A Room With Two Views (Five Squares IV)

Today we look at tilings that utilize just the four other squares. The first step in classifying these is again a simplification, making the split corner squares uniformly green. This leaves us with two tiles:


Ignoring the pink triangles for the moemnt, we recognize the problem we solved last time: The green squares need to occur in shifted rows or columns, like in the example below. Here we have four rows of green squares. Rows 1 and 2 are shifted, as are rows 2 and 3, but rows 3 and 4 are alined.


To add the pink triangles, note that two pink triangles fit together to a pink diamond, and each grid cell needs to have one of those, but we can only use those edges that are not already adorned with a green square. This leaved us with the following possibilities: If two consecutive rows of squares are aligned, we have place two diamonds in the square space between four squares, and we can do this horizontally or vertically. This can be done independently of neighboring squares, as shown between the two bottom rows below.


If the rows are shifted, we also have two possibilities to place the diamonds, but each choice affects the entire row, again as show above in the top rows.

Finally, we need to undo the merging of the orange and blue triangles into green squares, and we can do so by splitting each square either way and independently.


Below is an example how teh corresponding polyhedral surfaces will look like. The horizontal squares correspond to the green squares of the tiling. They are the floors and ceilings of rooms that have two opposing walls and two openings. I start seeing applications to randomly generated levels of video games here…


More about Decorated Squares (Five Squares III)

In order to classify surfaces that have five coordinate squares around each vertex, we were led to consider planar tilings with six different colored squares. Today we will discuss a special case of this, namely tilings that use just two of these squares. The only rule to follow is that colors of tiles need to match along edges. Here is an example:


To classify all tilings by these two squares (and their rotations), we first simplify by solely focussing  on the gray color (making it dark green), and considering the blue, orange, green as a single color, namely light green. This way we get away with just one tile. Of course we hope that understanding how this single tile can fill the plane will help us with the two tiles above.


We first note that placing the tile determines three of its neighbors around the dark green square. So instead of tiling the plane with copies of this squares, we can as well place dark green squares on the intersections of a line grid so that for each cell of the grid, precisely one corner is covered by a dark green square, like so:


We first claim that if we do this to the complete grid, we must have a complete row of squares or a complete column of squares. Below is a complete row (given the limitations of images). The red dots indicate where we cannot place green squares anymore, because the grid squares have all their green needs covered.


If we do not have such a row, there must be a square without left or right neighbor. Let’s say a square is missing its right neighbor, as indicated in the left figure below by the rightmost red dot.

Notice how the two grid squares to the right of the right dark square have only one free corner. We are forced to fill these with dark squares, as shown in the middle. This argument repeats, and we are forced to place consecutively more squares above and below, completing eventually two columns.

As soon as we know that we have (say) a complete horizontal row, directly above and below that row we will need to have again complete rows of squares, as in the example above. These rows can be shifted against each other, but that’s it. So any tiling of the plane by the dark/light green tile consists of complete rows or columns with arbitrary horizontal or vertical shifts, respectively.

Finally we have to address the question whether this tells us everything about tilings with the two tiles above. This is easy: Each dark green square represents a light gray square that is necessarily either surrounded by blue or orange tiles. So we can just replace each dark green square by an arbitrary choice of such a blue or orange cluster. The final image shows such a choice for the example above.


It is now easy to stack several such tiled planes on top of each other, thus creating infinite polyhedral surfaces that have five coordinate squares at each corner.

More Examples (Five Squares II)

To review, let’s start with the following tiling

New 4 tile

Now use the dictionary below to replace each tile by the corresponding 3-dimensional shape. Each tile from the bottom row is an abstraction of an idealized top view (top row) of a rotated version of five coordinate squares that meet around a vertex (middle row).


By using the top left quarter, we get the top layer of the polygonal surface below. The bottom layer uses the same pattern as above with blue and orange exchanged. This is a fundamental piece under translations, and we can see that the quotient has genus 4. This also follows from the Gauss-Bonnet formula, which says that a surface of genus g uses 8(g-1) of our tiles (12 for the top and bottom each in this case.

Genus 4

Similarly, this tiling

Genus5 tile

encodes one layer of the following surface of genus 5:

Genus 5

To make things more complicated, the next surface (of genus 4 as well)

Plus 1

needs four layers until it repeats itself. Two of them are shown below.


These tilings exhibit holes bordered by gray edges which complicates matters, as we will now also have to understand partial tilings (with gray borders).


A while ago I tried to start a blog about games and puzzles, which failed, mainly due to time constraints.
I will recycle some of the posts here.

Here are a few crafts of varying difficulty that you can do just with card stock and scissors. The idea is always the same: Use several copies of a simple shape with slits to build paper sculptures. They all make nice holiday ornaments.


The simplest such shape is an equilateral triangle that has been slit as shown below.


Using four such triangles, you can build the following star.

Folds 2

With eight triangles and a bit more patience, you get the following shape, which is Kepler’s Stella Octangula, a stellation of the octahedron, or the compound of two tetrahedra.

Folds 6

I like to curl the tips of the triangles to make them look like flower petals.

You can of course also build other objects.

Pentagonal Stars

Folds 1

Using 12 copies of the slit pentagrams below, one can build Kepler’s Small Stellated Dodecahedron.


This requires a bit patience. Start with one pentagram, and insert five pentagrams successively in all of its slits, thereby also linking the inserted pentagrams together as well. Then insert another five pentagrams into neighboring pairs of the first ring of pentagrams, again linking the pentagrams from the new ring together. Finally, insert the twelfth pentagram into the free slits of the pentagrams from the second ring.

The last steps require some heavy bending of the pentagrams, and careful adjustment at the end.

Triangles and Squares

Folds 7

Using properly slit triangles and squares, one can build a stellation of the cuboctahedron.


The slits in the squares and triangles must have the same length.

This is a bit easier than the previous example. During assembly, the model falls easily apart, but it is quite sturdy when done.

Irregular Hexagons

Folds 4

Twenty of the regular hexagons below can be used to create one of the stellations of the Icosahedron, the Small Triambic Icosahedron.


Escher’s Solid

Folds 3

This is the first stellation of the rhombic dodecahedron, also called Escher’s Solid. It tiles space. You need 12 of the non-convex hexagons below.


A simpler version is a stellation of a rhomboid, using 6 hexagons.

Folds 5

Final Comments

The strategy to design these models is to look for regular polyhedral shapes with few kinds of faces that intersect in a relatively simple way. Then, each intersection of two faces leads to slits on both faces half way along the intersection, so that the two faces can be slid into each other.

There are of course limits to this, but I am sure there are many more models one can assemble.