Consider $ \operatorname{GF}(16) $; denote by $ \mathfrak A $ its primitive element.

Input stripe $$ (Y_0,Y_1,\dots,Y_{14} ) , \quad Y_j\in \operatorname{GF}(16) \ . $$ $$ G(\mathfrak x):=Y_0+Y_1 \mathfrak x + \dots + Y_{14} \mathfrak x^{14} $$ $$ G(\mathfrak e)=\mathfrak o,\ G(\mathfrak A)= \mathfrak o,\ G(\mathfrak A^2)= \mathfrak o,\ G(\mathfrak A^3)= \mathfrak o \, . $$ (possibility to correct up to $ 2_{} $ SDCs).

Output stripe $$ (\widetilde Y_0, \widetilde Y_1,\dots, \widetilde Y_{14}) \ . $$

Theorem 1. There exists a unique polynomial $ H(\mathfrak x) $ over $ \operatorname{GF}(16) $ such that $$ H(\mathfrak A^{-1})=Y_{14},\ H(\mathfrak A^{-2})=Y_{13},\dots, H(\mathfrak A^{-15})=H(\mathfrak e)=Y_{0} \, . $$

It is claimed the existence of the «interpolation» polynomial with the number of indeterminate coefficients equal to $ 11 $, i.e. the $ 15 $ equalities from the theorem are evidently redundant.

$$ \left(\begin{array}{lllll} \mathfrak e & \mathfrak e & \mathfrak e & \dots & \mathfrak e \\ \mathfrak e & \mathfrak A^{-14} & (\mathfrak A^{-14})^2 & \dots & (\mathfrak A^{-14})^{10} \\ \mathfrak e & \mathfrak A^{-13} & (\mathfrak A^{-13})^2 & \dots & (\mathfrak A^{-13})^{10} \\ \vdots & & & & \vdots \\ \mathfrak e & \mathfrak A^{-2} & (\mathfrak A^{-2})^2 & \dots & (\mathfrak A^{-2})^{10} \\ \mathfrak e & \mathfrak A^{-1} & \mathfrak A^{-2} & \dots & \mathfrak A^{-10} \end{array} \right)_{15\times 11} \left(\begin{array}{l} H_0 \\ H_1 \\ \vdots \\ H_{10} \end{array} \right)= \left(\begin{array}{l} Y_0 \\ Y_1 \\ \vdots \\ Y_{14} \end{array} \right) \ . $$

Theorem 2. The following equalities are fulfilled $$ H_{10}=G(\mathfrak A^5),\ H_9=G(\mathfrak A^6),\dots, H_0= G(\mathfrak A^{15})= G(\mathfrak e) \ , $$ i.e. the coefficients of the «interpolation» polynomial $ H(\mathfrak x) $ turn out to be the values of the code polynomial $ G(\mathfrak x) $ on the set $ \mathfrak A^5,\mathfrak A^6, \dots, \mathfrak A^{15} $.

Proof. Multiply the system of linear equations from the previous theorem from the left-hand side by the matrix $$ \left(\begin{array}{lllll} \mathfrak e & \mathfrak e & \mathfrak e & \dots & \mathfrak e \\ \mathfrak e & \mathfrak A^{14} & (\mathfrak A^{14})^2 & \dots & (\mathfrak A^{14})^{14} \\ \mathfrak e & \mathfrak A^{13} & (\mathfrak A^{13})^2 & \dots & (\mathfrak A^{13})^{14} \\ \vdots & & & & \vdots \\ \mathfrak e & \mathfrak A^{6} & (\mathfrak A^{6})^2 & \dots & (\mathfrak A^{6})^{14} \\ \mathfrak e & \mathfrak A^{5} & (\mathfrak A^{5})^2 & \dots & (\mathfrak A^{5})^{14} \end{array} \right)_{11\times 15} \ . $$ The obtained matrix equation has in its right-hand side the column $$ \left(\begin{array}{c} Y_0+Y_1+\dots+Y_{14} \\ Y_0+Y_1\mathfrak A^{14}+ Y_2 (\mathfrak A^{14})^2 + \dots + Y_{14} (\mathfrak A^{14})^{14} \\ Y_0+Y_1\mathfrak A^{13}+ Y_2 (\mathfrak A^{13})^2 + \dots + Y_{14} (\mathfrak A^{13})^{14} \\ \vdots \\ Y_0+Y_1\mathfrak A^{5}+ Y_2 (\mathfrak A^{5})^2 + \dots + Y_{14} (\mathfrak A^{5})^{14} \end{array} \right)= \left(\begin{array}{c} G(\mathfrak e) \\ G(\mathfrak A^{14}) \\ G(\mathfrak A^{13}) \\ \vdots \\ G(\mathfrak A^{5}) \end{array} \right) \ ; $$ i.e. the column entries coincide with the right-hand sides of the equalities from the theorem statement.

Perform now the matrix multiplication $$ \left(\begin{array}{lllll} \mathfrak e & \mathfrak e & \mathfrak e & \dots & \mathfrak e \\ \mathfrak e & \mathfrak A^{14} & (\mathfrak A^{14})^2 & \dots & (\mathfrak A^{14})^{14} \\ \mathfrak e & \mathfrak A^{13} & (\mathfrak A^{13})^2 & \dots & (\mathfrak A^{13})^{14} \\ \vdots & & & & \vdots \\ \mathfrak e & \mathfrak A^{6} & (\mathfrak A^{6})^2 & \dots & (\mathfrak A^{6})^{14} \\ \mathfrak e & \mathfrak A^{5} & (\mathfrak A^{5})^2 & \dots & (\mathfrak A^{5})^{14} \end{array} \right) \left(\begin{array}{lllll} \mathfrak e & \mathfrak e & \mathfrak e & \dots & \mathfrak e \\ \mathfrak e & \mathfrak A^{-14} & (\mathfrak A^{-14})^2 & \dots & (\mathfrak A^{-14})^{10} \\ \mathfrak e & \mathfrak A^{-13} & (\mathfrak A^{-13})^2 & \dots & (\mathfrak A^{-13})^{10} \\ \vdots & & & & \vdots \\ \mathfrak e & \mathfrak A^{-2} & (\mathfrak A^{-2})^2 & \dots & (\mathfrak A^{-2})^{10} \\ \mathfrak e & \mathfrak A^{-1} & \mathfrak A^{-2} & \dots & \mathfrak A^{-10} \end{array} \right) $$ Compute the entries of the fisrt row: $$ \underbrace{\mathfrak e+\mathfrak e+\dots+\mathfrak e}_{15}=\mathfrak e . $$ $$ \mathfrak e + \mathfrak A^{-14} + \mathfrak A^{-13} + \dots + \mathfrak A^{-1} = \frac{(\mathfrak A^{-1})^{15}-\mathfrak e}{\mathfrak A^{-1}-\mathfrak e}=\mathfrak o $$ since, due to generalized Fermat theorem, $ (\mathfrak A^{-1})^{15}=\mathfrak e $. The third entry equals $$ \mathfrak e + (\mathfrak A^{-2})^{14} + (\mathfrak A^{-2})^{13} + \dots + \mathfrak A^{-2}= \mathfrak e + (\mathfrak A^{-14})^2 + (\mathfrak A^{-13})^2 + \dots + \mathfrak A^{-2} = \frac{(\mathfrak A^{-2})^{15}-\mathfrak e}{\mathfrak A^{-2}-\mathfrak e}=\mathfrak o $$ due to similar reason. And so forth. All the entries of the first row of the matrix, except for the first one, equal $ \mathfrak o $, i.e. it is just $$ [\mathfrak e,\mathfrak o,\mathfrak o,\dots,\mathfrak o] . $$ One can make a hypothesis that the second row of the matrix product equals $$ [\mathfrak o,\mathfrak e,\mathfrak o,\dots,\mathfrak o] , $$ and extend this hypothesis to the other rows. That means: the product of two non-square matrices is a square identity matrix. It can be verified by a direct computation, while for the general case it follows from the theory of discrete Fourier transform.

Wherefrom follows the statement of the theorem. ♦

The results of Theorems 1 and 2 state that the polynomials $ G(\mathfrak x) $ and $ H(\mathfrak x) $ reciprocally define each other.

If the output stripe $$ (\widetilde Y_0, \widetilde Y_1,\dots, \widetilde Y_{14}) $$ meats the conditions $$ \widetilde G(\mathfrak e)=\mathfrak o,\ \widetilde G(\mathfrak A)= \mathfrak o,\ \widetilde G(\mathfrak A^2)= \mathfrak o,\ \widetilde G(\mathfrak A^3)= \mathfrak o \, . $$ for $$ \widetilde G(\mathfrak x):= \widetilde Y_0+ \widetilde Y_1 \mathfrak x + \dots + \widetilde Y_{14} \mathfrak x^{14} , $$ we do not trouble. If at least one of the equalities fails, we, under the assumption that the number of SDCs $ \le 2_{} $, address Theorems 1 and 2 from the previous section. The blocks $$ \widetilde Y_0, \widetilde Y_1,\dots, \widetilde Y_{14} $$ are considered as the values of some polynomial $ H(\mathfrak x), \deg H \le 10 $. Two errors from the interpolation table constructed for a polynomial can be corrected via the rational interpolation.