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By other words, the polynomial e(x) includes no more t non-zero coefficients. From the expression

            (7) and (9) follows the equality

                                                                                          −
                                                              1    X  0    X 0 2  ...  X 0 N 1  T
                                                              1    X 1     X 1 2  ...  X 1 N 1
                                                                                          −
                           ( s , s ,..., s n− k 1 ) = ( e , e , e ,..., e N 1 ) ⋅  1  X  2  X 2 2  ...  X 2 N 1−  ,
                                                 1
                                1
                             0
                                              0
                                       −
                                                          −
                                                   2
                                                             ...   ...      ...   ...   ...
                                                                            2
                                                              1   X n− k 1−  X n− k 1−  ...  X n− k 1−
                                                                                        N 1−
            which is equivalent to the following system of equations:
                                                                           n −1
                                                                                 i
                                                                              e
                                       s 0  = e 0  + Xe 1  0  + e 2 X 0 2  + ... + e n −1 X 0 n −1  =  ∑ i X ,
                                                                                 0
                                                                           i =0
                                                                         n −1
                                     s 1  = e 0  + Xe 1  1  + e 2 X 1 2  + ... + e n −1 X 1 n −1  =  ∑ i X ,                                       (11)
                                                                                i
                                                                            e
                                                                               1
                                                                         i =0
                                                           …
                                                                               n −1
                                                                         −1
                                                                                  e
                               s n −k −1  = e 0  + Xe 1  n −k  −1  + e 2 X  n 2 −k  −1  + ... + e n −1 X n n −k −1  =  ∑ i X n i −k −1  .
                                                                               i =0
                                        *
                  Vector decoding task c  consists in finding all elements ei, i = 0,…, n – 1 by the known elements
            of the vector s = (s0, s1, …, sn-k-1). Equations system (11) includes n – k non-linear equations from n
            variables (unknowns), and any direct methods of its solution for this system (11) heaven knows. An
            artificial technique can be used for a finding the elements of the vector e = (e0, e1, …, en-1)  in the
            algebraic coding theory [38-40], which consists in a consideration of the error locator polynomial

            Λ(x), radicals of which are non-zero elements of the vector e, i.e.

                                                   Λ x)(  = ∏  x (  + X ) ,                                                         (12)
                                                                    j
                                                            j
            where j is an index of the non-zero elements of the vector e, Xj is the so-called error locator, which

            occurred in j  symbol of the code word.
                        th
                  Expand the brackets in the expression (12), we will get

                                              Λ(x)  = x  + λu-1x  + … + λ1x + λ0,                                       (13)
                                                       u
                                                               u-1
            where a degree u of the polynomial Λ(x) specifies the number of occurred errors on the block of n
            symbols, u ≤ t, i.e. a number of the non-zero elements of the vector e.

                  Set of coefficients (λ0, λ1, …, λu-1) of the polynomial (13) uniquely determines  its radicals,

            which clearly indicate (localize) the location of the occurred errors.
                  To multiple the polynomial (13) to eiX   and will calculate its volume in Xj, we will get:
                                                        i
                                      e i X  u j +i  + e λ − 1 X  u j  +i − 1  + ...+ e λ 1 X  j + i  1  + e λ 0 X  i j  = 0 ,
                                                 u
                                               i
                                                                          i
                                                                i
                             m
            where Xj ∈ GF(q ), i.e. Xj = α  (where α is a primitive element of the field GF(q )) for any Jj.
                                          Jj
                                                                                           m
                                  a+b    a+b+Jj   b
                  Consequently, Xj   = α       = X j+ a, i.e. the following expression is valid
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