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If the automaton is supposed to work as a cellular automaton, where each cell has four possible values, and steps into the next value if it connects with a neighbor, then, no, it is not possible to have a life-like blinker.
 
If there were such a blinker, it would have a minimum area of interaction; such an area would have corner cells that have two neighboring edges pointing out of the area; the cell in a corner position would eventually turn so that it both its legs would point out, and would no longer contribute to the blinker - thus, there would be an even smaller area of interaction. This leads to a contradiction.
 
Note 1) The applet does not seem to behave this way at all times.
Note 2) On an unbounded plane, such as a torus, you can do blinkers. A 1x2 torus suffices.
 
Thanks,
JP Ikaheimonen
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Hi Ed,

Just a question, what's the rule for this CA? I thought it was, "If one of
your endpoints it hit by a rotating tile, rotate 90 degrees clockwise,"
but if that's the rule, the implementation is a little broken (sometimes
tiles aren't activated when they should be).

BTW, _if_ that is the rule, I can prove there are no blinkers of finite
size, unless the board has non-trivial topology.

Assume there is a blinker. Consider the set of all active tiles in
its repeating sequence. Call this set S. Because S is finite and the
board has trivial topology, there must be a tile in S with no neighbors
to the west or south that are in S. This tile has to be activated in each
of its 4 positions at least once in the blinker's cycle. But there is no
way to activate this tile when it is pointing to the southwest. Contradiction.

I'm pretty sure you can set up blinkers on boards of non-trivial topology,
though.

Incidentally, if the rule is "rotate 90 degrees clockwise if you are
activated at one endpoint, 180 degrees if activated at two endpoints," the
same proof holds -- since there's no way to get rotated past the SW
position.

-Craig Helfgott
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I believe an oscillator is impossible if the truchet tiling automaton is on 
a plane.

By contradiction, assuming we have an oscillator: consider the set S of all 
tiles that move during its period.  Each of them must cycle through all 4 
positions at least once during the period.  Now let TL be the topmost 
leftmost tile in S.  At some point during the period TL must face up and 
left, at which point it can only be triggered by its neighbors above and to 
the left.  If TL does get triggered, then one of those neighbors moved, and 
therefore TL is not the topmost leftmost tile in S; if TL does not get 
triggered, then it never completes a cycle.

I have no idea whether it's possible to build an oscillator on a torus, 
though, or maybe an infinitely expanding pattern (like a glider gun) on a 
plane...
Michael Nelson