Cryptography Reference
In-Depth Information
Hu and Tzeng [10] algorithm for minimal access structure
Input:
1. 0 on a set of n participants.
2. Let C p and C p be the collection of basis Boolean matrices E p and
E p , where 1 p j 0 j.
3. Let C p be the collection of matrix F p .
Output:
The reconstructed secret image U.
Distribution phase:
The dealer encodes each transparency ti i as j 0 j subtransparecies t i;p and
each subblock consists of one secret image. For 1 p j 0 j, each white
(resp. black) pixel on subblock t i;p is encoded using n2 k p 1 matrices E p
(resp. E p ). To share a white (resp. black) pixel, the dealer performs the
following steps:
Step 1: Randomly choose a matrix S p in C p (resp. S p in C p ), and a
matrix A p = [a i;j ] in C p .
Step 2: For each participant i, put a white (resp. black) pixel on the
subblock t i;p if is i;j = 0 (resp. s i;j = 1).
Step 3: For each participant i, put a white (resp. black) pixel on the
subblock A i;p if a i;j = 0 (resp. a i;j = 1).
Reconstruction phase:
Let Q p = fi 1 ; ;i k p g be the minimal qualied set in 0 , participants
in Q p reconstruct the secret image by:
Step 1: XORing all the shares t j and stacking all the shares A j for j =
1; ;k p and obtain T and A: T = t 1 t k p , A = A 1 + A 2 ++ A k p
respectively.
Step 2: Computing U = (T + A) A.
Lemma 11 [10] The (k;k)-VC scheme [16] is an ideal contrast (k;k)-VC
scheme with reversing.
Proof k participants perform XOR operations on the k transparencies by
computing t 1 t 2 t k . It is easy to see that the white pixels are all white
since S 0 has an even number of 1's; whereas the black pixels are all black since
S 1 has an odd number of 1's.
Theorem 10 [10] Let = (P;Q;F) be an access structure on a set of n
participants. Then the basis matrices B 0 , B 1 , and A 0 constitute a compatible
ideal contrast VC scheme with reversing in two runs.
Proof It is obvious that the VC scheme is security. The basis matrix A 0 also
reveals absolutely no information about the secret image since no secret is
encoded into the shares A j for j = 1; ;k p .
 
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