IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_01
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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
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= 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
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= 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
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= 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)
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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)
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= 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)
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= 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)
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= 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)
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= 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)
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= 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)
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= 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)
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= 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
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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.
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