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Reed-Muller and Preparata Codes

Reed-Muller and Preparata Codes. OUTLINE. [1] Reed-Muller codes [2] Decoding Reed-Muller codes [3] Extended Preparata Codes [4] Encoding Extended Preparata Codes [5] Decoding Extended Preparata Codes. Chap 3 Perfect and Related Codes. Recall [3.8]. Reed-Muller codes

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Reed-Muller and Preparata Codes

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  1. Reed-Muller and Preparata Codes

  2. OUTLINE • [1] Reed-Muller codes • [2] Decoding Reed-Muller codes • [3] Extended Preparata Codes • [4] Encoding Extended Preparata Codes • [5] Decoding Extended Preparata Codes

  3. 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

  4. Perfect and Related Codes • 3. Generator matrix G of RM(r, m) • 4. Eg 3.8.4 Find G(1,3)

  5. Perfect and Related Codes • 5. The properties of RM(r,m)

  6. 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) 0 1 2 3 0 1 2 3 4 5 6 7

  7. Reed-Muller Codes • Any function f: Km {0, 1} has a unique vector v = (f(u0), f(u1), …, f(u2m-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 f  v

  8. Reed-Muller Codes • Eg. 9.1.2 Let 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

  9. Reed-Muller Codes • I={1,2} • I={0} • I= • I={1,2}{0} • I={1,2} • Two facts about fI • fI(x0, x1,…, xm-1) = 1 iff xi=0 for all iI • wt(vI)=2m-|I| • For each ui  Km fI(ui)fJ(ui) = fIJ(ui) and 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

  10. 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. 9.1.8 The generating matrix for RM(4,4) in canonical form (see next page) • Encoding

  11. Canonical form vI 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. 9.1.11 Encoding 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

  12. 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 = {uKm | 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 iI HI is a subspace of Km pf: x, y  HI  xi=yi=0  xi+yi=0  x+y HI

  13. 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

  14. I={1} s=010 J={0,1} Jc={2} t=001 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 Reed-Muller Codes • Lemma 9.2.1 Let I, J  Zm and |I||J|, for any sHIC and tHJ vI,s  vJC,t = 1 iff I = J pf: is even unless IJc=Zm I={1} s=010 J={1} Jc={0,2} t=101 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

  15. Reed-Muller Codes • Corollary 9.2.2 If c is a codeword in RM(r,m) and if |J| = r , then mJ = c  vJC,t ,  tHJ pf: ( In above, we choose s = (0,0,…,0) since (0,0,…,0) HJC for any J, |J| = r )

  16. Reed-Muller Codes J={0,1} HJ={00xx} HJc={xx00} • Lemma 9.2.3 Let J  Zm, for any word e (of length 2m) e  vJC,t = 1 forat most wt(e)values of tHJ pf: if t1,t2HJ , t1t2 then (HJc + t1)  (HJc + t2)= so vJc,t1 and vJc,t2 have no position where both digits are 1 Thus each 1 in e affects at most one of the values of e  vJC,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

  17. Reed-Muller Codes • Idea of decoding algorithm Let w=c+e where c is a codeword in RM(r,m) Let |J|=r by lemma 9.2.3, e  vJC,t = 0 for at least |HJ|-wt(e) values of tHJ for such t we have w  vJC,t = c  vJC,t + e  vJC,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)

  18. Reed-Muller Codes • Lemma 9.2.4 The 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

  19. 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 tHJ 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

  20. Reed-Muller Codes • Eg. 9.2.6 Decode 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 = 0OUTPUT 0 1000 000010

  21. 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

  22. w(1)=1111011100000000

  23. 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 GF(23) using 1+x+x3 word power of  000 0 100 0 010 1 001 2 110 3 011 4 111 5 101 6

  24. Preparata Codes GF(23) using 1+x+x3 word power of  000 0 100 0 010 1 001 2 110 3 011 4 111 5 101 6 • U  GF(2r) ,  GF(2r) (1) U +  = {u+|u  U} (2) U = {u| u U} (3) UV = {x| x UV but xUV } (U) + (V) = (UV) • 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 = {22, 25, 2 6}={4, 0, 1} (U) + (V) = 00100110 + 00000101 = 00100011 = ({2, 6, 0}) = (UV)

  25. Preparata Codes • Def.9.3.3 The extended Preparata code P(r) is the set of codewords of the form [(U), (V)], where U and V  GF(2r) satisfy • Eg.9.3.4 Construct 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) GF(23) using 1+x+x3 word power of  000 0 100 0 010 1 001 2 110 3 011 4 111 5 101 6

  26. Preparata Codes • Lemma 9.3.5 Suppose [(U), (V)] and [(A), (B)] are codewords in P(r). Let . Then [(UA + ), (VB)] is also a codeword Pf: check (i)(ii)(iii) of Def.9.3.3 (i) |U|, |V|,|A|,|B| are even |UA + | = |UA| = |U|+|A|-2|UA| is even |VB| = |V|+|B|-2 |VB| is even (ii)

  27. Preparata Codes (iii)

  28. 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.9.3.7 P(r) is distance invariant Pf: Let [(U), (V)] and [(A), (B)] be codewords that are distance i apart. By Lemma 9.3.5 [(UU + ), (VV)] and [(UA + ), (VB)] are both codewords and also distance i apart. Since [(UU + ), (VV)] is the zero word and [(UA + ), (VB)] has weight i

  29. Preparata Codes • Lemma 9.3.8 Suppose [(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.9.3.3 similar as the proof of lemma 9.3.5

  30. Preparata Codes • Lemma 9.3.13 If  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)=1 and if r is odd then 2r-1 = 1 mod 3 if r is even then 2r-1 = 0 mod 3 • Coro.9.3.14 If r is odd then for each nonzero element x of GF(2r), there is a unique y such that y3=x

  31. Preparata Codes • Thm.9.3.15 P(r) has min. distance 6 Pf: Since P(r) is distance invariant, it contains a codeword [(U), (V)] of weight d d = wt((U)) + wt((V)) = |U| + |V| d is even, so we only need to show that d2,4 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}, x0 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

  32. 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

  33. Preparata Codes Then w is not equal to x,y or z (since w=x imply y=z) and u0 (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.

  34. Preparata Codes • Thm.9.3.18 P(r) is not a linear code Pf: we know that [(U), (V)] + [(A), (B)] = [(UA), (VB)] from the proof of thm.9.3.15, we can construct them with U={0, u1} , V={w1, x1, y1, z1} A={0, u2} , B={w2, x2, y2, z2} Then by lemma 9.3.5, [(UA + u1), (VB)] is a codeword Since |UA + u1|  2, so d ([(UA + u1), (VB)] , [(UA), (VB)])  4 But min. distance is 6. So [(U), (V)] + [(A), (B)] is not a codeword, P(r) is not a linear code.

  35. 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 rows H’ : the submatrix of H formed by deleting the last 2r rows from H

  36. GF(23) using 1+x+x3 word power of  000 0 100 0 010 1 001 2 110 3 011 4 111 5 101 6 Preparata Codes • Eg.9.4.1 Construct GF(23) using 1+x+x3, we have

  37. 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

  38. Preparata Codes • Thm. 9.4.2 Let 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 = [vVv, vVv3] Similarly, mL() = uUu and mL(3) + (mL())3 = uUu3 + (uUu)3 Therefore Def.9.3.3(ii)(iii) hold, and clearly (i)(iv) hold.

  39. 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]

  40. 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

  41. 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 digits Then 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 digits By lemma 9.3.8(i) , we can check that (wR() + wL())3 = wR(3) +(wR())3 + wL(3) and i = wL() + wR()

  42. (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

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