ELSEVIER Discrete Mathematics 195 (1999) 287-293

DISCRETE MATHEMATICS

N o t e

Connected graphs which are not mod sum graphs

M a r t i n S u t t o n a, M i r k a M i l l e r b'*, J o s e p h R y a n c, S l a m i n a

aDepartment of Computer Science, University of Newcastle, NSW 2308, Australia bDepartment of Computer Science and Software Engineering, University of Newcastle,

NSW 2308, Australia CSchool of Management, University of Newcastle, NSW 2308, Australia

dDepartment of Computer Science and Software Engineering, University of Newcastle, NSW 2308, Australia

Received 5 May 1997; received in revised form 26 January 1998; accepted 6 May 1998

Abstract

In this paper we prove that no wheel with the exception of W4 can be a mod sum graph. We also give the unique (up to multiplication by positive integers) mod sum labelling of W4. We also prove that the symmetric complete bipartite graph K,,,, is not a mod sum graph. (~) 1999 Elsevier Science B.V. All rights reserved

Keywords: Mod sum labelling; Mod sum graphs; Sum graphs; Sum graph labelling; Wheels

1. Introduction

A sum graph is a graph G = (V,E) and a labelling o f the vertices o f G with distinct

elements o f the finite subset S c Z + such that vivj E E if and only if the sum of the

labels assigned to vi and vj is the label o f a vertex o f G. It is obvious that a sum graph cannot be connected. There must always be at least one isolated vertex, namely

the vertex with the highest label. The sum number a (H) of a connected graph H is

the least number r o f isolated vertices K'r such that G = H + K'r is a sum graph. For more information about sum graphs see [1 ,3 ,5-10] .

Mod sum graphs (MSGs) were introduced by Bolland et al. [2] in 1990. A mod

sum graph is a sum graph with S=Zm\{O} and all arithmetic performed modulo m

where m >~ IV[ + 1.

* Corresponding author. E-mail: mirka@cs.newcastle.edu.au.

0012-365X/99/$-see front matter (~) 1999 Elsevier Science B.V. All fights reserved PII: S0012-365X(98)00184-8

288 M. Sutton et al./ Discrete Mathematics 195 (1999) 287-293

Although there are no connected sum graphs, it is possible to have MSGs which

are connected. For example, paths on n vertices, n >~ 3 trees on n/> 3 vertices, cycles

on n >/4 vertices, cocktail party graphs and some complete bipartite graphs have been shown to be MSGs [2].

On the other hand, there are connected graphs which are not MSGs, for example

complete graphs K, for n >/2 [2]. Consequently, it is reasonable to define the 'mod

sum number'. The rood sum number p(H) of a connected graph H is the least number r of isolated vertices K'r such that G = H + K'r is a mod sum graph.

A wheel W~ is a graph G = ( V , E ) with a vertex set V={vc , vl,v2 . . . . ,v,} such that

Vc, Vi E E for i = 1,2 . . . . ,n, vivi+ 1 c E for i = 1,2 . . . . . n - 1 and v, vl EE. The edges Vc, Vi for i = 1,2 . . . . . n are called spokes.

In [2], it is conjectured that the wheel on n + 1 vertices W,,, n >f 4, is a MSG. Recently, Ghoshal et al. [4] disproved the conjectured by showing that the wheel on p + 1 vertices, Wp, p >. 5, p prime, is not a MSG. Their proof cannot be extended to

W,, for n nonprime.

Using a different technique, in the next section, we disprove the conjecture for all n~>5.

In Section 3 we prove that the symmetric complete bipartite graph K,,, is not a MSG.

To simplify notations, throughout this paper, we assume that the vertices of G are

already identified by their labels.

2. Wheels

It is known that Wz (= K3) and W3 (=K4) are not mod sum graphs [2]. In this section we prove that W4 is the only wheel which is a mod sum graph.

Theorem 1. For n ¢ 4 , W~ is not a mod sum 9raph for any modm.

Let n ~> 5. Let vc be the label of the centre and {Vl,V2 . . . . . Vn-l,Vn} be the labels of the rim vertices of W~ (mod some suitable m). To prove the theorem, we shall make use of the following lemmas.

Lemma 2. In a mod sum labellin9 o f W~, the set o f labels o f the rim vertices is the

same as the set o f labels 9enerated by addin 9 the labels o f the end vertices o f each spoke.

Proof. We need to show that

{v , ,v2 . . . . . v . _ l , v ° } = + vo, 2 + . . . . . + v ,v. +

This is obvious since vi + vc are n distinct labels from {1,2 . . . . . m - l } and are of course, all different from Vc. []

M. Sutton et al./Discrete Mathematics 195 (1999) 287-293 289

Lemma 3. Let Vc be the label o f the centre o f W~. In a mod sum labelling o f IV,, the

labels o f the rim vertices can be partitioned into sets o f equal size l such that the

elements o f each set form a l-cycle under addition o f v~.

Proof. This follows immediately from Lemma 2 and may be achieved in the following

manner. Start with the set o f all labels belonging to rim vertices. Select the smallest

(modm) label, vi say, remove it from the set and place it as the first label of a new

set representing the first partition.

It is obvious from Lemma 2 that a rim vertex VJ must exist with a label equal to

vg + Vc. Remove this label from the set and place in the first partition.

Repeat the process until the next label in the series cannot be found in the original

set. This can only occur when the label has previously been selected for the first parti-

tion since the label must have originally been presented by Lemma 2. This completes

the selection of the first partition set.

I f there are labels still remaining in the original label set then the process is repeated

for a second partition set until once again the required label is not found in the original

set. This label cannot be a member o f the first partition set as this would imply that a

label from the first partition set is equal to a label from the second partition set which

contradicts the requirement that all labels be distinct.

This completes the selection o f the second partition set. Repeat the selection process

until all labels are placed in a partition set and the original set is empty.

Since the consecutive elements of each cycle differ by the same quantity vc, it is

obvious that all cycles must be o f the same size, that is, we have vi + vc = vg+l for

i = 1 . . . . . l - 1 and Vl + v~=vl where 1 is the size of the sets. []

In this paper we shall refer to these partition sets as/-cycles. We denote the vertices

whose labels form /-cycle Ci by vii, vi2 . . . . . vii.

Lemma 4. In a mod sum labelling o f W~, the sum o f the labels o f any two vertices

from a l-cycle cannot be equal to the label o f a vertex from the same l-cycle.

Proof. Assume that Vxi + vxj = v~k for some i , j , k E {1 . . . . . I}, i # j .

LHS = [Vxl + (i-1)Vc] + [V~l + ( j -1)Vc] ,

RHS = Vxl + ( k - 1 ) V o

Therefore, vxl = tVc where t E {0, 1 . . . . . l}. Then for some i E { 1 . . . . . l} we have Vxi = Vc,

a contradiction since all labels must be distinct. []

Lemma 5. In a mod sum labelling o f W~, the sum o f the labels o f any two vertices from the same l-cycle cannot be equal to the label o f the central vertex.

Proof. Suppose that v I Jr-v 2 = V c . Then, since v I and v 2 are in the same /-cycle, there

is a vertex vt in the /-cycle such that vt = 0, a contradiction. []

290 M. Sutton et al./Discrete Mathematics 195 (1999) 287-293

Lemma 6. In a mod sum labellin 9 o f W~, the sum o f the labels o f a vertex f rom C1,

and a vertex f rom C2 cannot be equal to the label o f a vertex f rom either C1 or C2.

Proof. Suppose Vli-[-V2j=Vlk. Then v2j=Vlk--Vli=tVc. Therefore, there exists s E {1 . . . . . l} such that vzs = vc, a contradiction. []

Lemma 7. In a rood sum labelling o f IV,, there must be at least two distinct l-cycles.

Proof. This is a direct consequence of Lemmas 3 and 4. []

Corollary 8. 2 ~ I <~ n/2.

Lemma 9. A t least one l-cycle in a mod sum labelling o f W,, n >>. 5, contains a pair

o f adjacent vertices.

Proof. Assume that all adjacent rim vertices have labels from different /-cycles if

v~iVvj E E then x # y, Vx, y = 1 . . . . . n/l, and i , j = 1 . . . . . l. Without loss of generality,

we can assume that vliv2j E E. By Lemma 6, the vertex with the label vii + v2j cannot be in Cj or C2.

Case 1: Assume for adjacent vertices from cycles Cl and C2 that vii + vej=v3k. I f this is true for any particular pair o f vertices (one from C1 and the other from C2) then it will be true for every pair o f vertices (one from Cl and the other from C2),

since the labels in all /-cycles differ by an integral number of Vc. This implies that

every vertex in Ci is adjacent to every vertex in C2, VliV2j E E, Vi, j = 1 . . . . . l. This is clearly impossible in a wheel if the vertex labels partition into three or more l-cycles.

Case 2: I f vii + v2j = vc then consider the other rim vertex adjacent to v2j, say Vxk

with vejvxk E E. Clearly, v2j + Vxk # Vc since vii # Vxk and so we have a special example of Case 1 which is impossible in a wheel if the vertex labels partition into three or

more /-cycles. (Note: it is possible if the labels partition into exactly two /-cycles). I f v2j + vxk # v~, then Vxk = Vlk and v2j + v~k = Vr, for some rim vertex Yr. This, in turn, implies that one of the labels in C2 is Vc which is impossible as it contradicts the

definition of a mod sum graph. []

Lemma lO. The labels o f the rim vertices o f any mod sum labellin 9 o f W~ can be

partitioned into 2-cycles with respect to addition o f re.

Proof. Assume that the size of the /-cycles is greater than 2, i.e., l /> 3. By Lemma 9, there exist two adjacent vertices whose labels are from the same /-cycle. Without loss of generality, we can assume vlivlj c E . By Lemma 5, vii + v U # v c . By Lemma 4,

vii + vii -7/= Vlk, SO that vii + vii = Vxk where x # 1. This implies that the sum of every pair o f the labels in C~ will be a label in Cx. This, in turn, implies that every vertex in Ci is adjacent to every other vertex in C~ and the vertices of C1 induce a complete subgraph K/ so that vlivlj E E, Vi, j = 1 . . . . . 1, i # j . This can only be true in a wheel when l = 2. []

M. Sutton et al./Discrete Mathematics 195 (1999) 287-293 291

Corollary 11. IV. is not a mod sum graph for n odd.

Corollary 12. Vc = m/2.

Proof of Theorem 1. If n >~ 5 then suppose that Cl = { V l l , V l 2 } , C2 ~-{v21,v22} and,

without loss of generality, assume that vl i is adjacent to v21, so that v l i ~- v21 z W ( s a y ) .

By Lemma 6, w is not in Cl or C2. Suppose w E C~, C~ ~ Cl, C2. This implies that

every vertex in Cl is adjacent to every vertex in C2, which is impossible. Hence, w = vc

and vii ~- v21 = Pc. Since vii ~- v c = VI2 and v2~ + Vc = v22, we also have v12 + v22 =/3 c. Using Lemma 9, we can assume that V~l is adjacent to v~2, so that v~t + v~2=u.

Clearly u p Vc, because v12 ¢v21. We now suppose that u E C~', cy ¢ c i , c 2 and let

u=vy~. Then v~ +v~2 = Vy~. But we have Vl~ +v21 =Vc, thus Vvl +v2~ =VlZ+Vc=V~w.

It immediately follows that v,,2 + v21 = Vl2.

This implies that vzl is adjacent to both t~,l and vy2, which is impossible in a wheel

W,, n > 4, as vzl is adjacent to vii and vc. []

Theorem 13. The wheel W4 & a mod sum graph and has a unique (up to multipfication

by a positive integer) labelling.

Proof. Using Lemma 7 and the results in Proof o f Theorem 1, we have VllVl2,VllV21,

v~2v22 E E and can deduce that v21v22 EE. We know that in any mod sum labelling of W4, the rim vertices must be partitioned into two 2-cycles.

Let VII =x . Then v l 2 = m / 2 + x, v21 = m / 2 - x and v 2 2 = m - x . Since VII + Vlz=V21

or vii + v12 =Vz2 we have 2x + m / 2 = m / 2 - x or 2x + m / 2 = m - x .

Clearly we can reject the first possibility as this gives x = 0. Therefore, 2 x + ( m / 2 ) =

m - x which gives x = m/6. Thus, for a given mod, which must be a multiple o f 6,

there is a unique mod sum labelling o f the 4-wheel. For example, if the rood is 6 then

the unique mod sum labelling o f W4 is {1,2,3,4,5}. Thus, W4 is a mod sum graph

with a labelling that is unique up to multiplication by a positive integer of both the

labels and the rood (a multiple o f 6) used. []

3. Symmetric complete bipartite graphs

As mentioned earlier, Bolland et al. [2] showed that complete bipartite graphs Kin.,,

for certain m and n, are mod sum graphs. Here we prove that for m----n, the complete bipartite graph K,,, is not a rood sum graph.

Theorem 14. For n >~ 3, the symmetric complete bipartite graph K,. , is not a rood sum graph.

Proof. Suppose Kn, n is a mod sum graph. Let V be a set o f vertices of Kn,,,, with

partite sets V1 and V2, I Vl I = ]/I21 = n. We assume that a C Vi and b C V2. Then there

292 M. Sutton et al./D&crete Mathematics 195 (1999) 287-293

must exist a vertex a + b E V. Without loss of generality, we suppose a + b E 111. Then

there must exist a vertex a + 2b E V. This leads to the following two cases:

Case 1. a + 2b E V2. Then V must contain vertices 2a + 2b and 2a + 3b. Clearly

2a + 2b ~ V2, because there is a vertex 2a + 3b such that 2a + 2b is adjacent to b,

which is impossible as 2a + 2b and b are in the same partite set I"2.

Hence 2 a + 2 b E Vi. Then there must exist vertices 2 a + 3 b and 3 a + 4 b in V. Again,

clearly 2a + 3b ~ V2, because this gives a vertex 3a ÷ 3b such that 2a + 2b is adjacent

to a + b, which is impossible as 2a + 2b and a + b are in the same partite set Vi.

Thus 2a + 3b E I11. But this is impossible, since V also contains vertex 3a + 4b which

implies that 2a + 3b and a + b are adjacent.

Case 2. a + 2b E II1. Then V must contain a vertex a ÷ 3b. Clearly a + 3b ~V2,

because otherwise there would be a vertex 2a + 3b which leads to two vertices in the

same partite set Vl, i.e., a + b and a + 2b are adjacent. Thus a + 3b E Vl and it follows

that there is a vertex a + 4b which must be in 111. This implies that all a + ib, for

i -- 0, 1 . . . . . n - 1, must be in Vl, and so V contains a vertex a + nb which must be also

in Vl.

Since [Vi i=n, a ÷ n b = a ÷ ibmodm, for some i E { 0 , 1 . . . . . n - l } . Thus m - -

(a + n b ) - ( a + i b ) = j b , for some j E {1,2 . . . . . n}. It is obvious that m =nb.

Now, we consider the vertices in II2 other than b. Let c E 112. Then V must contain

vertices a ÷ c + kb, for all k = 0, 1 . . . . . n - 1. Clearly they cannot be in II2 (otherwise b

will be adjacent to a + c + kb, for some k E {0, 1 . . . . . n - 1 }). Hence a + c + kb must be

in Vt, so that a + c + k b = a + i b , i.e., c = l b , for certain I E {0 ,1 ,2 ,3 . . . . . n - l } . We see

that l cannot be 0 or 1, s o / - - - 2 , 4 , 7 . . . . . ( n - 2 ) + ( n - 1 ) + l . But ( n - 2 ) + ( n - 1 ) + l > n,

for n >~ 3, which is impossible since modm =nb. We conclude that K, , , , for n >/3, is

not a mod sum graph. []

Since complete graphs K, , n /> 2, wheels IV,, n ~ 4, and symmetric complete bipartite

graphs K, , , , n /> 3, are not MSGs, we have the following open problems.

Problem 15. What is the mod sum number o f K, , n >~ 2?

Problem 16. What is the mod sum number o f W~, n ~ 4?

Problem 17. What is the mod sum number o f K, , , , n ~> 3?

References

[1] D. Bergstrand, F. Harary, K. Hodges, G. Jennings, L. Kuklinski, J. Wiener, The sum number of a complete graph, Bull. Malaysian Math. Soc. 12 (1989) 25-28.

[2] J. Bolland, R. Laskar, C. Turner, G. Domke, On mod sum graphs, Congr. Numer. 70 (1990) 131-135. [3] Mark N. Ellingham, Sum graphs from trees, Ars Combin. 35 (1993) 335-349. [4] J. Ghoshal, R. Laskar, D. Pillone, G. Fricke, Further results on mod sum graphs, Congr. Numer. 101

(1994) 201-207.

M. Sutton et al./Discrete Mathematics 195 (1999) 287-293 293

[5] R. Gould, V. R6dl, Bounds on the number of isolated vertices in sum graphs, in: Y. Alevi, G. Chartrand, O.R. Oellermann, A.J. Schwenk (Eds.), Graph Theory, Combinatorics and Applications, Wiley, New York, 1991, 553-562.

[6] F. Harary, Sum graphs and difference graphs, Congr. Numer. 72 (1990) 101-108. [7] F. Harary, Sum graphs over all the integers, Discrete Math. 124 (1994) 99-105. [8] Nora Hartsfield, W.F. Smyth, The sum number of complete bipartite graphs, in: Rolf Rees (Ed.), Graphs

and Matrices, Marcel Dekker, New York, 1992, pp. 205-211. [9] Nora Hartsfield, W.F. Smyth, A family of sparse graphs of large sum number, Discrete Math. 141

(1995) 163-171. 10] Mirka Miller, Joseph Ryan, Slamin, W.F. Smyth, Labelling wheels for minimum sum number, Austral.

J. Combin., to appear.