Cryptography Reference
In-Depth Information
When k is odd, it is necessary to exchange x
(W)
i
by x
(B)
i
and vice versa.
nj
i
nj
i
nj
nj
X
X
x
(W)
i
x
(W)
ni
=
i=0
i=0
nj
ni
X
x
(W)
i
=
i=j
nj
ij
X
x
(W)
i
=
i=j
nj
ij
X
x
(B)
i
=
i=j
nj
i
nj
X
x
(B)
i
=
;
i=0
i.e., the variables x
(W
i
m x
(B
i
satisfy (8.16).
By the same argument we get
nk
ih
nk
ih
n
X
n
X
x
(B
i
+ (1)
h
=
x
(W)
i
(8.17)
i = 0
i = 0
for h k.
For h = k this shows that x
(W)
i
m x
(B)
i
is an optimal solution of the linear
program.
Let x
(W
i
, x
(B
i
, x
(W)
and x
(B)
i
be two solutions of the linear program, then
i
x
(W)
i
2
(x
(W)
+ x
(W
i
), x
(B)
2
(x
(B
i
+ x
(B
i
) is also a solution of the linear
1
1
=
=
i
i
program.
After this transformations we have a solution that satisfies (a) and (b).
Assume that we have a scheme that does not satisfy (c), i.e., the matri-
ces M
W
and M
B
have columns in common. Since the corresponding subpix-
els occur independent of the encoded color they can be omitted. Deleting all
columns that occur in M
W
and M
B
we get a scheme with smaller pixel expan-
sion and higher contrast. That proves that every optimal k-out-of-n scheme
must satisfy (c).
2
Now we are ready to determine the optimal contrast of a 3-out-of-n visual
cryptography scheme.
Theorem 14 ([2] Theorem 4.7) The
contrast of
a 3-out-of-n visual
cryptography scheme satisfies
(n 2b
n+
4
c)b
n+
4
c
2(n 1)(n 2)
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