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Section8.4Efficient Decoding

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.

Example8.35

Given the binary matrix

\begin{equation*} H = \begin{pmatrix} 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \end{pmatrix} \end{equation*}

and the 5-tuples \({\mathbf x} = (11011)^{\rm t}\) and \({\mathbf y} = (01011)^{\rm t}\text{,}\) we can compute

\begin{equation*} H{\mathbf x} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \qquad \text{and} \qquad H{\mathbf y} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}. \end{equation*}

Hence, \({\mathbf x}\) is a codeword and \({\mathbf y}\) is not, since \({\mathbf x}\) is in the null space and \({\mathbf y}\) is not. Notice that \(H{\mathbf y}\) is identical to the first column of \(H\text{.}\) In fact, this is where the error occurred. If we flip the first bit in \({\mathbf y}\) from 0 to 1, then we obtain \({\mathbf x}\text{.}\)

If \(H\) is an \(m \times n\) matrix and \({\mathbf x} \in {\mathbb Z}_2^n\text{,}\) then we say that the syndrome of \({\mathbf x}\) is \(H{\mathbf x}\text{.}\) The following proposition allows the quick detection and correction of errors.

The proof follows from the fact that

\begin{equation*} H{\mathbf x} = H({\mathbf c} +{\mathbf e}) = H{\mathbf c} + H{\mathbf e} = {\mathbf 0} + H{\mathbf e} = H{\mathbf e}. \end{equation*}

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 \(H{\mathbf e}\) is the \(i\)th column of the matrix \(H\text{.}\)

Example8.38

Consider the matrix

\begin{equation*} H = \begin{pmatrix} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{pmatrix} \end{equation*}

and suppose that the 6-tuples \({\mathbf x} = (111110)^{\rm t}\text{,}\) \({\mathbf y} = (111111)^{\rm t}\text{,}\) and \({\mathbf z} = (010111)^{\rm t}\) have been received. Then

\begin{equation*} H{\mathbf x} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, H{\mathbf y} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, H{\mathbf z} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}. \end{equation*}

Hence, \({\mathbf x}\) has an error in the third bit and \({\mathbf z}\) has an error in the fourth bit. The transmitted codewords for \({\mathbf x}\) and \({\mathbf z}\) must have been \((110110)\) and \((010011)\text{,}\) respectively. The syndrome of \({\mathbf y}\) does not occur in any of the columns of the matrix \(H\text{,}\) so multiple errors must have occurred to produce \({\mathbf y}\text{.}\)

SubsectionCoset Decoding

We can use group theory to obtain another way of decoding messages. A linear code \(C\) is a subgroup of \({\mathbb Z}_2^n\text{.}\) Coset or standard decoding uses the cosets of \(C\) in \({\mathbb Z}_2^n\) to implement maximum-likelihood decoding. Suppose that \(C\) is an \((n,m)\)-linear code. A coset of \(C\) in \({\mathbb Z}_2^n\) is written in the form \({\mathbf x} + C\text{,}\) where \({\mathbf x} \in {\mathbb Z}_2^n\text{.}\) By Lagrange's Theorem (Theorem 6.10), there are \(2^{n-m}\) distinct cosets of \(C\) in \({\mathbb Z}_2^n\text{.}\)

Example8.39

Let \(C\) be the \((5,3)\)-linear code given by the parity-check matrix

\begin{equation*} H = \begin{pmatrix} 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 1 \end{pmatrix}. \end{equation*}

The code consists of the codewords

\begin{equation*} (00000) \quad (01101) \quad (10011) \quad (11110). \end{equation*}

There are \(2^{5-2} = 2^3\) cosets of \(C\) in \({\mathbb Z}_2^5\text{,}\) each with order \(2^2 =4\text{.}\) These cosets are listed in Table 8.40.

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)
Table8.40Cosets of \(C\)

Our task is to find out how knowing the cosets might help us to decode a message. Suppose that \({\mathbf x}\) was the original codeword sent and that \({\mathbf r}\) is the \(n\)-tuple received. If \({\mathbf e}\) is the transmission error, then \({\mathbf r} = {\mathbf e} + {\mathbf x}\) or, equivalently, \({\mathbf x} = {\mathbf e} + {\mathbf r}\text{.}\) However, this is exactly the statement that \({\mathbf r}\) is an element in the coset \({\mathbf e} + C\text{.}\) In maximum-likelihood decoding we expect the error \({\mathbf e}\) to be as small as possible; that is, \({\mathbf e}\) 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 \({\mathbf r} + {\mathbf e}\) to obtain \({\mathbf x}\text{.}\)

Example8.41

In Table 8.40, notice that we have chosen a representative of the least possible weight for each coset. These representatives are coset leaders. Now suppose that \({\mathbf r} = (01111)\) is the received word. To decode \({\mathbf r}\text{,}\) we find that it is in the coset \((00010) + C\text{;}\) hence, the originally transmitted codeword must have been \((01101) = (01111) + (00010)\text{.}\)

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.

Syndrome Coset Leader
(000) (00000)
(001) (00001)
(010) (00010)
(011) (10000)
(100) (00100)
(101) (01000)
(110) (00110)
(111) (10100)
Table8.42Syndromes for each coset

Two \(n\)-tuples \({\mathbf x}\) and \({\mathbf y}\) are in the same coset of \(C\) exactly when \({\mathbf x} - {\mathbf y} \in C\text{;}\) however, this is equivalent to \(H({\mathbf x} - {\mathbf y}) = 0\) or \(H {\mathbf x} = H{\mathbf y}\text{.}\)

Example8.44

Table 8.42 is a decoding table for the code \(C\) given in Example 8.39. If \({\mathbf x} = (01111)\) is received, then its syndrome can be computed to be

\begin{equation*} H {\mathbf x} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}. \end{equation*}

Examining the decoding table, we determine that the coset leader is \((00010)\text{.}\) 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 \(2^{n - k}\) cosets of \(C\text{.}\) Suppose that we have a \((32, 24)\)-block code. We have a huge number of codewords, \(2^{24}\text{,}\) yet there are only \(2^{32 - 24} = 2^{8} = 256\) cosets.