MOD SUM NUMBER OF COMPLETE BIPARTITE GRAPHS. A Thesis. East Tennessee State University. In Partial Fulfillment. of the Requirements for the Degree

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1 MOD SUM NUMBER OF COMPLETE BIPARTITE GRAPHS A Thesis Presented to the Faculty of the Department of Mathematics East Tennessee State University In Partial Fulfillment of the Requirements for the Degree Master of Science in Mathematical Sciences by Christopher D. Wallace May, 1999

2 APPROVAL This is to certify that the Graduate Committee of Christopher D. Wallace met on the 1st day of April, The committee read and examined his thesis, supervised his defense of it in an oral examination, and decided to recommend that his study be submitted to the Graduate Council, in partial fulfillment of the requirements for the degree of Master of Science in Mathematics. James W. Boland Chair, Graduate Committee Robert B. Gardner Debra J. Knisley Linda M. Lawson Signed on behalf of the Graduate Council Dr. Brown, Dean School of Graduate Studies ii

3 ABSTRACT MOD SUM NUMBER OF COMPLETE BIPARTITE GRAPHS by Christopher D. Wallace A graph G = (V, E) is a mod sum graph if there exists a positive integer Z and a labelling of vertices with distinct elements of {1, 2,..., Z 1} such that {u, v} E if and only if u v and u + v (mod Z) V. First we discuss conditions which K n1,n 2,...,n m must satisfy to be a mod sum graph and then we determine the minimum number of isolated vertices such that K n,m m n 3 is a mod sum graph except when 2n m < 3n 3 and m is odd. iii

4 Copyright by Christopher D. Wallace 1999 iv

5 DEDICATION To my wife Olivia, my parents Billie and Bill, and my sister Sheena. Their support, encouragement, and love made this thesis possible. v

6 ACKNOWLEDGEMENTS I would like to thank Jay Boland who helped with assistance in making this thesis a reality. vi

7 Contents APPROVAL ii ABSTRACT iii COPYRIGHT iv DEDICATION v ACKNOWLEDGEMENTS vi LIST OF FIGURES viii 1. INTRODUCTION Definitions Survey on Sum Graphs Mod Sum Graph Definitions and Survey Overview MAIN RESULTS Notation Complete Partite Graphs CONCLUSIONS and CONJECTURES Conclusions Conjectures BIBLIOGRAPHY 21 VITA 24 vii

8 List of Figures 1 ρ(ik 5,11 ) 11 (mod 110) ρ(ik 4,7 ) = 4 (mod 40) K 4,8 (mod 32) IK 6,15 (mod 60) viii

9 CHAPTER 1 INTRODUCTION Definitions To begin, we first must define a simple undirected graph. A simple undirected graph G = (V, E) consists of a set V which is a nonempty, finite set of elements called vertices, and a set E which is a set of unordered pairs {u, v} called edges, where u, v are distinct elements of V. V is called the vertex set of G, and E is called the edge set of G. A simple graph can have at most one edge between any two vertices. From now on we will use the word graph to mean a simple undirected graph. We now define some classes of graphs. The complete graph, K n, has n vertices and every pair of vertices share an edge. The complete m-partite graph, K n1,n 2,...,n m, has partite sets V 1, V 2,..., V m with V i = n i having the property that if u V i, v V j, i j then {u, v} E and {u, v} E when i = j. One class of complete m- partite graphs is H n,m which is a m-partite graph which each partite set containing n vertices. A cycle containing n vertices is denoted C n. Also a wheel W n contains a cycle of length n and an additional vertex which is adjacent to every vertex on the cycle. A tree T n is a acyclic connected graph containing n vertices, while a forest F n is a acyclic graph containing n vertices. The concept of sum graph was introduced by Harary at the Nineteenth Southeastern Conference on Combinatorics at Baton Rouge in A graph, G = (V, E), is a sum graph if there exist a set F ZZ + and a bijection f : F V such that if x, y are 1

10 2 distinct elements of F, then x + y F if and only if {f(x), f(y)} E. In this paper, we will refer to vertices by their corresponding label in F. The first obvious class of graphs which are not sum graphs are connected graphs. Then not all graphs are sum graphs and so it is a natural extension to define sum number of a graph. The sum number of graph G, denoted σ(g), is the smallest number of isolated vertices that can be added to G to yield a sum graph. Survey on Sum Graphs Most papers thus far on sum graphs have involved finding the sum number for a particular class of graphs. These include σ(k n ) = 2n 3 for n 4 [1] σ(k m,n ) = 3m+n 3 2 for n m 2 [6] σ(h 2,n ) = 4n 5 for n 2 [10] σ(c n ) = 2 for n 4 [5] σ(w n ) = n for even n 4, σ(w n) = n for odd n 5 [7] σ(f n ) = 1 for all n 2 [3] Mod Sum Graph Definitions and Survey Boland, Laskar, Turner and Domke introduced in [2] a generalization of sum graph. A graph is a mod sum graph if there exist a positive integer Z and a labelling of the

11 3 vertices with distinct elements of {1, 2,..., Z 1} such that {u, v} E if and only if u v and u + v (mod Z) F. We similarly define the mod sum number of a graph ρ(g) as the smallest number of isolated vertices that must be added to G such that the resulting graph is a mod sum graph (MSG). So far the classifications on graphs are ρ(k n ) = n for n 4 [8] ρ(k 2,n ) = 0 and ρ(k 1,n ) = 0 for n 2 [2] ρ(k p+1,q ) = 0 for q 1 and p r q + r q 1 1 where r 1 = 1, r 2 = 2, and r j = r j 2 + r j for j 3 [2] ρ(k n,n ) > 0 for n 3 [9] ρ(h 2,n ) = 0 for n 2 [2] ρ(h m,n ) > 0 for m > n 2 [8] ρ(c n ) = 0 for n 4 [2] ρ(w n ) > 0 for n 5 [9] ρ(t n ) = 0 for all n 3 [2] Overview In this paper we determine conditions for when the complete m-partite graph is not a mod sum graph. Also we show K n,m is a mod sum graph when m is even and m 2n or when m is odd and m 3n 3 and ρ(k n,m ) = n when 3 n m < 2n.

12 In Chapter 2 we prove these results and we present conjectures and conclude with Chapter 3. 4

13 CHAPTER 2 MAIN RESULTS Notation For K n1,n 2,...,n m the complete m-partite graph, there exist a set of labels F, some nonnegative integer r, and positive integer Z, such that K n1,n 2,...,n m K r is a mod sum graph (MSG) modulo Z. Denote IK n1,n 2,...,n m = K n1,n 2,...,n m K r. Let V (K n1,n 2,...,n m ) = V, V (K r ) = R, E(K n1,n 2,...,n m ) = E. Denote partite sets V i for 1 i m. We refer to vertices by their label, so V (IK n1,n 2,...,n m ) = F. For easier notation, if G = (V, E) is a MSG, then v V means v (mod Z) V, and {u, v} E means u + v (mod Z) F. Define x y if and only if x = y (mod Z). Complete Partite Graphs We use the following observation in Lemma 2.6 and Lemma 2.7. Observation 2.1 Let Z ZZ + and let 0 < a i < Z for 1 i n be integers. Let d = gcd(a 1, a 2,..., a n ). Then there exist nonnegative integers L i for 1 i n such that n i=1 L i a i d. Discussion. Since d = gcd(a 1, a 2,..., a n ), there exist K i ZZ such that d = ni=1 K i a i. Let f : ZZ {0, 1,..., Z 1} by f(x) = x (mod Z). Then 0 f(x) < Z, f(xy) f(x)f(y), and f(x + y) f(x) + f(y). Since 0 < a i < Z, for 1 i n, 5

14 6 then f(a i ) = a i. Also 0 < d < a i for all a i. So f(d) = d. Therefore d = f(d) f( n i=1 K i a i ) n i=1 f(k i )f(a i ) n i=1 f(k i )a i, where f(k i ) are nonnegative integers for 1 i n. The following lemma is needed for Corollary 2.3. Lemma 2.2 For IK n1,n 2,...,n m, with a 1, a 2 V 1, b V V 1, and V 1 3, if a 1 + b V V 1, then a 2 + b a 1. Proof. If a 2 = a 1, then a 2 + b = a 1 + b V V 1, and the result follows. Suppose for the sake of contradiction, a 2 + b a 1. Since V 1 3, there exist a 3 V 1 such that a 3 / {a 1, a 2 }. Now {a 3, b + a 1 } E, since a 3 V 1 and b + a 1 V V 1. Thus a 3 + b + a 1 F. Notice a 3 + b F and a 3 + b a 2 + b a 1. So {a 3 + b, a 1 } E which implies a 3 + b V V 1. Hence {a 3 + b, a 2 } E and so a 3 + b + a 2 F. By assumption a 3 a 1 b + a 2, therefore {a 3, b + a 2 } E, a contradiction. The following corollary is cited in Lemma 2.4, Lemma 2.5, and Lemma 2.6. Corollary 2.3 Let V 1 be a partite set of IK n1,n 2,...,n m with V 1 3. If a 1 V 1, b V V 1, then either for all a i V, a i +b / V V 1, or for all a i V, a i +b V V 1 Proof. If for all a i V 1, a i + b / V V 1, the result follows. Assume there exist a a i V such that a i + b V V 1. Pick a j V 1. Then {a j, a i + b} E since a j V 1 and a i +b V V 1. So a j +a i +b F. Also a j +b F, since a j V and b V V 1. By Lemma 2.2, a j + b a i. So {a i, a j + b} E. This implies a j + b V V 1. The next lemma is used in Lemma 2.5. Lemma 2.4 Let V 1 and V 2 be partite sets of IK n1,n 2,...,n m, with V 1 = n 1, V 2 = n 2 and n 2, n 1 3. Suppose a V 1, b V 2. If a + b V 1, then for all b V 2, a + b V 1.

15 7 Proof. Denote V 1 = {a 1, a 2,..., a n1 }, V 2 = {b 1, b 2,..., b n2 }, and suppose a 1 + b 1 V 1. By Corollary 2.3, since a 1 + b 1 V V 2, then for every b i V 2, a 1 + b i V V 2. Likewise since a 1 + b 1 / V V 1, for every a j V 1, a j + b 1 / V V 1. If for every b i, a 1 + b i V 1, the result follows. Suppose there exist a b 2 such that a 1 + b 2 / V 1. Now b 2 + a 1 V V 2. So b 2 + a 1 V V 2 V 1 V V 1. Thus b 2 + a i V V 1, for all a i V 1 by Corollary 2.3. Pick a i V 1. Since b 1 + a 1 V 1 and b 2 + a i V V 1, then {b 1 + a 1, b 2 + a i } E. Hence b 1 + a 1 + b 2 + a i F. Now we showed b 1 + a i / V V 1, then b 1 + a i V 1 or b 1 + a i R. Well b 2 + a 1 V V 1, so b 1 + a i b 2 + a 1. Hence {b 1 +a i, b 2 +a 1 } E which implies b 1 +a i V 1. Therefore b 1 +a i V 1 for all a i V 1. So a 1, b 1 +a 1, b 1 +b 1 +a 1,..., nb 1 +a 1,... V 1. Hence there exist a smallest J Z + such that b 1 (J +1) 0 and kb 1 +a 1 V 1 for 0 k J. Since b 2 +a i V V 1 for all a i V 1, then b 2 +kb 1 +a 1 V V 1 for 0 k J. Then {b 2 +kb 1 +a 1, a 1 } E which implies b 2 + kb 1 + 2a 1 F for 0 k J. Hence for 0 k < J, b 2 + kb 1 + 2a 1 0, so b 2 + (k + 1)b 1 + 2a 1 b 1 for 1 k J. Also b 2 + 2a 1 b 1, since b 2 + Jb 1 + 2a 1 0. Notice b 2 + kb 1 + 2a 1 + b 1 F for 0 k J. So {b 2 + kb 1 + 2a 1, b 1 } E implies {b 2 + kb 1 + 2a 1, b 2 } E implies 2b 2 + kb 1 + 2a 1 F for 0 k J. Suppose for l > 0, lb 2 + kb 1 + 2a 1 F for 0 k J. So by the previous argument for 0 k J, lb 2 + kb 1 + 2a 1 b 1 and lb 2 + (k + 1)b 1 + 2a 1 F. Hence {lb 2 +kb 1 +2a 1, b 1 } E implies {lb 2 +kb 1 +2a 1, b 2 } E implies (l+1)b 2 +kb 1 +2a 1 F for 0 k J. Whence lb 2 + b 1 + 2a 1 F for all l 0. So there must exist K > N such that Kb 2 + b 1 + 2a 1 Nb 2 + b 1 + 2a 1. So (K N)b 2 0. Therefore (K N)b 2 + b 1 + 2a 1 b 1 + 2a 1 F. Now b 1 + a 1 a 1 since b 1 0. So {b 1 + a 1, a 1 } E, a contradiction.

16 The following result is an extension to a result by Hartfield and Smyth in [6], the lemma is referenced in Lemma 2.6, T heorem 2.10, and T heorem Lemma 2.5 Let V 1 and V 2 be partite sets of IK n1,n 2,...,n m, with V 1 = n 1, V 2 = n 2 and n 2 n 1 3. Suppose b 1 V 2. Then either for all a i V 1, a i + b 1 R, or for all a i V 1, a i + b 1 V V 1. Proof. Suppose a 1 V 1, b 1 V 2, and a 1 + b 1 V V 1. So by Corollary 2.3 for all a i V 1, a i + b 1 V V 1. Now suppose a 1 + b 1 V 1. By Lemma 2.4 for every b i V 2, a 1 + b i V 1. So a 1, a 1 + b i V 1 for 1 i n 2, but a 1 a 1 + b i. Also a 1 + b i a 1 + b j for i j. Thus a 1, a 1 + b i for 1 i n 2 are n distinct elements in V 1, a contradiction since n 2 V 1. Finally, observe that if a 1 + b 1 R, then by the conclusion of the previous paragraph, there exists no vertex a i V 1 such that a i + b 1 V. This completes the proof. The next lemma is needed in Lemma 2.7. Lemma 2.6 Suppose K n1,n 2,...,n m is a MSG, i.e. V = F, with V 1 and V 2 partite sets and V 2 = n 2 V 1 = n 1 3. If V 1 = {a 1, a 2,..., a n1 } with gcd(a 1, a 2,..., a n1 ) = d, then there exist L ZZ + such that (L + 1)d Z and for all b V 2, i ZZ, b + id V V 1. Proof. Denote V 2 = {b 1, b 2,..., b n2 }. By assumption R = 0, so by Lemma 2.5, b i + a j V V 1 for 1 i n 2, 1 j n 2. Pick b V 2.

17 Assume there exist nonnegative integers C i for 1 i n 1 such that b+ n 1 i=1 C ia i V V 1 for 0 C i C i. For sake of contradiction suppose there exist k and 0 K i C i for i k, such that b + n 1 i=1 K i a i + (C k + 1)a k V 1. Now by i k assumption b + n 1 i=1 K i a i + (C k )a k V V 1. Also since b + a k / V 1 either C k > 0 i k or there exist K i > 0 for some 1 i n 1, i k, denote K c. Case 1: (C k > 0) Since b + n 1 i=1,i k K ia i + C k a k V V 1 and b + n 1 i=1,i k K ia i + 9 C k a k + a k V 1, by Corollary 2.3, b + n 1 i=1,i k K ia i + (C k )a k + a j V 1 for 1 j n 1 and notice these are n 1 distinct vertices of V 1. So there exist l such that a k b + n 1 i=1 i k K i a i + C k a k + a l. Hence b + n 1 i=1 K i a i + (C k 1)a k + a l 0, i k but since C k > 0 by our assumption b + n 1 i=1 K i a i + (C k 1)a k V V 1. i k b + n 1 i=1 K i a i + (C k 1)a k + a l F, a contradiction. i k Thus Case 2: (C k = K k = 0) So K c > 0 for some c k. So by Corollary 2.3, b + n 1 i=1 K i a i + K c a c + a j V 1 for 1 j n 1, and also notice they are n 1 distinct i c elements. Since V 1 = n 1 and a c V 1, there exist l such a c b+ n 1 i=1 K i a i +K c a c +a l. i c Therefore b+ n 1 i=1 K i a i +(K c 1)a c +a l 0, a contradiction by the previous argument. i c Hence b + n 1 i=1 C i a i V V 1 for all C i 0. Now since d = gcd(a 1,..., a n1 ), by Observation 2.1, there exist nonnegative integers K i for 1 i n 1 such that d n 1 i=1 K i a i. Then it follows that b + n 1 i=1 K i a i b + d V V 1. Since b + n 1 i=1 C i a i V V 1 for all C i 0, then for all l 0, b + l n 1 i=1 K i a i b + ld V V 1. So there must exist i,j where i j such that b + id b + jd, since otherwise V V 1 =. From whence we conclude that there exist L ZZ + such that (L + 1)d Z, and so for any i ZZ, b + id b + ld for some nonnegative integer l, thus b + id (mod Z) V V 1.

18 The following lemma is the major result in this paper. With it, Lemma 2.8 shows K n,m with m > n 3 is not a MSG when m < 2n. 10 Lemma 2.7 Suppose K n1,n 2,...,n m is a MSG, i.e. V = F, with V 1 and V 2 partite sets, and V 2 = n 2 V 1 = n 1 3. Let V 1 = {a 1, a 2,..., a n1 } with gcd(a 1, a 2,..., a n1 ) = d. If n 2 > n 1, there exist b V 2 such that for all i ZZ, b + id V 2. Proof. Since for all b V 2, 0 < b < Z, then 2b 0 implies b = Z. Since n 2 2 3, we can chose a b V 2 such that 2b 0. Assume V 1 = {C 1 d, C 2 d,..., C n1 d}. By Lemma 2.6, for all i ZZ, b+id V V 1 and there exist a L ZZ + such that (L+1)d 0. Let L be the smallest such positive integer. Suppose there exist j such that b + jd / V 2. THEN WE ARE GOING TO SHOW n 1 = n 2. We first establish 2b + C 1 d V 1, b + C 1 d / V 2, and b + 2C 1 d, 3b + 2C 1 d V 2. Since R = 0, {b, b + jd} E. So 2b + jd V. Assume for k ZZ +, there exist an i such that kb + id V. Now either kb + id V 2 or kb + id / V 2. Hence either {kb + id, b + jd} E or {kb + id, b} E. Thus either (k + 1)b + (i + j)d V or (k + 1)b + id V. Therefore by induction for all k ZZ + there exist a i such that kb + id V. This implies that there must exist K > N > 0, and i, j ZZ such that Kb + id Nb + jd. We wish to show that there exist a J ZZ + such that (J +1)b 0. So (K N)b (j i)d. If j = i, since b 0, we have a J > 0 such that (J + 1)b 0. If j i, b+k(k N)b b+k(j i)d V V 1 for all k ZZ. Hence there must exist r, s ZZ + with r > s, such that b + r(k N)b b + s(k N)b. Thus (r s)(k N)b 0. Since r > s and K > N, there a exist a J Z + such that (J + 1)b 0. Let J be the smallest such positive integer. By the above argument there exist a k such that Jb + kd V. Hence (J + 1)b (L + 1)d 0.

19 Suppose for any l Z, that Jb + ld V 1. Now b + (C 1 + C 2 l)d V V 1. Then {Jb + ld, b + (C 1 + C 2 l)d} E which implies (C 1 + C 2 )d V, a contradiction. So Jb + ld V V 1. Thus Jb + kd / V 1. Whence {Jb + kd, C 1 d} E, and Jb+kd+C 1 d V. By the previous argument Jb+kd+C 1 d V V 1. Repeating one can see that Jb + kd + n 1 i=1 K i C i d V V 1 for all K i 0. Since d = gcd(a 1,..., a n1 ), 11 by Observation 2.1, there exist nonnegative integers K i for 1 i n 1 such that d n 1 i=1 K ia i. Therefore for l 0, Jb + kd + l n 1 i=1 K ia i Jb + kd + ld V V 1. Since we have L as the smallest positive integer such that (L+1)d 0, then Jb+id V V 1 for 0 i L are L + 1 distinct elements. We wish to show for all C i d, C j d V 1, that b + C i d, J + C j d are in the same partite set distinct from V 2. Denote V 3, the partite set containing b + C 1 d. Now if b + C 1 d Jb + C i d for some i then it follows that Jb + C i d V 3. Suppose b + C 1 d Jb+C i d for i 1. Then Jb+C i d V 3 since otherwise {Jb+C i d, b+c 1 d} E implies C i d + C 1 d F for i 1. Hence Jb + C i d V 3 for i 1. Since n 1 3, for b + C j d where j 1, there exist C l d such that C l d C j d, C l d C 1 d. So Jb + C l d V 3, by the previous argument, implies b + C j d V 3. Hence b + C j d V 3 for all C j d V 1. Finally, we have b + C 2 d V 3 implies Jb + C 1 d V 3. Now there exist an i such that Jb + C i d b. Since Jb + C i d + b C i d, it follows that {Jb + C i d, b} E. So V 3 is a different partite set then V 2 and V 3 contains b + C i d, Jb + C j d for all C i d, C j d V 1. Therefore we have {b, b + C i d} E implying that 2b + C i d V. Now {b + C i d, b + C j d} / E for i j. So 2b + C i d + C j d / V, for i j. So 2b + C i d V 1 for all C i d V 1. The vertices Jb, Jb + (C i + C j )d V 2 for i j, since these vertices must not be

20 12 adjacent to b. So b + C i d + C j d V 2 for i j, because they can not be adjacent to Jb. Now we established b+c 1 d / V 1. Then {b+c 1 d, C 1 d} E, and so b+2c 1 d V. Since 2b 0, then C 1 d 2b + C 1 d. Notice 2b + 2C 1 d / F implies {b + 2C 1 d, b} / E implying b + 2C 1 d V 2. Also since C 1 d 2b + C 1 d, we have b + (C 1 d) + (2b + C 1 d) 3b + 2C 1 d V 2. So we have established 2b + C 1 d V 1, b + C 1 d / V 2, and b + 2C 1 d, 3b + 2C 1 d V 2. We wish to establish n 1 = n 2 = V 3. By assumption n 2 n 1. Let f : V 2 V 1 by f(x) = x+b+c 1 d. Pick x V 2. Then {x, b+c 1 d} E, and so x+b+c 1 d V. Now if x + b + C 1 d / V 1, then {x + b + C 1 d, C 1 d}, {x + b + C 1 d, 2b + C 1 d} E. Therefore x + b + 2C 1 d, x + 3b + 2C 1 d V. Well 2b 0 implies b + 2C 1 d 3b + 2C 1 d. Hence either {x, b + 2C 1 d} E or {x, 3b + 2C 1 d} E, which can not happen. Therefore x + b + C 1 d V 1. If f(x) f(y) then x + b + C 1 d y + b + C 1 d, and so x y. So f is an injection. Thus n 1 = V 1 V 2 = n 2, a contradiction. The next lemma is cited in T heorem 2.12 and T heorem These theorems ρ(k n,m ) when m > n 3 and m is even. Lemma 2.8 If K n1,n 2,...,n m is a MSG, i.e. V = F, with V 1 and V 2 partite sets V 2 = n 2 > V 1 = n 1 3, then n 2 2n 1. If a 1, 2a 1 V 1, then n 2 3n 1 3. Proof. Denote V 1 = {C 1 d, C 2 d,..., C n1 d} where d = gcd(c 1 d, C 2 d,..., C n1 d). By Lemma 2.7, there exist b V 2 such that b + id V 2 for 0 i L where L is the smallest positive integer such that (L + 1)d 0. Thus b + id V 2 for 0 i L are L + 1 distinct vertices implying n 2 L + 1. Since L is the smallest positive integer such that (L + 1)d 0 we have at most L + 1 distinct multiples of d modulo Z. Now {C 1 d, C i d} / E, so C 1 d + C i d / F

21 for 2 i n 1. If C 1 d + C i d C 1 d + C j d, then C i d C j d. So C j d, {C 1 + C i }d 13 for 1 j n 1, 2 i n are 2n 1 1 distinct multiples of d modulo Z. So n 2 L + 1 2n 1 1. If C 1 d+c j d 2C 1 d, then C j d C 1 d. Hence if 2C 1 d / V 1, then n 2 L+1 2n 1. Suppose 2C 1 d C 2 d V 1. We will establish C 1 d, 2C 1 d, 3C 1 d, C i d, C 1 d + C i d, 2C 1 d + C i d for 3 i n 1 are 3n 1 3 distinct multiples of d mod Z. Now C 1 d C i d, 2C 1 d C i d for 3 i n 1. Also 3C 1 d, C 1 d + C i d, 2C 1 d + C i d / F for 3 i n 1. Likewise C 1 d, 2C 1 d, C i d / {3C 1 d, C 1 d + C j d, 2C 1 d + C j d} (mod Z) for 3 j n 1, 3 i n 1. If 3C 1 d C 1 d + C i d, then 2C 1 d C i d. Notice if 3C 1 d 2C 1 d + C i d then C 1 d C i d. So 3C 1 d / {C 1 d + C i d, 2C 1 d + C i d} (mod Z) for 3 i n 1. Finally C 1 d + C i d 2C 1 d + C j d, since C i d C 1 d + C j d. Hence we have (n 1 2) + (n 1 2) + (n 1 2) = 3n 1 3 distinct multiples of d mod Z. So n 2 L + 1 3n 1 3 2n 1 when n 1 3. Corollary 2.9 If K n1,n 2,...,n m is a MSG with n 1 > n 2 >... > n m 1 > n m 3, then n 1 2n n m 2 n m 1 2 m 1 n m. Proof. The result follows by induction using Lemma 2.8. The next theorem is a special case needed for T heorem This result was first proved by Miller, Ryan, Sutton, and Slamin in [9]. Theorem 2.10 K n,n is not a MSG for n 3. Proof. Denote partitions V 1 = {a 1, a 2,..., a n } and V 2 = {b 1, b 2,..., b n }. By Lemma 2.5, if K n,n is a MSG, with a j V 1 then for all b i V 2, a j + b i V V 2. Likewise

22 14 for b j V 2, for all a i V 1, a i + b j V V 1. Since both can not be true a j + b i R, for all a j V 1, b i V 2. The following theorem is used in T heorem Theorem 2.11 If K n,m is not a MSG, with n m then n ρ(ik n,m ) m. Proof. Denote partitions V 1 = {a 1, a 2,..., a n } and V 2 = {b 1, b 2,..., b m }. By assumption there exist a i, b j such that a i + b j R. Now by Lemma 2.5, b j + a i R for 1 i n. So ρ(ik n,m ) n. We now give a labelling for which R = m. Let a i = (i 1)10+7, b i = (i 1)10+9, r i = (i 1)10+6 with Z = 10m. Then R = {6, 16,..., 10(m 1)+6}. Notice a i +b j = (i+j 2)10+16 R, a i +a j = (i+j 2)10+14 / F, b i +b j = (i+j 2)10+18 / F, r i + r j = (i + j 2) / F, a i + r j = (i + j 2) / F, and finally, b i + r j = (i + j 2) / F. Thus ρ(ik n,m ) m. The next three theorems, T heorem 2.12, T heorem 2.13, and T heorem 2.14, show K n,m is a mod sum graph when m is even and m 2n or when m is odd and m 3n 3 and ρ(k n,m ) = n when 3 n m < 2n. Theorem 2.12 ρ(k n,m ) = n, if 3 n m < 2n. Proof. By Lemma 2.8, K n,m is not a MSG since m < 2n. By T heorem 2.11, ρ(k n,m ) n. Denote partitions V 1 = {a 1, a 2,..., a n } and V 2 = {b 1, b 2,..., b m }. We now give a labelling for which R = n. Let a i = (i 1)10+9, b i = (i 1)10+6 for 1 i n, b i = (i 1) for n < i m. Let r i = (i 1) with Z = 10n. Then R = {5, 15,..., 10(n 1)+5}. Notice if j n, then a i +b j = (i+j 2)10+15 R, and if j > n, then a i + b j = (i + j 2) V 2. Now for any i,j, a i + a j =

23 Figure 1: ρ(ik 5,11 ) 11 (mod 110) (i + j 2) / F, for i, j n, b i + b j = (i + j 2) / F, for i n, j > n, b i + b j = (i + j 2) / F, and for i, j > n, b i + b j = (i + j 2) / F. Also, r i + r j = (i + j 2) / F, a i + r j = (i + j 2) / F, and finally, for i n, b i + r j = (i + j 2) / F, for i > n, b i + r j = (i + j 2) / F. Thus ρ(ik n,m ) n. Theorem 2.13 If m is even and m 4, then K n,m with m n is a MSG if and only if m 2n. Proof. = By T heorem 2.10, K n,m is not a MSG if n = m. By Lemma 2.8, if K n,m is MSG with m > n, then m 2n. = Assume m 2n with m even. Let V 1 and V 2 be partite sets. Let V 1 {d, 3d,..., (m 1)d}, V 2 = {1, 1 + d, 1 + 2d,..., 1 + (m 1)d}, d = 4, and Z = dm = 4m. Pick (2i + 1)d V 1 and 1 + jd V 2, then {(2i + 1)d, 1 + jd} E since

24 Figure 2: ρ(ik 4,7 ) = 4 (mod 40) (2i + 1)d jd = 1 + (jd + 2(i + 1)d) (mod dm) = 1 + kd for some k. Pick (2i + 1)d, (2j + 1)d V 1. Now [(2i + 1)d + (2j + 1)d] = 2(i + j + 1)d (mod dm) = 2kd when m is even. Since 2kd / F, {(2i + 1)d, (2j + 1)d} / E. Pick 1 + id, 1 + jd V 2. Now (1 + id jd) (mod dm) = 2 + kd for some k. Also 2 + kd ld, since 2 = (k l)d (mod dm) implies d < 4. Likewise 2+kd 1+ld, since 1 = (k l)d (mod dm) implies d < 4. Thus 1 + id jd / F which in turn means {1 + id, 1 + jd} / E. Theorem 2.14 If m is odd and m 3, then K n,m with m 3n 3 is a MSG. Proof. Let L = m + 1. Then L is the largest integer such that 3L 3 m. So 3 if 3n 3 m, then 3n 3 3L 3, which implies n L. Either L = 2K + 1 or L = 2K. Case 1: (L=2K+1) Let V 1 {d, 3d,..., (l 2)d, (m l)d, (m l + 2)d,..., (m 3)d, (m 1)d}, and V 2 = {1, 1+d, 1+2d,..., 1+(m 1)d} where d = 4 and dm = Z. So for any s V 1, r V 2, s+r V 2, and for any s, r V 2, s+r / V. Denote a i = (2i 1)d

25 Figure 3: K 4,8 (mod 32) for 1 i K, b i = (m 2i + 1)d for 1 i K + 1. Then the a i are odd multiples of d and the b i are even multiples of d. Now a 1 < a 2 <... < a K < b K+1 < b K <... < b 1. Therefore for any a i, a j V 1 for i j, 2(i+j 1)d = a i +a j a K 1 +a K (2L 6)d < (2L 2)d (m L + 1)d < dm. So a i + a j = 2(i + j 1)d a h = (2h 1)d for any h, and a i + a j < (m L)d = b K+1. Hence a i + a j / V 1, and a i + a j / V 2. Pick a i, b j V 1. Now a i + b j = (2i 1)d + (m 2j + 1)d (m 2(i j))d. If a i + b j < dm, then a i +b j is and odd multiple of d, and a K < a i +b j, so a i +b j / V 1. If a i +b j > dm, then a i + b j < a i (mod dm) < b K+1, and a i + b j is an even multiple. So a i + b j / V 1, and a i + b j / V 2. Pick b i, b j V 1 with i j. Then dm < (2m 2L + 2)d = (2m 4K)d = b K + b K+1 b i + b j < 2md. So (2m 2L + 2)d (m 2L + 2)d, and a K = (L 2)d < (L 1)d (m 2L + 2)d < dm. So b i + b j / V 1, and b i + b j / V 2. Case 2: (L=2K) Let V 1 {d, 3d,..., (l 1)d, (m l + 1)d, (m l + 3)d,..., (m 3)d, (m 1)d}, and V 2 = {1, 1 + d, 1 + 2d,..., 1 + (m 1)d} where d = 4 and dm = Z. Denote a i = (2i 1)d for 1 i K, b i = (m 2i + 1)d for 1 i K. Now

26 a 1 < a 2 <... < a K < b K < b K <... < b 1. Similarly to the previous argument, we can show the desired properties results Figure 4: IK 6,15 (mod 60)

27 CHAPTER 3 CONCLUSIONS and CONJECTURES Conclusions In this paper we have extended the classes of graphs for which we know the mod sum number. These are ρ(k n1,n 2,...,n m ) > 0 if there exist n i and n j such that n i < n j < 2n i. ρ(k n,m ) = 0 when m 2n, n 3 and m even. ρ(k n,m ) = 0 when m 3n 3, n 3 and m odd. ρ(k n,m ) = n when 2n > m n 3. So we conclude with conjectures that our research has shown may be possible, but for lack of time we were not able to pursue. Conjectures Lemma 2.8 and T heorem 2.11 might be used to show the following. Conjecture 3.15 ρ(k n,m ) = m when 3n 3 > m n 3. The following would be an extension to T heorem Conjecture 3.16 If K n1,n 2,...,n m with n 1 > n 2 >... > n m is not a MSG, then (m 1)n m ρ(k n1,n 2,...,n m ) (m 1)n 1. 19

28 20 The following shows good promise and would be an extension of Lemma 2.8. Conjecture 3.17 If K n1,n 2,...,n m with n 1 > n 2 >... > n m is a MSG, then n i 2n i+1 + m j=i+2 n j. The following would be extension to work by [7] and [9]. Conjecture 3.18 ρ(w n ) = 3 when for n > 4 and n even. Conjecture 3.19 ρ(w n ) = n when for n > 4 and n odd. The following conjectures appear to be true, but seem difficult to prove or disprove. Conjecture 3.20 Let G = (V, E) be a graph with V = n, then ρ(g) n. Conjecture 3.21 Let G = (V, E) be a graph, then finding the mod sum number is np-complete. Conjecture 3.22 Let G = (V, E) be a graph. There exist L based on n such that if G is not MSG modulo Z L then G is not a MSG.

29 BIBLIOGRAPHY 21

30 [1] D. Bergstrand, F. Harary, K. Hodges, G. Jennings, L. Kuklinski, and J. Wiener, The sum number of a complete graph, Bull. Malaysain Math. Soc. 12 (1989) 22 [2] J. Boland, R. Laskar, C. Turner, and G. Domke, On mod sum graphs, Congressus Numeratium 70 (1990) [3] M. N. Ellingham, Sum graphs from trees, Ars Combinatoria 35 (1993) [4] J. Ghoshal, R. Laskar, D. Pillone, and G. Fricke, Further results on mod sum graphs, Congressus Numerantium 101 (1994) [5] F. Harary, Sum graphs and difference graphs, Congressus Numerantium 72 (1990) [6] N. Hartsfield and W. F. Smyth, The sum number of complete bipartite graphs, Graphs and Matrices (edited by Rolf Rees), Marcel Dekker (1992) [7] N. Hartsfield and W. F. Smyth, A family of sparse graphs of large sum number, Discrete Mathematics 141 (1995) [8] M. Miller, M. Sutton, Mod Sum Graph Labelling of H m,n and K n, to appear. [9] M. Miller, J. Ryan, Slamin, M. Sutton, Connected Graphs which are not Mod Sum Graphs, Discrete Mathematics 195 (1999) [10] J. Ryan, M. Miller, and W. F. Smyth, The Sum Labelling for the Cocktail Party Graph, Bulletin of the ICA 22 (1998)

31 [11] W. F. Smyth, Sum graphs of small sum numbers, Colloq. Math. Soc. Jonos Bolyai 60 (1991) [12] W. F. Smyth, Sum graphs: New results, new problems, Bulletin of the ICA 2 (1991)

32 VITA Christopher D. Wallace 111 Brown Road #26 Johnson City, TN Phone: (423) PROFESSIONAL EXPERIENCE: Math Tutor, ETSU, Johnson City, TN, 1/95-8/98 Math Lab Coordinator, ETSU, Johnson City, TN, 1/98-8/98 Math Instructor, ETSU, Johnson City, TN, 8/98-present EDUCATION B.S. with majors in Computer Science and Math, and concentration in Physics, ETSU, 5/97 M.S. in Mathematical Science, ETSU, 5/99 Graduate Student in Computer Science, ETSU, 8/97-present PRESENTATIONS: Progress on Mod Sum Graphs, at the 936 th AMS Meeting, Wake Forest University, October 9-10, 1998 HONORS AND AWARDS Hall of Fame in Computer Science, ETSU,

33 Outstanding Mathematics Senior, 1996, ETSU Outstanding Mathematics Senior, 1997, ETSU Runner Up for Outstanding Undergraduate in Computer Science, 1997, ETSU Graduated 1997, ETSU Graduating 1999, ETSU Member of Gamma Beta, Kappa Mu Epsilon, Phi Kappa Phi, and Upsilon Pi Upsilon Honor Societies