Proof about Octagonal Disk Periodicity

[brad klee]

Fig. "Necklaces" from Harm.On.ica @ Open.ai. 

In a previous thread: 

https://groups.google.com/g/seqfan/c/S61aFQk_alY

We drilled down through two lattice models and found a double 
Ɐpex (or Vpex if you'd rather) for producing chaotic evolution. 

I still don't know in general what happens for two hard body 

non-rotating octagons in square container when they don't 

have lattice-constrained initial conditions. 

For the 4.8.8 tiling in the "clock orientation" with sublattices along 

x and y (and right-aligned squares), the following constraints: 

1. Full size octagons starting centered on either sublattice
2. unit velocity in cardinal directions  

allow us to time step from any widely separate initial condition to 

local conditions only. Then we get a finite case check problem. 

Let's introduce notation H horizontal, V vertical and D for diagonal. 

H + H and V + V collisions are trivial, as are non-collisions. The H+V

case splits into D contact, H contact and V contact. D contact is 

another trivial "two elbows" periodic case, while H and V contact 

leave one disk frozen and the other takes away a diagonal velocity.

(again, assuming equal masses.) 

It's easy to notice by checking finite conditions that all H+V collisions 

are along diagonal or point-like. Point-like are rejected, so once we've

done the compute work, we've already proven all-periodic for the clock 

orientation. The tiling makes it easy. 

We also considered square grids (also equivalent to rotating the clock 

orientation of 4.8.8 by 45 degrees) with half-diameter octagons starting 

central to any unit square. As we've already seen, with unite cardinal 

direction velocity, these cases can collide H + V along an H or V facet, 

thus producing D+F motion with F for Freeze. 

The possible outcomes for re-collision of D+F are F+D along D surface 

and V+H meeting along either V or H surface. 

This is where things begin to get difficult again, and actually I'm not sure 

that induction through container sizes will immediately yield a proof. 

Already for the first few container sizes, measurements of period in terms

of collision counts before recurrence are apparently growing unbounded: 

2: 0 2 4 24
3: 0 2 4 12 14 42
4: 0 2 4 18 20 22 104 232
5: 0 2 4 24 26 28 30 140 184 192
6: 0 2 4 30 32 34 36 38 100 144 158 160 168 176

Some of this growth is due to repeat F+D:D collisions happening. I was 

also surprised that H+V:D can repeat on alternating D facets. 

If we delete duplicates in both X:D cases, then we get reduced periods: 

2: 0 1 4 16

3: 0 1 4 6 10 28

4: 0 1 4 6 10 14 18 76 164

5: 0 1 4 6 14 18 22 26 76 100

6: 0 1 4 6 14 22 26 30 34 36 48 68 92 114 124

Even the reduced sequence is still seeming to grow unboundedly, and 

there are two distinct T = 176 conditions at N = 6 (picture above). 

For the second geometry the best "proof" (not really) we have so far 

is by waving our hands and saying: "Well if we get to 6x6 and we 

haven't found chaos, then we just do an induction to higher NxN's 

that preserves the width=3 form of the periodic annulus". 

I asked Harm.On.ica to hand wave even more, and the return was 

pretty funny actually: 

<<

Yes, but the natural induction step is N↦N+6N\mapsto N+6N↦N+6: 
translate the old orbit by (3,3)(3,3)(3,3), then surround it with a three-cell 
annulus on all four sides.

There is real exact evidence for an annular transfer block. For the class-B 
family, the centered lifts give:

[ [ fabricated data omitted ] ] 

>>

The moral of the story is: If we want to prove the "half-diameter" model has 

only periodic orbits for two bodies (an interesting proposition, imo), 

then we might need the tables above or something similar--once they've 

been written down in easy-to-verify certificates and then checked a few

different ways. 

Exciting times!

--Brad

 

  

 

[brad klee]

On second thought: 

The "two elbows" case can end up going to multiple elbows 

and then breaking on a corner case. 

A better way to say this part of the proof is that the output 

of D contact goes back onto sublattice centroids. 

From the output to the next input we don't know what happens 

but staying on sublattice centroids is strong enough anyways. 

 

Clock orientation might have one or two difficulties, but this 

case is not the difficult one. If the suggested proof is correct

it should have no freeze cases. 

If anyone was writing this seriously, "No freeze" would actually 

be a good lemma. 

All the best, 

--Brad