Alan Schoen’s Cubons (Solitaire XIV)

Alan Schoen is best known for the discovery of the gyroid, but he has also invented an enormous number of puzzles. The one I will discuss today he named cubons.

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Above you see his model of the 24 cubons which I currently have on loan for exploration.


Cubons of order n are obtained from a regular cube by dividing each edge into n+1 equal segments, and choosing on each edge one of the n subdivision points. These are then joined with the vertices on the same edge, the face centers on the adjacent faces, and the center of the cube. Adding faces results in eight polyhedra that can be assembled into a cube, obviously.

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There are n^2+n(n-1)(n-2)/2 different cubons of order n, which gives the sequence 1, 4, 11, 24, 45, 76, … 

The 24 cubons for n=4 are particularly interesting because they might be used to assemble three cubes, using every cubon just once.  Finding a single solution is not so easy, because there are 735471 ways to select 8 from the 24 cubons, but only 18844 will allow themselves to be assembled into a cube, many of them in several different ways.

Still, there are a mind-blowing 1050759 different solutions to partition the 24 cubons into three groups of 8, each of which can be put together into a cube. One might want to put additional restrictions on the solution. For instance, one could ask that each cube has an equator, i.e. four consecutive unbroken segments.

Cuboncolor straight

The picture shows front and back side of each cube on top of each other, the back is obtained by rotating the fron by 180º about a horizonta axis parallel the screen. There are still 1887 solutions of this case. One can also aim for the opposite, namely insisting that there is no straight dividing edge on any face. This allows 6361 many solutions.

Cuboncolor broken

An esthetically pleasing limitation asks to have parallel edges being subdivided the same way. There are only two different solutions, using 16 cubons. Unfortunately, the remaining 8 cubons do not fit together.



One can also us the color of the faces to impose restrictions. I am using here Alan’s color scheme (which employs 5 colors, one for two of the twelve possible cubon faces that are visible in a cube), but I am sure there are many possibilities.

For instance, below is a solution so that each face uses two different colors. There are 2544 of those…


Cuboncolor two


Quadrons (Solitaire XIV – From the Pillowbook XI)

After discussing trions and hexons, it’s time for the simplest variety, the quadrons. They are obtained by cutting from a square a quadrilateral that has as its vertices the square center, a vertex, and two points on the edges adjacent to the chosen vertex. 

Quadron gen2 01

The points on the edges are chosen from the n possible points that divide the edge into n+1 equal segments. I call the number n the generation of the quadron. Above are the four generation 2 quadrons, which fit nicely into a single square, and can be regrouped to form the six square pillows I introduced a long time ago.

Quadron gen3 01

For generation 3, there are 9 quadrons, so they don’t fit together into squares. But we can leave one out, and try to assemble the remaining ones into two squares. There are 9 pairs of such squares, but not all quadrons can be left behind. Which ones can? The solution has to do something with the area (which color codes the quadrons above).

Quadrons gen4

The 16 generation 4 quadrons fit nicely into four squares as shown above. There are 48 individual squares, which will make nice cards for another puzzle…

Quadrons gen4squares

Then there are 75 eye-straining ways to select four of these 48 squares to obtain a complete set of generation 4 quadrons.

Quadrons gen4sols

Higher generation quadrons become very tedious. For generation 6, there are 36 quadrons, and 2139255 many ways to fit them into a set of 9 squares.


I promised a solution for last week’s puzzle, here it is. Now you can guess the second one yourself.


Trions (Solitaire X – From the Pillowbook X)

A trion is obtained by taking an equilateral triangle, dividing the edges into n segments of equal length, and cutting from the center of the triangle to two subdivision points on different edges. This will give a particular quadrilateral. If you divide the edges into just 3 segments, there are three different trions, which fit nicely into a single triangle:


This is the single puzzle piece from a previous post. We have also seen this mechanism (explained to me by Alan Schoen) to produce what I called hexons. Today we will look what trions we get when we divide the triangle sides into four segments.


There are six such trions, which fit nicely into two triangles. They can, as we did with the hexons, also be arranged in groups of six around a (former) triangle vertex, to create hexagonal pillows, i.e. hexagons whose edges can remain straight or possess inward or outward kinks.Trions3

There are too many of those for my taste, but there is only one (not counting its mirror) that uses each trion exactly once, namely the  one to the right. In the spirit of perfect solitairity, this makes an engaging single puzzle piece.


Can you extend the tiling above so that it tiles the plane? Using it as is gets a bit dizzying (but notice the triangle pattern on the left), so I have replaced it by a simpler version that contains all the essentials.


Two hexagons match along an edge if either both sides have no arrow, or you can keep following the arrow, as in the example. Below are two simple examples of periodic tilings:Trions7

There is much more one can do with this piece, but for now let’s end with a homework puzzle: Can you fill the board below so that everything matches?




Hexons (Solitaire VI – Pillowbook XI)

Hexons template 01

Take a regular hexagon, and mark points on each edge. Connect these points with the center of the hexagon to obtain six quadrilaterals. If we choose the marked points so that they divide an edge in one of the proportions 1:3, 2:2, or 3:1, there are exactly six such quadrilaterals (not counting mirror symmetric copies), and as you can see, they can tile a hexagon. I will call these hexons, in analogy of a puzzle by Alan Schoen that I will discuss in the near future. Above are mirror images of this solution, print & cut them out, then glue them together back to back.

Flipme 01

Here is a first puzzle: Suppose you flip one of the three pieces over (like here the blue one) that is not mirror symmetric, can you still tile the hexagon?

7Hexons 01

More complicated is the challenge to use seven hexons of each type to tile the seven hexagons above so that corners only meet corners, not just edges of neighboring hexons.


Triangle 01

Likewise, can you properly tile the above shape, so that the vertices of the tiles only meet at other vertices?

After playing with these hexons for a while, you will find it tempting and useful to group three of them together along so that they meet at their 120º-vertex. This can be done in 11 possible ways:



You will get inflated equilateral triangles, where an edge is either straight, or has a kink inside or outside. I have discussed some of these triangles before, and the square shaped analogies (which I called pillows) in a long sequence of blog posts. Understanding how these 11 triangles can fit together will help to create and solve many more puzzles for the hexons. We will begin this next week.


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.

Inside or Outside?

The last minimal surface that made it into Alan Schoen’s NASA report is the F-RD surface. It has genus 6 and looks fairly simple.


A fundamental decision one has to make these days is to choose the side one wants to live on. If, for instance, we decide on the orange side, we will have the impression to live in a network of tetrahedrally or cubically shaped rooms with connecting tunnels at the vertices of each. Not too bad, but, as things stand, we will never know what life on the other side looks like.


Luckily, our imagination is still free, and we can think about the other, green side. What we can hopefully see from the pictures above and below is that the rooms of the green world are all cubical, with tunnels towards the edges of each cube. Alternatively, we can also think of the rooms as rhombic dodecahedra, with tunnels towards the faces. That’s where F-RD got its name from: Faces – Rhombic Dodecahedron.


Incidentally, the conjugate of the F-RD surface is again one of those discussed by Berthold Steßmann, with the polygonal contours having been classified by Arthur Moritz Schoenfließ

A simple deformation of F-RD maintains the reflectional symmetries of a box over a square, but allows to change the height of the box. It turns out that there are two ways to squeeze the box together.


In both cases we get horizontal planes joined by catenoidal necks, but differently placed in each case.


Alan Schoen’s I6-Surface

Wp1After Alan Schoen was fired from NASA at the end of 1969, he moved back to California and continued to experiment with soap film. In October 1970, he used two identical wireframes bent into figure 8 curves consisting of two squares meeting at a vertex. When he dipped them into soapy water at a small distance from each other and pulled them out, he could poke the flat disks between the two figure 8s and create a minimal surface that looks like the top half in the picture above. It extends triply periodically to a surface of genus 5.


Several pages of notes with descriptions of successful experiments made it to Ken Brakke, who used his marvelous Surface Evolver to make 3D models of the surface. It was named I6, because it happened to be the 6th surface on page I of the notes. Hermann Karcher later called it Figure 8 surface. When you move the two figure 8s close to each other, you will get a surface that looks like a periodic arrangement of single periodic Scherk surfaces:


Note that these Scherk surfaces are vertically shifted in a subtle pattern. More interestingly, there is a second, unstable surface you won’t get as a soap film:


What you see here are Translation Invariant Costa Surfaces (or Callahan-Hoffman-Meeks surfaces) we looked at last time. So Alan Schoen’s I6 surface can be considered as a triply periodic version of the Costa surface, which Celso José da Costa discovered  about 10 years later.

Of course you can poke more handles into I6, as you can with the translation invariant Costa surface. Below is an example of genus 7:



Steßmann’s Surface (Wrapped Packages II)

In the paper Periodische Minimalflächen, published by the Mathematische Zeitschrift in 1934, Berthold Steßmann discusses the minimal surfaces that solve the Plateau problem for those spatial quadrilaterals for which rotations about the edges generate a discrete group. 



Arthur Moritz Schoenflies had classified these quadrilaterals, there are precisely six of them, up to similarity. For the three most symmetric cases, Hermann Amandus Schwarz had found the solutions to the Plateau problem in terms of elliptic integrals, and Steßmann treats the remaining cases. One of them is shown above. It is easier to describe the contour for three copies: Take a cubical box. Then the contour above consists of two (non-parallel) diagonals of top and bottom face, to vertical edges of the box, and two horizontal edges that lie diametrically across.



Extending the surface further produces the appealing triply periodic surface above. Below is a top view. This would make a nice design for a jungle gym. Unfortunately, this surface will not stay embedded; you see this at the corners where three pairwise orthogonal edges meet. 



However, the conjugate surface is embedded, and concludes the story from a few weeks back. The surface introduced there is the I-WP surface of Alan Schoen, and he mentions in the appendix of his NASA report on triply periodic minimal surfaces, that the conjugate of his I-WP surface had been discussed by Steßmann. Below is a more traditional view of the I-WP surface.

I WP cube

Its name (explains Schoen), stands for Wrapped Package, because a translational fundamental piece of its skeletal graph looks like four sticks wrapped together into a package:



The internet knows little about Berthold Steßmann. There is a short biographical note by the German Mathematical Society, telling that he was born on August 4, 1906 in Hüllenberg, Germany, studied in Göttingen and Frankfurt to become a high school teacher, which he completed in 1933. Then, a year later, he received his PhD about periodic minimal surfaces, with Carl Ludwig Siegel as advisor. The same year, the Mathematische Zeitschrift published a paper of Steßmann, covering the same topic. The note also mentions that Steßmann was Jewish. This leaves little hope.

The Gyroids (Algorithmic Geometry III)

When we use squares bent by 90 degrees about one diagonal and extend by the rotate-about-edges rule, we get Petrie’s triply periodic skew polyhedron {4,6|4} which has six squares about each vertex. The two tunnel systems it divides space into are another crude approximation of the primitive surface of Schwarz.


Coxeter observed that this polyhedron can be used to construct Laves’ remarkable chiral triply periodic graph as follows. Choose any diagonal of any of the squares of {4,6|4}. Take an end point of the diagonal, adjacent to which are six squares. Look at the six diagonals of the squares that share the end point as a vertex, and take every other of them, starting with the already chosen diagonal. Keep extending the emerging graph like this.


You obtain the 3-valent Laves graph. At each vertex, the edges meet 120 degree angles. It turns out a mirror symmetric copy fits onto the {4,6|4} without intersections. These two graphs are the skeletons of the two components of the Gyroid, a triply periodic minimal surface discovered by Alan Schoen. You can read all about the discovery at his Geometry Garret.


The Laves graph also lies on the dual skeleton of the tiling of space of rhombic dodecahedra. That means that you can get a solid neighborhood of the Laves graph consisting of rhombic dodecahedra:


This can be done both for the Laves graph and its mirror still leaving a gap in which one can fit the gyroid. Alan Schoen also discovered a uniform polyhedral approximation of the gyroid, consisting of squares and star hexagons. To build it, take a star, attach a square to every other edge, bending the squares alternatingly up and down. Then attach six more stars to the free edges of the first star, fitting them to one free edge of one of the squares each:


Two copies of this piece (without the downward pointing stars and and squares) make a translational fundamental piece of the uniform gyroid.


Images of larger portions are hard to parse, but it makes a wonderful model.