Cryptography Reference
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2. S N [r] ∈ B and index r is touched by at least one of the r − 1 many
j G values during the first r−1 rounds of the PRGA. Further, after the
swap(s), the value S N [r] remains in the set B. This will happen with
probability ( N + ǫ)
1−( N−1
N
b−1
) r−1
N−1 .
3. S N [r] ∈ B and index r is touched by at least one of the r− 1 many j G
values during the first r− 1 rounds of the PRGA. Due to the swap(s),
the value S N [r] comes to the set B. This will happen with probability
(1− N
1−( N−1
N
N .
Adding these contributions, we get the total probability as given by δ.
) r−1
−ǫ)
b
N + 2 N ,
b
Lemma 6.2.14. If P(S r−1 [r] ∈ B) =
N + δ, then P(z r
∈ C) =
where C = {c | c = r−b where b ∈ B}, r ≥ 1.
Proof: The event (z r
∈C) can happen in two ways.
1. S r−1 [r] ∈ B and z r = r − S r−1 [r]. From Corollary 5.2.2, we have
P(z r = r−S r−1 [r]) =
2
N
for r ≥ 1. Thus, the contribution of this part
N ( N + δ).
2. S r−1 [r] ∈ B and still z r
2
is
∈ C due to random association. The contribu-
tion of this part is (1− N ) N .
Adding these two contributions, we get the result.
b
Theorem 6.2.15. If P(S N [r] ∈B) =
N + ǫ, then P(z r
∈ C)
r−1
r−1
b
N +
2
N
b
N + ǫ
N −1
N
N −1
N
=
+
1−
r−1
b−1
N −1
b
N
b
N
N −1
N
,
where C = {c | c = r−b where b ∈ B}, r ≥ 1.
Proof: The proof immediately follows by combining Lemma 6.2.13 and
Lemma 6.2.14.
The above results imply that for a single value v, if
1
N + ǫ,
P(S N [r] = v) =
then
N +
1
N ,
where the value of δ can be computed by substituting b = 1 in Lemma 6.2.14.
This reveals a non-uniform distribution of the initial keystream output bytes
z r for small r.
P(z r = r−v) =
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