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next g rounds (t r−1 unchanged in rounds r,...,r + g − 1 and t r unchanged
in rounds r + 1,...,r + g) with probability e 2 N , given that the index i G in
these g rounds does not touch them. But each of t r−1 and t r does not fall in
the i G -affected area with probability 1 − N
and hence the probability of the
values at t r−1 and t r to remain in place is
1− g
N
2 e 2 N ≈e 4 N
(for small g).
Thus the overall probability of all indices to be in place is at least
e 4(g−2)
e 4 N = e 8−8 N .
N
Now, using the above two results, we are going to prove the main result
on the digraph repetition bias.
Theorem 6.3.3. For small values of ∆, the probability of the pattern ABTAB
in RC4 keystream, where T is a ∆-word string is
N 2 (1 + e −8−8∆
1
N
N
).
−i r−1 as defined in Lemma 6.3.2, the real gap between
the digraphs is ∆ = g−2. The digraph repetition can occur in two ways.
Proof: If g = j r−1
1. When the conditions of Lemma 6.3.2 are satisfied, the digraph repeti-
tion occurs with probability e 8−8g
= e −8−8∆
. The probability of the
N
N
conditions being satisfied is
P(S r−1 [i r ] = 1)P(j r−1 = i r+∆+1 )P(j r+∆+1 = i r−1 ) =
1
N 3 .
Thus, the contribution of this part is e −8−8∆
N
N 3
.
2. When the conditions of Lemma 6.3.2 are not satisfied (the probability of
which is 1− N 3 ), it may be assumed that the digraph repetition occurs
due to random association (the probability of which is
1
N 2 ). Hence the
contribution of this part is 1− N 3
N 2 .
Adding the above two components, we get the total probability to be approx-
imately
N 2 (1 + e −8−8∆
1
N
N ).
Note that in uniformly random stream, the probability of digraph repeti-
tion is
1
N 2 , irrespective of the length ∆ of T. Whereas in RC4, the probability
of such an event is more than
1
1
1
N 3 .
Using information theoretic arguments [21], it is claimed in [109] that the
number of samples required to distinguish RC4 from uniform random stream
are 2 29 ,2 28 ,2 26 for success probabilities 0.9,0.8 and 3 respectively.
N 2 and little less than
N 2 +
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