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2
m
[
n
−
rm
2
]
rm
n
+
2
h
h
=
0
.
(7.24)
rm
+
2
h
h
n
Combining (
7.23
)and(
7.24
) we can assert that the cardinality of
F
m
is given by
,
(
7.22
), which finishes the proof.
=
Remark 1.
(a) Theorem
4
and its proof are also valid when
m
3, therefore in this
case the value of expression (
7.22
) is equal to 3
n
.
(b) Due to the symmetry it is clear that the number of functions belonging to
n
F
,
m
n
,
m
divided by
m
. In a similar
for which
A
(
1
)=
r
is equal to the cardinality of
F
n
way, the number of functions belonging to
F
m
for which
A
(
1
)=
r
is equal to
,
3
n
,
m
divided by
m
.
the cardinality of
F
Example 1.
In a warlike situation the guerrilla wants to place
n
mines around the
perimeter of a zone. There is a weak point to penetrate in the ring where the mines
are going to be placed and it will be used by the guerrilla to do the incursion. The
ring is discretized in a grid with
n
columns and
m
rows, considering column 1 as
next to column
n
, and the weak point situated in the midst of them. The grid can be
represented by
L
=
{
1
,
2
,...,
n
}×{
1
,
2
,...,
m
},
and column
i
by the set
L
i
=
{
. The guerrilla will set a mine in
each column. In
L
1
he will set the mine at any of its
m
points, but he moves with
difficulty therefore if one mine is set at point
i
}×{
1
,
2
,...,
m
}
in column
L
i
the next mine
has to be placed in column
L
i
+
1
at one of the points
(
i
,
j
)
(
i
+
1
,
j
−
1
)
,
(
i
+
1
,
j
)
,
(
. Bearing in mind that column 1 is next to column
n
, it follows that,
if the first mine was placed in the point
i
+
1
,
j
+
1
)
(
1
,
j
)
the last mine has to be placed at
one of the points
. Furthermore, the perimeter is pro-
tected by the army, which tries to deactivate as many set mines as it can, beginning
by the the first column and ending in column
L
n
. The payoff to the army is max-
imum if it deactivates all the mines. In the rest of the cases the payoff is equal
to the number of deactivated mines from column
L
1
to the first column contain-
ing a mine that has not been deactivated, because the risk of an accident in this
column is very big. This situation can be modeled by the two-person zero-sum
game
(
n
,
j
−
1
)
,
(
n
,
j
)
,
(
n
,
j
+
1
)
2
n
,
m
(
X
,
Y
,
M
)
where player I is the army, player II the guerrilla,
X
=
Y
=
F
=
{
A
∈ F
m
:
A
(
i
)
∈{
0
,
1
,−
1
},
i
=
1
,...,
n
}
and the payoff function
n
,
h
if
A
(
i
)=
B
(
i
)
for
i
=
1
,
2
,...,
h
,
A
(
h
+
1
)
=
B
(
h
+
1
)
,
M
(
A
,
B
)=
kn
if
A
(
i
)=
B
(
i
)
for
i
=
1
,
2
,...,
n
.
where,
k
is a constant,
k
1.
Figure
7.2
shows a representation of the strategy
>
{
(
1
,
4
)
,
(
2
,
3
)
,
(
3
,
2
)
,
(
4
,
3
)
,
(
5
,
4
)
,
(
6
,
3
)
,
(
7
,
2
)
,
(
8
,
3
)
,
(
9
,
4
)
,
(
10
,
5
)
,
(
11
,
6
)
,
(
12
,
5
)(
13
,
4
)(
14
,
5
)
}
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