Cryptography Reference
In-Depth Information
1. Consider the first cases of both the items (2), (3) from Theorem 5.4.1.
Thus,
P(j
G
= z | i
G
odd)
P(j
G
= m)P(z = m | j
G
= m,i
G
odd)
=
m
1
N
2
2(N −1)!−(N −2)!
N!
+(N −2)
(N −1)!−(N −2)! + 2(N −3)!
N!
=
1
N
+
2
N
2
−
1
2
N
2
(N −1)
=
N(N −1)
+
≥
1
1
N
2
.
N
+
2. Consider the first case of item (1) from Theorem 5.4.1. Thus,
(N −1)!−(N −2)!
N!
P(j
G
= z | i
G
even)
=
≤
1
N
−
1
N
2
.
3. Consider the first and second cases of item (2) and the first case of item
(3) from Theorem 5.4.1. Note that,
=
P(j
G
= z ∧ z =
i
G
+1
j
G
= z | z =
i
G
+ 1
2
)
2
P
.
P(z =
i
G
+1
2
)
Now
j
G
= z ∧ z =
i
G
+ 1
2
1
N
2(N −1)!−(N −2)!
N!
P
=
and
z =
i
G
+ 1
2
P
1
N
2(N −1)!−(N −2)!
N!
+
(N −1)!−2(N −2)!
=
N!
+(N −2)
(N −1)!−(N −2)! + 2(N −3)!
N!
.
Thus,
j
G
= z | z =
i
G
+ 1
2
2
N
−
1
P
=
N(N −1)
.
