Automorphism groups and new constructions of maximum additive rank metric codes with restrictions
aa r X i v : . [ m a t h . C O ] A ug AUTOMORPHISM GROUPS AND NEW CONSTRUCTIONS OFMAXIMUM ADDITIVE RANK METRIC CODES WITHRESTRICTIONS
G. LONGOBARDI, G. LUNARDON, R. TROMBETTI, Y. ZHOU
Abstract.
Let d, n ∈ Z + such that 1 ≤ d ≤ n . A d -code C ⊂ F n × nq isa subset of order n square matrices with the property that for all pairs ofdistinct elements in C , the rank of their difference is greater than or equal to d . A d -code with as many as possible elements is called a maximum d -code.The integer d is also called the minimum distance of the code. When d < n ,a classical example of such an object is the so-called generalized Gabidulincode , [7]. In [2], [16] and [13], several classes of maximum d -codes made uprespectively of symmetric, alternating and hermitian matrices were exhibited.In this article we focus on such examples.Precisely, we determine their automorphism groups and solve the equivalenceissue for them. Finally, we exhibit a maximum symmetric 2-code which is notequivalent to the one with same parameters constructed in [16]. Introduction
Let F q be the finite field with q elements and denote by F n × nq the set of order n matrices with entries in F q . It is easy to verify that the map d defined by d ( A, B ) = rank( A − B ) , for A, B ∈ F n × nq , is a distance function on F n × nq , which is often called the rankdistance or the rank metric on F n × nq .Given any integer 1 ≤ d ≤ n , we consider here subsets C ⊂ F n × nq with theproperty that, for all distinct matrices M and M ∈ C , the rank of M − M isgreater than or equal to d . These sets are usually called rank metric codes withminimum distance d , and in some context also d -codes. Also, we say that a d -code C ⊂ F n × nq is additive if C is a subgroup of ( F n × nq , +). An F q -linear d -code is asubspace of F n × nq viewed as an n -dimensional vector space over F q .For the applications in classical coding theory, given n and d , it is desirable tohave d -codes which are maximum in size. In the general case, which means if it isnot required that all elements in the set must possess specific restrictions, Delsarteproved that this bound is q n ( n − d +1) (the so-called Singleton-like bound for rankdistance codes) [3]. If the cardinality of the code C meets this bound, we say that C is a Maximum Rank Distance code , (
MRD-code , for short), or maximum d-codes .Let F q n be a finite field of order q n , q a prime power. Let L q [ x ] = ( f ( x ) = k X i =0 c i x q i : c i ∈ F q n , k ∈ Z + ) , i.e., the set of so-called linearized polynomials over F q n (or q -polynomials ). If k isthe largest integer such that c k = 0, we say that k is the q -degree of f . ank metric codes consisting of order n square matrices can be considered alsoin q -polynomial representation.Indeed it is well known that L ( n,q ) [ x ] = L q [ x ] / ( x q n − x ) is equivalent to End F q ( F q n ),i.e., the set of all endomorphisms of F q n seen as a vector space over F q . Hence, thealgebraic structure ( L ( n,q ) [ x ] , + , ◦ , · ), where + is addition of maps, ◦ is the compo-sition of maps (mod x q n − x ) and · is the scalar multiplication by elements of F q ,is isomorphic to the algebra F n × nq .Let Tr q n /q be the trace function of F q n over F q . The map(1) T : ( x, y ) ∈ F q n × F q n → Tr q n /q ( xy ) ∈ F q , is a non-degenerate F q -bilinear form of F q n .Let f ( x ) = P n − i =0 a i x q i be an F q -linear map of F q n . Using the terminology of[17], we denote by f ⊤ the adjoint map of f with respect to T ; i.e., f ⊤ ( x ) = n − X i =0 a q i n − i x q i . If f = f ⊤ , we say that f is self-adjoint with respect to the bilinear form definedin (1). If C is a code consisting of q -polynomials, then the adjoint code of C is C ⊤ = { f ⊤ : f ∈ C} . In fact, the adjoint of f is equivalent to the transpose of thematrix in F n × nq derived from f .In the literature, codes in the rank metric context are studied up to severaldefinitions of equivalence relation; see [1, 11]. For what is needed here we may saythat two sets of q -polynomials over F q n , say C and C ′ , are equivalent if there existtwo permutation q -polynomials g , g and ρ ∈ Aut( F q ) such that(2) C ′ = { g ◦ f ρ ◦ g ( x ) + h ( x ) : f ∈ C} , where h ( x ) ∈ L ( n,q ) [ x ], and ( P a i x q i ) ρ := P a ρi x q i . Although, in general isometricequivalence covers the possibility when C ′ = { g ◦ f ⊤ ρ ◦ g ( x ) + h ( x ) : f ∈ C} ;see for instance [19].We indicate the fact that C and C ′ are equivalent codes by the symbol C ≃ C ′ ,and denote by [ C ] ≃ the equivalence class of C with respect to relevant equivalencerelation.Let g , ρ, g , h be as above. In the following we will use the symbol Φ g ,ρ,g ,h todenote the map of L ( n,q ) [ x ] defined by f ( x ) g ◦ f ρ ◦ g ( x ) + h ( x ) . The automorphism group of C consists of all Φ g ,ρ,g ,h fixing C .In this paper we will be mainly interested in the case when the sets C and C ′ areadditive. It is not difficult to see that if this is the case, we may assume h ( x ) to bethe null map in the definitions above.If n = d , then C ≤ q n . When the equality holds such a set consists of q n invertible endomorphisms of F q n . Hence, C is a spread set of End F q ( F q n ), and if C is additive this is also equivalent to a semifield spread set of End F q ( F q n ). For moreresults on semifields and related structures, we refer to [6], [12].In the case when d < n , the most important example of additive M RD -codeof L ( n,q ) [ x ], is the so-called Generalized Gabidulin code . This family was foundby Kshevetskiy and Gabidulin in [7]. It appeared as a generalization of the family iscovered many years before independently by Gabidulin [4] and Delsarte [3], whoseelements are nowadays known with the name of Delsarte-Gabidulin codes.Precisely, let k, n be positive integers and let s be an integer coprime with n ; aGeneralized Gabidulin code with stated parameters is the set of linearized polyno-mials(3) G n,k,s = ( k − X i =0 a i x q si : a , a , . . . , a k − ∈ F q n ) . The code G n,k,s is an F q -subspace of L ( n,q ) [ x ] of dimension kn , hence it hassize q nk , and any non-zero element in G n,k,s has rank greater than or equal to d = n − k + 1. Hence, G n,k,s is an F q -linear MRD-code with minimum distance d .In [2], [16] and [13], constructions of this sort have been exhibited for sets oflinearized polynomials with prescribed restrictions. Precisely, for polynomials as-sociated with symmetric, alternating and hermitian forms. In all such settings aheavy use of the theory of association schemes led to the determination of boundson the size of such d -codes. Moreover, in the additive case such bounds are provento be tight by exhibiting families of F q -linear examples attaining these bounds.In this article we elaborate on such maximum F q -linear codes. Precisely, inSection 3 we determine their automorphisms group and solve the equivalence issuefor them. In Section 4, we characterize relevant d -codes as the intersection of theirambient space with a suitable code which is equivalent to a generalized Gabidulincode with minimum distance d . Finally, in Section 5 we exhibit a symmetric 2-codeof order q m , which is not equivalent to the one with same parameters constructedin [16]. 2. Preliminaries
We start this section by giving a description of the known examples of maximumadditive d -codes presented in [3], [16] and [13], in terms of q -polynomials.In order to do that we first remind the following very well known fact, which inthe symmetric setting is stated for instance in [15, Lemma 13]: Proposition 2.1.
Let ℓ be an arbitrary integer.(1) For each m -dimensional F q -subspace U of F q n , every bilinear form B : U × F q n → F q can be written in the following form B ( x, y ) = Tr q n /q m − X j =0 a j yx q ( j − ℓ ) ! , for some uniquely determined a , a , . . . , a m − ∈ F q n .(2) For each m -dimensional F q -subspace U of F q n , every Hermitian form H : U × F q n → F q can be expressed in the form H ( x, y ) = Tr q n /q m − X j =0 a j y q x q j − ℓ ) ! , for some uniquely determined a , a , . . . , a m − ∈ F q n . In particular, each bilinear form say B ( · , · ) defined over F q n , seen as a vectorspace over F q , can be written in the following shape: B ( x, y ) = Tr q n /q ( f ( x ) y ) , here f ( x ) ∈ L ( n,q ) [ x ].2.1. Known constructions in the symmetric and alternating setting. A symmetric F q -bilinear form B of F q n is a bilinear form such that for each x, y ∈ F q n ,(4) B ( y, x ) = B ( x, y ) . By Proposition 2.1, there is a q -polynomial f ( x ) such that B ( x, y ) = Tr q n /q ( f ( x ) y ) , and by (4) we must have for all x, y ∈ F q n , Tr q n /q ( f ( y ) x ) = Tr q n /q ( f ( x ) y ). It isroutine to verify thatTr q n /q ( f ( y ) x ) = Tr q n /q ( f ( x ) y ) = Tr q n /q ( xf ⊤ ( y )) , which means that f is a self-adjoint map with respect to T given in (1).Therefore, by suitably choosing an F q -basis of F q n , we can identify the set ofsymmetric bilinear forms over F q n , with the n ( n +1)2 -dimensional subspace S n ( q ) ⊂ End F q ( F q n ) of self-adjoint F q -linear maps of F q n . Precisely,(5) S n ( q ) = ( n − X i =0 c i x q i : c n − i = c q ( n − i ) i for i ∈ { , , . . . , n − } ) . An alternating F q -bilinear form B of F q n instead is a bilinear form such that forall x ∈ F q n , (6) B ( x, x ) = 0;from which the additional property(7) B ( x, y ) + B ( y, x ) = 0follows.By Proposition 2.1, Equations (6) and (7), and again properly choosing an F q -basis of F q n , the set of alternating bilinear form with entries running over F q canbe seen as the following subset of q -polynomials:(8) A n ( q ) = ( n − X i =1 c i x q i : c n − i = − c q ( n − i ) i for i ∈ { , , . . . , n − } ) . Clearly, A n ( q ) is an n ( n − -dimensional subspace of End F q ( F q n ) and it is well knownthat the rank of each element of A n ( q ), is necessarily even.Denote by the symbol X n either the subspace S n ( q ) or A n ( q ). It is readilyverified that for given a ∈ F ∗ q , ρ ∈ Aut( F q ), g a permutation q -polynomial over F q n ,and r ∈ X n , the map Ψ : X n → X n defined by(9) Ψ a,g,ρ,r ( f ) = ag ◦ f ρ ◦ g ⊤ ( x ) + r ( x ) , preserves the rank distance on X n . In fact, the converse statement is also trueexcept when q = 2 and n = 3 if X n = S n ( q ), and except when n ≤ X n = A n ( q );see [19].For two subsets C and C of X n , if there exists a map Ψ a,g,ρ,r defined as inEquation (9) for certain a , g , ρ and r such that C := { Ψ a,g,ρ,r ( f ) : f ∈ C } , hen we say that C and C are equivalent in X n , and to distinguish this relationfrom the one defined in Section 1, we write C ∼ = C .Regarding upper bounds for such d -codes, parts of the following results can befound in [16, Theorem 3.3] and [15, Corollary 7, Remark 8], and the last open casethat q and d both even was proved in [14]. Theorem 2.2. [14]
Let C be a d -code in S n ( q ) , where C is required to be additiveif d is even. Then (10) C ≤ (cid:26) q n ( n − d +2) / , if n − d is even; q ( n +1)( n − d +1) / , if n − d is odd. Recall that in the alternating setting, the rank of matrices are always even. Wehave a result of the same sort due to Delsarte and Goethals; precisely,
Theorem 2.3. [2]
Let m = ⌊ n ⌋ and assume that C is any e -code in A n ( q ) , then C ≤ q n ( n − m ( m − e +1) . Also in [2], Delsarte and Goethals exhibited a class of F q -linear maximal codesin A n ( q ) for any characteristic, and any odd value of n .Precisely, let 2 ≤ d = 2 e ≤ n −
1, and let s be an integer coprime with n . Thenthe set of q -polynomials(11) A n,d,s = ( n − X i = e (cid:18) b i x q si − ( b i x ) q s ( n − i ) (cid:19) : b e , . . . , b n − ∈ F q n ) is a maximum alternating d -code [2, Theorem 7].In [16], Kai-Uwe Schmidt presented the following class of additive (in fact, F q -linear) codes in S n ( q ). For any integer 1 ≤ d ≤ n such that n − d is even and s coprime with n , consider the following subset of S n ( q ):(12) S n,d,s = ( b x + n − d X i =1 (cid:16) b i x q si + ( b i x ) q s ( n − i ) (cid:17) : b , b , . . . , b n − d ∈ F q n ) . The set S n,d,s turns out to be a maximum d -code [16, Theorem 4.4]. Also in[16] the author showed that for any such d , it is always possible to construct amaximal d -code of S n ( q ) with n − d an odd integer; in fact, by simply puncturing the ( d + 2)-code S n +1 ,d +2 ,s of S n +1 ( q ) [16, Theorem 4.1].2.2. Known constructions in the Hermitian setting.
Let F q n be the finitefield of order q n equipped with the involuntary automorphism a a q of F q .A Hermitian form on F q n , is a map H : F q n × F q n → F q which is F q -linear in the first coordinate and satisfies the following property(13) H ( y, x ) = H ( x, y ) q , for all x, y ∈ F q n .It is easy to check that for all x ∈ F q n , Tr q n /q ( x ) q = Tr q n /q ( x q ).Also, the map S : ( x, y ) ∈ F q n × F q n → Tr q n /q ( xy q ) s a non-degenerate sesquilinear form of F q n with companion automorphism a a q .Again by Proposition 2.1 (b), every such a sesquilinear form can be written inthe following fashion: H ( x, y ) = S ( f ( x ) , y ) = Tr q n /q ( f ( x ) y q ) , where f ( x ) ∈ L ( n,q ) [ x ] is a q -polynomial with coefficients in F q n .Now, let f ( x ) = P n − i =0 a i x q i be an element of L ( n,q ) [ x ] (which can be viewed asan element of End F q ( F q n )). It is easy to show that S ( f ( x ) , y ) q = S ( ˜ f ( y ) , x ) forall x, y ∈ F q n where ˜ f ( x ) = f ⊤ q ( x q ) = n − X i =0 a q n − i +1 i x q n − i +1) . Here f ⊤ denotes the adjoint map of f as an F q -linear map, i.e., f ⊤ = P n − i =0 a q i n − i x q i .It is routine to verify that ˜( · ) is involutionary on each F q -linear map.Then by (13), we obtain S ( f ( y ) , x ) = H ( y, x ) = H ( x, y ) q = S ( f ( x ) , y ) q = S ( ˜ f ( y ) , x )for all x, y ∈ F q n .Hence, we may identify the set of Hermitian forms defined on F q n with the setof q -polynomials(14) H n ( q ) = ( n − X i =0 c i x q i : c n − i +1 = c q n − i +1 i , i ∈ { , , , . . . , n − } ) , where the indices of the c i ’s are taken modulo n . The set H n ( q ) is an n -dimensional F q -vector subspace of End F q ( F q n ). We explicitly note that if f ( x ) = P n − i =1 c i x q i ∈ H n ( q ) with n odd, then c ( n +1) / ∈ F q n .For given a ∈ F ∗ q , ρ ∈ Aut( F q ), g a permutation q -polynomial over F q n , and r ∈ H n ( q ), the map Θ : H n ( q ) → H n ( q ) defined by(15) Θ a,g,ρ,r ( f ) = ag ◦ f ρ ◦ g ⊤ q n − ( x ) + r ( x ) , preserves the rank distance. The converse statement is also true, see [19] .In this context if for C , and C ∈ H n ( q ), there exists a map Θ a,g,ρ,r defined asin Equation (15) for certain a , g , ρ and r such that C := { Θ a,g,ρ,r ( f ) : f ∈ C } , then we say that C and C are equivalent in H n ( q ), and write C ∼ = C .Regarding upper bounds for codes in this context, we may state the followingresult. Theorem 2.4. [13, Theorem 1]
Assume that C is an additive d -code in H n ( q ) ,then C ≤ q n ( n − d +1) . Moreover, when d is odd, this upper bound also holds for non-additive d -codes. et s be an odd integer coprime with n . The following two classes of F q -linearcodes in H n ( q ), were presented in [13] only for s = 1. However, the case with s ∈ Z , s = 1 can be easily proved with the same technique used for generalizingGabidulin codes in [7, 17, 8].Suppose that n and d are integers with opposite parity such that 1 ≤ d ≤ n − H n,d,s = (cid:26) n − d +12 X j =1 (cid:18) ( b j x ) q s ( n − j +1) + b q s j x q sj (cid:19) : b , b , . . . , b n − d +12 ∈ F q n (cid:27) , is a maximum F q -linear Hermitian d -code [13, Theorem 4].Also, suppose that n and d are both odd integers such that 1 ≤ d ≤ n and s asabove; then, the set(17) E n,d,s = (cid:26) ( b x ) q s ( n +1) + n − d X j =1 (cid:18) ( b j x ) q s ( n +2 j +1) + b q s j x q s ( n − j +1) (cid:19) : b ∈ F q n and b , . . . , b ( n − d ) / ∈ F q n (cid:27) is a maximum F q -linear Hermitian d -code [13, Theorem 5].3. Automorphism groups of known constructions
Recall that the symbol X n denotes here one of the subspaces S n ( q ) and A n ( q ) ofEnd F q ( F q n ). Instead the symbol H n ( q ) is used to denote the n -dimensional F q -subspace of End F q ( F q n ) associated with a Hermitian form defined on F q n , withcompanion automorphism a a q .The aim here is determining the automorphism group of examples introduced inprevious section.We start by giving an alternative description of such d -codes in terms of the inter-section of their ambient space with suitable subspaces of L ( n,q ) [ x ] (or of L ( n,q ) [ x ],when dealing with the Hermitian setting). Precisely, Proposition 3.1.
Let n, s and d be integers such that ≤ d ≤ n and gcd( s, n ) = 1 .Let G = G n,n − d +1 ,s ⊂ L ( n,q ) [ x ] be the generalized Gabidulin code with minimumdistance d , then we have the following(1) S n,d,s = G ′ ∩ S n ( q ) , where G ′ = G ◦ x q s ( n + d .(2) A n,d,s = G ′ ∩ A n ( q ) , where G ′ = G ◦ x q s d .Moreover, let G = G n,n − d +1 ,s ⊂ L ( n,q ) [ x ] be the generalized Gabidulin code withminimum distance d , then we have the following(3) H n,d,s = G ′ ∩ H n ( q ) , where G ′ = G ◦ x q s ( n + d +1) .(4) E n,d,s = G ′ ∩ H n ( q ) , where G ′ = G ◦ x q s ( d +1) .Proof. Let f ( x ) = P n − di =0 a i x q si be an element of G n,n − d +1 ,s . Each element in G ′ = G n,n − d +1 ,s ◦ x q s ( n + d has the following form: n − d X i =0 a i x q s ( n + d i ) = n − d − X i =0 a i x q s ( n + d i ) + n − d X i = n − d a i x q s ( n + d i ) = − d X j =0 a j + n − d x q sj + n − X j = n + d a j − n + d x q sj =(18) a n − d x + n − d X j =1 (cid:0) a n − d + j x q sj + a n − d − j x q s ( n − j ) (cid:1) . It is clear that G ′⊤ = G ′ , and by intersecting G ′ with S n ( q ), we get the followingconditions a n − d − i = a q s ( n − i ) n − d + i , i = 1 , , . . . , n − d . Hence, each element in G ′ ∩ S n ( q ) has the following shape:(19) a n − d x + n − d X i =1 (cid:0) a n − d + i x q si + ( a n − d + i x ) q s ( n − i ) (cid:1) . This proves ( i ). Point ( ii ) is obtained arguing in the same way.Regarding ( iii ) and ( iv ), let f ( x ) = P n − di =0 a q s i x q si be any element in G n,n − d +1 ,s ⊂L ( n,q ) [ x ]. Composing f ( x ) on the right with the monomial x q s ( n + d +1) , we obtain a q s x q s ( n + d +1) + n − d − X i =1 a q s i x q s (cid:0) n + d +12 + i (cid:1) + n − d X i = n − d +12 a q s i x q s (cid:0) n + d +12 + i (cid:1) = n − d +12 X j =0 a q s j + n − d − x q sj + n − X j = n + d +12 a q s j − n + d +12 x q sj = n − d +12 X j =1 (cid:18) a q sn − d +12 − j x q s (2 n − j +2) + a q sn − d − + j x q sj (cid:19) . By intersecting G ′ with H n ( q ), we get the following conditions c q s (2 n − j +1) j = a q s (2 n − j +2) n − d − + j = c n − j +1 = a q sn − d +12 − j , for j = 1 , , . . . , n − d + 12 . Hence, we get G ′ ∩ H n ( q ) = H n,d,s . In a similar way, by composing an element f ( x ) ∈ G n,n − d +1 ,s with x x q s ( d +1) ,we obtain n − d X i =0 a q s i x q s (cid:0) d +12 + i (cid:1) = a q sn − d x q s ( n +1) + n − d − X i =0 a q s i x q s (2 i + d +1) + n − d X i = n − d +1 a q s i x q s (2 i + d +1) = a q sn − d x q s ( n +1) + n − d X i =1 a q s i − x q s (2 i + d − + n − d X j = n − d +1 a q s j x q s (2 j + d +1) . Setting i = n − d − l + 1 and j = n − d + m , we obtain q sn − d x q s ( n +1) + n − d X l =1 a q sn − d − l x q s ( n − l +1) + n − d X m =1 a q sn − d + m x q s ( n +2 m +1) = a q sn − d x q s ( n +1) + n − d X j =1 (cid:0) a q sn − d − j x q n − j +1 + a q sn − d + j x q s ( n +2 j +1) (cid:1) . Again by intersecting G ′ with the Hermitian space H n ( q ), we get: a q sn − d ∈ F q n c q s (2 n − j ) n +12 + j = a q s (2 n − j +1) n − d + j = c n +12 − j = a q sn − d − j , which finally gives the result. (cid:3) Regarding the punctured set obtained from S n +1 ,d +2 ,s , we can consider F q n +1 ≃ V ⊕ K , where K = h η i q with η ∈ F ∗ q n +1 and V is an n -dimensional F q -subspace of F q n +1 .Let s be a positive integer coprime with n + 1, let 1 ≤ d ≤ n − n − d isodd, and consider the F q -vector space U ′ η of G ′ = G n +1 ,n − d +2 ,s ◦ x q s (cid:0) n + d +12 (cid:1) definedas follows(20) U ′ η = ( n − d +12 X i =1 (cid:0) c i ( x q si − xη q si − ) + c n +1 − i ( x q s ( n +1 − i ) − xη q s ( n +1 − i ) − ) (cid:1) : c i , c n +1 − i ∈ F q n +1 , i ∈ (cid:26) , , . . . , n − d + 12 (cid:27) ) . We notice that U ′ η has dimension ( n + 1)( n − d + 1), and it is made up of allmaps f ∈ G ′ such that K ⊆ Ker f . Let S n +1 ,d,s ∩ U ′ η = ( n − d +12 X i =1 (cid:16) b i ( x q si − xη q si − ) + b q s ( n +1 − i ) i ( x q s ( n +1 − i ) − xη q s ( n +1 − i ) − ) (cid:17) : b , . . . , b n − d +12 ∈ F q n +1 ) . Clearly each polynomial f in this set has at most q n − d +1 roots in F q n +1 . Fur-thermore, since f is a linearized polynomial, we can write f ( x + u ) = f ( x )+ f ( u ) forall x, u ∈ F q n +1 . But K ⊆ Ker f which implies that, if f ( x ) = 0, then f ( x + u ) = 0for all u ∈ K . For each x ∈ V and each u ∈ K ∗ , we have x + u V , so the numberof roots of the polynomial f in V is at most q n − d , i.e.dim(Ker f ∩ V ) ≤ n − d. Hence, for each f ∈ S n +1 ,d,s ∩ U ′ η , the rank of the symmetric bilinear form on V B f | V : ( x, y ) ∈ V × V → Tr q n /q ( f ( x ) y )is at least d and the set T n,d,s ( η ) = ( S n +1 ,d,s ∩ U ′ η ) | V = (cid:8) B f | V : f ∈ S n +1 ,d,s ∩ U ′ η (cid:9) s a symmetric F q -linear maximum d -set of size q ( n +1) n − d +12 .By Proposition 3.1 (i), we have the following. Corollary 3.2.
Let ( n + 1 , s ) = 1 , and ≤ d ≤ n − . Let η ∈ F ∗ q n +1 and let V be an n -dimensional F q -subspace of F q n +1 such that F q n +1 = V ⊕ h η i q . Then the d -code (21) T n,d,s ( η ) = ( U ′ η ∩ S n +1 ( q )) | V , is maximum, where U ′ η is the F q -subspace in (20). Clearly, if η and η are linearly dependent over F q , then T n,d,s ( η ) = T n,d,s ( η ).Furthermore, we notice that U ′ η ∩ S n +1 ( q ) ⊂ G ′ ∩ S n +1 ( q ) = S n +1 ,d +2 ,s , while(22) T n,d,s ( η ) = ( U ′ η ∩ S n +1 ( q )) | V = ( S n +1 ,d +2 ,s ) | V . In the rest part of this section we prove that the subspace G ′ ⊂ L ( n,q ) [ x ] ( G ′ ⊂ L ( n,q ) [ x ]) defined in Proposition 3.1, is the unique element in [ G n,n − d +1 ,s ] ≃ satisfying properties ( i ) and ( ii ) of Proposition 3.1. More precisely, we have thefollowing Theorem 3.3.
Let n, s and d be integers such that d ≥ and ( s, n ) = 1 .(i) Let W ⊂ L ( n,q ) [ x ] be an ( n − d + 1) n -dimensional subspace of L ( n,q ) [ x ] suchthat W ∈ [ G n,n − d +1 ,s ] ≃ , and W ∩ S n ( q ) = S n,d,s (respectively, W ∩ A n ( q ) = A n,d,s ).Then W = G n,n − d +1 ,s ◦ x q s n + d (respectively, W = G n,n − d +1 ,s ◦ x q s d ).(ii) Let W ⊂ L ( n,q ) [ x ] be an ( n − d + 1) n -dimensional F q -subspace such that W ∈ [ G n,n − d +1 ,s ] ≃ and W ∩ H n ( q ) = H n,t,s (respectively, W ∩ H n ( q ) = E n,d,s ).Then, W = G n,n − d +1 ,s ◦ x q s ( n + d +1) (respectively, W = G n,n − d +1 ,s ◦ x q s ( d +1) ).Proof. ( i ) Since W is equivalent to G n,n − d +1 ,s , there exists a rank-preserving mapΦ g,ρ,h such that Φ g,ρ,h ( G n,n − d +1 ,s ) = W. As G ρn,n − d +1 ,s = G n,n − d +1 ,s for all ρ ∈ Aut( F q ), we may assume that ρ is theidentity. Hence, the elements of W are g ◦ n − d X j =0 α j x q sj ! ◦ h = n − d X j =0 ( g ◦ α j x q sj ◦ h ) = n − d X j =0 n − X m =0 c m,j ( α j ) x q sm ! = n − X m =0 n − d X j =0 c m,j ( α j ) ! x q sm , with α j ∈ F q n for all j ∈ J = { , , . . . , n − d } and c m,j ( α j ) = n − X i =0 g i h q si m − i − j α q si j . The indices here are taken modulo n . uppose that W ∩ S n ( q ) = S n,d,s . By (5) and (12), we have that L m ( α ) = P n − dj =0 c m,j ( α j ) is equal to zero for each α = ( α , α , . . . , α n − d ), m ∈ M = { n − d + 1 , n − d + 2 , . . . , n − ( n − d + 1) } .In particular L m ( α ) = 0 when α = (0 , . . . , , α j , , . . . , α j ∈ F q n , m ∈ M and j ∈ J . Then c m,j ( α ) = 0 for all α ∈ F q n and m ∈ M, j ∈ J. Hence, we obtain the following conditions:(23) ( g i h q si m − i − j = 0 i ∈ I := { , , ..., n − } , j ∈ J, m ∈ M. As g is an invertible q -polynomial, there exists at least an integer i ∈ I suchthat g i = 0. It is straightforward to verify that (cid:26) m − j + n − d j ∈ J and m ∈ M (cid:27) = { , , ..., n − } . Hence, we get that for each given i ∈ I , by letting j varying in J , the element m − i − j may equal, modulo n , all elements in I with the only exception of n + d − i .But this finally implies that there exists a unique index i between 0 and n − g i = 0 and h n + d − i = 0; while all others g i and h i are zero.Hence, g ( x ) = γx q si and h ( x ) = δx q s ( n + d − i with γ, δ ∈ F q n .On the other hand if W ∩ A n ( q ) = A n,d,s , by (8) and taking into account (11), we may conclude that L m ( α ) = n − d X j =0 c m,j ( α j )is equal to zero for each α = ( α , α , . . . , α n − d ) ∈ F n − d +1 q n , m ∈ M = M ∪ M = { , , . . . , d − } ∪ { n − ( d − , . . . , n − } . In particular L m ( α ) = 0 forall (0 , . . . , α j , . . . , α j ∈ F q n , m ∈ M .Then c m,j ( α ) = 0 for all α ∈ F q n and m ∈ M, j ∈ J. Hence, we obtain an analogous set of conditions; i.e., ( g i h q si m − i − j = 0 i ∈ I, j ∈ J, m ∈ M. As g is an invertible q -polynomial, there exists i ∈ I such that g i = 0, andsince d ≤ m − j − d ≤ n − j ∈ J . Again, one easily verifies that (cid:26) m − j − d j ∈ J and m ∈ M ∪ M (cid:27) = { , , ..., n − } . Again for each given i ∈ I , by letting j varying in J , the element m − i − j maybe equal, modulo n to all elements of I , except d − i . Arguing as in the previouspart this leads to prove that there exists a unique index i between 0 and n − g i = 0 and h d − i = 0; while all others g i and h i are zero. ence we have that g ( x ) = γx q si and h ( x ) = δx q s ( i d with γ, δ ∈ F ∗ q n . Thisconclude the proof.( ii ) The proof of this point is similar to that of previous one. For this reason weomit here computations. (cid:3) As a direct consequence of Theorem 3.3, we may state the following result.
Corollary 3.4.
Let d and s be integers such that < d < n and gcd( n, s ) = 1 . Let C ∈ X n be a d -code.(i) If either C = S n,d,s or C = A n,d,s . Then we have Aut( C ) = (cid:8) Ψ a,γx qr : a ∈ F ∗ q , γ ∈ F ∗ q n , r ∈ { , ..., n − } (cid:9) . (ii) If C ∈ H n ( q ) and either C = H n,d,s or C = E n,d,s . Then we have Aut( C ) = n Θ a,γ q x q r : a ∈ F ∗ q , γ ∈ F ∗ q n , r ∈ { , ..., n − } o . Proof. ( i ) We first observe that Aut( G ′ ) = Aut( G n,n − d +1 ,s ), whenever G ′ = G ◦ x q s ( n + d or G ′ = G ◦ x q s ( d . Nonetheless, in [17] it was proven that if 0 ≤ r ≤ n − G n,n − d +1 ,s ) = (cid:8) Φ αx qr ,id,βx qn − r | α, β ∈ F ∗ q n (cid:9) . Now, assume that either C = S n,d,s or C = A n,d,s .Since each element in the set(24) A = { Φ aγx qr ,id,γ qn − r x qn − r | a ∈ F ∗ q , γ ∈ F ∗ q n } fixes both S n ( q ) and A n ( q ), then as a consequence of Proposition 3.1, we get that A is a subgroup of Aut( C ). Conversely, let Φ ∈ Aut( C ). Of course, by points ( i )and ( ii ) of Proposition (3.1), we getΦ( G ′ ) ∩ Φ( X n ) = C , whenever X n = S n ( q ) and G ′ = G ◦ x q s ( n + d , or X n = A n ( q ) and G ′ = G ◦ x q s ( d ,respectively. This also means that D = Φ( G ′ ) ∩ X n ⊇ C . Now, assume that
D ⊃ C . Then D would be a d -code in X n with D > n ( n − d +2)2 ,which, since n − d is even, by Theorem 2.2 is clearly not possible. Hence,(25) Φ( G ′ ) ∩ X n = C . However, above Equation (25) contradicts Theorem 3.3, unless we have Φ( G ′ ) = G ′ ,which implies that Φ is an element of A . This conclude the proof of point ( i ).( ii ) Assume now that C ⊂ H n ( q ) is either H n,d,s or E n,d,s . Again, it is trivial tosee that, in both cases, each element in the set A = { Φ aγ q x q r ,id,γ q n − r +1 x q n − r : a ∈ F ∗ q , γ ∈ F ∗ q n } , fixes C . Moreover, an easy computation also shows that if either C = H n,d,s and G ′ = G ◦ x q s ( n + d +1) , or C = E n,d,s and G ′ = G ◦ x q s ( d +1) ; we have(26) Aut( G ′ ) ∩ Aut( C ) = A. Now, let Φ = Φ f,ρ,g where f and g are two invertible q -polynomials in L ( n,q ) [ x ],and ρ ∈ Aut( F q ), be an element of Aut( C ), and suppose that Φ does not belong to . Then by (26), Φ( G ′ ) = G ′ and this leads again to a contradiction by Theorem3.3. (cid:3) We end this section by proving the following equivalence results.
Theorem 3.5.
Let d ≥ . Two maximum d -codes S n,d,s and S n,d,s ′ (respectively, A n,d,s and A n,d,s ′ ), where s and s ′ are integers satisfying gcd( s, n ) = gcd( s ′ , n ) = 1 ,or, two maximal d -codes H n,d,s and H n,d,s ′ (respectively, E n,d,s and E n,d,s ′ ), where s and s ′ are integers satisfying gcd( s, n ) = gcd( s ′ , n ) = 1 , are equivalent if andonly if s ≡ ± s ′ (mod n ) .Proof. We give the proof only in symmetric and alternating setting. Similar argu-ments leads to the result for the two known constructions in the hermitian setting.For this reason we omit here the details.Suppose that s ≡ ± s ′ (mod n ). Let G ′ s = G s ◦ x q s ( n + d and G ′ s ′ = G s ′ ◦ x q s ′ ( n + d (respectively, G ′ s = G s ◦ x q s ( d and G ′ s ′ = G s ′ ◦ x q s ′ ( d ).By Proposition 3.1 points ( i ) and ( ii ), S n,d,s = G ′ s ∩ S n ( q ) and S n,d,s ′ = G ′ s ′ ∩ S n ( q ) , (respectively, A n,d,s = G ′ s ∩ A n ( q ) and A n,d,s ′ = G ′ s ′ ∩ A n ( q )).Since s ≡ ± s ′ (mod n ), by [8, Theorem 4.4 and 4.8, ( a )], we have that G s ′ = Φ ux qr ,id,vx qn − r ( G s ) = ux q r ◦ G s ◦ vx q n − r , for two given elements u, v ∈ F q n . Hence, G ′ s ′ = ( ux q r ◦ G s ◦ vx q n − r ) ◦ x q ( ± s + kn )( n + d , (respectively, G ′ s ′ = ( ux q r ◦ G s ◦ vx q n − r ) ◦ x q ( ± s + kn )( d ),If s ′ ≡ s (mod n ), from equation above we get(27) G ′ s ′ = ux q r ◦ G ′ s ◦ v q s n − d x q n − r , (respectively, G ′ s ′ = ux q r ◦ G ′ s ◦ v q − s ( d x q n − r ).If otherwise s ′ ≡ − s (mod n ), we have(28) G ′ s ′ = ux q r ◦ ( G s ◦ x q − s ( n + d ) ◦ v q s n + d x n − r , (respectively, G ′ s ′ = ux q r ◦ ( G s ◦ x q − s ( d ) ◦ v q s ( d x q n − r ).Since G ′ ⊤ s ′ = G ′ s ′ , by comparing coefficients in Equation (27) we get that it mustnecessarily be u = av ′ with a ∈ F ∗ q and v ′ = v q s n − d (respectively, u = av ′ with a ∈ F ∗ q and v ′ = v q − s ( d ).In a similar way, by comparing coefficients in Equation (28), we find u = av ′ with a ∈ F ∗ q , where v ′ = v q s n + d (respectively, u = av ′ with a ∈ F ∗ q and v ′ = v q − s ( d ).Hence, Φ av ′ x qr ,id,v ′ x qn − r ( S n,d,s ) = Ψ a,v ′ x qr ( S n,d,s ) = S n,d,s ′ , (respectively, Φ av ′ x,ρ,v ′ x ( A n,d,s ) = A n,d,s ′ ), where s ′ ≡ s (mod n ).Conversely, suppose that S n,d,s and S n,d,s ′ (respectively, A n,d,s and A n,d,s ′ ) areequivalent. Denote by Ψ = Ψ a,g,ρ the map such that Ψ( S n,d,s ) = S n,d,s ′ (respec-tively, Ψ( A n,d,s ) = A n,d,s ′ ). s gcd( s, n ) = gcd( s ′ , n ) = 1, we may assume that s ′ ≡ es (mod n ). In theremaining part of the proof we will write down computation only in the symmetriccontext. Similar arguments may be applied in the alternating case leading to thesame achievement.Each element f ∈ S n,d,s has the following shape: f ( x ) = b x + n − d X i =1 (cid:16) b i x q si + ( b i x ) q s ( n − i ) (cid:17) . Let g = P n − i =0 a i x q si ∈ F q n [ x ].Arguing as in the proof of Theorem 3.3 we have that each element in Ψ( S n,d,s )can be written as follows(29) n − X k =0 n − X i =0 b q s ( n − i ) a q s ( n − i ) i a q s ( n − i ) k + i + n − d X r =1 (cid:16) b q s ( n − i − r ) r a q s ( n − i ) i a q s ( n − i − r ) k + i + r + b q s ( n − i ) r a q s ( n − i ) i a q s ( r − i ) k + i − r (cid:17) x q sk . By comparing the coefficients of the term x q ks in Ψ( S n,d,s ) and in S n,d,s ′ we get(30) n − X i =0 b q s ( n − i ) a q s ( n − i ) i a q s ( n − i ) k + i + n − d X r =1 (cid:16) b q s ( n − i − r ) r a q s ( n − i ) i a q s ( n − i − r ) k + i + r + b q s ( n − i ) r a q s ( n − i ) i a q s ( r − i ) k + i − r (cid:17) = 0 , for each k ∈ { je : n − d < j < n + d } and all λ = ( b , · · · , b n − d ) ∈ F n − d +1 q n .By taking b = 0 and b j = 0 for j = 0, from above Equation (30) we have(31) a i a k + i = 0for i = 0 , , . . . , n −
1. Similarly, for each r ∈ { , . . . , n − d } , letting b r be the uniquenonzero elements among all b j , from (30) we can derive n − X i =0 (cid:16) a q s ( r − i ) i − r a q s ( n − i ) k + i + a q s ( n − i ) i a q s ( r − i ) k + i − r (cid:17) b q s ( n − i ) r = 0 . As the above equation holds for any b r ∈ F q n , it implies a q s ( r − i ) i − r a q s ( n − i ) k + i + a q s ( n − i ) i a q s ( r − i ) k + i − r = 0for every i , which means(32) a q sr i − r a k + i + a i a q sr k + i − r = 0 . Since g is a permutation q -polynomial, there must be at least one coefficient a i , i ∈ { , . . . , n − } which is different from zero. Denote such a coefficient with a i .By letting i = i in (31), we get a je + i = 0 for n − d < j < n + d . By taking i = i and i = r + i in (32) respectively, together with the aboveequation, we can derive a i + je − r = a i + je + r = 0 for n − d < j < n + d ≤ r ≤ n − d . ence, a je + i + i = 0 for n − d < j < n + d − n − d ≤ i ≤ n − d . As a i = 0, the equation je + i ≡ n )should have no solution for n − d < j < n + d and − n − d ≤ i ≤ n − d . As thereare d − { je (mod n ) : n − d < j < n + d } and n − d + 1 elements in { i : − n − d ≤ i ≤ n − d } , a je + i + i = 0 implies all a j = 0 for j = i .Thus g ( x ) = a i x q i . However, if e
6≡ ± n ), i.e. s
6≡ ± s ′ (mod n ), byCorollary 3.4. it is obvious that Ψ a,a i x qi ,ρ ( S n,d,s ) is not in S n,d,s ′ . Therefore, wemust have s ≡ ± s ′ (mod n ). (cid:3) A Characterization of known additive constructions
In this section we show that the property stated in Proposition 3.1 characterizingthe known examples of maximum d -codes in restricted setting, up to the equivalencerelation which we have denominated with the symbol ∼ =. More precisely, we provethe following Theorem 4.1.
Let n, s be two integers such that n ≥ and ( s, n ) = 1 , let d be aninteger such that ≤ d ≤ n − . Let D ⊂ X n be a maximum d -code. ( i ) If X n = S n ( q ) , then D ∈ [ S n,d,s ] ∼ = if and only if there is a unique subspace V of L ( n,q ) [ x ] , such that ( a ) V ∈ [ G ′ ] ≃ where G ′ = G n,n − d +1 ,s ◦ x q s n + d ; ( b ) V = V ⊤ , where V ⊤ = { f ⊤ : f ∈ V } ; ( c ) V ∩ S n ( q ) = D . ( ii ) If X n = A n ( q ) , then D ∈ [ A n,d,s ] ∼ = if and only if there is a unique subspace V of L ( n,q ) [ x ] , such that ( a ) V ∈ [ G ′ ] ≃ where G ′ = G n,n − d +1 ,s ◦ x q s d ; ( b ) V = V ⊤ , where V ⊤ = { f ⊤ : f ∈ V } ;( c ) V ∩ A n ( q ) = D .Proof. Let us prove the sufficiency first. Assume
D ∈ [ C ] ∼ = where either C is S n,d,s or C is A n,d,s . Hence, there exists a rank-preserving map of type Ψ = Ψ a,g,ρ , with a ∈ F ∗ q , ρ ∈ Aut( F q ) and g a permutation q -polynomial, such that Ψ( C ) = D .Let V = Φ ag,ρ,g ⊤ ( G ′ ), where G ′ = G ◦ x q s ( n + d if X n = S n ( q ) and C = S n,d,s , and G ′ = G ◦ x q s ( d if X n = A n ( q ) and C = A n,d,s .In both cases it is easy to see that G ′⊤ = G ′ . Hence, V satisfies the properties( a ) and ( b ). Moreover, as Φ ag,ρ,g ⊤ fixes X n , applying ( i ) of Proposition 3.1, weobtain that V ∩ X n = Φ ag,ρ,g ⊤ ( G ′ ) ∩ X n = Ψ( G ′ ∩ X n ) = Ψ( C ) = D , Hence V satisfies ( c ).Next, let us show the uniqueness. To this aim suppose that V and V ′ are twosubspaces of L ( n,q ) [ x ] both satisfying conditions ( a ), ( b ) and ( c ). n particular we have that V ∩ X n = D = V ′ ∩ X n . By hypothesis D = Ψ( C ) and Ψ fixes X n . This means that there is an elementsin [ G n,n − d +1 ,s ] ≃ different from G ′ , intersecting X n in C . Indeed, Φ − ag,ρ,g ⊤ ( V ′ ). This,by Theorem 3.3 ( i ), is a contradiction.Now, let us prove the necessity. By ( a ), G ′ and V are equivalent, more preciselythere exists a map Φ = Φ g,ρ,h such that V = Φ( G ′ ). Since again G ′⊤ = G ′ , by usingcondition ( b ), we have(33) Φ ⊤ ( G ′ ) = Φ h ⊤ ,ρ,g ⊤ ( G ′ ) = V. Now, from (33) and taking into account that V = Φ( G ′ ), we getΦ g − ◦ h ⊤ , id ,g ⊤ ◦ h − ( G ′ ) = G ′ . Hence, by Theorem [17, Theorem 4], we get g − ◦ h ⊤ = αx q r and g ⊤ ◦ h − = βx q n − r , with α, β ∈ F ∗ q n .In particular, r ≡ mod n ) and consequently β = α − , g = h ⊤ ◦ βx, andΦ = Φ h ⊤ ◦ βx,ρ,h .We show that Φ( C ) ∩ D contains at least one element which is different from thenull map. In fact, by ( c ), we have thatdim (Φ( C ) ∩ D ) ≥ dim Φ( C ) + dim D − dim V = n. Hence, let f be an element of C such that Φ( f ) ∈ D . Since Φ( f ) ∈ D ⊂ S n ( q ), wehave that Φ ⊤ ( f ) = Φ( f ). Consequently, f ρ ( βx ) = βf ρ ( x ) for each x ∈ F q n . Hence β ∈ F q and D = Φ( G ′ ) ∩ X n = Φ( G ′ ∩ X n ) = Ψ β,h ⊤ ,ρ ( C ) . Hence
D ∈ [ C ] ∼ = . (cid:3) A similar result can be stated also for the two known constructions of maximum d -codes in H n ( q ). Following is the precise statement Theorem 4.2.
Let n, s be two integers such that ( s, n ) = 1 , and let d be an integersuch that d > . Then we have the following(i) C ∈ [ H n,d,s ] ∼ = if and only if there is an unique subspace V of L ( n,q ) [ x ] , suchthat(a) V ∈ [ G ′ ] ≃ where G ′ = G n,n − d +1 ,s ◦ x q s ( n + d +1) ;(b) V = ˜ V , where ˜ V = { ˜ f : f ∈ V } ; (c) V ∩ H n ( q ) = C .(ii) C ∈ [ E n,d,s ] ∼ = if and only if there is an unique subspace V of L ( n,q ) [ x ] , suchthat(a) V ∈ [ G ′ ] ≃ where G ′ = G n,n − d +1 ,s ◦ x q s ( d +1) ;(b) V = ˜ V , where ˜ V = { ˜ f : f ∈ V } ;(c) V ∩ H n ( q ) = C . roof. The proof is similar to that of previous Theorem 4.1; in fact, by simply takinginto account that in this case we have ˜ G ′ = G ′ , whenever G ′ = G n,n − d +1 ,s ◦ x q s ( n + d +1) or G ′ = G n,n − d +1 ,s ◦ x q s ( d +1) . (cid:3) A new additive symmetric -code Let q be an odd prime power, m and s two integers such that m ≥ s, m ). Let η be an element of F q m such that N q m /q ( η ) = η q m − − q − , is not asquare.The set D k,s ( η ) = ax + k − X j =1 c i x q js + ηbx q ks : c , · · · , c k − ∈ F q m , a, b ∈ F q m is a maximum rank distance code with minimum distance d = 2 m − k + 1 [18].Now, let us consider the following set of q -polynomials S = ( a x + m − X j =1 a j x q sj + ηbx q s ( m − + ax q sm + η q s ( m +1) b q s x q s ( m +1) + m − X j =1 ( a j x ) q s (2 m − j ) : a , a , . . . , a m − ∈ F q m and a, b ∈ F q m ) . It is straightforward to see that, if we set D ′ = D m − ,s ( η ) ◦ x q sm then S = D ′ ∩ S m ( q ) and S = D ′⊤ ∩ S m ( q ) . In what follows we will show that any map in S has rank strictly greater thanone. In fact, let f m := f ◦ x q sm , where f ∈ S . Then the coefficients of terms x and x q s (2 m − ’ of f m are c and ηb , respectively. As a consequence of [5] (see also [17,Lemma 3]), the rank of f is then at least two. Hence, S is a maximum 2-code of S m ( q ). Theorem 5.1.
The -code S ∈ S m ( q ) is not equivalent to S m, ,s .Proof. Assume by way of contradiction that S is equivalent to S m, ,s . Then theremust be a map Ψ = Ψ a,g ⊤ ,ρ such that Ψ( S ) = S m, ,s , where a ∈ F q , ρ ∈ Aut( F q )and g ( x ) = P m − i =0 g i x q is is a permutation q -polynomial with coefficients in F q m .Consider g ⊤ ◦ α ρ x ◦ g , where α ∈ F q m . By computation the coefficient of x q ms is(34) a m ( α ) = m − X i =0 g q si m − i g q si m − i α ρq si where indices are taken modulo 2 m . Since the coefficient of the term with q -degree ms of S m, ,s is zero, we obtain g m − i g m − i = 0 for each i = 1 , , . . . , m. Without loss of generality, we can suppose that g m − i = 0 for i = 1 , , . . . , m. Let c ∈ F q m , in the same way the coefficient of degree q ms of the composition g ⊤ ◦ c ρ x q ms ◦ g is equal to m ( c ) = m − X i =0 g q s (2 m − i ) i g q s ( m − i ) i c ρq s (2 m − i ) = m − X i =0 g q s (2 m − i ) i g q s ( m − i ) i c ρq s ( m − i ) . Obviously, since ( g q s (2 m − i ) i g q s ( m − i ) i ) q m = g q s (2 m − i ) i g q s ( m − i ) i , the polynomial above has coefficients in F q m . On the other hand, as the coefficientof the term with q -degree ms in S m, ,s is zero, a m ( c ) = 0 for all c ∈ F q m . Thisimplies that g i = 0 for i = 0 , , · · · , m −
1. Therefore g is the null polynomial whichcontradicts the permutation property of g . (cid:3) Acknowledgment
This work is supported by the Research Project of MIUR (Italian Office for Uni-versity and Research) “Strutture geometriche, Combinatoria e loro Applicazioni”2012.
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