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
shown in Figure 11.8(a) and (b). By using M reg ( Figure 11.7(a)), we encrypt
 
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