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Error Correcting Codes: Combinatorics, Algorithms and Applications
(Fall 2007)
Lecture 5: Linear Codes
September 7, 2007
Lecturer: Atri Rudra
Last lecture we talked about the Hamming bound (for the special case of
d
= 3
) and informally
deﬁned binary linear codes. In today’s lecture, we will present the general form of Hamming
bound and deﬁne (general) linear codes. The latter will involve looking at ﬁnite ﬁelds and vector
spaces over ﬁnite ﬁelds.
1
General Family of Hamming Codes
We start with a new notation.
Deﬁnition 1.1.
A code
C
⊆
Σ
n
with dimension
k
and distance
d
will be called a
(
n, k, d
)
Σ
code.
We will also refer it to as a
(
n, k, d
)

Σ

code.
We now state the Hamming bound for
d
= 3
that we proved in the last lecture.
Theorem 1.2
(Hamming Bound for
d
= 3
)
.
For every
(
n, k,
3)
2
code
k
≤
n

log
2
(
n
+ 1)
.
We now proceed to generalize the above theorem to any distance
d
.
Theorem 1.3
(Hamming Bound for any
d
)
.
For every
(
n, k, d
)
q
code
k
≤
n

log
q
b
(
d

1)
2
c
X
i
=0
±
n
i
²
(
q

1)
i
.
Proof.
The proof is a straightforward generalization of the proof of Theorem 1.2 that we did last
lecture. For notational convenience, let
e
=
j
(
d

1)
2
k
. Given any two codewords,
c
1
6
=
c
2
∈
C
, the
following is true (as
C
has distance
1
d
):
B
(
c
1
, e
)
∩
B
(
c
2
, e
) =
∅
,
(1)
where recall that for any vector
x
∈
[
q
]
n
,
B
(
x
, e
) =
{
y
∈
[
q
]
n

Δ(
x
,
y
)
≤
e
}
.
1
Assume that
y
∈
B
(
c
1
, e
)
∩
B
(
c
2
, e
)
, that is
Δ(
y
, c
1
)
≤
e
and
Δ(
y
, c
2
)
≤
e
. Thus, by the triangle inequality,
Δ(
c
1
, c
2
)
≤
2
e
≤
d

1
, which is a contradiction.
1
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View Full DocumentWe claim that for all
x
∈
[
q
]
n
,

B
(
x
, e
)

=
e
X
i
=0
±
n
i
²
(
q

1)
i
.
(2)
Indeed any vector in
B
(
x
, e
)
must differ from
x
in exactly
0
≤
i
≤
e
positions. In the summation
(
n
i
)
is the number of ways of choosing the differing
i
positions and in each such position, a vector
can differ from
x
in
q

1
ways.
Now consider the union of all Hamming balls centered around some codeword. Obviously
their union is a subset of
[
q
]
n
. In other words,
³
³
³
³
³
[
c
∈
C
B
(
c, e
)
³
³
³
³
³
≤
q
n
.
(3)
As (1) holds for every pair of distinct codewords,
³
³
³
³
³
[
c
∈
C
B
(
c, e
)
³
³
³
³
³
=
X
c
∈
C

B
(
c, e
)

=
q
k
e
X
i
=0
±
n
i
²
(
q

1)
i
,
(4)
where (4) follows from (2) and the fact that
C
has dimension
k
. Combining (4) and (3) and taking
log
q
of both sides we will get the desired bound:
k
≤
n

log
q
´
e
X
i
=0
±
n
i
²
(
q

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