Page 186 - ISCI’2017

2 ;
= ( 0·385 0·616 2·1100 1·980 mod 1540 100)+ + + = < M = 40


·
A 2SRC = (0 ||0 || 2 ||1 = ) A 2PNS = (а 1 ·В + 12 аВ + 2 · 22 а 3 ·В + 32 аВ 42 ) mod M = 2
4
2 ;
+
+
+
= ( 0·385 0·231 2·330 1·210 mod 1155 870) = > M = 40


·
A 3SRC = (0 ||0 || 2 ||1 = ) A 3PNS = (а 1 ·В + 13 аВ + 2 · 23 а 3 ·В + 33 аВ 43 ) mod M = 3
4
2 ;
= ( 0·616 0·693 2·792 1·672 mod 924)+ + + = 418 < M = 40


A 4SRC = ( 0 ||0 ||0 ||1) = A 4PNS = (аВ + 1 · 14 а 2 ·В + 24 а 3 ·В + 34 а 4 ·В 44 ) mod M = 4
2 ;
= ( 0·220 0·165 2·396 1·540 mod660 540)+ + + = > M = 40


A 5SRC = ( 0 ||0 ||0 || 2) = A 5PNS = (аВ + 1 · 15 а 2 ·В + 25 а 3 ·В + 35 а 4 ·В 45 ) mod M = 5
2 .
)
+
+
+
= ( 0·280 0·105 2·336 1·120 mod 420 = 240 < M = 40

As a result of calculations of values A j PNS and comparing them with the value M = 420 of the
interval length [0,420) of processing of correct numbers A SRC in SRC we obtain the following.
Set of residuals a = , a = is correct (residuals are not distorted), and the residuals a = ,
0
0
0
2
1
4

a = 0 and a = 1 of the wrong number A 3180 = (0 ||0 ||0 || 2 ||1) may be distorted (may be wrong).
3
5

III. Correction is possible distorted residuals a , a и a of the number A 3180 .
3
1
5
0
It is necessary to be corrected, possibly distorted residuals a = , a = and a = by the
0
1
1
3
5
  m ⋅ (1 rm ) A   
+ ⋅
formula a = i   a + i  i n+ 1 −     mod m . Then we have that
i
  m n+ 1 ⋅ m i B i  
  m ⋅ (1 rm ) A      3 (1 r ⋅ 11) 3180  
+ ⋅
⋅
+
a = 1   a + 1  1 n+ 1 −   mod m = 1  0 +   −   mod 3 =
  m n+ 1 ⋅ m 1 B 1     11 1 ⋅ 1540  
])
])
)
−
+
= (0 + [3,27 2,06 mod 3 = (0 + [1,21 mod 3 = (0 1 mod 3 1.
=
1
In this way a = . For value a we have
1
3
  m ⋅ (1 rm ) A      5 (1 r ⋅ 11) 3180  
+
⋅
+ ⋅
a = 3   a + 3  3 n+ 1 −   mod m = 3  0 +   −   mod5 =
  m n+ 1 ⋅ m 3 B 3     11 4 ⋅ 3696  
])
)
0,5
= (0 + [1,36 0,86 mod5 = (0 + [ ]) mod5 = (0 0 mod5 0 .
=
−
+
0
In this way a = . To obtain the value of the residual a
3
5
  m ⋅ (1 rm ) A      11 (1 r ⋅ 11) 3180  
+ ⋅
⋅
+
a = 5   a + 5  5 n+ 1 −   mod m = 5  1+    −   mod11 =
  m n+ 1 ⋅ m 5 B 5     11 6 ⋅ 2520  
])
= (1+ [2 1,26 mod 11− ]) = (1+ [0,74 mod11 = (1 0 mod 11 1+ ) = .
1
0
We have that a = 1. According to the obtained values a = , a = and a = 1 of the recovered
5
3
1
5
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