A sufficient condition for pre-Hamiltonian cycles in bipartite digraphs

TL;DR

This paper establishes a sufficient degree condition for the existence of pre-Hamiltonian cycles in strongly connected balanced bipartite digraphs, identifying specific cycle lengths or a unique exceptional case.

Contribution

It introduces a new degree condition involving dominating pairs that guarantees the presence of cycles of all even lengths up to a certain size or a specific exceptional digraph.

Findings

If the maximum degree of vertices in dominating pairs is at least 2a-2, the digraph contains all even cycles up to length 2a-2.

The result applies to strongly connected balanced bipartite digraphs of order at least 10.

An exceptional digraph of order ten is characterized where the condition does not imply the cycle existence.

Abstract

Let $D$ be a strongly connected balanced bipartite directed graph of order $2a\geq 10$ other than a directed cycle. Let $x,y$ be distinct vertices in $D$. $\{x,y\}$ dominates a vertex $z$ if $x\rightarrow z$ and $y\rightarrow z$; in this case, we call the pair $\{x,y\}$ dominating. In this paper we prove: If $ max\{d(x), d(y)\}\geq 2a-2$ for every dominating pair of vertices $\{x,y\}$, then $D$ contains cycles of all lengths $2,4, \ldots , 2a-2$ or $D$ is isomorphic to a certain digraph of order ten which we specify.

Equations

$$d(x_2)=d(x_3)=d(y_0)=d(y_1)=7andd(x_0)=d(x_1)=d(y_2)=d(y_3)=3,$$

$$d(y_C^{-},P_1)=d(u_s,C^{\prime })=0.$$

$$d(x_2,\{u_1,u_3\})=d(u_2,\{y_1,y_C^{-}\})=0.$$

$$d(u_2,\{y_1,y_C^{--}\})=a(y_C^{-},u_1)=0,$$

$$a(u_1,x_2)=a(u_2,y_1)=0,$$

$$max\{d(u_1),d(x_2),d(u_2),d(y_1)\}\le 2a-2.$$

$$a(x_1,y_0)=d^{-}(x_1,\{y\})=d^{-}(y,\{x_2\})=0$$

$$y\leftrightarrow \{x_1,x_2,\dots ,x_m\}.$$

$$d(x_i)\ge 2a-2,for\; alli\in [1,m].$$

$$d(y_0,\{x_1,x_2,\dots ,x_m\})=0.$$

$$d^{-}(y_0,\{x_1,x_2,\dots ,x_m\})=0.$$

$$d^{-}(x_0,\{y_1,y_2,\dots ,y_m\})=0.$$

$$A(V(C)\cup \{x,y\}\to \{x_0,y_0\})=\varnothing ,$$

$$d(x)\le 2a-3,d(x_1)\ge 2a-2andy\mapsto X\cap R\setminus \{x\}.$$

$$y\leftrightarrow \{x_1,x_3,x_4,\dots ,x_m\}andy\mapsto x_2.$$

$$d^{-}(x,\{y_1,y_2,\dots ,y_m\})=d^{+}(x,\{y_1,y_2,\dots ,y_m\}-\{y_2\})=0.$$

$$d^{+}(x_0,\{y_1,y_2,\dots ,y_m\}-\{y_2\})=0.$$

$$d(x,\{y_1,y_2,\dots ,y_m\}-\{y_2\})=0,$$

$$max\{d(x_0),d(x),d(y_j)\}\le 2a-3,for\; allj\notin \{0,2\}.$$

$$d^{-}(u,\{y_1,y_m\})\le 1and{d}^{-}(v,\{x_0,x\})\le 1.$$

$$x_1\to yand{x}_i\leftrightarrow yfor\; alli\in [2,m]$$

$$d^{+}(x_0,\{y,y_1,y_2,\dots ,y_m\})=d(x,\{y_1,y_2,\dots ,y_{m-1}\})=0andx{y}_m\notin A(D).$$

$$y\leftrightarrow \{x_1,x_2,\dots ,x_m\}.$$

$$d^{-}(y_0,\{x_1,x_2,\dots ,x_m\})=d^{-}(x_0,\{y_1,y_2,\dots ,y_m\})=d^{+}(x,\{y_1,y_2,\dots ,y_m\})=0.$$

$$a(x,y)=d^{-}(x,\{y_1,y_2,\dots ,y_{r-2}\})=0andx{y}_r\notin A(D).$$

$$d(y_{r-2})\ge 2a-2.$$

$$A(\{x,x_0\}\to \{y,y_{r-1},y_r\})=\varnothing .$$

$$max\{d(y),d(y_0),d(x),d(x_0)\}\le 2a-3,$$

$$max\{d^{-}(u,\{y,y_0\}),d^{-}(v,\{x,x_0\})\}\le 1for\; allu\in Xandv\in Y.$$

$$max\{d(x),d(x_0),d(y),d(y_0),d(x_2)\}\le 2a-3.$$

Full text

**A sufficient condition for pre-Hamiltonian cycles in bipartite digraphs

**

**Samvel Kh. Darbinyan and Iskandar A. Karapetyan

**

Institute for Informatics and Automation Problems, Armenian National Academy of Sciences

E-mails: samdarbin@ipia.sci.am, isko@ipia.sci.am

Abstract

Let $D$ be a strongly connected balanced bipartite directed graph of order $2a\geq 10$ other than a directed cycle. Let $x,y$ be distinct vertices in $D$. $\{x,y\}$ dominates a vertex $z$ if $x\rightarrow z$ and $y\rightarrow z$; in this case, we call the pair $\{x,y\}$ dominating. In this paper we prove:

*If $max\{d(x),d(y)\}\geq 2a-2$ for every dominating pair of vertices $\{x,y\}$, then $D$ contains cycles of all lengths $2,4,\ldots,2a-2$ or $D$ is isomorphic to a certain digraph of order ten which we specify.

Keywords: Digraphs, pre-Hamiltonian cycles, bipartite balanced digraphs, even pancyclic.

1 Introduction

It is sequel to the paper [13] by the first author. We consider digraphs (directed graphs) in the sense of [4], and use standard graph theoretical terminology and notation (see Section 2 for details). A cycle passing through all the vertices of a digraph is called Hamiltonian. A digraph containing a Hamiltonian cycle is called a Hamiltonian digraph. A digraph $D$ of order $n$ is called pancyclic if it contains cycles of every lengths $3,4,\ldots,n$. Various sufficient conditions for a digraph to be Hamiltonian have been given in terms of the vertex degree of the digraph. Here we recall some of them which are due to Ghouila-Houri [16], Nash-Williams [23], Woodall [27], Meyniel [22], Thomassen [25] and Darbinyan [11]. The Meyniel theorem is a generalization Nash-Williams’, Ghouila-Houri’s and Woodall’s theorems.

Bondy suggested (see [9] by Chvátal) the following metaconjecture:

Metaconjecture. *Almost any non-trivial condition of a graph (digraph) which implies that the graph (digraph) is Hamiltonian also implies that the graph (digraph) is pancyclic. (There may be a ”simple” family of exceptional graphs (digraphs)).

In fact various sufficient conditions for a digraph to be Hamiltonian are also sufficient for the digraph to be pancyclic. Namely, in [20, 24, 10, 12], it was shown that if a digraph $D$ satisfies one of the above mentioned conditions for hamiltonicity digraphs, then the digraph $D$ also is pancyclic (unless some extremal cases which are characterized). For additional information on Hamiltonian and pancyclic digraphs, see, e.g, the book by Bang-Jensen and Gutin [4] and the surveys [7] by Bermond and Thomassen, [21] by Kühn and Ostus and [17] by Gutin.

Each of aforementioned theorems imposes a degree condition on all vertices (or, on all pairs of nonadjacent vertices). In [5] and [3], it was described a type of sufficient conditions for a digraph to be Hamiltonian, in which a degree condition requires only for some pairs of nonadjacent vertices. Let us recall only the following theorem of them.

Theorem 1.1 (Bang-Jensen, Gutin, H.Li [5]). *Let $D$ be a strongly connected digraph of order $n\geq 2$. Suppose that $min\{d(x),d(y)\}\geq n-1$ and $d(x)+d(y)\geq 2n-1$ for any pair of non-adjacent vertices $x,y$ with a common in-neighbour. Then $D$ is Hamiltonian.

A digraph $D$ is called a bipartite digraph if there exists a partition $X$, $Y$ of its vertex set into two partite sets such that every arc of $D$ has its end-vertices in different partite sets. It is called balanced if $|X|=|Y|$.

There are analogies results to the Nash-Williams, Ghouila-Houri, Woodall, Meyniel and Thomassen theorems for balanced bipartite digraphs (see e.g., [2] and the papers cited there).

An analogue of Theorem 1.1 for bipartite digraphs was given by R. Wang [26] and recently a different result was given by Adamus [1].

Theorem 1.2 (R. Wang [26]). Let $D$ be a strongly connected balanced bipartite digraph of order $2a$, where $a\geq 1$. Suppose that, for every dominating pair of vertices $\{x,y\}$, either $d(x)\geq 2a-1$ and $d(y)\geq a+1$ or $d(y)\geq 2a-1$ and $d(x)\geq a+1$. Then $D$ is Hamiltonian.

Theorem 1.3 (Adamus [1]). Let $D$ be a strongly connected balanced bipartite digraph of order $2a$, where $a\geq 3$. If $d(x)+d(y)\geq 3a$ for every pair of vertices $\{x,y\}$ with a common in-neighbour or a common out-neighbour, then $D$ is Hamiltonian.

Let $D$ be a balanced bipartite digraph of order $2a$, where $a\geq 2$. For integer $k\geq 0$, we say that $D$ satisfies condition $B_{k}$ when $max\{d(x),d(y)\}\geq 2a-2+k$ for every pair of dominating vertices $x$ and $y$.

Before stating the next theorems we need to define three digraphs.

Example 1. Let $D(10)$ be a bipartite digraph with partite sets $X=\{x_{0},x_{1},x_{2},x_{3},x_{4}\}$ and $Y=\{y_{0},y_{1},y_{2},y_{3},y_{4}\}$ satisfying the following conditions: The induced subdigraph $\langle\{x_{1},x_{2},x_{3},y_{0},y_{1}\}\rangle$ is a complete bipartite digraph with partite sets $\{x_{1},x_{2},x_{3}\}$ and $\{y_{0},y_{1}\}$; $\{x_{1},x_{2},x_{3}\}\rightarrow\{y_{2},y_{3},y_{4}\}$; $x_{4}\leftrightarrow y_{1}$; $x_{0}\leftrightarrow y_{0}$ and $x_{i}\leftrightarrow y_{i+1}$ for all $i\in[1,3]$. $D(10)$ contains no other arcs.

It is easy to check that the digraph $D(10)$ is strongly connected and satisfies condition $B_{0}$, but the underlying undirected graph of $D(10)$ is not 2-connected and $D(10)$ has no cycle of length 8. (It follows from the facts that $d(x_{0})=d(x_{4})=2$ and $x_{0}$ ($x_{4}$) is on 2-cycle). It is not difficult to check that any digraph obtained from $D(10)$ by adding a new arc whose one end-vertex is $x_{0}$ or $x_{4}$ contains no cycle of length eight. Moreover, if to $A(D)$ we add some new arcs of the type $y_{i}x_{j}$, where $i\in[2,4]$ and $j\in[1,3]$, then always we obtain a digraph which does not satisfy condition $B_{0}$.

Example 2. Let $K_{2,3}^{*}$ be a complete bipartite digraph with partite sets $\{x_{1},x_{2}\}$ and $\{y_{1},y_{2},y_{3}\}$. Let $H(8)$ be the bipartite digraph obtained from the digraph $K_{2,3}^{*}$ by adding three new vertices $x_{0},y_{0},x_{3}$ and the following new arcs $x_{0}y_{0}$, $y_{0}x_{0}$, $x_{0}y_{1}$, $y_{1}x_{0}$, $x_{3}y_{3}$ and $y_{3}x_{3}$.

It is not difficult to check that the digraph $H(8)$ is strongly connected and satisfies condition $B_{0}$, but the underlying undirected graph of $H(8)$ is not 2-connected and $H(8)$ has no cycle of length 6. Hence, the bound on order of $D$ in Theorem 3.4 is sharp.

Example 3. Let $D(8)$ be a bipartite digraph with partite sets $X=\{x_{0},x_{1},x_{2},x_{3}\}$ and $Y=\{y_{0},y_{1},y_{2},y_{3}\}$, and the arc set $A(D(8))$ contains exactly the following arcs $y_{0}x_{1}$, $y_{1}x_{0}$, $x_{2}y_{3}$, $x_{3}y_{2}$ and all the arcs of the following 2-cycles: $x_{i}\leftrightarrow y_{i}$, $i\in[0,3]$, $y_{0}\leftrightarrow x_{2}$, $y_{0}\leftrightarrow x_{3}$, $y_{1}\leftrightarrow x_{2}$ and $y_{1}\leftrightarrow x_{3}$.

It is easy to see that

[TABLE]

and the dominating pairs in $D(8)$ are: $\{y_{0},y_{1}\}$, $\{y_{0},y_{2}\}$,$\{y_{0},y_{3}\}$,$\{y_{1},y_{2}\}$, $\{y_{1},y_{3}\}$, $\{x_{0},x_{2}\}$, $\{x_{0},x_{3}\}$, $\{x_{1},x_{2}\}$, $\{x_{1},x_{3}\}$ and $\{x_{2},x_{3}\}$. Note that every dominating pair satisfies condition $B_{1}$. Since $x_{0}y_{0}x_{3}y_{2}x_{2}$ $y_{1}x_{0}$ is a cycle of length 6 in $D(8)$, it is not difficult to check that $D(8)$ is strong.

Observe that $D(8)$ is not Hamiltonian. Indeed, if $D(8)$ contains a Hamiltonian cycle, say $C$, then $C$ would contain the arcs $x_{1}y_{1}$ and $x_{0}y_{0}$. Therefore, $C$ must contain the path $x_{1}y_{1}x_{0}y_{0}$ or the path $x_{0}y_{0}x_{1}y_{1}$, which is impossible since $N^{-}(x_{0})=N^{-}(x_{1})=\{y_{0},y_{1}\}$.

For $a\geq 5$ Theorem 1.2 is an immediate consequence of the following theorem by the first author [14].

Theorem 1.4 (Darbinyan [14]). Let $D$ be a strongly connected balanced bipartite digraph of order $2a$, where $a\geq 4$. Suppose that, for every dominating pair of vertices $\{x,y\}$, either $d(x)\geq 2a-1$ or $d(y)\geq 2a-1$. Then either $D$ is Hamiltonian or isomorphic to the digraph $D(8)$.

A balanced bipartite digraph of order $2m$ is even pancyclic if it contains a cycle of length $2k$ for any $2\leq k\leq m$. A cycle of a balanced bipartite digraph $D$ is called pre-Hamiltonian if it contains all the vertices of $D$ except two.

Characterizations of even pancyclic bipartite tournaments was given in [6] and [28]. A characterization of pancyclic ordinary $k$-partite ($k\geq 3$) tournaments (respectively, pancyclic ordinary complete multipartite digraphs) was established in [18] (respectively, in [19]).

Motivated by the Bondy’s metaconjecture, it is natural to set the following problem:

Problem. *Characterize those digraphs which satisfy the conditions of Theorem 1.2 (or, 1.3 or 1.4) but are not even pancyclic.

In [15], the first author have proved the following Theorems 1.5 and 1.6.

Theorem 1.5 ([15]. *Let $D$ be a strongly connected balanced bipartite digraph of order $2a\geq 8$ other than a directed cycle. If $max\{d(x),d(y)\}\geq 2a-1$ for every dominating pair of vertices $\{x,y\}$, then $D$ contains a cycles of all even lengths less than equal $2a$ or $D$ is isomorphic to the digraph $D(8)$.

Theorem 1.6. ([15]. *Let $D$ be a strongly connected balanced bipartite digraph of order $2a\geq 8$ which contains a cycle of length $2a-2$. If $max\{d(x),d(y)\}\geq 2a-2$ for every dominating pair of vertices $\{x,y\}$, then for any $k$, $1\leq k\leq a-1$, $D$ contains a cycle of length $2k$.

In view of Theorem 1.6 it seems quite natural to ask whether a balanced bipartite digraph of order $2a$, which satisfies condition $B_{0}$ contains a pre-Hamiltonian cycle (i.e., a cycle of length $2a-2$). In this paper we prove the following theorems.

Theorem 1.7. *Let $D$ be a strongly connected balanced bipartite digraph of order $2a\geq 10$ other than the directed cycle of length $2a$. Suppose that $D$ satisfies condition $B_{0}$, i.e., $max\{d(x),d(y)\}\geq 2a-2$ for every dominating pair of vertices $\{x,y\}$. Then $D$ contains a cycle of lengths $2a-2$ unless $D$ is isomorphic to the digraph $D(10)$.

From Theorem 1.6 and 1.7 it follows the following theorem.

Theorem 1.8. *Let $D$ be a balanced bipartite digraph of order $2a\geq 10$ other than the directed cycle of length $2a$. Suppose that $D$ satisfies condition $B_{0}$, i.e., $max\{d(x),d(y)\}\geq 2a-2$ for every dominating pair of vertices $\{x,y\}$. Then $D$ contains cycles of all lengths $2,4,\ldots,2a-2$ unless $D$ is isomorphic to the digraph $D(10)$.

2 Terminology and Notation

In this paper we consider finite digraphs without loops and multiple arcs. The vertex set and the arc set of a digraph $D$ are denoted by $V(D)$ and by $A(D)$, respectively. The order of $D$ is the number of its vertices. For any $x,y\in V(D)$, we also write $x\rightarrow y$, if $xy\in A(D)$. If $xy\in A(D)$, then we say that $x$ dominates $y$ or $y$ is an out-neighbour of $x$, and $x$ is an in-neighbour of $y$. The notation $x\leftrightarrow y$ denotes that $x\rightarrow y$ and $y\rightarrow x$ ($x\leftrightarrow y$ is called a 2-cycle). We denote by $a(x,y)$ the number of arcs with end-vertices $x$ and $y$. For disjoint subsets $A$ and $B$ of $V(D)$ we define $A(A\rightarrow B)$   as the set $\{xy\in A(D)|x\in A,y\in B\}$ and $A(A,B)=A(A\rightarrow B)\cup A(B\rightarrow A)$. If $x\in V(D)$ and $A=\{x\}$ we write $x$ instead of $\{x\}$. If $A$ and $B$ are two disjoint subsets of $V(D)$ such that every vertex of $A$ dominates every vertex of $B$, then we say that $A$ dominates $B$, denoted by $A\rightarrow B$. $A\mapsto B$ means that $A\rightarrow B$ and there is no arc from $B$ to $A$.

We let $N^{+}(x)$, $N^{-}(x)$ denote the set of out-neighbours, respectively the set of in-neighbours of a vertex $x$ in a digraph $D$. If $A\subseteq V(D)$, then $N^{+}(x,A)=A\cap N^{+}(x)$ and $N^{-}(x,A)=A\cap N^{-}(x)$. The out-degree of $x$ is $d^{+}(x)=|N^{+}(x)|$ and $d^{-}(x)=|N^{-}(x)|$ is the in-degree of $x$. Similarly, $d^{+}(x,A)=|N^{+}(x,A)|$ and $d^{-}(x,A)=|N^{-}(x,A)|$. The degree of a vertex $x$ in $D$ is defined as $d(x)=d^{+}(x)+d^{-}(x)$ (similarly, $d(x,A)=d^{+}(x,A)+d^{-}(x,A)$).

For integers $a$ and $b$, $a\leq b$, let $[a,b]$ denote the set of all integers which are not less than $a$ and are not greater than $b$.

The subdigraph of $D$ induced by a subset $A$ of $V(D)$ is denoted by $D\langle A\rangle$ or $\langle A\rangle$ for brevity. The path (respectively, the cycle) consisting of the distinct vertices $x_{1},x_{2},\ldots,x_{m}$ ( $m\geq 2$) and the arcs $x_{i}x_{i+1}$, $i\in[1,m-1]$ (respectively, $x_{i}x_{i+1}$, $i\in[1,m-1]$, and $x_{m}x_{1}$), is denoted by $x_{1}x_{2}\cdots x_{m}$ (respectively, $x_{1}x_{2}\cdots x_{m}x_{1}$). We say that $x_{1}x_{2}\cdots x_{m}$ is a path from $x_{1}$ to $x_{m}$ or is an $(x_{1},x_{m})$-path. If $P$ is a path containing a subpath from $x$ to $y$ we let $P[x,y]$ denote that subpath. Similarly, if $C$ is a cycle containing vertices $x$ and $y$, $C[x,y]$ denotes the subpath of $C$ from $x$ to $y$. A digraph $D$ is strongly connected (or, just, strong) if there exists an $(x,y)$-path in $D$ for every ordered pair of distinct vertices $x,y$ of $D$. Given a vertex $x$ of a path $P$ or a cycle $C$, we denote by $x^{+}$ (respectively, by $x^{-}$) the successor (respectively, the predecessor) of $x$ (on $P$ or $C$), and in case of ambiguity, we precise $P$ or $C$ as a subscript (that is $x^{+}_{P}$ …). Two distinct vertices $x$ and $y$ are adjacent if $xy\in A(D)$ or $yx\in A(D)$ (or both).

Let $C$ be a non-Hamiltonian cycle in a digraph $D$. An $(x,y)$-path $P$ is a $C$-bypass if $|V(P)|\geq 3$, $x\not=y$ and $V(P)\cap V(C)=\{x,y\}$. The length of the path $C[x,y]$ is the gap of $P$ with respect to $C$.

The underlying undirected graph of a digraph $D$ is the unique graph that contains an edge $xy$ if $x\to y$ or $y\to x$ (or both).

3 Preliminaries

Lemma 3.1 (Bypass Lemma 3.17, Bondy [8]). *Let $D$ be a strongly connected digraph, and let $H$ be a non-trivial proper subdigraph of $D$. If the underlying undirected graph of $D$ is 2-connected, then $D$ contains a $H$-bypass.

Remark. One can prove Bypass Lemma using the proof of Theorem 5.4.2 [4].

Lemma 3.2 ([14]). *Let $D$ be a strongly connected balanced bipartite digraph of order $2a\geq 8$ other than the directed cycle of length $2a$. If $D$ satisfies condition $B_{0}$, then $D$ contains a non-Hamiltonian cycle of length at least 4.

Lemma 3.3 ([15]). *Let $D$ be a strongly connected balanced bipartite digraph of order $2a\geq 8$ with partite sets $X$ and $Y$. Assume that $D$ satisfies condition $B_{0}$. Let $C=x_{1}y_{1}x_{2}y_{2}\ldots x_{k}y_{k}x_{1}$ be a longest non-Hamiltonian cycle in $D$, where $k\geq 2$, $x_{i}\in X$ and $y_{i}\in Y$, and $P$ be a $C$-bypass. If the gap of $P$ with respect to $C$ is equal to one, then $k=a-1$, i.e., the longest non-Hamiltonian cycle in $D$ has length $2a-2$.

Theorem 3.4 ([13]). *Let $D$ be a strongly connected balanced bipartite digraph of order $2a\geq 10$. Assume that $D$ satisfies condition $B_{0}$. Then either the underlying undirected graph of $D$ is 2-connected or $D$ contains a cycle of length $2a-2$ unless $D$ is isomorphic to the digraph $D(10)$.

4 Proof of the main result

**Proof of Theorem 1.7. ** Suppose, on the contrary, that a digraph $D$ is not a directed cycle and satisfies the conditions of the theorem, but contains no cycle of length $2a-2$. Let $C=x_{1}y_{1}x_{2}y_{2}\ldots x_{m}y_{m}x_{1}$ be a longest non-Hamiltonian cycle in $D$, where $x_{i}\in X$ and $y_{i}\in Y$ (all subscripts of the vertices $x_{i}$ and $y_{i}$ are taken modulo $m$, i.e., $x_{m+i}=x_{i}$ and $y_{m+i}=y_{i}$).

By Lemma 3.2, $D$ contains a non-Hamiltonian cycle of length at least 4, i.e., $2\leq m\leq a-2$. By Theorem 3.4, the underlying undirected graph of $D$ is 2-connected. Therefore, by Bypass Lemma, $D$ contains a $C$-bypass. We choose a cycle $C$ and a $C$-bypass $P:=xu_{1}\ldots u_{s}y$ such that

(i) $C$ is a longest non-Hamiltonian cycle in $D$;

(ii) the gap of $C$-bypass $P$ with respect to $C$ is minimum subject to (i);

(iii) the length of $P$ is minimum subject to (i) and (ii).

Without loss of generality, we assume that $x:=x_{1}$. Let $R:=V(D)\setminus V(C)$, $P_{1}:=P[u_{1},u_{s}]$ and $C^{\prime}:=V(C[y_{1},y^{-}_{C}])$. Note that $|R|\geq 4$. From Lemma 3.3 it follows that $|V(C[x,y])|\geq 3$. Then $|C^{\prime}|\geq s$ since $C$ is a longest non-Hamiltonian cycle in $D$.

We first prove that $|V(P_{1})|=1$, i.e., $s=1$.

Proof of $s=1$. Suppose, on the contrary, that $s\geq 2$. By the minimality of the gap $|C^{\prime}|+1$, we have

[TABLE]

It is not difficult to see that $s\leq 3$. Indeed, if $s\geq 4$, then from (1) it follows that $d(y^{-}_{C})\leq 2a-4$ and $d(u_{s})\leq 2a-4$, since each of $P_{1}$ and $C^{\prime}$ contains at least two vertices from each partite set. So, we have a contradiction, since $\{u_{s},y^{-}_{C}\}\rightarrow y$. Therefore, $s=3$ or $s=2$.

First consider the case $s=3$, i.e., $P_{1}=u_{1}u_{2}u_{3}$. From our assumption that $x=x_{1}$ it follows that $u_{1},u_{3}\in Y$ and $y\in X$. Therefore, $u_{2}\in X$ and $y^{-}_{C}\in Y$. Note that $x_{2}\in C^{\prime}$ since $|C^{\prime}|\geq 3$. By the minimality of the gap $|C^{\prime}|+1$, we have

[TABLE]

Therefore, $d(x_{2})\leq 2a-4$. From the minimality of the gap $|C^{\prime}|+1$ and $P_{1}$ it follows that $x_{1}u_{3}\notin A(D)$. Therefore, by (2), $d(u_{3})\leq 2a-3$. This together with condition $B_{0}$ and $\{y^{-}_{C},u_{3}\}\rightarrow y$ imply that $d(y^{-}_{C})\geq 2a-2$. Therefore, by (2), the vertex $y^{-}_{C}$ and every vertex of $X\setminus\{u_{2}\}$ form a 2-cycle, i.e., $y^{-}_{C}\leftrightarrow X\setminus\{u_{2}\}$. In particular, $y^{-}_{C}\leftrightarrow\{x_{2},z\}$, where $z$ is an arbitrary vertex of $X\cap R\setminus\{u_{2}\}$. By the minimality of the gap, we have $a(z,u_{3})=d^{-}(z,\{u_{1}\})=0$. Therefore, $d(z)\leq 2a-3$. This and $d(x_{2})\leq 2a-4$ contradict condition $B_{0}$ since $\{z,x_{2}\}\rightarrow y^{-}_{C}$.

Now consider the case $s=2$, i.e., $P_{1}=u_{1}u_{2}$. Notice that $u_{1},y\in Y$ and $y^{-}_{C}\in X$ since $x_{1}\in X$. It is easy to see that $|C^{\prime}|$ is even.

Assume first that $|C^{\prime}|\geq 4$. By the minimality of the gap $|C^{\prime}|+1$,

[TABLE]

where $y^{--}_{C}$ denotes the predecessor of $y^{-}_{C}$ on $C$. Therefore, $d(u_{2})\leq 2a-4$ and $d(y^{-}_{C})\leq 2a-2$. Since $\{u_{2},y^{-}_{C}\}\rightarrow y$, i.e., $\{u_{2},y^{-}_{C}\}$ is a dominating pair, by condition $B_{0}$ we have, $d(y^{-}_{C})=2a-2$. This together with the second equality of (3) imply that $y^{-}_{C}$ and every vertex of $Y\setminus\{u_{1}\}$ form a 2-cycle. In particular, $y\rightarrow y^{-}_{C}\rightarrow y_{1}$ and $y^{-}_{C}\leftrightarrow z$, where $z$ is an arbitrary vertex of $Y\cap R\setminus\{u_{1}\}$. From the minimality of the gap $|C^{\prime}|+1$ it follows that $a(z,u_{2})=d^{-}(z,\{x_{2}\})=0$. Hence, $d(z)\leq 2a-3$. Now we consider the vertex $y^{--}_{C}$. It is easy to see that $y^{--}_{C}y^{+}_{C}\notin A(D)$ (for otherwise, the cycle $x_{1}u_{1}u_{2}yy^{-}_{C}y_{1}\ldots y^{--}_{C}y^{+}_{C}\ldots x_{1}$ is longer than $C$). From this and the first equality of (3) it follows that $d(y^{--}_{C})\leq 2a-3$. Thus, we have $\{z,y^{--}_{C}\}\rightarrow y^{-}_{C}$ and $max\{d(z),d(y^{--}_{C})\}\leq 2a-3$, which contradict condition $B_{0}$.

Assume then that $|C^{\prime}|=2$, i.e., $C^{\prime}=\{y_{1},x_{2}\}$. Then $y=y_{2}$. By the minimality of the gap $|C^{\prime}|+1$,

[TABLE]

i.e., the vertices $u_{1}$ and $x_{2}$ (respectively, $u_{2}$ and $y_{1}$) are not adjacent. Therefore,

[TABLE]

Without loss of generality, we may assume that $d(x_{2})=2a-2$, since $\{u_{2},x_{2}\}\rightarrow y_{2}$ (for otherwise, $d(u_{2})=2a-2$ and we will consider the cycle $x_{1}u_{1}u_{2}y_{2}\ldots y_{m}x_{1}$). Since $u_{1}$ and $x_{2}$ are non-adjacent and $d(x_{2})=2a-2$, it follows that $x_{2}$ and every vertex of $Y\setminus\{u_{1}\}$ form a 2-cycle, i.e., $x_{2}\leftrightarrow Y\setminus\{u_{1}\}$. In particular, $y_{2}\rightarrow x_{2}\rightarrow y_{1}$. Let $z$ be an arbitrary vertex in $Y\cap R\setminus\{u_{1}\}$. By the minimality of the gap $|C^{\prime}|+1$, $a(z,u_{2})=0$ and $x_{1}z\notin A(D)$. Hence, $d(z)\leq 2a-3$. If $y_{1}\rightarrow x_{3}$ (possibly, $x_{3}=x_{1}$), then, since $y_{2}\rightarrow x_{2}\rightarrow y_{1}$, we see that $x_{1}u_{1}u_{2}y_{2}x_{2}y_{1}x_{3}\ldots x_{1}$ is a cycle of length $|C|+2$, a contradiction. We may therefore assume that $y_{1}x_{3}\notin A(D)$. This together with $a(y_{1},u_{2})=0$ (by (4)) gives $d(y_{1})\leq 2a-3$. Thus we have that $\{y_{1},z\}\rightarrow x_{2}$ and $max\{d(y_{1}),d(z)\}\leq 2a-3$, which contradict condition $B_{0}$. This contradiction completes the proof of $s=1$.

From $s=1$ it follows that $u_{1}\in Y$ and $y\in X$ since $x_{1}\in X$. Without loss of generality, we may assume that $y:=x_{r}$. From now on, let $y:=u_{1}$.

Now we divide the proof of the theorem into two parts: $|C^{\prime}|=1$ and $|C^{\prime}|\geq 2$.

Part I. $|C^{\prime}|=1$, i.e., $r=2$ and $x_{1}\rightarrow y\rightarrow x_{2}$.

By condition $B_{0}$, $max\{d(y),d(y_{1})\}\geq 2a-2$ since $\{y,y_{1}\}\rightarrow x_{2}$. Without loss of generality, assume that $d(y)\geq 2a-2$. For this part we first prove Claims 1-5 below.

Claim 1. If $x\in R\cap X$ and $x\leftrightarrow y$, then $d(x)\leq 2a-3$ and $d(x_{1})\geq 2a-2$.

Proof of Claim 1. Assume that $x\in R\cap X$ and $x\leftrightarrow y$, but $d(x)\geq 2a-2$. It is easy to see that $a(x,y_{1})=0$ since $C$ is a longest non-Hamiltonian cycle in $D$. This and $d(x)\geq 2a-2$ imply that the vertex $x$ and every vertex of $Y\setminus\{y_{1}\}$ form a 2-cycle. In particular, $x\leftrightarrow\{y_{0},y_{2},y_{m}\}$ (possibly, $y_{2}=y_{m}$), where $y_{0}$ is an arbitrary vertex of $Y\cap R\setminus\{y\}$. Using this, it is easy to check that

[TABLE]

since $C$ is a longest non-Hamiltonian cycle in $D$. From the last equalities we have $d(x_{1})\leq 2a-3$, and $y\leftrightarrow x_{0}$, where $x_{0}$ is an arbitrary vertex of $X\cap R\setminus\{x\}$, since $d(y)\geq 2a-2$. Since $C$ is a longest non-Hamiltonian cycle in $D$ and since $x\leftrightarrow\{y_{0},y_{2},y_{m}\}$, it follows that $d(x_{0},\{y_{1},y_{0}\})=0$. Therefore, $d(x_{0})\leq 2a-4$, which contradicts condition $B_{0}$ since $max\{d(x_{1}),d(x_{0})\}\leq 2a-3$ and $\{x_{0},x_{1}\}\rightarrow y$. This contradiction proves that $d(x)\leq 2a-3$. From this and condition $B_{0}$ it follows that $d(x_{1})\geq 2a-2$ since $\{x,x_{1}\}\rightarrow y$. Claim 1 is proved.

Claim 2 follows immediately from Claim 1 and condition $B_{0}$.

Claim 2. There are no two distinct vertices $x,x_{0}\in R\cap X$ such that $x\leftrightarrow y$ and $x_{0}\leftrightarrow y$.

Claim 3. If $y\leftrightarrow x$ for some $x\in R\cap X$, then $a(y,z)=1$ for all $z\in R\cap X\setminus\{x\}$.

Proof of Claim 3. Let $x\leftrightarrow y$ for some $x\in R\cap X$. Then Claim 2 implies that $a(y,z)\leq 1$ for all $z\in R\cap X\setminus\{x\}$.

We want to show that $a(y,z)=1$ for all $z\in R\cap X\setminus\{x\}$. Assume that this is not the case. Then $a(y,x_{0})=0$ for some $x_{0}\in R\cap X\setminus\{x\}$. This together with Claims 1 and 2 thus imply that $|R|=4$ since $d(y)\geq 2a-2$ by our assumption. Then $R\cap X=\{x,x_{0}\}$. Let $R\cap Y=\{y,y_{0}\}$. Since $a(y,x_{0})=0$ and $d(y)\geq 2a-2$, it follows that

[TABLE]

Since $C$ is a longest non-Hamiltonian cycle in $D$, (5) implies that $d(x,\{y_{1},y_{2},\ldots,y_{m}\})=0$, which in turn implies that the vertices $x$ and $y_{0}$ are adjacent and $d(x)\leq 2a-4$ since $UG(D)$ is 2-connected and $m\geq 2$. By the assumption of Claim 3 and (5), $\{x,x_{i}\}\rightarrow y$ for all $i\in[1,m]$. Now using condition $B_{0}$, we obtain

[TABLE]

The remainder of the proof of Claim 3 is divided into two subcases depending on the value of $a(x,y_{0})$.

Case 1. $a(x,y_{0})=2$, i.e., $x\leftrightarrow y_{0}$.

Since $C$ is a longest non-Hamiltonian cycle in $D$, from (5) and $y\leftrightarrow x$, $x\leftrightarrow y_{0}$ it follows that

[TABLE]

Therefore, since $UG(D)$ is 2-connected, the vertices $y_{0}$ and $x_{0}$ are adjacent. Using (5), it is not difficult to see that:

If $x_{0}\rightarrow y_{0}$, then $d^{-}(x_{0},\{y_{1},y_{2},\ldots,y_{m}\})=0$, and if $y_{0}\rightarrow x_{0}$, then $d^{+}(x_{0},\{y_{1},y_{2},\ldots,y_{m}\})=0$.

In both cases we have $a(y_{i},x_{0})\leq 1$ for all $i\in[1,m]$. Together with $a(y_{i},x)=0$ this implies that $d(y_{i})\leq 2a-3$. On the other hand, from (6) and (7) it follows that every vertex $x_{i}$, $i\in[1,m]$, and every vertex of $Y\setminus\{y_{0}\}$ form a 2-cycle. In particular, $\{y_{1},y_{2}\}\rightarrow x_{2}$, which contradicts condition $B_{0}$, since $max\{d(y_{1}),d(y_{2})\}\leq 2a-3$.

Case 2. $a(x,y_{0})=1$, i.e., $y_{0}\mapsto x$ or $x\mapsto y_{0}$.

Let $y_{0}\mapsto x$. Since $C$ is a longest non-Hamiltonian cycle in $D$ and (5), we have

[TABLE]

Hence $x_{0}\rightarrow y_{0}$ since $D$ is strong and $y_{0}\mapsto x$. Now using (5), we obtain

[TABLE]

The last two equalities together with $a(y,x_{0})=0$ and $y_{0}\mapsto x$ imply that

[TABLE]

which contradicts that $D$ is strong.

Let now $x\mapsto y_{0}$. Again using (5), it is easy to see that $d^{+}(y_{0},\{x,x_{1},x_{2},\ldots,x_{m}\})=0$. Therefore, $y_{0}\rightarrow x_{0}$ since $D$ is strong. Together with (5) this implies that $d^{+}(x_{0},\{y_{1},y_{2},\ldots,y_{m},y\})=0$. Therefore $x_{0}\rightarrow y_{0}$ since $D$ is strong, and $d(x_{0})\leq 2a-4$ since $m\geq 2$. Thus, $d(x)\leq 2a-3$ (Claim 1) and $d(x_{0})\leq 2a-4$, which contradicts condition $B_{0}$ since $\{x,x_{0}\}\rightarrow y_{0}$. This contradiction completes the proof of Claim 3.

Claim 4. If $y\leftrightarrow x$ for some $x\in R\cap X$, then $d^{-}(y,R\cap X\setminus\{x\})=0.$

Proof of Claim 4. Assume that the claim is not true, i.e., there exist vertices $x\in R\cap X$ and $u\in R\cap X\setminus\{x\}$ such that $y\leftrightarrow x$ and $u\rightarrow y$. Then, by Claim 2, $yu\notin A(D)$. Notice that $\{x,u\}\rightarrow y$. Since $d(x)\leq 2a-3$ (Claim 1), condition $B_{0}$ implies that $d(u)\geq 2a-2$. It is clear that $y_{1}u\notin A(D)$ (if $y_{1}\rightarrow u$, then $x_{1}y_{1}uyx_{2}y_{3}\ldots y_{m}x_{1}$ is a cycle of length $2m+2$, a contradiction). Using this, $yu\notin A(D)$ and $d(u)\geq 2a-2$ we conclude that $uy_{1}\in A(D)$, $d(u)=2a-2$ and the vertex $u$ together with every vertex of $Y\setminus\{y,y_{1}\}$ forms a 2-cycle. In particular, $u\leftrightarrow\{y_{2},y_{m},v\}$, where $v\in Y\cap R\setminus\{y\}$ (possibly, $y_{2}=y_{m}$). Using this, it is not difficult to show that $a(x_{1},v)=0$ and $yx_{1}\notin A(D)$. Indeed, if $x_{1}\rightarrow v$, then $x_{1}vuy_{1}x_{2}\ldots y_{m}x_{1}$ is a cycle of length $2m+2$; if $v\rightarrow x_{1}$, then $y_{m}uvx_{1}y_{1}\ldots$ $x_{m}y_{m}$ is a cycle of length $2m+2$; if $y\rightarrow x_{1}$, then $y_{m}uyx_{1}y_{1}\ldots x_{m}y_{m}$ is a cycle of length $2m+2$. In each case we obtain a contradiction. Hence, $a(x_{1},v)=0$ and $yx_{1}\notin A(D)$. From this it follows that $d(x_{1})\leq 2a-3$, which contradicts that $d(x_{1})\geq 2a-2$ (Claim 1). Claim 4 is proved.

Claim 5. There is no $x\in X\cap R$ such that $y\leftrightarrow x$, i.e., in subdigraph $D\langle R\rangle$ through the vertex $y$ there is no cycle of length two.

Proof of Claim 5. Assume that the claim is not true, i.e., there exists a vertex $x\in X\cap R$ such that $y\leftrightarrow x$. By Claims 1, 3 and 4 we have

[TABLE]

Case 1. $x_{2}y\notin D$.

Then from the last expression of (8) and $d(y)\geq 2a-2$ we conclude that $|R|=4$, i.e., $m=a-2\geq 3$, and the vertex $y$ and every vertex of $\{x_{1},x_{2},x_{3},x_{4},\ldots,x_{m}\}\setminus\{x_{2}\}$ form a 2-cycle, i.e.,

[TABLE]

Put $R=\{y,x,y_{0},x_{0}\}$, where $x_{0}\in X$ and $y_{0}\in Y$. From (8) it follows that $y\mapsto x_{0}$. Now using (9), it is not difficult to see that

[TABLE]

Indeed, if $y_{i}\rightarrow x$ and $i\in[1,m]$ (respectively, $x\rightarrow y_{j}$ and $j\in[1,m]\setminus\{2\}$), then $x_{i}y_{i}xyx_{i+1}\ldots x_{i}$ (respectively, $x_{j}yxy_{j}\ldots y_{j-1}x_{j}$) is a cycle of length $2a-2$, a contradiction.

Similarly,

[TABLE]

In particular, (10) implies that

[TABLE]

i.e., the vertices $x$ and $y_{i}$, $i\notin\{0,2\}$, are not adjacent. From this, (11) and $y\mapsto x_{0}$ it follows that

[TABLE]

Using (12) and condition $B_{0}$, we obtain that for all $u\in X$ and $v\in Y$ the following holds

[TABLE]

Now we divide this case into four subcases.

Subcase 1.1. $x\leftrightarrow y_{0}$.

Since $x\leftrightarrow y$ and $x\leftrightarrow y_{0}$, using (9) and the fact that $m\geq 3$, it is not difficult to check that the vertices $x_{1}$ and $y_{0}$ are not adjacent (for otherwise, $D$ would contain a cycle of length $2a-2$, a contradiction). Together with the first inequality of (13) (when $u=x_{1}$) this implies that $d(x_{1})\leq 2a-3$, which contradicts Claim 1.

Subcase 1.2. $x\mapsto y_{0}$.

Then, by the second inequality of (13) when $v=y_{0}$, $x_{0}y_{0}\notin A(D)$. This together with $x_{0}y\notin A(D)$ and (11) imply that $x_{0}\rightarrow y_{2}$ since $D$ is strong. It is easy to see that $y_{0}x_{2}\notin A(D)$ (for otherwise, $x_{1}yxy_{0}x_{2}y_{2}\ldots y_{m}x_{1}$ is a cycle of length $2a-2$, a contradiction). Combining this with $x_{2}y\notin A(D)$ and $d^{-}(x_{2},\{y_{1},y_{m}\})\leq 1$ (by (13)) we obtain $d(x_{2})\leq 2a-3$. Thus, $d(x_{2})\leq 2a-3$ and $d(x_{0})\leq 2a-3$ (by (12)) and $\{x_{0},x_{2}\}\rightarrow y_{2}$, which contradict condition $B_{0}$.

Subcase 1.3. $y_{0}\mapsto x$.

If $x_{i}\rightarrow y_{0}$ and $i\in[1,m]$, then using (9) we obtain that the cycle $x_{i}y_{0}xyx_{i+1}\ldots x_{i}$ has length $2a-2$, which is a contradiction. We may therefore assume that $d^{-}(y_{0},\{x,x_{1},x_{2},\ldots,x_{m}\})=0$. Hence, $x_{0}\rightarrow y_{0}$ since $D$ is strong and $y_{0}\mapsto x$. Since $x_{m}\rightarrow y\rightarrow x_{2}$ and $C$ is a longest non-Hamiltonian cycle in $D$, it follows that the vertices $x_{1}$ and $y_{0}$ are not adjacent. Combing this with $d^{-}(x_{1},\{y_{1},y_{m}\})\leq 1$ (by (13)) we obtain that $d(x_{1})\leq 2a-3$, a contradiction to Claim 1.

Subcase 1.4. The vertices $x$ and $y_{0}$ are not adjacent.

Since the underlying undirected graph of $D$ is 2-connected, from $a(x,y_{0})=0$ and (10) it follows that $x\rightarrow y_{2}$. Together with (13) this imply that $x_{0}y_{2}\notin A(D)$. Therefore, by (11), we have that $d^{+}(x_{0},\{y,y_{1},y_{2},\ldots,y_{m}\})=0$. Hence, $x_{0}\rightarrow y_{0}$ since $D$ is strong. If $y_{0}\rightarrow x_{2}$, then $x_{1}yx_{0}y_{0}x_{2}y_{2}\ldots y_{m}x_{1}$ is a cycle of length $2a-2$, which is a contradiction. We may therefore assume that $y_{0}x_{2}\notin A(D)$. Combining this with $x_{2}y\notin A(D)$ and (13) we obtain that $d(x_{2})\leq 2a-3$, which contradicts condition $B_{0}$ since $\{x_{2},x\}\rightarrow y_{2}$ and $d(x)\leq 2a-3$ (Claim 1). The discussion of Case 1 is completed.

Case 2. $x_{2}\rightarrow y$.

Since $y\mapsto X\cap R\setminus\{x\}$ (by (8)) and $d(y)\geq 2a-2$, it follows that the vertices $y$ and $x_{i}$, where $i\in[1,m]$, are adjacent and $y\rightarrow\{x_{1},x_{2},\ldots,x_{m}\}$ or $\{x_{1},x_{2},\ldots,x_{m}\}\rightarrow y$. Therefore, without loss of generality, we may assume that

[TABLE]

(for otherwise we will have the considered Case 1). Let $x_{0}$ be an arbitrary vertex in $X\cap R\setminus\{x\}$. Then by (8) we have $y\mapsto x_{0}$. It is clear that

[TABLE]

This implies that $d(x_{0})\leq 2a-3$. Since $D$ is strong and (15), there is a vertex $y_{0}\in Y\cap R\setminus\{y\}$ such that $x_{0}\rightarrow y_{0}$. Now, since $y\rightarrow x_{0}\rightarrow y_{0}$ and $x_{i}\rightarrow y$ for all $i\in[1,m]$ (by (14)), we conclude that $d^{+}(y_{0},\{x_{1},x_{2},\ldots,x_{m}\})=0$ (for otherwise, for some $i\in[1,m]$, $y_{0}\rightarrow x_{i}$ and $x_{i-1}yx_{0}y_{0}x_{i}\ldots y_{i-2}x_{i-1}$ is a cycle of length $2m+2$, a contradiction). The last equality together with $d^{+}(x_{0},\{y,y_{1},y_{2},\ldots,y_{m}\})=0$ (by (15)) imply that $y_{0}\rightarrow x$ since $D$ is strong (for otherwise, $A(\{x_{0},y_{0}\}\rightarrow V(D)\setminus\{x_{0},y_{0}\})=\emptyset$, which contradicts that $D$ is strong). Then using the facts that $y\rightarrow x_{0}\rightarrow y_{0}$ and $y_{0}\rightarrow x\rightarrow y$, it is not difficult to show that $x_{1}$ and $y_{0}$ are not adjacent. Indeed, by (14) we have that: If $x_{1}\rightarrow y_{0}$, then $x_{1}y_{0}xyx_{2}y_{2}\ldots y_{m}x_{1}$ is a cycle of length $2m+2$; and if $y_{0}\rightarrow x_{1}$, then $x_{m}yx_{0}y_{0}x_{1}y_{1}\ldots$ $y_{m-1}x_{m}$ is a cycle of length $2m+2$, in both cases we have a contradiction. So, $x_{1}$ and $y_{0}$ are not adjacent. Together with $d(x_{1})\geq 2a-2$ (Claim 1) this implies that $y_{1}\rightarrow x_{1}$. On the other hand, since $d(x,\{y_{1},y_{m}\})=d^{+}(x_{0},\{y_{1},y_{m}\})=0$, we have that $max\{d(y_{1}),d(y_{m})\}\leq 2a-3$, which contradicts condition $B_{0}$, because of $\{y_{m},y_{1}\}\rightarrow x_{1}$. Claim 5 is proved.

Now we can finish the discussion of Part I.

From Claim 5 it follows that in $D\langle R\rangle$ there is no cycle of length two through the vertex $y$. Then, since $d(y)\geq 2a-2$ and $|R|\geq 4$, it follows that $|R|=4$. Put $X\cap R=\{x,x_{0}\}$ and $Y\cap R=\{y,y_{0}\}$. Then $a(y,x)=a(y,x_{0})=1$ and the vertex $y$ and every vertex of $X\cap C$ form a 2-cycle, i.e.,

[TABLE]

First consider the case $d^{+}(y,\{x,x_{0}\})\geq 1$. Assume, without loss of generality, that $y\mapsto x$. Then, by (16), $d^{+}(x,\{y_{1},y_{2},\ldots,y_{m}\})=0$. Together with $xy\notin A(D)$ this implies that $x\rightarrow y_{0}$ since $D$ is strong. Therefore $d(x)\leq 2a-3$ since $|Y\cap C|\geq 2$. By (16), it is clear that $d^{+}(y_{0},\{x_{1},x_{2},\ldots,x_{m}\})=0$. If $y\rightarrow x_{0}$, then analogously we obtain that $d^{+}(x_{0},\{y,y_{1},y_{2},\ldots,y_{m}\})=0$ and $x_{0}\rightarrow y_{0}$, $d(x_{0})\leq 2a-3$, which contradicts condition $B_{0}$ since $max\{d(x),d(x_{0})\}\leq 2a-3$ and $\{x,x_{0}\}\rightarrow y_{0}$. We may assume therefore that $yx_{0}\notin A(D)$. Then $x_{0}\rightarrow y$ (by $a(y,x_{0})=1$), $d^{-}(x_{0},\{y,y_{1},y_{2},\ldots,y_{m}\})=0$ (by (16)) and hence, $y_{0}\rightarrow x_{0}$ since $D$ is strong. Now it is not difficult to show that

[TABLE]

Therefore $d(x_{0})\leq 2a-3$ and $d(y_{i})\leq 2a-3$. Since for all $i\in[1,m]$, $\{x_{i},x_{0}\}\rightarrow y$ and $d(x_{0})\leq 2a-3$, from condition $B_{0}$ it follows that $d(x_{i})\geq 2a-2$ for all $i\in[1,m]$. This together with $a(x_{i},y_{0})=0$ (by (16) and the last equalities) imply that $y_{i}\leftrightarrow x_{i}$. Thus, $\{y_{i-1},y_{i}\}\rightarrow x_{i}$ and $max\{d(y_{i-1}),d(y_{i})\}\leq 2a-3$, which is a contradiction.

Now consider the case $d^{+}(y,\{x,x_{0}\})=0$. Then $\{x,x_{0}\}\rightarrow y$, because of $a(y,x)=a(y,x_{0})=1$, i.e., $\{x,x_{0}\}$ is a dominating pair. It is clear that $d^{-}(x,\{y_{1},y_{2}\})=d^{-}(x_{0},\{y_{1},y_{2}\})=0$. This together with $d^{+}(y,\{x_{0},x\})=0$ imply that $max\{d(x),d(x_{0})\}\leq 2a-3$, which is a contradiction because of $\{x,x_{0}\}\rightarrow y$. This completes the discussion of the part $|C^{\prime}|=1$.

Part 2. $|C^{\prime}|\geq 2$.

Then $|C^{\prime}|\geq 3$ since $|C^{\prime}|$ is odd. For this part we first will prove Claims 6-8.

Claim 6. If $|C^{\prime}|\geq 3$, then the following holds:

(a). $d(y)\leq 2a-3$ and $d(y_{r-1})\geq 2a-2$;

(b). There is no $x\in X\cap R$ such that $x\leftrightarrow y_{r-1}$, i.e., $a(y_{r-1},x)\leq 1$ for all $x\in X\cap R$;

(c). $a(y_{r-1},x)=1$ for all $x\in X\cap R$, $|R|=4$, $d(y_{r-1})=2a-2$ and the vertex $y_{r-1}$ together with every vertex of $X\cap V(C)$ forms a 2-cycle. In particular, $x_{r}\leftrightarrow y_{r-1}$ and $y_{r-1}\leftrightarrow x_{2}$;

(d). $d^{-}(y_{r-1},\{x,x_{0}\})=0$ and $y_{r-1}\mapsto\{x,x_{0}\}$;

(e). $max\{d(x_{0}),d(x)\}\leq 2a-3$ and $d^{-}(v,\{x_{0},x\})\leq 1$ for all $v\in Y$;

(f). $|C^{\prime}|=3$, i.e., $r=3$.

Proof of Claim 6.

(a). Suppose on the contrary, that $d(y)\geq 2a-2$. Then, since $d(y,C^{\prime})=0$, we have that $r=3$ and the vertex $y$ together with every vertex of $X\setminus\{x_{2}\}$ forms a 2-cycle. In particular, $y\leftrightarrow x_{1}$ and $y\leftrightarrow x_{3}$. It follows that for some $i\in[3,m]$, $x_{i}\rightarrow y\rightarrow x_{i+1}$, which contradicts that $C$-bypass $P$ has the minimum gap among the gaps of all $C$-bypasses. Therefore, $d(y)\leq 2a-3$. Now, since $\{y,y_{r-1}\}\rightarrow x_{r}$, from condition $B_{0}$ it follows that $d(y_{r-1})\geq 2a-2$.

(b). Suppose that Claim 6(b) is falls, i.e., there is a vertex $x\in X\cap R$ such that $x\leftrightarrow y_{r-1}$. From the minimality of the gap $|C^{\prime}|+1$ it follows that

[TABLE]

From this we have $d(x)\leq 2a-4$. This together with condition $B_{0}$ and $\{x,x_{r-1}\}\rightarrow y_{r-1}$ imply that $d(x_{r-1})\geq 2a-2$. Therefore, since $a(x_{r-1},y)=0$, it follows that

(i) *the vertex $x_{r-1}$ and every vertex of $Y\setminus\{y\}$ form a 2-cycle.

Let $y^{\prime}$ be an arbitrary vertex in $Y\cap R\setminus\{y\}$. By (i), $x_{r-1}\leftrightarrow y^{\prime}$. By the minimality of the gap $|C^{\prime}|+1$, $d^{+}(y^{\prime},\{x,x_{r}\})=d^{-}(y^{\prime},\{x_{1}\})=0$. Therefore, $d(y^{\prime})\leq 2a-3$. Since $\{y^{\prime},y_{r-2}\}\rightarrow x_{r-1}$, condition $B_{0}$ implies that

[TABLE]

First consider the case $r\geq 4$. Then $x_{r-2}\in C^{\prime}$ and it is not difficult to see that $x_{r-2}y^{\prime}\notin A(D)$ and $y_{r-2}x\notin A(D)$. Now, since $a(x_{r-2},y)=0$, we have $d(x_{r-2})\leq 2a-3$. Using condition $B_{0}$ and the fact that $d(x)\leq 2a-4$, we conclude that $xy_{r-2}\notin A(D)$. Therefore, by (17), $x$ and $y_{r-2}$ are not adjacent. This together with $d(y_{r-2})\geq 2a-2$ (by (18)) imply that $y_{r-2}\leftrightarrow x_{0}$, where $x_{0}$ is an arbitrary vertex in $R\cap X\setminus\{x\}$. By the minimality of the gap $|C^{\prime}|+1$, we have $x_{0}y_{r-1}\notin A(D)$, $y_{r-3}x_{0}\notin A(D)$ and $x_{0}y^{\prime}\notin A(D)$. Therefore, $d(x_{0})\leq 2a-3$, which contradicts condition $B_{0}$, since $\{x_{0},x_{r-2}\}\rightarrow y_{r-2}$ and $d(x_{r-2})\leq 2a-3$.

Now consider the case $r=3$. By (i), $x_{2}\leftrightarrow y_{3}$ and $x_{2}\leftrightarrow y_{2}$. If $y_{2}\rightarrow x_{4}$ (possibly, $x_{4}=x_{1}$) (respectively, $x_{3}\rightarrow y_{2}$), then the cycle $Q:=x_{1}yx_{3}y_{3}x_{2}y_{2}x_{4}\ldots y_{m}x_{1}$ (respectively, $Q:=x_{1}yx_{3}y_{2}x_{2}y_{3}x_{4}\ldots y_{m}x_{1}$) has length $2m$, the vertex $y_{1}$ is not on this cycle and $x_{1}\rightarrow y_{1}\rightarrow x_{2}$ is a $Q$-bypass whose gap with respect to $Q$ is equal to 4 and $d(y_{1})\geq 2a-2$, (by (18) since $r-2=1$), which contradicts Claim 6(a). We may therefore assume that $y_{2}x_{4}\notin A(D)$ and $x_{3}y_{2}\notin A(D)$. Combining this with $d(y_{2})\geq 2a-2$ (by Claim 6(a) and $r=3$) we obtain that $d(y_{2})=2a-2$ and $y_{2}$ together with every vertex of $X\setminus\{x_{3},x_{4}\}$ forms a 2-cycle. In particular, $y_{2}\leftrightarrow x_{0}$. On the other hand, from the minimality of the gap $|C^{\prime}|+1$ it follows that $x_{0},y$ are not adjacent and $y_{1}x_{0}\notin A(D)$. Therefore, $d(x_{0})\leq 2a-3$, which is a contradiction, since $d(x)\leq 2a-3$ and $\{x,x_{0}\}\rightarrow y_{2}$.

(c). Claim 6(c) is an immediate corollary of Claims 6(a) and 6(b).

From now on, we assume that $X\cap R=\{x,x_{0}\}$ and $Y\cap R=\{y,y_{0}\}$.

(d). Suppose, on the contrary, that there exists a vertex in $\{x,x_{0}\}$, say $x$, such that $x\rightarrow y_{r-1}$. From the minimality of the gap $|C^{\prime}|+1$ it follows that $d^{-}(x,\{y,y_{1}\})=0$. This together with $y_{r-1}x\notin A(D)$ (Claim 6(b)) imply that $d(x)\leq 2a-3$. Therefore, using Claim 6(c) and condition $B_{0}$, we obtain that $d(x_{2})\geq 2a-2$ and $d(x_{r-1})\geq 2a-2$ (possibly, $x_{2}=x_{r-1}$) since $\{x,x_{2},x_{r-1}\}\rightarrow y_{r-1}$. Together with $d(y,\{x_{2},x_{r-1}\})=0$ this implies that $x_{r-1}\rightarrow y_{r}$ and $x_{2}\leftrightarrow y_{0}$. Because of gap minimality, $d^{-}(y_{0},\{x_{1}\})=d^{+}(y_{0},\{x,x_{r}\})=0$. Therefore, $d(y_{0})\leq 2a-3$, and hence, by condition $B_{0}$, $d(y_{1})\geq 2a-2$ since $\{y_{1},y_{0}\}\rightarrow x_{2}$. Since $x_{r}\rightarrow y_{r-1}$ and $y_{r-1}\rightarrow x_{2}$ (Claim 6(c)) and $x_{r-1}\rightarrow y_{r}$, the cycle $Q:=x_{1}yx_{r}y_{r-1}x_{2}\ldots x_{r-1}y_{r}\ldots y_{m}x_{1}$ has length equal to $2a-4$. Observe that $y_{1}\notin V(Q)$ and $x_{1}\rightarrow y_{1}\rightarrow x_{2}$ is a $Q$-bypass, whose gap with respect to $Q$ is equal to 4, but $d(y_{1})\geq 2a-2$, which contradicts Claim 6(a).

(e). By Claim 6(d), $y_{r-1}\mapsto\{x,x_{0}\}$. Therefore, because of gap minimality, we have

[TABLE]

Therefore, $max\{d(x_{0}),d(x)\}\leq 2a-3$ and, by condition $B_{0}$, the vertices $x$, $x_{0}$ does not form a dominating pair, i.e., $d^{-}(v,\{x_{0},x\})\leq 1$ for all $v\in Y$.

(f). Suppose, on the contrary, that is $|C^{\prime}|>3$. Then $|C^{\prime}|\geq 5$, i.e., $r\geq 4$, since $|C^{\prime}|$ is odd. By Claim 6(c), $x_{i}\leftrightarrow y_{r-1}$ for all $i\in[1,m]$. This together with condition $B_{0}$ imply that $d(x_{i})\geq 2a-2$ for all $i\in[1,m]$ maybe except one. In particular, $d(x_{r-1})\geq 2a-2$ or $d(x_{r-2})\geq 2a-2$.

Fist consider the case $d(x_{r-1})\geq 2a-2$. From this and $a(y,x_{r-1})=0$ it follows that $x_{r-1}\leftrightarrow y_{r}$. Using this and the fact that $x_{r}\rightarrow y_{r-1}\rightarrow x_{2}$ (Claim 6(c)), we see that the cycle $Q:=x_{1}yx_{r}y_{r-1}x_{2}y_{2}\ldots x_{r-1}y_{r}$ $x_{r+1}\ldots x_{1}$ has length equal to $2a-4$. Observe that $x_{1}\rightarrow y_{1}\rightarrow x_{2}$ is a $Q$-bypass whose gap with respect to $Q$ is equal to 4, which contradicts the choice of the cycle $C$ and $C$-bypass $P$.

Now consider the case $d(x_{r-1})\leq 2a-3$. Then $d(x_{r-2})\geq 2a-2$. Observe that the vertex $x_{r-2}$ and every vertex of $Y$ other than $y$ form a 2-cycle (since $x_{r-2}$ and $y$ are not adjacent), in particular, $y_{0}\leftrightarrow x_{r-2}$. It is easy to see that $x_{r-2}=x_{2}$, i.e., $r=4$. Indeed, if $r-2\geq 3$, then $x_{r-3}\in C^{\prime}$ and $x_{r-3}y_{0}\notin A(D)$ because of the minimality of the gap $|C^{\prime}|+1$ and $y_{0}\leftrightarrow x_{r-2}$. This together with $a(y,x_{r-3})=0$ gives $d(x_{r-3})\leq 2a-3$. Thus we have that the vertices $x_{r-3}$ and $x_{r-1}$ both have degree less than $2a-2$, which contradicts the fact that at most one vertex $x_{i}$, $i\in[1,m]$ maybe has degree less that $2a-2$. Thus, $r=4$. By the above observations, $y_{4}\rightarrow x_{2}$ and $y_{3}\rightarrow x_{5}$. Therefore, $Q:=x_{1}yx_{4}y_{4}x_{2}y_{2}x_{3}y_{3}x_{5}\ldots x_{1}$ is a cycle of length $2a-4$. Notice that $y_{1}\notin V(Q)$ and the $Q$-bypass $x_{1}\rightarrow y_{1}\rightarrow x_{2}$ has gap with respect to $Q$ equal to 4. This contradicts the choice of $C$ and $C$-bypass $P$.

From Claims 6(c), 6(d) and 6(f)1 it follows that

(ii) *If $|C^{\prime}|\geq 3$, then $|C^{\prime}|=3$, $|R|=4$, $y_{2}\mapsto\{x,x_{0}\}$, $d(y)\leq 2a-3$, $d(y_{2})=2a-2$ and the vertex $y_{2}$ together with every vertex of $X\cap V(C)$ forms a 2-cycle.

Claim 7. If $|C^{\prime}|=3$, then $d(y_{0})\leq 2a-3$, (recall that $\{y_{0}\}=Y\cap R\setminus\{y\}$).

Proof of Claim 7. By Claim 6((e), $d^{-}(y_{0},\{x,x_{0}\})\leq 1$. Therefore, if $x_{2}$ and $y_{0}$ are not adjacent, then $d(y_{0})\leq 2a-3$. We may therefore assume that $x_{2}$ and $y_{0}$ are adjacent. Then $y_{0}\rightarrow x_{2}$ or $x_{2}\rightarrow y_{0}$. It is easy to see that if $y_{0}\rightarrow x_{2}$, then $x_{1}y_{0}\notin A(D)$; and if $x_{2}\rightarrow y_{0}$, then $y_{0}x_{3}\notin A(D)$. Therefore, $d(y_{0},\{x_{1},x_{2},x_{3}\})\leq 4$. This and $d^{-}(y_{0},\{x_{0},x\})\leq 1$ imply that $d(y_{0})\leq 2a-3$.

Combining Claims 6(a), 6(e) and 7 we obtain that if $|C^{\prime}|=3$, then

[TABLE]

in particular, by condition $B_{0}$, we have

[TABLE]

Claim 8. If $|C^{\prime}|=3$, then the following holds:

(a). $d(x_{2})\leq 2a-3$ and $d(x_{3})\geq 2a-2$;

(b). $d(y_{1})\geq 2a-2$;

(c). If $x\rightarrow y_{k}$ or $x_{0}\rightarrow y_{k}$, where $k\in[3,m]$, then $x_{k}y_{1}\notin A(D)$.

Proof of Claim 8.

(a). Suppose, on the contrary, that $|C^{\prime}|=3$ and $d(x_{2})\geq 2a-2$. This and $a(x_{2},y)=0$ imply that $d(x_{2})=2a-2$ and $x_{2}$ together with every vertex of $Y\setminus\{y\}$ form a 2-cycle. In particular, $x_{2}\leftrightarrow y_{0}$ and $x_{2}\leftrightarrow y_{3}$. Since $d(y_{0})\leq 2a-3$ (by (19)) and $\{y_{1},y_{0}\}\rightarrow x_{2}$, from condition $B_{0}$ it follows that $d(y_{1})\geq 2a-2$. On the other hand, using the facts that $x_{3}\rightarrow y_{2}\rightarrow x_{2}$ (Claim 6(c)) and $x_{2}\rightarrow y_{3}$ we see that $Q:=x_{1}yx_{3}y_{2}x_{2}y_{3}\ldots x_{1}$ is a cycle of length $2a-4$. Notice that $x_{1}\rightarrow y_{1}\rightarrow x_{2}$ is a $Q$-bypass, whose the gap with respect to $Q$ is equal to 4. This contradicts the minimality of the gap $|C^{\prime}|+1$ or Claim 6(a) since $d(y_{1})\geq 2a-2$. Therefore, $d(x_{2})\leq 2a-3$. Together with condition $B_{0}$ this implies that $d(x_{3})\geq 2a-2$ since $\{x_{2},x_{3}\}\rightarrow y_{2}$.

(b). Suppose, on the contrary, that $|C^{\prime}|=3$ and $d(y_{1})\leq 2a-3$. By condition $B_{0}$, $y_{1}x_{3}\notin A(D)$ since $d(y)\leq 2a-3$ (Claim 6(a)) and $y\rightarrow x_{3}$. From (20) we have that $y_{0}x_{3}\notin A(D)$ since $yx_{3}\in A(D)$. Thus, the arcs $y_{1}x_{3}$ and $y_{0}x_{3}$ are not in $A(D)$. This together with $d(x_{3})\geq 2a-2$ (Claim 8(a)) imply that $x_{3}\rightarrow y_{1}$ and $y_{3}\rightarrow x_{3}$. By (ii), $y_{2}\rightarrow x_{4}$ and hence, $Q:=x_{1}yx_{3}y_{1}x_{2}y_{2}x_{4}\ldots x_{1}$ is a cycle of length $2a-4$, which does not contain the vertices $x_{0},y_{0},x$ and $y_{3}$. The path $x_{3}y_{3}x_{4}$ is a $Q$-bypass whose the gap with respect to $Q$ is equal to 4. Therefore, by the minimality of the gap $|C^{\prime}|+1$ and Claim 6(a), $d(y_{3})\leq 2a-3$ but this is a contradiction since $\{y,y_{3}\}\rightarrow x_{3}$ and $d(y)\leq 2a-3$ (Claim 6(a)).

(c). Suppose that the claim is not true. Without loss of generality, assume that for some $k\in[3,m]$, $x\rightarrow y_{k}$ and $x_{k}\rightarrow y_{1}$. Then, by (ii) we have that $y_{2}\rightarrow\{x,x_{0}\}$ and hence, the cycle $x_{1}yx_{3}\ldots x_{k}y_{1}x_{2}y_{2}xy_{k}$ $\ldots y_{m}x_{1}$ is a cycle of length $2a-2$, a contradiction.

Now we are ready to complete the proof of Theorem 1.7. Combining (19) and Claim 8(a), we obtain

[TABLE]

This and condition $B_{0}$ imply

[TABLE]

Therefore, since $d(y_{1})\geq 2a-2$ (Claim 8(b)), the vertex $y_{1}$ and every vertex of $X\setminus\{x,x_{0},x_{2}\}$ form a 2-cycle, i.e.,

[TABLE]

Now using Claim 8(c), we obtain $A(\{x,x_{0}\}\rightarrow\{y_{2},y_{3},\ldots,y_{m}\})=\emptyset$. From $y_{2}\mapsto\{x,x_{0}\}$ (Claim 6(d)) and the minimality of the gap $|C^{\prime}|+1$ it follows that $d^{-}(y,\{x,x_{0}\})=0$. The last two equalities imply that

[TABLE]

By (20), in particular, we have

[TABLE]

Since $D$ is strong, from (23) and (24) it follows that $x\rightarrow y_{0}$ or $x_{0}\rightarrow y_{0}$. Again using (24), we obtain that if $x\rightarrow y_{0}$, then $x_{0}y_{0}\notin A(D)$ and $x_{0}\rightarrow y_{1}$; if $x_{0}\rightarrow y_{0}$, then $xy_{0}\notin A(D)$ and $x\rightarrow y_{1}$.

Because of the symmetry between the vertices $x$ and $x_{0}$, we can assume that $x_{0}\rightarrow y_{0}$, $x\rightarrow y_{1}$ and $x_{0}y_{1}\notin A(D)$, $xy_{0}\notin A(D)$. It is not difficult to show that $x_{2}$ and every vertex $y_{i}$ with $i\in[3,m]$ are not adjacent. Indeed, if $y_{i}\rightarrow x_{2}$, then, by (22), $y_{1}\rightarrow x_{i+1}$ and hence, $x_{1}yx_{3}y_{3}\ldots y_{i}x_{2}y_{2}xy_{1}x_{i+1}\ldots x_{1}$ is a cycle of length $2a-2$; if $x_{2}\rightarrow y_{i}$, then, by (ii), $x_{i}\rightarrow y_{2}$ and hence, $x_{1}yx_{3}y_{3}\ldots x_{i}y_{2}xy_{1}x_{2}y_{i}\ldots y_{m}x_{1}$ is a cycle of length $2a-2$. Thus, in both cases we have a contradiction.

Therefore, $a(y_{i},x_{2})=0$ for all $i\in[3,m]$. This and (20) imply that $d(y_{i})\leq 2a-3$ for all $i\in[3,m]$. From $y\rightarrow x_{3}$, $d(y)\leq 2a-3$ (Claim 6(a)) and condition $B_{0}$ it follows that $d^{-}(x_{3},\{y_{0},y_{3},y_{4},\ldots y_{m}\})=0$. Now frome $d(x_{3})\geq 2a-2$ (Claim 8(a)), we have that $m=3$, i.e., $a=5$. Since $x\rightarrow y_{1}$, $x_{0}\rightarrow y_{0}$ and (21), it follows that $d^{+}(x_{2},\{y_{0},y_{1}\})=0$. Because of $d(y_{1})\geq 2a-2$ (Claim 8(b)) and $d^{-}(y_{1},\{x_{0},x_{2}\})=0$ we have $y_{1}\rightarrow x_{0}$, $x_{3}\leftrightarrow y_{1}$ and $y_{1}\rightarrow x_{1}$. From this it follows that $y_{0}x_{2}\notin A(D)$ (for otherwise, $y_{1}x_{0}y_{0}x_{2}y_{2}\ldots x_{1}y_{1}$ is a cycle of length $2a-2$, a contradiction). Therefore, $a(x_{2},y_{0})=0$. If $y_{0}\rightarrow x_{1}$, then the cycle $x_{1}yx_{3}y_{1}x_{2}y_{2}x_{0}y_{0}x_{1}$ is a cycle of length $2a-2=8$, a contradiction. Therefore, $y_{0}x_{1}\notin A(D)$. So, we have $d^{+}(y_{0},\{x_{1},x_{2},x_{3}\})=0$. Then $d^{+}(y_{0},\{x,x_{0}\})\geq 1$ since $D$ is strong. It is easy to see that $y_{0}x\notin A(D)$ (for otherwise, $x_{1}yx_{3}y_{2}x_{0}y_{0}xy_{1}x_{1}$ is a cycle of length 8, a contradiction. Therefore, $d^{+}(y_{0},\{x_{1},x_{2},x_{3},x\})=0$. On the other hand, from (23) and $x_{0}y_{1}\notin A(D)$ we have $N^{+}(x_{0})=\{y_{0}\}$. Now it is not difficult to see that there is no path from $x_{0}$ to any vertex of $V(C)$ since $N^{+}(y_{0})=\{x_{0}\}$, which contradicts that $D$ is strong. So, the discussion of the case $|C^{\prime}|\geq 3$ is completed. Theorem 1.7 is proved.

5 Concluding remarks

In view of Theorem 1.3 it is natural to set the following problem.

Problem. Characterize those strongly connected balanced bipartite digraphs of order $2a\geq 6$ in which $d(x)+d(y)\geq 3a$ for every pair of vertices $x$, $y$ with a common in-neighbour or a common out-neighbour but are not even pancyclic.