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?
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.
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.
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.
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.
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.
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:
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?