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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 .
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Figure 2. A 2RDF of P 3 P n .
Figure 2. A 2RDF of P 3 P n .
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