IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_01 ______________________________________________________________________________________ Further Results on 3−equitable Labeling Gaurang V. Ghodasara, Sunil G. Sonchhatra Abstract—A mapping f from the vertex set of a graph G to the set {0, 1, 2} is called 3-equitable labeling, if the edge labels are produced by the absolute difference of labels of end vertices such that the absolute difference of number of vertices of G labeled with 0, 1 and 2 differ by atmost 1 and similarly the absolute difference of number of edges of G labeled with 0, 1 and 2 differ by atmost 1. A graph which admits 3-equitable labeling is called 3-equitable graph. In this paper we prove that the graph obtained by joining two copies of fan graph by a path of arbitrary length is 3-equitable. We also prove similar results for wheel, helm, gear and cycle with one pendant edge. Index Terms—3-equitable graph, Fan, Wheel, Helm, Gear. A graph G is 3-equitable if it admits 3-equitable labeling. The concept of 3-equitable labelings was introduced by Cahit[2] in 1990. Seoud and Abdel Maqsoud[6] proved that all fans except P2 + K1 are 3-equitable. Bapat and Limaye[1] proved that Helm Hn is 3-equitable for n ≥ 4. Youssef[11] proved that Wheels Wn = Cn + K1 are 3-equitable for n ≥ 4. In this paper we prove that the graph obtained by joining two copies of fan graphs by a path of arbitrary length is 3-equitable. We also prove similar results for wheel, helm, gear and cycle with one pendant edge. I. INTRODUCTION II. MAIN RESULTS E consider simple, finite, undirected graph G = (V, E). In this paper Fn denotes fan graph with n+1 vertices, Wn denotes the wheel graph with n + 1 vertices, Hn denotes helm graph with 2n + 1 vertices and Gn denotes gear graph with 2n + 1 vertices. For all other terminology and notations we follow Gross and Yellen[5]. If the vertices of the graph are assigned values subject to certain conditions is known as graph labeling. A survey on graph labeling is given by Gallian[4]. Definition 1.1 A fan graph, denoted by Fn , is the graph with n + 1 vertices which is the join of the graphs Pn and K1 . i.e. Fn = Pn + K1 . Definition 1.2 A wheel graph, denoted by Wn , is the join of the graphs Cn and K1 . i.e. Wn = Cn + K1 Here vertices correspond to Cn are called rim vertices and Cn is called rim of Wn and the vertex corresponding to K1 is called apex vertex. Definition 1.3 A helm graph, denoted by Hn , is the graph obtained from wheel Wn by adding a pendant edge at each vertex on rim of Wn . Definition 1.4 A gear graph, denoted by Gn , is obtained from the wheel Wn by adding a vertex between every pair of adjacent vertices of the rim of wheel. Definition 1.5 Let G = (V, E) be a graph. A mapping f : V (G) → {0, 1, 2} is called ternary vertex labeling of G and f (v) is called label of the vertex v of G under f . Let f ∗ : E(G) → {0, 1, 2} be the induced edge labeling function defined by f ∗ (e) = |f (u) − f (v)|, for an edge e = uv of G. Let us denote vf (i) = the number of vertices of G with label i under f and ef (i) = the number of edges of G with label i under f ∗ , 0 ≤ i, j ≤ 2. Definition 1.6 A ternary vertex labeling of a graph G is called 3-equitable labeling if |vf (i)−vf (j)| ≤ 1 and |ef (i)− ef (j)| ≤ 1, 0 ≤ i, j ≤ 2. Theorem 1 The graph obtained by joining two copies of fan graph Fn by a path of arbitrary length is 3-equitable. Proof: Let G be the graph obtained by joining two copies of fan graph Fn = Pn + K1 by a path Pk of length k − 1. Let us denote the successive vertices of first copy of fan graph by u1 , u2 , . . . , un+1 (where u1 is the vertex corresponding to K1 ) and the successive vertices of second copy of fan graph by v1 , v2 , . . . , vn+1 (where v1 is the vertex corresponding to K1 ). Let w1 , w2 , . . . , wk be the vertices of path Pk with w1 = u1 and wk = v1 . We define labeling function f : V (G) → {0, 1, 2} as follows. Case 1: n ≡ 0(mod6). Subcase I: k ≡ 0(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n + 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (wk ) = 2. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k − 1. The remaining vertices are labeled same as in Subcase I. Subcase III: k ≡ 2(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n + 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 0, 5(mod6) = 2; if j ≡ 2, 3(mod6), 1 ≤ j ≤ k. Subcase IV: k ≡ 3(mod6). f (v2 ) = 1, f (v4 ) = 0, f (vn+1 ) = 1. f (vi ) = 0; if i ≡ 2, 5(mod6) W Manuscript received June 4, 2014; revised December 25, 2014. Gaurang V. Ghodasara is an assistant professor of Mathematics in H. & H. B. Kotak Institute of Science, Rajkot, Gujarat. E-mail: gaurang enjoy@yahoo.co.in. Sunil G. Sonchhatra is the Ph.D. Scholar in School of Science, R.K. University, Rajkot. He is working as an assistant professor of Mathematics in Lukhdhirji Engineering College, Morbi, Gujarat. E-mail: sonchhabdasunil.20@gmail.com (Advance online publication: 17 February 2015) IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_01 ______________________________________________________________________________________ = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n, i ̸= 2, i ̸= 4. The remaining vertices are labeled same as in Subcase III. Subcase V: k ≡ 4(mod6). f (vn+1 ) = 1, f (wk ) = 2. The remaining vertices are labeled same as in Subcase III. Subcase VI: k ≡ 5(mod6). f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n + 1. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k. Case 2: n ≡ 1(mod6). Subcase I: k ≡ 0(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n + 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n + 1. The remaining vertices are labeled same as in Subcase I. Subcase III: k ≡ 2(mod6). f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n + 1. f (wk ) = 0. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 4, 5(mod6) = 2; if j ≡ 1, 2(mod6), 1 ≤ j ≤ k − 1. Subcase IV: k ≡ 3(mod6). f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n + 1. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 4, 5(mod6) = 2; if j ≡ 1, 2(mod6), 1 ≤ j ≤ k. Subcase V: k ≡ 3(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n + 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 0, 5(mod6) = 2; if j ≡ 2, 3(mod6), 1 ≤ j ≤ k. Subcase VI: k ≡ 5(mod6). f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n + 1. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k. Case 3: n ≡ 2(mod6). Subcase I: k ≡ 0(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n + 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n + 1. The remaining vertices are labeled same as in Subcase I. Subcase III: k ≡ 2, 3(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n + 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 0, 5(mod6) = 2; if j ≡ 2, 3(mod6), 1 ≤ j ≤ k. Subcase IV: k ≡ 4(mod6). f (v2 ) = 1, f (v4 ) = 0, f (wk ) = 2. The remaining vertices are labeled same as defined in Subcase I. Subcase V: k ≡ 5(mod6). f (v2 ) = 1, f (v4 ) = 0. The remaining vertices are labeled same as defined in Subcase I. Case 4: n ≡ 3(mod6). Subcase I: k ≡ 0(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n + 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) (Advance online publication: 17 February 2015) IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_01 ______________________________________________________________________________________ = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n + 1. The remaining vertices are labeled same as in Subcase I. Subcase III: k ≡ 2(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n + 1. f (vn+1 ) = 2. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. f (wk ) = 0. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k − 1. Subcase IV: k ≡ 3(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n + 1. f (vn−1 ) = 2, f (vn−3 ) = 0. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n + 1, i ̸= n − 3, i ̸= n − 1. f (wk ) = 2. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k − 1. Subcase V: k ≡ 4(mod6). f (wk ) = 2. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k − 1. The remaining vertices are labeled same as in Subcase I. Subcase VI: k ≡ 5(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n + 1. f (wk ) = 2. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 0, 5(mod6) = 2; if j ≡ 2, 3(mod6), 1 ≤ j ≤ k − 1. Case 5: n ≡ 4(mod6). Subcase I: k ≡ 0(mod6). f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n + 1. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 4, 5(mod6) = 2; if j ≡ 1, 2(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n + 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 0, 5(mod6) = 2; if j ≡ 2, 3(mod6), 1 ≤ j ≤ k. Subcase III: k ≡ 2(mod6). f (wk ) = 0. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 0, 5(mod6) = 2; if j ≡ 2, 3(mod6), 1 ≤ j ≤ k − 1. The remaining vertices are labeled same as in Subcase I. Subcase IV: k ≡ 3(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n + 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 0, 5(mod6) = 2; if j ≡ 2, 3(mod6), 1 ≤ j ≤ k. Subcase V: k ≡ 4(mod6). f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n + 1. f (w3 ) = 1, f (wk ) = 0. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 4, 5(mod6) = 2; if j ≡ 1, 2(mod6), 1 ≤ j ≤ k − 1, j ̸= 3. Subcase VI: k ≡ 5(mod6). f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k. The remaining vertices are labeled same as defined in Subcase I. Case 6: n ≡ 5(mod6). Subcase I: k ≡ 0(mod6). f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n + 1. f (v6 ) = 0. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n + 1, i ̸= 6. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (un−1 ) = 2, f (wk ) = 0. The remaining vertices are labeled same as in Subcase I. Subcase III: k ≡ 2(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n + 1. f (wk ) = 0. f (wj ) = 0; if j ≡ 1, 4(mod6) (Advance online publication: 17 February 2015) IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_01 ______________________________________________________________________________________ = 1; if j ≡ 0, 5(mod6) = 2; if j ≡ 2, 3(mod6), 1 ≤ j ≤ k − 1. Subcase IV: k ≡ 3(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n + 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 0, 5(mod6) = 2; if j ≡ 2, 3(mod6), 1 ≤ j ≤ k. Subcase V: k ≡ 4(mod6). f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n + 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n + 1. f (wk ) = 0. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 4, 5(mod6) = 2; if j ≡ 1, 2(mod6), 1 ≤ j ≤ k − 1. Subcase VI: k ≡ 5(mod6). f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n + 1. f (vn−2 ) = 0, f (vn−4 ) = 2. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n + 1, i ̸= n − 2, i ̸= n − 4. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k. The graph G under consideration satisfies the conditions |vf (i) − vf (j)| ≤ 1 and |ef (i) − ef (j)| ≤ 1, 0 ≤ i, j ≤ 2 in each case. Hence the graph G under consideration is 3-equitable graph. Theorem 2 The graph obtained by joining two copies of wheel graph by a path of arbitrary length is 3-equitable. Proof: Let G be the graph obtained by joining two copies of wheel graph Wn by path Pk of length k − 1. Let us denote the successive vertices of first copy of wheel graph by u0 , u1 , . . . , un (where u0 is apex vertex) and the successive vertices of second copy of wheel graph by v0 , v1 , . . . , vn (where v0 is apex vertex). Let w1 , w2 , . . . , wk be the vertices of path Pk with w1 = u1 and wk = v1 . We define labeling function f : V (G) → {0, 1, 2} as follows. Case 1: n ≡ 0(mod6). f (u0 ) = 0, f (v0 ) = 2. Subcase I: k ≡ 0(mod6). f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 4, 5(mod6) = 2; if j ≡ 1, 2(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1, 2(mod6). f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as in Subcase I. Subcase III: k ≡ 3(mod6). f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (wk−1 ) = 1. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 0, 1(mod6) = 2; if j ≡ 3, 4(mod6), 1 ≤ j ≤ k, j ̸= k − 1. Subcase IV: k ≡ 4(mod6). f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 0, 5(mod6) = 2; if j ≡ 2, 3(mod6), 1 ≤ j ≤ k. Subcase V: k ≡ 5(mod6). f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as in Subcase I. Case 2: n ≡ 1(mod6). Subcase I: k ≡ 0(mod6). f (u0 ) = 2. f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. f (v0 ) = 0, f (v1 ) = 2, f (vn−2 ) = 0. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 2 ≤ i ≤ n, i ̸= n − 2. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (u0 ) = 0, f (un−1 ) = 1. f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n, i ̸= n − 1. f (v0 ) = 2. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (wk ) = 0. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k − 1. Subcase III: k ≡ 2(mod6). The vertices are labeled same as in Subcase II except for f (wk ) = 0. (Advance online publication: 17 February 2015) IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_01 ______________________________________________________________________________________ Subcase IV: k ≡ 3(mod6). f (wk−1 ) = 1. The remaining vertices are labeled same as in Subcase II except for f (un−1 ) = 1. Subcase V: k ≡ 4(mod6). f (vn ) = 1. The remaining vertices are labeled same as in Subcase II except for f (un−1 ) = 1. Subcase VI: k ≡ 5(mod6). The vertices are labeled same as in Subcase II except for f (wk ) = 0 and f (un−1 ) = 1. Case 3: n ≡ 2(mod6). f (u0 ) = 0, f (v0 ) = 2. Subcase I: k ≡ 0(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (vn ) = 1. f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n − 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6), = 2; if j ≡ 0, 5(mod6) 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (un ) = 1, f (wk ) = 2. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as in Subcase I. Subcase III: k ≡ 2(mod6). f (wk−4 ) = 1. The remaining vertices are labeled same as in Subcase II. Subcase IV: k ≡ 3(mod6). f (wk−5 ) = 1. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as in Subcase I. Subcase V: k ≡ 4(mod6). f (wk ) = 1. The remaining vertices are labeled same as in Subcase IV except for f (wk−5 ) = 1. Subcase VI: k ≡ 5(mod6). f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (w3 ) = 1, f (wk ) = 0. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 1, 2(mod6) = 2; if j ≡ 4, 5(mod6), 1 ≤ j ≤ k − 1, j ̸= 3. Case 4: n ≡ 3(mod6). Subcase I: k ≡ 0(mod6). f (un ) = 2, f (u0 ) = 0. f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n − 1 f (v0 ) = 2. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 4, 5(mod6) = 2; if j ≡ 1, 2(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n f (v0 ) = 0. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase III: k ≡ 2(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (v0 ) = 2. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ j ≤ n. f (wk−3 ) = 1. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k, j ̸= k − 3. Subcase IV: k ≡ 3(mod6). f (wk−4 ) = 1. The remaining vertices are labeled same as in Subcase III except for f (wk−3 ) = 1. Subcase V: k ≡ 4(mod6). f (u0 ) = 2, f (v0 ) = 0 and f (wk ) = 2. The remaining vertices are labeled same as in Subcase II except for f (u0 ) = 0 and f (v0 ) = 2. Subcase VI: k ≡ 5(mod6). The vertices are labeled same as defined in Subcase V except for f (wk ) = 2. Case 5: n ≡ 4(mod6). f (u0 ) = 0, f (v0 ) = 2. Subcase I: k ≡ 0(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (vn ) = 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n − 1. f (wk ) = 0. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k − 1. Subcase II: k ≡ 1(mod6). The vertices are labeled same as in Subcase I except for f (wk ) = 0. Subcase III: k ≡ 2(mod6). f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) (Advance online publication: 17 February 2015) IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_01 ______________________________________________________________________________________ = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (wk ) = 2, f (wk−1 ) = 1. The remaining vertices are labeled same as in Subcase I except for f (wk ) = 0. Subcase IV: k ≡ 3, 4(mod6). The vertices are labeled same as in Subcase III except for f (wk−1 ) = 1. Subcase V: k ≡ 5(mod6). The vertices are labeled same as in Subcase IV except for f (wk ) = 2. Case 6: n ≡ 5(mod6). Subcase I: k ≡ 0(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (v0 ) = 2. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (wk−2 ) = 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k, j ̸= k − 2. Subcase II: k ≡ 1(mod6). f (wk−3 ) = 1, f (wk ) = 2. The remaining vertices are labeled same as in Subcase I except for f (wk−2 ) = 1. Subcase III: k ≡ 2(mod6). f (u0 ) = 2, f (v0 ) = 0, f (wk ) = 2, f (wk−4 ) = 1. The remaining vertices are labeled same as in Subcase I except for f (wk−2 ) = 1. Subcase IV: k ≡ 3(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (vn ) = 2, f (v0 ) = 2. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n − 1. f (wk ) = 2. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 0, 5(mod6) = 2; if j ≡ 2, 3(mod6), 1 ≤ j ≤ k − 1. Subcase V: k ≡ 4(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (v0 ) = 2. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. f (wk ) = 0. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k − 1. Subcase VI: k ≡ 5(mod6). f (u0 ) = 2. f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. f (v0 ) = 0. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. f (wk ) = 0, f (wk−2 ) = 1. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 1, 2(mod6), = 2; if j ≡ 4, 5(mod6), 1 ≤ j ≤ k − 1, j ̸= k − 2. The graph G under consideration satisfies the conditions |vf (i) − vf (j)| ≤ 1 and |ef (i) − ef (j)| ≤ 1 , 0 ≤ i, j ≤ 2 in each case. Hence the graph G under consideration is 3-equitable graph. Theorem 3 The graph obtained by joining two copies of helm graph Hn by a path of arbitrary length is 3-equitable. Proof: Let G be the graph obtained by joining two copies of helm graph Hn by a path Pk of length k − 1. Let u0 be the apex vertex, u1 , u2 , . . . , un be the rim vertices and u′1 , u′2 , . . . , u′n be the pendant vertices of first copy of helm Hn . Similarly let v0 be the apex vertex, v1 , v2 , . . . , vn be the rim vertices and v1′ , v2′ , . . . , vn′ be the pendant vertices of second copy of helm Hn . Let w1 , w2 , . . . , wk be the vertices of path Pk with w1 = u1 and wk = v1 . We define labeling function f : V (G) → {0, 1, 2} as follows. Case 1: n ≡ 0(mod6). f (u0 ) = 0, f (v0 ) = 2. Subcase I: k ≡ 0(mod6). f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (u′i ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (vi′ ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 4, 5(mod6) = 2; if j ≡ 1, 2(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1, 2(mod6). f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n f (vi′ ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as in Subcase I. Subcase III: k ≡ 3(mod6). f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n. f (u′i ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) (Advance online publication: 17 February 2015) IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_01 ______________________________________________________________________________________ = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (vi′ ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (wk−1 ) = 1. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 0, 1(mod6) = 2; if j ≡ 3, 4(mod6), 1 ≤ j ≤ k, j ̸= k − 1. Subcase IV: k ≡ 4(mod6). f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (u′i ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n. f (vi′ ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 0, 5(mod6) = 2; if j ≡ 2, 3(mod6), 1 ≤ j ≤ k. Subcase V: k ≡ 5(mod6). f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. f (vi′ ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as in Subcase I. Case 2: n ≡ 1(mod6). Subcase I: k ≡ 0(mod6). f (u0 ) = 2. f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. f (u′i ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (v0 ) = 0, f (v1 ) = 2, f (vn−2 ) = 1. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 2 ≤ i ≤ n, i ̸= n − 2. f (vi′ ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (u0 ) = 0, f (un−1 ) = 1. f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n, i ̸= n − 1. f (u′n ) = 0. f (u′i ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n − 1. f (v0 ) = 2. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (v2′ ) = 2, f (vn′ ) = 0. f (vi′ ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n − 1, i ̸= 2. f (wk ) = 0. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k − 1. Subcase III: k ≡ 2(mod6). The vertices are labeled same as in Subcase II except for f (v2′ ) = 2, f (wk ) = 0. Subcase IV: k ≡ 3(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (u′i ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (v0 ) = 2. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (vi′ ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. f (wk ) = 0,f (wk−1 ) = 1. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k − 2. Subcase V: k ≡ 4(mod6). f (vn ) = 1, f (v3′ ) = 0. The remaining vertices are labeled same as in Subcase IV except for f (wk−1 ) = 1. Subcase VI: k ≡ 5(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (u′n ) = 0, f (u′n−2 ) = 2. f (u′i ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n − 1, i ̸= n − 2. f (v0 ) = 2. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. ′ f (vn′ ) = 0, f (vn−1 ) = 1. ′ f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n − 2. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k. Case 3: n ≡ 2(mod6). f (u0 ) = 0, f (v0 ) = 2. Subcase I: k ≡ 0(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) (Advance online publication: 17 February 2015) IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_01 ______________________________________________________________________________________ = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (u′i ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. f (vn ) = 1. f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n − 1. f (v3′ ) = 0, f (v4′ ) = 1. f (vi′ ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n, i ̸= 3, i ̸= 4. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (un ) = 1. f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n − 1. f (u′3 ) = 2, f (u′n−2 ) = 1. f (u′i ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n,i ̸= 3, i ̸= n − 2. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (v4′ ) = 1. f (vi′ ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n, i ̸= 4. f (wk ) = 2. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k − 1. Subcase III: k ≡ 2(mod6). f (wk−4 ) = 1. The remaining vertices are labeled same as in Subcase II. Subcase IV: k ≡ 3(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (u′i ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n f (vi′ ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n. f (wk−5 ) = 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k, j ̸= k − 5. Subcase V: k ≡ 4(mod6). ′ ) = 1. f (v1′ ) = 0, f (vn−1 ′ f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 2 ≤ i ≤ n, i ̸= n − 1. The remaining vertices are labeled same as in Subcase IV except for f (wk−5 ) = 1. Subcase VI: k ≡ 5(mod6). f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. f (u′i ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (vn′ ) = 1. f (vi′ ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n − 1. f (w3 ) = 1, f (wk ) = 0. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 1, 2(mod6) = 2; if j ≡ 4, 5(mod6), 1 ≤ j ≤ k − 1, j ̸= 3. Case 4: n ≡ 3(mod6). Subcase I: k ≡ 0(mod6). f (un ) = 2, f (u0 ) = 0. f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n − 1. f (u′i ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (v0 ) = 2. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. f (vi′ ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 4, 5(mod6) = 2; if j ≡ 1, 2(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (u′n ) = 2. f (u′i ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n − 1. f (v0 ) = 2. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. f (vi′ ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase III: k ≡ 2(mod6). f (u0 ) = 2. f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) (Advance online publication: 17 February 2015) IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_01 ______________________________________________________________________________________ = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (u′i ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (v0 ) = 0. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. f (vi′ ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (wk−3 ) = 1. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k, j ̸= k − 3. Subcase IV: k ≡ 3(mod6). f (u0 ) = 0, f (v0 ) = 2, f (w5 ) = 1, f (wk ) = 0. The remaining vertices are labeled same as in Subcase III except for f (wk−3 ) = 1. Subcase V: k ≡ 4(mod6). f (u0 ) = 2, f (v0 ) = 0, f (wk ) = 2. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (v5′ ) = 1. f (vi′ ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n, i ̸= 5. The remaining vertices are labeled same as in Subcase II except for f (u0 ) = 0, f (u′n ) = 2. Subcase VI: k ≡ 5(mod6) The vertices are labeled same as in Subcase V except for f (wk ) = 2. Case 5: n ≡ 4(mod6). f (u0 ) = 0, f (v0 ) = 2. Subcase I: k ≡ 0(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (u′i ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (vn ) = 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n − 1. f (v1′ ) = 0. f (vi′ ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 2 ≤ i ≤ n. f (wk ) = 0. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k − 1. Subcase II: k ≡ 1(mod6). f (u′1 ) = 1. f (u′i ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 2 ≤ i ≤ n. The remaining vertices are labeled same as in Subcase I except for f (v1′ ) = 0, f (wk ) = 0. Subcase III: k ≡ 2(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (u′1 ) = 1. f (u′i ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 2 ≤ i ≤ n. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (v1′ ) = 0. f (vi′ ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 2 ≤ i ≤ n. f (wk ) = 2, f (wk−1 ) = 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k − 2. Subcase IV: k ≡ 3(mod6). f (u′i ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as in Subcase III except for f (v1′ ) = 0, f (wk−1 ) = 1. Subcase V: k ≡ 4(mod6). f (v1′ ) = 1. The remaining vertices are labeled same as in Subcase IV. Subcase VI: k ≡ 5(mod6). The vertices are labeled same as in subcase V except for f (wk′ ) = 2. Case 6: n ≡ 5(mod6). Subcase I: k ≡ 0(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (u′i ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (v0 ) = 2. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (v2′ ) = 0. f (vi′ ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n, i ̸= 2. f (wk−2 ) = 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k,j ̸= k − 2. Subcase II: k ≡ 1(mod6). f (u0 ) = 2, f (v0 ) = 0, f (wk−3 ) = 1, f (wk ) = 2. The remaining vertices are labeled same as in Subcase I except for f (v2′ ) = 0 and f (wk−2 ) = 1. Subcase III: k ≡ 2(mod6). f (u0 ) = 2. f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) (Advance online publication: 17 February 2015) IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_01 ______________________________________________________________________________________ = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (u′i ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. f (v0 ) = 0. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (vi′ ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (w4 ) = 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k, j ̸= 4. Subcase IV: k ≡ 3(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (u′i ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n f (v0 ) = 2, f (vn ) = 2. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n − 1, f (vn′ ) = 0. f (vi′ ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n − 1. f (wk ) = 2. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 0, 5(mod6) = 2; if j ≡ 2, 3(mod6), 1 ≤ j ≤ k − 1. Subcase V: k ≡ 4(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (u′i ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (v0 ) = 2. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. f (vi′ ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k. Subcase VI: k ≡ 5(mod6). f (u0 ) = 2. f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. f (u′i ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (v0 ) = 0. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. f (vi′ ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n. f (wk−2 ) = 1. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 1, 2(mod6) = 2; if j ≡ 4, 5(mod6), 1 ≤ j ≤ k, j ̸= k − 2. The graph G under consideration satisfies the condition |vf (i) − vf (j)| ≤ 1 and |ef (i) − ef (j)| ≤ 1, 0 ≤ i, j ≤ 2 in each case. Hence the graph G under consideration is 3-equitable graph. Theorem 4 The graph obtained by joining two copies of gear graph Gn by a path of arbitrary length is 3equitable. Proof: Let G be the graph obtained by joining two copies of gear graph Gn by path Pk of length k − 1. Let us denote the successive vertices of first copy of gear graph by u0 , u1 , . . . , u2n , where u0 is apex vertex, u1 , u3 , . . . , u2n−1 are rim vertices of wheel and u2 , u4 , . . . , u2n are the vertices inserted between two consecutive rim vertices corresponding to u1 , u3 , . . . , u2n−1 respectively. Similarly let v0 , v1 , . . . , v2n be the successive vertices of second copy of gear graph, where v0 is apex vertex, v1 , v3 , . . . , v2n−1 are rim vertices of wheel and v2 , v4 , . . . , v2n are the vertices inserted between two consecutive rim vertices corresponding to v1 , v3 , . . . , v2n−1 respectively. Let w1 , w2 , . . . , wk be the vertices of path Pk with w1 = u1 and wk = v1 . We define labeling function f : V (G) → {0, 1, 2} as follows. Case 1: n ≡ 0, 3(mod6). f (u0 ) = 0, f (v0 ) = 2. Subcase I: k ≡ 0(mod6). f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ 2n. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ 2n. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 4, 5(mod6) = 2; if j ≡ 1, 2(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1, 2(mod6). f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ 2n. The remaining vertices are labeled same as in Subcase I. Subcase III: k ≡ 3(mod6). f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ 2n. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ 2n. f (wk−1 ) = 1. f (wj ) = 0; if j ≡ 2, 5(mod6) (Advance online publication: 17 February 2015) IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_01 ______________________________________________________________________________________ = 1; if j ≡ 0, 1(mod6) = 2; if j ≡ 3, 4(mod6), 1 ≤ j ≤ k, j ̸= k − 1. Subcase IV: k ≡ 4(mod6). f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ 2n. f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ 2n. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k. Subcase V: k ≡ 5(mod6). f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ 2n. The remaining vertices are labeled same as in Subcase I. Case 2: n ≡ 1, 4(mod6). Subcase I: k ≡ 0(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ 2n. f (v0 ) = 2, f (v2n ) = 1, f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ 2n − 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ 2n. f (v0 ) = 2, f (v2 ) = 1. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ 2n, n ̸= 2. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k. Subcase III: k ≡ 2(mod6). f (u0 ) = 0, f (u2n ) = 1. f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ 2n − 1. f (v0 ) = 2. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ 2n. f (w4 ) = 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase IV: k ≡ 3(mod6). f (u0 ) = 2,f (v0 ) = 0, f (v2n ) = 1. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ 2n − 1. The remaining vertices are labeled same as in Subcase I. Subcase V: k ≡ 4(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ 2n. f (v0 ) = 2, f (v2n−1 ) = 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ 2n, i ̸= 2n − 1. f (wk ) = 0. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 4, 5(mod6) = 2; if j ≡ 1, 2(mod6), 1 ≤ j ≤ k − 1. Subcase VI: k ≡ 5(mod6). f (u0 ) = 2. f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ 2n. f (v0 ) = 0. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ 2n. f (w3 ) = 1, f (wk ) = 0. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 1, 2(mod6) = 2; if j ≡ 4, 5(mod6), 1 ≤ j ≤ k − 1, j ̸= 3. Case 3: n ≡ 2, 5(mod6). f (u0 ) = 0, f (v0 ) = 2. Subcase I: k ≡ 0(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ 2n. f (v2n ) = 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ 2n − 1. f (wk ) = 0. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k − 1. Subcase II: k ≡ 1(mod6). f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ 2n. f (v2n−3 ) = 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ 2n, i ̸= 2n − 3. f (wk ) = 0. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k − 1. Subcase III: k ≡ 2(mod6). All the vertices are labeled same as in Subcase II except for f (wk ) = 0. Subcase IV: k ≡ 3(mod6). f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ 2n. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) (Advance online publication: 17 February 2015) IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_01 ______________________________________________________________________________________ = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ 2n. f (wk−1 ) = 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 0, 5(mod6) = 2; if j ≡ 2, 3(mod6), 1 ≤ j ≤ k, j ̸= k − 1. Subcase V: k ≡ 4(mod6). f (v2n ) = 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ 2n − 1. The remaining vertices are labeled same as in Subcase IV. Subcase VI: k ≡ 5(mod6). f (wj ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ j ≤ k. The remaining vertices are labeled same as in Subcase IV. The graph G under consideration satisfies the conditions |vf (i) − vf (j)| ≤ 1 and |ef (i) − ef (j)| ≤ 1 , 0 ≤ i, j ≤ 2 in each case. Hence the graph G under consideration is 3-equitable graph. Theorem 5 The graph obtained by joining two copies of cycle with one pendant edge by a path of arbitrary length is 3-equitable. Proof: Let G be the graph obtained by joining two copies of cycle with one pendant edge by path Pk of length k − 1. Let us denote the successive vertices of first copy of cycle by u1 , u2 , . . . , un , e = u0 u1 be the pendant edge and u0 be the pendant vertex. Similarly let v1 , v2 , . . . , vn be the successive vertices of second copy of cycle, e′ = v0 v1 be the pendant edge and v0 be the pendant vertex. Let w1 , w2 , . . . , wk be the successive vertices of path Pk with w1 = u1 and wk = v1 . We define labeling function f : V (G) → {0, 1, 2} as follows. Case 1: n ≡ 0(mod6). Subcase I: k ≡ 0(mod6). f (u0 ) = 2. f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (v0 ) = 0. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1, 4(mod6). f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as in Subcase I. Subcase III: k ≡ 2(mod6). f (v0 ) = 1. f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as in Subcase I. Subcase IV: k ≡ 3(mod6). f (v0 ) = 2. The remaining vertices are labeled same as in Subcase III. Subcase V: k ≡ 5(mod6). f (v0 ) = 2. The remaining vertices are labeled same as in Subcase Case 2: n ≡ 1(mod6). Subcase I: k ≡ 0(mod6). f (u0 ) = 2. f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (v0 ) = 0. f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (u0 ) = 1, f (v0 ) = 2. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as in Subcase Subcase III: k ≡ 2(mod6). f (v0 ) = 2. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as in Subcase Subcase IV: k ≡ 3(mod6). f (v0 ) = 0. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (wk−1 ) = 1. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 0, 5(mod6) = 2; if j ≡ 2, 3(mod6), 1 ≤ j ≤ k, j ̸= k − 1. The remaining vertices are labeled same as in Subcase Subcase V: k ≡ 4(mod6). f (u0 ) = 1, f (v0 ) = 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as in Subcase Subcase VI: k ≡ 5(mod6). f (v0 ) = 2. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as in Subcase Case 3: n ≡ 2(mod6). Subcase I: k ≡ 0(mod6). f (u0 ) = 1. f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (v0 ) = 0. f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 1, 4(mod6) (Advance online publication: 17 February 2015) I. I. I. I. IV. V. IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_01 ______________________________________________________________________________________ = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (u0 ) = f (v0 ) = 0. f (v1 ) = f (wk ) = 2. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as Subcase III: k ≡ 2(mod6). f (v0 ) = 2. f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as Subcase IV: k ≡ 3(mod6). f (u0 ) = 2. f (ui ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (v0 ) = 1. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 0, 3(mod6) = 1; if j ≡ 4, 5(mod6) = 2; if j ≡ 1, 2(mod6), 1 ≤ j ≤ k. Subcase V: k ≡ 4(mod6). f (u0 ) = 1, f (v0 ) = 2. f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as Subcase VI: k ≡ 5(mod6). f (v0 ) = 1. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as Case 4: n ≡ 3(mod6). Subcase I: k ≡ 0(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (v0 ) = 1. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as Subcase III: k ≡ 2, 3(mod6). f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) in Subcase I. in Subcase I. in Subcase IV. in Subcase IV. in Subcase I. = 2; if i ≡ 3, 4(mod6), 1 ≤ j ≤ n. The remaining vertices are labeled same as Subcase IV: k ≡ 4(mod6). f (v0 ) = 2. The remaining vertices are labeled same as Subcase V: k ≡ 5(mod6). f (v0 ) = 2. The remaining vertices are labeled same as Case 5: n ≡ 4(mod6). Subcase I: k ≡ 0(mod6). f (u0 ) = 1. f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (v0 ) = 0. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (v0 ) = 2. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 2, 3(mod6) = 2; if i ≡ 0, 5(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as Subcase III: k ≡ 2(mod6). f (v0 ) = 1. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as Subcase IV: k ≡ 3(mod6). f (u0 ) = 2, f (v0 ) = 0. f (vi ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as Subcase V: k ≡ 4(mod6). f (u0 ) = 2. The remaining vertices are labeled same as Subcase VI: k ≡ 5(mod6). f (v0 ) = 2. The remaining vertices are labeled same as Case 6: n ≡ 5(mod6). Subcase I: k ≡ 0(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. f (v0 ) = 2. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 1, 4(mod6) = 1; if j ≡ 2, 3(mod6) = 2; if j ≡ 0, 5(mod6), 1 ≤ j ≤ k. Subcase II: k ≡ 1(mod6). f (u0 ) = 0. f (ui ) = 0; if i ≡ 0, 3(mod6) (Advance online publication: 17 February 2015) in Subcase I. in Subcase II. in Subcase I. in Subcase I. in Subcase I. in Subcase I. in Subcase II. in Subcase I. IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_01 ______________________________________________________________________________________ 1 0 1 1 0 2 2 0 1 0 0 1 Theorem 3, 3-equitable labeling of the graph G obtained by joining two copies of helm graph H6 by path P6 is shown in Fig. 3. It is the case related to n ≡ 0(mod6) and k ≡ 0(mod6). 2 2 2 2 2 0 1 1 2 0 2 0 0 2 0 0 1 2 0 1 0 1 1 2 0 1 0 1 1 1 Fig. 3. : 3-equitable labeling of the graph G obtained by joining two copies of H6 by P6 . Illustration 4 As an illustration of labeling pattern defined in Theorem 4, 3-equitable labeling of the graph G obtained by joining two copies of gear graph G6 by path P6 is shown in Fig. 4. It is the case related to n ≡ 0(mod6) and k ≡ 0(mod6). 1 0 1 1 2 2 2 0 1 2 0 1 1 2 2 1 0 2 0 0 1 0 0 2 2 0 2 1 0 1 Fig. 4. : 3-equitable labeling of the graph G obtained by joining two copies of G6 by P6 . Illustration 5 As an illustration of labeling pattern defined in Theorem 5, 3-equitable labeling of the graph G obtained by joining two copies of cycle C6 with one pendant edge by path P6 is shown in Fig. 5. It is the case related to n ≡ 0(mod6) and k ≡ 0(mod6). 2 0 1 1 2 2 1 0 2 1 1 0 11 0 2 2 01 1 2 2 2 0 Fig. 2. : 3-equitable labeling of the graph G obtained by joining two copies of W8 by P6 . 2 2 0 1 1 0 2 2 0 01 1 0 2 20 2 1 2 2 2 0 Illustration 1 As an illustration of Theorem 1, 3-equitable labeling of the graph G obtained by joining two copies of fan graph F7 by path P9 is shown in Fig. 1. It is the case related to n ≡ 1(mod6) and k ≡ 3(mod6). 2 0 III. ILLUSTRATIONS 0 0 = 1; if i ≡ 4, 5(mod6) = 2; if i ≡ 1, 2(mod6), 1 ≤ i ≤ n. f (v0 ) = 2. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 3, 4(mod6) = 2; if j ≡ 0, 1(mod6), 1 ≤ j ≤ k. Subcase III: k ≡ 2(mod6). f (u0 ) = 1, f (v0 ) = 0. f (vi ) = 0; if i ≡ 1, 4(mod6) = 1; if i ≡ 0, 5(mod6) = 2; if i ≡ 2, 3(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as in Subcase II. subcase IV: k ≡ 3, 4(mod6). f (v0 ) = 1. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 0, 1(mod6) = 2; if i ≡ 3, 4(mod6), 1 ≤ i ≤ n. The remaining vertices are labeled same as in Subcase II. Subcase V: k ≡ 5(mod6). f (u0 ) = 1. f (ui ) = 0; if i ≡ 0, 3(mod6) = 1; if i ≡ 1, 2(mod6) = 2; if i ≡ 4, 5(mod6), 1 ≤ i ≤ n. f (v0 ) = 0. f (vi ) = 0; if i ≡ 2, 5(mod6) = 1; if i ≡ 3, 4(mod6) = 2; if i ≡ 0, 1(mod6), 1 ≤ i ≤ n. f (wk ) = 2. f (wj ) = 0; if j ≡ 2, 5(mod6) = 1; if j ≡ 0, 1(mod6) = 2; if j ≡ 3, 4(mod6), 1 ≤ j ≤ k − 1. The graph G under consideration satisfies the conditions |vf (i) − vf (j)| ≤ 1 and |ef (i) − ef (j)| ≤ 1 , 0 ≤ i, j ≤ 2 in each case. Hence the graph G under consideration is 3equitable graph. 1 Fig. 5. : 3-equitable labeling of the graph G obtained by joining two copies of cycle C6 with one pendant edge by P6 . Fig. 1. : 3-equitable labeling of the graph G obtained by joining two copies of F7 by P9 . IV. CONCLUSION Illustration 2 As an illustration of labeling pattern defined in Theorem 2, 3-equitable labeling of the graph G obtained by joining two copies of wheel graph W8 by path P6 is shown in Fig. 2. It is the case related to n ≡ 2(mod6) and k ≡ 0(mod6). Illustration 3 As an illustration of labeling pattern defined in The research work presented here provide five new results in the theory of 3-equitable labeling of graphs. The entire work is focused on joining two copies of some graph by a path of arbitrary length. In this work two copies of fans, wheels helms, gears and cycle with one pendant edge are considered. (Advance online publication: 17 February 2015) IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_01 ______________________________________________________________________________________ ACKNOWLEDGEMENT The authors are grateful to the anonymous referee for valuable suggestions and comments. R EFERENCES [1] Bapat M. V., Limaye N. B., ”Some families of 3-equitable graphs”, Journal of Combinatorial Mathematics and Combinatorial Computing, 48(2004), pp. 179-196. [2] Cahit I., ”On Cordial and 3-equitable labelings of graphs”, Utilitas Mathematica, 37(1990), pp. 189-198. [3] Chitre A. M. and Limaye N. B., ”On edge-3-equitability of Kn-union of gears”, Journal of Combinatorial Mathematics and Combinatorial Computing, 83 (2012), pp. 129-150. [4] Gallian J. A., ”A dynemic survey of graph labeling”, The Electronics Journal of Combinatorics, ♯DS6, 16(2013), pp. 1-308. 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