We define ${\bf X}[0,t]=\{{\bf X}(\tau):0<\tau\leq t\}$ to be a sample path of the infection process from $0$ to $t.$ In addition, we define function $F(\cdot)$ such that $$F(X_v(t))=\left\{
  \begin{array}{ll}
    1, & \hbox{if $X_v(t)=I$;} \\
    0, & \hbox{otherwise.}
  \end{array}
\right.$$ We say ${\bf F}(\mathbf{X}[t])={\bf Y}$ if $F(X_v(t))=Y_v$ for all $v.$ Identifying the information source can be formulated as a maximum likelihood detection problem as follows: $$v^{\dag}\in\hbox{arg} \max_{v\in {\cal V}} \sum_{{\bf X}[0,t]: {\bf
F}({\bf X}(t))={\bf Y}} \Pr({\bf X}[0,t]|v^*=v),$$ where $\Pr({\bf X}[0,t]|v^*=v)$ is the probability to obtain sample path ${\bf X}[0,t]$ given the information source is node $v.$

We note the difficulty of solving this maximum likelihood problem is the curse of dimensionality. For each $v$ such that $Y_v=0,$ we need to decide its infection time and recovery time (the node is in susceptible state if the infection time is $>t$), i.e., $O(t^2)$ possible choices; for each $v$ such that $Y_v=1,$ we need to decide the infection time, i.e., $O(t)$ possible choices. Therefore, even for a fixed $t,$ the number of possible sample paths is at least at the order of $t^N,$ where $N$ is the number of nodes in the network. This curse of dimensionality makes it computationally expensive, if not impossible, 