$2\pi^{-}$, from working hypothesis 3.2 (\[work2\]), is the clock $O_{1}$ at vertex $A$. The situation before $O_{1}$ receives its kick of energy is when the phase of this clock is $2\pi^{-}$, i.e., the cycles is complete.

At this point we are able to describe what happens to the phases after a complete cycle of the reference clock.

We have $$%
\begin{cases}
\psi_{1}^{6} & ={2\pi}^{-}\\
\psi_{2}^{6} & =\psi_{2}^{5}+2\pi-\psi_{1}^{5}\\
& \simeq2\pi+2\pi-\left(  2\pi-{\psi}_{2}^{0}-2P\left(  {\psi}_{2}^{0}\right)
-P\left(  {\psi}_{3}^{0}\right)  -P\left(  {\psi}_{2}^{0}-{\psi}_{3}%
^{0}\right)  \right) \\
& =2\pi+{\psi}_{2}^{0}+2P\left(  {\psi}_{2}^{0}\right)  +P\left(  {\psi}%
_{3}^{0}\right)  +P\left(  {\psi}_{2}^{0}-{\psi}_{3}^{0}\right)  ;\\
\psi_{3}^{6} & =\psi_{3}^{5}+2\pi-\psi_{1}^{5}\\
& \simeq2\pi-{\psi}_{2}^{0}+{\psi}_{3}^{0}-P\left(  {\psi}_{2}^{0}\right)
+P\left(  {\psi}_{3}^{0}\right)  -2P\left(  {\psi}_{2}^{0}-{\psi}_{3}%
^{0}\right)  +2\pi\\
& -\left(  2\pi-{\psi}_{2}^{0}-2P\left(  {\psi}_{2}^{0}\right)  -P\left(
{\psi}_{3}^{0}\right)  -P\left(  {\psi}_{2}^{0}-{\psi}_{3}^{0}\right)  \right)
\\
& =2\pi+{\psi}_{3}^{0}+P({\psi}_{2}^{0})+2P({\psi}_{3}^{0})-P({\psi}_{2}%
^{0}-{\psi}_{3}^{0}).
\end{cases}$$

Now, we compute the phase differences after the first cycle of $O_{1}$.

We have $$\begin{aligned}
(BA)_{I}  &  =-(AB)_{I}=\psi_{2}^{6}-\psi_{1}^{6}\\
&  \simeq2\pi+{\psi}_{2}^{0}+2P\left(  {\psi}_{2}^{0}\right)  +P\left(  {\psi
}_{3}^{0}\right)  +P\left(  {\psi}_{2}^{0}-{\psi}_{3}^{0}\right)  -2\pi\\
&  ={\psi}_{2}^{0}+2P\left(  {\psi}_{2}^{0}\right)  +P\left(  {\psi}_{3}%
^{0}\right)  +P\left(  {\psi}_{2}^{0}-{\psi}_{3}^{0}\right) \\
&  =(BA)_{0}+2P((BA)_{0})+P((CA)_{0})+P((BA)_{0}-(CA)_{0})\\
&\end{aligned}$$ and$$\begin{aligned}
&  (CA)_{I}\\
&  =-(AC)_{I}=\psi_{3}^{6}-\psi_{1}^{6}\\
&  =2\pi+{\psi}_{3}^{0}+P({\psi}_{2}^{0})+2P({\psi}_{3}^{0})-P({\psi}_{2}%
^{0}-{\psi}_{3}^{0})-2\pi\\
&  ={\psi}_{3}^{0}+P({\psi}_{2}^{0})+2P({\psi}_{3}^{0})-P({\psi}_{2}^{0}%
-{\psi}_{3}^{0})\\
&  =({(CA)}_{0})+P({(BA)}_{0})+2P({(CA)}_{0})-P((BA)_{0}-(CA)_{0})\\
&\end{aligned}$$ Hence, if we set $x=BA$ and $y=CA$, we obtain the system $$\left\{
\begin{array}
[c]{c}%
x_{1}=x_{0}+2P(x_{0})+P(y_{0})+P(x_{0}-y_{0})\\
y_{1}=x_{0}+P(x_{0})+2P({y}_{0})-P(x_{0}-y_{0}).
\end{array}
\right.$$

**THE MODEL**

By iterating the argument above, we get, for $n$ equal to the number of 