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chemia - Studia

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COMPUTATION OF THE FIRST EDGE WIENER INDEX OF A COMPOSITION OF GRAPHS<br />

1<br />

1 1 1 1 1 1 1 1 1 1 1 1<br />

1 1<br />

G1<br />

+ min{ d ( u , v G ), d(<br />

u , v G ), d(<br />

u , v G ), d(<br />

u , v G )} = 1+<br />

d(<br />

u , v ) ,<br />

so the proof is completed<br />

In follow, we define five subsets B<br />

1<br />

, B2<br />

, B3,<br />

B4<br />

and B<br />

5<br />

of the set B .<br />

B1 = {{ e,<br />

f } ∈ B : e = [( u1,<br />

u2),(<br />

v1,<br />

v2)],<br />

f = [( u1,<br />

u2),(<br />

v1,<br />

z2)],<br />

u1,<br />

v1<br />

∈V<br />

( G1<br />

),<br />

u v , z ∈ V ( )}<br />

B<br />

B<br />

B<br />

B<br />

2, 2 2<br />

G2<br />

2<br />

= {{ e,<br />

f } ∈ B : e = [( u1,<br />

u2),(<br />

v1,<br />

v2)],<br />

f = [( u1,<br />

z2),(<br />

v1,<br />

t2)],<br />

u1,<br />

v1<br />

∈V<br />

( G1<br />

),<br />

u2, v2,<br />

z2,<br />

t2<br />

∈ V ( G2),<br />

z2<br />

≠ u2,<br />

t2<br />

≠ v2}<br />

3<br />

= e,<br />

f } ∈ B : e = [( u1,<br />

u2),(<br />

v1,<br />

v2)],<br />

f = [( u1,<br />

u2),(<br />

z1,<br />

z2)],<br />

u1,<br />

v1,<br />

z1<br />

∈V<br />

( G<br />

u2, v2,<br />

z2<br />

∈ V ( G2),<br />

z1<br />

≠ v1}<br />

4<br />

= {{ e,<br />

f } ∈ B : e = [( u1,<br />

u2),(<br />

v1,<br />

v2)],<br />

f = [( u1,<br />

t2),(<br />

z1,<br />

z2)],<br />

u1,<br />

v1,<br />

z1<br />

∈V<br />

( G1<br />

u2, v2,<br />

t2,<br />

z2<br />

∈ V ( G2),<br />

z1<br />

≠ v1,<br />

t2<br />

≠ u2}<br />

5<br />

= {{ e,<br />

f } ∈ B : e = [( u1,<br />

u2),(<br />

v1,<br />

v2)],<br />

f = [( z1,<br />

z2),(<br />

t1,<br />

t2)],<br />

u1,<br />

v1<br />

∈V<br />

( G1<br />

),<br />

z1, t1<br />

∈ V ( G1<br />

) −{<br />

u1,<br />

v1},<br />

u2,<br />

v2,<br />

z2,<br />

t2<br />

∈V<br />

( G2)}<br />

{{<br />

1),<br />

It is clear that, each pair of the above sets is disjoint and<br />

U 5 B = B i<br />

.<br />

i=<br />

1<br />

The next Proposition, characterizes d<br />

0( e,<br />

f G1[<br />

G2])<br />

for all{ e,<br />

f } ∈ B .<br />

Proposition 2. Let { e,<br />

f } ∈ B .<br />

(i) If { e,<br />

f } ∈ B1<br />

, then d<br />

0<br />

( e,<br />

f G1[<br />

G2])<br />

= 1<br />

(ii) If { e,<br />

f } ∈ B2<br />

, then d<br />

0<br />

( e,<br />

f G1[<br />

G2])<br />

= 2<br />

(iii) If { e,<br />

f } ∈ B3<br />

, then d<br />

0( e,<br />

f G1[<br />

G2])<br />

= d0([<br />

u1,<br />

v1<br />

],[ u1,<br />

z1]<br />

G1<br />

) ,<br />

where e = [( u1,<br />

u2),(<br />

v1,<br />

v2)],<br />

f = [( u1,<br />

u2),(<br />

z1,<br />

z2)]<br />

(iv) If { e,<br />

f } ∈ B4<br />

, then d<br />

0<br />

( e,<br />

f G1[<br />

G2])<br />

= d0([<br />

u1,<br />

v1],[<br />

u1,<br />

z1]<br />

G1<br />

) + 1,<br />

where e = [( u1,<br />

u2),(<br />

v1,<br />

v2)],<br />

f = [( u1,<br />

t2),(<br />

z1,<br />

z2)]<br />

(v) If { e,<br />

f } ∈ B5<br />

, then d<br />

0( e,<br />

f G1[<br />

G2])<br />

= d0([<br />

u1,<br />

v1<br />

],[ z1,<br />

t1]<br />

G1<br />

) ,<br />

where e = [( u1,<br />

u2),(<br />

v1,<br />

v2)],<br />

f = [( z1,<br />

z2),(<br />

t1,<br />

t2)]<br />

Proof. (i) Let { e,<br />

f } ∈ B1<br />

and e = [( u1,<br />

u2<br />

),( v1,<br />

v2<br />

)], f = [( u1,<br />

u2<br />

),( v1,<br />

z2<br />

)].<br />

Using the definition of d ( e,<br />

) , we have:<br />

0<br />

f<br />

d<br />

0( e,<br />

f G1[<br />

G2])<br />

= 1+<br />

min{ d((<br />

u1,<br />

u2),(<br />

u1,<br />

u2)<br />

G1[<br />

G2]),<br />

d((<br />

u1,<br />

u2),(<br />

v1,<br />

z2)<br />

G1[<br />

G2]),<br />

d(( v1,<br />

v2),(<br />

u1,<br />

u2)<br />

G1[<br />

G2]),<br />

d((<br />

v1,<br />

v2),(<br />

v1,<br />

z2)<br />

G1[<br />

G2])} =<br />

+ min{0,1,1, d((<br />

v , v ),( v , z ) G [ ])} = 1 + 0 = 1.<br />

1<br />

1 2 1 2 1<br />

G2<br />

),<br />

187

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