Locating-Total Domination in Grid Graphs

Abstract

Let $G=(V,E)$ be a graph with no isolated vertex. A subset $S\subseteq V(G)$ is a total dominating set of graph G if every vertex in V(G) is adjacent to at least one vertex in S. A total dominating set S of graph G is a locating-total dominating set if for every pair of distinct vertices $u_1$ and $u_2$ in $V(G)-S$, $N(u_1)\cap S\ne N(u_2)\cap S$. The locating-total domination number of graph G, denoted by $\gamma _t^L(G)$, is the minimum cardinality of a locating-total dominating set of G. In this paper, we investigate the bounds of locating-total domination number of grid graphs.

1 Introduction

The concept of the locating-total dominating set was introduced by Haynes et al. [6], which has been studied in [1,2,3,4, 8,9,10,11] and elsewhere. In a monitoring system, the problem of placing monitors such that each site (including the monitors themselves) in the system is adjacent to at least one monitor can be regarded as the total domination in graphs. The location of every monitor can be uniquely identified by the set of monitors in a monitoring system, which can regarded as the combination of total domination and locating in graphs. In this paper, we consider a different variation of this combination.

Given a graph $G=(V, E)$, the degree of vertex v in G, denoted by d(v), is the number of edges incident with v. The maximum degree of G, denoted by $\Delta (G)$ or $\Delta$, is equal to $\max \{d(v)\mid v\in V\}$. G is simple, if G has neither loops nor parallel edges. G is connected, if for any two vertices x and y, there is an xy-path in G. If A and B are two disjoint subsets of V, then $[A,B]=\{uv\in E(G)|u\in A,v\in B\}$.

Let $G=(V,E)$ be a simple graph on n vertices. For a vertex v in G, the set $N(v)=\{u\in V\mid uv\in E\}$ is called the open neighborhood of v and $N[v]=N(v)\cup \{v\}$ is the closed neighborhood of v. For a subset $S\subseteq V$, $N(S)=\bigcup _{v\in S}N(v)$ is the open neighborhood of S and $N[S]=N(S)\cup S$ is the closed neighborhood of S. A subset S of V is a dominating set of G if $N[S]=V$ and S is a total dominating set of G if $N(S)=V$. A total dominating set S is a locating-total dominating set (LTD-set) if for every pair of distinct vertices u and v in $V-S$, $N(u)\cap S\ne N(v)\cap S$. Every graph G with no isolated vertex has a LTD-set, since V is such a set. The locating-total domination number of G, denoted by $\gamma _t^L(G)$, is the minimum cardinality of a LTD-set of G. A LTD-set of cardinality $\gamma _t^L(G)$ is called a $\gamma _t^L(G)$-set.

A path of order n is denoted by $P_n$. For graphs G and H, the Cartesian product $G\times H$ is the graph with vertex set $V(G)\times V(H)$ where two vertices $(u_1, v_1)$ and $(u_2, v_2)$ are adjacent if and only if either $u_1 = u_2$ and $v_1v_2\in E(H)$ or $v_1 = v_2$ and $u_1u_2\in E(G)$. We call $P_m\times P_n$ the grid graph.

Locating-total dominating sets in graphs have been studied in several papers, see [2,3,4,5,6,7]. For grid graph, Henning and Rad [7] showed that if $n \equiv r (mod~ 5)$, where $0\le r < 5$, then

$$\begin{aligned} \gamma _t^L(P_2\times P_n)=\left\{ \begin{array}{ll} 4\lfloor \frac{n}{5}\rfloor +r &{} \hbox { if}\ r\not =1\\ 4\lfloor \frac{n}{5}\rfloor +2 &{} \hbox {if }r=1. \end{array} \right. \end{aligned}$$

They also showed that if $n\equiv 0 (mod~ 11)$, then $\gamma _t^L(P_3\times P_n)\le \frac{13}{11}n$. In this paper, we investigate the bounds of locating-total domination number of grid graphs $P_m\times P_n$ for $m=3,4,5$ and the case $m,n\ge 6$.

2 Notations and Lemmas

In the next, we will use $v_{ij}$ to denote the vertex of $P_m\times P_n$, where $i=1,\ldots ,m$ and $j=1,\ldots ,n$. Let S be a LTD-set of $P_m\times P_n$. For $k=0,1,2,3$, denote $X_k(S)=\{v_{ij}|v_{ij}\in S,~1\le i\le m,~1\le j\le n,~|N(v_{ij})\cap \overline{S}|=k\}$ and $|X_k(S)|=x_k(S)$, where $\overline{S}=V(P_m\times P_n)\setminus S$. We have the following lemmas which will be used in the next sections.

Lemma 1

If $m,n\ge 3$ and S is a LTD-set of $P_m\times P_n$, then $\sum _{k=0}^{3}kx_k(S)\ge 2nm-3|S|.$

Proof

Let $P=\{v_{ij}|v_{ij}\in \overline{S},~1\le i\le m,~1\le j\le n,~|N(v_{ij})\cap S|=1\}$. Since S is a LTD-set, $|P|\le |S|$. Denote $R=\overline{S}\setminus P$. Since $\Delta (P_m\times P_n)= 4$, we have $|[S,\overline{S}]|=\sum _{k=0}^{3}kx_k(S).$ On the other hand, $|[S,\overline{S}]|\ge |P|+2|R|\ge |P|+2(mn-|S|-|P|)\ge 2mn-3|S|.$ Thus $\sum _{k=0}^{3}kx_k(S)\ge 2mn-3|S|.$$\square$

Lemma 2

If $m,n\ge 3$ and S is a LTD-set of $P_m\times P_n$, then

• (1) $|S|\ge \frac{2mn}{5}-\frac{1}{5}x_3(S)$;

• (2) $|S|\ge \frac{mn}{3}+\frac{1}{6}y(S)$, where $y(S)=x_0(S)+x_1(S)+x_2(S)$.

Proof

Note that $x_0(S)+x_1(S)+x_2(S)+x_3(S)=|S|$. By Lemma 1, we have

$$\begin{aligned} 2|S|+x_3(S)\ge & {} (x_1(S)+x_2(S)+x_3(S))+(x_2(S)+x_3(S))+x_3(S)\\= & {} \sum _{k=0}^{3}kx_k(S)\ge 2nm-3|S|. \end{aligned}$$

Thus (1) holds.

On the other hand, by Lemma 1,

$$\begin{aligned} 6|S|\ge 2mn+3(x_0(S)+x_1(S)+x_2(S)+x_3(S))-\sum _{i=0}^{3}ix_i(S)\ge 2mn+y(S). \end{aligned}$$

Thus (2) holds obviously. $\square$

3 Locating-Total Domination Number of $P_3\times P_n$

Now we consider the lower and upper bounds of the locating-total domination number of $P_3\times P_n$.

Theorem 3

If $n\ge 3$, then $\lceil \frac{11n}{10}\rceil \le \gamma _t^L(P_3\times P_n)\le \lceil \frac{6n}{5}\rceil$.

Proof

Let $n=5k+r$, where $k\ge 0$ and $0\le r\le 4$. For $k\ge 1$, denote $S_0$ to be the empty set and

$$\begin{aligned} \begin{array}{rcl} S_5&{}=&{}\cup _{i=1}^3\{v_{i(l+2)},v_{i(l+4)}|~l=5s,\hbox { and }s=0,\ldots ,k-1\},\\ S_4&{}=&{}\{v_{1(5k+4)},v_{2(5k+1)},v_{2(5k+2)},v_{2(5k+4)},v_{3(5k+2)}\},\\ S_3&{}=&{}\{v_{1(5k+2)},v_{2(5k+2)},v_{2(5k+3)},v_{3(5k+2)}\},\\ S_2&{}=&{}\{v_{1(5k+1)},v_{2(5k+1)},v_{3(5k+1)}\},\\ S_1&{}=&{}\{v_{2(5k+1)},v_{3(5k+1)}\}. \end{array} \end{aligned}$$

Let $S=S_5\cup S_r$ for $n=5k+r$, where $k\ge 1$ and $0\le r\le 4$. If $k=0$, let $S=S_r$, where $3\le r\le 4$. It is not difficult to check that S is a LTD-set of $P_3\times P_n$. Thus $\gamma _t^L(P_3\times P_n)\le \lceil \frac{6n}{5}\rceil$.

Now we consider the lower bound of $\gamma _t^L(P_3\times P_n)$. Let S be a $\gamma _t^L$-set of $P_3\times P_n$ such that $x_3(S)$ is as small as possible. By Lemma 2(1), we have $|S|\ge \frac{6n}{5}-\frac{1}{5}x_3(S)$. Next we will show that $x_3(S)\le \frac{n}{2}$. Let $V_i=\{v_{ij}~|~j=1,\ldots ,n\}$, $1\le i\le 3$. Then $X_3(S)\subseteq V_2$ and if $v_{2i}\in X_3(S)$, then $i\not =1,n$.

Claim 1

If there is some i with $2\le i\le n-1$ such that $v_{2i}\in X_3(S)$, then $v_{2(i-1)},v_{2(i+1)}\not \in X_3(S)$.

Proof of Claim 1

Suppose there is some i with $2\le i\le n-1$ such that $v_{2i},v_{2(i+1)}\in X_3(S)$. Then $i\le n-2$ and $v_{2(i-1)},v_{2(i+2)},v_{1i},v_{1(i+1)},v_{3i},v_{3(i+1)}\notin S$ (see Fig. 1).

Fig. 1

Illustration of the structure of S if $v_{2i},v_{2(i+1)}\in X_3(S)$

If $v_{3(i+2)}\notin S$, then $i\le n-3$ and $v_{3(i+3)}\in S$. Since $N(v_{3(i+1)})\cap S=\{v_{2(i+1)}\}$, $v_{2(i+1)}\in N(v_{1(i+1)})\cap S$ and $v_{2(i+1)}\in N(v_{2(i+2)})\cap S$, we have $v_{1(i+2)},v_{2(i+3)}\in S$ by the definition of LTD-set. To total dominate $v_{1(i+2)}$, we have $v_{1(i+3)}\in S$ (see Fig. 2a).

By the same argument, we have the following result: If $v_{1(i+2)}\notin S$, then $v_{1(i+3)},v_{2(i+3)},$$v_{3(i+2)},v_{3(i+3)}\in S$ (see Fig. 2b). If $v_{1(i-1)}\notin S$, then $v_{1(i-2)},v_{2(i-2)},v_{3(i-2)},v_{3(i-1)}\in S$ (see Fig. 2c). If $v_{3(i-1)}\notin S$, then $v_{1(i-2)},v_{1(i-1)},v_{2(i-2)},v_{3(i-2)}\in S$ (see Fig. 2d). Now we consider the following four cases.

Fig. 2

Illustration of the structure of S

Case 1$v_{1(i-1)},v_{1(i+2)}\notin S$ or $v_{3(i-1)},v_{3(i+2)}\notin S$.

Assume, without loss of generality, that $v_{1(i-1)},v_{1(i+2)}\notin S$. In this case, we have $v_{1(i-2)},v_{2(i-2)},v_{3(i-2)},v_{3(i-1)},v_{3(i+2)},v_{3(i+3)},v_{2(i+3)},v_{1(i+3)}\in S$ (see Fig. 3a). Let $S'=(S\setminus \{v_{3(i-1)},v_{3(i+2)}\})\cup \{v_{3i},v_{3(i+1)}\}$ (see Fig. 3b). Then $S'$ is a $\gamma _t^L$-set of $P_3\times P_n$ but $x_3(S')<x_3(S)$, a contradiction.

Fig. 3

Illustration of the structure of case 1

Case 2$v_{1(i-1)},v_{3(i+2)}\notin S$ or $v_{3(i-1)},v_{1(i+2)}\notin S$.

Fig. 4

Illustration of the structure of case 2

Assume, without loss of generality, that $v_{1(i-1)},v_{3(i+2)}\notin S$. In this case, we have $v_{1(i-2)},v_{2(i-2)},v_{3(i-2)},v_{3(i-1)},v_{1(i+2)},v_{1(i+3)},v_{2(i+3)},v_{3(i+3)}\in S$ (see Fig. 4a). Let $S'=(S\setminus \{v_{3(i-1)},v_{1(i+2)}\})\cup \{v_{3i},v_{1(i+1)}\}$ (see Fig. 4b). Then $S'$ is a $\gamma _t^L$-set of $P_3\times P_n$ but $x_3(S')<x_3(S)$, a contradiction.

Case 3$v_{1(i-1)}\notin S$ and $v_{1(i+2)},v_{3(i+2)}\in S$, or $v_{1(i+2)}\notin S$ and $v_{1(i-1)},v_{3(i-1)}\in S$, or $v_{3(i-1)}\notin S$ and $v_{1(i+2)},v_{3(i+2)}\in S$, or $v_{3(i+2)}\notin S$ and $v_{1(i-1)},v_{3(i-1)}\in S$.

Fig. 5

Illustration of the structure of case 3

Fig. 6

Illustration of the structure of case 3

Assume, without loss of generality, that $v_{1(i-1)}\notin S$ and $v_{1(i+2)},v_{3(i+2)}\in S$. Then $v_{1(i-2)},v_{2(i-2)},v_{3(i-1)},v_{3(i-2)}\in S$ (see Fig. 5a). Let $S'=(S\setminus \{v_{3(i-1)}\})\cup \{v_{3i}\}$ (see Fig. 5b). Then $S'$ is a $\gamma _t^L$-set of $P_3\times P_n$ but $x_3(S')<x_3(S)$, a contradiction.

Case 4$v_{1(i-1)},v_{1(i+2)},v_{3(i-1)},v_{3(i+2)}\in S$.

To total dominate this four vertices, we should have $v_{1(i-2)},v_{1(i+3)},v_{3(i-2)},v_{3(i+3)}\in S$ (see Fig. 6a). Let $S'=(S\setminus \{v_{2i},v_{2(i+1)}\})\cup \{v_{2(i-1)},v_{2(i+2)}\}$ (see Fig. 6b). Then $S'$ is a $\gamma _t^L$-set of $P_3\times P_n$ but $x_3(S')<x_3(S)$, a contradiction. $\square$

By Claim 1, we have $x_3(S)\le \frac{n}{2}$. Thus $\gamma _t^L(P_3\times P_n)=|S|\ge \frac{6n}{5}-\frac{1}{5}x_3(S)\ge \lceil \frac{11n}{10}\rceil$. $\square$

4 Locating-Total Domination Number of $P_4\times P_n$

In this section, we give the lower and upper bounds of the locating-total domination number of $P_4\times P_n$.

Theorem 4

If $n\ge 3$, then $\lceil \frac{7n+1}{5}\rceil \le \gamma _t^L(P_4\times P_n)\le \lceil \frac{8n}{5}\rceil$.

Proof

Let $n=5k+r$, where $k\ge 0$ and $0\le r\le 4$. For $k\ge 1$, denote $S_0$ to be the empty set and

$$\begin{aligned} \begin{array}{rcl} S_5&{}=&{}\cup _{i=1}^4\{v_{i(l+2)},v_{i(l+4)}|~l=5s,\hbox { and }s=0,\ldots ,k-1\},\\ S_4&{}=&{}\{v_{1(5k+2)},v_{1(5k+4)},v_{2(5k+2)},v_{2(5k+4)},v_{3(5k+2)},v_{4(5k+2)},v_{4(5k+3)}\},\\ S_3&{}=&{}\{v_{1(5k+2)},v_{2(5k+1)},v_{2(5k+2)},v_{4(5k+2)},v_{4(5k+3)}\},\\ S_2&{}=&{}\{v_{2(5k+1)},v_{2(5k+2)},v_{3(5k+1)},v_{3(5k+2)}\},\\ S_1&{}=&{}\{v_{2(5k+1)},v_{3(5k+1)}\}. \end{array} \end{aligned}$$

Let $S=S_5\cup S_r$ for $n=5k+r$, where $k\ge 1$ and $0\le r\le 4$. If $k=0$, let $S=S_r$, where $3\le r\le 4$. It is not difficult to check that S is a LTD-set of $P_4\times P_n$. Thus $\gamma _t^L(P_4\times P_n)\le \lceil \frac{8}{5}n\rceil$.

Now we consider the lower bound of $\gamma _t^L(P_4\times P_n)$. Let $V_i=\{v_{ij}~|~j=1,\ldots ,n\}$, $1\le i\le 4$ and S a LTD-set of $P_4\times P_n$. By Lemma 2(1), $|S|\ge \frac{8n}{5}-\frac{1}{5}x_3(S)$. By the definition of LTD-set, $X_3(S)\subseteq V_2\cup V_3$. Note that if $v_{ij}\in X_3(S)$, then $j\not =1,n$.

Claim 2

For each $v_{2i}\in X_3(S)$, there is $v_{3i'}\notin X_3(S)$ and if $v_{2i},v_{2j}\in X_3(S)$ with $i\not =j$, then there are $v_{3i'},v_{3j'}\notin X_3(S)$ such that $i'\not =j'$, where $i,j\in \{2,\ldots ,n-1\}$ and $i',j'\in \{1,\ldots ,n\}$.

Proof of Claim 2

Suppose $v_{2i}\in X_3(S)$, $v_{lt}v_{2i}\in E(P_4\times P_n)$ and $v_{lt}\in S$. If $l\not = 3$, then $v_{3i}\notin S$ and let $v_{3i'}=v_{3i}$. Suppose $l=3$. Then $t=i$. If $v_{3i}\notin X_3(S)$, then let $v_{3i'}=v_{3i}$. If $v_{3i}\in X_3(S)$, then $v_{3(i-1)}\notin X_3(S)$ and let $v_{3i'}=v_{3(i-1)}$.

Suppose $v_{2i},v_{2j}\in X_3(S)$ with $i\not =j$. If $|i-j|=1$, then $i'=i,j'=j$ with $v_{3i'},v_{3j'}\notin X_3(S)$. If $|i-j|>1$, then $i'\in \{i-1,i\}$ and $j'\in \{j-1,j\}$. Since $\{i-1,i\}\cap \{j-1,j\}=\emptyset$, $v_{3i'},v_{3j'}\notin X_3(S)$ with $i'\not =j'$. Thus the claim holds. $\square$

By Claim 2, we have $x_3(S)\le n-1$. Thus $\gamma _t^L(P_4\times P_n)\ge \frac{8n}{5}-\frac{1}{5}x_3(S)\ge \lceil \frac{7n+1}{5}\rceil$. $\square$

5 Locating-Total Domination Number of $P_5\times P_n$

In this section, we give the lower and upper bounds of the locating-total domination number of $P_5\times P_n$.

Theorem 5

If $n\ge 3$, then $\lceil \frac{25n+3}{15}\rceil \le \gamma _t^L(P_5\times P_n)\le 2n$.

Proof

Let $n=5k+r$, where $k\ge 0$ and $0\le r\le 4$. For $k\ge 1$, denote $S_0$ to be the empty set and

$$\begin{aligned} \begin{array}{rcl} S_5&{}=&{}\cup _{i=1}^5\{v_{i(l+2)},v_{i(l+4)}|~l=5s,\hbox { and }s=0,\ldots ,k-1\},\\ S_4&{}=&{}\{v_{2(5k+1)},v_{2(5k+2)},v_{2(5k+3)},v_{2(5k+4)},v_{4(5k+1)},v_{4(5k+2)},v_{4(5k+3)},v_{4(5k+4)}\},\\ S_3&{}=&{}\{v_{2(5k+1)},v_{2(5k+2)},v_{2(5k+3)},v_{4(5k+1)},v_{4(5k+2)},v_{4(5k+3)}\},\\ S_2&{}=&{}\{v_{2(5k+1)},v_{2(5k+2)},v_{4(5k+1)},v_{4(5k+2)}\},\\ S_1&{}=&{}\{v_{2(5k-4)},v_{2(5k-2)},v_{2(5k)},v_{2(5k+1)},v_{4(5k-4)},v_{4(5k-2)},v_{4(5k)},v_{4(5k+1)}\}. \end{array} \end{aligned}$$

If $n=5k+1$ and $k\ge 1$, let $S=S_5\cup S_1-\{v_{1(5k-3)},v_{1(5k-1)},v_{3(5k-3)},v_{3(5k-1)},v_{5(5k-3)},v_{5(5k-1)}\}$. For $n=5k+r$ with $k\ge 1$ and $0\le r\le 4,~r\ne 1$, let $S=S_5\cup S_r$. If $k=0$, let $S=S_r$, where $3\le r\le 4$. It is not difficult to check that S is a LTD-set of $P_5\times P_n$. Thus $\gamma _t^L(P_5\times P_n)\le 2n$.

Now we consider the lower bound of $\gamma _t^L(P_5\times P_n)$. Let $V_i=\{v_{ij}~|~j=1,\ldots ,n\}$, $1\le i\le 5$ and S a LTD-set of $P_5\times P_n$. By Lemma 2(1), $|S|\ge 2n-\frac{1}{5}x_3(S)$. By the definition of LTD-set, $X_3(S)\subseteq V_2\cup V_3\cup V_4$. Note that if $v_{ij}\in X_3(S)$, then $j\not =1,n$.

Claim 3

For each $v_{3i}\in X_3(S)$, there is $v_{4i'}\notin X_3(S)$ and if $v_{3i},v_{3j}\in X_3(S)$ with $i\not =j$, then there are $v_{4i'},v_{4j'}\notin X_3(S)$ such that $i'\not =j'$, where $i,j\in \{2,\ldots ,n-1\}$ and $i',j'\in \{1,\ldots ,n\}$.

Proof of Claim 3

Suppose $v_{3i}\in X_3(S)$, $v_{lt}v_{3i}\in E(P_5\times P_n)$ and $v_{lt}\in S$. If $l\not = 4$, then $v_{4i}\notin S$ and let $v_{4i'}=v_{4i}$. Suppose $l=4$. Then $t=i$. If $v_{4i}\notin X_3(S)$, then let $v_{4i'}=v_{4i}$. If $v_{4i}\in X_3(S)$, then $v_{4(i-1)}\notin X_3(S)$ and let $v_{4i'}=v_{4(i-1)}$.

Suppose $v_{3i},v_{3j}\in X_3(S)$ with $i\not =j$. If $|i-j|=1$, then $i'=i,j'=j$ with $v_{4i'},v_{4j'}\notin X_3(S)$. If $|i-j|>1$, then $i'\in \{i-1,i\}$ and $j'\in \{j-1,j\}$. Since $\{i-1,i\}\cap \{j-1,j\}=\emptyset$, $v_{4i'},v_{4j'}\notin X_3(S)$ with $i'\not =j'$. Thus the claim holds. $\square$

By Claim 3, we have $|X_3(S)\cap (V_3\cup V_4)|\le n-1$. For any $j\in \{1,\ldots ,n-2\}$, by the definition of LTD-set, $|\{x_{2j},x_{2(j+1)},x_{2(j+2)}\}\cap X_3(S)|\le 2$. Hence $|V_2\cap X_3(S)|\le \frac{2n}{3}$ and $x_3(S)\le n-1+\frac{2n}{3}=\frac{5n}{3}-1$. Thus $\gamma _t^L(P_5\times P_n)\ge 2n-\frac{1}{5}x_3(S)\ge \lceil \frac{25n+3}{15}\rceil$. $\square$

6 Locating-Total Domination Number of $P_m\times P_n$

Finally, we present the lower and upper bounds of the locating-total domination number of $P_m\times P_n$.

Theorem 6

If $m\ge 6$ and $n\ge 6$, then $\gamma _t^L(P_m\times P_n)\le \lfloor \frac{mn+n+2m}{3}\rfloor .$

Proof

We will complete the proof by consider the following two cases.

Case 1m is even.

Let $m=2l$ and $n=6k+s$, where $l\ge 3$, $k\ge 1$ and $0\le s\le 5$. For $0\le t\le k-1$, denote

$$\begin{aligned} \begin{array}{rcl} A_{6t+1}&{}=&{}\{v_{(2i)(6t+1)}|i=1,2,\ldots ,l\},\\ B_{6t+2}&{}=&{}\{v_{(2i)(6t+2)}|i=1,2,\ldots ,l\}\cup \{v_{1(6t+2)}\},\\ C_{6t+4}&{}=&{}\{v_{(2i-1)(6t+4)}|i=1,2,\ldots ,l\},\\ D_{6t+5}&{}=&{}\{v_{(2i-1)(6t+5)}|i=1,2,\ldots ,l\}\cup \{v_{(2l)(6t+5)}\}. \end{array} \end{aligned}$$

Let $S'=\cup _{t=0}^{k-1}(A_{6t+1}\cup B_{6t+2}\cup C_{6t+4}\cup D_{6t+5})$ and

$$\begin{aligned} S=\left\{ \begin{array}{ll} S'\cup \{v_{(2i-1)(6k)}|i=1,2,\ldots ,l\} &{} \hbox {if }n=6k,\\ S'\cup \{v_{i(6k)}|i=1,2,\ldots ,2l\} &{} \hbox {if }n=6k+1,\\ S'\cup A_{6k+1}\cup B_{6k+2} &{} \hbox {if }n=6k+2,\\ S'\cup A_{6k+1}\cup B_{6k+2}\cup \{v_{(2i)(6k+3)}|i=1,2,\ldots ,l\} &{} \hbox {if }n=6k+3,\\ S'\cup A_{6k+1}\cup B_{6k+2}\cup \{v_{i(6k+4)}|i=1,2,\ldots ,2l\} &{} \hbox {if }n=6k+4,\\ S'\cup A_{6k+1}\cup B_{6k+2}\cup C_{6k+4}\cup D_{6k+5} &{} \hbox {if } n=6k+5. \end{array}\right. \end{aligned}$$

It is not difficult to check that S is a LTD-set of $P_m\times P_n$ when m is even. Note that $|S'|=k(4l+2)$ and then $|S|\le \lfloor \frac{mn+n+2m}{3}\rfloor$ in each case.

Case 2m is odd.

Let $m=2l+1$ and $n=6k+s$, where $l\ge 3$, $k\ge 1$ and $0\le s\le 5$. For $0\le t\le k-1$, denote

$$\begin{aligned} \begin{array}{rcl} A'_{6t+1}&{}=&{}\{v_{(2i)(6t+1)}|i=1,2,\ldots ,l\},\\ B'_{6t+2}&{}=&{}\{v_{(2i)(6t+2)}|i=1,2,\ldots ,l\}\cup \{v_{1(6t+2)},v_{(2l+1)(6t+2)}\},\\ C'_{6t+4}&{}=&{}\{v_{(2i-1)(6t+4)}|i=1,2,\ldots ,l+1\},\\ D'_{6t+5}&{}=&{}\{v_{(2i-1)(6t+5)}|i=1,2,\ldots ,l+1\}. \end{array} \end{aligned}$$

Let $S'=\cup _{t=0}^{k-1}(A'_{6t+1}\cup B'_{6t+2}\cup C'_{6t+4}\cup D'_{6t+5})$ and

$$\begin{aligned} S=\left\{ \begin{array}{ll} S'\cup \{v_{(2i-1)(6k)}|i=1,2,\ldots ,l+1\} &{} \hbox {if }n=6k,\\ S'\cup \{v_{i(6k)}|i=1,2,\ldots ,2l+1\} &{} \hbox {if }n=6k+1,\\ S'\cup A'_{6k+1}\cup B'_{6k+2} &{} \hbox {if }n=6k+2,\\ S'\cup A'_{6k+1}\cup B'_{6k+2}\cup \{v_{(2i)(6k+3)}|i=1,2,\ldots ,l\} &{} \hbox {if }n=6k+3,\\ S'\cup A'_{6k+1}\cup B'_{6k+2}\cup \{v_{i(6k+4)}|i=1,2,\ldots ,2l+1\} &{} \hbox {if }n=6k+4,\\ S'\cup A'_{6k+1}\cup B'_{6k+2}\cup C'_{6k+4}\cup D'_{6k+5} &{} \hbox {if }n=6k+5. \end{array}\right. \end{aligned}$$

It is not difficult to check that S is a LTD-set of $P_m\times P_n$ when m is odd. Note that $|S'|=k(4l+4)$ and then $|S|\le \lfloor \frac{mn+n+2m}{3}\rfloor$ in each case.

Thus $\gamma _t^L(P_m\times P_n)\le \lfloor \frac{mn+n+2m}{3}\rfloor .$$\square$

Theorem 7

If $m\ge 9$ and $n\ge 9$, then

$$\begin{aligned} \gamma _t^L(P_m\times P_n)\ge \frac{mn}{3}+\frac{2}{3}\max \left\{ \left\lfloor \frac{m-4}{7}\right\rfloor +\left\lfloor \frac{n}{7}\right\rfloor ,\left\lfloor \frac{m}{7}\right\rfloor +\left\lfloor \frac{n-4}{7}\right\rfloor \right\} . \end{aligned}$$

Proof

Let S be a $\gamma _t^L$-set of $P_m\times P_n$. By Lemma 2(2), we have $|S|\ge \frac{1}{3}mn+\frac{1}{6}y(S)$, where $y(S)=x_0(S)+x_1(S)+x_2(S)$. For $i\in \{1,2,m-1,m\}$ and $j\in \{1,2,\ldots ,n-6\}$, denote $A_j^{(i)}=\{v_{ij},v_{i(j+1)},\ldots ,v_{i(j+6)}\}$. For $i\in \{1,2,n-1,n\}$ and $j\in \{1,2,\ldots ,m-6\}$, denote $B_j^{(i)}=\{v_{ji},v_{(j+1)i},\ldots ,v_{(j+6)i}\}$. We first have the following result.

Claim 4

For any $1\le j\le n-6$, $|(A_j^{(1)}\cup A_j^{(2)})\cap (X_0(S)\cup X_1(S)\cup X_2(S))|\ge 2$ and $|(A_j^{(m-1)}\cup A_j^{(m)})\cap (X_0(S)\cup X_1(S)\cup X_2(S))|\ge 2$.

Proof of Claim 4

By symmetry, we just show that $|(A_j^{(1)}\cup A_j^{(2)})\cap (X_0(S)\cup X_1(S)\cup X_2(S))|\ge 2$. Suppose there is $j_0$ ($1\le j_0\le n-6$) such that $|(A_{j_0}^{(1)}\cup A_{j_0}^{(2)})\cap (X_0(S)\cup X_1(S)\cup X_2(S))|\le 1$. Then $|A_{j_0}^{(1)}\cap (X_0(S)\cup X_1(S)\cup X_2(S))|\le 1$.

If $|A_{j_0}^{(1)}\cap (X_0(S)\cup X_1(S)\cup X_2(S))|=0$, then $A_{j_0}^{(2)}\setminus \{v_{2j_0},v_{2(j_0+6)}\}\subseteq X_0(S)\cup X_1(S)\cup X_2(S)$ by the definition of LTD-set. Thus $v_{2(j_0+2)},v_{2(j_0+3)},v_{2(j_0+4)}\in X_0(S)\cup X_1(S)\cup X_2(S)$, a contradiction. Hence we have $|A_{j_0}^{(1)}\cap (X_0(S)\cup X_1(S)\cup X_2(S))|=1$ and then $(A_{j_0}^{(2)})\cap (X_0(S)\cup X_1(S)\cup X_2(S))=\emptyset$. Note that if $v_{1j}\in S$, then $v_{1j}\in X_0(S)\cup X_1(S)\cup X_2(S)$ by the definition of LTD-set.

Assume, without loss of generality, that $v_{1i}\notin S$, where $i\in \{j_0,j_0+1,j_0+2\}$. If $v_{1(j_0+3)},v_{1(j_0+4)}\notin S$, then $v_{2(j_0+1)},v_{2(j_0+2)},v_{2(j_0+3)}\in S$ which implies $v_{2(j_0+2)}\in X_0(S)\cup X_1(S)\cup X_2(S)$, a contradiction. So we have $v_{1(j_0+3)}\in S$ or $v_{1(j_0+4)}\in S$.

Suppose $v_{1(j_0+3)}\in S$. By the definition of LTD-set, $v_{2(j_0+3)}\in S$. Since $v_{1(j_0+3)}\in N(v_{1(j_0+2)})\cap S$ and $v_{1(j_0+3)}\in N(v_{1(j_0+4)})\cap S$, we have $v_{2(j_0+2)}\in S$ or $v_{2(j_0+4)}\in S$. Thus $v_{2(j_0+3)}\in X_0(S)\cup X_1(S)\cup X_2(S)$, a contradiction. If $v_{1(j_0+4)}\in S$, then $v_{2(j_0+4)}\in X_0(S)\cup X_1(S)\cup X_2(S)$ by the same argument, a contradiction. $\square$

By the same argument, we have the following claim.

Claim 5

For any $1\le j\le m-6$, $|(B_j^{(1)}\cup B_j^{(2)})\cap (X_0(S)\cup X_1(S)\cup X_2(S))|\ge 2$ and $|(B_j^{(n-1)}\cup B_j^{(n)})\cap (X_0(S)\cup X_1(S)\cup X_2(S))|\ge 2$.

By Claims 4 and 5, we have $y(S)\ge 4\max \{\lfloor \frac{m-4}{7}\rfloor +\lfloor \frac{n}{7}\rfloor ,\lfloor \frac{m}{7}\rfloor +\lfloor \frac{n-4}{7}\rfloor \}$. Since $\gamma _t^L(P_m\times P_n)=|S|\ge \frac{1}{3}mn+\frac{1}{6}y(S)$, the result holds immediately. $\square$