Information Technology Reference
In-Depth Information
Since 2
n−r
is a period of
S
1
and
S
2
,wehave
d
H
(
S
1
,
S
2
)=2
r
d
H
(
S
1
,
S
2
)=
·
2
r
+1
. This completes the proof of the theorem.
t
·
B Proof of Theorem 3
We assume
2
n
0
<L
≤
(49)
since the result holds trivially for
L
=0.
Suppose (
A
(
L
)+
E
i
1
,··· ,i
t
1
)
∩
(
A
(
L
)+
E
j
1
,··· ,j
t
2
)
=
∅
. So there exists sequences
S
,
S
∈
=
S
+
E
j
1
,··· ,j
t
2
. This implies that
A
(
L
) such that
S
+
E
i
1
,··· ,i
t
1
=
S
.
S
+
E
i
1
,··· ,i
t
1
+
E
j
1
,··· ,j
t
2
(50)
Consider the corresponding polynomials of
S
and
S
given by
x
)
2
n
x
)
2
n
−L
a
(
x
)and
S
(
x
)=(1
−L
a
(
x
)
,
S
(
x
)=(1
−
−
(51)
where
a
(1) =
a
(1) = 1. From equations (49) and (51) we have
x
2
n
)
,
S
(
x
)+
S
(
x
)))
>
2
n
deg(gcd((1
−
−
L.
(52)
From equations (50) and (52) we have
x
2
n
)
,x
i
1
+
+
x
i
t
1
+
x
j
1
+
+
x
j
t
2
))
>
2
n
deg(gcd((1
−
···
···
−
L.
(53)
To prove the theorem we first show that every sequence in
A
(
L
)+
E
i
1
,··· ,i
t
1
is in
A
(
L
)+
E
j
1
,··· ,j
t
2
.Considerany
R
∈A
(
L
) with the corresponding polynomial
x
)
2
n
−L
b
(
x
)
,
R
(
x
)=(1
−
where
b
(1) = 1
.
(54)
Then let
R
=
R
+
E
i
1
,··· ,i
t
1
+
E
j
1
,··· ,j
t
2
with the corresponding polynomial
R
(
x
). By equations (53) and (54) we have
deg(gcd((1
− x
2
n
)
,
R
(
x
)))
= deg(gcd((1
x
)
2
n
,
R
(
x
)+
x
i
1
+
+
x
i
t
1
+
x
j
1
+
+
x
j
t
2
))
(55)
−
···
···
=2
n
−
L.
From equation (55) using the definition of linear complexity we have
R
∈A
(
L
),
which implies
A
(
L
)+
E
i
1
,··· ,i
t
1
⊆A
(
L
)+
E
j
1
,··· ,j
t
2
. By symmetry
A
(
L
)+
E
j
1
,··· ,j
t
2
⊆A
(
L
)+
E
i
1
,··· ,i
t
1
, which proves the theorem.
