Cryptography Reference
In-Depth Information
Therefore, the reconstructed pixel = AND(
1
;:::;
mh+1
) results in all
white subpixels. On the other hand, in case a black pixel has been
s
hared,
i
contains all black subpixels, for each i = 1;:::;mh+ 1, thus also perfectly
reconstructs the pixel.
The security of the scheme directly follows from the security of the under-
lying VCS.
The reconstruction phase requires a qualified set of p participants to per-
form exactly mh + 2 reversing operations and (mh + 1)p 1 stacking
operations. Finally, notice that in the construction of
Figure 9.5
there is a loss
of resolution, since each pixel in the original image corresponds to m subpix-
els in the reconstructed image, where m denotes the pixel expansion of the
underlying VCS.
n
f1; 2g;f2; 3g;f3; 4g
o
Example 7 Let P = f1; 2; 3; 4g and
0
=
. The basis
matrices S
0
and S
1
in a VCS realizing the access structure
Qual
whose basis
is
0
are:
2
3
2
3
0
1
1
0
1
0
0
1
4
5
4
5
:
0
1
1
1
1
1
1
0
S
0
=
S
1
=
0
1
1
1
1
1
0
1
0
1
0
1
1
0
1
0
The collections C
0
and C
1
are obtained by permuting the columns of the corre-
sponding basis matrix (S
0
for C
0
, and S
1
for C
1
) in all possible ways. Let S
0
be the matrix chosen by the dealer to share a white pixel. Notice that m = 4,
h = 1, and mh+1 = 4. The shares generated for participant 1 are s
1
= 0110,
s
1
= 0011, s
1
= 1001, and s
1
= 1100, whereas, the shares generated for the
participant 2 are s
2
= 0111, s
2
= 1011, s
2
= 1101, and s
2
= 1110. During
the reconstruction phase, the two participants stack their shares and retrieve
1
= OR(s
1
;s
2
) = 0111,
2
= OR(s
1
;s
2
) = 1011,
3
= OR(s
1
;s
2
) = 1101,
and
4
= OR(s
1
;s
2
) = 1110. By applying the
reve
r
sing opera
ti
on t
o each
j
and stacking the results, th
ey
obtain = OR(0111; 1011; 1101; 1110) = 1111.
By reversing , they obtain = 0000, and reconstruct a white pixel.
Let S
1
be the matrix chosen by the dealer to share a black pixel. The
shares generated for participant 1 are s
1
= 1001, s
1
= 1100, s
1
= 0110, and
s
1
= 0011, whereas, the shares generated for the participant 2 are s
2
= 1110,
s
2
= 0111, s
2
= 1011, and s
2
= 1101. During the reconstruction phase, the
two participants stack their shares and retrieve
1
= OR(s
1
;s
2
) = 1111,
2
=
OR(s
1
;s
2
) = 1111,
3
= OR(s
1
;s
2
) = 1111, and
4
= OR(s
1
;s
2
) = 1111.
By applying the
rever
s
ing operat
i
on to
each
j
and stacking the results, they
ob
tain = OR(1111; 1111; 1111; 1111) = 0000. By reversing , they obtain
= 1111, and reconstruct a black pixel.
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