Cryptography Reference
In-Depth Information
are independent and with the same 0-or-
1 distrib
ution. Let
z
be the probability that
N
i
=
=
√
LP
c
a
p
c
1. W
e
also define
θ
=
2
z
−
1
,
b
. We thus mean to prove that
|
−
θ
√
n
.Wehave
p
0
|≤
2
n
u
z
u
(1
p
c
z
)
n
−
u
=
−
u
∈
A
and
z
u
(1
n
u
1
2
n
p
c
z
)
n
−
u
−
p
0
=
−
−
.
u
∈
A
We would like to upper-bound
|
p
c
−
p
0
|
over all possible
A
depending on
z
. Since
1
z
and 1
−
z
play a symmetric role, we assume without loss of generality that
z
≥
2
.
1
2
, the result is trivially true, so from now on we assume that
z
1
2
. Since
=
>
For
z
z
u
(1
−
z
)
n
−
u
is an increasing function in terms of
u
we have
z
u
(1
n
u
n
1
2
n
z
)
n
−
u
p
c
|
−
p
0
|≤
−
−
u
=
k
where
k
is the smallest integer
u
such that the difference in parenthesis is nonnegative,
i.e.
n
log
2
−
log(1
−
z
)
k
=
1
+
.
log
z
−
log(1
−
z
)
n
2
Replacing
u
by
in the same expression in parenthesis we obtain a negative difference.
n
+
1
Hence
k
z
, the expression in parenthesis turns out
to be an increasing function in terms of
z
, which is 0 for
z
≥
. Similarly, replacing
u
by
n
.
2
1
2
. Since
z
1
2
=
>
we obtain
n
−
1
2
that
k
≤
n
.
z
. Therefore
≤
k
−
1
≤
(
n
−
1)
z
+
z
.
p
c
1
2
If
n
=
1, we have
k
=
1; thus max
A
|
−
p
0
|=
z
−
and so the result holds. If
3
n
=
2, we have
k
≥
2
; thus
k
=
2 and
z
z
z
1
2
1
2
3
2
1
2
p
c
max
A
|
−
p
0
|=
−
+
≤
−
≥
and so the result holds as well. We now concentrate on
n
3.
We use the following identity.
3
z
n
u
z
u
(1
k
n
k
n
z
)
n
−
u
t
k
−
1
(1
t
)
n
−
k
dt
−
=
−
.
0
u
=
k
3
This identity was found in Ref. [155]. We can easily prove it by taking the derivative in terms of
z
.
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