SOME MORE RESULTS ON FACE EDGE PRODUCT CORDIAL GRAPHS 1A. Muthaiyan and 2P. Pugalenthi 1 Assistant Professor, P.G. and Research Department of Mathematics, Govt. Arts College, Ariyalur - 621713. 2 P.G. and Research Department of Mathematics, Govt. Arts College, Ariyalur - 621713. ABSTRACT In this paper, we investigate the face edge product cordial labeling of the planar graphs K1,nK2, K1,nK3, (PnK1)K2, (PnK1)K3 and the planar graph obtained from joining the graphs fn and Wn by a path of arbitrary length. Keywords : Edge product cordial labeling, Edge product cordial graph, Face edge product cordial labeling, Face edge product cordial graph. 1. Introduction Throughout this paper we consider only finite, planar, undirected and simple graphs. Let G be a graph with p vertices and q edges. For all terminologies and notations related to graph theory, we follow Harary [2]. For standard terminology and notations related to graph labeling, we refer to Gallian [1]. In [4], Vaidya et al. introduced the concept of edge product cordial labeling of graph. The concept of face edge product cordial labeling is introduced by Lawrence et al. [3]. The brief summaries of definition which are necessary for the present investigation are provided below. Definition : 1.1 The corona G1G2 of two graphs G1(p1, q1) and G2(p2, q2) is defined as the graph obtained by taking one copy of G1 and p1 copies of G2 and then joining the ith vertex of G1 to all the vertices in the ith copy of G2. Definition : 1.2 A wheel Wn, n 3, with n spokes is a graph that has a center vertex connected to all n vertices in cycle Cn. Definition : 1.3 A fan graph fn is constructed from a wheel Wn by deleting one edge in Cn. Definition : 1.4 [4] For a graph G = (V(G),E(G)), an edge labeling function f : E(G)→{0,1} induces a vertex labeling function f* :V(G)→{0,1} defined as f (v) f(e i ) for {ei E(G)/ei is incident to v}. Now denoting the number of vertices of G having label i under f* as 1 vf(i) and the number of edges of G having label i under f as ef(i). Then f is called edge product cordial labeling of graph G if |vf(0) − vf(1)| 1 and |ef(0) − ef(1)| 1. A graph G is called edge product cordial if it admits edge product cordial labeling. Definition : 1.5 [3] For a planar graph G, the edge labeling function is defined as g : E(G){0,1} and g(e) is called the label of the edge e of G under g, induced vertex labeling function g* : V(G) {0,1} is given as if e1,e2,…,em are the edges incident to vertex v, then g*(v) = g(e1)g(e2)…g(em) and induced face labeling function g** : F(G) {0,1} is given as if v1, v2, …, vn and e1, e2, …, em are the vertices and edges of face f then g**(f) = g*(v1)g*(v2)…g*(vn) g(e1)g(e2)…g(em). Let us denote vg(i) is the number of vertices of G having label i under g*, eg(i) is the number of edges of G having label i under g and fg(i) is the number of interior faces of G having label i under g** for i = 1,2. g is called face edge product cordial labeling of graph G if |vg(0)−vg(1)|1, |eg(0)−eg(1)| 1 and | fg(0) − fg(1)| 1. A graph G is face edge product cordial if it admits face edge product cordial labeling. 2. Main Theorems Theorem : 2.1 The planar graph G is obtained from joining the graphs fn and Wn by a path of arbitrary length is face edge product cordial graph. Proof. Let G be the planar graph is obtained from joining of fn and Wn by a path of arbitrary length is face edge product cordial graph. Let u,u1, u2,…, un be the vertices, e1, e2,…, e2n–1 be the edges and f1, f2, …, fn–1 be the interior faces of fn. Let v, v1,…,vn be the vertices, e2n, e2n+1,…,e4n–1 be the edges and fn,fn+1,…,f2n–1 be the interior faces of Wn. Let w1,w2,…,wk be the vertices of path Pk and e4n, e4n+1, … , e4n+k–2 be the edges of path Pk. In G, u = w1 and wk = v1. Then |V(G)| = 2n+k, |E(G)| = 4n + k – 2 and |F(G)| = 2n–1. Define edge labeling g : E(G) → {0,1} as follows g(ei) = 1, for 1 i 2n–1 g(e2n–1+i ) = 0, for 1 i 2n Case (i) : k is odd g(e4n–1+i ) = 1, for 1 i k 1 2 k 1 +1 i k – 1 2 In view of the above defined labeling pattern we have g(e4n–1+i ) = 0, for 4n k 1 2n k 1 , vg(0) = vg(1)+1 = and fg(0) = fg(1)+1= n. 2 2 Then |vg(0) − vg(1)| 1, |eg(0) − eg(1)| 1 and | fg(0) − fg(1)| 1 eg(0) = eg(1)+1 = Case (ii) : k is even g(e4n–1+i ) = 1, for 1 i k 2 2 k +1 i k – 1 2 In view of the above defined labeling pattern we have g(e4n–1+i ) = 0, for 4n k 2 2n k , vg(0) = vg(1) = and fg(0) = fg(1)+1= n. 2 2 Then |vg(0) − vg(1)| 1, |eg(0) − eg(1)| 1 and | fg(0) − fg(1)| 1 eg(0) = eg(1) = Therefore, the planar graph G is obtained from joining the graphs fn and Wn by a path of arbitrary length is face edge product cordial graph. Example : 2.1 The planar graph G is obtained from joining the graphs f5 and W5 by a path P6 and its face edge product cordial labeling is given in figure 2.1. 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Figure 2.1 Theorem : 2.2 K1,n K2 is face edge product cordial graph. Proof. Let G be the graph K1,n K2. The vertex set V(G) = {ui,vij :1 i n+1,1j 2}, edge set E(G) = {ei, ejk: 1 i n, 1 j n+1, 1 k 3} and interior face set F(G) = { fi:1 i n+1}, where ei = uiun+1 for 1 i n, eik = uivik for 1 i n+1 and 1 k 2, ei3 = vi1vi2 for 1 i n+1. Then |V(G)| = 3n + 3, |E(G)| = 4n+3 and |F(G)| = n+1. Define edge labeling g : E(G) → {0,1} as follows Case (i) : n is odd n 1 g(ei) = 1, for 1 i 2 n3 g(ei) = 0, for in 2 n 1 g(eij) = 1, for 1 i and 1 j 3 2 n3 g(eij) = 0, for i n+1 and 1 j 3 2 In view of the above defined labeling pattern we have 3n 3 n 1 and fg(0) = fg(1) = . 2 2 Then |vg(0) − vg(1)| 1, |eg(0) − eg(1)| 1 and | fg(0) − fg(1)| 1. eg(1) = eg(0) + 1 = 2n+2, vg(0) = vg(1) = 3 Case (ii) : n is even g(ei) = 1, for 1 i n 2 n +1 i n 2 n g(eij) = 1, for 1 i and 1 j 3 2 n g(eij) = 0, for +1 i n and 1 j 3 2 g(ei1) = 1, g(ei2) = 0 and g(ei3) = 1, for i = n +1 In view of the above defined labeling pattern we have g(ei) = 0, for 3n 4 n2 and fg(0) = fg(1) + 1 = . 2 2 Then |vg(0) − vg(1)| 1, |eg(0) − eg(1)| 1 and | fg(0) − fg(1)| 1 Therefore, the graph K1,n K2 is face edge product cordial graph. eg(1) = eg(0)+1 = 2n+2, vg(0) = vg(1)+1 = Example : 2.2 The graph K1,4K2 and its face edge product cordial labeling is given in figure 2.2. 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 Figure 2.2 Theorem : 2.3 K1,n K3 is face edge product cordial graph for n is odd. Proof. Let G be the graph K1,n K3. The vertex set V(G) = {ui,vij :1 i n+1,1j 3}, edge set E(G) = {ei, ejk: 1 i n, 1 j n+1, 1 k 6} and interior face set F(G) = { fi:1 i 3n+3}, where ei = uiun+1 for 1 i n, eik = uivik for 1 i n+1 and 1k 3, ei4 = vi1vi2 , ei5 = vi2vi3, ei6 = vi1vi3 for 1 i n+1. Then |V(G)| = 4n + 4, |E(G)| = 7n+6 and |F(G)| = 3n+3. Define edge labeling g : E(G) → {0,1} as follows Case (i) : n is odd n 1 g(ei) = 1, for 1 i 2 n3 g(ei) = 0, for in 2 4 g(eij) = 1, for 1 i n 1 and 1 j 6 2 n3 i n+1 and 1 j 6 2 In view of the above defined labeling pattern we have g(eij) = 0, for 7n 7 3n 3 , vg(0) = vg(1) = 2n+2 and fg(0) = fg(1) = . 2 2 Then |vg(0) − vg(1)| 1, |eg(0) − eg(1)| 1 and | fg(0) − fg(1)| 1 Case (ii) : n is even In order to satisfy the edge condition for G, it is essential to assign label 1 and 0 7n 6 to exactly edges. 2 Assigning any pattern of edge labels which satisfying the edge condition will induce vertex labels for 4n+4 number of vertices in such a way that |vg(0)–vg(1)| 2, that is vertex condition for G is violated. Thus the graph G under consideration is not a face edge product cordial graph when n is even. Therefore, the graph K1,n K3 is face edge product cordial graph for n is odd. eg(1) = eg(0) + 1 = Example : 2.3 The graph K1,3 K3 and its face edge product cordial labeling is given in figure 2.3. 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 Figure 2.3 Theorem : 2.4 (Pn K1) K2 is face edge product cordial graph. Proof. Let G be the graph (PnK1)K2. The vertex set V(G) = {ui,vij :1i2n,1j 2}, edge set E(G) = {ei, ejk: 1 i 2n–1, 1 j 2n and 1 k 3} and interior face set F(G) = {fi:1 i 2n}, where ei = uiui+1 for 1 i n–1, en–1+i = uiun+i for 1 i n, eik = uivik for 1 i 2n and 1 k 2, ei3 = vi1vi2 for 1 i 2n. Then |V(G)| = 6n, |E(G)| = 8n–1 and |F(G)| = 2n. Define edge labeling g : E(G) → {0,1} as follows 5 Case (i) : n is odd g(ei) = 1, for 1 i n 1 2 n 1 i n–1 2 n 1 g(en–1+i) = 1, for 1 i 2 n3 g(en–1+i) = 0, for in 2 n 1 g(eij) = 1, for 1 i and 1 j 3 2 n 1 g(eij) = 0, for i n and 1 j 3 2 n 1 g(e(n+i)j) = 1, for 1 i and 1 j 3 2 n3 g(e(n+i)j) = 0, for i n and 1 j 3 2 In view of the above defined labeling pattern we have g(ei) = 0, for eg(1) = eg(0) + 1 = 4n, vg(0) = vg(1) = 3n and fg(0) = fg(1) = n. Then |vg(0) − vg(1)| 1, |eg(0) − eg(1)| 1 and | fg(0) − fg(1)| 1 Case (ii) : n is even g(ei) = 1, for 1 i n 2 n +1 i n–1 2 n g(en–1+i) = 1, for 1 i 2 n g(en–1+i) = 0, for +1 i n 2 n g(eij) = 1, for 1 i and 1 j 3 2 n g(eij) = 0, for +1 i n and 1 j 3 2 n g(e(n+i)j) = 1, for 1 i and 1 j 3 2 n g(e(n+i)j) = 0, for +1 i n and 1 j 3 2 In view of the above defined labeling pattern we have g(ei) = 0, for eg(1) = eg(0) + 1 = 4n, vg(0) = vg(1) = 3n and fg(0) = fg(1) = n. Then |vg(0) − vg(1)| 1, |eg(0) − eg(1)| 1 and | fg(0) − fg(1)| 1 Therefore, the graph (Pn K1) K2 is face edge product cordial graph. 6 Example : 2.4 The graph (P4 K1) K2 and its face edge product cordial labeling is given in figure 2.4. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 0 0 0 1 0 1 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 1 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 Figure 2.4 Theorem : 2.5 (Pn K1) K3 is face edge product cordial graph. Proof. Let G be the graph (PnK1)K3. The vertex set V(G) = {ui,vij :1i2n,1j3}, edge set E(G) = {ei, ejk: 1 i 2n–1, 1 j 2n and 1 k 6} and interior face set F(G) = {fi:1 i 6n}, where ei = uiui+1 for 1 i n–1, en–1+i = uiun+i for 1 i n, eik = uivik for 1 i 2n and 1 k 3, ei4 = vi1vi2 , ei5 = vi2vi3, ei6 = vi1vi3 for 1 i 2n. Then |V(G)| = 8n, |E(G)| = 14n–1 and |F(G)| = 6n. Define edge labeling g : E(G) → {0,1} as follows Case (i) : n is odd n 1 g(ei) = 1, for 1 i 2 n 1 g(ei) = 0, for i n–1 2 n 1 g(en–1+i) = 1, for 1 i 2 n3 g(en–1+i) = 0, for in 2 n 1 g(eij) = 1, for 1 i and 1 j 6 2 n 1 g(eij) = 0, for i n and 1 j 6 2 n 1 g(e(n+i)j) = 1, for 1 i and 1 j 6 2 n3 g(e(n+i)j) = 0, for i n and 1 j 6 2 In view of the above defined labeling pattern we have eg(1) = eg(0) + 1 = 4n, vg(0) = vg(1) = 3n and fg(0) = fg(1) = n. Then |vg(0) − vg(1)| 1, |eg(0) − eg(1)| 1 and | fg(0) − fg(1)| 1 Case (ii) : n is even n g(ei) = 1, for 1 i 2 7 n +1 i n–1 2 n g(en–1+i) = 1, for 1 i 2 n g(en–1+i) = 0, for +1 i n 2 n g(eij) = 1, for 1 i and 1 j 6 2 n g(eij) = 0, for +1 i n and 1 j 6 2 n g(e(n+i)j) = 1, for 1 i and 1 j 6 2 n g(e(n+i)j) = 0, for +1 i n and 1 j 6 2 In view of the above defined labeling pattern we have eg(1) = eg(0) + 1 = 4n, vg(0) = vg(1) = 3n and fg(0) = fg(1) = n. Then |vg(0) − vg(1)| 1, |eg(0) − eg(1)| 1 and | fg(0) − fg(1)| 1 Therefore, the graph (Pn K1) K2 is face edge product cordial graph. g(ei) = 0, for Example : 2.5 The graph (P3 K1) K3 and its face edge product cordial labeling is given in figure 2.5. 1 1 1 1 1 1 1 1 1 1 0 1 0 0 0 0 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 Figure 2.5 References: [1] [2] [3] [4] J. A. Gallian, A dynamic survey of graph labeling, The Electronic Journal of Combinatorics, 16, # DS6, 2014. F. Harary, Graph theory, Addison Wesley, Reading, Massachusetts, 1972. P. Lawrence Rozario Raj and R. Lawrence Joseph Manoharan, Face and Total face edge product cordial graphs, International Journal of Mathematics Trends and Technology, Vol. 19, No. 2, pp 136-149, 2015. S. K. Vaidya and C. M. Barasara, Edge Product Cordial Labeling of Graphs, J. Math. Comput. Sci. Vol 2, No. 5, pp 1436-1450, 2012. 8