0 7 8 9 1 3
6 1 7 8 9 2
5 0 2 7 8 9
4 6 1 3 7 8
6 5
1 0
7 2
8 7
1 9 8
3 2 9
5 4 3
2 3 4
4 5 6
6 0
0
7
8
9
4 9
6 5
1 0
3 2
7 4
8 7
9 8
5 6
0 1
2 3
5 2 4 6
4 3 5 0
3 4 6 1
9 5 0 2
8 0 1 3
8 7 1 2 3
17 0 2 4
9 8 2 3 4
6 1 3 5
6 9 3 4 5
0 7 8 9
1 0 4 5 6
1 9 7 8
3 2 5 6 0
2 8 9 7
5 4 6 0 1
7 6 0 1 2
0 1 7 8 9
2 3 8 9 7
4 5 9 7 8
COM BIN A TOR I A L
MAT HEM A T I C S
YEA R
February 1969 - June 1970
A CLASS OF TRI-WEIGHT CYCLIC CODES
by
THOMAS A. DOWLING
University of North Carolina at Chapel Hill
Department of Statistics
Institute of Statistics Mimeo Series No. 600.3
JANUARY 1969
This research was partially sponsored by the Air Force Office of Scientific
Research, United States Air Force, under Grant No. AF-AFOSR-68-l4l5; the
National Science Foundation, under Grant No. GP-8624; and the National Aeronautices and Space Administration - American Society of Electrical Engineers
Summer Faculty Fellowship Program.
A CLASS OF TRI-WEIGHT CYCLIC CODES
INTRODUCTION
Ie
1, k
Let
Let
fO(x)
1
be positive integers such that
be a primitive polynomial of degree
be the minimum polynomial of
cyclic code
0.
241 ,
where
A with recursion polynomial
k
<
k
and
over
(2
1
GF(2),
is a root of
a
fO(x)fl(x)
k
+ 1, 2
is a
and let
fO(x).
(2
k
- 1)
= 1.
flex)
Then the
- 1, 2k) code
representing the direct sum of the maximal-length shift register codes with
recursion polynomials
In the case
that the code
k
fO(x), flex),
respectively.
(1, k) = 1,
odd and
Gold [5] and Solomon [11] have shown
A has three non-zero weights.
They give the weight distribution
and an algebraic characterization of the code words of each weight.
k :: 2 (mod 4)
obtained a similar result for
these results to
(2
arbitrary
1 + 1, 2k - 1) = 1.
k
and
1
and
1
= (k + 2)/2.
We generalize
1
subject to the conditions
Theorem 1 below shows that
Gold [4]
<
k,
A has three non-zero weights
and gives the weight distribution, and Theorem 2 gives an algebraic characterization of the code words of each weight.
Remark.
After completing this work, the author learned that the weight dis-
tribution (Theorem 1) had been obtained earlier by Kasami [6, 7].
Kasami's
proof is purely algebraic and is based on the fact (which he proves) that
is a sub code of a second-order modified Reed-Muller code of length
2
k
A
- 1.
Our proof involves geometric as well as algebraic methods, and is sufficiently
dissimilar to Kasami's to merit its inclusion here.
To the author's knowledge,
Theorem 2 is new.
2.
STATEMENT OF PRINCIPAL RESULTS
Throughout we shall assume
k
a fixed positive integer and
1
an integer
such that 1 $ 1 $ k-l. This latter restriction is merely a convenience, since
This research was partially sponsored by the Air Force Office of Scientific
Research, United States Air Force, under Grant Ncr. AF-AFOSR-68-l4l5; the National
Science Foundation, under Grant No. GP-8624; and the National Aeronautices and Space
Admiristration - American Society of Electrical Engineers Su~nmer Faculty Fellowship
Program.
z
m
2
l
k
+ 1 = Z + 1 (mod Z - 1)
since we require
fO(x)
if
and
m
=l
flex)
(mod k).
to be distinct.
=
l
The case
0
is omitted
We begin by characterizing
the set
{l:
L(k)
of integers
l
for which
For any integer
LEMMA 1:
=
met)
1
(a, c)
Zl _ 1,
=
c
1:::;
-
1,
only if
mel)
~
L (k)
COROLLARY 2:
If
Write
k
and distinct from
a, b, c
m(k)}.
are integers, and
(a, b)
=
Zl + 1 ,
(b, c).
But
Setting
=1
i f and only i f
(Zr - 1, Zs _ 1)
1
i f and only if
(Zl, k)
~
if and only if
k
a
1,
= (l,
=
k) ,
Z(r,s)
-
l.
i. e. , i f and
m(k).
COROLLARY 1:
Proof:
~
(Zl + 1, Zk _ 1)
we have
(Zl + 1, Zk _ 1)
mel)
(ab, c)
( l l _ 1, Zk _ 1) = (Zl _ 1, Zk _ 1) .
Hence
zk - 1) = l }
as the exponent of two in the prime
l :::; k-1,
i f and only i f
Zk
(Zl + 1,
t.
It is easily verified that i f
Proof:
b
k-l,
is primitive of degree
define
{l:
L(k)
~l :::;
flex)
t,
power factorization of
then
1
k
l
Zm s,
k
E
L(k),
where
(l, k) _ k (mod Z)
m
=
m(k),
and
l
for some
and
Zm t.
m ~
o.
k/(l, k) - 1 (mod Z).
Then
Zm(s + ( s, t » ,
+ (l, k)
k/(l, k)
Zm
=
siCs, t),
and the right-hand expressions are even and odd, respectively, since
s
is odd.
3
THEOREM 1:
the cyclic code
m
k f 2
for any
Suppose
m ~ 0,
A with recursion polynomial
code with three non-zero weights.
l
and let
fO(x)f1(x)
E
is a
(2
L(k).
k
Then
- 1, 2k)
A is given below:
The weight distribution of
Weight
Number of Code Words
k-l
2
(2 k _ 1) (2 k _ 2k -(l,k) + 1)
k-l
2(k+(l,k)-2)/2
2
-
(2 k _ 1) (2 k - a ,k)-1 + 2(k-a,k)-2)/2)
2k - l + 2 (k+a,k)-2) /2
(2 k _ 1) (2 k - a,k)-l _ 2(k- a,k)-2) /2)
The proof of Theorem 1, and Theorem 2 below, will be given in Section 5.
Before stating Theorem 2, we require some preliminary remarks on the algebraic
A, following Solomon [11] and Gold [5].
characterization of the code words of
Let
a
= (aO' aI' •.. , a k )
2 -2
A.
be a code word of
Then
a
is charac-
terized by its Mattson-Solomon polynomial [8]
1
Tr cx + Tr dx
g (x)
a
where
c, d
k-1
k
GF(2 ),
E
and
Tr x
2 +1
,
2i
is the trace of
x
l:
k
GF(2 )/GF(2).
i=O
The polynomial
satisfies
g (x)
a
i
g (a )
a
=
where
is a root of
Thus the mapping
ljJ:
a
-+
(c, d)
is a one-one linear mapping of the code
A onto
w(c, d)
a
as the weight of the code word
a
If
maps
a
clearly
a
f
E
A,
T (a)
s
+
a
E
AO'
ljJ(a) = (c, d) ,
and
so
(c, d)
-+
If
ljJ
-1
x
k
GF(2 ).
We define
(c, d).
to the right of length
a cyclic shift T
s
s l
s
(a c, a (2 +l)d). If c 1= 0, d
0, then
k-l
w(c, 0) = 2
.
k-l
.
Ai' w(O, d) = 2
=
k
GF(2 )
Similarly, i f
c 1= 0, d 1= 0,
c =
we can choose
°, d 1= 0,
s
then
above so that
s
4
a
s
d
1
2 +1
-~
since
l·
+ 1
is prime to
Since
T
s
is obviously
weight-preserving,
1
= w(cd-~, 1).
w(c, d)
Thus we can restrict attention to code words
where
~(a)
such that
=
(c, 1),
C:f O.
Let
where
a
x
t = k/(l, k).
THEOREM 2:
(i)
if
Let
=
if
denote the trace of
We then have
Then
C E
Tr k / (l,k) c
=
w(c, 1)
(ii)
t-l 2i (l,k)
E x
i=O
T
r k / (l, k)
2
1,
:f:
k-l
C
=
1,
l
=
w(c, 1)
where
S
r
if
Tr(8 + 82 +1)
k
2 - w(l, 1)
if
Tr(8 + 82 +1)
~
l
x
Remarks:
Hence
Note that since
w(c, 1)
w(c, 1) = w(l, 1)
1
= 0,
l
1,
is any solution of
(1)
or
w(l, 1)
t
22l
2l
+ x
= k/(l, k)
2k - l ± 2(k+(l,k)-2)/2
or
2
k
- w(l, 1)
(c + 1)
is odd by Corollary 2,
if
Trk/(l,k) c
=
1.
=
Trk/(l,k) 1
The proof that
in this case according as
will not be given below, as it is entirely analogous to the proof given
by Solomon [11] and Gold [5] for the case
(l, k) = 1.
If
a
1.
5
theorem of McEliece [9] shows that the
GF(2 k ),
Eq. (1) has
the solutions forming a coset of
S
Hence, if
is any solution of (1),
y ~ GF(2(f,k».
GF(2 k )
2(l,k)
solutions in
with respect to
GF(2(f,k».
8 + Y is a solution for any
But
f
Zf 1
zf+ l
Zf
f
Tr(s + 8 +) + Tr(y + y
) + Tr(s y + SyZ )
Tr[(S + y) + (B + y)Z +1]
f
Since
y
E
GF(Z(f,k»,
so
Tr(S
Tr(y + y
f
Zf
y + Sy2 )
2 +l
)
= 0,
and
f
Tr(s2
+ S)y .
f
(SZ
Tr (l,k) /1 [Trk / (l,k)
+ 8)Y]
Zf
Tr(l,k)/l [y Trk/(f,k)(8
o
f
since
(Ref. 1, pp. 118-119).
f
+ 8) = (Trk/(f,k) 8)Z
Trk/(f,k) (SZ
+ 8)]
+ (Trk/(f,k) S) = O.
Hence, any two solutions to (1) have the same value of
Tr (x + x
Zf+ l
) ,
so
the weight is well-defined in (ii).
3.
TRANSLATION OF THE PROBLEM
The remainder of this report will be devoted to the proofs of Theorem 1
and part (i) of Theorem 2.
sider code words
a
E
By our earlier remarks, it is sufficient to con-
A such that
wee, 1) is the number of
x
g (x)
a
E
~(a)
= (c, 1), where
GF(Zk) - {a}
such that
c # O.
g (x) = 1,
a
Then
where
6
Define
as the number of
p (c)
Tr ex
a
l+l
a
Tr x
hold simultaneously.
k l
2 - _ 1
such that
X E
Then, since each equation is satisfied separately by
k
GF(2 ) - {a},
elements of
we have
k l
2[2 - - 1 - p(c)] •
w(c, 1)
(2)
Hence we can consider the function
k
GF(2 )
Regard
p(c)
as a vector space over
GF(2),
and let
Then any element
representation in the form
We associate the vector
Y = L y.1 wi'
where
X.
1
x
E
k
GF(2 )
GF (2) ,
E
(xO' xl' •.. , ~-l)
x'
w(c, 1).
in order to determine
with
i
has a unique
... ,
= 0, 1,
x.
k-l.
Then if
we have
k-l k-l
=
Tr xy
l:
x. y. Tr (w . w .)
l:
J
1
i=O j=O
1
J
or
x' T Y ,
Tr xy
where
T = (t
ij
)
is the
k x k
matrix over
GF(2)
with
t .. = Tr w.w.
1J
1 J
c'
such that
k
GF(2 ),
which
is non-singular, for otherwise there exists a non-null vector
c' T = -0' ,
implies
the null vector.
c = 0,
The mapping
S = (sij) ,
Tr ex =
a for all x
E
T
a contradiction.
a:
exists a non-singular
(Take
Then
.
x
+
x
2l
k x k
where
s ..
1J
is an automorphism of
matrix
s~.
J1
S
and
over
GF(2)
2l
w. = l: s ~ .
1
j 1J
k
GF(2);
hence there
such that
W.
J
.)
a (x) = Sx.
Thus we can write,
7
Tr xy
(3)
for any
x,y
E
k
GF(2 ).
l-
In particular, if
Xl
x o' xl' ... ,
is a quadratic form in
x
Xl
x
l.
TS x
Hence we see that
p(c)
is the
such that the equations
cl
(5)
y
~-l'
number of non-null binary k-vectors
(4)
x I TSy
T x
TS x
o
= 0
hold simultaneously.
We now associate the elements
the finite projective geometry
x =
~
Xi wi'
quadric
~I
by the correspondence
in
PG(k-l, 2).
Hence the points
satisfy (4) and (5) are the points of a quadric
Since
T
~
X if
~k-2
and Eq. (5) a
X whose coordinate vectors
Qk-2
= Qk-l
n ~k-2
in
~k-2'
is non-singular, Eq. (4) establishes a one-one correspondence between
the elements
c
E
k
GF(2 ) - {a}
thus determine the number of
of hyperplanes
Qk-2
x
of
= (x O' Xl' ..• , x k _ l ), where
Eq. (4) then represents a hyperplane
Qk-l
x
wi th the points
E
PG(k-l, 2)
X is
the coordinate vector of
X
~k-2
in
= Qk-l n ~k-2 and
and the hyperplanes of
c
such that
PG(k-l, 2)
1Qk-21
p(c) = p
such that
PG(k-l, 2).
by determining the number
1Qk-21
= p, where
is the number of points of
we shall require some results on quadrics in
We can
PG(n, 2).
Qk-2'
For this,
Proofs not given
below may be found in Bose [3] or Ray-Chaudhuri [10].
4.
RESULTS ON QUADRICS IN
A quadric
vectors
~I
Q
n
in
PG(n, 2)
PG(n, 2)
= (xO' Xl' ..• , x n )
c.onsists of all points
satisfy an equation
X whose coordinate
8
x' A x
A = (a, ,)
where
Q
n
(n + 1)
is an
1J
x
0 ,
(n + 1)
is defined to be the minimum number
matrix over
n + 1 - r
U.
such that
x' A x
The rank of
Q
n
If
If
r
>
r = 0, i. e. , i f
cone of order
r.
A + A'
by
2 [1. rank Q 1
2
n
rank Q = n + 1,
n
is degenerate of order
Q
n
0,
UO' U , ... , U + - .
l
n 1 r
is related to the rank of the symmetric matrix
is the greatest integer not exceeding
[xl
of linear forms
is expressible as a quadratic form in
rank (A + A')
where
The rank of
0, 1, ... , n+l-r
i
1
GF(2).
Q
n
x.
is a non-degenerate quadric.
In this case,
r.
Q
n
is called a
We shall find i t convenient to regard a non-degenerate
quadric as a cone of order zero.
If
n
2 - 1
n
is even, non-degenerate quadrics are of one type.
points and flat spaces of dimension
are two types of non-degenerate quadrics:
tain
2n + 2(n-l)/2 _ 1
(n-3)/2.
If
n
is odd, there
hyperbolic (ruled) quadrics con-
points and flat spaces of dimension
2n _ 2(n-l) /2 - 1
'
.
( unru 1 e d) quad'
.
e 111pt1C
r1CS conta1n
of dimension
(n-2)/2.
These contain
(n-l)/2;
points and flat spaces
(By convention, a flat space of dimension
-1
is
empty.)
If
Q
n
is a cone of order
called the vertex of
Q
n'
r,
there exists a unique
such that the points of
where
Q
n
Er - l
quadric in
n - r
dimensions obtained by intersecting
(n-r)-flat
E
n-r
to points of
which is skew to
Qn-r'
Er- 1"
If
n - r
Er- l'
are those of the lines
Q
n-r
joining points of
( r-l)-flat
is a non-degenerate
Q
n
with any
is odd,
Qn
is an
9
elliptic or hyperbolic cone according as
Q
n-r
quadric.
Q
n
We
The number
therefore
IQnl
of points on
is given by
have
2n _ 1
(6)
is an elliptic or hyperbolic
n
1
if
n - r
odd,
Qn-r
hyperbolic,
2 n _ 2 (n-r-l)/2 - 1
if
n - r
odd,
Qn-r
elliptic.
2
IQnl
A point
B
-
if
n
(n-r-l) /2
+ 2
of
PG(n, 2)
even,
r
with coordinate vector
ular (regular) with respect to
if
Q
n
_b t (A
+
b
is said to be irreg-
is null (non-null),
At)
The
set of irregular points clearly form a flat space, called the nucleus of po-
Qn .
1arity of
odd.
B
If
~
n
Q
n
If
is non-degenerate, every point is regular if
is even, the nucleus of polarity consists of a single point
More generally, i f
Qn "
1arity is the vertex
L
r-1
Q
n
if
r,
is odd.
n - r
L
r
U
l:
containing
n
Qn
Q
n
Q
n-1
~
is non-degenerate,
then the quadric
PG(n, 2),
B
L _ ,
n 1
B
where
Qn-r"
LEMMA 2:
B
and define
Let
Q
n
n
Q
n-1
is even,
=
Q n l:
n
n-l
where
B
is
In this case,
= Qn
Ln - l
is non-degenerate if and only
If
Qn '
be a non-degenerate quadric in
n l: n- 1·
is any hyperplane of
BEl: n- l'
We use these results to prove
be the nucleus of polarity of
Qn- 1
and
is the nucleus of polarity of
is a cone of order one.
and let
B,
l: r- l'
If
if
is even, the nu-
and
r-l
the nucleus of polarity of the non-degenerate quadric
r
the nucleus of pa-
is a cone of order
n - r
cleus of polarity is the r-flat
l:
is
n
Then if
Qn '
Let
BEl: n- l'
PG(n,2),
L _
n l
n even,
be a hyperplane,
10
L
B
if
1!;
E
n-
Qn-l is a cone of order one',
l' then either
2.
Qn-l
is a non-degenerate hyperbolic quadric,
3.
Qn-l
is a non-degenerate elliptic quadric.
The numbers
Nt'
or
type
= 1, 2, 3,
t
=
2
Proof:
n
for which
n-l
Q
n-l
n-l
+ 2(n-2)/2 ,
2
N
n-l
2
- 2 (n-2) /2.
The first part is simply a restatement of the above results.
termine the
point of
Q
n
PG(n-l, 2)
Nt'
we count the number of pairs
and
E
n-l
(P, En-I)
is a hyperplane containing
is contained in
2n _ 1
P.
where
n
(2 _l)2.
Counting these pairs in
=
N
l
and
= 2n - 1,
= 2n+l - 1,
the total number of hyperplanes, and
the number of hyperplanes containing
B,
is a
2 n _ 1 points
or
N + N + N
2
3
l
To de-
Since each point of
hyperplanes and there are
by (6), the number of such pairs is
P
a second way, using (6), we obtain
Since
is of
- 1,
N
2
3
Q
n
L
tare
N
l
on
of hyperplanes
we have
11
Solving these equations for
LEMMA 3:
even.
Let
L:
vertex.
Let
Then if
L:
if
L:
Let
r
L:
r-l
N , N ,
2
3
we have
be a cone of order
r
in
PG(n, 2) ,
be the nucleus of polarity of
Qn
and
Q
n
be a hyperplane and define
n-l
Is; L:
Qn-l
is a cone of order
r - 1,·
1.
Q
n-1
is a cone of order
r + 1·,
C
L
but
n-l
r-l
= L:
r
n-r
n Q
n
is
the
Qn-l = Qn n L n-l'
n-l'
o.
r-1
~
where
L:
then either
<j: Ln- 1;
r
2.
Qn-1
is a hyperbolic cone of order
3.
Q
n-l
is an elliptic cone of order
r,
or
The number
Nt'
of type
is
t
NO
=
2n+l
N
l
=
2
N
2
=
2
N
3
Proof:
L:
r-l·
If
2
2
n-r
n-l
for which
L
r-2
n-r-l
= L r-l n L: n-l·
is
,
- 2
(n-r-2)/2
then
L
n-l
.
contains an
(n-r)-flat
= Qn n Ln-r is non-degenerate in
L:
clearly consists of all points on the lines joining points of
of
Q
n-l
(n-r-2) /2
,
+ 2
n-r-1
Qn-r
n-r+l
L:
- 1,
L r-l ¢ L: n-l'
The quadric
of hyperplanes
0, 1, 2, 3,
=
t
ro
Hence,
Qn-l
is a cone of order
L
n-r
and
n-r'
Qn-r
r - 1
skew to
Qn-l
to points
with vertex
L r- 2.
Suppose now
Lr- l'
and let
L
_
r l
C
L
- •
n l
Q
- Q n L:
n-r - n
n-r
Let
L
n-r
be a fixed
(n-r)-flat skew to
There is a one-one correspondence between
12
hyperplanes
such that
E
n-l
containing
En- r - l - En _ l
E
r-l
and (n-r-l)-flats
L:
Ln - r - l = Ln_ l n L: n- r
if and only if
E
in
n-r-l
E
n-r
E
n-r'
Since
is non-degenerate, we can apply Lenuna 2 to obtain the numbers
(n-r-l) -flats
of
n-r-l
N , N , N
1
2
3
Qn-r-l = Qn-r n En-r-l
for which
Q
n-r
of
is a cone
of order one, a non-degenerate hyperbolic quadric, and a non-degenerate elliptic
quadric, respectively.
s
order
in
vertex
E
vertex of
r+s-l
L:
then
n-r'
of
r
COROLLARY 3:
sect
Q
n
n-l - L n-r-l'
E
and
r
c
Qn-r-l'
then
B
= L: r
n L:
n-r'
Let
in exactly
M(p)
p
points.
5.
and
s'
L
The
is the
s
is the nu-
B
and hence
n-l'
B
E:
L:
n-r-l
n-I
which inter-
M(p)
2n+ l _ 2n- r
n-l 1
2
n-l
2(n+r-2)/2 _
2
I
2
n - 1
L:
Then we have
n-l
+ 2 (n+r-2) /2 _ I
2
Apply (6) with
0
L:
where
E
be the number of hyperplanes
2
types
in
E
•
n-l
.e.
Proof:
r + s
is a cone of order
Qn-l
is a cone of
Qn-r-l
The proof is completed by noting that i f
cleus of polarity of
E
and
E
is the join of
Qn-l
Qn-r- I"
i f and only i f
But i f
replacing
n
n-r-l
n-r-l
+ 2
- 2
-
I
(n-r-2)!2
(n-r-2) /2
and note that the quadrics of
1 have the same number of points.
PROOF OF THEOREMS 1 AND 2
We now consider the quadric
Qk-l
x' TS x
in
O.
PG(k-l, 2)
with equation
13
The points of
for which
Tr x
Qk-l'
we recall, correspond to the elements
241
o.
Then it is clear that
and by comparison with Eq. (6), we see that
where
k - 1 - r
p(c)
is even.
is three-valued.
Qk-l
Qk-l
k - 1 - r
intersecting
E
r
LEMMA 4:
c l eus
0
Proof:
-
{o}
The points
0'
Qk-l
in the vertex
Qk-l
~
U(l,k)-l
of
= (1, k) - 1.
r
r
th e e 1emen t s
is an r-flat
The points of the nu0
f
GF(2(1,k»
b
are those whose coordinate vectors
- {o}.
of
b
satisfy
-0' ,
(TS)')
Equivalently,
b' TS x
x.
Qk-l
= O.
Tr b
E
t0
CIS
Qk-l'
correspond to the elements
denotes the null vector.
for all binary k-vectors
E
is
L(1,k)-2
for which
B of
0,
We have
Er- 1.
c orrespond
(7)
X
r
is even, the nucleus of polarity of
1:' (TS +
where
~
In order to specify the values and the number of
The points of the vertex
GF(2(1,k»
r
The previous corollary then shows that the function
The order of
f po 1 arl' t y
points,
has
is a cone of order
mapped into each value, we must determine the order
Since
X E
BEL
r
if and only if
x' TS b
Using Eq. (2), (7) holds if and only if, for all
k
GF(2 ) ,
1
2
Tr bx
Tr b
21
x
or
-1
Tr b
2
x
=
Tr b
21
x.
14
Thus
e 12
b
2l
-l
2
b.
Since
- l.
2(l,k) _ 1
eJ2
k
b
e
- 1,
l E L(k).
22l
for
= blll
E
L
Hence
GF(2(l,k».
b,
then
Conversely, if
2(l,k) - 1
contains
r
L(l,k)-2
The points of the vertex
GF(2
b
is the order of
(22l_ 1 , 2k _l) = 2(2l,k) - 1 =
divides
Thus
e
points, and
r = (l, k) - 1.
therefore
of
Hence, if
(l
)
,k) - {a}
thus correspond to the elements
l
such that
Tr b
2 +1
= O.
b
But
[2(i~~5_1](2(t,k)-1) + 2
,
l
so that
Tr b
and
and only if
b
E
GF(2(l,k»
-
2 +1
p
and
and
{a}
M(p)
Tr b
n = k - 1
On substituting the values
expressions for
Tr b.
Thus
if
O.
and
r = (l, k) - 1
into the
of the corollary, and using Eq. (2), we obtain
the weight distribution of the words
a
E
A for which
~(a)
= (c, 1), c
~
0:
No. of Code Words
Weight
l.
k-l
2
2k _ 2k- (l, k) - 1
2.
k-l
2 (k+(l,k)-2) /2
2
-
2k -(l,k)-1 + 2 (k- (l,k) -2) /2
3.
2k - l + 2 (k+(l,k)-2) /2
2k-(l,k)-1 _ 2(k-(l,k)-2)/2
We obtain the weight distribution of Theorem 1 by multiplying the numbers
in the right-hand column by
(c, 1) words, and adding
2
k
- 1
k
2(2 -l)
to include all cyclic shifts of the type
to the number so obtained for the weight
15
k-l
2
words to include the words of type
(c, 0)
and
(0, d).
This completes
the proof of Theorem 1.
Returning now to the words of type
the words of weight
2k - 1 ± 2(k+(l,k)-2)!2
6k_ 2
x
E
~!
T x = O.
But
Thus
n LCl,k)-l = L(l,k)-2.
where
Qk-2
wee,
c
~
0,
are those for which
Qk-2 = Qk-l n L _ 2
k
type 2 or 3 in Lemma 3, where
plane with equation
(c, 1),
and
r k_2
we note that
Qk-2
is of
is the hyper-
is of type 2 or 3 if and only if
k-1
1) ~ 2
if and only if, for all
GF(2(l,k»,
Tr x
0,
Tr x
1,
Tr cx
i.e., if and only if
Tr(c + l)x
o
for all
x
E
GF(2(l,k».
But
Tr(c+l)x
Tr (l,k) 11 [Trkl (l,k) (c+l)x]
=
for
x
~
x
E
GF(2(l,k»,
GF(2(l,k»
by Corollary 2,
Trk/(l,k)c
= 1.
Tr (l,k) 11 [x Trk / (l,k) (c+l)]
[1, pp.118-119].
if and only if
Tr k /(l,k)l
= 1.
Hence
Tr(c+1)x
Trk/(l,k) (c+l)
Hence
w(c, 1)
O.
~
=0
Since
k-l
2
for all
k/(l, k)
is odd
if and only 1f
This completes the proof of part (i) of Theorem 2.
16
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1.
Albert, A.A.,
Fundamental Concepts
of Chicago Press
2.
Berlekamp, E.,
3.
Bose, R.C.,
(Chicago),
~
Higher Algebra,
The University
1956.
Algebraic Coding Theory,
Combinatorial Problems
~
McGraw-Hill
(New York),
1968.
Experimental Design, Vol. I.,
John Wiley & Sons (New York), 1969.
4.
Gold, R.
"Optimal Binary Sequences for Spread-Spectrum Multiplexing",
IEEE Transactions on Information Theory, IT-13 (1967), 619-621.
5.
Gold, R.,
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Correlation Functions",
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Kasami, T.,
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7.
Kasami, T.,
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Chapter 20 in
Combinatorial Mathematics and Its Applications:
Proceedings
the Conference held at the University of North Carolina
~
at Chapel Hill,
April 10-14, 1967, (R.C. Bose, T.A. Dowling, eds.)
University of North Carolina Press.
8.
Mattson, H.F. and Solomon,
J. Soc. Ind.
9.
McEliece, R.,
~.
G.,
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Math.,
2.
(1961), 654-669.
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11.
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