THE ORIGINAL VIEW OF REED-SOLOMON CODES THE ORIGINAL ...

L.R.Welch
THE ORIGINAL VIEW
OF
REED-SOLOMON CODES
THE ORIGINAL VIEW
[Polynomial Codes over Certain Finite Fields,
I.S.Reed and G. Solomon,
Journal of SIAM, June 1960]
Parameters:
Let GF(2 n ) be the eld with 2 n elements.
The number of message symbols encoded into a
codeword is M.
The numberofcodesymbols transmitted is
N=2 n .
Message symbols are elements of GF(2 n ).
Code symbols are elements of GF(2 n ).
1

The original view of Reed and Solomon is that
the codeword symbols are the VALUES of a polynomial
whose COEFFICIENTS are the message
symbols.
2

MATHEMATICALLY:
Let (x 1 ;x 2 ;;x N )beanenumeration of the
elements of the eld.
Let (m 1 ;m 2 ;;m M ) be a message where
m i 2 GF (2 n ).
Dene a polynomial by
P (Z) =m 1 +m 2 Z+m 3 Z 2 ++m M Z M,1
Then the codeword is
(P (x 1 );P(x 2 );:::;P(x N ))
The enumeration used by Reed and Solomon was
as follows.
Let be a primitive element in GF(2 n ). Then the
enumeration is
(x 1 ;x 2 ;;x N )=(0;; 2 ;; N,2 ;1)
3

Suppose the received word is
(R(x 1 );R(x 2 );:::;R(x N ))
At the receiver, N equations can be formed, some
of which are wrong if there are transmission errors.
R(0) = m 1
R() = m 1 + m 2 + m 3 2 + +m M M,1
R( 2 ) = m 1 +m 2 +m 3 22 ++m M 2(M,1)
.
R(1) = m 1 + m 2 + m 3 + +m M
Any M of these equations have full rank and can
be solved. By trying many sets of M equations,
the correct answer will appear more often than any
wrong answer provided the number of errors is less
than (2 n , M +1)=2.
Choosing M equations and solving those equations
is equivalent to Lagrange interpolation.
4

THE CLASSICAL VIEW
Reed Solomon Codes
are a sub type of
BCH Codes
I don't know who rst made this observation but
it leads to a practical decoding algorithm developed
by Wes Peterson.
Parameters:
Let GF(2 n ) be the eld with 2 n elements.
Let be a primitive element in GF(2 n ).
The number of message symbols encoded into a
codeword is M.
The numberofcodesymbols transmitted is
N =2 n ,1.
Let G(Z) be the polynomial whose roots are
(; 2 ;; N,M )
Message symbols are elements of GF(2 n ).
The BCH view is that the symbols of a codeword
are the COEFFICIENTS of a polynomial in
Z which is divisible by G.
5

MATHEMATICALLY:
Let
G(Z) = N,M Y
i=1
(Z , i )
Let (m 1 ;m 2 ;;m M ) be a message where
m i 2 GF (2 n ).
Dene a polynomial by
P (Z) =m 1 +m 2 Z+m 3 Z 2 ++m M Z M,1
and dene
C(Z) = P (Z)G(Z)
X
= N ,1
i=0
c i Z i
Then the codeword is
(c 0 ;c 1 ;;c N,1 )
6

Alternatively, dividing Z N ,M P (Z) byG(Z)to
get
Z N ,M P (Z) =Q(Z)G(Z)+R(Z)
Then
C(Z) =Z N,M P(Z),R(Z)
is divisible by G(Z) and its coecients form a systematic
codeword.
7

DECODING AS PRESENTED
IN
THE REED-SOLOMON PAPER
Reed and Solomon described the following decoding
procedure.
Receive (r 1 ;r 2 ;;r N )
Select M indices in all possible ways
(Lagrange Interpolation)For each selection,
nd P (Z) with deg(P )=M,1
P(x i )=r i at these indices.
The coecients of P form a potential message.
As all possible selections are made, the message
which gives a codeword closest to the received word
occurs more often than any other.
8

An alternative is to form a codeword corresponding
to each potential message and stop when the
result disagrees with the received message in at
most (N , M)=2 places. If none is found, failure
to decode occurs.
These procedures are very inecient.
return to this later.
We will
9

CLASSICAL DECODING
[W.W. Peterson, IRE IT-6 Sept 1960]
Receive (r 1 ;r 2 ;;r N )
and form
R(Z) =
C( i ) = 0
N X
i=1
r i Z i,1
= C(Z)+E(Z)
S i = R( i ) = E( i )
for 1 i N , M
for 1 i N , M
Peterson observed that the S i satisfy a linear recursion.
0= e X
i=0
q i S k+i
for 1 k N , M , e
where Q(Z) = P e
i=0
q i Z i has the error locations as
its roots.
This gives a simple system of linear equations to
nd the error locations. Another system solves for
the error values.
10

Later Berlekamp found a
faster algorithm for nding
the recursion satised by the S i .
11

Let us return to the Reed & Solomon's
Original View and see if there is another
decoding method.
We begin in the same way
Reed and Solomon did
by selecting a set of M indices, S,
and nding P (Z) for which
degree of P is at most M , 1
and P (x i )=r i for i 2 S
The tool is Lagrange interpolation.
We ask what is the eect of errors
at the selected indices.
12

LAGRANGE INTERPOLATION
For each i 2 S we need a polynomial, i (Z), of
degree M , 1 which has the value 1 at x i and the
value 0 at x j for j 2 S; j 6= i
With these, the solution for P is
P (Z) = X
i2S
r i i (Z)
If r i includes an error e i then e i i (Z) is added
to the correct P (Z).
13

SYNDROMES
Let T (Z) be the message polynomial
and e k be the error at position k
Then
r k = T (x k )+e k
Given an index selection, S, the Lagrange Interpolation
polynomial evaluated at x k will be
P (x k ) =
X r i i (x k )
i2S
= X T (x i ) i (x k )+ X e i i (x k )
i2S
i2S
14

Subtracting the interpolated values from the received
values:
r k , P (x k ) = T (x k ) , X
+e k , X
i2S
i2S
T (x i ) i (x k )
e i i (x k )
Since T has degree at most M , 1,
and we have
T (x k )= X
i2S
T(x i ) i (x k )
k =r k ,P(x k )=e k , X
for k 2 S c
i2S
which is a new family of syndromes.
e i i (x k )
Given k for k 2 S c and assuming the number
of errors is within the R-S bound, we would like to
solve for the e j 's.
15

INTERPOLATION POLYNOMIALS
Given S, the 's are constructed as follows:
Let
G(Z) = Y (Z,x j )
j2S
and let G 0 (Z) be its formal derivative.
Then
i (Z) =
G(Z)
(Z,x i )G 0 (x i )
It is clear that i (x j ) = 0 for i; j 2 S and
j 6= i. That i (x i ) = 1 follows from the rule of
L'Hospital which isvalid for ratios of polynomials
over any eld.
16

The syndromes can now be expressed as:
k = e k , X
i2S
e i
G(x k )
(x k , x i )G 0 (x i )
or
k
G(x k ) =
e k
G(x k ) , X
i2S
e i 1
G 0 (x i ) (x k , x i )
This looks simpler if we dene
k =
k
G(x k )
e k =
e k
G(x k )
for k 2 Sc
and
e i =
e i
G 0 (x i )
then
k = e k , X
i2S
e i
(x k , x i )
17

Not all of the e's are non-zero.
Let E be the set of indices for which e i 6=0
Dene
Q S (Z) =
Y
i2E\S
(Z,x i )
and
Q C (Z) =
Y
i2E\S c (Z,x i )
The roots of Q S (Z) are those x i for which i 2 S
and e i 6= 0 While the roots of Q C (Z) are those x i
for which i 2 S c and e i 6=0
The summation expression in the syndrome equations
can now be put over a common denominator
to give
k = e k , A(x k)
Q S (x k )
where the degree of Q S is the number of errors at
selected indices and the degree of A is less
18

For k 2 S c , either e k =0orQ C (x k ) = 0 so the
product is 0 for k 2 S c . Multiplying the previous
syndrome expression by Q(x k ) Q S (x k )Q C (x k )
gives
Q(x k ) k = Q C(x k )A(x k ) P (x k )
Q(x k ) k = P (x k )
This expression has, as unknowns, the coecients
of Q and P and provides a system of linear
equations for their solution.
At rst glance this looks the same has the Classical
case but is not. The classical equations are
statements about the relationship between COEF-
FICIENTS of polynomials while the above expression
is a relation about the VALUES of polynomials.
19

AFAST ALGORITHM
FOR SOLVING
THE WELCH-BERLEKAMP EQUATIONS
This algorithm was developed jointly
by Elwyn Berlekamp
Tze Hua Liu
Po Tong
Lloyd Welch
The Key WB Equations
are
Q(x k ) k = P (x k ) for k 2 S c (1)
These equations will be used sequentially
in some order depending on auxillary
information such assymbol reliability
We will assume that the x k 's are renumbered
so that the equations are
Q(x k ) k = P (x k) for k =1;;N ,M
20

We introduce an auxillary
pair of polynomials,
W (Z);V(Z)
and recursively compute an integer
and four polynomials as follows
J k ;Q k (Z);P k (Z);W k (Z);V k (Z)
The pair
(Q k ;P k )
will constitute a \minimal"
solution to the rst k equations
and the pair
(W k ;V k )
will also solve the rst k
equations but, in a certain sense,
will not be minimal.
21

Initialization:
J 0 =0;
Q 0 (Z)=1;P 0 (Z)=0;W 0 (Z)=1;V 0 (Z)=1
Since k = 0, there are
no equations to verify.
22

When we have generated
Q k ;P k ;W k ;V k
we will have two solutions
to the rst k equations.
(Q k ;P k ) and (W k ;V k )
However the pair (W k ;V k ) will be
"less suitable" than (Q k ;P k )
in a certain sense.
But the pair will be useful
in constructing (Q k+1 ;P k+1 )
23

STEP k
At the beginning of step k we have
J k,1 and
Q k,1 (Z);P k,1 (Z);W k,1 (Z);V k,1 (Z)
We form
d k = Q k,1 (x k ) k , P k,1 (x k )
IF d k =0
J k = J k,1 +1
Q k (Z) = Q k,1 (Z)
P k (Z) = P k,1 (Z)
W k (Z) = (Z,x k )W k,1 (Z)
V k (Z) = (Z,x k )V k,1 (Z)
In this case
(Q k,1 ;P k,1 )
already satisfy the k'th equation
and
multiplication by (Z,x k )
forces the pair
W k (Z);V k (Z)
to satisfy the k'th equation.
24

IF d k 6=0
We form another quantity,
c k = d ,1
k (W k,1 (x k ) k , V k,1(x k ))
and set
Q k (Z) = (Z , x k )Q k,1 (Z)
P k (Z) = (Z , x k )P k,1 (Z)
W k (Z) = W k,1 , c k Q k,1 (Z)
V k (Z) = V k,1 , c k P k,1 (Z)
It is readily veried that the two
pair of polynomials satisfy the rst
k equations.
HOWEVER, we are not done with this case.
IF J k,1 = 0 then
swap the two pair.
Q k (Z) $ W k (Z)
P k (Z) $ V k (Z)
and set J k =0
OTHERWISE set J k = J k,1 , 1
and do not swap pairs.
25

AMATRIX DESCRIPTION
The four polynomials can be written
as components of a matrix.
0
B
@ Q k(Z) P k (Z)
W k (Z) V k (Z)
1
C
A
In this form the k'th step can be written
26

IF d k =0
0
B
@ Q k(Z) P k (Z)
W k (Z) V k (Z)
1
C
A=
0
B
@ 1 0
0 (Z,x k )
10
C
A
B
@ Q k,1(Z) P k,1 (Z)
W k,1 (Z) V k,1 (Z)
1
C
A
IF d k 6= 0 and J k,1 6=0
0
B
@ Q k(Z) P k (Z)
W k (Z) V k (Z)
1
C
A=
0
B
@ (Z,x k) 0
c k 1
10
C
A
B
@ Q k,1(Z) P k,1 (Z)
W k,1 (Z) V k,1 (Z)
1
C
A
IF d k 6= 0 and J k,1 =0
0
B
@ Q k(Z) P k (Z)
W k (Z) V k (Z)
1
C
A=
0
B
@
,c k 1
(Z,x k ) 1
10
C
A
B
@ Q k,1(Z) P k,1 (Z)
W k,1 (Z) V k,1 (Z)
1
C
A
27

The fact that each pair,
(Q k (Z);P k (Z)) and (W k (Z);V k (Z))
satisfy the rst k equations
can be expressed as
0
B
@ Q k(x i ) P k (x i )
W k (x i ) V k (x i )
1 0
C
A
B
@ i
,1
1
C
A =
0
B
@ 0 0
1
C
A
for i k
28

WHAT ABOUT MINIMALITY
This is the dicult part of the theorem
and I will just outline the proof
with a few comments.
I will pattern this proof after that in
Tze Hwa Liu's dissertation ( 1984)
29

First we are going to replace the decision
criterion using J k by something
more intuitive. ( It is equivalent)
Dene the length of an ordered pair
of polynomials by
L(Q(Z);P(Z)) = max(deg(Q),1+deg(P ))
and dene
L k = min L(Q(Z);P(Z))
where the minimum is taken over all
pairs of polynomials satisfying
the rst k equations.
30

THEOREM:
The algorithm described above
gives a sequence of pairs, (Q k (Z);P k (Z))
for which
L k = L(Q k (Z);P k (Z))
That is: (Q k (Z);P k (Z)) is minimal.
I will just list the lemmas, with
a comment about some of them.
Lemma 1
L k is monotone increasing.
Lemma 2
If 2L k k then there is
a unique minimum pair satisfying
the rst k equations.
31

Lemma 3
If 2L k k and d k+1 6=0
when computed for a minimal pair at
level k then
L k+1 = L k +1
***
Lemma 4
Q k (Z) P k (Z)
W k (Z) V k (Z)
= Y k
i=1
(Z , x i )
(determinant of product is product of
determinants)
32

Lemmas 5,6
L(Q k (Z);P k (Z)) + L(W k (Z);V k (Z)) = k +1
THE SWAP PAIRS DECISION
The swap pairs decision based on the
value of J k is really the comparison
of L(Q k (Z);P k (Z)) and L(W k (Z);V k (Z))
and picking L(Q k (Z);P k (Z)) to be the smaller.
33

We restate the theorem:
THEOREM:
The algorithm described above
gives a sequence of pairs, (Q k (Z);P k (Z))
for which
L k = L(Q k (Z);P k (Z))
That is: (Q k (Z);P k (Z)) is minimal.
34