deletions | additions       diff --git a/section_Convergence_label_sec_convergence__.tex b/section_Convergence_label_sec_convergence__.tex  index d6d8964..45ad524 100644  --- a/section_Convergence_label_sec_convergence__.tex  +++ b/section_Convergence_label_sec_convergence__.tex 
...
This brings us back to convergence result for general (not necessarily Boolean) DeGroot processes. Indeed, for functional and serial influence graphs, a closed connected component is aperiodic if and only if its cycle is of length $1$.   \begin{lemma}\label{lemma:influence}  Let $\G=(G_{p_1},\dots,G_{p_m})$ $\G$  be an influence profile. Then the following are equivalent: \begin{itemize}  \item[] The BDP converges for any opinion profile $\O$ on $\G$   \item[] For all $j\in \{1,\ldots,m\}$, $G_{p_j}$ $p\in \Atoms$, $G_{p}$  contains no cycle of length $\geq 2$. \item[] For all $j\in \{1,\ldots,m\}$, $p\in \Atoms $,  all closed connected components of $G_{p_j}$ $G_{p}$  are aperiodic. \end{itemize}  \end{lemma}  \begin{proof}  Let $j\in\{1,\dots,m\}$ $p\in\Atoms$  and assume that $G_{p_j}$ $G_{p}$  contains no cycle of length $\geq 2$ and has diameter $k$. Let $C_{p_j}$ $C_{p}$  be a connected component of $G_{p_j}$. $G_{p}$.  By \ref{fact:uniquecycle}, $C_{p_j}$ $C_{p}$  contains a unique cycle, which, by assumption, is of length $1$. Hence, $C_{p_j}$ $C_{p}$  is aperiodic. Let $i$ be the node in the cycle. The opinion of $i$ will spread to all nodes in $C_{p_j}$ $C_{p}$  after at most $k$ steps. Therefore, all BDPs on $G$ will converge after at most $l$ steps, where $l$ is the maximum within the set of diameters of $G_{p_j}$ $G_{p}$  for all $j\in \{1,\ldots,m\}$. $p\in\Atoms$.  Assume that for some $j\in\{1,\dots,m\}$, $p\in\Atoms$,  a connected component $C_{p_j}$ $C_{p}$  of $G_{p_j}$ $G_{p}$  contains a cycle of length $k\geq 2$. By \ref{fact:uniquecycle}, this cycle is unique, and therefore the greatest common divisor of the cycles lengths of $C_{p_j}$ $C_{p}$  is $k$, so $C_{p_j}$ $C_{p}$  is not aperiodic. Let $S$ be the set of nodes in the cycle.   Let $\O$ be such that for some $i,j\in S$ with distance $d$ from $i$ to $j$, $O_i{p_j}\neq O_j{p_j}$. $O_i{p}\neq O_j{p}$.  Then $O_i{p_j}$ $O_i{p}$  will not converge, but enter a loop of size $k$: for all $x\in\mathbb{N}$, $O^{x\times k}_i{p_j}=\neq k}_i{p}=\neq  O^{(x\times k)+d}_i{p_j}$. k)+d}_i{p}$.  \end{proof}  Note that that one direction of the above result also follows both from the DeGroot convergence result stated earlier (since graphs with only cycles of length $1$ are trivially aperiodic) and from the propositional opinion diffusion sufficient condition above (since the unanimity rule is trivially monotonic).