The classification of all (145,7,72) binary linear codes

The classification of all (145,7,72) binary linear codes

Citation for published version (APA):

Tilborg, van, H. C. A., & Helleseth, T. (1980). The classification of all (145,7,72) binary linear codes. (EUT report. WSK, Dept. of Mathematics and Computing Science; Vol. 80-WSK-01). Technische Hogeschool Eindhoven.

Document status and date: Published: 01/01/1980

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TECHNISCHE HOGESCHOOL EINDHOVEN NEDERLAND

ONDERAFDELING DER WISKUNDE

TECHNOLOGICAL UNIVERSITY EINDHOVEN THE NETHERLANDS

DEPARTMENT OF MATHEMATICS

The classification of all (145,7,72) binary linear codes

by

Tor Helleseth and Henk C.A. van Tilborg

T.H.-Report BO-WSK-01 April 1980

1. Introduction

We let an (n,k,d) code denote a binary linear k-dimensional code of length n and minimum distance at least d.

We let

(1.1 ) n(k,d) := min{n

I

there exist an (n,k,d)code}.

Important results about n(k,d) which we will use repeatedly are:

Theorem 1.1. (Griesmer [2]). Let rxl denote the smallest integer greater than or equal to x. Then n(k,d) ~ g(k,d) := k-l

L

r~l

i=O 21.

..,

Theorem 1.2. (Logacev [4]). If 3 ~ d ~ 2k-2 _ 2, then

n(k,d) ~ g(k,d) + 1 .

Theorem 1.3. (van Tilborg [6] ). If 2k-2 + 3

~

d

~

2k-2 + 2k-3 - 4, then

n(k,d) ~ g(k,d) + 1 .

I

2Ui -1, i=l k-l and s·2 - d > O. >u p > u > 2 where k > u 1

Theorem 1.4. (Belov [1]). Let s = r--d--l k-l 2 I f min (p, s+l)

L

i=l u. ~ s·k 1. or u i+1 ui - 1, for s ~ i ~ P - 1 and up E {1,2} , then n(k,d) = g(k,d) •

Belov gave constructions of codes with n(k,d) = g(k,d) for all para-meters satisfying the conditions of Theorem 1.4. An open problem is to

2

-We will show the existence of new codes meeting the Griesmer bound without satisfying Belov's conditions. These codes are (145,7,72) codes. We will give a thorough analysis of codes with these parameters. We have several motivations for this. First of all the methods in this paper can in principle be applied to classify any (n,k,d) code with n(k,d) = g(k,d), and thus we can either prove that such a code does not exist or (as in our case) find new codes meeting the Griesmer bound. Further, in a follo-wing paper [3J we will construct infinite sequences of (n,k,d) codes based on our (145,7,72) codes. To prove uniqueness results for these sequences of (n,k,d) codes, as well as to prove nonexistence results for other (n,k,d) codes with s > 1 meeting the Griesmer bound, we will need a classification of all (145,7,72) codes.

(i)

2. Preliminaries

Let G denote a generator matrix of a binary (n,k,d) code c.

Definition 2.1. The residual code of C with respect to c E C is the code

a

C generated by the restriction of G to the columns where c has a zero.

We let w(c) denote the Hamming weight of £. We denote cO by Res(Ci£) or Res(Ciw) if only the weight w of £ is important.

If cO E cO we let c E C denote a word whose restriction to cO is ca. Also the restriction of c E C to cO is denoted by £0.

We denote Res(Res(Ci£l)i

£~»

by

ReS(Ci£l'£~)

or Res(Ciw,WO) if only

a

a

a

the weights of £1 E C and £2 E C are important, and w(£l)

=

w, w(£2)

=

w

_o'

etc.

Let lxJ denote the greatest integer smaller than or equal to x. From Definition 2.1 we get:

Lemma 2.2. We have that Res(Ciw) is a (n-w,k-1, d-LIJ) code.

Lemma 2.3. Let £1 E C have weight w. Let

£~

E cO

=

Res(Ci£l) have weight w00 Then {w(£2),w(£1 + £2)}

=

{w

O + l~J - i, W

o

+ r~l + i} for some non-negative

integer i .

Lemma 2.4. (van Tilborg [6J). Suppose that C meets the Griesmer bound. (i) There exists a generator matrix for C such that all row vectors have

weight d.

(ii) If s

=

r

k~ll

, the no column vector occurs more than s times in G.

2

Lemma 2.5. (van Tilborg [6J). The following (n,k,d) codes are unique:

C(i)

=

(2k_2u,k,2k-l_2u-l), 1

~

u

~

k - 1 • The weight distribution is

kiU

A

=

1 A k-1 u-l

=

2k _2k - u A k-1

=

2k -U_l

a ' 2 -2 ' 2

(ii) c(i)

=

(2k_2k-2_3,k,2k-1_2k-3_2), k ;;:: 5. The weight distribution is k;k-2,2 k-2 A O = 1,A2k-1_2k-3_2

=

3(2 -1), A2k-1_2k-3 (1.' 1.' 1.') Ck(ii) = (2k-1+k,k,2k-2+2), k ;;:: 3, k ~r 5.

=

2k-2_1,A k-l 2

=

3. 2

4

-Lemma 2.6. Let A,B denote the number of codewords of weight w in a

w w

binary linear code C and in its dual code respectively. Then

B w where 1 n

=

TCT I

i=O A.K (i) 1 w K (i) w

, a

~ i, w ~ n In particular, K O(i) 1, K 1(i)

=

n - 2i, K 2(i) (n) - 2ni + 2i 2

_.

2

3. Weight distribution of (145,7,72) codes

Let C denote a (145,7,72) code. We let G be its generatormatrix. The number of codewords of weight ware A .

w

Our approach can briefly be sketched as follows. We first show that the only possible occuring nonzero weights of Care 72,80, or 88. Next we show that C has only 5 possible weight distributions. Finally we prove that all (145,7,72) codes with a given weight distribution are isomorphic.

Lemma 3.1. If A ~ 0, then w E {0,72,80,82,84,86,88,90,92,96,98,100,112,114}. w

Proof. From Lemma 2.4(i) we get A

w

°

for w odd. By Lemma 2.4(ii) we con-clude A

=

°

for w > 128 since s

=

2 for C.

w

Suppose C contains a codeword of weight w ~ 128 and w even. The

resi-°

dual code C of C w.r.t. a word of weight w has parameters given by Lemma 2.2, and these contradicts Theorem 1.1-1.3, except for the values of w given

° ° °

°

in Lemma 3.1. The parameters of the (n ,k ,d ) code C are given in Table 1.

Table 1 (no,kO,dO) contra-w dicts 72 73,6,36 74 71,6,35 T.1.1. 76 69,6,34 T.1.1. 78 67,6,33 T.1.1. 80 65,6,32 82 63,6,31 84 61,6,30 86 59,6,29 BB 57,6,28 90 55,6,27 92 53,6,26 94 51,6,25 T.1.1. (nO,kO,dO) contra-w dicts 96 49,6,24 9B 47,6,23 100 45,6,22 102 43,6,21 T.1.1. 104 41,6,20 T.1.3. 106 39,6,19 '1'.1.1. lOB 37,6,18 T.1.1. 110 35,6,17 T.1.1. 112 33,6,16 114 31,6,15 116 29,6,14 T.1.2. l1B 27,6,13 T.1.1. -(nO,kO,dO) contra--w dicts 120 25,6,12 T.1.2. 122 23,6,11 T.1.1. 124 21,6,10 T.1.1. 126 19,6, 9 T.1.1. 128 17,6, 8 T.1.2.

o

6

-Lemma 3.2. If A ~ 0, then w E {0,72,80,82,84,88,92,96,112}.

w

Proof. In view of Lemma 3.1 we must prove that A

86 = A90 = Ag8 = AIOO = Al14 = O. ro illustrate the method we prove that A

86

=

O. Suppose

£

E C has weight 86.

Then cO

=

Res(C;~I)

is a (59,6,29) code. The extended code of cO is the unique

c~~;

code of Lemma 2.5. Since

c6(~;

has 48 codewords of weight

, 0 '

30 i t follows that C has a codeword ~g of weight 30. Using Lemma 2.3. with w

=

86 and W

o

=

30 we get

for some integer i ~ O. This contradicts that A

=

0 for 72 < w < 80 •

w

With the following choices of wand W

_o

the other cases can be proved in a similar way. cO _Res(Ciw) 0 w = C (extended) W

_o

90 (55,6,27) C(i) 28 6i3 98 (47,7,23) C (i) 24 6;4 100 (45,6,22) """- C (i)

-

24 6i 4 ,2 114 (31,6,15)

c

(i) 16 6i5 Lemma 3.3. If A

t

0, then w E {0,72,80,88,96,112} . w

Proof. According to Lemma 3.2 we have to prove that A

82 = A84 = A92 = O. We illustrate the method by proving that A

84 O. Suppose £1 E C has weight

a

84. Then C

=

Res(C i £l) is a (61,6,30) code. Since a (61,6,31) code would

o

0

contradict the Griesmer bound, C has a codeword £2 of weight 30.

00 0

a

00

Then C

=

Res (C'~2) is a (31,5,15) code. We claim that C has a codeword

00 00

of weight 16. If not, C must have a word c of weight 15 since a (31,5,17) cot'le contradicts the r::riec;mer r.ound. Then cOOO

=

Res(COO;151 is a (16,4,8) code which has a codeword of weight 8 since a (16,4,9) code

ld ' d' h ' b d 00 00 h ' ht 8 h

wou aga1n contra 1ct t e Gr1esmer oun. Let ~ E C a v e we1g w en

000 00 00 00

restrictedtoC . Then {w(£ ),w(£ .+~ )} = {15,16} by Lemma 2.3.

By symme ry we may assume w £t ( 00) = 16• Then aga1n e1t er £. . h

°

or c

° °

+ £2 has weight 31 or 32, and we assume w.l.o.g. that it is cO. Then c or £ + £1 has weight w such that 72 < w < 80, a contradiction.

The proof that A

S2 = A92 = 0 is similar.

Lemma 3.4. If A ~ 0, then w E {0,72,80,88,96} •

w

Proof. We have to prove that A

l12

=

0. Suppose £ E C has weight 112. Then cO = Res(Cic) is a (33,6,16) code. Let AO denote the number of codewords

- w

of weight w in cO. Since a

(33-w,5,16-L~J)

code will contradict at least one of the Theorems 1.1-1.3 for all w such that 19 ~ w ~ 31 we get that

°

A

=

°

for 19 ~ w ~ 31. Using Lemma 2.3 with w

=

112 and W

o

=

17 or 18 we

w O O get A

17 = A18 0. Since a (33,6,17) code does not exist (contradicts the

°

°

Griesmer bound), A

16 >

° .

Therefore A33 = 0, otherwise the complement of a word of weight 16 would have weight 17 which is excluded. From Lemma 2.6

°

we get, since A 32 ~ 1 that

o

°

31A 32

=

64

leading to

B~ =

1. This means that one column of the generator matrix of cO is the zero vector, and therefore this is also the case for G. This contra-dicts that C meets the Griesmer bound, and we conclude that A

l12

=

° .

0

Lemma 3.5. We have the following equations: An A₈₀ A₈₈ A₉₆ 1

°

°

1 _~(B2 + 195)

°

1

°

-3 42 - B 2

°

°

1 3 _~(B2 - 25) where B

2 is the number of pairs of repeated columns of G. Proof. This follows from Lemma 2.6. Note that B

1

=

°

since C meets the Griesmer bound. Further B

2 is the number of codewords in the dual code of C which is equal to the number of repeated columns of G since no column

8

-a

Lemma 3.6. Let A_w denote the number of codewords of weight w in the (49,6,24) code Res(Ci96). If AO

~

0, then w E {0,24,28,321 and we have

w

the following equations,

a

a

a

A 24 A28 A32 1

a

-1 49

a

1 2 14 0

a

1 B

a

2

where

B~

is the number of repeated columns in a generator matrix for Res(Ci96) .

Proof. Using the same techniques as in Lemma 3.1-3.4 it is easily seen that the only possible weights are 0,24,28, and 32. The lemma follows then

from Lemma 2.6.

0

a

Lemma 3.7. Let A denote the number of codewords of weight w in the

w 0

(57,6,28) code Res(Ci88). If A ~ 0, then w E {0,28,32,36,40} and we have w

the following equations,

(i) A

a

a

a

a

28 A32 A36 A40 1

a

-1 -2 48

a

1 2 3 15

a

a

1 3 BO_3 2

where

B~

is the number of repeated columns in a generator matrix for Res(Ci88) •

(ii)

Proof. (i) This goes along the same lines as the previous lemma.

0 0 0

(ii) Suppose £ E C

=

Res(Ci88) has weight 36, and that

A40

> O. According to van Tilborg [6J there are two non-isomorphic (21,5,10) codes with the weight distributions,

00 Ala 21

a

(3.1 ) 20 10

a

1 40(resp. 36} least 14} , have we get BO

=

7, 2 40 its gene-where AOO denotes the number of codewords of weight w in a (21,5,10)

w

code. Since Res(COi£O} is a (21,5,10) code it has one of the two weight distributions in (3.1). We note that if

~O

E cO has weight

then its restriction to Res(COi£O} has weight 16 (resp. at

a

a

a a

since otherwise w(v + £ ) < 28. Therefore Res(C i£ } must

A~~

=

1 and

A~~

=

0, and therefore

A~6

=

A~O

=

1. From (i) but since cO is a (57,6,28) code with a codeword of weight

rator matrix must have at least 8 repeated columns. This contradicts

that

B~

=

7 and (ii) is proved.

o

We next will prove that A

96

=

O. This is the hardest case and can not be done by the methods of Lemma 3.1-3.4 alone. We first introduce

(a 1, ••• ,an) and denote innerproduct (£1'£2) be~ween £1

=

defined as (£1'£2) :=

.r

aibi • We ~=1 some notations. The

£2

=

(b₁,···,b_n) is

+r++s-+ r

c

=

(1 •••1 0 •••a) by

t-I---+1

s •

Lemma 3.8. If A ~ 0, then w E {0,72,80,88} •

w

Proof. According to Lemma 3.4 we must show that A

96

=

O. Suppose ~1 E C has weight 96. We have A

96= 1 since if v E C is another codeword of weight 96, then

w(~

+ £1} < 72 since the restriction of v to cO

=

Res(Ci~1}

by Lemma 3.6 has weight at most 32.

From Lemma 3.5 we get B

2

=

31 + 2A88• Since C has a codeword of weight 96 there are at least 32 pairs of repeated columns and therefore A

88 ~ 1. We next show that A

88 ~ 2. Suppose A88

=

1 and let £2 E C have weight 88. Then

(£~'£2)

=

56 since

W(£~} ~

32 and w(£1 + £2)

~

72. Lemma 3.5 gives A

80

=

12 and B2

=

33. Therefore we can choose £3 E C of weight 80, such that we have,

o

0 0

Since B

2

=

33 and A32~1 we get from Lemma 3.6 that B2

=

A32

=

1, and therefore

00 0 0 1

(~1'~3)

=

52. The code C

=

Res(C ;~2) has the possible weights 8,10,12, or 6,

as can be proved using the methods of Lemma 3.1-3.4. Since YO is the weight of a

o

codeword in the (17,5,8) code Res(C ;3~and YO + _Y1 = 28 it follows from Lemma 3.6 that YO

=

10 or 12. If YO

=

10, then Y_l

=

18 leading to

B~ ~

2, a contra-diction.

Hence YO

=

12 and Yl

=

16. The (57,6,28) code Res(Ci£2) has a word of weight 40, namely the restriction of £1. By Lemma 3.7 (i) we get Y₂

=

16,20,24, or 28. If Y

2

=

16 or 24, then the restriction of £3 or £1 + £3 have weight 36, contradicting Lemma 3.7 (ii). Further Y₂

=

28

is impossible since this leads to B

2 > 33. Hence Y2

=

20 and Y3

=

32.

We choose ~ E C of weight 80 such that ~

i

< £1'£2'£3 >. This is possible since A

80

=

12. Therefore,

From the above discussions we get Y

7 + Y6

=

32 and therefore Y7

=

Y6

=

16, or a column of G will be repeated more than twice, contradicting Lemma

2.4 (ii). Also Y₁ + YO

=

12 gives Y₁ = 8 and Y

8

= 4 since YO is the

o

00

weight of a codeword in the (5,4,2) code Res(C ;£3). The Res(C;£1 + £2) is a (73,6,36) code and the restriction of c

2 + c + cA has weight 32,

- -3

-a contr-adiction. We conclude A

88 = 1 is impossible.

We next show A88 ~ 2. Suppose £1'£2'£3 E C where w(£l)

=

96, w(£2)

=

w(£3)

=

88. We have

96 49 56 40 32 17 (3.2)

Yo

t---ll • Since Y

1 + YO

=

32 and

ReS(c;£l'£~)

has only weights S,10,12, or 16. we get YO

=

12 or 16, otherwise

w(£~

+

£~)

< 24. If YO

=

12, then Y3

=

20 and w(cg + £g)

=

24 which means that w(£2 + £3)

=

72 and therefore Y3

=

32 and Y₂

=

24. Then the restrictions of £1 and £3 to Res(c;£2) have weights 40 and 36 respectively, contradicting Lemma 3.7 (ii). We conclude that

Yo

=

16. This also means that every

£ i

< £1,c

2'£3 > of weight 88 would

have weight 16 when restricted to Res(c;£l'£~). Since this is a (17,5,S) code this code can not contain more than one word of weight 16. We there-fore conclude ASS

=

2.

96 49 56 40 32 17 32

I

24

I

24 16 16 16 16

I

1 Y _{Y6 Y5 Y4 Y3 Y2 Y1 YO}

~I

~

t---=:--1

~ ~

r---:=---1

f---=--1

~

Finally we prove that ASS

=

2 is impossible. Suppose ASS

=

2 and let

< £1'£2'£3 > , w(~)

=

_{80 where £1,c 2 '£3 are as in (3.2). We have}

3.5 that B

2

=

35 and ASO

=

10, so ~ exists. Therefore, ~ E C -by Lemma £1 £2 £3 ~

We have (c ,c )

=

52 and therefore Y. + YO

=

10 for i

=

1,2,3, which means

-1 -4 1.

that YO

=

l'Y₁

=

Y

2

=

Y3

=

9, otherwise we contradict the occurring weights of cO or Res(CO;32). Further by Lemma 2.4 (ii) we have Y

7 = 16. Also Y₄ = S since otherwise B

2 ~ 36, which is impossible. Since £3 + ~ restricted to

ReS(COi£~)

has weight S, the restriction of £3 +

~

to

Res(ci£2) has weight 28. Therefore Y₅

=

12 and Y

6

=

16. We now get a contradiction since w(£2 +~)

=

6S. We therefore have proved that

A

12

-Lemma 3.9. A (145,7,72) code C must have one of the following 5 weight distributions, I II III IV V 1 1 1 1 113 112 111 110 11 13 15 17 3 2 1

o

Proof. By Lemma 3.5 it is sufficient to prove that A

88 E {0,1,2,3,5} • Suppose A₈₈ ~ 2 and let £1'£2 E C have weight 88. Then (£1'£2)

=

48

or 52. We will exclude (£1'£2) = 48. If (£1'£2) = _{48 then B2} ~ 32 and therefore Lemma 3.5 gives A

88 ~ 4. We can therefore choose £3 E C - <£1' £2 > , w('£3)= 88 such that, 88 57 .£1 48 40 40 17 (3.3) £2 Y3 Y2 Y1

I

YO

I •

.£3

I

By Lemma 3.7 (ii) we get (£1'£3)

=

(£2'£3)

=

48. The (17,5,8) code

o

ReS(Ci£1'£2) has only 8,10,12, or 16 as possible weights and therefore we get the unique solution YO

=

16, Y1

=

Y₂

=

Y₃

=

24. In < £1'£2'£3 > there are 3 words of weight 88. The argument above proves that any further code-word of weight 88 has weight 16 when restricted to

Res(Ci£1'£~).

Since

o

Res(Ci£1'£2) has at most one word of weight 16 this means that A

88 ~ 3, a contradiction. We conclude that any two codewords of weight 88 have innerproduct 52.

52 88

36 36

57

21 (3.4)

The (21,5,10) code Res(Ci88,36) has one of the two weight distributions

in (3.1). Since Y

i + YO = 36 for i = 1,2 we get YO = 14 or 16, otherwise

we contradict the minimum distance of Res(Ci88).

Case 1. (Yo = 16).

Then we have the unique solution Y

1

=

Y2

=

20, Y3

=

32, and < £1'£2'£3 >

contains three words of weight 88. Further it is impossible to choose another word £ E C - < £1'£2'£3 > of weight 88 since £ restricted to

ReS(Ci£l'£~)

would have weight 14 or 16, contradicting (3.1). Hence we conclude that A

88

=

3 in this case.

Case 2. (YO

=

14).

Then we have the unique solution Y

1

=

Y2

=

22, Y3

=

30. Inspection of (3.4) reveals at least 32 pairs of repeated columns of G. Lemma 3.5

gives A

88 ~ 4 and therefore there is a codeword ~ E C - < £1'£2'£3 > of

weight 88 since < £1'£2'£3 > only contains 3 words of weight 88.

We have, 88 57 £1

I

52 36 36 21 £2 (3.5) 30 22 22 14

_I

22

_I

14

I

14 7 c 3

I

~ Y7 Y6 Y5 Y4 Y3 Y2 Y1 YO ~ ~~ ~ ~ ~

J---=.1

~.

Here YO = 3,4, or 7 since YO is the weight of a codeword in Res(Ci88,36,14) which extended is a unique (8,4,4) code by Lemma 2.5 (i). Since Y

1 + YO = 14

we have YO

=

7, otherwise the weight of £3 + ~ when restricted to

Res(Ci£l,£g) gives a contradiction. Therefore we get YO = Yl = 7. By symmetry (interchanging £2 and £3) we obtain Y2

=

7 and therefore Y

14

-(interchanging £1 and £2) we get Y4

=

7, Y₅

=

15. Finally since

(£3'~)

=

52 we obtain Y

7

=

15 and therefore Y6

=

15. We observe that < £1'£2'£3~ > contains exactly 5 codewords of weight 88, namely

£1'£2'£3'~'

and £1 + £2 + £3 +

~.

Since

ReS(C;£1'£~)

by (3.1) only contains 3 words of weight 14 it follows that there are not more code-words of C of weight 88 since a further codeword of weight 88 would

o

have weight at least 14 when restricted to Res(C;£1'£2). Hence we have A

4. The final classification

In this section we will prove that there are exactly 5 non-isomorphic (145,7,72) codes. We will show that for each of the 5 listed weight distri-butions in Lemma 3.9, there is a unique code.

Let (AlB) be a partitioning of a matrix G. Then we shall often denote

*

B by G - A. Further F2~ denotes the set of all ~-dimensionalbinary column

*

vectors and F2~ := F2~ - {a}

Theorem 4.1. Let C be a (145,7,72) code with wei~ht distribution,

Then C is isomorphic to the code generated by ~1' where

11. .• 1 11. •• 1 11. .. 1 11. •. 1 00 ...

a

11. .• 1 11. .. 1 11. .. 1 00 ...

a

11. .• 1 11. •• 1 11. .. 1 00 ...

a

11. .. 1 11. .. 1

*

11. .• 1 00 ••. 1 11. •• 1 11. •• 1 11. .. 1 F 27 F 23 F 23 F 23 F 23 F 23 I I ~1

=

+ 127 + + 8 + + 8 + + 8 + + 8 + + 8 + + 15 + + 7 + I

.,

*

10

F 4 2

I

I

0/

_{F 3}

*

2

Proof. According to the proof of Lemma 3.9 (case 2) the first 4 rows of G can be chosen as in (3.5) i. e. , 88 57 £1 I 52 36 36 21 £2 1 £3 I 30 22 22 14 22 14 14 7

~~15~7

15 7 7 7 15 7

~

7

~

7 7 I---l !----l l----1 1----+

- 16

-Since A

80

=

7 and < ~1'~2'~3'~4 > contains no codewords of weight 80 we

can choose ~5 E C - < c

1,c2,c3,c4 > of weight 80. Inspection of the first

4 rows gives that B

2 ~ 35. By Lemma 3.5 we have B2

=

35 and therefore

Y_i for 1 < i ~ 15 must be chosen such that no more pairs of repeated columns are introduced, and also such that no columns occurs more than twice. 'rhis means Y1 'y2'Y3'Y4'Y5'Y6'Y8'Y9' Y10'y12 ~ 4 and y7'Y11 'Y13'Y14'Y15~ 8. Since

15

w(c_S)

=

.r

1

y.

=

80, equality must hold in all cases. We next note that

- ~= ~

< ~1 '~2'~3'~4'~5 >contains only one word of weight 80, namely ~5. Therefore we can choose ~6 E C - < ~1'~2'~3'~4'~5 > of weight 80. Exactly as above it follows that ~6 is uniquely determined. Further< ~1 '~2'~3'~4'~5'~6 > has 3

codewords of weight 80, namely ~5'~6' and ~5 + ~6' and we can therefore choose ~7 E C - < ~1'~2'~3'~4'~5'~6 > of weight 80. Again we are led to a unique choice of ~7 by the arguments above. Therefore a (145,7,72) code with A

88

=

5 must be unique. Since the code generated by ~1 in Theorem 4.1

is easily seen to be a (145,7,72) code with A

88

=

5, this must be the one.

0

Theorem 4.2. Let C be a (145,7,72) code with weight distribution,

Then C is isomorphic to the code generated by ~2' where

+ 127 -+ + 16 -+ + 5 -+ + 5 -+ + 5 -+ + 5 -+ + 3 -+ + '15 -+ ~2

=

11. .. 1 11111 11111 00000 00000 11. •. 1 11111 00000 11111 00000

*

11. .. 1 00000 11111 11111 00000 F 7 11110 11110 2 11110 11110 F 4 11101 11101 11101 11101 2 11011 11011 11011 11011 10111 10111 10111 10111 110 101

0

011

0

_{F 4}

*

2

Proof. First, note that the code generated by ~2 is a (145,7,72) code with the required weight distribution. It is therefore sufficient to

Since A

80

=

11 we can choose ~ E C - < £1'£2'£3 > of weight 80. Let CO be the (57,6,28) code cO

=

Res(Ci£l)' and define the (21,5,10) code

00 0 0 O .

C

=

Res(C i£2). We first observe that w(~) ~ 32 otherw~sew(£l +~) < 72.

o 0 00

By Lemma 3.7 we get w(~)

=

32 or 36. If w(~)

=

36 then w(~ ) ~ 14 contradicting (3.1). Therefore

w(~)

= 32 which gives

w(~O) ~

12,

other-. 0 0 00 000

w~se w(£3 +~) < 28. By (3.1) we get w(~ )

=

12. The code C

=

R_es(COO_i~00)._{~s t e}h _un~que. (5 4 2)_{" even we~ght}. _{co e. T er.e ore we}d h f h_ave

YO

=

w(~O)

=

4, since

w(~o)

=

2 gives

w(£~O +~~O)

< 10. Hence Y 1

=

8, and the symmetry in £1'£2' and £3 leads to Y

3 Y5

=

12, Y1

=

Y2

=

Y4

=

8. Therefore Y

6 + Y7 = 28 and since no column of G occurs more than twice according to Lemma 2.4 (ii) we get Y

7

=

16 and Y6

=

12. Hence ~ is uniquely determined.

We next note that < £1'£2'£3'~ >contains less than 11 words of weight 80. Therefore we can choose £5 E C - < £1'£2'£3'~ > of weight 80. Then,

£ 1 - 1 ;:;:.;88 -1

1 57

£2 1 - 1_ _--=5:.:::2 __

36 36 21

- 18

-We will show that ~ is uniquely determined. Since W(~)

=

w(£5)

=

80, and

~

is uniquely determined we have

W(~)

= 32,

W(£~

) = 12, and

000 . 000 000

w(c S ) 4. Sl.nce w(c₄ +c₅ ) ~ 2 we have y

=

1 and y

=

3 •. Further

- - - 00 0 1

Y2

=

Y3 4, otherwise w(~O + ~5 ) < 10 or w(~~O + ~~o + ~gO) < 10. sy symmetry

UU

we obtain Y

4

=

Y5 = Y8

=

Yg

=

4. Next, w(~ + ~O)

=

10 leau~ ~U

w(~

+

~)

28, otherwise

w(£~

+

~

+

£~)

<28. Therefore since Y

6+ Y7

=

12 we have Y

6 = 5 and Y7 = 7. Again by symmetry we get Y11 = Y13 = 7

and Y

10

=

Y12

=

5. Finally Y14

=

Y15 = 8, otherwise a column of G occurs more than twice contradicting Lemma 2.4 (ii).

With this unique choice of £5' inspection of these 5 rows gives at least 31 pairs of repeated columns. Lemma 3.5 gives B

2

=

31. Since

< £1'£2'£3'~'~ > contains less than 11 words of weight 80 we can choose

~ E C - < £1'£2'£3'~'~ > of weight 80. The same argument as we used to prove the uniqueness of ~,~ and £7 in Theorem 4.1 will now prove the existence of unique words ~ and £7 of weight 80. We therefore conclude that there is at most one (145,7,72) code with the weight distribution given in Theorem 4.2. Since the code generated by ~2 is a (145,7,72) code with this weight distribution, this must be the unique such code.

0

Theorem 4.3. Let C be a (145,7,72) code with weight distribution,

Then C is isomorphic to the code generated by ~3' where

+- 127 + +- 8 + +- 8 + +- 8 ++- 8 + +- 7 + +- 7 + 00 ...

a

00 . . . 0 00 ... 0 11111111 11. .. 1 11. .. 1 11. .. 1 00001111 11. .. 1 11. .. 1 00 ...

a

11001100 11. .. 1 00 ... 0 11. .. 1 10101010 F* 27

0

F 3 F 3 F 3 2 2 2 00 •.. 0 00 .•.

a

F 3

0

2

0

F* 23

Proof. By the proof of Lemma 3.9 it follows from (3.4) that the first 3 rows of "G can be taken to be as follows,

88 57 .£1 52 36 36 21 .£2 Y3 Y2

f--

Y1 YO .£3 (4.1)

where'£3 E C - < '£1''£2 > is any codeword of weight 80. From Lemma 3.5

o

and Lemma 3.7 we get ('£1''£3)'('£2''£3) E {44,48}. Let C denote the

(57,6,28) code cO

=

Res(c i'£l) and define the (21,5,10) code cOO

=

ReS(Ci.£l'~~).

We first prove that ('£1''£3)

=

('£2''£3) 48. By symmetry it is sufficient to show that ('£1''£3)

=

48. Suppose ('£1''£3)

=

44, i.e.

w(.£~) =

36. From (3.1) it follows that YO =

w(.£~o)

= 14 or 16, otherwise

w(.£~

+

.£~)

< 28. If

YO = 14, then Y1 = 22 which leads to at least 30 pairs of repeated columns, contradicting Lemma 3.5 which says B

2 29. Therefore YO

=

16 and Y1

=

20. If ('£2''£3) = 44 we get Y

2 = 20 and Y3 = 24. If ('£2''£3) = 48 we get Y2

=

16 and Y

3

=

28. In both cases the two (57,6,28) codes Res(Ci'£l) and Res(Ci'£2) contain at least two codewords of weight 36, as inspection of (4.1)

reveals. From Lemma 3.7 this leads to at least 5 pairs of repeated columns in these two co es.d S ·1nce COO.1S a (21 5 10)" co e mee 1ngd t' the G .r1esmer bound, it follows from Lemma "2.4 (ii) that it has no pairs of repeated columns. Therefore this leads to at least 30 pairs of repeated columns of ~, contradicting B

2 = 29. We therefore conclude that every codeword of weight 80 has innerproduct 48 with .£1 and '£2'

00 00 , 0 0

Suppose.£ E C has we1ght 16. Then by Lemma 3.7 w(.£ + '£2) ~ 36

o

0

or w(.£) ~ 36 so we can assume w.l.o.g. that w(.£ ) ~ 36. Then by Lemma 3.5 w(.£ + '£1) ~ 80 or w(.£) ~ 80 so we can assume w.l.o.g. that w(.£) ~ 80. Since A

88

=

2 and .£ ~ ~1 and .£ ~ ~2 we get that w(£)

=

80. Further since (~'~1)

=

('£''£2)

=

48 we get with .£

=

.£3 that YO

=

Y

1

=

Y2

=

16, Y3

=

32. Again this leads to two codewords of weight 36 in the two (57,6,28) codes Res(Ci'£l) and Res(Ci'£2) which as above contradicts B

2

=

29. Hence (3.1) shows that COO has 3 words of weight 14.

Let .£00 E cOO have weight 14. As above we show that there is a

there 20 there

-fore w.l.o.g. choose £3

=

£. Since (£1'£3)

=

(£2' ~)

=

48 we have the unique solution Y

l

=

Y2

=

18, and Y3

=

30. Since cOO has 3 words of weight 14 we can similarly w.l.o.g. choose ~ such that

00

~ E C - < £1'£2'£3 > , _w(~)

=

80, and w(~ )

=

14. Then we have,

£1 I 88 ₅₇ £2 , 52 ₃₆ 36 ₂₁ I 30 22 18 18 I 18 I 14 I £3 I 18 7 Y7 Y6 Y5 Y4

_{I~ ~}

_~

_~

~ ~ ~ ~ ~ 00 00 Since YO is the weight of a codeword in a (7,4,3) code Res(C ;~), we have YO

=

3,4, or 7. Since YO + Y_{1 0 1}

=

14 we get Y

=

Y

=

7, other-wise

w(c~O

+

~O)

< 10. From Lemma 3.7 it is straightforward to show that a (25,5,12) code Res(C;88,32) has only even weights. Therefore since Y 2 + Y3

=

Y4 + Y5

=

18 we get Y2

=

Y3

=

Y4

=

Y5

=

9, otherwise we contradict B 2

=

29. Since w(£3 + ~) E {72,80} we get Y6

=

Y7

=

15 i f w(c 3 + ~)

=

80 and Y6

=

11, Y7

=

19 if w(£3 + ~)

=

72. But Y7

=

19 is impossible since this contradicts Lemma 2.4 (ii) • Hence we have Y₆

=

Y

7

=

15. Inspection of these 4 rows reveals at least 29 pairs of repeated columns. Since B₂ = 29 and < £1'£2'£3'~ > contains less than 13 words of weight 80, the arguments used to prove the uniqueness of a

£5'~' and £7 in Theorem 4.1 can also be applied here. We conclude that all (145,7,72) codes with A

88

=

2 are isomorphic to the code generated

by G₃, which generates such a code.

0

To treat the two remaining cases we need the following two lemmas, whose proofs follow from Lemma 2.6 and the techniques in Lemma 3.1-3.4 and will be omitted here.

Lemma 4.4. Let AO denote the number of codewords of weight w in the

w 0

(65,6,32) code Res(C;80). If A ~ 0, then w E {0,32,36,40,44,48} and

w

a

a

a

a

a

A 32 A36 A40 A44 A48 1

a

-1 -2 -3 47

a

1 2 3 4 16

a

a

1 3 6 BO_7 2

where BO is the number of repeated columns in a generator matrix for

2

Res(Ci80).

Lemma 4.5. Let AOdenote the number of codewords of weight w in the

w

a

(73,6,36) code Res(Ci72). If A

#

0, then w E {0,36,40,44,48,52} and we

w

have the following equations,

a

a

0

a

a

A 36 A40 A44 A48 A52 1

a

-1 -2 -3 46

a

1 2 3 4 17

a

a

1 3 6 BO-12 2

where BO is the number of repeated columns in a generator matrix for

2

Res(Ci72) •

Theorem 4.6. Let C be a (145,7,72) code with weight distribution,

A

O A72 A80 A88

1 111 15 1 .

Then C is isomorphic to the code generated by

24'

where

+ 127 ~ + 4 ~ + 4 ~ + 4 ~ + 4 ~ + 8 ~ + 4 ~ 0000 0000 0000 0000 11111111 0000 1111 1111 1111 1111 00000000 0000 1111 1111 0000 0000 00000000 1100 1111 0000 1111 0000 00000000 1010

*

F 7 2 0100 0100 0100 0100 1111 0010 0010 0010 0010 F 3 1111 2 0001 0001 0001 0001 1111 + 3 ~+ 7 ~

0 0

*

0

F 2 2

0

_{F 3}

*

2

22

-Lemma 3.5 which says B 2 the weight distribution,

Proof. Let £1 be the codeword of weight SS. The (57,6,2S) code cO

=

a

a

Res (Ci£l) has B

2 ~ 3 according to Lemma 3.7. Therefore B2

=

3, other-wise we get at least 2S pairs of repeated columns of G, contradicting

=

27. From Lemma 3.7 i t fOllO:S that cO has

a

A 32 15 .

In particular if £ E C and w(£)

=

SO, then

A

SO

=

A~2

=

15 we note that if cO E cO and

32. Also since

a

32, then c is the restriction of some c E C of weight SO.

Let £2 E c, w(£2)

=

SO and define cOO as the (25,5,12) code cOO

=

a a

cOO E cOO and 00

a

a

Res(C ;C

2). I f w(£ ) > 16, then w(£ + £2) > 32 or

a

.

a

w(£ ) > 32, wh1ch contradicts the weight distribution of C • Further if w(cOO)

=

13 or 15 then again w(£O) or w(£O +

£~)

would contradict the weight distribution of COO. Therefore the only possible occurring weights in cOO are 12,14, and 16. From Lemma 2.6 we obtain the

equa-(4.2)

columns in a generator matrix for tions, AOO AOO AOO 12 14 16 1 1 1 31

a

1 2 14

a

a

1 2BOO + 1 2 00 is the number of where B 2 repeated 00

(4.2) we observe that cOO

C • From has a codeword of weight 16. This

a

must be the restriction of a word of weight 32 in C and therefore there is a £3 E C of weight SO such that

w(£~o) =

16. We can therefore assume

w.l.o.g. that the first rows of G can be chosen as in one of the two cases,

Case 2. _{(w(£2 + £3)}

₌

80) 88 57 £1 48 40 32 25 £2 24 124 1 24 1161 16 116

I

16

I

9 £3

We next show that we can transfer Case 1 to Case 2. Suppose we have the situation in Case 1. Since

B~

= 3 the generator matrix of at least one of the (65,6,32) codes Res(Ci£2)' Res(Ci c₃) or Res(Ci£l + £2 + £3) has at least 9 pairs of repeated columns. By symmetry we can assume it is Res(Ci£2). This code has a word of weight 40, namely the restriction of £1. Since the existence of a word of weight 44 or 48 in Res(Cic2) would lead to a codeword of weight at least 88, different from £1 it follows from Lemma 4.4 that Res(Ci£2) contains another word of weight 40. Let c E C be the codeword with this restriction to Res(Ci£2)' then w(c)

=

80 and, c 48 88 40 57 32 25 where YO + Y

2

=

40. We have Y2 ~ 24, otherwise we contradict B2

=

27. Further since YO ~ 16 by (4.2) we get YO

=

Y1

=

16, Y2

=

24, and Y₃

=

24. which with £3

=

£ gives Case 2.

We therefore only have to consider Case 2. Since Res(Ci£2) has two words of weight 40 we get from Lemma 4.4 that i t has 9 pairs of repeated columns. Since

B~

=

3 we have

B~O ~

1 otherwise we contradict B₂

=

27. Therefore

A~~ ~

3 follows from (4.2) and we can choose

~

E C-<£1 '£2,c

3 >

00

3 which by (4.2) leads to - 24 -£1 I 88 ₅₇ £2 1 48 40 32 25 (4.3) £3 I 24 I ₂₄ I 24 I 16 I 16 ₁₆ I 16 9 Y Y _{Y5 Y4 Y3} Y _Y1 YO ~I~

I---LI

_~ ~ 1--:4

I~

~ ~

D f '_{e 1ne} th_e (9 4 4)_{" co e}d COOO

₌

_{Res C}(00_;£300) _{. A co ewor}d d 0f we1g t' h 9 '1n

000 , . 00

C would lead to a word of we1ght greater than 16 1n C . Further 000

Res(C ;7) would have parameters (2,3,1) which is impossible. Hence cOOO has 4,5,6, and 8 as only possible weights. Suppose cOOO has no codeword of weight 8. Since

W(£~O)

=

W(~O)

=

16 this means YO

=

6 and

. 00 00

Y₁

=

10, otherw1se w(£3 + ~ ) < 12. Further Res(C;£3) has also two codewords of weight 40. We can therefore repeat the above arguments

*

with £2 and £3 interchanged and obtain the existence of a ~ E C of

* .----

0

weight 80 such that Y

2= 10. But then inspection of (4.3) leads to B2~ 4, 000

a contradiction. We therefore conclude that C has a codeword of weight 8. We can therefore assume w. Lo.g. that YO=

w(~~O)

= 8. Since no codeword of cO has weight greater than 32 we get Y

1

=

Y2

=

Y3a 8 .Since B2

=

27 we have Y4

=

8 , and all Y

5,Y6,Y7

s

16. From Lemma 4.4 applied to Res(C;~2)' Res(C;~3)' and Res(C;£2 + £3) i t follows Y

5'Y6'Y7 E {12,16}. Since Y5 +Y6 +Y7 = 40 we can assume w.l.o.g. that Y5

=

16 and Y

6

=

Y7

=

12. Inspection of (4.3) now reveals that Res (C;£2) has at least 4 words of weight 40,

namely the restrictions of £1'£3'~' and £1 + £3 +~. From Lemma 4.4 we conclude that Res(C;c ) has at least 11 pairs of repeated columns

-2

, . 00 h BOO

1n a generator matr1x. Hence C as 2

A₁₆OO

=

7_{. We can} th_{ere ore c oose}f h

£S

E C - < _{£1 '£2'£3'-=-4}c > s ch that_u

00

w(£5)

=

80 and w(£5 ) = 16, and we have,

88 ₅₇ ~1 I I 48 40 32 25 ~2 24 24 16 16 16 9 ~3 24 16 -I ~4

~

121~

12~

8

~-l

8

~

8

~

..

~

8

8

8

~

1

We immediately get Y₄

=

Y₅ = Y₆

=

Y₇

=

Y₈

=

Yg

=

Y₁₀

=

4, otherwise we contradict B2 = 27. From Lemma 2.4 (ii) we obtain Y

11 = 8. Since YO + Y

I + Y2 + Y3

=

16 we are lead to Yi + YO = 6 for i

=

1,2,3, so

o

0 0

8₂0 = 3 gives YO

=

1 and Y₁

=

Y₂

=

Y₃ = 5. Also w(~ + ~) = 28 meanS that w(~ +~) = 72 and therefore Y₁₂ + Y₁₄ ,= _Y13 + Y15 = 14.

Similarly

w(c~

+

~)

=

28 gives Y 14 + Y 15

=

Y 12 + Y 13

=

14. Finally since

w(£~

+

~ +~)

=

28 we get Y₁₂ + Y₁₅

=

Y₁₃ + Y₁₄

=

14. Hence we have Y₁₂

=

Y₁₃ = Y₁₄

=

Y₁₅

=

7.

Now inspection of the first 5 rows above gives at least 27 pairs of repeated columns. Since < £1,c2,c3'~'~ > contains less than 15 words of weight 80, there exist a % E C - < £1'£2,c:3'~'~ > of weight 80. To prove the uniqueness of % we use the same method as for~ in the proof of Theorem 4.1. Similarly we also prove the unique-ness of c₇ E C of weight 80 such that C = < £1,c2,c3,~,~,%,C7 > •

It is straightforward to check that the code generated by ~ is a (145,7,72) code with A88

=

1, and therefore all other (145,7,72) code.s with the weight distribution in Theorem 4.6 are isomorphic to this code.

0

Theorem 4.7. Let C be a (145,7,72) code with weight distribution,

Then C is isomorphic with the code generated by ~, where

+ 127 -+ + 8 -++ 8 -++ 9 + +1++3++3-+ 11111111 00000000

0

00000000 11111111

0

0 0

00001111

0

000111111

*

00110011 111000111 F 7 2

0

*

0

F 2 2

0

00001111 011011011 00110011 101101101

0

0

*

F 2 2 01010101 01010101 111111111 1

0

0

- 26

-Proof. If ~,~ E C, u # ~, and w(~)

=

w(~)

=

80, then (~,~)

=

40 or 44

since w(~ +~) E {72,80}. In the first part of the proof we will show the existence of two codewords of weight 80 with innerproduct 40.

Suppose the contrary, i.e. (~,~) = 44 for all ~,~ E C, ~

#

~, and w(u) = w(~) = 80. We will show this to be impossible. We let the first 3

the (65,6,32) code cO = Res(Ci'£l)' From Lemma 4.4 we obtain

7 since

A~O ~

0 by the assumption above. Let cOO be the (29,5,141

o

0 00 00

Res(C i'£2)' Then all.£ E C have even weights, otherwise will contradict Lemma 4.4. Also

A~O

=

0 for i

~

20 since

l.

we get the equations,

o

Let C be that B 30

=

code C =

o

0 w(.£ + '£2) BO =

i.

From Lemma 2.6 2 AOO 00 AOO 14 A16 18 1 1 1 31 (4.4) 0 1 2 15 0 0 1 2 (BOO-1) 2

where

B~O

is the number of pairs of repeated columns in a generator matrix

00 00 00

for C . If A

18 ~ 1 then we can assume w.l.o.g. that w('£3 ) = 18, and since ('£1''£3)

=

('£2''£3)

=

44 we get the unique solution y

=

y

=

y

=

18

o

0 1 0 2

and Y3

=

26. Since B3

=

7 one of the codes cOO Res(C

;.£~)'

Res(C

i.£~)'

or

ReS(CO;.£~

+

.£~)

has exactly 3 pairs of repeated columns in its generator matrix. By symmetry we can assume

B~O

=

3. Then (4.4) gives

A~~

=

4 and

00 we can choose ~ E C - < '£1,c

We have,

£1 I 80 65

£2 I 44 ₃₆ I· 36 ₂₉

£3 26 I ₁₈ I 18 ₁₈ 18 ₁₈ I 18 ₁₁

The (11,4,5) code cOOO

=

Res(COOiCOO) reach the Griesmer bound, and we

-3

get from Lemma 2.4 (ii) and Lemma 2.6 the equations,

AOOO_{5 .} A₆OOO 1 O·

o

1 1

o

2 1 3 3 15 13 5 • 000 000 . 000 000

Hence there is a c E C of we~ght 7. The (4,3,2) code Res(C i£ )

is unique and i t has a codeword of weight 4. One of the two codewords of cOOO whose restriction to Res(COOOi£OOO) has weight 4 must have

000 000 . 00 .

weight 8. We conclude that A

7

=

2 and A8

=

1. S~nce A18

=

4 1t follows that every word of weight 7 or 8 in cOOO is the restriction of some word of weight 18 in cOO. We can therefore assume w.l.o.g. that

. 00 00

a

0

Yo

=

7 and _Yl

=

11. S1nce w(£3 + ~ )

=

14 we get w(£3 + ~)

=

32 and therefore Y

2

=

Y3

=

9. Similarly we obtain Y4

=

Y5

=

9, Y6

=

11, and Y7 = 15. We further can choose ~ E C - < £1,c2'£3'~ > of weight 80,

00

and w.l.o.g. we assume w(~ ) = 18, and w(~OO)

=

7. Then we have, 80 ~1 44 ~2

1-.---1

36 36 65 29 ~3 26 18 18 18 18 18 18 11 ~5

A1187H9H9~9~9~7~4

Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y I IS.

H

14 I13₁ I 12,

H

11 ,10₁ 9 8 7 6 5 4 3 2 1 0 r---l r---"1r---1 ,...-,l--lHHI--I~HI--lI--lI----1I---4

2S

-000 000 Since YO is even we have YO = 4 and Y1 = 3. Further since w(~ +

.£s

)

= S it follows from the remark above that

w(~o

+

~o)

=

18 or

00 00 00

w(~3 + ~ + ~ )

=

lS. Since Y

2 + Y3

=

11 we have Y2

=

4 or 5. Here Y

2 5 is impossible because this contradicts

B~O

= 3. We conclude that Y₂

=

4 and Y

3

=

7. Similar arguments give Y12

=

4 and Y13

=

7. Since Y₁₄ + Y15

=

15 we obtain from Lemma 2.4 (ii) that Y

14, Y15 E {7,S} • If Y₁₄

=

7 and Y₁₅

=

8 then the restriction of £3 + ~ + ~ to the

(73,6,36) code Res(Ci~l + £2) has weight 38, contradicting Lemma 4.5. If Y₁₄

=

8 and Y₁₅

=

7, then since Y

i E {4,5} for 4 ~ i ~ 11 and (~,~)

=

44 we get Y₄

=

Y₆

=

Y_S

=

Y₁₀

=

4 and Y

_s

= Y₇ = Y₉ = Y₁₁ 5. We now get a contradiction since w(~l + ~2 + ~)

=

w(£S)

=

SO and

(~1 + £2 + ~'£5) = 40.

We can therefore assume that the codewords in Res(Ci80,36) have 14 and 16 as only possible non zero weights. Therefore we can choose

£1 SO 65

£2 _I 44 36 36 29

£3

_I

24 _{20 I} 20 ₁₆ 20 16 ~ 16 13

~ ~Y7 ~Y6 ~Y

s

~Y4

~

~Y2

~

~

a

00

a a

000

where w(~)

=

80. Let C

=

Res(C;~l)' C

=

Res(C i£2)' and C

00 00 00

Res(C i£3). By (4.4) we have Ai

=

0 for i > 16, and therefore

A9

00

₌

°

_{for i > 8. Hence the (13,4,6) code cOO}O_{has only 6,7, and 8}

~

as possible weights. From Lemma 2.6 we obtain in particular that

A₈000

~

3. We can therefore w.l.o.g. assume

w(~OO)

= 8. Since (c.,c,)=44

-- -~ -J

80 65 ~1 44 36 36 29 ~2 ~3 24 20 20 16 20 16 16 13

~4 ~12 i~

8

~

8

~I

8 11.L.j 8

I~

a

~

8

~

5 Y15 Y14 Y13 Y12 ;11, ~101 Y9 Ya Y7 Y6 Y5 Y4 Y3 Y2 Y1 Yo

~ 5 ~ 1--1 ~ ~ ~ I---l J--I I---l I---i ~ 1--1

H

1--1 1--1

. 0000

The (5,3,2) code def~ned as C Res(C 000 i~ 000 ) has A4 000 ~ 1 accor-ding to Lemma 2.6. Therefore we can assume w(c 00)

=

4, and using the

-S

uniqueness of ~ combined with the facts that Res(C;80,36) is a two-weight code and that (c.,c.) = 44 for 1 ~ i , j ~ 4, we get as above that

-~ - ]

Y1

=

Y2

=

Y3

=

Y4

=

Y5

=

Y₆

=

Y

_a

=

Yg 4 and

Y

7

=

Y11

=

Y13

=

Y14

=

8. We now get a contradiction by observing that w(£1 + £2 + £3 + ~ +

£S)

< 72. Hence we have shown the existence of two codewords in C of weight 80 whose innerproduct is 40.

We can therefore choose £1'£2 E C of weight 80 such that (£1,c_Q)=40.

o

00 0 0 00 00

Let C

=

Res(C;£1) and C = Res(C ;£2)' Every £ E C must have even weight otherwise w(£O +

£~)

or w(£O) will contradict Lemma 4.4. Further

A~O

=

0 for i

~

18 since B2

=

25 according to Lemma 3.5. Therefore

Lemma 2.6 gives the same equations as (4.2). Since

A~~ ~

1 we can choose 00

£3 E C - < £1'£2 > such that w(c] )

=

16 and w(£3)

=

80. Then, 80

65

40

40 40 25

£3

We have

B~

=

8 or 9 since B2

=

25. Lemma 4.4 gives

A~4

= a

so therefore

Y

1'Y2'Y3 E {20,24} which combined with Y1 + Y2 + Y3

=

64 gives w.l.o.g.

Y₁

=

24 and Y

2

=

Y3

=

20.

30 -£1 ' 80 65 £2 40 ₄₀ 40 25 £3 20 I 20 I 20 ₂₀ 1_ _24 I 16 ~ 16 9 ~ ~Y7 ~Y6 ~Yr: ~Y4

~

~Y2

~

~

(4.5) 15 6 B 000 2 and this contradicts B

2

AOOO AOOO AOOO

4 6 8

1 1 1

a

1 2

a

a

1

000 00 00 000

We define the (9,4,4) code C

=

Res(C i'£3). The codewords of C

have all even weights since if

w(~OO)

is odd, then Y

1 and Y2 are odd

=

25. From Lemma 2.6 we get the equations:

where BOOO is the number of repeated columns in a generator matrix for

2

COOO. In part '~cuaIr A000 >- l 's~nce B000 / 1 th f

2 ~ . We can ere ore assume

6 000

w.l.o.g. that w(~) = 80 and w(~ ) = 6. From Lemma 4.4 we get that

B~ =

9. Therefore YO

=

6 leads to Y1

=

Y

2

=

8 and Y3

=

14. We have Y_{4 +} Y₅ + Y

6 + Y7 = 44 but since ('£1 + £2'~)' (~2'~4) € {40,44} we obtain

Y

4 + Y5, Y6 + Y7 E {18,22} and therefore Y4 + Y5

=

Y6 + Y7

=

22. By Lemma 4.4 the restrictions of £3 + ~ to Res(Ci£2) and Res(Ci'£1 + £2) has weight 32 or 36 since 40 is excluded because then the restriction of £1 + £3 + ~ would have weight less than 32. Therefore

Y

4'Y5'Y6'Y7 € {10,12}. Because of the symmetry in .£2 and .£1 + .£2 we can assume Y

4

=

10. Then Lemma 4.4 applied to Res(Ci~3) gives the solution Y

5

=

12, Y6

=

12, and Y7

=

10. We therefore have,

~1 80 65 ~2

I

40 ₄₀ 40 ₂₅ 20 20 20 20 24 16 16 j ₉ ~3

:::5

~4 ~ 10~

8

~

8

I~

10

~II0 ~

8

~

8

~

3 Y Y Y Y Y Y Y Y

s

Y Y Y Y Y Y Y1 YO

~~A8~~~H~~~~~~~~.

000 000 where ~5 c C - <~1 '~2'~3'~4 > • _{We next show that Aa} = _{O. If Aa} ~ 1, then we can let w(~~OO) =a and w(~5) = ao. We get YO = 3 and Y₁ = 5 and since 8

2 = 25 it follows that Y2 = Y3 = Y4 = Y5 = Y10 = Y12 = 4. Since

o 0 0

8

2 = 9 we have A40 = 2 by Lemma 4.4 and therefore w(~5) = 36. Further

000 000 0 0

W(~4 + ~5 ) = 4 gives w(=4 + =5) = 32 which leads to Y_{6 = 5 and Y7 = 7.} The restriction of ~5 to Res(C;~4) has weight at least 36, and since 8

2

=

25 we get Y

a = Y14

=

6. From the discussion of ~3 we have Yg+Y11 = 14 and Y13 +Y

15= 10 or vica versa. If Yg+Y11 = 14 then Yg = 6 and Y11=8, and by Lemma 4.4. applied to Res(C;~3) we get Y

13E {4,a} • The restriction of of ~4 + ~5 to Res(Ci=3) has weight 30 or 34, a contradiction. The case Y9 + Y11= 10 gives a similar contradiction. We conclude A~OO'=,0i and.

000 (4.5) gives A

6 = 6.

We define cOOOO = Res(CoOO;cOOO). Then cOOOO is a (3,3,1) code and we 000 -4

can assume w.l.o.g. that w(~5) 2 and w(~5) = ao. From (4.2) and (4.5) we obtain Y₁ = Y

2

=

Y3

=

Y4

=

Y5 4 because

8g

0 O. Since w(~~OO + ~~OO) 4 one gets w(~~ + ~g)

=

32, leading to Y

6

=

6 and Y7 = a. Further 82 25 gives Y

10 = Y12

=

4 and also that Ya

=

Y14 = 6 because the restriction of

~5 to Res(c;~4) has weight at least 36. From the discussion of ~4 we get Yg = _{4 or 6. If Yg} = 4 we get Y

11 = Y13 =a and Y15 = 4 but this is im-possible since w(~1 + ~2 + ~3 + ~4 + ~5) < _{72. Therefore Yg = 6 which gives}

Y11

=

Y₁₃

=

Y 15

=

6. We therefore have, ~'1I 80 115 £2 I 40 40 41! 25 £3 I 20 20 20 20 24 111 16 9 ~ i---l..L-t 10 l---!L.-l e ~ e ,~ 10 ~ 10 ~ B ~ 8 ~3 %~ 4 ~ 4 ~ 6 I~ 4 ~ 6

f--!-l

4 ~ 4 ~ ..

t-!--l

6

rLt

4 ~.. r!-l:4'1~ 4~4 ~ :I ~

HI V30 V29 V28 Y27 Y26 Y2S Y24 Y23 Y22 Y21 Y20 YI9 YI8 YI7 YI8 Yl5 Y14 VI3YI2 VII yto Y9 Y8 Y7 Y6 Ys V4 Yg '12 yt YO

32

-000

We can assume w.l.o.g. that w(~ )

=

2 and w(~)

=

80. Since B

2

=

25,

B~

=

9, and

B~O

=

a

we immediately get from the discus-sion of £5 that YO

=

Y1

=

1, Y i 2 for 2 ~ i ~ 11, Y16

=

Y₁₈

=

Y₂₀

=

Y₂₁

=

Y₂₄

=

Y 25 Y28

=

Y30 2, and_{000 000 000) 4} Y₁₄

=

Y 15

=

Y17

=

Y19

=

Y29

=

Y31 4· We have w(~4 + ~5 + ~6

=

and thus the restriction of ~4 + ~5 + ~6 to Res(C;~i)' 1 ~ i ~ 3, has weight 32. Hence we get Y

12

=

Y13

=

Y22

=

Y23

=

Y26

=

Y27

=

3.

Finally we can use the same method as in Theorem 4.1 to prove

the existence of a unique £7 E C - < ~1'~2'£3'~'~'~ > of weight 80. Since ~5 can be shown to generate a (145,7,72) code w±thA

88

=

0,

REFERENCES

[lJ B.I. Belov, A conjecture on the Griesmer bound, Optimalization methods and their applications (All-Union Summer Sem. Khakusy, Lake Baikal, 1972) (Russian), 100-106, 182. Sibirsk Energet. Inst. Sibirsk, Otdel, Akad. Nauk SSSR, Irkutsk, 1974.

[2J J.H. Griesmer, A bound for error - correcting codes, IBM J. Res. and Develop., 4 (1960), 532-542.

[3J T. Helleseth and H.C.A. van Tilborg, A new class of codes meeting the Griesmer bound, to appear.

[4] V.W. Loga~ev, An improvement of the Griesmer bound in the case of small code distances, Optimalization methods and their applications

(All-Union Summer Sem. Khakusy, Lake Baikal, 1972) (Russian),

107-111,182. Sibirsk Energet. Inst. Sibirsk, Otdel, Akad. Nauk SSSR, Irkutsk, 1974.

[5J F.J. MacWilliams and N.J.A. Sloane, The theory of error correcting codes, North Holland, Amsterdam 1977.

[6J H.C.A. van Tilborg, On the uniqueness resp. nonexistence of certain codes meeting the Griesmer bound, Information and Control, 44 (1980), 16-35.