chemia - Studia

COMPUTATION OF THE FIRST EDGE WIENER INDEX OF A COMPOSITION OF GRAPHS
C3 = {{ e,
f } ∈ C : e = [( u1,
u2),(
u1,
t2)],
f = [( v1,
v2),(
z1,
z2)],
u1,
v1,
z1
∈V
( G1
),
where u t , v , z ∈ V ( G ), v ≠ u , z ≠ }
2, 2 2 2 2 1 1 1
u1
Clearly, every pair of the above sets is disjoint and U 3 C = C i
.
i=
1
In the following Proposition, we find d e,
f G [ ]) for all { e,
f } ∈ C .
0( 1
G2
Proposition 3. Let { e,
f } ∈ C .
(i) If { e,
f } ∈ C1
, then d
0
( e,
f G1[
G2])
= 1
(ii) If { e,
f } ∈ C2
, then d
0
( e,
f G1[
G2])
= 2
(iii) If { e,
f } ∈ C3
, then
d e,
f G [ ]) = + min{ d(
u , v G ), d(
u , z )}
,
0( 1
G2
1
1 1 1 1 1
G1
= [( u1,
u2
),( u1,
t2
)], f [( v1,
v2
),( z1,
z2
where e =
)]
Proof. (i) Let { e,
f } ∈ C1
and e = [( u1,
u2
),( u1,
v2
)], f = [( u1,
u2
),( z1,
z2
)] .
By definition of d ( e,
) , we have:
0
f
d
0( e,
f G1[
G2])
= 1+
min{ d((
u1,
u2),(
u1,
u2)
G1[
G2]),
d((
u1,
u2),(
z1,
z2)
G1[
G2]),
d(( u , v2),(
u1,
u2)
G1[
G2]),
d((
u1,
v2),(
z1,
z2)
G1[
G2])}
= 1+
min{0,1,1,1} = 1+
0
1
=
(ii) Let { e,
f } ∈ C2
and e = [( u1,
u2
),( u1,
v2
)], f = [( u1,
t2
),( z1,
z2
)] .
By definition of C 2
, t2 ≠ u2
, t2
≠ v2
. Thus, due to the distance between
two vertices in G [ G 1 2]
, the distances d (( u , ),( , ) [
1
u2
u1
t2
G1
G2])
and
d (( u1,
v2),(
u1,
t2)
G1[
G2])
are either 1 or 2. So
d
0( e,
f G1[
G2])
= 1+
min{ d((
u1,
u2),(
u1,
t2)
G1[
G2]),
d((
u1,
u2),(
z1,
z2)
G1[
G2]),
d (( u1,
v2),(
u1,
t2)
G1[
G2]),
d((
u1,
v2),(
z1,
z2)
G1[
G2])} =
1
1 2 1 2 1 2
1 2 1 2 1 2
=
+ min{ d (( u , u ),( u , t ) G [ G ]),1, d((
u , v ),( u , t ) G [ G ],1} = 1+
1 2 .
(iii) Let { e,
f } ∈ C3
and e = [( u1,
u2),(
u1,
t2)],
f = [( v1,
v2),(
z1,
z2)]
. By
definition of C
3
, v1 ≠ u1, z1
≠ u1
. Therefore
d e,
f G [ G ]) = 1+
min{ d((
u , u ),( v , v ) G [ G ]), d((
u , u ),( z , z ) G [ ]),
0( 1 2
1 2 1 2 1 2 1 2 1 2 1
G2
d (( u1,
t2),(
v1,
v2)
G1[
G2]),
d((
u1,
t2),(
z1,
z2)
G1[
G2])} =
1
1 1 1 1 1 1 1 1 1 1 1
G1
+ min{ d ( u , v G ), d(
u , z G ), d(
u , v G ), d(
u , z )} =
1
1 1 1 1 1
G1
+ min{ d(
u , v G ), d(
u , z )} , and the proof is completed
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