deletions | additions       diff --git a/section_Convergence_label_sec_convergence__.tex b/section_Convergence_label_sec_convergence__.tex  index 1ed2e33..fd52a06 100644  --- a/section_Convergence_label_sec_convergence__.tex  +++ b/section_Convergence_label_sec_convergence__.tex 
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\fbox{$1) \Rightarrow 3)$} We proceed by contraposition.   Assume that for some $p\in\Atoms$, a connected component $C_p$ of $G_p$ 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 $C_p$ is $k$, so $C_p$ 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 $j$, $\O_i(p)\neq \O_j(p)$. $O_i(p)\neq O_j(p)$.  Then $\O_i(p)$ will not converge, but enter a loop of size $k$: for all $x\in\mathbb{N}$, $\O^{x\times $O^{x\times  k}_i(p) \neq \O^{(x\times O^{(x\times  k)+d}_i(p)$. Hence, $\O$ does not converge. \fbox{$3) \Rightarrow 2)$} Trivial.  \end{proof}  It is worth noticing that one direction (namely from $3$ to $1$) of the above result is actually a corollary of both the convergence result for DeGroot processes stated at the beginning of this section (cf. \cite{jackson08social}), and of a known convergence result for propositional opinion diffusion \cite[Th. 2]{Grandi:2015:POD:2772879.2773278}, also stated earlier. 
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Let $\G$ be an influence profile and $\O$ be an opinion profile. Then the following statements are equivalent:  \begin{enumerate}  \item The BDP converges for $\O$ on $\G$.  \item For all $p\in \Atoms$, there is no set of agents $C\subseteq\N$ such that: $C$ is a cycle in $G_{p}$ and there are two agents $i,j\in C$ such that $\O_i(p)\neq \O_j(p)$. $O_i(p)\neq O_j(p)$.  \end{enumerate}  \end{theorem}  \begin{proof}  \fbox{$1) \Rightarrow 2)$} We proceed by contraposition.  Let $p\in\Atoms$ and assume $C\subseteq\N$ be such that $C$ is a cycle in $G_p$, $i,j\in C$, and $\O_i(p)\neq \O_j(p)$. $O_i(p)\neq O_j(p)$.  Let $k$ be the length of the cycle and $d$ be the distance from $i$ to $j$. Then $\O_i(p)$ $O_i(p)$  will enter a loop of size $k$: for all $x\in\mathbb{N}$, $\O^{x\times $O^{x\times  k}_i(p)\neq \O^{(x\times O^{(x\times  k)+d}_i(p)$. \fbox{$2) \Rightarrow 1)$}  Assume $C\subseteq\N$ be such that $C$ is a cycle in $G_p$, and for all $i,j\in C$, $\O_i(p)=\O_j(p)$. $O_i(p)=O_j(p)$.  Then, for all $j\in C$, and all $x\in\mathbb{N}$, $\O^{x}_j(p)=\O_i(p)$ $O^{x}_j(p)=O_i(p)$  and for all $f\in\N\notin C$ with distance $d$ from $f$ to $i$, for all $x\in\mathbb{N}$, such that $x\geq d$, $\O^{x}_f(p)=\O_i(p)$. $O^{x}_f(p)=O_i(p)$.  \end{proof}  This trivially implies that the class of opinion profiles which guarantees convergence for {\em any} influence profile, is the one where everybody agrees on everything already. Note that the only stable distributions of opinions are the ones where, in each connected component in $G$, all members have the same opinion, i.e, on BDPs, converging and reaching a consensus (within each connected component) are equivalent, unlike in the stochastic case. Moreover, for an influence profile where influence graphs have at most diameter $d$ and the smallest cycle in components with diameter $d$ is of length $c$, it is easy to see that if a consensus is reached, it will be reached in at most $d-c$ steps, which is at most $n-1$.