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\section*{\it {Supplementary materials}}


%%%%%%%%%%%%%                           %Mutated TPM decomposition               %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection*{Mutated TPM decomposition}
Let $\mathbf{P}$ be the probability transition matrix
of a BNp defined on $n$ genes with perturbation probability $p$ and suppose
the network undergoes a single mutation, $(i,k)$. Then the transition
probability matrix $\tilde{\mathbf{P}}$ of the mutated BNp possesses the
decompostion 
\begin{equation}
\tilde{\mathbf{P}}= \mathbf{P}%
+(1-p)^{n}\mathbf{Q}(\mathbf{T}_{i,k}-\mathbf{I}) = \mathbf{P}+(1-p)^{n}.\sum_{j=1}^{u}{\mathbf{a}_{j}.%
\mathbf{b}_{j}^{T}},  \label{decompositions}
\end{equation}%
for some vectors $\mathbf{a}_{j}$ and $\mathbf{b}_{j}$ satisfying $\mathbf{b}%
_{j}^{T}.\mathbf{e}=0$, where $u\leq 2^{n-1}$ is a positive integer. 

Proof: Let $(x_1,...,x_n)$ denote the GAP of the network. We note that there is a natural bijection between GAP and its decimal representation, which takes values in $\{0,1,...,2^{n}-1\}$ which for the sake of simplicity is changed to $\{1,...,2^n\}$. Let $\mathcal S = \left\{{1,...,{2^n}}\right\} $ denote the state space. Based on the given mutation $(i,k)$, we can partition state space as follows: 

\[ \mathcal {P_S} = \left\{ s_1,...s_m\right\}\] 

where $m$ and $s_j$s are defined as

\[
\begin{array}{c}

m = \frac{2^n}{r}\\
s_j = \left\{{(j-1)r+1},...,{jr}\right\}, j=1,...,m\\
\end{array}
\]
where $r = 2^{n-i+1}.$ Now, suppose we define the transformation matrix as ${\bf T}_{i,k} = \left[{\bf t}_1,...,{\bf t}_{2^n}\right]$. In order to find ${\bf T}_{i,k}$, we split it in two cases:
\begin{itemize}
\item $k=0$: we define a bijection mapping from $s_{2j}$ to $s_{2j-1}$ for all $j=1,...,\frac{m}{2}$ as 

\[ {(2j-2)r+l} \leftarrow {(2j-1)r+l}, \forall l \in\left\{{1,...,r}\right\}, \]
which means that the column corresponding to state ${(2j-1)r+l}$ in matrix $Q$ should be added to the column  ${(2j-2)r+l}$ of this matrix, while the ${(2j-1)r+l}$th column should be changed to the all-zero vector. 

So, assuming ${\bf T}_{i,k}=[{\bf t}_1,{\bf t}_2,...,{\bf t}_{2^n}]$, we would have

\begin{equation}\label{t-for-zero} {\bf t}_u = \begin{cases} {\bf 0} &; \forall u \in s_{2j}\\
\begin{cases} t_{u,u}=1& \\t_{v,u}=1;&\forall v \in s_{2j}\\t_{v,u}=0;& \text{otherwise} \end{cases} 
&; \forall u \in s_{2j-1}\end{cases} ; \forall j=1,...,\frac{m}{2}. \end{equation}

\item $k=1$: in this case we have another bijection mapping from $s_{2j-1}$ to $s_{2j}$ for all $j=1,...,\frac{m}{2}$ as follows

\[ {(2j-2)r+l} \rightarrow {(2j-1)r+l}, \forall l \in\left\{{1,...,r}\right\}. \]

then we would have

\begin{equation} \label{t-for-one}{\bf t}_u = \begin{cases} \bf{0}&; \forall u \in s_{2j-1}\\
\begin{cases} t_{u,u}=1& \\t_{v,u}=1;&\forall v \in s_{2j-1}\\t_{v,u}=0;& \text{otherwise} \end{cases} 
&; \forall u \in s_{2j}\end{cases} ; \forall j=1,...,\frac{m}{2}. \end{equation}
\end{itemize}
Hence, the overall altered transition matrix (i.e. under a given mutation $(i,k)$) of a given BNp with transition matrix ${\bf P}= (1-p)^n.{\bf Q}+{\bf H}$ can be written as 

\begin{equation} \label{additivepart} \tilde {\bf P} = {\bf P} + \underbrace{(1-p)^n.{\bf Q}.({\bf T}_{i,k}-{\bf I})}_{\text {additive  perturbation}}.\end{equation}

As we know, in general for any two matrices ${\bf A}$ and ${\bf B}$ we have \[Rank({\bf AB})\leq \min \left\{Rank({\bf A}),Rank({\bf B})\right\}.\] Now we prove the following:

For every transformation matrix, ${\bf T}_{i,k}$, related to the mutation $(i,k)$ in an n-gene network, we have 
\begin{equation}\label{rank} Rank({\bf T}_{i,k}-{\bf I})= 2^{n-1}.\end{equation}

proof: It is enough to show that ${\bf T}_{i,k}-{\bf I}$ has exactly $2^{n-1}$ nonzero eigenvalues. Every ${\bf T}_{i,k}$ can be considered as a transition matrix of a BN itself which exactly has $2^{n-1}$ singleton attractors. It shows that ${\bf T}_{i,k}$ has $2^{n-1}$ eigenvalues with value of $1$ and $2^{n-1}$ with value $0$, causing ${\bf T}_{i,k}-{\bf I}$ to have exactly $2^{n-1}$ zero eigenvalues and $2^{n-1}$ eigenvalues with value $-1$.  

From \eqref{additivepart} and \eqref{rank} we obtain that rank of perturbation is less than $2^{n-1}$, so 
\begin{equation*}
Rank(\text{additive perturbation})\leq \min \left\{2^{n-1},Rank(\mathbf{Q}%
)\right\},
\end{equation*}
showing that the additive perturbation can be decomposed into at most $%
2^{n-1}$ rank-one matrices. Now, we want to show that these
can be in the form of rank-one perturbations. Let ${\mathbf{b}_j}^T=\left[b_j^{(1)},...b_j^{(2^n)}%
\right]$, and $\mathbf{a}_j$ make the $j$th rank-one perturbation. It can be
shown that the corresponding components of these rank-one
perturbation are as follow

\begin{itemize}
\item $k=0$
\begin{equation}\label{general-b-k-0}
b_{(y-1)r+z}^{(w)}=
\begin{cases} 1& \text{for $w= (2y-1)r+z-1$}\\
-1& \text{for $w=2yr+z-1$}\\
0& \text{otherwise}
\end{cases} ; \forall z \in \left\{1,...,r\right\}  \forall y \in \{1,...,\frac{m}{2}\}
\end{equation}

and 

\begin{equation}\begin{array}{cccc}
{\bf a}_{(y-1)r+z}={\bf q}_{{l_z}};& \forall l_z  \in s_{2w},&  \forall z \in \left\{1,...,r\right\},& \forall y \in \left\{1,...,\frac{m}{2}\right\}.
\end{array}
\end{equation}

\item $k=1$
\begin{equation}\label{general-b-k-1}
b_{(y-1)r+z}^{(w)}= 
\begin{cases} -1& \text{for $ w= (2y-1)r+z-1$,}\\
1& \text{for $w=2yr+z-1$,}\\
0& \text{otherwise}
\end{cases} ; \forall z \in \{1,...,r\}  \forall y \in \{1,...,\frac{m}{2}\}
\end{equation}

and 


\begin{equation}
\begin{array}{cccc}
{\bf a}_{(y-1)r+z} = {\bf q}_{{l_z}};& \forall l_z  \in s_{2w-1},&  \forall z \in \left\{1,...,r\right\}, &\forall y \in \left\{1,...,\frac{m}{2}\right\}.
\end{array}
\end{equation}

As \eqref{general-b-k-0} and \eqref{general-b-k-1} show, ${\bf b}_j$s satisfy ${\bf b}_j^T.{\bf e}=0$. 


\end{itemize}

\subsection*{Commutativity property of transformation matrices}

For transformation matrices representing gene mutations we have 

\begin{equation}\label{commutative} {\bf T}_{i_1,k_1}.{\bf T}_{i_2,k_2} = {\bf T}_{i_2,k_2}.{\bf T}_{i_1,k_1} , 
\forall i_1 \neq i_2, \forall k_1,k_2 \in \{0,1\}. \end{equation}

Proof: Let $\mathbf{X} = (x_1,...,x_n)$ determine the GAP. From \eqref{t-for-zero} and \eqref{t-for-one} one can see that each ${\bf T}_{i,k}$ corresponds to a BN itself which is shown here by $N_{i,k}$. We start our proof from the right hand side of \eqref{commutative}: ${\bf T}_{i_1,k_1}.{\bf T}_{i_2,k_2}$ is an altered copy of $N_{i_1,k_1}$ after the mutation $(i_2,k_2)$. It can be easily seen that, $N_{i_1,k_1}$ has ${2^{n-1}}$ basins in the general following form: 

\[\begin{array}{cc} & \mathbf{X} = (x_1,...x_{i_1-1},1-k_1,x_{i_1+1},...,x_n)\rightarrow {\hat{s}} = (x_1,...x_{i_1-1},k_1,x_{i_1+1},...,x_n)\rightarrow {\hat{\mathbf{X}}} 
,\\ 
&\forall (x_1,...x_{i_1-1},x_{i_1+1},...,x_n)\in \{0,1\}^{n-1},\end{array}\]

which contains both the following transitions

\begin{equation} \label{transitions-in-the-first-matrix}\begin{array}{c}\mathbf{X} =(x_1,...x_{i_1-1},1-k_1,...,x_{i_2-1},1-k_2...,x_n)\rightarrow {\tilde{\mathbf{X}}}= (x_1,...x_{i_1-1},k_1,...,x_{i_2-1},1-k_2,...,x_n)\rightarrow {\tilde{\mathbf{X}}}\\ \mathbf{X} =(x_1,...x_{i_1-1},1-k_1,...,x_{i_2-1},k_2...,x_n)\rightarrow \tilde{\mathbf{X}} = (x_1,...x_{i_1-1},k_1,...,x_{i_2-1},k_2,...,x_n)\rightarrow {\tilde{\mathbf{X}}},\end{array}\end{equation} 
where without loss of generality we assumed $i_1 \leq {i_2 - 1}$. (we already know that $i_1\neq i_2.$) 
Now, we apply the matrix ${\bf T}_{i_2,k_2}$ to the transitions in \eqref{transitions-in-the-first-matrix} which produces the following transitions:

\begin{equation} \label{transitions-final}\begin{array}{cc}
& (x_1,...x_{i_1-1},1-k_1,...,x_{i_2-1},1-k_2...,x_n)\rightarrow {\hat{\mathbf{X}}}\rightarrow {\hat{\mathbf{X}}} \\
& (x_1,...x_{i_1-1},k_1,...,x_{i_2-1},1-k_2,...,x_n)\rightarrow {\hat{\mathbf{X}}}\rightarrow {\hat{\mathbf{X}}}\\ 
& (x_1,...x_{i_1-1},1-k_1,...,x_{i_2-1},k_2...,x_n)\rightarrow {\hat{\mathbf{X}}}\rightarrow {\hat{\mathbf{X}}}
,\end{array}\end{equation} 

where ${\hat{\mathbf{X}}}= (x_1,...x_{i_1-1},k_1,...,x_{i_2-1},k_2,...,x_n)$, and $(x_1,...x_{i_1-1},x_{i_1+1},...,x_{i_2-1},x_{i_2+1}...,x_n)\in \{0,1\}^{n-2}$. So, the final altered BN would have $2^{n-2}$ basins in the form of \eqref{transitions-final}. 

Now, suppose that one wants to compute ${\bf T}_{i_2,k_2}.{\bf T}_{i_1,k_1}$: in this case we should alter $N_{i_2,k_2}$ by the mutation $(i_1,k_1)$. In a similar way, $N_{i_2,k_2}$ has ${2^{n-1}}$ basins in the general following form: 

\[\begin{array}{cc} &\mathbf{X} = (x_1,...x_{i_2-1},1-k_2,x_{i_2+1},...,x_n)\rightarrow {\hat{\mathbf{X}}} = (x_1,...x_{i_2-1},k_2,x_{i_2+1},...,x_n)\rightarrow {\hat{\mathbf{X}}} 
,\\ &\forall (x_1,...x_{i_2-1},x_{i_2+1},...,x_n)\in \{0,1\}^{n-1},\end{array}\]

which contains both the following transitions 
\begin{equation} \label{transitions-in-the-first-matrix-2}\begin{array}{c}\mathbf{X} =(x_1,...x_{i_1-1},1-k_1,...,x_{i_2-1},1-k_2...,x_n)\rightarrow {\tilde{\mathbf{X}}}= (x_1,...x_{i_1-1},1-k_1,...,x_{i_2-1},k_2,...,x_n)\rightarrow {\tilde{\mathbf{X}}}\\ \mathbf{X} =(x_1,...x_{i_1-1},k_1,...,x_{i_2-1},1-k_2...,x_n)\rightarrow \tilde{\mathbf{X}} = (x_1,...x_{i_1-1},k_1,...,x_{i_2-1},k_2,...,x_n)\rightarrow {\tilde{\mathbf{X}}}.\end{array}\end{equation} 
Now, we apply the matrix ${\bf T}_{i_1,k_1}$ to the transitions in \eqref{transitions-in-the-first-matrix} which exactly produces the transitions in \eqref{transitions-final}. This shows that the produced BNs with the both multiplication orderings are the same and so are their transition matrices meaning that  ${\bf T}_{i_1,k_1}.{\bf T}_{i_2,k_2} = {\bf T}_{i_2,k_2}.{\bf T}_{i_1,k_1}.$


%%%%%%%%%%%%%%                     Approximated probability of algorithm failure                       %%%%%%%%%%%%%%%

\subsection*{Approximated probability of algorithm failure}
Suppose that the probability of algorithm failure is defined as 
 \begin{equation}  \label{p-miss-definition appendix}
P_{miss} = \Pr(\left[N_{normal};Path_c\right]\notin {\mathcal{BN}}^M).
\end{equation}

Define
\begin{equation} \label{corresponding}
p_i := {F_D}^{(M-i,{p_{cancer}})}(\beta_i),\forall i=1,...,M \newline
\end{equation}
where ${F_D}^{(M-i,{p_{cancer}})}(d)$ is the CDF of the distance $d$ between a BNp and its $(M-i)$-fold mutated counterpart. 

Defining 

\begin{align} \label{error approximate}
& p_{miss}^{(i+1)} =
(1-p_{i+1})^{M-i}(1-p_{miss}^{(i)})+p_{miss}^{(i)}; 1 \leq i\leq {M-1} \\
\nonumber &p_{miss}^{(1)} = (1-p_1)^M
\end{align}

we will have the following approximation 

\begin{equation} \label{missing error}
P_{miss} \approx p_{miss}^{(M)}.
\end{equation}


Proof: We start proving \eqref{error approximate} by finding the probability of missing in the first step of the algorithm. As we mentioned before, $Path_c$ is a {\it set} of alterations meaning that the occurrence ordering does not matter. Hence the probability of missing in the first step can be found as follows

\begin{equation}\label{appendix-p-miss-first}\begin{array}{cc} p^{(1)}_{miss}  & = Pr(\nexists alt \in Path_c \ni [N_{normal};alt] \in \mathcal{BN}^1)\\
&\\
 &= \Pr( \nexists alt \in Path_c \ni R([N_{normal};alt]) \leq \beta_1) \\
&\\
 &\stackrel{A}{=} \Pr( \forall alt \in Path_c : \rho(SSD_{cancer},SSD([N_{normal};alt]))> \beta_1)\\
 &\\
&\stackrel{B}{\approx} \left\{{\Pr}( \rho(SSD_{cancer},SSD([N_{normal};alt])) > \beta_1)\right\} ^{M}\\
 &\\
 &= {\{1- ({F_D}^{(M-1,p)}(\beta_1))\}}^M 
 , \end{array}\end{equation}
 
 
where in $B$, we assume independent which may not happen in general. \begin{comment}Generally, $p_{cancer}$ is unknown and has to be estimated from the data (e.g. gene expression); however, in our approximation we assume exact estimation.  \end{comment}
Since $Path$ does not have any order, there are many possibilities that can happen for the second step. We approximate this probability as
 
\begin{align} p^{(2)}_{miss} = \sum_{i=0}^{M}{\left(\begin{array}{c}M\\i\end{array}\right)\Pr\left\{miss|E_i\right\}P(E_i)},\end{align}
 
where $E_i$ is an {\it event} in which only $M-i$ mutations from $Path_c$ make the residual less than $\beta_1$ in the first iteration. We obtain 

\begin{equation}\begin{array}{cc} p^{(2)}_{miss}&=\sum_{i=0}^{M-1}{\left(\begin{array}{c}M\\i\end{array}\right)\Pr\left\{{miss}|E_i\right\}p(E_i)}+\Pr\left\{miss|E_M\right\}p(E_M)\\
&\\
 &= \sum_{i=0}^{M-1}{\left(\begin{array}{c}M\\i\end{array}\right)\Pr\left\{miss|E_i\right\}p(E_i)} + p^{(1)}_{miss}\\
 &\\
 &\le \Pr\left\{miss|E_{M-1}\right\}\sum_{i=0}^{M-1}{\left(\begin{array}{c}M\\i\end{array}\right)p(E_i)} +  p^{(1)}_{miss}\\
 &\\
&= \Pr\left\{{miss}|E_{M-1}\right\}(1-{p_{miss}}^{(1)})+p^{(1)}_{miss}\end{array} \end{equation}

which similar to \eqref{appendix-p-miss-first} can be written as 

\[\begin{array}{c}
p^{(2)}_{miss} \le \Pr\left\{miss|E_{M-1}\right\}(1-{p_{miss}}^{(1)}) +  p^{(1)}_{miss}\\
\\
\approx (1-{p_{miss}}^{(1)}){\{1- ({F_D}^{(M-2,p)}(\beta_2)\}}^{M-1} +  p^{(1)}_{miss}
\end{array}\]
This procedure should be done for the next steps as before which completes the proof. 

        
    
    
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