Cryptography Reference
In-Depth Information
9.4.2.3
The Scheme by Yang, Wang, and Chen
Yang, Wang, and Chen [14] proposed a different method to construct a VCS
with reversing starting from any perfect black VCS with pixel expansion m.
In their scheme each participant receives m h + 1 shares, where the rst
one corresponds to the one obtained by the underlying perfect black VCS,
whereas the i-th share is obtained by cyclically shifting the (i 1)-th share
one bit to the right, for each i = 2;:::;mh + 1. Notice that the right shift
operation can be implemented by means of OR and NOT operations, since
OR and NOT represent a complete basis for Boolean functions.
The scheme is shown in Figure 9.5.
Assume there exists a (
Qual
;
Forb
)-VCS with a perfect reconstruction
of black pixels. Let () be a one-bit cyclical right-shift function.
Distributionphase.For each pixel of the secret image, the dealer:
runs the distribution phase of the underlying (
Qual
;
Forb
)-VCS; let
s
1
i
be the share distributed to participant i, for i = 1;:::;n;
for each ` = 2;:::;mh + 1, compute s
`
i
= (s
`
i
);
distributes the (mh + 1)-tuple (s
1
i
;:::;s
mh+
i
) to participant i.
Reconstructionphase. A qualified set fi
1
;:::;i
p
g of participants
reconstructs the secret pixel as follows:
superimpose their shares to get
`
= OR(s
`
i
1
;:::;s
`
i
p
), for ` =
1;:::;mh + 1;
reverse the results of the previous step to obtain
`
, for ` = 1;:::;m
h + 1;
supe
ri
mpose
the
results
of
the
previous
step
to
get =
OR(
1
;:::;
mh+1
);
nally, reverse the result of the previous step to obtain , which is
the reconstructed pixel.
FIGURE 9.5
Yang, Wang, and Chen's ideal contrast VCS with reversing.
The reconstruction phase is the same as that of the schem
e s
hown in
Figure
9.1.
Recall that the computation of the reconstr
u
cte
d pixel corresponds
to
performing m h + 1 AND operations, since = OR(
1
;:::;
mh+1
) =
AND(
1
;:::;
mh+1
). It is easy to see that the construction of Figure 9.5
achieves a perfect reconstruction of both white and black pixels. Indeed, in the
case a white pixel has been shared,
1
contains h white subpixels and mh
black ones. The maximum interval between two 0s in
1
is mh, thus, by
shifting right one bit (mh) times, there is at least an
i
having a subpixel
equal to 0 at position j, for each i = 1;:::;m h + 1 and j = 1;:::;m.
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