SUPER MEAN NUMBER OF A GRAPH

Transcription

1 Kragujevac Journal of Mathematics Volume Number (0), Pages 9 0. SUPER MEAN NUMBER OF A GRAPH A. NAGARAJAN, R. VASUKI, AND S. AROCKIARAJ Abstract. Let G be a graph and let f : V (G) {,,..., n} be a function such that the label of the edge uv is f(u)+f(v) or f(u)+f(v)+ according as f(u) + f(v) is even or odd and f(v (G)) {f (e) : e E(G)} {,,..., n}. If n is the smallest positive integer satisfying these conditions together with the condition that all the vertex and edge labels are distinct and there is no common vertex and edge labels, then n is called the super mean number of a graph G and it is denoted by S m (G). In this paper, we find the bounds for super mean number of some standard graphs.. Introduction Throughout this paper, by a graph we mean a finite, undirected graph (simple graph) with p vertices. For notation and terminology, we follow []. Path on n vertices is denoted by P n and a cycle on n vertices is denoted by C n. K,m is called a star and it is denoted by S m. The union of m disjoint copies of a graph G is denoted by mg. The bistar B m,n is the graph obtained from K by identifying the central vertices of K,m and K,n with the end vertices of K respectively. C m, K,n is the graph obtained from C m and K,n by identifying any one of the vertices of C m with the central vertex of K,n. C m K,n is the graph obtained from C m and K,n by identifying any one of the vertices of C m with a pendant vertex of K,n (that is a non-central vertex of K,n ). The concept of super mean labeling was introduced by D. Ramya et al. []. They have studied in [,, ], the super mean labeling of some standard graphs. Further, some more results on super mean graphs are discussed in [, ]. Let V (G) and E(G) be the vertex set and edge set of a graph G, respectively, and V (G) = p, E(G) = q (the order and the size of G, respectively). Let f : Key words and phrases. Labeling, super mean graph, super mean number. 00 Mathematics Subject Classification. 0C. Received: December 0, 00. Revised: April, 0. 9

2 9 A. NAGARAJAN, R. VASUKI, AND S. AROCKIARAJ V (G) {,,..., p + q} be injective. For a vertex labeling f, the induced edge labeling f (e = uv) is defined by { f(u)+f(v) if f(u) + f(v) is even, f (e) = f(u)+f(v)+ if f(u) + f(v) is odd. Then, f is called super mean labeling if f(v (G)) {f (e) : e E(G)} = {,,,..., p+ q}. A graph that admits a super mean labeling is called a super mean graph. A super mean labeling of the graph K, is shown in Figure. 9 0 Figure. The concept of mean number of a graph was introduced by M. Sundaram and R. Ponraj [] and they have found the mean number of some standard graphs. Motivated by these work, we introduce the concept of super mean number of a graph. Let f : V (G) {,,..., n} be a function such that the label of the edge uv is f(u)+f(v) or f(u)+f(v)+ according as f(u) + f(v) is even or odd and f(v (G)) {f (e) : e E(G)} {,,..., n}. If n is the smallest positive integer satisfying these conditions together with the condition that all the vertex and edge labels are distinct and there is no common vertex and edge labels, then n is called the super mean number of a graph G and it is denoted by S m (G). For example, S m (K, ) = 0 is shown in the following Figure. 0 Figure. It is observed that S m (G) p + q, where p is the order and q is the size of the graph G. Clearly, the equality holds for a super mean graph. In this paper, we prove that S m (G) p for any graph G. Also, we find an upper bound of the super mean number of the graphs K,n, n, tk,n for n, t >, B(p, n) for p > n+, n, C m, K,n for n, m and C m K,n

3 SUPER MEAN NUMBER OF A GRAPH 9 for n, m. Further, we obtain the super mean number of the graphs K p, p, K,n for n, tk,, t >, B(p, n) for p = n, n + and any cycle C n. We use the following results in the subsequent theorems. Theorem.. [] A complete graph K n is a super mean graph if n. Theorem.. [] K n is not a super mean graph if n >. Theorem.. [] The star K,n is a super mean graph for n. Theorem.. [] K,n is not a super mean graph for n >. Theorem.. [] nk,, n >, is a super mean graph. Theorem.. [] The bistar, B m,n is a super mean graph for m = n or m = n +. Theorem.. [] If G is a super mean graph then mg is also a super mean graph. Theorem.. [] C n+, n, is a super mean graph. Theorem.9. [] C is not a super mean graph. Theorem.0. [] C n is a super mean graph for n. Based on the above theorems, we observe the following: Observation.. S m (K p ) = p(p+) if p and S m (K p ) p(p+) + if p >. A labeling of K in Figure shows that the above bound is attained for p = and S m (K ) =. 0 9 Figure. Observation.. S m (K,n ) = n + for n and S m (K,n ) n + (n > ). Observation.. S m (tk, ) = 9t for t >. Observation.. S m (B(p, n)) = n + when p = n and S m (B(p, n)) = n + when p = n +. Observation.. S m (C n+ ) = n + for n and S m (C n ) = n for n. Observation.. S m (C ) = 9, since C is not a super mean graph and from a labeling of C in Figure.

4 9 A. NAGARAJAN, R. VASUKI, AND S. AROCKIARAJ 9 Figure. Observation.. For any super mean graph G, S m (tg) = t(p + q), t > where p is the order and q is the size of the graph G.. Super mean number of some standard graphs The existence of the super mean number for any graph G is guaranteed by the following theorem. Theorem.. S m (G) p. Proof. It is enough if we prove that S m (K p ) p. Let v, v,..., v p be the vertices of K p. Define f : V (K p ) {,,..., n} by f(v i ) = i for i p and f(v p ) = p. Clearly all the vertex labels are distinct. Let us prove the same for edge labels. For i, j, s, t p, we consider the following two cases. Suppose f (v i v j ) = f (v s v t ); then i + j = s + t. This implies i + j = s + t. Case(i) Assume that the edges v i v j and v s v t have one vertex in common. Take i = s and j t. Since j t, we have j t then i + j s + t. Hence if two edges have one vertex in common their edge values are distinct. Case(ii) Assume that the edges v i v j and v s v t have no vertex in common. Then i s, i t, j s and j t. Suppose f (v i v j ) = f (v s v t ). Without loss of generality, assume that i is the smallest integer. Let j = i + k, s = i + k, t = i + k, k, k, k > 0. Then i + j = s + t implies i + i+k = i+k + i+k. Then i (+ k ) = i ( k + k ), which gives + k = k + k. This is a contradiction. Case(iii) Suppose one of the vertices is v p. Subcase(i) Assume that v i v j and v s v t have the common vertex v p. Without loss of generality, assume that i = s = p and j t. Suppose f (v p v j ) = f (v p v t ). This implies p + j + = p + t +, that is p + j = p + t. Then j = t which gives j = t. This is a contradiction. Subcase(ii) Suppose v i v j and v s v t have a common vertex other than v p.

5 SUPER MEAN NUMBER OF A GRAPH 9 For i, s, t p, take i = s and j = p. Then f (v i v p ) = f (v s v t ) implies i + p + = i + t. Then we have i + p = i + t, that is p = t which gives p = t. This is a contradiction. Subcase(iii) Suppose the edges v i v j and v s v t have no vertex in common. For i, s, t p and j = p f (v i v p ) = f (v s v t ) implies i + p + = s + t i.e. i + p = s + t. Then, we get a contradiction as in Case (ii). Let us prove now that for any v V (G) and e E(G), f(v) f (e). Let us take any edge v i v j in K p, i < j, where i and j p. Now, f (v i v j ) = i + j = i + j = i + j = i ( + j i ) k for any k. Also f (v i v j ) p. This implies that f (v i v j ) / f(v (G)). If j p and i =, f (v i v j ) = +j = j / f(v (G)). If i and j = p, f (v i v p ) = (i + p )+ = i + p = i + p k for any k. Also f (v i v p ) p. This implies that f (v i v p ) / f(v (G)). If i =, j = p, f (v v p ) = (+p )+ = p / f(v (G)). Thus, all the vertex and edge labels are distinct and no vertex and edge labels are equal. Hence S m (K p ) p. Theorem.. S m (K,n ) = n + for n =,,. Proof. Let V (K,n ) = {v, v, v,..., v n } and E(K,n ) = {vv i ; i n}. Define f on V (K, ) and V (K, ) as follows: f(v) =, f(v i ) = i, i, f(v i ) = +(i ), i n and f(v n ) = n+, for n =,. The vertex labeling f on V (K, ) is defined by f(v) =, f(v i ) = i, i, f(v ) =, f(v ) =, f(v ) = and f(v ) =. Clearly, the vertex labels and the induced edge labels are distinct. Hence, S m (K,n ) n + for n =,,. Then by Observation., the result follows. Theorem.. S m (K,n ) n 0, n. Proof. Let V (K,n ) = {v, v, v,..., v n } and E(K,n ) = {vv i ; i n}. Define f on V (K,n ) as follows: f(v) =, f(v i ) = i, i, f(v ) =, f(v i ) = i +, i, f(v i ) = + (i ), i n, f(v n ) = n 0. Clearly, the vertex labels and the edge labels are distinct and no vertex and edge labels are equal. Hence, S m (K,n ) n 0 for n. Theorem.. S m (tk,n ) (n + )t + for n =, and t >. Proof. Let v 0j, v ij, j t, i n be the vertices and v 0j v ij, j t, i n be the edges of tk,n. Define f on V (tk,n ), n =, as follows: When t = and n =, define f(v 0 ) =, f(v i ) = i, i, f(v i ) = + (i ), i, f(v ) =.

6 9 A. NAGARAJAN, R. VASUKI, AND S. AROCKIARAJ When t = and n =, define f(v 0 ) =, f(v i ) = i, i, f(v ) =, f(v ) =, f(v ) = and f(v ) =. For t >, label the vertices of tk, and tk, as follows : f(v 0j ) = f(v 0 ) + (n + )(j ), j t, f(v ) = f(v ) + n, f(v j ) = f(v ) + (n + )(j ), j t and f(v ij ) = f(v i ) + (n + )(j ), j t, i n. Clearly, the vertex labels are distinct. Also, the vertex labeling f induces distinct edge labels and f(e(g)) {,,,..., (n + )t + } f(v (G)). Hence S m (tk,n ) (n + )t +. According to Theorem., in the following Figure, the labeling of K, shows that S m (K, ) Figure. K, Theorem.. When t is an even integer, S m (tk,n ) t(n + ) for n >. Proof. Let v 0j, v ij, i n, j t be the vertices and v 0j v ij, i n, j t be the edges of tk,n. Define f on V (tk,n ) as follows: f(v 0j+ ) = (n + )j +, 0 j t, f(v 0j ) = (n + )j + j, j t, f(v ij+ ) = (n + )j + i, 0 j t, i n and f(v ij ) = (n + )(j ) + i +, j t, i n. It can be verified that all the vertex and edge labels are distinct and there is no common vertex and edge labels. Hence, S m (tk,n ) t(n + ) for n > and t is an even integer. Theorem.. When t is an odd integer, S m (tk,n ) t(n + ) + for n >. Proof. Let V (tk,n ) = {v 0j, v ij : i n, j t} and E(tK,n ) = {v 0j v ij : i n, j t}. Let t = k + for some k Z +. Define f on V (tk,n ) as follows:

7 SUPER MEAN NUMBER OF A GRAPH 99 For j k, When j = k +, f(v 0j+ ) = (n + )j +, 0 j k, f(v 0j ) = (n + )j + j, j k, f(v ij+ ) = (n + )j + i, 0 j k, i n and f(v ij ) = (n + )(j ) + i +, j k, i n. f(v 0k+ ) = + (n + )k, f(v ik+ ) = i + (n + )k, i, f(v k+ ) = + (n + )k, f(v ik+ ) = i + (n + )k +, i, f(v ik+ ) = i + (n + )k 0, i n and f(v nk+ ) = (n + )k + n. Clearly, the vertex labels and the induced edge labels are distinct and further f(v (G)) f(e(g)) =. Hence, S m (tk,n ) t(n + ) + for n > and t is an odd integer. Theorem.. S m (B(p, n)) p if p > n +. Proof. Let V (B(p, n)) = {u, v, u i, v j : i p, j n} and E(B(p, n)) = {uv, uu i, vv j : i p, j n}. Define f on V (B(p, n)) as follows: f(u) =, f(u i ) = i, i p, f(v) = p and f(v j ) = +(j ), j n. It can be verified that the vertex and edge labels are distinct and f(v (G)) f(e(g)) =. Thus, S m (B(p, n)) p, p > n +. Theorem.. C m, K,n is a super mean graph for n and m. Proof. Let V ( C m, K,n ) = {u, u,..., u m, v, v,..., v n, v = u } and E( C m, K,n ) = {u u, u u,..., u m u, u v i : i n}. For m =, the super mean labeling of the graphs C, K,, C, K,, C, K, and C, K, are shown in Figure. 9 0 C, K, C, K, 0 9

8 00 A. NAGARAJAN, R. VASUKI, AND S. AROCKIARAJ 9 0 C, K, C, K, 0 9 Figure. Case(i) n =,. Define f on V ( C m, K,n ), n =,, m and m as follows: Subcase(i) When m is odd. Let m = k +, k Z +. Subcase(ii) When m is even. Let m = k, k Z +. f(v i ) = i, i n, f(u ) = n +, f(u j ) = n + j, j k + and f(u k++j ) = n + k j +, j k. f(v i ) = i, i n, f(u ) = n +, f(u j ) = n + j, j k, f(u k+j ) = n + k (j ), j and f(u k++j ) = n + k j, j k. Clearly, f induces distinct edge labels and it can be verified that f induces a super mean labeling and hence C m, K,n, n =,, m and m is a super mean graph. Case (ii) n =,. Define f on V ( C m, K,n ), n =,, m and m as follows: Label the vertices of K,n, n =, as f(v i ) = i, i, f(v ) = and f(v ) = in the case of n =. Label the vertices of C m as follows: Subcase(i) When m is odd.

9 SUPER MEAN NUMBER OF A GRAPH 0 Let m = k + for some k Z +. f(u ) =, f(u j ) = n + j, j k, f(u k+j ) = n + k j +, j k and f(u k+ ) = n +. Subcase(ii) When m is even. Let m = k for some k Z +. f(u ) =, f(u j ) = n + i, j k, f(u k +j ) = n + k (j ), j, f(u k++j ) = n + k j, j k and f(u k ) = n +. Clearly, f induces distinct edge labels and it is easy to check that f generates a super mean labeling and hence C m, K,n, n =,, m and m is a super mean graph. Thus, C m, K,n is a super mean graph for n and m. For example, the super mean labelings of C, K, and C 9, K, are shown in the following Figure Figure. 0 Corollary.. S m ( C m, K,n ) = m + n for n and m. 9 Theorem.9. S m ( C m, K,n ) m + n for n, m and m and S m ( C, K,n ) n + for n. Proof. Let V ( C m, K,n ) = {u, u,..., u m, v, v,..., v n, v = u } and E( C m, K,n ) = {u u, u u,..., u m u, u v i : i n}. Define f on V ( C m, K,n ) as follows:

10 0 A. NAGARAJAN, R. VASUKI, AND S. AROCKIARAJ For n, label the vertices of K,n as follows: f(v ) = m +, f(v i ) = m + + (i ), i, f(v i ) = m + + (i ), i n, f(v n ) = m + n and f(v = u ) = m. Label the vertices of C m as follows: Case(i) When m is odd. Let m = k +, k Z +. The vertex labeling f is given by Case(ii) When m is even. Let m = k for some k Z +. f(u ) = m, f(u j ) = m j +, j k +, f(u k+ ) = and f(u k++j ) = + (j ), j k. f(u ) = m, f(u j ) = m j +, j k, f(u k+ ) =, f(u k++j ) = + (j ), j k and f(u k ) = m. Clearly, the vertex labels and the induced edge labels are distinct and f(e(g)) {,,,..., m + n } f(v (G)). Hence, S m ( C m, K,n ) m + n, n, m and m. Define f on V ( C, K,n ), n as follows: f(v ) =, f(v i ) = + (i ), i, f(v i ) = + (i ), i n, f(v n ) = n +, f(v = u ) = 9, f(u ) =, f(u ) = and f(u ) =. Clearly, the vertex labels and the edge labels are distinct and no vertex and edge labels are equal. Hence, S m ( C, K,n ) n + for n. Theorem.0. C m K,n is a super mean graph for n and m. Proof. Let V ( C m K,n ) = {u, u,..., u m, v = u, v,..., v n, v} and E( C m K,n ) = {u u, u u,..., u m u, u v, vv i : i n }. For m =, the super mean labelings of the graphs C, K,n, n =,,,,, are shown in Figure.

11 SUPER MEAN NUMBER OF A GRAPH C K, C K, C K, C K, C K, C K, Figure. Case (i) n =,,. Define f on V ( C m K,n ), n =,,, m and m as follows: Subcase(i) When m is odd, say m = k +, k Z +. f(v) = n, f(u = v ) = n +, f(v n+ i ) = i, i n, n =,, f(u j ) = n + j, j k + and f(u k++j ) = n + k j +, j k.

12 0 A. NAGARAJAN, R. VASUKI, AND S. AROCKIARAJ Subcase(ii) When m is even, say m = k, k Z +. f(v) = n, f(u = v ) = n +, f(v n+ i ) = i, i n, n =, f(u j ) = n + j, j k, f(u k+j ) = n + k (j ), j and f(u k++j ) = n + k j, j k. Clearly, the vertex labeling f induces distinct edge labels and it is easy to check that f is a super mean labeling. Hence, C m K,n, n =,,, m and m is a super mean graph. Case (ii) n =,,. Define f on V ( C m K,n ), n =,,, m and m as follows: Label the vertices of K,n, n =,, as given below: For n =,, f(v) =, f(v = u ) = n + and { i, i f(v n+ i ) = + (i ), i n. For n =, f(v) =, f(v = u ) =, f(v ) =, f(v ) =, f(v ) =, f(v ) = and f(v ) =. Label the vertices of C m as follows: Subcase(i) When m is odd, take m = k +, k Z +. f(u ) = n +, f(u ) = n +, f(u j ) = n + j, j k + and f(u k++j ) = n + k j +, j k. Subcase(ii) When m is even, take m = k, k Z + f(u ) = n +, f(u ) = n +, f(u j ) = n + j, j k, f(u k+j ) = n + k + (j ), j and f(u k++j ) = n + k j, j k. Clearly, the edge labels are distinct. It can be easily verified that f is a super mean labeling. Hence, C m K,n, n =,,, m and m is a super mean graph. Thus, C m K,n for n, m is a super mean graph.

13 SUPER MEAN NUMBER OF A GRAPH 0 For example, the super mean labelings of C 0 K, and C 9 K, are shown in the Figure Figure 9. Corollary.. S m ( C m K,n ) = m + n for n and m. Theorem.. S m ( C m K,n ) m + n for n, m and m and S m ( C K,n ) n for n. Proof. Let V ( C m K,n ) = {v i : i n, v, u = v, u,..., u m } and E( C m K,n = {u u, u u,..., u m u, u v, vv i, i n }. Define f on V ( C m K,n ) as follows: For n label the vertices of K,n by f(v = u ) = m, f(v ) = m +, f(v ) = m +, f(v i ) = m (i ), i, f(v i ) = m + + (i ), i n and f(v n ) = m + n. Now label the vertices of C m as follows: Case (i) When m is odd, say m = k +, k Z +. f(u ) = m, f(u j ) = m j +, j k +, f(u k+ ) = and f(u k++j ) = + (j ), j k. Case (ii) When m is even, say m = k, k Z +. f(u ) = m, f(u j ) = m j +, j k, f(u k+ ) =, f(u k++j ) = + (j ), j k and f(u k ) = m.

14 0 A. NAGARAJAN, R. VASUKI, AND S. AROCKIARAJ Clearly, the vertex labels and the induced edge labels are distinct and f(v (G)) f(e(g)) =. Hence, S m ( C m K,n ) m + n for n, m and m. Define f on V ( C K,n ), n as follows: f(v = u ) = 9, f(v ) = 0, f(v ) =, f(v i ) = 9 + (i ), i, f(v i ) = + (i ), i n, f(v n ) = n, f(u ) =, f(u ) = and f(u ) =. Clearly, the vertex labels and edge labels are distinct and f(e(g)) {,,,..., n } f(v (G)). Hence, S m ( C K,n ) n for n. Theorem.. For any graph G, if k is a super mean number of the graph G, then S m (tg) kt. Proof. Let V (G) = {u, u,..., u p } and V (tg) = V (G) {u i, u i,..., u pi : i t}. Let f be the vertex labeling of G which yields k as the super mean number. Define g on V (tg) by g(u j ) = f(u j ) for j p and g(u ji ) = f(u j ) + (i )k, j p, i t. Clearly, g induces distinct edge labels and hence S m (tg) kt.. Conclusion We proved that the graph C m, K,n for n, m and the graph C m K,n for n, m are super mean graphs. We found an upper bound of the super mean number of the graphs K,n, n, tk,n for n, t >, B(p, n) for p > n +, n, C m, K,n for n, m and C m K,n for n, m. It is also established that S m (G) p for any graph G. Further, we obtained the super mean number of the graphs K p for p, K,n for n, tk, for t >, B(p, n) for p = n, n + and C n. Acknowledgement: The authors are thankful to the referees for their helpful suggestions which lead to substantial improvement in the presentation of the paper. References [] F. Harary, Graph Theory, Addison Wesley, Reading Mass., (9). [] P. Jeyanthi, D. Ramya and P. Thangavelu, On super mean graphs, AKCE Int. J. Graphs. Combin., () (009), 0. [] R. Ponraj and D. Ramya, On super mean graphs of order, Bulletin of Pure and Applied Sciences, (Section E Maths and Statistics) E (00),. [] D. Ramya, R. Ponraj and P. Jeyanthi, Super mean labeling of graphs, Ars Combin., (To appear). [] M. Sundaram and R. Ponraj, Mean number of a graph, Pure and Applied Mathematika Sciences, LXV, (-) (00), 9 0. [] R. Vasuki and A. Nagarajan, Some results on super mean graphs, International Journal of Mathematical Combinatorics, (009), 9. [] R. Vasuki and A. Nagarajan, On the construction of new classes of super mean graphs, Journal of Discrete Mathematical Sciences and Cryptography, () (00), 90.

15 SUPER MEAN NUMBER OF A GRAPH 0 Deparment of Mathematics V.O. Chidambaram College Thoothukudi-00, Tamil Nadu, INDIA. address: nagarajan.voc@gmail.com Department of Mathematics Dr. Sivanthi Aditanar College of Engineering Tiruchendur-, Tamil Nadu, INDIA. address: vasukisehar@yahoo.co.in Department of Mathematics Mepco Schlenk Engineering College Sivakasi, Tamil Nadu, INDIA. address: