The (mod, integral and exclusive) sum number of

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1 Global Journal of Pure and Applied Mathematics. ISSN Volume 12, Number 1 (2016), pp Research India Publications The (mod, integral and exclusive) sum number of graph Knn, n 2 K2 1 R. K. Samal and 2 D. Mishra 1 Department of Mathematics, I. T. E. R, Siksha O Anusandhan University, Khandagiri, Bhubaneswar , Odisha, India. 2 Department of Mathematics, C. V. Raman College of Engineering & Technology, Bhubaneswar, Odisha, India. Abstract In this paper, we developed some new formulae for exclusive sum number of the complete bipartite graph K nm,. Also we show that for K E n 2 K, n 6 2n 4, 2n 6, n 3and 2n 4. Keywords: (mod, integral, exclusive) sum graph; Complete Bipartite graph K E nk K E n 2 K. K ; Graph and Graph nm, Introduction All graphs we considered in this paper are finite, simple, undirected graphs. We follow in general the graph-theoretic notation and terminology of Ref. [1] unless otherwise specified. The notion of a sum graph was introduced by Harary in 1990 [2]. In 1994 Harary introduced the notion of an integral sum graph [3]. Let N(Z) be the set of all positive integers (integers). The sum graph G + (S) of a finite subsets N(Z) is the graph (S, E) with uv E if and only if u + v S. A graph G is said to be an (integral) sum graph if it is isomorphic to the sum graph of some N(Z). The (integral) sum number σ(g) ( (G)) of G is the smallest number of isolated vertices which when added to G result in an (integral) sum graph. If G has a (integral) sum labeling then G is called a (integral) sum graph. The notion of an exclusive sum labeling is introduced in [4]. A mapping L is called a sum labeling of a graph H(V(H), E(H)) if it is an injection from V(H)to a set of positive integers, such that xy E(H) if and only if there exists a vertex w V(H)

2 298 R. K. Samal and D. Mishra such that L(w) = L(x) + L(y). In this case, w is called a working vertex. We define L as an exclusive sum labeling of G if it is sum labeling of G rk 1 for some nonnegative integer r and G contain no working vertex. In general a graph G will require some isolated vertices to be labeled exclusively. The least numbers of isolated vertices is called exclusive sum number of G,denoted by (G). An exclusive sum labeling of a graph G is said to be optimum if it labels G exclusively by using (G) isolated vertices. In case (G) = (G), where (G) denotes the maximum degree of vertices in G, the labeling is called optimum exclusive sum labeling. Mod sum graph was introduced by Bolland, Laskar, Turner and Domke in 1990, as a generalization of sum labeling [5]. A graph G = (V, E) is a mod sum graph if there exists a positive integer z and a labeling L of vertices of G with distinct elements from {1,2, z 1} so that uv E(G) if and only if the sum, modulo z, of the labels assigned to u and v is the label of a vertex of G. The mod sum number ρ(g)of a connected graph G is the smallest nonnegative r such that G rk 1 is a mod sum graph. Any sum graph can be considered as a mod sum graph by choosing a sufficiently large modulus z. The converse is not true. Unlike in the case of sum graphs, there exist mod sum graphs that are connected. New formula on Exclusive sum labeling of K n,m Let V Knm, A, Bbe the bipartition of Knm, and A a1, a2,... a n B b1, b2,... n n, C V K 1 be the isolated vertices, S V Knm, K1. Let K nm, Snm, max n, m for n 2 and 2, S n m. Since K 2n 3 n for n 2, 3 for 4 Theorem 2. 1 (K n,m ) = n + m 1, for n 2 and m 2. C n for m 3, n. P n n, 2 Proof. We consider the following labeling of the graph (K n,m ) (n + m 1)K 1 : a i = (i 1)N + 1, i = 1,2,, n, b j = (j 1)N + 3, j = 1,2,, m, c k = (k 1)N + 4, k = 1,2,, n + m 1, where N 8 is an integer. It is obvious that the vertex is distinct for each vertex of (K n,m ) (n + m 1)K 1. Let A a, a,..., a, B b, b,..., b, 1 2 n 1 2 m C c1, c2,..., cnm 1, S A B C. In fact it is easy to see that ai, bj, ck strict increase, the remainder of dividing N is 1, (3, 4). Thus, we verify that the following assertions are true. a i + a j S for any a i, a j A(i j). b i + b j S for any b i, b j B(i j). c i + c j S for any c i, c j C(i j). a i + c j S for any a i A and any c j C.

3 The (mod, integral and exclusive) sum number of graph 299 b i + c j S for any b i B and any c j C. a i + b j S for any a i A and any b j B. Thus the above labeling is an exclusive sum labeling of (K n,m ) (n + m 1)K 1 for n 2 and m 2. Exclusive sum labeling of K n,n E(nK 2 ) Let V (K n,n E(nK 2 )) = (A, B) be the bipartition of K n,n E(nK 2 ), and A = {a 1, a 2,, a n }, B = {b 1, b 2,, b n }, {a 1 b 1, a 2 b 2,, a n b n } = E(nK 2 ), C = V( K 1 ) be the isolated vertices and S = V ((K n,n E(nK 2 )) K 1 ). Let = (K n,n E(nK 2 )), for n 6, we will give some properties of the exclusive sum graph (K n,n E(nK 2 )), for n 6. It is clear that K n,n E(nK 2 ) is two independent edges for n = 2 and K n,n E(nK 2 ) is a 6-cycle for n = 3. In this paper, we only consider the case of n 6. Theorem 3. 1 (K n,n E(nK 2 )) = 2n 3, for n 6. Proof. We consider the following labeling of the graph (K n,n (nk 2 )) (2n 3)K 1 : a i = (i 1)N + 1, i = 1,2,, n, b j = (j 1)N + 3, j = 1,2,, n, c k = (k 1)N + 4, k = 1,2,, n 2, c k = kn + 4, k = n 1,,2n 3, where N 8 is an integer. Let V (K n,n E(nK 2 )) = (A, B) be the bipartition of K n,n E(nK 2 ), and A = {a 1, a 2,, a n }, B = {b 1, b 2,, b n }C = V((2n 3)K 1 ) = {c 1, c 2,, c 2n 3 }, S = V ((K n,n E(nK 2 )) (2n 3)K 1 ) = A B C. It is easy to verify that the following assertions are true. S Z m \{0}. a i + a j S for any a i, a j A(i j). b i + b j S for any b i, b j B(i j). c i + c j S for any c i, c j C(i j). a i + c j S for any a i A and any c j C. b i + c j S for any b i B and any c j C. a i + b j S if and only if i + j = n or i = j = n. So a 1 b n 1, a 2 b n 2,, a n 1 b 1, a n b n is E(nK 2 ).

4 300 R. K. Samal and D. Mishra Thus the above labeling is an exclusive sum labeling of (K n,n E(nK 2 )) (2n 3)K 1 for n 6. The sum number and integral sum number of K n,n E((n + 2)K 2 ) Let V (K n,n E((n + 2)K 2 )) = (A, B) be the bipartition of K n,n E((n + 2)K 2 ), and A = {a 1, a 2,, a n }, B = {b 1, b 2,, b n }, {a 1 b 1, a 2 b 2,, a n b n, a n+1 b n+1, a n+2 b n+2 } = E((n + 2)K 2 ), C = V(σK 1 ) be the isolated vertices, S = V ((K n,n E((n + 2)K 2 )) σk 1 ). Let σ = σ (K n,n E((n + 2)K 2 )), for n 6,we will give some properties of the (integral) sum graph (K n,n E((n + 2)K 2 )), for n 6. It is clear that K n,n E((n + 2)K 2 ) is four independent edges for n = 2 and K n,n E((n + 2)K 2 ) is a 10- cycle for n = 3. Since σ (K 2,2 E(2K 2 )) = 1, (K 2,2 E(2K 2 )) = ρ (K 2,2 E(2K 2 )) = 0 and σ(c 6 ) = 2 and ρ(c 6 ) = 0. In this paper, we only consider the case of n 6. Lemma 4. 1 In a (integral) sum labeling of (K n,n E((n + 2)K 2 )) sk 1 (s r) if there exists a p A and b q B(q p) such that a p + b q B then a i + b q B for any a i A(i q). Proof. Since a p + b q B,we may assume without loss of generality that a p + b q = b r. It is obvious that a i + b q S for any a i A and i q. Since a p + b q = b r Banda i A, we have that a i + b r = a p + (a i + b q ) S for i r, q. Thus we can obtain that a i + b q B {a p } for i r, q. We will prove that for a i + b q A for any a i A and i q. If a i + b q A we may assume without loss of generality thata i + b q = a s. Then a s + a j = a i + (a j + b q )S for j s, q. So a j + b q A {b i } Cforj s, q. Hence we have that a j + b q (A {b i } C) (B {a p }) = {a p, b i } for j r, s, q. So n 5, contradicting the fact that n 6. Therefore, a i + b q A for any a i A and i q. Thus we have that a r + b q B C, a i + b q B for i r, q. So we only need to prove thata r + b q C. If a r + b q C then a i + (a r + b q ) = a r + (a i + b q ) S for i q. So a i + b q (A {b r } C) fori q. Hence we have that a i + b q = b r for j r, q. So n 3, contracting the fact that n 6. From the above we have that the lemma holds. In a similar way we have the following lemma. Lemma In a (integral) sum labeling of (K n,n E((n + 2)K 2 )) sk 1 (s r) if there exists a p A and b q B(q p) such that a p + b q A then a p + b i A for any b i B(i p).

5 The (mod, integral and exclusive) sum number of graph 301 Lemma In a (integral) sum labeling of (K n,n E((n + 2)K 2 )) sk 1 (s r) if there exists a p A and b q B(q p) such that a p + b q B then a p + b i B C for any b i B(i p). Proof. By contradiction, If there exists r p such that a p + b r A then we have that a p + b i A for any b i B and i p by Lemma So a p + b q A, contradicting the fact that a p + b q B. Lemma 4. 4 K E n 2 K 2n 4 and for n 6. K E n 2 K 2n 6 Proof. If a i + b j C for any a i A and any b j B (j i) we assume that b b b and a i is the least integer of A {a n }. Then we have that n n n n n K E n 2 K K E n 2 K 2n 4 by the distinctness of p, 2, 2 ai b1, ai b2,... ai bi 1, ai bi 1,..., ai bn, a1 bn,..., ai 1 bn,..., ai 1 bn,..., an2 bn. If there exists a p A and b q B such that a p + b q A B.We assume without loss of generality that a p + b q B. Then we have that a i + b q B for any a i A(i q), a b B C for any b Bi p and B { bq, a1 bq,..., aq 1 bq, aq 1 bq, an 1 bq} i i. Suppose that a 1 < a 2 < < a n and b t is the largest integer of A {b q }.Then b t = a n + b q. Firstly we will prove that a i + b j C for any a i A(i p) and for any b j B(j i, q). If there exists a r A and b s B (s r, q) such that a r + b s B then a i + b s B for any a i A(i s)and B = {b s, a 1 + b s,, a s 1 + b s, a s+1 + b s,, a n + b s }. we may assume without loss of generality that b q < b s. Then since a 1 + b q < a 1 + b s a k + b s for s k then a 1 + b q = b s. So 0 < a 1 < a 2 < < a n and b q < a 1 + b q = b s < a 1 + b s a k + b s for s k. So b q B = {b s, a 1 + b s,., a s 1 + b s, a s+1 + b s,, a n + b s }, which is a contradiction. If there exists a r A (r q) and b s B(s r, q) such that a r + b s A then a r + b i A for any b i B(i r). So a r + b q A, contracting the fact a r + b q B. So a i + b j C for any a i A(i q) and any b j B(j i, q) and a q +b j B for any b j B(j i, q). Hence if q 1 then we have that (K n,n E((n + 2)K 2 )) 2n 6 by the distinctness of {a 1 + b 2, a 1 + b 3,, a 1 + b q 1, a 1 + b q+1,, a 1 + b n, a 2 + b t,..., a n 1 + b t } { a q + b t } { a t + b t }, if q = 1 then we have that (K n,n E((n + 2)K 2 )) 2n 6 by the distinctness of {a 2 + b 3,, a 2 + b n, a 3 + b t,, a n 1 + b t } { a t + b t }. Thus, we only need to prove that σ (K n,n E((n + 2)K 2 )) 2n 4. Hence we may assume that a i, b i > 0 for any 1 i n in the following proof. If there exists b s B (s q) such that a q + b s A then a q + b i A for any b j B(i q) and A = {a q, a q +

6 302 R. K. Samal and D. Mishra b 1,, a q + b q 1, a q + b q+1,, a q + b n }. Hence B = {b q, a q + b q + b 1,, a q + b q + b q 1, a q + b q + b q+1,, a q + b q + b n }. We have that (n 1)( a q + b q ) = 0. However, a q + b q > 0 for a q, b q > 0, which is a contradiction. So a i + b j C for any b j B(j q) and any a i A(i j). Since a 1 + b 1, a 1 + b 2,, a 1 + b q 1, a 1 + b q+1,, a 1 + b n, a 2 + b t,, a n 1 + b t are distinct. Let R = { a 1 + b 1, a 1 + b 2,, a 1 + b q 1, a 1 + b q+1,, a 1 + b n, a 2 + b t,, a n 1 + b t }. Then R = {2a 1 + b q, a 1 + a 2 + b q,, a 1 + a q 1 + b q, a 1 + b q+1 + b q,, a 1 + a n + b q, a 2 + a n + b q,, a n 1 + a n + b q, } where 2a 1 + b q < a 1 + a 2 + b q < < a 1 + a q 1 + b q < a 1 + b q+1 + b q, < a 1 + a n + b q < a 2 + a n + b q < < a n 1 + a n + b q and R { a 1 + b 1 } { a t + b t } C. Since a 1 + b q b q we have that a q + ( a 1 + b q ) = a 1 + a q + b q C. By a 1 + a q 1 + b q < a 1 + a q + b q < a 1 + a q+1 + b q we can obtain that a 1 + a q + b q R. So C R { a 1 + b 1 } { a t + b t } + 1 2n 4. Lemma σ (K n,n E((n + 2)K 2 )) 2n 4 for n 6. Proof. We consider the following labeling of the graph (K n,n E((n + 2)K 2 )) (2n 4)K 1 : a i = (i 1)N + 1, i = 1,2,, n, b j = (j 1)N + 3, j = 1,2,, n, c k = (k 1)N + 4, k = 1,2,, n 2, c k = kn + 4, k = n 1, n,,2n 4, where N 8 is an integer. Let V (K n,n E((n + 2)K 2 )) = (A, B) be the bipartition of K n,n E((n + 2)K 2 ), and A = {a 1, a 2,, a n }, B = {b 1, b 2,, b n }C = V((2n 4)K 1 ) = {c 1, c 2,, c 2n 4 }, S = V ((K n,n E((n + 2)K 2 )) (2n 4)K 1 ) = A B C. It is easy to verify that the following assertions are true. S Z m \{0}. a i + a j S for any a i, a j A(i j). b i + b j S for any b i, b j B(i j). c i + c j S for any c i, c j C(i j). a i + c j S for any a i A and any c j C. b i + c j S for any b i B and any c j C. a i + b j S if and only if i + j = n or 2 n 1 or i = j = n. So a 1 b n 1, a 2 b n 2,, a n 1 b 1, a n b n 1, a n 1 b n, a n b n is E((n + 2)K 2 ).

7 The (mod, integral and exclusive) sum number of graph 303 Thus the above labeling is a sum labeling of (K n,n E(n + 2K 2 )) (2n 4)K 1 for n 6. Lemma (K n,n E((n + 2)K 2 )) 2n 6 for n 6. Proof. We consider the following labeling of the graph(k n,n E((n + 2)K 2 )) (2n 6)K 1 : a i = (i 1)N + 7, i = 1,2,, n 1, a n = 3, b j = (j 1)N + 4, j = 1,2,, n 1, b n = 3, c k = (k 1)N + 11, k = 1,2,, n 3, c k = kn + 11, k = n 2, n 1,,2n 6, where N 30 is an integer. Let V (K n,n E((n + 2)K 2 )) = (A, B) be the bipartition of K n,n E((n + 2)K 2 ), and A = {a 1, a 2,, a n }, B = {b 1, b 2,, b n }C = V((2n 6)K 1 ) = {c 1, c 2,, c 2n 6 }, S = V ((K n,n E((n + 2)K 2 )) (2n 6)K 1 ) = A B C. It is easy to verify that the following assertions are true. S Z m \{0}. a i + a j S for any a i, a j A(i j). b i + b j S for any b i, b j B(i j). c i + c j S for any c i, c j C(i j). a i + c j S for any a i A and any c j C. b i + c j S for any b i B and any c j C. a i + b j S if and only if i + j = n 1 or 2n 3 or n 1 i = j n. E n 2 K. a 1 b n 2, a 2 b n 3,, a n 2 b 1, a n 2 b n 1 a n 1 b n 2,, a n 1 b n 1, a n b n is Thus the above labeling is a sum labeling of (K n,n E((n + 2)K 2 )) (2n 6)K 1 for n 6. We have the following theorem by Lemmas Theorem 4.1. K E n 2 K 2n 6 and for n 6. The mod sum number of K n,n E((n + 2)K 2 ) K E n 2 K 2n 4 Let ρ = ρ (K n,n E((n + 2)K 2 )), for n 6,we will give some properties of the mod sum graph (K n,n E((n + 2)K 2 )) ρk 1, for n 6. Let V (K n,n E((n + 2)K 2 )) = (A, B) be the bipartition of K n,n E((n + 2)K 2 ) and A = {a 1, a 2,, a n }, B = {b 1, b 2,, b n }, {a 1 b 1, a 2 b 2,, a n+2 b n+2 } = E((n + 2)K 2 ), C = 2

8 304 R. K. Samal and D. Mishra V(ρK 1 ) be the isolated vertices, S = V ((K n,n E((n + 2)K 2 )) ρk 1 ) and the modulus be m. Lemmas have been established for (integral) sum graph labeling of K n,n E((n + 2)K 2 ) in section 4. Lemma If there exists a p A and b q B(q p) such that a p + b q B then a i + b q B for any a i A(i q). Lemma If there exists a p A and b q B(q p) such that a p + b q A then a p + b i A for any b i B(i p). Lemma If there exists a p A and b q B(q p) such that a p + b q B then a p + b i B C for any b i B(i p). Lemma K n,n E((n + 2)K 2 )is not a mod sum graph for n 6. Proof. By contradiction. If K n,n E((n + 2)K 2 ) is a mod sum graph then C = and a i + b j A Bfor any a i A and any b j B(i j). We may assume without loss of generality that there exist a p A and b q B(q p) such thata p + b q B. Then a i + b q B for any a i A(i q) by Lemma Hence a i + b j B for any a i A(i q) and any b j B(i j) by Lemma if there exists b r B(r q)such that a q + b r A then a q + b k A for any b k Band k q. Thus A = {a q, a q + b 1,, a q + b q 1, a q + b q+1,, a q + b n }. So there exists an integer l i q such that a i = a q + b li for any i q. Since a i + a k = (a k + b li ) + a q S for any k i, l i we have that a k + b li A {b q } C, k i, l i. From the above, we know that a k + b li {b q }, k q, i, l i. So n 4, contradicting the fact that n 6. Hence a i + b j B for any a i A and any b j B(j i). Let R = {a i + b 1,, a i + b i 1, a i + b i+1,, a i + b n } B. Since B\R = 1 we may assume without loss of generality that B\R = {b i }. Thus B = {b t, a i + b 1,, a i + b i 1, a i + b i+1,, a i + b n }. Hence b t + (n 1)a i = b i. If ka i 0 for any k(2 k n 1) then B = {b t, b t + a i,, b t + 2a i, b t + (n 1)a i } = {a 1 + b t,, a t 1 + b t, a t+1 + b t,, a n + b t, b t }. Hence {a 1,, a t 1, a t+1. a n } = {a i, 2a i, (n 1)a i }. Since n 6 we have that a i, 2a i, 3a i are distinct vertices of A and a i + 2a i = 3a i. So a i is adjacent to 2a i, which is a contradiction. If there exists a least integer l i (2 l i n 1) such that l i a i = 0. Suppose that there exist integers s i and r i (0 r i l i ) such that 1 = s i l i + r i. We have that elements of B can be partitioned into {b t, b t + a i,, b t + r i a i } and s i ( 1) sets of equal size l i such that the elements of each set from an l i cycle under addition of a i. Assume that one of the l i cycles is {b k, b k + a i,, b k + (l i 1)a i }. Since B = {a 1 + b k,, a k 1 + b k, a k+1 + b k,, a n + b k, b k } we have that {a i, 2a i, (l i 1)a i } A. If l i 4 then a i, 2a i, 3a i are distinct vertices of A and a i + 2a i = 3a i. So a i is adjacent to 2a i, which is a contradiction. Hence 2 l i 3. If l i = 2 then a i =

9 The (mod, integral and exclusive) sum number of graph 305 m/2. If l i = 3 then a i = m/3 or 2m/3. So n = A 3, contradicting the fact that n 6. Lemma K E n 2 K n 3 for n 6. Proof. By Lemma 5. 4 we have that there exists a p A and b q B such that a p + b q C. If a i + b q C for any a i A (i q) then the lemma holds by the distinctness of a 1 + b q,, a q 1 + b q, a q+1 + b q,, a n + b q. If there exists a r A (r p, q) such that a r + b q C then a r + b q A B. If a r + b q B then by Lemma 5. 1 we have that a i + b q B for any a i A (i q), contradicting the fact that a p + b q C. If a r + b q A then a r + b j A for any b j B (j r) by Lemma Hence a i + b j B for any b j B (j r) and any a i A (i j)( If there exists a s A(s j) such that a s + b j B then we have that a k + b j B for any a k A(k j), contradicting the fact that a r + b j A. ). So a i + b j A C for any a i A and any b j B (j r, i). If there exists a p + b t A then a p + b j A for any b j B (j p) by Lemma 5. 2, contradicting the fact that a p + b q C. Hence a p + b j C for any b j B (j r, p). So { a p + b 1,, a p + b p 1, a p + b p+1,, a p + b n 1 } { a p + b r } C. The lemma holds. Theorem K E n 2 K n 3 for n 6. Proof. By Lemma 5. 5 we only need to prove that ρ (K n,n E((n + 2)K 2 )) n 3. nn 2 3 Mod sum labeling of the graph K E n K n K are as follows., 2 1 a i = (i 1)N + 7, i = 1,2,, n 1, a n = 3, b j = (j 1)N + 4, j = 1,2,, n 1, b n = m 3, c k = (k 1)N + 11, k = 1,3,4,, n 3, And take the modulus m = (n 1)N, Where N 30 is an integer. Let V (K n,n E((n + 2)K 2 )) = (A, B) be the bipartition of K n,n E((n + 2)K 2 ), and A = {a 1, a 2,, a n }, B = {b 1, b 2,, b n } C = V((n 3)K 1 ) = {c 1, c 2,, c n 3 }, S = V ((K n,n E((n + 2)K 2 )) (n 3)K 1 ) = A B C. It is easy to verify that the following assertions are true. S Z m \{0}. a i + a j S for any a i, a j A(i j). b i + b j S for any b i, b j B(i j). c i + c j S for any c i, c j C(i j). a i + c j S for any a i A and any c j C. b i + c j S for any b i B and any c j C. a i + b j S if and only if i + j = 3 or n or i = j = n.

10 306 R. K. Samal and D. Mishra So a 1 b 2, a 2 b 1, a 1 b n 1, a 2 b n 2,, a n 1 b 1, a n b n is E((n + 2)K 2 ). Thus the above labeling is a sum labeling of (K n,n E((n + 2)K 2 )) (n 3)K 1 for n 6. Exclusive sum labeling of K n,n E((n + 2)K 2 ) Let V (K n,n E((n + 2)K 2 )) = (A, B) be the bipartition of K n,n E((n + 2)K 2 ), and A = {a 1, a 2,, a n }, B = {b 1, b 2,, b n }, {a 1 b 1, a 2 b 2,, a n b n, a n+1 b n+1, a n+2 b n+2 } = E((n + 2)K 2 ), C = V( K 1 ) be the isolated vertices, S = V ((K n,n E((n + 2)K 2 )) K 1 ). Let = (K n,n E((n + 2)K 2 )), for n 6,we will give some properties of an exclusive sum graph (K n,n E((n + 2)K 2 )), for n 6. Theorem 6. 1 K E n 2 K 2n 4 for n 6. Proof. We consider the following labeling of the graph (K n,n ((n + 2)K 2 )) (2n 4)K 1 : a i = (i 1)N + 1, i = 1,2,, n, b j = (j 1)N + 3, j = 1,2,, n, c k = (k 1)N + 4, k = 1,2,, n 2, c k = kn + 4, k = n 1,,2n 4, where N 8 is an integer. Let V (K n,n E((n + 2)K 2 )) = (A, B) be the bipartition of K n,n E((n + 2)K 2 ), and A = {a 1, a 2,, a n }, B = {b 1, b 2,, b n }C = V((2n 4)K 1 ) = {c 1, c 2,, c 2n 4 }, S = V ((K n,n E((n + 2)K 2 )) (2n 4)K 1 ) = A B C. It is easy to verify that the following assertions are true. S Z m \{0}. a i + a j S for any a i, a j A(i j). b i + b j S for any b i, b j B(i j). c i + c j S for any c i, c j C(i j). a i + c j S for any a i A and any c j C. b i + c j S for any b i B and any c j C. a i + b j S if and only if i + j = n or 2n 1 or i = j = n. So a 1 b n 1, a 2 b n 2,, a n 1 b 1, a n b n 1, a n 1 b n, a n b n is E((n + 2)K 2 ). Thus the above labeling is an exclusive sum labeling of (K n,n E((n + 2)K 2 )) (2n 4)K 1 for n 6.

11 The (mod, integral and exclusive) sum number of graph 307 References [1] J. A. Bondy, U. S. R. Murty. Graph Theory with application [M]. American Elsevier Publishing Co., Inc., New York, [2] F. Harary, Sum graphs and difference graphs, Congressus Numerantium 72 (1990) [3] F. Harary, Sum graphs over all integers, Discrete Mathematics124(1994) [4] M. miller, J. Ryan, Slamin, K. Sugeng, M. Tuga, Exclusive sum labeling of Graphs, preprint, [5] J. Bolland, R. Laskar, C. Turner, G. Domke, (1990) On mod sum graphs Congressus Numerantium 70; [6] F. Fernau, J. F. Ryan, K. A. Sugeng (2009) A sum labeling for the generalized friendship graph, Discrete Math 308: [7] J. Wu, J. Mao, D. Li, New types of integral sum graphs, Discrete Mathematics 260, (2003)

12 308 R. K. Samal and D. Mishra

13 The (mod, integral and exclusive) sum number of graph 309