chemia - Studia

MAHDIEH AZARI, ALI IRANMANESH, ABOLFAZL TEHRANIAN
(i) If { e,
f } ∈ A1
, then d
0
( e,
f G1[
G2])
= 1
(ii) If { e,
f } ∈ A2
, then d
0
( e,
f G1[
G2])
= 3
(iii) If { e,
f } ∈ A3
, then d
0
( e,
f G1[
G2])
= 2
(iv) If { e,
f } ∈ A4
, then d
0( e,
f G1[
G2])
= 1+
d(
u1,
v1
G1
) ,
where e = [( u1,
u2
),( u1,
v2
)], f = [( v1,
z2
),( v1,
t2
)]
Proof. (i) Let { e,
f } ∈ A1
and e = [( u1,
u2
),( u1,
v2
)], f = [( u1,
u2
),( u1,
z2
)] .
Due to distance between two vertices in G [ G 1 2]
and by definition of d ( , 0
e f ) , we
have:
d e,
f G [ G ]) = 1+
min{ d((
u , u ),( u , u ) G [ G ]), d((
u , u ),( u , z ) G [ ]),
0( 1 2
1 2 1 2 1 2 1 2 1 2 1
G2
d(( u1 , v2),(
u1,
u2)
G1[
G2]),
d((
u1,
v2),(
u1,
z2)
G1[
G2])}
= 1+
min{0,1,1, d(
v2,
z2
G2)}
= 1+
0 = 1
(ii) Let { e,
f } ∈ A2
and e = [( u1,
u2),(
u1,
v2)],
f = [( u1,
z2),(
u1,
t2)]
.
By definition of the set A
2
, z
2
is adjacent neither to u
2
nor to v
2
in G
2
and this
is also true for t 2
. Therefore,
d e,
f G [ G ]) = 1+
min{ d((
u , u ),( u , z ) G [ G ]), d((
u , u ),( u , t ) G [ ]),
0( 1 2
1 2 1 2 1 2 1 2 1 2 1
G2
d(( u1 , v2
),( u1,
z2
) G1[
G2
]), d((
u1,
v2
),( u1,
t2
) G1[
G2
])} = 1+
min{2,2,2,2} = 3.
(iii) Let { e,
f } ∈ A3
and e = [( u1,
u2
),( u1,
v2
)], f = [( u1,
z2
),( u1,
t2
)]. By
definition of the set A
3
, z2 ∉ { u2
, v2},
t2
∉{
u2
, v2}
.
On the other hand { e,
f } ∉ A2
, so at least one of the following situations occurs:
[ u2 , z
2
] ∈ E(
G2
), [ u2
, t
2
] ∈ E(
G2
), [ v2
, z2
] ∈ E(
G2
) or
v , t ] ∈ E(
) .
[
2 2
G2
This means that, at least one of the distances d (( u1,
u2
),( u1,
z2
) G1[
G2
]),
d (( u1,
u2),(
u1,
t2)
G1[
G2]),
d (( u1,
v2
),( u1,
z2
) G1[
G2
])
or
d (( u1,
v2),(
u1,
t2)
G1[
G2])
is equal to 1. Therefore,
d
0( e,
f G1[
G2])
= 1+
min{ d((
u1,
u2),(
u1,
z2)
G1[
G2]),
d((
u1,
u2),(
u1,
t2)
G1[
G2]),
d (( u1 , v2),(
u1,
z2)
G1[
G2]),
d((
u1,
v2),(
u1,
t2)
G1[
G2])}
= 1+
1 = 2 .
(iv) Let { e,
f } ∈ A4
and e = [( u1,
u2
),( u1,
v2
)], f = [( v1,
z2
),( v1,
t2
)] .
Thus v1 ≠ u1
and
d e,
f G [ G ]) = 1+
min{ d((
u , u ),( v , z ) G [ G ]), d((
u , u ),( v , t ) G [ ]),
0( 1 2
1 2 1 2 1 2 1 2 1 2 1
G2
d (( u1,
v2),(
v1,
z2)
G1[
G2]),
d((
u1,
v2),(
v1,
t2)
G1[
G2])} =
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