Cryptography Reference
In-Depth Information
1 we use the following probability table that defines [
F
]
1
.
For
d
=
y
=
0
y
=
1
x
=
0
1
/
4
3
/
4
x
=
1
3
/
4
1
/
4
x
=
2
1
/
4
3
/
4
2 we use the following probability table which defines [
F
]
2
.
For
d
=
(
y
1
,
y
2
)
=
(0
,
0)
(
y
1
,
y
2
)
=
(0
,
1)
(
y
1
,
y
2
)
=
(1
,
0)
(
y
1
,
y
2
)
=
(1
,
1)
(
x
1
,
x
2
)
=
(0
,
0)
1
/
4
0
0
3
/
4
(
x
1
,
x
2
)
=
(1
,
0)
0
3
/
4
1
/
4
0
(
x
1
,
x
2
)
=
(2
,
0)
0
1
/
4
1
/
4
1
/
2
(
x
1
,
x
2
)
=
(0
,
1)
0
1
/
4
3
/
4
0
(
x
1
,
x
2
)
=
(1
,
1)
3
/
4
0
0
1
/
4
(
x
1
,
x
2
)
=
(2
,
1)
1
/
4
1
/
2
1
/
4
0
(
x
1
,
x
2
)
=
(0
,
2)
1
/
4
1
/
4
1
/
2
0
(
x
1
,
x
2
)
=
(1
,
2)
1
/
4
1
/
2
0
1
/
4
(
x
1
,
x
2
)
=
(2
,
2)
1
/
4
0
0
3
/
4
Therefore if we write rows and columns in this order we have
⎛
⎝
⎞
⎠
1/4
0
0
3/4
0
3/4
1/4
0
0
1/4
1/4
1/2
⎛
⎞
1/4
3/4
0
1/4
3/4
0
[
F
]
1
⎝
⎠
,
[
F
]
2
=
3/4
1/4
=
3/4
0
0
1/4
1/4
3/4
1/4
0
1/2
1/4
0
1/4
1/4
1/2
1/4
1/2
0
1/4
1/4
0
0
3/4
We can compare
F
with a uniformly distributed
F
∗
for which
⎛
⎝
⎞
⎠
1/2
0
0
1/2
1/4
1/4
1/4
1/4
1/4
1/4
1/4
1/4
⎛
⎞
1/2
1/2
1/4
1/4
1/4
1/4
[
F
∗
]
1
⎝
⎠
,
[
F
∗
]
2
=
1/2
1/2
=
1/2
0
0
1/2
1/2
1/2
1/4
1/4
1/4
1/4
1/4
1/4
1/4
1/4
1/4
1/4
1/4
1/4
1/2
0
0
1/2
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