Reed-Muller and Preparata Codes

p2.

OUTLINE

[1] Reed-Muller codes

[2] Decoding Reed-Muller codes [3] Extended Preparata Codes

[4] Encoding Extended Preparata Codes

[5] Decoding Extended Preparata Codes

p3.

Chap 3 Perfect and Related Codes

Recall [3.8]. Reed-Muller codes

1. r-th order, length 2m, 0≦r≦m, RM(r, m)

RM(0, m)={00…0, 11…1}, RM(m, m)=

RM(r, m) =

2. Eg 3.8.1

{( , ) | ( , 1), ( 1, 1)},0x x y x RM r m y RM r m r m

2m

K

2

4

(0,0) {0,1}

(0,1) {00,11} (1,1) {00,01,10,11}

(0,2) {0000,1111} (2,2)

(1,2) {( , ) | {00,01,10,11}, {00,11}}

RM

RM RM K

RM RM K

RM x x y x y

p4.

Perfect and Related Codes

3. Generator matrix G of RM(r, m)

4. Eg 3.8.4 Find G(1,3)

( , 1) ( , 1)i) 0<r<m ( , )

0 ( 1, 1)

ii) r=0, (0, ) [11....1]

( 1, )iii)r=m, ( , )

0........01

G r m G r mG r m

G r m

G m

G m mG m m

11 11

(1,2) 01 01

00 11

1111 1111

(1,2) (1,2) 0101 0101(1,3) (1,3)

0011 0011 0 (0,2)

0000 1111

G

G GG G

G

p5.

Perfect and Related Codes

5. The properties of RM(r,m)

m

m-r

r

i=0

(1) length n=2

(2) distance d=2

(3) dimension k= ( )

(4) RM(r-1,m) is contained in RM(r,m), r>0

(5) dual code RM(m-1-r, m), r<m

mi

p6.

Reed-Muller Codes

[1] Reed-Muller codes RM(r,m) linear (n, k, d) code with

n = 2m

Dimension k = Min. distance d = 2m-r

Standard ording of Km

Label position i with binary representation ui Km in reverse

order (low order digit first) Eg. 9.1.1

Standard ordering for K2 = (00, 10, 01, 11)

Standard ordering for K3 = (000, 100, 010, 110, 001, 101, 011, 111)

r

i i

m0

0 1 2 3 4 5 6 7

0 1 2 3

p7.

Reed-Muller Codes

Any function f : Km {0, 1} has a unique vectorv = (f(u0), f(u1), …, f(u2

m-1)) Kn , n=2m

where ui Km and follows standard ordering

Consider fI : Km {0, 1} as follow Given I {0, 1, …, m-1}

<def> vI is the corresponding vector form of fI

I if ,1

I if ),1(),...,,( 110

Ii imI

xxxxf

f v

p8.

Reed-Muller Codes

Eg. 9.1.2Let m = 3, so n = 23

(a) If I = {1, 2} then fI(x0, x1, x2) = (x1+1) (x2+1)

under standard ordering f{1,2}(0,0,0) = 1 f{1,2}(0,0,1) = 0

f{1,2}(1,0,0) = 1 f{1,2}(1,0,1) = 0

f{1,2}(0,1,0) = 0 f{1,2}(0,1,1) = 0

f{1,2}(1,1,0) = 0 f{1,2}(1,1,1) = 0

so vI = 11000000

(b) If I = {0} then fI(x0, x1, x2) = (x0+1)

so vI = 10101010

(c) If I = then fI(x0, x1, x2) = 1

so vI = 11111111

x0x1x2 (a) (b) (c) 0 0 0 1 1 1 1 0 0 1 0 1 0 1 0 0 1 1 1 1 0 0 0 1 0 0 1 0 1 1 1 0 1 0 0 1 0 1 1 0 1 1 1 1 1 0 0 1

p9.

Reed-Muller Codes

Two facts about fI

fI(x0, x1,…, xm-1) = 1 iff xi=0 for all iI wt(vI)=2m-|I|

For each ui Km

fI(ui)fJ(ui) = fIJ(ui) and

2) (mod )(

)(

)()(

12

0

12

0

JI

iiJI

iJi

iIJI

vwt

uf

ufufvv

m

m

x0x1x2 (a) (b) (c) (d) (e) 0 0 0 1 1 1 1 1 1 0 0 1 0 1 0 1 0 1 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 0 1 0 1 1 0 0 1 0 1 0 0 1 0 0 0 1 1 0 1 1 0 0 1 1 1 0 0 1 0 0

(a) I={1,2}(b) I={0}(c) I=(d) I={1,2}{0}(e) I={1,2}

I if ,1

I if ),1(),...,,( 110

Ii imI

xxxxf

p10.

Reed-Muller Codes

The Reed-Muller code RM(r,m) can be defined as the linear code <{vI | I Zm, |I| r}>

S = {vI | I Zm, |I| r} : linearly indep. and thus a basis for RM(r,m)

Generating matrix Gr,m in canonical form

Eg. 9.1.8The generating matrix for RM(4,4) in canonical

form (see next page)

Encodingm

I II Z ,|I| r

c mv

3,2,1,0

2,1,0

3,1,0

3,2,0

3,2,1

1,0

2,0

2,1

3,0

3,1

3,2

0

1

2

3

4,4

0000001000000000

0000001000000010

0000001000100000

0000001010000000

0000001100000000

0010001000100010

1000001010000010

0000001100000011

0000001010101000

0000001100110000

0000001111000000

1010101010101010

0011001100110011

1100001111000011

0000001111111100

1111111111111111

v

v

v

v

v

v

v

v

v

v

v

v

v

v

v

v

G

Canonical formvI comes before vJ if (1) |I|<|J| or (2) |I|=|J|, j such that fI(uj)<fJ(uj) and fI(ui)=fJ(ui) i > j

Eg. 9.1.11Encoding the following m using G2,4

(a) If m = 1 0000 001000 (so m = m0,3 = 1) then

c = v + v0,3 = 0101010111111111 (b) If m = 0 0101 001001 (so m2 = m0 = m0,3 = m0,1 = 1) then

c = v2 + v0 + v0,3 + v0,1 = 0111100011010010

p12.

Reed-Muller Codes

[2] Decoding Reed-Muller codes <Def>

IC = Zm\I : the complement of I in Zm

Eg. I = {1, 3, 5} Zm , then IC = {0, 2, 4, 6, 7}

HI = {uKm | fI(u)=1}

Eg. If I = {1, 2} then fI(x0, x1, x2) = (x1+1) (x2+1)

f{1,2}(0,0,0) = 1 f{1,2}(0,0,1) = 0

f{1,2}(1,0,0) = 1 f{1,2}(1,0,1) = 0

f{1,2}(0,1,0) = 0 f{1,2}(0,1,1) = 0

f{1,2}(1,1,0) = 0 f{1,2}(1,1,1) = 0

so HI = {000, 100}

fI(x0, x1,…, xm-1) = 1 iff xi=0 for all iI

HI is a subspace of Km

pf: x, y HI

xi=yi=0 xi+yi=0 x+y HI

p13.

Reed-Muller Codes

For u = (x0, x1, …, xm-1) Km , t = (t0, t1, …, tm-1) Km

Define fI,t(x0, x1, …, xm-1) = fI(x0+t0, x1+t1, …, xm-1+tm-1)

= fI(x+t)

Eg. If I = {1, 2} then fI(x0, x1, x2) = (x1+1) (x2+1)

Assume t = 010 fI,t(0,0,0) = fI(0+0, 0+1, 0+0) = 0

fI,t(1,0,0) = fI(1+0, 0+1, 0+0) = 0

fI,t(0,1,0) = fI(0+0, 1+1, 0+0) = 1

fI,t(1,1,0) = fI(1+0, 1+1, 0+0) = 1

fI,t(0,0,1) = fI(0+0, 0+1, 1+0) = 0

fI,t(1,0,1) = fI(1+0, 0+1, 1+0) = 0

fI,t(0,1,1) = fI(0+0, 1+1, 1+0) = 0

fI,t(1,1,1) = fI(1+0, 1+1, 1+0) = 0 so vI,t = 00110000

p14.

Reed-Muller Codes

Lemma 9.2.1Let I, J Zm and |I||J|, for any sHI

C and tHJ

vI,s vJC

,t = 1 iff I = J

pf:c

cI ,s i i j jJ ,t

i I j J

f (u)f (u) (x s 1) (x t 1)

c

cm |I J |

I ,s J ,tso |{u| f (u)f (u) 1}| 2

x0x1x2 f{1},010 f{2},001 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 1 1 0 1 0 0 0 0 1 0 1 0 1 0 1 0 1 0 0 1 1 1 1 1 1 1 1 1 1 1

I={1} s=010J={0,1} Jc={2} t=001

is even unless IJc=Zm

x0x1x2 f{1},010 f{0,2},101 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 1 1 0 1 0 0 0 0 1 0 0 0 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1 1 1

I={1} s=010J={1} Jc={0,2} t=101

p15.

Reed-Muller Codes

Corollary 9.2.2If c is a codeword in RM(r,m) and if |J| = r , then

mJ = c vJC

,t , tHJ

pf:

( In above, we choose s = (0,0,…,0) since (0,0,…,0) HJ

C for any J, |J| = r )

c c c

m

I I J J JJ ,t J ,t J ,tI Z ,|I| r

c v mv v mv v m

m

I II Z ,|I| r

c mv

p16.

Reed-Muller Codes

Lemma 9.2.3Let J Zm, for any word e (of length 2m)

e vJC

,t = 1

for at most wt(e) values of tHJ

pf:if t1,t2HJ , t1t2

then (HJc + t1) (HJ

c + t2)=

so vJc,t1 and vJ

c,t2 have no position

where both digits are 1Thus each 1 in e affects at most one of the values of e vJ

C,t

x0x1x2x3 v{2,3},0010 v{2,3},0001

0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 0 1 1 0 1 0 1 1 1 0 1 0 0 0 0 1 0 1 1 0 0 1 0 1 0 1 0 1 0 1 1 1 0 1 0 1 0 0 1 1 0 0 1 0 1 1 0 0 0 1 1 1 0 0 1 1 1 1 0 0

J={0,1}HJ={00xx} HJ

c={xx00}

p17.

Reed-Muller Codes

Idea of decoding algorithmLet w=c+e where c is a codeword in RM(r,m)Let |J|=rby lemma 9.2.3, e vJ

C,t = 0 for at least |HJ|-wt(e) values of tHJ

for such t we have w vJ

C,t = c vJ

C,t + e vJ

C,t

= c vJC

,t

= mJ (by cor. 9.2.2)

So if 2wt(e) < |HJ|, more than half of the w vJC

,t will be mJ

(i.e. more than |HJ|/2 will be mJ and less than |HJ|/2 will be mJ +1)

Now we can partially decode w by w(r-1) = w + |J|=rmJvJ

and apply similar method to w(r-1)

p18.

Reed-Muller Codes

Lemma 9.2.4The minimum distance of RM(r,m) is 2m-r

pf:The decoding algorithm can correct error

patterns of weight < |HJ|/2 where |J| r

Thus RM(r,m) has min. distance at least 2m-r

( |HJ| = wt(vJ) = 2m-|

J| )We know vI with |I|=r is a codeword in RM(r,m)

and wt(vI) = 2m-r

So min. distance of RM(r,m) is 2m-r

p19.

Reed-Muller Codes

Algorithm 9.2.5 (Majority logic decoding)Goal : decode a received word w encoded by RM(r,m)

1. i r, w(r) w 2. J Zm with |J|= i

compute w(i) vJC

,t for each tHJ

determine the majority mJ

3. If i > 0 then w(i-1) w(i) + mJvJ , where |J|=i

if wt(w(i-1)) 2m-r-1-1 then

set mJ=0 for all J with |J| r OUTPUT message m

else i i-1 goto step 2 If i = 0 then

OUTPUT message m

p20.

Reed-Muller Codes

Eg. 9.2.6Decode w = 0101011110100000 that was encoded using G2,4

1. i = r = 2 , w(2) = w m2,3 = 0 m1,3 = 0 m0,3 = 0

m1,2 = 0 m0,2 = 1 m0,1 = 0

Then w(1) = w(2) + v0,2 = 1111011100000000 2. i = 1

m3 = 1 m2 = 0 m1 = 0 m0 = 0 Then w(0) = w(1) + v3 = 0000100000000000

3. Since wt(w(0)) = 1 2m-r-1-1, so m = 0 OUTPUT 0 1000 000010

x0x1x2x3 v{2,3},0000

0 0 0 0 1 1 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 0 1 0 0 1 0 1 0 0 0 1 1 0 0 1 1 1 0 0 0 0 0 1 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 0 1 1 0 1 0 1 1 0 0 1 1 1 0 1 1 1 1 0

w(1)=1111011100000000

p23.

Preparata Codes

[3] Extended Preparata Codes U GF(2r), is a primitive element of GF(2r)

Define

Eg. 9.3.1 is a primitive element of GF(23)({0})=00000001({2, 5, 6})=00100110()=00000000

r

r

r

( ) be word of length 2 which is

1 in position i if (for 0 i 2 2)

1 in position 2 1 if 0 U

0 otherwise

i

U

U

GF(23) using 1+x+x3

word power of 000 0100 0

010 1

001 2

110 3

011 4

111 5

101 6

p24.

Preparata Codes

U GF(2r) , GF(2r) (1) U + = {u+|u U}(2) U = {u| u U}(3) UV = {x| x UV but xUV }

(U) + (V) = (UV) Eg.9.3.2

U={2, 5, 6} , V={5, 0}

Then U+ 2 = {2 + 2 , 5 + 2 , 6 + 2} = {0, 3, 0} 2 U = {22, 25, 2 6}={4, 0, 1}

(U) + (V) = 00100110 + 00000101= 00100011= ({2, 6, 0})= (UV)

GF(23) using 1+x+x3

word power of 000 0100 0

010 1

001 2

110 3

011 4

111 5

101 6

p25.

Preparata Codes

Def.9.3.3The extended Preparata code P(r) is the set of codewords of the form [(U), (V)], where U and V GF(2r) satisfy

Eg.9.3.4Construct GF(23) using 1+x+x3

U = {1, 2, 5, 0} , V = {0, 1, 2, 3, 6, 0}

[(U), (V)] = 01100101 11110011 is a codeword in P(3)

3 3 3

(i) | | and | | are even

(ii) =

(iii) ( )

(iv) is odd

u U v V

u U u U v V

U V

u v

u u v

r

GF(23) using 1+x+x3

word power of 000 0100 0

010 1

001 2

110 3

011 4

111 5

101 6

p26.

Preparata Codes

Lemma 9.3.5Suppose [(U), (V)] and [(A), (B)] are codewords in P(r).Let . Then [(UA + ), (VB)] is also a codeword

Pf:check (i)(ii)(iii) of Def.9.3.3 (i) |U|, |V|,|A|,|B| are even

|UA + | = |UA| = |U|+|A|-2|UA| is even |VB| = |V|+|B|-2 |VB| is even

(ii)

u U

u

ifor any , , since 2 0

x I J x I x J

I J

x x x

( ) | |

=

x U A y U A y U A

y U y A

y V y B y V B

x y y U A

y y

y y y

3 3 3

(i) | | and | | are even

(ii) =

(iii) ( )

(iv) is odd

u U v V

u U u U v V

U V

u v

u u v

r

p27.

Preparata Codes

(iii)

3 3 3 3

3 3 3

3 2 2 3

( ) ( ) ( ) (by (ii))

= ( ) ( ) ( )

| |

x U A x U A y U A y V B

y U y A y V y B

y U y U y U

x x y y

y y y y

y y y U

3 2 2 3

3 2 2 3

+ | |

+( ) ( ) ( ) ( )( ) ( )

y A y A y A

y V y V y B y V y B y B

y y y A

y y y y y y

2 2

(1)

(2)( )

(3) | | | |

y V y U

y V y V

y y

y y

U and V are even

3 3 3

y V y B y V B

y y y

p28.

Preparata Codes

Def A code is distance invariant if for any pair of codeword c1 and c2, the number of codewords distance i from c1

equals the number of codewords distance i from c2 for 1<=i<n.

Coro.9.3.7P(r) is distance invariant

Pf: Let [(U), (V)] and [(A), (B)] be codewords that are distance i apart.By Lemma 9.3.5 [(UU + ), (VV)] and [(UA + ), (VB)] are both codewords and also distance i apart.Since [(UU + ), (VV)] is the zero word

and [(UA + ), (VB)] has weight i

p29.

Preparata Codes

Lemma 9.3.8Suppose [(U), (V)] is a codeword in P(r) Then P(r) also contains the following codewords:(i) [(V), (U)](ii) [(U + ), (V + )](iii) [(U), (V)] , 0

Pf:Check (i)(ii)(iii) of Def.9.3.3similar as the proof of lemma 9.3.5

p30.

Preparata Codes

Lemma 9.3.13If is a primitive element of GF(2r) then 3 is a primitive element if r is odd and is not primitive if r is even

Pf: i is primitive iff gcd(i, 2r-1)=1and if r is odd then 2r-1 = 1 mod 3

if r is even then 2r-1 = 0 mod 3

Coro.9.3.14If r is odd then for each nonzero element x of GF(2r),there is a unique y such that y3=x

p31.

Preparata Codes

Thm.9.3.15P(r) has min. distance 6

Pf:Since P(r) is distance invariant, it contains a codeword [(U), (V)] of weight dd = wt((U)) + wt((V)) = |U| + |V|d is even, so we only need to show that d2,4 and there is a codeword of weight 6(1) suppose d=2, by lemma 9.3.8(i), assume |U|=2 and |V|=0

By lemma 9.3.8(ii), we can assume U={0, x}, x0 Then Def.9.3.3(ii) does not hold.

(2) suppose d=4, by lemma 9.3.8(i), either |U|=4 and |V|=0 or |U|=2 and |V|=2

p32.

Preparata Codes

(a) |U|=4 and |V|=0, assume U={0, x, y, z}, x,y,z are distinct and non-zero Then Def.9.3.3(iii) gives that 03+x3+y3+z3+(0+x+y+z)3=0 (x+y)(x+z)(y+z)=0 ( impossible !!)

(b) |U|=2 and |V|=2, assume U={0, x} and V={y,z} y,z are distinct Then Def.9.3.3(iii) gives that

03+x3+(0+x)3= y3+z3=0 By coro.9.3.14, if y3=z3 then y=z (-><-)

(3) Find a codeword of weight 6 Let x,y,z are distinct nonzero element

w be the unique element such that w3 = x3+y3+z3

and let u = w+x+y+z

p33.

Preparata Codes

Then w is not equal to x,y or z (since w=x imply y=z)and u0 (since w3 + (x+y+z)3 = (x+y)(x+z)(y+z) 0, so w x+y+z)Now let U={0,u} , V={w, x, y, z}Then [(U), (V)] is a word of weight 6 and we can check that it is a codeword.

p34.

Preparata Codes

Thm.9.3.18P(r) is not a linear code

Pf:we know that [(U), (V)] + [(A), (B)] = [(UA), (VB)]from the proof of thm.9.3.15, we can construct them withU={0, u1} , V={w1, x1, y1, z1}

A={0, u2} , B={w2, x2, y2, z2}

Then by lemma 9.3.5, [(UA + u1), (VB)] is a codeword

Since |UA + u1| 2, so

d ([(UA + u1), (VB)] , [(UA), (VB)]) 4

But min. distance is 6.So [(U), (V)] + [(A), (B)] is not a codeword,P(r) is not a linear code.

p35.

Preparata Codes

[4] Encoding extended Preparata codes Parity-check matrix of 2 error-correcting BCH code

Define

A : the submatrix of H formed by the last 2r rowsH’ : the submatrix of H formed by deleting the last 2r rows from H

0 0

1 3

2 6

2 2 3(2 2)

m m

H

p36.

Preparata Codes

Eg.9.4.1 Construct GF(23) using 1+x+x3,

we have

1 3

2 6

3 2

4 5

5 1

6 4

010 110 001 101

110 001

011 111 111 010 101 011

A

1

001 011

111 010

011 101 and

110 100

101 110

111 001

A

GF(23) using 1+x+x3

word power of 000 0100 0

010 1

001 2

110 3

011 4

111 5

101 6

p37.

Preparata Codes

Let m = mL,mR be any binary word of length 2r+1 - 2r – 2,

where mLis of length 2r - 1

mRis of length 2r- 2r – 1

then by using polynomial notation for mLand mR

we have [mL(),mL(3)] mLH

[mR(),mR(3)] mRH’

Define vR = [mL() + mR(), mL(3) + (mL())3 + mR(3)]A-1

p38.

Preparata Codes

Thm. 9.4.2Let r be odd. For any m of length 2r+1

- 2r – 2,

if (U)=[mL, pL] and (V)=[mR, vR, pR] , where pL and pR are parity check digits.Then [(U), (V)] is a codeword in P(r)

Pf: [mR, vR]H = [mR]H’ + [vR]A

= [mR(), mR(3)] + [mL() + mR(), mL(3) + (mL())3 + mR(3)]

= [mL(), mL(3) + (mL())3]

But [mR, vR]H = [vVv, vVv3]

Similarly, mL() = uUu and mL(3) + (mL())3 = uUu3 + (uUu)3

Therefore Def.9.3.3(ii)(iii) hold, and clearly (i)(iv) hold.

p39.

Preparata Codes

Coro.9.4.3 P(r) has 2^(2r+1

- 2r – 2) codewords.

Algo.9.4.4 (Encoding P(r))

Let mL and mR be words of length 2r-1 and 2r-2r-1 resp.

Let vR be as defined in Thm 9.4.2.

Then [mL, pL, mR, vR, pR] is a codeword of m=[mL, mR]

p40.

Preparata Codes

[5] Decoding extended Preparata codes w = [wL, pL, wR, pR] be a received word

where wL and wR are both of length 2r-1

Then we can compute [wL(), wL(3)] = wLH

[wR(), wR(3)] = wRH

consider 6 cases(1) If error only occurred on parity check digits

wL() = wR()

wL(3) +(wL())3 = wR(3)

and is easily checked

p41.

Preparata Codes

(2) If there are no errors in wL, one error in position i of wR and at most one error in the parity check digitsThen

wL() = wR() + i

wL(3) +(wL())3 = wR(3) + 3i

so (wL() + wR())3 = wL(3) +(wL())3 + wR(3)

and i = wL() + wR()

(3) If there are no errors in wR, one error in position i of wL and at most one error in the parity check digitsBy lemma 9.3.8(i) , we can check that

(wR() + wL())3 = wR(3) +(wR())3 + wL(3)

and i = wL() + wR()

(4) If two errors occur in wR, say in the position i and j

wL() = wR() + i + j

wL(3) +(wL())3 = wR(3) + 3i + 3j

so i + j and 3i + 3j are known.

i and j can be found in the same method in BCH decoding

(5) If two errors occur in wL, say in the position i and j

same as (4)

(6) If one error in wL and one in wR, say in position i and j resp.

Then

wL() + i = wR() + j

wL(3) + 3i +(wL() + i )3 = wR(3) + 3j

First, Let j = wL() + i + wR()

wL(3) + 3i +(wL() + i )3 = wR(3) + (wL() + i + wR())3

so (i + wR())3 = (wL(3) + wR(3)) + (wL() + wR())3 + wL()3 + wR()3 =

Therefore i = wR() + 1/3

j = wL() + 1/3