is $$\bm{\vec x} \times \bm{\vec w}^T = (x_1 w_1, x_2 w_2,...,x_n w_n).$$

[c c![width 1.5pt]{}c|c|c|c|c]{} $a$ & $b$ & $a\wedge b$ & $a\vee b$ & $a\oplus b$ & $a\equiv b$ & $a \rightarrow b$\
$0$ & $0$ & $0$ & $0$ & $0$ & $1$ & $1$\
$1$ & $0$ & $0$ & $1$ & $1$ & $0$ & $0$\
$0$ & $1$ & $0$ & $1$ & $1$ & $0$ & $1$\
$1$ & $1$ & $1$ & $1$ & $0$ & $1$ & $1$\

Second, the nonlinear calculation of BDNN is defined as follows: we count the number of “$-1$” and “$1$” in $(x_1 w_1, x_2 w_2,...,x_n w_n)$ and let the calculation return the one with larger number. Assume $f$ is the basic function of BDNN, with the linear and nonlinear calculation defined above, then we have $$y=f(\bm{\vec x},\bm{\vec w}) = \begin{cases} 1,& if \quad \sum_{i=1}^{n}x_i w_i \geq 0 \\ -1,& if \quad \sum_{i=1}^{n}x_i w_i < 0.
\label{f}
\end{cases}$$

Figure \[BDNNexample\] shows an example of the calculation of . The inputs $\bm{\vec x}$ are $(1,-1,-1)$ and the corresponding weights $\bm{\vec w}$ are $(1,1,-1)$. Then, we use Boolean operation as linear calculation and the results $(1,-1,1)$ are obtained. Finally, 