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I'd just like to know if the following model has received any attention:

A state at discrete time $t$ consists of a function $S_t:{\Bbb Z}^2\rightarrow S^1$.

So view each cell $c$ (element of ${\Bbb Z}^2$) as having eight neighbors in the usual way.

The update rule: set $S_{t+1}(c)$ equal that value $S_{t}(c')$, for $c'$ a neighbor of $c$, occurring nearest to (but not coincident with) $S_t(c)$ when moving clockwise around $S^1$.

Computer simulation suggests that an random initial state quickly evolves to a pattern reminiscent of lichens provided one represents locations on $S^1$ with shades of gray (despite the unfortunate spurious discontinuity). In time some of the "lichens" (distinguished visually one from another by their prevailing shade) grow while others eventually get absorbed.

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  • $\begingroup$ I have no idea if that has been done before, but you could use hue instead of gray-scale to avoid the discontinuity. $\endgroup$
    – user5810
    Commented Apr 14, 2011 at 4:39
  • $\begingroup$ @Ricky Right, have done, but actually the psychological effect is similar as the map from hues to qualia seems to break the symmetry. $\endgroup$
    – David Feldman
    Commented Apr 14, 2011 at 4:43
  • $\begingroup$ Reminds me of the voter models studied in cellular automata: each person does a local poll of his $k$-neighbourhood and adjusts his vote accordingly. $\endgroup$
    – Anthony Quas
    Commented Apr 14, 2011 at 15:51
  • $\begingroup$ I don't know if this has been looked at before either, but it seems to me that it's won't lead to particularly interesting dynamics. Once two neighboring cells have the same shade, they will never change again. Anything adjacent to such a cell can only move counterclockwise as far as the frozen shade and as long as you have only finitely many different shades to begin with, after finitely many steps, that cell will freeze as well. So either everything eventually freezes, or nothing freezes ever. $\endgroup$
    – Aubrey da Cunha
    Commented Apr 14, 2011 at 17:55
  • $\begingroup$ On second thought, if you pin down the cases where nothing freezes ever (rare though I expect those cases to be), that would be interesting. $\endgroup$
    – Aubrey da Cunha
    Commented Apr 14, 2011 at 17:57

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