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0
24
6
24
8
24
12
16
24
18
24
24
24
24
Fig. 9.3
A Defender optimal strategy when
a
=
6
/
24 and
b
=
8
/
24
or
(
1
−
3
ε
)
/
3 can contain at most two of these points; in addition, if an interval of
length
(
1
−
3
ε
)
/
3 contains two of them, the two points must be either
(
1
−
2
ε
)
/
3
and
Using symmetry and a little intuition,
it is not too difficult to deduce that an optimal Infiltrator strategy is given by play-
ing the points 0
(
1
+
ε
)
/
3or
(
2
−
ε
)
/
3and
(
2
+
2
ε
)
/
3
.
,
(
1
−
2
ε
)
/
3
,
(
1
+
ε
)
/
3
,
(
2
−
ε
)
/
3
,
(
2
+
2
ε
)
/
3
,
1 with probabilities
/
,
/
,
/
,
/
,
/
,
/
10
36
3
36
5
36
5
36
3
36
10
36 respectively.
9.6 The Value of the Game
In Sect.
9.4
we showed that an optimal Defender strategy gives rise to a number
of chains of intervals and used this fact to obtain a lower bound for
v
(
a
,
b
)
.
The
lower bound involved only two non-negative integers,
s
and
q
,
which satisfy
s
∈
Λ
+
,
∈
Λ
−
and
G
This suggests that, if the lower bound is attained,
an optimal Defender strategy might be constructible from a number,
x
s
,
q
(
s
,
q
)=
G
.
of chains
having precisely
s
intervals of length
b
and a number,
x
q
,
of chains having precisely
q
intervals of length
b
.
The total number of intervals of length
b
in these chains is
sx
s
+
From the
proof of Theorem
1
, we would expect these two quantities to be equal which implies
x
s
=
α
(
λ
q
−
qx
q
whereas the total number of intervals of length
a
is
λ
s
x
s
+
λ
q
x
q
.
q
)
and
x
q
=
α
(
s
−
λ
s
)
for some positive integer
α
.
For most cases
α
=
1
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