11 Minimal Distance and the Parity Check Matrix

Transcription

1 MATH32031: Coding Theory Part 12: Hamming Codes 11 Minimal Distance and the Parity Check Matrix Theorem 23 (Distance Theorem for Linear Codes) Let C be an [n, k] F q -code with parity check matrix H. Then d(c) = d if and only if all sets of d 1 columns of H are linearly independent and some set of d columns of H is linearly dependent. Remark. This condition implies that column rank (H) d 1. Proof. We know from the theorem in Part 5 that d(c) is the minimal weight of the non-zero codewords. Recall that for x V : x C if and only if xh T = 0 if and only if xi h i = 0 where h i denotes the ith column of H. Thus we have a linear relation with all coefficients non-zero between w columns of H if and only if there exists a codeword x of weight w. Since d is the minimal such non-zero weight, and is achieved, d is the minimal number of columns required for linear dependence, and there is such a set of columns. Example. The [4, 2] F 2 -code C with generator matrix [ ] G = has standard form [ and so has parity check matrix [ So d = 2 in this case, as we have two equal columns. We can come to the same conclusion using other means C = and apply the fact that d(c) = w(c) (from a Theorem in Part 5). 39 ], ].

2 12 Hamming codes These, together with cyclic codes, form the high-point of the course. These two techniques generate between them nearly all the codes that we shall meet. We start with the definition of proective (n 1)-space over a field K, which we denote by P n 1 (K). Consider all vectors v K (n), v 0. We introduce a relation on K (n) \ {0} by saying This is an equivalence relation, since: (1) v = 1 v (reflexive) v u if and only if v = λu for some λ K. (2) v = λu implies u = λ 1 v (symmetric) (3) v = λu, u = µw implies v = λµw (transitive). We define P n 1 (K) = (K (n) \ {0})/ (the set of equivalence classes). Geometrically, P n 1 (K) can be thought of as the set of (undirected) lines in K (n). Examples. (a) P 1 (R) is the unit circle with antipodes identified. P 2 (R) is the unit sphere with antipodes identified. (b) P n 1 (F 2 ) = F (n) 2 \ {0} (since 1 is the only non-zero scalar in F 2 ). In particular, (c) P 1 (F 3 ) = {(0, 1), (1, 0), (1, 2), (1, 1)}. Lemma 24 P n 1 (F q ) = qn 1 q 1. Proof. P 1 (F 2 ) = {(0, 1), (1, 1), (1, 0)}. F (n) q \ {0} = q n 1 and each equivalence class contains (F q ) = q 1 distinct elements (where (F q ) is the multiplicative group of F q as usual). Definition of Hamming Codes. Let s 1 be given. We let Ham(s, q) denote the F q -linear code whose parity check matrix has columns corresponding to the elements P s 1 (F q ), each column being a representative of the equivalence class. 40

3 Remarks. (1) Clearly Ham(s, q) is defined only up to linear equivalence, for we can alter the order of the columns and the representatives are only well defined only up to scalar multiplication. Thus we should strictly say a member of the set of linearly equivalent Hamming codes of certain parameters, rather than the Hamming code of those parameters. (2) Such a parity check matrix has dimensions (see Lemma 24). Set (length Ham(s, q)). s qs 1 q 1 n = qs 1 q 1 (3) Since the columns of the parity check matrix H can be reordered to begin λ λ with λ i λ s... we see that the rows of H are linearly independent. Since H has row rank s, it follows that Ham(s, q) has dimension n s. The principal properties of Hamming codes are summarised in the next theorem. Theorem 25 (a) Ham(s, q) is an F q -code with parameters [n, n s, 3] where n = qs 1 q 1. (b) Ham(s, q) is a perfect, single error correcting code. Remark. Since Ham(s, q) has length n, but is n s dimensional, we see that Ham(s, q) has redundancy n (n s) = s, (which is to say that there are s check symbols). Proof. (a) By the above remarks it only remains to show that d(ham(s, q)) = 3. By the Distance Theorem for Linear Codes (Theorem 23),it suffices to prove that no two columns in the parity check matrix H are linearly dependent but that some 3 columns are linearly dependent. To see this consider two distinct columns x, y. If they are linearly dependent then x = λy, i.e. they represent the same proective point in P s 1 (F q ). But this is a contradiction. On the other hand, the vectors (a00 0), (0b0 0) and (cc0 0) all appear as columns, for some choice of a, b, c. They are clearly linearly dependent but define three distinct points in proective space. 41

4 (b) By the result relating the error-correcting power of a code with its minimal distance (Theorem 2), we see that since d(ham(s, q)) = 3 it corrects single errors. Recall that to show that this code is perfect, we need to show that the Hamming bound is attained; i.e. (see Lecture 5) { t ( ) } n M (q 1) i = q n, i i=0 i.e., since t = 1 and the code is linear, q (n s) {1 + n(q 1)} = q n. But so n = qs 1 q 1, q (n s) {1 + q s 1} = q n. Recall that S t (0) denotes the set of vectors of weight at most t. Proposition 26 For a perfect code with d = 2t + 1, the coset leaders are exactly the points in S t (0). Proof. A linear code is perfect if and only if q k S t (0) = q n, in our usual notation, so S t (0) = q n k, which is equal to the number of cosets. By the pigeonhole principle, we only need to show that each element of S t (0) is in a different coset. But if x and y are vectors satisfying w(x) t, w(y) t then d(x, y) d(0, x) + d(0, y) = w(x) + w(y) 2t < d, so x = y. Corollary. For C = Ham(s, q) the coset leaders for the non-trivial cosets are precisely those vectors with exactly one non-zero co-ordinate. Example. matrix For this we reconsider the binary code Ham(3, 2). By definition it has parity H = , so from the relations between the generator and parity check matrices (Theorem 21 in Part 10), we see that G = Then by Theorem 25 this is a [7, 4, 3]-binary code. In particular, we know that d(c) = 3 straight away without examining the weights of the 15 non-zero vectors. 42

5 Example. Ham(2, 2). By the corollary this is a [3, 1, 3] code. By definition it has parity check matrix [ ] H =, so G = [ ], i.e. C is a repetition code. 13 Decoding Not only do Hamming codes provide an elegant way of generating perfect codes with minimal distance 3, they also have great advantages for syndrome decoding. We briefly recall that the syndrome decode scheme is. 1. For a received vector y, calculate S(y) = yh T. 2. Find the coset leader a with S(y) = S(a). Thus 3. Decode to y a. However, for a Hamming code, if a 0 then, by Proposition 26, a = (0,..., 0, λ, 0,..., 0), λ 0. S(a) = ah T = λh T where h is the th column of H. So for a Hamming code we may proceed as follows. We do not need to do any work to calculate the syndromes and coset leaders. 1. Calculate S(y). 2. Check S(y) against the columns of H and find S(y) = λh for some, some λ F. We then know by ( ) that S(y) = S(a) for a the coset leader ( ) a = (0,..., 0, λ, 0,..., 0), 3. In order to decode ust subtract λ from the -th co-ordinate of y. 43

6 14 Construction of codes with large distance We can obviously try and use the criterion provided by the Distance Theorem for Linear Codes to generate codes with large distance. Here we content ourselves with the following sufficient condition. Theorem 27 Let n > k, d n k and suppose that ( n 1 (q 1) i i i=0 Then there exists an [n, k, d ] F q -code with d d. ) < q n k. Proof. For ease of notation we write r = n k, W = F q (r). We let the first r columns of H be given by I r. This ensures that H has row rank r and so defines an n r = k dimensional code. We need to show that we can find k = n r further column vectors with the property that no d 1 subset of these columns are linearly dependent. We now consider the problem of choosing an i-th column (for i > r) given that i 1 columns have been chosen and have the desired property. So here we are not allowed to choose any of the linear combinations (with all non-zero coefficients) of the d 2 (and fewer) selections of the i 1 columns already chosen (Of course in general not all these combinations are distinct so we are overestimating). We denote this number of forbidden choices by N(i): for each : 1 d 2 we count that there are ( ) i 1 (q 1) non-zero linear combinations of the possible selections from the (i 1) columns. So ( ) i 1 ( ) i 1 N(i) = 1 + (q 1) = (q 1). =1 (1 for 0, which is also forbidden). Notice that ( i 1 =0 ( ) n 1 N(i) (q 1). =0 ) ( n 1 ), provided i n, so Thus, by our hypothesis, N(i) < q (n k) = q r, which is the total number of column vectors of the right length, so a choice of ith column is always possible for r < i n. By induction we see that we can choose the n columns that we need. 44