Biology Reference
In-Depth Information
This fact guarantees that the constructions below add the maximum number of
vertices allowed by the result.
Case 1: Suppose that T ( w 1 )=
Then, (2 |T ( w 1 ) |−|N ( w 1 ) | ) + =0,and
.
ˆ
( w 1 )
N ( w 1 )
|
V
|
is either 0 or 1 depending on whether
|
|
is even or odd,
N ( w 1 )
respectively. If
|
|
is even (odd), then no nodes are (a single node
ˆ
V ,
E ).Let
N ( w 1 )
( w 1 )
is) added to (
=2 p for the natural
number p .Let k be a natural number so that 2 k > 2 p . List the elements
of
W
,
|
V
|
2 k .
k lexicographically as h
1 , h
2 ,..., h
{−
1 , 1
}
Then, denoting the
ˆ
elements of N ( w 1 )
( w 1 ) as v i for i =1 , 2 ,..., 2 p ,wedefine η and
V
γ by
( η ( v ) , 1 , 1 ,..., 1) ,
∈V
v
i +1 ) ,
v = v i ,
i =1 , 2 ,...p
(1 , 1 ,..., 1 , h
η : ¯
n + k : v
V→{−
1 , 1
}
2 k −i + p ) ,v = v i ,
i = p +1 ,... 2 p
(1 , 1 ,..., 1 , h
and
( γ ( v ) , 2 , 2 ,..., 2) ,
∈W
(2 , 2 ,... 2 , 0 , 0 ,... 0) ,w = w 1 ,
w
n + k : w
γ :
W→{−
2 , 0 , 2
}
where γ ( w 1 ) has k zeros. We mention that Lemma 6.1 is used to guar-
antee that the last k elements of η ( v i ),for i =1 , 2 ,..., 2 p , can be paired
to satisfy the definition of a diversity graph.
Case 2: Suppose T ( w 1 )
=
. The difficulty with this case lies in the fact that
η ( v ) is defined for v
T ( w 1 ). Notice that
N ( w 1 )
T ( w 1 )
T ( w 1 )
N ( w 1 )
(
|
|−
2
|
|
)+(2
|
|−|
|
) +
N ( w 1 )
T ( w 1 )
=(
|
|−
2
|
|
) +
0 ,
in ( N ( w 1 )
which
guarantees
that
there
are
enough
nodes
ˆ
( w 1 ))
T ( w 1 ) to be uniquely paired with the nodes in T ( w 1 ).Let
V
\
ˆ
Z,Z C
be a two set partition of ( N ( w 1 )
( w 1 ))
T ( w 1 ) so that
{
}
V
\
. Notice that the definition of ˆ
T ( w 1 )
( w 1 ) guarantees that both
|
Z
|
=
|
|
V
T ( w 1 )
Z C
|
are even.
List the elements of T ( w 1 ), Z and Z C so that
Z
|
and
|
|
v 1 ,v 2 ,...,v |T ( w 1 ) | }
T ( w 1 )=
{
,
(6.3)
v |T ( w 1 ) | +1 ,v |T ( w 1 ) | +2 ,...,v 2 |T ( w 1 ) | }
Z =
{
, and
(6.4)
v 2 |T ( w 1 ) | +1 ,v 2 |T ( w 1 ) | +2 ,...,v | V ( w 1 ) | }
Z C =
{
.
(6.5)
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