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Coding new pentagon tiling

Ed Pegg, Wolfram Research
Posted 9 years ago
POSTED BY: Ed Pegg
3 Replies
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POSTED BY: Moderation Team

I decided to simplify and compress the motif. The offset got simplified to {{3 (1 + Sqrt[3])/2, 0}, {0, 1}} in the process.

p15 = {
{{5-5 s,1+s},{5-5 s,5-3 s},{1-s,-7+s},{-1+s,-1-s},{1-s,1+s}},
{{-5-3 s,7-s},{-7-s,1+s},{5-5 s,5-3 s},{5-5 s,1+s},{1-5 s,5+s}},
{{-1-3 s,11-s},{5-5 s,13-3 s},{1-s,1+s},{5-5 s,1+s},{1-5 s,5+s}},
{{-1-3 s,3-s},{1-5 s,-3+s},{-11-s,1-3 s},{-11-s,-3+s},{-7-s,1+s}},
{{-5-3 s,7-s},{-11-s,9-3 s},{-7-5 s,-3+s},{-11-s,-3+s},{-7-s,1+s}},
{{-11-s,-3+s},{-11-s,1-3 s},{-7-5 s,-11+s},{-5-7 s,-5-s},{-7-5 s,-3+s}}}/8/.s->Sqrt[3];

This allows a more compact presentation.

motif = Polygon /@ Join[p15, -p15];
colors = {{Red, Blue, Brown}, {Yellow, Orange, Purple}, {Green, Cyan, White}};
Graphics[{EdgeForm[Black], Table[{colors[[Mod[x, 3, 1], Mod[y, 3, 1]]], 
Translate[motif, {x 3 (1 + Sqrt[3])/2, y }]}, {x, -1, 1}, {y, -3, 3}]},
 PlotRange -> {{-4.5, 4.5}, {-1.5, 1.5}}, ImageSize -> {660, 220}]

the 15th tiling pentagon
POSTED BY: Ed Pegg

The Demonstration Ed mentioned for Tetartoid is very interesting. It driven by a multiset {a,b,c} has quite a few "degenerate" states for multiset like {n, m, n} when polyhedron collapses into a cube or pyramid. See this animation:

enter image description here

The new pentagonal tiling is striking. I'd like to note that is was found by using a computer algorithm (I wonder which):

We discovered the tile using a computer to exhaustively search through a large but finite set of possibilities,” ~ Casey Mann.

Here is a bit changed Ed's code to see how it is formed via translations of main motif ( do you have any other related visualizations ? - please share! ):

motif = Polygon[pent15];
Rotate[Graphics[{EdgeForm[Directive[Black, Opacity[.3]]], Table[{RandomColor[], 
Translate[motif, x offset15[[1]] + y offset15[[2]]]}, {x, 0, 2}, {y, 0, 8}]}], .85]

enter image description here
POSTED BY: Vitaliy Kaurov
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