Cryptography Reference
In-Depth Information
and thus we can set =
h
m
. The security property is immediate.
Example 1 Consider as starting point the deterministic scheme (3; 4) having
basis matrices M
W
and M
B
reported below, whose parameters are m = 6,
h = 5, l = 4, = 1=6:
2
3
2
3
0
0
1
1
1
0
1
1
0
0
0
1
4
5
4
5
:
0
0
1
1
0
1
1
1
0
0
1
0
M
W
=
M
B
=
0
0
1
0
1
1
1
1
0
1
0
0
0
0
0
1
1
1
1
1
1
0
0
0
1
To
illustrate
the
construction,
consider
the
following
6
-probabilistic
(3; 4; 4; 5; 1)-VCS S.
8
<
9
=
2
4
3
5
;
2
4
3
5
;
2
4
3
5
;
2
4
3
5
;
2
4
3
5
;
2
4
3
5
0
0
0
0
0
0
0
0
1
1
1
0
1
1
0
1
1
0
1
1
0
1
1
1
C
W
=
:
;
8
<
2
3
2
3
2
3
2
3
2
3
2
3
9
=
1
1
1
1
1
1
1
1
0
0
0
1
0
0
1
0
0
1
0
0
1
0
0
0
4
5
;
4
5
;
4
5
;
4
5
;
4
5
;
4
5
C
B
=
:
:
;
Let S be the (n;n; 0; 0; 1)-VCS obtained applying Lemma 3 to scheme S
D
.
For scheme S we have that p
bjb
= 5=6, p
wjb
= 1=6, p
wjw
= 1=3, p
bjw
= 2=3, =
(S) = 1=6.
Lemma 3 shows how starting from a deterministic scheme with base ma-
trices it is possible to obtain a corresponding probabilistic scheme. A more
general version of the above lemma we report below has been given in [11]
showing the transformation from a general (also with no base matrices) de-
terministic scheme to a probabilistic one.
Lemma 4 Let S be a deterministic (k;n;`;h;m)-VCS. Then, there exists a
canonical -probabilistic (k;n; 0; 1; 1)-VCS scheme with = (S).
Proof Let S be a deterministic (k;n;`;h;m)-VCS. Construct a probabilistic
scheme S
0
, by letting C
B
(S
0
) (resp. C
W
(S
0
)) consists of all the n 1 vectors
that appear in all the matrices of C
B
(S) (resp. C
W
(S)).
We need to prove that S
0
is a -probabilistic (k;n;`
0
;h
0
;m
0
)-VCS scheme
with `
0
= 0;h
0
= 1;m
0
= 1 and = (S) = (h`)=m. Obviously S
0
has pixel
expansion m
0
= 1. We set `
0
= 0 and h
0
= 1 and we have to prove that this
results in a -probabilistic scheme.
Let r = jC
B
(S)j. By the properties of S, we know that when the secret
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