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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 ) {\displaystyle b_{i}E(a_{i})=E(a_{i})F(a_{i})}

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():

Q ( a i ) = E ( a i ) F ( a i ) {\displaystyle Q(a_{i})=E(a_{i})F(a_{i})}

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.

b i E ( a i ) = Q ( a i ) {\displaystyle b_{i}E(a_{i})=Q(a_{i})} b i E ( a i ) − Q ( a i ) = 0 {\displaystyle b_{i}E(a_{i})-Q(a_{i})=0} b i ( e 0 + e 1 a i + e 2 a i 2 + ⋯ + e e a i e ) − ( q 0 + q 1 a i + q 2 a i 2 + ⋯ + q q a i q ) = 0 {\displaystyle 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 ee is constrained to be 1, the equations become:

b i ( e 0 + e 1 a i + e 2 a i 2 + ⋯ + e e − 1 a i e − 1 ) − ( q 0 + q 1 a i + q 2 a i 2 + ⋯ + q q a i q ) = − b i a i e {\displaystyle 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 ) {\displaystyle 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:

[ 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 ] [ e 0 e 1 q 0 q 1 q 2 q 3 q 4 ] = [ − 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 ] {\displaystyle {\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}}} [ 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 ] [ e 0 e 1 q 0 q 1 q 2 q 3 q 4 ] = [ 0 2 2 2 1 6 5 ] {\displaystyle {\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}}} [ 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 ] [ e 0 e 1 q 0 q 1 q 2 q 3 q 4 ] = [ 4 2 4 3 3 1 3 ] {\displaystyle {\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 e2 = 1:

Q ( a i ) = 3 x 4 + 1 x 3 + 3 x 2 + 3 x + 4 {\displaystyle Q(a_{i})=3x^{4}+1x^{3}+3x^{2}+3x+4}

E ( a i ) = 1 x 2 + 2 x + 4 {\displaystyle E(a_{i})=1x^{2}+2x+4}

F ( a i ) = Q ( a i ) / E ( a i ) = 3 x 2 + 2 x + 1 {\displaystyle F(a_{i})=Q(a_{i})/E(a_{i})=3x^{2}+2x+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

• Reed–Solomon error correction

External links

• MIT Lecture Notes on Essential Coding Theory – Dr. Madhu Sudan
• University at Buffalo Lecture Notes on Coding Theory – Dr. Atri Rudra
• Algebraic Codes on Lines, Planes and Curves, An Engineering Approach – Richard E. Blahut
• Welch Berlekamp Decoding of Reed–Solomon Codes – L. R. Welch
• US 4,633,470, Welch, Lloyd R. & Berlekamp, Elwyn R., "Error Correction for Algebraic Block Codes", published September 27, 1983, issued December 30, 1986  – The patent by Lloyd R. Welch and Elewyn R. Berlekamp