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May 24, 2005STOC 2005, Baltimore1 Limits to List Decoding Reed-Solomon Codes Venkatesan Guruswami Atri Rudra (University of Washington)

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May 24, 2005 STOC 2005, Baltimore 2 Error-Correcting Codes Linear Code C : GF(q) k ! GF(q) n Hamming Distance or (u,v) for u,v 2 GF(q) n Number of positions u and v differ Distance of code C, d=min x,y 2 GF(q k ) (C(x),C(y)) C is an [n,k,d] GF(q) code Relative distance =d/n This talk is about Reed-Solomon (RS) Codes

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May 24, 2005 STOC 2005, Baltimore 3 List Decoding Given r 2 GF(q) n and 0 · e · 1 Output all codewords c 2 C such that (c,r) · en Combinatorial Issues How big can the list of codewords be ? LDR(C) largest e such that list size is poly(n) Algorithmic issues Can one find list of codewords in poly(n) time ? Cannot have poly time algo beyond LDR(C) errors

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May 24, 2005 STOC 2005, Baltimore 4 List Recovery Related to List Decoding Problem Given C: GF(q) k ! GF(q) n and L i µ GF(q), 1 · i · n Find all codewords c= h c 1, ,c n i s.t. c i 2 L i 8 i |L i | · s LRB(C) largest s for which # of codewords is poly(n) L1L1 L2L2 L3L3 LnLn 12n3

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May 24, 2005 STOC 2005, Baltimore 5 Reed-Solomon Codes RS [n,k+1] GF(q) Message P a poly. of degree · k over GF(q) S µ GF(q) RS(P) = h P(a) i a 2 S n= |S| d = n – k For this talk S = GF(q) 9 poly time algo for list decoding of RS codes till error bound J( )=1-(1- ) 1/2= 1- (k/N) 1/2

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May 24, 2005 STOC 2005, Baltimore 6 The Big Picture for RS Polynomial Reconstruction RS List RecoveryRS List Decoding 9 poly time algo for error bound · J( ) Negative Result ) above algo optimal

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May 24, 2005 STOC 2005, Baltimore 7 Talk outline Our main result is about combinatorial limitation of List Recovery of Reed Solomon Codes Motivation of the problem Main Result and Implications Proof of the main result

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May 24, 2005 STOC 2005, Baltimore 8 Combinatorial Limitations- I For any C LDR(C ) ¸ /2 Unique decodability Relative Distance ( ) Error Bound Half Distance

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May 24, 2005 STOC 2005, Baltimore 9 Combinatorial Limitations- II LDR(C) ¼ is the best one can hope for e ¸ can’t detect errors Lots of “good” codes with LDR(C) ¼ Random Linear Codes 2x improvement over unique decoding Difficulty: getting explicit codes Relative Distance ( ) Error Bound Half Distance Full Distance

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May 24, 2005 STOC 2005, Baltimore 10 Combinatorial Limitations- III Johnson Bound For any code C LDR(C) ¸ J( )=(1-(1- ) 1/2 ) Exists codes for which Johnson Bound is tight Non-linear codes [GRS00] Linear codes [G02] Relative Distance ( ) Error Bound Half Distance Full Distance Johnson Bound

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May 24, 2005 STOC 2005, Baltimore 11 Going beyond the Johnson Bound Go beyond Johnson Bound Choice of code matters Random Linear codes get there What about well studied codes like RS codes ? Motivation of our work Relative Distance ( ) Error Bound Half Distance Full Distance Johnson Bound

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May 24, 2005 STOC 2005, Baltimore 12 Algorithmic Status of RS Unique decoding [Peterson60] List Decoding Johnson Bound [Sud97, GS99] Unknown beyond JB Some belief that LDR(RS)=(1-(1- ) 1/2 ) ? ? ? ? ? ? ? ? Half Distance Full Distance Johnson Bound

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May 24, 2005 STOC 2005, Baltimore 13 General setup for GS algorithm Polynomial Reconstruction Pairs of numbers {(a i,b i )}, i=1..N Finds all degree k poly P at most N-(Nk) 1/2 indices i, P(a i ) b i a i distinct ) List Decoding of RS a i not necessarily distinct ) List Recovery of RS a1a1 a2a2 a3a3 aiai anan b2b2 b3b3 bibi bnbn LiLi b1b1

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May 24, 2005 STOC 2005, Baltimore 14 Main Result of this talk Version of Johnson Bound implies LRB(RS) ¸ d n/k e -1 (GS algo works in poly time in this regime) We show LRB(RS) = d n/k e -1 ) For Polynomial reconstruction GS algo is optimal

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May 24, 2005 STOC 2005, Baltimore 15 Implication for List Decoding RS Polynomial Reconstruction In List Recovering setting N=n ¢d n/k e Number of disagreements = N-n w (Nk) 1/2 With (little more than) N-(Nk) 1/2 disagreement have super poly RS codewords GS algo works for disagreement · N-(Nk) 1/2 Improvement “must” use near distinctness of a i s a1a1 a2a2 a3a3 aiai anan d n/k e

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May 24, 2005 STOC 2005, Baltimore 16 Main Result n=q=p m D=(p m -1)/(p-1) =p m-1 +p m-2 + +p+1 Consider RS [n,k=D+1] GF(q m ) * For each i=1, ,n the list L i = GF(p) d n/k e = p Number of deg D polys over GF(p m ) which take values in GF(p) is p 2 m a1a1 a2a2 a3a3 aiai anan d n/k e

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May 24, 2005 STOC 2005, Baltimore 17 Explicit Construction of Polys P b (z) = i=0 b i ( za i + 1) D where b i 2 GF(p) a is a generator of GF(p m ) D=(p m -1)/(p-1)=p m-1 + +p+1 Poly over GF(p m ) Takes values in GF(p) Norm function: for all x 2 GF(p m ), x D 2 GF(p) Will now prove for distinct b, P b (z) are distinct polys over GF(p m ) 2 m -1

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May 24, 2005 STOC 2005, Baltimore 18 Proof Idea By Linearity,need to show P b (z) = i=0 b i ( za i + 1) D 0 ) b = b = = b 2 m -1 =0 Coefficients of all z j must be 0 ( ) i=0 b i (a i ) j =0 for j=0..D D+1 eqns and 2 m vars (some of them trivial) D j 2 m -1 D=p m-1 + +p+1

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May 24, 2005 STOC 2005, Baltimore 19 Lucas’ Lemma p prime and integers a and b a=a 0 +a 1 p+ +a r p r b=b 0 +b 1 p+ +b r p r ( ) ( ) ( ) ( ) mod p D=1+p+ p m-1, j=j 0 +j 1 p+ +j m p m-1 ( ) 0 iff for all i, j i 2 {0,1} ) 2 m equations and 2 m var a bb0b0 a0a0 a1a1 b1b1 arar brbr D j

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May 24, 2005 STOC 2005, Baltimore 20 Wrapping up the proof 2 m equations in 2 m variables T= { j 0 +j 1 p+ j m-1 p m-1 | j i 2 {0,1} } i=0 b i (a i ) j =0 for j 2 T Coefficient matrix is Vandermonde 2 m a j 0 (a j0 ) 2 … (a j0 ) 2 m -1 a j 1 (a j1 ) 2 … (a j1 ) 2 m -1 b0b0 b1b1 = 0

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May 24, 2005 STOC 2005, Baltimore 21 Other Results in the Paper Use connection with BCH codes to get an exact estimate Show existence of explicit received word with super poly “close by” RS codewords for certain parameters Uses ideas from [CW04]

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May 24, 2005 STOC 2005, Baltimore 22 Open Questions Is Johnson Bound the true list decoding radius of Reed Solomon codes ? Show RS of rate 1/L cannot be list recovered using lists of size L which are not prime powers. What RS codes on prime fields ?

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