## 4 Decoding Procedure
Consider a received vector \(r=\left(0,\,0,\,\beta^{2}\right)\), and \(S=r\mathcal{H}^{T}=\left(\beta^{2},\,\beta^{3}\right)\) are two syndromes corresponding to the received vector \(r\), which is non-zero; this means that \(r\) is not a codeword. Secondly, find the error locator polynomial \(\sigma(z)\) for the weight of the received vector \(r\), which is 1. Let \(t=1\), because only one component in \(r\) is non-zero. So, \(r_{i_{1}}=r_{3}\), for \(i_{1}=3\); this means the error is occurring at the third place in \(r\). The location of the error is \(X_{1}=\eta_{i_{1}}=\beta\), with error magnitude \(Y_{1}=r_{i_{1}}=\beta^{2}\). The error locator polynomial is \(\sigma(z)=1-\beta z=1+\beta z\). Furthermore, the error evaluator polynomial is \(\omega(z)=S(z).\sigma(z)\)\(\left(mod\,\,z^{2}\right)=\beta^{2}\), where \(S(z)=\beta^{2}+\beta^{3}z\). Finally, the error magnitude is \(\overline{Y}_{1}=\beta^{2}\). Hence, \(c=r-e=\left(0,\,0,\,0\right)\) is a corrected codeword of the alternant code over \(\mathbb{Z}_{2}[i]\)._  
**Example 5**. : _From Illustration 3.2, the cyclic group of order 80. Let us set \(\eta=\left(\alpha^{2},\,\alpha,\,\alpha^{55},\,0,\,0,\,\ldots,0\right)_{1\times 1 0}=\left(\eta_{1},\,\eta_{2},\,\eta_{3},\,\ldots,\,\eta_{10}\right)\), \(y=\left(1,\,1,\,1,\,\ldots,\,1\right)_{1\times 10}\) and \(r=4\), so \(t=2\) then we have length 10, and the alternant code \(\mathcal{C}(\eta,\,y)\) over \(\mathbb{Z}_{3}[i]\) with minimum Hamming distance is at least \(5\). Let_  
\[H= \begin{pmatrix}1&1&1&1&\ldots&1\\ \alpha^{2}&\alpha&\alpha^{55}&0&\ldots&0\\ \alpha^{4}&\alpha^{2}&\alpha^{30}&0&\ldots&0\\ \alpha^{6}&\alpha^{3}&\alpha^{5}&0&\ldots&0\end{pmatrix} ,\]  
_be the parity-check matrix. Let a received vector \(r=\left(1,\alpha^{2},0,\ldots,0\right)_{1\times 10}\) and \(S=a\mathcal{H}^{T}= \begin{pmatrix}1+\alpha^{2}\\ \alpha^{2}+\alpha^{3}\\ \alpha^{4}+\alpha^{4}\\ \alpha^{6}+\alpha^{5}\end{pmatrix} = \begin{pmatrix}\alpha^{77}\\ \alpha^{51}\\ \alpha^{44}\\ \alpha^{54}\end{pmatrix} = \begin{pmatrix}S_{1}\\ S_{2}\\ S_{3}\\ S_{4}\end{pmatrix} \) are the four syndromes corresponding to the received vector \(r\), which are non-zero; this means that \(r\) is not a codeword. Secondly, find the error locator polynomial \(\sigma(z)\). Let \(t=2\) because there are two non-zero components in \(r\). So, \(r_{i_{1}}=r_{1}\), and \(r_{i_{2}}=r_{2}\), for \(i_{1}=1\), and \(i_{2}=2\); this means the error occurs in the 1 st and 2 nd places in \(r\). The error locations are \(X_{1}=\eta_{i_{1}}=\eta_{1}=\alpha^{2}\) and \(X_{2}=\eta_{i_{2}}=\eta_{2}=\alpha\), with error magnitudes \(Y_{1}=a_{i_{1}}=1\) and \(Y_{2}=a_{i_{2}}=\alpha^{2}\). Let the error locator polynomial \(\sigma(z)=1+\alpha^{10}z+\alpha^{3}z^{2}\), and the error evaluator polynomial is \(\omega(z)=\alpha^{28}z+\alpha^{77}\). Further polynomials \(\sigma(z)\) and \(\omega(z)\) satisfy \(\sigma(z)=\omega(z)S(z)\) mod \(\left(z^{4}\right)\). Finally, the error magnitudes are_  
\[\overline{Y}_{1}=\frac{\omega(X_{1})}{Y_{1}\left(1-X_{2}X_{1}^{-1 }\right)}=\frac{\omega(\alpha^{28})}{1\left(1-X_{2}X_{1}^{-1}\right)}=\frac{ \frac{2+\alpha+i+\alpha}{2+\alpha+i+\alpha}}{2+\alpha+i+\alpha}=1.\] \[\overline{Y}_{2}=\frac{\omega(X_{2})}{Y_{1}\left(1-X_{1}X_{2}^{-1 }\right)}=\frac{\omega(\alpha^{29})}{1\left(1-X_{1}X_{2}^{-1}\right)}=\frac{ \frac{2+\alpha+i+\alpha}{1+2\alpha}}{1+2\alpha}=1+2\alpha+2i=\alpha^{2}.\]  
Therefore, the error pattern is \(e=\left(1,\,\alpha^{2},\,0,\,0\right)\) and the corrected codeword is \(c=\left(0,\,0,\,0,\,0\right)\).