Page 185 - ISCI’2017
P. 185
⋅
A 1PNS = (5 980)mod 1540 = 280 420 = M ;
<
A 2PNS = (2 385 5 231)mod1155 1925 (mod 1155 = > M ;
⋅
+ ⋅
) 770 420 =
=
A 3PNS = (2 616 5 672)mod924⋅ + ⋅ = 4592 (mod924 ) 896 420= > = M ;
A 4PNS = (2 220 5 540)mod660 3140 (mod660 = > M ;
) 500 420 =
=
+ ⋅
⋅
A 5PNS = 2 280(mod 420) 560 (mod 420 = ) 140 420 = M .
<
=
⋅
0
0
So as A 2ПСС , A 3ПСС and A 4ПСС > 420, then it is concluded that the residuals a = , a = and
2
3
a = 0 of number A = (2 ||0 ||0 ||0 ||5) is not distorted. Distortions a = and a = can be only
2
5
1
5
5
4
residuals a and a . At first spend correcting of residual a = . We have that
2
1
1
5
m ⋅ (1 rm ) A 3 (1 11) 1820 ⋅ +
+ ⋅
a = 1 a + 1 1 n+ 1 − mod m = 1 2 + − mod 3 =
m n+ 1 ⋅ m 1 B 1 11 1 ⋅ 1540
= (2 [3,27 1,18] mod 3+ − ) = (2 [2,09] mod 3 (2 2)mod 3 4(mod 3)1+ ) = + = = .
Thus the corrected residual by modular m is equal a = .
1
1
1
The similar way we obtain the value a = 5. According to the resulting residual a , a we correct
5
1
5
the wrong number A 1820 = (2 ||0 ||0 ||0 ||5). Eventually, in the process correcting we obtain the correct
number A 280 = (1||0 ||0 ||0 ||5).
Example 3. Implement number control A SRC = (0 ||0 ||0 || 2 ||1). In the case of distortion diagnose
and correct the data.
I. Data check A SRC = (0 ||0 ||0 || 2 ||1). In accordance with a known procedure of control determine
A PNS by the formula
n+ 1
A = aB i ∑ mod M = (0 1540 0 3465 0 3696 2 2640 +
+ ⋅
⋅
+⋅
+⋅
⋅
PNS i= 1 i 0
+⋅ 7800(mod 4620) = 3180 > 420 .
1 2520)mod 4620 =
This number is wrong A 3180 .
II. Data diagnostic A 3180 = (0 ||0 ||0 || 2 ||1). We construct all possible projections A of number
j
A 3180 : A = (0 ||0 || 2 ||1) , A = (0 ||0 || 2 ||1) , A = (0 ||0 || 2 ||1) , A = (0 ||0 ||0 ||1) and
4
1
3
2
A = (0 ||0 ||0 || 2) .
5
We define the values of all five projections A in the PNS:
j
A 1SRC = (0 ||0 || 2 ||1 = ) A 1PNS ( ·аВ + = 1 11 а 2 ·В + 21 а 3 ·В + 31 аВ 41 )mod M = 1
·
4
185
