Cryptography Reference
In-Depth Information
We obtain
z
k
n
k
p
c
t
k
−
1
(1
t
)
n
−
k
dt
max
A
|
−
p
0
|≤
−
(4.1)
1
2
and thus
z
max
t
k
n
k
1]
t
k
−
1
(1
t
)
n
−
k
.
1
2
p
c
|
−
p
0
|≤
−
−
∈
[0
,
k
1
n
−
1
; hence,
−
=
The maximum is obtained for
t
z
(
k
k
n
k
1)
k
−
1
(
n
k
)
n
−
k
1
2
−
−
p
c
|
−
p
0
|≤
−
.
(
n
−
1)
n
−
1
2
k
−
1
n
−
1
2
n
−
1
2
Let
x
=
n
−
1
−
1. We have
k
−
1
=
(1
+
x
) and
n
−
k
=
(1
−
x
). We have
0
≤
x
≤
1 and
z
1
k
n
k
2
n
−
1
(1
x
)
1
−
x
1
2
n
−
1
2
p
c
x
)
1
+
x
(1
|
−
p
0
|≤
−
+
−
.
k
−
1
and the Stirling approximation, we obtain that this
bound is asymptotically equal to
θ
√
n
n
n
−
1
Note that by using
k
k
=
and so the bound we want to prove is not so
√
2
π
loose.
2
2
x
2
. Hence,
+
x
)
1
+
x
(1
−
x
)
1
−
x
≤
We can easily prove that (1
z
1
2
n
−
1
2
(
n
−
1)
x
2
k
n
k
1
2
p
c
|
−
p
0
|≤
−
.
θ
1
n
θ
+
1
−
≤
−
+
≤
θ
+
n
−
1
+
n
−
1
=
Since
k
1
(
n
1)
z
z
,wehave
x
. Thus,
n
−
1
k
n
k
1
2
n
2
(
n
θ
+
1)
2
p
c
|
−
p
0
|≤
θ
×
×
.
n
−
1
3, we have
k
k
2
n
≤
3
For
n
=
4
; thus,
θ
√
n
1
2
√
3
×
3
4
×
2
(3
θ
+
1)
2
p
c
|
−
p
0
|≤
2
×
.
n
−
1
θ
√
n
and this remains true even for
1
p
c
1
For
θ
≤
2
√
3
, we obtain
|
−
p
0
|≤
2
θ>
2
√
3
.We
now concentrate on
n
≥
4.
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