A new sufficient condition for a Digraph to be Hamiltonian-A proof of Manoussakis Conjecture

TL;DR

This paper proves Manoussakis's conjecture that a 2-strongly connected digraph with a specific degree sum condition is Hamiltonian, and additionally shows such digraphs contain cycles of all lengths under certain degree constraints.

Contribution

It confirms Manoussakis's conjecture and establishes the existence of cycles of all lengths in certain degree-satisfying digraphs.

Findings

Proves the conjecture that degree sum conditions imply Hamiltonicity.

Shows existence of cycles of all lengths in specific digraphs.

Provides a new sufficient condition for Hamiltonian digraphs.

Abstract

Y. Manoussakis (J. Graph Theory 16, 1992, 51-59) proposed the following conjecture. \noindent\textbf{Conjecture}. {\it Let $D$ be a 2-strongly connected digraph of order $n$ such that for all distinct pairs of non-adjacent vertices $x$, $y$ and $w$, $z$, we have $d(x)+d(y)+d(w)+d(z)\geq 4n-3$. Then $D$ is Hamiltonian.} In this paper, we confirm this conjecture. Moreover, we prove that if a digraph $D$ satisfies the conditions of this conjecture and has a pair of non-adjacent vertices $\{x,y\}$ such that $d(x)+d(y)\leq 2n-4$, then $D$ contains cycles of all lengths $3, 4, \ldots , n$.

Equations

$$d^{-}(y_j,V(C))+d^{+}(y_{j-1},V(C))=\sum _{i=1}^{n-k}(\overrightarrow{a}[x_i,y_j]+\overrightarrow{a}[y_{j-1},x_{i+1}]))\le n-k.$$

$$d(y_1,V(C))+\cdots +d(y_s,V(C))=\sum _{j=1}^s(d^{-}(y_j,V(C))+d^{+}(y_{j-1},V(C)))\le s(n-k).$$

$$d(z,A)\le k,d(y_1,A)\le 2k-2andd(y_j,A)\le 2(k-2)+1=2k-3.$$

$$d(y_1,A)+\cdots +d(y_s,A)\le (s-2)(2k-3)+k+2k-2=2ks-3s-k+4.$$

$$d(y_1)+\cdots +d(y_s)\le ns+ks-3s-k+4.$$

$$d(y_1)+\cdots +d(y_s)+sd(y)\le 2ns-3s-k+4.$$

$$s(4n-3)/2\le d(y_1)+\cdots +d(y_s)+sd(y)\le 2ns-3s-k+4.$$

$$(s-1)(4n-3)/2+2n+k-2\le d(y_1)+\cdots +d(y_{s-1})+d(y_s)+sd(y)$$

$$\le 2ns-3s-k+4.$$

$$2n(s-1)-1.5(s-1)+2n+k-2\le 2ns-3s-k+4.$$

$$A(\{x_1,\dots ,x_{k-1}\}\to \{x_{k+1},\dots ,x_{n-2}\})\ne \varnothing .$$

$$d(y,\{x_1,\dots ,x_{p-1}\})=d(y,\{x_{r+1},\dots ,x_{n-2}\})=0.$$

$$d^{-}(z,\{x_1,\dots ,x_{p-1}\})=d^{+}(z,\{x_{r+1},\dots ,x_{n-2}\})=0$$

$$if{x}_i\to y,theny{x}_j\notin A(D)and\; if{x}_i\to z,thenz{x}_j\notin A(D).$$

$$A(\{x_1,\dots ,x_{k-1}\}\to \{x_{k+1},\dots ,x_{n-2}\})=\varnothing .$$

$$d^{+}(y,\{x_{m+1},\dots ,x_{n-2}\})=d^{+}(z,\{x_{l+1},\dots ,x_{n-2}\})=0.$$

$$d^{-}(y,\{x_1,\dots ,x_{m-1}\})=d^{-}(z,\{x_1,\dots ,x_{l-1}\})=0.$$

$$A(\{x_1,x_2,\dots ,x_{k-1}\}\to \{y,z,x_{k+1},x_{k+2},\dots ,x_{n-2}\})=\varnothing$$

$$A(\{y,z,x_1,x_2,\dots ,x_{k-1}\}\to \{x_{k+1},x_{k+2},\dots ,x_{n-2}\})=\varnothing ,$$

$$d^{+}(z,\{x_{a+1},\dots ,x_f\})=d^{-}(z,\{x_s,\dots ,x_{b-1}\})=0,$$

$$d(z)=d(z,\{x_1,\dots ,x_{s-1}\})+d(z,\{x_{f+1},\dots ,x_{n-2}\})\le n+s-f-2.$$

$$d^{-}(x_s,\{x_1,\dots ,x_{l-1}\})=d^{+}(x_s,\{x_{t+1},\dots ,x_{n-2}\})=0.$$

$$d(x_s)=d^{+}(x_s,\{x_1,\dots ,x_{l-1}\})+d^{-}(x_s,\{x_{t+1},\dots ,x_{n-2}\})+d(x_s,\{x_l,\dots ,x_t\})+d(x_s,\{y\})$$

$$\le l-1+n-2-t+2(t-l+1)=n+t-l-1.$$

$$d(y,\{x_{l+1},\dots ,x_{s-1}\})=d(y,\{x_{f+1},\dots ,x_{t-1}\})=0.$$

$$d(y)=d(y,\{x_1,\dots ,x_l\})+d(y,\{x_s,\dots ,x_f\})+d(y,\{x_t,\dots ,x_{n-2}\})$$

$$\le l+f-s+1+n-2-t+2=n+l+f-s-t+1.$$

$$4n-3\le d(y)+2d(z)+d(x_s)\le n+l+f-s-t+1+2n+2s-2f-4+n+t-l-1$$

$$=4n-4-(f-s)\le 4n-4,$$

$$d(y,\{x_1,\dots ,x_{p-1}\})=d(y,\{x_{q+1},\dots ,x_{k-1}\})=d(y,\{x_{r+1},\dots ,x_{n-2}\})$$

Full text

Conjecture. Let $D$ be a 2-strongly connected digraph of order $n$ such that for all distinct pairs of non-adjacent vertices $x$, $y$ and $w$, $z$, we have $d(x)+d(y)+d(w)+d(z)\geq 4n-3$. Then $D$ is Hamiltonian.

In this paper, we confirm this conjecture. Moreover, we prove that if a digraph $D$ satisfies the conditions of this conjecture and has a pair of non-adjacent vertices $\{x,y\}$ such that $d(x)+d(y)\leq 2n-4$, then $D$ contains cycles of all lengths $3,4,\ldots,n$.

1 Introduction

In this paper, we consider finite digraphs (directed graphs) without loops and multiple arcs. Every cycle and path are assumed simple and directed; its length is the number of its arcs. A digraph $D$ is Hamiltonian if it contains a cycle passing through all the vertices of $D$. There are many conditions that guarantee that a digraph is Hamiltonian (see, e.g., Bang-Jensen and Gutin (Springer-Verlag, London, 2000), Bermond and Thomassen (1981), Kühn and Osthus (2012), Manoussakis (1992), Meyniel (1973)). Manoussakis (1992) the following theorem was proved.

Theorem 1.1 (Manoussakis (1992)). *Let $D$ be a strong digraph of order $n\geq 4$. Suppose that $D$ satisfies the following condition for every triple $x,y,z\in V(D)$ such that $x$ and $y$ are non-adjacent: If there is no arc from $x$ to $z$, then $d(x)+d(y)+d^{+}(x)+d^{-}(z)\geq 3n-2$. If there is no arc from $z$ to $x$, then $d(x)+d(y)+d^{-}(x)+d^{+}(z)\geq 3n-2$. Then $D$ is Hamiltonian.

Definition 1.2. *Let $D$ be a digraph of order $n$. We say that $D$ satisfies condition $(M)$ when $d(x)+d(y)+d(w)+d(z)\geq 4n-3$ for all distinct pairs of non-adjacent vertices $x,y$ and $w,z$.

Manoussakis (1992) proposed the following conjecture. This conjecture is an extension of Theorem 1.1.

Conjecture 1.3 (Manoussakis (1992)). *Let $D$ be a 2-strong digraph of order $n$ such that for all distinct pairs of non-adjacent vertices $x$, $y$ and $w$, $z$ we have $d(x)+d(y)+d(w)+d(z)\geq 4n-3$. Then $D$ is Hamiltonian.

Manoussakis (1992) gave an example, which showed that if this conjecture is true, then the minimum degree condition is sharp. Notice that another examples can be found in a paper by Darbinyan (1983), where for any two integers $k\geq 2$ and $m\geq 1$, the author constructed a family of $k$-strong digraphs of order $4k+m$ with minimum degree $4k+m-1$, which are not Hamiltonian. This result improves a conjecture of Thomassen (see Bermond and Thomassen (1981) Conjecture 1.4.1: Every 2-strong $(n-1)$-regular digraph of order $n$, except $D_{5}$ and $D_{7}$, is Hamiltonian). Moreover, when $m=1$, then from these digraphs we can obtain $k$-strong non-Hamiltonian digraphs of order $n=4k+1$ with minimum degree equal to $n-1$ and the minimal semi-degrees equal to $(n-3)/2$. Thus, if in Conjecture 1.3 we replace $4n-3$ with $4n-4$, then for every $n$ there are many digraphs of order $n$ with high connectivity and high semi-degrees, for which Conjecture 1.3 is not true.

The cycle factor in a digraph $D$ is a collection of pairwise vertex disjoint cycles $C_{1},C_{2},\dots,C_{l}$ such that $\bigcup^{l}_{i=1}V(C_{i})=V(D)$. It is clear that the existence of a cycle factor in a digraph $D$ is a necessary condition for a digraph to be Hamiltonian. The following theorem gives a necessary and sufficient condition for the existence of a cycle factor in a digraph.

Theorem 1.4 (Yeo (1999)). *Let $D$ be a digraph. Then $D$ has a cycle factor if and only if $V(D)$ cannot be partitioned into subsets $Y$, $Z$, $R_{1}$, $R_{2}$ such that $A(Y\rightarrow R_{1})=A(R_{2}\rightarrow R_{1}\cup Y)=\emptyset$, $|Y|>|Z|$ and $Y$ is an independent set.

Using theorem Theorem 1.4, it is not difficult to construct 2-strong digraphs satisfying the condition that $d(x)+d(y)+d(w)+d(z)\geq 4n-4$ for every distinct pairs $\{x,y\}$, $\{w,z\}$ of non-adjacent vertices, but these digraphs do not even contain a cycle factor.

Thomassen suggested (see Bermond and Thomassen (1981)) the following two conjectures:

1. Conjecture 1.6.7. Every 3-strong digraph of order $n$ and with minimum degree at least $n+1$ is strongly Hamiltonian-connected.

2. Conjecture 1.6.8. Let $D$ be a 4-strong digraph of order $n$ such that the sum of the degrees of any pair of non-adjacent vertices is at least $2n+1$. Then $D$ is strongly Hamiltonian-connected.

Investigating these conjectures, Darbinyan (1990) disproved the first conjecture (proving that for every integer $n\geq 9$ there exists a 3-strong non-strongly Hamiltonian-connected digraph of order $n$ with the minimum degree at least $n+1$) and for the second proved the following two theorems.

Theorem 1.5. Any $k$-strong ($k\geq 1$) digraph $D$ of order $n\geq 8$ satisfying the condition that the sum of degrees of any pair of non-adjacent vertices $x,y\in V(D)\setminus\{z\}$ at least $2n-1$, where $z$ is some vertex in $V(D)$, is Hamiltonian if and only if any $(k+1)$-strong digraph of order $n+1$ satisfying the condition that the sum of degrees of any pair of non-adjacent vertices at least $2n+3$ is strongly Hamiltonian-connected.

Theorem 1.6. *Let $D$ be a strong digraph of order $n\geq 3$. Suppose that $d(x)+d(y)\geq 2n-1$ for every pair of non-adjacent vertices $x,y\in V(D)\setminus\{z\}$, where $z$ is some vertex of $V(D)$. Then $D$ contains a cycle of length at least $n-1$.

It is easy to see that if a digraph $D$ satisfies the condition $(M)$, then it contains at most one pair of non-adjacent vertices $x,y$ such that $d(x)+d(y)\leq 2n-2$. From this and Theorem 1.6, the following corollary immediately follows.

Corollary 1.7. *Let $D$ be a strong digraph of order $n$ satisfying condition $(M)$. Then $D$ contains a cycle of length at least $n-1$ (in particular, $D$ contains a Hamiltonian path).

Corollary 1.7 was also later proved by Ning (2015).

It is worth to note that Darbinyan (2017), Darbinyan (2015) and Darbinyan and Karapetyan (2015) studied some properties in digraphs with the conditions of Theorem 1.1. They obtained the following results (in first two results $D$ is a digraph of order $n$ satisfying the degree condition of Theorem 1.1).

(i) (Darbinyan and Karapetyan (2015)). If $D$ is strong, then it contains a cycle of length $n-1$ or $D$ is isomorphic to the complete bipartite digraph $K^{*}_{n/2,n/2}$.

(ii) (Darbinyan (2015)). If $D$ is strong, then it contains a Hamiltonian path in which the initial vertex dominates the terminal vertex or $D$ is isomorphic to one tournament of order 5.

(iii) (Darbinyan (2017)). Let $D$ be a digraph of order $n$ and let $Y$ be a non-empty subset of $V(D)$. Suppose that for every triple of the vertices $x,y,z\in Y$ such that $x$ and $y$ are non-adjacent: If there is no arc from $x$ to $z$, then $d(x)+d(y)+d^{+}(x)+d^{-}(z)\geq 3n-2$. If there is no arc from $z$ to $x$, then $d(x)+d(y)+d^{-}(x)+d^{+}(z)\geq 3n-2$. If there is a path from $u$ to $v$ and a path from $v$ to $u$ in $D$ for every pair of distinct vertices $u,v\in Y$, then $D$ has a cycle which contains at least $|Y|-1$ vertices of $Y$.

The last result is best possible in some situations and gives an answer to a question posted by Li et al. (2007).

Theorem 1.8 (Meyniel (1973)). *Let $D$ be a strong digraph of order $n\geq 2$. If $d(x)+d(y)\geq 2n-1$ for all pairs of non-adjacent vertices $x,y$ in $D$, then $D$ is Hamiltonian.

For a short proof of Theorem 1.8, see Bondy and Thomassen (1977). Darbinyan (1985) characterized those digraphs which satisfy Meyniel’s condition, but are not pancyclic. Before stating the main result obtained by Darbinyan (1985), we need to define a family of digraphs.

Definition 1.9. *For integers $n$ and $m$, $(n+1)/2<m\leq n-1$, let $\Phi_{n}^{m}$ denote the set of digraphs $D$, which satisfy the following conditions: (i) $V(D)=\{x_{1},x_{2},\ldots,x_{n}\}$; (ii) $x_{n}x_{n-1}\ldots x_{2}x_{1}x_{n}$ is a Hamiltonian cycle in $D$; (iii) for each $k$, $1\leq k\leq n-m+1$, the vertices $x_{k}$ and $x_{k+m-1}$ are not adjacent; (iv) $x_{j}x_{i}\notin A(D)$ whenever $2\leq i+1<j\leq n$ and (v) the sum of degrees for any two distinct non-adjacent vertices is at least $2n-1$.

Theorem 1.10 (Darbinyan (1979), Darbinyan (1985)). Let $D$ be a strong digraph of order $n\geq 3$. Suppose that $d(x)+d(y)\geq 2n-1$ for all pairs of distinct non-adjacent vertices $x$, $y$ in $D$. Then either (a) $D$ is pancyclic or (b) $n$ is even and $D$ is isomorphic to one of $K^{*}_{n/2,n/2}$, $K^{*}_{n/2,n/2}\setminus\{e\}$, where $e$ is an arbitrary arc of $K^{*}_{n/2,n/2}$, or (c) $D\in\Phi^{m}_{n}$ $($in this case $D$ does not contain a cycle of length $m$).

Later on, Theorem 1.10 was also proved by Benhocine (1986). Darbinyan (2019) investigated the pancyclicity of digraphs with the condition $(M)$. Using Theorem 1.10 and the Moser theorem for a strong tournament to be pancyclic (see Harary and Moser (1966)), we proved the following theorem.

Theorem 1.11 (Darbinyan (2019)). Let $D$ be a 2-strong digraph of order $n\geq 6$ satisfying condition $(M)$. Suppose that there exists a pair of non-adjacent vertices $x$, $y$ in $D$ such that $d(x)+d(y)\leq 2n-4$. Then $D$ contains cycles of all lengths $3,4,\ldots,n-1$.

In this paper we confirm Conjecture 1.3.

Theorem 1.12. Let $D$ be a 2-strong digraph of order $n\geq 3$ satisfying condition $(M)$. Then $D$ is Hamiltonian.

Theorem 1.12 also has the following immediate corollaries.

Corollary 1.13 (Woodall (1972)). A digraph of order $n$ is Hamiltonian if, for any two vertices $x$ and $y$, either $x\rightarrow y$ or $d^{+}(x)+d^{-}(y)\geq n$.

Corollary 1.14 (Nash-Williams (1969)). Let $D$ be a digraph of order $n\geq 2$. If for every vertex $x$, $d^{+}(x)\geq n/2$ and $d^{-}(x)\geq n/2$, then $D$ is Hamiltonian.

Note that Corollary 1.14 immediately follows from well-known theorem of Ghouila-Houri Ghouila-Houri (1960).

Corollary 1.15 ( Ore (1960)). Let $G$ be a simple graph of order $n\geq 3$, in which the degree sum of any two non-adjacent vertices is at least $n$. Then $G$ is Hamiltonian.

As an immediate corollary of Theorems 1.12 and 1.11, we obtain the following theorem.

Theorem 1.16. Let $D$ be a 2-strong digraph of order $n\geq 6$ satisfying condition $(M)$. Suppose that $D$ contains a pair of non-adjacent vertices $x$, $y$ such that $d(x)+d(y)\leq 2n-4$. Then $D$ is pancyclic.

In view of Theorem 1.16, it is natural to set the following problem.

Problem 1.17. Let $D$ be a 2-strong connected digraph of order $n$ satisfying condition $(M)$. Suppose that $\{x,y\}$ is a pair of non-adjacent vertices in $D$ such that $2n-3\leq d(x)+d(y)\leq 2n-2$. Whether $D$ is pancyclic?

2 Terminology and notation

In this paper we consider finite digraphs without loops and multiple arcs. We shall assume that the reader is familiar with the standard terminology on digraphs and refer to the book Bang-Jensen and Gutin (Springer-Verlag, London, 2000) for terminology and notations not defined here. The vertex set and the arc set of a digraph $D$ are denoted by $V(D)$ and $A(D)$, 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 $xy\in A(D)$. We use the notations $\overrightarrow{a}[x,y]=1$ if $xy\in A(D)$ and $\overrightarrow{a}[x,y]=0$ if $xy\notin A(D)$. If $xy\in A(D)$, $y$ is an out-neighbour of $x$ and $x$ is an in-neighbour of $y$. If $x\rightarrow y$ and $y\rightarrow z$, we write $x\rightarrow y\rightarrow z$. Two distinct vertices $x$ and $y$ are adjacent if $xy\in A(D)$ or $yx\in A(D)$ (or both). If there is no arc from $x$ to $y$, we shall use the notation $xy\notin A(D)$.

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 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 subdigraph of $D$ induced by a subset $A$ of $V(D)$ is denoted by $D\langle A\rangle$. If $z$ is a vertex of a digraph $D$, then the subdigraph $D\langle V(D)\setminus\{z\}\rangle$ is denoted by $D-z$.

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 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. Let $x$ and $y$ be two distinct vertices of a digraph $D$. Cycle that passing through $x$ and $y$ in $D$, we denote by $C(x,y)$.

A cycle (respectively, a path) that contains all the vertices of $D$, is a Hamiltonian cycle (respectively, is a Hamiltonian path). A digraph is Hamiltonian if it contains a Hamiltonian cycle. A digraph $D$ is strongly Hamiltonian-connected if, for every ordered pair $\{x,y\}$ of distinct vertices of $D$ there is a Hamiltonian path from $x$ to $y$. A digraph $D$ of order $n\geq 3$ is pancyclic if it contains cycles of all lengths $m$, $3\leq m\leq n$. For a cycle $C=x_{1}x_{2}\cdots x_{k}x_{1}$ of length $k$, the subscripts considered modulo $k$, i.e., $x_{i}=x_{s}$ for every $s$ and $i$ such that $i\equiv s\,(\hbox{mod}\,k)$. If $P$ is a path containing a subpath from $x$ to $y$, we let $P[x,y]$ denote that subpath. Similarly, if $C$ is a cycle containing vertices $x$ and $y$, $C[x,y]$ denotes the subpath of $C$ from $x$ to $y$. If $j<i$, then $\{x_{i},\ldots,x_{j}\}=\emptyset$.

A digraph $D$ is strongly connected (or just strong), if there exists a path from $x$ to $y$ and a path from $y$ to $x$ for every pair of distinct vertices $x,y$. A digraph $D$ is $k$-strongly connected (or $k$-strong), where $k\geq 1$, if $|V(D)|\geq k+1$ and $D\langle V(D)\setminus A\rangle$ is strongly connected for any subset $A\subset V(D)$ of at most $k-1$ vertices.

For a pair of disjoint subsets $A$ and $B$ of $V(D)$, we define $A(A\rightarrow B)=\{xy\in A(D)\,|\,x\in A,y\in B\}$ and $A(A,B)=A(A\rightarrow B)\cup A(B\rightarrow A)$.

3 Auxiliary known results

Lemma 3.1 (Häggkvist and Thomassen (1976)). *Let $D$ be a digraph of order $n\geq 3$ containing a cycle $C$ of length $m$, $m\in[2,n-1]$. Let $x$ be a vertex not contained in this cycle. If $d(x,V(C))\geq m+1$, then $D$ contains a cycle of length $k$ for all $k\in[2,m+1]$ *.

It is not difficult to prove the following lemma.

Lemma 3.2. *Let $D$ be a digraph of order $n$. Assume that $xy\notin A(D)$ and the vertices $x$, $y$ in $D$ satisfy the degree condition $d^{+}(x)+d^{-}(y)\geq n-2+k$, where $k\geq 1$. Then $D$ contains at least $k$ internally disjoint $(x,y)$-paths of length two.

The following results were proved by Darbinyan (2019) and its preliminary version presented at Emil Artin International Conference (Darbinyan (2018)).

Theorem 3.3. Let $D$ be a 2-strong digraph of order $n\geq 3$ satisfying condition $(M)$. Suppose that $\{x,y\}$ is a pair of non-adjacent vertices in $V(D)$ such that $d(x)+d(y)\leq 2n-2$. Then $D$ is Hamiltonian if and only if $D$ contains a cycle through the vertices $x$ and $y$.

Theorem 3.4. *Let $D$ be a 2-strong digraph of order $n\geq 3$. Suppose that $D$ contains at most one pair of non-adjacent vertices. Then $D$ is Hamiltonian.

Remark. *There is a strong non-Hamiltonian digraph of order $n\geq 5$, which is not 2-strong and has exactly one pair of non-adjacent vertices.

Using Lemma 3.2, it is not difficult to prove the following lemma.

Lemma 3.5. Let $D$ be a 2-strong digraph of order $n\geq 3$ and let $u$, $v$ be two distinct vertices in $V(D)$. If $D$ contains no cycle through $u$ and $v$, then $u$, $v$ are not adjacent and there is no path of length two between them. In particular, $d(u)+d(v)\leq 2n-4$.

Theorem 3.6. *Let $D$ be a 2-strong digraph of order $n\geq 3$ satisfying condition $(M)$. Suppose that $\{u,v\}$ is a pair of non-adjacent vertices in $V(D)$ such that $d(u)+d(v)\leq 2n-2$. Then $D$ is Hamiltonian or $D$ contains a cycle of length $n-1$ passing through $u$ and avoiding $v$ (passing through $v$ and avoiding $u$).

As an immediate corollary of Theorems 3.6, 3.3 and Lemma 3.1, we obtain

Corollary 3.7. *Let $D$ be a 2-strong non-Hamiltonian digraph of order $n\geq 3$ satisfying condition $(M)$. Suppose that $\{u,v\}$ is a pair of non-adjacent vertices in $V(D)$ such that $d(u)+d(v)\leq 2n-2$. Then $d(u)\leq n-1$, $d(v)\leq n-1$ and $D$ contains at most one cycle of length two passing through $u$ ($v$) *.

4 Preliminaries

Lemma 4.1. Let $D$ be a 2-strong digraph of order $n\geq 3$ satisfying condition $(M)$. Suppose that $\{y,z\}$ is a pair of non-adjacent vertices in $V(D)$ such that $d(y)+d(z)\leq 2n-2$ and $C=x_{1}x_{2}\ldots x_{n-k}x_{1}$ is a cycle in $D$ passing through $y$ and avoiding $z$, where $2\leq n-k\leq n-2$. If the subdigraph $D\langle V(D)\setminus V(C)\rangle$ contains a cycle passing through $z$ and $d(y,V(D)\setminus V(C))=0$, then $D$ is Hamiltonian.

Proof 4.1**.**

Suppose, on the contrary, that $D\langle V(D)\setminus V(C)\rangle$ contains a cycle passing through $z$, but $D$ is not Hamiltonian. Since $D$ contains at most one cycle of length two passing through $y$ (Corollary 3.7), from $d(y,V(D)\setminus V(C))=0$ it follows that $d(y)\leq n-k$. Let $y_{1}y_{2}\ldots y_{s}y_{1}$ be a cycle through $z$ in $D\langle V(D)\setminus V(C)\rangle$, where $s\in[2,k]$.

By Theorem 3.3 we have that $D$ contains no cycle through $y$ and $z$. Therefore, for each pair of integers $i$ and $j$, where $i\in[1,n-k]$ and $j\in[1,s]$, $\overrightarrow{a}[x_{i},y_{j}]+\overrightarrow{a}[y_{j-1},x_{i+1}]\leq 1$ (here, $y_{0}=y_{s}$ and $x_{n-k+1}=x_{1}$). This implies that for every $j\in[1,s]$ we have

[TABLE]

Hence,

[TABLE]

Since there is at most one cycle of length two through $z$ ($y$) (Corollary 3.7), it follows that for $A:=V(D)\setminus V(C)$ and for every $y_{j}\in\{y_{1},\ldots,y_{s}\}\setminus\{z,y_{1}\}$ (we may assume that $y_{1}\not=z$) the following holds:

[TABLE]

Therefore,

[TABLE]

Combining this with (1), we obtain

[TABLE]

The last inequality together with $d(y)\leq n-k$ implies that

[TABLE]

Notice that $\{y,y_{1}\}$, $\ldots$ , $\{y,y_{s}\}$ are $s$ distinct pairs of non-adjacent vertices. We will consider the cases when $s$ is even and $s$ is odd separately.

Assume first that $s$ is even. Using condition $(M)$ and (2), we obtain

[TABLE]

Therefore, $2ns-1.5s\leq 2ns-3s-k+4$, i.e., $1.5s+k\leq 4$. The last inequality is impossible, since $k\geq s\geq 2$.

Assume next that $s$ is odd. Then $s\geq 3$. Since $d(y)\leq n-k$, and $d(z)\leq n-1$ by Corollary 3.7 (we may assume that $z\not=y_{s}$), from condition $(M)$ it follows that $d(y)+d(y_{s})\geq 2n+k-2$. Now, by condition $(M)$ and (2) we have,

[TABLE]

[TABLE]

Hence,

[TABLE]

This means that $1.5s+2k\leq 4.5$, which is a contradiction. This contradiction completes the proof of Lemma 4.1.

Lemma 4.2. Let $D$ be a 2-strong digraph of order $n\geq 3$ satisfying condition $(M)$. Suppose that $\{y,z\}$ is a pair of non-adjacent vertices in $V(D)$ such that $d(y)+d(z)\leq 2n-2$ and $C=x_{1}x_{2}\ldots x_{n-2}zx_{1}$ is a cycle of length $n-1$ passing through $z$ and avoiding $y$ in $D$. Then either $D$ is Hamiltonian or for every $k\in[2,n-3]$, the following holds:

[TABLE]

Proof 4.2**.**

Suppose that $D$ is not Hamiltonian. Since $D$ is 2-strong, $n\geq 5$. Then by Theorem 3.3, there is no cycle through $y$ and $z$. Therefore, we have that if $x_{i}\rightarrow y$ with $i\in[1,n-3]$, then $d^{+}(y,\{x_{i+1},\ldots,x_{n-2}\})=0$ (for otherwise, $x_{1}\ldots x_{i}yx_{j}\ldots x_{n-2}zx_{1}$, where $j\in[i+1,n-2]$, is a cycle through $y$ and $z$, a contradiction). Let $x_{r}\rightarrow y\rightarrow x_{p}$, $1\leq p<r\leq n-2$, and $p$, $r$ be chosen so that $p$ is minimal and $r$ is maximal with these properties. Then

[TABLE]

If $p=1$ and $r=n-2$, then by a similar argument as above, we conclude that if $x_{i}\rightarrow z$ with $i\in[1,n-3]$, then $d^{+}(z,\{x_{i+1},\ldots,x_{n-2}\})=0$. Assume that $p\geq 2$ or $r\leq n-3$. Observe that $Q:=yx_{p}\ldots x_{r}y$ is a cycle through $y$ which does not contain $z$, and $d(y,V(D)\setminus V(Q))=0$ because of (3). Therefore by Lemma 4.1, the subdigraph $D\langle V(D)\setminus V(Q)\rangle$ contains no cycle through $z$ since $D$ is not Hamiltonian. This implies that

[TABLE]

since $x_{n-2}\rightarrow z\rightarrow x_{1}$. From the last equalities it follows that if there are $i,j$ such that $x_{i}\rightarrow z$ and $z\rightarrow x_{j}$ with $i<j$, then $i\geq p$, $j\leq r$ and $yx_{p}\ldots x_{i}zx_{j}\ldots x_{r}y$ is a cycle passing through $y$ and $z$, a contradiction. Thus, we may assume that for every pair of integers $i$ and $j$, $1\leq i<j\leq n-2$,

[TABLE]

Now suppose that the theorem is not true. Then $D$ is not Hamiltonian and there is an integer $k\in[2,n-3]$ such that

[TABLE]

It is easy to see that there are vertices $x_{m}$ and $x_{l}$ such that $y\rightarrow x_{m}$, $z\rightarrow x_{l}$ and

[TABLE]

Then by (4),

[TABLE]

Assume first that $m\leq l$. Since $D$ is 2-strong, (4) and (7) imply that $2\leq m\leq l\leq n-3$. Now from (5), (6) and (7) it follows that:

(i) if $k\leq m$ or $k\geq l$, then (respectively)

[TABLE]

or

[TABLE]

(ii) if $m<k<l$, then $A(\{y,x_{1},x_{2},\ldots,x_{k-1}\}\rightarrow\{z,x_{k+1},x_{k+2},\ldots,x_{n-2}\})=\emptyset$. Thus, in each case we have that $D-x_{k}$ is not strong, which contradicts the condition that $D$ is 2-strongly connected.

Assume next that $m>l$. This case is similar to the first case and we omit the details. Lemma 4.2 is proved.

The following lemma is proved by Darbinyan (2019). We present its proof for completeness.

Lemma 4.3. Let $D$ be a 2-strong digraph of order $n\geq 3$ satisfying condition $(M)$. Suppose that $\{y,z\}$ is a pair of non-adjacent vertices in $V(D)$ such that $d(y)+d(z)\leq 2n-2$ and $C=x_{1}x_{2}\ldots x_{n-2}zx_{1}$ is a cycle of length $n-1$ passing through $z$ and avoiding $y$ in $D$. If $x_{a}\rightarrow x_{b}$ and there are integers $l,s,f,t$ such that $1\leq l\leq a<s\leq f<b\leq t\leq n-2$ and $\{x_{f},x_{t}\}\rightarrow y\rightarrow\{x_{l},x_{s}\}$, then $D$ is Hamiltonian.

Proof 4.3**.**

Suppose, on the contrary, that $D$ is not Hamiltonian. By Theorem 3.3, $D$ contains no cycle through $y$ and $z$. Therefore, there are no integers $i$ and $j$, $1\leq i<j\leq n-2$, such that $x_{i}\rightarrow y\rightarrow x_{j}$ (for otherwise, $x_{1}\ldots x_{i}yx_{j}\ldots x_{n-2}zx_{1}$ is a cycle through $y$ and $z$). Since the arcs $yx_{l}$, $yx_{s}$, $x_{f}y$, $x_{t}y$ are in $D$ and $l\leq a<s\leq f<b\leq t$, it is easy to check that:

(i) if $z\rightarrow x_{i}$ with $i\in[a+1,f]$, then $C(y,z)=yx_{l}\ldots x_{a}x_{b}\ldots x_{n-2}zx_{i}\ldots x_{f}y$;

(ii) if $x_{j}\rightarrow z$ with $j\in[s,b-1]$, then $C(y,z)=x_{1}\ldots x_{a}x_{b}\ldots x_{t}yx_{s}\ldots x_{j}zx_{1}$. Thus, in both cases we have a contradiction. Therefore,

[TABLE]

in particular, $d(z,\{x_{s},\ldots,x_{f}\})=0$ and the vertices $z$ and $x_{s}$ ($z$ and $x_{f}$) are not adjacent. The last equality together with the fact that $D$ contains at most one cycle of length two passing through $z$ (Corollary 3.7) implies that

[TABLE]

Now we consider the vertex $x_{s}$. It is not difficult to check that:

(iii) if $x_{i}\rightarrow x_{s}$ with $i\in[1,l-1]$, then $C(y,z)=x_{1}\ldots x_{i}x_{s}\ldots x_{f}yx_{l}\ldots$ $x_{a}x_{b}\ldots x_{n-2}zx_{1}$;

(iv) if $x_{s}\rightarrow x_{j}$ with $j\in[t+1,n-2]$, then $C(y,z)=x_{1}\ldots x_{a}x_{b}\ldots x_{t}yx_{s}$ $x_{j}\ldots x_{n-2}zx_{1}$. In both cases we have a contradiction. Therefore, we may assume that

[TABLE]

This implies that

[TABLE]

[TABLE]

Without loss of generality, we may assume that $l$, $f$ are chosen as maximal as possible and $s$, $t$ are chosen as minimal as possible, i.e.,

[TABLE]

This, since $D$ contains at most one cycle of length two passing through $y$, implies that

[TABLE]

[TABLE]

Since $\{y,z\}$ and $\{x_{s},z\}$ are two distinct pairs of non-adjacent vertices, from (8), (9), the last inequality and condition $(M)$ it follows that

[TABLE]

[TABLE]

which is a contradiction. Lemma 4.3 is proved.

5 Proof of Theorem 1.12

Recall the statement of Theorem 1.12.

Theorem 1.12. Let $D$ be a 2-strong digraph of order $n\geq 3$ satisfying condition $(M)$. Then $D$ is Hamiltonian.

Proof 5.1**.**

By Theorem 3.4, the theorem is true if $D$ contains at most one pair of non-adjacent vertices. We may therefore assume that $D$ contains at least two distinct pairs of non-adjacent vertices. If the degrees sum of any two non-adjacent vertices at least $2n-1$, then by Meyniel’s theorem, the theorem is true. We may therefore assume that $D$ contains a pair of non-adjacent vertices, say $y,z$, such that $d(y)+d(z)\leq 2n-2$. By Theorem 3.3, to prove the theorem, it suffices to prove that $D$ contains a cycle through $y$ and $z$. If $d(y)+d(z)\geq 2n-3$, then by Lemma 3.5 we have that $D$ contains a cycle trough $y$ and $z$, which, in turn, implies that $D$ is Hamiltonian (by Theorem 3.3). Thus, we may assume that $d(y)+d(z)\leq 2n-4$. By Theorem 3.6 we have that either $D$ is Hamiltonian or $D$ contains a cycle of length $n-1$ passing through $z$ and avoiding $y$ (passing through $y$ and avoiding $z$).

Suppose that $D$ is not Hamiltonian, i.e., $D$ contains no cycle through $y$ and $z$. Let $C:=x_{1}x_{2}\ldots x_{n-2}zx_{1}$ be a cycle of length $n-1$ in $D$, which does not contain $y$. Let $q$ be the maximum integer such that $y\rightarrow x_{q}$ and $k$ be the minimum integer such that $x_{k}\rightarrow y$. Since $D$ is 2-strong and contains no cycle passing through $y$ and $z$, it follows that $k\geq q$ and there are some integers $p$, $r$, $1\leq p<q\leq k<r\leq n-2$, such that $x_{r}\rightarrow y\rightarrow x_{p}$ and

[TABLE]

[TABLE]

Note that if $D$ contains a cycle of length two passing trough $y$, then $k=q$, otherwise $k>q$, $yx_{k}\notin A(D)$ and $x_{q}y\notin A(D)$. Therefore, it is not difficult to see that

[TABLE]

In order to prove the theorem, it is convenient for the digraph $D$ and the cycle $C$ to prove the following claims.

Claim 5.1.* If $p\geq 2$, then $d^{-}(x_{n-2},\{z,x_{1},\ldots,x_{p-1}\})=0$.*

Proof 5.2**.**

Notice that $Q:=yx_{p}\ldots x_{r}y$ is a cycle passing through $y$ and avoiding $z$. By (10) we have that $d(y,V(D)\setminus V(Q))=0$. Now by Lemma 4.1 , the induced subdigraph $D\langle V(D)\setminus V(Q)\rangle$ contains no cycle through $z$. Then, since $x_{n-2}\rightarrow z\rightarrow x_{1}$, we have

[TABLE]

The first equality together with 2-connectedness of $D$ implies that there is an integer $t\in[p,n-3]$ such that $x_{t}\rightarrow z$. The last equality means that if $r\leq n-3$, then $d^{-}(x_{n-2},\{z,x_{1},\ldots,x_{p-1}\})=0$. Assume that $r=n-2$, i.e., $x_{n-2}\rightarrow y$. In this case, we have that if $x_{i}\rightarrow x_{n-2}$ with $i\in[1,p-1]$ (respectively, $z\rightarrow x_{n-2}$), then $C(y,z)=x_{1}\ldots x_{i}x_{n-2}yx_{p}\ldots x_{t}zx_{1}$ (respectively, $C(y,z)=yx_{p}\ldots x_{t}zx_{n-2}y$), which is a contradiction. This proves that $d^{-}(x_{n-2},\{z,x_{1},\ldots,x_{p-1}\})=0$.

Claim 5.2.* Suppose that $k\geq q+1$ and $x_{h}\rightarrow x_{l}$, where $h\in[q,k-1]$ and $l\in[k+1,n-2]$. Then $d^{-}(x_{k},\{x_{1},\ldots,x_{q-1}\})=0$.*

Proof 5.3**.**

Assume that Claim 5.2 is not true. Then for some $i\in[1,q-1]$, $x_{i}\rightarrow x_{k}$. Then, since the arcs $yx_{q}$, $x_{k}y$, $x_{h}x_{l}$ are in $D$ and $i<q\leq h<k<l$, we have a cycle $C(y,z)=x_{1}\ldots x_{i}x_{k}yx_{q}\ldots x_{h}x_{l}\ldots x_{n-2}zx_{1}$, which contradicts our initial supposition.

Claim 5.3.* Suppose that $k\geq q+1$, $x_{h}\rightarrow x_{l}$ with $h\in[q,k-1]$ and $l\in[k+1,r]$ (possibly, $r=n-2$). Then there is an integer $f\geq 0$ such that $l+f\leq r$, $x_{l+f}\rightarrow y$, $d(y,\{x_{l},\ldots,x_{l+f-1}\})=0$ (possibly, $\{x_{l},\ldots,x_{l+f-1}\}=\emptyset$). Moreover, either there is a vertex $x_{g}$ with $g\in[l+f+1,n-2]$ such that $x_{k}\rightarrow x_{g}$ or there is a vertex $x_{c}$ with $c\in[k,l-1]$ such that $x_{c}\rightarrow z$.*

Proof 5.4**.**

By Claim 5.2,

[TABLE]

*Since $l\leq r$ and $x_{r}\rightarrow y$, obviously there is an integer $f\geq 0$ such that $l+f\leq r$, $x_{l+f}\rightarrow y$, $d^{-}(y,\{x_{l},\ldots,x_{l+f-1}\})=0$ $($possibly $\{x_{l},\ldots,x_{l+f-1}\}=\emptyset$). This together with

$d^{+}(y,\{x_{l},\ldots,x_{l+f-1}\})=0$ implies that*

[TABLE]

Now suppose that the claim is not true. Then

[TABLE]

The second equality of (14) together with $d^{+}(y,\{x_{k},\ldots,x_{l-1}\})=0$ and the fact that there is no path of length two between $y$ and $z$ (Lemma 3.5) implies that the vertices $x_{k}$, $z$ are not adjacent and

[TABLE]

This together with (13), (10) and the fact that there is at most one cycle of length two through $z$ (Corollary 3.7) implies that

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Now consider the vertex $x_{k}$. Note that $d(x_{k},\{y\})=1$ since $k\geq q+1$. Using (12) and the first equality of (14), we obtain

[TABLE]

[TABLE]

Combining the last two inequalities, $d(z)\leq n-1$ (Corollary 3.7) and $r\leq n-2$, we obtain

[TABLE]

which contradicts condition $(M)$, since $\{y,z\}$, $\{z,x_{k}\}$ are two distinct pairs of non-adjacent vertices. This contradiction completes the proof of Claim 5.3.

Claim 5.4.* If $p\geq 2$, then $A(\{x_{1},\ldots,x_{p-1}\}\rightarrow\{x_{k+1},\ldots,x_{n-2}\})=\emptyset.$*

Proof 5.5**.**

Suppose, on the contrary, that $p\geq 2$ and $x_{a}\rightarrow x_{b}$ with $a\in[1,p-1]$ and $b\in[k+1,n-2]$. Let $b$ be the maximum with these properties, i.e.,

[TABLE]

Notice that $Q:=yx_{p}\ldots x_{r}y$ is a cycle in $D$ and $d(y,V(D)\setminus V(Q))=0$ by (10). Therefore by Lemma 4.1, the subdigraph $D\langle V(D)\setminus V(Q)\rangle$ does not contain a cycle through $z$. In particular,

[TABLE]

and if $r\leq n-3$, then

[TABLE]

By Claim 5.1, we have

[TABLE]

From (17) and (18) it follows that $b\leq r$ and, if $r=n-2$, then $b\leq n-3$. In both cases we have that $b\leq n-3$.

If $x_{i}\rightarrow z$ with $i\in[p,b-1]$, then $C(y,z)=x_{1}\ldots x_{a}x_{b}\ldots x_{r}yx_{p}\ldots x_{i}zx_{1}$, a contradiction. We may therefore assume that $d^{-}(z,\{x_{p},\ldots,x_{b-1}\})=0$. This together with (16) implies that

[TABLE]

Applying Lemma 4.2 to the vertex $x_{b}$, we obtain that

[TABLE]

Let $x_{s}\rightarrow x_{t}$, where $s\in[1,b-1]$ and $t\in[b+1,n-2]$. Choose $t$ maximal with these properties, i.e.,

[TABLE]

*From (15) it follows that $s\geq p$, i.e., $s\in[p,b-1]$. If $x_{i}\rightarrow y$ with $i\in[b,t-1]$, then $C(y,z)=x_{1}\ldots x_{a}x_{b}\ldots x_{i}yx_{p}\ldots x_{s}x_{t}\ldots x_{n-2}zx_{1}$, a contradiction. We may therefore assume that

$d^{-}(y,\{x_{b},\ldots,x_{t-1}\})=0$. This together with $d^{+}(y,\{x_{b},\ldots,x_{t-1}\})=0$ implies that*

[TABLE]

In particular, the vertices $x_{b}$ and $y$ are not adjacent, $t\leq r$ and $b\leq r-1$ since $b+1\leq t\leq r$ (i.e., $A(\{x_{p},\ldots,x_{b-1}\}\rightarrow\{x_{r+1},\ldots,x_{n-2}\})=\emptyset$). Using Lemma 4.3, we obtain

[TABLE]

Then, since $t\leq r$ and (20), we have that $A(\{x_{p},\ldots,x_{q-1}\}\rightarrow\{x_{b+1},\ldots,x_{n-2}\})=\emptyset$. This together with (15) implies that

[TABLE]

Therefore, $s\geq q$, i.e., $s\in[q,b-1]$. Since $b\leq r-1$, and $x_{b}$, $y$ are not adjacent, there is an integer $f\geq 0$ such that $d^{-}(y,\{x_{b},\ldots,x_{b+f}\})=0$ and $x_{b+f+1}\rightarrow y$. Then, since (21) and $d^{+}(y,\{x_{b},\ldots,x_{b+f}\})=0$ we have that $t\leq b+f+1$ and

[TABLE]

This together with (10) implies that

[TABLE]

[TABLE]

From (19), $d^{+}(y,\{x_{k},\ldots,x_{b-1}\})\leq 1$ and the fact that there is no path of length two between $y$ and $z$ (Lemma 3.5) it follows that

[TABLE]

This together with (10), (23) and the fact that there is at most one cycle of length two through $z$ (Corollary 3.7) implies that

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Since $t\leq b+f+1$ and (20), it follows that

[TABLE]

In particular, from $b\geq k+1$ and (26) it follows that

[TABLE]

*We will consider the cases $b\geq k+2$, $b=k+1$ separately.

Case 1.* $b\geq k+2$.*

Then by the first equality of (22) we have

[TABLE]

Using the fact that there is no path of length two between $y$ and $z$ (Lemma 3.5) and (19), we obtain that $d(x_{b-1},\{y,z\})\leq 1$. This together with $d^{+}(x_{b-1},\{x_{b+f+2},\ldots,x_{n-2}\})=0$ (by (26)) and (28) implies that

[TABLE]

[TABLE]

[TABLE]

*Now we divide this case into the following subcases.

Subcase 1.1.* The vertices $x_{b-1}$ and $y$ are not adjacent.*

Then $\{y,x_{b-1}\}$ and $\{y,z\}$ are two distinct pairs of non-adjacent vertices. Since $p\geq 2$, $r\leq n-2$, $f\geq 0$ and $k\geq q$, combining (25), (24) and (29), we obtain

[TABLE]

[TABLE]

*which contradicts condition $(M)$.

Subcase 1.2.* The vertices $x_{b-1}$ and $y$ are adjacent.*

Then $x_{b-1}\rightarrow y$. Therefore by Lemma 3.5 and (19), the vertices $z$ and $x_{b-1}$ are not adjacent. Since $d(z)\leq n-2$ (because of $d(z,\{y,x_{b-1}\})=0$ and Corollary 3.7) and $r\leq n-2$, from (25) and (29) it follows that

[TABLE]

[TABLE]

*which contradicts condition $(M)$. The discussion of Case 1 is completed.

Case 2.* $b=k+1$.*

*We divide this case into the following subcases.

Subcase 2.1.* $s\leq k-1$.*

Then $k\geq q+1$ since $s\geq q$. Then $yx_{k}\notin A(D)$ by the definition of $q$ and $k$. Recall that the vertices $z$, $x_{k}$ are not adjacent by (19) and Lemma 3.5. Now it is easy to see that $d(z)\leq n-2$. Since $x_{s}\rightarrow x_{t}$ with $s\in[q,k-1]$ and $t\in[b+1,n-2]$, by Claim 5.2 we have that $d^{-}(x_{k},\{x_{1},\ldots,x_{q-1}\})=0$. This together with (27) and $b=k+1$ implies that

[TABLE]

[TABLE]

This together with (24) and $d(z)\leq n-2$, we obtain

[TABLE]

[TABLE]

*which is a contradiction since $\{y,z\}$ and $\{x_{k},z\}$ are two distinct pairs of non-adjacent vertices.

Subcase 2.2.* $s=k$.*

From $b=k+1$, $t\in[b+1=k+2,b+f+1]$ and (23) it follows that

[TABLE]

in particular, the vertices $y$ and $x_{k+1}$ are not adjacent. Observe that $R:=yx_{p}\ldots x_{k}x_{t}\ldots\\ x_{r}y$ is a cycle in $D$ passing through $y$, avoiding $z$ and $d(y,V(D)\setminus V(R))=0$. By Lemma 4.1, the induced subdigraph $D\langle V(D)\setminus V(R)\rangle$ contains no cycle through $z$. In particular, this means that

[TABLE]

*for otherwise, if $x_{i}\rightarrow x_{j}$ with $i\in[k+1,t-1]$ and $j\in[r+1,n-2]$, then $H:=x_{1}\ldots x_{a}x_{k+1}\ldots x_{i}x_{j}\\ \ldots x_{n-2}zx_{1}$ is a cycle in $D\langle V(D)\setminus V(R)\rangle$ through $z$, a contradiction.

Subcase 2.2.1.* There is an integer $l\in[b+f+2,n-2]$ such that $x_{k+1}\rightarrow x_{l}$ and*

[TABLE]

*Then $b+f+2\leq n-2$, and $l\leq r$ because of the first equality of (31). Recall that $t\leq b+f+1\leq l-1$. Hence, $l\geq t+1$. If $x_{i}\rightarrow z$ with $i\in[t,l-1]$, then $C(y,z)=x_{1}\ldots x_{a}x_{k+1}x_{l}\ldots x_{r}yx_{q}\ldots x_{k}x_{t}\ldots x_{i}zx_{1}$, a contradiction. We may therefore assume that $d^{-}(z,\{x_{t},\ldots,x_{l-1}\})=0$. This together with *

$d^{+}(y,\{x_{t},\ldots,x_{l-1}\})=0$* and the fact that there is no path of length two between $y$ and $z$ implies that*

[TABLE]

Combining this, (10) and (30), we obtain

[TABLE]

[TABLE]

[TABLE]

[TABLE]

For the vertex $x_{k+1}$, using (32) and the second equality of (22), we obtain

[TABLE]

[TABLE]

[TABLE]

This together with (33), (24), $r\leq n-2$, $k\geq q$ and $p\geq 2$ implies that

[TABLE]

[TABLE]

which contradicts condition $(M)$ since $\{y,z\}$ and $\{y,x_{k+1}\}$ are two distinct pairs of non-adjacent vertices.

Subcase 2.2.2.* There is no $l\in[b+f+2,n-2]$ such that $x_{k+1}\rightarrow x_{l}$.*

Then $d^{+}(x_{k+1},\{x_{b+f+2},\ldots,x_{n-2}\})=0$. This together with the second equality of (22) implies that

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Combining this, $b=k+1$, (24) and $d(z)\leq n-2$, we obtain

[TABLE]

[TABLE]

which contradicts condition $(M)$. In each case we obtain a contradiction and hence the discussion of Case 2 is completed. This completes the proof of Claim 5.4.

Now we are ready to complete the proof of the main result.

By Claim 5.4, if $p\geq 2$, then $A(\{x_{1},\ldots,x_{p-1}\}\rightarrow\{x_{k+1},\ldots,x_{n-2}\})=\emptyset.$ Similarly, if $r\leq n-3$, then $A(\{x_{1},\ldots,x_{q-1}\}\rightarrow\{x_{r+1},\ldots,x_{n-2}\})=\emptyset$. Using Lemma 4.3, we obtain $A(\{x_{p},\ldots,x_{q-1}\}\rightarrow\{x_{k+1},\ldots,x_{r}\})=\emptyset.$ From the last three equalities it follows that

[TABLE]

From (34) and Lemma 4.2 it follows that $k\geq q+1$. Applying Lemma 4.2 to the vertices $x_{q}$ and $x_{k}$, we obtain

[TABLE]

Let $x_{a}\rightarrow x_{b}$ and $x_{h}\rightarrow x_{l}$ with $a\in[1,q-1]$, $b\in[q+1,n-2]$, $h\in[1,k-1]$ and $l\in[k+1,n-2]$. Choose $b$ maximal and $h$ minimal with these properties, i.e.,

[TABLE]

From (34) it follows that $b\leq k$ and $h\geq q$, i.e., $b\in[q+1,k]$ and $h\in[q,k-1]$. If $h\leq b-1$, then $C(y,z)=x_{1}\ldots x_{a}x_{b}\ldots x_{k}yx_{q}\ldots x_{h}x_{l}\ldots x_{n-2}zx_{1}$, a contradiction. We may therefore assume that $h\geq b$, which in turn implies that $k\geq q+2$. By Lemma 4.2, $A(\{x_{1},\ldots,x_{b-1}\}\rightarrow\{x_{b+1},\ldots,x_{n-2}\})\not=\emptyset$. Let $x_{s}\rightarrow x_{t}$, where $s\in[1,b-1]$ and $t\in[b+1,n-2]$. Choose $t$ maximal with this property, i.e.,

[TABLE]

From (35) it follows that $s\geq q$ and $t\leq k$, i.e., $s\in[q,b-1]$ and $t\in[b+1,k]$. We may assume that $l$ (recall that $x_{h}\rightarrow x_{l}$, $l\geq k+1$) is chosen so that

[TABLE]

*We consider the cases $l\leq r$ and $l\geq r+1$ separately.

Case 1.* $l\leq r$.*

For this case, it is not difficult to check that the conditions of Claim 5.3 hold. Therefore, there is an integer $f\geq 0$ such that $l+f\leq r$, $x_{l+f}\rightarrow y$, $d(y,\{x_{l},\ldots,x_{l+f-1}\})=0$ (possibly, $\{x_{l},\ldots,x_{l+f-1}\}=\emptyset$), and either there is a vertex $x_{g}$ with $g\in[l+f+1,n-2]$ such that $x_{k}\rightarrow x_{g}$ or there is a vertex $x_{c}$ with $c\in[k,l-1]$ such that $x_{c}\rightarrow z$.

Assume first that $t\geq h+1$. Then, since the arcs $yx_{q}$, $x_{a}x_{b}$, $x_{s}x_{t}$, $x_{h}x_{l}$, $x_{k}y$, $x_{l+f}y$ are in $D$ and $1\leq a\leq q-1<s<b\leq h<t\leq k<l\leq l+f\leq r\leq n-2$, we have that $C(y,z)=x_{1}\ldots x_{a}x_{b}\ldots x_{h}x_{l}\ldots x_{l+f}yx_{q}\ldots x_{s}x_{t}\ldots x_{c}zx_{1}$, or $C(y,z)=x_{1}\ldots x_{a}x_{b}\ldots x_{h}x_{l}\ldots x_{l+f}yx_{q}\ldots\\ x_{s}x_{t}\ldots x_{k}x_{g}\dots x_{n-2}zx_{1}$ when $x_{c}\rightarrow z$ or when $x_{k}\rightarrow x_{g}$ respectively. In each case we have a contradiction.

Assume next that $t\leq h$. By Lemma 4.2, $A(\{x_{1},\ldots,x_{t-1}\}\rightarrow\{x_{t+1},\ldots,x_{n-2}\})\not=\emptyset$. Let $x_{s_{1}}\rightarrow x_{t_{1}}$, where $s_{1}\in[1,t-1]$ and $t_{1}\in[t+1,n-2]$. Choose $t_{1}$ maximal with this property, i.e.,

[TABLE]

From (36) (respectively, from (35)) it follows that $s_{1}\geq b$, i.e., $s_{1}\in[b,t-1]$ (respectively, $t_{1}\leq k$, i.e., $t_{1}\in[t+1,k]$). If $t_{1}\geq h+1$, then $C(y,z)=x_{1}\ldots x_{a}x_{b}\ldots x_{s_{1}}x_{t_{1}}\ldots x_{k}yx_{q}\ldots x_{s}x_{t}\ldots x_{h}x_{l}\dots x_{n-2}zx_{1}$, a contradiction. We may therefore assume that $t_{1}\leq h$. By Lemma 4.2,

[TABLE]

Let $x_{s_{2}}\rightarrow x_{t_{2}}$, where $s_{2}\in[1,t_{1}-1]$ and $t_{2}\in[t_{1}+1,n-2]$. Choose $t_{2}$ maximal with this property, i.e.,

[TABLE]

From (38) (respectively, from (35)) it follows that $s_{2}\geq t$, i.e., $s_{2}\in[t,t_{1}-1]$ (respectively, $t_{2}\leq k$, i.e., $t_{2}\in[t_{1}+1,k]$).

Assume first that $t_{2}\geq h+1$. Then it is not difficult to see that $C(y,z)=x_{1}\ldots x_{a}x_{b}\ldots x_{s_{1}}x_{t_{1}}\ldots\\ x_{h}x_{l}\ldots x_{l+f}yx_{q}\ldots$ $x_{s}x_{t}\ldots x_{s_{2}}x_{t_{2}}\ldots x_{c}zx_{1}$ or $C(y,z)=x_{1}\ldots x_{a}x_{b}\ldots x_{s_{1}}x_{t_{1}}\ldots x_{h}x_{l}\ldots x_{l+f}y\\ x_{q}\ldots x_{s}x_{t}\ldots x_{s_{2}}x_{t_{2}}\ldots x_{k}x_{g}\dots$ $x_{n-2}zx_{1}$ when $x_{c}\rightarrow z$ or when $x_{k}\rightarrow x_{g}$, respectively. In each case we have a contradiction.

Continuing this process, we finally conclude that for some $m\geq 0$, $t_{m}\in[h+1,k]$ (here, $t_{0}=t$) since all the vertices $x_{t},x_{t_{1}},\ldots,x_{t_{m}}$ are distinct and in $\{x_{q+1},\ldots,x_{k}$}. We already have constructed a cycle $C(y,z)$ when $m\in\{0,1,2\}$. Assume that $m\geq 3$. By the above arguments we have that:

If $m\geq 3$ is odd, then $C(y,z)=x_{1}\ldots x_{a}x_{b}\ldots x_{s_{1}}x_{t_{1}}\ldots x_{s_{m}}x_{t_{m}}\ldots x_{k}yx_{q}\ldots x_{s}x_{t}\ldots x_{s_{2}}x_{t_{2}}\ldots\\$ $x_{s_{m-1}}x_{t_{m-1}}\ldots x_{h}x_{l}\ldots x_{n-2}zx_{1}$.

*If $m\geq 4$ is even, then $C(y,z)=x_{1}\ldots x_{a}x_{b}\ldots x_{s_{1}}x_{t_{1}}\ldots x_{s_{m-1}}x_{t_{m-1}}\dots x_{h}x_{l}\ldots x_{l+f}yx_{q}\ldots x_{s}x_{t}\\ \ldots x_{s_{2}}x_{t_{2}}\ldots x_{s_{m}}x_{t_{m}}\ldots x_{c}zx_{1}$ or $C(y,z)=x_{1}\ldots x_{a}x_{b}\ldots x_{s_{1}}x_{t_{1}}\ldots$ $x_{s_{m-1}}x_{t_{m-1}}\ldots x_{h}x_{l}\ldots\\$ $x_{l+f}yx_{q}\ldots x_{s}x_{t}\ldots x_{s_{2}}x_{t_{2}}\ldots x_{s_{m}}x_{t_{m}}\ldots x_{k}x_{g}\ldots x_{n-2}zx_{1}$ when $x_{c}\rightarrow z$ or when $x_{k}\rightarrow x_{g}$, respectively. In all cases we have a cycle through $y$ and $z$, which contradicts our supposition and hence the discussion of Case 1 is completed.

Case 2.* $l\geq r+1$.*

*Then $r\leq n-3$. Recall that $h\in[b,k-1]$, $x_{h}\rightarrow x_{l}$ and $x_{s}\rightarrow x_{t}$, where $l\leq n-2$, $s\in[q,b-1]$ and $t\in[b+1,k]$. Note that $\{y,x_{h}\}$, $\{y,z\}$ are two distinct pairs of non-adjacent vertices.

Subcase 2.1.* $t\geq h+1$.*

Since $s\in[q,b-1]$ and $t\in[h+1,k]$, we have that $Q:=yx_{p}\ldots x_{s}x_{t}\ldots x_{r}y$ is a cycle in $D$ and $d(y,V(D)\setminus V(Q))=0$. If $a\leq p-1$, then $H:=x_{1}\ldots x_{a}x_{b}\ldots x_{h}x_{l}\ldots x_{n-2}zx_{1}$ is a cycle in $D\langle V(D)\setminus V(Q)\rangle$ passing through $z$, which contradicts Lemma 4.1. We may therefore assume that $a\geq p$, i.e., $a\in[p,q-1]$.

Assume first that $b\leq h-1$. Then $q+1\leq b\leq h-1\leq k-2$ and $k\geq q+3$. From the first equality of (35) it follows that $d^{-}(x_{h},\{x_{1},\ldots,x_{q-1}\})=0$. This equality together with (37) implies that

[TABLE]

[TABLE]

[TABLE]

This together with (11) and $d(z)\leq n-1$ implies that

[TABLE]

[TABLE]

which contradicts condition $(M)$.

Assume that $b=h$, i.e., $x_{a}\rightarrow x_{h}$. We may assume that $a$ is chosen so that $d^{-}(x_{h},\{x_{1},\ldots,x_{a-1}\})=0$. This and (37) imply that

[TABLE]

[TABLE]

[TABLE]

Since $a\geq p$, it is not difficult to check that if $z\rightarrow x_{i}$ with $i\in[a+1,s]$, then $C(y,z)=yx_{p}\ldots x_{a}x_{h}x_{l}\ldots\\ x_{n-2}zx_{i}\ldots x_{s}x_{t}\ldots x_{k}y$, which is a contradiction. Therefore, $d^{+}(z,\{x_{a+1},\ldots,x_{s}\})=0$. This together with $d^{-}(y,\{x_{a+1},\ldots,x_{s}\})=0$ and the fact that there is no path of length two between $y$ and $z$ implies that

[TABLE]

Using this and (10), we obtain

[TABLE]

[TABLE]

[TABLE]

Combining this, (11) and (39), we obtain

[TABLE]

[TABLE]

*which contradicts condition $(M)$ and hence the discussion of Subcase 2.1 is completed.

Subcase 2.2.* $t\leq h$.*

Then $b\leq h-1$ since $h\geq t\geq b+1$.

Assume first that $t=h$. Then $x_{s}\rightarrow x_{h}\rightarrow x_{l}$. By Lemma 4.2,

[TABLE]

Let $x_{i}\rightarrow x_{j}$, where $i\in[1,h-1]$ and $j\in[h+1,n-2]$. From the second equality of (35) it follows that $j\leq k$, i.e., $j\in[h+1,k]$. By (36) we have that $i\geq b$, i.e., $i\in[b,h-1]$. Therefore, $C(y,z)=x_{1}\ldots x_{a}x_{b}\ldots x_{i}x_{j}\ldots x_{k}yx_{q}\ldots x_{s}x_{h}x_{l}\ldots x_{n-2}zx_{1}$, a contradiction.

Assume next that $t\leq h-1$. From the maximality of $b$ and $t$ it follows that $d^{-}(x_{h},\{x_{1},\ldots,x_{b-1}\})=0$. This last equality together with (37) implies that

[TABLE]

[TABLE]

[TABLE]

This together with (11), $d(z)\leq n-1$ and $r\leq n-3$ implies that

[TABLE]

[TABLE]

which contradicts condition $(M)$, since $k-q\geq 0$, $b-q\geq 1$. The discussion of Case 2 is completed. Theorem 1.12 is proved.

Acknowledgements.

The author would like to thank the anonymous referees for thoroughly review and many helpful comments and suggestions which improved substantially the rewriting of this paper. We also thank PhD P. Hakobyan for formatting the manuscript of this paper.