## Dual pillows (Solitaire XXIV – From the Pillowbook XVIII)

Today we introduce the dual pillows:

They arise as follows: Take a tiling by pillows, and practice social distancing. The gaps between the pillows will instantly become occupied by the dual pillows, like so:

While the dual pillow don’t look anything like the standard pillows (which also don’t really look like pillows, I just wish they would), they are nothing but real pillows, which might be the quintessence of any duality. To see this in this instance, let’s switch to the arrow representation of pillows I introduced to make Alan Schoen’s cubons accessible to flatlanders:

A pillow is replaced by a cross where an arrow pointing away from (resp. towards to) the center represents a pillow bulge inwards (resp. outwards). A tiling by pillows becomes an oriented grid graph. Rotating every edge of this grid graph about the center of teh edge by 90º and changing its color from orange to green creates the oriented dual grid graph, which we can now interpret pillows or dual pillows.

Purists will suffer under the different appearance of pillows and dual pillows. So above is a more sober but completely symmetric representation of pillows and their duals. They, as well as their more baroque counterparts, can be used for puzzles and games.

Above to the left is a 3×3 board game. It is easy to fill it with the abstract pillows / dual pillows so that colored squared are correspondingly covered and gray triangles never overlap. But you can do this competitively, green and red taking turns until one player can’t place a tile anymore.

Or, if you still prefer isolation, can you tile a larger 6×6 board with 6 sets of pillows and dual pillows?

## 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?

## Colored Pillows (Solitaire XXI – From the Pillowbook XVI)

Another lonely day gave me the idea to spice up the six standard pillows into domino type puzzle pieces by coloring them like so:

Below is a less pretty but easier to cut version, assembled in a 5×2 rectangle,

which solves what a mathematician would call a boundary value problem:

Here is another puzzlable boundary contour for your solitary enjoyment. The rule is to place the puzzle pieces so that they fit & match in color when they meet:

These are nice little puzzles, not too easy, not too hard, but 10 is an awkward number (why don’t we have 16 fingers, like everybody else out there?), and it is somewhat annoying to have to turn around some of the pieces by 180º to see whether they finally fit, so I decided to modify the coloring a bit, like so:

The 16 puzzle tiles above must not be rotated anymore. Equivalently, the horizontal and vertical color gradients need to match with adjacent fitting pieces. The pieces above all fit together to form a 4×4 square which periodically tiles the plane, moreover, this square is symmetric across one of its diagonals. Can you find other such square, tiled by using each of the 16 colored pillows exactly once? There are a few, but not too many.

These 16 puzzle pieces would make great 2-person games, for horizontal and vertical players… Their time will come. For now, enjoy the two boundary value problems above. One of them is very very very hard.

## 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.

## Above and Below (Solitaire XVII – From the Pillowbook XII)

I wrote the first Solitaire post in March, exactly four months ago, being almost certain that after maybe two months I could safely move on to two person games. Now it looks like this will have to continue for a while. At least I can assure you that by the time I run out of topics, the pandemic will be over, one way or the other…

Today’s puzzle is concerned with the Heawood map. This is a map consisting of seven hexagons arranged as up above in the Heawood tile to the left, with edge-zigzags matching in pairs of equal color. This matching can be used to periodically tile the plane as to the right, or to interpret this map as a map on a torus, thus showing that one needs at least 7 shades of gray to shade a general map on a torus. (7 is indeed optimal).

After the square pillows and triangular pillows, it is now finally time to introduce the 14 hexagonal pillows above. That’s all there is with curvy edges only — if you allow for straight edges, you get (too) many more. It is (for some of us) tempting to replace the hexagons of the Heawood tile by seven pillow tiles, so that the entire Heawood tile can be used to periodically tile the plane. If you only use one type or pillow, there are only two possibilities:

With two different pillows, it gets more interesting (and prettier). Below are two (slightly different) solutions using the same two pillows.

And here are two more, again using the same two pillows, which are less similar. You can find four more by reflecting all these, but that’s it with two pillows.

Now let’s jump ahead and try to use seven different pillows. Here is a simple example:

How hard is this? There are 3432 ways to select 7 different pillows from the 14, but only 380 of these choices allow you to form a Heawood tile. That’s maybe not too hard. But, of course, you (I) would want  to assemble the remaining 7 pillows also into a Heawood tile. That’s today’s challenge, and I think it’s rather difficult (there are still many different solutions). The hint below may not be that useful. It merely shows the contours of the Heawood tiles for one particular solution to this problem. But at least it now becomes a very concrete puzzle: Just tile the two regions using all of the 14 pillows exactly once.

## 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.