Editing BCH code (section)

==== Decoding with unreadable characters with a small number of errors ====
Let us show the algorithm behaviour for the case with small number of errors. Let the received word is [ 1 {{color|red|0}} 0 ? 1 1 ? 0 0 0 1 0 1 0 0 ].

Again, replace the unreadable characters by zeros while creating the polynomial reflecting their positions <math>\Gamma(x) = \left(\alpha^{8}x - 1\right)\left(\alpha^{11}x - 1\right).</math>
Compute the syndromes <math>s_1 = \alpha^{4}, s_2 = \alpha^{-7}, s_3 = \alpha^{1}, s_4 = \alpha^{1}, s_5 = \alpha^{0},</math> and <math>s_6 = \alpha^{2}.</math>
Create syndrome polynomial

:<math>\begin{align}
           S(x) &= \alpha^{4} + \alpha^{-7}x + \alpha^{1}x^2 + \alpha^{1}x^3 + \alpha^{0}x^4 + \alpha^{2}x^5, \\
  S(x)\Gamma(x) &= \alpha^{4} + \alpha^{7}x + \alpha^{5}x^2 + \alpha^{3}x^3 + \alpha^{1}x^4 + \alpha^{-1}x^5 + \alpha^{-1}x^6 + \alpha^{6}x^7.
\end{align}</math>

Let us run the extended Euclidean algorithm:

:<math>\begin{align}
  \begin{pmatrix}
    S(x)\Gamma(x) \\
    x^6
  \end{pmatrix}
    &= \begin{pmatrix}
         \alpha^{4} + \alpha^{7}x + \alpha^{5}x^2 + \alpha^{3}x^3 + \alpha^{1}x^4 + \alpha^{-1}x^5 + \alpha^{-1}x^6 + \alpha^{6}x^7 \\
         x^6
       \end{pmatrix} \\
    &= \begin{pmatrix}
         \alpha^{-1} + \alpha^{6}x & 1 \\
                                 1 & 0
       \end{pmatrix} \begin{pmatrix}
         x^6 \\
         \alpha^{4} + \alpha^{7}x + \alpha^{5}x^2 + \alpha^{3}x^3 + \alpha^{1}x^4 + \alpha^{-1}x^5 + 2\alpha^{-1}x^6 + 2\alpha^{6}x^7
       \end{pmatrix} \\
    &= \begin{pmatrix}
         \alpha^{-1} + \alpha^{6}x & 1 \\
                                 1 & 0
       \end{pmatrix} \begin{pmatrix}
         \alpha^{3} + \alpha^{1}x & 1 \\
                                1 & 0
       \end{pmatrix} \begin{pmatrix}
         \alpha^{4} + \alpha^{7}x + \alpha^{5}x^2 + \alpha^{3}x^3 + \alpha^{1}x^4 + \alpha^{-1}x^5 \\
         \alpha^{7} + \left(\alpha^{-5} + \alpha^{5}\right)x + 2\alpha^{-7}x^2 + 2\alpha^{6}x^3 +  2\alpha^{4}x^4 + 2\alpha^{2}x^5 + 2x^6
       \end{pmatrix} \\
    &= \begin{pmatrix}
         \left(1 + \alpha^{2}\right) + \left(\alpha^{0} + \alpha^{-6}\right)x + \alpha^{7}x^2 & \alpha^{-1} + \alpha^{6}x \\
         \alpha^{3} + \alpha^{1}x & 1
       \end{pmatrix} \begin{pmatrix}
         \alpha^{4} + \alpha^{7}x + \alpha^{5}x^2 + \alpha^{3}x^3 + \alpha^{1}x^4 + \alpha^{-1}x^5 \\
         \alpha^{7} + \alpha^{0}x
       \end{pmatrix}
\end{align}</math>

We have reached polynomial of degree at most 3, and as
: <math>
  \begin{pmatrix}
                          -1 & \alpha^{-1} + \alpha^{6}x \\
    \alpha^{3} + \alpha^{1}x & -\left(\alpha^{-7} + \alpha^{7}x + \alpha^{7}x^2\right)
  \end{pmatrix} \begin{pmatrix}
    \alpha^{-7} + \alpha^{7}x + \alpha^{7}x^2 & \alpha^{-1} + \alpha^{6}x \\
                     \alpha^{3} + \alpha^{1}x & 1
  \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},
</math>
we get
: <math>
  \begin{pmatrix}
                          -1 & \alpha^{-1} + \alpha^{6}x \\
    \alpha^{3} + \alpha^{1}x & -\left(\alpha^{-7} + \alpha^{7}x + \alpha^{7}x^2\right)
  \end{pmatrix}\begin{pmatrix}
    S(x)\Gamma(x) \\ x^6
  \end{pmatrix} = \begin{pmatrix}
    \alpha^{4} + \alpha^{7}x + \alpha^{5}x^2 + \alpha^{3}x^3 + \alpha^{1}x^4 + \alpha^{-1}x^5 \\
    \alpha^{7} + \alpha^{0}x
  \end{pmatrix}.
</math>

Therefore,
: <math>S(x)\Gamma(x)\left(\alpha^{3} + \alpha^{1}x\right) - \left(\alpha^{-7} + \alpha^{7}x + \alpha^{7}x^2\right)x^6 = \alpha^{7} + \alpha^{0}x.</math>

Let <math>\Lambda(x) = \alpha^{3} + \alpha^{1}x.</math> Don't worry that <math>\lambda_0 \neq 1.</math> The root of <math>\Lambda(x)</math> is <math>\alpha^{3-1}.</math>

Let 
:<math>\begin{align}
     \Xi(x) &= \Gamma(x)\Lambda(x) = \alpha^{3} + \alpha^{-7}x + \alpha^{-4}x^2 + \alpha^{5}x^3, \\
  \Omega(x) &= S(x)\Xi(x) \equiv \alpha^{7} + \alpha^{0}x \bmod{x^6}
\end{align}</math>

Let us look for error values using formula <math>e_j = -\Omega\left(\alpha^{-i_j}\right)/\Xi'\left(\alpha^{-i_j}\right),</math> where <math>\alpha^{-i_j}</math> are roots of polynomial <math>\Xi(x).</math>
: <math>\Xi'(x) = \alpha^{-7} + \alpha^{5}x^2.</math>

We get
:<math>\begin{align}
  e_1 &= -\frac{\Omega\left(\alpha^4\right)}{\Xi'\left(\alpha^{4}\right)}
       =  \frac{\alpha^{7} + \alpha^{4}}{\alpha^{-7} + \alpha^{-2}}
       =  \frac{\alpha^{3}}{\alpha^{3}}
       =  1 \\
  e_2 &= -\frac{\Omega\left(\alpha^7\right)}{\Xi'\left(\alpha^{7}\right)}
       =  \frac{\alpha^{7} + \alpha^{7}}{\alpha^{-7} + \alpha^{4}}
       =  0 \\
  e_3 &= -\frac{\Omega\left(\alpha^2\right)}{\Xi'\left(\alpha^2\right)}
       =  \frac{\alpha^{7} + \alpha^{2}}{\alpha^{-7} + \alpha^{-6}}
       =  \frac{\alpha^{-3}}{\alpha^{-3}}
       =  1
\end{align}</math>

The fact that <math>e_3 = 1</math> should not be surprising.

Corrected code is therefore [ 1 {{color|green|1}} 0 {{color|green|1}} 1 1 {{color|green|0}} 0 0 0 1 0 1 0 0].