Biomedical Engineering Reference
In-Depth Information
i
12
( )
Based.on.the.deinition.of. Q i .in.Lemma.4.7,.one.can.have. a
( )
i
11
=
Q
,. a
=
Q
,
1
3
( )
i
21
( )
i
22
a
=
Q
,.and. a
=
Q
..Hence,
4
2
P
(
,
t
1)
Q P
2
(
, )
t
(
Q Q P
) (
, ) (
t P
, )
t Q P
2
(
, )
t
ξ
+
=
ξ
+
+
ξ
ξ
+
ξ
1
1
1
3
4
1
2
2
2
2
2
=
Q P
(
ξ
, )
t
+
(
Q Q P
+
) (
ξ
, )(1
t
P
(
ξ
, ))
t
+
Q
(1
P
(
ξ
, ))
t
1
1
3
4
1
1
2
1
. (4.64)
2
=
(
Q Q Q Q P
+
)
(
ξ
, )
t
+
(
Q Q
+
2
Q P
) (
ξ
, )
t Q
+
1
2
3
4
1
3
4
2
1
2
1
2 (1
=
(
C
+
C P
) (
ξ
, )
t
+
C
C
)
row
col
1
row
col
.
By.iterating.backward,.one.has
t
1
2 (1
1 (
C
+
C
)
t
row
col
P
(
ξ
, )
t
=
(
C
+
C
)
P
(
ξ
, 0)
+
C
C
)
×
1
row
col
1
row
col
(1 (
C
+
C
)
row
col
. (4.65)
1
2
1
2
t
=
+
(
C
+
C
)
P
(
ξ
, 0)
row
col
1
.
t
Based.on.Lemma.4.6,.one.has. C
(
+
C
)
0
.when. t ..From. Equation.
row
col
(4.65) ,. (
.and. (
..Therefore,.the.theorem.is.true.
P ξ
,
∞ =
)
1
2
P
ξ
,
∞ =
) 1
P
(
ξ
,
∞ =
)
1
2
1
2
1
for.all. S ξ .with. o (ξ) = 1.
Now,.it.is.assumed.that.the.theorem.is.true.for.any. S ξ .with. o (ξ).= n .and. M
=
2
n
,.
that.is,.
Y
× .or.
P
(
)
S
ξ ,.where. P ( ξ .and. Y .represent. P i
(
ξ
,
∞ .
)
=
1
1
ξ
=
1
∀ξ ∈
M M
1
i
i
M
and. Y ( ∞ ,.respectively,.for.simplicity.
Consider. any. primary. schemata. competition. set. S ϕ . where. (
. It.
o
ϕ
)
=
n
+
1.
i
is.easily.proved.that.there.exists. n + 1.subsets.of. S ξ .with. o (ξ) = n .and. M
= .
such.that.the.locations.of.the.actual.bits.in. S ξ .are.also.the.locations.of.actual.
bits.in. S ϕ ..Mathematically,.
n
ξ ⊂ ϕ .
S ϕ .can.be.divided.into.two.subgroups.
f
f
( )
( )
L
L
S ϕ .and.
S ϕ ,.which.are.formed.by.
,0
,1
the.following.rules:
The.actual.bits.of.the.first.element.in.
S ϕ .are.all.zeros,.while.the.other.
elements.are.found.by.flipping.the.last.actual.bit.and.changing.the.
other.bits.according.to.Gray.coding.
The. actual. bits. of. the. irst. element. in.
,0
S ϕ . are. all. zeros. except. the. last.
actual. bit,. while. the. other. elements. are. found. by. lipping. the. last.
actual.bit.and.changing.the.other.bits.according.to.Gray.coding.
,1
Remark4.7
Obviously,.it.can.be.proved.that.
ϕ .and.
S
S = ∅
..Moreover,.
S
=
S
S
ϕ
,0
ϕ
,1
ϕ
,0
ϕ
,1
size S
(
)
=
size S
(
)
=
M
.
ϕ
,0
ϕ
,1
 
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