Limits to List Decoding Reed-Solomon Codes

May 24, 2005 STOC 2005, Baltimore 1

Limits to List Decoding Reed-Solomon Codes

Venkatesan Guruswami

Atri Rudra (University of Washington)


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=minx,y2GF(qk) (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


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


List Recovery

Related to List Decoding Problem Given C: GF(q)k! GF(q)n and Liµ GF(q), 1· i· n

Find all codewords c=hc1,,cni s.t. ci2 Li 8 i

|Li|· s LRB(C) largest s for which # of codewords

is poly(n)

L1

L2

L3

Ln

1 2 n3


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) ia2S

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


The Big Picture for RS

Polynomial Reconstruction

RS List Recovery RS List Decoding

9 poly time algo for error bound · J()

Negative Result ) above algo optimal


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


Combinatorial Limitations- I

For any C LDR(C ) ¸ /2

Unique decodability

Relative Distance ()

Err

or

Bo

un

d

Half Distance


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 codesRelative Distance ()

Err

or

Bo

un

d

Half Distance

Full Distance


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]


Err

or

Bo

un

d

Half Distance

Full Distance

Johnson Bound


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


Err

or

Bo

un

d

Half Distance

Full Distance

Johnson Bound


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


General setup for GS algorithm Polynomial Reconstruction

Pairs of numbers {(ai,bi)}, i=1..N Finds all degree k poly P at most N-(Nk)1/2

indices i, P(ai) bi

aidistinct ) List Decoding of RS

ai not necessarily distinct ) List Recovery of RSa1 a2 a3 ai an

b2 b3 bi bn

Li

b1


Main Result of this talk

Version of Johnson Bound implies

LRB(RS) ¸ dn/ke -1

(GS algo works in poly time in this regime)

We show LRB(RS) = dn/ke-1

) For Polynomial reconstruction GS algo is optimal


Implication for List Decoding RS

Polynomial Reconstruction In List Recovering setting N=n¢dn/ke 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 ais

a1 a2 a3 ai an

dn/ke


Main Result

n=q=pm

D=(pm-1)/(p-1) =pm-1+pm-2++p+1 Consider RS [n,k=D+1]GF(qm)*

For each i=1,,n the list Li= GF(p)

dn/ke = p

Number of deg D polys over GF(pm) which take values in GF(p) is p2m

a1 a2 a3 ai an

dn/ke


Explicit Construction of Polys

Pb(z) = i=0 bi( zai + 1)D where bi2 GF(p)

a is a generator of GF(pm) D=(pm-1)/(p-1)=pm-1++p+1

Poly over GF(pm) Takes values in GF(p)

Norm function: for all x2 GF(pm), xD2 GF(p)

Will now prove for distinct b, Pb(z) are distinct polys over GF(pm)

2m-1


Proof Idea

By Linearity,need to show

Pb(z) = i=0 bi( zai + 1)D 0

) b= b== b2m-1 =0

Coefficients of all zj must be 0

( ) i=0 bi (ai)j =0 for j=0..D

D+1 eqns and 2m vars (some of them trivial)

Dj

2m-1

2m-1

D=pm-1++p+1


Lucas’ Lemma

p prime and integers a and b a=a0+a1p++arpr

b=b0+b1p++brpr

( ) ( ) ( )( ) mod p

D=1+p+ pm-1, j=j0+j1p+ +jmpm-1

( ) 0 iff for all i, ji 2 {0,1}

) 2m equations and 2m var

ab b0

a0 a1

b1

ar

br

Dj


Wrapping up the proof 2m equations in 2m variables

T= { j0+j1p+ jm-1pm-1 | ji2 {0,1} }

i=0 bi (ai)j =0 for j 2 T

Coefficient matrix is Vandermonde

2m-1

1

1...1

aj0 (aj0)2 … (aj0)2m-1

aj1 (aj1)2 … (aj1)2m-1

b0

b1 = 0


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]


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 ?

Limits to List Decoding Reed-Solomon Codes

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