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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
p4.
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.)
p6.
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)
p7.
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)
p8.
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}
p10.
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
p13.
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
p14.
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
p17.
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
p31.
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