## Ninety-Six (Collaboration Games VI)

Today Alan Schoen is celebrating his 96th birthday (below you can see him at the youthful age of 93), and this gives me the opportunity to congratulate him with a puzzle. Its mechanics is motivated by the genus 3 Riemann surface which is the double cover branched over the vertices of a cube, the Riemann surface that underlies the Schwarz P surface, the Diamond surface, and Alan’s Gyroid, his best known discovery. Not quite incidentally, this surface has 96 symmetries. Today’s puzzle is also motivated by Alan’s puzzle RotoTiler. Above is the first version of the puzzle, consisting of 12 squares. You are allowed to put two of them next to each other if the adjacent edges have the same color and the same orientation of arrows, but opposite shading. It is also prohibited to place two tiles next to each other that use the same pair of edge colors. All that is forbidden is shown below (wrong edge colors, wrong edge direction, wrong shading, same edge pairs). It’s not that hard to lay out all 12 squares in a single connected way. You will notice that it is impossible to place four squares around a single vertex. As a simple puzzle: The above layout fits in a 4×7 rectangle. Can you find a solution that fits into a smaller rectangle?

And, if you have a partner to play: print say 8 sets of the tiles so that you have 96. One player gets all the dark squares, the other the light squares, and you take turns creating a connected layout of all tiles. It is quite easy to get stuck, so the players have to collaborate. The two crosses above give a hint how this is related to cubes. They can each be wrapped around a cube, one from the inside, the other from the outside, so that the two tiles on each face differ only in shading. Next we have a rhombic version of the same puzzle, now with 24 different tiles. This more truthfully can be used to represent a walk on the Schwarz P-surface. We have one additional rule: 60º vertices are not allowed to meet with 120º vertices. All that’s forbidden is listed below: Below are a few partial layouts. None of them can be extended. Can you lay out all 24 tiles in a single, connected figure? That you can’t has to do with the Schwarz P surface having congruent insides and outsides, and it being impossible to reach the inside from the outside. More down to earth is the following explanation: You can lay out all 24 tiles in the two rings above, but no tile from the left ring can be legally placed next to a tile from the right ring. To remedy this, we can relax the rules a bit. If we allow that two tiles that use the same pair of colored edges can be placed next to each other, a single chain can be found: This was not so easy. Can you find smaller displays? And again, you can play with two player, using four sets of tiles, and take turns trying to complete a single layout.

Happy Birthday, Alan!

## Alan Schoen’s Octons (Solitaire XXIII – From the Pillowbook XVII)

After introducing the six small octons, let’s talk about the larger octons a bit, that allow to divide the octahedron edges in the proportions 1:3, 2:2, or 3:1. There are 24 of them, suggesting to assemble four octahedra with them. This is Alan Schoen’s original version. These octons can be represented as decorations of the vertices of a planar graph using the 24 coins above. Above you see the octahedral graph, decorated on the left with the six chiral octons (or coins). The rule for placing them is simple: All edges must have either arrows with matching directions, are no arrows attached to them. Besides the one solution shown above for the chiral coins, there are three more, up to symmetry, which are indicated below in the octahedral subdivision: They are not so easy to find. To avoid duplicating a solution, just place one coin exactly as in the decorated graph on the left, and complete its in a different way.

We have already seen the 10 different ways to place the six cubons without 2:2 ratio (i.e. with four arrows in a coin) the last time. What about the remaining 12 cubons? There are 12 different ways to split them into six so that one can assemble them into two octahedra. Below is an example how to split them, assembling them is another puzzle, and finding the remaining 11 ways is really hard and tedious. One can of course also start from scratch. There are  134596 ways to select 6 octons of the 24, but only 5427 can be assembled into an octahedron. Then there are 11417 different ways to group the 24 octons into four groups of 6 that all can be assembled into four octahedrons.

## Alan Schoen’s Small Octons (Solitaire XXII – From the Pillowbook XVII) Above you can see how Alan decomposes a regular octahedron into octons. The recipe is the same as for the cubons and tetrons: Divide the edges of the octahedron in suitable ratios, connect the subdivision points to the face centers and the center of the octahedron. If you allow as subdivisions the proportions (1:2) and (2:1), there are six different octons that you can get this way. Up to rotations, they can be used to assemble an octahedron in 10 different ways, as indicated above. A better representation of these solutions is shown below. The octahedron is represented by its skeletal graph, and each octon becomes a vertex with arrows pointing towards or away from the vertex, indicating what proportion the edge is to be divided in (2:1 or 1:2). Equivalently, we can think of these six “small” octons as the six pillows I have discussed four years ago. So we can view the puzzle to assemble the six octons into an octahedron also as the puzzle to decorate the vertices of an octahedral graph with the six pillows so that the two arrow at each edge point the same way. More generally, one can ask to decorate any quartic graph this way, with any choice of the octons/pillows. Below is the graph of the cuboctahedron, with two sets of “coins” for decoration. This suggests a game for two players (gasp): Take turns to place the coins of your color, again so that arrows along each edge are pointing the same way. The player who moves last wins. Can you find a winning strategy for one of the two players? ## Inflation (Solitaire XX – From the Pillowbook XV)

In order to have a planar realization of Alan Schoen’s tetrons and cubons, we need to add a few coins to our currency. Here are the 24 coins you’ll need: Below is an example how to decorate three cube skeletons with them. This doesn’t look as pretty as the 3D-cubons, but one can much more easily play with them. Coins need to be placed on the vertices of a graph so that the two arrows that share an edge have the same color and point in consistent directions. What other pretty cubic graphs are out there? The Foster census lists one cubic symmetric graph with 24 vertices (the Nauru-graph), and it is a challenge quite in Alan’s spirit to try to decorate it with his 24 cubon-coins. This is indeed possible: The above representation of the Nauru graph lives in a hexagonal torus. It is also the dual graph of the Octahedral 3¹² polyhedral surface, which is the genus 4 quotient of a triply periodic polyhedron by its period lattice. Here is a picture of it. Note that this is twice a fundamental piece, as the period lattice is spanned by the half main diagonals of the cube. Octahedral 3¹² is intimately related to Alan Schoen’s I-WP surface. Everything we do at a certain depth is connected to everything else, it seems.

## Decorated Maps (Solitaire XIX – From the Pillowbook XIV)

We continue decorating cubic graphs with our four coins: Below is the complete bipartite graph K₃₃, realized as the edge graph of the Heawood map of the torus. Remember that opposite edges are identified. Here is a little puzzle to warm up. If we order the four coins as up above, we can for each decoration of the Heawood map make a tally like 0330 which lists for each coin how often it occurs. The 0330 is the tally for the simple solution to the the left. Besides that, the following tallies are possible: 1221, 1140, 0411, 1302, 2112, 2031, 3003. Find one decoration in each case. A more interesting map is the Möbius-Kantor map on a genus 2 surface, represented by an octagon with opposite edges identified. The map consist of 6 octagonal regions. Can you decorate it so that the boundary of each octagon uses each type of coin exactly twice? Here is a hint: This map is the double cover of the cube map, branched over the centers of the faces. So you if you can first decorate the cube such that each square uses each coin once, you can lift this decoration… Finally for today, here is the Pappus map on a torus. Again identify opposite edges of the diamond, matching all three coins on one side with the three corresponding three coins on the other side. Can you decorate this map using one blue coin, and for the rest only use purple and brown coins? It will help to remember what we learned about deficits for pillows a long time ago.

## Plane and Simple (Solitaire XVIII – From the Pillowbook XIII)

Alan Schoen’s Cubons and Tetrons make beautiful and interesting puzzles, but few people will have the patience to build them. So here is a workaround. I begin with simplified tetrons where the edges are divided either 1:2 or 2:1. There are just 4 of them, and they nicely fit together into a single tetrahedron. Here are three views of the same tetrahedron. We now represent the tetrahedron by its edge graph K₄, and each cubons becomes a disk with three arrows placed on the vertices of this graph. The graph on the left represents the tetrahedron above. An arrow pointing away from the center of the disk means that the corresponding edge of the cubon is long, and short otherwise. So instead of elaborately assembling tetrahedra, we can just place one of the four types of coins on the vertices of the graph so that the two arrows at the end points of an edge point in the same direction. As an exercise, try to find the tetrahedron below among the four graphs above: Here is a little worksheet so that you can cut out coins in our new currency. You will have realized that these four coins correspond to the four rounded trillows. In essence, we are doing nothing but decorating the vertices of cubic graphs with trillows. The same procedure works for simplified cubons. There are of course again just four of them, represented by the same set of coins. Instead of trying to parse a 3D image, we decorate the edge graph of the cube with our coins. Below you see what the cube on the left above looks like. Try to find decorations of the graph that correspond to the other solutions. Next time we will look into decorating other cubic graphs.

## Alan Schoen’s Tetrons (Cubons IV and Solitaire XVII)

After Alan’s cubons, now his tetrons. There is no end to it… They are constructed like the cubons: Take a tetrahedron, divide each edge into fifths, pick one of the four inner subdivision points on each edge, connect them to face centers and tetrahedron center to decompose the tetrahedron into four tetrons. Up to motions, each tetron is determined by the choice of three numbers from 1 to 4 (up to cyclic permutation), and, as for the cubons, there are 24 possible choices. In fact, the tetrons are just squished cubons. The natural question for anybody obsessed with puzzles is whether these 24 tetrons can be used to assemble six tetrahedra. The answer is yes, you can group them in 11417 different ways into six sets of four so that this is possible. Above is one of them, and it is clear that this image is lacking, so below is the same solution, unfolded into nets. You will recognize what I discussed earlier as trions (albeit there with fewer subdivision points): The tetrons are computationally much simpler than the cubons. For instance, we can again separate the 24 tetrons into 8 chiral and 16 achiral ones. Surprisingly, the 16 achiral ones can be assembled into four tetrahedra in exactly five different ways (up to rotations). Here they are, unfolded: For the 8 chiral ones, the situation is a bit more complicated. There are two ways they can be grouped into two sets of 4, and in each case, there are two ways to assemble each set into tetrahedra. If we denote a single tetron by a list if three numbers that give the number of the chosen subdivision point as seen from the tetrahedron vertex of the tetron, then the partition of the 8 tetrons for the first solution can be denoted like so: {(1, 2, 3), (4, 1, 3), (2, 1, 4), (3, 2, 4)} and {(1, 2, 4), (4, 2, 3), (1, 4, 3), (3, 2, 1)}. Prettier are the nets. The first way to assemble them into tetrahedra is on the left, the second on the right. And here is the other partition: {(1, 2, 3), (4, 2, 3), (2, 1, 3), (3, 2, 4)} and {(1, 2, 4), (4, 2, 1), (1, 4, 3), (3, 4, 1)}.

Again there are two different ways to assemble each quartet of tetrons into tetrahedra. I’ll show the solution next week.

## Pillars (Cubons III and Solitaire XVI)

This (for now) last past on Alan Schoen’s Cubons is dedicated to what Alan calls pillars. Above you see a page from one of several notebooks of Alan, introducing the pillars. A cubon solution has a pillar structure if all four horizontal faces are cut by unbroken lines. The pictures should make clear what this means. There are 456 ways to partition the 24 cubons into 3 groups of 8 so that one can assemble 3 pillar cubes. If we restrict our attention to those that in addition have top and bottom face each cut into unbroken lines, there is only one such pair, consisting of all 16 symmetric cubons. They are shown above, in front and back view, and below as nets. The remaining 8 chiral pillars can also be assembled into pillar cubes, in 8 different ways (not counting symmetries): Last week I asked about polarity, which divides the set of 24 cubons into polar pairs, which use a complementary subdivision of the cube edges. It turns out that there is no solution to the problem to divide the 24 cubons into three sets of 8 so that each set can be assembled into a cube and consist of four pairs of polar cubons. On the other hand, the eight achiral cubons obviously form four polar pairs (and can be assembled into a single cube). The remaining 16 symmetrical cubons can then be divided in four different ways into two sets of eight that are polar to each other, and that can both be assembled (in several different ways) into cube. Above are 3D solutions (one pair each column), and the nets are below. ## Chirality (Cubons II & Solitaire XV)

Alan Schoen’s 24 cubons possess a lot of structure. To get an idea why, let’s encode a cubon by a triple (abc) of numbers between 1 and 4 that indicate on which edge subdivision points its vertices are. For instance, the cubon below on the left would be encoded by (243). Cyclic permutations (432) and (324) encode the same cubon, but (342) is chirally different. The cubon in the right is achiral, as can be seen from its encoding (244). This makes it easy to count: there are 8 chiral and 16 achiral cubons, suggesting that one might be able to assemble a single cube just with the chiral cubons. This is indeed possible, in exactly 32 essentially different ways, i.e. up to rotations. Below is a representation of the same solution set as nets: Similarly, the 16 achiral cubons can be divided in 50 different ways into two groups of 8, each of which can be assembled in (several) different ways into cubes. Most of these have only few ways to be assembled but one of them has 27 essentially different ways to accomplish this for each of the two cubes. Here is one set and here the second one. Notice the striking color separation. There is more structure on the set of solutions. For instance, there is a polar “inversion” that changes the subdivision point of each edge from a to 5-a. This turns any cubon (a,b,c) into the cubon (n-a,n-b,n-c). Following Schoen, we’ll call a decomposition of a cube obtained this way from another decomposition its polar. A decomposition is self-polar if it is congruent to its polar.

Can you assemble the 24 cubons into three cubes that are self-polar, or so that one is self-polar and the second is the polar of the third?

## 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. 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. 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. 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. 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… 