Intructor's Solution Manual Introduction to Coding Theory, Second Edition [2nd Ed] Juergen Bierbraue

SOLUTION MANUAL

Jürgen Bierbrauer

Introduction to Coding Theory, Second Edition Solutions Manual

CRC PRESS Boca Raton London

New York

Washington, D.C.

Contents

I

An elementary introduction to coding

7

1 The concept of coding 1.1 Bitstrings and binary operations . . . . . . . . . . . . . . . .

1.2 The Hamming distance . . . . . . . . . . . . . . . . . . . . .

1.3 Binary codes . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.4 Error-correcting codes in general . . . . . . . . . . . . . . . .

1.5 The binary symmetric channel . . . . . . . . . . . . . . . . .

1.6 The sphere packing bound . . . . . . . . . . . . . . . . . . .

9 9 10 11 11 12 14

2 Binary linear codes 2.1 The concept of binary linear codes . . . . . . . . . . . . . . .

2.2 Block coding . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.3 The effect of coding . . . . . . . . . . . . . . . . . . . . . . .

2.4 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.5 Binary Hamming and Simplex codes . . . . . . . . . . . . . .

2.6 The principle of duality for binary linear codes . . . . . . . .

17 17 18 19 20 21 22

3 General linear codes 3.1 Prime fields . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3.2 Finite fields . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3.3 Linear codes over finite fields . . . . . . . . . . . . . . . . . .

3.4 Duality and orthogonal arrays . . . . . . . . . . . . . . . . .

3.5 Weight distribution . . . . . . . . . . . . . . . . . . . . . . .

3.6 The game of SET . . . . . . . . . . . . . . . . . . . . . . . .

3.7 Syndrome decoding . . . . . . . . . . . . . . . . . . . . . . .

25 25 25 28 29 31 34 35

4 Reed-Solomon codes

39

5 Recursive constructions I 5.1 Shortening and puncturing . . . . . . . . . . . . . . . . . . .

5.2 Concatenation . . . . . . . . . . . . . . . . . . . . . . . . . .

41 41 44

6 Universal Hashing

47

7 Designs and the binary Golay code

49

8 Shannon entropy

53

3

4 9 Asymptotic results

57

10 3-dimensional codes, projective planes

61

11 Summary and outlook

63

II

65

Theory and applications of codes

12 Subfield codes and trace codes 12.1 The trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12.2 Trace codes and subfield codes . . . . . . . . . . . . . . . . .

12.3 Galois invariant codes . . . . . . . . . . . . . . . . . . . . . .

12.4 Automorphism groups . . . . . . . . . . . . . . . . . . . . . .

67 67 70 70 72

13 Cyclic codes 13.1 Some primitive cyclic codes of length 15 . . . . . . . . . . . .

13.2 Theory of cyclic codes . . . . . . . . . . . . . . . . . . . . . .

13.3 Decoding BCH-codes . . . . . . . . . . . . . . . . . . . . . .

13.4 Constacyclic codes . . . . . . . . . . . . . . . . . . . . . . . .

75 75 75 79 79

14 Recursive constructions, covering radius 14.1 Construction X . . . . . . . . . . . . . . . . . . . . . . . . .

14.2 Covering radius . . . . . . . . . . . . . . . . . . . . . . . . .

83 83 85

15 The linear programming method

87

16 OA in statistics and computer science 16.1 OA and independent random variables . . . . . . . . . . . .

16.2 Linear shift register sequences . . . . . . . . . . . . . . . . .

16.3 Cryptography . . . . . . . . . . . . . . . . . . . . . . . . . .

16.4 Two-point based sampling . . . . . . . . . . . . . . . . . . .

16.5 Resilient functions . . . . . . . . . . . . . . . . . . . . . . . .

16.6 Derandomization of algorithms . . . . . . . . . . . . . . . . .

16.7 Authentication and universal hashing . . . . . . . . . . . . .

91 91 91 93 93 93 97 97

17 The geometric description of codes 17.1 Linear codes as sets of points . . . . . . . . . . . . . . . . . .

17.2 Quadratic forms, bilinear forms and caps . . . . . . . . . . .

17.3 Caps: constructions and bounds . . . . . . . . . . . . . . . .

99 99 105 108

18 Additive codes 18.1 Basic constructions and applications . . . . . . . . . . . . . .

18.2 The cyclic theory of additive codes . . . . . . . . . . . . . . .

18.3 Additive quaternary codes: the geometric approach . . . . .

18.4 Quantum codes . . . . . . . . . . . . . . . . . . . . . . . . .

18.5 Network codes, subspace codes . . . . . . . . . . . . . . . . .

113 113 115 115 117 118

5

III

Codes and algebraic curves

121

19 Introduction 19.1 Polynomial equations and function fields . . . . . . . . . . .

19.2 Places of the rational function field . . . . . . . . . . . . . .

123 123 123

20 Function fields, their places and valuations 20.1 General facts . . . . . . . . . . . . . . . . . . . . . . . . . . .

20.2 Divisors and the genus . . . . . . . . . . . . . . . . . . . . .

20.3 The Riemann-Roch theorem . . . . . . . . . . . . . . . . . .

20.4 Some hyperelliptic equations . . . . . . . . . . . . . . . . . .

125 125 125 125 125

21 Determining the genus 21.1 Algebraic extensions of function fields . . . . . . . . . . . . .

21.2 The hyperelliptic case . . . . . . . . . . . . . . . . . . . . . .

21.3 The Kloosterman codes and curves . . . . . . . . . . . . . .

21.4 Subrings and integrality . . . . . . . . . . . . . . . . . . . . .

21.5 The Riemann-Hurwitz formula . . . . . . . . . . . . . . . . .

127 127 127 127 127 127

22 AG-codes, Weierstraß points and universal hashing 22.1 The basic construction . . . . . . . . . . . . . . . . . . . . .

22.2 Pole orders . . . . . . . . . . . . . . . . . . . . . . . . . . . .

22.3 Examples of function fields and projective equations . . . . .

22.4 The automorphism group . . . . . . . . . . . . . . . . . . . .

22.5 AG-codes and universal hashing . . . . . . . . . . . . . . . .

129 129 129 130 131 131

23 The last chapter 23.1 List decoding . . . . . . . . . . . . . . . . . . . . . . . . . . .

23.2 Expander codes . . . . . . . . . . . . . . . . . . . . . . . . .

23.3 tms-nets . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

23.4 Sphere packings . . . . . . . . . . . . . . . . . . . . . . . . .

23.5 Permutation codes . . . . . . . . . . . . . . . . . . . . . . . .

133 133 133 133 134 134

Part I

An elementary introduction to coding

7

Chapter 1 The concept of coding

1.1

Bitstrings and binary operations

Exercises 1.1 1.1.1. Compute the sum of 11001 and 01110

11001 + 01110 = 10111 1.1.2. Assume 000000 was sent and two errors occurred. List all possible received messages.

There are ones.

6 2

= 15 possibilities: all bitstrings of length 6 with precisely two

1.1.3. Let x = 1101 be the message to be sent. Encode x using the repetition code of length 3.

Encode x = 1101 −→ 111 111 000 111.

1.1.4. Assume the repetition code of length 3 is used and 000110111101 is received. What is the result of decoding?

000110111101 −→ 0111.

1.1.5. Why does the Morse code represent letters E and T by bitstrings of length 1, whereas letters like Q,V,Z are represented by longer bitstrings?

Because E and T are frequent letters in standard text, whereas Q,V,Z typically occur less often.

9

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Introduction to Coding Theory, Second Edition

1.2

The Hamming distance

Solutions Manual

Exercises 1.2 1.2.1. Compute d(11001, 01110) and d(0000, 0110). d(11001, 01110) = 4, d(0000, 0110) = 2 1.2.2. Find wt(00110) and wt(10111). wt(00110) = 2, wt(10111) = 4.

1.2.3. List all vectors in F62 at distance 3 from 111000.

The list consists of 63 = 20 vectors: all the vectors which have the same weight in the first and in the last half of the coordinates.

1.2.4. The alphabet has 26 letters. If we want to represent all possible words of length ≤ 3 (all letters, pairs of letters and triples of letters) as bitstrings of the same length n, what is the smallest n such that this is possible?

There are 263 = 17, 576 words of length 3, 262 = 676 words of length 2 and of course 26 words of length 1. We have to choose n such that 2n ≥ 263 + 262 + 26 = 18, 278. The minimal value of n for which this is satisfied is n = 15.

1.2.5. Prove that the Hamming distance is a metric.

Only the triangle inequality is in doubt. For arbitrary x, y, z we have to prove d(x, z) ≤ d(x, y) + d(y, z).

Addition of a fixed string does not change the distance: d(x, y) = d(x+c, y+c).

This follows directly from the definition of the Hamming distance. We can therefore assume that y = 0 is the 0-string. What we have to prove is d(x, z) ≤ wt(x) + wt(z).

This is clear from the definition.

1.2.6. Assume x is sent and y = x + e is received. What can we say about d(x, y) and about wt(e) if not more than 3 errors have occurred?

If x is sent and y = x + e, then d(x, y) = wt(e). This is ≤ 3 if no more than 3 errors have occurred.

The concept of coding

1.3

11

Binary codes

1.3.1. If we want to correct 8 bit errors, what would the minimum distance of the code have to be?

In order to correct 8 bit errors, distance d ≥ 17 is needed.

1.3.2. Using our code (6, 8, 3)2 , decode the following received vectors: 111100, 111011, 000001, 011110.

Decode: 111100 7→ 111000, 111011 7→ 110011, 000001 7→ 000000, 011110 7→ 011110.

1.3.3. Does a code (5, 6, 3)2 exist?

It does not exist. In fact we can prove the stronger statement that a (5, 5, 3)2 -code cannot exist.

Assume such a code C does exist. Assume at first there are two codewords at distance 5. Adding the same word to all codewords we can assume that 00000 and 11111 belong to C. There is no further word at distance ≤ 3 from both, contradiction.

Assume next there are two words at distance 4 in C. These can be chosen without restriction as x = 00000 and y = 11110. Every word ending in 0 has distance < 3 from either x or y. All remaining words in C must therefore end in 1 and they must have weight 3. We can choose without restriction z = 11001 as a word of C. The fourth word is 00111, and there is no fifth word. Contradiction.

We have shown that any two different words in C must be at distance 3.

This allows us to choose as codewords x = 00000, y = 11100. There is no further word of weight 3 at distance 3 from y.

1.4

Error-correcting codes in general

Exercises 1.4 1.4.1. Find the smallest length n such that an (n, 27, 2)3 exists.

We have seen a (4, 27, 2)3 . Can a (3, 27, 2)3 exist? As it has 27 codewords it would have to consist of all ternary triples. This is impossible as the minimum distance is only 1. The answer is 4.

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Introduction to Coding Theory, Second Edition

Solutions Manual

1.4.2. Prove the following: if there is an (n, M, d)q , then there is an (n + 1, M, d)q .

Write out the words of the (n, M, d)q , lengthen each codeword by appending an arbitrary (n + 1)-st coordinate.

1.4.3. Prove: If there is an (n, M, d)q−1 , then there is an (n, M, d)q .

An (n, M, d)q−1 can be seen as an (n, M, d)q . The entries are from an alphabet of size q − 1, which can be seen as a subset of an alphabet of size q.

1.5

The binary symmetric channel

Exercises 1.5 1.5.1. Compute the probability that no more than 1 error occurs in the transmission of a bitstring of length 10, when the bit error probability is p = 10−3 , or p = 10−4 , or p = 10−5 .

The probability of no errors is (1 − p)10 , the probability of exactly one error is 10p(1 − p)9 . The probability of at most one error is therefore P = (1 − p)10 + 10p(1 − p)9 = (1 + 9p)(1 − p)9 = (1 + 9p)(1 − 9p + 36p2 − + . . . ), which is P = 1 − 45p2 + higher terms For p = 10−3 we have P ∼ 1 − 0.000045 = 0.999955. For p = 10−4 this is P ∼ 1 − 0.00000045 = 0.99999955. In case p = 10−5 finally P ∼ 1 − 0.0000000045 = 0.9999999955.

1.5.2. Describe a generalization of the BSC from the binary to the general q-ary case.

The alphabet A has q elements. The most obvious generalization of the BSC is the following: fix a probability p, the error probability. When x ∈ A is sent, define the probability that x0 is received as 1 − p if x0 = x, as p/(q − 1) if x0 6= x.

1.5.3. Sketch a formal picture of the binary erasure channel, analogous to Figure 2.3 for the BSC.

The concept of coding

0

1−p

13

0 p E p

1

1−p FIGURE 1.1:

1

The erasure channel

1.5.4. Show that if a code of minimum distance d is used for the erasure channel, then any d − 1 errors can be corrected.

Assume the 0-word was sent and errors occur in the first d − 1 coordinates.

The erasure channel is so kind to tell us that the final n − (d − 1) zeroes are correct whereas there may be problems with the first d − 1 bits. The 0-word is then the only codeword which explains this behaviour as any other codeword would have to have at least one entry 1 in the last coordinate section.

1.5.5. Use the subset interpretation of the binomials to prove the binomial formula n X n i n−i n (a + b) = ab . i i=0 Let S = {1, 2, . . . , n} and imagine (a + b)n as a product of n copies of (a + b) where factor number j corresponds to j ∈ S. Multiplying out yields 2n products. Associate to the choice of a from factor number j the choice of j ∈ S, to the choice of b to not choosing j ∈ S. In this way we obtain a one-to-one correspondence between subsets U ⊆ S and products of n copies of a and b. The factor ai bn−i results if and only if the corresponding subset U of

S has cardinality i. The formula results as there are ni subsets of cardinality i.

1.5.6. Prove (1 − b)n =

n X n i (−1)i b. i i=0

This follows from the preceding formula by substituting a 7→ −b, b 7→ 1.

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Introduction to Coding Theory, Second Edition

1.6

The sphere packing bound

Solutions Manual

Exercises 1.6 1.6.1. Construct a code (4, 2, 3)2 (this is really trivial).

We need only two codewords. Choose them as 0000 and 1111. This even is a (4, 2, 4)2 , obviously.

1.6.2. Show that there is no (4, 3, 3)2 -code If there are two words at distance 4, we can choose them as 0000 and 1111.

The third codeword cannot be found. So we can choose 0000 and 1110 as the first two words. The third word needs last entry 1, but the string of first three bits needs to have distance ≥ 2 from both 000 and 111, which is not possible.

1.6.3. Construct a code (5, 4, 3)2 (hint: use our code (6, 8, 3)2 )).

In our code (6, 8, 3) there are 4 words ending in 0. Write out these words and omit the final 0. This is a (5, 4, 3)2 .

1.6.4. Show that there is no (5, 5, 3)2 .

Assume a (5, 5, 3)2 exists. Either we can find 3 words ending in 0 or we can find 3 words ending in 1. Imagine these 3 words written out, with the last bit cancelled. This is a (4, 3, 3)2 . We know from Exercise 2.6.2 that it cannot exist.

1.6.5. Using the preceding exercise, show that there is no (6, 9, 3)2 .

Assume there is a (6, 9, 3)2 . Either there are 5 words ending in 0 or there are 5 words ending in 1. Erasing the last entry we obtain (5, 5, 3)2 , contradiction by the preceding exercise.

1.6.6. What does the sphere-packing bound tell us about the length n of a binary code (n, 27 , 5)2 ?

Assume (n, 27 , 5)2 exists. The sphere packing bound says

n 7 n M = 2 ≤ 2 /(1 + n + ), 2 equivalently

n + n + 1 ≤ 2n−7 .

2 For n = 13 this is not satisfied (13 + 78 ≤ 64 is not true), for n = 14 it is satisfied (15 + 91 ≤ 128 is true). The answer is: n ≥ 14.

The concept of coding

15

1.6.7. 6 candidates are examined by 9 referees. Each referee assigns a passfail grade to each candidate. Any two referees assign the same grade to not more than 3 of the candidates. Can this really happen?

No. Interpreting the candidates as coordinates, the grades (pass-fail) as 0 or 1 and the referees as codewords of a binary code, the question is if an (6, 9, 3)2 -code exists. We proved in Exercise 1.6.5 that this is not the case.

1.6.8. Show that the minimum distance of a perfect code must be odd Assume d = 2e. Pick a vector at distance e from two codewords. Such a vector is in the intersection of the spheres of radius e centered at the codewords, contradicting the definition of a perfect code.

1.6.9. Use the sphere-packing bound to show the nonexistence of (5, 6, 3)2 .

We have q = 2, n = 5, V1 (1, 5) = 6. It follows M ≤ 32/6 < 6.

1.6.10. Does a (7, 9, 3)2 -code exist?

In fact we will see in the next chapter that a (7, 16, 3)2 -code exists. This is a perfect code.

Chapter 2 Binary linear codes

2.1

The concept of binary linear codes

Exercises 2.1 2.1.1. Show that our code [7, 4, 3]2 is perfect.

We have V2 (1, 7) = 8. The sphere packing bound says M · 8 ≤ 27 , which is satisfied with equality.

2.1.2. Try to decide if an [8, 4, 4]2 exists.

Start from a generator matrix of the [7, 4, 3]2 . Append a new bit to each row such that the weight of each row is even. The result is a generator matrix of an [8, 4, 4]2 .

2.1.3. Give an example showing that the basis of a code is not uniquely determined.

Any old example will do.

2.1.4. Determine the parameters of the binary linear code generated by the rows of the matrix   1001101 0 1 0 1 0 1 1 0010111 Simply write out all codewords: 0000000 1001101 0101011 0010111 1100110 1011010 0111100 1110001

17

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Introduction to Coding Theory, Second Edition

Solutions Manual

These are 8 = 23 different codewords. Each nonzero codeword has weight 4. In particular d = 4. The parameters are [7, 3, 4]2 .

2.1.5. Compute the parameters [n, k, d]2 of the binary linear code generated by



 1100001101 G = 0 0 1 1 0 0 1 0 1 1 0000110111

Find a nonzero codeword of minimum weight.

Again write out all codewords:

0000000000 1100001101 0011001011 0000110111 1111000110 1100111010 0011111100 1111110001

Again we have 8 = 23 codewords. The smallest nonzero weight is d = 5.

The parameters are [10, 3, 5]2 . The second word in the list has weight 5.

2.2

Block coding

Exercises 2.2 2.2.1. Follow through all stages of block coding for the input string 00101110 when transmission errors occur in coordinates 3, 7 and 14.

Nothing to say here. Just do it.

Binary linear codes

2.3

19

The effect of coding

Exercises 2.3 2.3.1. Compute the block error probability of the repetition code of length 3, seen as a code encoding blocks of length 4 into messages of length 12.

For example, 0010 is encoded 000000111000. No decoding error means no error in each of the four blocks of length 3. For each block the probability of no error is (1 − p)3 + 3p(1 − p)2 ≈ 1 − 3p2 . The total probability of no decoding error is therefore ≈ (1 − 3p2 )4 ≈ 1 − 18p2 and the error probability is P ≈ 18p2 . This is very close to the case of the binary Hamming code, which has a much higher information rate.

2.3.2. Compute information rate and block error probability for a code [9, 5, 3]2 (it exists).

Information rate R = 5/9, probability of correct decoding (1 − p)9 + 9p(1 − p)8 ≈ 1 + 36p2 − 72p2 ≈ 1 − 36p2 .

Block error probability P ≈ 36p2 .

2.3.3. Compute information rate and block error probability for a code [23, 12, 7]2 (the binary Golay code).

R = 12/23. Probability 1 − P of no error

23 2 23 3 (1 − p)23 + 23p(1 − p)22 + p (1 − p)21 + p (1 − p)20 2 3 4 4 which is ≈ 1 − 23 4 p . The error probability is ≈ 8855p .

2.3.4. Show that the block error of a [n, k, 2e + 1]2 -code probability

n is bounded (approximately) by pe+1 . e+1 This generalizes our 1 − P of correct

Peearlier calculations. The probability decoding is at least i=0 ni pi (1 − p)n−i (there are ni error patterns with precisely i errors, the probability of each such pattern is pi (1−p)n−i ). See this expression as a polynomial in p. The leading term (the term) is of

Peconstant course 1. Fix some exponent j. The coefficient of pj is i=0 ni (−1)j−i n−i j−i .

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Introduction to Coding Theory, Second Edition

Solutions Manual

Our interpretation of binomial numbers in terms of subsets shows the following identity:

n n−i n j = i j−i j i (these are two ways of counting pairs of disjoint subsets, one of cardinality i, the other of cardinality j − i, of an n-set). Our coefficient is therefore X

e n j j−i (−1) j i=0 j−i For j = 0 the result is 1 confirming that the leading term is 1. For 1 ≤ j ≤ e j the sum simply is the binomial expansion of (1 − 1)P = 0, showing that those

e e+1 j powers p do not occur. For j = e + 1 we have i=0 (−1)e+1−i e+1−i = e+1 (1 − 1) − 1 = −1.

2.4

Duality

Exercises 2.4 2.4.1. When is the all-1-word orthogonal to itself ?

The all−1-word is orthogonal to itself if the length n is even.

2.4.2. A code is self-dual if it equals its dual. Is there a self-dual [6, 3, 3]2 ?

There is no self-dual [6, 3, 3]2 .

As in particular each codeword must be orthogonal to itself, all weights must be even. We would have a code [6, 3, 4]2 . The all−1-word is orthogonal to the code and therefore contained in the code. All other codewords have weight 4. However, the sum of a word of weight 4 and the all−1-word has weight 2, contradiction.

2.4.3. Find a (4, 8)-matrix in standard form (starting with the unit matrix I) which generates a self-dual code (C ⊥ = C) with parameters [8, 4, 4]2 .   10000111 0 1 0 0 1 0 1 1   0 0 1 0 1 1 0 1 00011110 2.4.4. Find a generator matrix of the Hamming code [7, 4, 3]2 in standard form. Use the P-transform to find a check matrix.

Binary linear codes

21

In order to find a generator matrix in standard form simply pick those codewords of H starting with the quadruples of weight 1. This yields the generator matrix   1000011 0 1 0 0 1 0 1  G= 0 0 1 0 1 1 1 0001110 The P-transform yields the check matrix   0111100 1 0 1 1 0 1 0 1110001 2.4.5. Is there a self-dual [12, 6, 6]2 -code?

Write a generator matrix in standard form G = (I|P ) where P is a (6, 6) − matrix. Because of self-duality the rows of P have odd weight, because of d = 6 they have weight 5. The sum of two rows of G has weight ≤ 4, contradiction.

2.4.6. Is there a [12, 6, 6]2 -code?

No. After puncturing (cancel the last coordinate in each codeword) an [11, 6, 5]2 -code is obtained. It contradicts the sphere-packing bound.

2.5

Binary Hamming and Simplex codes

Exercises 2.5 2.5.1. Using matrix M3 , find at least 5 different codewords in the Hamming code [7, 4, 3]2 .

The codewords in the Hamming code generated by matrix M3 : the rows 1001101, 0101011, 0010111, the pairwise sums 1100110, 1011010, 0111100 and 0000000, 1110001.

2.5.2. Use M4 and find at least 5 different codewords in H4 (2).

Similarly using M4 : there are 16 codewords.

2.5.3. Use the binary Hamming code [7, 4, 3]2 . Decode the received vectors y1 = (1, 1, 0, 1, 1, 0, 0), y2 = (1, 1, 1, 1, 1, 1, 1), y3 = (1, 1, 1, 0, 0, 0, 0).

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Solutions Manual

Decoding: y1 = (1, 1, 0, 1, 1, 0, 0) 7→ (1, 1, 0, 1, 0, 0, 0). y2 = (1, 1, 1, 1, 1, 1, 1) 7→ y2 . y3 = (1, 1, 1, 0, 0, 0, 0) 7→ (1, 1, 1, 0, 0, 0, 1).

2.5.4. Prove by induction on r that each nontrivial linear combination of the rows of Mr (each nonzero word of the Simplex code Sr (2)) has weight 2r−1 .

For r = 1 and r = 2 this is immediately checked. Let r > 2. Order the columns of Mr (this ordering is immaterial for our problem) such that in the first row we have all zeroes on the left, the ones on the right. Then Mr has the form

0 1 1 Mr = Mr−1 0 Mr−1 (in the middle there is a column (1, 0)t ). Each linear combination not involving the first row has weight 2r−2 + 2r−2 = 2r−1 . Adding the first row yields weight 2r−2 + 1 + (2r−1 − 1 − 2r−2 ) = 2r−1 . Fortunately also the first row has weight 2r−1 .

2.5.5. Show that the binary Simplex codes Sr (2), r ≥ 3 are self-orthogonal (contained in their orthogonals).

Consider the generator matrix Mr . Each codeword is orthogonal to itself as it has even weight 2r−1 . Compare row i and row j of Mr . The number of coordinates where both have entry 1 is 2r−2 , which is even.

2.6

The principle of duality for binary linear codes

Exercises 2.6 2.6.1. Define q-ary orthogonal arrays for arbitrary q.

An array with n columns and entries from a q-set is a q-ary orthogonal array of strength t if in the projection onto any set of t columns each t-tuple of entries occurs the same number λ of times. We write the parameters as OAλ (t, n, q).

2.6.2. Show that a binary orthogonal array of strength t > 1 also has strength t − 1.

Binary linear codes

23

Let A be a binary orthogonal array of strength t > 1. Consider any set of t − 1 columns and embed it in a set of t columns. Because of strength t there is a number λ such that every t-tuple occurs λ times in the projection onto the t columns. This shows that every (t − 1)-tuple occurs 2λ times in the projection onto the t − 1 columns.

2.6.3. Find a check matrix of S3 (2) by applying the P-transform to M3 . Use this check matrix to prove that S3 (2) has minimum weight 4.

Application of the P-transform to M3 yields as generator matrix of the Hamming code, check matrix of the Simplex code S3 (2), the matrix   1101000 1 0 1 0 1 0 0  H= 0 1 1 0 0 1 0 1110001 In order to prove that S3 (2) has minimum distance 4 we have to see that no 3 or less columns of H add up to 0. There is no 0-column, and there are no two equal columns. Assume 3 columns add to 0. Clearly they are not all in the right section, corresponding to the unit matrix. If two are on the right, the one column on the left must have weight 2, contradiction. If one is on the right, two on the left must add to a weight 1 column. This is not the case. The remaining case is that all are on the left, but the sum of the 3 left columns is (0, 0, 0, 1)t , contradiction.

2.6.4. Show that each perfect binary linear code of distance d = 3 has the parameters of one of the binary Hamming codes.

We have 2n = 2k (1 + n). This shows that the length n has the form n = 2 − 1. This shows r + k = n, k = 2r − 1 − r. r

2.6.5. Show that each perfect binary linear code of distance d = 3 is equivalent to one of the binary Hamming codes.

Continuing from the previous exercise we see that a check matrix is an (r, 2r −1)-matrix H which has no zero column (because d > 1) and no repeated columns (as d > 2). It follows that the columns of H are precisely all nonzero r-tuples in some order.

2.6.6. Describe an OA1 (n − 1, n, 2) for arbitrary length n.

This is the linear sum zero code: use as rows all binary n-tuples of even weight.

Chapter 3 General linear codes

3.1

Prime fields

Exercises 3.1 3.1.1. Show that Z/6Z is not a field (of 6 elements).

Z/6Z is not a field as I can find elements 2 and 3, both of which are nonzero whose product 2 · 3 = 6 = 0.

3.1.2. Find the multiplicative inverse of 16 mod 101.

16−1 = 19 (mod 101).

3.1.3. Find 3100 (mod 17).

34 = 81 ≡ 13, 35 ≡ 39 ≡ 5, 36 ≡ 15 ≡ −2, 38 ≡ −1, 316 ≡ 1, 3100 = 3 ≡ 34 ≡ 13 (everything mod 17).

6·16+4

3.1.4. Determine all elements a ∈ Fp satisfying a2 = 1. a2 ≡ 1 (mod p) means a2 − 1 = (a − 1)(a + 1) ≡ 0. Calculating in the integers this means that the prime p divides (a − 1)(a + 1). This implies that p either divides a + 1 or divides a − 1, hence a ≡ ±1 (mod p).

3.2

Finite fields

Exercises 3.2 3.2.1. Find all irreducible polynomials of degree 3 with coefficients in F2 . x3 + x2 + 1 and x3 + x + 1.

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3.2.2. Construct F8 .

Use such that 3 = 2 + 1. The nonzero elements of F8 are then the i , i = 0, . . . , 6. This determines the multiplication. The addition is given by 1 + 3 + 2 = 1 + + 5 = 1 + 4 + 6 = 0.

3.2.3. Is it true that F4 ⊂ F8 ?

No, F4 is not contained in F8 . If it were, then F8 would be a vector space over F4 , and 8 would have to be a power of 4, but this is not the case.

3.2.4. Determine the sum of all elements in Fq .

Let S denote this sum. Let x ∈ Fq , such that x 6= 0, x 6= 1. Then S = xS, therefore 0 = (x − 1)S. As x − 1 6= 0, we conclude S = 0.

This argument does not work in F2 , and in this case we have S = 1.

3.2.5. Construct the field C of complex numbers as a quadratic extension of the field R of real numbers. What is the natural choice of a quadratic irreducible polynomial?

The irreducible polynomial is X 2 + 1, and the complex number corresponding to X is denoted i. All complex numbers z therefore have a unique representation z = x + iy, where x, y are reals. Addition is componentwise, multiplication is determined by the rule i2 = −1 : (x1 + iy1 )(x2 + iy2 ) = (x1 x2 − y1 y2 ) + i(x1 y2 + x2 y1 ).

3.2.6. Construct F9 , using the irreducible polynomial X 2 − X − 1. f (X) = X 2 − X − 1 is irreducible over F3 as f (0) = 2, f (1) = 2, f (2) = 1 are all 6= 0. Let be the image of X. We have 2 = + 1, 3 = − + 1, 4 = −1.

This determines the additive and multiplicative structure of F9 . All nonzero elements of F9 are powers of .

3.2.7. Let α ∈ F = Fqr . Show that there is polynomial f (X) of degree ≤ r with coefficients in Fq such that f (α) = 0.

As F is a vector space of dimension r over Fq , the powers 1, α, . . . , αr cannot be linearly independent.

3.2.8. Let α ∈ F = Fqr . Show that there is a uniquely determined monic (highest coefficient = 1) polynomial f (X) of smallest degree s > 0 such that f (α) = 0. The polynomial f (X) is the minimal polynomial of α. It is irreducible.

It follows from the previous exercise that there is a polynomial f (X) of degree ≤ r such that f (α) = 0. Let s > 0 be the smallest degree of such a polynomial and choose f (X) to be monic by multiplying with a suitable

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27

constant. If there were two different such polynomials we could find a linear combination of smaller degree, contradicting the minimality of s. If f (X) was not irreducible, one of its factors g(X) would have to satisfy g(α) = 0, again contradicting the minimality of s.

3.2.9. Let α ∈ F = Fqr . Let s be the degree of the minimal polynomial of α, equivalently the smallest number such that the powers 1, α, . . . , αs are linearly dependent over Fq . Show that the smallest subfield Fq (α) of F containing Fq and α has 1, α, . . . , αs−1 as basis. Conclude that s must divide r.

The minimal polynomial f (X) of α is irreducible (see the previous exercise).

It follows that 1, α, . . . , αs−1 form a basis of an extension field Fqs of Fq , by our main construction of extension fields. As Fqr is an extension field, in particular a vector space, over Fqs , the number q r of its elements must be a power of q s , equivalently r is a multiple of s.

3.2.10. In the following exercises the readers are encouraged to prove some of the basic facts concerning finite fields. Start with the following: let F be a finite field. For every natural number n define n · 1 = 1 + 1 + · · · + 1 as the sum of n copies of 1 in F. Prove that there must exist some n > 0 such that n · 1 = 0.

As F is finite the a · 1 cannot all be different. Let a · 1 = b · 1 for some a > b.

Then (a − b) · 1 = 0.

3.2.11. Let F be a finite field. Define the characteristic of F as the smallest natural number p such that p · 1 = 0.

Prove that the characteristic is a prime number.

The distributive law in F shows (ab) · 1 = (a · 1)(b · 1). As F has no zero divisors it follows that the smallest number p > 0 such that p · 1 = 0 cannot be composite. It is a prime number.

3.2.12. Let p be the characteristic of the finite field F. Prove that {0, 1, 2 · 1, . . . , (p − 1) · 1} is a subfield of F and that this subfield is the prime field Fp .

There are exactly p different sums of 1 in F, and they are closed under addition and multiplication. It follows that they form a subfield. Clearly this is a copy of Fp .

3.2.13. Let F be a finite field of characteristic p. Prove that F is a vector space over Fp . Conclude: |F | must be a power of p.

A field of characteristic p contains Fp as a subfield. The axioms of a field are stronger than the axioms of a vector space over a subfield. In particular F is a vector space over Fp . Accept from basic linear algebra that each vector space has a basis, and therefore a dimension. If F has dimension r over Fp , then it has pr elements.

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3.3

Linear codes over finite fields

Solutions Manual

Exercises 3.3 3.3.1. Find a generator matrix of a ternary (q = 3) linear code [4, 3, 2]3 .  1200 0 1 2 0 0012 

3.3.2. Find a generator matrix of a 5-ary code [5, 3, 3]5 . 

 100ab Start with the unit matrix in the left corner: G =  0 1 0 c d  As each 001ef row must have weight at least 3 the constants a, b, . . . , f must be 6= 0. A linear combination involving all 3 rows is OK because of the unit matrix. It suffices to consider linear combinations involving precisely 2 of the rows. In order for those to have weight ≥ 3 it suffices that the pairs (a, b), (c, d), (e,  f ) are not 10011 scalar multiples of one another. We can choose G =  0 1 0 1 2  00113 3.3.3. Find a generator matrix of a [5, 2, 4]4 code.

0111 1 Use G = where F4 = {0, 1, ω, ω}.

101ωω 3.3.4. Find a generator matrix of an [8, 2, 7]7 code.

01111111 G= 10123456 3.3.5. Describe a generator matrix of a [q + 1, 2, q]q code.

Use as first row (1, 0, 1, 1, . . . ), as second row (0, 1, x1 , x2 , . . . , xq−1 ) where {x1 , . . . , xq−1 } = F∗q .

3.3.6. Show that a [q + 2, 2, q + 1]q -code cannot exist.

There are at least q coordinates where the first and second row both have nonzero entries. It follows that at least two of the corresponding fractions are identical. This shows that we can find a non-trivial linear combination of the rows of the generator matrix with two zeroes, contradiction.

General linear codes

3.4

29

Duality and orthogonal arrays

Exercises 3.4 3.4.1. Prove that the dual of the linear repetition code of length n is the sum zero-code.

By definition, a vector (x1 , x2 , . . . , xn ) is in the dual of the repetition code if x1 + x2 + · · · + xn = 0.

3.4.2. Prove the following generalization of the P-transform: Let a generator matrix for a code [n, k]q be written in the form G = (I|P ). Then a check matrix is H = (−P T |I). Here I in G is the (k, k)-unit matrix, the I in H is the (n − k, n − k)-unit matrix and P T is the transpose of P. Let G = (I|P ), H = (−P t |I), where G has k rows and n columns and H therefore has n − k rows and n columns. The presence of the unit matrix shows that G has rank k and H has rank n − k. Let C be the code generated by the rows of G. As C ⊥ has dimension n − k, it suffices to show that each row of G has dot product 0 with each row of H, equivalently that the matrix GH t is the 0-matrix. This is true: GH t = −P t + P t = 0.

3.4.3. Consider the following quaternary matrix H :   10011 100 011 111 1 1 1 1 1 1 1 0 1 0 1 ω ω 1 1 1 0 0 0 1 1 1 ω ω ω ω ω ω 00100 01ωω1ωω1ωω 1ωω 1ωω Determine the strength of the orthogonal array generated by H and the parameters of the quaternary code with H as check matrix.

Here F4 = {0, 1, ω, ω 2 = ω}. What is the name of this code?

There is no zero column in H, so the code generated by H does have strength 1. Also, none of the columns of H is a scalar multiple of some other column, so t = 2. Columns 1, 2, 4 are linearly dependent, so strength 3 certainly is not satisfied. The precise strength is therefore t = 2. This means that the code which has H as check matrix has d = 3.

As H has 21 columns and 21 = (43 − 1)/3, we see that 21 is the maximum number of columns of a quaternary code with 3 rows, which generates a code of strength 2. The code with H as check matrix is the Hamming code H3 (4).

3.4.4. Determine the parameters of the binary code with the following check matrix (this is easy):

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Introduction to Coding Theory, Second Edition   11001111110 0 0 1 1 0 0 1 1 0 1 1 00001111101

Solutions Manual

More interesting is the code with this matrix as generator matrix.

Determine its parameters.

Let the matrix be G. it has 11 columns and three independent rows. Let C be the code generated by G. We have dim(C) = 3, dim(C ⊥ ) = 8. As the first two columns of G are identical, d(C ⊥ ) = 2. So C ⊥ is an [11, 8, 2]2 -code (not interesting). In order to find d(C) we have to write out the remaining four codewords of C. The result is d = 4. The sum of row 1 and row 3 is 11000000011.

3.4.5. Determine the parameters of the code with the following check matrix:   0011ωω 0 ω 0 ω ω ω 1111 1 1 The result is [6, 3, 4]4 . This code is known as the hexacode.

3.4.6. Same problem as before. The check matrix is   1001ωω 0 1 0 1 ω ω 00111 1 Prove your result.

Same result as in the previous exercise.

3.4.7. What is the smallest length n such that a code [n, 7, 3]3 exists.

(Hint: think of the Hamming codes). Show your argument.

A check matrix of [n, 7, 3]3 has n − 7 rows and n columns, such that no two columns are scalar multiples of each other. For n = 9 this is impossible as n > (32 − 1)/2 = 4. For n = 10 we have 10 < (33 − 1)/2 = 13, so this can be constructed. The answer is n = 10. In fact, we can start from the check matrix of H3 (3) (this is our matrix M3 (3)) and delete three columns.

3.4.8. Construct 3 pairwise orthogonal Latin squares of order 4 and the corresponding OA of index 1. Determine its parameters when seen as a code.   1234 2 1 4 3  As first Latin square we use L1 =   3 4 1 2  An orthogonal partner is L2 = 4321     1234 1234 4 3 2 1 3 4 1 2      2 1 4 3  A third Latin square, orthogonal to both, is L3 =  4 3 2 1  .

3412 2143

General linear codes

31

The corresponding OA1 (2, 5, 4) is 11111 12222 13333 14444 21243 22134 23421 24312 31324 32413 33142 34231 41432 42341 43214 44123 Its rows form a (5, 16, 4)4 -code.

3.5

Weight distribution

Exercises 3.5 3.5.1. Determine the weight polynomial of the repetition code [n, 1, n]q and the sum-0-code [n, n − 1, 2]q .

The repetition code has A0 = 1, An = q − 1. Its weight polynomial is X n + (q − 1)Y n .

Its dual, the sum-0-code, therefore has weight polynomial 1 {(X + (q − 1)Y )n + (q − 1)(X − Y )n } q 3.5.2. Let D be a binary linear code with weight distribution (Bi ) and C the code obtained by adding a parity check bit.

Express the weight distribution of C in terms of (Bi ).

The weight distribution of the code obtained by adding a parity check bit is given by A0 = 1 (of course) and A2i = B2i + B2i−1 for all i > 0.

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3.5.3. The Simplex code [7, 3, 4]2 is a constant weight code. Its weight distribution is therefore A0 = 1, A4 = 7. Use the Mc Williams formula to compute the weight distribution of its dual, the binary Hamming code [7, 4, 3], and of the extended Hamming code, the [8, 4, 4]-code obtained by adding a parity check bit.

The Simplex code S3 (2) has weight enumerator X 7 + 7X 3 Y 4 . The weight enumerator of the Hamming code is therefore 1 {(X + Y )7 + 7(X + Y )3 (X − Y )4 } = X 7 + 7X 4 Y 3 + 7X 3 Y 4 + Y 7 .

8 The weight distribution of the Hamming code is therefore B0 = 1, B3 = B4 = 7, B7 = 1, confirming that the minimum distance is 3 and that the all-1-vector is contained in the Hamming code. The code [8, 4, 4]2 has weight distribution C0 = 1, C4 = 7 + 7 = 14, C8 = 1.

3.5.4. Show that each self-dual (equal to its dual) [8, 4, 4]2 -code has the same weight distribution as the extended Hamming code: A0 = A8 = 1, A4 = 14.

Let C = C ⊥ be an [8, 4, 4]2 -code. All codewords have even weight. As the all-1-vector is orthogonal to each word of C, it is contained in C, so A8 = 1.

The Mac Williams formula shows X X 16 Ai X 8−i Y i = (X + Y )8−i (X − Y )i , i

i

equivalently 16(X 8 +

8 X

Ai X 8−i Y i ) = (X + Y )8 +

i=4

8 X

Ai (X 2 − Y 2 )8−i (X − Y )2i−8

i=4

Comparing coefficients of X 7 Y shows 0=8−

8 X

(2i − 8)Ai = 8 − (4A6 + 8A8 ).

i=4

As A8 = 1 it follows A6 = 0. The remaining 14 codewords must have weight 4.

3.5.5. Show that the extended Hamming code is the only self-dual [8, 4, 4]2 -code.

After permuting coordinates and Gauß elimination we can arrange that a generator matrix has the form G = (I | P ), where P is a (4, 4)-matrix.

Because of the preceding problem each row of G has weight 4, so each row of P has weight 3. No two rows of P can be identical as otherwise addition of

General linear codes

33

the two corresponding rows of G would of weight 2. We  produce a codeword  10000111 0 1 0 0 1 0 1 1  can choose notation such that G =  0 0 1 0 1 1 0 1.

00011110 3.5.6. The weight polynomial in one variable is defined as X Ai Y i .

PC (Y ) = WC (1, Y ) = i

Find the Mc Williams formula for the one variable weight polynomial.

As WC (X, Y ) is homogeneous of degree n we have WC (λX, λY ) = λn WC (λX, λY ) for any constant λ. In particular WC (X, Y ) = X n WC (1, Y /X). Application to the right side of the Mc Williams formula yields the Mc Williams formula for weight polynomials in one variable: PC ⊥ (Y ) =

1 1−Y (1 + (q − 1)Y )n PC ( ). |C| 1 + (q − 1)Y

3.5.7. Let [n, k, d]q be the parameters of a perfect linear code.

Determine the number Ad of minimum weight codewords

n We have d = 2e+1 by Exercise 1.6.8. Consider the e+1 (q−1)e+1 vectors of weight e+1. Each such vector is in a ball of radius e centered at a codeword of

weight d. As each such ball contributes precisely 2e+1 such vectors it follows e

2e + 1 n Ad = (q − 1)e+1 . e e+1 3.5.8. Show that the weight distribution of a perfect [n, k, d]q -code is uniquely determined by n, k, d, q.

Ad has been determined in the previous exercise. Use induction. Assume Ad , . . . , Ai−1 are determined for some i > d. Consider the vectors of weight j = i − e. They are partitioned by the balls of radius e centered at codewords of weight ≤ i. The contribution of codewords of weight < i can be calculated because of the induction hypothesis. As the contribution of each codeword of weight i can be computed as well, Ai is uniquely determined.

3.5.9. Let f (j) be the number of j-tuples with nonzero entries in Fq with j j−1 }. nonzero sum. Prove that f (j) = q−1 q {(q − 1) + (−1) Clearly f (1) = q − 1, f (j + 1) = (q − 1)j+1 − f (j) for j > 1 and the formula is true for j = 1, 2, 3. Assume it is true for j. Then f (j + 1) = (q − 1)j+1 − f (j) = =

q−1 {q(q − 1)j − (q − 1)j + (−1)j } = q

q−1 {(q − 1)j+1 + (−1)j }. q

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3.6

The game of SET

Solutions Manual

Exercises 3.6 3.6.1. How many SETS are there? Show your counting method.

There are 81 2 = 81·80/2 = 81·40 pairs of cards. Each such pair determines exactly one SET. Each SET contains precisely 3 pairs. The number of SETS is 81 · 40/3 = 27 · 40 = 1080.

3.6.2. Consider the ternary (4, 10)-matrix A in the upper left corner of the matrix above:   0111111111  1011112222   A=  0001021221  0011220102 Determine the parameters of the code generated by A and the parameters of its dual.

A generates a code [10, 4, 6]3 , whose dual is a [10, 6, 4]3 -code. In fact, it is easier to work with a generator matrix in standard form, starting with a unit matrix:   1000222120  0100221012   G=  0010202211  0001021221 The last row of G is the third of A, the third row of G is the sum of third and fourth of A and so on. Each row of this matrix has weight 6. In particular it is orthogonal to itself. It is easy to check that each two rows of G are orthogonal. This shows that any two codewords are orthogonal. In particular each codeword is orthogonal to itself: its weight is a multiple of 3. It suffices therefore to show that there is no codeword of weight 3.

Assume v is such a codeword. How many rows of G does it involve? Certainly not only one (the rows have weight 6) and not all four. A linear combination of two rows of G has weight 2 in the first section corresponding to the unit matrix and 2 more entries 6= 0 in the latter section, where the 0 entry of one row meets a nonzero in the other. This is a contradiction. Only linear combinations of exactly 3 rows of G remain to be considered. As the first section already gives weight 3 we have to show that no 3 of the right segments of rows of G are linearly dependent. This is easy to see. There are only 4 cases to be considered.

General linear codes

35

3.6.3. Consider the following ternary matrix:   100000021212 0 1 0 0 0 0 1 0 2 1 1 2   0 0 1 0 0 0 2 2 0 1 1 1   0 0 0 1 0 0 1 1 1 0 1 1   0 0 0 0 1 0 1 2 2 2 0 1 000001212210 Determine the parameters of the ternary code with this check matrix.

What are the parameters after erasing a coordinate?

Compare with the sphere-packing bound.

This is a self-dual [12, 6, 6]3 -code. It is known as the extended ternary Golay code. Erasing a column leads to the ternary Golay code [11, 6, 5]3 , which is perfect.

3.7

Syndrome decoding

Exercises 3.7 3.7.1. Use the decoding scheme for [8, 4, 4]2 given in this section to decode the following received vectors: y = 00110011, y = 11111010, y = 10001001. y = 00110011 → σ(y) = 0000 7→ x = y. y = 11111010 → σ(y) = 0101σ 7→ x = y − 00000101 = 11111111. y = 10001001 → σ(y) = 0001σ 7→ x = y − 00010000 = 10011001.

3.7.2. Find coset leaders for the sum-0-codes [n, n − 1, 2]q .

The number of cosets of [n, n − 1, 2]q is q, a check matrix has only one row, and all the entries are = 1. It follows σ(x) = x1 + x2 + · · · + xn . As coset leaders we can take the (c, 0, 0, . . . , 0), of weight 0 if c = 0 (this coset is the code itself), of weight 1 if c 6= 0.

3.7.3. Find coset leaders for the Hamming code [7, 4, 3]2 .

There are 23 = 8 cosets, defined by the check matrix   1001101 H = 0 1 0 1 0 1 1 0010111

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This leads to the following coset leaders coset leader syndrome 0000000 000 1000000 100 0100000 010 0010000 001 0001000 110 0000100 101 0000010 011 0000001 111

All the coset leaders are uniquely determined, because the code is perfect.

3.7.4. Find a check matrix of a quaternary code [6, 3, 4]4 . Find coset leaders for all cosets.

The number of cosets is 43 = 64. As a check matrix we can use  1 0 0 1 ω2 ω  0 1 0 1 ω ω2  0011 1 1 

As d = 4, the vectors of weights ≤ 1 are unique leaders of their respective cosets. Up to scalar multiples this yields coset leader syndrome (0, 0, 0, 0, 0, 0) (0, 0, 0) (1, 0, 0, 0, 0, 0) (1, 0, 0) (0, 1, 0, 0, 0, 0) (0, 1, 0) (0, 0, 1, 0, 0, 0) (0, 0, 1) (0, 0, 0, 1, 0, 0) (1, 1, 1) (0, 0, 0, 0, 1, 0) (ω, ω, 1) (0, 0, 0, 0, 0, 1) (ω, ω, 1) So far we have 19 cosets. Each vector of weight 2 is a coset leader, but it is not necessarily uniquely determined. Vectors of weight 2 with intersecting supports certainly are leaders of different cosets. If we choose 1 in the first coordinate and an arbitrary nonzero entry in an arbitrary coordinate > 1, we obtain, after multiplication with scalars, 5 · 3 · 3 = 45 different further cosets.

However, as 45 + 19 = 64, in this way we obtain leaders for all cosets. Up to scalar multiples the following is obtained:

General linear codes coset leader syndrome (1, 1, 0, 0, 0, 0) (1, 1, 0) (1, ω, 0, 0, 0, 0) (1, ω, 0) (1, ω, 0, 0, 0, 0) (1, ω, 0) (1, 0, 1, 0, 0, 0) (1, 0, 1) (1, 0, ω, 0, 0, 0) (1, 0, ω) (1, 0, ω, 0, 0, 0) (1, 0, ω) (1, 0, 0, 1, 0, 0) (0, 1, 1) (1, 0, 0, ω, 0, 0) (ω, ω, ω) (1, 0, 0, ω, 0, 0) (ω, ω, ω) (1, 0, 0, 0, 1, 0) (ω, ω, 1) (1, 0, 0, 0, ω, 0) (0, ω, ω) (1, 0, 0, 0, ω, 0) (ω, 1, ω) (1, 0, 0, 0, 0, 1) (ω, ω, 1) (1, 0, 0, 0, 0, ω) (ω, 1, ω) (1, 0, 0, 0, 0, ω) (0, ω, ω)

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Chapter 4 Reed-Solomon codes

Exercises 4 4.1. Prove the following slight generalization of Lagrange interpolation: Let a ≥ 0 be a natural number, x1 , x2 , . . . , xn different nonzero ∈ F and y1 , . . . , yn arbitrary in F. Then there is exactly one polynomial p(X) of degree < n with coefficients in F, which satisfies xai p(xi ) = yi , i = 1, 2, . . . , n.

Lagrange interpolation is the case a = 0. However, what we need is a polynomial if degree < k satisfying p(xi ) = yi /xai , so all we need is Lagrange interpolation again.

4.2. Let a ≥ 0 be a natural number and q a prime-power. Define a q-ary code of length q − 1 with coordinates indexed by 0 6= u ∈ Fq and with codewords indexed by the polynomials p(X) of degree < k, where k < q. The entry of the codeword indexed by p(X) in coordinate u is ua p(u).

Determine the dimension, minimum distance and strength of this code.

The dimension is k, the length is q − 1 and by the preceding exercise the strength is k. It follows that the dual code is a [q − 1, q − 1 − k, k + 1]q -code, an MDS-code. The next exercise shows that the original code is MDS as well, therefore an [q − 1, k, q − k]q -code.

4.3. Prove: The dual of a linear MDS-code is an MDS-code.

Let C be an [n, k, n − k + 1]q -code. We want to show that C ⊥ has minimum distance k + 1, equivalently that C has strength k. Assume C does not have strength k. This means there must be a nonzero codeword v ∈ C which has entries 0 in some k coordinates. However, it follows wt(v) ≤ n − k, a contradiction.

4.4. Let C be an [n, k, d]q -code. Determine the parameters of C ⊥ as an orthogonal array. Derive the Singleton bound (in the case of linear codes).

C ⊥ has strength d − 1 and q n−k codewords. Its parameters are therefore OAλ (d − 1, n, q), where λ = q (n+1)−(d+k) . As λ is a non-negative integer it follows k + d ≤ n + 1, the Singleton bound in the linear case.

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4.5. Give another simple proof of the Singleton bound for linear [n, k, d]q -codes based on Gauß elimination (no calculations needed).

We can find a generator matrix with the unit matrix in some k coordinates.

When this is done each row has at least k − 1 zeroes, so d ≤ n − (k − 1).

4.6. In order to understand why covering arrays are also known as qualitatively independent partitions consider the CA(10; 3, 5, 2) which we gave as an example. Interpret the rows as the ground set X (of 10 elements).Then every column describes a partition of X into two parts.

Express the defining property of a strength 2 covering array in terms of this family of partitions.

Any two parts from different partitions intersect nontrivially.

4.7. Prove the following basic fact: if f (X) is a polynomial with coefficients in the field F, and u ∈ F such that f (u) = 0, then f (X) can be written in the form f (X) = (X − u)g(X), where g(X) is a polynomial.

We can always divide with remainder (long division); Dividing by X − u yields f (X) = (X − u)g(X) + r(X), where the degree of r(X) is less than the degree of X − u, which is 1. It follows that r(X) = r is a constant (in F ).

Evaluating at u shows r = f (u) = 0.

4.8. The finite geometric series is the identity (X − 1)

n−1 X

X i = X n − 1.

i=0

Prove this identity.

Simply multiply out the left side: (X + X 2 + · · · + X n ) − (1 + X + · · · + X n−1 ) = X n − 1.

P∞ 4.9. Let q be a complex number of absolute value |q| < 1. Prove n=0 q n = 1 + q + · · · = 1/(1 − q). This is the infinite geometric series.

Use the preceding exercise. Summing up to some number N we obtain (1 − q N +1 )/(1 − q). The infinite sum is defined as the limit of this expression when N −→ ∞. As q N +1 −→ 0 the result is 1/(1 − q).

4.10. Show the identity (X − u)

n−1 X

X i un−1−i = X n − un .

i=0

Use Exercise 4.8 with X/u in the place of X, multiply both sides of the resulting equation by un .

4.11. Solve Exercise 4.7 using the geometric series.

Pn P Let f (X) = i=0 ai X i . We have f (X) = f (X) − f (u) = i ai (X i − ui ).

In the preceding exercise we saw that X i − ui is divisible by X − u.

Chapter 5 Recursive constructions I

5.1

Shortening and puncturing

Exercises 5.1 5.1.1. Let C be a binary linear code with odd minimum distance d and D ⊂ C the set of codewords of even weight. Prove that D is a subspace of codimension 1 in C. First, D is a subspace as it is closed under addition. As d is odd, D 6= C. If v, w are in C, but not in D, then v + w has even weight (odd+odd=even), therefore is in D. It follows that the dimension of D is at most one less than the dimension of C. We conclude codim(D) = dim(C) − dim(D) = 1.

5.1.2. Given a linear code [192, 11, 92]2 (it exists), make a list of all parameters of codes of lengths 188 and above that can be constructed using iterated shortening and projection.

All [192 − a, 11 − b, 92 − c]2 , where a, b, c ≥ 0, a ≤ 4 and a = b + c.

5.1.3. Prove the following: if a code is MDS, then every code derived from it by shortening or projection is MDS as well.

An [n, k, d]q -code is MDS if k + d = n + 1. Shortening and puncturing decrease n by 1 and decrease either k or d by 1. The equality is preserved.

5.1.4. Determine the parameters of the 3-dimensional binary code generated by   111000000111100 G =  000111000110011  000111111111100 Add a parity check bit to this generator matrix. What are the parameters of the resulting extended code?

Clearly d = 7. As the rank is 3 we have a [15, 3, 7]2 -code. Adding a parity check bit yields a [16, 3, 8]2 -code.

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5.1.5. Let X = {1, 2, . . . , 8}. Define a binary matrix M of size (8, 24) as follows: the rows of M are indexed by the elements x ∈ X. The columns of M are indexed by unordered pairs {i, j} ⊂ X, where {i, j} is none of {1, 2}, {3, 4}, {5, 6}, {7, 8} (these forbidden pairs form what is known as a onefactor). The entry in row x and column {i, j} is 0 is x = i or x = j, it is 1 otherwise. Observe that every column of M has weight 6 and every row has weight 18.

Prove that M has rank 7 and that its row-space is a code [24, 7, 10]2 .

P Denote the rows by rx . We have x rx = 0. Assume there is a another linear dependence among the rows. Then there is such a dependence involving at most 4 rows. If the number of rows involved is even, choose a coordinate {i, j} such that ri is one of the rows involved whereas rj is not. The entry of the linear combination at coordinate {i, j} is 1, contradiction. If the number of rows involved is odd (and ≤ 4), it must be 3. Choose a coordinate {i, j} such that neither i nor j indexes one of the three involved columns. The same contradiction as before is obtained. It follows that there is only one non-trivial linear combination of the rows, hence rk(M ) = 7, equivalently dim(C) = 7.

Each rx has weight 18, the sum of two rows has weight 12. Consider the sum of three rows. Let A be the set of indexes of the three rows involved, B the complement. The weight equals the number of admissible pairs in A plus the number of admissible pairs in B. As there is at most

one pair from the 1-factor in A, and at most two in B the weight is ≥ 52 + 32 − 3 = 10. In case of a sum of four rows we have |A| = |B| = 4 and the weight is the number of admissible pairs having one endpoint in A and the other in B. This number is ≥ 16 − 4 = 12.

5.1.6. Given a linear code [24, 7, 10]2 , which parameters of 6-or 7-dimensional codes can be constructed using iterated shortening and projection?

[24, 7, 10] [23, 7, 9] [22, 6, 9] [22, 7, 8] [23, 6, 10].

5.1.7. Consider the matrix 

 0111111111  1011112222   A=  0001021221  0011220102 which generates a code [10, 4, 6]3 . Find generator matrices for the codes derived by shortening and by projection.

For puncturing erase the first column of A (or any other column), for shortening it is easiest to erase the second row and the leading 0 in each of the remaining rows.

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5.1.8. Let C be a code [n, k, d]q and x = (x1 , x2 , . . . , xw , 0, 0, . . . ) ∈ C ⊥ a codeword of weight w in the dual code. Apply shortening to C with respect to the first w coordinates. We know that the resulting code has parameters (at least) [n − w, k − w, d]q .

Show that this shortened code has in fact better parameters.

This is known as construction Y1 .

Imagine having shortened w − 1 times already. This code consists of all words (yw , yw+1 , . . . , yn ) such that (0, 0, . . . , 0, yw , yw+1 , . . . , yn ) ∈ C. Because of the orthogonality to x (we have xw 6= 0) we have automatically yw = 0.

This shows that the shortened code is an [n − w, k − w + 1, d]q -code.

5.1.9. Determine the parameters of the codes obtained by applying construction Y1 to the Simplex codes.

The Simplex codes have parameters [ d = 3. Construction Y1 yields [

qk − 1 , k, q k−1 ]q and their duals have q−1

qk − 1 − 3, k − 2, q k−1 ]q . q−1

5.1.10. Show that a code with minimum distance d > q cannot be MDS. Consider an MDS-code [n, n − d + 1, d]q and assume d > q. The residual code has parameters [n − d, n − d, dd/qe], which is impossible when d > q.

5.1.11. Prove that the Simplex codes meet the Griesmer bound with equality. qk − 1 and d = q k−1 . q−1 The Griesmer bound applied to q, k, d yields n ≥ q k−1 + q k−2 + · · · + 1 = (q k − 1)/(q − 1).

The Simplex codes have dimension k, length n =

5.1.12. Prove the following generalization of the residual code construction: Let C be an [n, k, d]q -code and v ∈ C a codeword of weight w. Assume d − w + dw/qe > 0. Then there is an [n − w, k − 1, d − w + dw/qe]q -code.

Let v = v1 , v2 , . . . , vk be a basis of C. We can choose notation such that the first w coordinates are the coordinates where v1 has nonzero entries. It can also be assumed that the nonzero entries of v1 are = 1. The code D we wish to construct is generated by the projection of v2 , . . . , vk to the right coordinates (where v1 has entry 0).

Let 0 6= x ∈ hv2 , . . . , vk i and w0 the weight of the right part of x. Let δi be the frequency of entry i in the left part of x. Then wt(x − iv1 ) = w0 + w − δi .

As x − iv1 is a nonzero codeword of C we have wt(x − iv1 ) ≥ d. It follows w0 ≥ d − w + δi . As the average value of the δi is w/q, we obtain the desired bound w0 ≥ d − w + dw/qe.

Assume dim(D) < k − 1. Then we can find x as above such that w0 = 0.

This contradicts an assumption.

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Concatenation

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Exercises 5.2 5.2.1. Use RS(4, 8) and [7, 3, 4]2 in a concatenation scheme. What are the parameters of the resulting concatenated code?

[8, 4, 5]8 as outer code and [7, 3, 4]2 as inner code give a concatenated code [56, 12, 20]2 .

5.2.2. We have already met the following matrix in Exercise 3.4.6. It generates a code [6, 3, 4]4 , which is known as the hexacode. We want to apply concatenation with the hexacode as outer code. List at least two parameters of binary codes that can be used as inner codes, and the parameters of the resulting concatenated codes.   1001ωω 0 1 0 1 ω ω 00111 1 The outer code is [6, 3, 4]4 . Any 2-dimensional binary code can be used as inner code. [2, 2, 1]1 as inner code yields [12, 6, 4]2 , the inner code [3, 2, 2]2 yields the concatenated code [18, 6, 8]2 .

5.2.3. Use concatenation with Reed-Solomon codes over F2r as outer codes and binary sum-0-codes as inner codes. Determine the parameters of the resulting binary codes.

The inner codes must be r-dimensional binary codes, so they are [r+1, r, 2]2 codes. Concatenation with the outer code [2r , k, 2r + 1 − k]2r yields a concatenated code [(r + 1)2r , kr, 2r+1 + 2 − 2k]2 , where k ≤ 2r .

5.2.4. Consider concatenation with Reed-Solomon codes as outer codes and a ternary [10, 4, 6]3 as inner code. What are the resulting parameters?

The outer Reed-Solomon codes is defined over F81 . They are [81, k, 82−k]81 codes. Concatenation with a [10, 4, 6]3 -code yields [810, 4k, 492 − 6k]3 -codes, where k ≤ 81.

5.2.5. Construct a 48-dimensional binary linear code by applying concatenation to a Reed-Solomon code as outer code and [23, 12, 7]2 as inner code. What are the parameters of this concatenated code?

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45

The inner code is the binary Golay code, of dimension 12. In order to obtain a 48-dimensional code by concatenation, the outer code must be 4-dimensional over F212 . This Reed-Solomon code is a [4096, 4, 4093]4096 -code. The resulting concatenated code has parameters [23 × 4096, 48, 7 × 4093]2 .

5.2.6. Show how to construct a code [128, 12, 56]2 by concatenation. What do you use as outer and inner codes?

Outer [16, 3, 14]16 , inner [8, 4, 4]2 .

5.2.7. Show how to construct a code [27, 10, 10]3 by concatenation. What do you use as outer and inner codes?

Outer [9, 5, 5]9 (Reed-Solomon), inner [3, 2, 2]3 .

5.2.8. An [11, 4, 5]2 was promised earlier. Construct it by concatenation.

An [11, 4, 5]2 is equivalent to [12, 4, 6]2 , which can be constructed by concatenation from [4, 2, 3]4 (a Reed-Solomon code) and [3, 2, 2]2 .

5.2.9. The Reed-Muller codes RM(a, r) are binary codes of length 2r , a X r r−a minimum distance 2 and dimension . i i=0 Use the (u, u + v)-construction to obtain codes with these parameters. Here RM(0, r) is the repetition code and RM(r, r) is the ambient space.

RM(a, r) is defined inductively by applying the (u, u + v)-construction to RM(a, r − 1) and RM(a − 1, r − 1). The familiar properties of the binomial coefficients show that the dimension is right.

5.2.10. Give several parameters of good short binary linear codes among the family constructed in the third exercise above.

Some examples are [12, 4, 6]2 (case r = k = 2), [32, 9, 12]2 , [32, 12, 10]2 , [32, 15, 8]2 , [32, 18, 6]2 (cases r = 3, 3 ≤ k ≤ 6) and [80, 28, 20]2 , [80, 32, 18]2 , [80, 44, 12]2 , [80, 48, 10]2 (here r = 4).

5.2.11. Give several examples of Reed-Muller codes with good parameters.

Here are some good codes among the RM(a, r). In case r = 3 we obtain the extended Hamming code [8, 4, 4]2 again. Some more good code parameters are [16, 5, 8]2 , [32, 6, 16]2 , [32, 16, 8]2 , [64, 7, 32]2 , [64, 42, 8]2 , [128, 8, 64]2 .

5.2.12. Construct [2i × 14, 4 + i, 2i−1 × 14]2 codes for all i ≥ 1. This yields in particular [28, 5, 14]2 , [56, 6, 28]2 , [112, 7, 56]2 .

We just saw an [16, 5, 8]2 -code. It implies a [14, 4, 7]2 -code. Apply the (u, u + v)-construction with the repetition code as second ingredient.

5.2.13. Construct a [240, 8, 120]2 -code.

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The (u, u + v)-construction shows that such a code would be implied by a [120, 7, 60]2 -code, which in turn would be implied by a [60, 6, 30]2 -code, which would be the consequence of a [30, 5, 15]2 -code. We know that such a code exists. It follows from a Reed-Muller [32, 6, 16]2 -code which in turn is a (u, u + v)-result of a [16, 5, 8]2 -code.

Chapter 6 Universal Hashing

Exercises 6 6.1. Determine the parameters of Reed-Solomon codes when interpreted as −universal hash families. [q, k, q + 1 − k]q is equivalent to an -universal (q, q k )q -hash family, where = (k − 1)/q.

6.2. Describe an (8, 16)2 hash family, which is 12 -universal.

The extended Hamming code [8, 4, 4]2 is an (8, 16)2 -hash family, which is 1 2 -universal.

6.3. Find a 25 -universal (10, 81)3 -hash family. Hint: this code has been constructed in Chapter ??. It is closely related to the ternary Golay code.

A linear [10, 4, 6]3 -code is a 52 -universal (10, 81)3 -hash family.

6.4. Does a 12 -universal (6, 9)2 -hash family exist?

An 12 -universal (6, 9)2 -hash family is an (6, 9, 3)2 -code, which does not exist.

The nonexistence was proved in Exercise 1.6.5.

1 6.5. Construct a (256, 25617 )256 hash family, which is 16 -universal.

1 -universal (256, 25617 )256 A Reed-Solomon code [256, 17, 256−16]256 is an 16 hash family.

6.6. Explain why a code [4000, 5, 3996]4096 exists and determine the parameters of the universal hash family derived from it.

A code [4000, 5, 3996]4096 arises by puncturing from a Reed-Solomon code.

It describes a 10−3 -universal (4000, 260 )4096 -hash family.

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Chapter 7 Designs and the binary Golay code

Exercises 7 7.1. Show that the all-1-word belongs to the extended binary Golay code [24, 12, 8]2 .

The all-1-word is orthogonal to G24 . As G24 is self-dual, it is a word of G24 .

7.2. Let C be a binary linear code, C ⊆ C ⊥ . Assume there is a basis of C all of whose elements have weights divisible by 4 (in other words, a generator matrix all of whose rows have weight divisible by 4). Show that all weights of codewords are multiples of 4.

It suffices to show the following: if the binary codewords v, w both have weight a multiple of 4 and satisfy v · w = 0, then v + w has weight divisible by 4. Why does it suffice? The rows of a generator matrix have weights ≡ 0( mod4). Then the sums of two basis vectors have the property, and so on.

As each codeword is a linear combination of rows of the generator matrix, the statement follows by induction.

Let a be the number of coordinates i where vi = 1, wi = 1, b the number of coordinates where vi = 1, wi = 0, c the number of coordinates where vi = 0, wi = 1. The weight of v + w is b + c. Both a + b and a + c are multiples of 4, and a is even as v · w = 0. It follows b + c ≡ 0( mod4).

7.3. Find a generator matrix of the form M = (I|P ) for G24 .

Starting from the generator matrix G of G23 , adding a last coordinate with all entries 1, we obtain a generator matrix of G24 with a triangular matrix as left half. Use Gauß elimination to transform the left half into the unit matrix. This is simplified by the cyclicity on the first 23 coordinates. We obtain M = (I|P ), where

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

 110001110101 0 1 1 0 0 0 1 1 1 0 1 1   1 1 1 1 0 1 1 0 1 0 0 0   0 1 1 1 1 0 1 1 0 1 0 0   0 0 1 1 1 1 0 1 1 0 1 0   1 1 0 1 1 0 0 1 1 0 0 1   P =  0 1 1 0 1 1 0 0 1 1 0 1 0 0 1 1 0 1 1 0 0 1 1 1   1 1 0 1 1 1 0 0 0 1 1 0   1 0 1 0 1 0 0 1 0 1 1 1   1 0 0 1 0 0 1 1 1 1 1 0 100011101011 7.4. Determine the weight distribution of G24 .

A0 = A24 = 1, A8 = A16 = 759, A12 = 2576.

7.5. Use the value A8 = 759 to show the following: for each set of 5 of the 24 coordinates of G24 there is exactly one codeword of weight 8 (a so-called octad) with entries 1 in those 5 coordinates.

Each octad contains 85 = 83 = 7 × 8 subsets of 5 coordinates. As octads intersect pairwise in at most 4 coordinates, each set of 5 is contained in at most one octad. The number of sets of 5 which are contained in an octad is 7 × 8 × 759 = 3 × 7 × 8 × 11 × 23. This number happens to equal 24 5 . It follows that each 5-set is contained in precisely one octad.

7.6. Find a picture of the Fano plane P G(2, 2).

A picture of the Fano plane is in Chapter 11.

7.7. For every prime-power q there is a design S(2, q, q 2 ) (the affine plane of order q). Determine the number of blocks.

Each point is on q + 1 blocks, each block has q points. Let b be the number of blocks. Count pairs (P, l), where P is a point and l a block such that P ∈ l.

The result is q 2 (q + 1) = bq. It follows b = q(q + 1).

7.8. The SETS form a 2-design. Find its parameters.

2 − (81, 3, 1).

7.9. The extended binary Golay code [24, 12, 8]2 has A12 = 2576 (2576 codewords have weight 12). The corresponding subsets are called dodecads. The 2576 dodecads form the blocks of a design 2 − (24, 12, λ). What is λ? In fact, this is a 5-design. What is λ now?

Designs and the binary Golay code

51

We have 24 points and b = 2576 = 16 × 7 × 23 blocks, each containing 12 points. Count triples (P, Q, l), where P, Q are different points, both contained in l. The result is 24 × 23 × λ2 = b × 12 × 11, so λ2 = 11 × 7 × 8. The parameters as a 2-design are 2 − (24, 12, 616).

What is λ = λ5 when we consider our design as a 5-design? Count pairs (T, l), where T is a set of 5 points contained in block l. The result is

24 12 λ5 = b .

5 5 It follows λ5 = 48. We have a 5 − (24, 12, 48)-design.

Chapter 8 Shannon entropy

Exercises 8 8.1. What is the relationship between H(p1 , p2 , p3 , p4 ) and p1 p1 p2 p2 p3 p3 p4 p4 H( , , , , , , , )?

2 2 2 2 2 2 2 2 H(

p1 p1 p2 p2 p3 p3 p4 p4 , , , , , , , ) = 1 + H(p1 , p2 , p3 , p4 ).

2 2 2 2 2 2 2 2

1 1 1 1 8.2. What is larger (contains more information), H( , , , ) or 2 4 8 8 1 1 2 1 H( , , , )? Compute both entropies and compare.

3 3 9 9 1 1 1 1 1 1 1 1 H( , , , ) = + · 2 + · 3 + · 3 = 1.75 and 2 4 8 8 2 4 8 8 1 2 1 4 2 1 1 2 1 H( , , , ) = 2 · log(3) + log(9/2) + log(3) = log(3) − .

3 3 9 9 3 9 9 3 9 This is the larger value.

8.3. Compare the following experiments: • A: Three possible events occur with probabilities 0.5, 0.25, 0.25 • B: Three possible events occur each with probability 1/3.

Which of these experiments yields a larger amount of information?

Compute the entropies and compare.

The uniform distribution has entropy log(3), which is larger than 1 1 1 1 1 1 H( , , ) = + + = 1.5.

2 4 4 2 2 2 8.4. Same type of problem as before, with • A: Five possible events occur with probabilities 14 , 14 , 41 , 18 , 18 .

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• B: Three possible events occur with probabilities 12 , 13 , 16 .

1 1 9 1 1 1 1 1 H( , , , , ) = 3 · · 2 + 2 · · 3 = , 4 4 4 8 8 4 8 8 1 1 1 2 1 1 1 1 H( , , ) = + log(3) + (1 + log(3)) = + log(3), 2 3 6 2 3 6 3 2 which is larger than 9/8.

8.5. Prove the statement concerning the case of equality in Jensen’s inequality.

This is clear.

8.6. Prove A(sm ) = A(sm−1 ) + A(s) for natural numbers s, m and conclude A(sm ) = mA(s).

Consider sm equiprobable choices. Divide them into blocks of s each.

The basic property states A(sm ) = A(sm−1 ) + A(s). By induction we obtain A(sm ) = mA(s).

8.7. Given natural numbers s, t, n define m such that sm ≤ tn < sm+1 .

With this notation prove m A(t) m+1 ≤ < n A(s) n and also m log(t) m+1 ≤ < n log(s) n Conclude A(t) = log(t), where log is the binary logarithm.

The monotonicity of A and the previous exercise show mA(s) ≤ nA(t) < (m + 1)A(s). Divide by n to obtain the first inequality. Using log instead of A the same argument applies. Let n go to infinity. This shows A(t)/A(s) = log(t)/log(s). The choice s = 2 yields the claim A(t) = log(t).

8.8. Prove the formula for H in the case of rational probabilities. It is clear i that because of continuity this suffices. So let pi = m m , where mi and m are Pr natural numbers, i=1 mi = m.

Hint: Use the central property of H in an appropriate situation to obtain A(m) = H(p1 , . . . , pr ) +

r X i=1

pi A(mi ).

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55

Consider at first m equally probable events. The corresponding entropy is A(m). Divide them into r blocks of sizes m1 , m2 , . . . , mr . The basic property yields X A(m) = H(p1 , . . . , pr ) + pi A(mi ). i

Using A(t) = log(t) and solving for H(p1 , . . . , pr ) the claim is obtained.

(here 8.9. The multinomial coefficient m1 ,m2n,...,mk is defined as m1 !mn!

2 !...mk ! m1 + m2 + · · · + mk = n). Prove that

n lim(log( )/n) = H(p1 , p2 , . . . , pk ) m1 , m2 , . . . , mk where n, mi −→ ∞ such that mi /n −→ pi .

For k = 2 we know the result. Use induction on k. As

n n n − m1 = × m1 , m2 , . . . , mk m1 m2 , . . . , mk it follows that our limit is H2 (p1 , 1 − p1 ) + Hk−1 (p2 /(1 − p1 ), . . . , pk /(1 − p1 )), which by the basic property of the entropy equals Hk (p1 , . . . , pk ).

Chapter 9 Asymptotic results

9.1. Prove the additivity of the expectation: if f and g are real-valued random variables defined on the same probability space, then E(f + g) = E(f ) + E(g).

This is trivial. We have by definition X E(f + g) = pi (f (i) + g(i)) = E(f ) + E(g). i

9.2. Let F = F1 + · · · + Fn be the real-valued random variable considered in the proof of the channel coding theorem. We have E(F ) = np.

Prove that the variance of F is σ 2 = npq, where q = 1 − p.

2 The formula V ar(F ) = E(F 2 ) − E(F )2 shows that it suffices σ = Pn to 2prove n i n−i 2 2 2 E(F ) = npq + (np) = np(np + q). By definition E(F ) = i=1 i i p q .

As ni = ni n−1 i−1 we can factor out n and obtain

n X n − 1 i n−i E(F ) = n i pq . i−1 i=1 2

With j = i − 1 this becomes E(F 2 ) = np

n−1 X

(j + 1)

j=0

n − 1 j n−1−j p q . j

The inner sum corresponding to the summand 1 from j + 1 is = 1, so E(F 2 ) = np +

n−1 X j=1

j

n − 1 j n−1−j p q . j

Repeating this process, letting k = j − 1, yields E(F 2 ) = np+np(n−1)p

n−2 X k=0

n − 2 k n−2−k p q = np(1+p(n−1)) = np(np+q), k

as claimed.

9.3. Determine the maximum dimension k such that the GV-bound guarantees the existence of a code [8, k, 4]2 .

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We have V2 (3, 8) = 1 + 8 + 82 + 83 = 93 < 128, so 7 = 9 − k. Dimension k = 2 is the largest.

9.4. Determine the maximum dimension k such that the GV-bound guarantees the existence of a code [6, k, 4]3 .

We have V3 (3, 6) = 233 < 35 , so 5 = 7 − k and k = 2 is the maximum.

9.5. Let C be a code [n, k, d]q . Prove that nd ≤ (q −1)q k−1 /(q k −1). This can be done by a simple counting argument. Conclude that if (δ, R) is asymptotically reachable by linear codes and R > 0, then δ ≤ (q − 1)/q.

Consider a linear [n, k, d]q -code. Imagine all the codewords written as rows of a (q k , n)-array. Count the triples (x, y, i), where x and y are different codewords (rows), and i is a coordinate such that xi 6= yi . Because of minimum distance d a lower bound is q k (q k − 1)d. Now count columnwise. There are n columns. In each column we find at most q different entries in q k rows.

The maximum number of pairs of rows with different entries is obtained when each entry occurs the same number q k−1 of times. An upper bound for our number of triples is therefore nq k (q k − q k−1 ). We have q k (q k − 1)d ≤ nq k q k−1 (q − 1), which is equivalent to the claim. d q − 1 qk We have ≤ . When k → ∞ the last factor goes to 1, so n q qk − 1 δ ≤ (q − 1)/q if R > 0.

9.6. Let h(x) be the binary entropy function. Determine the derivative h0 (x).

We have h(x) = −xlog(x) − (1 − x)log(1 − x), where log is the binary log. Differentiation simply using the product rule 1−x shows h0 (x) = log( ), which is 0 when x = 1/2, negative for 0 < x < 1/2 x and positive for 1/2 < x < 1. So h(x) is increasing from h(0) = 0, until it reaches the maximum h(1/2) = 1, then decreases to h(1) = 0.

9.7. The q-ary entropy function is defined as Hq (δ) = logq (2)h(δ) + δlogq (q − 1).

In particular H2 = h. The right side of the asymptotiv GV-bound is 1 − Hq (δ).

Prove that Hq is increasing for 0 ≤ δ ≤ (q − 1)/q. Determine Hq ((q − 1)/q).

We have H(x) = logq (2)h(x) + logq (q − 1)x.

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It follows H 0 (x) = logq (2)log(

(q − 1)(1 − x) 1−x ) + logq (q − 1) = logq ( ). x x

It follows H 0 (x) ≥ 0 for 0 ≤ x ≤ (q − 1)/q.

Hq (

q−1 ) = 1. q

9.8. Determine the maximum dimension k such that the Gilbert-Varshamov bound guarantees the existence of a code [10, k, 5]4 . Show your calculation.

V4 (4, 10) = 20, 686 < 48 , so 8 = 11 − k. The answer is k = 3.

9.9. Let the natural number n and the positive real number Pn S be given. Consider nonnegative numbers x , i = 1, 2, . . . , n such that i i=1 xi = S. Pn Prove: the expression i=1 x2i is minimized by the choice xi = S/n.

Determine this minimum.

Hint: Let some n-tuple (x1 , x2 , . . . , xn ) be given such that not all the xi equal S/n. Pick i such that xi > S/n and j such that xj < S/n. Show that you can replace xi by xi − and xj by xj + for some > 0 and obtain a tuple with a smaller sum of squares.

Follow the hint. Without restriction x1 > S/n, x2 < S/n. For some small P> 0 choose P 2 y1 = x1 − , y2 = x2 + and yi = xi for i > 2. It follows 2 y = i i i xi − 2 (x1 − x2 − ). For < x1 − x2 this is a smaller value than P 2 i xi .

9.10. Prove an asymptotic version of the sphere-packing bound (for q-ary linear codes): if (δ, R) is asymptotically reachable, then R ≤ 1 − Hq (δ/2).

Consider an [n, k, d]q -code. With established notation the sphere-packing bound states q k Vq (e, n) ≤ q n . Take base q logarithms on both sides, divide by e k n. Use only the asymptotically dominating terms. This yields + logq (q − n n e δ δ 1) + h( )logq (2) ≤ 1, equivalently R ≤ 1 − ( logq (q − 1) + logq (2)h( )), n 2 2 equivalently R ≤ 1 − Hq (δ/2).

9.11. Prove the following version of the Gilbert-Varshamov bound: If Vq (d − 2, n − 1) < q n−k , then an [n, k, d]q -code exists. Hint: assume at first that [n − 1, k − 1, d]q exists and use a check matrix.

Assume at first an [n − 1, k − 1, d]q -code exists. This is equivalent to an (n−k, n−1)q -matrix of rank n−k, which is (d−1)-independent (a check matrix H). If we can find an additional column, which is not linearly dependent of any d − 2 columns of H, then the matrix obtained by adding the new column

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to H is again (d−1)-independent, hence the check matrix of an [n, k, d]q -code.

The additional column is an element of Fn−k . The number of elements of Fn−k q Pd−2 qn−1 which cannot be used is at most i=0 i (q − 1)i = Vq (d − 2, n − 1).

Just as in the case of the original GV-bound we get rid of the assumption that an [n − 1, k − 1, d]q -code exists.

9.12. Prove that the version of the GV-bound proved in the preceding exercise is stronger than the original GV-bound.

It suffices to prove that qVq (d − 2, n − 1) < Vq (d − 1, n). . Adding a Consider the Vq (d − 2, n − 1) vectors of weight ≤ d − 2 in Fn−1 q coordinate with all possible extries in the extra coordinate yields qVq (d − 2, n − 1) different vectors of weight ≤ d − 1 in Fnq .

9.13. Use the method of the GV-bound to prove a Gilbert-Varshamov bound for not necessarily linear q-ary codes.

An (n, M, d)q -code exists if (M − 1)Vq (d − 1, n) < q n .

Chapter 10 -dimensional codes, projective planes 3

Exercises 10 10.1. Construct a code [14, 3, 8]2 by the geometrical method.

Assign weight 2 to each point of the Fano plane. There are 8 points outside each line (counted with multiplicity). This yields [14, 3, 8]2 .

10.2. Describe the construction of [q 2 , 3, q 2 − q]q by the geometric method.

Assign weight 0 to all points on a line, weight 1 to all points outside that line. This yields [q 2 , 3, q 2 − q]q .

10.3. Under which condition on the weight function w will the code really have dimension 3?

The dimension of the code will indeed be 3 if there are some three points of weight > 0, which form a triangle. Equivalently: the dimension will be < 3 if all points of positive weight are on a line.

10.4. The projective line P G(1, q) consists of the q + 1 one-dimensional subspaces (points) of F2q . Describe the parameters of codes [n, 2, d]q using geometric language.

In the projective line each point is a hyperplane. Let the point weights be w1 ≤ w2 · · · ≤ wq+1 . Then d = w1 + w2 + · · · + wq .

10.5. Given d, determine the smallest n such that a code [n, 2, d]q exists.

With notation as in the preceding exercise d equals n− the largest weight of a point. Let the point weights be w1 ≤ w2 · · · ≤ wq+1 . We must have w1 +· · ·+wq ≥ d, so we can choose w1 +· · ·+wq = d and all these wi ≤ dd/qe.

Choose wq+1 = wq = dd/qe, so n = d + dd/qe. This meets the Griesmer bound with equality.

10.6. Write the generator matrix of a code [10, 2, 8]7 and show that the parameters are as claimed.

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A [10, 2, 8]7 is described by assigning weight 1 to every points of P G(1, 7) with the exception of two points which receive weight 2. A generator matrix is

0011111111 1100123456 10.7. Find a description of 4-dimensional codes, which is analogous to our geometric description of 3-dimensional codes.

A 4-dimensional code is defined by assigning non-negative integer weights w(P ) to the points P ∈ P G(3, q). A generator matrix is obtained by writing P w(P ) columns which are generators of the point P. The length is n = P w(P ). The dimension is indeed 4 if and only if the points with positive weight are not contained in a plane. The minimum distance d is n− the maximum sum of weights over a plane.

10.8. Find a generator matrix of a code [17, 3, 9]2 .

In order to obtain [17, 3, 9]2 assign weight 3 to each of three points of the Fano plane forming a triangle. The four points outside the triangle receive weight 3 each. A generator matrix is   11100000011110 0 11 0 0 0 1 1 1 0 0 0 1 1 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 0 0 1 1 1 1∗1 1 10.9. Imagine you have 28 points in P G(2, 8) such that no line contains more than 4 of these points (they exist). Use these points as columns of a generator matrix.

What are the parameters of the code? Give a brief explanation.

Outside each line there are 24 points of P G(2, 8). It follows that we have a code [28, 4, 24]8 . It meets the Griesmer bound with equality.

10.10. Show that the maximum minimum distance of a [9, 3, d]-code is d = 4.

Obviously the optimum choices are, either one point of the Fano plane with weight 3, the others with weight 1, or two points each with weight 2, the others with weight 1. In each case the maximum weight of a line is 5, hence d = 9 − 5 = 4.

Chapter 11 Summary and outlook

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Theory and applications of codes

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Chapter 12 Subfield codes and trace codes

12.1

The trace

Exercises 12.1 12.1.1. What is the dimension of F212 as a vector space over F8 ?

As 212 = 84 the dimension of F212 over F8 is 4.

12.1.2. Show that every irreducible polynomial of degree 5 over F2 must be primitive.

For every irreducible polynomial of degree 5 over F2 the element ∈ F32 corresponding to X has an order > 1 and dividing 31. As 31 happens to be prime is a primitive element.

12.1.3. Let be a primitive element of Fq , where q is odd. How can you read off from the exponent j if j is a square in Fq ?

How many nonzero squares are there in Fq ? j is a square if and only if j is even. Exactly half of the nonzero elements of Fq for odd q are squares.

12.1.4. The norm N from Fqr to Fq is defined by N (x) = x · g1 (x) · · · · · gr−1 (x) =

Y

g(x).

g∈G

Prove that indeed N (x) ∈ Fq for every x ∈ Fqr .

Apply g1 to N (x). When considering the trace we saw already that g1 permutes the g(x), g ∈ G. It follows that the product of these terms is unchanged: g1 (N (x)) = N (x). This implies N (x) ∈ Fq .

12.1.5. Let N : Fq2 → Fq be the norm. How many elements x ∈ Fq2 have N (x) = 1?

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We have N (x) = xq+1 , and N (x) = 1 if and only if the order of x divides q + 1. As the multiplicative group of Fq2 is cyclic there are precisely q + 1 elements of norm 1.

12.1.6. Show that there are exactly q r−1 elements in Fqr whose trace is 0.

Continue to show that this kernel ker(tr) of the trace is an (r−1)-dimensional subspace of F when considered as a vector space over Fq .

This follows from basic principles of linear algebra. The trace is an Fq -linear mapping from Fqr to Fq . As Fqr is r-dimensional and tr is not identically zero, we have that ker(tr) has dimension r − 1.

12.1.7. Determine the kernel ker(tr) in the cases tr : F4 → F2 and tr : F8 → F2 .

When tr : F4 → F2 we have ker(tr) = F2 . In case tr : F4 → F2 we have ker(tr) = {0, 3 , 5 , 6 }.

12.1.8. Let tr : Fqr → Fq . Under which conditions on q, r is it true that Fq ⊆ ker(tr)? When do we have ker(tr) = Fq ?

We have Fq ⊂ ker(tr) if and only if tr(1) = 0. This is true if and only if p divides r. Equality can hold only if r = 2. It follows that ker(tr) = Fq if and only if p = r = 2.

12.1.9. Determine all values tr(x) for tr : F9 → F3 .

We have described F9 = F3 ( ), where 2 = + 1. It follows 3 = − + 1, 4 = −1. In particular is a primitive element. We have tr(1) = 1 + 1 = −1, tr(−1) = 1, tr( ) = tr( 3 ) = + 3 = 1. Clearly ker(tr) = {0, ± 2 }. Also tr(x) = 1 for x ∈ {−1, , 3 } and tr(x) = −1 for x ∈ {1, − , − 3 }.

12.1.10. Let α ∈ Fqr . We studied the minimal polynomial in a series of exercises.

Prove that α and αq have the same minimal polynomial.

Let f (X) = X s + as−1 X s−1 + · · · + a1 X + a0 be the minimal polynomial of α. As g1 is a field automorphism and g1 (ai ) = ai for all i we have f (αq ) = f (g1 (α)) = g1 (f (α)) = 0. The claim follows.

12.1.11. Prove tr(αq ) = tr(α), where tr : Fqr → Fq .

This follows from the fact that the images of α and of αq = g1 (α) under the elements of G are the same.

12.1.12. Let α ∈ Fqr . Show that Fq (α) = Fqr if and only if the images g(α) under the elements g ∈ G are pairwise different.

Let Fq (α) = Fqs , where s < r. It follows gs (α) = α.

Vice versa assume Fq (α) = Fqr . Assume gi (α) = gj (α) for some i > j. It follows gi−j (α) = α. Let i − j = s. It follows α ∈ Fqs .

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12.1.13. Why is the trace called the trace? The reason is that tr(α) is the trace of the Fq -linear mapping : Fqr → Fqr defined by multiplication with α.

We want to prove this in a special case.

Let F = Fqr = Fq (α). Prove that the Fq -linear mapping : F → F defined by x 7→ αx has trace tr(α).

We know that α has minimal polynomial f (X) of degree r and that the g(α) Q are different. As all these elements are roots of f (X) we have f (X) = g∈G (X − g(α)). In particular the second highest coefficient of f (X) is −tr(α) : f (X) = X r − tr(α)X r−1 + . . .

Consider the matrix representing the mapping x 7→ αx with respect to the basis 1, α, . . . , αr−1 . In the first r − 1 columns the entry in the main diagonal is 0. The last column is obtained by expanding αr . Comparison with f (X) shows that the southeast entry is indeed tr(α).

12.1.14. Show that X 3 − X − 1 is an irreducible polynomial over F3 . Decide if it is primitive.

Let f (X) = X 3 − X − 1. As f (0) = f (1) = f (2) = 2 the polynomial f (X) is irreducible over F3 . Let F27 = F3 ( ), where 3 = + 1. Computing successive powers we obtain 4 = 2 + , 5 = 2 + + 1, 6 = 2 − + 1, 7 = − 2 − + 1, 8 = − 2 − 1, 9 = − 1, 10 = 2 − , 11 = − 2 + + 1, 12 = 2 − 1, 13 = 1.

It follows that has order 13.

12.1.15. Show that X 3 − X 2 + 1 is an irreducible polynomial over F3 . Use it to construct F27 . Determine ker(tr), where tr is the trace down to F3 .

Let f (X) = X 3 − X 2 + 1. As f (0) = f (1) = f (2) = 1 it follows that f (X) is irreducible. Let F27 = F3 ( ), where 3 = 2 − 1. Computing successive powers: 4 = 2 − − 1, 5 = − − 1, 6 = − 2 − , 7 = 2 + 1, 8 = 2 + − 1, 9 = − 2 − − 1, 10 = 2 − + 1, 11 = − 1, 12 = 2 − , 13 = −1.

It follows that has order 26. We have tr(1) = 0, tr( ) = tr( 2 ) = 1, so ker(tr) is generated by 1 and 2 − e − 1 = 4 . The general rule is tr(a2 2 + a1 + a0 ) = a2 + a1 (where ai ∈ F3 ).

12.1.16. Prove the transitivity of the trace: if Fq ⊂ L = Fqs ⊂ F = Fqr , then trF |Fq (x) = trL|Fq (trF |L (x)) for all x ∈ F. Let r = sd, G = {g0 = id, g1 , . . . gsd−1 } the Galois group of F over Fq .

By definition trF |Fq (x) = g0 (x) + g1 (x) + · · · + gsd−1 (x). A partial sum is y = g0 (x) + gs (x) + · · · + gs(d−1) (x). As {g0 , gs , . . . , gs(d−1) } is the Galois group of F over L = Fqs , it follows y = trF |L (x). The total sum describing trF |Fq (x) is y + g1 (y) + · · · + gs−1 (y) = trL|Fq (y).

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Trace codes and subfield codes

Exercises 12.2 12.2.1. Determine trace code and subfield code of RS(2, 4) (clearly down to F2 ).

The trace code of RS(2, 4) is the sum 0 code [4, 3, 2]2 , the subfield code is the repetition code [4, 1, 4]2 . As RS(2, 4) is self-dual, Delsarte’s theorem promises that trace code and subfield code are duals of one another. This is indeed the case.

12.2.2. Determine trace code and subfield code of RS(2, 9) (down to F3 ).

As usual describe F9 as F3 ( ), where 2 = +1, 3 = − +1, 4 = −1. Recall from Exercise 12.1.9 that ker(tr) = {0, ± 2 }, tr(x) = 1 for x ∈ {−1, , 3 } and tr(x) = −1 for x ∈ {1, − , − 3 }.

Write the elements of F9 in the order 0, 1, , 2 , 3 , 4 = −1, 5 = − , 6 = − 2 , 7 = − 3 . Then RS(2, 9) has as basis z1 = (1, 1, 1, 1, 1, 1, 1, 1, 1) and z2 = (0, 1, , 2 , 3 , −1, − , − 2 , − 3 ). The F9 -multiples of z1 yield the ternary repetition code as subcode of the subfield code and of the trace code. Also tr(z2 ) = (0, 2, 1, 0, 1, 1, 2, 0, 2) and tr( z2 ) = (0, 1, 0, 1, 1, 2, 0, 2, 2). We claim that tr(RS(2, 9)) has as basis v1 = (1, 1, 1, 1, 1, 1, 1, 1, 1), v2 = tr(z2 ), v3 = tr( z2 ). Clearly v1 , v2 , v3 are linearly independent. Also tr( i z2 ) arises from v2 by a cyclic shift on the 8 last coordinates (this is how v3 is obtained from v2 ).

It is easy to see that all later such shifts are linear combinations of v1 , v2 , v3 .

As in v2 + v3 and in v2 − v3 each entry occurs precisely 3 times we conclude that tr(RS(2, 9)) has minimum distance 6. It is a [9, 3, 6]3 -code. The subfield code is the repetition code.

12.3

Galois invariant codes

Exercises 12.3 12.3.1. Determine the Galois closure of RS(2, 4) (over F2 ).

The standard basis of RS(2, 4) consists of the evaluations at each element of F4 of the monomials 1 and X. As an application of g1 = φ corresponds

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to multiplication by 2 in the exponent, the Galois closure of RS(2, 4) has as basis the evaluations of 1, X and X 2 , in other words RS(2, 4) = RS(3, 4).

12.3.2. Determine the Galois closure of RS(2, 9) (over F3 ).

RS(2, 9) has as basis the monomials 1, X, its Galois closure has as basis the monomials 1, X, X 3 .

12.3.3. The situation is Fq ⊂ F = Fqr . Prove the following general bound: if dimF (C) = k, then the Galois closure C˜ has dimension at most rk.

The Galois group G has r elements. The Galois closure of C is the code generated by the g(C), where g ∈ G. As each of these codes has dimension k, the Galois closure C has dimension ≤ kr.

12.3.4. Let C be Galois invariant over F = Fqr of minimum distance d. Show that each codeword of weight d in C is a scalar multiple of a codeword of tr(C).

Let x ∈ C of weight d. Without restriction x1 , . . . , xd 6= 0 and x1 = 1. Let H ⊂ F be the hyperplane consisting of the elements of trace 0. Consider cx, where c ∈ H. As tr(c) = tr(cx1 ) = 0 it follows from the minimality of d that tr(cx) is the 0-word, hence tr(cxi ) = 0 for all i and all c ∈ H. This shows that xi is orthogonal, with respect to the trace form, to the hyperplane H. As H has Fq -dimension r − 1, its orthogonal has dimension 1, is therefore Fq . We have shown that x ∈ CFq .

12.3.5. Let C be linear over F = Fqr and U ⊂ C a Galois invariant subcode.

Prove dimFq (tr(C)) ≤ r · dim(C) − (r − 1)dim(U) Write C as direct sum of U and a subcode D. We know that dimFq (tr(C)) is the dimension of the Galois closure of C, which equals dim(U) plus the dimension of the Galois closure of D. Apply the bound from Exercise 12.3.3.

12.3.6. Show that codewords of the subfield code which are linearly independent over the subfield Fq are also linearly independent over the large field F = Fq r . (j)

(j)

Let x(j) = (x1 , . . . , xn ) ∈ Fnq , j = 1, 2, . . . , m be elements of the subfield code which are linearly independent over Fq . Assume they are linearly dependent over F, with corresponding coefficients αj ∈ F, j = 1, . . . , m : Pm (j) α x = 0. Choose those coefficients such that for some j we have j=1 j Pm (j) trF |Fq (αj ) 6= 0. Then j=1 αj xi = 0 for all i = 1, . . . , n. Applying the trace Pm (j) shows j=1 trF |Fq (αj )xi = 0 for all i. As the x(j) are linearly independent over Fq it follows that trF |Fq (αj ) = 0 for all j which contradicts our choice.

12.3.7. Let Fq ⊂ F = Fqr . Let D be an Fq -linear code. Define the tensor product D ⊗Fq F by extension of constants: a basis of D is seen as a basis of

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D ⊗Fq F over F. Prove that the mapping D 7→ D ⊗Fq F defines a one-to-one correspondence between the Fq -linear codes and the Galois closed F -linear codes.

Why is D ⊗Fq F Galois closed? Let the code length be n and v1 , . . . , vk a (i) (i) (i) basis of D. Here vi = (v1 , . . . , vn ) and vj ∈ Fq . This is also a basis of D⊗Fq . (i)

(i)

The images under the Frobenius mapping are viq = ((v1 )q , . . . , (vn )q ) = vi .

The remaining claims follow from the preceding theorem. For example, as C = D ⊗Fq F is Galois closed its subfield code has the same dimension over Fq as C has over F. The subfield code contains D. It follows CFq = D.

12.4

Automorphism groups

Exercises 12.4 12.4.1. Find an example of two linear quaternary codes, which are monomially equivalent but have subfield codes and trace codes (down to F2 ) of different dimension.

110 Consider the quaternary code C1 with generator matrix . This is 101 the sum zero code [3, 2, 2]4 . Its trace code coincides with its subfield code, the binary sum zero code [3, 2, 2]2 .

1ω 0 The code C2 with generator matrix is monomially equivalent to 10ω C1 . Its trace code is the whole space [3, 3, 1]2 , its subfield code is the repetition code [3, 1, 3]2 .

12.4.2. Find all cyclic binary codes of length 3 (with respect to the cyclic permutation σ = (1, 2, 3)).

If C contains a codeword of weight 1 it contains all of them and C is the whole space [3, 3, 1]2 . If C contains a word of weight 2 it contains them all, and C is either the sum zero code [3, 2, 2]2 or the whole space. The remaining cyclic codes are the 0-code and the repetition code [3, 1, 3]2 .

12.4.3. Show that the rule σ2 σ1 = σ1 σ2 is not true in general for permutations σ1 , σ2 .

Choose for example σ1 = (1, 2), σ2 = (2, 3). Then σ1 σ2 = (1, 3, 2), σ2 σ1 = (1, 2, 3).

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12.4.4. Can a permutation on n objects have order larger than n?

Yes, this is possible. For example, the permutation (1, 2, 3, 4)(5, 6, 7) on 7 objects has order 4 × 3 = 12.

12.4.5. Determine the automorphism group of the extended binary Hamming code [8, 4, 4]2 , in particular its order and its degree of transitivity.

Let H be the extended binary Hamming code with parameters [8, 4, 4]2 and generator matrix   1000 0111  0100 1011     0010 1101  0001 1110 whose columns we denote by 1, 2, . . . , 8. The automorphism group Aut(H) ⊂ S8 consists of those permutations of the columns which map H to itself: Aut(H) is the stabilizer of H in the symmetric group S8 .

The transitivity of Aut(H) is demonstrated by the elementary-abelian group U = hu1 , u2 , u3 i, where u1 = (1, 5)(2, 6)(3, 7)(4, 8), u2 = (1, 2)(3, 8)(4, 7)(5, 6), u3 = (1, 3)(2, 8)(4, 6)(5, 7).

The double transitivity follows from the element σ = (2, 3, 7, 4, 6, 8, 5) of order 7. The triple transitivity follows from the elements (3, 7, 6)(4, 5, 8) and (46)(57), generating a group which is transitive on {3, 4, 5, 6, 7, 8}.

Each element of the stabilizer of 1, 2, 3 also stabilizes 8 as s8 is the only of the remaining columns which is linearly dependent on the first three. In particular Aut(H) is not 4-transitive.

The four-group generated by (4, 5)(6, 7) and (4, 6)(5, 7) is transitive on {4, 5, 6, 7}. Clearly the pointwise stabilizer of 1, 2, 3, 4 in Aut(H) is the identity group. It follows |Aut(H)| = 8 × 7 × 6 × 4 = 26 3.7 = 8 × 168.

12.4.6. Consider the extended binary Hamming code [8, 4, 4]2 .

Count the unordered 4-sets of coordinates with the property that the corresponding columns of a generator matrix sum to 0.

Prove that they form a Steiner system S(3, 4, 8).

We saw in the preceding exercise that Aut(H) is 3-transitive. In particular any set of 3 columns is linearly independent, and a set of 4 columns is linearly dependent if and only if the corresponding columns of a generator matrix sum to 0. This is independent of the choice of the generator matrix.

Aut(H) has two orbits of 4-subsets with representatives {1, 2, 3, 4} (those columns are linearly independent) and {1, 2, 3, 8} (dependent columns). The

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linearly dependent columns form an orbit, of length 14.

Because of 3-transitivity these subsets form a 3-design 3 − (8, 4, λ). As it has 14 blocks we must have λ = 1. The blocks of this Steiner quadruple system S(3, 4, 8) are the following:   1238 1 2 4 7   1 2 5 6   1 3 4 6   1 3 5 7   1 4 5 8   1 6 7 8   2 3 4 5   2 3 6 7   2 4 6 8   2 5 7 8   3 4 7 8   3 5 6 8 4567

Chapter 13 Cyclic codes

13.1

Some primitive cyclic codes of length 15

Exercises 13.1 13.1.1. Which powers of the primitive element α ∈ F16 are in F4 ?

The elements in F4 \ F2 have order 3. As α has order 15 the powers of α which have order 3 are α5 and α10 . It follows F4 = {0, 1 = α0 , α5 , α10 }.

13.1.2. Find the parameters of the code obtained when A = {1, 3}.

Follow the procedure in the example. Our target is C(A) = tr(B(A))⊥ .

As the Galois closure of A has 8 elements (1, 2, 4, 8, 3, 6, 12, 9) it follows that tr(B(A)) has dimension 8, so C(A) has dimension 15 − 8 = 7. The presence of {1, 2, 3, 4} in à shows that B(A) and tr(B(A)) have strength 4. It follows that C(A) is an [15, 7, 5]2 -code. Here we are sure about the dimension but the minimum distance might conceivably be even larger.

13.2

Theory of cyclic codes

Exercises 13.2 13.2.1. Determine the cyclotomic cosets in case q = 2, n = 31.

The cyclotomic cosets are obtained by multiplication by 2. They are {0}{1, 2, 4, 8, 16}{3, 6, 12, 17, 24}{5, 9, 10, 18, 20}{7, 14, 19, 25, 28} {11, 13, 21, 22, 26}{15, 23, 27, 29, 30}

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13.2.2. Determine the parameters of binary cyclic codes of length 31 in cases A = {1, 3, 5} and A = {−1, 0, 1}.

[31, 16, 7]2 and [31, 20, 6]2 .

13.2.3. Determine the parameters of the binary cyclic code of length 63 with A = {1, 3, 5, 7, 9} The minimum distance is d ≥ 11 as {1, 2, . . . , 10} ⊂ Ã. The cyclotomic cosets C63 (1), C63 (3), C63 (5), C63 (7) have length 6 each, whereas C63 (9) = {9, 18, 36}. It follows that the dimension is 63 − (6 + 6 + 6 + 6 + 3) = 36. We have a [63, 36, 11]2 -code.

13.2.4. Determine the cyclotomic cosets in case q = 3, n = 26.

The cyclotomic cosets in case q = 3, n = 26 are {0}{1, 3, 9}{2, 6, 18}{4, 10, 12}{5, 15, 19} {7, 11, 21}{8, 20, 24}{13}{14, 16, 22}{17, 23, 25} 13.2.5. Show that the binary primitive cyclic code (length 2r −1) with defining set A = {1} is equivalent to the binary Hamming code.

The binary cyclic code of length 2r − 1 with {1} as defining set is a [2r − 1, 2r − 1 − r, 3]2 -code, just as the binary Hamming code. It is easy to see that codes with these parameters are all equivalent.

Observe that in the binary case there is no difference between monomial equivalence and permutation equivalence. A check matrix of a code with the above parameters is an (r, 2r − 1)-matrix whose columns are nonzero and pairwise different. There are precisely 2r − 1 such columns.

13.2.6. Find a cyclic code C with parameters [26, 8, 13]3 , containing two cyclic codes D1 , D2 of parameters [26, 7, 14]3 such that D1 ∩D2 is a cyclic [26, 6, 15]3 code.

Use the interval A = {1, . . . , 12} for C. Then à avoids {0}, {13}, {14, 16, 22} and {17, 23, 25}, so C has dimension 8. The defining exponents for C1 , C2 are A ∪ {0} and A ∪ {13}.

13.2.7. Prove: if r is odd, then (3r − 1)/2 is odd.

Let r be odd. Then 3r is 3 (mod 4) and 3r − 1 is 2 (mod 4). Consequently (3 − 1)/2 is odd. r

13.2.8. Construct a generator matrix of a cyclic [13, 7, 5]3 -code.

We construct a generator matrix of a cyclic code [13, 7, 5]3 . The underlying field is F27 . We construct it as F = F3 ( ), where the primitive element satisfies 3 = 2 − 1. Observe tr(a2 2 + a1 + a0 ) = a2 + a1 , where the ai ∈ F3 .

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We know that the defining set can be chosen as A = {+1, −1}. The cyclotomic cosets are {0}{1, 3, 9}{2, 5, 6}{4, 10, 12}{7, 8, 11} The set B, the negative of the complement of the Galois closure of A, is B = C(0) ∪ C(2) ∪ C(−2).

The coordinates of the code are indexed by the nonzero squares of F, the powers of α = 2 . A basis of F over F3 is 1, , 2 . It follows that the monomials indexing the rows of the generator matrix can be chosen as , X 2 , X 2 , 2 X 2 , X −2 , X −2 , 2 X −2 .

The entries are the traces of the evaluations at the αj , j = 0, . . . , 12. The following generator matrix is obtained: j : 0 1 2 3 4 5 6 7 8 9 10 11 12 →1111111111 1 1 1 X2 → 0 0 2 0 2 2 2 1 1 0 2 1 2 X 2 → 1 2 1 0 0 1 0 1 1 1 2 2 0 2 2 X →1102120020 2 2 2 X −2 → 0 2 1 2 0 1 1 2 2 2 0 2 0 X −2 → 1 0 2 2 1 1 1 0 1 0 0 1 2 2 −2 X →1222020021 2 0 1 13.2.9. Let W = hαi the subgroup of order n of F ∗ and C(i) a cyclotomic coset mod n. Prove that the length |C(i)| = s is determined by Fq (αi ) = Fqs , the degree of the field generated by αi over Fq .

The length s of Cn (i) is the smallest positive number s such that iq s ≡ s s i (mod n). This is equivalent to αiq = αi . The elements fixed under x 7→ xq are precisely the elements of Fqs .

13.2.10. Prove the Roos bound for cyclic codes in its general form: Let I be an interval and J contained in an interval J0 such that |J0 | − |J| < |I| and I + J ⊆ Ã. Then d(C(A)) ≥ |I| + |J|.

What we proved is the special case when J = J0 is an interval. Following the proof of this theorem we see d > |I| as I is an interval. Assume now d ≤ |J0 |.

As J0 is an interval we have dim(Bs (J0 )) = d and therefore dim(BS (J) > (d − |I|). This contradicts our theorem for the special case of two intervals.

We conclude d > |J0 |. It follows dim(Bs (J0 )) = |J0 | and dim(Bs (J)) = |J|.

This proves the theorem.

13.2.11. Determine the binary cyclic codes of length 9 and their parameters.

The cyclotomic cosets are C(0) = {0}, C(1) = {1, 2, 4, 5, 7, 8}, C(3) = {3, 6}.

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There are therefore 8 cyclic codes four of which are boring (the 0-code, the ambient space F92 , the repetition code, the sum zero code). Here are the remaining four cyclic codes: C({1}) is a [9, 3, 3]2 -code. In fact we know d ≥ 3 because of the interval {1, 2} and d ≤ 4 as we saw in Chapter 10. The description of a generator matrix shows that all codewords have weights multiples of 3. It follows d = 3.

C({3}) is a [9, 7, 2]2 -code, C({0, 3}) is a [9, 6, 2]2 -code. In both cases d = 3 is impossible because of the sphere packing bound. Finally C({0, 1}) has parameters [9, 2, 6]2 as shown by the interval {7, 8, 0, 1, 2} and the optimality of the parameters.

13.2.12. Show that the maximum minimum distance of a [17, 9, d]2 -code is d = 5.

We saw a cyclic [17, 9, 5]2 -code. Assume a [17, 9, 6]2 -code exists. The residual code has parameters [11, 8, 3]2 which contradicts the sphere-packing bound.

13.2.13. Determine the binary cyclic codes of length 17 and their parameters.

The cyclotomic cosets are C(0) = {0}, C(1) = {1, 2, 4, 8, 9, 13, 15, 16}, C(3) = {3, 5, 6, 7, 10, 11, 12, 14}.

Aside of the boring cases we have C({1}) with optimal parameters [17, 9, 5]2 (see the preceding exercise), likewise C({3}). Finally C({0, 1}) has parameters [17, 8, 6]2 as shown by the interval {15, 16, 0, 1, 2}. Observe that these are optimal parameters as shown by the Griesmer bound (likewise for C({0, 3})).

13.2.14. Show that all three parameters of [13, 7, 5]3 are optimal.

It suffices to show that [12, 7, 5]3 does not exist. This follows from the sphere-packing bound.

13.2.15. Show that the [q 2 − q + 1, q 2 − q − 6, 6]q -codes have optimal d for q ≥ 5.

Let n = q 2 − q + 1 where q ≥ 5 and assume an [n, n − 7, 7]q exists. The sphere-packing bound shows q 7 ≥ Vq (3, n) > (q − 1)3 n(n − 1)(n − 2)/6. This is not satisfied.

Cyclic codes

13.3

Decoding BCH-codes

13.4

Constacyclic codes

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Exercises 13.4 13.4.1. Let K be an [n, k, d]q -code with the property that for every (a0 , a1 , . . . , an−1 ) ∈ K we also have (a1 , a2 , . . . , an−1 , γa0 ) ∈ C, where γ ∈ Fq such that γ u = 1.

Prove the following: there is a cyclic [un, k, ud]q -code C which possesses K as a projection (contraction). Here the term cyclic is to be interpreted as follows: there is a cyclic permutation on the coordinates, which is an automorphism of C. The natural isomorphism from K to C is given by mapping x = (x0 , x1 , . . . , xn−1 ) to (x|γx| . . . |γ u−1 x) ∈ Fun q .

13.4.2. Let K be a constacyclic code constructed according to Theorem 14.28 from a cyclic code with set of zeroes B. Prove the following: K⊥ is the contraction of the cyclic code with set of zeroes B 0 , where B 0 is the complement of −B in the set of numbers having modulus (mod u) the negative of the elements of B. We have N = un, so there are n numbers mod N with a given modulus mod u. As K has length n and dimension |B|, its dual must have dimension n − |B|. It follows that the dimension is right.

It remains to prove that the codes are orthogonal. This reduces to orthogonality of the P F -linear parent codes in the usual way. Let i ∈ B, i0 ∈ B 0 . We 0 n−1 have to show j=0 α(i+i )j = 0. Here α is a primitive N -th root of unity. As 0 i + i0 is a multiple of u bot not a multiple of N, αi+i is an n-th root of unity but 6= 1. The statement follows from the often used fact that the sum of all powers of root of unity 6= 1 vanishes.

13.4.3. When is a constacyclic code self-dual (equal to its dual)?

The criterion which decides when a constacyclic code is self-dual follows from the preceding exercise. Let the elements of the set of zeroes B of the underlying cyclic code be ≡ i mod u. The first condition is that 2i is a multiple of u. The second condition is that the set of all elements with modulus i (mod u) in Z/N Z is the disjoint union of B and −B. 13.4.4. Prove that the constacyclic [28, 14, 9]3 -code constructed in the present section is self-dual.

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The criterion derived in the preceding exercise applies to the [28, 14, 9]3 code. In fact, the set of zeroes of the underlying cyclic code is B = C(1) ∪ C(5) ∪ C(7), and B ∪ (−B) is the set of all odds mod 56.

13.4.5. Let F = Fp2f and W the subgroup of order pf + 1 in F ∗ . Show that W is not contained in a proper subspace of F when F is considered as a 2f -dimensional vector space over Fp .

Assume at first W is contained in L = Fpi ⊂ F. We have that i divides 2f.

As W ⊆ L∗ , we have (pf + 1) | 2i − 1. This implies i > f. It follows i = 2f.

Let now V be the smallest Fp -subspace of F containing W and assume V 6= F. We claim that V must be a subfield, contradicting the first part of the proof. It suffices to prove P that V is closed under multiplication. By definition V is the set of all w∈W λw w, where λw ∈ Fp . As W is a multiplicative subgroup it is clear that the product of two such expressions has this form again.

13.4.6. Prove the following converse of Theorem 14.28: Let C be a cyclic q-ary code of length N = nu and γ ∈ Fq such that γ u = 1, let K be a code of length n and dim(K) = dim(C) such that C = {(v, γv, . . . , γ u−1 v) | v ∈ K}.

Then K is constacyclic.

Let v = (v1 , v2 , . . . , vn ) ∈ K, hence (v, λv, . . . , λu−1 v) ∈ C. Let v 0 = (v2 , . . . , vn , λv1 ). As C is cyclic, we have (v 0 , λv 0 , . . . , λu−1 v 0 ) ∈ C (here we use that C is cyclic in the sense that cyclic shifts of codewords are codewords, and also the fact that λu = 1). It follows v 0 ∈ K, which therefore is constacyclic.

13.4.7. Prove that the check matrix H of Rf is indeed 4-independent.

It has to be shown that the check matrix H is 4-independent.

2-independence is obvious. Assume (w1 , 1/w12 ) = a2 (w2 , 1/w22 ) + a3 (w3 , 1/w32 ) with nonzero coefficients a2 , a3 ∈ F4 . Use the first coordinate, to express w1 in terms of w2 , w3 , substitute in the second coordinate. As w12 = a22 w22 + a23 w32 it follows a3 a2 1 = (a22 w22 + a23 w32 )( 2 + 2 ) = a22 a3 w + a2 a23 /w, w2 w3 where w = (w2 /w3 )2 . There is therefore c ∈ F4 such that 1 = cw + c2 /w, equivalently (cw)2 + cw + 1 = 0. As cw satisfies a quadratic equation with coefficients in F2 it follows aw ∈ F4 , hence w ∈ F4 . This is impossible as w2 , w3 had been chosen in different cosets mod F∗4 .

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Assume now a1 (w1 , 1/w12 ) + a2 (w2 , 1/w22 ) = a3 (w3 , 1/w32 ) + a4 (w4 , 1/w42 ) Square the equation from the first coordinate. This yields the system a1 /w12 + a2 /w22 = a3 /w32 + a4 /w42 a1 w12 + a2 w22 = a3 w32 + a4 w42 Multiplication yields cw + c2 /w = dw0 + d2 /w0 , where w = (w1 /w2 )2 , w0 = (w3 /w4 )2 . Let x = cw, y = dw0 . We have equation x+1/x = y +1/y, equivalently (x+y)(1+xy) = 0. It follows that either x = y or xy = 1, equivalently w1 w4 ∼ w2 w3 or w1 w3 ∼ w2 w4 , where ∼ indicates that the elements are in the same coset mod F∗4 . It is clear that in either of these cases it is impossible that also w1 w2 ∼ w3 w4 , as it would follow that two of the wi are in the same coset. However, we started from an arbitrary subdivision of the wi in two pairs and we saw that only one of the three equations can hold. Choose notation such that w1 w2 ∼ w3 w4 is true, and the other two therefore not true. None of the conditions x = y or xy = 1 can be satisfied, contradiction.

13.4.8. Prove that there is no cyclic [10, 4, 6]3 -code.

The cyclotomic cosets are {0}, {5} and C(1) = {1, 3, 7, 9}, {2, 4, 6, 8} A 4-dimensional code is obtained in two cases: A is the union of the cosets of length 1 and either C(1) or C(2). The set B of zeroes is either C(1) or C(2).

The 4-dimensional codes are trace codes (=subfield codes) of C(1) and of C(2).

As the elements of C(1) are odd and the elements of C(2) are even we would obtain a contraction with impossible parameters [5, 4, 3]3 . As the minimum distance of those 4-dimensional cyclic codes is even (because of contraction) and > 2 because of the BCH-bound, those codes have parameters [10, 4, 4]3 and their contractions are [5, 4, 2]3 -codes, not very interesting.

13.4.9. Does a [10, 4, 6]3 -code exist?

Yes, of course. This follows from the ternary Golay code [12, 6, 6]3 .

Chapter 14 Recursive constructions, covering radius

14.1

Construction X

Exercises 14.1 14.1.1. Show that the parity check theorem is a special case of construction X. Let C be an [n, k, d]2 -code, where d is odd. The intersection D of C with the sum-zero code [n, n − 1, 2]2 is an [n, k − 1, d + 1]2 -code. Construction X with [1, 1, 1]2 as auxiliary code produces [n + 1, k, d + 1]2 .

14.1.2. Use construction X to obtain a [12, 4, 6]2 -code.

The extended binary Hamming code [8, 4, 4]2 contains the all-1-word. Apply construction X to the chain [8, 4, 4]2 ⊃ [8, 1, 8]2 , with auxiliary code [4, 3, 2]2 .

14.1.3. Prove the following rule for cyclic codes: C(A1 )∩C(A2 ) = C(A1 ∪A2 ).

This follows from the definition of the cyclic codes via duality. C(A1 )∩C(A2 ) consists of all q-ary n-tuples which are orthogonal to B(A1 ) and to B(A2 ), equivalently to B(A1 ∪ A2 ).

14.1.4. Show that [8, 4, 4]4 exists.

It is easy to see that the generator matrix   10000111 0 1 0 0 1 0 1 1  G = (I|P ) =  0 0 1 0 1 1 0 1 00011110 of [8, 4, 4]2 also generates [8, 4, 4]4 when the entries are interpreted as elements of F4 . In fact, each row has weight 4, each linear combination of two rows of P has weight ≥ 2 (because of the zeroes), and no three rows of P are linearly dependent.

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14.1.5. Construct a [73, 8, 48]4 −code using construction X. Apply construction X with [8, 4, 4]4 as auxiliary code to the chain [65, 8, 44]4 ⊃ [65, 4, 48]4 .

14.1.6. Show that all codewords of C2 \ C3 (see Proposition 15.3) have weight q 2 − q.

The chain C2 ⊃ C3 has parameters [q 2 , 3, q 2 − q]q ⊃ [q 2 , 1, q 2 ]q and C3 is in fact the repetition code. Let x ∈ C2 \ C3 and δi the frequency of i ∈ Fq as P entry of x. As wt(x − 1) = q 2 − δi it follows δi ≤ q. As i δi = q 2 it follows δi = q for all i. In particular wt(x) = q 2 − δ0 = q 2 − q.

14.1.7. Show that the nonzero weights of C1 from Proposition 15.3 are q 2 − q − 1, q 2 − q, q 2 − 1, q 2 .

The repetition code shows that weight q 2 occurs. It has been observed that because of the Griesmer bound weight q 2 − q − 1 must occur. Subcode C2 shows that q 2 − q is a weight. Only weight q 2 − 1 is in doubt.

Choose x of weight q 2 −q −1 and δi as in the previous exercise. In particular δ0 = q + 1. Assume q 2 − 1 is not a weight. Then δi ∈ {0, q, q + 1} for all i.

Let α be the number of elements i such that δi = q + 1 and β the number of i such that δi = q. Then α(q + 1) + βq = q 2 . Calculating mod q the first equation implies that α is a multiple of q. As α 6= 0 it follows α ≥ q, which is impossible.

14.1.8. Show that all three parameters of [28, 8, 15]3 are optimal.

We gave several constructions for [28, 8, 15]3 -codes. In order to prove optimality it suffices to show that [27, 8, 15]3 does not exist. Assume it does.

The residual code has then parameters [12, 7, 5]3 which is impossible because of the sphere-packing bound.

14.1.9. Combine construction X with the Varshamov bound to prove the following construction: Assume Vq (d−2, n−1) < q n−k and the existence of an [n−1, k−1, d+δ]q -code.

Then an [n + δ, k, d + δ]q -code exists.

The Varshamov bound shows that [n, k, d]q exists. In fact, start from an [n−1, k−1, d]q -code and its check matrix H. Then H is an (n−k, n−1)-matrix which is (d − 1)-independent (any set of d − 1 of its columns is independent).

The assumption shows that we can find an additional column such that the resulting (n − k, n)-matrix is (d − 1)-independent. This is therefore the check matrix of an [n, k, d]q -code which contains the [n − 1, k − 1, d]q -code as a shortened code. Now assume the existence of an [n − 1, k − 1, d + δ]q -code. We just saw that it can be embedded in an [n, k, d]q -code. Now apply construction X to this chain of codes, using the repetition code [δ, 1, δ]q as auxiliary code.

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14.1.10. A [126, 36, 34]2 -code is known to exist. Use the preceding exercise and the fact that V2 (26, 126) < 290 to show the existence of an [133, 37, 34]2 code.

This is the case n = 127, d = 28, k = 37, δ = 6 of the preceding exercise.

14.2

Covering radius

Exercises 14.2 14.2.1. Formulate and prove a sphere-packing bound for covering codes.

Let C be a q-ary code of length n with covering radius ρ. The balls of radius ρ centered at the codewords must cover the ambient space. It follows |C| ≥

qn .

Vq (ρ, n)

14.2.2. Characterize the case of equality in the sphere-packing bound for covering codes.

Equality happens if the ambient space is the disjoint union of the balls of radius ρ. The covering code C must be perfect of minimum distance d = 2e+1, and ρ = e.

14.2.3. Prove the asymptotic sphere-packing bound for covering functions: Let α(x) be the lim sup of the information rate n/N for families of covering functions COV (ρ, N, n)q whose change rate is ρ/N ≤ x. Then α(x) ≤ Hq (x).

Here Hq (x) = xlogq (q − 1) + logq (2)h(x) is the q-ary entropy function. It was introduced in Exercise 9.8.

The sphere-packing bound for covering codes implies the following bound on covering functions COV (ρ, N, n)q : ρ X N (q − 1)i ≥ q n . i i=0

Take base q logarithms, divide by N, take the limit for N −→ ∞. The binomial term yields h(x)logq (2), the (q − 1)-power contributes xlogq (q − 1), the right side the information rate. It follows α(x) ≤ h(x)logq (2)+xlogq (q−1) = Hq (x).

14.2.4. Show that the existence of COV (ρ, N, n)q implies the existence of COV (ρ, N 0 , n)q for all N 0 ≥ N.

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Let f be a COV (ρ, N, n)q . We have to show the existence of COV (ρ, N + 1, n)q . Define g : AN +1 −→ An as g(x1 , . . . , xN +1 ) = f (x1 , . . . , xN ).

14.2.5. Show that the existence of COV (ρ, N, n)q implies the existence of COV (cρ, cN, cn)q for every natural number c.

Define f : AcN −→ Acn componentwise.

14.2.6. Determine the covering radius of the repetition code [n, 1, n]q .

Let C be the repetition code [n, 1, n]q . Let x ∈ Fnq and λi the number of entries i in x (i ∈ Fq ). We have that d(x, C) is n − M ax{λi }. The minimum value of M ax{λi } is dn/qe. The covering radius is ρ = n − dn/qe.

14.2.7. Determine the covering radius of the [8, 4, 4]2 -code.

The [8, 4, 4]2 is a lengthening of the (perfect) Hamming code [7, 4, 3]2 , which has ρ = 1. It follows that [8, 4, 4]2 has ρ ≤ 2, clearly = 2.

Alternatively consider the list of coset leaders in Section 4.7 on syndrome decoding. The maximum weight of a coset leader is 2. This is therefore the covering radius.

Chapter 15 The linear programming method

Exercises 15 15.1. Explain why in the standard form of the LP-problem it is usually assumed that m ≤ n.

If M = 6 ∅ then each x ∈ M is a solution of a system of m linear equations for n variables. It can be assumed that those equations are linearly independent, equivalently rk(A) = m. It follows m ≤ n.

15.2. Consider the LP-problem in standard form where 2 1 1/2 m = 2, n = 3, A = and −5 −1 0 b = (2, 1), p = (1, 0, 5).

Determine the region M of admissible points and solve the problem.

The second equation is equivalent to x2 = −(1 + 5x1 ). As x1 ≥ 0 it follows x2 < 0, so M = ∅.

15.3. Draw the convex hull of points P1 = (2, 0), P2 = (2, 1), P3 = (2, 2), P4 = (3, 1), P5 = (0, −2).

Write (1.5, 0) as a convex combination of P1 , . . . , P5 .

(1.5, 0) = (1/4)P3 + (1/3)P4 + (5/12)P5 = (1/2)P1 + (1/4)P3 + (1/4)P5 .

The solution −(3/4)P3 + P4 − (1/4)P5 is not correct: this is not a convex combination.

15.4. Determine the vertices of the first toy example.

The columns of A are c1 = (2, −5), c2 = (0, 1), c3 = (1/2, 0), and the right side b = (3, 5/3). It follows M = {(x1 , 5/3 + 5x1 , 6 − 4x1 )|0 ≤ x1 ≤ 1.5} The vertices are obtained when x1 = 0 and x1 = 3/2. They are (0, 5/3, 6) and (3/2, 55/6, 0).

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15.5. Let

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

 01 2 20 0 3  1 1 5 1 1 1 −5   A=  0 1 7 1 0 1 −2  1 0 −2 1 0 −3 −1

Then I = {4, 5, 6, 7} describes a basis. Express columns c1 , c2 , c3 in terms of c4 , c5 , c6 , c7 .  3/26 8/13 71/26 1 0 0 0  10/13 −3/13 −71/13 0 1 0 0   A∼  −7/26 3/13 51/26 0 0 1 0  −1/13 −1/13 −15/13 0 0 0 1 

15.6. Use Gauß elimination to determine the rank of   0 2 1 4 0 −3  −2 3 5 6 −7 2   A=  3 −1 −4 0 8 −2  1 0 0 2 1 3   100 2 1 3  0 1 0 10/7 5/7 −23/7   A∼  0 0 1 8/7 −10/7 25/7  000 0 0 0 The rank is 3.

15.7. Consider the linear program x1 + x2 − 3x3 ≤ 5 x1 − 2x2 + 2x3 ≥ 6 x1 , x2 , x3 ≥0 with objective function Q(x) = x1 + x2 + x3 → max.

Write it in standard form and determine the dual LP. In standard form there are 5 non-negative variables x1 , x2 , x3 , x4 , x5 , with conditions x1 + x2 − 3x3 + x4 = 5, x1 − 2x2 + 2x3 − x5 = 6.

It follows

A=

1 1 −1 1 0 1 −2 2 0 −1

, b = (5, 6)t , p = (−1, −1, −1, 0, 0)t .

The dual has variables w1 , w2 (no sign restriction) such that At w ≤ (−1, −1, −1, 0, 0).

Concretely: w1 ≤ 0, w2 ≥ 0 and w1 + w2 ≤ −1, w1 − 2w2 ≤ −1, − 3w1 + 2w2 ≤ −1.

The objective function is 5w1 + 6w2 and has to be maximized.

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15.8. Show how n unconstrained variables can be replaced by n+1 constrained variables.

Let xi , i = 1, . . . , n be the unconstrained variables. Write them as xi = P x − x for x , x ≥ 0.

Let y = x ≥ 0. Then xi = y − (xi,2 + i,1 i,2 i,1 i,2 i,1 i P x ).

Let the new constrained variables be i,1 j6=i y = y0 , yi = xi,2 +

X

xi,1 .

j6=i

Then yi ≥ 0 and the yi determine the xi as xi = y − yi , i 6= 0.

Maybe easier the other way around: let y0 , . . . , yn be variables yi ≥ 0. As unconstrained variables use xi = yi − y0 for i = 1, . . . , n.

15.9. Consider the LP with unknowns x1 , x2 , x3 and conditions. x1 , x2 ≥ 0, 2x1 − 3x3 ≥ 3, x2 + 5x3 ≤ 6, x1 + x2 + x3 = 1 where x1 − x3 has to be minimized.

Determine the dual problem.

By the general mechanism the dual problem has unknowns w1 , . . . , w5 where w1 , w2 , w3 ≥ 0, w4 ≤ 0. The matrix describing the dual problem is   10 2 01 At =  0 1 0 1 1  0 0 −3 5 1 and the main condition is At w ≤ p = (1, 0, −1). Maximize 3w3 + 6w4 + w5 .

15.10. Find a vertex of the polyhedron Ax = b, x ≥ 0 where     101 2 2 A =  0 1 1 −1  and b =  1  001 3 1 (1, 0, 1, 0) and (4/3, 4/3, 0, 1/3) are vertices.

15.11. Let



   1 0 0 0 2 −2 1 0 1 0 0 4 2  2    A= 0 0 1 0 3 0 b = 3 00011 7 4

Find the vertex X with basis {1, 2, 3, 4} and use it to find a vertex with basis I 0 ⊂ {1, 2, 3, 4, 6}, I 0 6= I. X = (1, 2, 3, 4), X 0 = (15/7, 6/7, 3, 0, 0, 4/7).

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15.12. Find the minimum of x1 + 2x2 subject to 2x1 − 5x2 ≥ −40 5x1 + 8x2 ≤ 720 −5x1 + 3x2 ≥ −830 8x1 + 10x2 ≥ 440 At (130, −60) the minimum value is 10.

15.13. Apply the Simplex algorithm to Q(x) = −x1 − x2 − x3 − x4 → min subject to x1 , x2 , x3 , x4 ≥ 0 −x1 + 3x2 + 7x3 + 11x4 ≤ 13 6x1 + 2x2 − 18x3 − 54x4 ≤ 78 x3 + 4x4 ≤ 4 The final vertex is (24, 3, 4, 0, 0, 0, 0), of value −31.

15.14. x five story and y two story buildings are to be constructed on a swampy plot, where the need for adequate foundations greatly increases the cost of a taller building. The work produced by one person in one month will be denoted a ”person-month”. The remaining information is in the following table: number of stories

costs in $

personmonths

area in yd2

5 2 available:

600, 000 200, 000 18, 000, 000

120 60 4, 500

800 600 42, 000

number of number of occupants buildings per building 30 x 12 y

How should x and y be chosen if we want to maximize the number of people who can live on this plot of land? x = 15, y = 45 for a total of 990 people.

Chapter 16 OA in statistics and computer science

16.1

OA and independent random variables

16.2

Linear shift register sequences

Exercises 16.2 16.2.1. Let f (X) be irreducible of degree r ≥ 2 over Fq . When is it true that every LSRS with f (X) as characteristic polynomial has maximum length?

The length of the LSRS is the order of an element in the multiplicative group of the field. This group is cyclic of order q r − 1. The question is when all elements 6= 1 of this group have order exactly q r − 1. This is equivalent with q r − 1 being a prime. For q > 2 this is clearly impossible. The primes of the form 2r − 1 are known as Mersenne primes. A necessary condition is that r itself must be prime. The answer is: q = 2, 2r − 1 prime.

16.2.2. Let F = Fqr and tr : F −→ Fq the trace. Show that for every Fq linear mapping f : F −→ Fq there is exactly one u ∈ F such that f (x) = tr(xu) for all u.

Each function gu defined by gu (x) = tr(xu) is a linear mapping. As tr is not the zero mapping each gu for u 6= 0 is a nonzero mapping. As gu − gv = gu−v the gu are pairwise different. As there are precisely q r linear mappings : F −→ Fq these are precisely the gu .

16.2.3. Show that the linear complexity of the termwise sum of two q-ary LSRS is at most the sum of the complexities.

This is trivial. If (an ) has period r1 and (bn ) has period r2 , then an+r1 r2 = an and bn+r1 r2 = bn .

16.2.4. Construct a maximal length binary LSRS of depth (linear complexity) 4

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We used the primitive polynomial f (X) = X 4 + X + 1 to describe F16 as an extension field of F2 . Interpreted as characteristic polynomial we see c1 = c4 = 1. The next coefficient of the LSRS is the sum of the first and the last register. Start with 1000 as initial values: 100011110101100|1000 As soon as the initial segment repeats we realize that the period has come to an end. Its length is 15 as it has to be.

16.2.5. Let x = (xi ), y = (yi ) ∈ Fn2 Prove that the correlation n1 is given by n1 (n − 2wt(x + y)).

P

i (−1)

xi +yi

Let w1 = wt(x), w2 = wt(y). Let a be the number of coordinates where xi = 1, yi = 0 and b the number of coordinates where yi = 1, xi = 0, so that a+b = wt(x+y). The number of coordinates where 1 meets 1 is w1 −a = w2 −b and the number of coordinates where 0 meets 0 is n − w1 − b. n times the correlation is therefore (w1 − a) + (n − w1 − b) − (a + b) = n − 2(a + b).

16.2.6. Use the Roos bound to prove that the binary cyclic code C({1, −1}) of length n = 2r − 1 has minimum distance 5 when r is odd.

As r is odd we have that 3 is coprime to n. It follows that both {0, 3, 6} and {−4, +4} are intervals. Their sum is contained in the closure of A = {+1, −1}.

16.2.7. What are the smallest code parameters of the family of binary codes constructed in the previous exercise?

After adding a parity check bit we obtain [32, 21, 6]2 (case r = 5) and [128, 113, 6]2 . The former code is good but we know a [33, 22, 6]2 .

16.2.8. The parameters [128, 113, 6]2 could be obtained from a [129, 114, 6]2 code, by shortening. Does such a code exist?

Yes. This is the special case m = 7 of the family of [2m + 1, 2m − 2m, 6]2 codes constructed in the chapter on cyclic codes.

OA in statistics and computer science

16.3

Cryptography

16.4

Two-point based sampling

16.5

Resilient functions

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Exercises 16.5 16.5.1. Let C be an [n, m, t + 1]q -code. Show how to obtain an RF (n, m, t)q .

Let G be a generator matrix of the [n, m, t+1]q -code C. Define f : Fnq → Fm q as f (x) = xGt (see x as a row vector). Then Ker(f ) is C ⊥ , an OA of strength t, and each f −1 (y) is a coset of C ⊥ , hence an OA of the same parameters.

16.5.2. Show that a systematic OA2k−t (t, n, 2) can be embedded in a partition of Fn2 into OA2k−t (t, n, 2).

Assume without restriction that the first k parameters are the systematic parameters. Write each row of the OA A in the form (a|b), where a ∈ Fk2 .

Observe that for each a there is exactly one row (a|b) of the OA, so we can write b = α(a) for some α : Fk2 −→ Fn−k . Define the partition of Fn2 as the 2 cosets with representatives (0|c), where c varies over Fn−k . The typical row 2 of A + (0|c1 ) is (a|α(a) + c1 ). It follows that A + (0|c1 ) and A + (0|c2 ) have no row in common when c1 6= c2 , and we obtain the desired partition.

16.5.3. Give a definition of RF (n, m, t)q for arbitrary q and prove m + t ≤ n.

Why do we call this the Singleton bound?

A resilient function RF (n, m, t)q is a balanced mapping f : X = An → Y = Am (here A is an alphabet, a set of size q) with the property that whenever y ∈ Y and some t input bits of f are prescribed there are exactly q n−m−t elements x ∈ X which have those prescribed input values and satisfy f (x) = y.

An RF (n, m, t)q exists if and only if An can be partitioned into q m parts each of which is an OAqn−m−t (t, n, q). As the index λ is an integer we must have n − (m + t) ≥ 0. In the linear case the OA is the dual of an [n, m, t + 1]q code, and the Singleton bound yields m + t ≤ n.

16.5.4. Show that there is no linear [16, 8, 6]2 -code

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The residual code of an [15, 8, 5]2 -code is a [10, 7, 3]2 -code. A check matrix of such a code has 10 different nonzero columns of length 3, which is impossible.

In fact, it has been shown in Exercise 3.4.7 that 11 is the smallest n such that [n, 7, 3]2 exists, using the same argument.

16.5.5. Determine the automorphism group of the octacode As the binary image of the octacode N is the extended Hamming code H, each automorphism of N yields an element of Aut(H). This group has been determined in Exercise 12.4.5. This defines a homomorphism : Aut(N ) −→ Aut(H). It is clear that the kernel of this mapping is −1, the automorphism of N mapping everything to its negative.

An important observation is that the homomorphism : Aut(N ) −→ Aut(H) is onto. Each automorphism of the extended Hamming code lifts to an automorphism of the octacode. We give a list of elements of Aut(N ) whose images in Aut(H) comprise a set of generators of Aut(H). The notation is such that the permutation has to be applied first, followed by mapping the entries in the coordinates of the m-list to their negatives: (2, 3, 4)(6, 7, 8) maps z1 7→ z1 and (z2 , z3 , z4 ), (1, 2)(5, 6)m(3, 8) maps z1 ↔ z2 , z3 ↔ −z3 and fixes z4 .

In particular the group hα, βi is 2-transitive on the subspaces Z4 zi . u1 = (1, 5)(2, 6)(3, 7)(4, 8)m1 m5 , u2 = (1, 2)(3, 8)(4, 7)(5, 6)m1 m2 , u3 = (1, 3)(2, 8)(4, 6)(5, 7)m1 m3 .

The group U = hu1 , u2 , u3 i is an elementary abelian normal subgroup of Aut(N ). σ = (2, 3, 7, 4, 6, 8, 5)m(1, 2, 3, 8) and α generate a Frobenius group of order 21 stabilizing coordinate 1. We have σ α = σ 2 .

We have (3, 7, 6)(4, 5, 8)m(1, 2, 5, 6) and the four-group generated by (4, 5)(6, 7)m(1, 4, 5, 8) and (4, 6)(5, 7)m(2, 4, 6, 8).

In particular |Aut(N )| = 2 × |Aut(H)| = 2688 and Aut(N )/h−1i ∼ = Aut(H).

16.5.6. Determine the weight distribution of the Nordstrom-Robinson code We saw that the weights of N R are even and ≥ 4. Denote the rows of the generator matrix of the octacode by z1 , . . . , z4 . We can work with the codewords of the octacode if we define the weight of a codeword as the sum of the weights of its entries, where wt(0) = 1, wt(±1) = 1 and wt(2) = 2.

The fact that the automorphism group of N is 3-transitive on the coordinates comes in helpful.

Assume there is a word x of weight 4. Then x is a word of weight 4 in the Hamming code and x is a word which does not have 2 as an entry. The unit matrix shows that x is a linear combination of the zi with 0, 1, 3 as coefficients.

It does not have the form ±zi as we have chosen wisely to use 2 as an entry in each zi . The structure of the submatrix P yields a contradiction.

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e = 2(z1 + z2 + z3 + z4 ) = (2, 2, 2, 2|2, 2, 2, 2). We have wt(e) = 16, so A16 = A0 = 1. Let x have a entries ±1, b entries 2 and c entries 0. Then x + e has a entries ±1, b entries 0 and c entries 2. As wt(x) + wt(x + e) = (a + 2b) + (a + 2c) = 2(a + b + c) = 16 it follows that the vectors in N R come in complementary pairs. In particular A16−i = Ai for all i. The only weights that can occur are 0, 6, 8, 10, 16 and we have A10 = A6 . It suffices to determine A8 in order to find the weight distribution. The automorphism group simplifies the determination.

The words of weight 8 are • 2zi (4 words) • 2zi + 2zj (6 words) • 2zi + 2zj + 2zk (4 words) • ±z1 ± z2 ± z3 ± z4 (16 words) We count A8 = 30. This yields the weight distribution: A0 = A16 = 1, A6 = A10 = 112, A8 = 30.

Here is another approach to the Nordstrom-Robinson code: Consider the doubling operation: x 7→ 2x. It transforms the generator matrix of N into the standard generator matrix of the extended binary Hamming code, with 1 replaced by 2 and therefore all weights doubled. Denote this code by H (it has one word of weight 16 and 14 words of weight 8 each). We have H ⊂ N and H is also the kernel of the doubling operation (N /H ∼ = H, where the homomorphism is given by doubling). We view N as union of the 16 cosets of H. One of these cosets is H itself. Another one has the word of weight 16 as image. This coset consists of words in N all of whose entries are ±1. It must therefore consist of all combinations of the rows of G with ±1 as coefficients, and each element of this coset has weight 16.

Each of the remaining cosets has as image a word of weight 8 in H. All those elements have 4 entries ±1. In order to see that 6 is the minimum weight it has to be shown that N contains no element with 4 entries ±1 and 4 entries 0.

P4 Let x be such a word, x = i=1 ai zi . We have ai 6= 2 as otherwise 2 would be an entry of x. Let w be the weight of the tuple (a1 , a2 , a3 , a4 ). It is impossible that w = 1 as each row of G contains an entry 2. Assume w = 2.

Two of the entries of x in the back section must be 0. This must happen in the coordinates where none of the participating rows of G has entry 2. However we took care that our rows differ in one of those entries and are equal in the other. This shows that x cannot vanish in both those coordinates. Assume

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w = 3. The dot product of x with the fourth row of G (the one not involved in the linear combination) is not 0, contradiction. Finally, if w = 4 then a linear combination of the rows of P with coefficients ±1 vanishes. This is not possible as different rows of P are orthogonal whereas each row has product −1 with itself, implying that the rows of P generate Z44 .

This argument simplifies the determination of the weight distribution, it seems: each of the 14 cosets with image of weight 8 has equally many words of weight 6 and of weight 10 (because of addition of the all-2-vector), therefore 8 each. It follows A6 = A10 = 14 × 8 = 112. The coset with the all-2-vector as image consists of 16 weight 8 vectors. Together with H itself we obtain A0 = 1 = A16 , A8 = 30.

16.5.7. Let G be the binary Golay code [24, 12, 8]2 and choose notation such that the first 8 coordinates from an octad. Let D ⊂ G be the subcode consisting of the codewords with entries 0 in the first 8 coordinates. Show: dim(D) = 5 and the image of the projection from G to the first 8 coordinates is the sum0-code [8, 7, 2]2 .

Let π : G −→ F82 be the projection on the first eight coordinates. Subcode D is the kernel of π. As G is self-dual, each element in the image of π is contained in the sum-0-code. It follows dim(D) ≥ 12 − 7 = 5. Assume dim(D) ≥ 6.

Projection on the last 16 parameters produces a [16, ≥ 6, 8]2 -code, which contradicts the Griesmer bound. We have dim(D) = 5 and consequently the image of the projection on the first coordinate segment is the [8, 7, 2]2 -code.

16.5.8. In the situation of the preceding exercise, define D1 = D and for i = 2, . . . 8 the subcode Di ⊂ G as the set of codewords having entry 1 in coordinates 1 and i, entry 0 in the remaining among the first eight coordinates.

Let E be the projection of D1 ∪ · · · ∪ D8 on the last 16 coordinates.

Determine the code parameters of E.

We know from the preceding exercise that G contains codewords with the required projections. It follows that each D1 is a coset of the linear subcode D of dimension 5. In particular E has 8 × 32 = 28 codewords. As any two of the underlying codewords of G have a distance of at most 2 in the first coordinate section, code E has a minimum distance of 8 − 2 = 6. It is a (16, 28 , 6)2 -code.

OA in statistics and computer science

16.6

Derandomization of algorithms

16.7

Authentication and universal hashing

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Exercises 16.7 16.7.1. Let A be an (N, n)m -hash family, which is -SU. Show that A is -U. Let an -SU family be given. In terms of probabilities: we have that P r(f (x) = z | f (y) = z) is bounded by . The probability P r(f (x) = f (y)) is an average over these conditional probabilities. It is therefore also bounded by .

16.7.2. Prove that the codes from Theorem 16.26 are OAqa−b (2, q a+b , q b ), where a ≥ b.

Consider two different columns u1 , u2 and corresponding entries c1 , c2 ∈ Fbq .

How many rows (x, y) are there with those entries in those columns? The condition is Φ(u1 x) + y = c1 , Φ(u2 x) + y = c2 .

After subtraction and using the additivity of Φ the equivalent condition Φ((u1 − u2 )x) = c1 − c2 is obtained (for each solution of this equation there is precisely one y). Each element of Fbq has precisely q a−b preimages under Φ. As u1 − u2 6= 0 this is also the number of solutions x. We have an OA of strength 2 with λ = q a−b .

Chapter 17 The geometric description of codes

17.1

Linear codes as sets of points

Exercises 17.1 17.1.1. Express the assumption that a generator matrix of the linear code C has no column of zeroes in terms of the dual distance d0 .

This is equivalent to d0 > 1, obviously.

17.1.2. Give a geometric description of [q 2 , 3, q 2 − q]q -codes.

Use all affine points of P G(2, q), all points outside a certain line. These are q 2 points, intersected by hyperplanes (=lines) in at most q points. The result is a [q 2 , 3, q 2 − q]q -code.

17.1.3. Show that a [q 2 − q − 1, 3, q 2 − 2q]q −code cannot exist.

Hint: use the geometric method. Show at first that such a code would have to be projective. Then show that each line of P G(2, q) would have to intersect the corresponding point set in either 0 or q − 1 points.

An [q 2 − q − 1, 3, q 2 − 2q]q -code is equivalent to a multiset of n = q 2 − q − 1 points in P G(2, q), which intersects each line in at most q − 1 points. Assume point P occurs with multiplicity 2. As P is on q + 1 lines the cardinality of the multiset is at most 2 + (q + 1)(q − 3) = q 2 − 2q − 1 < n, contradiction.

The same argument works if P has higher multiplicity.

It follows that we have a set of points (the code is projective). The same argument yields a bound of 1+(q +1)(q −2) = q 2 −q −1 = n on the cardinality of the set. We have equality. This shows that each line intersects the point set either in 0 or in precisely q − 1 points.

Let Q be a point not in the point set. The lines through Q show that n must be a multiple of q − 1, which is not the case, contradiction.

17.1.4. Prove the following: a [n, 3, d]q -code meeting the Griesmer bound with equality such that d ≤ q 2 is a multiple of q is necessarily projective.

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Let d = iq, i ≤ q. Then n = iq + q + 1 = i(q + 1) + 1. Each line contains at most i + 1 points. Assume P occurs with multiplicity 2. Considering the distribution of points on the lines through P yields 2 + (q + 1)(i − 1) = n − q ≥ n, contradiction. An even worse contradiction is obtained if P has higher multiplicity. The code must be projective. q 2m − 1 , 3m, q 2m−1 − q m−1 ]q for all q and m. q−1 Hint: use concatenation with an oval code (an extended 3-dimensional ReedSolomon code) as outer code.

17.1.5. Construct [

Let Q = q m . The oval code is a [Q + 1, 3, Q − 1]Q -code. As inner code m −1 use the Simplex code [ qq−1 , m, q m−1 ]q . Concatenation yields a code with the parameters as claimed.

17.1.6. Show that the code constructed in the previous exercise has only 3 nonzero weights. Determine the weight distribution.

The only weights occuring in the oval code are Q − 1, Q, Q + 1, obviously.

There are Q+1 = Q(Q + 1)/2 lines interseting the oval in 2 points, Q + 1 2 tangents and therefore Q(Q − 1)/2 exterior lines. This means that the weight distribution in the oval code is A0 = 1, AQ−1 = Q(Q2 − 1)/2, AQ = Q2 − 1, AQ+1 = Q(Q − 1)2 /2. As the inner code has constant weight q m−1 this yields the following weight distribution of the concatenated code: A0 = 1, Aq2m−1 −qm−1 = Q(Q2 − 1)/2, Aq2m−1 = Q2 − 1, Aq2m−1 +qm−1 = Q(Q − 1)2 /2.

17.1.7. Construct the Barlotti arcs (and with it the generator matrices of the corresponding 3-dimensional codes) over F5 and F7 .

At first q = 5, where ±1 are the squares. The points of the quadric are (0 : 0 : 1) and (1 : y : y 2 ), the interior points are those with x = 1 and y 2 −z = ±2. This yields as generator matrix for the first Barlotti code [11, 3, 8]5   01111111111 G = 0 2 3 0 2 3 0 1 4 1 4 11122233344 and for the second Barlotti code [16, 3, 12]5   0111111111111111 G = 0 0 1 2 3 4 2 3 0 2 3 0 1 4 1 4 1014411122233344 In case q = 7 the squares are {1, 2, 4}. The interior points are those with x = 1 and y 2 − z ∈ {3, 5, 6}. The first Barlotti code is a [22, 3, 18]7 -code with

The geometric description of codes

101

generator matrix 

 0111111111111111111111 G = 0 0 2 5 0 1 6 1 6 3 4 0 3 4 1 6 2 5 3 4 2 5 1111222333344455556666 and finally the second Barlotti code [29, 3, 24]7 has a generator matrix whose 8 first columns represent the points of the quadric, the remaining columns represent the 21 interior points.

17.1.8. The Griesmer bound is a consequence of the residual code construction, which states that from an [n, k, d]q code with precise minimum distance d we can construct an [n − d, k − 1, dd/qe]q -code, the residual code.

The reader is urged to give a geometric proof of this construction.

Hint: imagine the code described by a multiset P of cardinality n in P G(k − 1, q). By assumption there is a hyperplane H such that P ∩ H has cardinality n − d (in the multiset sense, including weights). It has to be proved that P ∩ H generates H, not a proper subspace, and that for every hyperplane L of H the intersection P ∩ L has cardinality at most . . . what? This is a simple counting exercise. Observe that L is contained in precisely q + 1 hyperplanes of P G(k − 1, q), one of them being H. Let C be an [n, k, d]q -code, described by a multiset of n points in P G(k − 1, q), and let H be a hyperplane intersecting the multiset in n − d points. The hyperplanes of hyperplanes are also known as secunda. Let S ⊂ H be a secundum, intersecting the multiset in x points. As S is in q + 1 hyperplanes, the cardinality of S is bounded by x + (q + 1)(n − d − x), which therefore must be ≥ n, equivalently qx ≤ qn − (q + 1)d or x ≤ n − q+1 q d. The intersection with H describes a code of length n − d, dimension k − 1 and distance d0 ≥ (n − d) − (n − q+1 q d) = d/q.

17.1.9. Give a geometric proof for the nonexistence of [13, 7, 6]4 -codes and of [13, 6, 7]4 -codes.

Assume at first [13, 6, 7]4 exists, equivalently a multiset of 13 points in P G(5, 4) intersecting each hyperplane in ≤ 6 points. As each secundum is in 5 hyperplanes it follows that the multiset intersects secunda in ≤ 4 points.

In case of equality the secundum is contained in four 6-hyperplanes and in one 5-hyperplane. The usual counting argument shows that the multiset of projective points is in fact a set (otherwise there would 3 points on some line, 4 points on some plane and more than 4 points on some secundum, contradiction). The same counting argument shows that any 5 points of our set generate a hyperplane. Let b be the number of 6-hyperplanes. Counting incident pairs of 6-hyperplanes and 4-sets of points the equation

13 b × 15 = × 4, 4

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a contradiction as the right side is not divisible by 3.

We have seen that [13, 6, 7]4 does not exist. Assume [13, 7, 6]4 exists. We claim that its dual is a [13, 6, 7]4 . This follows from the usual counting argument: the multiset describing the [13, 7, 6]4 intersects each secundum (a P G(4, 4)) in at most 5 points. This forces the multiset to be a set, a cap, and so on, finally any 6 points to be in general position (generating a hyperplane).

This implies that the dual code has d = 7.

17.1.10. Give a geometric proof for the fact that the dual of an [11, 6, 5]4 -code is an [11, 5, 6]4 -code.

Use this and the cyclic [11, 6, 5]4 from the section on constacyclic codes to construct an [12, 6, 6]4 -code.

Let an [11, 6, 5]4 -code C be given. We have to prove that its dual has minimum distance 6, equivalently that C has strength 5. Let P be a multiset of points in P G(5, 4) describing C. As each secundum P G(3, 4) is contained in 5 hyperplanes, P intersects S in at most 4 points. The usual counting argument shows that P is a set and that any 5 points from P generate a hyperplane. This is equivalent with C having strength 5.

The cyclotomic cosets in case q = 4, n = 11 are {0}, {1, 3, 4, 5, 9}, {2, 6, 7, 8, 10} We saw that C({1}) is an [11, 6, 5]4 -code. Its dual has as defining set the negative of the complement of C(1), which is {0} ∪ C(1), and is therefore contained in C({1}). By the preceding exercise the dual of an [11, 6, 5]4 -code is an [11, 5, 6]4 -code. Application of construction X to this chain yields a [12, 6, 6]4 -code.

17.1.11. Determine the weight distribution of the Denniston codes.

The Denniston codes are described geometrically by the Denniston arcs.

Let q = 2a . The Denniston 2i -arc in P G(2, q) is a Steiner system S(2, k, n), where k = 2i , n = 2a+i + 2i − 2a = 1 + (q + 1)(k − 1). The number of blocks of this Steiner system, which is the number of lines meeting the Denniston arc in k points, equals b = n(q + 1)/k = (q + 1)(q + 1 − 2a−i ) (each of the n points is on q + 1 blocks, the product n(q + 1) counts each block exactly k times). It follows that the number of lines avoiding the Denniston arc is q 2 + q + 1 − (q + 1)(q + 1 − 2a−i ) = (q + 1)2a−i − q = 22a−i − 2a + 2a−i .

The blocks are responsible for words of weight n − k = 2a+i − 2a = q(k − 1), the exterior lines contribute words of weight n. The weight distribution of the Denniston code is therefore A0 = 1, An−k = (q 2 − 1)(q + 1 − 2a−i ), An = (q − 1)(22a−i − 2a + 2a−i ).

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17.1.12. Give a geometric proof for the fact that the dual of an [11, 5, 6]3 -code is an [11, 6, 5]3 -code.

Let C be an [11, 5, 6]3 -code and P a multiset of points in P G(4, 3) describing C. It intersects each hyperplane in ≤ 5 points. The standard counting argument shows that a plane P G(2, 3) intersects P in at most )3 points. This implies that P is a set and that any 4 points of P generate a hyperplane. The claim follows.

17.1.13. Construct the ternary Golay codes [11, 6, 5]3 and [12, 6, 6]3 using cyclic codes of length 11.

The cyclotomic cosets in case q = 3, n = 11 are {0}, {1, 3, 4, 5, 9}, {2, 6, 7, 8, 10}, precisely the same as in the quaternary length 11 case. We have C({0, 1}) ⊂ C({1}) = C({0, 1})⊥ . The interval of length 3 in C(1) shows that the smaller of these codes has d ≥ 4. As it is contained in its dual it has d = 6. The previous exercise shows that C({1}) is an [11, 6, 5]3 -code. Application of construction X to this chain yields a [12, 6, 6]3 -code.

17.1.14. Fix disjoint subspaces P G(u − 1, q) and P G(k − u − 1, q) in P G(k − 1, q). Find the parameters of the q-ary k-dimensional linear codes described by the points of P G(k − 1, q), which are not contained in the union of those subspaces. Compare with the Griesmer bound.

We have n = (q k − 1)/(q − 1) + (q u − 1)/(q − 1) + (q k−u − 1)/(q − 1) = (q − 1)(q k−u − 1)/(q − 1) and d = q k−1 − q u−1 − q k−u−1 . These codes meet the Griesmer bound with equality. u

17.1.15. Complete the proof of the theorem from Belov et.al.

It remains to prove that the code has dimension k, equivalently that the points with positive weight are not contained in a hyperplane. For x > l this is clear as all points have positive weight. It is clear that it suffices to prove this statement for x = l = 1. We have that V = V (k, 2) is written a a direct sum of at least two nonzero subspaces and we have to prove that the vectors which are not contained in any of those subspaces generate all of V. Denote by W the subspace generated by those vectors. It suffices to prove W = V when there are only two subspaces: V = U1 ⊕ U2 , where dim(Ui ) = ui > 0. For k = 2 this is definitely not true. We assume k > 2. Write V = Fk2 , U1 defined by x1 = · · · = xu1 = 0 and U2 by xu1 +1 = · · · = xk = 0. We have k = u1 + u2 and choose u2 ≥ u1 , hence u2 ≥ 2. (1|110 . . . ) + (1|010 . . . ) = (0|100 . . . ) shows U2 ⊂ W. This implies W = V. 17.1.16. Show that the codes constructed in the theorem from Belov et.al. meet the Griesmer bound with equality if and only if the ui are pairwise different.

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P P We have n = x(2k − 1) − i (2ui − 1), d = x2k−1 − i 2ui −1 and the Pk−1 Griesmer bound is satisfied with equality if j=0 dd/2j e = n. Replace n by n0 = n − d and d by d0 = dd/2e. Then n0 , d0 have the same form as n, d, with k 7→ k − 1 and ui 7→ ui − 1. The Griesmer bound is reached with equality by n, d, k if and only if this is true of n0 , d0 , k − 1. Assume two of the ui are equal. Eventually P then we reach a situation where u1 = u2 = 0 and therefore d0 ≤ x2k−1 − i,ui ≥1 2ui −1 − 1. The Griesmer bound is not reached. If the original values ui are all different, then this situation will never arise and the Griesmer bound is reached.

17.1.17. Use the theorem from Belov et.al. to construct a [12, 4, 6]2 -code This is the case k = 4, u1 = 3 of the theorem. Here is the generator matrix:   001100011101 0 0 0 0 1 1 0 1 1 0 1 1   1 0 1 0 1 0 1 1 0 1 1 1 010101101111 17.1.18. The Reed-Muller code RM(1, r) is an [2r , r + 1, 2r−1 ]2 -code.

Give a geometric description for such a code.

This is the affine geometry AG(r, 2), the points in P G(r, 2) outside a hyperplane.

17.1.19. Show that the subgroup H of the Mathieu group M12 , the stabilizer of {1, 2, 3, 4, 5, 6}, is sharply 3-transitive on {1, 2, 3, 4, 5, 6}.

We know from the text that H has order 120 and is a subgroup of S6 , in its action on {1, 2, 3, 4, 5, 6}. The group H is generated by the permutation parts of h1 , h2 , h3 , h4 given in the text. h4 , h1 show that H is transitive on {1, 2, 3, 4, 5, 6}. Now h1 shows that the stabilizer of 1 is transitive on the remaining symbols. Finally h3 shows that the stabilizer of 1 and 2 is transitive on {3, 4, 5, 6}. All in all we have that H is 3-transitive. Its order is therefore a multiple of 6 × 5 × 4 = 120. As this equals the order of H, we have that H is sharply 3-transitive.

17.1.20. Find a set of generators of M12 .

We know the subgroup H of order 120, the stabilizer of {1, 2, 3, 4, 5, 6} and of {7, 8, 9, 10, 11, 12}. In order to obtain a permutation of M12 which is outside H consider the stabilizer of 1, 2, 3, 4. It is a group of order 8 and must contain an involution g. Then g must be fixed-point-free on the remaining 8 elements. A moment’s thought shows that it must map 5 7→ 12, 6 7→ 11. In fact g = (5, 12)(6, 11)(7, 8)(9, 10)m(1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2) in its action on the Golay code. Let M be the subgroup of M12 generated by H and g. We claim M = M12 . In fact it is immediately clear that M is transitive. The stabilizer of 1 contains h1 , h2 , h3 , g and is transitive on the remaining points.

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This shows that M is 2-transitive. The stabilizer of 1 and 2 contains g and h3 which implies that M is 3-transitive. Continuing like that we see that M is 5-transitive which shows M = M12 .

17.1.21. Show that the automorpism group of the hexacode is isomorphic to Z 3 × A6 .   111 100 Let the hexacode [6, 3, 4]4 be described by the generator matrix G =  0 1 ω ω 0 1  01ωω10 whose rows we denote by z1 , z2 , z3 . The permutation (34)(56) is an automorphism. The rows of the permuted matrix are z10 = z1 , z20 = z3 , z30 = z2 . Also (12)(34) is an automorphism, and z10 = z1 , z20 = z1 + z2 , z30 = z1 + z3 . An element of order 5 is described by (1, 2, 3, 4, 5)m where the constants in the monomial multiplication are m = (ω, ω, 1, ω, ω, 1). Here z10 = ωz3 , z20 = z2 + z3 , z30 = ω(z1 + z3 ). Clearly the center of order 3 is generated by multiplication by ω.

It is immediately checked that (1, 2)m can never be an automorphism.

17.1.22. Do we know the quaternary Denniston code?

This is case a = 2, i = 1, the hexacode [6, 3, 4]4 .

17.1.23. Determine the parameters of the F8 -linear Denniston codes.

[10, 3, 8]8 (the hyperoval in P G(2, 8)) and [28, 3, 24]8 .

17.1.24. Determine the parameters of the F16 -linear Denniston codes.

[18, 3, 16]16 , [52, 3, 48]16 , [120, 3, 112]16 .

17.1.25. Construct a [196, 9, 96]2 -code.

Apply concatenation to [28, 3, 24]8 (a Denniston code), with [7, 3, 4]2 as inner code.

17.2

Quadratic forms, bilinear forms and caps

Exercises 17.2 

 102 17.2.1. Consider the bilinear form with Gram matrix A =  µ 1 0  in odd λ01 characteristic. For which values of λ, µ is (, ) degenerate?

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The bilinear form is degenerate if and only if the Gram matrix has determinant 0. In our example this is equivalent (use the middle column) to λ = 1/2.

17.2.2. Let Q(x1 , x2 , x3 ) = x21 − x23 + 2x1 x3 + 3x2 x3 over Fp , p 6= 2.

Determine the corresponding bilinear form.

In which characteristic is this non-degenerate?  The Grammatrix of the corresponding symmetric bilinear form is A = 1 0 1  0 0 3/2  , of determinant −9/4. This is non-degenerate when p 6= 3.

1 3/2 −1 17.2.3. Consider the preceding example in characteristic 5.

Find a standard basis.   1 0 1 In case p = 5 we have A =  0 0 −1  . The Gram matrix with respect  1 −1−1 010 to basis e2 , e1 − e3 + e2 , e1 is  1 0 0  . We have a standard basis.

001   0111 1 0 1 1  17.2.4. Consider the bilinear form with Gram matrix A =   1 1 0 a  in odd 11a1 characteristic. When is it degenerate?

The bilinear form is degenerate if and only if det(A) = 0. Standard operations show det(A) = −(a2 − 4a + 2). The condition for degeneracy is a2 − 4a + 2 = 0. Let b = a − 2. The condition is b2 = 2. We have degeneracy if and only if the following conditions are satisfied: in Fp number 2 must be a square. Let b2 = 2. Then a = b + 2 or a = −b + 2.

17.2.5. Consider the preceding problem with a = 1. Over which fields is the bilinear form hyperbolic (respectively elliptic)?

Let a = 1 in the preceding exercise. Then b = a − 2 = −1 must satisfy b2 = 2. This is never the case. It follows that we have a non-degenerate symmetric bilinear form on V (4, q) for all odd q.

We have det(A) = 1, a square. The discriminant of A is therefore +1 and the form is hyperbolic.

17.2.6. Consider the preceding problem over F5 .

Find a totally isotropic line.

Let p = 5, a = 1 in the preceding problems. As the form is hyperbolic a totally isotropic line must exist. One example is he1 , 2e2 − e3 − e4 i.

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17.2.7. Let Q(x1 , x2 , x3 ) = x21 + x1 x2 + ωx2 x3 , a quadratic form on V (3, 4).

Determine the Gram matrix of the underlying symplectic form and its radical.

Is Q degenerate?   01 0 The Gram matrix of the underlying symplectic form is  1 0 ω  , clearly 0ω 0 of rank 2. The radical is generated by ωe1 + e3 . As Q(ω, 0, 1) = ω 6= 0 our quadratic form is non-degenerate.

17.2.8. Find the number of totally isotropic lines in P G(3, q) of a nondegenerate symplectic bilinear form on V (4, q).

Let P be a point. We have that P ⊥ has vector space dimension 3 and the factor space P ⊥ /P has dimension 2. It follows that P is on q + 1 totally isotropic lines. As each such line contains q + 1 points it follows that the number of totally isotropic lines with respect to a symplectic bilinear form on V (4, q) equals the number of points in P G(3, q), which is (q 2 + 1)(q + 1).

17.2.9. Consider V (4, 2) with a non-degenerate symplectic form.

Count the symplectic bases.

As first element of an ordered symplectic basis of V (4, 2) we can choose an arbitrary nonzero vector v1 (15 choices). The second vector v2 must be outside v1⊥ (8 choices for v2 ). We have that H1 = hv1 , v2 i is a hyperbolic plane. The remaining basis vectors must be chosen in H2 = H1⊥ . The number of choices is clearly 3 × 2. The total number of ordered symplectic bases is 15 × 8 × 3 × 2, which happens to be 6!

17.2.10. Define a design as follows: the points are the points of Q− (3, q) ⊂ P G(3, q), the blocks are the intersections with non-tangent planes. Determine the parameters of this design.

There are q 2 + 1 points. Each block has q + 1 points. As Q− (3, q) is a cap any 3 of its points determines a plane of P G(3, q). This plane is then a secant plane, meeting Q− (3, q) in q + 1 points (a block). It follows that we have a Steiner 3-design S(3, q + 1, q 2 + 1).

17.2.11. Show that the trace tr : F = Fqm −→ K = Fq defines a nondegenerate symmetric bilinear form, the trace form hi on F (seen as a vector space over K) by hx, yi = tr(xy).

This is obvious. Symmetry and bilinearity are evident. If x 6= 0 was orthogonal to all of F, then tr would be the 0-mapping, which is not the case.

17.2.12. Show that a quadric meets a line in either 0 or 1 or 2 or in q + 1 points.

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The intersection is a quadric defined on this line (= 2-dimensional vector space). If the line is not contained in the quadric, then it has at most 2 points on the line (we are talking about quadratic equations after all).

17.2.13. Show that a quadric in P G(2, q) which is not identically zero has at most 2q + 1 points.

We know the non-degenerate case: the parabolic quadric has q + 1 points (an oval). Let the radical have vector space dimension 1. The underlying V (3, q) is direct sum of the radical (a point) and a line with a non-degenerate quadric on this line. The quadric has therefore either 0 or 2 points on the line. In the former case we have only one point, the radical, on the quadric.

In the latter points the number is 1 + 2q. If finally the radical is a line, then the intersection size is q + 1.

17.2.14. Describe the parameters of the codes geometrically described by the hyperbolic quadrics on P G(3, q), q odd.

The number of points of Q+ (3, q) is (q + 1)2 . By the previous exercise the intersection with a plane (=hyperplane) has at most 2q +1 points. This shows that the code parameters are [(q + 1)2 , 4, q 2 ]q . Small examples are [16, 4, 9]3 and [36, 4, 25]5 .

17.2.15. Describe the parameters of the codes geometrically described by the elliptic quadrics on P G(3, q), q odd.

The elliptic quadric has q 2 + 1 points. The intersection with a plane has at most q +1 points (as the quadric does not contain lines). The code parameters are therefore [q 2 + 1, 4, q 2 − q]q . The Griesmer bound is met with equality.

17.3

Caps: constructions and bounds

Exercises 17.3 17.3.1. Show that lines in AG(k, q) have q points.

A line in P G(k, q) has q + 1 points. The affine points are the points outside some fixed hyperplane H of P G(k, q). The lines of AG(k, q) correspond to the lines of P G(k, q), which are not contained in H. Each such line meets H in precisely one point.

17.3.2. Determine the weight distribution of an ovoid code.

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An ovoid is a (q 2 + 1)-cap in P G(3, q). It has q 2 + 1 tangent planes. All q + q remaining planes meet the ovoid in q + 1 points each. This shows 3

A0 = 1, Aq2 −q = (q − 1)(q 3 + q), Aq2 = (q − 1)(q 2 + 1).

17.3.3. Prove that P (0) = (1, 0) extends the cap Kq ⊂ AG(4, q) to a (3q 2 +4)cap (here q = 2f , f odd). a1 w1 ∈ / Fq . Assume P (0) a2 w2 is collinear with P (a1 , w1 ) and P (a2 , w2 ). This means that the (3, 5)-matrix with those points as rows has rank 2. Using Gauß elimination with respect to row P (0) = (0, 0, 0, 0, 1) shows that a1 w1 and a2 w2 must be scalar multiples, which is not the case.

If (a1 , w1 ) 6= (a2 , w2 ) (0 6= ai ∈ F4 , wi ∈ W ), then

17.3.4. Describe the cap Kq ⊂ AG(4, q) in terms of cyclic codes.

Let q = 2f , f odd. In the general mechanism of q-ary cyclic codes consider length n = 3(q 2 + 1) | (q 4 − 1), defining set A = {0, 1}. The points 6= P (0) are the columns of a generator matrix of tr(B(A)) (a check matrix of C(A)). The fact that those columns form a cap is equivalent to C(A) having minimum distance d ≥ 4. The BCH-bound shows only d ≥ 3.

17.3.5. Show that AG(k, q) is the disjoint union of q copies of AG(k − 1, q).

AG(k, q) is by definition the complement of a hyperplane H0 in P G(k, q).

Let S ⊂ H0 be a hyperplane of H0 . Then S is a P G(k − 2, q) contained in our P G(k, q). Such a subspace is also called a secundum. It is contained in q + 1 hyperplanes, one of which is H0 . The remaining q hyperplanes through S partition our AG(k, q) into q subspaces AG(k − 1, q).

17.3.6. Show that the hyperplanes of AG(k, q) are represented by pairs (y, c) where y = (y1 , . . . , yk ) ∈ Fkq varies over representatives of 1-dimensional subspaces (in other words points of P G(k − 1, q)) and c ∈ Fq . Point x = (x1 , . . . , xk ) ∈ AG(k, q) is on the hyperplane described by the pair (y, c) if x · y = c.

Represent the points of AG(k, q) as (x1 : x2 : · · · : xk : 1) ∈ P G(k, q).

Those are the points outside the hyperplane H0 defined by xk+1 = 0. The hyperplanes of P G(k, q) are parametrized by [y1 : y2 : · · · : yk+1 ], where point Pk+1 x = (x1 : · · · : xk+1 ) is on the hyperplane if i=1 xi yi = 0. Hyperplane H0 corresponds to (0 : · · · : 0 : 1). The hyperplanes 6= H0 are therefore described by pairs (y, yk+1 = −c), where y = (y1 , . . . , yk ) 6= 0 is determined up to scalar multiples, c ∈ Fq and the affine point (x, 1) belongs to the hyperplane described by pair (y, c) if x · y = c.

17.3.7. Prove that the maximum size of a cap in AG(3, q) (q > 2) is q 2 .

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We know that q 2 + 1 the maximum size of a cap in P G(3, q) and that each plane of P G(3, q) meets a (q 2 + 1)-cap either in 1 point or in q + 1 points.

As there is no plane avoiding the (q 2 + 1)-cap, AG(k, q) does not contain a cap of that size. The affine part with respect to a tangent plane is a q 2 -cap in AG(3, q).

17.3.8. Prove the Cauchy-Schwartz inequality for real vectors: x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) satisfy |x · y| ≤ kxkkyk Let x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) be real vectors. We can assume that all the entries xi , yi are non-negative. Scalar multiplication shows that we can also assume kxk = kyk = 1 and we need to show x · y ≤ 1. This follows from the fact that (x − y) · (x − y) ≥ 0.

17.3.9. Derive the bound used in the proof of the Plotkin bound from the Cauchy-Schwartz inequality.

Let x like in the preceding exercise, M = x1X + · · · + xn , and y = (1, . . . , 1).

2 We have M = x · y ≤ kxkkyk, hence M /n ≤ x2i . i

17.3.10. Prove Theorem 17.58 for q > 3.

Let q > 3, k ≥ 3. As C3 (q) = q 2 is the size of the largest cap in AG(3, q), k it follows c3 (q) = 1/q ≤ 94 . Assume now ck−1 (q) ≤ (k−1) 2 . The main bound shows k + (k − 1)2 q −k ck (q) ≤ . k + (k − 1)2 2 k This is ≤ k+1 k2 if and only if (k(k − 1)) ≤ q , which is always true when q > 3.

In the ternary case we need to know the values Ck (3) for the first values of k to complete the verification of the theorem.

17.3.11. Prove that there is no 57-cap in P G(5, 3). In the proof use the fact that there is no 21-cap in P G(4, 3).

Assume K ⊂ P G(5, 3) is a 57-cap. Use representatives for the points of K as columns of a (6, 57)-matrix and consider the code C generated by this matrix. As the maximum size of a cap in P G(4, 3) is 20, it follows that K meets each hyperplane in ≤ 20 points. This means that C is a [57, 6, 37]3 -code, contradicting the Griesmer bound.

17.3.12. Prove that the following form the points of a 236-cap in AG(7, 3) : (1, 0, D) (1, 0, R) (1, 1, D (1, 1, R) (1, 2, U ).

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The two first types, with prefix 0, form a 112-cap, likewise the two types with prefix 1. The last type consists of the 12 vectors with prefix 2 followed by a vector of weight 1. In order to see that we have an affine cap we have to show that no three of our vectors sum to 0. Assume x, y, z are from our collection and x + y + z = 0. The prefixes cannot be all 0 or all 1. It is also clear that the prefixes cannot all be 2, as the sum of two vectors of weight 1 can never have weight 1. We conclude that the prefixes of x, y, z are 0, 1, 2.

The sum of two vectors of type R never has weight 1. The sum of a vector of type R and a vector of type D or D has weight ≥ 3. As it is also clear that the sum of a vector of type D and a vector of type D has weight > 1 we conclude that we have a 236-cap in AG(7, 3).

17.3.13. Show that the union of the Calderbank-Fishburn cap from the previous exercise and the points of type (0, 1, U ) form a 248-cap in P G(7, 3).

It is clear that the 12 new points contained in the hyperplane x1 = 0 form a cap. It remains to show that the difference between two points of the Calderbank-Fishburn cap never has the form (0, 1, U ). Assume this is the case.

The presence of the entry 1 in the second coordinate restricts the possibilities.

Consider the last coordinate section. It is obvious that a vector of weight 1 cannot be involved in the difference. Also, the difference between type R and type D or D has weight ≥ 3. We have already used the fact that the sum or difference of two vectors of type R cannot have weight 1. The only case remaining is the difference between type D and type D. Obviously, this results in weight ≥ 2.

17.3.14. Show λ(t, 2) = 1/bt/2c.

We know λ(t, 2) ≤ 1/bt/2c by the sphere-packing bound. In order to obtain a matching lower bound consider the cyclic code tr(B(A)) of length 2r − 1, where A = {0, 1, 3, . . . , 2(i − 1) + 1 = t − 2} when t = 2i + 1. The length is 2r − 1, dimension k ≤ 1 + ir and strength t as the Galois closure of A contains {0, 1, . . . , t − 1}. For r → ∞ we have log2 (2r − 1) log2 (n) = → 1/i = 1/bt/2c. k 1 + ir This is also OK when t = 2i.

17.3.15. Show λ(2, q) = 1 for all q.

The maximal length of a linear OA of strength 2 and dimension k is n = logq (n) (q k −1)/(q −1) (see the Simplex and Hamming codes) and we have → k 1.

17.3.16. Accepting the fact that 20 is the largest size of a cap in P G(4, 3), prove that the largest intersection of the Hill cap with a hyperplane is 20.

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As the intersection with a hyperplane is a cap in P G(4, 3) this intersection size cannot be larger than 20. Assume it is always smaller than 20. Then the geometric description yields a [56, 6, d]3 -code with d ≥ 37. This contradicts the Griesmer bound.

17.3.17. Show that the Hill cap defines a [56, 6, 36]3 -code and compare with the Griesmer bound.

This follows from the previous exercise. The Griesmer bound is met with equality.

17.3.18. Use the generator matrix of the Hill cap to obtain a 20-cap in P G(4, 3).

Take the 20 last columns of the generator matrix of the Hill cap, remove the leading 0 in each column. This is the intersection of the Hill cap with the hyperplane x1 = 0, a 20-cap in P G(4, 3).

Chapter 18 Additive codes

18.1

Basic constructions and applications

Exercises 18.1 18.1.1. Let C be a q-linear [n, k]qm -code whose dual has minimum weight > t.

Determine the parameters of C as an orthogonal array.

A q-linear [n, k]qm -code of strength t is an OAλ (t, n, q m ), where λ = q m(k−t) .

18.1.2. Formulate a generalization of the method of concatenation which applies to additive q-linear codes as outer codes.

In the definition of concatenation use the same inner code (a linear [n, r, d]q ) and as outer code a q-linear q r -ary code of length N, dimension K (this is the q r -dimension) and minimum distance D. The definition of the concatenated code is unchanged. It is a linear q-ary code with parameters [nN, rK, dD]q .

18.1.3. Prove: if there exists a q-linear [n, k]qm -code of strength t and m > 1, k−1 then there exists a q-linear [n, k + ] m−1 -code of strength t. m−1 q This is simply Chen projection, expressed in terms of the strength. We are given an [n, k]qm -code of strength t. Its dual is an [n, n − k, t + 1]qm -code. k−1 , t + 1]qm−1 -code, whose dual is an Chen projection yields an [n, n − k − m−1 k−1 [n, k + m−1 ]qm−1 -code of strength t.

18.1.4. Construct q-linear q b -ary codes of length q a + 1, strength 2 and dimension (a + b)/b for every b ≤ a.

Start from the extended 2-dimensional Reed-Solomon code over Fqa . This is in particular a q-linear [q a + 1, 2]qa -code of strength 2. Chen projection (see the preceding problem) yields a q a−1 -ary code, still of strength 2, with Fq -dimension 2a − 1 (one less than before). After a − b applications we have a q-linear q b -ary code of length q a + 1 and strength 2, with Fq -dimension 2a − (a − b) = a + b, which coresponds to the q b -ary dimension (a + b)/b.

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18.1.5. Let Dq (a, b) (where a ≥ b) the q-linear q b -ary code of dimension a/b and length q a constructed in this section. Determine its strength and minimum distance.

Recall that the entry of codeword c(u, 0) in coordinate x is Φ(ux). For fixed u 6= 0 this entry is 0 if ux ∈ Ker(Φ). As Ker(Φ) has q a−b elements it follows that c(u, 0) has weight q a −q b . This is therefore the minimum weight. Dq (a, b) is a constant weight code.

All codewords of Dq (a, b) have entry 0 in the coordinate corresponding to x = 0. This shows that Dq (a, b) does not even have strength 1.

18.1.6. Let a ≥ b and l = ba/bc − 1. Show that l X

q a−ib = (q a+b − q a−lb )/(q b − 1).

i=0

This simply is the geometric series. After factoring out q a−lb we obtain the geometric series 1 + q b + · · · + q bl = (q b(l+1) − 1)/(q b − 1). Multiplication by q a−lb (in the numerator) yields the result.

18.1.7. Determine the parameters of the OA resulting from the Bose-Bush construction when 2b ≤ a < 3b. Give at least two examples.

We have a = 2b + i, where 0 ≤ i < b, consequently l = 1, λ = q a−b = q b+i and n = 1 + (q 3b+i − q b+i )/(q b − 1) = 1 + q b+i (q b + 1) = q 2b+i + q b+i + 1.

The parameters are therefore OAqb+i (2, q 2b+i + q b+i + 1, q b ) where 0 ≤ i < b.

As examples we obtain OA8 (2, 41, 4), OA16 (2, 145, 8), OA32 (2, 289, 8).

18.1.8. Determine the parameters of the codes constructed in Theorem 18.9 for m = 1, r0 = 2 and show that they are d-optimal for q ≤ 4.

In case r0 = 2, m = 1 we obtain k = 6, d = q 4 − q 3 , n = q(q 2 − 1)(q 2 − q + 1)/(q − 1) = q 4 + q. We obtain parameters [q 4 + q, 6, q 3 (q − 1)]q . The Griesmer bound shows that for q ≤ 4 these parameters are d-optimal. The optimal parameters are [18, 6, 8]2 (but the binary Golay code is stronger),[84, 6, 54]3 , [260, 6, 192]4 .

Even the [68, 5, 48]4 obtained as residual code from the quaternary code is known to be d-optimal (but we know a [70, 7, 48]4 -code).

18.1.9. Construct linear codes [70, 9, 32]2 and [196, 9, 96]2 .

These are special cases of the family of 3r0 -dimensional 2-weight codes.

They correspond to q = 2, r0 = 3 and m = 1, m = 2, respectively.

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18.2

The cyclic theory of additive codes

18.3

Additive quaternary codes: the geometric approach

Exercises 18.3 18.3.1. Use geometric language to confirm the (obvious) fact that the largest minimum distance of an additive [n, 1, d]4 -code is d = n.

The ambient space is a line. Length n means that we use it with multiplicity n. A hyperplane is a point. No codeline is contained in a hyperplane, therefore all codeword weights are n.

18.3.2. Determine the maximum minimum distance d of additive [n, 1.5, d]4 codes, for all n.

The ambient space is the Fano plane. If n = 7a the optimal choice is to use each line with multiplicity a. Hyperplanes are lines. Each hyperplane has 7a − a = 6a codelins not contained in it. The parameters are [7a, 1.5, 6a]4 . Let n = 7a + b, 0 < b < 7. The optimal choice is to use b lines with multiplicity a+1, the others with multiplicity a. The parameters are [7a+b, 1.5, 6a+b−1]4 .

18.3.3. Let P G(2, q) ⊂ P G(2, q 2 ).

Show that each line of P G(2, q 2 ) intersects the Baer subplane either in one point or in q + 1 points.

Clearly each line meeting the Baer subplane in at least two points is a line of the subplane and therefore meets it in q + 1 points. The number of lines meeting the Baer subplane in at least one point is therefore (q 2 +q +1)+(q 2 +q +1){(q 2 +1)−(q +1)} = (q 2 +q +1)(q 2 −q +1) = q 4 +q 2 +1 which happens to be the number of all lines of P G(2, q 2 ).

18.3.4. Describe a construction of a [14, 3, 10]4 -code.

As the dimension is an integer it may be possible to find a linear such code.

As the dimension is 3 such a linear code is described by a point multiset in P G(2, 4). Choose the point set be the complement of a Baer subplane. As the Baer subplane is a blocking set by the previous exercise, the parameters of the code are indeed [14, 3, 10]4 . Alternatively: start from the elliptic quadric which gives a linear [17, 4, 12]4 , apply puncturing and shortening.

18.3.5. Describe a construction of a [13, 4, 8]4 -code.

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Start from the elliptic quadric and the corresponding [17, 4, 12]4 -code, apply puncturing. We see that indeed there is a linear quaternary code with those parameters.

18.3.6. Show that the binary linear codes obtained by concatenation from the additive [2f , 1 + f /2, 3 × 2f −2 ]4 -codes meet the Griesmer bound with equality.

Concatenation with [3, 2, 2]2 as inner code yields parameters [3 × 2f , f + 2, 3×2f −1 ]2 . Given this dimension k and distance d the Griesmer lower bound on the length is 3 × 2f −1 + 3 × 2f −2 + · · · + 3 × 20 + 2 + 1 = 3 × (2f −1 + · · · + 1 + 1) = 3 × 2f .

18.3.7. Give a geometric proof for the non-existence of a linear [10, 3, 7]4 code.

We work in P G(2, 4) (21 points, 21 lines, each line has 5 points). We are looking for a multiset of 10 points such that each line contains at most 10 − 7 = 3 of those points. No point can have multiplicity > 1 (otherwise the bundle of 5 lines through the point limits the cardinality to 2 + 5 = 7), so we are talking of a point set M. The bundle of lines through a point P ∈ M shows that it is on precisely one line meeting M in two points. There are therefore precisely 5 such lines. The x be the number of lines meeting M in 3 points. Then 10 2 = 5 + 3x, contradiction.

18.3.8. Give a geometric proof for the non-existence of an additive [10, 3, 7]4 code.

This is more general than the previous exercise, so the proof will be harder.

It is given in the Blokhuis-Brouwer paper. Let L be the corresponding multiset of lines in P G(5, 2). Assume the projection to some two codelines has binary dimension < 4. Then shortening with respect to those two lines would yield an [8, 1.5, 7]4 -code which cannot exist (we would have to choose 8 different lines from the Fano plane). It follows that the codelines form a partial spread. Let M be the union of the points on codelines. Then M is a set of 30 points, the codepoints. As each hyperplane H of P G(5, 2) contains at most 3 codelines we have |H ∩ M | ≤ 3 × 3 + 7 = 16. In fact, let H contain i codelines. Then |H ∩ M | = 3i + (10 − i) = 10 + 2i. Let hi be the number of hyperplanes that contain precisely i codelines. By elementary counting each secundum V4 meets M in at most 9 points, each plane E = V3 meets M in at most 5 points and each line of P G(5, 2) meets M in at most 3 points (which is clear anyway).

Let now L be a codeline and work in the factor space Π = V /L, which is a P G(3, 2). Call a point of Π a j-point if its preimage (a plane) contains 3 + j codepoints. Again by elementary counting we see that Π has three 1-points (they are on a line), the remaining points of Π are 2-points. This implies that there are no hyperplanes containing precisely one codeline (hence h1 = 0).

The three obvious linear equations relating the hi therefore have only one

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solution: h0 = 8, h2 = 15, h3 = 40. Let now H be one of the hyperplanes without a codeline (hence |H ∩ M | = 10) and let sk be the number of solids S ⊂ H (hyperplanes of H) which contain precisely k codepoints. Call S then a k-solid. S is of type (10, b, c) if the three hyperplanes containing S meet M in 10, b, c points, respectively. Then b, c ∈ {10, 14, 16} and b + c = 20 + 2k.

This shows that for each k the type of S is uniquely determined: k = 0 is type (10, 10, 10), k = 1 is impossible, k = 2 has type (10, 10, 14), k = 3 type (10, 10, 16), k = 4 type (10, 14, 14), k = 5 type (10, 14, 16) and k = 6 type (10, 16, 16). The intersection with the remaining 7 hyperplanes without codelines yields the equation 2s0 + s2 + s3 = 7. Eliminating s4 , s5 , s6 from the four standard equations that relate the sk yields 20s0 + 4s2 + s3 = 20 which together with the equation above yields a contradiction: 18s0 + 3s2 = 13.

18.3.9. Show the following: if E is a plane in P G(2m, 2) then P G(2m, 2) \ E can be partitioned into lines.

For m = 2 this is a special case of the proposition as a secundum is a plane.

The general case follows by induction.

18.3.10. Show that P G(2m, 2) possesses a partial spread of 1+8(22m−2 −1)/3 lines.

Use the partial spread of the preceding exercise and add a line of the Fano plane E. 18.3.11. Show that 9 is the maximal number of lines in P G(4, 2) which form a partial spread.

We know that there is a partial spread of 9 lines. Assume there is one with 10 lines. It covers all points but one point P. Any two of the ten spread lines generate a hyperplane H. Assume H contains a third spread line. Then the number of points on H is ≥ 9 + 7 > 15, contradiction. It follows that H contains those two spread lines, one point from each of the remaining 8 spread lines and necessarily then also P. This gives us 10 2 = 45 different hyperplanes containing P. This is a contradiction as P is on only 15 hyperplanes.

18.4

Quantum codes

Exercises 18.4 18.4.1. Show that there is no linear [6, 3, 4]4 -code which is self-dual with respect to the dot product.

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Let G = (I | P ) be a generator matrix of C in standard form. All entries of P are nonzero. Assume the first row of P is 111. Because of the dot product with the first row the remaining rows of P must have each nonzero entry exactly once. They cannot have dot product 0 with themselves.

It follows that each row of P contains entry 1 once, and two identical entries, ω or ω. The entries 1 of the three rows of P are in different positions, as otherwise the minimum weight would be < 4. Choose (1, ω, ω) as first row.

The second row cannot be (ω, 1, ω) as the dot product with the first row does not vanish. It is not (ω, 1, ω) either, as the condition of minimum weight 4 would be violated.

18.4.2. Give a geometric description of a pure [[5, 1, 3]]2 -quantum code.

This is a spread of P G(3, 2).

18.4.3. Show that the notion of a pure [[n, n − r, 3]]2 -quantum code does not make sense if r < 4.

We need lines, so r < 2 does not make sense. In case r = 2 the entire space is a line. In case r = 3 we are in the Fano plane, where all lines pairwise intersect, so again this does not make sense.

18.4.4. Show that P G(r − 1, 2) for even r possesses a spread of lines and determine the parameters of the corresponding binary quantum code.

2m Let r = 2m and consider Fm and points 4 . This is a representation of F2 in P G(m − 1, 4) are lines in P G(2m − 1, 2). As P G(m − 1, 2) is the disjoint union of its points, this yields a spread in P G(2m − 1, 2). Those spreads are also known as Desarguesian spreads.

18.5

Network codes, subspace codes

Exercises 18.5 18.5.1. Let Ω be the set of m-dimensional vector subspaces of a fixed ldimensional space Flq . Show that the subspace distance defines a metric on Ω.

As usual only the triangle inequality is in doubt. We have to show ds (A, C) ≤ ds (A, B) + ds (B, C), equivalently dim(A ∩ B) + dim(B ∩ C) − dim(A ∩ C) ≤ l.

In fact, we have dim(A ∩ B) + dim(B ∩ C) ≤ l + dim(A ∩ B ∩ C) which shows the inequality.

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18.5.2. Find the maximum length n of a subspace code Sq (l, n, 1, m).

This is the case of subspace distance ≥ 1 which means that the subspaces must be different. This means that the optimal code simply is the space Ω of all vector m-subspaces of our Flq . The maximal length is n = |Ω| =

(q l − 1)(q l − q) . . . (q l − q m−1 ) (q m − 1)(q m − q) . . . (q m − q m−1 )

These are the Gauß coefficients.

18.5.3. Show that the coding problem for subspace codes in case m = 2 is essentially trivial.

In case m = 2 the only distances occurring are 0, 1, 2. Subspace distance ds = 1 is trivial as we saw in the previous problem. In case ds = 2 the condition is that the codelines are pairwise skew, so the code describes a partial spread. The corresponding problem therefore asks what is the maximum number of lines in P G(l − 1, q) which form a partial spread.

18.5.4. Show that different linearized polynomials yield different linear mappings : F −→ F. The difference such linearized polynomials is a linearized polynomial Pn−1 of two i f (X) = i=0 ai X q where not all of the coefficients ai ∈ F vanish. Assume f (x) = 0 for all x ∈ F. Then the nonzero polynomial f (X) ∈ F [X] of degree q n−1 has q n roots, which contradicts one of the basic properties of polynomials in one variable with coefficients in fields.

18.5.5. Use double counting to show for S2 (6, n, 2, 3) we have n ≤ 81.

Count the pairs (P, E) where P is a point, E one of our n code planes and P ∈ E. Counting from the right shows that the number of pairs is 7n. Count from the left: each of the 63 points P is contained in at most 9 code planes as the E/P form a partial spread of lines in P G(4, 2). This yields the inequality 7n ≤ 9 × 63.

18.5.6. Show that the lines contained in the planes of the lifted Gabidulin code are precisely the lines which are skew to the special plane S. The lines of the 64 planes which form the lifted Gabidulin code are by construction different and skew to S. There are 7 × 64 such lines. The number of lines in P G(5, 2) is 63 × 62/6 = 21 × 31. Seven of them are in S and 7 × (63 − 7)/2 = 7 × 28 meet S in a point. The number of lines skew to S is therefore 21 × 31 − 7 − 7 × 28 = 7 × 64.

Part III

Codes and algebraic curves

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Chapter 19 Introduction

19.1

Polynomial equations and function fields

19.2

Places of the rational function field

Exercises 19.2 19.2.1. List the 24 solutions of the Klein quartic over F8 .

Describe F8 by F8 = F2 ( ) where 2 + 3 + 1 = 0. The solutions having some entry = 0 are (1 : 0 : 0), (0 : 1 : 0), (0 : 0 : 1). Observe that the equation x + x3 + 1 = 0 has solutions 3 , 5 , 6 . Let now (x : y : 1) be a solution where xy 6= 0. The equation is x3 y + y 3 + x = 0, equivalently x2 y + (x2 y)3 + 1 = 0, hence x2 y ∈ { 3 , 5 , 6 }. This yields 3 + 7 × 3 = 24 solutions.

19.2.2. Prove the additive version of Hilbert’s theorem 90 for finite fields We saw in Exercise 12.1.6 that the kernel of the trace map is an (r − 1)dimensional Fq -subspace of F = Fqr . On the other hand the elements of the form b = x − xq also form a vector space of the same dimension, and they do have trace 0.

19.2.3. Show that for a hyperelliptic equation y 2 = f (x) it can be assumed without restriction that f (x) is a polynomial without quadratic factors This is obvious: assume f (x) = g(x)2 h(x). Then the equation can be rewritten as (y/g(x))2 = h(x).

19.2.4. Show that the hyperelliptic equation y 2 = f (x) where f (x) is a polynomial of degree ≥ 3 has (0 : 1 : 0) as only point at infinity Let f (x) = an xn + . . . where an 6= 0. The homogenization is y 2 z n−2 = an xn + . . . When z = 0 it follows an xn = 0, hence x = 0. We obtain (0 : 1 : 0) as unique solution at infinity.

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19.2.5. Explain why the equation y 2 = f (x) is not interesting in characteristic 2 In this case p squaring is a field automorphism. Each value of x yields a unique y = f (x).

Chapter 20 Function fields, their places and valuations

20.1

General facts

20.2

Divisors and the genus

20.3

The Riemann-Roch theorem

20.4

Some hyperelliptic equations

125

Chapter 21 Determining the genus

21.1

Algebraic extensions of function fields

21.1.1. Use the generalized Eisenstein criterion to prove that the Artin-Schreier polynomial Y q − Y − f (X) is irreducible as a polynomial in Y provided the degree of the polynomial f (X) is not a multiple of the characteristic p.

In the language of the generalized Eisenstein criterion we have n = q a power of the characteristic p and aq = 1, a1 = −1, a0 = −f (X), the remaining coefficients 0. We see that the second item of the Eisenstein criterion applies, with F = K(x) the rational function field and P the place at infinity.

21.1.2. Find instances when the Artin-Schreier polynomial is not irreducible.

When p = q = 3 we have (Y + a(X))(Y 2 − a(X)Y + a(X)2 − 1) = Y 3 − Y + a(X)3 − a(X) and for p = 2, q = 4 we have (Y 2 + cY + c2 + a(X))(Y 2 + cY + a(X)) = Y 4 + Y + c2 a(X) + a(X)2 where 0 6= c ∈ F4 .

21.2

The hyperelliptic case

21.3

The Kloosterman codes and curves

21.4

Subrings and integrality

21.5

The Riemann-Hurwitz formula

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Chapter 22 AG-codes, Weierstraß points and universal hashing

22.1

The basic construction

22.2

Pole orders

Exercises 22.2 22.2.1. Show that the pole orders form a semigroup: a subset of the natural numbers which is closed under addition.

We are given a rational point P. We have that 0 is a pole order. In general a natural number m is a pole order if there exists x ∈ F such that (x)∞ = mP. Let now m1 , m2 be pole orders and (x)∞ = m1 P, (y)∞ = m2 P. Then (xy)∞ = (m1 + m2 )P which shows that m1 + m2 is a pole-order.

22.2.2. Let S be the semigroup generated by q and q + 1. Show that there are precisely 2q natural numbers which are not in S. By definition S consists of the numbers iq + j(q + 1) where i, j ≥ 0. The natural numbers not in S consist of the interval from 1 to q − 1, the interval from 2q + 3 to 2q − 1 . . . . The lengths of those interval keep decreasing by 1,

so the total number of numbers outside S is (q − 1) + (q − 2) + · · · + 1 = 2q .

22.2.3. Let S be the semigroup of the preceding exercise. Order its elements: u1 = 0, u2 = q, u3 = q + 1, u4 = 2q, u5 = 2q + 1, u6 = 2q + 2, u7 = 3q, . . . Show that u(i+1)+1 = iq provided that i ≤ q.

2

We just saw that the statement is true for i = 1, 2, 3. The claim is obvious.

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Solutions Manual

Examples of function fields and projective equations

Exercises 22.3 22.3.1. Show that the number of rational places of F | Fq (x) defined by the Klein quartic in characteristic 6= 7 equals the number of points in P G(2, q) satisfying Y 7 Z − XY 7 − X 3 Z 5 = 0.

22.3.2. Find a basis of L(4B).

We know dim(4B) = 10 and we have a basis of the 7-dimensional subspace L(3B). This is extended to a basis of L(4B) by x2 , 1/y 2 , y 2 /x.

22.3.3. Find a generator matrix of a [21, 4, 15]8 -code.

We have dim(L(2B)) = 4, with basis 1, x, 1/y, y/x. A generator matrix of the code has rows indexed by those basis elements, columns indexed by the rational points different from those used in B with the evaluations as entries.

The points are the P (x, y) = (x, y, 1) where xy 6= 0 and x2 y ∈ { 3 , 5 , 6 }.

There is no need to specify in which order they are written as the second row of the matrix gives the x-value, the third row gives the reciprocal of the y-value of the point in question.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 1→ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x → 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1/y → 4 6 3 5 1 2 2 4 6 3 5 1 3 5 1 2 4 6 y/x → 3 1 4 5 2 6 5 1 2 6 3 1 e4 6 3 1 4 5 2 22.3.4. Show that dim(B) = 1.

We know that 1, x, 1/y and y/x form a basis of L(2B). Then L(B) is contained in this space and contains the constants. Assume it is larger. This leads to a contradiction as the values of the basis elements at P0 for example are all different.

22.3.5. Prove the existence of a [65, 5, 52]8 -code.

This is a simple case of Construction X, using the chain [64, 4, 52]8 ⊂ [64, 5, 51]8 .

22.3.6. Show that a [70, 5, 56]8 -code exists.

This is another appliation of Construction X, using [64, 3, 56]8 ⊂ [64, 5, 51]8 and the auxiliary code [6, 2, 5]8 .

AG-codes, Weierstraß points and universal hashing

22.4

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The automorphism group

Exercises 22.4 22.4.1. Let K be a field containing an element of order 7. Show that the automorphisms σ of order 7 and ρ of order 3 of the Klein quartic together generate a group of order 21. Determine the conjugate ρ−1 σρ.

22.4.2. Show that the Klein quartic over F8 has GL(3, 2) as a group of automorphisms.

22.5

AG-codes and universal hashing

Exercises 22.5 22.5.1. Construct linear codes [27, 4, 21]9 and [27, 7, 18]9 .

These are the special cases of the Hermitian one-point codes [q 3 , i+1 + 2 1, q 3 − iq]q2 when q = 3 and i = 2, i = 3.

22.5.2. Which F16 -linear codes are obtained? q = 4 and i = 2, 3, 4 yields codes [64, 4, 56]16 , [64, 7, 52]16 , [64, 11, 48]16 .

Chapter 23 The last chapter

23.1

List decoding

Exercises 23.1 23.1.1. Show that the number of solutions (i, j) in non-negative integers to i + kj ≤ l is (bl/kc + 1)(2l + 2 − kbl/kc)/2.

j is between 0 P and t = bl/kc. The number of solutions is (l + 1)(t + 1) − k j j. The last sum is (t + 1)t/2.

23.2

Expander codes

23.3

tms-nets

P

j l + 1 − kj =

23.3.1. Prove that α can be chosen as required in this section.

We have q = 2r , L = Fq ⊂ F = efq2 and W ⊂ F the subgroup of order s = q + 1. Let T : F → L be the trace. For w ∈ W we have T (w) = w + 1/w.

The element α ∈ L \ F2 has to be chosen such that α2 + 1 does not have the form T (w) for w ∈ W. As T (1) = 0 and T (w) = T (1/w) there are q/2 elements of this form in L∗ . It follows that α can be chosen as required.

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23.4

Sphere packings

23.5

Permutation codes

Solutions Manual

Exercises 23.5 23.5.1. The alternating group A4 is sharply 2-transitive on 4 letters. Use it to construct a projective plane of order 4.

A4 consists of the identity permutation id, eight 3-cycles and three fixedpoint-free involutions (elements of order 2). The 16 lines not through ∞ are then (P1 , Q1 , id, (234), (243)), (P1 , Q2 , (123), (124), (12)(34)), (P1 , Q3 , (132), (134), (13)(24)), (P1 , Q4 , (142), (143), (14)(23)) (P2 , Q1 , (132), (142), (12)(34)), (P2 , Q2 , id, (134), (143)), (P2 , Q3 , (123), (234), (14)(23)), (P2 , Q4 , (124), (243), (13)(24)) (P3 , Q1 , (123), (143), (13)(24)), (P3 , Q2 , (132), (243), (14)(23)), (P3 , Q3 , id, (124), (142), ), (P3 , Q4 , (134), (234), (12)(34)) (P4 , Q1 , (124), (134), (14)(23)), (P4 , Q2 , (142), (234), (13)(24)), (P4 , Q3 , (143), (243), (12)(34)), (P4 , Q4 , id, (123), (132)) The lines through ∞ are the lines we started with: (∞, P1 , P2 , P3 , P4 ), (∞, Q1 , Q2 , Q3 , Q4 ), as well as (∞, id, (12)(34), (13)(24), (14)(23)), (∞, (123), (142), (243), (134)), (∞, (132), (124), (234), (143)).

Observe that pairs of permutations in one of the three last lines are at distance 4, so those lines correspond to Latin squares.