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Let:
⎛
2
⎞
⎠
,
1
4
2
1
+
+
1
2
T
2
T
−
1
2
2
T
T
⎝
M
=
−
(16.20)
and:
⎛
2
⎞
⎠
,
1
4
2
1
1
+
2
T
2
T
+
1
2
+
2
T
⎝
N
=
−
(16.21)
T
The solution to our recurrence relation is:
C
1
M
n
C
2
N
n
q
n
=
+
,
(16.22)
where
C
1
and
C
2
are constants. To determine their value, we use our initial con-
ditions, equations (
16.16
)and(
16.17
). We thus obtain the constants as functions
of
T
:
T
1
+
4
T
T
2
T
2
1
+
4
T
T
2
1
+
T
−
+
C
1
=
2
T
2
1
+
4
T
T
2
T
1
+
4
T
T
2
4
,
(16.23)
−
1
+
+
−
T
1
+
4
T
T
2
T
2
1
+
4
T
T
2
1
+
T
+
−
2
T
2
1
+
4
T
T
2
1
+
4
T
T
2
C
2
=(
−
1
)
T
4
,
(16.24)
1
+
+
+
1. So, though we will not write it out in full, we have obtained
an expression in terms of
T
to express the probability of the hider moving or searcher
searching in any particular period. Using the notation above, and substituting into
∑
Note that
C
1
+
C
2
=
L
i
1
q
i
=
1, we conclude:
=
L
i
=
1
M
n
L
i
=
1
N
n
C
1
+
C
2
=
1
,
(16.25)
C
1
1
C
2
1
M
L
+
1
N
L
+
1
−
−
⇒
+
−
−
=
,
C
1
C
2
1
(16.26)
1
−
M
1
−
N
Since
C
1
,
C
2
,
M
and
N
are all functions of
L
or
T
, this equation implicitly solves for
T
as a function of
L
. It can be shown that
C
1
M
C
2
N
is always positive over the
range of values for
T
we are interested in, and thus
q
n
is always positive and we do
indeed have a valid probability distribution. Since
C
1
,
C
2
,
M
and
N
are all functions
of
L
or
T
, we conclude the following:
Theorem 4.
With C
1
,C
2
, M and N as defined previously, the value, T , of the silent
search-ambush game as a function of L (where K
+
=
1
) is given implicitly by the
C
1
1
C
2
1
equation:
M
L
+
1
N
L
+
1
−
−
+
=
2
,
(16.27)
−
−
1
M
1
N
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