Cryptography Reference
In-Depth Information
235
45
239
Input: the secret image A =
188
103
234
Output: the final shadows S
1
;S
2
;S
3
;S
4
Construction:
Step 1: Generate random image B
1
=
105
14
208
.
228
90
2
105
235
14
208
45
239
Step 2: Generate B
2
= B
1
A =
=
228
90
2
188
103
234
130
35
63
.
88
61
232
n
k 1
4
2
. Construct a 4 6 shadows-assignment
matrix H, each column of H have exactly 2 zeros and 2 ones, so
Step 3: m =
=
0
1
0
0
0
1
1
1
@
A
0
1
1
0
0
1
H =
1
0
1
0
1
0
1
1
0
1
0
0
Step 4: Partition B
1
into 6 nonoverlapping blocks C
11
= 105;C
21
=
14;C
31
= 208;C
41
= 228;C
51
= 90;C
61
= 2, Partition into 6 nonoverlap-
ping blocks C
12
= 130;C
22
= 35;C
32
= 63;C
42
= 88;C
52
= 61;C
62
= 232.
Step 5: C
= C
11
C
21
C
61
= 10514208228902 = 11.
Compute C
13
= C
12
C
= 130 11 = 137, similarly C
23
= 40;C
33
=
52;C
43
= 83;C
53
= 54;C
63
= 227.
Step 6: Construct 6 temporary shadows C
1
=
105
137
14
40
;C
2
=
;C
3
=
208
52
228
83
90
54
2
227
;, where the upper half of each
C
i
is the block C
i1
and the lower half is the block Ci
i3
.
Step
;C
4
=
;C
5
=
;C
6
=
7:
According
to
the
shadows-assignment
matrix H,
assign
C
4
;C
5
;C
6
! S
1
, C
2
;C
3
;C
6
! S
2
, C
1
;C
3
;C
5
! S
3
, C
1
;C
2
;C
4
! S
4
.
Revealing:
Step 1: Suppose 3 participants using their shares S
1
;S
2
;S
3
to reconstruct
the secret image. Referring to H, all 6 temporary shadows C
1
=
105
137
;C
2
=
14
40
208
52
228
83
90
54
2
227
;C
3
=
;C
4
=
;C
5
=
;C
6
=
, can be extracted
from S
1
;S
2
;S
3
.
Step 2: The upper half of each Ci
i
is the block C
i1
and the lower half of
each C
i
is the block C
i3
. So we get all C
11
= 105;C
21
= 14;C
21
= 14;C
31
=
208;C
41
= 228;C
51
= 90;C
61
= 2 and C
13
= 137;C
23
= 40;C
33
= 52;C
43
=
83;C
53
= 54;C
63
= 227.
Step 3: Compute C
= C
11
C
21
C
61
= 10514208228902 =
11, then C
12
= C
13
C
= 137 11 = 130. Similarly, C
22
= 35;C
32
=
63;C
42
= 88;C
52
= 61;C
62
= 232.
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