Cryptography Reference
In-Depth Information
2.
for (each pixel B[i;j], 1 6 i 6 h and 1 6 j 6 w) do
2.1 f a
1
= R
1
[i;j]
2.2
for (2 6 k 6 n 1) do
f a
k
= f(R
k
[i;j];a
k1
)
g
2.3
if (B[i;j] = 0) then R
n
[i;j] =random pixel()
else R
n
[i;j] = f(B[i;j];a
n1
) (= f(1;a
n1
) = a
n1
g
3.
output(R
1
;R
2
;:::;R
n
)
Theorem 3 declares the validity of Algorithm 5 and 6 in producing (n;n)-
VCRG.
Theorem 3 Given a secret binary image B,
= fR
1
;R
2
;:::;R
n
g produced
by Algorithms 5 or 6 with respect to B is a set of (n;n)-VCRG of B.
E
The statements in the proof of Theorem 2 can be easily applied to prove
Theorem 3. We omit the details here.
In Algorithm 5, if b = 0, r
n
= f(b;a
n1
) = f(0;a
n1
) = a
n1
, otherwise
(b = 1) r
n
=random pixel( ). Consequently, we obtain
(S
E
[B(0)]) = 1=2
n1
T
by Corollary 3 and
(S
E
[B(1)]) = 1=2
n
because all pixels in R
k
[B(1)]
are random pixels for 1 6 k 6 n. Regarding Algorithm 6, if b
=
0,
r
n
=random pixel(); otherwise (b = 1), r
n
= f(p;a
n1
) = f(1;a
n1
) = a
n1
.
Thus, we have
T
(S
E
[B(0)]) = 1=2
n
since
(R
k
) = 1=2 for
1 6 k 6 n, andT(S
E
[B(1)]) = 0 since the only condition for r
1
r
2
r
n
to let through the light (i.e., r
1
= r
2
= = r
n1
= r
n
= 0) would never
occur
(
once r
1
= r
2
= = r
n1
= 0, we have a
n1
= 0; but b = 1 causes
r
n
= a
n1
= 1). We obtain the following corollary.
T
T
(S
k
[B(0)]) =
T
(S
E
[B(0)]);
(S
E
[B(1)]))
(1=2
n1
; 1=2
n
) or (1=2
n
; 0)
Corollary 4 (
T
T
=
where
= fR
1
;R
2
;:::;R
n
g is produced by Algorithms 5 or 6, respectively,
with respect to B and S
E
= R
1
R
2
R
n
.
E
The light contrasts of Algorithms 4{6 in point of Denition 3 are summa-
rized in
Table 7.6.
It is seen from Table 7.6 that among the three approaches,
the light contrast obtained by Algorithm 4 is the best.
Search WWH ::
Custom Search