Cryptography Reference
In-Depth Information
and
2
3
11 00 11 00 11 00
11
4
5
:
M
1
=
0
0
00
1
1
0
0
11
| {z }
a
0
| {z }
b
0
| {z }
c
| {z }
d
| {z }
e
| {z }
e
Let m be the pixel expansion, then it is obvious that m = a
0
+ b
0
+ c + d + 2e.
The collections C
0
and C
1
contain all the permutations of the basis matrices
M
0
and M
1
, and hence each has m! share matrices.
The shifted scheme is generated as follows:
Shift the second row of the m! share matrices in C
0
(resp. C
1
) to the left
(resp. right) by r subpixels, and let c
1
;c
2
; ;c
r
be the r-bit string that is
shifted in, where each ci
i
2f0; 1g represents a subpixel. By the above discus-
sion, we get m! shifted share matrices for C
0
(resp. C
1
). Take the share matrix
M
0
2 C
0
as an example, then the shifted share matrix, denoted by M
(r)
, is
0
as follows:
2
3
11 00 11 00 11 00
1
1
M
(r)
0
4
5
;
=
00
0
0
1
1
11
0
0
c
1
c
r
| {z }
a
0
| {z }
b
0
| {z }
c
| {z }
d
| {z }
e
| {z }
e
| {z }
r
where c
1
c
r
of share 2 are the adjacent subpixels of the right pixel that are
shifted in.
By going through all m! share matrices of C
0
and C
1
and all the possible
string of subpixels c
1
c
r
2f0; 1g
r
, where f0; 1g
r
is the set of all the binary
strings of length r, the shifted scheme is generated. Hence, we have:
Theorem 1 The shifted scheme of a DVCS is a PVCS, where the average
contrast of the shifted scheme is =
(mr)e
m
2
(m1)
, where 1 r m 1 is the
number of subpixels by which the share 2 (the second share) is shifted.
Proof: Without loss of generality, we only prove the case when the share 2 is
shifted to the left, which is equivalent to the case when the share 1 (the first
share) is shifted to the right. Note that, swapping rows 1 and 2 corresponds to
swapping the parameters c and d. We observe that, since the share matrices
of the DVCS satisfy condition 3 of Definition 1 and the shifting operation is
the same for matrices in C
0
and C
1
, the share matrices of the shifted scheme
satisfy condition 3 of Definition 2.
First, we prove the case that the share 2 is shifted by one subpixel, and
then we extend it to the case when it is shifted by r subpixels.
When share 2 is shifted by one subpixel, the adjacent right subpixel of
share 2 is shifted in. Let p
1
be the probability that a 1 is shifted in, and p
0
be the probability that a 0 is shifted in. Because the share matrices are all
the permutations of the basis matrices, so p
1
and p
0
have fixed values and
p
1
+ p
0
= 1, i.e., the shifted in subpixel is either 1 or 0. More precisely, by the
above discussion we have p
1
=
a
0
+d+e
m
and p
0
=
b
0
+c+
m
. In the general case
when r subpixels are shifted in, denote p
c
as the probability that a string of
subpixels c is shifted in, we also have
P
c2f0;1g
r
p
c
= 1.
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