11.2 Label the vectors in the lexicode in the order in which they are generated so that c 0 is the zero vector.

(i) Lis a linear code and the vectorsc2i are a basis ofL.

(ii) After c2i is chosen, the next 2i−1 vectors generated are c₁+c₂i, c₂+c₂i, . . . ,

c2i₋₁+c₂i.

(iii) LetLi = {c0,c1, . . . ,c2i₋₁}. ThenL_iis an[n,i,d]linear code.

Proof: If we prove (ii), we have (i) and (iii). The proof of (ii) is by induction oni. Clearly it is true fori =1. Assume the first 2ivectors generated are as claimed. ThenLiis linear with basis{c1,c2, . . . ,c2i−1}. We show thatLi₊1is linear with the next 2i vectors generated in

order by addingc2i to each of the previously chosen vectors inL_i. If not, there is anr<2i

such thatc2i_+r =c₂i+cr. Chooserto be the smallest such value. As d(c2i +cr,c2i +cj)= d(cr,cj)≥d for j <r,c2i +c_r was a possible vector to be chosen. Since it was not, it

must have come too late in the lexicographic order; so

c2i_+r <c₂i +c_r. (2.46)

As d(cr+c2i_+r,c_j)=d(c₂i_+r,c_r+c_j)≥dfor j <2i by linearity ofL_i,c_r+c₂i_+r could

have been chosen to be in the code instead ofc2i (which it cannot equal). So it must be that

c2i <c_r+c₂i_+r. (2.47)

If j <r, then c2i_+j =c₂i +c_j by the assumption on r. Hence d(c₂i_+r +c₂i,c_j)=

d(c2i_+r,c₂i_+j)≥d for j<r. So c₂i_+r +c₂i could have been chosen to be a codeword

instead ofcr. The fact that it was not implies that

cr <c2i_+r+c₂i. (2.48)

But then (2.47) and (2.48) with Lemma 2.11.1 imply c2i +c_r <c₂i_+r contradicting

(2.46).

The codesLisatisfy the inclusionsL1⊂L2⊂ · · · ⊂Lk =L, wherekis the dimension ofL. In general this dimension is not known before the construction. Ifi <k, the left-most coordinates are always 0 (exactly how many is also unknown). If we punctureLion these zero coordinates, we actually get a lexicode of smaller length.

Exercise 145 Do the following:

(a) Construct the codesLi of length 5 and minimum distance 2.

(b) Verify that these codes are linear and the vectors are generated in the order described by Theorem 2.11.2.

(c) Repeat (a) and (b) for length 5 and minimum distance 3. Exercise 146 Find an ordering ofF5₂so that the greedy algorithm does not produce a linear

code.

Exercise 147 Prove that the covering radius ofLisd−1 or less. Also prove that the lexicodes meet the Gilbert Bound. Hint: See Exercise 132. The lexicodeLis the largest of theLi constructed in Theorem 2.11.2. We can give a parity check matrix forLprovidedd ≥3, which is reminiscent of the parity check matrix constructed in the proof of the Varshamov Bound. IfCis a lexicode of lengthnwithd≥3, construct its parity check matrixH=[hn· · ·h1] as follows (wherehiis a column vector). Regard the columnshias binary numbers where 1↔(· · ·001)T, 2↔(· · ·010)T, etc. Leth1

be the column corresponding to 1 andh2the column corresponding to 2. Oncehi−1, . . . ,h1

are chosen, choosehi to be the column corresponding to the smallest number which is not a linear combination ofd−2 or fewer ofhi₋1, . . . ,h1. Note that the length of the columns

does not have to be determined ahead of time. Wheneverhi corresponds to a number that is a power of 2, the length of the columns increases and zeros are placed on the tops of the columnshjfor j<i.

99 2.11 Lexicodes

Example 2.11.3 Ifn=7 andd =3, the parity check matrix for the lexicodeLis H=  11 11 10 10 01 01 00 1 0 1 0 1 0 1  _.

So we recognize this lexicode as the [7,4,3] Hamming code. As this example illustrates, the Hamming code H3 is a lexicode. In fact, all binary Hamming and Golay codes are lexicodes [39].

Exercise 148 Compute the parity check matrix for the lexicode of length 5 and minimum distance 3. Check that it yields the code produced in Exercise 145(c). Exercise 149 Compute the generator and parity check matrices for the lexicodes of length

For deeper analysis and construction of linear codes we need to make use of the basic theory of finite fields. In this chapter we review that theory. We will omit many of the proofs; for those readers interested in the proofs and other properties of finite fields, we refer you to [18, 196, 233].

3.1

Introduction

Afieldis a setFtogether with two operations:+, called addition, and·, called multiplication, which satisfy the following axioms. The setFis an abelian group under+with additive identity called zero and denoted 0; the set F∗ of all nonzero elements of F is also an abelian group under multiplication with multiplicative identity calledoneand denoted 1; and multiplication distributes over addition. We will usually omit the symbol for multiplication and writeabfor the producta·b. The field isfiniteifFhas a finite number of elements; the number of elements inFis called theorderofF. In Section 1.1 we gave three fields denotedF2,F3, andF4of orders 2, 3, and 4, respectively. In general, we will denote a field withq elements byFq; another common notation is GF(q) and read “the Galois field with qelements.”

If p is a prime, the integers modulo pform a field, which is then denotedFp. This is not true ifpis not a prime. These are the simplest examples of finite fields. As we will see momentarily, every finite field contains someFpas a subfield.

Exercise 150 Prove that the integers modulondo not form a field ifnis not prime. The finiteness ofFq implies that there exists a smallest positive integer p such that 1+ · · · +1 (p1s) is 0. The integerpis a prime, as verified in Exercise 151, and is called thecharacteristicofFq. Ifa is a positive integer, we will denote the sum ofa 1s in the field bya. Also if we wish to write the sum ofa αs whereαis in the field, we write this as eitheraα ora·α. Notice that pα=0 for allα∈Fq. The set of p distinct elements {0,1,2, . . . ,(p−1)}ofFq is isomorphic to the fieldFp of integers modulo p. As a field isomorphic toFp is contained inFq, we will simplify terminology and say thatFp is a subfield ofFq; this subfield Fp is called theprime subfieldof Fq. The fact thatFp is a subfield ofFq gives us crucial information aboutq; specifically, by Exercise 151, the field

Fqis also a finite dimensional vector space overFp, say of dimensionm. Thereforeq =pm as this is the number of vectors in a vector space of dimensionmoverFp.

101 3.2 Polynomials and the Euclidean Algorithm

Although it is not obvious, all finite fields with the same number of elements are iso- morphic. Thus our notationFq is not ambiguous;Fq will be any representation of a field withqelements. As we did with the prime subfield ofFq, if we say thatFr is a subfield of

Fq, we actually mean thatFqcontains a subfield withrelements. IfFq has a subfield with r elements, that subfield is unique. Hence there is no ambiguity when we say thatFr is a subfield ofFq. It is important to note that although all finite fields of orderqare isomorphic, one field may have many different representations. The exact form that we use for the field may be crucial in its application to coding theory. We summarize the results we have just given in a theorem; all but the last part are proved in Exercise 151.

Theorem 3.1.1 LetFqbe a finite field with q elements. Then: