Cryptography Reference
In-Depth Information
of AB, A
120
B, and A
240
B) reconstruct (p
1
)
1
(), (p
2
)
1
(), and
(p
3
)
1
(), respectively.
Based upon the experience above, Table 3.8 summarizes the encoding
scheme for the blocks in the third chord of B for sharing 3 secrets. We can
see from Table 3.8 that given a set of corresponding pixels (p
1
;p
2
;p
3
)
j
, the
elementary block of b
j
is chosen to be s
btod(p
2
p
3
p
1
)
.
B
TABLE 3.8
Encoding a set of corresponding pixels (p
1
;p
2
;p
3
)
j
into a
j
(a
j
and a
j
)
and b
j
in terms of s
3
A
(s
1
A
, s
2
A
, respectively) and s
B
in the first chords of
A and B, respectively for visual 3-secret sharing.
p
2
p
3
p
1
s
A
s
A
s
A
s
B
s
A
s
B
s
A
s
B
s
A
s
B
Following the previous example in
Figure 3.11,
consider the particular case
(p
1
;p
2
;p
3
)
1
= (;;). Since btod(p
2
p
3
p
1
) = btod(110) = 6, we encode b
1
as
permute(s
6
B
;
1
) (
) as show in
Figure 3.14.
It is clearly seen that
a
1
b
1
= permute(s
3
A
;
1
) permute(s
6
B
;
1
)
= s
3
A
s
6
B
= =
a
1
b
1
= permute(s
1
A
;
1
) permute(s
6
B
;
1
)
= s
1
A
s
6
B
= =
a
1
b
1
= permute(s
2
A
;
1
) permute(s
6
B
;
1
)
= s
2
A
s
6
B
= =
We have that a
1
b
1
, a
1
b
1
, and a
1
b
1
(the first blocks in the third
chords of AB, A
120
B, and A
240
B) recover (p
1
)
1
(), (p
2
)
1
(), and
(p
3
)
1
(), respectively.
Figure 3.15
depicts the results of the first three blocks in the three chords
of AB, A
120
B and A
240
B, which reconstruct (p
1
)
1
in P
1
(;;)
(Figure 3.15(a) vs. Figure 3.11(a)), (p
2
)
1
in P
2
(;;) (Figure 3.15(b) vs.
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