On non-antipodal binary completely regular codes

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1 Discrete Mathematics 308 (2008) On non-antipoal binary completely regular coes J. Borges a, J. Rifà a, V.A. Zinoviev b a Department of Information an Communications Engineering, Universitat Autònoma e Barcelona, Bellaterra, Spain b Institute for Problems of Information Transmission of the Russian Acaemy of Sciences, Bol shoi Karetnyi per. 19, GSP-4, Moscow , Russia Receive 20 December 2005; receive in revise form 27 June 2007; accepte 4 July 2007 Available online 10 August 2007 Abstract Binary non-antipoal completely regular coes are characterize. Using a result on nonexistence of nontrivial binary perfect coes, it is conclue that there are no unknown nontrivial non-antipoal completely regular binary coes with minimum istance 3. The only such coes are halves an puncture halves of known binary perfect coes. Thus, new such coes with covering raius ρ = 6 an 7 are obtaine. In particular, a half of the binary Golay [23, 12, 7]-coe is a new binary completely regular coe with minimum istance = 8 an covering raius ρ = 7. The puncture half of the Golay coe is a new completely regular coe with minimum istance = 7 an covering raius ρ = 6. The new coe with = 8 isproves the known conjecture of Neumaier, that the extene binary Golay [24, 12, 8]-coe is the only binary completely regular coe with 8. Halves of binary perfect coes with Hamming parameters also provie an infinite family of binary completely regular coes with = 4 an ρ = 3. Puncturing of these coes also provie an infinite family of binary completely regular coes with = 3 an ρ = 2. Both these families of coes are well known, since they are uniformly packe in the narrow sense, or extene such coes. Some of these completely regular coes are new completely transitive coes Elsevier B.V. All rights reserve. Keywors: Binary coe; Completely regular coe; Antipoality; Perfect coe; Golay coe 1. Introuction Let F n be the n-imensional vector space of all n-tuples over the finite fiel F=GF (2). The Hamming weight,wt(v), of a vector v F n is the number of its nonzero coorinates. The Hamming istance between two vectors v, u F n is (v, u) = wt(v + u). A(binary) (n, N, )-coe C is a subset of F n where n is the length, is the minimum istance an N = C is the carinality of C. For the case when C is a k-imensional linear subspace of F n, the coe C becomes a linear coe, enote by [n, k, ], where N = 2 k. Given any vector v F n, its istance to the coe C is (v,c)= min{(v, x)} x C This work was partially presente in the workshop Optimal Coes an Relate Topics. OC 05 Bulgarian Acaemic of Sciences, Pomporovo, Bulgary, June This work has been partially supporte by the Spanish MEC an the European FEDER MTM Grant, by the PNL UAB Grant an also by the Russian fun of funamental researches (Project No ). aress: joaquim.borges@autonoma.eu (J. Borges) X/$ - see front matter 2007 Elsevier B.V. All rights reserve. oi: /j.isc

2 an the covering raius of the coe C is J. Borges et al. / Discrete Mathematics 308 (2008) ρ(c) = ρ = max v F n{(v,c)}. Given two sets X, Y F n, efine their minimum istance (X,Y) = min{(x, y) : x X, y Y }. We write X + x instea of X +{x}. For a given vector x F n let x be the complementary vector, i.e., (x, x) = n. For a given set X F n efine the complementary set X ={ x : x X}. We write 1 (respectively, 0) for the all one (respectively, all zero) vector in F n. For a given coe C with covering raius ρ = ρ(c) efine C(i) ={x F n : (x,c)= i}, i = 1, 2,...,ρ. We assume that coe C always contains the zero vector 0, unless state otherwise. Let D = C + x be a translate of C. The weight wt(d) of D is the minimum weight of the coewors of D. For an arbitrary translate D of weight i = wt(d) enote by μ(d) = (μ 0 (D), μ 1 (D),...,μ n (D)) its weight istribution (μ i (D) is the number of wors of D of weight i). So μ(c) = (μ 0 (C),...,μ n (C)) is the weight istribution of C. If this vector μ(c) is the same for any translate of C by a coewor, then C is istance invariant. Denote by C j (respectively, D j an C(i) j ) the subset of C (respectively, of D an C(i)), forme by all wors of the weight j. In this terminology μ i (D) = D i. A (n, N, ) coe C with minimum istance = 2e + 1weexten to (n + 1,N,+ 1) coe C, aing one overall parity check symbol to coewors of C, an puncture to (n 1,N, 1) coe C (1), eleting any one position of coewors of C. We say that two vectors x an y are neighbors if (x, y) = 1. For any vector x F n enote by W(x) the set of all its neighbors, i.e. W(x) is the sphere of raius 1 near x W(x) ={y F n : (x, y) = 1}. Delsarte [7], as a generalization of linear coes over a finite fiel alphabet in a Hamming scheme, efines the concept of aitive coes as subgroups of the unerlying Abelian group in a translation association scheme. For the special case of a binary Hamming scheme where the unerlying Abelian group is of size 2 n the only structures for an Abelian group of this size which gives rise to a translation association scheme are of the form Z α 2 Zβ 4, with α + 2β = n (see [15,8]). Let C be an aitive coe, so a subgroup of Z α 2 Zβ 4 an let C = Φ(C), where Φ is the extension of the usual Gray map Φ : Z α 2 Zβ 4 Zn 2, n = α + 2β, given by Φ(x, y) = (x, (y 1 ),..., (y β )) x Z α 2, y = (y 1,...,y β ) Z β 4, where : Z 4 Z 2 2 is the usual Gray map, i.e., (0) = (0, 0), (1) = (0, 1), (2) = (1, 1), (3) = (1, 0). Definition 1. Let C be an aitive coe, i.e., a subgroup of Z α 2 Zβ 4 an let C = Φ(C) be the binary coe obtaine from C after using the Gray map. When α = 0 the binary coe C will be calle Z 4 -linear coe an when α = 0 coe C will be calle Z 2 Z 4 -linear coe. Definition 2. A coe C is completely regular if, for all l 0, every vector x C(l)has the same number c l of neighbors in C(l 1) an the same number b l of neighbors in C(l + 1). Also, efine a l = n b l c l an note that c 0 = b ρ = 0. Define by {b 0,...,b ρ 1 ; c 1,...,c ρ } the intersection array of C an by L the intersection matrix of C: a 0 b c 1 a 1 b c..... L = bρ c ρ a ρ

3 3510 J. Borges et al. / Discrete Mathematics 308 (2008) For a binary coe C let Perm(C) be its permutation stabilizer group. For any θ Perm(C) an any translate D = C + x of C efine the action of θ on D as: θ(d) = C + θ(x). Definition 3 (Solé [19]). Let C be a binary linear coe with covering raius ρ. The coe C is calle completely transitive, if the set {C + x : x F n } of all ifferent cosets of C is partitione uner action of Perm(C) into exactly ρ + 1 orbits. Since two cosets in the same orbit shoul have the same weight istribution, it is clear, that any completely transitive coe is completely regular. It has been conjecture for a long time that if C is a completely regular coe an C > 2, then e 3. Moreover, in [14] it is conjecture that the only completely regular coe C with C > 2 an 8 is the extene binary Golay [24, 12, 8]-coe with ρ = 4. As we know from [20,22] for ρ = e an [21] (see also [18,11]) for ρ = e + 1, any such nontrivial unknown coe shoul have a covering raius ρ e + 2. For the special case of completely regular coes, for linear completely transitive coes [19], the problem of existence is solve: we prove in [5,6] that for e 4 such nontrivial coes o not exist. In this paper we give a complete characterization of binary nontrivial, non-antipoal, completely regular coes with istance 3. The only such coes are forme by halves of binary perfect coes. In particular, a half of the binary Golay [23, 12, 7]-coe is a new non-antipoal completely regular [23, 11, 8]-coe with covering raius ρ = 7 an intersection array (23, 22, 21, 20, 3, 2, 1; 1, 2, 3, 20, 21, 22, 23). This result implies that the conjecture of Neumaier [14] is not vali. The puncture half of the Golay coe is a new non-antipoal completely regular [22, 11, 7]-coe with covering raius ρ = 6 an intersection array (22, 21, 20, 3, 2, 1; 1, 2, 3, 20, 21, 22). Halves of binary perfect (n, N, 3) coes also give an infinite family of completely regular coes with = 4, ρ = 3 an intersection array (n, n 1, 1; 1,n 1,n). The puncture halves of binary perfect (n, N, 3) coes are uniformly packe in the narrow sense [18] an therefore, these coes are completely regular with = 3, ρ = 2 an intersection array (n, 1; 1,n). The same results are vali for q-ary perfect coes, uner certain conitions on the original coes (see [16]). In particular, from the ternary Golay coe we obtain new ternary completely regular coe with minimum istance 6, with covering raius 5 an with intersection array (22, 20, 18, 2, 1; 1, 2, 9, 20, 22). The present paper is organize as follows. In Section 2 we give some preliminary results concerning completely regular coes. In Section 3 we prove that the covering set C(ρ) of non-antipoal completely regular binary coe C is its translate by 1. This permits us to lower an upper boun the covering raius of non-antipoal completely regular coes. In Section 4 we prove that the only non-antipoal completely regular coes are forme either by even (or o) coewors of binary perfect coes, or the coes, obtaine by puncturing these coes. 2. Preliminary results We give some efinitions, an results which we will nee later. Definition 4. Let C be any binary coe of length n an let ρ be its covering raius. We say that such a coe is uniformly packe in the wie sense, i.e., in the sense of [1], if there exist rational numbers α 0,...,α ρ such that for any v F n ρ α k f k (v) = 1, (1) k=0 where f k (v) is the number of coewors at istance k from v. We say that such a coe is strongly uniformly packe (or uniformly packe in the narrow sense, or in the sense of [18]), if ρ = e + 1 an α e = α e+1, where e = ( 1)/2. The support of v F n, v = (v 1,...,v n ) is supp(v) ={l v l = 0}. Say that a vector v covers a vector z if supp(z) supp(v).

4 J. Borges et al. / Discrete Mathematics 308 (2008) Definition 5. A t-esign T (n, w, t, β) is a set of binary vectors of length n an weight w such that for any binary vector z of weight t,1 t w, there are precisely β vectors v i,i= 1,...,β, of T (n, w, t, β) each of them covering z. If β = 1 the esign T (n, w, t, 1) is a Steiner system S(n, w, t). Definition 6. Say that a binary coe C is even (respectively, o) if all its coewors have even (respectively, o) weights. The next fact follows from the efinition of completely regular coe. Lemma 7. Let C be a completely regular coe with minimum istance an containing the zero coewor. Then any nonempty set C j, j n, is a t-esign, where t = e, if = 2e + 1 an t = e + 1, if = 2e + 2. Lemma 8 (Neumaier [14]). If C is completely regular with covering raius ρ, then C(ρ) is also completely regular, with reverse intersection array. Definition 9. The coe C is calle antipoal, if for any c C the complementary vector c = c + 1 is also a coewor of C. It is clear that a istance invariant coe C, containing 0, is antipoal if it contains 1. Lemma 10. Let C be any binary coe. Then C an C(ρ) are antipoal or not simultaneously. Proof. Let C be any binary coe, an let C(ρ) be the corresponing covering set of C. Assume that C is antipoal. To see that C(ρ) is antipoal we take v C(ρ) an prove that 1 + v C(ρ). In orer to o this we observe that (1 + v,c)= ρ, since (1 + v,c)= (v, 1 + C) = (v,c)= ρ. The statement follows now since the antipoality of C(ρ) implies the antipoality of C by reversing of C an C(ρ). 3. The covering raius The natural question is: oes any completely regular coe contain the vector 1? Theorem 11. Let C be a completely regular coe with covering raius ρ, with minimum istance 3. If 0 C, but 1 / C, then 1 C(ρ) an C + 1 = C(ρ). Furthermore, { 2e if = 2e + 1, ρ 2e + 1 if = 2e + 2. Proof. Let C be a completely regular coe an let 1 / C. First we prove that 1 C(ρ). Assume to the contrary, i.e. 1 / C(ρ). Consier the subset C w of C with largest weight w an the subset C(ρ) v of C(ρ) with largest weight v. As C an C(ρ) o not contain 1, we have clearly: 1 n w ρ 1 an 1 n v ρ 1. Now we claim that (n w) + (n v) = ρ. (2) Inee, C is a completely regular coe in the Hamming space F n, which is a metric association scheme [7]. In particular, this means that for any vector x from F n there exist two vectors c C an v C(ρ) such that (c, x) + (x, v) = ρ. (3) Taking the vector 1 as x we immeiately obtain (2), since we have that (1, C w ) = n w an (1, C(ρ) v ) = n v.

5 3512 J. Borges et al. / Discrete Mathematics 308 (2008) By Lemma 7, the set C w is a t-esign, say T 1 (n, w, t, β 1 ) with t =e or e +1, where e = ( 1)/2. By the conition of theorem 3 an, therefore, t 1. By efinition, (C, C(ρ)) = ρ. Hence, these sets C w an C(ρ) v are at istance ρ at least from each other. Consier the complementary sets: S 1 = C w with vectors of weight w = n w an S 2 = C(ρ) v with vectors of weight v = n v, where 1 w,v ρ 1. From (2) we euce that w + v = ρ. (4) But the set S 1 (which is complementary of C w )isat-esign also [17], say T (n, w,t,α 1 ) with t 1. Taking any wor z from S 2 we can always fin x from S 1 such that supp(z) supp(x) t. Taking into account this last fact, we conclue that uner the conition (4) the two sets S 1 an S 2 have minimum istance (S 1,S 2 ) ρ 2. Thus, we obtain a contraiction an so 1 shoul belong to C(ρ). Now we claim that C + 1 belongs to C(ρ). This comes from the fact that C is completely regular, an therefore, the istance istribution is the same for all its coewors. An this istance istribution says that for any coewor c C the complementary vector c belongs to C(ρ). We conclue, therefore, that C + 1 is a subset of C(ρ). But C + 1 is a translate of C of weight ρ, an any such translate has the same weight istribution. But there is only one vector 1 of weight n. So, we can have only one such translate. This means that C(ρ) = C an, therefore, C + 1 = C(ρ). (5) This last property implies immeiately limitations for the possible values of ρ. Inee, since 1 belongs to C(ρ), it follows from (5) that the sets C n ρ an C(ρ) ρ are nonempty. As C n ρ is a t-esign (Lemma 7) the set C(ρ) ρ is a t-esign too [17], say T 2 (n, ρ,t,β 2 ). By (5) we euce that C(ρ) ρ is a constant weight coe with minimum istance (T 2 ) 2e + 2. If = 2e + 1, we have t = e (Lemma 7). This implies that ρ 2e,ifβ 2 = 1 an, ρ 2e + 1, if β 2 > 1. If = 2e + 2, we have t = e + 1 (Lemma 7). This implies that ρ 2e + 1, if β 2 = 1 an, ρ 2e + 2, if β 2 > 1. Lemma 12. Let C an its (even or o) extension C be completely regular coes of lengths n an n+1 with covering raii ρ an ρ + 1, respectively. Then C an C are antipoal or not simultaneously. Proof. Let 1 C. Assume to the contrary that 1 / C. Then by Theorem 11, 1 C (ρ + 1), an therefore, 1 C(ρ), which is a contraiction. Therefore, if C is antipoal, it follows that C is antipoal. Next two main theorems upper boun the covering raius of any non-antipoal nontrivial completely regular binary coe. Before these theorems we will go to some useful results. Assume that C is a nontrivial non-antipoal completely regular coe with covering raius ρ, with minimum istance = 2e an containing the zero coewor. Let J enote the coorinate set of C, i.e. J ={1, 2,...,n}. Let v C(ρ) ρ be a given vector of weight ρ. Let y C(ρ e) be a vector of weight ρ e which is covere by v an let x C(e) be a given vector of weight e covere by y an v. Note that x is well efine since from Theorem 11 we have ρ 2e. Define the set S c as the set of all the vectors c i C which are at istance e + 1 from x. Let J (1) be the union of supp(c)\supp(x ) of all coewors c from C, covering x an let J (2) be the union of supp(c)\supp(y ) of all vectors c from C, having at least e nonzero positions in supp(y ). By Lemma 7 the set C is an e-esign, say T (n, 2e + 1,e,λ ) an the set C(ρ) ρ (since it is a complementary [17] of C n ρ which is also an e-esign by Lemma 7) is an e-esign too, say T ρ (n, ρ,e,λ ρ ). With all this notation we establish the following lemmas: Lemma 13. Let C be a nontrivial non-antipoal completely regular coe with minimum istance = 2e + 1 3, covering raius ρ 2e + 1 containing the zero coewor. Let y C(ρ e) an x C(e). Then there are exactly λ vectors c i C, i = 1,...,λ which are at istance e + 1 from x an λ vectors v i C(ρ) which are at istance e + 1 from y. Proof. We know x C(e). Without loss of generality, we can assume that wt(x) = e. By efinition of T (n,, e, λ ), there are exactly λ vectors c i, i = 1,...,λ, which are at istance e + 1 from x.now,ify C(ρ e) an taking into account that C(ρ) = C + 1 we obtain the secon statement.

6 J. Borges et al. / Discrete Mathematics 308 (2008) Lemma 14. Let C be a nontrivial non-antipoal completely regular coe with minimum istance = 2e + 1 3, covering raius ρ 2e + 1 an containing the zero coewor. Using the notation introuce before we have the following results: J (1) = J (2) ; J (1) (2) = J =(e + 1)λ ; λ 2 an C +1 =. Proof. Clearly, the vectors c i S c intersect just in the support of x, so from the e-esign T (n, 2e + 1,e,λ ) we get J (1) =(e + 1)λ. Let l = J (2). It is easy to see that J (1) J (2) an so l (e + 1)λ, but we want to establish l = (e + 1)λ. Let W(y ) be the set of all vectors at istance one from y. We consier all l vectors z W(y ) with one nonzero position in the set J (2). Since y C(ρ e), by Lemma 13 shoul be exactly λ vectors from C(ρ), say v i, i=1,...,λ, which are at istance e + 1 from y (i.e. all these v i cover this vector y ). It is easy to see that v i cannot be from C(ρ) ρ since (v i, v ) 2e + 1 (recall that C(ρ) ρ is a coe with minimum istance 2e + 2). Hence these vectors v i shoul be from C(ρ) ρ+1. Furthermore, any such vector v i is at istance e from e + 1 vectors z, corresponing its nonzero positions in J (2). Since all these λ vectors v i have isjoint supports (i.e. supp(v i ) supp(v j ) = ), we euce that l (e + 1)λ. But, if l >(e+ 1)λ, then some position of J (2), say i, oes not belong to any of the sets supp(v i ) for i = 1,...,λ.Taking the vector z W(y ) with i th nonzero position, we obtain the contraiction with the arguments above: there is a vector c C at istance ρ e from z, but there is no any vector from C(ρ) at istance e from z, which is impossible. Let c i be any of the λ vectors from C covering fixe e positions of y. These vectors have isjoint supports on the. The same property have the λ vectors v i from C(ρ) ρ+1, consiere above. Since any two such vectors c i an v i can have only one common nonzero position at J (2) we euce that λ e + 1. For the rest of the statement, notice that by Lemma 13 the conition x C(e) implies that there are exactly λ coewors c i C at istance e + 1 from x. Since C is completely regular, this property shoul be satisfie for any set J (2) vector z from C(e). Assume that z of weight e + 1, covering x, has one nonzero position at the set J (1). This means that there is some vector c C such that (z, c) = e. Hence z C(e) an shoul be exactly λ coewors, say c i C, i = 1,...,λ, which are at istance e + 1 from z. One such a wor is the zero wor, an it is easy to see that the other λ 1 coewors c i might be only from C +1. Thus, we have that if λ > 1, then the set C +1 is not empty. We nee one more result. Fix any i from J (2) an let x i be a vector of weight e + 1 covering x, such that the ith coorinate is x i = 1. Define the set S u (i) of all the vectors u C +1 such that (x i,u)= e + 1, for some i. Lemma 15. For any i J (2) there are exactly λ 1 vectors in S u (i) an for any u j S u (i) we have supp(u j )\supp(x i ) J (2). Proof. Fix any vector x of weight e, covere by y C(ρ e) of weight ρ e an consier the λ corresponing vectors c i from C, covering x. By efinition, all supports of these vectors belong to supp(x ) J (2). Take any position i, out of supp(x ), covere by c i. The vector x i of weight e + 1, covering x with nonzero ith position is at istance e from c i. By Lemma 13 we conclue that there are exactly λ wors from C at istance e + 1 from x i. One such a wor is the zero coewor. It is easy to see that the other λ 1 wors, say u j, j = 1,...,λ 1, might be only from C +1. Now we claim that for any such wor u j its support belongs to supp(x ) J (2). Assume that it is not the case. Let s j supp(u j ) such that s j is out of supp(x ) an s j / J (2). Take the vector x s of weight e + 1, covering x, with nonzero s j th position. Since x s is not covere by coewors from C, this vector belongs to C(e + 1) an so it is covere by some y C(ρ e) (remark that we are assuming ρ 2e + 1). From Lemma 14 we have J (1) = J (2) an we conclue that J (2) oes not epen on the specific vector y we are consiering in our construction. Hence, we can assume that x s is covere by some y C(ρ e) an for this y by Lemma 13 there are exactly λ vectors v i from C(ρ) ρ+1 which are at istance e + 1 from y, i.e. every v i covers y. Now from the arguments, which we use in the proof of Lemma 14, we euce that vectors v i cover all the positions of J (2) (this last set epens only on x ). Therefore, for this position s j we can fin some v i which intersect u j on some position

7 3514 J. Borges et al. / Discrete Mathematics 308 (2008) from J (2). Then we will have (v i, u j ) = wt(v i ) + wt(u j ) 2(e + 2) = (ρ + 1) + 2(e + 1) 2(e + 2) = ρ 1 (inee, vectors v i an u j share the e coorinates in x, the position s j on supp(y )\supp(x ), an one position on J (2) ). So, this last equality contraicts our initial assumption. Theorem 16. Let C be a nontrivial non-antipoal completely regular coe with minimum istance = 2e + 1 3, covering raius ρ an containing the zero coewor. Then ρ = 2e. Proof. From Theorem 11 we have that ρ 2e. Now assume that ρ 2e +1. For the fixe vector x of weight e, covere by y, choose the λ vectors c j whose supports belong to supp(x ) J (2) an the λ vectors v i C(ρ) which are at istance e + 1 from y. For any fixe sth position in J (2) choose the λ 1 vectors u s,k, k = 1,...,λ 1 from C +1 covering x, having nonzero sth position. All these l (λ 1)/(e + 2) vectors u s,k cannot have more than one common nonzero positions at J (2) (note that we ivie l (λ 1) into (e + 2), since any wor u s,k is counte (e + 2) times). They also can coincie with the λ vectors v i an c j not more than in one nonzero position at J (2). This means that any two positions of J (2) can be covere by the all vectors u s,k, c j an v i above not more than once. Hence, on the set J (2) we can consier a packing of the pairs of nonzero positions of all the aforementione vectors c j, v i an u s,k. Thus, the following inequality shoul be vali: ( ) ( ) ( ) l e + 2 e + 1 e + 2 (λ l 1) + 2 λ 2. (6) 2 2 This inequality (taking into account that l =(e +1)λ by Lemma 14) reuces to the inequality e 0, i.e. we obtain a contraiction. This means that there are no any such coe C with istance 2e+1 3 an with covering raius ρ 2e+1. We conclue that the only possibility is ρ = 2e. Lemma 17. Let C be a nontrivial non-antipoal completely regular coe with minimum istance = 2e + 1 3, covering raius ρ an containing the zero coewor an let C(ρ) be its covering set. Denote by C (ρ) ρ (respectively, C (ρ) ρ+1 ) the coe of length n + 1 obtaine from C(ρ) ρ after aing one extra coorinate with value 1(respectively, with value 0). Then the union C (ρ) ρ+1 C (ρ) ρ results in a Steiner system S(n + 1, ρ + 1,e+ 1). Proof. It is straightforwar. Now we consier non-antipoal coes with even istance. Unfortunately, this proof is more technical, since in this case there is no such nice partition of the coorinate set J as above. Theorem 18. Let C be a nontrivial non-antipoal completely regular coe with minimum istance 4, covering raius ρ an containing the zero coewor. If = 2e + 2 4, then ρ = 2e + 1. Furthermore, C(ρ) ρ is a Steiner system S(n, 2e + 1,e+ 1). Proof. Assume to the contrary that ρ 2e + 2. From Theorem 11 we have that ρ 2e + 1. Note that C is an (e + 1)- esign, say T (n,, e + 1, λ ), an C(ρ) ρ is also (e + 1)-esign, say T ρ (n, ρ,e+ 1, λ ρ ). Compute the intersection numbers (a i,b i,c i ) for i = e + 1 an for i = ρ e 1. For a fixe vector x of weight e + 1 enote J x = supp(x). Define three subsets of J \J x. The set J (1) is forme by the supports of λ coewors c from C covering x, the set J ρ (1) is forme by the supports of λ ρ vectors v from C(ρ) ρ covering x, an the set J (1) +1 : J (1) +1 = J \(J x J (1) J (1) ρ ).

8 J. Borges et al. / Discrete Mathematics 308 (2008) Since any c C an v C(ρ) ρ can have not more than e + 1 common nonzero positions, i.e. supp(c) supp(v) e + 1, (7) we conclue that J (1) an J ρ (1) are isjoint. Therefore, J is partitione into four isjoint subsets J x, J (1), J ρ (1) an J (1) +1, respectively. Denote such a partition by P(x), since it is uniquely efine by x. Now we claim that J (1) +1 is forme by all λ +1 coewors from C +1 covering x. First, note that any such b C +1 (if it exists), which covers x, oes not have any nonzero positions on J (1) an J ρ (1) (inee, C is a coe with = 2e + 2 an ρ 2e + 2). To see that any element of J (1) +1 is containe in some b C +1, assume that it is not the case. Let a vector y of weight e + 2 cover x an be not covere by any wor from J (1), J ρ (1) or J (1) +1. This means that y is at istance e + 2 from the zero coewor, C, an C +2 (if it is nonempty), at istance e + 3 from C +1, at istance ρ e from C(ρ) ρ, an at istance ρ e 1 from C(ρ) ρ+1 (if it is nonempty also). Hence, y has istance e + 2 from C an istance ρ e 1 from C(ρ), which is impossible. The only possibility is that y is covere by some wor from C +1. We conclue also that C +1 an C(ρ) ρ+1 are empty or not simultaneously. Since C is an (e + 1)-esign, we know the carinality of J (1). Inee, the λ coewors c from C, which cover x, have isjoint supports on J (1) (1). Taking into account that the sets J, J ρ (1) an J (1) +1 are isjoint, we conclue that J (1) =(e + 1)λ. (8) Using this partition P(x), we have for the case i = e + 1: a e+1 = J (1) +1, b e+1 = J ρ (1), c e+1 = e J (1) =(e + 1)(λ + 1). (9) Nowwefixanyv C(ρ) ρ an any y F n of weight ρ e 1 which is covere by v. We efine on J \supp(v ) three sets J (2), J ρ (2) an J (2) (2) +1. The set J is forme by all vectors c from C such that supp(c) supp(y) =e + 1. (10) The set J (2) ρ is forme by supp(v) of wors v from C(ρ) ρ such that supp(v) supp(y) =ρ e 1. (11) The set J (2) +1 is the rest of J \supp(v ): J (2) +1 = J \(supp(v ) J (2) J (2) ρ ). The sets J (2) an J ρ (2) are isjoint. Inee, if we assume that there is an element i J such that i J (2) J ρ (2), then we obtain two vectors c C an v C(ρ) ρ, with e + 2 common nonzero positions, which is impossible. Denote such a partition by P(v, y). Having these sets, we have for the case i = ρ (e + 1): a ρ e 1 = J (2) +1, b ρ e 1 = e J ρ (2), c ρ e 1 = ρ e 1 + J (2). (12) By Lemma 8 we shoul have a e+1 = a ρ e 1, b e+1 = c ρ e 1 an c e+1 = b ρ e 1, which means (using (9) an (12)), that J (1) (2) (1) (2) (2) +1 = J +1, J = J ρ an J +ρ e 1 = J ρ (1). (13) Denote l = J (1) an l ρ = J ρ (1). From (8) we have that l = λ (e + 1). Denote by W(x) the sphere of raius one with center at x.forz W(x) with position i supp(z) enote by ξ (i) (respectively, by ξ ρ (i)) the number of vectors from C (respectively, from C(ρ) ρ ) covering z. From (8) we euce that ξ (i) = 1 for any i J (1).

9 3516 J. Borges et al. / Discrete Mathematics 308 (2008) Our first step is to obtain the exact expressions for λ ρ, ξ ρ (i) an l ρ. In all lemmas below the conitions of Theorem 18 are satisfie an ρ 2e + 2. Lemma 19. We have that ( ) ρ e 1 λ ρ = λ e + 1 (14) an for any i J (1) ρ, ξ ρ (i) = ξ ρ = ρ e 1 l ρ λ ρ = ( ρ e 2 e ) λ. (15) Proof. The partition P(v, y) becomes P(v + y) if we translate it by the vector v. Uner this translate the roles of C an C(ρ) interchange (inee, C is the translate of C(ρ)). The vectors with supports on J (2) supp(y) will be the vectors from J ρ (1). The vectors c i from J (2) supp(y) correspon to the vectors u j of weight ρ e 1onJ ρ (1). They are exactly the vectors c i, having e + 1 nonzero positions on supp(y). But the number of such vectors is equal to ( ) ρ e 1 λ ρ = λ. e + 1 Inee, for any choice of e + 1 positions from ρ e 1, there are exactly λ ifferent vectors c i from C, with these fixe e + 1 positions on supp(y) an the rest e + 1 positions on J (2). This gives expression (14) for λ ρ. By the arguments above the number ξ ρ (i) oes not epen on i supp(y) for a chosen y. To fin this number, we fix one position on supp(y) an choose the other e positions from the remaining ρ e 2 positions in all possible ways ( ) ρ e 2 ξ ρ (i) = λ. e It is clear, that we will have the same expression for ξ ρ (i), if we choose as v any other vector v from C(ρ) ρ, covering x. Thus, ξ ρ (i) is the same (i.e. ξ ρ (i)=ξ ρ ) for all i from J ρ (1). Counting in two ifferent ways all the numbers of nonzero positions of vectors v C(ρ) ρ, covering x, we obtain that ξ ρ l ρ = (ρ e 1)λ ρ, which gives (15). Lemma 20. We have that an l ρ = (ρ e 1)2 e + 1 (16) l ρ l + ρ e 1. (17) Proof. Returning to the proof of the previous lemma, we have that l ρ = λ ρ ξ ρ (ρ e 1). Now the expression for l ρ follows, if we take into account expressions for λ ρ an ξ ρ from the lemma above. To prove the inequality we euce from (13) that l ρ = J ρ (1) = J (2) +ρ e 1. The boun follows now from a simple observation that J (2) J (1) =l (inee, in the partition P(v, y) for any e + 1 fixe nonzero positions of y, we shoul have exactly λ isjoint vectors of weight e + 1onJ (2) ). Now we have to consier the cases ρ = 2e + 2 an ρ 2e + 3 separately. We start from the case ρ 2e + 3. In the partition P(v, y) let z i W(y ), i = 1, 2. Denote by ξ 1 (respectively, by ξ 2 ) the number of wors from C which are at

10 J. Borges et al. / Discrete Mathematics 308 (2008) istance ρ e 2 from z 1 (respectively, from z 2 ), where z 1 is covere by y (respectively, z 2 has one nonzero position on J (2) ). Lemma 21. Let ρ 2e + 3. Then ( ) ρ e 2 ξ 1 λ + 1 (18) e + 1 an e + 1 ξ 2 λ ρ l ρ ρ + e + 1. (19) Proof. Since wt(z 1 ) = ρ e 2, it is at istance ρ e 2 from the zero coewor. Now, for any choice of e + 1 positions in supp(z 1 ), there are exactly λ coewors from C at the istance ρ e 2. There might be also some coewors from C +2 at the same istance from z 1, which we cannot evaluate. Hence, we conclue that ξ 1 is not less than expression (18) of the lemma. Similarly, for the number ξ 2, for any ρ e 1 nonzero positions of y, there are exactly λ coewors from C where each one has exactly e + 1 nonzero positions on J (2). We can lower boun the number ξ 2 taking average contribution of these coewors from C to one position of J (2). This gives the following lower boun (again we o not know the number of possible coewors from C +2 ): ( ) ρ e 1 ξ 2 λ e + 1 But by (13), e + 1 J (2). J (2) +ρ e 1 = J (1) ρ. Recalling that J (1) ρ =l ρ, we obtain from these two expressions above the secon inequality of the lemma. Now we return to the proof of Theorem 18 for the case ρ 2e + 3. Consier the partition P(x). Let z 3 W(x) contain one nonzero position on J ρ (1). Then we know that there are exactly ξ ρ vectors from C(ρ) ρ at istance ρ e 2 from z 3, an there are no any vectors from C(ρ) at this istance (see Lemma 19). But C is completely regular coe an, since C(ρ) is a translate of C (Theorem 11), all these numbers ξ ρ, ξ 1 an ξ 2 shoul be equal. This implies the following inequality: ( ( ) ) ρ e 2 (e + 1)λ ρ max λ + 1, ξ e + 1 ρ. (20) l ρ ρ + e + 1 Consier the first inequality ( ) ρ e 2 λ + 1 ξ e + 1 ρ, which is equivalent to the following one: ( ) ρ e 2 λ < ξ e + 1 ρ. Taking into account (15), the last inequality is reuce to the following one: (ρ e 1)2 l ρ < ρ 2e 2. (21) The secon inequality (e + 1)λ ρ l ρ ρ + e + 1 ξ ρ,

11 3518 J. Borges et al. / Discrete Mathematics 308 (2008) implies that (ρ e 1)2 l ρ ρ 2e 2. (22) Comparing (21) an (22), we obtain a contraiction. We conclue that for the case ρ 2e + 3 there is no such coe C, which satisfies the conitions of the theorem. Now we continue the proof of theorem for the case ρ = 2e + 2. Consier the intersection numbers a i,b i,c i of C for i = ρ/2 = e + 1. As we mentione alreay, by Lemma 8 the intersection numbers a ρ i,b ρ i,c ρ i are the reverse of a i,b i,c i, i.e. b i = c ρ i an c i = b ρ i. For the case ρ = 2e + 2 an i = e + 1 all these numbers shoul be equal, since C(ρ) = C + 1 by Theorem 11, i.e. we shoul have b e+1 = c e+1. Using (9), we euce that l ρ = l + e + 1. But for the case ρ = 2e + 2, expression (16) gives l ρ = e + 1, i.e. we obtain that l = 0, which is impossible, since J (1) is nonempty. Thus, we obtain a contraiction. Therefore, such coe C with ρ = 2e + 2 cannot exist for any e 1. This means, that if such coe exists it shoul have ρ 2e + 1. Now combining this with Theorem 11, we conclue that ρ = 2e + 1. Hence C(ρ) ρ is a (e + 1)-esign T (n, 2e + 1,e+ 1, λ ρ ) an a constant weight coe with minimum istance 2e + 2, which is possible if an only if λ ρ = 1. Thus, C(ρ) ρ is a Steiner system S(n, 2e + 1,e+ 1). The theorem is prove. 4. Non-antipoal completely regular coes an binary perfect coes The next two statements give a characterization of all nontrivial binary non-antipoal completely regular coes with o or even minimum istance. Theorem 22. Let C be a nontrivial (i.e. C > 2) completely regular coe with parameters n, = 2e + 1 3, an ρ. If 0 C an 1 / C, then C is a puncture half of a perfect coe C an C = C C (ρ), where C is obtaine from C by extension with even parity check, C (ρ) is the covering set of C, an C is a binary perfect coe with parameters n = n + 1, = 2e + 1 an ρ = e. Proof. Let C be a coe obtaine from C by even parity check, i.e. it is a coe with = 2e + 2. From Theorem 16 we have that ρ = 2e. Denote by C (ρ) the covering set of C, obtaine from C(ρ) by o parity check. Then C has covering raius ρ = ρ + 1 = 2e + 1. It is easy to see also that C (ρ) is a translate of C by 1. Define a new coe C as a union of C an C (ρ). By efinition of covering set the coe C has minimum istance = ρ = 2e + 1. Now we have to show only that this new coe has the covering raius ρ = e. The lower boun ρ e is trivial. To see that ρ e, note that from Lemma 17 the union C (ρ) ρ+1 C (ρ) ρ form the Steiner system S(n + 1, 2e + 1,e+ 1).This means that any vector z of weight e + 1 is covere by exactly one vector from S(n+ 1, 2e + 1,e+ 1),orbyC (ρ) ρ+1. This implies that ρ e. Thus, ρ = e. Theorem 23. Let C be a nontrivial completely regular coe with parameters n, = 2e an ρ. If 0 C an 1 / C, then C is a half of a perfect coe C an C = C C(ρ),

12 J. Borges et al. / Discrete Mathematics 308 (2008) i.e. a union of C an its covering set C(ρ), is a binary perfect coe with parameters n = n, = 2e + 1 an ρ = e where e = ( 1)/2. Proof. By Theorem 18 we have that ρ = 2e + 1. Define a new coe C (with minimum istance an covering raius ρ ), taking a union of C an C(ρ). Since C(ρ) is a translate of C, we euce that = ρ = 2e + 1. We claim that C is a perfect coe. To have it we have to show that ρ = e. First, it is clear that ρ e (inee, ρ = 2e + 1 an so C an C(ρ) are coes with minimum istance 2e + 2). To see that ρ e, recall the proof of Theorem 18. In terms of partition P(x), inuce by any vector x of weight e + 1, the inequality ρ e is the same as existence of some v C(ρ) ρ covering x. But this follows from the fact that C(ρ) ρ is a (e + 1)-esign. Thus, C is a perfect coe, an C is a half of a perfect coe. The following example shows that for trivial completely regular coes with C =1 this last theorem is not vali. Example 24. Consier a trivial coe C, consisting of one vector in F n, which is completely regular non-antipoal coe with ρ = n for n multiple of 4. Let C ={(0, 0,...,0)}. The intersection array of C looks as follows: a i = 0, b i = n i, c i = i, i = 0, 1,...,n. By Theorem 11 above the set C(ρ) is the complementary vector 1 = (1, 1,...,1). The mile row for i = n/2 is symmetric b n/2 = c n/2 = n/2 as it shoul be, since to this coe we can a the complementary vector (1, 1,...,1) an obtain a completely regular coe again with two coewors an with even covering raius ρ = n/2 (but not perfect coe with o covering raius, as we have in Theorem 22). Now we have the following natural question: which half of a perfect coe C is a coe C? Since 0 oes belong to C, it is quite natural to suggest that it is an even subcoe of C. The next statement answer this question for known binary perfect coes, i.e. for coes with Hamming parameters an for the binary Golay coe (since these are the only nontrivial binary perfect coes [20,22]). Theorem 25. Let C be a nontrivial completely regular binary coe with parameters n, = 2e an ρ. Assume that 0 C, an 1 / C. Then ρ = 2e + 1 an C is the even half part of a perfect coe C with minimum istance (C ) = 2e + 1. Proof. By Theorem 22, the coe C is a half of an e-perfect coe C an the minimum istance of C is = 2e + 2. First consier 1-perfect coes (i.e. coes with = 3). Let (μ 0, μ 1,...,μ n ) be the weight istribution of 1-perfect coe C containing the zero coewor. It is well known that μ i = 0 for all region from 0 to n, except when i = 1, 2,n 1,n 2. The following two properties follow from the efinition of a perfect binary coe. For any neighbor sets C i an C i+1 where i = 3, 4,...,n 4: (Q.1) for any c C i there are coewors from C i+1 at istance 3 from c; (Q.2) for any c C i+1 there are coewors from C i at istance 3 from c. It is clear, that the even half of C is the coe C with carinality C = C /2 an with minimum istance 4, as well as, the remaining part C = C \C, which is a translate of C. Now we want to prove that it is the only possibility. Since 0 C, we euce that C cannot contain any wor from C 3. Hence we choose for C all wors from C 4. If not, the wors which are not chosen will have istance 3 from C (property (Q.2)). But now, since C contains all wors from C 4,we cannot choose any wor from C 5 (property (Q.2)). Continuing in this way we obtain that C contains of all coewors of C of even weight. For the Golay [23, 12, 7]-coe C the proof is similar. Thus, we euce from Theorems 16, 18 an 25 that any nontrivial non-antipoal completely regular coe with 3 is a half of a perfect coe, or a puncture half of it. But the only nontrivial binary perfect coes are the binary Golay [23,12,7]-coe an (n = 2 m 1,N= 2 n m, 3) coes with parameters of Hamming coes [20,22]. We have, therefore, from the results above the following result.

13 3520 J. Borges et al. / Discrete Mathematics 308 (2008) Theorem 26. Let C be a nontrivial non-antipoal completely regular binary coe with parameters n, 3 an ρ. Then there are exactly four cases: If = 2e + 2, then: (1) C is a half of binary perfect Golay coe an n = 23, = 8 an ρ = 7. (2) C is a half of binary perfect coe with Hamming parameters, i.e. n = 2 m 1, = 4, ρ = 3, where m = 3, 4,.... If = 2e + 1, then: (3) C is a puncture half of binary perfect Golay coe an n = 22, = 7 an ρ = 6. (4) C is a puncture half of binary perfect coe with Hamming parameters, i.e. n = 2 m 2, = 3, ρ = 2, where m = 3, 4,.... Proof. Let C be a non-antipoal completely regular coe with minimum istance. If = 2e + 1, then by Theorem 16 we euce that ρ = 2e, an by Theorem 22, we obtain that C is a puncture half of a perfect coe. In particular, by Theorem 25, we see that the coe C (obtaine by extension of C) which contain the zero coewor is the even part of a perfect coe. This means that the original coe C is either the puncture half of the Golay [23, 12, 7]-coe, i.e. case (3), or the puncture halves of the perfect coes with Hamming parameters, i.e. case (4), foun in [18]. If C has = 2e + 2, by Theorem 18 we euce that ρ = 2e + 1, an by Theorem 23, we obtain that C is a half of a perfect coe. In this case we obtain, either a half of Golay coe, i.e. case (1), or halves of binary perfect coes with Hamming parameters, i.e. case (2). Thus, from known binary perfect coes we obtain new completely regular coes taking halves of these coes. It comes from the following statement. Theorem 27. Let C be a binary nontrivial perfect coe of (o) length n, with minimum istance 2e + 1 an such that 0 C. Denote by C the subcoe of C, forme by all coewors of even (respectively, o) weight. Then C is completely regular with covering raius ρ = 2e + 1 an with intersection array (n,...,n e,e,...1; 1,...,e,n e,...,n). Furthermore, if C is completely transitive, then C is completely transitive too. Proof. Let C be the even subcoe of C, i.e. 0 C. Since C has minimum istance 2ρ + 1, clearly we have that ρ(c) = ρ = 2e + 1. Write out the intersection numbers a i,b i an c i of C. Since = 2e + 2 an ρ = 2e + 1, we have immeiately for i = 0, 1,...,e : a i = 0, b i = n i, c i = i. But C(ρ) is the translate of C by 1, so the numbers a i, b i an c i of C for i = ρ,...,e + 1 are inverse of those values for i = 0, 1,...,e (Lemma 8), i.e. a i = 0, b i = c ρ i, c i = b ρ i. Clearly these numbers o not epen on the choice of x an coincie with numbers given by the theorem. Thus, C is completely regular with intersection array given in the statement. For the secon statement consier the coset D of C of some weight i: say D = C + x where wt(x) = i an i ρ. Clearly this coset consists of two subsets: D = D D(ρ), where D is a coset of C of weight i an D(ρ) is the translate of C(ρ) by vector x of weight i. We note that, if D is even (respectively, o), then D(ρ) is o (respectively, even). This means that two sets D an D(ρ) consist of coewors of ifferent parities. We conclue, that these two sets are fixe uner action of any permutation automorphism of C (inee, permutations o not change the weight of wors). So, if D runs over all ifferent cosets of weight i, then uner action of the same automorphisms, the coset D of C runs over the all ifferent cosets of C of weight i. Assuming that C is completely transitive, we obtain the same for C, taking into account that Perm(C) contains Perm(C ) as a subgroup. Lemma 28. Let H be the binary Hamming coe of length n = 2 m 1. Let H be the even half of H an let H (1) be the puncture coe after eleting the x 1 coorinate. Then the coe H (1) is completely transitive.

14 J. Borges et al. / Discrete Mathematics 308 (2008) Proof. The coe H (1) is the puncture half of H. Hence, the covering raius of H (1) is two an H (1) (ρ) = H (1) + 1 (see Theorem 26). The cosets of H (1) are H (1) itself, n 2 cosets of weight 1 an one coset, H (1) (ρ), of weight 2. Any permutation in Perm(H (1) ) fix the all-ones vector 1 an so, also fix the coset H (1) + 1. Therefore, to prove that H (1) is completely transitive it is enough consiering the permutations in Perm(H ) which fix a coorinate (for instance, x 1 ) an looking for the transitivity on the other coorinates. But Perm(H (1) ) contains the stabilizer of one coorinate of Perm(H ) which is the general linear group GL(m, 2) [13]. Since GL(m, 2) is oubly transitive (see [13, p. 399]), it is clear that Perm(H (1) ) is transitive. Lemma 29. Let G be a binary perfect Golay [23, 12, 7]-coe. Denote by G its even half subcoe an by G (1) the puncture coe after eleting the x 1 coorinate in G. Then G (1) is a completely transitive coe. Proof. Since the covering raius of G (1) is 6, by Theorem 11, we have that G (1) + 1 is a coset with minimum weight 6. Thus, for any vector x such that w(x) = 1, the coset G (1) x has minimum weight 5. Similarly, G (1) y has minimum weight 4, for any vector y such that w(y) = 2. Taking into account that G (1) is a 3-error correcting coe, we have that the cosets of G (1) are {G (1) + x} x:w(x)<3 {G (1) + z} z:w(z)=3 {G (1) + x + 1} x:w(x)<3. Perm(G ) is the Mathieu group M 23 which is 4-transitive (see [13]). Therefore, Perm(G (1) ) is clearly 3-transitive (since it contains the stabilizer of one coorinate of M 23, namely M 22 ). If two cosets G (1) i an G (1) j are in the same orbit, then the cosets G (1) i + 1 an G (1) j + 1 are also in the same orbit. The conclusion is that all cosets are in seven orbits an hence G is completely transitive. Now taking halves of the binary perfect [23, 12, 7] Golay coe, we obtain new completely regular, completely transitive an uniformly packe in the wie sense (or in the sense of [1]) coes. Corollary 30. Let G be a binary perfect Golay [23, 12, 7]-coe. Denote by G its subcoe, forme by all coewors of even (respectively, o) weight. Then: (i) G is a completely regular (23, 2 11, 8)-coe with ρ = 7 an intersection array (23, 22, 21, 20, 3, 2, 1; 1, 2, 3, 20, 21, 22, 23). (ii) G is a completely transitive coe. (iii) G is a uniformly packe coe in the wie sense with parameters α i, i = 0, 1,...,7: α 0 = 1, α 1 = 7 23, α 2 = 3 11, α 3 = , 29 α 4 = , α 47 5 = , α 6 = , α 7 = Proof. The statements (i) an (ii), incluing the intersection array coming from Theorem 27; inee, it is known (see [19]) that the Golay coe is completely transitive. We explain shortly how to fin the parameters α i of the new uniformly packe coe. We use the efining equation (1). Recall that subsets G w of G for w = 7, 8, 11, 12, 15, 16 are 4-esigns T w (23,w,4, λ(w)) with values [13] λ(7) = 1, λ(8) = 4, λ(11) = 48, λ(12) = 72. (23) But any t-esign with given λ is also j-esign for any j = 1,...,t with λ j which is easily compute [17]. For the values λ(w) given in (23), enote by λ j (w) the corresponing values λ j. We start with α 0. Since (G) = 8 we have that α 0 = 1. For the case v G(1) Eq. (1) becomes α 1 f 1 (v) + α 7 f 7 (v) = 1,

15 3522 J. Borges et al. / Discrete Mathematics 308 (2008) where f j (v) enotes the number of wors of G which are at the istance j from v. Clearly f 1 (v) = 1 an f 7 (v) = λ 1 (8), implying that α α 7 = 1. (24) For the case v G(7) Eq. (1) becomes α 7 f 7 (v) = 1. Since G(ρ) ρ = G 7, we have that f 7(v) = 253, implying that α 7 = 253 1, an hence, α 1 = 23 7 by (24). For the case v G(2), Eq. (1) is α α 6 f 6 (v) = 1, (25) where f 2 (v) = 56, since G 8 is a 2-esign with λ 2 (8) = 56. Continuing in the same way, we obtain: For the case v G(3): α α 5 λ 3 (8) + α 7 (3(λ 2 (8) λ 3 (8))) = 1. (26) For the case v G(4): α 4 (1 + λ 4 (8)) + α 6 (4(λ 3 (8) λ 4 (8))) = 1. (27) For the case v G(5): α 5 (1 + 5λ 4 (8)) + α 7 f 7 (v) = 1. (28) For the case v G(6): α 6 f 6 (v) = 1. (29) For the values f 5 (v) an f 6 (v) we have clearly ( ) 6 f 6 (v) = 1 + λ 4 (8) + λ 6 (12) = 77 (30) 2 an f 7 (v) = 1 2 ( ) 5 λ 3 (8) + λ 5 (12) = 112. (31) 2 In the last two expressions we use the intersection arrays of the corresponing esigns (see [13]). From (29) we have that α 6 = Using this value for (25) (28), we obtain the other values α i, given in the statement. We remark, that this new uniformly packe coe G is the first example (known to the authors) of such coe with eight ifferent values of parameters α i. Similarly, we obtain the following result for a puncture half of a binary Golay coe. Corollary 31. Let G be a binary perfect Golay [23, 12, 7]-coe. Denote by G its subcoe, forme by all coewors of even (respectively, o) weight an by G (1) the coe, obtaine by puncturing one position of G. Then: (i) G (1) is a completely regular (22, 2 11, 7)-coe with ρ = 6 an intersection array (22, 21, 20, 3, 2, 1; 1, 2, 3, 20, 21, 22). (ii) G (1) is a completely transitive coe.

16 J. Borges et al. / Discrete Mathematics 308 (2008) (iii) G (1) is a uniformly packe coe in the wie sense with parameters α (1) i, i = 0, 1,...,6: α (1) 0 = 1, α (1) 1 = α (1) 2 = 3 11, α (1) 3 = α (1) 29 4 = , α(1) 5 = α (1) 6 = Proof. G (1) is completely transitive by Lemma 29. We explain shortly (iii). Remark that the (23, 2 11, 8) coe G, consiere in Corollary 30, is the extension of the (22, 2 11, 7) coe G (1). Since G is uniformly packe in the wie sense, the coe G (1) is also uniformly packe in the wie sense with parameters α (1) i, i = 0, 1,...,6, which satisfy the following property [2]: α (1) ρ 2i = α(1) ρ 2i 1, i = 0, 1,..., (ρ 1)/2. Now the values of parameters α (1) i, i = 0, 1,...,6 come (see [2, Theorem 2]) from the known parameters α i, i = 0, 1,...,7 of the coe G, which we know from Corollary 30. Corollary 32. Let H be a binary 1-perfect coe of length n = 2 m 1 7. Denote by H its even (respectively, o) subcoe. Then: (i) H is completely regular with covering raius ρ = 3 an with intersection array (n, n 1, 1; 1,n 1,n). (ii) H is uniformly packe in the wie sense with parameters α i : α 0 = 1, α 1 = 3 n, α 2 = 2 n 1, α 6 3 = n(n 1). (iii) If H is completely transitive, then H is completely transitive too. Proof. The fact that H is completely regular comes from [18] (see also [10]), since H is the extension of the coe H (1) (see Corollary 33) which is a uniformly packe in the narrow sense coe. Similarly to Corollary 30, we show here only how to fin the parameters α i. Since = 4 an ρ = 3wehaveα 0 = 1 an α 3 = 1 H 3 (32). For α 1 we have immeiately α 1 + α 3 4 n H 4 =1, implying that α 1 = 3/n. Finally for α 2 we have ( α n 3 ) = 1, 2 implying that α 2 = 2/(n 1). Finally, H is completely transitive by Theorem 27. We have a similar result for puncture halves of perfect single error correcting coes. These coes are also well known [18] (see also [10]). Corollary 33. Let H be a binary 1-perfect coe of length n = 2 m 1 7. Denote by H its even (respectively, o) subcoe an by H (1) the coe, obtaine by puncturing one position of H. Then: (i) H (1) is completely regular with covering raius ρ = 2 an with intersection array (n 1, 1; 1,n 1).

17 3524 J. Borges et al. / Discrete Mathematics 308 (2008) (ii) H (1) is uniformly packe in the narrow sense, i.e. in the sense of [18] with parameters α (1) 0 = 1 an α (1) 1 = α (1) 2/(n 1). (iii) If H is linear, then H (1) is completely transitive. Proof. Except for the point (iii) the statement comes from [18]. Regaring (iii), note that H (1) is completely transitive by Lemma 28. As it is known from [18,10] the binary coes which are closest to binary perfect coes are Preparata-like coes. Unfortunately, a half of such coe of length 15 is not even uniformly packe coe in the wie sense. From the results of [12,3] we know that for any m 4, there exist (m + 1)/2 non-equivalent extene 1-perfect Z 4 -linear coes of length n + 1 = 2 m. Also, from [4,3] we know that there exist (m + 2)/2 non-equivalent extene 1-perfect Z 2 Z 4 -linear with length n + 1 = 2 m 16. Hence, we can obtain completely regular coes taking halves of the corresponing puncture coes. Most of them will be non-equivalent coes ue to the following theorem. Theorem 34. Let C an B two non-equivalent 1-perfect coes of length n = 2 m 1. Let C an B their halves which are completely regular coes with = 4, ρ = 3 an with intersection array (n, n 1, 1; 1,n 1,n). Then C an B are non-equivalent coes. Proof. Assume in contrary that B an C are equivalent. This means that there is a vector h F n an a permutation θ in the group of S n (all permutations of n elements) such that: B + h = θ(c). (33) By Theorems 22 an 25 we have that C = C C(ρ) an B = B B(ρ). Now from (33) we obtain that B h = θ(c) + 1 = θ(c + 1), (34) where the last equality follows since 1 is fixe by any permutation from S n. But C(ρ) = C + 1 as well as B(ρ) = B + 1 (Theorem 11). Doing the union of the sets in (33) an (34) we obtain that the coes C an B are equivalent, i.e. a contraiction. We remark that if the starting coe C is extene 1-perfect Z 2 Z 4 -linear then we can puncture in such a way that the resulting coe is also Z 2 Z 4 -linear [3]. It is also easy to see that taking the half coe which have a zero at the elete coorinate we obtain a Z 2 Z 4 -linear coe again. This is not true for the Z 4 -linear case. In [9] it is prove that puncturing an extene 1-perfect Z 4 -linear coe gives a 1-perfect coe which is not Z 2 Z 4 -linear except for the case of the extene Hamming coe of length 16 (which has a Z 4 -linear structure). Therefore, for m>4 (avoiing the mentione exception) we can obtain (m + 2)/2 non-equivalent Z 2 Z 4 -linear completely regular coes an (m + 1)/2 non-equivalent (an neither Z 2 Z 4 -linear, nor Z 4 -linear) non-equivalent completely regular coes, starting from extene 1-perfect Z 2 Z 4 -linear an from extene 1-perfect Z 4 -linear coes, respectively, of length n + 1 = 2 m. References [1] L.A. Bassalygo, G.V. Zaitsev, V.A. Zinoviev, Uniformly packe coes, Problems Inform. Transmission 10 (1) (1974) [2] L.A. Bassalygo, V.A. Zinoviev, Remark on uniformly packe coes, Problems Inform. Transmission 13 (3) (1977) [3] J. Borges, K. Phelps, J. Rifa, The rank an kernel of extene 1-perfect Z 4 -linear an aitive non-z 4 -linear coes, IEEE Trans. Inform. Theory 49 (8) (2003) [4] J. Borges, J. Rifà, A characterization of 1-perfect aitive coes, IEEE Trans. Inform. Theory 45 (1999) [5] J. Borges, J. Rifa, On the nonexistence of completely transitive coes, IEEE Trans. Inform. Theory 46 (1) (2000) [6] J. Borges, J. Rifa, V.A. Zinoviev, Nonexistence of completely transitive coes with error-correcting capability e>3, IEEE Trans. Inform. Theory 47 (4) (2001) [7] P. Delsarte, An algebraic approach to the association schemes of coing theory, Philips Research Reports Suppl. 10 (1973). [8] P. Delsarte, V.I. Levenshtein, Association schemes coing theory, IEEE Trans. Inform. Theory 44 (1998) =