Burning Numbers of Path forests and Spiders

Abstract

Graph burning is a deterministic discrete-time graph process that can be interpreted as a model for the spread of influence in social networks. The burning number of a graph G is the minimum number of steps in a graph burning process for G. It is shown that the graph burning problem is NP-complete even for trees and path forests. In this paper, we first determine the burning numbers of path forests with at most three components and then use these results to determine the burning numbers of all spiders with three arms.

1 Introduction

Suppose that we have a message or a piece of information (such as a new idea or a gossip) that we want to distribute among all the users in a network such as Facebook or Twitter. Our assumptions, similar to those in the recent study on the spread of emotional contagion in Facebook [12], are that in-person interaction and agents in the network spread the contagion to their friends or followers, and the contagion propagates overtime. Our goal is to accomplish this task as quickly as possible with the minimum cost. Bonato et al. [5] defined a new graph process called graph burning to model these sort of real-world phenomenon. It is motivated by two other well-known models: the competitive diffusion game [1] and Firefighter [9].

Given a graph G, the burning process on G is a discrete-time process defined as follows. Initially, at time step $t=0$ all vertices are unburned. At each time step $t\ge 1$, one new unburned vertex is chosen to burn (if such a vertex is available); such a vertex is called a source of fire. If a vertex is burned, then it remains in that state until the end of the process. Once a vertex is burned in step t, in step $t+1$ each of its unburned neighbors becomes burned. The process ends when all vertices of G are burned. A graph is called k-burnable if it can be burned in at most k steps. The burning number of a graph G, denoted by b(G), is the minimum number of steps needed for the process to end. The vertices that are chosen to be burned are referred to as a burning sequence; a shortest such sequence is called optimal. Note that optimal burning sequences have length b(G). For example, for the path $P_6=v_1v_2 v_3v_4v_5v_6$, the sequence $(v_2, v_5, v_6)$ is an optimal burning sequence.

The burning number can be used as a measure for the speed of spreading fire on the vertex set of graphs. Burning also can be viewed as a simplified model for the spread of social contagion in a social network such as Facebook or Twitter. The lower the value of b(G), the easier it is to spread such contagion in the graph G. It turned out that graph burning is related to several other graph theory problems such as graph bootstrap percolation [2] and graph domination [11].

Let $G=(V(G),E(G))$ be a graph. For two vertices u and v, the distance between u and v, denoted by d(u, v), is the number of edges in a shortest path joining u and v. For any nonnegative integer k and a vertex u, the kth closed neighborhood of u is the set of vertices whose distance from u is at most k, and is denoted by $N_k[u]$. From the definition, a graph G is k-burnable if there is a burning sequence $(x_1, . . . , x_k)$ of vertices such that

$$\begin{aligned} \text{ for } \text{ any } u\in V(G), \text{ there } \text{ exists } \text{ some } 1\le i\le k \text{ satisfying } d(u,x_i)\le k-i~ \end{aligned}$$

(*)

and $d(x_i,x_j)\ge j-i~ \text{ for } \text{ all } 1\le i<j\le k.$

Note that ($*$) is equivalent to $N_{k-1}[x_1]\cup N_{k-2}[x_2]\cup \cdots \cup N_{1}[x_{k-1}]\cup N_{0}[x_k]=V(G)$.

Theorem 1.1

[5, 6, 18] In a graph G, $(x_1, x_2, \ldots , x_k)$ 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: $N_{k-1}[x_1]\cup N_{k-2}[x_2]\cup \cdots \cup N_{1}[x_{k-1}]\cup N_{0}[x_k]=V(G)$.

Bonato et al. [5] studied the burning number of paths and cycles.

Theorem 1.2

[5] For a path $P_n$ or a cycle $C_n$ of order n, we have that $b(P_n)=b(C_n) =\lceil \sqrt{n}\rceil$.

Based on this result, they made the following conjecture.

Conjecture 1.3

[5] For a connected graph G of order n, $b(G)\le \lceil \sqrt{n}\rceil$.

Several bounds on the burning number of graphs are given in [4, 5, 13, 18]. It is shown that the graph burning problem is NP-complete even for trees and path forests (that is, disjoint unions of paths), see [3, 18]. Bonato and Lidbetter [7] derived a $\frac{3}{2}$-approximation algorithm for computing the burning numbers of path forests and showed that Conjecture 1.3 is true for spiders (which are trees with exactly one vertex of degree strictly greater than two). Since the burning number is a relatively new parameter, it is natural to consider the burning number of special graphs, see [10, 14, 15, 17, 19]. Probabilistic results and random variations on the burning process were presented in [16, 18]. For more on graph burning and graph searching, see the recent book [8].

In this paper, we first determine the exact value of the burning numbers of path forests with at most three components and then determine the burning numbers of spiders with three arms.

2 Preliminaries

Let $P_n$ denote the path of order n. A spider $S_u$ is a tree with exactly one vertex u of degree more than 2, called the center of the spider. An arm of a spider is a path from the center to a vertex of degree 1. A spider $S_u$ with $r\ge 3$ arms can be thought of as $P^{(1)}\cup P^{(2)}\cup \cdots \cup P^{(r)}$, where the $P^{(i)}$’s are edge-disjoint paths that share one common end vertex u. Let $a_i$ represent the number of vertices (excluding the center) in path $P^{(i)}$, then $S_u$ is also denoted as $S_u(a_1,a_2,\ldots ,a_r)$. Assume, without loss of generality, that $a_1\ge a_2\ge \cdots \ge a_r$. Note that $S_u(a_1,a_2,\ldots ,a_r)$ has $1+a_1+a_2+\cdots +a_r$ vertices.

A rooted tree T(x) is a tree with one vertex x designed as the root. The depth of a vertex in a rooted tree is the number of edges in a shortest path from the vertex to the tree’s root. The height of a rooted tree T is the greatest depth in T. 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).

The following theorem provides an alternative characterization of the burning number.

Theorem 2.1

[6] Burning a graph G in k steps is equivalent to finding a rooted tree partition into k trees $T_1, T_2, . . . , T_k$, with heights at most $(k -1), (k-2), . . . , 0$, 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 $|i-j|$.

Let G be a graph, and let $(x_1, x_2, \ldots , x_k)$ be a burning sequence of G. By Theorem 2.1, G has a rooted tree partition $T_1(x_1), T_2(x_2), . . . , T_k(x_k)$ with heights at most $(k -1), (k-2), . . . , 0$, respectively. If $G=P_{a_1}\cup P_{a_2}\cup \cdots \cup P_{a_l}$ is a path forest of order n, then $T_i(x_i)$ is a path for $1\le i\le k$, i.e., $|V(T_i(x_i))|\le 2(k-i)+1$, and thus, $n=\sum _{i=1}^k|V(T_{i}(x_{i}))|\le \sum _{i=1}^{k}[2(k-i)+1]=k^2$. Specially, if $a_l=2$, then there exists some $i_0$ $(1\le i_0\le k-1)$ such that $T_{k-i_0}(x_{k-i_0})\subseteq P_{a_l}$, and then, $|V(T_{k-i_0}(x_{k-i_0}))|\le a_l=2$. Thus,

$$\begin{aligned} \sum _{i=1}^k|V(T_i(x_i))| \le \sum _{i=0}^{k-1}(2i+1)-(2i_0+1)+2=k^2-(2i_0-1)\le k^2-1. \end{aligned}$$

Therefore, we have the following result.

Lemma 2.2

Let $G=P_{a_1}\cup P_{a_2}\cup \cdots \cup P_{a_{l-1}}\cup P_{a_{l}}$ be a path forest of order n. Suppose that $(x_1, x_2, \ldots , x_k)$ is a burning sequence of G. Then, $n\le k^2$, that is, $k\ge \lceil \sqrt{n}\rceil$. Furthermore, if $a_l=2$, then $n \le k^2-1$, that is, $k\ge \lceil \sqrt{n+1}\rceil$.

Recently, Bonato and Lidbetter [7] presented the following upper bound for the burning numbers of path forests.

Lemma 2.3

[7] If G is a path forest of order n with $w\le \lceil \sqrt{n}\rceil$ components, then $b(G)\le \lceil \sqrt{n}+\frac{w-1}{2}\rceil$.

By Lemmas 2.2 and 2.3 , we have the following bounds for the burning numbers of path forests with at most three components.

Lemma 2.4

Let G be a path forest of order n with $(t-1)^2+1\le n\le t^2$. If G has at most three components, then $t\le b(G)\le t+1$. Furthermore, if $n=t^2$ and G has a $P_2$-component, then $b(G)=t+1$.

In the following, we first define some sets and then give some lemmas that will be useful in the proof of our main results.

Let,

$$\begin{aligned} D_1= & {} \{(2,2)\},\\ D_2= & {} \{(3,2)\},\\ D_3= & {} \{(1,1),(3,3),(4,2),(5,5)\},\\ D_4= & {} \{(2,1),(4,1),(4,3),(4,4),(6,1),(6,4),(6,5),(6,6),(7,7),(8,4),(8,6),(10,4)\},\\ D_5= & {} \{(11,10,4),(13,11,1),(11,11,3),(22,13,1),(19,13,4),(17,13,6),(15,13,8),\\&\quad (13,13,10),(17,15,4),(15,15,6),(30,15,4),(28,15,6),(26,15,8),(19,15,15),\\&\quad (28,17,4),(26,17,6),(17,17,15),(26,19,4),(43,17,4),(41,17,6),(30,17,17),\\&\quad (41,19,4),(30,30,4),(58,19,4)\}. \end{aligned}$$

Lemma 2.5

Let $G=P_{a_1}\cup P_{a_2}\cup P_{a_3}$ with $a_1\ge a_2\ge a_3\ge 1$. Suppose that $(x_1, x_2, \ldots , x_k)$ is a burning sequence of G, then $|V(G)|\le k^2-j$ for $(a_2,a_3)\in D_{5-j}$, that is, $k\ge \lceil \sqrt{|V(G)|+j}\rceil$, where $1\le j\le 4$.

Proof

Since $G=P_{a_1}\cup P_{a_2}\cup P_{a_3}$, $T_i(x_i)$ is a path for all $1\le i\le k$. Thus, $|V(T_{k-i}(x_{k-i}))|\le 2i+1$, and so $|V(G)|=\sum _{i=1}^k|V(T_i(x_i))| \le \sum _{i=0}^{k-1}(2i+1)=k^2$.

• 1. We show that $\sum _{i=1}^k|V(T_i(x_i))| \le k^2-4$ for $(a_2,a_3)\in D_1$. Since $a_2=a_3=2$, there exist $i_0$, $j_0$ $(1\le i_0<j_0\le k-1)$ such that $T_{k-i_0}(x_{k-i_0})\cup T_{k-j_0}(x_{k-j_0})\subseteq P_{a_2}\cup P_{a_3}$, which implies $|V(T_{k-i_0}(x_{k-i_0}))|+|V(T_{k-j_0}(x_{k-j_0}))|\le 4\le 2i_0+2j_0-2$, and thus,

  $$\begin{aligned} \sum _{i=1}^k|V(T_i(x_i))| \le \sum _{i=0}^{k-1}(2i+1)-(2i_0+1)-(2j_0+1)+(2i_0+2j_0-2)=k^2-4. \end{aligned}$$

• 2. By a similar argument as that of (1), we may show that $\sum _{i=1}^k|V(T_i(x_i))| \le k^2-3$ for $(a_2,a_3)\in D_2$.

• 3. We show that $\sum _{i=1}^k|V(T_i(x_i))| \le k^2-2$ for $(a_2,a_3)\in D_3$. If $(a_2,a_3)=(4,2)$, then by an argument similar to (1), we have $\sum _{i=1}^k|V(T_i(x_i))| \le \sum _{i=0}^{k-1}(2i+1)-2=k^2-2$. So in the following, assume $a_2=a_3=2t+1$ for $0\le t \le 2$. Assume, without loss of generality, that $x_{k-t}\notin V(P_{a_2})$; then, there exists $i_0$ $(1\le i_0\ne t\le k-1)$ such that $T_{k-i_0}(x_{k-i_0})\subseteq P_{a_2}$ and $|V(T_{k-i_0}(x_{k-i_0}))|\le 2i_0-1$, and thus,

  $$\begin{aligned} \sum _{i=1}^k|V(T_i(x_i))|\le \sum _{i=0}^{k-1}(2i+1)-(2i_0+1)+(2i_0-1)=k^2-2. \end{aligned}$$

• 4. We show that $\sum _{i=1}^k|V(T_i(x_i))|\le k^2-1$ for $(a_2,a_3)\in D_4$. Suppose to a contradiction that $\sum _{i=1}^k|V(T_i(x_i))|\ge k^2$. Then, $\sum _{i=1}^k|V(T_i(x_i))|= k^2$, and hence, we have the following fact.

Fact 1 $|V(T_i(x_i))|=2(k-i)+1$ for $1\le i\le k$, i.e., $T_{k-i}(x_{k-i})=P_{2i+1}$ for $0\le i\le k-1$.

We consider four cases:

Case 1 $(a_2,a_3)\in \{(2,1),(4,1),(6,1)\}$.

In this case, $P_{a_3}=T_{k}(x_{k})$ by Fact 1. Then, there exists $i_0$ ($1\le i_0\le k-1$) such that $T_{k-i_0}(x_{k-i_0})\subseteq P_{a_2}$ as $a_2\in \{2,4,6\}$, and thus, $|V(T_{k-i_0}(x_{k-i_0}))|\le 2i_0$, which is a contradiction to Fact 1.

Case 2 $(a_2,a_3)\in \{(4,3),(6,5),(7,7)\}$.

In this case, $a_3=2t+1$, where $t\in \{1,2,3\}$. By Fact 1, we have $P_{a_3}=T_{k-t}(x_{k-t})$. Note that $a_2=a_3+1$ or $a_2=a_3$; then, there exists $i_0$ ($t+1\le i_0\le k-1$) such that $T_{k-i_0}(x_{k-i_0})\subseteq P_{a_2}$ and $|V(T_{k-i_0}(x_{k-i_0}))|\le 2i_0$, which is a contradiction to Fact 1.

Case 3 $(a_2,a_3)\in \{(4,4),(6,4),(8,4),(10,4)\}$.

By Fact 1, we have $P_{a_3}=T_{k}(x_{k})\cup T_{k-1}(x_{k-1})$, and then, there exists $i_0$ ($2\le i_0\le k-1$) such that $T_{k-i_0}(x_{k-i_0})\subseteq P_{a_2}$ and $|V(T_{k-i_0}(x_{k-i_0}))|\le 2i_0$, which is a contradiction to Fact 1.

Case 4 $(a_2,a_3)\in \{(6,6),(8,6)\}$.

This case can be proved by a similar argument as that of Case 3.

Therefore, we complete the proof of Lemma 2.5. $\square$

Lemma 2.6

Let $G=P_{a_1}\cup P_{a_2}\cup P_{a_3}$ with $(a_1,a_2,a_3)\in D_5$. Then, $b(G)= \sqrt{|V(G)|}+1$.

Proof

Note that $|V(G)|=k^2$ for some $5\le k\le 8$. By Lemma 2.4, $k\le b(G)\le k+1$. Suppose for a contradiction that $b(G)=k$. Let $(x_1, x_2, \ldots , x_k)$ be an optimal burning sequence of G. Since $G=P_{a_1}\cup P_{a_2}\cup P_{a_3}$, $T_i(x_i)$ is a path for $1\le i\le k$. Thus, $|V(T_{k-i}(x_{k-i}))|\le 2i+1$, and $k^2=|V(G)|=\sum _{i=1}^k|V(T_i(x_i))| \le \sum _{i=0}^{k-1}(2i+1)=k^2$. Hence, we have the following facts.

Fact 2 $|V(T_i(x_i))|=2(k-i)+1$ for $1\le i\le k$, i.e., $T_{k-i}(x_{k-i})=P_{2i+1}$ for $0\le i\le k-1$.

Fact 3 (i) If $a_j$ is even for some $1\le j\le 3$, then there exist at least $i_1,i_2$ such that $V(T_{i_1}(x_{i_1}))\cup V(T_{i_2}(x_{i_2}))\subseteq V(P_{a_j})$.

(ii) If $a_j$ is odd and $a_j>2k-1$ for some $1\le j\le 3$, then there exist at least $i_1,i_2,i_3$ such that $\cup _{l=1}^{3}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_j})$.

(iii) If $a_j$ is even and $a_j> (2k-1)+(2k-3)=4k-4$ for some $1\le j\le 3$, then there exist at least $i_1,i_2,i_3,i_4$ such that $\cup _{l=1}^{4}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_j})$.

(iv) If $a_j$ is odd and $a_j> (2k-1)+(2k-3)+(2k-5)=6k-9$ for some $1\le j\le 3$, then there exist at least $i_1,i_2,\ldots ,i_5$ such that $\cup _{l=1}^{5}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_j})$.

(v) If $a_j$ is even and $a_j> (2k-1)+(2k-3)+(2k-5)+(2k-7)=8k-16$ for some $1\le j\le 3$, then there exist at least $i_1,i_2,\ldots ,i_6$ such that $\cup _{l=1}^{6}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_j})$.

We consider the following four cases:

Case 1 $(a_1,a_2,a_3)\in \{(11,10,4),(13,11,1),(11,11,3),(22,13,1),(19,13,4),(17,13,6)$, $(15,13,8),(13,13,10),(17,15,4),(15,15,6)\}$.

In this case, $5\le k\le 6$. First, we assume $(a_1,a_2,a_3)=(11,10,4)$, and then, $k=5$. Since $a_1>2k-1$ is odd, $a_2,a_3$ are even; then by Fact 3(i)(ii), there exist at least $i_1,i_2,\ldots ,i_7$ such that $\cup _{l=1}^{3}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_1})$, $\cup _{l=4}^{5}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_2})$, and $\cup _{l=6}^{7}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_3})$, where $i_{j_1}\ne i_{j_2}$ for $1\le j_1\ne j_2\le 7$. Hence, $25=|V(G)|\ge 1+3+\cdots +13=49$ by Fact 2, which is a contradiction.

Then, we assume $(a_1,a_2,a_3)=(22,13,1)$. In this case, $k=6$. Since $a_1>4k-4$ is even, $a_2>2k-1$ is odd; then by Fact 3(ii)(iii), there exist at least $i_1,i_2,\ldots ,i_8$ such that $\cup _{l=1}^{4}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_1})$, $\cup _{l=5}^{7}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_2})$, and $V(T_{i_8}(x_{i_8}))\subseteq V(P_{a_3})$, where $i_{j_1}\ne i_{j_2}$ for $1\le j_1\ne j_2\le 8$. Hence, $36=|V(G)|\ge 1+3+\cdots +15=64$ by Fact 2, which is a contradiction.

Now, we assume $(a_1,a_2,a_3)\notin \{(11,10,4),(22,13,1)\}$. Note that $a_1,a_2$ are odd and $a_1\ge a_2> 2k-1$. By Fact 3(ii), there exist at least $i_1,i_2,\ldots ,i_7$ such that $\cup _{l=1}^{3}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_1})$, $\cup _{l=4}^{6}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_2})$, and $V(T_{i_7}(x_{i_7}))\subseteq V(P_{a_3})$, where $i_{j_1}\ne i_{j_2}$ for $1\le j_1\ne j_2\le 7$. Hence, $36\ge |V(G)|\ge 1+3+\cdots +13=49$ by Fact 2, which is a contradiction.

Case 2 $(a_1,a_2,a_3)\in \{(30,15,4),(28,15,6),(26,15,8),(19,15,15),(28,17,4),(26,17,6),(17,17,15),(26,19,4)\}$.

In this case, $k=7$. First, we assume $(a_1,a_2,a_3)\in \{(19,15,15),(17,17,15)\}$. Note that $a_1,a_2,a_3$ are odd and $a_1\ge a_2\ge a_3>2k-1$. By Fact 3(ii), there exist at least $i_1,i_2,\ldots ,i_9$ such that $\cup _{l=1}^{3}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_1})$, $\cup _{l=4}^{6}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_2})$, and $\cup _{l=7}^{9}V(T_{i_l}(x_{i_l})) \subseteq V(P_{a_3})$, where $i_{j_1}\ne i_{j_2}$ for $1\le j_1\ne j_2\le 9$. Hence, $49=|V(G)|\ge 1+3+\cdots +17=81$ by Fact 2, which is a contradiction.

Now, we assume $(a_1,a_2,a_3)\notin \{(19,15,15),(17,17,15)\}$. Note that $a_1>4k-4$ is even, $a_2>2k-1$ is odd, and $a_3$ is even. By Fact 3(i)(ii)(iii), there exist at least $i_1,i_2,\ldots ,i_9$ such that $\cup _{l=1}^{4}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_1})$, $\cup _{l=5}^{7}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_2})$, and $\cup _{l=8}^{9}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_3})$, where $i_{j_1}\ne i_{j_2}$ for $1\le j_1\ne j_2\le 9$. Hence, $49=|V(G)|\ge 1+3+\cdots +17=81$ by Fact 2, which is a contradiction.

Case 3 $(a_1,a_2,a_3)\in \{(43,17,4),(41,17,6),(30,17,17),(41,19,4),(30,30,4)\}$.

In this case, $k=8$. First, we assume $(a_1,a_2,a_3)=(30,17,17)$. Note that $a_1>4k-4$ is even, $a_2,a_3$ are odd and $a_2=a_3>2k-1$. By Fact 3(ii)(iii), there exist at least $i_1,i_2,\ldots ,i_{10}$ such that $\cup _{l=1}^{4}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_1})$, $\cup _{l=5}^{7}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_2})$, and $\cup _{l=8}^{10}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_3})$, where $i_{j_1}\ne i_{j_2}$ for $1\le j_1\ne j_2\le 10$. Hence, $64=|V(G)|\ge 1+3+\cdots +19=100$ by Fact 2, which is a contradiction.

Then, we assume $(a_1,a_2,a_3)=(30,30,4)$. Note that $a_1,a_2,a_3$ are even, and $a_1=a_2>4k-4$. By Fact 3(i)(iii), there exist at least $i_1,i_2,\ldots ,i_{10}$ such that $\cup _{l=1}^{4}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_1})$, $\cup _{l=5}^{8}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_2})$, and $\cup _{l=9}^{10}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_3})$, where $i_{j_1}\ne i_{j_2}$ for $1\le j_1\ne j_2\le 10$. Hence, $64=|V(G)|\ge 1+3+\cdots +19=100$ by Fact 2, which is a contradiction.

Now, we assume $(a_1,a_2,a_3)\in \{(43,17,4),(41,17,6),(41,19,4)\}$. Note that $a_1>6k-9$ is odd, $a_2>2k-1$ is odd, and $a_3$ is even. By Fact 3(i)(ii)(iv), there exist at least $i_1,i_2,\ldots ,i_{10}$ such that $\cup _{l=1}^{5}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_1})$, $\cup _{l=6}^{8}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_2})$, and $\cup _{l=9}^{10}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_3})$, where $i_{j_1}\ne i_{j_2}$ for $1\le j_1\ne j_2\le 10$. Hence, $64=|V(G)|\ge 1+3+\cdots +19=100$ by Fact 2, which is a contradiction.

Case 4 $(a_1,a_2,a_3)=(58,19,4)$.

In this case, $k=9$. Note that $a_1>8k-16$ is even, $a_2>2k-1$ is odd, and $a_3$ is even. By Fact 3(i)(ii)(v), there exist at least $i_1,i_2,\ldots ,i_{11}$ such that $\cup _{l=1}^{6}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_1})$, $\cup _{l=7}^{9}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_2})$, and $\cup _{l=10}^{11}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_3})$, where $i_{j_1}\ne i_{j_2}$ for $1\le j_1\ne j_2\le 11$. Hence, $81=|V(G)|\ge 1+3+\cdots +21=121$ by Fact 2, which is a contradiction. $\square$

Lemma 2.7

$b(P_{11}\cup P_{11}\cup P_{2})=6$.

Proof

By Lemma 2.4, we have $5\le b(P_{11}\cup P_{11}\cup P_{2})\le 6$. Suppose to the contrary that $b(P_{11}\cup P_{11}\cup P_{2})=5$. Let $(x_1,x_2,\ldots ,x_5)$ be a burning sequences of $G=P_{a_1}\cup P_{a_2}\cup P_{a_3}$. Note that $T_i(x_i)$ is a path for $1\le i\le 5$. Then, $|V(T_{k-i}(x_{k-i}))|\le 2i+1$, and $k^2-1=|V(G)|=\sum _{i=1}^k|V(T_i(x_i))| \le \sum _{i=0}^{k-1}(2i+1)=k^2$. Hence, we have the following fact.

Fact 4 There exists $1\le i_0\le k$ such that $|V(T_{i_0}(x_{i_0}))|=2(k-{i_0})$ and $|V(T_i(x_i))|=2(k-i)+1$ for $i\ne i_0$.

Note that $a_3=2$; there exists $1\le j\le 4$ such that $V(T_{j}(x_{j}))\subseteq V(P_{a_3})$. If $j\le 3$, then $|V(T_{j}(x_{j}))|=2\le 2(k-j)-2$, a contradiction to Fact 4. Therefore, $j=4$. Hence, $|V(T_i(x_i))|=2(k-i)+1$ for $i\ne 4$. Note that $a_1=a_2=11$; there exist at least $i_1,i_2,\ldots ,i_{6}$ such that $\cup _{l=1}^{3}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_1})$, and $\cup _{l=4}^{6}V(T_{i_l}(x_{i_l}))\subseteq V(P_{a_2})$, where $i_{j_1}\ne i_{j_2}$ for $1\le j_1\ne j_2\le 6$. Hence, $24=|V(G)|> 1+3+\cdots +11=36$ by Fact 4, which is a contradiction. $\square$

3 Burning Numbers of Path forests

In this section, we will determine the burning numbers of all path forests with at most three components.

3.1 Burning Number of $P_{a_1}\cup P_{a_2}$

Lemma 3.1

[15] Let $G=P_{a_1}\cup P_{a_2}$ with $a_1\ge a_2\ge 1$ and $a_1+a_2\le t^{2}$.

• 1. If $a_1+a_2\le t^{2}-1$, then $b(G)\le t$;

• 2. If $a_1+a_2=t^{2}$ and $a_2\ne 2$, then $b(G)\le t$.

Let $J(t)=\{(t^2-2,2)$: $t\ge 2$ is an integer}. Then, $\left\lceil \sqrt{a_1+a_2}\right\rceil =t$ if $(a_1,a_2)\in J(t)$.

By Lemmas 2.4 and 3.1 , we can determine the burning number of $P_{a_1}\cup P_{a_2}$.

Theorem 3.2

Let $G=P_{a_1}\cup P_{a_2}$ with $a_1\ge a_2\ge 1$, then

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

3.2 Burning Number of $P_{a_1}\cup P_{a_2}\cup P_{a_3}$

The following lemma will be helpful.

Lemma 3.3

[15] Let $P_{a_j}$ be a path with $a_j\le t^2-\sum _{i=1}^l(2s_i+1)$, where $s_1>s_2>\cdots >s_l\ge 0$. Then, there exist $y_1,\ldots ,y_{t-s_1-1},y_{t-s_1+1}\ldots ,y_{t-s_l-1},y_{t-s_l+1},\ldots ,y_{t-1}, y_t\in V(P_{a_j})$ such that $V(P_{a_j})\subseteq N_{t-1}[y_{1}]\cup \cdots \cup N_{s_1+1}[y_{t-s_1-1}]\cup N_{s_1-1}[y_{t-s_1+1}]\cup \cdots \cup N_{s_l+1}[y_{t-s_l-1}]\cup N_{s_l-1}[y_{t-s_l+1}]\cup \cdots \cup N_{1}[y_{t-1}]\cup N_{0}[y_{t}]$.

By Lemma 3.3 and Theorem 1.1, we have the following result.

Lemma 3.4

Let $G=P_{a_1}\cup P_{a_2}\cup P_{a_3}$ with $a_1\ge a_2\ge a_3\ge 1$ and $a_1+a_2+a_3\le t^2$. Suppose there exist $y_{s_1},y_{s_2},\ldots ,y_{s_l}$ such that $a_1\le t^2-\sum _{i=1}^l(2s_i+1)$, $V(P_{a_2})\subseteq \cup _{i=1}^{j}N_{s_i}[y_{t-s_i}]$ and $V(P_{a_3})\subseteq \cup _{i=j+1}^{l}N_{s_i}[y_{t-s_i}]$ for some $1\le j\le l$, then $b(G)\le t$.

Lemma 3.5

Let $G=P_{a_1}\cup P_{a_2}\cup P_{a_3}$ with $a_1\ge a_2\ge a_3\ge 1$ and $a_1+a_2+a_3\le t^2$. Suppose that $t\ge 5$, $a_2=2t+1$ and $a_3\le 2t-2$, we have the following:

• 1. $a_1+a_2+a_3\le t^2-1$ and $(a_1,a_2,a_3)\ne (11,11,2)$, and then, $b(G)\le t$.

• 2. $a_1+a_2+a_3=t^2$, $a_3\ne 2$ and $(a_1,a_2,a_3)\notin D_5$, and then, $b(G)\le t$.

Proof

We consider the following two cases according to the parity of $a_3$:

Case 1 $a_3$ is even, say $a_3=2s$ for $1\le s\le t-1$.

Case 1.1 $s=1$, i.e., $a_3=2$.

In this case, $a_1+a_2+a_3\le t^2-1$. First, we assume $a_1+a_2+a_3=t^2-1$. If $t=5$, then $(a_1,a_2,a_3)=(11,11,2)$. Therefore, $t\ge 6$, and hence, $t-3\ge 3$. Assume $V(P_{a_3})= N_1[y_{t-1}]$, $V(P_{a_2})= N_0[y_{t}]\cup N_2[y_{t-2}]\cup N_{t-3}[y_{3}]$. Note that $a_1=(t^2-1)-(2t+1)-2=t^2-(2t-5)-5-3-1$; then by Lemma 3.4, $b(G)\le t$.

Then, we assume $a_1+a_2+a_3\le t^2-2$. Note that $t\ge 5$; then, $t-2\ge 3$. Assume $V(P_{a_3})= N_1[y_{t-1}]$, $V(P_{a_2})= N_2[y_{t-2}]\cup N_{t-2}[y_{2}]$. Note that $a_1\le (t^2-2)-(2t+1)-2=t^2-(2t-3)-5-3$; then by Lemma 3.4, $b(G)\le t$.

Case 1.2 $s=2$, i.e., $a_3=4$.

In this case, $t^2\ge a_1+a_2+a_3\ge 4t+6$, and then, $t\ge 6$. First, we assume $a_1+a_2+a_3=t^2$. If $t=6$, then $(a_1,a_2,a_3)=(19,13,4)\in D_5$. If $t=7$, then $(a_1,a_2,a_3)=(30,15,4)\in D_5$. If $t=8$, then $(a_1,a_2,a_3)=(43,17,4)\in D_5$. If $t=9$, then $(a_1,a_2,a_3)=(58,19,4)\in D_5$. Therefore, $t\ge 10$, and hence, $t-6\ge 4$. Assume $V(P_{a_3})= N_{0}[y_{t}]\cup N_1[y_{t-1}]$, $V(P_{a_2})= N_2[y_{t-2}]\cup N_3[y_{t-3}]\cup N_{t-6}[y_{6}]$. Note that $a_1=t^2-(2t+1)-4=t^2-(2t-11)-7-5-3-1$; then by Lemma 3.4, $b(G)\le t$.

Now, we assume $a_1+a_2+a_3\le t^2-1$. Note that $t\ge 5$; then, $t-2\ge 3$. Assume $V(P_{a_3})= N_0[y_{t}]\cup N_1[y_{t-1}]$, $V(P_{a_2})= N_2[y_{t-2}]\cup N_{t-2}[y_{2}]$. Note that $a_1\le (t^2-1)-(2t+1)-4=t^2-(2t-3)-5-3-1$; then by Lemma 3.4, $b(G)\le t$.

Case 1.3 $s=3$, i.e., $a_3=6$.

In this case, $t^2\ge a_1+a_2+a_3\ge 4t+8$; then, $t\ge 6$. First, we assume $a_1+a_2+a_3=t^2$. If $t=6$, then $(a_1,a_2,a_3)=(17,13,6)\in D_5$. If $t=7$, then $(a_1,a_2,a_3)=(28,15,6)\in D_5$. If $t=8$, then $(a_1,a_2,a_3)=(41,17,6)\in D_5$. Therefore, $t\ge 9$ and hence $t-5\ge 4$. Assume $V(P_{a_3})= N_{0}[y_{t}]\cup N_2[y_{t-2}]$, $V(P_{a_2})= N_1[y_{t-1}]\cup N_3[y_{t-3}]\cup N_{t-5}[y_{5}]$. Note that $a_1=t^2-(2t+1)-6=t^2-(2t-9)-7-5-3-1$; then by Lemma 3.4, $b(G)\le t$.

Now, we assume $a_1+a_2+a_3\le t^2-1$. Note that $t\ge 5$; then, $t-1\ge 3$. Assume $V(P_{a_3})= N_0[y_{t}]\cup N_2[y_{t-2}]$, $V(P_{a_2})= N_1[y_{t-1}]\cup N_{t-1}[y_{1}]$. Note that $a_1\le (t^2-1)-(2t+1)-6=t^2-(2t-1)-5-3-1$; then by Lemma 3.4, $b(G)\le t$.

Case 1.4 $s=4$, i.e., $a_3=8$.

In this case, $t^2\ge a_1+a_2+a_3\ge 4t+10$; then, $t\ge 6$. First, we assume $a_1+a_2+a_3=t^2$. If $t=6$, then $(a_1,a_2,a_3)=(15,13,8)\in D_5$. If $t=7$, then $(a_1,a_2,a_3)=(26,15,8)\in D_5$. Therefore, $t\ge 8$ and hence $t-4\ge 4$. Assume $V(P_{a_3})= N_{0}[y_{t}]\cup N_3[y_{t-3}]$, $V(P_{a_2})= N_1[y_{t-1}]\cup N_2[y_{t-2}]\cup N_{t-4}[y_{4}]$. Note that $a_1=t^2-(2t+1)-8=t^2-(2t-7)-7-5-3-1$; then by Lemma 3.4, $b(G)\le t$.

Now, we assume $a_1+a_2+a_3\le t^2-1$. Note that $t\ge 5$; then, $t-1\ge 4$. Assume $V(P_{a_3})= N_0[y_{t}]\cup N_3[y_{t-3}]$, $V(P_{a_2})= N_1[y_{t-1}]\cup N_{t-1}[y_{1}]$. Note that $a_1\le (t^2-1)-(2t+1)-8=t^2-(2t-1)-7-3-1$; then by Lemma 3.4, $b(G)\le t$.

Case 1.5 $s\ge 5$, i.e., $a_3\ge 10$.

In this case, $t^2\ge a_1+a_2+a_3\ge 4t+12$; then, $t\ge 6$. Moreover, if $t=6$, then $a_1=a_2=2t+1$, and hence, $(a_1,a_2,a_3)=(13,13,10)\in D_5$. Therefore, $t\ge 7$. If $s\le t-2$, assume $V(P_{a_3})= N_1[y_{t-1}]\cup N_{s-2}[y_{t-s+2}]$, $V(P_{a_2})= N_0[y_{t}]\cup N_2[y_{t-2}]\cup N_{t-3}[y_{3}]$. Note that $a_1\le t^2-(2t+1)-2s=t^2-(2t-5)-(2s-3)-5-3-1$; then by Lemma 3.4, $b(G)\le t$. Therefore, $s=t-1$, and hence, $a_3=2t-2$. Assume $V(P_{a_3})= N_0[y_{t}]\cup N_{t-2}[y_{2}]$, $V(P_{a_2})= N_1[y_{t-1}]\cup N_2[y_{t-2}]\cup N_{t-4}[y_{4}]$. Note that $a_1\le t^2-(2t+1)-(2t-2)=t^2-(2t-3)-(2t-7)-5-3-1$; then by Lemma 3.4, $b(G)\le t$.

Case 2 $a_3$ is odd, say $a_3=2s+1$ for $0\le s\le t-2$.

Case 2.1 $s=0$, i.e., $a_3=1$.

First, we assume $a_1+a_2+a_3=t^2$. If $t=5$, then $(a_1,a_2,a_3)=(13,11,1)\in D_5$. If $t=6$, then $(a_1,a_2,a_3)=(22,13,1)\in D_5$. Therefore, $t\ge 7$, and hence, $t-4\ge 3$. Assume $V(P_{a_3})= N_0[y_{t}]$, $V(P_{a_2})= N_1[y_{t-1}]\cup N_2[y_{t-2}]\cup N_{t-4}[y_{4}]$. Note that $a_1= t^2-(2t+1)-1=t^2-(2t-7)-5-3-1$; then by Lemma 3.4, $b(G)\le t$.

Now, we assume $a_1+a_2+a_3\le t^2-1$. Note that $t\ge 5$; then, $t-1\ge 4$. Assume $V(P_{a_3})= N_0[y_{t}]$, $V(P_{a_2})= N_1[y_{t-1}]\cup N_{t-1}[y_{1}]$. Note that $a_1\le (t^2-1)-(2t+1)-1=t^2-(2t-1)-3-1$; then by Lemma 3.4, $b(G)\le t$.

Case 2.2 $s=1$, i.e., $a_3=3$.

First, we assume $a_1+a_2+a_3=t^2$. If $t=5$, then $(a_1,a_2,a_3)=(11,11,3)\in D_5$. Therefore, $t\ge 6$, and hence, $t-3\ge 3$. Assume $V(P_{a_3})= N_1[y_{t-1}]$, $V(P_{a_2})= N_0[y_{t}]\cup N_2[y_{t-2}]\cup N_{t-3}[y_{3}]$. Note that $a_1\le t^2-(2t+1)-3=t^2-(2t-5)-5-3-1$; then by Lemma 3.4, $b(G)\le t$.

Now, we assume $a_1+a_2+a_3\le t^2-1$. Note that $t\ge 5$; then, $t-2\ge 3$. Assume $V(P_{a_3})= N_1[y_{t-1}]$, $V(P_{a_2})= N_2[y_{t-2}]\cup N_{t-2}[y_{2}]$. Note that $a_1\le (t^2-1)-(2t+1)-3=t^2-(2t-3)-5-3$; then by Lemma 3.4, $b(G)\le t$.

Case 2.3 $s\ge 2$ and $s\ne t-2$.

Assume $V(P_{a_3})= N_s[y_{t-s}]$, $V(P_{a_2})= N_0[y_{t}]\cup N_1[y_{t-1}]\cup N_{t-2}[y_{2}]$. Note that $a_1\le t^2-(2t+1)-(2s+1)=t^2-(2t-3)-(2s+1)-3-1$; then by Lemma 3.4, $b(G)\le t$.

Case 2.4 $s=t-2$, i.e., $a_3=2t-3$.

In this case, $t^2\ge a_1+a_2+a_3\ge 6t-1$, then $t\ge 6$, and hence $t-3\ge 3$. Assume $V(P_{a_3})= N_{t-2}[y_{2}]$, $V(P_{a_2})= N_0[y_{t}]\cup N_2[y_{t-2}]\cup N_{t-3}[y_{3}]$. Note that $a_1\le t^2-(2t+1)-(2s+1)=t^2-(2t-3)-(2t-5)-5-1$; then by Lemma 3.4, $b(G)\le t$. $\square$

In [15], Liu et al. considered the following upper bound for the burning number of $G=P_{a_1}\cup P_{a_2}\cup P_{a_3}$ with $a_1+a_2+a_3\le t^{2}$ and $a_2\le 2t-1$.

Lemma 3.6

[15] Let $G=P_{a_1}\cup P_{a_2}\cup P_{a_3}$ with $a_1\ge a_2\ge a_3\ge 1$ and $a_1+a_2+a_3\le t^{2}$. Suppose that $a_2\le 2t-1$, then

• 1. If $a_1+a_2+a_3\le t^{2}-2$ and $(a_2,a_3)\notin D_1\cup D_2$, then $b(G)\le t$;

• 2. If $a_1+a_2+a_3=t^{2}-1$ and $(a_2,a_3)\notin \cup _{i=1}^{3}D_i$, then $b(G)\le t$;

• 3. If $a_1+a_2+a_3=t^{2}$, $a_3\ne 2$ and $(a_2,a_3)\notin \cup _{i=1}^{4}D_i$, then $b(G)\le t$.

Lemma 3.7

Let $G=P_{a_1}\cup P_{a_2}\cup P_{a_3}$ with $a_1\ge a_2\ge a_3\ge 1$ and $a_1+a_2+a_3\le t^{2}$. Suppose that $a_2\le 2t-1$, we have the following:

• 1. If $a_1+a_2+a_3\le t^{2}-4$, then $b(G)\le t$;

• 2. If $a_1+a_2+a_3=t^{2}-3$ and $(a_2,a_3)\notin D_1$, then $b(G)\le t$;

• 3. If $a_1+a_2+a_3=t^{2}-2$ and $(a_2,a_3)\notin D_1\cup D_2$, then $b(G)\le t$.

Proof

• 1. By Lemma 3.6, we only need to consider the case that $(a_2,a_3)\in \{(2,2),(3,2)\}$. Assume $V(P_{a_3})=N_{1}[y_{t-1}]$, $V(P_{a_2})=N_{2}[y_{t-2}]$. Recall that $a_2\ge 2$ and $a_1+a_2+a_3\le t^{2}-4$. Then, $a_1\le t^{2}-4-2-2=t^{2}-5-3$. By Lemma 3.4, $b(G)\le t$.

• 2. We only need to consider $(a_2,a_3)=(3,2)$. Assume $V(P_{a_3})=N_{1}[y_{t-1}]$, $V(P_{a_2})=N_{2}[y_{t-2}]$. Note that $a_1+a_2+a_3=t^{2}-3$, and hence, $a_1= t^{2}-5-3$. By Lemma 3.4, $b(G)\le t$.

• 3. This statement holds by Lemma 3.6(1).

$\square$

Now, we need to consider the upper bound for the burning number of $G=P_{a_1}\cup P_{a_2}\cup P_{a_3}$ with $a_1+a_2+a_3\le t^{2}$ and $a_2\ge 2t$.

Lemma 3.8

Let $G=P_{a_1}\cup P_{a_2}\cup P_{a_3}$ with $a_1\ge a_2\ge a_3\ge 1$, and $a_1+a_2+a_3\le t^{2}$. Suppose that $a_2\ge 2t$, then

• 1. If $a_1+a_2+a_3\le t^{2}-1$ and $(a_1,a_2,a_3)\ne (11,11,2)$, then $b(G)\le t$;

• 2. If $a_1+a_2+a_3=t^{2}$, $a_3\ne 2$ and $(a_1,a_2,a_3)\notin D_5$, then $b(G)\le t$.

Proof

Suppose to the contrary that G is a minimal counterexample.

Note that $a_2\ge 2t$, then $t^2\ge a_1+a_2+a_3\ge 2t+2t+1=4t+1$, and hence $t\ge 5$.

Case 1 $a_3\ge 2t$.

In this case, $t^2\ge a_1+a_2+a_3\ge 6t$, $t\ge 6$.

Let w be a leaf of $P_{a_3}$, and let $v\in V(P_{a_3})$ such that $d(v,w)=t-1$. Set $P_{a_3'}=P_{a_3}-N_{t-1}[v]$, and set $G'=P_{a_1}\cup P_{a_2}\cup P_{a_3'}$. Then, $a_3'=a_3-(2t-1)\ge 1$ and $|V(G')|\le t^2-(2t-1)=(t-1)^2$. Note that $a_2\ge 2t\ge 12$. If $|V(G')|=(t-1)^2$, $a_3'\ne 2$ and $(a_1,a_2,a_3')\notin D_5$, or $|V(G')|\le (t-1)^2-1$, then $b(G')\le t-1$ by the minimality of G. But by Theorem 1.1, $b(G)\le 1+b(G')=t$, a contradiction. Therefore, $|V(G')|=(t-1)^2$, and $a_3'=2$ or $(a_1,a_2,a_3')\in D_5$. Hence, $a_1+a_2+a_3=t^2$.

First, we assume $(a_1,a_2,a_3')\in D_5$. Since $a_2-a_3'\ge 2t-1$, we have $(a_1,a_2,a_3')\in \{(26,19,4),(30,30,4)\}$. Then, $(a_1,a_2,a_3)\in \{(26,19,19),(30,30,21)\}$. Let u be a leaf of $P_{a_1}$, and let $v\in V(P_{a_1})$ such that $d(v,u)=t-1$. Set $P_{a_1'}=P_{a_1}-N_{t-1}[u]$, and set $G''=P_{a_2}\cup P_{a_3}\cup P'_{a_1}$. Now, $(a_2,a_3,a_1')\in \{(19,19,11),(30,21,13)\}$. By the minimality of G, we have $b(G'')\le t-1$. But by Theorem 1.1, $b(G)\le 1+b(G'')=t$, a contradiction.

Now, we assume $a_3'=2$. Then, $a_3=a_3'+2t-1=2t+1$. By a similar argument, we have $a_2=2t+1$. Then, $t^2=a_1+a_2+a_3\ge 6t+3$, and hence, $t\ge 7$. If $t=7$, then $(a_1,a_2,a_3)=(19,15,15)\in D_5$, and if $t=8$, then $(a_1,a_2,a_3)=(30,17,17)\in D_5$. Therefore, $t\ge 9$. Assume $V(P_{a_2})\subseteq N_{0}[y_{t}]\cup N_{2}[y_{t-2}]\cup N_{t-3}[y_{3}]$, $V(P_{a_3})\subseteq N_{1}[y_{t-1}]\cup N_{3}[y_{t-3}]\cup N_{t-5}[y_{5}]$. Note that $a_1= t^{2}-(2t+1)-(2t+1)=t^{2}-(2t-5)-(2t-9)-7-5-3-1$; then by Lemma 3.4, $b(G)\le t$, a contradiction.

Case 2 $a_3=2t-1$.

In this case, $t^2\ge a_1+a_2+a_3\ge 6t-1$, $t\ge 6$.

Let w be a leaf of $P_{a_3}$, and let $v\in V(P_{a_3})$ such that $d(v,w)=t-1$. Then, $V(P_{a_3})=N_{t-1}[v]$. Set $G'=P_{a_1}\cup P_{a_2}$. Then, $|V(G')|\le t^2-(2t-1)=(t-1)^2$. Note that $a_1\ge a_2\ge a_3=2t-1\ge 11$. Then, $b(G')\le t-1$ by Lemma 3.1. But by Theorem 1.1, $b(G)\le 1+b(G')=t$, a contradiction.

Case 3 $a_3\le 2t-2$.

Let w be a leaf of $P_{a_2}$, and let $v\in V(P_{a_2})$ such that $d(v,w)=t-1$. Set $P_{a_2'}=P_{a_2}-N_{t-1}[v]$, and set $G'=P_{a_1}\cup P_{a_2'}\cup P_{a_3}$. Then, $a_2'=a_2-(2t-1)\ge 1$ and $|V(G')|\le t^2-(2t-1)=(t-1)^2$.

Case 3.1 $a_2'\ge 2(t-1)$.

In this case, $a_1\ge a_2=a_2'+2t-1\ge 4t-3$. Since $t^2=a_1+a_2+a_3\ge 8t-5$, $t\ge 8$. Note that $a_2'\ge 2(t-1)\ge 14$. If $|V(G')|=(t-1)^2$, $a_3\ne 2$ and $(a_1,a_2',a_3)\notin D_5$, or $|V(G')|\le (t-1)^2-1$, then $b(G')\le t-1$ by the minimality of G. But by Theorem 1.1, $b(G)\le 1+b(G')=t$, a contradiction. Therefore, $|V(G')|=(t-1)^2$ and $(a_1,a_2',a_3)\in D_5$. Note that $a_1\ge 4t-3\ge 29$ and $a_1-a_2'\ge 2t-1=15$, then $(a_1,a_2',a_3)\in \{(43,17,4),(41,17,6),(41,19,4),(58,19,4)\}$, and thus $(a_1,a_2,a_3)\in \{(43,34,4),(41,34,6),(41,36,4),(58,38,4)\}$. Let u be a leaf of $P_{a_1}$, and let $v\in V(P_{a_1})$ such that $d(v,u)=t-1$. Set $P_{a_1'}=P_{a_1}-N_{t-1}[u]$, and set $G''=P'_{a_1}\cup P_{a_2}\cup P_{a_3}$. Now, $(a_1',a_2,a_3)\in \{(26,34,4),(24,34,6),(24,36,4),(39,38,4)\}$. Then, $b(G'')\le t-1$ by the minimality of G; we have $b(G'')\le t-1$. But by Theorem 1.1, $b(G)\le 1+b(G'')=t$, a contradiction.

Case 3.2 $a_2'\le 2(t-1)-1$.

In this case, $a_2=a_2'+2t-1\le 4t-4$. If $a_2=2t+1$, then $b(G)\le t$ by Lemma 3.5, which is a contradiction. Therefore, $a_2\ne 2t+1$. Then, $a_2'\ne 2$, and hence, $(a_2',a_3)\notin \{(2,1),(2,2),(2,3),(2,4)\}$.

Fact 5 Suppose that $a_3=1$, then $a_2\notin \{2t,2t+3,2t+5\}$.

Proof of Fact 5

Assume $V(P_{a_3})=N_0[y_{t}]$. If $a_2=2t$, assume $V(P_{a_2})=N_{1}[y_{t-1}]\cup N_{t-2}[y_{2}]$. Noting that $a_1\le t^2-2t-1=t^2-(2t-3)-3-1$, then $b(G)\le t$ by Lemma 3.5, which is a contradiction. If $a_2=2s+1$ with $t+1\le s\le t+2$, then $t\ge 6$ as $t^2\ge a_1+a_2+a_3\ge 4t+7$, and hence, $s-4\ge t-3\ge 3$. Assume $V(P_{a_2})=N_{1}[y_{t-1}]\cup N_{2}[y_{t-2}]\cup N_{s-4}(y_{t-s+4})$. Noting that $a_1\le t^2-(2s+1)-1=t^2-(2s-7)-5-3-1$, then $b(G)\le t$ by Lemma 3.5, which is a contradiction. $\square$

By Fact 5, we have $(a_2',a_3)\notin \{(1,1),(4,1),(6,1)\}$.

Fact 6 Suppose that $a_3=2$, then $a_2\notin \{2t,2t+2,2t+3\}$.

Proof of Fact 6

If $a_3=2$, then $a_1+a_2+a_3\le t^2-1$. Assume $V(P_{a_3})=N_1[y_{t-1}]$. If $a_2=2t$, assume $V(P_{a_2})=N_{0}(y_{t})\cup N_{t-1}[y_{1}]$. Noting that $a_1\le (t^2-1)-2t-2=t^2-(2t-1)-3-1$, then $b(G)\le t$ by Lemma 3.5, which is a contradiction. If $a_2=2t+2$, assume $V(P_{a_2})=N_{2}[y_{t-2}]\cup N_{t-2}[y_{2}]$. Noting that $a_1\le (t^2-1)-(2t+2)-2=t^2-(2t-3)-5-3$, then $b(G)\le t$ by Lemma 3.5, which is a contradiction. If $a_2=2t+3$, assume $V(P_{a_2})=N_{0}(y_{t})\cup N_{2}[y_{t-2}]\cup N_{t-2}[y_{2}]$. Noting that $a_1\le (t^2-1)-(2t+3)-2=t^2-(2t-3)-5-3-1$, then $b(G)\le t$ by Lemma 3.5, which is a contradiction. $\square$

By Fact 6, we have $(a_2',a_3)\notin \{(1,2),(3,2),(4,2)\}$.

By a similar argument, we have the following results.

Fact 7 Suppose that $a_3=3$, then $a_2\notin \{2t+2,2t+3\}$, and hence, $(a_2',a_3)\notin \{(3,3),(4,3)\}$.

Fact 8 Suppose that $a_3=4$, then $a_2\notin \{2t,2t+2,2t+3,2t+5,2t+7,2t+9\}$, and hence, $(a_2',a_3)\notin \{(1,4),(3,4),(4,4),(6,4),(8,4),(10,4)\}$.

Fact 9 Suppose that $a_3=5$, then $a_2\notin \{2t+4,2t+5\}$, and hence, $(a_2',a_3)\notin \{(5,5),(6,5)\}$.

Fact 10 Suppose that $a_3=6$, then $a_2\notin \{2t,2t+3,2t+4,2t+5,2t+7\}$, and hence, $(a_2',a_3)\notin \{(1,6),(4,6),(5,6),(6,6),(8,6)\}$.

Fact 11 Suppose that $a_3=7$, then $a_2\ne 2t+6$, and hence, $(a_2',a_3)\notin \{(7,7)\}$.

Fact 12 Suppose that $a_3=8$, then $a_2\notin \{2t+3,2t+5\}$, and hence, $(a_2',a_3)\notin \{(4,8),(6,8)\}$.

Fact 13 Suppose that $a_3=10$, then $a_2\ne 2t+3$, and hence, $(a_2',a_3)\ne (4,10)$. By the fact that $a_2'\ne 2$ and Facts 5–13, we have $a_1+a'_2+a_3\le (t-1)^2-1$ and $(a'_2,a_3),(a_3,a_2')\notin \cup _{i=1}^{3}D_i$, or $a_1+a'_2+a_3=(t-1)^2$, $a_2',a_3\ne 2$, and $(a'_2,a_3),(a_3,a_2')\notin \cup _{i=1}^{4}D_i$. By Lemma 3.6, we have $b(G')\le t-1$. But by Theorem 1.1, $b(G)\le 1+b(G')=t$, a contradiction. $\square$

Let $J^{1}=\{(a_1,a_2,a_3):(a_2,a_3)\in D_1$, $a_1+a_2+a_3= t^2-3$ for some t},

$J^{2}=\{(a_1,a_2,a_3):(a_2,a_3)\in D_1\cup D_2$, $a_1+a_2+a_3= t^2-2$ for some t},

$J^{3}=\{(a_1,a_2,a_3):(a_2,a_3)\in \cup _{i=1}^3D_i$, $a_1+a_2+a_3= t^2-1$ for some t},

$J^{4}=\{(a_1,a_2,a_3):a_3=2$ or $(a_2,a_3)\in \cup _{i=1}^{4}D_i$, $a_1+a_2+a_3= t^2$ for some t},

$J^{5}=D_5\cup \{11,11,2\}$.

By Lemmas 2.6 and 2.7 , we have $b(P_{a_1}\cup P_{a_2}\cup P_{a_3})=\lceil \sqrt{a_1+a_2+a_3}\rceil +1$ if $(a_1,a_2,a_3)\in J^{5}$. By Lemma 2.5, we have $b(P_{a_1}\cup P_{a_2}\cup P_{a_3})\ge \lceil \sqrt{a_1+a_2+a_3}\rceil +1$ if $(a_1,a_2,a_3)\in J^{1}\cup J^{2}\cup J^{3}\cup J^{4}$. By Lemmas 3.6, 3.7, 3.8, we have $b(P_{a_1}\cup P_{a_2}\cup P_{a_3})\le \lceil \sqrt{a_1+a_2+a_3}\rceil$ if $(a_1,a_2,a_3)\notin J^{1}\cup J^{2}\cup J^{3}\cup J^{4}\cup J^{5}$. Combining Lemma 2.4, we can determine the burning number of $P_{a_1}\cup P_{a_2}\cup P_{a_3}$.

Theorem 3.9

Let $G=P_{a_1}\cup P_{a_2}\cup P_{a_3}$ with $a_1\ge a_2\ge a_3\ge 1$. Then,

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

4 Burning Number of $S_u(a_1,a_2,a_3)$

In this section, we will determine the burning numbers of spiders with three arms. Bonato et al. [6] showed that Conjecture 1.3 is true for spiders.

Theorem 4.1

[6] Let G be a spider of order n, then $b(G)\le \lceil \sqrt{n}\rceil$.

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 $d_H(u, v) = d_G(u, v)$. The burning number is not monotonic even on the isometric subgraphs of a graph. For example, $b(C_5)=3$, but $b(W_5)=2$, where $W_5$ is the wheel graph obtained from a 5-cycle by adding a new vertex join to each vertex of the 5-cycle. However, Bonato et al. [6, 18] showed that the burning number is monotonic on the isometric subtrees.

Lemma 4.2

[6, 18] For any isometric subtree H of a graph G, we have that $b(H)\le b(G)$.

The following lemma will be useful in the proof of our main results.

Lemma 4.3

[15] Let $S_x$ be a spider with three arms, and let $w\in V(S_x)$ with $d(x,w)=p$. If the height of $S_x$ with w as the root is at most i, then $|V(S_x)|\le 3i+1-p$. Moreover, if $|V(S_x)|=3i+1$, then $x=w$ and $S_x-x=P_i\cup P_i\cup P_i$, and if $|V(S_x)|=3i$, then $x=w$ and $S_x-x=P_i\cup P_i\cup P_{i-1}$, or $xw\in E(G)$ and $S_x-w=P_i\cup P_{2i-1}$.

Suppose that $(x_1, x_2,\ldots , x_k)$ is an optimal burning sequence for $S_u(a_1,a_2,a_3)$. By Theorem 2.1, there exists a rooted tree partition $T_1(x_1), T_2(x_2), \ldots , T_k(x_k)$ of $S_u(a_1,a_2,a_3)$, with heights at most $(k-1), (k -2), \ldots , 0$, respectively. By the definition of spiders, $T_i(x_i)$ is a spider or a path. Let $I_1=\{i:T_i(x_i)$ is a spider, $1\le i\le k\}$. Then, $|V(T_{k-i}(x_{k-i}))|\le 2i+1$ for any $k-i\notin I_1$. Note that if $T_i(x_i)$ is a spider, then $u\in V(T_i(x_i))$, and thus, $|I_1|\le 1$.

Lemma 4.4

Let $S_u(a_1,a_2,a_3)$ be a spider of order $n=q^2+r$ with $1\le r\le 2q+1$, then

$$\begin{aligned} b(S_u(a_1,a_2,a_3))\ge \left\lceil \sqrt{n+\frac{5}{4}}-\frac{1}{2}\right\rceil . \end{aligned}$$

Proof

Suppose that $(x_1, x_2,\ldots , x_k)$ is an optimal burning sequence for $S_u(a_1,a_2,a_3)$. Then, $b(S_u(a_1,a_2,a_3))=k\ge 2$. We will show that $(k+\frac{1}{2})^2\ge n+\frac{5}{4}$, i.e., $n\le k^2+k-1$. Recall that $|V(T_{k-i}(x_{k-i}))|\le 2i+1$ for any $k-i\notin I_1$. We consider two cases:

Case 1 $|I_1|=0$. In this case, $I_1=\emptyset$; then,

$$\begin{aligned} n=\sum _{i=1}^k|V(T_i(x_i))| \le \sum _{i=0}^{k-1}(2i+1)=k^2\le k^2+k-1, \end{aligned}$$

where the last inequality follows from $k\ge 2$.

Case 2 $|I_1|=1$, say $k-i_0\in I_1$.

In this case, $i_0\le k-1$. By Lemma 4.3, $|V(T_{k-i_0}(x_{k-i_0}))|\le 3i_0+1$, and thus,

$$\begin{aligned} n=\sum _{i=1}^k|V(T_i(x_i))|\le \sum _{i=0}^{k-1}(2i+1)-(2i_0+1)+(3i_0+1)=k^2+i_0\le k^2+k-1. \end{aligned}$$

Therefore, we complete the proof of Lemma 4.4. $\square$

Theorem 4.5

Let $S_u(a_1,a_2,a_3)$ be a spider of order $n=q^2+r$ with $1\le r\le 2q+1$. Then,

$$\begin{aligned} q\le b(S_u(a_1,a_2,a_3))\le q+1. \end{aligned}$$

Proof

Note that

$$\begin{aligned} \sqrt{n+\frac{5}{4}}-\frac{1}{2}=\sqrt{q^2+r+\frac{5}{4}}-\frac{1}{2}>\sqrt{q^2-q+\frac{1}{4}}-\frac{1}{2}=q-1, \end{aligned}$$

and hence, by Lemma 4.4, we have $k\ge \lceil \sqrt{n+\frac{5}{4}}-\frac{1}{2}\rceil \ge q$. By Theorem 4.1, we have $k\le \lceil \sqrt{n}\rceil =q+1$. Thus, $q\le b(S_u(a_1,a_2,a_3))\le q+1$. $\square$

4.1 Spiders with Burning Number $q+1$

Let $S_u(a_1,a_2,a_3)$ be a spider of order $n=a_{1}+a_{2}+a_{3}+1=q^2+r$ with $1\le r\le 2q+1$. Then, $\lceil \sqrt{n}\rceil =q+1$. In this subsection, we will characterize the spiders with burning number $q+1$. Recall that $(x_1, x_2,\ldots , x_k)$ is an optimal burning sequence for $S_u(a_1,a_2,a_3)$. By Theorem 4.5, we only need to give some sufficient conditions for a spider to satisfy $k\ge q+1$.

Lemma 4.6

If $I_1= \emptyset$, then $k\ge q+1$.

Proof

Since $I_1= \emptyset$, $|V(T_{i}(x_{i}))|\le 2(k-i)+1$ for any $1\le i\le k$. Thus,

$$\begin{aligned} n=\sum _{i=1}^k|V(T_i(x_i))|\le \sum _{i=1}^k(2(k-i)+1)=k^2, \end{aligned}$$

that is, $k\ge \lceil \sqrt{n}\rceil =q+1$. $\square$

Lemma 4.7

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 Lemma 4.4, $k\ge q+1$. $\square$

Lemma 4.8

If $a_3\le r-1$, then $k\ge q+1$.

Proof

Since $a_3\le r-1$, we have $a_1+a_2+1=q^2+r-a_3\ge q^2+1$; thus, $b(P_{a_1+a_2+1})\ge q+1$ by Theorem 1.2. Note that $P_{a_1+a_2+1}$ is an isometric subtree of $S_u(a_1,a_2,a_3)$; then, $b(S_u(a_1,a_2,a_3))\ge b(P_{a_1+a_2+1})$ by Lemma 4.2, which implies $k\ge q+1$. $\square$

Denote $A:=\{(q,q),(q+2,q+1),(q+2,q+2),(q+4,q+4)\}$.

Lemma 4.9

If $a_3\ge q$, $r= q-2$ and $(a_2,a_3)\in A$, then $k\ge q+1$.

Proof

By Lemma 4.6, we may assume $I_1\ne \emptyset$, say $k-i_0\in I_1$ with $T_{k-i_0}(x_{k-i_0})\cong S_u$. Then, $i_0\le k-1$ and $|V(T_{k-i_0}(x_{k-i_0}))|\le 3i_0+1-p$ by Lemma 4.3, where $p:=d(u,x_{k-i_0})$. So,

$$\begin{aligned} ~~~~n=q^2+q-2= & {} \sum _{i=1}^k|V(T_i(x_i))|\le \sum _{i=0}^{k-1}(2i+1)-(2i_0+1)+(3i_0+1-p)\\ {}= & {} k^2+i_0-p.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(\dagger ) \end{aligned}$$

If $\sum _{i=1}^k|V(T_i(x_i))|\le k^2+k-3$, then $q^2+q-2=n=\sum _{i=1}^k|V(T_i(x_i))|\le k^2+k-3$, i.e., $(k+q+1)(k-q)\ge 1$, and thus, $k\ge q+1$. So, we have the following assumption.

Assumption 1

$\sum _{i=1}^k|V(T_i(x_i))|\ge k^2+k-2$. By Assumption 1 and ($\dagger$), $i_0-p\ge k-2$. Moreover, if $i_0-p= k-2$, then $\sum _{i=1}^k|V(T_i(x_i))|= k^2+k-2$, and then, all the inequalities in ($\dagger$) should be equalities, that is, $|V(T_{k-i_0}(x_{k-i_0})|=3i_0+1-p$ and $T_{k-i}(x_{k-i})=P_{2i+1}$ for all $i\ne k-i_0$. We consider two cases:

Case 1 $p=0$. In this case, $i_0\in \{k-2,k-1\}$.

Subcase 1.1 $i_0=k-2$.

In this subcase, $\sum _{i=1}^k|V(T_i(x_i))|= k^2+k-2$ and $T_{k-i}(x_{k-i})=P_{2i+1}$ for all $i\ne k-2$. Then by Lemma 4.3, $x_2=u$ and $T_2(x_2)=S_u(k-2,k-2,k-2)$, that is, $V(T_2(x_2))=N_{k-2}[u]$. Let $T'=S_u(a_1,a_2,a_3)-N_{k-2}[u]$, then $T'\cong P_{a_1'}\cup P_{a_2'}\cup P_{a_3'}$, where $a_i'=a_i-k+2$ for $i=1,2,3$. Suppose that $k= q$. Then, $(a_2',a_3')\in \{(2,2), (4,3),(4,4),(6,6)\}$ as $(a_2,a_3)\in A$.

Note that $\bigcup _{i\ne k-2}T_i(x_i)= P_{a_1'}\cup P_{a_2'}\cup P_{a_3'}$. Since $T_{k-i}(x_{k-i})=P_{2i+1}$ for all $i\ne k-2$, we have $a_2'\ne 2$. Moreover, if $a_2'=2t$ $(t\in \{2,3\})$, then $P_{a_2'}=T_{k-t+1}(x_{k-t+1})\cup T_k(x_k)$, and then, there exists some $j\ge t$ and $j\ne k-2$ such that $T_{k-j}(x_{k-j})\subseteq P_{a_3'}$, which implies $|V(T_{k-j}(x_{k-j}))|\le a_3'\le a_2'=2t\le 2j$, a contradiction with $T_{k-j}(x_{k-j})=P_{2j+1}$. So $k\ge q+1$.

Subcase 1.2 $i_0=k-1$.

In this subcase, $x_1=u$, and $k^2+k-2\le \sum _{i=1}^k|V(T_i(x_i))|\le k^2+k-1$. Then, $T_1(x_1)=S_u(k-1,k-1,k-1)$ or $T_1(x_1)=S_u(k-1,k-1,k-2)$.

Suppose that $k= q$. Let $T'=S_u(a_1,a_2,a_3)-V(T_1(x_1))$, then $T'=P_{a_1'}\cup P_{a_2'}\cup P_{a_3'}$, where $a_i'=a_i-q+2$ for at most one $1\le i\le 3$, and $a_j'=a_j-q+1$ for $1\le j\ne i\le 3$. Since $(a_2,a_3)\in A$, we have

$$\begin{aligned} (a_2',a_3')\in \left\{ \begin{array}{ll} \{(2, 1),(4,2),(4,3),(6,5)\}, &{} \text{ if } a_2'=a_2-q+2,~a_3'=a_3-q+1;\\ \{(1, 1),(3,2),(3,3),(5,5)\}, &{} \text{ if } a_2'=a_2-q+1,~a_3'=a_3-q+1;\\ \{(1, 2),(3,3),(3,4),(5,6)\}, &{} \text{ if } a_2'=a_2-q+1,~a_3'=a_3-q+2. \end{array}\right. \end{aligned}$$

Furthermore, $(a_2',a_3')\in \{(1, 1),(3,2),(3,3),(5,5)\}$ if $T_1(x_1)=S_u(k-1,k-1,k-1)$. Note that $(x_2,\ldots ,x_k)$ is a burning sequence of $P_{a_1'}\cup P_{a_2'}\cup P_{a_3'}$. By Lemma 2.5, $\sum _{i=2}^{k}|V(T_i(x_i))|\le (k-1)^2-1$. Moreover, $\sum _{i=2}^{k}|V(T_i(x_i))|\le (k-1)^2-2$ if $(a_2',a_3')\in \{(1, 1),(3,2),(3,3),(5,5)\}$. Then,

$$\begin{aligned} \sum _{i=1}^{k}|V(T_i(x_i))|\le \left\{ \begin{array}{ll} 3(k-1)+(k-1)^2-1,&{} \text{ if } T_1(x_1)=S_u(k-1,k-1,k-2);\\ 3k-2+(k-1)^2-2, &{} \text{ if } T_1(x_1)=S_u(k-1,k-1,k-1). \end{array}\right. \end{aligned}$$

In either case, $\sum _{i=1}^{k}|V(T_i(x_i))|\le k^2+k-3$, a contradiction to Assumption 1. Therefore, $k\ge q+1$.

Case 2 $p=1$. In this case, $i_0=k-1$ (i.e., $i_0-p=k-2$), $d(x_1,u)=1$. Suppose that $k= q$, then $N_{k-1}[x_1]=V(S_u(k,k-2,k-2))$ as $a_3\ge q=k$, and thus, $|V(T_1(x_1))|=3k-3=3q-3$. Let $T'=S_u(a_1,a_2,a_3)-N_{k-1}[x_1]$, then $T'\cong P_{a_1'}\cup P_{a_2'}\cup P_{a_3'}$. If $x_1\in V(P^{(1)})$, then $a_2'=a_2-q+2$ and $a_3'=a_3-q+2$, and thus, $(a_2',a_3')\in \{(2,2), (4,3), (4,4),(6,6)\}$ as $(a_2,a_3)\in A$. Similarly, if $x_1\in V(P^{(2)})$, then $a_2'=a_2-q$, $a_3'=a_3-q+2$ and $(a_2',a_3')\in \{(0,2), (2,3), (2,4),(4,6)\}$, and if $x_1\in V(P^{(3)})$, then $a_2'=a_2-q+2$, $a_3'=a_3-q$ and $(a_2',a_3')\in \{(2,0), (4,1), (4,2),(6,4)\}$.

Note that $(x_2,x_2,\ldots ,x_k)$ is a burning sequence of $P_{a_1'}\cup P_{a_2'}\cup P_{a_3'}$. Then by Lemmas 2.2 and 2.5 , $\sum _{i=2}^{k}|V(T_i(x_i))|\le (k-1)^2-1$ in either case. Thus, $\sum _{i=1}^{k}|V(T_i(x_i))|\le 3k-3+(k-1)^2-1=k^2+k-3$, a contradiction to Assumption 1. So, $k\ge q+1$.

Therefore, we complete the proof of Lemma 4.9. $\square$

Let

$$\begin{aligned} B= & {} \{(q,q),(q+1,q),(q+1,q+1),(q+2,q+2),(q+3,q),(q+3,q+2),(q+3,q+3),\\&(q+4,q+4),(q+5,q),(q+5,q+3),(q+5,q+4),(q+5,q+5),(q+6,q+6),\\&(q+7,q+3),(q+7,q+5),(q+9,q+3)\}.\\ S= & {} \{(18,16,6),(16,16,8),(28,19,7),(25,19,10),(23,19,12),(21,19,14),(19,19,16),\\&(23,21,10),(21,21,12),(37,22,11),(35,22,13),(33,22,15),(26,22,22),(35,24,11),\\&(33,24,13),(24,24,22),(33,26,11),(51,25,12),(49,25,14),(38,25,25),(49,27,12),\\&(38,38,12),(67,28,13)\}. \end{aligned}$$

Lemma 4.10

Suppose that $r=q-1$. Then, we have the following:

• 1. If $a_2=q+1$ and $a_3=q-1$, then $k\ge q+1$;

• 2. If $a_3=q+1$, then $k\ge q+1$;

• 3. If $a_3\ge q$, and $(a_2,a_3)\in B$ or $(a_1,a_2,a_3)\in S$, then $k\ge q+1$.

Proof

By Lemma 4.6, we can assume $I_1\ne \emptyset$, say $k-i_0\in I_1$ with $T_{k-i_0}(x_{k-i_0})\cong S_u$. Then, $i_0\le k-1$ and $|V(T_{k-i_0}(x_{k-i_0}))|\le 3i_0+1-p$ by Lemma 4.3, where $p:=d(u,x_{k-i_0})$. So,

$$\begin{aligned} q^2+q-1=n=\sum _{i=1}^k|V(T_i(x_i))|\le \sum _{i=0}^{k-1}(2i+1)+i_0-p=k^2+i_0-p.~~~~~~~(\ddagger ) \end{aligned}$$

If $\sum _{i=1}^k|V(T_i(x_i))|\le k^2+k-2$, then $q^2+q-1=n=\sum _{i=1}^k|V(T_i(x_i))|\le k^2+k-2$, and thus, $k\ge q+1$. Therefore, we have the following assumption. $\square$

Assumption 2

$\sum _{i=1}^k|V(T_i(x_i))|\ge k^2+k-1$.

By Assumption 2 and ($\ddagger$), $i_0-p\ge k-1$. Since $i_0\le k-1$ and $p\ge 0$, $p=0,i_0=k-1$. Suppose that $k= q$, then $\sum _{i=1}^k|V(T_i(x_i))|= k^2+k-1$, and thus, all the inequalities in ($\ddagger$) should be equalities, that is, $|V(T_{1}(x_{1}))|=3(k-1)+1$, and $T_{k-i}(x_{k-i})=P_{2i+1}$ for all $1\le i\le k-2$. Thus, $x_1=u$ and $T_1(x_1)=S_u(k-1,k-1,k-1)$ (i.e., $|V(T_1(x_1))|=3k-2$). Let $T'=S_u(a_1,a_2,a_3)-V(T_1(x_1))$, then $T'\cong P_{a_1'}\cup P_{a_2'}\cup P_{a_3'}$, where $a_i'=a_i-k+1$ for $i=1,2,3$. Note that $a_1'+a_2'+a_3'=q^2+r-(3q-2)=(q-1)^2$.

If $a_2=q+1$ and $a_3=q-1$, then $a_2'=2$ and $a_3'=0$. But now, $(x_2,\ldots ,x_k)$ is a burning sequence of $P_{a_1'}\cup P_{2}$, which is a contradiction to Theorem 3.2. If $a_3=q+1$, then $a_3'=2$; if $(a_2,a_3) \in B$, then $(a_2',a_3')\in \{(1,1),(2,2),(5,5)\}\cup D_4$; if $(a_1,a_2,a_3)\in S$, then $a_1'+a_2'+a_3'=(q-1)^2$ and $(a_1',a_2',a_3')\in J^{5}$. But now, $(x_2,\ldots ,x_k)$ is a burning sequence of $P_{a_1'}\cup P_{a_2'}\cup P_{a_3'}$, which is a contradiction to Theorem 3.9. Hence, $k\ge q+1$.

$\square$

By Lemmas 4.7–4.10 and Theorem 4.5, we have the following result.

Theorem 4.11

Suppose that $S_u(a_1,a_2,a_3)$ ($a_1\ge a_2\ge a_3\ge 1$) is a spider of order $n=q^2+r$ with $1\le r\le 2q+1$. If $q\le r\le 2q+1$, or $1\le a_3\le r-1$, or $(r,a_2,a_3)=(q-1,q+1,q-1)$, then $b(S_u(a_1,a_2,a_3))= \left\lceil \sqrt{n}\right\rceil$. Moreover, if $a_3\ge q$, $r= q-2$, and $(a_2,a_3)\in A$; or $a_3= q+1$ and $r= q-1$; or $a_3\ge q$, $r= q-1$, and $(a_2,a_3)\in B$ or $(a_1,a_2,a_3)\in S$, then $b(S_u(a_1,a_2,a_3))= \left\lceil \sqrt{n}\right\rceil$.

4.2 Spiders with Burning Number q

Let $S_u(a_1,a_2,a_3)$ be a spider of order $n=a_1+a_2+a_3+1$, where $a_1\ge a_2\ge a_3\ge 1$. In this subsection, we will give some sufficient conditions for $S_u(a_1,a_2,a_3)$ to have burning number q. Recall that $n=q^2+r$ with $1\le r\le 2q+1$. If $q \le r\le 2q+1$, then $b(S_u(a_1,a_2,a_3))=q+1$ by Lemma 4.7. So, we always assume $1\le r\le q-1$ in this subsection.

Lemma 4.12

Suppose that $r\le a_3\le q-1$. If $(r,a_2,a_3)\ne (q-1,q+1,q-1)$, then $b(S_u(a_1,a_2,a_3))\le q$.

Proof

Let $x\in P^{(1)}$ such that $d(u,x)=q-a_3-1$ and set $x_1=x$. Then, $G'=S_u(a_1,a_2,a_3)-N_{q-1}[x_1]=P_{a_1'}\cup P_{a_2'}$ as $a_3\le q-1$. Note that $a_1'=a_1+a_3-2q+2$ and $a_2'=a_2-a_3$. Thus, $a_1'+a_2'=q^2+r-2q+1-a_3=(q-1)^2+r-a_3\le (q-1)^2$ as $r\le a_3$.

If $b(G')\le q-1$, then $b(G)\le 1+b(G')=q$ by Theorem 1.1. Therefore, $b(G')\ge q$. By Theorems 1.2 and 3.2 , $a_1'+a_2'=(q-1)^2$, and $a_1'=2$ or $a_2'=2$. Then, $r=a_3$.

If $d(u,x)=q-a_3-1=0$, then $r=a_3=q-1$. Note that $(r,a_2,a_3)\ne (q-1,q+1,q-1)$, then $a_2\ne q+1$, thus $a_2'\ne 2$, and hence $a_1'=2$. Recall that $a_1'=a_1+a_3-2q+2$; then, $a_1=q+1$. Since $a_1\ge a_2\ge a_3$ and $a_2\ne q+1$, $a_2=q$ or $q-1$. Then, $a_2'=0$ or 1. But now, $a_1'+a_2'=2$ or 3, a contradiction to $a_1'+a_2'=(q-1)^2$.

Therefore, $d(u,x)=q-a_3-1\ge 1$. Let $x'\in P^{(1)}$ such that $d(u,x')=q-a_3-2$ and set $x_1=x'$. Then, $G''=S_u(a_1,a_2,a_3)-N_{q-1}[x_1]=P_{a_1''}\cup P_{a_2''}$. Note that $a_1''=a_1'+1$, $a_2''=a_2'-1$, and $a_1''+a_2''=(q-1)^2$. If $a_1''=2$, then $a_1'=1$, and thus, $a_2'=2$, but now $a_1''+a_2''=3$, a contradiction to $a_1''+a_2''=(q-1)^2$. Therefore, $a_1''\ne 2$. If $a_2''=a_2'-1=2$, then $a_2'=3$, and thus, $a_1'=2$, but now $a_1''+a_2''=5$, a contradiction to $a_1''+a_2''=(q-1)^2$. Therefore, $a_2''\ne 2$. By Theorems 1.2 and 3.2 , $b(G')\le q-1$, and thus, $b(G)\le 1+b(G')=q$ by Theorem 1.1. $\square$

Lemma 4.13

Suppose that $a_3\ge q$.

• 1. If $1\le r\le q-3$, then $b(S_u(a_1,a_2,a_3))\le q$;

• 2. If $r=q-2$ and $(a_2,a_3)\notin A$, then $b(S_u(a_1,a_2,a_3))\le q$;

• 3. If $r=q-1$, $a_3\ne q+1$, $(a_2,a_3)\notin B$ and $(a_1,a_2,a_3)\notin S$, then $b(S_u(a_1,a_2,a_3))\le q$.

Proof

(1) In this case, $1\le r\le q-3$. First, we assume $(a_2,a_3)\notin \{(q+1,q+1),(q+2,q+1)\}$. Let $x_1=u$, then $G'=S_u(a_1,a_2,a_3)-N_{q-1}[x_1] =P_{a_1'}\cup P_{a_2'}\cup P_{a_3'}$. Note that $a_i'=a_i-q+1$ for $i=1,2,3$, and $a_1'+a_2'+a_3'=q^2+r-3q+2\le (q-1)^2-2$ as $r\le q-3$. Since $(a_2,a_3)\notin \{(q+1,q+1),(q+2,q+1)\}$, we have $(a_2',a_3')\notin \{(2,2),(3,2)\}$. By Lemma 3.6(1), $b(G')\le q-1$, and hence, $b(S_u(a_1,a_2,a_3))\le q$ by Theorem 1.1.

Now, we assume $(a_2,a_3)\in \{(q+1,q+1),(q+2,q+1)\}$. Let $x\in P^{(3)}$ such that $d(u,x)=2$ and set $x_1=x$. Then, $G''=S_u(a_1,a_2,a_3)-N_{q-1}[x_1]=P_{a_1''}\cup P_{a_2''}$. Note that $a_1''=a_1-(q-3)\ge 4$, $a_2''=a_2-(q-3)\ge 4$ and $a_1''+a_2''=q^2+r-2(q-1)-1-(q-3)=q^2+r-3q+4\le (q-1)^2$, and so, by Lemma 3.1, $b(G'')\le q-1$. Hence, $b(S_u(a_1,a_2,a_3))\le q$ by Theorem 1.1.

(2) In this case, $r= q-2$ and $(a_2,a_3)\notin A$. First, we assume $(a_2,a_3)\notin \{(q+1,q+1),(q+3,q+1)\}$. Let $x_1=u$, then $G'=S_u(a_1,a_2,a_3)-N_{q-1}[x_1] =P_{a_1'}\cup P_{a_2'}\cup P_{a_3'}$. Note that $a_i'=a_i-q+1$ for $i=1,2,3$, and $a_1'+a_2'+a_3'=q^2+r-3q+2=(q-1)^2-1$ as $r= q-2$. Since $(a_2,a_3)\notin A \cup \{(q+1,q+1),(q+3,q+1)\}$, we have $(a_2',a_3')\notin D_1\cup D_2\cup D_3$. If $(a_1',a_2',a_3')\ne (11,11,2)$, then by Theorem 3.9, $b(G')\le q-1$ and hence $b(S_u(a_1,a_2,a_3))\le q$ by Theorem 1.1. Therefore, $(a_1',a_2',a_3')=(11,11,2)$ and hence $q=6$ and $(a_1,a_2,a_3)=(16,16,7)$. Let $x\in P^{(3)}$ such that $d(u,x)=1$ and set $x_1=x$. Then, $G''=S_u(a_1,a_2,a_3)-N_{5}[x_1]=P_{a_1''}\cup P_{a_2''}\cup P_{a_3''}$. Note that $a_1''=a_1-4=12$, $a_2''=a_2-4=12$ and $a_2''=a_3-6=1$. By Lemma 3.1, $b(G'')\le 5$. Hence, $b(S_u(a_1,a_2,a_3))\le 6$ by Theorem 1.1.

Now, we assume $(a_2,a_3)\in \{(q+1,q+1),(q+3,q+1)\}$. Let $x\in P^{(3)}$ such that $d(u,x)=1$ and set $x_1=x$. Then, $G'''=S_u(a_1,a_2,a_3)-N_{q-1}[x_1]=P_{a_1'''}\cup P_{a_2'''}\cup P_{a_3'''}$. Note that $a_1'''=a_1-(q-2)$, $a_2'''=a_2-(q-2)$ and $a_3'''=a_3-q=1$. Then, $a_1'''+a_2'''+a_3'''=q^2+r-2(q-1)-1-(q-2)=(q-1)^2$ and $(a_2''',a_3''')\in \{(3,1),(5,1)\}$. By Lemma 3.6(3), $b(G''')\le q-1$ and hence $b(S_u(a_1,a_2,a_3))\le q$ by Theorem 1.1.

(3) In this case, $r= q-1$, $a_3\ne q+1$, $(a_2,a_3)\notin B$ and $(a_1,a_2,a_3)\notin S$. Let $x_1=u$, then $G'=S_u(a_1,a_2,a_3)-N_{q-1}[x_1] =P_{a_1'}\cup P_{a_2'}\cup P_{a_3'}$. Note that $a_i'=a_i-q+1$ for $i=1,2,3$, and $a_1'+a_2'+a_3'=q^2+r-3q+2=(q-1)^2$ as $r= q-1$. Since $a_3\ne q+1$, $(a_2,a_3)\notin B$ and $(a_1,a_2,a_3)\notin S$, then $a_3'\ne 2$, $(a_2',a_3')\notin \cup _{i=1}^{4}D_i$ and $(a_1',a_2',a_3')\notin J^{5}$. By Theorem 3.9, $b(G')\le q-1$ and hence $b(S_u(a_1,a_2,a_3))\le q$ by Theorem 1.1. $\square$

By Lemmas 4.12–4.13 and Theorem 4.5, we have the following result.

Theorem 4.14

Suppose that $S_u(a_1,a_2,a_3)$ ($a_1\ge a_2\ge a_3\ge 1$) is a spider of order $n=q^2+r$ with $1\le r\le 2q+1$. If $1\le r\le a_3\le q-1$ and $(r,a_2,a_3)\ne (q-1,q+1,q-1)$, or $a_3\ge q$ and $1\le r\le q-3$, then $b(S_u(a_1,a_2,a_3))= \left\lceil \sqrt{n}\right\rceil -1$. Moreover, if $a_3\ge q$, $r= q-2$, and $(a_2,a_3)\notin A$; or $a_3\ge q$, $r= q-1$, $a_3\ne q+1$, $(a_2,a_3)\notin B$ and $(a_1,a_2,a_3)\notin S$, then $b(S_u(a_1,a_2,a_3))= \left\lceil \sqrt{n}\right\rceil -1$.

5 Conclusion

The burning number measures how rapidly social contagion spreads in a given graph. Determining the burning number remains open for many classes of graphs, including trees and disconnected graphs. It is shown that the graph burning problem is NP-complete even for trees and path forests.

Here, we determine the burning numbers of path forests with at most three components. Let $G=P_{a_1}\cup P_{a_2}$ with $a_1\ge a_2\ge 1$, then the burning number of G is $\lceil \sqrt{a_1+a_2}\rceil +1$ if $(a_1,a_2)\in J(t)$, and $\lceil \sqrt{a_1+a_2}\rceil$ otherwise. Let $G=P_{a_1}\cup P_{a_2}\cup P_{a_3}$ with $a_1\ge a_2\ge a_3\ge 1$, then the burning number of G is $\lceil \sqrt{a_1+a_2+a_3}\rceil +1$ if $(a_1,a_2,a_3)\in J^{1}\cup J^{2}\cup J^{3}\cup J^{4}\cup J^{5}$, and $\lceil \sqrt{a_1+a_2+a_3}\rceil$ otherwise.

Table 1 Burning number of spiders

Based on those results, we establish the burning numbers of spiders with exactly three arms. Let $S_u(a_1,a_2,a_3)$ be a spider of order $n=1+a_1+a_2+a_3$ with three arms, where $a_1\ge a_2\ge a_3\ge 1$. Denote $n=q^2+r$, $1\le r\le 2q+1$. Our results are listed in Table 1.