Resolving the Alignment Problem in Visual Cryptography - Visual Cryptography and Secret Image Sharing - page 308

l whitearea is the average white area for a white pixel, which contains two
cases (a1) and (a2), i.e.,

A W;a1 + A W;a2

= A 1
2

+ A 2
2

+ A 3
4

l whitearea =

+ A 4 (11.12)

2

According to Definition 1 (definition of a deterministic VCS), in order to
deterministically recover the secret image by its original color, the values of
l blackarea and h blackarea should satisfy l blackarea < h blackarea . Together with
Equations (11.1), (11.5), and (11.6), we get (d x ;d y ) to satisfy (sd x )(sd y ) >
s 2 =2, i.e., (d x ;d y ) falls in the region R 1 . And the contrast R 1 is
R 1 = h blackarea l blackarea

2s 2 = s 2 2(d x + d y )s + 2d x d y 2s 2

Similarly, in order to recover deterministically by its complementary color,
the values of h whitearea and l whitearea should satisfy h whitearea > l whitearea .
However, according to Equations (11.9) and (11.10), we get that h whitearea >
l whitearea does not hold. Hence, the secret image cannot be recovered deter-
ministically by its complementary color.
According to Definition 2 (definition of a probabilistic VCS), in order
to probabilistically recover the secret image by its original color, the val-
ues of l blackarea and h blackarea should satisfy l blackarea < h blackarea . To-
gether with Equations (11.1), (11.7), and (11.8), we get (d x ;d y ) to satisfy
0 d x < 2s=3; 0 d y < s. By excluding the region R 1 we get to know that
(d x ;d y ) falls in the region R 2 . And the contrast R 2 is
R 2 = h blackarea l blackarea

2s 2 = (2s 3d x )(sd y ) 4s 2

Similarly, in order to probabilistically recover the secret image by its
complementary color, the values of h whitearea and l whitearea should satisfy
h whitearea > l whitearea . Together with Equations (11.1), (11.11), and (11.12),
we get (d x ;d y ) to satisfy 2s=3 < d x s; 0 d y < s, i.e. (d x ;d y ) fall in the
region R 3 . And the contrast R 3 is
R 3 = h blackarea l blackarea

= (3d x 2s)(sd y )
4s 2 2

2s 2

The above Theorem 2 is consistent with Theorem 1. According to The-
orem 1 and Theorem 2, for the deviation (d x ;d y ) = (s; 0), the secret image
can be probabilistically recovered by its complementary color with average
contrast = 1=4.
The above Theorem 2 considers the deviations (d x ;d y ) less than one sub-
pixel. In fact, when the value of d x is between s and 2s for a misaligned
(2; 2)-VCS, the secret image can also be recovered. The proof is left to the
interested readers.