Cryptography Reference
In-Depth Information
Lemma 3 (One-dimensional deviation) The darkness and the whiteness
ratios always satisfy R
0
W;a1
R
W;a1
, R
0
W;a2
R
W;a2
R
0
B;b1
R
B;b1
and
R
0
B;b2
R
B;b2
for the deviations (d
x
; 0) and (0;d
y
) where 0 < d
x
;d
y
< s
1
.
Proof: We only consider the case with deviation (d
x
; 0) since the proof is the
same for the case with deviation (0;d
y
). According to equation (11.1), when
d
y
= 0, we have
A
1
= A
0
1
= 0
A
2
= A
0
2
= 0
A
3
= A
0
3
= d
x
s
A
4
= A
0
4
= (sd
x
)s
A
3
+ A
4
= s
2
(11.13)
We consider the relation of R
C;c
and R
0
C;c
for C 2 fW;Bg and c 2 (a1),
(a2), (b1) and (b2). Recall that s
1
< s
2
, and R
C;c
and R
0
C;c
can be calculated
by the following formula:
R
C;c
=
A
C;c
2s
1
and R
0
C;c
=
A
C;c
(11.14)
2s
2
We have Table 11.4 to show the relation of R
C;c
and R
0
C;c
for different
values of C and c.
TABLE 11.4
The relation of the ratios R
C;c
and R
0
C;c
.
Stacked
pixel
Area(A
C;c
) Ratio(R
C;c
) Ratio(R
0
C;c
) Relation
White,a1
A
W;a1
=A
4
R
W;a1
=
s
1
d
x
2s
1
R
0
W;a1
=
s
2
d
x
2s
2
R
0
W;a1
>R
W;a1
A
W;a1
=A
3
+A
4
R
W;a1
= 1=2 R
0
W;a1
= 1=2 R
0
W;a1
=R
W;a1
White,a2
A
W;a2
=A
0
4
R
W;a2
=
s
1
d
x
2s
1
R
0
W;a2
=
s
2
d
x
2s
2
R
0
W;a2
>R
W;a2
Black,b1
A
B;b1
=A
3
+A
4
+A
0
4
R
B;b1
= 1
d
x
2s
1
R
0
W;a1
= 1
d
x
2s
2
R
0
B;b1
>R
B;b1
A
B;b2
=A
4
+A
0
3
+A
0
4
R
B;b2
= 1
d
x
2s
1
R
0
W;a2
= 1
d
x
2s
2
R
0
B;b2
>R
B;b2
Black,b2
A
B;b2
=A
3
+
A
4
+A
0
3
+A
0
4
R
B;b2
= 1 R
0
W;a2
= 1 R
0
B;b2
=R
B;b2
The proofs of the relations in Table 11.4 are quite similar. We only take
the proof of the relation between R
0
W;a1
and R
W;a1
, for example.
According to equation (11.3), for the stacked pixel of
Figure 11.3
(a1),
there are two cases of A
W;a1
, i.e., I: A
W;a1
= A
4
and II: A
W;a1
= A
3
+ A
4
.
For the Case I, we have R
W;a1
=
A
W;a1
2s
1
=
A
4
2s
1
=
s
1
d
x
2s
1
and R
0
W;a1
=
s
2
d
x
,
2s
2
by subtraction we have R
0
W;a1
R
W;a1
=
s
2
d
x
2s
2
s
1
d
x
=
d
x
(s
2
s
1
)
2s
1
s
2
> 0, i.e.
2s
1
R
0
W;a1
> R
W;a1
.
For the Case II, we have R
0
W;a1
R
W;a1
=
s
2
2s
2
s
1
= 0, i.e. R
0
W;a1
= R
W;a1
.
2s
1
According to Table 11.4, the lemma follows.
2
Search WWH ::
Custom Search