$K$, say $K\geq10$, as long as ${\boldsymbol{M}}$ satisfies that each column contains at least $K$ zeros, and the positions of the zeros appear relatively random, it is very likely that it is sufficiently scattered, and thus can be uniquely recovered via solving .

Algorithm
=========

Our identifiability analysis based on the sufficiently scattered condition poses an interesting non-convex optimization problem . First of all, the given co-occurrence probability ${\boldsymbol{\varOmega}}$ may not be exact, therefore it may not be a good idea to put  as a hard constraint. For algorithm design, we propose the following modification to problem . $$\begin{aligned}
\label{prob:alg}
\operatorname*{\textrm{minimize}}_{{\boldsymbol{\varTheta}},{\boldsymbol{M}}}~~~ & 
\sum_{n,\ell=1}^{N}-\varOmega_{n\ell}\log\!\!\sum_{k,j=1}^{K}\!\!M_{nk}\varTheta_{kj}M_{\ell j} 
 + \lambda|\det{\boldsymbol{\varTheta}}| \nonumber\\
\textrm{subject to}~~~ & {\boldsymbol{M}}\geq0, {\boldsymbol{\mathit{1}}}^{{\!\top\!}}{\boldsymbol{M}}={\boldsymbol{\mathit{1}}}^{{\!\top\!}}, \\
& {\boldsymbol{\varTheta}}\geq0, {\boldsymbol{\varTheta}}{\boldsymbol{\mathit{1}}}={\boldsymbol{\varTheta}}^{{\!\top\!}}{\boldsymbol{\mathit{1}}}, {\boldsymbol{\mathit{1}}}^{{\!\top\!}}{\boldsymbol{\varTheta}}{\boldsymbol{\mathit{1}}}=1. \nonumber\end{aligned}$$ In the loss function of , the first term is the Kullback-Leibler distance between the empirical probability ${\boldsymbol{\varOmega}}$ and the parameterized version ${\boldsymbol{M}}{\boldsymbol{\varTheta}}{\boldsymbol{M}}^{{\!\top\!}}$ (ignoring a constant), and the second term is our identifiability-driven regularization. We need to tune the parameter $\lambda$ to yield good estimation results. However, intuitively we should use a $\lambda$ with a relatively small value. Suppose ${\boldsymbol{\varOmega}}$ is sufficiently accurate, then the priority is to minimize the difference between ${\boldsymbol{\varOmega}}$ and ${\boldsymbol{M}}{\boldsymbol{\varTheta}}{\boldsymbol{M}}^{{\!\top\!}}$; when there exist equally good fits, then 