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On the other hand, if
n
α
<
β
holds, we have
β
α
+
β
,
α
α
+
β
a
k
b
k
(
,
0
,
1
)
.
If
n
r
k
α
and
β
α
hold, we have
(
n
2
n
+
1
)
)
,
(
n
−
1
)(
2
n
+
1
)
+
1
2
n
3
n
a
k
b
k
,
1
,
.
(
+
(
+
)
+
+
n
3
n
1
n
3
n
1
3
n
1
8.4.2 One Boat Case
If the ship can dispatch only one boat, the payoff matrix is shown in Table
8.4
.
Amine
No mine
α
−
β
n
n
Reconnaissance
r
k
(
)+
v
k
−
1
1
−
n
+
v
k
−
1
n
+
1
+
1
No reconnaissance
r
k
(
−
β
)+
v
k
−
1
1
+
v
k
−
1
Tabl e 8. 4
Payoff matrix for Γ
(
k
)
—one boat case
Tab le
8.4
also differs from Table
8.2
in the left upper part. Similar to the explana-
tion above, if the ships dispatch a reconnaissance boat, a stage contains
n
1 slots.
Since there is only one boat, it can find a mine only if it is laid at the first slot of the
stage. Thus, the reward of the ships is
+
. On the contrary, one of the ships
strikes the mine if it is laid at other slots. Thus, the loss of the ships is
α
/
(
n
+
1
)
β
n
/
(
n
+
1
)
.
Next, the game
Γ
(
m
)
for
m
>
1 can be expressed as follows.
r
m
(
α
−
β
n
n
1
)+
Γ
(
m
−
1
)
1
−
n
+
Γ
(
m
−
1
)
Γ
(
m
)=
+
(
−
β
)+
Γ
(
−
)
+
Γ
(
−
)
r
m
m
1
1
m
1
Γ
(
)
=
−
(
β
+
)
/
(
(
α
+
β
)
/
(
+
)+
)
The game value
v
m
for
m
,where
v
1
1
n
r
1
1
r
1
n
1
n
,
can be solved as follows.
val
r
m
(
α
−
β
n
n
+
1
)+
v
m
−
1
1
−
n
+
v
m
−
1
v
m
=
r
m
(
−
β
)+
v
m
−
1
1
+
v
m
−
1
val
r
m
(
α
−
β
n
)
1
−
n
=
v
m
−
1
+
n
+
1
r
m
(
−
β
)
1
1
n
(
r
m
β
+
1
)
=
v
m
−
1
+
−
r
m
(
α
+
β
)
/
(
n
+
1
)+
n
m
k
=
1
n
(
r
k
β
+
1
)
=
m
−
n
.
r
k
(
α
+
β
)
/
(
+
)+
n
1
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