Cryptography Reference
In-Depth Information
Let L
0
= E
1
kkE
t
, L
1
= E
1
kkE
t
and A
0
= F
1
kkF
t
be the basis
matrices for a VC scheme with reversing, constructed using the previously
described technique. Without loss of generality, let
0
= fQ
0
; ;Q
t
g and
X = Q
1
, X be a subset of qualified participants. Since the secret image is
reconstructed by computing (T +A)A, we will prove that the general access
structure has an ideal contrast, i.e., H((E
1
+ F
1
) F
1
) = 0, H((E
1
+ F
1
)
F
1
) = 2
jQ
1
j1
and H((E
i
+ F
i
) F
i
) for i = 2; ;jj and b = 0; 1. It
results that H((E
1
+ F
1
) F
1
) = H((E
1
+ 0) 0) = H(E
1
0) = H(E
1
) =
0 by Lemma 11, and H((E
1
+ F
1
) F
1
) = H((E
1
+ 0) 0) = H(E
1
0) =
H(E
1
) = 2
jQ
1
j1
by Lemma 11, whereas, H((E
i
+ F
i
) F
i
) = H((E
i
+ 1)
1)1 = w(11) = 0 for i = 2; ;j
0
j and b = 0; 1.
We give the following example to illustrate the construction method above.
Example 5 Let p = f1; 2; 3; 4g and
0
= ff1; 2g;f1; 3g;f2; 3gg. Then the
basis matrices L
0
, L
1
, and A are constructed as follows according to the method
above. B
0
and B
1
are basis matrices of a (2; 2) V C scheme.
10
10
10
01
B
0
=
;B
1
=
:
2
3
2
3
2
3
2
3
2
3
2
3
10
10
10
11
10
10
10
11
10
10
01
11
11
10
01
10
11
01
4
5
;E
2
=
4
5
;E
3
=
4
5
;E
1
=
4
5
;E
2
=
4
5
;E
3
=
4
5
:
E
1
=
2
3
2
3
101110
101011
111010
101110
011011
110101
4
5
;L
1
=
4
5
:
L
0
=
2
3
2
3
2
3
2
3
00
00
11
11
00
00
00
11
00
001100
000011
110000
4
5
;F
2
=
4
5
;F
3
=
4
5
;A
0
=
4
5
:
F
1
=
The distribution phase is shown in Table 6.4 and the reconstruction phase
is shown in
Tables 6.5,
6.6,
and
6.7.
Distribution phase
TABLE 6.4
Distribution phase of the visual cryptography with minimal access
structure
0
= ff1; 2g;f1; 3g;f2; 3gg using Hu and Tzeng [10] algorithm.
Pixel
First run
Second run
Black t
1
= (101110)
A
1
= (001100)
Black t
2
= (011011)
A
2
= (000011)
Black
t
3
= (110101)
A
3
= (110000)
White t
1
= (101110)
A
1
= (001100)
White t
2
= (101011)
A
2
= (000011)
White
t
3
= (111010)
A
3
= (110000)
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