Hamiltonicity and Connectivity in Distance-Colored Graphs

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Dissertations Graduate College

1-2011

Kyle C. Kolasinski Western Michigan University

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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]. Hamiltonicity and Connectivity in Distance-Colored Graphs

by Kyle C. Kolasinski

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

Western Michigan University Kalamazoo, Michigan June 2011 Hamiltonicity and Connectivity in Distance-Colored Graphs

Kyle Kolasinski, Ph.D.

Western Michigan University, 2011

For a connected graph G and a positive integer fc, the fcth power Gk of G is the graph with V(Gk) — V(G) where uv E E(Gk) if the distance dc{u^v) between u and v is at most fc. The edge coloring of Gk defined by assigning each edge uv of

Gk the color da(u,v) produces an edge-colored graph Gk called a distance-colored graph.

A distance-colored graph Gk is Hamiltonian-colored if Gk contains a properly colored Hamiltonian cycle. The minimum k for which Gk is Hamiltonian-colored is the Hamiltonian coloring exponent hce(G). It is shown that for each pair fc, d of integers with 4 < k < d, there exists a tree T with hce(T) = k and diam(T) = d.

Hamiltonian coloring exponents are determined for several well-known classes of graphs. It is also shown that for each integer k > 2, there exists a tree T/~ with hce(T/c) = k such that every properly colored Hamiltonian cycle in the fcth power of T/c must use all colors 1, 2,..., fc. For a grid G — Pn • Pm with n^m > 2, it is shown that G2 is Hamiltonian-colored if and only if nm = 0 (mod 4) and a complete solution is presented for restricted Hamiltonian-colored cycles in G3.

A distance-colored graph is properly p-connected if two distinct vertices u and v in the graph are connected by p internally disjoint properly colored u — v paths.

For a connected graph G and an integer fc > 2, the color-connectivity of Gk is the maximum positive integer p for which Gk is properly p-connected. The color- connectivities are determined for some well-known classes of graphs. It is shown that G2 is properly 2-connected for every 2-connected graph that is not complete,

a double star is the only tree T for which T2 is properly 2-connected and G3 is properly 2-connected for every connected graph G of diameter at least 3. All pairs

fc, n of positive integers for which Pk is properly fc-connected are determined and other related results are presented.

Various color-related distance parameters and rainbow concepts in distance-

colored graphs are introduced and studied. Results and open problems are also presented in this area of research.

UMI Number: 3467827

INFORMATION TO ALL USERS The quality of this reproduction is dependent upon the quality of the copy submitted.

In the unlikely event that the author did not send a complete manuscript and there are missing pages, these will be noted. Also, if material had to be removed, a note will indicate the deletion. UMT Dissertation Publishing

First and foremost, I would like to express my gratitude towards Dr. Ping

Zhang. Without her constant guidance and inspiration, I would have been lost. I truly appreciate the kind of person that she is, and feel honored to have her as my academic mother.

I would next like to thank the other members of my dissertation committee. I always received constructive feedback and support on my thesis and the members of my committee always helped me along the way. I was privileged to work under such brilliant minds who have really given so much to the world of mathematics.

I would specifically like to thank Dr. Gary Chartrand. He always went the extra mile for me and I am very grateful for his efforts.

I also would like to thank the Mathematics Department at Western Michigan

University. Whether I wanted to attend a conference, or needed summer support, they were always there for me. It was nice to know that I had the support and encouragement from many staff and faculty members who wished for my success.

Another person who deserves recognition is Dr. Ryan Jones. We went through all of graduate school together and having him as a friend always seemed to make things better.

Special thanks go to my family and friends, who had to deal with the constant rambling of my dissertation. Specifically, I would like to thank my parents, Steve and Laurie, my brothers, Steve and Jake, and my sisters, Karissa and Jessica for

ii Acknowledgments-Continued their continuous moral support. I could not have done this without your encour agement. I know it was a long and boring ride from Howell, but I appreciate how many times you drove it just to see me.

Lastly, I would like to thank my fiance, Jessica. She has been my rock through out my entire college experience, and I really could not have done this without her.

She was always the one who told me to keep going, and knowing that she believed in me made all the difference in the world. She really is a special person.

Kyle Kolasinski

in TABLE OF CONTENTS

ACKNOWLEDGMENTS ii

LIST OF FIGURES vi

CHAPTER

1. Introduction 1

1.1 History and Motivation 1

1.2 Definitions and Notation 8

2. Color-Hamiltonicity 12

2.1 Distance-Colored Graphs 12

2.2 Hamiltonian Coloring Exponents of Graphs 21

2.3 Graphs with Hamiltonian Coloring Exponent 2 26

2.4 Hamiltonian Coloring Exponents of Trees 29

2.5 Some Well-Known Classes of Trees 32

3. Colored Bridge Problems 49

3.1 The Konigsberg Bridge Problem 49

3.2 A Colored Bridges Problem 52

3.3 The Colored Bridges Problem for Grids 55

3.4 Restricted Colored Bridge Problems 59

iv Table of Contents-Continued

CHAPTER

4. Color Distance and Connectivity 82

4.1 Color-Connection Exponent and Color-Distance 82

4.2 Color-Eccentricity, Color-Radius and Color-Diameter 88

4.3 Color-Center and Color-Periphery 97

4.4 Color Connectivities of Graphs 103

4.5 On //-Colored and //-Chromatic Graphs 123

5. Rainbow Connected Graphs 128

5.1 Introduction 128

5.2 Rainbow Pancyclic Graphs 135

5.3 Rainbow Connectivity 142

6. Topics for Further Study 150

6.1 Questions on Color-Hamiltonicity 150

6.2 Questions on Color-Distance 152

6.3 Questions on Rainbow Subgraphs 153

6.4 Questions on Rainbow Connectivity 154

BIBLIOGRAPHY 156

v LIST OF FIGURES

1 1 A connected graph whose square is not Hamiltonian

2 1 A triangle in K7

2 2 A 3-edge coloring of K7

2 3 A properly colored Hamiltonian cycle m K7

2 4 Properly colored Hamiltonian cycles in C| and C%

2 5 A properly colored Hamiltonian cycle in P2

2 6 A connected grapg G of order 6 and A(G) = 4

such that G3 contains a properly colored Hamiltonian cycle

2 7 Properly colored Hamiltonian cycles in P^ for 4 < n < 6

2 8 Properly colored Hamiltonian cycles in P3 for 7 < r? < 16

2 9 A step in the proof of Theorem 2 13

3 2 10 Hamiltonian r2 — rn paths in P for 3 < r? < 8 2 11 Illustrating the graphs in the proof of Theorem 2 2 2 for k=A and k=5

2 12 Labeling the vertices of G = Spq

3 2 13 A colored Hamiltonian cycle in S(K\ 4)

2 14 Illustrating the proof of Proposition 2 5 3

2 15 Properly colored Hamiltonian cycles in T$ for 4 < d < 11

2 16 A step in the proof of Theorem 2 5 4

2 17 A step in the proof of Theorem 2 5 4

2 18 The graphs Gk for k E {4,5, 6} in the proof of Proposition 2

VI List of Figures-Continued

2.19 Illustrating the properly colored Hamiltonian cycles of Tk(4 < k < 7) in the proof of Theorem 2.5.8 46

3.1 A map of Konigsberg 50

3.2 a multigraph representing Konigsberg 51

3.3 A portion of a town 53

3.4 The graph representation of the town in Figure 3.3 55

3.5 A properly colored Hamiltonian cycle in the graph Gs of Figure 3.4(c) 55 3.6 Properly colored Hamiltonian cycle in

the cubes of PnDPm when n, m E {2, 3, 5, 6} 57 3.7 Properly colored Hamiltonian cycles in the distance-colored graph P3 for n E {4, 7,8, 9,10} 58

3.8 A properly colored Hamiltonian cycle in

3 the distance-colored graph (P7DP6) 59

3.9 A step in the proof of Theorem 3.4.1 61

3.10 Properly colored Hamiltonian cycles in

the squares of PnDP4 and PnOP$ for n E {3,9} 65 3.11 Properly colored Hamiltonian paths in

the squares of PnDP4 for n E {4,6, 8,10} 66 3.12 Properly colored Hamiltonian cycles in

the squares of PnDP6 for n E {4,10} 67

3.13 The graph P2DP7 68

3.14 The situation where C contains the edge x\x± 68

vn List of Figures-Continued

3.15 The situation where C contains the edges

#52/7, zij/3,2/12/3 and yby7 69

3.16 The graph P2DP5 70

3 3.17 Properly colored Hamiltonian cycles in (P2DPm) for m E {4, 5, 6,11}

containing v^mv2,m, vi,m-2Vi,m-i and ?;2,m_2?;2,m_i 70

3 3.18 Constructing a properly colored Hamiltonian cycle in (P2DPm+4) dotfill 71

3.19 The vertices of a graph P2DPm 71

3 3.20 Properly colored Hamiltonian cycles in (PnDPm) using only the colors 1 and 3 and containing r the edge Vn,m-2^n,m-\ f° four pairs (n, ra) 73

3 3.21 Properly colored Hamiltonian cycles in (P2DPm) for m = 4, 6 using only the colors 2 and 3 75

3.22 Constructing a properly colored Hamiltonian cycle 3 in (P2DPm+2) using only the colors 2 and 3 and

containing the edges ^1,1^2,2 and ^i,m+i^2,m-f2 77 3.23 A properly colored Hamiltonian cycle in 3 (P3DP4) using only the colors 2 and 3 and containing the edges ^1,1^2,2 and ^1,3^2,3 79 4.1 A color-eccentric vertex of u that is not an eccentric vertex of u 90

4.2 The color-eccentricities of Pg 93

4.3 A graph with a given color-center 98

4.4 A graph H with given center and color-center 101

4.5 A step in the proof of Theorem 4.4.4 108

vin List of Figures-Continued

4.6 Showing that P| is properly 3-connected 112

5.1 A bi-graceful labeling of C5 and a rainbow C5 in C\x 134

5.2 Illustrating the proof of Theorem 5.2.1 138

7 5.1 Rainbow 6-cycle and 5-cycle in P8 139

5.2 Rainbow cycles of length d — 1 in P^+1 for d — 11 and d = 15 140

5.3 Rainbow cycles of length d — 2 in P$+1 for d = 11 and d — 15 141

5.4 Illustrating the proof of Theorem 5.3.4 for k = 3 146

5.5 Three internally disjoint rainbow paths in P^ 149

6.1 Decomposing K7 into properly colored Hamiltonian cycles 151

IX Chapter 1 Introduction

1.1 History and Motivation

A fundamental problem in graph theory deals with whether a given graph G con tains a Hamiltonian cycle or a Hamiltonian path or possesses various other Hamil tonian properties such as G being Hamiltoman-connected, where every two vertices of G are connected by a Hamiltonian path

Over the years, a number of operations defined on a graph G have been studied

Among the best known of these are.

(1) The subdivision graph 5(G), where each edge of G is subdivided exactly once,

(2) The kth power Gk where V(Gk) = V(G) and uv E E(Gk) if the distance

dc{u,v) between u and v (the length of a shortest u — v path in G) is at

most k The graph G2 is called the square of G and G3 is the cube of G,

(3) The line graph L(G) of G whose vertices can be put in one-to-one correspon

dence with the edges of G where two vertices of L(G) are adjacent if the

1 2

corresponding edges of G are adjacent;

(4) The total graph T[G) of G whose vertices can be put in one-to-one corre

spondence with the vertices and the edges of G where two vertices of T[G)

are adjacent if the corresponding elements of G are adjacent or incident.

When each of the operations (2)-(4) is applied to a Hamiltonian graph G, a Hamil tonian graph results; while S{G) is Hamiltonian if and only if G = Cn for some n>3.

In [6] Chartrand and Kronk showed that every path in a Hamiltonian graph G of order n > 3 with a given but arbitrary vertex v can be extended to a Hamiltonian cycle with initial vertex v of G if and only if G = Gn, G = Kn or G = Kn/2jTl/2 where n is even.

The terms Hamiltonian path, Hamiltonian cycle and Hamiltonian graph are of course named for the famous mathematician Sir William Rowan Hamilton who considered spanning cycles on the dodecahedron in connection with the Icosian

Game he developed in 1856. Months earlier, however, the Reverend Thomas Kirk- man had considered spanning cycles on a variety of polyhedra. Hamilton not only observed that the dodecahedron contains a spanning cycle but in the instructions accompanying his Icosian Game, Hamilton wrote that every path of order 5 could be extended to a spanning cycle on the dodecahedron, that is, for every path

P = (?;l5 v2, i?3, ?;4, ?;5) in the graph of the dodecahedron, there is a spanning cycle

G = (i>i, ?;2, v3, ?;4, v5, uu u2,..., 7ii5, uie = vi) in this graph. Hamilton's observation led Chartrand to define a related concept. Let G be 3 a Hamiltonian graph of order n and let k be an integer with 3 < k < n. Then

G is k-ordered Hamiltonian if for every sequence s : v\, v2,..., i;fc of k vertices of G, there is a Hamiltonian cycle of G encountering these k vertices of s in the order listed. While every Hamiltonian graph is 3-ordered Hamiltonian, not every

Hamiltonian graph is 4-ordered Hamiltonian. Chartrand defined a Hamiltonian- connected graph G to be k-ordered Hamiltonian-connected (fc > 3) if for every sequence s : Vi,v2)..., Vk of fc vertices of G there exists a Hamiltonian v\ — v^ path encountering the vertices of s in the given order.

Milan Sekanina (1931-1987) was a mathematician from Czechoslovakia whose research interests comprised many fields of mathematics including algebra, topol ogy and graph theory. Much of Sekanina's research concerned ordered sets. In 1960

Sekanina [27] proved that for every two distinct vertices u and v in a connected graph G of order n, there is an ordering u = ?;i, ?;2,..., vn — v of the vertices of G such that dc{vu ^i+i) < 3 for i = 1, 2,..., n — 1. Using current terminology, he showed that the cube of every connected graph of order at least 3 is Hamiltonian- connected and therefore Hamiltonian as well. The square G2 is not necessarily

Hamiltonian however. For example, for the graph G of Figure 1.1, its square G2

is not Hamiltonian. In 1968 Karaganis [21] independently proved that the cube of every connected graph is Hamiltonian-connected. The square, cube and powers of

a graph were evidently terms coined by Ian Ross and Frank Harary [26] in 1960.

Three years later, while attending the 1963 Graph Theory Symposium in

Smolenice (now in Slovakia), Sekanina [28] suggested the research problems of

determining the class of graphs G such that G2 is Hamiltonian and the class of 4

Figure 1.1: A connected graph whose square is not Hamiltonian graphs G such that G2 is Hamiltonian-connected. The smallest graph G of order

2 2 at least 3 such that G is not Hamiltonian is S(Kl;3), that is, (S{Kh3)) = T(Klj3) is not Hamiltonian (see Figure 1.1). In 1964, Sekanina's student Frantisek Neu- man [23] determined those trees T and pairs ?i, v of vertices of T for which the graph T2 has a Hamiltonian u — v path. The following theorem is a consequence of Neuman's work but was later proved more directly and more simply by Frank

Harary and Allen Schwenk [18].

Theorem 1.1.1 The square T2 of a tree T of order at least 3 is Hamiltonian if

and only ifT is a caterpillar.

The graph G of Figure 1.1 contains cut-vertices, that is, vertices v for which G — v is disconnected. A connected graph without cut-vertices is a 2-connected graph.

At the 1966 Graph Theory Conference in Tihany, Hungary, Crispin Nash-Williams

[22] asked whether the square of every 2-connected graph is Hamiltonian. Michael

D. Plummer had independently made the same conjecture. Herbert Fleischner [11]

and Hansjoachim Walther [29] independently showed, using a theorem of Petersen

(guaranteeing the existence of a 1-factor), that if G is a cubic bridgeless graph, 5 then G2 is Hamiltonian

During the period September 1970 - January 1971, Herbert Fleischner worked on this conjecture and was finally successful m proving it [12]

Theorem 1.1.2 (Fleischner's Theorem) The square of every 2-connected graph

is Hamiltonian

Fleischner's proof was contained in four papers, all of which were submitted to the Journal of Combinatorial Theory According to Fleischner [14], the referee

"gave hell to the author" and the four papers never appeared in their original form but in two revised papers that were much clearer and much shorter than the original (thanks to the referee and, later, to Nash-Williams) These two papers were finally published in 1974 [13]

Prior to the publication of Fleischner's two papers, his theorem had become widely known The technique and the theorem itself were useful in proving that the square of a 2-connected graph had even stronger properties In 1973 Hobbs

[20] proved that the square of every 2-connected graph is in fact vertex-pancychc

In 1974 Chartrand, Hobbs, Jung, Kapoor and Nash-Williams [5] showed that the square of every 2-connected graph is Hamiltonian-connected In 1976 Faudree and

Schelp [10] proved that the square of every 2-connected graph is panconnected

More recently, shorter proofs of Fleischner's theorem have been given - in 1991 by Riha [25] and in 2009 by Georgakopoulos [16] In fact, Georgakopoulos showed that for every 2-connected graph G and every vertex v of G, the square G2 has a

Hamiltonian cycle whose two edges incident with v belong to G 6

In 1964 Chartrand [3] showed (see also [4]) that if G is a connected graph that is not a path, then for sufficiently large integers r, the iterated line graph Lr(G) is Hamiltonian, where L2{G) = L(L(G)) and for r > 3, Lr(G) = L(Lr~l(G)).

In 1973 Chartrand and Wall [8] defined the Hamiltonian index h(G) of G as the minimum r for which Lr(G) is Hamiltonian. They showed that if 8(G) > 3, then h(G) < 2. In 1965 Harary and Nash-Williams [17] characterized those graphs having a Hamiltonian line graph. A dominating circuit in a connected graph G is a circuit C in G in which every edge of G is incident with a vertex of C.

Theorem 1.1.3 (Harary and Nash-Williams) Let G be a connected graph. Then

L(G) is Hamiltonian if and only if G = K\^n for some n > 3 or G contains a dominating circuit.

The first paper on fc-ordered Hamiltonian graphs was authored by Ng and

Schultz [24] in 1997. They gave sufficient conditions for a graph to be fc-ordered

Hamiltonian. Several other papers dealing with this concept have been written since then. The following results also appear in the paper of Ng and Schultz.

(1) Every (fc + l)-ordered Hamiltonian-connected graph, fc > 3, is fc-ordered

Hamiltonian.

(2) For each integer fc > 4, there is a fc-ordered Hamiltonian-connected graph

that is not fc-ordered Hamiltonian.

In 2008, Chebikin [9] showed that for each integer fc > 4 and every connected graph G of order at least fc, the graph GL3/c/2J-2 is fc-ordered Hamiltonian. Chebikin 7 also showed that for n > 5, the graph G3 is 5-ordered Hamiltonian He also showed that for r > 3 and r? sufficiently large, there exists a sequence s of 2r vertices in Cn

r such that no cycle in C n contains the vertices of s encountered in the given order

In 2009 Hartke and Ponto [19] showed for a graph G that if Lr(G) is fc-ordered

Hamiltonian for a sufficiently large integer r, then LrJtl(G) is (fc + l)-ordered

Hamiltonian They also showed that if G is a connected graph that is not a path, cycle or K\j, then there exists an integer N such that LN+^k~6\G) is fc-ordered

Hamiltonian for fc > 3 Hartke and Ponto also showed that for fc = 4 and fc = 5 that if G is a fc-ordered Hamiltonian graph, then so is L(G) They asked if this is true for fc > 6

In the past few decades, graph labehngs have been growing in popularity be cause of their applications to many other areas of research in graph theory The origin of the study of graph labehngs as a major area of graph theory can be traced to a 1967 research paper by Alexander Rosa [30] Among the labehngs he intro duced was a vertex labeling he referred to as a /?-valuation Let G be a graph of order n and size m A one-to-one function / V(G) —» {0,1, 2, ,ra} is called a

(5-valuation (or a ^-labeling) of G if {\f(u) - f(v)\ uv E E(G)} = {1, 2, , m}

In order for a graph to possess a /3-labelmg, it is necessary that m > n — 1 Five years later (in 1972), Golomb [?] referred to a /3-labehng as a graceful labeling and a graph possessing a graceful labeling as a graceful graph Eventually, it was this terminology that became standard Galhan [15] has written a survey on la behngs of graphs that includes an extensive discussion of graceful labehngs and their applications One of the major conjectures in graph theory concerns graceful 8 graphs and is due to Kotzig (see Rosa [30]) which is known as the Graceful Tree

Conjecture.

The Graceful Tree Conjecture Every tree is graceful.

1.2 Definitions and Notation

In this section, we introduce some basic definitions and notation in graph theory.

We refer to the book [7] for graph theory notation and terminology not described in this work. We assume that all graphs under consideration are connected.

• In a connected graph G, the distance dc(u, v) (or simply d(u, v) if the graph

G under consideration is clear) between two vertices u and v in G is the

length of a shortest u — v path in G. Au — v geodesic is a u — v path of length

d(u, v). The eccentricity e(v) of a vertex v in a connected graph G is

e(v) = msx{d(v,x) : x E V(G)}.

The radius of a connected graph G is

rad(G) = min{e» : v E V(G)},

while the diameter of G is

diam(G) = max{e» : v E V(G)}.

Thus, the diameter of G is the greatest distance between any two vertices of

G. The center C(G) of a connected graph G is the subgraph of G induced 9

by those vertices of G having eccentricity rad(G), while the periphery P(G)

of G is the subgraph of G induced by the vertices of G having eccentricity

diam(G).

• A proper vertex coloring (or simply a coloring) of the vertices of a graph G

is an assignment of colors (the elements of some set) to the vertices of G,

one color to each vertex, such that adjacent vertices are assigned distinct

colors The minimum number of colors in a proper vertex coloring of G is

the chromatic number x(G) °f G

• A coloring of the edges of a graph G is an assignment of colors (the elements

of some set) to the edges of G, one color to each edge A graph with a

coloring of its edges is called an edge-colored graph If adjacent edges are

assigned distinct colors, then the coloring is a proper edge coloring The

minimum number of colors in a proper edge coloring of G is the chromatic

index x'(G) of G

• Recall that a cycle containing every vertex of a graph G (necessarily exactly

once) is called a Hamiltonian cycle of G A graph G is Hamiltonian if G

contains a Hamiltonian cycle A path containing every vertex of a graph

G is called a Hamiltonian path of G A graph G is traceable if G contains

a Hamiltonian path A graph G is Hamiltoman-connected if for every pair

u, v of vertices of G, there is a Hamiltonian u — v path in G Necessarily,

every Hamiltonian-connected graph of order 3 or more is Hamiltonian (but

the converse is not true) 10

• A circuit G of G (a cyclic walk which never repeats an edge) that contains

every edge of G (necessarily exactly once) is an Eulenan circuit. A connected

graph G is called Eulenan if G contains an Eulerian circuit.

• A graph G of order n > 3 is called pancyclic if G contains a cycle of every

possible length, that is, G contains a cycle of length ( for each ( with 3 <

( < n. Thus if G is pancyclic, then G is Hamiltonian. A connected graph G

of order n is said to be panconnected if for each pair fi, v of distinct vertices

of G, there exists a u — v path of length ( for each integer f satisfying

d(u,v) < ( < n — 1. If a graph is panconnected, then it is Hamiltonian-

connected.

• The connectivity K(G) of a graph G is the minimum number of vertices whose

removal from G results in a disconnected or trivial graph. If K(G) > t for

some positive integer £, then G is ^-connected. By a well-known theorem of

Whitney [31], a graph G is ^-connected if and only if G contains t internally

disjoint u — v paths for every two distinct vertices u and v of G. That is, G

contains u — v paths Pi, P2,..., Pt such that V(Pt) Pi V(P3) — {u, v} (and

so E(P%) n E(P3) = 0) for all distinct integers i and j with 1 < ?,, j < t. In particular, if K(G) = I, then G contains ( internally disjoint u — v paths

for every pair u, v of vertices of G, but G does not contain ( + 1 internally

disjoint x — y paths for some pair x, y of vertices of G.

• A graph G of size m is graceful if it is possible to label the vertices of G with

distinct elements from the set {0,1,..., m} in such a way that the induced 11

edge labeling, which prescribes the integer |? — j\ to the edge joining vertices

labeled ? and j, assigns the distinct labels from the set {1,2, ,ra} to the

edges of G A labeling that accomplish this is a graceful labeling (This was

called a /3-valuation by Rosa [30] )

• A graph G of size m is bi-graceful if the vertices of G can be labeled with

distinct elements from the set {0,1, ,2m} in such a way that the set S

of induced edge labels that prescribes the integer |? — j\ to the edge joining

vertices labeled ? and j has the property that for each set St = {7, 2m +1 — 1}

with 1 < 1 < m, \S H St\ = 1 A labeling with this property is called a bi graceful labeling That is, a bi-graceful labeling of a graph G is a one-to one

function / V(G) -> {0,1, 2, , 2m} such that the set S = {\f(u) - f(v)\

uv E E(G)} has the property that \S H Sz\ = 1 for all ? with 1 < 1 < m (This was called a p-valuation by Rosa [30] ) A graph possessing a bi graceful

labeling is a bi-graceful graph If G is a graceful graph and / is a graceful

labeling of G Then SnSl = {1} for all 1 (1 < 1

2.1 Distance-Colored Graphs

We have mentioned that Sekanina proved that if G is a connected graph of order at least 3, then G3 is Hamiltonian. Consequently, if G is any connected graph of order n > 3, then there is a cyclic ordering

s : vu v2,..., vn_u vn, ?;n+i = vx

of the vertices of G such that 1 < dc(vz, v%+i) < 3 for every integer i with 1 < i < n.

Thus for each consecutive pair vu v1+i of vertices in s, the distance dc(vZ) ?;2+i) is 1,

2 or 3. This of course says that for each integer i with 1 < i < n, vzvl+i is an edge

3 3 of G . This also suggests the idea of assigning to the edge v%vl+i of G the distance dc(vz, I>H-I), which is either 1, 2 or 3. We can interpret such an assignment of the integers 1, 2 and 3 to the edges of G3 as a 3-edge coloring of G3, which may not be a proper edge coloring of G3.

Such an assignment of the integers 1, 2 and 3 to the edges of G3 is not entirely new. For example, consider the triangle constructed on three of the seven vertices

12 13 labeled 0,1, ,6 shown in Figure 2 1, where the seven vertices are those of a regular 7-gon and the edges of the triangle are assigned the integers 1, 2 and 3

6o \^pl

4 3

Figure 2 1 A triangle in K7

Of course, with the vertices of the triangle (having order r? = 3 and size m — 3) in

Figure 2 1 labeled with three of the four labels 0,1, 2, 3 = r? and each edge incident with vertices labeled ? and j is itself labeled |? — j|, producing the edge labels

1, 2, 3 = ??7,, we have obtained the well-known fact that K$ is graceful

If this triangle is rotated clockwise through an angle of 27r/7 radians a total six times, where the color of each rotated edge is maintained, then we obtain the

3-edge coloring of K7 in Figure 2 2 This illustrates another well-known fact for a graceful graph G of size m, namely that there is a decomposition of K2m+i into

2m + 1 copies of G In fact, there is a rainbow /^-decomposition of K7 (that is, all three edges of K$ are colored differently)

The 3-edge coloring of K7 shown in Figure 2 2 can be obtained in another man

3 ner Suppose that G — C7 By Sekanma's theorem, G = K7 contains a Hamilto nian cycle Of course, this is a rather trivial observation as we are beginning with a cycle Also, Kn is Hamiltonian for every integer r? > 3 (In fact, C7 itself is a 14

Figure 2.2: A 3-edge coloring of K7

Hamiltonian cycle of G3.) For distinct vertices u and v in G, assign uv the color dc(u,v). Thus every edge of K7 is colored with one of the colors 1, 2 and 3. This edge coloring of K7 suggests the problem of determining whether K7 contains a properly colored Hamiltonian cycle. It does, as Figure 2.3 shows. This of course is equivalent to the statement that there is a cyclic sequence s : vX) v2,..., v7, v$ = v\ of the seven vertices of G = C7 such that dG(vt) vl+i) ^ dG(vl+i,vul+2) for every integer i with 1 < i < 6.

Figure 2.3: A properly colored Hamiltonian cycle in K7

3 More generally, suppose that G = Gn, where n > 3, and we consider G . An 15

3 edge uv of G is colored ? E {1,2,3} if dc(u,v) = ? Since diam(Gn) = |_?^/2j ? it

3 3 follows that G = /Cn if 77 < 7 The question here then is whether G contains a properly colored Hamiltonian cycle If G = Cs, then every edge of K^ is colored 1, while if G = G5, then every edge of K§ is colored 1 or 2 Since ^(C^) — yf(C$) — 3, there is no properly colored Hamiltonian cycle in G3 for either n = 3 or 77 = 5

There is a properly colored Hamiltonian cycle in both Gj and C%, however, as is shown in Figure 2 4

\. 1 o 2 C 3

2 S\

Figure 2 4 Properly colored Hamiltonian cycles in C\ and C\

Since the diameter of G4 is 2, not only is G| = /Q but C\ — K4 Consequently, every edge of K4 is colored 1 or 2 and there is a properly colored Hamiltonian cycle in /Q, using the colors 1 and 2 Of course, it is only possible for C\ to have a properly colored Hamiltonian cycle if 77 is even While such a Hamiltonian cycle exists in C\, no such Hamiltonian cycle exists in C2 In fact, we have the following result

Theorem 2.1.1 Let G = Cn where n > 4 and n is even Assign the edge uv m G2 the color 1 E {1,2} if dc(u,v) = 1 where 1 < 1 < 2 Then G2 has a properly colored Hamiltonian cycle if and only ifn = 0 (mod 4) 16

2 Proof. Let Cn = (i>i, v2,..., vn) v{) and suppose that G has a properly colored Hamiltonian cycle G. We may assume that the initial (and terminal) vertex of

G is v\ and the initial edge incident with V\ is colored 2 and that this edge is v\v>$. The next two vertices following V\v<$ thus have even subscripts. Hence the vertices of C can be described in terms of a sequence of 2-element subscripts, namely E\, E2,..., Ek, E\ where E% consists of two odd integers if i is odd and E% consists of two even integers if i is even. Hence fc is even, say fc = 2( and each set

Ex (1 < i < 21) is a 2-element set; so n = 2(2f) = 0 (mod 4). If 77, = 0 (mod 4), say n = 4f, then the cyclic sequence

1, 3, 2, 4, 5, 7, 6, 8,..., n - 3, n - 1, n - 2, n, 1 describes a properly colored Hamiltonian cycle in G2. •

Before proceeding further, we introduce a more general concept. Let G be a connected graph of order n with diameter d and let fc be an integer with 1 < fc < d.

An edge uv of Gh is colored i (1 < i < fc) if dc(u,v) — i. This edge coloring of

Gh produces what is called a distance-colored graph Gk. Suppose that Gk contains a subgraph isomorphic to a graph //. We say that Gk is H-colored if Gk contains a properly colored copy of //. Certainly, this is only possible if x!(H) < fc. If

k // = Gn, then G is called a Hamiltonian-colored graph. By Theorem 2.1.1 the distance-colored graph G2 is Hamiltonian-colored if and only if n = 0 (mod 4).

As an illustration, let P be the Petersen graph containing two disjoint 5-cycles

(uu u2, ?i3, ?i4, ^5, ^i) and (vu ?;3, vh, v2, v4) vx) 17

2 such that uxv% E E(P) for 1 < ? < 5 Then P contains a properly colored Hamiltonian cycle as shown in Figure 2 5 and so P2 is Hamiltonian-colored

«i

^4 Ws

Figure 2 5 A properly colored Hamiltonian cycle in P2

If G is a connected graph of order at least 3 such that G contains a vertex u that is adjacent to every vertex of G, then d(u,v) = 1 for all v E V(G) — {u}) which implies that u does not belong to any properly colored cycle in Gk for every integer fc > 2 Thus we have the following observation

Observation 2.1.2 If G is a connected graph of order n > 3 with maximum degree A(G) = 77 — 1, then the distance-colored graph Gk is not Hamiltonian-

k colored for any integer fc > 2 In particular, if G — Kn or G = Kijn-i, then G is not Hamiltonian-colored for any integer fc > 2

Figure 2 6 shows a connected graph G of order r? > 3 with A(G) =77—2 such that G3 contains the properly colored Hamiltonian cycle

(vu v2) 7;3, ?;4, 7;5, vb, vx) 18 and so G3 is Hamiltonian colored For this graph G, its square G2 is not Hamiltonian colored, however, for such a properly colored Hamiltonian cycle G would have to contain the path

(vuvb,v^v2),

say, implying that v2V\ is an edge of G, which is impossible

V6 I

Figure 2 6 A connected graph G of order 6 and A(G) = 4 such that G3 contains a properly colored Hamiltonian cycle

We now consider the case where G is the path Pn of order 77 A properly colored

3 Hamiltonian cycle is shown in each of distance colored graphs P$, P^ and P6 in Figure 2 7, where only the edges of the cycle are shown While the Hamiltonian

cycle in P4 is properly colored, this Hamiltonian cycle is also properly colored in

P2 There is no integer 77 > 3 with 77 / 4 such that P2 has a properly colored

Hamiltonian cycle The situation for P* is different, however

n — 4 n = 5 n = b

Figure 2 7 Properly colored Hamiltonian cycles in P3 for 4 < 77 < 6

Theorem 2.1.3 For each integer n > A, the distance colored graph P3 is Hamiltonian

colored 19

Proof. We have already seen that P3 is Hamiltonian-colored for 77 = 4,5,6.

Thus, we now assume that 77 > 7. Let Pn = (vi,v2,..., vn). We show by induction on 77 > 7 that the distance-colored graph P3 has a properly colored Hamiltonian cycle G containing an edge e incident with vn that is colored 1 such that the two edges of G adjacent to e are both colored 2. That the statement is true when

7 < 77 < 16 is shown in Figure 2.8.

Figure 2.8: Properly colored Hamiltonian cycles in P3 for 7 < 77 < 16

Assume that the statement is true for every integer 77 with 7 < 77 < fc, where

fc > 16, and consider the distance-colored graph P^+1. By the induction hypothesis, there exists a properly colored Hamiltonian cycle G' in the distance-colored graph 20

P^_9 such that an edge e of G' incident with vk_9 is colored 1 and the two edges of C' adjacent to e are colored 2. Thus e = Vk-ioVk-9-

3 Let P be the properly colored Hamiltonian path in P 0 shown in Figure 2.9 where

P = (Vfc-7, Vfc_8, Vfc_5, . . . , 7;fc_6).

Delete the edge e = ?;&_ 10^-9 from C' and add the edges Vk-gVk-t an-d ^fc—lo^fc—7-

This produces a properly colored Hamiltonian cycle in the distance-colored graph

P^+1 in which the edge / = v^+iVk is colored 1 and the two edges of G adjacent to / are colored 2. •

Figure 2.9: A step in the proof of Theorem 2.1.3

For 3 < 77 < 5, G2 = G3 and so C± is Hamiltonian-colored and G3 and C% are not Hamiltonian-colored. For each 77 > 6, the properly colored Hamiltonian cycle of P^ described in the proof of Theorem 2.1.3 is also a properly colored

Hamiltonian cycle of G3.

Theorem 2.1.4 For each integer n > 6, the distance-colored graph G3 is Hamiltonian-

colored. 21

2.2 Hamiltonian Coloring Exponents of Graphs

We've seen that for some connected graphs G, there is no positive integer fc such that the distance-colored graph Gk is Hamiltonian-colored For example, for every graph G of odd order and diameter 2, the distance-colored graph Gk is not Hamiltonian-colored for any fc On the other hand, if for a given connected graph G, there is a positive integer fc for which the distance-colored graph Gk is

Hamiltonian-colored, then it is natural to ask for the smallest integer fc for which

Gk is Hamiltonian-colored This gives rise to the following concept Let G be a

connected graph for which Gk is Hamiltonian-colored for some positive integer fc

The Hamiltonian coloring exponent hce(G) of G is the minimum fc for which Gk

is Hamiltonian-colored Thus if hce(G) = fc, then Gk is Hamiltonian-colored but

Qk-i 1S noj. Hamiltonian-colored In this section, we show that for every integer fc > 2, there exists a connected graph G such that hce(G) = fc Of course, we have

already seen that there are graphs with Hamiltonian coloring exponents 2 and 3,

namely G4 and G5, respectively We begin with the following lemma

Lemma 2.2.1 For each integer n > 3; let Pn — (7*1, r2, , xn) be a path of order 71 Then the distance-colored graph P3 contains a properly colored Hamiltonian

?2 — Tn path m which the edge incident with x2 is colored 1 and the edge incident

with xn is colored 1 or 2

Proof. We show that the distance-colored graph P3 has a properly colored

Hamiltonian x2 — xn path Q such that (1) the edge incident with r2 is colored 1 22

and (2) if 77 > 3 is odd, then the edge incident with xn in Q is colored 2 and Q ends with the subpath (.xn_i, .xn_2,.xn), while if 77 > 4 is even, then the edge inci dent with xn in Q is colored 1 and Q ends with the subpath (.xn_2, .xn_3, xn_l5 xn).

Figure 2.10 shows a properly colored Hamiltonian x2 — xn path having the desired properties in each distance-colored graph P3 for 3 < 77 < 8, where only the edges of the path are shown.

3 Figure 2.10: Hamiltonian x2 — xn paths in P for 3 < 77 < 8

The properly colored Hamiltonian x2 — xn paths Q of P^ shown in Figure 2.10 for 3 < 77 < 8 illustrate a general pattern. For 77 > 9, we consider the path Q in P* that begins with (.x2, X\, x4, .x3, xb, .x5,..., .xr_i, .xr_2), where r — 2 is the largest odd integer that is less than 77 — 1. Thus r — 2 = 77 — 2 if 77 is odd, while r — 2 — n — 3 if 77 is even. The path Q is then completed as follows:

Q = (.x2,.Xi,.X4,.X3,.x6,.x5,...,.xn_i,.Tn_2,.xn) if 77 is odd,

Q = (.x2, .xi, ,x4, .x3, .xb, .x5,..., .xn_2, xn_3, .xn_i, xn) if r? is even.

In both cases, Q is a properly colored Hamiltonian .x2 — xn path whose initial edge is colored 1 and whose final edge is (i) colored 2 if 77 is odd and (ii) colored 1 if 77 23 is even •

We are now prepared to present the following result

Theorem 2.2.2 For every integer fc > 2, there ensts a connected graph G such that

hce(G) = fc

Proof. Since C2 is Hamiltonian-colored if and only if r? = 0 (mod 4) by Theo

3 rem 2 11, hce(Gn) = 2 if 77 = 0 (mod 4) Also, G is Hamiltonian colored for all

77 > 6 by Theorem 2 14 and so hce(Gn) = 3 if r? > 6 and 77 ^ 0 (mod 4) Thus we may assume that fc > 4

For fc > 4, let F and H be two copies of Kk, where

V(F) = {u,uuU2, ,uk 1} and V(H) = {w,wuw2, ,wk 1},

and let P = (7;0,^i,^2, ,^fc-3,^/c 2) be a path of order fc — 1 Let Gk be the graph obtained from P, H and P where u and ?;0 are identified and w and vk 2 are identified Figure 2 11 shows the graphs Gk for fc = 4 and fc = 5 Observe that diam(Gfc) = fc

Let G = Gk We first show that Gk is Hamiltonian colored By Lemma 2 2 1, for the path P = (u,Vi,V2, ,7;^_3,7i;) of order fc — 1 > 3, the distance colored graph P3 contains a properly colored Hamiltonian V\ —w path Q in which the edge incident with ?;x is colored 1 and the edge incident with w is colored 2 if fc — 1 is odd and colored 1 if fc — 1 is even 24

Figure 2.11: Illustrating the graphs in the proof of Theorem 2.2.2 for fc = 4 and fc = 5

Next, we construct a properly colored v\ — w path Q* of Gk such that V(Q) U

V(Q*) = V(G) and V(Q)nV(Q*) = {vuw}.

If fc — 1 is odd, then let

Q* = (Vu Wu uuu2, w2) ws,uS)u4, w4, wb,..., wk_2) wk-i, uk_u w).

The color of ViWi is fc — 2 > 2 and the color of uk_iw is fc — 1 > 3. For example, if fc = 4, then the path

Q* = (vuwuuuu2, w2, ^3, u3) w)

4 for the distance-colored graph G of the graph G — G4 is shown in Figure 2.11(a).

If fc — 1 is even, then let

Q* = (Vu Ul, wu w2) u2) ?i3, w3, w4,..., ufc_3, ?i/c_2,wfc_2, wk_uuk-i,w). 25

The color of V1U1 is 2 and the color of uk-\w is fc — 1 > 3. For example, if fc = 5, then the path

Q* = (Vu uuwuw2, u2) ?i3,w3,7i;4, u4, w) for the distance-colored graph G5 of the graph G = G$ is shown in Figure 2.11(b).

The union of the v\ — w paths Q* and Q in Gfc produces a properly colored

Hamiltonian cycle in Gk. Therefore, Gk is Hamiltonian-colored.

It remains to show that Gk~l is not Hamiltonian-colored. Assume, to the contrary, that Gk~l contains a properly colored Hamiltonian cycle G. Let

U = V(F) - {u} = {uuu2,..., uk_i] and

W = V(H)-{w} = {w1,w2,...,wk.1}.

As we proceed cyclically about G in some direction, we encounter at most two consecutive vertices belonging to U or to W. Let X\, X2,..., Xp be the maximal subsets of consecutive vertices in U or in W that are encountered on G in the given order. Then 1 < \Xt\ < 2 for 1 < i < p. Since \U\ = \W\ = fc - 1, it follows that

\l%=1Xt\ = \UUW\=2k-2.

Thusp > fc —1. Since (1) X% cannot be immediately followed by X%+i \i X%\JXl+i C

U or Xz U Xl+i C W from the maximality of these sets and (2) X% cannot be immediately followed by X%+\ if one of Xx and Xz+i is a subset of U and the other is a subset of W as no vertex in U is adjacent to a vertex of W in G^-1, it follows that a vertex of P must lie between Xz and X2+i for each i (1 < i < p) where 26

Xp+i = X\. Since the order of P is fc — 1 and p > fc — 1, it follows that p = fc — 1 and that \XZ\ — 2 for i — 1, 2,... ,p = fc — 1. This implies that exactly one vertex of P lies between Xx and Xz+i for i = 1, 2,..., fc — 1. This, however, implies that the vertex u (as well as w) lies between X3 and X3+i for some j (1 < j < k — 1). If either X3 C JJ or A^+i C [/, then, since |J^| = |^+i| = 2, two consecutive edges on G are colored 1, a contradiction. Otherwise, both X3 and X3+i are subsets of W, implying that two consecutive edges on C are colored fc — 1, again a contradiction.

Therefore, Gfc_1 is not Hamiltonian-colored. •

2.3 Graphs with Hamiltonian Coloring Exponent 2

There is a concept that is closely related to powers of a graph. For a connected graph G of diameter d and each integer i with 1 < % < d, the i-step graph G^ of G is the graph whose vertex set is V(G) and two vertices u and v of G are adjacent in G^ if dc(u, v) = i. In particular, G^ = G. Thus if i and fc are integers with 1 < i < k < d, then the i-step graph G^ is a spanning subgraph of the distanced-colored graph Gk and each edge of G^ is colored i. Furthermore, the set

{P(GW), P(GPl),..., P(GW)} of color classes of Gk is a partition of E(Gk).

If G is a connected graph of order n > 3 such that G2 is Hamiltonian-colored

(so hce(G) = 2), then every properly colored Hamiltonian cycle of G2 gives rise to two 1-factors Pi and F2 such that P2 is a 1-factor of G^ for ?, = 1,2. The converse of this statement is not true, however. For example, if G = P4/~ for some integer 27

fc > 2, then each of G and G® has a 1-factor but hce(G) = 3 by Theorem 2 1 3

The following two observations related to a connected graph G with hce(G) = 2 will be useful to us

Observation 2.3.1 If G is a connected graph of order n > 3 such that hce(G) =

2, then n is even

Observation 2.3.2 If G is a connected graph of order at least 3 such that hce(G) —

2, then G cannot contain two vertices u and w such that there is a unique vertex v for which d(u, 7;) = d(w, 7;) = ? for 1 = 1 or 1 = 2

As an immediate consequence of Observation 2 3 2, if G is a star of order at

least 3 or a double star (a tree of diameter 3) of order at lease 5, then G2 is not

Hamiltonian-colored

By Observation 2 3 1, if G is a connected graph of odd order having diameter 2, then Gk is not Hamiltonian-colored for each integer fc > 2 A well known class of graphs of diameter 2 are the complete multipartite graphs Let Kni n2 Uk be the

complete fc-partite graph of 77 > 5, where r?i < r?2 < < nk and 77 = X]2=1772 Then it is known that

n Kni n2 nk is Hamiltonian if and only if nk < Yli=i i (2 1)

By (2 1), if 77i, 772, , nk are fc > 2 positive even integers, then

Kni n2 nk is Hamiltonian if and only if Kn± n^ ^ is Hamiltonian (2 2) 28

Theorem 2.3.3 Let G — KnuU2^ iUk be the complete k-partite graph of order

T1 2 Yln=i i ^ 5, where r?i < 772 < ••• < nk The distance-colored graph G ?s

Hamiltonian-colored if and only ifnl is even for 1 < ? < fc and G is Hamiltonian

Proof. First, suppose that the distance-colored graph G2 is Hamiltonian-colored

Then G2 has a properly colored Hamiltonian cycle G whose edges are alternately

2 colored 1 and 2 We saw that G produces two 1-factors Pi and F2 in G such

2 that P2 is a 1-factor of the ?-step graph G^ of G for ? = 1,2 Since G' ' =

Kni U Kn2 U • • U Knk, it follows that each Kn% (1 < i < fc) is 1-factorable and so

772 is even Next, we show that G is Hamiltonian Contracting each edge colored 2 in the properly colored Hamiltonian cycle G in G2 produces a Hamiltonian cycle in the complete fc-partite graph H = K?± 22 HL Thus H is Hamiltonian It then 2 ' 2 ' '2 follows by (2 2) that G is Hamiltonian

For the converse, suppose that 772 is even for 1 < 1 < fc and G is Hamiltonian

[2] [2] Then G = Kni U KU2 U • • • U KUk contains 1-factors Let P be a 1-factor of G We now construct the complete fc-partite graph H — K?± 22. ^ by identifying every pair 7i, v of vertices of G if uv is an edge in F and labeling the resulting vertex by u Since G is Hamiltonian, it follows by (2 2) that H is Hamiltonian

Let G' be a Hamiltonian cycle of H Replacing each vertex u of C' by the vertices u and v and the edge 7i7; in F produces a properly colored Hamiltonian cycle in

G2 Therefore, G2 is Hamiltonian-colored •

Corollary 2.3.4 Let G = Kr^s be a complete bipartite graph of order r + s > 3 The distance-colored graph G2 is Hamiltonian-colored if and only if r — s — 2t for 29 some positive integer t

Let G be a connected graph of order 77 and diameter 2 Then hce(G) exists if and only if 77 is even and hce(G) = 2 Note that if the diameter of G is 2, then

G'2l = G Thus, if G is a connected graph of diameter 2 and hce(G) = 2, then each of G and G contains 1-factors

2.4 Hamiltonian Coloring Exponents of Trees

In this section, we investigate Hamiltonian coloring exponents of trees It is known that for each integer fc > 2, there exists a connected graph whose Hamiltonian coloring exponent is fc (see Theorem 2 2 2) For each integer fc > 4, the graph G constructed in the proof of Theorem 2 2 2 has clique number fc and diam(G) = hce(G) We now show that not only does there exist a tree with Hamiltonian coloring exponent fc but the tree can have an arbitrarily large diameter We begin with a lemma, which is a consequence of the proof of Theorem 2 13

Lemma 2.4.1 For each integer n > 4, let Pn be a path of order 77 Then the distance-colored graph P^ has a properly colored Hamiltonian cycle C in which an edge e incident with one of the end vertices of Pn is colored 1 such that the two edges of C adjacent to e are both colored 2

Theorem 2.4.2 For each pair fc, d of integers with 4 < fc < d, there ensts a tree

T with hce(T) = fc and diam(T) = d

Proof. Let F and H be two copies of the star K\ k-\ where 30

V(F) = {u, uuu2,..., 7ifc_i}, V(H) = {w, wuw2,..., Wfc_i}, u is the central vertex of F and w is the central vertex of H. Next, let P =(vo,

7;i, 7;2, ..., 7;fc_3, 7^-2) be a path of order fc — 1. Construct a tree Vk from P, // and P by (i) identifying u and 7;0 and labeling the identified vertex by v0 and (ii) identifying w and 7;^_2 and labeling the identified vertex by vk_2. If d = fc, let

Tj^fc = T£, while if d > fc + 1, then Tkjd is obtained from Tk and the path (2:1, z2) ..., 2^-fc) of order d — fc by joining z\ to the vertex wk_i in T^. Observe that diam(Tfc>d) = d.

k Let T = Tkjd- We first show that T is Hamiltonian-colored. By Lemma 2.2.1, for the path P = (?;0, t>i, ?;2,..., 7^-3,7;^_2) of order fc — 1 > 3, the distance-colored

3 graph P contains a properly colored Hamiltonian vi — vk_2 path Q in which the edge incident with v\ is colored 1 and the edge incident with vk^2 is colored 1 or

fc 2. Next, we construct a properly colored v\ — vk_2 path Q* of T such that

(i) V(Q) UV(Q*) = K(T) and 7(Q)n7(Q*) = {vuvk_2} and

(ii) the edge incident with v\ in Q* is colored a where a > 2 and the edge incident

with vk-2 in Q* is colored 6 where b > 3.

k We begin with a v\ — vk_2 path in T according to whether fc is even or fc is odd. If fc is even, then let

Q0 = (Vl, Wi, 771, U2, W2, 7i;3, 7i3, ^4> W4, W5, . . . , Wfc_2, Wfc_i, 7i/e_i, 7;/c_2). (2.3)

If fc is odd, then let

Qi = (7;i,771,7i;i,7i;2,u2,773, ws, w4,..., wfc_3,ufc_2,7i;fc_2, wk_x,uk_X) vk_2). (2.4) 31

In (2.3) and (2.4), the color of v\W\ is fc — 2 > 2 and the color of uk_ivk_2 is fc — 1 > 3. In (2.4), the color of viU\ is 2 and the color of uk_ivk_2 is fc — 1 > 3. We now consider four cases, according to the values of d and fc.

Case 1. d = fc. If fc is even, then let Q* — Q0 as described in (2.3); while if fc is odd, then let Q* = Qi as described in (2.4).

Case 2. d = fc + 1. If fc is even, then let Q* be the v\ — vk^2 path obtained from the path QQ in (2.3) by replacing the edge wk_2wk_i in Qo by the path

(wk_2, z\, 7i;/c_i); while if fc is odd, then let Q* be the v\ — vk^2 path obtained from the path Q\ in (2.4) by replacing the edge wk_2u)k_i in Qi by the path

(Wfc-2,2l,Wfc-l).

Case 3. d = fc + 2. If fc is even, then let Q* be the v\ — vk_2 path obtained from the path QQ in (2.3) by replacing the edge wk_2wk_i in Q0 by the path

(wk_2) Zi,z2, wk-i)\ while if fc is odd, then let Q* be the v\ — vk_2 path obtained from the path Qi in (2.4) by replacing the edge ivk^2wk_i in Qi by the path

Case 4. d > fc + 3. In this case, the order of the path P^-^+i =(wk_i, z\, z2)

..., Zd-k) is d — k + 1 > 4. By Lemma 2.4.1, the distance-colored graph Pl_k+l has a properly colored Hamiltonian cycle G in which (z2,wk_i, Zi) z3) is a subpath of

G. Let P* = C — wk-\Z\ be the zi~wk_i subpath of G. Then P* is a Hamiltonian z\ — wk_i path of PjL/,+1 whose initial and terminal edges (the edges incident with z\ and itffc-i) are both colored 2. If fc is even, then let Q* be the vi — vk_2 path 32

m obtained from the path Q0 (2.3) by replacing the edge wk_2wk__i in Q0 by the path (wk-2) P*); while if fc is odd, then let Q* be the v\ — vk_2 path obtained from the path Qi in (2.4) by replacing the edge iuk_2wk_i in Qi by the path (wk_2) P*).

Since d(wk_2, Zi) — 3 and d(wk_i,uk_i) = k > 4, the path Q* is properly colored.

In each case, the Vi~vk_2 path Q* has the prescribed properties. Thus the union

k of the 7;i — vk_2 paths Q* and Q in T produces a properly colored Hamiltonian cycle in Tk. Therefore, Tk is Hamiltonian-colored.

It remains to show that Tk~l is not Hamiltonian-colored. Assume, to the contrary, that Tk~l contains a properly colored Hamiltonian cycle G. Let U =

u {7ii, 77-2, • • •, k-i} and let V — {7;0, t?i,..., 7;/c_2}. Since at most two vertices from U can be consecutive vertices in G, it follows that in G, each vertex in U is adjacent to a vertex not in U. Furthermore, NTk-i(u) = U U V(P) for each u E U and |V(P)| = |t/| = fc — 1. Thus, we may assume, without loss of generality, that

7v;z_i G P(G) for 1 < i < fc — 1. Since dT(u2, v\) = 2, it follows that u2u% £ E(C) for all i with 1 < i < fc — 1 and i =^ 2. This implies that 7i27;; G P(G) for some j with 2 < j < k — 2. However then, either (7i,2, V3, u3) or (u3,v3,u2) is a subpath of G, each of whose two edges is colored 2, which is impossible. Therefore, Tk~~l is

not Hamiltonian-colored, as claimed. •

2.5 Some Well-Known Classes of Trees

We continue our study of Hamiltonian-colored powers of trees by determining the

Hamiltonian coloring exponents of trees belonging to some much-studied classes. 33

We saw in Observation 2.1.2 that if T is a star (a tree of diameter 2), then Tk is

Hamiltonian for no integer fc > 2. In this section, we determine the Hamiltonian coloring exponents of double stars and the so-called cubic caterpillars.

2.5.1 Double Stars

If T is a double star, then T has exactly two non-end-vertices called the central vertices of T. For integers p, q > 2, let SPiq be the double star having central vertices A and B such that deg A = p and deg B — q. Then the order of SVA is 77 = p + q > 4. We saw by Observation 2.3.2 that the square of a double star of order n > 4 is Hamiltonian-colored if and only if n = 4. Next, we determine all double stars whose cube is Hamiltonian-colored. In order to do this, we present two additional definitions. For a Hamiltonian cycle

G = (vuv2,...,vn,vi) in a graph G, the distance sequence of G is defined as

sc = (do^v2),dG(v2,vz),... .do^^i.v^.dG^Vx)).

Thus if C is a properly colored Hamiltonian cycle, then every two consecutive terms in sc are distinct.

Theorem 2.5.1 Let G be the double star SPiq where p, q > 2. Then the distance- colored graph G3 is Hamiltonian-colored if and only if

|p-g|

Proof. Let G = SPiq whose central vertices are A and P, where deg A = p and degP = q and n = p + q is the order of G. Let VA = N(A) — {B} and VB = N(B) — {A}. We denote any vertex in VA by a and any vertex in VB by

6 (see Figure 2.12). For 77,7; E V(G), it follows that d(u,v) = 3 if and only if

{77,7;} = {a, 6} and d(u, v) < 2 if {77,7;} n {A, B} ^ 0.

Figure 2.12: Labeling the vertices of G = 5^

Suppose first that G3 contains a properly colored Hamiltonian cycle

G =(vu v2) ..., 7;n, 7;i).

We consider three cases, according to the distance dc(A,B) of A and B on the cycle G.

Case 1. dc(A B) = 1. We may assume that A = 7^ and B = ?;i+i, where 1 < 7 < n — 1. Then the sequence 6, A, P, a appears clockwise on G, the sequence

2,1, 2 appears clockwise on sc and each of other terms in sc is either 2 or 3. This implies that

G = (6, A, P, a, 6, 6, a, a, fc, fc,a , a,..., fc, fc,a , a, fc) and so the distance sequence of C is sc = (2,1, 2, 3,2, 3,..., 2, 3). Since at least one subsequence fc, fc,a , a must appear in G, it follows that |V^l = |V/?| > 3 and so

P = q > 4. Therefore, \p - q\ < 1 and {p, g} / {2, 3}, {3, 3}. 35

Case 2. dc(A,B) = 2. We may assume that A = Vi and B — vi+2, where 1 < i < n — 2. Then either the sequence a, A, 6, P, a or the sequence fc, A, a, P, fc appears clockwise on G. Assume, without loss of generality, that a, A) fc, P, a ap pears clockwise on C. Then the sequence 1,2,1,2 appears clockwise on sc and each of the remaining terms in sc is either 2 or 3. Thus either

G = (a, A, fc, P, a, fc, fc,a , a, fc, fc,a , a,..., fc, fc,a , a) or

G = (a, A, fc, P, a, fc, fc,a , a, fc, fc, a, a,..., fc, fc, a, a, fc, fc, a), where at least one subsequence fc, fc,a , a must appear in G. Hence either

sc = (l,2,l,2,3,2,...,3,2)orsc = (l,2,l,2,3,2,...,3,2,3).

Thus either \VA\ = |Vb| + 1 > 3 or \VA\ = \VB\ > 3, that is, either p = q + 1 > 4 or p = q > 4. Therefore, \p — q\ < 1 and {p, q} ^ {2, 3}, {3, 3}.

Case 3. dc(A,B) > 3. We may assume that A = Vi and P = v3, where

1 < i < J < n, and that a, A) b appears clockwise on G. Thus either fc, P, a or a,B,b appears clockwise on C and exactly two terms in SQ are 1. We consider these two subcases.

Subcase 3.1. The sequence b,B,a appears clockwise on C. Let

G = (a, A,fc, Si,fc,P,a,s2,a),

where «i and s2 are subsequences of G. The length of each of Si and s2 at least 2.

In fact, si is either 36

a, a, fc, fc,a , a, fc, fc,..., a, a, fc, fc,a , a, fco r a, a, fc, fc,a , a, fc, fc,..., a, a, fc, fc,a , a,

where it is possible that s\ is a, a, fco r a, a. Similarly, s2 is either

fc, fc,a , a, b, fc, a, a, ..., fc, fc, a, a, fc, fc, a or fc, fc,a , a, fc, fc,a , a, fc, fc, ..., a, a, fc, fc,

where it is possible that s2 is fc, 6, a or fc, fc. In each case then, \VA\ > 4, |Vz?| > 4, and HVAI - |V^|| < 1. Therefore, \p - q\ < 1 and {p, q} ^ {2, 3}, {3, 3}.

Subcase 3.2. The sequence a,B,b appears clockwise on C. Let

C = (a,A,b,s1)a,B,b,s2,a),

where sx and s2 are subsequences of G. Then

Si : a, a, fc, fc,a , a, fc, fc,..., a, a, fc, fc, where the subsequence a, a, fc, fc must appear at least once in Si. Thus the length of Si is 4£ for some integer £ >1. On the other hand, s2 is one of the following:

(1) a,

(2) b,

(3) 6, a,

(4) 6, a, a, 6, 6, a, a, 6, /;,..., a, a, 6,6, a,

(5) b, a, a, b, b, a, a, 6,6,..., a, a, 6,6 or

(6) a, a, 6,6, a, a, b, b,..., a, a, 6,6. 37

In each of (4), (5) and (6), the subsequence a, a,fc, fcmus t appear at least once. If

(3), (4) or (6) occurs, then p = q\ while if (1), (2) or (5) occurs, then \p — q\ — 1.

In each case, \VA\ > 4, \VD\ > 4. Therefore, \p - q\ < 1 and {p, q} / {2, 3}, {3,3}.

For the converse, assume that \p — q\ < 1 and {p, q} ^ {2,3}, {3, 3}. lip — q, then

G = (fc, A, P, a, fc, fc,a , a, fc, fc, a, a,..., fc, fc, a, a, fc) is a properly colored Hamiltonian cycle in G3; while if p = q + 1, then

G = (a, A, fc, P, a, fc, fc,a , a, fc, fc,a , a,..., fc, fc,a , a) is a properly colored Hamiltonian cycle in G3. Thus G3 is Hamiltonian-colored. •

Since a double star has diameter 3, the following is an immediate consequence of Theorem 2.5.1.

Corollary 2.5.2 Let G be the double star SPjq where p, q > 2. Then hce(G) exists if and only if

|p-g|

The subdivision graph S(F) of a connected graph F is obtained from F by subdividing each edge of P exactly once. By Observation 2.3.2, if F is a star or a double star, then the square of S(F) is not Hamiltonian-colored. On the other hand, the cube of S(F) is, in fact, Hamiltonian-colored. 38

Proposition 2.5.3 If F is a star of order at least 3 or a double star of order at

least 4 and G = S(F), then the distance-colored graph G3 ?9 Hamiltonian-colored

Proof. First, let P = K\it for some integer / > 2 with V(F) ={u, v\, v2, , vt}, where deg 77 = t Suppose that G = S(F) is obtained from F by subdividing the edges of F with t new vertices w\, w2, , wt such that wz is adjacent to u and v% for 1 < ? < t Then G3 contains the properly colored Hamiltonian cycle

G =(vu w2l v2, 7i;3, 7;3, , wt-u Vt-u wu u, vx), where the resulting cyclic distance sequence of G is sc = (3,1,3,1, ,3,1,2)

Figure 2 13 shows such a properly colored Hamiltonian cycle G for / = 4, where only edges on the cycle G are drawn

Figure 2 13 A colored Hamiltonian cycle in S^Z^i^)3

Next, let F = SPjq for some integers p, q > 2 We may assume that p > q If p = q = 2, then the result follows by Theorem 2 13 Thus, we may assume that p > 3 Suppose that the central vertices of SPjq are 77 and v such that u is

adjacent to 7;, i^, 772, , 77p_i and v is adjacent to 77, ?;x, 7;2, , vq_i Let G be the subdivision graph of P obtained by subdividing the edges of F with new vertices w, Ti, r2, , rp_i and T/I,T/2, . , J/^-i such that (1) 7i; is adjacent to 7i and 7;, (2) 39

rt is adjacent to u and ux for 1 < i < p — 1 and (3) y3 is adjacent to v and v3 for 1 < j < ^ — 1 If g = 2, then G3 contains the properly colored Hamiltonian cycle

C = (771,777, 7/1, V\, 7;, Tp_l, 7ip_l, Tp_2, 7ip_2 , , T2, 712, Ti, 14, Hi),

(see Figure 2 14(a) for p = 3), while if q > 3, then G3 contains the properly colored

Hamiltonian cycle

C = (77,1,77;, 7/1,7;i, 7/2,7;2, , yq_u vq_u 7;, rp_i, 77p_x,

Tp-2,Up-2, ,T2,772,Ti,77,7ii), as desired •

Figure 2 14 Illustrating the proof of Proposition 2 5 3

2.5.2 Cubic Caterpillars

A tree T is called r7/fc?r if the degree of each non-end-vertex of T is 3 A caterpillar is a tree of order 3 or more, the removal of whose end-vertices produces a path called 40 the spine of the caterpillar For each integer d > 2, let T^ be the cubic caterpillar of diameter d Thus T2 is the star Ki^ and T3 is the double star Ss,s We have seen that Tk is not Hamiltonian-colored for each k > 2 by Observation 2 12, while Tk

is Hamiltonian-colored when k > 3 by Theorem 2 5 1 Thus, we may assume that

d > 4 By Observation 2 3 2, for each integer d > 4, the distance-colored graph

Tj is not Hamiltonian-colored On the other hand, Tj is Hamiltonian-colored for each integer d > 4, as we show next Figure 2 15 shows a properly colored

Hamiltonian cycle in Tj for each integer d with 4 < d < 11 and so these graphs

are Hamiltonian-colored

Theorem 2.5.4 For each integer d > 4, the distance-colored graph Tj is Hamiltonian-

colored

Proof. Let P = (?;i, 7;2, ., 7Vfi) be a geodesic of length d in 7^, where the vertex vx is adjacent to the end-vertex 77z for 2 < 7 < d We have seen (in Figure 2 15)

3 3 that T4 and T5 are Hamiltonian-colored We now show by induction on d > 6 that the distance-colored graph Tj has a properly colored Hamiltonian cycle G such that C contains one of the two paths shown in Figure 2 16 That the statement is true when 6 < d < 11 is verified m Figure 2 15

Assume that the statement is true for every integer d' with 6 < df < d, where

d > 11, and consider the distance-colored graph T^+1 By the induction hypothesis, there exists a properly colored Hamiltonian cycle C' in the distance-colored graph

T^_4 such that C' contains one of the two paths as shown in Figure 2 16 in which the subscript i of each vertex (77z or v%) is replaced by ? — 4 We consider two cases 41

d = 4

d = b

d = 7

d = 8

d = 9 rKX

d=10

d = ll • * v12

Figure 2.15: Properly colored Hamiltonian cycles in Tj for 4 < d < 11

Case 1. The cycle C' contains the path as shown in Figure 2.16(a). Let

P = (Vd-4, Vd-i,Vd+i,Vd+2, Ud+UUd, Vd, Ud_!,Ud_2, Vd_U 77^_3, 7^_3)

be the properly colored path in T'd+1. (See Figure 2.17(a).) Replace the edge vd-4^d-3 from C' by the path P. This produces a properly colored Hamiltonian cycle G in the distance-colored graph T^+1 such that G contains the path in Fig ure 2.16(a) in which the subscript % of each vertex (ut or v%) is replaced by 7 + 1. 42

u Ud-2 Ud_i Ud Ud-2 d-l Ud

l Vd-2 Vd-l^/Vd Vd+l Vd_^^^ Vd_2 Vd-l^^Vd yd+1

(a) (b)

Figure 2.16: A step in the proof of Theorem 2.5.4

V>d-4 ^d-i Ud_2 Ud_i Ud Ud+l

Vd-4

Vd-4 Vd+2

Figure 2.17: A step in the proof of Theorem 2.5.4

Case 2. The cycle C' contains the path as shown in Figure 2.16(b). Let

Q = {vd-4, Vd_2, Vd+U Vd+2, Ud+U Ud, Vd, Ud_u Vd_u Ud_2) 77^_3, 7^_3

be the properly colored path in Tj+i. (See Figure 2.17(b).) Replace the edge Vd-4Vd-3 from C1 by the path Q. This produces a properly colored Hamiltonian cycle G in the distance-colored graph Tj+1 such that G contains the path in Fig ure 2.16(6), in which the subscript i of each vertex (ut or v%) is replaced by i + 1. • 43

The following is an immediate consequence of Observation 2 3 2 and Theo rem 2 5 4

Corollary 2.5.5 For each integer d>3, hce(Tj) = 3

If G is a connected graph with hce(G) = /r, then necessarily every properly colored Hamiltonian cycle in Gk must use the color k On the other hand, it is possible that Gk contains a properly colored Hamiltonian cycle that does not use

all k colors 1, 2, , k

Theorem 2.5.6 For each integer k > 3, there ensts a connected graph Gk with

hce(Gfc) = k and a properly colored Hamiltonian cycle in the kth power of Gk

whose edges are colored with fewer than k colors

Proof. For k = 3, let Gs = G6 = (7;i,7;2, ,i?b = vi) We have seen that

hce(G3) = 3 and the properly colored Hamiltonian cycle (vi, v2, 7;5, vb, 7;3, 7;4, Vi) in the cube of G3 uses two colors, namely 1 and 3 For each integer k > 4,

let Gk be the tree T^k constructed in Case 1 when diam(Gfc) = k in the proof

of Theorem 2 4 2 and so hce(Gfc) = k To simplify notation, let G — Gk For

k E {4,5,6}, Figure 2.18 shows a properly colored Hamiltonian cycle C in each

graph Gk whose edges are colored with less than k colors In Figure 2 18, the edges

of G are drawn as dashed lines, the edges of G are drawn as solid lines and the

remaining edges in Gk are not drawn (Notice that the color 3 is not used in any

of these cases and that the cycle in Gb uses only four colors ) 44

Figure 2.18: The graphs Gk for k E {4, 5, 6} in the proof of Proposition 2.5.6

For k > 7, let C be the properly colored Hamiltonian cycle in Gk described in the proof of Theorem 2.4.2; that is, C consists of the Vi — vk-2 paths Q and Q*, where the path Q is described in Lemma 2.2.1 and the path Q* is described in

(2.3) or (2.4) of Theorem 2.4.2 according to the parity of k. The path Q uses the colors 1, 2, 3 and the path Q* uses the colors 1, 2, k and at most two other colors

(namely, k — 2 and k — 1). Observe that for k > 5, the color \k/2] is never used and for k > 8, none of the colors in the set {5, 6,..., k — 3} are used. Thus, the cycle C uses at most six different colors for each integer k > 7. •

Theorem 2.5.6 suggests another concept. For a connected graph G with hce(G) =

/c, let n(G) denote the minimum number of colors used in a properly colored 45

Hamiltonian cycle of Gk Thus, if hce(G) = 2, then fi(G) = hce(G) = 2, while if hce(G) = 3 and the order of G is odd, then fi(G) = hce(G) = 3 Since fi(G) < hce(G) for every connected graph G, it follows that » 4, the graph Gk constructed in the proof of Theorem 2 5 6 is a tree Thus the following corollary is an immediate consequence of the proof of

Theorem 2 5 6

Corollary 2.5.7 There exists a sequence {7/c}£L4 of trees such that

fc->oo hce(Tfc)

Next we show for each integer k > 2, that there exists a tree Tk such that

/i(Tfc) = hce(Tfc) and so the upper bound in (2 5) holds

Theorem 2.5.8 For each integer k > 2, there exists a tree Tk with hce(T^) = k such that every properly colored Hamiltonian cycle in the kth power ofTk must use all of the colors 1,2, , k

Proof. For k = 2, let Tk = T2 = P4, while for k = 3, let Tk = Ts = P5 For each fixed integer k > 4, let P = (ui,u2, ^uk) be a path of order k and let H be a copy of star K\ k-i with V(H) = {v,v\,v2, ,7;/c_i}, where v is the central vertex of H The tree Tk is constructed from P and H by identifying u\ and v and labeling the identified vertex by u^ We claim that hce(Tk) = /r Let T = Tk We first show that Tk is Hamiltonian-colored For each even integer k > 4, let 46

; ; u C = (u\,Vi, 722, 713, 7;2, 7;3, 714, 725, ^4,^5, > ^fc-2, ^fc-1> ? fc-2, ? fc-l, k, ^l)

Figures 2 19(a), (b) show the cycle G for /r = 4 and k = 6, respectively, where only the edges on C are drawn Since 5, let

C = (721, 7;i, 7;2, 724, 712, 723, ^3, ^4, 725, 726, 7;5, 7J6, , 72/c_2, 72fc-l, Vh-2, Vk-\ , 72fc, 72l)

Figures 2 19(c), (d) show the cycle G for k = 5 and /r = 7, respectively, where only the edges on G are drawn Since dT(ui,vi) = 1, dT(vi,v2) = 2, dr^s^vs) = 3, dT(u4,i)2) = 4 and ^(1^,7^+1) = 7 + 1 for 4 < ? < /r — 1, it follows that G is a properly colored Hamiltonian cycle in Tk

(a) fc = 4 (b) fc = 6 Vb w> ^4 Q- VA

U*Ov Vi ^2 Q- V2

ViO- ^P Vi U\ U4 UK, 'U(> urs >w2

Figure 2 19 Illustrating the properly colored Hamiltonian cycles of Tk (4 < k < 7) in the proof of Theorem 2 5 8 47

Next, we show that Tk l is not Hamiltonian-colored Assume, to the contrary, that Tfc_1 contains a properly colored Hamiltonian cycle G Let

U = {ui,u2, , uk} and let V = {vuv2, , vk-i]

Since at most two vertices from V can be consecutive vertices in G, it follows that each vertex in V is adjacent to a vertex in U in G Furthermore, no two vertices in

V can be adjacent to a common vertex in U and no vertex in V can be adjacent to

k 1 the vertex uk E U in T ~~ Thus, we may assume, without loss of generality, that

7;27i2 E E(C) for 1 < i < k — 1 In particular, 7;27i2 E E(C) Since dT(v2,u2) — 2, it follows that 7;27;7 ^ E(C) for all 1 < j < k — 1 and j / 2 This implies that

7;27i£ E P(G) for some f with 1 < j < k — 1 and f ^ 2 However then, either

(v£, 7i£, v2) or (v2, U£, vg) is a subpath of G, each of whose edges is colored ^, which is impossible Therefore, Tk~l is not Hamiltonian-colored

Therefore, hce(T) = Zr, as claimed It remains to show that every properly colored Hamiltonian cycle in Tk must use all colors 1,2, , /r, that is, fi(T) = k

Assume, to the contrary, that there is a properly colored Hamiltonian cycle G in

Tk such that C avoids the color j for some j with 1 < j < k — 1 If j =2, then no two vertices in V can be consecutive vertices in G Thus for each integer ? with 1 < ? < k — 1, the vertex 7;z must be adjacent to two vertices in U Since k > 4, \V\ = k — 1 and |[/| = A*, it follows that some vertex of U is adjacent to two vertices of V in G, which is a contradiction Thus G must use the color 2 and so j y£ 2 This implies that no vertex in V can be adjacent to the vertex u3 E C/ Furthermore, recall that in G, each vertex in V is adjacent to a vertex in [/, no two 48 vertices in V can be adjacent to a common vertex in U and no vertex in V can be adjacent to the vertex Uk E U. Since \V\ — fc — 1 and \U — {UJ, Uk}\ — fc — 2, this is impossible. Therefore, as claimed, every properly colored Hamiltonian cycle in

Th must use all colors 1, 2,..., fc and so fi(T) = fc. •

The following is an immediate consequence of the proof of Theorem 2.5.8.

Corollary 2.5.9 There exists a sequence {Pfc}£L2 of trees such that

hm -—\_\ = 1. fc-^oo hce(Tk) Chapter 3 Colored Bridge Problems

3.1 The Konigsberg Bridge Problem

The city of Konigsberg was founded in 1255 when it was the capital of German

East Prussia and home of the Prussian Royal Castle. The River Pregel flowed through Konigsberg, separating it into four land regions. Seven bridges were built over the river. A map of Konigsberg, showing the four land regions (labeled A, B,

C, D) and the location of the river and the seven bridges (labeled a, b, c, d, e, f, g), is shown in Figure 3.1.

Konigsberg played an interesting role in the origin of graph theory. The story goes that during the 1730s the citizens of Konigsberg enjoyed strolling about the city. Some citizens wondered whether it was possible to go for a walk in the city and pass over each bridge exactly once. This problem eventually became known as the Konigsberg Bridge Problem.

Undoubtedly, the greatest mathematician of the 18th century was Leonhard

49 50

C d g

Figure 3.1: A map of Konigsberg

Euler from Switzerland. Euler was known to correspond with many people, math ematicians and non-mathematicians alike. One of the people with whom Euler corresponded was Carl Leonhard Gottlieb Ehler, who was the mayor of the city of

Danzig in Prussia. Danzig is located approximately 80 miles west of Konigsberg.

Ehler was aware of the Konigsberg Bridge Problem, which had become known to many people in the area. It is believed that it was Ehler who wrote to Euler men tioning this problem to him. In fact, in a letter dated March 9, 1736, Ehler asked

Euler if he would send him a solution of the problem.

Initially, Euler did not think that this problem was particularly mathematical in nature but when he discovered a method for solving not only the Konigsberg

Bridge Problem but a generalization of the problem, he wrote a paper on this subject, describing the problem and its generalization and presenting his proof. In particular, Euler showed, mathematically, that it was not possible to walk about

Konigsberg and pass over each bridge exactly once.

Euler's proof, indeed Euler's paper, neither mentioned graphs nor even hinted at 51 graphs. Nevertheless, Euler's reasoning was graph theoretic in nature. In fact, the term "graph" as used in this context was evidently not introduced until 1878 when the British mathematician James Joseph Sylvester first used this word. That the map of Konigsberg can be represented by a graph (actually a multigraph since some pairs of vertices are joined by more than one edge) is apparent (see Figure 3.2).

Figure 3.2: A multigraph representing Konigsberg

In 1736 Euler proved the following result, which we state in the language of graph theory. Suppose that there is a city consisting of land regions, some pairs of which are joined by one or more bridges. Then this town can be represented by a graph or multigraph G whose vertices are the land regions where every two vertices of G are joined by a number of edges equal to the number of bridges joining the corresponding land regions. Then there is a walk in the town passing over each bridge exactly once if and only if G is connected and every vertex of G has even degree. 52

3.2 A Colored Bridges Problem

There is another problem concerning traveling about a town with bridges that also has a graph theory representation and was inspired by a problem in geographical analysis and transportation networks and the Konigsberg Bridge Problem.

The Colored Bridges Problem A town has a network of walkways running through it, each of which is colored gray. A red bridge is constructed between every two walkway intersections that are two blocks apart and a blue bridge is constructed between every two intersections that are three blocks apart. Is it possible to take a round trip about the town that passes through every intersection in the town exactly once and such that during this trip the walkway or bridge used to enter every intersection is of a different color than that used to exit it?

As one would probably expect, the answer to the question posed in the Colored

Bridges Problem depends on the structure of the network of walkways in the town being considered. For example, the business area of a town might consist of a collection of rectangular blocks with two parallel walkways running east and west and three parallel walkways running north and south. Figure 3.3(a) shows such a situation, where A, B, C, D, E, F are the six intersections in this case. Figure 3.3(b) shows the six red bridges that are constructed between each pair of intersections that are two blocks apart and Figure 3.3(c) shows the two blue bridges that are constructed between each pair of intersections that are three blocks apart. At the end, the bridges shown in Figure 3.3(b) and in Figure 3.3(c) will all be present in 53 the town. J

(b)

(c)

Figure 3.3: A portion of a town 54

This problem can be stated as a problem in graph theory. Consider a network

N of walkways in a town. This can be represented by a connected graph G whose vertices are the intersections in N and where two vertices u and v of G are joined by an edge if u and v correspond to two intersections connecting a walkway passing through no intermediate intersection. Color each edge of G gray. If the distance between two vertices u and v of G is 2, then add an edge between u and v colored red. If the distance between u and v is 3, then add an edge between u and v colored blue. Hence the resulting graph is G3, where each edge of G3 is colored gray, red or blue. For this town, the Colored Bridges Problem has a solution if and only if the distance-colored graph G3 has a properly colored Hamiltonian cycle.

Perhaps the most common structure for a network of walkways in a town is when all blocks in town are rectangular, having n walkways running north and south, say, and m walkways running east and west, that is, the walkways in the town form a grid. In the particular case of the town shown in Figure 3.3(a), n — 3 and m = 2. The graph G representing the network of walkways in this town is shown in Figure 3.4(a), while Figure 3.4(b) shows the graph G2 representing the town with the six red bridges in Figure 3.3(b). Figure 3.4(c) shows the graph G3 representing the town with both the red and blue bridges. Here, g, r and b indicate the colors gray, red and blue, respectively.

The Colored Bridges Problem has a solution for the town in Figure 3.3 since there is a properly colored Hamiltonian cycle in the graph of Figure 3.4(c). Such a cycle is shown in Figure 3.5. 55

A B C O- -O- -o

E (a)

^3 .

Figure 3.4: The graph representations of the town in Figure 3.3

Figure 3.5: A properly colored Hamiltonian cycle in the graph G3 of Figure 3.4(c)

3.3 The Colored Bridges Problem for Grids

The Colored Bridges Problem has a connection with distance-colored graphs. We have already seen that the graph G3 in Figure 3.4(c) is Hamiltonian-colored and 56 so the Colored Bridges Problem has a solution for the town in Figure 3 3(a) The graph G in Figure 3 4(a) is often denoted by Ps • P2 and called the Cartesian product of the paths P3 and P2 We now show that the Colored Bridges Problem always has a solution whenever the walkways in the town form a grid and so the graph representation of the town is Pn • Pm for some integers 77, ra > 2

3 Before verifying that the distance-colored graph (Pn • Pm) is Hamiltonian- colored for all integers 77, ra > 2, we first observe (m Figure 3 6) that this is the case when 77, ra E {2, 3, 5, 6}, where as in Figure 3 6, a bold line represents a gray edge, a solid line represents a red edge and a dashed line represents a blue edge

(We have already observed this in Figure 3 5 when 77 = 3 and ra = 2 )

We have seen that for each integer 77 > 4, the distance-colored graph P3 is

Hamiltonian-colored In the following lemma, we provide an alternative proof of this result for 77 > 7 that shows that P3 has a properly colored Hamiltonian cycle possessing four specified edges

Lemma 3.3.1 For each integer n > 7, let the path Pn — (v\,v2, , ?;n) Then the distance-colored graph P3 has a properly colored Hamiltonian cycle containing the four edges vxv2j vn^vn_i, vn_2vn and vn_ivn

Proof. We begin by observing that for each 77 E {4, 7, 9,10}, the graph P3 has a properly colored Hamiltonian cycle containing the four edges v\v2, vn_$vn-i,vn_2vn

3 and 7;n_i7;n (see Figure 3 7(a)) Next, observe that P8 has this property as well

This can be seen by adding the four vertices 7;5, i>6, v7 and 7;8 to the properly colored 57

X Z7./H Z/t i XKI X

n = m = 2 n = 5,m = 2 n = 6, m = 2 n = m = 3

C^X n = 6, m = 3 !£* n = m = 5 i :\_\N

n = 6, m = 5 n = m = 6

Figure 3.6: Properly colored Hamiltonian cycles in the cubes of Pn • Pm when n, ra E {2, 3, 5,6}

3 Hamiltonian cycle in P in Figure 3.7(b), deleting the edge 7;37;4 and adding the edges 7;37;6, 7;47;5, v^v7, v^v8 and 7;77;8.

3 3 3 Thus the distance-colored graphs P , P%, P9 and P 0 possess properly colored Hamiltonian cycles with the specified edges. Assume, for an integer n > 11, that all distance-colored graphs P^, where 7 < k < n possess properly colored Hamiltonian cycles containing the specified edges. Consider the distance-colored graph P3. By

3 the induction hypothesis, P _4 possesses a properly colored Hamiltonian cycle 58

n = 7 (^ n = 10

Figure 3.7: Properly colored Hamiltonian cycles in the distance-colored graph P3 for n E {4, 7, 8, 9,10}

G containing the four edges ViV2,7;n_77;n_5,7;n_(,7;n_4 and 7;n_57;n_4. Proceeding as above, we construct a properly colored Hamiltonian cycle C' in the distance-colored

3 graph P by deleting the edge 7;n_57;n_4 from G and adding the edges 7;n_57;n_2,

^n-4^n-3, vn_3vn_i, 7;n_27;n and 7;n_i7;n. •

We now verify the Colored Bridges Problem for all towns represented by a

graph Pn • Pm for all integers n, ra > 2.

Theorem 3.3.2 For every two integers n,m > 2, the distance-colored graph

3 (Pn • Pm) is Hamiltonian-colored.

Proof. We have already seen that this is true when n, ra E {2,3,5,6}. Thus,

we may assume that at least one of n and ra does not belong to {2,3,5,6}, say

77 ^ {2,3,5,6}. Therefore, either n — 4 or n > 7. Hence Pn • Pm contains ra

3 copies of Pn = (t>i, v2)..., vn) and (Pn D Pm) contains 777 copies Gx (1 < i < ra)

3 of P . By Lemma 3.3.1, G2 possesses a properly colored Hamiltonian cycle Cz 59

containing the four edges 7;!7;2,7;n_37;n_i, 7;n_27;n and 7;n_x7;n. Next, we delete the edge 7;i7;2 from each cycle G2, unless ra is odd in which case the edge 7;i7;2 is not deleted from Gm. In addition, the edge 7;n_i7;n is deleted from Gz for 1 < i < m and from Gm as well if ra is odd. For each odd integer i with 1 < i < ra, the two vertices 7;i and the two vertices 7;2 are joined by an edge in G2 and G2+i; while for each even integer i with 2 < i < ra, the two vertices vn_i and the two vertices vn are joined by an edge in Gz and G2+i. (See Figure 3.8 for the cases when n = 7

3 and ra = 6.) This produces a properly colored Hamiltonian cycle in (Pn • Pm) . •

Figure 3.8: A properly colored Hamiltonian cycle in 3 the distance-colored graph (P7 • P^)

3.4 Restricted Colored Bridge Problems

As a consequence of Theorem 3.3.2, whenever a town can be represented by the

graph Pn • Pm for some n, ra > 2, then it is always possible to take a round trip about the town that encounters each intersection in the town exactly once such 60 that the walkway or bridge entering every intersection is of a different color than that used to exit it. In fact there are some circumstances when no blue bridges, or no red bridges or only bridges are needed to solve the Colored Bridges Problem.

In this section, we present necessary and sufficient conditions for such a round trip to exist when (1) no blue bridge is used, (2) no red bridge is used and (3) only bridges are used.

3.4.1 Restricted Colored Bridge Problem I

We now show that if a town can be represented by a certain grid, then it is always possible to take a round trip about the town that encounters each intersection in the town exactly once such that the walkway or bridge entering every intersection is of a different color than that used to exit it and no blue bridges are needed. We begin with a lemma.

Lemma 3.4.1 For each even integer n > 2; the square of Pn • P2 is Hamiltonian- colored.

Proof. Let Gn = Pn • P2. Suppose that Gn is constructed from two copies (u\, u2) ..., un) and (7^, 7;2, ..., vn) of Pn by adding the edges uxv% for 1 2 that Gn contains a properly colored Hamiltonian cycle G such that either

(7xn_i, un, 7;n_i, vn) or (un, 7in_i, 7;n, 7;n_i) is a subpath of G. This statement is true for 2 < n < 6, as shown in Figure 3.9. 61

Vi V2 V3 V4 V5 VQ

Figure 3.9: A step in the proof of Theorem 3.4.1

Assume for some even integer k > 6 that the statement is true for G\ and consider the distance-colored graph G\+2. By the induction hypothesis, G\ con tains a properly colored Hamiltonian cycle C' such that either (uk-\, uk, 7^-1, vk) or (uk, 7ifc_i, Vk, Vk-i) is a subpath of C'. We now construct a properly colored

Hamiltonian cycle G of G\+2 from the cycle C' as follows:

(a) If (uk-i, Uk, Vk-\, Vk) is a subpath of G', then G can be constructed from Cf

by replacing the edge ukvk-\ by the path (uk, uk+2, uk+u vk+2, vk+u vk-i).

Then (uk+2, uk+i, vk+2, vk+i) is a subpath of G.

(b) If (uk, Uk-i, Vk, Vk-i) is a subpath of G', then G can be constructed from C'

by replacing the edge uk-\Vk by the path (uk-\, uk+u uk+2, vk+u 7;fc+2, vk).

Then (uk+i, uk+2) vk+i, vk+2) is a subpath of G.

Therefore, G^+2 is Hamiltonian-colored. •

Theorem 3.4.2 For integers n and ra with 77, ra > 2, the square of Pn • Pm is Hamiltonian-colored if and only if nm = 0 (mod 4). 62

2 Proof. Let G = Pn D Pm. First, we show that if nra ^ 0 (mod 4), then G is not Hamiltonian-colored. Since nra ^ 0 (mod 4), it follows that

(i) n=£0 (mod 4) and ra ^ 0 (mod 4) and

(ii) at least one of n and 777 is odd.

If n and ra are both odd, then the order nm of G (and G2) is odd. Since the chromatic index of an odd cycle is 3, it follows that G2 is not Hamiltonian-colored.

Thus, exactly one of n and 777 is even, say n is odd and 777 is even. Then n = 2p+1 and ra = 4q + 2 where p > 1 and q > 0. Then nm = 8pq + 4p + 4q + 2. Assume, to the contrary, that G2 is Hamiltonian-colored. Let G be a properly colored

Hamiltonian cycle in G2. Thus the edges of G are alternately colored 1 and 2.

Since G is bipartite, the vertex set of G can be partitioned into two sets X and Y", where then

\X\ = \Y\ =4pq + 2p + 2q+l is odd. Suppose that

G = (^1, z2,..., znm, znm±i = z\).

We may assume that z\ E X and Z\Z2 is colored 2. Thus z%zl+i is colored 2 if i is odd and zlzl+i is colored 1 if i is even, where 1 < i < nm. This implies that zz e X if i = l,2 (mod 4) and 2:, e Y if 2 = 3,4 (mod 4). Since nra = 8pq + 4p + 4q + 2, it follows that nra + 1=3 (mod 4) and so znm+i E F, contradicting the fact that ^nm+i = Zi E X. Therefore, G2 is not Hamiltonian-colored. 63

For the converse, suppose that nra = 0 (mod 4). We show that G2 is Hamiltonian- colored. Since nra = 0 (mod 4), at least one of n and ra is even, say ra is even.

Thus either ra = 0 (mod 4) or ra =. 2 (mod 4). By Lemma 3.4.1, we may assume that ra > 4. Hence either ra = 4q or ra = 4q + 2 for some positive integer q. We consider these two cases.

Case 1. m — 4q where q > 1. By Lemma 3.4.1, we may assume that n > 3.

Suppose that G = Pn D Pm is constructed from ra copies Pi, P2,..., Fm of Pn where

Ft = (7;z,i,7;2?2,.. .,7;2?n) for 1 < i < ra by adding the edges

^1,1^+1,1, vli2vl+h2, ..., vhnvl+1}n

for 1 < i < ra — 1. For each integer j with 1 < j < p — 1, let H3 = Pn O P4p be the induced subgraph of G with

V(H3) = V(F43) U V(F4J+1) U V(F4]+2) U F(F4j+3).

For each integer j with 1 < j < p — 1, define a (4n)-cycle Go in the square of H3 by

?; C3 = C%,1, ^4J,2J ^4J + 1,1J ^4j + l,2) 4j+2,l, ^4^+2,2, ^4j+3,l J ^4j+3,2>

^+2,3? ^4.7+3,3> ^4j+2,4, ^4.7+3,4? • • • > ^4.7+2,71 j ^4j+3,n?

v ^4j + l,nj ?%n, ^4j + l,n-l? ?%n-l, • • • , %-fl,3, %,35 4j,l)- 64

2 Then C3 is a properly colored Hamiltonian cycle in H for 1 < j < p — 1. Fig ure 3.10(a) shows such a properly colored Hamiltonian cycle in the square of

Pn • P4 for each n E {3,9}. Then a properly colored Hamiltonian cycle G in

2 G can be constructed from these p — 1 cycles Gi, G2,..., Gp_i by

v (1) deleting the edges V43^s,2 43-\-2,s and 7;4?_f4jl7;4:7+4)3 for 1 < j < p — 1 and

(2) adding the edges 7;47+4ji7;4o+3j2 and 7;4o+4j37;4o+2j3 for 1 < j < p - 1.

For each n E {3,9}, Figure 3.10(b) shows such a properly colored Hamiltonian cycle in the square of Pn • P8 that is constructed from the two properly colored

Hamiltonian cycles in the squares of the two copies of Pn • P4. In Figure 3.10(b), the dashed lines indicate the deleted edges and the bold lines indicate the added edges between the squares of the two copies of Pn • P4 for each n E {3, 9}.

Case 2. ra = 4q + 2 where q > 1. Since nra = 0 (mod 4), it follows that n is even. By Lemma 3.4.1, we may assume that n > 4. Suppose that G = Pn • Pm is constructed from ra copies Pi,P2,..., Pm of Pn, as described in Case 1. Let

H = Pn D P2 be the induced subgraph of G with V(H) = V(FX)UV(F2). Construct

2 a properly colored Hamiltonian path P in H with the initial vertex 7;2j2 and the terminal vertex 7;2>4 such that

(a) if n = 4, then P = (v2y2, v2fU 7;i>2, ^1,1, ^1,3, ^2,3, ^1,4, ^2,4); 65

Figure 3.10: Properly colored Hamiltonian cycles in the squares of Pn • P4 and Pn • P8 for n E {3, 9}

(b) if n > 6, then

P = (^2,2, ^2,1, ^1,2, ^1,1, vl,3, ^2,3, ^1,4, ^1,5, ^2,6, ^2,7,

7;ij8, 7;i}9, 7;2>io, V2,ll, . . . , Vl,n-2, ^l,n-l, V2,n, ^l,n

^2,71-1,^2,71-2, ^l,n-3, ^l,n-4, • • • , ^1,7, ^1,6, ^2,5, ^2,4)-

Figure 3.11 shows such a properly colored Hamiltonian path P in the square of

Pn D P2 for nE {4, 6, 8,10}. 66 "-*•• *ZM

-»= ZlOCX!

Figure 3.11: Properly colored Hamiltonian paths in the squares of Pn • P2 for n E {4, 6, 8,10}

Next, let H' = Pn • P4p be the induced subgraph of G with

V(ff') = V(F3) U 7(F4) U • • • U K(F4p+2) and let G be the properly colored Hamiltonian cycle in the square of H' that is constructed as described in Case 1. Then a properly colored Hamiltonian cycle in

2 G can be constructed from P and G by (1) deleting the edge 7;3>i7;3;3 and (2) adding the two edges 7;3)i7;2>2 and 7;3j37;2)4. For each n E {4,10}, Figure 3.12 shows such a properly colored Hamiltonian cycle in the square of Pn • P6 that is constructed from the properly colored Hamiltonian path P in the square of Pn • P2 and the properly colored Hamiltonian cycle G in the square of Pn • P4. In Figure 3.12, the dashed lines indicates the deleted edges and the bold lines indicate the added edges.Therefore, G2 is Hamiltonian-colored. • 67

n = 4 n = 10

Figure 3.12: Properly colored Hamiltonian cycles in the squares of Pn • P6 for n E {4,10}

3.4.2 Restricted Colored Bridge Problem II

We now show that whenever a town can be represented by the graph Pn • Pm for some n, ra > 2 under certain conditions, then it is possible to take a round trip about the town that encounters each intersection in the town exactly once such that the walkway or bridge entering every intersection is of a different color than that used to exit it and no red bridges are needed.

3 Theorem 3.4.3 For integers n,m > 2, the distance-colored graph (Pn • Pm) contains a properly colored Hamiltonian cycle using only the colors 1 and 3 if and only if

nm is even unless n = m = 2 or {n, ra} is {2,3} or {2, 7}.

3 Proof. First, observe that if nra is odd, then no Hamiltonian cycle in (Pn • Pm) can be properly colored with only two colors. If n = ra = 2, then the diameter

3 of Pn • Pm is 2 and no edge in (Pn D Pm) is colored 3. If {n, ra} = {2,3}, 68

3 then there are only two edges colored 3 in (Pn • Pm) and so no properly colored

3 Hamiltonian cycle using the colors 1 and 3 exists in (Pn • Pm) .

3 Next, we show the distance-colored graph (P2 • P7) does not contains a prop erly colored Hamiltonian cycle using only the colors 1 and 3. Assume, to the contrary, that such a properly colored Hamiltonian cycle G exists. Let the vertices of P2 • P7 be labeled as shown in Figure 3.13. Necessarily, each vertex of G is incident with an edge colored 1 and an edge colored 3. We consider two cases, according to the edge of G colored 3 that is incident with the vertex x\.

X[\ X4 x X(y X XI x2 5 7 Q- •o— O— -o- -o- -o— -o

O- -o- -o 2/2 2/1 V\ 2/4 2/5 2/7

Figure 3.13: The graph P2 • P7

Case 1. The edge x\x± is on C. (See Figure 3.14, where we always represent an edge colored 1 by a solid edge and an edge colored 3 by a dotted edge.) Since the only possible edges on G that are colored 3 and incident with y2 are y2.x4 and

7/27/5, the cycle G must contain y2y$. Since the only possible edges on G that are colored 3 and incident with x7 are .x4.x7 and y$x7) a contradiction is produced.

Xl X2 x:\ X4 X*, Xb x7 O O O O O O O

O O O O O O O 2/1 2/2 V:i 2/4 2/5 2/6 2/7

Figure 3.14: The situation where G contains the edge x\x± 69

Since Case 1 cannot occur, none of the edges x\x±, yiy^ x4x7 and T/4T/7 can belong to G.

Case 2. All of the edges rcij/3,7/1.T3, x$y7 andyhx7 belong to C. (See Figure 3.15.)

Since the only possible edges on G that are colored 3 and incident with x2 are .T2.X5 and .x2y4, the edge .x2y4 must belong to G. Similarly, the only possible edges on

G that are colored 3 and incident with x§ are .T3.X6 and y4.x6. Therefore, the edge y4.x6 must belong to G, which produces a contradiction. •

Xl X2 £4 £5 £7 0 0 0 0 0 0 O

0 0 0 0 0 0 O 2/1 2/2 2/3 2/4 2/5 2/6 2/7

Figure 3.15: The situation where G contains the edges £52/7, £i2/3,2/12/3, and 2/52/7

For the converse, assume that n > 2 and ra > 2 are integers such that nra is even and such that neither n = ra = 2 nor {n, ra} = {2,3} or {2, 7} occurs.

3 We show that the distance-colored graph (Pn • Pm) contains a properly colored Hamiltonian cycle using only the colors 1 and 3. Since nra is even, we may assume that n is even. The graph Pn • Pm consists of n paths Pm which we denote by

= ?; Pm,i ( z,l, ^2,2, • • •, vhm)

for 1 < i < n such that vlft is adjacent to v3it (1 < t < m) when \i — j\ = 1. With this labeling, the graph P2 • P5 is shown in Figure 3.16.

We first show by induction on ra > 4 with ra ^ 7 that the distance-colored

3 graph (P2 • Pm) has a properly colored Hamiltonian cycle using only the colors 70

^1.2 vi.-\ ^1,4 o— -o— -O— -o— -o

^2,1

Figure 3.16: The graph P2 • P5

1 and 3 and containing the edges vljmv2,m, ^i,m-2^i,m-i and 7;2?m_27;2?m_i. First, observe that this is true for ra E {4, 5, 6,11} as shown in Figure 3.17. (Again, the solid edges are colored 1 and dotted edges are colored 3.)

i r. i .-•. i• i• •—~ • „•— • i•

ft DP4 P2 • ft P2 • ft

* • • • • • • • P2 D Pii : I 1 1 . I 1 3 Figure 3.17: Properly colored Hamiltonian cycles in (P2 • Pm) for ra E {4,5,6,11} containing 7;l5m7;2>m, ^1,^-2^1,^-1 and 7;2?m_27;25m_i

3 For ra > 4 with ra 7^ 7, assume that (P2 D Pm) has a properly colored

Hamiltonian cycle C' using the colors 1 and 3 and containing the edges 7;ljm7;2?m, ^i,m-2^i,m-i and ^2,m-2^2,m-i- For the properly colored Hamiltonian cycle C" in

3 (P2 D P4) shown in Figure 3.18, we delete the edge vi^mv2^m from C' and the edge 7;i}m+i7;2,m+i from C" and add the edges i>i>m?;i)m+i and 7;2>m7;2jm+i to produce

3 a properly colored Hamiltonian cycle G in (P2 • Pm+4) containing the edges

^l,m-f4^2,m+4, ^l,m+2^1,m4-3 and 7;2,m+2^2,m+3- 71

Vl,m+1 vl.m+2 Ul.m+'i ^l,m4-4 o o o O

O O - O V2.m4-1 •W2.m4-2 ^2,m + 'i ^2,m+4

vl.m-2 Vl,m-l vl,m ^l.m+1 vl ,m+2 ^l,m-M ^l.m+4 o o o o o- o o o o o o o-

^2,m-2 V2.m-1 V2,m ^.m+l V2 n+2 V2,Tn + '\ ^2,m+4

3 Figure 3.18: Constructing a properly colored Hamiltonian cycle in (P2 • Pm+4)

3 Therefore, (P2 • Pm) contains a properly colored Hamiltonian cycle using the colors 1 and 3 and containing the edges 7;i?m7;2,m, V\^m^2vi^m_i and 7;2jm_27;2>m_i for every integer ra > 4 with ra / 7.

Next, we show by induction on even integers n > 2 that except for n = ra = 2

3 and {n, ra} = {2,3}, {2, 7} the distance-colored graph (Pn • Pm) has a prop erly colored Hamiltonian cycle using the colors 1 and 3 and containing the edge

Vn,m-2Vn,m-i- We have seen that this is true for n = 2. Assume first that this is true for an even integer n > 2 and an integer ra > 2 where ra ^ {2,3, 7}. Then

3 there exists a properly colored Hamiltonian cycle C' in (Pn • Pm) using the colors

1 and 3 and containing the edge t>n,m-2^n,m-i- Consider the graph P2 • Pm whose vertices are labeled as in Figure 3.19

Un + 1,1 vn + l,2 Un + 1,1 Vn + l,m-2 ^n + l.m-l ^n + l.m O O O ' ' ' O O o

o o o ... o o o Vn + 2,1 ^n + 2,2 Vn + 2,3 ^n + 2,ro-2 ^n+2,m-l t>n + 2.ra

Figure 3.19: The vertices of a graph P2 • Pr 72

We have seen that when ra ^ {2,3, 7}, there is a properly colored Hamiltonian

3 cycle C" in (P2 • Pm) using only the colors 1 and 3 and containing the edges

^n+i,m-2?Wi,m-i and 7;n+2,m-2^n+2,m-i . We now delete the edge vn,m-2vn,m-i of

G' and the edge vn+i,m-2Vn+i,m-i of C" and add the two edges 7;njm_27;n+i)m_2 and

3 Vn,m-iVn+i,m-i, producing a properly colored Hamiltonian cycle G in (Pn+2 • Pm) using only the colors 1 and 3 and containing the edge 7;n+25m_27;n+25m_i. Conse quently, for every even integer n > 2 and every integer ra > 2 with ra ^ {2, 3, 7}, there is a properly colored Hamiltonian cycle using only the colors 1 and 3 in

(Pn D Pmf.

3 3 To complete the proof, it remains to show that (Pn • P3) and (Pn • P7) contain properly colored Hamiltonian cycles using only the colors 1 and 3 for each even integer n > 4. We verify by induction on even integers n > 4 that such a cycle exists containing the edge vn,m-2Vn,m-i for ra = 3 and ra = 7. That this is true for n = 4 and n = 6 is shown in Figure 3.20 (where the edges 7;n?m_27;n?m_i are drawn in bold).

3 3 Assume for an even integer k > 6 that (P3 • P3) and (P3 • P7) contain properly colored Hamiltonian cycles using only the colors 1 and 3 and containing the edge v3im-.2v3jm-i for every even integer j with 4 < j < k and ra E {3, 7}. We

3 3 show that (P/c+2 • P3) and (P/c+2 • P7) contain properly colored Hamiltonian cycles using only the colors 1 and 3 and containing the edge Vk+2im-2Vk+2jm-i for ra E {3,7}. From the induction hypothesis, it follows that (Pk-2 D P3)3 and

3 (Pk-2 0 ^V) contain properly colored Hamiltonian cycles C[ and G2, respectively, 73

• • • •- -• • • • ^ J J l y! i • • •—• (n,m) = (4,7) (n,m) = (4,3) • • • t ~ 1 I • • • • ..11 i ~ : •J 1 ~ ~i i (n, m) = (6,3) (n, m) = (6,7)

3 Figure 3.20: Properly colored Hamiltonian cycles in (Pn • Pm) using only the colors 1 and 3 and containing

the edge vnyrn_2vn,m-i for four pairs (n, ra)

using only the colors 1 and 3 and containing the edge Vk-2jm-2Vk-2,m-i where me {3,7}.

3 3 Each of the graphs (P4 • P3) and (P4 • P7) consists of four paths Pm (ra E {3, 7}) which we denote by

Pm,i — (^fc+2,l> ^fc+2,2, • • • , ^fc+2,mj

for — 1 < i < 2 such that vht is adjacent to v3jt (1 < t < ra) when \i — j\ = 1. Then

3 3 (P4 D P3) and (P4 • P7) contain properly colored Hamiltonian cycles G" and G2, respectively, using only the colors 1 and 3 and containing the edge ^fc_i,m-2^fc-i,m-i and 7;/c+2,m-2^+2,m-i for rn E {3, 7}. Deleting the edge 7;/c_2,m-2^-2,m-i of C[ and the edge ^-1^-2^-1^-1 of C" for i = 1, 2 and adding the edges 7;/c_2)m_2^/c-i,m-2 74

and Vk-2,m-iVk-i,m-i result in properly colored Hamiltonian cycles C\ and G2 in

3 3 (P/c+2 D P3) and (P^+2 D P7) , respectively, using only the colors 1 and 3 and containing the edge 7;fc+2>m_27;/c+2jm_i for ra E {3, 7}, completing the proof. •

3.4.3 Restricted Colored Bridge Problem III

We now show that if a town can be represented by the graph Pn • Pm for some n, ra > 2 under certain conditions, then it is possible to take a round trip about the town that encounters each intersection in the town exactly once such that the walkway or bridge entering every intersection is of a different color than that used to exit it and only bridges are used.

3 Theorem 3.4.4 For integers n,m > 2, the distance-colored graph (Pn • Pm) contains a properly colored Hamiltonian cycle using only the colors 2 and 3 if and only if nm = 0 (mod 4) unless n = ra = 2.

3 Proof. Suppose first that (Pn • Pm) contains a properly colored Hamiltonian cycle G using only the colors 2 and 3. Since the diameter of P2 • P2 is 2, it is impossible that n — m = 2. Let

G = {vuv2,..., 7;nm_i, 7;nm, vx).

Since Pn • Pm is bipartite, it has two partite sets A and B. We may assume that vi E A and that 7;i7;2 on G is colored 2. Thus the distance between vx and 7;2 in

Pn • Pm is 2 and 7;2 E A as well. Since 7;27;3 is colored 3 and 7;37;4 is colored 2, both 7;3 and 7;4 belong to B. Because 7;47;5 is colored 3, it follows that 7;5 E A. This 75

implies that the nra vertices of Pn • Pm are encountered cyclically on G in groups of 4, the first pair of which belongs to A and the second pair of which belongs to

B. Thus nra = 0 (mod 4).

For the converse, let n, ra > 2 be integers with nra = 0 (mod 4) such that

3 (n, ra) ^ (2,2). We show that the distance-colored graph (Pn • Pm) contains a properly colored Hamiltonian cycle using only the colors 2 and 3. Since nra is even, we may assume that n is even. The graph Pn • Pm consists of n paths of order ra, which we denote by

Pm,i — (^2,1, ^2,2, • • • , Vhm)

for 1 < i < n such that vht is adjacent to v3jt (1 < t < ra) when |z — j| = 1.

We first show by induction on the even integers ra > 4 that the distance-colored

3 graph (P2 • Pm) contains a properly colored Hamiltonian cycle using only the colors 2 and 3 and containing the edges ^1,1^2,2 and vijm_iv2jm colored 2. That this is true for m = 4 and 777 = 6 is shown in Figure 3.21 (where the solid edges are colored 2 and dotted edges are colored 3).

3 Figure 3.21: Properly colored Hamiltonian cycles in (P2 • Pm) for ra = 4, 6 using only the colors 2 and 3

Assume for an even integer ra > 6 that for every even integer t with 4 < t < ra,

3 the distance-colored graph (P2 • Pt)' contains a properly colored Hamiltonian cy- 76

cle using only the colors 2 and 3 and containing the edges ^1,1^2,2 and Vi^iV2^t.

3 We claim that the distance-colored graph (P2 • Pm+2) contains a properly col ored Hamiltonian cycle using only the colors 2 and 3 and containing the edges

^1,1^2,2 and 7;i>m+i7;2?m+2- By the induction hypothesis, the distance-colored graph

3 (P2 • Pm_2) contains a properly colored Hamiltonian cycle C' using only the colors 2 and 3 and containing the edges ^1,1^2,2 and 7;i)m_37;2?m-2 (see Figure 3.22).

3 The graph (P2 • P4) consists of two paths P4 which we denote by

(^i,m-i, ^i,m, ?>L,m+i, vhm+2) and (7;2?m_i, 7;2>m, v2jm+u 7;2?m+2),

3 respectively, as shown in Figure 3.22 where the vertices of (P2 • P4) are drawn as

3 solid vertices. Let C" be the cycle in (P2 • P4) also shown in Figure 3.22. We now construct a properly colored Hamiltonian cycle G from Cf and C" by removing the edge 7;ijm_37;2>m_2 from C' and the edge ^i,m,—i^2,m from C" and adding the edges

^i,m-3^2,m-i and 7;2jm_27;2,m- The cycle G uses only colors 2 and 3 and contains the edges ^1,1^2,2 and 7;i5m+i7;2,m+2- This verifies the claim.

Next, we show by induction on the even integers n > 2 that for every even in

3 teger ra > 2, except for ra = 2 when n = 2, the distance-colored graph (Pn • Pm) contains a properly colored Hamiltonian cycle using only the colors 2 and 3 and containing the edge vn,ivn;z colored 2. We have already seen that this is true for n = 2. Assume that this is true for an even integer n > 2 and every even integer ra > 2 where (n, ra) 7^ (2,2). Then for each even integer ra > 2, there is a prop

3 erly colored Hamiltonian cycle C' in (Pn • Pm) using only the colors 2 and 3 and 77

^2,1 ^2.2 V2y7n-:\ V2,m-2 ~ ^2,m+l V2,m+2 V2,m-1 V2,m

Figure 3.22: Constructing a properly colored Hamiltonian cycle 3 in (P2 • Pm+2) using only the colors 2 and 3 and containing the edges Vit\V2j2 and 7;i)m+i7;2?m+2

containing the edge 7;nji7;nj3. For this integer ra,le t P2 D Pm consist of two paths

Pm which we denote by

; (?Vfl,l, ? n+l,2, • • • , Vn+l,m) and (^n+2,1) ^n+2,2, • • • , ^n+2,m),

respectively, and 7;n+i?z is adjacent to 7;n+2,z for 1 < i < m. We have seen that

3 there is a properly colored Hamiltonian cycle C" in (P2 • Pm) using only the colors 2 and 3 and containing the edges 7;n+ij27;n+i)4 and 7;n+2ji7;n+2,3- The cycle in

3 ! (Pn+2 • Pm) obtained by deleting the edges 7;n5i7;nj3 and Vn+1,2^71+1,4 from C and

hr C , respectively, and adding the edges 7;n?i7;n+ij2 and ^n,3^n+i,4 produces a properly colored Hamiltonian cycle using only the colors 2 and 3 and containing the edge

^71+2,1^71+2,3 • 1^ therefore follows that for every two even integers n > 2 and ra > 2

3 with (n, 7Ti) / (2,2), the distance-colored graph (Pn • Pm) contains a properly colored Hamiltonian cycle using only the colors 2 and 3.

To complete the proof, it remains to show that the distance-colored graph

S (Pn • Pm) contains a properly colored Hamiltonian cycle using only the colors 2 78 and 3 when n = 0 (mod 4) and ra > 3 is odd or equivalently, when n > 3 is odd and ra = 0 (mod 4). We verify this latter formulation of the statement.

First, we show by induction on the integers ra > 4 with ra = 0 (mod 4) that

3 the distance-colored graph (P3 • Pm) has a properly colored Hamiltonian cycle using only the colors 2 and 3. We use the same notation as before for the vertex set

3 of P3 • Pm. In fact, we show that (P3 • Pm) has a properly colored Hamiltonian cycle using only the colors 2 and 3 and containing the two edges Vi^v2y2 and ^i,m-i^2,m) both colored 2. That this is true for ra = 4 is shown in Figure 3.23. 79

v 1)1,1 l,2 V\,3 ViA

7J2,1 • • • * V2A

^2,2 7J2.'} .

«3,1 V'i,2 V'\/l VXA

Figure 3.23: A properly colored Hamiltonian cycle 3 in (P3 • P4) using only the colors 2 and 3 and containing the edges ^1,1^2,2 and ^1,3^2,4

Assume, for an integer ra > 4 with ra = 0 (mod 4) that the distance-colored

3 graph (P3 • Pm) has a properly colored Hamiltonian cycle C' using only the colors

2 and 3 and containing the edges 1^1,1^2,2 and 7;ljm_i7;2m. Let P3 D P4 consist of three paths P4 which we denote by

(/?;2,m+l, vi,m+2i ^z,m+3) ^i,m+4j

3 for i = 1,2,3. We have seen that the distance-colored graph (P3 • P4) has a properly colored Hamiltonian cycle C" using only the colors 2 and 3 and contain

3 ing the edges vi>m+iV2>m+2 and 7;i)m+37;2,m+4. Then the cycle G in (P3 • Pm+4) constructed by deleting the edges 7;i?m_i7;2?m and 7;i?m+i7;2)m+2 from C' and G", re spectively, and adding the edges ^1,^—i^i,m-hi and 7;2jm7;2jm+2 (both colored 2) is a properly colored Hamiltonian cycle using only the colors 2 and 3 and containing the edges 7;i7i7;2?2 and 7;i?m+37;2>m+4. This verifies the statement. Furthermore, observe that the cycle G constructed also contains the edge ^3,1^3,3, which is colored 2.

We now show by induction on the odd integers n > 3 that for every integer ra > 4 with m = 0 (mod 4) there exists a properly colored Hamiltonian cycle 80

3 in the distance-colored graph (Pn • Pm) that uses only the colors 2 and 3 and contains the edge 7;n?i7;n53. We have already seen that this is true when n = 3. Assume for an odd integer n > 3 and for every integer ra > 4 with ra = 0 (mod 4) there exists a properly colored Hamiltonian cycle G in the distance-colored graph

3 (Pn • Pm) that uses only the colors 2 and 3 and contains the edge 7;n5i7;n?3.

For an integer ra > 4 with ra = 0 (mod 4), let P2 • Pm consist of two paths

Pm which we denote by

Pm,i — (7;2,1, ^z,2, • • • , vi,m)

3 for i = n +1, n + 2. We have seen that (P2 D Pm) has a properly colored Hamilto nian cycle C" using only the colors 2 and 3 and containing the edges Un+1,2^71+1,4 and

^7i+2,i^n+2,3- By deleting the edges 7;n>i7;n)3 and 1^+1,2^71+1,4 from C' and G", respec tively, and adding the edges 7;nji7;n+ij2 and 7;nj37;n+i)4 (both colored 2), we obtain a

3 properly colored Hamiltonian cycle G in the distance-colored graph (Pn+2 d Pm) using only the colors 2 and 3 and containing the edge 7;n+2,i^n+2,3- This completes the proof. •

In Theorem 3.4.2, all pairs n, ra > 2 of integers are determined for which the

3 2 distance-colored graphs (Pn • Pm) and (Pn • Pm) contain a properly colored Hamiltonian cycle. In the second case, this is equivalent to determining all pairs

3 n, ra > 2 of integers for which the distance-colored graph (Pn • Pm) has a properly colored Hamiltonian cycle the colors of whose edges follow the permutation (1 2).

Theorems 3.4.3 and 3.4.4 answered the question for the permutations (1 3) and

(2 3) respectively. This might suggest another question: For which pairs n, ra > 2 81

3 of integers, does the distance-colored graph (Pn • Pm) contain a properly colored Hamiltonian cycle the colors of whose edges follow the permutation (12 3)? Of course, this is only possible if nra = 0 (mod 3). Suppose that such a cycle G =

(7;i,7;2,... ,7;nm_i,7;nm,7;i) exists. Since Pn • Pm is bipartite, it has two partite sets A and B. If nra is even, then |>1| = |5|; while if nra is odd, then one of \A\ and \B\ exceeds the other by 1, say \B\ = \A\ + 1. We may assume that v\ E A and that viv2 is colored 1, which implies that 7;2 E B. Thus 7;27;3 is colored 2, 7;37;4 is colored 3 and 7;47;5 is colored 1. Therefore, 7;3 E P, 7;4 E A and 7;5 E B. This implies that two-thirds of the vertices of Pn • Pm belong to B and one-third to A, that is, \B\ = 2\A\. This is impossible if \A\ = \B\. If \B\ = \A\ + 1, then

3 \B\ = 2, \A\ — 1 and (P2 • Pi) is not Hamiltonian. Hence there are no integers

3 n, ra > 2 for which the distance-colored graph (Pn • Pm) has a properly colored Hamiltonian cycle the colors of whose edges follow the permutation (1 2 3). Chapter 4 Color Distance and Connectivity

4.1 Color-Connection Exponent and Color-Distance

For a connected graph G and a positive integer fc, the distance-colored graph Gk is called properly color-connected, or simply color-connected, if every two vertices u and v in Gk are connected by a properly colored u — v path in Gk. If the diameter of G is d, then Gd is complete and so every two vertices u and v in Gd are connected by the path (u, v) of length 1 in Gd and so Gd is trivially color-connected. It is natural, therefore, to seek the smallest integer fc for which Gk is color-connected.

This gives rise to the following concept. The color-connection exponent cce(G) of

G is defined as the minimum fc for which Gk is color-connected. Thus cce(G) = 1 if and only if G is a complete graph; while if cce(G) = fc > 2, then Gk is color- connected but Gk~l is not color-connected. First, we make a useful observation.

Observation 4.1.1 Let G be a connected graph that is not complete. If a distance-

82 83 colored graph Gk is Hamiltonian-colored, then Gk is color-connected and so

2 < cce(G) < hce(G).

In particular, if the diameter of G is 2, then cce(G) = 2.

By Observation 4.1.1, if the diameter of a connected graph G is 2, then cce(G) =

2. In fact, we show that cce(G) = 2 for all connected graphs G of diameter at least 2. In order to do this, we first verify this fact for paths by establishing a stronger property possessed by the square of Pn.

Proposition 4.1.2 For each integer n > 3, the distance-colored graph P% con tains a properly colored Hamiltonian path.

Proof. By Observation 4.1.1, we may assume that n > 4. Let G = Pn =

(ui, 7;2,..., vn). We consider four cases, according to whether n is congruent to 0, 1, 2 or 3 modulo 4.

Case 1. n = 0 (mod 4). Then n = 4fc for some positive integer fc. We

2 show by induction on fc that G contains a properly colored Hamiltonian v\ — vn path, in which each of v\ and vn is incident with an edge colored 2. If fc = 1, then (7;i,7;3,7;2,7;4) has the desired properties. Assume, for H = P4fc, where fc > 1, that the distance-colored graph H2 contains a properly colored Hamiltonian v\ — v4k path Q such that each of v\ and vn is incident with an edge colored 2.

2 Now for G = P4fc+4, the distance-colored graph G contains the properly colored

Hamiltonian vi — 7;4fc+4 path (Q, i^fc+i, 7;4^+3,7;4fc+2,7;4/~+4), in which each of v\ and

7;4fc+4 is incident with an edge colored 2. 84

2 Case 2. n = 1 (mod 4). Let F = G — vn. By Case 1, P contains a properly colored Hamiltonian v\ — vn-\ path Q such that each of v\ and 7;n_i is incident with an edge colored 2. Then (Q,vn) has the desired properties.

2 Case 3. n = 2 (mod 4). Let F = G — {vi,vn}. By Case 1, P contains a properly colored Hamiltonian 7;2 — 7;n_i path Q such that each of 7;2 and vn-\ is incident with an edge colored 2. Then (vi,Q,vn) has the desired properties.

2 Case 4. n = 3 (mod 4). Let F = G — J7;n_2,7;n_x, 7;n}. By Case 1, P contains a properly colored Hamiltonian V\ — vns path Q such that each of v\ and 7;n_3 is incident with an edge colored 2. Then the properly colored Hamiltonian v\ — vn~\ path (Q,7;n_2,7;n,7;n_!) has the desired properties. •

The following is then a consequence of Proposition 4.1.2.

Corollary 4.1.3 If G is a non-complete connected graph, then cce(G) = 2.

Proof. For two nonadjacent vertices u and v of G, let P be a u — v geodesic in G.

By Proposition 4.1.2, the distance-colored graph P2 contains a properly colored

Hamiltonian path and so P2 is color-connected. Since each properly colored u — v path in P2 is also a properly colored u — v path in G2, it follows that G2 is color- connected. •

By Corollary 4.1.3, we know that for every two vertices u and v in a connected graph G, there is a properly colored u — v path in G2. In fact, there may be several properly colored u — v paths in G2. A properly colored u — v geodesic is a properly 85 colored u — v path of minimum length in G2 and its length is the properly colored distance between u and v in G2 or simply the color-distance between u and v and is denoted by cd(7i, 7;). Once we know the distance d(u, v) between two vertices u and v in a connected graph G, the color-distance cd(7i,7;) between two vertices u and v in G2 can be immediately determined.

Proposition 4.1.4 Let u and v be two vertices in a connected graph G. If d{u, v) = £, then cd(u, 7;) - \%f±].

Proof. First, we show that cd(iz, 7;) > [pp]. Let P be a u—v geodesic (of length

£) in G and let Q be a properly colored u — v geodesic in G2. Then £ = 3q + r, where 0 < r < 2. Suppose first that r = 0. Since every two consecutive edges of Q are colored 1 and 2, these two edges produce a subpath of Q of length 2 in G2 that covers at most three edges of P in G. Thus Q contains at least 2q edges of G2 and at least 2q edges of G2 are needed to produce Q; that is, cd(u,7;) >2q= [pp] • If r > 0, then Q contains more than 2q edges of G2 and so cd(?i, v) >2q + l = [pp].

It remains to show that cd(?z, 7;) < |~pp]. To verify this, it suffices to exhibit a

2 properly colored u—v path of length |~pp] in G . Suppose that P = (v0,7;x,..., 7;^) is a 72 —7; geodesic in G. If £ = 3q) then the properly colored u — v path Qo =(^0) ^i>

1 7;3, 7;4,7;6, ..., 7;3,, v3i+1) 7;3i+3, ..., 7;3

7;6, ..., v3i, 7;3i+2, 7;3i+3, ..., 7;3

It is well known that the distance d is a metric on the vertex set of a connected graph G. This, however, is not the case for the color-distance cd. While cd(?i, 7;) > 0 for all u, v E V(G), cd(u, v) — 0 if and only if u = v and cd(7i, 7;) = cd(v, 71) for all u,v E V(G), the color-distance cd does not satisfy the triangle inequality. For example, suppose that P is a u — w geodesic of length 10 in a connected graph G and v is the central vertex of P. Then d(u, v) — d(v, w) = 5 and d(u, w) — 10. By

Proposition 4.1.4, cd(u,v) = cd(v,w) = [^-p1] = 3 while cd(u,w) = [^p1] = 7.

So cd(7i, 7;) + cd(7;, w) < cd(u, w).

Theorem 4.1.5 For every three vertices u,v and w in a connected graph G,

cd(7i, w) < cd(u, v) + cd(7;, w) + 1. (4.1)

Equality holds in (4.1) if and only if d(u,v) = 2 (mod 3), d(v,w) = 2 (mod 3) and d(u, w) = d(u, v) + d(v, w).

Proof. Suppose that d(u, v) = a, d(v, w) = b and d(u, w) = c. We first establish

(4.1). Since a + b > c, it follows by Proposition 4.1.4 that

2a -1 26 - 1 2a - 1 26-1 cd(u, v) + cd(u, w) = + > 3 + 3

2(a + 6) - 2 2c - 2 2c -4"

2c-1 — 1 = cd(u,w) — 1 and so (4.1) holds.

Next, assume that a = 2 (mod 3), 6 = 2 (mod 3) and c = a+b. Then a = 3p+2 and 6 = 3g + 2 for some nonnegative integers p and q. By Proposition 4.1.4, 87 cd(u, v) = 1^} = 2p + 1, cd(u, w) = [^] = 2g + 1 and

"2c-1" 2(a + 6) - 1 2(3p + 3g + 4)-l cd(u,w) = 3

"6p + 6g + 7" 2p + 2q + 3 = cd(7i, v) + cd(?;, w) + 1. 3

For the converse, first suppose that a ^ 2 (mod 3) or 6 ^ 2 (mod 3), say the former. Since c < a + 6, it follows by Proposition 4.1.4 that

2c-1 2(a + 6) -1 cd(u,w) = < (4.2)

It suffices to show that '2(a + 6) - 1 [2a-1] [26-11 < + (4.3) 3 3

r Let a = 3gi + rx and 6 = 3g2 + 2 > where 0 < r^, r2 < 2 and at least one of ri and r2 is not 2. Without loss of generality, we consider the following cases.

2{a+b)- -1 • If n = 0 and r2 = 0, then \*2=±] = 2qi and \®=±] = 2g2 and 3

2gi + 2q2.

• If n = 0 and r = 1, then |"22pl] = 2 and [^±] = 2q +l and 2(a+6)--1 2 9l 2 3

2g1 + 2g2 + l.

2{a+b)--1 • If ri = 0 and r2 = 2, then [^±] = 2gi and [^] = 2g2 + l and 3

2q1+2q2 + l.

If n = 1 and r2 = 1, then [^] = 2^ + 1 and |"^1] = 2g2 + 1 and

2(q+6)-l 3 = 2?! +292 + I. If n = 1 and r2 - 2, then [2^1] =2^ + 1 and |"^] = 2g2 + 1 and

2(a+fe)-l 3 - 2gi + 2q2 + 2.

In each case, cd(^, w) < cd(u,v) + cd(v,w). Finally, suppose that a = 2

(mod 3) and b = 2 (mod 3) but c < a + b — 1. Then a = 3gi + 2 and ft = 3g2 + 2.

Thus [2^1] = 2gi + 1 and \&=±] = 2

2c-1 2(3q1+3q2 + 3)-l cd(u,w) = < = 2qi +2q2 + 2

2a-l 2b-1 + = cd(u,v) + cd(v,w). Thus, strict inequality holds in (4.1).

4.2 Color-Eccentricity, Color-Radius and Color- Diameter

For a vertex v in a connected graph G, the color-eccentricity ce(v) of v is the color-distance between v and a vertex farthest from v in G2. The minimum color eccentricity among the vertices of G is its color-radius and the maximum eccentric ity is its color-diameter, which are denoted by crad(G) and cdiam(G), respectively.

A vertex in G whose color-eccentricity equals crad(G) is a color-central vertex of G and a vertex whose color-eccentricity equals cdiam(G) is a color-peripheral vertex of G. Two vertices u and v of G with cd(u, 7;) = cdiam(G) are color-antipodal vertices of G. Necessarily, if u and v are color-antipodal vertices in G, then each of u and v is a color-peripheral vertex. 89

Proposition 4.2.1 For every nontrivial connected graph G,

crad(G) < cdiam(G) < 2crad(G) + 1.

Proof. The inequality crad(G) < cdiam(G) is immediate from the definitions.

Let u and w be two vertices such that cd(7i, w) — cdiam(G) and let v be a color- central vertex of G. Therefore, ce(7;) = crad(G). By Theorem 4.1.5,

cdiam(G) = cd(7i, w) < cd(u, v) + cd(7;, w) + 1 < 2 ce(7;) + 1 = 2 crad(G) + 1, as desired. •

For a vertex u in a connected graph G, a vertex v is an eccentric vertex of u if v is a vertex that is farthest from u in G and the set of all eccentric vertices of u in G is denoted by Ecc(7i). As expected, a vertex w is a color-eccentric vertex of u if w is a vertex that is farthest from u in G2, and the the set of all color-eccentric vertices of u in G2 is denoted by CEcc(7i). First, we present three preliminary results on the eccentric vertices and color-eccentric vertices of a given vertex in a graph.

Lemma 4.2.2 Let G be a nontrivial connected graph and u E V(G). If x,y E

CEcc(7i); then

\d(u,x) -d(u,y)\ < 1.

Proof. Suppose that d(u, x) = a and d(u, y) = b. Since ce(7i) = \^f^} = \^j^~\ > it follows that la — 61 < 1. • 90

Lemma 4.2.3 For each vertex u in a nontrivial connected graph G,

Ecc(u) C CEcc(7i).

Proof. Let x be a vertex of G such that x £ CEcc(7i). Let y E CEcc(7i). Thus cd(7i, y) = ce(u). Then cd(tx, y) > cd(u, x). Suppose that d(u, x) = a and d(u, y) =

6. Since ce(u) = f^1] > cd(u,x) = f"2^], it follows that d(u,y) = 6 > a = e(u) and .T ^ CEcc(7i). •

The reverse set inclusion of that stated in Lemma 4.2.3 does not hold in general.

For example, consider the graph G = P\Q in Figure 4.1. Since d(u,v) — 4 and d(u, w) — 5, it follows that ce(?z) = cd(7i, 7;) = cd(7i, w) = 3. In this case, e(u) — 5 and so v is not an eccentric vertex of u. While v and w are both color-eccenti ic vertices of 7i, only w is an eccentric vertex of u.

P10 : • O O O • O O O O • v u w

Figure 4.1: A color-eccentric vertex of u that is not an eccentric vertex of u

Lemma 4.2.4 For two vertices u and v in a nontrivial connected graph,

• if e(u) > e(v), then ce(u) > ce(v),

• if ce(u) > ce(v), then e(u) > e(v) — 1.

Proof. Let x E Ecc(7i) and y E Ecc(7;). Suppose that e(u) — d(u,x) = a and e(v) = d(v,y) — 6. By Lemma 4.2.3, x E CEcc(7i) and y E CEcc(7;). Then ce(u) = 91 cd(7i, r) = [2^1] and ce(v) = cd(v,y) = \^f±] li a > 6, then \*f±] > [2*-T|, while if [^-^] > [^y^] 5 then either a > 6 or 6 = a + 1 In particular, if a = 1 (mod 3) and 6 = 2 (mod 3), then 6 = a + 1 and |"==±] = [^] •

For a nontrivial connected graph G and 7i, 7; E V^(G), let Q be a properly colored u — v path in G2 An underlying walk WQ of Q is a u — v walk in G constructed from Q by replacing each edge of Q that is colored 2 by a path of length 2 in G

A u — v path of minimum length in WQ IS called an underlying path of Q and is denoted by PQ

Theorem 4.2.5 Le/ G be a nontrivial connected graph

(a) If u and v are adjacent vertices of G2, then | ce(u) — ce(v)\ < 2

(6) If u and v are adjacent vertices of G, then \ ce(u) — ce(v)\ < 1

such that ce(w) = k

(d) If k is an integer with crad(G) < k < cdiam(G), then there are at least two

vertices of G with color-eccentricity k

Proof. First, we verify (a) Let u and v be adjacent vertices of G2 Assume, without loss of generality, that ce(u) > ce(v) Suppose that ce(u) = cd(n, r) By

Theorem 4 15,

ce(?i) = cd(7i, r) < cd(7i, 7;) + cd(7;, r) + 1 < 2 + ce(v) 92

Hence ce(u) < 2+ce(7;) and so 0 < ce(u)—ce(v) < 2. Therefore, | ce(u)—ce(v)\ < 2.

Second, we verify (b). Let u and v be adjacent vertices of G. Let x E Ecc(7i) and y E Ecc(7;). Suppose that e(u) = d(u,x) — a and e(v) — d(v,y) = b.

By Lemma 4.2.3 and Proposition 4.1.4, ce(7i) = cd(u,x) = [^^] and ce(7;) = cd(?;,y) = ["^j^~|. Since \e(u) — e(v)\ < 1, either a = 6, a = b + 1 or 6 = a + 1.

If a = 6, then ce(7i) = ce(v); while if a = b + 1 or 6 = a + 1, say the former, then

2 ce(u) = [2^1] = \^f^\ = p±i] . Since \ -^] - \%=±] < 1, it follows that 0 < ce(u) — ce(7;) < 1 and so | ce(7x) — ce(7;)| < 1.

Next, we verify (c). Let u and v be adjacent vertices of G such that ce(7;) = crad(G) and ce(7i) = cdiam(G) and let P be a u — v path in G, say P = (u =

Vi,v2, • • • ,Vk = v). By (b), the color-eccentricities of Vi and vi+i (1 < i < k — 1) differ by at most 1. Hence some vertex vt of P has color-eccentricity k.

Finally, we verify (d). Let k be an integer with crad(G) < k < cdiam(G). By

(c), there is a vertex w such that ce(7i;) = k. Let u be a vertex with cd(7i;,7i) = ce(w) = k. For a color-central vertex v of G, let Q be a properly colored u — v path of length cd(7i, 7;) in G2. Thus cd(u,v) < ce(v) = crad(G). Since cd(7i,7L>) = /c, it follows that ce(7;) < k < ce(u). Let P be the underlying 71 — 7; path of Q in G where say P = (u — Vi,v2, • • • ,V£ = v) where £ > k. By (b), the color-eccentricities of

Vi and vi+i (1 < i < £ — 1) differ by at most 1. There is a vertex x on P such that ce(.x) = fc. Because cd(u,x) < cd(u,v) < k and cd(7i;,7i) — fc, it follows that x ^ w. m 93

For two adjacent vertices u and v in G2, it is possible that | ce(u) — ce(v) \ = i for each i E {0,1, 2} even when u and v are not adjacent in G. This fact is illustrated by the graph G = P8 = (vi, 7;2,7;3,..., 7;8) of Figure 4.2, where each vertex of G is labeled by its color-eccentricity. In particular, Vi and 7;3 are adjacent vertices

2 of G . Since d(7;i,7;8) = 7 and d(v3)v8) = 5, it follows that ce(^i) = cd(7;i,7;8) =

[2^1] = 5 and ce(7;3) - cd(7;3,7;8) - [~^] = 3. Thus | ce(vx) - ce(v3)\ = 2.

Vi V2 V3 V4 ?;5 V(y Vj Vg P : O O O O O O O O 8 54 333345

Figure 4.2: The color-eccentricities of P±8

Next, we present a necessary and sufficient condition for two adjacent vertices u and v in G2 to satisfy the equality | ce(u) — ce(i>)| = 2.

Theorem 4.2.6 Let G be a nontrivial connected graph and let u and v be adjacent vertices of G2. Suppose that max{e(ii),e(i;)} = a and min{e(i/),e(t>)} = 6. Then

| ce(7i) — ce(7;)| = 2 if and only if a and 6 satisfy one of the following conditions:

(?;) a = 6 + 2 and 6 = 2 (mod 3);

(ii) a = b + 3,

(in) a = b + 4 and 6=1 (mod 3).

Proof. Assume, without loss of generality, that ce(u) > ce(v). By Lemma 4.2.4, e(u) > e(v) or e(v) = e(ix) + 1. Let e(u) = a and e(v) — 6.

First, assume that ce(?z) — ce(v) = 2. By Theorem 4.2.5(b), u and 7; are not adjacent vertices of G. Let x E Ecc(7i) and y E Ecc(v). Then e(u) = d(u,x) = a 94 and e(v) = d(v,y) = 6. By Lemma 4.2.3, ce(u) = cd(u,r) = \^j^-~\ and ce(i;) = cd(v,y) = \^f^~\- We consider five cases, according to the value of a — 6.

Case 1. \a — b\ < 1. It then follows by the proof of Theorem 4.2.5(b) that

0 < ce(u) — ce(v) < 1. Thus Case 1 is impossible.

2(&+2)-l Case 2. a-b = 2. Then ce(u) = \^=1] = 3 = [f ] + 1 and so

26' 26-1 ce(u) — ce(w) = + 1. ~3~ Suppose that 6 = 3A- + ? where ? = 0,1,2. If b = 3k, then [f ] = \&j±] = 2k, if b = 3k + l, then [f] = [2*=I] = 2A- + 1, and if 6 = 3fr + 2, then [f] = 2A- + 2and

[2^i] = 2fr + 1. Thus Case 2 is possible only when 6 = 2 (mod 3).

1 2(b+3)-l Case 3. a-b = 3. Then ce(u) = [^f ] = 3 = [^1 = P¥l+2 = ce(v) + 2. Thus Case 3 is possible.

2(b+4)-l Case 4. a - b = 4. Then ce(u) = pjpl] = 3 = [^1 + 2 and so

26+1 26-1 ce(u) — ce(v) = + 2.

Suppose that 6 = 3k + ? where ? = 0,1,2. If 6 = 3fr, then [^±1] = 2k + 1 and

r«=i] = 2fr, if 6 = 3k + 1, then \®$±] = [^i] = 2fr + 1, and if b = 3k + 2, then

[?±±I] = 2k + 2 and [^=1] = 2fr + 1. Thus Case 4 is possible only when 6 = 1

(mod 3).

2(6+5)-l Case 5. a-b > 5. Then ce(u) = [^] = 3 > [2ti]+3 = Ce(,;) + 3. Thus Case 5 is impossible. 95

For the converse, if a and /; satisfy one of the conditions (?)-(???), then, by the arguments in Cases 1 4, we see that ce(u) — ce(v) — 2 •

By Theorem 4 2 1, if G is a nontrivial connected graph such that crad(G) = r and cdiam(G) = d, then r < d < 2r+l Next, we determine all pairs r, d of positive integers with r < d < 2r +1 for which there is a connected graph with color radius r and color-diameter d First we present a well-known realization result on the radius and diameter of a graph (see [1])

Lemma 4.2.7 For each pair a, b of positive integers with a

a connected graph G with rad(G) = a and diam(G) = b

Theorem 4.2.8 Let r and d be positive integers with r < d <2r + 1 Then there

exists a connected graph G with crad(G) = r and cdiam(G) = d if and only if (i) r < d <2r or (??) d = 2r + 1 and r is odd

Proof. Suppose that r and d are positive integers with r < d < 2r + 1 that satisfies (I) or (ii) First, we make an observation If G is a connected graph with rad(G) = a and diam(G) = 6, then by Lemma 4 2 3 and Proposition 4 14, crad(G) = \^f±] and cdiam(G) = [^y1] By Lemma 4 2 7, it suffices to show that there exist positive integers a and b such that a < b < 2a, r = \^f^] and d = \^f^~\ We consider four cases

Case 1 r and d are both even Then r < d <2r Let a = y and b = y Then a

Case 2 r is even and d is odd Then r < d < 2r — 1 Let a = y and

6 = &±± Observe that a < 6 = ^1 < 3(2r~1)+1 = 3r - 1 = 2a - 1 Furthermore, Rf1! = P^l = r and [ ^J = [f ] = d

Case 3 r is odd and d is even Then r < d < 2r Let a = 3-r±i and 6 = =y

3(d 1)+1 Since r < d - 1, it follows that a = &±± < "2 = ^ < f = 6 and 6=^<^i = 3r = 2a-K2a Furthermore, [ ^] = r and \%=±] = d

CW 4 r andd are both odd Then r < d < 2r + 1 If d = 2r + 1, let a = ^±1,

6 = 3r + 1 Then 6 = 2a Furthermore, [2s=I] = r and [^1] = d Thus we may assume that r < d < 2r - 1 Then a = ^ and 6 = ^±i Then a < b < 2a - 2, [VI = r and [^] = d

For the converse, we show that there is no connected graph G with crad(G) = r and cdiam(G) = d such that r is even and d = 2r + 1 Assume, to the contrary, that such a graph G exists Then r = 2k for some positive integer k and so d = 2r + l=4k + l Suppose that u and v are color-antipodal vertices of G Then cd(7i, 7;) = 4k + 1 Suppose that d(u, 7;) = a Then

2a -1 cd(7i,7;) = = 4^ + 1 (44)

Now \**f±] is either ^, f or 2a±l If [^i] = ^i, then a = 6k + 2, if

p^I] = ^, then 2a = 12k + 3, which is impossible, if p^l] = ?a±i) then a = 6k + 1 Hence

either a = 6k+ 2 or a = 6k + 1 (4 5) 97

Let w be a color-central vertex of G. Then ce(w) — crad(G) = r = 2k and so cd(u, w) < 2k and cd(7;, w) < 2k. If cd(u, w) < 2k or cd(7;, w) < 2k, it then follows by Theorem 4.1.5 that cd(u, v) < cd(u, w) + cd(v, w) + 1 < 2k + 2fc + 1 = 4k + 1, which contradicts (4.4). Thus

cd(7i, w) = 2k and cd(7;,7i;) = 2k. (4.6)

Suppose that d(u,w) = 6. Then cd(u, M) = f^1] = 2fc by (4.6). However, f"^1] is one of ^, f and ^. If \^\ = 2^1, then 26- 1 = 6fc, which is impossible; if [2^1] = |, then 2b = 6k and 6 - 3fc; if [^i] = 2^1, then 26 + 1 = 6fc, which is impossible. Therefore, d(u,w) = b = 3k. Similarly, d(v,w) = b = 3k. However then, by the triangle inequality, d(u,v) < d(u,w) + d(v,w) < 3k + 3k — 6fc, which contradicts (4.5). Therefore, as claimed, there is no connected graph G with crad(G) = r and cdiam(G) = d such that r is even and d = 2r + 1. m

4.3 Color-Center and Color-Periphery

The subgraph of a connected graph G induced by its color-central vertices is the color-center CC(G) of G and the subgraph induced by the color-peripheral vertices of G is the color-periphery of G and is denoted by CP(G).

In an observation first made by Hedetniemi, there is no restriction on which graphs can be the center of some graph.

Theorem 4.3.1 [2] Every graph is the center of some connected graph.

There is an analogous result for color-centers, the proof of which uses the same construction as employed by Hedetniemi. 98

Theorem 4.3.2 Every graph is the color center of some connected graph

Proof. Let G be a graph We construct a graph H from G by first adding two new vertices u and v to G and joining them to every vertex of G but not to each other The construction of H is completed by adding two additional vertices u\ and vi and the two new edges U\U and V\V (This construction is illustrated in

Figure 4 3 ) Since ce(?i) = ce(7;) = 2, ce(?ii) = ce(i?i) = 3 and ce(r) = 1 for every vertex r in G, it follows that V(G) is the set of the color central vertices of H and so CC(H) =G m

Figure 4 3 A graph with a given color-center

If every vertex of G is a color-central vertex, then CC(G) = G and G is self

color centered Thus, if G is self-color-centered, then crad(G) = cdiam(G)

Theorem 4.3.3 Let G be a connected graph with rad(G) = a and diam(G) = 6

Then G is self color-centered if and only if (?) b — a or (??) b = a + 1 and a = 1

(mod 3)

Proof. First, suppose that a and 6 satisfy (I) or (n) By Lemma 4 2 3, crad(G) =

P^i] and cdiam(G) = \^} If a = 6, then \**=±] = \&=±] If 6 = a + 1 and

a = 1 (mod 3), then let a = 3k + 1 for some integer k Then [2^1] = f^L-I] -

2k + 1 Thus G is self-color centered 99

For the converse, suppose that neither (1) nor (n) holds We consider two cases

Case 1 b = a + 1 and a ^ 1 (mod 3) Thus a — 3k + 1 where ? = 0 or

? = 2 If ? - 0, then [^i] = 2k and [^] = p*±l] = 2k + 1, if i = 2, then [^_2i] =2k + l and \&f±] - [^±1] = 2£- + 2 Thus crad(G) < cdiam(G)

CW2 b>a + 2 Then [26dL] > T^MJ = [^±3] = |y|+l> P^]+l and so crad(G) < cdiam(G)

In either case, G is not self-color centered •

Let v be any vertex in the center of G Then by Lemma 4 2 3, Ecc(7;) C CEcc(7;)

Hence, it follows that Cen(G) C CC(G) for every nontrivial connected graph G

The graph H constructed in the proof of Theorem 4 3 2 has the property that

Cen(i7) = CC(H) = G This gives rise to the following question

Problem 4.3.4 Suppose that F andG are two graphs where F is a proper induced subgraph of G Under what conditions does there exist a connected graph H such that Cen(H) = F and CC(H) = G?

Although the answer to Problem 4 3 4 is not known, we present two necessary conditions for such a connected graph H

Proposition 4.3.5 Suppose that F and G are two graphs where F is a proper induced subgraph of G If R is a connected graph such that Cen(H) = F and

CC(H) = G, then H must satisfy the following two conditions 100

(i) There is a positive integer a with a = 1 (mod 3) such that e#(.x) = a for all

x G V(F), eH(y) = a + 1 for all y G V(G) - V(F), and eH(z) >a + 2 for all z G V(H) - V(G)

(ii) No vertex in F is adjacent to any vertex in V(H) — V(G).

Proof. We first verify (i). Since Cen(H) = F, there is a positive integer a such that eH(x) = a for all x G V(F). Let y G V(G) - V(F). Then eH(y) = b for some integer b > a + 1. Since CC(H) = G, there is a positive integer r such that ceH(x) = r for all x G V(G). Since r = \*=±] = [2^1], it follows that a = 1

(mod 3) and b = a + 1. Thus eH(y) = a + 1 for all y G V(G) - V(F). If there is z G V(H) — V(G) such that a < e#(z) < a + 1, then ce#(z) = r. On the other hand, ceH(z) > r + 1 for all z G V^i?) — V'(G), which is impossible.

We next verify (ii). If there is x G V(F) such that x is adjacent to some vertex z in V(H) - V(G), then |e#(.T) - eH(z)\ < 1. Thus a < eH(z) < a + 1 and so cejff (z) = r, which is impossible. •

By Proposition 4.3.5, if F = Pk and G = Pfc+i, then there is no connected graph 77 such that Cen(H) = F and CC(H) = G. To see this, assume that there is a connected graph H such that Cen(i7) = Pk and CC(H) = P*;+i- Let

G = PM = (vi, 7;2,..., vfc+i) and P = Ph = Pfc+i - vfc+1. By Proposition 4.3.5(h), no vertex 7^ (1 eH(vk) for each i with 1 < i < k — 1, which contradicts Proposition 4.3.5(i). 101

Suppose that P and G are two graphs where P is a proper induced subgraph of G. The F-augmenting graph G of G is obtained from G by adding two new vertices x and y and joining each of x and y to every vertex of P.

Theorem 4.3.6 If F and G are two graphs where F is a proper induced subgraph

of G, then there is a connected graph H such that Cen(H) = F and CC(H) = G.

Proof. Let U = V(G) — V(F) = {ui,u2,..., up}. We begin with two copies of

Kp with vertex sets V = {vi, 7;2,..., vp} and W — {wi, 7i;2,..., 7i;p}, respectively.

Let Qi = (.Xi, .T2, .T3) and Q2 = (7/1, y2, y3) be two copies of the path P3. The graph P is constructed by (1) joining x to each vertex in V(F) U 1/ U {#i}, (2) joining y to each vertex in V(F) UWU {7/1} and (3) joining Ui to vt and ^ for 1 < i < p. (See Figure 4.4).

2/1 2/2 2/3 o o o

V U W

Figure 4.4: A graph H with given center and color-center

Since eH(v) = 4 if 7; G V(P), eH(v) = 5 if v G C/ U {x,y}, eH(v) = 6 if

7; G V U V^ and e^(^) = eH(yl) = 5 + i for 1 < i < 3, it follows that Cen(P) = P. 102

Furthermore, ceH(v) = 3 if v G V(G), ce#(7;) = 4 if v G V U VK, ce^(^) = ceH(yi) = 3 + i for 2 = 1, 2 and ce#(.x3) = ceH(y3) — 5. Thus CC(P) = G. •

Bielak and Syslo [1] verified that not every graph is the periphery of some graph.

Theorem 4.3.7 A nontrivial graph G is the periphery of some graph if and only if every vertex of G has eccentricity 1 or no vertex of G has eccentricity 1.

We now present a necessary and sufficient condition for a graph to be the color-periphery of some graph.

Theorem 4.3.8 A nontrivial graph G is the color-periphery of some graph if and only if every vertex of G has color-eccentricity 1 or no vertex of G has color- eccentricity 1.

Proof. First, suppose that every vertex of G has color-eccentricity 1. Then

G2 is complete and CP(G) = G. Next, assume that no vertex of G has color- eccentricity 1. Let V(G) = {ui,u2)... ,7in} and let F — Kn with V(F) —

{7;!,7;2,... ,7;n}. The graph H is obtained from G and P by joining Vi to Ui for 1 < i < n. Since ce(7;^) = 1 for 1 < i < n and ce(7^) — 2 for 1 < i < n. Thus

CP(P) = G.

For the converse, suppose that there is a graph G for which some but not all vertices have color-eccentricity 1 and G is the color-periphery of a connected graph

H. Then G is a proper induced subgraph of H and cdiam(P) > 2. Let u be a vertex 103 of G having color eccentricity 1 in G, that is, cea(u) = 1. Then cdc(u, w) — 1 for all w G V(G) — {u}. Consequently, u is adjacent in G2 to all vertices w G V(G) — {u}.

Let v be a vertex of H such that ceH(u) = cdH(u, v) = cdiam(P) > 2. Therefore, ceH(v) = cdiam(P) > 2 and so v is a color peripheral vertex of H. On the other

2 hand, since cdH(v,u) > 2, the vertex v is not adjacent to u in P and so v is not a vertex of G, which is a contradiction. •

4.4 Color Connectivities of Graphs

By Theorem 4.1.3, the distance-colored graph G2 is color-connected and so for every two vertices u and v of G2, there is at least one properly colored u — v path in G2. This gives rise to another concept. If G is a connected graph with connectivity K(G) = ft, then it follows from a well-known theorem of Whitney [31] that for every two distinct vertices u and v of G, the graph G contains K internally disjoint u — v paths. A graph G is p-connected if K(G) > p. For a connected graph G and an integer fc > 2, the distance-colored graph Gk is properly colored p-connected or simply properly p-connected if for every two distinct vertices u and v of Gfc, there are p internally disjoint properly colored u — v paths in Gk. If

G is a complete graph, then Gk is not properly p-connected for all integers fc,p with fc > 1 and p > 2. Thus we consider only non-complete connected graphs.

For a connected graph G and an integer fc > 2, the color-connectivity of Gk is the maximum positive integer p for which Gk is properly p-connected. In Section 4.4.1, we study the color-connectivities of the square and the cube of a connected graph, 104 while in Section 4.4.2 we determine all pairs fc, n of positive integers for which Pk is properly fc-connected.

4.4.1 Properly 2-Connected Graphs

For a connected graph G of order 3 or more, it is well known that G2 is 2-connected and so every two distinct vertices u and v of G are connected by two internally disjoint u — v paths in G2. By Theorem 4.1.3, the distance-colored graph G2 is color-connected. However, G2 need not be properly 2-connected, that is, it is possible that for some pair u,v of vertices of G2, two internally disjoint properly

2 colored u — v paths do not exist in G . For example, if G = Pi,n-i where n > 3,

2 then G = Kn is 2-connected but not properly 2-connected (as any two end-vertices of G are not connected by two internally disjoint properly colored paths in G2).

On the other hand, if G is 2-connected, then G2 is properly 2-connected.

Theorem 4.4.1 // G is a 2-connected graph that is not complete, then G2 is properly 2-connected.

Proof. Let u and v be two distinct vertices of G. Since G is 2-connected, G contains at least two internally disjoint u — v paths. Hence, u and v lie on a common cycle. Among all cycles in G that contain u and 7;, let G be one of shortest length. Construct two internally disjoint u — v paths in G, P and P', by traversing G in opposite directions. Say P = (u — .XI,.T2, ..., xs = 7;) and

P' = (u — 7/!, y2,..., yt = v). By the defining property of P and P', it follows that dP(xu Xi+2) = 2 = dG(xu xi+2) for 1 (yh yH2) = 2 = dG(%, yj+2) 105 for 1 < j < t — 2. By Lemma 4.1.2, there is a properly colored u — v path Q in the square of P and a properly colored u — v path Qr in the square of P'. Since P and

P' are internally disjoint, so are Q and Qf. Therefore, G2 is properly 2-connected. •

For a connected graph G, its square can be properly 2-connected without G be ing 2-connected, however. In the case of trees, for example, we know precisely those trees whose square is properly 2-connected. A double star is a tree of diameter 3.

The double stars are the only trees T for which T2 is properly 2-connected.

Theorem 4.4.2 Let T be a tree of order at least 3. Then T2 is properly 2-

connected if and only ifT is a double star.

Proof. First, suppose that T is a double star whose central vertices are x and y.

Let X = {.Ti, .x2,..., xr} and Y = {yi, y2, ..., ys} be the sets of end-vertices of T such that x is adjacent to every vertex in X and y is adjacent to every vertex in y, where then r, s > 1. Let 7i, v G V(T). We show that u and v are connected by two internally disjoint properly colored paths. If {u,v} = {.x,y}, say u — x and v — y, then (72,7;) and (u^x^v) are two internally disjoint properly colored u — v paths in T2. If {72,7;} n {.x, y} — 0, then we may assume that u G X. If v G X, say v — .x2, then (72,7;) and (72, .x,yl5 y,7;) are two internally disjoint properly colored

2 u — v paths in T . If 7; G Y\ say 7; = yl5 then (7i, y, 7;) and (71, .x, 7;) are two internally disjoint properly colored u — v paths in T2. If |{7i, 7;} n {.x,y}| = 1, then we may assume that 7i = x and 7; 7^ y. If 7; G X, say v = .Xi, then (u,v) and (u,y,v) are

2 two internally disjoint properly colored 7i — v paths in T . If v G V, say v = yl5 106 then (u, 7;) and (71, .xi, y, 7;) are two internally disjoint properly colored u — v paths inT2.

For the converse, assume that T is not a double star. Let d = diam(T) and so d 7^ 3. If d = 2, then T is a star and we saw that T2 is not properly 2-connected.

Thus we may assume that d > 4. Let u be an end-vertex of T whose eccentricity er(u) is d and let v be a vertex of T such that dT(7i, 7;) = 4. We show that there are no two internally disjoint properly colored u — v paths in T2. Suppose that

2 (72,7;i,7;2,7;3,7;) is the u — v path in T. Assume to the contrary that T contains two properly colored u — v paths, Pi and P2. By an extensive case-by-case analysis, it can be shown that each properly colored u — v path in T2 must use at least two vertices in {y\, 7;2,7;3} Hence, P\ and P2 must each contain at least two vertices from {7;i,7;2,7;3}. But this is impossible since we assumed that Pi and P2 were internally disjoint. •

We have seen that if G is a connected graph, then G2 is 2-connected. Since

(G2)2 = G4 for each connected graph G, the following is an immediate consequence of Theorem 4.4.1.

Corollary 4.4.3 If G is a connected graph such that G2 is not complete, then G4 is properly 2-connected.

By Corollary 4.4.3, if G is a connected graph of diameter at least 3, then G4 is properly 2-connected. This gives rise to a natural question: For a connected graph G of diameter at least 3, what is the minimum fc such that Gk is properly 107

2-connected? By Theorem 4.4.2 and Corollary 4.4.3, either fc = 3 or fc = 4. Next, we show that fc = 3.

Theorem 4.4.4 If G is a connected graph of diameter at least 3, then G3 is properly 2-connected.

Proof. Since diam(G) > 3, it follows that the order of G is at least 4. Let u and v be two distinct vertices of G. We show that u and v are connected by two internally disjoint properly colored paths in G3. We consider two cases, according to whether 1 < dG(u, v) < 2 or dG{u, v) > 3.

Case 1. 1 < dG(u,v) < 2. Suppose that u has a neighbor x(^ v) that is not a neighbor of v. Then (?z, 7;) and (?x, .x, 7;) are two internally disjoint properly colored u — v paths in G3. Similarly, G3 contains two internally disjoint u — v paths in v has a neighbor x(^ u) that is not a neighbor of u. We may therefore assume that NG(u) — v — NG(v) — u. Since diam(G) > 3, it follows that rad(G) > 2 and

^G(V) > 2 for all y G V(G). Ii eG(u) > 3 or eG(v) > 3, say the former, then there is w G V(G) such that dG(u, w) — 3. Let (72, .xi, .x2, w) be a u — w geodesic in G. Since

NQ(U) —v = NG(v) — 7i, it follows that v is adjacent to x\ (and v is not adjacent to

.x2). Then (72,7;) and (u, w, x\,v) are two internally disjoint properly colored u — v paths in G3. Thus we may assume that ec(u) = ec(v) = 2. Then rad(G) = 2 and

3 < diam(G) < 4. Since NG(u) —v = NG(v) — u and eG(u) = eG(v) = 2, it follows that G contains a subgraph isomorphic to one of the four graphs in Figure 4.5 such that d(w,wf) = 3. A dashed line between u and v in Figure 4.5 indicates that u 108 and v are not adjacent. Thus neither u nor v is adjacent to u/. Then (71,7;) and

(71, 7i;, 7i/, 7;) are two internally disjoint properly colored u — v paths in G3.

O w'

Ow'

Figure 4.5: A step in the proof of Theorem 4.4.4

Case 2. dG(u, v) > 3. Let d — dG(u, v) and let P =(u = 7;0, t>i, 7;2, ..., vd = v) be a 71 — 7; geodesic in G. Thus dP(x,y) = dG(x,y) for all .x,y G V(P). If d = 3, then (71,7;) and (71,7;2,7;) are two internally disjoint properly colored u — v paths in G3. If d = 4, then (71,771,7;) and (71,7^,7;) are two internally disjoint properly colored u — v paths in G3. Thus, we may assume that d > 5. Suppose that d = i

(mod 4) where i = 0,1,2, 3 and let d — 4k + i for some positive integer fc. In each

3 case, G contains two internally disjoint properly colored 71 — v paths Qi and Q2 as follows: For d — 4k,

Qi = (71,7;i, 7;4,7;5,7;8,7;9,..., 7;4/c_3, v4k)

Q2 = (71,7;2,7;3,7;6,7;7, vw,..., 7;4fc_2, v4k). 109

For d = 4fc + l,

Qi = 71,7;i, 7;4,7;5,7;8,7;9,..., 7;4/c, v4k+i)

Q2 = 71,7;2,7;3,7;6,7;7, v10,..., 7;4fc_i, 7;4fc+i).

For d = 4fc + 2,

Qi = 71, vu 7;4,7;5,7;8,7;9,..., 7;4/c, 7;4/c+2)

g2 = 71,7;2,7;3,7;6,7;7, v10,..., v4k_i, v4k+2).

For d = 4fc + 3,

Qi = 7i, 7;x, 7;4,7;5,7;8,7;9,..., v±k+\, v±k+$)

Q2 = 71,7;2,7;3,7;6,7;7,7;i0,..., 7;4/c+2,7;4/c+3).

Therefore, G3 is properly 2-connected. •

Theorem 4.4.4 brings up another question: For a connected graph G of diameter at least 3, is G3 properly 3-connected? We will see in the next section that this is not true in general.

4.4.2 Color-Connectivities of the Powers of a Path

We have seen in the proofs of Theorems 4.4.1 and 4.4.4 that the color-connectivity

k of the distance-colored graph P of a path Pn of order n plays an important role in determining the color-connectivity of a connected graph. Thus in this section, we investigate the color-connectivities of the distance-colored graph Pk for integers n and fc with n > fc + 1 > 3. By Theorem 4.4.2, the distance-colored graph P2 110 is properly 2-connected if and only if n = 4. By Theorem 4.4.4, P3 is properly

2-connected for all n > 4. However, P3 is not properly 3-connected in general. In fact, for n > 4, it can be verified that the distance-colored graph P3 is properly

3-connected if and only if n = 6. Next, we determine all pairs fc, n of integers with n > fc + 1 > 4 for which Pk is properly fc-connected. In order to do this, we first present two lemmas.

Lemma 4.4.5 For each even integer k > 2, the distance-colored graph Pk+2 is properly k-connected.

2 Proof. We proceed by induction on even integers fc > 2. By Theorem 4.4.2, P4 is properly 2-connected and so the statement holds for fc = 2. Suppose that P^+2 is properly fc-connected for some even integer fc > 2. Let P/c+4 = (7;0, t>i, 7;2,..., 1^+3) and let 71 and v be two distinct vertices of P/c+4. We show that there are fc + 2 internally disjoint properly colored 71 — 7; paths in Pk+l* We consider two cases.

Case 1. 71 and v are not end-vertices of G. Then 71,7; G \y\, v2,..., 7;^+2}. Let

Pk+2 = (i?1? 7;2,..., 7;/c+2). Since Pk+2 is properly fc-connected by the induction hy pothesis, there are fc internally disjoint properly colored 71 — 7; paths Qi, Q2)..., Qk in P£+2. Let Qk+1 = (71,7;0,7;) and Qk+2 = (u, vk+s, v). Therefore, Qu Q2,..., Qk+2 are fc + 2 internally disjoint properly colored 71 — v paths in Pk+l.

Case 2. At least one of u and v is an end-vertex of G, say u = VQ. If v = vk+$, then let Qi = (v0, vt, 7;fc+3) for 1 < i < fc + 2, producing fc + 2 internally disjoint properly colored v0 — vk+$ paths in Pk+%. Thus, we may assume that 7; ^ vk+3. Ii Ill

v — v2£+i for some nonnegative integer £, where 1<2£ + I

(vo,Vi,V2£+i) if 1

If v = v2£ for some positive integer £ where 2 < 2£ < fc + 2, then construct fc + 2 internally disjoint properly colored VQ — v2£ paths Q1? Q2,..., Qfc+2 in P/f+4 as follows: ' fa), vi. ^) if 1 < i < fc + 2 and ?; 7^, 2^

Qi = \ (vo,v2e) Hi = 2£ < (v0,ve,vk+3,V2£) Hi = L

Therefore, P%+% is properly (fc + 2)-connected. •

Lemma 4.4.6 For each odd integer k > 3, the distance-colored graph P^+3 is properly k-connected.

Proof. We proceed by induction on odd integers fc > 3. Figure 4.6 shows that

P| is properly 3-connected.

Assume that Pk+3 is properly fc-connected for some odd integer fc > 3. We show that Pk+l is properly (fc + 2)-connected. Let Pfc+5 =(t>o, ^i, • • •, ^fc+4) and let

71 and 7; be two distinct vertices of Pk+$. We may assume that 71 = vs and v — vt where 0 < s < t < k + 4. We consider two cases.

Case 1. u and v are not end-vertices of Pfc+5. Then 71,7; G {v\, v2) ..., Ufc+3}.

First, suppose that 71 7^ v\ and v / vk+3. By the induction hypothesis, there are 112

t^T^*^*

Figure 4.6: Showing that P| is properly 3-connected

fc internally disjoint properly colored u — v paths Qi, Q2,..., Qk whose vertices belong to {vx,v2,... ,vk+s}. Let QM = (u,vQ,v) and Qk+2 = (71, vk+4) v). Then

k 2 Qi5 Q2, • • •, Qfc+2 are fc + 2 internally disjoint properly colored 71 — v paths in P ^ and so Pk+$ is properly (fc + 2)-connected. Next, suppose that either 71 = v\ or 7; = 7^+3. We may assume, without loss of generality, that 71 G {7;i,7;2,... ,7;fc+2} and 7; = vk+$. Let dPk+5(u,v) = p. Suppose that 71 = 7;2, where 1 < i < fc + 2, and so p = fc + 3 — ?;. If p is odd, then define

= ( (7;,, 7;;, 7;fc+3) if j G {1, 2,..., fc + 2} - {?;} J \ K,^+3) if i = i. If p is even, then define

f (7;2,7;„7;,+3) if j G {1, 2,..., fc + 2} - {*, ^} Q, = < (7^,7^3) if j - i { (^,^+4,^+3) ifj^1^.

In each case, Qi, Q2,..., Qk±2 are fc + 2 internally disjoint properly colored 71 — 7; paths in P^2 and so Pk+2 is properly (fc + 2)-connected.

Case 2. At least one of u and v is an end-vertex of Pk+$, say u = v0. Observe that the neighborhood of v0 in Pk+l is {v\, 7;2,..., vk+2}. 113

• If v = vt and 1 < t < fc + 2, then define

f (v0,vuvt) if 7 G {1,2, , fc + 2}-{/,§} Qz = I {VQ, Vt) if 7 = f [ (^0,^,^+3,^) if 7 = I

• If 7; = 7;fc+3, then define Q2 = (v0, v2,7^+3) for 1 < ? < fc + 2 and ? 7^ ^ and

• If 7; = vk+4, then define Qi = (7;0,7;i,7;/c+3,7;fc+4) and Qz =(v0) vt, 7;fc+4) for 2

In each case, Qi, Q2, , Qk+2 are fc + 2 internally disjoint properly colored 71 — 7; paths in Pfc+52 and so P£+2 is properly (fc + 2) connected •

We are now prepared to determine all pairs fc, 77 of integers with 77 > fc + 1 > 4 for which Pk is properly fc-connected

Theorem 4.4.7 Let fc and 77 be integers inhere 77 > fc+1 > 4 Then Pk is properly k-connected if and only if n is even and either (?) fc is odd and fc + 3 < 77 < 2fc or

(11) fc is even and fc + 2 < 77 < 2fc

Proof. First, suppose that 77 is even and we show that Pk is properly fc-connected if either (1) fc is odd and fc + 3 < 77 < 2fc or (11) fc is even and fc + 2 < 77 < 2fc We consider two cases

Case 1 fc is odd and fc + 3 < 77 < 2fc We proceed by a finite induction on integers 77 to show that, for each odd integer fc > 3, if fc + 3 < 77 < 2fc, then Pk 114 is properly fc-connected. By Lemma 4.4.6, this statement is true for n — k + 3.

Assume, for each odd integer fc > 3, that Pk is properly fc-connected for an even

k integer 77 with fc + 3<77<2fc — 2. We show that P ±2 is properly fc-connected.

Let Pn+2 = (^0,^1, • •. ,?Vfi) and let 71 and 7; be two distinct vertices of Pn+2

First, suppose that dpn+2(u,v) < n. We may assume, with out loss of generality, that 71,7; G {7;0, t>i,..., 7;n_i}. By the induction hypothesis, there are fc internally

k disjoint properly colored 71 — v paths in P where Pn = (VQ, VI, ..., ?;n-i) and so in

k P +2 as well. Next, suppose that 77 < dpn+2(u,v) < 77 + 1. We may assume that

71 = v0. If 7; = 7;n, then define

(vo, v%, vl+k, vn) if 1 < 7 < r? - fc and ? 7^ |

Qz= i (v0, V%, vn) if 77 - fc < ? < fc and ? 7^ |

, (vQ,v^,vn+1,vn) if ? = f.

If 7; — 7;n+1, then define

(7;0, vt, 7;z+/c, 7;n+i) if 1 < ? < 77 - fc 2 ^ (vQ, VX, vn+i) if 71 - fc < ? < fc.

In each case, Qi, Q2,..., Qk are fc internally disjoint properly colored 71 — 7; paths

k k in P +2 and so P +2 is properly fc-connected.

Gasp 2. fc 75 e7;e77 and fc + 2 < 77 < 2fc. We proceed by a finite induction on integers 77 to show that, for each even integer fc > 4, if fc + 2 < 77 < 2fc, then Pk is properly fc-connected. By Lemma 4.4.5, this statement is true for 77 = fc + 2.

Assume for each even integer fc that Pk is properly fc-connected for an even integer

77 with fc + 2<77 < 2fc — 2. An argument similar to the one in Case 1 shows that

Pn+2 is properly fc-connected. 115

To verify the converse, we show that if Pk is properly fc-connected where n > fc + 1 > 4, then n is even and either (i) k is odd and fc + 3 < n < 2fc or (ii) k is even and fc + 2 < n < 2k. We consider two cases, according to n is odd or n is even. In the case when n is odd, we show that Pk is not properly fc-connected for all fc and n with fc > 2 and n > fc + 1. In the case when n is even, we show that either (i) k is odd and fc + 3 < n < 2fc or (ii) fc is even and fc + 2 < n < 2fc.

Case 1. n is odd. We show that Pk is not properly fc-connected for all fc and n with fc > 2 and n > fc + 1. Assume, to the contrary, that for some integer fc > 2 there is an odd integer n > k + 1 such that Pk is properly fc-connected.

By Theorem 4.4.2, fc > 3. Let n = 2£ + 1 for some integer £ > 2 and let Pn =

k (7;0,7;!,... ,7;2^). Since P is properly fc-connected, there are fc internally disjoint

k properly colored 7;0 — v2£ paths Qi, Q2,..., Qk in P . Since degpfc v0 = fc and v0 is adjacent to v\, v2,..., vk in P^, we may assume that 7;0 is adjacent to Vi in Q^ for 1 < i < fc. Since these fc paths are internally disjoint, this implies that Vi G V(Qi) must be adjacent to vk+i in Qi and so 7;2 G V^(Q2) must be adjacent to vk+2 in Q2.

Continuing in this manner, we obtain that v^ G V(Qi) must be adjacent to vk+i in

Qi for 1 < i < fc. However then, Q£ = (7;0,7;^, 7;2^) is not properly colored, which is a contradiction.

Case 2. n is even. In this case, for an odd integer fc > 3, we show that if n = fc + 1 or n > 2fc + 2, then Pk is not properly fc-connected; while for an even integer fc > 2, we show that if n > 2fc + 2, then Pk is not properly fc-connected.

Therefore, it suffices to show that if n = fc + 1 or n > 2fc + 2, then Pk is not 116 properly fc-connected. Assume, to the contrary, that Pk is properly fc-connected for some integers fc > 2 and 77, such that n = fc + lorn>2fc + l. We consider these two subcases.

Subcase 2.1. n = k + 1. Let Pk+\ = (VQ, VI, ..., vk). Since P^+1 is prop erly fc-connected, there are fc internally disjoint properly colored 7;0 — 7;2 paths

Qi? Q25 • • • ? Qk in Pfc+i- Since degpfc v0 = fc and 7;0 is adjacent to 7;i, 7;2,..., vk in Pfc+1, it follows that 7;0 is adjacent to 7;z (1 < i < fc) in exactly one of these fc paths. We may assume that 7;0 is adjacent to v% in Q% for 1 < i < fc. However then, Qi = (7;0,7;i, 7;2) is not properly colored, which is a contradiction.

k Subcase 2.2. 77, > 2fc + 2. Let Pn = (7;0,7;x,... ,7;n_i). Since P is prop erly fc-connected, there are fc internally disjoint properly colored v0 — v2k paths

Qi, Q2, • • •, Qk in P^. Since degpfc v0 = fc and v0 is adjacent to 7;l5 7;2,..., vk in P^, we may assume that v0 is adjacent to v% in Q2 for 1 < i < fc. Thus, vt G V(QZ) must be adjacent to vk+l in Q2 for 1 < 7 < fc. However then, 7^7;^ and vkv2k are adjacent edges in Qk) which is impossible. •

While it follows from Theorem 4.4.7 that the distance-colored graph Pk is not properly fc-connected whenever 77, > 2fc + 1 > 7, the following is believed to be true.

Conjecture 4.4.8 For each integer k > 3, the distance-colored graph Pk+1 is properly k-connected for all n > 2fc + 1. 117

Conjecture 4 4 8 is true for fc = 3, however, as we now verify First, we introduce some additional definitions and notation For a path Q of a connected graph H and v G V(H) — V(Q), let f (Q, 7;) denote the distance between v and the terminal vertex of Q For a set Q of paths in a connected graph P, let V(Q) be the set of vertices of H that belong to some path in Q For each v G V(H) — V(Q), define T(Q, v) = max{f (Q, 7;) Q G Q} That is, T(Q, v) is the maximum distance between v and the terminal vertex of a path in Q

For a connected graph G, let Q be a properly colored 71 — 7; path in Gk for some integer fc > 2 and let w be a vertex of Gk that does not belong to Q Then Q is

extendable to w if (Q, 7i>) is a properly colored 71 — w path in Gk In this case, we can extend Q to w in Gfc and the path (Q, it;) is then called the extension of Q to w in Gfc

Let Pn = (^0,^1, j^n-i) be a path of order 77 > 7 Next, we present an algorithm that produces three internally disjoint properly colored paths Qi, Q2 and

Qs in P^ with initial vertex 7;0 such that the distance between vn_i and the terminal vertex of each path Q% (1 < ? < 3) is at most 4 That is, if Q = {Q\, ^2,^3}, then

T(Q,vn_i)<4

Algorithm 1 For Pn = (^0, ^1, ,7;n_i) ?i;/iere 77 > 7, this algorithm produces

three internally disjoint properly colored paths Qi,Q2 and Q3 in P^ with initial

vertex 7;0 such that the distance between vn_i and the terminal vertex of each path

Qi (1 < 7 < 3) 79 at most 4

Input An integer 77 > 7 a77^ a path Pn = (?;0, v\, , 7;n_i) 118

Step 1 The first path begins at v0 and moves to Vi, the second path begins

at v0 and moves to v2 and the third path begins at v0 and moves to v4

At the end of Step 1, we obtain three paths

Q\ = (v0,vi), Q\ = {vQ,v2)} Ql = (vo,v4)

LetQ} = {QlQlQ\]

Step 2 IfT(Q\vn_i) < 4, then stop, while ifT(Q\vn_i) > 4, then do the following

1 l Let Ql G Q (1 < p < 3) such that t(Q p,vn_i) = T(Q\vn_i) Suppose

that j is the smallest integer such that v3 does not belong to any path in

1 2 Q and Q* is extendable to v3 in P^ Let Q — (Q*, v3) and rename the

x l f l remaining two paths Q q and Q ql (where q, q G {1, 2, 3} — {p}) ?77 Q as Q\ and Q\,, respectively Let Q2 = {QlQlQj}

2 2 If T(Q ,7;n_i) < 4, then stop, while if T(Q , vn_i) > 4, then repeat this procedure above In general, at the step ? for an integer ? > 1, let Q —

Step 1 + 1 IfT{Q\vn_l) < 4, then stop, while t/T(Q\t;n_i) > 4, then do the following

l Let Q;eQ'(l

that j is the smallest integer such that v3 does not belong to any path in 119

1 +1 Q and Qp is extendable to v3 in P* Let Qp = (Qp)Vj) and rename

l l the remaining two paths Qq and Q q, (where q) q' G {1, 2,3} — {p}) ??7 Q

1 l +1 l 1 as Q? and Q qt\ respectively Let Q* - {Q\+\ Q +\ Q^ }

Output Three internally disjoint properly colored paths Qi, Q2 and Qj in P^ with

initial vertex 7;0 such that T({Qi, Q2, Q3}, 7;n_i) < 4

k For a properly colored path Q = (ri,r2, , xi) in G ) the distance sequence of Q is defined as d(Q) di) d2, , d^i where dx — dc(x%^ r2+1) for 1 < ? < ( — 1

Lemma 4.4.9 For each integer n > 7, let Pn = (7;0,7;i, ,7;n_i) Then the

distance colored graph P% contains three internally disjoint properly colored v0 — vn_i paths

Proof. First, suppose that 7 < 77 < 9

• For 77 = 7, let Qx = (v0,vi,v3,vt), Q2 = (v0,v2,vb,vt) and Q3 = (v0,7;4,7;6),

• For 77 = 8, let Q1 = (v0,vuv3,v7), Q2 = (v0,v2)vb,v7) and Qs = (vQ, 7;4,7;7),

• For 77 = 9, let Q1 = (v0, vu 7;3,7;7,7;8), Q2 = (7;0,7;2, vb, v$) and Q3 = (v0,7;4,7;5,7;8) We now assume that 77 > 10 By Algorithm 1, we obtain three internally disjoint properly colored paths Q1? Q2 and Qs in P^ for an arbitrarily large integer 77 such that T({Qi, Q2, Q3}, 7;n_i) < 4 and the distance sequences of these three paths are

d(Qi) 1,2,3,2,4,3,4,3,2,4,3,4,3,2,

d(Q2) 2,3,4,1,3,4,1,3,4,1,3,4,1,3,4, 120

Let

si : 4, 3, 4,3, 2, s2 : 1,3,4,1, 3, 4 and s3 : 4, 3, 2, 4,3. (4.7)

Then the three internally disjoint properly colored paths Qi, Q2 and Q3 obtained by Algorithm 1 have the distance sequences as follows: d(Qi) : 1,2,3,2,*!,*!,,...

d(Q2) : 2,3,4,s2,s2,...

d(03) : 4,3,53,53,...

Observe that the sum of integers in sz in (4.7) is 16 for 1 < i < 3. Let

Qi = (7;0,7;i,7;3,7;6,7;8), Q°2 = (v0,v2)v5)v9) and Q?3 = (v0,v4)v7).

Then Q\, Q2 and Q3 are internally disjoint properly colored paths in P^ for a

hl +1 sufficiently large integer n. For each integer i > 0, the three paths Q\ ) Q2 and

+1 Q3r are constructed from Q\, Q\ and Q3, respectively, as follows:

(1) the path Q1^1 is obtained from Q\ and the path

; ; Xz = (7;8+16x, ^8+16z+4, ? 8+16z+7) ^8+162+11, ? 8+162+145 ^8+16(2+1))

by identifying the the vertex 7;8+i62 in Q\ and X%, respectively.

(2) the path Q1^1 is obtained from Q\ and the path

y% = (t>9+i62> 7;9+i62+i, 7;9+i6z+4,7;9+i6z4.8,ug+i^+g,7;9+i62+i2,7;9+16(i+1))

by identifying the the vertex 7;9+i62 in Q2 and YL^ respectively.

(3) the path Q^1 is obtained from Q3 and the path

Zi = (l>7+i62, ^7+16x4-4? 7;7+i62+7, 7;7+i6z+9, ^7+16z+13? ^7+16(i+l))

by identifying the the vertex 7;7+i6z in Q\ and Z2, respectively. 121

Observe that Q\, Q2 and Q\ are internally disjoint properly colored paths in P^ for all i > 0 (where n is sufficiently large). We now verify the following claim.

Claim. For each n > 10, three internally disjoint properly colored v0 — 7;n_i

2 paths Qi, Q2 and Q3 can be constructed from the paths Q X, Q\ and Q\ for some integer i with i > f22^] by an appropriate modification.

Proof of Claim. The distance sequences of Q\, Q2 and Q3 are

d(Q°i) : 1, 2, 3, 2, d(Q°2) : 2, 3, 4, and d(Q§) : 4, 3.

In general, for each integer i > 0,

+1 +1 +1 d(Q\ ) : d(Q\), su d(Ql ) : d(Q\), s2, and d^ ) : d(Q\), s3

where sz are shown in (4.7) for 1 < i < 3. Since (i) the sum of integers in st is 16

1 for 1 0, it follows that, to verify the claim, it suffices to show that for each 77, with

10 < 77, < 25, three internally disjoint properly colored 7;0 — 7;n_i paths Qi, Q2 and

Q3 can be obtained from Q}, Q2 and Q3 in P^. This is verified by the following table, where a path (vn, 7;22,..., vls) is denoted by (i±, i2)..., is) 122

n = 10 Qi (0, 1, 3, 7, 9) n = 18 Qi (0, 1, 3 6, 8, 12, 15, 17) Q2 (0, 2, 6, 9) Q2 (0, 2, 5 9, 10, 13, 17) Qs (0, 4, 5, 9) Qs (0, 4, 7 11, 14, 16, 17) n = ll Qi (0, 1, 3, 6, 10) n = 19 Qi (0, 1, 3 6, 8, 12, 15, 16, 18)

Q2 (0, 2, 5, 9, 10) Q2 (0, 2, 5 9, 10, 13, 17, 18) Qs (0, 4, 7, 8, 10) Qs (0, 4, 7 11, 14, 18 ) n - 12 Qi (0, 1, 3, 6, 8, 11) n = 20 Qi (0, 1, 3 6, 8, 12, 15, 19) Q2 (0, 2, 5, 9, 11) Q2 (0, 2, 5 9, 10, 13, 17, 19) Qs (0, 4, 7, 11) Qs (0, 4, 7 11, 14, 16, 19) n = 13 Qi (0, 1, 3, 6, 8, 12) n = 21 Qi (0, 1, 3 6, 8, 12, 15, 19, 20)

Q2 (0, 2, 5, 9, 12) Q2 (0, 2, 5 9, 10, 13, 17, 20) Qs (0, 4, 7, 11, 12) Qs (0, 4, 7 11, 14, 16, 20) n - 14 Qi (0, 1, 3, 6, 8, 12, 13) n - 22 Qi (0, 1, 3 6, 8, 12, 15, 19, 21) Q2 (0, 2, 5, 9, 10, 13) Q2 (0, 2, 5 9, 10, 13, 17, 18, 21) Qs (0, 4, 7, 11, 13) Qs (0, 4, 7 11, 14, 16, 20, 21) n - 15 Qi (0, 1, 3, 6, 8, 12, 14) n = 23 Qi (0, 1, 3 6, 8, 12, 15, 19, 22) Q2 (0, 2, 5, 9, 10, 13, 14) Q2 (0, 2, 5 9, 10, 13, 17, 18, 21, 22) Qs (0, 4, 7, 11, 14) Qs (0, 4, 7 11, 14, 16, 20, 22) n = 16 Qi (0, 1, 3, 6, 8, 12, 15) n = 24 Qi (0, 1, 3 6, 8, 12, 15, 19, 22, 23) Q2 (0, 2, 5, 9, 10, 13, 15) Q2 (0, 2, 5 9, 10, 13, 17, 18, 21, 23) Qs (0, 4, 7, 11, 14, 15) Qs (0, 4, 7 11, 14, 16, 20, 23) n = 17 Qi (0, 1, 3, 6, 8, 12, 15, 16) n = 25 Qi (0, 1, 3 6, 8, 12, 15, 19, 21, 24) Q2 (0, 2, 5, 9, 10, 14, 16) Q2 (0, 2, 5 9, 10, 13, 17, 18, 22, 24) Qs (0, 4, 7, 11, 13, 16) Qs (0, 4, 7 11, 14, 16, 20, 23, 24)

The result then follows from the claim. •

We are now prepared to show that P^ is properly 3-connected for each n > 7.

Theorem 4.4.10 For each integer n > 7, the distance-colored graph P% is prop

erly 3-connected.

Proof. Let Pn = (t?o,^i, • • • ,^n-i)- We proceed by induction on n. For n = 7, it is straightforward to verify that P^ is properly 3-connected. Assume that P^

is properly 3-connected for some integer k > 7. Now let u and v be two distinct vertices of P£+v First, suppose that {71,7;} ^ {^o, ^/c}- We may assume, without

loss of generality, that Vk £ {71,7;}. Let Pk = Pfc+i — vk- Then 71,7; G V(Pk) and 123

dpk (71,7;) = dPk+l (71,7;) By the induction hypothesis, P^ contains three internally disjoint properly colored 71—7; paths and these three paths are also properly colored

paths in P^+1 Next, suppose that {71,7;} = {7;0,v^} It then follows by Lemma 4 4 9

that P/f+1 contains three internally disjoint properly colored VQ — Vk paths •

4.5 On iJ-Colored and i7-Chromatic Graphs

In Section 4 4, we investigated connected graphs G of diam(G) — d for which Gh is

properly p-connected for some integers /r, p > 2 If Gh is properly p-connected with

k p < k < d, then G contains a properly colored subdivision of K2p as a subgraph In fact, there is no restriction on what properly colored subgraphs that Gk can

possess

Theorem 4.5.1 For every connected graph H, there exists a connected graph G

and a positive integer k such that the distance-colored graph Gk contains a properly

colored copy of H using x'(P) colors

Proof. Since the result is obvious if H — K2, we may assume that H has order at least 3 Let x'(P) — X > 2 and let c be a proper X-edge coloring of H using

the colors 1,2,. , x We consider two cases

Case 1 x ~ 2 For each e G E(H) such that c(e) = 2, subdivide the edge e

exactly once Denote the resulting graph by G Then the distance-colored graph

G2 contains a copy of H and a proper 2-edge coloring of H using colors 1 and 2 124

Case 2. x > 3. Let M = x — 3. For each e G E(H), subdivide the edge e a total of M + c(e) — 1 times, resulting in a path of length M + c(e). Denote the resulting graph by G. Suppose that e joins 71 and v in P. Since there is a path of length M + c(e) connecting 71 and v in G, it follows that dc(u, v) < M + c(e). We claim that dc(u,v) = M + c(e). Suppose that dc(u,v) < M + c(e). Then there exists a 7i —7; path P in G of length less than M + c(e). From the way in which G is constructed, the length of such a path must be at least (M +p) + (M + q) for some positive integers p and q such that p + q > 3. Since (M+p) + (M + q) > 2M + 3, it follows that 2M + 3 < M + c(e) < M + x and so M < x — 3, which is impossible.

Thus GM+X contains a copy of H and a proper x~edge coloring of H using colors

M+l,M + 2,...,M + x. •

Theorem 4.5.1 raises the question of determining, for a given connected graph

P, the smallest order of a connected graph G such that there is a positive integer k for which the distance-colored graph Gk contains a properly colored copy of H and a proper edge coloring of H using ^(H) colors. Let H be a given connected graph. A connected graph G is H-colored if there is a positive integer k such that the distance-colored graph Gk contains a properly edge-colored copy of H with

X'(P) colors. If the properly edge-colored copy of P in an P-colored graph G is produced by a x'(P)-edge coloring c, then G is an H-colored graph with respect to c. The minimum order of such a graph G is called the H-color-order of G or the color-order of P, denoted by co(P). An P-colored graph of order co(P) is called an H-chromatic graph. Thus if H is a nontrivial connected graph of order n and 125

G is an P-colored graph of order 77', then 77 < co(P) < 77'

Let H be a nontrivial connected graph of size ra with E(H) ={ei, e2, , em} and x'(H) — X F°r a X_e(lge coloring r of P, define the

m a(c) = £(r(e.) - 1)

2=1

Let C(P) be the set of all x colorings of H The minimum a-number (or simply a-number) of P is defined by

a(P) = mm {a(c) ceC(H)}

First, we establish bounds for the color order of a connected graph in terms of its order, size, chromatic index and minimum cr-number

Theorem 4.5.2 If H is a nontrivial connected graph of order n with x'(P) > 3,

then

n + a(H) < co(P) < n + m(^(H) - 3) + a(H) (4 8)

Proof. The upper bound for co(P) is a consequence of the proof of Theo rem 4 5 1, in which the P-colored graph has order 77 + m(x'(H) — 3) + cr(H)

if we choose a x-edge coloring whose

the lower bound Let c be a x_e(lge coloring of P using the colors 1,2, , x and let

G be an P-colored graph with respect to c Thus there is a positive integer k such

that the distance-colored Gk contains a properly edge-colored copy of P produced

by the coloring c Let P(P) = {^1,^2, ,<°m} For each ? with 1 < 7 < ra, let 126

ex = x%yx and c(el) = c%. Since dc(xl)yl) — cu there is an xz — y% geodesic Q in

k G . Suppose that Q = (xz = v0, v\,..., vc% = yz). Since ez is an edge of P, each of the vertices vu v2,..., 7;Cz_i belongs to V(G) — V(H). This implies that each edge ex (1 n + °(H), giving the desired result. •

Next, we describe a class of connected graphs H for which co(P) = n + cr(H).

The girth g(H) of a graph H having a cycle is the length of a smallest cycle in P.

Proposition 4.5.3 Let H be a nontrivial connected graph of order n. If H is a

tree or x'(P) < L3(.9(P) ~ 1)/2J, then co(H) =n + a(H).

Proof. Suppose that x!(H) = x« Let c be a x_edge coloring of P using the colors

1, 2,..., x such that a(c) = cr(H). For each e G E(H), subdivide the edge e a total of c(e) — 1 times, resulting in a path of length c(e). Denote the resulting graph by

G. Thus the order of G is n + cr(c). It remains to show that G is P-colored with respect to the coloring c. Suppose that e joins 71 and v in P. Since there is a 7i — v path of length c(e) connecting 7i and v in G, it follows that dc(u,v) < c(e). We claim that dc(u, v) = c(e).

First, suppose that P is a tree. Since G is a subdivision of a tree, G is a tree

as well. Thus the 71 — v path of length c(e) is the only 7i — v path in G. Hence dc(u,v) — c(e) if P is a tree. Next, suppose that P is not a tree. Assume, to the contrary, that dc(u,v) < c(e). Then there exists a 7i — v path Q in G of

length less than c(e). Let P be a 7i — v path in P. Since the girth of P is g 127 and 717; G E(H), it follows that the length of P is either 1 or is at least g — 1. Ii the length of P is 1, then P = (71,7;) and P gives rise to a 71 — 7; path of length c(e) in G. Thus, we may assume that the length of P is £ and so £ > g — 1. Let

P = (71 = .x0, .TI, .T2, ..., xi = v). Since P is properly colored by c, it follows from the way in which G is constructed that P gives rise to a 71 — v path in G whose length is at least

[5-1] .9-1 = 3(.9 - 1) Y^c(xiXi+1) > + 2 2 2=0 L 2 J [ 2 J

This implies that the length of Q is at least |_3(.g — 1)/2J. However then,

3(.9-l) c(e) > dG(u,v) > > X'(H) > c(e), which is a contradiction. Therefore, Gx contains a properly edge-colored copy of

H using x colors and G is P-colored. •

The following is an immediate consequence of Theorem 4.5.2 and Proposi tion 4.5.3.

Corollary 4.5.4 If H is a nontrivial connected graph of order n such that x'(P) =

2 or x'(P) - 3; then co(H) =n + a(H). Chapter 5 Rainbow Connected Graphs

5.1 Introduction

As we described in Chapter 1, graph labelings have grown in popularity in recent decades because of their applications to many other areas of research in graph the ory. The origin of the study of graph labelings as a major area of graph theory can be traced to a 1967 research paper by Alexander Rosa [30]. Among the labelings he introduced was a vertex labeling he referred to as a /3-valuation. Let G be a graph of order n and size 772. A one-to-one function / : V(G) —> {0,1, 2,..., ra} is called a ^-valuation (or a ^-labeling) of G if (|/(7i) — /(^)| : uv G E(G)} = {1,2,..., ra}.

In order for a graph to possess a /3-labeling, it is necessary that ra > n — 1. In 1972,

Golomb [?] referred to a /3-labeling as a graceful labeling and a graph possessing a graceful labeling as a graceful graph. Eventually, it was this terminology that became standard. Gallian [15] has written a survey on labelings of graphs that includes an extensive discussion of graceful labelings and their applications. One of the major conjectures in graph theory concerns graceful graphs and is due to

128 129

Kotzig (see Rosa [30]) which is known as the Graceful Tree Conjecture

The Graceful Tree Conjecture Every tree is graceful

The following is a consequence of Rosa's work [30]

Theorem 5.1.1 If G is a graceful graph of size m, then the complete graph K2m+i is cyclically G decomposable

There is a connection between distance-colored graphs and graceful trees and it can be made even more specific in the following setting For a nontrivial connected graph G and a positive integer /r, a subgraph P in the distance colored graph Gk is a rainbow subgraph if no two edges of P are colored the same First, we show that every connected graph P can be a rainbow subgraph for some distance-colored graph

Theorem 5.1.2 For every connected graph H, there exists a connected graph G

and a positive integer k such that the distance-colored graph Gk contains a rainbow

Proof. Since the result is obvious if H = K2, we may assume that P has order

e at least 3 Suppose that V(H) = {7;i,7;2, ,7;n} and E(H) = {ei,e2, > m}

So 77 > 3 and ra > 2 For ? = 1,2, ,ra, subdivide the edge e% a total of ra — 3 + 7 times, resulting in a path of length ra — 2 + i Denote the resulting graph by G Suppose that the edge ex (1 < i < ra) joins vr and 7;s in H Since there is a path of length ra — 2 + ? connecting vr and vs in G, it follows that 130

d>G(vr)vs) < ra — 2 + ? We claim that dG(vr,vs) — m — 2 + i Suppose that dc(vr, vs) < ra — 2 + ? Then there exists a vr — vs path in G of length less than ra — 2 + ? From the way in which G is constructed, the length of such a path must be at least (ra — 2 + p) + (ra — 2 + q) for some positive integers p and q such that p + q > 3 Since (ra — 2 + p) + (ra — 2 + q) > 2ra — 1, it follows that

2ra — 1 < ra — 2 + 7 and so 7 > ra + 2, which is impossible Thus G2m 2 contains a rainbow P in which each edge ex of P is colored ra — 2 + 7forl<7

The following result illustrates a connection between rainbow subgraphs in distance colored graphs and graceful trees

Theorem 5.1.3 A nontrivial tree T of size ra is graceful if and only if there exists a rainbow T in the distance-colored graph P™+1

Proof. Let T be a nontrivial tree of size 777, and let Pm+i = (7;o,7;i, ,7;m) First, suppose that a tree T is graceful and that / V(T) —» {0,1,2, , ra} is a graceful labeling of T Since the order of T is ra + 1, it follows that / is in fact a bijection We show that P™+1 contains a rainbow tree T Define a function

/* V(T) -> V(Pm+i) by f*(v) = Vf(v) for each v G V(G) Since / is one-to-one, it follows that /* is also one-to-one Furthermore, for each edge 717; G V'(T),

d Pm+i(/* Mi/*(*>)) = dPrn+l(uf(uhvf{v))

= \f(u)-f(v)\=f(uv)

Since / is a graceful labeling, /' is one-to-one and so

Rwi (/»./») ™e£(T)} = {/'H m;e£(T)} = {l,2, ,m} 131

This implies that /* induces a rainbow tree T in the distance-colored graph P™+v

For the converse, suppose that the distance-colored graph P™+1 contains a rainbow tree T of size ra. Thus V(T) = V(Pm+i) = {7;0, vu ..., 7;m}. If 7;^ G JS(T) where 0 < i / j, then 7;^ is colored by

d v v ?; Pm+i( »> j) = l -J'l e {l,2,...,ra} in P™+1. Since T is a rainbow tree of size ra, it follows that |{|i — j\ : vtv3 G E(T)}\ =m and so

{\i-j\:vlv,eE(T)} = {l,2,...,m}. (5.1)

Now define a labeling g : V(T) —> {0,1,2,..., ra} by p(v2) = i for 0 {1,2,..., ra} of 5 defined by

9'favj) = |.g(7;,)-.g(7;J| - |i-j| is one-to-one by (5.1). Therefore, g is graceful and so T is graceful. •

By Theorems 5.1.1 and 5.1.3, we have the following corollary.

Corollary 5.1.4 Let T be a nontrivial tree of size ra. If there exists a rainbow T in the distance-colored graph P^+1, then K2m+i is cyclically T-decomposable.

Next we illustrates a connection between distance-colored graphs and another well-known graph labeling. Let G be a graph of size ra and let S% = {2, 2ra + 1 — i} for 1

{0,1,2,..., 2ra} that induces the edge labeling /' : E(G) -> {1, 2,..., 2ra} defined by 132

f(e) = |/(7i) — f(v)I for each edge e = uv of G

such that the set S = {f'(e) e G E(G)} has the property that |5 H 5Z| = 1 for all 7 (1 < ? < 772) A bi-graceful labeling was called a p-valuation by Rosa in [30]

A graph admitting a bi-graceful labeling is a bi-graceful graph In particular, if

S n 5Z = {?} for all 7 (1 < ? < ra), then G is graceful and so every graceful graph is bi-graceful Rosa [30] proved the following

Theorem 5.1.5 A nonempty graph G of size ra is bi-graceful if and only if the

complete graph K2m+i is cyclically G-decomposable

We now show a connection between rainbow subgraphs in distance-colored graphs and bi-graceful labelings

Theorem 5.1.6 Let G be a nonempty graph of size ra Then G is bi-graceful if

and only if there exists a rainbow G in the distance-colored graph G^+1

Proof. Let G2m+i = (i>o, ^1, , t;2m, 7;0) be a (2ra + l)-cycle First, suppose that G is bi-graceful Then there is an mjective function / V(G) —> {0,1, 2, , 2ra} such that the induced edge labeling /' E(G) —> {1,2, , 2ra} defined by f'(e) =

|/(7i) — f(v)I for each edge e = uv of G uses exactly one label in {?, 2ra + 1 — 7} for each? withl V{C2m+i) by f*(v) = Vf(v) for each 7; G V(G) If 717; G E(G) such that \f(u) - f(v)\ G {7, 2771 + 1 — 7} for some ? with 1 < ? < 771, then

^+i(/*(w),/*(v)) = dC2m+1(uf{u),vf{v)) = 1 133

This implies that /* describes a rainbow copy of G in the distance-colored graph

m C 2m+l-

For the converse, suppose that G^+1 contains a rainbow G. We show that

G is bi-graceful. Since diam(G2m+i) = ra, each edge of G is assigned a color from the set {1,2,..., ra}. Let V(G) = {vtl,vl2,..., vln} C V(C2m+i), where

0 < ?i < ?2 < • • • < vln < 2m. Define a labeling g : V(G) -» {0,1, 2,..., 2ra} by 9(vij) — h f°r aH J with 1 < j < n. For each pair 7;^,7v of vertices of G where

1 < J7 7^ j' < ^, since

rf E c2m+1K,^) {1,2,...,m},

either dc2m+1K,^;) = |^ ~ ^| or dc2m+1(vvv,/) = (2ra - 1) - |^ - ?;^|. Now consider the induced edge labeling g' : E(G) —» {1, 2,..., 2ra} of g defined by

5'K^;) = l.9K)-ffK;)l

for each edge vxv%i of G where 1 < j'^ f < n. Since G is a rainbow subgraph of

size ra in G^+1, it follows that

d u v \{ c2rn+1( , ) :u,v€V{G)}\ =m

and so

{dc2m+1(u,v) : 71,7; G V(G)} = {1,2,... ,ra}.

This implies that each edge of G uses exactly one label in {?, 2ra + 1 — ?} for each

i with 1 < i < ra. Therefore, 5 is a bi-graceful labeling and so G is bi-graceful. • 134

To illustrate Theorem 5.1.6, consider the 5-cycle G5 = (u,v,w,x,y,u) of size ra = 5. A bi-graceful labeling of the cycle G5 is shown in Figure 5.1(a). In this case, the set of induced edge labels is S = {1, 4,6, 8,9}, Si = {1,10}, S2 = {2, 9},

S3 = {3,8}, £4 = {4, 7} and 55 = {5, 6}. Observe that \S n Sz\ = 1 for 1 < i < 5.

Thus G5 is bi-graceful. By Theorem 5.1.6, the distance-colored graph C\x contains a rainbow G5 induced by the labeling /* as described in the proof of Theorem 5.1.6.

In this case, f*{u) = 7;0, f*(v) = vu f*(w) = 7;10, f*{x) = v2, f*{y) = v6. The 5- cycle in C\x is (?;0,7;i, i;10,7;2,7;6,7;0) as shown in Figure 5.1(b). The colors assigned to the edges of this 5-cycles are 1,2, 3,4, 5 and so this 5-cycle is a rainbow G5 in

uio Cd 2 ^ >^l - V9 O ^T^ )V2 5/ v* O Ol>3 /4

V7 O 0-u4 O U6 ^n (a) (b)

Figure 5.1: A bi-graceful labeling of G5 and a rainbow G5 in Gfx

The following is an immediate consequence of Theorems 5.1.5 and 5.1.6.

Corollary 5.1.7 Let G be a nonempty graph of size ra. Then the complete graph

Kzm+i is cyclically G-decomposable if and only if there exists a rainbow G in the distance-colored graph C™m+i. 135

5.2 Rainbow Pancyclic Graphs

Recall that a graph G of order 77 > 3 is called pancyclic if G contains a cycle of every possible length, that is, G contains a cycle of length £ for each integer £ with

3 < £ < 77 Let G be a connected graph of order 77 > 4 and diameter d > 3

d d The distance colored graph G = Kn is called rainbow pancyclic if G contains a rainbow cycle of length £ for every £ with 3 < £ < d First, we show that rainbow pancyclic graphs exist

Theorem 5.2.1 For each integer d > 3, there is a connected graph G of diame ter d such that Gd is rainbow pancyclic

Proof. For d G {3,4}, let Gd = Pd+i = (v0, vu , vd) If d = 3, then (v0,7;x, 7;3, vQ) is a rainbow triangle in G3, while if d — 4, then (7;0, v\, 7;3,7;0) and (7;0,7;2, ?;i, 7;4,7;0) are rainbow 3-cycle and 4-cycle, respectively, in G\ Thus the result holds for d G {3,4} Furthermore, the order of a connected graph of diameter d is at least d+1

For each integer d > 5, let Gd be the graph obtained from Pd+i = (i;0, t>i, , ?;<*) by adding a new vertex ?i and joining 71 to the vertices v0 and v\ in P^+i Then G^ has order d + 2 and diameter d We proceed by induction on d > 5 to show that

G^ is rainbow pancyclic Since

(?;0,7;i,7;3,7;0), (v0,v2, vu 7;4,7;0) and (v0,7;3, vu 7;5,7i,?;0) are rainbow /r-cycles for /r = 3, 4, 5, respectively, in G|, the result holds for d = 5

Assume that the distance-colored graph Gk is rainbow pancyclic for some integer 136 k > 5, that is, G£ contains a rainbow cycle of length £ for every £ with 3 < £ < k

We show that G^\ is rainbow pancyclic Since Gk is a subgraph of GJT^, it follows that G^} contains a rainbow cycle of length £ for every £ with 3 < £ < k Thus, it remains to find a rainbow (k + l)-cycle in G^J Suppose that k + 1 = i (mod 4) where ? G {0,1, 2,3} We consider these four cases

Case 1 k + 1 = 0 (mod 4) Let k + 1 = 4p ior some positive integer p Then

v G = (7;0,7;2p, 7;2p_i, 7;2p+i, ^2p-2, 2p+2, , 7;p+i, 7;3p_i, 7;p, 7;3p+1,

Vp-1, %,+2, Vp-2, ^3p+3 J J ^1» ^4p, ^0 )

Note that G does not contain the vertices 71 and 7;3p (See Figure 5 2(a) for k + 1 =

Case 2 k + 1 = 1 (mod 4) Let k + 1 = 4p + 1 for some positive integer p

Then

G = (vQ, ?i, 7;4p+i, 7;2p_i, 7;2p+i, 7;2p_2,7;2p+2, , 7;p+i, 7;3p_i, ?;p, 7;3p,

Vp-U V3p+2, Vp-2, %>+3, > ?;1, %>, VQ)

Note that G does not contain the vertices 7;2p and 7;3p+1 (See Figure 5 2(b) for k+1=9)

Case 3 k + 1 = 2 (mod 4) Let k + 1 = 4p + 2 for some positive integer p

Then

C = (VQ , 7;2p, 7;2p+2,7;2p_ 1,7;2p+3,7;2p_2,7;2p+4, , 7;p+2,7;3p,

v Vp+l > 3p+2, , ^2, %>+i, ?;i, 7;4p+2, 71, v0) 137

Note that G does not contain the vertices 7;2p+i and 7;3p+i (See Figure 5 2(c) for k + 1 = 10)

Case 4 k + 1 = 3 (mod 4) Let k + 1 = 4p + 3 for some positive integer p

Then

C — (v0,7;3p, ?;3p+i, 7;3p_i, ?;3p+2,7;3p_2, , 7;p+2,7;3p,

Up, %H-I, , v2,7;4p+2, vu 7;4p+3, v0)

Note that G does not contain the vertices u and 7;p+i (See Figure 5 2(d) for k+1 = 11 )

In each case, G^\ contains a rainbow (k + 1) cycle and so G^{ is rainbow pancyclic •

Next we determine the minimum order of a connected graph G of diameter d for which Gd is rainbow pancyclic

Theorem 5.2.2 For each integer d > 3, the minimum order nd of a connected graph G of diameter d for which Gd is rainbow pancyclic is

_(d+l ifd = 0 (mod 4) or d = 3 (mod 4) Vd~ \ d + 2 ifd=l (mod 4) or d = 2 (mod 4)

Proof. Figure 5 3 shows that 77^ = d + 1 for d G {3, 4} Thus we may assume d > 5 We consider two cases

Case 1 d = 0 (mod 4) or d = 3 (mod 4) We show that nd = d+1 in this case

It suffices to show that Pd+X is rainbow pancyclic In the proof of Theorem 5 2 1, 138

u 9

V() Vi v2 v3

(b) 9 - fc + 1 = 0 (mod 4)

(c) 10 = fc + 1 = 0 (mod 4) (d) 11 = fc + 1 = 0 (mod 4)

Figure 5.2: Illustrating the proof of Theorem 5.2.1

we saw that Pd+X contains a rainbow cycle of length d for d = 0 (mod 4) or d = 3

k (mod 4). Since the distance-colored graph P +i is a subgraph of the distance- colored graph P/f/+1 for every two positive integers k and k! where k! > fc, we only need show that if d = 3 (mod 4), then P$+1 contains rainbow cycles of length d—1

7 and d — 2, respectively. For d = 7, Figure 5.4(a) shows a rainbow 6-cycle in P8

7 and Figure 5.4(b) shows a rainbow 5-cycle in P8 . Thus the result holds for d = 7. Thus, we may assume that d > 8. 139

Figure 5.3: nd = d+l iord G {3, 4}

7 Figure 5.4: Rainbow 6-cycle and 5-cycle in P8

Let d — 4fc + 3 for some integer fc > 2 and let

P4/C+3 = (vo, 7;i, 7;2,..., 7;4/c+3).

The distance-colored graph P^+1 contains a rainbow cycle of length d — 1 > 7, namely

Gd-l — (^0) ^4fc-f3? ^15 ^4fc+27 v2, • • • , ^3fc+3> ?;fc+l, v3fc+lj ^fc+2> ^3fc> ^fc+3)

• •• ,U2k+3,U2k,U2k+uU0).

Observe that the distance sequence of G^-i is

4fc + 3,4fc + 2,4fc + 1,..., 2fc + 2, 2fc, 2fc - 1,..., 4, 3,1, 2fc + 1.

So the color 2 is the only color that is not used in Cd-i- Figure 5.5 shows a rainbow

Gio in P^2 and a rainbow Gi4 in P^^ respectively. 140

rf=ll d = 15

Figure 5.5: Rainbow cycles of length d — 1 in P^+1 for d — 11 and d — 15

The distance-colored graph P^+1 contains a rainbow cycle of length d — 2 > 6, namely

v C^-2 = (^0) 4k+3, ^i, 7;4fc+2,7;2,..., ?;fc, 7;3fc+3,Vfc+2,7;3fc+2,7;fc+3,7;3fc+i,

v v • • • > ^2fc+5) ^2fc? 2k-\-4, V2k+2i o)-

Observe that the distance sequence of Cd-2 is

4fc + 3,4fc + 2,4fc + 1,..., 2fc + 3,2fc + 1, 2fc,..., 6,5,4,2, 2fc + 2.

So the colors 1 and 3 are the only colors that are not used in Cd-2. Figure 5.6 shows a rainbow Cg in P^2 and a rainbow Gi3 in P/g, respectively.

Case 2. d = 1 (mod 4) or d = 2 (mod 4). We show that n^ = d + 2 for each d > 5. Since the graph G^ described in the proof of Theorem 5.2.1 has order d + 2, it follows that nd < d + 2. Thus it suffices to show that Pd+i is not rainbow 141

d=ll d=15

Figure 5.6: Rainbow cycles of length d — 2 in P^+1 for d = 11 and d = 15 pancyclic. Assume, to the contrary, that there is an integer d > 5 such that d = l

(mod 4) or d = 2 (mod 4) for which P^+1 is rainbow pancyclic. Thus there is a rainbow cycle G of length d in Pd+V Let Pd+i = (7;0,7;i, 7;2,..., i;d). Observe that G must contain an edge colored i for each color i with 1 < i < d. Necessarily, vo belongs to G. We start at 7;0, traverse around G, say clockwise, and partition the edge set G into two sets Si and S2 as follows: Let e = v%Vj G P(G), where

0 < i,j < d + 1. If i < j, then let e G Sx; while if i > j, then e G S2. Since G starts and ends at 7;0, it follows that the sum of colors of edges in Si equals the sum of colors of edges in S2, that is,

(5.2)

uv^Si uveS2

Since d = 1 (mod 4) or d = 2 (mod 4), the number of odd numbers between 1 and d is odd. Thus either there is an odd number of odd colors used in the edges of Si 142

or there is an odd number of odd colors used in the edges of S2, say the former.

Then the number of odd colors used in the edges of S2 is even. This contradicts

(5.2). Therefore, nd = d + 2 for d = 1 (mod 4) or d = 2 (mod 4). •

5.3 Rainbow Connectivity

Let G be a nontrivial connected graph of order n and diameter d. For an integer fc with 1 < fc < d, a path P in the distance-colored graph Gk is a rainbow path if no two edges of P are colored the same. The graph Gk is called rainbow-connected if every two vertices u and v in Gk are connected by a rainbow u — v path in

k d d G . Since G — Kn, the graph G is certainly rainbow-connected. Thus there is a smallest positive integer fc such that Gk is rainbow-connected. This gives rise to the following concept. The rainbow-connection exponent rce(G) of G is defined as the minimum fc for which Gk is rainbow-connected. In order to determine the rainbow-connection exponent of an arbitrary connected graph, we first present a lemma.

Lemma 5.3.1 Let k > 2 be an integer. If £ is an integer with k < £ < (k^), then there exist integers fci, fc2,..., kPi, where fc > k\ > fc2 > • • • > kPe > 1 and

Pt > 1; such that £ = k + ki + k2 + ... + kPf.

Proof. We proceed by induction on fc. If fc = 2, then 2 < £ < (^) = 3 and so

^ = 3 = 2 + 1. Suppose that the result is true for all integers £ with fc < £ < (k^) for some integer fc > 2. Let £ be an integer with fc + 1 < £ < (k^2)^ First, suppose 143 that fc + 1 < £ < (fcJ1) Then fc < £ < (k^) By the induction hypothesis,

£ = fc + fc1 + fc2 + + kPf where fc > fci > fc2 > > kPf > 1 for some positive integer p£ If kPf > 2, then £ = (fc + 1) + kx + k2 + + fcP/_i + (fcp, - 1), where fc + l>fci>fc2> > fcp^_i > fcp, - 1 If fcP/ = 1, then since £ > fc + 1, it follows that pi > 2 and fcp,_i > 2 Thus £ = (k+ 1)+ kx + k2+ + fcp,_i, where

k 2 fc+1 > fci > fc2 > > fcp,_i andp^-1 > 1 Next, suppose that (*+*) < £ < ( + )

fc 2 li£= ( + )? then £ = (k + l) + k+ + 2 + 1 and the result follows Thus we may assume that £ < (k+2) Since (kf) = (k+l) + (fc + 1), it follows that £ = (k+x) +p where 1 < p < fc and so

1 £ = r+ \+p = k + (k-l) + (k-2)+ +2 + l+p

= (k + l) + k+ +[k-(p-l) + l} + (k-p) + (k-p-l)+ +2 + 1, producing the desired result •

Theorem 5.3.2 If G is a connected graph of diameter d > 2, then the rainbow connection exponent rce(G) of G is the unique positive integer fc for which

(l)

Proof. We first show that rce(G) < fc For two vertices 71 and v of G, we show that there is a rainbow 71 — 7; path in Gk Suppose that d(u, v) — £ where 1 < £ < d

Let P = (u — 7;0,7;i, 7;2, , V£ = v) be a 71—7; geodesic in G If 1 < £ < fc, then 717; is an edge in Gk and so 71 and 7; are connected by a rainbow path in Gfc, namely (71,7;)

Thus we may assume that £ > k Since d < (k^), it follows that fc < £ < (k^) By 144

Lemma 5.3.1, £ = k + fci + fc2 + ... + kPt such that fc > fci > fc2 > • • • > kPf > 1 for some positive integer pi. Let 7ii = 7;fc and, for 2 < r < p^ let ur = Vk+hx+k2+ +fcr-i-

Then Q = (71,711,7i2,..., upnv) is a rainbow 71 — v path in which the edge 71711 is colored fc, the edge urur+i is colored kr for 1 < r < pt — 1 and the edge uPfv is colored fcp,.

Next, we show that rce(G) > fc. Assume, to the contrary, that the distance- colored graph GJ is rainbow-connected for some positive integer j with j < fc. Let x and y be two antipodal vertices of G, that is, d(x, y) = d. Since GJ is rainbow- connected, there exists a rainbow x — y path in G3 using each color in {1, 2,..., j} at most once. This implies that d = d(x,y) < 1 + 2-1 \-j < 1 + 2-1 h(fc —1) =

(2) < d, which is impossible. Therefore, rce(G) > fc and so rce(G) = fc. •

The following is an immediate consequence of Theorem 5.3.2.

Corollary 5.3.3 If G is a connected graph of diameter d>2, then

-1 + VT+M' rce(G) =

Proof. Let fc be the minimum positive integer such that ( ^) > d. Then fc2 + fc - 2d > 0, which implies that fc > ~1+y+^. The result then follows by Theorem 5.3.2. •

By Corollary 5.3.3, if G is a connected graph with rce(G) = fc > 2, then every two vertices 71 and v in G are connected by at least one rainbow 71 — v path in

Gk. In Chapter 4, the concept of color-connectivity was introduced in terms of 145 internally disjoint properly colored paths in distance-colored graphs In the view of rainbow paths, this gives rise to another concept

For a connected graph G and positive integers fc and £, the distance-colored

k k graph G is rainbow £-connected if for every two distinct vertices 71 and v of G ) there are £ internally disjoint rainbow 71 — 7; paths in Gk In particular, if £ = 1, then Gk is rambow-connected Thus, we now assume that £ > 2 If G is a complete graph, then Gk is not rainbow £ connected for all positive integers fc and £ with

£ > 2 Hence we only consider non complete connected graphs For a connected graph G and an integer fc > 2, the rainbow connectivity of Gk is the maximum positive integer £ for which Gk is rainbow /'-connected We begin by considering rainbow 2-connected graphs Gk where G is a path

Theorem 5.3.4 For each integer k > 3, let P^+n+1 = (i>o, ^1, ^2 > ,V(k+i\) be the path of order (k^) + 1 Then the distance colored graph P/L+n ?s properly 2- rainbow connected Furthermore, for each pair vt,v3 of vertices of Prk+i\, p where

0<7 3; there are two internally disjoint rainbow v% — v3

suc paths Qi and Q2 in P/\+i\ h that

V(Qi)UV(Q2) C{7;z,7;z+i, ,M

Proof. We proceed by induction on fc > 3 If fc - 3, then (*+*) + 1 = (*) + 1 =

7 Figure 5 3 4 shows that the statement is true for P^ Now assume that the statement is true for the distance colored graph P(M\+I for some integer fc > 3

h 2 Let m = ( + ) and Pm+i = (v0, vuv2, , vm) Suppose that 7;2, v3 G V(Pm+i), where 0 < 1 < j < m If 1 < j — ? < 2, then the result is true (see Figure 5 3 4) 146

o o o o -• o o

o • o o o o

o o o

Figure 5.7: Illustrating the proof of Theorem 5.3.4 for fc = 3

Thus we may assume that j -— ?i; > 3.

k First, assume that i = 0 and 3 < j < m. Ii j < ( ^): then let

k l P( + )+i = \v0,vuv2,...,v,k+i\y

Since P/k+n is a subgraph of P^+\, the result follows by the induction hypothesis. \ 2 J+1 Thus we may assume that j > (k^) +1. Now let t = j- (k + 1). Since (k+l) +1 <

fc 2 j < ( 2 ), it follows that *<(T) "(* + !) = (T) and since k > 3,

*>(*+i) + i-(fc+i) = ®ffi > 3.

Thus 3 < t < (/c^1). By the induction hypothesis there are two internally disjoint rainbow 7;0 — vt paths Q[ and Q'2 in Pfk+i\ (and so in P^+\) such that v 2 J+1

nQi)UF(Q'2)C{Vo,Vl,...,t;t}.

Let Q'l = S^pn'W'-.V = «t)

#2 = (v0,Vqi,Vqa,...,Vq.=Vt). 147

Since j — / = fc + l,we can construct two internally disjoint rainbow v0 — v3 paths

Qi and Q2 from Q[ and Q2 by

v Qi = (^o, vPl, 7;P2, , 7;Pr_x, 7;Pr 1+(fc+i), V+(fc+i) = j)

Q2 = (v0, vqi, vq2, , vqs_x, 7;9s_1+(fc+1), 7;gs+(/c+i) = 7;^)

Next, assume that 1 < 7 < j < m We first construct two internally disjoint

wl rainbow v0 — v3_% paths Q\ and Q2 in P^Xi ^h the desired properties, where say

Ql = (v0, vai, 7;a2, , var = ^_2)

Q2 = (^0,^n^2> »^s=^-x)

Then

v V V v v Wl — ( 0-^-i) ai+ii a2-\-ii i ar^i — j)

Q2 = (v0+i, vbl+2,7;62+2, , 7;6s+2 = v3) have the desired properties •

The proof of Theorem 5 3 4 also establishes the following result

Theorem 5.3.5 For each integer m > 3, let fc be the smallest positive integer for which m + 1 < ( ^ ) Then the distance-colored graph P^+i is properly 2-rambow connected

Theorem 5.3.6 Let G be a connected graph with rad(G) > 3 and diam(G) = d

Ifk is the minimum positive integer for which d < ( * ), then the distance-colored graph Gk is 2-rambow connected. 148

Proof. Let 71 and v be two distinct vertices of G. We show that there are two

k internally disjoint rainbow 71 — 7; paths in G . If dG(u,v) = p > 3, then let

P — (u — 7;0,7;i,..., 7;p = 7;) be a 71 — 7; geodesic in G. By Theorem 5.3.5, there are

k k two internally disjoint rainbow 71 — 7; paths Qi and Q2 in the subgraph P of G

k and so Qi and Q2 are internally disjoint rainbow 71 — 7; paths in G as well. Thus we may assume that dc(u,v) < 2. Since rad(G) > 3, there is a vertex w G V(G) such that G?G(7I, W) = 3. Let P' = (71, .T, T/, W) be 71 — w geodesic in G. We consider two cases, according to dc(u,v) = 1 or dc(u,v) = 2.

Case 1. dc(u,v) = 1. Let Qi = (71,7;). If 7;y ^ E(G), then let Q2 = (71, y,7;), while if 7;T/ G E(G), then let Q2 = (71,7i;,7;).

Gase 2. dc(u,v) = 2. Let Qi = (71,7;). To construct Q2, we consider two subcases.

Subcase 2.1. .T7; G E(G). Necessarily, (71, .T, 7;) is a, u — v geodesic, livy G E(G), then let Q2 = (71,^,7;), while if 7;y ^ P(G), then let Q2 = (71, y,w,v), whose edges are colored 1, 2 and 3.

Subcase 2.2. xv fi E(G). Let 2 G V(G) such that (71, z, 7;) is a 71 — 7; geodesic in

G. Since .X7; ^ E(G)) let Q2 = (71, x, 7;) is a rainbow path whose edges are colored 1 and 2 or 1 and 3 (depending on whether xz G E(G) or not).

In each case, there are two internally disjoint rainbow 71 — v paths in Gk and so Gk is properly 2-rainbow connected. •

In general, we have the following conjecture. 149

2£ 2 Conjecture 5.3.7 Let £ > 2 be any integer and let G = P/2/-i\+r Then G ~ is £-rainbow connected.

Figure 5.8 shows that P^ is a properly 3-rainbow connected and so Conjec ture 5.3.7 is true for £ = 3.

4 Figure 5.8: Three internally disjoint rainbow paths in PX X Chapter 6 Topics for Further Study

6.1 Questions on Color-Hamiltonicity

1. The complete graph K7 can be decomposed into properly colored Hamilto

nian cycles Gi, G2 and G3 where

Gi = (vu v2,7;5, v7,7;4,7;6,7;3, vx)

C2 = (vu 7;5,7;6,7;2,7;4,7;3, v7, vx)

G3 = (vi, 7;4,7;5,7;3,7;2, v7,7;6,7;x)

Such a deposition is shown in Figure 6.1.

This gives rise to the following question: Which complete graphs of odd order

can be decomposed into properly colored Hamiltonian cycles?

2. Let G be a connected graph for which Gk is Hamiltonian-colored for some

positive integer fc. Recall that the Hamiltonian coloring exponent hce(G) of

G is the minimum fc for which Gk is Hamiltonian-colored. For a connected

150 151

1"5 V4

Figure 6 1 Decomposing K7 into properly colored Hamiltonian cycles graph G with hce(G) = fc, let ii(G) denote the minimum number of colors used in a properly colored Hamiltonian cycle of Gk The following results were presented in Chapter 2

Theorem 6.1.1 For each integer k > 2, there exists a tree Tk with hce(Tfc) = fc such that every properly colored Hamiltonian cycle in the kth power ofTk must use all of the colors 1,2, , fc

Theorem 6.1.2 For each integer fc > 3, there exists a connected graph Gk with hce(Gfc) = fc and a properly colored Hamiltonian cycle in the kth power of Gk whose edges are colored with fewer than fc colors

Corollary 6.1.3 There exists a sequence {P/c}£L4 of trees such that

k->oo hce(Tfc) 152

Corollary 6.1.4 There exists a sequence {7fc}£L2 of trees such that >- M, = i. fc-4oo hce(Tfc)

Problem 6.1.5 For which pairs a, b of integers with 2 < a < 6, does there

exist a tree T with fi(T) = a and hce(P) = b?

Corollaries 6.1.3 and 6.1.4 give rise to the following concepts. For a connected

graph G with hce(G) = fc, the color range p(G) of G is defined as the set

of all colors j that are used in some properly colored Hamiltonian cycle of

Gfc, while the color spectrum a(G) of G is defined as the set of all colors

j that are used in every properly colored Hamiltonian cycle of Gk. Thus

(j(G) C p(G) C {1,2,..., fc} and fc G a(G) n /9(G). In particular, if fc - 2,

then cr(G) = p(G) = {1,2}. For each integer fc > 3, the graph Gk constructed

in Theorem 6.1.2 has a(Gk) $i p(Gk), while the graph Hk constructed in

Theorem 6.1.1 has a(Hk) = p(Pfc).

Problem 6.1.6 Study the relationship between a(G) and p(G).

6.2 Questions on Color-Distance

1. For which rational numbers r with 0 < r < 1, does there exist a graph G

such that the order of CC(G) is fc, the order of Cen(G) = £ and | = r?

2. Let P and H be graphs where every vertex of H has color-eccentricity 1 or

none do. Does there exist a graph G such that CC(G) = P and CP(G) ^ P? 153

3 If a graph G has a sufficiently large diameter and fc is sufficiently large, is Gk

properly 3-connected?

6.3 Questions on Rainbow Subgraphs

Let G be a connected graph of order 77 and diameter d Then each edge of the

d distance-colored graph G = Kn is colored with one of the colors 1,2, , d

d ? 1 Which subgraphs P of size d in G — Kn are rainbow subgraph of Pn

For each rainbow subgraph H of Pn, define the rainbow quotient rq(H) of P as the maximum number of pairwise edge-disjoint copies of a rainbow H in

Kn Thus

? Find examples of graphs H for which rq(P) = |_(2)/^J

2 Let G be a connected graph of diameter d > 3 and let 3 < fc < d be an

integer For each unicychc graph H of size fc, does there exist a rainbow H

in Gdr? If this is not true, then which connected graphs have this property7

d 3 Let 7; be a vertex of G = Kn Let P be a path with initial vertex 7; con structed by taking any edge incident with v colored 1, say vv\, followed by

any edge incident with 7;i whose color is the smallest color greater than 1,

etc Among all paths constructed in this manner, let r£(v) denote the maxi

mum length of such a path Determine the minimum and maximum values 154

of r£(v) over all vertices v in Kn. [Note: Instead of encountering the edges whose colors in increasing order, we could just demand that the colors be

different.]

d 4. Does the distance-colored graph G = Kn contain a rainbow path P of length d the colors of whose edges follow in the order 1, 2, 3,..., d?

For each ordering a of the elements in the set {1, 2, 3,..., d}, is there a path

of length d the colors of whose edge follow a?

d 5. Study decompositions of distance-colored graphs G = Kn into prescribed rainbow subgraphs.

6.4 Questions on Rainbow Connectivity

Recall that the rainbow-connection exponent rce(G) of a connected graph G is defined as the minimum fc for which the distance-colored graph Gk is rainbow-

connected in Chapter 5. We have seen that if G is a connected graph of diame ter d > 2, then \-l + Vl + 8d] rce(G) = . (6.1)

Suppose that G is a connected graph with rce(G) = fc > 2. By (6.1), we know that for every two vertices 7i and v in a connected graph G, there is a rainbow u — v path in Gk. In fact, there may be several rainbow u — v paths in Gk. A rainbow

7i — v path of minimum length in Gk is a rainbow u — v geodesic in Gk and its

length is the rainbow-distance between u and v in Gk and is denoted by rd(u,v). 155

We have the following conjecture

Conjecture 6.4.1 Suppose that G is a connected graph with rce(G) = fc > 2 //

7i and v are two vertices in G such that d(u,v) = £, then rd(u,v) is the maximum positive integer p for which (k^) — (ty > £

For a fixed integer £, define the number rce(G, £) to be the minimum positive

integer fc for which Gk is ^-rainbow connected Thus rce(G, 1) = rce(G) Recall the following result in Chapter 5

Theorem 6.4.2 Let G be a connected graph with rad(G) > 3 and diam(G) = d

Ifkis the minimum positive integer for which d < ( ^), then the distance colored

graph Gk is 2 rainbow connected

The following is a consequence of Theorem 6 4 2

Corollary 6.4.3 Let G be a connected graph with rad(G) > 3 and diam(G) = d

/c 1 Ifk is the minimum positive integer for which d < ( ^ ); then rce(G, 2) = fc

Problem 6.4.4 Study rce(G, £) in general

d Let G be a connected graph of order 77 and diameter d A tree T in G — Kn is a rainbow tree if no two edges of T are colored the same For an integer fc with

2 < fc < 77, the distance-colored graph Gd is called k-rambow tree-connected if for

every set S of fc vertices of G, there exists a rainbow tree in Gd containing the

vertices of S

Problem 6.4.5 Determine conditions under which Gd is k-rambow tree connected 156 Bibliography

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