Cryptography Reference
In-Depth Information
∪
∪
∩
∩
∩
∩
4
1
2
1
3
2
3
Figure 11.4.
Example for an access structure term.
as drawn in Fig. 11.4. We start by attaching
S
to the whole expression
. It is written as
∪
1
∩
2
)
∪
1
∩
3
) and (
2
∩
3
)
∩
4
. So we attach
S
to both terms.
a
between (
(
Similarly, in the former term, we attach
S
to
1
∩
2
and
1
∩
3
. We have thus three
terms to which
S
is attached. For the first one
1
∩
2
, we attach
W
to
1
and
S
−
W
to
2
. For the second one
1
∩
3
, we attach
X
to
1
and
S
−
X
to
3
. For the third
one (
2
∩
3
)
∩
4
we attach
Y
to
2
∩
3
and
S
−
Y
to
4
. It thus remains to attach
Z
to
2
and
Y
−
Z
to
3
. To summarize, we have two occurrences of
1
to which
are attached
W
and
X
, we have two occurrences of
2
to which are attached
S
−
W
and
Z
, we have two occurrences of
3
to which are attached
S
−
X
and
Y
−
Z
, and
we have one occurrence of
4
to which is attached
S
−
Y
. Therefore we define the
share
S
1
=
,
(
W
X
)
S
2
=
−
,
(
S
W
Z
)
S
3
=
(
S
−
X
,
Y
−
Z
)
S
4
=
S
−
Y
where
W
Z
are independent uniformly distributed random variables. We can,
for instance, check that
P
1
and
P
2
can reconstruct
S
because they have
W
and
S
,
X
,
Y
,
−
W
.
But
P
2
and
P
3
, for instance, cannot reconstruct
S
because (
S
−
W
,
Z
,
S
−
X
,
Y
−
Z
)
is equivalent to (
S
−
W
,
X
−
W
,
Y
,
Z
) which gives no clue about
S
.
11.3
Special Purpose Digital Signatures
In this section we list a few important variants of digital signature schemes.
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