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Let us first recall some vocabulary.   Let $G = \tuple{\N, R}$ be a graph and $R^*$ be the transitive and symmetric closure of $R$.   A \emph{path} is a sequence of nodes $$, such that, for all $l\in\{1,\dots,k\}$, $i_lRi_{l+1}$.   The \emph{distance} between two nodes $i,j$ is the length of the shortest path $$ between them.   The \emph{diameter} of a graph is the maximal distance between any two of its nodes.  A \emph{cycle} is a path of length $k$ such that $i_1=i_k$.   A set of nodes $S\subseteq \N$ is said to be:  \begin{itemize} 
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\item[]\emph{aperiodic} if the greatest common divisor of the lengths of its cycles is $1$.  \end{itemize}  Note that the class of graphsour  BDPs come with rely on  (finite, serial and functional) come with present  a special shape: \begin{fact} \label{fact:uniquecycle}  Let $G$ be an influence graph and $C$ be a connected component of $G$.   Then $C$ contains exactly one cycle, and this cycle forms a closed set.  
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Assume that $C$ contains more than one cycle. Since each cycle forms a closed set, there is no path connecting any node inside the cycle to any node outside the cycle, which contradicts connectedness. So $C$ contains a unique cycle, which forms a closed set.   \end{proof}  This means that influence graphs will look like confluent chains aiming together towards cyclical tails. \subsection{Convergence of BDPs} 
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\begin{lemma}\label{lemma:influence}  Let $\G=(G_{p_1},\dots,G_{p_m})$ be an influence profile. Then the following are equivalent:  \begin{itemize}  \item[] The BDP converges for any opinion profile $\O$ on $\G$. $\G$   %\item[] The BDP converges for any opinion profile $\O$ on $\G$ in at most $k$ steps, where $k$ is the maximum in the set of diameters of $G_{p_j} for all $j\in \{1,\ldots,m\}$.  \item[] For all $j\in \{1,\ldots,m\}$, $G_{p_j}$ contains no cycle of length $\geq 2$.  \item[] For all $j\in \{1,\ldots,m\}$, all closed connected components of $G_{p_j}$ are aperiodic.  \end{itemize}  \end{lemma}  \begin{proof}  Let $j\in\{1,\dots,m}$ and assume that  $G_{p_j}$ contain contains  no cycle of length $\geq 2$. 2$ and has diameter $k$. Let $C_{p_j}$ be a connected component of $G_{p_j}$.  ByFact  \ref{fact:uniquecycle}, we know $C_{p_j}$ contains a unique cycle, which, by assumption, is of length $1$. Hence, $C_{p_j}$ is aperiodic. Let $i$ be the node in the cycle. The opinion of $i$ will spread to all nodes in $C_{p_j}$ 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} for all $j\in \{1,\ldots,m\}$.  Assume  that each for some $j\in\{1,\dots,m}$, a  connected component $C_{p_j}$  of $G_{p_j}$ contains a unique cycle, which has to be cycle  of length $1$. $k\geq 2$. By \ref{fact:uniquecycle}, this cycle is unique, and therefore the greatest common divisor of the cycles lengths of $C_{p_j}$ is $k$, so $C_{p_j}$ 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 $d$, $O_i{p_j}\neq O_j{p_j}$. Then $O_i{p_j}$ will not converge, but enter a loop of size $k$: for all $x\in\mathbb{N}$, $O^{x\times k}_i{p_j}=\neq O^{(x\times k)+d}_i{p_j}$.  \end{proof}  The above gives a characterization of the class of influence profiles on which \emph{all} opinion streams converge.