# Constructing 2D and 3D Jerusalem Cube Fractal

Posted 2 years ago
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| Consider constructing a Jerusalem Cube Fractal. It seems the logic is if there's a square of certain size (dependent on the level) available, then punch out the shape. I did not notice a nice pattern for this, so I took a brute force approach which means this solution most likely suboptimal.First let's get the shape we'd like to punch out. To get the connectivity, I'll import the image and turn it to a mesh. To get the correct coordinates, I'll map the pixel approximations to the appropriate values.The connectivity: im = Import["https://i.stack.imgur.com/9uvoz.png"]; mesh = RegionResize[ConnectedMeshComponents[ImageMesh[ColorNegate[im]]][], {{0, 1}, {0, 1}}]; Find the correct coordinates: s = Last[x /. NSolve[x + x^4 + x^3 + x^2 + x == 1, x, Reals]]; nf = Nearest[Union[#, 1 - #] &[{0., s^4, s^3, s^2, s, s + s^4, s + s^4 + s^3}]]; seed = RegionDifference[ BoundaryDiscretizeGraphics[Rectangle[]], BoundaryMeshRegion[Map[First@*nf, MeshCoordinates[mesh], {2}], MeshCells[mesh, 1]] ] Now for the code that looks for valid squares of with side lengths s^lev and punches out the shape. Note that I didn't take the time to refine makeRects or to make it readable. iJerusalemCube[mr_, lev_] := Block[{coords, nn, rects, sub}, coords = MeshCoordinates[mr]; nn = Nearest[coords]; rects = Union @@ makeRects[mr, s^lev] /@ coords; sub = RegionUnion @@ (RegionResize[seed, Transpose[#] + s^(4+lev){{1, -1}, {1, -1}}]& /@ rects); RegionDifference[mr, sub] ] makeRects[mr_, lev_][p:{x_, y_}] := First[Reap[ If[Or[Count[Max[Chop[Abs[First[nn[#]]-#]]]& /@ Tuples[Transpose@{p, p+lev}], 0] > 1, Nor @@ RegionMember[mr, {p + {lev/2, -s^(6+lev)}, p + {-s^(6+lev), lev/2}}]] && RegionWithin[mr, Rectangle[p, p+lev]], Sow[{p, p+lev}]]; If[Or[Count[Max[Chop[Abs[First[nn[#]]-#]]]& /@ Tuples[Transpose@{{x, y-lev}, {x+lev, y}}], 0] > 1, Nor @@ RegionMember[mr, {p + {lev/2, s^(6+lev)}, p + {-s^(6+lev), -lev/2}}]] && RegionWithin[mr, Rectangle[{x, y-lev}, {x+lev, y}]], Sow[{{x, y-lev}, {x+lev, y}}]]; If[Or[Count[Max[Chop[Abs[First[nn[#]]-#]]]& /@ Tuples[Transpose@{{x-lev, y}, {x, y+lev}}], 0] > 1, Nor @@ RegionMember[mr, {p + {-lev/2, -s^(6+lev)}, p + {s^(6+lev), lev/2}}]] && RegionWithin[mr, Rectangle[{x-lev, y}, {x, y+lev}]], Sow[{{x-lev, y}, {x, y+lev}}]]; If[Or[Count[Max[Chop[Abs[First[nn[#]]-#]]]& /@ Tuples[Transpose@{p-lev, p}], 0] > 1, Nor @@ RegionMember[mr, {p + {-lev/2, s^(6+lev)}, p + {s^(6+lev), -lev/2}}]] && RegionWithin[mr, Rectangle[p-lev, p]], Sow[{p-lev, p}]]; ][[-1]], {}] Here, iJerusalemCube finds all rectangles that have side lengths s^lev have a corner that's a coordinate of mr either has multiple corners that are a coordinate of mr (the nn part) is tucked into a corner, i.e. does not just meet the boundary at a single point (the Nor @@ RegionMember part) is fully contained within mr (the RegionWithin part). Then it punches out the shape in all valid rectangles.The main function will iterate this: JerusalemCubeList[lev_Integer?NonNegative] := FoldList[ iJerusalemCube, BoundaryDiscretizeGraphics[Rectangle[]], Range[0, lev-1] ] JerusalemCubeList Multicolumn[MapIndexed[ BoundaryMeshRegion[#1, MeshCellStyle -> {1 -> Black, 2 -> ColorData[First[#2]]}] &, JerusalemCubeList], 3, Appearance -> "Horizontal"] And unsurprising that the number of boundary edges grows exponentially: cnts = MeshCellCount[#, 1] & /@ JerusalemCubeList  {4, 32, 144, 704, 3504, 17504}  FindSequenceFunction[Rest@cnts, n]  4/5 (5 + 7 5^n)  To get a cube in 3D, we can intersect the 2D mesh extruded in 3 different directions: mr = Last[JerusalemCubeList]; bmr = BoundaryMesh[RegionProduct[mr, Line[{{0.}, {1.}}]]]; RegionIntersection @@ ( BoundaryMeshRegion[MeshCoordinates[bmr][[All, #]], MeshCells[bmr, 2]] & /@ {{1, 2, 3}, {3, 2, 1}, {1, 3, 2}}) Original post can be found HERE. Answer
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Posted 2 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] bmr = BoundaryMesh[RegionProduct[wallis2D, Line[{{0.}, {1.}}]]]; RegionIntersection @@ (BoundaryMeshRegion[ MeshCoordinates[bmr][[All, #]], MeshCells[bmr, 2]] & /@ {{1, 2, 3}, {3, 2, 1}, {1, 3, 2}})  Answer
Posted 2 years ago
 It's kind of neat to see the tunnel structure here: BoundaryMeshRegion[wallis3D, MeshCellStyle -> {1 -> None, 2 -> Opacity[.5]}, PlotTheme -> "Default"]  Answer
Posted 2 years ago - Congratulations! This post is now a Staff Pick as distinguished by a badge on your profile! Thank you, keep it coming! Answer
Posted 2 years ago
 Can this method make the Wallis sponge? Answer
Posted 2 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.) Answer
Posted 1 year 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 Answer
Posted 1 year ago
 Hear, hear, Eric's description is way better than mine at https://robertdickau.com/jerusalemcube.html and https://robertdickau.com/spongeslices.html#asterisk. Answer
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