Counting Additive Z2Z4 Codes Steven T. Dougherty Department of Mathematics, University of Scranton Scranton, PA 18510, USA Email: [email protected] and Eseng¨ ul Salt¨ urk Department of Mathematics, Yildiz Technical University 34210 Istanbul, Turkey Email: [email protected] January 2, 2013 Abstract It is well known, from the work of Delsarte, that the abelian groups in the binary Hamming scheme correspond to additive Z2 Z4 codes. These codes give rise to propelinear binary codes via a Gray map. In this paper, we define free codes in this space and count the number of these codes. We then count the number of arbitrary Z2 Z4 linear codes of type (α, β; γ, δ; κ).

Key words: Additive Z2 Z4 codes, translation-invariant propelinear codes. MSC: 94B05

1

Introduction

Let Z2 and Z4 denote the rings of integers modulo 2 and modulo 4, respectively. The spaces of k-tuples over these rings are denoted by Zk2 and Zk4 . In [5], Delsarte defined additive codes as subgroups of the underlying abelian group in a translation association scheme. In the binary Hamming scheme, that is when the underlying abelian group is of order 2n , the only structures for the abelian group are those of the form Zα2 × Zβ4 , with α + 2β = n ([6]). This gives that the subgroups C of Zα2 × Zβ4 are the only additive codes in a binary Hamming 1

scheme. These codes are called translation-invariant propelinear codes. They are extremely important in that they behave similarly to linear codes without necessarily being binary linear codes. For fundamental results for codes in this space see [1, 2, 3, 4, 9]. We write v ∈ Zα2 × Zβ4 as v = (v1 |v2 ), where v1 = (x1 , . . . , xα ) ∈ Zα2 and v2 = (y1 , . . . , yβ ) ∈ Zβ4 . There is a Gray map defined for Z2 Z4 -additive codes, by taking an extension of the usual Gray map. The usual quaternary Gray map is defined as φ(0) = (0, 0), φ(1) = (0, 1), φ(2) = (1, 1), φ(3) = (1, 0). We define Φ : Zα2 × Zβ4 −→ Zn2 , where n = α + 2β, by Φ(v1 |v2 ) = (v1 |φ(y1 ), . . . , φ(yβ )). For any Z2 Z4 code in Zα2 Zβ4 we shall say that n = α + 2β. Let C be a subgroup of Zα2 × Zβ4 , then this code is also isomorphic to an abelian structure Zγ2 × Zδ4 . We have that the code C has |C| = 2γ+2δ codewords and the number of order two codewords in C is 2γ+δ . Denote by X and Y the set of Z2 and Z4 coordinate positions respectively. It follows that |X| = α and |Y | = β. We let X correspond to the first α coordinates and Y corresponds to the last β coordinates. Define CX to be the punctured code of C formed by deleting the coordinates outside X and define CY to be the punctured code of C formed by deleting the coordinates outside Y . Denote by Cb the subcode of C which contains all order two codewords and let κ be the dimension of (Cb )X . The code Cb is a binary linear code. When α = 0, we will write κ = 0. For a Z2 Z4 code we say that C is of type (α, β; γ, δ; κ), where γ, δ and κ are defined as above. In [4], it is shown that a Z2 Z4 -additive code is permutation equivalent to a Z2 Z4 -additive code with standard generator matrix of the form:   Iκ Tb 2T2 0 0   CGS =  0 0 2T1 2Iγ−κ 0  , (1) 0 Sb Sq R Iδ where Ik is the identity matrix of size k ×k; Tb , Sb are matrices over Z2 ; T1 , T2 , R are matrices over Z4 with all entries in {0, 1} ⊂ Z4 ; and Sq is a matrix over Z4 . We say that a Z2 Z4 code C is decomposable if C is the direct product of the binary code D2 and the quaternary code D4 . A code that is not decomposable is said to be indecomposable. As usual, we define the inner product for any two vectors u, v ∈ Zα2 × Zβ4 as α+β α X X [u, v] = 2( ui vi ) + uj vj ∈ Z4 . i=1

j=α+1

Note that the inner-product is an element of Z4 . We denote by C ⊥ the Z2 Z4 -additive dual code of C, i.e. C ⊥ = {v ∈ Zα2 × Zβ4 | [u, v] = 0 for all u ∈ C}. 2

¯κ If C is a Z2 Z4 code with parameters (α, β; γ, δ; κ) then C ⊥ has parameters (α, β; γ¯ , δ; ¯) where γ¯ = α + γ − 2κ, δ¯ = β − γ − δ + κ, κ ¯ = α − κ. Definition 1. We say that a code C is free if for every vector v in C of order 2 there exists an order 4 vector w in C with v = w + w. Notice, that unlike for codes over chain rings, we do not define free codes in terms of independent vectors. In general, it is important to determine the number of linear codes given a set of parameters. For finite fields, the result is well known. Specifically, the number of codes of dimension k that are subspaces of an n dimensional space over a field of order q is " # (q k − 1)(q k − q) . . . (q k − q r−1 ) n = r . (2) (q − 1)(q r − q) . . . (q r − q r−1 ) k q

These numbers are called Gaussian binomial coefficients. Moreover, we have the following two recursions: " # " # " # n n−1 n − 1 = + qk (3) k k−1 k q

and

"

n k

q

#

" = q n−k

q

n−1 k−1

q

"

# + q

n−1 k

# .

(4)

q

In [8], generalizations of this result were found for codes over chain rings and principal ideal rings. In this work we shall count the number of free codes and the number of arbitrary codes of a given parameter set for Z2 Z4 codes.

2

Free codes

From the standard form given in Equation 1, a free Z2 Z4 additive code has a generator matrix of the following form ! 0 Sb |Sq R Iδ where Sb is a binary matrix, Sq and R are quaternary matrices and Iδ is the identity matrix. The following definition is only for quaternary vectors, not for Z2 Z4 vectors. 3

Definition 2. Let v1 , v2 , . . . , vk be vectors over Z4 . The vectors are said to be linearly P independent if α v = 0 implies αi = 0 for all i. The vectors are said to be modular P i i independent if αi vi = 0 implies αi ∈ {0, 2} for all i. It has been shown in [10] that every linear quaternary code has a minimal generating set that is modular independent. There is not necessarily a minimal generating set that is linearly independent. However, we define that a quaternary code is a free code if and only if it has a linearly independent generating set. Lemma 2.1. Vectors v1 , v2 , . . . , vk generate a free code if and only if (v1 )Y , (v2 )Y , . . . , (vk )Y generate a quaternary free code. Proof. Assume c is a Z2 Z4 order 4 vector. Then c + c 6= 0 and cX + cX = 0 since cX is a binary vector. Hence 2cY 6= 0 and its quaternary part is of order 4. It follows that a Z2 Z4 vector is order 4 if and only if its quaternary part is of order 4. Assume w1 , w2 , . . . , wk are quaternary linearly independent vectors generating a free code. Then, if w is an order two vector in this code, we have that w = α1 w1 + α2 w2 + · · · + αk wk , for αi ∈ Z4 . Since the vector is of order 2 we have: 2w = 2(α1 w1 + α2 w2 + · · · + αk wk ) = 0. This implies that each αi is either 2 or 0 since the vectors are linearly independent. Let αi0 = α2i . Then (α10 w1 + α20 w2 + · · · + αk0 wk ) is an order 4 vector in the code. This proves that a quaternary code generated by linearly independent vectors has no order 2 vectors that are not a multiple of an order 4 vector. A free code generated by s vectors has type (α, β, 0, s, κ). Lemma 2.2. The number of ways of choosing s vectors that generate a free Z2 Z4 code in Zα2 Zβ4 is β

β

α

β

β

α

β

β 2

α

β

β s−1

(4 − 2 )2 (4 − 2 2)2 (4 − 2 2 )2 . . . (4 − 2 2

α

)2 =

s−1 Y

(4β − 2β 2i )2α .

(5)

i=0

Proof. In Lemma 2.3 in [8], (when applied to Z4 ) it is shown that the number of ways of choosing s linearly independent vectors is: (4β − 2β )(4β − 2(2β ))(4β − 22 (2β )) . . . (4β − (2s−1 )2β ).

(6)

However, in this case, each choice of a vector gives an arbitrary choice of an accompanying binary vector. That is, by Lemma 2.1, it does not matter what (vi )X is. Hence the number of ways of choosing s vectors to generate a free code is (4β − 2β )(2α )(4β − 2(2β ))(2α )(4β − 22 (2β ))(2α ) . . . (4β − (2s−1 )2β )(2α ).

4

(7)

Lemma 2.3. The number of ways of finding a minimal generating set of a free Z2 Z4 code generated by s vectors in Zα2 Zβ4 is s

s

s

s

s

s 2

s

s s−1

(4 − 2 )(4 − 2 2)(4 − 2 2 ) . . . (4 − 2 2

)=

s−1 Y

(4s − 2s 2i ).

(8)

i=0

Proof. The counting is the same as in Lemma 2.2 except that there are no free choices for the binary part. Theorem 2.4. The number of free Z2 Z4 codes generated by s vectors in Zα2 Zβ4 is " # β 2s(β+α−s) . s

(9)

2

Proof. The numerator comes from Lemma 2.2 and the denominator from Lemma 2.3. Then we have (4β − 2β )(4β − 2β 2)(4β − 2β 22 ) . . . (4β − 2β 2s−1 )2sα (4s − 2s )(4s − 2s 2)(4s − 2s 22 ) . . . (4s − 2s 2s−1 )   (2β )s 2sα (2β − 1)(2β − 2) . . . (2β − 2s−1 ) = (2s )s (2s − 1)(2s − 2) . . . (2s − 2s−1 ) " # β = 2s(β+α−s) . s 2

Example 1. The number of free Z2 Z4 additive codes with α = 1, β = 2 generated by 1 vector 2 −22 )21 in Zα2 Zβ4 is 12, since (4 (4−2) = 12. These codes are generated by the following matrices: !

! ,

1 |0 1

0 |0 1

,

1 |1 1

!

!

,

0 |1 1

,

1 |2 1

!

!

,

0 |2 1

,

1 |3 1

!

!

,

0 |3 1

,

1 |1 0

!

!

! ,

0 |1 0

,

1 |1 2 !

,

0 |1 2

.

The number of free Z2 Z4 additive codes in the same space generated by 2 vectors is 4, 2 2 )(42 −22 2)22 since (4(4−2 2 −22 )(42 −22 2) = 4. These codes are generated by the following matrices:                 0 |1 0 , 1 |1 0 , 0 |1 0 , 1 |1 0 . 0 |0 1 0 |0 1 1 |0 1 1 |0 1

5

( Define

α, β s

) to be the number of free Z2 Z4 codes generated by s vectors in Zα2 Zβ4 ,

that is

(

α, β s

)

" = 2s(β+α−s)

β s

# . 2

Theorem 2.5. We have the following recurrence relation: ( ) ( ) ( ) α, β α, β − 1 α, β − 1 = 2α+β−s + 22s . s s−1 s Proof. We add the right side of the equation: (

= = = =

)

) α, β − 1 2α+β−s + 22s s " # " # β − 1 β − 1 2α+β−s 2(s−1)(β−1+α−s+1) + 22s 2s(β−1+α−s) s−1 s 2 2 " # " # β−1 β−1 + 2s ) 2s(β+α−s) ( s s−1 2 2 " # β 2s(β+α−s) s 2 ( ) α, β . s α, β − 1 s−1

(

The end of this proof follows from the recursion in Equation 3. We also have the following different recursion. Theorem 2.6. We have the following recurrence relation: ( ) ( ) ( ) α, β − 1 α, β α, β − 1 = 2α+2β−2s + 2s . s s−1 s Proof. We add the right side of the equation:

6

(

= = =

(

)

α, β − 1 s " # " # β−1 β−1 2α+2β−2s 2(s−1)(β−1+α−s+1) + 2s 2s(β−1+α−s) s−1 s 2 2 " # " # β−1 β−1 2s(β+α−s) (2β−s + ) s−1 s 2 2 " # β s(β+α−s) 2 s 2 ( ) α, β . s 2α+2β−2s

=

α, β − 1 s−1

)

+ 2s

The end of this proof follows from the recursion in Equation 4. For α there is also a recurrence relation which is more evident, namely ( ) ( ) α, β α − 1, β = 2s . s s Unlike free quaternary codes, the orthogonal of a free code need not be a free code. For example, consider the free code (0, Zβ4 ). Its orthogonal is (Zα2 , 0) which is not free.

3

Additive Codes

In this section, we shall count the number of arbitrary additive codes of type (α, β; γ, δ; κ). Lemma 3.1. The number of ways of choosing vectors to generate a Z2 Z4 additive code of type (α, β; γ, δ; κ) is δ−1 Y i=0

α

β

β i

2 (4 − 2 2 )

κ−1 Y

γ−κ−1 β

α

j

2 (2 − 2 )

(2β − 2δ+r ).

(10)

r=0

j=0

Proof. Recall the generating matrix given in  Iκ Tb  CGS =  0 0 0 Sb

Y

Equation 1:  2T2 0 0  2T1 2Iγ−κ 0  . Sq R Iδ

(11)

We shall split the count by examining the three major parts of the generating matrix. We begin with the bottom part, namely counting the number of elements that have order 4. 7

This part of the proof is similar to the counting of free codes done earlier. There are 2α 4β vectors in total, so we choose the first one of order 4 in 2α 4β − 2α 2β = 2α (4β − 2β ) different ways. Then we choose the remaining δ − 1 vectors of order 4 in a similar way as before. Hence the count of order 4 vectors is 2α (4β − 2β )2α (4β − 2β 2)2α (4β − 2β 22 ) . . . 2α (4β − 2β 2δ−1 ) =

δ−1 Y

2α (4β − 2β 2i ).

(12)

i=0

Assume that we have chosen δ vectors of order 4. Next, we need to choose κ vectors of order 2 that are not in the space generated by the δ vectors of order 4 and that are not all zero in the binary part. We also want to eliminate the vectors of order 2 in the code which have trivial binary part. We have 2α 2β vectors of order 2 in total. Of these vectors, 2β vectors have trivial binary parts. Hence, the first choice of the vectors of order 2 which have a non-trivial binary part is 2α 2β − 2β = 2β (2α − 1). Then, we continue choosing the remaining κ − 1 vectors in the same way. Then we have that the number of ways of choosing the top part of the generating matrix is: β

α

β

α

β

α

2

β

α

κ−1

2 (2 − 1)2 (2 − 2)2 (2 − 2 ) . . . 2 (2 − 2

)=

κ−1 Y

2β (2α − 2j ).

(13)

j=0

Finally, we need to pick vectors for the middle part of the generating matrix, namely we need to choose the order 2 vectors which have a trivial binary part. There are 2β vectors to choose from without considering those that might already be in the generated code. We have to eliminate the vectors which are in the code generated by the δ vectors of order 4. Hence, for the first choice there are 2β − 2δ vectors. Since there are γ − κ vectors, we need to choose the remaining γ − κ − 1 vectors in a similar way. Hence the number of ways to choose vectors for the middle part of the generating matrix is γ−κ−1 β

δ

β

δ

β

δ γ−κ−1

(2 − 2 )(2 − 2 2) . . . (2 − 2 2

)=

Y

(2β − 2δ+r ).

(14)

r=0

Therefore, the number of ways of choosing vectors to generate a code of type (α, β; γ, δ; κ) is the product of the results in Equation 12, Equation 13, and Equation 14. This gives that there are γ−κ−1 δ−1 κ−1 Y Y Y 2α (4β − 2β 2i ) 2β (2α − 2j ) (2β − 2δ+r ) i=0

r=0

j=0

ways to generate a code of type (α, β; γ, δ; κ). The next lemma will count the number of ways of finding a generator matrix for a given code.

8

Lemma 3.2. The number of distinct generating sets of an additive Z2 Z4 code with type (α, β; γ, δ; κ) is δ−1 Y

γ

δ

δ l

2 (4 − 2 2 )

κ−1 Y

γ−κ−1 δ

γ

Y

γ−κ m

2 (2 − 2

2 )

(15)

n=0

m=0

l=0

2δ (2γ−κ − 2n ).

Proof. Since C is of type 2γ 4δ as a group, it has |C| = 2γ 4δ codewords. We do the enumeration in a manner similar to the counting that was done in Lemma 3.1. First, we choose δ elements of order 4. The first choice is done in 2γ 4δ − 2γ 2δ = 2γ (4δ − 2δ ) ways since 2γ 2δ gives the number of order 2 elements. Continuing in the same way, the number of ways of choosing δ elements of order 4 is γ

δ

δ

γ

δ

δ

γ

δ

δ 2

γ

δ

δ δ−1

2 (4 − 2 )2 (4 − 2 2)2 (4 − 2 2 ) . . . 2 (4 − 2 2

)=

δ−1 Y

2γ (4δ − 2δ 2l ).

(16)

l=0

Our next step is to choose κ vectors of order 2 which have non-trivial binary part. We shall choose elements from the set of elements whose orders are not 4 as we did in Lemma 3.1. There are γ − κ vectors which have trivial binary part. Since the number of order 2 vectors in C is 2γ 2δ , we choose the first vector in 2γ 2δ − 2δ−κ 2δ = 2δ (2γ − 2δ−κ ) different ways. Here the eliminated part 2δ−κ 2δ stands for order 2 vectors which have trivial binary parts. Note that the 2δ−κ vectors may be seen in the standard form of generator matrix and the 2δ vectors are those vectors already generated by the order 4 vectors. Hence the number of ways of choosing κ vectors is δ

γ

γ−κ

2 (2 −2

δ

γ

γ−κ

)2 (2 −2

δ

γ

γ−κ 2

2)2 (2 −2

δ

γ

γ−κ κ−1

2 ) . . . 2 (2 −2

2

)=

κ−1 Y

2δ (2γ − 2γ−κ 2m ). (17)

m=0

Finally, we choose the elements whose orders are 2 and that have trivial binary part. We will choose γ − κ elements. There are 2γ−κ 2δ ways to choose vectors without considering those that might already be in the generated code. We shall remove those elements which are already generated. Hence, the first choice has 2γ−κ 2δ −2δ = 2δ (2γ−κ −1) possible vectors. Continuing, the number of ways of choosing γ − κ elements of order 2 which have trivial binary part, is γ−κ−1 δ

γ−κ

2 (2

δ

γ−κ

− 1)2 (2

δ

γ−κ

− 2)2 (2

2

δ

γ−κ

− 2 ) . . . 2 (2

γ−κ−1

−2

)=

Y

2δ (2γ−κ − 2n ).

(18)

n=0

Then the number of ways of generating the code is the product of the results in Equation 16, Equation 17, and Equation 18. Theorem 3.3. The number of distinct Z2 Z4 additive codes of type (α, β; γ, δ; κ) is " # " # " # β α β − δ 2(α+β−γ−δ)δ+(β−δ−γ+κ)κ . δ κ γ−κ 2

9

2

2

(19)

Proof. Lemma 3.1 gives the numerator and Lemma 3.2 gives the denominator. This gives: Qδ−1 α β Qκ−1 β α Qγ−κ−1 β β i j 2 (4 − 2 2 ) 2 (2 − 2 ) (2 − 2δ+r ) i=0 j=0 r=0 Qδ−1 γ δ Q Q κ−1 δ γ γ−κ−1 δ γ−κ δ l γ−κ 2m ) 2 (2 − 2n ) l=0 2 (4 − 2 2 ) m=0 2 (2 − 2 n=0 Qδ−1 α β Q Q κ−1 β α γ−κ−1 δ β−δ β i j 2 (2 − 2r ) i=0 2 (4 − 2 2 ) j=0 2 (2 − 2 ) r=0 = Qδ−1 Qκ−1 δ+γ−κ κ Q γ δ δ l (2 − 2m ) γ−κ−1 2δ (2γ−κ − 2n ) l=0 2 (4 − 2 2 ) m=0 2 n=0 Qδ−1 α β β Qκ−1 β α Qγ−κ−1 δ β−δ i j 2 (2 − 2r ) i=0 2 2 (2 − 2 ) j=0 2 (2 − 2 ) r=0 = Qδ−1 Qκ−1 δ+γ−κ κ Q γ δ δ l (2 − 2m ) γ−κ−1 2δ (2γ−κ − 2n ) l=0 2 2 (2 − 2 ) m=0 2 n=0 # " Qδ−1 Q Qγ−κ−1 β−δ κ−1 β i α j r (2 − 2 ) (2 − 2 ) (2 − 2 ) 2αδ+βδ+βκ+δγ−δκ i=0 j=0 r=0 = γδ+δ2 +δκ+γκ−κ2 +δγ−δκ Qδ−1 Qκ−1 κ Qγ−κ−1 γ−κ δ l m 2 (2 − 2 ) m=0 (2 − 2 ) n=0 (2 − 2n ) " Ql=0 # " Qκ−1 #"Q # α j δ−1 γ−κ−1 β i β−δ r (2 − 2 ) (2 − 2 ) (2 − 2 ) j=0 r=0 = 2(α+β−γ−δ)δ+(β−δ−γ+κ)κ Qi=0 Qκ−1 κ Qγ−κ−1 δ−1 m δ l (2 − 2 ) (2γ−κ − 2n ) (2 − 2 ) m=0 n=0 # " l=0 # " # " β α β−δ = 2(α+β−γ−δ)δ+(β−δ−γ+κ)κ . δ κ γ−κ 2

2

2

Example 2. The number of Z2 Z4 additive codes of type (2, 2; 2, 0; 1) is 18, since 18. These codes are generated by the following generator matrices:          0 1 |y 0 1 x |0 0 x 1 |0 2 0 1 |0 1 x |y 0          0 0 |y 2 , 0 0 |y 2 , 0 0 |2 0 ,  0 0 |2 0 , 0 0 |2

22 (22 −1)(22 −1) (22 −2)(2−1)

=

   0 1 0 |0 2    0 , 0 0 |2 0

where x ∈ {0, 1} and y ∈ {0, 2}. Thus we obtain the 18 codes. Corollary 3.4. The number of additive Z2 Z4 codes of type (α, β; γ, δ; κ) is equal to the number of additive Z2 Z4 codes of type (α, β; α + γ − 2κ, β − γ − δ + κ; α − κ). Proof. The map that sends a code C to its orthogonal C ⊥ is a bijection. Given a code of type (α, β; γ, δ; κ) its orthogonal has type (α, β; α + γ − 2κ, β − γ − δ + κ; α − κ). Hence the numbers of codes of these two types are equal. Example 3. Continuing Example 2, where we showed that number of Z2 Z4 additive codes of type (2, 2; 2, 0; 1) was 18, we consider codes of type (2, 2; 2, 1; 1) since γ¯ = 2 + 2 − 2 = 2, δ¯ = 2 − 2 − 0 + 1 = 1, κ ¯ = 2 − 1 = 1. Namely, these are the parameters of its dual. Then we have that the number of Z2 Z4 additive codes of type (2, 2; 2, 1; 1) is the same as the number of Z2 Z4 additive codes of type (2, 2; 2, 0; 1), namely: 22 (42 − 22 )22 (22 − 1)(22 − 21 ) = 18. 22 (41 − 21 )21 (22 − 21 )21 (21 − 1) 10

Example 4. The number of Z2 Z4 additive codes of type (2, 4; 2, 3; 1) is: 22 (44 − 24 )22 (44 − 25 )22 (44 − 26 )24 (22 − 1)(24 − 23 ) = 360. 22 (43 − 23 )22 (43 − 24 )22 (43 − 25 )23 (22 − 21 )23 (21 − 1) It is equal to the number of Z2 Z4 additive codes of type (2, 4; 2, 0; 1), that is their duals: 24 (22 − 1)(24 − 1) = 360. (22 − 2)(2 − 1) ( Define

α, β γ, δ, κ (

) to be the number of Z2 Z4 additive codes of type (α, β; γ, δ; κ), that is

α, β γ, δ, κ

)

" (α+β−γ−δ)δ+(β−δ−γ+κ)κ

=2

β δ

# " 2

α κ

# " 2

β−δ γ−κ

# . 2

Theorem 3.5. We have the following recurrence relation: (

α, β γ, δ, κ

)

( = 2β−γ−δ+κ

α − 1, β γ − 1, δ, κ − 1

)

( + 2δ+κ

α − 1, β γ, δ, κ

) .

Proof. We add the right side of the equation: (

) α − 1, β 2 +2 γ, δ, κ " # " # " # β α−1 β−δ β−γ−δ+κ (α−1+β−γ+1−δ)δ+(β−δ−γ+1+κ−1)(κ−1) 2 2 + δ κ−1 γ−κ 2 2 2 " # " # " # β α−1 β−δ δ+κ (α−1+β−γ−δ)δ+(β−δ−γ+κ)κ 2 2 δ κ γ−κ 2 2 " # " # " 2 # " # β β − δ α − 1 α − 1 ( + 2κ ) 2(α+β−γ−δ)δ+(β−δ−γ+κ)κ δ γ−κ κ−1 κ 2 2 " #2 " #2 " # β β−δ α 2(α+β−γ−δ)δ+(β−δ−γ+κ)κ δ γ−κ κ 2 2 2 ( ) α, β . γ, δ, κ β−γ−δ+κ

=

= = =

α − 1, β γ − 1, δ, κ − 1

)

(

δ+κ

The end of this proof follows from the recursion in Equation 3.

We also have the following different recursion. 11

Theorem 3.6. We have the following recurrence relation: (

α, β γ, δ, κ

)

( = 2α+β−γ−δ

α − 1, β γ − 1, δ, κ − 1

)

( + 2δ

α − 1, β γ, δ, κ

) .

Proof. We add the right side of the equation: ( 2α+β−γ−δ =

= = =

α − 1, β γ − 1, δ, κ − 1

(

) + 2δ

α − 1, β γ, δ, κ "

)

# " # " # β α − 1 β − δ 2α+β−γ−δ 2(α−1+β−γ+1−δ)δ+(β−δ−γ+1+κ−1)(κ−1) + δ κ−1 γ−κ 2 2 " # " # " 2 # β α−1 β−δ 2δ 2(α−1+β−γ−δ)δ+(β−δ−γ+κ)κ δ κ γ−κ 2 2 " # " 2 " # # " # β α − 1 β − δ α − 1 2(α+β−γ−δ)δ+(β−δ−γ+κ)κ (2α−κ + ) δ κ γ−κ κ−1 2 2 " #2 " #2 " # β β−δ α 2(α+β−γ−δ)δ+(β−δ−γ+κ)κ δ γ−κ κ 2 2 2 ( ) α, β . γ, δ, κ

The end of this proof follows from the recursion in Equation 4.

Lemma 3.7. A decomposable Z2 Z4 code of type (α, β; γ, δ; κ) is the direct product of a binary code of dimension κ in an α dimensional space and a quaternary code in Zβ4 of quaternary type (δ, γ − κ). Proof. In this case, the generator matrix  Iκ  CGS =  0 0

has the following form:  Tb 0 0 0  0 2T1 2Iγ−κ 0  . 0 Sq R Iδ

Then the result follows. Lemma 3.8. The number of decomposable codes of type (α, β; γ, δ; κ) is " # Pa−1 Q Q a −1 β 2βk0 1a=0 ki=0 (2 − 2 b=0 kb 2i ) α ( )( k2 +2k k Q1 ), κ 2 0 0 1 i=0 (2ki − 1)(2ki − 2) . . . (2ki − 2ki −1 ) 2

where k0 = δ, k1 = γ − κ. 12

(20)

Proof. The number of binary codes of dimension κ in an α dimensional space is well known. The number of quaternary codes in Zβ4 of quaternary type (δ, γ − κ) is given in [8]. Theorem 3.9. The number of indecomposable codes of type (α, β; γ, δ; κ) is " # " # " # β α β − δ 2(α+β−γ−δ)δ+(β−δ−γ+κ)κ δ κ γ−κ 2 2 2 " # Pa−1 Q1 Qka −1 β βk0 k 2 (2 − 2 b=0 b 2i ) α − ( )( k2 +2k k Q1 a=0 i=0 ), ki − 1)(2ki − 2) . . . (2ki − 2ki −1 ) 0 1 0 κ 2 (2 i=0 2 where k0 = δ, k1 = γ − κ. Proof. The number of total codes is given in Theorem 3.3 and the number of decomposable codes is given in Lemma 3.8. Example 5. The number of Z2 Z4 additive codes of type (2, 2; 2, 0; 1) was 18 in Example 2. The number of decomposable" codes # of "type#(2, 2; 2, 0; 1) is 9. Because the number of binary α 2 codes of dimension κ = 1 is = = 3, and the number of quaternary codes in Z24 κ 1 2 2 of quaternary type (δ, γ − κ) = (0, 1) is 3 from the formula given in [8]. Then the product, 3 × 3 = 9, gives the number of decomposable codes. They are generated by the following matrices:        0 1 |0 0 1 x |0 0 0 1 |0 0 1 x |0 0         0 0 |y 2 , 0 0 |y 2 , 0 0 |2 0 , 0 0 |2 0 

where x ∈ {0, 1} and y ∈ {0, 2}. The remaining 9 codes are indecomposable ones. Codes which are the Gray image of Z2 Z4 codes are similar to binary codes in that they have a group structure. They may or may not be linear. In general, this adds a large number of codes which are of interest"in Z#n2 . For example, the number of ordinary binary codes of n length n and dimension k is . To find the number of Gray images of Z2 Z4 additive k 2 codes which have images in Zn2 we count the codes of types (α, β; γ, δ; κ) where n = α + 2β, k = γ + 2δ, κ ≤ min{α, γ} and 0 < γ + δ ≤ β + κ. Notice that there are numerous cases for these parameters. It may be true however that the Gray images of codes with different parameters may give the same code. For example, any purely quaternary code with β = n2 that has a linear Gray image is the same as a purely binary code with α = n. Hence it is not an easy task in general to count the number of codes which are the Gray images of Z2 Z4 codes.

13

5 We can consider an example. Consider n = 6, for"codes # with 2 elements. The number 6 of binary linear codes of length 6 and dimension 5 is = 63. 5 2 In order to find the number of Gray images of additive codes we count the number of Z2 Z4 additive codes of type (α, β; γ, δ; κ) where n = α + 2β = 6 and k = γ + 2δ = 5. There are 6 sets of parameters that give codes in this case. Summing the numbers of additive codes in this case gives 1876 codes in total. Of course, some of their Gray images may in fact be equal. However, we note that this number is quite large in comparison with 63.

References [1] Bilal, M., Borges, J., Dougherty, S. T., and Fernndez-Crdoba, C.: Maximum distance separable codes over Z4 and Z2 × Z4 . Des. Codes Cryptogr. Vol. 61, No. 1, 31 - 40, (2011). [2] Borges, J., Dougherty, S.T. and Fern´andez-C´ordoba, C.: Self-Dual codes over Z2 × Z4 . To appear in Adv. Math. Commun. [3] Borges, J., Fern´andez, C., Pujol, J., Rif`a, J., and Villanueva, M.: On Z2 Z4 -linear codes and duality. V Jornades de Matem`atica Discreta i Algor´ısmica, Soria (Spain), Jul. 11-14, 171-177, (2006). [4] Borges, J., Fern´andez-C´ordoba, C., Pujol, J., Rif`a, J., and Villanueva, M.: Z2 Z4 -linear codes: generator matrices and duality. Des. Codes Cryptogr. Vol. 54 , No. 2, 167-179, (2010). [5] Delsarte, P.: An algebraic approach to the association schemes of coding theory. Philips Res. Rep. Suppl., Vol. 10, (1973). [6] P. Delsarte, P. and Levenshtein, V.: Association schemes and coding theory. IEEE Trans. Inform. Theory. Vol. 44, No. 6, 2477-2504, (1998). [7] Dougherty, S.T. and Liu, H.: Independence of vectors in codes over rings. Designs, Codes, Cryptogr. Vol. 51, No. 1, 55-68 (2009). [8] Dougherty, S.T. and Salt¨ urk, E.: Counting codes over rings. In submission. [9] Fernndez-Crdoba, C., Pujol, J., Villanueva, M.: Z2 Z4 -linear codes: rank and kernel. Des. Codes Cryptogr. Vol. 56, No. 1, 43 - 59, (2010). [10] Park, Y. H.: Modular independence and generator matrices for codes over Zm , Des. Codes Cryptogr. Vol. 50, No. 2, 147-162, (2009). 14