Consecutive radio labelings and the Cartesian product of graphs

Transcription

1 University of Iowa Iowa Research Online Theses and Dissertations Summer 2013 Consecutive radio labelings and the Cartesian product of graphs Amanda Jean Niedzialomski University of Iowa Copyright 2013 Amanda Jean Niedzialomski This dissertation is available at Iowa Research Online: Recommended Citation Niedzialomski, Amanda Jean "Consecutive radio labelings and the Cartesian product of graphs" PhD (Doctor of Philosophy) thesis, University of Iowa, Follow this and additional works at: Part of the Mathematics Commons

2 CONSECUTIVE RADIO LABELINGS AND THE CARTESIAN PRODUCT OF GRAPHS by Amanda Jean Niedzialomski A thesis submitted in partial fulfillment of the requirements for the Doctor of Philosophy degree in Mathematics in the Graduate College of The University of Iowa August 2013 Thesis Supervisor: Associate Professor Maggy Tomova

3 Graduate College The University of Iowa Iowa City, Iowa CERTIFICATE OF APPROVAL PHD THESIS This is to certify that the PhD thesis of Amanda Jean Niedzialomski has been approved by the Examining Committee for the thesis requirement for the Doctor of Philosophy degree in Mathematics at the August 2013 graduation Thesis Committee: Maggy Tomova, Thesis Supervisor Frauke Bleher Victor Camillo Isabel Darcy Colleen Mitchell

4 To Robert, My Favorite ii

5 ACKNOWLEDGEMENTS There are many people who have contributed to this work First I would like to thank my advisor, Maggy Tomova, who inspired me to keep going I would also like to thank Professors Anderson and Tymoczko for their guidance as advisors I thank my wonderful Iowa friends; their encouragement and companionship through this unfamiliar territory of graduate school have saved me innumerable times Among these, I thank my best friend, my husband Robert This thesis would most definitely not exist without all the many ways he has helped me I thank my family, whose love and support share responsibility for every one of my accomplishments And I thank my God, without whom I could do nothing iii

6 ABSTRACT For k Z + and G a simple connected graph, a k-radio labeling f : V G Z + of G requires all pairs of distinct vertices u and v to satisfy f(u) f(v) k+1 d(u, v) When k = 1, this requirement gives rise to the familiar labeling known as vertex coloring for which each vertex of a graph is labeled so that adjacent vertices have different colors We consider k-radio labelings of G when k = diam(g) In this setting, no two vertices can have the same label, so graphs that have radio labelings of consecutive integers are one extreme on the spectrum of possibilities; graphs that can be labeled with such a labeling are called radio graceful In this thesis, we give four main results on the existence of radio graceful graphs, which focus on Hamming graphs (Cartesian products of complete graphs) and a generalization of the Petersen graph In particular, we prove the existence of radio graceful graphs of arbitrary diameter, a result previously unknown Two of these main results show that, under certain conditions, the t th Cartesian power G t of a radio graceful graph G is also radio graceful We will also speak to occasions when G t is not radio graceful despite G being so, as well as some partial results about necessary and sufficient conditions for a graph G so that G t is radio graceful iv

7 TABLE OF CONTENTS LIST OF TABLES vi LIST OF FIGURES vii CHAPTER 1 INTRODUCTION 1 2 EXISTENCE RESULTS 9 21 Powers of Complete Graphs Definition of x 1, x 2,, x n t The x i s Form an Ordering The Ordering Induces a Consecutive Radio Labeling Alternate Proof Powers of Incomplete Graphs Other Hamming Graphs A Generalization of the Petersen Graph 50 3 NON-EXISTENCE RESULTS 62 4 TOWARDS NECESSARY AND SUFFICIENT CONDITIONS 66 5 REMARKS AND FUTURE WORK 71 REFERENCES 74 v

8 LIST OF TABLES Table 11 All induced radio labelings of P The list of vertices x 1,, x 27 for K All nonequivalent induced radio labelings of K All G pi graphs for p = 3, 5, 7, The construction of the 7-Petersen graph C 7 and its diameter complement 68 vi

9 LIST OF FIGURES Figure 11 The Petersen graph with a consecutive radio labeling 7 21 Cartesian product example: K 3 P The matrix A for K4 4 with blocks A consecutive radio labeling of K Visual description of Theorem 12 and Corollary vii

10 1 CHAPTER 1 INTRODUCTION For a fixed graph, the problem of finding an optimal labeling (a function that maps vertices of our graph into some set, most commonly the set of positive integers) satisfying certain conditions has been of great interest in graph theory The notion of graph labeling was first introduced in 1966 by Rosa in [29], and since then many different graph labelings have been defined and studied Probably the best known labeling is vertex coloring, where we assign a color to each vertex of a graph in such a way that adjacent verices differ in color Other labelings widely studied include graceful labeling, harmonious labeling, and k-radio labeling The abundance of different labelings is rooted in the idea that existence of a certain optimal labeling has application to problems in computer science, information theory, and coding theory, among others For a survey of graph labeling we recommend a heroic effort by Gallian ([14]) and its updates (the most recent version was published in 2012 and counts 260 pages) In this thesis we study radio labeling, which has its historical roots in the optimal assignment of radio frequencies (or channels) to transmitters This so-called Channel Assignment Problem addresses the issue that interference can be created when geographically close radio transmitters have an insufficient difference in their radio frequencies The goal, then, is to assign frequencies to transmitters in a way that minimizes that interference The problem was put into a graph-theoretic context by Hale in 1980 ([20]); transmitters and frequency assignment were modeled by vertices

11 2 in a graph and a labeling of them There have since been a number of variations of Hale s labeling; most notably L(2, 1)-labeling and radio labeling, which both lie under the umbrella of k-radio labeling (first defined by Chartrand, Zhang, and Ping in [8]) Given a connected graph G with vertex set V G and k Z +, a labeling f : V G Z + is a k-radio labeling if f satisfies the inequality f(u) f(v) k + 1 d(u, v) (11) for all distinct u, v V G In particular, this inequality guarantees that any two vertices distance k or less apart are assigned different labels As k has a natural relationship to distance in the graph, we typically bound k above by diam(g) If k = 1, (11) simply requires adjacent vertices to have different labels, which makes 1-radio labeling equivalent to vertex coloring An optimal vertex coloring of a fixed graph G is one that uses the smallest number of labels/colors (ie, a coloring f is optimal if it has the smallest possible range with respect to G) The number of labels used in an optimal coloring of G is called the chromatic number of G, denoted χ(g) The study of k-radio labeling will likewise be centered on finding optimal labelings, and the chromatic-number-analogs associated to them Originally introduced by Griggs and Yeh in [19], 2-labeling has several aliases including L(2, 1)-labeling, and has been extensively studied The λ-number of a graph G is the smallest number m such that there exists a (not necessarily surjective) L(2, 1)- labeling f : V G {1,, m} Results about the λ-numbers of various families of graphs include [3], [5], [12], [13], [15], [17], [18], and [22]; see [32] for a survey L(2, 1)-

12 3 labeling has been the inspiration of several variants of its own, including L(j, k)- labeling, which requires adjacent vertices to have label differences of at least j and vertices distance two apart to have label differences of at least k ([6], [10], [16]) Our focus in this paper will be on radio labeling, first introduced in [7], which is k-radio labeling when k is maximized at diam(g) Definition 1 For a connected graph G, a labeling f : V G Z + is a radio labeling of G if it satisfies f(u) f(v) diam(g) + 1 d(u, v) (12) for all distinct vertices u, v V G Inequality (12) is called the radio condition The span of a labeling f is the largest element in the range of f; the minimal possible span of a radio labeling of a fixed graph G is called the radio number of G, and is denoted rn(g) Much attention has been given to the pursuit of bounds for, and, more desirable, formulas for the radio numbers of graphs Two of the first of these results came in [26], the paper of Liu and Zhu, with paths and cycles; progress has been made for a number of other types of graphs: [1], [9], [14], [23], [24], [25], [27], [28], [30], and [31] The computational complexity of k-radio labeling has connections to the Travelling Salesman and Hamiltonian path problems, and has been studied with partial results in [2], [3], [5], [18], and [11] In general, the computational complexity is unknown for radio labeling The size of the computation currently prohibits the discovery of radio numbers of fixed graphs of modest size by any known general algorithm One basic algorithm involves considering permutations of the vertices of

13 4 a graph: Given a radio labeling f of a graph with vertex set {v 1,, v n }, we can rename (order) the vertices as x 1,, x n so that f(x i ) < f(x j ) if and only if i < j On the other hand, we can produce a radio labeling of a graph by picking an ordering of the vertices and then forcing the labels to increase with respect to that order As we are interested in minimizing span, the smallest label that satisfies all requirements at the given moment is always chosen, so that an ordering of the vertices uniquely determines a radio labeling when we enforce this increasing condition In this way, an ordering of the vertices of G induces a radio labeling of G Finding a radio labeling of G with span rn(g) can be accomplished by finding the appropriate ordering of V G that induces the optimal labeling For example, Table 11 shows all possible orderings of the vertices of P 4 (the path with four vertices), as well as the corresponding induced radio labelings The optimal labelings in which the span is rn(p 4 ) = 6 are distinguished by filled vertices Clearly symmetry could be used to reduce the number of orderings considered in this case Improving this algorithm, however, has not in general sufficiently helped the practicality of computing the radio numbers of larger examples Instead, techniques and results have been tailored to graphs of some specified structure Radio labeling differs from all other k-radio labeling in an immediate way: if f is a radio labeling, then f is necessarily injective We will be interested in graphs G for which a surjective radio labeling f : V G {1, 2,, rn(g)} exists The injectivity of a radio labeling f implies rn(g) V G When rn(g) = V G, there is a radio labeling of G of consecutive integers 1,, V G, which leads to the following definition:

14 x 1 x 2 x 3 x 4 x 1 x 2 x 4 x 3 x 1 x 3 x 2 x x 1 x 3 x 4 x 2 x 1 x 4 x 2 x 3 x 1 x 4 x 3 x x 2 x 1 x 3 x 4 x 2 x 1 x 4 x 3 x 2 x 3 x 1 x x 2 x 3 x 4 x 1 x 2 x 4 x 1 x 3 x 2 x 4 x 3 x x 3 x 1 x 2 x 4 x 3 x 1 x 4 x 2 x 3 x 2 x 1 x x 3 x 2 x 4 x 1 x 3 x 4 x 1 x 2 x 3 x 4 x 2 x x 4 x 1 x 2 x 3 x 4 x 1 x 3 x 2 x 4 x 2 x 1 x x 4 x 2 x 3 x 1 x 4 x 3 x 1 x 2 x 4 x 3 x 2 x 1 Table 11: All induced radio labelings of P 4 Definition 2 A radio labeling f of a graph G is a consecutive radio labeling of G if f(v G ) = {1, 2,, V G } A graph for which a consecutive radio labeling exists is called radio graceful The term radio graceful was introduced by Sooryanarayana and Ranghunath in [30] We will use the language G has a consecutive radio labeling and G is radio graceful interchangeably

15 6 Labelings which are descendants of the Channel Assignment Problem that map surjectively onto Z n for some n have been of recent interest Their study is meaningful for modeling optimal frequency assignment in cases where all frequencies in a range need to be used Full colorings and no-hole colorings (defined by Fishburn and Roberts in [13] and [12] for L(2, 1)-labeling and later extended to L(j, k)-labeling) are two examples; for related results see [4], [10], and [22] Less work has been done with radio labeling; Sooryanarayana and Ranghunath made some progress in [30] In this thesis we will give several results concerning the existence of consecutive radio labelings The complete graph K n with n vertices is a trivial example of a radio graceful graph; since diam(k n ) 1, the radio condition (12) is automatically satisfied, and any labeling of distinct integers is a radio labeling of K n Another well-known example is that of the Petersen graph, denoted P ; since diam(p ) = 2, this example is nontrivial That is, unlike K n, not every injective labeling of P is a radio labeling, but there is some radio labeling of P consisting of consecutive integers 1,, 10 (Figure 11) There are two properties of P that allow it to have a consecutive radio labeling: it has diameter 2, and there is a Hamiltonian path in P, its complement The Hamiltonian path in the graph complement in this case corresponds to an ordering x 1, x 10 of the vertices of P such that d(x i, x i+1 ) = diam(p ) = 2 for all i {1,, 9} This condition will produce a graph with a consecutive radio labeling only if it has diameter 2, however The higher the diameter of the graph, the more

16 Figure 11: The Petersen graph with a consecutive radio labeling specialized the graph must be in order for it to possess a consecutive radio labeling In general, for a graph G with V G = {v 1,, v n } to have a consecutive radio labeling, we need an ordering x 1,, x n of its vertices such that d(x i, x i+c ) diam(g) c + 1 for all i {1,, n c} Stated this way, the difficulty of finding high diameter examples becomes more clear One result we will see here shows in particular that Kn n (which we will define) is radio graceful for n 3 (Theorem 12); since diam(kn) n = n, this result shows there are indeed radio graceful graphs of arbitrary diameter, which had not been known before It is worth mentioning that there is some potential for a symbiotic relationship between the studies of consecutive radio labeling and Hamiltonian or semi- Hamiltonain (traceable) graphs We will prove in Chapter 4 that there is a strong connection between the existence of a consecutive radio labeling of a graph and the existence of a Hamiltonian path (and sometimes a Hamiltonian cycle) in its comple-

17 8 ment a fact we have already observed for the Petersen graph The existence problem for Hamiltonian graphs is another area of study where examples are widely sought A tool we will use throughout is the Cartesian product of graphs, where the notation G H means the Cartesian product of graphs G and H (see Definition 3) We will in large part be focusing on the Cartesian product of t copies of a graph G, denoted G t : G t = G} G {{ G} t copies of G When G has a consecutive radio labeling, we will look to G t for new examples of graphs with such a labeling Since diam(g t ) increases as t does, we have some hope of finding high diameter examples by looking here Before moving on, we define Hamming graphs, an important example of the Cartesian product both within the context of this thesis and without; Hamming graphs are strongly linked to coding theory, for example ([6], [33]) A Hamming graph is any graph of the form K n1 K n2 K nd where n 1,, n d are (not necessarily distinct) integers with both d 2 and n i 2 for all i We have already mentioned Theorem 12 which gives conclusions for such a Hamming graph when n 1 = = n d 3 and d n 1 Theorem 17 is another related result; it says that a Hamming graph with n 1,, n d 3 and n 1,, n d relatively prime is radio graceful

18 9 CHAPTER 2 EXISTENCE RESULTS In this chapter, the central portion of the thesis, we will see four main results that speak to the existence of consecutive radio labelings The first three results use some chosen radio graceful graphs as building blocks, and the Cartesian product as cement, to produce new examples The focus in these three results is almost exclusively on complete graphs and Hamming graphs (Cartesian products of complete graphs) The fourth result looks to one of the few preexisting nontrivial examples of a radio graceful graph: the Petersen graph In Section 24, we will be able to construct a family of graphs that are the same in spirit as the Petersen graph, and show that they too share the title radio graceful The proofs of these results are collectively similar in flavor In each case, we will begin by defining some well-chosen ordering of the vertices of the graph in question As none of these definitions is a priori well-defined, the first subsequent hurdle is to prove as much We will then prove that the given ordering induces a consecutive radio labeling; this task is summarized by Lemma 4 Before diving into the actual mathematics, we must pause for a few preliminaries Radio labeling is defined for simple, connected graphs, and the graphs in this thesis will be assumed to have those properties unless otherwise indicated The distance between vertices u, v in a graph G will be denoted d G (u, v), or, when G is clear from context, d(u, v); diameter of G is denoted diam(g) We will use the convention that a (mod b) {1,, b} As mentioned in the Introduction, the following

19 10 operation on graphs will be used heavily throughout (see Figure 21 for an example): Definition 3 Given two graphs G and H, the Cartesian product of G and H, denoted G H, is a graph defined by the following: (1) The vertex set V G H is given by V G V H (2) Two vertices (u, v), (u, v ) V G H are adjacent if the vertices in one coordinate are the same, and the vertices in the other coordinate are adjacent in their respective original graph That is, if (i) u = u and v is adjacent to v in H, or (ii) v = v and u is adjacent to u in G a b c See Fig = (a, 1) (b, 1) (a, 2) (c, 1) (b, 2) (a, 3) (b, 3) (c, 2) (c, 3) Figure 21: Cartesian product example: K 3 P 3 One very nice property of the Cartesian product of graphs is that distance is inherited from the original graphs: d G H ((u, v), (u, v )) = d G (u, u ) + d H (v, v )

20 11 This allows us to investigate the influence that rn(g) and rn(h) have over rn(g H) The Cartesian product of t copies of a graph G is denoted G t : G t = G} G {{ G} t copies of G Vertices of G t can be represented as t-tuples with entries in V G, and there is an edge between two vertices (v i1, v i2,, v it ) and (v j1, v j2,, v jt ) if, for some a {1,, t}, v ia is adjacent to v ja in G, and v ik = v jk for all k a There are V G t vertices and t V G t 1 E G edges in G t (where E G denotes the set of edges in G) Distance is given by d G t((v i1, v i2,, v it ), (v j1, v j2,, v jt )) = t d G (v ik, v jk ) Diameter is then easily computed: diam(g t ) = t k=1 diam(g) = t diam(g) The following lemma is a simple rewording of the definition of consecutive radio labeling We make special mention of it here because this vantage point of the k=1 definition will be found repeatedly useful for our purposes Lemma 4 Let G be a graph with n vertices An ordering x 1,, x n of the vertices of G induces a consecutive radio labeling of G if and only if the following is satisfied: d(x i, x i+c ) diam(g) + 1 c (21) for all c {1,, diam(g)}, i {1,, n c} Proof Let G be a graph with n vertices, and suppose x 1,, x n induces a consecutive radio labeling f Then f(x i ) < f(x j ) if and only if i < j This together with the assumption that f is a consecutive radio labeling puts f(x i ) = i for all {1,, n}

21 12 And, as f must satisfy the radio condition (12), we have f(x i ) f(x j ) diam(g) + 1 d(x i, x j ) i j diam(g) + 1 d(x i, x j ) d(x i, x j ) diam(g) + 1 i j for all distinct vertices x i, x j Since this inequality is trivially satisfied for any i, j such that i j > diam(g), we can restrict j to i + c where c {1,, diam(g)}, i {1,, n c} So we get (21): d(x i, x i+c ) diam(g) + 1 c Instead, suppose x 1,, x n is an ordering of the vertices that satisfies (21) for all c {1,, diam(g)}, i {1,, n c} Define a labeling f : V G Z + by f(x i ) = i for all {1,, n} Then d(x i, x i+c ) diam(g) + 1 c i + c i diam(g) + 1 d(x i, x i+c ) f(x i ) f(x i+c ) diam(g) + 1 d(x i, x i+c ), which is the radio condition So f satisfies the radio condition for all integers c {1,, diam(g)}, i {1,, n c} If c > diam(g), then the above f satisfies the radio condition trivially because then c diam(g) + 1 d(x i, x i+c ) regardless of the value of d(x i, x i+c ) Hence, f is a consecutive radio labeling

22 13 21 Powers of Complete Graphs We will encounter our first significant result in Theorem 12, which asserts that Kn t has a consecutive radio labeling if n 3 and 1 t n This result is consequential because it proves for the first time that there are radio graceful graphs of arbitrary diameter In fact, with it we are able to construct infinitely many radio graceful graphs of any specified diameter We begin by defining a list (with possible repetition and exclusion) of the vertices of Kn, t then we prove that this definition is actually an ordering of V Kn (ie, our list has no repetition or exclusion after all), and finally we show that the ordering induces a consecutive radio labeling of Kn t We will be able to add to this result in Section 22; see Corollary 15 At the end of this section we give an alternate proof of Theorem 12; it is shorter, with less emphasis on construction details 211 Definition of x 1, x 2,, x n t Let V Kn = {v 1,, v n } be the vertices of K n, n 3 Consider Kn t where t Z n We describe the desired ordering of the vertices of Kn t in groups of n vertices at a time The first n vertices are the t-tuples x 1 = (v 1, v 1,, v 1 ) x 2 = (v 2, v 2,, v 2 ) x n = (v n, v n,, v n )

23 14 For the sake of easier reference, we will think of these n vertices as an n t matrix A 1 where a 1 i,j is the j th coordinate of x i That is, v 1 v 1 v 1 v 2 v 2 v 2 A 1 = v n v n v n Let σ S VKn be the n-cycle (v 1 v 2 v n ) We produce a second matrix A 2 where a 2 i,j = σ(a 1 i,j) a 1 i,j if j = t otherwise We then form subsequent matrices by the following rule To produce A k, we first determine p, the largest integer such that k 1 (mod n p ) Then A k is the n t matrix made up of entries a k i,j = σ(a k 1 a k 1 i,j i,j ) if j = t p otherwise Using this definition we create n t 1 matrices of dimensions n t: A 1, A 2,, A nt 1 Each matrix A k is identical to the one that came before (A k 1 ) except for a single column which differs by an application of σ The n rows of each matrix (which total n t rows over all n t 1 matrices) correspond to vertices of Kn, t and tell us our ordering, n vertices at a time Formally, if i = bn + c, c {1, 2,, n}, then x i = x bn+c = (a b+1 c,1, a b+1 c,2,, a b+1 c,t ), (22)

24 15 the c th row of matrix A b+1 Table 211 displays this definition of x 1,, x 27 for the example K 3 3 x 1 : (v 1, v 1, v 1 ) x 10 : (v 1, v 2, v 3 ) x 19 : (v 1, v 3, v 2 ) x 2 : (v 2, v 2, v 2 ) x 11 : (v 2, v 3, v 1 ) x 20 : (v 2, v 1, v 3 ) x 3 : (v 3, v 3, v 3 ) x 12 : (v 3, v 1, v 2 ) x 21 : (v 3, v 2, v 1 ) x 4 : (v 1, v 1, v 2 ) x 13 : (v 1, v 2, v 1 ) x 22 : (v 1, v 3, v 3 ) x 5 : (v 2, v 2, v 3 ) x 14 : (v 2, v 3, v 2 ) x 23 : (v 2, v 1, v 1 ) x 6 : (v 3, v 3, v 1 ) x 15 : (v 3, v 1, v 3 ) x 24 : (v 3, v 2, v 2 ) x 7 : (v 1, v 1, v 3 ) x 16 : (v 1, v 2, v 2 ) x 25 : (v 1, v 3, v 1 ) x 8 : (v 2, v 2, v 1 ) x 17 : (v 2, v 3, v 3 ) x 26 : (v 2, v 1, v 2 ) x 9 : (v 3, v 3, v 2 ) x 18 : (v 3, v 1, v 1 ) x 27 : (v 3, v 2, v 3 ) Table 21: The list of vertices x 1,, x 27 for K The x i s Form an Ordering Our first goal is to show that this definition of x 1,, x n t is actually an ordering of the vertices of Kn t Certainly {x 1,, x n t} V Kn This coupled with the fact that V Kn = n t means we only need to show that our proposed ordering has no repetition: x i x j for all i j We can make this task a bit easier by noticing some structure of the A k s A 1 has a structure that is inherited by A k for all k: a k i,j = σ(a k i 1,j) for

25 16 all i {2,, n}, j {1,, t} The fact that σ is an n-cycle tells us immediately that, for any fixed index k {1,, n t 1 }, the matrix A k has no duplicate rows Also, since k n t 1, and since p is the largest integer such that k 1 (mod n p ), we see that p t 2 Consequently, σ is never applied to the first column during the construction of the matrices A 1, A 2,, A nt 1 We conclude that all of these matrices have the same first column These two observations about the matrices A 1, A 2,, A nt 1 imply that any row of A k determines all of A k For this reason, it suffices to show that no two matrices have the same first row Since we are narrowing our scope to the first rows of each matrix, we make a new matrix A comprised of them Let A be the n t 1 t matrix defined by a i,j = a i 1,j We can give an equivalent definition by using the definition of A k Let q be the largest integer such that i 1 (mod n q ) Then v 1 if i = 1 a i,j = σ(a i 1,j ) if j = t q a i 1,j otherwise is the entry of matrix A in the i th row and j th column Now we show that A has no identical rows We will start by studying the specific and repetitive structure of A To do this, we break up each column into blocks what we ll call a certain collection of column vectors that, when adjoined, produce the original column Fix a column j We wish to break up the column into groups of n t j entries which will

26 17 form the blocks So, the vectors a 1,j, a n t j +1,j,, a (n j 1)n t j +1,j (23) a n t j,j a 2n t j,j a n t,j are the blocks of column j Note that the blocks are of uniform size, but that size is dependent on the column The first column of A has only one block, while in the last (t th ) column of A each entry is itself a block In order to easily reference which column a block comes from, we will commonly call a block from column j a j-block In summary: Definition 5 A j-block is any one of the vectors listed in line (23) We will call the set of rows that are associated to the entries of a block the scope of the block Definition 6 The scope of a j-block a cn t j +1,j a (c+1)n t j,j is the set with rows cn t j + 1 through (c + 1)n t j of A as its elements: {[a cn t j+1,h] 1 h t,, [a (c+1)n t j,h] 1 h n } The scope of multiple j-blocks is the union of the scopes of the individual j-blocks Claim 7 If a j-block of A has entries a cn t j +1,j,, a (c+1)n t j,j, then a cn t j +1,j = = a (c+1)n t j,j Proof The j = t case is immediate; assume j < t Based on the definition of A, a i,j = a i 1,j unless j = t q where q is the largest integer such that i 1

27 18 (mod n q ) Equivalently, a i,j = a i 1,j unless q = t j The only way to get any distinct entries in the block is for one of the entries (excluding the first) to be the image under σ of the previous entry So the question is, is it possible for q = t j when i {cn t j + 2, cn t j + 3,, (c + 1)n t j }? One condition for q = t j is that i 1 (mod n t j ) It s clear that this cannot be the case for any of the possibilities for i: cn t j (mod n t j ) cn t j (mod n t j ) (c + 1)n t j 0 (mod n t j ) Therefore, since j t q for all i {cn t j + 2, cn t j + 3,, (c + 1)n t j }, each entry of the vector must be the same as the first entry a cn t j +1,j As a result of this claim, along with the definition of A, we know that adjacent blocks in a column can only have two possible relationships: either they are identical, or the second block has entries which are the image under σ of the first block s entries We d like to know when adjacent blocks are identical Claim 8 The adjacent blocks in column j of A a cn t j +1,j, a (c+1)n t j +1,j are identical if and only if n divides c + 1 a (c+1)n t j,j a (c+2)n t j,j Proof It follows from Claim 7 that the blocks are identical if and only if a (c+1)n t j,j =

28 19 a (c+1)n t j +1,j We know from our definition of A that this happens only when j t q, where q is computed for i = (c + 1)n t j + 1 So what is q in this case? Certainly t j is a candidate since (c + 1)n t j (mod n t j ) In fact, q t j if and only if q > t j, which is equivalent to requiring that n divides c + 1 The blocks have now given us a really detailed picture of what s happening with our matrix A If we look down an arbitrary column j, the column is partitioned into n j 1 blocks, each with n t j entries The entries of any fixed block are identical The first block in each column has all entries of v 1 When changes do happen down a column, they change according to σ; so v 1 can be followed by v 2 then v 3 and so on And we now know by Claim 8 exactly when those changes will not occur due to the repetition of adjacent blocks To see an example of this, Figure 22 depicts the matrix A for K4; 4 blocks are separated in each column by lines, and different entries are shaded differently to better display the pattern The 64 rows of the matrix prohibit displaying it in one piece, so it is broken into four pieces with arrows signifying how they should be mentally attached Now we want to examine how the structure of the different columns relate It happens that the scope of a (j 1)-block is equal to the scope of a collection of j-blocks the appropriate n of them in a row This is immediate because of the size of the blocks A (j 1) block has n t j+1 entries, and j-blocks have n t j entries So, for every block in column j 1, there are n t j+1 /n t j = n blocks in column j Interestingly, those n j-blocks with the same scope as the single (j 1)-block must be distinct (in the sense that no two of these j-blocks can have the same entries)

29 20 Figure 22: The matrix A for K 4 4 with blocks We will prove this in the following claim Note that the blocks listed in the claim are in column j, of the correct size (n t j entries), they are adjacent, and there are n of them Also note that they cover rows bn t j through (b + 1)n t j+1 This is significant because a bn t j+1 +1,j+1 a (b+1)n t j+1,j+1 is a (j 1)-block Claim 9 The following j-blocks in A are all distinct: a bn t j+1 +1,j a bn t j+1 +n t j,j, a bn t j+1 +n t j +1,j a bn t j+1 +2n t j,j,, a bn t j+1 +(n 1)n t j +1,j a (b+1)n t j+1,j

30 21 Proof Let s begin by proving that the first two j-blocks a bn t j+1 +1,j a bn t j+1 +n t j,j, a bn t j+1 +n t j +1,j a bn t j+1 +2n t j,j are distinct In order to easily reference Claim 8, we rewrite the indices of the entries in the form that appears there: a bn t j+1 +1,j a bn t j+1 +n t j,j = a (bn)n t j +1,j a (bn+1)n t j,j, a bn t j+1 +n t j +1,j a bn t j+1 +2n t j,j = a (bn+1)n t j+1 +1,j a (bn+2)n t j+1,j Now we are already done because (by Claim 8) showing these two blocks are distinct is equivalent to showing n does not divide bn + 1, which is immediate With the same argument we can show that, in this collection of n blocks, adjacent blocks are distinct We proceed by similarly rearranging the remaining indicies to be compatible with the previous claim a bn t j+1 +2n t j +1,j a bn t j+1 +3n t j,j = a (bn+2)n t j +1,j a (bn+3)n t j,j a bn t j+1 +(n 1)n t j +1,j a (b+1)n t j+1,j = a (bn+n 1)n t j +1,j a (bn+n)n t j,j

31 22 Showing that adjacent blocks are distinct is equivalent to showing that n does not divide bn + 2, bn + 3,, bn + n 1, which is again immediate Because of this we can now say the following: if the first block has entries equal to v k, then the second has entries of σ(v k ), and the third has entries of σ 2 (v k ),, and the n th has entries of σ n 1 (v k ) Since σ is an n-cycle, v k σ(v k ) σ 2 (v k ) σ n 1 (v k ), which confirms that the n blocks are distinct We are now able to prove matrix A has no repeated rows by showing that an arbitrary row can only exist in one place in A Lemma 10 The matrix A has no two rows with identical entries Proof We begin by proving the following claim using an induction argument Claim: If two rows share their first k entries, then they both belong to the scope of the same k-block Proof of Claim: As we ve already mentioned, the first column of A is itself one block, so the k = 1 case is trivial Suppose the claim is true for k, and let the x th and y th rows share their first k + 1 entries Then they must share their first k entries, so by assumption rows x and y belong to the scope of a single k-block By Claim 3, this scope is equal to that of a collection of n distinct (k + 1)-blocks Since rows x and y also share the (k + 1)st entry, they can only be in the scope of one of those (k + 1)-blocks (the one with entries equal to the (k+1)st entry of the x th and y th rows), which proves the claim This result tells us that if rows x and y share all t of their entries, then they

32 23 belong to the scope of the same t-block But a t-block consists of n t t = 1 entry, so its scope contains only one row Then x = y and no two rows of A can have identical entries Theorem 11 Let n Z 3, t Z n As defined in line (22), x 1,, x n t is an ordering of the vertices of K t n Over the course of this section, we have already made the arguments proving Theorem 11: We defined the x i s to be the rows of the matrices A 1,, A nt 1 We needed only to show that no two of those rows are identical For each k, A k has the characterisitic that a k i,j = σ(a k i 1,j), which implies that fixing a row fixes all of A k (It also implies that A k cannot have any repeated rows itself) So if any two rows across all the rows of A 1,, A nt 1 are identical, that would imply two of the A k s are identical Then of course the first rows of those two matrices are identical, which makes two rows in A identical, but Lemma 10 proved that this cannot happen So our definition of x 1,, x n t is indeed an ordering of V K t n 213 The Ordering Induces a Consecutive Radio Labeling To prove that our ordering induces a consecutive radio labeling, we need to keep track of instances where close vertices have common entries Let e i,j be the number of coordinates where x i and x j agree If x i = (v i1, v i2,, v it ) and x j = (v j1, v j2,, v jt ), then d K t n (x i, x j ) = t d Kn (v ik, v jk ) = t e i,j k=1

33 24 Let f be the radio labeling induced from our ordering The radio condition becomes f(x i ) f(x j ) diam(k t n) + 1 d(x i, x j ) = t + 1 (t e i,j ) = 1 + e i,j Equivalently, e i,j f(x i ) f(x j ) 1 In order for f to be consecutive, (ie, f(x i ) = i for all i), e i,j must then satisfy e i,j i j 1 (24) Since e i,j counts coordinates, 0 e i,j t Then e i,i+s trivially satisfies (24) if s t + 1 Also, if s = t, then (24) is equivalent to requiring that x i and x i+s are not identical, which Theorem 11 asserts We will show that e i,i+s satisfies (24) for all s Theorem 12 Let n Z 3, t Z n Then K t n is radio graceful Proof We use the ordering x 1,, x n t of the vertices of K t n from Theorem 11, and we let f : V K t n Z + be defined as f(x i ) = i We will prove that f satisfies the radio condition From the above discussion, we need to show that e i,i+s satisfies e i,i+s i (i s) 1 = s 1 for all 1 s t 1 We will be referring again to the A k s the matrices from Section 211 used to define x 1,, x n t Let x i be a row in matrix A k i If x i+s is also a row in matrix A k i, then e i,i+s = 0 since a k i i,j = σ(ak i i 1,j ) Suppose x i+s is not a row in A k i By assumption s t 1 < n Since the A k s each have n rows, x i+s must be a row in A k i+1 Note that, because s < n, x i and x i+s cannot have the same placement as rows in A k i and A k i+1 (ie, if x i is the third row of A k i, then x i+s cannot be the third row of A k i+1 ) We have seen that A k i and A k i+1

34 25 are identical except for a single column, and that column differs by an application of σ These last two observations together imply that e i,i+s 1 With this, there is only one case left to consider: s = 1 We need to show e i,i+1 = 0 In this situation, x i must be the n th row of A k i row of A k i+1 So x i+1 is identical to the first row of A k i and x i+s must be the first except for one coordinate, let s call this the c th coordinate That is, x i+1 = (a k i 1,1,, σ(a k i 1,c),, a k i 1,t) We also know that x i = (σ n 1 (a k i 1,1),, σ n 1 (a k i 1,c),, σ n 1 (a k i 1,t)) We can observe then that x i and x i+1 share no common coordinates (because σ is an n-cycle, n 3) In particular, we check that the c th coordinates are different: indeed, σ(a k i 1,c) σ n 1 (a k i 1,c) We have shown e i,i+1 = 0 Therefore, e i,i+s s 1 for all s, and K t n is radio graceful Figure 23 shows the consecutive radio labeling induced by the ordering given in Table 211, completing the example of K3 3 in reference to Theorem 12 We note that Kn 2 = C 4 does not have a consecutive radio labeling (hence the n 3 hypothesis) There are only three nonequivalent orderings of the vertices of K2 2 Table 22 shows these orderings, along with the radio labelings induced by them, none of which are consecutive 214 Alternate Proof We also give the following more concise proof of Theorem 12 This proof provides less intuition about the construction of the ordering of the vertices it uses

35 26 Figure 23: A consecutive radio labeling of K 3 3 to induce a consecutive radio labeling, which is the price of its brevity Lemma 13 Let n Z 3, t Z n Then this recursive definition is an ordering on V K t n : x 1 i = v i, and, for k = 1,, n, x t mn+k = ( ) x t 1 m/n n+k, v k+m m/n (mod n) where we use the convention that a (mod n) {1,, n} Proof We must first to confirm 1 m n n + k n t 1 to check that x t 1 m/n n+k is

36 27 x 1 x 2 x 1 x 2 x 1 x 3 x 4 x 3 x 3 x 4 x 4 x Table 22: All nonequivalent induced radio labelings of K 2 2 well-defined Since 1 mn + k n t, and 1 k n, then 0 m n t 1 1 Then 0 m n n t m n n n t 1 n k m n n + k n t 1 n + k 1 k m n n + k n t 1 n + k n t 1 It remains to be shown that {x t 1,, x t n t } = V K t n We will do this by inducting on t If t = 1, {x t 1,, x t n t } = V K t n by definition Case: t = 2 Suppose x 2 mn+k, x2 Mn+K {x2 1,, x 2 n 2 } where 1 k, K n, such that ( x 1 m/n n+k, v k+m m/n (mod n) ) = ( x 1 M/n n+k, v K+M M/n (mod n) ) (25)

37 28 Then x 1 m/n n+k = x 1 M/n n+k v m/n n+k = v M/n n+k m n n + k = M n n + K k K (mod n) k = K since 1 k, K n This equality also tells us that m n = M n Since we are assuming (25), we also know v k+m m/n (mod n) = v K+M M/n (mod n) k + m m n K + M M n (mod n) m M (mod n) Together, m n = M n and m M (mod n) tell us that m = M So, if k = K and m = M, then mn + k = Mn + K This shows that x 2 i, x 2 j {x 2 1,, x 2 n 2 } are the same if and only if i = j It s clear that {x 2 1,, x 2 n 2 } V K 2 n, and now we know that both sets contain n 2 distinct elements We conclude {x 2 1,, x 2 n 2 } = V K 2 n Case: Suppose the statement is true for t = j: {x j 1,, x j n j } = V K j n Then let x j+1 mn+k, xj+1 Mn+K {xj+1 1,, x j+1 n j+1 } where 1 k, K n, such that ( ) ( ) x j m/n n+k, v k+m m/n (mod n) = x j M/n n+k, v K+M M/n (mod n) Then x j m/n n+k = xj M/n n+k, and by assumption that can only be true if m n n+k =

38 29 M n n + K Just as before, we will conclude that k = K, m n = M n, and m = M So we get {x t 1,, x t n t } = V K t n For the convenience of the reader, we restate Theorem 12 before giving its alternate proof Theorem 12 Let n Z 3, t Z n Then K t n is radio graceful Proof To show that the ordering given in Lemma 13 induces a consecutive radio labeling on Kn, t we must check that e i,j i j 1 (where e i,j is the number of coordinates with common entries between x i and x j ) Note that x i and x j have t coordinates and are distinct, so the statement e i,j t 1 is automatically satisfied Thus we need only concern ourselves with vertices of the form x i and x i+ where 0 < t 1 Since t n, we will let n 1 We d like to know the value of e i,i+, but first we ask the following question: For fixed γ {1,, t}, when do x i and x i+ share the γ th coordinate? Let γ(x i ) denote the entry of the γ th coordinate of x i, where i = mn+k Then γ(x i ) = v k+ m/n t γ m/n t γ+1 Suppose γ(x i ) = γ(x j ), where j = i + Case 1: k + n j = mn + (k + ) Then we show that γ(x i ) γ(x j ): (mod n) 0 m m k + + n t γ n t γ+1 (mod n) k + m n t γ γ(x mn+(k+ ) ) (mod n) γ(x mn+k ) m n t γ+1

39 30 Case 2: k + > n and n t γ does not divide m + 1 Then j = (m + 1)n + δ where δ = k + n, m+1 n = m t γ n, and m+1 t γ n = t γ+1 γ(x i ) γ(x j ): m n Again we can show that t γ+1 (mod n) 0 n (mod n) 0 k δ (mod n) 0 k (mod n) δ m m m m k + n t γ n t γ+1 (mod n) δ + n t γ n t γ+1 m m m + 1 m + 1 k + n t γ n t γ+1 (mod n) δ + n t γ n t γ+1 γ(x mn+k ) (mod n) γ(x (m+1)n+δ ) Case 3: k + > n, n t γ divides m + 1, and n t γ+1 does not divide m + 1 Then j = (m + 1)n + δ where δ = k + n, m+1 n = m t γ n + 1, and m+1 t γ n = t γ+1 m n t γ+1 In this case, it is possible for γ(x i ) to equal γ(x j ) The following are equivalent

40 31 statements: γ(x mn+k ) (mod n) γ(x (m+1)n+δ ) m m m + 1 m + 1 k + n t γ n t γ+1 (mod n) δ + n t γ n t γ+1 m m m m k + n t γ n t γ+1 (mod n) δ n t γ n t γ+1 k (mod n) δ + 1 k (mod n) k + n + 1 (mod n) n 1 = n 1 Case 4: k + > n and n t γ+1 does not divide m + 1 Then j = (m + 1)n + δ where δ = k + n, m+1 n = m t γ n + 1, and m+1 t γ n = t γ+1 computation yields: m n +1 A similar t γ+1 (mod n) 0 n (mod n) 0 k δ (mod n) 0 k (mod n) δ m m m m k + n t γ n t γ+1 (mod n) δ + n t γ n t γ+1 m m m ( m k + n t γ n t γ+1 (mod n) δ n t γ n t γ+1 m m m + 1 m + 1 k + n t γ n t γ+1 (mod n) δ + n t γ n t γ+1 γ(x mn+k ) (mod n) γ(x (m+1)n+δ ) ) + 1

41 32 By the inspection of these cases, γ(x mn+k ) = γ(x (mn+k)+ ) if and only if = n 1, k + > n, n t γ divides m + 1, and n t γ+1 does not divide m + 1 We conclude e i,i+ = 0 if < n 1 Now we only need to confirm e i,i+n 1 n 2 Suppose that x i and x i+n 1 share entries in both the γ th and ϕ th coordinates, such that γ < ϕ Then, in particular, that would mean n t γ divides m+1 and n t ϕ+1 does not divide m + 1 But, since t γ t ϕ + 1, this isn t possible So, x i and x i+n 1 can have the same entries in only one coordinate Then e i,i+n 1 1 n 2 We have shown this ordering induces a consecutive radio labeling for Kn t 22 Powers of Incomplete Graphs A natural question follows Theorem 12: Are there incomplete graphs G such that G t is radio graceful for some t? Theorem 14 will describe such a G, and Corollary 15 will show show the existence of graphs satisfying that description Interestingly, Theorem 14 about powers of incomplete graphs will allow us to extend our results in Theorem 12 about powers of complete graphs (see Corollary 15) We will need the following definition: The distance modulo n between i, j {1,, n} is equal to min{ i j, n i j } and is denoted i j n It is helpful to note that i j n n/2; since i j + n i j = n, it is always the case that min{ i j, n i j } n/2 Theorem 14 Let G be an incomplete connected graph with n vertices, let t Z +, and let y 1,, y n be an ordering of the vertices of G such that d(y i, y j ) diam(g) i j n 1 t

42 33 for all distinct i, j {1,, n} Then G t is radio graceful Proof Let G be an incomplete connected graph with such an ordering y 1,, y n of its vertices for some fixed t Z + We begin by showing that t n Since G is diam(g) connected, it must have at least n 1 edges Because of this, there must be at least n 1 pairs {i, j} {1,, n} {1,, n} such that diam(g) i j n 1 t 1, (26) corresponding to the at least n 1 pairs {y i, y j } V G V G of adjacent vertices Solving for t produces diam(g) i j n 1 t 1 t diam(g) i j n + 1 t t diam(g) t i j n 1 t i j n 1 diam(g) 1 Note that the division in the last step is valid because G is incomplete By the definition of i j n, it follows that i j n n/2 So t n/2 1 diam(g) 1

43 34 And, since diam(g) > 1, Therefore, t 0 (diam(g) 2)n + 2 diam(g) 0 n diam(g) + 2 diam(g) 2n n diam(g) 2 diam(g) 2n diam(g) 2n n diam(g) (n 2)diam(G) 2n(diam(G) 1) n 2 diam(g) 1 n/2 1 diam(g) 1 2n diam(g) n diam(g) We will define an ordering x 1,, x n t of the vertices of G t in a nearly identical way as in the beginning of Section 21 The first n vertices are the t-tuples x 1 = (y 1, y 1,, y 1 ) x 2 = (y 2, y 2,, y 2 ) x n = (y n, y n,, y n ), and the subsequent vertices in the list will be defined again by the A m matrices (m {1,, n t 1 }) where σ will now be the n-cycle (y 1 y 2 y n ) S VG The proof that this definition of x 1,, x n t is an ordering of V G t is identical to the one found in Section 21 Let f : V G t Z + be the labeling defined by f(x i ) = i We wish to show f is a radio labeling That is, we wish to show f satisfies the radio condition for all

44 35 i, j {1,, n t } with i j: f(x i ) f(x j ) diam(g t ) + 1 d G t(x i, x j ), or this equivalent inequality d G t(x i, x j ) t diam(g) + 1 i j (27) If we let x i = (y i1,, y it ) and let x j = (y j1,, y jt ), then we have d G t(x i, x j ) = t k=1 d G(y ik, y jk ) And we are assuming d(y i, y j ) diam(g) i j n 1 t for i j We can use this to bound d G t(x i, x j ) and eventually prove (27) Since x 1,, x n t is an ordering, d G t(x i, x j ) 1 for all i, j {1,, n t } with i j With this, it remains to show (27) when t diam(g)+1 i j 2, which will only happen when i and j are sufficiently close We can manipulate this inequality to make sufficiently close precise: t diam(g) + 1 i j 2 t diam(g) 1 i j Since we showed t n, we can simplify this bound on i j : diam(g) i j t diam(g) 1 ( ) n diam(g) 1 diam(g) = n 1 Therefore we must show (27) for all i, j {1,, n t } with 0 < i j n 1 So consider x i = (y i1,, y it ) and x i+s with 1 s n 1 Suppose x i and x i+s are both

45 36 rows in A m i Then x i+s = (σ s (y i1 ),, σ s (y it )) Now, for each coordinate k, d G (y ik, σ s (y ik )) = d G (y ik, y ik +s (mod n)) diam(g) i k (i k + s (mod n)) n 1 t diam(g) s 1 t We can justify the last step by proving i k (i k + s (mod n)) n s, which we will now do Recall i k {1,, n}, s {1,, n 1} If i k + s < n, then i k (i k + s (mod n)) n = a (a + s) n = min{ a (a + s), n a (a + s) } = min{s, n s} s If i k + s n, then (i k + s (mod n)) = i k + s n since i k {1,, n} and s {1,, n 1} So, i k (i k + s (mod n)) n = a (a + s n) n = n s n = min{ n s, n n s } Now, we can even ignore any relationship between n and s for the moment; if s < n, then n n s = s, and if s n, then n s < s So in all cases min{ n s, n n s } s Therefore, i k (i k + s (mod n)) n s Now that we have the bound d G (y ik, σ s (y ik )) diam(g) s 1 t which is inde-

46 37 pendent of the coordinate k, we can bound d G t(x i, x i+s ): ( d G t(x i, x i+s ) t diam(g) s 1 ) t = t diam(g) (s 1) = t diam(g) s + 1 = t diam(g) + 1 i (i + s) So in this case (the case where x i and x i+s are both rows in A m i ), (27) is satisfied Now we suppose x i is a row in A m i and x i+s is not Since we are assuming s n 1, we can conclude x i+s is a row in A mi+1 We know from their definitions that A m i and A mi+1 are identical except for a single column which differs by an application of σ Say the column in this case is column c That is, if a i represents the i th column of A m i, then A mi+1 = [a 1 a c 1 σ(a c ) a c+1 a t ] So if x i = (y i1,, y it ) then x i+s = (σ s (y i1 ),, σ s (y ic 1 ), σ s+1 (y ic ), σ s (y ic+1 ),, σ s (y it )) We proceed in a similar manner as the previous case, but we will need more precision So instead of manipulating the inequality d(y i, y j ) diam(g) i j n 1, t we will use i j n 1 d(y i, y j ) diam(g) t

47 38 These two inequalities give the exact same bound on d(y i, y j ) since d(y i, y j ) Z + (If a, b, c, d, e, f Z +, f < d, and c/d = e + f/d, then we get a b c/d a b (e + f/d) a b e f/d a b e a b c/d ) We have d G t(x i, x i+s ) = and we know ) ( t ) d G (y ik, σ s (y ik )) + d G (y ic, σ s+1 (y ic )) + d G (y ik, σ s (y ik )), ( c 1 k=1 k=c+1 (28) d G (y ik, σ s (y ik )) = d G (y ik, y ik +s (mod n)) ik (i k + s (mod n)) n 1 diam(g) t s 1 diam(g) t Using justifications similar to those given earlier, it is also true that d G (y ic, σ s+1 (y ic )) = d G (y ic, y ic+s+1 (mod n)) ic (i c + s + 1 (mod n)) n 1 diam(g) t s diam(g) t s = diam(g) t

48 39 Using these bounds in conjunction with (28) gives s 1 If = t ( d G t(x i, x i+s ) (t 1) diam(g) ) s 1 s + diam(g) t t s 1 = t diam(g) (t 1) t s t, then we can further conclude s 1 d G t(x i, x i+s ) t diam(g) t t ( ) s 1 t diam(g) t t = t diam(g) s + 1 = t diam(g) + 1 i (i + s) (29) s t (210) s 1 Again we have satisfied (27) We have left to consider the possibility that t s s, in which case t s So = s s 1 t t t and = s 1 Now the bound starting t t in line (29) becomes ( s ) d G t(x i, x i+s ) t diam(g) (t 1) t 1 s t ( s ) = t diam(g) t + (t 1) t = t diam(g) s + t 1 = t diam(g) + (t 1) i (i + s) t diam(g) + 1 i (i + s) This finally shows that (27) is satisfied for all distinct i, j {1,, n t }, and therefore f is a consecutive radio labeling of G t

49 40 This leads to the following result, which, in conjunction with Theorem 12, increases the value of t for which K t n has a consecutive radio labeling Note that p n/p > n when p n Corollary 15 Let n Z + and let p {2, 3,, n 2} such that p n Then K n/p p n has a consecutive radio labeling Proof Let n Z + and let p {2, 3,, n 2} such that p n We will show that Kn n/p ( K n/p n satisfies the hypotheses of Theorem 14 for t = p, and therefore conclude ) p = n/p p K has a consecutive radio labeling n Our goal is to exhibit an ordering y 1,, y n n/p of the vertices of K n/p n such that d n/p K (y i, y j ) n/p i j n 1 n/p n p (211) for all distinct i, j {1,, n n/p } Let {v 1,, v n } be V Kn, and let {v 1,, v n } {v 1,, v n } {v 1,, v n } }{{} n/p copies of V Kn represent the vertices of K n/p n Consider the ordering x 1,, x n n/p of V K n/p n defined in Section 21 Recall that e i,j denotes the number of coordinates over which ) ) x i and x j agree Say x i = (v i1,, v i n/p and x j = (v j1,, v j n/p Because n/p d n/p K (x i, x j ) = d Kn (v ik, v jk ) = n/p e i,j, (212) n k=1 as we again will be exclusively concerned with the value of e i,j If x 1,, x n n/p is to be an ordering that satisfies (211), then the following would need to be true (from

50 41 (212)): n/p e i,j n/p i j n 1 n/p p So our goal is to prove that e i,j is bounded thusly: e i,j i j n 1 n/p (213) p As e i,j counts coordinates within n/p -tuples, e i,j n/p We also know x i x j when i j, so e i,j n/p 1 when i j Therefore, we will be interested in proving (213) when i j n n/p 1 p n/p 2 i j n n/p p n/p 2p + 1 i j n n/p p (n/p + 1) 2p + 1 i j n n/p n p + 1 i j n n/p n 1 since p 2 So let i, j be distinct elements of {1,, n n/p } with i j n n/p n 1 Without loss of generality, j = i + s (mod n n/p ) for some s {1,, n 1} As we have seen several times now, the value of e i,i+s (mod n n/p ) will depend on the placement of x i and x i+s (mod n n/p ) as rows in the matrices used to define the ordering ) (Section 21) Let x i = (v i1,, v i n/p be a row in the matrix A m i Case 1: Suppose x i+s (mod n n/p ) is also a row in A m i Then x i+s (mod n n/p ) = ( ) σ s (v i1 ),, σ s (v i n/p )

51 42 where σ is the n-tuple (v 1 v 2 v n ) S VKn As s {1,, n 1}, v ik σ s (v ik ) for all k We conclude e i,i+s (mod n n/p ) = 0, and (213) is satisfied Case 2: Suppose x i+s (mod n n/p ) is not a row in A m i, and suppose i + s n n/p Then, because s < n, x i+s (mod n n/p ) = x i+s is a row in A m i+1 Matrices A m i and A m i+1 are identical except for a single column which differs by an application of σ Say the column in this case is column c Then x i+s = ( ) σ s (v i1 ),, σ s (v ic 1 ), σ s+1 (v ic ), σ s (v ic+1 ),, σ s (v i n/p ) As in Case 1, v ik σ s (v ik ) because s {1,, n 1}, which means 1 if v ic = σ s+1 (v ic ) e i,i+s =, 0 otherwise or in other words 1 if s = n 1 e i,i+s = 0 otherwise So if s n 1, 213 is satisfied If s = n 1, then i j n n/p 1 p = i (i + n 1) n n/p 1 p = n 1 n n/p 1 p = n 2 p The final equality is due to the constraints on p Since p {2, 3,, n 2} such that p n, we know that n > 4 and n/p 2, in which case n 1 < n n/p (n 1)

52 43 Finally, since p n 2, in this case, so we get (213) i j n n/p 1 p = n 2 p 1 = e i,j Case 3: Suppose x i+s (mod n n/p ) is not a row in A m i, and suppose i + s > n n/p Then A m i = A n n/p 1 and x i+s (mod n n/p ) is a row in A 1 To simplify notation here, let t = n n/p Comparing these matrices will require us to determine the entries of A nt 1 Consider A, the n t 1 t matrix defined in Section 21 Determining the (n t 1 ) th row of A determines A nt 1, as we have seen In pursuit of that goal, we will describe A by describing the j-blocks that comprise it Recall that we were able to give a description of the j-blocks in Section 21: The j th column of A is partitioned into n j 1 blocks, each with n t j entries Those blocks are the vectors We will refer to a 1,j a n t j,j, a (i 1)n t j +1,j a in t j,j a n t j +1,j a 2n t j,j,, a (n j 1)n t j +1,j a n t,j as the i th j-block The entries of any fixed block are identical The first block in each column has entries of v 1 By Claim 4 which was proven in Section 21, the i th and (i + 1) th j-blocks are identical if and only if n i Otherwise, if the entries of the i th j-block are all equal to v k, then the entries of the

53 44 (i + 1) th j-block are all equal to σ(v k ) While this collection of conditions do uniquely describe the entries of the j-blocks, the following is more useful for our purposes: Claim 16 If i = bn + d where b Z 0 and d {1,, n}, then the i th j-block of A has entries of σ b(n 1)+(d 1) (v 1 ) Proof of Claim 16 We need to check that this description of the entries of the j-blocks matches the one we already have, which can be stated in three parts: 1 The first j-block has entries of v 1 If i = 1, then i = 0 n + 1, and σ 0 (n 1)+(1 1) (v 1 ) = σ 0 (v 1 ) = v 1 2 If n i then the entries in the i th j-block are identical to those in the (i + 1) th j-block If n i, then i = bn + n for some b Z 0, and σ b(n 1)+(n 1) (v 1 ) = σ (b+1)(n 1) (v 1 ) 3 If n i and the entries in the i th j-block are equal to v k, then the entries in the (i + 1) th j-block are equal to σ(v k ) If n i, then i = bn + d for some d {1,, n 1}, and σ(σ b(n 1)+(d 1) (v 1 )) = σ b(n 1)+d (v 1 ) = σ b(n 1)+[(d+1) 1] (v 1 ) Therefore, if i = bn + d where b Z 0 and d {1,, n}, then the i th j-block of A has entries of σ b(n 1)+(d 1) (v 1 ) because the descriptions match (Claim 16)

54 45 Using Claim 16, we can determine the (n t 1 ) th row of A As the (n t 1 ) th row is the last row in A, each of its entries are contained in the last j-block for each j (ie, a n t 1,j is an entry in the (n j 1 ) th j-block) Notice that if j = 1, then (n j 1 ) = 1 = 0 n + 1 If j = 2, then (n j 1 ) = n = 0 n + n And if j > 2, then (n j 1 ) = (n j 2 1)n + n It follows that σ 0 (n 1)+(1 1) (v 1 ) = σ 0 (v 1 ) = v 1 if j = 1 a n t 1,j = σ 0 (n 1)+(n 1) (v 1 ) = σ n 1 (v 1 ) = v n if j = 2 (214) σ (nj 2 1)(n 1)+(n 1) (v 1 ) = σ nj 2 (n 1) (v 1 ) = v 1 if j > 2 Now this row of A is the first row of A nt 1 ; by its definition we have A nt 1 = v 1 v n v 1 v 1 σ(v 1 ) σ(v n ) σ(v 1 ) σ(v 1 ) σ n 1 (v 1 ) σ n 1 (v n ) σ n 1 (v 1 ) σ n 1 (v 1 ), which simplifies to A nt 1 = v 1 v n v 1 v 1 v 2 v 1 v 2 v 2 v n v n 1 v n v n

55 46 Now we observe the result of applying σ to the second column of A nt 1 : v 1 σ(v n ) v 1 v 1 v 1 v 1 v 1 v 1 v 2 σ(v 1 ) v 2 v 2 v 2 v 2 v 2 v 2 = = A 1 v n σ(v n 1 ) v n v n v n v n v n v n Therefore, matrices A nt 1 and A 1 are related in the same way that the matrices were related in Case 2: they are identical except for a single column which differs by an application of σ The argument given in Case 2 is hence sufficient to show (213) in this case as well We have now shown that Kn n/p ( t = p, and therefore K n/p n ) p = K n/p p n satisfies the hypotheses of Theorem 14 for is radio graceful 23 Other Hamming Graphs In Section 21, we constructed examples of graphs that have consecutive radio labelings of arbitrary diameter In this section, we will focus on another basic property of a graph its order (number of vertices) The complete graphs K n of order n immediately show that there are graphs of any order that have consecutive radio labelings Corollary 18 of Theorem 17 gives a partial result to the next logical question: Are there nontrivial examples such graphs of arbitrary order? First we prove the following result about Hamming graphs (any Cartesian product of complete graphs with multiple vertices), which are meaningful in their own right due to their close connections to coding theory

56 47 Theorem 17 If n 1, n 2,, n s Z 2 are relatively prime, then K n1 K n2 K ns is radio graceful Proof Let n 1, n 2,, n s Z 2 be relatively prime, and G = K n1 K n2 K ns For each i, let the vertex set of K ni be V Kni = {v i1, v i2,, v ini }, and let f i : V Kni Z + be the labeling defined by f i (v j ) = j Then, to simplify notation, we will use the s-tuple (a 1, a 2,, a s ) {1, 2,, n 1 } {1, 2,, n 2 } {1, 2,, n s } to refer to the vertex (v 1a1, v 2a2, v sas ) in G Let n = n 1 n 2 n s For k {1, 2,, n}, define x k = ( ) k (mod n 1 ), k (mod n 2 ),, k (mod n s ) where we use the convention that k (mod n i ) {1, 2,, n i } We will show that x 1, x 2,, x n is an ordering of the vertices of G Each x k is a vertex of G by construction, and G has n vertices These facts together tell us that we only need to show x j x k whenever j k By way of contradiction, suppose x j = x k for some pair of distinct j, k

57 48 {1, 2,, n} Then ( ) j (mod n 1 ), j (mod n 2 ),, j (mod n s ) = ( ) k (mod n 1 ), k (mod n 2 ),, k (mod n s ) Without loss of generality, j = k + c for some integer c < n So for each i {1, 2,, s} j k (mod n i ) k + c k (mod n i ) c 0 (mod n i ) But, since n 1, n 2,, n s are relatively prime, this implies n divides c, a contradiction since c < n Therefore, the x k s are distinct and x 1, x 2,, x n form an ordering of the vertices of G Let f : V G Z + be defined by f(x k ) = k We will show f is a radio labeling of G For f to be a radio labeling, it needs to satisfy the radio condition for each pair of distinct vertices x k, x k+c : f(x k ) f(x k+c ) diam(g) + 1 d(x k, x k+c ) If e i,j is the number of coordinates x i and x j have in common, then we have the

58 49 following for c > 0: If e k,k+c c 1, then c 1 + e k,k+c k (k + c) s + 1 (s e k,k+c ) f(x k ) f(x k+c ) diam(g) + 1 d(x k, x k+c ) So showing e k,k+c c 1 will show that f is a radio labeling Note that e k,k+c s 1 since x k x k+c, so we must prove the inequality for 1 c s 1 Let 1 c s 1 By definition, x k and x k+c agree over e k,k+c coordinates, so k k +c (mod n i ) for e k,k+c different values of i Then n i divides c for e k,k+c different values of i Since the number of distinct prime divisors of c is bounded above by c 1, any collection of relatively prime divisors of c must be less than or equal to c 1 in number Then e k,k+c c 1 Then f gives a consecutive radio labeling of G Corollary 18 If n Z + has at least s distinct prime divisors, then there is a radio graceful graph with n vertices and diameter s Proof Let n Z + have at least s distinct prime divisors If s = 1 then K n is a graph with n vertices and diameter s that has a consecutive radio labeling Consider s > 1 Then n has a prime factorization n = p α 1 1 p α 2 2 p αt t where t s Set n i = p α i i for i {1, 2,, s 1}, and let n s = p αs s p α s+1 s+1 p αt t Then n 1, n 2,, n s are relatively prime By Lemma 1, G = K n1 K n2 K ns

59 50 has a consecutive radio labeling Note that G = n 1 n 2 n s = n and diam(g) = diam(k n1 ) + diam(k n2 ) + + diam(k ns ) = s 24 A Generalization of the Petersen Graph We end this chapter with a family of graphs (Definition 21) whose invention was inspired by the structure of one of its members: the Petersen graph A small collection of related definitions and facts preceded the main result (Theorem 23), which cultivates a necessary understanding of these new graphs Let p be a prime with p 3 We will construct graphs, (p 1)/2 in number, each of size p For i Z with 1 i (p 1)/2, we define the graph G pi by the following two requirements: (1) G pi has vertex set {v i,1, v i,2,, v i,p }, and (2) v i,j and v i,k are adjacent in G pi if and only if j k p = i See Table 23 Claim 19 For each vertex v i,j V Gpi, deg(v i,j ) = 2 Proof Let v i,j be a vertex in G pi, and let k {1,, p} such that j k p = i Case 1 Let j k p = j k = i Then either j k = i k = j i, or k j = i k = j + i Case 2 Let j k p = p j k = i Then either p (j k) = i k = j + i p, or p (k j) = i k = j i + p Since k {1,, p}, k = k (mod p) Coupled with the facts that j i j i +

60 51 p = 3 p = 5 p = 7 p = 11 i = 1 i = 2 i = 3 i = 4 i = 5 Table 23: All G pi graphs for p = 3, 5, 7, 11 p (mod p) and j + i j + i p (mod p), this shows that there are only two possible values for k, namely j i (mod p) and j + i (mod p) Therefore deg(v i,j ) 2 As both of these values satisfy the hypotheses for k in the theorem, which correspond precisely to the conditions for the existence of edges in the definition of

61 52 G pi, deg(v i,j ) = 2 Claim 20 For each i, G pi = Cp Proof Let i, j {1,, p}, i (p 1)/2 Since j (j + i (mod p)) p = min{ j (j + i (mod p)), p j (j + i (mod p)) } = i, G pi contains the following walk of length p: v i,1, v i,1+i (mod p), v i,1+2i (mod p),, v i,1+pi (mod p) (215) Clearly v i,1+pi (mod p) = v i,1, and therefore the given walk is closed Let j, k {1,, p 1} be distinct If 1 + ji (mod p) = 1 + ki (mod p), then p i(j k) But since i, j, k < p, j k, and p is prime, this is not possible Therefore the vertices v i,1, v i,1+i (mod p), v i,1+2i (mod p),, v i,1+(p 1)i (mod p) are distinct, and the the closed walk in line (215) is a Hamiltonian cycle in G pi This together with Claim 19 shows G pi = Cp Definition 21 Let p be a prime, p 3 The p-petersen graph, denoted P p, is defined by } 1 The vertex set V Pp = {v i,j i {1,,(p 1)/2} j {1,,p} 2 If i 1 i 2, then v i1,j 1 is adjacent to v i2,j 2 if and only if j 1 = j 2 3 For fixed i {1,, (p 1)/2}, the subgraph of P p generated by {v i,1, v i,p } is G ip See Table 24 at the end of this section for a visual construction of P 7

62 53 Notice that P 3 = K3 and P 5 = P ; the latter observation is, of course, the reason for the name p-petersen graph All p-petersen graphs other than the two just mentioned share the following property Claim 22 Let p be a prime with p > 5 Then diam(p p ) = 3 Proof Step 1 We show diam(p p ) 3 If j 1 = j 2, then d Pp (v i1,j 1, v i2,j 2 ) 1 by part (2) of the definition of P p Then consider j 1 j 2 ; we make an observation: If j 1, j 2 {1,, p} with j 1 j 2, then d Pp (v j1 j 2 p,j 1, v j1 j 2 p,j 2 ) = 1 This is because the vertices { v j1 j 2 p,1,, v j1 j 2 p,p} generate G ( j1 j 2 p) p, in which v j1 j 2 p,j 1 and v j1 j 2 p,j 2 are adjacent by definition This tells us that, for any distinct j 1, j 2 {1,, p}, there is a value of i such that v i,j1 and v i,j2 are adjacent (namely i = j 1 j 2 p ) Equipped with this information, and part (2) of the definition of P p, we see that there is always the following walk of (not necessarily distinct) vertices: v i1,j 1 v j1 j 2 p,j 1 v j1 j 2 p,j 2 v i2,j 2 Therefore, for any v i1,j 1, v i2,j 2 V Pp, d Pp (v i1,j 1, v i2,j 2 ) 3 Step 2 We show diam(p p ) 3, by showing d Pp (v 1,1, v 1,4 ) = 3 By the definition of P p, if i 1 i 2, then v i1,j 1 and v i2,j 2 are adjacent if and only if j 1 = j 2 Then any path from v 1,1 to v 1,4 containing a vertex v i,j with i 1 must

63 54 have length at least 3, because (whatever the value of j) v i,j cannot be adjacent to both v 1,1 and v 1,4 So, we consider all other paths from v 1,1 to v 1,4 A path made up of all vertices of the form v 1,j lies within the subgraph of P p generated by {v 1,1, v 1,p }, which is G p1 By the definition of G p1, vertices v 1,j1 and v 1,j2 are adjacent if and only if j 1 j 2 p = 1 Then, within this subgraph, there are only two possible paths from v 1,1 to v 1,4 : v 1,1 v 1,2 v 1,3 v 1,4, or (216) v 1,1 v 1,p v 1,p 1 v 1,4 (217) Since p > 5, the path in (216) is the shorter path, and it has length 3 Hence, d Pp (v 1,1, v 1,4 ) = 3 Steps 1 and 2 together show diam(p p ) = 3 With this Claim in hand, we are ready to show that this family of graphs that generalize the Petersen graph have consecutive radio labelings Theorem 23 For any odd prime p, the p-petersen graph P p is radio graceful Proof It is a simple exercise to show that P 3 = K3 and P 5 = P, which have both been shown to have consecutive radio labelings previously We will show P p is radio graceful for all primes p > 5 To do this, we will define an ordering of its 1 p(p 1) 2 vertices (Steps 1 and 2), one that we will prove induces a consecutive radio labeling (Step 3)

64 55 Let p be a prime with p > 5 Step 1 We define the following list of the vertices of P p Let k { 1,, 1 2 p(p 1)} Then there exist unique a Z 0 and b { 1,, 1 2 (p 1)} such that k = a ( 1 2 (p 1)) + b Then x k is the vertex v i,j in P p such that 1 i = b, and a ( 1 (p + 5)) + 1 (mod p) if b = j = a ( 1 (p + 5)) + b + 2 (mod p) otherwise 2 Recall that here, as elsewhere in this thesis, we use the convention that α (mod β) {1,, β} Step 2 We prove that the list x 1,, x p(p 1)/2 V Pp as defined in Step 1 is an ordering of the vertices of P p As it is clear from Definition 21 that x 1,, x p(p 1)/2 V Pp, and {x 1,, x p(p 1)/2 } = V Pp = 1 p(p 1), 2 we only need to show that no vertex is repeated in the list Suppose x k1 = x k2 for some k 1, k 2 { 1,, 1 2 p(p 1)} Say k 1 = a 1 ( 1 2 (p 1)) + b 1 and k 2 = a 2 ( 1 2 (p 1)) + b 2 for a 1, a 2 Z 0 and b 1, b 2 { 1,, 1 2 (p 1)} Then,

65 56 by the definition of x k, we can conclude b 1 = b 2 and ( a 1 1 (p + 5)) ( + 1 a (p + 5)) + 1 (mod p) if b 2 1 = b 2 = 1 (218) ( a 1 1 (p + 5)) ( + b a 1 2 (p + 5)) + b (mod p) otherwise Both equivalences in (218) are equivalent to ( ) ( ) 1 1 a 1 2 (p + 5) a 2 2 (p + 5) (mod p) (219) Given the constraints on k and b (namely, k 1p(p 1) and b 1 (p 1)), we see 2 2 that a is bounded: 0 a p 1 As such, (219) implies a 1 = a 2, or p divides ( 1 2 (p + 5)) Since p > 5, p cannot divide ( 1 2 (p + 5)), so a 1 = a 2 Since a 1 = a 2 and b 1 = b 2, we have shown k 1 = k 2 Hence x 1,, x p(p 1)/2 is an ordering of the vertices of P p Step 3 We prove that the ordering x 1,, x p(p 1)/2 of V Pp induces a consecutive radio labeling of the p-petersen graph Let f : V Pp Z + be defined as f(x k ) = k The goal is to show that f satisfies the radio condition for distinct x k and x m, which, by Claim 22, is f(x k ) f(x m ) diam(p p ) + 1 d Pp (x k, x m ) k m 4 d Pp (x k, x m ) d Pp (x k, x m ) 4 k m In other words, since we have already shown d Pp (x k, x m ) 1 always, we must show two additional things:

66 57 1 d Pp (x k, x k+1 ) = 3 for all k { 1,, 1 2 p(p 1) 1}, and 2 d Pp (x k, x k+2 ) 2 for all k { 1,, 1 2 p(p 1) 2} We begin by making a few notes that are useful moving forward Note 1: If x k = v i1,j 1, x k+1 = v i2,j 2, and x k+2 = v i3,j 3, then, by the construction of x 1,, x p(p 1)/2, and since p > 5, we know i 2 i 1 i 3 Note 2: By the definition of P p, if i 1 i 2, then v i1,j 1 and v i2,j 2 are adjacent if and only if j 1 = j 2 Note 3: Let i 1 i 2 In light of Note 2, there are at most two paths of length 2 from v i1,j 1 to v i2,j 2 : v i1,j 1 v i1,j 2 v i2,j 2, or (220) v i1,j 1 v i2,j 1 v i2,j 2 (221) The path in line (220) exists if and only if j 1 j 2 p = i 1, and the path in line (221) exists if and only if j 1 j 2 p = i 2 Therefore, d Pp (v i1,j 1, v i2,j 2 ) < 3 if and only if j 1 j 2 p {0, i 1, i 2 } Consider x k, x k+1, for k { 1,, 1 2 p(p 1) 1} Then k = a ( 1 2 (p 1)) + b for some a Z 0, b { 1,, 1 2 (p 1)} Using the definition given in Step 1, we can determine which vertices have been assigned to x k and x k+1 Case 11: b = 1 Then x k = v 1,j1 where ( ) 1 j 1 = a (p + 5) + 1 (mod p), 2

67 58 and x k+1 = v 2,j2 where ( ) 1 j 2 = a (p + 5) + 4 (mod p) 2 Since ( a ( 1 2 (p + 5)) + 4 ) ( a ( 1 2 (p + 5)) + 1 ) = 3 < 1 2 p, we can conclude j 1 j 2 p = 3 Therefore, j 1 j 2 p {0, i 1 = 1, i 2 = 2}, and d Pp (x k, x k+1 ) = 3 Case 12: b = 1 2 (p 1) Then x k = v (p 1)/2,j1 where ( ) 1 j 1 = a (p + 5) + 1 (p 1) + 2 (mod p), 2 2 and x k+1 = v 1,j2 where ( ) 1 j 2 = (a + 1) (p + 5) + 1 (mod p) 2 Since ( (a + 1) ( 1 (p + 5)) + 1 ) ( a ( 1 (p + 5)) + 1(p 1) + 2) = 2 < 1 p, we can conclude j 1 j 2 p = 2 Therefore, j 1 j 2 p { 0, i 1 = 1 2 (p 1), i 2 = 1 }, and d Pp (x k, x k+1 ) = 3 Case 13: 1 < b < 1 2 (p 1) Then x k = v b,j1 where ( ) 1 j 1 = a (p + 5) + b + 2 (mod p), 2 and x k+1 = v b+1,j2 where ( ) 1 j 2 = a (p + 5) + b + 3 (mod p) 2 Since ( a ( 1 (p + 5)) + b + 3 ) ( a ( 1 (p + 5)) + b + 2 ) = 1 < 1 p, we can conclude j 1 j 2 p = 1 Therefore, j 1 j 2 p {0, i 1 = b, i 2 = b + 1}, and d Pp (x k, x k+1 ) = 3

68 59 By Cases 11-13, we have proven d Pp (x k, x k+1 ) = 3 for all positive integers k { 1,, 1 2 p(p 1) 1} Now consider x k, x k+2, for k { 1,, 1 2 p(p 1) 2}, where k = a ( 1 2 (p 1)) + b for some a Z 0, b { 1,, 1 2 (p 1)} Case 21: b = 1 Then x k = v 1,j1 where ( ) 1 j 1 = a (p + 5) + 1 (mod p), 2 and x k+2 = v 3,j2 where ( ) 1 j 2 = a (p + 5) + 5 (mod p) 2 Since ( a ( 1 2 (p + 5)) + 5 ) ( a ( 1 2 (p + 5)) + 1 ) = 4 p, we conclude j 1 j 2 Therefore d Pp (x k, x k+2 ) 1 Case 22: b = 1 2 (p 1) Then x k = v (p 1)/2,j1 where ( ) 1 j 1 = a (p + 5) + 1 (p 1) + 2 (mod p), 2 2 and x k+2 = v 2,j2 where ( ) 1 j 2 = (a + 1) (p + 5) + 4 (mod p) 2 Since ( (a + 1) ( 1 2 (p + 5)) + 4 ) ( a ( 1 2 (p + 5)) (p 1) + 2) = 5 p, we conclude j 1 j 2 Therefore d Pp (x k, x k+2 ) 1

69 60 Case 23: b = 1 2 (p 1) 1 Then x k = v ((p 1)/2) 1,j1 where ( ) 1 j 1 = a (p + 5) + 1 (p 1) + 1 (mod p), 2 2 and x k+2 = v 1,j2 where ( ) 1 j 2 = (a + 1) (p + 5) + 1 (mod p) 2 Since ( (a + 1) ( 1 2 (p + 5)) + 1 ) ( a ( 1 2 (p + 5)) (p 1) + 1) = 3 p, we conclude j 1 j 2 Therefore d Pp (x k, x k+2 ) 1 Case 24: 1 < b < 1 2 (p 1) 1 Then x k = v b,j1 where ( ) 1 j 1 = a (p + 5) + b + 2 (mod p), 2 and x k+2 = v b+2,j2 where ( ) 1 j 2 = a (p + 5) + b + 4 (mod p) 2 Since ( a ( 1 2 (p + 5)) + b + 4 ) ( a ( 1 2 (p + 5)) + b + 2 ) = 2 p, we conclude j 1 j 2 Therefore d Pp (x k, x k+2 ) 1 By Cases 21-24, we have proven d Pp (x k, x k+2 ) 2 for all positive integers k { 1,, 1 2 p(p 1) 2} Therefore, f is a consecutive radio labeling

70 61 Table 24: The construction of the 7-Petersen graph

71 62 CHAPTER 3 NON-EXISTENCE RESULTS After the results of Chapter 2, it is tempting to wonder if K t n might have a consecutive radio labeling for all t Z + Theorem 24 gives the negative answer not only for K n but for any graph Theorem 24 Given a graph G, there is an integer s such that G t does not have a consecutive radio labeling for any t s In particular, if G has n vertices, is such a value s = 1 + n 1 i=diam(g) i (n i) diam(g) Proof Let V G = {v 1,, v n } Let s = 1 + n 1 i=diam(g) (n i) i diam(g), and let t Z, t s In search of contradiction, suppose x 1,, x n t is an ordering of the vertices such that f : V G t Z + defined by f(x i ) = i is a radio labeling Then f must satisfy the radio condition for all x i, x j V G t: d(x i, x j ) diam(g t ) f(x i ) f(x j ) + 1 = t diam(g) i j + 1 We can get an upper bound for d(x i, x j ) by counting the number of coordinates for which x i and x j have common entries If we define e i,j to be this number of coordinates where x i and x j agree, then d(x i, x j ) diam(g)(t e i,j )

72 63 Combining these two bounds on d(x i, x j ) yields t diam(g) i j + 1 diam(g)(t e i,j ) t i j 1 diam(g) t e i,j e i,j i j 1 diam(g) Then the maximum possible number of coordinates with common entries between x i and x j is { } i j 1 E i,j = min t, diam(g) Fix x i The maximum number of coordinates that x i can have in common with any prior vertex is i 1 j=1 E i,j Then the total number of coordinates in which any of the first n + 1 vertices in the ordering can agree is 1 j=1 E 2,j }{{} number of coordinates x 2 can have in common with any prior vertex n+1 i 1 = E i,j i=2 j=1 + 2 j=1 E 3,j }{{} number of coordinates x 3 can have in common with any prior vertex + + n j=1 E n+1,j }{{} number of coordinates x n+1 can have in common with any prior vertex

73 64 We can obtain with some computation that n+1 i 1 E i,j = n+1 i 1 i=2 j=1 j=1 i j 1 diam(g) i=2 j=1 1 n+1 2 j 1 i 1 = + j=1 diam(g) i=3 j=1 2 3 j 1 3 = + diam(g) j=1 j=1 2 2 j 3 = + diam(g) i=1 j=1 i j 1 diam(g) 4 j 1 diam(g) 3 j diam(g) n n + 1 j 1 j=1 n n j diam(g) j=1 diam(g) 1 i 2 n 1 i i = diam(g) diam(g) diam(g) i=1 i=1 i=1 1 2 n 1 = (n 1) + (n 2) + + diam(g) diam(g) diam(g) n 1 n 1 i i = (n i) = (n i) = s 1 < t diam(g) diam(g) i=diam(g) Since t is larger than the number of coordinates with this property, there is at least one coordinate in which none of the vertices x 1,, x n+1 agree This is impossible to accomplish, however, as we have only n possible entries: v 1,, v n So there is no ordering of the n t vertices of G t that induces a consecutive radio labeling Corollary 25 K t n does not have a consecutive labeling for any t 1 + n(n2 1) 6

74 65 Proof This is a simple computation: n 1 i n 1 s = 1 + (n i) = i=1 i=1 n 1 ni i 2 n 1 = 1 + n i i=1 i=1 i 2 ( ) n(n 1) = 1 + n 2 = 1 + n(n2 1) 6 n(n 1)(2n 1) 6

75 66 CHAPTER 4 TOWARDS NECESSARY AND SUFFICIENT CONDITIONS Much of this thesis thus far has been dedicated to the idea that, under certain conditions, G t is radio graceful; we also saw that G t will not be radio graceful for t sufficiently large As might be expected, we now wish to discover necessary and sufficient conditions for a graph G so that G t is radio graceful While that goal remains unattained, this chapter holds some partial results Sufficient conditions follow immediately by summarizing Theorem 12 and Theorem 14: Corollary 26 Let G K 2 be a connected graph with n vertices, let t Z + with t n, and let y diam(g) 1,, y n be an ordering of the vertices of G such that d(y i, y j ) diam(g) i j n 1 t (41) for all distinct i, j {1,, n} Then G t is radio graceful Note: While the condition that t n diam(g) is not a hypothesis of Theorem 14, it is a bound that is implied by the other hypotheses, and a bound necessary for the proof of Theorem 14 If we remove the hypothesis that G is incomplete, as we have done here, t is no longer bounded in the complete case, and the conclusion that G t has a consecutive radio labeling is no longer valid (if G = K n then (41) is true for all t Z +, but K t n is not consecutive for all t Z + by Theorem 24) Thus we include the condition that t n diam(g)

76 67 Proof If G is incomplete, then G t has a consecutive radio labeling by Theorem 14 If G is complete, then t by Theorem 12 n diam(g) = n, and therefore G has a consecutive radio labeling There are no known examples that have been proven to both be radio graceful and to fail these sufficient conditions, so it remains unknown whether the conditions given in Corollary 26 are also necessary While the Petersen graph satisfies the conditions for t = 2 if i j n is replaced with i j, the author conjectures that P does not satisfy the whole of Corollary 26 and might therefore be an example disproving the Corollary s necessity We now return to the discussion of the connection between consecutive radio labeling and Hamiltonian paths As we will see, every radio graceful graph gives rise to a semi-hamiltonian graph Definition 27 The diameter complement of a graph G, denoted G d, is a graph with vertex set V G such that two vertices u, v V Gd are adjacent if and only if d G (u, v) = diam(g) Note: G d is a subgraph of G unless G = K n for some n > 1 If A is a semi-hamiltonian subgraph of a graph B such that V A = V B, then B is also semi- Hamiltonian This simple point is made because the following result could also be made for the complement of G (if G is incomplete), but the given result is stronger and implies the result for G Theorem 28 If G is a radio graceful graph, then G d is semi-hamiltonian

77 68 C 7 (C 7 ) d Table 41: C 7 and its diameter complement Proof Let G be a radio graceful graph with n vertices Then G has a consecutive radio labeling: there exists an ordering x 1,, x n of the vertices of G such that the labeling f : V G Z + defined by f(x i ) = i is a radio labeling As such, f must satisfy the radio condition for x i, x i+1, i {1,, n 1}: d(x i, x i+1 ) diam(g) + 1 f(x i ) f(x i+1 ) = diam(g) + 1 i (i + 1) = diam(g) Therefore, x i is adjacent to x i+1 in G d for all i {1, n 1}, and x 1 x 2 x n is a Hamiltonian path in G d Theorem 29 If G satisfies the hypotheses of Corollary 26 for some t Z + with t n, then G diam(g) d is Hamiltonian Proof Let G be a graph that satisfies the hypotheses of Corollary 26 for t Then there exists y 1,, y n, an ordering of the vertices of G such that d(y i, y j ) diam(g) i j n 1 t n diam(g)

78 69 for all distinct i, j {1,, n} Then d(y i, y i+1 ) diam(g) i (i + 1) n 1 t = diam(g) 1 1 t = diam(g) Therefore, x i is adjacent to x i+1 in G d for all i {1, n 1}, and x 1 x 2 x n is a Hamiltonian path in G d In addition, d(y 1, y n ) diam(g) 1 n n 1 t = diam(g) 1 1 t = diam(g) So x 1 and x n are also adjacent in G d, and x 1 x 2 x n x 1 is a Hamiltonian cycle in G d In addition to marking the relationship between semi-hamiltonian graphs and radio graceful graphs, Theorem 28 has also given us necessary conditions for a graph G to have a consecutive radio labeling Indeed, when diam(g) = 2, those conditions are necessary and sufficient Corollary 30 Let G be a graph with diam(g) = 2 Then G is radio graceful if and only if G d is semi-hamiltonian Proof Let G be a graph with diam(g) = 2 and n vertices If G is radio graceful, then G d is semi-hamiltonian by Theorem 28 If G d is semi-hamiltonian, then in it there exists a Hamiltonian path x 1 x 2 x n By the definition of G d, this implies

79 70 d G (x i, x i+1 ) = diam(g) = 2 for all i {1,, n 1} Let f : V G Z + be defined by f(x i ) = i Then, for all i {1,, n 1}, f(x i ) f(x i+1 ) = i (i + 1) = = diam(g) + 1 d G (x i, x i+1 ), (42) and, if j > i + 1, f(x i ) f(x j ) = i j diam(g) + 1 d G (x i, x j ) (43) Lines (42) and (43) together show that f satisfies the radio condition for any pair of distinct vertices in G, so G is radio graceful This result gives another proof that K2 2 is not radio graceful, as (K2) 2 d is not connected, never mind semi-hamiltonian The diameter 2 assumption cannot be removed from Corollary 30 Take for example C 7, the cycle v 1 v 2 v 3 v 4 v 5 v 6 v 7 v 1 Then (C 7 ) d, the cycle v 1 v 4 v 7 v 3 v 6 v 2 v 5 v 1, is Hamiltonian (Table 41) However, no ordering will induce a consecutive radio labeling of C 7 : Without loss of generality, f(x 1 ) = f(v 1 ) = 1 and f(x 2 ) = f(v 4 ) = 2 (since we want d C7 (x 1, x 2 ) to be 3, x 2 can be v 4 or v 5, which are symmetric choices) Then x 3 is v 7 since this is the only remaining vertex diameter away from v 4, and f(v 7 ) = 3 But f(v 1 ) f(v 7 ) = 1 3 = 2, and diam(c 7 ) + 1 d(x 1, x 7 ) = = 3, so the radio condition is not satisfied In other words, the necessary conditions given in Theorem 28 are not in general sufficient

80 71 CHAPTER 5 REMARKS AND FUTURE WORK The (t) th Cartesian power of a complete graph (K t n) is an object that has surfaced repeatedly throughout the course of this thesis Together, Theorem 12 and Corollary 25 give us a lot of information about K t n If 1 t n, then K t n has a consecutive radio labeling, so rn(k t n) = V K t n = n t If t 1 + n(n2 1) 6 then K t n does not have a consecutive radio labeling, so rn(k t n) > n t This is illustrated in Figure 51 rn(k t n) >n t? (3, 0) rn(k t n)=n t t t Figure 51: Visual description of Theorem 12 and Corollary 25 In addition, Corollary 15 tells us K n/p p n is radio graceful for n Z + and p {2, 3,, n 2} such that p n One ambitious future goal: to be able to say, based on the values of any n and t, whether K t n is radio graceful Short of this, any improvement upon this current picture is desirable