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.
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.
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.
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?
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.
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.
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).
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…
Then there are 75 eye-straining ways to select four of these 48 squares to obtain a complete set of generation 4 quadrons.
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.
The puzzle from over a month ago is based on a curios 3-coloring of the edges of the dodecahedral graph:
Whenever you delete the edges of one color, the remaining edges form a Hamiltonian cycle. Incidentally, on the dodecahedral graph there are two different such paths, up to symmetries of the graph. The question arises whether there are more such graphs.
One such example is the Heawood graph drawn on a torus above (i.e. you identify opposite edges of the big shaded hexagon like in hexagonal astroids). There is a very symmetrical (and boring) edge coloring that colors all parallel edges the same way that is triply Hamiltonian, but there is a different one as well. You can think of this as living on the vertices of a map drawn inside a big hexagon, so that you can travel across a hexagon edge by continuing at the same position of the opposite edge. There are three types of roads, color coded, and each inhabitant of your country is issued two colored passes allowing them to travel on roads of those two colors only. Luckily, the colorings above allow each inhabitant to still travel everywhere, regardless of the passes they have been issued.
To turn this into a puzzle, we use the dual tiling by triangles, that gives us a single triangular puzzle piece (and its mirror) to place on the vertices of the graphs. For instance, below we place one triangle (with its translated copies) on three of hexagon vertices, choose another triangle elsewhere, and are tasked to complete this to a triply Hamiltonian tiling. The solution on the right corresponds to the first coloring of the Heawood graph up above.
For a decent puzzle, the Heawood graph is a little bit too small. Now, Hamiltonian cycles on trivalent graphs have been extensively studied (partially because of Tait’s conjecture, which turned out to be incorrect), so I suspect this phenomenon is known. There is a list of small symmetric cubic graphs, the Foster census, and it is no hard to write code that searches for more examples. Below is thus today’s puzzle, on Foster26A, also a map on a larger hexagonal torus. Complete it to make it triply Hamiltonian.
I continue last week’s discussion of tiling hexagons with the four trillows below.
Instead of curved triangles I am using the markings with curved arrows that have to match in direction when the tiles are put together. This creates directed graphs like so.
Last time we saw that one can systematically solve such puzzles by first drawing the undirected graphs on the region to be tiled, and then directing the edges. Today we will use the concept of deficits to look at this more closely. The goal is to solve puzzles that tile the hexagon of edge length two with an equal number of green and gray triangles. Above we see two solutions that use 12 and 6 green and gray triangles, respectively. Let’s try it with 10 each.
For this, we need to leave two purple and brown triangles. A solution is up above to the right. To the left is the undirected graph. Each edge comes with a deficit, which counts how many more left turns one takes than right turns when following that edge. This number is determined up to sign only as long as the edges aren’t oriented. The orientation needs to be chosen so that the deficits cancel, as left turns give not unexpectedly green triangles, and right turns grey ones. Below is another example with the brown purple triangles located differently.
Can we do it with 11 green and gray triangles each? Below is the complete graph, from which purple diamonds have to be removed until only two purple triangles are left. There are, up to symmetry, only two possibilities, and we see that the deficits cannot be combined to give 0, Therefore this version of the puzzle has no solution. Thanks for trying.
This can be continued, of course. Try it with 9 green and gray triangles each. Below are two solutions with six triangles of each kind.