C HAPTER 4: R ESULTS AND D ISCUSSIONS

4-1 Cyclotomic Cosets

By narrowing down our focus on prime numbers pas modulus in constructing cyclo- tomic cosets modulo p, we observed a certain pattern on two types of primes p. We consider the primespwhere2has orderp−1modulopor ^p−1₂ modulop. Examples ofp that has orderp−1modulop is3,5,11,13,19,29,37whereas7,17,23,41and 47have ^p−1₂ modulop. Based on the primesp, we observed a pattern on the number of elements for the cyclotomic cosets modulop² andp³. The following theorems are the result of our observations.

Theorem 4.1. Letp be an odd prime. Suppose2is a primitive root modulopand 2 has orderp²−pmodulop². Then, there are exactly two distinct nonzero2-cyclotomic cosets modulop².

Proof. The2-cyclotomic coset modulop²with coset representative1is C₁ ={1,2,2², ...,2^t1−1},

where2^t1 ≡ 1(modp²). Since2has orderp²−pmodulop², i.e. 2^p2−p ≡ 1(modp²), then we havet₁ =p² −p, and so|C₁|=p²−p.

Next, since all the elements in C₁ is either 1 or even, we see thatpbeing an odd prime is not inC₁. Thus, we construct the second 2-cyclotomic coset modulop² with coset representativep, that is,

C_p ={p,2p,2²p, ...,2^tp−1p},

where2^tpp≡p(modp²). The condition2^tpp≡p(modp²) is equivalent top²|2^tpp−p, that is,p|2^tp−1. Since2is a primitive root modulop, thentp =φ(p) =p−1where φis the Euler-phi function, and so|C_p|=p−1. Now we have

C1 ={1,2,2², ...,2^p2−p−1}, and Cp ={p,2p,2²p, ...,2^p−2p},

23

If y ∈ C₁ ∩ C_p, then y ∈ C₁ and y ∈ C_p, and so y = 2^j and y = 2^kp for some integersj, k(w.l.o.g we may assume thatj > k). Then, we have2^j = 2^kpthat implies 2^j−k = p, which is a contradiction as an odd prime pcannot be even. Therefore, we conclude thatC₁∩C_p =∅. Next, clearly for allz ∈Zp² ⊆C₁∪C_p∪ {0}as we have

|C₁ ∪C_p ∪ {0}| = p² −p+p−1 + 1 = p² = |Zp²|. Therefore, we conclude that Zp² =C₁∪C_p∪ {0}. The results follows directly.

Theorem 4.2. Letpbe an odd prime. Suppose2has orderp³−p² modulop³. Then, there are exactly three distinct nonzero 2-cyclotomic cosets modulop³.

Proof. Given2has orderp³−p²modulop³, then

2^p3−p2 ≡1(mod p³). (4.1) Thus, by Euler’s Theorem together with the definition of order, we have

2^p2−p ≡1(mod p²), (4.2)

and 2^p−1 ≡1(mod p). (4.3)

The first2-cyclotomic coset modulop³ with coset representative1is C₁ ={1,2,2², ...,2^t1−1},

where2^t1 ≡1(modp³). From (4.1), we see thatt₁ =p³−p² and so|C₁| =p³−p². Next, sincep /∈C₁, we construct

C_p ={p,2p,2²p, ...,2^tp−1p},

where2^tpp≡p(modp³) which is equivalent to2^tp ≡1(modp²) and sotp =p²−p=

|C_p|(follows from (4.2)). If p² ∈ C_p, then p² = 2ⁱpfor some i implies that p = 2ⁱ which is a contradiction. Hence, p² ∈/ Cp. Finally, as p² ∈/ C1 and p² ∈/ Cp, we consider

C_p² ={p²,2p²,2²p², ...,2^tp2−1p²},

where2^tp2p² ≡p²(modp³) which is equivalent to2^tp2 ≡1(modp) and sot_p² =p−1 (follows from (4.3)). Now, we have

|C₁|=p³−p²,

|C_p|=p²−p,

|C_p²|=p−1,

Chapter 4. Results and Discussions 25

and so

|C₁|+|C_p|+|C_p²|+|{0}|=p³−p²+p²−p+p−1 + 1 =p³ =|Zp³|.

Suppose that x ∈ C₁ ∩C_p. Then x ∈ C₁ andx ∈ C_p which impliesx = 2^j and x = 2^kp for some integersj, k (w.l.o.g assume j ≥ k). It follows that2^j = 2^kpand then 2^j−k = p, which is a contradiction as an odd primep cannot be even. Hence, C₁∩C_p =∅.

Next, suppose thaty ∈C₁∩C_p². Theny ∈C₁andy∈C_p² which impliesy= 2^j andy = 2^kp² for some integersj, k (w.l.o.g assumej ≥ k). It follows that2^j = 2^kp² implies2^j−k=p², is a contradiction as the square of odd primepis odd and cannot be even. Hence,C₁∩C_p² =∅.

Finally, suppose that z ∈ C_p ∩C_p². Then z ∈ C_p and z ∈ C_p² which implies z = p2^j and z = p²2^k for some integers j, k (w.l.o.g assume j ≥ k). It follows that p2^j = p²2^k and it implies 2^j−k = p since p 6= 0, is a contradiction. Hence, C_p∩C_p² =∅.

Therefore, we can conclude thatZp³ =C₁∪C_p∪C_p² ∪ {0}andC_i∩C_j =∅for alli, j ∈ {1, p, p²}. And hence,C₁, C_p andC_p² are the required2-cyclotomic cosets modulop³.

Theorem 4.3. Letpbe an odd prime andn ≥2. Suppose2is a primitive root modulo pⁿ. Then there are exactlynnonzero2-cyclotomic cosets modulopⁿwith

|C₁|=pⁿ−pⁿ⁻¹,

|C_p|=pⁿ⁻¹−pⁿ⁻², ...

|C_pⁿ⁻¹|=p−1.

Proof. Letpbe an odd prime and n ≥ 2. LetP(n)be the statement "Suppose 2is a primitive root modulopⁿ. Then there are exactlynnonzero cylcotomic cosets modulo pⁿ."

For the basis step, when n = 2, we shown that there are exactly2nonzero cyclo- tomic cosets modulop²by Theorem 4.1. Hence,P(2)is true.

Inductively, suppose P(k) is true for some k ≥ 2, that is, there are exactly k nonzero cyclotomic cosets modulop^kprovided2is a primitive root modulop^k. Hence,

whenn =k + 1, we suppose that2is a primitive root modulop^k+1. FromP(k), we havek nonzero cyclotomic cosets modulop^k, that isC₁, C_p, C_p², .... andC_p^k−1 and in total there arep^k−1elements. When we increase the modulus fromp^k to p^k+1, the same elements remain in the same cyclotomic cosets but the same cosets can now be filled with more elements fromZp^k+1.

Suppose there are exactlyk nonzero cyclotomic cosets modulo p^k+1. Then,C₁ ∪ C_p∪C_p²∪...∪C_p^k−1 =Zp^k+1 ={1,2, ..., p^k, p^k+ 1, ..., p^k+1−1}.Sincep^k+ 1is not a multiple ofpso it is not in C_p, C_p², ...,andC_p^k−1 and it is forced to be inC₁. Same goes for other elements in Zp^k+1 that is not a multiple ofp must be inC₁. Then, the remaining elements like{..., p^k+p, .., p^k+p², ..., p^k+p^k, ...}which are multiples ofp modulop^k+1will be in their respective cyclotomic cosets modulop^k+1. For example, p^k+p² ∈ C_p² but not inC_p because it is in the form of2^jp² instead of2ⁱpfor some integersiandj.

However,p^kis not in any of thesekcyclotomic cosets. Suppose we assume thatp^k is in one of these cosets. Then,p^k =p^l2^tfor some integersl andtwhere0≤ l < k, implies p^k−l = 2^t which is a contradiction. Hence, p^k must be the smallest integer that is in the new cyclotomic coset modulo p^k+1 that is C_p^k. This contradicts the assumption that there are exactly k nonzero cyclotomic cosets modulop^k+1. Hence, there are exactlyk+ 1nonzero cyclotomic cosets modulop^k+1.

Hence,P(k+1)is true. By mathematical induction,P(n)is true for alln ≥2.

Theorem 4.4. Let pbe an odd prime. Suppose 2has order ^p−1₂ modulop and2 has order ^p(p−1)₂ modulop². Then there are exactly4distinct nonzero 2-cyclotomic cosets modulop².

Proof. The cyclotomic coset modulop² with coset representative1is C₁ ={1,2,2², ...,2^t1−1},

where2^t1 ≡1(modp²). Since2has order ^p(p−1)₂ modulop², then we havet₁ = ^p(p−1)₂ and so|C1|= ^p(p−1)₂ .

Next, for all x ∈ C₁, x = 1 or x is even, we choose an odd prime q that is the smallest quadratic non-residue modulo pwhere q /∈ C1 andq < √

p+ 1. Thus, we next construct the second 2-cyclotomic coset modulo p² with coset representative q,

Chapter 4. Results and Discussions 27

that is,

C_q ={q,2q,2²q, ...,2^tq−1q},

where2^tqq ≡q(modp²). The condition2^tqq ≡q(modp²) can be reduced to2^tq ≡ 1 (modp²), which gives ust_q = ^p(p−1)₂ and also|C_q|= ^p(p−1)₂ .

Next, sincep /∈C₁andp /∈C_q, we consider

C_p ={p,2p,2²p, ...,2^tp−1p},

where2^tpp ≡ p (modp²) which is equivalent to 2^tp ≡ 1(modp). Since2has order

p−1

2 modulop, then we havet_p = ^p−1₂ and so|C_p|= ^p−1₂ .

Finally, aspq /∈ C₁ ∪C_q∪C_p, we construct the last cyclotomic coset modulo p² with coset representativepq, that is,

C_pq ={pq,2pq,2²pq, ...,2^tpq−1pq},

where2^tpqpq≡pq(modp²) which is equivalent to2^tpqq ≡q(modp). Similarly, it can be reduced to2^tpq ≡1(modp). Then we have thatt_pq = ^p−1₂ and so|C_pq|= ^p−1₂ . Now, we have

|C₁|=|C_q|= p(p−1) 2 ,

and |C_p|=|C_pq|= p−1

2 ,

and clearly,|C₁|+|C_q|+|C_p|+|C_pq|+|{0}| =|Zp²|. The proof that all cyclotomic cosets are mutually disjoint are similar to the previous theorems. Hence, C₁, C_q, C_p andC_pq are the required2-cyclotomic cosets modulop².

Theorem 4.5. Letpbe an odd prime. Suppose2 has order ^p2(p−1)₂ modulop³. Then there are exactly6distinct nonzero 2-cyclotomic cosets modulop³.

Proof. Given2has order ^p2(p−1)₂ modulop³, then

2^p2(p−1)2 ≡1(mod p³). (4.4)

Thus, by Euler’s Theorem together with the definition of order, we have

2^p(p−1)2 ≡1(mod p²), (4.5)

and 2^p−12 ≡1(mod p). (4.6)

The cyclotomic coset modulop³with coset representative1is C₁ ={1,2,2², ...,2^t1−1},

where2^t1 ≡1(modp³). From above, we havet1 = ^p2(p−1)₂ and so|C1|= ^p2(p−1)₂ . Next, we choose an odd primeqthat is the smallest quadratic non-residue modulo pwhereq <√

p+ 1. Obviously,q /∈C1, so we construct C_q ={q,2q,2²q, ...,2^tq−1q},

where2^tqq ≡ q (modp³) which can be reduced to2^tq ≡ 1(modp³), which gives us t_q = ^p2(p−1)₂ and so|C_q|= ^p2(p−1)₂ .

Moving forward, sincep /∈C₁∪C_q, we consider the third cyclotomic coset modulo p³with coset representativep, that is,

C_p ={p,2p,2²p, ...,2^tp−1p},

where 2^tpp ≡ p (mod p³) which is equivalent to2^tp ≡ 1 (mod p²). Then, we have t_p = ^p(p−1)₂ and so|C_p|= ^p(p−1)₂ (follows from (4.5)).

Similar to previous theorem, we considerpqwhich is not in any of the three cyclo- tomic cosets. We then construct

Cpq ={pq,2pq,2²pq, ...,2^tpq−1pq},

where 2^tpqpq ≡ pq (mod p³) which is equivalent to 2^tpqq ≡ q (mod p²). It can be reduced to2^tpq ≡1(modp). Then we have thatt_pq = ^p(p−1)₂ and so|C_pq|= ^p(p−1)₂ .

Next, asp² ∈/ C₁∪C_q∪C_p∪C_pq, we consider

C_p² ={p²,2p²,2²p², ...,2^tp2−1p²},

where 2^tp2p² ≡ p²(mod p³) which is equivalent to 2^tp2 ≡ 1(mod p) and so t_p² =

p−1

2 (follows from (4.6)). Finally, we consider p²q as it is also not in the other five cyclotomic cosets modulo p³ and hereby construct the cyclotomic coset modulo p³, that is,

C_p²_q ={p²q,2p²q,2²p²q, ...,2^tp2q−1p²q},

where2^tp2qp²q ≡ p²q(modp³) which is equivalent to 2^tp2qq ≡ q(modp). Later, it is reduced to2^tp2q ≡ 1(modp). Clearly, we see that t_p²_q = ^p−1₂ = |C_p²_q|(follows from (4.6)).

Chapter 4. Results and Discussions 29

The sum of all six cyclotomic cosets with |{0}| gives the size of Zp³. The ver- ification of all cyclotomic cosets to be mutually disjoint sets can be done using the technique in previous theorems. Hence, the required 2-cyclotomic cosets modulo p³ areC₁, C_q, C_p, C_pq, C_p², andC_p²_q.

4-2 Construction of Cyclic Codes

In this section, we will illustrate the construction of[9, k, d]and[25, k, d]cyclic codes using Theorem 4.1 while for[49, k, d] cyclic codes, we are using Theorem4.4 from previous section. Throughout this whole section, we are dealing with binary cyclic codes.

4-2-1 Construction of [9, k, d]-Cyclic Codes

Now, we construct the cyclic codes withn = 9. Using Theorem4.1, we verified that2 is a primitive root modulo3. Hence, there are exactly two nonzero cyclotomic cosets modulo9. The following are the 2-cyclotomic cosets modulo9.

C₀ ={0}, C₁ ={1,2,4,8,7,5}andC₃ ={3,6}.

To represent these cosets in term of group ring, we letΩ₁ = P

s∈C₁g^s ∈ F[Z⁹] and Ω₂ =P

r∈C3g^r ∈F[Z9]. In term of group ring, the cyclotomic cosets ofF[Z9]are Ω₀ =g⁰,

Ω₁ =g¹+g²+g⁴+g⁸ +g⁷+g⁵, Ω₂ =g³+g⁶.

We can see thatΩ²₀ = 1² = 1 = Ω₀ and easily verify thatΩ²₁ = Ω₁ andΩ²₂ = Ω₂ as well as the union of differentΩare also idempotent. These idempotent are called the generating idempotent as each of them can generate a cyclic code.

There are two ways to find the dimension and minimum distance of a cyclic code from each generating idempotent. The first way is to form a matrix by cyclic shifting the generating idempotent until it is the same as the original generating idempotent.

Then, we perform Gaussian elimination to obtain the row echelon form and get the

dimensionk. The second method to obtain the dimension is by using Theorem3.26to find the generator polynomial.

After obtaining k, we find the weight for all combination of the generator polyno- mial or idempotent; or we can list down all the elements for the cyclic code to find the minimum distance d. It is preferable to use the second method as we can avoid handling large matrix which consumes more time and effort for latter cases.

Clearly, the generator polynomial forΩ₀is1so <Ω₀> is a [9,9,1]-cyclic code.

ForΩ₁, the generator matrixGis found as follows:







1 0 1 1 0 1 1 0 1 1 1 0 1 1 0 1 1 0 0 1 1 0 1 1 0 1 1







R2→R₂+R1

−−−−−−−→







1 0 1 1 0 1 1 0 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1







R3→R₃+R2

−−−−−−−→







1 0 1 1 0 1 1 0 1 0 1 1 0 1 1 0 1 1 0 0 0 0 0 0 0 0 0





= G 0

!

From above, we found that the dimension of the cyclic code is2. So the only four code- words are the three codewords in each row of the frist matrix and the zero codeword.

Clearly, the minimum distance is6. <Ω₁> is a [9,2,6]-cyclic code.

Next, we find the generator polynomial for <Ω₂> using Theorem3.26.

g(x) =gcd(g⁴⁹−1, g³+g⁶) =gcd(g³+g⁶,1 +g³) = 1 +g³.

The dimension of the cyclic code is6and the minimum distance is2. Hence, <Ω₂> is a[9,6,2]-cyclic code.

SinceΩ₀+Ω₁is also a generating idempotent, we have the corresponding generator polynomial.

g(x) =gcd(g⁴⁹−1,1 +g¹+g²+g⁴+g⁸+g⁷+g⁵)

=gcd(1 +g¹+g²+g⁴ +g⁸+g⁷+g⁵, g³+g⁴+g⁶+g⁷)

=gcd(g³+g⁴+g⁶+g⁷,1 +g+g²)

=1 +g+g².

The dimension of this code is7and the minimum distance is2. <Ω0+Ω1> is a [9,7,2]- cyclic code. Later, we found that <Ω₀+ Ω₂> is a [9,3,3]-cyclic code as1 +g³+g⁶ divides g⁹ −1completely so the generating idempotent is the generator polynomial itself.

Chapter 4. Results and Discussions 31

Next, we obtained the generator polynomial for <Ω₁+ Ω₂> by the following steps:

g(x) =gcd(g⁴⁹−1, g¹+g²+g³+g⁴ +g⁵+g⁶+g⁷+g⁸)

=gcd(g¹+g²+g³+g⁴ +g⁵+g⁶+g⁷+g⁸,1 +g)

=1 +g.

Clearly, the dimension of the code is 9−1 = 8 while the minimum distance is 2.

Hence, <Ω₁+ Ω₂> is a [9,8,2]-cyclic code.

Finally, we have <Ω0+ Ω1+ Ω2>={000000000,111111111}which tells us that it is a [9,1,9]-binary cyclic code.

4-2-2 Construction of [25, k, d]-Cyclic Codes

Next, we consider the casep= 5. From Theorem4.1, we verified that2is a primitive root modulo5, so there are exactly two nonzero cyclotomic cosets modulo25. Hence, we letΩ₁ = P

h∈C1g^h ∈F[Z25]andΩ₂ =P

k∈C5g^k ∈ F[Z25]. Then, we see that all 2-cyclotomic cosets modulo25are as follows:

C₀ ={0}, C₅ ={5,10,20,15}and

C₁ ={1,2,4,8,16,7,14,3,6,12,24,23,21,17,9,18,11,22,19,13}.

In term of group ring, the cyclotomic cosets ofF[Z25]are Ω₀ =g⁰,

Ω₁ =g¹+g²+g⁴+g⁸+g¹⁶+g⁷+g¹⁴+g³+g⁶ +g¹²+ g²⁴+g²³+g²¹+g¹⁷+g⁹+g¹⁸+g¹¹+g²²+g¹⁹+g¹³, Ω₂ =g⁵+g¹⁰+g²⁰+g¹⁵.

Using Theorem3.26, we find the generator polynomial with respect to<Ω0 >to be g(x) =gcd(Ω₀, g²⁵−1) =gcd(1, g²⁵−1) = 1.

From Theorem3.21, this cyclic code has dimension 25. Clearly, the minimum distance here is 1as the minimum weight of all nonzero codewords for this cyclic code is 1.

Hence, we obtained a [25,25,1]-binary cyclic code.

Next, forΩ₁, we perform Gaussian elimination on the matrix formed by the gener- ating idempotent to obtaink.

















0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0

















R3→R₃+R2

R4→R4+R2

R5→R₅+R2

−−−−−−−→

















0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1

















R1→R₁+R5

R3→R3+R5

R4→R₄+R5

−−−−−−−→

















0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1

















R1→R₁+R3

−−−−−−−→

















0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1

















R1↔R₅

−−−−→

















0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1

















R1↔R2

R5→R₅+R4

−−−−−−−→

















1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0















 .

From the row echelon form of the matrix, the dimension is4. Then we have2⁴ = 16 codewords. We easily list down all the elements as follows:

<Ω₁ >={0000000000000000000000000,1101111011110111101111011, 1110111101111011110111101,1111011110111101111011110,

Chapter 4. Results and Discussions 33

0111101111011110111101111,1011110111101111011110111, 1010010100101001010010100,0101001010010100101001010, 0010100101001010010100101,1001010010100101001010010, 0100101001010010100101001,1000110001100011000110001, 1100011000110001100011000,0110001100011000110001100, 0011000110001100011000110,0001100011000110001100011}.

From these codewords, we can see that the minimum distance is10. Hence,< Ω₁ >

is a[25,4,10]-binary cyclic code.

SinceΩ₂ is a generating idempotent, we have

g(x) =gcd(g²⁵+ 1, g⁵+g¹⁰+g²⁰+g¹⁵)

=gcd(g⁵+g¹⁰+g²⁰+g¹⁵, g⁵+ 1)

=g⁵+ 1.

The corresponding cyclic code to <Ω₂> is a [25,20,2]-binary cyclic code. Similarly, Ω₀+ Ω₁ is a generating idempotent, we have the following generator polynomial.

g(x) =gcd(g²⁵+ 1,1 +g¹ +g²+g⁴+g⁸+g¹⁶+g⁷+g¹⁴+g³+g⁶+ g¹²+g²⁴+g²³+g²¹+g¹⁷+g⁹+g¹⁸+g¹¹+g²²+g¹⁹+g¹³)

=gcd(1 +g¹+g² +g⁴+g⁸+g¹⁶+g⁷+g¹⁴+g³+g⁶+g¹²+ g²⁴+g²³+g²¹+g¹⁷+g⁹+g¹⁸+g¹¹+g²²+g¹⁹+g¹³, g⁵+g⁶ +g¹⁰+g¹¹+g¹⁵+g¹⁶+g²⁰+g²¹)

=gcd(g⁵+g⁶+g¹⁰+g¹¹+g¹⁵+g¹⁶+g²⁰+g²¹,1 +g+g²+g³+g⁴)

=1 +g+g²+g³+g⁴.

Clearly, the dimension is 21. Minimum distance can be difficult to find by listing all the elements as there are 2²¹ of them. However, we can observe from the generator polynomial. By addinggⁱg(x)andgⁱ⁺¹g(x)for anyi, we still get the minimum weight of2. For example,g(x) +g·g(x) = 1 +g⁵. Hence, <Ω₀+ Ω₁> is a[25,21,2]-binary cyclic code.

SinceΩ₀ + Ω₂is a generating idempotent, we have

g(x) =gcd(g²⁵+ 1,1 +g⁵+g¹⁰+g²⁰+g¹⁵) = 1 +g⁵+g¹⁰+g²⁰+g¹⁵.

Clearly, the dimension is 5 and we can list all the elements as follows:

<Ω0+ Ω2 >={0000000000000000000000000,1000010000100001000010000, 0100001000010000100001000,0010000100001000010000100, 0001000010000100001000010,0000100001000010000100001, 0011000110001100011000110,0001100011000110001100011, 1000110001100011000110001,1100011000110001100011000, 0110001100011000110001100,0100101001010010100101001, 1010010100101001010010100,0101001010010100101001010, 0010100101001010010100101,1001010010100101001010010, 1110111101111011110111101,1111011110111101111011110, 0111101111011110111101111,1011110111101111011110111, 1101111011110111101111011,0011100111001110011100111, 1001110011100111001110011,1100111001110011100111001, 1110011100111001110011100,0111001110011100111001110, 0101101011010110101101011,1010110101101011010110101, 1101011010110101101011010,0110101101011010110101101, 1011010110101101011010110,1111111111111111111111111}.

Hence, <Ω₀+ Ω₂> is a[25,5,5]-binary cyclic code. Next, we haveΩ₁+ Ω₂ which is also a generating idempotent, we have the following

g(x) =gcd(g²⁵+ 1, g¹ +g²+g⁴+g⁸+g¹⁶+g⁷+g¹⁴+g³+g⁶+g¹²+g²⁴+ g²³+g²¹+g¹⁷+g⁹+g¹⁸+g¹¹+g²²+g¹⁹+g¹³+g⁵+g¹⁰+g²⁰+g¹⁵)

=gcd(g¹+g² +g⁴+g⁸+g¹⁶+g⁷+g¹⁴+g³+g⁶+g¹²+g²⁴+g²³+g²¹+ g¹⁷+g⁹+g¹⁸+g¹¹+g²²+g¹⁹+g¹³+g⁵+g¹⁰+g²⁰+g¹⁵,1 +g)

=1 +g.

We see that k = 24. Again, listing all the elements by hand will be very tedious.

However, we can observe from the generator polynomial and we see that the minimum distance is2. Hence, <Ω₁ + Ω₂> is a[25,24,2]-binary cyclic code. Finally, we have

<Ω0+ Ω1+ Ω2 >={0000000000000000000000000,1111111111111111111111111}

which tells us that it is a [25,1,25]-binary cyclic code.

Chapter 4. Results and Discussions 35

4-2-3 Construction of [49, k, d]-Cyclic Codes

From Theorem 4.4, we let Ω₁ = P

s∈C₁g^s ∈ F[Zp²], Ω₂ = P

r∈Cqg^r ∈ F[Zp²], Ω₃ = P

t∈Cpg^t ∈ F[Zp²] andΩ₄ = P

v∈Cpqg^v ∈ F[Zp²]. Consider the case p = 7.

Then, we see that all2-cyclotomic cosets modulo49are

C₀ ={0}, C₁ ={1,2,4,8,16,32,15,30,11,22,44,39,29,9,18,36,23,46,43,37,25}, C₃ ={3,6,12,24,48,47,45,41,33,17,34,19,38,27,5,10,20,40,31,13,26},

C₇ ={7,14,28}andC₂₁ ={21,42,35}.

In term of group ring, the cyclotomic cosets ofF[Z49]are Ω0 =g⁰,

Ω1 =g¹+g²+g⁴+g⁸+g¹⁶+g³²+g¹⁵+g³⁰+g¹¹+g²²+g⁴⁴+ g³⁹+g²⁹+g⁹ +g¹⁸+g³⁶+g²³+g⁴⁶+g⁴³+g³⁷+g²⁵, Ω2 =g³+g⁶+g¹²+g²⁴+g⁴⁸+g⁴⁷+g⁴⁵+g⁴¹+g³³+g¹⁷+g³⁴+

g¹⁹+g³⁸+g²⁷+g⁵+g¹⁰+g²⁰+g⁴⁰+g³¹+g¹³+g²⁶, Ω₃ =g⁷+g¹⁴+g²⁸andΩ₄ =g²¹+g⁴²+g³⁵.

We can easily verify that all these cyclotomic cosets in term of group ring are idempo- tent and since they generate a cyclic code, they are known as the generating idempotent.

Clearly, forΩ₀ = 1, we have a[49,49,1]-binary cyclic code.

Next, we have the following generator polynomial for<Ω₁ >.

g(x) =gcd(g⁴⁹−1, g¹+g² +g⁴+g⁸+g¹⁶+g³²+g¹⁵+g³⁰+g¹¹+g²²+ g⁴⁴+g³⁹+g²⁹+g⁹ +g¹⁸+g³⁶+g²³+g⁴⁶+g⁴³+g³⁷+g²⁵)

=gcd(g¹+g²+g⁴+g⁸+g¹⁶+g³²+g¹⁵+g³⁰+g¹¹+g²²+g⁴⁴+g³⁹+ g²⁹+g⁹+g¹⁸+g³⁶+g²³+g⁴⁶+g⁴³+g³⁷+g²⁵,1 +g+g³+g⁷+ g⁸+g¹⁰+g¹⁴+g¹⁵+g¹⁷+g²¹+g²²+g²⁴+g²⁸+g²⁹+g³¹+ g³⁵+g³⁶+g³⁸+g⁴²+g⁴³+g⁴⁵)

=gcd(1 +g+g³+g⁷ +g⁸+g¹⁰+g¹⁴+g¹⁵+g¹⁷+g²¹+g²²+g²⁴+g²⁸+ g²⁹+g³¹+g³⁵+g³⁶+g³⁸+g⁴²+g⁴³+g⁴⁵,0)

=1 +g+g³+g⁷+g⁸+g¹⁰+g¹⁴+g¹⁵+g¹⁷+g²¹+g²²+g²⁴+g²⁸+ g²⁹+g³¹+g³⁵+g³⁶+g³⁸+g⁴²+g⁴³+g⁴⁵.

From above, we see thatk = 4and we have16codewords as follows:

<Ω1>=

{0000000000000000000000000000000000000000000000000,1111111111111111111111111111111111111111111111111, 1101000110100011010001101000110100011010001101000,0110100011010001101000110100011010001101000110100, 0011010001101000110100011010001101000110100011010,0001101000110100011010001101000110100011010001101, 1000110100011010001101000110100011010001101000110,0100011010001101000110100011010001101000110100011, 1010001101000110100011010001101000110100011010001,0101110010111001011100101110010111001011100101110, 0010111001011100101110010111001011100101110010111,1001011100101110010111001011100101110010111001011, 1100101110010111001011100101110010111001011100101,1110010111001011100101110010111001011100101110010, 0111001011100101110010111001011100101110010111001,1011100101110010111001011100101110010111001011100}.

From above, we found that the minimum distance is 21. Hence, we obtained a [49,4,21]-binary cyclic code.

The generator matrix based on the generating idempotentΩ₂ is as follows and we performed Gaussian Elimination on it.













1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1













R2→R2+R1

R3→R3+R1

−−−−−−−→













1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1













R3→R₃+R2

R5→R5+R2

−−−−−−−→













1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1













R4→R4+R3

R6→R₆+R3

R7→R7+R5

−−−−−−−→













1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0













R5→R5+R4

−−−−−−−→













1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0













=G₂ 0

.

Chapter 4. Results and Discussions 37

Once again, the dimension is 4so there are 16 codewords. If we perform cyclic shift on the first two rows, we will get14different codewords. Then we include 0 and 1, we will have exactly16codewords. Hence, from the matrix G₂, d = 21. Again, we obtained a[49,4,21]-binary cyclic code.

The generator polynomial for <Ω₃> using Theorem3.26is as follows:

g(x) =gcd(g⁴⁹−1, g⁷+g¹⁴+g²⁸)

=gcd(g⁷ +g¹⁴+g²⁸,1 +g⁷+g²¹)

=gcd(1 +g⁷+g²¹,0)

=1 +g⁷+g²¹.

Based on g(x), the Hamming distance is at least 3. So, we have a[49,28,3]-binary cyclic code.

The generator polynomial for <Ω4> using Theorem3.26is as follows:

g(x) =gcd(g⁴⁹−1, g²¹+g³⁵+g⁴²)

=gcd(g²¹+g³⁵+g⁴²,1 +g²¹+g²⁸+g³⁵)

=gcd(1 +g²¹+g²⁸+g³⁵, g⁷+g²¹+g²⁸)

=gcd(g⁷+g²¹+g²⁸,1 +g¹⁴+g²¹)

=1 +g¹⁴+g²¹.

Based ong(x), the minimum distance is3. So, we have a[49,28,3]-binary cyclic code.

The generator polynomial for <Ω₀+ Ω₁> using Theorem3.26is as follows:

g(x) =gcd(g⁴⁹−1,1 +g¹ +g²+g⁴+g⁸+g¹⁶+g³²+g¹⁵+g³⁰+g¹¹+g²²+ g⁴⁴+g³⁹+g²⁹+g⁹+g¹⁸+g³⁶+g²³+g⁴⁶+g⁴³+g³⁷+g²⁵)

=gcd(1 +g¹ +g²+g⁴+g⁸+g¹⁶+g³²+g¹⁵+g³⁰+g¹¹+g²²+g⁴⁴+g³⁹+ g²⁹+g⁹+g¹⁸+g³⁶+g²³+g⁴⁶+g⁴³+g³⁷+g²⁵, g⁷ +g⁸+g¹⁰+g¹⁴+g¹⁵+ g¹⁷+g²¹+g²²+g²⁴+g²⁸+g²⁹+g³¹+g³⁵+g³⁶+g³⁸+g⁴²+g⁴³+g⁴⁵)

=gcd(g⁷+g⁸+g¹⁰+g¹⁴+g¹⁵+g¹⁷+g²¹+g²²+g²⁴+g²⁸+g²⁹+ g³¹+g³⁵+g³⁶+g³⁸+g⁴²+g⁴³+g⁴⁵,1 +g+g²+g⁴)

=1 +g+g²+g⁴.

Based ong(x), the dimension is45while the minimum distance is4. Hence, we have a[49,45,4]-binary cyclic code.

The generator polynomial for <Ω₀+ Ω₂> using Theorem3.26is as follows:

g(x) =gcd(g⁴⁹−1,1 +g³ +g⁶+g¹²+g²⁴+g⁴⁸+g⁴⁷+g⁴⁵+g⁴¹+g³³+g¹⁷+ g³⁴+g¹⁹+g³⁸+g²⁷+g⁵+g¹⁰+g²⁰+g⁴⁰+g³¹+g¹³+g²⁶)

=gcd(1 +g³ +g⁶+g¹²+g²⁴+g⁴⁸+g⁴⁷+g⁴⁵+g⁴¹+g³³+g¹⁷+g³⁴+g¹⁹+ g³⁸+g²⁷+g⁵+g¹⁰+g²⁰+g⁴⁰+g³¹+g¹³+g²⁶, g+g³+g⁴+g⁵+g⁷+ g¹⁰+g¹¹+g¹²+g¹⁴+g¹⁷+g¹⁸+g¹⁹+g²¹+g²⁴+g²⁵+g²⁶+g²⁸+g³¹+ g³²+g³³+g³⁵+g³⁸+g³⁹+g⁴⁰+g⁴²+g⁴⁵+g⁴⁶+g⁴⁷)

=gcd(g+g³+g⁴ +g⁵+g⁷+g¹⁰+g¹¹+g¹²+g¹⁴+g¹⁷+g¹⁸+g¹⁹+g²¹+ g²⁴+g²⁵+g²⁶+g²⁸+g³¹+g³²+g³³+g³⁵+g³⁸+g³⁹+g⁴⁰+g⁴²+g⁴⁵+ g⁴⁶+g⁴⁷,1 +g²+g³+g⁴+g⁸+g¹⁰+g¹¹+g¹⁵+g¹⁷+g¹⁸+g²²+g²⁴+ g²⁵+g²⁹+g³¹+g³²+g³⁶+g³⁸+g³⁹+g⁴³+g⁴⁵+g⁴⁶)

=gcd(1 +g² +g³+g⁴+g⁸+g¹⁰+g¹¹+g¹⁵+g¹⁷+g¹⁸+g²²+g²⁴+g²⁵+ g²⁹+g³¹+g³²+g³⁶+g³⁸+g³⁹+g⁴³+g⁴⁵+g⁴⁶, g⁷+g⁹+g¹⁰+g¹⁴+g¹⁶+ g¹⁷+g²¹+g²³+g²⁴+g²⁸+g³⁰+g³¹+g³⁵+g³⁷+g³⁸+g⁴²+g⁴⁴+g⁴⁵)

=gcd(g⁷+g⁹+g¹⁰+g¹⁴+g¹⁶+g¹⁷+g²¹+g²³+g²⁴+g²⁸+g³⁰+g³¹+ g³⁵+g³⁷+g³⁸+g⁴²+g⁴⁴+g⁴⁵,1 +g²+g³+g⁴)

=1 +g²+g³+g⁴.

Based on the generator polynomial, the cyclic code has a dimension of 45and Ham- ming distance at least 4. So,<Ω₀+ Ω₂ >is a[49,45,4]-binary cyclic code.

Ω₀+ Ω₃ = 1 +g⁷+g¹⁴+g²⁸dividesg⁴⁹−1completely. So, the dimension is21 andd = 4. Hence, <Ω₀+ Ω₃> is a [49,21,4]-binary cyclic code.

Next, the generator polynomial for <Ω₀+ Ω₄> is g(x) =gcd(g⁴⁹−1,1 +g²¹+g³⁵+g⁴²)

=gcd(1 +g²¹+g³⁵+g⁴², g⁷+g²¹+g²⁸+g³⁵)

=gcd(g⁷+g²¹+g²⁸+g³⁵,1 +g¹⁴+g²¹+g²⁸)

=1 +g¹⁴+g²¹+g²⁸.

Similarly, we obtained a [49,21,4]-binary cyclic code.

Chapter 4. Results and Discussions 39

Next, the generator polynomial for <Ω₁+ Ω₂> is as follows:

g(x)

=gcd(g⁴⁹−1, g¹+g²+g⁴+g⁸+g¹⁶+g³²+g¹⁵+g³⁰+g¹¹+g²²+g⁴⁴+ +g³⁹+g²⁹+g⁹+g¹⁸+g³⁶+g²³+g⁴⁶+g⁴³+g³⁷+g²⁵+g³+g⁵+g⁶+ g¹⁰+g¹²+g¹³+g¹⁷+g¹⁹+g²⁰+g²⁴+g²⁶+g²⁷+g³¹+g³³+g³⁴+ g³⁸+g⁴⁰+g⁴¹+g⁴⁵+g⁴⁷+g⁴⁸)

=gcd(g¹+g²+g³+g⁴+g⁵ +g⁶+g⁸+g⁹+g¹⁰+g¹¹+g¹²+g¹³+g¹⁵+g¹⁶+ g¹⁷+g¹⁸+g¹⁹+g²⁰+g²²+g²³+g²⁴+g²⁵+g²⁶+g²⁷+g²⁹+g³⁰+g³¹+g³²

+g³³+g³⁴+g³⁶+g³⁷+g³⁸+g³⁹+g⁴⁰+g⁴¹+g⁴³+g⁴⁴+g⁴⁵+g⁴⁶+g⁴⁷+ g⁴⁸,1 +g+g⁷+g⁸+g¹⁴+g¹⁵+g²¹+g²²+g²⁸+g²⁹+g³⁵+g³⁶+g⁴²+g⁴³)

=gcd(1 +g+g⁷ +g⁸+g¹⁴+g¹⁵+g²¹+g²²+g²⁸+g²⁹+g³⁵+g³⁶+g⁴²+g⁴³,0)

=1 +g+g⁷+g⁸+g¹⁴+g¹⁵+g²¹+g²²+g²⁸+g²⁹+g³⁵+g³⁶+g⁴²+g⁴³. The dimension of the code is 6 so there are 2⁶ codewords in this cyclic code. The following are all64codewords.

<Ω1+ Ω2>

={0000000000000000000000000000000000000000000000000,1100000110000011000001100000110000011000001100000, 0110000011000001100000110000011000001100000110000,0011000001100000110000011000001100000110000011000, 0001100000110000011000001100000110000011000001100,0000110000011000001100000110000011000001100000110, 0000011000001100000110000011000001100000110000011,1000001100000110000011000001100000110000011000001, 0010100001010000101000010100001010000101000010100,0001010000101000010100001010000101000010100001010, 0000101000010100001010000101000010100001010000101,1000010100001010000101000010100001010000101000010, 0100001010000101000010100001010000101000010100001,1010000101000010100001010000101000010100001010000, 0101000010100001010000101000010100001010000101000,0100010010001001000100100010010001001000100100010, 0010001001000100100010010001001000100100010010001,1001000100100010010001001000100100010010001001000, 0100100010010001001000100100010010001001000100100,0010010001001000100100010010001001000100100010010, 0001001000100100010010001001000100100010010001001,1000100100010010001001000100100010010001001000100, 1110010111001011100101110010111001011100101110010,0111001011100101110010111001011100101110010111001, 1011100101110010111001011100101110010111001011100,0101110010111001011100101110010111001011100101110, 0010111001011100101110010111001011100101110010111,1001011100101110010111001011100101110010111001011, 1100101110010111001011100101110010111001011100101,1110100111010011101001110100111010011101001110100, 0111010011101001110100111010011101001110100111010,0011101001110100111010011101001110100111010011101, 1001110100111010011101001110100111010011101001110,0100111010011101001110100111010011101001110100111, 1010011101001110100111010011101001110100111010011,1101001110100111010011101001110100111010011101001, 0011110001111000111100011110001111000111100011110,0001111000111100011110001111000111100011110001111, 1000111100011110001111000111100011110001111000111,1100011110001111000111100011110001111000111100011,

1110001111000111100011110001111000111100011110001,1111000111100011110001111000111100011110001111000, 0111100011110001111000111100011110001111000111100,1101100110110011011001101100110110011011001101100, 0110110011011001101100110110011011001101100110110,0011011001101100110110011011001101100110110011011, 1001101100110110011011001101100110110011011001101,1100110110011011001101100110110011011001101100110, 0110011011001101100110110011011001101100110110011,1011001101100110110011011001101100110110011011001, 1010110101011010101101010110101011010101101010110,0101011010101101010110101011010101101010110101011, 1010101101010110101011010101101010110101011010101,1101010110101011010101101010110101011010101101010, 0110101011010101101010110101011010101101010110101,1011010101101010110101011010101101010110101011010, 0101101010110101011010101101010110101011010101101,1111101111110111111011111101111110111111011111101, 1111110111111011111101111110111111011111101111110,0111111011111101111110111111011111101111110111111, 1011111101111110111111011111101111110111111011111,1101111110111111011111101111110111111011111101111, 1110111111011111101111110111111011111101111110111,1111011111101111110111111011111101111110111111011}.

Based on the list of elements above, the minimum distance is14. <Ω₁ + Ω₂> is a [49,6,14]-binary cyclic code.

We constructed the generator polynomial for <Ω₁ + Ω₃> as follows:

g(x) =gcd(g⁴⁹−1, g¹ +g²+g⁴+g⁷+g⁸+g¹⁴+g¹⁶+g³²+g¹⁵+g³⁰+g¹¹+ g²²+g²⁸+g⁴⁴+g³⁹+g²⁹+g⁹ +g¹⁸+g³⁶+g²³+g⁴⁶+g⁴³+g³⁷+g²⁵)

=gcd(g¹+g²+g⁴+g⁷+g⁸ +g¹⁴+g¹⁶+g³²+g¹⁵+g³⁰+g¹¹+g²²+ g²⁸+g⁴⁴+g³⁹+g²⁹+g⁹+g¹⁸+g³⁶+g²³+g⁴⁶+g⁴³+g³⁷+g²⁵, 1 +g+g³+g²¹+g²²+g²⁴+g³⁵+g³⁶+g³⁸+g⁴²+g⁴³+g⁴⁵)

=gcd(1 +g+g³ +g²¹+g²²+g²⁴+g³⁵+g³⁶+g³⁸+g⁴²+g⁴³+g⁴⁵, g⁷+g⁸+g⁹+g¹¹+g¹⁴+g¹⁵+g¹⁶+g¹⁸+g²⁸+g²⁹+g³⁰+g³²)

=gcd(g⁷+g⁸+g⁹+g¹¹+g¹⁴+g¹⁵+g¹⁶+g¹⁸+g²⁸+g²⁹+g³⁰+g³², 1 +g+g³+g¹⁴+g¹⁵+g¹⁷+g²¹+g²²+g²⁴+g²⁸+g²⁹+g³¹)

=gcd(1 +g+g³ +g¹⁴+g¹⁵+g¹⁷+g²¹+g²²+g²⁴+g²⁸+g²⁹+g³¹, g+g² +g⁴+g⁷+g⁸+g⁹+g¹¹+g¹⁴+g²²+g²³+g²⁵+g²⁸)

=gcd(g+g²+g⁴+g⁷+g⁸+g⁹+g¹¹+g¹⁴+g²²+g²³+g²⁵+g²⁸, 1 +g+g²+g⁴+g⁷ +g⁸+g⁹+g¹¹+g²¹+g²²+g²³+g²⁵)

=gcd(1 +g+g² +g⁴+g⁷+g⁸+g⁹+g¹¹+g²¹+g²²+g²³+g²⁵,0)

=1 +g+g²+g⁴+g⁷ +g⁸+g⁹+g¹¹+g²¹+g²²+g²³+g²⁵.

From the degree of the generator polynomial, the dimension is 24and the Hamming distance should be at least12. <Ω₁+ Ω₃> is a[49,24,12]-binary cyclic code.

Chapter 4. Results and Discussions 41

Next, the <Ω₁+ Ω₄> has the following generator polynomial:

g(x) =gcd(g⁴⁹−1, g¹ +g²+g⁴+g⁸+g¹⁶+g³²+g¹⁵+g³⁰+g¹¹+g²¹+g²²+ g³⁵+g⁴²+g⁴⁴+g³⁹+g²⁹+g⁹ +g¹⁸+g³⁶+g²³+g⁴⁶+g⁴³+g³⁷+g²⁵)

=gcd(g¹+g²+g⁴+g⁸+g¹⁶+g³²+g¹⁵+g³⁰+g¹¹+g²¹+g²²+g³⁵+ g⁴²+g⁴⁴+g³⁹+g²⁹+g⁹+g¹⁸+g³⁶+g²³+g⁴⁶+g⁴³+g³⁷+g²⁵, 1 +g+g³+g⁷+g⁸ +g¹⁰+g¹⁴+g¹⁵+g¹⁷+g²⁸+g²⁹+g³¹)

=gcd(1 +g+g³ +g⁷+g⁸+g¹⁰+g¹⁴+g¹⁵+g¹⁷+g²⁸+g²⁹+g³¹, g+g² +g⁴+g⁷+g¹⁵+g¹⁶+g¹⁸+g²¹+g²²+g²³+g²⁵+g²⁸)

=gcd(g+g²+g⁴+g⁷+g¹⁵+g¹⁶+g¹⁸+g²¹+g²²+g²³+g²⁵+g²⁸, 1 +g+g²+g⁴+g¹⁴+g¹⁵+g¹⁶+g¹⁸+g²¹+g²²+g²³+g²⁵)

=gcd(1 +g+g² +g⁴+g¹⁴+g¹⁵+g¹⁶+g¹⁸+g²¹+g²²+g²³+g²⁵,0)

=1 +g+g²+g⁴+g¹⁴+g¹⁵+g¹⁶+g¹⁸+g²¹+g²²+g²³+g²⁵.

From the degree of the generator polynomial, the dimension is 24and the Hamming distance should be at least12. <Ω₁+ Ω₄> is a[49,24,12]-binary cyclic code.

Next, we haveΩ3 + Ω4 = g⁷ +g¹⁴+g²¹+g²⁸+g³⁵+g⁴². We constructed the generator polynomial as follows:

g(x) = gcd(g⁴⁹−1, g⁷+g¹⁴+g²¹+g²⁸+g³⁵+g⁴²)

=gcd(g⁷ +g¹⁴+g²¹+g²⁸+g³⁵+g⁴²,1 +g⁷)

=gcd(1 +g⁷,0)

= 1 +g⁷.

Clearly,<Ω₃+ Ω₄> is a [49,42,2]-binary cyclic code.

By addingΩ₀ to the previous generating idempotent, we have1 +g⁷+g¹⁴+g²¹+ g²⁸+g³⁵+g⁴²which is a factor ofg⁴⁹−1as we found that

(1 +g⁷+g¹⁴+g²¹+g²⁸+g³⁵+g⁴²)(1 +g⁴) = g⁴⁹−1.

Hence, <Ω₀+ Ω₃+ Ω₄> is a[49,7,7]-binary cyclic code.

We found that < Ω₂ + Ω₄ > is a [49,24,12]-binary cyclic code based on the

generator polynomial with degree25and weight12found below.

g(x) =gcd(g⁴⁹−1, g³+g⁵+g⁶+g¹⁰+g¹²+g¹³+g¹⁷+g¹⁹+g²⁰+g²¹+g²⁴+ g²⁶+g²⁷+g³¹+g³³+g³⁴+g³⁵+g³⁸+g⁴⁰+g⁴¹+g⁴²+g⁴⁵+g⁴⁷+g⁴⁸)

=gcd(g³+g⁵+g⁶+g¹⁰+g¹²+g¹³+g¹⁷+g¹⁹+g²⁰+g²¹+g²⁴+g²⁶+ g²⁷+g³¹+g³³+g³⁴+g³⁵+g³⁸+g⁴⁰+g⁴¹+g⁴²+g⁴⁵+g⁴⁷+g⁴⁸,1 +g³

+g⁴+g⁵+g⁷+g¹⁰+g¹¹+g¹²+g¹⁴+g¹⁷+g¹⁸+g¹⁹+g²²+g²⁴+g²⁵+ g²⁶+g²⁸+g³¹+g³²+g³³+g³⁶+g³⁸+g³⁹+g⁴⁰+g⁴³+g⁴⁵+g⁴⁶+g⁴⁷)

=gcd(1 +g³ +g⁴+g⁵+g⁷+g¹⁰+g¹¹+g¹²+g¹⁴+g¹⁷+g¹⁸+g¹⁹+g²²+ g²⁴+g²⁵+g²⁶+g²⁸+g³¹+g³²+g³³+g³⁶+g³⁸+g³⁹+g⁴⁰+g⁴³+g⁴⁵+ g⁴⁶+g⁴⁷, g+g³+g⁴+g⁸+g¹⁰+g¹¹+g¹⁵+g¹⁷+g¹⁸+g²¹+g²³+g²⁴+ g²⁵+g²⁹+g³¹+g³²+g³⁵+g³⁷+g³⁸+g³⁹+g⁴²+g⁴⁴+g⁴⁵+g⁴⁶)

=gcd(g+g³+g⁴ +g⁸+g¹⁰+g¹¹+g¹⁵+g¹⁷+g¹⁸+g²¹+g²³+g²⁴+g²⁵+ g²⁹+g³¹+g³²+g³⁵+g³⁷+g³⁸+g³⁹+g⁴²+g⁴⁴+g⁴⁵+g⁴⁶,

1 +g²+g³+g⁷+g⁹+g¹⁰+g¹⁴+g¹⁶+g¹⁷+g²⁸+g³⁰+g³¹)

=gcd(1 +g² +g³+g⁷+g⁹+g¹⁰+g¹⁴+g¹⁶+g¹⁷+g²⁸+g³⁰+g³¹, g+g³+ g⁴ +g⁷+g⁸+g⁹+g¹⁵+g¹⁷+g¹⁸+g²¹+g²³+g²⁴+g²⁵+g²⁸+g²⁹+g³⁰)

=gcd(g+g³+g⁴ +g⁷+g⁸+g⁹+g¹⁵+g¹⁷+g¹⁸+g²¹+g²³+g²⁴+g²⁵+ g²⁸+g²⁹+g³⁰,1 +g³ +g⁴+g⁵+g⁷+g⁸+g¹⁴+g¹⁷+g¹⁸+g¹⁹+g²²+ g²⁴+g²⁵+g²⁶+g²⁸+g²⁹)

=gcd(1 +g³ +g⁴+g⁵+g⁷+g⁸+g¹⁴+g¹⁷+g¹⁸+g¹⁹+g²²+g²⁴+g²⁵+g²⁶ +g²⁸+g²⁹, g³+g⁵+g⁶+g⁷+g¹⁷+g¹⁹+g²⁰+g²¹+g²⁴+g²⁶+g²⁷+g²⁸)

=gcd(g³+g⁵+g⁶+g⁷+g¹⁷+g¹⁹+g²⁰+g²¹+g²⁴+g²⁶+g²⁷+g²⁸, 1 +g³+g⁵+g⁶+g¹⁴+g¹⁷+g¹⁹+g²⁰+g²¹+g²⁴+g²⁶+g²⁷)

=gcd(1 +g³ +g⁵+g⁶+g¹⁴+g¹⁷+g¹⁹+g²⁰+g²¹+g²⁴+g²⁶+g²⁷, g+g³+g⁴+g⁵+g¹⁵+g¹⁷+g¹⁸+g¹⁹+g²²+g²⁴+g²⁵+g²⁶)

=gcd(g+g³+g⁴ +g⁵+g¹⁵+g¹⁷+g¹⁸+g¹⁹+g²²+g²⁴+g²⁵+g²⁶, 1 +g²+g³+g⁴+g¹⁴+g¹⁶+g¹⁷+g¹⁸+g²¹+g²³+g²⁴+g²⁵)

=gcd(1 +g² +g³+g⁴+g¹⁴+g¹⁶+g¹⁷+g¹⁸+g²¹+g²³+g²⁴+g²⁵,0)

=1 +g²+g³+g⁴+g¹⁴+g¹⁶+g¹⁷+g¹⁸+g²¹+g²³+g²⁴+g²⁵.

Chapter 4. Results and Discussions 43

Next, we constructed the generator polynomial based on the two generating idem- potentΩ₂+ Ω₃ and the result is as follows:

g(x) =gcd(g⁴⁹−1, g³+g⁵+g⁶+g⁷+g¹⁰+g¹²+g¹³+g¹⁴+g¹⁷+g¹⁹+g²⁰+ g²⁴+g²⁶+g²⁷+g²⁸+g³¹+g³³+g³⁴+g³⁸+g⁴⁰+g⁴¹+g⁴⁵+g⁴⁷+g⁴⁸)

=gcd(g³+g⁵+g⁶+g⁷+g¹⁰+g¹²+g¹³+g¹⁴+g¹⁷+g¹⁹+g²⁰+g²⁴+g²⁶ +g²⁷+g²⁸+g³¹+g³³+g³⁴+g³⁸+g⁴⁰+g⁴¹+g⁴⁵+g⁴⁷+g⁴⁸,1 +g³+ g⁴ +g⁵+g⁸+g¹⁰+g¹¹+g¹²+g¹⁵+g¹⁷+g¹⁸+g¹⁹+g²¹+g²⁴+g²⁵+ g²⁶+g²⁹+g³¹+g³²+g³³+g³⁵+g³⁸+g³⁹+g⁴⁰+g⁴²+g⁴⁵+g⁴⁶+g⁴⁷)

=gcd(1 +g³ +g⁴+g⁵+g⁸+g¹⁰+g¹¹+g¹²+g¹⁵+g¹⁷+g¹⁸+g¹⁹+g²¹+ g²⁴+g²⁵+g²⁶+g²⁹+g³¹+g³²+g³³+g³⁵+g³⁸+g³⁹+g⁴⁰+g⁴²+g⁴⁵+ g⁴⁶+g⁴⁷, g+g³+g⁴+g⁷+g⁹+g¹⁰+g¹¹+g¹⁴+g¹⁶+g¹⁷+g¹⁸+g²²+ g²⁴+g²⁵+g²⁸+g³⁰+g³¹+g³²+g³⁶+g³⁸+g³⁹+g⁴³+g⁴⁵+g⁴⁶)

=gcd(g+g³+g⁴ +g⁷+g⁹+g¹⁰+g¹¹+g¹⁴+g¹⁶+g¹⁷+g¹⁸+g²²+g²⁴ +g²⁵+g²⁸+g³⁰+g³¹+g³²+g³⁶+g³⁸+g³⁹+g⁴³+g⁴⁵+g⁴⁶,1 +g²+ g³ +g²¹+g²³+g²⁴+g³⁵+g³⁷+g³⁸+g⁴²+g⁴⁴+g⁴⁵)

=gcd(1 +g² +g³+g²¹+g²³+g²⁴+g³⁵+g³⁷+g³⁸+g⁴²+g⁴⁴+g⁴⁵, g⁷ +g⁹+g¹⁰+g¹¹+g¹⁴+g¹⁶+g¹⁷+g¹⁸+g²⁸+g³⁰+g³¹+g³²)

=gcd(g⁷+g⁹+g¹⁰+g¹¹+g¹⁴+g¹⁶+g¹⁷+g¹⁸+g²⁸+g³⁰+g³¹+g³², 1 +g²+g³+g¹⁴+g¹⁶+g¹⁷+g²¹+g²³+g²⁴+g²⁸+g³⁰+g³¹)

=gcd(1 +g² +g³+g¹⁴+g¹⁶+g¹⁷+g²¹+g²³+g²⁴+g²⁸+g³⁰+g³¹, g+g³+g⁴+g⁷+g⁹ +g¹⁰+g¹¹+g¹⁴+g¹⁵+g¹⁶+g²²+g²⁴+g²⁵+ g²⁸+g²⁹+g³⁰)

=gcd(g+g³+g⁴ +g⁷+g⁹+g¹⁰+g¹¹+g¹⁴+g¹⁵+g¹⁶+g²²+g²⁴+g²⁵+ g²⁸+g²⁹+g³⁰,1 +g³ +g⁴+g⁵+g⁸+g¹⁰+g¹¹+g¹²+g¹⁴+g¹⁵+g²¹+ g²⁴+g²⁵+g²⁶+g²⁸+g²⁹)

=gcd(1 +g³ +g⁴+g⁵+g⁸+g¹⁰+g¹¹+g¹²+g¹⁴+g¹⁵+g²¹+g²⁴+ g²⁵+g²⁶+g²⁸+g²⁹, g³+g⁵+g⁶+g⁷+g¹⁰+g¹²+g¹³+g¹⁴+g²⁴+ g²⁶+g²⁷+g²⁸)

=gcd(g³+g⁵+g⁶+g⁷+g¹⁰+g¹²+g¹³+g¹⁴+g²⁴+g²⁶+g²⁷+g²⁸, 1 +g³+g⁵ +g⁶+g⁷+g¹⁰+g¹²+g¹³+g²¹+g²⁴+g²⁶+g²⁷)

=gcd(1 +g³+g⁵+g⁶+g⁷+g¹⁰+g¹²+g¹³+g²¹+g²⁴+g²⁶+g²⁷, g+g³+g⁴+g⁵ +g⁸+g¹⁰+g¹¹+g¹²+g²²+g²⁴+g²⁵+g²⁶)

=gcd(g+g³ +g⁴+g⁵+g⁸+g¹⁰+g¹¹+g¹²+g²²+g²⁴+g²⁵+g²⁶, 1 +g²+g³ +g⁴+g⁷+g⁹+g¹⁰+g¹¹+g²¹+g²³+g²⁴+g²⁵)

=gcd(1 +g²+g³+g⁴+g⁷+g⁹+g¹⁰+g¹¹+g²¹+g²³+g²⁴+g²⁵,0)

=1 +g²+g³ +g⁴+g⁷+g⁹+g¹⁰+g¹¹+g²¹+g²³+g²⁴+g²⁵.

From theg(x), the dimension of the cyclic code is24and the minimum distance is12.

Hence,<Ω₂+ Ω₃ >is a[49,24,12]-binary cyclic code.

For <Ω₀+ Ω₁+ Ω₂>, the generator polynomial is as follows:

g(x)

=gcd(g⁴⁹−1,1 +g¹+g²+g⁴+g⁸+g¹⁶+g³²+g¹⁵+g³⁰+g¹¹+g²²+g⁴⁴+ +g³⁹+g²⁹+g⁹+g¹⁸+g³⁶+g²³+g⁴⁶+g⁴³+g³⁷+g²⁵+g³+g⁵ +g⁶+ g¹⁰+g¹²+g¹³+g¹⁷+g¹⁹+g²⁰+g²⁴+g²⁶+g²⁷+g³¹+g³³+g³⁴+g³⁸+ g⁴⁰+g⁴¹+g⁴⁵+g⁴⁷+g⁴⁸)

=gcd(1 +g¹+g²+g³+g⁴ +g⁵+g⁶+g⁸+g⁹+g¹⁰+g¹¹+g¹²+g¹³+g¹⁵+ g¹⁶+g¹⁷+g¹⁸+g¹⁹+g²⁰+g²²+g²³+g²⁴+g²⁵+g²⁶+g²⁷+g²⁹+g³⁰+g³¹

+g³²+g³³+g³⁴+g³⁶+g³⁷+g³⁸+g³⁹+g⁴⁰+g⁴¹+g⁴³+g⁴⁴+g⁴⁵+g⁴⁶+ g⁴⁷+g⁴⁸, g⁷+g⁸+g¹⁴+g¹⁵+g²¹+g²²+g²⁸+g²⁹+g³⁵+g³⁶+g⁴²+g⁴³)

=gcd(g⁷+g⁸ +g¹⁴+g¹⁵+g²¹+g²²+g²⁸+g²⁹+g³⁵+g³⁶+g⁴²+g⁴³, 1 +g +g²+g³+g⁴+g⁵+g⁶)

=gcd(1 +g+g²+g³+g⁴+g⁵+g⁶,0)

=1 +g +g²+g³+g⁴+g⁵+g⁶.

Clearly, the dimension is 43. It can be found that the minimum distance is 2. For example, we can useg(x) +g·g(x) = 1 +g⁷. Hence, the corresponding cyclic code is a[49,43,2]-cyclic code.

Chapter 4. Results and Discussions 45

For <Ω₀+ Ω₁+ Ω₃>, the generator polynomial is as follows:

g(x) =gcd(g⁴⁹−1,1 +g¹ +g²+g⁴+g⁷+g⁸+g¹⁴+g¹⁶+g³²+g¹⁵+ g³⁰+g¹¹+g²²+g²⁸+g⁴⁴+g³⁹+g²⁹+g⁹+g¹⁸+g³⁶+g²³+ g⁴⁶+g⁴³+g³⁷+g²⁵)

=gcd(1 +g¹+g²+g⁴+g⁷+g⁸+g¹⁴+g¹⁶+g³²+g¹⁵+g³⁰+g¹¹+ g²²+g²⁸+g⁴⁴+g³⁹+g²⁹+g⁹+g¹⁸+g³⁶+g²³+g⁴⁶+g⁴³+g³⁷+ g²⁵, g²¹+g²²+g²⁴+g³⁵+g³⁶+g³⁸+g⁴²+g⁴³+g⁴⁵)

=gcd(g²¹+g²²+g²⁴+g³⁵+g³⁶+g³⁸+g⁴²+g⁴³+g⁴⁵,1 +g+g² +g⁴ +g⁷+g⁸+g⁹+g¹¹+g¹⁴+g¹⁵+g¹⁶+g¹⁸+g²⁸+g²⁹+g³⁰+g³²)

=gcd(1 +g+g²+g⁴+g⁷+g⁸+g⁹+g¹¹+g¹⁴+g¹⁵+g¹⁶+g¹⁸+g²⁸ +g²⁹+g³⁰+g³², g⁷+g⁸ +g¹⁰+g²¹+g²²+g²⁴+g²⁸+g²⁹+g³¹)

=gcd(g⁷+g⁸+g¹⁰+g²¹+g²²+g²⁴+g²⁸+g²⁹+g³¹,1 +g+g²+ g⁴ +g⁷+g¹⁴+g¹⁵+g¹⁶+g¹⁸+g²²+g²³+g²⁵+g²⁸)

=gcd(1 +g+g²+g⁴+g⁷+g¹⁴+g¹⁵+g¹⁶+g¹⁸+g²²+g²³+g²⁵+g²⁸, g+g²+g⁴+g¹⁵+g¹⁶+g¹⁸+g²²+g²³+g²⁵)

=gcd(g+g²+g⁴ +g¹⁵+g¹⁶+g¹⁸+g²²+g²³+g²⁵, 1 +g+g³+g¹⁴+g¹⁵+g¹⁷+g²¹+g²²+g²⁴)

=gcd(1 +g+g³+g¹⁴+g¹⁵+g¹⁷+g²¹+g²²+g²⁴,0)

=1 +g+g³+g¹⁴+g¹⁵+g¹⁷+g²¹+g²²+g²⁴.

Clearly, the dimension is 25. The Hamming distance is at least 9 by looking at the generator polynomial. Hence, <Ω₀+ Ω₁+ Ω₃> is[49,25,9]-cyclic code.

Most of the calculations were done by hand but it became harder to perform poly- nomial division as the number of elements increased, so we used a Matlab code to help us calculate the remainder for the division of two polynomials as follows:

[q,r]=gfdeconv(g,u,2)

gis the numerator whileuis the denominator, the third parameter is the division over which finite field. By default, it is 2(binary). Then, it will produce two result :quotient and remainder. Since we just need the remainder, we just display the remainder. Ba-

sically, to get the results for all idempotent in one run, we loop the function multiple times using a for loop.

For <Ω₀+ Ω₁+ Ω₄>, the generator polynomial is as follows:

g(x)

=gcd(g⁴⁹−1,1 +g¹+g²+g⁴+g⁸+g¹⁶+g³²+g¹⁵+g³⁰+g¹¹+g²¹+g²²+ g³⁵+g⁴²+g⁴⁴+g³⁹+g²⁹+g⁹+g¹⁸+g³⁶+g²³+g⁴⁶+g⁴³+g³⁷+g²⁵)

=gcd(1 +g¹+g²+g⁴+g⁸ +g¹⁶+g³²+g¹⁵+g³⁰+g¹¹+g²¹+g²²+g³⁵+ g⁴²+g⁴⁴+g³⁹+g²⁹+g⁹+g¹⁸+g³⁶+g²³+g⁴⁶+g⁴³+g³⁷+g²⁵,

g⁷+g⁸+g¹⁰+g¹⁴+g¹⁵+g¹⁷+g²⁸+g²⁹+g³¹)

=gcd(g⁷+g⁸+g¹⁰+g¹⁴+g¹⁵+g¹⁷+g²⁸+g²⁹+g³¹,1 +g+g²+g⁴+g⁸+ g⁹+g¹¹+g¹⁴+g²¹+g²²+g²³+g²⁵+g²⁸)

=gcd(1 +g+g²+g⁴+g⁸+g⁹+g¹¹+g¹⁴+g²¹+g²²+g²³+g²⁵+g²⁸, g+g²+g⁴+g⁸ +g⁹+g¹¹+g²²+g²³+g²⁵)

=gcd(g+g² +g⁴+g⁸+g⁹+g¹¹+g²²+g²³+g²⁵, 1 +g+g³+g⁷+g⁸+g¹⁰+g²¹+g²²+g²⁴)

=gcd(1 +g+g³+g⁷+g⁸+g¹⁰+g²¹+g²²+g²⁴,0)

=1 +g+g³+g⁷+g⁸+g¹⁰+g²¹+g²²+g²⁴.

The dimension of the cyclic code is49−24 = 25. From the generator polynomial, the Hamming distance should be at least9. Hence, the code is a [49,25,9]-binary cyclic code.

Next, we found that < Ω₀ + Ω₂ + Ω₃ > is a [49,25,9]-binary cyclic code by constructing the generator polynomial as follows:

g(x)

=gcd(g⁴⁹−1,1 +g³+g⁵+g⁶+g⁷+g¹⁰+g¹²+g¹³+g¹⁴+g¹⁷+g¹⁹+g²⁰ +g²⁴+g²⁶+g²⁷+g²⁸+g³¹+g³³+g³⁴+g³⁸+g⁴⁰+g⁴¹+g⁴⁵+g⁴⁷+g⁴⁸)

=gcd(1 +g³+g⁵+g⁶ +g⁷+g¹⁰+g¹²+g¹³+g¹⁴+g¹⁷+g¹⁹+g²⁰+g²⁴+ g²⁶+g²⁷+g²⁸+g³¹+g³³+g³⁴+g³⁸+g⁴⁰+g⁴¹+g⁴⁵+g⁴⁷+g⁴⁸, g+g³ +g⁴+g⁵+g⁸+g¹⁰+g¹¹+g¹²+g¹⁵+g¹⁷+g¹⁸+g¹⁹+g²¹+g²⁴+g²⁵+ g²⁶+g²⁹+g³¹+g³²+g³³+g³⁵+g³⁸+g³⁹+g⁴⁰+g⁴²+g⁴⁵+g⁴⁶+g⁴⁷)

Chapter 4. Results and Discussions 47

=gcd(g+g³+g⁴+g⁵+g⁸+g¹⁰+g¹¹+g¹²+g¹⁵+g¹⁷+g¹⁸+g¹⁹+g²¹+ g²⁴+g²⁵+g²⁶+g²⁹+g³¹+g³²+g³³+g³⁵+g³⁸+g³⁹+g⁴⁰+g⁴²+g⁴⁵+ g⁴⁶+g⁴⁷,1 +g²+g³+g⁴+g⁷+g⁹+g¹⁰+g¹¹+g¹⁴+g¹⁶+g¹⁷+g¹⁸+ g²²+g²⁴+g²⁵+g²⁸+g³⁰+g³¹+g³²+g³⁶+g³⁸+g³⁹+g⁴³+g⁴⁵+g⁴⁶)

=gcd(1 +g²+g³+g⁴+g⁷+g⁹ +g¹⁰+g¹¹+g¹⁴+g¹⁶+g¹⁷+g¹⁸+g²²+ g²⁴+g²⁵+g²⁸+g³⁰+g³¹+g³²+g³⁶+g³⁸+g³⁹+g⁴³+g⁴⁵+g⁴⁶, g²¹+ g²³+g²⁴+g³⁵+g³⁷+g³⁸+g⁴²+g⁴⁴+g⁴⁵)

=gcd(g²¹+g²³+g²⁴+g³⁵+g³⁷+g³⁸+g⁴²+g⁴⁴+g⁴⁵,1 +g²+g³+g⁴+ g⁷+g⁹+g¹⁰+g¹¹+g¹⁴+g¹⁶+g¹⁷+g¹⁸+g²⁸+g³⁰+g³¹+g³²)

=gcd(1 +g²+g³+g⁴+g⁷+g⁹ +g¹⁰+g¹¹+g¹⁴+g¹⁶+g¹⁷+g¹⁸+g²⁸+ g³⁰+g³¹+g³², g⁷+g⁹+g¹⁰+g²¹+g²³+g²⁴+g²⁸+g³⁰+g³¹)

=gcd(g⁷+g⁹+g¹⁰+g²¹+g²³+g²⁴+g²⁸+g³⁰+g³¹,1 +g²+g³+g⁴+ g⁷+g⁸+g⁹+g¹⁴+g¹⁶+g¹⁷+g¹⁸+g²²+g²⁴+g²⁵+g²⁸+g²⁹+g³⁰)

=gcd(1 +g²+g³+g⁴+g⁷+g⁸ +g⁹+g¹⁴+g¹⁶+g¹⁷+g¹⁸+g²²+g²⁴+ g²⁵+g²⁸+g²⁹+g³⁰, g+g³+g⁴+g⁵+g⁷+g⁸+g¹⁵+g¹⁷+g¹⁸+g¹⁹+ g²¹+g²⁴+g²⁵+g²⁶+g²⁸+g²⁹)

=gcd(g+g³+g⁴+g⁵+g⁷+g⁸+g¹⁵+g¹⁷+g¹⁸+g¹⁹+g²¹+g²⁴+g²⁵+ g²⁶+g²⁸+g²⁹,1 +g³+g⁵+g⁶+g⁷+g¹⁴+g¹⁷+g¹⁹+g²⁰+g²⁴+g²⁶+ g²⁷+g²⁸)

=gcd(1 +g³+g⁵+g⁶+g⁷+g¹⁴+g¹⁷+g¹⁹+g²⁰+g²⁴+g²⁶+g²⁷+g²⁸, g³+g⁵+g⁶+g¹⁷+g¹⁹+g²⁰+g²⁴+g²⁶+g²⁷)

=gcd(g³+g⁵+g⁶ +g¹⁷+g¹⁹+g²⁰+g²⁴+g²⁶+g²⁷,1 +g³+g⁴ +g⁵+ g¹⁴+g¹⁷+g¹⁸+g¹⁹+g²¹+g²⁴+g²⁵+g²⁶)

=gcd(1 +g³+g⁴+g⁵+g¹⁴+g¹⁷+g¹⁸+g¹⁹+g²¹+g²⁴+g²⁵+g²⁶, g+g³+g⁴+g¹⁵+g¹⁷+g¹⁸+g²²+g²⁴+g²⁵)

=gcd(g+g³+g⁴+g¹⁵+g¹⁷+g¹⁸+g²²+g²⁴+g²⁵, 1 +g²+g³+g¹⁴+g¹⁶+g¹⁷+g²¹+g²³+g²⁴)

=gcd(1 +g²+g³+g¹⁴+g¹⁶+g¹⁷+g²¹+g²³+g²⁴,0)

=1 +g²+g³+g¹⁴+g¹⁶+g¹⁷+g²¹+g²³+g²⁴.

Similarly, we found that<Ω₀+Ω₂+Ω₄ >is a[49,25,9]-binary cyclic code by finding the generator polynomial as follows:

g(x)

=gcd(g⁴⁹−1,1 +g³+g⁵+g⁶+g¹⁰+g¹²+g¹³+g¹⁷+g¹⁹+g²⁰+g²¹+g²⁴ +g²⁶+g²⁷+g³¹+g³³+g³⁴+g³⁵+g³⁸+g⁴⁰+g⁴¹+g⁴²+g⁴⁵+g⁴⁷+g⁴⁸)

=gcd(1 +g³+g⁵+g⁶ +g¹⁰+g¹²+g¹³+g¹⁷+g¹⁹+g²⁰+g²¹+g²⁴+g²⁶+ g²⁷+g³¹+g³³+g³⁴+g³⁵+g³⁸+g⁴⁰+g⁴¹+g⁴²+g⁴⁵+g⁴⁷+g⁴⁸, g+g³+ g⁴+g⁵+g⁷+g¹⁰+g¹¹+g¹²+g¹⁴+g¹⁷+g¹⁸+g¹⁹+g²²+g²⁴+g²⁵+ g²⁶+g²⁸+g³¹+g³²+g³³+g³⁶+g³⁸+g³⁹+g⁴⁰+g⁴³+g⁴⁵+g⁴⁶+g⁴⁷)

=gcd(g +g³+g⁴+g⁵+g⁷+g¹⁰+g¹¹+g¹²+g¹⁴+g¹⁷+g¹⁸+g¹⁹+g²²+ g²⁴+g²⁵+g²⁶+g²⁸+g³¹+g³²+g³³+g³⁶+g³⁸+g³⁹+g⁴⁰+g⁴³+g⁴⁵+ g⁴⁶+g⁴⁷,1 +g²+g³+g⁴ +g⁸+g¹⁰+g¹¹+g¹⁵+g¹⁷+g¹⁸+g²¹+g²³+ g²⁴+g²⁵+g²⁹+g³¹+g³²+g³⁵+g³⁷+g³⁸+g³⁹+g⁴²+g⁴⁴+g⁴⁵+g⁴⁶)

=gcd(1 +g²+g³+g⁴ +g⁸+g¹⁰+g¹¹+g¹⁵+g¹⁷+g¹⁸+g²¹+g²³+g²⁴+ g²⁵+g²⁹+g³¹+g³²+g³⁵+g³⁷+g³⁸+g³⁹+g⁴²+g⁴⁴+g⁴⁵+g⁴⁶, g⁷+g⁹+g¹⁰+g¹⁴+g¹⁶+g¹⁷+g²⁸+g³⁰+g³¹)

=gcd(g⁷+g⁹+g¹⁰+g¹⁴+g¹⁶+g¹⁷+g²⁸+g³⁰+g³¹,1 +g²+g³+g⁴+ g⁸+g¹⁰+g¹¹+g¹⁴+g¹⁵+g¹⁶+g²¹+g²³+g²⁴+g²⁵+g²⁸+g²⁹+g³⁰)

=gcd(1 +g²+g³+g⁴ +g⁸+g¹⁰+g¹¹+g¹⁴+g¹⁵+g¹⁶+g²¹+g²³+g²⁴+ g²⁵+g²⁸+g²⁹+g³⁰, g+g³+g⁴+g⁵+g⁷+g¹⁰+g¹¹+g¹²+g¹⁴+g¹⁵+ g²²+g²⁴+g²⁵+g²⁶+g²⁸+g²⁹)

=gcd(g +g³+g⁴+g⁵+g⁷+g¹⁰+g¹¹+g¹²+g¹⁴+g¹⁵+g²²+g²⁴+g²⁵+ g²⁶+g²⁸+g²⁹,1 +g³+g⁵+g⁶+g¹⁰+g¹²+g¹³+g¹⁴+g²¹+g²⁴+ g²⁶+g²⁷+g²⁸)

=gcd(1 +g³+g⁵+g⁶ +g¹⁰+g¹²+g¹³+g¹⁴+g²¹+g²⁴+g²⁶+g²⁷+g²⁸, g³+g⁵+g⁶+g¹⁰+g¹²+g¹³+g²⁴+g²⁶+g²⁷)

=gcd(g³+g⁵+g⁶+g¹⁰+g¹²+g¹³+g²⁴+g²⁶+g²⁷,

1 +g³+g⁴+g⁵+g⁷+g¹⁰+g¹¹+g¹²+g²¹+g²⁴+g²⁵+g²⁶)

Chapter 4. Results and Discussions 49

=gcd(1 +g³+g⁴+g⁵+g⁷+g¹⁰+g¹¹+g¹²+g²¹+g²⁴+g²⁵+g²⁶, g+g³+g⁴+g⁸+g¹⁰+g¹¹+g²²+g²⁴+g²⁵)

=gcd(g+g³+g⁴+g⁸+g¹⁰+g¹¹+g²²+g²⁴+g²⁵,1 +g² +g³+g⁷+ g⁹+g¹⁰+g²¹+g²³+g²⁴)

=gcd(1 +g²+g³+g⁷+g⁹+g¹⁰+g²¹+g²³+g²⁴,0)

=1 +g²+g³+g⁷+g⁹+g¹⁰+g²¹+g²³+g²⁴.

For<Ω₁+ Ω₂+ Ω₃ >, the generator polynomial of this code is as follows:

g(x) =gcd(g⁴⁹−1, g¹+g²+g⁴+g⁷+g⁸+g¹⁴+g¹⁶+g³²+g¹⁵+g³⁰+g¹¹+ g²²+g²⁸+g⁴⁴+g³⁹+g²⁹+g⁹+g¹⁸+g³⁶+g²³+g⁴⁶+g⁴³+g³⁷+g²⁵+ g³ +g⁵+g⁶+g¹⁰+g¹²+g¹³+g¹⁷+g¹⁹+g²⁰+g²⁴+g²⁶+g²⁷+g³¹+ g³³+g³⁴+g³⁸+g⁴⁰+g⁴¹+g⁴⁵+g⁴⁷+g⁴⁸)

=gcd(g¹+g²+g³+g⁴+g⁵+g⁶+g⁷+g⁸ +g⁹+g¹⁰+g¹¹+g¹²+g¹³+g¹⁴ +g¹⁵+g¹⁶+g¹⁷+g¹⁸+g¹⁹+g²⁰+g²²+g²³+g²⁴+g²⁵+g²⁶+g²⁷+ g²⁸+g²⁹+g³⁰+g³¹+g³²+g³³+g³⁴+g³⁶+g³⁷+g³⁸+g³⁹+g⁴⁰+g⁴¹+ g⁴³+g⁴⁴+g⁴⁵+g⁴⁶+g⁴⁷+g⁴⁸,1 +g+g²¹+g²²+g³⁵+g³⁶+g⁴²+g⁴³)

=gcd(1 +g+g²¹+g²²+g³⁵+g³⁶+g⁴²+g⁴³, g⁷+g⁸+g⁹+g¹⁰+g¹¹+ g¹²+g¹³+g¹⁴+g¹⁵+g¹⁶+g¹⁷+g¹⁸+g¹⁹+g²⁰+g²⁸+g²⁹+g³⁰+ g³¹+g³²+g³³+g³⁴)

=gcd(g⁷+g⁸+g⁹+g¹⁰+g¹¹+g¹²+g¹³+g¹⁴+g¹⁵+g¹⁶+g¹⁷+g¹⁸+ g¹⁹+g²⁰+g²⁸+g²⁹+g³⁰+g³¹+g³²+g³³+g³⁴,1 +g+g¹⁴+g¹⁵+ g²¹+g²²+g²⁸+g²⁹)

=gcd(1 +g+g¹⁴+g¹⁵+g²¹+g²²+g²⁸+g²⁹, g+g²+g³+g⁴+g⁵+g⁶+g⁷+ g⁸ +g⁹+g¹⁰+g¹¹+g¹²+g¹³+g¹⁴+g²²+g²³+g²⁴+g²⁵+g²⁶+g²⁷+g²⁸)

=gcd(g+g²+g³ +g⁴+g⁵+g⁶+g⁷+g⁸+g⁹+g¹⁰+g¹¹+g¹²+g¹³+g¹⁴+ g²²+g²³+g²⁴+g²⁵+g²⁶+g²⁷+g²⁸,1 +g+g²+g³ +g⁴+g⁵+g⁶+g⁷+ g⁸ +g⁹+g¹⁰+g¹¹+g¹²+g¹³+g²¹+g²²+g²³+g²⁴+g²⁵+g²⁶+g²⁷)

=1 +g+g²+g³+g⁴+g⁵+g⁶+g⁷ +g⁸+g⁹+g¹⁰+g¹¹+g¹²+g¹³+ g²¹+g²²+g²³+g²⁴+g²⁵+g²⁶+g²⁷.

Clearly, the dimension of the code is 22. If we take g(x) + g · g(x), we will get 1 +g¹⁴+g²¹+g²⁸which is actually the generator polynomial of <Ω₀+ Ω₄>. Hence, the minimum distance is4. So, <Ω₁+ Ω₂+ Ω₃> is a[49,22,4]-binary cyclic code.

Next, we have <Ω₁+ Ω₂+ Ω₄> as a generating idempotent. By Theorem3.26, the generator polynomial is related as follows:

g(x) =gcd(g⁴⁹−1, g¹+g²+g⁴+g⁸+g¹⁶+g³²+g¹⁵+g³⁰+g¹¹+g²¹+g²²+ g³⁵+g⁴²+g⁴⁴+g³⁹+g²⁹+g⁹+g¹⁸+g³⁶+g²³+g⁴⁶+g⁴³+g³⁷+g²⁵+ g³+g⁵+g⁶+g¹⁰+g¹²+g¹³+g¹⁷+g¹⁹+g²⁰+g²⁴+g²⁶+g²⁷+g³¹+ g³³+g³⁴+g³⁸+g⁴⁰+g⁴¹+g⁴⁵+g⁴⁷+g⁴⁸)

=gcd(g¹+g²+g³+g⁴+g⁵+g⁶+g⁸+g⁹+g¹⁰+g¹¹+g¹²+g¹³+g¹⁵+ g¹⁶+g¹⁷+g¹⁸+g¹⁹+g²⁰+g²¹+g²²+g²³+g²⁴+g²⁵+g²⁶+g²⁷+ g²⁹+g³⁰+g³¹+g³²+g³³+g³⁴+g³⁵+g³⁶+g³⁷+g³⁸+g³⁹+g⁴⁰+ g⁴¹+g⁴²+g⁴³+g⁴⁴+g⁴⁵+g⁴⁶+g⁴⁷+g⁴⁸,1 +g+g⁷+g⁸+g¹⁴+ g¹⁵+g²⁸+g²⁹)

=gcd(1 +g+g⁷+g⁸+g¹⁴+g¹⁵+g²⁸+g²⁹, g+g²+g³+g⁴+g⁵+g⁶+ g⁷+g¹⁵+g¹⁶+g¹⁷+g¹⁸+g¹⁹+g²⁰+g²¹+g²²+g²³+g²⁴+g²⁵+ g²⁶+g²⁷+g²⁸)

=gcd(g+g²+g³+g⁴+g⁵+g⁶+g⁷+g¹⁵+g¹⁶+g¹⁷+g¹⁸+g¹⁹+g²⁰+ g²¹+g²²+g²³+g²⁴+g²⁵+g²⁶+g²⁷+g²⁸,1 +g+g²+g³+g⁴+g⁵+ g⁶+g¹⁴+g¹⁵+g¹⁶+g¹⁷+g¹⁸+g¹⁹+g²⁰+g²¹+g²²+g²³+g²⁴+ g²⁵+g²⁶+g²⁷)

=gcd(1 +g+g²+g³+g⁴+g⁵+g⁶+g¹⁴+g¹⁵+g¹⁶+g¹⁷+g¹⁸+g¹⁹+ g²⁰+g²¹+g²²+g²³+g²⁴+g²⁵+g²⁶+g²⁷,0)

=1 +g+g² +g³+g⁴+g⁵+g⁶+g¹⁴+g¹⁵+g¹⁶+g¹⁷+g¹⁸+g¹⁹+g²⁰+ g²¹+g²²+g²³+g²⁴+g²⁵+g²⁶+g²⁷.

Clearly, the dimension is22. If we sum upg(x)andg·g(x), we get1 +g⁷+g¹⁴+g²⁸ again. Hence, <Ω₁ + Ω₂+ Ω₄> is a [49,22,4]-binary cyclic code.