Burning Numbers of t-unicyclic Graphs

Abstract

Given a graph G, the burning number of G is the smallest integer k for which there are vertices $x_1, x_2,\ldots ,x_k$ such that $(x_1,x_2,\ldots ,x_k)$ is a burning sequence of G. It has been shown that the graph burning problem is NP-complete, even for trees with maximum degree three, or linear forests. A t-unicyclic graph is a unicycle graph in which the unique vertex in the graph with degree greater than two has degree $t + 2$. In this paper, we first present the bounds for the burning number of t-unicyclic graphs, and then use the burning numbers of linear forests with at most three components to determine the burning number of all t-unicyclic graphs for $t\le 2$.

1 Introduction

Let $G=(V(G),E(G))$ be a simple graph, and we simply write $|V(G)|=|G|$ and $|E(G)|=||G||$. For any nonnegative integer k and a vertex u, the kth closed neighborhood of u, denoted by $N_k[u]$, is the set of vertices whose distance from u is at most k; we denote $N_1[u]$ simply by N[u], and $d(u)=|N(u)|$ is the degree of u. Let $\Delta (G)=\max \{d(u):u\in V(G)\}$. A vertex of degree k is called a k-vertex. For two vertices u and v in G, the distance between u and v, denoted by $d_G(u,v)$, and simply write d(u, v), is the number of edges in a shortest path joining u and v. The eccentricity of a vertex v is the greatest distance from v to any other vertices.

A tree is a connected acyclic graph. A caterpillar is a tree where deleting all vertices of degree 1 leaves a path. Let $SP_s (x) (s\ge 3)$ be a generalized star which is a tree with exactly one s-vertex x of degree more than 2, and a path can be seen as an $SP_2(x)$. A rooted tree T(x) is a tree with one vertex x designated as the root. The height of a rooted tree T(x) is the eccentricity of x. A rooted tree partition of G is a collection of rooted trees that are subgraphs of G, with the property that the vertex sets of the trees partition V(G).

A unicyclic graph G is a connected graph with $||G||=|G|$. A t-unicyclic graph $U_t(x)$ ($t\ge 1$) is a unicyclic graph in which there exists exactly one $(t+2)$-vertex x of degree more than 2. A cycle can be seen as a 0-unicyclic graph. We denote a path of order n and a cycle of order n by $P_n$ and $C_n$, respectively.

Let $G_1,G_2,\ldots ,G_k$ be k vertex disjoint graphs. The union of graphs $G_i ~(1\le i\le k)$, denoted by $G_1 +\cdots +G_k$ , is the graph with vertex set $\bigcup _{i=1}^{k}V (G_i)$ and edge set $\bigcup _{i=1}^{k}E(G_i)$. For any positive integer k, let [k] denote the set $\{0,1,2,\ldots ,k-1\}$.

Bonato, Janssen, and Roshanbin [6] defined graph burning involving a graph process as follows. Initially, at time round $t=0$ all vertices are unburned. During each time round $t\ge 1$, one new unburned vertex is chosen to burn (if such a vertex exists): burned vertex remains burned until the end of the process and its unburned neighbors becomes burned. The process ends when all vertices are burned. A graph is called k-burnable if it can be burned by at most k rounds. The burning number of a graph G, denoted by b(G), is the smallest integer k such that G is k-burnable. The vertices that are chosen to be burned are referred to as a burning sequence, and a shortest such sequence is called optimal. Note that any optimal burning sequences have length b(G). Figure 1 illustrates an optimal burning sequence the path $C_4$ (resp., $C_5$) with vertices $\{v_1, v_2, v_3, v_4\}$ (resp., $\{v_1, v_2, v_3, v_4,v_5\}$).

Fig. 1

Burning $C_4$ and $C_5$ (the open circles represent burned vertices)

Burning can be viewed as a simplified model for the spread of social influence in a social network such as Facebook or Twitter [15], which is motivated by two other well-known models: the competitive diffusion game [1] and Firefighting [10]. It turned out that graph burning is related to several other graph theory problems such as graph bootstrap percolation [2] and graph domination [12]. Recently, the graph burning problem attracted the attention of many researchers and some results are achieved, see [11, 14, 16, 18, 20,21,22, 24]). For more on graph burning and graph searching, see [5, 9]. The graph burning problem is NP-complete, even for trees with maximum degree three, or linear forests (i.e., a disjoint union of paths), see [3, 4, 23]. So it is interesting to compute the burning number of some special classes of graphs.

Roshanbin [23] gave the following criterion for a graph to have burning number 2.

Theorem 1.1

[23] A graph G of order n satisfies $b(G) =2$ if and only if $|G|\ge 2$ and $n-2\le \Delta (G)\le n-1$.

Bonato at al. in [6] studied the burning number of paths and cycles.

Theorem 1.2

[6] For a path $P_n$ or a cycle $C_n$, we have that $b(P_n)=b(C_n) =\lceil \sqrt{n}\rceil$. Moreover, if a graph G has a Hamiltonian path, then $b(G) \le \lceil \sqrt{|G|}\rceil$.

A graph G of order n is well-burnable if $b(G)\le \lceil \sqrt{n}\rceil$. Bonato at al. [6] conjectured that all connected graphs are well-burnable, and they showed that $P_n$ and $C_n$ are well-burnable. As shown in [17] and independently in [13], the caterpillar is well-burnable. Bonato and Lidbetter [8] proved that the generalized star is also well-burnable. Moreover, Tan and Teh [25] showed that the bound $\lceil \sqrt{n}\rceil$ is tight for generalized stars.

Theorem 1.3

[8, 25] Let G be a generalized star of order n, then $b(G)\le \lceil \sqrt{n}\rceil$.

Recently, Liu et al. [19] determined the burning numbers of linear forests with at most three components.

Theorem 1.4

[19] Let $G=P_{a_1}+P_{a_2}$ with $a_1\ge a_2\ge 1$, and let $J(t)=\{(t^2-2,2):t\ge 2$ is an integer$\}$, then

$$\begin{aligned} b(G)= \left\{ \begin{array}{ll} \left\lceil \sqrt{a_1+a_2}\right\rceil +1, &{} \quad \hbox {if } (a_1,a_2)\in J(t) ;\\ \lceil \sqrt{a_1+a_2}\rceil , &{} \quad \hbox {otherwise}. \end{array}\right. \end{aligned}$$

Let

$$\begin{aligned} J^{1}= & {} \{(a_1,a_2,a_3):(a_2,a_3)\in D_1, \sum _{i=1}^3a_i= t^2-3\hbox { for some }t\},\\ J^{2}= & {} \{(a_1,a_2,a_3):(a_2,a_3)\in D_1\cup D_2, \sum _{i=1}^3a_i= t^2-2\hbox { for some }t\},\\ J^{3}= & {} \{(a_1,a_2,a_3):(a_2,a_3)\in \cup _{i=1}^3D_i\hbox { and }\sum _{i=1}^3a_i= t^2-1\hbox { for some }t\}\cup \{(11,11,2)\},\\ J^{4}= & {} \{(a_1,a_2,a_3):a_3=2 or (a_2,a_3)\in \cup _{i=1}^{4}D_i, \sum _{i=1}^3a_i= t^2\hbox { for some }t\}.\\ J^{5}= & {} \{(13,11,1),(11,11,3),(22,13,1),(19,13,4),(17,13,6),(15,13,8),\\&\quad (13,13,10),(17,15,4),(15,15,6),(30,15,4),(28,15,6),(26,15,8),\\&\quad (19,15,15),(28,17,4),(26,17,6),(17,17,15),(26,19,4),(43,17,4),\\&\quad (41,17,6),(30,17,17),(41,19,4),(30,30,4),(58,19,4)\}, \end{aligned}$$

where $D_1=\{(2,2)\}$, $D_2=\{(3,2)\}$, $D_3=\{(1,1),(3,3),(4,2),(5,5)\}$ and $D_4=$ $\{(2,1),(4,1),(4,3),(4,4),(6,1),(6,4),(6,5),(6,6),(7,7),(8,4),(8,6),(10,4)\}$.

Theorem 1.5

[19] Let $G=P_{a_1}+P_{a_2}+P_{a_3}$ with $a_1\ge a_2\ge a_3\ge 1$. Then

$$\begin{aligned} b(G)= \left\{ \begin{array}{ll} \lceil \sqrt{a_1+a_2+a_3}\rceil +1, &{} \quad \hbox {if } (a_1,a_2,a_3)\in J^{1}\cup J^{2}\cup J^{3}\cup J^{4}\cup J^{5};\\ \lceil \sqrt{a_1+a_2+a_3}\rceil , &{} \quad \hbox {otherwise}. \end{array}\right. \end{aligned}$$

Let $U_{g}^{a_1,\ldots , a_t}$ be a unicyclic graph of order n obtained from a cycle $C_{g}=v_1v_2\cdots v_gv_1$ by attaching t paths of lengths $a_{1},a_{2},\ldots ,a_{t}$ at the vertex $v_g$, where $a_1\ge a_2\ge \cdots \ge a_{t}\ge 1$ (see Fig. 2). Then $P_{a_i}$ ($1\le i\le t$) are called the arms. Denote ${\mathcal {U}}_{n,g}^{(t)}=\{U_{g}^{a_1,\ldots , a_t}:n=g+\sum _{i=1}^ta_i\}$ for $1\le t\le n-g$.

Fig. 2

$U_{g}^{a_{1},\ldots ,a_{t}}$

In this paper, we show that the burning number of all unicyclic graphs in ${\mathcal {U}}_{n,g}^{(t)}$ of order $n=q^2+r$ with $1\le t\le 2$ is either q or $q+1$, where $1\le r\le 2q+1$. Furthermore, we find all the sufficient conditions for all unicyclic graphs in ${\mathcal {U}}_{n,g}^{(t)}$ ($1\le t\le 2$) to have burning number q or $q+1$, respectively.

2 Preliminaries

We first cite the following results about graph burning.

Theorem 2.1

[6, 7, 23] In a graph G, the sequence $(x_1, x_2, \ldots , x_k)$ forms a burning sequence if and only if, for each pair i and j, with $1 \le i <j \le k$, $d(x_i, x_j)\ge j -i$, and the following set equation holds: $N_{k-1}[x_1]\cup N_{k-2}[x_2]\cup \cdots \cup N_{1}[x_{k-1}]\cup N_{0}[x_k]=V(G)$.

Theorem 2.2

[7] Burning a graph G in k steps is equivalent to finding a rooted tree partition $\{T_1,T_2,\ldots , T_k\}$, with heights at most $(k-1), (k-2),\ldots , 0$, respectively, such that for every $1 \le i, j\le k$, the distance between the roots of $T_i$ and $T_j$ is at least $|i-j|$.

Lemma 2.3

[7] For a graph G, we have $b(G)=\min \{b(T): T~{\textit{is}}~a~{\textit{spanning}}~{\textit{tree}}~ {\textit{of}}~\;G\}$.

A subgraph H of a graph G is called an isometric subgraph if for every pair of vertices u and v in H, we have that $d_H(u, v) = d_G(u, v)$.

Lemma 2.4

[7, 23] For any isometric subtree H of a graph G, we have $b(H)\le b(G)$.

The following assertions will be used in the proof of our main results.

Proposition 2.5

Let $w\in V(SP_s(x))$ ($s\ge 3$) with $d(x,w)=t$. If the height of $SP_s(x)$ with w as the root is at most i, then $|SP_s(x)|\le 2i+1+(s-2)(i-t)$. Moreover,

• (i) if $|SP_s(x)|=si+1$, then $x=w$ and $SP_s(x)-x=sP_i$;

• (ii) if $|SP_s(x)|=si$, then $x=w$ and $SP_s(x)-x=(s-1)P_i+P_{i-1}$, or $d(x,w)=1$, $s=3$ and $SP_s(x)-\{x,w\}=P_i+2P_{i-1}$;

• (iii) if $|SP_s(x)|=si-1$, then either $x=w$ and $SP_s(x)-x\in \{(s-1)P_{i}+P_{i-2},(s-2)P_i+2P_{i-1}\}$, or $d(x,w)=2$, $s=3$ and $SP_s(x)-\{x,y,w\}=P_i+2P_{i-2}$ where xyw is a path of length 2, or $d(x,w)=1$, $s=4$ and $SP_s(x)-\{x,w\}=P_i+3P_{i-1}$, or $d(x,w)=1$, $s=3$ and $SP_s(x)-\{x,w\}\in \{P_i+P_{i-1}+P_{i-2}, 3P_{i-1}\}$.

Proof

Let $x_j$ be a vertex of degree 1 of $SP_s(x)$, where $1\le j\le s$. Note that $d(w,z)\le i$ for any $z\in V(SP_s(x))$. Hence,

$$\begin{aligned} |SP_s(x)|\le i+t+1+(s-1)(i-t)=2i+1+(s-2)(i-t). \end{aligned}$$

(*)

• (i) If $|SP_s(x)|=si+1$, then by ($*$), $t=0$, and then $d(w,x_j)=i$ for $1\le j\le s$, that is, $x=w$ and $SP_s(x)-x= sP_i$.

• (ii) If $|SP_s(x)|=si$, then by ($*$), $(s-2)t\le 1$ which implies $t\le 1$ as $s\ge 3$, moreover, if $t=1$, then $s=3$, that is $d(x,w)=1$, and then $SP_s(x)-\{x,w\}=P_i+2P_{i-1}$. If $t=0$, that is $x=w$, then there exists some $j_0$ with $1\le j_0\le s$ such that $d(w,x_{j_0})= i-1$ and $d(w,x_j)= i$ for $1\le j\le s$ and $j\not =j_0$, thus $SP_s(x)-x=(s-1)P_{i}+P_{i-1}$.

• (iii) If $|SP_s(x)|=si-1$, then by ($*$), $(s-2)t\le 2$ which implies $t\le 2$ as $s\ge 3$. If $t=2$, then $s=3$, and then $d(x,w)=2$ and $SP_s(x)-\{x,y,w\}=P_i+2P_{i-2}$, where xyw is a path of length 2. If $t=1$, then $3\le s\le 4$, that is $d(x,w)=1$, and then $SP_s(x)-\{x,w\}=P_i+3P_{i-1}$ for $s=4$, and $SP_s(x)-\{x,w\}\in \{P_i+P_{i-1}+P_{i-2}, 3P_{i-1}\}$ for $s=3$. If $t=0$, then $x=w$ and $SP_s(x)-x\in \{(s-1)P_{i}+P_{i-2},(s-2)P_i+2P_{i-1}\}$.

$\square$

For any $G\in {\mathcal {U}}_{n,g}^{(t)}$ with $\Delta (G)=d(v_g)=t+2$ and $d(v)\le 2$ for any $v\ne v_g$, we always suppose that $(x_1, x_2,\ldots , x_k)$ is an optimal burning sequence for G. By Theorem 2.2, there exists a rooted tree partition $\{T_1(x_1), T_2(x_2), \ldots , T_k(x_k)\}$ of G with heights at most $(k-1), (k -2), \ldots , 0$, respectively. Then $T_i(x_i)= P_{l_i}$ with $l_i\le 2(k-i)+1$, or $T_i(x_i)= SP_s(v_g)$ with $3\le s\le t+2$. Assume $v_g\in V(T_{k-i_0}(x_{k-i_0}))$ for some $i_0\in [k]$. Then by Proposition 2.5, $|T_{k-i_0}(x_{k-i_0})|\le 2i_0+1+(s-2)(i_0-p_{i_0})$ with $s\le t+2$ and $d(v_g,x_{k-i_0})=p_{i_0}\ge 0$, and then $|T_{k-i}(x_{k-i})|\le 2i+1$ for $i\in [k]\backslash \{i_0\}$. Hence,

$$\begin{aligned} n= & {} \sum _{i=1}^k|T_i(x_i)|\le \sum _{i=0}^{k-1}(2i+1)-(2i_0+1)+2i_0+1+(s-2)(i_0-p_{i_0})\nonumber \\= & {} k^2+(s-2)(i_0-p_{i_0})\le k^2+t(k-1), \end{aligned}$$

(1)

where the last inequality follows from $p_{i_0}\ge 0$, $s\le t+2$ and $i_0\le k-1$.

Note that $C_g$ is the unique cycle of G, and hence $H:=G-e$ is a generalized star for any $e\in E(C_g)$. By Lemma 2.3 and Theorem 1.3, $b(G)\le b(H)\le \left\lceil \sqrt{n}\right\rceil$. So, by (1), we have the following result.

Theorem 2.6

For any $G\in {\mathcal {U}}_{n,g}^{(t)}$, we have

$$\begin{aligned} \left\lceil \sqrt{n+\frac{t^2+4t}{4}}-\frac{t}{2}\right\rceil \le b(G)\le \lceil \sqrt{n}\rceil . \end{aligned}$$

(2)

If $t\le 2$, then by (1), $b(G)\ge \left\lceil \sqrt{n+3}\right\rceil -1\ge \left\lceil \sqrt{n}\right\rceil -1.$ Therefore, we have the following result.

Corollary 2.7

Let $G\in {\mathcal {U}}_{n,g}^{(t)}$ with $n=q^2+r$ ($1\le r\le 2q+1$). If $1\le t\le 2$, then $q\le b(G)\le q+1$.

3 Burning Number of Unicyclic Graphs in ${\mathcal {U}}_{n,g}^{(1)}$

In this section, ${\mathcal {U}}_{n,g}^{(1)}=\{U_{g}^{n-g}\}$. Let $V(U_{g}^{n-g})=\{v_1,\ldots , v_g, w_1,\ldots w_{n-g}\}$ with $w_0=v_g$ and $d(v_g,w_i)=i$ for $0\le i\le n-g$. By Theorem 1.1, if $4\le n\le 5$, $b(G)=2$ for any $G\in \mathcal {U}_{n,g}^{(1)}$. So we can assume $n\ge 6$. Then $b(G)\ge 3$ for $G\in {\mathcal {U}}_{n,g}^{(1)}$.

3.1 Unicyclic Graphs in ${\mathcal {U}}_{n,g}^{(1)}$ with Burning Number q

In this subsection, we will give some sufficient conditions for the unicyclic graphs in ${\mathcal {U}}_{n,g}^{(1)}$ to have burning number $q\ge 3$. Recall that $d(v_g) =3$ and $n=q^2+r$ with $1\le r\le 2q+1$. Denote ${\mathcal {A}}=\{(2q+1,q^{2}-q-2),(q^{2}-2,q+1)\}$.

Lemma 3.1

Suppose that $1\le r\le q-1$. If $2r+1\le g\le q^{2}$ and $(g,n-g)\notin {\mathcal {A}}$, then $b(U_{g}^{n-g})\le q$.

Proof

Let $G_w:=U_{g}^{n-g}-N_{q-1}[w]$ for $w\in V(U_{g}^{n-g})$. If $b(G_w)\le q-1$, then $b(U_{g}^{n-g})\le q$ by Theorem 2.1. Hence, we can assume $b(G_w)\ge q$ for any $w\in V(U_{g}^{n-g})$.

Let $g':=\lfloor \frac{g-1}{2}\rfloor$. Then $r\le g'$ as $g\ge 2r+1$. If $g\le 2q-1$, then we can assume $n-g'\ge 2q$. Otherwise, $V(U_{g}^{n-g})\subseteq N_{q-1}[w_{q-g+g'}]$, then $b(U_{g}^{n-g})\le q$ by Theorem 2.1. So $G_{w_{q-g+g'}}= P_{n-g'-2q+1}$ with $n-g'-2q+1= q^2+r-g'-2q+1\le (q-1)^2$ as $r\le g'$, and thus, by Theorem 1.2, $b(G_{w_{q-g+g'}})\le q-1$, a contradiction. Therefore, $g\ge 2q$.

If $n-g\le q-1$, then $G_{v_g}= P_{g-2q+1}$ and $|G_{v_g}|\le (q-1)^2$ as $g\le q^2$, and then by Theorem 1.2, $b(G_{v_g})\le q-1$, a contradiction. So $n-g\ge q$, and then $G_{v_g}= P_{g-2q+1}+ P_{n-g-q+1}$ with $|G_{v_g}|= q^2+r-3q+2\le (q-1)^2$ as $r\le q-1$. Note that $(g,n-g)\notin {\mathcal {A}}$, and hence $2\notin \{n-g-q+1,g-2q+1\}$. By Theorem 1.4, $b(G_{v_g})=q-1$, a contradiction. $\square$

By Corollary 2.7 and Lemma 3.1, we have the following result.

Theorem 3.2

Suppose that $U_{g}^{n-g}$ is a unicyclic graph of order $n=q^2+r$ with $1\le r\le q-1$. If $2r+1\le g\le q^{2}$ and $(g,n-g)\notin {\mathcal {A}}$, then $b(U_{g}^{n-g})= \left\lceil \sqrt{n}\right\rceil -1=q$.

3.2 Unicyclic Graphs in ${\mathcal {U}}_{n,g}^{(1)}$ with Burning Number $q+1$

In this subsection, we will characterize the graphs in ${\mathcal {U}}_{n,g}^{(1)}$ with burning number $q+1$. Recall that ${\mathcal {U}}_{n,g}^{(1)}=\{U_{g}^{n-g}\}$ and $(x_1, x_2,\ldots , x_k)$ is an optimal burning sequence for $U_{g}^{n-g}$. By Corollary 2.7, we only need to give some sufficient conditions for $U_{g}^{n-g}$ to satisfy $k\ge q+1$.

Lemma 3.3

If $r\ge q$, then $k\ge q+1$.

Proof

Since $r\ge q$, we have

$$\begin{aligned} n+\frac{5}{4}=q^2+r+\frac{5}{4}\ge q^2+q+\frac{5}{4}=(q+\frac{1}{2})^2+1. \end{aligned}$$

Then $\sqrt{n+\frac{5}{4}}>q +\frac{1}{2}$, i.e., $\left\lceil \sqrt{n+\frac{5}{4}}-\frac{1}{2}\right\rceil \ge q+1$, and so, by (2), $k\ge q+1$.

Lemma 3.4

If $g\le 2r$, then $k\ge q+1$.

Proof

Note that $H:=P_{n-\lfloor \frac{g-1}{2}\rfloor }$ is an isometric subtree of $U_{g}^{n-g}$. Since $g\le 2r$, we have $\lfloor \frac{g-1}{2}\rfloor \le r-1$. Then $|H|=q^2+r-\lfloor \frac{g-1}{2}\rfloor \ge q^2+1$, and hence, by Theorem 1.2, $b(H)\ge q+1$. By Lemma 2.4, we have $k\ge b(H)\ge q+1$. $\square$

Lemma 3.5

If $g\ge q^{2}+1$, then $k\ge q+1$.

Proof

Since $g\ge q^{2}+1$, $n-g=q^2+r-g\le r-1$. Recall that $v_g\in V(T_{k-i_0}(x_{k-i_0}))$. If $x_{k-i_0}\in V(C_g)$, then $|T_{k-i_0}(x_{k-i_0})|\le 2i_{0}+1+(n-g)$; and if $x_{k-i_0}\notin V(C_g)$, then $|T_{k-i_0}(x_{k-i_0})|\le 2(i_{0}-p_{i_{0}})+1+(n-g)\le 2i_{0}+1+(n-g)$. Hence, by (1), $n=q^2+r\le k^2+n-g\le k^2+r-1,$ that is $k^{2}\ge q^2+1$. Thus, $k\ge q+1$. $\square$

Lemma 3.6

If $(g,n-g)\in {\mathcal {A}}$, then $k\ge q+1$.

Proof

Since $(g,n-g)\in {\mathcal {A}}$, $g\in \{2q+1,q^2-2\}$ and $n=q^2+q-1$. If $n\le k^2+k-2$, then $k\ge q+1$ clearly. So we can assume $n\ge k^2+k-1$. Then by (1), $n=q^2+q-1= k^2+k-1$, which implies $k=q$, $i_0=k-1=q-1$, $d(x_{k-i_0},v_g)=0$ (that is $x_1=v_g$), $|T_{k-i_0}(x_{k-i_0})|=3(q-1)+1$ and $|T_{k-i}(x_{k-i})|=2i+1$ for all $i\not =q-1$. Hence, by Proposition 2.5, $T_1(x_1)-v_g=3P_{q-1}$, and thus, $U_{g}^{n-g}-N_{q-1}[v_g]= P_{g-2q+1}+ P_{n-g-q+1}$. Note that $g\in \{2q+1,q^2-2\}$, and hence $2\in \{g-2q+1,n-g-q+1\}$, which is a contradiction with $|T_{k-i}(x_{k-i})|=2i+1$ for all $i\not =q-1$. Hence, $k\ge q+1$. $\square$

By Corollary 2.7 and Lemmas 3.3–3.6, we have the following result.

Theorem 3.7

Suppose that $U_{g}^{n-g}$ is a unicyclic graph of order $n=q^2+r$. If $q\le r\le 2q+1$, or $g\ge q^{2}+1$, or $g\le 2r$, or $(g,n-g)\in {\mathcal {A}}$, then $b(U_{g}^{n-g})=q+1$.

Note 1 By Theorems 3.2 and 3.7, $b(U_{g}^{n-g})$ are listed in Table 1.

Table 1 Burning number of $U_{g}^{n-g}$

4 Burning Number of Unicyclic Graphs in ${\mathcal {U}}_{n,g}^{(2)}$

Recall that ${\mathcal {U}}_{n,g}^{(2)}=\{U_{g}^{a_1,a_2}:a_1+a_2=n-g\}$ with $a_1\ge a_2$, and $\Delta (G)=d(v_g) =4$ for $G\in {\mathcal {U}}_{n,g}^{(2)}$. By Theorem 1.1, $b(G)=2$ if $G\in \{{\mathcal {U}}_{5,g}^{(2)},{\mathcal {U}}_{6,g}^{(2)}\}$. Hence, we may assume $n\ge 7$ and $b(G)\ge 3$ for all $G\in {\mathcal {U}}_{n,g}^{(2)}$. Let $V(U_{g}^{a_1,a_2})=\{v_1,\ldots , v_g,u_1,\ldots ,u_{a_1},w_1,\ldots w_{a_2}\}$ with $d(v_g,u_i)=i$ for $0\le i\le a_1$ and $d(v_g,w_j)=j$ for $0\le j \le a_2$, where $u_0=w_0=v_g$. Let $n=q^2+r$ with $1\le r\le 2q+1$. Then $\lceil \sqrt{n}\rceil =q+1$.

4.1 Unicyclic Graphs in ${\mathcal {U}}_{n,g}^{(2)}$ with Burning Number q

In this subsection, we will give some sufficient conditions for the unicyclic graphs in ${\mathcal {U}}_{n,g}^{(2)}$ to have burning number q. By Theorem 2.1, we only need to show that $b(U_{g}^{a_{1},a_{2}})\le q$, or find a burning sequence $(x_1,x_2,\ldots ,x_q)$ so that $V(U_{g}^{a_{1},a_{2}})\subseteq N_{q-1}[x_1]\cup \cdots \cup N_1[x_{q-1}]\cup N_0[x_q]$. Denote ${\mathcal {B}}:=\{(2q-2,q^2-q-2,q+1),(2q-1,q^2-q-2,q+1)\}$.

Let $v\in V(U_{g}^{a_{1},a_{2}})$, and let $H(v):=U_{g}^{a_{1},a_{2}}-N_{q-1}[v]$. If $b(H(v))\le q-1$, then $b(U_{g}^{a_{1},a_{2}})\le q$ by Theorem 2.1. Hence, we can assume $b(H(v))\ge q$ for any $v\in V(U_{g}^{a_{1},a_{2}})$.

Lemma 4.1

Suppose that $1\le r\le 2q-2$. If $r+1\le g\le 2q-1$, $a_{1}\le q^{2}-\lfloor \frac{g}{2}\rfloor -1$ and $(g,a_{1},a_{2})\notin {\mathcal {B}}$, then $b(U_{g}^{a_{1},a_{2}})\le q$.

Proof

If $a_2\le \lfloor \frac{g}{2}\rfloor$, then $H({u_{q-1-\lfloor \frac{g}{2}\rfloor }})= P_{a_1-2(q-1)+\lfloor \frac{g}{2}\rfloor }$ and $|H({u_{q-1-\lfloor \frac{g}{2}\rfloor }})|\le q^{2}-\lfloor \frac{g}{2}\rfloor -1-2q+2+\lfloor \frac{g}{2}\rfloor =(q-1)^2$, and then by Theorem 1.2, $b(H(u_{q-1-\lfloor \frac{g}{2}\rfloor }))\le q-1$, a contradiction.

So $a_{2}\ge \lfloor \frac{g}{2}\rfloor +1$. Then $H(u_{q-1-\lfloor \frac{g}{2}\rfloor })=P_{a_{1}'}+ P_{a_{2}'}$ with $a_1'=a_1-2(q-1)+\lfloor \frac{g}{2}\rfloor$, $a_2'=a_2-\lfloor \frac{g}{2}\rfloor$ and $a_{1}'+a_{2}'=q^{2}+r-g-(2q-2)\le (q-1)^2$ as $r+1\le g$. By our assumption and Theorem 1.4, we have $\min \{a_1',a_2'\}= 2$, $g=r+1$ and $a_{1}'+a_{2}'=(q-1)^2$.

If $a_1'=2$, then $a_{2}'=q^2-2q-1$, i.e., $a_1=2q-\lfloor \frac{g}{2}\rfloor$ and $a_2=q^2-2q-1+\lfloor \frac{g}{2}\rfloor$. Since $a_1\ge a_2$, we have $q^2-4q-1+2\lfloor \frac{g}{2}\rfloor \le 0$, which implies $q=3$ and $g\le 5$. Since $(g,a_{1},a_{2})\notin {\mathcal {B}}$, i.e., $(g,a_{1},a_{2})\notin \{(4,4,4),(5,4,4)\}$ for $q=3$. Hence, $g=3$, and then $n=11$, $a_1=5$ and $a_2=3$. Then, $H(w_1)= P_4$, and then by Theorem 1.2, $b(H(w_1))=2=q-1$, a contradiction.

If $a_2'=2$, then $a_1=q^2-3-\lfloor \frac{g}{2}\rfloor$ and $a_2=\lfloor \frac{g}{2}\rfloor +2$. If $g\le 2q-3$, then $q-2-\lfloor \frac{g}{2}\rfloor \ge 0$, and then $H({u_{q-2-\lfloor \frac{g}{2}\rfloor }})=P_{a_1''}+ P_{a_2''}$ with $a_1''=a_1-(2q-3-\lfloor \frac{g}{2}\rfloor )=(q-1)^2-1$ and $a_2''=1$. By Theorem 1.4, $b(H(u_{q-2-\lfloor \frac{g}{2}\rfloor }))=q-1$, a contradiction. Therefore, $2q-2\le g\le 2q-1$, furthermore, $a_1=q^2-q-2$ and $a_2=q+1$, a contradiction with $(g,a_{1},a_{2})\notin {\mathcal {B}}$. $\square$

Let

$$\begin{aligned} B_1:= & {} \{(q^2-2,q,q),(q^2-6,q+2,q+2),(q^2-10,q+4,q+4),(q^2-3,q+1,q),\\&\quad (q^2-5,q+3,q),(q^2-7,q+3,q+2),(q^2-8,q+3,q+3),(q^2-7,q+5,q),\\&\quad (q^2-10,q+5,q+3),(q^2-11,q+5,q+4),(q^2-12,q+5,q+5),(q^2-14,\\&\quad q+6,q+6),(q^2-12,q+7,q+3),(q^2-14,q+7,q+5),(q^2-14,q+9,q+3)\}.\\ B_2:= & {} \{(2q,q^2-q-2,q),(2q+2,q^2-q-6,q+2),(2q+4,q^2-q-10,q+4),\\&\quad (2q+1,q^2-q-3,q),\\&\quad (2q+3,q^2-q-5,q),(2q+3,q^2-q-7,q+2),(2q+3,q^2-q-8,q+3),\\&\quad (2q+5,q^2-q-7,q),(2q+5,q^2-q-10,q+3),(2q+5,q^2-q-11,q+4),\\&\quad (2q+5,q^2-q-12,q+5),(2q+6,q^2-q-14,q+6),(2q+7,\\&\quad q^2-q-12,q+3),(2q+7,q^2-q-14,q+5),(2q+9,q^2-q-14,q+3)\}.\\ B_3:= & {} \{(2q,q^2-q-3,q+1),(2q,q^2-q-5,q+3),(2q+2,q^2-q-7,q+3),\\&\quad (2q,q^2-q-7,q+5),(2q+3,q^2-q-10,q+5),(2q+4,q^2-q-11,q+5),\\&\quad (2q +3,q^2-q-12,q+7),(2q+5,q^2-q-14,q+7),(2q+3,q^2-q-14,q+9)\}.\\ B_4:= & {} \{(24,16,6),(22,16,8),(35,19,7),(32,19,10),(30,19,12),(28,19,14),\\&\quad (26,19,16),(30,21,10),(28,21,12),(45,22,11),(43,22,13),(41,22,15),\\&\quad (34,22,22),(43,24,11),(41,24,13),(32,24,22),(41,26,11),(60,25,12),\\&\quad (58,25,14),(47,25,25),(58,27,12),(47,38,12),(77,28,13)\}.\\ B_5:= & {} \{(22,18,6),(22,16,8),(26,28,7),(26,25,10),(26,23,12),(26,21,14),\\&\quad (26,19,16),(28,23,10),(28,21,12),(30,37,11),(30,35,13),(30,33,15),\\&\quad (30,26,22),(32,35,11),(32,33,13),(32,24,22),(34,33,11),(34,51,12),\\&\quad (34,49,14),(34,38,25),(36,49,12),(47,38,12),(38,67,13)\}.\\ B_6:= & {} \{(12,18,16),(14,16,16),(14,28,19),(17,25,19),(19,23,19),(21,21,19),\\&\quad (23,19,19),(17,23,21),(19,21,21),(19,37,22),(21,35,22),(23,33,22),\\&\quad (30,26,22),(19,35,24),(21,33,24),(30,24,24),(19,33,26),(21,51,25),\\&\quad (23,49,25),(34,38,25),(21,49,27),(21,38,38),(23,67,28)\}. \end{aligned}$$

Denote

$$\begin{aligned} {\mathcal {C}}_1= & {} \{(2q+1,q^{2}-q-2,r-q+1),(q^{2}-2,q+1,r-q+1)\};\\ {\mathcal {C}}_2= & {} \{(q^{2}-7,q+2,q+1),(2q+1,q^{2}-q-6,q+1),(2q+1,q^{2}-q-7,q+2)\};\\ {\mathcal {C}}_3= & {} \{(q^2-3,q,q),(q^2-5,q+1,q+1),(q^2-6,q+2,q+1),(q^2-7,q+2,q+2),\\&\quad (q^2-7,q+3,q+1),(q^2-11,q+4,q+4),(2q+4,q^2-q-11,q+4),\\&\quad (2q+1,q^2-q-5,q+1),(2q+1,q^2-q-6,q+2),(2q+1,q^2-q-7,q+3),\\&\quad (2q+2,q^2-q-6,q+1),(2q+2,q^2-q-7,q+2),(2q+3,q^2-q-7,q+1),\\&\quad (2q,q^2-q-3,q),(22,16,7),(13,16,16)\};\\ {\mathcal {C}}_{4}= & {} \{(g,a_{1},a_{2}):n=q^{2}+2q-2,g-q\ge a_{1}\ge a_{2}=q+1\}\cup B_{1}\cup B_{4};\\ {\mathcal {C}}_{5}= & {} \{(g,a_{1},a_{2}):n=q^{2}+2q-2,a_{1}\ge g-q\ge a_{2}=q+1\}\cup B_{2}\cup B_{5};\\ {\mathcal {C}}_{6}= & {} \{(g,a_{1},a_{2}):n=q^{2}+2q-2,a_{1}\ge a_{2}\ge g-q=q+1\}\cup B_{3}\cup B_{6}. \end{aligned}$$

If $g\ge 2q+2$, $a_1\ge a_2\ge q$, we let $\{m_1,m_2,m_3\}=\{g-2q+1,a_1-q+1,a_2-q+1\}$ with $m_1\ge m_2\ge m_3\ge 1$, then $m_3\not =a_1-q+1$ as $a_{1}\ge a_{2}$.

Lemma 4.2

Suppose that $1\le r\le 2q-2$. If $2q\le g\le q^{2}$, $a_{2}\ge r-q+1$ and $(g,a_{1},a_{2})\notin \bigcup _{i=1}^6 {\mathcal {C}}_i$, then $b(U_{g}^{a_{1},a_{2}})\le q$.

Proof

If $a_{2}\le q-1$, then $a_{1}\ge q$ (otherwise, $H(v_g)= P_{g-2q+1}$ with $|H(v_g)|\le (q-1)^{2}$ as $2q\le g\le q^{2}$, then by Theorem 1.2, $b(H(v_g))\le q-1$, a contradiction). So $H(v_g)= P_{g-2q+1}+ P_{a_{1}-q+1}$ with $|H(v_g)|=q^{2}+r-3(q-1)-1-a_2\le (q-1)^{2}$ as $a_2\ge r-q+1$. By our assumption and Theorem 1.4, $a_2=r-q+1$ and $2\in \{g-2q+1,a_{1}-q+1\}$, which implies $(g,a_{1},a_{2})\in {\mathcal {C}}_1$, a contradiction. Thus, $a_{2}\ge q$. Then $H(v_g) =P_{m_1}+ P_{m_2}+ P_{m_3}$ with $|H(v_g)|=q^2+r-4q+3$. Note that $|H(w_1)|=|H(v_g)|+2$.

If $r\le 2q-6$, then $|H(v_g)|\le (q-1)^2-4$, and then $b(H(v_g))=q-1$ by Theorem 1.5, a contradiction.

If $r=2q-5$, then $|H(v_g)|= (q-1)^{2}-3$, and $(m_2,m_3)=(2,2)$ by our assumption and Theorem 1.5. Then $m_1\ge 2$ and $a_2-q+1=2$, and then $H(w_1) \in \{P_{m_1+2}+ P_{3}+ P_{1},P_{4}+P_{m_1+1}+ P_{1}\}$ with $|H(w_1)|= (q-1)^{2}-1$, and thus, by Theorem 1.5, $b(H(w_1))= q-1$, a contradiction.

If $r=2q-4$, then $|H(v_g)|= (q-1)^{2}-2$, and $(m_2,m_3)\in \{(2,2),(3,2)\}$ by our assumption and Theorem 1.5. If $m_3=g-2q+1=2$, then $m_2=a_2-q+1\in \{2,3\}$ as $a_1\ge a_2$, which implies $(g,a_{1},a_{2})\in {\mathcal {C}}_{2}$, a contradiction. Therefore, $m_3=a_2-q+1$. If $m_2=a_1-q+1$, then $m_2=2$ (otherwise, $a_1-q+1=3$, and then $(g,a_{1},a_{2})\in {\mathcal {C}}_{2}$, a contradiction), and then $m_1=(q-1)^{2}-6\ge 3$. Note that $H(w_1) =P_{m_1+2}+ P_{3}+ P_{1}$ with $|H(w_1)|= (q-1)^{2}$, and hence, by Theorem 1.5, $b(G)= q-1$, a contradiction. So $m_2=g-2q+1$, furthermore, $g-2q+1=3$ (otherwise, $(g,a_{1},a_{2})\in {\mathcal {C}}_{2}$, a contradiction), and thus $H(w_1)=P_{m_1+1}+ P_{5}+ P_{1}$ with $m_1=(q-1)^{2}-7\ge 9$ and $|H(w_1)|=(q-1)^{2}$. By Theorem 1.5, $b(H(w_1))= q-1$, a contradiction.

If $r=2q-3$, then $|H(v_g)|= (q-1)^{2}-1$, by our assumption and Theorem 1.5, $(m_1,m_2,m_3)\in J^3$, which implies $(g,a_{1},a_{2})\in {\mathcal {C}}_3$, a contradiction.

If $r=2q-2$, then $|H(v_g)|= (q-1)^{2}$, and then by our assumption and Theorem 1.5, we have $(m_1,m_2,m_3)\in J^4\cup J^5$, which implies $(g,a_{1},a_{2})\in {\mathcal {C}}_{4}\cup {\mathcal {C}}_{5}\cup {\mathcal {C}}_{6}$, a contradiction. $\square$

By Corollary 2.7 and Lemmas 4.1–4.2, we have

Theorem 4.3

Suppose that $U_{g}^{a_{1},a_{2}}$ $(a_{1}\ge a_{2})$ is a unicyclic graph of order $n=q^2+r$ ($1\le r\le 2q+1$). If $1\le r\le 2q-2$, $r+1\le g\le 2q-1$, $a_{1}\le q^{2}-\lfloor \frac{g}{2}\rfloor -1$, $(g,a_{1},a_{2})\notin {\mathcal {B}}$ or $2q\le g\le q^{2}$, $a_{2}\ge r-q+1$ and $(g,a_{1},a_{2})\notin \bigcup _{i=1}^6 {\mathcal {C}}_i$, then $b(U_{g}^{a_{1},a_{2}})= \left\lceil \sqrt{n}\right\rceil -1=q$.

4.2 Unicyclic Graphs in ${\mathcal {U}}_{n,g}^{(2)}$ with Burning Number $q+1$

In this subsection, we will characterize the graphs in ${\mathcal {U}}_{n,g}^{(2)}$ with burning number $q+1$. By Corollary 2.7, we only need to give some sufficient conditions for a unicyclic graph to satisfy $k\ge q+1$.

Recall that $(x_1, x_2,\ldots , x_k)$ is an optimal burning sequence for $U_{g}^{a_{1},a_{2}}$, where $n=g+a_{1}+a_{2}$. Suppose that $v_g\in V(T_{k-i_0}(x_{k-i_0}))$ for some $i_0\in [k]$. Then $T_{k-i}(x_{k-i})$ is a path with $|T_{k-i}(x_{k-i})|\le 2i+1$ for any $i\ne i_{0}$.

Lemma 4.4

If $r\ge 2q-1$, then $k\ge q+1$.

Proof

Since $r\ge 2q-1$, we have $n+3=q^2+r+3\ge q^2+2q+2=(q+1)^2+1$. Then $\left\lceil \sqrt{n+3}\right\rceil -1\ge q+1$, and then by (2), $k\ge q+1$. $\square$

By Lemma 4.4, we can assume $r\le 2q-2$ in the following.

Lemma 4.5

If $3\le g\le r$ or $a_{1}\ge q^{2}-\lfloor \frac{g}{2}\rfloor$, then $k\ge q+1$.

Proof

If $3\le g\le r$, then $a_{1}+a_{2}+1=q^2+r-g+1\ge q^{2}+1$; and if $a_{1}\ge q^{2}-\lfloor \frac{g}{2}\rfloor$, then $a_{1}+\lfloor \frac{g}{2}\rfloor +1\ge q^2+1$. Note that $P_{a_{1}+a_{2}+1}$ and $P_{a_{1}+\lfloor \frac{g}{2}\rfloor +1}$ are isometric subtrees of $U_{g}^{a_{1},a_{2}}$, and hence, by Lemma 2.4 and Theorem 1.2, $k\ge b(P_{a_{1}+a_{2}+1})=\lceil \sqrt{a_{1}+a_{2}+1}\rceil \ge q+1$ and $k\ge b(P_{a_{1}+\lfloor \frac{g}{2}\rfloor +1})=\left\lceil \sqrt{a_{1}+\lfloor \frac{g}{2}\rfloor +1}\right\rceil \ge q+1$. $\square$

Lemma 4.6

If $g\ge q^{2}+1$, then $k\ge q+1$.

Proof

Note that $|T_{k-i_0}(x_{k-i_0})|\le 2i_{0}+1+a_1+a_2$. Hence, $n\le \sum _{i=0}^{k-1}(2i+1)+a_{1}+a_{2}=k^2+n-g\le k^2+n-q^2-1,$ that is $k^{2}\ge q^2+1$. Thus, $k\ge q+1$. $\square$

Lemma 4.7

Let $2q\le g\le q^{2}$. If $a_{2}\le r-q$ or $(g,a_{1},a_{2})\in {\mathcal {C}}_1$, then $k\ge q+1$.

Proof

Since $2q\le g\le q^{2}$ and $a_{2}\le r-q+1$, we have $a_1=q^2+r-g-a_2\ge q-1$. Then $|T_{k-i_0}(x_{k-i_0})|\le 2i_{0}+1+(i_{0}-p_{i_{0}})+a_2$, where $p_{i_{0}}=d(v_g,x_{k-i_0})$. Hence,

$$\begin{aligned} n=q^2+r=\sum _{i=1}^k|V(T_{i}(x_i))|\le k^2+i_0-p_{i_{0}}+a_2\le k^2+k-1+a_2. \end{aligned}$$

(3)

If $a_{2}\le r-q$, then by (3), $k^2+k\ge q^2+q+1$, i.e., $k\ge q+1$.

If $(g,a_{1},a_{2})\in {\mathcal {C}}_1$, then $a_{2}=r-q+1\le q-1$ as $r\le 2q-2$, and then by (3), $n=q^2+r\le k^2+k+r-q$. If $n=q^2+r\le k^2+k-1+r-q$, then $k\ge q+1$. So we assume $n\ge k^2+k+r-q$. Then $n=q^2+r=k^2+k+r-q$, which implies $k=q$ and all inequalities in (4) should be equal, that is, $i_0=k-1$, $p_{i_0}=0$ and $|T_{k-i}(x_{k-i})|=2i+1$ for any $i\not =i_0$. So $x_1=v_g$ and $H(v_g)\cong P_{2}+P_{q^2-2q-1}$, a contradiction. $\square$

Lemma 4.8

If $(g,a_{1},a_{2})\in {\mathcal {B}}$, then $k\ge q+1$.

Proof

Since $(g,a_{1},a_{2})\in {\mathcal {B}}$, we have $2q-3\le r=g-1\le 2q-2$ and $a_1\ge a_2= q+1$. Then $|T_{k-i_0}(x_{k-i_0})|\le 2i_{0}+g$, and hence, $n=q^2+r=\sum _{i=1}^k|T_{i}(x_i)|\le k^2+g-1=k^2+r$. Then $n=q^2+r=k^2+r$ (otherwise $n=q^2+r\le k^2+r-1$, then $k\ge q+1$), which implies $k=q$, $|T_{k-i_0}(x_{k-i_0})|=2i_{0}+g$ and $|T_{k-i}(x_{k-i})|=2i+1$ for any $i\not =i_0$, then $i_0=q-1$ and $|T_1(x_1)|=2i_0+g\in \{4(q-1),4(q-1)+1\}$. By Proposition 2.5 we can find that $x_1=v_g$, and $|T_{1}(x_{1})|=|U_{g}^{q-1,q-1}|$, then $H(v_g)=P_{q^2-2q-1}+P_{2}$, a contradiction with $|T_{k-i}(x_{k-i})|=2i+1$ for any $i\not =i_0$. $\square$

Lemma 4.9

If $(g,a_{1},a_{2})\in \bigcup _{i=2}^{6}{\mathcal {C}}_i$, then $k\ge q+1$.

Proof

Since $k \ge q$, we shall assume, to the contrary, that $k = q$. Let $G^*:=U_{g}^{a_{1},a_{2}}-V(T_{1}(x_{1}))$, then by Theorem 2.1, $b(G^*)\le q-1$.

By Proposition 2.5, $|T_{k-i_0}(x_{k-i_0})|\le 2i_{0}+1+(s-2)(i_{0}-p_{i_{0}})$ with $s\le 4$ and $p_{i_{0}}=d(v_g,x_{k-i_0})$. Hence, by (1),

$$\begin{aligned} n=q^2+r\le k^2+2(i_0-p_{i_{0}})\le k^2+2k-2=q^2+ 2q-2. \end{aligned}$$

(4)

If $(g,a_{1},a_{2})\in \bigcup _{i=4}^{6}{\mathcal {C}}_i$, i.e., $n=q^2+ 2q-2$, then all inequalities in (4) should be equal, that is $p_{i_0}=0$, $i_0=q-1$, $|T_{1}(x_{1})|=4(q-1)+1$, and then $x_1=v_g$, $T_{1}(x_{1})-v_g=4P_{q-1}$ by Proposition 2.5, which implies $G^*= P_{m_1}+P_{m_2}+P_{m_3}$ and $|G^*|=(q-1)^2$, where $\{m_1,m_2,m_3\}=\{g-2q+1,a_1-q+1,a_2-q+1\}$ and $m_1\ge m_2\ge m_3\ge 1$. On the other hand, since $(g,a_{1},a_{2})\in \bigcup _{i=4}^{6}{\mathcal {C}}_i$, we can check that $(m_1,m_2,m_3)\in J^4\cup J^5$, and thus, by Theorem 1.5, $b(G^*)=q$, a contradiction.

If $(g,a_{1},a_{2})\in {\mathcal {C}}_{3}$, then $n=q^2+2q-3$, and then by (4), $p_{i_0}=0$ (that is $x_1=v_g$) and $i_0=q-1$, furthermore, $|T_{1}(x_{1})|=4(q-1)$ and $T_{1}(x_{1})-v_g=3P_{q-1}+P_{q-2}$ by Proposition 2.5, or $|T_{1}(x_{1})|=4(q-1)+1$ and $T_{1}(x_{1})-v_g=4P_{q-1}$, which implies $G^*= P_{m}+P_{m'}+P_{M}$ for some $M\ge m'\ge m\ge 1$. In the former case, $G^*\in \{P_{g-2q+2}+P_{a_1-q+1}+P_{a_2-q+1}, P_{g-2q+1}+P_{a_1-q+2}+P_{a_2-q+1}, P_{g-2q+1}+P_{a_1-q+1}+P_{a_2-q+2}\}$, $|G^*|=(q-1)^2$ and we can check that $(M,m',m)\in J^4\cup J^5$. In the later case, $G^*= P_{g-2q+1}+P_{a_1-q+1}+P_{a_2-q+1}$, $|G^*|=(q-1)^2-1$ and $(M,m',m)\in J^3$. Then in either case, $b(G^*)=q$ by Theorem 1.5, a contradiction.

If $(g,a_{1},a_{2})\in {\mathcal {C}}_{2}$, then $n=q^2+2q-4$, and then by (4), $4q-5\le |T_{1}(x_{1})|\le 4q-3$. If $|T_{1}(x_{1})|=4q-3$, then $x_1=v_g$, $T_{1}(x_{1})-v_g=4P_{q-1}$ by Proposition 2.5, which implies $G^*\in \{ P_{q^2-2q-6}+P_3+P_2,P_{q^2-2q-5}+P_2+P_2\}$ and $|G^*|=(q-1)^2-2$, and then $b(G^*)=q$ by Theorem 1.5, a contradiction. If $|T_{1}(x_{1})|=4(q-1)$, then $x_1=v_g$, $T_{1}(x_{1})-v_g=3P_{q-1}+P_{q-2}$ by Proposition 2.5, which implies $G^*\in \{ P_{q^2-2q-5}+P_3+P_2,P_{q^2-2q-6}+P_4+P_2,P_{q^2-2q-6}+P_3+P_3, P_{q^2-2q-4}+P_2+P_2\}$ and $|G^*|=(q-1)^2-1$, and then $b(G^*)=q$ by Theorem 1.5, a contradiction. So $|T_{1}(x_{1})|=4q-5$. Then $|G^*|=(q-1)^2$ and $p_{i_{0}}=d(v_g,x_{1})\le 1$. If $x_1=v_g$, then $T_{1}(x_{1})-v_g\in \{3P_{q-1}+P_{q-3}, 2P_{q-1}+2P_{q-2}\}$ by Proposition 2.5, which implies $G^*\in \{ P_{q^2-2q-6}+P_4+P_3,P_{q^2-2q-5}+P_4+P_2,P_{q^2-2q-6}+P_5+P_2, P_{q^2-2q-3}+P_2+P_2, P_{q^2-2q-4}+P_3+P_2, P_{q^2-2q-6}+P_5+P_2, P_{q^2-2q-5}+P_3+P_3\}$, and then $b(G^*)=q$ by Theorem 1.5, a contradiction. So $d(x_1,v_g)=1$, then $x_1\in \{w_1,u_1,v_1\}$ and $T_{1}(x_{1})-\{v_g,x_1\}=3P_{q-2}+P_{q-1}$. If $x_1=w_1$, then $G^*\in \{ P_{q^2-2q-4}+P_4+P_1,P_{q^2-2q-5}+P_4+P_2\}$; if $x_1=u_1$, then $G^*\in \{ P_{q^2-2q-4}+P_3+P_2,P_{q^2-2q-6}+P_4+P_3,P_{q^2-2q-7}+P_4+P_4\}$; and if $x_1=v_1$, then $G^*\in \{ P_{q^2-2q-4}+P_3+P_2,P_{q^2-2q-6}+P_4+P_3,P_{q^2-2q-5}+P_4+P_2\}$, and thus $b(G^*)=q$ by Theorem 1.5, a contradiction. $\square$

By Corollary 2.7 and Lemmas 4.4–4.9, we have

Theorem 4.10

Suppose that $U_{g}^{a_{1},a_{2}}$ ($a_{1}\ge a_{2}$) is a unicyclic graph of order $n=q^2+r$ with $1\le r\le 2q+1$. If $r\ge 2q-1$ or $3\le g\le r$ or $g\ge q^{2}+1$ or $a_{1}\ge q^{2}-\lfloor \frac{g}{2}\rfloor$, or $2q\le g\le q^{2}$ and $a_{2}\le r-q$, or $(g,a_{1},a_{2})\in \bigcup _{i=1}^{6}{\mathcal {C}}_{i}\cup {\mathcal {B}}$, then $b(U_{g}^{a_{1},a_{2}})= \left\lceil \sqrt{n}\right\rceil =q+1$.

Note 2. By Theorems 4.3 and 4.10, $b(U_{g}^{a_1,a_2})$ are listed in Table 2, where $g+a_1+a_2=q^2+r$ and $1\le r\le 2q+1$.

Table 2 Burning number of $U_{g}^{a_{1},a_{2}}$