1. Cyclic Codes 1. Definition Linear: 𝑎 0 ,…, 𝑎 𝑛−1 ∈𝐶, 𝑏 0 ,…, 𝑏 𝑛−1 ∈𝐶
⟹α 𝑎 0 ,…, 𝑎 𝑛−1 +β 𝑏 0 ,…, 𝑏 𝑛−1 ∈𝐶
Cyclic: 𝑐 0 ,…, 𝑐 𝑛−1 ∈𝐶⟹ 𝑐 𝑛−1 , 𝑐 0 ,…, 𝑐 𝑛−2 ∈𝐶

2. 𝑭𝒐𝒓 𝒆𝒙𝒂𝒎𝒑𝒍𝒆:
C= 𝟎 𝟎 𝟎 , 𝟎 𝟏 𝟏 , 𝟏 𝟎 𝟏 , 𝟏 𝟏 𝟎 is a 𝒏=𝟑, 𝒌=𝟐 cyclic code 𝒐𝒗𝒆𝒓 𝑮𝑭 𝟐 with 𝒒 𝒌 =𝟒 𝒄𝒐𝒅𝒆𝒘𝒐𝒓𝒅𝒔;
C= 𝟎 𝟎 𝟎 , 𝟎 𝟏 𝟐 , 𝟐 𝟎 𝟏 , 𝟏 𝟐 𝟎 , 𝟎 𝟐 𝟏 , 𝟏 𝟎 𝟐 , 𝟐 𝟏 𝟎 , 𝟏 𝟏 𝟏 ,(𝟐,𝟐,𝟐)
is a 𝒏=𝟑, 𝒌=𝟐 𝒄𝒚𝒄𝒍𝒊𝒄 𝒄𝒐𝒅𝒆 𝒐𝒗𝒆𝒓 𝑮𝑭 𝟑 with 𝒒 𝒌 =𝟗 𝒄𝒐𝒅𝒆𝒘𝒐𝒓𝒅𝒔.

3. 2. Generating function 𝑐 0 ,…, 𝑐 𝑛−1 ∈𝐶→
𝑐 0 ,…, 𝑐 𝑛−1 ∈𝐶→
𝑐 0 + 𝑐 1 𝑥+ 𝑐 2 𝑥 2 +…+ 𝑐 𝑛−1 𝑥 𝑛−1 ∈𝐶
which is a corresponding polynomial for the codeword,
Note that
generating function is for codeword, not for code, and cannot generate the whole cyclic code !
from now on, codeword and polynomial are the same !

4. Example:
C= 𝟎 𝟎 𝟎 , 𝟎 𝟏 𝟐 , 𝟐 𝟎 𝟏 , 𝟏 𝟐 𝟎 , 𝟎 𝟐 𝟏 , 𝟏 𝟎 𝟐 , 𝟐 𝟏 𝟎 , 𝟏 𝟏 𝟏 ,(𝟐,𝟐,𝟐)
is a 𝟑,𝟐 𝒄𝒚𝒄𝒍𝒊𝒄 𝒄𝒐𝒅𝒆 𝒐𝒗𝒆𝒓 𝑮𝑭 𝟑 with 𝒒 𝟐 =𝟗 𝒄𝒐𝒅𝒆𝒘𝒐𝒓𝒅𝒔.
The generating functions of the codewords are: 0,
x+2 𝒙 𝟐 , 2+ 𝒙 𝟐 , 1+2x,
2x+ 𝒙 𝟐 , 1+2 𝒙 𝟐 , 2+x,
1+x+ 𝒙 𝟐 , 2+2x+ 𝟐𝒙 𝟐

5. 3. Re-define Cyclic code by using generating function
⟹ 𝐶 𝑅 (𝑥)∈𝐶
Linear:
𝑎 𝑥 =𝑎 0 + 𝑎 1 𝑥+…+ 𝑎 𝑛−1 𝑥 𝑛−1 ∈𝐶,
𝑏 𝑥 =𝑏 0 + 𝑏 1 𝑥+…+ 𝑏 𝑛−1 𝑥 𝑛−1 ∈𝐶
⟹α𝑎(𝑥)+β𝑏(𝑥)∈𝐶

6. 4. Generator polynomial (lowest degree of nonzero codewords)

7. combine the conditions on “cyclic and linear” together
Example:
C= 𝟎 𝟎 𝟎 , 𝟎 𝟏 𝟐 , 𝟐 𝟎 𝟏 , 𝟏 𝟐 𝟎 , 𝟎 𝟐 𝟏 , 𝟏 𝟎 𝟐 , 𝟐 𝟏 𝟎 , 𝟏 𝟏 𝟏 ,(𝟐,𝟐,𝟐)
is a 𝟑,𝟐 𝒄𝒚𝒄𝒍𝒊𝒄 𝒄𝒐𝒅𝒆 𝒐𝒗𝒆𝒓 𝑮𝑭 𝟑 with 𝒒 𝟐 =𝟗 𝒄𝒐𝒅𝒆𝒘𝒐𝒓𝒅𝒔.
The generating functions of the codewords are:
0, x+2 𝒙 𝟐 , 2+ 𝒙 𝟐 , 1+2x, 2x+ 𝒙 𝟐 , 1+2 𝒙 𝟐 , 2+x, 1+x+ 𝒙 𝟐 , 2+2x+ 𝟐𝒙 𝟐
Generator polynomial: g(x)=1+2x or 2+x ? We need monic: 2+x
need monic !
The 3rd definition of cyclic code by using generating function:
combine the conditions on “cyclic and linear” together
Theorem 8.2 implies C = 𝑡 𝑥 𝑔(𝑥) 𝑛 :𝑡(𝑥)∈𝐹[𝑥]

8. 4.2 Property of Factor: Part I
𝒈(𝒙) is unique in this way
𝒈 𝒙 |𝑷 𝒙 ,
Otherwise 𝑃 𝑥 =𝑢 𝑥 𝑔 𝑥 +𝑣 𝑥 where
0<𝜕𝑣 𝑥 <𝜕𝑔 𝑥 ≤𝑛−1,
which implies that
𝑣 𝑥 = [𝑣 𝑥 ] 𝑛 = 𝑃 𝑥 −𝑢 𝑥 𝑔 𝑥 𝑛 = [𝑃 𝑥 ] 𝑛 − [𝑢 𝑥 𝑔 𝑥 ] 𝑛
is a nonzero codeword, i.e. a contradiction. Corollary:
𝒈 𝒙 |𝒄 𝒙 where 𝒄 𝒙 is any codeword, 𝒈 𝒙 | 𝒙 𝒏 −𝟏.

9. 4.3 Property of Factor: Part II,
which is the 4th definition of cyclic code, i.e. Th. 8.3 (1)
C = 𝑡 𝑥 𝑔(𝑥) 𝑛 :𝑡(𝑥)∈𝐹[𝑥]
= 𝑡 𝑥 𝑔 𝑥 :𝑡 𝑥 ∈𝐹 𝑥 , 𝜕𝑡 𝑥 ≤𝑛−𝑟−1
where 𝑟=𝜕𝑔 𝑥
Proof. 𝑡 𝑥 𝑔 𝑥 :𝑡 𝑥 ∈𝐹 𝑥 , 𝜕𝑡 𝑥 ≤𝑛−1−𝑟
⊇𝐶 because of Lemma 2
⊇ 𝑡 𝑥 𝑔(𝑥) 𝑛 :𝑡(𝑥)∈𝐹[𝑥] because of Theorem 8.2
⊇ 𝑡 𝑥 𝑔 𝑥 :𝑡 𝑥 ∈𝐹 𝑥 , 𝜕𝑡 𝑥 ≤𝑛−1−𝑟 𝐨𝐛𝐯𝐢𝐨𝐮𝐬𝐥𝐲 #
So 𝒌= 𝒍𝒐𝒈 𝒒 𝑪 = 𝒍𝒐𝒈 𝒒 𝒒 𝒏−𝒓 =𝒏−𝝏𝒈 𝒙

10. 5. One to one correspondence
between cyclic code with length n and monic divisor of xn-1
(b) i.e. if 𝒈 𝒙 | 𝒙 𝒏 −𝟏
𝒕 𝒙 𝒈 𝒙 :𝒕 𝒙 ∈𝑭 𝒙 , 𝝏𝒕 𝒙 ≤𝒏−𝒓−𝟏 is a cyclic code.

11. 𝒕 𝒙 𝒈(𝒙) 𝒏 = 𝒕 𝒙 𝒈 𝒙 +v 𝒙 𝒙 𝒏 −𝟏 , and 𝒈 𝒙 | 𝒙 𝒏 −𝟏,
(b) If 𝒈 𝒙 | 𝒙 𝒏 −𝟏, 𝐂= 𝒕 𝒙 𝒈 𝒙 :𝒕 𝒙 ∈𝑭 𝒙 , 𝝏𝒕 𝒙 ≤𝒏−𝒓−𝟏 is a cyclic code.
Proof. 𝐰𝐞 𝐨𝐧𝐥𝐲 𝐧𝐞𝐞𝐝 𝐭𝐨 𝐩𝐫𝐨𝐯𝐞 𝐭𝐡𝐚𝐭
𝑡 𝑥 𝑔 𝑥 :𝑡 𝑥 ∈𝐹 𝑥 , 𝜕𝑡 𝑥 ≤𝑛−𝑟−1 = 𝑡 𝑥 𝑔(𝑥) 𝑛 :𝑡(𝑥)∈𝐹[𝑥]
since 𝑡 𝑥 𝑔(𝑥) 𝑛 :𝑡 𝑥 is a cyclic code. O𝐧𝐥𝐲 𝐧𝐞𝐞𝐝 𝐭𝐨 𝐩𝐫𝐨𝐯𝐞 𝑔 𝑥 | 𝑡 𝑥 𝑔(𝑥) 𝑛 .
From the definition of mod,
𝒕 𝒙 𝒈(𝒙) 𝒏 = 𝒕 𝒙 𝒈 𝒙 +v 𝒙 𝒙 𝒏 −𝟏 , and 𝒈 𝒙 | 𝒙 𝒏 −𝟏,
It is easy to see that 𝑔 𝑥 | 𝑡 𝑥 𝑔(𝑥) 𝑛 . #
For example, for 𝒏=𝟑 𝐨𝐯𝐞𝐫 𝑮𝑭(𝟑), 𝑥 𝑛 −1= 𝑥−1 3 has 𝟒 factors：
(1) 𝑔 𝑥 =1
(2) 𝑔 𝑥 =𝑥−1
(3) 𝑔 𝑥 = 𝑥−1 2 = 𝑥 2 +𝑥+1
(4) 𝑔 𝑥 = 𝑥−1 3 = 𝑥 3 −1=0 𝑚𝑜𝑑 𝑥 𝑛 −1. see follows.

12. 𝑔 𝑥 =1 corresponds to
C= 𝑡 𝑥 :𝑡 𝑥 ∈𝐹 𝑥 , 𝜕𝑡 𝑥 ≤2
={(0 0 0), (1 0 0), (2 0 0), …, (2 2 2)} totally 27 codewords, i.e. whole space;
(2) 𝑔 𝑥 =𝑥−1 corresponds to
C= 𝑡 𝑥 (𝑥−1):𝑡 𝑥 ∈𝐹 𝑥 , 𝜕𝑡 𝑥 ≤1
={(0 0 0), (2 1 0), (0 2 1), (1 0 2), (1 2 0), (0 1 2), (2 0 1), (1 1 1), (2 2 2)}
totally 9 codewords;
(3) 𝑔 𝑥 = 𝑥−1 2 = 𝑥 2 +𝑥+1 corresponds to
C= 𝑡 𝑥 ( 𝑥 2 +𝑥+1):𝑡 𝑥 ∈𝐹 𝑥 , 𝜕𝑡 𝑥 ≤0
={(0 0 0), (1 1 1), (2 2 2)} totally 3 codewords;
(4) 𝑔 𝑥 = 𝑥−1 3 = 𝑥 3 −1=0 𝑚𝑜𝑑 𝑥 𝑛 −1 corresponds to
C={(0 0 0)}.

13. 6. Parity-check polynomial

14. 7. Generator matrix and Parity-check matrix
Example: for C ={(0 0 0), (2 1 0), (0 2 1), (1 0 2), (1 2 0), (0 1 2), (2 0 1), (1 1 1), (2 2 2)}
over 𝐺𝐹(3) We have 𝒈 𝒙 =𝟐+𝒙, and 𝐺=

15. Note that the previous form of 𝐺 1 , and the relation 𝑔 𝑥 ℎ 𝑥 = 𝑥 𝑛 −1, imply that one parity check matrix can be 𝐻 1 .
Example cont'd: 𝒉 𝒙 = 𝒙 𝟑 −𝟏 𝟐+𝒙 = 𝒙 𝟐 +𝒙+𝟏, and 𝒉 𝒙 = 𝒙 𝟐 +𝒙+𝟏
𝐻= 1 1 1

16. Encoding example by polynomials: 𝒄 𝒙 =𝒖 𝒙 𝒈(𝒙)
𝑢 𝑥 = 𝑢 0 + 𝑢 1 𝑥+…+ 𝑢 𝑘−1 𝑥 𝑘−1
𝑐 𝑥 = 𝑐 0 + 𝑐 1 𝑥+…+ 𝑐 𝑛−1 𝑥 𝑛−1 … + + +
𝑐 0 , 𝑐 1 ,…, 𝑐 𝑛−1 … ×
𝑔 𝑛−𝑘 ×
𝑔 𝑛−𝑘−1 ×
𝑔 1 ×
𝑔 0
𝑢 0 , 𝑢 1 ,…, 𝑢 𝑘−1 …
Begin of output: 𝑐 𝑛−1 =𝑢 𝑘−1 𝑔 𝑛−𝑘
End of output: 𝑐 0 = 𝑢 0 𝑔 0

17. 7. Dual code of cyclic code
Dual code definition. 𝐶 ⊥ ={ 𝛼 0 , 𝛼 1 ,…, 𝛼 𝑛−1 : 𝛼 0 𝛽 0 + 𝛼 1 𝛽 1 +…+ 𝛼 𝑛−1 𝛽 𝑛−1 =0 𝑓𝑜𝑟 𝑎𝑙𝑙 ( 𝛽 0 , 𝛽 1 ,…, 𝛽 𝑛−1 )∈𝐶}
Proposition 1. The dual code of cyclic code is also cyclic.
Proof.

18. For example: 𝑔 𝑥 =𝑥−1 corresponds to a [3,2] cyclic code over GF(3)
={(0 0 0), (2 1 0), (0 2 1), (1 0 2), (1 2 0), (0 1 2), (2 0 1), (1 1 1), (2 2 2)}.
We have [3,1] dual cyclic code 𝐶 ⊥ ={(0 0 0), (1 1 1), (2 2 2)} = 𝑡 𝑥 (𝑥−1) 2 :𝑡 𝑥 ∈𝐹 𝑥 , 𝜕𝑡 𝑥 ≤0 .

19. Proposition 2. C is a [n, k] cyclic code
Proposition 2. C is a [n, k] cyclic code. The generator polynomial of is
Proof. Step 1: Since generator matrix of is parity check matrix of C, so
is a generator matrix of

20. Proposition 3. C is a cyclic code. The parity check polynomial of is
Step 2: is a generator polynomial of the dual code:
𝑔 (𝑥) ℎ (𝑥)
Proposition 3. C is a cyclic code. The parity check polynomial of is
𝑔 𝑥 ℎ 𝑥 = 𝑥 𝑛 −1
⟹𝑔 𝑥 −1 ℎ 𝑥 −1 =( 𝑥 −𝑛 −1)
⟹ [𝑥 𝑛−𝑘 𝑔 𝑥 −1 ]× [𝑥 𝑘 ℎ 𝑥 −1 ]= 𝑥 𝑛 ( 𝑥 −𝑛 −1)
⟹ 𝑔 (𝑥) ℎ (𝑥)= −(𝑥 𝑛 −1)
Proof.
From Prop.2, ℎ 𝑥 is generator of 𝑪 ⊥ , so 𝑔 (𝑥) is the parity check of 𝑪 ⊥ .