Cryptography Reference
In-Depth Information
Proof For any pixel r
1
2 R
1
and its corresponding pixel r
2
2 R
2
, we have
P
rob
(r
1
= 0) =
1
andP
rob
(r
2
= 0) =
2
. r
1
r
2
is transparent if and only
if both of r
1
and r
2
are transparent, that is,
P
rob
(r
1
r
2
= 0) =P
rob
(r
1
=
P
rob
(r
2
= 0) =
1
2
. Therefore,T(R
1
R
2
) =T(R
1
)T(R
2
) =
1
2
.
The theme of Lemma 3 can be extended for more than two random grids.
0)
Lemma 4 IfU= fR
1
;R
2
;:::;R
u
g is a set of u > 2 independent random
grids withT(R
i
) = for 1 6 i 6 u, then S
U
is a random grid withT(S
U
) =
u
where S
U
= R
1
R
2
R
u
.
Proof We prove by mathematical induction. Consider the case of
V
=
(S
V
) =
2
) is true by Lemma 3. Assume
that the statement holds for the case of
fR
1
;R
2
g. The statement (i.e.,
T
= fR
1
;R
2
;:::;R
u1
g, that is, S
V
V
(S
V
) =
u1
where S
V
= R
1
R
2
R
u1
.
is a random grid with
T
[fR
u
g. S
U
is simply
the superimposed result of S
V
and R
u
, i.e., S
U
= S
V
R
u
. By Lemma 3, we
have
Then, we further consider
U
= fR
1
;R
2
;:::;R
u
g =
V
(R
u
) = (
u1
) =
u
.
Since the superimposition operation is commutative, it is not hard to
obtain that the results of different superimposing orders of R
1
;R
2
;:::;R
u
are
all the same (i.e., S
U
). As a matter of fact, the superimposed result of any
subset of
(S
U
) =
(S
V
R
u
) =
(S
V
)
T
T
T
T
U
is also a random grid as indicated in Lemma 5.
Lemma 5 Let
U
= fR
1
;R
2
;:::;R
u
g be a set up u > 2 independent random
grids and
V
= fR
i
1
;R
i
2
;:::;R
i
v
g
U
where 1 6 i
1
< i
2
< < i
v
6 u
and
T
(R
i
) = for 1 6 i 6 u. S
V
is a random grid with
T
(S
V
) =
v
where
S
V
= R
i
1
R
i
2
:::R
i
v
.
The proof of Lemma 4 can be easily applied to prove Lemma 5. Con-
sider a set of u independent random gridsU= fR
1
;R
2
;:::;R
u
g where
T(R
k
) = 1=2 for 1 6 k 6 u. We may generate a set of u binary images
A= fA
1
;A
2
;:::;A
u
g with respect toUin such a way that each pixel a
k
2 A
k
is defined by formula (7.12), that is,
a
k
= f
r
1
if k = 1;
f(r
k
;a
k1
)
otherwise,
where a
k
2 A
k
and r
k
2 R
k
are corresponding pixels for 1 6 k 6 u. Lemma
6 claims that all of A
1
;A
2
;:::;A
u
are random grids with
T
(A
k
) = 1=2 for
1 6 k 6 u.
Lemma 6 ForA= fA
1
;A
2
;:::;A
u
g generated by formula (7.12) with re-
spect toU= fR
1
;R
2
;:::;R
u
g whereT(R
k
) = 1=2, A
k
is a random grid with
T
(A
k
) = 1=2 for 1 6 k 6 u:
Proof We prove by mathematical induction. For the case of j
j = 1 (i.e.,
u = 1), since a
1
2 A
1
is equal to r
1
2 R
1
by formula (7.12), A
1
is exactly the
A
Search WWH ::
Custom Search