5.1 BCH Codes—Part I
67
contradiction. If i = r − 3 then the congruence r (q − 3) ≡ (r − 3)(q + 1) mod n
holds, that is, 4r ≡ 3(q + 1) mod n holds. Since r | (q + 1) and q + 1 > r hold,
it implies that q + 1 ≥ 2r so, 3(q + 1) − 4r ≥ 2r > 0. Moreover, the inequality
3(q + 1) − 4r < n holds, which is a contradiction. Therefore, C is Euclidean dualcontaining.
Let C
be the cyclic code generated by
M
(r )
(x)M
(r +1)
(x) · . . . · M
(2r −4)
(x).
C
is an enlargement of C; C
has dimension k
= n − 2(r − 4) − 1 and minimum
distance d
≥ r − 2. Since m = 2 then k
− k = 2, where k
denotes the dimension
of C
and k is the dimension of C. We know that
q+1
q
d
≥ r − 1. Thus, applying
the Steane’s construction one has an [[n, n − 4(r − 3), d ≥ r − 1]] q quantum code,
as required.
Recall that an [[n, k, d]] q code C satisfies the quantum Singleton bound given
by k + 2d ≤ n + 2. If C attains the quantum Singleton bound, i.e., k + 2d = n + 2,
then it is called a quantum maximum distance separable (MDS) code. In the following
two examples we construct quantum MDS-BCH codes:
Example 5.1.5 Applying Theorem 5.1.7 for q = 9 and n = 40 one has r = 5. Thus
there exists an [[40, 36, 3]] 9 quantum MDS-BCH code. Analogously, applying Theorem 5.1.7 for q = 11 and n = 60 one obtains an [[60, 56, 3]] 11 quantum MDS-BCH
code. Additionally, an [[60, 48, d ≥ 5]] 11 and an [[60, 52, d ≥ 4]] 11 quantum codes
can be constructed.
5.1.4 Construction IV
In this subsection we present the fourth proposed construction, which is based on
finding good Hermitian dual-containing BCH codes. Let us recall some useful concepts.
Lemma 5.1.4 ([4, Lemma 13]) Assume that gcd(q, n) = 1. A cyclic code of length n
over F q 2 with defining set Z contains its Hermitian dual code if and only if Z ∩ Z
−q
=
∅, where Z
−q
= {−qz mod n | z ∈ Z }.
Lemma 5.1.5 ([4, Lemma 17c]) (Hermitian Construction) If there exists a classical
linear [n, k, d] q 2 code D such that D
⊥ H ⊂ D, then there exists an [[n, 2k − n, ≥ d]] q
stabilizer code that is pure to d. If the minimum distance d
⊥ H of D
⊥ H exceeds d,
then the stabilizer code is pure and has minimum distance d.
Let us start with an example of how Lemma 5.1.1 can be applied together the Hermitian construction in order to construct good codes. Assume that q = 7, n = 144,
m = 3 and r = 3; the q
2 -ary cosets C 3 , C 6 , C 9 and C 12 contain only one element. The
67
contradiction. If i = r − 3 then the congruence r (q − 3) ≡ (r − 3)(q + 1) mod n
holds, that is, 4r ≡ 3(q + 1) mod n holds. Since r | (q + 1) and q + 1 > r hold,
it implies that q + 1 ≥ 2r so, 3(q + 1) − 4r ≥ 2r > 0. Moreover, the inequality
3(q + 1) − 4r < n holds, which is a contradiction. Therefore, C is Euclidean dualcontaining.
Let C
be the cyclic code generated by
M
(r )
(x)M
(r +1)
(x) · . . . · M
(2r −4)
(x).
C
is an enlargement of C; C
has dimension k
= n − 2(r − 4) − 1 and minimum
distance d
≥ r − 2. Since m = 2 then k
− k = 2, where k
denotes the dimension
of C
and k is the dimension of C. We know that
q+1
q
d
≥ r − 1. Thus, applying
the Steane’s construction one has an [[n, n − 4(r − 3), d ≥ r − 1]] q quantum code,
as required.
Recall that an [[n, k, d]] q code C satisfies the quantum Singleton bound given
by k + 2d ≤ n + 2. If C attains the quantum Singleton bound, i.e., k + 2d = n + 2,
then it is called a quantum maximum distance separable (MDS) code. In the following
two examples we construct quantum MDS-BCH codes:
Example 5.1.5 Applying Theorem 5.1.7 for q = 9 and n = 40 one has r = 5. Thus
there exists an [[40, 36, 3]] 9 quantum MDS-BCH code. Analogously, applying Theorem 5.1.7 for q = 11 and n = 60 one obtains an [[60, 56, 3]] 11 quantum MDS-BCH
code. Additionally, an [[60, 48, d ≥ 5]] 11 and an [[60, 52, d ≥ 4]] 11 quantum codes
can be constructed.
5.1.4 Construction IV
In this subsection we present the fourth proposed construction, which is based on
finding good Hermitian dual-containing BCH codes. Let us recall some useful concepts.
Lemma 5.1.4 ([4, Lemma 13]) Assume that gcd(q, n) = 1. A cyclic code of length n
over F q 2 with defining set Z contains its Hermitian dual code if and only if Z ∩ Z
−q
=
∅, where Z
−q
= {−qz mod n | z ∈ Z }.
Lemma 5.1.5 ([4, Lemma 17c]) (Hermitian Construction) If there exists a classical
linear [n, k, d] q 2 code D such that D
⊥ H ⊂ D, then there exists an [[n, 2k − n, ≥ d]] q
stabilizer code that is pure to d. If the minimum distance d
⊥ H of D
⊥ H exceeds d,
then the stabilizer code is pure and has minimum distance d.
Let us start with an example of how Lemma 5.1.1 can be applied together the Hermitian construction in order to construct good codes. Assume that q = 7, n = 144,
m = 3 and r = 3; the q
2 -ary cosets C 3 , C 6 , C 9 and C 12 contain only one element. The