OJDMOpen Journal of Discrete Mathematics2161-7635Scientific Research Publishing10.4236/ojdm.2021.113005OJDM-109708ArticlesPhysics&Mathematics A Note on <i>n</i>-Set Distance-Labelings of Graphs RogerK. Yeh1*Department of Applied Mathematics, Feng Chia University, Taiwan08062021110355602, March 20215, June 2021 8, June 2021© Copyright 2014 by authors and Scientific Research Publishing Inc. 2014This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

This note is considered as a sequel of Yeh . Here, we present a generalized (vertex) distance labeling (labeling vertices under constraints depending the on distance between vertices) of a graph. Instead of assigning a number (label) to each vertex, we assign a set of numbers to each vertex under given conditions. Some basic results are given in the first part of the note. Then we study a particular class of this type of labelings on several classes of graphs.

Graph Distance Labeling
1. Introduction

Inspired by a channel assignment problem proposed by Roberts  in 1988, Griggs and Yeh  formulated the L(2,1)-labeling problem for graphs. There are considerable amounts of articles studying this labeling and its generalizations or related problems. Readers can see  or  for a survey. In this note, we like to consider a generalization of the L(2,1)-labeling. Let A and B be two subsets of natural numbers. Define ‖ A − B ‖ = min { | a − b | : a ∈ A , b ∈ B } . Denote the set

[ k ] = { 0 , 1 , ⋯ , k } and ( [ k ] n ) the collection of alln-subsets of [ k ] .

Motivated by the article , we propose the following labeling on a graph.

Let G = ( V , E ) be a graph and n be a positive integer. Given non-negative integers δ 1 ≥ δ 2 an L ( n ) ( δ 1 , δ 2 ) -labeling is a function f : V ( G ) → ( [ k ] n ) for

some k ≥ 1 such that | f ( u ) − f ( v ) | ≥ δ i whenever the distance between u andv inG isi, for i = 1 , 2 . (The minimum value and the maximum value of ∪ v ∈ V ( G ) f ( v ) is 0 and k, respectively.) The value kis called the span of f. The smallest k so that there is an L ( n ) ( δ 1 , δ 2 ) -labeling f with span k, is denoted by λ ( n ) ( G ; δ 1 , δ 2 ) and called the L ( n ) ( δ 1 , δ 2 ) -labeling number of G. An L ( n ) ( δ 1 , δ 2 ) -labeling with span λ ( n ) ( G ; δ 1 , δ 2 ) is called an optimal L ( n ) ( δ 1 , δ 2 ) -labeling. If n = 1 then notations L ( 1 ) and λ ( 1 ) will be simplified as L and λ , respectively.

Note: 1) The elements in [ k ] are called “numbers” and f ( u ) is called the “label” of u. So, a label is a set in this problem. 2) Using our notation, the labeling in  is the L ( δ 1 , 0 ) -labeling for δ 1 ≥ 1 .

Previously, we have studied the L ( 2 ) ( 2 , 1 ) -labeling problem (cf.  ). In this note, we will first investigate properties of the L ( n ) ( δ 1 , δ 2 ) for n ≥ 1 . Then, we study the case of ( δ 1 , δ 2 ) = ( 1 , 1 ) .

2. Preliminarily

Let G be a graph and n an positive integer. Now, we construct a new graph G ( n ) by replacing each vertex v in G by n vertices v i , 1 ≤ i ≤ n and u i is adjacent to v j for all i , j , in G ( n ) , whenever u and v is adjacent in G. That is, u i and v j , for all i , j , induces a complete bipartite graph K n , n . Note that G ( 1 ) = G .

It is easy to verify that λ ( n ) ( G ; δ 1 , 1 ) = λ ( G ( n ) ; δ 1 , 1 ) . Thus, for example, λ ( n ) ( K m ; 2 , 1 ) = λ ( K n , n , ⋯ , n ; 2 , 1 ) = n m + m − 2 , where m ≥ 2 , by previous result on complete m-partite graph K n , n , ⋯ , n (cf.  ).

Next, we consider the relation between the labeling numbers for n = 1 and n ≥ 1 . In the following, λ ( G ; 1 , 1 ) and λ ( n ) ( G ; 1 , 1 ) are denoted by λ 1 ( G ) and λ 1 ( n ) ( G ) , respectively, for short.

Proposition 2.1. Let n ≥ 1 , δ 1 ≥ δ 2 be nonnegative integers and Δ be the maximum degree of G. Then

1) ( n − 1 ) ( Δ + 1 ) + δ 1 + ( Δ − 1 ) δ 2 ≤ λ ( n ) ( G ; δ 1 , δ 1 ) .

2) λ ( n ) ( G ; δ 1 , δ 1 ) ≤ λ ( G ; n + δ 1 − 1 , n + δ 2 − 1 ) + n − 1 .

Proof.

1) A vertexu with the maximum degree Δ in a graphG is called a major vertex of G. By counting the numbers for the labels of a major vertex and its neighbors and numbers need to separate each label (the ( δ 1 , δ 2 ) condition), we shall have the trivial lower bound.

2) Let λ ( G ; n + δ 1 − 1 , n + δ 2 − 1 ) = k and f an optimal L ( n + δ 1 − 1 , n + δ 2 − 1 ) -labeling. Define sets L i = { i , i + 1 , ⋯ , i + n − 1 } , i = 0 , 1 , ⋯ , k and function

g f : V ( G ) → ( [ k + n − 1 ] n ) by g f ( u ) = L i whenever f ( u ) = i for i = 0 , 1 , ⋯ , k .

Let u and v be distinct vertices with d G ( u , v ) = j for j = 1 , 2 in G. Suppose f ( u ) = i and f ( v ) = i + n + δ ′ j − 1 for δ ′ j ≥ δ j for j = 1 , 2 . Then g f ( u ) = { i , i + 1 , ⋯ , i + n − 1 } and g f ( v ) = { i + n + δ ′ j − 1 , i + n + δ ′ j , ⋯ , i + n + δ ′ j − 1 + n − 1 } . Hence ‖ g f ( u ) − g f ( v ) ‖ = ( i + n + δ ′ j ) − ( i + n − 1 ) = δ ′ j ≥ δ j for j = 1 , 2 . Thus g f is an L ( n ) ( δ 1 , δ 2 ) -labeling with span k + n − 1 . Therefore

λ ( n ) ( G ; δ 1 , δ 1 ) ≤ λ ( G ; n + δ 1 − 1 , n + δ 2 − 1 ) + n − 1 . ∎

The following is the direct consequence of Proposition 2.1 when ( δ 1 , δ 1 ) = ( 1 , 1 ) . Also notice that λ ( G ; d δ 1 , d δ 2 ) = d λ ( G ; δ 1 , δ 2 ) .

Corollary 2.2. Let Δ be the maximum degree of G. Then

( Δ + 1 ) n − 1 ≤ λ 1 ( n ) ( G ) ≤ n λ 1 ( G ) + n − 1 . ∎

By Corollary 2.2, we know that whenever λ 1 ( G ) = Δ , the lower bound and the upper bound are equal and hence λ 1 ( n ) ( G ) = ( Δ + 1 ) n − 1 . There are several well-known classes of graphs whose λ 1 values are all Δ (see  ). For example, tree T, wheel W m (with m rims), the square lattice Γ S (4-regular infinite plane graph), the hexagonal lattice Γ H (3-regular infinite plane graph), and the triangular lattice Γ Δ (6-regular infinite plane graph) are all with λ 1 = Δ . We summarize as follows.

Theorem 2.3.

1) λ 1 ( n ) ( T ) = ( Δ ( T ) + 1 ) n − 1 .

2) λ 1 ( n ) ( W m ) = ( m + 1 ) n − 1 .

3) λ 1 ( n ) ( Γ S ) = 5 n − 1 .

4) λ 1 ( n ) ( Γ H ) = 4 n − 1 .

5) λ 1 ( n ) ( Γ Δ ) = 7 n − 1 . ∎

3. Cycles

We know that the maximum degree of a cycle Cm of order m ≥ 3 is 2. However, λ 1 ( C m ) is not necessary 2. It depends on m. In this section, we will consider L ( n ) ( 1 , 1 ) -labelings on cycles.

Proposition 3.1. Let Cm be a cycle of order m ≥ 3 . Then λ 1 ( n ) ( C m ) = 3 n − 1 if m ≡ 0 ( mod 3 ) .

Proof. Since the maximum degree of Cm is 2, the trivial lower bound is 3 n − 1 by Corollary 2.2. On the other hand, we use { 0 , 1 , ⋯ , n − 1 } , { n , n + 1 , ⋯ , 2 n − 1 } and { 2 n , 2 n + 1 , ⋯ , 3 n − 1 } consecutively to label vertices of Cm where m ≡ 0 ( mod 3 ) , to obtain an L ( n ) ( 1 , 1 ) -labeling of Cm with span 3 n − 1 . Thus, we have the exact value of λ 1 ( n ) ( C m ) in this case. ∎

Lemma 3.2. Let Cm be a cycle of order m where m ≡ 0 ( mod 3 ) . Then λ 1 ( n ) ( C m ) ≥ 3 n .

Proof. Let V ( C m ) = { v 1 , v 2 , ⋯ , v m } where v i is adjacent to v i + 1 for i = 1 , 2 , ⋯ , m where v m + 1 = v 1 . Suppose λ 1 ( n ) ( C m ) ≤ 3 n − 1 . Letf be an L ( n ) ( 1 , 1 ) -labeling with span 3 n − 1 . Let f ( v 1 ) = A , f ( v 2 ) = B and f ( v 3 ) = C . Since, by definition, f ( v 1 ) , f ( v 2 ) and f ( v 3 ) are distinct, that is, | A ∪ B ∪ C | = 3 n and A ∪ B ∪ C = [ 3 n − 1 ] . Now, f ( v 4 ) ∩ ( B ∪ C ) = ∅ and f ( v 4 ) ⊆ [ 3 n − 1 ] . Hence f ( v 4 ) = A . Consider f ( v 5 ) . Again, we have f ( v 5 ) ∩ ( A ∪ C ) = ∅ and f ( v 5 ) ⊆ [ 3 n − 1 ] . Hence f ( v 5 ) = B . In general, we have 1) f ( v i ) = A if i ≡ 1 ( mod 3 ) , 2) f ( v i ) = B if i ≡ 2 ( mod 3 ) and 3) f ( v i ) = C if i ≡ 0 ( mod 3 ) , for i = 1 , 2 , ⋯ , m .

If m ≡ 1 ( mod 3 ) then f ( v m ) = A . But v m is adjacent to v 1 , where

f ( v 1 ) = A . This violates the condition on adjacent vertices. If m ≡ 2 ( mod 3 ) then f ( v m ) = B = f ( v 2 ) while the distance between v m and v 2 is 2. Again, this violates the condition on distance 2 vertices. We have a contradiction on each case. Therefore, λ 1 ( n ) ( C m ) ≥ 3 n for m ≡ 0 ( mod 3 ) . ∎

Proposition 3.3. If 1) m ≡ 1 ( mod 3 ) and m ≥ 3 n + 1 or 2) m ≡ 2 ( mod 3 ) and m ≥ 6 n + 2 then λ 1 ( n ) ( C m ) = 3 n .

Proof. Let V ( C m ) = { v 1 , v 2 , ⋯ , v m } .

1) Suppose m = 3 n + 1 and Define A 0 = { 0 , 1 , ⋯ , n − 1 } , A 1 = { n , n + 1 , ⋯ , 2 n − 1 } , A 2 = { 2 n , 2 n + 1 , ⋯ , 3 n − 1 } and A 3 = { 3 n , 0 , ⋯ , n − 2 } . Denote X − i ( mod k ) to be that set { x − i ( mod k ) : x ∈ X } . Then we use A 1 , A 2 , A 3 , A 1 − 1 , A 2 − 1 , A 3 − 1 , A 1 − 2 , A 2 − 2 , A 3 − 2 , ⋯ , A 1 − ( n − 1 ) , A 2 − ( n − 1 ) , A 3 − ( n − 1 ) to label v 1 , v 2 , ⋯ , v 3 n . The last vertex v 3 n + 1 is labeled by A 0 . We see that this is an L ( n ) ( 1 , 1 ) -labeling with span 3n of C m .

Suppose m > 3 n + 1 . Then we label first 3 n + 1 vertices as we did above. And then we repeatedly use A 0 , A 1 and A 2 to label remaining vertices.

2) First consider m = 6 n + 2 . We use the sequence presented in (1) for m = 3 n + 1 twice to label vertices of C 6 n + 2 . Obviously, it is still an L ( n ) ( 1 , 1 ) -labeling for C 6 n + 2 with span 3n.

For m > 6 n + 2 , we label the first 6 n + 1 vertices (namely, v 1 , v 2 , ⋯ , v 6 n + 1 ) using the same sequence as above and then repeat using A 0 , A 1 and A 2 to label remaining vertices. Thus λ 1 ( n ) ( C m ) ≤ 3 n in each case. On the other hand, by Lemma 3.2, we have the equality. ∎

Lemma 3.4. Let G be a diameter two graph with order p. Then λ 1 ( n ) ( G ) = n p − 1 .

Proof. SinceG is a diameter two graph, every vertex must receive distinct label. Thus, we need at least np numbers, i.e., λ 1 ( n ) ( G ) = n p − 1 . On the other hand, we can use { i n , i n + 1 , ⋯ , i n + n − 1 } for i = 0 , 1 , ⋯ , p − 1 to label vertices of G in any order. Hence λ 1 ( n ) ( G ) ≤ n p − 1 . ∎

Corollary 3.5.

λ 1 ( n ) ( C m ) = { 5 m ≡ 0 ( mod 3 ) , 6 m ≡ 1 ( mod 3 ) , m ≥ 7     or     m ≡ 2 ( mod 3 ) , m ≥ 14 , 7 m = 4 , 8 , 11 , 9 m = 5.

Proof. Let V ( C m ) = { v 1 , v 2 , ⋯ , v m } where v i is adjacent to v i + 1 for i = 1 , 2 , ⋯ , m where v m + 1 = v 1 .

Claim 1. λ 1 ( 2 ) ( C 8 ) = 7 .

Suppose λ 1 ( 2 ) ( C 8 ) ≤ 6 . Let f be an L ( 2 ) ( 1 , 1 ) -labeling with span 6. Since m = 8 , there must have three consecutive vertices, say v 1 , v 2 and v 3 , be labeled without using 6; and let 6 ∈ f ( v 4 ) . Also let f ( v 1 ) = { a 1 , a 2 } , f ( v 2 ) = { b 1 , b 2 } and f ( v 3 ) = { c 1 , c 2 } . Then f ( v 4 ) = { a , 6 } where a = a 1 or a 2 . Suppose a = a 1 . Hence f ( v 5 ) ⊆ { b 1 , b 2 , a 2 } and f ( v 8 ) ⊂ { c 1 , c 2 , 6 } . Since ( f ( v 6 ) ∪ f ( v 7 ) ) ∩ ( f ( v 5 ) ∪ f ( v 8 ) ) = ∅ , we left only 3 numbers for f ( v 6 ) ∪ f ( v 7 ) , (that is two from { b 1 , b 2 , a 2 , c 1 , c 2 , 6 } \ ( f ( v 5 ) ∪ f ( v 8 ) ) plus a 1 ). It is not enough. The case for a = a 2 is similar.

Thus, λ 1 ( 2 ) ( C 8 ) ≥ 7 . On the other hand, we can use { 0 , 1 } , { 2 , 3 } , { 4 , 5 } , { 6 , 7 } consecutively to label v 1 , v 2 , ⋯ , v 8 to obtain an L ( 2 ) ( 1 , 1 ) -labeling with sapn 7. Hence the claim holds.

Claim 2. λ 1 ( 2 ) ( C 11 ) = 7 .

Let f be an L ( 2 ) ( 1 , 1 ) -labeling with span 6. Similar to Claim 1, we may assume that f ( v 1 ) = { a 1 , a 2 } , f ( v 2 ) = { b 1 , b 2 } , f ( v 3 ) = { c 1 , c 2 } and f ( v 4 ) = { 6 , a } where a ∈ { a 1 , a 2 } and 6 ∉ { a 1 , a 2 , b 1 , b 2 , c 1 , c 2 } .

Again, we have f ( v 11 ) ⊂ { c 1 , c 2 , 6 } . Consider the following cases:

1) f ( v 8 ) ⊂ { c 1 , c 2 , 6 } . Since f ( v 4 ) = { 6 , a } (as indicated above), the discussion on f ( v 4 ) , f ( v 5 ) , f ( v 6 ) , f ( v 7 ) and f ( v 8 ) is the same as Claim 1.

2) f ( v 8 ) ⊂ { a 1 , a 2 , b 1 , b 2 , 6 } . Since f ( v 1 ) = { a 1 , a 2 } , f ( v 10 ) ∩ { a 1 , a 2 } = ∅ . Let c ∈ { c 1 , c 2 , 6 } \ f ( v 11 ) . Hence f ( v 9 ) ⊂ { a 1 , a 2 , c } . So f ( v 8 ) ⊂ { b 1 , b 2 , c } . Thus, there is only one number left available for f ( v 10 ) . This is a contradiction.

3) Suppose f ( v 8 ) consists of one number of f ( v 1 ) and one number of f ( v 3 ) . Without loss of generality, say f ( v 8 ) = { a 1 , c 1 } . Then f ( v 5 ) ⊂ { b 1 , b 2 , a ′ } where a ′ = a 2 if a = a 1 and vice versa. Then there only three numbers available for f ( v 6 ) ∪ f ( v 7 ) and they are one from { b 1 , b 2 , a ′ } \ f ( v 5 ) , c 1 and 6. That is not enough.

Therefore, λ 1 ( 2 ) ( C 11 ) ≥ 7 . On the other hand, we can use { 0 , 1 } , { 2 , 3 } , { 4 , 5 } , { 6 , 7 } consecutively to label v 1 , v 2 , ⋯ , v 11 to obtain an L ( 2 ) ( 1 , 1 ) -labeling with span 7. Hence the claim holds. Finally, we have

1) m ≡ 0 ( mod 3 ) .

By Proposition 3.2, λ 1 ( 2 ) ( C m ) = 5 .

2) m ≡ 1 ( mod 3 ) .

By Proposition 3.3, λ 1 ( 2 ) ( C m ) = 6 if m ≥ 7 . Since C4 is diameter 2 graph, by Lemma 3.4, λ 1 ( 2 ) ( C 4 ) = 7 .

3) m ≡ 2 ( mod 3 ) .

By Proposition 3.3, λ 1 ( 2 ) ( C m ) = 6 if m ≥ 14 . Case for m = 11 and 8 are obtained by Claim 1 and Claim 2. Since C5 is also a diameter 2 graph, by Lemma 3.4, λ 1 ( 2 ) ( C 5 ) = 9 . ∎

4. Concluding Remark

We have obtained values of λ 1 ( 2 ) ( C m ) for all m and λ 1 ( n ) ( C m ) for some m where n ≥ 3 . Otherwise, the labeling numbers are still unknown. It is known that λ 1 ( C m ) = 4 if m ≠ 0 ( mod 3 ) (cf.  ). Hence an upper bound is 4 n − 1 in this case. On the other hand, the lower bound we have in Lemma 3.2 is 3n. Thus, there is still a gap between 3n and 4 n − 1 for n > 1 .

Acknowledgements

The author would like to thank the referee for valuable editorial suggestions.

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

Cite this paper

Yeh, R.K. (2021) A Note on n-Set Distance-Labelings of Graphs. Open Journal of Discrete Mathematics, 11, 55-60. https://doi.org/10.4236/ojdm.2021.113005

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