Cryptography Reference
In-Depth Information
a
k
= f
r
1
if k = 1;
(7.12)
f(r
k
;a
k1
)
otherwise.
Then, we nd r
n
according to b and a
n1
by
r
n
= f(b;a
n1
):
(7.13)
It is noticed that formula (7.11) is a special case of formula (7.13) by set-
ting n = 2. After all r
n
's 2 R
n
corresponding to all b's 2 B are computed, we
obtain R
n
. Then,
= fR
1
;R
2
;:::;R
n
g is reported as a set of (n;n)-VCRG
of B. The whole idea is formally illustrated in Algorithm 4.
E
Algorithm 4. Encrypting a secret image into a set of (n;n)-VCRG
Input: an hw binary image B and an integer n
Output:
E
= fR
1
;R
2
;:::;R
n
g constituting (n;n)-VCRG of B
1.
for (1 6 k 6 n 1) do
f generate R
k
as a random grid,
T
(R
k
) = 1=2
g
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
)
//f(x;s) is dened in formula (7.10)
g
2.3
R
n
[i;j] = f(B[i;j];a
n1
)
g
3.
output(R
1
;R
2
;:::;R
n
)
= fR
1
;R
2
;:::;R
n
g generated by Algorithm 4
with respect to B is indeed a set of (n;n)-VCRG of B, we explore more useful
features among multiple random grids in the following.
It is easy to see that two independent random grids produce another ran-
dom grid as they are superimposed even when their light transmission is not
1=2. We formalize this in Lemma 3, which is a generalized version of Lemma
1.
Before we prove that
E
Lemma 3 Given two independent random grids R
1
and R
2
with
T
(R
1
) =
1
and
T
(R
2
) =
2
, R
1
R
2
is a random grid with
T
(R
1
R
2
) =
1
2
.
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