Cryptography Reference
In-Depth Information
TABLE 7.2
Encoding b into r
1
and r
2
and results of s = r
1
r
2
by Algorithms 1, 2,
and 3.
b Probability r
1
r
2
s = r
1
r
2
P
rob
(s = 0) (
t
(s))
1
2
1
2
1
2
Algorithm 1
1
2
0
1
2
1
2
1
2
1
2
Algorithm 2
1
4
1
4
1
4
1
4
1
4
1
4
1
4
1
4
Algorithm 3
1
4
1
4
1
2
0
1
2
Proof To prove that R
1
and R
2
constitute a set of (2, 2)-VCRG, we should
validate whether the following two conditions (setting n = 2 in Definition 2)
hold:
1
(1)T(R
1
) =T(R
2
) =
2
; and
(2)
T
(S[B(0)]) >
T
(S[B(1)]) where S = R
1
R
2
.
Let us examine Algorithm 1 first. Let R
1
[B(b)] denote the area of
pixels in R
1
that corresponds to B(b) for b = 0 or 1. Note that R
1
=
R
1
[B(0)] [ R
1
[B(1)] where R
1
[B(0)] \ R
1
[B(1)] = . Since Step 1 in Al-
gorithm 1 composes R
1
as a random grid with
T
(R
1
)
=1/2, we have
T
(R
1
) =
T
(R
1
[B(0)]) =
T
(R
1
[B(1)]) = 1=2. When B[i;j] = 0, we have
R
2
[i;j] = R
1
[i;j]. That means
(R
1
[B(0)]) = 1=2. Moreover,
when B[i;j] = 1;R
2
[i;j] = R
1
[i;j]. By Lemma 2, we have
T
(R
2
[B(0)]) =
T
T
(R
2
[B(1)]) =
( R
1
[B(1)]) = 1
T
(R
1
[B(1)]) = 1 1=2 = 1=2. Due to the facts that
R
2
= R
2
[B(0)] [R
2
[B(1)] and
T
T
(R
2
[B(0)]) =
T
(R
2
[B(1)]) = 1=2, we have
T
(R
2
) = 1=2 by the principle of combination. Therefore both of R
1
and R
2
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