Visual Cryptography and Random Grids - Visual Cryptography and Secret Image Sharing - page 187

Algorithm 2.

1. Generate R 1 as a random grid, T (R 1 ) = 1=2

2. for (each pixel B[i, j ], 16i6w and 16j6h) do

2.1 f if (B[i, j ] = 0) then R 2 [i;j] = R 1 [i;j]
else R1[i, 2 [i;j] = random pixel (0, 1)

g

3. output(R 1 , R 2 )

Algorithm 3.

1. Generate R 1 as a random grid, T (R 1 ) = 1=2

2. for (each pixel B[i, j ], 16i6w and 16j6h) do

2.1 f if (B[i, j ] = 0) then R 2 [i;j] = random pixel (0, 1)
else R1[i, 2 [i;j] = R 1 [i;j]

g

3. output(R 1 , R 2 )

The three algorithms are capsulated into a generic procedure named En-
cryption so that when Encryption(B) is called, each of Algorithms 1, 2, or 3
can be applied onto B. Also note that in this paper, 0 (1) denotes a white
(black) pixel in the secret binary image or a transparent (opaque) pixel in the
encrypted share interchangeably.

Table 7.2 summarizes the encoding process of pixel b in secret image B
into r 1 and r 2 by Algorithms 1, 2, and 3 respectively, the result of r 1 r 2 and
its average light transmission.
Let B(0) (B(1)) denote the area of all of the transparent (opaque) pixels
in B, that is, pixel b is in B(0) (B(1)) if and only if b = 0 (b = 1) where
B = B(0) [B(1) and B(0) \B(1) =?. We denote the area of pixels in ran-
dom grid R corresponding to B(0) (B(1)) by R[B(0)] (R[B(1)]), that is, pixel
r is in R[B(0)] (R[B(1)]) if and only if r's corresponding pixel b is in B(0)
(B(1)). Surely, R = R[B(0)] [R[B(1)] and R[B(0)] \R[B(1)] =?.

Based upon the above notations, we have the following theorem.

Theorem 1 Given a binary image B, fR 1 , R 2 g produced by Algorithms 1{3,
respectively, is a set of (2; 2)-VCRG of B.