# lect5 - Error Correcting Codes Combinatorics Algorithms and...

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 -
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 10/16/2011 for the course ELECTRICAL EE251202 taught by Professor Rejaei during the Spring '10 term at Sharif University of Technology.

### Page1 / 5

lect5 - Error Correcting Codes Combinatorics Algorithms and...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online