The Outer-Planar Anti-Ramsey Number of Matchings

Abstract

A subgraph H of an edge-colored graph G is called rainbow if all of its edges have different colors. Let $ar(G,H)$ denote the maximum positive integer t, such that there is a t-edge-colored graph G without any rainbow subgraph H. We denote by $k{K}_2$ a matching of size k and ${\mathcal{O}}_n$ the class of all maximal outer-planar graphs on n vertices, respectively. The outer-planar anti-Ramsey number of graph H, denoted by $ar({\mathcal{O}}_n,H)$, is defined as $max\left\{ar(O_n,H)\right|O_n\in {\mathcal{O}}_n\}$. It seems nontrivial to determine the exact values for $ar({\mathcal{O}}_n,H)$ because most maximal outer-planar graphs are asymmetry. In this paper, we obtain that $ar({\mathcal{O}}_n,k{K}_2)\le n+3k-8$ for all $n\ge 2k$ and $k\ge 6$, which improves the existing upper bound for $ar({\mathcal{O}}_n,k{K}_2)$, and prove that $ar({\mathcal{O}}_n,k{K}_2)=n+2k-5$ for $n=2k$ and $k\ge 5$. We also obtain that $ar({\mathcal{O}}_n,6K_2)=n+6$ for all $n\ge 29$.

1. Introduction

In this paper, all graphs considered are finite, simple and undirected. Let G be a graph with vertex set $V\left(G\right)$ and edge set $E\left(G\right)$. Let $e\left(G\right)$, $v\left(G\right)$ and $\delta \left(G\right)$ denote the number of edges, number of vertices and minimum degree of G, respectively. The circumference of graph G, denoted by $\ell \left(G\right)$, is the length of a longest cycle in G. Denote by $d_G\left(v\right)$ and $N_G\left(v\right)$ the degree and neighborhood of the vertex v in G respectively. For any subset $A\subseteq V\left(G\right)$, let $G\left[A\right]$ denote the subgraph of G induced by A, and $N_G\left(A\right)=\{v\in V\left(G\right)\backslash A|uv\in E\left(G\right),u\in A\}$. For a set B, we denote the cardinality of B by $\left|B\right|$. For two disjoint subsets $A_1$, $A_2$ of $V\left(G\right)$, let $e_{{}_G}(A_1,A_2)$ denote the number of edges in G satisfying one end in $A_1$ and the other in $A_2$. A graph G is called a planar graph if it can be drawn in the plane such that its edges intersect only at their ends, and such a drawing is called a planar embedding of G. For convenience, a planar embedding of G is still represented by G. A graph G is outer-planar if it admits a planar embedding such that all vertices lie on the boundary of its outer face. An outer-planar graph G is maximal if $G+uv$ is not outer-planar for any two non-adjacent vertices u and v of G. A graph G is bipartite if its vertex set can be partitioned into two subsets X and Y so that every edge has one end in X and the other in Y. We denote a bipartite graph G with bipartition $(X,Y)$ by $G[X,Y]$. If any two edges of M are not adjacent in G, where $M\subseteq E\left(G\right)$, then M is called a matching of graph G. The number of edges in a maximum matching of a graph G is called the matching number of G, denoted by $\alpha \left(G\right)$. Let M be a matching of graph G, if $v\left(G\right)=n$ and $\left|M\right|=\frac{n}{2}$, then M is called a perfect matching of G. A graph G is called factor-critical if $G-v$ contains a perfect matching for every vertex $v\in V\left(G\right)$. We call a graph G an H-minor if H may be obtained from G by means of a sequence of vertex deletions, edge deletions or edge contractions. A component of a graph G is odd component (even component) if the order of the component is odd (even). The number of odd components in G is denoted by $o\left(G\right)$. Let $G\cup H$ denote the vertex disjoint union of graphs G and H. Denote by $G+H$ the graph obtained from $G\cup H$ by adding all edges joining each vertex of G and each vertex of H. For a positive integer k and a graph G, denote by $kG$ the vertex disjoint union of k copies of G. For any positive integer t, let $\left[t\right]:=\{1,2,\dots ,t\}$. The terminology and notation used but undefined in this paper can be found in [1].

If a subgraph H of an edge-colored graph G contains no two edges of the same color, then we say that G contains a rainbow H. Let $K_n$, $P_n$ and $C_n$ be the complete graph, path and cycle on n vertices, respectively. The anti-Ramsey number of H, denoted by $ar(K_n,H)$, is the maximum positive integer t such that there is a t-edge-colored $K_n$ without any rainbow H. In 1975, Erdős et al. [2] introduced anti-Ramsey numbers, and showed that these are closely related to Turán numbers. In the following discussion, the subgraph induced by a matching is still called a matching, and let $k{K}_2$ denote a matching of size k. In 2004, Schiermeyer [3] considered the anti-Ramsey number of matchings and determined the exact values of $ar(K_n,k{K}_2)$ for all $k\ge 2$ and $n\ge 3k+3$. Chen et al. [4] also studied $ar(K_n,k{K}_2)$ and completely determined the exact values of the anti-Ramsey number of matchings. When replacing $K_n$ by other graph G, let $ar(G,H)$ denote the maximum positive integer t such that there is a t-edge-colored G without any rainbow H. The researchers studied $ar(G,k{K}_2)$ when G is a bipartite graph [5,6,7], complete split graph [8], hypergraph [9] and so on. For more results on anti-Ramsey numbers, we refer the readers to [10,11,12,13,14,15,16,17].

Let ${\mathcal{T}}_n$ be the family of all plane triangulations on n vertices. The planar anti-Ramsey number of H is denoted by $ar({\mathcal{T}}_n,H)=max\left\{ar(T_n,H)\right|T_n\in {\mathcal{T}}_n\}$. In 2014, $\mathrm{Jendrol}$’ et al. [18] investigated the planar anti-Ramsey number of $k{K}_2$, in which the upper and lower bounds of $ar({\mathcal{T}}_n,k{K}_2)$ for all $k\ge 5$ and $n\ge 2k$ were established, and the exact values of $ar({\mathcal{T}}_n,k{K}_2)$ for $2\le k\le 4$ and $n\ge 2k$ were determined. Qin et al. [19] improved the upper bound of $ar({\mathcal{T}}_n,k{K}_2)$ in [18] and determined the exact value of $ar({\mathcal{T}}_n,5K_2)$ for all $n\ge 11$. Later, Chen et al. [20] improved the upper and lower bounds of $ar({\mathcal{T}}_n,k{K}_2)$ for $k\ge 6$ and $n\ge 3k-6$ existing in [18,19], and determined the exact value of $ar({\mathcal{T}}_n,6K_2)$ for all $n\ge 30$. Recently, Qin et al. [21] determined the exact values of $ar({\mathcal{T}}_n,k{K}_2)$ for all $k\ge 7$ and $n\ge 9k+3$.

Let ${\mathcal{O}}_n$ be the family of all maximal outer-planar graphs on n vertices. For $n\ge 3$, let ${\mathcal{O}}_n^{-}$ (${\mathcal{O}}_n^{=}$) denote the family of all outer-planar graphs with n vertices and $2n-4$ ($2n-5$) edges. The outer-planar anti-Ramsey number of H is denoted by $ar({\mathcal{O}}_n,H)=max\left\{ar(O_n,H)\right|$ $O_n\in {\mathcal{O}}_n\}$. It seems non-trivial to determine the exact values for $ar({\mathcal{O}}_n,H)$ because most maximal outer-planar graphs are asymmetry. There are two lemmas about the properties of maximal outer-planar graphs as follows.

Lemma 1

([22]). Let $O_n$ be a maximal outer-planar graphs on n vertices. If $n\ge 3$, then $e\left(O_n\right)=2n-3$ and $\delta \left(O_n\right)\ge 2$.

Lemma 2

([22]). Any maximal outer-planar graph contains neither a $K_{2,3}$-minor nor a $K_4$-minor.

In 2018, Jin et al. [23] studied the outer-planar anti-Ramsey numbers of $k{K}_2$, which were further studied by Pei et al. [24] in 2022. We summarize their results as follows.

Theorem 1

([23]). Let n and k be positive integers. Then

(1)

$ar({\mathcal{O}}_n,2K_2)=\left\{\begin{array}{cc}3,\hfill & n=4;\hfill \\ 1,\hfill & n\ge 5.\hfill \end{array}\right.$

(2)

$ar({\mathcal{O}}_n,3K_2)=\left\{\begin{array}{cc}7,\hfill & n=6;\hfill \\ n,\hfill & n\ge 7.\hfill \end{array}\right.$

(3)

$ar({\mathcal{O}}_n,4K_2)=\left\{\begin{array}{cc}11,& n=8;\\ n+2,& n\ge 9.\end{array}\right.$

(4)

for all $k\ge 5$ and $n\ge 2k$, we have $n+2k-6\le ar({\mathcal{O}}_n,k{K}_2)\le n+14k-25$.

Theorem 2

([24]). Let n and k be positive integers. Then

(1)

for all $k\ge 2$ and $n\ge 3k-3$, we have $ar({\mathcal{O}}_n,k{K}_2)\le n+4k-9$.

(2)

for all $n\ge 15$, we have $ar({\mathcal{O}}_n,5K_2)=n+4$.

By Theorem 1, when $3\le k\le 4$, if $n=2k$, then $ar({\mathcal{O}}_n,k{K}_2)$ is the lower bound given by Theorem 1(4) plus 1; if $n\ge 2k+1$, then $ar({\mathcal{O}}_n,k{K}_2)$ is exactly the lower bound given by Theorem 1(4). By Theorem 2, $ar({\mathcal{O}}_n,k{K}_2)$ is exactly the lower bound given by Theorem 1(4) when $k=5$ and $n\ge 2k+5$.

2. Main Results

It is non-trivial to determine the exact values for $ar({\mathcal{O}}_n,k{K}_2)$ for all $n\ge 2k$. The previous best upper bound for $ar({\mathcal{O}}_n,k{K}_2)$ is $n+4k-9$. Here, we improve the existing upper bound of $ar({\mathcal{O}}_n,k{K}_2)$ to $n+3k-8$.

Theorem 3.

For all $n\ge 2k$ and $k\ge 6$, we have $ar({\mathcal{O}}_n,k{K}_2)\le n+3k-8$.

Also, we obtain that the exact value of $ar({\mathcal{O}}_n,k{K}_2)$ when $n=2k$, which is equal to the lower bound given by Theorem 1(4) plus 1.

Theorem 4.

For all $k\ge 5$ and $n=2k$, we have $ar({\mathcal{O}}_n,k{K}_2)=n+2k-5$.

Finally, we attain that the exact value of $ar({\mathcal{O}}_n,k{K}_2)$ for $k=6$ and $n\ge 2k+17$, which is exactly the lower bound given by Theorem 1(4).

Theorem 5.

For all $n\ge 29$, we have $ar({\mathcal{O}}_n,6K_2)=n+6$.

The following two lemmas are useful in the proofs of Theorems 3 and 5.

Lemma 3

(Tutte-Berge Lemma [25]). If G is a graph with n vertices, then there exists a subset $S\subset V\left(G\right)$ satisfying $\left|S\right|\le \alpha \left(G\right)$, such that $\alpha \left(G\right)=\frac{1}{2}(n-o(G-S)+|S\left|\right)$. Furthermore, each odd component of $G-S$ is factor-critical and each even component of $G-S$ has a perfect matching.

Lemma 4

([24]). Let $G=G[X,Y]$ be a bipartite outer-planar graph on n vertices. If $\left|Y\right|\ge \left|X\right|\ge 1$, then $e\left(G\right)\le n+\left|X\right|-2$.

3. Proof of Theorem 3

The outer-planar anti-Ramsey number is closely related to the outer-planar Turán number of graphs. The outer-planar Turán number of H, denoted by $e{x}_{op}(n,H)$, is the maximum number of edges of an outer-planar graph on n vertices that does not contain H as a subgraph. To get Theorem 3, we first prove the following two lemmas.

Lemma 5.

For all $n\ge v\left(H\right)$, $ar({\mathcal{O}}_n,H)\le e{x}_{op}(n,H)$.

Proof. 

Let $ar({\mathcal{O}}_n,H)=t$. Then there exists an $O_n\in {\mathcal{O}}_n$, such that $O_n$ does not contain any rainbow H under a given t-edge-coloring. Let $G\subset O_n$ be a rainbow spanning subgraph with t edges. Thus G is an outer-planar graph on n vertices that does not contain H as a subgraph. It follows that $e{x}_{op}(n,H)\ge t$. Therefore, $ar({\mathcal{O}}_n,H)\le e{x}_{op}(n,H)$ for all $n\ge v\left(H\right)$. □

Lemma 6.

For all $n\ge 2k$ and $k\ge 6$, $e{x}_{op}(n,k{K}_2)\le min\{2n-3,n+3k-8\}$.

Proof. 

The proof will be conducted by induction on n. Since $n\ge 12$, then $e{x}_{op}(n,k{K}_2)\le 2n-3$ by Lemma 1. Thus $e{x}_{op}(n,k{K}_2)\le 2n-3=min\{2n-3,n+3k-8\}$ when $2k\le n\le 3k-6$. Now we assume that $n\ge 3k-5$. Next we will prove that $e{x}_{op}(n,k{K}_2)\le n+3k-8$ for $k\ge 6$ and $n\ge 3k-5$ by contradiction. Suppose $e{x}_{op}(n,k{K}_2)\ge n+3k-7$. Then there exists an outer-planar graph G such that $v\left(G\right)=n$ and $e\left(G\right)\ge n+3k-7$, and G does not contain $k{K}_2$ as a subgraph. Notice that $\alpha \left(G\right)\le k-1$. By Lemma 3, there exists a subset $S\subset V\left(G\right)$ satisfying $\left|S\right|\le \alpha \left(G\right)\le k-1$, such that $o(G-S)=n+\left|S\right|-2\alpha \left(G\right)$. Let $s=\left|S\right|$ and $p=o(G-S)$. Then $s\le k-1$ and $p\ge n+s+2-2k$. Denote by $B_1,B_2,\dots ,B_p$ all the odd components of $G-S$. We may assume that $v\left(B_1\right)\ge v\left(B_2\right)\ge \cdots \ge v\left(B_p\right)$. Let $w=0$ when $v\left(B_1\right)=1$, otherwise let $w=max\left\{i\right|v\left(B_i\right)>1\}$. Let $V\left(B_j\right)=\left\{v_j\right\}$ for any $j>w$. Let $I=\{v_{w+1},v_{w+2},\dots ,v_p\}$. Since $n=v\left(G\right)\ge \left|S\right|+v\left(B_1\right)+v\left(B_2\right)+\cdots +v\left(B_p\right)\ge s+3w+p-w=2w+s+p\ge 2w+s+(n+s+2-2k)=n+2s+2w-2k+2$, then $w\le k-s-1$.

We first prove that $s\le 1$. Suppose $s\ge 2$. Then $\left|I\right|=p-w\ge (n+s+2-2k)-(k-s-1)=n+2s-3k+3\ge (3k-5)+2s-3k+3=2s-2\ge s=\left|S\right|$. Therefore, $e_{{}_G}(S,I)\le \left(\right|S|+\left|I\right|)+\left|S\right|-2=2s+p-w-2$ by Lemma 4. Since $s\ge 2$, then $v(G-I)\ge 2$. So $e(G-I)\le 2(n-(p-w\left)\right)-3=2n-2p+2w-3$. Therefore, $e\left(G\right)=e_{{}_G}(S,I)+e(G-I)\le (2s+p-w-2)+(2n-2p+2w-3)=2n+2s-p+w-5\le 2n+2s-(n+s+2-2k)+(k-s-1)-5=n+3k-8$. But $e\left(G\right)\ge n+3k-7$, a contradiction. Thus $s\le 1$.

Let $H_1,H_2,\dots ,H_{\ell }$ be all components of $G-S$, where $\ell \ge 1$. Then $v\left(H_i\right)\ge 1$ for any $i\in [\ell ]$. We next prove that $\ell =1$. Suppose $\ell \ge 2$. If there exists $j\in [\ell ]$, such that $v\left(H_j\right)=1$, then $e_{{}_G}(S,V\left(H_j\right))\le 1$ since $s\le 1$. Therefore, $G-V\left(H_j\right)$ is an outer-planar graph with $n-1$ vertices containing no $k{K}_2$, and $e(G-V\left(H_j\right))=e\left(G\right)-e_{{}_G}(S,V\left(H_j\right))\ge n+3k-7-1=n+3k-8>min\{2(n-1)-3,(n-1)+3k-8\}$. But $e{x}_{op}(n-1,k{K}_2)\le min\{2(n-1)-3,(n-1)+3k-8\}$ by induction hypothesis, a contradiction. Therefore, we have $v\left(H_i\right)\ge 2$ for any $i\in [\ell ]$. Then $p=w$, and $e\left(G[S\cup V\left(H_i\right)]\right)\le 2(s+v\left(H_i\right))-3$ for any $i\in [\ell ]$. Thus, $e\left(G\right)=e\left(G[S\cup V\left(H_1\right)]\right)+\cdots +e\left(G[S\cup V\left(H_{\ell }\right)]\right)\le 2(s+v\left(H_1\right))-3+\cdots +2(s+v\left(H_{\ell }\right))-3=2(n-s)+2\ell s-3\ell$. Since $\ell \ge 2$ and $s\le 1$, then $e\left(G\right)\le 2(n-s)+2\ell s-3\ell =2n-3+(2s-3)(\ell -1)\le 2n-3+(2s-3)=2n+2s-6$. On the other hand, we have $n\le 3k-2s-3$ since $n+s+2-2k\le p=w\le k-s-1$. Thus $e\left(G\right)\le 2n+2s-6\le n+2s-6+(3k-2s-3)=n+3k-9$. But $e\left(G\right)\ge n+3k-7$, a contradiction. Therefore, $\ell =1$. Then $G-S$ has only one component $H_1$. Since $k\ge 6$, then $n\ge 3k-5>2k$. Combining $s\le 1$, we have $v\left(H_1\right)\ge 2k$. Thus, $H_1$ must contain a $k{K}_2$ by Lemma 3, which contradicts to the fact that G contains no $k{K}_2$. Thus, $e{x}_{op}(n,k{K}_2)\le n+3k-8=min\{2n-3,n+3k-8\}$ when $n\ge 3k-5$ and $k\ge 6$. Therefore, $e{x}_{op}(n,k{K}_2)\le min\{2n-3,n+3k-8\}$ for all $n\ge 2k$ and $k\ge 6$. □

Now we prove Theorem 3.

Proof of Theorem 3.

Since $n\ge 2k$, then $ar({\mathcal{O}}_n,k{K}_2)\le e{x}_{op}(n,k{K}_2)$ by Lemma 5. By Lemma 6, $e{x}_{op}(n,k{K}_2)\le n+3k-8$ for all $n\ge 2k$ and $k\ge 6$. Therefore, $ar({\mathcal{O}}_n,k{K}_2)\le n+3k-8$ for all $n\ge 2k$ and $k\ge 6$. □

4. Proof of Theorem 4

By Theorem 1(1–3), we observe that the outer-planar anti-Ramsey number of $k{K}_2$ when $n=2k$ is different from the case when $n\ge 2k+1$. It is not hard to see that the outer-planar anti-Ramsey number of $k{K}_2$ when $n=2k$ is equal to the lower bound given in Theorem 1(4) plus 1 for $2\le k\le 4$. In this section, we will prove that it is also equal to the lower bound given in Theorem 1(4) plus 1 when $k\ge 5$.

Now we are ready to prove Theorem 4.

Proof of Theorem 4.

We will first prove $ar({\mathcal{O}}_n,k{K}_2)\ge n+2k-5$ for $k\ge 5$ and $n=2k$. Construct a graph $G^{*}$ as follows: choose a maximal outer-planar graph G on k vertices, and the vertices of the outer face in a planar embedding of G are $v_1,v_2,\dots ,v_k$ in order; add vertex set $\{u_1,u_2,\dots ,u_k\}$ such that $u_i$ is only adjacent to $v_i$ and $v_{i+1}$ for each $i\in \left[k\right]$ (here $v_{k+1}$ is identified as $v_1$). Then $G^{*}$ is an outer-planar graph with $2k$ vertices, and $e\left(G^{*}\right)=e\left(G\right)+2k=(2k-3)+2k=4k-3$ combining Lemma 1. Therefore, by the definition of maximal outer-planar graphs, $G^{*}$ is a maximal outer-planar graph on n vertices, where $n=2k$.

Suppose that H is any matching $k{K}_2$ of $G^{*}$. Then we have $v\in V\left(H\right)$ for any $v\in V\left(G^{*}\right)$ since $v\left(G^{*}\right)=2k$. Note that $N_{G^{*}}\left(u_i\right)=\{v_i,v_{i+1}\}$ for each $i\in \left[k\right]$. Then for $u_k\in V\left(G^{*}\right)$, we have $u_k{v}_k\in E\left(H\right)$ or $u_k{v}_1\in E\left(H\right)$. If $u_k{v}_k\in E\left(H\right)$, then $u_i{v}_i\in E\left(H\right)$, $i\in [k-1]$; If $u_k{v}_1\in E\left(H\right)$, then $u_i{v}_{i+1}\in E\left(H\right)$, $i\in [k-1]$. Therefore, either $E\left(H\right)=\{u_1{v}_1,u_2{v}_2,\dots ,u_k{v}_k\}$ or $E\left(H\right)=\{u_1{v}_2,u_2{v}_3,\dots ,u_{k-1}{v}_k,u_k{v}_1\}$.

Let $\phi$ be an edge-coloring of $G^{*}$ as follows: $\phi \left(u_1{v}_1\right)=\phi \left(u_2{v}_2\right)=1$, $\phi \left(u_1{v}_2\right)=\phi \left(u_2{v}_3\right)=2$, and color all the remaining edges of $G^{*}$ with different new colors. Then the number of colors used for $\phi$ is $(4k-3)-2=4k-5=n+2k-5$. Since H is not a rainbow $k{K}_2$ under the $(n+2k-5)$-edge-coloring $\phi$, then $G^{*}$ does not contain any rainbow $k{K}_2$. Therefore, $ar({\mathcal{O}}_n,k{K}_2)\ge n+2k-5$. The graph $G^{*}$ and the edges of coloring 1 and 2 under its $(n+2k-5)$-edge-coloring $\phi$ when $k=6$ are depicted in Figure 1.

Next we will prove $ar({\mathcal{O}}_n,k{K}_2)\le n+2k-5$ for $k\ge 5$ and $n=2k$ by contradiction. Suppose $ar({\mathcal{O}}_n,k{K}_2)\ge n+2k-4$. Then there exists an $O_n\in {\mathcal{O}}_n$, such that $O_n$ does not contain any rainbow $k{K}_2$ under a given t-edge-coloring, where $t\ge n+2k-4=2n-4$. Let $G^{\prime }$ be a rainbow spanning subgraph of $O_n$ and $e\left(G^{\prime }\right)=t$. By Lemma 1, $e\left(O_n\right)=2n-3$. Then $2n-4\le t\le 2n-3$. Thus either $G^{\prime }\in {\mathcal{O}}_n^{-}$ or $G^{\prime }\in {\mathcal{O}}_n$. Note that $O_n$ contains a cycle $C_n$, which means that $G^{\prime }$ contains a $P_n$. It follows that $O_n$ must contain a rainbow $k{K}_2$, a contradiction.

This completes the proof of Theorem 4. □

5. Proof of Theorem 5

It is easy to see, from the previous results, that the exact value of the outer-planar anti-Ramsey number of $k{K}_2$ is equal to the lower bound given in Theorem 1(4) when n is large and $3\le k\le 5$. It is natural to ask for whether it is also equal to the lower bound given in Theorem 1(4) when $k\ge 6$. We verify it is true when $k=6$.

Now we shall prove Theorem 5.

By Theorem 1(4), $ar({\mathcal{O}}_n,6K_2)\ge n+6$. Next it suffices to prove that $ar({\mathcal{O}}_n,6K_2)\le n+6$. By contradiction, suppose that $ar({\mathcal{O}}_n,6K_2)\ge n+7$. Then there exists an $O_n\in {\mathcal{O}}_n$, such that $O_n$ contains no rainbow $6K_2$ under an edge-coloring c with k colors, where $k\ge n+7$. It follows from Theorem 2(2) that $O_n$ contains a rainbow $5K_2$. Now let $G\subset O_n$ be a rainbow spanning subgraph with k edges which contains a $5K_2$. Then $\alpha \left(G\right)=5$. Thus by Lemma 3, there exists a subset $S\subset V\left(G\right)$ satisfying $\left|S\right|\le 5$ such that $o(G-S)=n+\left|S\right|-10$. Let $s=\left|S\right|$ and $p=o(G-S)$, we have $0\le s\le 5$ and $p=n+s-10$. Denote by $B_1,B_2,\dots ,B_p$ all the odd components of $G-S$. We may assume that $v\left(B_1\right)\ge v\left(B_2\right)\ge \cdots \ge v\left(B_p\right)$. Let $w=0$ when $v\left(B_1\right)=1$, otherwise let $w=max\left\{i\right|v\left(B_i\right)>1\}$. Let $V\left(B_j\right)=\left\{v_j\right\}$ for any $j>w$. Without loss of generality, we assume that $d_G\left(v_{w+1}\right)\ge \cdots \ge d_G\left(v_p\right)$. Let $I=\{v_{w+1},v_{w+2},\dots ,v_p\}$. Let $Q=V\left(B_1\right)\cup V\left(B_2\right)\cup \cdots \cup V\left(B_w\right)$, where $w\ge 1$; otherwise let $Q=\varnothing$. Let $R=V\left(G\right)-(S\cup I\cup Q)$. Let $n_{{}_I}^q=\left|\{v\in I:d_G\left(v\right)=q\}\right|$ and $n_{{}_I}^{q^{+}}=\left|\right\{v\in I:$$d_G\left(v\right)\ge q\left\}\right|$. For convenience, we replace $e_{{}_G}(A_1,A_2)$ by $e(A_1,A_2)$ in the following proof. We first present several useful claims, which shall be proved in Section 6.

Claim 1. If $M_1$ and $M_2$ are two edge-disjoint $5K_2$ of G, then

$$E\left(O_n\left[V\left(G\right)-V\left(M_1\cup M_2\right)\right]\right)=\varnothing .$$

Claim 2. $s\ge 1$. Especially, $s\ge 2$ when $R\ne \varnothing$.

Claim 3. $R=\varnothing$.

Claim 4. If $G\left[S\right]$ is a connected graph, $e\left(O_n\left[I^{\prime }\right]\right)=0$ and $|I^{\prime }|>2b$, then $O_n$ contains a $K_{2,3}$-minor.

Claim 5. $s\ge 2$.

Claim 6. $v\left(B_1\right)\le 5$.

Claim 7. $v\left(B_1\right)\ge 3$.

By Claim 3, we get $n=v\left(G\right)=v\left(B_1\right)+\cdots +v\left(B_p\right)+s=v\left(B_1\right)+\cdots +v\left(B_w\right)+n+2s-w-10$ when $w\ge 1$. So the following result holds for any $w\ge 1$:

$$v\left(B_1\right)+\cdots +v\left(B_w\right)=w+10-2s.$$

(1)

Let $b=w+n_{{}_I}^{2^{+}}$, and $I^{\prime }=\{v_{b+1},v_{b+2},\dots ,v_p\}$. By Claims 6 and 7, $3\le v\left(B_1\right)\le 5$. If $v\left(B_i\right)=3$, then $B_i\cong K_3$ by Lemma 3. Since $w\ge 1$, then by $\left(1\right)$, $v\left(B_1\right)+\cdots +v\left(B_w\right)=w+10-2s$. Combining $3w\le v\left(B_1\right)+\cdots +v\left(B_w\right)\le 5w$, we have $(5-s)/2\le w\le 5-s$. It follows from Claims 5 and 7 that $2\le s\le 4$. We next distinguish the following three cases to finish the proof of Theorem 5.

5.1. $s=2$

In this case, $p=n-8$ and $n_{{}_I}^{3^{+}}=0$. Since $(5-s)/2\le w\le 5-s$, we see $2\le w\le 3$. We consider the following two situations according to w.

Case A.1.$w=2$.

Then $v\left(B_1\right)=5$, $v\left(B_2\right)=3$ and $\left|I\right|=p-2=n-10$. Since $n_{{}_I}^2\le 2$, then $b\le 4$. By Lemma 1, we have $e(G-I)\le 2\times 10-3=17$. By Lemma 4, $e(S,I)\le \left(\right|I|+|S\left|\right)+\left|S\right|-2=n-8$, which implies that $e(G-I)=e\left(G\right)-e(S,I)\ge 15$. Therefore, $15\le e(G-I)\le 17$.

If $e(G-I)=17$, then $G-I\in {\mathcal{O}}_{10}$. Thus $N_G\left(V\left(B_1\right)\right)=N_G\left(V\left(B_2\right)\right)=S$. Therefore, $G\left[S\right]$ is connected and $G-I$ contains two edge-disjoint $5K_2$.

If $e(G-I)=16$, then $e(S,I)=e\left(G\right)-e(G-I)\ge n-9$. Thus $d_G\left(v_3\right)=2$. Then $G[S\cup Q\cup \left\{v_3\right\}]\in {\mathcal{O}}_{11}^{-}$, which implies that $G[S\cup Q\cup \left\{v_3\right\}]$ contains $P_{11}$. Therefore, $G-I^{\prime }$ contains two edge-disjoint $5K_2$. On the other hand, combining $d_G\left(v_3\right)=2$, we get $|N_G\left(V\left(B_i\right)\right)|\le 1$ for some $i\in \left[2\right]$, then $G\left[S\right]$ is connected.

If $e(G-I)=15$, then $e(S,I)\ge n-8$. Thus $d_G\left(v_3\right)=d_G\left(v_4\right)=2$. Then $G[S\cup Q\cup \{v_3,v_4\}]\in {\mathcal{O}}_{12}^{=}$, which implies that $G-I^{\prime }$ contains two edge-disjoint $5K_2$. On the other hand, combining $d_G\left(v_3\right)=d_G\left(v_4\right)=2$, we get $|N_G\left(V\left(B_i\right)\right)|\le 1$ for each $i\in \left[2\right]$, then $G\left[S\right]$ is connected.

From the above discussion of $e(G-I)$ and combining Claim 1, we have $e\left(O_n\left[I^{\prime }\right]\right)=0$. Since $n\ge 29$ and $b\le 4$, then $|I^{\prime }|=p-b\ge (n-8)-4>2b$. Therefore, by Claim 4, $O_n$ contains a $K_{2,3}$-minor, which contradicts to Lemma 2.

Case A.2.$w=3$.

Then $v\left(B_1\right)=v\left(B_2\right)=v\left(B_3\right)=3$ and $\left|I\right|=p-3=n-11$. Since $n_{{}_I}^2\le 2$, then $b\le 5$. By Lemma 2, there exists at least one $i\in \left[3\right]$ such that $|N_G\left(V\left(B_i\right)\right)|\le 1$. Thus combining Lemma 1, we have $e(G-I)\le 2\left(\right|Q|+s)-4=2\times 11-4=18$. By Lemma 4, we have $e(S,I)\le \left(\right|I|+|S\left|\right)+\left|S\right|-2=n-9$, which implies that $e(G-I)\ge 16$. Therefore, $16\le e(G-I)\le 18$.

If $e(G-I)=18$, then $G-I\in {\mathcal{O}}_{11}^{-}$, which implies that $G-I$ contains $P_{11}$. Therefore, $G-I$ contains two edge-disjoint $5K_2$. On the other hand, since there exists at least one $i\in \left[3\right]$ such that $|N_G\left(V\left(B_i\right)\right)|\le 1$, we have $G\left[S\right]$ is connected.

If $e(G-I)=17$, then $e(S,I)=e\left(G\right)-e(G-I)\ge n-10$. Thus $d_G\left(v_4\right)=2$. Therefore, $G[S\cup Q\cup \left\{v_4\right\}]\in {\mathcal{O}}_{12}^{=}$, which implies that $G-I^{\prime }$ contains two edge-disjoint $5K_2$. On the other hand, combining $d_G\left(v_4\right)=2$, we get that there exist at least two $i\in \left[3\right]$ such that $|N_G\left(V\left(B_i\right)\right)|\le 1$, thus $G\left[S\right]$ is connected.

If $e(G-I)=16$, then $e(S,I)\ge n-9$. Thus $d_G\left(v_4\right)=d_G\left(v_5\right)=2$. Therefore, $G[S\cup Q\cup \{v_4,v_5\}]$ is the graph obtained from a maximal outer-planar graph with 13 vertices by deleting 3 edges, then $G-I^{\prime }$ contains two edge-disjoint $5K_2$. On the other hand, combining $d_G\left(v_4\right)=d_G\left(v_5\right)=2$, we get $|N_G\left(V\left(B_i\right)\right)|\le 1$ for each $i\in \left[3\right]$, thus $G\left[S\right]$ is connected.

From the above discussion of $e(G-I)$ and combining Claim 1, we have $e\left(O_n\left[I^{\prime }\right]\right)=0$. Since $n\ge 29$ and $b\le 5$, then $|I^{\prime }|=p-b\ge (n-8)-5>2b$. Therefore, by Claim 4, $O_n$ contains a $K_{2,3}$-minor, which contradicts to Lemma 2.

5.2. $s=3$

In this case, $p=n-7$ and $n_{{}_I}^{4^{+}}=0$. Since $(5-s)/2\le w\le 5-s$, then $1\le w\le 2$. We consider the following two situations according to w.

Case B.1.$w=1$.

Then $v\left(B_1\right)=5$ and $\left|I\right|=p-1=n-8$. By Lemma 1, we have $e(G-I)\le 2\times 8-3=13$. By Lemma 4, $e(S,I)\le \left(\right|I|+|S\left|\right)+\left|S\right|-2=n-4$, which implies that $e(G-I)\ge 11$. Therefore, $11\le e(G-I)\le 13$.

If $e(G-I)=13$, then $G-I\in {\mathcal{O}}_8$. Thus $d_G\left(v_2\right)\le 2$ and $|N_G\left(V\left(B_1\right)\right)|\ge 2$. Since $e(S,I)=e\left(G\right)-e(G-I)\ge n-6$, then $d_G\left(v_2\right)=d_G\left(v_3\right)=2$. Thus $G[S\cup V\left(B_1\right)\cup \{v_2,v_3\}]\in {\mathcal{O}}_{10}$, which means that $G-I^{\prime }$ contains two edge-disjoint $5K_2$. On the other hand, we have $e(S,V\left(B_1\right))\le 4$ because $d_G\left(v_2\right)=d_G\left(v_3\right)=2$. So $e\left(G\left[S\right]\right)=e(G-I)-e(S,V\left(B_1\right))-e\left(B_1\right)\ge 13-4-7=2$, which implies that $G\left[S\right]$ is connected.

If $e(G-I)=12$, then $e(S,I)\ge n-5$. Thus either $d_G\left(v_2\right)=3$ and $d_G\left(v_3\right)=2$ or $d_G\left(v_2\right)=d_G\left(v_3\right)=d_G\left(v_4\right)=2$. Therefore, we have $G[S\cup V\left(B_1\right)\cup \{v_2,v_3\}]\in {\mathcal{O}}_{10}$; or $G[S\cup V\left(B_1\right)\cup \{v_2,v_3,v_4\}]\in {\mathcal{O}}_{11}^{-}$. Then we always get that $G-I^{\prime }$ contains two edge-disjoint $5K_2$. From the degree situation of the vertices of I in G, we have $e(S,V\left(B_1\right))\le 3$. So $e\left(G\left[S\right]\right)=e(G-I)-e(S,V\left(B_1\right))-e\left(B_1\right)\ge 12-3-7=2$, which implies that $G\left[S\right]$ is connected.

If $e(G-I)=11$, then $e(S,I)\ge n-4$. Thus either $d_G\left(v_2\right)=3$ and $d_G\left(v_3\right)=d_G\left(v_4\right)=2$ or $d_G\left(v_2\right)=d_G\left(v_3\right)=\cdots =d_G\left(v_5\right)=2$. Therefore, we have $G[S\cup V\left(B_1\right)\cup \{v_2,v_3,v_4\}]\in {\mathcal{O}}_{11}^{-}$; or $G[S\cup V\left(B_1\right)\cup \{v_2,v_3,\dots ,v_5\}]\in {\mathcal{O}}_{12}^{=}$. Then we always get that $G-I^{\prime }$ contains two edge-disjoint $5K_2$. From the degree situation of the vertices of I in G, we have $e(S,V\left(B_1\right))\le 2$. So $e\left(G\left[S\right]\right)=e(G-I)-e(S,V\left(B_1\right))-e\left(B_1\right)\ge 11-2-7=2$, which implies that $G\left[S\right]$ is connected.

From the above discussion of $e(G-I)$ and combining Claim 1, we have $e\left(O_n\left[I^{\prime }\right]\right)=0$. Since $s=3$, then $n_{{}_I}^2\le 4$. So $b\le 5$. Then $|I^{\prime }|=p-b\ge (n-7)-5>2b$ because $n\ge 29$ and $b\le 5$. Therefore, by Claim 4, $O_n$ contains a $K_{2,3}$-minor, which contradicts to Lemma 2.

Case B.2.$w=2$.

Then $v\left(B_1\right)=v\left(B_2\right)=3$ and $\left|I\right|=p-2=n-9$. By Lemma 1, we have $e(G-I)\le 2\times 9-3=15$. By Lemma 4, $e(S,I)\le \left(\right|I|+|S\left|\right)+\left|S\right|-2=n-5$, which implies that $e(G-I)=e\left(G\right)-e(S,I)\ge 12$. Therefore, $12\le e(G-I)\le 15$.

If $e(G-I)=15$, then $G-I\in {\mathcal{O}}_9$. Thus $d_G\left(v_3\right)\le 2$, $|N_G\left(V\left(B_1\right)\right)|\ge 2$ and $|N_G\left(V\left(B_2\right)\right)|\ge 2$. Since $e(S,I)=e\left(G\right)-e(G-I)\ge n-8$, then $d_G\left(v_3\right)=2$. Thus $G[S\cup Q\cup \left\{v_3\right\}]\in {\mathcal{O}}_{10}$, which implies that $G-I^{\prime }$ contains two edge-disjoint $5K_2$. On the other hand, we have $e(S,Q)\le 7$ because $d_G\left(v_3\right)=2$. Therefore, $e\left(G\left[S\right]\right)=e(G-I)-e(S,Q)-e\left(B_1\right)-e\left(B_2\right)\ge 15-7-3-3=2$, which means that $G\left[S\right]$ is connected.

If $e(G-I)=14$, then $e(S,I)\ge n-7$. Thus either $d_G\left(v_3\right)=3$ or $d_G\left(v_3\right)=d_G\left(v_4\right)=2$. Therefore, we have $G[S\cup Q\cup \left\{v_3\right\}]\in {\mathcal{O}}_{10}$; or $G[S\cup Q\cup \{v_3,v_4\}]\in {\mathcal{O}}_{11}^{-}$. Then we always get that $G-I^{\prime }$ contains two edge-disjoint $5K_2$. From the degree situation of the vertices of I in G, we have $e(S,Q)\le 6$. So $e\left(G\left[S\right]\right)=e(G-I)-e(S,Q)-e\left(B_1\right)-e\left(B_2\right)\ge 14-6-3-3=2$, which means that $G\left[S\right]$ is connected.

If $e(G-I)=13$, then $e(S,I)\ge n-6$. Thus either $d_G\left(v_3\right)=3$ and $d_G\left(v_4\right)=2$ or $d_G\left(v_3\right)=d_G\left(v_4\right)=d_G\left(v_5\right)=2$. Therefore, we have $G[S\cup Q\cup \{v_3,v_4\}]\in {\mathcal{O}}_{11}^{-}$; or $G[S\cup Q\cup \{v_3,v_4,v_5\}]\in {\mathcal{O}}_{12}^{=}$. Then we always get that $G-I^{\prime }$ contains two edge-disjoint $5K_2$. From the degree situation of the vertices of I in G, we have $e(S,Q)\le 5$. So $e\left(G\left[S\right]\right)=e(G-I)-e(S,Q)-e\left(B_1\right)-e\left(B_2\right)\ge 13-5-3-3=2$, which means that $G\left[S\right]$ is connected.

If $e(G-I)=12$, then $e(S,I)\ge n-5$. Thus either $d_G\left(v_3\right)=3$ and $d_G\left(v_4\right)=d_G\left(v_5\right)=2$ or $d_G\left(v_3\right)=d_G\left(v_4\right)=\cdots =d_G\left(v_6\right)=2$. Therefore, we have $G[S\cup Q\cup \{v_3,v_4,v_5\}]\in {\mathcal{O}}_{12}^{=}$; or $G[S\cup Q\cup \{v_3,v_4,\dots ,v_6\}]$ is the graph obtained from a maximal outer-planar graph with 13 vertices by deleting 3 edges. Then we always get that $G-I^{\prime }$ contains two edge-disjoint $5K_2$. From the degree situation of the vertices of I in G, we have $e(S,Q)\le 4$. So $e\left(G\left[S\right]\right)=e(G-I)-e(S,Q)-e\left(B_1\right)-e\left(B_2\right)\ge 12-4-3-3=2$, which means that $G\left[S\right]$ is connected.

From the above discussion of $e(G-I)$ and combining Claim 1, we have $e\left(O_n\left[I^{\prime }\right]\right)=0$. Since $n_{{}_I}^2\le 4$, we have $b\le 6$. Then $|I^{\prime }|=p-b\ge (n-7)-6>2b$ because $n\ge 29$ and $b\le 6$. Therefore, by Claim 4, $O_n$ contains a $K_{2,3}$-minor, which contradicts to Lemma 2.

5.3. $s=4$

In this case, $p=n-6$, $w=1$ and $n_{{}_I}^{5^{+}}=0$. Then $v\left(B_1\right)=3$ and $\left|I\right|=p-1=n-7$. Since $s=4$, we see $n_{{}_I}^2\le 6$. So $b\le 7$. Then we get $|I^{\prime }|=p-b\ge (n-6)-7>2b$ because $n\ge 29$ and $b\le 7$. By Lemma 1, $e(G-I)\le 2\times 7-3=11$. By Lemma 4, $e(S,I)\le \left(\right|I|+|S\left|\right)+\left|S\right|-2=n-1$, which implies that $e(G-I)=e\left(G\right)-e(S,I)\ge 8$. Therefore, $8\le e(G-I)\le 11$. We consider the following four situations according to $e(G-I)$.

Case C.1.$e(G-I)=11$.

Then $G-I\in {\mathcal{O}}_7$. Thus $d_G\left(v_2\right)\le 2$. Since $e(S,I)=e\left(G\right)-e(G-I)\ge n-4$, then $d_G\left(v_2\right)=d_G\left(v_3\right)=d_G\left(v_4\right)=2$. Thus $G[S\cup V\left(B_1\right)\cup \{v_2,v_3,v_4\}]\in {\mathcal{O}}_{10}$, which implies that $G-I^{\prime }$ contains two edge-disjoint $5K_2$. Therefore, by Claim 1, $e\left(O_n\left[I^{\prime }\right]\right)=0$. On the other hand, we have $e(S,V\left(B_1\right))\le 5$ because $d_G\left(v_2\right)=d_G\left(v_3\right)=d_G\left(v_4\right)=2$. Thus $e\left(G\left[S\right]\right)=e(G-I)-e(S,V\left(B_1\right))-e\left(B_1\right)\ge 11-5-3=3$. We next prove that $G\left[S\right]$ is connected. If $e\left(G\right[S\left]\right)=3$ and $G\left[S\right]$ contains a cycle, then $n_{{}_I}^{3^{+}}=0$ and $n_{{}_I}^2\le 2$ because $S\cup V\left(B_1\right)\in {\mathcal{O}}_7$. So $e(S,I)\le n-5$, which contradicts to $e(S,I)\ge n-4$. Therefore, either $e\left(G\right[S\left]\right)\ge 4$ or $e\left(G\right[S\left]\right)=3$ and $G\left[S\right]$ contains no cycle. Then we clearly get that $G\left[S\right]$ is connected. Thus, by Claim 4, $O_n$ contains a $K_{2,3}$-minor, which contradicts to Lemma 2.

Case C.2.$e(G-I)=10$.

Then $d_G\left(v_2\right)\le 3$ and $e(S,I)=e\left(G\right)-e(G-I)\ge n-3$. Thus either $d_G\left(v_2\right)=3$ and $d_G\left(v_3\right)=d_G\left(v_4\right)=2$ or $d_G\left(v_2\right)=d_G\left(v_3\right)=\cdots =d_G\left(v_5\right)=2$. Therefore, we have $G[S\cup V\left(B_1\right)\cup \{v_2,v_3,v_4\}]\in {\mathcal{O}}_{10}$; or $G[S\cup V\left(B_1\right)\cup \{v_2,v_3,\dots ,v_5\}]\in {\mathcal{O}}_{11}^{-}$. Then we always get that $G-I^{\prime }$ contains two edge-disjoint $5K_2$. Thus, by Claim 1, $e\left(O_n\left[I^{\prime }\right]\right)=0$. We next prove that $G\left[S\right]$ is connected. If $\ell \left(G\right[S\left]\right)=4$, then it is obvious that $G\left[S\right]$ is connected. If $\ell \left(G\right[S\left]\right)=3$, then we have $|N_G\left(V\left(B_1\right)\right)|\le 2$ combining Lemma 2, thus $e(S,V\left(B_1\right))\le 3$. Then $e\left(G\left[S\right]\right)=e(G-I)-e(S,V\left(B_1\right))-e\left(B_1\right)\ge 10-3-3=4$, which implies that $G\left[S\right]$ is connected. If $\ell \left(G\right[S\left]\right)\le 2$, that is, $G\left[S\right]$ contains no cycle, then we get $|N_G\left(V\left(B_1\right)\right)|\le 3$ by Lemma 2. Thus $e(S,V\left(B_1\right))\le 4$. So $e\left(G\left[S\right]\right)=e(G-I)-e(S,V\left(B_1\right))-e\left(B_1\right)\ge 10-4-3=3$, which means that $G\left[S\right]$ is connected.

Therefore, by Claim 4, $O_n$ contains a $K_{2,3}$-minor, which contradicts to Lemma 2.

Case C.3.$e(G-I)=9$.

Then $e(S,I)\ge n-2$. Combining Lemma 2, the degree situation of the vertices of I in G satisfies one of the following: (1) $d_G\left(v_2\right)=4,d_G\left(v_3\right)=d_G\left(v_4\right)=2$; (2) $d_G\left(v_2\right)=d_G\left(v_3\right)=3,d_G\left(v_4\right)=2$; (3) $d_G\left(v_2\right)=3,d_G\left(v_3\right)=d_G\left(v_4\right)=d_G\left(v_5\right)=2$; (4) $d_G\left(v_2\right)=d_G\left(v_3\right)=\cdots =d_G\left(v_6\right)=2$. Thus, we have $G[S\cup V\left(B_1\right)\cup \{v_2,v_3,v_4\}]\in {\mathcal{O}}_{10}$; or $G[S\cup V\left(B_1\right)\cup \{v_2,v_3,\dots ,v_5\}]\in {\mathcal{O}}_{11}^{-}$; or $G[S\cup V\left(B_1\right)\cup \{v_2,v_3,\dots ,v_6\}]\in {\mathcal{O}}_{12}^{=}$. Obviously, we always get that $G-I^{\prime }$ contains two edge-disjoint $5K_2$. Thus, by Claim 1, we have $e\left(O_n\left[I^{\prime }\right]\right)=0$.

We first claim that $\ell \left(G\right[S\left]\right)\le 3$. Since otherwise combining Lemma 2, we have $n_{{}_I}^{3^{+}}=0$ and $n_{{}_I}^2\le 4$, which means $e(S,I)\le n-3$. Next we will prove $G\left[S\right]$ is connected. If $\ell \left(G\right[S\left]\right)=3$, then one of (3)–(4) above is satisfied. By Lemma 2, we have $|N_G\left(V\left(B_1\right)\right)|\le 1$. Thus $e(S,V\left(B_1\right))\le 2$. So $e\left(G\left[S\right]\right)=e(G-I)-e(S,V\left(B_1\right))-e\left(B_1\right)\ge 9-2-3=4$, which implies that $G\left[S\right]$ is connected. If $\ell \left(G\right[S\left]\right)\le 2$, that is, $G\left[S\right]$ contains no cycle, then one of (1)–(4) above is satisfied. Thus by Lemma 2, we have $|N_G\left(V\left(B_1\right)\right)|\le 2$. Then $e(S,V\left(B_1\right))\le 3$. So $e\left(G\left[S\right]\right)=e(G-I)-e(S,V\left(B_1\right))-e\left(B_1\right)\ge 9-3-3=3$, which means that $G\left[S\right]$ is connected.

Therefore, by Claim 4, $O_n$ contains a $K_{2,3}$-minor, which contradicts to Lemma 2.

Case C.4.$e(G-I)=8$.

Then $e(S,I)\ge n-1$. Combining Lemma 2, the degree situation of the vertices of I in G satisfies one of the following: (1) $d_G\left(v_2\right)=4,d_G\left(v_3\right)=d_G\left(v_4\right)=d_G\left(v_5\right)=2$; (2) $d_G\left(v_2\right)=d_G\left(v_3\right)=3,d_G\left(v_4\right)=d_G\left(v_5\right)=2$; (3) $d_G\left(v_2\right)=3,d_G\left(v_3\right)=d_G\left(v_4\right)=\cdots =d_G\left(v_6\right)=2$; (4) $d_G\left(v_2\right)=d_G\left(v_3\right)=\cdots =d_G\left(v_7\right)=2$. Therefore, we have $G[S\cup V\left(B_1\right)\cup \{v_2,v_3,\dots ,v_5\}]\in {\mathcal{O}}_{11}^{-}$; or $G[S\cup V\left(B_1\right)\cup \{v_2,v_3,\dots ,v_6\}]\in {\mathcal{O}}_{12}^{=}$; or $G[S\cup V\left(B_1\right)\cup \{v_2,v_3,\dots ,v_7\}]$ is the graph obtained from a maximal outer-planar graph with 13 vertices by deleting 3 edges. Then we always get that $G-I^{\prime }$ contains two edge-disjoint $5K_2$. Thus, by Claim 1, $e\left(O_n\left[I^{\prime }\right]\right)=0$.

We claim that $G\left[S\right]$ is connected. If $G\left[S\right]$ contains a cycle, then $e(S,I)\le n-7+5=n-2$, a contradiction. Thus $G\left[S\right]$ does not contain any cycle. Then one of (1)–(4) above is satisfied. Thus $e(S,V\left(B_1\right))\le 2$. So $e\left(G\left[S\right]\right)=e(G-I)-e(S,V\left(B_1\right))-e\left(B_1\right)\ge 8-2-3=3$, which means that $G\left[S\right]$ is connected.

Therefore, by Claim 4, $O_n$ contains a $K_{2,3}$-minor, which contradicts to Lemma 2.

This completes the proof of Theorem 5.

6. Proof of Claims 1–7

In this section, we shall prove the seven claims used in the proof of Theorem 5.

6.1. Proof of Claim 1

Suppose that there exists some $e\in E\left(O_n\left[V\left(G\right)-V\left(M_1\cup M_2\right)\right]\right)$. If there exists an $e^{\prime }\in E\left(M_1\right)$ such that $c\left(e\right)=c\left(e^{\prime }\right)$, then $M_2\cup \left\{e\right\}$ is a rainbow $6K_2$ in $O_n$, a contradiction. Thus, $c\left(e\right)\ne c\left(e^{\prime }\right)$ for any $e^{\prime }\in E\left(M_1\right)$. But then $M_1\cup \left\{e\right\}$ is a rainbow $6K_2$ in $O_n$, a contradiction. This completes the proof of Claim 1.

6.2. Proof of Claim 2

We first prove $s\ge 1$. Suppose that $s=0$. Then $p=n-10$. Hence, combining Lemma 1, we have $e\left(G\right)={\sum }_{i=1}^{w}e\left(B_i\right)+e\left(G\left[R\right]\right)\le {\sum }_{i=1}^w(2v\left(B_i\right)-3)+\left(2\right|R|-3)=2({\sum }_{i=1}^{w}v\left(B_i\right)+\left|R\right|)-3(w+1)=2(n-(p-w))-3(w+1)\le 17-w$. Therefore, we have $e\left(G\right)<n+7$ because $w\ge 0$ and $n\ge 29$, a contradiction.

We next prove $s\ge 2$ when $R\ne \varnothing$. Suppose that $s\le 1$ when $R\ne \varnothing$. Since $s\ge 1$, then $s=1$. So $p=n-9$. Hence, combining Lemma 1, we have $e\left(G\right)={\sum }_{i=1}^{w}e\left(G[S\cup V\left(B_i\right)]\right)+e\left(G[S\cup R]\right)+e(S,I)\le {\sum }_{i=1}^w(2(1+v\left(B_i\right))-3)+\left(2\right(1+\left|R\right|)-3)+(p-w)=2({\sum }_{i=1}^{w}v\left(B_i\right)+\left|R\right|)+p-2w-1=2(n-(p-w)-1)+p-2w-1=n+6<n+7$, a contradiction. This completes the proof of Claim 2.

6.3. Proof of Claim 3

Suppose that $R\ne \varnothing$. Since $\left|R\right|\le n-s-p$ and $p=n+s-10$, we have $2\le \left|R\right|\le 10-2s$. Then $s\le 4$. Thus combining Claim 2, we have $2\le s\le 4$. We distinguish the following three cases to finish the proof of this claim.

Case 3.1.$s=4$.

In this case, $p=n-6$ and $\left|R\right|=2$. Then $\left|I\right|=p$. By Lemma 1, $e(G-I)\le 2\left(v\right(G)-|I\left|\right)-3=9$. Since $|I\cup R|=n-4>\left|S\right|$, then $e(S,I\cup R)\le n+4-2=n+2$ by Lemma 4. Note that $e\left(G\right[R\left]\right)=1$. Thus, $e\left(G\right[S\left]\right)=e\left(G\right)-e(S,I\cup R)-e\left(G\right[R\left]\right)\ge n+7-(n+2)-1=4$. Therefore, $G\left[S\right]$ must contain a cycle, that is, $3\le \ell \left(G\right[S\left]\right)\le 4$. We consider the following two subcases.

Subcase 3.1.1.$\ell \left(G\right[S\left]\right)=4$.

By Lemma 2, $n_{{}_I}^{3^{+}}=0$ and $n_{{}_I}^2\le 4$. Thus, $e(S,I)\le n-6+4=n-2$. If $e(S,I)<n-2$, then $e\left(G\right)=e(S,I)+e(G-I)<n-2+9=n+7$, a contradiction. If $e(S,I)=n-2$, then $n_{{}_I}^2=4$, which implies that $e(S,R)\le 2$. Therefore, $e(G-I)=e\left(G\left[S\right]\right)+e(S,R)+e\left(G\left[R\right]\right)\le 5+2+1=8$. But $e(G-I)=e\left(G\right)-e(S,I)\ge 9$, a contradiction.

Subcase 3.1.2.$\ell \left(G\right[S\left]\right)=3$.

Then $e\left(G\left[S\right]\right)=4$. By Lemma 2, $n_{{}_I}^{4^{+}}=0$, $n_{{}_I}^3=1$ and $n_{{}_I}^2\le 3$; or $n_{{}_I}^{3^{+}}=0$ and $n_{{}_I}^2\le 5$. Therefore, $e(S,I)\le n-6+5=n-1$. If $e(S,I)=n-1$, then $n_{{}_I}^{4^{+}}=0$, $n_{{}_I}^3=1$ and $n_{{}_I}^2=3$; or $n_{{}_I}^{3^{+}}=0$ and $n_{{}_I}^2=5$. So $e(S,R)\le 2$, which implies that $e(G-I)=e\left(G\left[S\right]\right)+e(S,R)+e\left(G\left[R\right]\right)\le 4+2+1=7$. But $e(G-I)=e\left(G\right)-e(S,I)\ge 8$, a contradiction. If $e(S,I)\le n-2$, because $e(S,I)=e\left(G\right)-e(G-I)\ge n+7-9=n-2$, then $e(S,I)=n-2$. Thus combining Lemma 2, we have $n_{{}_I}^{4^{+}}=0$, $n_{{}_I}^3=1$ and $n_{{}_I}^2\ge 2$; or $n_{{}_I}^{3^{+}}=0$ and $n_{{}_I}^2\ge 4$. Then $e(S,R)\le 3$. Therefore, $e(G-I)=e\left(G\left[S\right]\right)+e(S,R)+e\left(G\left[R\right]\right)\le 4+3+1=8$. But $e(G-I)=e\left(G\right)-e(S,I)\ge 9$, a contradiction.

Case 3.2.$s=3$.

In this case, $p=n-7$ and $n_{{}_I}^{4^{+}}=0$. Then $w\le 1$ because $\left|R\right|\ge 2$. We consider the following two subcases based on w.

Subcase 3.2.1.$w=0$.

Then $\left|I\right|=p$ and $\left|R\right|=4$. By Lemma 1, we have $e(G-I)\le 11$.

If $e(G-I)=11$, then $G-I\in {\mathcal{O}}_7$. So $n_{{}_I}^3=0$ and $|N_G\left(R\right)|\ge 2$. When $|N_G\left(R\right)|=2$, we have $G\left[S\right]\cong K_3$. Combining Lemma 2, we have $n_{{}_I}^2\le 2$. Thus $e(S,I)\le n-5$. But $e(S,I)=e\left(G\right)-e(G-I)\ge n-4$, a contradiction.

If $e(G-I)=10$, then $G-I\in {\mathcal{O}}_7^{-}$ and $|N_G\left(R\right)|\ge 1$. When $|N_G\left(R\right)|=1$, we have $G\left[S\right]\cong K_3$. Combining Lemma 2, we have $n_{{}_I}^3=0$ and $n_{{}_I}^2\le 3$. Thus $e(S,I)\le n-4$. But $e(S,I)=e\left(G\right)-e(G-I)\ge n-3$, a contradiction.

If $e(G-I)\le 9$, then $e(S,I)=e\left(G\right)-e(G-I)\ge n-2$. But by Lemma 4, we have $e(S,I)\le \left(\right|I|+|S\left|\right)+\left|S\right|-2=n-3$, a contradiction.

Subcase 3.2.2.$w=1$.

Then $v\left(B_1\right)=3$, $\left|R\right|=2$ and $\left|I\right|=p-1=n-8$. Clearly, we have $B_1\cong K_3$ by Lemma 3. By Lemma 1, we have $e(G-I)\le 13$.

If $e(G-I)=13$, then $G-I\in {\mathcal{O}}_8$. Then $n_{{}_I}^3=0$, $|N_G\left(V\left(B_1\right)\right)|\ge 2$ and $|N_G\left(R\right)|\ge 2$. When both of the above equalities hold, we have $G\left[S\right]\cong K_3$. Combining Lemma 2, we have $n_{{}_I}^2\le 1$. Therefore, $e(S,I)\le n-7$. But $e(S,I)=e\left(G\right)-e(G-I)\ge n-6$, a contradiction.

If $e(G-I)=12$, then $G-I\in {\mathcal{O}}_8^{-}$. Combining Lemma 2, if $e\left(G\right[S\left]\right)=3$, then $|N_G\left(V\left(B_1\right)\right)|+|N_G\left(R\right)|\ge 3$, $n_{{}_I}^3=0$ and $n_{{}_I}^2\le 2$; if $e\left(G\right[S\left]\right)\le 2$, then $|N_G\left(V\left(B_1\right)\right)|+|N_G\left(R\right)|\ge 4$, and either $n_{{}_I}^3=1$ and $n_{{}_I}^2=0$, or $n_{{}_I}^3=0$ and $n_{{}_I}^2\le 2$. In both cases of $e\left(G\right[S\left]\right)$, we have $e(S,I)\le n-6$. But $e(S,I)=e\left(G\right)-e(G-I)\ge n-5$, a contradiction.

If $e(G-I)=11$, then $G-I\in {\mathcal{O}}_8^{=}$. If $e\left(G\right[S\left]\right)=3$, then $|N_G\left(V\left(B_1\right)\right)|+|N_G\left(R\right)|\ge 2$. If $e\left(G\right[S\left]\right)\le 2$, then $|N_G\left(V\left(B_1\right)\right)|+|N_G\left(R\right)|\ge 3$. In both cases of $e\left(G\right[S\left]\right)$ and combining Lemma 2, we have $n_{{}_I}^3=1$ and $n_{{}_I}^2\le 1$; or $n_{{}_I}^3=0$ and $n_{{}_I}^2\le 3$. Thus $e(S,I)\le n-5$. But $e(S,I)=e\left(G\right)-e(G-I)\ge n-4$, a contradiction.

If $e(G-I)\le 10$, then $e(S,I)=e\left(G\right)-e(G-I)\ge n-3$. But by Lemma 4, we have $e(S,I)\le \left(\right|I|+|S\left|\right)+\left|S\right|-2=n-4$, a contradiction.

Case 3.3.$s=2$.

In this case, $p=n-8$ and $n_{{}_I}^{3^{+}}=0$. Since $\left|R\right|\ge 2$, we see $w\le 2$. We consider the following two subcases based on w.

Subcase 3.3.1.$w=0$.

Then $\left|I\right|=p$ and $\left|R\right|=6$. By Lemma 1, we have $e(G-I)\le 13$. If $e(G-I)=13$, then $G-I\in {\mathcal{O}}_8$. Combining Lemma 2, we have $n_{{}_I}^2\le 1$. Therefore, $e(S,I)\le n-7$. But $e(S,I)=e\left(G\right)-e(G-I)\ge n-6$, a contradiction. If $e(G-I)\le 12$, then $e(S,I)=e\left(G\right)-e(G-I)\ge n-5$. But $e(S,I)\le \left(\right|I|+|S\left|\right)+\left|S\right|-2=n-6$ by Lemma 4, a contradiction.

Subcase 3.3.2.$w=1$.

Then $\left|I\right|=p-1=n-9$. By Lemma 1, $e(G-I)\le 15$.

If $14\le e(G-I)\le 15$, then $G-I$ is the graph obtained from a maximal outer-planar graph with 9 vertices by deleting $15-e(G-I)$ edges. Hence, when $e(G-I)=14$, we have at least one of $N_G\left(V\left(B_1\right)\right)$ and $N_G\left(R\right)$ is S; when $e(G-I)=15$, we have $N_G\left(V\left(B_1\right)\right)=N_G\left(R\right)=S$. Then $n_{{}_I}^2\le 15-e(G-I)$. Therefore, $e(S,I)\le n-9+15-e(G-I)$. Then $e\left(G\right)=e(G-I)+e(S,I)\le n+6$, a contradiction.

If $e(G-I)\le 13$, then $e(S,I)=e\left(G\right)-e(G-I)\ge n-6$. But $e(S,I)\le \left(\right|I|+|S\left|\right)+\left|S\right|-2=n-7$ by Lemma 4, a contradiction.

Subcase 3.3.3.$w=2$.

Then $v\left(B_1\right)=v\left(B_2\right)=3$, $\left|R\right|=2$ and $\left|I\right|=p-2=n-10$. Obviously, we have $B_1\cong B_2\cong K_3$ by Lemma 3. By Lemma 1, $e(G-I)\le 17$.

If $e(G-I)=17$, then $G-I\in {\mathcal{O}}_{10}$. Thus, $N_G\left(V\left(B_1\right)\right)=N_G\left(V\left(B_2\right)\right)=N_G\left(R\right)=S$. But then G contains a $K_{2,3}$-minor, one part of which is S, and the other part is $\{V\left(B_1\right),V\left(B_2\right),R\}$. This contradicts to Lemma 2.

If $15\le e(G-I)\le 16$, then $G-I$ is the graph obtained from a maximal outer-planar graph with 10 vertices by deleting $17-e(G-I)$ edges. Hence, when $e(G-I)=15$, we have at least one of $N_G\left(V\left(B_1\right)\right)$, $N_G\left(V\left(B_2\right)\right)$ and $N_G\left(R\right)$ is S; when $e(G-I)=16$, we have at least two of $N_G\left(V\left(B_1\right)\right)$, $N_G\left(V\left(B_2\right)\right)$ and $N_G\left(R\right)$ are S. Then $n_{{}_I}^2\le 16-e(G-I)$. Therefore, $e(S,I)\le n-10+16-e(G-I)$. Then $e\left(G\right)=e(G-I)+e(S,I)\le n+6$, a contradiction.

If $e(G-I)\le 14$, then $e(S,I)=e\left(G\right)-e(G-I)\ge n-7$. But $e(S,I)\le \left(\right|I|+|S\left|\right)+\left|S\right|-2=n-8$ by Lemma 4, a contradiction. This completes the proof of Claim 3.

6.4. Proof of Claim 4

Clearly, $d_G\left(v\right)\le 1$ for any $v\in I^{\prime }$. Since $e\left(O_n\left[I^{\prime }\right]\right)=0$, $\delta \left(O_n\right)\ge 2$, and combining Claim 3, we have $N_{O_n}\left(v\right)\cap V\left(B_i\right)\ne \varnothing$ for some $i\in \left[b\right]$ and any $v\in I^{\prime }$. Since $|I^{\prime }|>2b$, then $I^{\prime }$ contains at least three vertices, say $v_{b+1},v_{b+2}$ and $v_{b+3}$, such that $d_G\left(v_{b+j}\right)=1$, $N_{O_n}\left(v_{b+j}\right)\cap V\left(B_{\ell }\right)\ne \varnothing$ for any $j\in \left[3\right]$ and some $\ell \in \left[b\right]$. Since $G\left[S\right]$ is connected, then $O_n$ contains a $K_{2,3}$-minor, one part of which is $\{S,V\left(B_{\ell }\right)\}$, and the other part is $\{v_{b+1},v_{b+2},v_{b+3}\}$. This contradicts to Lemma 2. This completes the proof of Claim 4.

6.5. Proof of Claim 5

By Claim 2, we just need to prove that $s\ge 2$ when $R=\varnothing$. Suppose that $s\le 1$ when $R=\varnothing$. Again combining Claim 2, we have $s=1$. Then $p=n-9$ and $n_{{}_I}^{2^{+}}=0$. Thus $b=w$. If $w=0$, then $n=v\left(G\right)=v\left(B_1\right)+\cdots +v\left(B_p\right)+s=p+s=n-8$ by Claim 3, a contradiction. So $w\ge 1$. Hence, by Claim 3 and Lemma 1, $n+7\le e\left(G\right)={\sum }_{i=1}^{w}e\left(G[S\cup V\left(B_i\right)]\right)+e(S,I)\le {\sum }_{i=1}^w(2(1+v\left(B_i\right))-3)+(p-w)=2(n-(p-w)-1)+p-2w=n+7$, which implies that $G[S\cup V\left(B_i\right)]$ is a maximal outer-planar graph for each $i\in \left[w\right]$. By $\left(1\right)$, we get $v\left(B_1\right)+\cdots +v\left(B_w\right)=w+8$. Then we have $w\le 4$ because $v\left(B_1\right)+\cdots +v\left(B_w\right)\ge 3w$. So $1\le w\le 4$. If $w=1$, then $v\left(B_1\right)=9$. If $w=2$, then either $v\left(B_1\right)=7$ and $v\left(B_2\right)=3$, or $v\left(B_1\right)=v\left(B_2\right)=5$. If $w=3$, then $v\left(B_1\right)=5$ and $v\left(B_2\right)=v\left(B_3\right)=3$. If $w=4$, then $v\left(B_1\right)=v\left(B_2\right)=\cdots =v\left(B_4\right)=3$. From the above four cases of w, we always get that $G\left[S\cup Q\right]$ contains two edge-disjoint $5K_2$. Thus by Claim 1, $e\left(O_n\left[I^{\prime }\right]\right)=0$. Since $n\ge 29$ and $b=w\le 4$, then $|I^{\prime }|=p-b\ge (n-9)-4>2b$. Therefore, $O_n$ contains a $K_{2,3}$-minor by Claim 4, which contradicts to Lemma 2. This completes the proof of Claim 5.

6.6. Proof of Claim 6

Suppose that $v\left(B_1\right)\ge 7$. Then $w\ge 1$. Thus by $\left(1\right)$ and $v\left(B_1\right)+\cdots +v\left(B_w\right)\ge 7+3(w-1)=3w+4$, we have $s+w\le 3$. Then $w=1$ and $s=2$ by Claim 5. Therefore, $v\left(B_1\right)=7$, $p=n-8$ and $\left|I\right|=p-1=n-9$. By Lemma 2, $n_{{}_I}^{3^{+}}=0$ and $n_{{}_I}^2\le 2$. Then $e(S,I)\le n-9+2=n-7$. Thus $e(G-I)\ge 14$. On the other hand, we have $e(G-I)\le 15$ by Lemma 1. So $14\le e(G-I)\le 15$.

If $e(G-I)=14$, then $e(S,I)=e\left(G\right)-e(G-I)\ge n-7$. Thus $d_G\left(v_2\right)=d_G\left(v_3\right)=2$. Then $|N_G\left(V\left(B_1\right)\right)|\le 1$. Since $e(S\cup V\left(B_1\right))=14$, then $G\left[S\cup V\left(B_1\right)\cup \{v_2,v_3\}\right]\in {\mathcal{O}}_{11}^{-}$. On the other hand, combining $|N_G\left(V\left(B_1\right)\right)|\le 1$, we can obtain that $G\left[S\right]$ is connected.

If $e(G-I)=15$, then $S\cup V\left(B_1\right)\in {\mathcal{O}}_9$. Thus $N_G\left(V\left(B_1\right)\right)=S$. Since $e(S,I)=e\left(G\right)-e(G-I)\ge n-8$, then $d_G\left(v_2\right)=2$. Therefore, $G[S\cup V\left(B_1\right)\cup \left\{v_2\right\}]\in {\mathcal{O}}_{10}$ and $G\left[S\right]$ is connected.

From the above two cases of $e(G-I)$, we always get that $G-I^{\prime }$ contains two edge-disjoint $5K_2$. Thus by Claim 1, $e\left(O_n\left[I^{\prime }\right]\right)=0$. Since $n_{{}_I}^2\le 2$, we see $b\le 3$. Then $|I^{\prime }|=p-b\ge (n-8)-3>2b$ because $n\ge 29$ and $b\le 3$. Therefore, by Claim 4, $O_n$ contains a $K_{2,3}$-minor, which contradicts to Lemma 2. This completes the proof of Claim 6.

6.7. Proof of Claim 7

Suppose that $v\left(B_1\right)<3$. Then $\left|I\right|=p$. Combining Claim 3, we have $p+s=n$. Then $s=5$ because $p=n+s-10$. Thus $n_{{}_I}^{6^{+}}=0$ and $p=n-5$. In the following proof, let $U_1$ denote the graph obtained by hanging a path of length 2 at one vertex of $C_3$; let $U_2$ denote the graph obtained by hanging an edge at each of two vertices of $C_3$; let $T_1$ denote the tree with 5 vertices and diameter 3.

We will prove that $G\left[S\right]$ does not contain any cycle. Suppose not, that is, $3\le \ell \left(G\right[S\left]\right)\le 5$. We consider the following three cases according to $\ell \left(G\right[S\left]\right)$.

If $\ell \left(G\right[S\left]\right)=5$, then $G\left[S\right]$ is clearly connected, and we have $n_{{}_I}^{3^{+}}=0$ and $n_{{}_I}^2\le 5$ by Lemma 2. Thus $e(S,I)\le n$. By Lemma 1, we have $e\left(G\right[S\left]\right)\le 7$, which implies that $e(S,I)=e\left(G\right)-e\left(G\right[S\left]\right)\ge n$. Therefore, $e(S,I)=n$. Then $n_{{}_I}^2=5$ and $e\left(G\right[S\left]\right)=7$. It follows that $b=5$ and $G\left[S\right]\in {\mathcal{O}}_5$. Thus, $G[S\cup \{v_1,v_2,\dots ,v_5\}]\in {\mathcal{O}}_{10}$, which means that $G[S\cup \{v_1,v_2,\dots ,v_5\}]$ contains two edge-disjoint $5K_2$. Then we obtain $e\left(O_n\left[I^{\prime }\right]\right)=0$ by Claim 1. Then $|I^{\prime }|=p-b=(n-5)-5>2b$ because $n\ge 29$ and $b=5$. Therefore, by Claim 4, $O_n$ contains a $K_{2,3}$-minor, which contradicts to Lemma 2.

If $\ell \left(G\right[S\left]\right)=4$, then $e\left(G\right[S\left]\right)\le 6$. By Lemma 2, $n_{{}_I}^{4^{+}}=0$, $n_{{}_I}^3=1$ and $n_{{}_I}^2\le 4$; or $n_{{}_I}^{3^{+}}=0$ and $n_{{}_I}^2\le 6$. So $e(S,I)\le n+1$. Since $n+7\le e\left(G\right)=e(S,I)+e\left(G\right[S\left]\right)\le n+7$, then $e\left(G\right[S\left]\right)=6$ and $e(S,I)=n+1$. Therefore, $G\left[S\right]$ is connected, and either $n_{{}_I}^3=1$ and $n_{{}_I}^2=4$ or $n_{{}_I}^2=6$. Thus, $5\le b\le 6$ and $G[S\cup \{v_1,v_2,\dots ,v_b\}]$ contains two edge-disjoint $5K_2$. Then by Claim 1, $e\left(O_n\left[I^{\prime }\right]\right)=0$. Since $n\ge 29$ and $b\le 6$, then $|I^{\prime }|=p-b\ge (n-5)-6>2b$. Therefore, by Claim 4, $O_n$ contains a $K_{2,3}$-minor, which contradicts to Lemma 2.

If $\ell \left(G\right[S\left]\right)=3$, then we also have $e\left(G\right[S\left]\right)\le 6$. If $e\left(G\right[S\left]\right)=6$, then $e(S,I)=e\left(G\right)-e\left(G\right[S\left]\right)\ge n+1$ and $G\left[S\right]\cong K_1+2K_2$. By Lemma 2, $n_{{}_I}^{4^{+}}=0$, $n_{{}_I}^3=1$ and $n_{{}_I}^2=4$; or $n_{{}_I}^{3^{+}}=0$ and $n_{{}_I}^2=6$. It follows that $5\le b\le 6$ and $G[S\cup \{v_1,v_2,\dots ,v_b\}]$ contains two edge-disjoint $5K_2$. Then by Claim 1, $e\left(O_n\left[I^{\prime }\right]\right)=0$. We have $|I^{\prime }|=p-b\ge (n-5)-6>2b$ because $n\ge 29$ and $b\le 6$. Therefore, by Claim 4, $O_n$ contains a $K_{2,3}$-minor, which contradicts to Lemma 2. Therefore, $e\left(G\right[S\left]\right)\le 5$. By Lemma 2, $n_{{}_I}^5=0$, $n_{{}_I}^4=1$ and $n_{{}_I}^2\le 4$; or $n_{{}_I}^{4^{+}}=0$, $n_{{}_I}^3=2$ and $n_{{}_I}^2\le 3$; or $n_{{}_I}^{4^{+}}=0$, $n_{{}_I}^3=1$ and $n_{{}_I}^2\le 5$; or $n_{{}_I}^{3^{+}}=0$ and $n_{{}_I}^2\le 7$. From the degree situation of the vertices of I in G, we can obtain $e(S,I)\le n-5+7=n+2$. Since $n+7\le e\left(G\right)=e(S,I)+e\left(G\right[S\left]\right)\le n+7$, $e\left(G\right[S\left]\right)=5$ and $e(S,I)=n+2$. It follows that $G\left[S\right]$ is connected, and $G\left[S\right]\in \{U_1,U_2,(K_2\cup 2K_1)+K_1\}$. On the other hand, we can also get that the degree situation of the vertices of I in G satisfies one of the following: (1) $n_{{}_I}^5=0$, $n_{{}_I}^4=1$ and $n_{{}_I}^2=4$; (2) $n_{{}_I}^{4^{+}}=0$, $n_{{}_I}^3=2$ and $n_{{}_I}^2=3$; (3) $n_{{}_I}^{4^{+}}=0$, $n_{{}_I}^3=1$ and $n_{{}_I}^2=5$; (4) $n_{{}_I}^{3^{+}}=0$ and $n_{{}_I}^2=7$. If $G\left[S\right]\in \{U_1,U_2\}$, then one of (1)–(4) is satisfied. If $G\left[S\right]\cong (K_2\cup 2K_1)+K_1$, then one of (2)–(4) is satisfied. Thus, from the above three structures of $G\left[S\right]$, we always find that $b\le 7$ and $G[S\cup \{v_1,v_2,\dots ,v_b\}]$ contains two edge-disjoint $5K_2$. Then by Claim 1, $e\left(O_n\left[I^{\prime }\right]\right)=0$. Then $|I^{\prime }|=p-b\ge (n-5)-7>2b$ because $n\ge 29$ and $b\le 7$. Therefore, by Claim 4, $O_n$ contains a $K_{2,3}$-minor, which contradicts to Lemma 2.

From the above discussion of $\ell \left(G\right[S\left]\right)$, we get that $G\left[S\right]$ contains no cycle. Then $e\left(G\right[S\left]\right)\le 4$. By Lemma 4, we have $e(S,I)\le \left(\right|I|+|S\left|\right)+\left|S\right|-2=n+3$ because $\left|I\right|>\left|S\right|$, which implies that $e\left(G\right[S\left]\right)\ge 4$. Thus $e\left(G\right[S\left]\right)=4$. Then $G\left[S\right]$ is connected, and $G\left[S\right]\in \{P_5,T_1,K_{1,4}\}$. Since $e\left(G\right)\ge n+7$, we have $e(S,I)=e\left(G\right)-e\left(G\right[S\left]\right)\ge n+3$. Combining $e(S,I)\le n+3$, we have $e(S,I)=n+3$ and $e\left(G\right)=n+7$. Thus by Lemma 4, we have $d_G\left(v_i\right)\ge 1$ for each $v\in I^{\prime }$. Then the degree situation of the vertices of I in G satisfies one of the following: (1) $n_{{}_I}^5=1$ and $n_{{}_I}^2=4$; (2) $n_{{}_I}^5=0$, $n_{{}_I}^4=1$, $n_{{}_I}^3=1$ and $n_{{}_I}^2=3$; (3) $n_{{}_I}^5=0$, $n_{{}_I}^4=1$ and $n_{{}_I}^2=5$; (4) $n_{{}_I}^{4^{+}}=0$, $n_{{}_I}^3=3$ and $n_{{}_I}^2=2$; (5) $n_{{}_I}^{4^{+}}=0$, $n_{{}_I}^3=2$ and $n_{{}_I}^2=4$; (6) $n_{{}_I}^{4^{+}}=0$, $n_{{}_I}^3=1$ and $n_{{}_I}^2=6$; (7) $n_{{}_I}^{3^{+}}=0$ and $n_{{}_I}^2=8$. If $G\left[S\right]\cong P_5$, then one of (1)–(7) is satisfied. If $G\left[S\right]\cong T_1$, then one of (2)–(7) is satisfied. If $G\left[S\right]\cong K_{1,4}$, then one of (4)–(7) is satisfied. From the above three structures of $G\left[S\right]$, we can get that $b\le 8$ and $G[S\cup \{v_1,v_2,\dots ,v_b\}]$ contains two edge-disjoint $5K_2$. Then by Claim 1, $e\left(O_n\left[I^{\prime }\right]\right)=0$.

Note that $b\le 8$ and $n\ge 29$. If $b<8$ and $n\ge 29$, or $b=8$ and $n>29$, then it can be found that $|I^{\prime }|=p-b\ge (n-5)-8>2b$. Then by Claim 4, $O_n$ contains a $K_{2,3}$-minor, which contradicts to Lemma 2. So we only need to consider the case of $b=8$ and $n=29$. If there exists some $i\in \left[b\right]$ such that $|N_{O_n}\left(v_i\right)\cap I^{\prime }|\ge 3$, without loss of generality, we assume that $N_{O_n}\left(v_i\right)\cap I^{\prime }=\{v_{b+1},v_{b+2},v_{b+3}\}$. Then $O_n$ contains a $K_{2,3}$-minor, one part of which is $\{S,\left\{v_i\right\}\}$, and the other part is $\{v_{b+1},v_{b+2},v_{b+3}\}$. It contradicts to Lemma 2. If $|N_{O_n}\left(v_i\right)\cap I^{\prime }|\le 2$ for each $i\in \left[b\right]$, then $e_{{}_{O_n}}(\{v_1,v_2,\dots ,v_b\},I^{\prime })\le 2b$. It follows that $e\left(O_n\right)=e\left(G\right)+e_{{}_{O_n}}(\{v_1,v_2,\dots ,v_b\},I^{\prime })\le n+7+2b<2n-3$, which contradicts to Lemma 1. This completes the proof of Claim 7.

7. Concluding Remarks

Theorem 4 determines the exact value of $ar({\mathcal{O}}_n,k{K}_2)$ for $n=2k$. It seems non-trivial to determine the exact value of $ar({\mathcal{O}}_n,k{K}_2)$ when $n\ge 2k+1$. Theorem 3 gives a better upper bound of $ar({\mathcal{O}}_n,k{K}_2)$ for all $n\ge 2k+1$. However, we conjecture that the exact value of $ar({\mathcal{O}}_n,k{K}_2)$ for $n\ge 2k+1$ is equal to the lower bound given in Theorem 1(4). In [23,24], the authors proved that the conjecture holds when $3\le k\le 4$, and $k=5$ and $n\ge 2k+5$. Theorem 5 verifies the conjecture holds when $n\ge 2k+17$ and $k=6$. The conjecture is wide open when $k\ge 7$.