An Ore-type condition for existence of two disjoint cycles

TL;DR

This paper improves a degree sum condition for guaranteeing two disjoint cycles of specified lengths in a graph, extending previous results and confirming a conjecture about the minimal degree sum needed.

Contribution

It establishes that a degree sum condition of at least n+2 suffices for the existence of two disjoint cycles of given lengths, generalizing prior theorems.

Findings

Improves the degree sum condition to d(u)+d(v) ≥ n+2

Confirms the conjecture posed by Yan et al.

Generalizes El-Zahar's result for odd n1 and n2.

Abstract

Let $n_{1}$ and $n_{2}$ be two integers with $n_{1},n_{2}\geq3$ and $G$ a graph of order $n=n_{1}+n_{2}$. As a generalization of Ore's degree condition for the existence of Hamilton cycle in $G$, El-Zahar proved that if $\delta(G)\geq \left\lceil\frac{n_{1}}{2}\right\rceil+\left\lceil\frac{n_{2}}{2}\right\rceil$ then $G$ contains two disjoint cycles of length $n_{1}$ and $n_{2}$. Recently, Yan et. al considered the problem by extending the degree condition to degree sum condition and proved that if $d(u)+d(v)\geq n+4$ for any pair of non-adjacent vertices $u$ and $v$ of $G$, then $G$ contains two disjoint cycles of length $n_{1}$ and $n_{2}$. They further asked whether the degree sum condition can be improved to $d(u)+d(v)\geq n+2$. In this paper, we give a positive answer to this question. Our result also generalizes El-Zahar's result when $n_{1}$ and $n_{2}$ are both odd.

Equations

$${\sigma }_2(G)=\min \{d(x)+d(y):x,y\in V(G),xy\notin E(G)\}.$$

$$e(\{s,t\},W_2)\ge n_2+3$$

$$d_{G[V_2]}(x)+d_{G[V_2]}(y)\le n_2+2\mathrm{and}e(\{x,y\},V_1)\ge n_1.$$

$$e(W_1^{\prime })+e(W_2^{\prime })$$

$$d_G(s_1)+d_G(v_3)\le \left(\frac{1}{2}(n_1-1)+\frac{1}{2}(n_2+3)\right)+\left(\frac{1}{2}(n_1-3)+\frac{1}{2}(n_2+1)\right)<n+2,$$

Full text

An Ore-type condition for existence of two disjoint cycles

Maoqun Wang and Jianguo Qian***Corresponding author. E-mail: jgqian@xmu.edu (J.G. Qian)

School of Mathematical Sciences, Xiamen University, Xiamen 361005, PR China

Abstract. Let $n_{1}$ and $n_{2}$ be two integers with $n_{1},n_{2}\geq 3$ and $G$ a graph of order $n=n_{1}+n_{2}$. As a generalization of Ore’s degree condition for the existence of Hamilton cycle in $G$, El-Zahar proved that if $\delta(G)\geq\left\lceil\frac{n_{1}}{2}\right\rceil+\left\lceil\frac{n_{2}}{2}\right\rceil$ then $G$ contains two disjoint cycles of length $n_{1}$ and $n_{2}$. Recently, Yan et. al considered the problem by extending the degree condition to degree sum condition and proved that if $d(u)+d(v)\geq n+4$ for any pair of non-adjacent vertices $u$ and $v$ of $G$, then $G$ contains two disjoint cycles of length $n_{1}$ and $n_{2}$. They further asked whether the degree sum condition can be improved to $d(u)+d(v)\geq n+2$. In this paper, we give a positive answer to this question. Our result also generalizes El-Zahar’s result when $n_{1}$ and $n_{2}$ are both odd. Keywords: degree sum condition; disjoint cycles

1 Introduction

We consider only finite simple graphs. For terminologies and notations not defined here we refer to [3]. Let $G$ be a graph with $n$ vertices. We denote by $V(G)$ and $E(G)$ the vertex set and edge set of $G$, respectively. For two vertices $u$ and $v$ in $G$, we denote by $uv$ the edge joining $u$ and $v$. A neighbour of a vertex $v$ is a vertex adjacent to $v$. The set of all neighbours of $v$ is denoted by $N_{G}(v)$, or simply by $N(v)$, and the degree of $v$ is defined to be $d_{G}(v)=|N_{G}(v)|$. We use $\delta(G)$ to denote the minimum degree of vertices in $G$. A path (resp., cycle) is called a Hamilton path (resp., Hamilton cycle) if it has length $n-1$ (resp., $n$). If $G$ has a Hamilton cycle, then $G$ is called Hamiltonian. In general, if $G$ contains a cycle of length $k$ for every $k$ with $3\leq k\leq n$, then $G$ is called pancyclic. If, for any two distinct vertices $u$ and $v$, $G$ contains a Hamilton path with $u$ and $v$ as the two endvertices, then $G$ is called Hamilton-connected. Define

[TABLE]

The degree condition for the existence of cycle(s) with specified length(s) is one of the most elementary concerns in graph theory. A classic result should be the one given by Dirac in 1952, which says that every graph of order $n$ with minimum degree at least $n$ has a Hamilton cycle. Since then, this result has been generalized to various forms in terms of degree condition or degree sum condition. We recall some typical results on this subject.

Theorem 1.1**.**

(Dirac, [5]) For a graph $G$ of order $n$, if $n\geq 3$ and $\delta(G)\geq\frac{n}{2}$ then $G$ is Hamiltonian.

Theorem 1.2**.**

(Ore, [12]) For a graph $G$ of order $n$, if $n\geq 3$ and $\sigma_{2}(G)\geq n$ then $G$ is Hamiltonian.

Theorem 1.3**.**

(Bondy, [4]) For a graph $G$ of order $n$, if $n\geq 3$ and $\sigma_{2}(G)\geq n$ then $G$ is either pancyclic or else is the complete bipartite graph $K_{\frac{n}{2},\frac{n}{2}}$.

Theorem 1.4**.**

(El-Zahar, [6]) Let $n_{1},n_{2}$ be two integers with $n_{1},n_{2}\geq 3$ and $G$ a graph of order $n$ with $n=n_{1}+n_{2}$. If $\delta(G)\geq\left\lceil\frac{n_{1}}{2}\right\rceil+\left\lceil\frac{n_{2}}{2}\right\rceil$ then $G$ has two disjoint cycles of length $n_{1}$ and $n_{2}$, where, for a real number $r$, $\lceil r\rceil$ is the least integer not less than $r$.

The degree condition in Theorem 1.4 is sharp since the complete bipartite graph $K_{n/2,n/2}$ does not have any odd cycle. In general, El-Zahar posed the following conjecture:

Conjecture 1.5**.**

(El-Zahar, [6]) Let $G$ be a graph of order $n=n_{1}+n_{2}+\cdots+n_{k}$ with $n_{i}\geq 3$ for each $i\in\{1,2,\cdots,k\}$. If \delta(G)\geq\left\lceil\frac{n_{1}}{2}\right\rceil+\left\lceil\frac{n_{2}}{2}\right\rceil$$+\cdots+\left\lceil\frac{n_{k}}{2}\right\rceil then $G$ has $k$ disjoint cycles of length $n_{1}$, $n_{2},\cdots,n_{k}$.

In [1], Abbasi confirmed the conjecture for sufficiently large graphs by using the regularity lemma. For the special case when $n_{1}=n_{2}=\cdots=n_{k}=4$, the conjecture was posed earlier by Erdős [7] and was proved later by Wang [15]. In general, the conjecture still remains open. Recently, instead of degree condition, Yan et. al [17] considered the problem from the view point of degree sum condition and proved the following result:

Theorem 1.6**.**

(Yan. et.al,[17]) Let $G$ be a graph on $n$ vertices. For any two integers $n_{1}$ and $n_{2}$ with $n_{1},n_{2}\geq 3$ and $n=n_{1}+n_{2}$, if $\sigma_{2}(G)\geq n+4$, then $G$ has two disjoint cycles of length $n_{1}$ and $n_{2}$.

As an extension of two disjoint cycles with specified lengths, Kostochka and Yu [10] showed that if $\sigma_{2}(G)\geq\frac{4}{3}n-1$ (not sharp in general) then the graph $G$ contains every 2-factor. On the other hand, the sharpness of Theorem 1.4 implies that the degree sum $\sigma_{2}(G)$ is at least $n+2$ for Theorem 1.6. Even so, Yan et. al [17] pointed that it might not be easy to get a better degree sum condition than the one in Theorem 1.6 and therefore, posed the following question.

Question (Yan. et.al, [17]). Let $G$ be a graph of order $n=n_{1}+n_{2}$ with $n_{1},n_{2}\geq 3$. Can $\sigma_{2}(G)\geq n+2$ guarantee that $G$ has two disjoint cycles of length $n_{1}$ and $n_{2}$?

In this paper, we give a positive answer to this question.

Theorem 1.7**.**

Let $n_{1},n_{2}$ be two integers with $n_{1},n_{2}\geq 3$ and $G$ a graph of order $n$ with $n=n_{1}+n_{2}$. If $\sigma_{2}(G)\geq n+2$ then $G$ has two disjoint cycles of length $n_{1}$ and $n_{2}$.

2 Lemmas

In order to prove the main theorem, in this section we introduce some necessary lemmas.

Let $W$ and $W^{\prime}$ be two disjoint subsets of $V(G)$. We denote by $e(W,W^{\prime})$ (resp., $e(W)$) the number of the edges in $G$ that lie between $W$ and $W^{\prime}$ (resp., lie in $G[W]$), where $G[W]$ denotes the subgraph of $G$ induced by $W$. The graph obtained from $G$ by removing the vertices in $W$ is denoted by $G-W$. In particular, if $W$ consists of a single vertex $w$, then we simply write $e(\{w\},W^{\prime})$ and $G-\{w\}$ as $e(w,W^{\prime})$ and $G-w$, respectively.

For a cycle $C$ in $G$, we always give a direction on $C$ and use $C^{-}$ to denote the cycle $C$ with the opposite direction. For a vertex $v\in V(C)$, we use $v^{i-}$ and $v^{i+}$ to represent the $i$-th predecessor and $i$-th successor of $v$ along the direction of $C$, respectively. For simplicity, we write $v^{-}$ and $v^{+}$ instead of $v^{1-}$ and $v^{1+}$, respectively. For two vertices $u,v\in V(C)$, we denote by $C[u,v]$ the section of $C$ from $u$ to $v$ along the direction of $C$.

Lemma 2.1**.**

[13]** Let $u$ and $v$ be the two endvertices of a Hamilton path in a graph $G$ of order $n$. If $d(u)+d(v)\geq n$, then $G$ is Hamiltonian.

Lemma 2.2**.**

[13]** Let $C$ be a Hamilton cycle of a graph $G$ of order $n$ with a given direction and let $u$ and $v$ be two vertices on $C$. If $d_{G}(u^{+})+d_{G}(v^{+})\geq n+1$ or $d_{G}(u^{-})+d_{G}(v^{-})\geq n+1$, then $G$ contains a Hamilton path with endvertices $u$ and $v$.

Lemma 2.3**.**

[16]** If $P$ is a path of length $k$ in a graph $G$ and $u,v$ are two vertices in $G-V(P)$ such that $e(\{u,v\},V(P))\geq k+2$, then $G[V(P)\cup\{u,v\}]$ has a Hamilton path.

Lemma 2.4**.**

[8]** If $G$ is a graph on $n$ vertices and $\sigma_{2}(G)\geq n+1$, then $G$ is Hamilton-connected.

Lemma 2.5**.**

[17]** Let $n_{1}$ and $n_{2}$ be two integers with $n_{1},n_{2}\geq 5$ and let $G$ be a graph on $n=n_{1}+n_{2}$ vertices with $\sigma_{2}(G)\geq n+2$. Suppose that $(G_{1},G_{2})$ is a partition of $G$ such that $V(G)=V(G_{1})\cup V(G_{2})$ and $|G_{i}|=n_{i}$ for i=1,2, and $G_{2}$ contains a Hamilton cycle. For two nonadjacent vertices $u,v\in V(G_{2})$, if $G_{2}-\{u,v\}$ does not contain a Hamilton path, then there exist two vertices $u^{\prime},v^{\prime}\in V(G_{2})$ such that $G_{2}-\{u^{\prime},v^{\prime}\}$ contains a Hamilton path and the following holds: $d_{G_{2}}(u^{\prime})+d_{G_{2}}(v^{\prime})\leq n_{1}$, $e(\{u^{\prime},v^{\prime}\},V(G_{1}))\geq n_{2}+2$.

The following lemma was obtained independently by Ainouche and Christofides [2], Jung [9], Nara [11], and Schmeichel and Hayes [14].

Lemma 2.6**.**

[2, 9, 11, 14]** Let $G$ be a graph of order $n\geq 3$. If $\sigma_{2}(G)\geq n-1$, then one of the following holds:

(i)

$G$* contains a Hamilton cycle.*

(ii)

$K_{m,m+1}$* $\subseteq G\subseteq K_{m}+(m+1)K_{1}$, where $m=\frac{n-1}{2}$ and $n$ is odd with $n\geq 5$.*

(iii)

$G\cong K_{1}+(K_{p}\cup K_{q})$* for some positive integers $p$ and $q$ with $p+q=n-1$.*

Lemma 2.7**.**

Let $n_{1}$ and $n_{2}$ be two integers with $n_{1},n_{2}\geq 5$ and $G$ a graph on $n=n_{1}+n_{2}$ vertices. If $\sigma_{2}(G)\geq n+2$ and $G$ has no pair of two disjoint cycles of length $n_{1}$ and $n_{2}$, then $V(G)$ has a partition $V(G)=V_{1}\cup V_{2}$ such that $|V_{1}|=n_{1}-2$ and $G[V_{1}]$ has a Hamilton path; $|V_{2}|=n_{2}+2$, $\sigma_{2}(G[V_{2}])\geq n_{2}+3$ and moreover, one of the following holds:

(i)

For any two vertices $x,y\in V_{2}$, $G[V_{2}]-\{x,y\}$ has a Hamilton cycle;

(ii)

$K_{m+2,m+1}$* $\subseteq G[V_{2}]\subseteq K_{m+2}+(m+1)K_{1}$, where $n_{2}$ is odd and $m=\frac{1}{2}(n_{2}-1)$;*

(iii)

$3K_{1}+(K_{p}\cup K_{q})\subseteq G[V_{2}]\subseteq K_{3}+(K_{p}\cup K_{q})$, where $p$ and $q$ are two positive integers with $p+q=n_{2}-1$.

Proof.

Let $W_{1}\cup W_{2}$ be a partition of $V(G)$ with $|W_{1}|=n_{1},|W_{2}|=n_{2}$ such that $G[W_{i}]$ contains a Hamilton path for each $i\in\{1,2\}$ and $e(W_{1})+e(W_{2})$ is maximum among all such partitions. Since $\sigma_{2}(G)\geq n+2$, Theorem 1.2 implies that $G$ is Hamiltonian and therefore, the partition above exists. Since $G$ has no pair of two disjoint cycles of length $n_{1}$ and $n_{2}$, one of $G[W_{1}]$ and $G[W_{2}]$, say $G[W_{1}]$, is not Hamiltonian. Let $s$ and $t$ be the two endvertices of a Hamilton path in $G[W_{1}]$. Since $G[W_{1}]$ is not Hamiltonian, $s$ and $t$ are not adjacent and, moreover, by Lemma 2.1 we have $d_{G[W_{1}]}(s)+d_{G[W_{1}]}(t)\leq n_{1}-1$ and, therefore,

[TABLE]

because $\sigma_{2}(G)\geq n+2$. Let $V_{1}=W_{1}\setminus\{s,t\}$ and $V_{2}=W_{2}\cup\{s,t\}$. It is clear that $G[V_{1}]$ contains a Hamilton path since $s$ and $t$ are the two endvertices of a Hamilton path in $G[W_{1}]$. Further, by Lemma 2.3 and (1), $G[V_{2}]$ has a Hamilton path.

Claim 1. If $\sigma_{2}(G[V_{2}])\leq n_{2}+2$ then $V_{2}$ has two vertices $x$ and $y$ such that $G[V_{2}]-\{x,y\}$ contains a Hamilton path and the following two inequalities hold:

[TABLE]

Proof.

Assume first that $G[V_{2}]$ is not Hamiltonian. Choose $x$ and $y$ to be the two endvertices of a Hamilton path in $G[V_{2}]$. Since $G[V_{2}]$ is not Hamiltonian, we have $xy\notin E(G)$ and, by Lemma 2.1, $d_{G[V_{2}]}(x)+d_{G[V_{2}]}(y)\leq n_{2}+1$. Thus, $e(\{x,y\},V_{1})\geq n_{1}+1$, as desired. Now assume that $G[V_{2}]$ is Hamiltonian. Let $x^{\prime}$ and $y^{\prime}$ be arbitrary two nonadjacent vertices in $G[V_{2}]$. Since $\sigma_{2}(G[V_{2}])\leq n_{2}+2$, we may choose $x^{\prime}$ and $y^{\prime}$ to be such that satisfy (2). If $G[V_{2}]-\{x^{\prime},y^{\prime}\}$ contains a Hamilton path then we are done by letting $x=x^{\prime}$ and $y=y^{\prime}$. Otherwise, in Lemma 2.5 we replace $G_{1},G_{2}$ and $u,v$ with $G[V_{1}],G[V_{2}]$ and $x^{\prime},y^{\prime}$, respectively. Then $u^{\prime},v^{\prime}$ clearly satisfy our requirement for $x,y$, which completes the proof of the claim. ∎

Let $x$ and $y$ be defined as in Claim 1 and let $W^{\prime}_{1}=V_{1}\cup\{x,y\},W^{\prime}_{2}=V_{2}\setminus\{x,y\}$. Then $G[W^{\prime}_{2}]$ contains a Hamilton path. By Lemma 2.3 and (2), $G[W^{\prime}_{1}]$ has a Hamilton path. Moreover, recall that $V_{1}=W_{1}\setminus\{s,t\},V_{2}=W_{2}\cup\{s,t\}$ and therefore, $|W_{1}|=|W^{\prime}_{1}|,|W_{2}|=|W^{\prime}_{2}|$. Further, by (1) and Claim 1, if $\sigma_{2}(G[V_{2}])\leq n_{2}+2$ then

[TABLE]

This contradicts the maximality of $e(W_{1})+e(W_{2})$. Thus, we have $\sigma_{2}(G[V_{2}])\geq n_{2}+3$.

Suppose now that (i) does not hold, that is, there are two vertices $x,y\in V_{2}$ such that $G[V_{2}]-\{x,y\}$ is not Hamiltonian. Consider the graph $H=G[V_{2}]-\{x,y\}$. Recalling that $\sigma_{2}(G[V_{2}])\geq n_{2}+3$, we have $\sigma_{2}(H)\geq n_{2}-1=|V(H)|-1$. So by Lemma 2.6 we have $K_{m,m+1}\subseteq H\subseteq K_{m}+(m+1)K_{1}$, where $n_{2}$ is odd and $m=\frac{1}{2}(n_{2}-1)$; or $H\cong K_{1}+(K_{p}\cup K_{q})$ for some positive integers $p$ and $q$ with $p+q=n_{2}-1$. For the former case, we notice that $H\subseteq K_{m}+(m+1)K_{1}$ means that $H$ contains at least $m+1$ pairwise nonadjacent vertices, which is of course the case for $G[V_{2}]$. Further, the condition $\sigma_{2}(G[V_{2}])\geq n_{2}+3$ implies that each of these $m+1$ vertices is adjacent to both $x$ and $y$. Thus, $K_{m+2,m+1}\subseteq G[V_{2}]\subseteq K_{m+2}+(m+1)K_{1}$. If the latter holds, it is clear that $3K_{1}+(K_{p}\cup K_{q})\subseteq G[V_{2}]\subseteq K_{3}+(K_{p}\cup K_{q})$ for some positive integers $p$ and $q$ with $p+q=n_{2}-1$. Thus, (ii) or (iii) holds, which completes our proof. ∎

3 Proof of Theorem 1.7

The main idea in our proof comes from [6] and [17]. In the following, when we say that a graph $G$ has a pair of disjoint $(n_{1},n_{2})$-cycles we always mean that $G$ has two disjoint cycles of length $n_{1}$ and $n_{2}$. Further, for more clarity we always write a cycle $C=v_{1}v_{2}\cdots v_{n}$ as the form $v_{1}v_{2}\cdots v_{n}v_{1}$, instead of its standard form.

Firstly, we prove the following proposition.

Proposition 1. If $\sigma_{2}(G)\geq n+2$, then $G$ has a pair of disjoint $(3,n-3)$-cycles and a pair of disjoint $(4,n-4)$-cycles.

Proof.

Since $\sigma_{2}(G)\geq n+2$, by Theorem 1.3, $G$ has a 3-cycle, denoted by $C=uvwu$. Let $G^{\prime}=G-\{u,v,w\}$. Obviously, $\sigma_{2}(G^{\prime})\geq(n+2)-6=|V(G^{\prime})|-1$. By Lemma 2.6, we distinguish among three cases.

Case 1. $G^{\prime}$ contains a Hamilton cycle.

In this case, we need only to find a pair of disjoint $(4,n-4)$-cycles in $G$ for $n\geq 8$. Let $C^{\prime}$ be a Hamilton cycle in $G^{\prime}$ (with a given direction).

Subcase 1.1. $G^{\prime}$ has two nonadjacent vertices $x$ and $y$ with $d_{G^{\prime}}(x)+d_{G^{\prime}}(y)\leq n-2$.

In this case, $e(\{x,y\},V(C))\geq 4$. This implies that $x$ or $y$ is adjacent to at least two vertices on $C$, say $uy,vy\in E(G)$. It is clear that $u,v,w,y$ form a 4-cycle. If $y^{+}y^{-}\in E(G)$, then $G^{\prime}-y$ has an $(n-4)$-cycle. Now assume that $y^{+}y^{-}\notin E(G)$. If $d_{G^{\prime}}(y^{+})+d_{G^{\prime}}(y^{-})\geq n-2$ then by Lemma 2.1, $G^{\prime}-y$ is Hamiltonian. Now we consider the case that $d_{G^{\prime}}(y^{+})+d_{G^{\prime}}(y^{-})\leq n-3$. In this case, again by $\sigma_{2}(G)\geq n+2$ we have $e(\{y^{-},y^{+}\},V(C))\geq 5$. This means that at most one of $uy^{-},vy^{-},wy^{-},uy^{+},vy^{+},wy^{+}$ is not an edge in $E(G)$, say $uy^{-}\notin E(G)$. Replacing the roles $y^{-},y^{+}$ by $y,y^{2+}$ and repeating the discussion above, we can see that at most one of $uy,vy,wy,uy^{2+},vy^{2+},wy^{2+}$ is not an edge in $E(G)$. Thus, $zC^{-}[y^{-},y^{2+}]z$ is an $(n-4)$-cycle and $(\{u,v,w\}\setminus\{z\})\cup\{y,y^{+}\}$ induces a 4-cycle, where $z$ is a common neighbour of $y^{-}$ and $y^{2+}$ on $C$.

Subcase 1.2. $d_{G^{\prime}}(x)+d_{G^{\prime}}(y)\geq n-1$ for any two nonadjacent vertices $x,y\in V(G^{\prime})$.

We first assume that $G^{\prime}$ contains a vertex $z$ that is adjacent to at least two vertices on $C$, say $uz,vz\in E(G)$. If $z^{-}$ and $z^{+}$ are adjacent then we are done. If $z^{-}$ and $z^{+}$ are not adjacent, then by Lemma 2.1, $G^{\prime}-z$ is Hamiltonian since $d_{G^{\prime}}(z^{-})+d_{G^{\prime}}(z^{+})\geq n-1$, i.e., $d_{G^{\prime}-z}(z^{-})+d_{G^{\prime}-z}(z^{+})\geq(n-1)-2>|V(G^{\prime}-z)|$. Thus, we get a pair of disjoint $(4,n-4)$-cycles $uwvzu$ and the Hamilton cycle in $G^{\prime}-z$.

We now assume that every vertex in $V(G^{\prime})$ is adjacent to at most one vertex on $C$. Let $z$ be an arbitrary vertex in $G^{\prime}$ and, without loss of generality, assume that $uz,vz\notin E(G)$. If $d_{G^{\prime}}(z)\leq\frac{1}{2}(n+1)$, then $d_{G}(u)\geq\frac{1}{2}(n+1)$ and $d_{G}(v)\geq\frac{1}{2}(n+1)$ since $\sigma_{2}(G)\geq n+2$. This means that $u$ and $v$ have a common neighbour in $G^{\prime}$, which contradicts that $e(z,C)\leq 1$. Thus, $\delta(G^{\prime})\geq\frac{1}{2}(n+2)$. Let $ux$ be an arbitrary edge with $x\in V(G^{\prime})$. Since $e(x,C)\leq 1$, $x$ and $v$ are not adjacent and therefore, have a common neighbour $y$ in $G^{\prime}-x$. Notice that $u$ and $v$ are the only neighbours of $x$ and $y$ on $C$, respectively. Therefore, $w$ is neither adjacent to $x$ nor to $y$. So again by $\sigma_{2}(G)\geq n+2$, $w$ is adjacent to at least two vertices $x^{\prime}$ and $y^{\prime}$ in $G^{\prime}-\{x,y\}$. Further, since $\delta(G^{\prime})\geq\frac{1}{2}(n+2)$, so by Lemma 2.4, $V(G^{\prime})\setminus\{x,y\}$ is Hamilton-connected. Thus, we get a pair of disjoint $(4,n-4)$-cycles $vuxyv$ and $wPw$, where $P$ is a Hamilton path in $G^{\prime}-\{x,y\}$ with endvertices $x^{\prime}$ and $y^{\prime}$.

Case 2. $K_{m,m+1}$ $\subseteq G^{\prime}\subseteq K_{m}+(m+1)K_{1}$, where $m=\frac{1}{2}(n-4)$ and $n\geq 8$ is even.

In this case $G^{\prime}$ has a set of $m+1$ pairwise non-adjacent vertices. We denote this set by $S$ and denote $T=V(G^{\prime})\setminus S$. It is clear that $|S|=m+1=\frac{1}{2}(n-2)$ and $|T|=m=\frac{1}{2}(n-4)$. Since $\sigma_{2}(G)\geq n+2$, each vertex in $S$ must be adjacent to all of the vertices on $C$ and the vertices in $T$. Choose a vertex $s\in S$ and two vertices $t_{1},t_{2}\in T$. If $t_{1}$ and $t_{2}$ are not adjacent then, again by $\sigma_{2}(G)\geq n+2$, either $t_{1}$ or $t_{2}$ is adjacent to one of $u,v,w$, say $wt_{1}\in E(G)$. In either two cases that $t_{1}$ and $t_{2}$ are adjacent or not, one can see that $\{u,v,s\}$ forms a 3-cycle and $V(G)\setminus\{u,v,s\}$ has an $(n-3)$-cycle while $\{u,v,w,s\}$ forms a 4-cycle and $V(G)\setminus\{u,v,w,s\}$ has an $(n-4)$-cycle.

Case 3. $G^{\prime}\cong K_{1}+(K_{p}\cup K_{q})$ for two positive integers $p$ and $q$ with $p+q=n-4$.

Without loss of generality, we assume that $p\geq q$. If $n=6$, then $G^{\prime}$ is a path $P=xyz$ where $y\in V(K_{1})$. Since $\sigma_{2}(G)\geq n+2$, $y$ is adjacent to two vertices in $\{u,v,w\}$, say $vy,wy\in E(G)$. Thus, $G$ has a pair of disjoint $(3,3)$-cycles $vxyv$ and $uwzu$. We now consider the case that $n\geq 7$. Again by the assumption that $\sigma_{2}(G)\geq n+2$, each vertex in $K_{p}$ and $K_{q}$ is adjacent to all of the vertices $u,v,w$ because the vertices between $K_{p}$ and $K_{q}$ are not adjacent. In this case, it is not difficult to find a pair of disjoint $(3,n-3)$-cycles and a pair of disjoint $(4,n-4)$-cycles in $G$. The proposition follows. ∎

By Proposition 1, we need only to consider the case that $n_{1},n_{2}\geq 5$. Throughout the following, we assume that $V(G)$ has a partition $V(G)=V_{1}\cup V_{2}$ such that

1)

$|V_{1}|=n_{1}-2$ and $G[V_{1}]$ has a Hamilton path;

2)

$|V_{2}|=n_{2}+2$, $\sigma_{2}(G[V_{2}])\geq n_{2}+3$ and therefore, $G[V_{2}]$ has a Hamilton cycle, denoted by $C_{2}$.

Thus, to prove Theorem 1.7, it suffices to prove that if $G$ satisfies any one of (i),(ii),(iii) in Lemma 2.7 then $G$ has a pair of disjoint $(n_{1},n_{2})$-cycles. We will distinguish among three claims. The following two propositions are necessary for our further discussion.

Proposition 2. $G[V_{1}]$ is Hamiltonian or $G[V_{1}\cup\{x,y\}]$ is Hamiltonian for some two consecutive vertices $x$ and $y$ on $C_{2}$.

Proof.

Let $u$ and $v$ be the two endvertices of a Hamilton path in $G[V_{1}]$. If $u$ and $v$ are adjacent then clearly $G[V_{1}]$ is Hamiltonian. If $d_{G[V_{1}]}(u)+d_{G[V_{1}]}(v)\geq n_{1}-2$ then by Lemma 2.1, again $G[V_{1}]$ is Hamiltonian. Now assume that $u$ and $v$ are not adjacent and $d_{G[V_{1}]}(u)+d_{G[V_{1}]}(v)\leq n_{1}-3$. In this case we have $e(\{u,v\},V_{2})\geq n_{2}+5$, which implies that there are two consecutive vertices $x$ and $y$ on $C_{2}$ such that $e(\{u,v\},\{x,y\})$ $\geq 3$. Thus, $G[V_{1}\cup\{x,y\}]$ is Hamiltonian. ∎

Proposition 3. Let $u\in V_{1}$ and $x,y\in V_{2}$. If $G[V_{1}\cup\{x,y\}]$ is not Hamiltonian but has a Hamilton path with $u$ and $x$ as the two endvertices, then $u$ and $x$ have three common neighbours in $V_{2}\setminus\{x,y\}$.

Proof.

Let $\Gamma=G[V_{1}\cup\{x,y\}]$. If $ux\in E(G)$, or $ux\notin E(G)$ but $d_{\Gamma}(u)+d_{\Gamma}(x)\geq n_{1}$ then $\Gamma$ would be Hamiltonian because of Lemma 2.1. Now assume that $ux\notin E(G)$ and $d_{\Gamma}(u)+d_{\Gamma}(x)\leq n_{1}-1$. Then $e(\{u,x\},V_{2}\setminus\{x,y\})\geq n_{2}+3$. This means that $u$ and $x$ have at least three common neighbours in $V_{2}\setminus\{x,y\}$. ∎

Claim 1. If $G[V_{2}]-\{x,y\}$ is Hamiltonian for any two vertices $x,y\in V_{2}$ then $G$ has a pair of disjoint $(n_{1},n_{2})$-cycles.

Proof.

It suffices to prove that $G[V_{1}\cup\{x,y\}]$ is Hamiltonian for some $x,y\in V_{2}$. If $G[V_{1}\cup\{x,y\}]$ is Hamiltonian for two consecutive vertices $x$ and $y$ on $C_{2}$ then the claim follows directly. Now assume that $G[V_{1}\cup\{x,y\}]$ is not Hamiltonian for any two consecutive vertices $x$ and $y$ on $C_{2}$. By Proposition 2, let $C_{1}$ be a Hamilton cycle in $G[V_{1}]$.

Case 1. $\sigma_{2}(G[V_{1}])<n_{1}$.

Let $u$ and $v$ be two nonadjacent vertices in $V_{1}$ and $d_{G[V_{1}]}(u)+d_{G[V_{1}]}(v)<n_{1}$. Since $\sigma_{2}(G)\geq n+2$, we have $e(\{u,v\},V_{2})\geq n_{2}+3$. This implies that $C_{2}$ has two consecutive vertices $x$ and $y$ such that $e(\{u,v\},\{x,y\})\geq 3$ and therefore, $x$ or $y$ is adjacent to both $u$ and $v$, say $xu,xv\in E(G)$. Since $G[V_{1}\cup\{x,y\}]$ is not Hamiltonian, $\{u,v\}$ is not the pair of the two endvertices of any Hamilton path in $G[V_{1}]$. This implies that $u^{-}v^{-}\notin E(G)$ and, moreover, by Lemma 2.2, $d_{G[V_{1}]}(u^{-})+d_{G[V_{1}]}(v^{-})\leq n_{1}-2$. Thus, $e(\{u^{-},v^{-}\},V_{2})\geq n_{2}+4$ and therefore, $u^{-}$ and $v^{-}$ have a common neighbour $x^{\prime}$ in $V_{2}$ with $x^{\prime}\not=x$. So $C_{1}^{-}[u^{-},v]xC_{1}[u,v^{-}]x^{\prime}u^{-}$ is a Hamilton cycle in $G[V_{1}\cup\{x,x^{\prime}\}]$, again as desired.

Case 2. $\sigma_{2}(G[V_{1}])\geq n_{1}$.

Notice that $n_{1}=|V_{1}|+2$. So by Lemma 2.4, $G[V_{1}]$ is Hamilton-connected.

Let $ux\in E(G)$ with $u\in V(C_{1})$ and $x\in V(C_{2})$. Then $G[V_{1}\cup\{x,x^{+}\}]$ has a Hamilton path with two endvertices $u^{+}\in V(C_{1})$ and $x^{+}\in V(C_{2})$. Since $G[V_{1}\cup\{x,x^{+}\}]$ is not Hamiltonian, then by Proposition 3, $u^{+}$ and $x^{+}$ have a common neighbour $y\in V_{2}\setminus\{x,x^{+}\}$, i.e., $x^{+}y,u^{+}y\in E(G)$. Since $\sigma_{2}(G[V_{1}])\geq n_{1}=|V_{1}|+2$, we have $d_{G[V_{1}]-u}(w)+d_{G[V_{1}]-u}(w^{\prime})\geq n_{1}-2=|V_{1}\setminus\{u\}|+1$ for any two nonadjacent vertices $w$ and $w^{\prime}$ in $V_{1}\setminus\{u\}$. Thus, by Lemma 2.4, $G[V_{1}]-u$ is Hamilton-connected.

If $e(x,V_{1})\geq 2$, then $x$ is adjacent to a vertex $v$ in $V_{1}$ other than $u$. Moreover, we may choose a direction of $C_{1}$ such that $v\neq u^{+}$. Then $G[V_{1}]-u$ has a Hamilton path $P$ with endvertices $v$ and $u^{+}$ because $G[V_{1}]-u$ is Hamilton-connected. Thus, $G[V_{1}\cup\{x,y\}]$ has a Hamilton path $uxPy$ with endvertices $u$ and $y$. If $G[V_{1}\cup\{x,y\}]$ is Hamiltonian, then we are done. If not, by Proposition 3, $u$ and $y$ have a common neighbour $z\in V_{2}\setminus\{x,y\}$. Thus, $G[V_{1}\cup\{z,y\}]$ has a Hamilton cycle $yC_{1}[u^{+},u]zy$.

Now consider the case that $e(x,V_{1})\leq 1$ for any vertex $x\in V_{2}$. First assume that there is a vertex $x\in V_{2}$ such that $d_{G[V_{2}]}(x)\leq\lfloor\frac{n_{2}+5}{2}\rfloor$, i.e. $d_{G}(x)\leq\lfloor\frac{n_{2}+5}{2}\rfloor+1$. Let $u$ and $v$ be two distinct vertices in $G[V_{1}]$ such that $ux\notin E(G)$, $vx\notin E(G)$. By the degree sum condition, we have $d(u)+d(v)+2d_{G}(x)\geq 2(n+2)$ and therefore, $d(u)+d(v)\geq 2n_{1}+n_{2}-3$. Since $|V_{1}|=n_{1}-2$, we have $d_{G[V_{1}]}(u)+d_{G[V_{1}]}(v)\leq 2(n_{1}-3)$, and then $e(\{u,v\},V_{2})\geq n_{2}+3$. Thus, there is a vertex $y\in V_{2}$ such that $uy,vy\in E(G)$, a contradiction.

Assume now that $d_{G[V_{2}]}(x)\geq\lfloor\frac{n_{2}+5}{2}\rfloor+1\geq\frac{n_{2}+6}{2}$ for any vertex $x\in V_{2}$. Let $ux\in E(G)$ with $u\in V(C_{1})$ and $x\in V(C_{2})$. Then $u^{+}$ and $x^{+}$ are the endvertices of the Hamilton path $C_{1}[u^{+},u]xx^{+}$ in $G[V_{1}\cup\{x,x^{+}\}]$. If $G[V_{1}\cup\{x,x^{+}\}$ is not Hamiltonian then by Proposition 3, $u^{+}$ and $x^{+}$ have three common neighbours, say one of which is $w$. Similarly, $u^{-}$ and $x^{+}$ have three common neighbours. Further, notice that $C^{-}_{1}[u,u^{+}]wx^{+}$ is a Hamilton path with endvertices $u$ and $x^{+}$. If $G[V_{1}\cup\{w,x^{+}\}]$ is not Hamiltonian then $u$ and $x^{+}$ have three common neighbours. Consequently, we can choose three distinct vertices $y_{1},y_{2},y_{3}\in V_{2}$ such that $u^{+}y_{1},uy_{2},u^{-}y_{3},x^{+}y_{1},x^{+}y_{2},x^{+}y_{3}\in E(G)$. Thus, $G[(V_{1}-u)\cup\{y_{1},x^{+},y_{3}\}]$ has a Hamilton cycle $C_{1}[u^{+},u^{-}]y_{3}x^{+}y_{1}u^{+}$. Write $\Gamma=G[V_{2}-\{y_{1},x^{+},y_{3}\}]$. Notice that, for any pair of nonadjacent vertices $a$ and $b$ in $\Gamma$, we have $d_{\Gamma}(a)+d_{\Gamma}(b)\geq(n_{2}+6)-6=|\Gamma|+1$, i.e., $\sigma_{2}(\Gamma)\geq|\Gamma|+1$. So by Lemma 2.4, $\Gamma$ is Hamilton-connected. Thus, $\Gamma$ has a Hamilton path $P$ with endvertices $x$ and $y_{2}$, and hence we obtain a desired $n_{2}$-cycle $uPu$. ∎

Claim 2. If $K_{m+2,m+1}$ $\subseteq G[V_{2}]\subseteq K_{m+2}+(m+1)K_{1}$, then $G$ has a pair of disjoint $(n_{1},n_{2})$-cycles.

Proof.

It is clear that $G[V_{2}]$ has a set of $m+1$ pairwise non-adjacent vertices. We denote this set by $S$ and denote $T=V_{2}\setminus S$. Thus, $|S|=m+1=\frac{1}{2}(n_{2}+1)\geq 3$ and $|T|=m+2=\frac{1}{2}(n_{2}+3)\geq 4$. Since $\sigma_{2}(G[V_{2}])=n_{2}+3$, $G[T]$ has at most one isolated vertex.

Let $P$ be a Hamilton path in $G[V_{1}]$ and let $u$ and $v$ be its two endvertices. If $d_{G[V_{1}]}(u)+d_{G[V_{1}]}(v)\leq n_{1}-3$ and $uv\notin E(G)$, then $e(\{u,v\},V_{2})\geq n_{2}+5$, say $e(u,V_{2})\geq\frac{1}{2}(n_{2}+5)$. Thus, $e(u,T)\geq 2$ and $e(u,S)\geq 1$, say $us,ut_{1},ut_{2}\in E(G)$ where $s\in S,t_{1},t_{2}\in T$. Since $G[T]$ has at most one isolated vertex, one of $G[T]-t_{1}$ and $G[T]-t_{2}$ contains an edge, say $E(G[T]-t_{1})\not=\emptyset$. If there is a vertex $s_{1}\in S$ such that $vs_{1}\in E(G)$, then $G$ has a pair of disjoint $n_{1}$-cycles $t_{1}Ps_{1}$ and an $n_{2}$-cycles in $G[V_{2}]-\{t_{1},s_{1}\}$. Otherwise, we have $e(v,S)=0$ and therefore, $e(v,T)\geq 3$ because $e(\{u,v\},V_{2})\geq n_{2}+5$. Thus, there is a vertex $t\in T$ such that $E(G[T]-t)$ is not empty. Hence, $G$ has a pair of disjoint $n_{1}$-cycle $sPt$ and an $n_{2}$-cycle in $G[V_{2}]-\{t,s\}$.

We now assume that $uv\in E(G)$ or $d_{G[V_{1}]}(u)+d_{G[V_{1}]}(v)\geq n_{1}-2$ holds for the two endvertices $u,v$ of any Hamilton path in $G[V_{1}]$. Similar to the proof of Proposition 2, $G[V_{1}]$ is Hamiltonian. Let $C_{1}=v_{1}v_{2}\cdots v_{n_{1}-2}v_{1}$ be a Hamilton cycle in $G[V_{1}]$.

Since $\sigma_{2}(G[V_{2}])=n_{2}+3$ and $|S|\geq 3$, there are two distinct vertices $s_{1},s_{2}\in S$ such that $e(s_{1},V_{1})\geq\frac{1}{2}(n_{1}-1)$ and $e(s_{2},V_{1})\geq\frac{1}{2}(n_{1}-1)$. Thus, each of $s_{1}$ and $s_{2}$ is adjacent to a pair of two consecutive vertices on $C_{1}$, say $\{v_{i},v_{i+1}\}$ and $\{v_{j},v_{j+1}\}$, respectively.

Claim 2.1. If $G$ has no pair of disjoint $(n_{1},n_{2})$-cycles and $G[T]$ contains a $P_{4}$ then the pairs $\{v_{i},v_{i+1}\}$ and $\{v_{j},v_{j+1}\}$ can be properly chosen to be distinct.

Proof.

If $s_{1}$ or $s_{2}$ is adjacent to at least two pairs of two consecutive vertices on $C_{1}$ then the assertion clearly holds. Assume now that each of $s_{1}$ and $s_{2}$ is adjacent to exactly one pair of two consecutive vertices and moreover, they are adjacent to the same pair, say $\{v_{1},v_{2}\}$. In this case, recall that $e(s_{1},V_{1})\geq\frac{1}{2}(n_{1}-1)$ and $e(s_{2},V_{1})\geq\frac{1}{2}(n_{1}-1)$. This implies that $s_{i}v_{4},s_{i}v_{6},\cdots,s_{i}v_{n_{1}-3}\in E(G)$ and $s_{i}v_{3},s_{i}v_{5},\cdots,s_{i}v_{n_{1}-2}\notin E(G)$ for each $i\in\{1,2\}$ (in this case $n_{1}$ must be odd). Since $G[T]$ contains a $P_{4}$, $G[T]-\{t\}$ has at least one edge for any $t\in T$. Thus, if $tv_{3}\in E(G)$ for some $t\in T$ then $G$ has a pair of disjoint $n_{1}$-cycle $s_{1}C_{1}[v_{4},v_{3}]t$ and an $n_{2}$-cycle in $G[V_{2}]-\{t,s_{1}\}$, a contradiction. Similarly, if $v_{3}$ is adjacent to some $v_{i}$ with $i$ odd then $G$ has a pair of disjoint $n_{1}$-cycle $s_{1}v_{2}s_{2}C^{-}_{1}[v_{i-1},v_{3}]v_{i}C^{-}_{1}[v_{i},v_{1}]s_{1}$ and an $n_{2}$-cycle in $G[V_{2}]-\{s_{1},s_{2}\}$. This is again a contradiction. Therefore, $v_{3}$ is adjacent to neither vertex in $T$ nor vertex in $\{v_{5},v_{7},\cdots,v_{n_{1}-2}\}$. Therefore,

[TABLE]

which contradicts our assumption that $\sigma_{2}(G)\geq n+2$ since $s_{1}$ and $v_{3}$ are not adjacent. ∎

If $G[T]$ contains a $P_{4}$ then by Claim 2.1, $G[V_{1}\cup\{s_{1},s_{2}\}]$ is Hamiltonian and moreover, it is not difficult to find a Hamilton cycle in $G[V_{2}]-\{s_{1},s_{2}\}$.

If $G[T]$ does not contain $P_{4}$, then each component in $G[T]$ is a star (i.e., $K_{1,q}$) or has at most three vertices. Thus, we may choose two nonadjacent vertices $t_{1},t_{2}\in T$ such that $d_{G[T]}(t_{1})\leq 2$ and $d_{G[T]}(t_{2})\leq 2$. Moreover, each of $G[T]-t_{1}$ and $G[T]-t_{2}$ has at least one edge because, except one possible isolated vertex, every vertex in $G[T]$ has degree at least one. Therefore, $d_{G[V_{2}]}(t_{1})+d_{G[V_{2}]}(t_{2})\leq n_{2}+5$ and hence, $e(\{t_{1},t_{2}\},V_{1})\geq n_{1}-3$, say $e(t_{1},V_{1})\geq\frac{1}{2}(n_{1}-3)$. Choose a Hamilton cycle $C_{2}$ in $G[V_{2}]$ such that $t_{1}^{+},t_{1}^{-}\in S$. Since $d_{G[V_{2}]}(t_{1}^{+})+d_{G[V_{2}]}(t_{1}^{-})=n_{2}+3$, we have $e(\{t_{1}^{+},t_{1}^{-}\},V_{1})\geq n_{1}-1$ and therefore, $t_{1}^{-}$ or $t_{1}^{+}$ is adjacent to two consecutive vertices on $C_{1}$, say $t_{1}^{-}v_{i},t_{1}^{-}v_{i+1}\in E(G)$. If $t_{1}v_{i-1}\in E(G)$, then $G$ has a pair of disjoint $n_{1}$-cycle $t_{1}C_{1}^{-}[v_{i-1},v_{i}]t_{1}^{-}t_{1}$ and an $n_{2}$-cycle in $G[V_{2}]-\{t_{1},t_{1}^{-}\}$ since $G[T]-t_{1}$ has at least one edge. Similarly, if $t_{1}$ is adjacent to one of $v_{i},v_{i+1},v_{i+2}$ then we can get a pair of disjoint $(n_{1},n_{2})$-cycles. Now assume that $t_{1}c_{i-1},t_{1}c_{i},t_{1}c_{i+1},t_{1}c_{i+2}\notin E(G)$. Then $t_{1}$ is adjacent to two consecutive vertices $v_{j},v_{j+1}$ on $C_{1}$ as $e(t_{1},V_{1})\geq\frac{1}{2}(n_{1}-3)$. Therefore, $G$ has a pair of disjoint $n_{1}$-cycle $t_{1}C_{1}[v_{j+1},v_{i}]t_{1}^{-}C_{1}[v_{i+1},v_{j}]t_{1}$ and an $n_{2}$-cycle in $G[V_{2}]-\{t_{1},t_{1}^{-}\}$. The claim follows. ∎

Claim 3. If $3K_{1}+(K_{p}\cup K_{q})\subseteq G[V_{2}]\subseteq K_{3}+(K_{p}\cup K_{q})$ for some positive integers $p$ and $q$ with $p+q=n_{2}-1$, then $G$ has a pair of disjoint $(n_{1},n_{2})$-cycles.

Proof.

If $n_{2}=5$, then $K_{4,3}$ $\subseteq 3K_{1}+(K_{p}\cup K_{q})\subseteq 3K_{1}+K_{4}$. So by Claim 2, Claim 3 holds. Thus, we assume that $n_{2}\geq 6$. Let $C_{2}$ be a Hamilton cycle in $3K_{1}+(K_{p}\cup K_{q})$ and let $M=3K_{1}$. It is clear that $C_{2}$ is also a Hamilton cycle in $G[V_{2}]$. So by Proposition 2, $G[V_{1}\cup\{x,y\}]$ is Hamiltonian for some two consecutive vertices $x$ and $y$ on $C_{2}$ or $G[V_{1}]$ is Hamiltonian. Noticing that at most one of $x$ and $y$ is in $M$, $(3K_{1}+(K_{p}\cup K_{q}))\setminus\{x,y\}$ is Hamiltonian. So if the former holds, then we are done. We now assume that $G[V_{1}]$ has a Hamilton cycle $C_{1}$.

For any vertex $x\in K_{p}$ and $y\in K_{q}$, since $xy\notin E(G)$ and $\sigma_{2}(G[V_{2}])=n_{2}+3$, we have $e(\{x,y\},V_{1})\geq n_{1}-1$. Therefore, $x$ or $y$ is adjacent to two consecutive vertices $w,w^{+}$ on $C_{1}$, say $xw,xw^{+}\in E(G)$. Thus, $G[V_{1}\cup\{x,x^{+}\}]$ has a Hamilton path $C_{1}[w^{2+},w^{+}]xx^{+}$ with endvertices $w^{2+}$ and $x^{+}$. If $G[V_{1}\cup\{x,x^{+}\}]$ is Hamiltonian, then we are done. If not, then by Proposition 3, $w^{2+}$ and $x^{+}$ have a common neighbour $z\in V_{2}\setminus\{x,x^{+}\}$. It is clear that $G[V_{2}]-\{x,z\}$ is Hamiltonian because $x\notin M$. Hence, if $G[V_{1}\cup\{x,z\}]$ is Hamiltonian then we are done. If not, again by Proposition 3, $w^{+}$ and $z$ have a common neighbour in $V_{2}\setminus\{x,z\}$, say $t\in V_{2}\setminus\{x,z\}$. Noticing that at most one of $z$ and $t$ is in $M$, $G[V_{2}]-\{z,t\}$ is Hamiltonian. Thus, we get a pair of an $n_{1}$-cycle $C_{1}[w^{2+},w^{+}]tzw^{2+}$ and an $n_{2}$-cycle (Hamilton cycle) in $G[V_{2}]-\{z,t\}$. ∎

By Lemma 2.7 and the three claims above, Theorem 1.7 follows.

Remark. Since $\sigma_{2}(G)\leq\delta(G)$, Theorem 1.7 gives a generalization of Theorem 1.4 when $n_{1}$ and $n_{2}$ are both odd. This remains a natural question: Can $\sigma_{2}(G)\geq n+1$ guarantee that $G$ has a pair of disjoint $(n_{1},n_{2})$-cycles if at least one of $n_{1}$ and $n_{2}$ is even?

4 Acknowledgements

This work was supported by the National Natural Science Foundation of China [Grant numbers, 11471273, 11561058].