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BCH Codes
Hsin-Lung Wu
NTPU
OUTLINE





[1] Finite fields
[2] Minimal polynomials
[3] Cyclic Hamming codes
[4] BCH codes
[5] Decoding 2 error-correcting BCH codes
p2.
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 irreducible
f(x)=1+x+x2+x3=(1+x)(1+x2) is not irreducible
f(x)=1+x+x4 is irreducible
p3.
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
p4.
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.)
p5.
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)
p6.
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 10
n
 1 x a = a x 1 = a,  a  K
The operation x is distributive over +
 a x ( b + c ) = ( a x b ) + ( a x c )
It is communicative
n
 a x b = b x a,  a,b  K
All non-zero elements have multiplicative inverses
Galois Fields: GF(2r)
m
 For every prime power order p , there is a unique finite field
of order pm
m
 Denoted by GF(p )
p7.
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
1
010
x
001
x2
110
x3  1+x
011
x4  x+x2
111
x5  1+x+x2
101
x6  1+x2
p8.
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 ji iff i = j-i i
 j-i

Kn\{0}={i | i = 0,1,…,2n-2}
=1
p9.
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
p10.
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
p11.
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
p12.
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
p13.
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
r
2
 m(x) is a factor of 1  x 1




p14.
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).
p15.
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
p16.
BCH Codes
 Useful facts: f(x)2=f(x2)
n
n
n
(  ai x )   a ( x )   ai ( x 2 )i
i 2
i 0
i 0
2
i
i 2
i 0
 If f()=0, then f(2)=(f())2=0
 If  is a root of f(x), so are , 2, 4,…,  2r 1
 The degree of m(x) is |{, 2, 4,…,  2r 1 }|
p17.
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+a22+a33+a44
=a00+a13+a26+a39+a412
0000=a0(1000)+a1(0001)+a2(0011)+a3(0101)+a4(1111)
 a0=a1=a2=a3=a4=1 and
m(x)=1+x+x2+x3+x4
p18.
BCH Codes

Example
 Let m5(x) be the minimal polynomials of =5, 5GF(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
p19.
BCH Codes

Table 5.2: Minimal polynomials in GF(24)
constructed using 1+x+x4
Element of GF(24)
0
1
, 2, 4, 8
3, 6, 9, 12
5, 10
7, 11, 13, 14
Minimal polynomial
x
1+x
1+x+x4
1+x+x2+x3+x4
1+x+x2
1+x3+x4
p20.
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 matrix of a Hamming code of
length n=2r-1
 1 





H   2 


  
 2r 2 
 
p21.
BCH Codes

2. Generator polynomial
 For any received word w=w0w1…wn-1
wH=w0+w1+…+wn-1n-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
p22.
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
1
100 
 
010
 


2
 
 001
 3




110
 

H
 4 
 011
 5



111
 


6
  
101
p23.
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
p24.
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
p25.
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
p26.
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)= m ( x )m  3 ( x ) , r  4
0 
 0


3
 
 
H   2
6 


 
 
 2 r  2 3( 2 r  2 ) 



The generator polynomial:
g(x)=m1(x) m3(x)
Degree(g(x))=2r, the code has
dimension n-2r=2r-1-2r
p27.
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
p28.
BCH Codes

3. The parity check matrix of C15 (distance d=5)
 1


2
 3

4
 5

6
 7

8
 9

  10
 11

  12

  13
 14

1  1000
 3  0100
 
6
  0010
 9   0001
 12  1100
 
1  0110
 3   0011
 
 6   1101
 9  1010
 
 12   0101
1  1110
 
 3   0111
 6  1111
 
9
  1011
 12  1001
1000
0001

0011

0101
1111

1000
0001

0011  H
0101

1111
1000

0001
0011

0101

1111
(Table 5.3)
p29.
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], s 3  s
1
3
p30.
BCH Codes
 If two errors occurred, say in positions i and j, ij,
e(x)=xi+xj, wH=eH=[e(), e(3)]
=[i+j, 3i+3j]=[s1,s3]
s3   3i   3 j  (  i   j )(  2i   i  j   2 j )  s1 ( s12   i  j )

s3
i
j
 s12   i  j , let ( x   )( x   )  0
s1
 The error locator polynomial:
x 2  s1 x  (
s3
 s12 )  0
s1
p31.
BCH Codes
 Example:
Let ww(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
s3
 s12   8  11   22   12   7   2
s1
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
p32.
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 s30, ask for retransmission
If (s1)3=s3, a single error at position i, where s1=i
From the quadratic equation:
s3
x  s1 x  (  s12 )  0
s1
2
(*)
 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
p33.
BCH Codes
 Example:
Assume w is received and the syndrome is
wH=01111010  [11,8]. Now
s13  (  11 )3   33   3   8  s3
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.
p34.
BCH Codes
 Example
Assume w=110111101011000 is received.
The syndrome is
wH=01110110  [11, 5]= [s1,s3].
Now
s13  (  11 )3   33   3  s3   5
s3
 s12   5  11   22   9   7  0101  1101  1000   0
s1
So in this case, (*) becomes x2+11x+0=0.
p35.
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+117+0=14+3+0
1001+0001+1000=0000
Now 7j=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.
p36.
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
s13  (  10 )3   30  1  s3   3
s3
 s12   3  10   20   8   5  1010  0110  1100   4
s1
(*) becomes x2+10x+4=0, no roots in GF(24). Therefore
IMLD for C15 concludes correctly, that at least three errors
occurred.
p37.