decompositions and extremal type ii codes over z4

800 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 44, NO. 2, MARCH 1998

Decompositions and Extremal Type II Codes over

W. Cary Huffman

Abstract—In previous work by Huffman and by Yorgov, a decomposi-tion theory of self-dual linear codesC over a finite field Fq is given whenC has a permutation automorphism of prime order r relatively prime toq: We extend these results to linear codes over the Galois ring4 andapply the theory to 4-codes of length24. In particular we obtain 42inequivalent [24; 12] 4-codes of minimum Euclidean weight16 whichlead to 42 constructions of the Leech lattice.

Index Terms— 4-codes, extremal codes, self-dual codes.

I. BACKGROUND MATERIAL

In this section we present some preliminary results needed for ourdecompositions. In particular, we describe the structure of Galoisrings and related rings including4[X]=(Xr

� 1): General resultson Galois rings are found in [2], [8], [14], and [15]. In Section II,we present the code decomposition. In Section III, we present theself-dual 4-codes of length24 and minimum Euclidean weight16with an automorphism of prime order5 or more, including partialinformation on their inequivalence. In Section IV, we present theLee and Euclidean distributions of these codes and complete thediscussion of their equivalence or inequivalence.

There are similarities and analogies between finite fields and finiteGalois rings. There are also important differences. We denote byFq = GF(q) the finite field withq elements and by n the ring ofintegers modulon: The finite field GF(pm) of characteristicp is the(unique) Galois extension of degreem of the finite field p wherepis a prime. The Galois ring GR(qm) of characteristicq is the (unique)Galois extension of degreem of the Galois ring q whereq is a primepower. In this correspondence, we concentrate on the caseq = 4:

One method of constructing the Galois ring GR(4m) is as follows.Let R[X] be the ring of polynomials over a ringR whereX is anindeterminate. Define�: 4[X] ! F2[X] by

�(f(X)) = f(X) (mod 2)

that is,� is determined by�(0) = �(2) = 0; �(1) = �(3) = 1; and�(X) = X: A polynomial f(X) 2 4[X] is a basic irreducibleif�(f(X)) is irreducible inF2[X]; it is monicif its leading coefficientis 1. If f(X) is any monic basic irreducible polynomial of degreem, then 4[X]=(f(X)) ' GR(4

m) where(f(X)) is the principal

ideal of 4[X] generated byf(X):

The nonzero elements of a finite field form a cyclic group. Thisis not the case with Galois rings, but an equally useful structure ispresent. The Galois ringR = GR(4

m) contains an element� of

order2m � 1, called aprimitive element. Every elementc 2 R canbe uniquely expressed in the formc = a + 2b wherea and b areelements of

T (R) = f0; 1; �; �2; � � � ; �2 �2g:

This is the 2-adic representationof c: R has only three ideals:(0); (1); and (2). Of course,(0) = f0g and (1) = R: The ideal

Manuscript received April 5, 1997; revised July 15, 1997. The material inthis correspondence was presented at the 35th Annual Allerton Conferenceon Communication, Control, and Computing, Allerton House, University ofIllinois, September 29–October 1, 1997.

The author is with the Department of Mathematical Sciences, LoyolaUniversity, Chicageo, IL 60611 USA.

Publisher Item Identifier S 0018-9448(98)00842-6.

(2) = 2R = f2tjt 2 T (R)g consists of0 together with the divisorsof 0 in R: The elements ofR n 2R are precisely the invertibleelements ofR: The quotient ringR=2R is isomorphic to the fieldGF(2m) with the coset�+2R corresponding to a primitive elementof GF(2m):

The Galois group of GR(4m) over 4 is cyclic of order mgenerated by theFrobenius map�2 defined by

�2(c) = a2

+ 2b2

wherea + 2b is the 2-adic representation ofc:Our decomposition results for codes will be based on the decom-

position of the ring<r = 4[X]=(Xr� 1) where r is odd. The

polynomialXr� 1 can be factored in 4[X] (uniquely up to order

and multiplication by units) as a product

Xr� 1 = m0(X)m1(X) � � �mg(X)

wheremi(X) are monic basic irreducible polynomials for0 � i � g

with m0(X) = X � 1: Furthermore,

�(m0(X))�(m1(X)) � � ��(mg(X))

is the factorization ofXr� 1 into irreducible polynomials overF2:

Let mj(X) = (Xr�1)=mj(X), and letdj denote the degree of both

mj(X) and�(mj(X)): In what follows, (f) will denote the idealgenerated byf in 4[X], while hfi will denote the ideal generatedby f in <r: In particular, letJ denote the ideal

hX � 1i = hm1(X)m2(X) � � �mg(X)i

of <r:

Lemma 1.1: For 0 � j � g

(mj) + (mj) = 4[X] and hmji � hmji = <r:

Proof: The ring 4[X]=(mj(X)) is a Galois ring with onlythree ideals, and therefore if(mj) + (mj) 6= 4[X], then

(mj) + (mj) � (mj) + 2 4[X]:

In particular,

mj(X) = �(X)mj(X) + 2�(X)

for some�(X); �(X) 2 4[X]: But then

�(mj(X)) = �(�(X))�(mj(X))

a contradiction as�(mj(X)) and�(mj(X)) are relatively prime inF2[X] sinceXr

� 1 has distinct roots in an extension field ofF2:This verifies the first equality andhmji+hmji = <r: That the lattersum is direct follows from the fact thathmji has order4r�d andhmji has order4d , while <r has order4r:

Let Ij = hmj(X)i for 0 � j � g: Notice that

I0 = fa(1 +X + � � �+Xr�1

)ja 2 4g ' 4

under the mapa(1+X+� � �+Xr�1)! ar: For any positive integeru relatively prime tor, define the map�u: <r ! <r by

�u

r�1

i=0

aiXi

=

r�1

i=0

aiXui: (1)

Since(Xr� 1)j(Xru

� 1) andgcd (r; u) = 1; �u is clearly a well-defined ring automorphism. Furthermore,�u = �v if u � v (mod r)

as we can simply read the exponents in the right-hand side of (1)modulo r: Using this observation, the definition of�u extends tonegative values ofu relatively prime tor: Parts of the followinglemma are analogous to [10, Lemma 1] and [11, Lemma 1].

0018–9448/98$10.00 1998 IEEE

IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 44, NO. 2, MARCH 1998 801

Lemma 1.2: The following hold:

i) <r = I0 � I1 � � � � � Ig:

ii) J = hX � 1i = I1 � I2 � � � � � Ig:

iii) For 0 � j � g

Ij = hmj(X)i ' 4[X]=(mj(X)) ' GR(4d):

iv) �u is a ring automorphism of<r:

v) �u is the identity onI0:vi) �u permutesI1; � � � ; Ig and if �u(Ii) = Ij ; then�u is a ring

isomorphism fromIi onto Ij :

Proof:i) First note that

i6=j

Ii \ Ij � hmj(X)i \ hmj(X)i = f0g

by Lemma 1.1. Therefore, the sumI0 + I1 + � � � + Ig is directand hence has�g

j=0 4d = 4r elements, the same as<r, proving

i). Part ii) follows sinceIi � J for 1 � i � g while bothJ andI1 � I2 � � � � � Ig have order4r�1:

By Lemma 1.1, in<r;

1 = ej(X) + ej(X) (modXr � 1)

where ej(X) and ej(X) are pairwise-orthogonal idempotents inhmj(X)i and Ij , respectively. Since

mj(X) = mj(X)ej(X) + mj(X)ej(X) = mj(X)ej(X)

ej(X) is the identity ofIj : The map�: 4[X]=(mj(X)) ! Ijgiven by

�(f(X) + hmj(X)i) = f(X)ej(X) (modXr � 1)

is the desired isomorphism in iii).Part iv) follows from the discussion prior to the lemma and v) is

obvious as

�u(1 +X + � � �+Xr�1

) = 1 +X + � � �+Xr�1

:

For part vi), �u(Ii) is an ideal of<r: Let

1 =

g

j=0

ej(X) (modXr � 1); where ej(X) 2 Ij :

So

�u(Ii) = ej(X)�u(Ii)

where the latter sum is over valuesj with ej(X)�u(Ii) 6= f0g: Aseachej(X)�u(Ii) is an ideal inIj and�u(Ii) is a Galois ring, therecan be only one suchj because Galois rings are not the nontrivialdirect sum of two or more ideals. Hence�u(Ii) is an ideal ofIj ;and since�u(Ii) has invertible elements,�u(Ii) = Ij :

The polynomiala0 + a1X + � � � + ar�1Xr�1 in <r is denoted

a0a1 � � � ar�1:

Example 1.3: We illustrate the decomposition of<r for r = 7:

First

X7 � 1 = (3 +X)(3 + 2X + 3X

2+X

3)(3 +X + 2X

2+X

3)

is the factorization ofX7 � 1 into monic basic irreducibles.I0 = h1 + X + � � � + X6i has identity

e0 = 3 + 3X + � � �+ 3X6:

I0 ' 4: The idealsI1 andI2 are both isomorphic to GR(43): Iihas a seventh root of unity�i such that

Ii = fa+ 2bja; b 2 f0; �0i ; � � � ; �6i gg:

Table I gives enough information to construct the ideals.

TABLE IPRIMITIVE ELEMENTS FOR r = 7

The map�6 = ��1 interchangesI1 andI2 interchanging�1 and�2: The maps�1; �2; and�4 = � 22 make up the Galois group of bothI1 and I2: Notice that�j1 is obtained by cyclically shifting�01 j

places to the right;�j2 is obtained by cyclically shifting�02 j placesto the left. Also

X = 0100000 = 3333333+ 3122323+ 2133232 = e0 + �1 + �62 :

So multiplying an element inI1 by �1 and an element inI2 by �62cyclically shifts the element one to the right.

II. THE CODE DECOMPOSITION

In this section, we will decompose self-dual4-codes. This decom-position will involve codes over Galois rings GR(4m): We beginwith the necessary terminology.

Let R = GR(4m): A linear R-code of length n is an R-submodule ofRn: In particular, alinear 4-code of length n isa 4-submodule of n

4 ; that is, an additive subgroup ofn4 : Let � bea permutation of the coordinates ofRn: If x 2 Rn hasith coordinatexi, definex� 2 Rn by (x�)i = xi� : Two linearR-codesC1 andC2 arepermutation equivalentprovided there exists a permutation�such thatC1� = C2: Recall thatR = fa + 2bja; b 2 T (R)g whereT (R) = f0; 1; �; � � � ; �2 �2g with � an element of order2m�1: AlinearR-code of lengthn is permutation equivalent to a code whosegenerator matrixG has the form

G =Ik A B

0 2Ik 2C(2)

whereIk is the k � k identity matrix,A andC have entries onlyfrom T (R), andB has entries fromR: Every codeword is uniquelydetermined by(v1; v2)G where v1 2 Rk and v2 2 Rk : Thusthe code contains4mk 2mk codewords, and we say it hastype4mk 2mk :

Let S be a ring. Define theordinary inner producthu; viS ofvectorsu = (u1; � � � ; un) andv = (v1; � � � ; vn) in Sn by

hu; viS =

n

i=1

uivi:

We denoteh�; �iS by h�; �i4 whenS = 4: In particular, for vectorsu and v in n

4 ,

hu; vi4 =

n

i=1

uivi (mod4):

The dual C? of the linearR-codeC is defined by

C? = fu 2 Rnjhu; viR = 0 for all v 2 Cg:

A straightforward “linear algebra” type argument using matrixinvariants shows that the dual of a linearR-code is again a linearR-code. Furthermore, ifC has generator matrix (2), thenC?, underh�; �iR, has generator matrix

G?=

�BT � CTAT CT In�k �k2AT 2Ik 0

: (3)

ThusC? is of type4m(n�k �k )2mk : As usual,C is self-orthogonalif C � C? and self-dual if C = C?:


Let Mn be the group of all invertiblen � n monomial matriceswith entries in 4: So every matrixM 2 Mn can be written as theproductM = PD, whereP is ann � n permutation matrix, calledthe permutation partof M; and D = diag (a1; a2; � � � ; an) is ann�n diagonal matrix, called thediagonal partof M , with diagonalentriesai 2 f�1g: We say that two linear 4-codesC1 and C2 oflengthn areequivalentprovided there exists anM 2Mn such thatC1 = C2M:

Aut(C) = fM 2MnjCM = Cg

is the automorphism groupof the 4-codeC:For v 2 n

4 ; the Lee weightof v is the integern�1+2n2, and theEuclidean weightwtE(v) of v is the integern�1 +4n2, wheren�1is the total number of components ofv which equal�1 andn2 is thenumber of components ofv which equal2. A self-dual 4-codeC isof Type II provided8jwtE(c) for all c 2 C: It is easy to show that ifC is self-dual with a basis of vectors whose Euclidean weights are alldivisible by8, thenC is Type II. Type II codes exist only for lengthsa multiple of8. There are four inequivalent Type II codes of length8

all with minimum Euclidean weight8 [4]. There are 133 inequivalentType II codes of length16 all with minimum Euclidean weight8 [5].A complete enumeration of such codes of length24 seems infeasible.

For the remainder of the section letC be a linear 4-code of lengthn which has an automorphism� of odd prime orderr: By applying asimilar argument to that of [13, Lemma 1], we see, by replacingC byan equivalent code, that without loss of generality� is a permutationwith c r-cycles andf = n � cr fixed points where

� = (1; 2; � � � ; r) � � � ((c� 1)r+ 1; (c� 1)r+ 2; � � � ; cr): (4)

(Later we will assume thatC is self-dual; note that any codeequivalent to a self-dual code is also self-dual.) Denote the orbitsof � by

i = f(i� 1)r+ 1; (i� 1)r+ 2; � � � ; irg

for 1 � i � c and

c+j = fcr + jg

for 1 � j � f . Let vj denote the restriction ofv 2 n4 to i: For

1 � i � c; vj can be viewed as an element

a0 + a1X + � � �+ ar�1Xr�1

2 <r

where

v�j = (a0 + a1X + � � �+ ar�1Xr�1

)X 2 <r:

Define

C(�) = fv 2 C j v� = vg

and

E(�) = fv 2 C j vj 2 J for 1 � j � c

andvj = 0 for c+ 1 � j � c+ fg:

For 1 � i � g, let

Ei(�) = fv 2 C j vj 2 Ii for 1 � j � c

andvj = 0 for c+ 1 � j � c+ fg:

The following theorem gives our code decomposition.

Theorem 2.1: Let C be a linear 4-code with automorphism� ofodd prime orderr as in (4). Then

C = C(�)� E(�)

and

E(�) = E1(�)� � � � � Eg(�):

Proof: Let v 2 C and

w =

r�1

i=0

v�i:

Clearly,w 2 C(�): Let x = v � (1=r)w 2 C, which we claim is inE(�), noting thatr is invertible in 4: If c + 1 � i � c + f , thenwj = rvj implying thatxj = 0: If 1 � i � c and

vj =

r�1

j=0

vi;jXj

thenwj = ai(1 + X + � � � + Xr�1) where

ai =

r�1

j=0

vi;j :

So

xj =

r�1

j=0

(vi;j � (1=r)ai)Xj

which when divided byX � 1 has remainderr�1

j=0

(vi;j � (1=r)ai) = 0:

Hencexj 2 J yieldingx 2 E(�): SoC = C(�)+E(�): Notice thatif x 2 C(�), then for1 � i � c; xj 2 I0: With this observationandI0\J = f0g by Lemma 1.2 i) and ii), clearlyC(�)\E(�) = f0g

and soC = C(�) � E(�):

Let ej(X) be the identity ofIj and

e(X) =

g

j=1

ej(X)

which is the identity ofJ : For

x = (xj ; � � � ; xj ; 0; � � � ; 0) 2 E(�)

and for 1 � j � g, let

x(j)

= (xj ej(X); � � � ; xj ej(X);0; � � � ; 0):

Note thatx� 2 E(�) and

x�j = xj X 2 J ; for 1 � i � c

asJ is an ideal of<r: Therefore, eachx(j) is in Ej(�): So as

x =

g

j=1

x(j)

we have

E(�) = E1(�) + � � �+ Eg(�):

The sum is direct asIi \j 6=i

Ij = f0g:

This theorem shows thatC can be decomposed as

C(�)� E1(�)� � � � � Eg(�):

It is easy to see that the converse also holds; namely, ifC can bedecomposed in the obvious way, then� is an automorphism ofC:Each Ei(�) can be viewed as a linear code over the Galois ringIi: Therefore, letEi(�)� denote the code formed fromEi(�) bypuncturing on the fixed pointsc+1 [ � � � [ c+f and consideringthe codewords asc-tuples with components inIi: Let

E(�)�= E1(�)

�� Et(�)

�� J

c

and fora 2 E(�), denote bya� the associatedc-tuple inE(�)� withcomponents fromJ :


For the remainder of the section assumeC is also self-dual. A greatdeal can be said in this case about the codesC(�) andEi(�): Beforewe give the main result, we need three lemmas.

Lemma 2.2: If a; b 2 E(�) � n4 , then

ha�; ��1(b

�)iJ =

r�1

h=0

ha�h; bi4X

�h:

Proof: Let

a�=

r�1

j=0

a1;jXj; � � � ;

r�1

j=0

ac;jXj

and defineb� analogously. Then

ha�h; bi4 =

c

i=1

r�1

j=0

ai;j�hbi;j ; for 0 � h � r � 1 (5)

where the second subscript ofai;j�h is read modulor: Then

ha�; ��1(b

�)iJ =

c

i=1

r�1

k=0

ai;kXk

r�1

j=0

bi;jX�j

=

c

i=1

r�1

j=0

r�1

k=0

ai;kbi;jXk�j

=

c

i=1

r�1

j=0

j

h=j�r+1

ai;j�hbi;jX�h

=

c

i=1

r�1

j=0

r�1

h=0

ai;j�hbi;jX�h

=

r�1

h=0

c

i=1

r�1

j=0

ai;j�hbi;j X�h

:

The result follows from (5).

By Lemma 1.2 (vi),��1 defines a permutation� of 1; 2; � � � ; ggiven by

��1(Ii) = I�(i): (6)

Lemma 2.3: With � defined as in (6),E(�) � E(�)? underh�; �i4if and only if

E�(i)(�)�� (��1(Ei(�)

�))?

under h�; �iJ for all i with 1 � i � g:

Proof: First suppose

E(�) � E(�)?:

Choosea 2 E�(i)(�) and b 2 Ei(�) with associated vectorsa� 2 E�(i)(�)

� and b� 2 Ei(�)�: Then ha�h; bi4 = 0 for all h

with 0 � h � r�1: By Lemma 2.2,ha�; ��1(b�)iJ = 0, and henceE�(i)(�)

�� (��1(Ei(�)

�))?:

Conversely, suppose

E�(i)(�)�� (��1(Ei(�)

�))?

for all i with 1 � i � g: If a� 2 E�(i)(�)� and b� 2 Ei(�)

�, thenha�; ��1(b

�)iJ = 0 by assumption. Now considera� 2 E�(i)(�)�

andb� 2 Ej(�)� with j 6= i: Then��1(b

�) has entries inI�(j): Soha�; ��1(b

�)iJ is a sum of productsxy with x 2 I�(i); y 2 I�(j);hence

xy 2 I�(i)I�(j) � I�(i) \ I�(j) = f0g

implying ha�; ��1(b�)iJ = 0: Thus, ha�; ��1(b�)iJ = 0 for all

a�; b� 2 E(�)� which implies ha; bi4 = 0 for all a; b 2 E(�) byLemma 2.2. ThusE(�) � E(�)?:

Define�: C(�) ! c+f4 by

� a1

r�1

i=0

Xi; � � � ; ac

r�1

i=0

Xi; ac+1; � � � ; ac+f

= (ra1; � � � ; rac; ac+1; � � � ; ac+f ):

Lemma 2.4: If C is self-dual,j�(C(�))j = jC(�)j � 2c+f :

Proof: Clearly, j�(C(�))j = jC(�)j: Let a; b 2 C(�): Then

0 = ha; bi4 = r

c

i=1

aibi +

c+f

i=c+1

aibi

= r

c

i=1

�(a)i�(b)i +

c+f

i=c+1

aibi

sincer2 � 1 (mod 4) asr is odd. So vectors in�(C(�)) satisfy thesystem of equations

r

c

i=1

�(a)ixi +

c+f

i=c+1

aixi = 0; for all a 2 C(�):

By linear algebra type arguments, there are4c+f=jC(�)j such solu-tions. Therefore,

jC(�)j �4c+f

jC(�)j

yielding jC(�)j � 2c+f :

Theorem 2.5: AssumeC is a self-dual 4-code with automorphism� as in (4). Let� be the permutation defined by (6). ThenC(�) is self-orthogonal with2c+f codewords andE�(i)(�)

� = (��1(Ei(�)�))?

underh�; �iJ for all i with 1 � i � g:

Proof: By Lemma 2.4

jC(�)j � 2c+f (7)

and by Lemma 2.3E�(i)(�)�� (��1(Ei(�)

�))?: But

j��1(Ei(�)�)?j = j(Ei(�)

�)?j =

jIijc

jEi(�)�j:

Therefore,

jE�(i)(�)�j �

jIijc

jEi(�)�j; for all 1 � i � g: (8)

The theorem is proved if we show equality in (7) and in (8) for alli with 1 � i � g: Now

2cr+f

= jCj = jC(�)j

g

i=1

jEi(�)�j: (9)

But by (8)g

i=1

jEi(�)�j =

g

i=1

jE�(i)(�)�j �

g

i=1

jIijc

jEi(�)�j

which implies, together withg

i=1

jIij = 4r�1

thatg

i=1

jEi(�)�j �

g

i=1

jIij

c=2

= (4r�1

)c=2 (10)


with strict inequality if and only if there is a strict inequality in (8)for somei: Combining (7), (9), and (10) gives

2cr+f

� 2c+f

(4r�1

)c=2

with strict inequality if and only if there is a strict inequality in either(7) or (8) for somei: But

2c+f

(4r�1

)c=2

= 2cr+f

and strict inequality can never occur.

The converse of this theorem also holds.

Theorem 2.6: Let C be a linear 4-code with automorphism� asin (4). AssumeC(�) is self-orthogonal with2c+f codewords and

E�(i)(�)�= (��1(Ei(�)

�))?

underh�; �iJ for all i with 1 � i � g: ThenC is self-dual.Proof: The argument in the proof of Theorem 2.5 can be

reversed to show thatjCj = 2cr+f : By Lemma 2.3E(�) � E(�)?,and by assumptionC(�) � C(�)?: We only need to show thatC(�) � E(�)?: Let a 2 C(�) and b 2 E(�): Note thatha; bi4 =

ha0; b0i4 wherea0; b0 area; b with fixed points deleted sinceb is zeroon the fixed points. By the proof of Lemma 2.2

ha�; ��1(b

�)i< =

r�1

h=0

ha0�h; b0i4X

�h=

r�1

h=0

ha�h; bi4X

�h

where

a�= a1

r�1

i=0

Xi; � � � ; ac

r�1

i=0

Xi

2 Ic0

andb� 2 J c: As ha�; ��1(b�)i< is a sum of productsxy with

x 2 I0; y 2 J ; xy 2 I0 \ J = f0g

implying ha�; ��1(b�)i< = 0: Thus ha�h; bi4 = 0 implying

ha; bi4 = 0:

If two self-dual codes have automorphism�, it is sometimes rathereasy to decide their equivalence or inequivalence. We describe whenthis occurs.

Let � be as in (4); recall thatr is an odd prime. Ifi 2 , defineir 2 by ir � i (mod r) where0 � ir <r: For 1 � u<r, definefu to be the permutation(sr + 1 + i)fu = sr + 1 + (ui)r for0 � i< r and 0 � s< c, andxfu = x for cr + 1 � x � cr + f:

Let F = ffuj1 � u<rg, which is isomorphic to the cyclicmultiplicative group of the field r: The effect of applyingfu to C isto replaceX byXu in eachr-cycle of�. Notice thatfu is merely thepermutation associated to�u defined in Section I extended to eachr-cycle of�: For 0 � s< c define�s = (sr+1; sr+2; � � � ; (s+1)r)

and

W = f��0 � � ��

�

c�1 j0 � �s<r for 0 � s< cg:

Application of an element ofW to C cycles the entries of ther-cyclesseparately. Let�a denote the symmetric group onf1; � � � ; ag: If � 2�c, define�� to be the permutation((s�1)r+ i)�� = (s��1)r+ i

for 1 � s � c and1 � i � r andx�� = x for cr+1 � x � cr+ f:

Let ��c = f��j� 2 �cg: If � 2 �f , define�0 to be the permutationx�0 = x for 1 � x � cr and(cr+i)�0 = cr+i� for 1 � i � f: Let�0f = f�0j� 2 �fg: Application of an element of��c to C permutesthe c r-cycles, while applying an element of�0f permutes the fixedpoints. Let

D = fdiag (a1; � � � ; an)jasr+1 = asr+2 = � � � = a(s+1)r

for 0 � s< cg;

that is,D consists of the diagonal matrices that are constant on eachr-cycle. Finally, let

N = fN 2 MnjNh�iN�1

= h�ig

be the normalizer ofh�i in Mn: The following results can be provedin the same manner as [11, Theorems 2 and 3].

Theorem 2.7Let C and C0 be 4-linear codes of lengthn bothhaving automorphism� as in (4). Assume thath�i is a Sylow r-subgroup ofAut (C): Then,C and C0 are equivalent if and only ifC0 = CM for someM 2 N : FurthermoreN =W �0f �

�c DF :

Theorem 2.8: Let C and C0 be 4-linear codes of lengthn bothhaving automorphism� as in (4). Let

C = C(�)� E1(�)� � � � � Eg(�)

and

C0= C

0(�)� E

0

1(�)� � � � � E0

g(�)

be the code decompositions of Theorem 2.1. LetM 2 W �0f ��c DF

whereC0 = CM: Then,C0(�) = C(�)M andE 0�(i)(�) = Ei(�)M;

where� is a permutation of1; � � � ; g:

We can apply Theorem 2.7 whenh�i is a Sylowr-subgroup ofAut (C): The following lemma gives a condition under which this istrue. For its proof, see [12, Lemma 1].

Lemma 2.9: Let C be a 4-linear code of lengthn with automor-phism � of odd prime orderr as in (4). If eitherr > f or r2>n

and every element of orderr in Aut (C) has permutation part withexactlyc r-cycles andf fixed points, thenh�i is a Sylowr-subgroupof Aut (C):

We now describe the natural actions thatW;�0f ;��c ;D; andF

have on�(C(�)) and Ei(�)�: First, W acts trivially on�(C(�));application of �j 2 W to Ei(�)

� scales thejth coordinate bymultiplying by the element ofIi corresponding toX: �0f permutesthe fixed-point coordinates of�(C(�)) and acts trivially onEi(�)�,while ��c permutes ther-cycle coordinates of both�(C(�)) andEi(�)

�: SinceD is constant on cycles, application of an element ofD

to either�(C(�)) or Ei(�)� scales each coordinate by�1: Finally,F acts trivially on�(C(�)), and onEi(�)� it induces a permutationof the Galois ringsI1; � � � ; Ig by Lemma 1.2 asfu restricted toan r-cycle is �u: If M 2 W �0f �

�c DF , denote these actions by

M: Notice that ifM 2 �0f ��c D the mapM jC(�) ! M j�(C(�)) is

one-to-one. This implies the following:

Corollary 2.10: Let C and C0 be self-dual 4-codes of lengthnboth having automorphism� as in (4). LetC0(�) = C(�) and supposeC0 = CM whereM 2 �0f �

�c D: ThenM 2 Aut (�(C(�))):

III. CODES OF LENGTH 24

In this section we find all the Type II 4-codes of length24 andminimum nonzero Euclidean weight16 with an automorphism ofprime orderr � 5: (The symmetrized weight enumerator for a self-dual 4-code belongs to a certain polynomial ring [1], [4]. Usingthis ring it is easy to show that no such Type II code exists whoseminimum nonzero Euclidean weight is greater than16. Thus the codesin this section areextremal in the sense that they have the highestpossible minimum Euclidean weight.) As discussed in Section II,if such a code has a monomial automorphism of orderr, we mayassume the automorphism is a permutation as given in (4). The firstlemma reduces the number of cases that we have to check.


TABLE IIPRIMITIVE ELEMENTS FOR r = 5 AND 11

Lemma 3.1: Let C be a Type II 4-code of lengthn = 24 with anautomorphism� given by (4) of prime orderr � 5: If the minimumnonzero Euclidean weight inC is 16, then the only possibilities areas follows:

i) r = 23; c = f = 1,ii) r = 11; c = f = 2,iii) r = 7; c = f = 3, andiv) r = 5; c = f = 4:

Proof: Consider first the casec = 1 and1 +X + � � � +Xr�1

irreducible. Then<r = I0 � I1 and E1(�)� is an ideal ofI1

of order 2r�1 by Theorem 2.5; thusE1(�)� contains 2 + 2X

which corresponds to a codeword inC of Euclidean weight8. Thiseliminatesr = 13 and 19 completely andr = 5 and 11 whenc = 1: If r = 17;�(C(�)) has order44 by Theorem 2.5, and so thecode obtained by shortening�(C(�)) on the17-cycle yields a self-orthogonal code of length7 with at least43 codewords, minimumEuclidean weight16, and Euclidean weights divisible by8. A simplecalculation shows this is impossible, eliminatingr = 17:

Next we eliminate the casesr = 7 with c = 2 and r = 5 withc = 3: If r = 7 and c = 2; �(C(�)) has order46 by Theorem2.5, and so the code obtained by shortening�(C(�)) on the two7-cycles and one fixed point yields a self-orthogonal code of length9 with at least43 codewords, minimum Euclidean weight16, andEuclidean weights divisible by8. If r = 5 andc = 3; �(C(�)) alsohas order46, and the code obtained by shortening�(C(�)) on thethree5-cycles also yields a self-orthogonal code of length9 with atleast43 codewords, minimum Euclidean weight16, and Euclideanweights divisible by8. It is straightforward to show no such length9 code can exist.

The remaining case to be eliminated isr = 5 with c = 2: Then<5 = I0�I1 and Table II gives the primitive element� and identity�0 of I1: By Theorem 2.5,E1(�)� and��1(E1(�)

�) are dual codesof length 2 over I1; thus

gen (E1(�)�

) = [�0

�] or2�0 0

0 2�0:

The latter yields the element(2 + 2X; 0) 2 E1(�)� which is

associated to a word of Euclidean weight8 in C: In the former case,by duality,

�0+ ��1(�)� = 0: (11)

To the generator matrix we may apply elements ofW, which areobtained by column multiplication by�3i (inducing a cyclic shiftto the right by i positions), and elements ofF , generated by�2applied componentwise (which sends� to �2 ). By (11), � 62 2I1:

By examining the action ofW andF on the generator matrix, wemay assume that

� 2 f�0; �

0+ 2�

0; �

0+ 2�; �

0+ 2�

3; �

0+ 2�

5; �

0+ 2�

7g

or � = � + 2�j : The only� which satisfies (11) is�0 + 2�7: But

(�0; �

0+ 2�

7) = (03333;01133)

which leads to a codeword inC of Euclidean weight8.

In what follows, the all-one vector of lengthr will be denoted1: Recall also thata0 + a1X + � � � + ar�1X

r�1 will be denoted

a0a1 � � � ar�1: A total of 44 codes will occur which we labelC1; � � � ; C44: We begin with the caser = 23:

Theorem 3.2: Up to monomial equivalence, there is only one TypeII 4-codeC = C1 of lengthn = 24 and minimum Euclidean weight16 with an automorphism� of order 23:

C1 = C(�)� E1(�)� E2(�)

with

gen (C(�)) = [1j1] gen (E1(�)�

) = [�] gen (E2(�)�

) = [0]

where

� = 12102300311110000000000:

Also, � may be replaced by the idempotent generator ofI1 given by

�0= 12222323223322332323333:

This code is equivalent to the4 extended quadratic residue code.Proof:

�1 +X23=(�1 +X)m1(X)m2(X)

where

m1(X)=3 + 2X +X2+X

4+X

5+X

6+ 2X

7+ 3X

10+X

11

and

m2(X)=3+X+2X4+3X

5+3X

6+3X

7+3X

9+2X

10+X

11:

Also I1 = h(�1+X)m2(X)i = h�i; where� is as in the statementof the theorem, andI2 = h(�1 + X)m1(X)i;��1 interchangesI1 and I2: By Theorem 2.5,jC(�)j = 4 and must be as given;also E1(�)� and ��1(E2(�)

�) are duals overI1. By applying ��1;

we may assumejE1(�)�j � jE2(�)�j: Thus we have either the

form in the statement orgen (E1(�)�) = [2�] and gen (E2(�)�) =[2��1(�)]: However, the latter case yields the codeword2 + 2X inE1(�)

� � E2(�)� of Euclidean weight8. By [1] the 4 extended

quadratic residue code has an automorphism of order23 and hencemust be the remaining code. The idempotent generator ofI1 is foundin [15].

The quadratic residue code of Theorem 3.2 is discussed thoroughlyin [1].

The next three theorems examine the casesr = 11; r = 7; andr = 5: In each case, when deciding equivalence, by Lemma 3.1,Lemma 2.9 applies and hence the conditions of Theorem 2.7 hold.If r = 11 or 7, we will see that up to equivalence there is only onechoice forC(�); if r = 5, there are two inequivalent choices. Theresults of Theorems 2.7 and 2.8 and Corollary 2.10 can, therefore, beapplied. In each case we need check equivalence only by applyingelements ofW �0

f ��

c DF where elementsM 2 �0

f ��

c D are suchthat M 2 Aut(�(C(�))): We let

G(C) = fM jM 2 �0

f ��

c Dg \Aut(�(C(�))):

Theorem 3.3: There are exactly six monomially inequivalent TypeII 4-codes C = C2; � � � ; C7 of length n = 24 and minimumEuclidean weight16 with an automorphism� of order 11: C =

C(�) � E1(�) with

gen (C(�)) =1 0 1 2

0 1 2 1

and

gen (E1(�)�

) = [�0

�]


where�0 = 21111111111 and the inequivalent codes are determinedby the six choices for� = �

i+ 2�

j given by

code i j � = �i+ 2�

j

C2 0 73 01113131333

C3 0 157 01131313133

C4 0 341 01311133313

C5 31 46 32111200321

C6 31 104 10133022121

C7 31 188 30111020103

Proof: The ring <11 = I0 � I1, where the identity�0 andprimitive element� of I1 are given in Table II. By Theorem 2.5,C(�)has 16 codewords and the form stated is clearly the only possibilityup to equivalence. The only possibility forgen (E1(�)�) is the onestated or

2�0

0

0 2�0 :

However, in this case, the codeword(2 + 2X; 0) is in E1(�)�,

which has Euclidean weight8. Hencegen (E1(�)�) is of the formstated. When testing for equivalence, the groupG(C) writtenas maps acting on thec + f = 4 coordinates is given byh(1;2)(3;4);diag (1; 3; 1; 3)i of order 8. Elements ofW acting onE1(�)

� are induced by column multiplication by powers of�93 while

elements ofF are componentwise application of�u; 0 � u � 10:

As E1(�)� is dual to �

�1(E1(�)�

); �0+ ��

�1(�) = 0: Upto equivalence there are only eight choices for� satisfying thiscondition. The six listed yield codes of minimum Euclidean weight16.

Theorem 3.4: There are precisely nine monomially inequivalentType II 4-codesC = C8; � � � ; C16 of lengthn = 24 and minimumEuclidean weight16 with an automorphism� of order7.

C = C(�)� E1(�)� E2(�)

with

gen (C(�)) =

1 0 0 1 2 2

0 1 0 2 1 2

0 0 1 2 2 1

:

The nine inequivalent codes are determined by the following:i) One codeC8:

gen (E1(�)�

) =

�0

1 0 �0

1

0 �0

1 �0

1

0 0 2�0

1

gen (E2(�)�

) = [2�0

2 2�0

2 2�0

2 ]

ii) Four codesC9; � � � ; C12:

gen (E1(�)�

) =�0

1 0 �

0 �0

1 �gen (E2(�)

�

) = [ � �0

2 ]

wherecode � � �

C9 �0

1 �0

1 + 2�1

1 �0

2 + 2�0

2 �0

2 + 2�5

2

C10 2�0

1 �0

1 + 2�1

1 2�0

2 �0

2 + 2�5

2

C11 2�0

1 2�0

1 2�0

2 2�0

2

C12 �0

1 + 2�1

1 �0

1 + 2�2

1 �0

2 + 2�5

2 �0

2 + 2�3

2

iii) One codeC13:

gen (E1(�)�

) =

�0

1 �0

1 �0

1

0 2�0

1 0

0 0 2�0

1

gen (E2(�)�

) =2�

0

2 2�0

2 0

2�0

2 0 2�0

2

:

iv) Three codesC14; C15; C16:

gen(E1(�)�

) =�0

1 � �

0 2�0

1 �

gen(E2(�)�

) = � �

0

2

� 2�0

2 0

where

code � � � � �

C14 0 �0

1 + 2�1

1 2�0

1 �0

2 + 2�5

2 �0

2 0

C15 0 2�0

1 2�0

1 2�0

2 �0

2 0

C16 �0

1 �0

1 2�1

1 �5

2 + 2�2

2 �1

2 2�0

2

The relevant elements ofI1 and I2 for these generator matricesare as follows:

I1 I2

�0

2 1332322

�0

1 1223233 �1

2 3323221

2�0

1 2002022 2�0

2 2220200

2�1

1 2200202 �0

2 + 2�0

2 3112122

�0

1 + 2�1

1 3023031 �0

2 + 2�3

2 1132100

�0

1 + 2�2

1 3003213 �0

2 + 2�5

2 1310120

�5

2 + 2�2

2 0233301

Proof: The ring<7 is discussed in Example 1.3 with pertinentdata in Table I. By Theorem 2.5,C(�) has43 codewords. AsC(�)clearly cannot contain a nonzero codeword which is zero on all7-cycle coordinates, the only possibility forgen (C(�)) is the onepresented.G(�) is a group of order48 given by

h(1;2)(4;5); (2; 3)(5;6);diag (3; 1; 1; 3; 1; 1)i:

Since ��1 interchangesI1 and I2, we may assumejE1(�)�j �

jE2(�)�j: Again E1(�)

� and ��1(E2(�)

�

) are duals. SinceG(�)

contains the full symmetric group acting on the7-cycles, there areonly five possible forms forgen (E1(�)�):

a)�0

1 0 0

0 �0

1 0

0 0 �0

1

b)�0

1 0 �

0 �0

1 �

0 0 2�0

1

c)�0

1 0 �

0 �0

1 �

d)�0

1 � �

0 2�0

1 0

0 0 2�0

1

e)�0

1 � �

0 2�0

1 2 :

Since m1 = 1231100 2 I1 has Euclidean weight8, a) isimpossible. The elements ofI1 are of the forma + 2b wherea; b 2 f0; �01 ; �1; � � � ; �

6

1g: In case b), by adding a multiple of row3to rows1 and2, we may assume�; � are0 or powers of�1: As incase a), neither� nor � is 0. Applying elements ofW is the same asscaling columns by powers of�1: By scaling columns2 and 3 andrescaling row2, we may assume� = � = �

0

1 in b). Similarly, byhand, usingG(C);W; andF ; we can reduce the number of casesfor c), d), and e). Fifteen possibilities resulted. They were testedfor low-weight codewords and the nine possibilities of the theoremsurvived.

Theorem 3.5: There are exactly 28 monomially inequivalent TypeII 4-codesC = C17; � � � ; C44 of length n = 24 and minimumEuclidean weight16 with an automorphism� of order 5. C =


C(�) � E1(�) with gen (C(�)) = A or B where

A =

1 0 0 1 1 1 2 00 1 0 1 1 0 1 20 0 1 1 1 2 0 10 0 0 2 0 2 2 20 0 0 0 2 2 2 2

and

B =

1 0 0 1 0 1 1 20 1 0 1 0 3 2 10 0 1 1 0 2 3 30 0 0 2 1 1 3 1

:

The 28 codes are determined by the following generator matrices forgen (E1(�)

�):i) Four codes with

gen(E1(�)�) =

�0

�0

�a

�b + 2�c

0 2�0 0 2�d

0 0 2�0 2�0

where

code=gen(C(�)) a b c d

C17=A; C29=B 0 0 5 0C18=A; C30=B 1 1 2 11

ii) Twenty four codes with

gen (E1(�)�) =

�0 0 �

i + 2�j �k + 2�`

0 �0

�m + 2�n �

p + 2�q

where (see the first table at the bottom of this page).Here the element��1 represents0.The relevant elements ofI1 for these generator matrices are as

follows (in the second table at the bottom of this page).

Proof: The proof is similar to that of Theorem 3.4. Twopossibilities arise forC(�) as stated. These are inequivalent, asthey are of different types. Lemma 2.9 applies to this case byLemma 3.1, and by Theorems 2.7 and 2.8, two codesC and C

0

havinggen (C(�)) = A andgen (C0(�)) = B must be inequivalent.<5 = I0 � I1, and a primitive element and identity ofI1 are foundin Table II. The codeE1(�)� is dual to�

�1(E1(�)�): General forms

for E1(�)� were set up. To test equivalence (done in this case bycomputer), in the casegen (C(�)) = A

G(�) = h(1; 2)(6; 7)diag (3; 3; 3; 3; 1; 1; 1; 1); (1; 2; 3; 4)(5;6; 7; 8)

� diag (1; 1; 1; 1; 1; 1; 1; 3; 3);diag (3; 1; 1; 1; 1; 3; 1; 1)i

a group of order16�24, was used. In the case wheregen (C(�)) = B

G(�)=h(1;2)(7;8)diag (1; 1; 1; 1; 3; 3; 1; 1); (1; 2; 3; 4)(5;8; 7; 6);

diag (3; 3; 3; 3; 3; 3; 3; 3);

(5; 6)(7;8)diag (1; 1; 3; 3; 3; 3; 1; 1)i

a group of order8 � 24, was used. Inequivalent codes were tested forminimum Euclidean weight. The 28 codes of the theorem resulted.

IV. WEIGHT DISTRIBUTIONS AND EQUIVALENCES

In Table III, we give the Lee weight distribution of the codesC1; � � � ; C44 that arise in Theorems 3.2–3.5. These Lee weight dis-tributions are Hamming weight distributions of the binary imagesof the 4-codes under the Gray map [8]. All weights not listedhave no codewords of that weight. Table IV contains the Euclideandistributions of these codes.

A question remains regarding the equivalence of codes that arisefrom automorphisms with different prime orders. From the tables, we

code=gen(C(�)) i j k ` m n p q

C19=A; C31=B �1 0 0 7 0 7 �1 0C20=A; C32=B �1 0 0 7 0 11 �1 0

C33=B �1 0 0 7 0 13 �1 0C21=A; C34=B �1 1 0 7 0 11 �1 4

C35=B �1 1 0 7 0 14 �1 4C22=A; C36=B �1 1 0 11 0 11 �1 4

C37=B �1 1 0 11 0 14 �1 4C23=A; C38=B 1 �1 2 �1 2 �1 13 13C24=A; C39=B 1 �1 2 �1 2 2 13 �1

C25=A; C40=B 1 �1 2 2 2 �1 13 �1

C26=A; C41=B 1 0 2 4 2 4 13 11C27=A; C42=B 1 0 2 4 2 8 13 9

C43=B 1 0 2 4 2 10 13 12C28=A; C44=B 1 1 2 �1 2 �1 13 �1

�0 03333 2�4 22000 �

0 + 2�14 03131 �2 + 2�8 30212

xi1 10201 2�11 02020 �

1 + 2�0 12023 �2 + 2�10 10232

�2 10012 �

0 + 2�5 01331 �1 + 2�1 30203 �

13 + 2�9 20213�13 02011 �

0 + 2�7 01133 �1 + 2�2 30221 �

13 + 2�11 000312�0 02222 �

0 + 2�11 01313 �2 + 2�2 30032 �

13 + 2�12 202312�1 20002 �

0 + 2�13 03311 �2 + 2�4 32012 �

13 + 2�13 02033


TABLE IIILEE DISTRIBUTIONS

see that the only possible equivalences are those where two or morecodes have the same distributions. If two codes were constructedfrom the same automorphism, they would not be equivalent by thetheorems of the previous section. Therefore, the only possible equiv-alences are among the codesfC1; C5; C6; C7; C9; C12g or betweenC4andC37: If � is the primitive element ofI1 for r = 11 from TableII, then �4(�) = �4 and

f4(�0; �

0+ 2�

341) = (�4(�

0); �4(�

0+ 2�

341)) = (�

0; �

0+ 2�

341):

Therefore, the codesC4 and C37 are indeed equivalent asf4 is anautomorphism ofC4 of order 5.

We now consider possible equivalences amongfC1; C5; C6; C7;

C9; C12g: By Theorems 3.3 and 3.4,C5; C6; andC7 are inequivalent asareC9 andC12: By [1], Aut (C1) contains a subgroup, which moduloits center, isPSL2(23): As PSL2(23) has an element of order11,

TABLE IVEUCLIDEAN DISTRIBUTIONS

C1 is equivalent to precisely one ofC5; C6; or C7: The map

(1; 24; 23; 19; 9; 5; 11; 21; 15; 22; 13; 7; 2)(3;6; 16; 12; 14)

�(4; 8; 17; 10)(18;20)diag (111; 313)

sendsC1 to C6:We now show that none ofC5; C6; or C7 is equivalent toC9 or

C12: Suppose two of these are equivalent. Then there is a codeCwith automorphisms of order7 and 11. Let P be the permutationparts of the automorphisms ofC: By Lemma 3.1,P is a permutationgroup on 24 points containing�; which has three7-cycles and threefixed points, and� , which has two11-cycles and two fixed points. LetG be the subgroup ofP generated by� and�: By the orbit structureof � and � , either a)G is transitive, b)G has orbit sizes23 and1,or c) G has orbit sizes22, 1, and1. In a), letG1 be the stabilizer ofsome point. By transitivity ofG; �; and� have conjugates�1 and�1


in G1: ReplacingG by h�1; �1i; we may assume b) or c) holds. Inb) a similar argument using the stabilizer of a point in the orbit ofsize 23 allows us to reduce to case c).

AssumeG satisfies c). SinceG has two fixed points, letG� beG with the fixed points deleted. SoG� is transitive on 22 points.By Lemmas 3.1 and 2.9 and the fact thatC does not have anautomorphism of order5 by the distribution tables,jG�j has order2a3b � 7 � 11: Let N be a minimal normal subgroup ofG�: By [6,Theorem 1.5],N is either an elementary Abelianp-group for someprimep or a direct product of isomorphic non-Abelian simple groups.Let O1; � � � ;Ot be the orbits ofN : As N is normal inG�, whichis transitive,G� permutes the orbits ofN transitively and hence allorbits have the same size, namely2, 11, or 22. If jOij = 2 or 11,since� has a fixed point,� fixes someOi, which is clearly impossibleby the cycle structure of�: HenceN is transitive.

Assume first thatN is an elementary Abelianp-group. If only theidentity of N fixes a point, then by transitivityjN j = 22, which isimpossible. Let� be a nonidentity element ofN fixing a point i:Let j be any point moved by�: By transitivity of N , there exists� 2 N such thati� = j: Then j��1�� = j: But ��1�� = �

asN is Abelian, and hence� fixes j, a contradiction. ThusN isa direct product of isomorphic non-Abelian simple groups. AsN istransitive,11 j jN j; but 112 6 j jN j, and soN must be simple. ThusN is a simple transitive subgroup ofA22 (the alternating group on22 points) of order either2c3d � 7 � 11 or 2c3d � 11: A check of thefinite simple groups (see, for example, [7, Table 2.4]) showsN doesnot exist. Therefore, we have the following theorem.

Theorem 4.1: The codesC1; � � � ; C44 are inequivalent with thefollowing exceptions:C1 andC6 are equivalent, andC4 andC37 areequivalent.

Even unimodular lattices can be constructed from linear4-codesC of length n as follows. Let

�(C) = fx 2 njx � c (mod 4) for somec 2 Cg:

By [1, Theorem 4.1],�(C)=2 is an even unimodular lattice wheneverC is a self-dual 4-code with all Euclidean weights divisible by8.The 42 inequivalent codesCi of Theorem 4.1 have minimum nonzeroEuclidean weight16; hence each�(Ci)=2 is an even unimodularlattice of minimum norm4 in 24: By [3, Ch. 12], these lattices areall the Leech lattice�24:

REFERENCES

[1] A. Bonnecaze, P. Sole, and A. R. Calderbank, “Quaternary quadraticresidue codes and unimodular lattices,”IEEE Trans. Inform. Theory,vol. 41, pp. 366–377, 1995.

[2] A. R. Calderbank and N. J. A. Sloane, “Modular andp-adic cycliccodes,”Des., Codes Cryptogr., vol. 6, pp. 21–35, 1995.

[3] J. H. Conway and N. J. A. Sloane,Sphere Packings, Lattices, andGroups. New York: Springer-Verlag, 1988.

[4] , “Self-dual codes over the integers modulo 4,”J. Comb. TheoryA, vol. 62, pp. 30–45, 1993.

[5] J. Fields, J. S. Leon, and V. Pless, “All4 codes of Type II and length16 are known,”J. Comb. Theory A, vol. 78, pp. 32–50, 1997.

[6] D. Gorenstein,Finite Groups. New York: Harper & Row, 1968.[7] , Finite Simple Groups. New York: Plenum Press, 1982.[8] A. R. Hammons, Jr., P. V. Kumar, A. R. Calderbank, N. J. A. Sloane,

and P. Sol´e, “The 4-linearity of Kerdock, Preparata, Goethals, andrelated codes,”IEEE Trans. Inform. Theory, vol. 40, pp. 301–319, 1994.

[9] W. C. Huffman, “Automorphisms of codes with applications to extremaldoubly even codes of length 48,”IEEE Trans. Inform. Theory, vol.IT-28, pp. 511–521, 1982.

[10] , “On the [24; 12;10] quaternary code and binary codes with anautomorphism having two cycles,”IEEE Trans. Inform. Theory, vol.34, pp. 486–493, 1988.

[11] , “On extremal self-dual quaternary codes of lengths 18–28, I,”IEEE Trans. Inform. Theory, vol. IT-36, pp. 651–660, 1990.

[12] , “On the equivalence of codes and codes with an automorphismhaving two cycles,”Discr. Math., vol. 83, pp. 265–283, 1990.

[13] , “On extremal self-dual ternary codes of lengths 28–40,”IEEETrans. Inform. Theory, vol. 38, pp. 1395–1440, 1992.

[14] B. R. McDonald, Finite Rings with Identity. New York: MarcelDekker, 1974.

[15] Z. Qian, “Cyclic codes over 4,” Ph.D. dissertation, University ofIllinois at Chicago, 1996.

[16] V. Y. Yorgov, “Binary self-dual codes with automorphisms of oddorder,” Probl. Inform. Transm., vol. XIX, pp. 11–24, 1983.

Some New Extremal Self-Dual Codeswith Lengths and

Ilya Boukliev and Stefka Buyuklieva

Abstract—New extremal self-dual codes with lengths44; 50; 54; and58 are constructed. They have weight enumerators for which extremalcodes were previously not known to exist.

Index Terms—Heuristic algorithm, self-dual codes

I. INTRODUCTION

A binary linear[n; k] codeC is a k-dimensional subspace ofFn

2

whereFn

2 is then-dimensional vector space over the binary fieldF2.The number of the nonzero coordinates of a vector inFn

2 is calledits weight. An[n; k; d] code is an[n; k] linear code with minimumnonzero weightd. An automorphism of the codeC is a permutationof the coordinates ofC which preservesC.

Let

(u; v) =

n

i=1

uivi 2 F2;

for u = (u1; � � � ; un); v = (v1; � � � ; vn) 2 Fn

2

be the inner product inFn

2 . Then if C is an [n; k] code overF2,C? = fu 2 Fn

2 : (u; v) = 0 for all v 2 Cg. If C � C?, C istermed self-orthogonal and ifC = C?, C is self-dual. A binaryself-dual code in which all weights are divisible by four is termeddoubly-even. If not all weights are divisible by four the code is singly-even. Self-dual codes with the largest minimum weight for a givenlength are called extremal. A list of possible weight enumerators of

Manuscript received February 28, 1997; revised July 9, 1997. This workwas supported in part by the Bulgarian National Science Fund under ContractI-618/1995 and Contract I-602/1996.

I. Boukliev is with the Institute of Mathematics, Bulgarian Academy ofSciences, 5000 Veliko Tarnovo, Bulgaria.

S. Buyuklieva is with the Faculty of Mathematics and Information, Univer-sity of Veliko Tarnovo, 5000 Veliko Tarnovo, Bulgaria.

Publisher Item Identifier S 0018-9448(98)01629-0.

0018–9448/98$10.00 1998 IEEE

decompositions and extremal type ii codes over z4

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