# Finite-Time Set Reachability of Probabilistic Boolean Multiplex Control Networks
## 3 Finite-Time Set Reachability
### Finite-Time Set Reachability with Probability One
\begin{array}{ll}\hat{Q}=(P^{\hat{x}}W_{[2^{KN},2^{m}]})^{\gamma}W_{[2^{m}, 2^{KN}]},\\ \tilde{Q}=(P^{\hat{x}}W_{[2^{KN},2^{m}]})\left(P^{\hat{x}}W_{[2^{KN},2^{m}]} \right)^{\gamma-1}W_{[2^{m},2^{KN}]}.\end{array}  \tag{16}\]  
The \(\gamma\)-step nonzero state overall transition probability matrix \(\hat{Q}\) and the \(\gamma\)-step nonzero state global transition probability matrix \(\hat{Q}\) of RLDS (7) are divided into \(2^{m\gamma}\) matrices as follows:  
\[ \begin{array}{ll}\hat{Q}=[\hat{Q}_{(1)}\ \hat{Q}_{(2)}\ \cdots\ \hat{Q}_{(2^{m \gamma})}],\\ \tilde{Q}=[\hat{Q}_{(1)}\ \hat{Q}_{(2)}\ \cdots\ \hat{Q}_{(2^{m\gamma})}]\end{array}  \tag{17}\]  
The symbol \(\bigwedge\limits_{s}\) can be defined as follows:  
\[\left\{ \begin{array}{ll}Col_{j}(\hat{Q}_{q})\bigwedge\limits_{s}\delta^{0}_{2 ^{KN}}&=\delta^{0}_{2^{KN}},\\ Col_{j}(\hat{Q}_{q})\bigwedge\limits_{s}\delta^{k}_{2^{KN}}&=Col_{j}(\hat{Q}_{q}), \ \ \ k\in[1,2^{KN}]\end{array} \right.\]In the same way, we can obtain  
\[\left\{ \begin{array}{ll}Col_{j}(\tilde{Q}_{q})\bigwedge_{s}\delta_{2^{n}}^{0}= \delta_{2^{n}}^{0},\\ Col_{j}(\tilde{Q}_{q})\bigwedge_{s}\delta_{2^{n}}^{k}=Col_{j}(\tilde{Q}_{q}),&k \in[1,2^{n}]\end{array} \right.\]  
**Theorem 1**. : _For PBMCNs (4), suppose that \(\mathcal{M}_{0}=\{\delta_{2^{KN}}^{j}\mid j\in\Lambda_{0}\}\), where \(\Lambda_{0}\) is a subset of \([1:2^{KN}]\). Then, the following two statements hold._  
1. _An input sequence_ \(\{\boldsymbol{u}(t),t=0,1,\ldots,\gamma-1\}\) _exists such that_ \(\mathcal{M}_{d}\) _is overall reachable with probability one from_ \(\mathcal{M}_{0}\) _on_ \([0:\gamma-1]\) _if and only if_ \[Col_{j}\biggl{[}\bigwedge_{s}^{2^{m\gamma}}\hat{Q}_{(q)}\biggr{]}=\delta_{2^{ KN}}^{0},\qquad\forall j\in\Lambda_{0}.\]
2. _An input sequence_ \(\{\boldsymbol{u}(t),t=0,1,\ldots,\gamma-1\}\) _exists such that_ \(\widetilde{\mathcal{M}}_{d}\) _is global reachable with probability one from_ \(\mathcal{M}_{0}\) _on_ \([0:\gamma-1]\) _if and only if_ \[Col_{j}\biggl{[}\bigwedge_{s}^{2^{m\gamma}}\hat{Q}_{(q)}\biggr{]}=\delta_{2^{ n}}^{0},\qquad\forall j\in\widetilde{\Lambda}_{0}.\]  
Proof. : First, we demonstrate that the necessary condition is satisfied. Known as \(\mathcal{M}_{d}\), it is overall reachable with probability one from \(\mathcal{M}_{0}\) on \([0,\gamma-1]\). Then, for any \(x_{0}=\delta_{2^{KN}}^{j}\) and \(t\geq 0\), there exists an input sequence \(\{\boldsymbol{u}(t),t=0,1,\ldots\gamma-1\}\) such that  
\[ \begin{array}{ll}&\Pr\{\exists\xi\in[0:t-1],s.t.\;\hat{x}(\xi;\eta,x_{0})\in \mathcal{M}_{d}\}\\ &=\Pr\{\exists\xi\in[1:t],s.t.\;\hat{x}(\xi;\eta,x_{0})=\delta_{2^{KN}}^{0}\} \\ &=\Pr\{\hat{x}(t;\eta,x_{0})=\delta_{2^{KN}}^{0}\}\\ &=1\end{array} \]  
Since  
\[\Pr\{\hat{x}(t;\eta,x_{0})=\delta_{2^{KN}}^{0}\}=1-\sum_{i=1}^{2^{KN}}\biggl{[} \hat{Q}_{(q)}\biggr{]}_{ij}=1\]  
for which there exists an input sequence \(\{\boldsymbol{u}(t),t=0,1,\ldots\gamma-1\}\) such that  
\[\sum_{i=1}^{2^{KN}}\biggl{[}\hat{Q}_{(q)}\biggr{]}_{ij}=0\]  
Then, we have  
\[Col_{j}\biggl{[}\hat{Q}_{(q)}\biggr{]}=\delta_{2^{KN}}^{0}\]  
Further, we obtain  
\[Col_{j}\biggl{[}\bigwedge_{s\;q=1}^{2^{m\gamma}}[\hat{Q}_{(q)}]\biggr{]}=\delta _{2^{KN}}^{0}\qquad\forall j\in\Lambda_{0}\]  
Next, we prove the sufficiency condition holds. For any \(j\in\Lambda_{0}\), we have \(Col_{j}\biggl{[}\bigwedge_{s\;q=1}^{2^{m\gamma}}[\hat{Q}_{(q)}]\biggr{]}=\delta _{2^{KN}}^{0}\). Then, we are able to easily verify that there exists an input sequence \(\boldsymbol{u}\) such that  
\[Col_{j}\biggl{[}\hat{Q}_{(q)}\biggr{]}=\delta_{2^{KN}}^{0}\]which is  
\[1-\sum_{i=1}^{2\kappa N}\left[(\hat{Q}_{(q)})\right]_{ij}=1 \tag{18}\]  
Then, (18) becomes  
\[\Pr\{\hat{x}(\gamma+1)=\delta_{2^{KN}}^{0}\mid x(0)=\delta_{2^{KN}}^{j}\}=1\]  
Consequently, there exists an input sequence \(\{\mathbf{u}(t),t=0,1,\ldots,\gamma-1\}\) such that \(\mathcal{M}_{d}\) is overall reachable with probability one from \(\mathcal{M}_{0}\) on \([0,\gamma-1]\). Similarly, we can prove Conclusion 2, no need to repeat the arguments.