Unpacking the Hypercube

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:

HexaSwap2

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.

Rhomb1

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.

Rhomb2

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.

Rhomb3

 

 

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.

Rhomb4

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…

 

Advertisements

Roto-Tiler

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.

Roto0

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.

Roto3

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.

Roto2

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:

Roto1

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.

Vertical Flux (The pleasures of Failure IV)

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

 

 

Fr1

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.

Fr2

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.

Spcat1

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.

 Spcat2

 

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.

Dpcat1

 

Dihedralization (The Pleasures of Failure II)

W1 2Up 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

W1 5

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.

W2 2

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: W2 3

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).

W2 2 cut

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.W2 2 9 cut

So I tend to think that this is the first case where dihedralization fails.

Death of Proof (The Pleasures of Failure I)

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.

 

Almost

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.

Notquite

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.

Fail

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?

Side

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.

Top

Three Planes

When you take two non-parallel planes, they will intersect in a line. The singly periodic Scherk surfaces are the only minimal way to “desingularize” this, in the sense that they are the only known minimal surfaces asymptotic to these two planes. To show this is one of the many famous open problems about minimal surfaces.

Scherk6sym

The situation gets vastly more complicated with three planes. Nobody has yet succeeded in constructing a minimal surface that is asymptotic to the three coordinate planes. That is another open problem. A case where we do know something is that of three (or more) vertical planes. Martin Traizet has shown in 1994 that in case the planes are reasonably general one can wiggle them a little bit and desingularize them by gluing in singly periodic Scherk surfaces. The concrete and very symmetric example above was known before that.

 

Otherscherk

The only requirement on the Scherk surfaces is that they have the same translational period and share a horizontal reflectional symmetry plane to ground them. But nothing prevents us from shifting one of the Scherk surfaces by a half-period, like up above. To make the image, I assumed another reflectional symmetry at a vertical plane (roughly parallel to the screen). This still left me with a 1-parameter family, whose existence is truly only guaranteed near the limit that looks like three Scherk surfaces (with one of them shifted). But nothing keeps us from looking at the other surfaces in this family.

Middle

Above I have turned it around so that one can appreciate the handles better. What emerges becomes clear when one pushed the parameter further:

Costa

A singly periodic Costa surface! There is a similar one constructed by Bastista and Martín where the Costa-necks are rotated by 45 degrees. It then loses its reflectional symmetries but gains straight lines.

Closing the Gaps

In 1982, Chi Cheng Chen and Fritz Gackstatter published a paper that described the surface below.
Cg 1

Like some of the classical examples of minimal surfaces, this surface is complete and has finite total curvature. A famous theorem of Osserman from 1964 asserts that any such surface can be defined on a punctured Riemann surface. In the classical examples, this had always been a sphere, but here we have a torus with one puncture.  There were some earlier examples, but this one, while not embedded, was surprisingly simple. From far away, it looks just like the Enneper surface.Cg 1 no

How does one make such an example? One problem is illustrated above: While Osserman’s theorem also guarantees that the derivative of a conformal parametrization has a meromorphic extension to the compact surface, the integration of these so-called Weierstrass data might leave gaps.

Flat 01

To close the gap, we use the help of symmetries: Two vertical planes cut the surface into four congruent pieces, each represented by the upper half plane. The Weierstrass forms \phi_1 and  \phi_2 then turn out to be Schwarz-Christoffel integrands. The corresponding integrals map the upper half plane to (infinite) Euclidean polygons, shown above. The left extends to cover a bit more than a quarter plane, the right a bit less than a three quarter plane.Torus 01

Incidentally, we can see the torus by fitting four copies of the right polygon together. We obtain the plane with a square missing. Identifying opposite edges of the missing square creates a torus with one puncture.

Now the condition that makes the gaps disappear is just that the two polygons fit together, which can be achieved by scaling. It’s really that simple. Similarly one can have more symmetric versions by just changing the angles in the polygons. Below is an example with sevenfold symmetry.

Cg 1 7