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Mathematics 2019, 7(2), 203; https://doi.org/10.3390/math7020203

Article
k-Rainbow Domination Number of P3Pn
1
Department of network technology, South China Institute of Software Engineering, Guangzhou 510990, China
2
Institute of Computing Science and Technology, Guangzhou University, Guangzhou 510006, China
3
South China Business College, Guang Dong University of Foreign Studies, Guangzhou 510545, China
4
Department of Mathematics and Computer Science, Sirjan University of Technology, Sirjan 7813733385, Iran
5
Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz 5375171379, Iran
6
School of Mathematics and Physics, Anhui Jianzhu University, Hefei 230601, China
*
Author to whom correspondence should be addressed.
Received: 12 January 2019 / Accepted: 19 February 2019 / Published: 21 February 2019

## Abstract

:
Let k be a positive integer, and set $[ k ] : = { 1 , 2 , … , k }$. For a graph G, a k-rainbow dominating function (or kRDF) of G is a mapping $f :$$V ( G ) → 2 [ k ]$ in such a way that, for any vertex $v ∈ V ( G )$ with the empty set under f, the condition $⋃ u ∈ N G ( v ) f ( u ) = [ k ]$ always holds, where $N G ( v )$ is the open neighborhood of v. The weight of kRDF f of G is the summation of values of all vertices under f. The k-rainbow domination number of G, denoted by $γ r k ( G )$, is the minimum weight of a kRDF of G. In this paper, we obtain the k-rainbow domination number of grid $P 3 □ P n$ for $k ∈ { 2 , 3 , 4 }$.
Keywords:
k-rainbow dominating function; k-rainbow domination number; grids

## 1. Introduction

For a graph G, we denote by $V ( G )$ and $E ( G )$ the vertex set and the edge set of G, respectively. For a vertex $v ∈ V ( G )$, the open neighborhood of v, denoted by $N G ( v )$, is the set ${ u ∈ V ( G ) : u v ∈ E ( G ) }$ and the closed neighborhood of v, denoted by $N G [ v ]$, is the set $N G ( v ) ∪ { v }$. The degree of a vertex $v ∈ V ( G )$, denoted by $d G ( v )$, is defined by $d G ( v ) = | N G ( v ) |$. We let $δ ( G )$ and $Δ ( G )$ denote the minimum degree and maximum degree of a graph G, respectively.
Let k be a positive integer, and $[ k ] : = { 1 , 2 , … , k }$. For a graph G, a k-rainbow dominating function (or kRDF) of G is a mapping $f :$$V ( G ) → 2 [ k ]$ in such a way that for any vertex $v ∈ V ( G )$ with the empty set under f, the condition $⋃ u ∈ N G ( v ) f ( u ) = [ k ]$ always holds. The weight of a kRDF f of G is the value $ω ( f ) : = ∑ v ∈ V ( G ) | f ( v ) |$. The k-rainbow domination number of G, denoted by $γ r k ( G )$, is the minimum weight of a kRDF of G. A kRDF f of G is a $γ r k$-function if $ω ( f ) = γ r k ( G )$. The k-rainbow domination number was introduced by Brešar, Henning, and Rall [1] was studied by several authors (see, for example [2,3,4,5,6,7,8,9,10,11,12,13,14,15]).
For graphs F and G, we let $F □ G$ denote the Cartesian product of F and G. Vizing [16] conjectured that for arbitrary graphs F and G, $γ ( F □ G ) ≥ γ ( F ) γ ( G )$. This conjecture is still open, and the domination number or its related invariants of $F □ G$ are extensively studied with the motivation from Vizing’s conjecture.
Concerning the k-rainbow domination number of $F □ G$, one problem naturally arises: Given two graphs F and G under some conditions, determine $γ r k ( F □ G )$ for all k. In [3], the authors determined $γ r k ( P 2 □ P n )$ for $k = 3 , 4 , 5$.
In this paper, we examine grid graphs $P 3 □ P n$, and determine the value $γ r k ( P 3 □ P n )$ for $k ∈ { 2 , 3 , 4 }$ and all n, where $P m$ is the path of order m.

## 2. 2-Rainbow Domination Number of $P 3 □ P n$

We write $V ( P 3 □ P n ) = { v i , u i , w i ∣ 0 ≤ i ≤ n − 1 }$ and let $E ( P 3 □ P n ) = { v i u i , u i w i ∣ 0 ≤ i ≤ n − 1 } ∪ { v i v i + 1 , u i u i + 1 , w i w i + 1 ∣ 0 ≤ i ≤ n − 1 }$ (see Figure 1). A 2RDF f is given in three lines, where in the first line there are values of the function f for vertices ${ v 0 , v 1 , … , v n − 1 }$, in the second line of the vertices ${ u 0 , u 1 , … , u n − 1 }$, and in the third line of the vertices ${ w 0 , w 1 , … , w n − 1 }$ (see Figure 2). Furthermore, we use $0 , 1 , 2 , 3$ to encode the sets $∅ , { 1 } , { 2 } , { 1 , 2 }$.
To provide a complete answer, we need the following fact that can easily be proved as an exercise.
Fact 1.
$γ r 2 ( P 3 □ P 3 ) = 4 , γ r 2 ( P 3 □ P 4 ) = 6 , γ r 2 ( P 3 □ P 5 ) = 7 , γ r 2 ( P 3 □ P 6 ) = 8 , γ r 2 ( P 3 □ P 7 ) = 10 .$
Theorem 1.
For $n ≥ 8$, $γ r 2 ( P 3 □ P n ) = 5 n + 3 4$.
Proof.
First, we present constructions of a 2RDF of $P 3 □ P n$ of the desired weight.
• $n ≡ 0 ( mod 8 )$:
$0200 30010200 ⋯ 30010200 3001$
$1010 00200010 ⋯ 00200010 0020$
$0202 01003002 ⋯ 01003002 0101$
• $n ≡ 1 ( mod 8 )$:
$0200 30010200 ⋯ 30010200 30010$
$1010 00200010 ⋯ 00200010 00202$
$0202 01003002 ⋯ 01003002 01010$
• $n ≡ 2 ( mod 8 )$:
$0200 30010200 ⋯ 30010200 300101$
$1010 00200010 ⋯ 00200010 002020$
$0202 01003002 ⋯ 01003002 010101$
• $n ≡ 3 ( mod 8 )$:
$0200 30010200 ⋯ 30010200 3001001$
$1010 00200010 ⋯ 00200010 0020220$
$0202 01003002 ⋯ 01003002 0101001$
• $n ≡ 4 ( mod 8 )$:
$0200 30010200 ⋯ 30010200 30010010$
$1010 00200010 ⋯ 00200010 00202202$
$0202 01003002 ⋯ 01003002 01010010$
• $n ≡ 5 ( mod 8 )$:
$0200 30010200 ⋯ 30010200 300102020$
$1010 00200010 ⋯ 00200010 002000101$
$0202 01003002 ⋯ 01003002 010030020$
• $n ≡ 6 ( mod 8 )$:
$0200 30010200 ⋯ 30010200 30$
$1010 00200010 ⋯ 00200010 01$
$0202 01003002 ⋯ 01003002 01$
• $n ≡ 7 ( mod 8 )$:
$0200 30010200 ⋯ 30010200 301$
$1010 00200010 ⋯ 00200010 002$
$0202 01003002 ⋯ 01003002 010$
To show that these are also lower bounds, we prove there is a $γ r 2 ( P 3 □ P n )$-function, f such that for every $0 ≤ i ≤ n − 1$, $ω ( f i ) = | f ( v i ) | + | f ( u i ) | + | f ( w i ) | ≥ 1$. Let $n ≥ 8$ and f be a $γ r 2 ( P 3 □ P n )$-function such that the cardinality of $S = { i ∣ 0 ≤ i ≤ n − 1 and ω ( f i ) = 0 }$ is as small as possible. We claim that $| S | = 0$. Suppose, to the contrary, that $| S | ≥ 1$ and let s be the smallest positive integer for which $ω ( f s ) = 0$. Then, $ω ( f s − 1 ) + ω ( f s + 1 ) ≥ 6$. Then, we consider the following cases.
Case 1.
$s = 1$ (the case $s = n − 1$ is similar).
Then, we have $f ( v 1 ) = f ( u 1 ) = f ( w 1 ) = { 1 , 2 }$ and the function g defined by $g ( u 0 ) = { 1 }$, $g ( v 1 ) = g ( w 1 ) = { 2 }$, $g ( u 2 ) = f ( u 2 ) ∪ { 1 }$, $g ( v 0 ) = g ( w 0 ) = g ( u 1 ) = ∅$ and $g ( x ) = f ( x )$ otherwise, is a 2RDF of $P 3 □ P n$ of weight at most $ω ( f )$, which contradicts the choice of f.
Case 2.
$s = 1$ ($s = n − 2$ is similar).
Then, $ω ( f 0 ) + ω ( f 2 ) ≥ 6$ and the function g defined by $g ( u 0 ) = g ( u 2 ) = { 1 }$, $g ( v 1 ) = g ( w 1 ) = { 2 }$, $g ( v 3 ) = f ( v 3 ) ∪ { 2 }$, $g ( w 3 ) = f ( w 3 ) ∪ { 2 }$, $g ( v 0 ) = g ( w 0 ) = g ( u 1 ) = g ( v 2 ) = g ( w 2 ) = ∅$ and $g ( x ) = f ( x )$ otherwise, is an 2RDF of $P 3 □ P n$ of weight at most $ω ( f )$, which contradicts the choice of f.
Case 3.
$2 ≤ s ≤ n − 3$.
Since $ω ( f s − 2 ) ≥ 1$, then $| f ( v s − 2 ) | + | f ( u s − 2 ) | + | f ( w s − 2 ) | ≥ 1$. First, let $| f ( u s − 2 ) | ≥ 1$. We may assume that ${ 1 } ⊆ f ( u s − 2 )$. It is easy to see that the function g defined by $g ( v s − 1 ) = g ( v s + 1 ) = g ( w s − 1 ) = g ( w s + 1 ) = { 2 }$, $g ( u s ) = { 1 }$, $g ( u s + 2 ) = f ( u s + 2 ) ∪ { 1 }$, $g ( u s − 1 ) = g ( v s ) = g ( w s ) = g ( u s + 1 ) = ∅$ and $g ( x ) = f ( x )$ otherwise, is an 2RDF of $P 3 □ P n$ of weight at most $ω ( f )$, which contradicts the choice of f. Now, let $| f ( w s − 2 ) | ≥ 1$ ($| f ( v s − 2 ) | ≥ 1$ is similar). We may assume that ${ 1 } ⊆ f ( w s − 2 )$. Hence, the function g defined by $g ( v s − 2 ) = f ( v s − 2 ) ∪ { 1 }$, $g ( v s + 1 ) = g ( u s − 1 ) = g ( w s + 1 ) = { 2 }$, $g ( u s ) = { 1 }$, $g ( u s + 2 ) = f ( u s + 2 ) ∪ { 1 }$, $g ( u s − 1 ) = g ( v s ) = g ( w s ) = g ( w s − 1 ) = g ( u s + 1 ) = ∅$ and $g ( x ) = f ( x )$ otherwise, is an 2RDF of $P 3 □ P n$ of weight $ω ( f )$, which is contradicting the choice of f. Therefore, $| S | = 0$.
We can see that for every $0 ≤ i ≤ n − 2$, if $ω ( f i ) = ω ( f i + 1 ) = ω ( f i + 2 ) = 1$, then $ω ( f i − 1 ) , ω ( f i + 3 ) > 1$. In addition, there is the function f such that, if $ω ( f 0 ) = 1$ ($ω ( f n − 1 ) = 1$ is similar), then $ω ( f 1 ) > 1$ and $ω ( f 1 ) + ω ( f 2 ) + ω ( f 3 ) + ω ( f 4 ) ≥ 6$ and if $ω ( f 0 ) = 2$ ($ω ( f n − 1 ) = 2$ is similar), then $ω ( f 0 ) + ω ( f 1 ) + ω ( f 2 ) + ω ( f 3 ) ≥ 6$.
If $ω ( f 0 ) = 1$ and $ω ( f n − 1 ) = 1$, then
$4 ω ( f ) = 4 ∑ 0 ≤ i ≤ n − 1 ω ( f i ) = [ 3 ω ( f 0 ) + 2 ω ( f 1 ) + ω ( f 2 ) ] + [ 3 ω ( f n − 1 ) + 2 ω ( f n − 2 ) + ω ( f n − 3 ) ] + ∑ i ∈ { 0 , … , n − 4 } − { 1 , n − 5 } ω ( f i ) + ω ( f i + 1 ) + ω ( f i + 2 ) + ω ( f i + 3 ) + [ ω ( f 1 ) + ω ( f 2 ) + ω ( f 3 ) + ω ( f 4 ) ] + [ ω ( f n − 5 ) + ω ( f n − 4 ) + ω ( f n − 3 ) + ω ( f n − 2 ) ] ≥ 8 + 8 + 5 ( n − 5 ) + 12 = 5 ( n − 3 ) + 18 .$
If $ω ( f 0 ) = 1$ and $ω ( f n − 1 ) = 2$, then
$4 ω ( f ) = 4 ∑ 0 ≤ i ≤ n − 1 ω ( f i ) = [ 3 ω ( f 0 ) + 2 ω ( f 1 ) + ω ( f 2 ) ] + [ 3 ω ( f n − 1 ) + 2 ω ( f n − 2 ) + ω ( f n − 3 ) ] + ∑ i ∈ { 0 , … , n − 4 } − { 1 } ω ( f i ) + ω ( f i + 1 ) + ω ( f i + 2 ) + ω ( f i + 3 ) + [ ω ( f 1 ) + ω ( f 2 ) + ω ( f 3 ) + ω ( f 4 ) ] ≥ 8 + 9 + 5 ( n − 4 ) + 6 = 5 ( n − 3 ) + 18 .$
If $ω ( f 0 ) = 2$ and $ω ( f n − 1 ) = 2$, then
$4 ω ( f ) = 4 ∑ 0 ≤ i ≤ n − 1 ω ( f i ) = [ 3 ω ( f 0 ) + 2 ω ( f 1 ) + ω ( f 2 ) ] + [ 3 ω ( f n − 1 ) + 2 ω ( f n − 2 ) + ω ( f n − 3 ) ] + ∑ i ∈ { 0 , … , n − 4 } − { 1 } [ ω ( f i ) + ω ( f i + 1 ) + ω ( f i + 2 ) + ω ( f i + 3 ) ] + [ ω ( f 1 ) + ω ( f 2 ) + ω ( f 3 ) + ω ( f 4 ) ] ≥ 9 + 9 + 5 ( n − 3 ) = 5 ( n − 3 ) + 18 .$
Thus, $ω ( f ) = 5 n + 3 4$. □

## 3. 3-Rainbow Domination Number of $P 3 □ P n$

As in the previous section, a 3RDF is given in three lines and we use $0 , 1 , 2 , 3$ to encode the sets $∅ , { 1 } , { 2 } , { 3 }$.
To provide a complete answer, we need the following fact.
Fact 2.
$γ r 3 ( P 3 □ P 3 ) = 5 , γ r 3 ( P 3 □ P 4 ) = 8 .$
Theorem 2.
For $n ≥ 5$,
$γ r 3 ( P 3 □ P n ) = ( 3 n + 1 ) / 2 if n ≡ 1 ( mode 2 ) , ( 3 n + 2 ) / 2 if n ≡ 0 ( mode 2 ) ,$
Proof.
First, we present constructions of a 3RDF of $P 3 □ P n$ of the desired weight.
• $n ≡ 0 ( mod 4 )$:
$2010 ⋯ 2010 2201$
$0303 ⋯ 0303 0030$
$1020 ⋯ 1020 1102$
• $n ≡ 1 ( mod 4 )$:
$2010 ⋯ 2010 2$
$0303 ⋯ 0303 0$
$1020 ⋯ 1020 1$
• $n ≡ 2 ( mod 4 )$:
$2010 ⋯ 2010 201201$
$0303 ⋯ 0303 030030$
$1020 ⋯ 1020 102102$
• $n ≡ 3 ( mod 4 )$:
$2010 ⋯ 2010 201$
$0303 ⋯ 0303 030$
$1020 ⋯ 1020 102$
To show that these are also lower bounds, we prove there is a $γ r 3 ( P 3 □ P n )$-function, f that satisfies the following conditions:
• For every $0 ≤ i ≤ n − 1$, $ω ( f i ) = | f ( v i ) | + | f ( u i ) | + | f ( w i ) | ≥ 1$,
• For every $1 ≤ i ≤ n − 2$, if $ω ( f i ) = 1$, then $ω ( f i − 1 ) + ω ( f i + 1 ) ≥ 4$. In particular, if $ω ( f i ) = 1$, then $( ω ( f i − 1 ) + ω ( f i ) ) + ( ω ( f i ) + ω ( f i + 1 ) ) ≥ 6$,
• $ω ( f 0 ) ≥ 2$ and $ω ( f n − 1 ) ≥ 2$.
First, we show that for every $γ r 3 ( P 3 □ P n )$-function f, $ω ( f i ) = | f ( v i ) | + | f ( u i ) | + | f ( w i ) | ≥ 1$ when $0 ≤ i ≤ n − 1$. Let $n ≥ 5$ and f be a $γ r 3 ( P 3 □ P n )$-function and $S = { i ∣ 0 ≤ i ≤ n − 1 and ω ( f i ) = 0 }$. We claim that $| S | = 0$. Assume to the contrary that $| S | ≥ 1$. Then, we consider the following cases.
Case 1.
$0 ∈ S$ (the case $n − 1 ∈ S$ is similar).
Then, we have $f ( v 1 ) = f ( u 1 ) = f ( w 1 ) = { 1 , 2 , 3 }$ and it is easy to see that the function g defined by $g ( v 0 ) = { 1 }$, $g ( u 1 ) = { 3 }$, $g ( w 0 ) = { 2 }$, $g ( v 2 ) = f ( v 2 ) ∪ { 2 }$, $g ( w 2 ) = f ( w 2 ) ∪ { 1 }$, $g ( u 0 ) = g ( v 1 ) = g ( w 1 ) = ∅$ and $g ( x ) = f ( x )$ otherwise, is an 3RDF of $P 3 □ P n$ of weight less than $ω ( f )$, which is a contradiction.
Let s be the smallest positive integer for which $ω ( f s ) = 0$. Then, $s ≥ 1$ and $ω ( f s − 1 ) + ω ( f s + 1 ) ≥ 9$.
Case 2.
$s = 1$ ($s = n − 2$ is similar).
Then, the function g defined by $g ( v 0 ) = g ( u 0 ) = g ( w 0 ) = { 1 }$, $g ( v 1 ) = { 2 }$, $g ( w 1 ) = { 1 }$, $g ( u 2 ) = { 3 }$, $g ( v 3 ) = f ( v 3 ) ∪ { 1 }$, $g ( w 3 ) = f ( w 3 ) ∪ { 2 }$, $g ( u 1 ) = g ( v 2 ) = g ( w 2 ) = ∅$ and $g ( x ) = f ( x )$ otherwise, is an 3RDF of $P 3 □ P n$ of weight less than $ω ( f )$, which is a contradiction.
Case 3.
$2 ≤ s ≤ n − 3$.
The function g defined by $g ( u s − 1 ) = g ( u s + 1 ) = { 3 }$, $g ( v s ) = { 2 }$, $g ( w s ) = { 1 }$, $g ( v s − 2 ) = f ( v s − 2 ) ∪ { 1 }$, $g ( w s − 2 ) = f ( w s − 2 ) ∪ { 2 }$, $g ( v s + 2 ) = f ( v s + 2 ) ∪ { 1 }$, $g ( w s + 2 ) = f ( w s + 2 ) ∪ { 2 }$, $g ( v s − 1 ) = g ( v s + 1 ) = g ( u s ) = g ( w s − 1 ) = g ( w s + 1 ) = ∅$ and $g ( x ) = f ( x )$ otherwise, is an 3RDF of $P 3 □ P n$ of weight less than $ω ( f )$, which is a contradiction. Therefore, $| S | = 0$.
Now, let f be a $γ r 3 ( P 3 □ P n )$-function. It is easy to see that, if $ω ( f i ) = 1$, then $ω ( f i − 1 ) + ω ( f i + 1 ) ≥ 4$ when $1 ≤ i ≤ n − 2$.
Finally, we show that there is $γ r 3 ( P 3 □ P n )$-function f such that $ω ( f 0 ) ≥ 2$ ( $ω ( f n − 1 ) ≥ 2$ is similar). Let f be a $γ r 3 ( P 3 □ P n )$-function such that $ω ( f 0 ) = 1$. If $| f ( v 0 ) | = 1$ ($| f ( w 0 ) | = 1$ is similar), then $| f ( w 0 ) | = | f ( u 0 ) | = 0$, $| f ( u 1 ) | ≥ 2$ and $| f ( w 1 ) | = 3$. We may assume that ${ 1 , 2 } ⊆ f ( u 1 )$. It is easy to see that the function g defined by $g ( w 0 ) = { 3 }$, $g ( w 2 ) = { 3 }$, $g ( w 1 ) = ∅$ and $g ( x ) = f ( x )$ otherwise, is an 3RDF of $P 3 □ P n$ of weight less than $ω ( f )$, which is a contradiction. Now, let $| f ( u 0 ) | = 1$. Then, $| f ( w 0 ) | = | f ( v 0 ) | = 0$, $| f ( v 1 ) | ≥ 2$ and $| f ( w 1 ) | ≥ 2$. It is easy to see that the function g defined by $g ( w 0 ) = { 1 }$, $g ( w 2 ) = { 2 }$, $g ( u 1 ) = { 3 }$, $g ( v 2 ) = f ( v 2 ) ∪ { 1 }$, $g ( w 2 ) = f ( w 2 ) ∪ { 2 }$, $g ( u 1 ) = g ( u 2 ) = ∅$ and $g ( x ) = f ( x )$ otherwise, is an 3RDF of $P 3 □ P n$ of weight $ω ( f )$.
Hence, there is a $γ r 3 ( P 3 □ P n )$-function, f that satisfies the following conditions:
• For every $0 ≤ i ≤ n − 1$, $ω ( f i ) ≥ 1$;
• For every $1 ≤ i ≤ n − 2$, if $ω ( f i ) = 1$, then $ω ( f i − 1 ) + ω ( f i + 1 ) ≥ 4$; and
• $ω ( f 0 ) ≥ 2$ and $ω ( f n − 1 ) ≥ 2$.
If n is odd, then
$2 ω ( f ) = 2 ∑ 0 ≤ i ≤ n − 1 ω ( f i ) = ω ( f 0 ) + ω ( f n − 1 ) + ∑ 0 ≤ i ≤ n − 2 ( ω ( f i ) + ω ( f i + 1 ) ) ≥ 4 + 3 ( n − 1 ) .$
Then, $ω ( f ) = 3 n + 1 2$ when n is odd. Now, let n is even. Then, there is $s ≠ n − 1$ such that $ω ( f s ) + ω ( f s + 1 ) ≥ 4$. Hence,
$2 ω ( f ) = 2 ∑ 0 ≤ i ≤ n − 1 ω ( f i ) = ω ( f s ) + ω ( f s + 1 ) + ω ( f 0 ) + ω ( f n − 1 ) + ∑ 0 ≤ i ≤ n − 2 , i ≠ s ( ω ( f i ) + ω ( f i + 1 ) ) ≥ 8 + 3 ( n − 2 ) .$
Therefore, $ω ( f ) = 3 n + 2 2$ when n is even. □

## 4. 4-Rainbow Domination Number of $P 3 □ P n$

As above, a 4RDF is given in three lines and we use $0 , 1 , 2 , 5$ to encode the sets $∅ , { 1 } , { 2 } , { 3 , 4 }$.
To provide a complete answer, we need the following fact.
Fact 3.
$γ r 4 ( P 3 □ P 3 ) = 6 , γ r 4 ( P 3 □ P 4 ) = 9 .$
Theorem 3.
For $n ≥ 5$, $γ r 4 ( P 3 □ P n ) = 2 n$.
Proof.
First, we show that $γ r 4 ( P 3 □ P n ) ≤ 2 n$. To do this, we present constructions of a 4RDF of $P 3 □ P n$ of the desired weight.
• $n ≡ 0 ( mod 4 )$:
$2010 ⋯ 2010 2201$
$0505 ⋯ 0505 0050$
$1020 ⋯ 1020 1102$
• $n ≡ 1 ( mod 4 )$:
$2010 ⋯ 2010 2$
$0505 ⋯ 0505 0$
$1020 ⋯ 1020 1$
• $n ≡ 2 ( mod 4 )$:
$2010 ⋯ 2010 201201$
$0505 ⋯ 0505 050050$
$1020 ⋯ 1020 102102$
• $n ≡ 3 ( mod 4 )$:
$2010 ⋯ 2010 201$
$0505 ⋯ 0505 050$
$1020 ⋯ 1020 102$
To prove the inverse inequality, we show that every $γ r 4 ( P 3 □ P n )$-function f satisfies the following conditions:
• For every $0 ≤ i ≤ n − 1$, $ω ( f i ) = | f ( v i ) | + | f ( u i ) | + | f ( w i ) | ≥ 1$;
• For every $1 ≤ i ≤ n − 2$, if $ω ( f i ) = 1$, then $ω ( f i − 1 ) + ω ( f i + 1 ) ≥ 6$; and
• $ω ( f 0 ) ≥ 2$ and $ω ( f n − 1 ) ≥ 2$.
First, we show that for every $γ r 4 ( P 3 □ P n )$-function f, $ω ( f i ) = | f ( v i ) | + | f ( u i ) | + | f ( w i ) | ≥ 1$ when $0 ≤ i ≤ n − 1$. Let $n ≥ 5$ and f be a $γ r 4 ( P 3 □ P n )$-function and $S = { i ∣ 0 ≤ i ≤ n − 1 and ω ( f i ) = 0 }$. We claim that $| S | = 0$. Assume to the contrary that $| S | ≥ 1$. Then, we consider the following cases.
Case 1.
$0 ∈ S$ (the case $n − 1 ∈ S$ is similar).
Then, we have $f ( v 1 ) = f ( u 1 ) = f ( w 1 ) = { 1 , 2 , 3 , 4 }$ and the function g defined by $g ( v 0 ) = { 1 }$, $g ( u 1 ) = { 3 , 4 }$, $g ( w 0 ) = { 2 }$, $g ( v 2 ) = f ( v 2 ) ∪ { 2 }$, $g ( w 2 ) = f ( w 2 ) ∪ { 1 }$, $g ( u 0 ) = g ( v 1 ) = g ( w 1 ) = ∅$ and $g ( x ) = f ( x )$ otherwise, is an 4RDF of $P 3 □ P n$ of weight less than $ω ( f )$, which is a contradiction.
Let $ω ( f s ) = 0$. Then, $s ≥ 1$ and $ω ( f s − 1 ) + ω ( f s + 1 ) ≥ 12$.
Case 2.
$s = 1$ ($s = n − 2$ is similar).
The function g defined by $g ( v 0 ) = g ( u 0 ) = g ( w 0 ) = { 1 }$, $g ( v 1 ) = { 2 }$, $g ( w 1 ) = { 1 }$, $g ( u 2 ) = { 3 , 4 }$, $g ( v 3 ) = f ( v 3 ) ∪ { 1 }$, $g ( w 3 ) = f ( w 3 ) ∪ { 2 }$, $g ( u 1 ) = g ( v 2 ) = g ( w 2 ) = ∅$ and $g ( x ) = f ( x )$ otherwise, is an 4RDF of $P 3 □ P n$ of weight less than $ω ( f )$, which is a contradiction.
Case 3.
$2 ≤ s ≤ n − 3$.
Then, it is easy to see that the function g defined by $g ( u s − 1 ) = g ( u s + 1 ) = { 3 , 4 }$, $g ( v s ) = { 2 }$, $g ( w s ) = { 1 }$, $g ( v s − 2 ) = f ( v s − 2 ) ∪ { 1 }$, $g ( w s − 2 ) = f ( w s − 2 ) ∪ { 2 }$, $g ( v s + 2 ) = f ( v s + 2 ) ∪ { 1 }$, $g ( w s + 2 ) = f ( w s + 2 ) ∪ { 2 }$, $g ( v s − 1 ) = g ( v s + 1 ) = g ( u s ) = g ( w s − 1 ) = g ( w s + 1 ) = ∅$ and $g ( x ) = f ( x )$ otherwise, is an 4RDF of $P 3 □ P n$ of weight less than $ω ( f )$, which is a contradiction. Therefore, $| S | = 0$.
Now, let f be a $γ r 4 ( P 3 □ P n )$-function. It is easy to see that, if $ω ( f i ) = 1$, then $ω ( f i − 1 ) + ω ( f i + 1 ) ≥ 6$ when $1 ≤ i ≤ n − 2$.
We show that for every $γ r 4 ( P 3 □ P n )$-function f $ω ( f 0 ) ≥ 2$ ($ω ( f n − 1 ) ≥ 2$ is similar). Let f be a $γ r 4 ( P 3 □ P n )$-function such that $ω ( f 0 ) = 1$. If $| f ( v 0 ) | = 1$ ($| f ( w 0 ) | = 1$ is similar), then $| f ( w 0 ) | = | f ( u 0 ) | = 0$, $| f ( u 1 ) | ≥ 3$ and $| f ( w 1 ) | = 4$. We may assume that ${ 1 , 2 , 3 } ⊆ f ( u 1 )$. The function g defined by $g ( w 0 ) = { 4 }$, $g ( w 2 ) = { 4 }$, $g ( w 1 ) = ∅$ and $g ( x ) = f ( x )$ otherwise, is an 4RDF of $P 3 □ P n$ of weight less than $ω ( f )$, which is a contradiction. Now, let $| f ( u 0 ) | = 1$. Then, $| f ( w 0 ) | = | f ( v 0 ) | = 0$, $| f ( v 1 ) | ≥ 3$ and $| f ( w 1 ) | ≥ 3$. The function g defined by $g ( w 0 ) = { 1 }$, $g ( w 2 ) = { 2 }$, $g ( u 1 ) = { 3 , 4 }$, $g ( v 2 ) = f ( v 2 ) ∪ { 1 }$, $g ( w 2 ) = f ( w 2 ) ∪ { 2 }$, $g ( u 1 ) = g ( u 2 ) = ∅$ and $g ( x ) = f ( x )$ otherwise, is an 4RDF of $P 3 □ P n$ of weight less than $ω ( f )$, which is a contradiction.
Hence, every $γ r 4 ( P 3 □ P n )$-function f satisfies the following conditions:
• For every $0 ≤ i ≤ n − 1$, $ω ( f i ) ≥ 1$;
• For every $1 ≤ i ≤ n − 2$, if $ω ( f i ) = 1$, then $ω ( f i − 1 ) + ω ( f i + 1 ) ≥ 6$. In particular $( ω ( f i − 1 ) + ω ( f i ) ) + ( ω ( f i ) + ω ( f i + 1 ) ) ≥ 8$; and
• $ω ( f 0 ) ≥ 2$ and $ω ( f n − 1 ) ≥ 2$.
Hence,
$2 ω ( f ) = 2 ∑ 0 ≤ i ≤ n − 1 ω ( f i ) = ∑ 0 ≤ i ≤ n − 2 ( ω ( f i ) + ω ( f i + 1 ) ) + ω ( f 0 ) + ω ( f n − 1 ) ≥ 4 ( n − 1 ) + 4 .$
Hence, $ω ( f ) = 2 n$. □

## Author Contributions

R.K. contributes for supervision, methodology, validation, project administration and formal analysing. N.D., J.A., Y.W., J.-B.L. contribute for resources, some computations and wrote the initial draft of the paper which were investigated and approved by Y.W., X.W., J.-B.L., and J.A. wrote the final draft.

## Funding

This research was funded by the National Natural Science Foundation of China (Grant No. 11701118), Guangdong Provincial Engineering and Technology Research Center ([2015]1487), Guangdong Provincial Key Platform and Major Scientific Research Projects (Grant No. 2016KQNCX238), Key Supported Disciplines of Guizhou Province - Computer Application Technology (Grant No. QianXueWeiHeZi ZDXK [2016]20), and the Specialized Fund for Science and Technology Platform and Talent Team Project of Guizhou Province (Grant No. QianKeHePingTaiRenCai [2016]5609), the China Postdoctoral Science Foundation under Grant 2017M621579; the Postdoctoral Science Foundation of Jiangsu Province under Grant 1701081B; Project of Anhui Jianzhu University under Grant no. 2016QD116 and 2017dc03.

## Conflicts of Interest

The author declares no conflict of interest.

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Figure 1. The grid graph $P 3 □ P 16$.
Figure 1. The grid graph $P 3 □ P 16$.
Figure 2. A 2RDF of $P 3 □ P n$.
Figure 2. A 2RDF of $P 3 □ P n$.