288 14 ERROR-CORRECTING CODES

This determinant is nonzero because α, α2, . . . , α2t are all different if t < 2m−1. [The expression for the Vandermonde determinant can be verified as follows. When the j th column is subtracted from the ith column, each term contains a factor (αsi − αsj ); hence the determinant contains this factor. Both sides are polynomials in αs2 , . . . , αs2t of the same degree and hence must differ by a multiplicative constant. By looking at the leading diagonal, we see that this constant is 1.]

There is, for example, a BCH (127,92)-code that will correct up to five errors. This code adds 35 check digits to the 92 information digits and hence contains 235 syndromes. It would be impossible to store all these syndromes and their coset leaders in a computer, so decoding has to be done by other methods. The errors in BCH codes can be found by algebraic means without listing the table of syndromes and coset leaders.

In fact, any code with a relatively high information rate must be long and consequently, to be useful, must possess a simple algebraic decoding algorithm. Further details of the BCH and other codes can be found in Roman [46], Lidl and Pilz [10], and Lidl and Niederreiter [34].

EXERCISES

14.1.Which of the following received words contain detectable errors when using the (3, 2) parity check code?

110, 010, 001, 111, 101, 000.

14.2.Decode the following words using the (3, 1) repeating code to correct errors:

111, 011, 101, 010, 000, 001.

Which of the words contain detectable errors?

14.3.An ancient method of detecting errors when performing the arithmetical operations of addition, multiplication, and subtraction is the method known as casting out nines. For each number occurring in a calculation, a check digit is found by adding together the digits in the number and casting out any multiples of nine. The original calculation is then performed on these check digits instead of on the original numbers. The

answer obtained, after casting out nines, should equal the check digit of the original answer. If not, an error has occurred. For example, check the following:

9642 × (425 − 163) = 2526204.

Add the digits of each number; 9 + 6 + 4 + 2 = 21 = 2 × 9 + 3, 4 + 2 + 5 = 9 + 2, 1 + 6 + 3 = 9 + 1. Cast out the nines and perform the calculation on these check digits:

14.4.Find the redundancy of the English language. Copy a paragraph from a

book leaving out every nth letter, and ask a friend to try to read the paragraph. (Try n = 2, 3, 4, 5, 6. If a passage with every fifth letter missing can usually be read, the redundancy is at least 15 or 20%.)

14.5.Each recent book, when published, is given an International Standard Book Number (ISBN) consisting of ten digits, for example, 0-471-29891- 3. The first digit is a code for the language group, the second set of digits is a code for the publisher, and the third group is the publisher’s number for the book. The last digit is one of 0, 1, 2, . . . , 9, X and is a check digit.

Have a look at some recent books and discover how this check digit is calculated. What is the 1 × 10 parity check matrix? How many errors does this code detect? Will it correct any?

14.6.Is 1 + x3 + x4 + x6 + x7 or x + x2 + x3 + x6 a code word in the (8,4) polynomial code generated by p(x) = 1 + x2 + x3 + x4?

14.7. Write down all the code words in the (6,3)-code generated by p(x) =

1 + x2 + x3.

14.8.Design a code for messages of length 20, by adding as few check digits as possible, that will detect single, double, and triple errors. Also give a shift register encoding circuit for your code.

14.9.Decode the following, using the (6,3)-code given in Table 14.9:

000101, 011001, 110000.

14.10. A (7,4) linear code is defined by the equations

u1 = u4 + u5 + u7, u2 = u4 + u6 + u7, u3 = u4 + u5 + u6,

where u4, u5, u6, u7 are the message digits and u1, u2, u3 are the check digits. Write down the generator and parity check matrices for this code. Decode the received words 0000111 and 0001111 to correct any errors.

14.11.Find the minimum Hamming distance between the code words of the code with generator matrix G, where

Discuss the error-detecting and error-correcting capabilities of this code, and write down the parity check matrix.

14.12.Encode the following messages using the generator matrix of the (9,4)- code of Example 14.11:

1101, 0111, 0000, 1000.

For Exercises 14.13 to 14.15, find the generator and parity check matrices for the polynomial codes.

14.13.The (4,1)-code generated by 1 + x + x2 + x3.

14.14.The (7,3)-code generated by (1 + x)(1 + x + x3).

14.15.The (9,4)-code generated by 1 + x2 + x5.

14.16. Find the syndromes of all the received words in the (3,2) parity check code.

14.17.Using the parity check matrix in Example 14.14 and the syndromes in Table 14.10, decode the following words:

101110, 011000, 001011, 111111, 110011.

14.18.Using the parity check matrix in Example 14.11 and the syndromes in Table 14.12, decode the following words:

110110110, 001001101, 111111111, 000000111.

For Exercises 14.19 to 14.22, construct a table of coset leaders and syndromes for each code.

14.19.The (3,1)-code generated by 1 + x + x2.

14.20.The (7,4)-code with parity check matrix

14.21.The (9,4)-code generated by 1 + x2 + x5.

14.22.The (7,3)-code generated by (1 + x)(1 + x + x3).

14.23.Consider the (63,56)-code generated by (1 + x)(1 + x + x6).

(a)What is the number of digits in the message before coding?

(b)What is the number of check digits?

(c)How many different syndromes are there?

(d)What is the information rate?

(e)What sort of errors will it detect?

(f)How many errors will it correct?

14.24.One method of encoding a rectangular array of digits is to add a parity check digit to each of the rows and then add a parity check digit to each of the columns (including the column of row checks). For example, in the array in Figure 14.8, the check digits are shaded and the check on checks is crosshatched. This idea is sometimes used when transferring information to and from magnetic tape. The same principle is used in accounting. Show that one error can be corrected and describe how to correct that error. Will it correct two errors? What is the maximum number of errors that it will detect?

Figure 14.8

14.25.Let V be a vector space over Zp , where p is a prime. Show that every subgroup is a subspace. Is this result true for a vector space over any Galois field?

14.26. Show that the Hamming distance between vectors has the following properties:

(1)d(u, v) = d(v, u).

(2)d(u, v) + d(v, w) d(u, w).

(3)d(u, v) 0 with equality if and only if u = v. (This shows that d is a metric on the vector space.)

14.27.We can use elements from a finite field GF (q), instead of binary digits,

Ik

H = (In−k | − P ) is the parity check matrix.

14.28.Using elements of Z5, find the parity check matrix of the (7,4)-code generated by 1 + 2x + x3 Z5[x].

14.29.Find the generators of the twoand three-error-correcting BCH codes

of length 15 by starting with the primitive element β in GF (16), where

β4 = 1 + β3.

14.30.Find the generator polynomial of a single-error-correcting BCH code of length 7.

14.31.Let α be a primitive element in GF (32), where α5 = 1 + α2. Find an irreducible polynomial in Z2[x] with α3 as a root.

14.32.Find the generator polynomial of a double-error-correcting BCH code of length 31.

Note that it is important in Example 2 that every integer n is even or odd, so that the two cases considered cover every possibility. Of course, a proof can proceed by reduction to more than two cases, at least one of which must always hold.

The statements used in mathematics are chosen so that they are either true or false. This leads to another method of proof of an implication p q called proof by contradiction. Since q is either true or false, the idea is to show that it cannot happen that both p is true and q is false. We accomplish this by showing that the assumption that p is true and q is false leads to a contradiction.

Example 3. If r is a rational number (that is, a fraction), show that r 2 = 2.

both even. The statement r 2 = 2 leads to m2 = 2n2, so m2 is even. Hence m is even (by Example 1), say m = 2k, where k is an integer. But then the equation m2 = 2n2 becomes 4k2 = 2n2, so n2 = 2k2 is even. Hence n is even (again by Example 1), so we have shown that m and n are both even, contradicting the choice of m and n. This completes the proof.

As in Example 3, proof by contradiction often provides the simplest verification of an implication.

To provide another example, we need the following concept. An integer greater than 1 is called a prime (and we say that it is a prime number) if it cannot be factored as the product of two smaller integers both greater than 1. Hence the first few primes are 2, 3, 5, 7, 11, 13, . . ., but 6 = 2 · 3 and 35 = 5 · 7 are not primes.

Example 4. If 2n − 1 is a prime number, show that n is prime.

Proof. We must show that p q where p is the statement “2n − 1 is prime” and q is the statement that “n is prime.” Suppose that q is false, so that n = ab where a 2 and b 2 are integers. For convenience, write k = 2a . Then 2n = 2ab = (2a )b = kb , and we verify that

2n − 1 = kb − 1 = (k − 1)(kb−1 + kb−2 + · · · + k2 + k + 1).

Since k 4 this is a factorization of 2n − 1 as a product of integers greater than 2, a contradiction.

The next example illustrates one way to verify that an implication is not valid.

Example 5. Show that the implication “n is a prime” “2n − 1 is a prime” is false.

We say that n = 11 is a counterexample to the (proposed) implication in Example 5. Note that it is enough to find even one example in which an implication is not valid to show that the implication is false. Hence it is in a sense easier to disprove an implication than to prove it.

The implications in Examples 4 and 5 are closely related. They have the form p q and q p, respectively, where p is the statement “2n − 1 is a prime” and q is the statement “n is a prime.” In general, each of the statements p q and q p is called the converse of the other, and these examples show that an implication can be valid even though its converse is not valid.

If both p q and q p are valid, we say that p and q are logically equivalent. We write this as

p q,

and read it as “p if and only if q.” Many of the most satisfying theorems assert that two statements, ostensibly quite different, are in fact logically equivalent.

Example 6. If n is an integer, show that “n is odd” “n2 is odd.”

Every mathematics book is full of examples of proofs of implications. This is because of the importance of the axiomatic method. This procedure arises as follows: In the course of studying various examples, it is observed that they all have certain properties in common. This leads to the study of a general, abstract system where these common properties are assumed to hold. The properties are then called axioms in the abstract system, and the mathematician proceeds by deducing other properties (called theorems) from these axioms using the methods introduced in this appendix. These theorems are then true in all the concrete examples because the axioms hold in each case. The body of theorems is called a mathematical theory, and many of the greatest mathematical achievements take this form. Two of the best examples are number theory and group theory, which derive a wealth of theorems from 5 and 4 axioms, respectively.

The axiomatic method is not new: Euclid first used it in about 300 B.C.E. to derive all the propositions of (euclidean) geometry from a list of 10 axioms. His book, The Elements, is one of the enduring masterpieces of mathematics.