Partition of a Subset into Two Directed Cycles with Partial Degrees

TL;DR

This paper proves that under certain degree conditions, a directed graph contains two disjoint cycles covering a specified subset, and conjectures a similar property for multiple cycles with a partition of the subset.

Contribution

It establishes a degree condition ensuring two disjoint cycles cover a subset with a given partition, and proposes a conjecture for multiple cycles with sharper conditions.

Findings

Degree condition guarantees two disjoint cycles covering a subset

Conjecture extends the result to multiple cycles with larger partitions

Degree condition is shown to be sharp in general

Abstract

Let $D=(V,A)$ be a directed graph of order $n\geq 6$. Let $W$ be a subset of $V$ with $|W|\geq 6$. Suppose that every vertex of $W$ has degree at least $(3n-3)/2$ in $D$. Then for any integer partition $|W|=n_1+n_2$ with $n_1\geq 3$ and $n_2\geq 3$, $D$ contains two disjoint directed cycles $C_1$ and $C_2$ such that $|V(C_1)\cap W|=n_1$ and $|V(C_2)\cap W|=n_2$. We conjecture that for any integer partition $|W|=n_1+n_2+\cdots +n_k$ with $k\geq 3$ and $n_i\geq 3(1\leq i\leq k)$, $D$ contains $k$ disjoint directed cycles $C_1,C_2,\ldots , C_k$ such that $|V(C_i)\cap W|=n_i$ for all $1\leq i\leq k$. The degree condition is sharp in general.

Equations

$$x_1\dots x_{i}y{x}_j{x}_{j-1}\dots x_{i+1}z{x}_{j+1}\dots x_k$$

$$d(x,G)$$

$$\ge$$

$$a(u,Y)$$

$$|Y|$$

$$\ge$$

$$l(P_1)+l(P_2)\mathrm{is}minimal.$$

$$e(G_1)+e(G_2)\mathrm{is}maximal.$$

$$d(x,G_1)+d(y,G_2)\ge d(x,G_2)+d(y,G_1)-2\mu (xy).$$

$$e(G_1-x+y)+e(G_2-y+x)$$

$$d(xy,G_1)+d(uv,G_2)\ge d(xy,G_2)+d(uv,G_1)-2d(xy,uv).$$

$$d(vw,H)$$

$$\ge$$

$$a(x_1{x}_s,G_2)\ge 3n-2-{\lambda }_n-a(x_1{x}_s,G_1+R)\ge 3t+s+r-{\lambda }_n;$$

$$d(x_1{x}_s,G_2)\ge 3t+s+r-{\lambda }_n-|N_D(x_1,G_2)|-|N_D(x_s,G_2)|\ge t+s+r-{\lambda }_n.$$

$$d(x_1,G_2)$$

$$\ge$$

$$=$$

$$t-2\ge d(x_1,G_2-\{b_{i-1},b_{i+1}\})\ge (n-{\lambda }_n)/2\mathrm{and}\mathrm{so}t\ge (n+4-{\lambda }_n)/2.$$

$$d(x_1,G_2-V(h_i))\ge (t+s+r-{\lambda }_n)/2-|V(h_i)|.$$

$$\mathrm{Either}a(b_{i-1}{b}_{i+1},C-b_i)\ge 2(l(C)-1)\mathrm{or}a(b_{i-1}{b}_{i+1},R^{\prime })\ge 2r^{\prime }+1.$$

$$d(y_1{y}_t,G_2)\le t-1,I(y_1{y}_t,R)=\varnothing \mathrm{and}d(y_1{y}_t,G_2+R)\le r+t-1.$$

$$d(y_1{y}_t,G_1)$$

$$\ge$$

$$=$$

$$\mathrm{If}d(y_1{y}_t,G_1)=s,\mathrm{then}{\lambda }_n=1\mathrm{and}|N_D(y_1)|=|N_D(y_t)|=n-1;$$

$$\mathrm{If}d(y_1{y}_t,G_1)=s+1,\mathrm{then}{y}_1{y}_t\in A^{\prime }\mathrm{and}|N_D(y_1)|+|N_D(y_t)|\ge 2n-3.$$

$$d(x_1{x}_s,y_1{y}_t)\le 2,d(y_1{y}_t,G_1)\le s+1\mathrm{and}{y}_1{y}_t\in A^{\prime }.$$

$$d(x_1{x}_s,G_2)+d(y_1{y}_t,G_1)-d(x_1{x}_s,G_1)-d(y_1{y}_t,G_2)$$

$$\ge (t+s+r-{\lambda }_n)+(s+2-{\lambda }_n)-(s-1)-(t-1)$$

$$=s+4+r-2{\lambda }_n\ge 5.$$

$$d(y_1{y}_t,G_2)\mathrm{is}\mathrm{minimal}.$$

$$d(x_s,G_2)\le (t-2)/2\mathrm{and}\mathrm{if}{x}_2\in W\mathrm{then}d(x_2,G_2)\le (t-1)/2.$$

$$d(y_t,G)\le (t-s-r+{\lambda }_n)/2+(s-1)/2+r=(t+r-1+{\lambda }_n)/2<(n-{\lambda }_n)/2.$$

Full text

Partition of a Subset into Two Directed Cycles

with Partial Degrees

Hong Wang

Department of Mathematics

The University of Idaho

Moscow, Idaho, USA 83844

Let $D=(V,A)$ be a directed graph of order $n\geq 6$. Let $W$ be a subset of $V$ with $|W|\geq 6$. Suppose that every vertex of $W$ has degree at least $(3n-3)/2$ in $D$. Then for any integer partition $|W|=n_{1}+n_{2}$ with $n_{1}\geq 3$ and $n_{2}\geq 3$, $D$ contains two disjoint directed cycles $C_{1}$ and $C_{2}$ such that $|V(C_{1})\cap W|=n_{1}$ and $|V(C_{2})\cap W|=n_{2}$. We conjecture that for any integer partition $|W|=n_{1}+n_{2}+\cdots+n_{k}$ with $k\geq 3$ and $n_{i}\geq 3(1\leq i\leq k)$, $D$ contains $k$ disjoint directed cycles $C_{1},C_{2},\ldots,C_{k}$ such that $|V(C_{i})\cap W|=n_{i}$ for all $1\leq i\leq k$. The degree condition is sharp in general.

1 Introduction

We discuss only finite simple graphs and strict directed graphs. The terminology and notation concerning graphs is that of [2], except as indicated. A set of graphs or directed graphs is said to be disjoint if no two of them have any common vertices. El-Zahar [5] conjectured that if $G$ is a graph of order $n=n_{1}+\cdots+n_{k}$ with $n_{i}\geq 3(1\leq i\leq k)$ and the minimum degree of $G$ is at least $\lceil n_{1}/2\rceil+\cdots+\lceil n_{k}/2\rceil$ then $G$ contains $k$ disjoint cycles $C_{1},\ldots,C_{k}$ of orders $n_{1},\ldots,n_{k}$, respectively. This conjecture has been verified for a number of cases (for example, see, [3], [5] and [10]). Sauer and Spencer in their work [8] conjectured that if the minimum degree of $G$ is at least $2n/3$ then $G$ contains every graph of order $n$ with maximum degree of at most 2. This conjecture was proved by Aigner and Brandt [1]. Shi [9] showed that if $G$ is 2-connected, $W\subseteq V(G)$ and $d(x)\geq n/2$ for each $x\in W$ then $G$ contains a cycle passing through all the vertices of $W$. This generalizes the classic result of Dirac [4].

In [13], we proposed a conjecture on disjoint cycles of a graph $G$ if we only know a degree condition on vertices in a subset of $V(G)$:

Conjecture $A$ [13] Let $G$ be a graph of order $n\geq 3$. Let $W$ be a subset of $V(G)$ with $|W|\geq 3k$ where $k$ is a positive integer. Suppose that $d(x)\geq 2n/3$ for each $x\in W$. Then for any integer partition $|W|=n_{1}+\cdots+n_{k}$ with $n_{i}\geq 3(1\leq i\leq k)$, $G$ contains $k$ disjoint cycles $C_{1},\ldots,C_{k}$ such that $|V(C_{i})\cap W|=n_{i}$ for all $1\leq i\leq k$.

In [13], we showed that $G$ contains $\lfloor|W|/3\rfloor$ disjoint cycles such that each of them contains at least three vertices of $W$. This conjecture was verified in [14] for the case $k=2$, i.e., $|W|=n_{1}+n_{2}$. This result further supports Conjecture $A$. In [7] and [11], we proved the following theorem.

**Theorem $B$ ** $($[7],[11]$)$ Let $D$ be a directed graph of order $n\geq 4$. Suppose that the minimum degree of $D$ is at least $(3n-3)/2$. Then for any two integers $s$ and $t$ with $s\geq 2$, $t\geq 2$ and $s+t\leq n$, $D$ contains two disjoint directed cycles of orders $s$ and $t$, respectively.

We proposed the following conjecture [12].

Conjecture $C$ [12] Let $D$ be a directed graph of order $n\geq 4$. Suppose that the minimum degree of $D$ is at least $(3n-3)/2$. Then for any directed graph $H$ with $\Delta^{+}(H)\leq 1$ and $\Delta^{-}(H)\leq 1$, $D$ contains a directed subgraph isomorphic to $H$.

To support this conjecture, we proved the following theorem:

**Theorem $D$ ** [12] If $D$ is a directed graph of order $n\geq 4$ with minimum degree at least $(3n-3)/2$, then $D$ contains $\lfloor n/3\rfloor$ disjoint directed triangles.

Continuing the work along this line, we propose the following conjecture:

Conjecture $E$ *Let $D=(V,A)$ be a directed graph of order $n\geq 6$. Let $W$ be a subset of $V$. Suppose that every vertex of $W$ has degree at least $(3n-3)/2$ in $D$. Then for any integer partition $|W|=n_{1}+n_{2}+\cdots+n_{k}$ with $k\geq 2$ and $n_{i}\geq 3(1\leq i\leq k)$, $D$ contains $k$ disjoint directed cycles $C_{1},C_{2},\ldots,C_{k}$ such that $|V(C_{i})\cap W|=n_{i}$ for all $1\leq i\leq k$. *

To demonstrate that the degree condition in Conjecture $E$ is sharp in general for each $n\geq 6$, we construct the following directed graph $B_{n}$ of order $n$. For each positive integer $k$, we use $K_{k}^{*}$ to denote the complete directed graph of order $k$, i.e., $K_{k}^{*}$ contains both $(x,y)$ and $(y,x)$ for any two distinct vertices $x$ and $y$ of $K_{k}^{*}$. The directed graph $B_{n}$ consists of two disjoint complete directed subgraphs $D^{\prime}$ and $D^{\prime\prime}$ of order $\lfloor n/2\rfloor$ and $\lceil n/2\rceil$, respectively, and all the arcs $(x,y)$ with $x\in V(D^{\prime})$ and $y\in V(D^{\prime\prime})$. Then the minimum degree of $B_{n}$ is $\lfloor(3n-4)/2\rfloor$. Clearly, $B_{n}$ does not contain a directed cycle of order greater than $\lceil n/2\rceil$.

In [15], we showed that if $|W|\geq 3k$ then $D$ contains $k$ disjoint directed cycles covering $W$ such that each of them contains at least three vertices of $W$. In this paper, we prove the following result:

Theorem $F$ *Let $D=(V,A)$ be a directed graph of order $n\geq 6$. Let $W$ be a subset of $V$. Suppose that every vertex of $W$ has degree at least $(3n-3)/2$ in $D$. Then for any integer partition $|W|=n_{1}+n_{2}$ with $n_{1}\geq 3$ and $n_{2}\geq 3$, $D$ contains two disjoint directed cycles $C_{1}$ and $C_{2}$ such that $|V(C_{1})\cap W|=n_{1}$ and $|V(C_{2})\cap W|=n_{2}$. *

For convenience, we need some special terminology and notation. Let $D$ be a digraph and $G$ a graph with the same vertex set, i.e., $V=V(D)=V(G)$. We use $V(D)$ and $A(D)$ to denote the vertex set and the arc set of $D$, respectively. Use $E(G)$ to denote the edge set of $G$ and let $e(G)=|E(G)|$. Let $x\in V$. The out-degree, in-degree and degree of $x$ in $D$ are denoted by $a^{+}(x)$, $a^{-}(x)$ and $a(x)$, respectively. Let each of $X$ and $Y$ be a subset of $V$ or a sequence of some vertices of $V$ or a subgraph of $G$ or a subdigraph of $D$. We use $N_{D}^{+}(x,Y)$ denote the set of all the vertices $y$ that are contained in $Y$ with $(x,y)\in A(D)$. We similarly define $N_{D}^{-}(x,Y)$ and let $N_{D}(x,Y)=N_{D}^{+}(x,Y)\cup N_{D}^{-}(x,Y)$. We also define $a^{+}(x,Y)=|N^{+}(x,Y)|$, $a^{-}(x,Y)=|N^{-}(x,Y)|$ and $a(x,Y)=a^{+}(x,Y)+a^{-}(x,Y)$. Thus $a^{+}(x,D)=a^{+}(x)$, $a^{-}(x,D)=a^{-}(x)$ and $a(x,D)=a(x)$. Define $a(X,Y)=\sum_{u}a(x,Y)$, where $u$ runs over all the vertices of $X$.

We define $N_{G}(x,Y)$ to be the set of all the vertices $y$ in $Y$ with $xy\in E$ and let $d(x,Y)=|N_{G}(x,Y)|$. Hence $d(x,G)=d(x,D)=d(x,V)=d(u)$ which is the degree of $u$ in $G$. For a subset $U\subseteq V$, $G[U]$ is the subgraph of $G$ induced by $U$ and $D[U]$ is the subdigraph of $D$ induced by $U$. Define $d(X,Y)=\sum_{u}d(x,Y)$, where $u$ runs over all the vertices of $X$.

We define $I(xy,X)=N_{G}(x,X)\cap N_{G}(y,X)$ and let $i(xy,X)=|I(xy,X)|$. We define $I(\vec{xy},X)=N_{D}^{+}(x,X)\cap N_{D}^{-}(y,X)$ and let $i(\vec{xy},X)=|I(\vec{xy},X)|$.

A graph or a digraph is said to be traceable if it contains a hamiltonian path or directed hamiltonian path, respectively. A graph or a digraph is called hamiltonian if it contains a hamiltonian cycle or directed hamiltonian cycle, respectively.

If $C$ is a dicycle of $D$ or a cycle of $G$ and $x$ is a vertex of $C$, we use $x^{+}$ and $x^{-}$ to denote the successor and the predeccssor of $x$ on $C$, respectively. If $a$ and $b$ are two vertices of $C$, we use $C[a,b]$ to represent the path of $C$ from $a$ to $b$ along the direction of $C$. We adopt the notation $C(a,b]=C[a,b]-a$, $C[a,b)=C[a,b]-b$ and $C(a,b)=C[a,b]-a-b$.

For two vertices $x$ and $y$ of $D$, we say that $x$ is adjacent to $y$ in $D$ if either $(x,y)$ or $(y,x)$ is an arc of $D$. For any two vertices $x$ and $y$ of $G$, we define $\mu(xy)=1$ if $xy$ is an edge of $G$ and $\mu(xy)=0$ otherwise. For any integer $m$, let $\lambda_{m}=1$ if $m$ is odd and otherwise $\lambda_{m}=0$.

2 Lemmas

Let $D=(V,A)$ be a digraph. Let $G=(V,E)$ be the underlining graph of $D$, i.e., $V(G)=V$ and $E(G)=\{xy|(x,y)\in A\ {\rm and}\ (y,x)\in A\}$. We will use the following lemmas which are generally well known.

Lemma 2.1

Let $P=x_{1}x_{2}\ldots x_{k}$ be a path of $G$ and $\{y,z\}\subseteq V-V(P)$. The following two statements hold:

$(a)$* If $d(y,P)>k/2$ then $\{x_{i},x_{i+1}\}\subseteq N_{G}(y,P)$ for some $i\in\{1,2,\ldots,k-1\}$ unless $k$ is odd, $d(y,P)=(k+1)/2$ and $N_{G}(y,P)=\{x_{1},x_{3},x_{5},\ldots,x_{k}\}$;*

$(b)$* If $d(y,P)\geq k/2$, then $P+y$ is traceable;*

$(c)$* If $d(x_{1}y,P)\geq k$, then $P+y$ is traceable;*

$(d)$* If $d(yz,P)\geq k+2$ then $P+y+z$ is traceable.*

Proof. To show $(a)$, it is easy to see that if $d(y,x_{i}x_{i+1})\leq 1$ for all $i\in\{1,\ldots,k-1\}$, then $k$ must be odd and $\{x_{1},x_{k}\}\subseteq N_{G}(y,P)$ as $d(y,P)>k/2$ and it follows that $N_{G}(y,P)=\{x_{1},x_{3},x_{5},\ldots,x_{k}\}$.

To show $(b)$, it is easy to see that if $\{x_{1},x_{k}\}\cap N_{G}(y,P)=\emptyset$, then there exists $i\in\{2,\ldots,k-2\}$ such that $\{x_{i},x_{i+1}\}\subseteq N_{G}(y,P)$. Thus the path $P^{\prime}=x_{1}\ldots x_{i}yx_{i+1}\ldots x_{k}$ satisfies the requirement.

To show $(c)$, it is easy to see that if $d(x_{k},x_{1}y)\geq 1$ then $(c)$ holds. So we may assume that $d(x_{k},x_{1}y)=0$. As $d(x_{1}y,P)\geq k$, it follows that $\{x_{i}y,x_{i+1}x_{1}\}\subseteq E$ for some $i\in\{1,\ldots,k-2\}$. Therefore $yx_{i}x_{i-1}\ldots x_{1}x_{i+1}x_{i+1}\ldots x_{k}$ is a hamiltonian path of $P+y$.

To show $(d)$, we see that $\{yx_{i},yx_{j},zx_{i+1},zx_{j+1}\}\subseteq E$ for some $\{i,j\}\subseteq\{1,2,\ldots,k\}$ with $i<j$, where $x_{k+1}=x_{1}$. If $j=k$, $zPy$ is a path of $G$ and if $j<k$ then

[TABLE]

is a hamiltonian path of $P+y+z$.

Lemma 2.2

Let $C$ be a dicycle of $D$ or a cycle of $G$. Let $y\in V-V(C)$. Suppose that $d(y,C)\geq l(C)/2$. Then either $D[V(C)\cup\{y\}]$ is hamiltonian if $C$ is a cycle of $D$ and $G[V(C)\cup\{y\}]$ is hamiltonian if $C$ is a cycle of $G$, or $l(C)$ is even and there is a list $x_{1},\ldots,x_{l(C)}$ of vertices of $C$ along the direction of $C$ such that $N_{G}(y,C)=\{x_{2i-1}|1\leq i\leq l(C)/2\}$.

Proof. If the last statement does not hold, then there exists $x\in V(C)$ such that $\{x,x^{+}\}\subseteq N_{G}(y)$ and the lemma follows.

Lemma 2.3

Let $P=x_{1}x_{2}\ldots x_{k}$ be a path of $G$ with $k\geq 3$. If either $a^{+}(x_{1},P)+a^{+}(x_{k},P)\geq k$ or $a^{-}(x_{1},P)+a^{-}(x_{k},P)\geq k$, then $D[V(P)]$ is hamiltonian. Therefore if $a(x_{1},P)+a(x_{k},P)\geq 2k-1$ then $D[V(P)]$ is hamiltonian.

Proof. If $a^{+}(x_{1},P)+a^{+}(x_{k},P)\geq k$, then $\{(x_{1},x_{i+1}),(x_{k},x_{i})\}\subseteq A$ for some $i\in\{1,\ldots,k-1\}$ and so $(x_{1},x_{i+1},x_{i+2},\ldots,x_{k},x_{i},x_{i-1},\ldots,x_{1})$ is a dicycle of $D[V(P)]$. Similarly, we can easily see that the lemma holds if $a^{-}(x_{1},P)+a^{-}(x_{k},P)\geq k$. Since $a(x_{1},P)+a(x_{k},P)\geq 2k-1$ implies either $a^{+}(x_{1},P)+a^{+}(x_{k},P)\geq k$ or $a^{-}(x_{1},P)+a^{-}(x_{k},P)\geq k$, the lemma follows.

Lemma 2.4

[6]* Let $P=x_{t}x_{t-1}\ldots x_{1}$ be a longest path starting at $x_{t}$ in $G$. Let $c>t/2$. Suppose that for each $v\in V(G)$, if there exists a longest path starting at $x_{t}$ in $G$ such that the path ends at $v$ then $d_{G}(v)\geq c$. Then there exists $r\geq c+1$ such that $N(x_{i})\subseteq\{x_{1},x_{2},\ldots,x_{r}\}$, $G[P]$ has an $x_{t}$-$x_{i}$ hamiltonian path containing $x_{t}x_{t-1}\ldots x_{r}$ and $d_{G}(x_{i})\geq c$ for all $i\in\{1,2,\ldots,r-1\}$. Moreover, if $t>r$ then $x_{r}$ is a cut-vertex of $G$.*

3 Proof of Theorem $F$

Let $D=(V,A)$ be a directed graph of order $n$. Let $W\subseteq V$ with $|W|\geq 6$. Suppose that $a(x)\geq(3n-2-\lambda_{n})/2$ for all $x\in W$. Suppose, for a contradiction, that for some integer partition $|W|=n_{1}+n_{2}$ with $n_{1}\geq 3$ and $n_{2}\geq 3$, $D$ does not contain two disjoint dicycles $C_{1}$ and $C_{2}$ such that $|V(C_{1})\cap W|=n_{1}$ and $|V(C_{2})\cap W|=n_{2}$. For the proof, we may assume that $D-W$ has no arcs. Let $G=(V,E)$ be the underlining graph of $D$.

For any subgraph $G^{\prime}$ of $G$, we use $W(G^{\prime})$ to denote $W\cap V(G^{\prime})$. For convenince, let $A^{\prime}=\{xy|\ \{x,y\}\subseteq V\ {\rm and\ either}\ (x,y)\in A\ {\rm or}\ (y,x)\in A\}$, i.e., $x$ and $y$ are adjacent in $D$ if and only if $xy\in A^{\prime}$. For a dicycle or a cycle $C$, we use $l(C)$ to denote the length of $C$ and use $l_{w}(C)$ to denote $|V(C)\cap W|$ which is called the $W$-length of $C$. Similarly, we define $l(P)$ and $l_{w}(P)$ for a dipath or a path $P$.

For each $x\in W$, we have

[TABLE]

For each partition $(X,Y)$ of $V$ and $u\in X$, if $a(u,X)\leq|X|-1$, then we have

[TABLE]

Lemma 3.1

There is a path $P$ in $G$ such that $V(P)\supseteq W$.

Proof. We choose a path $P$ in $G$ with $l_{w}(P)$ maximal. Subject to this, we choose $P$ with $l(P)$ minimal. Assume there exists $x_{0}\in W-V(P)$. Say $P=x_{1}x_{2}\ldots x_{t}$. Then $x_{0}x_{1}\not\in E$, $I(x_{1}x_{0},V-V(P))=\emptyset$ and so $d(x_{1}x_{0},V-V(P))\leq n-t-1$. By (2), $d(x_{1}x_{0},P)\geq(n-\lambda_{n})-(n-t-1)\geq t$. By Lemma 2.1$(c)$, $P+x_{0}$ is traceable, a contradiction.

By Lemma 3.1, $G$ has two disjoint paths $P_{1}$ and $P_{2}$ with $l_{w}(P_{1})=n_{1}$ and $l_{w}(P_{2})=n_{2}$. Let $G_{1}=G[V(P_{1})]$ and $G_{2}=G[V(P_{2})]$. We choose $P_{1}$ and $P_{2}$ such that

[TABLE]

Subject (7), we choose $P_{1}$ and $P_{2}$ such that

[TABLE]

Say $P_{1}=x_{1}x_{2}\ldots x_{s}$, $P_{2}=y_{1}y_{2}\ldots y_{t}$, $R=V-V(G_{1}\cup G_{2})$ and $r=|R|=n-s-t$. Let $H=G[V(G_{1}\cup G_{2})]$, $D_{1}=D[V(G_{1})]$ and $D_{2}=D[V(G_{2})]$. By (8), we obtain the following lemma.

Lemma 3.2

Let $x$ be an endvertex of a hamiltonian path of $G_{1}$ and $y$ an endvertex of a hamiltonian path of $G_{2}$. If both $G_{1}-x+y$ and $G_{2}-y+x$ are traceable, then

[TABLE]

In particular, $(9)$ holds if $d(x,G_{2})\geq(t+1)/2$ and $d(y,G_{1})\geq(s+1)/2$ hold.

Proof. We have

[TABLE]

If both $G_{1}-x+y$ and $G_{2}-y+x$ are traceable, then $e(G_{1})+e(G_{2})\geq e(G_{1}-x+y)+e(G_{2}-y+x)$ by (8) and it follows that (9) holds. Note that by Lemma 2.1$(b)$, if $d(x,G_{2})\geq(t+1)/2$ and $d(y,G_{1})\geq(s+1)/2$, both $G_{1}-x+y$ and $G_{2}-y+x$ are traceable.

Similarly, by (8), one can easily see that the following Lemma 3.3 holds.

Lemma 3.3

Let $x$ and $y$ be the two endvertices of a hamiltonian path of $G_{1}$ and $u$ and $v$ be the two endvertices of a hamiltonian path of $G_{2}$. Suppose that both $G_{1}-x-y+u+v$ and $G_{2}-u-v+x+y$ are taceable. Then

[TABLE]

Lemma 3.4

For some $i\in\{1,2\}$, $D_{i}$ does not have a dicycle of $W$-length $n_{i}$ and if $x$ and $y$ are the two endvertices of a hamiltonian path of $G_{i}$, then $I(\vec{xy},R)=\emptyset$, $I(\vec{yx},R)=\emptyset$ and so $a(xy,R)\leq 2r$.

Proof. If $D_{1}$ contains a dicycle of $W$-length $n_{1}$, then $D[V(D_{2})\cup R]$ does not contain a dicycle of $W$-length $n_{2}$, and this implies that the lemma holds with $i=2$. Therefore we may assume that $D_{1}$ does not contain a dicycle of $W$-length $n_{1}$. Similarly, we may assume that $D_{2}$ does not contain a dicycle of $W$-length $n_{2}$. In particular, $x_{1}x_{s}\not\in A^{\prime}$ and $y_{1}y_{t}\not\in A^{\prime}$. On the contrary, say the lemma fails. Say w.l.o.g. that $I(\vec{x_{1}x_{s}},R)\not=\emptyset$ and $I(\vec{y_{1}y_{t}},R)\not=\emptyset$. Since $D$ does not contain the two required dicycles, we see that for some $u\in R$, $I(\vec{x_{1}x_{s}},R)=I(\vec{y_{1}y_{t}},R)=\{u\}$. Moreover, $I(\vec{x_{s}x_{1}},R)\subseteq\{u\}$ and $I(\vec{y_{t}y_{1}},R)\subseteq\{u\}$. This implies that $a(x_{1}x_{s},R)\leq 2r+2$ and $a(y_{1}y_{t},R)\leq 2r+2$. Then for each $vw\in\{x_{1}x_{s},y_{1}y_{t}\}$, $a(vw,H)\geq 3n-2-\lambda_{n}-2r-2=3s+3t+r-4-\lambda_{n}$. Therefore for each $vw\in\{x_{1}x_{s},y_{1}y_{t}\}$, we have

[TABLE]

Since both $G_{1}$ and $G_{2}$ are not hamiltonian, $d(x_{1}x_{s},G_{1})\leq s-1$ and $d(y_{1}y_{t},G_{2})\leq t-1$. Therefore $d(x_{1}x_{s},G_{2})\geq s+t+r-\lambda_{n}-(s-1)\geq t+1+r-\lambda_{n}$ and $d(y_{1}y_{t},G_{1})\geq s+t+r-\lambda_{n}-(t-1)\geq s+1+r-\lambda_{n}$. Let $\sigma=d(x_{1}x_{s},G_{2})+d(y_{1}y_{t},G_{1})-d(x_{1}x_{s},G_{1})-d(y_{1}y_{t},G_{2})$. Then $\sigma\geq 4+2(r-\lambda_{n})$.

First, assume that there exist two independent edges between $\{x_{1},x_{s}\}$ and $\{y_{1},y_{t}\}$ in $G$. Say w.l.o.g. $\{x_{1}y_{t},x_{s}y_{1}\}\subseteq E$. By Lemma 3.2, $d(x_{1},G_{1})+d(y_{1},G_{2})\geq d(x_{1},G_{2})+d(y_{1},G_{1})-2\mu(x_{1}y_{1})$ and $d(x_{s},G_{1})+d(y_{t},G_{2})\geq d(x_{s},G_{2})+d(y_{t},G_{1})-2\mu(x_{s}y_{t})$. Thus $2\mu(x_{1}y_{1})+2\mu(x_{s}y_{t})\geq\sigma$. Since $4\geq 2\mu(x_{1}y_{1})+2\mu(x_{s}y_{t})$ and $\sigma\geq 4+2(r-\lambda_{n})$, it follows that equality hold in (11) and (12) with $r=\lambda_{n}=1$. In particular, $|N_{D}(z,H)|=s+t-2$ for all $z\in\{x_{1},x_{s},y_{1},y_{t}\}$. Hence $x_{1}y_{2}\in A^{\prime}$ and $y_{1}x_{2}\in A^{\prime}$. Consequently, $D_{1}-x_{1}+y_{1}$ and $D_{2}-y_{1}+x_{1}$ are hamiltonian, a contradiction.

Therefore $G$ does not contain two independent edges between $\{x_{1},x_{s}\}$ and $\{y_{1},y_{t}\}$ and so $d(x_{1}x_{s},y_{1}y_{t})\leq 2$. Assume for the moment that $d(x_{1}x_{s},G_{2})\geq(t+2)$ and $d(y_{1}y_{t},G_{1})\geq s+2$. Then $d(x_{1}x_{s},G_{2}-y_{1}-y_{t})\geq(t-2)+2$ and $d(y_{1}y_{t},G_{1}-x_{1}-x_{s})\geq(s-2)+2$. By Lemma 2.1$(d)$ and Lemma 3.3, we shall have that $d(x_{1}x_{s},G_{1})+d(y_{1}y_{t},G_{2})\geq d(x_{1}x_{s},G_{2})+d(y_{1}y_{t},G_{1})-2d(x_{1}x_{s},y_{1}y_{t})$. Thus $s-1+t-1\geq s+2+t+2-4$, a contradiction. Therefore we may assume w.l.o.g.that $d(x_{1}x_{s},G_{2})\leq t+1$. By (12), $d(x_{1}x_{s},G_{2})\geq s+t+r-\lambda_{n}-(s-1)=t+1+r-\lambda_{n}$. It follows that $d(x_{1}x_{s},G_{1})=s-1$, $d(x_{1}x_{s},G_{2})=t+1$, $r=1$, $\lambda_{n}=1$ (i.e., $n$ is odd), and equality hold in (11) and (12) with $\{v,w\}=\{x_{1},x_{s}\}$ and $N_{D}(x_{1})=N_{D}(x_{s})=V-\{x_{1},x_{s}\}$. Moreover, $a(x_{1}x_{s},R)=a(x_{1}x_{s},u)=4$, i.e., $d(x_{1}x_{s},R)=d(x_{1}x_{s},u)=2$. As $x_{1}x_{s}\not\in A^{\prime}$ and by (1), we see that $d(x_{1},G)\geq(n+1)/2$ and $d(x_{s},G)\geq(n+1)/2$. As $x_{1}x_{s-1}\in A^{\prime}$, $D_{1}-x_{s}$ is hamiltonian and so $d(x_{s},G_{1})\leq(s-1)/2$ by Lemma 2.2 because $D_{1}$ is not hamiltonian. Similarly, $d(x_{1},G_{1})\leq(s-1)/2$. It follows that $s$ and $t$ are odd and $d(x_{i},G_{1})=(s-1)/2$ and $d(x_{i},G_{2})=(t+1)/2$ for $i\in\{1,s\}$. Suppose that $d(y_{1}y_{t},G_{1})\leq s+1$. Then this argument allows us to see that $d(y_{j},G_{2})=(t-1)/2$ and $d(y_{j},G_{1})=(s+1)/2$ for $j\in\{1,t\}$. For each $i\in\{1,s\}$ and $j\in\{1,t\}$, applying Lemma 3.2 with $x_{i}$ and $y_{j}$ in place of $x$ and $y$, we see that $\mu(x_{i}y_{j})=1$. Thus $d(x_{1}x_{s},y_{1}y_{t})=4$, a contradiction.

Therefore $d(y_{1}y_{t},G_{1})\geq s+2$. Say w.l.o.g. $d(y_{1},G_{1})\geq\lceil(s+2)/2\rceil=(s+3)/2$. Applying Lemma 3.2 with $x_{1}$ and $y_{1}$ in place of $x$ and $y$, we see that $d(y_{1},G_{2})\geq(t+1)/2$. Thus $d(y_{t},G_{2})\leq t-1-(t+1)/2=(t-3)/2$. By (1), $d(y_{t},G)\geq(n+1)/2$. Consequently, $d(y_{t},G_{1})\geq(n+1)/2-(t-3)/2-\mu(y_{t}u)\geq(s+3)/2$. Applying Lemma 3.2 with $x_{1}$ and $y_{t}$ in place of $x$ and $y$, we obtain a contradiction to (9).

We now may assume that Lemma 3.4 holds for $i=1$. As $D_{1}$ is not hamiltonian, $a(x_{1}x_{s},P_{1})\leq 2(s-1)$ by Lemma 2.3 and so $a(x_{1}x_{s},G_{1}+R)\leq 2(s-1)+2r$. Thus

[TABLE]

We assume w.l.o.g. that $a(x_{1},G_{1}+R)\leq a(x_{s},G_{1}+R)$. Then $a(x_{1},G_{1}+R)\leq s+r-1$ and $a(x_{1},G_{2})\geq(3n-2-\lambda_{n})/2-(s+r-1)=(3t+s+r-\lambda_{n})/2$. Thus

[TABLE]

It follows that $t\geq(n-\lambda_{n})/2$. Clearly, $(n-\lambda_{n})/2-1>(t-1)/2$. By Lemma 2.1$(b)$ and Lemma 2.2, we obtain the following corollary:

Corollary 3.5

*For each $v\in V(G_{2})$, if $G_{2}-v$ is traceable then $G_{2}-v+x_{1}$ is traceable and if $D[V(G_{2}-v)]$ is hamiltonian then $D[V(G_{2}-v+x_{1})]$ is hamiltonian.  *

Lemma 3.6

$G_{2}$* is not hamiltonian and $G_{2}+z$ is not hamiltonian for each $z\in R$.*

Proof. On the contrary, say the lemma fails. If $G_{2}$ is a hamiltonian, let $C$ be a hamiltonian cycle and otherwise let $C$ be a hamiltonian cycle of $G_{2}+z_{0}$ for some $z_{0}\in R$. Let $b_{1},b_{2},\ldots,b_{n_{2}}$ be a list of vertices in $W(C)$ along the direction of $C$. Let $h_{i}=C[b_{i},b_{i+1})$ for each $i\in\{1,\ldots,n_{2}\}$, where the operation in the subscripts is taken in $\{1,2,\ldots,n_{2}\}$ by modulo $n_{2}$. Note that as $D-W$ has no arcs, each $h_{i}$ has at most two vertices. Let $R^{\prime}=R-\{z_{0}\}$ if $l(C)=t+1$ and otherwise let $R^{\prime}=R$. Say $r^{\prime}=|R^{\prime}|$.

Let $x^{*}=x_{2}$ if $x_{2}\in W$ and otherwise let $x^{*}=x_{3}$ with $x_{3}\in W$. Suppose that $a(x^{*}x_{s},h_{i})\leq 2|V(h_{i})|$ for all $i\in\{1,2,\ldots,n_{2}\}$. Then $a(x^{*}x_{s},C)\leq 2l(C)$. Thus $a(x^{*}x_{s},D-V(C))\geq 3n-2-\lambda_{n}-2l(C)=3s+3r^{\prime}+l(C)-2-\lambda_{n}$. As $x_{1}x_{s}\not\in A^{\prime}$, we obtain that $2(s+r^{\prime}-1)+2(s+r^{\prime}-2)\geq a(x^{*}x_{s},D-V(C))\geq 3s+3r^{\prime}+l(C)-2-\lambda_{n}$ and so $s+r^{\prime}\geq l(C)+4-\lambda_{n}$. This implies that $n\geq 2l(C)+4-\lambda_{n}$, i.e., $l(C)\leq(n-4+\lambda_{n})/2$. But $l(C)\geq t\geq(n-\lambda_{n})/2$, a contradiction.

Therefore $a(x^{*}x_{s},h_{i})\geq 2|V(h_{i})|+1$ for some $i\in\{1,\ldots,n_{2}\}$. Then either $a^{+}(x^{*},h_{i})+a^{-}(x_{s},h_{i})\geq|V(h_{i})|+1$ or $a^{-}(x^{*},h_{i})+a^{+}(x_{s},h_{i})\geq|V(h_{i})|+1$. As $|V(h_{i})|\leq 2$, it follows that $D[V(P_{1}[x^{*},x_{s}])\cup V(h_{i})]$ has a hamiltonian dicycle $C_{1}$ of $W$-length $n_{1}$. Therefore $D-V(C_{1})$ does not contain a dicycle of $W$-length $n_{2}$. In particular, $d(x_{1},b_{i-1}b_{i+1})\leq 1$ and $a(x_{1},b_{i-1}b_{i+1})\leq 2$. Recall that $a(x_{1},G_{1}+R)\leq s+r-1$. By (3) to (6) with $Y=V(G_{2})-\{b_{i-1},b_{i+1}\}$ and $x_{1}$ in place of $u$, we obtain

[TABLE]

By (16), we now have

[TABLE]

Note that if $C$ is a hamiltonian cycle of $G_{2}$, then by (7), $V(G_{2})\subseteq W$, i.e., $n_{2}=t$. We claim

[TABLE]

To see (19), say, for a contradiction, that $a(b_{i-1}b_{i+1},R^{\prime}\cup V(C)-\{b_{i}\})\leq 2r^{\prime}+2(l(C)-1)-1$. As $a(x_{1},b_{i-1}b_{i+1})\leq 2$, $a(b_{i-1}b_{i+1},R^{\prime}\cup V(C)\cup\{x_{1}\}-\{b_{i}\})\leq 2r^{\prime}+2l(C)-1=2(t+r-1)+1$. This implies that $a(u,R^{\prime}\cup V(C)\cup\{x_{1}\}-\{b_{i}\})\leq t+r-1$ for some $u\in\{b_{i-1},b_{i+1}\}$. By (3) to (6) with $Y=V(G_{1})\cup\{b_{i}\}-\{x_{1}\}$, we obtain $s\geq(n-\lambda_{n})/2$. By (17), $s+t\geq(n-\lambda_{n})/2+(n+4-\lambda_{n})/2\geq n+1$, a contradiction. Hence (19) holds.

We break into the following four cases. Cases 2 to 4 are similar to Case 1 with more subtle details.

Case 1. $b_{i}^{+}=b_{i+1}$ and $b_{i}^{-}=b_{i-1}$.

Then $h_{i}=b_{i}$. Let $P=C-b_{i}$. If the latter of (19) holds, $a(z^{\prime},b_{i-1}b_{i+1})\geq 3$ for some $z^{\prime}\in R^{\prime}$ and we let $C^{\prime}$ be a hamiltonian dicycle of $D[V(P)\cup\{z^{\prime}\}]$. Otherwise the former holds and we let $C^{\prime}$ be a hamiltonian dicycle of $D[V(P)]$ by Lemma 2.3. By (18), $d(x_{1},C^{\prime})\geq(t+s+r-\lambda_{n})/2-1=(t+s-2+r-\lambda_{n})/2>l(C^{\prime})/2$. Thus $D[V(C^{\prime})\cup\{x_{1}\}]$ is hamiltonian by Lemma 2.2, a contradiction.

Case 2. $b_{i}^{+}=b_{i+1}$ and $b_{i}^{--}=b_{i-1}$.

Then $h_{i}=b_{i}$. In this case, as $b_{i}^{-}\not\in W$, $C$ is not a hamiltonian cycle of $G_{2}$ by (7). Thus $l(C)=t+1$ with $V(C)\cap R=\{z_{0}\}$ and so $r\geq 1$ . As in Case 1, we see that we may assume that $D[V(C)-\{b_{i}\}]$ is not hamiltonian. Thus $b_{i}^{-}b_{i+1}\not\in A^{\prime}$. If the former of (19) holds, then $a(b_{i-1}b_{i+1},C-b_{i}-b_{i}^{-})\geq 2(l(C)-2)$ and we let $C^{\prime\prime}$ be a hamiltonian dicycle of $D[V(C-b_{i}-b_{i}^{-})]$ by Lemma 2.3. Otherwise $D[V(C-b_{i}-b_{i}^{-})\cup\{z^{\prime\prime}\}]$ is hamiltonian for some $z^{\prime\prime}\in R^{\prime}$ with $a(b_{i-1}b_{i+1},z^{\prime\prime})\geq 3$ and we let $C^{\prime\prime}$ be one of its hamiltonian dicycles. If $x_{1}b_{i}^{-}\not\in E$, then $d(x_{1},C^{\prime\prime})\geq(t+s+r-\lambda_{n})/2-1\geq(l(C^{\prime\prime})+1)/2$ and so $D[V(C^{\prime\prime})\cup\{x_{1}\}]$ is hamiltonian, a contradiction. Therefore $x_{1}b_{i}^{-}\in E$. Then $x_{1}b_{i+1}\not\in A^{\prime}$ for otherwise $D[V(C)\cup\{x_{1}\}-\{b_{i}\}]$ is hamiltonian. By (14), we see that $d(x_{1},C^{\prime\prime})\geq d(x_{1},G_{2})-2\geq(3t+s+r-\lambda_{n})/2-(t-1)-2=(t+s-2+r-\lambda_{n})/2\geq(l(C^{\prime\prime})+1)/2$. Again, $D[V(C^{\prime\prime})\cup\{x_{1}\}]$ is hamiltonian, a contradiction.

Case 3. $b_{i}^{++}=b_{i+1}$ and $b_{i}^{-}=b_{i-1}$.

Then $h_{i}=b_{i}b_{i}^{+}$. In this case, $l(C)=t+1$ with $V(C)\cap R=\{z_{0}\}$. Let $P^{\prime}=C-b_{i}-b_{i}^{+}$. First, assume that the latter of (19) holds. Then $D[V(P^{\prime})\cup\{u\}]$ has a hamiltonian dicycle $L$ for some $u\in R^{\prime}$ with $a(b_{i-1}b_{i+1},u)\geq 3$. Thus $r^{\prime}\geq 1$ and so $r\geq 2$. Clearly, $l(L)\leq t$ and $d(x_{1},L-u-z_{0})=d(x_{1},G_{2}-b_{i}-b_{i}^{+})\geq(t+s+r-\lambda_{n})/2-2=(t+s+r-4-\lambda_{n})/2\geq l(L)/2$. As $D[V(L)\cup\{x_{1}\}]$ is not hamiltonian and by Lemma 2.2, it follows that $l(L)=t$, $d(x_{1},G_{2}-b_{i}-b_{i}^{+})=d(x_{1},L)=t/2$, $d(x_{1},N_{L}(z_{0}))=2$ and so $D[V(L)\cup\{x_{1}\}-\{z_{0}\}]$ is hamiltonian, a contradiction. Hence the latter of (19) does not hold. Note that this argument allows us to see that $D[V(P^{\prime})]$ does not have a hamiltonian dicycle in both Case 3 and Case 4.

Therefore $a(b_{i-1}b_{i+1},R^{\prime})\leq 2r^{\prime}$, and by Lemma 2.3, $a(b_{i-1}b_{i+1},P^{\prime})\leq 2(|V(P^{\prime})|-1)$. As $a(x_{1},b_{i-1}b_{i+1})\leq 2$, it follows that $a(b_{i-1}b_{i+1},R^{\prime}\cup V(C)\cup\{x_{1}\}-\{b_{i},b_{i}^{+}\})\leq 2(t+r-2)$. Thus $a(v,R^{\prime}\cup V(C)\cup\{x_{1}\}-\{b_{i},b_{i}^{+}\})\leq t+r-2$ for some $v\in\{b_{i-1},b_{i+1}\}$. By (3) to (6) with $Y=V(G_{1})\cup\{b_{i},b_{i}^{+}\}-\{x_{1}\}$, we obtain $s+1\geq(n-\lambda_{n})/2$. By (17), $n\geq s+t+1\geq(n-\lambda_{n})/2+(n+4-\lambda_{n})/2\geq n+1$, a contradiction.

Case 4. $b_{i}^{++}=b_{i+1}$ and $b_{i}^{--}=b_{i-1}$.

In this case, $h_{i}=b_{i}b_{i}^{+}$, $l(C)=t+1$ and $r\geq 1$. As noted in Case 3, we may assume that $D[V(P^{\prime})]$ is not hamiltonian for otherwise $D[V(P^{\prime})\cup\{x_{1}\}]$ is hamiltonian, where $P^{\prime}=C-b_{i}-b_{i}^{+}$. Similarly, $D[V(C)-\{b_{i},b_{i}^{-}\}]$ is not hamitonian. Thus $b_{i+1}b_{i}^{-}\not\in A$ and $b_{i-1}b_{i}^{+}\not\in A^{\prime}$. Set $P^{\prime\prime}=C-b_{i}^{-}-b_{i}-b_{i}^{+}$.

If the former of (19) holds, then $a(b_{i-1}b_{i+1},P^{\prime\prime})\geq 2(l(C)-1)-4=2|V(P^{\prime\prime})|$ and by Lemma 2.3, we let $Q$ be a hamiltonian dicycle of $D[V(P^{\prime\prime})]$. Otherwise $D[V(P^{\prime\prime})\cup\{c\}]$ is hamiltonian for some $c\in R^{\prime}$ with $a(c,b_{i-1}b_{i+1})\geq 3$, and we let $Q$ be a hamiltonian dicycle of $D[V(P^{\prime\prime})\cup\{c\}]$. If $x_{1}b_{i}^{-}\not\in E$, $d(x_{1},Q)\geq(t+s+r-\lambda_{n})/2-2=(t+s-4+r-\lambda_{n})/2>l(Q)/2$ and so $D[V(Q+x_{1})]$ is hamiltonian by Lemma 2.2, a contradiction. Therefore $x_{1}b_{i}^{-}\in E$. In this situation, $x_{1}b_{i+1}\not\in A^{\prime}$. By (14), $d(x_{1},G_{2})\geq(t+s+r+2-\lambda_{n})/2$. Thus $d(x_{1},Q)\geq(t+s+r+2-\lambda_{n})/2-3=(t+s-4+r-\lambda_{n})/2>l(Q)/2$ and so $D[V(Q+x_{1})]$ is hamiltonian, a contradiction.

By Lemma 3.6, we obtain

[TABLE]

Since we have $d(y_{1}y_{t},G)\geq 3n-2-\lambda_{n}-|N_{D}(y_{1})|-|N_{D}(y_{t})|$, it follows that

[TABLE]

From (21) and (22), it follows that

[TABLE]

We claim

[TABLE]

Proof of $(27)$. On the contray, say (27) fails. Since $G_{1}$ is not hamiltonian, $d(x_{1}x_{s},G_{1})\leq s-1$. First, assume that $d(x_{1}x_{s},y_{1}y_{t})\geq 3$. Then there exist two independent edges of $G$ between $\{x_{1},x_{s}\}$ and $\{y_{1},y_{t}\}$. Say w.l.o.g. $\{x_{1}y_{t},x_{s}y_{1}\}\subseteq E$. If $y_{t}y_{2}\in A^{\prime}$ then $D[V(G_{2})\cup\{x_{1}\}-\{y_{1}\}]$ is hamiltonian by Corollary 3.5 and so $y_{1}x_{2}\not\in A^{\prime}$. That is, either $y_{t}y_{2}\not\in A^{\prime}$ or $y_{1}x_{2}\not\in A^{\prime}$. By (22), we obtain $d(y_{1}y_{t},G_{1})\geq s+2-\lambda_{n}$. With (13) and (20), we obtain

[TABLE]

Thus $d(x,G_{2})+d(y,G_{1})-d(x,G_{1})-d(y,G_{2})\geq 3$ for some $(x,y)\in\{(x_{1},y_{1}),(x_{s},y_{t})\}$. This contradicts Lemma 3.2. Hence $d(x_{1}x_{s},y_{1}y_{t})\leq 2$. Thus by (13), $d(x_{1}x_{s},G_{2}-y_{1}-y_{t})\geq t+s+r-\lambda_{n}-2\geq(t-2)+2$. Next, assume $d(y_{1}y_{t},G_{1})\geq s+2$. Then $d(y_{1}y_{t},G_{1}-x_{1}-x_{s})\geq s+2-2=(s-2)+2$. By Lemma 2.1$(d)$, both $G_{1}-x_{1}-x_{s}+y_{1}+y_{t}$ and $G_{2}-y_{1}-y_{t}+x_{1}+x_{s}$ are traceable. By Lemma 3.3, we would have that $(s-1)+(t-1)\geq(t+s+r-\lambda_{n})+(s+2)-4$, which is impossible. Hence $d(y_{1}y_{t},G_{1})\leq s+1$. Together with (24), (25) and (26), we obtain $y_{1}y_{t}\in A^{\prime}$. Hence (27) holds. $\Box$

We now choose $P_{2}=y_{1}y_{2}\ldots y_{t}$ in $G_{2}$ such that

[TABLE]

Lemma 3.7

For each $v\in\{y_{1},y_{t}\}$, if $d(v,G_{2})\leq(t-1)/2$ then $d(v,G_{1})\leq(s-1)/2$.

Proof. On the contrary, say $d(y_{1},G_{2})\leq(t-1)/2$ and $d(y_{1},G_{1})\geq s/2$. By Corollary 3.5, $G_{2}-y_{1}+x_{1}$ is traceable. First, assume that $G_{1}-x_{1}+y_{1}$ is tracable. Then by Lemma 3.2, $(s-2)+(t-1)/2\geq(t+s+r-\lambda_{n})/2+\lceil s/2\rceil-2\mu(x_{1}y_{1})$. This implies that $r=0$, $\lambda_{n}=1$, $d(x_{1},G_{1})=s-2$, $d(y_{1},G_{2})=(t-1)/2$, $d(y_{1},G_{1})=s/2$ and $x_{1}y_{1}\in E$. Thus $s$ is even, $s\geq 4$ and $N_{G}(x_{1},G_{1})=\{x_{2},\ldots,x_{s-1}\}$. It follows that $d(y_{t},G_{2})\leq(t-1)/2$ by (20), and by (24), $d(y_{t},G_{1})\geq s/2$. Clearly, $d(x_{s},G_{1})=1$ as $d(x_{1}x_{s},G_{1})\leq s-1$, and $G_{1}-x_{s}+y_{t}$ is traceable. As $x_{1}x_{s}\not\in A^{\prime}$ and by (1), $d(x_{s},G)\geq(t+s+1)/2$ and so $d(x_{s},G_{2})\geq(t+s-1)/2$. Thus $G_{2}-y_{1}+x_{s}$ is traceable. Applying Lemma 3.2 with $x_{s}$ and $y_{1}$ in place of $x$ and $y$, we obtain $1+(t-1)/2\geq(t+s-1)/2+s/2-2\mu(x_{s}y_{1})$, which is impossible. Hence $G_{1}-x_{1}+y_{1}$ is not traceable. Therefore $d(y_{1},x_{2}x_{s})=0$ and $d(y_{1},P_{1}-\{x_{1},x_{2},x_{s}\})\leq\lfloor(s-2)/2\rfloor$ by Lemma 2.1$(a)$. As $d(y_{1},G_{1})\geq s/2$, it follows that $d(y_{1},G_{1})=s/2$ with $N_{G}(y_{1},G_{1})=\{x_{1},x_{3},x_{5},\ldots,x_{s-1}\}$. Then $N_{G}(x_{2},G_{1}-x_{1})\subseteq\{x_{3},x_{5},\ldots,x_{s-1}\}$ and $N_{G}(x_{s},G_{1})\subseteq\{x_{3},x_{5},\ldots,x_{s-1}\}$ for otherwise $G_{1}-x_{1}+y_{1}$ is traceable. Thus $d(x_{2},G_{1})\leq s/2$ and $d(x_{s},G_{2})\leq(s-2)/2$. We claim

[TABLE]

To see (29), we have that $d(x_{2},G_{1})\leq s/2$, $d(y_{1},x_{2}x_{s})=0$ and $G_{1}-x_{i}+y_{1}$ is traceable for each $i\in\{2,s\}$. If $d(x_{s},G_{2})\geq(t-1)/2$, then $G_{2}-y_{1}+x_{s}$ is traceable by Lemma 2.1$(b)$ and so $(s-2)/2+(t-1)/2\geq(t-1)/2+s/2$ by Lemma 3.2, which is impossible. Hence $d(x_{s},G_{2})\leq(t-2)/2$. If $x_{2}\in W$ and $d(x_{2},G_{2})\geq t/2$, then $G_{2}-y_{1}+x_{2}$ is traceable by Lemma 2.1$(b)$ and so $s/2+(t-1)/2\geq t/2+s/2$ by Lemma 3.2, which is impossible. Therefore (29) holds.

Then $d(x_{s},R)\geq(n-\lambda_{n})/2-(s-2)/2-(t-2)/2=(r+4-\lambda_{n})/2$. We also have that $d(y_{1},R)\geq(n-\lambda_{n})/2-s/2-(t-1)/2=(r+1-\lambda_{n})/2$. We obtain that $d(x_{s}y_{1},R)\geq r+5/2-\lambda_{n}$ and so $i(x_{s}y_{1},R)\geq 2$. Note that $r\geq 2$. If $x_{2}\in W$ and $a(x_{2},N_{G}(y_{1},R))=0$ then $d(x_{2},G)\geq(n+4-\lambda_{n})/2$ by (1) and so $i(x_{2}y_{1},R)\geq 2$, a contradiction. Therefore if $x_{2}\in W$ then $a(x_{2},N_{G}(y_{1},R))>0$. It follows that $D_{1}-x_{1}+y_{1}+R$ contains a dicycle of $W$-length $n_{1}$. Thus $D[V(G_{2}-y_{1}+x_{1})]$ does not contain a hamiltonian dicycle. By Corollary 3.5, $D[V(G_{2}-y_{1})]$ is not hamiltonian. In particular, $y_{2}y_{t}\not\in A^{\prime}$. By (24), $d(y_{t},G_{1})\geq s/2$ as $d(y_{1},G_{1})=s/2$. If $d(y_{t},G_{2})\leq(t-1)/2$, then with $y_{t}$ in place of $y_{1}$ in the above argument, we see that $d(y_{t},G_{1})=s/2$ and so $y_{t}y_{2}\in A^{\prime}$ by (25), a contradiction. Therefore $d(y_{t},G_{2})\geq t/2$. With (28), we apply Lemma 2.4 to $y_{2}y_{3}\ldots y_{t}$ in $G_{2}-y_{1}$, we see that there exists $a\in\{2,\ldots,t-2\}$ such that for each $i\in\{a+1,a+2,\ldots,t\}$, there exists a hamiltonian path of $G_{2}$ from $y_{1}$ to $y_{i}$ containing $y_{1}y_{2}\ldots y_{a}$, $d(y_{i},G_{2})\geq t/2$ and $N(y_{i},G_{2})\subseteq\{y_{a},y_{a+1},\ldots,y_{t}\}$. Clearly $G[\{y_{a},y_{a+1},\ldots,y_{t}\}]$ is hamiltonian. Thus $a\not=2$ and so $a\geq 3$. By Lemma 3.6, $y_{1}y_{i}\not\in E$ for all $i\in\{a+1,a+2,\ldots,t\}$. As $y_{t}y_{2}\not\in A^{\prime}$ and by (20), (21), (22) and (27), we obtain that $d(y_{1}y_{t},G_{1})=s+1$ and $d(y_{1}y_{t},G_{2})=t-1$. It follows that $N_{G}(y_{1},G_{2})=\{y_{2},y_{3},\ldots,y_{a}\}$. Thus $G_{2}$ has a hamiltonian path from $y_{2}$ to $y_{t}$. By (27) with $y_{2}$ in place of $y_{1}$, we obtain that $y_{2}y_{t}\in A^{\prime}$, a contradiction.

We are now in a position to complete our proof. By (24) , we may assume w.l.o.g. that $d(y_{1},G_{1})\geq s/2$. By Lemma 3.6 and Lemma 3.7, $d(y_{1},G_{2}-y_{t})\geq t/2$. With (28), applying Lemma 2.4 to $y_{t-1}y_{t-2}\ldots y_{1}$, there exists $p\geq\lceil t/2\rceil$ such that for each $i\in\{1,\ldots,p\}$, $N_{G}(y_{i},G_{2})\subseteq\{y_{1},\ldots,y_{p+1}\}$, $G_{2}$ has a hamiltonian path from $y_{t}$ to $y_{i}$ containing $y_{t}y_{t-1}\ldots y_{p+1}$ and $d(y_{i},G_{2})\geq d(y_{1},G_{1})$. By Lemma 3.6, $N_{G}(y_{t},G_{2})\subseteq\{y_{p+1},y_{p+1},\ldots,y_{t-1}\}$ and so $d(y_{t},G_{2})\leq t-1-p\leq\lfloor(t-2)/2\rfloor$. By Lemma 3.7, $d(y_{t},G_{1})\leq\lfloor(s-1)/2\rfloor$. Thus $d(y_{1},G_{1})\geq d(y_{1}y_{t},G_{1})-d(y_{t},G_{1})\geq\lceil(s+1)/2\rceil$ by (24). By (27) with $y_{i}$ in place of $y_{1}$, $y_{t}y_{i}\in A^{\prime}$ for all $i\in\{1,2,\ldots,p\}$. Thus for all $j\in\{1,2,\ldots,p\}$, $D[V(G_{2})-y_{j}]$ is hamiltonian and so $D[V(G_{2})-y_{j}+x_{1}]$ is hamiltonian by Corollary 3.5. We conclude that $D[V(G_{1}-x_{1}+y_{i}+R)]$ does not contain a dicycle of $W$-length $n_{1}$ for all $i\in\{1,2,\ldots,p\}$. This implies that $a(y_{i},x_{2}x_{s})\leq 2$ for all $i\in\{1,2,\ldots,p\}$ and so $a(x_{2}x_{s},y_{1}y_{2}\ldots y_{p}\})\leq 2p$.

Assume that $x_{2}\not\in W$. By Corollary 3.5, $G_{2}-y_{1}+x_{1}$ is traceable. By (7), $y_{1}x_{3}\not\in E$ and $y_{1}x_{s}\not\in E$. By (7) and Lemma 2.1$(a)$, $d(y_{1},P_{1}-\{x_{1},x_{2},x_{3},x_{s}\})\leq(s-3)/2$. As $d(y_{1},G_{1})\geq(s+1)/2$, it follows that $d(y_{1},x_{1}x_{2})=2$ and $d(y_{1},P_{1}-\{x_{1},x_{2},x_{3},x_{s}\})=(s-3)/2$. Thus $y_{1}x_{s}\not\in A^{\prime}$ for otherwise $D[V(G_{1}-x_{1}+y_{1})]$ is hamiltonian. By (20), (21), (22) and (27), we obtain that $d(y_{1}y_{t},G_{1})=s+1$ and so $d(y_{t},G_{1})=(s+1)/2$, a contradiction since we have $d(y_{t},G_{1})\leq\lfloor(s-1)/2\rfloor$ in the above paragraph.

Therefore $x_{2}\in W$. We claim that $a(x_{2}x_{s},R)\leq 2r$ and $x_{2}x_{s}\not\in A^{\prime}$. If this is not true, then either $e(z,x_{2}x_{s})\geq 3$ for some $z\in R$ or $x_{2}x_{s}\in A^{\prime}$. Since $D[V(G_{1}-x_{1}+R+y_{1})]$ does not contain a dicycle of $W$-length $n_{2}$, we see that $d(y_{1},x_{2}x_{s})\leq 1$ and $d(y_{1},x_{i}x_{i+1})\leq 1$ for all $i\in\{2,\ldots,s-1\}$. By Lemma 2.1$(a)$, this yields that $d(y_{1},P_{1}-x_{1})\leq(s-1)/2$. It follows that $d(y_{1},G_{1})=(s+1)/2$, $y_{1}x_{1}\in E$ and $d(y_{1},x_{2}x_{s})=1$. Thus either $y_{1}x_{2}\not\in A^{\prime}$ or $y_{1}x_{s}\not\in A^{\prime}$. By (20), (21), (22) and (27), it follows that $d(y_{1}y_{t},G_{1})=s+1$ and so $d(y_{t},G_{1})=(s+1)/2$, a contradiction. Therefore $a(x_{2}x_{s},R)\leq 2r$ and $x_{2}x_{s}\not\in A^{\prime}$. Thus $s\geq 4$.

Next, we claim that $a(x_{2}x_{s},P_{1}-x_{1})\leq 2(s-2)$. If this is not true, then by Lemma 2.3, $D[V(G_{1}-x_{1})]$ has a hamiltonian dicycle $C$. By Lemma 2.2, $d(x_{1},C)\leq(s-1)/2$ and $d(y_{1},C)\leq(s-1)/2$. It follows that $d(y_{1},C)=(s-1)/2$ and $y_{1}x_{1}\in E$. Since $d(y_{1},G_{1})=(s+1)/2$, $G_{1}-x_{1}+y_{1}$ is traceable by Lemma 2.1$(b)$. By Corollary 3.5, $G_{2}-y_{1}+x_{1}$ is traceable. By Lemma 3.2, $(s-1)/2+d(y_{1},G_{2})\geq d(x_{1},G_{2})+(s+1)/2-2$. Thus $p\geq d(y_{1},G_{2})\geq d(x_{1},G_{2})-1\geq(t+s+r-\lambda_{n})/2-1$. Then $d(y_{t},G_{2})\leq t-1-p\leq(t-s-r+\lambda_{n})/2$. By Lemma 3.7, $d(y_{t},G_{1})\leq(s-1)/2$. Thus

[TABLE]

This contradicts (2).

Therefore $a(x_{2}x_{s},P_{1}-x_{1})\leq 2(s-2)$. Clearly, $a(x_{2}x_{s},x_{1})=2$ as $x_{1}x_{s}\not\in A^{\prime}$. Let $X=R\cup V(G_{1})\cup\{y_{1},\ldots,y_{p}\}$. We obtain that $a(x_{2}x_{t},X)\leq 2r+2(s-2)+2+2p=2(|X|-1)$. By (6), $n-|X|=t-p\geq(t+s+r-\lambda_{n})/2$. Thus $t\geq 2p+s+r-\lambda_{n}\geq t+s+r-\lambda_{n}>t$, a contradiction. This proves the theorem.