Cryptography Reference
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and therefore,
0
1
0
1
X
X
@
2
jT
0
jjTj
A
+ 1 =
@
2
jT
0
jjTj
A
+ (1)
jT
0
j+jSj
:
ST
(
T
0
jT
0
jjTj mod 2
St3T
(
T
0
jT
0
j6jTj mod 2
(Note that (1)
jT
0
jjSj
= (1)
jT
0
j+jSj
.) Division by 2 gives
0
1
0
1
X
X
(1)
jT
0
j+jSj
1
2
@
2
jT
0
j1jTj
A
=
@
2
jT
0
j1jTj
A
+
ST
(
T
0
jT
0
jjTj mod 2
ST
(
T
0
jT
0
j6jTj mod 2
as required for equation (8.33).
Suppose that h
T
(for ;6= T f1;:::;ng) satisfy (8.30) with equality, too.
If inequality (8.30) is satisfied with equality for all subsets S, we can solve
these equations recursively and get
h
S
= h
f1;:::;ng
+ F
S
(
T
j T f1;:::;ng) ;
for some function F
S
. Since inequality (8.30) is satisfied with equality for the
contrast values hT
T
and h
0
T
this yields
h
S
= h
S
+ h
f1;:::;ng
h
f1;:::;ng
:
h
f1;:::;ng
= h
f1;:::;ng
and therefore
But for S = ; inequality (8.30) yields
h
T
= h
T
for all ;6= T f1;:::;ng.
This proves that (8.32) is the only solution of (8.30) that satisfies all
inequalities with equality.
Thus, we nd
X
X
1
2
(3
n
1)
T
0
2
jT
0
j1
2
jT
0
j1
=
m h
f1;:::;ng
;6=T
0
f1;:::;ng
;6=T
0
f1;:::;ng
what proves the theorem.
2
Without proof we mention the following result about the contrast of the
different images.
Result 20 ([13] Theorem 3.6) For ;6= T f1;:::;ng let
T
=
h
T
l
T
m
be
the contrast of the image IT
T
. The contrast levels of the images satisfy
X
2
jTj1
T
1 :
(8.34)
;6=Tf1;:::;ng
Further, let
0
T
0 (for ;6= T f1;:::;ng) satisfy (8.34). Then for every
" > 0 there exists a generalized visual cryptography scheme with contrast levels
T
(for ;6= T f1;:::;ng) where j
T
0
T
j < " for all nonempty subsets T
of f1;:::;ng.
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