Cryptography Reference
In-Depth Information
Step 2:
Use p
B
and p
S
to generate the M
ran
mask or choose the M
reg
mask
for the case p
B
= p
S
= 50%; let the chosen mask M = M
ran
or
M
reg
.
Step 3:
For each block in the mask M, do the following:
Step 3-1: If the block is the large block
B
, for the secret pixel "s" in
the corresponding block of the large-scaled image I
B
, do the
following:
Assign (bp)
j
= D
B
(s) for i 2 [1;n] and j 2 [1;m];
Use A
m
x
;m
y
((bp)
j
) to create n (m
x
m
y
)-sized rectangles,
where m = m
x
m
y
, and deliver them in the corresponding
block to O
(1)
, O
(2)
,:::, O
(n)
, re
spe
ctively.
Step 3-2: If the block is the large block
S
, for the secret pixel "s" in
the corresponding block of the large-scaled image I
S
, do the
following:
Assign (bp)
j
= D
S
(s) for i 2 [1;n] and j 2 [1;m];
Use A
m
x
;m
y
((sp)
j
) to create n (m
x
m
y
)-sized rectangles,
where m = m
x
m
y
, and deliver them in the corresponding
block to O
(1)
, O
(2)
,:::, O
(n)
, respectively.
Decrypting Algorithm:
Input: Any k shares from O(i),
(i)
, i 2 [1;n].
Output: I
0
.
Step 1:
Print out k shares on transparencies. Stack and align them by hand
with the approximate accuracy.
/* Note that, when stacking shares the tradition VCS needs the pre-
cise alignment; the proposed scheme has the misalignment tolerance
and so that one just aligns them roughly. */
Step 2:
Decrypt the secret directly by human eyes.
Align shares gradually to get a refined secret image of I
0
.
/* One can first align transparencies precisely in the vertical direc-
tion, and then move gradually by hand without losing the secret in
the horizontal direction, finally the recovered secret image with no
deviation can be obtained.*/
Step 3:
For easily understanding the encrypting algorithm, we herein give an ex-
ample to show how to encode a secret image.
Example 4 Construct two shares by the (2; 2) misalignment tolerant VCS,
where p
B
= p
S
= 50%,
2
= 4, and M
reg
are used. We use a 16 16-pixel
secret image I, one black horizontal on a white background
(Figure 11.8).
Since I is black and white now, so I
S
= I, and I
B
has the size of 88 pixels
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