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There are possible some other mechanisms contributing to the probability of
the two-round characteristic and corresponding to generation the differences
4
at the output of the operation
G
(Case 5). Besides, due to the use of the mutually
P 1
32
inverse CP boxes
P 32 / 80 and
80 there are possible significantly contributing
/
cases when the box
P 32 / 80 generates an additional pair of active bits and the
P 1
box
32 / 80 annihilates this pair of active bits (Case 6).
Probability
2 =Pr 2 |i,i +1 δ 1 |i can be calculated considering the
avalanche effect corresponding to the operations
( i,i +1)
p
G
. Each input bit
l i of
G
influ-
ences several output bits
y i (except the 32nd input bit influences only the 32nd
output bit). The following formulas describe the avalanche caused by inverting
the bit
l i (3
≤ i ≤
32):
Pr(
∆y i =1
/∆l i =1)=1
,
Pr(
∆y i +1 =1
/∆l i =1)=1
,
Pr(
∆y i + k =1
/∆l i =1)=1
/
2
,
for
k
=2
,
3
,
4
,
5;
i
+
k ≤
32
.
One can see that alteration of the input bit
l i , where 3
≤ i ≤
27, causes de-
terministic alteration of two output bits
y i and
y i +1 and probabilistic alteration
of the output bits
y i +2 ,
y i +3 ,
y i +4 ,
y i +5 which change with probability
p
=0
.
5.
Note that for
i
=1
,
2 alteration of
l i causes deterministic alteration of three out-
put bits
y i ,
y i +1 , and
y i +3 . When passing trough the operation
G
the difference
1 |i
2 ,
can be transformed with certain probability in the output differences
3 , ...,
6 .
)=Pr (
0 ,∆ 1 ) be the probability that the (
0 ,∆ 1 ) r
p
r
0 ,∆ 1 )
Let
(
(
0 ,∆ 1 ) output difference when passing
input difference transforms into the (
1 denotes arbitrary difference with one active bit
in the right data subblock, i.e.
through
r
rounds. Note that
1 denotes a batch of the one bit differences.
Probability
(2) of the two-round characteristic can be calculated using the
following formula:
P
(2) = i p
( i )
2 13
)=2 5
P
p
(
i
)=1
.
15
·
,
where
p
(
i
is the
probability that after the fist round active bit moves to
i
th digit.
we have performed the statistic test ”1,000
keys and 100,000 pairs of plaintexts” including 10 8
For each value
i ∈{
1
,
2
, ...,
32
}
experiments in order to de-
(
i
) that
0 passes the right branch of the
termine the experimental probability
p
procedure
Crypt
in the case when in the left data subblock we have the differ-
1 |i
(
i
) =
10 8 .We
ence
. Let
ν i be the number of such events. Then we have ˆ
p
ν i /
( i ) taking into account the mechanisms of
the formation of the two-round characteristic described above. For all
have also calculated the probabilities
p
i
theoretic
( i )
( i ) . For example we
values
p
matches suciently well the experimental ones ˆ
p
(21) =1
2 9
(21) =1
2 9 ;
(18) =1
2 10
(18) =1
2 10 ;
have
p
.
25
·
p
.
22
·
p
.
5
·
p
.
62
·
2 11 . This demonstrates that the most impor-
tant mechanisms of the formation of the two-round differential characteristic
correspond to Cases 1-6.
The performed analysis has shown that 21st and 18th digits contribute to
(17)
2 11
(17)
p
=1
.
0
·
p
=1
.
11
·
P
(2) about 12%. Contribution
of other digits is very small. Such strongly non-uniform dependence
(2) about 88% and 17th digit contributes to
P
( i )
i
is caused by several lacks in the structure of the extension box the E box,
p
on
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