Simply Sequentially Additive Labeling of Some Special Trees

6506 K. Manimekalai, J. Baskar Babujee and K. Thirusangu
In the following theorem we prove is a 1-sequentially additive
graph.
Theorem 2.2: The tree is 1-sequentially additive.
Proof: Let V be the vertex set and E be the edge set of . Then |V| =
2n+3, |E| = 2n+2. We define bijection f : V ∪ E → {1, 2, …, 4n+5} as follows.
Case 1: n is even.
Let V = V1 ∪ V2 ∪ V3 , where V1 = {v1, v2, u0}, V2 = {v1,i ; 1 ≤ i ≤ n},
V3 = {v2,j ; 1 ≤ j ≤ n}, deg (v1) = dev(v2) = n+1 and deg(u0) = 2 and all the other
vertices are of degree one and E = E1 ∪ E2 ∪ E3, where E1 = {v1u0, v2u0}, E2 =
{v1v1,i ; 1 ≤ i ≤ n}, E3 = {v2v2,j ; 1 ≤ j ≤ n}.
Define
f(v1) = 1, f(v2) = 2, f(u0) = 3, f(v1,i) = 2(i+2) ; 1 ≤ i ≤ n,
n
⎧2(n
+ 2j+
1) ; 1≤
j ≤ 2
f(v2,
j)
= ⎨
n
⎩3
+ 4j;
( 2 + 1)
≤ j ≤ n
f(v1u0) = 4, f(v2u0) = 5,
f(v1v1,i) = 2i +5 ; 1 ≤ i ≤ n,
n
⎧2(n
+ 2j + 2) ; 1 ≤ j ≤ 2
f(v 2 v 2, j)
= ⎨
n
⎩4j
+ 5;
( + 1)
2 ≤ j ≤ n
Case 2: n is odd.
Subcase 2.1: n ≡ 0 (mod 3)
Let V = V1 ∪ V2 ∪ V3 ∪ V4, where V1 = {v1, v2, u0}, V2 = {v1,i ; 1 ≤ i ≤ n},
n n
V3 = {ui,j ; 1 ≤ i ≤ [] 6 & 1 ≤ j ≤ 6}, V4 = {ui,j ; i = [ 6 ] + 1 & 1 ≤ j ≤ 3} and
E = E1 ∪ E2 ∪ E3 ∪ E4 where E1 = {v1u0, v2u0}, E2 = {v1v1,i ; 1 ≤ i ≤ n},
n n
E3 = {v2ui,j ; 1 ≤ i ≤ [] 6 & 1 ≤ j ≤ 6}and E4 = {v2ui,j ; i =[ 6 ] +1 & 1 ≤ j ≤ 3}.
Define
f(v1) = 1, f(u0) = 4, f(v2) = 6,
⎧2
; i = 1
⎪
f(v1, i ) = ⎨11;
i = 2
⎪
⎩2(n
+ i + 2) ; 3 ≤ i ≤ n
n
⎧12i
+ j+
3 ; 1≤
i ≤ [] 6 & 1≤
j ≤ 6
f(u i, j)
= ⎨
n
⎩j
+ 6 ; i = [] 6 + 1 & 1≤
j ≤ 3
f(v1u0) = 5, f(v2u0) = 10,
⎧3
; i = 1
⎪
f(v1 v1,
i ) = ⎨12
; i = 2
⎪
⎩2(n
+ i) + 5 ; 3 ≤ i ≤ n
n
f(v2ui,j) = 12i +j + 9 ; 1≤ i ≤ [] 6 & 1 ≤ j ≤ 6
f(v2ui,j) = j + 12 ; i = []
6
n +1 & 1 ≤ j ≤ 3.
Subcase 2.2: (i) n ≡ 1 (mod 3) and n ≠1.
Let V and E be defined as in case 1.