Resolving the Alignment Problem in Visual Cryptography - Visual Cryptography and Secret Image Sharing - page 296

where (a) is the recovered secret image without being shifted, and (b) is the
recovered secret image with the second share being shifted to the left by one
subpixel, and (c) is the recovered secret image with the second share being
shifted to the left by 2 subpixels. From the experimental results we notice
that, the original secret image can be recovered visually by shifting one or two
subpixels for the (2; 2)-DVCS with pixel expansion m = 3. And the clearness
of the recovered secret image with two subpixels being shifted is not as clear
as the one with one subpixel being shifted. Furthermore, the shifted scheme
is not a DVCS anymore; we give an example to show this point:

010
010

100
010

Example 2 Take the share matrices S 0 and S 1

as an
example, which are chosen from the permutations of the basis matrices M 0
and M 1 in Example 1. By shifting the second share by one subpixel to the left,
we get the following four matrices:

= =

010
0

010
0

100
0

100
0

S 0 = S 1 = S 0 = S 1 =

100 101 100 101

where S j , i = 0; 1, j = 0; 1, is the shifted share matrix that a subpixel j is
shifted into Si i . The left bottom subpixel of Si i is shifted out, and the asterisk
in the S j can be either 1 or 0, which belongs to the pixel on the left of the
pixel that we considered. So, here and hereafter, we no longer consider the two
subpixels anymore, i.e., we only need to consider the right 3 columns in the
shifted share matrix S j .
For the above four shifted share matrices, the stacking Hamming weights
are 2, 3, 1, and 2. In particular, the stacking Hamming weight of S 0 and
S 1 are the same. Here and hereafter, the stacking Hamming weight means
the Hamming weight of the resulting vector generated by stacking the shares.
Hence, the shifted scheme is not a DVCS anymore.

Generally, we aim at proving the conclusion that, the shifted scheme can
visually recover the original secret image based on the (k;n)-DVCS. However,
it is noticed that this proof can be reduced to the proof based on the (2; 2)-
DVCS in the case that only one share is shifted. The reason is as follows:

(k;k)-DVCS. For a set of k shares, if
no share is shifted, then the k shares can recover the secret image obviously.
And because we only consider the case when only one of the n shares is shifted,
we only need to consider the k shares that contain the shifted share, i.e., we
only need to prove our conclusion based on a (k;k)-DVCS.
Second, denote the k shares of a (k;k)-DVCS as s 1 ;s 2 ; ;s k , without loss
of generality, let is k be the share that is shifted, and let is 0 k be the resulting
image of stacking the remaining k 1 shares s 1 ;s 2 ; ;s k1 together. Then
the scheme becomes a (2; 2)-DVCS, where the two shares are s 0 k and s k . Note
that, the stacking result of this (2; 2)-DVCS is the same as that of the previous
(k;k)-DVCS. The previous (k;k)-DVCS can visually recover the secret image

First, a (k;n)-DVCS consists of k