# Construction Method of Probabilistic Boolean Networks Based on Imperfect Information
## 2 Preliminaries
### Matrix-Based Representation for PBNs  
In this subsection, we explain the outline of the matrix-based representation for PBNs [27]. Instead of this representation, we may use the matrix-based representation proposed in [5; 6; 14], where the semi-tensor product (STP) of matrices is used. In this paper, we do not need manipulation of matrices using the STP. Hence, we use a simple matrix-based representation based on truth tables. First, we define the notation. Binary variables \(x_{i}^{0}(k)\) and \(x_{i}^{1}(k)\) are introduced. If \(x_{i}(k)=0\) holds, then \(x_{i}^{0}(k)=1\) holds, otherwise \(x_{i}^{0}(k)=0\) holds. If \(x_{i}(k)=1\) holds, then \(x_{i}^{1}(k)=1\) holds, otherwise \(x_{i}^{1}(k)=0\) holds. Then, the equality \(x_{i}^{0}(k)+x_{i}^{1}(k)=1\) is satisfied. Using \(x_{i}^{0}(k)\) and \(x_{i}^{1}(k)\), consider transforming the BN (1) into a matrix-based representation. Define  
\[\overline{x}_{i}(k):=\left[ \begin{array}{c}x_{i}^{0}(k)\\ x_{i}^{1}(k)\end{array} \right]\left(=\left[ \begin{array}{c}1-x_{i}(k)\\ x_{i}(k)\end{array} \right]\right).\]  
Then, the matrix-based representation for \(x_{i}(k+1)\) is given by  
\[\overline{x}_{i}(k+1)=A^{(i)}\bigotimes_{j\in\mathcal{N}^{(i)}}\overline{x}_{ j}(k), \tag{3}\]  
where \(A^{(i)}\in\{0,1\}^{2\times 2^{|\mathcal{N}^{(i)}|}}\) and \(\bigotimes_{j\in\mathcal{N}^{(i)}}\overline{x}_{j}(k)\in\{0,1\}^{2^{|\mathcal{ N}^{(i)}|}}\). The matrix \(A^{(i)}\) can be derived from the following procedure [24]. **Procedure for deriving \(A^{(i)}\) in (3):**  
**Step 1:** Derive a truth table for \(x_{i}(k+1)\). **Step 2:** Based on the obtained truth table, assign \(x_{i}(k+1)=0\) or \(x_{i}(k+1)=1\) for each element of \(\bigotimes_{j\in\mathcal{N}^{(i)}}\overline{x}_{j}(k)\). **Step 3:** Express the assignment obtained in Step 2 by a row vector. Denote the obtained row vector by \(\overline{A}^{(i)}\in\{0,1\}^{1\times 2^{|\mathcal{N}^{(i)}|}}\). **Step 4:** Derive \(A^{(i)}\) as  
\[A^{(i)}=\left[ \begin{array}{c}1_{1\times 2^{|\mathcal{N}_{i}|}}-\overline{A}^{ (i)}\\ \overline{A}^{(i)}\end{array} \right].\]  
Next, consider extending the matrix-based representation of BNs to that of PBNs. Noting that the probability distribution of each \(\overline{x}_{i}\) is independent, \(E[\overline{x}_{i}(k+1)]\) can be obtained as  
\[E[\overline{x}_{i}(k+1)]=\left(\sum_{l=1}^{q(i)}c_{l}^{(i)}A_{l}^{(i)}\right) \bigotimes_{j\in\mathcal{N}_{i}}E[\overline{x}_{j}(k)], \tag{4}\]  
where \(A_{l}^{(i)}\in\{0,1\}^{2\times 2^{|\mathcal{N}_{i}|}}\) and \(\bigotimes_{j\in\mathcal{N}_{i}}\overline{x}_{j}(k)\in\{0,1\}^{2^{|\mathcal{N} _{i}|}}\). The matrix \(A_{l}^{(i)}\) can be derived from the above procedure. We present a simple example. **Example 1**. : _Consider the PBN with two states. Boolean functions and those selection probabilities are given as follows:_  
\[f^{(1)}=\begin{cases}f_{1}^{(1)}=x_{1}(k)\wedge\
eg x_{2}(k),&c_{1}^{(1)}=0. 8,\\ f_{2}^{(1)}=\
eg x_{1}(k),&c_{2}^{(1)}=0.2,\\ f^{(2)}= \begin{cases}f_{1}^{(2)}=x_{1}(k)\lor x_{2}(k),&c_{1}^{(2)}=0.5,\\ f_{2}^{(2)}=x_{1}(k),&c_{2}^{(2)}=0.3,\\ f_{3}^{(2)}=\
eg x_{2}(k),&c_{3}^{(2)}=0.2,\end{cases} \]_where \(\mathcal{N}_{1}=\mathcal{N}_{2}=\{1,2\}\), \(q(1)=2\), and \(q(2)=3\). Using the truth table, we derive the matrix-based representation for each Boolean function. As an example, consider the Boolean function \(f_{1}^{(1)}\). From the truth table, we can obtain \(f_{1}^{(1)}=0\) if \((x_{1},x_{2})=(0,0)\), \((0,1),(1,1)\) and \(f_{1}^{(1)}=1\) if \((x_{1},x_{2})=(1,0)\). Then, we can obtain the following matrix representation:_  
\[A_{1}^{(1)}= \begin{bmatrix}1&1&0&0\\ 0&0&1&0\end{bmatrix} .\]  
_The second row in this matrix corresponds to the truth table. In a similar way, we can obtain other matrices as follows:_  
\[A_{2}^{(1)}= \begin{bmatrix}0&0&1&1\\ 1&1&0&0\end{bmatrix} ,\;\;A_{1}^{(2)}= \begin{bmatrix}1&0&0&0\\ 0&1&1&1\end{bmatrix} ,\;\;A_{2}^{(2)}= \begin{bmatrix}1&1&0&0\\ 0&0&1&1\end{bmatrix} ,\;\;A_{3}^{(2)}= \begin{bmatrix}0&1&0&1\\ 1&0&1&0\end{bmatrix}