lect5 - Error Correcting Codes: Combinatorics, Algorithms...

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Error Correcting Codes: Combinatorics, Algorithms and Applications (Fall 2007) Lecture 5: Linear Codes September 7, 2007 Lecturer: Atri Rudra Last lecture we talked about the Hamming bound (for the special case of d = 3 ) and informally defined binary linear codes. In today’s lecture, we will present the general form of Hamming bound and define (general) linear codes. The latter will involve looking at finite fields and vector spaces over finite fields. 1 General Family of Hamming Codes We start with a new notation. Definition 1.1. A code C Σ n with dimension k and distance d will be called a ( n, k, d ) Σ code. We will also refer it to as a ( n, k, d ) | Σ | code. We now state the Hamming bound for d = 3 that we proved in the last lecture. Theorem 1.2 (Hamming Bound for d = 3 ) . For every ( n, k, 3) 2 code k n - log 2 ( n + 1) . We now proceed to generalize the above theorem to any distance d . Theorem 1.3 (Hamming Bound for any d ) . For every ( n, k, d ) q code k n - log q b ( d - 1) 2 c X i =0 ± n i ² ( q - 1) i . Proof. The proof is a straightforward generalization of the proof of Theorem 1.2 that we did last lecture. For notational convenience, let e = j ( d - 1) 2 k . Given any two codewords, c 1 6 = c 2 C , the following is true (as C has distance 1 d ): B ( c 1 , e ) B ( c 2 , e ) = , (1) where recall that for any vector x [ q ] n , B ( x , e ) = { y [ q ] n | Δ( x , y ) e } . 1 Assume that y B ( c 1 , e ) B ( c 2 , e ) , that is Δ( y , c 1 ) e and Δ( y , c 2 ) e . Thus, by the triangle inequality, Δ( c 1 , c 2 ) 2 e d - 1 , which is a contradiction. 1
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We claim that for all x [ q ] n , | B ( x , e ) | = e X i =0 ± n i ² ( q - 1) i . (2) Indeed any vector in B ( x , e ) must differ from x in exactly 0 i e positions. In the summation ( n i ) is the number of ways of choosing the differing i positions and in each such position, a vector can differ from x in q - 1 ways. Now consider the union of all Hamming balls centered around some codeword. Obviously their union is a subset of [ q ] n . In other words, ³ ³ ³ ³ ³ [ c C B ( c, e ) ³ ³ ³ ³ ³ q n . (3) As (1) holds for every pair of distinct codewords, ³ ³ ³ ³ ³ [ c C B ( c, e ) ³ ³ ³ ³ ³ = X c C | B ( c, e ) | = q k e X i =0 ± n i ² ( q - 1) i , (4) where (4) follows from (2) and the fact that C has dimension k . Combining (4) and (3) and taking log q of both sides we will get the desired bound: k n - log q ´ e X i =0 ± n i ² ( q -
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lect5 - Error Correcting Codes: Combinatorics, Algorithms...

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