Variations on a Graph Coloring Theme

Western Michigan University ScholarWorks at WMU

Dissertations Graduate College

6-2012

Bryan Phinezy Western Michigan University

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Recommended Citation Phinezy, Bryan, "Variations on a Graph Coloring Theme" (2012). Dissertations. 3400. https://scholarworks.wmich.edu/dissertations/3400

This Dissertation-Open Access is brought to you for free and open access by the Graduate College at ScholarWorks at WMU. It has been accepted for inclusion in Dissertations by an authorized administrator of ScholarWorks at WMU. For more information, please contact [email protected]. Variations on a Graph Coloring Theme

Bryan Phinezy

A Dissertation Submitted to the Faculty of The Graduate College in partial fulfllment of the requirements for the Degree of Doctor of Philosophy Department of Mathematics Advisor: Ping Zhang, Ph.D.

I would like to thank my doctoral committee consisting of Dr. Ping Zhang, Dr. Gary Chartrand, Dr. Garry Johns, Dr. Allen Schwenk, and Dr. Arthur White. Without them, none of this would have happened. I would specificallylike to thank my committee chair, Dr. Ping Zhang for all of the time she spent with me to further my research endeavors. I would like to thank the staff in the math office for all of the help they have given me over the years. My classmates have been a big help for me to get through the program as well. Specifically, the people I share an office with have been around to listen to me as I worked on these problems. I would also like to thank my family for all of the support they have given me during this process.

Bryan Phinezy

11 Variations on a Graph Coloring Theme

Bryan Phinezy, Ph.D.

Western Michigan University, 2012

Historically, the subject of graph colorings has been the most popular research area in graph theory. There are many problems in mathematics and in real life that can be represented by a graph and whose solution involves finding a specific coloring of this graph. Our research consists of two parts: (1) combinatorial problems and vertex colorings and (2) distance-defined colorings. In this research, we show that certain combinatorial puzzles and problems can be placed in a graph coloring setting and graph colorings can be defined in terms of distance in graphs that are useful in applications. Two vertices u and v in a nontrivial connected graph G are twins if u and v have the same neighbors in V (G) - { u,v}. If u and v are adjacent, they are referred to as true twins; while if u and v are nonadjacent, they are false twins. For a positive integer k, V(G) let c: -+ Zk be a vertex coloring where adjacent vertices may be assigned the V(G) same color. The coloring c induces another vertex coloring c' : -+ Zk defined by V(G), c'(v) = I::uEN[v] c(u) for each v E where N[v] is the closed neighborhood of v.

Then c is called a closed modular k-coloring if c'( u) #- c'( v) in Zk for all pairs u, v of adjacent vertices that are not true twins. The minimum positive integer k for which G has a closed modular k-coloring is the closed modular chromatic number mc(G) of G. These concepts were inspired by two combinatorial problems.

For an ordered set W = { w1,w 2, ... ,w k} of k distinct vertices in a nontrivial con nected graph G, the metric code of a vertex v of G with respect to W is the k-vector

code(v) = (d(v,w1),d(v,w2), .. · ,d(v,wk)) where d( v, wi) is the distance between v and Wi for 1 � i � k. The set W is a local metric set of G if code( u) #- code( v) for every pair u, v of adjacent vertices of G. The minimum positive integer k for which G has a local metric k-set is the local metric dimension lmd(G) of G. A local metric set of G of cardinality lmd( G) is a local metric basis of G. These concepts were inspired by the well-studied concepts of metric sets and metric dimension in graphs and their applications. TABLE OF CONTENTS

ACKNOWLEDGEMENTS ...... ii

LIST OF TABLES...... V

LIST OF FIGURES...... VI

CHAPTER

1. Introduction ...... 1

1.1. History and Motivation ...... 1

1.2. A Checkerboard Problem ...... 3

1.3. The Lights Out Problem...... 4

1.4. Distance-Defned Neighbor-Distinguishing Sets...... 6

2. Preliminary Results on Closed Modular Colorings ...... 9

2.1. Terminology and Notation...... 9

2.2. Paths, Cycles and Complete Bipartite Graphs...... 14

2.3. Regular Complete Multipartite Graphs...... 20

2.4. Closed k-Modular Graphs...... 29

3. Closed Modular Colorings in 'ees ...... 34

3.1. Preliminary Results...... 34

3.2. A Four Color Theorem...... 41

3.3. Rooted Tees ...... 49

3.4. Odd Rooted Tees ...... 51

iii Table of Contents - Continued

CHAPTER

4. Closed Modular Colorings and Graph Operations ...... 62

4.1. Cartesian Products ...... 62

4.2. Coronas of Graphs...... 75

4.3. Joins of Graphs...... 78

5. Local Metric Dimension...... 90

5.1. Terminology and Notation...... 90

5.2. Graphs with Prescribed Order and Local Metric Dimension...... 92

5.3. Bounds for Local Metric Dimension...... 98

5.4. Uniqueness and Non-Uniqueness of Local Metric Bases...... 104

5.5. Joins and Compositions of Graphs...... 107

5.6. Vertex and Edge Deletions...... 118

6. Upper Local Metric Dimension ...... 124

6.1. Introduction ...... 124

6.2. Comparing the Two Local Metric Dimensions...... 126

6.3. Realization Results ...... 131

6.4. Local Metric Dimensions of Graphs with Prescribed Order ...... 135

7. Topics for Further Study ...... 140

BIBLIOGRAPHY...... 143

iv LIST OF TABLES

6.1 Values of lmd(G) and lmd+ (G) of all connected graphs of order 6...... 136

V LIST OF FIGURES

1.1 A coin placement on a 3 x 6 checkerboard...... 4

1.2 Lights Out Game...... 5

2.1 A graph G with x(G) = 2 and mc(G) = 3 ...... ·...... 10

2.2 A closed modular 3-coloring of the Petersen graph...... 12

3.1 A tree T with mc(G)=3...... 35

3.2 A bipartite graph with me(G) = 2...... 37

3.3 The tree Tk for k = 2...... 47

3.4 A tree T with mc(T) = 3...... 55

3.5 A tree T* of type S = {2, 5} with mc(T*) = 2 ...... 56

3.6 A tree T with me( G) = 3...... 57

4.1 An illustration of the coloring c in case 1...... 64

4.2 An illustration of the coloring c in case 2...... 64

4.3 An illustration of the coloring c in case 3 ...... 65

4.4 An illustration of the coloring c in case 4...... 66

4.5 A closed modular 3-coloring of CnDK2, for n = 5, 9...... 67

4.6 A closed modular 3-coloring of CnDK2, for n = 7, 11 ...... 68

4.7 A closed modular 2-coloring of CsDK2...... 68

4.8 A closed modular 3-coloring of C4□K2 and C6□K2 ...... 69

4.9 A closed modular 4-coloring of Cn + K1 for n = 5, 9...... 87 + 4.10 A closed modular 4-coloring of C1 3 K1 ...... 88

vi List of Figures - Continued

4. 11 A closed modular 4-coloring of Cn + K1 for n = 7, 11...... 89

5.1 Metric sets...... 91

5.2 A graph with metric dimension 3 and local metric dimension 2 ...... 92

6.1 A graph G with lmd(G) = 3 and dim(G) = lmd+ (G) = 4...... 125

6.2 Graphs F and H with lmd(F) < dim(F) < lmd+ (F) and lmd(H) < lmd+ (H) < dim(H)...... 126

6.3 A graph G with lmd(G) = 40 and lmd+ (G) = 224...... 134

6.4 Graphs G1 , Fi and H1 of order 6...... 135

vu Chapter 1

Introduction

1.1 History and Motivation

In 1986, at the 250th Anniversary of Graph Theory Conference held at Indiana University Purdue University Fort Wayne, a weighting (or edge labeling with positive integers) of a connected graph G was introduced fr the purpose of producing a weighted graph whose degrees ( obtained by adding the weights of the incident edges of each vertex) were distinct. Such a weighted graph was called irregular. This concept could be looked at in another manner, however. In particular, let N denote the set of positive integers and let

Ev denote the set of edges of G incident with a vertex v. An edge coloring c: E(G) - N, where adjacent edges may be colored the same, is said to be vertex-distinguishing if the coloring c': V(G) - N induced by c and defned by

c'(v) = L c(e) eEEv has the property that c' ( x) # c' (y) for every two distinct vertices x and y of G. A paper [9] on this concept was contained in the proceedings of this confrence. The main emphasis of this research dealt with minimizing the largest color assigned to the edges of the graph to produce an irregular graph.

Two decades earlier, in 1968, Rosa [60] introduced a vertex labeling that induces an edge-distinguishing labeling defned by subtracting labels. In particular, fr a graph G of size m, a vertex labeling (an injective function) f: V(G) -t {0,1, ... ,m} was called a /-valuation by Rosa if the induced edge labeling f' : E(G) - {1,2,... ,m} defned by f'(uv) = lf(u) - f(v)I is bijective. In 1972 Golomb [33] called a /-valuation a grceful labeling and a graph possessing a gracefl labeling a graceful graph. It is this

1 terminology that became standard. Much research has been done on graceful graphs. A popular conjecture in graph theory, due to Anton Kotzig and Gerhard Ringel, is the fllowing.

The Gracefl Tree Conjecture Every nontrivial tree is graceful.

In 1991 Gnana Jothi [32] introduced a concept that, in a certain sense, reverses the roles of vertices and edges in graceful labelings ( also see [31]). For a connected graph

G of order n 2 3, let f : E(G) -+ Zn be an edge labeling of G that induces a bijective function f' : V ( G) -+ Zn defned by

J'(v) = L f(e) eEEv for each vertex v of G. Such a labeling f is called a modular edge-grceful labeling, while a graph possessing such a labeling is called modular edge-graceful. Verifying a conjecture by Jothi on trees, Jones, Kolasinski and Zhang [34] showed not only that every tree of order n 2 3 is modular edge-graceful if and only if n ,t.2 (mod 4) but a connected graph of order n 2 3 is modular edge-graceful if and only if n ,t. 2 (mod 4). These concepts have been studied further in [?].

Many of these weighting or labeling concepts were later interpreted as coloring con cepts with the resulting vertex-distinguishing labeling becoming a vertex-distinguishing coloring. A neighbor-distinguishing coloring is a coloring in which every pair of adjacent vertices are colored diferently. Such a coloring is more commonly called a proper col oring. The minimum number of colors in a proper vertex coloring of a graph G is its chromatic number x(G). In 2004 a neighbor-distinguishing edge coloring c: E(G) -+ {1, 2, ... , k} of a graph G was introduced (see [25, p.385]) in which an induced vertex colorings: V(G) -+ N is defned by s(v) = L c(uv), uEN(v) where N(v) denotes the neighborhood of v (the set of vertices adjacent to v). The minimum k for which such a neighbor-distinguishing coloring exists is called the sum distinguishing index, denoted by sd(G) of G. This is therefore the proper coloring ana logue of the irregular weighting mentioned earlier. It wa shown in [45] that if x(G) : 3, then sd(G) ::; 3. In [1] it was shown for every connected graph G of order at least 3 that

2 sd( G) : 4. In fact, Karonski, Luczak and Thomason [45] made the fllowing conjecture, which has acquired a catchy name used by many.

The 1-2-3 Conjecture If G is a connected graph of order 3 or more, then sd(G) : 3.

Consequently, if the 1-2-3 Conjecture is true, then fr every connected graph G of order 3 or more, it is possible to assign each edge of G one of the colors 1, 2, 3 in such a way that if u and v are adjacent vertices of G, then the sums of the colors of the incident edges of u and v are different.

A number of neighbor-distinguishing vertex colorings diferent from standard proper colorings have been introduced in the literature (see [25, p.379-385], for example). In 2010 a neighbor-distinguishing vertex coloring of a graph [22] was introduced based on sums of colors. For a nontrivial connected graph G, let c : V(G) - N be a vertex coloring of G where adjacent vertices may be colored the same. If k colors are used by c, then c is a k-coloring of G. The color sum a(v) of v is the sum of the colors of the vertices in N(v). If a(u) c a(v) for every two adjacent vertices u and v of G, then c is neighbor-distinguishing and is called a sigma coloring of G. The minimum number of colors required in a sigma coloring of a graph G is called the sigma chromatic number of G and is denoted by a(G). It was shown in [22] that fr each pair a, b of positive integer with a :Sb, there is a connected graph G with a(G) = a and x(G) = b.

1.2 A Checkerboard Problem

In 2010 a problem appeared in [56] dealing with the placement of coins on the squares of a checkerboard. Specifcally, let there be given an m x n checkerboard (m rows and n columns) where m + n 2 3 whose squares are alterately colored white and black. Also, let there be given a collection of mn coins. The problem asked fr which pairs m, n, coins could be placed on some squares, at most one coin per square, such that fr each square of one of the colors white and black, the total number of coins on adjacent squares is even, while fr each square of the other color, the total number of coins on adjacent squares is odd. For example, as illustrated in Figure 1.1, such a coin placement is possible for a 3 x 6 checkerboard.

In [56] it was shown that this problem could be placed in a graph theory setting. For example, let G be the graph whose vertices are the squares of the checkerboard and where two vertices of G are adjacent if the corresponding squares are adjacent. Then G is the grid (bipartite graph) Pm □ Pn (or Pm x Pn ) which is the Cartesian product of

3 -;- --;-i I 0 I 0 3 0 ■-I• - ■ 0 I 0 3 0 3 ■ ■ •■ I 0 I 0 3 0 --■�� Figure 1.1: A coin placement on 3 x 6 checkerboard

the paths Pm and Pn. This suggests a function (coloring) con G = Pm □ Pn , where c: V(G) ➔ Z2 such that

if v corresponds to a square with no coin c(v) = { � if v corresponds to a square containing a coin.

This induces another coloring c' : V(G) ➔ Z2 defined by

c'(v) = L c(u). uEN(v)

If c' is a proper coloring, then the checkerboard problem has a solution on the checker board represented by G. This checkerboard problem gave rise to a new coloring. For a connected graph G,

a modular k-coloring is a function c : V(G) ➔ Zk that induces a proper coloring c'

V ( G) ➔ Zk defined by c'(v) = L c(u). uEN(v)

The minimum k for which G has a modular k-coloring is called the modular chromatic

number mc(G) of G. If mc(Pm □ Pn) = 2, then the checkerboard problem has a solution

for the m x n checkerboard. In [57] it was verified that mc(Pm □ Pn) = 2 for all m and n with m + n 2: 3, and so the checkerboard problem has solutions for all checkerboards.

1.3 The Lights Out Problem

Another recreational problem concerns the electronic game of "Lights Out", consisting of a cube, each of whose six faces contains 9 squares in 3 rows and 3 columns. Thus there are 54 squares in all. Figure 1.2 shows the "front" of the cube as well as the faces on the top, bottom, left and right. The back of the cube is not shown.

4 0 0 0 0 0 0 0 • 0 0 0 0 • • • 0 0 0 0 0 0 0 • 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Figure 1.2: Lights Out Game

A button is placed on each square of a Lights Out cube containing a light which is either on or off. When a button is pushed, the light on that square changes from on to off or from off to on. Moreover, not only is the light on that square affected when its button is pushed, but the lights on its four neighboring squares (top, bottom, left, right) are reversed as well. The four neighboring squares of the middle square of a face lie on the same face as the middle square. Only three neighboring squares of a "side square" ( top middle, bottom middle, left middle, right middle) lie on the same face of a such a square, with the remaining neighboring square lying on an adjacent face as the middle square. For example, if all 54 lights are on initially and the button on the top middle square on the front face is pushed, then this light goes off as well as the lights on its four neighboring squares (see Figure 1.2). Only two neighboring squares of a "corner" square lie on the same face as that square; the other two neighboring squares lie on two other faces.

One goal of the game "Lights Out" is to begin with such a cube where all lights are on and to push a set of buttons so that, at the end, all lights are out. Two observations are (1) no button need be pushed more than once and (2) the order in which the buttons are pushed is immaterial.

This game also has a setting in graph theory. Let each square be a vertex and join each such vertex to the vertices corresponding to its neighboring squares. Thus a 4- regular graph G of order 54 results. The goal is to find a collection S of vertices of G, which correspond to the buttons to be pushed, such that every vertex of G is in the closed neighborhood of an odd number of vertices of S. This says that each vertex v, corresponding to a lit square, will have its light reversed an odd number of times, resulting in the light being turned out.

5 The Lights Out Game can in fact be played on any connected graph G on which there is a light at each vertex of G. The game has a solution for the graph G if all lights are on initially and there exists a collection S of vertices which, when the button on each vertex of S is pushed, all lights of G will be out. This problem has another interpretation. A vertex v of a graph G dominates a vertex u if u belong to the closed neighborhood N[v] of v (consisting of v and the vertices in N(v)). The Lights Out Game has a solution fora given graph G if G contains a set S of vertices.such that every vertex of G is dominated by an odd number of vertices of S. In [54] Sutner showed that every graph has this property and so the Lights Out Game is solvable on every graph.

The Lights Out Game is also equivalent to beginning with a connected graph G where every vertex of G is initially assigned the color 1 in Z2 ( corresponding to its light being on) and finding a set S of vertices of G and a coloring c : V ( G) --+ Z2 each that

if XE S c(v) = { � if X (/_ 8.

A new coloring c' : V(G) --+ Z2 induced by c is defined by

c'(v) = 1 + L c(u). uEN[v]

The goal of the Lights Out Game is therefore to have c' ( v) = 0 for all v E V ( G).

The problem of lights out can be generalized to one where an on-off light is replaced by a light that can be bright (high beam), can be low (low beam), or can be off; that is, Z2 can be replaced by Z.3 or Z2 can be replaced by Zk for some integer k 2: 3

Modular colorings and the Lights Out Game suggests other colorings and other col oring problems. We study one such coloring in Chapters 2, 3 and 4.

1.4 Distance-Defned Neighbor-Distinguishing Sets

Distance in graphs has been used to distinguish all of the vertices of a graph. The distance d( u, v) between two vertices u and v in a nontrivial connected graph G is the length of a shortest path between these two vertices. Suppose that W = { w1, w2, ... , wk} is an ordered set of vertices of a nontrivial connected graph G. For each vertex v of G, there is associated a k-vector called the metric code, or simply the code of v (with respect

6 to W), which is denoted by codew(v) and defned by

codew(v) = (d(v, w1), d(v, w2), · · · , d(v, wk)) ( or simply code( v) if the set W under consideration is clear). If code( u) i=code( v) for every pair u, v of distinct vertices of G, then W is called a metric set, or a resolving set or a locating set. The minimum k fr which G has a metric k-set is the global metric dimension, or simply the metric dimension of G, which is denoted by dim(G). Resolving sets and metric dimensions of graph were introduced, independently, by Harary and Melter [35] and Slater [52, 53], although, as indicated in [7], these concepts in hypercubes were studied earlier under the guise of a coin-weighing problem. In recent years, this concept has been studied widely in a number of papers (see [7, 10, 35, 42, 46, 51, 52, 53], fr example) with a variety of applications.

One of the numerous reasons fr studying resolving sets in graphs deals with distin guishing all the structures in a collection of discrete structures in some manner. If we represent each discrete structure by a graph, then the goal is to distinguish all the graphs in a collection of graphs in some way. While the importance of determining whether two graphs are isomorphic is well known, it can also be useful to know how diferent or how similar two graphs are. A number of integer-valued metrics have been introduced on clases of graphs (see [14], for example), often with restrictions on the orders or sizes of the graphs. Each such metric emanates from a certain graph transfrmation, where an "elementary" transformation of a graph G produces a (non-isomorphic) graph H where the distance between G and His 1. Each such metric then produces a meaure on how close or how far two graphs are fom being isomorphic.

In addition to their applications in mathematics, these metrics have found important applications in chemistry. For example, suppose that two graphs G1 and G2 correspond to two chemical compounds. If the distance between G1 and G2 is k (by some metric), then a graph G lies between G1 and G2 if the sum of the distance between G and G1 and the distance between G and G2 is k. In the context of medical chemistry, the two compounds corresponding to G1 and G2 may have been designed to treat a particular illness. Each such compound is characterized by its efcacy relative to its undesired side efects.

By studying compounds corresponding to graphs lying between G1 and G2, a com pound with an improved efcacy-to-side-efects ratio may be discovered. In particular, it is potentially valuable to be able to distinguish every pair of "nearby" chemical com pounds in as simple a manner as possible.

7 A collection C of graphs under investigation on which a certain metric has been defined itself gives rise to a graph G, whose vertex set is C and where two vertices are adjacent in G if the corresponding two graphs can be transformed into one another by an elementary transformation. The standard distance between the vertices of G then corresponds to the metric defined on C. In this context, a possible goal is to locate a set W of vertices in G such that every two adjacent vertices in G can be distinguished by knowing their distances to the vertices of W.

The foregoing discussion then gives rise to a local version of resolving sets. In this case, we consider those ordered sets W of vertices of G for which any two vertices of G having the same code with respect to Ware not adjacent in G. We study these concepts in Chapters 5 and 6.

We refer to [18] for graph theory notation and terminology not described in this paper. All graphs under consideration here are nontrivial connected graphs.

8 Chapter 2

Preliminary Results on Closed Modular Colorings

2.1 Terminology and Notation

k G, V(G) For a positive integer and a connected graph let c: -+ '11.k be a vertex coloring where adjacent vertices may be assigned the same color. Then c induces another vertex V coloring c' : ( G) -+ '11.k where c'(v) = L c(u) uEN[v] for each v E V(G). We refer to c'(v) as the (closed) color sum of the vertex v with respect to the coloring c.

Two vertices u and v in a connected graph G are twins if u and v have the same neighbors in V ( G) - { u, v}. If u and v are adjacent, they are referred to as true twins; while if u and v are nonadjacent, they are false twins. If u and v are adjacent vertices of a graph G such that N[u] = N[v], that is, if u and v are true twins, then c'(u) = c'(v) for G. V(G) k every vertex coloring c of Define a coloring c: -+ '11.k to be a closed modular coloring if c'(u) -=/ c'(v) in '11.k for all pairs u, v of adjacent vertices for which N[u]-=/ N[v] in G ( or u and v are not true twins in G). A vertex coloring c is a closed modular coloring of G if c is a closed modular k-coloring of G for some positive integer k. That is, in a closed modular coloring c of a graph, c'(u) = c'(v) if u and v are true twins, c'(u) -=/ c'(v) if u and v are adjacent vertices that are not true twins and no condition is placed on c' ( u) and c' (v) otherwise. The minimum k for which G has a closed modular k-coloring is called the closed modular chromatic number of G and is denoted by me( G). If c is a closed modular mc(G)-coloring of G, then c is called a minimum closed modular

9 coloring. A graph G is closed modular k-colorable if mc(G) � k. Since every two vertices in a nontrivial complete graph Kn are true twins, it follows that mc(Kn) = 1 for all n 2 2. In fact, for each integer n 2 2, the complete graph Kn of order n is the only connected graph G of order n for which me(G) = 1.

Proposition 2.1.1 Let G be a nontrivial connected graph. Then mc(G) 1 if and only if G = Kn for some n 2 2.

Proof. We saw that mc(Kn) = 1. Thus, it remains to verify the converse. Let G be a connected graph of order n that is not complete. Then n 2 3 and G contains au- w geodesic (u, v, w) of length 2 for some pair u, w of vertices. Since w E N[v] - N[u], it follows that N[u] # N[v]. Since the closed color sums of u and v in every closed modular coloring of G are different, me(G) 2 2. •

To illustrate these concepts, consider the bipartite graph G of Figure 2.1. Since

G # Kn, it follows that mc(G) 2 2 by Proposition 2.1.1. We show that mc(G) = 3. Figure 2.1 shows a closed modular 3-coloring of G (where the color of a vertex is placed within the vertex) together with the color c'(v) for each vertex v of G (where the color c'(v) of a vertex is placed next to the vertex). Thus mc(G) � 3. Assume, to the contrary, that there exists a closed modular 2-coloring c of G. Since U = { u1, u2, u3, u4} and V = { v1 , v2} are the partite sets of G and the induced coloring c' is a proper 2-coloring of G, the color classes of c' are U and V.

Figure 2.1: A graph G with x(G) = 2 and mc(G) = 3

Assume first that c'(x) = 0 for all x EU and c'(x) = 1 for all x EV. Since c'(u1) = 0, either c(u1) = c(v1) = 0 or c(u1) = c(v1) = 1. In either case, it follows because c'(v1) = 1 that {c(u2), c(u3)} = {0, 1} and so c(u2) # c(u3). However then, since c(u2) # c(u3) and N(u2) = N(u3), we have c'(u2) # c'(u3), a contradiction. Consequently, c'(x) = 1 for all x EU and c'(x) = 0 for all x EV. Since c'(u1) = 1, it followsthat {c(u1), c(v1)} = {0, 1}. Because c'(v1) = 0, we have {c(u2), c(u3)} = {0, 1}. Again, c(u2) # c(u3) and so c'(u2) # c'(u3), once again a contradiction. Therefore, mc(G) = 3, as claimed. Observe

10 for the false twins u2 and u3 in the graph G of Figure 2.1 that if c(u2) = c(u3), then c'(u2) = c'(u3). This example illustrates the following �seful result.

Proposition 2.1.2 Let c be a closed modular coloring of a connected graph G and let u and v be false twins in G. Then c(u) = c(v) if and only if c'(u) = c'(v).

Proof. Observe that

c'(u) = c(u) + L c(x) and c'(v) = c(v) + L c(x). xEN(u) xEN(v)

Since N ( u) = N ( v), the result follows. •

Proposition 2.1.3 If G is a nontrivial connected graph, then mc(G) exists. Further more, if G contains no true twins, then

mc(G) � x(G). (2.1)

Proof. Let V(G) = { vi, v2, ... , vn } where n � 2 and

n-1 k = L 2i = 2n - l. i=O

i Defne a coloring c: V(G)-+ Zk of G by c(vi) = 2 -l fr 1 :i :n. Then 1 : c'(vi) : k for all i (1 : i : n). If Vi and v1 are adjacent vertices that are not true twins, then N[vi] = N[v1] and c' (vi) = c' (v 1). Hence c is a closed modular k-coloring of G and so me(G) exists. Since this is a proper vertex coloring of G using at most me(G) colors, it fllows that x(G) : mc(G). ■

To illustrate (2.1), we consider the Petersen graph P. Since P contains no true twins, mc(P) � x(P) = 3. Figure 2.2 shows a closed modular 3-coloring of P together with the corresponding induced color fr each vertex of P and so mc(P) = 3.

For an edge uv of a graph G, the graph G/uv obtained fom G by contracting the edge uv has the vertex set V(G) in which u and v are identifed. If we denote the vertex u = v in G/uv by w, then V(G/uv) = (V(G)U{w})-{u,v} and the edge set of G/uv is

E(G/uv) = {xy: xy E E(G), x,y E V(G) - {u,v}} U {wx: ux E E(G) or vx E E(G),x E V(G) - {u,v}}. 11 1 2

Figure 2.2: A closed modular 3-coloring of the Petersen graph

The graph G / uv is a minor of G and is also referred to as an elementary contraction of G.

Proposition 2.1.4 Let u and v be true twins of a nontrivial connected graph G. Then G has a closed modular k-coloring if and only if G /uv has a closed modular k-coloring.

Proof. Suppose that w is the vertex of G/uv obtained by identifying u and v in G.

Let co : V(G) ➔ Zk be a closed modular k-coloring of G. Since u and v are true twins of G, it fllows that

c�(u) = c�(v) = co(u) + co(v) + co(x). xEN(u)-{v}

Defne the coloring c: V(G/uv) ➔ Zk by c(x) = co(x) if x -wand c(w) = co(u) + co(v). Then c'(w) = c(w) + co(x) = c�(u) = c�(v).

Furthermore, if z - w in G / uv, then c' ( z) = c� ( z). Thus c is a closed modular k-coloring of G/uv.

For the converse, let co : V(G/uv) ➔ Zk be a closed modular k-coloring of G/uv.

Defne the coloring c: V(G) ➔ Zk by c(x) = co(x) if x 'I- {u,v}, c(u) = co(w) and c(v) = 0. Then c'(u) = c'(v) c(u) + c(v) + L c(x) xEN(u)-{v}

co(w) + co(x) = c�(w). xEN(u)-{v}

12 Furthermore, if x ( { u, v} in G, then c' (x) = c� (x). Thus c is a closed modular k-coloring of G. ■

For a nontrivial connected graph G, defne the true twins closureTC(G) of G to be the graph obtained from G by a sequence of elementary contractions of pairs of true twins in G until no such pair remains. In particular, if G contains no true twins, then TC(G) = G. Thus TC(G) is a minor of G. The fllowing then is a consequence of Proposition 2.1.4.

Corollary 2.1.5 For a nontrivial connectedgraphG, mc(G) = mc(TC(G)).

By Proposition 2.1.3, if G is a nontrivial connected graph that contains no true twins, then mc(G) 2 x(G). On the other hand, if G contains true twins, then it is possible that mc(G) < x(G). In fact, more can be said.

Proposition 2.1.6 For each pair a,b ofpositive integers witha :S b andb2 2, there is a connected graphG such thatme(G) = a andx( G) = b.

Proof. First, suppose that a = b 2 2. Let G = K2,2, ...,2 be the regular complete a :S :S partite graph with partite sets U1, U2, ... , Ua such that IUil = 2 fr 1 i a. Then x(G) = a. Since G contains no true twins, it fllows by Proposition 2.1.3 that mc(G) 2 a. i - :Si Next, defne the coloring c: V(G) ➔ Za by c(x) = 1 if x E Ui fr 1 :Sa. Thus, if :Si Ui E ui (1 :Sa), then

c'(u;) (t,12u - 1)]) - 2(i - 1) + (i - 1)

= 2 ( �) - (i - 1) = -( i - 1) in Za .

i :S :S Hence c' (u i ) i c' (U j ) if Ui E Ui and Uj E Uj where # j and 1 i, j a. Thus c' is a proper vertex coloring of G. Since c is a closed modular a-coloring of G, it follows that mc(G) = a. Next, assume that a < b. If a = l, then let G = Kb and the result follows by Proposition 2.1.1. For a 2 2, let G be the graph obtained from the regular complete a-partite graph K2,2,... ,2 by replacing a vertex v of K2,2, ...,2 by the complete graph Kb-a+l and joining each vertex of Kb-a+l to all the neighbors of v in K2,2, ... ,2. Thus G contains

13 Kb as a subgraph. Since TC(G) = K2,2, ... ,2, it follows by Corollary 2.1.5 that mc(G) = mc(K2,2, ... ,2) = a. Furthermore, x(G) = x(Kb) = b. ■

By Corollary 2.1.5, it sufces to consider nontrivial connected graphs containing no true twins.

2.2 Paths, Cycles and Complete Bipartite Graphs

In this section we determine the closed modular chromatic numbers of paths, cycles and complete bipartite graphs. First, we present an observation that will be useful in seeking a closed modular coloring of a graph.

Proposition 2.2.1 Let u and v be two adjacent vertices in a connected graph G and let X be the set of edges incident with u or v. If the subgraph induced by X is P4 = (uo, u, v, vo), then c(uo) -f. c(vo) for any closed modular coloring c of G.

Proof. Let c: V(G) - Zk be a closed modular k-coloring of G fr some integer k 2: 2. Assume, to the contrary, that c(uo) = c(vo) = a. Then c'(u) = a+ c(u) + c(v) = c'(v), which is impossible. Thus c(uo) - c(vo). ■ For the path Pn = (v1, v2, ... , vn) of order n 2: 3 and a coloring c : V(Pn) -t Zk where k 2: 2, defne the color sequence Sc of c as Sc= (c(v1), c(v2), ... , c(vn)).

Proposition 2.2.2 For each integer n 2: 3

if n is even and n = 2 (mod 3) otherwise.

Proof. Let G = Pn = (v1, v2, ... , vn)- First, assume that n 2: 3 is odd. Defne a coloring c: V(G) -t Z2 of G by c(vi) = 1 if i is odd and 1 'i 'n and c(vi) = 0 if i is even and 2 ' i ' n - l. Then c' ( vi) = c(vi) in Z2 fr 1 ' i ' n and so c' is a proper coloring of G. Hence c is a closed modular 2-coloring of G and me(G) = 2 if n is odd by (2.1).

Next, assume that n > 4 is even. We consider three cases, according to n r (mod 3) fr r E {O, 1, 2}.

14 Case l. n = 0 (mod 3). Since n is even, n = 6k fr some integer k 2 1. Defne a coloring c : V ( G) - Z2 by

ifi:0,1,2 (mod c(vi) = { � 6) (2.2) ifi = 3, 4, 5 (mod 6).

Then the color sequence of c is

Sc= (0,0, 1, 1, 1,0, 0, 0, 1,1, 1,0, . .. , 0,0, 1, 1, 1, 0), that is, this sequence consists of k subsequences (0,0, 1, 1, 1,0 ). Thus the color sequence of the induced coloring c' is

sd (0,1,2,3,2,1, 0,1,2,3,2,1,. .. ,0,1,2,3,2,1) in N ( 0, 1, 0, 1,... , 0, 1) in Z2 .

Hence c is a closed modular 2-coloring of G and so mc(G) = 2 by (2.1) in this case. + Case 2. n = l (mod 3). Since n is even, n = 6k 4 fr some integer k 2 0. Defne a coloring c : V ( G) - Z2 by

ifi:0,1,5 (mod 6) (2.3) if i = 2, 3, 4 (mod 6).

Then the color sequence of c is

Sc = (0,1,1,1,0,0,0,1,1,1, 0,0,0,1,1,1,.. . ,0,0,0,1,1,1).

that is, this sequence consists of the subsequence ( 0, 1,1, 1 ) and k subsequences (0, 0, 0, 1, 1,1 ). Thus the color sequence of the induced coloring c' is

Sc' (1,2, 3,2, 1, 0, 1, 2, 3,2, 1,0, 1, 2, 3, 2, ... , 1,0, 1, 2, 3, 2) in N ( 1, 0, 1, 0,... , 1, 0) in Z2 .

Hence c is a closed modular 2-coloring of G and so mc(G) = 2 by (2.1) in this case. + Case 3. n = 2 (mod 3). Since n is even, n = 6k 2 for some integer k 2 1. We show me(G) = 3 in this case. Assume, to the contrary, that me( G) < 3 and so me(G) = 2. Then there is a closed modular 2-coloring c of G. Thus c'(v1) = 1 or c'(v1) = 0 We consider these two subcases.

15 Subcase 3.1. c'(v1) = 0. In this case, the color sequence of c' is (0, 1, 0, ... , 1,0, 1) in Z2, and either (c(v1), c(v2)) = (0,0) or (c(v1), c(v2)) = (1, 1). Thus there are two subcases.

Subcase 3.1.l. (c(v1),c(v2)) = (0,0). Since c'(v2) = 1, it fllows that c(v3) = 1. However, c'(v3) = 0 implies that c(v4) = 1. Again, c'(v4) = 1 and so c(v5) = 1. Since c'(vs) = 0, we must have c(v6) = 0. Now c'(v6) = 1 implies that c(v1) = 0. Since c'(v1) = 0, we must have c(vs) = 0. Repeating this procedure, we obtain that c(vi) = 0 if i = 0, 1,2 (mod 6) and c(vi) = 1 if i = 3,4, 5 (mod 6). That is, c is the coloring described in (2.2) in Case 1. Since n = 6k + 2 for some integer k 2 1, it follows that

(c(vn-2), c(vn-1),c(v n)) = (0,0, 0). However then, c'(vn-1) = c'(vn) = 0, which is a contradiction.

Subcase 3.1.2. (c(v1),c (v2)) = (1, 1). Since c'(v2) = 1, it follows that c(v3) = 1. However, c'(v3) = 0 implies that c(v4) = 0. Again, c'(v4) = 1 and so c(vs) = 0. Since c'(vs) = 0, we have c(v6) = 0. However then, c'(v6) = 1 implies that c(v1) = 1. Repeating this procedure, we obtain that c(vi) = 0 if i = 0, 4, 5 (mod 6) and c(vi) = 1 if i = 1, 2, 3 (mod 6). Since n = 6k + 2 fr some integer k 2 1, it fllows that (c(vn-2), c(vn-1), c(vn)) = (0,1, 1). However then, c'(vn-1) = c'(vn) = 0, which is a contradiction.

Subcase 3.2. c'(v1) = 1. In this cae, the color sequence of c' is (1,0, 1, 0, ... , 1, 0) in Z2 and either (c(v1),c(v2)) = (1, 0) or (c(v1),c(v2)) = (0, 1). Thus there are two subcases.

Subcase 3.2.l. (c(v1), c(v2)) = (1, 0). Since c'(v2) = 0, it fllows that c(v3) = l. However, c'(v3) = 1 implies that c(v4) = 0. Again, c'(v4) = 0 and so c(v5) = l. Repeating this procedure, we obtain that c(vi) = 1 if i is odd and 1 : i : n - l and

c(vi) = 0 if i is even and 2 : i :Sn. However then, c'(vn-1) = c'(vn) = 1, which is a contradiction.

Subcase 3.2.2. (c(v1),c(v2)) = (0, 1). Since c'(v2) = 0, it fllows that c(v3) = 1. However, c'(v3) = 1 implies that c(v4) = 1. Again, c'(v4) = 0 and so c(v5) = 0. Since c'(vs) = 1, we have c(v6) = 0. However then, c'(v6) = 0 implies that c(v7) = 0. Repeating this procedure, we obtain that c(vi) = 0 if i = 0, 1, 5 (mod 6) and c(vi) = 1 if i = 2, 3,4 (mod 6). That is, c is the coloring described in (2.3) in Case 2. Since n = 6k + 2 fr k some integer 2 1, it follows that (c(vn-2), c(vn-i), c(vn)) = (0, 0,1). However then, c'(vn-1) = c'(vn) = 1, which is a contradiction.

16 Therefore,mc(G) 2'. 3. To show that mc(G) S 3, define a coloring co : V(G) -t Z3 by if i = 1 (mod 4) if i = 0, 2 (mod 4) (v ) = 4). � ; u if i = 3 (mod Then the color sequence of co is

(0, 1,2,1, 0,1,2,1,... ,0, 1,2, 1,0, 1 ) ifn= 2 (mod4) Seo = { (0, 1, 2, 1, 0, 1, 2, 1,... , 0, 1, 2, 1 ) ifn = 0 (mod4) .

Thus the color sequence of the induced coloring c� is

( 1,0, 1,0, 2,0, 1,0, 2,0, 1,0,... ,2,0, 1,0,2, 1 ) if n 2 (mod4) Sc'= { = 0 ((1,0,1,D, 2,0, 1,0, 2,0, 1,0,. .. ,2,0, 1,0) if n = 0 (mod4).

Hence co is a closed modular 3-coloring of G and so me(G) = 3, as claimed. ■

While mc(Pn ) > x(Pn ) only if n 2'. 3 and n = 2 (mod 6), there is no value of n for which mc(Cn ) > x(Cn )- For the cycle Cn (v1, v2 , ... , Vn,v1) of order n 2'. 4 = and a coloring c : V(Cn) -t Zk where k 2'. 2, define the color sequence Sc of c as

Sc= (c(v1),c(v2), ... ,c(vn )).

Theorem 2.2.3 For each integer n 2'. 4, mc(Cn ) = x(Cn )-

Proof. Let G = Cn = (vi, v2, ... , Vn, Vn+l = v1). First, assume that n 2'. 4is even.

Define a coloring c : V(G) -t Z2 of G by c(vi) = 0 if i is odd and 1 S i S n - l and

c(vi) = 1 if i is even and 2 Si Sn. Then c'(vi) = c(vi) in Z2 for 1 Si Sn and so c' is a proper coloring of G. Hence c is a closed modular 2-coloring of G and mc(G) = 2 if n is even by (2.1). Next, we assumethat n 2'. is5 odd. By (2.1), mc(G) 2'. x(G) = 3 for each odd integer n 2'. 5. Thus it remains only to show that mc(G) S 3. For n = 5, define c: V(G) -t Z3 (0, such that the color sequence of c is Sc = 2,2, 1, 1). Thus the color sequence of the (0, induced coloring c' is Sc' = 1,2, 1,2). Thus c is a closed modular 3-coloring of G. For n 2'. 7, define c: V(G) -t Z3 by

if i E { 1,2, 3} c , = if i = 0, 4, 5 (mod 6) (v ) {! if i = 1,2,3 (mod 6) and i 2'. 4. 17 For n = 7, Sc = (0, 0, 0, 1, 1, 1, 2) and so Sc' = (2, 0, 1, 2, 0, 1, 0). Then c is a closed modular 3-coloring of G for n = 7. Thus, we may assume that n � 9. Let So (0, 0,0, 1,1, 1, 2,2, 2, 1,1, 1, 2,2, 2, ... , 1, 1, 1, 2,2, 2) . S1 (0, 0, 0, 1, 1, 1, 2, 2,2, 1, 1, 1, 2,2, 2, ..., hh_!, 2,2, 2, 1, 1, 1). Then the color sequence of c is

so ifn = 3k and k is qdd s ifn=3k+ andk.isodd Sc= { ( o,1,1) 2 (s1, 2) if n= 3k + 1and k is even.

For example then,

(0,0, 0, 1, 1, 1,2, 2, 2) if n = 9 if n Sc = { (0,0,0,1,1,1,2,2,2,1,1) = 11 (0,0,0,1,1,1,2,2,2,1,1,1,2) ifn = 13.

If n = 9, then Sc' = (2, 0, 1, 2, 0, 1, 2, 0, 1). For n � 11, let s� (0,1,2,0,1, 2,0,2, 1,0,1, 2,0,2, 1,0,1, ...,2,0) s� (0,1,2,0,1, 2,0,2, 1,0,1, 2,0,2, 1,0,1, ...,2, o,2,1,o).

Then the color sequence of c' is

(2,s�,1) if n = 3k and k is odd (1,s�,2,1,2) if n = 3k + 2 and k isodd (2, s�,1, 0) ifn = 3k +1 and k is even.

For example then,

(2,0,1,2,0,1,2,0,2,1,0,1,2,0,1) ifn = 15 (1,0,1,2,0,1,2,0,2,1,0,1,2,0,2,1,2) ifn=17 (2,0,l,2,0,l,2,0,2,1,0,1,2,0,2,1,0,1,0) if n = 19.

Hence c is a closed modular 3-coloring of G and so me( G) = 3, as claimed. ■

By Theorems 2.2.2 and 2.2.3, mc(Pn ) = 2 if n � 3 is odd and mc(Cn ) = 2 if n � 4 is even. In both cases, these are bipartite graphs, at least one of whose partite sets consists only of even vertices. These are special cases of the following lemma. A vertex v in a graph G is even if deg v is even and v is odd if deg v is odd. 18 Lemma 2.2.4 If G is a connected bipartite grph at least one of whose partite sets consists only of even vertices, then mc(G) = 2.

Proof. Let G be connected bipartite graph with partite sets U and V, where say each vertex in U is even. Defne c: V(G) ➔ Z2 by c(u) = 0 fr each u E U and c(v) = 1 fr each v E V. Then c'(u) = degu = 0 (mod 2) for u E U and c'(v) = 1 fr each v E V. Thus c is a closed modular 2-coloring of G and so me(G) = 2. ■

The converse of Lemma 2.2.4 is not true, however. For example, if n 2 4 and n = 0 (mod 6) or n = 4 (mod 6) , then each partite set of Pn contains an end-vertex but mc(Pn) = 2 by Theorem 2.2.2. In the case of complete bipartite graphs the condition stated in Lemma 2.2.4 is both necessary and sufcient fr mc(G) = 2.

Theorem 2.2.-5 For positive integers r and s with r + s 2 3,

if at least one of r and s is even mc(K ,s) r = { � if both r and s are odd.

Proof. By Lemma 2.2.4, it sufces to show that mc(Kr,s) = 3 if both r and s are odd.

Assume, to the contrary, that mc(Kr,s) = 2. Let G = Kr,s and let c: V(G) ➔ Z2 be a closed modular 2-coloring of G. Since the induced coloring c' is a proper 2-coloring of G, we may assume, without loss of generality, that c'(u) = 0 for all u E U and c'(v) = 1 fr each v E V. There are two cases.

Case l. There is u E U such that c(u) = 0. Since c'(u) = 0, it fllows that u is adjacent to an even number of vertices in V that are colored 1. Since u is adjacent to every vertex in U, an even number of vertices in V that are colored 1. If there is uo E U such that c(uo) = 1, then c'(uo) = 1 as well since uo is adjacent to an even number of vertices in V that are colored 1, which contradicts that c'(uo) = 0. Thus c assigns Oto every vertex in U. Since IV I is odd and there is an even number of vertices in V are colored 1, there is v E V such that c(v) = 0. However then, c'(v) = 0 in Z2, which contradicts that c' (v) = 1. Case 2. c(u) = 1 for all u E U. Assume there is a vertex v E V such that c(v) = 1. Since I UI is odd and c(u) = 1 fr all u E U, it fllows that c' ( v) = 0 in Z2, a contradiction. Thus c(v) = 0 for all v EV. However then, c'(u) = 1 for each u E U, which is impossible.

Therefre, me(G) 2 3. To show that me( G) : 3, we defne a closed modular 3- coloring of G. Since r+s 2 3 and rands are both odd, we may assume that IVI = s 2 3.

19 Let V = { v1,v2, ... , v8 }. Define c: V(G) -+ Z3 by c(v1) = c(v2) = 1 and c(x) = 0 for all x E V(G)- {v1,v2}. Then c'(u) = 2 for all u EU, c'(vi) = 1 for i = 1,2 and c'(vi) = 0 for 3 :Si :Ss. Therefore, mc(G) = 3 if rands are both odd. ■

2.3 Regular Complete Multipartite Graphs

For every graph G without true twins that we have encountered thus far, either mc(G) = x(G) or me(G) = x( G) + 1. We will see that this is not the case in general, however. In fact, we will see that there is no positive integer constant c for which me(G) '.S x( G) + c for every graph G.

By Theorem 2.2.5, the regular complete bipartite graph Kr,r has mc(Kr,r ) = 2 if r is even while mc(Kr,r ) = 3 if r is odd. We now turn our attention to the closed modular chromatic numbers of regular complete k-partite graphs in general. First, we present two lemmas.

Lemma 2.3.1 Let G = Kn1 ,n2, ... ,nk be a complete k-partite graph (k 2: 2) with partite sets Vi,Vi, ... , Vk such that I¼ I = ni for 1 '.S i '.S k. If c is a closed modular k-coloring of G, then for every set¼ (1 '.Si '.Sk) and for every pair x,y of vertices of¼, c(x) = c(y).

Proof. Suppose that c is a closed modular k-coloring of G. Let x, y E ¼ for some i ( 1 :S i :Sk). Since ( i) x( G) = k, (ii) the induced coloring c' is a proper k-coloring of G and (iii) the graph G is uniquely k-colorable, the color classes of c' are the partite sets of G. Thus c'(x) = c'(y). Because N(x) = N(y), it follows by Proposition 2.1.2 that c(x) = c(y). ■ We denote the regular complete k-partite graph each partite set of which has r vertices by Kk(r)·

For integers r, k let G be a regular complete k-partite graph Lemma 2.3.2 2: 2, = Kk(r) with partite sets Vi,Vi, ... , Vk. If c is a closed modular k-coloring of G, then c( x) = c(y) if and only if x,y E ¼ for some i with 1 '.S i '.S k; that is, c is a proper vertex coloring of G.

Proof. Let x,y E V(G). If x,y E ¼ (1 :S i :S k), then by Lemma 2.3.1 c(x) = c(y). For the converse, suppose that c(x) = c(y) = f. Assume, to the contrary, that x E ¼

20 and y E ½ where i =I= j. By Lemma 2.3.1, c(v) =£for all v E ¼ U ½ and so

c'(x) = L c(v) + (r + 1)£ vEN(x)-Vj

c'(y) L c(v) + (r + vEN(y)-V; 1)£.

Since N(x) - ½ N(y) - ¼, it follows that c'(x) c'(y), which is impossible as xy E E(G). ■

Since x(Kk(r) ) = k for all positive integers k and r, it follows that mc(Kk(r) ) 2: k. k. The following result provides a necessary and sufficient condition for mc(Kk(r) ) =

Theorem 2.3.3 For integers r, k 2: 2, let G = Kk(r)· Then mc(G) =k if and only if 1 -r and k are relatively prime.

Proof. Let the partite sets of G be denoted by Vi,Vi , .. . , Vi, where l¼I = r S: (l i S: k). First, assume that mc(G) = k. Then there exists a closed modular k-coloring V(G) c: -+ Zk. By Lemma 2.3.2, c(x) =I= c(y) if x and y belong to different partite sets. We may assume that c(x) = i - 1 if x E ¼ for 1 S: i S: k. Thus, if x E ¼ (1 S: i S:k), then

c'(x) + + � t[r(j - l)] - r(i - 1) (i - 1) � rG) (1 - r)(i - 1). (2.4)

If k and 1-r are not relatively prime, then gcd(k, 1-r) = d 2: 2and d I k and d I (1-r). Since k (1- --r) k k), (1 - r)d = d = O (mod

it followsthat k/d = i-1 =I= for0 some element i E Zk and so (l-r)(i-1) = 0 E Zk . This implies that there are vertices u E Vi and v E ¼ , where i =/= l, such that c'(u) = c'(v) by (2.4), which is impossible. Therefore, k and 1 -r are relatively prime.

For the converse, assume that k and 1 - r are relatively prime. This implies .that

1 -r has a multiplicative inverse in Zk and that the elements (1 -r) (i -1) for 1 S: i S: k V(G) are distinct. Hence if we define the coloring c : -+ Zk by c(x) = i - 1 for x E ¼ (1 S: i S: k), then the induced coloring c' defined by (2 .4) has the property that c' ( u) =I=

21 c'( v) if u and v belong to distinct partite sets in G. Consequently, c is a closed modular k-coloring of G and so mc(G) :Sk. Since x(G) = k, we have mc(G) = k. ■

By Theorem 2.3.3, it thereforefollows that if the integers 1-r and k are not relatively prime, then mc(Kk(r)) > x(Kk(r)) = k. We now look at some conditions under which certain regular complete multipartite graphs cannot have closed modulars-coloring for various values ofs. We begin with the following lemma.

Lemma 2.3.4 For integersr, k 2'. 2, let G = Kk(r) withpar tite sets Vi,V2, ... , Vk . For an integers 2'. k, if c : V(G) ➔ Zs is a closed modul ars-coloring for whic h c(½) = {c(x): x E ½} consists of ni 2'. 1 el em ents in Zs for l :Si :Sk, then the induc ed coloring Z . c' has n1 + n2 + · · · + n k dist inctvalu es in s

Proof. If x and y belong to the same partite set, then c'(x) = c'(y) if and only if c(x) = c(y). This implies that c'(½) = {c'(x) : x E ½} consists of ni elements for each i with 1 :S i k.:S Since c' (½) n c'(½) = 0 for all i =/= j, it follows that c' has n1 + n2 + · · · + n k values. ■

An immediate consequence of Lemma 2.3.4 is the following lemma.

k Lemma 2.3.5 For integersr, 2'. 2, let G= Kk(r) with partite sets Vi,V 2, ... , Vk . For an integers 2'. k, if c: V(G) ➔ Zs is a coloring with l{c(x): x E ½}I= ni for l :Si :Sk su c h that n1 + n2 + · · · + n k > s, then c is not a closed modul ars-coloring of G.

Proposition 2.3.6 For integersr, k 2'. 2, let G = Kk(r)· (a ) If s is an integer su c h that k :S s :S2k - 2 and r = 1 (mods), then G do es not have a closed modul ars-coloring.

( b) If r is an odd integer, then for eachpo sitive integert witht < k, the gr aph G do es not have a closed modul ar 2t-coloring.

Proof. Suppose that the partite sets of G are Vi,Vi, ... , Vk . We first verify (a). As sume, to the contrary, that there is a closed modular s-coloring c : V(G) ➔ Zs of G. Sinces :S2k- 2, it followsthat at least two of the sets c(½) = {c(x) : x E ½} (1 :S i k):S are singletons; for otherwise, the induced coloring c' has at least 2(k - 1) + 1 = 2k - 1 distinct values in Zs by Lemma 2.3.5, which is impossible. We may assume that c(x) = a for each x E Vi and c(x) = b for each x E Vi- Since r = 1 (mods), it follows

22 c(x) that 0-1 = LxEVi = ra = a E Zs and 0-2 = LxEVi c(x) = rb = b E Zs. Let o- = LxEV(G)-(ViUVi) c(x). However then c'(x) = o- +a+ b in Zs for all x E Vi U V2, which is impossible. Next, we verify(b). Assume, to the contrary, that G has a closed modular 2t-coloring V(G) c(x) c: --t Z2t for some positive integer t with t < k. Let O"i= LxEV; for 1 � i � k. Since l¼I = r is odd for each i with 1 � i � k, there must be Xi E ¼ such that c(xi) and O"i are of the same parity. For i, j E {1, 2, ... , k} with i -=I= j, we claim that c'(xi) and c'(xj) are of the same parity. Without loss of generality, consider c'(x1) and c'(x2 ).

Leto-= L�=3 O"i- Then c'(x1) = c(x1) + 0-2 + o- and c'(x2 ) = 0-1+ c(x 2 ) + o-. Since c(x1) and 0-1 are of the same parity and c(x2) and 0-2 are of the same parity, it follows that c'(x1) andc'(x 2 ) are of the same parity. Therefore, c'(x1),c'(x 2 ), ... , c'(xk) are k distinct elements, every two of which are of the same parity in Z2t- Since there are no more than t t elements of the same parity in Z2t, it follows that k > produces a contradiction. ■

By Proposition 2.3.6(a), if s is an integer such that k � s � 2k - 2 and r = 1

(mods), then G = Kk(r) does not have a closed modulars-coloring. This, however, does not imply that me( G) 2:: s + 1 since it may occur that G has a closed modular s' -coloring for some positive integer s' � s - 1. In fact, we can only conclude that me( G) 2:: s + 1 if it can be shown that G has no closed modular s'-coloring for alls' � s. Similarly, if r 2: 3 is an odd integer, then, by Proposition 2.3.6(b), G does not have a closed modular 2t-coloring for each positive integer t with 2t � 2k - 2. Again, this does not imply that mc(G) 2:: 2k - 1. Part (b) of the following result is an extension of Theorem 2.3.3.

k k Proposition 2.3.7 For integers r, 2: 2, let G = Kk(r) and lets 2: be an integer.

(a) If r = 0 (mods), then G has a clos ed modulars-c oloring. (b) If 1 - r and s are relatively prime, then G has a clos ed modulars-c oloring.

Proof. Suppose that the partite sets of G are Vi, Vi,... , Vk. For an integer s 2:: k, V k. define a coloring c : ( G) --t Zs by c( x) = i - 1 for 1 � i � We claim that if s satisfies the conditions in either (a) or (b), then c is a closed modulars-coloring. First, suppose r c(x) r( i that = 0 (mods). Then O"i = LxEV; = - 1) = 0 in Zs for 1 � i � k. Thus c'(x) = c(x) forall x E V(G) and soc is a closed modulars-coloring. Next, suppose that

23 1 -r and s are relatively prime. For each Vj E ½, it follows that

k k(k - 1) c'(vj) [ r(i-l) -r(j-l)+(j-l)=r -rj+r+j-l � l 2

Since 1-rands are relatively prime, c'(vp) -=/ c'(vq) if Vp E Vp and vq E Vq where p-=/ q. Thus c is a closed modular s-coloring, as claimed. ■

Corollary 2.3.8 For integers r, k 2: 2, let G = Kk(r) and let s 2: k be an integer. If either r = 0 (mods) or l - r ands are relatively prime, mc(G) :S s.

For G = Kk(r), we can now present upper bounds for mc(G) in terms of k.

Theorem 2.3.9 For integers r,k 2: 2, let G = Kk(r)·

(a) If r 2: 2k + 1, then mc(G) :S 2k- 1.

(b) If r 2: 2 is even, then me(G) :S 2k -2.

Proof. We firstverify (a) . Writer = (2k-l)t+p, where tis the maximum nonnegative integer for which p2: 2 is even, say p= 2s for some positive integers. Let Vi,Vi, ... , Vk be the partite sets of G. Define a coloring c: V(G) ➔ Z2k-l by (1) assigning Oto all vertices of Vi and (2 ) for each i with 2 :S i :S k, assigning i -1 to r -s vertices of½ and c(x) assigning 2k -i to the remaining s vertices of½. Let CTi= �xE¼ for 1 :S i :S k. Since (1) CT1 = 0 and (2 ) for 2 :S i :S k,

CTi = (i- l)(r - s) + (2k- i)s= (i - l)(r - 2s) + (2k - l)s= 0 in Z2k-l,

it follows that c'(x) = c(x) for all x E V(G). Therefore, c is a closed modular (2k- 1) coloring and so me( G) :S 2k- 1. Next, we verify (b) . Lets= 2k-2 2: k and r = p (mod s), where O:Sp :S s -1 and pis even. If p= 0, then the result follows by Proposition 2.3.7. Thus, we may assume

that p 2: 2. Define c : V(G) ➔ Zs by (1) assigning the color O to all vertices of Vk-l,

24 (2) assigning the color s/2 to all vertices of Vk and (3) for each i with 1 :::; i :::; k - 2, assigning the color s - i to p/2 vertices of½, and assigning the color i to the remaining r c(x) - p/2 vertices of½. Let O'i = Z:xEV; for 1 :::;. i :::; k. Clearly, O'k-1 = O'k = 0. For 1 :::; i:::; k - 2, p)i O'i = �(s - i) + (r - �) i = (r - = 0 in Zs. Hence c'(x) = c(x) for all x E V(G) and soc is a closed modulars-coloring. Therefore, me( G) : s = 2k - 2.

Finally, we verify (c). Since r and k � 4 are both even, we can write s = 2k - 4 = 4a and r = st + p, where a, t,p are integers, a � 2, t � 0, 0 :::; p :::; s - 1 and pis even.

Define a coloring c : V ( G) -+ Zs by

• assigning the color a to all vertices of Vi,

• assigning the color s - a = 3a to all vertices of Vi,

• assigning the color O to all vertices of Vi,

• assigning the color s /2 = 2a to all vertices of Vi,

• for each integer i with 1 :::; i :::; a - 1, assigning the color s - i to p/2 vertices of

Vi+i and the color i to the remaining vertices of Vi+i,

• for each integer i with for a :::; i :::; k - 4, assigning the color s - (i + 1) to p/2

vertices of Vi+i and the color i + 1 to the remaining vertices of Vi+i.

c(x) for 1 :::; i:::; k. In Z then, 0'1 a(st+p) ap , 0'2 3a(st+p) 3ap, Let O'i = Z:xEV; s = = = = + 0'3 = 0, 0'4 = 2a(st p) = 2ap and O'j = 0 for 5:::; j :::; k. If 4 Ip, then O'i = 0 in Zs for 1 :::; i:::; k and so c'(x) = c(x) for all x E V(G), which implies thatc is a closed modular s-coloring. Hence we may assume that 4 f p. Since pis even, we can write p = 2q for some positive integer q, where then 2 f q. Since qis odd, it follows that 0'1 = ap = 2aq = s/2 (mods) and 0'2 = 3ap = 3(2a)q = s/2 (mods). Furthermore, O'i = 0 for 3 :::; i :::; k. This implies that c'(x) =a + s/2 = s - a in Zs if x E Vi, c'(x) = (s - a)+ s/2 = a in Zs if x E Vi, and c'(x) = c(x) if x E ½ for3:::; i:::; k. Thusc is a closed modulars-coloring. In each case then, me( G) : 2k - 4. ■ According to Theorem 2.3.9(a), the closed modular chromatic number of every regular complete k-partite graph is at most 2k - 1. This bound is sharp.

For every integer k � and r + the graph has Theorem 2.3.10 4 = (2k - 3)! 1, Kk(r) closed modular chromatic number 2k - 1.

25 Proof. Let G K · Since r 1 (mods) fr every integers with k � s � 2k - 2, it = k(r) = fllows by Proposition 2.3.6(a) that G does not have a closed modulars-coloring and so mc(G) 2'. 2k - 1. By Theorem 2.3.9(a), mc(G) � 2k - 1. Therefore, mc(G) = 2k - 1. ■

We have already determined mc(Kk(r)) when k = 2. We now determine exact values of mc(Kk(r)) fr all values of r 2'. 2 when k E {3, 4, 5}.

Theorem 2.3.11 For each integer r 2'. 2,

if 1 - r and 3 are relatively prime if r is even and l - r and 3 are not relatively prime if r is odd and l - r and 3 are not relatively prime.

G K U, V W. r Proof. Let = 3(r) with partite sets and If 1 - and 3 are relatively prime, then mc(G) = 3 by Theorem 2.3.3. Thus, we may assume that 1 - r and 3 are not relatively prime. Hence me(G) 2'. 4. If r is even, then me(G) � 4 by Theorem 2.3.9(b) and so me(G) = 4. If r is odd, then G does not have a closed modular 4-coloring by Proposition 2.3.6(b) and so mc(G) 2'. 5. It remains to show that mc(G) � 5. Since 1 - r and 3 are not relatively prime, 3 I (r-1). Because r is odd, it fllows that r = 1 (mod 6) and so r = 6£ + 1 fr some positive integer £. This implies that r 2'. 7 = 2k + 1. Thus, by Theorem 2.3.9 (a), we have mc(G) � 2k - 1 = 5. The result follows. ■

Theorem 2.3.12 For each integer r 2'. 2,

if r is even if r = 3, 5, 7, 9 (mod 10) if r = 1 (mod 10).

G K Proof. Let = 4(r) with partite sets v for 1 � i � 4. If r 2'. 2 is an even integer, then 1 - r and 4 are relatively prime and so mc(G) = 4 by Theorem 2.3.3. Thus, we may asume that r 2'. 3 is an odd integer. Then r = p (mod 10) where p E {1, 3, 5, 7, 9}. By Theorem 2.3.3, me( G) 2'. 5. First, suppose that r = 3, 5, 7, 9 (mod 10). If r = 5 (mod 10), then r = 0 (mod 5) and so mc(G) � 5 by Corollary 2.3.8. If r = 3, 7, 9 (mod 10), then 1 - r and 5 are relatively prime, and by Proposition 2.3. 7 (b) there is a closed modular 5-coloring. Thus mc(G) = 5 if r = 3, 5, 7, 9 (mod 10).

26 Next, suppose that r = 1 (mod 10). Then r = 10£ + 1 fr some positive integer £. Since r = 1 (mod 5), it fllows by Proposition 2.3.6(a) that G does not have a closed modular 5-coloring. Because r is odd, there is no closed modular 6-coloring by Proposition 2.3.6(b). Hence mc(G) 2 7. Since r = 10£ + 1, we must have r 2 11 > 2k + 1 = 9. Thus, by Theorem 2.3.9 (a), mc(G) : 7. ■

Theorem 2.3.13 For each integer r 2 2,

5 if r "t 1, 6, 11, 16, 21, 26 (mod 30) 6 if r = 6, 26 (mod 30) ) 7 if r 1, 11, 16, 21 (mod 30) and r 1 (mod 7). mc(K5(r) = = 't 8 if r = 16 (riod 30) and r = 1 (mod 7). 9 if r = 1, 11, 21 (mod 30) and r = 1 (mod 7).

Proof. If r 2 2 is an integer such that r "t 1, 6, 11, 16, 21, 26 (mod 30), then 1 - r and 5 are relatively prime and so mc(G) = 5 by Theorem 2.3.3. Thus, we may asume that r 1, 6, 11, 16, 21, 26 (mod 30) and so mc(G) 2 6 by Theorem 2.3.3. Let G K = = 5(r) with the partite sets V, Vi, ... , Vi- We consider two cases, according to whether r is even or r is odd. Case 1. r is even. Then r = 6, 16, 26 (mod 30). We consider these three subcases. Subcase 1.1. r = 6 (mod 30). Since r = 0 (mod 6) in this case, G has a closed modular 6-coloring by Proposition 2.3.7(a). Thus mc(G) = 6. Subcase 1.2. r = 16 (mod 30). Let r = 30£ + 16 fr some integer £ 2 0. We frst show that there is no closed modular 6-coloring of G. Assume, to the contrary, that G has a closed modular 6-coloring c : V(G) � Zf. Let c(¼) = {c(x) : x E ¼} for 1 : i : 5. By Lemmas 2.3.4 and 2.3.5, there exists at most one i E {1, 2, 3, 4, 5} such that ic(¼)I 2 2. Hence we may assume that ic(¼)I fr 1 :i : 4. We consider any two c(V2), c(Vi) c(Vi), of the fur sets c(V), and say V and V- Let Oi = LxEV c(x) fr i 1, 2 and a= c(x). = LxEV(G)-(VUV) • If c(V) = {O} and c(V) = {2}, then 0-1 = 0 and 0-2 = 2. However, then c'(x) = a+ 2 fr all x E V U V2, a contradiction.

• If c(V) = {0} and c(V2) = { 4}, then 0-1 = 0 and 0-2 = 4. However, then c'(x) = a+ 4 fr all x E V U V, a contradiction.

• If c(V1) = {2} and c(V2) = { 4}, then 0-1 = 2 and 0-2 = 4. However, then c'(x) = a fr all x E V1 U V2, a contradiction.

27 • If c(V) = {1} and c(Vi) = {3}, then 0-1 = 4 and 0-2 = 0. However, then c'(x) = a- + 1 for all x E V U Vi, a contradiction.

This is the same for all possible pairs c(¼), c(½) where 1 � i =I= j � 4. Thus implies that c(¼) = c(½) for some pair i, j with 1 � i =I=j � 4. Suppose that c(x) = ai if x E ¼ for 1 � i � 4, where a1, a2, a3, a4 must be distinct elements in Zfi. However, at most one of 0, 2, 4 can be in {a1, a2, a3, a4} and at most one of 1, 3 can be in {a1, a2, a3, a4}. This implies that at least two of a1, a2, a3, a4 are same, which is a contradiction. Therefore, mc(G) 2: 7.

First, assume that r ¢ 1 (mod 7). Since 7 is prime, this implies that 1 - rand 7 are relatively prime and by Proposition 2.3.7 (b) there is a closed modular 7-coloring. Thus mc(G) = 7 if r = 30£ + 16 and r ¢ 1 (mod 7). Next, we assume that r = 1 (mod 7). By Proposition 2.3.6(a), there is no closed modular 7-coloring of G and so mc(G) 2: 8. Also, by Theorem 2.3.9 (b) mc(G) � 8. Thus mc(G) = 8. Subcase 1.3. r = 26 (mod 30). Since 1 - r and 6 are relatively prime, there is a closed modular 6-coloring by Proposition 2.3.7 (6). Thus mc(G) = 6 in this case. Case 2. r is odd. Then r = 1, 11, 21 (mod 30). Since r is odd, mc(G) =I= 6 and mc(G) =I= 8by Proposition 2.3.6(6). Since r 2'. 2k + 1 = 11, it follows by Theorem 2.3.9 that mc(G) � 9. Thus either mc(G) = 7 or mc(G) = 9. If r = 0 (mod 7), then mc(G) = 7 by Corollary 2.3.8. If r = 1 (mod 7), then it follows by Proposition 2.3.6(a) that G does not have a closed modular 7-coloring and so mc(G) = 9. Thus we may assume that r ¢ 0, 1 (mod 7). Since 7 is prime and r ¢ 1 (mod 7), it follows that 1 - r and 7 are relatively prime, and so there is a closed modular 7-coloring by Proposition 2.3. 7 (b). Thus me(G) = 7 in this case.

■

We present a class of complete multipartite graphs G for which me(G) = x( G).

Proposition 2.3.14 For each integer k 2'. 2, let G = Kn1 ,n2, ... ,nk be a complete k partite graph with partite sets U1, U2, ... , Uk such that IUi l = ni for 1 � i � k. If ni¢ nj (mod k) for each pair i,j with 1 � i =I=j � k, then mc(G) = k.

Proof. Since ni ¢ nj (mod k) for each pair i, j with 1 � i =I= j � k, we may assume k) k. V(G) that ni = i - 1 (mod for 1 :::; i :::; Define c : -+ Zk by c(x) = 1 for all

28 x E V(G). If x E Ui where 1 :i : k, then

c'(x) + + (t nj) - ni 1 = [tu - (i - 1) 1 J=l J=l l)l -

(�) + 2 - i in Zk,

Thus c is a closed modular k-coloring of G and so mc(G) = k. ■

2.4 Closed k-Modular Graphs

For a positive integer k, a graph G is k-colorable if there is a proper coloring of G using k colors. Similarly, fr an integer k 2 2, a graph G is closed k-modular or simply k-modular if there is a closed modular k-coloring of G. It is clear that if G is a k-chromatic graph of order n, then a proper k-coloring of G can induce a proper k' -coloring of G fr each integer k' with k : k' : n by introducing a new color to a vertex of G. Therefore, it is easy to see every graph G of order n is k-colorable for all k with x( G) : k : n.

In the case of closed modular colorings, the situation is quite diferent. First, fr positive integers k and k' where k' > k, a closed modular k-coloring of a graph G may not induce a closed modular k' -coloring of G by introducing a new color to a vertex of G. Secondly, if a connected graph G has a closed modular k-coloring fr some positive integer k, it is in general not true that G has a closed modular (k + 1)-coloring. For example, K4(5) = 5 by Theorem 2.3.12 and so K4(5) ha a closed modular 5-coloring. On the other hand, since r = 5 is odd and t = 3 < k = 4, it follows by Proposition 2.3.6(b) that K4(5) does not have a closed modular 6-coloring. Therefore, K4(5) is 5-modular but not 6-modular. Hence it is much more challenging to determine whether a connected graph G is k-modular if k > mc(G). In this section we show that paths, stars, double stars and cycles are k-modular fr each integer k greater than its closed modular chromatic number. First, we present an additional defnition. For a coloring c: V(G)-+ Zk, the (closed) color sum of a vertex v in Z is defined as o(v) = L c(x). xEN[v]

For an ordering (v1, v2, ..., vn) of the vertices of G, the color sequence Sa of the color

29 sums of vertices in Z is defined as

Proposition 2.4.1 For every integer n 2:: 3, the path Pn is k-modular for all k > mc(Pn)-

Proof. Let Pn = ( v1,v2, ... , vn) where n 2:: 3. For a vertex coloring c of Pn, the ( ( ) ) color sequence of c is defined as Sc = c v1 , c(v2 , ... , c(vn)). First, assume that n is odd. By Proposition 2.2.2, mc(Pn) = 2 and so Pn has a closed modular 2-coloring. We show that Pn has a closed modular k-coloring Ck for all integers k 2:: 3. For k 2:: 3, let

Ck : V(G) ➔ Zk be the coloring with color sequence S ek = (1, 0, 1,0, . .. ,1, 0, 1). Then

c�(vi) = 1 in Zk if i is odd and c�(vi) = 2 in Zk if i is even for1 :Si :Sn, that is, the color

sequence Sc� of c� is Sc� = (1,2 ,1, 2 ,... , 1, 2,1 ) for k 2:: 3. Hence Ck is a closed modular k-coloring ck of Pn for all k 2:: 3. Next, assume that n is even. Let n = i (mod 3) where i = 0, 1,2. We consider these three cases. Case l. n = 0 (mod 3). In this case, n = 0 (mod 6) as n is even. By Proposi tion 2.2.2, mc(Pn) = 2 and so Pn has a closed modular 2-coloring. For k 2:: 3, define the coloring ck : V ( G) ➔ Zk such that

= S ek (0,0,1,1,1,0, 0,0,1,1,1,0, ... ,0,0,1,1,1,0).

The color sequence Sa of the color sums of vertices of Pn in Z is

Sa= (0,1,2,3,2,1, 0,1,2,3,2,1, .. . ,0,1,2,3,2,1).

Now, considering the color sum of each vertex in Zk for k 2:: 3, we obtain

sc; = (0, 1, 2, 0, 2,1, 0, 1,2 , 0, 2, 1, ... , 0, 1, 2, 0, 2, 1)

and Sc� = Sa for k 2:: 4. Thus ck is a closed modular k-coloring ck of Pn for all k 2:: 3.

Case 2. n = l (mod 3). By Proposition 2.2.2, mc(Pn) = 2 and so Pn has a closed k V G) modular 2-coloring. For 2:: 3, define the coloring Ck : ( ➔ Zk such that

S ek = (0,1,1,1,0,0,0,1,1,1, 0,0,0,1,1,1, ... ,0,0,0,1,1,1).

The color sequence of a of the vertices of Pn in Z is

30 Sa= (1,2, 3 ,2, 1,0, 1, 2,3 ,2, 1, 0,1, 2, 3, 2, ... ,1, 0, 1, 2, 3,2 ).

Now,considering the color sum of each vertex in Zk fr k 2 3, we obtain

Sc�= (1,2,0,2,1,0, 1,2,0,2, 1,0, 1,2,0,2, ... ,1,0,1,2,0,2 )

and Sc� = Sa fr k 2 4. Thus ck is a closed modular k-coloring ck of Pn fr all k 2 3. Case 3. n = 2 (m od 3). In this case,mc(P n) = 3 by Proposition 2.2.2 and so Pn has a closed modular 3-coloring. For k 2 4, we consider two subcases. V(G) Subcase 3.1. n = 0 (m od 4). Defne the coloring ck : -+ Zk such that

Sq = (0,1,2,1, 0,1,2,1, ... ,0,1,2,1).

Then the color sequence of a of the vertices of Pn in Z is

Sa= (1, 3, 4, 3, 1,3, 4 , 3, ..., 1, 3, 4,3).

Thus Sc� = (1, 3,0, 3 , 1,3 ,0, 3, ..., 1, 3, 0, 3) and Sc� = Sa fr k 2 5. Therefore, ck is a closed modular k-coloring ck of Pn fr all k 2 4. V(G) Subcase 3.2. n = 2 (mod 4). Defne the coloring Ck : -+ Zk such that

= Sek (0,1,2,1, 0,1,2,1, ... ,0,1,2,1, 0,1).

Then the color sequence of a of the vertices of Pn in Z is

Sa= (1,3,4,3, 1,3,4,3, ... ,1,3,4,3, 2,1).

Thus Sc� = (1,3 , 0, 3, 1,3 , 0, 3, ..., 1, 3 ,0, 3 , 2,1 ) and Sc� = Sa fork 2 5. Therefre, Ck is a closed modular k-coloring Ck of Pn for all k 2 4. ■

Proposition 2.4.2 For positive integers r and s with r + s 2 3, the complete bipartite

graph Kr,s is k-modular for all k 2 mc(Kr,s)-

Proof. By Theorem 2.2.5, mc(Kr,s) = 2 if at least one of rands is even and mc(Kr,s) =

3 if both rands are odd. Thus,it sufces to show that Kr,s is k-modular for all k 2 3. Let U V = { u1,u2 ,... ; Ur} and = { vi,v2 ,... ,v 5 }. We may assume that r: s and sos 2 2. Defne c: V(G)-+ Zk by c(v1) = c(v2) = 1 and c(x) = 0 fr all x E V(G) - {v1,v2}. Then c'(u) = 2 fr all u E U, c'(vi) = 1 for i = 1,2 and c'(vi) = 0 fr 3: i: s. Thus, c is a closed modular k-coloring Ck of Kr,s for all k 2 3. ■

3 1 Proposition 2.4.3 Every double star G is k-modular for all integers k 2:: me(G).

Proof. Let G be a double star with central vertices u and v such that deg u = a and deg v = b, where a,b 2:: 2. Let U = {ui,u2,ua-i} be the set of end-vertices adjacent to u and V = {vi,v2,... ,vb-i} the set of end-vertices adjacent to v. If a= b = 2, then

G = P4 and the result follows by Proposition 2.4.1. Thus, we may assume that one of a and b is at least 3, say a 2:: 3. Define the coloring c : V(G) -+ Zk by c(x) = 1 if x E {u,ui,u2,vi} and c(x) = 0, otherwise. Then the color sums of vertices of Gin Z are o=(u) = 3,o=(ui) = 2 for i = 1, 2, a'(ui) = 1 for 3::; i::; a - l,o=(v) = 2,o=(vi) = 1 and o=(vj) = 0 for 2::; j::; b-1. If k = 3,then o=(x) = c'(x) if x E V(G)- {u} and c'(u) = O; while if k 2:: 4, then o=(x) = c'(x) if x E V(G). In either case, c' is proper and so c is a closed modular k-coloring Ck of G for all k 2:: 3. ■

Proposition 2.4.4 For every integer n 2:: 4, the cycle Cn is k-modular for all k > mc(Cn)-

n n Proof. Let Cn = (vi,v 2,... ,V n,vi) where 2:: 4. First, assume that is even. By

Proposition 2.2.3, me(C n) = 2 and so Cn is 2-modular. Thus, we may assume that k 2:: 3.

Define the coloring c: V(Cn) -+ Zk such that the color sequence of c is

Sc= (c(vi),c(v 2),.. . , c(vn)) = (1,0, 1, 0, ... , 1, 0).

The color sequence of the induced coloring c' of c is

Sc'= (c'(vi),c'(v2),... ,c '(vn)) = (1,2,1,2,... ,1,2)

k in Zk. Thus c is a closed modular k-coloring for all 2:: 3and so Cn is k-modular for even integers n 2:: 4. n Next, assume that is odd. By Proposition 2.2.3, mc(Cn) = 3 and so Cn is 3-

modular. We show that Cn has a closed modular k-coloring ck : V(Cn) -+ Zk for all integers k 2:: 4.

n 5, (0, • For = let Ck be the coloring with the color sequence Sek = 2, 2,1, 1). If k k k = 4, then Sc� = (3, 0,1, 0, 2), if = 5,then Sc� = (3, 4, 0, 4,2) and if 2:: 6, then (3, 4, 5, 4, 2). Sc'k =

• For n = 7,let ck be the coloring with the color sequence Sek = (0,0, 0,1, 1, 1, 2). If k k = 4,then Sc� = (2, 0, 1,2, 3, 0, 3) and if 2:: 5,then Sc� = (2,0, 1, 2, 3, 4,3).

32 • For n 2 9, let Ck be the coloring with the color sequence

1, 1, 1, Sq = (0,0, 0, 1, 2, 2,2, 1,1, 2, 2, 2,... ).

We show that Ck is a closed modular k-coloring. There are three cases.

Case l. n = 0 (mod 3). For k 2 4, the color sequence of the induced coloring c� is

if k = 4 (2,0,1,2,3,0, 1,2,1,0,3,0,1,2,1,0,3,0,... ,1,2,1,0,3,0,1,2,0)1,4) (2,o,1,2,3,4,o,1,o,4,3,4,o,1,o,4,3,4,... ,o,1,0,4,3,4,o, if k = 5 (2,0,1,2,3,4,5,0,5,4,3,4,5,0,5,4,3,4,... ,5,0,5,4,3,4,5,0,4) if k = 6 (2,0,1,2,3,4,5,6,5,4,3,4,5,6,5,4,3,4,... ,5,6,5,4,3,4,5,6,4) if k 2 7.

Case 2. n = l (mod 3). For k 2 4, the color sequence of the induced coloring c� is

(2,0,1,2,3,0,1,2,1,0,3,0,1,2, 1,0,3,0,... ,1,2,1,0,3,0,3) if k = 4 (2,0,1,2,3,4,0,1,0,4,3,4,0,1,0,4,3,4,... ,0,1,0,4,3,4,3) if k = 5 (2,0,1,2,3,4,5,0,5,4,3,4,5,0,5,4,3,4,... ,5,0,5,4,3,4,3) if k = 6 (2,0,1,2,3,4,5,6,5,4,3,4,5,6,5,4,3,4,... ,5,6,5,4,3,4,3) if k 2 7.

Case 3. n = 2 (mod 3). Fork 2 4, the color sequence of the induced coloring c� is

(1,0,1,2,3,0,1,2,1,0,3,0, 1,2,1,0,3,0,... ,0,3,0,1,2,1,0,2) if k = 4 (1,0,1,2,3,4,0, 1,0,4,3,4,0,1,0,4,3,4,... ,4,3,4,0,1,0,4,2) if k = 5 (1,0,1,2,3,4,5,0,5,4,3,4,5,0,5,4,3,4,... ,4,3,4,5,0,5,4,2) if k = 6 (1,0,1,2,3,4,5,6,5,4,3,4,5,6,5,4,3,4,... ,4,3,4,5,6,5,4,2) if k 2 7.

In each case, Ck is a closed modular k-coloring and so Cn is k-modular for all k > ■

33 Chapter 3

Closed Modular Colorings Ill Trees

3 .1 Preliminary Results

In this chapter we study the closed modular chromatic numbers of trees. By Theo rem 2.2.5, for a star K1,n-1 of order n 2: 3,

if n is odd -mc(K1,n-1) 2 (3.1) = { 3 if n is even.

If T = S(K1,n-i) is the subdivision graph of K1,n-1 (obtained by subdividing each edge of K1,n-l exactly once), then mc(T) = 2 by Lemma 2.2.4. Next, we consider double stars. A double star is a tree whose diameter is 3.

Proposition 3.1.1 For integers a, b 2: 2, let Sa,b be the double star of order a+b whose central vertices have degrees a and b, respectively. Then

if at least one of a and b is even -mc(S ,b) = 2 a { 3 if a and b are both odd.

Proof. Let G = Sa,b be a double star with central vertices u and v such that deg u = a and deg v = b, where U is the set of end-vertices adjacent to u and V is the set of end-vertices adjacent to v. We consider two cases.

Case 1. At least one of a and b is even. Since x( G) = 2, it remains only to show that

G has a closed modular 2-coloring. If a and b are both even, then define c: V(G)-+ Z2

34 by c( x) = 0 fr all x EV and c( x) = 1 fr x ff.V. Then c'( x) = 0 if x EU U { v} and c' (x) = 1 if x E{ u} U V. If a and b are of opposite parity, say a is odd and b is even, V(G)-+ U U. then defne c: Z2 by c(x) = 0 if x E and c(x) = 1 if x ff. Then c'(x) = 0 if x E{ u} U V and c'( x) = 1 if x EU U{ v}. In each case, c is a closed modular 2-coloring of G and so mc(G) = 2.

Case 2. Both a and bare odd. Let U = { u1,u 2,... ,U a-1} and V = { vi,v 2,... ,vb_i}. V(G)-+ Defne co: Z3 by co(x) = 0 if x E(U-{ui})UVand co(x) = 1 if x E{u,u 1,v}. Then c�(u) = 0, c�(u1) = c�(v) = 2 and c�(x) = 1 for x E(U -{ui}) UV. Since co is a closed modular 3-coloring of G, it follows that mc(G) : 3. Assume, to the contrary, V(G) G. that mc(G) = 2. Let c : - Z2 be a closed modular 2-coloring of Then the induced vertex coloring c' is a proper 2-coloring of G. Thus, we may assume, without loss of generality, that c'(x) = 0 if x EU U {v} and c'(x) = 1 if x E{u} UV. First, assume that c(u) = 0. Since c'(u) = 1 and c'(ui) = 0 for 1 :i :a - l, it follows that c(ui) = 0 (1 :i :a - l) and c(v) = 1. Since c'(vj) = 1 for 1 : j :b- 1 and c(v) = 1, we have c(vj) = 0 fr 1 : j :b - 1. However then, c'(v) = 1, a contradiction. Next, assume that c(u) = 1. Since c'(u) = 1 and c'(ui) = 0 fr 1 :i :a - 1, it fllows that c(ui) = 1 (1 :i :a - l) and c(v) = 0. Since c'(vj) = 1 fr 1 : j :b- 1 and c(v) = 0, it fllows that c(Vj) = 1 fr 1 : j : b - 1. However then, c' (v) = deg v = 1 in Z2, a contradiction. Therefore, me(G) = 3. ■ By (3.1) and Proposition 3.1.1, if T is a tree of diameter 2 or 3 containing only odd vertices,then mc(T) = 3. Figure 3.1 shows a tree of diameter 4, all of whose vertices are odd. We show that mc(T) = 3. Since there is a closed modular 3-coloring of T shown in Figure 3.1, mc(T) : 3.

U1 V1 W1 2(D 1@

u I V w U2 11 W2 0 =-----

Next, we show that mc(T) 2 3. Assume,to the contrary,that there exists a modular

2-coloring c of T. The partite sets of Tare V= {u,v1,w} and V2 = {u1,u2,v,w1,w2} First, assume that c'(x) = 0 for all x EV and c'(x) = 1 fr all x EV 2. If c(u) = 0, then c(ui) = 1 fr i = 1,2 and c(v) = c( v1) = 0. This implies that c(w) = 1. However then, c( wi) = 0 fr i = 1,2 and so c'( w) = 1,which is a contradiction. Thus c( u) = 0. Similarly, c(w) = 0. Hence c(u) = c(w) = 1. Since c'(v) = 1, it fllows that (c(v),c(v 1)) = (0,1)

35 or (c(v), c(v1)) = (1, 0), which is impossible as c'(v1) = 0. Next,assume that c'(x) = 1 fr all x E V and c'(x) = 0 fr all x E V- If c(u) = 1, then c(ui) = 1 for i = 1, 2 and c(v) = 0 and c(v1) = 1. This implies that c(w) = 0. However then, c(wi) = 0 fr i = 1, 2 and so c'( w) = 0, which is a contradiction. Thus c( u) = 1. Similarly, c(w) = 1 and so c(u) = c(w) = 0. Since c'(v) = 0, either c(v) = c(v1) = 0 or c(v) = c(v1) = 1, which is impossible in either case. Therefre, mc(T) = 3, as claimed. In fact, the argument used to show that mc(T) = 3 fr the tree T of Figure 3.1 can be extended to show the fllowing.

Theorem 3.1.2 If T is a tree of order at least 4 each of whose vertices is odd, then mc(T) = 3.

Proof. First, we show that mc(T) 2 3. Assume, to the contrary, that this statement is false. Among all trees each of whose vertices is odd and having closed modular chromatic number 2, let T be one of minimum order n. Thus n is even. By (3.1) and Proposi tion 3.1.1, Tis neither a star nor a double star and son 2 8. Let v be an end-vertex of T with maximum eccentricity and let u be the vertex adjacent to v in T that is not an end-vertex of T. Since T contains no vertex of degree 2, it fllows that u is adjacent to at least two end-vertices in T. Suppose that u1, u2, ... , uk (k 2 2) are end-vertices of T that are adjacent to u. Let c : V(T) - Z2 be a closed modular 2-coloring of T. Since the induced vertex coloring c' is a proper 2-coloring of T, it fllows that either c' (ui) = 0 for 1 '.S i '.S k or c' ( ui) = 1 for 1 '.S i '.S k. This implies that either c(ui) = 0 fr 1 '.S i '.S k or c(ui) = 1 for 1 '.S i '.S k. However then, the restriction of c to the tree T' = T - u1 - u2 is a closed modular 2-coloring of T' and so mc(T') = 2. Since each vertex of T' has odd degree, this contradicts the defning property of T. To show that mc(T) '.S 3, we proceed by induction on the even order of trees each of whose vertices is odd. By (3.1) and Proposition 3.1.1, the result is true for all trees of order 4 and 6. Suppose that if T' is a tree of order n for some even n 2 6 each of whose vertices is odd, then mc(T') '.S 3. Let T be a tree of order n + 2 each of whose vertices is odd. Note that T contains a vertex u that is adjacent to at least two end-vertices of T, say u is adjacent to the end-vertices x and y in T. Then To = T - x - y is a tree of order n each of whose vertices is odd. By the induction hypotheses, mc(To) = 2 or mc(To) = 3. We consider these two cases. Case 1. mc(To) = 2. Let co : V(To) --- Z2 be a closed modular 2-coloring of To. We extend co to a closed modular 2-coloring c of T. There are two subcases, according to whether c�(u) = 0 or c�(u) = 1.

36 Subcase 1.1. c0(u) = 0. If co(u) = 0, then defne c(v) = co(v) fr v E V(To) and c(x) = c(y) = l. Thus c'(v) = c0(v) fr v E V(To) and c'(x) = c'(y) = l. If co(u) = 1, then defne c(v) = co(v) fr v E V(To) and c(x) = c(y) = 0. Thus c'(v) = c0(v) fr v E V(To) and c'(x) = c'(y) = l. Subcase 1.2. c0(u) = l. If co(u) = 0, then defne c(v) = co(v) fr v E V(To) and c(x) = c(y) = 0. Thus c'(v) = c0(v) fr v E V(To) and c'(x) = c'(y) = 0. If co(u) = 1, then defne c(v) = co(v) fr v E V(To) and c(x) = c(y) = l. Thus c'(v) = c0(v) fr v E V(To) and c'(x) = c'(y) = 0.

Case 2. mc(To) = 3. Let co : V(T) -+ Z3 be a closed modular 3-coloring of T0. We extend co to a closed modular 3-coloring c of T. Since c0(u) E {0, 1, 2}, we consider these subcases. As in Case 1, in each of these subcases, defne c(v) = co(v) for v E V(T0). Subcase 2.1. c0(u) = 0. If co(u) = 0, then defne c(x) = 1 and c(y) = 2. Thus c'(v) = co(v) for fr v E V(To) and c'(x) = 1 and c'(y) = 2. If co(u) = i where i = 1, 2 then defne c(x) = c(y) = 0. Thus c'(v) = c0(v) fr v E V(To) and c'(x) = c'(y) = i fr i = 1, 2. Subcase 2.2. c0(u) = l. If co(u) = i fr i = 0, 2, then defne c(x) = c(y) = 0. Thus c'(v) = c0(v) fr v E V(To) and c'(x) = c'(y) = i for i = 0, 2. If co(u) = 1, then defne c(x) = 1 and c(y) = 2. Thus c'(v) = c0(v) fr v E V(To) and c'(x) = 2 and c'(y) = 0. Subcase 2.3. c0(u) = 2. If co(u) = i fr i = 0, 1, then defne c(x) = c(y) = 0. Thus c'(v) = c0(v) fr v E V(To) and c'(x) = c'(y) = i fr i = 0, l. If co(u) = 2, then defne c(x) = 1 and c(y) = 2. Thus c'(v) = c0(v) fr v E V(To) and c'(x) = 0 and c'(y) = l. ■ By Theorem 2.2.2, if n 2 8 is even and n = 2 (mod 3), then mc(Pn) = 3. Thus the converse of Theorem 3.1.2 is not true. Furthermore, Theorem 3.1.2 does not hold fr bipartite graphs in general. For example, each vertex is odd in the bipartite graph G of Figure 3.2 but me( G) = 2. A closed modular 2-coloring of G is also shown in Figure 3.2.

0 0 0 0 0 1

1 0 1 1

Figure 3.2: A bipartite graph G with mc(G) = 2

A caterpillar is a tree of order 3 or more, the removal of whose end-vertices produces a path, called the spine of the caterpillar. Thus every path and star (of order at least 3)

37 and every double star is a caterpillar.

Proposition 3.1.3 Every caterpillar of order at least 3 is closed modular 3-colorable.

Proof. Let T be a caterpillar of order at least 3 and let Pk = ( v1,v 2,... ,v k) be the spine of T. Since stars and double stars are closed modular 3-colorable,we may assume that k � 3. Define a coloring c : V(T) -+ Z3 by c(vi) = 1 if i is odd and 1 � i � k, c(vi) = 2 if i is even and 2 � i � k and c(x) = 0 for all end-vertices x of T. The color

sequence of the induced coloring c' on the spine Pk of Tis

Sc' (0,1,2,1,2,... , 1,2,1,0) if k is odd (0,1, 2, 1,2, ... , 1, 2, 0) if k is even.

Let x and y be two adjacent vertices of T. If x, y E V(Pk), then c'(x) i- c'(y). Thus, we may assume that x is an end-vertex of T and y = vi for some i with 1 � i � s. If i E {1,k}, then c'(x) = c(vi) i- 0 and c'(vi) = 0; if 2 � i � k - 1 and i is even, then c'(x) = c(vi) = 2 and c'(y) = 1; if 3 � i � k - 1 and i is odd, then c'(x) = c(vi) = 1 and c'(y) = 2. In any case, c'(x) i- c'(y). Hence c is a closed modular 3-coloring of T and so mc(G) � 3. ■

Theorem 3.1.4 Let T be a caterpillar of order at least 3 where Pk = (v1,v 2,... ,v k) is the spine of T. For each i with 1 � i � k, let Wi be the set of end-vertices adjacent to vi in T.

(a) If IWil is even for 1 � i � k, then

3 if k is even and k 2 (mod 3) mc(T) = { = 2 otherwise.

(b) If IWil is odd for 1 � i � k, then

if k 1 ( mod 4) mc(T) = { � = otherwise.

Proof. We first verify(a) which says that mc(T) = mc(Pk) by Proposition 2.2.2. Since

mc(Pk) = 2 or mc(Pk) = 3, we consider these two cases.

38 Case l. mc(Pk) = 2. Let cpk be a closed modular 2-coloring of Pk. Define a coloring 1 (1 c : V(T) ➔ Z2 by c(V i) = Cpk ( Vi) for 1 S i S k, c( w) = if w E wi S i S k) such that cpk (vi) = c�k (vi) in Z2 and c(w) = 0 if w E Wi (1 S i S k) such that cpk (vi)-/- c�k (vi) in Z2. We show that c is a closed modular 2-coloring of T. Let x and y be two adjacent vertices of Since c'( v T. i) = c�k ( vi) for 1 S i S k, it follows that c'(x) -/- c'(y) if x, y E V(Pk )- Thus we may assume that x is an end-vertex and y = Vi for some i with 1 S i S k. It follows by the definition_of c that (1) if c(x) = 0, then c(x) = c(vi) = cpk (vi) -/- c�k (vi) = c'(vi) in Z2 and (2) if c(x) = 1, then c(x) = c(vi) + 1 = cpk ( vi) + 1 -/- c�k ( vi) + 1 -/- c�k ( vi) = c' ( vi) in Z2. Thus c is a closed modular 2-coloring of T and so mc(T) = 2.

Case 2. mc(Pk) = 3. Assume, to the contrary, that mc(T) = 2. Let c be a closed modular 2-coloring of T. Since the induced vertex coloring c' is a proper 2-coloring of T, it follows that for each i with 1 S i S k, if x, y E Wi, then c'(x) = c'(y). By 1 Proposition 2.1.2, c(x) = c(y) for all x, y E Wi where S i S k. Thus c'(vi) = c(vi-1)+c(vi)+c(vi+i)+ LwEW; c(w), where we definec(vi-1) = 0 ifi = 1 and c(vi+1) = 0 if i = k. Since IWil = 0 (mod 2), it follows that LwEW; c(w) = 0 in Z2 and so c'(vi ) = c(vi-1) + c(vi) + c(vi+1) for 1 S i S k. However then, this implies that if we restrict c to Pk, then we obtain a closed modular 2-coloring of Pk, which is a contradiction. Therefore, mc(T) = 3. Next, we verify (b). First, suppose that k = 1 (mod 4). Since the result is true for stars by (3.1), we may assume that k 2: 5. Assume, to the contrary, that there is a closed modular 2-coloring c of T. Since the induced vertex coloring c' is a proper

2-coloring, it follows that for each i with 1 Si S k, if x, y E Wi, then c'(x) = c'(y). By

Proposition 2.1.2, c(x) = c(y) for all x, y E Wi where 1 Si S k. First, we claim that

c(v1) -/- c(v1) + c(v2) = c*(v1)

c(vi) -/- c(vi-1) + c(vi) + c(vH1) = c*(vi) for 2 Si S k - 1 + c(vk) -/- c(vk-1) c(vk) = c*(vk)- 1 For otherwise, suppose that there is i with S i S k such that c( vi) = c*(v i). If c(w) = 0 for all w E Wi , then c'(w) = c(vi) = c*(vi) = c'(vi) in Z2, which is a contradiction. Thus, we may assume that c(w) = 1 for all w E Wi. Since IWil is odd, c'(w) = c(vi) + 1 = c*(vi) + 1 = c'(vi) in Z2, which is a contradiction. Thus, as claimed, c(vi) -/- c*(vi) for 1 Si S k.

Let s = (c(v1), c(v2), ... , c(vk )) where c(vi) E Z2 for 1 S i S k. First, suppose that c(v1) = 0. Then c*(v1) = 1. This implies that c(v2) = 1 and c*(v2) = 0. Since c(v1) = 0,

39 c(v2) = 1 and c*(v2) = 0, it fllows that c(v3) = 1 and c*(v3) = 0. Continuing in this manner, we obtain that

s=(0,1,1, 0,0,1,1, o,0,1,1, ... ,o,o,1,1).

However, this implies that k = 3 (mod 4), a contradiction. Next, suppose that c(v1) = 1. Then c*(v1) = 0. This implies that c(v2) = 1 and c*(v2) = 0. Then c(v3) = 0 and c*(v3) = 1, which implies that c(v4) = 0 and c*(v4) = 1. Continuing in this manner, we obtain that s must be one of the following two sequences

...,l,1,0,0, 1,1) s1 (1,1,0,0, 1,1,0,0,

s2 (1,1,0,0, 1,1,0,0,. ..,1,1,0,0, 1,1,0).

However then, this implies that k = 2 (mod 4) or k = 3 (mod 4), a contradiction. Therefore, mc(T) = 2 and so mc(T) = 3 by Proposition 3.1.3. Next, suppose that k "t l (mod 4). We show mc(T) = 2 by providing a closed modular 2-coloring of T.

• If k = 0 (mod 4), then let c : V(T) - Z2 be defned by c(vi) = 0 if i = 0, 1 (mod 4), c(vi) = 1 if i = 2, 3 (mod 4), c(w) = 0 if w E Wi and i = 0, 3 (mod 4) and c(w) = 1 if w E Wi and i = 1, 2 (mod 4). The color sequence of the induced =(0, 0,1, ... ,0, 1). = coloring c' on the spine Pk of Tis Sc' 1, Furthermore, c'(w) 1 if w E Wi and i is odd and c'(w) = 0 if w E Wi and i is even. Thus c'(x) = c'(y) fr every pair x, y of adjacent vertices of T. Hence c is a closed modular 2-coloring of T.

• If k = 2 (mod 4) or k = 3 (mod 4), then let c: V(T) - Z2 be defned by c(vi) = 0 if i = 0, 3 (mod 4), c(vi) = 1 if i = 1, 2 (mod 4), c(w) = 0 if w E Wi and i = 2, 3 (mod 4) and c(w) = 1 if w E Wiand i = 0, 1 (mod 4) . Fork= 2 (mod 4), the color

sequence of the induced coloring c' on the spine Pk of Tis Sc' = (1, 0, 1, 0, ..., 1, 0); while fork= 3 (mod 4), the color sequence of the induced coloring c' on the spine = (1,0, 1, 0, ... ,1, 0, 1). = 0 Pk of Tis Sc' In each case, c'(w) if w E Wi and i is odd and c'(w) = 1 if w E Wi and i is even. Thus c'(x) = c'(y) fr every pair x, y of adjacent vertices of T. Hence c is a closed modular 2-coloring of T.

Therefore, mc(T) = 2 if k "t l (mod 4). ■

40 3.2 A Four Color Theorem

While the closed modular chromatic number of every tree considered thus far is either 2 or 3, no upper bound for mc(T) has been determined for trees of order at least 3 in general. We now show that mc(T) :S 4 for every such tree. In fact, we show that every tree of order at least 3 has a special type of closed modular 4-coloring. A closed modular k-coloring c : V(T) ➔ Zk of a tree T of order 3 or more is a nowhere-zero coloring if c(x) =I= 0 for each vertex x of T.

Lemma 3.2.1 Every star of order at least 3 has a nowhere-zero closed modular 4- coloring.

Proof. Let T = K1 ,k be a star with V(T) = {v,v1,v2, ... ,vk } where vis the central vertex of T and k � 2. First assume that k = 0, 2, 3 (mod 4). Define the coloring c : V(G) ➔ Z4 by c(x) = 1 for each x E V(T). Then c'(vi) = 2 for 1 :S i :S k. If k = 0 (mod 4), then c'(v) = 1; if k = 2 (mod 4), then c'(v) = 3; and if k = 3 (mod 4), then c'(v) = 0. Next assume that k = 1 (mod 4). Define the coloring c : V(G) ➔ 'E,4 by c(v1) = c(v2) = 2 and c(x) = 1 for each x E V(T) - {v1,v2}. Then c'(v) = 0 and c'(v1) = c'(v2) = 3 and c'(vi) = 2 for 3 :Si :Sk. In each case, c is a nowhere-zero closed modular 4-coloring. ■

Theorem 3.2.2 Every tree of order at least 3 has a nowhere-zero closed modular 4- coloring.

Proof. We proceed by strong induction on the order of a tree. By Lemma 3.2.1, the base step of induction holds. Assume for some integer n � 4 that every tree of order at least 3 and at most n has a nowhere-zero closed modular 4-coloring. Let T be a tree of order n + 1. We show that Thas a nowhere-zero closed modular 4-coloring. By Lemma 3.2.1, we may assume that Tis not a star. Let z be a peripheral vertex of T and so z is an end-vertex of T. Suppose that z is adjacent to the vertex u in T. Hence each vertex adjacent to u is an end-vertex of T with exactly one exception. Let V = {z = v1, v2, ... , vk } be the set of end-vertices of T that are adjacent to u. Then T* = T - V is a tree of order at least 3 and u is an end-vertex of T*. By the induction hypothesis, T* has a nowhere-zero closed modular

4-coloring c : V(T*) ➔ Z4. Next, we show that the coloring c can be extended to a nowhere-zero closed modular 4-coloring er : V(T) ➔ Z4 of T; that is, cr(x) = c(x) for

41 each x E V(T*) and so c;,(x) = c'(x) fr each x E V(T*) - {u}. Suppose that u is adjacent to the vertex w in T*. Since c is a nowhere-zero closed modular coloring, it fllows that c(u), c(w) E {1, 2, 3}. W consider three cases, according to the values of c(u).

Case 1. c(u) = 1. If c(w) = 1, then c'(u) = 2; if c(w) = 2, then c'(u) = 3; and if c(w) = 3, then c'(u) = 0. Hence there are three subcases.

Subcase 1.1. c'(u) = 2. Then c'(w) E {O, 1, 3}. We consider these three subcases. Subcase 1.1.1. c'(w) = 0. Defne the coloring er on Vas fllows: If k = 1, 3 (mod 4), then er assigns the color 1 to each vertex in V. Hence c� ( v) = 2 fr each v E V and c�(x) = c'(x) for all x E V(T*) - {u}. If k = 1 (mod 4), then c�(u) = 3, while if k = 3 (mod 4), then c� ( u) = l. If k = 2, 0 (mod 4), then er assigns the color 2 to each vertex in V. Hence c;.,(v) = 3 fr each v EVand c�(x) = c'(x) fr all x E V(T*).

Subcase 1.1.2. c' ( w) = 1. Defne the coloring er on V as fllows: If k = 3 ( mod 4), then defne er as in Subcae 1.1.1 (since c;(u) i 1 in this coloring). If k = 3 (mod 4), then defne the coloring er on V by assigning 2 to each vertex in V. Hence c;.,v) ( = 3 for each v EV, c;.(u) = 0 and cHx) = c'(x) fr all x E V(T*) - {u}. Subcase 1.1.3. c'(w) = 3. Defne the coloring er on V as fllows: If k = 1 (mod 4), then er assigns the color 2 to each vertex in V. Hence c;.,v) ( = 3, c� ( u) = 0 and c;.(x) = c' (x) for all x E V (T*)-{ u }. If k = 1 (mod 4), then defne er as in Subcase 1.1.2 ( since c� ( u) i 3 in this coloring).

Subcase 1.2. c'(u) = 0. Then c'(w) E {1, 2, 3}. We consider these three subcases. Subcase 1.2.1. c'(w) = 1. Defne the coloring er on Vas fllows: If k = 1 (mod 4), then er assigns the color 3 to each vertex in V. Hence c� ( v) = 0 for each v E V, c;,(u) = 3 and c�(x) = c'(x) fr all x E V(T*) - {u}. If k = 2 (mod 4), then er assigns the color 2 to each vertex in V. Hence c� ( v) = 3 fr each v E V and c� ( x) = c' (x) fr all x E V(T*). If k = 3, 0 (mod 4), then er asigns the color 1 to each vertex in V. Hence c;,(v) = 2 fr each v E V. If k = 3 (mod 4), then c�(u) = 3 and c;,(x) = c'(x) fr all x E V(T*) - {u}; while if k = 0 (mod 4), then c;,(x) = c'(x) fr all x E V(T*). Subcase 1.2.2. c'(w) = 2. Defne the coloring er on V as the same in Subcase 1.2.1 (since cHu) i 2 in this coloring). Subcase 1.2.3. c'(w) = 3. Defne the coloring er on Vas fllows: If k = 1 (mod 4), then er assigns the color 1 to each vertex in V. Hence c;.,v) ( = 2 for each v E V, c�(u) = 1 and c;.(x) = c'(x) for all x E V(T*) - { u}. If k = 3 (mod 4), then er

42 assigns the color 2 to each vertex in V. Hence c� ( v) = 3 for each v E V, c� ( u) = 2 and c�(x) = c'(x) fr all x E V(T*) - {u}. If k = 0,2 (mod 4), then defne CT as in Subcase 1.2.1 (since c'(u) = 3 in this coloring).

Subcase 1.3. c'(u) = 3. Then c'(w) E {0,1, 2} and so we consider these three subcases. Subcase 1.3.1. c'(w) = 0. Defne the coloring CT on Va fllows: If k = l,3 (mod 4), then CT asigns the color 2 to each vertex in V. Hence c� (v) = 3 fr each v E V, c�(u) = 1 and c�(x) = c'(x) fr all x E V(T*) - {u}. If k = 2 (mod 4), then CT assigns the color 2 to each vertex in V. Hence c�(v) = 3 fr each v E V and �(x) = c'(x) for all x E V(T*). If k = 0 (mod 4), then CT assigns the color 1 to each vertex in V. Hence c�(v) = 2 fr each v E V and c�(x) = c'(x) for all x E V(T*). Subcase 1.3.2. c'(w) = 1. Defne the coloring CT on V as follows: If k = l (mod 4), then CT assigns the color 1 to each vertex in V. Hence �( v) = 2 fr each v E V, c�(u) = 0 and cHx) = c'(x) fr all x E V(T*) - {u}. If k = 3 (mod 4), then CT assigns the color 1 to v1 and assigns the color 2 to each vertex in V - {vi}. Hence c� ( v1) = 2, �(v) = 3 fr each v EV - {vi}, �(u) = 0 and c�(x) = c'(x) fr all x E V(T*)- {u}. If k = 0, 2 (mod 4), then define CT a in Subcae 1.2.1 (since �(u) = l in this coloring). Subcase 1.3.3. c'(w) = 2. Defne CT on Vas in Subcase 1.2.2 (since c�(u) = 2 in this coloring).

Case 2. c(u) = 2. If c(w) = 1, then c'(u) = 3; if c(w) = 2, then c'(u) = 0; and if c(w) = 3, then c'(u) = 1. W consider these three subcases, according to the values of c'(u). Subcase 2.1. c'(u) = 0. Then c'(w) E {1,2, 3}. There are three subcaes. Subcase 2.1.1. c'(w) = 1. Defne the coloring CT on Vas follows. If k = l (mod 4), then CT assigns the color 3 to each vertex in V. Hence c� (v) = 1 fr each v E V, �(u) = 3 and �(x) = c'(x) fr all x E V(T*) - {u}. If k = 2 (mod 4), then CT assigns the color 1 to each vertex in V. Hence c� (v) = 3 for each v EV, c� ( u) = 2 and �(x) = c'(x) for all x E V(T*) - {u}. If k = 3 (mod 4), then CT assigns the color 2 to each vertex in V. Hence �(v) = 0 fr each v E V, c�(u) = 2 and c�(x) = c'(x) for all x E V(T*) - {u}. If k = 0 (mod 4), then CT asigns the color 1 to each vertex in V. Hence c�(v) = 3 fr each v EV and �(x) = c'(x) fr all x E V(T*). Subcase 2.1.2. c'(w) = 2. Defne the coloring CT on Vas fllows. If k = l,0 (mod 4), then defne CT as in Subcase 2.1.1 (since�= 2 in this coloring). If k = 2 (mod 4), then CT assigns the color 1 to v1 and 3 each vertex in V - {v 1}. Hence c� (v 1) = 3,� ( v) = 1 for each v E V - {v1} and �(x) = c'(x) fr all x E V(T*). If k = 3 (mod 4), then CT

43 assigns the color 1 to v1 and the color 2 to each vertex in V - {vi}. Hence c� ( v1) = 3, c�(v) = 0 for each v EV - {vi}, c�(u) = 1 and c�(x) = c'(x) fr all x E V(T*) - {u}. Subcase 2.1.3. c'(w) = 3. Defne the coloring CT on V as fllows. If k = l (mod 4), then CT assigns the color 1 to each vertex in V. Then c� ( v) = 3 fr each v E V, c�(u) = 1 and c�(x) = c'(x) fr all x E V(T*) - {u}. If k "t l (mod 4), then defne CT as in Subcase 2.1.2 (since c�(u) i 3 in this coloring).

Subcase 2.2. c'(u) = 1. Then c'(w) E {0,2,3}. There·are three subcases. Subcase 2.2.1. c'(w) = 0. Defne the coloring CT on V as follows. If k = l (mod 4), then CT assigns the color 2 to each vertex in V. Hence c� ( v) = 0 for each v E V, c� ( u) = 3 and c� ( x) = c' ( x) fr all x E V ( T*) - { u}. If k = 2 ( mod 4), then CT assigns the color 1 to each vertex in V. Hence c� ( v) = 3 fr each v E V, c� ( u) = 3 and �(x) = c'(x) fr all x E V(T*)- {u}. If k = 3 (mod 4), then CT assigns the color 2 to each vertex in V. Hence c�(v) = 0 fr each v EV, c�(u) = 3 and c�(x) = c'(x) fr all x E V(T*) - {u}. If k = 0 (mod 4), then CT assigns the color 1 to each vertex in V. Hence c�(v) = 3 fr each v EV and c�(x) = c'(x) fr all x E V(T*). Subcase 2.2.2. c'(w) = 2. Defne the coloring CT on V as fllows as in Subcase 2.2.1 (since c� ( u) / 2 in this coloring). Subcase 2.2.3. c'(w) = 3. Defne the coloring CT on Vas fllows. If k = l (mod 4), then CT asigns the color 3 to each vertex in V. Hence c� ( v) = 1 fr each v E V, c�(u) = 0 and c�(x) = c'(x) for all x E V(T*) - {u}. If k = 2 (mod 4), then CT assigns the color 2 to each vertex in V. Hence c�(v) = 0 fr each v E V and c�(x) = c'(x) fr all x E V(T*). If k = 3 (mod 4), then CT assigns the color 1 to each vertex in V. Hence c�(v) = 3 for each v E V, c�(u) = 0 and c�(x) = c'(x) fr all x E V(T*) - {u}. If k = 0 (mod 4), then CT assigns the color 1 to each vertex in Vas in Subcase 2.2.l(since � ( u) / 3 in this coloring).

Subcase 2.3. c'(u) = 3. Then c'(w) E {0,1, 2}. There are three subcases. Subcase 2.3.1. c'(w) = 0. Defne the coloring CT assigns the color 2 to each vertex in V. Then c�(v) = 0 for each v E V. If k = l, 3 (mod 4), c�(u) = 1 and c�(x) = c'(x) for all x E V(T*) - {u}. If k = 0,2 (mod 4), then c�(x) = c'(x) fr all x E V(T*). Subcase 2.3.2. c'(w) = 1. Define the coloring CT as fllows. If k = l (mod 4), the coloring CT assigns the color 3 to each vertex in V. Hence c� ( v) = 1 fr each v E V, �(u) = 2 and c�(x) = c'(x) fr all x E V(T*) - {u}. If k = 3 (mod 4), then CT assigns the color 1 to each vertex in V. Hence c� ( v) = 3 fr each v E V, � ( u) = 2

44 and c�(x) = c'(x) fr all x E V(T*) - {u}. If k = 0, 2 (mod 4), then defne CT as Subcase 2.3.1 (since 4(u) = 1 in this coloring).

Subcase 2.3.3. c'(w) = 2. Defne the coloring CT as in Subcase 2.3.l. (since c�(u) = 2 in this coloring). Case 3. c(u) = 3. If c(w) = 1, then c'(u) = 0; if c(w) = 2, then c'(u) = 1 and if c(w) = 3, then c'(u) = 2. We consider these three subcases, according to the values of c'(u). Subcase 3.1. c'(u) = 0. Then c'(w) E {1, 2, 3}. There are three subcases. Subcase 3.1.l. c'(w) = l. Defne the coloring CT on Vas follows. If k = l, (mod 4), then CT assigns the color 3 to each vertex in V. Hence 4 ( v) = 2 for each v E V, 4(u) = 3 and c�(x) = c'(x) fr all x E V(T*) - {u}. If k ¢ l (mod 4), then CT asigns the color 2 to each vertex in V. Hence 4(v) = 1 fr each v E V. If k = 2, 0 (mod 4), then c�(x) = c'(x) fr all x E V(T*). If k = 3 (mod 4), then c�(u) = 2 and 4(x) = c(x) for all x E V(T*)- {u}. Subcase 3.1.2. c'(w) = 2. Defne the coloring CT on Vas fllows. If k - 3 (mod 4), then define CT as in Subcase 3 .1.1 (since 4 ( u) = 2 in this coloring). If k = 3 (mod 4), then CT assigns the color 1 to each vertex in V. Then c� ( v) = 0 fr each v EV, c� ( u) = 3 and 4(x) = c'(x) for all x E V(T*) - {u}. Subcase 3.1.3. c'(w) = 3. Define the coloring CT on V as fllows. If k = l (mod 4), then CT assigns the color 1 to each vertex in Vi . Hence c� (v) = 0 fr each v E V, c�(u) = 1 and 4(x) = c'(x) fr all x E V(T*) - {u}. If k ¢ l (mod 4), then defne CT as in Subcase 3.1.1 (since c�(u) = 3 in this coloring).

Subcase 3.2. c'(u) = l. Then c'(w) E {0, 2, 3}. There are three subcases. Subcase 3.2.l. c'(w) = 0. Defne the coloring CT on Vas fllows. If k = l (mod 4), then CT assigns the color 2 to each vertex in V. Hence 4 ( v) = 1 fr each v E V, c�(u) = 3 and 4(x) = c'(x) for all x E V(T*) - {u}. If k = 2 (mod 4), then CT assigns the color 1 to each vertex in V. Hence c� (v) = 0 fr each v E V, c� ( u) = 3 and 4(x) = c'(x) for all x E V(T*) - {u}. If k = 3 (mod 4), then CT assigns the color 2 to each vertex in V. Hence 4(v) = 1 fr each v E V, 4(u) = 3 and 4(x) = c'(x) fr all x E V(T*) - {u}. If k = 0 (mod 4), then CT assigns the color 1 to each vertex in V. Hence c�(v) = 0 for each v EV and c�(x) = c'(x) fr all x E V(T*). Subcase 3.2.2. c'(w) = 2. Defne the coloring CT on V as Subcase 3.2.1 (since 4 ( u) = 2 in this coloring).

45 Subcase 3.2.3. c'(w) = 3. Defne the coloring cr on V as follows. If k = l (mod 4), then cr assigns the color 1 to each vertex in V. Hence c� ( v) = 0 fr each v E V, c� ( u) = 2 and c�(x) = c'(x) fr all x E V(T*) - {u}. If k = 2 (mod 4), then er assigns the color 1 to v1 and assigns the color 3 each vertex in V - {v1}- Hence c�(v1) = 0, c�(v) = 2 fr each v E V - {vi} and 4(x) = c'(x) fr all x E V(T*). If k = 3 (mod 4), then er assigns the color 2 to v1 and v2 and asigns the color 3 to each vertex in V - {v1, v2}. Hence c� ( v1) = c� ( v2) = 1, c� ( v) = 2 fr each v E V - {v1, v2}, c� ( u) = 0 and c� ( x) = c' ( x) fr all x E V(T*) - {u}. If k = 0 (mod 4), then the coloring er on Vas Subcase 3.2.1 (since c� ( u) = 3 in this coloring).

Subcase 3.3. c'(u) = 2. Then c'(w) E {0, 1,3}. Again, there are three subcases. Subcase 3.3.1. c'(w) = 0. Defne the coloring er on Vas fllows. If k = l (mod 4), then er assigns the color 1 to each vertex in V. Hence c� ( v) = 0 fr each v E V, c� ( u) = 3 and c�(x) = c'(x) for all x E V(T*) - {u}. If k = 2,0 (mod 4), then er assigns the color 2 to each vertex in V. Hence 4(v) = 1 fr each v E V and c�(x) = c'(x) fr all x E V(T*). If k = 3 (mod 4), then er assigns the color 1 to each vertex in V. Hence 4(v) = 0 fr each v EV, c�(u) =land c�(x) = c'(x) fr all x E V(T*) - {u}.

Subcase 3.3.2. c'(w) = 1. Defne the coloring er on Vas follows. If k ¢ 3 (mod 4), then defne er as in Subcase 3.3.1 (since 4(u) = 1 in this coloring). If k = 3 (mod 4), then er assigns the color 3 to each vertex in V. Hence c� (v) 2 fr each v E V, c�(u) = 3 and c�(x) = c'(x) for all x E V(T*) - {u}. Subcase 3.3.3. c'(w) = 3. Defne the coloring er on Vas follows. If k = l (mod 4), then er assigns the color 2 to each vertex in V. Hence c� (v) = 1 for each v E V, c�(u) = 1 and c�(x) = c'(x) for all x E V(T*) - {u}. If k ¢ l (mod 4), then defne er as in Subcase 3.3.1 (since c�(u) = 3 in this coloring).

In each case, c is a nowhere-zero closed modular 4-coloring. ■

Although every path and every caterpillar of diameter at most 5 has a nowhere-zero closed modular 3-coloring, this is not true in general. In fact, there is an infnite class of trees that do not have a nowhere-zero closed modular 3-coloring and so Theorem 3.2.2 cannot be improved. In order to show this, we frst present a lemma.

Lemma 3.2.3 Suppose that T is a tree that contains a vertex u of degree 3 such that u is adjacent to two end-vertices u1 and u2 and one non-end-vertex w. If c is a nowhere-zero closed modular 3-coloring of T, then c(u1) = c(u2) and c'(w) = c(u).

46 Proof. Assume, to the contrary, that there is a nowhere-zero closed modular 3-coloring c: V(G) -t Z3 - {O} such that either c(u1) =I= c(u2) or c'(w) = c(u). First, suppose that c(u1) =/= c(u2), where say c(u1) = 1 and c(u2) = 2. Then c'(ui) = c(u) + i fori = 1, 2 and c'(u) = c(u) + c(w). Since c(w) E {1,2}, it follows that either c'(u) = c'(u1) or c'(u) = c'(u2), which is impossible. Thus, c(u1) = c(u2). Next, suppose that c'(w) = c(u). Since c(u1) = c(u2) E {1, 2}, there are two cases. Case l. c(ui) = 1 for i = 1,2. Hence c'(ui) = 1 + c(u) for i = 1,2 and c'(u) =

2 + c(u) + c(w) in Z3. If c(w) = 1, then c'(u) = c(u) = c'(w), which is impossible; while if c(w) = 2, then c'(u) = c(u) + 1 = c'(ui) for i = 1,2, which again is impossible. Case 2. c(ui) = 2 for i = 1,2. Hence c'(ui) = 2 + c(u) fori = 1,2 and c'(u)

1 + c(u) + c(w) in Z3. If c(w) = 1, then c'(u) = c(u) + 2 = c'(ui) for i = 1, 2, which is impossible; while if c(w) = 2, then c'(u) = c(u) = c'(w), which again is impossible. ■

Theorem 3.2.4 There is an infinite class of trees that do not have a nowhere-zero closed modular 3-coloring.

Proof. For each integer k � 2 with k = 2 (mod 3), we construct a tree Tk that does not have a nowhere-zero closed modular 3-coloring. We begin with the star K1 ,k+1 with the central vertex v. Then the graph Tk is obtained by (1) subdividing exactly k edges of Ki,k+l exactly once and (2) adding two pendant edges at each end-vertex of K1,k+l· Suppose that N(v) = {u,w1,w2,... ,wk } where degu = 3, degwi = 2 for 1 Si S k, u is adjacent to two end-ve�tices u1 and u2 and k non-end-vertices w1, w2, ... , w2, each vertex Wi is adjacent to v and Xi and each vertex Xi is adjacent to exactly two end-vertices for

1 Si S k. Figure 3.3 shows the tree Tk and the labeling of the vertices of Tk for k = 2.

Figure 3.3: The tree Tk for k = 2

Assume, to the contrary, that for some k � 2 with k = 2 (mod 3) the tree Tk has a nowhere-zero closed modular 3-coloring c: V(Tk) -t Z3 - {0}. For each i with 1 S i S k, since deg Xi = 3 and Xi is adjacent to two end-vertices and one non-end-vertex Wi, it

47 follows by Lemma 3.2.3 that

c'(wi) -=I- c(xi) for 1 � i � k. (3.2)

Furthermore, since deg u = 3 and u is adjacent to two end-vertices u1 and u2 and one non-end-vertex v, it follows by Lemma 3.2.3 that c(u1) = c(u2) E {1, 2}. We consider two cases. Case 1. c(u1) = c(u2) = 1. Since c(u) E {1, 2}, there are two possibilities. Subcase 1.1. c( u) = 1. Then c' ( ui) = 2 for i = 1, 2. This forces c(v) = 1 and so c'(u) = 1. First, suppose that c(wi) = 2 for some i with 1 � i � k, then c'(wi) = c(v)+c(wi)+c(xi) = 1+2+c(xi) = c(xi) in :l.3, which contradicts (3.2). Thus c(wi) = 1 for all i with 1 � i � k. Since k = 2 (mod 3), it follows that c'(v) = 1 = c'(u), which is a contradiction. Subcase 1.2. c(u) = 2. Then c'(ui) = 0 for i = 1, 2. This forces c(v) = 1 and so c'(u) = 2. By (3.2), an argument similar to the one in Subcase 1.1 shows that c(wi) = 1 for all i with 1 � i � k. Since k = 2 (mod 3), it follows that c'(v) = 2 = c'(u), which is a contradiction. Case 2. c(u1) = c(u2) = 2. Since c(u) E {1, 2}, there are two possibilities. Subcase 2.1. c(u) = 1. Then c'(ui) = 0 for i = 1, 2. This forces c(v) = 2 and so c'(u) = 1. First, suppose that c(wi) = 1 for some i with 1 � i � k, then c'(wi) = c(v)+c(wi)+c(xi) = 1+2+c(xi) = c(xi) in :l.3, which contradicts (3.2). Thus c(wi) = 2 for all i with 1 � i � k. Since k = 2 (mod 3), it follows that c'(v) = 1 = c'(u), which is a contradiction. Subcase 2.2. c(u) = 2. Then c'(ui) = 1 for i = 1, 2. This forces c(v) = 2 and so c'(u) = 2. By (3.2), an argument similar to the one in Subcase 1.1 shows that c(wi) = 2 for all i with 1 � i � k. Since k = 2 (mod 3), it follows that c'(v) = 2 = c'(u), which is a contradiction. ■ In fact, Lemma 3.2.3 can be extended to the following.

Lemma 3.2.5 For an integer k � 2 with k = 2 (mod 3), suppose that T is a tree that contains a vertex u of degree k + 1 such that u is adjacent to k end-vertices u1, u2, ... , Uk and one non-end-vertex w. If c is a nowhere-zero closed modular 3-coloring of T, then (1) c must assign the same color to each vertex Ui for 1 � i � k and (2) c'(w) -=I- c(u).

With the aid of Lemma 3.2.3, it can be shown that the tree Tk constructed in the proof of Theorem 3.2.4, where k � 2 with k = 2 (mod 3), can be extended by assuming

48 that each vertex Xi (1 � i � k) in Tk is adjacent to exactly k end-vertices. Then the resulting tree has no nowhere-zero closed modular 3-coloring as well. We know of no trees T for which mc(T) = 4. Therefore, we close this section with the following conjecture.

Conjecture For every tree T of order at least 3, mc(T) � 3.

In fact, there is a more general question related to the closed modular chromatic number of a connected bipartite graph; that is,

Is there a constant c for which me( G) � c for all connected bipartite graphsG?

3.3 Rooted Tees

Let T be a rooted tree of order at least 3 having the root v. For each integer i with 0 � i � e ( v), where e ( v) is the distance between v and a ·vertex farthest from v, let

11; = {x E V(T) : d(v, x) = i}.

If x Ev; where O � i � e(v), then xis at level i. If x Ev; (0 � i � e(v) - 1) is adjacent toy E ¼+1, then x is the parent of y andy is a child of x. A vertex z is a descendant of x (and x is an ancestor of z) if the x - z path in T lies below x. In this section, we show that if T is a rooted tree of order at least 3 such that the numbers of children of all vertices of T have the same parity and no vertex of T has exactly one child, then either mc(T) = 2 or mc(T) = 3. A caterpillar is a tree of order 3 or more, the removal of whose end-vertices produces a path called the spine of the caterpillar. Thus every path and star (of order at least 3) and every double star (a tree of diameter 3) is a caterpillar.

Theorem 3.3.1 If T is a caterpillar of order at least 3, then mc(T) � 3.

Theorem 3.3.2 Let T be a rooted tree of order at least 3.

(a) If each vertex of T has an even number of children, then mc(T) = 2.

(b) If each vertex of T has either no child or an odd number of children and no vertex has exactly one child, then mc(T) � 3.

Proof. Suppose that v is the root of T. For each integer i with O � i � e( v), let

¼ = {x E V(T): d(v, x) = i}.

49 To verif (a), defne the coloring c: V ( G) ➔ Z2 by

if x E ½ where i 0, 1 (mod 4) c(x) = { � = (3.3) if x E ½ where i = 2, 3 (mod 4).

Then c'(x) = 1 if x E ½ and i is even and c'(x) = 0 if x E ½ and i is odd. Thus c is a closed modular 2-coloring and so mc(T) = 2 if each vert�x of T has an even number of children. To verif (b), we proceed by strong induction. If T is a star, then mc(T) s 3 by Theorem 3.3.l. Assume fr an integer n � 4 that if each vertex of a tree of order at most n has either no child or an odd number of children and no vertex has exactly one child, then the closed modular chromatic number of the tree at most 3. Let T be a tree of order n + l such that each vertex of T has either no child or an odd number of children and no vertex has exactly one child. We may assume that T is not a star. Let x be a peripheral vertex of T and so x is an end-vertex of T. Suppose that x is a child of the vertex y in T. Since each vertex of T ha either no child or an odd number of children and no vertex of T has exactly one child, it fllows that y has an odd number r � 3 children, say x = x1, x2, ... , Xr are children of y. Then each child of y is an end-vertex of T. Let X = { x = x1, x2, ..., Xr }. Consider T* = T - X which is a tree of order less than n + l such that each vertex of T* ha either no child or an odd number of children and no vertex of T* ha exactly one child. By the induction hypothesis, T* has a closed modular 3-coloring c: V(T*) ➔ :3. Next, we show that T has a closed modular 3-coloring CT : V(T) ➔ :3 such that cT(u) = c(u) and 4(u) = c'(u) fr each u E V(T*). Since r is odd, r = l, 3, 5 (mod 6). We consider these three cases. Case l. r = l (mod 6). In this case, r � 7. We defne CT on X such that c�(y) = c'(y). If c(y) # c'(y), then CT assigns the color O to Xi fr 1 S i S r. Hence c�(xi) = c(y) # c' (y) fr 1 S i S r. If c(y) = c' (y), then CT assigns the color 2 to x1 and x2 and the color 1 to Xi for 3 S i S r. Hence 4(xi) = c'(y) + 2 # c'(y) fr i = 1, 2 and 4(xi) = c'(y) + 1 # c'(y) fr 3 Si Sr. Case 2. r = 3 (mod 6). We defne CT on X such that 4(y) = c'(y). If c(y) # c'(y), then CT assigns the color O to Xi fr 1 S i S r. Hence 4(xi) = c(y) # c'(y) fr 1 S i S r. If c(y) = c'(y), then CT assigns the color 1 to Xi fr 1 S i S r. Hence 4(xi) = c'(y) + 1 # c'(y) fr 1 Si Sr. Case 3. r = 5 (mod 6). We defne CT on X such that c�(y) = c'(y). If c(y) # c'(y), then CT assigns the color Oto Xi fr 1 Si Sr. Hence c�(xi) = c(y) # c'(y) fr 1 Si Sr.

50 If e(y) = e' (y), then er asigns the color 2 to x1 and assigns the color 1 to Xi for 2 � i � r. Hence e�(x1) = e'(y) + 2 and �(xi)= c!(y) + 1 fr 2 � i � r.

In each case, er is a closed modular 3-coloring of T and so mc(T) � 3. ■

Theorem 3.3.2 provides the closed modular chromatic numbers for a well-known class of rooted trees. A rooted tree T is a complete r-ary tree fr some integer r � 2 if every vertex of T ha either r children or no child. The following is a consequence of Theorem 3.3.2.

Corollary 3.3.3 For an integer r � 2, let T be a complete r-ary tree.

(a) If r is even, then mc(T) = 2 .

(b) If r is odd, then mc(T) � 3.

In the view of Theorem 3.3.2, it would be useful to introduce an additional terminol ogy. A rooted tree T of order at least 3 is even if every vertex of T has an even number of children; while T is odd if every vertex of T has an odd number of children. It then follows by Theorem 3.3.2 that mc(T) = 2 if T is an even rooted tree and mc(T) � 3 if T is an odd rooted tree and no vertex of T has exactly one child.

3.4 Odd Rooted Tees

In this section we investigate the closed modular colorings of odd rooted trees of order at least 3. We will see that if the locations of leaves of an odd rooted tree T are given, then in some cases it is possible to determine the exact value of mc(T). For each integer p E {O, 1, 2, 3, 4, 5}, an odd rooted tree T of order at least 3 having root v is said to be of type p if d(v, u) = p (mod 6) for every leaf u in T. We now determine all odd rooted trees of type p were O � p � 5 that have closed modular chromatic number 2.

Theorem 3.4.1 For each integer p E {O, 1, 2, 3, 4, 5}, let T be an odd rooted tree of order at least 3 that is of type p. Then mc(T) = 2 if and only if p = 1.

Proof. Suppose that v is the root of T. For each integer i with O � i � e(v), let ¼ = {x E V(T) : d(v, x) = i}. First, suppose that O � p � 5 and p = ". We show mc(T) = 2. Since x(T) = 2 for every nontrivial tree T, it sufces to construct a closed modular 2-coloring e: V(T) -+ Z2 of T. We consider three cases, according to the values of p.

51 Case 1. p = 0. In this case, a coloring c: V(T) -+ Z2 is defned by

if x Ev and i 0, 1,5 (mod 6) c(x) = { � = if x Ev and i = 2, 3, 4 (mod 6).

Then the induced coloring c' : V(T) -+ Z2 is defned as

if x E v;and i is even c'(x) = � (3.4) { if x E v;and i is odd.

Case 2. p = 2, 3, 4 (mod 6). In this cae, a coloring c: V(T)- Z2 is defned by

if x E v; and i 0, 1,2 (mod 6) c(x) = { = t if x Ev; and i = 3,4,5 (mod 6).

Then the induced coloring c': V(T)- Z2 is defned as in (3.4). Case 3. p = 5 (mod 6). In this case, a coloring c: V(T) -+ Z2 is defned by

if x E v; and i 0,4, 5 (mod 6) c(x) = { � = if x Ev and i = 1,2,3 (mod 6).

Then the induced coloring c': V(T) -+ Z2 is defned as

if x E v;and i is even c'(x) = { t if x E v;and i is odd.

Thus c is a closed modular 2-coloring of T and so mc(T) = 2. For the converse, suppose that T is an odd rooted tree of order at least 3 that is of type 1. Thus if u is a leaf of T, then u E Vk for some integer k, where then 1 : k : e(v) and k = 1 (mod 6). We show that mc(T) - 2. Assume, to the contrary, that there is a closed modular 2-coloring c: V(T) -+ Z2 of T. Thus c'(v) = 0 or c'(v) = 1. We consider these two caes. Case 1. c'(v) = 0. Thus c'(x) = 0 if x Ev; and i is even and c'(x) = 1 if x E v;and i is odd. Since c(v) E {O, 1}, there are two subcases. Subcase 1.1. c( v) = 0. Since c' ( v) = 0 and c( v) = 0, there is v1 E Vi such that c(v1) = 0. Since c'(v1) = 1 and c(v) = c(v1) = 0, there is v2 E V such that c(v2) = 1.

52 Since c'(v2) = 0, c(vi) = 0 and c(v2) = 1, there is v3 E V such that c(v3) = l. Observe fr each i � 3 that c(vi) is uniquely determined by c'(vi-i), c(vi-2) and c(vi-i)

Repeating this procedure, we obtain a path Pk = (vi, v2, ... , vk) in T such that (1) vk is a leaf of T, d(v, vi) = i fr 1 � i � k and k = 1 (mod 6) and (2) the color sequence Sc= (c(vi), c(v2), ... , c(vk)) of the coloring con the path Pk is

Sc= (0, 1,1, 1, 0,0, 0, 1, 1,1, 0, 0, 0, ... ,1, 1, 1, 0, 0, 0).

Hence (c(vk-2), c(vk_i), c(vk)) = (0,0, 0). However then, c'(vk-i) = c'(vk) = 0, which is a contradiction. Subcase 1.2. c(v) = l. By the same argument in Subcase 1.2, we conclude that there must be a path Pk = (vi, v2, ... , vk) in T such that (1) Vk is a leaf of T, d(v, vi) = i fr 1 � i � k and k = 1 (mod 6) and (2) the color sequence Sc = (c(vi), c(v2), ... , c(vk)) of the coloring c on the path Pk is

Sc= (1,1,0,0,0,1,1,1,0,0,0, 1,1, ... ,1,0,0,0,1,1).

Hence (c(vk-2), c(vk-i), c(vk)) = (0, 1, 1). However then, c'(vk-i) = c'(vk) = 0, which is a contradiction.

Case 2. c'(v) = l. Thus c'(x) = 1 if x E ¼ and i is even and c'(x) = 0 if x E ¼ and i is odd. Since c(v) E {O, 1}, there are two subcases.

Subcase 2.1. c(v) = 0. Since c'(v) = 1 and c(v) = 0, there is vi E V such that c(vi) = l. Since c'(vi) = 0, c(v) = 0 and c(vi) = 1, there is v2 E V such that c(v2) = l. Since c'(v2) = 1, c(vi) = c(v2) = 1, there is v3 E V such that c(v3) = l. Observe fr each i � 3 that c(vi) is uniquely determined by c'(vi-i), c(vi-2) and c(vi-i).

Repeating this procedure, we obtain a path Pk = (vi, v2, ... , vk) in T such that (1) vk is a leaf of T, d(v, vi) = i for 1 � i � k and k = 1 (mod 6) and (2) the color sequence Sc = ( c(vi), c(v2), ... , c(vk)) of the coloring c on the path Pk is

= Sc (1,l,1,0,0,0,1,1,1,0,0,0,1,... ,1,1,0,0,0,1).

Hence (c(vk-2), c(vk-d, c(vk)) = (0, 0, 1). However then, c'(vk-i) = c'(vk) = 1, which is a contradiction. Subcase 2.2. c(v) = l. By the same argument in Subcase 2.1, we conclude that there must be a path Pk= (vi, v2, ... , vk) in T such that (1) Vk is a leaf of T, d(v, vi) = i for 1 � i � k and k = 1 (mod 6) and (2) the color sequence Sc = (c(vi), c(v2), ... , c(vk)) of the coloring c on the path Pk is

Sc= (0,1,0,1,0,1,0,1,0,1,0, 1,0,... ,1,0,1,0,1,0). 53 Hence (c(vk-2), c(vk-1), c(vk)) = (0,1, 0). However then, c'(vk-1) = c'(vk) = 1, which is a contradiction. • By Theorems 3.4.1, if T is an odd rooted tree of order at least 3 that is of type 1, then mc(T) 2': 3. On the other hand, every odd rooted tree of order at least 3 we have encountered that is of type 1 has closed modular chromatic number 3. Furthermore, the following is a consequence of Theorems 3.3.2 and 3.4.1.

Corollary 3.4.2 If T is an odd rooted tree of order at least 3 that is of type 1 such that no vertex has exactly one child, then mc(T) = 3.

By Theorem 3.4.1, if pis an integer with O :Sp : 5 and p = 1, then every odd rooted tree of order at least 3 that is of type p has closed modular chromatic number 2. This gives rise to the following question:

If S � {0, 1, 2, 3, 4, 5} where ISi 2': 2 and 1 t S and T is an odd rooted tree of order at least 3 having root v such that for every leaf u in T, d(v, u) = p (mod 6) fr some p E S, then is it necessary that mc(T) = 27

The answer to this question is no, as we show next. First, it will be convenient to introduce an additional defnition. For a nonempty subset S � {0, 2, 3, 4, 5}, an odd rooted tree T having root v is said to be of type S if fr every leaf u in T, d(v, u) = p (mod 6) fr some p E S and for each p E S, there is at least one leaf u in T such that d(v,u) = p (mod 6). In particular, if S = {p} where p E {0,2,3,4,5}, then Tis of type p. We frst consider odd rooted trees of type S, where S = {2, 5} or S = {0, 3}. In the next two results, we show that if S = {2, 5} or S = {0, 3}, then it is possible for an odd rooted tree T of type S to have mc(T) = 3.

Theorem 3.4.3 For S = {2, 5}, there are odd rooted trees of type S such that mc(T) = 3.

Proof. Consider the tree Tin Figure 3.4 each of whose leaves are at level 2 or at level 5. W show mc(T) = 3. For each integer i with O :i : 5, let v;= {x E V(T): d(v,x) = i}.

Thus if x is a leaf of T, then x E V2 or x E V5 . By Corollary 3.3.3, mc(T) : 3. Thus it remains to show that mc(T) = 2. Assume, to the contrary, that there is a closed

modular 2-coloring c: V(T) -- Z2 of T. Thus c(v) = 0 or c(v) = 1. We consider these two cases.

54 V

T mc(T) Figure 3.4: A tree with = 3

Case 1. c(v) = 0. Since either c'(v) = 0 or c'(v) = 1, there are two subcases.

Subcase 1.1. c'(v) = 0. Thus c'(w) = 0 if w E ½ and i is even and c'(w) = 1 if w E ½ and i is odd. Furthermore, c(u1) = 0 or c(u1) = l. First, assume that c(u1) = 0. Since c'(u1) = 1 and c(v) = 0, there is a child w of u1 such that c(w) = l. However then, c'(w) = c'(u1) = 1, a contradiction. Next, assume that c(u1) = 1. Since c'(u1) = 1 and c(v) = 0, there is a child w of u1 such that c(w) = 0. However then, c'(w) = c'(u1) = 1, a contradiction.

Subcase 1.2. c'(v) = 1. Thus c'(w) = 0 if w E ½ and i is odd and c'(w) = 1 if w E ½ and i is even. Furthermore, c(u1) = 1 or c(u1) = 0. First, assume that c(u1) = 1. Since c'(u1) = 0 and c(v) = 0, there is a child w of u1 such that c(w) = 1. However then, c'(w) = c'(u1) = 0, a contradiction. Next, asume that c(u1) = 0. Since c'(u1) = 0 and c(v) = c(u1) = 0, there is a child w of u1 such that c(w) = 0. However then, c'(w) = c'(u1) = 0, a contradiction.

Case 2. c(v) = 1. Since either c'(v) = 0 or c'(v) = 1, there are two subcases.

Subcase 2.1. c'(v) = 0. Then c(u) = 1 or c(u) = 0. First, assume that c(u) = 1. Since c'(u) = 1 and c(v) = 1, there is a child w of u such that c(w) = 1. We claim that c(y) # 0 and c(z) # 0, fr otherwise, say c(y) = 0. Then c'(u) = d(y) = 1, a contradiction. Thus c(y) = c(z) = 1, as claimed, which implies that c(x) = 1. Since c'(x) = 0, there is a child w of x such that c(w) = 0. Since c'(w) = 1, there is a child w1 of w such that c(wi) = 0. Since c'(w1) = 0, there is a child w2 of w1 such that c(w2) = 0. However then, c'(w1) = c'(w2) = 0, a contradiction. Next, assume that c(u) = 0. We saw that c(y) -/ 1 and c(z) -/ 1 and so c(y) = c(z) = 0. Since c'(u) = 1, it fllows that c(x) = 0. Since c'(x) = 1, there is a child w of x such that c(w) = 0. Since c'(w) = 1, there is a child w1 of w such that c(w1) = 1. Since c'(w1) = 0, there is a child w2 of w1

55 such that c(w2) = l. However then, c'(w1) = c'(w2) = 0, a contradiction.

Subcase 2.2. c'(v) = l. Then c(u) = 1 or c(u) = 0. We consider these two possibili ties.

Subcase2.2.l. c(u) = l. Now either c(x) = 0 or c(x) = l. First assume that c(x) = 0. Since c'(x) = 1 and c(u) = 1, there is a child w of x such that c(w) = 0. Since c'(w) = 0 and c(x) = 0, there is a child w1of wsuch that c( w1) = 0. Since c' ( w1) = 1 and c( w1) = 0, there is a child w2 of w1 such that c(w2) = l. However· then, c'(w1) = c'(w2) = 1, a contradiction. Next, asume that c(x) = l. Since c(u) = 1 and c'(u) = 0, one of y and z must be colored 1, say c(y) = l. However then, c'(y) = c'(u) = 0, a contradiction.

Subcase 2.2.2. c(u) = 0. Now either c(x) = 0 or c(x) = l. First assume that c(x) = 0. Since c'(u) = 0, exactly one of y and z is colored 1, say c(y) = 1 and c(z) = 0. However then c'(z) = c'(u) = 0, a contradiction. Next, asume that c(x) = l. Since c'(x) = c(x) = 1, there is a child w of x such that c(w) = 0. Since c'(w) = 0, c(x) = 1 and c(w) = 0, there is a child w1 of w such that c(w1) = l. Since c'(w1) = 1, there is a child w2 of w1 such that c(w2) = 0. However then, c'(w1) = c'(w2) = 1, a contradiction.■ By Theorem 3.4.3, despite the fact that every odd rooted tree of type 2 or type 5 has closed modular chromatic number 2, there are odd rooted trees Tof type S = {2, 5} fr which mc(T) = 3. On the other hand, there are odd rooted trees of type S = {2, 5} having closed modular chromatic number 2. For example, we start with the tree T in Figure 3.4. Let T' be the subtree of T whose vertex set consists of x and all descendants of x. Then the tree T* is constructed from T of Figure 3.4 by replacing y with a copy of T' (see Figure 3.5). The coloring c : V(T) - Z2 defned by asigning the color Oto each vertex in { u,u1, u2, x,y} and assigning the color 1 to the remaining vertices of T* is a closed modular 2-coloring. Therefore, mc(T*) = 2.

Figure 3.5: A tree T* of type S = {2, 5} with mc(T*) = 2

Theorem 3.4.4 For S = {0, 3}, there are oddtreesT oftype S such that mc(T) = 3.

Proof. Consider the tree T of Figure 3.6, each of whose leaves are at level 3 or at level 6. We show that mc(T) = 3. For each integer i with 0 : i : 6, let ¼ = { x E

56 V(T) : d(v, x) = i}. Thus if xis a leaf of T, then x E Vi or x E Vi- By Corollary 3.3.3, mc(T) ::::; 3. Thus it remains to show that mc(T) -=I= 2. Assume, to the contrary, that there is a closed modular 2-coloring c: V(T)-+ Z2 of T. Thus c(v) = 0 or c(v) = 1. We consider these two cases.

Figure 3.6: A tree T with mc(T) = 3

Case l. c(v) = 0. Since either c'( v) = 0 or c'( v) = 1, there are two subcases. Subcase l. 1. c'( v) = 0. In this case, there is a child of v that is colored 0. First, assume that c(u1) = 0. Since c'( u1) = 1, there is a child w of u1 such that c( w) = 1. Since c'( w) = 0 and c(w) = 1, there is a child w1 of w such that c(w1) = 1. However then, c'( w1) = c' ( w) = 0, a contradiction. Thus c( u1) = 1 and similarly, c( u2) = 1. This implies that c(u) must be 0. Note that c(x1) = 1 or c(x1) = 0. If c(x1) = 1, then there is a child w of x1 such that c(w) = 1. However then, c'(x1) = c'(w) = 0, a contradiction. If c(x1) = 0, then there is a child w of x1 such that c(w) = 0. However then, c'(xi) = c'(w) = 0, a contradiction. Subcase 1.2. c'(v) = 1. Since c(v) = 0, either exactly one or exactly three children of v must be colored 1. First, suppose that c(u1) = 0. Then there is a child w of u1 such that c(w) = 0. Since c'( w) = 1, there is a child w1 of w such that c(w 1) = 1. However then, c'(w1) = c'(w) = 1, a contradiction. Thus c(u1) = 1 and similarly, c(u2) = 1. This implies that c(u) must be 1. Note that c(x) = 1 or c(x) = 0. We consider these two subcases.

Subcase 1.2.1. c(x) = 1. In this case, either exactly one or exactly three children of x must be colored 1. Since c'(x) = 1, it follows c(y1) = 1 (for otherwise, c'(y1) = 1). Similarly c(y2) = 1. Thus c(y) must be 1. Since c' (y) = 0, there is a child w of y such that c(w) = 0. Since c' (w) = 1, there is a child w1 of w such that c(w1) = 0.

This in turn implies that there is a child w2 of w1 such that c(w 2) = 0. However then,

57 c'(w1) = c'(w2) = 0, a contradiction. Subcase 1.2.2. c(x) = 0. Since c'(x) = 1, it fllows that c(y1) = c(y2) = 0, which implies that c(y) = 0. Since c' (y) = c(y) = 0, there is a child w of y such that c( w) = 0. Since c'( w) = 1, there is a child w1 of w such that c(w1) = 1. This in turn implies that there is a child w2 of w1 such that c(w2) = 1. However then, c'(wi) = c'(w2) = 0, a contradiction.

Case 2. c( v) = 1. Since either c' ( v) = 0 or c' ( v) = 1, there are two subcases. Subcase 2.1. c'(v) = 0. Note that c(u1) = 0 or c(u1) = 1. First, assume that c(u1) = 0. Since c'(u1) = 1, there is a child w of u1 such that c(w) = 0. Since c'(w) = 0, there is a child w1 of w such that c(w1) = 0. However then, c'(w1) = c'(w) = 0, a contradiction. Thus c(u1) = 1 and similarly, c(u2) = 1. This implies that c(u) must be 1. Note that c(x) = 0 or c(x) = 1. There are two subcases. Subcase 2.1.1. c(x) = 0. If c(y1) = 0, then c'(y) = c'(x) = 0, a contradiction. Thus c(y1) = 1 and similarly, c(y2) = 1. This implies that c(y) must be 1. Since c'(y) = 1, there is a child w of y such that c(w) = 0. Since c'(w) = 0 and c(y) = 1, there is a child w1 of w such that c(w1) = 1. This in turn implies that there is a child w2 of w1 such that c(w2) = 0. However then, c'(w1) = c'(w2) = 1, a contradiction. Subcase 2.1.2. c(x) = 1. If c(y1) = 1, then c'(y) = c'(x) = 0, a contradiction. Thus c(y1) = 0 and similarly, c(y2) = 0. This implies that c(y) must be 0. Since c'(y) = 1 and c(x) = 1, there is a child w of y such that c(w) = 0. Since c'(w) = 0, there is a child w1 of w such that c(w1) = 0. This in turn implies that there is a child w2 of w1 such that c(w2) = 0. However then, c'(w1) = c'(w2) = 1, a contradiction. Subcase 2.2. c'(v) = 1. Note that c(u1) = 0 or c(u1) = 1. First, assume that c(u1) = 0. Since c'(u1) = c(u1) = 0 and c(v) = 1, there is a child w of u1 such that c(w) = 1. Since c'(w) = 1, there is a child w1 of w such that c(w1) = 0. However then, c'(w1) = c'(w) = 1, a contradiction. Thus c(u1) = 1 and similarly, c(u2) = 1. This implies that c( u) must be 0. Note that c(x1) = 0 or c(x1) = 1. Furthermore, c'(x1) = 1. If c(x1) = 0, then there is a child w of x1 such that c(w) = 1. However then, c'(x1) = c'(w) = 1, a contradiction. If c(x1) = 1, then there is a child w of x1 such that c(w) = 0. However then, c'(x1) = c'(w) = 1, a contradiction. ■ As with the case when S = {2, 5}, there are odd rooted trees of ty pe S = {0, 3} having closed modular chromatic number 2. For example, we start with the tree T in Figure 3.6. Let T' be the subtree of T whose vertex set consists of y and all descendants of y. Then the tree T* is constructed from T of Figure 3.6 by replacing Y1 with a copy

58 of T' in the fashion as we did in the case when S = {2, 5} (see Figure 3.5). The coloring c : V(T) ➔ Z2 defned by assigning the color 0 to each vertex in { v, y, Y1} U V5 and assigning the color 1 to the remaining vertices of T* is a closed modular 2-coloring. Therefore, mc(T*) = 2. Next, we show that if S is a nonempty subset of {0, 2, 3,4, 5} such that S contains at most one of 2 and 5 and at most one of 0 and 3, then every odd rooted tree of type S has closed modular chromatic number 2.

Theorem 3.4.5 Let S be a nonempty subset of {0, 2, 3, 4, 5} such that S contains at most one of 2 and 5 and at most one of O and 3. If T is an odd rooted tree of order at least 3 that is of type S, then mc(T) = 2.

Proof. By Theorem 3.4.1, we may assume that ISi � 2. Since IS n {2, 5}1 � 1 and ISn{0,3}1 � 1, it follows that ISi � 3. Thus we consider two cases, according to whether ISi = 3 or ISi = 2. Case 1. ISi = 3. Then Sis one of the sets {0,2, 4}, {0, 4, 5},{2, 3, 4},{3, 4, 5}. Since x(T) = 2 for every nontrivial tree T, it sufces to show that there is a closed modular 2-coloring by Proposition 2 .1.3. For each integer i with 0 � i � e (v), let

¼ = {x E V(T): d(v,x) = i}.

First, suppose that S = {0,2, 4}. Defne a coloring c: V(T) ➔ Z2 by

if x E ¼ and i is odd c(x) = { � if x E ¼ and i is even.

Then c'(x) = c(x) fr each x E V(T). Next, suppose that S is one of {0, 4, 5} and {2,3, 4}. If S = {0,4, 5}, then defne a coloring c: V(T) ➔ Z2 by

if x E ¼ where i 0, 1, 5 (mod 6) c(x) = { � = if x E ¼ where i = 2,3,4 (mod 6).

If S = {2,3, 4}, then defne a coloring c: V(T) ➔ Z2 by

if x E ¼ where i 3, 4, 5 (mod 6) c(x) = { � = if x E ¼ where i = 0, 1,2 (mod 6).

59 In either case, c'(x) = 0 if x E ¼ and i is even and c'(x) = 1 if x E ¼ and i is odd.

Finally, suppose that S = {3, 4, 5}. Define a coloring c: V(T) -+ Z2 by

if x E ¼ where i 0, 4, 5 (mod 6) c(x) = { � = if x E ¼ where i = 1,2,3 (mod 6).

Then c'(x) = 0 if x E ¼ and i is odd and c'(x) = 1 if x E. ¼ and i is even. In each case, c is a closed modular 2-coloring of T and so mc(T) = 2. Case 2. ISi = 2. Then Sis a 2-element subset of one of the sets {0, 2, 4}, {0, 4, 5}, {2, 3, 4}, {3, 4, 5} in Case 1. Observe that those closed modular 2-colorings described in Case 1 will provides closed modular 2-colorings for this case. For example, if S is a 2-element subset of S' = {0, 2, 4}, then a closed modular 2-coloring of a tree of type S' described in Case 1 provides a closed modular 2-coloring of T. Therefore, mc(T) = 2 in this case as well. ■

Theorem 3.4.6 For S = {0, 1}, there are odd rooted trees of type S such that mc(T) = 3.

Proof. Let T be an odd rooted tree of type S = { 0, 1} having root v and height 6 such that v has exactly three children and exactly one of these three children is a leaf. Suppose that x is the child of v that is a leaf. We show mc(T) = 3. For each integer i with O � i � 6, let ¼ = {x E V(T) : d(v, x) = i}. Thus if w is a leaf of T, then w E Vi or w E V6. Assume, to the contrary, that there is a closed modular 2-coloring c: V(T) -+ Z2 of T. Thus c(v) = 0 or c(v) = 1. We consider these two cases. Case 1. c(v) = 0. Since either c'(v) = 0 or c'(v) = 1, there are two subcases. Subcase 1.1. c'(v) = 0. Since c'(x) = 1, it follows that c(x) = 1. This implies that exactly one other child of v can be colored 1, say this child is u1. Since c'(u1) = 1, there is a child u2 of u1 such that c(u2) = 0. Since c'(u2) = 0, there is a child U3 of u2 such that c(u3) = 1. Similarly, the fact that c'(u3) = 1 implies that there is a child u4 of u3 such that c(u4) = 0, there is a child u5 of u4 such that c(us) = 1 and there is a child U6 of u5 such that c(u6) = 0. However then, c'(us) = c'(u6) = 1, a contradiction. Subcase 1.2. c'(v) = 1. Since c'(x) = 0, it follows that c(x) = 0. This implies that exactly one other child of v can be colored 1, say this child is u1 and c(u1) = 1. Since c'(u1) = 0, there is a child u2 of u1 such that c(u2) = 1. Since c'(u2) = 1, there is a child u3 of u2 such that c(u3) = 1. Similarly, there is a child u4 of u3 such that c(u4) = 0,

60 there is a child u5 of u4 such that c(u5) = 0 and there is a child Ufi of u5 such that c(uB) = 0. However then, c'(u5) = c'(uB) = 0, a contradiction.

Case 2. c(v) = 1. Since either c'(v) = 0 or c'(v) = 1, there are two subcases.

Subcase 2.1. c'(v) = 0. Since c'(x) = 1, it follows that c(x) = 0. This implies that exactly one other child of v can be colored 1, say this child is u1 and c(u1) = 1. Since c'(u1) = 1, there is a child u2 of u1 such that c(u2) = 1. Since c'(u2) = 0, there is a child u3 of u2 such that c(u3) = 0. Similarly, there is a child.u4 of u3 such that c(u4) = 0, there is a child u5 of u4 such that c(u5) = 0 and there is a child Ufi of u5 such that c(uB) = 1. However then, c'(u5) = c'(uB) = 1, a contradiction.

Subcase 2.2. c'(v) = 1. Since c'(x) = 0, it follows that c(x) = 1. This implies that exactly one other child of v can be colored 1, say this child is u 1 and c( u 1) = 1. Since c'(u1) = 0, there is a child u2 of u1 such that c(u2) = 0. Since c'(u2) = 1, there is a child u3 of u2 such that c(u3) = 0. Similarly, there is a child u4 of u3 such that c(u4) = 0, there is a child u5 of u4 such that c(u5) = 1 and there is a child Ufi of u5 such that c(uB) = 1. However then, c'(u5) = c'(uB) = 0, a contradiction. ■

Theorem 3.4. 7 For S = {0, 1}, there are odd rooted trees of type S such that mc(T) = 2.

Proof. Let T be an odd rooted tree of type S = {0, 1} having root v. For each integer i with 0 � i � e(v), let½= {x E V(T): d(v,x) = i}. Thus if w is a leaf of T, then w E ½ where i = 0, 1 (mod 6). We claim that if for each i with i = 1 (mod 6) the set½ contains an even number of leaves of T, then mc(T) = 2. Since T has no true twins and x(T) = 2, it suffices to find a closed modular 2-coloring. Let X be the set of all leaves that belong to½ where i = 1 (mod 6). Define the coloring c: V(G) -t Z2 by

if u E ½ - X and i is odd c(u) = � { if u E X or u E ½ and i is even.

Then c'(u) = 1 if u E ½ and i is even and c'(u) = 0 if u E ½ and i is odd. Thus c is a closed modular 2-coloring of T and so mc(T) = 2. ■

61 Chapter 4

Closed Modular Colorings and Graph Operations

In this chapter we determine the closed modular chromatic numbers of graphs that are obtained fom various graph operations on some well-known classes of graphs, namely the complete graphs, paths and cycles.

4.1 Cartesian Products

For two graphs G and H, the Cartesian product G □ H has vertex set V(G □ H) = V(G) x V(H) and two distinct vertices (u, v) and (x, y) are adjacent in G □ H if either (1) u = x and vy E E(H) or (2) v = y and ux E E(G). In particular if H = K2, then the graph G □ K2 is obtained from two copies G1 and G2 of G by joining the corresponding vertices in G1 and G2, respectively. First, we consider the Cartesian products G □ K2 of G and K2, where GE {Kn, Pn, Cn }-

Proposition 4.1.1 For each integer n 2 2, mc(Kn □ K2) = x(Kn □ K2) = n.

Proof. Let G = Kn □ K2. Since G has no true twins andx(G) = n, it fllows by (2.1) that mc(G) 2 n. Thus, it remains only to show that mc(G) : n. Suppose that

V(G) = UUV where U = {u1,u2,.. ,,un } and V = {v1,v2,... ,vn } such that the subgraph induced by each of U and Vis complete and UiViE E(G) fr 1 :i :Sn. Defne a coloring c: V(G) ➔ Zn by c(ui) = i - 1 for 1 :i :Sn and c(vi) = i fr 1 :i :Sn. Let n f E Zn such that f = I:;�.:}i = (�-l). Then c'(ui) = f + (i - 1) and c'(vi) = f + i fr 1 : i : n. Thus c'( ui) #- c'( vi) for 1 : i :n. Furthermore, if i #- j where 1 : i, j : n,

62 then£+ i =£ + j (mod n), which implies that c'(ui) i c'(uj) and c'(vi) i c'(vj) fr all i,j with i y j. Thus c is a closed modular n-coloring and so me( G) : n. •

Theorem 4.1. 2 For each integer n 2: 3

ifn = 0,2, 4, 6, 7 (mod 8) ifn=l,3,5 (mod8).

G P F H ,v ) Proof. Let = n □ K2 and let = (u1,u2,. .. ,un) and = (v1,v2,... n be the two copies of Pn in G such that UiVi E E(G) fr 1 :i :Sn. Suppose that c: V(G)-+ Z2 is a closed modular 2-coloring. Then {c'(u1), c'(vi)} = {O,1}. By symmetry, we may assume that c'( u1) = 0 and c'( v1) = l. Hence,· the color sequences of the induced coloring c' on F and H are

Sc',F (c'(u1), c'(u2),... , c'(un)) = (0,1, 0, 1, ...)

Sc',H (c'(v1), c'(v2), ... , c'(vn)) = (1, 0, 1, 0,. ..).

Since c'(u1) = 0, it fllows that c(u1) + c(u2) + c(v1) = 0 (mod 2). Thus, either all of c(u1), c(u2) and c(v1) are O or exactly one of c(u1), c(u2) and c(v1) is 0. We consider four cases, according to the values of c(u1), c(u2) and c(v1).

Case l. c(u1) = c(u2) = c(v1) = 0. Since c'(v1) = 1 and c(u2) = c(v1) = 0, it fllows that c(v2) = 1. Since c'(u2) = 1 and c'(v2) = 0 and N[u2] - {u3} and N[v2] - {v3} are colored, it frces that c(u3) = 0 and c(v3) = l. Similarly, since c'(u3) = 0 and c'(v3) = 1 and N[u3] - {u4} and N[v3] - {v4} are colored, we must have c(u4) = c(v4) = l. If we continue this procedure, then we see fr each integer i 2: 4, since c'( ui) and c'( vi) are fxed and N[ui] - {uH1} and N[vi] - {vi+i} are colored, the values c(ui+l) and c(vi+1) are uniquely determined. In fact, if c: V(G) - Z2 is a closed modular 2-coloring of G such that c(u1) = c(u2) = c(v1) = 0, then there is only one possibility fr c, namely

0 ifi:0,1,2,3 (mod8) c(ui) = { 1 ifi = 4, 5, 6, 7 (mod8) 0 ifi 0,1, 6, 7 (mod8) c(Vi) = = { 1 ifi = 2,3, 4, 5 (mod8).

The coloring c is partially shown in Figure 4.1 and this pattern repeats. Observe that c(ui) = c(vi) = 0 only when i = 0,1 (mod8). If n = 0,7 (mod8), then c is a closed

63 modular 2-coloring of G; while if n "¥:- 0, 7 (mod 8), thenc'(u n) = c'(vn) in each case and soc cannot be a closed modular 2-coloring of G. Therefre, ifc: V(G) - Z2 is a closed modular 2-coloring of G such that c(u1) = c(u2) = c(v1) = 0, then n = 0, 7 (mod 8).

Figure 4.1: Illustrating the coloring c in Case 1

Case 2. c(u1) = 0 and c(u2) = c(v1) = 1. Since c'(v1) = 1 and c(v1) = 1, it fllows that c(v2) = 0. Since c'(u2) = 1 and c'(v2) = 0, and N[u2] - { u3} and N[v2] - { v3} are colored, it frces that c(u3) = c(v3) = 0. An argument similar to the one in Cae 1 show that fr each integer i � 3, since c'(ui) and c'(vi) are fxed and N[ui] -{ui+l} and N[vi] - {vi+i} are colored, the values c(ui+1) and c(vi+1) are uniquely determined. Thus, if c : V(G) - Z2 is a closed modular 2-coloring of G such that c(u1) = 0 and c( u2) =c( v1) = 1, then there is only one possibility fr c, namely

0 if i 0, 1, 3, 6 (mod 8) c(ui) = { 1 if i = 2, 4, 5, 7 (mod 8) 0 if i 0, 2, 3, 5 (mod 8) c(Vi) = = { 1 if i = 1, 4, 5, 6, 7 (mod 8).

The coloring c is partially shown in Figure 4.2 and this pattern repeats. Observe that c(ui) = c(vi) = 0 only when i = 0, 3 (mod 8). If n = 2, 7 (mod 8), then c is a closed modular 2-coloring of G; while if n "¥:- 2, 7 (mod 8), then c'(un) = c'(vn) in each case and soc cannot be a closed modular 2-coloring of G. Therefre, if c : V(G) - Z2 is a closed modular 2-coloring of G such that c(u1) = 0 and c(u2) = c(v1) = 1, then n = 2, 7 (mod 8).

Figure 4.2: Illustrating the coloring c in Cae 2

64 Case 3. c(u1) = 1, c(u2) = 0 and c(v1) = l. This forces that c(v2) = 1 and there is only one one possibility for c, namely

0 if i 0, 2, 5, 7 (mod 8) c(Ui) = { 1 if i = 1, 3, 4, 6 (mod 8) 0 if i 0, 3, 5, 6 (mod 8) c(vi) = = { 1 if i = 1, 2, 4, 7 (mod 8).

The coloring c is partially shown in Figure 4.3 and this pattern repeats. Observe that c(ui) = c(vi) = 0 only when i = 0, 5 (mod 8). If n = 4, 7 (mod 8), then c is a closed modular 2-coloring of G; while if n ,/:. 2, 7 (mod 8), then c cannot be a closed modular 2-coloring of G. Therefore, if c : V(G) ➔ Z2 is a closed modular 2-coloring of G such that c(u1) = 1, c(u2) = 0 and c(v1) = 1, then n = 4, 7 (mod 8).

Figure 4.3: Illustrating the coloring c in Case 3

Case 4. c(u1) = c(u2) = 1 and c(v1) = 0. This forces that c(v2) = 0 and there is only one one possibility for c, namely

0 if i 0, 5, 6, 7 (mod 8) c(ui) = { 1 if i = 1, 2, 3, 4, (mod 8) 0 if i 0, 1, 2, 7 (mod 8) c(vi) = = { 1 if i = 3, 4, 5, 6, 7 (mod 8).

The coloring c is partially shown in Figure 4.4 and this pattern repeats. Observe that c(ui) = c(vi) = 0 only when i = 0, 7 (mod 8). If n = 6, 7 (mod 8), then c is a closed modular 2-coloring of G; while if n ,!. 6, 7 (mod 8), then c cannot be a closed modular 2-coloring of G. Therefore, if c : V(G) ➔ Z2 is a closed modular 2-coloring of G such that c(u1) = c(u2) = 1 and c(v1) = 0, then n = 6, 7 (mod 8). By Cases 1 - 4, if n = 0, 2, 4, 6, 7 (mod 8), then G has a closed modular 2-coloring and so mc(G) = 2; while if n = 1, 3, 5 (mod 8), then G does not have a closed modular

65 Figure 4.4: Illustrating the coloring c in Case 4

2-coloring and so mc(G) 2: 3. For an odd integer n, define a coloring c: V(G) -+ Z3 by c(ui) = 0 if i is even and c(ui) = 2 if i is odd for 1 :Si :Sn and c(vj) = 1 for 1 :Sj :Sn. Then c'(ui) = 2 if i is even and c'(ui) = 0 if i is odd where 1 :S i n.:S Furthermore, c'(v ) 1, c'(vj) 0 if j is even and c'(vj) 2 if j is odd and 2 :Sj :Sn - l. c'(v1) = n = = = Since c is a closed modular 3-coloring, mc(G) = 3 if n = l, 3, 5 (mod 8). ■

Next we determine the closed modular chromatic numbers of Cn □ K2 where n 2'. 3. K2) K2) We show that 2 :S mc(Cn □ :S 3 and mc(Cn D = 2 if and only if n 2'. 8and n = 0 (mod 8). We begin with odd cycles.

K2) Theorem 4.1.3 If n 2'. 3 is an odd integer, then mc(Cn □ = 3.

K2, .. ,v ,Vn Proof. For G = Cn □ let (u1,u2,.. . ,un,Un+1 = u1) and (v1,v2, . n +l = K2 E v1) be the two copies of Cn in Cn □ such that UiVi E( G) for 1 :S i n.:S By Theorem 2.1.3, mc(G) 2: 3 for each odd integer n 2'. 3 and so it remains to show that mc(G) :S 3. We consider two cases, according to whether n = l (mod 4) or n = 3 (mod 4). In each case, we define a closed modular 3-coloring of G. Z3 Case l. n = l (mod 4). For n = 5, define a coloring c: V(G)- such that (1, 0, 1, 0, 2) (c(u1),c(u2),c(u3),c(u4),c(u5)) (0, 2, 1, 0, 1). (c(v1),c(v2),c(v3),c(v4),c(v5)) See Figure 4.5. The induced coloring c' satisfies

(0, 1, 2, 0, 1) ( c' ( u1), c' ( u2),c' ( u3), c' ( U4), c'( u5)) (1, 0, 1, 2, 0). ( c'( v1), c'( v2), c' ( v3), c' ( V4), c' ( v5)) Z For n = 9, define a coloring c : V (G) -+ 3 such that (2,0,1,0,2,0,1,0,2) (c(u1), c(u2),... ,,c(ug)) (c(v1),c(v2),... , ,c(vg)) (1, 1,1, 1, 0, 1,1, 0, 0).

66 Proof. Define the coloring c : V (G) -+ Z2 by

0 ifi 0, 1,2, 7 (mod 8) c(ui) = { 1 ifi =3, 4, 5, 6, (mod 8) 0 ifi::::1,2,3,4 (mod 8) c( Vi) = { 1 ifi = 0, 5, 6, 7 (mod 8).

The� c'(ui) = 0 if i is odd and c'(ui) = 1 if i is even, while c'(vi) = 1 if i is odd and c' (Vi) = 0 if i is even (1 :S i :S n). Thus c is a closed modular2-coloring and so mc(G) =2. ■

Next we show if n �4 is an even integer and n ¢. 0 (mod 8), then mc(Cn D K2) -=I=2, beginning with n =4 or n = 6. Figure4.8 shows closed modular3-colorings of C4 D K2 and C6 D K2 and so mc(Cn D K2) :S 3 for n = 4 or n = 6. We show, in fact, that mc(Cn D K2) =3 for n =4 or n = 6 in the next two results.

2 0 ,,.,1 ,-----·1'

Figure4.8: Closed modular3-colorings for C4 D K2 and C6 D K2

Proposition 4.1.5 mc(C4 D K2) =3.

Proof. Since Q3 = C4 D K2 has a closed modular3-coloring, mc(Q3) :S3. Assume, to the contrary, that Q3 has a closed modular2-coloring c: V(G) -+ Z2. Let X and Y be the partite sets of Q3. Since the induced coloring c' is a proper2-coloring of Q3 and

Q3 is uniquely colorable, we may assume that c'(x) = 0 for all x EX and c'(y) = 1 for ally E Y. First, assume that c(x) = 0 for all x EX. Since c'(y) = 1 for each y E Y, it follows that c(y) = 1 for ally E Y. However then, c'(x) = 1 for all x E X, which is a contradiction. Next, assume that c(x) = 1 for all x E X. Then c(y) = 0 for all y E Y.

This implies that c'(x) = 1 for all x E X, a contradiction. Thus if n1 is the number of the vertices in X that are colored 1 by c, then n1 E {1,2, 3}. We consider these three cases. Let X = {x1 ,x2,x3,x4} and Y = {y1 ,Y2,y3,y4}.

69 Case l. ni =1, sayc(xi) =1 andc(xi) = 0 for i =1, 2, 3. Sincec'(xi) = 0, there is y E Y such thatc(y) =1. On the other hand, sincec' (y) =1, at least onex E X - {xi} must be colored1, which is a contradiction.

Case 2. ni = 2, sayc(xi) = c(x2 ) = 1 andc(x3) = c(x4) = 0. We may assume

thatYi andY 2 are adjacent to both X3 and X4 and to exactly one ofxi andx 2 , say Yi

is adjacent to xi and Y2 is adjacent to x2 . Since c'(yi) = c'(y2 ) = 1, it follows that c(yi) = c(y2 ) = 0. Nowy3 andy4 are adjacent to both-xi andx 2 and exactly one of X3 and X4 and so c(y3) = c(y4) = 1. However then, c'(x3) = c'(x4) = 1, which is a contradiction. Case 3. ni = 3, sayc(xi) = 1 for i =1, 2, 3 andc(x4) = 0. We may assume that

each ofYi, Y 2 , y3 is adjacent to exactly two vertices in {xi,x 2 , x3}. Thusc(yi) = 1 for i =1, 2, 3. However then, c'(x4) =1, which is a contradiction. ■

Proposition 4.1.6 mc(C6 D K2 ) = 3.

Proof. For G = C5 D K2 , let (ui,u 2 , ..., u5,u7 = ui) and (vi,v 2 , ..., v5, V7 = vi) E(G) b. G be the two copies ofC 5 in C5 □ K2 such that UiVi E for1 :S i :S Since has a closed modular 3-coloring, mc(G) :S 3. Assume, to the contrary, that G has a V(G) X Y G. closed modular 2-coloringc: ---+ Z2 . Let and be the partite sets of Since the induced coloringc' is a proper 2-coloring of G and G is uniquely colorable, we may assume that c'(x) = 0 for all x E X andc'(y) = 1 for ally E Y. First, assume that c(x) = 0 for allx E X. Sincec'(y) = 1 for eachy E Y, it follows that c(y) = 1 for all y E Y. However then, c'(x) = 1 for all x E X, which is a contradiction. Next, assume thatc( x) = 1 for allx E X. Then c(y) = 0 for ally E Y. This implies that c'( x) = 1 for all x E X, a contradiction. Thus if ni is the number of the vertices in X that are colored1 by c, then ni E {1, 2, 3, 4, 5}. We consider these five cases. Let X Y = {xi,x 2 , ... ,x 5 } and = {Yi,y 2 , ...,y 5 }. We may assume thatui =xi EX. Then X = {ui,v2,u3,v4,u5,v5} and Y = {vi,u2,v3,u4, V5,u5}. Case l. ni =1, sayc(xi) =1. Sincec'(xi) = 0, there isy E Y such thatc(y) =1. On the other hand, sincec' (y) =1, at least one x E X - {xi} must be colored1, which is a contradiction.

Case 2. ni = 2, sayc(xi) = c(x2 ) = 1. Then d(xi,x2) = 2 or d(xi,x2) = 4.

First, assume that d(xi,x 2 ) = 2. By symmetry and xi = ui, we may assume that x2 E {v2 ,u3}. Ifc(ui) = c(v2 ) = 1, then c(u3) = c(v4) = c(u5) = c(v5 ) = 0. Since c'(x) = 0 for allx E X and c'(y) = 1 for ally E Y, it follows that {c(x) : x E X}

70 uniquely determines each color c(y) for all y E Y. In particular, this forces c(vi) = 1, c(u2) = 1 and c(u6) = 0. However then, c'(u1) = 1, a contradiction. If c(u1) = c(u3) = 1, then c(v2) = c(v4) = c(u5) = c(v6) = 0. This forces c(u4) = 0, c( v3) = 0 and c( v5) = 1. However then, c'(v4) = 1, a contradiction. Next, assume that d(x1,x2) = 4. Then· x2 = v4. Thus c(x) = 0 for all x EX - {u1, v4}. This forces c(v1) = c(u2) = c(u6) = 0. However then, c'(u1) = 1, a contradiction. Case 3. n1 = 3, say c(xi) = 1 for 1 :S i :S 3. We may assume that x1, x2 E {u1,u2,... ,u6}- Thus d(x1,x2) = 2. By symmetry and x1 = u1, we can assume that x2 = u3. First, suppose that X3 E {u1,u2, ... , u6}- Then x3 = u5. Then c(v2) = c(v4) = c(v6) = 0. This forces c(u2) = 1, c(v1) = 0 and c(u6) = 1. However then, c'(u1) = 1, a contradiction. Next, suppose that x3 E {v1,v2,... ,v6}- Thus X3 E {v2,v4,v6}- If x3 = v2, then c(v1) = 1, c(u2) = 0 and c(v3) = 1 and so c'(v2) = 1, a contradiction. If x3 = v4, then c(v3) = 1, c(u4) = 1 and c(v5) = 0 and so c'(v4) = 1, a contradiction. If x3 = v6, then c(u6) = 1, c(v1) = 1 and c(v5) = 0 and so c'(v6) = 1, a contradiction. Case 4. n1 = 4, say c(x1) = c(x2) = 0. Then d(x1,x2) = 2 or d(x1,x2) = 4. First, suppose that d(x1,x2) = 2. Then we may assume that x2 E {v2,u3}. If x2 = v2, then c(v1) = c(u2) = 0 and c(u6) = 1. However then c'(u1) = 1, a contradiction. If x2 = u3, then c(v5) = 0 and c(u4) = c(u6) = 1. However then c'(u5) = 1, a contradiction. Next, suppose that d(x1,x2) = 4, then x2 = v4. This forces that c(v1) = c(u2) = c(u6) = 1 and so c'( u1) = 1, a contradiction. Case 5. n1 = 5, say c(x1) = 0. Then this forces c(vi) = c(u2) = c(u6) = 1 and so c' ( u1) = 1, a contradiction. ■

Theorem 4.1. 7 If n 2'. 4 is an even integer,

- {2 if n = 0 (mod 8) mc(Cn D K2) = 3 otherwise

Proof. As a consequence of Proposition 2.1.3, if n 2'. 4 is an even integer, then 2 :S mc(Cn D K2). By Proposition 4.1.4, it remains to show that if n 2'. 4 is even such that n ¢. 0 (mod 8), then mc(Cn D K2) = 3. For G = Cn □ K2, let (u1, u2,... , Un, Un+l = u1) and (v1, v2,... , Vn,V n+l = v1) be the two copies of Cn in Cn □ K2 such that UiVi E E(G) for 1 :S i :S n. We first show that mc(G) :S 3. Define a coloring c : V(G) -+ Z3 by c(ui) = 1 for 1 :S i :S n and c(vi) = 0 if i is odd and c(vi) = 2 if i is even (1 :S i :S n). Then c' ( ui) = 0 if i is odd and c'( ui) = 2 if i is even, while c'( vi) = 2 if i is odd and

71 c'(vi) = 0 ifi is even where 1 � i � n. Figure 4.8 shows the coloring c fr n = 4,6. Thus c is a closed modular 3-coloring and so me( G) � 3 fr all even integers n � 4.

Next, we show that mc(G) � 3 for all even integers n � 4. By Propositions 4.1.5 and 4.1.6, we may assume that n � 10. For G = Cn □ K2, let (u1 ,u2, ... ,U n+I = u1 ) and (v1 ,v2, ... ,V n+I = v1) be the two copies of Cn in Cn □ K2 such that UiVi E E(G) fr 1 � i � n. Since G ha a closed modular 3-coloring, me( G) � 3. Assume, to the contrary, that G has a closed modular 2-coloring c : V(G) -+ Z2. Let X and Y be the partite sets of G. Since the induced coloring c' is a proper 2-coloring of G and G is uniquely colorable, we may assume that c'(x) = 0 for all x E X and c'(y) = 1 for all y E Y. We may assume that u1 EX. Let

U1 { Ui : i is odd and 1 � i � n - 1} U2 { Ui i is even and 2 � i � n } Vi {vi : i is odd and 1 � i � n - 1} Vi { Vi : i is even and 2 � i � n }

Then X = U1 U V2 and Y = U2 U Vi- By assumption, c'(u1) = 0. We consider two cases, according to whether c( u1 ) = 0 or c( u1 ) = 1.

Case 1. c(u1 ) = 0. Since c'(u1) = 0, either all of the three neighbors u2,U n,v1 of u1 are colored 0 by c or exactly two of them are colored 1 by c. There are two subcases.

Subcase 1.1. All of u2,U n,V 1 are colored 0 by c. Since c' ( v1 ) = 1 and c(v1) = 0,exactly one of v2 and Vn is colored 0 by c. By symmetry, we can assume that c(v n) = 0 and c(v2) = 1. Since c'(u2) = 1,c'(v2) = 0 and each vertex in N[u2] - {u3} and N[v2] - {v3} are colored, it frces that c( u3) = 0 and c(v3) = 1. Similarly, since c'( u3) = 0 and c'(v3) = 1 and all vertices N[u3] - {u4} and N[v3] - {v4} are colored, we must have c(u4) = c(v4) = 1. If we continue this procedure, then we see fr each integer i � 4,since c' ( Ui) and c'(Vi) are fxed and all vertices in N [ui] - { ui+l} and N [ vi] - { vi+ 1} are colored, V(G) the values c(uH 1 ) and c(vi+ 1 ) are uniquely determined. In fact, if,c : -+ Z2 is a closed modular 2-coloring of G with the desired properties in this case,then there is only one possibility for c,namely

0 if i = 0,1, 2, 3 (mod 8) { 1 ifi = 4,5, 6, 7 (mod 8) 0 ifi:0,1,6,7 (mod 8) { 1 ifi = 2,3, 4, 5 (mod 8).

72 If n = 2 (mod 8), then (c(un),c(v n)) = (0, l); if n = 4 (mod 8), then (c(un),c(v n)) = (1,1); if n = 6 (mod 8), then (c(un), c(vn)) = (1, 0). Since (c(un),c(v n)) = (0,0) in this case, a contradiction is produced.

Subcase 1.2. Exactly two of u2, Un, v1 are colored 1 by c. We may assume, without loss of generality, that either c(u2) = c(un) = 1 or c(u2) = c(v1) = 1. First, suppose that c(u2) = c(un) = 1. By symmetry, we can assume that c(vn) = 0. With an argument similar to the one used in Subcase 1.1, we can show that if c is a closed modular 2- coloring of G with the desired properties in this case,then the values of c( ui) and c( vi) for i 2: 3 are

if i 1,4, 6, 7 (mod 8) c(ui) = if i 0, 2, 3, 5 (mod 8) � =

{ ifi 0, 1,3, 6 (mod 8) c( Vi) = = ifi 2, 4, 5, 7 (mod 8). { � = If n = 2 (mod 8), then (c(un),c(v n)) = (1,1); if n = 4 (mod 8), then (c(un), c(vn)) = (0, l); and if n = 6 (mod 8), then (c(un),c(v n)) = (0, 0). Since (c(un),c(v n)) = (1,0) in this case, a contradiction is produced when n = 2 (mod 8) or n = 4 (mod 8). Next, suppose that c(u2) = c(vi) = 1. Then either c(v2) = c(vn) = 0 or c(v2) = c(vn) = 1. If c(v2) = c(vn) = 0, then the values of c(ui) and c(vi) fori 2: 3 are

ifi::::0,1,3,6 (mod 8) c(ui) if i 2, 4, 5, 7 (mod 8) � =

{ if i 0, 2, 3, 5 (mod 8) c( Vi) = = if i 1,4, 6, 7 (mod 8). { � = If n = 2 (mod 8), then (c(un),c(v n)) = (1,0); if n = 4 (mod 8), then (c(un), c(vn)) = (1,1); while if n = 6 (mod 8), then (c(un),c(v n)) = (0,1). Since (c(un), c(vn)) = (0, 0) in this case,a contradiction is produced. If c(v2) = c(v n) = 1, then the values of c(ui) and c(vi) fori 2: 3 are

if i = 0, 1,6, 7 (mod 8) ifi 2, 3, 4, 5 (mod 8) { � = if i = 4, 5, 6, 7 (mod 8) � ifi::::0,1,2,3 (mod 8). { 73 If n = 2 (mod 8), then (c(un),c(v n)) = (1,1); if n = 4 (mod 8), then (c(un),c(v n)) = (1,O) ; if n = 6 (mod 8), then (c(un),c(v n)) = (0, 0). Since (c(un), c(vn)) = (0,1) in this case, a contradiction is produced.

Case 2. c(u1) = 1. Since c'(u1) = 0, either all of the three neighbors u2, Un,v1 of u1 are colored 1 by c or exactly one of them are colored 1 by c. There are two subcases.

Subcase 2.1. All of u2, Un,v1 are colored l by c. Since c'(v1) = 1, exactly one of v2 and Vn is colored Oby c, say c(vn) = 0 and c(v1) = 1. Then the values of c(ui) and c(vi) fr i � 3 are

if i 3,4, 5, 6 (mod 8) c( ui) � = { if i = 0, 1, 2, 7 (mod 8) if i 0, 5, 6, 7 (mod 8) c( Vi) = � = { if i = 1, 2, 3, 4 (mod 8). (1, 1); If n = 2 (mod 8), then (c(un),c(v n)) = if n = 4 (mod 8), then (c(un),c(v n)) = (0, 1); if n = 6 (mod 8), then ( c( un),c( v n)) = (0, 0). Since ( c( Un), c( vn)) = (1, 0) in this case, a contradiction is produced.

Subcase 2.2. Exactly one of u2, Un,v1 is colored l by c. We may assume, without loss of generality, c(v1) = 1 or c( u2) = 1. First, assume that c(v1) = 1. Since c' (v1) = 1, exactly one of v2 and Vn is colored Oby c, say c(vn) = 0 and c(v1) = 1. Then the values of c(ui) and c(vi) fr i � 3 are

if i 0, 2, 5, 7 (mod 8) c(ui) � = { if i = 1, 3, 4, 6 (mod 8) if i 0, 3, 5, 6 (mod 8) c( Vi) � = { if i = 1,2, 4, 7 (mod 8). If n = 2 (mod 8), then (c(un),c(vn)) = (0, 1); if n = 4 (mod 8), then (c(un),c(vn)) = (1, 1); if n = 6 (mod 8), then (c(un),c(v n)) = (1, 0). Since (c(un),c(v n)) = (0, 0) in this case, a contradiction is produced.

Next, assume that c(u2) = 1. Since c'(v1) = 1 and c(v1) = 0, either c(vn) = c(v2) = 0 or c( vn) = c(v2) = 1. If c( vn) = c(v2) = 0, then the values of c( ui) and c( vi) fr i � 3

74 are if i = 0, 5, 6, 7 (mod 8) { � if i = 1,2, 3, 4 (mod 8) if i = 0, 1, 2,7 (mod 8) if i = 3, 4, 5, 6 (mod 8).

If n = 2 (mod 8), then (c(un), c(vn)) = (1,0); if n = 4 (mod 8), then (c(un),c(v n)) = (1,1); if n = 6 (mod 8), then (c(un), c(vn)) = (0, 1). Since (c(un),c(v n)) = (0,0) in this case, a contradiction is produced. If c(vn) = c(v2) = 1, then the values of c(ui) and c(vi) fori 2: 3 are

if i 0, 3, 5, 6 (mod 8) c(ui) = { � ifi::::1,2,4,7 (mod 8) ifi = 1, 3,4, 6 (mod 8) c( Vi) { � ifi = 0, 2, 5, 7 (mod 8). If n = 2 (mod 8), then (c(un), c(vn)) = (1,1); if n = 4 (mod 8), then (c(un),c(v n)) = (1,0); if n = 6 (mod 8), then (c(un),c(vn)) = (0,0). Since (c(un),c(vn)) = (0, 1) in this case, a contradiction is produced. ■ Combing Theorem 4.1.3 and Theorem 4.1.7, we have the following result.

Corollary 4.1.8 For every integer n 2: 3,

if n 0 (mod 8) -mc(C D K2) { 2 = n = 3 otherwise.

4.2 Coronas of Graphs

For a graph H, the corona cor(H) of H is that graph obtained from H by adding a pendant edge to each vertex of H. By Theorem 3.1.4,

- 3 if n l ( mod 4) mc(cor(Pn)) = = { 2 otherwise.

Proposition 4.2.1 For each integer n 2: 3, mc(cor(Kn)) = n.

75 Proof. Let G = cor(Kn)- Since x(G) = n and G contains no true twins, mc(G) 2: n by Proposition 2.1.3. To show mc(G) :Sn, suppose that V(Kn) = {v1,v2, ... ,vn } and

V(G) = V(Kn) U { u1,u2, ... ,u n } where UiVi E E(G) for 1 :Si :Sn. Define a coloring c : V(G) ➔ Zn by c(vi) = 1 and c(ui) = i - 1 for 1 :S i :S n. Since c'(vi) = i - 1 and c' ( ui) = i for 1 :S i :Sn, it follows that c is a closed modular n-coloring of G and so me( G) :S n. Therefore, me( G) = n. ■ For a nontrivial connected graph H, let co : V(H) ➔ Zk be a coloring of H where k 2: 2 and c0 may not be a closed modular k-coloring. A vertex v of G is called a color-fi,r:ed vertex with respect to co if co(v) = dcJ(v).

Proposition 4.2.2 Let G be a connected graph of order at least 3 and let u be an end vertex of G. Suppose that u is adjacent to the vertex v in G. If v is a color-fixed vertex with respect to every coloring co : V ( G - v) ➔ Zk for some fixed integer k 2: 2, then G does not have a closed modular k-coloring.

Proof. Let H = G - v. Assume, to the contrary, that G has a closed modular k coloring c : V ( G) ➔ Zk. Let co be the restriction of c to H, that is, Co : V ( H) ➔ Zk is defined by co(x) = c(x) for all x E V(H). By assumption, vis a color-fixed vertex with respect to eo and so co(v) = c0(v). However then, c'(u) = c(v) + c(u) = co(v) + c(u) = Co ( V) + c(u ) = c' ( V), which is a contradiction. ■ The following corollaries are immediate consequences of Proposition 4.2.2.

Corollary 4.2.3 Let G = cor(H) for a graph H and let k 2: 2 be an integer. If H contains a color-fixed vertex with respect to every coloring co : V(H) ➔ Zk , then G does not have a closed modular k-coloring.

Corollary 4.2.4 Let G = cor(H) for a graph H and let k 2: 2 be an integer. If a coloring Co : V ( H) ➔ ;;zk such that co ( V) = Co ( V) for some V E V ( H)' then there is no closed modular k-coloring c of G such that the restriction of c to H is co.

Theorem 4.2.5 For each integer n 2: 4,

{ 2 ,ifn = 0 (mod 4) mc(cor(Cn)) = 3 otherwise.

76 Proof. Let G = cor(Cn)- Suppose that Cn = (v1,v2, ... , vn) and V(G) = V(Cn) U

{u1,u2,... ,un } where UiVi E E(G) for 1 :i :n. We consider two cases, according to whether n = 0 (mod 4) or n "# 0 (mod 4). Case l. n = 0 (mod 4). Defne a coloring c: V(G) ➔ Z2 by c(vi) = 1 if i = 1, 2 (mod 4) c(vi) = 0 if i = 0, 3 (mod 4) c(ui) = 1 if i = 0, 1 (mod 4) c(ui) = 0 if i = 2, 3 (mod 4).

For each i with 1 : i : n, c'(vi) = 1 and c'(ui) = 0 if i is odd and c'(vi) = 0 and c'(ui) = 1 if i is even. Thus c is a closed modular 2-coloring of G and so mc(G) = 2. Case 2. n "# 0 (mod 4). Thus either n is odd or n = 2 (mod 4). First, suppose that n is odd and so x(G) = 3. Since G contains no true twins, mc(G) � 3 by Proposition 2.1.3. To show that mc(G) : 3, defne a coloring c : V(G) ➔ Z3 by c(vi) = 1 fr 1 :i : n, c(un) = 0, c(ui) = 1 if i is odd and 1 : i : n - l and c(ui) = 2 if i is even and 2 : i :n - l. Then c'(vi) = c(ui) and c'(ui) = c(ui) + 1 fr 1 :i : n. Hence c is a closed modular 3-coloring of G and so mc(G) = 3. Next, suppose n = 2 (mod 4). We show mc(G) = 3 in this cae. First, we show mc(G) � 3 by Corollary 4.2.3. That is, we show that fr every coloring co : V(Cn) ➔ Z2, there is v E V(H) such that co(v) = c0(v). We proceed by two cases, according whether co(v1) = 0 or co(v1) = 1.

Subcase 2.1. co(v1) = 0. If co(v1) = 0, then let V = V1 and we have the desired result. Thus we may assume that c0(v1) = 1. Thus exactly one of co(v2) and co(vn) is 1, say co(v2) = 1 and co(vn) = 0. If co(v2) = 1, then let V = V2 and we have the desired result. Thus c0(v2) = 0 and so co(v3) = 1. Continuing in this manner and assuming that co(vi) = c0(vi) fr each i with 1 : i : n - l, we conclude that the color sequence sc0 must be

Seo = (0, 1,1, 0, 0, 1, 1, 0, 0, . .. , 1, 1,0, 0, co(v n)).

Since co(Vn-1) = co(Vn-2) = 0 and co(Vn-1) = co(Vn-i), it fllows that co(vn) must be 1, which contradicts to the fact that co(vn) = 0.

Subcase 2.2. co(v1) = 1. If co(v1) = 1, then let V = V1 and we have the desired result. Hence, we may asume that c0(v1) = 0. Thus exactly one of co(v2) and co(vn) is

77 1, say co(v2) = 1 and co(vn) = 0. Using an argument similar to the on in Subcase 2.1, we conclude that the color sequence Seo must be

Seo = (1,1,0,0,1,1,0,0, .. ., 1,1,0,0,1,co(Vn)).

However, in order for co(vn-1) =f c�(vn-1), the vertex Vn must be colored 1 by co. Since co(vn) = 0, this is a contradiction.

Hence, mc(G) � 3. To show that mc(G) :S 3, we define a coloring c: V(G) ➔ Z3 by c(vi) = 1 for 1 :Si :Sn, c(ui) = 0 if i is even and 2 :Si :Sn and c(ui) = 1 if i is odd and 1 :S i :Sn. Then c' ( vi) = c(ui) and c' ( ui) = c(ui) + 1 for 1 :S i :Sn. Thus c' is proper and soc is a closed modular 3-coloring of G. Therefore, mc(G) = 3. ■ There is a question related to the closed modular chromatic numbers of coronas of graphs in general.

Problem 4.2.6 If H is a nontrivial connected graph with mc(H) = k, then what can we say about mc(cor(H))?

4.3 Joins of Graphs

For two vertex-disjoint graphs G and H, the join G + H of G and H has vertex set V(G) U V(H) and edge set consisting of E(G) U E(H) and all edges joining a vertex of G and a vertex of H. In particular, the join of a graph G and the trivial graph K1 is

obtained by joining the vertex of K1 to all vertices of G. First, we show that mc(G) is + bounded above by mc(G K1 ) for every nontrivial connected graph G. + Proposition 4.3.1 If G is a nontrivial connected graph, then mc(G) :Smc(G K1 ).

Proof. Suppose that mc(G + K1 ) = k and letc: V(G + K1 ) ➔ Zk be a closed modular k-coloring of G + K 1 · We show that the restriction cc ofc to G is also a closed modular k-coloring of G. Let w E V(G + K1) and w (j. V(G). For each x E V(G),

c'(x) = ( I: c(y)) + c(w) = ( I: cc(y)) + c(w) = ca(x) + c(w). (4.1) yENc(x) yENc(x)

If u and v are two adjacent vertices of G, thenc' ( u) =/c' ( v) and so ca(u) =/ cc( v) by ( 4.1). Therefore, cc is a closed modular k-coloring of G and so mc(G) :Sk = mc(G + K1). ■ 78 Proposition 4.3.2 Let G be a nontrivial connected graph and let H = G + K1, where ) V( K 1 = { v}. If me( H) = k, then for each element a E Zk , there is a closed modular k-coloring Ca : V(H) ➔ Zk of H such that ca(v) = a.

Proof. Let c: V(H) ➔ Zk be a closed modular k-coloring of G. We may assume that c(v) = a. Defne the coloring Ca : V(H) ➔ Zk by ca(u) = c(u) fr all u E V(G) and ca(v) = a. Then c�(x) = c'(x) - c(v) + a for all x E V(H). (4.2) Let x, y E V(H) be two adjacent vertices of H. Since c is a closed modular k-coloring of H, it fllows that c'(x) = c'(y). Thus c�(x) = c�(y) by (4.2) and so Ca is a closed modular k-coloring of H. ■ + By Proposition 4.3.2, there is a closed modular k-coloring c: V(G K1 ) ➔ Zk such that c(v) = 0.

4.3.1 The Closed Modular Chromatic Number of Pn + K1

We now determine the closed modular chromatic number of Pn + K1 fr n � 3. If n = 3, then the true twin closure TC(P3 + K1 ) is P3 . It then fllows by Corollary 2.1.5 and

Proposition 2.2.2 that mc(P3 + K1 ) = mc(P3) = 2. In this subsection, we show that fr each integer n � 4

if n = l, 3 (mod 18) or n = 2 (mod 6) otherwise.

In order to do this, we frst we present three lemmas. Recall, fr the path Pn =

(v1, v2, ... , vn) of order n � 4 and a coloring c : V(Pn) ➔ Zk where k � 2, that the color sequence Sc of c is defned as Sc= (c(v1 ), c(v2), ... , c(vn)).

Lemma 4.3.3 For each integer n � 4, mc(Pn + K1 ): 4.

Proof. Let G = Pn + K1 where Pn = (v1 , v2, ... vn) and V(K1) = { v }. Let n = r (mod 4) where r E {0, 1, 2, 3}. We consider these fur cases. In each case, we provide a closed modular 4-coloring c: V(G) ➔ Z4 such that c(v) = 0. Hence to defne such a coloring c, it sufces to describe the color sequence Sc of c on the path Pn.

Case 1. n = 0 (mod 4). The color sequence Sc of con Pn is

Sc= (3, 2, 1, 2, 3, 2, 1, 2, ... , 3, 2, 1, 2).

79 Thus c'(v) = 0 and if n = 4, then Sc' = (1,2, 1,3 ); while if n 2 8, then

Sc' = (1, 2,1, 2, 3,2, 1, 2, 3,2, 1,2, ... ,3, 2, 1, 2, 3, 2, 1, 3).

Case 2. n = l (mod 4). The color sequence Sc of con Pn is

Sc = (1,2, 3,2, 1,2, 3, 2, ... ,1, 2, 3, 2, 0 ).

Thus c'(v) = 0 and if n = 5, then Sc' = (3,2, 3, 1,2 ); while if n 2 9, then

sd = (3,2,3,2,1,2,3,2,1,2,3,2,... ,1,2,3,2,1,2,3,1,2 ).

Case 3. n = 2 (mod 4). If n = 2 (mod 8), then the color sequence Sc of con Pn is

Sc= (3,0,3,0,3,0,3,0, ... ,3,0,3,0,3,0,1,1,2,0).

Then c'(v) = 1 and

Sd = (3,2,3,2,3,2,3,2, ... ,3,2,3,2,3,0,2,0,3,2).

If n = 6 (mod 8), then the color sequence Sc of con Pn is Sc = (3, 0, 3,1, 2, 0) if n = 6; while fr n 2 14,

Sc= (3,0,3,0,3,0,3,0, ... ,3,0,3,0,3,0,1,1,2,0).

Thus c'(v) = 1 and if n = 6, then Sc' = (3,2, 0, 2, 3, 2); while if n 2 14, then

sd = (3,2,3,2,3,2,3,2,... ,3,2,3,2,3,0,2,0,3,2 ).

Case 4. n = 3 (mod 4). If n = 3 (mod 8), then the color sequence Sc of con Pn is

Sc = (3,0,3,0,3,0,3,0, ... ,3,0,3,0,3,0,3,1,2,0,2).

Then c'(v) = 1 and

Sc' = (3,2, 3, 2, 3, 2,3, 2, ... , 3, 2, 3,2, 3, 2, 0, 2, 3, 0, 2).

If n = 7 (mod 8), then the color sequence Sc of con Pn is

Sc= (3,0, 3, 0, ... , 3, 0, 3).

In this case, c' ( v) = 0 and

80 Sc' = (3, 2,3, 2,... ,3, 2, 3).

In each case, c is a closed modular 4-coloring of G and so me( G) :S: 4. ■

Lemma 4.3.4 If n � 4 is an integer such that with n = 2 (mod 6), then Pn + K1 has no closed modular 3-coloring.

Proof. G + Let = Pn K1 where Pn = (v1,v2,... ,v n) arid V(K1 ) = {v}. Assume, to V(G) the contrary, that there is a closed modular 3-coloring c: -+ Z3 for some integer n � 4 such that n = 2 (mod 6). Since v is adjacent to every vertex of Pn , it follows that c' ( v) =/- c' ( vi) for all 1 :S: i :S: n and so the set C' = { c' ( vi) : 1 :S: i :S: n} consists of exactly two elements in {0,1,2}; that is, C' = {0,1}, C' = {0,2} or C' = {1,2}. We consider these three cases. First, we ma�e an observation. If C' = { a, b}, where b a< band a, E {0, 1,2}, then by the symmetry of Pn where n is even, we may assume b, b, ... , b). that c'(v1) = a and so Sc' = (a, a, a, Furthermore, we may also assume that c(v) = 0 by Proposition 4.3.2.

Case 1. C' = {0,1}. Then Sc'= (0,1,0,1, ... ,0,1). Since c(v1) E {0,1,2}, we consider these three possibilities for c(v1). First, assume c(v1) = 0. Since c'(v1) = 0, we must have c(v2) = 0. Since c'(v2) = 1, we must have c(v3) = 1. Observe that, for each i � 3, the color c(vi) is uniquely determined by c'(vi-1),c(vi-2) and c(vi-1). Thus by repeating the argument as above, we obtain the color sequence

Sc = (0,0,1,2,1,0,... ,0,0,1,2,1,0,0,0).

1, However then, c'(vn- 1) = c'(vn) = 0, which is a contradiction. Similarly, if c(v1 ) = then we must obtain the color sequence

Sc = (1,2,1,0,0,0, ... ,l,2,1,0,0,0,1,2).

However then, c'(vn-1) = c'(vn) = 0, which is a contradiction. Finally, if c(v1) = 2, then we must obtain the color sequence

Sc = (2, 1, 1, 1,2, 0,... ,2, 1, 1,1, 2,0, 2, 1).

However then, c'(vn-1 ) = c'(vn) = 0, which is a contradiction.

Case 2. C' = {0, 2}. Then Sc' = (0, 2, 0,2, ... ,0, 2). We apply an argument similar to the one used in Case 1. Since c(v1) E {O, 1, 2}, we consider these three choices for c(v1). If c(v1) = 0, then we obtain the color sequence 81 Sc= (0,0,2, 1,2,0,... ,0,0,2,1,2,0,0,0).

However then, c'(vn-1) = c'(vn) = 0, a contradiction. If c(v1) = 1, then we obtain the color sequence

Sc = (1,2,2,2,1,0,... ,l,2,2,2, 1,0,1,2).

However then, c'(vn-1) = c'(vn) = 0, which is a contradiction. If c(v1) = 2, then we must have the color sequence

Sc = (2,1,2,0,0,0, ... ,2,1,2,0,0,0,2,1).

However then, c'(vn-1) = c'(vn) = 0, which is a contradiction.

Case 3. C' = {1, 2}. Then Sc' = (1,2, 1, 2, ... ,1, 2). We again apply an argument similar to the one used in Case 1 and consider these three choices fr c(v1) E {0,1, 2 }. If c(v1) = 0, then we obtain the color sequence

Sc = (0, 1,1, 2, 2,0, ... ,0, 1, 1,2, 2,0, 0, 1).

However then, c(vn-1) = c(vn) = 1, which is a contradiction. If c(v1) 1, then we obtain the color sequence

Sc= (1,0, 1, 0, ... , 1,0, 1,0).

However then, c(vn-1) = c(vn) = 1, which is a contradiction. If c(v1) 2, then we obtain the color sequence

Sc = (2,2, 1, 1, 0, 0, ... ,2, 2,1, 1,0, 0, 2, 2).

However then, c(vn-1) = c(vn) = 1, which is a contradiction.

Therefre, there is no closed modular 3-coloring fr Pn + K1 fr n = 2 (mod 6). ■

Lemma 4.3.5 If n 2 4 is an integer such that either n = l (mod 18) or n 3 (mod 18), then Pn + K1 does not have a closed modular 3-coloring.

Proof. Let G = Pn +K1 where Pn = (vi,v2 ,... , vn) and V(K1) = { v }. Assume, to the V(G) contrary, that there is a closed modular 3-coloring c: -- Z3 fr some integer n 2 4

82 such that n = 1 (mod 18) or n = 3 (mod 18). Since v is adjacent to every vertex of Pn, it follows that c'(v) = c'(vi) fr all 1 :i :n and so C' = {c'(vi): 1 :i :n} consists of exactly two elements in {0, 1, 2}. As we have seen in the proof of Lemma 4.3.4, if c(v1) and c'(v1) are given, then the color of each vertex in Pn is uniquely determined by c(v1) and c'(v1). Furthermore, we can assume that c(v) = 0 by Proposition ·4.3.2. + First, suppose that n = 3 (mod 18). Then n � 21 and so n = 18k 3 fr k. some positive integer It can be shown that the colors in the color sequence Sc

(c(v1), c(v2), ..., c(vn)) of con the path Pn repeat every six vertices; that is,

(c(visk-5), c(visk-4), ..., c(v1sk)).

n + 3k Since = 18k 3, there are blocks ( c(V6i+1) , c( V6i+2), ... , c( V6i+6)) of six colors where 0 :i : 3k - l. Let a= c(v1) + c(v2) + ... + c(vB) be the sum of these six colors in such a block. Then

3k + + c'(v) = a + c(vn-2) c(vn-1) c(vn) = c'(vn-1), which is a contradiction. Thus, there is no closed modular 3-coloring when n 3 (mod 18). Next, suppose that n = 1 (mod 18). Since n is odd, if C' = {c'(vi) : 1 : i :n} = { a, b} where a, b E {0, 1, 2} and a < b, then c'(v1) = a and c'(v1) = b will be two diferent cases. Since C' is one of {0, 1}, {0, 2} and {1, 2}, there are six possible induced color sequences Sc' of the induced coloring c' on the path Pn. Thus, we consider these six cases. In each case, c(v1) E {0, 1, 2} and c'(v1) is given. Since c(v) = 0, the color of each vertex in Pn is uniquely determined by c( v1) and c'(v1) and so the color sequence

Sc = (c(v1), c(v2), ..., c(vn)) is uniquely determined by c(v1) and c'(v1).

Case 1. Sc' = (0, 1, 0, 1, ..., 0, 1, 0). In this case, c'(vi) = 0. If c(v1) = 0, then

Sc = (0, 0, 1, 2, 1, 0, 0, 0, 1, 2, 1, 0, ... , 0, 0, 1, 2, 1, 0, 0).

However then, c'(v) = 0, a contradiction. If c(v1) = 1, then

Sc = (1, 2, 1, 0, 0, 0, 1, 2, 1, 0, 0, 0, ... , 1, 2, 1, 0, 0, 0, 1).

83 ( ) In this case, c'(vn-1) = c'(vn) = 1, which is a contradiction If c v1 = 2, then

Sc= (2,1,1,1,2,0,2,1,1,1,2,0, ...,2,1,1,1,2,0,2 ).

However then, c'(vn) = 2, which is a contradiction.

( ) Case 2. Sc' = (0,2, 0, 2, ... ,0, 2, 0 ). If c v1 = 0, then

Sc= (0,0,2,1,2,0,0,0,2,1,2,0, ... ,0,0,2,1,2,0,0).

However then, c'(v) = 0, a contradiction. If c(v1) = 1, then

Sc = (1,2,2,2, 1,0,1,2,2,2,1,0, ... ,1,2,2,2,1,0,1).

( However then, c'(vn) = 1, which is impossible. If c v1) = 2, then

Sc = (2,1,2,0,0,0,2,1,2,0,0,0, ... ,2,1,2,0,0,0,2).

In this case, c'(vn-1) = c'(vn) = 2, a contradiction.

( ) Case 3. Sc' = (1,0, 1, 0, ... ,1, 0, 1 ). If c v1 = 0, then

Sc = (0,1,2,1,0,0,0,1,2,1,0,0, ... ,0,1,2,1,0,0,0).

( ) However then, c'(vn-1) = c'(vn) = 0, a contradiction. If c v1 = 1, then

Sc= (1,0,2,2,2,0,1,0,2,2,2,0, ...,1,0,2,2,2,0,1 ).

However then, c'(v) = 1, which is impossible. If c(v1) = 2, then

Sc = (2,2, 2, 0, 1, 0, 2, 2, 2,0, 1,0, ... , 2,2, 2, 0, 1,0, 2).

However then, c' ( vn) = 2, is a contradiction.

( ) Case 4. Sc' = (1,2, 1,2, ... ,1, 2, 1 ). If c v1 = 0, then

Sc= (0,1,1,2,2,0,0,1, 1,2,2,0, ... ,0,1,1,2,2,0,0).

( ) However then, c'(vn) = 0, a contradiction. If c v1 = 1, then

Sc= (1, 0,1, 0, ..., 1, 0, 1,0, 1 ).

However then, c'(v) = 1, which ·is a contradiction. If c(v1) = 2, then

Sc= (2, 2,1, 1, 0, 0, 2,2, 1, 1, 0, 0, ... , 2,2, 1, 1, 0, 0, 2). 84 However then, c'(vn-1) = c'(vn) = 2, a contradiction.

Case 5. Sc' = (2,0, 2,0, ... ,2, 0, 2 ). If c(v1) = 0, then

Sc= (0,2,1,2,0,0,0,2, 1,2,0,0,. .. ,0,2,1,2,0,0,0).

However then, c'(vn-1) = c'(vn) = 0, which is a contradiction. If c(v1) = 1, then

Sc= (1,1,1,0,2,0,1,1,1,0,2,0,... ,1,1�1,0,2,0,1).

However then, c'(vn) = 1, which is a contradiction. If c(vi) = 2, then

Sc= (2,0,1,1,1,0,2,0,1,1,1,0,. .. ,2,0,1,1,1,0,2).

However then, c'( v) = 2. This is a contradiction.

Case 6. Sc' = (2,1, 2, 1,. . . ,2, 1, 2 If c(v1) = 0, then

Sc = (0,2,2,1,1,0,0,2,2,1,1,0,. .. ,0,2,2,1,1,0,0 ), which gives c'(vn) = 0. This is a contradiction. If c(v1) = 1, then

Sc= (1,1,2,2,0,0,1,1,2,2,0,0,.. . ,1,1,2,2,0,0,1), which yields c'(vn-1) = c'(vn) = l. This isimpossible. If c(v1) = 2, then

Sc = (2,0,2,0,2,0,... ,2,0,2,0,2), which gives c' ( v) = 2, which is a contradiction.

Therefore, there is no closed modular 3-coloring of Pn + K1 for n = 1,3 (mod 18). ■

We are now prepared to present the main result of this subsection.

Theorem 4.3.6 For each integer n 2'. 4

ifn = 1,3 (mod 18) or n = 2 (mod 6) otherwise.

Proof. Let G = Pn + K1 where Pn = (vi, v2, ... , Vn) and V(K1) = { v }. Since Pn + K1 has no true twins for n 2'. 4, it follows by Proposition 2.1.3 that me(G) 2'. x( G) = 3. Furthermore, mc(G) :S 4 by Lemma 4.3.3. If n = 1,3 (mod 18) or n = 2 (mod 6), then it 85 follows by Lemmas 4.3.4 and 4.3.5 that mc(G) = 4. Thus, we may assume that n ¢ l, 3 (mod 18) and n ¢ 2 (mod 6). Thus, either n = 0, 4,5 (mod 6) or n = 7, 9,13 ,15 (mod 18). We consider these seven cases and in each case, we provide a closed modular 3-coloring c : V ( G) -+ Z3 such that c(v) = 0. Thus, it suffices to describe the color sequence Sc of c on the path Pn as follows:

• For n = 0 (mod 6), let Sc = (0,1, 1,2, 2, 0, 0, 1,1, 2, 2, 0, ... ,0, 1, 1, 2, 2,0).

Then c'(v) = 0 and Sc' = (1, 2,1, 2, ... ,1, 2, 1, 2).

• For n = i'lc (mod 6), let Sc = (2,2, 1, 1, 0, 0, 2, 2, 1,1, 0, 0, ... , 2, 2,1, 1, 0, 0, 2, 2,1, 1).

Then c'(v) = 0 and sc' = (1, 2,1, 2, ... , 1, 2,1, 2).

• For n = 5 (mod 6), let Sc = (1,0, 1, 0,1, 0, ... ,1, 0, 1, 0,1). Then c'(v) = 0 and Sc' = (1,2, 1, 2, 1, 2, ..., 1,2, 1).

• For n = 7 (mod 18), let Sc = (1, 0, 2, 2,2, 0, 1, 0, 2,2, 2,0, ... ,1, 0, 2, 2,2, 0, 1).

Then c'(v) = 2 and Sc' = (1,0, 1, 0,... ,1, 0, 1,0, 1).

• For n = 9 (mod 18), let Sc = (1, 2, 1,0, 0,0, 1, 2, 1, 0,0, 0, ... , 1,2, 1,0, 0, 0, 1, 2,1).

Then c'(v) = 2 and Sc' = (0,1, 0, 1, ..., 0,1, 0, 1, 0).

• For n = 13 (mod 18), let Sc = (0,0, 2,1, 2,0, 0, 0,2, 1, 2, 0,... ,0, 0, 2, 1,2, 0,0).

Then c'(v) = 1 and Sc' = (0, 2,0, 2,... ,0, 2, 0, 2, 0).

• For n = 15 (mod 18), let Sc = (2,2, 2, 0, 1, 0, 2,2, 2,0, 1,0, ... , 2,2, 2, 0, 1, 0, 2,2, 2).

Then c'(v) = 2 and Sc' = (1,0, 1, 0, ... , 1,0, 1, 0, 1).

In each case, c is a closed modular 3-coloring and so mc(G) = 3 if n ¢ l, 3 (mod 18) and n ¢ 2 (mod 6). ■

4.3.2 The Closed Modular Chromatic Number of Cn + K1

C We now determine the closed modular chromatic number of n + K1 for n 2:: 3. If n = 3,

then C3 + K1 = K4 and mc(K4) = 1. Thus, we may assume that n 2:: 4. For the cycle

Cn = (v1,v2,... ,vn,v1) of order n 2:: 4 and a coloring c: V(Cn)-+ Zk where k 2:: 2, recall that the color sequence Sc of c on Cn is defined as Sc = ( c( v1), c( v2), ... , c( Vn)).

86 Theorem 4.3. 7 For each integer n 2 4,

if n is even if n is odd.

Proof. Let Cn = (vi,v2,.. . ,vn,vi) and V(Ki) = {v} where then vis adjacent to Vi for 1 :Si :Sn. We first show that if n 2 4 is an even integer, then mc(Cn + Ki ) = 3. If n 2: 4 is even, then G has no true twins and so mc(G) 2: x(G) = 3 by Proposition 2.1.3. To show that me( G) :S 3, we provide a closed modular 3-coloring. Define a coloring c: V(G) -+ :l.3 by c(vi) = 2 for i is odd and c(vi) = 1 if i is even where 1 :Si :Sn and c(v) = 0. Then c'(vi) = 1 if i is odd and c'(vi) = 2 if i is even where 1 :S i :Sn and c'( v) = 0 in :l.3. Thus c is a closed modular 3-coloring and so me(G) = 3 for all even integers n 2 4. If n 2: 5 is odd, then G has no true twins and so me(G) 2: x( G) = 4. To show that me( G) :S4, it suffices to show that G has a closed modular 4-coloring. We consider two cases, according to whether n = 1 (mod 4) or n = 3 (mod 4). In each case, we define a closed modular 4-coloring c: V(G) -+ Z4. By Proposition 4.3.2, we may assume that c(v) = 0 and so it suffices to define the color sequence Sc of con Cn. Case 1. n = 1 (mod 4). For n = 5, define a coloring c : V(G) -+ :l.4 such that Sc = (1, 1,0, 3, 2) as shown in Figure 4.9. Then the color sequence Sc' of the induced coloring c' on Cn is Sc' = (0, 2,0, 1, 2) and c'(v) = 3. Thus c is a closed modular 4�coloring and so mc(G) = 4 for n = 5. For n 2: 9, define a coloring c: V(G) -+ :l.4,according to whether n = 1 (mod 8) or n = 5 (mod 8). 2

2\2\1 2

1 0

Figure 4.9: Closed modular 4-colorings for Cn + Ki for n = 5, 9

For n = 1 (mod 8), let n = 8£ + 1, where f 2 1. Define c such that

Sc = (2, 1, 2,1, ... , 2, 1, 3,0, 3).

87 Figure 4.9 shows such a coloring fr n = 9. Then the induced coloring c' satisfes c'(v) = 3

in Z4 and

Sc' = (2,1, 0,1, 0, 1, 0,1, 0,1, 0, 1, ..., 0, 1, 0, 1, 0, 2, 0, 2, 1). There are 2£ - 2 subsequences 0,1, 0, 1 in the sequence. Thus c is a closed modular 4-coloring and some( G) = 4. + For n = 5 (mod8) , let n = 8£ 5, where£ 2 1 and defne c such that

Sc = (1, 0,1, 0, ..., 1, 0, 2, 3,2).

Then the induced coloring c' satisfes c' ( v) = 0 in Z4 and Sd = (3,2,1,2,1,2,1,2,1,2,1,2, ...,1,2,1,2,1,3,1,3,2).

Figure 4.10 shows a closedmodular 4-coloring fr C13+K1. There are2 £-1 subsequences 1,2, 1,2 in the sequence. Thus c is a closed modular 4-coloring and somc(G) = 4. 3

Figure 4.10: A closed modular 4-coloring for C13 + K1

Case 2 . n = 3 (mod 4) . Defne a coloring c : V(G) -+ Z4 , according to whether n = 3 (mod 8) or n = 7 (mod 8). First suppose that n = 3 (mod 8). Defne c such that

Sc= (3,0, 3,0, 3, 0, 3,0 ... , 3,0, 3, 0, 1,2, 1, 3,0, 3, 1).

Figure 4.11 shows such a coloring fr n = ll. Then the induced coloring c' satisfes

c'(v) = 1 in Z4 and

Sc' = (0,2,3,2,3,2,3,2, ...,3,2,3,2,3,2,3,0,3,0,2,0,2,0,3 ).

Next suppose that n = 7 (mod 8). For n = 7, defne c such that Sc= (0,0, 0, 2, 3, 1,1) as shown in Figure 4.9. Then Sc'= (1,0,2,1,2,1,2) and c'(v) = 3.For n 2 15, defne c such that

Sc= (3,0,3,0,3,0,3,0, .. .,3,0,3,0,2,0, 1,0,1,2,2).

88 1 1

Figure 4.11: Closed modular 4-colorings fr Cn + K1 fr n = 7, 11

Then c'(v) = 0 and

Sc' = (1, 2, 3, 2, 3, 2, 3, 2, ... , 3, 2, 3, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3).

In each cae, c is a closed modular 4-coloring and so me( G) = 4. ■

89 Chapter 5 Local Metric Dimension

5.1 Terminology and Notation

The distance d( u, v) between two vertices u and v in a connected graph G is the length of a shortest path between these two vertices. Suppose that W = { w1, w2,... ,W k} is an ordered set of vertices of a nontrivial connected graph G. For each vertex v of G, there is associated a k-vector called the metric code, or simply the code of v (with respect to W), which is denoted by codew(v) and defined by

codew(v) = (d(v,w1),d(v,w2),··· ,d(v,wk))

(or simply code(v) if the set W under consideration is clear). If code(u) =I= code(v) for every pair u, v of distinct vertices of G, then W is called a metric set, a resolving set or a locating set. The minimum k for which G has a metric k-set is the global metric dimension, or simply the metric dimension of G, which is denoted by dim(G). A metric set in G of cardinality dim(G) is a minimum metric set. To illustrate these concepts, consider the graph G shown in Figure 5.1. The ordered set W1 = { v1,V3} is not a metric set for G since codew1 ( v2) = (1,1) = codew1 ( v4), that is, G contains two vertices with the same code. On the other hand, W2 = { v1, v2,v3} is a metric set for G since the codes for the vertices of G with respect to W2 are

codew2 (v1) = (0,1, 1), codew2(v2) = (1,0, 1), codew2(v3) = (1,1, 0),

codew2 ( v4) = (1,2, 1) and codew2 ( vs) = (2,1, 1).

However, W2 is not a minimum metric set for G since W3 = {v1,v2} is also a metric set.

The codes for the vertices of G with respect to W3 are

90 (0, 1), (1,0), (1, 1), codew3 (v1) = codew3 (v2) = codew3 (v3) = (1,2) (2,1). codew3 (v4) = and codew3 (v5) =

Since no single vertex constitutes a metric set fr G, it follows that W3 is a minimum metric set for this graph G and so dim(G) = 2.

Figure 5 .1: Metric sets

As we mentioned in Chapter 1, the idea of metric sets (and of minimum metric sets) has appeared in the literature previously. In [52], and later in [53], Slater introduced the concept of metric set (or locating set) and described the usefulness of these idea when working with U.S. sonar and Coast Guard Loran (Long range aids to navigation) stations. Other applications of this concept in communication networks appeared in [40, 41] fr example. In recent years, many mathematicians have shown interest in this subject, both theoretically and for its numerous applications (see [7, 10, 35, 42, 46, 51], fr example). In fact, there was a workshop (the Bordeaux Workshop) taking place in November 2011 in Bordeaux, France on this and related topics.

W now introduce a local version of metric sets by considering those ordered sets W of vertices of G fr which any two vertices of G having the same code with respect to W are not adjacent in G. If code(u) - code(v) fr every pair u, v of adjacent vertices of G, then W is called a loc�l metric set of G. The minimum k fr which G has a local metric k-set is the local metric dimension of G, which is denoted by lmd(G). A local metric set of cardinality lmd( G) in G is a local metric basis of G. The local metric dimension exists fr every nontrivial connected graph G. In fact, V ( G) is always a local metric set of G. Indeed, fr each independent set U of vertices in G, the set V (G) - U is a local metric set. Thus we have the fllowing observation. The independence number of a graph G is denoted by a(G) (the maximum cardinality of an independent set of vertices in G).

Observation 5.1.1 For every nontrivial connected graph G of order n,

lmd(G) � n - a(G).

91 The sharpness of this upper bound will be discussed in Section 5.3.

While each metric set of a nontrivial connected graph G is vertex-distinguishing (since every two vertices of G have distinct codes), each local metric set is neighbor distinguishing (since every two adjacent vertices of G have distinct codes). Thus every metric set is also a local metric set and so if G is a nontrivial connected graph of order n, then 1 � lmd(G) � dim(G) � n -1. (5.1)

To illustrate these concepts, consider the graph G of Figure 5.2. In this case, W1 = { v1, v4} is a local metric 2-set and W2 = { v1, v3, v5} is a metric 3-set. The corresponding codes for the vertices of G with respect to the sets W1 and W2 , respectively, are shown in Figure 5.2. In fact, lmd(G) = 2 and dim(G) = 3.

(0, 2) (0, 2, 2) V1 V1 (1, 2) V2

V4 V4 (2, 0) (2, 1, 1)

Figure 5.2: A graph with metric dimension 3 and local metric dimension 2

These examples illustrate the following useful observations.

Observation 5.1.2 When determining whether a given set W of vertices of a nontrivial connected graph G is a local metric set of G, one need only investigate the pairs of adjacent vertices in V ( G) - W since w E W is the only vertex of G whose distance from w is 0.

Observation 5.1.3 If W is a subset of the vertex set of a nontrivial connected graph G containing a local metric set of G, then W is also a local metric set of G.

5.2 Graphs with Prescribed Order and Local Metric Dimension

It is known that if G is a nontrivial connected graph of order n, then dim(G) = n - l if In and only if G = Kn and dim(G) = 1 if and only if G = Pn . the case of local metric

92 dimension, there is an analogous result. Befre stating this result, we present some additional terminology. Two vertices u and v in a connected graph G are refrred to as twins if u and v have the same neighbors in V ( G)-{ u, v}. If u and v are adjacent, they are refrred to as true twins or adjacent twins; while if u and v are nonadjacent, they are false twins or or nonadjacent twins. If u and v are true twins and v and w are true twins, then so too are u and w. Hence two vertices being true twins produces an equivalence relation on V ( G). The resulting equivalence classes are called true twin equivalence classes of G. True twins and flse twins can also be defned in terms of distance. Two vertices u and v in a connected graph G are distance similar if d(u, x) = d( v, x) for all x E V ( G) - { u, v}. Therefre, if u and v are distance similar vertices, then O : d(u, v) : 2. In particular, two nonadjacent vertices u and v are distance similar if and only if N(u) = N(v) (where N(x) denotes the neighborhood of the vertex x or the set of vertices adjacent to x in G), that is, two nonadjacent vertices u and v are distance similar if and only if they are false twins. Frthermore, two adjacent vertices u and v are distance similar if and only if N ( u) - { v} = N ( v) - { u}, that is, two adjacent vertices u and v are distance similar if and only if they are true twins. A relation R is defned on V ( G) by u is related to v by R if either (i) u = v or (ii) u and v are adjacent and u and v are distance similar. Then R is an equivalence relation on V(G). The equivalence classes resulting fom the relation R are called the distance equivalence classes or true twin equivalence classes of G. If the resulting true twin equivalence classes are U1, U2, ... , Ue, then every local metric set of G must contain at least IUil -1 vertices fom Ui fr each i with 1 :i :C. Thus we have the fllowing observation.

Observation 5.2.1 If G is a nontrivial connected graph of order n having C true twin equivalence classes, then lmd(G) 2 n - £.

The sharpness of this lower bound will be discussed in Section 5.3.

There is no connected graph having exactly two true twin equivalence classes. To see this, suppose that G is a nontrivial connected graph having U1 and U2 as its only true twin equivalence classes. Since G is connected, there exist two adjacent vertices x E U1 and y E U2. However, this implies that every vertex in U1 is adjacent to y, which in turn implies that every vertex in U2 is adjacent to every vertex in U1. Therefre, G is a complete graph, which is impossible. We state this observation below.

Observation 5.2.2 There is no nontrivial connected graph having exactly two true twin equivalence classes.

93 For a vertex v of a graph G, the eccentricity e(v) of vis the distance between v and a vertex farthest fom v. The diameter diam(G) of G is the largest eccentricity among all vertices of G. If G is a nontrivial connected graph of order n with diameter d and ( vo, v1, ... , vd) is a geodesic of length din G, then V ( G) -{ v1, v2, ... , vd} is a local metric set of G. This yields another observation.

Observation 5.2.3 If G is a nontrivial connected graph of order n and diameter d, then

lmd(G) : n - d.

The sharpness of this upper bound will be discussed in Section 5.3.

Theorem 5.2.4 Let G be a nontrivial connected graph of order n. Then lmd(G) = n-1 if and only if G = Kn and lmd(G) = 1 if and only if G is bipartite.

Proof. Since the complete graph Kn has only one distance equivalence class, lmd(Kn) 2 n - 1 (by Observation 5.2.1). It then fllows, by (5.1), that lmd(Kn) = n - 1. On the other hand, if G - Kn, then a(G) 2 2 and so lmd(G): n - 2 by Observation 5.1.1. It remains to show that lmd( G) = 1 if and only if G is bipartite. Suppose frst that G is a bipartite graph with partite sets U and V. Let W = { w}, where w E U say. Since d( u, w) is even fr each u E U and d( u, v) is odd fr each v E V, it fllows that W is a local metric basis and so lmd( G) = 1. To verify the converse, let G be a nontrivial W connected graph having local metric dimension 1 and let = { w} be a local metric basis of G. For O: i: e(w), let Ni= {v E V(G) : d(v, w) = i}. (Therefore, No= W and N1 = N(w).) Since W is a local metric basis, each set Ni is an independent set. Furthermore, if i and j are integers with O : i, j : e( w) and Ii - j I 2 2, then no vertex in Ni is adjacent to any vertex in Nj . Therefre, G is a bipartite graph with partite sets U = No U N2 U · · · U N2le(w)/2J and V = N1 U N3 U · · · U N2re(w)/21-1· ■

Next, we characterize all nontrivial connected graphs of order n 2 3 having local metric dimension n - 2. The clique number w(G) of a graph G is the order of a largest complete subgraph (clique) in G.

Theorem 5.2.5 A connected graph G of order n 2 3 has local metric dimension n - 2 if and only if w(G) = n - 1.

94 Proof. First, let G be a connected graph of order n 2: 3 with clique number n - l.

Since G # Kn , it follows that lmd(G) :n - 2. Let H = Kn-l be a clique in G and let V(G) V(H) d d v be the vertex of - where = deg v. Then 1 :S :n - 2. Observe that V(G) U1 = { v}, U2 = N(v) = { vi, v2, ... , vd} and U3 = - N[v] = { vd+l, Vd+2,... , Vn-d are the true twin equivalence classes. Therefore, lmd(G) 2: n - 3 by Observation 5.2.l.

Assume, to the contrary, that G contains a local metric set W' consisting of n - 3 V(G) vertices. Then again, by Observation 5.2.1, - W' = { v, x2, x3}, where Xi E Ui for i = 2, 3. However, then x2 and x3 are adjacent and d(x2, w) = d(x3, w) for all w E W'. Thus code(x2) = code(x3), which is a contradiction. Therefore,lmd(G) = n - 2. For the converse, let G be a connected graph of order n 2: 3 with lmd(G) = n-2. We show that w(G) = n-1. Since G # Kn (by Theorem 5.2.4), it follows that w(G) :S n-1.

The result follows immediately for n = 3 since G = P3. Since every connected graph G of order 4 with w(G) :S 2 is bipartite, it follows that lmd(G) = 1 (by Theorem 5.2.4). Therefore,w(G) = 3 for all connected graphs G of order 4 with lmd(G) = 2. Hence we may now assume that n 2: 5. Suppose that there is some graph G of order n 2: 5 for which lmd(G) = n - 2 and w(G) :S n - 2. By Observation 5.2.3, diam(G) = 2. Also, there exists a set X = { x1, x2, x3, x4} consisting of four distinct vertices such that x1x2,x 3x4 rJ. E(G). Consider the induced subgraph H = (X) by X.

If f(H) = 2, say degH(xi) = 2, then let W1 = V(G)-{x2, x3,x4}. Since d(x1, x3) = d(x1, X4) = 1 while d(x1, x2) = 2 and X3 and x4 are not adjacent in G, it follows that W1 is a local metric set. Similarly, if 8(H) = 0, say degH(x1) = 0, then W2 = V( G) - { x1, x3,x4} is a local metric set since the three vertices x1, x3, and x4 are mutually nonadjacent. Therefore, lmd(G) :S n - 3 in each case, which cannot occur. Therefore,we consider the case where f(H) = 8(H) = 1 and we may assume that E(H) = {x1x3,x2x4}. If there exists a vertex v* E V(G) - X such that d(x1,v*) # d(x3,v*) or d(x2,v*) # d(x4,v*), say the former,then codew2 (xi) # codew2 (x3) since v* E W2. Since X4 is adjacent to neither x1 nor X3, it follows that W2 is a local metric set. On the other hand, if d(x1,v) = d(x3,v) and d(x2,v) = d(x4,v) for every v E V(G)-X, then there exists a vertex v' in V(G) - X which is adjacent to every vertex in X since diam(G) = 2. Then let W3 = V(G)- {x3, X4,v'} and observe that d(v',xi) = d(v', x2) =

1 while d(x4,x1) = d(x3,x2) = 2. Therefore, codew3 (v') rJ. {codew3 (x3),codew3 (x4)}. Since X3 and x4 are not adjacent, it follows that W3 is a local met:r:ic set. Hence, lmd(G) :S n - 3, which is again a contradiction. ■

95 It is known fr a connected graph G of order n 2 3 that x( G) = n - 1 if and only if w(G) = n - 1. Thus the fll�wing corollary is a consequence of Theorem 5.2.5.

Corollary 5.2.6 A connected graph G of order n 2 3 has local metric dimension n - 2 if and only if x(G) = n - 1.

We have seen that if G is a nontrivial connected graph of order n with lmd(G) = k, then 1 : k : n - 1. In fact, every pair k, n of integers with 1 : k : n -1 is realizable as the local metric dimension and order, respectively, of a connected graph as we show next.

Theorem 5.2. 7 For each pair k, n of integers with 1 < k < n - 1, there exists a connected graph G of order n with lmd(G) = k.

Proof. By Theorems 5.2.4 and 5.2.5, there exists a connected graph G of order n with lmd(G) = k fork E {1,n -2,n -1}. Thus we may asume that 2 : k :Sn - 2. Let G be the graph obtained from Kk+I with vertex set V = { v1,v2, ... , Vk+d and the path (vk+2, Vk+3, ... , vn) by joining Vk+I and Vk+2· Since V - { Vk+1} is a local metric set, lmd( G) : k. Assume, to the contrary, that lmd(G) : k -1 and let W' be a local metric set of k - 1 vertices. Since every local metric set must contain at least k -1 vertices fom V- {vk+d by Observation 5.2.1, we may assume that W' = {v1,v2, ... ,vk_i}. However then codew,(vk) = codew,(vk+i) = (1, 1, · · · ,1), which is a contradiction. Thus lmd(G) = k. ■

We noted that if G is a nontrivial connected graph with lmd(G) = a and dim(G) = b, then a : b. On the other hand, every pair a, b of positive integers with a : b can be realized as the local metric dimension and metric dimension, respectively, of some connected graph. In order to verif this, we present two lemmas which provide the metric dimension and local metric dimension of all complete multipartite graphs.

.. Lemma 5.2.8 Let G = Kn1 ,n2, -,nk be a complete k-partite graph of order n, where k 2 2, n = n1 + n2 + · · · + nk, and n1 : n2 : · · · : nk. Define

* 1 if n2 - 1 p = { p if n2 = 1, np = 1 and either p = k or np+l 2 2.

Then dim(G) = n -k + p* -1.

96 Proof. If p* = k, then n1 = n2 = · · · = nk = 1, then G is complete and the result follows from Theorem 5.2.4. We may therefore assume that 1 � p* � k - l. Let Vi,Vi, ... , Vk be the partite sets of G, where l¼I= ni for 1 � i � k. Observe first that if W is a metric set of G, then I¼ n WI 2: ni - 1 for 1 � i � k. Therefore,

IWI 2: n -k. First, suppose that p* = 1 and let X1 = { xi, x2, ... , xk} be a subset of V(G), k. V(G) where Xi is a vertex of¼ for1 � i � Since the set W1 = -Xi is a metric set, it follows that dim(G) � IW1I = n - k and so dim(G) = n- k. Next suppose that p* = p 2: 2. Then every metric set must contain at least p - 1 vertices in the set ViU ViU · · · U Vp. Therefore,

IWI 2: (p - 1) + [(n- p) - (k -p)] = n - k + p - l.

V(G), Let X2 = {xp, Xp+l, ... , xk} be a subset of where Xi is a vertex in¼ forp � i � k, and let W2 = V(G) -X2. Since W2 is a metric set, dim(G) � IW2I = n- k + p - 1 and so dim(G)=n-k+p-1. ■

Lemma 5.2.9 For each complete k-partite graph G, where k 2: 2, lmd(G)= k - l.

Proof. Let Vi,Vi, ... , Vk be the partite sets of G and let W = { w1, w2, ... , Wk-l}, where Wi E ¼ for 1 � i � k - l. For each Vi E ¼ (1 � i � k - 1), where Vi =I= Wi, the i-th coordinate of code(vi) is 2 and the remaining coordinates of code(vi) are l; while all coordinates of code(vk) are 1 for each vk E Vk . Thus code(vi) =I= code(vj) for all pairs i, j with i =I= j and 1 � i, j � k. Thus W is a local metric set and so lmd(G) � k - l. If W' is a set of k - 2 vertices, then there are partite sets ¼ and ½, where i =I= j and 1 � i,j � k, such that W' n¼ = W' n ½ = 0. Then code(vi) = code(vj) = (1, 1, · · · , 1) for each Vi E ¼ and each Vj E ½, implying that W' is not a local metric set. Thus lmd(G) 2: k -1 and so lmd(G) = k - l. ■

Theorem 5.2.10 For each pair a, b of positive integers with a � b, there is a nontrivial connected graph G with lmd(G) = a and dim(G) = b.

Proof. Consider the complete (a+ 1)-partite graph G = K1,1, ... ,1,b-a+2 of order b + 2. Then lmd(G) = (a+ 1)-1 =a by Lemma 5.2.9 and dim(G) = (b+2)-(a+ 1)+a-1 = b by Lemma 5.2.8. ■

97 5.3 Bounds for Local Metric Dimensions

In this section we establish bounds for the local metric dimension of a nontrivial con nected graph in terms of its order and other well-known graphical parameters. Earlier bounds are revisited here as well. Other results involving the local metric dimension and the number of distance equivalence classes of a graph are also presented.

Theorem 5.3.1 If G is a nontrivial connected graph with clique number w, then

(5.2)

Furthermore, for ea ch integer w � 2, there exists a connected grph Gw with clique number w such that lmd(G w) = pog2 w l -

Proof. Let F = Kw be a clique in G with V(F) = { v1, v2,... , vw }- Suppose that lmd(G) = k and let W be a local metric basis. For each Vi E V(F), let code(vi) = (a1,i,a2,i,··· ,ak,i)- Since Jd(vi,x)-d(vi,,x)I : 1 for every two vertices Vi,Vi' E V(F) and every vertex x in G, it fllows that I { a1,i : 1 : i : w} I : 2 fr 1 : j : k. Therefre, there are at most 2k possible codes for the w vertices in F with respect to W. Since every k vertex in F must have a distinct code, it fllows that w : 2 or k � log2 w. Therefore, lmd(G) � flog2 w l -

We now construct a connected graph Gw with clique number w such that lmd(Gw) =

flog2 w l for each integer w � 2. If w = 2, then let G2 be a nontrivial tree and so lmd(G2) = 1 = flog2 w l Thus we may assume that w � 3. Then there exists a unique integer k � 2 such that 2k-l + 1 : w : 2k. Let w = 2k-l + p, where p is an integer with k 1 : p : 2 -l _ Construct the graph Gw from the complete graph Kw with vertex set

V(Kw) = { v1,v 2 , ..., V w } by adding the k new vertices in the set W = { w1, w2,... ,wk} k as follows: Let X = {x1,x2, ... , x2k} be the set of the 2 distinct ordered k-tuples whose coordinates are elements in { 1, 2}, where x1,x 2, ... , x2k are listed in lexicographic order.

Thus x1 = (1,1,··· ,1) and x 2k = (2,2,··· ,2). For each j with 1 : j : k, the vertex k w1 is joined to the vertex Vi (1 : i : 2 -l + p) if and only if the j-th coordinate of Xi is

1. Thus v1 is adjacent to every vertex in W while Vw is adjacent to none of the vertices k k in W if w = 2 . Since Gw is a connected graph with w(Gw) = w = 2 -l + p, it fllows k that lmd(Gw) � k by (5.2). On the other hand, code(vi) = Xi for 1 : i : 2 -l + p and so Wis a local metric set of Gw . Therefore, lmd(Gw) = k. ■

The graph Gw constructed in the proof of Theorem 5.3.1 illustrates an interesting fature of the local metric dimension; namely, if H is a subgraph of a graph G, then it

98 is possible that lmd(H) > lmd(G). For example, let H = Kw be the complete subgraph of order win the graph Cw . Then lmd(H) = w - 1 > flog2 wl = lmd(G) for each w 2 4. We now present another lower bound for the local metric dimension of a graph in terms of its order and clique number. This lower bound is particularly useful when n is large and n - w is small.

Theorem 5.3.2 If G is a nontrivial connected graph of order n with w = w(G), then

lmd(G) 2 n - 2n-w_

Furthermore, fo r each pair n, w of integers with 2n-w $ w $ n, there exists a connected n- graph G of order n whose clique number is w such that lmd(G) = n - 2 w.

Proof. The result is immediate if w E { n - 1, n} and so assume that 2 :S w $ n - 2. Suppose that H = Kw is a clique in G and let X = V(H) and Y = V(G) -X. Consider an arbitrary local metric set W. We show that [WI 2 n - 2n-w_ Let p = [Y n W[. Therefore, 0 :Sp$ n -w. Since d(x, w) = 1 for all x EX -X n W and w E X n W, there are at most 2P possible codes for the vertices in X - X n W. + Hence, [X - X n W[ :S 2P and so [X n WI 2 w - 2P. Therefore, [WI 2 p (w - 2P). If p = 0, then [WI 2 w - 1 = n - (n - w + 1) > n - 2n-w,

since n - w 2 2. If 1 :Sp $ n - w, then consider the function f from � to � defined by f(x) = x + (w - 2x). Observe that f'(x) = 1 - 2x ln 2 < 0 for x > 1 and so f(x) 2 f(n - w) = n - 2n-w for1 :S x $ n - w. Therefore,

[W[ 2 f(p) 2 n - 2n-w_

Next, let n and w be positive integers with 2n-w :Sw $ n. Clearly G = Kn possesses the desired property for w = n. Therefore, suppose that 2n-w :S w < n. Let X, Y, and Z be pairwise disjoint sets of vertices, where X = { x1, x2, ... , Xn-w }, Y = n {Y1' Y2' ... ' Y2n-w}' and = 0 if 2 -w = w and = { z1 ' z2 ' ... Z n-w} otherwise. z z ' w-2 Also, let P(A) = { S1 , S2, ... , S2n-w} be the power set of the set A = {1, 2, ... , n - w}. We construct a graph G with V(G) = XUYUZ such that (XU Z) � Kn_2n-w, (YU Z) � n Kw, and XiYj E E(G) if and only if i E S1 for 1 :Si :Sn-wand 1 :Sj $ 2 -w_ Therefore, n w + the order of G is n and since n-w 2 1, it followsthat w(G) = max{w, n-2 - 1} = w. Furthermore, X U Z is a local metric basis of G and so lmd( G) = IX U Z I = n - 2n-w. ■

The following is an immediate consequence of Theorem 5.3.2.

99 Corollary 5.3.3 If G is a nontrivial connected graph of order n and lmd(G) = n - k, then

We have seen in Observations 5.1.1 and 5.2.3 that if G is a nontrivial connected graph of order n having independence number a and diameter d, then lmd( G) : n - a and lmd( G) : n - d. In fact, more can be said.

Theorem 5.3.4 For each pair a, n of integers with 1 : a : n - 1, there exists a nontrivial connected graph G of order n and independence number a such that lmd( G) = n-a.

Proof. For integers a and n with 1: a: n -1, let G =Kn-a + Ka . Since V(Kn-a) is a local metric set, it fllows that lmd(G) : n - a. By Observation 5.2.1, every local metric set must contain at least n - a - 1 vertices fom the set V(Kn-a )- Let n a v W be any subset of V(Kn-a ) with IWI = - - l. If V(Kn-a ) - W = { }, then v) x) x code w ( = co dew ( for each vertex in Ka. Thus W is not a local metric set and so lmd( G) = n - a. ■

d, d Theorem 5.3.5 For ea ch pair n of integers with 1 : : n - 1, there exists a nontrivial connected graph G of order n and diameter d such that lmd( G) = n - d.

Proof. Let d and n be integers with 1 : d::; n - l. Ford= 1, the complete graph Kn has the desired property. Also, consider the path Pn fr d = n - 1. Hence suppose that d::; 2 : n - 2. Let G be the graph obtained fom a complete graph H = Kn-d of order n - d and a path P : v1, v2, ... , Vd of order d by joining every vertex in H to v1. Then diam(G) = d and lmd(G) : n - d by Observation 5.2.3. Also, observe that Ui = { vi} + fr 1 : i : d and Ud+1 = V(Kn-d) are the d 1 true twin equivalence classes. Hence, lmd(G) 2 n-d-1 by Observation 5.2.1 and furthermore, every local metric set contains d- d at leat n - 1 vertices in Ud+l. Assume, to the contrary, that lmd( G) = n - -1 and let W be a local metric basis. Then there exists a vertex x E Ud+l -W. However then, code(x) = code(v1), which contradicts the fact that W is a local metric set. Therefre, lmd(G) = n -d. ■

By Observations 5.2.1 and 5.2.2, if G is a nontrivial connected graph of order n f. having £ true twin equivalence classes, then £ f 2 and lmd( G) 2 n - For a fxed integer n 2 2, we next determine all possible values of£ for which there is a connected graph G of order n with £ true twin equivalence clases such that lmd(G ) = n - f.

100 Theorem 5.3.6 Let n and R be integers with 1 S £ S n - l. There exists a connected graph G of order n with R distance equivalence classes such that lmd(G) = n - R if and only if R = 1 or 3 S R S n - 2.

Proof. Let G be a connected graph of order n with £ true twin equivalence classes

U1 , U2, ..., Ue. Then £ = 1 or 3 S R S n - 1 by Observation 5.2.2. If R = n - 1, then we may assume, without loss of generality, that IUil = 1 for 1 S i S n - 2 and

Un-l = {x,y}. Then x and y are adjacent. Since G is connected and d(x,v) = d(y,v) fr every v E V( G) - { x, y}, there exists a vertex z that is adjacent to both x and y. This implies that G contains a triangle and so G is not bipartite. By Theorem 5.2.4, lmd(G) 2 2 and so lmd(G) = n - (n - 1) = n - R.

To verif the converse, first observe that G = Kn has the desired property fr R = l. If 3 S R S n - 2, then let G be the graph obtained fom vertex-disjoint complete graphs

Hi = Kn-e and H2 = K2 of orders n-R and 2, respectively, and a path P: v1 , v2, ..., W-2

of order R - 2 by joining (i) every vertex in H1 to V1 and (ii) the two vertices in H2 to

ve-2. Then Ui = { vi} fr 1 S i S £ - 2, Ue-i = V(H1 ), and Ue = V(H2) are the true twin equivalence classes. Let x E V(H1) and y E V(H2) and observe that the set W = V (G) - [V( P) U { x, y}] is a local metric set containing n - R vertices. Therefre, lmd(G) S n - R. Since lmd(G) 2 n - R by Observation 5.2.1, we obtain the desired result. ■

The following result presents a sharp upper bound for the local metric dimension of a nontrivial connected graph in terms of its order, the number of true twin equivalence classes and the number of singleton equivalence classes.

Theorem 5.3. 7 Let G be a nontrivial connected graph of order n having R true twin equivalence classes. If p of these R true twin equivalence classes consist of a single vertex, then lmd(G) Sn - R + p.

Proof. If £ = 1 or p E { £ - 1, R}, then the result immediately fllows by ( 5.1). Since R = 2 by Observation 5.2.2, we may assume that £ 2 3 and O S p S R - 2. Suppose

that U1 , U2, ..., Ue are the true twin equivalence classes and IUil 2 2 fr p + 1 S i S £.

For each i with p + 1 S i S £, let Ui E Ui. Now let U = { Up+l, Up+2, ... , ue} and W = V(G) - U. Then IWI = n - (R - p) = n - R + p. We show that W is a local metric set of G. Let x and y be two adjacent vertices of U. Then x and y belong to distinct true twin equivalence classes. Therefre, there exists z E V (G) - { x,y} such

101 that d(x, z) i d(y, z). If z E W, then codew(x) i codew(y), as desired. Thus we may assume that z (/. W and so z E U. Then z = Uj fr some j with p + 1 : j : e. Let z' E Uj -{z}. Then z' E Wand d(x, z') =d(x, z) i d(y, z) = d(y, z'). Thus Wis a local metric set and so lmd(G) : IWI =n - e +p. ■

The upper bound in Theorem 5.3.7 is sharp. To see this, for an integer k 2 3, let A= {1, 2,... ,k - 2} and let P(A) = {Si,S 2,... ,S 2k-2} be the power set of A. Defne the sets S2k-2+ 1 , S2k-2+2,... , S2k by

Si+zk-2 =Siu {k - 1}, Si+zk-1 = Si U {k}, Si+2k-1+zk-2 =Siu {k - 1, k}

k fr 1 :i : 2 -Z_ Thus {S1 ,S 2,... ,S 2k} is the power set of AU{k-1, k} = {1,2, ... , k}. k k Let H = K2 be a complete graph of order 2 with V(H) ={u 1 ,u2,... ,u2k}. We construct G fom H by adding k new vertices in the set Wo ={ w 1 ,w 2,... ,wk} and k joining Ui to Wj if and only if j E Si. Note that G = Cw (w = 2 ) is described in the proof of Theorem 5.3.l. Hence Wo is a local metric basis and lmd(G) = k. Frthermore, degwi = 2k-l fr 1 :i : k. We show that lmd(G -wi ) =k + 2k-I -1 for 1 : i : k. By symmetry, it sufces to show that lmd(G - wk) = k + 2k-I - l.

Since the set W = { u 1 ,u 2,... ,Uzk-i} U (Wo -{wk}) is a local metric set of G -wk containing k + 2k-l - 1 vertices, lmd(G - wk) : k + 2k-l - l. Observe that each set Ui = {u i , ui+zk-i} is a distance equivalence class in G - wk fr 1 : i : 2k-l. Thus, if there exists a local metric set W' in G -Wk containing at most k + 2k-I - 2 vertices, then we may asume that {u1, u 2,... ,Uzk-1} � W' and Wk-I f W'. On the other hand, de-wk (u i+zk-1, v) i de-wk ( ui+zk-1+zk-2,v) if and only if v = Wk-I fr k 1 : i : 2 -Z. Therefre, we may assume that {Uzk-1+ 1 ,Uzk-1+ 2,... ,Uzk-1+zk-2} � W' as well. However then, 2k-l +2k-Z : IW'I < 2k-l +k-1,which is impossible. Therefre, lmd(G - wk) = k + 2k-l -1 as claimed. The order of G is n = 2k + k - 1 and G has k e = 2 -l + k - 1 true twin equivalence classes, namely the 2-sets Ui (1 : i : 2k-l) and the singleton sets {W j } (1 : j : k - 1 ). It fllows then that p = k - 1 and so lmd(G - wk) =n -e + p, as desired. The corollary below is an immediate consequence of Observation 5.2.1 and Theo rem 5.3.7.

Corollary 5.3.8 If G is a nontrivial connected graph of order n withe distance equiv alence classes none of which is a singleton set, then lmd(G) =n -e.

Suppose that G is a nontrivial connected graph of order n having e true twin equiv alence classes,p of which consist of a single vertex. Theorem 5.3.6 provides all possible

102 values of £ for which lmd( G) = n - f. We now study the structures of such graphs. If p = 0, then lmd(G) = n - f by Corollary 5.3.8; while if p = f, then every distance equiv alence class is a singleton set and so n = f. Since G is nontrivial, lmd(G) > 0 = n - f and so lmd(G) i- n -£. Therefore, it remains to consider the case where 1 :Sp :S£ - 1. We first establish some additional definitions.

For two subsets X and Y of the vertex set V(G) of a connected graph G, define the distance d(X, Y) between X and Y by

d(X, Y) = min{d(x,y): x EX and y E Y}.

Thus d(X,Y) = 0 if and only if X n Y #- 0 and d(X,Y) = 1 if and only if X n Y = 0 X Y. S and some vertex in is adjacent a vertex in Suppose that = {U1, U2, ..., Ue} is the set of all true twin equivalence classes of a nontrivial connected graph G. For an ordered subset X = {X1,X2,... ,Xk} of Sand an element U of S, the code codex(U) of U (with respect to X) is defined as the ordered k-tuple

where if u i- xi if u = xi.

In other words, ai = 1 if and only if d(U,Xi) :S 1 for 1 :S i :S k. We are now prepared to present a necessary and sufficient condition for a nontrivial connected graph G of order n having£ distance equivalence classes (where p of these classes consist of a single vertex and 1 :Sp :S £ - 1) to have local metric dimension n -f.

Theorem 5.3.9 Let G be a nontrivial connected graph of order n and let S = {U1 ,U2,. . . ,Ue} be the set of distance equivalence classes of G. Suppose that I Ui I 2: 2 for l :S i :S £ - p and IUil = 1 for f - p + l :S i £, :S where l :Sp :S £ - 1. Let X = {U1,U2, ... , Ut-p} Then lmd(G) = n - f if and only if codex (U) #- codex(U') for every two elements U,U' in S with d(U,U') = 1.

Proof. Suppose first that lmd(G) = n-f and let Wo be a local metric basis of G. Then V(G) - Wo = { u1, u2, ... , ue}, where Ui E Ui for 1 :Si :S £ and Wo � u;:fui. Suppose that Ui, Uj E S and d(Ui, Uj) = 1. Then i #- j. Let codex(Ui) = ( a1,a2, · · · , at-p) and codex(Uj) = (b1, b2, · · · , bt-p)- Since UiUj E E(G) and codewo(ui) c/- codewo(uj), there exists a vertex w EWon U5 , where 1 :S s :S £ - p, such that d(Ui, w) i- d(U j, w). Observe

103 s that = i, j, since otherwise d( Ui, w) = d(U j ,w) = 1. Then Us = Ui, Uj and so as = d(Ui, Us)= d(ui,w) = d(uj,w) = d(Uj , Us)= bs, implying that codex(Ui) = codex(Uj )-

For the converse, let Ui E Ui fr 1 : i : f and consider the set W1 = V (G) - G. { u1, u2, ... , ue}. We show that W1 is a local metric set of Suppose that Ui and Uj are adjacent in G. Let codex(Ui) = (a1, a2, · · · , ae-p) and codex(Uj) = (b1, b2, · · · , be-p) These codes are diferent and so we may assume, without loss of generality, that a1 = b1. Since IU11 2 2, there exists a vertex w E U1 - {ui}. Then observe that d(ui,w) = G a1 = b1 = d(uj ,w). Since w E W1, it fllows that W1 is a local metric set of and so lmd(G) : IW1 1 = n - f. Therefore, lmd(G) = n -f by Observation 5.2.1. ■

Corollary 5.3.10 LetG be anontrivial connectedgraph ofordern havingf truetwin equivalenceclasses,p ofwhichconsistofasinglevertex. Thenlmd( G) = n-fifandonly if(i) p= 0 or(ii) 1 :p: f-1 andGsatisfestheconditionsdescribedin Theorem5.3.9.

5.4 Uniqueness and Non-Uniqueness of Local Metric Bases

We now turn our attention to determining those positive integers kfr which there exists a nontrivial connected graph G with local metric dimension k such that either (i) G has a single local metric basis or (ii) G contains two local metric bases that are arbitrarily fr apart. We begin with (i).

Theorem 5.4.1 There exists a nontrivial connected graphG withlmd(G) = k having auniquelocal metricbasisifand onlyifk2 2.

Proof. Let G be a nontrivial connected graph having local metric dimension k. If k = 1, then G is bipartite and any singleton set W � V(G) is a local metric basis. Therefre, if G has a unique local metric basis, then k 2 2.

To verif the converse, suppose that k 2 2. Consider the set A= {1, 2, ... , k} and let P(A) = {S1, S2, ... , S2k} be the power set of A. Let H = K2k be a complete graph of order 2k with V ( H) = { u1, u2, ..., u2k}. We construct G from H by adding the k vertices fom the set W = {w 1, w2, ..., wk} to H and joining Ui to Wj if and only if j E Si. Note k that G = Cw, where w = 2 , the graph described in the proof of Theorem 5.3.1. Hence lmd(G) = k and, furthermore, W is a local metric basis. W show that W is the only local metric basis of G. Let W' be a local metric basis and assume, to the contrary, that W' = W. By symmetry, we may assume that

104 � Wk( W'. Let P(B) = {S�, s , ... , s;k-1} be the power set of B = A - {k}. We may k also assume that Si = Si and Si+2k-1 = Si U { k} fr 1: i : 2 -l. Since W' is a local 1 metric set of G and Wk� W', this implies that fr each i (1: i: 2k- ), at least one k of U and u k-1 belongs to W'. Hence, k IW'I 2 2 -l. This is impossible if k 2 3. i i+2 = If, on the other hand, k = 2, then W' � V ( H), say W' = { u1,u2}. However then, codew,(u3) = codew,(u4), a contradiction. Therefre, if W' is a local metric basis of G, then W' = W. ■

To describe a solution to the problem stated in (ii), we frst present some prelim inary infrmation, beginning with a lemma, which gives the local metric dimension lmd( G □ H) of the Cartesian product G □ H of two graphs G and H in terms of lmd(G) and lmd(H). For the metric dimension of graphs, it was shown in [10] that

dim(G): dim(G □ K2): dim(G) + 1

fr every connected graph G. In [7] bounds ( or exact values) have been established on dim( G □ H) for many well-known classes of graphs G and H.

Lemma 5.4.2 For every two connected grphsG andH,

lmd(G □ H) = max{lmd(G), lmd(H)}.

Proof. Suppose that G and H are connected graphs of orders pand q, respectively,

with V(G) = {u1,u2,... ,up} and V(H) = {v1,v2,... ,vq} such that lmd(G) = k and lmd(H) = . where k 2 .. Rename the vertices of G and H, if necessary, so that We = { u1,u2, ... , uk} and WH = { v1,v2, ... , w} are local metric bases of G and H, respectively. Construct G H fom q disjoint copies G1, G2, ... , G of G with V(G ) □ q i = {u1,i,U2,i,···,u ,i} fr 1: q by joining Uj, and Uj', ' if and only if j j' and p i::; i i = ViVi' E E(H). Now consider the set W = { u1,1,u2,2, ... ,Ut,d U W', where W' = 0 if k = f and W' = { uH1,1, uH2,1,... ,u k,1} if k > .. We show that Wis a local metric set of G □ H. Let x and y be adjacent vertices in G □ H not belonging to W. We consider two cases.

Case l. Bothx andy belong toV(G a) for somea (1: a: q). Since x,y E V(Ga),

observe that x = Ua,a and y = ub,a fr some aand band UaUb E E(G). Since We is a

local metric basis of G, there exists a vertex Uc E We such that de(ua, uc) = de(ub,u c) W Then the set contains a vertex w1 = uc,/ fr some / and observe that

105 Therefore, codew ( x) =I= codew (y).

Case 2. x E V(G0 ) and y E V(G,a ) for some a and /3, wherea =I= /3 (1 Sa, /3 S q).

Then x = Ua,o and y = Ua,,B for some a and v0v,a E E(H). Since WH is a local metric

basis of H, there exists a vertex v, E WH such that dH(v0 , v,) =I= dH(v,a , v,). Then the

set W contains a vertex w2 = ub,, for some b and observe that

Therefore, codew ( x) =I= co dew (y).

Hence, every two adjacent vertices in G □ H have distinct codes with respect to W and so lmd(G □ H) S IWI = k = lmd(G) = max{lmd(G), lmd(H)}. To show that lmd(G) S lmd(G □ H), let W be a local metric basis of G □ H and

let W1 be the subset of V(G1) such that ui,1 E W1 if and only if Ui,j E W. We show that

W1 is a local metric set of G1 . Let x,y E V(G1) -W1 be two adjacent vertices in G1 .

Since W is a local metric set of G □ H, there exists a vertex w = Ua,o E W such that de w) de w). w' □ H(x, =I= □ H(Y, Therefore, W1 contains a vertex = ua,l· Then observe that

implying that codew1 (x) =I= codew1 (y). Hence lmd(G1) S IW1 I S IWI = lmd(G DH). ■

By Observation 5.2.1, if U1 , U2, ... , Ue are the true twin equivalence classes of a graph G, then every local metric basis of G must contain at least I Ui I - 1 vertices from Ui for each i (1 S i S f). Therefore, if G has two local metric bases W and W' that are disjoint, then 1 S d(W, W') S diam(G) and IUil S 2 for each i. Let d(W, W') = t. Then IWI + IW'I + (t - 1) Sn and sot S n - 2 lmd(G) + 1. Also, if IUil = 2 for some i, say U1 = { u, v}, then u E W and v E W', implying that d(W, W') = 1. Hence, if G contains two local metric bases W and W' such that d(W, W') 2: 2, then every distance equivalence class of G is a singleton set.

Theorem 5.4.3 For each pair k, t of positive integers, there exists a connected graph G with lmd(G) = k having two local metric bases W and W' such that d(W, W') = s for each s with 1 S s St.

Proof. Construct G = Kk+l □ Pt+l fromt+l copies Hi, H2, ... , Ht+l of Kk+l, where

V(Hi) = {u1 ,i, u2,i, ... , uk+ 1 ,i} for 1 Si St+ 1, by joining Uj,i to Uj,i+l for 1 Si St and

106 1 : j : k + l. By Lemma 5.4.2 and Theorem 5.2.4, lmd(G) = k. Furthermore, each set Wi = V(Hi) - { Uk+i,i} is a local metric basis fr 1 :i :t+ l. Since d(W1, Ws+1) = s for 1 : s : t, we obtain the desired result. ■

By the proofs of Theorems 5.4.1 and 5.4.3, there exists a connected graph with a local metric basis W for which the subgraph (W) induced by Wis an empty graph and there also exists a connected graph with a local metric basis W' fr which the subgraph (W') induced by W' is a complete graph. In fact, for every graph H, there is a connected graph with a local metric bais W such that (W) = H.

Theorem 5.4.4 For every graph H, there exists a connected graph G having local metric basis W such that (W) = H.

Proof. Suppose that His a graph of order k 2 1 with V(H) = { w1, w2, ... , wk}. Let P(A) = {S1 , S2, ... , S2k} be the power set of the set A = {l, 2, ... , k}. Construct G k from H by adding 2 new vertices in the set U = { u1, u2, ... , u2k} such that (U) = K2 k k k and joining Wi to Uj if and only if i E Sj fr 1 : i : and 1 : j : 2 . Observe that w(G) = 2k and so lmd(G) 2 k by Theorem 5.3.l. Since V(H) is a local metric set containing k vertices, it fllows that lmd(G) = k and V(H) is a local metric basis of G. ■

5.5 Joins and Compositions of Graphs

We have seen in Lemma 5.4.2 that for every two connected graphs G and H,

lmd(G DH)= max{lmd(G), lmd(H)}.

In this section, we study the lo�al metric dimension of the join G+ H and the composition G[H] of two connected graphs G and H.

5.5.1 Joins of Graphs

For two vertex-disjoint graphs G and H, recall that the join G + H of G and H has vertex set V(G) U V(H) and edge set consisting of E(G) U E(H) and all edges joining a vertex of G and a vertex of H. We frst present a lemma which determine the local metric dimensions of cycles.

107 Lemma 5.5.1 For each integer n 2 3,

if n is even lmd(C n ) = { � if n is odd.

Proof. By Theorem 5.2.4, lmd(C n ) = 1 if n is even and lmd(C n ) 2 2 if n is odd. On the other hand, any set consisting of two adjacent vertices of C n is a local metric set and so lmd(C n ) = 2 if n is odd. ■

Theorem 5.5.2 For every two connected graphs G and H,

lmd(G + H) 2 lmd(G) + lmd(H).

Proof. If G= H= K1 , then the result is true trivially. Thus we may assume that Gis a nontrivial connected graph. Let W be a local metric basisof G+Hand We= WnV(G) and WH = WnV(H). We claim that We# 0; forotherwise, let x and y be two adjacent vertices of G. Then codew(x) = codew(y) = (1, 1, · · · , 1), which is impossible. Thus

We # 0. Similarly, if His a nontrivial connected graph, then WH # 0. Next we show that We is a local metric set of G. If this is not the case, then there are adjacent vertices x and y such that codewa(x) = codewa(Y) in G. Let

We = {w1,w2, ... ,wk} � V(G). Thus de(x,wi) = de(y,wi) for 1::; i::; k. If

de(x,wi) = de(y,wi) = 1, where 1::; i ::; k, then de+H(x, wi) = de+H(Y, wi) = 1,

while if de(x, wi) = de(Y, wi) 2 2, then de+H(x, wi) = de+H(Y, wi) = 2. This implies that codew(x) = codew(y) in G + H, which is impossible. Similarly, if His a nontrivial connected graph, then WH is a local metric set of H. Thus regardless of H being trivial or nontrivial, lmd(G) + lmd(H)::; IWel + IWHI= IWI= lmd(G + H). ■

There are graphs G and H for which lmd(G + H) = lmd(G) + lmd(H). For example, lmd(Cs +Cs)= 4= 2 lmd(Cs), which we will see later in Theorem 5.5.6. Also, there are

graphs G and H for which lmd(G + H) > lmd(G) + lmd(H). For example, lmd(Kn 1 + ) Kn2 = lmd(Kn 1) + lmd(Kn2) + 1 for all integers n1,n2 2 1. In fact, the difference between lmd( G + H) and lmd( G) + lmd( H) can be arbitrarily large, as we will see soon. We now determine lmd( G + H) for several well-known classes of graphs G and H, namely complete graphs, cycles, and paths. In order to do this, we present a useful lemma.

Lemma 5.5.3 Let G and H be graphs and let W be a local metric set of G.

108 (a) If G =Kn + H where n 2 1, then IWn V(Kn)I 2 n - l.

(b) If G = Cn + H where n 2 4, then IWn V(Cn)I 2 I n/41- (c) If G = Pn + H where n 2 3, then IWn V(Pn)I 2 f(n - 1)/41.

Proof. The result in (a) follows by the proof of Theorem 5.5.2.

For (b), let Cn = (u1, u2,... , Un,u1) and construct· G = Cn + H. Consider an arbitrary local metric set W of G and let X be a set of fur consecutive vertices in Cn, say X = {u1,u2,u3,u4}. If X n W = 0, then observe that codew(u2) = codew(u3), which is a contradiction. This implies that IWn V(Cn)I 2 In/41-

For ( c), let Pn = ( u1, u2, ..., un) and construct G = Pn + H. Let W be an arbitrary local metric set of G. Observe that { ui, Ui+I, Ui+2, Ui+J} n W - 0 fr 1 :i :Sn - 3 and

frthermore, { u1, u2, u3} nW - 0 and { Un-2, Un-l, un} nW - 0. Hence, IWn V(Pn)I 2 f(n - 1)/41. ■

Theorem 5.5.4 For integers n1 2 4 and n2 2 1,

Proof. Let Cn1 = (u1,u2, ...,u n1 ,u1) and V(Kn2 ) = {v1,v2, ...,vn 2} and construct

G = Cn1 +Kn2• ByLemma 5.5.3, lmd(G) 2 fni/4l+n2-l. Consider the set U � V(Cn1) defned by

if n1 = 4, 6 if n1 = 5 or n1 2 7, and let W = U U [V(Kn2) - {vi}]. Then Wis a local metric set of G and so

lmd(G) : IWI = max{2, fni/41} + n2 -1.

If n1 = 4, then lmd(G) E { n2, n2 + 1}. Assume, to the contrary, that there exists a local metric set W' of G containing n2 vertices. Then IW'n V(C4)I = 1 and IW' n

V (K n2) I = n2 - l. Without loss of generality, assume that u1 E W' and v1 tJ. W'. However, this implies that codew,(u2) = codew,(v1), which is impossible since u2v1 E E(G). Therefre, lmd(G) = n2 + 1 for n1 = 4. This establishes the desired result. ■

109 As a consequence of Theorem 5.5.4, if Wn = Cn + K1 is the wheel of order n + 1, then

if n = 3 lmd(Wn) = { � if n = 4 (5.3) 1 n/41 if n 2 5.

Theorem 5.5.5 For integers n1 2 3 and n2 2 1,

Proof. Let Pn1 = (u1, u2,... ,U n1) and V(Kn2) = {vi, v2, ... , Vn2} and construct G = U � Pn1 + Kn2- By Lemma 5.5.3, lmd(G) 2 f(n1 - 1)/41 + n2 - 1. Let V(Pn1) be the set defned by

if 3 : n1 : 5 if n1 2 6 and n1 = 0, 1 (mod 4) if n1 2 6 and n1 = 2, 3 (mod 4), W U K W G and let = U [V ( n2) - { v1}]. Then is a local metric set of and so lmd(G) : IWI = max{2, f(n1 - 1)/41} + n2 - 1.

For n1 : 5, observe that lmd(G) E {n2,n2 + l}. By a similar argument used in the proof of Theorem 5.5.4, there is no local metric set containing n2 vertices. Therefre, lmd(G) = n2 + 1 if n1 : 5. ■

Theorem 5.5.6 For integers n1, n2 2 4,

Proof. Suppose that n1 : n2. Let

- be disjoint cycles and construct G = Cn1 + Cn2 By Lemma 5.5.3, lmd(G) 2 I ni/41 + I n2/4l Consider the sets W1 � V(CnJ and W2 � V(Cn2) defned by

{u1, u4}

{U 1,U5 , • • •, U4fn 1/4l-3}

{V 1, V4} if n2 = 4,6 {V 1, V5, • • •,V 4fn2/4l-3} if n2 = 5 or n2 2 7.

110 Then W = W1 U W2 is a local metric set of G and so

lmd(G): IWI = max{3, 1¥ l + 1¥1}-

If n2 = 4, then lmd(G) = 2 or lmd( G) = 3. By a similar argument used in the proof of Theorem 5.5.4, there is no local metric set containing two vertices. Therefre, lmd(G) = 3 in this case. ■

Theorem 5.5. 7 For integers n1,n 2 2 3,

Proof. Suppose that n1 : n2. Let

+ . be disjoint paths and construct G = Pn 1 Pn2 By Lemma 5.5.3, + lmd(G) 2 f(n1 - 1)/41 f(n2 - 1)/41-

{u3} if 3: n1 : 5 w,-{ {u3 , u7, ... , u4 lni/4J- 1 } if n1 2 6, n1 = 0, 1 (mod 4) { u3, U7, ... , u4 lni/4J- 1 } U { u4 lni/4J+2} if n1 2 6, n1 = 2,3 (mod 4)

{v2,v3} if 3 : n2 : 5 w,-{ { V3, V7, • • ·, V4 ln2/4J-1 } if n2 2 6, n2 = 0, 1 (mod 4) { V3, V7, · · ·, V4 ln2/4J-i} U { V4 ln2/4J+2} if n2 2 6, n2 = 2,3 (mod 4).

Then W = W1 U W2 is a local metric set of G and so

lmd(G) : W = ln1-l + n2-l I I max{3, 4 l l 4 l}.

If n2 : 5, then lmd(G) = 2 or lmd(G) = 3. By a similar argument used in the proof of Theorem 5.5.4, lmd(G) - 2 and so lmd(G) = 3 if n2 : 5. ■

Theorem 5.5.8 For integers n1 2 4 and n2 2 3,

111 n ) Proof. Let Cn 1 = (u1 , u2,... , Un 1, u1 ) and P 2 = (vi, v2, ... , Vn2 and construct G =

Cn1 + Pn2 • By Lemma 5.5.3,

lmd(G) 2: n I i/41 + 1(n2 - 1)/41-

If n1 = 4 and n2 :S 5, then let W = {u1, v2, v3}. Otherwise, consider the sets W1 �

V ( Cn1 ) and W2 � V ( Pn2) defined by

if 3 :S n2 :S 5 if n2 2: 6, n2 = 0, 1 (mod 4) if n2 2: 6, n2 = 2, 3 (mod 4),

and let W = W1 U W2. Then W is a local metric set of G and so

n2 lmd(G) :SI WI = max{ 3, I� l + f 4-l l}.

If n1 = 4 and n2 :S 5, then lmd(G) = 2 or lmd(G) = 3. Again by a similar argument used in the proof of Theorem 5.5.4, lmd(G) -=/ 2 and so lmd(G) = 3 in this case. ■

5.5.2 Compositions of Graphs

For graphs G and H, the composition G[H] is the graph with vertex set V(G) x V(H) such that (v1,u1 ) is adjacent to (v2,u2) if either (i) v1v2 E E(G) or (ii) v1 = v2 and u1u2 E E(H). Let V(G) = {v 1 , v2, ... ,vn}- Then G[H] is constructed by replacing each vertex Vi by a copy Hi of H and joining each vertex in Hi to every vertex in Hj if and only if vivj E V(G) with 1 :Si,j '.Sn. Let us always assume that V(G[H]) = ViUViU· · -UVn, where¼= {(vi,u): u E V(H)} for 1 :Si :Sn.

If G = K1 , then G[H] = H. Otherwise, consider two distinct vertices (vi, u) E ¼ and (vj, u') E ½ in G[H] where 1 :S i,j :Sn. Observe that

, min{dH(u,u'),2} if i = j dc[H]((vi, u ), (vj, u )) = { . · d G ( Vi, VJ ) if i # j.

Theorem 5.5.9 If G is a connected graph of order n and H is a graph, then

lmd(G[H]) 2: n - lmd(H).

112 Proof. The result is trivially true if G = Ki or H = Ki . Hence suppose that both G and H are nontrivial graphs. Suppose that lmd(H) = k and consider an arbitrary local metric set W of G[H]. Assume, to the contrary, that IWI � nk - l. Let Wi = W n ¼ for 1 � i � n and, without loss of generality, suppose that IWiI � k - l. Let

Since W H = { wi , w2,... ,w1w 1 I} is not a local metric set of H, there exist adjacent V(H) vertices x,y E such that dH(x,w) = dH(y,w) for every w E WH. Consider the two vertices ( vi ,x) and (vi, y) in Vi and observe that they are adjacent in G[ H]. Then

da[H] ((vi,x), (v i , w)) = min{dH(x, w),2} = min{dH(Y,w), 2}

= da[H]((vi,Y),(vi ,w)) for every (vi , w) E Wi , while

da[H]((vi, x),(v, u )) = da(vi, v) = da[H] ((vi,y), (v, u )) V(G[H]) for every (v,u) E - Vi. Therefore,code((v i,x)) = code((vi ,Y)), which is a contradiction. Therefore, IWI 2 nk and so lmd(G[H]) 2 nk. ■

The following is a consequence of Theorem 5.5.9.

Corollary 5.5.10 If G is a connected graph of order n each of whose distance equiva lence classes is a singleton set and H is a nontrivial connected graph with diam(H) � 2, then lmd(G[H]) = n - lmd(H).

Proof. If n = 1, then G[H] = H and the result is obvious. Suppose that n 2 2. Let lmd (H) = k and suppose that W H = { wi, w2,... , wk} is a local metric basis of H. Let

Wi = {(vi,wi), (vi,w2), ... , (vi,wk)} �¼for 1 � i � n and W = Wi U W2 U · · · U Wn. Hence IWI = nk. We show that Wis a local metric set of G[H]. Suppose that (vi, x),(vj, y) E V(G[H]) - W and (vi,x)(vj, y) E E(G[H]). Hence, x, y ¢ WH. We consider two cases.

Case 1. i = j, say i = j = 1. Since WH is a local metric set of H, there exists a vertex w E WH such that dn(x,w) i- dH(y,w). Since diam(H) � 2, it follows that

{dH(x,w),dH (y,w)} = {1,2}. Then observe that (vi,w) E W and

da[H] ((vi, x),(vi, w)) = min{dH (x,w), 2} i- min{dH (Y,w), 2}

= da[H]((vi,y),(vi,w)).

113 Case 2. i =I= j, say i = 1 and j = 2. Hence v1v2 E E(G). Since v1 and v2 belong to different distance equivalence classes in G, there exists a vertex v E V ( G) - { v1, v2} such that dc(v1, v) =I= dc(v2,v). Observe then that (v,w 1) E Wand

dc[H]((v1,x), (v,wi)) = dc(v1,v) =I= dc(v2,v) = dc[H]((v2,Y),(v,w 1)).

Thus, codew((vi,x)) =I= codew((vj, y)) in each case and so Wis a local metric set of G[H]. Consequently, lmd(G[H]) :Snk and the result now follows by Theorem 5.5.9. ■

The converse of Corollary 5.5.10 is false. To see this, consider a 7-cycle

Then diam(H) = 3 and the set WH = { u3, u6} is a local metric set and so lmd(H) = 2. G K V G G Let = 3 with ( G) = { v1, v2,v3} and construct [ H]. Of course, has only one distance equivalence class, which is not a singleton set. Let W = { (Vi, U3),(Vi, u6) : 1 :S i :S3} and observe that W is a local metric set of G[H]. Therefore, lmd(G[H]) = 6 = 3 · lmd(H) in this case. In fact, with this particular graph H, we have lmd(G[H]) = n · lmd(H) for every connected graph G of order n.

First, we make an important observation. The graph H described above has the property that there exists a local metric basis W such that

• for every vertex x E V(H) - W,

max{d(x,w): wEW}2:2,

that is, there is no vertex in V(H) - W that is adjacent to all vertices in W, and

• for every two adjacent vertices x, y E V(H) - W, there exists a vertex w E W such that {d(x,w), d(y, w)} = {1, 2}. H, H +H For every graph the join equals K2[H]. By Theorems 5.5.6 and 5.5.7 then,

if n = 4 lmd(K2[Cn]) Uril ifn 2: 5

if3:Sn:S5 lmd(K2[Pn]) = 1 { ! r·, 1 ifn2:6.

114 We next generalize these to obtain the local metric dimension of G[H], where one of G and H is a complete graph and the other is either a cycle or a path. First, we present a lemma, whose proof is similar to the proof of Lemma 5.5.3.

Lemma 5.5.11 Let G be a connected graph of order n1� 2. Let H be a graph and W a local metric set of G[H].

(a) If H = Kn2 where n2� 2, then IWn ½I� n2 - 1 for l::; i::; n1.

(b) If H = Cn2 where n2� 4, then IWn ½I� I n2/4l for l::; i::; n1.

Theorem 5.5.12 Suppose that n1� 2 is an integer.

(a) For n2� 4,

if n2 = 4 if n2� 5

(b) For n2� 3,

if 3::; n2::; 5 if n2� 6

G V(G) Proof. Let = Kn1 with = { v1, v2, ... , VnJ- We first verify (a). Let

and construct G[H]. For n2� 5, consider W � V(G[H]) such that

if n2 = 6 if n2 = 5 or n2� 7

for 1::; i::; n1. Then Wis a local metric set with IWI= n11 n2/4l By Lemma 5.5.11, lmd(G[H]) = n11n2/4l

115 If n2 = 4, then let W � V(G[H]) such that W n V = {(v1,u1)} and W n ¼ = {(vi,u1), (vi,u4)} fr 2 Si S n1. Then Wis a local metric set whose cardinality is 2n1 - l.

IfW' � V (K n1 [C4]) is a set containing 2n1 -2 vertices, then we may assume, without loss of generality, that IW' n Vl= IW' n V2l = 1. Suppose that W' n ¼ = {(vi,u1)} fr i = 1,2. Then codew,((v1,u2)) = codew,((v2, u2)) and so W' is not a local metric set of

Kn1 [C4]. Therefre, lmd(Kn1 [C4]) = 2n1 - l.

For Cn2 [Kn1] = H[G], consider the set W = V(H[G]) - {(ui,v1): 1 Si S n2} and observe that W is a local metric set of H[G]. By Lemma 5.5.11,

and the result now fllows.

To verif (b), let H P (u1, u2,... , U ) and construct G[H]. For n2 2 6, = n2 n2 consider W � V(G[H]) such that

if n2 = 0,1 (mod 4); while

if n2 = 2, 3 (mod 4) (where 1 S i s n1). Then W is a local metric set with IWI = n11(n2 - 1)/4l By Lemma 5.5.11, lmd(G[H]) = n1 f(n2 - 1)/4l

If 3 S n2 S 5, then let W � V(G[H]) such that W n V= {(v1,u3)} and W n ¼ = { (Vi,u2), (Vi, u 3)} for 2 Si S n1. Then W is a local metric set containing 2n1 -1 vertices. By a similar argument used for verifying (a), IW n ¼I = 1 fr at most one i for any local metric set W of G[H], implying that IWI2 2n1 - l. Therefre, lmd(G[H]) = 2n1 - l.

To prove that lmd(H[G]) = n2(n1 - 1) is straightforward. ■

Corollary 5.5.13 Let G be a connected graph of order n1 2 2. Then

]) (b) lmd(G[Cn2 = n1 I n2/4l for n2 2 5,

116 Proof. We first verify (a). Since the result is trivial for n2 = 1, suppose that n2 2: 2.

By Lemma 5.5.11, n1(n2 - 1) � lmd(G[Kn2 ]) and so we only show that lmd(G[Kn2 ]) � + n1(n2 - 1) lmd(G). Let lmd(G) = k and V(G) = {v1,v2, ... ,vn1 } such that We= { v1,v2, ... ,vk} is a local metric basis of G. Let W � V(G[Kn2]) such that

ifl�i�k if k +l � i � n1.

] + Then Wis a local metric set of G[Kn2 whose cardinality is n1(n2 - 1) k. ]) � For (b), we need only show that lmd(G[Cn2 n1 In2/4l by Lemma 5. 5.11. Let

]) and consider the set W � V(G[Cn2 such that

]) for 1 � i � n1. Consider two adjacent vertices (vi,x) and (vj, y) belonging to V(G[Cn2 W. If i = j, then there exists w E W n ¼ such that { d((vi, x), w), d(( Vj, y),w)} = { 1, 2}, that is, d((vi,x),w)-/= d((vj,Y),w). On the other hand, if i-/= j, then ViVj E E(G) and so

where Wi = W n ¼. Therefore, codew((vi, x)) -/= codew((vj, y)). Hence, W is a local ] metric set of G[Cn2 containing n1 I n2/ 41 vertices and we obtain the desired result. ]) � For (c), we again only show that lmd(G[Cn2 n11(n2 - 1)/41- Let

]) and consider the set W � V(G[Pn2 such that

if n2 = 0, 1 (mod 4); while

117 if n2 = 2, 3 (mod 4) (where 1 � i � n1). The result now followssince Wis a local metric set of G[Cn2 ] containing n1 f(n2 - 1)/41 vertices. ■

The upper and lower bounds in (a) are both sharp. For example,

for n1 2:'. 4, while

5. 6 Vertex and Edge Deletions

A common question in graph theory concerns how the value of a parameter is affected by making a small change in the graph. In this section, we study how the local metric dimension of a connected graph is affected by deleting a vertex or an edge from the graph, beginning with deleting a vertex. Observe that for the wheel Wn = Cn + K1 of order n + l 2:'. 4and a vertex v E V(Wn),

Wn - VE {Cn , Pn-l + K1},

For n 2:'. 3 we have seen that

3 if n = 3 lmd(Wn) -{ 2 if n = 4 ,�}l if n 2:'. 5.

1 if n is even lmd(Cn) = { 2 if n is odd.

2 if 3 � n � 6 K1) lmd(Pn-1 + - { n r :/1 if n 2:'. 7.

Thus fora connected graph G and a vertex v of G, it is possible that lmd(G-v) = lmd(G) or lmd( G -v) < lmd(G). In general, we have the following. For a vertex v in a nontrivial graph G, recall that N[v] = N(v) U {v} denotes the closed neighborhood of v.

118 Theorem 5.6.1 If v is a non-cut-vertex of a nontrivial connected graph G, then

lmd(G - v) '.S lmd(G) + degv.

Proof. Let W be a local metric basis of a connected graph G and let W' = [WU N[v]] { v }. Thus IW'l :S IWI + degv = lmd(G) + degv. We show that W' is a local metric set G V G W' of - v, fr otherwise, there exists a pair x, y E ( - v) - of -djacent vertices such that codew,( x) = codew, (y) in G - v. Since W is a local metric set of G, there exists a a= vertex w E W such that da(x, w) = dc(y,w) , say da(x, w) < dc(Y, w). Then since w E W' and codew,(x) = codew,(Y), it fllows that dc-v(x,w) = dc-v(y,w). Hence

a= da(x,w) < dc(y,w) '.S dc-v(y,w) = dc-v(x,w), implying that every x - w geodesic in G contains the vertex v. Let

p = ( X = UQ, U 1, ... , Ub, Ub+ 1 = V, Ub+ 2, ... , Ua = W) be an x - w geodesic in G. Thus, dc-v(x, ub) = da(x, ub) = b. Furthermore, since dc(Y, w) �a+ l, the path Q: y, x fllowed by Pis a y - w geodesic in G. This implies that dc-v(Y, ub) = dc(Y, ub) = b + l > dc-v(x, ub),

which is a contradiction since Ub E N(v) � W' and codew,(x) = codew,(y). Therefre, W' is a local metric set of G - v and so lmd(G - v) '.S IW'I :S lmd(G) + degv. ■

By the proof of Theorem 5.6.1, if there exists a local metric basis W of a connected graph G such that W n N[v] = 0, then W' = [WU N[v]] - {v} is a local metric set of G - v and IW'I :S lmd(G) + degv - 1. Thus

lmd(G - v) '.S IW'I: lmd(G) + degv - 1.

In fct, fr every local metric basis W of G,

lmd(G - v) '.S lmd(G) + degv - IW n N[v]I.

Although it is not known whether there exists a connected graph H containing a non cut-vertex v fr which lmd(H -v) = lmd(H) +deg v, there are infnitely many connected graphs G containing a non-cut-vertex v for which lmd( G - v) = lmd(G) + deg v - l.

Theorem 5.6.2 For every positive integer k, there exists a connected graph G contain ing a non-cut-vertex v such that G has local metric dimension k and

lmd(G - v) = lmd(G) + degv - 1.

119 Proof. For k = 1, let G be a tree of order at least 3 and v an end-vertex. Hence, assume that k 2:: 2 and consider the set A = {1, 2, ..., k - 2} for k 2:: 3 while A = 0 if k = 2, and let P(A) = {S1 , S2, ... , S2k-2} be the power set of A. Define the sets

S2k-2+1 , S2k-2+2, ... , S2k by

k 2 for 1 � i � 2 - and observe that { S1 , S2, ... , S2k} is the power set of A U { k -

k k 1, k} = {1, 2, ... , k}. Let H = K2 be a complete graph of order 2 with V(H) =

{ u1 , u2, ..., u2k}. We construct G from H by adding the k vertices in the set W = H G G , {w 1 , w2, ... , wk} to and joining Ui to Wj if and only if j E Si. Note that = w where w = 2k, th� graph described in the proof of Theorem 5.3.1, where we saw that W k is a local metric basis and lmd(G) = k. Furthermore, degwi = 2 -l for 1 � i � k. We show that lmd(G -wi) = k + 2k-l - 1 for 1 � i � k. By symmetry, it suffices to show that lmd(G -wk) = k + 2k-l - 1. Since the set

is a local metric set of G -wk containing k + 2k-l - 1 vertices,

lmd(G -wk) � k + 2k-l - 1.

Observe that each set Ui = { ui, ui+2k-1} is a distance equivalence class in G - Wk for 1 � i � 2k-l _ Thus, if there exists a local metric set W* containing at most k + 2k-l - 2 vertices, then we may assume that

On the other hand,

for 1 � i � 2k-2. Since Wk 1 W*, we further assume that - i

However then, 2k-l + 2k-2 � JW*J < 2k-l + k - 1, which is impossible. Therefore, lmd( G - Wk) = k + 2k-l - 1 as claimed. ■

The following result provides a sufficientcondition for a connected graph G containing a vertex v to have lmd( G - v) � lmd(G) + deg v - 1.

120 Theorem 5.6.3 Let v be a vertex with deg v � 2 that is not a cut-vertex in a connected graph G. If there exists a vertex v1 E N(v) such that dc-v(z, v1) :S: 2 for every z E N(v) - {v1}, then lmd(G - v) :S: lmd(G) + deg v - 1.

Proof. Let W be a local metric basis of G. We may assume that W n N[v] = 0. Let W' =WU N(v) and W{ = W' - {vi}. We show that W{ is a local metric set of G - v. Suppose that this is not the case. Since W' is a local metric set of G - v, it follows that there exists a pair x, y E V(G - v) - W{ of adjacent vertices such that

dc-v(x, w) -=/ dc-v(Y, w) if and only if w = v1 for each vertex w E W'. Also, since W is a local metric set of G, there exists a vertex w* E W < W' such that dc(x, w*) -=/ dc(Y, w*), say a= dc(x, w*) < dc(y, w*). Observe that dc-v(x, w*) = dc-v(Y, w*) by assumption since w*-=/v1. Then

dc(x, w*) < dc(y, w*) :S: dc-v(Y, w*) = dc-v(x, w*), implying that every x - w* geodesic in G contains v. Let

p = (x = uo, u1, ... , Ub, Ub+l = v, Ub+2, ... , Ua = w*)

be an x - w* geodesic in G. Observe that ub, ub+2 E N(v) < W'. Since dc(x, w*) < dc(y, w*), it fllows that dc-v(x, ub) -=/ dc-v(Y, ub) and so ub = v1. However then, dc-v(ub, ub+2) = dc-v(v1, ub+2) :S: 2, implying that there exists an x - w* path in G - v having length at most a. This is a contradiction.

Therefre, no such pair x, y exists and W{ is a local metric set of G - v. Consequently, lmd(G - v) :S: IW{I = lmd(G) + degv - 1. ■

It is not known whether there is a connected graph G containing a non-cut-vertex v of G such that lmd(G - v) = lmd(G) + degv. On the other hand, there are many connected graphs G with a vertex v such that lmd( G - v) = lmd( G). First, we state an observation.

Observation 5.6.4 If G is a nontrivial connected graph and v is an end-vertex of G, then G contains a local metric basis not containing v.

Proposition 5.6.5 If v is an end-vertex in a connected graph G, then

121 lmd(G - v) = lmd(G).

Proof. We first show that lmd(G-v) � lmd(G). By Observation 5.6.4 there is a local metric basis Wof G such that v (/.W. Since de-v(x, y) = de(x, y) forevery two vertices x, y E V(G -v), it follows that Wis a local metric set of G - v.

Next we verify that lmd(G) � lmd( G - v). Let v1 be the vertex adjacent to v in G and suppose that W' is a local metric basis of G - v. Consider a pair x, y of adjacent vertices in V ( G) -W'. If v E { x, y}, then { x, y} = { v, vi}. Since de (v, u) = de (v1, u) + 1 for every u E V ( G) - { v}, it follows that codew, (v) -::/- codew, (v1). Hence, assume that v (/. { x, y}. Since W' is a local metric set of G -v, it follows that there exists a vertex w E W' such that de-v(x,w)-::/- de-v(y,w) . On the other hand, de(x,w) = de-v(x,w) as well as de(Y, w) = de-v(Y, w) and so

de(x,w) = de-v(x,w) -::/- de-v(Y, w) = de(Y, w), that is, de(x, w) -::/- de(Y, w). Hence, W' is a local metric set of G. ■

Next, we investigate how the local metric dimension of a connected graph is affected by deleting an edge from the graph.

Theorem 5.6.6 If e is an edge that is not a bridge of a connected graph G, then + lmd(G - e) � lmd(G) 2.

Proof. Let W be a local metric basis of G, e = v1v2, and W' = W U { v1, v2}. Then IW' I � IWI+ 2 = lmd(G) + 2. We show that W' is a local metric set of G - e. If this is not the case, then there exists a pair x, y E V(G - e) - W' of adjacent vertices such that codew,(x) = codew,(Y) in G - e. Since Wis a local metric set of G, there exists a vertex w E W such that de(x, w) -::/- de(Y, w), say a= de(x, w) < de(Y, w). Also, since w E W' and codew,(x) = codew1 (y), it follows that de-e(x,w) = de-e(y,w). Hence

a= de(x, w) < de(y, w) � de-e(x, w) = de-e(Y, w), implying that every x -w geodesic in G contains the edge e. We may assume, therefore, that

p = (x = uo, U1,. .. , Ub = V1, Ub+l = V2, ... , Ua = w)

122 is an x - w geodesic in G. Observe that dc-e(x, v1) = dc(x,v1) = b. Furthermore, since dc(y,w) �a+ l, the path Q = (y,x) fllowed by Pis a y - w geodesic in G. Therefre, dc-e (y,v1) = de (y, v1) = b + l > dc-e( x, v1), which contradicts the fact that codew, ( x) = codew, (y) and v1 E W'. Therefore, W' is a local metric set of G - e and lmd(G - e) : IW'I : lmd(G) + 1. ■

By the proof of Theorem 5.6.6 if e = v1v2 is not a bridge of G and there exists a local metric basis W of G such that W n {v1, v2} -1- 0, then lmd(G - e) : lmd(G) + 1.

In fact, for every local metric basis W of G,

lmd(G - e) : lmd(G) + 2 - IW n {v1, v2}I-

We conclude this chapter with the fllowing two conjectures.

Conjecture 5.6.7 If vis a vertex that is not a cut-vertex of a connected graph G, then

lmd(G - v) � lmd(G) - degv.

Conjecture 5.6.8 If e is an edge that is not a bridge of a connected graph G, then

lmd(G - e) � lmd(G) - 2.

123 Chapter 6

Upper Local Metric Dimension

6.1 Introduction

While the local metric dimension of a graph G is the minimum number of vertices in a local metric set of G, there are certainly larger local metric sets as well. Indeed, V(G) is a local metric set. Many local metric sets contain proper subsets that are also local metric sets - but not all. A local metric set W of a graph G is said to be a minimal local metric set if no proper subset of W is a local metric set fr G. Every connected graph G has a minimal local metric set since every local metric basis is a minimal local metric set. The maximum cardinality of a minimal local metric set fr G is the upper local metric dimension lmd+ ( G) of G. Thus if G is a nontrivial connected graph of order n, then 1 � lmd(G) � lmd+ (G) � n - l. (6.1)

To illustrate these concepts, consider the graph G = W1 0 = C10 + K1 of Figure 6.1 obtained from a IO-cycle C1 0 = (v1, v2, ... , v10, v1) by adding a new vertex vo and joining = { vo to every vertex of C10. The set W1 v1, v5, v7} is a local metric set for G. The code of each vertex with respect to W1 is shown in Figure 6. l(a). Therefore, lmd(G) � 3. To see that lmd( G) 2 3, assume, to the contrary, that there exists a local metric 2-set W' for G. If vo E W', say W' = {vo,v1}, then codew,(vi) = (1,2) for 3 � i � 9, which = { is impossible. If Vo 1 W', then we may assume that W' v1, Vs} for some s with 2 � s � 6. Then codew,(vs) = codew,(vg) = (2, 2), which is also impossible. Therefre, = lmd( G) 3 and so W1 is a local metric basis fr G.

Since the set W2 = { v1, v3, v6, vs} is both a metric set and a minimal local metric set + for G (see Figure 6.l(b)), dim(G) � 4 � lmd (G). We show that dim(G) = 4. Assume, to the contrary, that there is a metric 3-set W' = { w1 , w2, W3} for G. Therefore, if

124 (a) (b) Vg

Figure 6.1: A graph G with lmd(G) = 3 and dim(G) = lmd+ (G) = 4 v E V(G)- W', then codew,(v) = (a,b,c), where {a,b,c} � {1,2}. This implies th�t there are 23 = 8 possible codes fr those vertices not belonging to W', that is, every possible code must be used since IV(G) - W'I = 8. Then degw1 2 4, since there are ( fur codes of the frm 1, b, c). Similarly, there are four codes of the form (a, 1, c) and so deg w2 2 4. However, this is a contradiction since vo is the only vertex whose degree exceeds 3. Thus, dim( G) = 4. Next, we show that lmd+(G) : 4. Let W be a local metric 5-set for G. We show that W contains a proper subset that is also a local metric set for G. Suppose, frst, that vo E W and let W' = W- {vo}. Since there is no vertex in G whose code with respect to W is (1, 1, 1, 1, 1), codew,(v) = (1, 1, 1, 1) if and only if v = vo. Since d(vi, vo) = 1 fr 1 : i : 10, every two adjacent vertices in G have distinct codes with respect to

W'. We may therefore assume that vo � W, say W v1 Vi , Vi Vi Vi Vis}, where = { = 1 2 , 3 , 4 , 1 = i1 < i2 < i3 < i4 < i5 : 10. If i3 : 5, then let W' = W-{vi2 }. If i3 2 6, then let W' = W -{viJ· In each case, W' is a local metric set fr G. This implies that no local metric set containing more than fur vertices is a minimal local metric set. Consequently, lmd+ (G) = 4.

In Figure 6.l(c) it is also shown that W3 = {v1,v3,v5,v1} = W1 U {v3} is a metric set that is not a minimal local metric set; while in Figure 6.l(d), W4 = {v1,v2,vB,v1} is

125 a minimal local metric set that is not a metric set. For the graph G of Figure 6.1, we have shown that lmd(G) = 3 and dim(G) = lmd+ (G) = 4. It is possible for all three of these metric parameters to be distinct. For example, lmd(P) = lmd+ (P) = dim(P) = 3 for the Petersen graph P. On the other hand, for the graphs F and H in Figure 6.2, it can be shown that lmd(F) = 3, dim(F) = 4 and lmd+(F) = 5, while lmd(H) = 3, lmd+(H) = 4 and dim(H) = 5.

H: --Ks

Figure 6.2: Graphs F and H with lmd(F) < dim(F) < lmd+(F) and lmd(H) < lmd+ (H) < dim(H)

6.2 Comparing the Two Local Metric Dimensions

In Figure 6.1 we saw a graph G for which lmd+(G) > lmd(G). In this section we consider those graphs G with the property that if Wis a local metric set with IWI > lmd(G), then G must also contain a proper subset of W that is a local metric set. These are precisely the graphs G for which lmd+ (G) = lmd(G). First, we restate a result established in Chapter 5.

Theorem 6.2.1 Let G be a nontrivial connected graph of order n. Then

(a) lmd(G) = n - 1 if and only if G = Kn ,

(b) lmd(G) = n - 2 if and only if w(G) = n - 1 and

126 For each local metric basis W ofa nontrivial connected graph G, the resulting codes with respect to W can be interpreted as colors ofthe vertices ofG. Thus each local metric basis of G gives �ise to a proper coloring of G using at most lmd(G) + diam(G)1rd (G) colors. Therefre, x(G) � lmd(G) + diam(G)1rd (G). Furthermore, there are graphs fr which equality holds a well a graphs fr which strict inequality holds by Theorem 6.2.1. On the other hand, the fllowing is another consequence of Theorem 6.2.1.

Corollary 6.2.2 Let G be a nontrivial connected graph of order n. For each k E {1,n -2,n -1}, lmd(G) = k if and only if x(G) = k+ 1.

We have seen in (6.1) that if G is a nontrivial connected graph of order n, then 1 � lmd(G) � lmd+ (G) � n -1. With the aid of Theorem 6.2.1, we next show fr k E {1,n -2,n -1} that lmd(G) =kifand only iflmd+ (G) = k.

Theorem 6.2.3 Let G be a nontrivial connected graph of order n. For each k E {1, n- 2,n - 1}, lmd(G) = k if and only if lmd+ (G) = k.

Proof. For k = 1 or k = n -1, the result is an immediate consequence of Theo rem 6.2.l(a), (c). Thus it remains to establish the result fr k = n - 2. First, suppose that G is a connected graph of order n with lmd(G) = n -2. We claim that every local metric set whose cardinality is n -1 contains a proper subset that is a local metric set. To see this, suppose that x E V ( G) and W = V (G) - {x} is a local metric set. If deg x = n - 1, then since G is not complete, there exists a vertex w1 E W whose degree is less than n-1. Then W - {w1} is a local metric set. Ifdeg x � n - 2, then let w2 E W such that xw2 (j. E( G). Then W -{w2} is a local metric set. Thus, a claimed, every (n - 1)-element local metric set W contains a proper subset that is a local metric set. Therefre, lmd+ (G)- n -1 and so lmd+ (G) = n - 2 by (6.1).

For the converse, suppose that G is a connected graph of order n with lmd+ ( G) = n - 2. We show that lmd(G) = n -2 as well. By Theorem 6.2.1, it sufces to show that w(G) = n -1. Since the result is obvious fr 2 � n � 4, we may assume that n � 5. We claim that ifw(G) � n-2, then every local metric set W fr G containing n -2 vertices has a proper subset that is also a local metric set fr G. Let W be a local metric set fr G containing n -2 vertices and suppose that V ( G) = W U {x, y}.

127 If there exists a vertex w1 in W such that W1 is adjacent to neither x nor y, then consider the set W - { wi}. If xy � E ( G), then { x, y,w 1} is an independent set and so W - { w1} is a local metric set. If xy E E (G), then since W is a local metric set and G is connected, there exists a vertex w2 E W - { wi} such that w2 is adjacent to only one of x and y. Since w1 i- w2, it follows that W - { wi} is still a local metric set for G. Thus, we may assume that every vertex in W is adjacent to at least one of x and y. Let X = N(x) n Wand Y = N(y) n W. Hence, W =XU Y. We consider three cases.

Case l. X = W or Y = W, say X = W. Since w(G) � n - 2, there are two vertices w1,w2 E W that are not adjacent. If Y = W, then since W is a local metric set, it follows that x and y are not adjacent. Then W - { w1} is a local metric set. Suppose next that Yi- W. If {w1,w2} � Y, then W - {w1} is a local metric set. We therefore assume that w1 � Y. If { xy,w2y} CZ E ( G), then W - { wi} is a local metric set. Otherwise, either (i) there exists a vertex w3 E [W- Y] - { wi} or (ii) there exist two vertices Wa, w1 E W - { w1} that are not adjacent. Let W' = W - { W1} if (i) occurs and W' = W - { wa} otherwise. Then in each case W' is a local metric set. This also implies that if X = 0 or Y = 0, then W always contains a proper subset that is a local metric set.

Case 2. X n Y = 0. We may assume that X, Y i- 0 by Case 1. If no vertex in X is adjacent to any vertex in Y, then xy E E( G) since G is connected and W - { w} is a local metric set for every w in W. Otherwise, suppose that w1 E X, w2 E Y and w1w2 E E ( G). Then observe that W - { w1} is a local metric set.

Case 3. Neither Case l nor Case 2 .ccurs. Then we may assume that none of X, Y and X n Y is empty. We may further assume that both X' = W - Y and Y' = W - X are not empty. Let w1, w2 and W3 be three vertices in W, where w1 E X', w2 E Y' and w3 EX n Y. If wiw3 E E(G) for i = 1,2 or wiw3 � E(G) for i = 1,2, then W - {w3} is a local metric set. Otherwise, assume that w1W3 E E( G) and w2w3 � E( G). Then W - { w2} is a local metric set.

Therefore,if w(G) � n - 2, then there is no minimal local metric set containing n - 2 vertices, that is, lmd+ (G) � n - 3. Therefore, if lmd+(G) = n - 2, then w(G) = n - l and so lmd(G) = n - 2 by Theorem 6.2.l(b). ■

The following result is an immediate consequence of Theorems 6.2.1 and 6.2.3, which characterizes all nontrivial connected graphs of order n with upper local metric dimen sion 1, n - 2 or n - l.

128 Corollary 6.2.4 Let G be a nontrivial connected grph of order n. Then

+ (a) lmd (G) = n - l if and only if G = Kn,

(b) lmd+(G) = n - 2 if and only if w(G) = n - land

By Corollary 6.2.4, we obtain the fllowing, which is similar to Corollary 6.2.2.

Corollary 6.2.5 Let G be a nontrivial connected graph of order n. For k E {1, n - 2,n -1}, lmd+ (G) = k if and only if x(G) = k + l.

By Theorem 6.2.3(c), if G is a nontrivial tree, then lmd(G) = lmd+(G) = 1. There + are other classes of graphs G for which lmd(G) = lmd (G). For example, lmd(Cn) = + lmd (C n ) = 2 for each cycle Cn of odd order n 2 3. We have seen that if G is a complete k-partite graph, where k 2 2, then lmd( G) = k - l. In fact, lmd+ (G) equals k - l as well, which we show next.

Proposition 6.2.6 If G is a complete k-partite graph for some k 2 2, then

lmd(G) = lmd+ (G) = k - l.

Proof. We may assume that k 2 3. Suppose that the partite sets of G are are V, Vi,... , Vk. If W is a subset of V(G) such that W n V = W n V2 = 0, then W is not a local metric set since the vertices in V1 U V2 have the same code with respect to W. Therefore, if W is a local metric set fr G containing at least k vertices, then we may assume that either (i) W n ¼ = 0 fr 1 : i : k or (ii) W n ¼ = 0 if and only if i = k and [W n Vil 2 2. Let w E W n V- Then in either cae, W - { w} is a local metric set for G, that is, W is not a minimal local metric set fr G. ■ We have seen that there are infnitely many graphs G fr which lmd(G) = lmd+ (G). In particular, if G is a complete multipartite graph, a tree, or a cycle, then lmd(G) = lmd+ ( G). On the other hand, there are also infnitely many graphs G fr which lmd+ (G) lmd(G) can be arbitrarily large. The following result on the wheel G = Cn + K1 of order n + l 2 4 is a consequence of Theorem 5.5.4 as we indicated in Chapter 5.

129 Theorem 6.2. 7 For every integer n 2'. 3,

ifn = 3 lmd(Cn + K1) = { � ifn = 4 r n/41 ifn 2'. 5.

We now determine the upper local metric dimension of wheels.

Theorem 6.2.8 For an integer n 2'. 3,

2 ifn = 4 3 ifn E {3, 5, 6} l 2n/5 J ifn 2'. 7.

+ Proof. LetG =C n K1, whereC n = (v1,v2, ... ,Vn, v1) and the vertex v ofG is adja cent to every vertex ofC n. The result immediately fllows fr n = 3, 4 by Corollary6.2.4. For n = 5,6, the set {v,v1,v3} is a minimal local metric set and so lmd+ (G) 2: 3. Sup pose that W is a local metric set with IWI 2: 4. If n = 5, then W contains at least three vertices belonging toC 5 . Hence, we may assume thatv1, v2 E W. Then Wis not a minimal local metric set since { v1, v2} is a local metric set. For n = 6, observe that if {Vi?Vi +3} � W fr some i ( 1 � i � 3), then W is not a minimal local metric set.

Otherwise, we may assume that either W = { v,v1, v2,v3} or W = { v,v1, V3,V 5}. Then W is not a minimal local metric set in either case since W - {v} is a local metric set. Therefre, lmd+ (G) = 3. For n 2'. 7, we frst claim that there is no minimal local metric set containing v. Let W be a local metric set containing v and consider a pair x, y E V(G) - W' of adjacent vertices, where W' = W - {v}. We show that codew,(x) = codew,(y). Since codew(x) = codew(Y), there exists a vertex w E W such that d(x, w) = d(y, w). If v � {x,y}, then d(x,v) = d(y,v) = 1 and sow = v. Therefre, w E W', implying that codew,(x) = codew,(y). If v E {x,y}, then we may assume that x = v and y = v1. If

{v3,V4,... ,Vn-1} n W' = 0, then codew(v4) = codew(v5) since n 2'. 7. However, this is a contradiction since W is a local metric set. Therefore, { v3, V4,... ,V n-d n W' = 0, implying that at least one of the coordinates of codew,(y) is 2. Since codew,(x) = (1, 1, · · · , 1), it fllows that codew,( x) = codew,(y).

Let W be a minimal local metric set. Then W � V(Cn) and so we may write W= = + Vip W2 Vi , .••,wk Vi }, k ik � n. {w1 = = 2 k where = lmd (G) and 1 � i1 < i2 < · · · <

130 k, Furthermore, assume that i1 = 1. We claim that ij+2-i j 2 5 for 1 ::; j '.S where ik+ l = n + i 1 = n + 1 and ik+2 = n + i2. Suppose that ij+2- ij '.S 4 for some j, say i 3 ::; 4+i 1 = 5. Then observe that W- { w2} is a local metric set, which is a contradiction since Wis a minimal local metric set. This implies that for every set U of fiveconsecutive vertices in C n, at most two vertices in U belongs to W, that is, IWI :'.S 2n/5. Therefore, lmd+ (G)::; l2n/5J. On the other hand, consider the set

Wo = { V5i+ 1 : 0 :'.S i :'.S l n / 5 J - 1} U { V5i+4 : 0 :'.S i :'.S l n / 5J - 1} and let Wo if n = 0, 1, 2 (mod 5) w-{- Wo U { v5ln/5J+ i} if n = 3,4 (mod 5).

Then W is a minimal local metric set and IWI = l 2n/5J. Therefore, lmd+ (G) 2 l 2n/5J, which is the desired result. ■

With the aid of Theorems 6.2.7 and 6.2.8, we show next that lmd+(G)-lmd(G) can be arbitrarily large.

Theorem 6.2.9 For every positive integer k, there exists a graph G such that

lmd+ (G) - lmd(G) = k.

Proof. Let k be a positive integer. Then k E {3q -2, 3q -1, 3q} for some positive integer q. Consider the graph

C2oq-3 + K1 if k = 3q - 2 G C2oq-1 + K1 if k = 3q - 1 = if k 3q. { C2oq + K1 =

Then lmd+ (G) - lmd(G) = k by Theorems 6.2.7 and 6.2.8. ■

6.3 Realization Results

We have seen that if G is a connected graph with lmd(G) = f and lmd+ (G) = L, then 1 :'.S f '.S L. This gives rise to the question of determining which pairs (.f_, L) of positive integers are realizable as the local metric dimension and upper local metric dimension, respectively, of some connected graph. We refer to such pairs of positive integers as realizable pairs. By Proposition 6.2.6, each pair (£, f) is a realizable pair for every positive

131 integer£. Also, for R, < L, there are infinitely many realizable pairs by Theorems 6.2.7 and 6.2.8. On the other hand, by Theorem 6.2.3, no pair (1,L) is realizable for L 2 2. Thus we have the following question for those pairs (£,L) with£ 2 2.

Problem 6.3.1 Which pairs(£, L) of positive integers with 2 � R, � L are realizable?

Next,we show that for each integer£ 2 2,every pair (£,L) is realizable for all integers L with 2£-l + R, -1 � L �£ 2 - 1. In order to do this, we present useful lower bounds for the local metric dimension and upper local metric dimension of a graph. Recall that if G is a nontrivial connected graph, then

(6.2)

By (6.1) and (6.2), if G is a nontrivial connected graph, then

(6.3)

Theorem 6.3.2 For every pair£, L of integers with R, 2 2 and 2£-l +£-1 � L � 2t -1, there exists a connected graph G such that lmd( G) = £ and lmd+ ( G) = L.

Proof. Let {S1 ,S2,... ,S 2e} be the power set of {1,2,... ,R,}. We first construct a £ graph Go of order 2 +£ from a graph H = K2t with V(H) = { v1 , v2,... , v2e} by adding

R, new vertices in the set U = { u1, u 2,... , Uf} and joining Ui to Vj if and only if i E Sj . If t L = 2 -1,then let G = G0. Otherwise, let G be the graph obtained fromGo by deleting £ the 2 -L - 1 vertices VL+2,VL+3,... ,v2t. Let V = V(Go) -U = {v1 ,v2,... ,VL+d· Since w(G) = L + 1 > 2e-i and U is a local metric set, lmd(G) =£ by (6.2). Next we show that lmd+ ( G) = L. First let v E V and consider the set Wo = V-{ v}. Then JN(v) n Wol = JWol = L while JN(u) n Wol � deg u � 2£-l < L for each u E U, implying that Wo is a local metric set forG. Furthermore,Wo is a minimal local metric set, implying that lmd+(G) 2 JWo[ = L.

To show that lmd+ ( G) � L, suppose that W is a local metric set containing L + 1 vertices. We show that W contains a vertex w such that W - { w} is also a local metric set forG. Since the result immediately follows if JW n U[ E {O, 1,£}, we may assume that 2 � [W n U[ = t � R, - 1. Renaming the vertices, if necessary, let

}. W = {u 1 ,u 2,... ,Ut,Vt+l,Vt+2,... ,V£+1 Furthermore, let ¼ = {v EV: codewnu(v) = codewnu(vi)}

132 for 1 :S i :S t. Then the sets Vi, Vi,... , ½ are pairwise disjoint and I¼ I :S 2e-t. Since

VI - I (L t e t e t 1 I u!=1 ¼I 2: + 1) - · 2 - 2: 2 -t (2 - - t) + £ 2: £ 2: 2, there exists a vertex v* EV - U!=l ¼ � W. We show that W* = W - { v*} is a local metric set. We need only show that if u E U and v E V are adjacent and u, v (/. W*, then their codes with respect to W* are different. Observe that IN(u) n W*I :S deg u - 1 :S 2e-i - 1

while IN( v) n W* I 2: IV n W* I = L - t 2: L - £ + 1 2: e2 -1. Therefore, codew• ( u) =I- codew• (v). Since W* is a proper subset of W, we obtain the desired result. Hence, lmd+ ( G) = L. ■

For a nontrivial connected graph G, define the (f, L)-ratio of G by

_ lmd(G) re ,L( G) - lmd+(G)'

It then follows by ( 6 .1) that 0 < re,L(G) :S 1.

With the aid of Theorem 6.3.2, we next show that for every rational number r E (0, 1] there exists a connected graph G such that re,L(G) = r. We denote the graph G with + L) (lmd(G), lmd (G)) = (£, constructed in the proof of Theorem 6.3.2 by Gl,L ·

Theorem 6.3.3 For every rational number r with O < r :S 1, there exists a connected graph G such that rl,L(G) = r.

Proof. Let r = a/b,where a and b are positive integers with a :S b. We consider two cases, according to whether a= l or a 2: 2.

l/b. + (K ) b Case l. r = Since lmd(Kn) = lmd n for n 2: 2, we may assume that 2: 2. "' Let f : JR ➔ JR be a function defined by f( x) = 2 ;1 . Then f is strictly increasing on (1, oo) and f(2) < 2 < f(3). Hence, there exists the unique integer Pb 2: 3 such that 2"' 1 J(fb - 1) < b :S f(fb)- Let g : JR ➔ JR be a function defined by g(x) = - ;x-l. For

x 2: 6, observe that g(x) < f(x - 1) and so g(f_b) < b :S f(fb) for b 2: 7. In fact, we can

verify that g(fb) :S b :S f(fb) for every b 2: 2. Therefore,

133 Then by Theorem 6.3.2 the graph Geb,b-eb has the desired property.

Case 2. r = a/b, where 2 ::; a< b and a f b. Let Go = Ge,e(b-a+l) as described in the

proofs of Theorem 6.3.2 and in Case 1, where f, = f,b-a+l � 3. Hence, lmd(Go) = f, and + f(b- H f(b- lmd (Go) = a+ 1). Let be the clique of order w(Go ) (= a+ 1) + 1) in Go.

Also let F = Ke(a-l)+l be a graph with V(Go) n V(F) = 0. We construct a graph G from Go and F by adding a new vertex x and joining x to every vertex in V(F) U V(H). Then

lmd(G) = lmd(Go) + £(a - 1) = fo

and

which gives us the desired result. ■

We now illustrate the proof of Theorem 6.3.3 with the following example. For r =

5/28, first obtain f2s-5+ 1 = £24 = 8. Consider the graph Go = Gs,8-24 = Gs,192 whose local metric dimension and upper local metric dimension are 8 and 192, respectively.

Adding a copy of K8(5-l)+l = K33 and a new vertex x to Go and joining x to every vertex except for the eight vertices u1, u2, ... , us, we obtain a graph G with lmd(G) = 8 + 32 = 40 and lmd+(G) = 192 + 32 = 224. (See Figure 6.3.)

Figure 6. 3: A graph G with lmd(G) = 40 and lmd+(G) = 224

134 6.4 Local Metric Dimensions of Graphs with Prescribed Order

In this section, we investigate the possible values of lmd(G) and lmd+ ( G) for a connected graph G of fxed order n. By Proposition 6.2.6, for each integer n 2 2 and an integer a with 1 : a:=; n - 1, the complete (a+ 1)-partite graph G = K1,1, ...,I,n-a of order n has lmd( G) = lmd+ ( G) = a. Thus we have the fllowing.

Proposition 6.4.1 Let n 2 2. For each integer a with 1 < a < n - 1, there is a connected graph G of order n with lmd(G) = lmd+ (G) = a.

For those connected graphs G of order n with lmd(G) < lmd+ (G), we have the fllowing consequence of Theorem 6.2.3.

Corollary 6.4.2 If G is a nontrivial connected graph of order n such that lmd(G) - lmd+ (G), then 2: lmd(G) < lmd+ (G): n - 3.

By Corollary 6.4.2, if G is a connected graph of order n such that lmd( G) - lmd+ (G), then n 2 6. In particular, observe that if G is a connected graph of order 6, then either (i) lmd(G) = lmd+ (G) or (ii) lmd(G) = 2 and lmd+ (G) = 3. If (ii) occurs, then G must contain an odd cycle and w(G) : 4. In fact, there are exactly eight such graphs, three of which are shown in Figure 6.4.

Figure 6.4: Graphs G1 , F1 and H1 of order 6

For each of G1 and G2 = G1 + v5v6, the set { vs, v6} is the unique local metric basis,

while any 3-set W � { v1 , v2, v3, V4} is a minimal local metric set. Therefre, each of G 1

and G2 has exactly fur minimal local metric 3-sets. Observe also that G1 and G2 are the only connected graph of order 6 with w(G) = 4 and lmd(G) = 2.

For each of Fi= Ps + K1 and F2 = F1 + v1vs =Cs+ K1 , the set {v1, vs} is a local

metric bais. Also, the set {v 2, V4, V6} is the unique minimal local metric 3-set fr Fi,

135 while every 3-set { Vfi,x, y} where xy €j. E(F2) is a minimal local metric set for F2. Thus,

F1 has only one minimal local metric 3-set and F2 has exactly five minimal local metric 3-sets.

Finally, consider the graphs H1 , H2 = H1 +v2v4 , H3 = Hi +v2 v6 and H4 = H2+v2Vfi.

Then the set {v3,v5} is a local metric basis for Hi (1 :Si :S4). Furthermore, let W1 ,

W2, W3 and W4 be 3-sets satisfying

and observe that W is a minimal local metric 3-set for Hi if and only if W = Wi ,

Therefore, the number of minimal local metric 3-sets for H1 , H2, H3 and H4 is four,two, three and two, respectively.

Let 9 = {G1 ,G2,F1 ,F2,H1 ,H2,H3,H4}. We summarize the result for connected graphs of order 6 in Table 6.1.

lmd(G) lmd-t- (G) 1 1 G is bipartite. 2 2 w(G) :S3, G €j. 9, G is not bipartite. 2 3 GE Q. 3 3 w(G) = 4 and G €j. {G1 ,G2}. 4 4 w(G) = 5. 5 5 w(G) = 6, i.e., G is complete.

Table 6.1: The values of lmd(G) and lmd+(G) of all connected graphs G of order 6

A triple (a, b,n) of positive integers is said to be realizable if there exists a connected graph G of order n for which lmd(G) = a and lmd+ (G) = b. Thus, if (a,b,n) is a realizable triple, then 1 :S a :S b :S n - l. Furthermore, every triple (a, a,n) with 1 :S a :S n - l is realizable; while if a i= b and a triple ( a, b,n) is realizable, then 2 :Sa < b :Sn - 3. We have seen that no triple (1, b, n) is realizable if b 2:: 2. Thus, we have the following question.

Problem 6.4.3 Which triples ( a, b,n) of integers with 2 :S a < b :Sn - 3 are realizable?

136 Next, we show that if a triple ( a, b, n) is realizable, then every triple ( a, b, n') is realizable for each integer n' 2:: n. In order to verify this, we first present some additional definitions. Recall that two vertices u and v in a connected graph G are twins if u and v have the same neighbors in V ( G) - { u, v}. Also, if u and v are adjacent, then they are referred to as true twins; while if u and v are nonadjacent, then they are false twins.

Theorem 6.4.4 Let a, b and n be integers with l '.S a '.S b '.S n - l. If (a, b, n) is a realizable triple, then so is (a, b, n') for every integer n' 2:: n.

Proof. Let G be a connected graph of order n such that lmd(G) = a and lmd+ ( G) = b. (Hence, a '.S b '.S n - l.) We need only show that there exists a connected graph of order n + l whose local metric dimension and upper local metric dimension are a and b, respectively. Let x1 E V ( G) and obtain a connected graph H of order n + l by adding a new vertex x2 to G and joining x2 to every vertex in N(x1). Therefore, x1 and x2 are false twins in H. First suppose that W is a local metric set for G. Then the code of a vertex v E V ( G) with respect to W in G equals the code of v with respect to W in H. Furthermore, co dew (x2) =I= co dew (v) in H for every v E N ( x1) in H. Therefore,

every local metric set for G is a local metric set for H. (6.4)

This implies that lmd(H) '.S lmd(G). On the other hand, suppose next that W is a local metric set for H. If x1, x2 E W, then W - { xi} is also a local metric set for H for i = 1, 2. Hence, we may assume that x2 (/. W. Then observe that W is a local metric set for G. That is,

every local metric set for H contains a local metric set for G. (6.5)

Therefore, lmd(H) 2:: lmd(G) and so lmd(H) = a.

+ We next show that lmd (H) = b. First suppose that W is a minimal local metric set for G. Then W is a local metric set for H by (6.4). Furthermore, W is a minimal local metric set for H by (6.5). Therefore, lmd+(H) 2:: lmd+(G). Finally, suppose that W is a minimal local metric set for H. Then at most one of x1 and x2 belongs to W and so we may assume that x2 (/. W. Therefore, W is a local metric set for G by (6.5). Moreover, W must be a minimal local metric set for G by + + + b, (6.4). Hence, lmd (H) '.S lmd (G) and so lmd (H) = which is the desired result. ■

137 By Theorems 6.3.2 and 6.4.4, if a 2 2 and 2a-l + a - l : b : 2a - 1, then there exists a positive integer N such that all triples (a, b, n) are realizable fr n 2 N.

As described in [23], if f is a parameter defned in terms of the vertices of a connected graph G of order n, then O < f(G) :Sn. There are numerous instances in the literature of two such parameters Ji and f being studied where O < Ji( G) : h (G) : n fr every graph G. A common problem concerns whether every two integers a and b with 0 < a : bare realizable as the values of Ji and fz, respectively, fr some graph. Normally, a considerably more challenging problem involves, fr a given integer n 2 2, determining those pairs a, b of integers with O < a : b : n for which there exists a graph G of order n such that fi(G) = a and f(G) = b. Often, only partial results of this nature exist. Of course, if fr some pair a, b of integers with O < a : b : n, say, there exists a graph G of order n such that fi(G) = a and f(G) = b, then O : a/n : b/n : 1. In this case, we say that the rational numbers a/n and b/n are realizable as the Ji-ratio and f-ratio, respectively, of some graph. This suggests a new, less restrictive problem when considering such pairs Ji, f of parameters. Thus, one can investigate problems of realizable ratios concering pairs of parameters such as local metric dimension and upper local metric dimension.

For a nontrivial connected graph G of order n, the local metric ratio rt(G) and upper local metric ratio r L ( G) of G are defned as

lmd(G) lmd+ (G) rt(G) = and rL(G) = . n n Thus rt(G) and rL(G) are rational numbers and O < rt(G) : rL(G) < 1. Certainly, there are infnitely many pairs s,t of rational numbers with O < s :St < l fr which

there exists a connected graph G with (rt(G), rL(G)) = (s, t). (6.6)

For example, fr every rational number s E (0, 1), the pair (s, s) satisfes (6.6) by Propo sition 6.4.1. Also, each pair (1/n, 2/n) satisfes (6.6) for every integer n 2 3. To see this, note frst that (3, 6, 10) is a realizable triple by the proof of Theorem 6.3.2. Then so is (3, 6, 3n) fr every integer n 2 4 by Theorem 6.4.4. That is, the result holds fr n 2 4. To verif that the pair (1/3, 2/3) also satisfes (6.6), obtain the graph G3,7 of order 11 with V (G3,7) = { u1, u2,u3} U { v1,v2, ... ,vs} as described in the proof of Theorem 6.3.2.

Let G be the graph obtained fom G3,7 by adding a new vertex Vg and joining v9 to Vi for 1 :i : 8. Hence, G is of order 12 and w(G) = 9. Let U = {u1,u2,u3} and V = {v1, v2, .. . , vg}. Since U U {vg} is a local metric set, lmd(G) = 4 by (6.2). Also,

138 V - {vi} is a minimal local metric set for 1 � i � 9, implying that lmd+ (G) 2: 8. As sume, to the contrary, that there exists a minimal local metric set W for G containing 9 vertices. Then IW n Uj � 2 and IW n VI � 7. Hence, renaming the vertices, if nec essary, we may assume that V ( G) - W = { u1, v1, v2}. Since co dew ( v1) i- co dew ( v2), we may further assume that v1u3 E E(G) and v2u3 (f. E(G). Now let W' = W - {u2} and observe that IN(ui) n W'I � degui = 4 while JN(vi) n W'I 2: 7 for i = 1, 2. This implies, however, that W' is a local metric set, which contradicts the fact that W is a minimal local metric set. Hence, lmd+ (G) = 8. Therefore, (re(G), rL(G)) = (1/3, 2/3). We conclude this chapter with the followingquestion.

Problem 6.4.5 For which pairs s, t of rational numbers with O < s � t < 1 does there exist a connected graph G such that re(G) = s and rL(G) = t?

139 Chapter 7

Topics for Further Study

The goal is to continue investigating open questions on closed modular colorings and local metric sets in graphs. Furthermore, we also plan to investigate some related topics.

1. Closed Sigma Colorings

Recall for a positive integer k and a connected graph G that we consider a vertex coloring c : V(G) - Zk where adjacent vertices may be asigned the same color.

Then c induces another vertex coloring c': V(G) - Zk where c'(v) = L c(u) uEN[v]

for each v E V( G). The coloring c is called a cl0sed modular k-coloring if c' (u) - c'(v) in Zk for all pairs u, v of adjacent vertices fr which N[u] - N[v] in G. The minimum k fr which G has a closed modular k-coloring is called the closed modular chromatic number of G and is denoted by mc(G). There is a variation of this concept where we use the set N of positive integers as colors rather than

Zk , for some k E N fr a coloring c and its induced coloring c'. This concept is closely related to the so-called sigma colorings in graphs, introduced in 2010 by Chartrand, Okamoto and Zhang [22]. For a nontrivial connected graph G, let c : V ( G) -+ N be a vertex coloring of G where adjacent vertices may be colored the same. If k colors are used by c, then c is a k-coloring of G. The color sum a( v) of v is the sum of the colors of the vertices in the neighborhood N(v) of v, that is, a(v) = L c(u). uEN(v)

140 Then the vertex coloring c induces another vertex coloring a : V(G) -+ N. If a(u) = a(v) fr every two adjacent vertices u and v of G, then c is neighbor distinguishing and is called a sigma coloring of G. The minimum number of colors required in a sigma coloring of a graph G is called the sigma chromatic number of G and is denoted by a(G). If G is a nontrivial connected graph with V(G) = { v1, v2, ..., vn}, then the coloring c of G defned by c(vi) = 2i-l fr 1 : i : n is a sigma coloring of G. Thus the sigma chromatic number a( G) exists fr every nontrivial connected graph G and 1 : a(G) : n fr every connected graph G of order n. Furthermore, a(Kn) = n fr every positive integer n and a(G) = 1 if and only if every two adjacent vertices of G have diferent degrees. It was shown in [22] that a sigma coloring may use fwer colors than a proper coloring of the graph.

For a nontrivial connected graph G, let c : V (G) -+ N be a vertex coloring of G where adjacent vertices may be colored the same. The closed color sum a( v) of a vertex v in G is defned as the sum of the colors of the vertices in the closed neighborhood N[v] of v, that is,

a(v) = L c(u). uEN[v]

Thus, the vertex coloring c induces another vertex coloring a : V(G) ➔ N. If a(u) = a(v) fr every two adjacent vertices u and v of G fr which N[u] = N[v] (that is, u and v are not true twins), then c is neighbor-distinguishing and is called a closed sigma coloring of G. That is, in a closed sigma coloring c of a graph, a(u) = a( v) if u and v are true twins, a(u) = a( v) if u and v are adjacent vertices that are not true twins, and no condition is placed on a(u) and a( v) otherwise. A closed sigma coloring c using k colors is called a closed sigma k-coloring. If a graph G ha a closed sigma k-coloring, then G is closed sigma k-colorble. The minimum number of colors required in a closed sigma coloring of a graph G is called the closed sigma chromatic number of G and is denoted by a(G). We plan to investigate closed sigma colorings of graphs.

141 2. Local Metric Sets with Prescribed Properties

For a nontrivial connected graph G, recall that an ordered set W = { w1, w2, ... , Wk} of k vertices of G is local metric set if code( u) - code( v) fr every pair u, v of ad jacent vertices of G. For a graphical property P, a local metric set W is a local P-metric set if the subgraph G[W] ha property P. For example, if Pis the property of being connected, then G[W] is a connected subgraph in G; if Pis the property of being independent, then G[W] is an empty graph; while if P is the property of being acyclic, then G[W] is a frest in G. In each case, we frst investigate the existence of such metric sets and then the minimum and maximum cardinality of local metric sets with prescribed properties.

142 Bibliography

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