Binary Cyclic codes with two primitive nonzeros

0.16 MB·English

by  Tao Feng

BINARY CYCLIC CODES WITH TWO PRIMITIVE NONZEROS
TAO FENG1, KA HINLEUNG, QINGXIANG
Abstract. In this paper, we make some progress towards a well-known conjecture on the
minimum weights of binary cyclic codes with two primitive nonzeros. We also determine the
3
1 WalshspectrumofTr(xd)overF2m inthecasewherem=2t,d=3+2t+1andgcd(d,2m−1)=1.
0
2

n
a
J 1. Introduction

1
In this paper, we are concerned with the weight distributions of binary cyclic codes with two
2
primitive nonzeros. Let q = 2m, where m ≥ 1 is an integer, and F= Fq, the finite field of size q.
] Let α be a primitive element of F, and let C be the binary cyclic code of length q−1 with two
T d
nonzeros α−1 and α−d, where d is an integer such that 1 ≤ d ≤ q−2, gcd(d,q −1) = 1. Then
I
s. Cd is a [q−1,2m]2 code, and its codewords are given by
c
[ c(a,b) = Tr(a+b),Tr(aαd+bα),...,Tr(aα(q−2)d +bαq−2) , a,b ∈ F,

1
v where Tr is the absolu(cid:0)te trace function defined on F. (cid:1)
3 Let us consider the Hamming weights of c(a,b), where a,b ∈ F. When exactly one of a,b is
7 0, the codeword c(a,b) has weight q/2. When a,b are both nonzero, c(a,b) has weight
7
4
q−2
1. 1 1−(−1)Tr(aαdi+bαi) = 1 q− (−1)Tr(xd+ba−d1x) , (1.1)
0 2 2
i=0 x∈F
3 X(cid:0) (cid:1) (cid:0) X (cid:1)
1 wherewe use1/d to denote the uniqueinteger j such that jd ≡ 1 (mod q−1) and 1≤ j ≤ q−2.
:
v Therefore, the weight distribution of C is completely determined by the Walsh spectrum of the
d
i function f : F→F , x 7→ Tr(xd), and vice versa. Here the Walsh coefficients of f are defined
X d 2 d
by
r
a
W (a) = (−1)Tr(xd+ax), a ∈ F.
d
x∈F
X
ThedistributionofW (a),a ∈ F,iscalled the Walsh spectrumoff . Theproblemofdetermining
d d
the Walsh spectrum of f is also equivalent to the problem of determining the crosscorrelations
d
of an m-sequence and its d-decimation. We refer the reader to the appendix in [9] for more
details on various formulations of this problem. A lot of work has gone into determining the
Walsh spectrum of f when d takes special forms, see [11], [4], [2], and [8]. There are a few
d
general conjectures on the Walsh spectrum of f which have proved to be quite challenging. We
d
refer the reader to the recent paper [1] for a list of these conjectures, and some recent progress
made on them.
In this paper, we are primarily interested in the following well-known conjecture due to Dilip
V. Sarwate, cf. [1]; see [3, p. 258] also.
1Research supported in part by Fundamental Research Fund for the Central Universities of China, Zhejiang
ProvincialNaturalScienceFoundation(LQ12A01019),NationalNaturalScienceFoundationofChina(11201418)
and Research Fund for Doctoral Programs from theMinistry of Education of China (20120101120089).
1
2 TAOFENG,KAHINLEUNG,QINGXIANG
Conjecture 1.1. Let m = 2t, and let C be the [2m−1,2m] binary cyclic code with two nonzeros
d
α−1 and α−d (gcd(d,2m −1) = 1), where α is a primitive element of F. Then the minimum
distance of C is ≤ 2m−1 −2t.
d
Using (1.1), the existence of a nonzero codeword of weight ≤ 2m−1 −2t is equivalent to the
existence of nonzero a ∈ F such that W (a) ≥ 2t+1. Charpin [3] showed that Conjecture 1.1 is
d
truewhend≡ 2j (mod 2t−1),forsomej,0 ≤ j ≤ t−1. (Suchd’sarecalledtheNihoexponents.)
In this paper, without putting any conditions on d (of course, gcd(d,2m − 1) = 1 is still
assumed),weshallproveanupperboundontheminimumdistanceofC ,whichisslightlyweaker
d
than the bound in Conjecture 1.1. Furthermore, we will determine the weight distributions of
C for two special classes of d; one of the two classes was previously considered by Cusick and
d
Dobbertin [4], the other class is new. Details are given in Section 3. Throughout the rest of
this paper, we shall fix m = 2t. We use Tr , Tr to denote the absolute traces defined on F and
m t
L := F , respectively. Also we use Tr (resp. N ) to denote the relative trace (resp. norm)
2t m/t m/t
from F to F . We shall drop the subscripts if we believe that no confusion will arise.
2t
2. An upper bound on the minimum weight of C
d
First,wegiveasummaryofsomewell-knownidentitiesinvolvingtheWalshcoefficientsW (a),
d
a ∈ F. We refer the reader to [7, 3, 9, 6] for the proof of these identities.
Lemma 2.1. (1) W (a) = q, W (a)2 = q2.
a∈F d a∈F d
P P
(2)
q, if u∈ F∗ ;
W (au) = 2t
d
aX∈F2t (0, if u ∈/ F2t.
Now we are ready to prove our first result.
Theorem 2.2. Let m = 2t, and let C be the [2m−1,2m] binary cyclic code with two nonzeros
d
α−1 and α−d (gcd(d,2m −1) = 1), where α is a primitive element of F. Then the minimum
distance of C is < 2m−1−2t−1−2⌊t/2⌋−1. That is, there is a nonzero a ∈ F such that W (a) >
d d
2t+2⌊t/2⌋.
Proof. For any nonzero b ∈ F\F , by direct calculations we have
2t
W (a) 1−(−1)Trm(ba)ǫ = 2m+2t|M |, (2.1)
d b b
aX∈F2t (cid:16) (cid:17)
where M = (−1)Trm((x+b)d) and ǫ = ±1 is chosen such that ǫ M = −|M |. For
b x∈F2t b b b b
b ∈ F\F , it will be convenient to introduce a function p on F defined by
2t P b 2t
p (a) := 1−(−1)Trm(ba)ǫ , ∀a ∈F .
b b 2t
Then for b ∈ F\F , we have p (a) = 2t, p (a) ≥ 0, and (2.1) can be rewritten as
2t a∈F2t b b
P
W (a)p (a) = 2m+2t|M |. (2.2)
d b b
aX∈F2t
WEIGHT DISTRIBUTIONS OF BINARY CYCLIC CODES 3
Next we compute
M2 = 2t (−1)Trm((x+b)d+bd)
b
Xb∈F Xb∈FxX∈F2t
= 2t|F|+2t (−1)Trm(xd((1+b)d+bd))
b∈Fx∈F∗
X X2t
= 2t|F|+2t 2t·|{b ∈ F |Tr ((1+b)d+bd) = 0}|−|F|
m/t
= 22t|{b ∈ F(cid:16)|(1+b)d+bd ∈ F }|. (cid:17)
2t
Since M = 2t if b ∈ F , we thus have
b 2t
M2 = 22t·|{b ∈ F\F | (1+b)d+bd ∈ F }|.
b 2t 2t
b∈XF\F2t
Let c ∈ F∗ be an element of order 2t+1. Then a system of coset representatives of (F ,+) in
2t
(F,+) is given by uc, u∈ F2t. Since Mb+x = Mb for any x ∈ F2t, and F\F2t = ∪u∈F∗ (uc+F2t),
2t
we get
M2 = 2t·|{b ∈ F\F |(1+b)d+bd ∈ F }|. (2.3)
uc 2t 2t
u∈F∗
X2t
If u∈ F∗ , we have
2t
M = (−1)Trm((x+uc)d)
uc
xX∈F2t
= (−1)Trt(cid:16)ud(cid:16)(x+c)d+(x+c2t)d(cid:17)(cid:17)
xX∈F2t
= ψ (z),
ud
zX∈Rd
whereR denotes themultiset “(x+c)d+(x+c2t)d, x ∈ F ”(each element of R indeed belongs
d 2t d
to F ), and ψ is the additive character of F defined by
2t ud 2t
ψ (x) = (−1)Trt(udx), x ∈ F .
ud 2t
We write the multiset R as a group ring element: R = a [g] ∈ Q[(F ,+)]. Then
d d g∈F2t g 2t
a = 2t, each a is a nonnegative integer, and for u∈ F∗ , M = ψ (R ). Furthermore
a∈F2t g g P2t uc ud d
note that each coefficient a of R must be even since (x+c)d +(x+c2t)d = ((x+c+c2t)+
P g d
c)d +((x+c+c2t)+c2t)d for any x ∈ F , and c+c2t 6= 0. We compute the coefficient of the
2t
identity (i.e., the zero element of F ) in R R(−1) in two ways, where R(−1) = a [−g]. In
2t d d d g∈F2t g
(−1)
fact, we have Rd = Rd here. On the one hand, this coefficient is equal to P
a2 ≥22 ·2t−1 =2t+1.
g
gX∈F2t
On the other hand, by the inversion formula (see, for example [6]), the coefficient of the identity
element in R R(−1) is equal to 1 ψ (R )2 = 1 M2 . It follows that
d d 2t u∈F2t ud d 2t u∈F2t uc
P M2 ≥ 22t+1.P
uc
uX∈F2t
4 TAOFENG,KAHINLEUNG,QINGXIANG
Using (2.3) we now obtain
(2t)2+2t·|{b ∈ F\F |(1+b)d+bd ∈ F }| ≥ 22t+1.
2t 2t
Therefore
|{b ∈ F\F | (1+b)d +bd ∈ F }| ≥ 2t,
2t 2t
with equality if and only if R has size 2t−1 as a set. As a consequence, there exists an element
d
u∈ F∗ such that
2t
|M | ≥ 22t/(2t −1) > 2⌊t/2⌋.
uc
Using the above element uc as b in Eqn. (2.2), we see that there is some a ∈ F such that
p 2t
W (a) > 2t +2⌊t/2⌋ by an averaging argument. The proof of the theorem is now complete. (cid:3)
d
Remarks. (1) In the case where d = 1+2i, for x ∈ F , we have Tr ((x+b)d) = Tr (xv)+
2t m t
Tr (bd), where v = Tr (b)2i +Tr (b)2−i. Choosing b ∈ F\F such that Tr (b) = 1, we
m m/t m/t 2t m/t
have v = 0, and |M |= 2t. We see that Conjecture 1.1 is true in this case by using (2.2).
b
(2) If d is a Niho exponent, then from [3, p. 253] we know that 2t|W (a) for all a ∈ F.
d
Combining this divisibility result with the conclusion of Theorem 2.2 that there is some a ∈ F
with W (a) > 2t +2⌊t/2⌋, we immediately get W (a) ≥ 2t+1. The same argument shows that
d d
more generally, for any d, 1 ≤ d ≤ q −2, gcd(d,q −1) = 1, such that 2t|W (a) for all a ∈ F,
d
Conjecture 1.1 is also true.
3. The Walsh spectrum of Tr(xd) with d = 1+2i+2i+t
Inthissection,weassumethatd = 1+2i+2i+t forsomei,0 < i < t−1,andgcd(d,2m−1) = 1.
Such a d is not a Niho exponent. First, we show that for any d of the aforementioned form,
Conjecture 1.1 is true. Secondly, specializing to the i = 1 case, i.e., d = 3+2t+1, we determine
the Walsh spectrum of Tr(xd) completely.
For a nonzero integer n, we use v (n) to denote the highest power of 2 dividing n.
2
Lemma 3.1. Let m = 2t and d = 1+2i+2i+t for some i, 0 < i< t−1, with gcd(d,2m−1)= 1.
Then v (i+1) ≥ v (t).
2 2
Proof. Since gcd(d,2m −1) = 1, we have gcd(2i+1 +1,2t −1) = 1. It follows that gcd(2i+1 −
1,2t−1) = gcd(22(i+1)−1,2t−1). Therefore gcd(i+1,t) = gcd(2(i+1),t), which is easily seen
to be equivalent to v (i+1) ≥ v (t). The proof is complete. (cid:3)
2 2
Let c be a fixed element of F∗ such that c 6= 1 and c2t+1 = 1. Then each element of F can
be written uniquely as x+yc with x,y ∈ L := F . We shall write c¯:= c2t, θ := c+c¯. Now we
2t
compute W (a+bc¯), where a,b ∈L. For x,y ∈L, we have
d
Tr((x+yc)d+(a+bc¯)(x+yc)) =Tr(xN (x+yc)2i +yN (x+yc)2ic+ax+by+ayc+bxc¯)
m/t m/t
=Tr (y(x2+xyθ+y2)2iθ)+Tr (ayθ+bxθ)
t t
=Tr (yx2i+1θ+y1+2iθ1+2ix2i)+Tr (y1+2i+1θ+ayθ+bxθ)
t t
=Tr (y2t−i−1θ2t−i−1 +y1+2t−iθ1+2t−i +bθ)x +Tr (y1+2i+1θ+ayθ).
t t
Therefore, (cid:16) (cid:17)
W (a+bc¯)= (−1)Trt(cid:16)(y2t−i−1θ2t−i−1+y1+2t−iθ1+2t−i+bθ)x(cid:17)+Trt(y1+2i+1θ+ayθ)
d
y∈Lx∈L
XX
= 2t (−1)Trt(y1+2i+1θ+ayθ),
y
X
WEIGHT DISTRIBUTIONS OF BINARY CYCLIC CODES 5
where the last sum is taken over
{y ∈ L|yθ+(yθ)2+2i+1 +(bθ)2i+1 = 0}.
After a change of variable, we have
W (a+bc¯) = 2t (−1)Trt(z1+2i+1θ−2i+1+az), (3.1)
d
zX∈Sb
where
S := {z ∈L|z+z2+2i+1 +(bθ)2i+1 = 0}.
b
When b = 0, we have S = {0,1} since gcd(2i+1+1,2t −1) = 1. It follows that
0
W (a) = 2t(1+(−1)Trt(θ−1+a)), ∀a∈ L.
d
Choosing a = θ−1, we have W (θ−1) = 2t+1. Thus we have proved the following:
d
Theorem 3.2. Conjecture 1.1 holds when d is of the form 1+2i +2i+t, 0 < i < t−1, and
gcd(d,2m −1) = 1.
In the case where b 6= 0, we need to solve the equation
z+z2i+1+2 = w, z ∈L,
for each w ∈ L∗. For general i, 0 < i < t−1, the solutions are complicated. We will consider
the i= 1 case below.
From now on, we assume that i = 1 (so d = 3+2t+1). By Lemma 3.1, v (t) ≤ 1; that is,
2
either t is odd or t ≡ 2 (mod 4). The equation we need to consder is now z6 +z = w, z ∈ L
and w ∈ L∗.
Assume that z ∈ L∗ is a solution to z6+z =w, w ∈ L∗. Suppose z +x is another solution
0 0
with x ∈ L∗. Now expanding (z +x)6+z +x = w gives
0 0
5 3
x x x 1
+ + = .
z z z z5
(cid:18) 0(cid:19) (cid:18) 0(cid:19) (cid:18) 0(cid:19) 0
The polynomial X5+X3+X ∈ F [X] is the Dickson polynomial D (X,1). For convenience of
2 5
the reader, we include the definition of general Dickson polynomials here. Let a ∈ F (here q
q
is an arbitrary prime power) and let n be a positive integer. We define the Dickson polynomial
D (X,a) over F by
n q
⌊n/2⌋
n n−j
D (X,a) = (−a)jXn−2j.
n
n−j j
j=0 (cid:18) (cid:19)
X
Itis well known[10] that theDickson polynomial D (X,a), a ∈ F∗, is apermutation polynomial
n q
of F if and only if gcd(n,q2−1)= 1. For more details about Dickson polynomials, we refer the
q
reader to [10].
We are now ready to determine the Walsh spectrum of Tr(xd) in the case where m = 2t, t is
odd, and d = 3+2t+1.
Theorem 3.3. Let m = 2t be a positive integer with t odd, and d = 3 + 2t+1. The Walsh
spectrum of Tr(xd) over F = F is given as follows.
2m
Proof. We have observed that X5 +X3 +X is the Dickson polynomial D (X,1). If t is odd,
5
then gcd(5,22t −1) = 1; consequently D (X,1) induces a permutation over L = F . Hence by
5 2t
the computations that we did above, |S |= 0 or 2 whent is odd and b 6=0. We already saw that
b
S = {0,1}. It follows that W (a+bc¯), a,b ∈ L, take three values only: 0, ±2t+1. Now denote
0 d
6 TAOFENG,KAHINLEUNG,QINGXIANG
Table 1.
W (·) multiplicity
d
0 3·22t−2
2t+1 22t−3 +2t−2
−2t+1 22t−3 −2t−2
by N ,N ,N the multiplicity of 0, 2t+1, −2t+1 in the Walsh spectrum of Tr(xd), respectively.
0 + −
From part (1) of Lemma 2.1, we have
N +N +N = 22t, 2t+1N −2t+1N = 22t, 22t+2N +22t+2N = 24t.
0 + − + − + −
Solving this system of equations, we get
N = 22t−22t−2, N = 22t−3 +2t−2, N = 22t−3 −2t−2.
0 + −
(cid:3)
Remarks. (1). Let t be an odd positive integer. The fact that z6+z = w, w ∈ F , has 0 or 2
2t
solutions in L is equivalent to the fact that D(6) = {(1,x,x6) | x ∈ F }∪{(0,1,0),(0,0,1)} is
2t
a hyperoval in PG(2,2t). See [5] for more details.
(2). Theorem 3.3 was first proved in [4] by a slightly different argument.
Next we consider the case where d = 3+2t+1 and t ≡2 (mod 4).
Theorem 3.4. Let m = 2t be a positive integer with v (t) = 1, t ≥ 6, and d = 3+2t+1. The
2
Walsh spectrum of Tr(xd) over F = F2m is given as follows.
Table 2.
W (·) multiplicity
d
0 22t−1 −22t−5−2t−1 +2t−3
2t 22t+2t
5
−2t 22t+2t
5
2t+1 22t−4 +2t−2
−2t+1 22t−4 −2t−2
2t+2 22t−6−2t−4
5
−2t+2 22t−6−2t−4
5
The remaining part of this paper is devoted to the proof of Theorem 3.4. From now on we
always assume that v (t) = 1 and t ≥ 6. Let G := {x ∈ F | x2t+1 = 1}. Furthermore we will
2
assume that the element c used in (3.1) to have order 5. Since t ≡ 2 (mod 4) by assumption,
we have 5|(2t +1). Thus c2t+1 = 1, i.e., c ∈ G (and c6∈ L).
Lemma 3.5. Let w ∈ L∗. Then the number of solutions z ∈ L to
z6+z = w
is 0, 1, 2 or 6.
Proof. The main difference from the t odd case is that X5+X3+X ∈ F [X] no longer induces
2
a permutation of L = F when t ≡ 2 (mod 4). We start in the same way as before. Assume
2t
WEIGHT DISTRIBUTIONS OF BINARY CYCLIC CODES 7
that z ∈L∗ is a solution to z6+z = w, w ∈ L∗. Supposez +x is another solution with x ∈ L∗.
0 0
Then expanding (z +x)6+z +x = w gives
0 0
5 3
x x x 1
+ + = , (3.2)
z z z z5
(cid:18) 0(cid:19) (cid:18) 0(cid:19) (cid:18) 0(cid:19) 0
which has 0, 1, or 5 solutions in L when v (t) = 1 and t ≥ 6. This can be seen as follows.
2
It is well known that each element y of L∗ can be written in the form u+ 1, with u ∈ L∗ or
u
u∈ G, according as Tr (1/y) is equal to 0 or 1 (see [10]). Now if u+ 1 ∈ L is a solution to (3.2),
t u
then so are γu+ 1 , γ ∈ F∗ and γ5 = 1, since D (u+ 1,1) = u5 + 1 . When u ∈ L∗, γu+ 1
γu 5 u u5 γu
is in L if and only if γ = 1. When u ∈ G, any choice of γ (γ5 = 1) will give γx+ 1 ∈ L. This
γx
proves the claim that (3.2) has 0, 1 or 5 solutions in L. The conclusion of the lemma follows as
a consequence. (cid:3)
From Lemma 3.5 and (3.1), we see that the Walsh coefficients of Tr(x3+2t+1) are in {±i·2t |
i = 0,1,2,4,6}. We use N to denote the number of a+bc¯∈ F such that W (a+bc¯) = i·2t, for
i d
i ∈ {0,±1,±2,±4,±6}.
3.1. Now, we examine for which w ∈ L∗, z6 +z = w, has six solutions in L. Assume that z
0
and x are as in the proof of Lemma 3.5. By the above analysis, there exists u ∈ G such that
x = u+1, and 1 = u5+ 1 , i.e., z5 = 1 . Since gcd(5,2t−1) = 1, we get z = 1 .
z0 u z05 u5 0 u−5+u5 0 (u−5+u5)1/5
The other five solutions are
1 1
1+uγ + , γ5 = 1.
(u−5+u5)1/5 uγ
(cid:18) (cid:19)
Therefore, z6+z = w, w ∈ L∗, has six solutions in L if and only if w is in the following set
1
T := {z6+z|z = , u ∈G, u5 6= 1}.
6 (u−5+u5)1/5
The set T has size 2t+1−5 = 2t−2−1 : the factor 5 in the denominator comes from the fact that
6 5·2·6 15
u 7→ u5 is 5-to-1 on G; the factor 6 comes from the fact z 7→ z6 + z is 6-to-1 on the set in
consideration; and the factor 2 comes from the fact u and u−1 give the same element. In this
case, with (bθ)4 =w, W (a+bc¯)∈ {±i·2t |i = 0,2,4,6}.
d
Next, we examine for which w ∈ L, z6+z = w has two solutions in L. Clearly, when w = 0,
this equation has two solutions in L. So in what follows we consider the case where w 6= 0.
Assume that z and x are as in the proof of Lemma 3.5. By the same analysis, there exists
0
u∈ L∗ such that x = u+ 1, and 1 = u5+ 1 , i.e., z5 = 1 . Therefore, z6+z = w, w ∈ L,
z0 u z05 u5 0 u−5+u5
has two solutions in L if and only if w is in the following set
1
T := {z6+z|z = , u∈ L\F }∪{0}.
2 4
(u−5+u5)1/5
Theset T has size 2t−4+1= 2t−2. Inthis case, with (bθ)4 = w, W (a+bc¯) ∈ {±i·2t :i = 0,2}.
2 2·2 d
Itnowfollowsthatthereare2t−2·2t−2−6·2t−4 = 2t+1+2 elements w ∈ Lsuchthatz6+z = w
60 5
has only one solution in L. Only these w will give the values W (a +bc¯) = ±2t (again with
d
(bθ)4 = w). We observe that the two values, 2t and −2t, occur for equally many a ∈ L, since
for the unique solution z ∈ L∗ to z6 +z = w, half of the a’s in L satisfy Tr (az ) = 0 and the
0 t 0
other half satisfy Tr (az ) = 1. Therefore we have
t 0
2t+1 +2 22t +2t
N = N = 2t−1 · = .
1 −1
5 5
8 TAOFENG,KAHINLEUNG,QINGXIANG
Finally we note that the number of w ∈ L such that z6 +z = w has no solutions in L at all
is equal to 2t− 2t−2−1 −2t−2− 2t+1+2 = 2t−1.
15 5 3
3.2. We now show that W (a + bc¯) 6= ±6 · 2t for all a,b ∈ L. As seen above, only when
d
z6+z = w, w = (bθ)4 ∈ L∗, has 6 solutions in L, could W (a+bc¯) possibly be equal to ±6·2t.
d
Let z = 1 ∈ L∗, u ∈ G, be a solution to z6 +z = w, w = (bθ)4 ∈ L∗. The other five
0 (u−5+u5)1/5
solutions are z = z +x ∈ L, with xj = uγj + 1 , 1 ≤ j ≤ 5, o(γ) = 5, u ∈ G. The fact that
j 0 j z0 uγj
±6·2t won’t occur as Walsh coefficients of Tr(xd) amounts to the fact that the following system
of equations does not have a solution a ∈ L:
Tr (z5θ−4+az ) = Tr (z5θ−4+az ), 1 ≤ j ≤ 5.
t j j t 0 0
We will prove the latter fact by way of contradiction. Assume that the above system has a
solution a ∈ L. With z = x +z , we get
j j 0
Tr x z4θ−4+z2t−2θ−1+a = Tr (x5θ−4), 1 ≤ j ≤ 5.
t j 0 0 t j
(cid:16) (cid:16) (cid:17)(cid:17)
Since xj = uγj + 1 =Tr (uγj), we have
z0 uγj m/t
Tr uγjz z4θ−4+z2t−2θ−1+a = Tr u5+u3γ3j z5θ−4 , 1 ≤ j ≤5.
m 0 0 0 m 0
(cid:16) (cid:16) (cid:17)(cid:17)
(cid:0)(cid:0) (cid:1) (cid:1)
Now, we rewrite the above equations as
Tr γjU = V +Tr γ3jW , 1 ≤ j ≤ 5.
4 4
where (cid:0) (cid:1) (cid:0) (cid:1)
U := Tr (uz (z4θ−4+z2t−2θ−1+a)) = Tr u θ−4+ u θ−1+uz a ,
m/4 0 0 0 m/4 u5+u−5 (u5+u−5)1/4 0
(cid:18) (cid:19)
u5
V := Tr (u5z5θ−4) = Tr θ−4 = Tr (θ−1),
m 0 m u5+u−5 t
(cid:18) (cid:19)
u3
W := Tr (u3z5θ−4) =Tr θ−4 .
m/4 0 m/4 u5+u−5
(cid:18) (cid:19)
Taking summation of the above equations over 1 ≤ j ≤ 5, we get V = 0. However, as we
stated before,Tr (θ−1) = 1sinceθ = c+c−1 withc ∈ G. Thiscontradiction completes theproof.
t
3.3. (1) We now compute N and N . As we have seen above, W (a+bc¯) = ±2t+2 if and only
4 −4 d
if z6+z = w, w = (bθ)4 ∈ L∗, has 6 solutions in L, and for some i ∈ {0,1,...,5} the following
0
equations hold:
Tr (z5θ−4+az )= Tr (z5θ−4+az )+1, 0 ≤j ≤ 5, j 6= i .
t j j t i0 i0 0
Without loss of generality we may assume that i = 0. Similar to the above computations, we
0
can rewrite the above equations as
Tr γjU = Tr γ3jW , 1 ≤ j ≤ 5,
4 4
where U,W are the same as abo(cid:0)ve. It(cid:1)follows(cid:0)that (cid:1)
Tr γjU = Tr γjW2 , 1 ≤ j ≤ 5.
4 4
(cid:0) (cid:1) (cid:0) (cid:1)
WEIGHT DISTRIBUTIONS OF BINARY CYCLIC CODES 9
Since γj, 1≤ j ≤ 5, span F , we obtain that U = W2, i.e.,
24
u
Tr (uz a) = Tr a
m/4 0 m/4 (u5+u−5)1/5
(cid:18) (cid:19)
u u u6
= Tr θ−4+ θ−1+ θ−8
m/4 u5+u−5 (u5+u−5)1/4 u10+u−10
(cid:18) (cid:19)
By assumption c has order 5, it follows that θ = c+c¯has order 3. We have
u u u6
Tr (uz a)= Tr θ2+ θ2+ (θ2+1)
m/4 0 m/4 u5+u−5 (u5+u−5)1/4 u10+u−10
(cid:18) (cid:19)
u u16 u6 u6
= θ2Tr + + +Tr
m/4 u5+u−5 u20+u−20 u10+u−10 m/4 u10+u−10
(cid:18) (cid:19) (cid:18) (cid:19)
u u−4 u3 2
= θ2Tr + +Tr
m/4 u5+u−5 u20+u−20 m/4 u5+u−5
(cid:18) (cid:19) (cid:18) (cid:19)
u+u−1 u−1 u3 2
= θ2Tr +θ2Tr +Tr
m/4 u5+u−5 m/2 u5+u−5 m/4 u5+u−5
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
u+u−1 u+u−1 u3 2
= θ2Tr +θ2Tr +Tr
t/2 u5+u−5 t/2 u5+u−5 m/4 u5+u−5
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
u3 2
= Tr .
m/4 u5+u−5
(cid:18) (cid:19)
2
Conversely, if Tr (uz a)= Tr u3 , a ∈ L, and z6+z = w, w = (bθ)4 ∈ L∗, has 6
m/4 0 m/4 u5+u−5
solutions in L, then W (a+bc¯)= ±2t+(cid:16)2. (cid:17)
d
Below we will count the number of solutions to
u3 2
Tr (uz a)= Tr , a ∈ L. (3.3)
m/4 0 m/4 u5+u−5
(cid:18) (cid:19)
2
Write Tr u3 = h+gγ with h,g ∈ F and
m/4 u5+u−5 22
(cid:16) (cid:17) u
uz = = α+βγ, α,β ∈L = F , o(γ) = 5.
0 (u5+u−5)1/5 2t
We claim thatα/β 6∈ F∗. Otherwise, uisinF∗ ·F∗ andthushasorderdividinglcm(15,2t−1) =
4 24 2t
5(2t −1). Noting that u has order dividing 2t +1, we have u5 = 1, which is a contradiction.
Now (3.3) becomes Tr (αa)+Tr (βa)γ = h+gγ, that is,
m/4 m/4
Tr (αa) = h, Tr (βa) = g.
t/2 t/2
Since α/β 6∈ F∗, this system of equations clearly has 2t−4 solutions a ∈ L.
4
We thus have
2t−2 −1 22t−5 −2t−3
N +N =6·2t−4· = .
4 −4
15 5
(2) Let b ∈ L∗ be such that z6+z = w, w = (bθ)4 ∈ L∗, has 6 solutions in L. Assume that the
six solutions are z , 0 ≤ j ≤ 5, as given above. We claim that for each i ∈ {0,1,...,5} there
j 0
exists an x ∈ L such that
Tr (uz x) = 0, Tr (z x) = 1, ∀j, 0 ≤ j ≤ 5. (3.4)
m/4 i0 t j
10 TAOFENG,KAHINLEUNG,QINGXIANG
An immediate consequence is that N = N ; this can be seen as follows: If W (a+bc¯)= 4·2t,
4 −4 d
a,b ∈ L, then W (x+a+bc¯) = −4·2t since every term in the sum on the right hand side of
d
2
(3.1) is negated and Tr (uz (x+a)) = Tr (uz a) = Tr u3 . We thus conclude
m/4 i0 m/4 i0 m/4 u5+u−5
that (cid:16) (cid:17)
22t−6 −2t−4
N = N = .
4 −4
5
Nowweprovetheclaimabouttheexistenceofsolutionof(3.4). Again, withoutlossofgenerality
we assume that i = 0. Multiplying both sides of Tr (uz x) = 0 by γj and taking trace to
0 m/4 0
F , we get
2
Tr (x x) = 0, ∀1 ≤ j ≤ 5.
t j
As above, writing uz = α+βγ, α,β ∈ L, o(γ) = 5, and noting that z = x +z , for 1 ≤ j ≤ 5,
0 j j 0
we see that the system of equations under consideration reduces to
Tr (αx) =0, Tr (βx) = 0, Tr (z x) = 1,
t/2 t/2 t 0
We prove that this system of equations has a solution by showing that z does not lie in the
0
F -linear span of α and β. Raising uz = α+βγ to the 2t-th power gives u−1z = α+βγ−1.
4 0 0
We solve that
uγ−1+u−1γ u+u−1
α= z , β = z .
γ +γ−1 0 γ+γ−1 0
Suppose to the contrary that there exist r,s ∈ F such that rα+sβ = z . After expansion we
4 0
get
u2(r+sγ−1)+u(γ+γ−1)+(r+sγ) =0.
This is a degree 2 equation with coefficients in F . Since u ∈ F and 2||t, we have u ∈ F∗ .
24 22t 16
Hence u5 = 1, which is impossible.
3.4. It remains to determine N , N , N . By Lemma 2.1, we have the following equations
0 2 −2
22t−5 −2t−3 22t +2t
N +N +N = 22t− −2· = 19·22t−5 −3·2t−3;
0 2 −2
5 5
2t+1(N −N ) = 22t;
2 −2
22t−5 −2t−3 22t +2t
22t+2(N +N ) = 24t− ·22t+4−2· ·22t = 24t−1.
2 −2
5 5
Solving these equations, we get
N = 22t−1 −22t−5 −2t−1+2t−3, N = 22t−4 +2t−2, N = 22t−4 −2t−2.
0 2 −2
The proof of Theorem 3.4 is now complete.
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