Simply Sequentially Additive Labeling of Some Special Trees

Simply sequentially additive labeling 6505
Subcase 2.2: n ≡ 1 (mod 3).
Let V = V1 ∪ V2 ∪ V3 ∪ V4, where
n− 1
V1 = {v1, v2}, V2 = {v1,i ; 1 ≤ i ≤ m}, V3 = {uk,j ; 1 ≤ k ≤ 3 and 1 ≤ j ≤ 3}
n+ 2
V4 = {uk,1 ; k = 3 } and E = E1 ∪ E2 ∪ E3 ∪ E4, where E1 = {v1v2},
n− 1
E2 = {v1v1,i ; 1 ≤ i ≤ m}, E3 = {v2uk,j ; 1 ≤ k ≤ 3 , 1 ≤ j ≤ 3}and
n+ 2
E4 = {v2uk,j ; k = 3 }.
Define
f(v1) = 1, f(v2) = 3, f(v1,i) = 2(n+i+1);1 ≤ i ≤ m,
n-1
⎧6k
+ j −1
; 1 ≤ k ≤ 3 & 1 ≤ j ≤ 3
f(u k, j)
= ⎨
n+
2
⎩2
; k = 3 & j = 1
f(v1v2) = 4, f(v1v1,i) = 2(n+i)+3 ;1 ≤ i ≤ m,
n-1
⎧6k
+ j+
2 ; 1 ≤ k ≤ 3 & 1 ≤ j ≤ 3
f(v2
u k, j)
= ⎨
n+
2
⎩5
; k = 3 & j = 1
The labels are distinct and satisfy the conditions f(uv) = f(u) + f(v) for each edge
uv ∈ E. Hence Bm,n admits 1-sequentially additive labeling, when m, n are odd
and n ≡ 1 (mod 3).
Subcase 2.3: n ≡ 2 (mod 3).
Let V and E be defined as in case 1.
Define
f(v1) = m+n+1, f(v2) = 1, f(v1,i) = i+1 ; 1 ≤ i ≤ m,
n-1
⎧m
+ 2j ; 1 ≤ j ≤ 2
f(v2,
j)
= ⎨
n+
1
⎩2(m
+ j + 1) ; 2 ≤ j ≤ n
f(v1v2) = m+n+2, f(v1v1,i)= (m+n+2) + i; 1 ≤ i ≤ m,
n-1
⎧m
+ 2j + 1;
1 ≤ j ≤ 2
f(v 2v
2, j)
= ⎨
n+
1
⎩2(m
+ j) + 3 ; 2 ≤ 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 m, n are
odd and n≡2 (mod 3).
In all the cases Bm,n has a 1-sequentially additive labeling. Hence Bm,n is
1-sequentially additive.
In Fig. 3 , we give the 1-sequentially additive labeling for the bistar B7,11.
2
11 v2,2
v2,1
3 v1,2 v1,1
13 v2,3
9
12 v2,4
4 v1,3 22 21
10 15
14
23
16 17 v2,5
v1,4 24 v1 20 v2 18 28
v2,6
5
19 1 29 30
25
31
v1,5
39 33 v2,7
6 26
32
27
37 35
34 v2,8
v1,6 7
38
v1,7 v2,11 36 v2,9
8
v2,10
Fig. 3: B7,11