On the Burning Number of Generalized Petersen Graphs

Abstract

The burning number b(G) of a graph G is used for measuring the speed of contagion in a graph. In this paper, we study the burning number of the generalized Petersen graph P(n, k). We show that for any fixed positive integer k, $\lim _{n\rightarrow \infty } \frac{b(P(n,k))}{\sqrt{\frac{n}{k}}}=1$. Furthermore, we give tight bounds for b(P(n, 1)) and b(P(n, 2)).

1 Introduction

Graph burning is a discrete-time process that can be used to model the spread of social contagion in social networks. It was introduced by Bonato et al. [2, 3, 8]. This process is defined on the vertex set of a simple finite graph. Throughout the process, each vertex is either burned or unburned. Initially, at time step $t=0$, all vertices are unburned. At the beginning of every time step $t\ge 1$, an unburned vertex is chosen to burn (if such a vertex is available). After that, if a vertex is burned in time step $t-1$, then in time step t, each of its unburned neighbours becomes burned. A burned vertex will remain burned throughout the process. The process ends when all vertices are burned, in which case we say the graph is burned.

Suppose a graph G is burned in m time steps in a burning process. For $1 \le i \le m$, we denote the vertex we choose to burn at the beginning of time step i by $x_i$. The sequence ($x_1, x_2, \ldots , x_m$) is called a burning sequence for G. Each $x_i$ is called a burning source of G. The burning number of a graph G, denoted by b(G), is the length of a shortest burning sequence for G. It is straightforward to see that $b(K_n)=2$. For paths and cycles, Bonato et al. [3] determined their burning numbers exactly.

Theorem 1.1

[3, Theorem 9 and Corollary 10] Let $P_n$ be a path with n vertices and $C_n$ be a cycle with n vertices. Then,

$$\begin{aligned} b(P_n)= \left\lceil n^{1/2} \right\rceil =b(C_n). \end{aligned}$$

For general graphs, they showed that the burning number of any graph G can be bounded by its radius r and diameter d, giving $\left\lceil (d+1)^{1/2} \right\rceil \le b(G) \le r+1$. In the same paper, they also gave an upper bound on the burning number of any connected graph G of order n, showing that $b(n) \le 2 \sqrt{n} -1$. This upper bound was later improved to roughly $\frac{\sqrt{6}}{2} \sqrt{n}$ by Land and Lu [5]. It was conjectured in [3] that $b(G) \le \lceil \sqrt{n} \rceil$ for any connected graph G of order n. Very recently, Bonato and Lidbetter [4] verified this conjecture for spider graphs, which are trees with exactly one vertex of degree at least 3.

Determining b(G) for general graphs is a non-trivial problem. It is known that computing the burning number of a graph is NP-complete [1]. The burning number of the hypercube $Q_n$ is asymptotically $\frac{n}{2}$ [7], but the exact value of $b(Q_n)$ is still unknown. Several other results on burning number of graphs have also been studied recently. For example, Mitsche, Pralat and Roshanbin investigated the burning number of graph products in [7] and they also focused on the probabilistic aspects of the burning number in [6].

In this paper, we are interested in the burning number of the generalized Petersen graphs. Let $n\ge 3$ and k be integers such that $1 \le k \le n-1$. The generalized Petersen graph P(n, k) is defined to be the graph on 2n vertices with vertex set

$$\begin{aligned} V(P(n,k)) = \left\{ u_i, v_i{:}~i=0,1,2,\ldots ,n-1\right\} \end{aligned}$$

and edge set

$$\begin{aligned} E(P(n,k)) =\lbrace u_iu_{i+1}, u_iv_i, v_iv_{i+k}{:}~i=0,1,2,\ldots , n-1\rbrace , \end{aligned}$$

where subscripts are taken modulo n. Let $D_1=\lbrace u_i: i=0, 1, 2, \ldots , n-1 \rbrace$ and $D_2=\lbrace v_i: i=0, 1, 2, \ldots , n-1 \rbrace$. The subgraph induced by $D_1$ is called the outer rim, while the subgraph induced by $D_2$ is called the inner rim. A spoke of P(n, k) is an edge of the form $u_iv_i$ for some $0 \le i \le n-1$.

The following are the main results of this paper.

Theorem 1.2

Let k be a fixed positive integer. Then,

$$\begin{aligned} \left\lceil \sqrt{\left\lfloor \frac{n}{k} \right\rfloor } \right\rceil \le b(P(n,k))\le \left\lceil \sqrt{\left\lfloor \frac{n}{k} \right\rfloor } \right\rceil + \left\lfloor \frac{k}{2} \right\rfloor +2. \end{aligned}$$

In particular,

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{b(P(n,k))}{\sqrt{\frac{n}{k}}}=1. \end{aligned}$$

Theorem 1.3

For $n \ge 3$,

$$\begin{aligned} \left\lceil \sqrt{n}\right\rceil \le b(P(n,1)) \le \left\lceil \sqrt{n}\right\rceil +1. \end{aligned}$$

Furthermore, the bounds are tight, and if n is a square, then $b(P(n,1)) =\sqrt{n} +1$.

Theorem 1.4

For $n \ge 3$,

$$\begin{aligned} \left\lceil \sqrt{\frac{n}{2}}\right\rceil +1 \le b(P(n,2)) \le \left\lceil \sqrt{ \frac{n}{2} } \right\rceil +2. \end{aligned}$$

Furthermore, the bounds are tight, and if $\frac{n}{2}$ is a square, then $b(P(n,2)) =\sqrt{\frac{n}{2}} +2$.

We use standard graph terminology throughout the paper. The distance between two vertices u and v in a graph G, denoted by $\text {dist}_G (u, v)$, is the length of a shortest path from u to v in the graph G. By convention, $\text {dist}_G (u, u)=0$. Furthermore, we shall write $\text {dist} (u, v)$ for $\text {dist}_G (u, v)$ if the graph in question is clear. Given a non-negative integer s, the s-th closed neighbourhood of a vertex u, denoted by $N^{G}_{s} [u]$, is the set of vertices whose distance from u is at most s, i.e.

$$\begin{aligned} N^{G}_{s} [u]= \lbrace v \in V(G){:}~\text {dist}_{G} (u,v) \le s \rbrace . \end{aligned}$$

Again, if the graph in question is clear, we shall write $N_{s} [u]$ for $N^{G}_{s} [u]$.

Let ($x_1,x_2,\ldots ,x_m$) be a burning sequence of a graph G. As in [3, Section 2], for each pair i and j, with $1\le i<j\le m$, we have $\text {dist} (x_i, x_j) \ge j-i$ and

$$\begin{aligned} V(G)= N_{m-1}[x_1] \cup N_{m-2}[x_2] \cup \cdots \cup N_{0}[x_m]. \end{aligned}$$

(1)

The plan of the paper is as follows. In Sect. 2, we provide bounds for the burning number of P(n, k) and show that b(P(n, k)) is asymptotically $\sqrt{\frac{n}{k}}$. In Sect. 3, we determine the exact values of b(P(n, k)) for $1 \le n \le 8$. Then, we prove Theorems 1.3 and 1.4 in Sect. 4.

2 General Case

Lemma 2.1

For $n \ge 3$ and $1 \le k < n$,

$$\begin{aligned} b(P(n,k))\ge \left\lceil \sqrt{\left\lfloor \frac{n}{k} \right\rfloor } \right\rceil . \end{aligned}$$

Proof

Let C be a cycle with $\left\lfloor \frac{n}{k} \right\rfloor$ vertices, $V(C)=\{0,1,2, \ldots , \left\lfloor \frac{n}{k} \right\rfloor -1\}$ and $E(C)=\{ i(i+1){:}~0,1,\ldots , \left\lfloor \frac{n}{k} \right\rfloor -1\}$, where the integers are taken modulo $\left\lfloor \frac{n}{k} \right\rfloor$. Recall that the outer rim and inner rim of P(n, k) are $D_1=\{ u_0,u_1,\ldots , u_{n-1}\}$ and $D_2=\{ v_0,v_1,\ldots ,v_{n-1}\}$, respectively.

For each $m\in \{0,1,2,\ldots , n-1\}$, let

$$\begin{aligned} f(m) = \left\{ \begin{array}{llll} p,&{} \text{ if }\quad m=pk+q,&{} 0 \le p< \left\lfloor \frac{n}{k} \right\rfloor ,&{} 0 \le q \le k-1; \\ \left\lfloor \frac{n}{k} \right\rfloor -1, &{} \text{ if }\quad m=\left\lfloor \frac{n}{k} \right\rfloor k +q,&{} 0 \le q < k-1&{} . \end{array}\right. \end{aligned}$$

(2)

Let $\varphi : V(P(n,k))\rightarrow V(C)$ be defined by

$$\begin{aligned} \varphi (u_i)=f(i)=\varphi (v_i),\quad \forall i\in \{0,1,2,\ldots , n-1\}. \end{aligned}$$

(3)

Clearly, $\varphi$ is surjective.

Let $(x_1, x_2, \ldots , x_s)$ be a burning sequence of P(n, k). We construct a burning sequence for C using the map $\varphi$ as follows:

• (a) At the beginning of time step 1, burn $y_1=\varphi (x_1)$;

• (b) At the beginning of time step t ($2\le t\le s$), if $\varphi (x_t)$ is still unburned, then burn $y_t=\varphi (x_t)$; otherwise, burn any unburned vertex $y_t\in V(C)$.

Note that in (b) above, if at the beginning of time step t ($2\le t\le s$), no unburned vertex can be found, then $(y_1,y_2,\ldots , y_{t-1})$ is a burning sequence of C. So, we may assume that such an unburned vertex can be found at the beginning of every time step. We shall show that $(y_1,y_2,\ldots , y_{s})$ is a burning sequence of C. This follows from $\varphi$ is surjective and the following claim.

Claim

If $z\in V(P(n,k))$ is burned at time step $t_0$, then its image $\varphi (z)$ in C is burned at time step $t_1\le t_0$.

Proof

If $z=x_1$, then it is burned at time step 1. Its image $\varphi (z)=y_1$ is also burned at time step 1. The claim is true. Assume that the claim is true for a $t_0<s$.

Suppose z is burned at time step $t_0+1$. If z is a burning source, then $z=x_{t_0+1}$. By (b), $\varphi (z)$ is burned at time step $t_0+1$ provided that $\varphi (x_{t_0+1})$ is unburned. If $\varphi (x_{t_0+1})$ is burned, then it must be burned at an earlier time step. So, the claim holds.

We may assume that $z \ne x_{t_0+1}$. Note that for any two distinct vertices $w_1, w_2 \in V(P(n,k))$ such that $\varphi (w_1), \varphi (w_2) \in V(C)$ and $\vert \varphi (w_1) -\varphi (w_2) \vert \le 1$ or $\vert \varphi (w_1) -\varphi (w_2) \vert = \lfloor \frac{n}{k} \rfloor -1$, then $\varphi (w_1)=\varphi (w_2)$ or $\varphi (w_1)$ and $\varphi (w_2)$ are adjacent in C. We shall distinguish two cases.

Case 1 Let $z=u_l$. Then, it is adjacent to $v_l, u_{l+1}$ and $u_{l-1}$ where the subscript are taken modulo n. Furthermore, either $v_l, u_{l+1}$ or $u_{l-1}$ is burned at time step $t_0$. So, by induction, $\varphi (v_l), \varphi (u_{l+1})$ or $\varphi (u_{l-1})$ is burned at time step $t_1\le t_0$ respectively. By Eqs. (2) and (3), $\vert \varphi (u_l) -\varphi (v_l) \vert =0$, $\vert \varphi (u_l) -\varphi (u_{l-1}) \vert \le 1$ and $\vert \varphi (u_l) -\varphi (u_{l+1}) \vert \le 1$ where $l=1,2, \ldots , n-2$ and $\vert \varphi (u_0) -\varphi (u_{n-1}) \vert = \lfloor \frac{n}{k} \rfloor -1$. This means that $\varphi (z) =\varphi (u_l)$ is burned at time step $t_1+1 \le t_0 +1$.

Case 2 Let $z=v_l$. It is adjacent to $u_l, v_{l+k}$ and $u_{l-k}$ where the subscript are taken modulo n. Either $u_l, v_{l-k}$ or $v_{l+k}$ is burned at time step $t_0$. Here, we denote $v_{-i}=v_{n-i}$ for a non-negative i. So, by induction, $\varphi (u_l), \varphi (v_{l+k})$ or $\varphi (v_{l-k})$ is burned at time step $t_1\le t_0$ respectively. By Eqs. (2) and (3), we have $\vert \varphi (v_l) -\varphi (u_l) \vert =0$,

$$\begin{aligned} \vert \varphi (v_l) -\varphi (v_{l-k}) \vert = \left\{ \begin{array}{ll} \lfloor \frac{n}{k} \rfloor -1, &{} \text{ if }\quad l=0,1,2, \ldots , k-1; \\ 1,&{} \text{ if }\quad l=k, k+1, \ldots , \lfloor \frac{n}{k} \rfloor k -1; \\ 0,&{} \text{ if }\quad l= \lfloor \frac{n}{k} \rfloor k, \lfloor \frac{n}{k} \rfloor k+1, \ldots , n-1. \end{array}\right. \end{aligned}$$

and

$$\begin{aligned}&\vert \varphi (v_l) -\varphi (v_{l+k}) \vert \nonumber \\&\quad = \left\{ \begin{array}{ll} 1,&{} \text{ if }\quad l=0,1,2, \ldots , \left( \lfloor \frac{n}{k} \rfloor -1 \right) k -1 ; \\ 0,&{} \text{ if }\quad l=\left( \lfloor \frac{n}{k} \rfloor -1 \right) k,\left( \lfloor \frac{n}{k} \rfloor -1 \right) k +1, \ldots , n-1-k; \\ \lfloor \frac{n}{k} \rfloor -1,&{}\text{ if }\quad l=n-k, n-k+1, \ldots , n-1. \end{array}\right. \end{aligned}$$

This means that $\varphi (z) =\varphi (v_l)$ is burned at time step $t_1+1 \le t_0 +1$.

This completes the proof of the claim. $\square$

Therefore, given any burning sequence of P(n, k), we can construct a burning sequence for C with shorter or the same length. Hence, $b(P(n,k))\ge b(C)= \left\lceil \sqrt{\left\lfloor \frac{n}{k} \right\rfloor } \right\rceil$, where the last equality follows from Theorem 1.1. $\square$

Lemma 2.2

For $n \ge 3$ and $1 \le k < n$,

$$\begin{aligned} b(P(n,k)) \le \left\lceil \sqrt{\left\lfloor \frac{n}{k} \right\rfloor } \right\rceil + \left\lfloor \frac{k}{2} \right\rfloor +2. \end{aligned}$$

Proof

Recall that the outer rim and inner rim of P(n, k) are $D_1=\{ u_0,u_1,\ldots , u_{n-1}\}$ and $D_2=\{ v_0,v_1,\ldots ,v_{n-1}\}$, respectively. Let $r=\left\lfloor \frac{n}{k} \right\rfloor$. We shall construct a burning sequence for P(n, k) of length at most $\left\lceil \sqrt{r} \right\rceil + \left\lfloor \frac{k}{2} \right\rfloor +2$. Note that a subgraph G induced by the vertices $v_0, v_{k}, v_{2k}, \ldots , v_{(r-1)k}$ in P(n, k) is a path or cycle of order r. By Theorem 1.1, $b(G)=\left\lceil \sqrt{r} \right\rceil$. So, there is a burning sequence $(x_1, x_2, \ldots , x_{\lceil \sqrt{r} \rceil })$ of G. We shall take $x_1, x_2, \ldots , x_{\lceil \sqrt{r} \rceil }$ as the first part of our burning sequence for P(n, k).

Note that at time step $\lceil \sqrt{r} \rceil$, all $v_0, v_{k}, v_{2k}, \ldots , v_{(r-1)k}$ are burned. If at time step $\lceil \sqrt{r} \rceil$, $u_{rk}$ is unburned, then we set $x_{\lceil \sqrt{r} \rceil +1}=u_{rk}$. Otherwise, we set $x_{\lceil \sqrt{r} \rceil +1}$ to be any unburned vertex. Since $u_{ik}$ is adjacent to $v_{ik}$ for $0\le i\le (r-1)$, at time step $\lceil \sqrt{r} \rceil +1$, all $u_0, u_{k}, u_{2k}, \ldots , u_{(r-1)k}, u_{rk}$ are burned. Furthermore, at most $k-1$ vertices are unburned in the path $u_{ik}u_{ik+1}u_{ik+2}\ldots u_{(i+1)k}$ in the outer rim (see Fig. 1).

Now, for $j\ge \lceil \sqrt{r} \rceil +2$, we can choose $x_j$ to be any unburned vertex. Note that at time step $\lceil \sqrt{r} \rceil +1+\left\lfloor \frac{k}{2} \right\rfloor$, all the vertices in the outer rim are burned. Since $u_i$ and $v_i$ are adjacent, at time step $\lceil \sqrt{r} \rceil +2+\left\lfloor \frac{k}{2} \right\rfloor$, all vertices in the inner rim are also burned. Hence, the lemma follows. $\square$

Fig. 1

Filled vertices are burned, whereas empty vertices are unburned

Proof of Theorem 1.2

By Lemmas 2.1 and 2.2, we have

$$\begin{aligned} \left\lceil \sqrt{\left\lfloor \frac{n}{k} \right\rfloor } \right\rceil \le b(P(n,k))\le \left\lceil \sqrt{\left\lfloor \frac{n}{k} \right\rfloor } \right\rceil + \left\lfloor \frac{k}{2} \right\rfloor +2. \end{aligned}$$

By noting that $\lim _{n\rightarrow \infty } \frac{\left\lceil \sqrt{\left\lfloor \frac{n}{k} \right\rfloor } \right\rceil }{\sqrt{\frac{n}{k}}}=1$ and $\lim _{n\rightarrow \infty } \frac{\left\lfloor \frac{k}{2} \right\rfloor +2}{\sqrt{\frac{n}{k}}}=0$, we conclude

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{b(P(n,k))}{\sqrt{\frac{n}{k}}}=1. \end{aligned}$$

$\square$

3 Case $1\le N\le 8$

We shall give the exact burning numbers for the case $1 \le n \le 8$ in this section. Note that P(n, k) is isomorphic to $P(n,n-k)$. So, we may assume that $1\le k\le \left\lfloor \frac{n}{2} \right\rfloor$. Recall that the sth closed neighbourhood of a vertex $x\in V(P(n,k))$ is

$$\begin{aligned} N_{s} [x]= \lbrace y \in V(P(n,k)): \ \text {dist} (y,x) \le s \rbrace , \end{aligned}$$

and the outer rim and inner rim of P(n, k) are $D_1=\{u_0,u_1,\ldots , u_{n-1}\}$ and $D_2=\{v_0,v_1,\ldots ,v_{n-1}\}$, respectively.

Proposition 3.1

Let $3 \le n \le 8$ and $1 \le k \le \left\lfloor \frac{n}{2} \right\rfloor$. Then,

$$\begin{aligned} b(P(n,k)) = \left\{ \begin{array}{ll} 3,&{} \mathrm{if}\quad 3 \le n \le 6~\mathrm{or}~n=7, k \ne 1, \\ 4,&{} \mathrm{if}\quad n=8~\mathrm{or}~n=7, k=1. \end{array}\right. \end{aligned}$$

Proof

Since each vertex $x\in V(P(n,k))$ is of degree 3, $\vert N_0 [x]\vert =1$, $\vert N_1 [x]\vert \le 4$ and $\vert N_2 [x]\vert \le 10$.

Let $3\le n\le 7$. If $(x _1, x_2)$ is a burning sequence of P(n, k), then by Eq. (1),

$$\begin{aligned} 2n\le \vert N_{1}[x_1]\vert +\vert N_{0}[x_2]\vert \le 4+1=5, \end{aligned}$$

implying that $n<3$, which is a contradiction. Hence, $b(P(n,k)) \ge 3$. Similarly, if $(x_1, x_2, x_3)$ is a burning sequence of P(8, k), then

$$\begin{aligned} 16\le \vert N_{2}[x_1]\vert +\vert N_{1}[x_2]\vert +\vert N_{0}[x_3]\vert \le 10+4+1=15, \end{aligned}$$

again is a contradiction. Hence, $b(P(8,k)) \ge 4$.

Note that for each $x\in V(P(7,1))$, $\vert N_2 [x]\vert =8$. So, if $(x _1, x_2, x_3)$ is a burning sequence of P(7, 1), then

$$\begin{aligned} 14\le \vert N_{2}[x_1]\vert +\vert N_{1}[x_2]\vert +\vert N_{0}[x_3]\vert \le 8+4+1=13, \end{aligned}$$

which is a contradiction. Hence, $b(P(7,1)) \ge 4$.

Now, the proposition can be verified easily from the burning sequences in the following table (see also Fig. 2).

Burning sequence	Graph
$(u_0,v_1,v_2)$	P(3, 1), P(4, 1), P(4, 2)
$(u_0,v_3,u_3)$	P(5, 1), P(6, 1), P(6, 2)
$(u_0,u_2,v_4)$	P(5, 2), P(6, 3)
$(u_0,u_2,u_4,v_4)$	P(7, 1)
$(u_0,u_3,v_4)$	P(7, 2)
$(v_0,v_2,u_5)$	P(7, 3)
$(u_0,u_2,v_4,u_4)$	P(8, 1), P(8, 2), P(8, 3), P(8, 4)

Fig. 2

Burning sequences

$\square$

4 Case $1\le K\le 2$

4.1 Proof of Theorem 1.3

Note that for each $x\in V(P(n,1))$, $\vert N_m [x]\vert \le 4m$ for $m\ge 1$ and $\vert N_0 [x]\vert =1$. So, if $(x _1, x_2,\ldots , x_l)$ is a burning sequence of P(n, 1), then by Eq. (1),

$$\begin{aligned} 2n\le \vert N_0 [x_l]\vert +\sum _{i=1}^{l-1} \vert N_{l-i}[x_i]\vert \le 1+\sum _{i=1}^{l-1} 4(l-i)=2l^2-2l+1. \end{aligned}$$

Since $l\ge 1$, by completing the square, we conclude that $l\ge \frac{2+\sqrt{4-8(1-2n)}}{4}=\frac{1}{2}+\sqrt{n-\frac{1}{4}}>\sqrt{n}$. Hence, $b(P(n,1))\ge \left\lceil \sqrt{n} \ \right\rceil$, and if n is a square, then $b(P(n,1))\ge \left\lceil \sqrt{n} \ \right\rceil +1$.

The subgraph C induced by the vertices in the outer rim $D_1=\{ u_0,u_1,\ldots , u_{n-1}\}$ is a cycle of length n. By Theorem 1.1, $b(C)=\left\lceil \sqrt{n} \ \right\rceil$. So, C has a burning sequence $(y_1,y_2,\ldots , y_{\left\lceil \sqrt{n} \ \right\rceil })$. We shall take $y_1, y_2, \ldots , y_{\lceil \sqrt{n} \rceil }$ as the first part of our burning sequence for P(n, 1). Note that at time step $\lceil \sqrt{n} \rceil$, all the vertices in the outer rim are burned. Choose any unburned vertex z in the inner rim. Let $y_{\lceil \sqrt{n} \rceil +1}=z$. Since $u_iv_i$ are adjacent for $1\le i\le n-1$, at time step $\lceil \sqrt{n} \rceil +1$ all vertices in the inner rim are also burned. Hence, $b(P(n,1))\le \left\lceil \sqrt{n} \ \right\rceil +1$, and if n is a square, then $b(P(n,1))= \sqrt{n} +1$.

Finally, by Proposition 3.1, $b(P(5,1))=3=\left\lceil \sqrt{5} \ \right\rceil$. So the bounds are tight. This completes the proof of Theorem 1.3. $\square$

4.2 Proof of Theorem 1.4

We shall first define an isomorphic graph of P(n, 2), say H(n). Let

$$\begin{aligned} W_1&=\left\{ s_i, s'_i, t_i, t'_i: i=1, 2, \ldots , \left\lfloor \frac{n}{2} \right\rfloor \right\} ;\\ W_2&=\left\{ t_it_{i+1}, t'_it'_{i+1}, s_is'_{i+1}, s_jt_j, s'_jt'_j, s_js'_j : 1 \le i \le \left\lfloor \frac{n}{2}\right\rfloor -1, 1 \le j \le \left\lfloor \frac{n}{2}\right\rfloor \right\} . \end{aligned}$$

If n is even, then let

$$\begin{aligned} V(H(n))&= W_1;\nonumber \\ E(H(n))&= W_2 \cup \left\{ t_1t_{\frac{n}{2}}, t'_1t'_{\frac{n}{2}}, s_{\frac{n}{2}} s'_1 \right\} . \end{aligned}$$

(4)

If n is odd, then let

$$\begin{aligned} V(H(n))&= W_1\cup \left\{ s_0, t_0 \right\} ;\nonumber \\ E(H(n))&= W_2 \cup \left\{ s_0s_{\frac{n-1}{2}}, s_0s'_1, s_0t_0, t_0t_1, t_0t'_{\frac{n-1}{2}}, t_{\frac{n-1}{2} } t'_1 \right\} . \end{aligned}$$

(5)

We now show that P(n, 2) is isomorphic to H(n) (see Figs. 3 and 4). Define $\phi : V(P(n,2)) \rightarrow V(H(n))$ as follows: Let $\phi (u_i)=s'_{\frac{i}{2}+1}$ if i is even and $i\ne n-1$; $\phi (u_i)=s_{\frac{i-1}{2}+1}$ if i is odd; $\phi (u_{n-1})=s_{0}$ if $n-1$ is even. Let $\phi (v_i)=t'_{\frac{i}{2}+1}$ if i is even and $i\ne n-1$; $\phi (v_i)=t_{\frac{i-1}{2}+1}$ if i is odd; $\phi (v_{n-1})=t_{0}$ if $n-1$ is even. Note that the subgraph induced by all the vertices $s_i, s'_i$ in H(n) is isomorphic to the outer rim in P(n, 2), and the subgraph induced by all the vertices $t_i, t'_i$ in H(n) is isomorphic to the inner rim in P(n, 2). Furthermore, $s_it_i, s'_it'_i$ are the spokes in P(n, 2). So P(n, 2) is isomorphic to H(n).

Let $T_1=\lbrace t_i: 1 \le i \le \lfloor \frac{n}{2}\rfloor \rbrace$, $T_2=\lbrace t'_i: 1 \le i \le \lfloor \frac{n}{2}\rfloor \rbrace$, and level $L_i=\lbrace s_i, s'_i, t_i, t'_i\rbrace$ for $i=1, 2, \ldots , \lfloor \frac{n}{2} \rfloor$.

Fig. 3

H(n) is isomorphic to P(n, 2) where n is even

Fig. 4

H(n) is isomorphic to P(n, 2) where n is odd

Lemma 4.1

For $n \ge 3$,

$$\begin{aligned} b(P(n,2))\ge \left\lceil \sqrt{\frac{n}{2}} \ \right\rceil +1. \end{aligned}$$

Furthermore, if $\frac{n}{2}$ is a square, then $b(P(n,2))\ge \sqrt{\frac{n}{2}} +2$.

Proof

Note that if $x \notin T_1 \cup T_2$, then $\vert N_0 [x]\vert =1$, $\vert N_1 [x]\vert \le 4$, $\vert N_2 [x]\vert \le 10$, $\vert N_3 [x]\vert \le 16$, $\vert N_4 [x]\vert \le 22$, $\vert N_5 [x]\vert \le 30$ and $\vert N_r [x]\vert \le 30+8(r-5)$ for $r\ge 6$ (see Fig. 5). After 5 steps, a maximum of 8 vertices are newly burned in each following step.

Fig. 5

Spreading of fire from $x\notin T_1 \cup T_2$. Filled vertices are burned, whereas empty vertices are unburned

If $x \in T_1 \cup T_2$, then $\vert N_0 [x]\vert =1$, $\vert N_1 [x]\vert \le 4$, $\vert N_2 [x]\vert \le 10$, $\vert N_3 [x]\vert \le 18$ and $\vert N_r [x]\vert \le 18 + 8(r-3)$ for $r\ge 4$ (see Fig. 6). After 4 steps, a maximum of 8 vertices are newly burned in each following step.

Fig. 6

Spreading of fire from $x \in T_1 \cup T_2$. Filled vertices are burned, whereas empty vertices are unburned

In either case, we have $\vert N_0 [x]\vert =1$, $\vert N_1 [x]\vert \le 4$, $\vert N_2 [x]\vert \le 10$, $\vert N_3 [x]\vert \le 18$ and $\vert N_r [x]\vert \le 18 + 8(r-3)=8r-6$ for $r\ge 4$.

By Proposition 3.1, $b(P(n,2))=3=\left\lceil \sqrt{\frac{n}{2}} \ \right\rceil +1$ for $3\le n\le 7$ and

$$\begin{aligned} b(P(8,2))=4=\sqrt{\frac{n}{2}}+2. \end{aligned}$$

Hence, the lemma holds for $3\le n\le 8$. So, we may assume $n\ge 9$. Suppose $9 \le n \le 16$, then $\left\lceil \sqrt{\frac{n}{2}}\right\rceil \le 3$. If P(n, 2) has a burning sequence of length 3, say $(x_1,x_2,x_3)$, then by Eq. (1), $18\le 2n\le \sum _{i=1}^{3} \vert N_{3-i}[x_i] \vert \le 1+4+10=15$, a contradiction. Suppose $17 \le n \le 32$, then $\left\lceil \sqrt{\frac{n}{2}} \ \right\rceil \le 4$. If P(n, 2) has a burning sequence of length 4, say $(x_1,x_2,x_3,x_4)$, then $34\le 2n\le \sum _{i=1}^{4} \vert N_{4-i}[x_i] \vert \le 1+4+10+18=33$, a contradiction. So, $b(P(n,2))\ge \left\lceil \sqrt{\frac{n}{2}} \ \right\rceil +1$ for $3\le n\le 32$.

Note that for $9\le n\le 32$, $\frac{n}{2}$ is a square if and only if $n=18, 32$. When $n=18$, $\sqrt{\frac{n}{2}} +2=5$. If P(18, 2) has a burning sequence of length 4, then $\sum _{i=1}^{4} \vert N_{4-i}[x_i] \vert \le 33$, but $\vert V(P(18,2))\vert =36$. When $n=32$, $\sqrt{\frac{n}{2}} +2=6$. If P(32, 2) has a burning sequence of length 5, then $\sum _{i=1}^{5} \vert N_{5-i}[x_i] \vert \le 1+4+10+18+26=59$, but $\vert V(P(32,2))\vert =64$. Thus, if $\frac{n}{2}$ is a square and $9\le n\le 32$, then $b(P(n,2))\ge \sqrt{\frac{n}{2}} +2$.

Suppose $n \ge 33$. If P(n, 2) has a burning sequence of length l, say $(x_1,x_2,\ldots ,x_l)$, then by Eq. (1),

$$\begin{aligned} 2n&\le \sum _{i=1}^{l} \vert N_{l-i}[x_i]\vert \le \vert N_0 [x_l] \vert + \vert N_1 [x_{l-1}] \vert + \vert N_2 [x_{l-2}] \vert +\sum _{i=1}^{l-3}\vert N_{l-i}[x_i]\vert \\&\le 1+4+ 10+ \sum _{r=3}^{l-1} (8r-6) \\&= 4l^2 -10l+9. \end{aligned}$$

Since $l\ge 1$, by completing the square, we conclude that

$$\begin{aligned} l\ge \frac{10+\sqrt{100-16(9-2n)}}{8}=\frac{5}{4}+\sqrt{\frac{n}{2}-\frac{11}{16}}>\sqrt{\frac{n}{2}}+1. \end{aligned}$$

Hence, $b(P(n,2))\ge \left\lceil \sqrt{\frac{n}{2}} \ \right\rceil +1$, and if $\frac{n}{2}$ is a square, then $b(P(n,2))\ge \sqrt{\frac{n}{2}}+2$. This completes the proof of the lemma. $\square$

Lemma 4.2

For $n \ge 3$,

$$\begin{aligned} b(P(n,2)) \le \left\lceil \sqrt{ \frac{n}{2} } \right\rceil +2. \end{aligned}$$

Proof

Let $l =\left\lceil \sqrt{ \frac{n}{2} } \right\rceil$. It is sufficient to show that there is a burning sequence $(x_1,x_2, \ldots , x_l, x_{l+1}, x_{l+2})$ in H(n).

Note that for $2\le j\le l$, the term $(2j-1)l-(j-1)^2$ is increasing. Let $m_0$ be the largest positive integer such that $(2m_0-1)l-(m_0-1)^2 \le \left\lfloor \frac{n}{2} \right\rfloor$. Since

$$\begin{aligned} (2l-1)l-(l-1)^2 =l^2+l-1\ge \left( \sqrt{ \frac{n}{2} } \right) ^2+\left( \sqrt{ \frac{n}{2} }-1\right) >\frac{n}{2}, \end{aligned}$$

we must have $m_0\le l-1$.

Now, we construct the first part of a burning sequence for H(n), say $x_1,x_2,\ldots ,x_l$, as follows:

• (a) Let $x_1=t_l$;

• (b) For each $2\le j\le m_0$, set $x_j=t_{(2j-1)l-(j-1)^2}$ if j is odd, or $x_{j}=t'_{(2j-1)l-(j-1)^2}$ if j is even;

• (c) For $j\ge m_0+1$:

  • (i) Suppose $m_0 \le l-2$. If $x_{m_0}=t_{(2m_0-1)l-(m_0-1)^2}$, then set $x_{m_0+1}=t'_{\left\lfloor \frac{n}{2} \right\rfloor }$, whereas if $x_{m_0}=t'_{(2m_0-1)l-(m_0-1)^2}$, then set $x_{m_0+1}=t_{\left\lfloor \frac{n}{2} \right\rfloor }$. For $m_0+2\le w \le l$, choose $x_w$ to be any unburned vertex (if possible).

  • (ii) Suppose $m_0=l-1$. If $x_{l-1}=t_{(2l-3)l-(l-2)^2}$, then set $x_l=t'_{\left\lfloor \frac{n}{2} \right\rfloor }$, whereas if $x_{l-1}=t'_{(2l-3)l-(l-2)^2}$, then set $x_l=t_{\left\lfloor \frac{n}{2} \right\rfloor }$.

In Fig. 7, the filled vertices are $N_{l-i} [x_i]$ and the shaded vertices are $N_{l+2-i} [x_i] \backslash N_{l-i} [x_i]$. In particular, $L_4\cup L_5\cup \cdots \cup L_l\subseteq N_{l-1}[x_1]$. So $\left( L_1\cup L_2\cup \cdots \cup L_l\right) \setminus \{t_1'\}\subseteq N_{l+1}[x_1]$ (see Fig. 7a).

Fig. 7

Construction

Suppose $2\le j\le m_0$. Note that $x_j$ is contained in level $L_{(2j-1)l-(j-1)^2}$ and $x_{j-1}$ is contained in level $L_{(2j-3)l-(j-2)^2}$. There are exactly $2l-2j+4=((2j-1)l-(j-1)^2)-((2j-3)l-(j-2)^2)+1$ levels between $L_{(2j-1)l-(j-1)^2}$ and $L_{(2j-3)l-(j-2)^2}$ (inclusive). All these levels are contained in $N_{l-j+3}[x_{j-1}]\cup N_{l-j+2}[x_j]$ (see Fig. 7b).

Suppose $m_0\le l-2$. By the choice of $m_0$, $(2m_0+1)l-m_0^2>\left\lfloor \frac{n}{2} \right\rfloor$. So, the number of levels between $L_{\left\lfloor \frac{n}{2} \right\rfloor }$ and $L_{(2m_0-1)l-(m_0-1)^2 }$ (inclusive) is at most

$$\begin{aligned} \left\lfloor \frac{n}{2} \right\rfloor -((2m_0-1)l-(m_0-1)^2)+1&<(2m_0+1)l-m_0^2\\&\quad -\,((2m_0-1)l-(m_0-1)^2)+1\\&= 2l-2m_0+2. \end{aligned}$$

All these levels are contained in $N_{l-m_0+2}[x_{m_0}]\cup N_{l-m_0+1}[x_{m_0+1}]$ (see Fig. 7b).

Suppose $m_0=l-1$. Then, $x_{l-1}$ is in level $L_{(2l-3)l-(l-2)^2}$ and $x_{l}$ is in level $L_{\left\lfloor \frac{n}{2} \right\rfloor }$. Note that

$$\begin{aligned} (2l-3)l-(l-2)^2+2=l^2+l-2>\frac{n}{2}-1\ge \left\lfloor \frac{n}{2} \right\rfloor -1. \end{aligned}$$

Hence, we have

$$\begin{aligned} (2l-3)l-(l-2)^2+2 \ge \left\lfloor \frac{n}{2} \right\rfloor . \end{aligned}$$

Therefore,

$$\begin{aligned} L_{(2l-3)l-(l-2)^2}\cup L_{(2l-3)l-(l-2)^2+1}\cup \cdots \cup L_{\left\lfloor \frac{n}{2} \right\rfloor }\subseteq N_{3}[x_{l-1}]\cup N_{2}[x_{l}], \end{aligned}$$

(see Fig. 7c).

If we set $x_{l+1}=t_1'$ and $x_{l+2}$ to be any unburned vertex at time step $l+1$ (if possible), then $(x_1,x_2, \ldots , x_l, x_{l+1}, x_{l+2})$ is a burning sequence of H(n) when n is even. If n is odd, it is also a burning sequence by noticing that $\lbrace s_0, t_0 \rbrace \in N_{l+1} [x_1]$ (see Figs. 4 and 7a). This completes the proof of the lemma. $\square$

The first part of Theorem 1.4 follows from Lemmas 4.1 and 4.2. Furthermore, if $\frac{n}{2}$ is a square, then $b(P(n,2)) =\sqrt{\frac{n}{2}} +2$. Finally, by Proposition 3.1, $b(P(3,2))=3=\left\lceil \sqrt{\frac{3}{2}} \ \right\rceil +1$. So the bounds are tight. This completes the proof of Theorem 1.4. $\square$