lect12 - Error Correcting Codes: Combinatorics, Algorithms...

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Error Correcting Codes: Combinatorics, Algorithms and Applications (Fall 2007) Lecture 12: Reed-Solomon Codes September 28, 2007 Lecturer: Atri Rudra Scribe: Michel Kulhandjian Last lecture we saw the proof of the Singleton bound which claims that for any ( n, k, d ) q code, k n d + 1 . In today’s lecture we will study Reed-Solomon codes. These codes meet the Singleton bound, i.e. satisfy k = n d + 1 (but have the unfortunate property that q n ). Note that this implies that the Singleton bound is tight, at least for q n . 1 Reed-Solomon Codes We begin with the definition of Reed-Solomon codes. Definition 1.1 (Reed-Solomon code) . Let F q be a finite field and F q [ x ] denote the F q -space of univariate polynomials where all the coefficients of x are from F q . Pick { α 1 , α 2 , ...α n } distinct elements (also called evaluation points ) of F q and choose n and k such that k n q . We define an encoding function for Reed-Solomon code as RS : F k q F n q as follows. A message m = ( m 0 , m 1 , ..., m k - 1 ) with m i F q is mapped to a degree k 1 polynomial. m m→ f m ( x ) , where f m ( x ) = k - 1 s i =0 m i x i . (1) Note that f m ( x ) F q [ x ] is a polynomial of degree k 1 . The encoding of m is the evaluation of f m ( x ) at all the α i ’s : RS ( m ) = a f m ( α 1 ) , f m ( α 2 ) , ..., f m ( α n ) A We call this image Reed-Solomon code or RS code after two inventors Irving Reed and Gus Solomon of this code [1]. A common special case is n = q 1 with the set of evaluation points being F * d F \ { 0 } . Notice that by definition, the entries in { α 1 , ..., α n } are distinct and thus, must have n q . We now turn to some properties of Reed-Solomon codes. Claim 1.2. RS codes are linear codes. 1
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Proof. The proof follows from the fact that if a F q and f ( x ) , g ( x ) F q [ x ] are polynomials of degree k 1 , then af ( x ) and f ( x ) + g ( x ) are also polynomials of degree k 1 . In particular, let messages m 1 and m 2 be mapped to f m 1 ( x ) and f m 2 ( x ) where f m 1 ( x ) , f m 2 ( x ) F q [ x ] are polynomials of degree k 1 and because of the mapping defined in (1), it is easy to verify that: f m 1 ( x ) + f m 2 ( x ) = f m 1 + m 2 ( x ) , and af m 1 ( x ) = f a m 1 ( x ) . Therefore,
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lect12 - Error Correcting Codes: Combinatorics, Algorithms...

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