equal since the network is connected, due to Assumption \[Ass:Network\]. Note that they can be written more compactly as $(B^\top \otimes I_n) \bar{x} = 0$, where $B \in \mathbb{R}^{P\times E}$ is the node arc-incidence matrix of the graph, $I_n$ is the identity matrix in $\mathbb{R}^n$, and $\otimes$ is the Kronecker product. Each column of $B$ is associated with an edge $\{i,j\} \in \mathcal{E}$ and has $1$ and $-1$ in the $i$th and $j$th entry, respectively; the remaining entries are zeros. We can partition $B$ into two matrices $B_1$ and $B_2$, where $B_1$ contains the first $C_1$ rows and $B_2$ contains the remaining rows. Therefore, $(B^\top \otimes I_n)\bar{x} = (B_1^\top \otimes I_n)\bar{x}_1 + (B_2^\top \otimes I_n)\bar{x}_2$, where $\bar{x}_1 = (x_1,\ldots,x_{C_1}) \in (\mathbb{R}^{n})^{C_1}$ and $\bar{x}_2 = (x_{C_1 + 1},\ldots,x_{C}) \in (\mathbb{R}^{n})^{C_2}$. This enables rewriting  as $$\label{Eq:BipClonedProb2}
    \begin{array}{ll}
      \underset{\bar{x}_1,\bar{x}_2}{\text{minimize}} & \sum_{p \in \mathcal{C}_1}f_p(x_p) + \sum_{p \in \mathcal{C}_2}f_p(x_p) \\
      \text{subject to} & \bar{x}_1 \in \bar{X}_1\,\,,\,\, \bar{x}_2  \in \bar{X}_2\\
                        & (B_1^\top \otimes I_n) \bar{x_1} + (B_2^\top \otimes I_n) \bar{x_2} = 0\,,
    \end{array}$$ where $\bar{X}_i = \bigcap_{p \in \mathcal{C}_i} X_p$, for $i=1,2$. Problem  can be solved with ADMM, explained next.

[[**ADMM.**]{}]{} The alternating direction method of multipliers (ADMM) was proposed in the seventies by [@Glowinski75-ADMMFirst; @Gabay76-ADMMFirst]. 