## Appendix A  
For 2D convolution by multiplying vectors, we convert matrices \(\mathbf{F}\) and \(\mathbf{G}\) into vectors \(\mathbf{f}\) and \(\mathbf{g}\) as follows:  
\[\mathbf{f}=\big{[}f_{-2,-2}\quad f_{-2,-1}\quad\cdots\quad f_{0,0}\quad\cdots \quad f_{2,1}\quad f_{2,2}\big{]}^{T}, \tag{1}\]  
\[\mathbf{g}=\big{[}g_{-1,-1}\quad g_{-1,0}\quad\cdots\quad g_{0,0}\quad\cdots \quad g_{1,0}\quad g_{1,1}\big{]}^{T}. \tag{2}\]  
Simultaneously, the 2D data \(a[j,k]\) and \(y[j,k]\) are converted into vector form as follows:  
\[\mathbf{a}=\big{[}a[j+1,k+1]\quad\dots\quad a[j,k]\quad\dots\quad a[j-1,k-1] \big{]}^{T}, \tag{3}\]  
\[\mathbf{y}=\big{[}y[j+2,k+2]\quad\dots\quad y[j,k]\quad\dots\quad y[j-2,k-2] \big{]}^{T}. \tag{4}\]  
In Figure 2, \(z[j,k]\) is the output of the convolution of signal \(y[j,k]\) and matrix \(\mathbf{F}\). \(d[j,k]\) is the output of the convolution of signal \(a[j,k]\) and matrix \(\mathbf{G}\). Therefore, based on (1)-(4), the output of the equalizer \(z[j,k]\) and GPR target \(d[j,k]\) are given as  
\[z[j,k]=\mathbf{f}^{T}\mathbf{y}, \tag{5}\]  
\[d[j,k]=\mathbf{g}^{T}\mathbf{a}. \tag{6}\]  
Subsequently, the error signal \(e[j,k]\) can be expressed as  
\[e[j,k]=z[j,k]-d[j,k]=\mathbf{f}^{T}\mathbf{y}-\mathbf{g}^{T}\mathbf{a}. \tag{7}\]  
Finally, the mean square error is defined as follows:  
\[E\Big{\{}\big{(}e[j,k]\big{)}^{2}\Big{\}} = E\bigg{\{}\Big{(}\mathbf{f}^{T}\mathbf{y}-\mathbf{g}^{T}\mathbf{ a}\Big{)}\Big{(}\mathbf{f}^{T}\mathbf{y}-\mathbf{g}^{T}\mathbf{a}\Big{)}^{T}\bigg{\}}\] \[= \mathbf{f}^{T}\mathbf{R}\mathbf{f}-2\mathbf{f}^{T}\mathbf{Tg}+ \mathbf{g}^{T}\mathbf{Ag},\]  
where \(\mathbf{R}=E[\mathbf{y}\mathbf{y}^{T}]\) is the autocorrelation matrix of the channel output data, \(\mathbf{T}=E[\mathbf{ya}^{T}]\) is the cross-correlation between the input data and the channel output data, and \(\mathbf{A}=E[\mathbf{aa}^{T}]\) is the autocorrelation of the input data, where \(E\) denotes the expectation and \(T\) is the transpose operator. Then, we minimize (A) using the Lagrange function. Because the trivial answer for minimizing (A) is \(\mathbf{f}=\mathbf{g}=0\), we must impose a constraint on \(\mathbf{f}\) or \(\mathbf{g}\). Here, we impose a constraint on \(\mathbf{g}\) and create cases of the \(N\)-way GPR target [21]. This constraint is represented by the following expression:  
\[\mathbf{E}^{T}\mathbf{g}=\mathbf{b}. \tag{8}\]We analyzed each \(N\)-way GPR target to define matrices \(\mathbf{E}\) and \(\mathbf{b}\). First, we set the constraint for the 2-way GPR target. From the desired form of the 2-way GPR target, we can obtain the vector \(\mathbf{g}\) as follows:  
\[\mathbf{g}= \begin{bmatrix}0&g_{-1,0}&0&0&1&0&0&g_{1,0}&0\end{bmatrix} ^{T}. \tag{10}\]  
We can deduce the matrix \(\mathbf{E}\) and vector \(\mathbf{b}\) as follows:  
\[\mathbf{b}= \begin{bmatrix}0&0&0&1&0&0&0\end{bmatrix} ^{T}, \tag{11}\]  
\[\mathbf{E}^{T}= \begin{bmatrix}1&0&0&0&0&0&0&0\\ 0&0&1&0&0&0&0&0&0\\ 0&0&0&1&0&0&0&0&0\\ 0&0&0&0&1&0&0&0&0\\ 0&0&0&0&0&1&0&0&0\\ 0&0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&0&1\end{bmatrix} . \tag{12}\]  
We set a constraint for the 4-way GPR target as  
\[\mathbf{g}= \begin{bmatrix}0&g_{-1,0}&0&g_{0,-1}&1&g_{0,1}&0&g_{1,0}&0\end{bmatrix} ^ {T}, \tag{13}\]  
\[\mathbf{b}= \begin{bmatrix}0&0&1&0&0\end{bmatrix} ^{T}, \tag{14}\]  
\[\mathbf{E}^{T}= \begin{bmatrix}1&0&0&0&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1\end{bmatrix} . \tag{15}\]  
Finally, we set the constraint for the 8-way GPR target as  
\[\mathbf{g}= \begin{bmatrix}g_{-1,-1}&g_{-1,0}&g_{-1,1}&g_{0,-1}&1&g_{0,1}&g_{1, -1}&g_{1,0}&g_{1,1}\end{bmatrix} ^{T}, \tag{16}\]  
\[\mathbf{b}=1, \tag{17}\]  
\[\mathbf{E}^{T}= \begin{bmatrix}0&0&0&0&1&0&0&0&0\end{bmatrix} .