Cryptography Reference
In-Depth Information
Example 3.6
Suppose that we have an LFSR with
=4
, and tap sequence
(
c
1
c
2
c
3
c
4
) = (0101)
and initial state
s
0
=(
k
(3
,
0)
k
(2
,
0)
k
(1
,
0)
k
(0
,
0)
) = (1101)
.
Then we calculate the following.
j k
(3
,j
)
k
(2
,j
)
k
(1
,j
)
k
(0
,j
)
0
1
1
0
1
1
0
1
1
0
2
1
0
1
1
3
1
1
0
1
For instance, for
j
=1
, the state after the first bit iteration is given by
s
1
=(
k
(3
,
1)
k
(2
,
1)
k
(1
,
1)
k
(0
,
1)
) = (0110)
,
where from
(3.6)
,
k
(3
,
1)
=
c
1
k
(3
,
0)
⊕
c
2
k
(2
,
0)
⊕
c
3
k
(1
,
0)
⊕
c
4
k
(0
,
0)
=
0
·
1
⊕
1
·
1
⊕
0
·
0
⊕
1
·
1=0
,
k
(2
,
1)
=
k
(3
,
0)
=1
,
(1
,
1)
=
k
(2
,
0)
=1
,
and
k
(0
,
1)
=
k
(1
,
0)
=0
.
The LFSR has period length
L
=3
, since
s
3
=
s
L
= (1101) =
s
0
.
The output bitstring consists of the rightmost entry in each of the above table's
rows for each
(
distinct
)
bit iteration
j
=0
,
1
,
2
, namely,
k
=(
K
2
K
1
K
0
) = (101)
.
There is a very palatable, simple, easy-to-understand matrix method of de-
scribing the above. Consider the following
tap matrix
derived from the tap
sequence, and
state matrix
derived from the states.
c
1
c
2
c
3
···
c
−
1
c
k
(
−
1
,i
)
k
(
−
2
,i
)
k
(
−
3
,i
)
.
.
.
k
(0
,i
)
100
···
00
010
···
00
C
=
and
S
i
=
,
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
000
···
1
0
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