The second bifoldable object Jiangmei showed me was this:
You can find a movie showing how this folds together in two ways here. To understand how and why this works, let’s first look at a simple saddle:
This is a polyhedron with a non-planar 8-gon as boundary. Its faces are precisely the four types of faces that are allowed in our polyhedra: All others have to be parallel to these four. The four edges that meet at the center of this saddle constitute the star I talked about the last time. Again, all edges that can occur must be parallel to one of these four. One can fold the saddle by moving the upwards pointing star edges further up (or down), and the downwards pointing edges further down (or up), thereby keeping the faces congruent. This works locally everywhere and therefore allows a global folding of anything built that way.
For instance, the hollow rhombic dodecahedron above can be bi-folded. Now note that this piece is also a polyhedron with boundary. In fact, its boundary is exactly the same octagon as the boundary of the saddle.
Observe also that at the center of this piece we have a vertex in saddle form. This suggests to subdivide all rhombi into four smaller rhombi, remove the saddle an the middle vertex of the doubled hollow dodecahedron, and replace it by a copy of the standard hollow dodecahedron. This gives you Jiangmei’s fractal. Repeating this is now easy. Below is the generation 2 fractal (animation):
And, just for fun, the generation 10 fractal:
You can see it being bifolded here. So far, the two completely folded states of our polyhedra looked very much the same. We will see next week that this doesn’t need to be the case.
We all know that cardboard cubes are rigid, which is why we get our packages in boxes. We also all know that if we remove two opposite faces from a cube, we can fold it together. This started to interest me when I noticed that the polyhedral approximation of the Schwarz P surface is surprisingly flexible. This summer, I showed this to our local Origami and Paper Folding expert, Jiangmei Wu from our School of Art and Design, and she became interested. A few days later she came with a paper model that looked like this:
She called it a simple variation of the polyhedral P-surface. Hmm. This is a triply periodic polyhedral surfaces tiled with rhombi. To understand it, we build it out of smaller units (which we called butterflies):
The really cool thing about it is that it can be folded together in two different ways, like so:
You can find an animation showing the continuous deformation here. We stared at this for a (long) while, until we realized that this has to do with rhombic dodecahedra. The structures up above are composed of the rhomboids from last week that tile a rhombic dodecahedron. The latter has, as the name hints, 12 faces, which occur in opposite pairs. Like the cube, it is rigid per se, but becomes foldable if we remove two pairs of parallel faces, leaving us with four faces to use, which are distinguished by color up above.
Above you can see the four hollow parallelepipeds (which we called hollowpeds). The almost trivial but nevertheless mind bending realization is that everything you build out of these hollowpeds becomes a structure foldable in two different ways. Next week I’ll show Jiangmei’s second model, a foldable fractal… If you can’t wait, check out this.
Last week we learned how Rototiler moves can unpack a cube. As a warmup, below are the moves for a 2x2x2 cube projected parallel along a cube diagonal onto hexagons:
We start at the left, remove the frontmost cube, and keep going. The solution is far from being unique, but not too complicated. Today, we do the same with a hypercube. The projection of a 1x1x1x1 into 3-space along a main diagonal is a rhombic dodecahedron, tiled by four rhomboids. These rhombic dodecahedra have 8 obtuse, 3-valent vertices at the corners of a cube, and 6 acute, 4-valent vertices at the corners of an octahedron.
There are two ways to tile the rhombic dodecahedron with these rhomboids, and changing one to the other corresponds to a rototiler move in space. Let’s do this with the 2x2x2x2 hypercube, whose projection is a rhombic dodecahedron tiled by 32 rhomboids.
At first it seems as if there is no swappable rhombic dodecahedron available, but if we remove three rhomboids and look inside (which is the direction of the fourth dimension, after all), we can see it. After swapping it, we also remove the frontmost rhomboid of the swapped dodecahedron.
We then see that the four removed rhomboids together make up another swappable dodecahedron. We replace it by its swap. The same can be done at three other places.
The next thing to do is to swap 6 more dodecahedra. One of them is the one which shows yellow and purple rhomboids in the right figure above, sitting between the red and blue “vertex”. All these dodecahedra correspond to the edges of the tetrahedron whose vertices are the already swapped four peripheral dodecahedra. Doing these six swaps leads to a tiling very much like the one above to the right, where now the other four obtuse vertices mark swappable dodecahedra. Swapping these and finally the hidden central dodecahedron completely unpacks the hypercube. It took us 1+4+6+4+6+1 = 32 swaps, as expected.
Next week we’ll see what this is good for…
Today we look at a puzzle invented by Alan Schoen that he calls Roto-Tiler. He explained this to me a few years ago, and when I showed him notes I made for a class, he denied that this is the puzzle he described. I insist it is, and it is quite certainly not mine.
Things happen on a hexagonal board like the one above (it can but doesn’t need to be regular), tiled by hexagonal rhombi of equal size. The acute angles are marked by 1/3-circles, which occasionally happen to close up when three acute angles meet. In that case, a move consists of rotating the three involved rhombi by 120º either way.
Above you can see the possible four moves from the central position. At this point it is not clear at all that a move is always possible. The puzzle consists of transforming one given tiling by rhombi to another given tiling of the same hexagon. For instance, a simple example asks to find the smallest number of moves that takes the left tiling to the right tiling.
The clue to solve this puzzle is to view the hexagons as the parallel projection of a box subdivided into smaller cubes, and the rhombi as the projections of the faces of the smaller cubes. This becomes visually more intuitive if we color the rhombi by their orientation so that parallel cube faces have the same color:
Then the hexagon above becomes the projection of a box partially filled with cubes, and a move consists of adding or removing a frontmost cube. This step into the third dimension explains everything: We see that we can solve every Roto-Tiler puzzle by emptying and filling boxes with cubes. Last week’s first example was a 1-dimensional version of this, next week we will try to grasp a 3-dimensional version and practice our 4-dimensional intuition.
One of the standard examples of a failed minimal surface construction is the Finite Riemann surface, named in reference to the rather successful infinite Riemann minimal surface
The attempt was to make an embedded minimal sphere with 3 ends. Two ends is easy: the catenoid will do. If you want more, the ends can only be catenoidal or planar, and the catenoidal ends need to point in the same direction, i.e. have the same limiting normal. Such a sphere will always have what is called vertical flux, which is disastrous in its usefulness.
Whenever you have vertical flux, you can deform the minimal surface in a very simple way, keeping the catenoidal ends vertical but tilting non-horizontal planar ends like above. This is nice but also disastrous because it allows to prove that embedded surface of this sort cannot exist.
This is a famous theorem by Francisco José Lópe and Antonio Ros. The proof consists of three steps: If the surface is embedded, its stays embedded for all parameter values of the deformation. Secondly, if such a surface has either a planar end or a finite point with vertical normal, it cannot stay embedded. These two steps imply that such a surface can only have catenoidal ends and no points with vertical normal. Being a sphere then implies that there can only two ends, and we have the catenoid.
It is amusing to make more examples of surfaces with vertical flux. Above are single periodic versions with vertical catenoidal and annular ends. The first one feels a little bit embedded because the only intersections that happen are those of a catenoidal end with a translational copy, and I am inclined not to count that.
Up above is Meinhard Wohlgemuth’s first surface. It is very similar to Costa’s surface, but has one more end and genus 2. In Mathematics, like in every experimental science, you gain intuition about truth through observation. For instance, every minimal surface person I know will probably believe in dihedralization.
Whenever you have a minimal surface that has two vertical symmetry planes making an angle π/2, there should be more symmetric version where the planes make the angle π/n. Like up above for n=5, observed for Wohlgemuth’s first surface.
Then there is Wohlgemuth’s second surface. It has genus 3, squeezing in a little suspicious handle between the two planar ends, so that it looks a little bit like two copies of the ill-fated Horgan surface stacked together. But this time it works, one (Meinhard) can prove this surface does indeed exist. So, let’s make it more symmetric:
This is what I get when I run the experiment. The parameter values are so extreme that I don’t think this actually exists. So, do we trust our intuition, or the experiment? Neither. The only way out is to dig deeper. Let’s cut Wohlgemuth’s surface open (in half, to be acurate).
We can now apply the dihedralization continously, slowly increasing n=2 to n=3, thereby tracking the numerical solutions. They show a clear deformation,up to approximately n=2.9 (below). After that, the periods don’t close accurately enough anymore to make me feel confident about the surface.
So I tend to think that this is the first case where dihedralization fails.
Die Wahrheit ist dem Menschen zumutbar.
Occasionally, after confronting students with evidence of fact (Euler Polyhedron Theorem is a great example), I ask them whether they want to see a proof or prefer to accept the statement as a miracle. The overwhelming majority is always happy with the miracle. Such are the times. Below is such an evidence of fact: A minimal surface with 3 ends and of genus 2.
Should we doubt its existence? In 1993, John Horgan published an article in Scientific American questioning whether proofs were about to become obsolete, in times where shear length and difficulty made validation next to impossible, and numerical experiments supplied by computers could be an acceptable substitute. For many reasons, large parts of the mathematical community were outraged.
Above is another example of that surface, for a different parameter value, but something seems off. There appears to be a little crack. Maybe I didn’t compute accurately enough? Changing the parameter a bit more widens the gap.
The question whether this surface does actually exist hinges on the possibility to truly close that gap, for at least one parameter value. It appears that we have done so in the top image. But the parameter value there is 1.01, pretty close to 1, where the surface will clearly break down. A more accurate computation shows that there still is a gap at 1.01, which we can’t see, or don’t want to see. But maybe 1.001 will do?
David Hoffman and Hermann Karcher analyzed this surface in 1993, the same year as Horgan’s article, and it became known as the Horgan surface. One can indeed prove that the gap cannot be closed, so, despite all the evidence, this minimal surface does not exist.