Сидельников В.М. Теория кодирования. Справочник по принципам и методам кодирования
Подождите немного. Документ загружается.
x
T
x
(j) = (x, x
(j)
) =
X
x∈F
q
µ
x(x + j)
p
¶
+
µ
j
p
¶
+
µ
−j
p
¶
.
³
−1
p
´
= −1 p = 4t − 1
³
j
p
´
+
³
−j
p
´
= 0
¤
p = 4t + 1
³
j
p
´
=
³
−j
p
´
T
x
(j) =
(
p, j ≡ 0 mod p ,
2
³
j
p
´
− 1, j 6≡ 0 mod n
.
p = 4t − 1
x
(−1, −1, . . . , −1)
p+1
2
p + 1
x
F
p
x
F
q
, q = p
l
q − 1
T r(x) : F
q
→ F
p
, F
p
r = p
a = (T r(aθ
0
), T r(aθ
1
), T r(aθ
2
), . . . , T r(aθ
q−2
)), a ∈ F
q
\ {0},
θ F
q
a
a F
q
p−
F
p
K
θ
l K
θ
a
(j)
= (T r(aθ
j
), T r(aθ
j+1
), . . . ,
T r(aθ
j+q−2
)) = (T r(a
0
θ
0
), T r(a
0
θ
1
), T r(a
0
θ
2
), . . . , T r(a
0
θ
q−2
)) a j
a
0
= θ
j
a
a
g(x) =
x
l
f(x
−1
) = x
l
+ g
l−1
x
l−1
+ ···+ g
1
x + g
0
f(x) θ
a = (a
0
, a
1
, . . . , a
q−1
)
a
j+l
= −(a
j+l−1
g
1
+ ··· + a
j−1
g
l−1
+ a
j
g
l
), j = 0, . . . , q −2,
a
l
a
(t)
= (a
t+l−1
, a
t+l−2
, . . . , a
t
) t
t+1 a
(t+1)
= (a
t+l
, a
t+l−12
, . . . , a
t+1
)
a
t+l
= −(a
t+l−1
g
1
+ ··· + a
t−1
g
l−1
+ a
t
g
l
), t = 0, . . . , q − 2,
L(x) =
−(x
l−1
g
1
+ ··· + x
1
g
l−1
+ x
0
g
l
) a
(t)
b
a =
µ
exp
2π i T r(aθ
0
)
p
, exp
2π i T r(aθ
1
)
p
, . . . , exp
2π i T r(aθ
q−2
)
p
¶
T
b
a
(j)
T
b
a
(j)
b
a
T
b
a
(j) =
(
q − 1, j ≡ 0 mod q −1 ,
−1, j 6≡ 0 mod q − 1
.
T
b
a
(j) =
q−2
X
s=0
exp
2π i (T r(aθ
s
) − T r(aθ
s+j
))
p
=
q−2
X
s=0
exp
2π i T r(a
0
θ
s
)
p
,
a
0
= a(1 −θ
j
) 6= 0 j 6≡ 0 mod q −1
−1 a
0
6= 0
α ∈ F
p
a
0
6= 0
T r(a
0
x) = 0,
F
q
q
p
a
0
6= 0 T r(a
0
x)
F
p
L
F
p
l −1
|L| =
q
p
= p
l−1
T r(a
0
x) F
p
T r(a
0
x)
T r(a
0
αx) = αT r(a
0
x), α ∈ F
p
T r(a
0
x) = α α ∈ F
p
p
l−1
θ
s
, s = 0, . . . , q − 2,
F
q
T r(a
0
θ
s
), s = 0, . . . , q − 2,
p
l−1
α α 6= 0 p
l−1
− 1 0
p−1
X
α=0
exp
2π i α
p
= 0
¤
d(x, y) =
1
2
(n−(
b
x,
b
y))
m = 2t − 1 θ
F
q
, q = 2
m
R
(c)
m
n = 2
m
− 1
2m
Θ(a, b) = (T r(aθ
3
+ bθ), T r(aθ
3·2
+ bθ
2
), . . . , T r(aθ
3·(2
m
−1)
+ bθ
2
m
−1
)), a, b ∈ F
2
m
, ε ∈ F
2
,
R
m
, |R
m
| = 2
m
+ 1 =
1
2
m
−1
(|R
(c)
m
|−1)
R
m
τ(R
m
) = η(R
(c)
m
) = 2
m+1
2
− 1.
5 6
p−
2 p
l
− 1, p
l
p
l
+ 1
K ⊂ F
n
q
t t
t + 1 F
n
q
d(a, K) = min
x∈K
d(a, x).
d(a, K) a K
K ⊂ F
n
q
d(K) ≥ 2t + 1 x ∈ F
n
q
t + 1
K
F
q
K d ≥ 2t + 1
k
n × k− n k B
a
T
k F
q
F
n
q
t + 1
B
k k a ∈ F
k
q
x ∈ F
n
q
a
T
=
B
x
T
K x
∈
F
n
q
y
t + 1 x + y ∈ K
a
T
= Bx
T
= −By
T
a
T
t+1
B
a
T
t + 1
B x
a
T
= Bx
T
K t + 1 K
¤
p− q
d 5
5
α
j
, j = 0, 1, 2, 3
α
j
, j = 0, 1, 2, 3
z
1
x
j
1
+ z
2
x
j
2
+ z
3
x
j
3
= α
j
, j = 0, 1, 2, 3, z
1
, z
2
, z
2
, α
0
∈ F
p
, x
1
, x
2
, x
2
∈ F
q
,
α
j
, j = 1, 2, 3, F
q
p > 3 α
j
α
j
= z
0
x
j
0
, z
0
∈ F
p
\ {0}, j = 0, 1, 2, α
3
= z
0
0
x
j
0
x
0
F
q
z
0
0
6= z
0
, z
0
0
6= 0
x
0
6∈ {x
1
, x
2
, x
3
} |{x
1
, x
2
, x
3
}| = 3
z
1
, z
2
, z
3
z
j
z
0
3
kx
j
i
k
i,j=0,...,3
x
0
= x
1
|{x
1
, x
2
, x
3
}| = 3
z
1
−z
0
, z
2
, z
3
z
1
−z
0
= 0, z
2
= 0, z
3
= 0 F
p
kx
j
i
k
i,j=0,1,2
|{x
1
, x
2
, x
3
}| < 3
α
j
α
j
p > 3
BCH
3,q
(n, 5) n = 3
l
−1
B BCH
3,q
(n, 5)
B =
θ
0
1
θ
0
2
··· θ
0
n
θ
1
θ
2
··· θ
n
θ
2
1
θ
2
2
··· θ
2
n
,
{θ
1
, θ
2
, . . . , θ
n
} = F
q
r {0} B
θ
3
1
, θ
3
2
, ··· , θ
3
n
3
z
1
+ z
2
+ z
3
= α
0
z
1
x
1
+ z
2
x
2
+ z
3
x
3
= α
1
, z
1
, z
2
, z
3
∈ F
3
, x
1
, x
2
, x
3
∈ F
q
z
1
x
2
1
+ z
2
x
2
2
+ z
3
x
2
3
= α
2
α
1
, α
3
∈ F
q
α
0
∈ F
3
F
q
(α
0
, α
1
, α
2
) 6= (0, 0, 0)
α
j
α
j
z
1
z
1
= 1
x
1
= −z
2
x
2
− z
3
x
3
, z
2
= z
3
= 1 x
2
6= 0
(1 + y)
2
+ 1 + y
2
= −y
2
− y − 1 =
α
2
x
2
,
y =
x
3
x
2
α
2
6= 0 x
2
∈ F
q
r{0}
α
2
= 0
y = 1 ¤
C
l
F
q
, q = 3
l
, 3 G
l
n =
q+1
2
F
∗
q
2
F
q
G
l
x
n
−1 = 0 ξ
G
l
l = 2r l = 2r + 1
l C
2r
n
B
2r
= (ξ
1
, ξ
2
, . . . , ξ
n
),
ξ
j
= ξ
j
C
2r+1
n
B
2r+1
= (ξ
1
, ξ
2
, . . . , ξ
n
),
ξ
j
= ξ
j
, i = 0, . . . ,
n
2
− 1, ξ
j
= θξ
j
, j =
n
2
, . . . , n − 1, θ
4 F
∗
q
2
θ
x
4
− 1 = 0 θ
2
= −1
C
l
, l ≥ 1
l C
l
C
l
3
X
i=0
x
i
ξ
s
i
= 0, x
i
∈ F
3
, s
0
< s
1
< s
2
< s
3
x
0
= ··· = x
3
= 0
2, 3
l
, 3
l+1
ξ
3
l
i
= ξ
−1
i
x
i
3
X
i=0
x
i
ξ
s
i
=
3
X
i=0
x
i
ξ
3
s
i
=
3
X
i=0
x
i
ξ
−1
s
i
=
3
X
i=0
x
i
ξ
−3
s
i
= 0.
∆ =
¯
¯
¯
¯
¯
¯
¯
¯
ξ
s
0
ξ
s
1
ξ
s
2
ξ
s
3
ξ
3
s
0
ξ
3
s
1
ξ
3
s
2
ξ
3
s
3
ξ
−1
s
0
ξ
−1
s
1
ξ
−1
s
2
ξ
−1
s
3
ξ
−3
s
0
ξ
−3
s
1
ξ
−3
s
2
ξ
−3
s
3
¯
¯
¯
¯
¯
¯
¯
¯
= (ξ
s
0
ξ
s
1
ξ
s
2
ξ
s
3
)
−3
¯
¯
¯
¯
¯
¯
¯
¯
ξ
2
s
0
ξ
2
s
1
ξ
2
s
2
ξ
2
s
3
ξ
6
s
0
ξ
6
s
1
ξ
6
s
2
ξ
3
s
6
ξ
4
s
0
ξ
4
s
1
ξ
4
s
2
ξ
4
s
3
ξ
0
s
0
ξ
0
s
1
ξ
0
s
2
ξ
0
s
3
¯
¯
¯
¯
¯
¯
¯
¯
0 ξ
2
s
0
, ξ
2
s
1
, ξ
2
s
2
, ξ
2
s
3
ξ
j
C
l
C
l
α ∈ F
q
2
, α 6= 0, ξ
s
1
, ξ
s
2
, ξ
s
3
x
1
, x
2
, x
3
∈ F
3
α = x
1
ξ
s
1
+ x
2
ξ
s
2
+ x
3
ξ
s
3
{±ξ
j
|j = 1, . . . , n} Π
x
3
l
+1
− 1 = 0,
π
1
+ π
2
+ π
3
= α, π
1
, π
2
, π
3
∈ Π.
α α = π · β
π ∈ Π β y
2(3
l
−1)
− 1 = 0
π
1
, π
2
, π
3
π
1
+ π
2
+ π
3
= β, π
1
, π
2
, π
3
∈ Π,
β y
3
l
−1
− 1 = 0 y
3
l
−1
+ 1 = 0
χ
F
q
l C
l
¤
C
l
l
C
G
l
n =
3
l
−1
2
C
l
2l C
l
n
0
=
4
l
−1
3
2l
H
l
n =
3
l
−1
2
F
∗
3
l
F
3
l
J
l
n
0
=
4
l
−1
3
F
∗
4
l
F
4
l
x
n
− 1 = 0
x
n
0
− 1 = 0 ξ H
l
θ
J
l
l n C
G
l
C
G
l
B
G
l
=
µ
ξ ξ
2
··· ξ
n
ξ
−1
ξ
−2
··· ξ
−n
¶
.
B
G
l
C
G
l
l ≥ 1 C
G
l
C
G
l
B
G
l
3
X
i=0
x
i
ξ
s
i
= 0,
3
X
i=0
x
i
ξ
−s
i
= 0, x
i
∈ F
3
, s
0
< s
1
< s
2
< s
3
.
3
x
0
, x
1
, x
2
, x
3
3
X
i=0
x
i
ξ
s
i
= 0,
3
X
i=0
x
i
ξ
3s
i
= 0,
3
X
i=0
x
i
ξ
−s
i
= 0,
3
X
i=0
x
i
ξ
−3s
i
= 0, x
i
∈ F
3
, s
0
< s
1
< s
2
< s
3
.
∆
0
=
¯
¯
¯
¯
¯
¯
¯
¯
ξ
s
0
ξ
s
1
ξ
s
2
ξ
s
3
ξ
3s
0
ξ
3s
1
ξ
3s
2
ξ
3s
3
ξ
−s
0
ξ
−s
1
ξ
−s
2
ξ
−s
3
ξ
−3s
0
ξ
−3s
1
ξ
−3s
2
ξ
−3s
3
¯
¯
¯
¯
¯
¯
¯
¯
= (ξ
s
0
ξ
s
1
ξ
s
2
ξ
s
3
)
−3
¯
¯
¯
¯
¯
¯
¯
¯
ξ
2s
0
ξ
2s
1
ξ
2s
2
ξ
2s
3
ξ
6s
0
ξ
6s
1
ξ
6s
2
ξ
6s
6
ξ
4s
0
ξ
4s
1
ξ
4s
2
ξ
4s
3
ξ
0
ξ
0
ξ
0
ξ
0
¯
¯
¯
¯
¯
¯
¯
¯
0
ξ
2s
0
, ξ
2s
1
, ξ
2s
2
, ξ
2s
3
l
H
l
2
C
G
l
¤
l B
G
l
B
2r+1
B
G
l
=
µ
γ γ
2
··· γ
n
γ
−1
γ
−2
··· γ
−n
¶
,
γ F
3
l
C
l
5
ϑ F
4
l
, n =
4
l
−1
3
B
G
l
=
µ
ϑ ϑ
2
··· ϑ
n
ϑ
−1·2
ϑ
−2·2
··· ϑ
−n·2
¶
,
C
G
l
C
G
l
n =
4
l
−1
3
F
4
log
4
(3n − 1)
x
0
, x
1
, x
2
, x
3
3
X
i=0
x
i
ϑ
s
i
= 0,
3
X
i=0
x
i
ϑ
4s
i
= 0,
3
X
i=0
x
i
ϑ
−s
i
= 0,
3
X
i=0
x
i
ϑ
−4s
i
= 0, x
i
∈ F
4
, s
0
< s
1
< s
2
< s
3
.
¤
C
G
l
C
G
l
ζ F
2
l
, n = 2
l
− 1
B
G
l
=
µ
ζ ζ
2
··· ζ
n
ζ
1·3
ζ
2·3
··· ζ
n·3
¶
,
K
l
K
l
2l
K
l
n = 2
l
− 1
3
X
i=1
y
i
= α,
3
X
i=1
y
3
i
= β, y
i
∈ F
2
l
α, β ∈ F
2
l
z
i
= y
i
+ α
3
X
i=1
z
i
= 0,
3
X
i=1
z
3
i
= γ, z
i
∈ F
2
l
, γ = α
3
+ β.
z
3
= z
1
+ z
2
z
1
z
2
(z
1
+ z
2
) = γ.
x =
z
1
z
2
, z
2
6= 0
F
2
l
x
2
+ x +
γ
z
3
2
= 0.
z
2
∈ F
∗
q
z
2
T r(
γ
z
3
2
) = 0
l > 2
¤
K
G
n = 2
l
B
l
=
µ
1
ϑ − a
1
,
1
ϑ − a
1
, . . . ,
1
ϑ − a
2
l
−1
¶
,
ϑ x
2
+ax + b, a, b ∈ F
2
l
{a
0
, a
1
, . . . , a
2
l
−1
} = F
2
l
K
G
n = 2
l