$i \in I \setminus \{2\}$), respectively. This property ensures that, when in the desynchronization set, the relative distance between the leading timer and each of the other timers is equal, before and after jumps. More precisely, $$\begin{aligned}
{\widetilde}\tau_{1}^{k} - \widetilde{\tau}_{i}^{k} = {\widetilde}\tau_{2}^{k}\null^{+} - {\widetilde}\tau_{\mbox{\scriptsize next}(i)}^{k}\hspace{-0.21in}\null^{+} \qquad
\qquad \forall \ i \in I \setminus \{1\} \label{eqn:ANcondition},\end{aligned}$$ where ${\widetilde}\tau^{k}\null^{+} = G({\widetilde}\tau^{k})$ and next$(i) = i + 1$ if $i + 1 \leq N$ and $1$ otherwise.[^5] Since ${\mathcal{X}}$ contains all points such that at least two or more timers are the same, we can consider the case when one component of ${\widetilde}\tau^k$ is equal to ${\bar{\tau}}$ at a time. For each such case, we have $(N - 1)!$ possible permutations of the other components and $N$ possible timer components equal to ${\bar{\tau}}$, leading to $N!$ total possible sets $\ell_{k}$.

For the $N$ case, the algorithm above results in the system of equations $\Gamma \tau_s = b$, where and $
b = \bar{\tau} {\bf 1},
$ where $\tau_s$ is the state ${\widetilde}\tau^{k}$ sorted into decreasing order. It can be shown that for any $\varepsilon \in (-1,0)$, a solution $\tau_{s}$ exists (see Lemma \[eqn:tauSsolution\]). Then, $\tau_s$ needs to be unsorted and becomes ${\widetilde}\tau^k$ in 