Editing BCH code (section)

==== Decoding with unreadable characters ====
Suppose the same scenario, but the received word has two unreadable characters [ 1 {{color|red|0}} 0 ? 1 1 ? 0 0 {{color|red|1}} 1 0 1 0 0 ]. We replace the unreadable characters by zeros while creating the polynomial reflecting their positions <math>\Gamma(x) = \left(\alpha^8x - 1\right)\left(\alpha^{11}x - 1\right).</math> We compute the syndromes <math>s_1=\alpha^{-7}, s_2=\alpha^{1}, s_3=\alpha^{4}, s_4=\alpha^{2}, s_5=\alpha^{5},</math> and <math>s_6=\alpha^{-7}.</math> (Using log notation which is independent on GF(2<sup>4</sup>) isomorphisms. For computation checking we can use the same representation for addition as was used in previous example. Hexadecimal description of the powers of <math>\alpha</math> are consecutively 1,2,4,8,3,6,C,B,5,A,7,E,F,D,9 with the addition based on bitwise xor.)

Let us make syndrome polynomial

:<math>S(x)=\alpha^{-7}+\alpha^{1}x+\alpha^{4}x^2+\alpha^{2}x^3+\alpha^{5}x^4+\alpha^{-7}x^5,</math>

compute

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

Run the extended Euclidean algorithm:

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

We have reached polynomial of degree at most 3, and as

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

we get

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

Therefore,

:<math>S(x)\Gamma(x)\left(\alpha^{3} + \alpha^{-5}x + \alpha^{6}x^2\right) - \left(\alpha^{7}x + \alpha^{5}x^2 + \alpha^{3}x^3\right)x^6 = \alpha^{-4} + \alpha^{4}x + \alpha^{2}x^2 + \alpha^{-5}x^3.</math>

Let <math>\Lambda(x) = \alpha^{3}+ \alpha^{-5}x+ \alpha^{6}x^2.</math> Don't worry that <math>\lambda_0\neq 1.</math> Find by brute force a root of <math>\Lambda.</math> The roots are <math>\alpha^2,</math> and <math>\alpha^{10}</math> (after finding for example <math>\alpha^2</math> we can divide <math>\Lambda</math> by corresponding monom <math>\left(x - \alpha^2\right)</math> and the root of resulting monom could be found easily).

Let

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

Let us look for error values using formula

:<math>e_j = -\frac{\Omega \left(\alpha^{-i_j} \right)}{\Xi' \left(\alpha^{-i_j} \right)},</math>

where <math>\alpha^{-i_j}</math> are roots of <math>\Xi(x).</math> <math>\Xi'(x)=\alpha^{2}x^2.</math> We get

:<math>\begin{align}
  e_1 &=-\frac{\Omega(\alpha^4)}{\Xi'(\alpha^{4})} = \frac{\alpha^{-4}+\alpha^{-7}+\alpha^{-5}+\alpha^{7}}{\alpha^{-5}} =\frac{\alpha^{-5}}{\alpha^{-5}}=1 \\
  e_2 &=-\frac{\Omega(\alpha^7)}{\Xi'(\alpha^{7})} = \frac{\alpha^{-4}+\alpha^{-4}+\alpha^{1}+\alpha^{1}}{\alpha^{1}}=0 \\
  e_3 &=-\frac{\Omega(\alpha^{10})}{\Xi'(\alpha^{10})} = \frac{\alpha^{-4}+\alpha^{-1}+\alpha^{7}+\alpha^{-5}}{\alpha^{7}}=\frac{\alpha^{7}}{\alpha^{7}}=1 \\
  e_4 &=-\frac{\Omega(\alpha^{2})}{\Xi'(\alpha^{2})} = \frac{\alpha^{-4}+\alpha^{6}+\alpha^{6}+\alpha^{1}}{\alpha^{6}}=\frac{\alpha^{6}}{\alpha^{6}}=1
\end{align}</math>

Fact, that <math>e_3=e_4=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 {{color|green|0}} 1 0 1 0 0].