BCH_code

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BCH Codes

Yunghsiang S. Han

Graduate Institute of Communication Engineering,

National Taipei University

Taiwan

E-mail: [email protected]

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Y. S. Han BCH codes 1

Description of BCH Codes

The Bose, Chaudhuri, and Hocquenghem (BCH) codes form a

large class of powerful random error-correcting cyclic codes.

This class of codes is a remarkable generalization of the

Hamming code for multiple-error correction.

We only consider binary BCH codes in this lecture note.

Non-binary BCH codes such as Reed-Solomon codes will bediscussed in next lecture note.

For any positive integers m 3 and t

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We call this code a t-error-correctingBCH code.

Let be a primitive element in GF(2m). The generator

polynomial g(x) of the t-error-correcting BCH code of length

2m 1 is the lowest-degree polynomialover GF(2) which has

, 2, 3, . . . , 2t

as its roots.

g(i) = 0 for 1 i 2t and g(x) has , 2, . . . , 2t and their

conjugates as all its roots.

Let i(x) be the minimal polynomial of i. Then g(x) must be

the least common multipleof1(x), 2(x), . . . , 2t(x), i.e.,

g(x) = LCM{1(x), 2(x), . . . , (x)2t}.

Ifi is an even integer, it can be expressed as i= i2, where i is

Graduate Institute of Communication Engineering, National Taipei University

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odd and > 1. Then i =

i2

is a conjugate of i

. Hence,

i(x) = i(x).

g(x) = LCM{1(x), 3(x), . . . , 2t1(x)}.

The degree ofg(x) is at most mt. That is, the number ofparity-check digits, n k, of the code is at most equal to mt.

Ift is small, n k is exactly equal to mt.

Since is a primitive element, the BCH codes defined are usually

called primitive(or narrow-sense) BCH codes.


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Example

Let be a primitive element ofGF(24) such that 1 + + 4 = 0.

The minimal polynomials of , 3, and 5 are

1(x) = 1 + x + x4,

3(x) = 1 + x + x2 + x3 + x4,

5(x) = 1 + x + x2,

respectively. The double-error-correcting BCH code of length

n= 24 1 = 15 is generated by

g(x) = LCM{1(x), 3(x)}

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

= 1 + x4 + x6 + x7 + x8.

n k= 8 such that this is a (15, 7, 5) code. Since the weight of


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the generator polynomial is 5, it is a (15, 7, 5) code.

The triple-error-correcting BCH code of length 15 is generated by

g(x) = LCM{1(x), 3(x), 5(x)}

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

= 1 + x + x2 + x4 + x5 + x8 + x10.

n k= 10 such that this is a (15, 5, 7) code. Since the weight

of the generator polynomial is 7, it is a (15, 5, 7) code.

The single-error-correcting BCH code of length 2m 1 is a

Hamming code.


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!"#$"%"&'(')*&% *, -./0123 #/42 5 416 4 6 7

89#*&"&')(:

;*'(')*&

)(:

;*'(')*&

?)&($=

;*'(')*&

@"A)>(:

;*'(')*&

B)&)>(:

)(:

C C CCCC C 9!

C 7 CCC7 7 9 6 7!

7 4 CC7C 0 916 9 6 7!0 40 C7CC 1 916 9 6 7!

D 4D 7CCC E 916 9D6 906 9 6 7

!1 4 6 7 CC77 D 916 9 6 7!

F 406 4 C77C G 906 9 6 7!

G 4D6 40 77CC 70 916 9D6 906 9 6 7

!H 4D6 4 6 7 7C77 77 916 9D6 7!

E 406 7 C7C7 F 916 9 6 7!

I 4D6 4 7C7C 7C 916 9D6 906 9 6 7

!7C 406 4 6 7 C777 H 906 9 6 7

!77 4D6 406 4 6 7 777C 71 916 9D6 7!

70 4D6 406 4 6 7 7777 7F 916 9D6 906 9 6 7!

7D 4D6 406 7 77C7 7D 916 9D6 7!

71 4D6 7 7CC7 I 916 9D6 7

!!0!1!E!7G!!!

D!G!70!01!1E!!D

!!I


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Examples of Finite Fields

"#$%&

' !"#$%'()*+)&

,),&- .$)' / )&

,),&01.23435678

92575623:28;32;

92575623?4@6

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BCH Codes of Lengths Less than 210 1 (1)

! " # $ ! " # $ ! " # $ " # $ " # $

% & ' ( "# )' & ()& *+ (% )** (,& - )** &( )-

' (* (( ( (, (+ '% (' (&- (+ .% %+

& ) (. (( %. (* (&( (( ** %(

* % (+ (% )- )( (.% () '& ')

* %( ). ( & (* )) )% (** (% '* '%

)( ) $ ()& ()+ ( (* )& ('& (' %& '*

(. % ((% ) , %( (%- (* )- '&

(( * (+. % , )** )'& ( (%( (, )( **

. & -- ' )%- ) ()% (- (% *-

. .% *& ( -) * )%( % ((* )( - .%

*( ) ,* . ))% ' (+& )) *(( *+) (

'* % &, & )(* * -- )% '-% )

%- ' &( - )+& . -( )* ',' %

%. * .' (+ (-- & ,& ). '&* '

%+ . *& (( (-( , &- )& '.. *

%&' ( )*+,,

- . / 0 *(


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BCH Codes of Lengths Less than 210 1 (2)

! # $ ! # $ ! # $ ! # $ ! # $

%&& '%( ) %&& *++ ++ %&& &,* '* %&& %- ,& &.+* ,** ,

''- ( *&* +* &-' '% ', ,* ,+* &.

'*, - *.' +% &(% ') '. ,% ,&* &&

'*. , +,% +) &)) '( *& &., ,.* &+

'+& &. +-) +( &%( %& +- &&& -,* &*

'&+ && +(( +- &'- %* &, &&, --* &'

'.* &+ +)- +, &*, %' &. &+& -(* &%

*,' &* +%, *. &*. %% &.&* & -)* &)

*-% &' +%. *& &+& %- &.+* &..* + -%- &(

*() &% +'& *) &&+ %, ,,* *

*)( &) +*- *( &.* )& ,-* '

*%- &- ++, *- ,' )+ ,(* %

*', &, ++. *, -% )* ,)* )

*'. +. +&& '& () -% ,%* (

**& +& +.+ '+ )( -( ,'* -


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Minimal Polynomials of the Elements in GF(26)


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Generator Polynomials of All BCH Codes of Length 63


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Parity-Check Matrix of a BCH Code

We can define a t-error-correcting BCH code of length

n= 2m 1 in the following manner: A binary n-tuple

v= (v0, v1, . . . , vn1) is a code word if and only if the polynomialv(x) =v0+ v1x + + vn1x

n1 has , 2, . . . , 2t as roots.

Since i is a root ofv(x) for 1 i 2t, then

v(i) =v0+ v1i + v2

2i + + vn1(n1)i = 0.


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This equality can be written as a matrix product as follows:

(v0, v1, . . . , vn1)

1

i

2i

...

(n1)i

= 0 (1)

for 1 i 2t.


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Let

H=

1 2 3 n1

1 (2) (2)2 (2)3 (2)n1

1 (3) (3)2 (3)3 (3)n1

......

1 (2t) (2t)2 (2t)3 (2t)n1

. (2)

From (1), ifv= (v0, v1, . . . , vn1) is a code word in the

t-error-correcting BCH code, then

vHT =0.

If an n-tuple v satisfies the above condition, i is a root of thepolynomial v(x). Therefore, v must be a code word in the

t-error-correcting BCH code.


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H is a parity-check matrix of the code.

If for some i andj, j is a conjugate of i, then v(j) = 0 if and

only ifv(i) = 0.

The Hmatrix can be reduced to

H=

1 2 3 n1

1 (3) (3)2 (3)3 (3)n1

1 (5) (5)2 (5)3 (5)n1

......

1 (2t1) (2t1)2 (2t1)3 (2t1)n1

.

If each entry ofHis replaced by its corresponding m-tuple overGF(2) arranged in column form, we obtain a binary parity-check

matrix for the code.


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BCH Bound

The t-error-correcting BCH code defined has minimum distance

at least 2t + 1.

Proof: We need to show that no 2t of fewer columns ofH sum

to zero. Suppose that there exists a nonzero code vector v with

weight 2t. Let vj1 , vj2 , . . . , vj be the nonzero components of

v. Then

0 = v HT

= (vj1 , vj2 , . . . , vj)

j1 (2)j1 (2t)j1

j2 (2)j2 (2t)j2

j3 (2)j3 (2t)j3

... ... ...

j (2)j (2t)j


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= (1, 1, . . . , 1)

j1 (j1 )2 (j1 )2t

j2

(j2

)

2

(j2

)

2t

j3 (j3 )2 (j3 )2t

......

...

j (j )2 (j)2t

.

The equality above implies the following equality:

(1, 1, . . . , 1)

j1 (j1 )2 (j1 )

j2 (j2 )2 (j2 )

j3 (j3 )2 (j3 )

... ... ...

j (j)2 (j )

=0,


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which the second matrix on the left is a square matrix. To

satisfy the above equality, the determinant of the matrix

must be zero. That is,

j1 (j1 )2 (j1 )

j2 (j2 )2 (j2 )

j3 (j3 )2 (j3 )

......

...

j (j)2 (j)

= 0.


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Then

j1+j2++j

1 j1 j1(1)

1 j2 j2(1)

1 j3 j3(1)

......

...

1 j j(1)

= 0.

The determinant in the equality above is a Vandermonde

determinantwhich is nonzero. Contradiction!

The parameter 2t + 1 is usually called the designed distanceof

the t-error-correcting BCH code.

The true minimum distance of the code might be larger than

2t + 1.


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Syndrome Calculation

Let

r(x) =r0+ r1x + r2x2 + + rn1x

n1

be the received vector and e(x) the error pattern. Then

r(x) =v(x) +e(x).

The syndrome is a 2t-tuple,

S= (S1, S2, . . . , S 2t) =r HT,

where His given by (2).

Si=r(i) =r0+ r1

i + r22i + + rn1

(n1)i

for 1 i 2t.


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Dividing r(x) by the minimal polynomial i(x) of i, we have

r(x) =ai(x)i

(x) +bi(x),

where bi(x) is the remainder with degree less than that ofi(x).

Since i(i) = 0, we have

Si =r(i) =bi(

i).

Since 1, 2, . . . , 2t are roots of each code polynomial, v(i) = 0

for 1 i 2t.

Then Si=e(i) for 1 i 2t.

We now consider a general case that is also good for non-binary

case.

Suppose that the error pattern e(x) has v errors at locations


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0 j1 < j2

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(3) can be rewritten as follows:

S1 = Y1X1+ Y2X2+ + YvXv

S2 = Y1X21 + Y2X

22 + + YvX

2v

S3 = Y1X31 + Y2X

32 + + YvX

3v

...

S2t = Y1X2t1 + Y2X

2t2 + + YvX

2tv . (4)

We need to transfer the above set of non-linear equations into a

set of linear equations.

Consider the error-locator polynomial

(x) = (1 X1x)(1 X2x) (1 Xvx)

= 1 + 1x + 2x2 + + vx

v. (5)

Multiplying (5) by YiXj+vi , where 1 j v, and set x= X

1i


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we have

0 =YiXj+vi 1 + 1X1i + 2X2i + + vXvi .

for 1 i v.

Summing all above v equations, we have

0 =v

i=1

Yi

Xj+vi + 1X

j+v1i + + vX

ji

=v

i=1

YiXj+vi + 1

vi=1

YiXj+v1i + + v

vi=1

YiXji

= Sj+v+ 1Sj+v1+ 2Sj+v2+ + vSj .

We have

1Sj+v1+ 2Sj+v2+ + vSj = Sj+v

for 1 j v.


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Putting the above equations into matrix form we have

S1 S2 Sv1 Sv

S2 S3 Sv Sv+1...

Sv Sv+1 S2v2 S2v1

v

v1

...

1

=

Sv+1

Sv+2...

S2v

. (6)

Since v t, S1, S2, . . . , S 2v are all known. Then we can solve for

1,2, . . . ,v.

We still need to find the smallest v such that the above system of

equations has a unique solution.


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Let the matrix of syndromes, M, be defined as follows:

M=

S1 S2 Su

S2 S3 Su+1...

......

Su Su+1 S2u1

.

M is nonsingular ifuis equal to v, the number of errors thatactually occurred. M is singular ifu > v.

Proof: Let

A=

1 1 1

X1 X2 Xu

......

...

Xu11 Xu12 X

u1u


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with Aij =Xi1j and

B =

Y1X1 0 00 Y2X2 0...

......

0 0 YuXu

with Bij =YiXiij , where

ij =

1 i= j

0 i =j.

We have

ABAT

ij

=u

=1

Xi1

uk=1

YXkXj1k


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=u

=1

Xi1

YXXj1

=u

=1

YXi+j1

=Mij .

Hence, M=ABAT. Ifu > v, then det(B) = 0 and then

det(M) = det(A)det(B)det(AT) = 0. Ifu= v, then det(B) = 0.

Since A is a Vandermonde matrix with Xi =Xj , i =j ,

det(A) = 0. Hence, det(M) = 0.


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The Peterson-Gorenstein-Zierler Algorithm


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Example

Consider the triple-error-correcting (15, 5) BCH code with

g(x) = 1 + x + x2 + x4 + x5 + x8 + x10. Assume that the received

vector is r(x) =x2 + x7. The operating finite field is GF(24). Then

the syndromes can be calculated as follows:

S1 = 7

+2

= 12

S2 = 14 +4 = 9

S3 = 21 +6 = 0

S4 = 28 +8 = 3

S5 =

35

+

10

=

0

= 1S6 =

42 +12 = 0.


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Set v= 3, we have

det(M) =

S1 S2 S3

S2 S3 S4

S3 S4 S5

=

12

9 0

9 0 3

0 3 1

= 0.

Set v= 2, we have

det(M) =

S1

S2

S2 S3

=

12

9

9 0

= 0.


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We then calculate

M1 = 0 6

6

9

.Hence,

2

1

= M1

0

3

=

9

12

and

(x) = 1 +12x +9x2

=

1 + 2x

1 + 7x

= 9 x 8

x 13

.Since 1/8 = 7 and 1/13 = 2, we found the error locations.


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