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Error Correcting Codes: Combinatorics, Algorithms and Applications
(Fall 2007)
Lecture 12: ReedSolomon Codes
September 28, 2007
Lecturer: Atri Rudra
Scribe: Michel Kulhandjian
Last lecture we saw the proof of the Singleton bound which claims that for any
(
n, k, d
)
q
code,
k
≤
n
−
d
+ 1
. In today’s lecture we will study ReedSolomon codes. These codes meet the
Singleton bound, i.e. satisfy
k
=
n
−
d
+ 1
(but have the unfortunate property that
q
≥
n
). Note
that this implies that the Singleton bound is tight, at least for
q
≥
n
.
1
ReedSolomon Codes
We begin with the definition of ReedSolomon codes.
Definition 1.1
(ReedSolomon code)
.
Let
F
q
be a finite field and
F
q
[
x
]
denote the
F
q
space of
univariate polynomials where all the coefficients of
x
are from
F
q
. Pick
{
α
1
, α
2
, ...α
n
}
distinct
elements (also called
evaluation points
) of
F
q
and choose
n
and
k
such that
k
≤
n
≤
q
. We
define an encoding function for ReedSolomon code as
RS
:
F
k
q
→
F
n
q
as follows. A message
m
= (
m
0
, m
1
, ..., m
k

1
)
with
m
i
∈
F
q
is mapped to a degree
k
−
1
polynomial.
m
m→
f
m
(
x
)
,
where
f
m
(
x
) =
k

1
s
i
=0
m
i
x
i
.
(1)
Note that
f
m
(
x
)
∈
F
q
[
x
]
is a polynomial of degree
≤
k
−
1
. The encoding of
m
is the evaluation
of
f
m
(
x
)
at all the
α
i
’s :
RS
(
m
) =
a
f
m
(
α
1
)
, f
m
(
α
2
)
, ..., f
m
(
α
n
)
A
We call this image ReedSolomon code or
RS
code after two inventors
Irving Reed
and
Gus
Solomon
of this code [1]. A common special case is
n
=
q
−
1
with the set of evaluation points
being
F
*
d
F
\ {
0
}
.
Notice that by definition, the entries in
{
α
1
, ..., α
n
}
are distinct and thus, must have
n
≤
q
.
We now turn to some properties of ReedSolomon codes.
Claim 1.2.
RS
codes are linear codes.
1
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View Full DocumentProof.
The proof follows from the fact that if
a
∈
F
q
and
f
(
x
)
, g
(
x
)
∈
F
q
[
x
]
are polynomials of
degree
≤
k
−
1
, then
af
(
x
)
and
f
(
x
) +
g
(
x
)
are also polynomials of degree
≤
k
−
1
. In particular,
let messages
m
1
and
m
2
be mapped to
f
m
1
(
x
)
and
f
m
2
(
x
)
where
f
m
1
(
x
)
, f
m
2
(
x
)
∈
F
q
[
x
]
are
polynomials of degree
≤
k
−
1
and because of the mapping defined in (1), it is easy to verify that:
f
m
1
(
x
) +
f
m
2
(
x
) =
f
m
1
+
m
2
(
x
)
,
and
af
m
1
(
x
) =
f
a
m
1
(
x
)
.
Therefore,
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