Erdos-Gallai Stability Theorem for Linear Forests

TL;DR

This paper establishes a stability version of the Erd ext{"o}s-Gallai Theorem for linear forests, linking minimum degree conditions to the guaranteed presence of specific disjoint paths in large connected graphs.

Contribution

It extends and strengthens existing results by providing minimum degree conditions that ensure the embedding of certain linear forests, with exceptions for specific graph classes.

Findings

Minimum degree condition guarantees linear forest embedding

Extends previous results for even and odd paths

Applicable to large connected graphs with specific path structures

Abstract

The Erd\H{o}s-Gallai Theorem states that every graph of average degree more than $l-2$ contains a path of order $l$ for $l\ge 2$. In this paper, we obtain a stability version of the Erd\H{o}s-Gallai Theorem in terms of minimum degree. Let $G$ be a connected graph of order $n$ and $F=(\bigcup_{i=1}^kP_{2a_i})\bigcup(\bigcup_{i=1}^lP_{2b_i+1})$ be $k+l$ disjoint paths of order $2a_1, \ldots, 2a_{k}, 2b_1+1, \ldots, 2b_l+1,$ respectively, where $k\ge 0$, $0\le l\le 2$, and $k+l\geq 2$. If the minimum degree $\delta(G)\ge \sum_{i=1}^ka_i+\sum_{i=1}^lb_i-1$, then $F\subseteq G$ except several classes of graphs for sufficiently large $n$, which extends and strengths the results of Ali and Staton for an even path and Yuan and Nikiforov for an odd path.

Equations

$$h=\sum _{i=1}^k{a}_i-1\ge 2a_k$$

$$2h+1=\sum _{i=1}^{k}2a_i-1\ge \sum _{i=1}^{k-1}2a_i,$$

$$h+1=\sum _{i=1}^k{a}_i+b_1\ge 2b_1+1$$

$$2h+1\ge \sum _{i=1}^{k}2a_i+2b_1-1\ge \sum _{i=1}^{k}2a_i.$$

$$h+1=\sum _{i=1}^k{a}_i+b_1\ge \sum _{i=1}^{k}2a_i$$

$$2h+1\ge \sum _{i=1}^{k}2a_i+2b_1-1\ge 2b_1+1.$$

$$h=\sum _{i=1}^k{a}_i+b_1-1\ge 2b_1+1,$$

$$2h+1=\sum _{i=1}^{k}2a_i+2b_1-1\ge \sum _{i=1}^{k}2a_i.$$

$$h=\sum _{i=1}^k{a}_i+b_1-1\ge \sum _{i=1}^{k}2a_i,$$

$$2h+1=\sum _{i=1}^{k}2a_i+2b_1-1\ge 2b_1+1.$$

$$h-1=\sum _{i=1}^{k}2a_i-2\ge 2a_k,$$

$$h=\sum _{i=1}^{k}2a_i-1>\sum _{i=1}^{k-1}2a_i,$$

$$h+2=\sum _{i=1}^k{a}_i+b_1+1=2b_1+1.$$

$$h=\sum _{i=1}^{k}2a_i-2\ge \sum _{i=1}^{k-1}2a_i,$$

$$h=\sum _{i=1}^{k}2a_i-2\ge 2a_k,$$

$$h+1=\sum _{i=1}^k{a}_i+b_1=2b_1+1.$$

Full text

AMS Classification: 05C35, 05C05

Key words: Erdős-Gallai Theorem; Stable problem; Linear forest; Path.

1 Introduction

1.1 Notation

Let $G$ be a finite simple undirected graph with vertex set $V(G)$ and edge set $E(G)$. Denote $e(G)$ by the number of edges of $G$. For $v\in V(G)$, the neighborhood $N_{G}(v)$ of $v$ is $\{u:uv\in E(G)\}$ and the degree $d_{G}(v)$ of $v$ is $|N_{G}(v)|$. For a subgraph $H$ of $G$, the $H$-neighborhood $N_{H}(v)$ of $v$ is $N_{G}(v)\bigcap V(H)$ and the $H$-degree $d_{H}(v)$ of $v$ is $|N_{H}(v)|$. Denote $\delta(G)$ by minimum degree of $G$. Denote $P_{n}$, $K_{n}$, and $K_{m,n}$ by a path of order $n$, a complete graph of order $n$, and a complete bipartite graph with partition of size $m$ and $n$, respectively. For two disjoint graphs $G$ and $H$, denote $G\bigcup H$ and $G\bigvee H$ by the disjoint union of $G$ and $H$, and the join of $G$ and $H$ which is obtained from $G\bigcup H$ by joining every vertex of $G$ to every vertex of $H$, respectively. Moreover, $kG$ denotes a graph consisting of $k$ disjoint copies of $G$ and $\overline{G}$ denotes the complement of $G$. For $X\subseteq V(G)$, $G[X]$ denotes the graph induced by $X$ and write $E(X)$ for $E(G[X])$. For $X,Y\subseteq V(G)$, $e(X,Y)$ denotes the number of the edges of $G$ with one end vertex in $X$ and the other in $Y$. For two graphs $G$ and $H$, write $H\subseteq G$ if $G$ contains $H$ as a subgraph, and $H\nsubseteq G$ otherwise. An odd (even) path is a path of odd (even) order. A graph is connected if there is a path between every pair of vertices. A connected graph $G$ is $k$-connected if $|V(G)|>k$ and $G-X$ is connected for every set $X\subseteq V(G)$ with $|X|<k$. A graph $G$ is $F$-free if it does not contain $F$ as a subgraph. A star forest is a forest whose components are stars and a linear forest is a forest whose components are paths. A cut vertex (edge) of a graph $G$ is a vertex (edge) whose removal increases the number of components of $G$. A block of is a maximal connected subgraph without any cut vertex. An end block of $G$ is block that contains precisely a cut vertex of $G$. For two end blocks $B_{1}$ and $B_{2}$ of a graph $G$, denote $p(B_{1},B_{2})$ by the order of a longest path between the unique cut vertex of $G$ in $V(B_{1})$ and the unique cut vertex of $G$ in $V(B_{2})$. For other notations not defined here, readers are referred to [10].

1.2 History and known results

The study of Turán-type problems is the center of extremal graph theory. Erdős and Gallai in [6] proved the following key result, which opens a new subject for degenerate graphs:

Theorem 1.1**.**

[6]** Let $G$ be a graph of order $n$. If $e(G)>\frac{(l-2)n}{2}$, where $l\geq 2$, then $P_{l}\subseteq G$.

Let $H_{n,l,a}:=K_{a}\bigvee(K_{l-2a}\bigcup\overline{K}_{n-l+a})$ with $h(n,l,a):=e(H_{n,l,a})=\binom{l-a}{2}+a(n-l+a)$ for $a\leq\lfloor\frac{l}{2}\rfloor.$ Recently, Füredi, Kostochka, and Verstraëte [7] obtained a stability theorem for paths.

Theorem 1.2**.**

[7*]** Let $G$ be a connected graph of order $n$, where $t\geq 2$ and $n\geq 3t-1$. If $e(G)>h(n+1,l+1,t-1)-n$, where $l\in\{2t,2t+1\}$, then $P_{l}\subseteq G$, unless one of following holds:

(i). $l=2t$, $l\neq 6$, and $G\subseteq H_{n,l,t-1}$;

(ii). $l=2t+1$ or $l=6$, and $G-A$ is a star forest for some $A\subseteq V(G)$ of size at most $l-1$.*

Another question in Turán-type problems which has received extensive attention is how large minimum degree of a graph $G$ is to guarantee $P_{l}\subseteq G$. Erdős and Gallai [6] and Andrasfai [1] proved the following result for an odd path.

Theorem 1.3**.**

[6, 1]** Let $G$ be a connected graph of order $n$, where $n\geq 2h+3\geq 3$. If $\delta(G)\geq h+1$, then $P_{2h+3}\subseteq G$.

Let $S_{n,h}:=K_{h}\bigvee\overline{K}_{n-h}$ and $S^{+}_{n,h}$ be a graph obtained by adding an edge to $S_{n,h}$, i.e., $S^{+}_{n,h}=K_{h}\bigvee(K_{2}\bigcup\overline{K}_{n-h-2})$ (see Fig.1).

K_{h}$$\overline{K}_{n-h}$$S_{n,h}$$K_{h}$$K_{2}\bigcup\overline{K}_{n-h-2}$$S^{+}_{n,h}Fig.1. Graphs $S_{n,h}$ and $S_{n,h}^{+}$.

Given $t_{1},t_{2}\geq 0$, let $L_{t_{1},t_{2},h,h+1}:=K_{1}\bigvee(t_{1}K_{h}\bigcup t_{2}K_{h+1})$ (see Fig.2). In particular, write $L_{t,0,h,h+1}:=L_{t,h}$ (see Fig.2). The center of $L_{t_{1},t_{2},h,h+1}$ is the vertex of maximum degree in $L_{t_{1},t_{2},h,h+1}$.

K_{h+1}$$K_{h+1}$$K_{h+2}$$K_{h+2}$$t_{1}$$t_{2}$$L_{t_{1},t_{2},h,h+1}$$K_{h+1}$$K_{h+1}$$t$$L_{t,h}Fig.2. Graphs $L_{t_{1},t_{2},h,h+1}$ and $L_{t,h}$.

For an even path, the question was answered by Ali and Staton [2].

Theorem 1.4**.**

[2]** Let $G$ be a connected graph of order $n$, where $n\geq 2h+2$. If $\delta(G)\geq h\geq 1$, then $P_{2h+2}\subseteq G$ unless either $G\subseteq S_{n,h}$ or $G=L_{t,h}$, where $n=th+1$.

Recently, Yuan and Nikiforov [9] gave a stability version of Theorem 1.3.

Theorem 1.5**.**

[9*]** Let $G$ be a connected graph of order $n$, where $n\geq 2h+3$. If $\delta(G)\geq h\geq 2$, then $P_{2h+3}\subseteq G$, unless unless one of following holds:

(i). $G\subseteq S_{n,h}^{+}$;

(ii). $G=L_{t,h}$, where $n=th+1$;

(iii). $G\subseteq L_{t,h,1,h+1}$, where $n=(t+1)h+2$;

(iv). $G$ is obtained by joining the centers of two disjoint graphs $L_{s,h}$ and $L_{t,h}$, where $n=(s+t)h+2$.*

Bushaw and Kettle [3] proved the maximum number of edges in a $kP_{l}$-free graph of sufficiently large order $n$ for $l\geq 3$ and determined all the extremal graphs. Recently, Yuan and Zhang [11] extended their result for $l=3$ and all possible $n$ and $k$. Lidický, Liu, and Palmer [8] extended Bushaw-Kettle result for linear forests. Their main result can be stated as follows.

Theorem 1.6**.**

[8*]** Let $F=(\bigcup_{i=1}^{k}P_{2a_{i}})\bigcup(\bigcup_{i=1}^{l}P_{2b_{i}+1})$ and $h=\sum\limits_{i=1}^{k}a_{i}+\sum\limits_{i=1}^{l}b_{i}-1$, where $k+l\geq 2$, $a_{1}\geq\cdots\geq a_{k}\geq 1$, and $b_{i}\geq 1$ for $1\leq i\leq l$. Let $G$ be an $F$-free graph of sufficiently large order $n$.

(i). If $k\geq 1$, then $e(G)\leq e(S_{n,h})$ with equality if and only if $G=S_{n,h}$;

(ii). If $k=0$ and there is at least one $b_{i}$ such that $b_{i}>1$, where $1\leq i\leq l$, then $e(G)\leq e(S^{+}_{n,h})$ with equality if and only if $G=S^{+}_{n,h}$.

It is easy to see that we have the following result from Theorem 1.6.

Corollary 1.7**.**

*Let $F=(\bigcup_{i=1}^{k}P_{2a_{i}})\bigcup(\bigcup_{i=1}^{l}P_{2b_{i}+1})$ and $h=\sum\limits_{i=1}^{k}a_{i}+\sum\limits_{i=1}^{l}b_{i}-1$, where $k+l\geq 2$, $a_{1}\geq\cdots\geq a_{k}\geq 1$, and $b_{i}\geq 1$ for $1\leq i\leq l$. Let $G$ be a graph of sufficiently large order $n$.

(i). If $k\geq 1$ and $\delta>2h-\frac{h^{2}+h}{n}$, then $F\subseteq G$;

(ii). If $k=0$, $\delta>2h-\frac{h^{2}+h-4}{n}$, and there is at least one $b_{i}$ such that $b_{i}>1$, where $1\leq i\leq l$, then $F\subseteq G$.*

The related results can be referred to [4] and references therein.

1.3 Our main results

In this paper, we obtain stability versions of Corollary 1.7 for a graph which does not contain a linear forest with at most two odd paths. The main results in this paper are Theorems 1.8-1.11.

Theorem 1.8**.**

*Let $F=\bigcup_{i=1}^{k}P_{2a_{i}}$ and $h=\sum_{i=1}^{k}a_{i}-1\geq 1$, where $k\geq 2$ and $a_{1}\geq\cdots\geq a_{k}\geq 1$. Let $G$ be a connected graph of order $n$, where $n\geq 2h+2$. If $\delta(G)\geq h$, then $F\subseteq G$, unless one of the following holds:

(i). $G\subseteq S_{n,h}$;

(ii). $F=2P_{2a_{1}}$ and $G=L_{t,h}$, where $n=th+1$.*

Remark 1. The minimum degree condition is best possible. For example, let $F=2P_{2a_{1}}$ and $h=2a_{1}-1$. If $G=L_{\frac{n-1}{h-1},h-1}$, where $n=h(h-1)q+1$ with positive integer $q$, then $\delta(G)=h-1$, $F\nsubseteq G$, $G\nsubseteq S_{n,h}$, and $G\ncong L_{\frac{n-1}{h},h}$. So Theorem 1.8 does not hold for $\delta(G)=h-1$. For another example, let $F=P_{2}\bigcup 2P_{4}$ and $h=4$. If $G=L_{\frac{n-1}{3},3}$ where $n=12q+1$ with positive integer $q$, then $\delta(G)=3$, $F\nsubseteq G$, and $G\nsubseteq S_{n,4}$. So Theorem 1.8 does not hold for $\delta(G)=h-1$. Therefore the condition $\delta(G)\geq h$ in Theorem 1.8 can not be weaken and is best possible.

Theorem 1.9**.**

*Let $F=(\bigcup_{i=1}^{k}P_{2a_{i}})\bigcup P_{2b_{1}+1}$ and $h=\sum_{i=1}^{k}a_{i}+b_{1}-1\geq 1$, where $k\geq 1$, $a_{1}\geq\cdots\geq a_{k}\geq 1$, and $b_{1}\geq 1$. Let $G$ be a connected graph of order $n$, where $n\geq 2h+3$. If $\delta(G)\geq h$, then $F\subseteq G$, unless one of the following holds:

(i). $G\subseteq S_{n,h}$;

(ii). $F=P_{6}\bigcup P_{3}$ and $G\subseteq K_{2}\bigvee\frac{n-2}{2}K_{2}$, where $n$ is even;

(iii). $F\in\{P_{2b_{1}}\bigcup P_{2b_{1}+1},P_{2b_{1}+2}\bigcup P_{2b_{1}+1}\}$ and $G=L_{t,h}$, where $n=th+1$.*

Remark 2. The minimum degree condition is best possible. For example, let $F=P_{2b_{1}}\bigcup P_{2b_{1}+1}$ and $h=2b_{1}$. If $G=L_{\frac{n-1}{h-1},h-1}$, where $n=h(h-1)q+1$ with positive integer $q$, then $\delta(G)=h-1$, $F\nsubseteq G$, $G\nsubseteq S_{n,h}$, and $G\ncong L_{\frac{n-1}{h},h}$. Thus Theorem 1.9 does not hold for $\delta(G)=h-1$. For another example, let $F=P_{4}\bigcup P_{2}\bigcup P_{3}$ and $h=3$. If $G=L_{\frac{n-1}{2},2}$, where $n=6q+1$ with positive integer $q$, then $\delta(G)=2$, $F\nsubseteq G$, and $G\nsubseteq S_{n,3}$. Thus Theorem 1.9 does not hold for $\delta(G)=h-1$. Therefore the condition $\delta(G)\geq h$ in Theorem 1.9 is necessary and best possible.

Theorem 1.10**.**

Let $F=(\bigcup_{i=1}^{k}P_{2a_{i}})\bigcup(\bigcup_{i=1}^{2}P_{2b_{i}+1})$, $h=\sum_{i=1}^{k}a_{i}+\sum_{i=1}^{2}b_{i}-1\geq 2$, and $G$ be a 2-connected graph of order $n$, where $k\geq 0$, $a_{1}\geq\dots\geq a_{k}\geq 1$, $b_{1}\geq b_{2}$, and $n\geq 4(2h+1)^{2}\binom{2h+1}{h}$.

*(a). If $\delta(G)\geq h$ and $k=0$, then $F\subseteq G$, unless one of the following holds:

(i). $G\subseteq S^{+}_{n,h}$;

(ii). $F=P_{7}\bigcup P_{3}$ and $G\subseteq K_{2}\bigvee\frac{n-2}{2}K_{2}$, where $n$ is even;

(iii). $F=P_{9}\bigcup P_{3}$ and $G\subseteq K_{3}\bigvee\frac{n-3}{2}K_{2}$, where $n$ is odd.*

*(b). If $\delta(G)\geq h$ and $k\geq 1$, then $F\subseteq G$, unless one of the following holds:

(iv). $G\subseteq S_{n,h}$;

(v). $F=P_{4}\bigcup 2P_{3}$ and $G\subseteq K_{2}\bigvee\frac{n-2}{2}K_{2}$, where $n$ is even;

(vi). $F=P_{6}\bigcup 2P_{3}$ and $G\subseteq K_{3}\bigvee\frac{n-3}{2}K_{2}$, where $n$ is odd.*

Let $H_{n}^{1}$ be a graph of order $n$, where $n\geq 7$, obtained from $S_{n,2}$ and $K_{3}$ by identifying a vertex with maximum degree of $S_{n,2}$ with a vertex of $K_{3}$ (see Fig.3). Let $H_{n}^{2}$ be a graph order $n$, where $n\geq 9$, obtained from $H_{n-2}^{1}$ and $K_{3}$ by identifying a vertex with the second largest degree of $H_{n-2}^{1}$ with a vertex of $K_{3}$ (see Fig.3).

Let $U_{3,h}$ be a graph of order $3h+3$ obtained from $3K_{h+1}$ and $K_{3}$ by identifying every vertex of $K_{3}$ with a vertex of $K_{h+1}$, respectively (see Fig.3).

H_{n}^{1}$$H_{n}^{2}$$K_{h+1}$$K_{h+1}$$K_{h+1}$$U_{3,h}Fig.3. Graphs $H_{n}^{1}$, $H_{n}^{2}$, and $U_{3,h}$.

For $h\geq 2$, let $F_{t_{1},t_{2},h,h+1}$ be a graph of order $t_{1}h+(t_{2}+1)(h+1)+1$ obtained from $L_{t_{1},t_{2},h,h+1}$ and $K_{h+1}$ by adding an edge joining the center of $L_{t_{1},t_{2},h,h+1}$ and a vertex of $K_{h+1}$ (see Fig.4). Moreover, for $h\geq 2$, let $T_{t_{1},t_{2},h,h+1}$ be a graph of order $t_{1}h+(t_{2}+2)(h+1)+1$ obtained from $L_{t_{1},t_{2},h,h+1}$ and $2K_{h+1}$ by adding an edge joining the center of $L_{t_{1},t_{2},h,h+1}$ and a vertex of each $K_{h+1}$, respectively (see Fig.4).

K_{h+1}$$K_{h+1}$$K_{h+2}$$K_{h+2}$$K_{h+1}$$t_{1}$$t_{2}$$F_{t_{1},t_{2},h,h+1}$$K_{h+1}$$K_{h+1}$$K_{h+2}$$K_{h+2}$$K_{h+1}$$K_{h+1}$$t_{1}$$t_{2}$$T_{t_{1},t_{2},h,h+1}Fig.4. $F_{t_{1},t_{2},h,h+1}$ and $T_{t_{1},t_{2},h,h+1}$.

Theorem 1.11**.**

Let $F=(\bigcup_{i=1}^{k}P_{2a_{i}})\bigcup(\bigcup_{i=1}^{2}P_{2b_{i}+1})$, $h=\sum_{i=1}^{k}a_{i}+\sum_{i=1}^{2}b_{i}-1$, and $G$ be a connected graph of order $n$ with at least one cut vertex, where $k\geq 0$, $a_{1}\geq\dots\geq a_{k}\geq 1$, $b_{1}\geq b_{2}\geq 1$, and $n\geq 2h+4$.

*(a). If $\delta(G)\geq h\geq 1$ and $k=0$, then $F\subseteq G$, unless one of the following holds:

(i) $F=P_{5}\bigcup P_{3}$ and either $G\subseteq H_{n}^{1}$ or $G\subseteq H_{n}^{2}$;

(ii) $F=P_{2b_{1}+1}\bigcup P_{2b_{1}-1}$ and $G=L_{t,h}$, where $n=th+1$;

(iii) $F=2P_{2b_{1}+1}$ and $G=U_{3,h}$, where $n=3h+3$;

(iv) $F=2P_{2b_{1}+1}$ and $G\subseteq L_{t_{1},t_{2},h,h+1}$, where $n=t_{1}h+t_{2}(h+1)+1$

(v) $F=2P_{2b_{1}+1}$ and $G\subseteq F_{t_{1},t_{2},h,h+1}$, where $n=t_{1}h+(t_{2}+1)(h+1)+1$

(vi) $F=2P_{2b_{1}+1}$ and $G\subseteq T_{t_{1},t_{2},h,h+1}$, where $n=t_{1}h+(t_{2}+2)(h+1)+1$.*

(b). If $\delta(G)\geq h\geq 2$ an $k\geq 1$, then $F\subseteq G$, unless $F=P_{2}\bigcup 2P_{2b_{1}+1}$ and $G=L_{t,h}$, where $n=th+1$.

The rest of this paper is organized as follows. In Section 2, some technical lemmas are provided. In Section 3, we present the proofs of Theorems 1.8 and 1.9. In Section 4, we present the proofs of Theorems 1.10 and 1.11.

2 Preliminary

In order to prove Theorems 1.8–1.11, we require several known and new technical lemmas.

Lemma 2.1**.**

[6*]** Let $G$ be a 2-connected graph of order $n$ with $u_{1}\in V(G)$. If $d(u)\geq h\geq 2$ for any vertex $u$ different from $u_{1}$, then there is a path $P_{\min\{n,2h\}}$ with end vertex $u_{1}$ and $P_{\min\{n,2h\}}\subseteq G$. *

Lemma 2.2**.**

[5]** Let $G$ be a 2-connected graph of order at least $2h$, where $h\geq 2$. If $\delta(G)\geq h$, then $C_{l}\subseteq G$, where $l\geq 2h$.

Lemma 2.3**.**

[8]** Let $G$ be a graph of order $n$ with a set $P$ of $p$ vertices, where $p\geq 2$ and $n\geq 4p^{2}\binom{p}{\lfloor\frac{p}{2}\rfloor}$. If $e(P,V(G)\backslash P)\geq(\lfloor\frac{p}{2}\rfloor-\frac{1}{2})n$, then $P$ contains a subset of $\lfloor\frac{p}{2}\rfloor$ vertices with a common neighborhood of $p$ vertices in $V(G)\backslash P$.

Lemma 2.4**.**

*Let $G$ be a connected graph of order $n$ with a longest cycle $C_{l}$ and $\delta(G)\geq h\geq 2$. Let $U=V(G)\backslash V(C_{l})$.

(i). If $l=2h$ and $U$ is an independent set, then $G\subseteq S_{n,h}$.

(ii). If $l=2h+1$ and $P_{2h+3}\nsubseteq G$, then $G\subseteq S^{+}_{n,h}$.

(iii). If $l=2h+2$ and $P_{2h+4}\nsubseteq G$, then $N_{C_{l}}(u_{1})=N_{C_{l}}(u_{2})$ and $h\leq d_{C_{l}}(u_{1})=d_{C_{l}}(u_{2})\leq h+1$ for every pair of vertices $u_{1},u_{2}\in U$.*

Proof. Let $C_{l}=v_{1}v_{2}\cdots v_{l}v_{1}$. Since $C_{l}$ is a longest cycle, none of vertices in $U$ is adjacent to any two consecutive vertices of $C_{l}$.

$(i)$. Since $l=2h$ and $U$ is an independent set, $d_{C_{l}}(u)=h$ for all $u\in U$. Furthermore either $N_{C_{l}}(u)=\{v_{1},v_{3},\ldots,v_{2h-1}\}$ or $N_{C_{l}}(u)=\{v_{2},v_{4},\ldots,v_{2h}\}$ for all $u\in U$. In fact, if there exist two distinct vertices $u_{1},u_{2}\in U$ such that $N_{C_{l}}(u_{1})=\{v_{1},v_{3},\ldots,v_{2h-1}\}$ and $N_{C_{l}}(u_{2})=\{v_{2},v_{4},\ldots,v_{2h}\}$, then a cycle $u_{1}v_{1}v_{2h}u_{2}v_{2}v_{3}v_{4}\cdots v_{2h-3}v_{2h-2}v_{2h-1}u_{1}$ is a longer cycle than $C_{l}$, a contradiction. Hence, we assume without loss of generality that $N_{C_{l}}(u)=\{v_{2},v_{4},\ldots,v_{2h}\}$ for all $u\in U$. Moreover, $\{v_{1},v_{3},\ldots,v_{2h-1}\}$ is an independent set. In fact, if there exists an edge $v_{2s+1}v_{2t+1}\in E(G)$ with $1\leq 2s+1<2t+1\leq 2h-1$, then a cycle $uv_{2t}v_{2t-1}v_{2t-2}\cdots v_{2s+2}v_{2s+1}v_{2t+1}v_{2t+2}\cdots v_{2h-1}v_{2h}v_{1}v_{2}\cdots v_{2s-1}v_{2s}u$ for $u\in U$ is a longer cycle than $C_{l}$, a contradiction. Hence $G\subseteq S_{n,h}$.

$(ii)$. If $l=2h+1$ and $P_{2h+3}\nsubseteq G$, then $U$ is an independent set and $d_{C_{l}}(u)=h$ for all $u\in U$. Furthermore, we claim that $N_{C_{l}}(u_{1})=N_{C_{l}}(u_{2})$ for any two vertices $u_{1},u_{2}\in U$. In fact, we suppose without loss of generality that there exists $v\in N_{C_{l}}(u_{2})\backslash N_{C_{l}}(u_{1})$. Since $P_{2h+3}\nsubseteq G$, neither of two neighbors of $v$ along $C_{l}$ belongs to $N_{C_{l}}(u_{1})$. Hence $N_{C_{l}}(u_{1})$ is a subset of a set consisting of $2h-2$ consecutive vertices of ${C_{l}}$. Furthermore, since ${C_{l}}$ is a longest cycle, $u_{1}$ is not adjacent to any two consecutive vertices of ${C_{l}}$. Hence we $d_{C_{l}}(u_{1})\leq h-1$, a contradiction. We assume without loss of generality that $N_{C_{l}}(u)=\{v_{2},v_{4},\ldots,v_{2h}\}$ for all $u\in U$. Moreover, $\{v_{1},v_{3},\ldots,v_{2h-1}\}$ is an independent set. Otherwise there exists an edge $v_{2s+1}v_{2t+1}\in E(G)$ with $1\leq 2s+1<2t+1\leq 2h-1$, which causes a longer cycle $uv_{2t}v_{2t-1}v_{2t-2}\cdots v_{2s+2}v_{2s+1}v_{2t+1}v_{2t+2}\cdots v_{2h}v_{2h+1}v_{1}\cdots v_{2s}u$ for $u\in U$ than $C_{l}$, a contradiction. Similarly, $\{v_{3},\ldots,v_{2h+1}\}$ is also an independent set. Hence G\big{[}U\bigcup\{v_{1},\ldots,v_{2h+1}\}\big{]} contains at most one edge $v_{1}v_{2h+1}$. Therefore $G\subseteq S^{+}_{n,h}$.

$(iii).$ If $l=2h+2$ and $P_{2h+4}\nsubseteq G$, then $U$ is an independent set and $h\leq d_{C_{l}}(u)\leq h+1$ for all $u\in U$. Furthermore, $N_{C_{l}}(u_{1})=N_{C_{l}}(u_{2})$ for any two vertices $u_{1},u_{2}\in U$. In fact, suppose without loss of generality that there exists a vertex $v\in N_{C_{l}}(u_{2})\backslash N_{C_{l}}(u_{1})$. Since $P_{2h+4}\nsubseteq G$, neither of two neighbors of $v$ along $C_{l}$ belongs to $N_{C_{l}}(u_{1})$. Hence $N_{C_{l}}(u_{1})$ is a subset of a set consisting of $2h-1$ consecutive vertices of ${C_{l}}$. Furthermore, since ${C_{l}}$ is a longest cycle, $u_{1}$ is not adjacent to any two consecutive vertices of $C_{l}$. Hence $d_{C_{l}}(u_{1})\leq h-1$, a contradiction. Hence $N_{C_{l}}(u_{1})=N_{C_{l}}(u_{2})$ and the assertion holds.  $\square$

Lemma 2.5**.**

*Let $G$ be a connected graph of order $n$ with a longest cycle $C_{l}$ and minimum degree $\delta(G)$, where $l\leq 2h+1$ and $\delta(G)\geq h\geq 2$. Let $U=V(G)\backslash V(C_{l})$.

(i). If $G[U]$ is $P_{3}$-free, then $N_{C_{l}}(u_{1})=N_{C_{l}}(u_{2})$ for every edge $u_{1}u_{2}\in E(U)$.

(ii). If $G[U]$ is $P_{4}$-free, then $N_{C_{l}}(u_{1})=N_{C_{l}}(u_{3})$ for every $P_{3}=u_{1}u_{2}u_{3}\subseteq G[U]$.*

Proof. $(i)$. Suppose that there exists a vertex $v\in N_{C_{l}}(u_{2})\backslash N_{C_{l}}(u_{1})$ for some edge $u_{1}u_{2}\in E(U)$. Since $C_{l}$ is a longest cycle, the distance along $C_{l}$ between $v$ and any vertex in $N_{C_{l}}(u_{1})$ is at least $3$. Thus $N_{C_{l}}(u_{1})$ is a subset of a set consisting of at most $2h-5$ consecutive vertices of ${C_{l}}$. Furthermore, since $C_{l}$ is a longest cycle, $u_{1}$ is not adjacent to any two consecutive vertices of $C_{l}$. These discussions imply that $d_{C_{l}}(u_{1})\leq h-2$. However, since $P_{3}\nsubseteq G[U]$, we have $d_{C_{l}}(u_{1})\geq d_{G}(u_{1})-1\geq h-1$, a contradiction. Hence the assertion holds.

$(ii)$. Suppose that there exists a vertex $v\in N_{C_{l}}(u_{3})\backslash N_{C_{l}}(u_{1})$ for some path $P_{3}=u_{1}u_{2}u_{3}\subseteq G[U]$. Since $C_{l}$ is a longest cycle, the distance along $C_{l}$ between $v$ and any vertex in $N_{C_{l}}(u_{1})$ is at least $4$. Hence $N_{C_{l}}(u_{1})$ is a subset of a set consisting of at most $2h-7$ consecutive vertices of $C_{l}$. Furthermore, since $C_{l}$ is a longest cycle, $u_{1}$ is not adjacent to any two consecutive vertices of $C_{l}$. These discussions imply that $d_{C_{l}}(u_{1})\leq h-3$. However, since $P_{4}\nsubseteq G[U]$, we have $d_{C_{l}}(u_{1})\geq d_{G}(u_{1})-2\geq h-2$, a contradiction. Hence the assertion holds.  $\square$

Lemma 2.6**.**

*Let $F=(\bigcup_{i=1}^{k}P_{2a_{i}})\bigcup(\bigcup_{i=1}^{2}P_{2b_{i}+1})$ and $h=\sum_{i=1}^{k}a_{i}+\sum_{i=1}^{2}b_{i}-1\geq 2$, where $k\geq 0$, $a_{1}\geq\cdots\geq a_{k}\geq 1$, and $b_{1}\geq b_{2}\geq 1$. Let $H$, i.e., $H=(X,Y;E)$, be a complete bipartite graph with partition size $|X|=h$ and $|Y|=h+2$.

$(i).$ If $k\geq 1$ and the graph $G$ is obtained from $H$ and $P_{3}$ by identifying a vertex $u\in X$ with an end vertex of $P_{3}$, then $F\subseteq G$.

$(ii).$ If the graph $G$ is obtained from $H$ and $P_{4}$ by identifying a vertex $u\in X$ with an end vertex of $P_{4}$, then $F\subseteq G$.

$(iii).$ If the graph $G$ is obtained from $H-v$ with $v\in X$ and $P_{6}$ by identifying $u\in X$ with an end vertex of $P_{6}$, then $F\subseteq G$.

Proof. $(i).$ Since $H$ is a complete bipartite graph with partition size $|X|=h$ and $|Y|=h+2$, there exist three disjoint complete bipartite subgraphs $H_{i}$ of $H$, i.e., $H_{i}=(X_{i},Y_{i}:E_{i})$ for $1\leq i\leq 3$, where $|X_{1}|=|Y_{1}|=\sum_{i=1}^{k}a_{i}-1$ with $u\in X_{1}$, $|X_{2}|=b_{1},|Y_{2}|=b_{1}+1$, $|X_{3}|=b_{2}$, and $|Y_{3}|=b_{2}+1$. Hence there exists a path $P_{\sum_{i=1}^{k}2a_{i}-2}$in $H_{1}$ with one end vertex $u$. In addition, there exist two disjoint paths $P_{2b_{1}+1}$ in $H_{2}$ and $P_{2b_{2}+1}$ in $H_{3}$. Hence $F\subseteq G$. So $(i)$ holds. By a similar argument in $(i),$ it is easy to see that $(ii)$ and $(iii)$ hold.  $\square$

Lemma 2.7**.**

*Let $H$ be a 2-connected graph of order $n$, where $n\geq 2h+1$ and $2\leq h\leq 3$, and $u_{1}\in V(H)$ such that $d_{H}(u)\geq h$ for every $u$ different from $u_{1}$. Let $G$ be a graph obtained from $H$ and $P_{t}$ by identifying $u_{1}$ with an end vertex of $P_{t}$, where $3\leq t\leq 4$.

$(i).$ If $h=2$ and $t=3$, then $2P_{2}\bigcup P_{3}\subseteq G$, $P_{4}\bigcup P_{3}\subseteq G$, and $P_{2}\bigcup P_{5}\subseteq G$. $(ii).$ If $h=2$ and $t=4$, then $P_{5}\bigcup P_{3}\subseteq G$.

$(iii).$ If $h=3$ and $t=4$, then $P_{7}\bigcup P_{3}\subseteq G$ and $2P_{5}\subseteq G$.

Proof. ($i$). Let $C_{l}$ be a longest cycle in $H$. Since $H$ be a 2-connected graph, $\delta(H)\geq 2$. By Lemma 2.2, $l\geq 4$. Since $H$ is connected, if $l\geq 5$ then $2P_{2}\bigcup P_{3}\subseteq G$, $P_{4}\bigcup P_{3}\subseteq G$, and $P_{2}\bigcup P_{5}\subseteq G$. Since $n\geq 5$, if $l=4$ then there exists a vertex $v\in V(H)\backslash V(C_{l})$ adjacent to some vertex of $C_{l}$. We have $2P_{2}\bigcup P_{3}\subseteq G$, $P_{4}\bigcup P_{3}\subseteq G$, and $P_{2}\bigcup P_{5}\subseteq G$. Thus the assertion holds.

($ii$). The proof of $(ii)$ is similar to that of $(i)$ and the detail is omitted.

$(iii)$. By Lemma 2.1, $H$ has a path $P_{6}$ with an end vertex $u_{1}$. Let $P_{6}=u_{1}u_{2}\cdots u_{6}$. Since $n\geq 7$, there exists a vertex $v\in V(H)\backslash V(P_{6})$ adjacent to some vertex of $P_{6}$. If $u_{1}$, $u_{3}$, or $u_{6}\in N_{P_{6}}(v)$, then $P_{7}\bigcup P_{3}\subseteq G$ and $2P_{5}\subseteq G$. Furthermore, if $\{u_{4},u_{5}\}\subseteq N_{P_{6}}(v)$, then $P_{7}\bigcup P_{3}\subseteq G$ and $2P_{5}\subseteq G$. Next we assume $N_{P_{6}}(v)\in\{\{v_{2},v_{4}\},\{v_{2},v_{5}\},\{u_{2}\},\{u_{4}\},\{u_{5}\}\}$. Since $d_{H}(v)\geq 3$, if $N_{P_{6}}(v)\in\{\{v_{2},v_{4}\},\{v_{2},v_{5}\}\}$ then there exists a vertex in $V(H)\backslash V({P_{6}})$ adjacent to $v$. Thus $P_{7}\bigcup P_{3}\subseteq G$ and $2P_{5}\subseteq G$. Moreover, since $d_{H}(v)\geq 3$, if $N_{P_{6}}(v)\in\{\{u_{4}\},\{u_{5}\}\}$ then there exists two distinct vertices in $V(H)\backslash V({P_{6}})$ adjacent to $v$. Thus $P_{7}\bigcup P_{3}\subseteq G$ and $2P_{5}\subseteq G$. Finally, since $H$ is 2-connected, if $N_{P_{6}}(v)=\{u_{2}\}$ then there exists a vertex in $V(H)\backslash V({P_{6}})$ adjacent to some vertex in $V({P_{6}})\backslash\{u_{2}\}$ ($\{u_{2}\}$ can not separate $V(P)\backslash\{u_{2}\}$ from the rest). By repeating the arguments above, $P_{7}\bigcup P_{3}\subseteq G$ and $2P_{5}\subseteq G$. Thus the assertion holds.  $\square$

Lemma 2.8**.**

Let $H$ be a 2-connected graph of order $n$ and $H\nsubseteq S_{n,2}$, where $n\geq 6$. If a graph $G$ is obtained from $H$ and $P_{3}$ by identifying a vertex $u_{1}\in V(H)$ with an end vertex of $P_{3}$, then $P_{5}\bigcup P_{3}\subseteq G$.

Proof. Suppose $P_{5}\bigcup P_{3}\nsubseteq G$. Let $C_{l}$ be a longest cycle in $H$. Since $H$ is 2-connected, $\delta(H)\geq 2$. By Lemma 2.2, $l\geq 4$.

Case 1: $l\geq 6$, or $l=5$ with $u_{1}\notin V(C_{l})$. According to the structure of $G$, we have $P_{5}\bigcup P_{3}\subseteq G$, a contradiction.

Case 2: $l=5$ with $u_{1}\in V(C_{l})$. Since $n\geq 6$, there exists a vertex $u\in V(H)\backslash V(C_{l})$ adjacent to some vertex of $C_{l}$. Furthermore, since $P_{5}\bigcup P_{3}\nsubseteq G$, we have $N_{C_{l}}(u)=\{u_{1}\}$. Moreover, since $\delta(H)\geq 2$, there exists a vertex in $V(H)\backslash V(C_{l})$ adjacent to $u$. Hence $P_{5}\bigcup P_{3}\subseteq G$, a contradiction.

Case 3: $l=4$. There exists a shortest path $P_{k}$ between $u_{1}$ and some vertex of $C_{l}$ such that $|V(P_{k})\bigcap V(C_{l})|=1$. Since $P_{5}\bigcup P_{3}\nsubseteq G$, we have $1\leq k\leq 2$.

Subcase 3.1: $k=2$. Let $C_{4}=v_{1}v_{2}v_{3}v_{4}v_{1}$, $V(C_{4})\bigcap V(P_{2})=\{v_{1}\}$, and $U=V(H)\backslash(V(C_{4})\bigcup\{u_{1}\})$. It is easy to see that $U\neq\varnothing$. First we assume that $U$ is an independent set. Since $C_{4}$ is a longest cycle in $H$ and $\delta(H)\geq 2$, $N_{H}(u)\in\{\{u_{1},v_{1}\},\{v_{1},v_{3}\},\{v_{2},v_{4}\}\}$ for every $u\in U$. However, in each case, $P_{5}\bigcup P_{3}\subseteq G$, a contradiction. Next we assume that $H[U]$ contains at least one edge. Then there exists one edge $w_{1}w_{2}$ such that one of $w_{1}$ and $w_{2}$ is adjacent to some vertex in $V(C_{4})\bigcup\{u_{1}\}$ and therefore $P_{5}\bigcup P_{3}\subseteq G$, a contradiction.

Subcase 3.2: $k=1$. Obviously $u_{1}\in V(C_{4})$. Let $C_{4}=u_{1}u_{2}u_{3}u_{4}u_{1}$ and $U=V(H)\backslash V(C_{4})$. Since $n\geq 6$, we have $|U|\geq 2$. First we assume that $U$ is an independent set. Since $C_{4}$ is a longest cycle and $\delta(H)\geq 2$, either $N_{H}(v)=\{u_{1},u_{3}\}$ or $N_{H}(v)=\{u_{2},u_{4}\}$ for $v\in U$. We claim that $N_{H}(v)=\{u_{1},u_{3}\}$ for all $v\in U$. Otherwise $P_{5}\bigcup P_{3}\subseteq G$, a contradiction. Moreover, since $P_{5}\bigcup P_{3}\nsubseteq G$, it follows that $u_{2}u_{4}\notin E(G)$. Hence $H\subseteq S_{n,2}$, a contradiction. Next we assume that $H[U]$ contains at least one edge $v_{1}v_{2}$. Furthermore, since $P_{5}\bigcup P_{3}\nsubseteq G$, we have $P_{3}\nsubseteq H[U]$. By Lemma 2.5 (i), $N_{C_{4}}(v_{1})=N_{C_{4}}(v_{2})$. Since $C_{4}$ is a longest cycle, $d_{C_{4}}(v_{1})=d_{C_{4}}(v_{2})=1$. This discussion implies that the unique common neighbor of $v_{1}$ and $v_{2}$ in $V(C_{4})$ is a cut vertex of $H$, which contradicts that $H$ is 2-connected. Thus the assertion holds.  $\square$

3 Proofs of Theorems 1.8 and 1.9

Now we are ready to prove Theorem 1.8, i.e.,

Theorem 3.1**.**

*Let $F=\bigcup_{i=1}^{k}P_{2a_{i}}$ and $h=\sum_{i=1}^{k}a_{i}-1\geq 1$, where $k\geq 2$ and $a_{1}\geq\cdots\geq a_{k}\geq 1$. Let $G$ be a connected graph of order $n$, where $n\geq 2h+2$. If $\delta(G)\geq h$, then $F\subseteq G$, unless one of the following holds:

(i). $G\subseteq S_{n,h}$;

(ii). $F=2P_{2a_{1}}$ and $G=L_{t,h}$, where $n=th+1$.*

Proof. Suppose $F\nsubseteq G$. Since $\sum_{i=1}^{k}2a_{i}=2h+2$, we have $P_{2h+2}\nsubseteq G$. We consider the following two cases.

Case 1: $G$ is 2-connected. It follows that $\delta(G)\geq 2$. Let $C_{l}$ be a longest cycle in $G$ and $U=V(G)\backslash V(C_{l})$. By Lemma 2.2, $l\geq 4$. First we claim that $h\geq 2$. In fact, if $h=1$, then $F=2P_{2}$ and thus $F\subseteq G$, a contradiction. By Lemma 2.2, $h\geq 2$ and therefore $l\geq 2h$. Since $n\geq 2h+2$ and $P_{2h+2}\nsubseteq G$, we have $l\leq 2h$ and therefore $l=2h$. Moreover, since $P_{2h+2}\nsubseteq G$, $U$ is an independent set. By Lemma 2.4 $(i)$, $G\subseteq S_{n,h}$.

Case 2: $G$ has at least one cut vertex. Since $k\geq 2$, if $h=1$ then $F=2P_{2}$. Since $F\nsubseteq G$, $G$ must be a star $K_{1,n-1}$, i.e., $G=L_{n-1,1}$. Next we assume that $h\geq 2$. Since $G$ has at least one cut vertex, there exist at least two end blocks. We claim that $|V(B)|=h+1$ for every end block $B$ of $G$. In fact, since $\delta(G)\geq h$, it follows that $|V(B)|\geq h+1$. Furthermore, since $P_{2h+2}\nsubseteq G$, Lemma 2.1 implies that $B$ has order at most $h+1$ and hence $|V(B)|=h+1$. Let $B_{i}$ be an end block of $G$ with a vertex $u_{i}$ which is a cut vertex of $G$ for $1\leq i\leq 2$. It follows that $|V(B_{i})|=h+1$ for $1\leq i\leq 2$. If there exists a path $P_{k}$ with $k\geq 2$ starting at $u_{1}$ and ending at $u_{2}$ such that $V(P_{k})\bigcap(V(B_{1})\bigcup V(B_{2}))=\{u_{1},u_{2}\}$, then Lemma 2.1 implies $P_{2h+2}\subseteq G$, a contradiction. It follows that $G=L_{t,h}$, where $n=th+1$. Moreover, since $n\geq 2h+2$, $G$ has at least three blocks, i.e., $t\geq 3$. By Lemma 2.1, $P_{h}\bigcup P_{2h+1}\subseteq G$. We claim that $k=2$ and $a_{1}=a_{2}$. Otherwise, since

[TABLE]

and

[TABLE]

we have $F\subseteq G$, a contradiction. Therefore $F=2P_{2a_{1}}$ and $G=L_{t,h}$, where $n=th+1$.  $\square$

Theorem 1.9 can be stated as follows.

Theorem 3.2**.**

*Let $F=(\bigcup_{i=1}^{k}P_{2a_{i}})\bigcup P_{2b_{1}+1}$ and $h=\sum_{i=1}^{k}a_{i}+b_{1}-1\geq 1$, where $k\geq 1$, $a_{1}\geq\cdots\geq a_{k}\geq 1$, and $b_{1}\geq 1$. Let $G$ be a connected graph of order $n$, where $n\geq 2h+3$. If $\delta(G)\geq h$, then $F\subseteq G$, unless one of the following holds:

(i). $G\subseteq S_{n,h}$;

(ii). $F=P_{6}\bigcup P_{3}$ and $G\subseteq K_{2}\bigvee\frac{n-2}{2}K_{2}$, where $n$ is even;

(iii). $F\in\{P_{2b_{1}}\bigcup P_{2b_{1}+1},P_{2b_{1}+2}\bigcup P_{2b_{1}+1}\}$ and $G=L_{t,h}$, where $n=th+1$.*

Proof. Suppose $F\nsubseteq G$. Since $\sum_{i=1}^{k}2a_{i}+2b_{1}+1=2h+3$, we have $P_{2h+3}\nsubseteq G$. We consider the following two cases.

Case 1: $G$ is 2-connected. It follows that $\delta(G)\geq 2$. Let $C_{l}$, denoted by $v_{1}v_{2}\cdots v_{l}v_{1}$, be a longest cycle in $G$ and $U=V(G)\backslash V(C_{l})$. By Lemma 2.2, $l\geq 4$. We claim that $h\geq 2$. In fact, if $h=1$, then $F=P_{2}\bigcup P_{3}$. Since $n\geq 5$ and $l\geq 4$, we have $F\subseteq G$, a contradiction. By Lemma 2.2, $l\geq 2h$. Furthermore, since $n\geq 2h+3$ and $P_{2h+3}\nsubseteq G$, we have $l\leq 2h+1$. So $2h\leq l\leq 2h+1$. We consider the following two subcases.

Subcase 1.1: $l=2h$. Since $P_{2h+3}\nsubseteq G$, we have $P_{3}\nsubseteq G[U]$. If $U$ is an independent set, then Lemma 2.4 $(i)$ implies $G\subseteq S_{n,h}$. Now we assume that $G[U]$ consists of $p$ disjoint edges and $q$ isolated vertices, i.e., $u_{1}u_{2},u_{3}u_{4},\ldots,u_{2p-1}u_{2p},w_{1},\ldots,w_{q}$, where $p\geq 1$ and $q\geq 0$. By Lemma 2.5 $(i)$, $N_{C_{2h}}(u_{2i-1})=N_{C_{2h}}(u_{2i})$ for $1\leq i\leq p$. Note that $d_{C_{2h}}(u_{j})=d_{G}(u_{j})-1\geq h-1$ for $1\leq j\leq 2p$. Since $C_{{2h}}$ is a longest cycle in $G$, the distance along $C_{{2h}}$ between two vertices in $N_{C_{2h}}(u_{1})$ is at least 3, which implies $3(h-1)\leq 2h$. Thus $h\leq 3$. Furthermore, $h=3$ (Otherwise $h=2$, which implies $d_{C_{4}}(u_{i})=1$ and $d_{G}(u_{i})=2$ for $1\leq i\leq 2$. Thus the common neighbor of $u_{1}$ and $u_{2}$ in $V(C_{4})$ is a cut vertex, which contradicts that $G$ is 2-connected). Now let $\mathcal{A}=\{P_{6}\bigcup P_{3},P_{4}\bigcup P_{5},P_{2}\bigcup P_{7},P_{4}\bigcup P_{2}\bigcup P_{3},2P_{2}\bigcup P_{5},3P_{2}\bigcup P_{3}\}$. Since $h=3$ and $k\geq 1$, we have $F\in\mathcal{A}$. Moreover, $d_{C_{6}}(u_{2i-1})=d_{C_{6}}(u_{2i})=2$ and $d_{G}(u_{2i-1})=d_{G}(u_{2i})=3$ for $1\leq i\leq p$. Since $C_{6}$ is a longest cycle in $G$, we assume without loss of generality that $N_{C_{6}}(u_{1})=N_{C_{6}}(u_{2})=\{v_{1},v_{4}\}$. Since $F\nsubseteq G$, we have $N_{C_{6}}(u_{2i-1})=N_{C_{6}}(u_{2i})=\{v_{1},v_{4}\}$ for $1\leq i\leq p$. Next we claim that $q=0$. In fact, if $q\geq 1$, then $d_{G}(w_{1})=d_{C_{6}}(w_{1})=3$, which implies $F\subseteq G$, a contradiction. So $n$ must be even. Hence $G\subseteq K_{2}\bigvee\frac{n-2}{2}K_{2}$. Moreover, $F=P_{6}\bigcup P_{3}$. The assertion holds.

Subcase 1.2: $l=2h+1$. By the proof of Lemma 2.4 $(ii)$, we have $N_{C_{2h+1}}(u)=\{v_{2},v_{4},\ldots,v_{2h}\}$ for every $u\in U$. Hence there exist two disjoint paths $uv_{2}v_{3}\cdots v_{2b_{1}+1}$ and $wv_{2b_{1}+2}\cdots v_{2h+1}v_{1}$ for two different vertices $u,w\in U$. Hence $F\subseteq G$ and it is a contradiction.

Case 2: $G$ has at least one cut vertex. If $h=1$, then $F=P_{2}\bigcup P_{3}$. Since $F\nsubseteq G$, $G$ must be a star $K_{1,n-1}$, i.e., $G=L_{n-1,1}$. Now we assume that $h\geq 2$. First we prove the following three Claims.

Claim 1: $P_{h+1}\bigcup P_{2h+1}\nsubseteq G$.

Suppose $P_{h+1}\bigcup P_{2h+1}\subseteq G$. If $\sum_{i=1}^{k}a_{i}\geq b_{1}+1$, then

[TABLE]

and

[TABLE]

Thus $F\subseteq G$, a contradiction. If $\sum_{i=1}^{k}a_{i}\leq b_{1}$, then

[TABLE]

and

[TABLE]

Thus $F\subseteq G$, a contradiction. So Claim 1 holds.

Claim 2: $\large|V(B)\large|=h+1$ for every end block $B$ of $G$.

Since $\delta(G)\geq h$, it follows that $|V(B)|\geq h+1$ for every end block $B$ of $G$. Next we prove that $\large|V(B)\large|\leq h+1$ for every end block $B$ of $G$. Suppose that there exists an end block $B_{1}$ of order at least $h+2$. If there exists another end block $B_{2}$ such that $V(B_{1})\bigcap V(B_{2})=\varnothing$, then Lemma 2.1 implies $P_{2h+3}\subseteq G$, a contradiction. If there exist another two end blocks $B_{2}$ and $B_{3}$ such that $B_{1},B_{2},$ and $B_{3}$ share a common cut vertex $u$ of $G$, then Lemma 2.1 implies that there exist three paths $P_{h+2}$, $P_{h+1}$, and $P_{h+1}$ in $B_{1},B_{2}$ and $B_{3}$ with an end vertex $u$, respectively. Hence $P_{h+1}\bigcup P_{2h+1}\subseteq G$, a contradiction. These discussions imply that there are exactly two blocks $B_{1}$ and $B_{2}$ sharing one common cut vertex $u$ of $G$, where $|V(B_{1})|\geq h+2$ and $|V(B_{2})|=n-|V(B_{1})|+1$. By Lemma 2.1, there exist two paths $P_{l_{1}}$ and $P_{l_{2}}$ with an end vertex $u$ in $B_{1}$ and $B_{2}$, respectively, where $l_{1}\geq\min\{|V(B_{1})|,2h\}$ and $l_{2}\geq\min\{|V(B_{2})|,2h\}$. Hence $P_{2h+3}\subseteq G$, a contradiction. Therefore Claim 2 holds.

Claim 3: $G=L_{t,h}$, where $n=th+1$ and $F\in\{P_{2b_{1}}\bigcup P_{2b_{1}+1},P_{2b_{1}+2}\bigcup P_{2b_{1}+1}\}$.

Choose two end blocks $B_{1}$ and $B_{2}$ of $G$ such that $p(B_{1},B_{2})$ is as large as possible, where $p(B_{1},B_{2})$ is the order of a longest path between the unique cut vertex $u_{1}$ of $G$ in $V(B_{1})$ and the unique cut vertex $u_{2}$ of $G$ in $V(B_{2})$. By Claim 2, $|V(B_{1})|=|V(B_{2})|=h+1$. If $p(B_{1},B_{2})\geq 3$, then Lemma 2.1 implies $P_{2h+3}\subseteq G$, a contradiction. Since $n\geq 2h+3$, if $p(B_{1},B_{2})=2$ then there exists another end block $B_{3}$ of $G$, sharing a common vertex $u_{1}$ with $B_{1}$ or a common vertex $u_{2}$ with $B_{2}$. By Lemma 2.1, $P_{h+1}\bigcup P_{2h+1}\subseteq G$, which contradicts Claim 1. Hence all blocks share a common cut vertex of $G$, i.e., $G=L_{t,h}$, where $n=th+1$. Moreover, $P_{h}\bigcup P_{2h+1}\subseteq G$, $P_{h-1}\bigcup P_{h}\bigcup P_{h+2}\subseteq G$, and $P_{h}\bigcup P_{h}\bigcup P_{h+1}\subseteq G$. If $\sum_{i=1}^{k}a_{i}\geq b_{1}+2$, then

[TABLE]

and

[TABLE]

Hence $F\subseteq G$, a contradiction. If $\sum_{i=1}^{k}a_{i}\leq b_{1}-1$, then

[TABLE]

and

[TABLE]

Hence $F\subseteq G$, a contradiction. If $\sum_{i=1}^{k}a_{i}=b_{1}$ and $k\geq 2$, then

[TABLE]

[TABLE]

and

[TABLE]

Hence $F\subseteq G$, a contradiction. If $\sum_{i=1}^{k}a_{i}=b_{1}+1$ and $k\geq 2$, then

[TABLE]

[TABLE]

and

[TABLE]

Hence $F\subseteq G$, a contradiction. So we have $F\in\{P_{2b_{1}}\bigcup P_{2b_{1}+1},P_{2b_{1}+2}\bigcup P_{2b_{1}+1}\}$ and $G=L_{t,h}$, where $n=th+1$.  $\square$

4 Proofs of Theorems 1.10 and 1.11

We will use the next several lemmas in Theorem 1.10.

Lemma 4.1**.**

*Let $G$ be a connected graph of order $n$.

(i). If $\delta(G)\geq 1$ and $n\geq 6$, then $2P_{3}\subseteq G$, unless either $G=U_{3,1}$, or $G\subseteq L_{t_{1},t_{2},1,2}$ for $n=t_{1}+2t_{2}+1$.

(ii). If $\delta(G)\geq 2$ and $n\geq 8$, then $P_{5}\bigcup P_{3}\subseteq G$, unless $G\subseteq S_{n,2}^{+}$, $G\subseteq H_{n}^{1}$, $G\subseteq H_{n}^{2}$, or $G=L_{t,2}$ for $n=2t+1$.

(iii). If $\delta(G)\geq 2$ and $n\geq 8$, then $P_{2}\bigcup 2P_{3}\subseteq G$, unless either $G\subseteq S_{n,2}$, or $G=L_{t,2}$ for $n=2t+1$.*

Proof. ($i$). Suppose $2P_{3}\nsubseteq G$. Let $P_{l}$ be a longest path in $G$, denoted by $v_{1}v_{2}\cdots v_{l}$. Since $n\geq 6$ and $2P_{3}\nsubseteq G$, we have $3\leq l\leq 5$. If $l=3$, then $G$ must be a star $K_{1,n-1}$, i.e., $G=L_{n-1,1}$. Since $n\geq 6$ and $2P_{3}\nsubseteq G$, if $l=4$ then $G\subseteq L_{n-3,1,1,2}$. Next we assume that $l=5$. Since $2P_{3}\nsubseteq G$, we have $N_{P_{5}}(u)\subseteq\{v_{3}\}$ for all $u\in V(G)\backslash V(P_{5})$. Let $X\subseteq V(G)\backslash V(P_{5})$ such that $N_{P_{5}}(u)=\{v_{3}\}$ for all $u\in X$ and $Y=V(G)\backslash(V(P_{5})\bigcup X)$. Since $n\geq 6$ and $2P_{3}\nsubseteq G$, we have $X\neq\varnothing$ and $P_{3}\nsubseteq G[X]$. So we can assume that $G[X]$ consists of $p$ disjoint edges and $q$ isolated vertices, i.e., $u_{1}u_{2},\ldots,u_{2p-1}u_{2p},w_{1},\ldots,w_{q}$, where $p+q\geq 1$. Since $P_{5}$ is a longest path in $G$, $Y$ must be an independent set if $|Y|\geq 1$. Furthermore, $N_{G}(u)=\{w_{i}\}$ and $N_{G}(v)=\{w_{j}\}$ for any two distinct vertices $u,v\in Y$ (it also holds for $|Y|=1$), where $1\leq i,j\leq q$ and $i\neq j$. Moreover, $v_{1}v_{4}$, $v_{1}v_{5}$, $v_{2}v_{5}\notin E(G)$, otherwise $2P_{3}\subseteq G$, a contradiction. Since $2P_{3}\nsubseteq G$, if $v_{2}v_{4}\in E(G)$ then $v_{1}v_{3}\notin E(G)$, $|Y|=p=0$, and $q=1$, which implies that $G=U_{3,1}$. If $v_{2}v_{4}\notin E(G)$, then $G\subseteq L_{t_{1},t_{2},1,2}$, where $n=t_{1}+2t_{2}+1$. Thus the assertion holds.

($ii$). Suppose $P_{5}\bigcup P_{3}\nsubseteq G$. Let $C_{l}$ be a longest cycle in $G$, denoted by $v_{1}v_{2}\cdots v_{l}v_{1}$, and $U=V(G)\backslash V(C_{l})$. Since $C_{l}$ is a longest cycle, none of vertices in $U$ is adjacent to any two consecutive vertices of $C_{l}$. Since $n\geq 8$ and $P_{5}\bigcup P_{3}\nsubseteq G$, it follows that $3\leq l\leq 6$. We consider the following four cases.

Case 1: $l=3$. This implies that all cycles in $G$ are triangles. Since $\delta(G)\geq 2$, $G$ has at least two triangles. Let $T_{1}$ and $T_{2}$ be two triangles in $G$ such that the longest path $P_{k}$, denoted by $u_{1}u_{2}\cdots u_{k}$, with $V(P_{k})\bigcap V(T_{1})=\{u_{1}\}$ and $V(P_{k})\bigcap V(T_{2})=\{u_{k}\}$, is as long as possible. Since $n\geq 8$, $\delta(G)\geq 2$, and $P_{5}\bigcup P_{3}\nsubseteq G$, it follows that $1\leq k\leq 2$. More precisely, $k=1$ (Otherwise $k=2$. $P_{5}\bigcup P_{3}\nsubseteq G$ together with $\delta(G)\geq 2$ implies that $d_{G}(w)=\{u_{1},u_{2}\}$ for every $w\in V(G)\backslash\bigcup_{i=1}^{2}V(T_{i})$. Then $|V(G)\backslash\bigcup_{i=1}^{2}V(T_{i})|=1$, i.e., $n=7$, which contradicts that $n\geq 8$.). So all triangles in $G$ share a common vertex. Since $\delta(G)\geq 2$, $G=L_{t,2}$ for $n=2t+1$.

Case 2: $l=4$. Let $P_{k}$, denoted by $u_{1}\cdots u_{k}$, be a longest path in $G[U]$. Since $P_{5}\bigcup P_{3}\nsubseteq G$, it follows that $1\leq k\leq 4$. we consider the following four subcases.

Subcase 2.1: $k=4$. Since $P_{5}\bigcup P_{3}\nsubseteq G$, it follows that $N_{C_{4}}(u_{1})=N_{C_{4}}(u_{4})=\varnothing$. Since $G$ is connected, either $N_{C_{4}}(u_{2})\neq\varnothing$ or $N_{C_{4}}(u_{3})\neq\varnothing$. If $u_{1}u_{4}\in E(G)$, then $P_{5}\bigcup P_{3}\subseteq G$, a contradiction. Since $\delta(G)\geq 2$, if $u_{1}u_{4}\notin E(G)$ then $u_{1}u_{3}\in E(G)$ and $u_{2}u_{4}\in E(G)$. Thus $P_{5}\bigcup P_{3}\subseteq G$, a contradiction.

Subcase 2.2: $k=3$. By Lemma 2.5 (ii), $N_{C_{4}}(u_{1})=N_{C_{4}}(u_{3})$. Since $C_{4}$ is a longest cycle, $d_{C_{4}}(u_{1})=d_{C_{4}}(u_{3})=1$. Note that $n\geq 8$. We have $P_{5}\bigcup P_{3}\subseteq G$, a contradiction.

Subcase 2.3: $k=2$. Obviously, $G[U]$ consists of $p$ disjoint edges and $q$ isolated vertices, i.e., $u_{1}u_{2},u_{3}u_{4},\ldots,u_{2p-1}u_{2p},w_{1},\ldots,w_{q}$, where $p\geq 1$, $q\geq 0$ and $2p+q=n-4$. Since $C_{4}$ is a longest cycle, Lemma 2.5 $(i)$ implies that $N_{C_{4}}(u_{2i-1})=N_{C_{4}}(u_{2i})$ and $d_{C_{4}}(u_{2i-1})=d_{C_{4}}(u_{2i})=1$ for $1\leq i\leq p$. So we assume without loss of generality that $N_{C_{4}}(u_{1})=N_{C_{4}}(u_{2})=\{v_{1}\}$. Since $C_{4}$ is a longest cycle and $\delta(G)\geq 2$, it follows that $N_{C_{4}}(w_{1})=\cdots=N_{C_{4}}(w_{r})=\{v_{1},v_{3}\}$ and $N_{C_{4}}(w_{r+1})=\ldots=N_{C_{4}}(w_{r+t})=\{v_{2},v_{4}\}$, where $r+t=q$. Moreover, since $P_{5}\bigcup P_{3}\nsubseteq G$, if $p\geq 2$ then $N_{C_{4}}(u_{3})=N_{C_{4}}(u_{4})=\{v_{3}\}$, $p=2$, and $t=0$. Thus $G\subseteq H_{n}^{2}$. In addition, since $P_{5}\bigcup P_{3}\nsubseteq G$ and $r+t=q\geq 2$, if $p=1$ then $t=0$. Thus $G\subseteq H_{n}^{1}$.

Subcase 2.4: $k=1$. Obviously, the graph $G[U]$ consists of $q$ isolated vertices, i.e., $w_{1},\ldots,w_{q}$, where $q=n-4\geq 4$. Since $C_{4}$ is a longest cycle and $\delta(G)\geq 2$, $N_{C_{4}}(w_{1})=\cdots=N_{C_{4}}(w_{r})=\{v_{1},v_{3}\}$ and $N_{C_{4}}(w_{r+1})=\cdots=N_{C_{4}}(w_{r+t})=\{v_{2},v_{4}\}$, where $r+t=q\geq 4$. We assume without loss of generality that $r\geq t$. It follows that $t=0$, otherwise $P_{5}\bigcup P_{3}\subseteq G$, a contradiction. Furthermore, since $P_{5}\bigcup P_{3}\nsubseteq G$, we also have $v_{2}v_{4}\notin E(G)$. Thus $G\subseteq S_{n,2}$.

Case 3: $l=5$. Since $P_{5}\bigcup P_{3}\nsubseteq G$, we have $P_{3}\nsubseteq G[U]$. Furthermore, $U$ is an independent set (Otherwise, the graph $G[U]$ contains an edge $u_{1}u_{2}$. Since $C_{5}$ is a longest cycle, Lemma 2.5 (i) implies that $N_{C_{5}}(u_{1})=N_{C_{5}}(u_{2})$ and $d_{C_{5}}(u_{1})=d_{C_{5}}(u_{2})=1$. Note that $n\geq 8$ and $\delta(G)\geq 2$. We have $P_{5}\bigcup P_{3}\subseteq G$, a contradiction.). Since $n\geq 8$ and $\delta(G)\geq 2$, the longest cycle $C_{5}$ together with $P_{5}\bigcup P_{3}\nsubseteq G$ implies that $N_{C_{5}}(u_{1})=N_{C_{5}}(u_{2})$ and $d_{C_{5}}(u_{1})=d_{C_{5}}(u_{2})=2$ for any two vertices $u_{1},u_{2}\in U$. We assume without loss of generality that $N_{C_{5}}(u)=\{v_{1},v_{3}\}$ for all $u\in U$. Since $P_{5}\bigcup P_{3}\nsubseteq G$, $G[\{v_{2},v_{4},v_{5}\}]$ contains exactly one edge $v_{4}v_{5}$. Hence $G\subseteq S_{n,2}^{+}$. Thus the assertion holds.

Case 4: $l=6$. Let $U=V(G)\backslash V(C_{6})$. Since $P_{5}\cup P_{3}\nsubseteq G$, we have $P_{8}\nsubseteq G$. By Lemma 2.4 ($iii$), $N_{C_{6}}(u)=N_{C_{6}}(v)$ for any two vertices $u,v\in U$ and $2\leq d_{C_{6}}(u)=d_{C_{6}}(v)\leq 3$. Since $P_{5}\bigcup P_{3}\nsubseteq G$, it follows that $d_{C_{6}}(u)=d_{C_{6}}(v)=2$ and the distance along $C_{6}$ between two vertices in $N_{C_{6}}(u)$ is at least $3$. However, we also have $P_{5}\bigcup P_{3}\subseteq G$, a contradiction.

$(iii)$. Suppose $P_{2}\bigcup 2P_{3}\nsubseteq G$. This implies $P_{5}\bigcup P_{3}\nsubseteq G$. By Lemma 4.1 ($ii$), either $G\subseteq S_{n,2}^{+}$, $G\subseteq H_{n}^{1}$, $G\subseteq H_{n}^{2}$, or $G=L_{t,2}$ for $n=2t+1$. However, if $G=S_{n,2}^{+}$, $G\subseteq H_{n}^{1}$, or $G\subseteq H_{n}^{2}$, then $P_{2}\bigcup 2P_{3}\subseteq G$, a contradiction. Hence either $G\subseteq S_{n,2}$, or $G=L_{t,2}$ for $n=2t+1$. Thus the assertion holds.  $\square$

Lemma 4.2**.**

Let $F=(\bigcup_{i=1}^{k}P_{2a_{i}})\bigcup(\bigcup_{i=1}^{2}P_{2b_{i}+1})$, $h=\sum_{i=1}^{k}a_{i}+\sum_{i=1}^{2}b_{i}-1\geq 3$, and $G$ be a connected graph of order $n$ with a longest cycle $C_{l}$, where $k\geq 0$, $a_{1}\geq\dots\geq a_{k}\geq 1$, $b_{1}\geq b_{2}\geq 1$, and $n\geq 2h+4$. If $\delta(G)\geq h$ and $F\nsubseteq G$, then $2h\leq l\leq 2h+1$.

Proof. Let $C_{l}=v_{1}v_{2}\cdots v_{l}v_{1}$. By Lemma 2.2, $l\geq 2h$. Since $\sum_{i=1}^{k}2a_{i}+\sum_{i=1}^{2}(2b_{i}+1)=2h+4$ and $F\nsubseteq G$, we have $P_{2h+4}\nsubseteq G$. Note that $G$ is connected and $n\geq 2h+4$. So $l\leq 2h+2$. We now prove $l\leq 2h+1$. Suppose that $l=2h+2$. Let $U=V(G)\setminus V(C_{2h+2})$. By Lemma 2.4 ($iii$), all vertices in $U$ have the same $C_{2h+2}$-neighborhood, denoted by $V_{1}$, and $h\leq|V_{1}|\leq h+1$. Let $V_{2}=V(C_{2h+2})\setminus V_{1}$. We consider the following two cases.

Case 1: $|V_{1}|=h+1$. Since $C_{2h+2}$ is a longest cycle, none of vertices in $U$ is adjacent to any two consecutive vertices in $V(C_{2h+2})$. So we assume without loss of generality that $V_{1}=\{v_{1},v_{3},\ldots,v_{2h+1}\}$. Let $H$, i.e., $H=(X,Y;E)$, be a bipartite subgraph of $G$, where $X=V_{1}\setminus\{v_{2h+1}\}$ and $Y\subseteq U$ with $|Y|=h+2$ and $|U|=|V(G)|-|V(C_{2h+2})|\geq h+2$. By definition, $H$ must be a complete bipartite graph. Moreover, let $G_{1}$ be the graph obtained from $H$ by identifying $v_{2h-1}$ with a path $P_{4}$, $v_{2h-1}v_{2h}v_{2h+1}v_{2h+2}$. By Lemma 2.6 $(ii)$, we have $F\subseteq G$, a contradiction.

Case 2: $|V_{1}|=h$. Since none of vertices in $U$ is adjacent to any two consecutive vertices in $V(C_{2h+2})$, there exist two edges $e_{1},e_{2}\in E(C_{2h+2})$ whose end vertices are all in $V_{2}$. If $e_{1}$ and $e_{2}$ share a common vertex, then we assume without loss of generality that $e_{1}=v_{2h}v_{2h+1}$ and $e_{2}=v_{2h+1}v_{2h+2}$. This implies that $V_{1}=\{v_{1},\ldots,v_{2h-1}\}$ and $v_{2h}v_{2h+1}v_{2h+2}\subseteq G[V_{2}]$. By Lemma 2.6 $(ii)$, $F\subseteq G$, a contradiction. So $e_{1}$ and $e_{2}$ share no vertex in common. We assume without loss of generality that $e_{1}=v_{1}v_{2}$ and $e_{2}=v_{2s+2}v_{2s+3}$, where $1\leq s\leq h-1$. Let $W=V_{2}\backslash\{v_{1},v_{2},v_{2s+2},v_{2s+3}\}$. Furthermore, we claim that $W$ is an independent set. In fact, if $v_{p}v_{q}\in E(G)$, where $v_{p},v_{q}\in W$, then $uv_{p-1}v_{p-2}\cdots v_{q}v_{p}v_{p+1}\cdots v_{q-1}u$ is a longer cycle than $C_{2h+2}$ with $u\in U$, a contradiction. Moreover, since $F\nsubseteq G$, Lemma 2.6 $(ii)$ implies that $G[V_{2}]$ contains exactly two disjoint edges $e_{1}$ and $e_{2}$. Thus for every $u\in\{v_{1},v_{2},v_{2s+2},v_{2s+3}\}$, $N_{G}(u)\bigcap V_{1}\geq h-1\geq 2$ and $|N_{G}(v_{1})\bigcap N_{G}(v_{2s+2})\bigcap V_{1}|\geq h-2\geq 1$. Choose $v\in N_{G}(v_{1})\bigcap N_{G}(v_{2s+2})\bigcap V_{1}$. Note that $v_{2}v_{1}vv_{2s+2}v_{2s+3}\subseteq G$ and $v_{2}$ has a neighbor different from $v$ in $V_{1}$. By Lemma 2.6 $(iii)$, $F\subseteq G$, a contradiction.

Hence the assertion holds.  $\square$

Lemma 4.3**.**

Let $F=\bigcup_{i=1}^{2}P_{2b_{i}+1}$, $h=\sum_{i=1}^{2}b_{i}-1\geq 3$ and $G$ be a connected graph of order $n$ with a longest cycle $C_{2h}$, where $b_{1}\geq b_{2}\geq 1$ and $n\geq 2h+4$. If $\delta(G)\geq h$ and $F\nsubseteq G$, then $P_{3}\nsubseteq G[U]$, where $U=V(G)\setminus V(C_{2h})$.

Proof. Let $C_{2h}=v_{1}v_{2}\cdots v_{2h}v_{1}$. Since $\sum_{i=1}^{2}(2b_{i}+1)=2h+4$ and $F\nsubseteq G$, we have $P_{2h+4}\nsubseteq G$. We first prove the following two Claims.

**Claim 1: ** $P_{4}\nsubseteq G[U]$.

Suppose $P_{4}\subseteq G[U]$. Since $P_{2h+4}\nsubseteq G$, there exists one path $P_{4}$, denoted by $u_{1}u_{2}u_{3}u_{4}$, such that $N_{C_{2h}}(u_{1})=N_{C_{2h}}(u_{4})=\varnothing$, and either $N_{C_{2h}}(u_{2})\neq\varnothing$ or $N_{C_{2h}}(u_{3})\neq\varnothing$. Furthermore, since $P_{2h+4}\nsubseteq G$, either $N_{G}(u_{1})=\{u_{2},u_{3}\}$ or $N_{G}(u_{4})=\{u_{2},u_{3}\}$. So $d_{G}(u_{1})=2$ or $d_{G}(u_{4})=2$, which contradicts that $\delta(G)\geq h\geq 3$. So Claim 1 holds.

Claim 2: $P_{3}\nsubseteq G[U]$.

Suppose $P_{3}\subseteq G[U]$. Let $P_{3}=u_{1}u_{2}u_{3}\subseteq G[U]$. By Claim 1, $P_{4}\nsubseteq G[U]$, and thus by Lemma 2.5 $(ii)$, $N_{C_{2h}}(u_{1})=N_{C_{2h}}(u_{3})$. Moreover, $d_{C_{2h}}(u_{1})\geq d_{G}(u_{1})-2\geq h-2$ and $d_{C_{2h}}(u_{3})\geq d_{G}(u_{3})-2\geq h-2$. Since $C_{2h}$ is a longest cycle in $G$, the distance along $C_{2h}$ between any two vertices in $N_{C_{2h}}(u_{1})$ is at least 4, which implies that $4(h-2)\leq 2h$. Thus $3\leq h\leq 4$. We consider the following two cases.

Case 1: $h=4$. Obviously, either $F=P_{9}\bigcup P_{3}$ or $F=P_{7}\bigcup P_{5}$. It is easy to see that $d_{C_{8}}(u_{1})=d_{C_{8}}(u_{3})=2$, $d_{G}(u_{1})=d_{G}(u_{3})=4$, and the distance along $C_{8}$ between two vertices in $N_{C_{8}}(u_{1})$ is exactly $4$. Hence we assume without loss of generality that $N_{C_{8}}(u_{1})=N_{C_{8}}(u_{3})=\{v_{1},v_{5}\}$. Since $P_{4}\nsubseteq G[U]$, we have $u_{1}u_{3}\in E(G)$. Note that $n\geq 12$ in this case. So $U\backslash\{u_{1},u_{2},u_{3}\}\neq\varnothing$. Since $F\nsubseteq G$ and $P_{4}\nsubseteq G[U]$, $u$ is adjacent to none of vertices in $V(C_{8})\bigcup\{u_{1},u_{2},u_{3}\}$ for all $u\in U\backslash\{u_{1},u_{2},u_{3}\}$, which contradicts that $G$ is connected.

Case 2: $h=3$. Obviously, either $F=P_{7}\bigcup P_{3}$ or $F=2P_{5}$. It is easy to see that $d_{C_{6}}(u_{1})=d_{C_{6}}(u_{3})=1$ and $d_{G}(u_{1})=d_{G}(u_{3})=3$. We assume without loss of generality that $N_{C_{6}}(u_{1})=N_{C_{6}}(u_{3})=\{v_{1}\}$. Note that $P_{4}\nsubseteq G[U]$. We have $u_{1}u_{3}\in E(G)$. Note that $n\geq 10$ in this case. So $U\backslash\{u_{1},u_{2},u_{3}\}\neq\varnothing$. Furthermore, since $F\nsubseteq G$, it follows that $N_{C}(u)\bigcap\{v_{1},v_{2},v_{3},v_{5},v_{6}\}=\varnothing$ for all $u\in U\backslash\{u_{1},u_{2},u_{3}\}$. Hence $N_{C_{6}}(u)\subseteq\{v_{4}\}$. Moreover, since $P_{4}\nsubseteq G[U]$, $u$ is not adjacent to any vertex in $\{u_{1},u_{2},u_{3}\}$. Thus the connectedness of $G$ implies that there exists $u_{0}\in U\backslash\{u_{1},u_{2},u_{3}\}$ such that $N_{C_{6}}(u_{0})=\{v_{4}\}$. However, since $\delta(G)\geq 3$, there exists another vertex $w\in U\backslash\{u_{1},u_{2},u_{3}\}$ such that $u_{0}w\in E(G)$. Therefore $F\subseteq G$, a contradiction.

Thus the assertion holds.  $\square$

Corollary 4.4**.**

Let $F=\bigcup_{i=1}^{2}P_{2b_{i}+1}$, $h=\sum_{i=1}^{2}b_{i}-1\geq 3$, and $G$ be a connected graph of order $n$ with a longest cycle $C_{2h}$, where $b_{1}\geq b_{2}\geq 1$ and $n\geq 2h+4$. If $\delta(G)\geq h$, then $F\subseteq G$, unless either $G\subseteq S_{n,h}$, or $F=P_{7}\bigcup P_{3}$ and $G\subseteq K_{2}\bigvee\frac{n-2}{2}K_{2}$ for even $n$.

Proof.

Suppose $F\nsubseteq G$. Let $C_{2h}=v_{1}v_{2}\cdots v_{2h}v_{1}$ and $U=V(G)\setminus V(C_{2h})$. By Lemma 4.2, $P_{3}\nsubseteq G[U]$. If $U$ is an independent set, then Lemma 2.4 $(i)$ implies $G\subseteq S_{n,h}$. Hence we assume that $G[U]$ consists of $p$ disjoint edges and $q$ isolated vertices, i.e., $u_{1}u_{2},u_{3}u_{4},\ldots,$ $u_{2p-1}u_{2p},w_{1},\ldots,w_{q}$, where $p\geq 1$ and $q\geq 0$. By Lemma 2.5 $(i)$, $N_{C_{2h}}(u_{2i-1})=N_{C_{2h}}(u_{2i})$ for $1\leq i\leq p$. Furthermore, $d_{C_{2h}}(u_{j})=d_{G}(u_{j})-1\geq h-1$ for $1\leq j\leq 2p$. Since $C_{2h}$ is a longest cycle, the distance along $C_{2h}$ between any two vertices in $N_{C_{2h}}(u_{1})$ is at least 3, which implies that $3(h-1)\leq 2h$. Thus $h=3$. It follows that either $F=P_{7}\bigcup P_{3}$ or $F=2P_{5}$, and $d_{C_{6}}(u_{j})=2$ for $1\leq j\leq 2p$. Moreover, the distance along $C_{6}$ between two vertices in $N_{C_{6}}(u_{1})$ is exactly 3. Hence we assume without loss of generality that $N_{C_{6}}(u_{1})=N_{C_{6}}(u_{2})=\{v_{1},v_{4}\}$. Since $F\nsubseteq G$, it follows that $N_{C_{6}}(u_{2i-1})=N_{C_{6}}(u_{2i})=\{v_{1},v_{4}\}$ for $1\leq i\leq p$. Next we prove that $q=0$. In fact, since $n\geq 10$, if $q\geq 1$ then either $p=1$ and $q\geq 2$, or $p\geq 2$ and $q\geq 1$. If $p=1$ and $q\geq 2$, then $d_{G}(w_{i})=d_{C_{6}}(w_{i})=3$ for $1\leq i\leq 2$ and thus $F\subseteq G$, a contradiction. If $p\geq 2$ and $q\geq 1$, then $d_{G}(w_{1})=d_{C_{6}}(w_{1})=3$, and thus $F\subseteq G$, a contradiction. So $n$ must be even. In addition, since $F\nsubseteq G$, $G[\{v_{2},v_{3},v_{5},v_{6}\}]$ consists of two disjoint edges $v_{2}v_{3}$ and $v_{5}v_{6}$. Therefore $G\subseteq K_{2}\bigvee\frac{n-2}{2}K_{2}$. Moreover, $F=P_{7}\bigcup P_{3}$. Thus the assertion holds. ∎

Corollary 4.5**.**

Let $F=(\bigcup_{i=1}^{k}P_{2a_{i}})\bigcup(\bigcup_{i=1}^{2}P_{2b_{i}+1})$, $h=\sum_{i=1}^{k}a_{i}+\sum_{i=1}^{2}b_{i}-1\geq 3$, and $G$ be a connected graph of order $n$ with a longest cycle $C_{2h}$, where $k\geq 1$, $b_{1}\geq b_{2}\geq 1$, and $n\geq 2h+4$. If $\delta(G)\geq h$, then $F\subseteq G$ unless, either $G\subseteq S_{n,h}$, or $F=P_{4}\bigcup 2P_{3}$ and $G\subseteq K_{2}\bigvee\frac{n-3}{2}K_{2}$ for even $n$.

Proof.

Suppose $F\nsubseteq G$. Let $F^{\prime}=\bigcup_{i=1}^{2}P_{2b_{i}^{\prime}+1}$, where $b_{1}^{\prime}=\sum_{i=1}^{k}a_{i}+b_{1}$ and $b_{2}^{\prime}=b_{2}$. We have $F^{\prime}\nsubseteq G$. By Corollary 4.4, either $G\subseteq S_{n,h}$, or $F^{\prime}=P_{7}\bigcup P_{3}$ and $G\subseteq K_{2}\bigvee\frac{n-2}{2}K_{2}$ for even $n$. Note $F\nsubseteq G$ whenever $G\subseteq S_{n,h}$. So we assume that $F^{\prime}=P_{7}\bigcup P_{3}$ and $G\subseteq K_{2}\bigvee\frac{n-2}{2}K_{2}$ for even $n$. Then we have $F\in\{P_{4}\bigcup 2P_{3},P_{2}\bigcup P_{5}\bigcup P_{3},2P_{2}\bigcup 2P_{3}\}$. In addition, $P_{2}\bigcup P_{5}\bigcup P_{3}\subseteq G$, $2P_{2}\bigcup 2P_{3}\subseteq G$, and $P_{4}\bigcup 2P_{3}\nsubseteq G$. Hence $F=P_{4}\bigcup 2P_{3}$. Thus the assertion holds. ∎

Lemma 4.6**.**

Let $F=\bigcup_{i=1}^{2}P_{2b_{i}+1}$, $h=\sum_{i=1}^{2}b_{i}-1\geq 3$, and $G$ be a connected graph of order $n$ with a longest cycle $C_{2h+1}$, where $b_{1}\geq b_{2}\geq 1$ and $n\geq 4(2h+1)^{2}\binom{2h+1}{h}$. If $\delta(G)\geq h$, then $F\subseteq G$, unless either $G\subseteq S^{+}_{n,h}$, or $F=P_{9}\bigcup P_{3}$ and $G\subseteq K_{3}\bigvee\frac{n-3}{2}K_{2}$ for odd $n$.

Proof. Suppose $F\nsubseteq G$. Since $\sum_{i=1}^{2}(2b_{i}+1)=2h+4$, we have $P_{2h+4}\nsubseteq G$. Let $C_{2h+1}=v_{1}v_{2}\cdots v_{2h+1}v_{1}$ and $U=V(G)\setminus V(C_{2h+1})$. We consider the following two cases.

Case 1: $U$ is an independent set. Since $C_{2h+1}$ is a longest cycle, none of vertices in $U$ is adjacent to any two consecutive vertices along $C_{2h+1}$. It follows that $d_{C_{2h+1}}(u)=h$ for all $u\in U$. By Lemma 2.3, there exists $V_{1}\subseteq V(C_{2h+1})$ and $U_{1}\subseteq U$ with $|V_{1}|=h$ and $|U_{1}|=h+2$ such that $H(V_{1},U_{1};E)$ is a complete bipartite graph. Hence $N_{C_{2h+1}}(u)=V_{1}$ for all $u\in U_{1}$. We assume without loss of generality that $V_{1}=\{v_{2},v_{4},\ldots,v_{2h}\}$. Moreover, $N_{C_{2h+1}}(u)=V_{1}$ for all $u\in U\backslash U_{1}$. Otherwise, there exists $u\in U\backslash U_{1}$ such that $N_{C_{2h+1}}(u)\bigcap(V(C_{2h+1})\backslash V_{1})\neq\varnothing$. Since $u$ is not adjacent to any two consecutive vertices along $C_{2h+1}$, we have $v_{1}$ or $v_{2h+1}\in N_{C_{2h+1}}(u)\bigcap(V(C_{2h+1})\backslash V_{1})$. Note $v_{2},v_{2h}\in V_{1}$ and $v_{1}v_{2h+1}\in E(G)$. By Lemma 2.6 $(ii)$, $F\subseteq G$, a contradiction. Furthermore, $\{v_{1},v_{3},\ldots,v_{2h-1}\}$ is an independent set (Othewise, let $v_{2s+1}v_{2t+1}\in E(G)$ and $u\in U$, where $1\leq 2s+1<2t+1\leq 2h-1$. It follows that $uv_{2s}v_{2s-1}\cdots v_{2}v_{1}v_{2h+1}v_{2h}\cdots v_{2t+2}v_{2t+1}v_{2s+1}v_{2s+2}\cdots v_{2t-1}v_{2t}u$ is a longer cycle than $C_{2h+1}$, a contradiction). Similarly, $\{v_{3},\ldots,v_{2h+1}\}$ is an independent set. Hence $G[\{v_{1},v_{3},\ldots,v_{2h+1}\}]$ contains exactly one edge $v_{1}v_{2h+1}$. Thus $G\subseteq S^{+}_{n,h}$.

Case 2: $G[U]$ contains at least one edge. Since $\delta(G)\geq 2$, we have $P_{3}\nsubseteq G[U]$. Hence we assume that $G[U]$ consists of $p$ disjoint edges and $q$ isolated vertices, i.e., $u_{1}u_{2},u_{3}u_{4},\ldots,u_{2p-1}u_{2p},w_{1},\ldots,w_{q}$, where $p\geq 1$ and $q\geq 0$. By Lemma 2.5 $(i)$, $N_{C_{2h+1}}(u_{2i-1})=N_{C_{2h+1}}(u_{2i})$ for $1\leq i\leq p$. Furthermore, $d_{C_{2h+1}}(u_{j})=d_{G}(u_{j})-1\geq h-1$ for $1\leq j\leq 2p$. Since $C_{2h+1}$ is a longest cycle in $G$, the distance along $C_{2h+1}$ between any two vertices in $N_{C_{2h+1}}(u_{1})$ is at least 3, which implies that $3(h-1)\leq 2h+1$. Thus $3\leq h\leq 4$. We claim that $h=4$ (Otherwise $h=3$ and thus either $F=P_{7}\bigcup P_{3}$ or $F=2P_{5}$. Moreover, the distance along $C_{7}$ between two vertices in $N_{C_{7}}(u_{1})$ is exactly 3. So we assume without loss of generality that $N_{C_{7}}(u_{1})=N_{C_{7}}(u_{2})=\{v_{1},v_{4}\}$. Since $F\nsubseteq G$, it follows that $N_{C_{7}}(u)\bigcap\{v_{1},v_{2},v_{3},v_{4},v_{5},v_{7}\}=\varnothing$ for all $u\in U\backslash\{u_{1},u_{2}\}$. Hence $N_{C_{7}}(u)\subseteq\{v_{6}\}$, which contradicts $d_{C_{7}}(u)\geq 2$.). Hence either $F=P_{9}\bigcup P_{3}$ or $F=P_{7}\bigcup P_{5}$. Moreover, the distance along $C_{9}$ between any two vertices in $N_{C_{9}}(u_{1})$ is exactly 3. So we assume without loss of generality that $N_{C_{9}}(u_{1})=N_{C_{9}}(u_{2})=\{v_{1},v_{4},v_{7}\}$. Since $F\nsubseteq G$, it follows that $N_{C_{9}}(u_{2i-1})=N_{C_{9}}(u_{2i})=\{v_{1},v_{4},v_{7}\}$ for $1\leq i\leq p$, and $q=0$. So $n$ must be odd. In addition, since $F\nsubseteq G$, $G[\{v_{2},v_{3},v_{5},v_{6},v_{8},v_{9}\}]$ consists of three disjoint edges $v_{2}v_{3}$, $v_{5}v_{6}$ and $v_{8}v_{9}$. Therefore $G=K_{3}\bigvee\frac{n-3}{2}K_{2}$. Moreover, $F=P_{9}\bigcup P_{3}$. Thus the assertion holds.  $\square$

Lemma 4.7**.**

Let $F=(\bigcup_{i=1}^{k}P_{2a_{i}})\bigcup(\bigcup_{i=1}^{2}P_{2b_{i}+1})$, $h=\sum_{i=1}^{k}a_{i}+\sum_{i=1}^{2}b_{i}-1\geq 3$, and $G$ be a connected graph of order $n$ with a longest cycle $C_{2h+1}$, where $k\geq 1$, $b_{1}\geq b_{2}\geq 1$, and $n\geq 4(2h+1)^{2}\binom{2h+1}{h}$. If $\delta(G)\geq h$, then $F\subseteq G$, unless $F=P_{6}\bigcup 2P_{3}$ and $G\subseteq K_{3}\bigvee\frac{n-3}{2}K_{2}$ for odd $n$.

Proof.

Suppose $F\nsubseteq G$. Let $F^{\prime}=\bigcup_{i=1}^{2}P_{2b_{i}^{\prime}+1}$, where $b_{1}^{\prime}=\sum_{i=1}^{k}a_{i}+b_{1}$ and $b_{2}^{\prime}=b_{2}$. We have $F^{\prime}\nsubseteq G$. By Lemma 4.6, either $G\subseteq S_{n,h}^{+}$, or $F^{\prime}=P_{9}\bigcup P_{3}$ and $G\subseteq K_{3}\bigvee\frac{n-3}{2}K_{2}$ for odd $n$. If $G\subseteq S_{n,h}^{+}$, then $F\subseteq G$, a contradiction. Thus we assume that $F^{\prime}=P_{9}\bigcup P_{3}$ and $G\subseteq K_{3}\bigvee\frac{n-3}{2}K_{2}$ for odd $n$. Hence $F\in\{P_{2}\bigcup P_{7}\bigcup P_{3},P_{2}\bigcup 2P_{5},P_{4}\bigcup P_{5}\bigcup P_{3},P_{6}\bigcup 2P_{3},2P_{2}\bigcup P_{5}\bigcup P_{3},P_{4}\bigcup P_{2}\bigcup 2P_{3},\\ 3P_{2}\bigcup 2P_{3}\}$. In addition, since $F\nsubseteq G$, it follows that $F=P_{6}\bigcup 2P_{3}$. Thus the assertion holds. ∎

Theorem 1.10 can be stated as follows.

Theorem 4.8**.**

Let $F=(\bigcup_{i=1}^{k}P_{2a_{i}})\bigcup(\bigcup_{i=1}^{2}P_{2b_{i}+1})$, $h=\sum_{i=1}^{k}a_{i}+\sum_{i=1}^{2}b_{i}-1\geq 2$, and $G$ be a 2-connected graph of order $n$, where $k\geq 0$, $a_{1}\geq\dots\geq a_{k}\geq 1$, $b_{1}\geq b_{2}$, and $n\geq 4(2h+1)^{2}\binom{2h+1}{h}$.

*(a). If $\delta(G)\geq h$ and $k=0$, then $F\subseteq G$, unless one of the following holds:

(i). $G\subseteq S^{+}_{n,h}$;

(ii). $F=P_{7}\bigcup P_{3}$ and $G\subseteq K_{2}\bigvee\frac{n-2}{2}K_{2}$, where $n$ is even;

(iii). $F=P_{9}\bigcup P_{3}$ and $G\subseteq K_{3}\bigvee\frac{n-3}{2}K_{2}$, where $n$ is odd.*

*(b). If $\delta(G)\geq h$ and $k\geq 1$, then $F\subseteq G$, unless one of the following holds:

(iv). $G\subseteq S_{n,h}$;

(v). $F=P_{4}\bigcup 2P_{3}$ and $G\subseteq K_{2}\bigvee\frac{n-2}{2}K_{2}$, where $n$ is even ;

(vi). $F=P_{6}\bigcup 2P_{3}$ and $G\subseteq K_{3}\bigvee\frac{n-3}{2}K_{2}$, where $n$ is odd.*

Proof. Suppose $F\nsubseteq G$. If $h=2$, then $F=P_{5}\bigcup P_{3}$. By Lemma 4.1 $(ii)$, either $G\subseteq S_{n,2}^{+}$, $G\subseteq H_{n}^{1}$, $G\subseteq H_{n}^{2}$, or $G=L_{t,2}$ for $n=2t-1$. However, $G$ is not 2-connected whenever $G\subseteq H_{n}^{1}$, $G\subseteq H_{n}^{2}$, or $G=L_{t,2}$. Hence $G\subseteq S_{n,2}^{+}$. Next we assume that $h\geq 3$. Let $C_{l}$ be the longest cycle in $G$. By Lemma 4.2, $2h\leq l\leq 2h+1$. It is easy to see that Corollary 4.4 together with Lemma 4.6 implies that $G\subseteq S_{n,h}^{+}$, $F=P_{7}\bigcup P_{3}$ and $G\subseteq K_{2}\bigvee\frac{n-2}{2}K_{2}$ for even $n$, or $F=P_{9}\bigcup P_{3}$ and $G\subseteq K_{3}\bigvee\frac{n-3}{2}K_{2}$ for odd $n$.

(b). Suppose $F\nsubseteq G$. If $h=2$, then $F=P_{2}\bigcup 2P_{3}$. By Lemma 4.1 $(iii)$, either $G\subseteq S_{n,2}$, or $G=L_{t,2}$ for $n=2t-1$. However, if $G=L_{t,2}$, then $G$ is not 2-connected. Hence $G\subseteq S_{n,2}$. Next we assume that $h\geq 3$. Let $C_{l}$ be the longest cycle in $G$. By Lemma 4.2, $2h\leq l\leq 2h+1$. It is easy to see that Corollary 4.5 together with Lemma 4.7 implies that $G\subseteq S_{n,h}$, $F=P_{4}\bigcup 2P_{3}$ and $G\subseteq K_{2}\bigvee\frac{n-2}{2}K_{2}$ for even $n$, or $F=P_{6}\bigcup 2P_{3}$ and $G\subseteq K_{3}\bigvee\frac{n-3}{2}K_{2}$ for odd $n$. Thus the assertion holds.  $\square$

We will use the next two lemmas in Theorem 1.11.

Lemma 4.9**.**

*Let $F=\bigcup_{i=1}^{2}P_{2b_{i}+1}$, $h=\sum_{i=1}^{2}b_{i}-1\geq 3$, and $G$ be a connected graph of order $n$ with at least one cut vertex, where $b_{1}\geq b_{2}\geq 1$ and $n\geq 2h+4$. If $\delta(G)\geq h$, then $F\subseteq G$, unless one of the following holds:

(i) $F=P_{2b_{1}+1}\bigcup P_{2b_{1}-1}$ and $G=L_{t,h}$, where $n=th+1$;

(ii) $F=2P_{2b_{1}+1}$ and $G=U_{3,h}$, where $n=3h+3$;

(iii) $F=2P_{2b_{1}+1}$ and $G\subseteq L_{t_{1},t_{2},h,h+1}$, where $n=t_{1}h+t_{2}(h+1)+1$;

(iv) $F=2P_{2b_{1}+1}$ and $G\subseteq F_{t_{1},t_{2},h,h+1}$, where $n=t_{1}h+(t_{2}+1)(h+1)+1$;

(v) $F=2P_{2b_{1}+1}$ and $G\subseteq T_{t_{1},t_{2},h,h+1}$, where $n=t_{1}h+(t_{2}+2)(h+1)+1$.*

Proof. Suppose $F\nsubseteq G$. Since $\sum_{i=1}^{k}2a_{i}+\sum_{i=1}^{2}(2b_{i}+1)=2h+4$, we have $P_{2h+4}\nsubseteq G$. Note that $h+2\geq 2b_{2}+1$ and $2h+1\geq 2b_{1}+1$. We have $P_{h+2}\bigcup P_{2h+1}\nsubseteq G$. Note that $h+1\geq 2b_{2}+1$ and $2h+1\geq 2b_{1}+1$ for $b_{1}\geq b_{2}+1$. We also have $P_{h+1}\bigcup P_{2h+1}\nsubseteq G$ for $b_{1}\geq b_{2}+1$. Note that $h\geq 2b_{2}+1$ and $2h+1\geq 2b_{1}+1$ for $b_{1}\geq b_{2}+2$. Moreover, $P_{h}\bigcup P_{2h+1}\nsubseteq G$ for $b_{1}\geq b_{2}+2$.

Since $G$ has at least one cut vertex, it has at least two end blocks. Furthermore, since $\delta(G)\geq h$, it follows that $|V(B)|\geq h+1$ for every end block $B$. Choose two end blocks $B_{1}$ and $B_{2}$ of $G$ such that $p(B_{1},B_{2})$ is as large as possible, where $p(B_{1},B_{2})$ is the order of a longest path between the cut vertex $u_{1}$ of $G$ in $V(B_{1})$ and the cut vertex $u_{2}$ of $G$ in $V(B_{2})$. Lemma 2.1 together with $P_{2h+4}\nsubseteq G$ implies that $p(B_{1},B_{2})\leq 3$. We assume without loss of generality that $|V(B_{1})|\leq|V(B_{2})|$. We consider the following three cases.

Case 1: $p(B_{1},B_{2})=1$. This implies that all blocks in $G$ are end blocks and they share a common cut vertex of $G$. We first prove the following Claim.

Claim: $G$ has at least three end blocks.

Suppose that $G$ has exactly two end blocks $B_{1}$ and $B_{2}$. Since $n\geq 2h+4$ and $|V(B_{1})|\leq|V(B_{2})|$, if $|V(B_{1})|\geq h+2$ then $|V(B_{2})|\geq h+3$. By Lemma 2.1, $P_{2h+4}\subseteq G$, a contradiction. Hence we assume that $|V(B_{1})|=h+1$ and thus $|V(B_{2})|\geq h+4$. We claim $h=3$ (Otherwise $h\geq 4$. By Lemma 2.1, $B_{2}$ has a path $P_{\mathrm{min}\{2h,h+4\}}$ with an end vertex $u_{2}$. Since $\mathrm{min}\{2h,h+4\}\geq h+4$, we have $P_{2h+4}\subseteq G$, a contradiction.). It follows that either $F=P_{7}\bigcup P_{3}$ or $F=2P_{5}$. Furthermore, by Lemmas 2.1 and 2.7 $(iii)$, $F\subseteq G$, a contradiction. This completes the claim.

By Claim above, $G$ has at least three end blocks. Since $P_{h+2}\bigcup P_{2h+1}\nsubseteq G$, all blocks have order at most $h+2$ and $P_{h}\bigcup P_{2h+1}\subseteq G$. Note $P_{h}\bigcup P_{2h+1}\nsubseteq G$ for $b_{1}\geq b_{2}+2$. Hence $b_{2}\leq b_{1}\leq b_{2}+1$. If $b_{1}=b_{2}$, then $F=2P_{2b_{1}+1}$ and $G\subseteq L_{t_{1},t_{2},h,h+1}$, where $n=t_{1}h+t_{2}(h+1)+1$. Next we assume that $b_{1}=b_{2}+1$. Note $P_{h+1}\bigcup P_{2h+1}\nsubseteq G$ for $b_{1}=b_{2}+1$. It follows that all blocks have order $h+1$. Thus $F=P_{2b_{1}+1}\bigcup P_{2b_{1}-1}$ and $G=L_{t,h}$, where $n=th+1$.

Case 2: $p(B_{1},B_{2})=2$. This implies that all blocks in $G$ are end blocks. Lemma 2.1 together with $P_{2h+4}\nsubseteq G$ implies that $|V(B_{1})|=h+1$ and $h+1\leq|V(B_{2})|\leq h+2$. Thus $G$ is a graph obtained by adding an edge to the centers of $L_{t_{1},t_{2},h,h+1}$ and $L_{t_{3},t_{4},h,h+1}$, where $n=(t_{1}+t_{3})h+(t_{2}+t_{4})(h+1)+2$, $t_{1}+t_{2}\geq 1$, and $t_{3}+t_{4}\geq 1$. Furthermore, since $P_{h+2}\bigcup P_{2h+1}\nsubseteq G$, either $t_{1}+t_{2}=1$ or $t_{3}+t_{4}=1$. We assume without loss of generality that $t_{1}+t_{2}=1$. Since $n\geq 2h+4$, it follows that $t_{3}+t_{4}\geq 2$. So $P_{h+1}\bigcup P_{2h+1}\subseteq G$. Note $P_{h+1}\bigcup P_{2h+1}\nsubseteq G$ for $b_{1}\geq b_{2}+1$. Then $b_{1}=b_{2}$ and thus $F=2P_{2b_{1}}$. Moreover, since $P_{h+2}\bigcup P_{2h+1}\nsubseteq G$, it follows that $t_{2}=0$ and thus $t_{1}=1$. Therefore, $F=2P_{2b_{1}+1}$ and $G\subseteq F_{t_{3},t_{4},h,h+1}$, where $n=t_{3}h+(t_{4}+1)(h+1)+1$.

Case 3: $p(B_{1},B_{2})=3$. Lemma 2.1 together with $P_{2h+4}\nsubseteq G$ implies that $|V(B_{1})|=|V(B_{2})|=h+1$. Note $P_{h+2}\bigcup P_{2h+1}\nsubseteq G$. The end block $B_{i}$ is the unique end block with $u_{i}\in V(B_{i})$ for $1\leq i\leq 2$. Denote $u_{1}uu_{2}$ by the path between $u_{1}$ and $u_{2}$. Since $\delta(G)\geq 3$, there exists an end block of order at least $h+1$ with the vertex $u$. Hence $P_{h+1}\bigcup P_{2h+1}\subseteq G$. Note $P_{h+1}\bigcup P_{2h+1}\nsubseteq G$ for $b_{1}\geq b_{2}+1$. So $b_{1}=b_{2}$ and thus $F=2P_{2b_{1}}$. Furthermore, since $p(B_{1},B_{2})=3$, there is exactly one path $P_{3}$ between $u_{1}$ and $u_{2}$. Since $P_{h+2}\bigcup P_{2h+1}\nsubseteq G$, if $u_{1}u_{2}\in E(G)$ then there is exactly one end block with the vertex $u$ and it has order $h+1$. Hence $F=2P_{2b_{1}+1}$ and $G=U_{3,h}$, where $n=3h+3$. Next we assume $u_{1}u_{2}\notin E(G)$. Obviously, $u_{1}u$ and $u_{2}u$ are both cut edges of $G$. Since $P_{h+2}\bigcup P_{2h+1}\nsubseteq G$, all blocks that contains $u$ have order at most $h+2$. Therefore $F=2P_{2b_{1}+1}$ and $G\subseteq T_{t_{1},t_{2},h,h+1}$, where $n=t_{1}h+(t_{2}+2)(h+1)+1$.

Thus the assertion holds.  $\square$

Lemma 4.10**.**

Let $F=(\bigcup_{i=1}^{k}P_{2a_{i}})\bigcup(\bigcup_{i=1}^{2}P_{2b_{i}+1})$, $h=\sum_{i=1}^{k}a_{i}+\sum_{i=1}^{2}b_{i}-1\geq 2$, and $G$ be a connected graph of order $n$ with at least one cut vertex, where $k\geq 1$, $a_{1}\geq\cdots\geq a_{k}\geq 1$, $b_{1}\geq b_{2}\geq 3$, and $n\geq 2h+4$. If $\delta(G)\geq h$, then $F\subseteq G$, unless $F=P_{2}\bigcup 2P_{2b_{1}+1}$ and $G=L_{t,h}$, where $n=th+1$.

Proof. Suppose $F\nsubseteq G$. If $h=2$, then $F=P_{2}\bigcup 2P_{3}$. By Lemma 4.1 $(ii)$, we have $G\subseteq S_{n,2}^{+}$, $G\subseteq H_{n}^{1}$, $G\subseteq H_{n}^{2}$, or $G=L_{t,2}$ for $n=2t+1$. However, $G$ is 2-connected whenever $G\subseteq S_{n,2}^{+}$ and $F\subseteq G$ whenever $G\subseteq H_{n}^{1}$ or $G\subseteq H_{n}^{2}$. Hence $G=L_{t,2}$, where $n=2t+1$. Next we assume that $h\geq 3$. Let $F^{\prime}=\bigcup_{i=1}^{2}P_{2b_{i}^{\prime}+1}$, where $b_{1}^{\prime}=\sum_{i=1}^{k}a_{i}+b_{1}$ and $b_{2}^{\prime}=b_{2}$. We have $F^{\prime}\nsubseteq G$. Note that $k\geq 1$ and $b_{1}\geq b_{2}$. It follows that $b_{1}^{\prime}>b_{2}^{\prime}$. By Lemma 4.9, $G=L_{t,h}$, where $n=th+1$ and $F^{\prime}=P_{2b_{1}^{\prime}+1}\bigcup P_{2b_{1}^{\prime}-1}$. Moreover, since $F^{\prime}=P_{2b_{1}^{\prime}+1}\bigcup P_{2b_{1}^{\prime}-1}$, it follows that $k=1$, $a_{1}=1$, and $b_{1}=b_{2}$. Hence $F=P_{2}\bigcup 2P_{2b_{1}+1}$. Thus the assertion holds.  $\square$

Theorem 1.11 can be stated as follows.

Theorem 4.11**.**

Let $F=(\bigcup_{i=1}^{k}P_{2a_{i}})\bigcup(\bigcup_{i=1}^{2}P_{2b_{i}+1})$, $h=\sum_{i=1}^{k}a_{i}+\sum_{i=1}^{2}b_{i}-1$, and $G$ be a connected graph of order $n$ with at least one cut vertex, where $a_{1}\geq\dots\geq a_{k}\geq 1$, $b_{1}\geq b_{2}\geq 1$, $k\geq 0$, and $n\geq 2h+4$.

*(a). If $\delta(G)\geq h\geq 1$ and $k=0$, then $F\subseteq G$, unless one of the following holds:

(i) $F=P_{5}\bigcup P_{3}$ and either $G\subseteq H_{n}^{1}$ or $G\subseteq H_{n}^{2}$;

(ii) $F=P_{2b_{1}+1}\bigcup P_{2b_{1}-1}$ and $G=L_{t,h}$, where $n=th+1$;

(iii) $F=2P_{2b_{1}+1}$ and $G=U_{3,h}$, where $n=3h+3$;

(iv) $F=2P_{2b_{1}+1}$ and $G\subseteq L_{t_{1},t_{2},h,h+1}$, where $n=t_{1}h+t_{2}(h+1)+1$;

(v) $F=2P_{2b_{1}+1}$ and $G\subseteq F_{t_{1},t_{2},h,h+1}$, where $n=t_{1}h+(t_{2}+1)(h+1)+1$;

(vi) $F=2P_{2b_{1}+1}$ and $G\subseteq T_{t_{1},t_{2},h,h+1}$, where $n=t_{1}h+(t_{2}+2)(h+1)+1$.*

(b). If $\delta(G)\geq h\geq 2$ an $k\geq 1$, then $F\subseteq G$, unless $F=P_{2}\bigcup 2P_{2b_{1}+1}$ and $G=L_{t,h}$, where $n=th+1$.

Proof. (a). Suppose $F\nsubseteq G$. If $h=1$, then $F=2P_{3}$. By Lemma 4.6 ($i$), either $G=U_{3,1}$ or $G\subseteq L_{t_{1},t_{2},1,2}$ for $n=t_{1}+2t_{2}+1$. Next we assume that $h=2$. It follows that $F=P_{5}\bigcup P_{3}$. By Lemma 4.6 ($ii$), $G\subseteq S_{n,2}^{+}$, $G\subseteq H_{n}^{1}$, $G\subseteq H_{n}^{2}$, or $G=L_{t,2}$ for $n=2t+1$. However, $G$ is 2-connected whenever $G\subseteq S_{n,2}^{+}$. Hence $G\subseteq H_{n}^{1}$, $G\subseteq H_{n}^{2}$, or $G=L_{t,2}$ for $n=2t+1$. Finally we assume that $h\geq 3$. The assertion follows by Lemma 4.9.

(b). The assertion follows by Lemma 4.10.  $\square$

Acknowledgements:

The authors would like to thank the anonymous referee for many helpful and constructive suggestions to an earlier version of this paper, which results in an great improvement.