## 2 Reverse Conversion for General View Moduli
Then the coefficients are \(k_{i}^{\prime}\):  
\[k_{1}^{\prime}=\left\lceil\frac{\left|P_{1}^{-1}\right|_{p_{1}}}{p_{1}}\cdot 2 ^{N}\right\rceil=\left\lceil\frac{4}{7}\cdot 2^{13}\right\rceil=4682,k_{2}^{ \prime}=7168,k_{3}^{\prime}=4552.\]  
From the Formula (5) for the number \(X=(3,2,1)=10\) we obtain:  
\[X^{\prime}=\left|4682\cdot 3+7168\cdot 2+4552\cdot 1\right|_{2^{13}}=\left|329 34\right|_{2^{13}}=166.\]  
Then \(X\) will be equal to:  
\[P_{1}=\frac{504}{7}=72,\qquad P_{2}=63,\qquad P_{3}=56,\]0  
Another method based on the CRT is a method called New CRT [8], which is given by the formula:  
\[P_{1}=\frac{504}{7}=72,\qquad P_{2}=63,\qquad P_{3}=56,\]1  
where  
\[P_{1}=\frac{504}{7}=72,\qquad P_{2}=63,\qquad P_{3}=56,\]2  
For the above considered RNS {7,8,9} the coefficients \(k_{i}\) of the new CRT will be:  
\[P_{1}=\frac{504}{7}=72,\qquad P_{2}=63,\qquad P_{3}=56,\]3  
Then, using the Formula (7) for the number \(X=(3,2,1)=10\) we obtain:  
\[P_{1}=\frac{504}{7}=72,\qquad P_{2}=63,\qquad P_{3}=56,\]4  
The article [16] proposes two modifications of the New CRT, which are denoted as New CRT I and New CRT II. Let us consider each of these methods separately. In New CRT I, the Formula (8) similarly calculates constants \(k_{i}\), which are used to find coefficients \(a_{i}\), \(i=0,\ldots,n-1\):  
\[P_{1}=\frac{504}{7}=72,\qquad P_{2}=63,\qquad P_{3}=56,\]5 \[P_{1}=\frac{504}{7}=72,\qquad P_{2}=63,\qquad P_{3}=56,\]6 \[P_{1}=\frac{504}{7}=72,\qquad P_{2}=63,\qquad P_{3}=56,\]7 \[P_{1}=\frac{504}{7}=72,\qquad P_{2}=63,\qquad P_{3}=56,\]8 \[P_{1}=\frac{504}{7}=72,\qquad P_{2}=63,\qquad P_{3}=56,\]9  
Then, for these coefficients we build a matrix of characteristics:  
\[A= \begin{pmatrix}a_{0,0}&0&\ldots&0&0\\ a_{0,1}&a_{1,1}&\ldots&0&0\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ a_{0,n-2}&a_{1,n-2}&\ldots&a_{n-2,n-2}&0\\ a_{0,n-1}&a_{1,n-1}&\ldots&a_{n-2,n-1}&a_{n-1,n-1}\end{pmatrix} \]  
which coefficients are obtained from the representation of \(a_{i}\) in the mixed-radix conversion:  
\[a_{0} =a_{0,0}+a_{0,1}p_{1}+\ldots+a_{0,n-1}p_{1}p_{2}\ldots p_{n-1}\] \[a_{1} =a_{1,1}+a_{1,2}p_{2}+\ldots+a_{1,n-1}p_{2}\ldots p_{n-1}\] \[\ldots\] \[a_{n-2} =a_{n-2,n-2}+a_{n-2,n-1}p_{n-1}\] \[a_{n-1} =a_{n-1,n-1}\]  
Then, from the matrix \(A\) and the value \(X=(x_{1},x_{2},\ldots,x_{n})\) a vector \(B=A\times X\) can be obtained, i.e. :  
\[B= \begin{pmatrix}B_{0}\\ B_{1}\\ \ldots\\ B_{n-2}\\ B_{n-1}\end{pmatrix} = \begin{pmatrix}a_{0,0}&0&\ldots&0&0\\ a_{0,1}&a_{1,1}&\ldots&0&0\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ a_{0,n-2}&a_{1,n-2}&\ldots&a_{n-2,n-2}&0\\ a_{0,n-1}&a_{1,n-1}&\ldots&a_{n-2,n-1}&a_{n-1,n-1}\end{pmatrix} \times \begin{pmatrix} x_{1}\\ x_{2}\\ \ldots\\ x_{n-1}\\ x_{n}\end{pmatrix} \]  
Then, the number \(X\) can be reconstructed by the formula:  
\[X=\left|B_{0}+B_{1}p_{1}+B_{2}p_{1}p_{2}+\ldots+B_{n-1}p_{1}p_{2}\ldots p_{n-1 }\right|_{p_{1}p_{2}\ldots p_{n}}\]  
Consider a similar example for which the coefficients of \(a_{i}\) are:  
\[a_{0} =\left|1-31\cdot 7\right|_{504}=288=1+1p_{1}+5p_{1}p_{2},\] \[a_{1} =\left|31-5\cdot 8\right|_{72}=63=7+7p_{2},\] \[a_{2} =\left|5\right|_{9}=5.\]  
For the number \(X=(3,2,1)=10\) we obtain the vector \(B\):  
\[B= \begin{pmatrix}1&0&0\\ 1&7&0\\ 5&7&5\end{pmatrix} \times \begin{pmatrix}3\\ 2\\ 1\end{pmatrix} = \begin{pmatrix}3\\ 17\\ 34\end{pmatrix} .\]  
From which we obtain:  
\[X=\left|3+17\cdot 7+34\cdot 7\cdot 8\right|_{504}=10.\]For the case of ordered RNS with moduli satisfying the expression \(p_{i}>p_{1}+\ldots+p_{i-1}\) or more strictly \(p_{i}>2p_{i-1}\), the reverse conversion can be obtained without taking into account the remainder modulo \(P\). In this case, the function \(Y<2P\) is given by  
\[Y=B_{0}+B_{1}p_{1}+B_{2}p_{1}p_{2}+\ldots+\left|B_{n-1}\right|_{p_{n}}p_{1}p_{2 }\ldots p_{n-1}\]  
and the required value of \(X\) is  
\[X= \begin{cases}Y&\text{if }Y<P,\\ Y-P&\text{if }Y\geq P.\end{cases}  \tag{9}\]  
The restriction of moduli in this case leads to an unbalanced RNS.