Three new classes of optimal frequency-hopping sequence sets
aa r X i v : . [ c s . I T ] M a y Three new classes of optimal frequency-hopping sequence sets ∗ Bocong Chen , , Liren Lin , San Ling , Hongwei Liu School of Mathematics, South China University of Technology, Guangzhou, Guangdong, 510641, China Division of Mathematical Sciences, School of Physical & Mathematical Sciences, Nanyang TechnologicalUniversity, Singapore 637616, Singapore School of Mathematics and Statistics, Central China Normal University, Wuhan, Hubei, 430079, China
Abstract
The study of frequency-hopping sequences (FHSs) has been focused on the establishment of theo-retical bounds for the parameters of FHSs as well as on the construction of optimal FHSs with respectto the bounds. Peng and Fan (2004) derived two lower bounds on the maximum nontrivial Hammingcorrelation of an FHS set, which is an important indicator in measuring the performance of an FHSset employed in practice.In this paper, we obtain two main results. We study the construction of new optimal frequency-hopping sequence sets by using cyclic codes over finite fields. Let C be a cyclic code of length n overa finite field F q such that C contains the one-dimensional subcode C = { ( α, α, · · · , α ) ∈ F nq | α ∈ F q } . Two codewords of C are said to be equivalent if one can be obtained from the other through applyingthe cyclic shift a certain number of times. We present a necessary and sufficient condition under whichthe equivalence class of any codeword in C \ C has size n . This result addresses an open questionraised by Ding et al. in [9]. As a consequence, three new classes of optimal FHS sets with respectto the Singleton bound are obtained, some of which are also optimal with respect to the Peng-Fanbound at the same time. We also show that the two Peng-Fan bounds are, in fact, identical. Keywords:
Frequency-hopping sequence set, cyclic code, maximum distance separable (MDS)code, cyclotomic coset.
Let ℓ be a positive integer and let F = { f , f , · · · , f ℓ − } be an alphabet of ℓ available frequencies. Asequence X = { x t } n − t =0 is called a frequency-hopping sequence (FHS) of length n over F if x t ∈ F for all0 ≤ t ≤ n −
1. For two FHSs X = { x t } n − t =0 and Y = { y t } n − t =0 of length n over F , if x t = y t for all0 ≤ t ≤ n −
1, then we say X = Y . Let S be the set of all FHSs of length n over F . Any subset of S iscalled an FHS set . For any
X, Y ∈ S , their
Hamming correlation is defined by H X,Y ( t ) = n − X i =0 h [ x i , y i + t ] , ≤ t ≤ n − h [ a, b ] = 1 if a = b and 0 otherwise, and the subscript addition is taken modulo n . For any distinct X, Y ∈ S , we have the following measures, H ( X ) = max ≤ t F ⊆ S be a set of N sequences of length n over an alphabetof size ℓ . Define I = ⌊ nN/ℓ ⌋ . Then M ( F ) ≥ (cid:24) ( nN − ℓ ) n ( nN − ℓ (cid:25) (1.1) and M ( F ) ≥ (cid:24) InN − ( I + 1) Iℓ ( nN − N (cid:25) . (1.2) Remark 1.2. Yang et al. compared the above two Peng-Fan bounds in [25]; it was shown that the Peng-Fan bound of (1.2) may be tighter than that of (1.1). However, the authors failed to find examples wherethe bound of (1.2) is strictly tighter than that of (1.1), and finally suggested that the exact relationshipbetween the bounds (1.1) and (1.2) needs to be studied further. In this paper, we show that the twoPeng-Fan bounds are, in fact, identical. (See Theorem 1.3 below. It is reasonable to assume that nN ≥ ℓ and its proof is deferred to the Appendix.) Theorem 1.3. Let F ⊆ S be a set of N sequences of length n over an alphabet of size ℓ . Define I = ⌊ nN/ℓ ⌋ . If nN ≥ ℓ , then M ( F ) ≥ (cid:24) ( nN − ℓ ) n ( nN − ℓ (cid:25) = (cid:24) InN − ( I + 1) Iℓ ( nN − N (cid:25) . Besides the bounds on the Hamming correlation, several bounds on the size of an FHS set were alsoestablished. Ding et al. in [9] obtained a number of bounds on the size of an FHS set from certainclassical bounds in coding theory. Lemma 1.4. (Sphere-packing bound on the size of FHS sets, [9]) For any ( n, N, λ ; ℓ ) FHS set F , where λ < n and ℓ > , we have N ≤ ℓ n n (cid:16) P ⌊ ( n − λ − / ⌋ i =0 (cid:0) ni (cid:1) ( ℓ − i (cid:17) . (1.3) Lemma 1.5. (Singleton bound on the size of FHS sets, [9]) For any ( n, N, λ ; ℓ ) FHS set F , where λ < n and ℓ > , we have N ≤ (cid:22) ℓ λ +1 n (cid:23) . (1.4)2n FHS set is called optimal if one of the bounds (1.1)-(1.4) is met. It is of great interest to constructoptimal FHS sets with respect to the bounds. In recent years, numerous constructions of optimal FHSsets have been proposed (e.g., see [1]-[14], [16], [20], [23], [26]-[31], and references therein). Ding et al. generalized the ideas in [24] to construct optimal FHS sets by using some special classes of cyclic codes[9]. This idea was further investigated in [11] to obtain more optimal FHS sets. As shown in [9], there isa natural equivalence relation defined on any cyclic code: two codewords of a cyclic code are said to beequivalent if one can be obtained from the other through applying the cyclic shift a certain number oftimes. A special class of cyclic codes C ( q,m ) of length n = ( q m − / ( q − 1) over a finite field F q containing C = { ( α, α, · · · , α ) ∈ F nq | α ∈ F q } was discussed in [9]. It was shown in [9] that if the code length n is aprime number, then the equivalence class of any codeword in C ( q,m ) \ C has size n ; an FHS set is thusobtained by taking exactly one element from every equivalence class of C ( q,m ) \ C , which turns out to beoptimal with respect to the sphere-packing bound (1.3). A natural open question posed in [9, p.3302] iswhether the prime-length constraint can be dropped without changing the situation that the equivalenceclass of any codeword in C ( q,m ) \ C has size n .In this paper, we further explore the above idea to construct more optimal FHS sets by using maximumdistance separable (MDS) cyclic codes. Let F q be the finite field with q elements and let n be a positiveinteger co-prime to q . Assume that C is a cyclic code of length n over F q containing C , where C = { ( α, α, · · · , α ) ∈ F nq | α ∈ F q } . In Section 3, we present a necessary and sufficient condition under whichthe equivalence class of any codeword in C \ C has size n . This result addresses the aforementioned openquestion. Actually, it turns out that the prime-length constraint is necessary and cannot be dropped (seeCorollary 3.3 in Section 3). In Section 4, using the results in Section 3, we obtain three new classes ofoptimal FHS sets with respect to the Singleton bound (1.4), some of which are also optimal with respectto the Peng-Fan bound (1.2) at the same time. More precisely, the parameters of the new FHS sets aregiven as follows:(i) (cid:16) q + 1 , q k +1 − qq + 1 , k ; q (cid:17) where q = 2 m with m > ≤ k ≤ min { p − , m − } with p being the smallest prime divisor of q + 1. This FHS set is optimal with respect to the Singletonbound (1.4).(ii) (cid:16) q + 1 , q ( q − , q (cid:17) where q is an odd prime power. The parameters of this FHS set meet both the Peng-Fan bound(1.2) and the Singleton bound (1.4) at the same time.(iii) (cid:16) n, ( q k +2 − /n, k + 1; q (cid:17) where n > q + 1 with q being a prime power, and where k is an integer suchthat 0 ≤ k ≤ ( n − / − M with M ≤ ( n − / M, n ) > k = 0,we have an FHS set with parameters (cid:16) n, ( q − /n, q (cid:17) which meet both the Peng-Fan bound (1.2) and the Singleton bound (1.4). In this section, we review some basic notation and results about cyclic codes over finite fields. For thedetails, the reader is referred to [19] or [21]. 3et F q be the finite field with q elements and let n be a positive integer co-prime to q . A linear code C of length n over F q is called cyclic if it is an ideal of F q [ x ] / h x n − i . It follows that any cyclic code C oflength n over F q is generated uniquely by a monic divisor g ( x ) ∈ F q [ x ] of x n − 1, which is referred to asthe generator polynomial , and h ( x ) = ( x n − /g ( x ) is called its parity-check polynomial . We then knowthat the irreducible factors of x n − F q [ x ] determine all cyclic codes of length n over F q . Theoretically,the irreducible factors of x n − F q [ x ] can be derived by the q -cyclotomic cosets modulo n . For anyinteger t , the q -cyclotomic coset C t of t modulo n is defined by C t = n tq j (mod n ) (cid:12)(cid:12)(cid:12) t = 0 , , · · · o . Take α (maybe in some extension field of F q ) to be a primitive n -th root of unity, which means that n isthe smallest positive integer such that α n = 1.Let C = { } , C i , C i , · · · , C i t be all the distinct q -cyclotomic cosets modulo n . It is well known that x n − (cid:0) x − (cid:1) M ( x ) M ( x ) · · · M t ( x )with M j ( x ) = Y s ∈ C ij (cid:0) x − α s (cid:1) , ≤ j ≤ t, all being monic irreducible in F q [ x ]. The defining set of C = h g ( x ) i is the subset of integers Z = n j (cid:12)(cid:12)(cid:12) ≤ j ≤ n − , g ( α j ) = 0 o . It is readily seen that the defining set Z is a union of q -cyclotomic cosets modulo n .A linear code of length n over F q is called an [ n, k, d ] code if its dimension is k and minimum (Hamming)distance is d . The following results are well known. Lemma 2.1. (BCH bound for cyclic codes) Let C be a cyclic code of length n over F q . Let α be a primitive n -th root of unity in some extension field of F q . Assume the generator polynomial of C has roots thatinclude the set { α i | i ≤ i ≤ i + d − } . Then the minimum distance of C is at least d . Proposition 2.2. (Singleton bound) If C is an [ n, k, d ] linear code over F q , then d ≤ n − k + 1 . A linear code achieving this Singleton bound is called a maximum distance separable (MDS) code . Aremark is in order at this point. Lemma 2.1 and Proposition 2.2 provide a useful method to constructMDS cyclic codes: If the generator polynomial of a cyclic code C has roots precisely equal to the set { α i | i ≤ i ≤ i + d − } , then the minimum distance of C is exactly equal to d . In particular, C is anMDS cyclic code with parameters [ n, n − d + 1 , d ]. Indeed, it follows from Lemma 2.1 that the minimumdistance of C is at least d . Since the dimension of C is equal to n − d + 1, then the minimum distance of C is no more than n − ( n − d + 1) + 1 = d , which implies that C is an MDS cyclic code with parameters[ n, n − d + 1 , d ] (e.g., see [2]). We will construct MDS cyclic codes based on this fact. Throughout this paper, C denotes the cyclic code of length n over F q with generator polynomial 1 + x + · · · + x n − , i.e., C = n α (cid:0) x + · · · + x n − (cid:1) (cid:12)(cid:12)(cid:12) α ∈ F q o . We always assume that C is a cyclic code of length n over F q with parity-check polynomial h ( x ) suchthat h (1) = 0. Note that C is contained in C if and only if h (1) = 0.Two codewords c ( x ) , c ( x ) of C are said to be equivalent if there exists an integer t such that x t c ( x ) ≡ c ( x ) (mod x n − C then are classified into equivalence classes. We say that acodeword c ( x ) ∈ C has size n if the equivalence class containing c ( x ) has size n .The following result establishes a necessary and sufficient condition under which any codeword in C \ C has size n . 4 heorem 3.1. Let C be a cyclic code of length n over F q with parity-check polynomial h ( x ) such that h (1) = 0 , say h ( x ) = ( x − h ′ ( x ) . Then the following statements are equivalent:(i) Any codeword of C \ C has size n .(ii) Any root of h ′ ( x ) is a primitive n -th root of unity.Proof. Assume first that ( i ) holds. If n is a prime number, then all the roots of ( x n − / ( x − 1) areprimitive n -th roots of unity. In particular, the roots of h ′ ( x ) are primitive n -th roots of unity, and weare done. Therefore, we can assume that n is a composite number. To get ( ii ), it is enough to prove thatthe roots of ( x n − / ( x − 1) which have multiplicative order less than n are roots of g ( x ) = ( x n − /h ( x ).Let α be a primitive n -th root of unity. Assume to the contrary that there exists 0 < i < n withgcd( i , n ) = 1 such that α i is not a root of g ( x ), i.e., α i is a root of ( x n − / ( x − 1) having order(say r ) less than n and g ( α i ) = 0. Let C i denote the q -cyclotomic coset modulo n containing i , andassume that C = { } , C i , C i , · · · , C i t , C i t +1 , · · · , C i t + s are all the distinct q -cyclotomic cosets modulo n . Without loss of generality, suppose that the defining set of C is C i S C i S · · · S C i t , i.e., g ( x ) = t Y j =1 M j ( x ) , where M j ( x ) = Q k ∈ C ij ( x − α k ) for 1 ≤ j ≤ t. Now let m ( x ) = s Y j =1 M t + j ( x ) , where M t + j ( x ) = Q k ∈ C it + j ( x − α k ) for 1 ≤ j ≤ s. It follows that c ( x ) = m ( x ) g ( x ) lies in C \ C . However, from our construction,( x r − m ( x ) g ( x ) ≡ x n − . This says that c ( x ) is a codeword of C \ C , and the size of the equivalence class containing c ( x ) is lessthan n . This is a contradiction.Assume that ( ii ) holds. Suppose otherwise that there exists a codeword c ( x ) ∈ C \ C such that theequivalence class containing it has size t < n . Thus x t c ( x ) ≡ c ( x ) (mod x n − . Let c ( x ) = m ( x ) g ( x ), then we have( x t − m ( x ) g ( x ) ≡ x n − , and hence ( x t − m ( x ) ≡ h ′ ( x )) . (3.1)Recall that all the roots of h ′ ( x ) are primitive n -th roots of unity. Since the roots of x t − n -th root of unity, h ′ ( x ) is a divisor of m ( x ). We have arrived at a contradiction since thisimplies c ( x ) = m ( x ) g ( x ) ∈ C . Example 3.2. Consider cyclic codes of length 9 over F . It is easy to verify that all the distinct 8-cyclotomic cosets modulo 9 are given by C = { } , C = { , } , C = { , } , C = { , } and C = { , } .Take α to be a primitive ninth root of unity in F . Then x − x − M ( x ) M ( x ) M ( x ) M ( x ) , with M i ( x ) = Y j ∈ C i ( x − α j ) , ≤ i ≤ x − F . It is readily seen that the roots of M ( x ) are primitivethird roots of unity. Let C and C be cyclic codes of length 9 over F with parity-check polynomials( x − M ( x ) M ( x ) M ( x ) and ( x − M ( x ) M ( x ), respectively. It follows from Theorem 3.1 that anycodeword of C i \ C has size 9, for i = 1 , 2. Note that C is an MDS cyclic code with parameters [9 , , et al. [9, p.3302].Let m be a positive integer such that gcd( m, q − 1) = 1. Take n = ( q m − / ( q − β be agenerator of F ∗ q m = F q m \ { } and γ = β q − . Define a cyclic code C ( q,m ) by C ( q,m ) = n c ( x ) (cid:12)(cid:12)(cid:12) c ( x ) ∈ F q [ x ] n and c ( γ ) = 0 o where F q [ x ] n consists of all polynomials of degree at most n − F q . It is known that C ( q,m ) is an[ n, n − m, 3] code with defining set C = n q i (mod n ) (cid:12)(cid:12)(cid:12) ≤ i ≤ m − o . (3.2)Assuming that n = ( q m − / ( q − 1) is a prime number, Ding et al. in [9, Theorem 12] used C ( q,m ) toconstruct an FHS set whose paramters meet the sphere-packing bound (1.3). The following open questionwas posed in [9, p.3302]: “If the condition that n = ( q m − / ( q − 1) is a prime number is dropped, is itstill true that any codeword of C ( q,m ) \ C has size n ?”By Theorem 3.1, we have a complete answer to this question. Corollary 3.3. Assume the same notation as previously defined. Then any codeword of C ( q,m ) \ C hassize n if and only if n is a prime number.Proof. It is known that C ( q,m ) is a cyclic code of length n over F q with defining set given by (3.2). Thusthe generator polynomial of C ( q,m ) is g ( x ) = Y i ∈ C ( x − γ i ) . Let h ( x ) = x n − g ( x ) = ( x − h ′ ( x ). Suppose that any codeword of C ( q,m ) \ C has size n . It follows fromTheorem 3.1 that all the roots of h ′ ( x ) are primitive n -th roots of unity. Suppose otherwise that 1 < d < n is a divisor of n . Then γ d must be a root of h ′ ( x ), contradicting Theorem 3.1.The following result can be proven in a fashion similar to Theorem 3.1. Proposition 3.4. Let D be a nonzero cyclic code of length n over F q with parity-check polynomial h ( x ) ,where n is a positive integer co-prime to q . Then any nonzero codeword of D has size n if and only if theroots of h ( x ) are primitive n -th roots of unity. Note that the result of Proposition 3.4 appeared previously in [17, Theorem 1]. We will apply Theorem3.1 and Proposition 3.4 to construct optimal FHS sets. In this section, three families of optimal FHS sets with respect to the Singleton bound (1.4) are obtained.It turns out that some of the FHS sets are also optimal with respect to the Peng-Fan bound (1.2) at thesame time. In the light of [9] and [11], we first present two basic facts.Fact 1. If one has found an [ n, k, n − k + 1] MDS cyclic code C over F q with n > q satisfying Theorem 3.1,then an FHS set is obtained by taking exactly one element from every equivalence class of C \ C ,which results in an ( n, q k − qn , k − q ) FHS set. It is straightforward to verify that this FHS set isoptimal with respect to the Singleton bound (1.4).Fact 2. Similarly, if one has found an [ n, k, n − k + 1] MDS cyclic code D over F q satisfying Proposition 3.4,then an FHS set is obtained by taking exactly one element from every equivalence class of D \ { } ,which results in an ( n, q k − n , k − q ) FHS set. This FHS set is optimal with respect to the Singletonbound (1.4). 6e explain the reason for which an [ n, k, n − k + 1] MDS cyclic code over F q with n > q satisfyingTheorem 3.1 gives rise to an ( n, N, λ ; ℓ ) = ( n, q k − qn , k − q ) FHS set, i.e., Fact 1 holds. It follows fromthe definition of λ and the Singleton bound (1.4) that λ ≤ n − ( n − k + 1) = k − N = q k − qn ≤ ⌊ q λ +1 n ⌋ ,which forces λ = k − 1. Fact 2 is also satisfied for a similar reason.In the following, three families of MDS cyclic codes satisfying Fact 1 or Fact 2 are constructed andtheir parameters are computed. Consequently, optimal FHS sets with respect to the Singleton bound(1.4) are derived from these MDS cyclic codes. These FHS sets are new in the sense that their parametershave not been covered in the literature. q + 1 Consider cyclic codes of length q + 1 over F q . We first study the case q = 2 m , where m > s = 2 m − . By [21, p.324], we know that all the distinct q -cyclotomic cosets modulo q + 1 aregiven by C = (cid:8) (cid:9) and C i = (cid:8) i, q + 1 − i (cid:9) , for 1 ≤ i ≤ s. Suppose p is the smallest prime divisor of q + 1. Let α ∈ F q be a primitive ( q + 1)-st root of unity. Forany integer k with 1 ≤ k ≤ min { p − , s } , let C be a cyclic code of length q + 1 over F q with parity-checkpolynomial h ( x ) = ( x − M ( x ) M ( x ) · · · M k ( x ) , M j ( x ) = Y t ∈ C j ( x − α t ) , ≤ j ≤ k. Equivalently, C is the cyclic code of length q + 1 over F q with defining set C k +1 [ C k +2 [ · · · [ C s . Thus C is a [ q + 1 , k + 1 , q − k + 1] MDS cyclic code satisfying Theorem 3.1. The above discussion leadsto the following result. Theorem 4.1. Let q = 2 m , where m > is a positive integer. Suppose p is the smallest prime divisor of q + 1 . For any integer k with ≤ k ≤ min { p − , m − } , we have a ( q + 1 , q k +1 − qq +1 , k ; q ) FHS set whoseparameters meet the Singleton bound (1.4). As an additional remark, we observe that if m is odd in Theorem 4.1, then p = 3. This is simplybecause 3 is always a divisor of 2 m + 1 for any odd m . Example 4.2. Take q = 2 = 8 and n = q + 1 = 9. Then p = 3 and 1 ≤ k ≤ 2. Thus we have a(9 , (8 k +1 − / , k ; 8) FHS set for k = 1 , 2. It is easy to verify that these parameters indeed meet theSingleton bound (1.4). Example 4.3. Take q = 2 = 16 and n = q + 1 = 17. We have p = 17 and min { p − , } = 8, thus1 ≤ k ≤ 8. Theorem 4.1 applies to give an optimal (17 , k +1 − , k ; 16) FHS set for any 1 ≤ k ≤ q + 1 over F q in the case where q is an odd prime power. Itis easy to verify that all the distinct q -cyclotomic cosets modulo q + 1 are given by C = (cid:8) (cid:9) , C q +12 = (cid:8) q + 12 (cid:9) and C i = (cid:8) i, q + 1 − i (cid:9) , for 1 ≤ i ≤ q +12 − . Let α ∈ F q be a primitive ( q + 1)-th root of unity. In this case, let C be a cyclic code of length q + 1 over F q with parity-check polynomial h ( x ) = ( x − M ( x ), where M ( x ) = ( x − α )( x − α − ).Thus C is a [ q + 1 , , q − 1] MDS cyclic code satisfying Theorem 3.1. From Fact 1, there is an optimal( q + 1 , q ( q − , q ) FHS set with respect to the Singleton bound (1.4), where q is an odd prime power.7e claim that these parameters also meet the Peng-Fan bound (1.2). To see this, note that M ( F ) = 2and I = ⌊ nN/ℓ ⌋ = q − 1. It is clear that (cid:24) InN − ( I + 1) Iℓ ( nN − N (cid:25) = & q ( q + 1) ( q − − q ( q + 1)( q − q ( q − (cid:0) q ( q + 1)( q − − (cid:1) ' . After simple computations we have2 q ( q + 1) ( q − − q ( q + 1)( q − q ( q − (cid:0) q ( q + 1)( q − − (cid:1) > , proving the claim.Summarizing the previous discussion, we arrive at the following result. Theorem 4.4. Let q be an odd prime power. Then there is a ( q + 1 , q ( q − , q ) FHS set whoseparameters meet both the Peng-Fan bound (1.2) and the Singleton bound (1.4). Example 4.5. Take q = 5 = 25 and n = 25 + 1 = 26. Theorem 4.4 applies to give a (26 , , 2; 25)FHS set. It is easy to check that both the Peng-Fan bound (1.2) and the Singleton bound (1.4) are met. q + 1 Let n > q + 1. Let α ∈ F q be a primitive n -th root of unity. It is easy to verifythat all the distinct q -cyclotomic cosets modulo n are given by C = (cid:8) (cid:9) and C i = { i, n − i } , for 1 ≤ i ≤ n − . Let M ≤ ( n − / M, n ) = 1; if no such integer exists,then M is assumed to be 0. Observe that gcd( n − , n ) = 1, thus M ≤ ( n − / 2. Fix a value k ,0 ≤ k ≤ ( n − / − M . Let D be a cyclic code of length n over F q with parity-check polynomial h ( x ) = M ( n − / ( x ) M ( n − / − ( x ) · · · M ( n − / − k ( x )where M j ( x ) = Q t ∈ C j ( x − α t ) , ( n − / − k ≤ j ≤ ( n − / 2. It follows that D is an [ n, k +1) , n − k − Theorem 4.6. Let n > be an odd divisor of q + 1 . Let M ≤ ( n − / be the largest integer such that gcd( M, n ) > . For any integer k with ≤ k ≤ ( n − / − M , we have an optimal (cid:0) n, ( q k +2 − /n, k +1; q (cid:1) FHS set with respect to the Singleton bound (1.4).In particular, by taking k = 0 , we have an (cid:0) n, ( q − /n, q (cid:1) FHS set, whose parameters meet boththe Peng-Fan bound (1.2) and the Singleton bound (1.4). Note that the parameters (cid:0) n, ( q − /n, q (cid:1) appeared previously in [11, Corollary 12]. In comparison,our method is quite neat and clearer for understanding. Remark 4.7. Theorems 4.1, 4.4 and 4.6 generate three new families of optimal FHS sets with respectto the Singleton bound (1.4), some of which are also optimal with respect to the Peng-Fan bound (1.1).Most previously known optimal FHS sets in the literature are constructed with respect to the Peng-Fanbound (e.g., see [25, TABLE I] or [27, TABLE II]). Optimal FHS sets with respect to the Singletonbound have been mainly studied in [9], [11] and [25]. [11, Corollary 12] gives the first class of FHSsets simultaneously meeting the Peng-Fan bound and the Singleton bound. Theorem 4.4 gives one moreexample of such FHS sets. Example 4.8. Take q = 2 = 512, then q + 1 = 513 = 3 × 19. Choose n = 27, thus M = 12 and k = 0.It follows from Theorem 4.6 that we have a (27 , , 1; 512) FHS set, whose parameters meet both thePeng-Fan bound (1.2) and the Singleton bound (1.4).8 xample 4.9. Take q = 2 = 32, then q + 1 = 33 = 3 × 11. Choose n = 11, thus M = 0 and 0 ≤ k ≤ , (32 k +2 − / , k + 1; 32) FHS set for any 0 ≤ k ≤ Appendix We give a detailed proof of Theorem 1.3. Proof of Theorem 1.3: Let P F nN − ℓ ) n ( nN − ℓ and P F InN − ( I + 1) Iℓ ( nN − N . We want to prove that ⌈ P F ⌉ = ⌈ P F ⌉ . If nN = ℓ , then I = 1, thus ⌈ P F ⌉ = 0 and ⌈ P F ⌉ = 0.Hereafter, we assume that nN > ℓ .Suppose nN = ℓI + J with 0 ≤ J < ℓ . [25, Proposition 13] says that ⌈ P F ⌉ ≥ ⌈ P F ⌉ , P F − P F InN − ( I + 1) Iℓ ( nN − N − ( nN − ℓ ) n ( nN − ℓ = 2 InN ℓ − ( I + 1) Iℓ ( nN − N ℓ − ( nN − ℓ ) nN ( nN − ℓN = 2( nN − J ) nN − ( nN − J + ℓ )( nN − J ) − ( nN − ℓ ) nN ( nN − ℓN = ( ℓ − J ) J ( nN − ℓN ≥ . (4.1)If J = 0, i.e., ℓ is a divisor of nN , then P F P F 2, and we are done. We can assume, therefore, that ℓ does not divide nN . In particular, ℓ > 1. Note that I = ⌊ nN/ℓ ⌋ ≥ 1. Since nN = Iℓ + J with J = 0, wehave nN > Iℓ . Clearly, 2 I ≥ I + 1, which implies that 2 InN − ( I + 1) Iℓ > ⌈ P F ⌉ ≥ n ≤ ℓ . In this case, ⌈ P F ⌉ = 1. Our task is thus to show that ⌈ P F ⌉ = 1.To this end, it is enough to show that nN − (cid:0) In + 1 (cid:1) N + I ℓ + Iℓ > . (4.2)This is obvious because, regarding the expression on the left hand side of (4.2) as a quadratic in thevariable N , its discriminant ∆ is less than 0, i.e.,∆ = 4 (cid:0) I n − I ℓn (cid:1) + 4 (cid:0) In − Inℓ (cid:1) + 1 < . Therefore, we have obtained that ⌈ P F ⌉ = ⌈ P F ⌉ = 1 in the case where nN > ℓ and n ≤ ℓ .Now let n = sℓ + r with s > < r ≤ ℓ − ℓ is not a divisor of n ). Let rN = tℓ + J with t ≥ < J ≤ ℓ − 1. We then see that P F nN − − ℓ ) n ( nN − ℓ = s + rℓ − ( ℓ − n ( nN − ℓ . (4.3)Combining Equations (4.1) and (4.3), P F P F ℓ − J ) J ( nN − ℓN = s + rℓ + ( ℓ − J ) J ( nN − ℓN − ( ℓ − n ( nN − ℓ . (4.4)We now consider two subcases separately.Subcase 1: rℓ ≤ ( ℓ − n ( nN − ℓ , i.e., r ( nN − ≤ ( ℓ − n. In this subcase, rnN − r = r ( nN − ≤ ( ℓ − n < ℓn − r, rN < ℓ. This means that t = 0 and J = rN . Clearly,( ℓ − n ( nN − ℓ − rℓ = ( ℓ − n − r ( nN − nN − ℓ . It is readily seen that ( ℓ − n − r ( nN − < ( nN − ℓ. Thus − < rℓ − ( ℓ − n ( nN − ℓ ≤ . This leads to ⌈ P F ⌉ = s. In order to show that ⌈ P F ⌉ = ⌈ P F ⌉ = s , by Equation (4.4), it suffices toprove that rℓ + ( ℓ − J ) J ( nN − ℓN − ( ℓ − n ( nN − ℓ = rN ( nN − 1) + ( ℓ − rN ) rN − ( ℓ − nN ( nN − ℓN ≤ , or equivalently, r ( nN − 1) + ( ℓ − rN ) r − ( ℓ − n = ( n − r )( rN − ℓ + 1) ≤ . We are done because rN < ℓ and n > r .Subcase 2: rℓ > ( ℓ − n ( nN − ℓ . In this subcase, it is clear that ⌈ P F ⌉ = s + 1 . We claim that rℓ + ( ℓ − J ) J ( nN − ℓN − ℓ − nN − ℓ ≤ , then ⌈ P F ⌉ = s + 1 = ⌈ P F ⌉ and this will complete the proof.Since ( ℓ − J ) J ( nN − ℓN = ( ℓ + tℓ − rN )( rN − tℓ )( nN − ℓN = rN ℓ + 2 rN tℓ − r N − tℓ − t ℓ ( nN − ℓN , it follows that rℓ + ( ℓ − J ) J ( nN − ℓN − ( ℓ − n ( nN − ℓ = rnN − rN + rN ℓ + 2 rN tℓ − r N − tℓ − t ℓ − ℓnN + nN ( nN − ℓN . We are left to prove that ℓnN ≥ rnN − rN + rN ℓ + 2 rN tℓ − r N − tℓ − t ℓ − ℓnN + nN + ℓN. (4.5)We prove Inequality (4.5) by induction on t ≥ 0. For the base step t = 0, ℓnN − (cid:0) rnN − rN + rN ℓ − r N − ℓnN + nN + ℓN (cid:1) = n ( ℓN − rN ) + rN − rN ℓ + r N + N ( ℓn − n − ℓ ) ≥ ( ℓ + 1)( ℓN − rN ) − rN ℓ + r N where the last inequality holds because n > ℓ > r . To obtain the desired result, it is enough to show that( ℓ + 1)( ℓN − rN ) − rℓ + r N ≥ . By a simple computation:( ℓ + 1)( ℓN − rN ) − rℓ + r N ≥ ℓ − ℓr + ℓ − r + r = (cid:0) ℓ − r (cid:1) + ℓ − r > . Now suppose Inequality (4.5) holds true for any t > 0. For the inductive step, rnN − rN + rN ℓ + 2( t + 1) rN ℓ − r N − ( t + 1) ℓ − ( t + 1) ℓ − ℓnN + nN + ℓN is equal to rnN − rN + rN ℓ + 2 trN ℓ − r N − tℓ − t ℓ − ℓnN + nN + ℓN + (2 rN ℓ − ℓ − tℓ ) . 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