deletions | additions       diff --git a/section_Unanimity_and_2_colorability__.tex b/section_Unanimity_and_2_colorability__.tex  index eef2e6e..d1e896d 100644  --- a/section_Unanimity_and_2_colorability__.tex  +++ b/section_Unanimity_and_2_colorability__.tex 
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Let $G$ be an influence profile and $\O$ be an opinion profile, such that, for some $p\in \Atoms $, for all $i,j\in\N$: if $i\in R_p(j)$, then $O_i(p)\neq O_j(p)$. Then $\O$ does not converge in UP.  \end{lemma}  Recall that a graph is properly colored if no node has a successor of the same color. color, and consider as two colors the two possible opinions on issue $p$.  The above result can be reformulated as follows: if for some $p\in\Atoms$, $\O$ properly colors $G_p$, then $\O$ does not converge. In fact, it is easy to see that in such a case, all agents will change their opinion on $p$ at every step, entering an oscillation of size $2$. Note that this limit case of opinion distribution is yet another special case of DeGroot processes, another example within the intersection between the two frameworks of propositional opinion diffusion and DeGroot. The possibility of this extreme distribution of opinions relies on the graph $G_p$ being $2$-colorable, which is equivalent to $G_p$ not containing any cycle of odd length. However, non $2$-colorability is not a sufficient condition for convergence of UPs in general: a simple cycle of three agents, for instance, is not $2$-colorable but does not guarantee convergence either (as illustrated above with the convergence conditions for BDPs).   Nevertheless, there is a class of influence profiles for which being $2$-colorable is a necessary condition of non-convergence of UPs, the \emph{symmetric} ones: