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Although the technical details are somewhat involved, the basic idea in the proof
of the next theorem is quite simple. Two sets of intervals
X
and
Y
are defined with
all the members of
X
having length b and all the members of
Y
having length a
.
For each interval J in
X
the set of all those sets in
Y
which are disjoint from J is
denoted by
B
.
It is shown that
B
J has a set of distinct representatives so that the
J
representative of
J together with J can be interpreted as a Defender pure strategy.
The strategy which chooses one of these pure strategies at random enables the result
to be established.
B
Λ + (
Λ (
Theorem 3. Let 2 b
<
1
,
s
a
,
b
)
and q
a
,
b
) ,
then Defender can restrict
Infiltrator to at most 1
G
(
s
,
q
)
where G
(
s
,
q
)
is given by ( 9.4 ).
Λ + ,
Λ .
X and
Y by
Proof. Let 2 b
<
1
,
s
and q
Define
X = {
B i
(
j
)
:1
i
q
,
1
j
s
λ s + λ q
q
}
B i
∪{
(
j
)
: q
+
1
i
s
,
1
j
λ q
q
}
and
Y = {
A i
(
j
)
:1
i
λ s ,
1
j
s
λ s + λ q
q
}
A i
∪{
(
j
)
:
λ s +
1
i
λ q ,
1
j
s
λ s }
X and
Y have the directions of the arrows reversed. It is
Definitions for
easy to see that
|X | = |X | =
λ s = |Y | = |Y |.
s
λ q
q
B 1
B s
A λ s ,
A λ s 1 (
A 1
Note that, for 1
j
λ q
q
,
(
j
) ,...,
(
j
) ,
j
) ,...,
(
j
)
is not
in general a chain in the sense of Sect. 9.4 because the definition of
λ
s in ( 9.2 )
means that B s
can have interior points in common. However it is
easy to generate a chain from it by moving the A i
(
j
)
and A λ s (
j
)
(
j
)
s an appropriate distance to the
right. Similar comments apply to B 1
B q
A λ q ,
A λ q 1 (
A 1
(
j
) ,...,
(
j
) ,
j
) ,...,
(
j
) .
This
B q
B s
Y ∪Y = Y
motivates the definitions for
(
j
)
and
(
j
)
below. Putting
let, for each B i
) ∈X ,
and denoting the interior of an interval I by int I
,
(
j
: B i
{
A
∈ Y
(
j
)
int A
=
0
}
if i
∈{
q
,
s
}
B i
(
j
)=
(9.6)
A λ i (
or B i
{
∈ Y
=
)
(
)
=
}
∈{
,
}.
A
: A
j
j
int A
0
if i
q
s
B i
The
are defined by reversing the directions of the arrows.
We will obtain the existence of a mixed optimal strategy by taking pure strate-
gies comprising B i
(
j
)
(respectively B i
B i
(
j
)
(
j
)
) and a member of
(
j
)
(respectively
B i
(
j
)
). To do this, we will show that, for most cases, the family
B
of all the
B i
B i
has a set of distinct representatives; by Remarks 1 and 2 these
cases will be sufficient to prove the theorem. It is easy to see that
(
j
)
and
(
j
)
|B| =
2
(
s
λ q
q
λ s ) .
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