Prime and Prime Cordial Labeling for Some Special Graphs

Transcription

1 Int. J. Contemp. Math. Sciences, Vol. 5, 2, no. 47, Prime and Prime Cordial Labeling for Some Special Graphs J. Baskar Babujee and L. Shobana Department of Mathematics Anna University Chennai, Chennai-6 25, India shobana Abstract A graph G(V,E) with vertex set V is said to have a prime labeling if its vertices are labeled with distinct integers, 2,..., V such that for each edge xy E the labels assigned to x and y are relatively prime. A prime cordial labeling of a graph G with vertex set V is a bijection f from V to {, 2,..., V } such that if each edge uv is assigned the label ifgcd(f(u),f(v)) = and if gcd(f(u),f(v)) >, and the number of edges labeled with and the number of edges labeled with differ by atmost. Mathematics Subject Classification: 5C78 Keywords: Graph, prime, functions, labeling Introduction The notion of a prime labeling originated with Entringer and was introduced in a paper in Tout et al. [8]. A graph with vertex set V is said to have a prime labeling if there exist a bijection f : V (G) {, 2,..., V } such that for each edge xy E(G), f(x) and f(y) are relatively prime. In other words the Greatest Common divisor of f(x) and f(y) denoted by g.c.d of (f(x),f(y)) =. Around 98 Entringer Conjectured, that all trees have a prime labeling. Paths, stars, caterpillar, complete binary trees, spider, np 2, P n, B n,n and <K,n :2> have prime labeling. Sundaram et al. [7] have introduced the notion of prime cordial labelings. A prime cordial labeling of a graph G with vertex set V is a bijection f from V to {, 2,..., V } such that if each edge uv is assigned the label if gcd(f(u),f(v)) = and if gcd(f(u),f(v)) >, then the number of edges labeled with and the number of edges labeled with differ by atmost.

2 2348 J. Baskar Babujee and L. Shobana The following graphs are proved as prime cordial labeling: C n if and only if n 6; P n if and only if n 3 or 5; K,n (n odd); the graph obtained by subdividing each edge of K ;n if and only if n 3; bistars; dragons; crowns; triangular snakes if and only if the snake has at least three triangles; ladders. we have proved the existence of prime cordial labeling for sun graph, kite graph and coconut tree [] and Y -tree, <K,n :2> (n ), Hoffman tree and K 2 ΘC n (C n ) in [6]. Definition.. Let G =(V,E) be a graph. Let e = uv be an edge of G and w is not a vertex of G. The edge e is subdivided when it is replaced by edges e = uw and e = wv. LetG =(V,E) be a graph. If every edge of graph G is subdivided, then the resulting graph is called barycentric subdivision of graph G. In other words barycentric subdivision is the graph obtained by inserting a vertex of degree two into every edge of original graph. Consider barycentric subdivision of cycle and join each newly inserted vertices of incident edges by an edge. We denote the new graph by c n (c n ) as it look like c n inscribed in c n. Definition.2. Let G be the graph obtained by joining two copies of c n (c n ) by a path of length one and it is denoted by K 2 ΘC n (C n ). Definition.3. Let graphs G,G 2,...,G n, n 2 be all copies of a fixed graph G. Adding an edge between G i to G i+ for i =, 2,...,n is called the path union of G. Definition.4. The double star K,n,n is a tree obtained from the star K,n by adding a new pendent edge to each of the existing n pendent vertices. Definition.5. K,m ΘK,n is a tree obtained by adding n pendant edges to each pendant vertices of K,m. It has totally m(n +)+vertices. Definition.6. A shell S n is the graph obtained by taking n 3 concurrent chords in a cycle C n on n vertices. The vertex at which all the chords are concurrent is called the apex vertex. The shell is also called fan F n. i.e., S n = F n = P n + K. Consider two shells S n () and S n (2) then graph G =< S n () : S n (2) > obtained by joining apex vertices of shells to a new vertex x. In this paper we obtain three results for barycentric subdivision graph, K,m ΘK,n for all m, n and S n () : S n (2) for n>2 and n, 2(mod 3) as prime labeling and the graphs S n () : S n (2), full binary trees from second level, K 2 ΘC n (C n ) and K,n,n for n 3 admits prime cordial labeling. 2 Main Results Algorithm 2.. Step : V = {u,u 2,u 3,...,u n } {u,u 2,u 3,...,u n }

3 Prime and prime cordial labeling 2349 Step 2: E = {u i u i : i n} {u j u j+ : j n } {u nu } Step 3: f(u )=, f(u )=2 Step 4: for i =to n, f(u i )=2i Step 5: for j =to n, f(u j)=2j Figure : Prime labeling for c 6 (c 6 ) Theorem 2.2. The barycentric subdivision of cycle C n is prime. Proof. Let {u,u 2,u 3,...,u n } be the vertices of cycle C n and {u,u 2,u 3,...,u n } be the newly inserted vertices to obtain barycentric subdivision of cycle C n. To define prime labeling f : V (G) {, 2, 3,...,2n} is defined in Algorithm 2. which gives us f(u ) = and f(u ) = 2. For any edge e = u i u i : i n, gcd(f(u i ),f(u i)) = since f(u ) and f(u i) and are labeled with consecutive positive integer which implies that they are relatively prime. For any edge e = u j u j+ : j n, gcd(f(u j ),f(u j+))=. For any edge e = u u n, gcd(f(u ),f(u n)) =, since (,x) = for any positive integer x. Hence the barycentric subdivision of the cycle C n is prime. Algorithm 2.3. Step : V = {u,u 2,u 3,...,u n } {u,u 2,u 3,...,u n } {v,v 2,v 3,...,v n } {v,v 2,v 3,...,v n } {w,w 2,w 3,...,w k } Step 2: E = {u iu i : i n} {u j u j+ : j n } {u n u } E 2 = {v i v i : i n} {v j v j+ : j n } {v nv } E 3 = {w i+ w i+2 : i k 2} E 4 = {u 2 w 2,w k v }

4 235 J. Baskar Babujee and L. Shobana Step 3: f(u i )=2i; i n f(u i )=2i ; i n Step 4: f(v i )=2n +2i; i n f(v i )=2n +2i ; i n Step 5: f(w j+ )=4n + j; j k Figure 2: Two copies of c 6 (c 6 ) joined by path P 5 and its prime labeling Theorem 2.4. The graph G obtained by two copies of c n (c n ) is prime only when n mod 3 joined by a path P k, k 3 and k is odd of arbitrary length is prime. Proof. Let G be a graph obtained by joining two copies of c n (c n ) by a path of arbitrary length k. Let {u,u 2,u 3,...,u n } be the vertices of cycle c n and {u,u 2,u 3,...,u n } be the corresponding vertices of the cycle which is obtained by joining newly inserted vertices of adjacent edges in cycle c n. Next denote the corresponding vertices in second copy of c n (c n )byv,v 2,v 3,...,v n and v,v 2,v 3,...,v n respectively. Let w,w 2,w 3,...,w n be the vertices of path P k with u = w and v = w k. To define prime labeling f : V (G) {, 2, 3,...,4n+k 2}. For any edge e = {u i u i : i n}, gcd(f(u i ),f(u )) = since 2i and 2i are consecutive positive integers. For any edge e = {v i v i : i n}, gcd(f(v ),f(v i)) =, since 2n +2i and 2n + 2i are consecutive positive integers. For any edge e = {u i u i+ : i n }, gcd(f(u i ),f(u i+)) = from the above algorithm. For any edge e = {v j v j+ : j n }, gcd(f(v j),f(v j+)) =. u and u n are relatively prime, since gcd(,x) = for any integer x. Similarly it holds for v and v n. For any edge e = {u 2 w 2,w k v },(f(u 2 ),f(w 2 )) and (f(w k ),f(v ))are relatively prime. Finally for any edge e = {w j+ w j+2 ;

5 Prime and prime cordial labeling 235 j k 2}, gcd(f(w j+ ),f(w j+2 )) =, since 4n+j and 4n+j + are relatively prime from the above Algorithm 2.3. Hence the proof Figure 3: Path union of three copies of c 4 (c 4 ) joined by path P 3 is not prime Theorem 2.5. The graph S () n : S (2) n n>2 is prime if n, 2(mod 3). Proof. Let v,v 2,v 3,...,v 2n+ be the vertices of the graph S n () : S n (2) and the edge set is defined as E = E E 2 E 3 where E = {v v 2,v 2 v m : m is the largest prime number 2n +} E 2 = {v v i,v i v i+ ;3 i n} {v v n+ } E 3 = {v m v j,v j v j+ ; n +2 j 2n,m j} {v m v 2n }. Let f : V (G) {, 2, 3,...,2n +} be a bijection map defined by f(v i )=i. For any edge e =(v v 2,v v n+,v v i ;3 i n), gcd(f(v ),f(v 2 )) = gcd(f(v ),f(v n+ )) = gcd(f(v ),f(v i )) = since gcd(,x) = where x is any positive integer. For any edge e =(v 2 v m,v m v 2n,v m v j ; n +2 j 2n ), gcd(f(v 2 ),f(v m )) = gcd(f(v m ),f(v 2n )) = gcd(f(v m ),f(v j )) = since gcd(m, x) = where m is the largest prime number. For any edge e =(v i v i+ ;3 i n, v j v j+ ; n +2 j 2n ), gcd(f(v i ),f(v i+ )) = gcd(f(v j ),f(v j+ )) = since i and i +, j and j + are consecutive positive integers. Hence the graph S n () : S n (2), n>2ifn, 2(mod 3) is prime. Theorem 2.6. The graph K,m ΘK,n for all m, n is prime labeling. Proof. Case (i): Let a be the root of the tree. Let a,a 2,...,a m be the children of the root. Each subtree a i, i m will have n number of vertices which are considered as leaves of the graph namely a i,a i2,...,a in ; i m leaves. Let a = 2. The internal vertices a,a 2,a 3,...,a m are labeled with consecutive largest prime numbers less than or equal to mn +

6 2352 J. Baskar Babujee and L. Shobana m +. The vertices that acts as a leaves of the graph K,m ΘK,n are labeled with the remaining numbers other than 2, a,a 2,...,a m in the consecutive order. The greatest common divisor of a with a,a 2,a 3,...,a m is one because a,a 2,a 3,...,a m all are prime numbers. The greatest common divisor of a i with a i,a i2,...,a in ; i m is one, since a i s are all consecutive largest prime numbers. Hence the the graph K,m ΘK,n for all m, n is prime. Case (ii): In K,m ΘK,n when m is too large and n is too small then the construction in the Theorem 2.6 need not hold. For example the above theorem holds for certain restriction for m i.e., When n =2,m 2; n =3,m 56; n =4,m 72; n =5,m 89. For further values of n, m can be calculated using program to list prime numbers and their count in C programming Figure 4: Theorem 2.7. The graph K 2 ΘC n (C n ) admits prime cordial labeling if n, 2(mod 3). Proof. Let G be the graph K 2 ΘC n (C n ) obtained by joining two copies of c n (c n ) by a path of length one. Let u,u 2,...,u n be the vertices of cycle c n and u,u 2,...,u n be the corresponding vertices of the cycle which is obtained by joining newly inserted vertices of adjacent edges in cycle c n. Next denote the corresponding vertices in second copy of c n (c n )byv,v 2,...,v n and v,v 2,...,v n respectively. Let u and v be the vertices of path p 2. Let the vertex and the edge sets are defined as follows: V = {{u,u 2,...,u n }, {u,u 2,...,u n }, {v,v 2,...,v n }, {v,v 2,...,v n }} E = E E 2 E where E = {u i u i; i n } E 2 = {u j u j+; j n } E 3 = {u u n} E 4 = {v i v i ; i n } E 5 = {v j v j+; j n } E 6 = {v v n}

7 Prime and prime cordial labeling 2353 E 7 = {u i u i+ ; i n} E 8 = {u u n } E 9 = {v iv i+; i n} E = {v v n } Let f : V {, 2, 3,..., v } be the bijective function defined as f(u i )=4i 2; i n f(u j )=4j; j n f(v i )=4i 3; i n f(v j )=4j ; j n g(u i u i) = ; i n g(u j u j+) = ; j n g(v i v i) = ; i n g(v j v j+) = ; j n g(u i u i+ ) = ; j n g(u u n)= g(v i v i+ ) = ; i n g(v v n )= g(u v )= g(u u n )= g(v v n)=. The total number of edges labeled with s are given by K =3n + and the total number of edges labeled with s are given by K 2 =3n. Therefore the total difference between s and s is given by K K 2 = (3n +) 3n and they differ by one. This proves that the graph K 2 ΘC n (C n ) is prime cordial labeling Figure 5: The graph K 2 ΘC 6 (C 6 ) is prime cordial Theorem 2.8. The graph <K,n,n > for n 3 admits prime cordial labeling.

8 2354 J. Baskar Babujee and L. Shobana Proof. Let the vertex set and edge set of < K,n,n > be defined as V = {v,v 2,v 3,...,v 2n+ } and E = {E E 2 E 3 } where E = {v v 2,v v 2n+,v 4 v 5,v 6 v 3 }, E 2 = {v 2 v 2i+2 ; i n } and E 3 = {v 2i+6 v 2i+5 ; i n 3}. Let f : V {, 2, 3,..., v } be the bijective function defined as f(v i )=i; i 2n +. We compute { the edge labeling defined by if (f(u),f(v)) > g(u, v) = if (f(u),f(v)) = and g(v v 2 )=g(v v 2n+ )=g(v 4 v 5 )= g(v 6 v 3 )= g(v 2 v 2i+2 ) = ; i n g(v 2i+6 v 2i+5 ) = ; i n 3 The total number of edges labeled with s is given by K = n and the total number of edges labeled with s are given by K 2 = n. Therefore the total difference between s and s is given by K K 2 = n n and they differ by zero. This proves that <K,n,n > for n 3 is a prime cordial labeling. for all n>2 admits prime cordial label- Theorem 2.9. The graph S n () : S n (2) ing. Proof. Let the vertex set and edge set of S n () : S n (2) be defined as V = {v,v 2,v 3,...,v 2n+ } and E = {E E 2 E 3 E 4 E 5 } where E = {v v 3,v 3 v 6,v 2 v 4,v 4 v 8,v 6 v 2,v 6 v 4 } E 2 = {v 6 v 2i ; i n} E 3 = {v 2i+6 v 2i+8 ; i n 4} E 4 = {v v 2i+3 ; i n } E 5 = {v 2i+3 v 2i+5 ; i n 2}. Let f : V {, 2, 3,..., v } be the bijective function defined as f(v i )=i; i 2n +. We compute { the edge labeling defined by if (f(u),f(v)) > g(u, v) = if (f(u),f(v)) = and g(v v 3 )= g(v 6 v 3 )=g(v 2 v 4 )=g(v 4 v 8 )=g(v 6 v 2 )=g(v 6 v 4 )= g(v 6 v 2i ) = ; 4 i n g(v 2i+6 v 2i+8 ) = ; i n 4 g(v v 2i+3 ) = ; i n g(v 2i+3 v 2i+5 ) = ; i n 2 The total number of edges labeled with s are given by K =2n 2and the total number of edges labeled with s are given by K 2 =2n 2. Therefore the total difference between s and s is given by K K 2 = (2n 2) (2n 2) and they differ by zero. This proves that the graph S n () : S n (2) for n>2asa

9 Prime and prime cordial labeling 2355 prime cordial labeling. Theorem 2.. The full binary tree admits prime cordial. Proof. The root a is called the special vertex in the tree. Let N denotes the number of levels in full binary tree. The root has edges to n other vertices called its children. The children of the root are said to be on level one. There are 2 N+ vertices and 2 N+ 2 edges in the full binary tree. Case (i): Let a and a 2 be the children of a. The vertices on the last level N have no children and are leaves. The vertices that are not leaves are said to be internal vertices. Let a =,a = 2 and a 2 = 3 are fixed. The vertices a and a 2 are divided into two sub trees. The leftmost subtree a are labeled with consecutive even numbers from top to bottom. The greatest common divisor of any two numbers on the leftmost subtree is greater than one and the edges are labeled with zero. In the second level the vertices on the rightmost subtree a 3 are labeled with 7 and 5 and the remaining levels of the rightmost subtree are labeled with consecutive odd numbers starting with the number 9. The rightmost vertices of level 2 and level 3 are connected by an edge with label zero since gcd(5, 5) > and all the other edges on the rightmost subtree a 3 are labeled with one. The root a with their children a and a 2 are connected by the edges labeled with one. The total number of edges labeled with is given by k =2 n and the total number of edges labeled with is given by k 2 =2 n. Therefore the total difference between s and s is given by K K 2 = (2 n ) (2 n ) and they differ by zero. This proves that the binary tree is prime cordial labeling. Case (ii): The full binary tree with the second level is prime cordial when a =2,a = 4 and a 2 = 6. In the second level the vertices are numbered as, 5, 3 and 7. The total number of edges labeled with is given by k = 3 and the total number of edges labeled with is given by k 2 = 3. Therefore the total difference between s and s is given by K K 2 = 3 3 and they differ by zero. References [] J. Baskar Babujee and L. Shobana, Prime Cordial Labeling, International Review of Pure and Applied Mathematics, Vol. 5, No. 2, (29). [2] I. Cahit, Cordial graphs: A weaker version of graceful and harmonious graphs, Ars Combin, 23, 2 27 (987). [3] J.A. Gallian, A Dynamic Survey of Graph Labeling, Electronic Journal of Combinatorics, 6, #DS6, (29).

10 2356 J. Baskar Babujee and L. Shobana [4] B. Gayathri and K. Amuthavalli, k-odd Mean Labeling of Graphs, Proceedings of the International Conference on Mathematics and Computer Science-2, pp. 2 5 (27), Sitech Publications. [5] S.M. Lee, I. Wui and J. Yeh, On the amalgamation of prime graphs, Bull. Malaysian Math. Soc. (Second Series), (988) [6] L. Shobana and J. Baskar Babujee, Prime Cordial Labeling for Some Special Graphs, Proceedings of the National Conference on Applied Mathematics (NCAM), Allied Publishers, 39 42, (2). [7] M. Sundaram, R. Ponraj, and S. Somasundram, Prime cordial labeling of graphs. Indian Acad. Math., 27, (25). [8] A. Tout, A. N. Dabboucy and K. Howalla, Prime labeling of graphs, Nat. Acad.Sci. Letters,, (982). [9] S.K. Vaidya and K.K. Kanani, Some Cordial Graphs in the context of Barycentric Subdivision, Int. J. Contemp. Math. Sciences, Vol. 4, No. 3, (29). [] W.D. Wallis, Magic graphs, Birkhauser, Buston, 2. Received: April, 2