italian journal of pure and applied mathematics – n.
33−2014 (191−200)
191
REVERSE MAGIC STRENGTH OF FESTOON TREES
S. Sharief Basha
K. Madhusudhan Reddy
Applied Algebra Division
School of Advanced Sciences
VIT University
Vellore 632 014, Tamilnadu
India
e-mails: Shariefbasha.s@vit.ac.in
drkmsreddy@yahoo.com
Abstract. In this paper, we prove that the reverse super edge-magic strength of some
different festoon trees.
Keywords. graph labeling, reverse super edge-magic labeling, Festoon trees.
1. Introduction
Consider a family of finite number of stars. Arrange them in an array and join one
centre of a star to that of the next one. The tree so obtained is called a f estoon
tree. In ([9]), we defined a reverse magic labeling of a graph G(V, E) as a bijection
f : V ∪ E → {1, 2, 3, ..., v + ²} such that for all edges xy, f (xy) − {f (x) + f (y)}
are the same where v and ² denote the order and size of the graph G. A graph
is said to be A reverse edge-magic labeling f is called reverse super edge-magic
if f (V ) = {1, 2, 3, ..., } and f (E) = {v + 1, v + 2, v + 3, ..., v + ²}. A graph
G is called reverse super edge-magic if there exists a reverse super edge-magic
labeling of G. In ([5]), Hugand introduced the concept of reverse super edge-magic
strength of a graph. A reverse edge-magic labeling of a graph G(V, E) is a bijection
f : V ∪ E → {1, 2, 3, ..., v + ²} such that for all edges xy, f (xy) − {f (x) + f (y)} is
a constant which is denoted by c(f ). The reverse edge-magic strength of a graph
G, rsm(G), is defined as the minimum of all c(f ) where the minimum is taken
over all reverse edge-magic labelings f of G. A reverse magic labeling of a graph
G(V, E) is called reverse super edge-magic labeling of G if f (V ) = {1, 2, ..., v} and
f (E) = {v + 1, v + 2, ..., v + ²}.
In ([5]), the following results have been proved.
192
s. sharief basha, k. madhusudhan reddy
1. rsm(P2n ) = n − 1 , rsm(P2n+1 ) = n
2. rsm(K1,n ) = n − 1
3. rsm(Bn,n ) = n
4. rsm(Pn2 ) = n − 2
5. rsm(C2n+1 ) = n
6. rsm(< K1,n : 2 >) = 2n
7. rsm[(2n + 1)P2 ] = n
Note 1. Let f be a reverse super edge-magic labeling of G with the constant
c(f ). Then, adding all the constants obtained at each edge, we get
X
X
ε c(f ) =
f (e) −
d(v)f (v)
e∈E
v∈V
In this paper, we determined the reverse super edge-magic strength of some different festoon trees.
2. Reverse super edge-magic strength of Festoon trees
In this paper, we obtain the reverse super edge-magic strength of festoon tree
α
X
Tα , n + α vertices of the stars K1,n , ni ≥ 1, 1 ≤ i ≤ α,
ni = n.
i=1
α
− 1.
2
Proof. We prove this theorem by assigning reverse super edge-magic labeling to
Tα , α is even.
Let ui be the center and vi be the set of pendent vertices of ith star. Each
vi has ni vertices, i = 1, 2, ..., α. Thus n1 + n2 + n3 + · · · + nα = n, v = n + α,
ε = n + α − 1. Then the following labeling f is a reverse super edge-magic labeling
of Tα .
Theorem 2.1. rsm(Tα ) = n + α − n1 − n3 − . . . − nα−1 −
α
2
α
f (u2i−1 ) = f (uα ) + n2 + n4 + n6 + · · · + n2i−2 + i, i = 1, 2, 3, ...,
2
f (v2i−1 ) = {n1 + n3 + · · · + n2i−3 + i, ..., n1 + n3 + · · · + n2i−1 + i − 1},
α
i = 1, 2, 3, ...,
2
f (v2i ) = {f (u1 ) + n2 + n4 + . . . + n2i−2 + i, ..., f (u1 ) + n2 + n4
α
+ · · · + n2i + i − 1}, i = 1, 2, 3, ...,
2
f (u2i ) = n1 + n3 + n5 plusn2i−1 + i,
i = 1, 2, 3, ...,
reverse magic strength of festoon trees
193
Figure 1:
This labeling of vertices gives f (x) + f (y) for all edges xy ∈ E to vary from
f (uα ) + 2 to f (uα ) + n + α. Thus, for each edge e = xy, if f (x) + f (y) = f (uα ) + i,
i = 2, 3, ..., n + α, then f (e) = n + α + i. By Note 1, if f is a reverse super
edge-magic labeling of Tα with constant c(f ).
X
X
ε c(f ) =
f (e) −
d(v)f (v)
e∈E
(n + α − 1) c(f ) =
X
(
f (e) −
e∈E
=
X
f (e) −
e∈E
v∈V
α
X
f (v) +
i=1,v∈Vi
α
n X
α
X
)
d(ui )f (ui )
i=1
f (v) + (n1 + 1)f (u1 )
i=1,v∈Vi
o
+ (n2 + 2)f (u2 ) + . . . + (nα−1 + 2)f (uα−1 ) + (nα + 1)f (uα )
=
X
f (e) −
e∈E
α
nh X
i
f (v) + f (u1 ) + f (u2 ) + . . . + f (uα )
i=1,v∈Vi
o
+ n1 f (u1 )+(n2 +1)f (u2 ) + . . . + (nα−1 +1)f (uα−1 )+nα f (uα )
=
X
e∈E
f (e) −
n n+α−1
X
i + n1 [f (uα ) + 1] + (n2 + 1)(n1 + 1)
i=1i
+ (n3 + 1)[f (uα ) + n2 + 2] + (n4 + 1)[n1 + n3 + 2]
¢¤
£
¡α
−1
+ . . . + (nα−2 + 1) n1 + n3 + . . . + nα−3 +
2
£
α¤
+ (nα−1 + 1) f (uα ) + n2 + n4 + . . . + nα−2 +
2
£
α ¤o
+ nα n1 + n3 + . . . + nα−1 +
2
194
s. sharief basha, k. madhusudhan reddy
(n + α − 1)[n + α + 1 + 2n + 2α − 1]
2
n (n + α − 1)(n + α)
£
α − 2¤
−
+ f (uα ) n1 + n3 + . . . + nα−1 +
2
2
+ n1 + (n2 + 1)(n1 + 1) + (n3 + 1)(n2 + 2) + (n4 + 1)(n1 + n3 + 2)
£
¡α
¢¤
+ (n5 + 1)(n2 + n4 + 3) + . . . + (nα−2 + 1) n1 + n3 + . . . + nα−3 +
−1
2 o
£
£
α¤
α¤
+ (nα−1 + 1) n2 + n4 + . . . + nα−2 +
+ nα n1 + n3 + . . . + nα−1 +
2
2
(n + α − 1)[3n + 3α] (n + α − 1)(n + α)
=
−
2
2
n¡
α ¢¡
α − 2¢
− n1 + n3 + . . . + nα−1 +
n1 + n3 + . . . + nα−1 +
2
2
+ n1 + (n3 + 1)(n2 + 2) + (n5 + 1)(n2 + n4 + 3) + . . .
£
α¤
+ (nα−1 + 1) n2 + n4 + . . . + nα−2 +
2
+ (n2 + 1)(n1 + 1) + (n4 + 1)(n1 + n3 + 2) + . . .
£
¡α
¢¤
+ (nα−2 + 1) n1 + n3 + . . . + nα−3 +
−1
2
£
α ¤o
+ nα n1 + n3 + . . . + nα−1 +
2
n¡
α−2 ¢
α ¢¡
n1 +n3 + . . . + nα−1 +
= (n+α−1)[n+α]− n1 +n3 + . . . + nα−1 +
2
2
+ n1 (n1 + n2 + n3 + . . . + nα ) + n3 (n1 + n2 + n3 + . . . + nα ) + . . .
¡α α
¢
α
+ (n1 + n2 + n3 + . . . + nα ) +
+ − 1 (n1 + n3 + . . . + nα−1 )
2
2
2
¢o
¡
α
α¢ ¡
+ 1 + 2 + 3 + ... + − 1
+ 2 + 3 + 4 + ... +
2
2
= (n + α − 1)[n + α]
n¡
α¢
− n1 + n3 + . . . + nα−1 +
(n1 + n3 + . . . + nα−1 + α − 1)
2
α α
α α
o
¢ α
( + 1)
( − 1)
α¡α
+
− 1 + (α − 1) + 2 2
+ 2 2
−1
2 2
2
2
2
n¡
o
α¢
= (n + α − 1)[n + α] − n1 + n3 + . . . + nα−1 +
+ (n + α − 1)
2
=
(n + α − 1)c(f ) = (n + α − 1)(n + α)
o
n¡
α¢
+ (n + α − 1)
− n1 + n3 + . . . + nα−1 +
2
¡
α¢
c(f ) = n + α − n1 + n3 + . . . + nα−1 +
2
α¢
∴ rsm(Tα ) = n + α − (n1 + n3 + . . . + nα−1 +
2
reverse magic strength of festoon trees
195
Note 2. rsm(Tα ) can be increased if the number of vertices in a star at any odd
position is less that the no. of vertices in a star at any even position.
For example, we have rsm(T6 ) = 17 in Fig. 2.
Figure 2: α = 6, n1 = 1, n2 = 4, n3 = 2, n4 = 5, n5 = 3, n6 = 6, rsm(T6 ) = 17
Note 3. rsm(Tα ) can also be reduced if the number of vertices in the star at
any odd position is greater than the number of vertices in the star at any even
position. For example, we have rsm(T6 ) = 8 in Fig. 3
Figure 3: α = 6, n1 = 4, n2 = 1, n3 = 5, n4 = 2, n5 = 6, n6 = 3, rsm(T6 ) = 8
µ
Theorem 2.2. rsm(Tα ) = n + α − n2 − n4 − . . . − nα−1 −
α+3
2
¶
Proof. We prove this theorem by assigning reverse super edge-magic labeling to
Tα , α is odd.
Let ui be the center and vi be the set of pendent vertices of the ith star.
Each vi has ni vertices, i = 1, 2, ..., α. Thus
n1 + n2 + n3 + · · · + nα = n, v = n + α, ε = n + α − 1.
Then, the following labeling f is a reverse super edge-magic labeling of Tα .
196
s. sharief basha, k. madhusudhan reddy
f (u1 ) = 1
(α − 1)
2
(α − 1)
f (u2i−1 ) = f (u1 ) + n2 + n4 + n6 + · · · + n2i−2 + i − 1, i = 2, 3, ...
2
f (v2i−1 ) = {f (uα ) + n1 + n3 + · · · + n2i−3 + i, ...,
α+1
f (uα ) + n1 + n3 + · · · + n2i−1 + i − 1},
i = 1, 2, 3, ...,
2
f (v2i ) = {f (u1 ) + n2 + n4 + · · · + n2i−2 + i, ...,
α−1
f (u1 ) + n2 + n4 + · · · + n2i + i − 1},
i = 1, 2, 3, ...,
.
2
f (u2i ) = f (uα ) + n1 + n3 + n5 + · · · + n2i−1 + i,
i = 1, 2, 3, ...
This labeling of vertices gives f (x) + f (y) for all edges xy ∈ E to vary from
f (uα ) + 2 to f (uα ) + n + α. Thus for each edge e = xy if f (x) + f (y) = f (uα ) + i,
i = 2, 3, ..., n + α then f (e) = n + α − 1.
For example, the reverse super edge-magic labeling of Tα , α = 7 is shown in
Fig. 4.
Figure 4: α=7, n1 =1, n2 = 2, n3 =3, n4 =5, n5 =6, n6 =5, rsm(T7 )=16
By Note 1, if f is a reverse super edge-magic labeling of Tα with constant c(f ).
ε c(f ) =
X
f (e) −
e∈E
(n + α − 1) c(f ) =
X
=
d(v)f (v)
v∈V
(
f (e) −
e∈E
X
X
α
X
f (v) +
f (e) −
e∈E
)
d(ui )f (ui )
i=1
i=1,v∈Vi
α
n X
α
X
f (v) + (n1 + 1)f (u1 ) + (n2 + 2)f (u2 ) + ...
i=1,v∈Vi
o
+ (nα−1 + 2)f (uα−1 ) + (nα + 1)f (uα )
=
X
e∈E
f (e) −
α
nh X
i=1,v∈Vi
i
f (v) + f (u1 ) + f (u2 ) + . . . + f (uα )
o
+ n1 f (u1 ) + (n2 + 1)f (u2 ) + · · · + (nα−1 + 1)f (uα−1 ) + nα f (uα )
reverse magic strength of festoon trees
=
X
e∈E
f (e) −
n n+α−1
X
197
i + n1 f (u1 ) + (n2 + 1)[f (uα ) + n2 + 1]
i=1i
+ (n3 + 1)[f (uα ) + n2 + 1]
+ (n4 + 1)[f (uα ) + n1 + n3 + 2] + (n5 + 1)[f (u1 ) + n2 + n4 + 2] + · · ·
£
¡ α − 1 ¢¤
+ (nα−1 + 1) f (uα ) + n1 + n3 + . . . + nα−2 +
2
£
α − 1 ¤o
+ nα f (u1 ) + n2 + n4 + . . . + nα−1 +
2
n (n + α − 1)(n + α)
X
£
α − 3¤
=
f (e) −
+ f (u1 ) n1 + n3 + . . . + nα +
2
2
e∈E
£
α − 1¤
+ f (uα ) n2 + n2 + . . . + nα−1 +
2
+ (n2 + 1)(n1 + 1) + (n3 + 1)(n2 + 1)
+ (n4 + 1)(n1 + n3 + 2) + (n5 + 1)(n2 + n4 + 2) + . . .
£
¡ α − 1 ¢¤
+ (nα−1 + 1) n1 + n3 + . . . + nα−2 +
2
£
α − 1 ¤o
+ nα n2 + n4 + . . . + nα−1 +
2
= {(n + α + 1) + (n + α + 2) + . . . + (2n + 2α − 1)}
n (n + α − 1)(n + α) ¡
(α − 3) ¢
+ n1 + n3 + . . . + nα +
−
2
2
¡
α + 1 ¢¡
α − 1¢
+ n2 + n4 + . . . + nα−1 +
n2 + n4 + . . . + nα−1 +
2
2
+ (n2 + 1)(n1 + 1) + (n4 + 1)(n1 + n3 + 2) + . . .
£
α − 1¤
+ (nα−1 + 1) n1 + n3 + . . . + nα−2 +
2
+ (n3 + 1)(n2 + 1) + (n5 + 1)(n2 + n4 + 2) + . . .
£
(α − 1) ¤o
+ nα n2 + n4 + . . . + nα−1 +
2
n+α−1
=
[n + α + 1 + 2n + 2α − 1]
2
n (n + α − 1)(n + α) ¡
(α − 3) ¢
−
− n1 + n3 + . . . + nα +
2
2
+ (n2 + n4 + . . . + nα−1 )(n2 + n4 + . . . + nα−1 + α)
¡ α + 1 ¢¡ α − 1 ¢ ¡ α − 1 ¢
+
[n1 + n2 + . . . + nα ]
+
2
2
2
¡
α − 3 ¢o
α − 1¢ ¡
+ 1 + 2 + 3 + ... +
+ 1 + 2 + 3 + ... +
2
2
n¡
n+α−1
(α − 3) ¢
=
[3n + 3α − n − α] − n1 + n3 + . . . + nα +
2
2
¡ α + 1 ¢¡ α − 1 ¢
¡α + 1¢
(n + α) +
+ (n2 + n4 + . . . + nα−1 )(n + α) +
2
2
2
α−1
α−3
α+1
α−1
o
¡ α + 1 ¢ ¡ ( 2 )( 2 ) ( 2 )( 2 ) ¢
+
+
+α
2
2
2
198
s. sharief basha, k. madhusudhan reddy
n¡
n+α−1
(α − 3) ¢
=
[2n + 2α] − n1 + n3 + . . . + nα +
2
2
+ (n2 + n4 + . . . + nα−1 )(n + α)
o
¡ α2 − 1 ¢
¡ α − 1 ¢ α2 − 1 α2 − 4α + 3 ¡ α + 1 ¢
+
+α
+
+
+
(n + α)
2
2
8
2
2
£
α + 1¤
= (n + α) n + α − 1 − n2 − n4 − . . . − nα−1 −
2
n¡
(α − 3) ¢
− n1 + n3 + . . . + nα +
+
2
o
1
+ [2α2 − 2 − 4α2 + 4α2 + α2 − 1 + α2 − 4α + 3]
2
£
α + 1¤
= (n + α) n + α − 1 − n2 − n4 − . . . − nα−1 −
2
£
(α − 3) ¤
− n1 + n3 + . . . + nα +
2
£
¡ α + 1 ¢¤
= (n + α − 1) n + α − 1 − n2 − n4 − . . . − nα−1 −
2
£
¡ α + 1 ¢¤
+ n + α − 1 − n2 − n4 − . . . − nα−1 −
2
£
(α − 3) ¤
− n1 + n3 + . . . + nα +
2
£
¡ α + 1 ¢¤
(n + α − 1)c(f ) = (n + α − 1) n + α − 1 − n2 − n4 − . . . − nα−1 −
2
¡α + 3¢
c(f ) = n + α − n2 − n4 − . . . − nα−1 −
2
¡α + 3¢
rsm[Tα ] = n + α − n2 − n4 − . . . − nα−1 −
2
Note 4. rsm(Tα ) is odd, will also be increased if the number of vertices in the
star at any even position is less than the number of vertices in the star at any odd
position as shown in Fig. 5.
Figure 5: α=7, n1 =4, n2 =1, n3 =5, n4 =2, n5 =5, n6 =3, n7 =6, rsm(T7 )=22
reverse magic strength of festoon trees
199
Note 5. rsm(Tα ) is odd, will also be decreased if the number of vertices in the
star at any even position is greater than the number of vertices in the star at any
odd position as shown in Fig. 6.
Figure 6: α=7, n1 =1, n2 =4, n3 =2, n4 =5, n5 =3, n6 =5, n7 =6, rsm(T6 )=14
3. Observation
Theorem 3.1. rsm (hK1,1 , K1,2 , K1,r i) = r − 2.
Proof. This is obtained by joining centres u, v, w of K1,1 , K1,2 , and K1,r to a
new X. The pendent vertex of the star K1,1 is denoted by u1 ; the pendent
vertices of K1,2 are denoted by v1 , v2 and the pendent vertices of K1,r are denoted
by w1 , w2 , ..., wr . The reverse super edge-magic labeling f of the graph is given
below: f (u1 ) = 1, f (v) = 2, f (w1 ) = 3, f (w) = 4, f (u) = 5, f (v1 ) = 6, f (x) = 7,
f (v2 ) = 8 and f (wi ) = i + 7, i = 2, 3, 4, ..., r, f (wu1 ) = r + 8, f (ww1 ) = r + 9,
f (vv1 ) = r + 10, f (xv) = r + 11, f (vv2 ) = r + 12, f (xw) = r + 13, f (xu) = r + 14,
f (wwi ) = f (wr ) + 3r + 1 + i, i = 2, 3, 4, ... Thus, f (xu) − {f (x) + f (u)} =
r + 14 − {7 + 5} = r + 2.
This observation is illustrated in Fig. 7 for r = 4.
Figure 7:
200
s. sharief basha, k. madhusudhan reddy
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Accepted: 21.04.2014