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1
m n
M
(
x F ,
B 0 )=
min
B
Y M
(
x F ,
B
)=
min
B
M
(
A
,
B
)
Y
A
∈F
1
m n
n
1
m n
n
=
|
||
| =
|
|
min
B
A ∈F
c i
B
L i
A
B
L i
min
B
A ∈F
c i s i
A
B
L i
i
=
1
i
=
1
n
i = 1 c i s i s i m n 1
n
i = 1 c i s i
1
m n
1
m
=
min
B
=
min
s 1 , s 2 ,..., s n
n
where s i = |
B
L i |
,andso
s i =
s . Therefore to obtain the solution of this game
i
=
1
we need to solve the following problem: Let c 1 , c 2 , ..., c n , be positive real numbers
satisfying ( 3.31 ), and s a positive integer. Minimize the quadratic function
n
i = 1 c i x i .
(
,
,...,
)=
f
x 1
x 2
x n
(3.32)
subject to
n
i = 1 x i = s and x i i = 1 , 2 ,..., n
non negative integers
(3.33)
, α
,..., α
,
=
,
,...,
If
n , thenan
optimal strategy for player II is the uniformly concentrated distribution in B 0 ,where
α
n is a solution for this problem and
α
m
i
1
2
i
1
2
B 0 = { (
i
,
j
)
:1
j
α i for i
=
1
,
2
,..,
n
}
and the value of the game is
n
i = 1 c i α
1
m
i
v
=
.
To solve the game of Example 3 we need to minimize ( 3.32 ) subject to ( 3.33 ),
which we do by taking advantage of the particular properties of this problem.
Lemma 5. The minimum of the function f , given by ( 3.32 ), for real values x i satis-
fying
i
1 x i =
s, is achieved when x i takes the following values r i
=
s
r i =
,
i
=
1
,
2
,...,
n
(3.34)
n
j = 1 c 1
c i
j
and for any other x 1 ,x 2 , ..., x n the following equality is satisfied
n
i = 1 c i x i =
n
i = 1 c i r i +
n
i = 1 c i ( x i r i )
2
.
(3.35)
Proof. It is sufficient to prove ( 3.35 ) because from this it follows that the minimum
condition is satisfied. Write
 
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