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Constructing 2D and 3D Jerusalem Cube Fractal

Posted 7 years ago
POSTED BY: Greg Hurst
7 Replies
Posted 7 years ago

Yes, a similar method can.

iWallis[n_] := With[{k = 2 n - 1},
  BoundaryDiscretizeGraphics[
   Table[Rectangle[{x, y}, {x, y} + 1/(k + 2)!!], 
      {x, (k + 1)/(2 (k + 2)!!), 1 - 1/(k + 2)!!, 1/k!!}, 
      {y, (k + 1)/(2 (k + 2)!!), 1 - 1/(k + 2)!!, 1/k!!}
    ]
  ]
]

init = BoundaryDiscretizeGraphics[Rectangle[]];

wallis2D = Fold[RegionDifference[#1, iWallis[#2]] &, init, Range[3]]

enter image description here

bmr = BoundaryMesh[RegionProduct[wallis2D, Line[{{0.}, {1.}}]]];

RegionIntersection @@ (BoundaryMeshRegion[
     MeshCoordinates[bmr][[All, #]], MeshCells[bmr, 2]] & /@ {{1, 2, 
     3}, {3, 2, 1}, {1, 3, 2}})

enter image description here

POSTED BY: Greg Hurst
Posted 7 years ago

It's kind of neat to see the tunnel structure here:

BoundaryMeshRegion[wallis3D, MeshCellStyle -> {1 -> None, 2 -> Opacity[.5]}, PlotTheme -> "Default"]

enter image description here

POSTED BY: Greg Hurst

enter image description here - Congratulations! This post is now a Staff Pick as distinguished by a badge on your profile! Thank you, keep it coming!

POSTED BY: EDITORIAL BOARD

Can this method make the Wallis sponge?

POSTED BY: Ed Pegg
Posted 7 years ago

Coincidentally, the Demonstration "Cross Menger (Jerusalem) Fractal" was published today, to give more details about the construction. Instead of punching holes through the shape, each iteration is constructed by shrinking and assembling copies of the previous two iterations. (See "Cross Menger (Jerusalem) Cube Fractal" for the difference between using one or two previous iterations.)

POSTED BY: Robert Dickau
Posted 6 years ago

Hi Chip!

FWIW, your model isn't the "official" Jerusalem Cube, it's a "houndstooth" variant (probably by Robert Dickau).

The "real thing" has crosses, whose arms have the proportions 1:root(2), and it has mirror-symmetry. The problem with the non-mirror-symmetrical variants is that they don't extend cleanly into three dimensions: if you punch a "right-handed" houndstooth through a cube, it comes out on the other side as left-handed. So you end up with three right-handed faces circling around one cube corner and three left-handed faces circling the opposite corner, and then you have another six corners whose three adjacents are either 2+1 or 1+2 left/right. It loses some of the original cube symmetry.

Regards, Eric

POSTED BY: Eric Baird
Posted 6 years ago

Hear, hear, Eric's description is way better than mine at https://robertdickau.com/jerusalemcube.html and https://robertdickau.com/spongeslices.html#asterisk.

POSTED BY: Robert Dickau
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