Section 9.2 Coding Theory
725
PrACtiCe 42 Example 26 shows four members of one coset. The parity-check matrix that generated the
code for this example is
h = E
1 0 1
1 1 1
1 0 0
0 1 0
0 0 1
U
Compute the syndrome for each member of the coset.
■
Now suppose that we somehow have available a list of the 2
r
coset leaders. It
is then easy to find which coset leader corresponds to X by making use of the fact
from the previous theorem that X and the coset leader of X’s coset will have the
same syndrome.
example 27
In the (5, 2) code of Example 26, n = 5, m = 2, and r = 3. The size of each coset
is 2
m
= 4, and there are 2
r
= 8 distinct cosets. Here is a list of the coset leaders
and their corresponding syndromes. The syndromes were computed using the matrix h of Practice 42, and the coset leaders were found by brute force.
Coset leaders
Syndromes
00000
000
00001
001
00010
010
00011
011
00100
100
10000
101
00110
110
01000
111
A received word of 10101 is decoded by computing its syndrome:
(10101)h = 000. The coset leader sharing this same syndrome is 00000. The
received word is decoded as
10101 + 2 00000 = 10101
The received word is a code word, so it is assumed that no errors have occurred.
A received word of 11000 is decoded by computing its syndrome:
(11000)h = 010.The coset leader sharing this same syndrome is 00010, so the
received word is decoded as
11000 + 2 00010 = 11010
assuming that a single error has occurred.
A received word of 10011 has a syndrome of 110, so it can be decoded as
10011 + 2 00110 = 10101
But because its coset leader has weight 2, we might instead generate a flag noting that
at least two errors have occurred and that decoding cannot be done with certainty.