BCH Codes

Hsin-Lung Wu

NTPU

p2.

OUTLINE

[1] Finite fields [2] Minimal polynomials [3] Cyclic Hamming codes [4] BCH codes [5] Decoding 2 error-correcting BCH codes

p3.

BCH Codes

[1] Finite fields

1. Irreducible polynomial

f(x)K[x], f(x) has no proper divisors in K[x]Eg. f(x)=1+x+x2 is irreduciblef(x)=1+x+x2+x3=(1+x)(1+x2) is not irreducible

f(x)=1+x+x4 is irreducible

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BCH Codes

2. Primitive polynomial

f(x) is irreducible of degree n > 1 f(x) is not a divisor of 1+xm for any m < 2n-1

Eg. f(x)=1+x+x2 is not a factor of 1+xm for m < 3 so f(x)

is a primitive polynomial f(x)= 1+x+x2+x3+x4 is irreducible but 1+x5=(1+x)

(1+x+x2+x3+x4) and m=5 < 24-1=15 so f(x) is not a primitive polynomial

p5.

BCH Codes

3. Definition of Kn[x] The set of all polynomials in K[x] having degree less than

n Each word in Kn corresponds to a polynomial in Kn[x] Multiplication in Kn modulo h(x), with irreducible h(x)

of degree n If we use multiplication modulo a reducible h(x), say, 1+x4

to define multiplication of words in K4, however: (0101)(0101)(x+x3)(x+x3)

= x2+x6

= x2+x2 (mod 1+x4)

= 0 0000 (K4-{0000} is not closed under multiplication.)

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BCH Codes

Furthermore each nonzero element in Kn can have

an inverse if we use irreducible h(x).

But if we use reducible h(x) then there exists nonzero

element, which has no inverse.

Why?

Let f(x) is nonzero element and h(x) is irreducible

then gcd(f(x),h(x))=1 and so exists

a(x)f(x)+b(x)h(x)=1 => a(x)f(x)=1 mod h(x)

and so a(x) is the inverse of f(x)

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BCH Codes

4. Definition of Field (Kn,+,x) (Kn,+) is an abelian group with identity denoted 0 The operation x is associative

a x ( b x c) = ( a x b ) x c There is a multiplicative identity denoted 1, with 10

1 x a = a x 1 = a, a Kn

The operation x is distributive over + a x ( b + c ) = ( a x b ) + ( a x c )

It is communicative a x b = b x a, a,b Kn

All non-zero elements have multiplicative inverses Galois Fields: GF(2r)

For every prime power order pm, there is a unique finite field of order pm

Denoted by GF(pm)

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BCH Codes

Example

Let us consider the construction of GF(23) using the primitive polynomial h(x)=1+x+x3 to define multiplication. We do this by computing xi mod h(x):

word xi mod h(x)100 1010 x001 x2

110 x3 1+x011 x4 x+x2

111 x5 1+x+x2

101 x6 1+x2

p9.

BCH Codes

5. Use a primitive polynomial to construct GF(2n)

Let Kn represent the word corresponding

to x mod h(x)

i xi mod h(x)

m 1 for m<2n-1

since h(x) dose not divide 1+xm for m<2n-1

Since j = i for ji iff i = j-i i j-i = 1

Kn\{0}={i | i = 0,1,…,2n-2}

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BCH Codes

6. GF(2r) is primitive

is primitive if m 1 for 1 m <2r-1

In other words, every non-zero word in GF(2r) can be

expressed as a power of

Example

Construct GF(24) using the primitive polynomial

h(x)=1+x+x4. Write every vector as a power of

x mod h(x)(see Table 5.1 below)

Note that 15=1.

(0110)(1101)= 5.7= 12=1111

p11.

BCH Codes Table 5.1 Construction of GF(24) using

h(x)=1+x+x4

word polynomial in x mod h(x)power of

0000 0 -

1000 1 0=1

0100 x

0010 x2 2

0001 x3 3

1100 1+x=x4 4

0110 x+x2=x5 5

0011 x2+x3=x6 6

p12.

BCH Codes Table 5.1(continue) Construction of GF(24)

using h(x)=1+x+x4

word polynomial in x mod h(x)power of

1101 1+x+x3=x7 7

1010 1+x2=x8 8

0101 x+x3=x9 9

1110 1+x+x2 =x10 10

0111 x+x2+x3 =x11 11

1111 1+x+x2+x3 =x12 12

1011 1+x2+x3 =x13 13

1001 1+x3 =x14 14

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BCH Codes

[2] Minimal polynomials

1. Root of a polynomial

: an element of F=GF(2r), p(x)F[x] is a root of a polynomial p(x) iff p()=0

2. Order of

The smallest positive integer m such that m=1 in GF(2r) is a primitive element if it has order 2r-1

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BCH Codes

3. Minimal polynomial of

The polynomial in K[x] of smallest degree having as root

Denoted by m(x) m(x) is irreducible over K If f(x) is any polynomial over K such that f()=0,

then m(x) is a factor of f(x) m(x) is unique m(x) is a factor of

121 r

x

p15.

BCH Codes

Example

Let p(x)=1+x3+x4, and let be the primitive element in GF(24) constructed using h(x)=1+x+x4(see Table

5.1):

p()=1+3+4=1000+0001+1100=0101=9

is not a root of p(x). However

p(7)=1+(7)3+(7)4=1+21+28=1+6+13

=1000+0011+1011=0000=0

7 is a root of p(x).

p16.

BCH Codes

4. Finding the minimal polynomial of

Reduce to find a linear combination of the vectors {1, , 2,…, r}, which sums to 0 Any set of r+1 vectors in Kr is dependent, such a solution exists Represent m(x) by mi(x) where =I

eg.

Find the m(x), =3, GF(24) constructed using

h(x)=1+x+x4

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BCH Codes

Useful facts: f(x)2=f(x2)

If f()=0, then f(2)=(f())2=0 If is a root of f(x), so are , 2, 4,…, The degree of m(x) is |{, 2, 4,…, }|

in

ii

n

i

ii

n

i

ii xaxaxa )()()(

0

22

0

22

0

12 r

12 r

p18.

BCH Codes

Example

Find the m(x), =3, GF(24) constructed using

h(x)=1+x+x4

Let m(x)= m3(x)=a0+a1x+a2x2+a3x3+a4x4 then we must find the value for a0,a1,…,a4 {0,1}

m()=0=a01+a1+a22+a33+a44

=a00+a13+a26+a39+a412

0000=a0(1000)+a1(0001)+a2(0011)+a3(0101)+a4(11

11)

a0=a1=a2=a3=a4=1 and

m(x)=1+x+x2+x3+x4

p19.

BCH Codes

Example

Let m5(x) be the minimal polynomials of =5,

5GF(24)

Since {, 2, 4, 8}={5 , 10}, the roots of m5(x) are

5 and 10 which means that degree (m5(x))=2. Thus

m5(x)=a0+a1x+a2x2:

0=a0+a1 5+a2 10

=a0(1000)+a1 (0110)+a2 (1110)

Thus a0=a1=a2=1 and m5(x)=1+x+x2

p20.

BCH Codes

Table 5.2: Minimal polynomials in GF(24) constructed using 1+x+x4

Element of GF(24) Minimal polynomial

01, 2, 4, 8

3, 6, 9, 12

5, 10

7, 11, 13, 14

x1+x1+x+x4

1+x+x2+x3+x4

1+x+x2

1+x3+x4

p21.

BCH Codes

[3] Cyclic Hamming codes

1. Parity check matrix

The parity check matrix of a Hamming code of length n=2r-1 has its rows all 2r-1 nonzero words of length r

is a primitive element of GF(2r)

H is the parity check ma-trix of a Hamming code oflength n=2r-1

22

2

1

r

H

p22.

BCH Codes

2. Generator polynomial

For any received word w=w0w1…wn-1

wH=w0+w1+…+wn-1n-1 w() w is a codeword iff is a root of w(x) m(x) is its generator polynomial

Theorem 5.3.1 A primitive polynomial of degree r is the generator polynomial of a cyclic Hamming code of length 2r-1

p23.

BCH Codes

Example:

Let r=3, so n=23-1=7. Use p(x)=1+x+x3 to construct GF(23), and 010 as the primitive element. Recall that i xi mod p(x). Therefore a parity

check matrix for a Hamming code of length 7 is

H

101

111

011

110

001

010

1001

6

5

4

3

2

p24.

BCH Codes

3. Decoding the cyclic Hamming code

w(x)=c(x)+e(x), where c(x) is a codeword,

e(x) is the error

w(β)=e(β)

e has weight 1, e(β)= βj, j is the position of the 1 in e

c(x)=w(x)+xj

p25.

BCH Codes

Example:

Suppose GF(23) was constructed using 1+x+x3.

m1(x)=1+x+x3 is the generator for a cyclic Hamming

code of length 7. Suppose

w(x)=1+x+x3+x6 is received. Then

w()=1+ 2+ 3+ 6

=100+001+110+101 =110 = 3

e(x)= x3 and c(x)=w(x)+x3=1+x2+x6

p26.

BCH Codes

[4] BCH codes

1. BCH: Bose-Chaudhuri-Hocquengham

Admit a relatively easy decoding scheme The class of BCH codes is quite extensive For any positive integers r and t with t 2r-1-1,

there is a BCH codes of length n=2r-1 which is t-error correcting and has dimension k n-rt

p27.

BCH Codes

2. Parity check matrix for the 2 error-correcting BCH

The 2 error-correcting BCH codes of length 2r-1 is the cyclic linear codes, generated by

g(x)= , r 4

)()( 3 xmxm

)22(3

6

3

0

22

2

0

rr

H

The generator polynomial:

g(x)=m1(x) m3(x)

Degree(g(x))=2r, the code has dimension n-2r=2r-1-2r

p28.

BCH Codes

Example:

is a primitive element in GF(24) constructed with p(x) = 1+x+x4. We have that m1(x)=1+x+x4 and m3(x) = 1+x+x2+x3+x4. Therefore

g(x)= m1(x) m3(x)= 1+x4+x6+x7+x8

is the generator for a 2 error-correcting BCH code of length 15

p29.

BCH Codes

3. The parity check matrix of C15 (distance d=5)

(Table 5.3)

H

11111001

01011011

00111111

00010111

10001110

11110101

01011010

00111101

00010011

10000110

11111100

01010001

00110010

00010100

10001000

1

1

11

1214

913

612

311

10

129

98

67

36

5

124

93

62

3

p30.

BCH Codes

[5] Decoding 2 error-correcting BCH codes

1. Error locator polynomial

w(x): received word syndrome wH=[w(),w(3)]=[s1,s3]

H is the parity check matrix for the (2r-1, 2r-2r-1, 5) 2 error-correcting BCH code with generator g(x)=m1(x) m3(x)

wH=0 if no errors occurred

If one error occurred, the error polynomial e(x)=xi

wH=eH=[e(), e(3)]=[i, 3i]=[s1,s3], 331 ss

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BCH Codes

If two errors occurred, say in positions i and j, ij, e(x)=xi+xj, wH=eH=[e(), e(3)]

=[i+j, 3i+3j]=[s1,s3]

The error locator polynomial:

)())(( 211

22333

jijjiijiji sss

jiss

s 21

1

3 0))((let , ji xx

0)( 21

1

31

2 ss

sxsx

p32.

BCH Codes

Example:Let ww(x) be a received word with syndromes s1=0111=w() and s3=1010= w(3), where w was

encoded using C15. From Table 5.1 we have that

s1 11 and s3 8. Then

We form the polynomial x2+11x+2 and find that it has roots 4 and 13. Therefore we can decide that the most likely errors occurred in positions 4 and

13, e(x)= x4+x13, the most likely error pattern is

0000100000000010

27122211821

1

3 ss

s

p33.

BCH Codes

2. Decoding algorithm of BCH codes Calculate the syndrome wH=[s1,s3]=[w(),w(3)] If s1=s3=0, no errors occurred If s1=0 and s30, ask for retransmission If (s1)3=s3, a single error at position i, where s1=i

From the quadratic equation:

(*)

If equation(*) has two distinct roots i and j, correct errors at positions i and j If equation(*) does not have two distinct roots in GF(2r), conclude that at least three errors

occurred

0)( 21

1

31

2 ss

sxsx

p34.

BCH Codes

Example: Assume w is received and the syndrome is

wH=01111010 [11,8]. Now

In this case equation(*) is x2+11x+2=0 which has roots 4 and 13. Correct error in positions i=4 and j=13.

Example: Assume the syndrome is wH=[w(),w(3)]=[3, 9].

Then (s1)3= (3)3=s3. A single error at position i=3.

e(x)=x3 is the error polynomial.

383333113

1 )( ss

p35.

BCH Codes

Example

Assume w=110111101011000 is received. The syndrome is

wH=01110110 [11, 5]= [s1,s3].

Now

So in this case, (*) becomes x2+11x+0=0.

53

33331131 )( ss

0792211521

1

3 100011010101 ss

s

p36.

BCH Codes

So in this case, (*) becomes x2+11x+0=0.

Trying the elements of GF(24) in turn as possible roots,

we come to x= 7 and find

(7)2+117+0=14+3+0

1001+0001+1000=0000

Now 7j=1=15, so j=8, is the other root. Correct error at positions i=7 and j=8;

u=000000011000000 is the most likely error pattern.

We decode v=w+u=110111110011000 as the word sent.

p37.

BCH Codes

Example: Assume a codeword in C15 is sent, and

errors occur in positions 2, 6 and 12. Then the syndrome wH is the sum of rows 2, 6, and 12 of H, where w is the word received. Thus

wH=00100011+00110001+11110011

= 11100001 [10, 3]= [s1,s3]

Now

(*) becomes x2+10x+4=0, no roots in GF(24).

Therefore IMLD for C15 concludes correctly, that at

least three errors occurred.

33

3031031 1)( ss

4582010321

1

3 110001101010 ss

s