Cryptography Reference
In-Depth Information
their elementary forms s
1
A
, s
2
A
, and s
3
A
, we design s
B
to be so that both
s
1
A
s
B
and s
3
A
s
B
reveal one white and five black subpixels, while s
2
A
s
B
show six black subpixels. Our eyes recognize s
1
A
s
B
and s
3
A
s
B
as white,
while is
2
A
s
B
as black. That means (p
1
;p
2
;p
3
) is recovered by is
1
A
s
B
,
s
2
A
s
B
, s
3
A
s
B
) in a visual sense.
In actual implementation, the set of three related blocks (a
j
;a
j
;a
j
) in A is
deliberately assigned as (permute(s
1
A
;
j
);permute(s
2
A
;
j
);permute(s
3
A
;
j
))
so that we only need to assign b
j
to be permute(s
B
;
j
) to preserve the su-
perimposition results designed in
Table 3.5.
Then, when we superimpose a
j
and b
j
, we identify (p
1
)
j
form a
j
b
j
. When we rotate A 120
counterclock-
wise, b
j
's corresponding block in A
120
turns out to be a
j
(i.e., A
120
[1, j],
Likewise, when rotating A 240
counterclockwise, b
j
's corresponding block in
A
240
is a
j
and a
j
b
j
reveals (p
3
)
j
. In general, when rotating A (i 1)
counterclockwise we recognize a
j
b
j
as (p
i
)
j
in the first chord of A
(i1)
B
by our visual system for 1 i 3(= x) and = 120
(= 360
=x).
We call the blocks in column s
B
of Table 3.5 the elementary blocks cir-
cle share B for sharing 3 secrets, which consists of three white and three
black subpixels. They are named by s
0
B
;s
1
B
;:::;s
7
B
in sequence as indicated
in Figure 3.10. When we denote as 0 and as 1, the superscript l of
s
l
B
is equal to the code formed by p
1
p
2
p
3
in binary, i.e. l = btod(p
1
p
2
p
3
)
where btod bi- is a function that returns the decimal representation of a bi-
nary number b. It means that based upon Table 3.5, once (a
j
;a
j
;a
j
) is
assigned to be (s
1
A
;s
2
A
;s
3
A
) and b
j
is encoded to be s
btod(p
1
p
2
p
3
B
(specifi-
cally, (a
j
;a
j
;a
j
) = (permute(s
1
A
;
j
);permute(s
2
A
;
j
);permute(s
3
A
;
j
)) and
b
j
= permute(s
btod(p
1
p
2
p
3
B
;
j
) in practical implementation with respect to
give (p
1
;p
2
;p
3
)
j
, (a
j
b
j
;a
j
b
j
;a
j
b
j
) recovers (p
1
;p
2
;p
3
)
j
.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
FIGURE 3.10
Elementary blocks of share B for sharing 3 secrets: (a) s
0
B
, (b) s
1
B
, (c) s
2
B
, (d)
s
3
B
, (e) s
4
B
, (f) s
5
B
, (g) s
6
B
, (h) s
7
B
.
Now, we take the instances in
Figure 3.11,
in which the first three pixels of
the three divided strips in Pi
i
are depicted for 1 i 3, as an example to show
how the corresponding blocks in B are encoded. From Figure 3.11, we have
(p
1
;p
2
;p
3
)
1
= (;;). According to Table 3.5, the elementary block for b
1
is chosen to be s
btod(p
1
p
2
p
3
)
B
= s
btod(010
B
= s
2
B
. Since in practical implemen-
tation, b
1
's corresponding block a
1
(a
1
, a
1
) in A (A
120
, A
240
, respectively)
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