Resolving the Alignment Problem in Visual Cryptography - Visual Cryptography and Secret Image Sharing - page 300

at the columns i and i + 1 in the collection C 1 is c+ m a 0

m1 . Similarly, for the
remaining three patterns, the results are shown in Table 11.2.

TABLE 11.2
The probability of the four patterns appearing at the columns i and
i + 1 in the collections C 0 and C 1 .

10
11

11
11

10
01

11
01

collectionsnpatterns

a 0
m

a 0
m

a 0 1
m1

a 0
m1

d+e
m1

c+e
m

d+e
m1

c+e
m

C 1

a 0 +e
m

a 0 +e
m

a 0 +e1
m1

a 0 +e
m1

d
m1

c
m

d
m1

c
m

C 0

The collection C 0 and C 1 contain all the column permutations of the basis
matrices in all possible ways, and there are only m 1 choices for the value
of i, so the total number of 1's that are ineective of the four patterns of the
collection C 1 is is 1;1 = s 1;0 = ( a 0

m1 + a 0

a 0 1 a 0

d+e

m1 + c+ m m1 + c+ m

d+e

m1 )(m1)m!, m m

and that of the collection C 0 is is 0;1 = s 0;0 = ( a 0 +e

m1 + a 0 +e

a 0 +e1
m1

d

+ m

d

m1 + m m

a 0 +e

c
m m1 )(m 1)m!.

Denote the total number of 1's in the share matrices in collections C 0 and
C 1 plus the number of 1's in the subpixel c that is shifted in as T 0;c and T 1;c ,
respectively. Then, when a 1 is shifted in, we have T 0;1 = T 1;1 = (2a 0 +c+d+
2e+1)m!, and when a 0 is shifted in, we have T 0;0 = T 1;0 = (2a 0 +c+d+2e)m!.
Denote T c;C as the total stacking Hamming weight of all the matrices of
the collection C (C 0 or C 1 ) when a string of subpixels c are shifted in. The
above discussion shows that when a 1 is shifted in, the total stacking Hamming
weight of all the matrices of the collection C 1 is

= T 1;1 s 1;1 s 1;1 s 1;1
=

T 1;C 1

[(2a 0 + c + d + 2e + 1) a 0 +d+e
m

a 0 +c+e
m

( a 0 (d+e)
m

+ a 0 (a 0 1)
m

(c+e)a 0
m

(c+e)(d+e)
m

+ + )]m!

and that of the collection C 0 is
T 1;C 0 = T 0;1 s 0;1 s 0;1 s 0;1

=

[(2a 0 + c + d + 2e + 1) a 0 +d+e
m

a 0 +c+e
m

( (a 0 +e)d
m

(a 0 +e)(a 0 +e1))
m

+ c m + c(a 0 +e)

+ )]m!

m

When a 0 is shifted in, the total stacking Hamming weight of the collection
C 1 is,

= T 1;0 s 1;0 s 1;0 s 1;0
=

T 0;C 1

[(2a 0 + c + d + 2e) a 0 +d+ m
( a 0 (d+e)
m

+ a 0 (a 0 1)
m

(c+e)a 0
m

(c+e)(d+e)
m

+ + )]m!

and that of the collection C 0 is