Cryptography Reference
In-Depth Information
h
y
, S
y−1
[y−1] and j
y
are respectively replaced by S
0
[y−1] and h
y−1
(S
0
,K).
Thus, we would have
y−1
π
y−1
.
N −1
N
π
y
=
Solving this recurrence, one can get
y
x−1
N −1
N
π
y
=
x=1
y(y−1)
2
N − 1
N
=
.
Example 3.1.11. For one more example, suppose
u(y,j
y
,S
y
[y],S
y
[j
y
],K[y],K[j
y
]) = j
y
+ yS
y
[j
y
] + K[j
y
].
As before,
h
0
(S
0
,K)
= u(0,0,S
0
[0],S
0
[0],K[0],K[0])
= 0 + 0S[0] + K[0]
= 0 + 0 + K[0]
= K[0]
and π
0
= P(j
1
= h
0
(S
0
,K)) = 1. For y ≥ 1,
h
y
(S
0
,K)
= u(y,h
y−1
(S
0
,K),S
0
[y],S
0
[y−1],K[y],K[h
y−1
(S
0
,K)])
= h
y−1
(S
0
,K)]) + yS
0
[y−1] + K[h
y−1
(S
0
,K)]
= h
y−1
(S
0
,K)]) + y(y−1) + K[h
y−1
(S
0
,K)].
As in the previous example, here also the recurrence relation for the probabil-
ities is
y−1
π
y−1
,
N −1
N
π
y
=
whose solution is
y
x−1
N −1
N
π
y
=
x=1
y(y−1)
2
N − 1
N
=
.
The above discussion shows that the design of RC4 KSA cannot achieve
further security by changing the update rule by any rule from a large class.
