J. Comp. & Math. Sci. Vol.4 (4), 274-281 (2013)
Mean Cordial Labeling of Certain Graphs ALBERT WILLIAM1, INDRA RAJASINGH2 and S. ROY1 1
Department of Mathematics, Loyola College, Chennai, INDIA. 2 School of Advanced Sciences, VIT University, Chennai, INDIA. (Received on: August 10, 2013) ABSTRACT
p vertices and q edges. A vertex labeling f : V (G ) → {0,1, 2} is said to be a mean cordial labeling of G
Let G be a graph with
if it induces an edge labeling
f (u ) + f (v) f * given by 2
suchthat v f ( i ) − v f ( j ) ≤ 1 and e f ( i ) − e f ( j ) ≤ 1 ,
i , j ∈ {0,1, 2} , where v f ( r ) and e f (r ) denote the number of vertices and edges respectively labeled with
r (r = 0,1, 2) . A
graph G is said to be a mean cordial graph if it admits a mean cordial labeling. In this paper, we establish the mean cordial labeling of caterpillar, S (PnٖK1), S ( Bn,n ) , S ( P2 × Pn ) and banana tree. Keywords: Mean cordial labeling, Mean Caterpillar, Banana tree, Path banana tree.
1. INTRODUCTION All graphs in this paper are finite, simple and undirected. The vertex set and edge set of a graph are denoted by V (G) and E (G) respectively. The concept of cordial labeling was introduced by Cahit1. Let f be
cordial
graph,
a function from V (G) to { 0 ,1} and let each edge u v be assigned the label f (u ) − f ( v ) . Then f is a cordial labeling of G if the number of vertices labeled with 0 and the number of vertices labeled with 1 differ by at most 1, and the number of edges labeled with 0 and the number of edges
Journal of Computer and Mathematical Sciences Vol. 4, Issue 4, 31 August, 2013 Pages (202-321)
275
Albert William, et al., J. Comp. & Math. Sci. Vol.4 (4), 274-281 (2013)
labeled with 1 differ by at most 1. For the survey of graph labeling one can refer2. Mean cordial labeling was introduced by Ponraj et al.4, and the definition goes as follows: A vertex labeling f : V ( G ) → {0,1, 2} is said to be a mean cordial labeling of G if it induces an edge labeling f * given by
f (u ) + f (v) such that v f (i) − v f ( j ) ≤ 1 2 and e f (i ) − e f ( j ) ≤ 1 , i , j ∈ {0,1, 2} , where v f ( r ) and e f ( r ) denote the number of
Since the caterpillars are having any number of leaves in its body, it is not easy to prove the mean cordial labeling of all types of caterpillars. So we deal with some particular types of caterpillars and their mean cordial labeling. We should note that not all caterpillars admit mean cordial labeling. The star graph is a caterpillar but not a mean cordial4. Let us label the vertices of caterpillars as it is shown in Fig. (1). Theorem 1: Let vt ∈V (CPn ) be a vertex in the body of CPn such that
vertices and edges respectively labeled with r (r = 0,1, 2) . A graph G is said to be a mean cordial graph if it admits a mean cordial labeling. The mean cordial labeling of path, cycle, star, comb, wheel and complete graph are discussed in4. In this paper, we obtain some new mean cordial graphs.
(i) deg(vt ) = t + 3 when n = 3t (n ≥ 12) (ii) deg(vt ) = t + 1 when n = 3t − 2( n ≥ 7) (iii) deg(vt ) = t + 2 when n = 3t − 1 (n ≥ 8) .
2. MEAN CORDIAL LABELING OF CATERPILLAR
Proof: Let v1 , v2 ,..., vn ∈V (CPn ) . Case 1: deg(vt ) = t + 3 Let n = 3t . Define f : V (CPn ) → {0,1, 2} by
Definition 1: A caterpillar is a path, called the body, where each vertex except the end in the path may have any number of single vertices, called leaves, connected to it. It is denoted by CPn .
If all the vertices are of degree 2 except the end vertices and vt in the body, then CPn is mean cordial.
f (vi ) = 0, t ≤ i ≤ 2t − 1 f (vi ) = 1, 2t ≤ i ≤ 3t − 2 f (vt −1 ) = 1 ; f (vi ) = 2, 1 ≤ i ≤ t − 2 and 3t − 1 ≤ i ≤ 3t . Then v f (0) = v f (1) = v f (2) = t and e f (0) = t − 1, e f (1) = e f (2) = t . Hence f is a mean cordial labeling.
Figure 1: Caterpillar CPn
Case 2: deg(vt ) = t + 1 Let n = 3t − 2 Define f : V (CPn ) → {0,1, 2} by
Journal of Computer and Mathematical Sciences Vol. 4, Issue 4, 31 August, 2013 Pages (202-321)
Albert William, et al., J. Comp. & Math. Sci. Vol.4 (4), 274-281 (2013)
f (vi ) = 0, t ≤ i ≤ 2t − 1 f (vi ) = 1, 2t ≤ i ≤ 3t − 3 ; f (vt −1 ) = 1 f (vi ) = 2, 1 ≤ i ≤ t − 2 ; f (v3t −2 ) = 2 . v f (0) = t , v f (1) = v f (2) = t − 1 and
Then
e f (0) = e f (1) = e f (2) = t . Hence f is a mean cordial labeling.
276
Case 1: n ≡ 0mod3 Let n = 3t Define f : V (CPn ) → {0,1, 2} by
f (vi ) = 0, 1 ≤ i ≤ t ; f (vi ) = 1, t + 1 ≤ i ≤ 2t f (vi ) = 2, 2t + 1 ≤ i ≤ 3t . Then v f (0) = v f (1) = v f (2) = t and e f (0) = t − 1, e f (1) = e f (2) = t . Hence f is a
Case 3: deg(vt ) = t + 2 Let n = 3t − 1 . Define f : V (CPn ) → {0,1, 2} by
mean cordial labeling. Case 2: n ≡ 1mod3 Let n = 3t − 2 . Define f : V (CPn ) → {0,1, 2} by
f (vi ) = 0, t ≤ i ≤ 2t − 1 f (vi ) = 1, 2t ≤ i ≤ 3t − 2 f (vt −1 ) = 1 ; f (vi ) = 2, 1 ≤ i ≤ t − 2 f (v3t −1 ) = 2 .
f (vi ) = 0, 1 ≤ i ≤ t f (vi ) = 1, t + 1 ≤ i ≤ 2t − 1 f (vi ) = 2, 2t ≤ i ≤ 3t − 2 . Then v f (0) = t , v f (1) = v f (2) = t − 1 and e f (0) = e f (1) = e f (2) = t . Hence f is a mean cordial labeling. Case 3: n ≡ 2mod3 Let n = 3t − 1 .
Figure 2: Mean cordial labeling of caterpillar when n = 8
Then
v f (0) = v f (1) = t , v f (2) = t − 1
and
e f (0) = t − 1, e f (1) = t , e f (2) = t − 1 . Hence f
Define f : V (CPn ) → {0,1, 2} by
f (vi ) = 0, 1 ≤ i ≤ t ; f (vi ) = 1, t + 1 ≤ i ≤ 2t f (vi ) = 2, 2t + 1 ≤ i ≤ 3t −1 .
is a mean cordial labeling. Theorem 2: Let CPn be a caterpillar of order n and body of length ≥ 3 suchthat all the vertices in the body have equal number of leaves except the end vertices. Then CPn is mean cordial. Proof: Let v1 , v2 ,..., vn ∈V (CPn ) .
Figure 3: Mean cordial labeling of Caterpillar when n = 11
Journal of Computer and Mathematical Sciences Vol. 4, Issue 4, 31 August, 2013 Pages (202-321)
277
Albert William, et al., J. Comp. & Math. Sci. Vol.4 (4), 274-281 (2013)
Then and
v f (0) = v f (1) = t , v f (2) = t − 1
e f (0) = t − 1, e f (1) = t , e f (2) = t − 1 .
Hence f is a mean cordial labeling. 3. MEAN CORDIAL LABELING OF SUBDIVISON OF GRAPHS
Figure 5: Mean cordial labeling of V ( S ( P4
Theorem 3: The graph S ( Pn cordial graph.
Case 2: m ≡ 1mod3 Let m = 3t − 2 . Define f : V ( S ( Pn K1 )) → {0,1, 2} by
Proof: Let V ( Pn the edges of ( Pn
V ( S ( Pn
K1 ) is a mean
K1 ) = 2n . Subdividing
K1 ) , we get
K1 )) = 4n − 1 = m . Let
v1 , v2 ,..., vm be the vertices of V (S (Pn
Label the vertices of V ( S ( Pn shown in Fig. (4).
K1 )) .
K1 )) as it is
K1 ))
f (vi ) = 0, 1 ≤ i ≤ t f (vi ) = 1, t + 1 ≤ i ≤ 2t − 1 f (vi ) = 2, 2t ≤ i ≤ 3t − 2 . Then v f (0) = t , v f (1) = v f (2) = t − 1 and e f (0) = e f (1) = e f (2) = t . Hence f is a mean cordial labeling. Case 3: m ≡ 2mod3 Let m = 3t − 1 .
Case 1: m ≡ 0mod 3 Let m = 3t . Define f : V ( S ( Pn K1 )) → {0,1, 2} by
K1 )) → {0,1, 2} by
Define f : V ( S ( Pn
f (vi ) = 0, 1 ≤ i ≤ t ; f (vi ) = 1, t + 1 ≤ i ≤ 2t f (vi ) = 2, 2t + 1 ≤ i ≤ 3t − 1 . Then v f (0) = v f (1) = t , v f (2) = t − 1 and
f (vi ) = 0, 1 ≤ i ≤ t ; f (vi ) = 1, t + 1 ≤ i ≤ 2t f (vi ) = 2, 2t + 1 ≤ i ≤ 3t .
e f (0) = t − 1, e f (1) = t , e f (2) = t − 1 . Hence f is a mean cordial labeling. Theorem 4: The graph S ( Bn,n ) is a mean cordial graph. Proof: Let V ( Bn ,n ) = 2n + 2 . Subdividing Figure 4: V ( S ( P4
K1 ))
the
edges
of
Bn,n ,
we
get
and
V ( S ( Bn ,n )) = 4n + 3 = m . Let v1 , v2 ,..., vm be
e f (0) = t − 1, e f (1) = e f (2) = t . Hence f is a
the vertices of V ( S ( Bn,n )) .Label the vertices
mean cordial labeling.
of V ( S ( Bn,n )) as it is shown in Fig. (6).
Then
v f (0) = v f (1) = v f (2) = t
Journal of Computer and Mathematical Sciences Vol. 4, Issue 4, 31 August, 2013 Pages (202-321)
Albert William, et al., J. Comp. & Math. Sci. Vol.4 (4), 274-281 (2013)
Case 1: m ≡ 0mod 3 Let m = 3t .
278
Theorem 5: The graph S ( P2 × Pn ), n ≥ 3 is mean cordial if m ≡ 1mod3 and m ≡ 2mod3 where V (S(P2 × Pn )) = 5n − 2 = m .
Define f : V ( S ( Bn ,n )) → {0,1, 2} by
f (vi ) = 0, 1 ≤ i ≤ t ; f (vi ) = 1, t + 1 ≤ i ≤ 2t f (vi ) = 2, 2t + 1 ≤ i ≤ 3t
Proof: Let V ( P2 × Pn ) = 2n . Subdividing the edges of ( P2 × Pn ) ,we get V(S(P2 ×Pn)) =5n−2 =m and
E(S(P2 ×Pn)) =6n−4.
Label the vertices of S ( P2 × Pn ) as it is in Fig.(7) . Let v1 , v2 ,..., vm be the vertices of S ( P2 × Pn ) . Figure 6: Graph V ( S ( B3,3 ))
Then
v f (0) = v f (1) = v f (2) = t
and
e f (0) = t − 1, e f (1) = e f (2) = t . Hence f is a mean cordial labeling.
Case 2: m ≡ 1mod3 Let m = 3t − 2 . Define f : V ( S ( Bn ,n )) → {0,1, 2} by
f (vi ) = 0, 1 ≤ i ≤ t f (vi ) = 1, t + 1 ≤ i ≤ 2t − 1 f (vi ) = 2, 2t ≤ i ≤ 3t − 2 . v f (0) = t , v f (1) = v f (2) = t − 1 and Then e f (0) = e f (1) = e f (2) = t . Hence f is a mean
Figure 7: Graph S ( P2 × P3 )
Case 1: m ≡ 1mod3 Let m = 3t − 2 Define f : V ( S ( P2 × Pn )) → {0,1,2} by
f (vi ) = 0, 1 ≤ i ≤ t f (vi ) = 1, t + 1 ≤ i ≤ 2t − 1 f (vi ) = 2, 2t ≤ i ≤ 3t − 2 .
cordial labeling.
Case 3: m ≡ 2mod3 Let m = 3t − 1 . Define f : V ( S ( Bn ,n )) → {0,1, 2} by
f (vi ) = 0, 1 ≤ i ≤ t ; f (vi ) = 1, t + 1 ≤ i ≤ 2t f (vi ) = 2, 2t + 1 ≤ i ≤ 3t − 1 . v f (0) = v f (1) = t , v f (2) = t − 1 and Then
Then
e f (0) = t − 1, e f (1) = t , e f (2) = t − 1 . Hence f
e f (0) = t − 1, e f (1) = e f (2) = t . Hence f is a
is a mean cordial labeling.
mean cordial labeling.
Figure 8: Mean cordial labeling of S ( P2 × P3 )
v f (0) = t , v f (1) = v f (2) = t − 1
Journal of Computer and Mathematical Sciences Vol. 4, Issue 4, 31 August, 2013 Pages (202-321)
and
279
Albert William, et al., J. Comp. & Math. Sci. Vol.4 (4), 274-281 (2013)
Case 2: m ≡ 2mod3 Let m = 3t − 1 Define f : V ( S ( P2 × Pn )) → {0,1,2} by
f (vi ) = 0, 1 ≤ i ≤ t ; f (vi ) = 1, t + 1 ≤ i ≤ 2t f (vi ) = 2, 2t + 1 ≤ i ≤ 3t − 1 . v f (0) = v f (1) = t , v f (2) = t − 1 and Then
If n1 = n2 = ... = nk and k = 6, then the total number of vertices n = 6( ni ) + 1, 1 ≤ i ≤ 6 in a banana tree is congruent to 1 modulo 6.
e f (0) = t , e f (1) = e f (2) = t + 1 . Hence f is a mean cordial labeling.
Theorem 6: The graph S ( P2 × Pn ), n ≥ 3 is not mean cordial if m ≡ 0mod 3 where V ( S ( P2 × Pn )) = 5n − 2 = m . Proof: Let V ( S ( P2 × Pn )) = 5n − 2 = m and E ( S ( P2 × Pn )) = 6n − 4 .
Let
m = 3t . Labeling t = m3 vertices of
S ( P2 × Pn ) with 0, we get e f (0) = t − 1 .
Since E ( S ( P2 × Pn )) = 6n − 4 ≥ 3t , it is a contradiction. Thus, the graph is not a mean cordial.
cordial labeling if n1 = n2 = ... = nk ; k = 6 .
BT (n1, n2 ,..., nk ) . Label as it is labeled in Fig. (9).
Definition 2: Let K1, n1 , K1,n2 ,..., K1,nk be a family of stars with the vertex sets V ( K1,ni ) = {ci , ai1 ,..., ain } and deg (ci ) = ni ,1 ≤ i ≤ k . i
A banana tree BT (n1 , n2 ,..., nk ) is a tree
a
Theorem 7: A banana tree BT (n1, n2 ,..., nk ) of order n admits a mean
Proof: Let v1 , v2 ,..., vn be the vertices of
4. MEAN CORDIAL LABELING OF BANANA TREE
obtained by adding a new vertex joining it to a11 , a21 , ..., ak 1 3.
Figure 9: Banana tree BT (4, 4,4,4,4, 4) with 25 vertices
and
In the following theorem, we discuss a particular case of banana tree where n1 = n2 = ... = nk ; k = 6; V (K1,ni ) ≥ 2, 1 ≤ i ≤ 6 .
Let n = 6t + 1 . Define f : V ( BT (n1 , n2 ,..., nk )) → {0,1, 2} by
f (vi ) = 0, 1 ≤ i ≤ 2t + 1 f (vi ) = 1, 2t + 3 ≤ i ≤ 3t + 1 and 3t + 3 ≤ i ≤ 4t + 1 f (vi ) = 2, 4t + 3 ≤ i ≤ 5t + 1 and 5t + 3 ≤ i ≤ 6t + 1
f (v4t + 2 ) = 1 ; f (v5t + 2 ) = 1 ; f (v2t + 2 ) = 2 ; f (v3t + 2 ) = 2
Journal of Computer and Mathematical Sciences Vol. 4, Issue 4, 31 August, 2013 Pages (202-321)
Albert William, et al., J. Comp. & Math. Sci. Vol.4 (4), 274-281 (2013)
280
In the following theorem, we show that a path banana tree is admitting mean cordial labeling if n1 = n2 = ... = nk ; k = 6 ; V (K1,ni ) ≥ 2, 1 ≤ i ≤ 6 . If n1 = n2 = ... = nk and
k = 6 , then the total number of vertices
n = 6((ni ) + 1) + 1, 1 ≤ i ≤ 6 in a path banana tree is congruent to 1 modulo 6. Theorem 8: A path banana tree PBT (n1 , n2 ,..., nk ) of order n admits a
Figure 10: Mean cordial labeling of
mean cordial if n1 = n2 = ... = nk ; k = 6 .
BT (4, 4,4,4,4, 4)
Proof. Let v1 , v 2 , ..., v n be the vertices of PBT ( n1 , n 2 ,..., n k ) . Label the path banana tree as it is labeled in Fig. (11). Let n = 6t + 1 . Define f : V ( PBT (n1 , n2 ,..., nk )) → {0,1, 2} by
Figure 11: Path banana tree PBT (4, 4,4,4,4,4) with 31 vertices
Then
v f (0) = t , v f (1) = v f (2) = t − 1
and
f (vi ) = 0, 1 ≤ i ≤ 2t + 1 f (vi ) = 1, 2t + 4 ≤ i ≤ 3t + 1 and 3t + 4 ≤ i ≤ 4t + 3 f (vi ) = 2, 4t + 4 ≤ i ≤ 5t + 1 and 5t + 4 ≤ i ≤ 6t + 1 f (vi ) = 1, 5t + 2 ≤ i ≤ 5t + 3 f (vi ) = 2, 2t + 2 ≤ i ≤ 2t + 3 f (vi ) = 2, 3t + 2 ≤ i ≤ 3t + 3
e f (0) = e f (1) = e f (2) = t . Hence f is a mean cordial labeling.
Definition 3: Let K1, n1 , K1,n2 ,..., K1,nk be a family of stars with the vertex sets V (K1,ni ) ={ci , ai1,..., aini } and deg(ci ) = ni ,1≤ i ≤ k . A path banana tree PBT (n1 , n2 ,..., nk ) is a tree obtained by adding a new vertex a and joining it by a subdivided edge (i.e a path of length two) to a11 , a21 , ..., ak 1 .
Figure 12: Mean cordial labeling of
PBT (4, 4,4,4,4,4)
Journal of Computer and Mathematical Sciences Vol. 4, Issue 4, 31 August, 2013 Pages (202-321)
281 Then
Albert William, et al., J. Comp. & Math. Sci. Vol.4 (4), 274-281 (2013)
v f (0) = t , v f (1) = v f (2) = t − 1
and
e f (0) = e f (1) = e f (2) = t . Hence f is a mean cordial labeling.
ACKNOWLEDGEMENT This work is supported by the University Grant Commission of India (UGC-MANF No: F1-17.1/2011/MANFCHR-TAM-2135).
REFERENCES 1. I. Cahit, Cordial graphs: A weaker
version of Graceful and Harmonious Graphs, Ars combinatorial, 23, 201-208 (1987). 2. J. A. Gallian, A dynamic survey of graph labeling, Electronic Journal of Combinatorics, 19 (2012). 3. M. Hussain, E. T. Baskoro and Slamin, On super edge-magic total labeling of banana trees, Utilitas Math., 79, 243-251 (2009). 4. R. Ponraj, M. Sivakumar, and M. Sundaram, Mean cordial labeling of graphs, Open Journal of Discrete Mathematics, 2, 145-148 (2012).
Journal of Computer and Mathematical Sciences Vol. 4, Issue 4, 31 August, 2013 Pages (202-321)