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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