ICT12 8. Linear codes. The Gilbert-Varshamov lower bound and the MacWilliams identities SXD

Transcription

1 1 ICT12 8. Linear codes. The Gilbert-Varshamov lower bound and the MacWilliams identities SXD 8.1. The Gilbert Varshamov existence condition 8.2. The MacWilliams identities

2 The Gilbert Varshamov existence condition Theorem. Fix positive integers n, k, d such that k n and 2 d n + 1. If the relation vol q (n 1, d 2) < q n k is satisfied (this is called the Gilbert Varshamov condition), then there exists a linear code of type [n, k, d ] with d d. Proof. A) It is sufficient to see that the condition allows us to construct a matrix H M n n k (F) of rank n k with the property that any d 1 of its columns are linearly independent. Indeed, if this is the case, then the code C defined as the orthogonal space of the rows of H has length n, dimension k (from n (n k) = k), and minimum weight d d (since there are no linear relations of length less than d among the columns of H, which is a check matrix of C).

3 3 B) Before constructing H, we first establish that the Gilbert Varshamov condition implies that d 1 n k (note that this is the Singleton bound for the parameters d and k). Indeed, d 2 vol q (n 1, d 2) = n 1 j=0 (q 1) j j d 2 d 2 j=0 (q 1) j j = 1 + (q 1) d 2 = q d 2 and hence q d 2 < q n k if the Gilbert Varshamov condition is satisfied. Therefore d 2 < n k, or d 1 n k. C) In order to construct H, we first select a basis c 1,, c n k F n k. Since d 1 n k, any d 1 vectors extracted from this basis are lin-

4 4 early independent. Also, any matrix H of type (n k) m with entries in F which contains the columns c j T (j = 1,, n k), has rank n k. Now assume that we have constructed, for some i [n k, n], vectors c 1,, c i F n k with the property that any d 1 of them are linearly independent. If i = n, it is sufficient to take H = (c 1 T,, c n T ) and by part A our question is answered. Otherwise we will have i < n. In this case, the number of linear combinations that can be formed with at most d 2 vectors from among c 1,, c i is not greater than vol q (i, d 2) (see P.2.6). Since i n 1, vol q (i, d 2) vol q (n 1, d 2). If the Gilbert Varshamov condition is satisfied, then there is a vector c i+1 F n k which is not a linear combination of any subset of d 2 vectors extracted from the list c 1,, c i, and our claim follows by induction.

5 5 Remark. Since A q (n, d) A q (n, d ) if d d, the Gilbert--Varshamov condition shows that A q (n, d) q k. This lower bound, called the bert Varshamov bound, often can be used to improve the Gilbert bound. Example. Gilbert(10,3)=19, as in this case q n /vol q (n, d 1) = 2 10 /vol 2 (10,2) = 2 10 / And GilbertVarshamov(10,3)=64, as vol 2 (9,1) = 10 and 10 < = 16 while = 8 < 10.

6 The MacWilliams identities The weight enumerator of a code C is defined as the polynomial A(t) = n i=0 A i t i, where A i is the number of elements of C that have weight i. It is clear that A(t) can be expressed in the form A(t) = x C t x for the term t i appears in this sum as many times as the number of solutions of the equation x = i, x C. Remark. A 0 = 1, since 0 n is the unique element of C with weight 0. On the other hand A i = 0 if 0 < i < d, where d is the minimum distance of C. The determination of the other A i is not easy in general and it is one of the basic problems of coding theory.

7 7 Theorem (F. J. MacWilliams). 1 The weight enumerator B(t) of the dual code C of a code C of type [n, k] can be determined from the weight enumerator A(t) of C according to the following identity: q k B(t) = (1 + (q 1)t) n A 1 t. 1+(q 1)t Example (Weight enumerator of the Hamming codes). We know that the dual Hamming code Ham q (r) is equidistant, with minimum distance q r 1. This means that its weight enumerator, say B(t), is the polynomial B(t) = 1 + (q r 1)t qr 1, because q r 1 is the number of non-zero vectors in Ham q (r) and each of these has weight q r 1. Now the MacWilliams identity allows us to determine the weight enumerator A(t) of Ham q (r). Setting q = q 1, we have: q r A(t) = (1 + q t) n B 1 t 1+q t

8 8 = (1 + q t) n + (q r 1)(1 + q t) n (1+t)qr 1 (1+q t) qr 1 = (1 + q t) n + (q r 1)(1 + q t) n qr 1 (1 + t) qr 1 = (1 + q t) n qr 1 (1 + q t) qr 1 + (q r 1)(1 t) qr 1 = (1 + q t) qr 1 1 q 1 (1 + q t) qr 1 + (q r 1)(1 t) qr 1 since n = (q r 1)/(q 1) and n q r 1 = (q r 1 1)/(q 1). In the binary case the previous formula yields that the weight enumerator A(t) of Ham(r) satisfies 2 r A(t) = (1 + t) (2r 1 1) (1 + t) (2r 1) + (2 r 1)(1 t) 2r 1 For r = 3, Ham(3) has type [7,4] and 8A(t) = (1 + t) 3 ((1 + t) 4 + 7(1 t) 4 ), A(t) = 1 + 7t 3 + 7t 4 + t 7. So A 0 = A 7 = 1, A 3 = A 4 = 7, A 1 = A 2 = A 5 = A 6 = 0.

9 9 Actually it is easy to find, using the description of this code, that the weight 3 vectors of Ham(3) are [1,1,0,1,0,0,0], [0,1,1,0,1,0,0], [1,0,1,0,0,1,0], [0,0,1,1,0,0,1], [1,0,0,0,1,0,1], [0,1,0,0,0,1,1], [0,0,0,1,1,1,0] and the weight 4 vectors are [1,1,1,0,0,0,1], [1,0,1,1,1,0,0], [0,1,1,1,0,1,0], [1,1,0,0,1,1,0], [0,1,0,1,1,0,1], [1,0,0,1,0,1,1], [0,0,1,0,1,1,1] Note that the latter are obtained from the former, in reverse order, by interchanging 0 and 1. Notes 1.