Cryptography Reference
In-Depth Information
depicted in Fig. 4.12) outputs 1 when the oracle implements the distributions of C and
C . We have
E LP C ( a
b ) +
n
C )
Adv ( C
,
3 3
n
.
,
3 3
1 .
2 m
Therefore, the attack is meaningless until the number of known plaintexts n reaches
the order of magnitude of
E LP C ( a
b ) .
/
,
n
1
Proof. We first notice that the advantage is zero when a
=
0or b
=
0, so the bound
=
=
holds. Let us now assume that a
0. We now take a random permutation
C with the corresponding Z and p C as in the previous lemma. Let
0 and b
δ =
1) 2 ).
E ((2 Z
E (LC C ( a
(Note that
δ =
,
b )).) When
|
2 Z
1
|≤ α
,wehave
× α n
p C
|
p 0 |≤
2
.
1) 2
is positive, the probability that
|
2 Z
1
|
is greater than
α
is less than
Since (2 Z
δ
α
2 . Hence
× α n
+ δ
α
|
p
p 0 |≤
2
2
for any
α
.
n
δ
1
3
Let us now fix
α =
. We obtain
|
p
p 0 |≤
3
×
3
n .
E LP C ( a
b ) . Since a
We recall that
δ =
,
=
0 and b
=
0 we finally note that
E LP C ( a
b ) =
1
2 m
,
and so we can have
1
n
p
|
p 0 |≤
3 3
1 .
2 m
|
p |≤|
p 0 |+|
p
p 0 |
Finally, we use the fact that
p
p
.
4.3.4
Ad hoc Construction
We define
a = 0 , b E DP C ( a
b )
EDP max =
,
max
0 E LP C ( a
b ) .
ELP max =
max
a
,
,
b
=
 
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