Cryptography Reference
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same as R
1
. Thus, A
1
is a random grid with
T
(A
1
) =
T
(R
1
) = 1=2. Assume
that the statement holds for the case of j
A
j = u1, that is, A
1
;A
2
;:::;A
u1
are random grids with
T
(A
k
) = 1=2 for 1 6 k 6 u 1. Due to the reasons
that r
u
(2 R
u
) = 0 or 1,
(A
u1
) = 1=2, and for each pixel a
u
2 A
u
, a
u
=
f(r
u
;a
u1
), we obtain that A
u
is also a random grid with
T
T
(A
k
) = 1=2 by
Corollary 2.
According to formula (7.12), for each pixel a
k
2 A
k
,
a
k
= f(r
k
;a
k1
) = f(r
k
; f(r
k1
;a
k2
))
= = f(r
k
; f(r
k1
; f(:::; f(r
2
;a
1
) :::)))
=
f(r
k
; f(r
k1
; f(:::; f(r
2
;r
1
) :::))):
That is, the value of a
k
is determined by its k corresponding pixels
r
1
;r
2
;:::;r
k
for 1 6 k 6 u. From Lemma 6, we know that a
k
2 A
k
for
1 6 k 6 u is a random pixel with
t
(a
k
) =
1
2
.
LetV= fR
i
1
;R
i
2
;:::;R
i
v
g be a set of v random grids randomly selected
from
) where 1 6 v 6 u. Let S
U
(0)(S(0)) denote the area
of transparent pixels in S
U
(S
V
) where S
U
= R
1
R
2
R
u
(S
V
=
R
i
1
R
i
2
R
i
v
). Now we would like to examine the light transmis-
sion of the area of pixels in A
u
corresponding to S
U
(0)(S(0)), referred to as
A
u
[S
U
(0)] (A
u
[S
V
(0)]).
U
, (i.e.
V
U
Lemma 7
(1)
(A
u
[S
U
(0)]) = 1;
T
(A
u
[S
V
(0)]) = 1=2:
(2)
T
Proof
(1) From Lemma 4, we have
(S
U
) = 1=2
U
. Consider pixel s = 0(or
s 2 S
U
(0)). The only chance for s = 0 is that its corresponding pixels
r
1
;r
2
;:::;r
u
should all be transparent, i.e., r
1
= r
2
= = r
u
= 0. Un-
der this circumstance, by formula (7.14) s's corresponding pixel a
u
in A
u
becomes
T
a
u
= f(r
u
;f(r
u1
;f(:::;f(r
2
;r
1
) :::)))
= f(0;f(0;f(:::;f(0; 0) :::)))
=
0:
That is, if s = 0 (or s 2 S
U
(0)). then a
u
= 0. As a result,
(A
u
[S
U
(0)]) = 1.
T
= fR
j
1
;R
j
2
;:::;R
j
w
g where w = u v (i.e.,
fj
1
;j
2
;:::;j
w
g = f1; 2;:::;ugfi
1
;i
2
;:::;i
v
g). From Lemma 5, we know
that both S
V
and S
W
are random grids with
(2) Let
U
V
=
W
(S
W
) =
1=2
uv
. Since the order of random grids does not affect the result of super-
imposition, without losing generality, let (j
1
;j
2
;:::;j
w
) = (1; 2;:::;w) and
(i
1
;i
2
;:::;i
v
) = (w + 1;w + 2;:::;u). That is,
(S
V
) = 1=2
v
and
T
T
W
= fR
1
;R
2
;:::;R
w
g and
V
= fR
w+1
;R
w+2
;:::;R
u
g. (This can easily be done by renaming all of the
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