One Weight Z2Z4 Additive Codes Steven T. Dougherty1 , Hongwei Liu2 , Long Yu2 1

Department of Mathematics, University of Scranton, Scranton PA 18510, USA 2 School of Mathematics and Statistics, Central China Normal University, Wuhan, Hubei 430079, China

Abstract We study one weight Z2 Z4 additive codes. It is shown that the image of an equidistant Z2 Z4 code is a binary equidistant code and that the image of a one weight Z2 Z4 additive code, with nontrivial binary part, is a linear binary one weight code. The structure and possible weights for all one weight Z2 Z4 additive codes are described. Additionally, a lower bound for the minimum distance of dual codes of one weight additive codes is obtained. Keywords: One weight additive codes, dual codes, Gray map. Mathematics Subject Classification (2010):

1

94B60, 94B05.

Introduction

A code of length n over a finite field Fq of size q is a subset of Fnq . If a code has the property that each nonzero codeword has the same Hamming weight, then we call this code a one weight code. A code is said to be equidistant if the distance between any two codewords is a constant. A linear equidistant code is necessarily a one weight code. The structure of linear one weight codes over finite fields has been determined by Bonisoli (see [2]). It was proven in [2] that every equidistant linear code is a sequence of simplex codes, where the simplex codes are the duals of Hamming codes. In [8], Carlet studied linear one weight codes over Z4 , where the weight used here is the Lee weight. Wood determined the structure of linear one weight codes over Zm for various weights in [16]. In [9], Delsarte gave foundational results on additive codes. He showed, in particular, that any abelian binary propelinear code has the form Zγ2 × Zδ4 for some nonnegative integers γ and δ. Hence it became important to study additive codes in Zα2 × Zβ4 for some Email addresses: [email protected]; [email protected]; [email protected]

1

nonnegative integers α and β. Foundational results on Z2 Z4 additive codes, including the generator matrix, the existence and constructions of self-dual codes and formally self-dual codes and several different bounds can be found in [4], [5], [6], [7] and [11]. In this paper, we introduce one weight codes over Z2 Z4 , that is, a subset of Zα2 × Zβ4 with only one weight for all nonzero vectors in the code. We prove that every one weight additive Z2 Z4 code has a linear image and use this to classify all additive one weight Z2 Z4 codes. We describe the structure and possible weights for all one weight Z2 Z4 additive codes. We obtain a lower bound for the minimum distance of dual codes of one weight additive codes. Finally, we show that the image of an equidistant Z2 Z4 code is a binary equidistant code.

2

Preliminaries

A binary code is a subset of Zn2 , and if it is a vector space, it is said to be linear. A quaternary code is a subset of Zn4 and it is said to be linear if it is a module. A Z2 Z4 code is a subset of Zα2 × Zβ4 , where α, β ≥ 0 are nonnegative integers such that α + β > 0. The code is said to be an additive code if it is a subgroup of the ambient group. Throughout the paper we use the calligraphic C to denote codes in Zα2 × Zβ4 and we use the standard C to denote codes over the binary field Z2 . Recall that the usual quaternary Gray map, from Z4 to Z22 , is given by ϕ(0) = (0, 0), ϕ(1) = (0, 1), ϕ(2) = (1, 1), ϕ(3) = (1, 0).

(2.1)

Let (v|w) ∈ Zα2 × Zβ4 , where v = (a1 , · · · , aα ) ∈ Zα2 , w = (b1 , · · · , bβ ) ∈ Zβ4 , then the is defined as follows: Gray map Φ : Zα2 × Zβ4 → Zα+2β 2 Φ((v|w)) = (v|ϕ(w)). Notationally, we let bj be an element of Z4 and write ϕ(bj ) = (bj,0 , bj,1 ). Then when applying the map ϕ to the vector w = (b1 , · · · , bβ ) we collect the first coordinates bj,0 as the first β coordinates and the second coordinates bj,1 as the last β coordinates. Namely we write ϕ(w) = (b1,0 , · · · , bβ,0 , b1,1 , · · · , bβ,1 ). Let (x|y) ∈ Zα2 × Zβ4 . We denote by wH (x|y) the Hamming weight of (x|y). For any two vectors (x|y), (x0 |y0 ) ∈ Zα2 × Zβ4 , the Hamming distance between (x|y) and (x0 |y0 ) is defined to be dH ((x|y), (x0 |y0 )) = wH ((x|y) − (x0 |y0 )) = wH (x − x0 |y − y0 ). We define the Lee weight of (x|y) as: wt((x|y)) = wH (Φ((x|y))) = wH (x) + wH (ϕ(y)). For any two vectors (v1 |w1 ), (v2 |w2 ) ∈ Zα2 × Zβ4 , we define the Lee distance between (v1 |w1 ) and (v2 |w2 ) as d((v1 |w1 ), (v2 |w2 )) = wt(v1 − v2 |w1 − w2 ) = wtH (v1 − v2 ) + wH (ϕ(w1 − w2 )). (2.2) It is well known, see [6], that the Gray map Φ : Zα2 × Zβ4 → ZN 2 is an isometry from β α N (Z2 × Z4 , d(·, ·)) to (Z2 , dH (·, ·)), where N = α + 2β. 2

Definition 2.1. Let (v1 |w1 ), (v2 |w2 ) ∈ Zα2 × Zβ4 , where v1 , v2 ∈ Zα2 and w1 , w2 ∈ Zβ4 . The inner product between (v1 |w1 ) and (v2 |w2 ) is defined as follows: h(v1 |w1 ), (v2 |w2 )i = 2hv1 , v2 i + hw1 , w2 i ∈ Z4 , where the computation of 2hv1 , v2 i is done in Z4 . Let C ⊆ Zα2 × Zβ4 be a code. We define the dual code C ⊥ of C as C ⊥ = {(x|y) | (x, y) ∈ Zα2 × Zβ4 , h(x|y), (v|w)i = 0 for all (v|w) ∈ C}. The code C is called self-orthogonal if C ⊆ C ⊥ and self-dual if C = C ⊥ . It is immediate that the dual code C ⊥ of C is an additive code. We denote the minimum Hamming weight of the code C by dH (C). We denote the minimum Lee distance of the code C by d(C). This may not be equal to the minimum Lee weight of the code and so we denote the minimum Lee weight of code C by wt(C). The MacWilliams relations for additive groups are given in [9] and [11]. In this case the MacWilliams relations for the Lee weight enumerator coincide with those of the Hamming weight enumerator for linear codes over the finite field Z2 = F2 . Namely we have the following theorem. P Theorem 2.2. [9] Let LC = c∈C xα+2β−wtL (c) y wtL (c) where wtL (c) denotes the Lee weight of c. Let C be an additive code in Zα2 × Zβ4 . Then LC ⊥ (x, y) =

1 LC (x + y, x − y). |C|

Let C be a code in Zα2 ×Zβ4 with minimum distance 2t+1, then C is a t-error-correcting code. Since the Gray map is an isometry, we know that Φ(C) is a t-error-correcting code of length α + 2β over Z2 as well. By the usual sphere packing bound over Z2 , we get |Φ(C)| ·

 t  X α + 2β i=0

i

≤ |Zα+2β | = 2α+2β . 2

Since |Φ(C)| = |C|, and |Zα2 × Zβ4 | = 2α+2β , we obtain the sphere packing bound for a code C in Zα2 × Zβ4 :  t  X α + 2β |C| · ≤ 2α+2β = |Zα2 × Zβ4 |. (2.3) i i=0 A code C in Zα2 × Zβ4 that meets the bound in Equation 2.3 is called a perfect code. Example 2.3. Let C = h(011|11), (110|20)i be an additive code in Z32 ×Z24 . The codewords of C are: C = {(000|00), (110|20), (011|11), (000|22), (011|33), (101|31), (110|02), (101|13)}.

3

The weight enumerator of C is LC (x, y) = x7 + 7x3 y 4 . By the MacWilliams relations (see Theorem 2.2), we get that the weight enumerator of C ⊥ is LC ⊥ (x, y) = x7 + 7x4 y 3 + 7x3 y 4 + y 7 . Hence the dual code C ⊥ of C has minimum distance 3. We have ⊥

|C | ·

1   X 7 i=0

i

= 16 · (1 + 7) = 128 = 23+2·2 .

This implies that the code C ⊥ meets the sphere packing bound and therefore C ⊥ is a perfect code in Z32 × Z24 .

3

One weight codes in Zα2 × Zβ4

In this section, we study the properties of one weight Z2 Z4 codes. Definition 3.1. A nonzero code C in Zα2 × Zβ4 is called a one weight code if all of its nonzero codewords have the same weight. In this case, if the weight is m, then the code is called a one weight code with weight m. For convenience, we adopt the convention that the zero code in Zα2 × Zβ4 , denoted by O, is a one weight code with weight 0. Definition 3.2. A code C in Zα2 ×Zβ4 is called an equidistant code if for any four codewords c 6= d, c0 6= d0 in C, we have d(c, d) = d(c0 , d0 ). The next theorem follows immediately. Theorem 3.3. If C is an equidistant code in Zα2 × Zβ4 with distance m, then Φ(C) is a binary equidistant code with the same distance m. Furthermore, if 0 ∈ C, then C is a one weight code in Zα2 × Zβ4 with weight m and Φ(C) is also a one weight binary code with weight m. Proof. Let Φ(c), Φ(c0 ) ∈ Φ(C) with Φ(c) 6= Φ(c0 ), where c, c0 ∈ C. Then, since the Gray map is an isometry, we have dH (Φ(c), Φ(c0 )) = d(c, c0 ) = m. This gives that Φ(C) is an equidistant code with the same distance m that is the distance of C. If 0 ∈ C, then for any nonzero codeword c ∈ C, we have wt(c) = d(c, 0) = m. Note that Φ(0) = 0, then the equality above gives that wH (Φ(c)) = dH (Φ(c), Φ(0)) = d(c, 0) = wt(c) = m. This finishes the proof. 4

It is easy to see that if C is an additive code, then C is a one weight code if and only if C is an equidistant code. Example 3.4. Let C = h(1|1)i be an additive code in Z12 × Z14 . Then C is a one weight additive code with weight 2, and its dual code C ⊥ = h(1|2)i, which is a one weight additive code with weight 3. This implies that C and C ⊥ are both one weight codes. The image is Φ(C) = h(101), (011)i ≤ Z32 , which is a one weight binary linear [3, 2, 2] code with weight 2. It is easy to check that Φ(C ⊥ ) = Φ(C)⊥ , which is the binary Hamming code with parameters [3, 1, 3]. The code C = h(011|11), (110|20)i in Z32 × Z24 , given in Example 2.3, is a one weight additive code with weight 4. The dual code C ⊥ of C is C ⊥ = h(100|13), (010|11), (001|02)i. The image is Φ(C) = h(0110101), (1101100), (1011001)i, which is the binary simplex code of length 7. Furthermore, we can verify that Φ(C ⊥ ) = Φ(C)⊥ . Example 3.5. The code C = h(0011011|1111), (0111100|0220), (1001101|0022)i is an additive code in Z72 × Z44 , which is a one weight code with weight 8. It can be verified that the image Φ(C) of the code C is a binary simplex code of length 15. If a code C over Fq meets the Plotkin bound then C must be an equidistant code, see [15] for a proof of this result. Applying the Gray map to a code in Zα2 ×Zβ4 which meets the Plotkin bound for Z2 Z4 codes and applying this known result to the binary image we have that any additive Z2 Z4 code meeting this bound is a one weight code. We have examples of additive codes that meets this bound. For example, the code C in Example 2.3 meets this bound, that is C ∼ = Z12 × Z14 as a group, with α = 3, β = 2, γ = 1 and δ = 1. In this particular case d(C) = 4 which is equal to the right side of the inequality. The following result gives a construction of one weight additive codes in Zα2 × Zβ4 . Theorem 3.6. Let C be a one weight additive code with weight m in Zα2 × Zβ4 . Then for tβ any positive integer t, there exists a one weight additive code of weight tm in Ztα 2 × Z4 . Proof. Let G be a generator matrix of the one weight code C with weight m. We can write G as G = (G1 |G2 ), where G1 , G2 are the binary and quaternary parts of the generator matrix G respectively. Let C˜ be an additive code generated by the following matrix t

t

z }| { z }| { ˜ = ( G1 , · · · , G1 | G2 , · · · , G2 ). G ˜ there exists a nonzero vector a such that c˜ = aG. ˜ Then for any nonzero codeword c˜ ∈ C, Note here that the vector a is an integer vector and multiplication of a row by an integer is simply the row added to itself that many times. Hence t

t

z }| { z }| { ˜ = wt( aG1 , · · · , aG1 | aG2 , · · · , aG2 ) = t · wH (aG1 ) + t · wt(aG2 ) wt(˜c) = wt(aG) = t · (wH (aG1 ) + wt(aG2 )) = t · wt(aG). Note that C is a one weight code with weight m, which implies that wt(aG) = m for all nonzero a. Then we get that wt(˜c) = tm. 5

Definition 3.7. An additive code in Zα2 × Zβ4 is called decomposable if it is of the form C × D, where C ⊆ Zα2 and D ⊆ Zβ4 . Otherwise it is called indecomposable. Sometimes in the literature, decomposable codes are called separable and indecomposable codes are called non-separable. Proposition 3.8. There do not exist decomposable one weight additive codes in Zα2 × Zβ4 with C 6= 0 and D 6= 0. Proof. Suppose C ×D is a decomposable one weight additive code with C 6= 0 and D 6= 0. Let 0 6= (c|d) ∈ C ×D. Note that C ×D is an additive code, we get that 0 ∈ C and 0 ∈ D. Then we can see that (c|0), (0|d) ∈ C × D. If c 6= 0 and d 6= 0, then wt(c|d) 6= wt(c) and wt(c|d) 6= wt(d). This is a contradiction, and we have the result. The next lemma is a direct generalization of an analogous results for codes over fields. Lemma 3.9. Let C be an additive code in Zα2 × Zβ4 with no all zero columns. Then the sum of the weights of all the codewords of C is equal to |C| (α + 2β). 2 Proof. Write the codewords of C as rows of a matrix G. Consider the first α coordinates of G, namely those that contain the binary part of the codewords. Since C is an additive code, the number of coordinates containing a 0 is equal to the number of coordinates containing a 1 for any column. Next consider the last β columns of G, namely those containing the quaternary part of the codewords. Any column in this part either contains an equal number of 0, 1, 2 and 3 or it contains an equal number of 0 and 2 and does not contain any coordinates with a 1 or 3 in it. Suppose there are s of these columns containing only 0 and 2 in the last β columns. Then there are β − s columns containing an equal number of 0, 1, 2 and 3. Hence the sum of the weights in C is X

wt(c) = α ·

c∈C

 |C|  2

+

 |C|

Then we have X c∈C

wt(c) = α ·

  |C| |C| |C|  ·2 ·s+ + ·2+ (β − s). 2 4 4 4

 |C|  2

+ β · |C| =

|C| · (α + 2β), 2

which gives the result. This leads to the following theorem. Theorem 3.10. Let C be a one weight additive code with weight m in Zα2 × Zβ4 such that γ there exists no zero columns in the generator matrix of C, and suppose C ∼ = Z2 × Zδ4 as an additive group. Then there exists a unique positive integer λ such that m = λ · 2γ+2δ−1 , where α and β satisfy α+2β = λ·(2γ+2δ −1). Furthermore, if m is an odd integer, then α is odd and C = {(0α |0β ), (1α |2β )}, where 1α = ( 1, · · · , 1 ) ∈ Zα2 , and 2β = ( 2, · · · , 2 ) ∈ Zβ4 .

6

Proof. By Lemma 3.9, we have that X

wt(c) =

c∈C

|C| · (α + 2β). 2

On the other hand, the sum of the weights of all codewords in C is (|C| − 1)m. This gives (|C| − 1)m =

|C| · (α + 2β). 2

Note that the cardinality of C is |C| = 2γ 4δ = 2γ+2δ , and gcd(|C| − 1, |C| ) = gcd(2γ+2δ − 2 = λ · 2γ+2δ−1 , 1, 2γ+2δ−1 ) = 1. Hence there exists a positive integer λ such that m = λ · |C| 2 and α + 2β = λ · (|C| − 1) = λ · (2γ+2δ − 1). Furthermore, if m is odd, then λ · 2γ+2δ−1 is odd. This implies that λ is odd and γ+2δ−1 2 = 1, which gives that γ = 1, δ = 0 and m = λ. In this case α + 2β = λ = m is odd, hence α is odd. Recall that C is a one weight additive code with weight m = α + 2β. Since (1α |2β ) is the only word with weight α + 2β, we get C = {(0α |0β ), (1α |2β )}.

4

Dual one weight codes

In this section, we study dual codes of one weight additive codes. We give a lower bound on the minimum distance for the dual codes of one weight additive codes. Since there exist one weight codes whose dual are also one weight, we give a complete classification for this class of one weight codes. γ Theorem 4.1. Let C be a one weight code in Zα2 × Zβ4 where C ∼ = Z2 × Zδ4 as a group. If there exists no zero columns in the generator matrix of C, then d(C ⊥ ) ≥ 2. Furthermore, d(C ⊥ ) ≥ 3 if and only if λ = 1. In the case when λ = 1, if |C| ≥ 4 then d(C ⊥ ) = 3.

Proof. If there exists no zero columns in the generator matrix of C, then by Theorem 3.10, there exists a positive integer λ such that m = λ·2γ+2δ−1 and N = α +2β = λ·(2γ+2δ −1). Since C is a one weight additive code, we get that the weight enumerator of C is LC (x, y) = xN + (|C| − 1)xN −m y m . By the MacWilliams relations (see Theorem 2.2), the weight enumerator of the dual code C ⊥ is as follows: LC ⊥ (x, y) =

1 ((x + y)N + (|C| − 1)(x + y)N −m (x − y)m ). |C|

The coefficient of xN −1 y on the right side of Equation 4.1 is       1 N N −m m 1 ( + (|C| − 1)( − )) = (N |C| − 2m(|C| − 1)). |C| 1 1 1 |C| Note that, |C| = 2γ+2δ , N = λ · (2γ+2δ − 1) = λ(|C| − 1), and 2m = λ|C|. Hence 1 (N |C| − 2m(|C| − 1)) = λ(|C| − 1) − λ(|C| − 1) = 0. |C| 7

(4.1)

This implies that there exists no codeword with weight 1 in C ⊥ , which gives that d(C ⊥ ) ≥ 2. Furthermore, the coefficient of xN −2 y 2 in the right side of Equation 4.1 is   N − m m N − mm 1  N ν = + (|C| − 1) + − |C| 2 2 2 1 1   2 2 1 N (N − 1) (|C| − 1)(N − 4N m + 4m − N ) + . = |C| 2 2 Substituting N = λ(|C| − 1), m = fying, we get

λ|C| 2

to the right side of the above equation and simpli-

(λ − 1)λ(|C| − 1) . 2 Since |C| ≥ 2 and λ ≥ 1, we get ν = 0 if and only if λ = 1, which gives the result. Now suppose |C| ≥ 4. Computing the coefficient of xN −3 y 3 on the right side of Equation 4.1, we have   N − m N − mm N − mm m 1  N + (|C| − 1) − + − . µ= |C| 3 3 2 1 1 2 3 ν=

For convenience, we let           N −m N −m m N −m m m = − + − . 3 2 1 1 2 3 Note that N = |C| − 1, m =

|C| , 2

then N − m =

|C| 2

− 1 = m − 1. Then we have

(N − m)(N − m − 1)(N − m − 2) (N − m)(N − m − 1)m − 6 2 (N − m)m(m − 1) m(m − 1)(m − 2) + − 2 6 N −m−2 m N −m m−2 − ) + m(m − 1)( − ) = (N − m)(N − m − 1)( 6 2 2 6 (−2m − 3) (2m − 1) = (m − 1)(m − 2) + m(m − 1) 6 6 (m − 1) = ((m − 2)(−2m − 3) + (2m − 1)m) = m − 1. 6

 =

Hence    1  N 1  N (N − 1)(N − 2) + 6(|C| − 1)(m − 1)  µ = + (|C| − 1)(m − 1) = |C| 3 |C| 6 (|C| − 1)(|C| − 2) = . 6 Note that |C| = 2γ+2δ ≥ 4. In this case, we can check that µ ≥ 1. This implies that d(C ⊥ ) = 3. 8

The following result is a characterization of additive codes all of whose codewords have even weights. Theorem 4.2. Let C be an additive code in Zα2 × Zβ4 . Then the weights of all codewords of C are even if and only if (1α |2β ) ∈ C ⊥ . Proof. Let (v|w) = (v1 , · · · , vα |w1 , · · · , wβ ) ∈ C, where v = (v1 , · · · , vα ) ∈ Zα2 and w = (w1 , · · · , wβ ) ∈ Zβ4 . We consider the following equality h(1α |2β ), (v|w)i =

α X i=1

2vi +

β X

2wj .

j=1

It is easy to see that h(1α |2β ), (v|w)i = 0 if and only if wt(v|w) is even, which gives the result. From this theorem, we get the following corollary immediately. Corollary 4.3. Let C be a one weight additive code in Zα2 × Zβ4 . Then the weight of C is even if and only if (1α |2β ) ∈ C ⊥ . Note that the code C and C ⊥ in Example 3.4 are both one weight codes. The code generated by (10|) in Z22 × Z04 is a one weight code with weight 1, its dual code is generated by (01|) which is also a one weight code with weight 1. The code D1 = D1⊥ = {(00|), (11|)} ⊆ Z22 × Z04 and D2 = D2⊥ = {(|0), (|2)} ⊆ Z02 × Z14 are also one weight codes. As a trivial example, the dual code O⊥ = Zα2 × Zβ4 , where O is the zero code in Zα2 × Zβ4 . We know that O⊥ is a one weight code if and only if α = 1, β = 0. In this case, O⊥ = Z12 × Z04 = {(0|), (1|)} = Z2 is a one weight code with weight 1. We shall use these examples in Theorem 4.4 when classifying all possible cases where this occurs. Suppose C and C ⊥ are both one weight codes with weights m1 and m2 respectively. Then the weight enumerators of C and C ⊥ are as follows: LC (x, y) = xN + (|C| − 1)xN −m1 y m1 ,

LC ⊥ (x, y) = xN + (|C ⊥ | − 1)xN −m2 y m2 .

By the MacWilliams relations (see Theorem 2.2), we have xN + (|C| − 1)xN −m1 y m1 =


1 N ⊥ N −m2 m2 (x + y) + (|C | − 1)(x + y) (x − y) . |C ⊥ |

(4.2)

If C is a one weight code with even weight, then by Corollary 4.3, (1α |2β ) ∈ C ⊥ . Since the code C ⊥ is also a one weight code, we have that |C ⊥ | = 2 and m2 = α + 2β = N . Then Equation 4.2 is equivalent to
1 (x + y)N + (x − y)N 2 (

xN + N2 xN −2 y 2 + N4 xN −4 y 4 + · · · + y N , N is even;

= N N xN + 2 xN −2 y 2 + 4 xN −4 y 4 + · · · + N xy N −1 , N is odd.

xN + (|C| − 1)xN −m1 y m1 =

9

Hence, N has only three possible values: N = 1, N = 2 or N = 3. If C and C ⊥ are both one weight codes with odd weights, then by Corollary 4.3, we have that (1α |2β ) ∈ / C ⊥ and (1α |2β ) ∈ / C. Comparing the coefficients of y N in both sides of Equation 4.2, and noting that m1 < N since (1α |2β ) ∈ / C, we have 0=

1 1 (1 + (|C ⊥ | − 1) · (−1)m2 ) = ⊥ (2 − |C ⊥ |). ⊥ |C | |C |

It follows that |C ⊥ | = 2. By a similar argument, it is easy to show that |C| = 2. Since |C| · |C ⊥ | = 2N , we have that N = 2. By the above discussion, we can prove the following theorem. Theorem 4.4. Assume C and C ⊥ are both one weight additive codes in Zα2 × Zβ4 with N = α + 2β. 1. If the weight of C is even and the weight of C ⊥ is odd, then N = 1 or 3. • If α = 1 and β = 0 then C = O = {(0|)} and C ⊥ = O⊥ = {(0|), (1|)} = Z2 . • If α = 1 and β = 1 then C = {(0|0), (1|1), (0|2), (1|3)} and C ⊥ = {(0|0), (1|2)}. • If α = 3 and β = 0 then C = {(000|), (110|), (101|), (011|)} and C ⊥ = {(000|), (111|)}. 2. If the weights of C and C ⊥ are both even, then N = 2 and C is self-dual. • If α = 2 and β = 0, then C = C ⊥ = {(00|), (11|)}. • If α = 0 and β = 1 then C = C ⊥ = {(|0), (|2)} = 2Z4 . 3. If the weights of C and C ⊥ are both odd, then N = 2. In this case, the only possibility for α and β is α = 2, β = 0 and C = {(00|), (10|)} and C ⊥ = {(00|), (01|)}. Proof. (1) Since C is a one weight code with even weight, then (1α |2β ) ∈ C ⊥ , and N = α + 2β = 1, 2, or 3 by the previous discussion. Note that C ⊥ is also a one weight code with odd weight N , hence N = 1 or N = 3. If N = 1, then α = 1, β = 0, and C ⊥ = {(0|), (1|)} with C = O = {(0|)}. If N = 3, then α = 3, β = 0, and C ⊥ = {(000|), (111|)} with C = {(000|), (110|), (101|), (011|)}, or α = 1, β = 1, and C ⊥ = {(0|0), (1|2)} ∈ Z12 × Z14 , with C = {(0|0), (1|1), (0|2), (1|3)}. (2) If the weights of C and C ⊥ are both even, then C = C ⊥ = h(1α |2β )i. Hence C is self-dual and N = 2. In this case, if α = 2, β = 0, then C = C ⊥ = {(00|), (11|)}. If α = 0 and β = 1, then C = C ⊥ = {(|0), (|2)} = 2Z4 . (3) If the weights of C and C ⊥ are both odd, then N = 2 by the previous discussion before the theorem. If α = 0 and β = 1, we know that there exists no one weight additive code with odd weight of length 1 in Z02 × Z14 . Hence α = 2 and β = 0 and in this case, C = {(00|), (10|)} and C ⊥ = {(00|), (01|)}. By this theorem, we can obtain the following corollary. Corollary 4.5. If N = α + 2β ≥ 4 then there exists no one weight additive code in Zα2 × Zβ4 such that its dual is also a one weight code. 10

5

The structure of one weight additive codes

In this section, we give a method to construct one weight codes in Zα2 × Zβ4 . Then we characterize the structure of one weight codes. We begin by noting that if C is a nontrivial one weight code with weight m, with a generator matrix that has no zero columns, then m = λ · 2γ+2δ−1 is even, where λ is a γ positive integer, and C ∼ = Z2 × Zδ4 as a group. Lemma 5.1. Let G be the generator matrix of a one weight code in Zα2 × Zβ4 with weight m. Then if c = (v|w) is a row of G, then the number of units in w is either 0 or m2 . Proof. Let c = (v|w) be a row of G. Then we have wtH (v) + wtL (w) = m. If c + c = 0, then w contains no units. Consider the vector c + c, where c + c 6= 0. Then v + v = 0 and c + c = (0|2w). This means that wtL (2w) = m. We have that wtL (2w) = 2|{i | (w)i = ±1}|. Then |{i | (w)i = ±1}| =

m 2

and we have our result. Consider the vector (1|1312). This generates the code {(0|0000), (1|1312), (0|2220), (1|3132)}. This code is a one weight code with weight 6 and the quaternary part of the vector has 3 units. Lemma 5.2. Let C be a one weight code in Zα2 × Zβ4 with weight m. If c1 = (v1 |w1 ) and c2 = (v2 |w2 ) are two distinct order 4 codewords of C, then {i | (w1 )i = ±1} = {i | (w2 )i = ±1}. Proof. Assume c1 = (v1 |w1 ) and c2 = (v2 |w2 ) are two distinct order 4 vectors in a one weight code C with weight m. We have that the number of units in w1 and w2 is m2 . Consider the vector c = c1 + c1 + c2 + c2 . The binary part of this vector is 0 and the quaternary part consists of elements that are either 0 or 2. If (w1 )i = ±1 whenever (w2 )i = ±1 then the number of coordinates in c that are 2 is 0. That is c = 0. So assume that there are coordinates in the quaternary part of c1 that have a unit where the corresponding coordinate in c2 does not have a unit. We need the number of units in w1 + w2 to be m2 . To accomplish this we need |{i | (w1 )i = ±1 = −(w2 )i }| =

m 2

and in all other coordinates where (w1 )i = ±1 we need that (w2 )i = 0 or 2, and in all other coordinates where (w2 )i = ±1 we need that (w1 )i = 0 or 2. Then consider d = c1 + c2 + c2 + c2 . This vector has the same binary part as c1 + c2 but it has a 2 in the coordinates where (w1 )i = ±1 = −(w2 )i and has coordinates of equal Lee weight in all other coordinates. Therefore d has weight greater than m. This gives our result. 11

As an example of the previous, consider the two quaternary vectors x = (111100) and y = (003311). They both have weight 4 with 4 units, and their sum has 4 weight 4 units. But x + 3y = (112233) has weight 8. Lemma 5.3. Let C be a one weight code in Zα2 × Zβ4 with weight m. If c1 = (v1 |w1 ) and c2 = (v2 |w2 ) are two distinct order 4 vectors in C, then |{i|(w1 )i = (w2 )i = ±1}| = |{i|(w1 )i = −(w2 )i = ±1}| =

m . 4

Proof. Assume c1 = (v1 |w1 ) and c2 = (v2 |w2 ) are two distinct order 4 vectors in C. Consider the vectors c = c1 + c2 and c0 = c1 + c2 + c2 + c2 . The binary parts of c and c0 are identical. In the quaternary part, we know that there is a unit in a coordinate of c1 if and only if there is a unit in that coordinate of c2 . In all other quaternary coordinates the values in c and c0 are the same. Hence the Lee weight in the coordinates where there are units in c1 must be the same in c and c0 . These have the same number of coordinates with a 2 in them if and only if |{i|(w1 )i = (w2 )i = ±1}| = |{i|(w1 )i = −(w2 )i = ±1}|. Then by Lemma 5.1 we know that there are m2 coordinates of c1 with a unit and we have that m |{i|(w1 )i = (w2 )i = ±1}| = |{i|(w1 )i = −(w2 )i = ±1}| = . 4 By identical reasoning we have that if x = (v1 |w1 ) and y = (v2 |w2 ) are distinct vectors of order 4 where v1 , v2 generate a one weight binary code of weight m2 , with |{i|(w1 )i = (w2 )i = ±1}| = |{i|(w1 )i = −(w2 )i = ±1}| =

m , 4

then x and y generate a one weight Z2 Z4 additive code of weight m. Example 5.4. Consider the code with α = 0 and β = 7 generated by the matrix   1 1 1 1 2 0 2 . 1 3 3 1 0 2 2 This code has 8 elements and is a one weight code with weight 8. Notice that each vector has 4 units and that in 2 places with units they coincide  and in two places with  units they 1 1 1 1 2 0 2 differ. This matrix can be put into standard form as . This does 0 2 2 0 2 2 0 not violate Proposition 2 in [8] since in our generator matrix we have repeated columns. It is not true that if G2 generates a nontrivial one weight quaternary code and G1 generates a nontrivial one weight binary code that G = (G1 |G2 ) generates a one weight code in Zα2 × Zβ4 . This follows from the fact that if c = (v|w) is a row of G then wtH (v) + wtL (w) = m. Then c + c = (0|2w). This gives that wtL (2w) = m, unless c + c = 0. But if G2 generates a one weight quaternary code then wtL (2w) = wtL (w). This gives that the weight of the binary part of every vector is 0. Hence this does not give a generator of a one weight code. 12

Lemma 5.5. If c = (v|w) ∈ C, where C is a one weight code and w contains no coordinates with a 2 in them, then wtH (v) = wtL (w) = m2 . Proof. Assume c = (v|w) ∈ C, where C is a one weight code. Then if w consists only of units and zeros then wtL (2w) = 2wtL (w) = m. Hence wtH (v) = wtL (w) = m2 , since wH (v) + wL (w) = m. Theorem 5.6. Let G = (G1 |G2 ) be a generator matrix of a code in Zα2 × Zβ4 , where G1 generates a one weight binary code with weight m2 and G2 consists only of units of number m where any two rows of G2 have an equal number of coordinates where they are equal 2 and where they differ. Then G generates a one weight Z2 Z4 code with weight m. Proof. Let C be an additive code generated by G = (G1 |G2 ), where G1 generates a m binary P one weight linear code with weight 2 . For anyPnonzero mcodeword cP∈ C, then c = i (vi |wi ), where (vi |wi ) is a row of G. Then wH ( i vi ) = 2 and wL ( i wi ) = m2 by the previous results. Hence wt(c) = 2 · ( m2 ) = m. Let D be the subcode of an additive code C in Zα2 × Zβ4 which contains all order two ˜ which is the punctured code codewords and let κ be the dimension of the binary code D, of D by deleting the coordinates outside of the first α coordinates. In this case, we will say that C is of type (α, β; γ, δ; κ). We have the following theorem. Theorem 5.7. If C is a one weight code in Zα2 ×Zβ4 isomorphic to Zγ2 ×Zδ4 as a group, then δ ≤ 1. In this case, the one weight additive code C with type (α, β; γ, δ; κ) is permutation equivalent to a one weight Z2 Z4 -additive code with a canonical generator matrix G, whose rows have same weight, of the following form: • If δ = 0 then  G=

Iκ Tb 2T2 0 0 0 2T1 2Iγ−κ

 ,

where Tb , T1 , T2 are matrices over Z2 . • If δ = 1 then  0 0 Iκ Tb 2T2 G =  0 0 2T1 2Iγ−κ 0  , 0 v1 v2 v3 1 

where Tb , T1 , T2 are matrices over Z2 , v1 and v3 are binary vectors and v2 is a quaternary vector. Proof. By Lemma 5.3 any two distinct order 4 vectors have their units in the same coordinates. Hence, by adding the first order 4 vector in a generator matrix to the others we have a new generator matrix with only one order 4 vector. This gives that δ ≤ 1. The form of the generator matrix of one weight additive codes then follows from the standard form given in [6]. Theorem 5.8. Let C be a one weight additive code in Zα2 ×Zβ4 then Φ(C) is a linear binary code. 13

Proof. It is well known that if 2v ∗ w ∈ C for all v, w in a generator matrix for C then Φ(C) is linear, where v = (a1 , a2 , · · · , aα+β ), w = (b1 , b2 , · · · , bα+β ) and v ∗ w = (a1 b1 , a2 b2 , · · · , aα+β bα+β ). Since δ ≤ 1 we have that 2v ∗ w ∈ C for all v, w in its generator matrix, therefore Φ(C) is linear. Example 5.9. Consider the generator matrix:   0 1 1 1 1 . 1 0 1 1 3 It generates a code with m = 4 and |C| = 8. Notice the code has type 41 21 . The image of this code under the Gray map is the Simplex code. Theorem 5.10. Every binary simplex code is the Gray image of a nontrivial Z2 Z4 one weight additive code. Proof. We shall construct a generator matrix (G1 |G2 ) of a Z2 Z4 code that gives the binary simplex code of length 2r −1. First take the generator for the binary simplex code of length 2r−1 − 1 as G1 . Construct the quaternary part of the generating matrix as follows. Take as G2 the matrix formed by writing the columns as all possible vectors with entries from the set {1, 3} of length r − 1. Then the generating matrix satisfies the conditions in Theorem 5.6. It is easy to see that the image has the parameters of the simplex code and is linear by Theorem 5.8. This theorem allows us to classify all abelian propelinear one weight binary codes. Theorem 5.11. If C is an abelian propelinear one weight binary code then C is either the multiple of simplex codes or it is the multiple of codes that are the image of a quaternary code of type 4δ 2γ such that no column in the generator matrix of the quaternary code is zero, no columns are repeated and the length is 2γ+2δ . Proof. Any abelian propelinear code is the image of a code in Zα2 Zβ4 . If α = 0 then the code is purely quaternary and the result for quaternary codes is found in [8]. If α is not 0, then we know the image is linear by Theorem 5.8, so it must be the multiple of simplex codes since these are precisely the linear binary one weight codes.

Acknowledgements The second author thanks the Department of Mathematics, University of Scranton for their hospitality and support, where he stayed from December 2013 to February 2014. The second author was supported by NSFC through Grant No. 11171370.

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