simultaneously solving for the best response of each player. In [@BKA10], an algorithm that finds the full set of Nash equilibria in exponential time is proposed. For a vector $%
\mathbf{x}$, let $S$ denote the set of active agents, i.e., $S=\{i:x_{i}>0\}$. Let $\mathbf{x}_{S}$ denote the vector of actions of the agents in $S$. The set of active players induce a subgraph $\mathcal{G}_{S}\subseteq 
\mathcal{G}$, with node-set $S\subseteq \mathcal{V}\left( \mathcal{G}\right) 
$ and a set of edges $\mathcal{E}_{S}\subseteq \mathcal{E}\left( \mathcal{G}%
\right) $ connecting active agents. We denote by $\mathcal{G}_{\mathcal{N}%
\backslash S,S}$ the subgraph of $\mathcal{G}$ whose edges connect active agents in $S$ to inactive agents in $\mathcal{N}\backslash S$. The adjacency matrices of $\mathcal{G}_{S}$ and $\mathcal{G}_{\mathcal{N}\backslash S,S}$ are denoted by $A_{S}\in \mathbb{R}^{\left\vert S\right\vert \times
\left\vert S\right\vert }$ and $A_{\mathcal{N}\backslash S,S}\in \mathbb{R}%
^{n-\left\vert S\right\vert \times \left\vert S\right\vert }$, respectively. Then, one can show the following [@BKA10]:

\[Nash Conditions\]A profile $\mathbf{x}$ with active agents $S$ is a Nash equilibrium if and only if: $$\begin{aligned}
(I_{\left\vert S\right\vert }+\delta A_{S})\mathbf{x}_{S} &=&\mathbf{1}%
_{\left\vert S\right\vert }\mathbf{,}\text{ and} \\
\delta A_{\mathcal{N}\backslash S,S}~\mathbf{x}_{S} &\geq &\mathbf{1}%
_{n-\left\vert S\right\vert }\mathbf{,}\end{aligned}$$where $I_{p}$ is the $p\times p$ identity matrix and $\mathbf{1}_{q}$ is the $q$-dimensional vector of ones.

Thus, in order to determine the complete set of 