Page 180 - ISCI’2017
P. 180

cases, able to correct single errors in SRC  d (SRC )  =  2 (at  k = 1).
                                                       min
               Really, if given the expression (3) and (5), for the orderly SRC, it can be draw a conclusion. When

                      1
            one ( k = ) control based  m n+ 1  of the  SRC, the  NCS  A =  (a 1  || a 2  ||...|| a i−  1  || aa i+  1  ||...|| a n  || a n+  1 )
                                                                                          ||
                                                                                         i
                                           (SRC
            may have different meanings  d min  ) . In this case, it depends on the magnitude of the control base
             m n+ 1 .


               If, for each individual module of SRC condition  m <  m n+ 1  (i = 1, n ), then, in accordance with the
                                                                i
                                                  (SRC
            expression (1), we can conclude that  d min  )  =  2, i. e., in accordance with equation (2) we obtain that
             t det.  = 1 . If the totality of  information  bases  {}m  for an  arbitrary pair of  modules condition
                                                              i
                                         j
                                                         (SRC
             mm⋅  i  j  <  m  ( , ij = 1, n ; i ≠ ), in this case  d min  )  =  3 и t det.  =  2 . Thus, for the NCS in the SRC with
                         1
                       n+
                           (SRC
             k = 1, MCD  d min  )  can be different depending on the magnitude of the control base  m n+ 1  of the SRC.
               Consider the ratio by which the error can be corrected in the residual  a   [1]. Let the wrong number
                                                                                  i
              
                       
                ≥
                                           
            ( AM )  A =   (а 1  || а 2  ||...|| a i−  1  || aa i+  1  ||...|| a n  || a n+  1 ) including error  a =  (a +∆ a i )mod m  reliably
                                             ||
                                                                                
                                                                                                    i
                                                                                 i
                                                                                      i
                                            i
            contained in the residue аi modulo mi.
               It is obvious
                                                
                                               A   ( А =  А +∆  )mod M  0 .                             (6)
               Given that the amount of error may be represented as  A∆ =  (0 ||0 ||...||0 || a∆  i  ||0 ||...||0 ||0) , when
            the correct ( AM<  ) number A can be determined as follows:
                       
                  А=    ( – А )mod M =  0   (a 1  || a 2  ||...|| a i−  1  || aa i+  1  ||...|| a n  || a n+  1 ) (0 ||0 ||...||0 || a i  ||0 ||...
                                                                               −
                                                             ||
                      A ∆
                                                                                              ∆
                                                           
                                        
                                                           i
                     ...||0 ||  ) 0 mod M = [a 1  || a 2  ||...|| a i−  1  ||(a  i  – a∆  i )mod  ||ma i+  1  ||...|| a n  || a n+  1 ]mod M .
                             
                             
                                                                         i
                                                                                                   0
                                     0
               Obtain a quantitative estimate of the value of A. Since the number A is correct, i.e. stored in the
            numerical range [0, M), then the following inequality must be fulfilled
                                             А   ( –A =    А ∆  ) mod M  0  M <   .                    (7)
               Given that the value  A∆  of the error value is equal  А∆ = ∆ aB⋅  i , then the inequality (7) will have
                                                                         i
            the following form:
                                             
                                            A −∆ ⋅    i  – rM <  M  or
                                                 a B
                                                          ⋅
                                                   i
                                                              0
                                    
                                    A −∆ a B⋅  i  – r M⋅  0  <  M 0  / m n+ 1 (r = 1, 2, 3,...) ,
                                          i
                                                                                                        (8)
                                        
                                        A –(a  i  – a ⋅  i  rM <  M 0  / m n+ 1 ,
                                                   ) B −⋅
                                                              0
                                                  i
                                        
                                        A –(a i  – a  i ) B⋅  i  −⋅  0  M 0  / m n+ 1 ,
                                                         rM <
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