Cryptography Reference
In-Depth Information
Proof Because the "" operation is associative and B
i
B
i
is a zero matrix
for any i, we have
A
0
= B
1
(B
1
B
2
) (B
n2
B
n1
) (B
n1
A)
= (B
1
B
1
) (B
n1
B
n1
) A
= A:
Theorem 12 [23] A
i1
A
i2
A
ik
6= A for any set of integers fi
1
; ;i
k
g
when k < n.
Proof We consider two cases. Case 1 is for n 2fi
1
; ;i
k
g and Case 2 is for
n 2fi
1
; ;i
k
g.
Case 1: n 2 fi
1
; ;i
k
g. In this case, A
n
(
j=s
A
j
) = A B
n1
(
j=s
A
j
) where
j=s
means A
s
A
t
with s; ;t being the indices in
fi
1
; ;i
k
g besides n. Since there are an odd number of n 2 term random
matrices involved, at least one of them cannot be absorbed into a zero matrix,
thus A
i
1
A
i
2
A
i
k
must be random, thus not equal to A.
Case 2: n 2fi
1
; ;i
k
g. Since no matrix A involved in A
i
1
A
i
2
A
i
k
to begin with, A
i
1
A
i
2
A
i
k
is constructed from the n1 term random
matrices only and it must be random.
Next, we analyze the steps of Wang et al.'s [23] (n;n) algorithm in the
context of the grayscale image below. It is trivially applicable to a binary
image and can be easily extended to a color image. Now we give an example
to demonstrate the computation steps in the construction/revealing process
using the above algorithm.
Example 6 Let n = 3 and the secret image A be a single pixel.
Input: A = (240)
10
= (1111 0000)
2
Construction:
Step 1:
B
1
= (125)
10
= (0111 1101)
2
,
B
2
= (10)
10
= (0001 0010)
2
Step 2:
A
1
= B
1
= (125)
10
= (0111 1101)
2
,
A
2
= B
1
B
2
= (0110 1111)
2
,
A
3
= B
2
A = (1110 0010)
2
Revealing: A
0
= A
1
A
2
A
3
= (1111 0000)
2
= (240)
10
.
From the above algorithm, it is known that the proposed (n;n) scheme
reconstructs the secret image exactly and when fewer than n shadows are
used, the original secret image A will not be revealed. Moreover, only simple
Boolean XOR operations are used and the size of each shadow image is the
same as the original image, thus no pixel expansion. The above (n;n) secret
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