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
APPROVAL
This is to certify that the Graduate Committee of
Christopher D. Wallace
met on the
1st day of April, 1999.
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
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 6= v and u + v (mod Z) ∈ V . First we discuss conditions which
Kn1 ,n2 ,...,nm must satisfy to be a mod sum graph and then we determine the minimum
number of isolated vertices such that Kn,m m ≥ n ≥ 3 is a mod sum graph except
when 2n ≤ m < 3n − 3 and m is odd.
iii
Copyright by Christopher D. Wallace 1999
iv
DEDICATION
To my wife Olivia, my parents Billie and Bill, and my sister Sheena. Their support,
encouragement, and love made this thesis possible.
v
ACKNOWLEDGEMENTS
I would like to thank Jay Boland who helped with assistance in making this thesis
a reality.
vi
Contents
APPROVAL
ii
ABSTRACT
iii
COPYRIGHT
iv
DEDICATION
v
ACKNOWLEDGEMENTS
vi
LIST OF FIGURES
1.
2.
3.
viii
INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Survey on Sum Graphs . . . . . . . . . . . . . . . . . . . . . . . . .
2
Mod Sum Graph Definitions and Survey . . . . . . . . . . . . . . . .
2
Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
MAIN RESULTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
Complete Partite Graphs . . . . . . . . . . . . . . . . . . . . . . . .
5
CONCLUSIONS and CONJECTURES . . . . . . . . . . . . . . . . .
19
Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
Conjectures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
BIBLIOGRAPHY
21
VITA
24
vii
List of Figures
1
ρ(IK5,11 ) ≤ 11 (mod 110) . . . . . . . . . . . . . . . . . . . . . . .
15
2
ρ(IK4,7 ) = 4 (mod 40) . . . . . . . . . . . . . . . . . . . . . . . . .
16
3
K4,8
(mod 32) . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
4
IK6,15
(mod 60) . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
viii
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, Kn , has n vertices
and every pair of vertices share an edge. The complete m-partite graph, Kn1 ,n2 ,...,nm ,
has partite sets V1 , V2 , ..., Vm with |Vi | = ni having the property that if u ∈ Vi , v ∈ Vj ,
i 6= j then {u, v} ∈ E and {u, v} 6∈ E when i = j . One class of complete mpartite graphs is Hn,m which is a m-partite graph which each partite set containing
n vertices. A cycle containing n vertices is denoted Cn . Also a wheel Wn contains a
cycle of length n and an additional vertex which is adjacent to every vertex on the
cycle. A tree Tn is a acyclic connected graph containing n vertices, while a forest Fn
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 1988. 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
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(Kn ) = 2n − 3 for n ≥ 4 [1]
• σ(Km,n ) =
d3m+n−3e
2
for n ≥ m ≥ 2 [6]
• σ(H2,n ) = 4n − 5 for n ≥ 2 [10]
• σ(Cn ) = 2 for n 6= 4 [5]
• σ(Wn ) =
n
2
+ 2 for even n ≥ 4,
σ(Wn ) = n for odd n ≥ 5 [7]
• σ(Fn ) = 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
3
vertices with distinct elements of {1, 2, ..., Z − 1} such that {u, v} ∈ E if and only if
u 6= 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(Kn ) = n for n ≥ 4 [8]
• ρ(K2,n ) = 0 and ρ(K1,n ) = 0 for n ≥ 2 [2]
• ρ(Kp+1,q ) = 0 for q ≥ 1 and p ≥ rq + rq−1 − 1 where r1 = 1, r2 = 2, and
rj = rj−2 + rj−1 + 1 for j ≥ 3 [2]
• ρ(Kn,n ) > 0 for n ≥ 3 [9]
• ρ(H2,n ) = 0 for n ≥ 2 [2]
• ρ(Hm,n ) > 0 for m > n ≥ 2 [8]
• ρ(Cn ) = 0 for n 6= 4 [2]
• ρ(Wn ) > 0 for n ≥ 5 [9]
• ρ(Tn ) = 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 Kn,m is a mod sum graph when m is even and
m ≥ 2n or when m is odd and m ≥ 3n − 3 and ρ(Kn,m ) = n when 3 ≤ n ≤ m < 2n.
4
In Chapter 2 we prove these results and we present conjectures and conclude with
Chapter 3.
CHAPTER 2
MAIN RESULTS
Notation
For Kn1 ,n2 ,...,nm the complete m-partite graph, there exist a set of labels F, some
nonnegative integer r, and positive integer Z, such that Kn1 ,n2 ,...,nm ∪Kr is a mod sum
graph (MSG) modulo Z. Denote IKn1 ,n2 ,...,nm = Kn1 ,n2 ,...,nm ∪Kr . Let V (Kn1 ,n2 ,...,nm ) =
V , V (Kr ) = R, E(Kn1 ,n2 ,...,nm ) = E. Denote partite sets Vi for 1 ≤ i ≤ m. We refer
to vertices by their label, so V (IKn1 ,n2 ,...,nm ) = 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 < ai < Z for 1 ≤ i ≤ n be integers. Let
d = gcd(a1 , a2 , ..., an ). Then there exist nonnegative integers Li for 1 ≤ i ≤ n such
that
Pn
i=1
Li ai ≡ d.
Discussion.
Pn
i=1
Since d = gcd(a1 , a2 , ..., an ), there exist Ki ∈ ZZ such that d =
Ki ai . 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 < ai < Z, for 1 ≤ i ≤ n,
5
6
then f (ai ) = ai . Also 0 < d < ai for all ai . So f (d) = d. Therefore d = f (d) ≡
f(
Pn
i=1
Ki ai ) ≡
Pn
i=1
f (Ki )f (ai ) ≡
Pn
i=1
f (Ki )ai , where f (Ki ) are nonnegative inte-
gers for 1 ≤ i ≤ n. 2
The following lemma is needed for Corollary 2.3.
Lemma 2.2 For IKn1 ,n2 ,...,nm , with a1 , a2 ∈ V1 , b ∈ V − V1 , and |V1 | ≥ 3, if a1 + b ∈
V − V1 , then a2 + b 6≡ a1 .
Proof. If a2 = a1 , then a2 + b = a1 + b ∈ V − V1 , and the result follows. Suppose
for the sake of contradiction, a2 + b ≡ a1 . Since |V1 | ≥ 3, there exist a3 ∈ V1 such
that a3 ∈
/ {a1 , a2 }. Now {a3 , b + a1 } ∈ E, since a3 ∈ V1 and b + a1 ∈ V − V1 . Thus
a3 + b + a1 ∈ F. Notice a3 + b ∈ F and a3 + b 6≡ a2 + b ≡ a1 . So {a3 + b, a1 } ∈ E
which implies a3 + b ∈ V − V1 . Hence {a3 + b, a2 } ∈ E and so a3 + b + a2 ∈ F . By
assumption a3 6≡ a1 ≡ b + a2 , therefore {a3 , b + a2 } ∈ E, a contradiction. 2
The following corollary is cited in Lemma 2.4, Lemma 2.5, and Lemma 2.6.
Corollary 2.3 Let V1 be a partite set of IKn1 ,n2 ,...,nm with |V1 | ≥ 3. If a1 ∈ V1 ,
b ∈ V −V1 , then either for all ai ∈ V , ai +b ∈
/ V −V1 , or for all ai ∈ V , ai +b ∈ V −V1
Proof. If for all ai ∈ V1 , ai + b ∈
/ V − V1 , the result follows. Assume there exist a
ai ∈ V such that ai + b ∈ V − V1 . Pick aj ∈ V1 . Then {aj , ai + b} ∈ E since aj ∈ V1
and ai + b ∈ V − V1 . So aj + ai + b ∈ F . Also aj + b ∈ F, since aj ∈ V and b ∈ V − V1 .
By Lemma 2.2, aj + b 6≡ ai . So {ai , aj + b} ∈ E. This implies aj + b ∈ V − V1 . 2
The next lemma is used in Lemma 2.5.
Lemma 2.4 Let V1 and V2 be partite sets of IKn1 ,n2 ,...,nm , with |V1 | = n1 , |V2 | = n2
and n2 , n1 ≥ 3. Suppose a ∈ V1 , b ∈ V2 . If a + b ∈ V1 , then for all b ∈ V2 , a + b ∈ V1 .
7
Proof. Denote V1 = {a1 , a2 , ..., an1 }, V2 = {b1 , b2 , ..., bn2 }, and suppose a1 + b1 ∈ V1 .
By Corollary 2.3, since a1 + b1 ∈ V − V2 , then for every bi ∈ V2 , a1 + bi ∈ V − V2 .
Likewise since a1 + b1 ∈
/ V − V1 , for every aj ∈ V1 , aj + b1 ∈
/ V − V1 . If for every bi ,
a1 + bi ∈ V1 , the result follows. Suppose there exist a b2 such that a1 + b2 ∈
/ V1 . Now
b2 + a1 ∈ V − V2 . So b2 + a1 ∈ V − V2 − V1 ⊂ V − V1 . Thus b2 + ai ∈ V − V1 , for all
ai ∈ V1 by Corollary 2.3. Pick ai ∈ V1 . Since b1 + a1 ∈ V1 and b2 + ai ∈ V − V1 , then
{b1 + a1 , b2 + ai } ∈ E. Hence b1 + a1 + b2 + ai ∈ F . Now we showed b1 + ai ∈
/ V − V1 ,
then b1 + ai ∈ V1 or b1 + ai ∈ R. Well b2 + a1 ∈ V − V1 , so b1 + ai 6≡ b2 + a1 . Hence
{b1 + ai , b2 + a1 } ∈ E which implies b1 + ai ∈ V1 . Therefore b1 + ai ∈ V1 for all ai ∈ V1 .
So a1 , b1 +a1 , b1 +b1 +a1 , ..., nb1 +a1 , ... ∈ V1 . Hence there exist a smallest J ∈ Z +
such that b1 (J + 1) ≡ 0 and kb1 + a1 ∈ V1 for 0 ≤ k ≤ J. Since b2 + ai ∈ V − V1 for all
ai ∈ V1 , then b2 +kb1 + a1 ∈ V − V1 for 0 ≤ k ≤ J. Then {b2 +kb1 + a1 , a1 } ∈ E which
implies b2 + kb1 + 2a1 ∈ F for 0 ≤ k ≤ J. Hence for 0 ≤ k < J, b2 + kb1 + 2a1 6≡ 0, so
b2 + (k + 1)b1 + 2a1 6≡ b1 for 1 ≤ k ≤ J. Also b2 + 2a1 6≡ b1 , since b2 + Jb1 + 2a1 6≡ 0.
Notice b2 + kb1 + 2a1 + b1 ∈ F for 0 ≤ k ≤ J. So {b2 + kb1 + 2a1 , b1 } ∈ E implies
{b2 + kb1 + 2a1 , b2 } ∈ E implies 2b2 + kb1 + 2a1 ∈ F for 0 ≤ k ≤ J.
Suppose for l > 0, lb2 + kb1 + 2a1 ∈ F for 0 ≤ k ≤ J. So by the previous
argument for 0 ≤ k ≤ J, lb2 + kb1 + 2a1 6≡ b1 and lb2 + (k + 1)b1 + 2a1 ∈ F . Hence
{lb2 +kb1 +2a1 , b1 } ∈ E implies {lb2 +kb1 +2a1 , b2 } ∈ E implies (l+1)b2 +kb1 +2a1 ∈ F
for 0 ≤ k ≤ J. Whence lb2 + b1 + 2a1 ∈ F for all l ≥ 0. So there must exist
K > N such that Kb2 + b1 + 2a1 ≡ N b2 + b1 + 2a1 . So (K − N )b2 ≡ 0. Therefore
(K − N )b2 + b1 + 2a1 ≡ b1 + 2a1 ∈ F .
{b1 + a1 , a1 } ∈ E, a contradiction. 2
Now b1 + a1 6≡ a1 since b1 6≡ 0.
So
8
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 2.11.
Lemma 2.5 Let V1 and V2 be partite sets of IKn1 ,n2 ,...,nm , with |V1 | = n1 , |V2 | = n2
and n2 ≥ n1 ≥ 3. Suppose b1 ∈ V2 . Then either for all ai ∈ V1 , ai + b1 ∈ R, or for all
ai ∈ V1 , ai + b1 ∈ V − V1 .
Proof. Suppose a1 ∈ V1 , b1 ∈ V2 , and a1 + b1 ∈ V − V1 . So by Corollary 2.3 for all
ai ∈ V1 , ai + b1 ∈ V − V1 .
Now suppose a1 + b1 ∈ V1 . By Lemma 2.4 for every bi ∈ V2 , a1 + bi ∈ V1 . So
a1 , a1 + bi ∈ V1 for 1 ≤ i ≤ n2 , but a1 6≡ a1 + bi . Also a1 + bi 6≡ a1 + bj for i 6= j.
Thus a1 , a1 + bi for 1 ≤ i ≤ n2 are n2 + 1 distinct elements in V1 , a contradiction since
n2 ≥ |V1 |.
Finally, observe that if a1 + b1 ∈ R, then by the conclusion of the previous paragraph, there exists no vertex ai ∈ V1 such that ai + b1 ∈ V . This completes the proof.
2
The next lemma is needed in Lemma 2.7.
Lemma 2.6 Suppose Kn1 ,n2 ,...,nm is a MSG, i.e. V = F, with V1 and V2 partite sets
and |V2 | = n2 ≥ |V1 | = n1 ≥ 3. If V1 = {a1 , a2 , ..., an1 } with gcd(a1 , a2 , ..., an1 ) = d,
then there exist L ∈ ZZ+ such that (L + 1)d ≡ Z and for all b ∈ V2 , i ∈ ZZ, b + id ∈
V − V1 .
Proof. Denote V2 = {b1 , b2 , ..., bn2 }. By assumption |R| = 0, so by Lemma 2.5,
bi + aj ∈ V − V1 for 1 ≤ i ≤ n2 , 1 ≤ j ≤ n2 . Pick b ∈ V2 .
9
Assume there exist nonnegative integers Ci for 1 ≤ i ≤ n1 such that b+
0
V − V1 for 0 ≤ Ci ≤ Ci .
Pn1
i=1
i6=k
i=1
0
Ci ai ∈
For sake of contradiction suppose there exist k and
0 ≤ Ki ≤ Ci for i 6= k, such that b +
assumption b +
Pn1
Pn1
Ki ai + (Ck + 1)ak ∈ V1 .
i=1
i6=k
Now by
Ki ai + (Ck )ak ∈ V − V1 . Also since b + ak ∈
/ V1 either Ck > 0
or there exist Ki > 0 for some 1 ≤ i ≤ n1 , i 6= k, denote Kc .
Case 1: (Ck > 0) Since b +
Pn1
i=1,i6=k
Ki ai + Ck ak ∈ V − V1 and b +
Ck ak + ak ∈ V1 , by Corollary 2.3, b +
Pn1
i=1,i6=k
Pn1
i=1,i6=k
Ki ai +
Ki ai + (Ck )ak + aj ∈ V1 for 1 ≤
j ≤ n1 and notice these are n1 distinct vertices of V1 . So there exist l such that
ak ≡ b +
Pn1
i=1
i6=k
Ki ai + Ck ak + al .
Hence b +
but since Ck > 0 by our assumption b +
b+
Pn1
i=1
i6=k
Pn1
i=1
i6=k
Pn1
i=1
i6=k
Ki ai + (Ck − 1)ak + al ≡ 0,
Ki ai + (Ck − 1)ak ∈ V − V1 . Thus
Ki ai + (Ck − 1)ak + al ∈ F, a contradiction.
Case 2: (Ck = Kk = 0) So Kc > 0 for some c 6= k.
b+
Pn1
i=1
i6=c
So by Corollary 2.3,
Ki ai + Kc ac + aj ∈ V1 for 1 ≤ j ≤ n1 , and also notice they are n1 distinct
elements. Since |V1 | = n1 and ac ∈ V1 , there exist l such ac ≡ b+
Therefore b+
Hence b +
Pn1
Pn1
i=1
i6=c
i=1
Pn1
i=1
i6=c
Ki ai +Kc ac +al .
Ki ai +(Kc −1)ac +al ≡ 0, a contradiction by the previous argument.
Ci ai ∈ V − V1 for all Ci ≥ 0.
Now since d = gcd(a1 , ..., an1 ), by Observation 2.1, there exist nonnegative integers Ki for 1 ≤ i ≤ n1 such that d ≡
b + d ∈ V − V1 . Since b +
b+l
Pn1
i=1
Pn1
i=1
Pn1
i=1
Ki ai . Then it follows that b +
Pn1
i=1
Ki ai ≡
Ci ai ∈ V − V1 for all Ci ≥ 0, then for all l ≥ 0,
Ki ai ≡ b + ld ∈ V − V1 . So there must exist i,j where i 6= j such that
b + id ≡ b + jd, since otherwise |V − V1 | = ∞. 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 − V1 . 2
10
The following lemma is the major result in this paper. With it, Lemma 2.8 shows
Kn,m with m > n ≥ 3 is not a MSG when m < 2n.
Lemma 2.7 Suppose Kn1 ,n2 ,...,nm is a MSG, i.e. V = F, with V1 and V2 partite sets,
and |V2 | = n2 ≥ |V1 | = n1 ≥ 3. Let V1 = {a1 , a2 , ..., an1 } with gcd(a1 , a2 , ..., an1 ) = d.
If n2 > n1 , there exist b ∈ V2 such that for all i ∈ ZZ, b + id ∈ V2 .
Proof. Since for all b ∈ V2 , 0 < b < Z, then 2b ≡ 0 implies b =
Z
.
2
Since n2 ≥ 3, we
can chose a b ∈ V2 such that 2b 6≡ 0. Assume V1 = {C1 d, C2 d, ..., Cn1 d}. By Lemma
2.6, for all i ∈ ZZ, b+id ∈ V −V1 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 ∈
/ V2 .
THEN WE ARE GOING TO SHOW n1 = n2 . We first establish 2b + C1 d ∈ V1 ,
b + C1 d ∈
/ V2 , and b + 2C1 d, 3b + 2C1 d ∈ V2 . 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 ∈ V2 or kb + id ∈
/ V2 . 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 ≡ N b + 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 6= 0, we have a J > 0 such that (J + 1)b ≡ 0. If j 6= i,
b+k(K −N )b ≡ b+k(j −i)d ∈ V −V1 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.
11
Suppose for any l ∈ Z, that Jb + ld ∈ V1 . Now b + (C1 + C2 − l)d ∈ V − V1 . Then
{Jb + ld, b + (C1 + C2 − l)d} ∈ E which implies (C1 + C2 )d ∈ V , a contradiction.
So Jb + ld ∈ V − V1 .
Thus Jb + kd ∈
/ V1 .
Whence {Jb + kd, C1 d} ∈ E, and
Jb+kd+C1 d ∈ V . By the previous argument Jb+kd+C1 d ∈ V −V1 . Repeating one
can see that Jb + kd +
Pn1
i=1
Ki Ci d ∈ V − V1 for all Ki ≥ 0. Since d = gcd(a1 , ..., an1 ),
0
by Observation 2.1, there exist nonnegative integers Ki for 1 ≤ i ≤ n1 such that
d≡
Pn1
i=1
0
Ki ai . Therefore for l ≥ 0, Jb + kd + l
Pn1
i=1
0
Ki ai ≡ Jb + kd + ld ∈ V − V1 .
Since we have L as the smallest positive integer such that (L+1)d ≡ 0, then Jb+id ∈
V − V1 for 0 ≤ i ≤ L are L + 1 distinct elements.
We wish to show for all Ci d, Cj d ∈ V1 , that b + Ci d, J + Cj d are in the same
partite set distinct from V2 . Denote V3 , the partite set containing b + C1 d. Now if
b + C1 d ≡ Jb + Ci d for some i then it follows that Jb + Ci d ∈ V3 . Suppose b + C1 d 6≡
Jb+Ci d for i 6= 1. Then Jb+Ci d ∈ V3 since otherwise {Jb+Ci d, b+C1 d} ∈ E implies
Ci d + C1 d ∈ F for i 6= 1. Hence Jb + Ci d ∈ V3 for i 6= 1. Since n1 ≥ 3, for b + Cj d
where j 6= 1, there exist Cl d such that Cl d 6= Cj d, Cl d 6= C1 d. So Jb + Cl d ∈ V3 , by
the previous argument, implies b + Cj d ∈ V3 . Hence b + Cj d ∈ V3 for all Cj d ∈ V1 .
Finally, we have b + C2 d ∈ V3 implies Jb + C1 d ∈ V3 . Now there exist an i such
that Jb + Ci d 6≡ b. Since Jb + Ci d + b ≡ Ci d, it follows that {Jb + Ci d, b} ∈ E.
So V3 is a different partite set then V2 and V3 contains b + Ci d, Jb + Cj d for all
Ci d, Cj d ∈ V1 . Therefore we have {b, b + Ci d} ∈ E implying that 2b + Ci d ∈ V . Now
{b + Ci d, b + Cj d} ∈
/ E for i 6= j. So 2b + Ci d + Cj d ∈
/ V , for i 6= j. So 2b + Ci d ∈ V1
for all Ci d ∈ V1 .
The vertices Jb, Jb + (Ci + Cj )d ∈ V2 for i 6= j, since these vertices must not be
12
adjacent to b. So b + Ci d + Cj d ∈ V2 for i 6= j, because they can not be adjacent to
Jb. Now we established b+C1 d ∈
/ V1 . Then {b+C1 d, C1 d} ∈ E, and so b+2C1 d ∈ V .
Since 2b 6≡ 0, then C1 d 6≡ 2b + C1 d. Notice 2b + 2C1 d ∈
/ F implies {b + 2C1 d, b} ∈
/E
implying b + 2C1 d ∈ V2 . Also since C1 d 6≡ 2b + C1 d, we have b + (C1 d) + (2b +
C1 d) ≡ 3b + 2C1 d ∈ V2 . So we have established 2b + C1 d ∈ V1 , b + C1 d ∈
/ V2 , and
b + 2C1 d, 3b + 2C1 d ∈ V2 .
We wish to establish n1 = n2 = |V3 |. By assumption n2 ≥ n1 . Let f : V2 → V1 by
f (x) = x+b+C1 d. Pick x ∈ V2 . Then {x, b+C1 d} ∈ E, and so x+b+C1 d ∈ V . Now
if x + b + C1 d ∈
/ V1 , then {x + b + C1 d, C1 d}, {x + b + C1 d, 2b + C1 d} ∈ E. Therefore
x + b + 2C1 d, x + 3b + 2C1 d ∈ V . Well 2b 6≡ 0 implies b + 2C1 d 6≡ 3b + 2C1 d. Hence
either {x, b + 2C1 d} ∈ E or {x, 3b + 2C1 d} ∈ E, which can not happen. Therefore
x + b + C1 d ∈ V1 . If f (x) ≡ f (y) then x + b + C1 d ≡ y + b + C1 d, and so x ≡ y. So f
is an injection. Thus n1 = |V1 | ≥ |V2 | = n2 , a contradiction. 2
The next lemma is cited in T heorem 2.12 and T heorem 2.13. These theorems
ρ(Kn,m ) when m > n ≥ 3 and m is even.
Lemma 2.8 If Kn1 ,n2 ,...,nm is a MSG, i.e. V = F, with V1 and V2 partite sets |V2 | =
n2 > |V1 | = n1 ≥ 3, then n2 ≥ 2n1 . If a1 , 2a1 ∈ V1 , then n2 ≥ 3n1 − 3.
Proof. Denote V1 = {C1 d, C2 d, ..., Cn1 d} where d = gcd(C1 d, C2 d, ..., Cn1 d). By
Lemma 2.7, there exist b ∈ V2 such that b + id ∈ V2 for 0 ≤ i ≤ L where L is the
smallest positive integer such that (L + 1)d ≡ 0. Thus b + id ∈ V2 for 0 ≤ i ≤ L are
L + 1 distinct vertices implying n2 ≥ 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 {C1 d, Ci d} ∈
/ E, so C1 d + Ci d ∈
/ F
13
for 2 ≤ i ≤ n1 . If C1 d + Ci d ≡ C1 d + Cj d, then Ci d ≡ Cj d. So Cj d, {C1 + Ci }d
for 1 ≤ j ≤ n1 , 2 ≤ i ≤ n are 2n1 − 1 distinct multiples of d modulo Z.
So
n2 ≥ L + 1 ≥ 2n1 − 1.
If C1 d+Cj d ≡ 2C1 d, then Cj d ≡ C1 d. Hence if 2C1 d ∈
/ V1 , then n2 ≥ L+1 ≥ 2n1 .
Suppose 2C1 d ≡ C2 d ∈ V1 . We will establish C1 d, 2C1 d, 3C1 d, Ci d, C1 d + Ci d,
2C1 d + Ci d for 3 ≤ i ≤ n1 are 3n1 − 3 distinct multiples of d mod Z. Now C1 d 6≡
Ci d, 2C1 d 6≡ Ci d for 3 ≤ i ≤ n1 .
Also 3C1 d, C1 d + Ci d, 2C1 d + Ci d ∈
/ F for
3 ≤ i ≤ n1 . Likewise C1 d, 2C1 d, Ci d ∈
/ {3C1 d, C1 d + Cj d, 2C1 d + Cj d} (mod Z)
for 3 ≤ j ≤ n1 , 3 ≤ i ≤ n1 . If 3C1 d ≡ C1 d + Ci d, then 2C1 d ≡ Ci d. Notice if
3C1 d ≡ 2C1 d + Ci d then C1 d ≡ Ci d. So 3C1 d ∈
/ {C1 d + Ci d, 2C1 d + Ci d} (mod Z)
for 3 ≤ i ≤ n1 . Finally C1 d + Ci d 6≡ 2C1 d + Cj d, since Ci d 6≡ C1 d + Cj d. Hence we
have 1 + 1 + 1 + (n1 − 2) + (n1 − 2) + (n1 − 2) = 3n1 − 3 distinct multiples of d mod
Z. So n2 ≥ L + 1 ≥ 3n1 − 3 ≥ 2n1 when n1 ≥ 3. 2
Corollary 2.9 If Kn1 ,n2 ,...,nm is a MSG with n1 > n2 > ... > nm−1 > nm ≥ 3, then
n1 ≥ 2n2 ≥ 22 n3 ≥ ... ≥ 2m−2 nm−1 ≥ 2m−1 nm .
Proof. The result follows by induction using Lemma 2.8. 2
The next theorem is a special case needed for T heorem 2.12. This result was first
proved by Miller, Ryan, Sutton, and Slamin in [9].
Theorem 2.10 Kn,n is not a MSG for n ≥ 3.
Proof. Denote partitions V1 = {a1 , a2 , ..., an } and V2 = {b1 , b2 , ..., bn }. By Lemma
2.5, if Kn,n is a MSG, with aj ∈ V1 then for all bi ∈ V2 , aj + bi ∈ V − V2 . Likewise
14
for bj ∈ V2 , for all ai ∈ V1 , ai + bj ∈ V − V1 . Since both can not be true aj + bi ∈ R,
for all aj ∈ V1 , bi ∈ V2 . 2
The following theorem is used in T heorem 2.12.
Theorem 2.11 If Kn,m is not a MSG, with n ≤ m then n ≤ ρ(IKn,m ) ≤ m.
Proof. Denote partitions V1 = {a1 , a2 , ..., an } and V2 = {b1 , b2 , ..., bm }. By assumption there exist ai , bj such that ai + bj ∈ R. Now by Lemma 2.5, bj + ai ∈ R for
1 ≤ i ≤ n. So ρ(IKn,m ) ≥ n.
We now give a labelling for which |R| = m. Let ai = (i−1)10+7, bi = (i−1)10+9,
ri = (i−1)10+6 with Z = 10m. Then R = {6, 16, ..., 10(m−1)+6}. Notice ai +bj =
(i + j − 2)10 + 16 ∈ R, ai + aj = (i + j − 2)10 + 14 ∈
/ F , bi + bj = (i + j − 2)10 + 18 ∈
/ F,
ri + rj = (i + j − 2)10 + 12 ∈
/ F , ai + rj = (i + j − 2)10 + 13 ∈
/ F , and finally,
bi + rj = (i + j − 2)10 + 15 ∈
/ F. Thus ρ(IKn,m ) ≤ m. 2
The next three theorems, T heorem 2.12, T heorem 2.13, and T heorem 2.14, show
Kn,m is a mod sum graph when m is even and m ≥ 2n or when m is odd and
m ≥ 3n − 3 and ρ(Kn,m ) = n when 3 ≤ n ≤ m < 2n.
Theorem 2.12 ρ(Kn,m ) = n, if 3 ≤ n ≤ m < 2n.
Proof. By Lemma 2.8, Kn,m is not a MSG since m < 2n. By T heorem 2.11,
ρ(Kn,m ) ≥ n. Denote partitions V1 = {a1 , a2 , ..., an } and V2 = {b1 , b2 , ..., bm }.
We now give a labelling for which |R| = n. Let ai = (i−1)10+9, bi = (i−1)10+6
for 1 ≤ i ≤ n, bi = (i − 1)10 + 7 for n < i ≤ m. Let ri = (i − 1)10 + 5 with Z = 10n.
Then R = {5, 15, ..., 10(n−1)+5}. Notice if j ≤ n, then ai +bj = (i+j−2)10+15 ∈ R,
and if j > n, then ai + bj = (i + j − 2)10 + 16 ∈ V2 . Now for any i,j, ai + aj =
15
7
17
6
16
27
26
37
47
57
67
77
9
19
29
39
49
36
46
56
66
76
87
97
107
86
96
106
Figure 1: ρ(IK5,11 ) ≤ 11 (mod 110)
(i + j − 2)10 + 18 ∈
/ F , for i, j ≤ n, bi + bj = (i + j − 2)10 + 12 ∈
/ F, for i ≤ n, j > n,
bi + bj = (i + j − 2)10 + 13 ∈
/ F, and for i, j > n, bi + bj = (i + j − 2)10 + 14 ∈
/ F.
Also, ri + rj = (i + j − 2)10 + 10 ∈
/ F, ai + rj = (i + j − 2)10 + 14 ∈
/ F , and finally,
for i ≤ n, bi + rj = (i + j − 2)10 + 11 ∈
/ F, for i > n, bi + rj = (i + j − 2)10 + 12 ∈
/ F.
Thus ρ(IKn,m ) ≤ n. 2
Theorem 2.13 If m is even and m ≥ 4, then Kn,m with m ≥ n is a MSG if and
only if m ≥ 2n.
Proof. =⇒ By T heorem 2.10, Kn,m is not a MSG if n = m. By Lemma 2.8, if
Kn,m is MSG with m > n, then m ≥ 2n.
⇐= Assume m ≥ 2n with m even. Let V1 and V2 be partite sets. Let V1 ⊂
{d, 3d, ..., (m − 1)d}, V2 = {1, 1 + d, 1 + 2d, ..., 1 + (m − 1)d}, d = 4, and Z = dm =
4m. Pick (2i + 1)d ∈ V1 and 1 + jd ∈ V2 , then {(2i + 1)d, 1 + jd} ∈ E since
16
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Figure 2: ρ(IK4,7 ) = 4 (mod 40)
(2i + 1)d + 1 + jd = 1 + (jd + 2(i + 1)d) (mod dm) = 1 + kd for some k. Pick (2i + 1)d,
(2j + 1)d ∈ V1 . 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 ∈ V2 .
Now (1 + id + 1 + jd) (mod dm) = 2 + kd for some k. Also 2 + kd 6= ld, since 2 =
(k −l)d (mod dm) implies d < 4. Likewise 2+kd 6= 1+ld, since 1 = (k −l)d (mod dm)
implies d < 4. Thus 1 + id + 1 + jd ∈
/ F which in turn means {1 + id, 1 + jd} ∈
/ E. 2
Theorem 2.14 If m is odd and m ≥ 3, then Kn,m with m ≥ 3n − 3 is a MSG.
Proof. Let L = b m3 c + 1. Then L is the largest integer such that 3L − 3 ≤ m. So
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 V1 ⊂ {d, 3d, ..., (l − 2)d, (m − l)d, (m − l + 2)d, ..., (m −
3)d, (m−1)d}, and V2 = {1, 1+d, 1+2d, ..., 1+(m−1)d} where d = 4 and dm = Z. So
for any s ∈ V1 , r ∈ V2 , s+r ∈ V2 , and for any s, r ∈ V2 , s+r ∈
/ V . Denote ai = (2i−1)d
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Figure 3: K4,8
1
5
9
13
17
21
25
29
(mod 32)
for 1 ≤ i ≤ K, bi = (m − 2i + 1)d for 1 ≤ i ≤ K + 1. Then the ai are odd multiples of
d and the bi are even multiples of d. Now a1 < a2 < ... < aK < bK+1 < bK < ... < b1 .
Therefore for any ai , aj ∈ V1 for i 6= j, 2(i+j−1)d = ai +aj ≤ aK−1 +aK ≤ (2L−6)d <
(2L − 2)d ≤ (m − L + 1)d < dm. So ai + aj = 2(i + j − 1)d 6≡ ah = (2h − 1)d for any
h, and ai + aj < (m − L)d = bK+1 . Hence ai + aj ∈
/ V1 , and ai + aj ∈
/ V2 . Pick ai ,
bj ∈ V1 . Now ai + bj = (2i − 1)d + (m − 2j + 1)d ≡ (m − 2(i − j))d. If ai + bj < dm,
then ai + bj is and odd multiple of d, and aK < ai + bj , so ai + bj ∈
/ V1 . If ai +bj > dm,
then ai + bj < ai (mod dm) < bK+1 , and ai + bj is an even multiple. So ai + bj ∈
/ V1 ,
and ai + bj ∈
/ V2 . Pick bi , bj ∈ V1 with i 6= j. Then dm < (2m − 2L + 2)d =
(2m − 4K)d = bK + bK+1 ≤ bi + bj < 2md. So (2m − 2L + 2)d ≡ (m − 2L + 2)d, and
aK = (L − 2)d < (L − 1)d ≤ (m − 2L + 2)d < dm. So bi + bj ∈
/ V1 , and bi + bj ∈
/ V2 .
Case 2: (L=2K) Let V1 ⊂ {d, 3d, ..., (l − 1)d, (m − l + 1)d, (m − l + 3)d, ..., (m −
3)d, (m − 1)d}, and V2 = {1, 1 + d, 1 + 2d, ..., 1 + (m − 1)d} where d = 4 and dm = Z.
Denote ai = (2i − 1)d for 1 ≤ i ≤ K, bi = (m − 2i + 1)d for 1 ≤ i ≤ K. Now
18
a1 < a2 < ... < aK < bK < bK−‘ < ... < b1 . Similarly to the previous argument, we
can show the desired properties results. 2
4
1
5
9
1
3
1
1
7
2
2
2
1
2
0
5
4
0
2
4
9
Figure 4: IK6,15
8
3
5
3
6
3
7
(mod 60)
4
1
4
5
4
9
5
3
5
7
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(Kn1 ,n2 ,...,nm ) > 0 if there exist ni and nj such that ni < nj < 2ni .
• ρ(Kn,m ) = 0 when m ≥ 2n, n ≥ 3 and m even.
• ρ(Kn,m ) = 0 when m ≥ 3n − 3, n ≥ 3 and m odd.
• ρ(Kn,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(Kn,m ) = m when 3n − 3 > m ≥ n ≥ 3.
The following would be an extension to T heorem 2.11.
Conjecture 3.16 If Kn1 ,n2 ,...,nm with n1 > n2 > ... > nm is not a MSG, then
(m − 1)nm ≤ ρ(Kn1 ,n2 ,...,nm ) ≤ (m − 1)n1 .
19
20
The following shows good promise and would be an extension of Lemma 2.8.
Conjecture 3.17 If Kn1 ,n2 ,...,nm with n1 > n2 > ... > nm is a MSG, then ni ≥
2ni+1 +
Pm
j=i+2
nj .
The following would be extension to work by [7] and [9].
Conjecture 3.18 ρ(Wn ) = 3 when for n > 4 and n even.
Conjecture 3.19 ρ(Wn ) = 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.
BIBLIOGRAPHY
21
22
[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)
[2] J. Boland, R. Laskar, C. Turner, and G. Domke, On mod sum graphs, Congressus Numeratium 70 (1990) 131-135
[3] M. N. Ellingham, Sum graphs from trees, Ars Combinatoria 35 (1993) 335339.
[4] J. Ghoshal, R. Laskar, D. Pillone, and G. Fricke, Further results on mod
sum graphs, Congressus Numerantium 101 (1994) 202-201.
[5] F. Harary, Sum graphs and difference graphs, Congressus Numerantium 72
(1990) 101-108.
[6] N. Hartsfield and W. F. Smyth, The sum number of complete bipartite
graphs, Graphs and Matrices (edited by Rolf Rees), Marcel Dekker (1992) 205211.
[7] N. Hartsfield and W. F. Smyth, A family of sparse graphs of large sum
number, Discrete Mathematics 141 (1995) 163-171.
[8] M. Miller, M. Sutton, Mod Sum Graph Labelling of Hm,n and Kn , to appear.
[9] M. Miller, J. Ryan, Slamin, M. Sutton, Connected Graphs which are not
Mod Sum Graphs, Discrete Mathematics 195 (1999) 287-293.
[10] J. Ryan, M. Miller, and W. F. Smyth, The Sum Labelling for the Cocktail
Party Graph, Bulletin of the ICA 22 (1998) 79-90.
23
[11] W. F. Smyth, Sum graphs of small sum numbers, Colloq. Math. Soc. Jonos
Bolyai 60 (1991) 669-678.
[12] W. F. Smyth, Sum graphs: New results, new problems, Bulletin of the
ICA 2 (1991) 79-81.
VITA
Christopher D. Wallace
111 Brown Road #26
Johnson City, TN 37615
Phone: (423)952-0545
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 936th AMS Meeting, Wake Forest University,
October 9-10, 1998
HONORS AND AWARDS
Hall of Fame in Computer Science, ETSU, 1994-1997
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Graduating 1999, ETSU
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