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IFS Generators for a Hinsley Irreptile and its Boundary

Background

In 2002, Stewart Hinsley posted a page illustrating several fractal irreptiles inspired by the Hokkaido[[?]] tiling of Shigeki Akiyama. In Feb 2018, Ed Pegg posted the StackExchange Mathematics query "Curve of a Fractal Triangle" asking if there were a good generator for the boundary of a certain Hinsley irreptile (as opposed to its interior).

This post shows the derivation of the IFS generators for both the irreptile of interest and its boundary.

The Irreptile of Interest

Here is an illustration of the subject irreptile showing its decomposition into two smaller copies of itself as well as labeling several notable points on its perimeter:

Fig 1: The Irreptile of Interest

Although this has been referred to as a fractal "triangle", it will be simpler to treat it as a quadrilateral []ABCD, due to the alignment requirements of the component copies at points C and F.

Component Transformations

The irreptile []ABCD is composed of two copies of itself, []BCFA and []CDEF, generated from the original by the transformations T1 and T2, respectively. As these are similarity transformations (translation, rotation, uniform scaling only), it will be convenient to use complex numbers to represent points in the plane.

Z[{x_, y_}] := x + I y ;
XY[z_] := {Re[z], Im[z]} ;

With this representation, the transformations become simple linear functions parameterized by a scaling/rotation value and a translation value.

T1[z_] := s1 z + t1 ;
T2[z_] := s2 z + t2 ;

We can then express the required mappings of the tile onto its component copies by the transformations T1 and T2 as the constraint equations E1 and E2.

E1 = T1 /@ {a,b,c,d} == {b,c,f,a} ;
E2 = T2 /@ {a,b,c,d} == {c,d,e,f} ;

To pin down the overall translation, rotation and scale, we place B at the origin and D at (1,0) via constraint equation E0.

E0 = b == Z[{0, 0}] && d == Z[{1, 0}] ;

We can now solve these constraint equations to give us the parameters of the transformations and the coordinates of the labeled vertices. (The symbolic results are rather lengthy, so will be suppressed here.)

Solns = Solve[{E0, E1, E2}] ;

This gives us three solutions. The first is a purely real solution which corresponds to a degenerate tile with all points lying on the x-axis. The second and third are complex conjugates of each other corresponding to a mirror images pair of tiles. We'll select the first of this pair, and evaluate it numerically.

Soln = Solns[[2]] // N
(* { s1 -> -0.232786 + 0.792552 I,  s2 -> -0.42605 + 0.368989 I, 
     t1 -> 0.602245 - 0.141188 I, t2 -> 1., a -> 0.369459 + 0.651364 I, b -> 0., 
     c -> 0.602245 - 0.141188 I, d -> 1., e -> 0.79551 + 0.282375 I, f -> 0.57395 + 0.368989 I } *)

Plotting the Resulting IFS's

To Be Continued...

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