Paired domination and 2- distance Paired domination of the flower graph $f_{n\times m}$

TL;DR

This paper determines the exact values of the paired domination number and 2-distance paired domination number for flower graphs $f_{n\times m}$ when $m,n \geq 3$, advancing understanding of domination parameters in specific graph classes.

Contribution

The paper provides the first exact formulas for paired and 2-distance paired domination numbers of flower graphs $f_{n\times m}$ for $m,n \geq 3$, filling a gap in graph domination theory.

Findings

Exact paired domination number for $f_{n\times m}$ with $m,n \geq 3$

Exact 2-distance paired domination number for $f_{n\times m}$ with $m,n \geq 3$

Advances understanding of domination parameters in flower graphs

Abstract

Let $G = (V, E)$ be a graph without an isolated vertex. A set $D\subseteq V(G)$ is a $k$-distance paired domination set of $G$ if $D$ is a $k$-distance dominating set of $G$ and the induced subgraph $\langle D \rangle$ has a perfect matching. The minimum cardinality of a $k$-distance paired dominating set for graph $G$ is the $k$-distance paired domination number, denoted by $\gamma_{p} ^{k}(G)$. In this paper, the $k$-distance paired domination of the flower graph $f_{n\times m}$ is discussed. For $m,n\geq 3$, the exact values for paired domination number and $2$-distance paired domination number of flower graph $f_{n\times m}$ are determined

Equations

$$E(f_{n\times m})=E_1\cup E_2\cup E_3,$$

$$\gamma_{p}(f_{n\times m})=\left\{\begin{array}[]{ll}2\lceil\frac{nm-2n}{4}\rceil,&if\,\,\,m\equiv 0\,\ (\textrm{mod}\ 4)\\ \\ 2\lceil\frac{nm-n}{4}\rceil,&if\,\,\,m\equiv 1,2\,\ (\textrm{mod}\ 4)\\ \\ 2\lceil\frac{3nm-5n}{12}\rceil,&if\,\,\,m\equiv 3\,\ (\textrm{mod}\ 4)\\ \end{array}\right.$$

$$\begin{array}{r}{\gamma }_p(f_{n\times m})\le 2\lceil \frac{nm-2n}{4}\rceil .\end{array}$$

$$\begin{array}{rl}q& \ge n\lceil \frac{m-4}{4}\rceil +\frac{n}{2}\\ & =\frac{nm-2n}{4}\end{array}$$

$$\begin{array}{r}{\gamma }_p(f_{n\times m})\ge 2\lceil \frac{nm-2n}{4}\rceil .\end{array}$$

$${\gamma }_p(f_{n\times m})=2\lceil \frac{nm-2n}{4}\rceil .$$

$$\begin{array}{r}{\gamma }_p(f_{n\times m})\le 2\lceil \frac{nm-n}{4}\rceil .\end{array}$$

$$\begin{array}{rl}q& \ge n\lceil \frac{m-4}{4}\rceil \\ & =n(\frac{m-1}{4})\\ & =\frac{nm-n}{4}\end{array}$$

$$\begin{array}{r}{\gamma }_p(f_{n\times m})\ge 2\lceil \frac{nm-n}{4}\rceil .\end{array}$$

$${\gamma }_p(f_{n\times m})=2\lceil \frac{nm-2n}{4}\rceil .$$

$$\begin{array}{r}{\gamma }_p(f_{n\times m})\le 2\lceil \frac{nm-n}{4}\rceil .\end{array}$$

$$\begin{array}{rl}|D_p|& \ge n\lceil \frac{m-4}{4}\rceil +\lceil \frac{n}{4}\rceil \\ & =n(\frac{m-2}{4})+\lceil \frac{n}{4}\rceil \\ & =\frac{nm-2n}{4}+\lceil \frac{n}{4}\rceil \\ & =\lceil \frac{nm-2n+n}{4}\rceil \\ & =\lceil \frac{nm-n}{4}\rceil \end{array}$$

$$\begin{array}{r}{\gamma }_p(f_{n\times m})\ge 2\lceil \frac{nm-n}{4}\rceil .\end{array}$$

$${\gamma }_p(f_{n\times m})=2\lceil \frac{nm-2n}{4}\rceil .$$

$$\begin{array}{r}{\gamma }_p(f_{n\times m})\le 2\lceil \frac{3nm-5n}{12}\rceil .\end{array}$$

$$\begin{array}{rl}|D_p|& \ge n\lceil \frac{m-4}{4}\rceil +\lceil \frac{n}{3}\rceil \\ & =n(\frac{m-3}{4})+\lceil \frac{n}{3}\rceil \\ & =\lceil \frac{n}{3}+\frac{nm-3n}{4}\rceil \\ & =\lceil \frac{3nm-5n}{12}\rceil \end{array}$$

$$\begin{array}{r}{\gamma }_p(f_{n\times m})\ge 2\lceil \frac{3nm-5n}{12}\rceil .\end{array}$$

$${\gamma }_p(f_{n\times m})=2\lceil \frac{3nm-5n}{12}\rceil .$$

$$\gamma_{p}^{2}(f_{n\times m})=\left\{\begin{array}[]{ll}2\lceil\frac{nm-3n}{6}\rceil,\,\,\,if\,\,\,m\equiv 0,5\,\ (\textrm{mod}\ 6)\\ \\ 2\lceil\frac{nm-n}{6}\rceil,\,\,\,if\,\,\,m\equiv 1,2\,\ (\textrm{mod}\ 6)\\ \\ 2\lceil\frac{5nm-9n}{30}\rceil,\,\,\,if\,\,\,m\equiv 3\,\ (\textrm{mod}\ 6)\\ \\ 2\lceil\frac{2nm-5n}{12}\rceil,\,\,\,if\,\,\,m\equiv 4\,\ (\textrm{mod}\ 6)\\ \end{array}\right.$$

$$\begin{array}{r}{\gamma }_p^2(f_{n\times m})\le 2\lceil \frac{nm-3n}{6}\rceil .\end{array}$$

$$\begin{array}{rl}q& \ge n\lceil \frac{m-6}{6}\rceil +\lceil \frac{n}{2}\rceil \\ & =\frac{nm-6n}{6}+\lceil \frac{n}{2}\rceil \\ & =\lceil \frac{nm-6n+3n}{6}\rceil \\ & =\lceil \frac{nm-3n}{6}\rceil \end{array}$$

$$\begin{array}{r}{\gamma }_p^2(f_{n\times m})\ge 2\lceil \frac{nm-3n}{6}\rceil .\end{array}$$

$${\gamma }_p^2(f_{n\times m})=2\lceil \frac{nm-3n}{6}\rceil .$$

$$\begin{array}{r}{\gamma }_p^2(f_{n\times m})\le 2\lceil \frac{nm-n}{6}\rceil .\end{array}$$

$$\begin{array}{rl}q& \ge n\lceil \frac{m-6}{6}\rceil \\ & =n(\frac{m-1}{6})\\ & =\frac{nm-n}{6}\end{array}$$

$$\begin{array}{r}{\gamma }_p^2(f_{n\times m})\ge 2\lceil \frac{nm-n}{6}\rceil .\end{array}$$

$${\gamma }_p^2(f_{n\times m})=2\lceil \frac{nm-n}{6}\rceil .$$

$$\begin{array}{r}{\gamma }_p^2(f_{n\times m})\le 2\lceil \frac{nm-n}{6}\rceil .\end{array}$$

$$\begin{array}{rl}q& \ge n\lceil \frac{m-6}{6}\rceil +\lceil \frac{n}{6}\rceil \\ & =n(\frac{m-2}{6})+\lceil \frac{n}{6}\rceil \\ & =\frac{nm-2n}{6}+\lceil \frac{n}{6}\rceil \\ & =\lceil \frac{nm-2n+n}{6}\rceil \\ & =\lceil \frac{nm-n}{6}\rceil \end{array}$$

$$\begin{array}{r}{\gamma }_p^2(f_{n\times m})\ge 2\lceil \frac{nm-n}{6}\rceil .\end{array}$$

Full text

Keywords:*domination number, paired domination number, flower graph

*Mathematics Subject Classification: 05C15, 05C65

1 Introduction

All the graphs considerd in this paper are finite and simple. Let $G=(V,E)$ be a graph without an isolated vertex. A set $D\subseteq V(G)$ is said to be a dominating set if every vertex in $V(G)-D$ is adjacent to at least one vertex in $D$. A paired dominating is a paired dominating set of G if it is dominating and the induced subgraph $\langle D\rangle$ has a perfect matching. This type of domination was introduced by Haynes and Slater in [1, 2] and is well studied, for example [3, 4, 5].

For two vertices $x$ and $y$, let $d(x,y)$ denote the distance between $x$ and $y$ in $G$. A set $D\subseteq V(G)$ is a k-distance dominating set of $G$ if every vertex in $V(G)-D$ is within distance k of at least one vertex in $D$. The k-distance domination number $\gamma^{k}(G)$ of G is the minimum cardinality among all k-distance dominating sets of $G$. The k-distance paired-domination was introduced by Joanna Raczek [6] as a generalization of paired-domination. For a positive integer $k$, a set $D\subseteq V(G)$ is a k-distance paired-dominating set if every vertex in $V(G)-D$ is within distance $k$ of a vertex in $D$ and the induced subgraph $\langle D\rangle$ has a perfect matching. The k-distance paired-domination number, denoted by $\gamma_{p}^{k}(G)$ is the minimum cardinality of a k-distance paired-dominating set. The distance paired domination number of different families of graph such as generalized Peterson graphs, circulant graphs were studies in [7, 8].

In this paper, paired domination number and $2$-distance paired domination number of $f_{n\times m}$ are studied. The exact values of the paired dominating number has been found for every value of $m$ and $n$. Throughout the paper, the subscripts are taken modulo $n$ when it is unambiguous.

2 Paired domination number of flower graph $f_{n\times m}$

A graph G is called an $f_{n\times m}$ flower graph if it has $n$ vertices which form an $n$-cycle and $n$ sets of $m-2$ vertices which from $m$-cycles around the $n$-cycle, so that each $m$-cycle uniquely intersects the $n$-cycle on a single edge. Let $C_{1,m},C_{2,m},C_{3,m},\dots,C_{n,m}$ are edge disjoint outer cycles of length $m$. Every two consecutive outer cycles has a common vertex of degree four. In each cycle, there are $m-2$ vertices of degree two and two vertices of degree four.

This graph will be denoted by $f_{n\times m}$. It is clear that $f_{n\times m}$ has $n(m-1)$ vertices and $nm$ edges. The $m$-cycles are called the petals and the $n$-cycles is called center of $f_{n\times m}$. The $n$ vertices which form the center are all of degree $4$ and all other vertices have degree $2$. The centered vertices are denoted by $u_{i}$, where $i=1,\dots,n$. The vertices of outer cycles are denoted by $v_{ij}$, where $1\leq i\leq n$ and $1\leq j\leq m-2$. Thus, the vertex and edge set of the flower graph $f_{n\times m}$ is

$V(f_{n\times m})=\{u_{i},v_{i,j}:1\leq i\leq n,1\leq j\leq m-2\}$

[TABLE]

where $E_{1}=\{u_{i}u_{i+1}:1\leq i\leq n\}$, $E_{2}=\{v_{i,j}v_{i,j+1}:1\leq i\leq n,1\leq j\leq m-3\}$ and $E_{3}=\{u_{i}v_{i,1},u_{i+1}v_{i,m-2}:1\leq i\leq n\}$.

Let $D_{p}=\{x_{i},y_{i}:i=1,2,...,q\}$ be an arbitrary paired dominating set of the flower graph $f_{n\times m}$. For convenience, let $V_{i}$ is the set of vertices of the outer cycles $C_{i,m}$, for each $i=1,\dots,n$ and $U$ is the set of vertices of the inner cycle. Thus,

$V_{i}=\{v_{i,j}$ $\in$ $V(f_{n\times m})$ $:$ $deg(v_{i,j})=2:1\leq i\leq n,1\leq j\leq m-2\}$

$U=\{u_{i}$ $\in$ $V(f_{n\times m})$ $:$ $deg(u_{i})=4:1\leq i\leq n\}$

and let

$D_{vv}=\{(x_{i},y_{i})$ $\in$ $D_{p}$ : $x_{i}$ $\in$ $V_{i}$, $y_{i}$ $\in$ $V_{i}\}$,

$D_{uu}=\{(x_{i},y_{i})$ $\in$ $D_{p}$ : $x_{i}$ $\in$ $U$, $y_{i}$ $\in$ $U\}$,

$D_{vu}=\{(x_{i},y_{i})$ $\in$ $D_{p}$ : $x_{i}$ $\in$ $V_{i}$, $y_{i}$ $\in$ $U\}$.

Obviously, $D_{p}=D_{vv}\cup D_{uu}\cup D_{vu}$.

Lemma 2.1**.**

Let $D_{p}$ be a paired dominating set of the graph $f_{n\times m}$ and $V_{i}$ be the set of vertices of degree $2$ of the outer m-cycles $C_{i,m}$. Then $D_{p}$ contain at least $2\lceil\frac{m-2(k+1)}{2(k+1)}\rceil$ vertices from each $V_{i}$.

Proof.

Since, each $C_{i,m}$ has $2$ vertices of degree $4$ and these vertices can dominate at most $2k$ $(\forall\,\ k\geq 1)$ vertices of each $V_{i}$. Therefore, to dominate the remaining $m-2(k+1)$ vertices of each $V_{i}$, we need at least $2\lceil\frac{m-2(k+1)}{2(k+1)}\rceil$ vertices in $D_{p}$ from each set $V_{i}$.

In the next theorem, the exact value of the paired dominating number of the graph $f_{n\times m}$ for $m\equiv 0,1,2,3$ $(\textrm{mod}\ 4)$ are given.

Theorem 2.2**.**

*For $m,n\geq 3$,

[TABLE]

Proof.

Let $t=\lfloor\frac{m}{4}\rfloor$. We prove this theorem by giving the following cases.

Case 1: $m\equiv 0\,\ (\textrm{mod}\ 4)$.

If $n$ is even, define $t^{{}^{\prime}}=\frac{n}{2}$. The set $D_{p}$ for $m=4$ is defined as follows:

$D_{p}=\{u_{2l-1},u_{2l}\ :\ 1\leq l\leq t^{{}^{\prime}}\}.$

If $m\equiv 0\,\ (\textrm{mod}\ 4)$ (where $m\neq 4$), define

$D_{p}=\{v_{i,4j-1},v_{i,4j}:\ 1\leq i\leq n,\,\,1\leq j\leq t-1\}\cup\{u_{2l-1},u_{2l}\ :\ 1\leq l\leq t^{{}^{\prime}}\}.$

If $n$ is odd, then define $t^{{}^{\prime}}=\frac{n-1}{2}$. The set $D_{p}$ for $m=4$ is defined as follows:

$D_{p}=\{u_{2l-1},u_{2l}\ :\ 1\leq l\leq t^{{}^{\prime}}\}\cup\{u_{n},v_{n,1}\}$

If $m\equiv 0\,\ (\textrm{mod}\ 4)$ (where $m\neq 4$), define

$D_{p}=\{v_{i,4j-1},v_{i,4j}:\ 1\leq i\leq n-1,\,\,1\leq j\leq t-1\}\cup\{v_{n,4j},v_{n,4j+1}\}\cup\{u_{2l-1},u_{2l}\>\,\,\,\ 1\leq l\leq t^{{}^{\prime}}\}\cup\{u_{n},v_{n,1}\}.$

In each case, it is easy to verify that $D_{p}$ is a paired dominating set. The cardinality of $D_{p}$ in each case is $2\lceil\frac{nm-2n}{4}\rceil$. Hence,

[TABLE]

To prove the lower bound for the paired dominating set $D_{p}$. Let $D_{p}=\{x_{i},y_{i}:1\leq i\leq q\}$ be a paired dominating set of $f_{n\times m}$. By Lemma 2.1, $D_{p}$ contain at least $2\lceil\frac{m-4}{4}\rceil$ pair of vertices from each $V_{i}$. With loss of generality, we can suppose that $v_{i,1}$ and $v_{i,m-2}$ are the vertices which are yet to be dominated in $V_{i}$. Then, to dominate $v_{i,1}$ and $v_{i}$ either $v_{i,1},v_{i,m-2}\in D_{p}$ or $u_{i},u_{i+1}\in D_{p}$. In both these cases, each $C_{i,m}$ has at least two vertices belong to $D_{p}$. Since, each vertex of degree $4$ belong to neighboring cycle, therefore each vertex of degree $4$ belong to $D_{p}$. Further, $\langle D_{p}\rangle$ is a perfect matching. Thus only edges that are not adjacent to each other can belong to $D_{p}$. There are $\lceil\frac{n}{2}\rceil$ non adjacent edges of type$(4,4)$ if $n$ is even and $\lceil\frac{n}{2}\rceil-1$ edges if $n$ is odd. In the later case, one edge of the type $(2,4)$ also belong to $D_{p}$. Thus

[TABLE]

which implies that $q\geq\lceil\frac{nm-2n}{4}\rceil$. Hence

[TABLE]

From Equation $1$ and $2$, it is clear that

[TABLE]

Case 2: $m\equiv 1\,\ (\textrm{mod}\ 4)$.

In this case, define the set $D_{p}$ as follows:

$D_{p}=\{v_{i,4j-3},v_{i,4j-2}:\ 1\leq i\leq n,\,\,1\leq j\leq t\}.$

It is easy to see that $D_{p}$ is a paired dominating set and the cardinality of paired dominating set is $2\lceil\frac{nm-n}{4}\rceil$. Hence,

[TABLE]

The lower bound of paired dominating set is proved in the following way.

Let $D_{p}=\{x_{i},y_{i}:1\leq i\leq q\}$ be a paired dominating set. By Lemma 2.1, $D_{p}$ contain at least $\lceil\frac{m-4}{4}\rceil$ vertices from each $V_{i}$ of $C_{i,m}$. If $m\equiv 1,$ $(\textrm{mod}\ 4)$, then $\lceil\frac{m-4}{4}\rceil$ pair of vertices dominate $m-1$ vertices from each $V_{i}$ of $C_{i,m}$, $where1\leq i\leq n.$ Therefore

[TABLE]

which implies that $q\geq\lceil\frac{nm-n}{4}\rceil$. Hence

[TABLE]

Equation $3$ and $4$ implies that

[TABLE]

Case 3: $m\equiv 2\,\ (\textrm{mod}\ 4).$

Let $t^{{}^{\prime}}=\lceil\frac{n}{4}\rceil$. If $n=5$, define

$D_{p}=\{u_{1},u_{2},u_{4},u_{5},v_{i,4j-2},v_{i,4j-1}:\ 1\leq i\leq n,\ 1\leq j\leq t\}.$

If $n\neq 5$, define

$D_{p}=\{v_{i,4j-2},v_{i,4j-1}:\ 1\leq i\leq n,\,\,1\leq j\leq t\}\cup\{u_{4l-3},u_{4l-2}:\ 1\leq l\leq t^{{}^{\prime}}\}.$

It is easy to see that $D_{p}$ is a paired dominating set and the cardinality of $D_{p}$ is $2\lceil\frac{nm-n}{4}\rceil$. Hence

[TABLE]

To prove the lower bound, let $D_{p}=\{x_{i},y_{i}:1\leq i\leq q\}$ be a paired dominating set. By Lemma 2.1, $D_{p}$ contain at least $\lceil\frac{m-4}{4}\rceil$ vertices from each $V_{i}$ of $C_{i,m}$.If $m\equiv 2,$ $(\textrm{mod}\ 4)$, then $\lceil\frac{m-4}{4}\rceil$ pair of vertices dominate $m-2$ vertices from each $V_{i}$ of $C_{i,m}$, $where1\leq i\leq n.$ The only vertices which are yet to be dominated are the vertices $u_{i}$ of degree $4$. Since there are $n$ vertices of degree $4$, therefore we need at least $\lceil\frac{n}{4}\rceil$ more pair of vertices in $D_{p}$. Thus

[TABLE]

which implies that $q\geq\lceil\frac{nm-n}{4}\rceil$. Hence

[TABLE]

Equation $5$ and $6$ implies that

[TABLE]

Case 4: For $m\equiv 3\,\ (\textrm{mod}\ 4)$.

For $n\equiv 0,2\,\ (\textrm{mod}\ 3)$, let $t^{{}^{\prime}}=\lceil\frac{n}{3}\rceil$ and for $n\equiv 1\,\ (\textrm{mod}\ 3)$, $t^{{}^{\prime}}=\lfloor\frac{n}{3}\rfloor$.

If $n=4$, define

$D_{p}=\{v_{1,4j-1},v_{1,4j},v_{2,4j-1},v_{2,4j},v_{3,4j-1},v_{3,4j},v_{4,4j-1},v_{4,4j}:\ 1\leq j\leq t\}\cup\{u_{1},u_{2},u_{3},u_{4}\}$

If $n=3t,\forall$ $t\geq 1$, then define

$D_{p}=\{v_{3i-2,4j-1},v_{3i-2,4j},v_{3i-1,4j-1},v_{3i-1,4j}:\ 1\leq i\leq t^{{}^{\prime}},\,\,1\leq j\leq t\}\cup\{v_{3l,4j-2},v_{3l,4j-1}:\ 1\leq i\leq t^{{}^{\prime}}\}\cup\{u_{3l^{{}^{\prime}}-2},u_{3l^{{}^{\prime}}-1}:\ 1\leq l^{{}^{\prime}}\leq t^{{}^{\prime}}\}.$

If $n=3t+1,\forall$ $t\geq 2$, then define

$D_{p}=\{v_{3i-2,4j-1},v_{3i-2,4j},v_{3i-1,4j-1},v_{3i-1,4j}:\ 1\leq i\leq t^{{}^{\prime}},\,\,1\leq j\leq t\}\cup\{v_{3l,4j-2},v_{3l,4j-1}:\ 1\leq i\leq t^{{}^{\prime}}-1\}\cup\{v_{n-1,4j-1},v_{n-1,4j},v_{n,4j-1},v_{n,4j}\}\cup\{u_{3l^{{}^{\prime}}-2},u_{3l^{{}^{\prime}}-1}:\ 1\leq l^{{}^{\prime}}\leq t^{{}^{\prime}}\}\cup\{u_{n-1},u_{n}\}.$

If $n=3t+2,\forall$ $t\geq 1$, then define

$D_{p}=\{v_{3i-2,4j-1},v_{3i-2,4j},v_{3i-1,4j-1},v_{3i-1,4j}:\ 1\leq i\leq t^{{}^{\prime}},\,\,1\leq j\leq t\}\cup\{v_{3l,4j-2},v_{3l,4j-1}:\ 1\leq i\leq t^{{}^{\prime}}-1\}\cup\{u_{3l^{{}^{\prime}}-2},u_{3l^{{}^{\prime}}-1}:\ 1\leq l^{{}^{\prime}}\leq t^{{}^{\prime}}\}.$

It is easy to see that $D_{p}$ is a paired dominating set in each case and the cardinality of $D_{p}$ is $2\lceil\frac{3nm-5n}{12}\rceil$. Hence,

[TABLE]

Now we prove the lower bound of paired dominating set.

Let $D_{p}=\{x_{i},y_{i}:1\leq i\leq q\}$ be a paired dominating set. By Lemma 2.1, $D_{p}$ contain at least $\lceil\frac{m-4}{4}\rceil$ pair of vertices from each $V_{i}$ of $C_{i,m}$. The graph $f_{n\times m}$ has vertices of degree $2$ and $4$. The $D_{p}$ can contain the edges of types $(2,2),(2,4)and(4,4)$. The edge of the type $(2,2),(2,4)$ and $(4,4)$ can dominate $4,6$ and $8$ vertices respectively of $f_{n\times m}$. Each $C_{i}$ contain $m-2$ vertices of degree $2$ and $2$ vertices of degree $4$. Since any pair of $D_{p}$ can dominate at most $4$ vertices of each $C_{i}$. Therefore, to dominate remaining $m-4$ vertices, we need at least $\lceil\frac{m-4}{4}$ pairs of adjacent vertices in the $D_{p}$. Since $m\equiv 3$ $(\textrm{mod}\ 4)$, therefore these $\lceil\frac{m-4}{4}\rceil$ pairs of adjacent vertices dominate $m-3$ vertices in each $C_{i}$. Also each edge of the type $(4,4)$ dominate $8$ vertices. So we have to choose at least one edge from $3$ consecutive copies of outer $C_{i,m}$. This implies that

[TABLE]

which implies that $q\geq\lceil\frac{3nm-5n}{12}\rceil$. Hence

[TABLE]

From Equation $7$ and $8$, it is clear that

[TABLE]

In Figure $1$, we show the paired dominating set of $f_{n\times m}$ for different values of $n$ and $m$, where the vertices of paired dominating set are in dark.

3 2-distance paired domination number of flower graph $f_{n\times m}$

In this section, the exact value of $2$-distance paired domination number of flower graph $f_{n\times m}$ is determined.

Theorem 3.1**.**

For $m,n\geq 3$,

[TABLE]

Proof.

Let $t=\lfloor\frac{m}{6}\rfloor$. We have the following cases.

Case 1: For $m\equiv 0\,\ (\textrm{mod}\ 6)$.

Define $t^{{}^{\prime}}=\lceil\frac{n}{2}\rceil$. The $2$-paired dominating set for $m=6$ and $n=2t+1$, $\forall$ $t\geq 1$ is defined as:

$D_{2,p}=\{u_{2i-1},u_{2i}:\ 1\leq i\leq t^{\prime}-1\}\cup\{u_{n},v_{n,1}\}.$

For $m=6$ and $n=2t$, $\forall$ $t\geq 2$, define

$D_{2,p}=\{u_{2i-1},u_{2i}:\ 1\leq i\leq t^{\prime}\}.$

If $m\neq 6$ and $n=2t+1$, $\forall$ $t\geq 1$, then define

$D_{2,p}=\{v_{i,6j-1},v_{i,6j}:\ 1\leq j\leq t-1,\,\ 1\leq i\leq n-1\}\cup\{v_{n,6j},v_{n,6j+1}:1\leq j\leq t-1\}\cup\{u_{n},v_{n,1}\}\cup\{u_{2l-1},u_{2l}\ :\ 1\leq l\leq t^{\prime}-1\}.$

If $m=6$ and $n=2t$, $\forall$ $t\geq 2$, then define

$D_{2,p}=\{v_{i,6j-1},v_{i,6j}:\ 1\leq j\leq t-1,\,\ 1\leq i\leq n\}\cup\{u_{2l-1},u_{2l}\ :\ 1\leq l\leq t^{\prime}\}.$

In all these possibilities, it is easy to verify that $D_{2,p}$ is a $2$-paired dominating set. Further, the cardinality of $D_{2,p}$ in each case is $2\lceil\frac{nm-3n}{6}\rceil$. Hence,

[TABLE]

Now to prove the lower bound for $2$-distance paired dominating set. Let $D_{2,p}=\{x_{i},y_{i}:1\leq i\leq q\}$ be a paired dominating set of $f_{n\times m}$. By Lemma 2.1, $D_{2,p}$ contains at least $\lceil\frac{m-6}{6}\rceil$ vertices from each $V_{i}$ $(m\neq 6)$. These $\lceil\frac{m-6}{6}\rceil$ vertices dominate $m-6$ vertices of degree $2$ in each $C_{i,m}$. Suppose that $v_{i,1},v_{i,2},v_{i,m-2}$ and $v_{i,m-3}$ are the vertices which are yet to be dominated. To dominate these vertices either vertices of degree $2$ belongs to $D_{2,p}$ or vertices of degree $4$ belongs to $D_{2,p}$. In both the cases each $C_{i}$ has at least two vertices which belongs to $D_{2,p}$. Since each vertex of degree $4$ belong to neighboring cycle, so it must belong to $D_{2,p}$. Also $\langle D_{2,p}\rangle$ has perfect matching which implies that the only non adjacent edges of type $(4,4)$ belong to $D_{2,p}$. There are $\lceil\frac{n}{2}\rceil$ non adjacent edges of type $(4,4)$ if $n$ is even and $\lceil\frac{n}{2}\rceil-1$ if $n$ is odd. In later case one edge of type $(2,4)$ also belong to $D_{2,p}$. Thus

[TABLE]

which implies that $q\geq\lceil\frac{nm-3n}{6}\rceil$. Thus

[TABLE]

From Equation $9$ and $10$, it is clear that

[TABLE]

Case 2: For $m\equiv 1\,\ (\textrm{mod}\ 6)$.

Define

$D_{2,p}=\{v_{i,6j-4},v_{i,6j-3}:1\leq i\leq n,\,\ 1\leq j\leq t\}.$

It is easy to verify that $D_{2,p}$ is a $2$-paired dominating set. Further, the cardinality of $D_{2,p}$ is

[TABLE]

Now to prove the lower bound of $2$-distance paired dominating set. Let $D_{2,p}=\{x_{i},y_{i}:1\leq i\leq q\}$ be a paired dominating set. By Lemma 2.1, $D_{2,p}$ contain at least $\lceil\frac{m-6}{6}\rceil$ vertices from each $V_{i}$ of $C_{i,m}$. If $m\equiv 1,$ $(\textrm{mod}\ 6)$, then $\lceil\frac{m-6}{6}\rceil$ pair of vertices dominate $m-1$ vertices from each $V_{i}$ of $C_{i,m}$, where $1\leq i\leq n.$ Therefore

[TABLE]

which implies that $q\geq\lceil\frac{nm-n}{6}\rceil$. Thus

[TABLE]

From Equation $11$ and $12$, it is clear that

[TABLE]

Case 3: $m\equiv 2\,\ (\textrm{mod}\ 6)$

Let $t^{{}^{\prime}}=\lceil\frac{n}{6}\rceil.$ For $n\equiv 0,2,3,4,5\,\ (\textrm{mod}\ 6)$, define

$D_{2,p}=\{v_{i,6j-3},v_{i,6j-2}:1\leq i\leq n,\,\ 1\leq j\leq t\}\cup\{u_{6l-5},u_{6l-4}:1\leq l\leq t^{\prime}\}.$

For $n\equiv 1\,\ (\textrm{mod}\ 6)$, define

$D_{2,p}=\{v_{i,6j-3},v_{i,6j-2}:1\leq i\leq n,\,\ 1\leq j\leq t\}\cup\{u_{6l-5},u_{6l-4},u_{n-1},u_{n}:1\leq l\leq t^{\prime}\}.$

In all these possibilities, it is easy to verify that $D_{2,p}$ is a $2$-paired dominating set. Further, the cardinality of $D_{2,p}$ in each case is $2\lceil\frac{nm-n}{6}\rceil$. Hence,

[TABLE]

Now we give the lower bound of $2$-distance paired dominating set. Let $D_{2,p}=\{x_{i},y_{i}:1\leq i\leq q\}$ be a paired dominating set of $f_{n\times m}$. If $m\equiv 2,$ $(\textrm{mod}\ 6)$, then $\lceil\frac{m-6}{6}\rceil$ pair of vertices dominate $m-2$ vertices from each $V_{i}$ of $C_{i,m}$, $where1\leq i\leq n.$ The only vertices which are yet to be dominated are the vertices $u_{i}$ of degree $4$. Since there are $n$ vertices of degree $4$, therefore we need at least $\lceil\frac{n}{6}\rceil$ more pair of vertices in $D_{2,p}$. Thus

[TABLE]

which implies that $q\geq\lceil\frac{nm-n}{6}\rceil$. Therefore

[TABLE]

From Equation $13$ and $14$, it is clear that

[TABLE]

Case 4: $m\equiv 3(\textrm{mod}\ 6)$.

let $t^{{}^{\prime}}=\lceil\frac{n}{5}\rceil.$ For $n=3$, define

$D_{2,p}=\{v_{1,6j-1},v_{1,6j},v_{2,6j-1},v_{2,6j},v_{3,6j-2},v_{3,6j-1}:1\leq i\leq t^{\prime},\,\ 1\leq j\leq t\}\cup\{u_{1},u_{2}\}.$

For $n=5$, define

$D_{2,p}=\{v_{1,6j-1},v_{1,6j},v_{2,6j-1},v_{2,6j},v_{3,6j-2},v_{3,6j-1},v_{4,6j-3},v_{4,6j-2},v_{5,6j-2},v_{5,6j-1}:1\leq j\leq t\}\cup\{u_{1},u_{2}\}.$

For $n=4,6$, define

$D_{2,p}=\{v_{4i-3,6j-1},v_{4i-3,6j},v_{4i-2,6j-1},v_{4i-2,6j},v_{3,6j-2},v_{3,6j-1},v_{4,6j-2},v_{4,6j-1}:1\leq i\leq t^{\prime},\,\ 1\leq j\leq t\}\cup\{u_{1},u_{2}\}.$

For $n=5t,\forall$ $t\geq 2$, define

$D_{2,p}=\{v_{5i-4,6j-1},v_{5i-4,6j},v_{5i-3,6j-1},v_{5i-3,6j},v_{5i-2,6j-2},v_{5i-2,6j-1},v_{5i,6j-2},\\ v_{5i,6j-1},v_{5i-1,6j-3},v_{5i-1,6j-2}:1\leq i\leq t^{\prime},\,\ 1\leq j\leq t\}\cup\{u_{5l-4},u_{5i-3}:1\leq l\leq t^{\prime}\}.$

For $n=5t+1,\forall$ $t\geq 1$, define

$D_{2,p}=\{v_{5i-4,6j-1},v_{5i-4,6j},v_{5i-3,6j-1},v_{5i-3,6j},v_{5i-2,6j-2},v_{5i-2,6j-1},v_{5p,6j-2},\\ v_{5p,6j-1},v_{5p-1,6j-3},v_{5p-1,6j-2},v_{n-2,6j-2},v_{n-2,6j-1},v_{n-1,6j-1},v_{n-1,6j},v_{n,6j-1},v_{n,6j}:1\leq i\leq t^{\prime}-1,\,\ 1\leq j\leq t,\,\ 1\leq p\leq t^{\prime}-2\}\cup\{u_{5l-4},u_{5i-3},u_{n-1},u_{n}:1\leq l\leq t^{\prime}-1\}.$

For $n=5t+2\,\ \forall$ $t\geq 1$, define

$D_{2,p}=\{v_{5i-4,6j-1},v_{5i-4,6j},v_{5i-3,6j-1},v_{5i-3,6j}:1\leq i\leq t^{\prime},\,\ 1\leq j\leq t\}\cup\{v_{5i-2,6j-2},v_{5i-2,6j-1},v_{5i-1,6j-3},v_{5i-1,6j-2},v_{5i,6j-2},v_{5i,6j-1}:1\leq i\leq t^{\prime}-1,\,\ 1\leq j\leq t\}\cup\{u_{5l-4},u_{5i-3}:1\leq l\leq t^{\prime}\}.$

For $n=5t+3\,\ \forall$ $t\geq 1$, define

$D_{2,p}=\{v_{5i-4,6j-1},v_{5i-4,6j},v_{5i-3,6j-1},v_{5i-3,6j},v_{5i^{\prime}-1,6j-3},v_{5i^{\prime}-1,6j-2},:1\leq i\leq t^{\prime},\,\ 1\leq i^{\prime}\leq t^{\prime}-1,\,\,1\leq j\leq t\}\cup\{v_{5i-2,6j-2},v_{5i-2,6j-1},\\ v_{5p,6j-2},v_{5p,6j-1}:1\leq i\leq t^{\prime},\,\ 1\leq j\leq t,\,\ 1\leq p\leq t^{\prime}-1\}\cup\{u_{5l-4},u_{5i-3}:1\leq l\leq t^{\prime}\}.$

For $n=5t+4\,\ \forall$ $t\geq 1$, define

$D_{2,p}=\{v_{5i-4,6j-1},v_{5i-4,6j},v_{5i-3,6j-1},v_{5i-3,6j},v_{5i^{\prime}-1,6j-3},v_{5i^{\prime}-1,6j-2}:1\leq i\leq t^{\prime},\,\,1\leq i^{\prime}\leq t^{\prime}-1,\,\ 1\leq j\leq t\}\cup\{v_{5i-2,6j-2},v_{5i-2,6j-1},v_{5p,6j-2},v_{5p,6j-1},v_{n,6j-2},v_{n,6j-1}:1\leq i\leq t^{\prime},\,\ 1\leq j\leq t,\,\ 1\leq p\leq t^{\prime}-1\}\cup\{u_{5l-4},u_{5i-3}:1\leq l\leq t^{\prime}\}.$

In all these possibilities, it is easy to verify that $D_{2,p}$ is a $2$-paired dominating set. Further, the cardinality of $D_{2,p}$ in each case is $2\lceil\frac{5nm-9n}{30}\rceil$. Hence,

[TABLE]

Now we prove the lower bound of $2$-distance paired dominating set.

Let $D_{2,p}=\{x_{i},y_{i}:1\leq i\leq q\}$ be a $2$-distance paired dominating set. According to Lemma 2.1, $D_{2,p}$ contain at least $\lceil\frac{m-6}{6}\rceil$ vertices from each $V_{i}$ of $C_{i,m}$. If $m\equiv\ 3,$ $(\textrm{mod}\ 6)$, then $\lceil\frac{m-6}{6}\rceil$ pair of vertices dominate $m-3$ vertices from each $V_{i}$ of $C_{i,m}$, $where1\leq i\leq n.$ The number of vertices which are yet to be dominated in each $C_{i}$ are $3$, from which $2$ vertices are of degree $4$ and one vertex of degree $2$. Suppose that these vertices are $v_{i,1},u_{i}$ and $u_{i+1}$. To dominate these vertices we choose pair of vertices of type $(4,4)$ from $5$ consecutive copies of $C_{i}$ because each pair of type $(4,4)$ dominates $16$ vertices of $f_{n\times m}$ Since $\langle D_{2,p}\rangle$ is a perfect matching, so it contains only the non adjacent edges. This implies that

[TABLE]

which implies that $q\geq\lceil\frac{5nm-9n}{30}\rceil$. Therefore

[TABLE]

From Equation $15$ and $16$, it is clear that

[TABLE]

Case 5: $m\equiv 4(\textrm{mod}\ 6)$.

Let $t^{{}^{\prime}}=\lceil\frac{n}{4}\rceil.$ If $n=4t\,\ \forall$ $t\geq 1$, then define

$D_{2,p}=\{v_{4i-3,6j-1},v_{4i-3,6j},v_{4i-2,6j-1},v_{4i-2,6j},v_{4i-1,6j-2},v_{4i-1,6j-1},v_{4i,6j-2},v_{4i,6j-1}:1\leq i\leq t^{\prime},\,\ 1\leq j\leq t\}\cup\{u_{4l-3},u_{4l-2}:1\leq l\leq t^{\prime}\}.$

If $n=5$, define

$D_{2,p}=\{v_{3i-2,6j-1},v_{3i-2,6j},v_{3i-1,6j-1},v_{3i-1,6j},v_{3,6j-2},v_{3,6j-1}:1\leq j\leq t:1\leq i\leq t^{\prime}\}\cup\{u_{1},u_{2},u_{4},u_{5}\}.$

If $n=4t+1\,\ \forall$ $t\geq 2$, define

$D_{2,p}=\{v_{4i-3,6j-1},v_{4i-3,6j},v_{4i-2,6j-1},v_{4i-2,6j},v_{n-1,6j-1},v_{n-1,6j},v_{n,6j-1},v_{n,6j}:1\leq i\leq t^{\prime}-1,\,\ 1\leq j\leq t\}\cup\{v_{4i-1,6j-2},v_{4i-1,6j-1},v_{4p,6j-2},v_{4p,6j-1}:1\leq i\leq t^{\prime}-1,\,\ 1\leq j\leq t,\,\ 1\leq p\leq t^{\prime}-2\}\cup\{u_{4l-3},u_{4l-2}:1\leq l\leq t^{\prime}-1\}\cup\{u_{n-1},u_{n}\}.$

If $n=4t+2\,\ \forall$ $t\geq 1$, define

$D_{2,p}=\{v_{4i-3,6j-1},v_{4i-3,6j},v_{4i-2,6j-1},v_{4i-2,6j}:1\leq i\leq t^{\prime},\,\ 1\leq j\leq t\}\cup\{v_{4i-1,6j-2},v_{4i-1,6j-1},v_{4i,6j-2},v_{4i,6j-1}:1\leq i\leq t^{\prime}-1,\,\,\,\ 1\leq j\leq t\}\cup\{u_{4l-3},u_{4l-2}:1\leq l\leq t^{\prime}\}.$

If $n=3$, then

$D_{2,p}=\{v_{1,6j-1},v_{1,6j},v_{2,6j-1},v_{2,6j},v_{3,6j-2},v_{3,6j-1}:1\leq i\leq t^{\prime},\,\ 1\leq j\leq t\}\cup\{u_{4l-3},u_{4l-2}:1\leq l\leq t^{\prime}\}.$

If $n=4t+3\,\ \forall$ $t\geq 1$, then

$D_{2,p}=\{v_{4i-3,6j-1},v_{4i-3,6j},v_{4i-2,6j-1},v_{4i-2,6j}:1\leq i\leq t^{\prime},\,\ 1\leq j\leq t\}\cup\{v_{4i-1,6j-2},v_{4i-1,6j-1},v_{4p,6j-2},v_{4p,6j-1}:1\leq i\leq t^{\prime},\,\ 1\leq j\leq t,\,\ 1\leq p\leq t^{\prime}-1\}\cup\{u_{4l-3},u_{4l-2}:1\leq l\leq t^{\prime}\}.$

In all these possibilities, it is easy to verify that $D_{2,p}$ is a $2$-paired dominating set. Further, the cardinality of $D_{2,p}$ in each case is $2\lceil\frac{2nm-5n}{12}\rceil$, Hence

[TABLE]

Now we prove the lower bound of $2$-distance paired dominating set.

Let $D_{2,p}=\{x_{i},y_{i}:1\leq i\leq q\}$ be a paired dominating set. According to Lemma 2.1, $D_{2,p}$ contain at least $\lceil\frac{m-6}{6}\rceil$ vertices from each $V_{i}$ of $C_{i,m}$. If $m\equiv 4,$ $(\textrm{mod}\ 6)$, then $\lceil\frac{m-6}{6}\rceil$ pair of vertices dominate $m-4$ vertices from each $V_{i}$ of $C_{i,m}$, $where1\leq i\leq n.$ The number of vertices which are yet to be dominated in each $C_{i}$ are $4$, from which $2$ vertices are of degree $4$ and other vertices of degree $2$. Suppose that these vertices are $v_{i,1},v_{i,2},u_{i}$ and $u_{i+1}$. To dominate these vertices either $v_{i,1},v_{i,2}$ belongs to $D_{2,p}$ or $u_{i},u_{i+1}$ belongs to $D_{2,p}$. If $v_{i,1},v_{i,2}$ belongs to $D_{2,p}$, then the only vertices which are dominated by these vertices are $u_{i}$ and $u_{i+1}$. If $u_{i},u_{i+1}$ belongs to $D_{2,p}$ then these vertices dominate $13$ vertices of $4$ consecutive copies of $C_{i}$. Thus we choose pair of vertices of type $(4,4)$ from $4$ consecutive copies of $C_{i}$. Since $\langle D_{2,p}\rangle$ is a paired dominating set and has perfect matching, so it contains only the non adjacent edges. This implies that

[TABLE]

which implies that $q\geq\lceil\frac{2nm-5n}{12}\rceil$. Therefore

[TABLE]

From Equation $17$ and $18$, it is clear that

[TABLE]

Case 6: $m\equiv 4(\textrm{mod}\ 6)$.

Let $t^{{}^{\prime}}=\lceil\frac{n}{3}\rceil.$

If $n=3$, define

$D_{2,p}=\{v_{1,6j-1},v_{1,6j},v_{2,6j-1},v_{2,6j},v_{3,6j-2},v_{3,6j-1}:1\leq j\leq t\}\cup\{u_{1},u_{2}\}.$

If $n=4$, define

$D_{2,p}=\{v_{i,6j-1},v_{i,6j}:1\leq i\leq n,\,\ 1\leq j\leq t\}\cup\{u_{1},u_{2},u_{3},u_{4}\}.$

If $n=5$, define

$D_{2,p}=\{v_{1,6j-1},v_{1,6j},v_{2,6j-1},v_{2,6j},v_{4,6j-1},v_{4,6j},v_{5,6j-1},v_{5,6j},v_{3,6j-2},v_{3,6j-1}:1\leq j\leq t\}\cup\{u_{1},u_{2},u_{3},u_{4}\}.$

For $n=3t\,\ \forall$ $t\geq 2$, define

$D_{2,p}=\{v_{3i-2,6j-1},v_{3i-2,6j},v_{3i-1,6j-1},v_{3i-1,6j},v_{3i,6j-2},v_{3i,6j-1}:1\leq i\leq t^{\prime},\,\ 1\leq j\leq t\}\cup\{u_{3l-2},u_{3l-1}:1\leq l\leq t^{\prime}\}.$

For $n=3t+1\,\ \forall$ $t\geq 2$, define

$D_{2,p}=\{v_{3i-2,6j-1},v_{3i-2,6j},v_{3i-1,6j-1},v_{3i-1,6j},v_{n-1,6j-1},v_{n-1,6j},v_{n,6j-1},v_{n,6j}:1\leq i\leq t^{\prime}-1,\,\ 1\leq j\leq t\}\cup\{v_{3i,6j-2},v_{3i,6j-1}:1\leq i\leq t^{\prime}-2,\,\ 1\leq j\leq t\}\cup\{u_{3l-2},u_{3l-1},u_{n-1},u_{n}:1\leq l\leq t^{\prime}-1\}.$

For $n=3t+2\,\ \forall$ $t\geq 2$, define

$D_{2,p}=\{v_{3i-2,6j-1},v_{3i-2,6j},v_{3i-1,6j-1},v_{3i-1,6j}:1\leq i\leq t^{\prime},\,\ 1\leq j\leq t\}\cup\{v_{3i,6j-2},v_{3i,6j-1}:1\leq i\leq t^{\prime}-1,\,\ 1\leq j\leq t\}\cup\{u_{3l-2},u_{3l-1}:1\leq l\leq t^{\prime}\}.$

In all these possibilities, it is not difficult to see that $D_{2,p}$ is a $2$-paired dominating set. Further, the cardinality of $D_{2,p}$ in each case is $2\lceil\frac{nm-3n}{6}\rceil$. Hence

[TABLE]

Now we give the lower bound of $2$-distance paired dominating set.

Let $D_{2,p}=\{x_{i},y_{i}:1\leq i\leq q\}$ be a paired dominating set. According to Lemma 2.1, $D_{2,p}$ contain at least $\lceil\frac{m-6}{6}\rceil$ vertices from each $V_{i}$ of $C_{i,m}$. If $m\equiv 4,$ $(\textrm{mod}\ 6)$, then $\lceil\frac{m-6}{6}\rceil$ pair of vertices dominate $m-5$ vertices from each $V_{i}$ of $C_{i,m}$, $where1\leq i\leq n.$ The number of vertices which are yet to be dominated in each $C_{i}$ are $5$, from which $2$ vertices are of degree $4$ and other vertices of degree $2$. Suppose that these vertices are $v_{i,1},v_{i,2},v_{i,m-1},u_{i}$ and $u_{i+1}$. To dominate these vertices either $v_{i,1},v_{i,m-2}\in D_{2,p}$ or $u_{i},u_{i+1}\in D_{2,p}$. In both these cases, each $C_{i,m}$ has at least two vertices from each $C_{i}$ belong to $D_{2,p}$. Since, each vertex of degree $4$ belong to neighboring cycle, therefore each vertex of degree $4$ belong to $D_{2,p}$. Further, $\langle D_{2,p}\rangle$ has perfect matching. Thus only edges that are not adjacent to each other can belong to $D_{2,p}$. Thus we choose pair of vertices of type $(4,4)$ from $3$ consecutive copies of $C_{i}$. Since $\langle D_{2,p}\rangle$ is a paired dominating set and has perfect matching, so it contains only the non adjacent edges. This implies that

[TABLE]

which implies that $q\geq\lceil\frac{nm-3n}{6}\rceil$. Therefore

[TABLE]

From Equation $19$ and $20$, it is clear that

[TABLE]

In Figure $2$, the vertices (dark) of $2$-distance paired dominating set of the graph $f{n\times m}$ are shown.