5.5 Quantum Synchronizable Codes
121
Let gcd(n, q) = 1. Recall that the q-coset, of s modulo n is defined as C s =
{s, sq, . . . , sq
ms−1
}, where sq
ms
≡ s mod n. Let α be a primitive nth root of unity
and M i (x) denotes the minimal polynomial of α
i .
Let us recall two results shown in [4].
Proposition 5.5.1 ([4, Theorems 3 and 10]) Let n be a positive integer such that
gcd(n, 2) = 1 and let m = ord n (2). If 2 ≤ δ ≤ δ max = =κ, where κ =
n
2 m −1
(2
m/2
− 1), then the narrow-sense BCH(n, 2, δ) code contains its Euclidean dual BCH
⊥
(n, 2, δ).
Lemma 5.5.1 ([4, Lemmas 8 and 9]) Let n ≥ 1 be an integer such that gcd(n, 2) = 1
and 2
m/2
< n ≤ 2
m
− 1, where m = ord n (2). Then the following hold:
(i) The 2-coset C x has cardinality m for all x in the range 1 ≤ x ≤ n2
m/2
/(2
m
− 1).
(ii) If x and y are distinct integers in the range 1 ≤ x, y ≤ min{{n2
m/2
/(2
m
−
1) − 1, n − 1} such that x, y ≡ 0 mod 2, then the 2-cosets of x and y modulo n are
disjoint.
We are now able to prove the first result of this subsection.
Theorem 5.5.4 Let n ≥ 3 be an integer such that gcd(n, 2) = 1 and assume that
2
m/2
< n ≤ 2
m
− 1, where m = ord n (2). Take integers a, b such that 1 ≤ a < b <
r = min{{n2
m/2
/(2
m
− 1) − 1, n − 1, κ}, where κ =
n
2 m −1
(2
m/2
− 1) and
a, b ≡ 0 mod 2. Then, for any pair of nonnegative integers (a l , a r ) satisfying
a l + a r < m(t − u), there exists an (a l , a r ) − [[n + a l + a r , n − 2m(t + 1)]] QSC
that corrects up to at least
d−1
2
phase errors and up to at least
d
∗ −1
2
bit errors,
where d ≥ b + 1, d
∗
≥ a + 1, t = (b − 1)/2 and u = (a − 1)/2.
Proof Let D be the binary BCH code of length n generated by
D = =M 1 (x)M 3 (x) · · · M a (x),
where a = 2u + 1 and u ≥ 0 is an integer. Furthermore, let C be the binary BCH
code of length n generated by
C = =M 1 (x)M 3 (x) · · · M b (x),
where b = 2t + 1 and t ≥ 1 is an integer. From construction, we have C ⊂ D, and by
Proposition 5.5.1, C is dual-containing. From Lemma 5.5.1 and from straightforward
computation, the dimension of D is equal to k 2 = n − m(u + 1) and the dimension
of C is k 1 = n − m(t + 1), which implies k 2 − k 1 = m(t − u) and 2k 1 − n = n −
2m(t + 1). From the BCH bound, because the defining set of D contains a sequence
of a consecutive integers, the minimum distance of D satisfies d 2 ≥ a + 1. Similarly,
as the defining set of C contains a sequence of b consecutive integers, the minimum
distance of C satisfies d 1 ≥ b + 1. The result follows from Theorem 5.5.1 and the
proof is complete.
We next show how to construct QSCs from the sum of BCH codes.
121
Let gcd(n, q) = 1. Recall that the q-coset, of s modulo n is defined as C s =
{s, sq, . . . , sq
ms−1
}, where sq
ms
≡ s mod n. Let α be a primitive nth root of unity
and M i (x) denotes the minimal polynomial of α
i .
Let us recall two results shown in [4].
Proposition 5.5.1 ([4, Theorems 3 and 10]) Let n be a positive integer such that
gcd(n, 2) = 1 and let m = ord n (2). If 2 ≤ δ ≤ δ max = =κ, where κ =
n
2 m −1
(2
m/2
− 1), then the narrow-sense BCH(n, 2, δ) code contains its Euclidean dual BCH
⊥
(n, 2, δ).
Lemma 5.5.1 ([4, Lemmas 8 and 9]) Let n ≥ 1 be an integer such that gcd(n, 2) = 1
and 2
m/2
< n ≤ 2
m
− 1, where m = ord n (2). Then the following hold:
(i) The 2-coset C x has cardinality m for all x in the range 1 ≤ x ≤ n2
m/2
/(2
m
− 1).
(ii) If x and y are distinct integers in the range 1 ≤ x, y ≤ min{{n2
m/2
/(2
m
−
1) − 1, n − 1} such that x, y ≡ 0 mod 2, then the 2-cosets of x and y modulo n are
disjoint.
We are now able to prove the first result of this subsection.
Theorem 5.5.4 Let n ≥ 3 be an integer such that gcd(n, 2) = 1 and assume that
2
m/2
< n ≤ 2
m
− 1, where m = ord n (2). Take integers a, b such that 1 ≤ a < b <
r = min{{n2
m/2
/(2
m
− 1) − 1, n − 1, κ}, where κ =
n
2 m −1
(2
m/2
− 1) and
a, b ≡ 0 mod 2. Then, for any pair of nonnegative integers (a l , a r ) satisfying
a l + a r < m(t − u), there exists an (a l , a r ) − [[n + a l + a r , n − 2m(t + 1)]] QSC
that corrects up to at least
d−1
2
phase errors and up to at least
d
∗ −1
2
bit errors,
where d ≥ b + 1, d
∗
≥ a + 1, t = (b − 1)/2 and u = (a − 1)/2.
Proof Let D be the binary BCH code of length n generated by
D = =M 1 (x)M 3 (x) · · · M a (x),
where a = 2u + 1 and u ≥ 0 is an integer. Furthermore, let C be the binary BCH
code of length n generated by
C = =M 1 (x)M 3 (x) · · · M b (x),
where b = 2t + 1 and t ≥ 1 is an integer. From construction, we have C ⊂ D, and by
Proposition 5.5.1, C is dual-containing. From Lemma 5.5.1 and from straightforward
computation, the dimension of D is equal to k 2 = n − m(u + 1) and the dimension
of C is k 1 = n − m(t + 1), which implies k 2 − k 1 = m(t − u) and 2k 1 − n = n −
2m(t + 1). From the BCH bound, because the defining set of D contains a sequence
of a consecutive integers, the minimum distance of D satisfies d 2 ≥ a + 1. Similarly,
as the defining set of C contains a sequence of b consecutive integers, the minimum
distance of C satisfies d 1 ≥ b + 1. The result follows from Theorem 5.5.1 and the
proof is complete.
We next show how to construct QSCs from the sum of BCH codes.