Information Technology Reference
In-Depth Information
s
i
(
m
−
s
h
)(
m
−
s
k
)
s
i
(
m
−
s
h
)(
m
−
s
k
)
∑
−
∑
i
P
P
m
−
s
i
m
−
s
i
i
=
h
,
k
=
h
,
k
s
i
=
−
∑
i
s
i
(
s
h
−
s
k
−
)
.
1
P
m
−
=
h
,
k
Then (
3.17
) can be rewritten as
s
i
−
∑
i
(
s
h
−
s
k
−
1
)
P
(
2
s
i
)
m
−
=
h
,
k
and the proof is finished.
To state the following lemma we will write
s
i
=
p
+
1
,
for
i
≤
q
,
(3.18)
s
i
=
p
for
i
>
q
.
where
p
=[
s
/
n
]
and
q
=
s
−
pn
,
(3.19)
Lemma 4.
Let s and m be integers,
0
<
s
<
2
m. The minimum of function f defined
by (
3.14
) for integers s
1
,
s
2
,...,
s
n
satisfying (
3.15
) and
n
1
i
=
2
−
s
i
s
i
≤
2
,
(3.20)
m
−
is achieved with the integers s
i
=
s
i
defined by (
3.18
) and it is equal to
q
−
1
(
n
−
q
−
1
.
(
ms
−
np
(
p
+
1
))(
m
−
p
−
1
)
m
−
p
)
(3.21)
with p and q defined by (
3.19
).
Proof.
Let
S
be the set defined by
2
n
i
=
1
s
i
=
s
,
n
1
i
=
2
−
s
i
S
=
{
s
1
,
s
2
,...,
s
n
}
:
s
i
integers
,
0
≤
s
i
≤
m
,
s
i
≤
m
−
We want to obtain the minimum
min
f
(
s
1
,
s
2
,...,
s
n
)
.
(
s
1
,
s
2
,...,
s
n
)
∈S
Given a set of integers
{
s
1
,
s
2
,...,
s
n
}∈S
, the symmetry of the function (
3.14
)
allows us to assume that
s
1
≥
s
2
≥ ...≥
s
n
.
If
s
1
−
s
n
=
0or
s
1
−
s
n
=
1, then
s
i
should be equal to
s
i
(for each index
i
). If
s
1
,
s
2
,...,
s
n
}∈S
s
1
−
s
n
≥
2, we can build the new set of integers
{
,by
Search WWH ::
Custom Search