Steganography in Halftone Images - Visual Cryptography and Secret Image Sharing - page 359

In the right half of R which is in W b , the y 1 (i;j) and y 2 (i;j) are favored to
be conjugate to each other (y 2 (i;j) y 1 (i;j)) such that y 1 (i;j) and y 2 (i;j)
tend not to be black at the same time. To investigate the behavior of y(i;j),
we consider two dierent cases: (1) 255 A > 127 and (2) 127 A 0.
For the case of 255 A > 127, there are more white pixels than black
pixels in R of Y 1 . The percentage of black pixels in the right half of R of Y 1
is about (255 A)=255 100%. For example, if A = 190, about 25% of the
pixels in the right half of R of Y 1 should be black. As y 1 (i;j) and y 2 (i;j)
tend not to be black at the same time, the black pixels in the right half of
R of Y 2 tend to be at different locations from those in Y 1 . Consequently, the
percentage of black pixels in the right half of R of Y tends to be doubled to
2(255A)=255100%. In our example of A = 190, the approximately 25%
black pixels in the right half of R of Y 1 and those of Y 2 tend to be at different
locations such that there are about 50% black pixels in the right half of R of
Y . Thus,

P[y(i;j) = 0] 2P[y 1 (i;j) = 0] = 2(255 A)=255 (13.27)

P[y(i;j) = 255] (2A 255)=255 (13.28)

such that

E[y(i;j)] = 0 P[y(i;j) = 0] + 255 P[y(i;j) = 255]
2A 255 = A (255 A) A (13.29)

In other words, the expected value should increase approximately linearly with
A, from 1 (for A = 128) to 255 (for A = 255). The difference between the
average intensity of the left and right halves of R for 255 A > 127 is

intensity = E[y(i;j)j(i;j) 2 W w ] E[y(i;j)j(i;j) 2 W b ]
A (2A 255) = 255 A (13.30)

For the case of 127 A 0, there are fewer white pixels than black pixels
in the right half of R of Y 1 . Again the y 1 (i;j) and y 2 (i;j) tend not to be black
at the same time in DHCED. This implies that the black pixels in the right
half of R of Y 2 tend to fill up all the white pixel locations in the right half of
R of Y 1 , leading to all y(i;j) being black. Thus, P(y(i;j) = 0) 1 and

E[y(i;j)] = 0 P[y(i;j) = 0] + 255 P[y(i;j) = 255] 0 (13.31)

for 127 A 0. And the dierence between the average intensity of the left
and right halves of R is, for 127 A 0,

intensity A 0 = A (13.32)

Consequently, the contrast between the left and right halves of R of Y can be