Lecture 12: November 6, 2013

Transcription

1 Information an Coing Theory Autumn 204 Lecturer: Mahur Tulsiani Lecture 2: November 6, 203 Scribe: Davi Kim Recall: We were looking at coes of the form C : F k q F n q, where q is prime, k is the message length, an n is the block length of the coe We also saw C (its range) as a set in F n q an efine the istance of the coe as (C) := min { (x, y)} x,y C,x y where (x, y) is the Hamming istance between x an y We showe that a coe C can correct t errors iff (C) 2t + Hamming Coe The following is an example of the Hamming Coe from F 4 2 to F7 2 : Example Let C : F 4 2 F7 2, where C(x, x 2, x 3, x 4 ) = (x, x 2, x 3, x 4, x 2 + x 3 + x 4, x + x 3 + x 4, x + x 2 + x 4 ) Note that each element of the image is a linear function of the x i s, ie, one can express C with matrix multiplication as follows: x C(x, x 2, x 3, x 4 ) = x 2 0 x 3 0 x 4 0 Definition 2 (Linear Coes) A coe C : F k q F n q is a linear coe if for all u, v F k q an α F q, C(αu + v) = αc(u) + C(v), an thus the image of C is a subspace of F n q Since a linear coe is a linear map from a finite imensional vector space to another, we can write it as a matrix of finite size That is, there is a corresponing G F n k q st C(x) = Gx for all x F k q If the coe has nonzero istance, then the rank of G must be k (otherwise there exist x, y F k q such that Gx = Gy) Hence, the null space of G T has imension n k, so let b,, b n k be a basis of the null space of G T Definition 3 (Parity Check Matrix) Let b,, b n k be a basis for the null space of G T corresponing to a linear coe C Then H F (n k) n q, efine by H T = [ ] b b 2 b n k is calle the parity check matrix of C

2 Since G T H T = 0 HG = 0, we have (HG)x = 0 for all x F k q, ie, Hz = 0 for all z C Moreover, since the columns of H T are a basis for the null-space of G T, we have that z C Hz = 0 So the parity check matrix gives us a way to quickly check a coewor, by checking the parities of some bits of z (each row of H gives a parity constraint on z) Also, one can equivalently efine a linear coe by either giving G or the parity check matrix H Example 4 The parity check matrix of our example Hamming Coe is: H = Note that the i th column is the integer i in binary One can easily check that HG = 0 Now suppose z = (z,, z 7 ) T is our coewor an we make a single error in the i th entry Then the output coewor with the error is z 0 z + e i = z i + z 7 0 an H(z + e i ) = Hz + He i = He i = H i, the i th column of H, which reas i in binary So this is a very efficient ecoing algorithm just base on parity checking Since the Hamming coe (C) can correct at least t = errors, we must have that (C) 2t + = 3 Verify that the istance is exactly 3 using the following characterization of istance for linear coes Exercise 5 For z F n q, let wt(z) = {i [n] z i 0} Prove that for a linear coe C (C) = min wt(z) z C One can generalize the Hamming coe to larger message an block lengths, we can create a parity matrix H F (n k) n q, where the i th column reas i in binary 2 Hamming Boun We now show an optimality boun on the size of the coe, starting with the case of istance-3 coes an then generalizing to istance- coes 2

3 Theorem 2 Let C : F k 2 Fn 2 be any istance-3 coe, ie, (C) = 3 Then C = 2 k 2n n + Proof: For each z C, let B(z) be the ball of size n + consisting of z an the n elements in F n 2 (not in C), each at istance from z Then the balls forme by the coewors in C are isjoint, if B(z) an B(z ) intersect, then (z, z ) 2 by triangle inequality For each coewor z C, we have B(z) = n + coes, so C (n + ) 2 n Note that our example hamming coe from F 4 2 to F7 2 satisfie C = 24 = 27, so it was an optimal 8 istance-3 coe Generalizing to istance- coes, we have: Theorem 22 (Hamming Boun) Let C : F k 2 Fn 2 be any istance- coe, ie, (C) = Then C = 2 k 2 n ( ) vol B 2 ( where vol B z C 2 ) = 2 ( n ) i= i is the number of coes at istance at most 2 from any fixe coewor Remark 23 The Hamming boun also gives us a boun on the rate of the coe in terms of entropy (recall: the rate of the coe is k n ) Let = δn for δ 2 Since l ( n i= i) 2 nh( l n ) for l n 2, we have: k H(δ/2) + o() n 3 Ree-Solomon Coe We now look at Ree-Solomon coes over F q These are optimal coes which can achieve a very large istance However, they have a rawback that they nee q n Definition 3 (Ree-Solomon Coe) Assume q n an fix S = {a,, a n } F q, istinct st S = n For each message (m 0,, m k ) F k q, consier the polynomial P (x) = m 0 + m x + + m k x k Then the Ree-Solomon Coe is efine as: C(m 0,, m k ) = (P (a ),, P (a n )) Remark 32 Ree-Solomon Coes can again be encoe using inner coes to create concatenate coes, which can work with smaller q However, we will not iscuss these Let s compute the istance of the Ree-Solomon Coe: Claim 33 (C) n k + 3

4 Proof: Consier C(m 0,, m k ) with P (x) = m 0 +m x+ +m k x k an C(m 0,, m k ) with P (x) = m 0 + m x + + m k xk, where we assume (m 0,, m k ) (m 0,, m k ) an hence P P Then P P is a non-zero polynomial of egree at most k an has at most k roots, ie, P an P agree on at most k points This implies that C(m 0,, m k ) = (P (a ),, P (a n )) an C(m 0,, m k ) = (P (a ),, P (a n )) iffer on at least n k + places Since our choice of the messages was arbitrary, (C) n k + We now show that this is optimal: Theorem 34 (Singleton Boun) Let C : F k q F n q be a istance- coe Then n k + Proof: Consier the map Γ : F k q F n + q, where Γ(x) is the first n + coorinates of C(x) Then for all x y, we have Γ(x) Γ(y), since (C(x), C(y)) Therefore, img(γ) = F k q F n + q Remark 35 If n = 2k, then for = k +, we can correct k 2 errors using the Ree-Solomon Coe, an this is optimal Moreover, the Ree-Solomon Coe is a linear coe: a a 2 a k C(m 0,, m k ) = a n a 2 n a k n 4 Berlekamp-Welch Decoing Algorithm m 0 m m k If the coewors output by the Ree-Solomon coe i not contain any errors, we coul simply use Lagrange interpolation to recover the messages However, we must be able to hanle noise, an the trick is to work with a hypothetical error-locator polynomial telling us where the error is Definition 4 (Error-locator Polynomial) An error-locator polynomial, E(x), is a polynomial which satisfies i [n] E(a i ) = 0 y i P (a i ) for all i =,, n, where y i is the i th element of the output Ree-Solomon Coe obtaine with noise Note that this irectly implies that y i E(a i ) = P (a i )E(a i ) for all a i S Let s enote Q as Q(x) := P (x)e(x) In the next lecture, we will iscuss the following algorithm by Berelekemp an Welch [], which uses the above intuition to ecoe a message with at most (n k + )/2 errors Algorithm 42 Fin Q, E with eg(e) t an eg(q) k + t st y i E(a i ) = Q(a i ) for all i =,, n Output Q E 4

5 References [] LR Welch an ER Berlekamp Error correction for algebraic block coes, December US Patent 4,633,470 5