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finds a new point, he is detected and cannot carry out his mission. This game has a
trivial solution when
s
In fact, player II can choose
B
filling up two columns,
which is an optimal strategy for him and the value of the game is zero. The game
satisfies hypothesis of Theorem
2
. Therefore, an optimal strategy for player I is the
uniform distribution on
X
≥
2
m
.
. To obtain an optimal strategy for player II we must
determine
B
0
in such a way that
=
F
1
m
n
M
(
x
F
,
B
0
)=
min
B
Y
M
(
x
F
,
B
)=
min
B
M
(
A
,
B
)
∑
∈
∈
Y
A
∈F
n
i
=
1
s
i
1
m
n
1
m
n
j
=
i
(
m
−
s
j
)
,
=
min
f
(
s
1
,
s
2
,...,
s
n
)=
min
s
1
,
s
2
,...,
s
n
s
1
,
s
2
,...,
s
n
n
∑
i
=
where
s
i
=
|
B
∩
L
i
|
and
s
i
=
s
. And the value of the game is given by
M
(
x
F
,
B
0
)
.
1
To solve the game when
s
<
2
m
, we have to solve the following problem. Given
integers
s
and
m
such that
s
<
2
m
, find the minimum of the function
n
i
=
1
s
i
j
=
i
(
m
−
s
j
)
,
f
(
s
1
,
s
2
,...,
s
n
)=
(3.14)
for
s
1
,
s
2
,...,
s
n
integers satisfying
n
i
=
1
s
i
=
s
.
0
≤
s
i
≤
m
(3.15)
In the following
[
x
]
denote the integer part of
x
. To obtain the solution for the
OEIG when
n
=
2 we will need the next result.
Lemma 1.
If n
=
2
, the minimum of the function f
=
f
(
s
1
,
s
2
)
defined by (
3.14
)is
obtained with the values
s
1
=
s
−
[
s
/
2
]
,
s
2
=[
s
/
2
]
.
Proof.
Given integers
s
1
,
s
2
satisfying (
3.15
), the symmetry of the function (
3.14
)
allows us to assume that
s
1
≥
s
2
We will prove that either
s
1
=
s
1
and
s
2
=
s
2
or
f
(
s
1
,
s
2
)
is not the minimum of
f
.
If
s
1
−
s
2
=
0or
s
1
−
s
2
=
1, then
s
i
should be equal to
s
i
for
i
=
1
,
2. If
s
1
−
s
2
≥
2,
we can build
s
1
=
s
2
=
−
,
+
s
1
1
s
2
1
and compute the difference
s
1
,
s
2
)=(
f
(
s
1
,
s
2
)
−
f
(
s
1
−
s
2
−
1
)
>
0
,
therefore
f
(
s
1
,
s
2
)
is not a minimum and the proof is complete.
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