BCH Codes HsinLung Wu NTPU OUTLINE n n

BCH Codes Hsin-Lung Wu NTPU

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

BCH Codes n [1] Finite fields n 1. Irreducible polynomial n f(x) K[x], f(x) has no proper divisors in K[x] Eg. f(x)=1+x+x 2 is irreducible f(x)=1+x+x 2+x 3=(1+x)(1+x 2) is not irreducible f(x)=1+x+x 4 is irreducible p 3.

BCH Codes n 2. Primitive polynomial n n f(x) is irreducible of degree n > 1 f(x) is not a divisor of 1+xm for any m < 2 n-1 Eg. f(x)=1+x+x 2 is not a factor of 1+xm for m < 3 so f(x) is a primitive polynomial f(x)= 1+x+x 2+x 3+x 4 is irreducible but 1+x 5=(1+x)(1+x+x 2+x 3+x 4) and m=5 < 24 -1=15 so f(x) is not a primitive polynomial p 4.

BCH Codes n 3. Definition of Kn[x] n The set of all polynomials in K[x] having degree less than n n Each word in Kn corresponds to a polynomial in Kn[x] n Multiplication in Kn modulo h(x), with irreducible h(x) of degree n n If we use multiplication modulo a reducible h(x), say, 1+x 4 to define multiplication of words in K 4, however: (0101) (x+x 3) = x 2+x 6 = x 2+x 2 (mod 1+x 4) = 0 0000 (K 4 -{0000} is not closed under multiplication. ) p 5.

BCH Codes n 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) p 6.

BCH Codes n 4. Definition of Field (Kn, +, x) n n n n (Kn, +) is an abelian group with identity denoted 0 The operation x is associative n a x ( b x c) = ( a x b ) x c There is a multiplicative identity denoted 1, with 1 0 n n 1 x a = a x 1 = a, a K The operation x is distributive over + n a x ( b + c ) = ( a x b ) + ( a x c ) It is communicative n n a x b = b x a, b K All non-zero elements have multiplicative inverses Galois Fields: GF(2 r) m n For every prime power order p , there is a unique finite field of order pm m n Denoted by GF(p ) p 7.

BCH Codes n Example n Let us consider the construction of GF(23) using the primitive polynomial h(x)=1+x+x 3 to define multiplication. We do this by computing xi mod h(x): word xi mod h(x) 100 1 010 x 001 x 2 110 x 3 1+x 011 x 4 x+x 2 111 x 5 1+x+x 2 101 x 6 1+x 2 p 8.

BCH Codes n 5. Use a primitive polynomial to construct GF(2 n) n Let Kn represent the word corresponding to x mod h(x) n i xi mod h(x) n m 1 for m<2 n-1 n since h(x) dose not divide 1+xm for m<2 n-1 n Since j = i for j i iff i = j-i i j-i n Kn{0}={ i | i = 0, 1, …, 2 n-2} =1 p 9.

BCH Codes n 6. GF(2 r) is primitive n is primitive if m 1 for 1 m <2 r-1 n In other words, every non-zero word in GF(2 r) can be expressed as a power of n Example Construct GF(24) using the primitive polynomial h(x)=1+x+x 4. 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 p 10.

BCH Codes n Table 5. 1 Construction of GF(24) using h(x)=1+x+x 4 word polynomial in x mod h(x) power of 0000 0 - 1000 1 0=1 0100 x 0010 x 2 2 0001 x 3 3 1100 1+x=x 4 4 0110 x+x 2=x 5 5 0011 x 2+x 3=x 6 6 p 11.

BCH Codes n Table 5. 1(continue) Construction of GF(24) using h(x)=1+x+x 4 word polynomial in x mod h(x) power of 1101 1+x+x 3=x 7 7 1010 1+x 2=x 8 8 0101 x+x 3=x 9 9 1110 1+x+x 2 =x 10 0111 x+x 2+x 3 =x 11 1111 1+x+x 2+x 3 =x 12 1011 1+x 2+x 3 =x 13 1001 1+x 3 =x 14 p 12.

BCH Codes § [2] Minimal polynomials § 1. Root of a polynomial § § § : an element of F=GF(2 r), 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(2 r) is a primitive element if it has order 2 r-1 p 13.

BCH Codes § 3. Minimal polynomial of § § § The polynomial in K[x] of smallest degree having as root Denoted by 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 p 14.

BCH Codes § Example Let p(x)=1+x 3+x 4, and let be the primitive element in GF(24) constructed using h(x)=1+x+x 4(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+ 28=1+ 6+ 13 =1000+0011+1011=0000=0 7 is a root of p(x). p 15.

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+x 4 p 16.

BCH Codes § § Useful facts: f(x)2=f(x 2) 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, …, }| p 17.

BCH Codes § Example § § Find the m (x), = 3, GF(24) constructed using h(x)=1+x+x 4 Let m (x)= m 3(x)=a 0+a 1 x+a 2 x 2+a 3 x 3+a 4 x 4 then we must find the value for a 0, a 1, …, a 4 {0, 1} m ( )=0=a 01+a 1 +a 2 2+a 3 3+a 4 4 =a 0 0+a 1 3+a 2 6+a 3 9+a 4 12 0000=a 0(1000)+a 1(0001)+a 2(0011)+a 3(0101)+a 4(1111) a 0=a 1=a 2=a 3=a 4=1 and m (x)=1+x+x 2+x 3+x 4 p 18.

BCH Codes § Example § Let m 5(x) be the minimal polynomials of = 5, 5 GF(24) Since { , 2, 4, 8}={ 5 , 10}, the roots of m 5(x) are 5 and 10 which means that degree (m 5(x))=2. Thus m 5(x)=a 0+a 1 x+a 2 x 2: 0=a 0+a 1 5+a 2 10 =a 0(1000)+a 1 (0110)+a 2 (1110) Thus a 0=a 1=a 2=1 and m 5(x)=1+x+x 2 p 19.

BCH Codes § Table 5. 2: Minimal polynomials in GF(24) constructed using 1+x+x 4 Element of GF(24) 0 1 , 2, 4, 8 3, 6, 9, 12 5, 10 7, 11, 13, 14 Minimal polynomial x 1+x+x 4 1+x+x 2+x 3+x 4 1+x+x 2 1+x 3+x 4 p 20.

BCH Codes § [3] Cyclic Hamming codes § 1. Parity check matrix § § § The parity check matrix of a Hamming code of length n=2 r -1 has its rows all 2 r-1 nonzero words of length r is a primitive element of GF(2 r) H is the parity check matrix of a Hamming code of length n=2 r-1 p 21.

BCH Codes § 2. Generator polynomial § § § For any received word w=w 0 w 1…wn-1 w. H=w 0 +w 1 +…+wn-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 2 r-1 p 22.

BCH Codes § Example: Let r=3, so n=23 -1=7. Use p(x)=1+x+x 3 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 p 23.

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 p 24.

BCH Codes § Example: Suppose GF(23) was constructed using 1+x+x 3. m 1(x)=1+x+x 3 is the generator for a cyclic Hamming code of length 7. Suppose w(x)=1+x+x 3+x 6 is received. Then w( )=1+ 2+ 3+ 6 =100+001+110+101 =110 = 3 e(x)= x 3 and c(x)=w(x)+x 3=1+x 2+x 6 p 25.

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 2 r-1 -1, there is a BCH codes of length n=2 r-1 which is t-error correcting and has dimension k n-rt p 26.

BCH Codes § 2. Parity check matrix for the 2 error-correcting BCH § The 2 error-correcting BCH codes of length 2 r-1 is the cyclic linear codes, generated by g(x)= , r 4 The generator polynomial: g(x)=m 1(x) m 3(x) Degree(g(x))=2 r, the code has dimension n-2 r=2 r-1 -2 r p 27.

BCH Codes n Example: is a primitive element in GF(24) constructed with p(x) = 1+x+x 4. We have that m 1(x)=1+x+x 4 and m 3(x) = 1+x+x 2+x 3+x 4. Therefore g(x)= m 1(x) m 3(x)= 1+x 4+x 6+x 7+x 8 is the generator for a 2 error-correcting BCH code of length 15 p 28.

BCH Codes § 3. The parity check matrix of C 15 (distance d=5) (Table 5. 3) p 29.

BCH Codes § [5] Decoding 2 error-correcting BCH codes § 1. Error locator polynomial w(x): received word syndrome w. H=[w( ), w( 3)]=[s 1, s 3] § § § H is the parity check matrix for the (2 r-1, 2 r-2 r-1, 5) 2 error-correcting BCH code with generator g(x)=m 1(x) m 3(x) w. H=0 if no errors occurred If one error occurred, the error polynomial e(x)=xi w. H=e. H=[e( ), e( 3)]=[ i, 3 i]=[s 1, s 3], p 30.

BCH Codes § § If two errors occurred, say in positions i and j, i j, e(x)=xi+xj, w. H=e. H=[e( ), e( 3)] =[ i+ j, 3 i+ 3 j]=[s 1, s 3] The error locator polynomial: p 31.

BCH Codes § Example: Let w w(x) be a received word with syndromes s 1=0111=w( ) and s 3=1010= w( 3), where w was encoded using C 15. From Table 5. 1 we have that s 1 11 and s 3 8. Then We form the polynomial x 2+ 11 x+ 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)= x 4+x 13, the most likely error pattern is 000010000010 p 32.

BCH Codes § 2. Decoding algorithm of BCH codes § § § Calculate the syndrome w. H=[s 1, s 3]=[w( ), w( 3)] If s 1=s 3=0, no errors occurred If s 1=0 and s 3 0, ask for retransmission If (s 1)3=s 3, a single error at position i, where s 1= 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(2 r), conclude that at least three errors occurred p 33.

BCH Codes § Example: Assume w is received and the syndrome is w. H=01111010 [ 11, 8]. Now In this case equation(*) is x 2+ 11 x+ 2=0 which has roots 4 and 13. Correct error in positions i=4 and j=13. § Example: Assume the syndrome is w. H=[w( ), w( 3)]=[ 3, 9]. Then (s 1)3= ( 3)3=s 3. A single error at position i=3. e(x)=x 3 is the error polynomial. p 34.

BCH Codes § Example Assume w=110111101011000 is received. The syndrome is w. H=01110110 [ 11, 5]= [s 1, s 3]. Now So in this case, (*) becomes x 2+ 11 x+ 0=0. p 35.

BCH Codes So in this case, (*) becomes x 2+ 11 x+ 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=11011111000 as the word sent. p 36.

BCH Codes § Example: Assume a codeword in C 15 is sent, and errors occur in positions 2, 6 and 12. Then the syndrome w. H is the sum of rows 2, 6, and 12 of H, where w is the word received. Thus w. H=00100011+00110001+11110011 = 11100001 [ 10, 3]= [s 1, s 3] Now (*) becomes x 2+ 10 x+ 4=0, no roots in GF(24). Therefore IMLD for C 15 concludes correctly, that at least three errors occurred. p 37.