Editing BCH code (section)

==== Decoding of binary code without unreadable characters ====
Consider a BCH code in GF(2<sup>4</sup>) with <math>d=7</math> and <math>g(x) = x^{10} + x^8 + x^5 + x^4 + x^2 + x + 1</math>. (This is used in [[QR code]]s.) Let the message to be transmitted be <nowiki>[1 1 0 1 1]</nowiki>, or in polynomial notation, <math>M(x) = x^4 + x^3 + x + 1.</math>
The "checksum" symbols are calculated by dividing <math>x^{10} M(x)</math> by <math>g(x)</math> and taking the remainder, resulting in <math>x^9 + x^4 + x^2</math> or <nowiki>[ 1 0 0 0 0 1 0 1 0 0 ]</nowiki>. These are appended to the message, so the transmitted codeword is <nowiki>[ 1 1 0 1 1 1 0 0 0 0 1 0 1 0 0 ]</nowiki>.

Now, imagine that there are two bit-errors in the transmission, so the received codeword is [ 1 {{color|red|0}} 0 1 1 1 0 0 0 {{color|red|1}} 1 0 1 0 0 ]. In polynomial notation:
:<math>R(x) = C(x) + x^{13} + x^5 = x^{14} + x^{11} + x^{10} + x^9 + x^5 + x^4 + x^2</math>
In order to correct the errors, first calculate the syndromes. Taking <math>\alpha = 0010,</math> we have <math>s_1 = R(\alpha^1) = 1011,</math> <math>s_2 = 1001,</math> <math>s_3 = 1011,</math> <math>s_4 = 1101,</math> <math>s_5 = 0001,</math> and <math>s_6 = 1001.</math>
Next, apply the Peterson procedure by row-reducing the following augmented matrix.
:<math>\left [ S_{3 \times 3} | C_{3 \times 1} \right ] =
\begin{bmatrix}s_1&s_2&s_3&s_4\\
s_2&s_3&s_4&s_5\\
s_3&s_4&s_5&s_6\end{bmatrix} =
\begin{bmatrix}1011&1001&1011&1101\\
1001&1011&1101&0001\\
1011&1101&0001&1001\end{bmatrix} \Rightarrow
\begin{bmatrix}0001&0000&1000&0111\\
0000&0001&1011&0001\\
0000&0000&0000&0000
\end{bmatrix}</math>
Due to the zero row, {{math|''S''<sub>3×3</sub>}} is singular, which is no surprise since only two errors were introduced into the codeword.
However, the upper-left corner of the matrix is identical to {{closed-closed|''S''<sub>2×2</sub> <nowiki>|</nowiki> ''C''<sub>2×1</sub>}}, which gives rise to the solution <math>\lambda_2 = 1000,</math> <math>\lambda_1 = 1011.</math>
The resulting error locator polynomial is <math>\Lambda(x) = 1000 x^2 + 1011 x + 0001,</math> which has zeros at <math>0100 = \alpha^{-13}</math> and <math>0111 = \alpha^{-5}.</math>
The exponents of <math>\alpha</math> correspond to the error locations.
There is no need to calculate the error values in this example, as the only possible value is 1.