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Ernst Haeckel (1834 - 1919) was a German philosopher, physician, artist and professor of zoology at the University of Jena. He promoted and popularised Charles Darwin's work in Germany.

enter image description here

Haeckel (left) with Nicholai Miklukho - Maklai, his assistant, in the Canaries, 1866

The published artwork of Haeckel includes over 100 detailed, multi-colour illustrations of animals and sea creatures, collected in his famous "Art Forms of Nature", which is still published in various languages. As I was recently browsing through the book, I was struck by his depiction of Scleratina corals, because their beautiful structures are so "mathematical".

enter image description here

The middle one, a "Diploria labyrinthiformis", also called "brain coral", seemed to be a good start for a "Mathematica replica".

enter image description here

From my collection of mathematical surfaces I picked up this Gyroid - like structure which looks like a coral skeleton:

gyro = 
  10 (Cos[x] Sin[y] + Cos[y] Sin[z] + Cos[z] Sin[x]) -
   0.5 (Cos[2 x] Cos[2 y] + Cos[2 y] Cos[2 z] + Cos[2 z] Cos[2 x]) - 12;

n = 8;
ContourPlot3D[gyro == 0, {x, -n, n}, {y, -n, n}, {z, -n, n},
 Axes -> False,
 Background -> GrayLevel[0],
 Boxed -> False,
 ContourStyle -> White,
 Lighting -> "Accent",
 Mesh -> 0,
 PlotPoints -> 15]

enter image description here

I was curious to see what it would look like if I gave him a rounded form:

ContourPlot3D[gyro, {x, y, z} \[Element] Ball[{0, 0, 0}, 8],
 Axes -> False,
 Background -> GrayLevel[0],
 Boxed -> False,
 ContourStyle -> White,
 Lighting -> "Accent",
 Mesh -> 0,
 PlotPoints -> 5,
 PlotRange -> All]

enter image description here

Disappointing, and certainly not the way to go!

Robert Fathauer, one of the leading mathematical artists, took a different way when he produced this ceramic brain coral sculpture:

enter image description here

His explanation:

"This sculpture, created in 2014, is based on the first three generations of a fractal curve that develops radially outward. The starting point is a simple saddle, and the final form has an envelope that is roughly hemispherical. The space curves were created by fitting a series of planar fractal curves to the surface of an octahedron. This fractal structure possesses two-fold rotational symmetry. The sculpture, which was partly inspired by brain coral, measures 21" in width and 11" in height."

Sounds frightening complicated to me, and so I would like to ask:

How would you create a brain coral?

Addendum 1

Similar to Silvia's comment:

enter image description here

American Mathematical Society

Addendum 2

Please also have a look at Repulsive Curves by Henrik Schumacher et al. (see his comment), especially 14:50, page 41, where the authors show Hilbert-curve-like patterns meandering on a sphere. Also interesting:

Interpolating the Hilbert Curve with a B-Spline

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    $\begingroup$ Well, my collaborators and I played a bit with the idea to maximize the length of the curve inscribed into a sphere under some self-avoidance constraint (which we built in via using the so-called tangent-point energy). See, e.g., Fig. 24 in dl.acm.org/doi/abs/10.1145/3439429. The numerics of solving this problem is quite involved. But I have some Mathematica code for it. (It is just to long to post it here, I am afraid.) $\endgroup$ Dec 6, 2023 at 15:27
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    $\begingroup$ Might be of interest for OP: Crocheting Adventures with Hyperbolic Planes. $\endgroup$
    – Silvia
    Dec 6, 2023 at 18:08
  • $\begingroup$ Thank you, Silvia, I added a similar thing to the question :) $\endgroup$
    – eldo
    Dec 7, 2023 at 10:34
  • 2
    $\begingroup$ Interesting and challenging question. I hope you will get some nice answers. But, please, do not close the question soon, some people may need time to produce good solutions. $\endgroup$
    – yarchik
    Dec 7, 2023 at 11:30
  • $\begingroup$ Thanks, Henrik for sharing your extremely interesting video. I added the link to my question. $\endgroup$
    – eldo
    Dec 7, 2023 at 17:58

3 Answers 3

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Not quite there, but this could be a basis for a better answer that does a line sweep along the B-Spline.

Remove["Global`*"];
SeedRandom[1];
n = 1000;
pts = Normalize /@ (SpherePoints[n] + RandomPoint[Sphere[{0, 0, 0}, 0.025], n]);
st = pts[[Last@FindShortestTour[pts]]];
Graphics3D[Tube[BSplineCurve[st], .08], Boxed -> False]

brain

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  • Still not so good.
ContourPlot3D[ (Cos[x] Sin[y] + Cos[y] Sin[z] + Cos[z] Sin[x] == 
     0 /. {x -> 2 x, y -> 2 y, z -> 2 z}) // Evaluate, {x, -8, 
  8}, {y, -8, 8}, {z, -8, 8}, 
 RegionFunction -> 
  Function[{x, y, z}, x^2 + y^2 + z^2 <= 8^2 && z >= 0], 
 ContourStyle -> White, Mesh -> None, Boxed -> False, Axes -> False, 
 BoxRatios -> Automatic, RegionBoundaryStyle -> None, 
 PlotTheme -> {"ThickSurface"}, Method -> {"Extrusion" -> .2}]

enter image description here

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An attempt at a coral like brain:

I have 8GB of RAM so choosing better settings for PlotPoints and MaxRecursion hangs up my system.

uhemi = RegionIntersection[Ellipsoid[{0, 0, 0}, {6, 8, 8}], 
   Cuboid[{-6, -8, 0}, {6, 8, 8}]];

ContourPlot3D[Cos[2 x  y] + Sin[2 y  z] + Sin[2 z x] == 0
 , {x, y, z} \[Element] uhemi
 , Boxed -> False
 , BoxRatios -> Automatic
 , Axes -> False
 , Mesh -> None
 , ContourStyle -> Directive[
   Lighter@ColorData["HTML"]["Chocolate"]
   , Opacity[0.5]
   , Specularity[1, 20]
   ]
 , PlotPoints -> 15
 , Extrusion -> 0.2
 , MaxRecursion -> 1
 , RegionBoundaryStyle -> None
 , BoundaryStyle -> Lighter@Brown
 , NormalsFunction -> None
 , Lighting -> "Neutral"
 , ImageSize -> 800
 ]

enter image description here

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  • $\begingroup$ You're surely worthy of and deserve more than 8 GiB, @Syed :) $\endgroup$ Dec 7, 2023 at 14:17

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