8.5: Efficient Decoding
( \newcommand{\kernel}{\mathrm{null}\,}\)
We are now at the stage where we are able to generate linear codes that detect and correct errors fairly easily, but it is still a time-consuming process to decode a received n-tuple and determine which is the closest codeword, because the received n-tuple must be compared to each possible codeword to determine the proper decoding. This can be a serious impediment if the code is very large.
Given the binary matrix
and the 5-tuples x=(11011)t and y=(01011)t,
Solution
we can compute
Hence, x is a codeword and y is not, since x is in the null space and y is not. Notice that Hy is identical to the first column of H. In fact, this is where the error occurred. If we flip the first bit in y from 0 to 1, then we obtain x.
If H is an m×n matrix and x∈Zn2, then we say that the syndrome of x is Hx. The following proposition allows the quick detection and correction of errors.
Let the m×n binary matrix H determine a linear code and let x be the received n-tuple. Write x as x=c+e, where c is the transmitted codeword and e is the transmission error. Then the syndrome Hx of the received codeword x is also the syndrome of the error e.
- Proof
-
The proof follows from the fact that
Hx=H(c+e)=Hc+He=0+He=He.
This proposition tells us that the syndrome of a received word depends solely on the error and not on the transmitted codeword. The proof of the following theorem follows immediately from Proposition 8.36 and from the fact that He is the ith column of the matrix H.
Let H∈Mm×n(Z2) and suppose that the linear code corresponding to H is single error-correcting. Let r be a received n-tuple that was transmitted with at most one error. If the syndrome of r is 0, then no error has occurred; otherwise, if the syndrome of r is equal to some column of H, say the ith column, then the error has occurred in the ith bit
Consider the matrix
and suppose that the 6-tuples x=(111110)t, y=(111111)t, and z=(010111)t have been received.
Solution
Then
Hence, x has an error in the third bit and z has an error in the fourth bit. The transmitted codewords for x and z must have been (110110) and (010011), respectively. The syndrome of y does not occur in any of the columns of the matrix H, so multiple errors must have occurred to produce y.
Coset Decoding
We can use group theory to obtain another way of decoding messages. A linear code C is a subgroup of Zn2. Coset or standard decoding uses the cosets of C in Zn2 to implement maximum-likelihood decoding. Suppose that C is an (n,m)-linear code. A coset of C in Zn2 is written in the form x+C, where x∈Zn2. By Lagrange's Theorem (Theorem 6.10), there are 2n−m distinct cosets of C in Zn2.
Let C be the (5,3)-linear code given by the parity-check matrix
Solution
The code consists of the codewords
There are 25−2=23 cosets of C in Z52, each with order 22=4. These cosets are listed in Table 8.40.
\(Table \text { } 8.40.\) Cosets of C
Coset | Coset |
Representative | |
C | (00000)(01101)(10011)(11110) |
(10000)+C | (10000)(11101)(00011)(01110) |
(01000)+C | (01000)(00101)(11011)(10110) |
(00100)+C | (00100)(01001)(10111)(11010) |
(00010)+C | (00010)(01111)(10001)(11100) |
(00001)+C | (00001)(01100)(10010)(11111) |
(10100)+C | (00111)(01010)(10100)(11001) |
(00110)+C | (00110)(01011)(10101)(11000) |
Our task is to find out how knowing the cosets might help us to decode a message. Suppose that x was the original codeword sent and that r is the n-tuple received. If e is the transmission error, then r=e+x or, equivalently, x=e+r. However, this is exactly the statement that r is an element in the coset e+C. In maximum-likelihood decoding we expect the error e to be as small as possible; that is, e will have the least weight. An n-tuple of least weight in a coset is called a coset leader. Once we have determined a coset leader for each coset, the decoding process becomes a task of calculating r+e to obtain x.
In Table 8.40, notice that we have chosen a representative of the least possible weight for each coset.
Solution
These representatives are coset leaders. Now suppose that r=(01111) is the received word. To decode r, we find that it is in the coset (00010)+C; hence, the originally transmitted codeword must have been (01101)=(01111)+(00010).
A potential problem with this method of decoding is that we might have to examine every coset for the received codeword. The following proposition gives a method of implementing coset decoding. It states that we can associate a syndrome with each coset; hence, we can make a table that designates a coset leader corresponding to each syndrome. Such a list is called a decoding table.
\(Table \text { } 8.42\). Syndromes for each coset
Syndrome | Coset Leader |
(000) | (00000) |
(001) | (00001) |
(010) | (00010) |
(011) | (10000) |
(100) | (00100) |
(101) | (01000) |
(110) | (00110) |
(111) | (10100) |
Let C be an (n,k)-linear code given by the matrix H and suppose that x and y are in Zn2. Then x and y are in the same coset of C if and only if Hx=Hy. That is, two n-tuples are in the same coset if and only if their syndromes are the same.
- Proof
-
Two n-tuples x and y are in the same coset of C exactly when x−y∈C; however, this is equivalent to H(x−y)=0 or Hx=Hy.
Table 8.42 is a decoding table for the code C given in Example 8.39. If x=(01111) is received,
Solution
then its syndrome can be computed to be
Examining the decoding table, we determine that the coset leader is (00010). It is now easy to decode the received codeword.
Given an (n,k)-block code, the question arises of whether or not coset decoding is a manageable scheme. A decoding table requires a list of cosets and syndromes, one for each of the 2n−k cosets of C. Suppose that we have a (32,24)-block code. We have a huge number of codewords, 224, yet there are only 232−24=28=256 cosets.