Simply Sequentially Additive Labeling of Some Special Trees

6504 K. Manimekalai, J. Baskar Babujee and K. Thirusangu
Theorem 2.1: The bistar Bm,n is 1-sequentially additive.
Proof: G(V,E) = Bm,n . Then G has (m+n+2) vertices and (m+n+1) edges. We
define the bijection f : V ∪ E → {1, 2, …, 2(m+n)+3} as follows:
Case 1: m or n or both m and n are even.
Let n be even.
Let V = V1 ∪ V2 ∪ V3, where V1 = {v1, v2}, V2 = {v1,i ; 1 ≤ i ≤ m},
V3 = {v2,j ; 1≤ j ≤ n}with degree of v1 is (m+1) and degree of v2 is (n+1)
and E = E1 ∪ E2 ∪ E3, where E1 = {v1v2}, E2 = {v1v1,i ; 1 ≤ i ≤ m},
E3 = {v2v2,j ; 1 ≤ j ≤ n}.
Define
f(v1) = 1, f(v2) = 2, f(v1,i) = 2(i+1) ; 1 ≤ i ≤ m
n
⎧2(m
+ 2j) ; 1 ≤ j ≤ 2
f(v2,
j)
= ⎨
n+
2
⎩2(m
− n + 2j) + 1;
2 ≤ j ≤ n
f(v1 v2) = 3
f(v1v1,i) = f(v1) + f(v1,i) = 2i +3 ; 1 ≤ i ≤ m
⎧2(m
+ 2j+
1) ; 1≤
j ≤
f(v v ) = ⎨
+
⎩2(m
− n + 2j) + 3 ; 2 ≤ j ≤ n.
2 n
n
2
2 2, j
The labels are distinct and satisfy the condition f(uv) = f(u) + f(v) for each uv ∈ E.
Hence Bm,n has 1-sequentially additive labeling, when m or n or both m, n are
even.
Case 2: Both m, n are odd.
Subcase 2.1: n ≡ 0 (mod 3).
Let V and E be defined as in case 1.
Define
f(v1) = 1, f(v2) = m+n+1,
m-1
⎧n
+ 2i ; 1 ≤ i ≤ 2
f(v1,
i ) = ⎨
m+
1
⎩2(n
+ i + 1)
; 2 ≤ i ≤ m
f(v2,j) = j+1 ; 1 ≤ j ≤ n.
f(v1 v2) = 3
m-1
⎧n
+ 1+
2i ; 1 ≤ i ≤ 2
f(v1
v1,
i ) = ⎨
m+
1
⎩2(n
+ i) + 3 ; 2 ≤ i ≤ m
f(v2v2,j) = (m+n+2) + j; 1 ≤ j ≤ n.
The labels are distinct and satisfy the conditions f(uv) = f(u) + f(v) for each edge
uv ∈ E. Hence the bistar Bm,n admits 1-sequentially additive labeling, when n is
odd and n ≡ 0 (mod 3). In Fig. 2 we give the 1-sequentially additive labeling for
the bistar B3,3.
v1,2
12
5
v1,3
v1,1
14
13
6
15
v1 8
v2
1 7
Fig. 2: B3,3
9
10
11
v2,1
2
v2,3 4
3
v2,2