chemia - Studia

MAHDIEH AZARI, ALI IRANMANESH, ABOLFAZL TEHRANIAN
(ii) Let { e,
f } ∈ B2
and e = [( u1,
u2
),( v1,
v2
)], f = [( u1,
z2
),( v1,
t2
)] . By
definition of B
2
, z2 ≠ u2, t2
≠ v2
. So due to distance between two vertices in
G [ G ] 1 2
, the distances d (( u , ),( , ) [
1
u2
u1
z2
G1
G2])
and d (( v1,
v2),(
v1,
t2)
G1[
G2])
are either 1 or 2. Therefore,
d e,
f G [ G ]) = 1+
min{ d((
u , u ),( u , 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 (( v1,
v2),(
u1,
z2)
G1[
G2]),
d((
v1,
v2),(
v1,
t2)
G1[
G2])} =
1+ min{ d (( u1 , u2),(
u1,
z2)
G1[
G2]),1,1,
d((
v1,
v2),(
v1,
t2)
G1[
G2]}
= 1+
1 = 2
(iii) Let { e,
f } ∈ B3
and e = [( u1,
u2
),( v1,
v2
)], f = [( u1,
u2
),( z1,
z2
)] .
By the definition of B3
we have z1 ≠ v1
and hence
d e,
f G [ G ]) = 1+
min{ d((
u , u ),( u , u ) G [ G ]), d((
u , u ),( z , z ) G [ ]),
0( 1 2
1 2 1 2 1 2 1 2 1 2 1
G2
d (( v1,
v2),(
u1,
u2)
G1[
G2]),
d((
v1,
v2),(
z1,
z2)
G1[
G2])} =
1
1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1
G1
Let { e,
f } ∈ B4
and e [( u1,
u2
),( v1,
v2
)], f = [( u1,
t2
),( z1,
z2
)]
of B
4
, z1 ≠ v1
, t2 ≠ u2
. So ( v1,
z1
G1
) ≥ 1 (( u1,
u2),(
u1,
t2)
G1[
G2])
≥
+ min{ d ( u , u G ), d(
u , z G ), d(
v , u G ), d(
v , z G )} = d ([ u , v ],[ u , z ] ) (iv)
= . By definition
d and d 1.
Therefore
d
0( e,
f G1[
G2])
= 1+
min{ d((
u1,
u2),(
u1,
t2)
G1[
G2]),
d((
u1,
u2),(
z1,
z2)
G1[
G2]),
d (( v1,
v2),(
u1,
t2)
G1[
G2]),
d((
v1,
v2),(
z1,
z2)
G1[
G2])} =
+ min{ d (( u , u ),( u , t ) G [ G ]),1,1, d(
v , z G )} = 1+
1 = d ([ u , v ],[ u , z ] G ) 1 (v)
1
1 2 1 2 1 2
1 1 1
0 1 1 1 1 1
+
Let { e,
f } ∈ B5
and e [( u1,
u2
),( v1,
v2
)], f = [( z1,
z2
),( t1,
t2
)]
definition of B 5
, z1 ≠ u1
, z1 ≠ v1
, t1 ≠ u1
and
1
v1
[ u 1,
v 1
and [ z , t 1 1]
of G
1
are distinct. Therefore
d
0( e,
f G1[
G2])
= 1+
min{ d((
u1,
u2),(
z1,
z2)
G1[
G2]),
d((
u1,
u2),(
t1,
t2)
G1[
G2]),
= . By the
t ≠ . So the edges ]
d(( v1,
v2),(
z1,
z2)
G1[
G2]),
d((
v1,
v2),(
t1,
t2)
G1[
G2])} =
1+
min{ d ( u1,
z1
G1
), d(
u1,
t1
G1
), d(
v1,
z1
G1
), d(
v1,
t1
G1
)} = d0([
u1,
v1
],[ z1,
t1]
G1
)
and the proof is completed
Now, we consider three subsets C ,C and 1 2
C
3
of the set C as follows:
C1 = {{ e,
f } ∈C
: e = [( u1,
u2),(
u1,
v2)],
f = [( u1,
u2),(
z1,
z2)],
u1,
z1
∈V
( G1
),
where u v , z ∈ V ( )}
2, 2 2
G2
C2 = {{ e,
f } ∈C
: e = [( u1,
u2),(
u1,
v2)],
f = [( u1,
t2),(
z1,
z2)],
u1,
z1
∈V
( G1
),
where u2, v2,
t2,
z2
∈ V ( G2),
t2
≠ u2,
t2
≠ v2}
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