CS 140
Class Notes 3
1 Error Correction and Detection
A variety of factors such as faulty components and inadequate design tolerances can result in
errors appearing in the information b eing processed by a computer. Such errors frequently occur
in information transmission between two relatively distant points.
One of the simplest and most widely used techniques for error control is using a single
parity
bit
C
0
. This parity bit is appended to an
n
bit input word
X
=(
X
0
X
1
:::X
n
1
) to form an (
n
+1)
bit word
X
=(
X
0
X
1
:::X
n
1
C
0
). The value assigned to bit
C
0
makes the total number of bits in
X
even (if we are using even parity) or odd (in the case of odd parity). Thus, at the receiving
end, an error can be detected by taking an
exclusive-or
of all the bits received (
X
).
Note that this simple technique of error control allows only detection of an error. It is not
possible to determine the bit in error (or the p osition of the error). Hence, it is not p ossible to
correct the error at the receiving end. This leads to retransmission of the information. In addition,
this parity only provides single error detection (or in fact, an odd number of errors can be detected).
It does not detect multiple errors (even in number).
The parity-checking concept can be easily extended to detection of multiple errors, or to the
location of single or multiple errors. These goals can be achieved by providing additional parity bits,
each of whichcheck the parity of a subset of the bits in
X
. For example, if we can deduce (from
the parity bits) that bit
X
i
is in error, then the error can be corrected by simply complementing
that bit, thus proving single error correction. Let
c
be the number of parity bits required to achieve
single error correction with
n
-bit data words. Clearly, the check bits must be able to distinguish
between (
n
+
c
) possible error location and the single error-free case. Hence,
2
c
n
+
c
+1
It can be proved that this code is
Single Error Correcting Double Error Detecting (SECDED)
code.
This is also known as
Hamming Code
.
The key question that remains to be answered is: How to generate the value of
c
parity bits
that are appended to
X
=(
x
0
x
1
:::x
n
1
) to obtain
X
? Also, how to obtain the location of error
from
X
? The process is illustrated in the following example:
Let
n
=4. Then,
c
= 3 (since, 2
3
4+3+1). Let
M
=(
m
0
m
1
m
2
m
3
) be the message bits and
let the three parity bits be (
p
0
p
1
p
2
). Organize the bits as follows:
POSITION 7 6 5 4 3 2 1
BIT
m
0
m
1
m
2
p
0
m
3
p
1
p
2
