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
.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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