## HOF+ (Tetrasticks I)

A polystick is a connected finite subgraph of the grid graph, and a tetrastick is a polystick with four edges. There are 25 of them, counting mirror copies. In other words, these are squiggles you can make with four strokes. They’d make a nice alphabet for people who are addicted to abstraction. Today we are focussing on six of them, fattened and colored above. They are denoted by the letters H, O, F, +, and the mirrors of H and F. For reasons to become clear later we consider O and + also as mirrors of each other in a certain sense. The goal is to tile rectangles with them, like in the 3×7 and 2×12 rectangles below. There are many constraints on what tiles one can use, and how many. For instance, an a x b rectangular grid has a(b+1) vertical and (a+1)b horizontal edges, for a total of 2ab+a+b edges. This number is divisible by 4 only if a-b is divisible by 4, so squares are good for tiling, as are the two rectangles above. They both consist of 52 segments and thus require 13 tetrasticks. Below is a different example. Note that all our 6 letter except for H and its mirror use two horizontal and two vertical segments. As the 3×7 rectangle has 4 more horizontal edges than vertical ones, we need at least four H-tetrasticks (or its mirrors) to tile this rectangle. We can use more, but then only an even number of them. Likewise, we need at least two H-tetrasticks to tile the 2×12 rectangle. This brings us to today’s puzzle: Tile the 3×7 rectangle with your choice of 13 tetrasticks from our selection of six, and then use the same set to tile the 2×12 rectangle. The examples on this page are attempts that require to flip an H or an F into its mirror (or an O into +). Can you find a perfect solution that doesn’t require flipping a tetrastick over?

## In Chains (Cooperation Games IV)

We continue with translucent trominoes, add an I-tromino, and select those that have two translucent squares (gray with red border). There are just three of them, shown below to the left. To the right are how they can be attached to each other, overlapping in exactly one translucent square: What we don’t allow today is that two trominoes overlap in both of their translucent squares, as below. This has the interesting consequence that the gray squares will necessarily form chains or loops, which adds useful structure. A tiling is complete if all translucent squares are covered twice. We will then have twice as many translucent squares visible than colored squared. Below is an example of a complete tiling of a 6×6 board. And here is our game: Each player gets the same number of tiles. For two players, is a good choice, say three of each kind. The players take turns placing one tile on the table so that

1. each new tile links with a previous tile
2. two linked tiles are of different color

Alternatively, a player can also remove the last played tile and replace it by another one.

The goal is to create a complete closed chain when all tiles are used up.

Below is an example of a successfully closed loop of length 8. ## Translucent Trominoes (Cooperation Games III)

After the translucent dominoes let’s continue with translucent L-trominoes. There are eight of them: The dark-gray one is a solid tromino, which will be a bit lonely, as it can’t connect to the others.

As before, a region is tiled if either any square is covered by a solid color or by two translucent (gray) squares. Let’s begin with a simple puzzle. Above you see an attempt to tile the 5×5 square just with copies of the purple tromino. This is of course impossible, because each purple tromino tiles 1+1/2+1/2 = 2 squares, so it can only tile regions with an even number of squares. Clearly one can tile a 2×2 square, and thus every 2n x 2n square. But can you also tile a say 6×6 square so that the tiles are all connected? Above is an example how you can tile a 5×5 square with two copies of each tromino except the orange and the dark gray one. There is an abundance of tiling problems like this. You usually begin by determining for a set of tiles the total number of squares they will cover, counting each gray piece as 1/2. As all eight trominoes cover 18 full squares, two sets of them should be enough to cover a 6×6 square. One solution is shown below. This suggests the following cooperative game for two players: Each gets a set of the seven trominoes without the dark gray one, shuffled, and stacked face up. They take turns by either

— placing the top tromino from their stack on a 6×6 board so that the newly placed tromino connects to the network of previously placed trominoes; or
— removing a previously placed tromino from the board so that the network remains connected, and placing it under their pile.

The goal is to leave at the end just room for the two dark gray trominoes. The example above is therefore no solution, because the network has too many components.

## Translucent Dominoes (Cooperation Games II)

Most tiling problems are strictly segregational, i.e. the tiles are not allowed to overlap. To change this, let’s consider tiles that are partially translucent, so that in order to really tile a part of the plane, one needs to cover it multiple times.

This is the first in a series of posts about partially translucent polyominoes, and we begin with translucent dominoes, of which there are three: The purple one is a regular (non-translucent) domino, and to the right you can see my feeble attempt to tile a 4×4 square of which two opposite squares have been removed (This is of course a very classical puzzle). The other two dominoes have one or two translucent squares, which are shown as gray. This translucency means that in order to properly tile a square with dominoes, we need to cover it either with a single solid color square, or with two translucent (gray) squares, i.e. the gray portions of two dominoes must overlap, like so: The left image shows two fully translucent dominoes that overlap in the middle square, while the left and right squares are still only covered once. By counting the small connector squares you can see how many gray squares sit on top of each other. In the middle is a chain of four dominoes, all gray tiles are doubled. And to the right we have covered the middle gray square three times, which is illegal for now.

If we use only the blue singly translucent domino to tile, two of them need to overlap to form a single classical (segregational) tromino, so tiling with these dominoes alone is equivalent to tiling with trominoes. You should try to tile the 4×4 square (with two opposite corners removed as above) with copies of either the singly or the doubly  translucent domino. In both cases, this is impossible (and impossible for larger squares as well, again with opposite corners removed. You will enjoy finding the arguments). Tiling becomes easier when you allow both types of translucent tiles, a simple solution to the 4×4 puzzle is shown below to the right. The left figure gives a hint what limitations you face when you try to tile with the doubly translucent domino alone. As a cooperative game, start with a 6×6 square, mark a few tiles as forbidden, and then take turns to place translucent dominoes on the board with the goal to tile the board completely, following the translucency rule.

## Poly-Worms (Solitaire III)

I have occasionally written about polyforms before. These are shapes obtained by putting simple shapes (like squares) together to form more complicated shapes. In the case of two squares, you ged dominoes, and more generally polyominoes. If you use other shapes, you get general polyforms. If we, in the insatiable desire for more, allow the shapes to change size, we get even more general polyforms. The ones we study today I will call poly-worms. We start with an isosceles right triangle, halve it, and attach the smaller copy to the larger, edge-to-edge. Then we keep going, halving and attaching restlessly. Above you see the first four generations, giving us eight 4-worms, which come in mirror symmetric pairs. Above is a template that allows you to print all eight 4-worms at once. Growing the polyworms further leads to problems with self-overlapping, but also to the tantalizing possibility of having polyworms with infinitely many sides. Maybe more about this in the finite future. Let’s practice tiling with 4-worms. Below are combinations of 4-worms that can be used to tile the plane periodically by translating them. There are also two 4-worms that tile the plane by themselves. It should be amusing to study this for general polyworms. Here are two puzzles, hopefully not too easy. The goal is to tile the left one with 8 and the right one with 16 4-worms.  You will need the same number of each kind. ## The Advantages of Different Viewpoints (From the Pillowbook IX)

It all started with a question about polysticks: I wanted to see how to tile parts of the square grid with 3-sticks so that there are always three 3-sticks touching. A 3-stick is nothing but a T with all segments the same length: it has 3 arms that end at the hands, and are joined together at a head. By tiling I mean that the arms align with the edges of the standard square grid and don’t overlap. And, as I said, I want always three of them to hold hands. Here is an example (that you should imagine continued periodically): The condition about holding hands in threes means that each such tiling has a dual tiling where the new 3-sticks have their heads wherever three hands come together, the same arms, and the previous heads are replaced by three hands. This also implies that a tiling and its dual will tile the same portion of the square grid, as below to the left and right. We can also combine a tiling and its dual into a single figure by centrally scaling each 3-stick by 50%, and taking the union: The gaps created by the scaling makes room for the 3-sticks from the other tiling. You can examine the result up above in the middle. The new skeleton will have all 3-sticks hold hands in pairs instead of in triples. Here is another, more complicated example. The original periodic 3-stick tiling: The inflated version that combines the original with its dual: In the inflated version we can replace each 3-stick by a square so that the sides touch when the 3-sticks hold hands. The unused edge of the square is pushed inwards, turning the square into the familiar 3/4 pillow we admired the last time (the next image shows only a quarter of the previous piece. It repeats itself using horizontal and vertical translations). As a final simplification, we can fill in the holes as follows: We replace the 3/4 pillows that border a hole by the polyomino they cover  (thereby filling in the hole). Below are the two simplest polyominoes that surround a hole: These polyominoes will tile the plane as before, because each 3/4 pillow must belong to exactly one hole. Our 3/4-pillow tiling now becomes a very simple polyomino tiling: Thus periodic 3-stick-tilings with triple hand holding, 3-stick tilings with hand holding in pairs, 3/4-pillow tilings with holes, and  tilings by polyominoes that surround holes are all the same thing.

You can use this to design  much more intricate patterns, with holes of any size, for instance.

## Three Quarters (From the Pillowbook VIII)

Today we will talk about a single, very neglected pillow, which I will call 3/4. To make things look pretty, we will color it depending on its orientation, as follows. Discriminated for centuries by the other, more curvy pillows, 3/4 only likes to hang out with other 3/4s and hide its straight edges as much as possible. Like so: These two examples are portions of periodic tilings of the entire plane. This way, all straight edges are hidden, and we have periodically placed holes. This quickly becomes more interesting. Here, for instance, is the only tiling that has just circular holes (up to symmetries, of course): The holes can be more complicated, like these double holes in subtly different periodic tilings. Next month, we will analyze these patterns a bit. For today, we end with an example that indicates that the holes can become really big (as we will learn). For today it’s enough to have learned that also the seemingly uninteresting can do pretty things.

## Double Parity (From the Pillowbook VII)

Here are the 36 pillowminoes introduced last time, arranged by their imbalance, i.e. according to how many more convex than concave edges they have. Isn’t that a pretty bell curve? This time we will focus on the pillowminoes near the border of existence, namely the six ones that have all but one edge either bulging in or out. They have an imbalance of +4 or -4. Gathered and recolored, here are the marginal pillowminoes: Let’s tile some curvy shapes with these. A curvy rectangle has odd dimensions, so cannot be entirely tiled by pillowminoes. If we decide to leave a round hole in order to fix that, the entire curvy shape will be balanced. This means that we will need the same number of brownish pillowminoes (with imbalance +4) as bluish ones (with imbalance -4). In particular, we will need an even number of these pillowminoes, so the total area of our shape needs to be divisible by 4. That’s our double parity argument.

The simplest example is that of a 5×5 square with a center hole (it’s easy to see that skinny rectangles with one edge of length 3 are not tilable with marginal pillowminoes). The example to the left is the only one I could find, up to the obvious symmetries. To the right you see how one can inflate it to make frames, proving:

Theorem: If you can tile an axb holy rectangle with marginal pillows, then you can also tile a holy (a+4)x(b+4) rectangle with marginal pillows.

We have seen this trick before, talking about ragged rectangles.

The next interesting case are 7×7 squares. Here is one example that also teaches us another trick:

Theorem: If you can tile an axb holy rectangle with marginal pillows, then you can also tile a holy ax(b+4) rectangle with marginal pillows. This second trick decenters the holes, however.

Finally, two examples that employ all six different marginals. First a 5×7 rectangle with center hole, then another 7×7 square that uses four marginals of each kind, nice and symmetrically. ## Pillow Puzzles (From the Pillowbook V)

After admitting a few pillows with straight edges, there is no end to it. Here are all 24 pillows based on a square that either have a straight, concave, or convex edge. We disregard rotational copies but keep mirrors. Usually, polyformists try to tile simple shapes using each polyform exactly once. The archetypical example is to tile a 6×10 rectangle with all 12 pentominoes. This is in most cases a tedious exercise that doesn’t teach you much more than backtracking. On the other hand, nothing is worse than not knowing, so here you go: Three puzzles that ask to tile the outlined region by using each of the 24 pillows exactly once. The grid is there to help placing the pillows. These puzzles are actually not so bad. The first one for instance requires to make economic use of the pillows with straight edges. I post the solutions below, mainly because nobody would do them anyway and to prevent future waste of time. Note in the solution above the second column consisting entirely of pillows with parallel straight edges. I think this has to appear in any solution of this puzzle. The one above is my favorite. Unfortunately, one could go ahead and ask to find solutions of similar puzzles where the shape of the hole in the center is any of the remaining 23 pillows. No. ## Triangles and Squares

There are two Archimedean tiling using triangles and squares. Both of them use twice as many triangles than squares. I find the first one is more interesting, maybe because it is chiral. There are still many other ways to tile the plane say periodically with just triangles and squares. There are three different ways to assemble two triangles and a square, and all of them give polyforms that can be used as a single subtile for the first Archimedean tiling: Among these three polyforms I like the middle one best, maybe because it cannot be used to subtile the second Archimedean tiling. It is an amusing exercise to doodle around and find other tilings of the plane with this tile. Here, for instance, are two small turtles and a giant caterpillar, all part of a big creation. I find it amusing how this simple polyform lends it self easily to organic shapes and abstract designs. There are (I think) 10 ways to combine two of them into a single polyform, not counting mirror images. At least two look like cats. Confusing as they look, almost all of them tile the plane. The two exceptions are shown below. It is not difficult to find an argument why these two do not tile. More interestingly the other eight tile, even though they look much more complicated. Typically one needs for each tile its mirror, suitably rotated. Here are two pretty examples. Homework is to find the others. 