Subdivisions of vertex-disjoint cycles in bipartite graphs

TL;DR

This paper proves that bipartite graphs with sufficiently high minimum degree contain subdivisions of certain disjoint even cycles, extending previous results in graph theory.

Contribution

It generalizes Wang's earlier theorem by establishing conditions for subdivisions of multiple even cycles in bipartite graphs.

Findings

Bipartite graphs with minimum degree at least n/2 - k + 1 contain subdivisions of specified disjoint even cycles.

The result applies to graphs with multiple components, each being an even cycle of length at least 6.

The theorem extends classical cycle subdivision results to more complex bipartite structures.

Abstract

Let $n\geq 6,k\geq 0$ be two integers. Let $H$ be a graph of order $n$ with $k$ components, each of which is an even cycle of length at least $6$ and $G$ be a bipartite graph with bipartition $(X,Y)$ such that $|X|=|Y|\geq n/2$. In this paper, we show that if the minimum degree of $G$ is at least $n/2-k+1$, then $G$ contains a subdivision of $H$. This generalized an older result of Wang.

Equations

$$\sum _{i=1}^{k-1}|C_i^{\ast }|\le |G|-|C_k|.$$

$$\beta (G_1)=\sum _{i=1}^{k-1}|E(G[V(C_i^{\ast })])|.$$

$$d(u^{\prime },G)+d(u^{\prime \prime },G)$$

$$d(u_1,G_2)=d(u_1,P)\le \frac{|C_k|}{2}-1$$

$$d(u_s,G_2)=d(u_s,P)\le \frac{|C_k|}{2}-1.$$

$$d(y_1,Q)\le \frac{|C_k|}{2}-1$$

$$d(y_2,Q)\le \frac{|C_k|}{2}-1.$$

$$d(y_1,P)=d(y_1,G_2)-d(y_1,G_2^{\prime })\ge |C_k|/2+1-d(y_1,Q)\ge 2$$

$$d(y_2,P)=d(y_2,G_2)-d(y_2,G_2^{\prime })\ge |C_k|/2+1-d(y_2,Q)\ge 2.$$

$$2min\{\frac{|C_k|}{2}+1-d(y_1,Q),\frac{|C_k|}{2}+1-d(y_2,Q)\}-1$$

$$2max\{d(y_1,Q),d(y_2,Q)\}$$

$$d(x,P_3)\le \frac{|C_k|}{2}-1$$

$$d(y,P_3)\le \frac{|C_k|}{2}-1.$$

$$d(x,P)\ge |C_k|/2+1-d(x,P_3)\ge 2$$

$$d(y,P)\ge |C_k|/2+1-d(y,P_3)\ge 2.$$

$$d(u_1,G_2)\le |C_k|/2-1,$$

$$d(u_2,G_2)\le |C_k|/2,$$

$$d(u_1,C_p^{\ast })+d(u_s,C_p^{\ast })\ge |C_p|-1$$

$$|C_p^{\ast }|=|C_p|.$$

$$d(z,G_2)+d(z,C_p^{\ast })\le \frac{|C_k|}{2}+\frac{|C_p|}{2}-1,$$

$$d(u_2,G_2)+d(u_2,C_p^{\ast })<\frac{|C_k|}{2}+\frac{|C_p|}{2}-1.$$

$$d(x^{\ast },G_2)+d(x^{\ast },C_p^{\ast })\le \frac{|C_k|}{2}+\frac{|C_p|}{2}-1.$$

$$d(y^{\ast },G_2)+d(y^{\ast },C_p^{\ast })\le \frac{|C_k|}{2}+\frac{|C_p|}{2}-1.$$

$$d(u_{s-1},G_2)+d(u_{s-1},C_p^{\ast })\le \frac{|C_k|}{2}+2\le \frac{|C_k|}{2}+\frac{|C_p|}{2}-1.$$

$$d(u_2,G_2)+d(u_2,C_p^{\ast })\le \frac{|C_k|}{2}+1<\frac{|C_k|}{2}+\frac{|C_p|}{2}-1.$$

$$\sum _{z\in S}d(z,C_q^{\ast })\ge 3|C_q|-5$$

$$\sum _{z\in S}d(z,C_p^{\ast })+\sum _{z\in S}d(z,G_2)<3|C_k|+3|C_p|-6.$$

$$\sum _{z\in S}d(z,G)$$

$$\sum _{z\in S}d(z,C_q^{\ast })\ge 3|C_q|-5.$$

$$d(u_1,C_q^{\ast })+d(u_s,C_q^{\ast })\ge |C_q|-1,$$

Full text

Keywords: subdivision; vertex-disjoint cycles; bipartite graph

1 Introduction

We use Bondy and Murty [2] for terminology and notation not defined here and consider finite simple graphs only.

Let $G$ be a graph. A set of subgraphs of $G$ is said to be vertex-disjoint if no two of them have a common vertex in $G$. Wang [3] considered the minimum degree condition for existence of vertex-disjoint large cycles in a bipartite graph, as follows.

Theorem 1** (Wang [3]).**

Let $G$ be a bipartite graph with bipartition $(X,Y)$ such that $|X|=|Y|\geq sk$, where $s\geq 3$ and $k\geq 1$ are two integers. If the minimum degree of $G$ is at least $(s-1)k+1$, then $G$ contains $k$ vertex-disjoint cycles of length at least $2s$.

A *subdivision * of a graph $H$ is a graph that can be obtained from $H$ by a sequence of edge subdivisions. Let $H$ be a graph on $n$ vertices and $k$ non-trivial components such that every component contains at most one cycle. A *cyclic subdivision * of $H$ is one such that only cyclic edges of $H$ are subdivided.

Recently, Babu and Diwan [1] gave the following result.

Theorem 2** (Babu and Diwan [1]).**

Let $H$ be a graph with $n$ vertices and $k$ non-trivial components such that every component contains at most one cycle. Let $G$ be a graph with at least $n$ vertices. If the minimum degree of $G$ is at least $n-k$, then $G$ contains a cyclic subdivision of $H$.

Motivated by Theorem 2, this paper focuses on the analogous problem for existence of subdivisions of vertex-disjoint cycles in bipartite graphs. Our result is as follows.

Theorem 3**.**

Let $H$ be a graph of order $n$ with $k$ components, each of which is an even cycle of length at least $6$. Suppose that $G$ is a bipartite graph with bipartition $(X,Y)$ such that $|X|=|Y|\geq n/2$. If the minimum degree of $G$ is at least $n/2-k+1$, then $G$ contains a subdivision of $H$.

It is easy to see that Theorem 3 generalized Theorem 1. In fact, we conjecture the following.

Conjecture 1**.**

Let $H$ be a graph of order $n$ with $k$ components, each of which is an even cycle. Suppose that $G$ is a bipartite graph with bipartition $(X,Y)$ such that $|X|=|Y|\geq n/2$. If the minimum degree of $G$ is at least $n/2-k+1$, then $G$ contains a subdivision of $H$.

If the Conjecture is true, then the minimum degree condition is sharp. Let $k$ be an even. Let $H$ be an union of $k$ cycles such that $k-1$ cycles of length 4 and a cycle of length 6. Let $G$ be a bipartite graph with bipartition $(X_{1}\cup X_{2}\cup\{u\},Y_{1}\cup Y_{2}\cup\{v\})$ such that the two induced subgraphs by $X_{1}\cup Y_{1}$ and $X_{2}\cup Y_{2}$ are isomorphic to $K_{k,k}$. Further, $u$ is adjacent to every vertex in $Y_{1}\cup\{v\}$ and $v$ is adjacent to every vertex in $X_{2}\cup\{u\}$. The vertices of $X_{1}$ are matched with vertices of $Y_{2}$ by $k$ independent edges of $G$. Clearly, $G$ contains exactly $4k+2$ vertices and the minimum degree of $G$ is $k+1<k+2$. It is not difficult to see that if $G$ contains $H$ then the edge $uv$ will be contained by the cycle of length exactly 6. Let $C$ be the cycle of length exactly 6 with contained the edge $uv$. But, $G-V(C)$ does not contain a spanning subgraph with a union of $k-1$ cycles of length 4.

We postpone the proof of Theorem 3 to the next section.

2 Proof of Theorem 3

We first introduce some further terminology and notations that will be used later.

Let $G$ be a graph. The order of $G$ is denoted by $|G|$. For a vertex $v$ and a subgraph $H$ of $G$, we use $N(v,H)$ and $d(v,H)$ to denote the set of vertices and the number of vertices in $H$ which are adjacent to $v$, respectively. Thus $d(v,G)$ is the degree of $v$ in $G$. By $\delta(G)$ we denote the minimum degree of $G$. For a subset $S$ of the vertices or the edges in $G$, we use $G[S]$ to denote the subgraph of $G$ induced by $S$. If $v\not\in V(H)$, then we denote $G[V(H)\cup\{v\}]$ by $H+v$. If $v\in V(H)$, then we denote $G[V(H)\setminus\{v\}]$ by $H-v$. By $G-H$ we denote $G[V(G)\setminus V(H)]$. Let $x,y$ be two nonadjacent vertices in $G$. We use $G+xy$ to denote the graph by adding the edge $xy$ in $G$. Let $C$ be a cycle with a given orientation. We use $v^{+}$ (resp. $v^{-}$) to denote the successor (resp. the predecessor) of $v$ along $C$ according to this orientation.

Proof of Theorem 3. By contradiction. Suppose that $G$ is a graph satisfying the conditions of the theorem but containing no subdivisions of $H$ such that the number of edges of $G$ is as large as possible. This implies that that $G$ is not a complete bipartite graph. Let $a\in X$ and $b\in Y$ be two nonadjacent vertices of $G$. Denote the components of $H$ by $C_{1},C_{2},\ldots,C_{k}$. It follows from the choice of $G$ that $G+ab$ contains a subdivision $H^{*}$ of $H$. Let $C^{*}_{i}$ be the subdivision of $C_{i}$ in $H^{*}$ for $1\leq i\leq k$. Without loss of generality, we assume that $ab\in E(C^{*}_{k})$. Then $G$ contains a subdivision of $H-C_{k}$ with

[TABLE]

Let $G_{1}$ be the subgraph of $G$ induced by $\bigcup_{i=1}^{k-1}V(C^{*}_{i})$ and

[TABLE]

Set $G_{2}=G-G_{1}$ and let $P$ be a longest path in $G_{2}$. We give an orientation to each $C^{*}_{i}$ for $1\leq i\leq k-1$.

We assume that $C^{*}_{1},C^{*}_{2},\ldots,C^{*}_{k-1}$ are chosen in $G$ such that

(i) $|G_{1}|$ is as small as possible;

(ii) $P$ is as long as possible, subject to (i);

(iii) $\beta(G_{1})$ is as large as possible, subject to (i) and (ii).

First, we prepare some claims.

Claim 1**.**

Let $u$ be a vertex of $G_{2}$ and $v$ be a vertex of $C^{*}_{i}$ with $1\leq i\leq k-1$. If $|C^{*}_{i}|>|C_{i}|$, then $d(u,C^{*}_{i})<|C_{i}|/2$ and $d(v,C^{*}_{i})<|C_{i}|/2$. If $d(u,C^{*}_{i})=|C_{i}|/2$ or $d(v,C^{*}_{i})=|C_{i}|/2$, then $|C^{*}_{i}|=|C_{i}|$.

Proof.

If $d(u,C^{*}_{i})\geq|C_{i}|/2$, then it is not difficult to see that $C^{*}_{i}+u$ contains a cycle of length less than $|C^{*}_{i}|$ but at least $|C_{i}|$, which can be seen as a subdivision of $C_{i}$. This contradicts the choice of $C^{*}_{1},C^{*}_{2},\ldots,C^{*}_{k-1}$ in (i). If $d(v,C^{*}_{i})\geq|C_{i}|/2$, then we can get a similar contradiction. So we have $d(u,C^{*}_{i})<|C_{i}|/2$ and $d(v,C^{*}_{i})<|C_{i}|/2$. The result $|C^{*}_{i}|=|C_{i}|$ if $d(u,C^{*}_{i})=|C_{i}|/2$ or $d(v,C^{*}_{i})=|C_{i}|/2$ follows immediately. ∎

By Claim 1, we can immediately obtain that for any vertices $u\in V(G_{2}),v\in V(C^{*}_{i})$, we have $d(u,C^{*}_{i})\leq|C_{i}|/2$ and $d(v,C^{*}_{i})\leq|C_{i}|/2$ for each $i$ with $1\leq i\leq k-1$.

Claim 2**.**

Let $u^{\prime}$ be a vertex in $V(G_{2})\cap X$ and $u^{\prime\prime}$ be a vertex in $V(G_{2})\cap Y$. If $d(u^{\prime},G_{2})+d(u^{\prime\prime},G_{2})<|C_{k}|$, then there exists an $C^{*}_{i}$ with $1\leq i\leq k-1$ such that $d(u^{\prime},C^{*}_{i})+d(u^{\prime\prime},C^{*}_{i})\geq|C_{i}|-1$. Moreover, we have $|C^{*}_{i}|=|C_{i}|$, and there exists a vertex $v\in N(u^{\prime},C_{i}^{*})$ such that $C^{*}_{i}+u^{\prime\prime}-v$ contains $C_{i}$.

Proof.

If $d(u^{\prime},C^{*}_{j})+d(u^{\prime\prime},C^{*}_{j})\leq|C_{j}|-2$ for all $j$ with $1\leq j\leq k-1$, then

[TABLE]

contradicting the condition $\delta(G)\geq n/2-k+1$. Thus, there exists an $C^{*}_{i}$ with $1\leq i\leq k-1$ such that $d(u^{\prime},C^{*}_{i})+d(u^{\prime\prime},C^{*}_{i})\geq|C_{i}|-1$. This implies that $d(u^{\prime},C^{*}_{i})=|C_{i}|/2$ or $d(u^{\prime\prime},C^{*}_{i})=|C_{i}|/2$. By Claim 1 we have $|C^{*}_{i}|=|C_{i}|$. If $d(u^{\prime\prime},C^{*}_{i})=|C_{i}|/2$, then any $v\in N(u^{\prime},C^{*}_{i})$ can be chosen as the required vertex. If $d(u^{\prime\prime},C^{*}_{i})=|C_{i}|/2-1$, then $d(u^{\prime},C^{*}_{i})=|C_{i}|/2$. Since $|C_{i}|\geq 6$, we can also find a vertex $v\in N(u^{\prime},C^{*}_{i})$ such that $C^{*}_{i}+u^{\prime\prime}-v$ contains $C_{i}$. ∎

Now let $P=u_{1}u_{2}\cdots u_{s}$ and assume that $u_{1}\in X$.

Claim 3**.**

$P$ is a Hamilton path of $G_{2}$.

Proof.

Suppose that $P$ is not a Hamilton path of $G_{2}$. Then from the choice of $P$ and the fact that $G_{2}$ does not contain a subdivision of $C_{k}$, we have

[TABLE]

and

[TABLE]

Let $G^{\prime}_{2}=G_{2}-V(P)$ and $M=\{a_{1}b_{1},a_{2}b_{2},\ldots,a_{m}b_{m}\}$ be a maximum matching in $G^{\prime}_{2}$ with $\{a_{1},a_{2},\ldots,a_{m}\}\subseteq X$. We distinguish two cases as follows.

Case 1. $M$ is not a perfect matching of $G^{\prime}_{2}$.

First note that $G_{2}$ has an even number of vertices. If $u_{s}\in Y$, then $P$ has also an even number of vertices. Thus, $|G^{\prime}_{2}|$ is even and $|V(G^{\prime}_{2})\cap X|=|V(G^{\prime}_{2})\cap Y|$. If $u_{s}\in X$, then $|V(G^{\prime}_{2})\cap Y|=|V(G^{\prime}_{2})\cap X|+1$. Since $M$ is not a perfect matching of $G^{\prime}_{2}$, we have $(V(G^{\prime}_{2})\cap Y)\setminus\{b_{1},b_{2},\ldots,b_{m}\}\neq\emptyset$.

Let $y_{0}$ be a vertex in $(V(G^{\prime}_{2})\cap Y)\setminus\{b_{1},b_{2},\ldots,b_{m}\}$. Choose $P_{1}$ as a longest $M$-alternating path in $G^{\prime}_{2}$ starting from $y_{0}$. Then $P_{1}$ must end at a vertex $y_{1}$ with $y_{1}\in\{y,b_{1},\ldots,b_{m}\}$. Let $M^{\prime}=M-E(P_{1})$ and $G^{\prime\prime}_{2}=G^{\prime}_{2}-V(P_{1})+y$. Choose $P_{2}$ as a longest $M^{\prime}$-alternating path in $G^{\prime\prime}_{2}$ starting from $y_{0}$. Then $P_{2}$ must end at a vertex $y_{2}\in\{y,b_{1},\ldots,b_{m}\}$. Therefore, $Q=P_{1}\cup P_{2}$ is a path from $y_{1}$ to $y_{2}$ in $G^{\prime}_{2}$ such that $d(y_{1},G^{\prime}_{2})=d(y_{1},Q)$ and $d(y_{2},G^{\prime}_{2})=d(y_{2},Q)$. Since $G_{2}$ does not contain a subdivision of $C_{k}$, we have

[TABLE]

and

[TABLE]

If $d(y_{1},G_{2})\leq|C_{k}|/2$, then by $d(u_{1},G_{2})+d(y_{1},G_{2})<|C_{k}|$ and Claim 2 we know that $d(u_{1},C^{*}_{i})+d(y_{1},C^{*}_{i})\geq|C_{i}|-1$ for some $i$ with $1\leq i\leq k-1$, and there exists a vertex $v\in N(u_{1},C^{*}_{i})$ such that $C^{*}_{i}+y_{1}-v$ contains $C^{*}_{i}$. This contradicts the choice of $C^{*}_{1},C^{*}_{2},\ldots,C^{*}_{k-1}$ in (ii). Hence, we have $d(y_{1},G_{2})\geq|C_{k}|/2+1$. Similarly we have $d(y_{2},G_{2})\geq|C_{k}|/2+1$. This implies that

[TABLE]

and

[TABLE]

It is not difficult to find a segment of $P$ with at least

[TABLE]

vertices such that the two end-vertices of this segment are adjacent to $y_{1}$ and $y_{2}$, respectively. Since $Q$ has at least

[TABLE]

vertices, we can obtain a cycle with at least $|C_{k}|$ vertices in $G[V(P\cup Q)]$, which can be seen as a subdivision of $C_{k}$. This contradicts our assumption that $G$ contains no subdivisions of $H$.

Case 2. $M$ is a perfect matching of $G^{\prime}_{2}$.

Choose $P_{3}$ as a longest $M$-alternating path in $G^{\prime}_{2}$ such that the first edge of $P_{3}$ is in $M$. Denote the initial vertex and the final vertex of $P_{3}$ by $x$ and $y$, respectively. Then we have $x\in\{a_{1},a_{2},\ldots,a_{m}\}$ and $y\in\{b_{1},b_{2},\ldots,b_{m}\}$. Since $G_{2}$ does not contain a subdivision of $C_{k}$, we know that

[TABLE]

and

[TABLE]

Clearly, $u_{s}\in Y$. If $d(x,G_{2})\leq|C_{k}|/2$, then it follows from $d(u_{s},G_{2})+d(x,G_{2})<|C_{k}|$ and Claim 2 that $d(u_{s},C^{*}_{i})+d(x,C^{*}_{i})\geq|C_{i}|-1$ for some $i$ with $1\leq i\leq k-1$, and there exists a vertex $v\in N(u_{s},C^{*}_{i})$ such that $C^{*}_{i}+x-v$ contains $C^{*}_{i}$. This contradicts the choice of $C^{*}_{1},C^{*}_{2},\ldots,C^{*}_{k-1}$ in (ii). Hence, we have $d(x,G_{2})\geq|C_{k}|/2+1$. Similarly, we can show that $d(y,G_{2})\geq|C_{k}|/2+1$. This means that

[TABLE]

and

[TABLE]

The rest of the proof of this case is just as same as that of Case 1. ∎

Since $G_{2}$ does not contain a subdivision of $C_{k}$ and $P$ is a Hamilton path of $G_{2}$, we can immediately get

[TABLE]

[TABLE]

Note $u_{s}\in Y$. Then, by (1) and Claim 2, we know that there exists a component $C^{*}_{p}$ for some $p$ with $1\leq p\leq k-1$ such that

[TABLE]

and

[TABLE]

Without loss of generality, we may assume $d(u_{1},C^{*}_{p})=|C_{p}|/2$ and $d(u_{s},C^{*}_{p})\geq|C_{p}|/2-1$. Let $x^{*}$ be a vertex of $V(C^{*}_{p})\cap X$ such that $u_{s}$ is adjacent to all vertices in $(V(C^{*}_{p})\cap X)\setminus\{x^{*}\}$. Let $y^{*}$ be a vertex of $V(C^{*}_{p})\cap Y$ such that $x^{*}$ and $y^{*}$ are not adjacent on $C^{*}_{p}$.

We have the following Claim 4 and Claim 5.

Claim 4**.**

For all $z\in\{u_{1},u_{s-1},u_{s},x^{*},y^{*}\}$ we have

[TABLE]

and for $u_{2}$ we have

[TABLE]

Proof.

The case $z=u_{1}$ or $z=u_{s}$ follows from (1) and the fact $|C^{*}_{p}|=|C_{p}|$ immediately.

Suppose $z=x^{*}$. Then we have $u_{1}x^{*}\not\in E(G)$. If $x^{*}$ is adjacent to at least $|C_{k}|/2$ vertices in $P$, then $P+x^{*}-u_{1}$ contains a subdivision of $C_{k}$ and $C^{*}_{p}+u_{1}-x^{*}$ contains $C_{p}$. This contradicts our assumption that $G$ contains no subdivisions of $H$. Thus, we have

[TABLE]

Similarly, we can prove

[TABLE]

Suppose $z=u_{s-1}$. If $u_{s-1}$ is adjacent to a vertex $v$ in $(V(C^{*}_{p})\cap Y)\setminus\{x^{*+},x^{*-}\}$, then from $u_{s-1}\in X$ and $|C^{*}_{p}|\geq 6$ we know that $G_{2}+v-u_{s}$ contains a subdivision of $C_{k}$ and $C^{*}_{p}+u_{s}-v$ contains $C_{p}$. This contradicts our assumption that $G$ contains no subdivisions of $H$. Together with (2), we can obtain

[TABLE]

Note that $u_{2}\in Y$. If $u_{2}$ is adjacent to a vertex $v$ in $(V(C^{*}_{p})\cap X)\setminus\{x^{*}\}$, then $G_{2}+v-u_{1}$ contains a subdivision of $C_{k}$ and $C^{*}_{p}+u_{1}-v$ contains $C_{p}$, a contradiction. Thus, together with (2) and $|C_{p}|\geq 6$, we have

[TABLE]

∎

Claim 5**.**

Let $S=\{u_{1},u_{2},u_{s-1},u_{s},x^{*},y^{*}\}$. Then there exists a $C^{*}_{q}$ with $q\neq p$ and $1\leq q\leq k-1$ such that

[TABLE]

and $|C^{*}_{q}|=|C_{q}|$.

Proof.

By Claim 4, we can get that

[TABLE]

If $\sum_{z\in S}d(z,C^{*}_{i})\leq 3|C_{i}|-6$ for all $i$ with $1\leq i\leq k-1$ and $i\neq p$, then we have

[TABLE]

This contradicts the condition $\delta(G)\geq n/2-k+1$. Hence, there exists a $C^{*}_{q}$ for some $q$ with $1\leq q\leq k-1$ and $q\neq p$ such that

[TABLE]

Then, at least one of the following holds.

[TABLE]

[TABLE]

and

[TABLE]

If $d(u_{1},C^{*}_{q})+d(u_{s},C^{*}_{q})\geq|C_{q}|-1$ or $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})\geq|C_{q}|-1$, then by Claim 2 we have $|C^{*}_{q}|=|C_{q}|$. If $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})\geq|C_{q}|-1$, then $d(x^{*},C^{*}_{q})=|C_{q}|/2$ or $d(y^{*},C^{*}_{q})=|C_{q}|/2$. Since $C^{*}_{p}+u_{1}-x^{*}$ or $C^{*}_{p}+u_{s}-y^{*}$ contains $C_{p}$, we can prove $|C^{*}_{q}|=|C_{q}|$ as in Claim 1. ∎

In order to complete the proof of Theorem 3, we will show that $G[V(G_{2})\cup V(C^{*}_{p})\cup V(C^{*}_{q})$ contains a subdivision of $C_{k}\cup C_{p}\cup C_{q}$, which contradicts our assumption that $G$ contains no a subdivision of $H$. Note that $d(u_{1},C^{*}_{q})+d(u_{s},C^{*}_{q})\leq|C_{q}|$, $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})\leq|C_{q}|$ and $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})\leq|C_{q}|$.

We consider the two cases as follows.

Case 1. $d(u_{1},C^{*}_{q})+d(u_{s},C^{*}_{q})\geq|C_{q}|-1$.

Suppose that $d(u_{1},C^{*}_{q})+d(u_{s},C^{*}_{q})=|C_{q}|$. Then by Claim 5 we have $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})\geq|C_{q}|-5\geq 1$. This implies that either $u_{1}$ and $u_{s-1}$ have at least one common neighbor in $C^{*}_{q}$ or $u_{2}$ and $u_{s}$ have at least one common neighbor in $C^{*}_{q}$. By the symmetry, we assume that $v\in V(C^{*}_{q})$ with $u_{1}v,u_{s-1}v\in E(G)$. It is easy to see that $G_{2}+v-u_{s}$ contains a subdivision of $C_{k}$ and $C^{*}_{q}+u_{s}-v$ contains a subdivision of $C_{q}$.

Suppose that $d(u_{1},C^{*}_{q})+d(u_{s},C^{*}_{q})=|C_{q}|-1$. By the symmetry, we assume that $d(u_{1},C^{*}_{q})=|C_{q}|/2$ and $d(u_{s},C^{*}_{q})=|C_{q}|/2-1$. Let $v$ be the vertex in $V(C^{*}_{q})\cap X$ with $u_{s}v\not\in E(G)$. If $|C_{q}|\geq 8$, then by Claim 5 we have $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})\geq|C_{q}|-4\geq 4$. So, we can deduce that either $u_{2}$ has at least one neighbor in $(V(C^{*}_{q})\setminus\{v\})\cap X$ or $u_{s-1}$ has at least one neighbor in $(V(C^{*}_{q})\setminus\{v^{+},v^{-}\})\cap Y$. If $u_{2}$ has at least one neighbor in $(V(C^{*}_{q})\setminus\{v\})\cap X$, say $v_{x}$, then $G_{2}+v_{x}-u_{1}$ contains a subdivision of $C_{k}$ and $C^{*}_{q}+u_{1}-v_{x}$ contains a subdivision of $C_{q}$. If $u_{s-1}$ has at least one neighbor in $(V(C^{*}_{q})\setminus\{v^{+},v^{-}\})\cap Y$, say $v_{y}$, then $G_{2}+v_{y}-u_{s}$ contains a subdivision of $C_{k}$ and $C^{*}_{q}+u_{s}-v_{y}$ contains a subdivision of $C_{q}$. Thus, we have $|C_{q}|=6$. If $u_{2}$ or $u_{s-1}$ has a neighbor in $V(C^{*}_{q})\setminus\{v^{-},v,v^{+}\}$, then the proof is as same as that of the case $|C_{q}|\geq 8$. We may assume that $N(u_{2},C^{*}_{q})\cup N(u_{s-1},C^{*}_{q})\subseteq\{v^{-},v,v^{+}\}$. Since $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})\geq|C_{q}|-4=2$, we have $u_{s-1}v^{-}\in E(G)$ or $u_{s-1}v^{+}\in E(G)$. Without loss of generality, we assume that $u_{s-1}v^{+}\in E(G)$. If $u_{2}v\not\in E(G)$, then $d(u_{2},C^{*}_{q})=0$ and $d(u_{s-1},C^{*}_{q})=|C_{q}|-4$. By Claim 5 we have $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})=|C_{q}|$. It follows from the choice of $y^{*}$ in $C^{*}_{p}$ that $G_{2}+v^{+}-u_{s}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{s}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+y^{*}-v^{+}$ contains a subdivision of $C_{q}$. If $u_{2}v\in E(G)$, then we can deduce that $v$ is adjacent to every vertex in $V(C^{*}_{q})\cap Y$. If not, then $P+v-u_{1}$ is a path of order $|P|$ and $C^{*}_{q}+u_{1}-v$ contains a subdivision of $C_{q}$ such that $|C^{*}_{q}+u_{1}-v|=|C^{*}_{q}|$ and $|E(G[V(C^{*}_{q}+u_{1}-v)])|>|E(G[V(C^{*}_{q})])|$, which contradicts the choice of $C^{*}_{1},C^{*}_{2},\ldots,C^{*}_{k-1}$ in (iii). Then it is not difficult to see that $G_{2}+v^{+}-u_{s}$ contains a subdivision of $C_{k}$ and $C^{*}_{q}+u_{s}-v^{+}$ contains a subdivision of $C_{q}$.

Case 2. $d(u_{1},C^{*}_{q})+d(u_{s},C^{*}_{q})\leq|C_{q}|-2$.

We will consider three subcases as follows.

Case 2.1. $d(u_{1},C^{*}_{q})+d(u_{s},C^{*}_{q})\leq|C_{q}|-4$.

By Claim 5 we have

[TABLE]

This implies that either $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})=|C_{q}|$ and $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})\geq|C_{q}|-1$ or $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})\geq|C_{q}|-1$ and $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})=|C_{q}|$. If $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})=|C_{q}|$ and $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})\geq|C_{q}|-1$, then there exists an edge $v_{x}v_{y}\in E(C^{*}_{q})$ such that $x^{*}$ is adjacent to every vertex in $(V(C^{*}_{q})\setminus\{v_{y}\})\cap Y$ and $y^{*}$ is adjacent to every vertex in $(V(C^{*}_{q})\setminus\{v_{x}\})\cap X$. Thus we can get that $G_{2}+v_{x}+v_{y}-u_{1}-u_{s}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{1}+u_{s}-x^{*}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+x^{*}+y^{*}-v_{x}-v_{y}$ contains a subdivision of $C_{q}$. The case $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})\geq|C_{q}|-1$ and $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})=|C_{q}|$ can be proved similarly.

Case 2.2. $d(u_{1},C^{*}_{q})+d(u_{s},C^{*}_{q})=|C_{q}|-3$.

By Claim 5, we have $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})+d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})\geq 2|C_{q}|-2$. This implies that $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})=|C_{q}|$, $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})=|C_{q}|$, or $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})=|C_{q}|-1$ and $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})=|C_{q}|-1$. We consider three subcases as follows.

Case 2.2.1. $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})=|C_{q}|$.

Then by Claim 5 we have $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})\geq|C_{q}|-2>|C_{q}|/2$. It is not difficult to see that there exists an edge $v_{x}v_{y}\in E(C^{*}_{q})$ with $u_{2}v_{x},u_{s-1}v_{y}\in E(G)$ such that $G_{2}+v_{x}+v_{y}-u_{1}-u_{s}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{1}+u_{s}-x^{*}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+x^{*}+y^{*}-v_{x}-v_{y}$ contains a subdivision of $C_{q}$.

Case 2.2.2. $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})=|C_{q}|$.

Then by Claim 5 we have $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})\geq|C_{q}|-2$.

First, we assume that $|C_{q}|\geq 8$. Then we can see that $d(u_{1},C^{*}_{q})\geq 1$. If $d(y^{*},C^{*}_{q})\geq|C_{q}|/2-1$, then we can deduce that there exists an edge $v_{x}v_{y}\in E(C^{*}_{q})$ with $u_{2}v_{x},u_{s-1}v_{y}\in E(G)$ such that $G_{2}+v_{x}+v_{y}-u_{1}-u_{s}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{s}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+u_{1}+y^{*}-v_{x}-v_{y}$ contains a subdivision of $C_{q}$. If $d(y^{*},C^{*}_{q})=|C_{q}|/2-2$, then $d(x^{*},C^{*}_{q})=|C_{q}|/2$ and we can also deduce that there exists an edge $v_{x}v_{y}\in E(C^{*}_{q})$ with $u_{2}v_{x},u_{s-1}v_{y}\in E(G)$ such that $G_{2}+v_{x}+v_{y}-u_{1}-u_{s}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{1}+u_{s}-x^{*}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+x^{*}+y^{*}-v_{x}-v_{y}$ contains a subdivision of $C_{q}$.

Next, we assume that $|C_{q}|=6$. If $d(u_{1},C^{*}_{q})=3$, then by $d(y^{*},C^{*}_{q})\geq 1$ and $u_{1}y^{*}\in E(G)$ we can deduce that there exists an edge $v_{x}v_{y}\in E(C^{*}_{q})$ such that $u_{2}v_{x},u_{s-1}v_{y}\in E(G)$ and $G_{2}+v_{x}+v_{y}-u_{1}-u_{2}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{s}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+u_{1}+y^{*}-v_{x}-v_{y}$ contains a subdivision of $C_{q}$. So we assume that $d(u_{1},C^{*}_{q})\leq 2$. Then $d(u_{s},C^{*}_{q})\geq 1$. Let $v^{\prime}_{x}$ be a vertex in $C^{*}_{q}$ with $u_{s}v^{\prime}_{x}\in E(G)$. We claim that if $d(x^{*},C^{*}_{q})=3$, then $x^{*}y^{*}\in E(G)$. If $x^{*}y^{*}\not\in E(G)$, then we can get that $P+v^{\prime}_{x}-u_{1}$ is a path of order $|P|$, $C^{*}_{q}+x^{*}-v^{\prime}_{x}$ contains a subdivision of $C_{q}$ with $|C^{*}_{q}+x^{*}-v^{\prime}_{x}|=|C_{q}|$ and $|E(G[V(C^{*}_{q}+x^{*}-v^{\prime}_{x})])|\geq|E(G[V(C^{*}_{q})])|$, and $C^{*}_{p}+u_{1}-x^{*}$ contains a subdivision of $C_{p}$ with $|C^{*}_{p}+u_{1}-x^{*}|=|C_{p}|$ and $|E(G[V(C^{*}_{p}+u_{1}-x^{*})])|>|E(G[V(C^{*}_{p})])|$, which contradicts the choice of $C^{*}_{1},C^{*}_{2},\ldots,C^{*}_{k-1}$ in (iii). Thus, if $d(x^{*},C^{*}_{q})=3$, then by $d(y^{*},C^{*}_{q})\geq 1$, it is not difficult to see that there exists an edge $v_{x}v_{y}\in E(C^{*}_{q})$ such that $u_{2}v_{x},u_{s-1}v_{y}\in E(G)$ and $G_{2}+v_{x}+v_{y}-u_{1}-u_{2}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{1}+u_{s}-x^{*}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+x^{*}+y^{*}-v_{x}-v_{y}$ contains a subdivision of $C_{q}$. If $d(y^{*},C^{*}_{q})=3$ and $d(u_{1},C^{*}_{q})\geq 1$, then we can also obtain that there exists an edge $v_{x}v_{y}\in E(C^{*}_{q})$ such that $u_{2}v_{x},u_{s-1}v_{y}\in E(G)$ and $G_{2}+v_{x}+v_{y}-u_{1}-u_{2}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{s}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+u_{1}+y^{*}-v_{x}-v_{y}$ contains a subdivision of $C_{q}$. If $d(y^{*},C^{*}_{q})=3$ and $d(u_{s},C^{*}_{q})=3$, then we claim that $x^{*}y^{*}\in E(G)$. If $x^{*}y^{*}\not\in E(G)$, then we can get that $P+v^{\prime}_{y}+x^{*}-u_{1}-u_{s}$ is a path of order $|P|$, where $v^{\prime}_{y}\in N(x^{*},C^{*}_{q})$, $C^{*}_{q}+u_{s}-v^{\prime}_{y}$ contains a subdivision of $C_{q}$ with $|C^{*}_{q}+u_{s}-v^{\prime}_{y}|=|C_{q}|$ and $|E(G[V(C^{*}_{q}+u_{s}-v^{\prime}_{y})])|\geq|E(G[V(C^{*}_{q})])|$, and $C^{*}_{p}+u_{1}-x^{*}$ contains a subdivision of $C_{p}$ with $|C^{*}_{p}+u_{1}-x^{*}|=|C_{p}|$ and $|E(G[V(C^{*}_{p}+u_{1}-x^{*})])|>|E(G[V(C^{*}_{p})])|$, which contradicts the choice of $C^{*}_{1},C^{*}_{2},\ldots,C^{*}_{k-1}$ in (iii). Note that $d(x^{*},C^{*}_{q})\geq 1$. So, there exists an edge $v_{x}v_{y}\in E(C^{*}_{q})$ such that $u_{2}v_{x},u_{s-1}v_{y}\in E(G)$ and $G_{2}+v_{x}+v_{y}-u_{1}-u_{2}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{1}+u_{s}-x^{*}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+x^{*}+y^{*}-v_{x}-v_{y}$ contains a subdivision of $C_{q}$. We may assume that $d(x^{*},C^{*}_{q})=d(y^{*},C^{*}_{q})=2$. Let $C^{*}_{q}=v_{1}v_{2}v_{3}v_{4}v_{5}v_{6}v_{1}$ with $v_{1}\in X$. Without loss of generality, we suppose that $v_{2}x^{*},v_{4}x^{*}\in E(G)$. If $v_{1}y^{*},v_{3}y^{*}\in E(G)$ or $v_{3}y^{*},v_{5}y^{*}\in E(G)$, by the symmetry, say $v_{1}y^{*},v_{3}y^{*}\in E(G)$, then $G_{2}+v_{5}+v_{6}-u_{1}-u_{2}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{1}+u_{s}-x^{*}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+x^{*}+y^{*}-v_{5}-v_{6}$ contains a subdivision of $C_{q}$. Thus let $v_{1}y^{*},v_{5}y^{*}\in E(G)$. If $u_{1}v_{2}$ or $u_{1}v_{4}\in E(G)$, by the symmetry, say $u_{1}u_{2}\in E(G)$, then $G_{2}+v_{3}+v_{4}-u_{1}-u_{2}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{s}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+u_{1}+y^{*}-v_{3}-v_{4}$ contains a subdivision of $C_{q}$. This implies that $d(u_{1},C^{*}_{q})\leq 1$ and $d(u_{s},C^{*}_{q})\geq 2$. If $u_{1}v_{6},u_{s}v_{1},u_{s}v_{5}\in E(G)$, then we can see that $G_{2}+v_{6}-u_{s}$ contains a subdivision of $C_{k}$ and $C^{*}_{q}+u_{s}-v_{6}$ contains a subdivision of $C_{q}$. Thus we assume that $u_{s}v_{1},u_{s}v_{3}\in E(G)$ or $u_{s}v_{3},u_{s}v_{5}\in E(G)$. By the symmetry, we suppose that $u_{s}v_{1},u_{s}v_{3}\in E(G)$. Then it is not difficult to see that $G_{2}+v_{5}+v_{6}-u_{1}-u_{2}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{1}-x^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+u_{s}+x^{*}-v_{5}-v_{6}$ contains a subdivision of $C_{q}$.

Case 2.2.3. $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})=|C_{q}|-1$ and $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})=|C_{q}|-1$.

If $|C_{q}|\geq 8$, or $|C_{q}|=6$ and $N(x^{*},C^{*}_{q})\cup N(y^{*},C^{*}_{q})\neq N(u_{2},C^{*}_{q})\cup N(u_{s-1},C^{*}_{q})$, then it is easy to deduce that there exists an edge $v_{x}v_{y}\in E(C^{*}_{q})$ with $u_{2}v_{x},u_{s-1}v_{y}\in E(G)$ such that $G_{2}+v_{x}+v_{y}-u_{1}-u_{s}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{1}+u_{s}-x^{*}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+x^{*}+y^{*}-v_{x}-v_{y}$ contains a subdivision of $C_{q}$. Now suppose that $|C_{q}|=6$ and $N(x^{*},C^{*}_{q})\cup N(y^{*},C^{*}_{q})=N(u_{2},C^{*}_{q})\cup N(u_{s-1},C^{*}_{q})$. Let $C^{*}_{q}=v_{1}v_{2}v_{3}v_{4}v_{5}v_{6}v_{1}$ with $v_{1}\in X$. Without loss of generality, we assume that $u_{2}v_{1},v_{1}y^{*}\not\in E(G)$. Then we have $u_{2}v_{3},u_{2}v_{5},v_{3}y^{*},v_{5}y^{*}\in E(G)$. If $d(u_{1},C^{*}_{q})=3$, then we can obtain that $G_{2}+v_{3}+v_{4}-u_{1}-u_{s}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{s}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+u_{1}+y^{*}-v_{3}-v_{4}$ contains a subdivision of $C_{q}$. If $d(u_{1},C^{*}_{q})\leq 2$, then $d(u_{s},C^{*}_{q})\geq 1$, say $v_{x}\in V(C^{*}_{q})$ with $u_{s}v_{x}\in E(G)$. We claim that $x^{*}y^{*}\in E(G)$. If not, then by $d(x^{*},C^{*}_{q})=3$ we can deduce that $P+v_{x}-u_{1}$ is a path of order $|P|$, $C^{*}_{q}+x^{*}-v_{x}$ contains a subdivision of $C_{q}$ with $|C^{*}_{q}+x^{*}-v_{x}|=|C_{q}|$ and $|E(G[V(C^{*}_{q}+x^{*}-v_{x})])|\geq|E(G[V(C^{*}_{q})])|$, and $C^{*}_{p}+u_{1}-x^{*}$ contains a subdivision of $C_{p}$ with $|C^{*}_{p}+u_{1}-x^{*}|=|C_{p}|$ and $|E(G[V(C^{*}_{p}+u_{1}-x^{*})])|>|E(G[V(C^{*}_{p})])|$, which contradicts the choice of $C^{*}_{1},C^{*}_{2},\ldots,C^{*}_{k-1}$ in (iii). Thus we can see that $G_{2}+v_{3}+v_{4}-u_{1}-u_{s}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{1}+u_{s}-x^{*}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+x^{*}+y^{*}-v_{3}-v_{4}$ contains a subdivision of $C_{q}$.

Case 2.3. $d(u_{1},C^{*}_{q})+d(u_{s},C^{*}_{q})=|C_{q}|-2$.

By the symmetry, we let $d(u_{1},C^{*}_{q})\geq|C^{*}_{q}|/2-1$. Note that $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})\leq|C_{q}|$. It follows from Claim 5 that $|C_{q}|-3\leq d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})\leq|C_{q}|$.

First, we assume that $|C_{q}|\geq 8$. If $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})\geq|C_{q}|-1$ and $d(u_{1},C^{*}_{q})=|C^{*}_{q}|/2$, then $d(u_{s},C^{*}_{q})=|C^{*}_{q}|/2-2\geq 2$ and there exists at least one vertex $v_{x}$ in $V(C^{*}_{q})\cap X$ such that $u_{2}v_{x},u_{s}v_{x}\in E(G)$. So we can obtain that $G_{2}+v_{x}-u_{1}$ contains a subdivision of $C_{k}$ and $C^{*}_{q}+u_{1}-v_{x}$ contains a subdivision of $C_{q}$. If $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})\geq|C_{q}|-1$ and $d(u_{1},C^{*}_{q})=|C^{*}_{q}|/2-1\geq 3$, then $d(u_{s},C^{*}_{q})=|C^{*}_{q}|/2-1\geq 3$. By the symmetry we let $d(u_{2},C^{*}_{q})=|C_{q}|/2$. Then there exists at least one vertex $v_{x}$ in $V(C^{*}_{q})\cap X$ such that $u_{2}v_{x},u_{s}v_{x},u_{1}v^{-}_{x},u_{1}v^{+}_{x}\in E(G)$. It is easy to see that $G_{2}+v_{x}-u_{1}$ contains a subdivision of $C_{k}$ and $C^{*}_{q}+u_{1}-v_{x}$ contains a subdivision of $C_{q}$. If $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})=|C_{q}|-3$, then by $|C_{q}|-3>|C^{*}_{q}|/2$, there exists at least an edge $v_{x}v_{y}\in E(C^{*}_{q})$ such that $u_{2}v_{x},u_{s-1}v_{y}\in E(G)$. By Claim 5, we can know that $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})=|C_{q}|$. It follows from the choices of $x^{*},y^{*}$ in $C^{*}_{p}$ that $G_{2}+v_{x}+v_{y}-u_{1}-u_{2}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{1}+u_{2}-x^{*}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+x^{*}+y^{*}-v_{x}-v_{y}$ contains a subdivision of $C_{q}$. If $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})=|C_{q}|-2$, then by Claim 5 we have $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})\geq|C_{q}|-1$. Since $|C_{q}|-2-|C^{*}_{q}|/2\geq 2$, we can get that there exists at least an edge $v_{x}v^{+}_{x}\in E(C^{*}_{q})$ such that $u_{2}v_{x},u_{s-1}v^{+}_{x},v^{-}_{x}x^{*},v^{++}_{x}y^{*}\in E(G)$. Further, we have $d(x^{*},C^{*}_{q}-v_{x}-v^{+}_{x})+d(y^{*},C^{*}_{q}-v_{x}-v^{+}_{x})\geq|C^{*}_{q}-v_{x}-v^{+}_{x}|-1$. Thus we can obtain that $G_{2}+v_{x}+v^{+}_{x}-u_{1}-u_{2}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{1}+u_{2}-x^{*}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+x^{*}+y^{*}-v_{x}-v^{+}_{x}$ contains a subdivision of $C_{q}$.

Next, we assume that $|C_{q}|=6$. Let $C^{*}_{q}=v_{1}v_{2}v_{3}v_{4}v_{5}v_{6}v_{1}$ with $v_{1}\in X$.

Suppose that $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})\geq|C_{q}|-1$ and $d(u_{1},C^{*}_{q})=|C^{*}_{q}|/2=3$. If there exists a vertex $v_{x}$ in $V(C^{*}_{q})\cap X$ such that $u_{2}v_{x},u_{s}v_{x}\in E(G)$, then we can obtain that $G_{2}+v_{x}-u_{1}$ contains a subdivision of $C_{k}$ and $C^{*}_{q}+u_{1}-v_{x}$ contains a subdivision of $C_{q}$. Note that $d(u_{s},C^{*}_{q})=1$. So we may assume that $d(u_{2},C^{*}_{q})=|C_{q}|/2-1$ and $N(u_{2},C^{*}_{q})\cap N(u_{s},C^{*}_{q})=\emptyset$. Without loss of generality, we let $u_{s}v_{1},u_{2}v_{3},u_{2}v_{5}\in E(G)$. We claim that $v_{1}v_{4}\in E(G)$. If not, then $P+v_{1}-u_{1}$ is a path of order $|P|$ and $C^{*}_{q}+u_{1}-v_{1}$ contains a subdivision of $C_{q}$ with $|C^{*}_{q}+u_{1}-v_{1}|=|C_{q}|$ and $|E(G[V(C^{*}_{q}+u_{1}-v_{1})])|>|E(G[V(C^{*}_{q})])|$, which contradicts the choice of $C^{*}_{1},C^{*}_{2},\ldots,C^{*}_{k-1}$ in (iii). By Claim 5 we have $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})\geq|C_{q}|-2=4$. If $y^{*}$ has at least two neighbors in $C^{*}_{q}$, then by $d(u_{1},C^{*}_{q})=|C^{*}_{q}|/2,d(u_{s-1},C^{*}_{q})=|C^{*}_{q}|/2$ and the choice of $y^{*}$ in $C^{*}_{p}$, there exists a vertex $v_{y}$ in $C^{*}_{q}$ such that $G_{2}+v_{y}-u_{s}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{s}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+y^{*}-v_{y}$ contains a subdivision of $C_{q}$. So we have $d(x^{*},C^{*}_{q})=3$ and $d(y^{*},C^{*}_{q})=1$. We claim that $x^{*}y^{*}\in E(G)$. If not, then $P+v_{1}-u_{1}$ is a path of order $|P|$, $C^{*}_{q}+x^{*}-v_{1}$ contains a subdivision of $C_{q}$ with $|C^{*}_{q}+x^{*}-v_{1}|=|C_{q}|$ and $|E(G[V(C^{*}_{q}+x^{*}-v_{1})])|=|E(G[V(C^{*}_{q})])|$, and $C^{*}_{p}+u_{1}-x^{*}$ contains a subdivision of $C_{p}$ with $|C^{*}_{p}+u_{1}-x^{*}|=|C_{p}|$ and $|E(G[V(C^{*}_{p}+u_{1}-x^{*})])|>|E(G[V(C^{*}_{p})])|$, which contradicts the choice of $C^{*}_{1},C^{*}_{2},\ldots,C^{*}_{k-1}$ in (iii). If $v_{1}y^{*}\in E(G)$, then we can get that $G_{2}+v_{2}+v_{3}-u_{1}-u_{s}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{1}+u_{s}-x^{*}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+x^{*}+y^{*}-v_{2}-v_{3}$ contains a subdivision of $C_{q}$. The cases $v_{3}y^{*}\in E(G)$ and $v_{5}y^{*}\in E(G)$ can be proved similarly.

Suppose that $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})\geq|C_{q}|-1$ and $d(u_{1},C^{*}_{q})=d(u_{s},C^{*}_{q})=|C^{*}_{q}|/2-1=2$. By the symmetry, we may assume that $d(u_{2},C^{*}_{q})=|C_{q}|/2=3$ and $d(u_{s-1},C^{*}_{q})\geq|C^{*}_{q}|/2-1=2$. Without loss of generality, we let $u_{1}v_{2},u_{1}v_{6},u_{s-1}v_{2}\in E(G)$. If $u_{s}v_{1}\in E(G)$, then it is easy to see that $G_{2}+v_{1}-u_{1}$ contains a subdivision of $C_{k}$ and $C^{*}_{q}+u_{1}-v_{1}$ contains a subdivision of $C_{q}$. Thus, we consider $u_{s}v_{1}\not\in E(G)$ and by $d(u_{s},C^{*}_{q})=|C^{*}_{q}|/2-1=2$, we have $u_{s}v_{3},u_{s}v_{5}\in E(G)$. We claim that $v_{1}v_{4}\in E(G)$. If not, then $P+v_{3}+v_{4}-u_{1}-u_{2}$ is a path of order $|P|$ and $C^{*}_{q}+u_{1}+u_{2}-v_{3}-v_{4}$ contains a subdivision of $C_{q}$ with $|C^{*}_{q}+u_{1}+u_{2}-v_{3}-v_{4}|=|C_{q}|$ and $|E(G[V(C^{*}_{q}+u_{1}+u_{2}-v_{3}-v_{4})])|>|E(G[V(C^{*}_{q})])|$, which contradicts the choice of $C^{*}_{1},C^{*}_{2},\ldots,C^{*}_{k-1}$ in (iii). It follows from Claim 5 that $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})\geq|C_{q}|-3=3$. If $v_{1}y^{*}$ or $v_{5}y^{*}\in E(G)$, then $G_{2}+v_{2}+v_{3}-u_{1}-u_{s}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{s}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+y^{*}+u_{1}-v_{2}-v_{3}$ contains a subdivision of $C_{q}$. If $v_{3}y^{*}\in E(G)$, then $G_{2}+v_{1}+v_{2}-u_{1}-u_{s}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{s}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+y^{*}+u_{1}-v_{1}-v_{2}$ contains a subdivision of $C_{q}$. So we let $d(y^{*},C^{*}_{q})=0$ and $d(x^{*},C^{*}_{q})=3$. Then we can see that $G_{2}+v_{1}+v_{2}-u_{1}-u_{s}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{1}-x^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+x^{*}+u_{s}-v_{1}-v_{2}$ contains a subdivision of $C_{q}$.

Suppose that $|C_{q}|-3\leq d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})\leq|C_{q}|-2$. From Claim 5, we can obtain that $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})\geq|C_{q}|-1$. First, we assume that $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})=|C_{q}|$. If there exists an edge $v_{x}v_{y}$ in $C^{*}_{q}$ with $u_{2}v_{x},u_{s-1}v_{y}\in E(G)$, then $G_{2}+v_{x}+v_{y}-u_{1}-u_{s}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{1}+u_{s}-x^{*}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+x^{*}+y^{*}-v_{x}-v_{y}$ contains a subdivision of $C_{q}$. So, we consider $d(u_{2},C^{*}_{q})=3$ or $d(u_{s-1},C^{*}_{q})=3$. By the symmetry, we let $d(u_{2},C^{*}_{q})=3$. Note that $d(u_{s},C^{*}_{q})>0$. This implies that there exists a vertex $v_{x}\in V(C^{*}_{q})$ such that $u_{2}v_{x},u_{s}v_{x}\in E(G)$. Then $G_{2}+v_{x}-u_{1}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{1}-x^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+x^{*}-v_{x}$ contains a subdivision of $C_{q}$. Next, we assume that $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})=|C_{q}|-1=5$. It follows from Claim 5 that $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})=|C_{q}|-2=4$. Since $d(x^{*},C^{*}_{q})+d(y^{*},C^{*}_{q})=|C_{q}|-1=5$, we have $d(x^{*},C^{*}_{q})=3$ or $d(y^{*},C^{*}_{q})=3$. If $d(x^{*},C^{*}_{q})=3$, then $d(y^{*},C^{*}_{q})=2$. We claim that $x^{*}y^{*}\in E(G)$. If not, then by $d(u_{s},C^{*}_{q})\geq 1$, say $u_{s}v_{1}\in E(G)$, we can get that $P+v_{1}-u_{1}$ is a path of order $|P|$, $C^{*}_{q}+x^{*}-v_{1}$ contains a subdivision of $C_{q}$ with $|C^{*}_{q}+x^{*}-v_{1}|=|C_{q}|$ and $|E(G[V(C^{*}_{q}+x^{*}-v_{1})])|\geq|E(G[V(C^{*}_{q})])|$, and $C^{*}_{p}+u_{1}-x^{*}$ contains a subdivision of $C_{p}$ with $|C^{*}_{p}+u_{1}-x^{*}|=|C_{p}|$ and $|E(G[V(C^{*}_{p}+u_{1}-x^{*})])|>|E(G[V(C^{*}_{p})])|$, which contradicts the choice of $C^{*}_{1},C^{*}_{2},\ldots,C^{*}_{k-1}$ in (iii). From $d(u_{2},C^{*}_{q})+d(u_{s-1},C^{*}_{q})=|C_{q}|-2=4$, we can deduce that there exists an edge $v_{x}v_{y}\in E(C^{*}_{q})$ such that $u_{2}v_{x},u_{s-1}v_{y}\in E(G)$ and $G_{2}+v_{x}+v_{y}-u_{1}-u_{s}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{1}+u_{s}-x^{*}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+x^{*}+y^{*}-v_{x}-v_{y}$ contains a subdivision of $C_{q}$. Now we let $d(y^{*},C^{*}_{q})=3$. If $d(u_{1},C^{*}_{q})\geq 2$ or $N(u_{s-1},C^{*}_{q})\setminus N(u_{1},C^{*}_{q})\neq\emptyset$, then we can obtain that there exists an edge $v_{x}v_{y}\in E(C^{*}_{q})$ such that $u_{2}v_{x},u_{s-1}v_{y}\in E(G)$ and $G_{2}+v_{x}+v_{y}-u_{1}-u_{s}$ contains a subdivision of $C_{k}$, $C^{*}_{p}+u_{s}-y^{*}$ contains a subdivision of $C_{p}$ and $C^{*}_{q}+u_{1}+y^{*}-v_{x}-v_{y}$ contains a subdivision of $C_{q}$. If $d(u_{1},C^{*}_{q})=d(u_{s-1},C^{*}_{q})=1$ and there exists the vertex $v_{y}\in V(C^{*}_{q})$ with $u_{1}v_{y},u_{s-1}v_{y}\in E(G)$, then by $d(u_{s},C^{*}_{q})=3$ we can get that $G_{2}+v_{y}-u_{s}$ contains a subdivision of $C_{k}$ and $C^{*}_{q}+u_{s}-v_{y}$ contains a subdivision of $C_{q}$.

The proof of Theorem 3 is complete. $\Box$