Cryptography Reference
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if and only if is
0
k
and s
k
can do so. Hence, it is sucient to prove the conclusion
based on a (2; 2)-DVCS.
We analyze the structure of the basis matrix of the (2; 2)-DVCS. Denote
M
0
and M
1
as the basis matrices of the (2; 2)-DVCS, then the M
0
and M
1
,
without loss of generality, are in the following form:
2
3
11 00 11 00
1
1
| {z }
a
4
5
;
M
0
=
00
| {z }
b
0
0
| {z }
c
1
1
| {z }
d
and
2
3
11 00 11 00
1
1
4
5
;
M
1
=
00
0
0
1
1
| {z }
a
0
| {z }
b
0
| {z }
c
0
| {z }
d
0
where a, b, c, d, a
0
, b
0
, c
0
, and d
0
are nonnegative integers satisfying a+c+d = l
and a
0
+ c
0
+ d
0
= h. According to the contrast and security property of
Definition 1, we have,
8
<
a + b + c + d = a
0
+ b
0
+ c
0
+ d
0
a + c = a
0
+ c
0
a + d = a
0
+ d
0
b > b
0
:
solving the above equations, we get a a
0
= b b
0
= c
0
c = d
0
d. Let
e = bb
0
, hence, by deleting identical columns of M
0
and M
1
, we get,
2
3
11 00
1
1
M
0
4
5
0
=
00
| {z }
e
| {z }
e
and,
2
3
11 00
0
0
| {z }
e
M
0
4
5
1
=
11
| {z }
e
where the number of columns in M
0
0
and M
0
1
is 2e.
Now we know that the basis matrices of an arbitrary (2; 2)-DVCS M
0
and M
1
contain the same number of identical columns
1
1
1
0
0
1
,
,
,
0
0
apart from the submatrices M
0
0
and M
0
and
1
. Hence, without loss of
generality, they can be represented as the following form:
2
3
11 00 11 00 11 00
1
1
| {z }
a
0
4
5
M
0
=
00
| {z }
b
0
0
0
| {z }
c
11
| {z }
d
1
1
| {z }
e
00
| {z }
e
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