# 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 $m_{1},\cdots ,m_{k}$ is used as coefficients of a polynomial $F(a_{i})$ or used with Lagrange interpolation to generate the polynomial $F(a_{i})$ of degree < k for inputs $a_{1},\cdots ,a_{k}$ and then $F(a_{i})$ is applied to $a_{k+1},\cdots ,a_{n}$ to create an encoded codeword $c_{1},\cdots ,c_{n}$.
The goal of the decoder is to recover the original encoding polynomial $F(a_{i})$, using the known inputs $a_{1},\cdots ,a_{n}$ and received codeword $b_{1},\cdots ,b_{n}$ with possible errors. It also computes an error polynomial $E(a_{i})$ where $E(a_{i})=0$ corresponding to errors in the received codeword.

## The key equations

Defining e = number of errors, the key set of n equations is
    $b_{i}E(a_{i})=E(a_{i})F(a_{i})$
Where E(a$_i$) = 0 for the e cases when b$_i$ ≠ F(a$_i$), and E(a$_i$) ≠ 0 for the n - e non error cases where b$_i$ = F(a$_i$) . These equations can't be solved directly, but by defining Q as the product of E and F:
    $Q(a_{i})=E(a_{i})F(a_{i})$
and adding the constraint that the most significant coefficient of E(a$_i$) = e$_e$ = 1, the result will lead to a set of equations that can be solved with linear algebra.
    $b_{i}E(a_{i})=Q(a_{i})$
    $b_{i}E(a_{i})-Q(a_{i})=0$
    $b_{i}(e_{0}+e_{1}a_{i}+e_{2}a_{i}^{2}+\cdots +e_{e}a_{i}^{e})-(q_{0}+q_{1}a_{i}+q_{2}a_{i}^{2}+\cdots +q_{q}a_{i}^{q})=0$
where q = n - e - 1. Since e$_e$ is constrained to be 1, the equations become:
    $b_{i}(e_{0}+e_{1}a_{i}+e_{2}a_{i}^{2}+\cdots +e_{e-1}a_{i}^{e-1})-(q_{0}+q_{1}a_{i}+q_{2}a_{i}^{2}+\cdots +q_{q}a_{i}^{q})=-b_{i}a_{i}^{e}$
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(a$_i$) are calculated for the locations where E(a$_i$) = 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: a$_i$ = i-1 : {0,1,2,3,4,5,6}. The message to be systematically encoded is {1,6,3}. Using Lagrange interpolation, F(a$_i$) = 3 x$^2$ + 2 x + 1, and applying F(a$_i$) for a$_4$ = 3 to a$_7$ = 6, results in the code word {1,6,3,6,1,2,2}. Assume errors occur at c$_2$ and c$_5$ resulting in the received code word {1,5,3,6,3,2,2}. Start off with e = 2 and solve the linear equations:
    ${\begin{bmatrix}b_{1}&b_{1}a_{1}&-1&-a_{1}&-a_{1}^{2}&-a_{1}^{3}&-a_{1}^{4}\\b_{2}&b_{2}a_{2}&-1&-a_{2}&-a_{2}^{2}&-a_{2}^{3}&-a_{2}^{4}\\b_{3}&b_{3}a_{3}&-1&-a_{3}&-a_{3}^{2}&-a_{3}^{3}&-a_{3}^{4}\\b_{4}&b_{4}a_{4}&-1&-a_{4}&-a_{4}^{2}&-a_{4}^{3}&-a_{4}^{4}\\b_{5}&b_{5}a_{5}&-1&-a_{5}&-a_{5}^{2}&-a_{5}^{3}&-a_{5}^{4}\\b_{6}&b_{6}a_{6}&-1&-a_{6}&-a_{6}^{2}&-a_{6}^{3}&-a_{6}^{4}\\b_{7}&b_{7}a_{7}&-1&-a_{7}&-a_{7}^{2}&-a_{7}^{3}&-a_{7}^{4}\\\end{bmatrix}}{\begin{bmatrix}e_{0}\\e_{1}\\q0\\q1\\q2\\q3\\q4\\\end{bmatrix}}={\begin{bmatrix}-b_{1}a_{1}^{2}\\-b_{2}a_{2}^{2}\\-b_{3}a_{3}^{2}\\-b_{4}a_{4}^{2}\\-b_{5}a_{5}^{2}\\-b_{6}a_{6}^{2}\\-b_{7}a_{7}^{2}\\\end{bmatrix}}$
    ${\begin{bmatrix}1&0&6&0&0&0&0\\5&5&6&6&6&6&6\\3&6&6&5&3&6&5\\6&4&6&4&5&1&3\\3&5&6&3&5&6&3\\2&3&6&2&3&1&5\\2&5&6&1&6&1&6\\\end{bmatrix}}{\begin{bmatrix}e_{0}\\e_{1}\\q0\\q1\\q2\\q3\\q4\\\end{bmatrix}}={\begin{bmatrix}0\\2\\2\\2\\1\\6\\5\\\end{bmatrix}}$
    ${\begin{bmatrix}1&0&0&0&0&0&0\\0&1&0&0&0&0&0\\0&0&1&0&0&0&0\\0&0&0&1&0&0&0\\0&0&0&0&1&0&0\\0&0&0&0&0&1&0\\0&0&0&0&0&0&1\\\end{bmatrix}}{\begin{bmatrix}e_{0}\\e_{1}\\q0\\q1\\q2\\q3\\q4\\\end{bmatrix}}={\begin{bmatrix}4\\2\\4\\3\\3\\1\\3\\\end{bmatrix}}$
Starting from the bottom of the right matrix, and the constraint e$_2$ = 1:
$Q(a_{i})=3x^{4}+1x^{3}+3x^{2}+3x+4$
$E(a_{i})=1x^{2}+2x+4$
$F(a_{i})=Q(a_{i})/E(a_{i})=3x^{2}+2x+1$ with remainder = 0.
E(a$_i$) = 0 at a$_2$ = 1 and a$_5$ = 4 Calculate F(a$_2$ = 1) = 6 and F(a$_5$ = 4) = 1 to produce corrected code word {1,6,3,6,1,2,2}.