Cryptography Reference
In-Depth Information
=−
√
r
Let us write this sum as
p
=
1
−
p
1
−
p
2
. In the second integral we let
x
+
is between 0 and
4
ρ
co
s
θ
and
y
=
ρ
sin
θ
. Variable
θ
in
T
. Then
ρ
goes from
√
2
r
/
(2 cos(
θ
+
π/
4)) to
+∞
. Hence
⎛
⎞
4
+∞
√
2
r
1
π
e
−
ρ
2
d
⎝
⎠
d
p
2
=
2 cos
(
θ
+
4
)
ρ
ρ
θ
0
4
1
π
r
4 cos
2
(
θ
+
4
)
d
e
−
=
θ.
0
=
√
r
In the first integral we let
x
+
ρ
cos
θ
and
y
=
ρ
sin
θ
. Variable
θ
is between
4
,
th
en
0 and
π
in
T
. When
θ
is between 0 and
ρ
goes from 0 to
+∞
. When
θ
is
−
√
2
r
between
4
and
π
, then
ρ
goes from 0 to
/
(2 cos(
θ
+
π/
4)).
−
d
π
√
2
r
2 cos
(
θ
+
4
)
1
4
+
1
π
e
−
ρ
2
d
p
1
=
ρ
ρ
θ
4
0
1
4 cos
2
(
θ
+
4
)
d
π
1
4
+
1
π
r
e
−
=
−
θ
4
π
r
4 cos
2
(
θ
+
4
)
d
1
π
e
−
=
1
−
θ.
4
θ
+
4
) around
θ
=
4
When adding
p
1
and
p
2
, by using the symmetry of cos
2
(
we
obtain
π
1
π
r
4 cos
2
(
θ
+
4
)
d
e
−
p
=
θ
2
4
1
π
r
4 cos
2
e
−
=
θ
d
θ
−
4
By using the symmetry of cos
θ
around
θ
=
0 we obtain
4
2
π
r
4 cos
2
e
−
p
=
θ
d
θ
0
≤
θ
≤
4
[
2
,
Clearly, when 0
we have cos
2
θ
∈
1]. Therefore
1
2
e
−
1
2
e
−
r
2
r
4
≤
p
≤
.
Having a number of bad candidates equal to
B
, we deduce that the expected rank
R
of
the right candidates is such that
B
2
e
−
λ
2
B
2
e
−
λ
4
≤
R
≤
.
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