Cyclic codes and some new entanglement-assisted quantum MDS codes
aa r X i v : . [ c s . I T ] S e p Cyclic codes and some newentanglement-assisted quantum MDS codes ∗ Wan Jiang , Shixin Zhu , Xiaojing Chen Abstract:
Entanglement-assisted quantum error correcting codes (EAQECCs)play a significant role in protecting quantum information from decoherence andquantum noise. Recently, constructing entanglement-assisted quantum maximumdistance separable (EAQMDS) codes with flexible parameters has received muchattention. In this work, four families of EAQMDS codes with a more generallength are presented. And the method of selecting defining set is different fromothers. Compared with all the previously known results, the EAQMDS codes weconstructed have larger minimum distance. All of these EAQMDS codes are newin the sense that their parameters are not covered by the quantum codes availablein the literature.
Keywords : EAQECCs, EAQMDS codes, Cyclic codes, Defining set
Quantum error correcting codes (QECCs) play an important role in quantumcomputing and quantum communications [1]-[2]. Usually, we use [[ n, k, d ]] q todenote a q -ary quantum error correcting code (QECC), whose length is n withsize q k and minimum distance d . It can detect up to d − ⌊ d − ⌋ quantum errors. Similar to classical linear codes, there existquantum Singleton bound k ≤ n − d + 2. If k = n − d + 2, the QECC is calleda quantum maximum-distance separable (QMDS) code. As we all know, one ofthe central topics in quantum coding theory is to construct quantum codes withgood parameters, especially QMDS codes. After Calderbank et al. [3] found thatQECCs can be constructed from classical self-orthogonal codes with certain innerproduct, there are many good works about QECCs and QMDS codes [4]-[7].The CSS construction and Hermitian construction are two famous construc-tion methods for QECCs. However, they both need classical codes to be dual-containing or self-orthogonal, which is not easy to satisfy. This limitation condi-tion form a gap between classical linear codes and QECCs. In 2006, Brun et al. [9] ∗ E-mail addresses: [email protected](W.Jiang), [email protected](S.Zhu), [email protected](X.Chen). roposed the concept of EAQECCs and solved this problem successfully. Theyproved that EAQECCs allows non-dual-containing classical codes to constructQECCs if the sender and receiver shared entanglement in advance.Let q be a prime power. A q -ary EAQECC, denoted as [[ n, k, d ; c ]] q , thatencodes k information qubits into n channel qubits with the help of c pairs ofmaximally entangled states and can correct up to ⌊ d − ⌋ errors, where d is theminimum distance of the code. Similar to QECCs, there is entanglement-assistedquantum Singleton bound in the following proposition. Proposition 1.1 [9, 11]
Assume that C is an entanglement-assisted quantumcode with parameters [[ n, k, d ; c ]] q . If d ≤ ( n +2) / , then C satisfies the entanglement-assisted Singleton bound n + c − k ≥ d − . If C satisfies the equality n + c − k = 2( d − for d ≤ ( n + 2) / , then it is calledan entanglement-assisted quantum MDS code. Although it is possible to construct EAQECCs from any classical linear codesin theory, it is not easy to calculate the number of entangled states c . Thisproblem gets solved after Li et al. proposed the concept of decomposing thedefining set [10]. Besides, they used this method to construct some EAQECCswith good parameters [12]. After that, this concept is generated to more generalcases, from cyclic codes to negacyclic codes and constacyclic codes [13]-[14]. AsEAQMDS codes are an important class of quantum codes, a great deal of effortshave been made to construct some new EAQMDS codes in recent years. Thereare many new EAQMDS codes with a small number of entangled states c havebeen constructed [15]-[21].Actually, if we want to correct more quantum errors, the EAQECCs we con-structed need to have larger minimum distance. And the larger the minimumdistance of EAQECCs are, the more the entangled states c will be employed.However, it is not an easy task to analyze the accurate parameters if the value ofentangled states c is too large or flexible. Recently, some scholars have obtainedgreat progress. Luo et al. obtained several new infinite families of EAQECCswith flexible parameters by using generalized ReedSolomon codes and extendedgeneralized ReedSolomon codes [22]. In [23], Qian and Zhang constructed somenew EAQMDS codes with length n = q + 1 and some new entanglement-assistedquantum almost MDS codes. Besides, Wang et al. got series of EAQECCs withflexible parameters of length n = q + 1 by using constacyclic codes [24].Based on the previous work, this paper dedicates to study cyclic codes toconstruct EAQECCs of length n = ( q + 1) /a with flexible parameters naturally,where a = m + 1 ( m ≥ q -ary EAQMDS codes with parameters as follows:(1) [[ n, n − α ( q − m − aα ) − mk, αq + mk ) + 1; 4 α ( aα + m )]] q , where q = 2 ak + m ( k ≥
1) is an odd prime power and 1 ≤ α ≤ k .
2) [[ n, n − α ( q − a − m − aα ) − a + m ) k − ( a + 2 m ) , αq + ( a + m ) k + a +2 m ] + 1; 4 α ( aα + a + m ) + a + 2 m ]] q , where q = 2 ak + a + m ( k ≥
1) is an oddprime power and 1 ≤ α ≤ k .(3) [[ n, n − α ( q − a + m − aα ) − a − m ) k − ( a − m ) , αq + ( a − m ) k + a − m ] + 1; 4 α ( aα + a − m ) + a − m ]] q , where q = 2 ak + a − m ( k ≥
1) is an oddprime power and 1 ≤ α ≤ k .(4) [[ n, n − α [ q − (2 a − m ) − aα ] − a − m ) k − a − m ) , αq + (2 a − m ) k +2( a − m )] + 1; 4 α ( aα + 2 a − m ) + 4( a − m )]] q , where q = 2 ak + 2 a − m ( k ≥
1) isan odd prime power and 1 ≤ α ≤ k .The main organization of this paper is as follows. In Sect.2, some basic back-ground and results about cyclic codes and EAQECCs are reviewed. In Sect.3,four classes of EAQMDS codes with length n = ( q + 1) /a are obtained, where a = m + 1 ( m ≥ In this section, we will review some relevant concepts and basic theory on cycliccodes and EAQECCs.Let F q be a finite field of q elements and F nq be the n -dimensional row vectorspace over F q , where q is a prime power, k and n are positive integer. Let C bea q -ary [ n, k, d ] linear code of length n with dimension k and minimum distance d , which is a k -dimensional linear subspace of F nq . Then C satisfies the followingSingleton bound: n − k ≥ d − . When the equality n − k = d − C is called an maximal distance separa-ble (MDS) code. Let vectors x = ( x , x , · · · , x n − ) and y = ( y , y , · · · , y n − ) ∈ F nq , their Hermitian inner product is defined as h x , y i h = n − X i =0 x iq y i = x q y + x q y + · · · + x n − q y n − . The Hermitian dual code of C is defined as C ⊥ h = { x ∈ F nq | h x , y i h = 0 for all y ∈ C} . If C ⊆ C ⊥ h , then C is called a Hermitian self-orthogonal code. If C ⊥ h ⊆ C , then C is called a Hermitian dual-containing code.Let C be a linear code over F nq with length n and gcd ( q, n ) = 1. If for anycodeword ( c , c , · · · , c n − ) ∈ C implies its cyclic shift ( c n − , c , · · · , c n − ) ∈ C ,then C is said to be a cyclic code. For a cyclic code C , each codeword c =( c , c , · · · , c n − ) is identified with its polynomial form c ( x ) = c + c x + · · · + c n − x n − and cyclic code C of length n is an ideal of F q [ x ] / h x n − i . Thus, C an be generated by a monic polynomial factor of x n −
1, i.e. C = h g ( x ) i and g ( x ) | ( x n − g ( x ) is called the generator polynomial of C and the dimensionof C is n − deg ( g ( x )).Let λ denote a primitive n -th root of unity in some extension field of F q .Hence, x n − Q n − i =0 ( x − λ i ). The defining set of cyclic code C = h g ( x ) i oflength n is the set Z = { ≤ i ≤ n − | g ( λ i ) = 0 } . For 0 ≤ i ≤ n −
1, the q -cyclotomic coset modulo n containing i is defined by the set C i = { i, iq , iq , ..., iq m i − } , where m i is the smallest positive integer such that iq m i ≡ i mod n . Each C i corresponds to an irreducible divisor of x n − F q . Let C be an [ n, k, d ]cyclic code over F q with defining set Z . Obviously, Z must be a union of some q -cyclotomic cosets modulo n and dim ( C ) = n − | Z | .For 0 ≤ b + δ − ≤ n −
1, if C has defining set Z δ = C b S C b +1 S · · · S C b + δ − ,then it is called cyclic BCH code with designed distance δ . Then the followingBCH bound is a lower bound for cyclic codes. Proposition 2.1 [25] (BCH Bound for Cyclic Codes)
Let C be a q -arycyclic code of length n with defining set Z . If Z contains d − consecutiveelements, then the minimum distance of C is at least d . Let the conjugation transpose of an m × n matrix H = ( x i,j ) entries in F q isan n × m matrix H † = ( x qj,i ). According to literatures [4, 26], EAQECCs can beconstructed from arbitrary classical linear codes over F q , which is given by thefollowing proposition. Proposition 2.2 If C is an [ n, k, d ] q classical code and H is its parity checkmatrix over F q , then there exist entanglement-assisted quantum codes with pa-rameters [[ n, k − n, d ; c ]] q , where c = rank ( HH † ) . Because it is not easy to determine the number of entangled states c by comput-ing the rank of HH † , there are scholars put forward the concept of decomposingthe defining set of C as follows. Definition 2.3
Let C be a q -ary cyclic code of length n with defining set Z .Assume that Z = Z T ( − qZ ) and Z = Z \ Z , where − qZ = { n − qx | x ∈ Z } .Then, Z = Z S Z is called a decomposition of the defining set of C .After decomposing the defining set of C , there is the following lemma whichgive a relative easy method to calculate the number of entangled states c . Lemma 2.4
Let C be a cyclic code with length n over F q , where gcd( n, q ) = 1 .Suppose that Z is the defining set of the cyclic code C and Z = Z S Z is adecomposition of Z . Then, the number of entangled states required is c = | Z | . Construction of EAQMDS Codes
In this section, we devote to derive four new classes of EAQMDS codes from cycliccodes over F q . Let q be an odd prime power with a | ( q + 1), where a is even. Inthis case, we always assume n = ( q + 1) /a , where a = m + 1 ( m ≥ n over F q . By Lemma 3.1 in [8], we havethe following results directly. Lemma 3.1
Let n = ( q +1) /a , where a = m +1 ( m is odd ) , then all cyclotomiccosets modulo n containing i are as follows: C i = { i, − i } = { i, n − i } , for ≤ i ≤ n − . Next, we give a useful lemma which will be used in later constructions.
Lemma 3.2
Let n = ( q + 1) /a, a = m + 1 ( m ≥ is odd ) and s = ( n − / , q be an odd prime power with the form a | ( q ± m ) . Let ≤ l ≤ ( m − / , when m ≥ and l = 0 , when m = 1 . Then we have the following results in four casesbellow: − qC uq + v = C vq − u .
1) When a | ( q − m ) and q = 2 ak + m , then ≤ v ≤ mk , if ≤ u ≤ k , or mk + 1 ≤ v ≤ mk , m + 1) k + l (2 mk + 1) + 2 ≤ v ≤ (3 m + 1) k + l (2 mk + 1) + 1 , q − l (2 mk + 1) − ( m + 1) k ≤ v ≤ q − l (2 mk + 1) − (2 k + 1) , if ≤ u ≤ k − .2) When a | ( q − m ) and q = 2 ak + a + m , then ≤ v ≤ (2 k +1) m , l (2 mk + m +1)+( m + 1)(2 k + 1) + 2 ≤ v ≤ l (2 mk + m + 1) + (2 k + 1)(3 m + 1) / , if ≤ u ≤ k ,or q − l (2 mk + m + 1) − (2 k + 1)( m + 1) / ≤ v ≤ q − l (2 mk + m + 1) − k + 1) ,if ≤ u ≤ k − .3) When a | ( q + m ) and q = 2 ak + a − m , then l (2 mk + m −
1) + 2 k + 1 ≤ v ≤ l (2 mk + m −
1) + (2 k + 1)( m + 1) / − , if ≤ u ≤ k , or q − l (2 mk + m − − (2 k +1)(3 m +1) / ≤ v ≤ q − l (2 mk + m − − ( m +1)(2 k +1)+1 , if ≤ u ≤ k − .4) When a | ( q + m ) and q = 2 ak + 2 a − m , then l [2 m ( k + 1) −
1] + 2 k + 2 ≤ v ≤ [2 m ( k + 1) − l + m ( k + 1) + k , q − l [2 m ( k + 1) − − m ( k + 1) − k + 1 ≤ v ≤ q − l [2 m ( k + 1) − − (2 m + 1)( k + 1) − k , q − m ( k + 1) + 1 ≤ v ≤ q − m ( k + 1) if ≤ u ≤ k , or q − m ( k + 1) + 1 ≤ v ≤ q if ≤ u ≤ k − . Proof
1) Note that C uq + v = { uq + v, − ( uq + v ) } for 0 ≤ v ≤ mk , if 0 ≤ u ≤ k ,or mk +1 ≤ v ≤ mk , 2( m +1) k + l (2 mk +1)+2 ≤ v ≤ (3 m +1) k + l (2 mk +1)+1, q − l (2 mk + 1) − ( m + 1) k ≤ v ≤ q − l (2 mk + 1) − (2 k + 1), if 0 ≤ u ≤ k − ≤ l ≤ ( m − /
2, when m ≥ l = 0, when m = 1. ince − q · ( − ( uq + v )) = uq + vq = u ( q + 1) + vq − u ≡ vq − u mod n . Thisgives that − qC uq + v = C vq − u .The proofs of 2), 3) and 4) are similar to case 1), so we omit it here. (cid:3) From Lemma 3.2, we can also obtain − qC jq − t = C tq + j in each case. Wherethe range of q , t and j is below:1) When a | ( q − m ) and q = 2 ak + m , let 0 ≤ l ≤ ( m − /
2, if m ≥ l = 0, if m = 1, then − qC jq − t = C tq + j , 0 ≤ j ≤ mk , if 0 ≤ t ≤ k , or mk + 1 ≤ j ≤ mk , 2( m + 1) k + l (2 mk + 1) + 2 ≤ j ≤ (3 m + 1) k + l (2 mk + 1) + 1, q − l (2 mk + 1) − ( m + 1) k ≤ j ≤ q − l (2 mk + 1) − (2 k + 1), if 0 ≤ t ≤ k − Case I q = 2 ak + m In order to obtain the number of entangled states c , we give the followinglemma for preparation. Lemma 3.3
Let n = ( q + 1) /a, a = m + 1 ( m ≥ is odd ) , s = ( n − / and q = 2 ak + m ( k ≥ be an odd prime power. For a positive integer ≤ α ≤ k ,let T = [ s +( m + t ) k + h + α ≤ v ≤ s +( m + t +2) k +( h − − α,if v ≤ s + mk, ≤ u ≤ α, else ≤ u ≤ α − − m ≤ t ≤ (2 m − m and t is odd C uq + v when − m ≤ t ≤ − , h = 1 .when ≤ t ≤ m − , h = 2 .when m + 1 ≤ t ≤ m − , h = 3 .. . . . . .when m − m + 1 ≤ t ≤ (2 m − m, h = m + 1 . Then T T − qT = ∅ . Proof
For a positive integer α with 1 ≤ α ≤ k , let T = [ s +( m + t ) k + h + α ≤ v ≤ s +( m + t +2) k +( h − − α,if v ≤ s + mk, ≤ u ≤ α, else ≤ u ≤ α − − m ≤ t ≤ (2 m − m and t is odd C uq + v when − m ≤ t ≤ − , h = 1 . hen ≤ t ≤ m − , h = 2 .when m + 1 ≤ t ≤ m − , h = 3 .. . . . . .when (2 m − m + 1 ≤ t ≤ (2 m − m, h = m + 1 . Then by Lemma 3.2, we have − qT = [ s +( m + t ) k + h + α ≤ v ≤ s +( m + t +2) k +( h − − α,if v ≤ s + mk, ≤ u ≤ α, else ≤ u ≤ α − − m ≤ t ≤ (2 m − m and t is odd C vq − u when − m ≤ t ≤ − , h = 1 .when ≤ t ≤ m − , h = 2 .when m + 1 ≤ t ≤ m − , h = 3 .. . . . . .when (2 m − m + 1 ≤ t ≤ (2 m − m, h = m + 1 . When − m ≤ t ≤ − , h = 1, then s + 1 + α ≤ v ≤ s + mk and 0 ≤ u ≤ α , itfollows that u q + v ≤ αq + s + mk, ( s + 1 + α ) q − α ≤ v q − u . When − m ≤ t ≤ − , h = 1, then s + mk + 1 ≤ v ≤ s + ( m + 1) k − α and0 ≤ u ≤ α −
1, it follows that u q + v ≤ ( α − q + s + ( m + 1) k − α, ( s + mk + 1) q − α + 1 ≤ v q − u . When 1 ≤ t ≤ m − , h = 2, then s +( m +1) k +2+ α ≤ v ≤ s +(3 m +1) k − α +1and 0 ≤ u ≤ α −
1, it follows that u q + v ≤ ( α − q + s +(3 m +1) k − α +1 , [ s +( m +1) k +2+ α ] q − α +1 ≤ v q − u . When 2 m + 1 ≤ t ≤ m − , h = 3, then s + (3 m + 1) k + 3 + α ≤ v ≤ s + (5 m + 1) k + 2 − α and 0 ≤ u ≤ α −
1, it follows that u q + v ≤ ( α − q + s +(5 m +1) k +2 − α, [ s +(3 m +1) k +3+ α ] q − α +1 ≤ v q − u .. . . . . . When 2( m − m + 1 ≤ t m +2 ≤ (2 m − m, h m +2 = m + 1, then s + [(2 m − m +1] k + m + 1 + α ≤ v m +2 ≤ s + 2( m + 1) k + m − α and 0 ≤ u m +2 ≤ α −
1, itfollows that u m +2 q + v m +2 ≤ ( α − q + s + 2( m + 1) k + m − α, s + (2 m − m + 1) k + m + 1 + α ] q − α + 1 ≤ v m +2 q − u m +2 . It is easy to check that u q + v < v q − u , u q + v < v q − u , . . . . . . , u q + v < v m +2 q + u m +2 .u q + v < v q − u , u q + v < v q − u , . . . . . . , u q + v < v m +2 q + u m +2 .. . . . . .u m +2 q + v m +2 < v q − u , . . . . . . , u m +2 q + v m +2 < v m +2 q + u m +2 . For the range of v , v , . . . . . . , v m +2 and u , u , . . . . . . , u m +2 , note that u i q + v i ≤ ( m + q ) k ( i = 1 , , . . . . . . , m + 2), the subscripts of C u i + v i is the biggestnumber in the set. Then T T − qT = ∅ . The desired results follows. (cid:3) Based on Lemma 3.3, we can determine the number of entangled states c inthe following theorem. Theorem 3.4
Let n = ( q + 1) /a , a = m + 1 ( m ≥ is odd ) and q = 2 ak + m ( k ≥ be an odd prime power. For a positive integer α with ≤ α ≤ k , let C bea cyclic code with defining set Z given as follows Z = C s +1 [ C s +2 [ · · · [ C s +( αq + mk ) . Then | Z | = 4 α ( aα + m ) . Proof
Let T = [ s +( m + t ) k + h + α ≤ v ≤ s +( m + t +2) k +( h − − α,if v ≤ s + mk, ≤ u ≤ α, else ≤ u ≤ α − − m ≤ t ≤ (2 m − m and t is odd C uq + v when − m ≤ t ≤ − , h = 1 .when ≤ t ≤ m − , h = 2 .when m + 1 ≤ t ≤ m − , h = 3 .. . . . . .when (2 m − m + 1 ≤ t ≤ (2 m − m, h = m + 1 . and ′ = [ s +1 ≤ v ≤ s + α, ≤ u ≤ α C uq + v [ s +2 tk +1 − α ≤ v ≤ s +2 tk + α, ≤ t ≤ m − , ≤ u ≤ α C uq + v [ s +2 tk + f +32 − α ≤ v ≤ s +2 tk + f +12 + α, fm +32 ≤ t ≤ ( f +2) m − , ≤ f ≤ m − is odd, ≤ u ≤ α − C uq + v [ s +2 tk + m +1 − α ≤ v ≤ s +2 tk + m + α, (2 m − m +32 ≤ t ≤ m , ≤ u ≤ α − C uq + v [ s +2 ak + m +1 − α ≤ v ≤ s + q, ≤ u ≤ α − C uq + v [ s +( gm +1) k + g +12 − α ≤ v ≤ s +( gm +1) k + g +12 + α, ≤ g ≤ m − is odd, ≤ u ≤ α − C uq + v . From Lemma 3.2, we have − qT ′ = [ s +1 ≤ v ≤ s + α, ≤ u ≤ α C vq − u [ s +2 tk +1 − α ≤ v ≤ s +2 tk + α, ≤ t ≤ m − , ≤ u ≤ α C vq − u [ s +2 tk + f +32 − α ≤ v ≤ s +2 tk + f +12 + α, fm +32 ≤ t ≤ ( f +2) m − , ≤ f ≤ m − is odd, ≤ u ≤ α − C vq − u [ s +2 tk + m +1 − α ≤ v ≤ s +2 tk + m + α, (2 m − m +32 ≤ t ≤ m , ≤ u ≤ α − C vq − u [ s +2 ak + m +1 − α ≤ v ≤ s + q, ≤ u ≤ α − C vq − u [ s +( gm +1) k + g +12 − α ≤ v ≤ s +( gm +1) k + g +12 + α, ≤ g ≤ m − is odd, ≤ u ≤ α − C vq − u . It is easy to check that − qT ′ = T ′ . From the definitions of Z , T and T ′ , wehave Z = T S T ′ . Then from the definition of Z , Z = Z \ ( − qZ ) = ( T [ T ′ ) \ ( − qT [ − qT ′ )= ( T \ − qT ) [ ( T \ − qT ′ ) [ ( T ′ \ − qT ) [ ( T ′ \ − qT ′ )= T ′ . Therefore, | Z | = | T ′ | = 4 α ( aα + m ). (cid:3) From Lemmas 3.2, 3.3 and Theorem 3.4 above, we can obtain the first con-struction of EAQMDS codes in the following theorem.
Theorem 3.5
Let n = ( q + 1) /a , a = m + 1 ( m ≥ is odd ) and q = 2 ak + m ( k ≥ be an odd prime power. There are EAQMDS codes with parameters [[ n, n − α ( q − m − aα ) − mk, αq + mk ) + 1; 4 α ( aα + m )]] q , here ≤ α ≤ k . Proof
For a positive integer 1 ≤ α ≤ k . Suppose that C is a cyclic code oflength n = ( q + 1) /a, a = m + 1 ( m ≥ Z = C s +1 [ C s +2 [ · · · [ C s +( αq + mk ) . Then the dimension of C is n − αq + mk ). Note that cyclic code C have 2( αq + mk )consecutive roots. By Proposition 2.1, the minimum distance of C is at least2( αq + mk ) + 1. From the Singleton bound, C is an MDS code with parameters[[ n, n − αq + mk ) , αq + mk ) + 1]] q . Then by Theorem 3.4, we have | Z | = | T ′ | = 4 α ( aα + m ). From Proposition 2.2 and Lemma 2.4, there are EAQECCswith parameters[[ n, n − α ( q − m − aα ) − mk, αq + mk ) + 1; 4 α ( aα + m )]] q . It is easy to check that n − k + c + 2 = 4( αq + mk ) + 2 = 2 d. By Proposition 1.1, it implies that the EAQECCs we constructed are EAQMDScodes. (cid:3)
TABLE I: SAMPLE PARAMETERS OF MDSEAQECCs OF THEOREM 3.5 m q n α
Parameters1 13 85 1 [[85 , ,
33; 12]] , ,
59; 40]] , ,
85; 84]] , ,
43; 12]] , ,
77; 40]] , , , , , ,
99; 52]] , , , , , , , , , , , , , , Case II q = 2 ak + a + m s for the case that n = ( q + 1) /a, a = m + 1 ( m ≥ q =2 ak + a + m ( k ≥
1) is an odd prime power, we can produce the followingEAQMDS codes. The proofs are similar to that in the Case I, so we omit it here.
Lemma 3.6
Let n = ( q + 1) /a , a = m + 1 ( m ≥ is odd ) , s = ( n − / and q = 2 ak + a + m ( k ≥ be an odd prime power. For a positive integer ≤ α ≤ k ,let T = [ s + t (2 k +1)+( h +1)+ α ≤ v ≤ s +( t +1)(2 k +1)+( h − − α,if v ≤ s +( a + m ) k + a +2 m , ≤ u ≤ α, else ≤ u ≤ α − h − m + γ ≤ t ≤ hm − γ , ≤ h ≤ m C uq + v when ≤ h ≤ ( m − / , γ = 0 , γ = 1; when h = ( m + 1) / , γ = 0 , γ = 0; when ( m + 1) / ≤ h ≤ m, γ = 1 , γ = 0 . Then T T − qT = ∅ . Theorem 3.7
Let n = ( q + 1) /a , a = m + 1 ( m ≥ is odd ) , s = ( n − / and q = 2 ak + a + m ( k ≥ be an odd prime power. For a positive integer mwith ≤ α ≤ k , let C be a cyclic code with defining set Z given as follows Z = C s +1 [ C s +2 [ · · · [ C s +[ αq +( a + m ) k + a +2 m ] . Then | Z | = 4 α ( aα + a + m ) + a + 2 m . Theorem 3.8
Let n = ( q + 1) /a , a = m + 1 ( m ≥ is odd ) , s = ( n − / and q = 2 ak + a + m ( k ≥ be an odd prime power. There are EAQMDS codeswith parameters [[ n, n − α ( q − a − m − aα ) − a + m ) k − ( a + 2 m ) , αq + ( a + m ) k + a + 2 m α ( aα + a + m ) + a + 2 m ]] q . where ≤ α ≤ k . m q n α Parameters1 11 61 1 [[61 , ,
39; 24]] , ,
61; 60]] , ,
67; 24]] , , , , , , , , , , , , , , , , , , , , , , , , Case III q = 2 ak + a − m we also have similar results for n = ( q + 1) /a, a = m + 1 ( m ≥ q = 2 ak + a − m ( k ≥
1) is an odd prime power, we can produce the followingEAQMDS codes. These results are given in the following lemma and theorems.Because the proofs of them are similar to that in Lemma 3.3 and Theorems 3.4,3.5, we omit it here.
Lemma 3.9
Let n = ( q + 1) /a , a = m + 1 ( m ≥ is odd ) , s = ( n − / and q = 2 ak + a − m ( k ≥ be an odd prime power. For a positive integer ≤ α ≤ k ,let T = [ s + t (2 k +1)+(2 − h )+ α ≤ v ≤ s + t (2 k +1)+2 k − (1 − h ) − α,if v ≤ s +( a − m ) k + a − m , ≤ u ≤ α, else ≤ u ≤ α − h − m + γ ≤ t ≤ hm − γ , ≤ h ≤ m C uq + v when ≤ h ≤ ( m − / , γ = 0 , γ = 1; when h = ( m + 1) / , γ = 0 , γ = 0; when ( m + 1) / ≤ h ≤ m, γ = 1 , γ = 0 . Then T T − qT = ∅ . Theorem 3.10
Let n = ( q + 1) /a , a = m + 1 ( m ≥ is odd ) , s = ( n − / and q = 2 ak + a − m ( k ≥ be an odd prime power. For a positive integer mwith ≤ α ≤ k , let C be a cyclic code with defining set Z given as follows Z = C s +1 [ C s +2 [ · · · [ C s +[ αq +( a − m ) k + a − m ] . hen | Z | = 4 α ( aα + a − m ) + a − m . Theorem 3.11
Let n = ( q + 1) /a , a = m + 1 ( m ≥ is odd ) , s = ( n − / and q = 2 ak + a − m ( k ≥ be an odd prime power. There are EAQMDS codeswith parameters [[ n, n − α ( q − a + m − aα ) − a − m ) k − ( a − m ) , αq + ( a − m ) k + a − m α ( aα + a − m ) + a − m ]] q . where ≤ α ≤ k . TABLE III: SAMPLE PARAMETERS OF MDSEAQECCs OF THEOREM 3.11 m q n α
Parameters1 29 421 1 [[421 , ,
73; 12]] , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , Case IV q = 2 ak + 2 a − m Similarly, for the case that n = ( q + 1) /a , a = m + 1 ( m ≥ q = 2 ak + 2 a − m ( k ≥
1) is an odd prime power, there are EAQMDS codes asfollows. We only list the results in the following lemma and theorems and omitproofs for simplification.
Lemma 3.12
Let n = ( q + 1) /a , a = m + 1 ( m ≥ is odd ) , s = ( n − / and q = 2 ak + 2 a − m ( k ≥ be an odd prime power. For a positive integer ≤ α ≤ k , let T = [ s +( m + t )( k +1)+ h + α ≤ v ≤ s +( m + t +2)( k +1)+( h − − α,if v ≤ s +(2 a − m ) k +2( a − m ) , ≤ u ≤ α, else ≤ u ≤ α − − m ≤ t ≤ (2 m − m and t is odd C uq + v hen − m ≤ t ≤ − , h = 2 .when ≤ t ≤ m − , h = 1 .when m + 1 ≤ t ≤ m − , h = 0 .. . . . . .when (2 m − m + 1 ≤ t ≤ (2 m − m, h = 2 − m. Then T T − qT = ∅ . Theorem 3.13
Let n = ( q + 1) /a , a = m + 1 ( m ≥ is odd ) and q =2 ak + 2 a − m ( k ≥ be an odd prime power. For a positive integer m with ≤ α ≤ k , let C be a cyclic code with defining set Z given as follows Z = C s +1 [ C s +2 [ · · · [ C s +[ αq +(2 a − m ) k +2( a − m )] . Then | Z | = 4 α ( aα + 2 a − m ) + 4( a − m ) . Theorem 3.14
Let n = ( q + 1) /a , a = m + 1 ( m ≥ is odd ) and q =2 ak + 2 a − m ( k ≥ be an odd prime power. There are EAQMDS codes withparameters [[ n, n − α [ q − (2 a − m ) − aα ] − a − m ) k − a − m ) , αq + (2 a − m ) k + 2( a − m )] + 1; 4 α ( aα + 2 a − m ) + 4( a − m )]] q , where ≤ α ≤ k . TABLE IV: SAMPLE PARAMETERS OF MDSEAQECCs OF THEOREM 3.14 m q n α
Parameters1 23 265 1 [[265 , ,
81; 24]] , , , , , , , , , , , , , , , , , , , , Remark
The required number of entangled states c of the EAQMDS codesobtained in the literatures (see for instance [10], [12], [13], [14]) is fixed. However,the EAQMDS codes we constructed in Theorems 3.5, 3.8, 3.11 and 3.14 withflexible parameters. And we list the parameters of these codes in Table I , II , III and IV . Conclusion
In this paper, we have utilized decomposing the defining set of cyclic codes to ex-tend the present results about EAQMDS codes and construct series of EAQMDScodes with good parameters of length n = ( q + 1) /a . They conclude all thelength with the form n = ( q + 1) /a , where a = m + 1, m ≥ q isan odd prime power with the form of a | ( q + m ) or a | ( q − m ). Compared withknown results, these EAQMDS codes have flexible parameters and much biggerminimum distance than the known quantum MDS codes and EAQECCs with thesame length. Therefore, our EAQMDS codes can detect and correct more errors.The study of EAQECCs is an interesting problem in coding theory. We believethat it will have tremendous application in the future. Acknowledgments
This research is supported by the National Natural Science Foundation ofChina (No.61772168).
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