Error Correcting Codes: Combinatorics, Algorithms and Applications
(Fall 2007)
Lecture 3: Error Correction and Distance
August 31, 2007
Lecturer: Atri Rudra
Scribe: Michael Pfetsch & Atri Rudra
The following topics were discussed in the last lecture:
•
Shannon and Hamming noise models.
•
The
C
3
,rep
repetition code can correct
≤
1
errors and has a rate of
R
=
1
3
.
•
The
C
⊕
parity code cannot correct even
1
error and has a rate of
R
=
4
5
.
The last two points confirmed our intuition that one can correct more errors if the underlying code
has more redundancy. In today’s lecture we will look a bit more closely at the parity code
C
⊕
,
which among other things will motivate another important parameter of codes called the
distance
.
1
A closer look at
C
⊕
Last lecture we saw that
C
⊕
cannot correct even
1
error. However, we will now see that
C
⊕
can
detect
one error. Consider the following algorithm. Let
y
= (
y
1
, y
2
, y
3
, y
4
, y
5
)
be the received
word– compute
b
=
y
1
⊕
y
2
⊕
y
3
⊕
y
4
⊕
y
5
and declare an error if
b
= 1
. Note that when no error
has occurred during transmission,
y
i
=
x
i
for
1
≤
i
≤
4
and
y
5
=
x
1
⊕
x
2
⊕
x
3
⊕
x
4
, in which
case
b
= 0
as required. If there is a single error then either
y
i
=
x
i
⊕
1
(for exactly one
1
≤
i
≤
4
)
or
y
5
=
x
1
⊕
x
2
⊕
x
3
⊕
x
4
⊕
1
. It is easy to check that in this case
b
= 1
. In fact, one can extend
this argument to to obtain the following result.
Proposition 1.1.
The parity code
C
⊕
can
detect
an odd number of errors.
Let us now revisit the example that showed that one cannot correct
1
error using
C
⊕
. Consider
the two codewords in
C
⊕
,
u
= 00000
and
v
= 10001
(which are codewords corresponding to mes
sages
0000
and
1000
respectively). Now consider the two scenarios when
u
and
v
are transmitted
and a single error occurs resulting in the received word
r
= 10000
in both cases. Now given the
received word
r
and the fact that at most one error can occur, the decoder has no way of knowing
whether the original transmitted codeword was
u
or
v
. Looking back at the example, it is clear that
the decoder gets “confused” because the two codewords
u
and
v
do not differ in a lot of positions.
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 Spring '10
 Rejaei
 Electromagnet, Distance, Yi, Coding theory, Hamming Code

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