasi-transitive digraph and u, v ∈ V ( D ) such that d ( u, v ) = k + 2 . Suppose that P = x x . . . x k +2 is a shortest ( u, v ) -path, where u = x and v = x k +2 . Theneach of the following holds: (1) x k +2 → x k − i , for every odd i such that ≤ i ≤ k . (2) x k +1 → x k − i , for every even i such that ≤ i ≤ k . From the above proposition, we have seen that for a longer path, thereare some structural property in a k -quasi-transitive digraph. There exist twovertices u, v such that d ( u, v ) = k + 2 in a strong digraph D with diam( D ) ≥ k + 2. So for the rest of this paper, let k be an odd integer with k ≥ D denote a strong k -quasi-transitive digraph with diam( D ) ≥ k + 2 and P = x x . . . x k +2 denote a shortest ( u, v )-path in D , where u = x and v = x k +2 . Denote O ( P ) = { x , x , . . . , x k +2 } and E ( P ) = { x , x , . . . , x k +1 } .First we show the following structural property on D [ V ( P )]. Lemma 2.3.
Let x t and x s be two arbitrary vertices in V ( P ) with s > t .The following three statements hold. D [ V ( P )] contains a semicomplete bipartite digraph as its subdigraph withbipartition ( O ( P ) , E ( P )) . (2) There is a path of length k − from x s to x t in D [ V ( P )] when the parityof s and t are different. Moreover, for any x, y ∈ V ( D ) \ V ( P ) , if y → x s and x t → x , then xy . (3) There is a path of length k − from x s to x t in D [ V ( P )] when the parityof s and t are same. Moreover, for any x ∈ V ( D ) \ V ( P ) , if x → x s ,then xx t and x → x t if s ≥ t + 4 ; if x t → x , then xx s and x s → x if s ≥ t + 4 .Proof. (1) It suffices to show that, for any x i ∈ E ( P ) and x j ∈ O ( P ), x i x j .By Proposition 2.2, x k +2 → { x , x , . . . , x k − } and x k +1 → { x , x , . . . , x k − } .So we assume 0 ≤ i ≤ k − ≤ j ≤ k . Clearly, x j − x j and x j +1 x j .Now assume that i ≤ j − i ≥ j + 3. First consider the case 0 ≤ i ≤ j −
3. For i = 0, we proof x j → x by induction on j . The length ofthe path P [ x , x k +2 ] x is k , which implies that x x and further x → x since P is minimal. For the inductive step, let us suppose that x j → x for 3 ≤ j ≤ k −
2. The path P [ x j +2 , x k +1 ] P [ x , x j ] x implies x j +2 x andfurther x j +2 → x . For i ≥
2, the path P [ x j , x k +1 ] P [ x i +1 , x j − ] P [ x , x i ]implies x i x j . Now consider the case i ≥ j + 3. By the above argument,we know that x i − → x . The path P [ x i , x k +1 ] P [ x j +2 , x i − ] P [ x , x j ] implies x i x j . Therefore, D [ V ( P )] contains a semicomplete bipartite digraph as itssubdigraph with bipartition ( O ( P ) , E ( P )).(2) Assume x s ∈ E ( P ) and x t ∈ O ( P ). If s − t = 1, then P [ x s , x p ] P [ x p − k +2 , x t ]is a path of length k −
2, where p = k + 1 when s = k + 1 and p = k − s < k + 1. If s − t = 3, then P [ x s , x k +1 ] P [ x , x t ] is a path of length k − s − t ≥
5, then P [ x s , x k +1 ] P [ x t +2 , x s − ] P [ x , x t ] is a path of length k − x s ∈ O ( P ) and x t ∈ E ( P ), we can find the desired path.Denote by Q the path from x s to x t of length k − D [ V ( P )]. Let x, y ∈ V ( D ) \ V ( P ). If y → x s and x t → x , then the path yQx implies xy .(3) Assume x s , x t ∈ O ( P ). If s ≤ k , then, by (2), there is a path Q of length k − x s +1 to x t in D [ V ( P )]. By the proof of (2), observethat we can find such a path Q so that x s / ∈ V ( Q ). So x s Q is the desiredpath. Now assume s = k + 2. For t = k , x k +2 P [ x , x k ] is the desiredpath. For t ∈ { , , . . . , k − } , x k +2 P [ x t +1 , x k − ] P [ x , x t ] is the desired path.Analogously, if x s , x t ∈ E ( P ), we can find the desired path.4enote by R the path from x s to x t of length k − D [ V ( P )]. Let x ∈ V ( D ) \ V ( P ) be arbitrary. If x → x s , then the path xR implies xx t and x → x t if s ≥ t + 4 as P is minimal; if x t → x , then the path Rx implies xx s and x s → x if s ≥ t + 4 as P is minimal. Lemma 2.4.
If there exist two vertices x i , x j ∈ E ( P ) or x i , x j ∈ O ( P ) suchthat x i x j , then D [ V ( P )] is a semicomplete digraph and for any ≤ t + 1 β + 1, then x α → x β by the minimality of P .First we show that x k +2 → { x , x , . . . , x k } . We claim that for some q ≥ x k +2 → x q , then x k +2 → x q − . In fact, the length of the path x k +2 P [ x q , x k +1 ] P [ x , x q − ] is k . So x k +2 x q − and x k +2 → x q − . Next we claimthat for some p ≥ x p → x , then x p +2 → x (if x p +2 exists). In fact, thelength of the path P [ x p +2 , x k +2 ] P [ x , x p ] x is k and so x p +2 → x . Supposethat x i , x j ∈ O ( P ). If j = k + 2 or i = 1, then, by above two claims, we canobtain that x k +2 → x . Now assume that j ≤ k and i ≥
3. Note that thelength of the path P [ x j +2 , x k +2 ] P [ x i +1 , x j ] x i P [ x , x i − ] is k . So x j +2 → x i − .Repeating this way and using above two claims, we can obtain x k +2 → x .Suppose that x i , x j ∈ E ( P ). Analogously, we can obtain x k +1 → x . Notethat the length of the path x k +2 P [ x , x k +1 ] x x is k . So x k +2 → x . Next weshow x k +2 → x k − p for 0 ≤ p ≤ k − p . For p = 0, thepath x k +2 P [ x , x k ] implies x k x k +2 and x k +2 → x k . For the inductive step, letus suppose that x k +2 → x k − p for even p and 0 ≤ p ≤ k −
5. Then the pathof x k +2 P [ x k − p , x k +1 ] P [ x , x k − ( p +2) ] implies x k +2 x k − ( p +2) and x k +2 → x k − ( p +2) .From now on, we have shown that x k +2 → { x , x , . . . , x k } .Now we show that for any 1 ≤ t + 1 < s ≤ k + 2, x s → x t . It sufficesto show that, for any x t , x s ∈ O ( P ) or x t , x s ∈ E ( P ), there exists a path oflength k from x s to x t since P is minimal. Suppose that x s , x t ∈ O ( P ). If s = k + 2, then we are done. Assume s ≤ k . If s − t = 2, then the lengthof the path P [ x s , x k +2 ] P [ x , x t ] is k . If s − t ≥
4, then the length of thepath P [ x s , x k +2 ] P [ x t +2 , x s − ] P [ x , x t ] is k . Suppose that x s , x t ∈ E ( P ). Firstconsider s = k + 1. If t = 0, then the length of the path x k +1 x k +2 P [ x , x k ] x is k . If t ≥
2, then the length of the path x k +1 x k +2 P [ x t +1 , x k ] P [ x , x t ] is k . Now consider that s ≤ k −
1. If s − t = 2, then the length of the5ath P [ x s , x k +1 ] P [ x , x t ] is k . If s − t ≥
4, then the length of the path P [ x s , x k +1 ] P [ x t +2 , x s − ] P [ x , x t ] is k .From now on, we have shown that D [ V ( P )] is a semicomplete digraphand x s → x t for 1 ≤ t + 1 < s ≤ k + 2.According to Lemmas 2.3 and 2.4, we can easily obtain the followingtheorem. Theorem 2.5.
The digraph D [ V ( P )] is either a semicomplete digraph or asemicomplete bipartite digraph. In the rest of the paper, we study the structure of D [ V ( D ) \ V ( P )]. Lemma 2.6.
Let H be a digraph and u, v ∈ V ( H ) such that d ( u, v ) = n with n ≥ in H . Let Q = x x . . . x n be a shortest ( u, v ) -path in H . If H [ V ( Q )] is a semicomplete digraph, then, for any x i , x j ∈ V ( Q ) with ≤ i < j ≤ n ,there exists a path of length p from x j to x i with p ∈ { , , . . . , n − } in H [ V ( Q )] .Proof. We proof the result by induction on n . For n = 4, it is not difficultto check that the result is true. Suppose n ≥
5. Assume j − i = n . Itmust be j = n and i = 0. Then the length of the path x n P [ x , x p ] x is p , with p ∈ { , , . . . , n − } . Now assume 1 ≤ j − i ≤ n −
1. Then x i , x j ∈ V ( H [ x , x , . . . , x n − ]) or x i , x j ∈ V ( H [ x , x , . . . , x n ]). Without lossof generality, assume that x i , x j ∈ V ( H [ x , x , . . . , x n − ]). By induction,there exists a path of length p from x j to x i with p ∈ { , , . . . , n − } . Nowwe only need to show that there exists a path of length n − x j to x i . If j − i = 1, then P [ x j , x n − ] P [ x , x i ] is the desired path. If j − i = 2,then P [ x j , x n ] P [ x , x i ] is the desired path. If 3 ≤ j − i ≤ n −
1, then P [ x j , x n ] P [ x i +2 , x j − ] P [ x , x i ] is the desired path.By Lemma 2.6, we can obtain the following lemma. Lemma 2.7.
Suppose that D [ V ( P )] is a semicomplete digraph. For any x ∈ V ( D ) \ V ( P ) and x i ∈ V ( P ) , if x → x i , then x and every vertexof { x , x , . . . , x i − } are adjacent; if x i → x , then x and every vertex of { x i +1 , x i +2 , . . . , x k +2 } are adjacent.Proof. If x → x i , then for any x j ∈ { x , x , . . . , x i − } , by Lemma 2.6, thereexists a path Q of length k − x i to x j . Then the path xQ implies xx j . If x i → x , then for any x j ∈ { x i +1 , x i +2 , . . . , x k +2 } , by Lemma 2.6, there6xists a path R of length k − x j to x i . Then the path Rx implies xx j . Lemma 2.8. [8] Let k be an integer with k ≥ and D be a strong k -quasi-transitive digraph. Suppose that C = x x . . . x n − x is a cycle of length n with n ≥ k in D . Then for any x ∈ V ( D ) \ V ( C ) , x and V ( C ) are adjacent. Lemma 2.9. [5] Let k be an integer with k ≥ and D be a k -quasi-transitivedigraph. Suppose that C = x x . . . x n − x is a cycle of length n with n ≥ k in D . For any x ∈ V ( D ) \ V ( C ) , if x → x i and x ⇒ V ( C ) , then x → x i +( k − ;if x i → x and V ( C ) ⇒ x , then x i − ( k − → x , where the subscripts are takenmodulo n . Note that x k +2 → x and so P contains a cycle x x . . . x k +2 x of length k + 3. Combining this with Lemma 2.8, every vertex of V ( D ) \ V ( P ) isadjacent to V ( P ). Hence we can divide V ( D ) \ V ( P ) into three sets: I = { x ∈ V ( D ) \ V ( P ) : x ⇒ V ( P ) } , W = { x ∈ V ( D ) \ V ( P ) : V ( P ) ⇒ x } and B = V ( D ) \ ( V ( P ) ∪ W ∪ I ). One of I , W and B may be empty. Lemma 2.10.
For any x ∈ V ( D ) \ V ( P ) , the following hold: (1) Suppose x ∈ I . If there exists a vertex x i ∈ E ( P ) such that x → x i , then x E ( P ) ; if there exists a vertex x i ∈ O ( P ) such that x → x i , then x O ( P ) . (2) Suppose x ∈ W . If there exists a vertex x i ∈ E ( P ) such that x i → x ,then E ( P ) x ; if there exists a vertex x i ∈ O ( P ) such that x i → x ,then O ( P ) x . (3) Suppose x ∈ B . If x and O ( P ) are adjacent, then either x and every ver-tex of O ( P ) are adjacent or there exist two vertices x s , x t ∈ O ( P ) with ≤ t < s ≤ k such that { x s , x s +2 , . . . , x k +2 } 7→ x
7→ { x , x , . . . , x t } ; If x and E ( P ) are adjacent, then either x and every vertex of E ( P ) areadjacent or there exist two vertices x s , x t ∈ E ( P ) with ≤ t < s ≤ k − such that { x s , x s +2 , . . . , x k +1 } 7→ x
7→ { x , x , . . . , x t } .Proof. (1) First assume that there exists a vertex x i ∈ E ( P ) such that x → x i . Recall that x x . . . x k +2 x is a cycle of length k + 3. From now on, thesubscripts are taken modulo k + 3. By Lemma 2.9, x → x i +( k − . Withoutloss of generality, assume i = 0. Denote U = { m ( k − ∈ Z k +3 , m ∈ Z } .Repeating using Lemma 2.9, we can obtain x → U . It is easy to see that7 = { md : m ∈ { , , , . . . , k +3 d − }} , where d = gcd( k − , k + 3) (here,gcd means the greatest common divisor.) Note that d ≥ k is odd. Sothere exist two integers a and b such that k − da and k + 3 = db . Thus,4 = ( k + 3) − ( k −
1) = db − da = d ( b − a ). From this, we know that d = 2or d = 4.If d = 2, then U = { , , . . . , k + 1 } . So x → E ( P ) and furthermore x E ( P ) by the definition of I . Now assume d = 4. It is easy to see that U = { , , . . . , k − } . So x
7→ { x , x , . . . , x k − } . Let x j ∈ { x , x , . . . , x k − } be arbitrary. Note that x x j +2 . The path xC [ x j +2 , x k +1 ] C [ x , x j ] implies xx j and so x x j . In addition, xC [ x , x k +1 ] implies that xx k +1 and so x → x k +1 . Analogously, if there exists x i ∈ O ( P ) such that x → x i , then wecan obtain x O ( P ).(2) By Remark 2.1, considering the converse of D , the statement is obvi-ous.(3) By the definition of B , we have ( x, V ( P )) = ∅ and ( V ( P ) , x ) = ∅ .First assume that x and O ( P ) are adjacent. Without loss of generality,assume that ( x, O ( P )) = ∅ , otherwise consider the converse of D . If x andevery vertex of O ( P ) are adjacent, then we are done. Now assume that thereexists x n ∈ O ( P ) such that x and x n are not adjacent. By Lemma 2.3(3), x ⇒ { x , x , . . . , x n − } and { x n +2 , x n +4 , . . . , x k +2 } ⇒ x . From this with( x, O ( P )) = ∅ , we have n ≥
3. Take t = max { i : x → x i and x i ∈ O ( P ) } .Then 1 ≤ t < n . By Lemma 2.3(3), x and every vertex { x , x , . . . , x t } areadjacent and furthermore x
7→ { x , x , . . . , x t } . By Lemma 2.9 and x → x ,we have xx k . If x → x k , then t ≥ k and so n = k + 2. But x → x andLemma 2.9 implies xx k +2 , a contradiction. Thus x k x . This together withLemma 2.3(3) implies xx k +2 and moreover x k +2 x . Take s = min { i : x i → x and x i ∈ O ( P ) } . Thus n < s ≤ k . By Lemma 2.3(3), x and every vertexof { x s , x s +2 , . . . , x k +2 } are adjacent and furthermore { x s , x s +2 , . . . , x k +2 } 7→ x . Then x x . . . x k +2 x implies that xx and so 3 < n . This also implies t ≥
3. The proof is similar to the above argument when x and E ( P ) areadjacent. So we omit it. Lemma 2.11. If D [ V ( P )] is a semicomplete digraph, then for any x ∈ B ,either x and every vertex of V ( P ) are adjacent or there exist two vertices x t , x s ∈ V ( P ) with ≤ t + 1 < s ≤ k − such that { x s , . . . , x k +2 } 7→ x x , . . . , x t } .Proof. If x and every vertex of V ( P ) are adjacent, then we are done. Supposenot. By the definition of B , ( x, V ( P )) = ∅ and ( V ( P ) , x ) = ∅ . Take t =8ax { i : x → x i } and s = min { j : x j → x } . By Lemma 2.7, x and everyvertex V ( P [ x , x t − ]) are adjacent and x and every vertex of V ( P [ x s , x k +2 ])are adjacent. Furthermore, since x and some vertex of V ( P ) are not adjacent,we can conclude that s > t + 1 and { x s , . . . , x k +2 } 7→ x
7→ { x , . . . , x t } .By Lemmas 2.6 and 2.10, we can obtain the following lemma. Lemma 2.12.
Suppose that D [ V ( P )] is a semicomplete digraph. For any x ∈ I , x V ( P ) and for any y ∈ W , V ( P ) y . Lemma 2.13. If D [ V ( P )] is a semicomplete digraph, then D [ V ( D ) \ V ( P )] is a semicomplete digraph.Proof. According to Lemmas 2.11 and 2.12, similar to the proof of Lemma2.11 in [9], the result can be shown.
Lemma 2.14. If D [ V ( P )] is a semicomplete bipartite digraph, then D [ B ] iseither a semicomplete bipartite digraph or an empty digraph.Proof. Now we divide B into two subsets. Denote B = { x ∈ B : x and E ( P )are adjacent } and B = { x ∈ B : x and O ( P ) are adjacent } . One of B and B may be empty.First we claim B ∩ B = ∅ . Suppose not. Let x ∈ B ∩ B be arbitrary.It is not difficult to obtain that there exist two vertices x i and x j in V ( P )such that x i → x → x j and the parity of i and j is different. Without lossof generality, assume that j is even and i is odd. Take t = max { i : x i → x and x i ∈ O ( P ) } and r = min { j : x → x j and x j ∈ E ( P ) } . By Lemma2.3(3), t = k or k + 2 and r = 0 or 2. If t = k , the path x k +1 x k +2 P [ x , x k ] xx r implies that x k +1 x r . Note that x k +1 , x r ∈ E ( P ). If t = k + 2, then the path P [ x , x k +2 ] xx r implies that x x r . Note that x , x r ∈ E ( P ). We have foundtwo vertices of E ( P ) such that they are adjacent. By Lemma 2.4, D [ V ( P )]is a semicomplete digraph, a contradiction. Hence B ∩ B = ∅ .Now we show that every vertex of B and every vertex of B are adjacent.Let x ∈ B and y ∈ B be arbitrary two vertices. Suppose, on the contrary,that x and y are not adjacent. Note that for any x i ∈ V ( P ), it is impossiblethat x → x i and x k − i → y both hold, otherwise, xx i x i +1 . . . x k − i y impliesthat xy , a contradiction, where the subscripts are taken modulo k + 3. ByLemma 2.3(2), if y → x j for some x j ∈ O ( P ), then x ⇒ { x , . . . , x j − } ∩ E ( P ). Analogously, if x → x i for some x i ∈ E ( P ), then y ⇒ { x , . . . , x i − } ∩ O ( P ). By Lemma 2.10, x and every vertex of { x , x , x k − , x k +1 } are adjacent9nd y and every vertex of { x , x , x k , x k +2 } are adjacent. Suppose y → x k .Then we have x ⇒ { x , x , . . . , x k − } ∩ E ( P ), in particular, x → x k − . Bythe definition of B and B ∩ B = ∅ , x k +1 → x . So it must be x → y ,otherwise yx x . . . x k +1 x implies xy , a contradiction. But it is impossibleas x → x k − . Therefore x k y . This implies x x and furthermore { x , x , . . . , x k , x k +2 } ⇒ y , in particular, x → y . By the definition of B and B ∩ B = ∅ , we have y → x . So x → x k − . But it is impossible as x → y .Thus xy . By the arbitrariness of x and y , every vertex of B and every vertexof B are adjacent.Next we show that B and B both are independent sets. Suppose not.Without loss of generality, assume that there exist two vertices x ′ , x ′′ ∈ B such that x ′ → x ′′ . By the definition of B and B ∩ B = ∅ , there exists x i ∈ E ( P ) such that x ′′ → x i . But the path x ′ x ′′ C [ x i , x i +( k − ] implies x ′ x i + k − . Note that i + k − x ′ ∈ B . Ifone of B and B is empty, then D [ B ] is an empty digraph, otherwise D [ B ]is a semicomplete bipartite digraph. Theorem 2.15.
The subdigraph induced by V ( D ) \ V ( P ) is either a semi-complete digraph, a semicomplete bipartite digraph or an empty digraph.Proof. By Theorem 2.5, D [ V ( P )] is either a semicomplete digraph or a semi-complete bipartite digraph. If D [ V ( P )] is a semicomplete digraph, then byLemma 2.13, we are done. Now consider that D [ V ( P )] is a semicompletebipartite digraph. From now on, all subscripts appearing in this proof aretaken modulo k + 3.By Lemma 2.14, D [ B ] is either a semicomplete bipartite digraph or anempty digraph. By Lemma 2.10, for any x ∈ I , either x E ( P ) or x O ( P ) or both and for any y ∈ W , either E ( P ) y or O ( P ) y or both.Hence, we divide I into two sets: I = { x ∈ I : x E ( P ) } and I = { x ∈ I : x O ( P ) } and divide W into two sets: W = { x ∈ W : E ( P ) x } and W = { x ∈ W : O ( P ) x } . B and B are defined similar to Lemma2.14. Now we show that D [ I ] is either a semicomplete bipartite digraph oran empty digraph. If | I | ≤
1, there is nothing to prove. Now assume | I | ≥ D is strong and ( V ( P ) , I ) = ∅ , we have B ∪ W = ∅ and ( B ∪ W, I ) = ∅ .Define e I = { x ∈ I : ( B ∪ W, x ) = ∅} . Clearly, e I = ∅ .To complete the proof of this theorem, we first give the following severalclaims. Claim 1.
Every vertex of I i is adjacent to every vertex of B − i ∪ W − i and10very vertex of W i is adjacent to every vertex of B − i for i = 1 , Proof.
Let x ∈ I and y ∈ B ∪ W be arbitrary. By the definition of B ∪ W ,there exists x i ∈ O ( P ) such that x i → y . Then the path xC [ x i +5 , x i ] y is apath of length k , which implies that xy . Analogously, we can show that everyvertex of I is adjacent to every vertex of B ∪ W and every vertex of W i isadjacent to every vertex of B − i for i = 1 , Claim 2.
Let Q = y y . . . y q be a path of V ( D ) \ V ( P ) with q ≤ k − x i ∈ V ( P ) such that x i → y . If i and q havethe same parity, then y q and E ( P ) are adjacent; if the parity of i and q aredifferent, then y q and O ( P ) are adjacent. Proof.
Note that x ( q + i ) − ( k − . . . x i y . . . y q is a path of length k , which implies x ( q + i ) − ( k − y q . If i and q have the same parity, then ( q + i ) − ( k −
1) is evenand so x ( q + i ) − ( k − ∈ E ( P ). If the parity of i and q are different, then( q + i ) − ( k −
1) is odd and so x ( q + i ) − ( k − ∈ O ( P ).Similar to Claim 2, we can obtain the following claim. Claim 3.
Let Q = y y . . . y q be a path of V ( D ) \ V ( P ) with q ≤ k − x j ∈ V ( P ) such that y q → x j . If j and q havethe same parity, then y and E ( P ) are adjacent; if j and q have the differentparity, then y and O ( P ) are adjacent. Claim 4. ( W, I ) = ∅ . Proof.
Suppose not. Let xy be an arc from W to I . By Lemma 2.10(1) and(2), there exist x i ∈ { x , x } and x j ∈ { x k +1 , x k +2 } such that x i → x and y → x j . Since P is minimal, we have d ( x i , x j ) ≥ k ≥
5. However x i xyx j is apath of length 3, a contradiction. Claim 5. ( B i , I i ) = ∅ and ( W i , B i ) = ∅ , for i = 1 , Proof.
Assume ( B , I ) = ∅ . Let xy be an arc from B to I . By Lemma2.10(1), y → x k +1 . Consider the path xy . By Claim 3, x and O ( P ) areadjacent. Since B ∩ B = ∅ , it is impossible. Thus ( B , I ) = ∅ . Analogously,we can show that ( B , I ) = ∅ and ( W i , B i ) = ∅ , for i = 1 , Claim 6. I ∩ I = ∅ and W ∩ W = ∅ .11 roof. Suppose I ∩ I = ∅ . Let z ∈ I ∩ I be arbitrary. By Lemma2.10(1), z V ( P ) and by Claim 1, z and every vertex of B ∪ W is adjacent.According to Claims 4 and 5, we can conclude that z B ∪ W . That is tosay, I ∩ I B ∪ W , which also implies that for any w ∈ e I , w / ∈ I ∩ I . Since D is strong, there exist v ∈ I ∩ I and u ∈ e I such that u → v . Accordingto Claim 3 and v → V ( P ), u ∈ I ∩ I , a contradiction. Hence I ∩ I = ∅ .Analogously, we can show that W ∩ W = ∅ .By Claims 3 and 6, we have the following claim. Claim 7. I i and W i are both independent sets, for i = 1 , Claim 8. ( I i , B i ) = ∅ and ( B i , W i ) = ∅ , for i = 1 , Proof.
Assume ( I , B ) = ∅ . Let uv be an arc from I to B . There exists x i ∈ E ( P ) such that v → x i . Then by Claim 4, u and O ( P ) are adjacent, acontradiction. Thus ( I , B ) = ∅ . Analogously, we can show that ( I , B ) = ∅ and ( B i , W i ) = ∅ , for i = 1 , Claim 9.
For any w ∈ I \ e I , d D [ I ] ( e I, w ) < ∞ . Proof.
Since D is strong, w is reachable from e I in D . Let Q = y y . . . y m be a shortest path from e I to w in D , where y ∈ e I , y m = w and m ≥ V ( Q ) ⊂ e I . Because y m ∈ I \ e I , we have y m − ∈ I . Denote r = min { j : y j , . . . , y m ∈ I } . If r ≥
1, then y r − / ∈ I . By the definition of e I , y r ∈ e I . Then y r . . . y m is a shorter path from e I to w than Q , a contradiction.Hence r = 0 and so V ( Q ) ⊂ I . Claim 10. ( I i , W i ) = ∅ for i = 1 , Proof.
Assume ( I , W ) = ∅ . Let uv be an arc from I to W . If u ∈ e I ,then by Claims 4 and 5, there exists z ∈ B such that z → u . There exists x i ∈ O ( P ) such that x i → z . Consider the path zuv . By Claim 3, v and O ( P ) are adjacent, a contradiction. Hence u / ∈ e I . By Claim 9, there existsa path Q = u u . . . u t from e I to u in D [ I ], where u ∈ e I and u t = u . If t ≥ k −
1, then u t − ( k − . . . u t v is a path of length k , which implies u t − ( k − v and furthermore u t − ( k − → v by Claim 4. Repeating using this way, thereexists an integer i such that u t − i ( k − → v and 0 ≤ t − i ( k − < k − u t − i ( k − ∈ I . So we assume, without loss of generality, that t ≤ k −
2. 12f u ∈ I , then, by Claim 7, t is even and so t ≤ k −
3. Since u ∈ e I and I ⇒ B ∪ W , there exists y ∈ B such that y → u and there exists x j ∈ O ( P )such that x j → y . Consider the path R = yu . . . u t v . Note that the lengthof R is even. By Claim 2, v and O ( P ) is adjacent, a contradiction. If u ∈ I ,we have t is odd. Since u ∈ e I and I ⇒ B ∪ W , there exists y ∈ B suchthat y → u and there exists x i ∈ E ( P ) such that x i → y . Consider the path R = yu . . . u t v . By Claims 5 and 8, we have that y and v are not adjacentand so t ≤ k −
4. By Claim 2, v and O ( P ) is adjacent, a contradiction.Therefore, ( I , W ) = ∅ . Analogously, we can show ( I , W ) = ∅ . Claim 11.
Every vertex of e I ∩ I i is adjacent to every vertex of I − i for i = 1 , Proof.
Let x ∈ e I ∩ I and x ′′ ∈ I be arbitrary. By Claims 4 and 5, thereexists y ∈ B such that y → x and furthermore there exists x j ∈ O ( P ) suchthat x j → y . Then x ′′ P [ x j − ( k − , x j ] yx implies x ′′ x . Analogously, we canshow that every vertex of e I ∩ I is adjacent to every vertex of I .Now we return the proof of the theorem. By Claim 7, I and I are bothindependent sets. If one of I and I is an empty set, then D [ I ] is an emptydigraph. Assume that I and I are both nonempty sets.Now we show that D [ I ] is a semicomplete bipartite digraph. Using Claim11, we only need to prove that D [ I \ e I ] is a semicomplete bipartite digraph.Let x ′ ∈ I \ e I and x ′′ ∈ I \ e I be arbitrary. We shall show that x ′ x ′′ . ByClaim 9, d D [ I ] ( e I, x ′ ) < ∞ and d D [ I ] ( e I, x ′′ ) < ∞ . Without loss of generality,assume that d D [ I ] ( e I, x ′ ) ≤ d D [ I ] ( e I, x ′′ ). Let R = z z . . . z m be a shortest pathfrom e I to x ′ in D [ I ], where z ∈ e I and z m = x ′ . By the minimality of R , z , . . . , z m ∈ I \ e I . In addition, x ′′ / ∈ V ( R ). By the definition of e I , thereexists y ∈ B such that y → z .Assume that z ∈ I . In this case y ∈ B and there exists x j ∈ O ( P )such that x j → y . By Claim 7 and x ′ ∈ I , we have m is even and so m ≥
2. By Claim 11, z x ′′ and x ′′ → z as d D [ I ] ( e I, x ′ ) ≤ d D [ I ] ( e I, x ′′ ). If m ≤ k −
3, then x ′′ P [ x j − ( k − − m ) , x j ] yR implies that x ′′ z m . If m = k −
1, then x ′′ R implies x ′′ z m . The proof for the case m ≥ k − m with the case m = k − x ′′ and every vertex of { z , z , . . . , z m − } are adjacent, in particular, x ′′ and z m − ( k − are adjacent.As d D [ I ] ( e I, x ′ ) ≤ d D [ I ] ( e I, x ′′ ), x ′′ → z m − ( k − . Note that x ′′ R [ z m − ( k − , z m ] isa path of length k , which implies that x ′′ z m .13ow consider the case z ∈ I . In this case y ∈ B and x ′′ → y . Thereexists x j ∈ E ( P ) such that x j → y . According to Claim 7 and x ′ ∈ I ,we have m is odd and m ≥
1. If m ≤ k −
3, then x ′′ P [ x j − ( k − − m ) , x j ] yR implies that x ′′ z m . If m = k −
2, then x ′′ yR implies that x ′′ z m . The prooffor the case m ≥ k is by induction on odd m with the case m = k − x ′′ and every vertex of { z , z , . . . , z m − } are adjacent,in particular, x ′′ and z m − ( k − are adjacent. As d D [ I ] ( e I, x ′ ) ≤ d D [ I ] ( e I, x ′′ ), x ′′ → z m − ( k − . Then the path x ′′ R [ z m − ( k − , z m ] implies x ′′ z m . From nowon, we have shown that D [ I ] is either a semicomplete bipartite digraph or anempty digraph. Analogously, we can show that D [ W ] is either a semicom-plete bipartite digraph or an empty digraph. Recalled that D [ B ] is either asemicomplete bipartite digraph or an empty digraph. Combining these withClaims 1,5,8 and 10, we have that D [ V ( D ) \ V ( P )] is either a semicompletebipartite digraph or an empty digraph.From Theorems 2.5 and 2.15, Theorem 1.3 holds. References [1] J. Bang-Jensen, J. Huang. Quasi-transitive digraphs, Journal of GraphTheory, 20 (1995) 141–161.[2] J. Bang-Jensen, G. Gutin. Digraphs: Theory, Algorithms and Applica-tions, Springer, London, 2000.[3] H. Galeana-S´anchez, I.A. Goldfeder, I. Urrutia. On the structure of 3-quasi-transitive digraphs, Discrete Mathematics, 310 (2010) 2495–2498.[4] H. Galeana-S´anchez, C. Hern´andez-Cruz, M.A. Ju´arez-Camacho. Onthe existence and number of ( k + 1)-kings in k -quasi-transitive digraphs,Discrete Mathematics, 313 (2013) 2582–2591.[5] C. Hern´andez-Cruz. 4-transitive digraphs I: the structure of strong4-transitive digraphs, Discussiones Mathematicae Graph Theory, 33(2013) 247–260.[6] C. Hern´andez-Cruz, H. Galeana-S´anchez. k -kernels in k -transitive and k -quasi-transitive digraphs, Discrete Mathematics, 312 (2012) 2522–2530.[7] R. Wang. ( k + 1)-kernels and the number of k -kings in k -quasi-transitivedigraphs, Discrete Mathematics, 338 (2015) 114–121.148] R. Wang, W. Meng. k -kings in k -quasitransitive digraphs, Journal ofGraph Theory, 79 (2015) 55–62.[9] R. Wang, H. Zhang. Hamiltonian paths in kk