Berlekamp–Welch algorithm

The Berlekamp–Welch algorithm, also known as the Welch–Berlekamp algorithm, is named for Elwyn R. Berlekamp and Lloyd R. Welch. This is a decoder algorithm that efficiently corrects errors in Reed–Solomon codes for an RS(n, k), code based on the Reed Solomon original view where a message is used as coefficients of a polynomial or used with Lagrange interpolation to generate the polynomial of degree < k for inputs and then is applied to to create an encoded codeword .

The goal of the decoder is to recover the original encoding polynomial , using the known inputs and received codeword with possible errors. It also computes an error polynomial where corresponding to errors in the received codeword.

The key equations

Defining e = number of errors, the key set of n equations is

Where E(ai) = 0 for the e cases when bi F(ai), and E(ai) 0 for the n - e non error cases where bi = F(ai) . These equations can't be solved directly, but by defining Q() as the product of E() and F():

and adding the constraint that the most significant coefficient of E(ai) = ee = 1, the result will lead to a set of equations that can be solved with linear algebra.

where q = n - e - 1. Since ee is constrained to be 1, the equations become:

resulting in a set of equations which can be solved using linear algebra, with time complexity O(n^3).

The algorithm begins assuming the maximum number of errors e = ⌊ (n-k)/2 ⌋. If the equations can not be solved (due to redundancy), e is reduced by 1 and the process repeated, until the equations can be solved or e is reduced to 0, indicating no errors. If Q()/E() has remainder = 0, then F() = Q()/E() and the code word values F(ai) are calculated for the locations where E(ai) = 0 to recover the original code word. If the remainder 0, then an uncorrectable error has been detected.

Example

Consider RS(7,3) (n = 7, k = 3) defined in GF(7) with α = 3 and input values: ai = i-1 : {0,1,2,3,4,5,6}. The message to be systematically encoded is {1,6,3}. Using Lagrange interpolation, F(ai) = 3 x2 + 2 x + 1, and applying F(ai) for a4 = 3 to a7 = 6, results in the code word {1,6,3,6,1,2,2}. Assume errors occur at c2 and c5 resulting in the received code word {1,5,3,6,3,2,2}. Start off with e = 2 and solve the linear equations:



Starting from the bottom of the right matrix, and the constraint e2 = 1:

with remainder = 0.

E(ai) = 0 at a2 = 1 and a5 = 4 Calculate F(a2 = 1) = 6 and F(a5 = 4) = 1 to produce corrected code word {1,6,3,6,1,2,2}.

See also

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