Visual Cryptography Schemes with Reversing - Visual Cryptography and Secret Image Sharing - page 270

9.4.2.4 Comparisons

The eciency of a VCS with reversing is evaluated according to the follow-
ing parameters: the contrast, the size (expansion) and the number of the
transparencies held by each participant, the number of stacking and reversing
operations, and the size (expansion) of the reconstructed image. In Table 9.1
we summarize and compare the parameters of the constructions described in
Sections 9.3 a nd 9.4.2., which are all based on perfect black VCSs for a general
access structure ( Qual ; Forb ).

TABLE 9.1
Comparison between VCSs with reversing based on perfect black VCSs.
Scheme Contrast Share Number Number Number Secret

exp. of shares OR oper. NOT oper. exp.

Fig. 9.1 Almost m c cp1 c+1 m

ideal

Fig. 9.3 Ideal NO m mp1 m+1 NO

Fig. 9.4 Ideal j 0 j 2 4p 1 4p NO

Fig. 9.5 Ideal m mh+1 (mh+1)p1 mh+2 m

9.4.3 A Construction Using a Nonperfect Black VCS
Yang, Wang and Chen [14] proposed a way to construct an ideal VCS with
reversing when the dierence h` is odd, starting from any VCS (i.e., not
necessarily a perfect black) with pixel expansion m. Each participant receives
m shares, where the first one corresponds to that obtained by the underlying
VCS, whereas, the i-th share is obtained by cyclically shifting the (i 1)-
th share one bit to right, for each i = 2;:::;m. 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.
The construction of Figure 9.5 achieves perfect reconstruction of both
white and black pixels. Indeed, in case a white pixel has been shared, 1
contains h white subpixels and m h black ones. By shifting right one bit
m times, 1 ;:::; m are constructed in such a way that exactly (mh) out
of m have a black subpixel at position j, for each j = 1;:::;m. Therefore,
= XOR( 1 ;:::; m ) is the reco ns tructed white pixel if (m h) is even,
otherwise the reconstructed pixel is . On the other hand, since h` is odd,
i.e. h and ` cannot be both odd or both even, if (mh) is even (odd, resp.)
it holds that m` is odd (even, resp.). Hence, in the case a black pixel has
been shared, 1 contains ` white subpixels and m` black ones. By shifting
right one bit m times, 1 ;:::; m are constructed in such a way that exactly
(m`) out of m have a black subpixel at position j, for each j = 1;:::;m.
Therefore, = XOR( 1 ;:::; m ) is the reconstructed black pix el if (mh)
is even, i.e., (m`) is odd, otherwise the reconstructed pixel is .