P

Summing up, if G = is the generator matrix of an (n, k)-code, then

Ik

|P ) is the parity check matrix. We encode a message m by calculating Gm, and we can detect errors in a received vector u by calculating H u. A linear code is determined by either giving its generator matrix or by giving its parity check matrix.

ERROR CORRECTING AND DECODING

We would like to find an efficient method for correcting errors and decoding. One crude method would be to calculate the Hamming distance between a received word and each code word. The code word closest to the received word would be assumed to be the most likely transmitted word. However, the magnitude of this task becomes enormous as soon as the message length is quite large.

be u = v + e. The decoder receives the vector u and has to determine the most likely transmitted code vector v by finding the most likely error pattern e. This error is e = −v + u = v + u. The decoder does not know what the code vector v is, but knows that the error e lies in the coset V + u.

The most likely error pattern in each coset of Zn2 by V is called the coset leader.

The coset leader will usually be the element of the coset containing the smallest number of 1’s. If two or more error patterns are equally likely, one is chosen arbitrarily. In many transmission channels, errors such as those caused by a stroke of lightning tend to come in bursts that affect several adjacent digits. In these cases, the coset leaders are chosen so that the 1’s in each error pattern are bunched together as much as possible.

dimensional vector H u is called the syndrome of u. (Syndrome is a medical term meaning a pattern of symptoms that characterizes a condition or disease.)

Every element of Zn2−k is a syndrome; thus there are 2n−k different cosets and different syndromes.

Theorem 14.12. Two vectors are in the same coset of Zn2 by V if and only if they have the same syndrome.

Proof. If u1, u2 Zn2 , then the following statements are equivalent:

1.Calculate the syndrome of the received word.

2.Find the coset leader in the coset corresponding to this syndrome.

3.Subtract the coset leader from the received word to obtain the most likely transmitted word.

4.Drop the check digits to obtain the most likely message.

For a polynomial code generated by p(x), the syndrome of a received polynomial u(x) is the remainder obtained by dividing u(x) by p(x). This is because the j th column of H is the remainder obtained by dividing xj −1 by p(x). Hence the syndrome of elements in a polynomial code can easily be calculated by means of a shift register that divides by the generator polynomial.

Example 14.13. Write out the cosets and syndromes for the (6,3)-code with parity check matrix

Solution. Each of the rows in Table 14.9 forms a coset with its corresponding syndrome. The top row is the set of code words.

The element in each coset that is most likely to occur as an error pattern is chosen as coset leader and placed at the front of each row. In the top row 000000 is clearly the most likely error pattern to occur. This means that any received word in this row is assumed to contain no errors. In each of the next six rows,

TABLE 14.9. Syndromes and All Words of a (6,3) Code

282 14 ERROR-CORRECTING CODES

there is one element containing precisely one nonzero digit; these are chosen as coset leaders. Any received word in one of these rows is assumed to have one error corresponding to the nonzero digit in its coset leader. In the last row, every word contains at least two nonzero digits. We choose 000110 as coset leader. We could have chosen 101000 or 010001, since these also contain two nonzero digits; however, if the errors occur in bursts, then 000110 is a more likely error pattern. Any received word in this last row must contain at least two errors. In decoding with 000110 as coset leader, we are assuming that the two errors occur in the fourth and fifth digits.

Each word in Table 14.9 can be constructed by adding its coset leader to the code word at the top of its column.

A word could be decoded by looking it up in the table and taking the code word at the top of the column in which it appears. When the code is large, this decoding table is enormous, and it would be impossible to store it in a computer. However, in order to decode, all we really need is the parity check matrix to calculate the syndromes, and the coset leaders corresponding to each syndrome.

Example 14.14. Decode 111001, 011100, 000001, 100011, and 101011 using Table 14.10, which contains the syndromes and coset leaders. The parity check matrix is

Solution. Table 14.11 shows the calculation of the syndromes and the decoding of the received words.

Example 14.15. Calculate the table of coset leaders and syndromes for the (9,4) polynomial code of Example 14.11, which is generated by p(x) = 1 + x2 +

TABLE 14.10. Syndromes and Coset Leaders for a (6,3) Code

Solution. There is no simple algorithm for finding all the coset leaders. One method of finding them is as follows.

We write down, in Table 14.12, the 25 possible syndromes and try to find their corresponding coset leaders. We start filling in the table by first entering the error patterns, with zero or one errors, next to their syndromes. These will be the most likely errors to occur. The error pattern with one error in the j th position is the j th standard basis vector in Z92 and its syndrome is the j th column of the parity check matrix H , given in Example 14.11. So, for instance, H (000000001) = 01101, the last column of H .

The next most likely errors to occur are those with two adjacent errors. We enter all these in the table. For example,

H (000000011) = H (000000010) + H (000000001)

=11010 + 01101, the last two columns of H

=10111.

This still does not fill the table. We now look at each syndrome without a coset leader and find the simplest way the syndrome can be constructed from the columns of H . Most of them come from adding two columns, but some have to