8.5: Efficient 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.

Example \(8.35\)

Given the binary matrix

\[ H = \begin{pmatrix} 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \end{pmatrix} \nonumber \]

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

Solution

we can compute

\[ H{\mathbf x} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \qquad \text{and} \qquad H{\mathbf y} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\text{.} \nonumber \]

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 \(1\text{,}\) 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.

Proposition \(8.36\)

Let the \(m \times n\) binary matrix \(H\) determine a linear code and let \({\mathbf x}\) be the received \(n\)-tuple. Write \({\mathbf x}\) as \({\mathbf x} = {\mathbf c} +{\mathbf e}\text{,}\) where \({\mathbf c}\) is the transmitted codeword and \({\mathbf e}\) is the transmission error. Then the syndrome \(H{\mathbf x}\) of the received codeword \({\mathbf x}\) is also the syndrome of the error \({\mathbf e}\text{.}\)

Proof

The proof follows from the fact that

\[ H{\mathbf x} = H({\mathbf c} +{\mathbf e}) = H{\mathbf c} + H{\mathbf e} = {\mathbf 0} + H{\mathbf e} = H{\mathbf e}\text{.} \nonumber \]

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 \(H\text{.}\)

Theorem \(8.37\)

Let \(H \in {\mathbb M}_{ m \times n} ( {\mathbb Z}_2)\) and suppose that the linear code corresponding to \(H\) is single error-correcting. Let \({\mathbf r}\) be a received \(n\)-tuple that was transmitted with at most one error. If the syndrome of \({\mathbf r}\) is \({\mathbf 0}\text{,}\) then no error has occurred; otherwise, if the syndrome of \({\mathbf r}\) is equal to some column of \(H\text{,}\) say the \(i\)th column, then the error has occurred in the \(i\)th bit

Example \(8.38\)

Consider the matrix

\[ H = \begin{pmatrix} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{pmatrix} \nonumber \]

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

Solution

Then

\[ H{\mathbf x} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, H{\mathbf y} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, H{\mathbf z} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\text{.} \nonumber \]

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 \({\mathbf y}\text{.}\)

Coset 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 \({\mathbb Z}_2^n\text{.}\)

Example \(8.39\)

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

\[ H = \begin{pmatrix} 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 1 \end{pmatrix}\text{.} \nonumber \]

Solution

The code consists of the codewords

\[ (00000) \quad (01101) \quad (10011) \quad (11110)\text{.} \nonumber \]

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 \(8.40\).

\(Table \text { } 8.40.\) 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 \({\mathbf x}\) 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 \({\mathbf x}\text{.}\)

Example \(8.41\)

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 \({\mathbf r} = (01111)\) 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.

\(Table \text { } 8.42\). Syndromes for each coset

Syndrome	Coset Leader
\((000)\)	\((00000)\)
\((001)\)	\((00001)\)
\((010)\)	\((00010)\)
\((011)\)	\((10000)\)
\((100)\)	\((00100)\)
\((101)\)	\((01000)\)
\((110)\)	\((00110)\)
\((111)\)	\((10100)\)

Proposition \(8.43\)

Let \(C\) be an \((n,k)\)-linear code given by the matrix \(H\) and suppose that \({\mathbf x}\) and \({\mathbf y}\) are in \({\mathbb Z}_2^n\text{.}\) Then \({\mathbf x}\) and \({\mathbf y}\) are in the same coset of \(C\) if and only if \(H{\mathbf x} = H{\mathbf y}\text{.}\) That is, two \(n\)-tuples are in the same coset if and only if their syndromes are the same.

Proof

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 \(H {\mathbf x} = H{\mathbf y}\text{.}\)

Example \(8.44\)

Table \(8.42\) is a decoding table for the code \(C\) given in Example \(8.39\). If \({\mathbf x} = (01111)\) is received,

Solution

then its syndrome can be computed to be

\[ H {\mathbf x} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\text{.} \nonumber \]

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.