Cryptography Reference
In-Depth Information
T
0;C
0
= T
0;0
s
0;0
s
0;0
s
0;0
=
[(2a
0
+ c + d + 2e)
a
0
+d+
m
(
(a
0
+e)d
m
(a
0
+e)(a
0
+e1))
m
+
c
m
+
c(a
0
+e)
+
)]m!
m
We now define the average stacking Hamming weight of each share ma-
trix to be the total stacking Hamming weight of all the share matrices being
divided by the number of the share matrices, i.e.,
T
c;
m!
. Then the difference
between the average stacking Hamming weight of each share matrix of the
shifted collections C
0
and C
1
, denoted by D
A
, is
D
A
= (T
1;C
1
T
1;C
0
)p
1
+ (T
0;C
1
T
0;C
0
)p
0
=
e(p
1
+ p
0
)
m
=
e
m
According to the definition of the average contrast in
Section 11.2,
with
h = T
1;C
1
p
1
+ T
0;C
1
p
0
and
l = T
1;C
0
p
1
+ T
0;C
0
p
0
, we get the value of the
hl
m
=
D
A
m
=
m
2
.
Now we consider the general case when the share 2 is shifted by r subpixels.
For this case there are 2
r
possible subpixels that can be shifted in. For example,
for r = 2, the shifted in strings of subpixels have four cases, 00, 01, 10 and 11.
Denote by p
00
, p
01
p
10
and p
11
the probabilities of these four cases to happen
respectively, then we have p
00
+ p
01
+ p
10
+ p
11
= 1.
We consider the string of subpixels c, let the Hamming weight of c be s,
i.e., w(c) = s. The total number of 1's that are ineective in the top right
corner of all the share matrices in C
0
and C
1
is is
0;c
= s
1;c
= s
a
0
+c+
m
m!, and
the total number of 1's that are ineective in the bottom left corner of all the
share matrices in C
0
and C
1
is is
0;c
= s
1;c
= r
a
0
+d+e
average contrast =
m!. For the third case,
1
1
m
the pattern
in the shifted share matrices is the shifted result of the
2
4
3
5
,
2
4
3
5
,
2
4
3
5
and
2
4
3
5
11
10
10
11
|{z}
r
01
|{z}
r
11
|{z}
r
01
|{z}
r
11
following four patterns,
|{z}
r
|{z}
r
|{z}
r
|{z}
r
in the collections C
0
and C
1
, and there are only mr choices for the value
of the position i, so the total number of the 1's that are ineective of the four
patterns of the collection C
1
is
(
a
0
m
+
a
0
m
a
0
1
m 1
a
0
m 1
)(mr)m!
d + e
m 1
+
c + e
m
d + e
m 1
+
c + e
m
and of the collection C
0
is
(
a
0
+ e
m
a
0
+ e
m 1
)(mr)m!
Hence, when a string of subpixels of c is shifted in, the total stacking Hamming
weight of all the matrices of the collection C
1
is
T
c;C
1
=
+
a
0
+ e
m
a
0
+ e 1
m 1
d
m 1
+
c
m
d
m 1
+
c
m
[(2a
0
+ c + d + 2e + s) r
a
0
+d+e
m
s
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
)
mr
+
+
m1
]m!
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