Chapter 5

Cyclic CodesDr.Mohammed Taha El Astal

Note: This PPT is an enhanced version of Dr.Musbah Shaat’s PPT.

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Definition

Definition: An (n,k) linear code C is cyclic if every

cyclic shift of a codeword in C is also a codeword in C.

• This mean that If c0 c1 c2 …. cn-2 cn-1 is a codeword, then

cn-1 c0 c1 …. cn-3 cn-2

cn-2 cn-1 c0 …. cn-4 cn-3

: : : : :

c1 c2 c3 …. cn-1 c0 are also codewords.

• Cyclic is a great feature. Why?

3

Example

• The (7,4) Hamming code discussed

before is cyclic:

1010001 1110010 0000000 1111111

1101000 0111001

0110100 1011100

0011010 0101110

0001101 0010111

1000110 1001011

0100011 1100101

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Code Polynomial

• Example:

1010001 ; 1+X2+X6

0101110 ; X+X3+X4+X5

• Each codeword is represented by a polynomial of

degree less than or equal n-1.

• .deg[ ]c(X) n 1

c = c0 c1 c2 …. cn-1. c(X) = c0 + c1X+ …. + cn-1 Xn-1

where the power of X corresponds to the bit position, and the coefficients are 0’s and 1’s.

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Implementing the Shift

• What is the relation between c(X) and c (i)(X)?

c = c0 c1 c2 …. cn-1. c(X) = c0 + c1X+ …. + cn-1 Xn-1

c(i) = cn-i cn-i+1… cn-1 c0 ..

...cn-i-1

c(i)(X) = cn-i + cn-i+1 X + …. + cn-1 Xi-1

+ c0Xi +…. +cn-i-1 X

n-1

i shifts to the right

Apparently, shifting a bit one place to the right is equivalent to multiplying

the term by X.

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• Rearranging,

Xic(X) = cn-i Xn +…+ cn-1 Xn+i-1 + c0X

i + c1X i+1 + …+ cn-i-1 Xn-1

• Add to the polynomial the zero-valued sequence:

(cn-i + cn-i ) + (cn-i+1 + cn-i+1 )X + …. + (cn-1 + cn-1 )Xi-1

• This leads to

Xic(X) = cn-i (Xn +1) + cn-i+1 X (Xn +1) +…. + cn-1 Xi-1 (Xn +1) +

cn-i + cn-i+1 X +…. + cn-1 Xi-1 +

c0Xi +c1X i+1 + …. + cn-i-1 X

n-1

• That is: Xic(X) = q(X)(Xn +1) + c(i)(X)

Where q(X)= cn-i + cn-i+1 X +…. + cn-1 Xi-1

Implementing the Shift (2)

• For i shifts:

Xic(X)= c0Xi +c1X

i+1 + ….+ cn-i-1 Xn-1 + cn-i Xn ….+ cn-1 Xn+i-1

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• Summary:

c(i)(X) is the remainder from dividing Xic(X) by (Xn +1).

c(i)(X) = Rem[Xic(X)/ (Xn +1)] = Xic(X) mod (Xn +1).

• Example: c = 0101110. c(3) = 1100101

Mathematically: c(X) = X + X3 + X4 + X5.

X3c(X) = X4 + X6 + X7 + X8

Rem[X3c(X)/ (X7 +1)] = 1 + X + X4 + X6

c(3) = 1100101

Implementing the Shift (3)

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The Code Polynomial of Minimum

Degree• TH 5.1: The nonzero code polynomial of minimum degree in

a cyclic code C is unique. Show?

• TH 5.2: Let g(X) = g0 + g1X +….+ gr-1Xr-1 +Xr be the nonzero

code polynomial of minimum degree in an (n,k) cyclic code.

Then the constant term g0 must be equal to 1.

• For the (7,4) code given in the Table 5.1, the nonzero code

polynomial of minimum degree is g(X) = 1 + X + X3

(If not, the sum of the two polynomials will be a code polynomial of

degree less than the minimum. Contradiction)

(If not, then one cyclic shift to the left (n-1 to right) will produce a code

polynomial of degree less than the minimum. Contradiction)

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Generator Polynomial

Let g(X), the minimum-degree polynomial in C, have degree r

• Since the code is cyclic:

Xg(X), X2g(X),…., Xn-r-1g(X) (Cyclic shift versions) are code

polynomials in C. (Note that deg[Xn-r-1g(X)] = n-1).

• Since the code is linear:

(a0 + a1X + …. + an-r-1 Xn-r-1)g(X) is also a code polynomial, where ai =

0 or 1.

• TH 5.3: A binary polynomial of degree ≤ n-1 is a code

polynomial if and only if it is a multiple of g(X). Show?

If a code polynomial c(X) is not a multiple of g(X), then Rem[c(X)/g(X)]

will be a code polynomial of degree less than the minimum.

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• All code polynomials are generated from the multiplication

c(X) = a(X)g(X).

deg[c(X)] n-1, deg[g(X)] = r, => deg[a(X)] ≤ n-r-1

There is a total of 2n-r polynomials of degree ≤ n-r-1

Therefore the number of codewords = 2n-r = 2k

Therefore, r = deg[g(X)] = n-k

• Since deg[a(X)] ≤ k-1, the polynomial a(X) may be taken

to be the information polynomial u(X) (a polynomial

whose coefficients are the information bits). Encoding is

performed by the multiplication c(X) = u(X)g(X).

• g(X), generator polynomial, completely defines the code.

Generator Polynomial (2)

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• TH 5.4:

In an (n,k) cyclic code, there exists one and only

one code polynomial of degree n-k,

1+ g1X + g2X2 + …+ gn-k-1X

n-k-1 +Xn-k

Every code polynomial is a multiple of g(X), and

every binary polynomial of degree n-1 or less that

is a multiple of g(X) is a code polynomial.

Generator Polynomial (3)

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(7,4) Code Generated by 1+X+X3

Infor. Code Code polynomials

0000 0000000 0 = 0 . g(X)

1000 1101000 1 + X + X3 = 1 . g(X)

0100 0110100 X + X2 + X4 = X . g(X)

1100 1011100 1 + X2 + X3 + X4 = (1 + X) . g(X)

0010 0011010 X2 + X3 + X5 = X2 . g(X)

1010 1110010 1 + X+ X2 + X5 = (1 + X2) . g(X)

0110 0101110 X+ X3 + X4 + X5 = (X+ X2) . g(X)

1110 1000110 1 + X4 + X5 = (1 + X + X2) . g(X)

0001 0001101 X3 + X4 + X6 = X3 . g(X)

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Infor. Code Code polynomials

1001 1100101 1 + X + X4 + X6 = (1 + X3) . g(X)

0101 0111001 X+ X2 + X3 + X6 = (X+ X3) . g(X)

1101 1010001 1 + X2 + X6 = (1 + X + X3) . g(X)

0011 0010111 X2 + X4 + X5 + X6 = (X2 + X3). g(X)

1011 1111111 1 + X + X2 + X3 + X4 + X5 + X6

= (1 + X2 + X3) . g(X)

0111 0100011 X + X5 + X6 = (X + X2 + X3). g(X)

1111 1001011 1 + X3 + X5 + X6

= (1 + X + X2 + X3) . g(X)

(7,4) Code Generated by 1+X+X3. (2)

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Properties of the Generator

Polynomial

• TH 5.5: The generator polynomial g(X) of an (n,k) cyclic

code is a factor of Xn+1.

• TH 5.6: If g(X) is a polynomial of degree n-k and is a

factor of Xn+1, then it generates an (n,k) cyclic code.

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Constructing g(X)

• To construct a cyclic code of length n,

1. find the factors of the polynomial Xn+1.

2. The factor (or product of factors) of degree n-k serves

as the generator polynomial of an (n,k) cyclic code.

• Clearly, a cyclic code of length n does not exist for

every k.

• The code generated this way is guaranteed to be

cyclic. But we know nothing yet about its minimum

distance. The generated code may be good or bad.

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Cyclic Codes of Length 7

Example (5.1-mod): What cyclic codes of length 7 can be

constructed?

X7+1 = (1 + X)(1 + X + X3)(1 + X2 + X3)

r=n-k k=n-r.

g(X) Code g(X) Code

(1 + X) (7,6) (1 + X)(1 + X + X3) (7,3)

(1 + X + X3) (7,4) (1 + X) (1 + X2 + X3) (7,3)

(1 + X2 + X3) (7,4) (1 + X + X3)(1 + X2 + X3) (7,1)

What if u=(1010) u= 1 + X2 v(x)= 1 + X + X2 + X5

• 1110010

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Encoding of Systematic Cyclic Codes

c(X) = b(X) + Xn-k u(X) a(X)g(X) = b(X) + Xn-k u(X)

Xn-k u(X)= a(X)g(X) + b(X)

Or b(X) = Rem[Xn-k u(X)/g(X)]

Encoding Procedure (Systematic code):

1. Multiply u(X) by Xn-k

2. Divide Xn-k u(X) by g(X), obtaining the remainder b(X).

3. Add b(X) to Xn-k u(X), obtaining c(X) in systematic form.

uk-1…. u2 u1u0 bn-k-1 …. b2 b1 b0

b(X) = b0 + b1X+….+bn-k-1Xn-k-1 u(X) = u0+u1X+ ….+uk-1X

k-1

, then c(X) = b(X) + Xn-k u(X)

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Example

Consider the (7,4) cyclic code generated by g(X) = 1 + X +

X3. Find the systematic codeword for the message 1001.

Encoding of Systematic Cyclic Codes (2)

Systematic

u(X) = 1 + X3

X3u(X) = X3 + X6

b(X) =

Rem[X3u(x)/g(X)] = X + X2

Therefore, c = 0111001

Not Systematic

u(X) = 1 + X3

c(X) = u(x)g(X) = 1+X + X4+X6

Therefore, c = 1100101

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Generator Matrix of Cyclic Codes

• Formed as follows:

– 1st row: coefficients of g(X) followed by k-1 zeros

Other rows are the k-1 cyclic shifts of first row

• The matrix can be arranged in systematic form

by row operations.

EELE 6338 21

Generator Matrix of Cyclic Codes (2)

Generator Matrix in Systematic form

EELE 6338 22

• This mean, you should determine the reminders of

Xn-k+i (for i=0,1,…k-1) over g(X) and then formulate

in G.

Example 5.4

EELE 6338 23

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Encoder Circuit

• Switch Down & Gate ON: k message bits are shifted into

the channel. The parity bits are formed in the register.

• Switch Up & Gate OFF: Contents of register are shifted

into the channel.

g2g1

++ + +

Gate

gr-1

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(7,4) Encoder Based on 1 + X + X3

Input 1 1 0 1

Register : 000 110 101 100

100initial 1st shift 2nd shift 3rd shift 4th shift

Codeword: 1 0 0 1 0 1 1

+ +

Gate

Check this: https://www.youtube.com/watch?v=JQ6Ad4fCGIo

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Decoding of Cyclic Codes

1. Syndrome computation

2. Associating the syndrome to the error pattern

3. Error correction• Refer to https://www.youtube.com/watch?v=d5u9hbVvT4o

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Syndrome Computation• Let assume that the received word is r0 r1 …. rn-1 :

– r(X) = r0 + r1X +….+ rn-1Xn-1

• If r(X) is a correct codeword, it is divisible by g(X).

Otherwise:

– r(X) = q(X)g(X) + s(X). and deg[s(X)] ≤ n-k-1.

– s(X) is called the syndrome polynomial.

• The syndrome polynomial depends on the error pattern

only. Show?

• s(X) is obtained by shifting r(X) into a divider-by-g(X)

circuit. The register contents are the syndrome bits.

s(X) = Rem[r(X)/g(X)] = Rem[ {a(X)g(X) + e(X)}/g(X)] = Rem[e(X)/g(X)]

Syndrome Circuit

EELE 6338 28

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Circuit for Syndrome Computation

++

Gate

r = 0010110

Shift Input Register contents

0 0 0 (initial state)

1 0 0 0 0

2 1 1 0 0

3 1 1 1 0

4 0 0 1 1

5 1 0 1 1

6 0 1 1 1

7 0 1 0 1 (syndrome s)

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More on Syndrome Computation

• TH5.8 Let s(X) be the syndrome of a received polynomial r(X)

= r0 + r1X +….+ rn-1Xn-1 . Then the remainder resulting from

dividing Xs(X) by g(X) is the syndrome of r(1)(X), which is a

cyclic shift of r(X). Proof?

• Proof: r(X) = r0 + r1X +….+ rn-1Xn-1

r(1)(X) = rn-1 + r0X +….+ rn-2Xn-1 = rn-1 + Xr(X) + rn-1X

n

= rn-1(Xn+1) + Xr(X)

c(X)g(X) + y(X) = rn-1 g(X)h(X)+ X{a(X)g(x) + s(X)}

where y(X) is the syndrome of r(1)(X) .

Xs(X) = {c(X) + Xa(X) + rn-1 h(X)}g(X) + y(X)

Therefore, Syndrome of r(1)(X)= Rem[Xs(X)/g(X)].

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• Note: for simplicity of notation, let Rem[Xs(X)/g(X)] bedenoted by s(1)(X).

– s(1)(X) is NOT a cyclic shift of s(X), but the syndrome of r(1)(X)which is a cyclic shift of r(X).

• Example:

r(X) = X2 + X4 + X5; g(X) = 1 + X + X3. Find s(1)(X) ?

s(X) = Rem[r(X)/g(X)] = 1 + X2

r(1)(X) = X3 + X5 + X6

s(1)(X) = Rem[r(1)(X)/g(X)] = 1 (polynomial)

Also, s(1)(X) = Rem[Xs(X)/g(X)] = 1.

• N.B. It follows from TH 5.8 that, in general, the remainders(i)(X) resulting from dividing Xis(X) by g(X) is the

syndrome of r(i)(X).

More on Syndrome Computation (2)

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Shift Input Register contents

0 0 0 (initial state)

1 0 0 0 0

2 1 1 0 0

3 1 1 1 0

4 0 0 1 1

5 1 0 1 1

6 0 1 1 1

7 0 1 0 1 (syndrome s)

8 (input gate off) - 1 0 0 (syndrome s(1)

)

9 - 0 1 0 (syndrome s(2)

)

++

Gate

Gater = 0010110

More on Syndrome Computation (3)

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More on Syndrome Computation (4)

• Let r(X) = r0 + r1X +….+ rn-1Xn-1 has the syndrome s(X).

• Then r(1)(X) = rn-1 + r0 X + ….+ rn-2Xn-1 has the syndrome:

s(1)(X) = Rem[r(1)(X)/g(X)].

• Define r1 (X) = r(X) + Xn-1 = r0 + r1X +….+ (rn-1+1)Xn-1

• The syndrome of r1 (X), call it s1 (X):

s1 (X)= Rem[{r(X)+ Xn-1}/g(X)] = s(X) + Rem[Xn-1/g(X)]

r1(1)(X), which is one cyclic shift of r1 (X), has the syndrome

s1(1)(X) = Rem[X s1 (X)/g(X)] = Rem[Xs(X)/g(X)+ Xn/g(X)]

= s(1)(X) + 1 (since Xn +1 = g(X)h(X))

• Therefore, adding “1” to the left of R removes the effect ofthe erroneous digit on the syndrome

• Important

• https://www.youtube.com/watch?v=dW3NhK2oWcc

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General Decoding Circuit

Important : Refer to the CyclicDecoding.pdf for the explanation and worked example

• Again , don’t forget, go to the

CyclicDecoding.pdf for the explanation and

worked example regarding the full-image of

the decoding that illustrated in the class

session.

Cyclotomic Coset • The cyclotomic coset of 𝑞 modulo 𝑛 that contain 𝑖 is

defined as follows:

𝐶𝑖 = 𝑖. 𝑞𝑖 𝑚𝑜𝑑 − 𝑛 ∈ 𝑍𝑛: 𝑗 = 0,1,⋯

• When 𝐶𝑖1 , ⋯ , 𝐶𝑖𝑡 are disjoint sets and 𝑗=1ڂ𝑡 𝐶𝑖𝑗 = 𝑍𝑛

the set {𝑖1, ⋯ , 𝑖𝑡} is called complete set of representatives

of cyclotomic coset.

37

• Example 1: Find the cyclotomic cosets of 2 mod-7. n=7

and q=2.

• 𝐶𝑖 = 𝑖. 𝑞𝑗 𝑚𝑜𝑑 − 𝑛 ∈ 𝑍𝑛: 𝑗 = 0,1,⋯ =

𝑖. [1,2,4,8,⋯ ] 𝑚𝑜𝑑 − 7

• 𝐶0 = 0 , 𝐶1 = 1,2,4 = 𝐶2 = 𝐶4, 𝐶3 = 3,5,6 = 𝐶5 = 𝐶6• Complete sets 0,1,3 𝑜𝑟 0,2,5 … . .

How to Factorize 𝒙𝒏 + 𝟏 over GF(2)

• Step 1: Find the complete set of representatives of

cyclotomic coset of 2 mod-n {𝑆𝑖1 , 𝑆𝑖2 , ⋯ , 𝑆𝑖𝑡}.

• Step2: Find the minimum number m such that

𝑅𝑒𝑚2𝑚−1

𝑛= 0.

• Step3: Construct the field GF( 2𝑚) using the

primitive polynomials from the Table 2.7 page 42.

Note that for m=3,4,6, and 15

primitive polynomial=1 + 𝑥 + 𝑥𝑚

EELE 6338 38

• Step 4: Obtain the minimal polynomials Φ2𝑚−1 .𝑆𝑖

𝑛

of the elements 𝛼2𝑚−1 .𝑆𝑖

𝑛 for every element in the

complete set found in step1.

Recall that : 𝛼𝑥 and its conjugates are roots of the

minimal polynomial.

• Step 5: Evaluate 𝑥𝑛 + 1 = ς𝑖=1𝑡 Φ

2𝑚−1 .𝑆𝑖𝑛 (𝑥)

EELE 6338 39

Factorization of 𝒙𝟕 + 𝟏 in EX.5.1

• Form Example 1 the complete set 𝑆 = 𝑠1, 𝑠2, 𝑠3= {0,1,3}.

• 𝑅𝑒𝑚2𝑚−1

7= 0 → 𝑚 = 3.

• We need to construct the field GF(8)=GF(23) using

the generator polynomial 1 + 𝑥 + 𝑥3 .

Let 𝛼 to be root fro the primitive polynomial 1+ 𝛼 + 𝛼3 = 0 𝛼3 = 1 + 𝛼.

Elements of the field : 0, 1, 𝛼, 𝛼2, 𝛼3 = 𝛼. 𝛼2 = 1+ 𝛼, 𝛼4 = 𝛼2 + 𝛼, 𝛼5 = 𝛼2 + 𝛼 + 1, 𝑎𝑛𝑑 𝛼6= 𝛼2 + 1.

EELE 6338 40

• We need to find Φ2𝑚−1 .𝑆𝑖

𝑛 (x)= Φ23−1 .𝑆𝑖

7

where 𝑆 = 𝑠1, 𝑠2, 𝑠3 = {0,1,3}.

Φ 0 (x), Φ 1 (x), Φ 3 (x)

Φ 0 (x)= x + 𝛼0 = x + 1

Φ 1 (x)= x + 𝛼1 x + 𝛼2 x + 𝛼4 = 1 + x + x3

Φ 3 (x)= x + 𝛼3 x + 𝛼5 x + 𝛼6 = 1 + x2 + x3

X7+1 = (1 + X)(1 + X + X3)(1 + X2 + X3)

EELE 6338 41

Factorization of 𝒙𝟕 + 𝟏 in EX.5.1 (2)

42

Error Trapping Decoders

• If the errors (that we intend to correct) can be

trapped in the last n-k locations, then it can be

shown that the contents of R is identical to the

error pattern, and therefore correction can be

done in one step.

• So, error trapping decoders can correct the

majority of correctable patterns at a faster rate.

EELE 6338

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Shortened Cyclic Codes

• Consider an (n,k) cyclic code.

• Select the codewords for which the l leading high-order

information digits are zeros.

• Delete those zero digits. This produces (n-l,k-l) linear

code, but not cyclic.

• A shortened cyclic code has at least the same error-

correcting capability of the original code.

• The same encoder circuit of the original code can be used.

• The same decoder circuit of the original code can be used

with little modification.

EELE 6338

Next Lecture

• Chapter 6 : Binary BCH Codes.

EELE 6338 44