Visual Cryptography for Multiple Secrets - Visual Cryptography and Secret Image Sharing - page 67

b j in B is superimposed with a j in area k and a k1 in k's previous area

j

k 1 when A is rotated 0 and clockwise where a k1

j = prev(a j ) (or a j =
next(a k j )). Note that a k j is the result of rotating a j 90 counterclockwise
(or a j is the result of rotating a k j 90 clockwise; see Figure 3.4). That means
the result of A B in Wu and Chang's scheme emulates that of S 90
1

S 2
in Wu and Chen's scheme. There is no restriction for to be 90 , 180 , or
270 merely. Yet there exist some inconsistent situations in some of the areas
in A B when = 360 = > 4. Interested readers refer to Ref. [13] for
details.
As mentioned above, square share S 1 in Wu and Chen's scheme is not
a totally random image. Strictly speaking, neither is circle share A in Wu
and Chang's scheme due to the reason that sector block a j is assigned as
next(a t j ) (i.e., the result of rotating a t j 90 clockwise) for 2 t ; that
is, only sector block a j in the first area is randomly determined from those
in Figure 3.4(a) for 1 j , while the other areas are not. Furthermore,
sector blocks in the first area of circle share A (see Figure 3.4(a)) in Wu and
Chang's scheme (or extended blocks in the rst area of square share S 1 (see
Figure 3.2(c)) contain only four patterns, instead of six, which is the number
of all possible combinations for four subpixels with two white and two black
subpixels (see Table 3.2) .

3.4 Visual Multiple-Secret Sharing Schemes
Both of the above-mentioned schemes accomplish visual secret sharing for
only two secrets in two shares. In this section we discuss two more generalized
visual secret sharing scheme for x 2 secrets in two shares by Shyu et al. [8]
and Feng et al. [3] in Sections 3.4.1 and 3.4.2, respectively.

3.4.1 Shyu et al.'s Scheme
3.4.1.1 Informal Description
Let us start by using a simple example. Assume that the number of secret
images to be shared is x = 3. Let P 1 , P 2 , and P 3 be the three binary secret
images with the same size hw. Let p 1 , p 2 , and p 3 denote the corresponding
pixels in P 1 , P 2 , and P 3 , respectively. Let A and B denote the two circle shares
encoded by the scheme. Our aim is to assure AB recovers P 1 , A 120 B
recovers P 2 , and A 240 B recovers P 3 .
Since there are three secrets, we decompose circle share A and B into three
(x = 3) chord-areas (chords for short), respectively, in which the angle of each
chord extends up to 120 (= 360 =x = 360 =3). Each chord is divided into a