## Over and Under

Time for a game. You will need one or more decks of the following 16 triangular cards.

The first game is a puzzle and asks to tile a triangle with all cards from a complete deck so that the tiles match along their edges, like so,

except that in my attempt above two triangles don’t match. No hints today.

Next we use one or more decks to play a 2-person game. All cards are shuffled and form a single deck, top card visible. You will need three special cards, each just marked with one of the three card colors on it. Both players draw one of these color cards and keep their color secret.

Now they take turns picking the top card from the deck and placing it on the table so that it matches previously placed cards. However, this time the matching has to happen along half-edges. For instance, after using a 16 card deck, the table might look like this:

Each newly placed card has to border a previously placed card, and match along half edges on all sides where it borders another card.

When all cards are played, the players reveal their secret colors and score. Note that the colored arcs form chains of equal color. Suppose that player A is orange and player B purple. A looks at all orange chains of length at least 2. There are three orange chains of length 2 and one of length 3. For each of these chains, A counts how often they go over a purple arc. This happens 7 times, and A scores as many points. Similarly, B looks at all purple chains of length at least two. There are three, of lengths 2, 3, and 4. They go over an orange arc six times, so A wins by one point.

The idea is to arrange cards in chains of your color that go often over the snakes of your opponent’s color. The problem is, of course, that in the beginning you won’t know your opponent’s color. So you might want to put cards that have your arc go under both other arcs not into chains but out of the way. This, however, might give away your color…

Finally, here is a 3 person game played on a hexagonal board of edge length 4.

The three players are dealt a color card each, and again the colors are being kept secret. You will need six decks of cards, for a total of 96 cards. Shuffle all cards and let each player grab 32. Then the players put their cards onto the board so that they meet at least one previously placed card along an entire edge, and match the colors of all cards they meet. After all cards are played, the board will look similar to the one below, except that there will be crossings.

When the board is completely tiled, the players reveal their colors and score: For each completed circle of their color, the player counts how often that circle stays above the other two colors. This can happen (for each circle) between 0 and 12 times. That number is squared, and all numbers for all full circles for each color are added up, giving the score for that player.

There is a little catch to be aware of: There are two colors with 12 full circles each, and one color with 13 circles (purple in the example). Clearly the player with 13 circles will have an advantage when scoring. The first player will decide which color has 13 circles, and is therefore likely to claim the advantage. But then the two opponents might unite…

## Spring Cleaning I

Spring Cleaning is played on a rectangular array of randomly placed dirt pieces. A sweep consists of removing a single row or column of consecutive dirt pieces.

Above are some example of legal sweeps, and below are illegal sweeps.

This is a game for two players, who take turns by doing exactly one sweep. The player who sweeps the last time is the winner. This is an impartial game which means that each position is equivalent to a single game of Nim. This is usually bad news, because playing Nim well requires us to perform exclusive or additions of binary numbers in our head, for which our brains are not (yet) well equipped.

The good news here is that many simple positions are equivalent to very small Nim piles, meaning that computations are easy. I will explain this using an example. No proofs (even though they are easy, too).

It’s your turn to find a winning move in the position above. You know (because I promise) that rectangles completely filled with dirt pieces are easy positions, so you will look for moves that separate the dirt pieces into such rectangles. Here is such a move:

After that, we are left with four separate rectangles, all completely dirty. This means that this game is equivalent to a game of Nim with four Nim piles. The question is what the pile sizes are. The answer is simple: Any rectangle both of whose dimensions are odd corresponds to a Nim pile of size 1, if both dimensions are even, the Nim pile is empty (size 0), and otherwise, the Nim pile has size 2. In our example, we have a 1×1 rectangle, a 1×3 rectangle, and two 1×2 rectangles. They correspond to Nim piles of sizes 1, 1, 2, and 2. The exclusive or sum of these numbers is 0. This is what we want, because it means that after this move, the game is equivalent to an empty Nim pile. From now on it’s easy. Suppose that our opponent performs a vertical swipe on the 1×3 rectangle. What do we do to return the game to Nim-value 0?

We can sweep away any of the isolated dirt pieces: From then on, the game is symmetrical and we can win easily without any Nim-theory. And we better leave the two 1×2 rectangles untouched. Suppose we remove one of them completely. Then we are left with three 1×1 rectangles and a single 1×2 rectangle, which exclusive or sums up to a Nim pile of size 3, in which case our opponent can win.

The winning move would be to reduce the remaining 1×2 rectangle to a 1×1 rectangle with a horizontal sweep.

So, if we know how to deal with Nim positions that consist of Nim piles of sizes 1 and 2, we will be able to win Spring Cleaning by dissecting a given position eventually into rectangles.

## 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)

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.

Similarly, this tiling

encodes one layer of the following surface of genus 5:

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

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).

## Columns

After looking at the intersections of symmetrically placed cylinders and obtaining curved polyhedra, it is tempting to straighten these intersections by looking at intersections of columns instead.

The simplest case is that of three perpendicular columns. The intersection is a cube. Fair enough. But what happens if we rotate all columns by 45 degrees about their axes?

Before we look, let’s make it more interesting. In both cases, we can shift the columns so that their cross sections tile a plane with squares. Surely, every point of space will then be in the intersection of a triplet of perpendicular columns. In other words, the intersection shapes will tile space.

Yes, right, we knew that in the first case. I find the second case infinitely harder to visualize. Fortunately, I have seen enough symmetrical shape to guess what the intersection of the three twisted columns looks like it is a rhombic dodecahedron.

But not all triplets of columns that meet do this in such a simple way, there is a second possibility, in which case the intersection is just a twelfth, namely a pyramid over the face of the rhombic dodecahedron.

Together with the center rhombic dodecahedron they form a stellation of the rhombic dodecahedron, or the Escher Solid, of which you have made a paper model using my slidables.

Above you can see a first few of Escher’s solids busy tiling space.

## Alchemy (From the Pillowbook X)

Here is a variation of the pillow theme. This time, the tiles are not based on squares as the regular pillows or on triangles as in an older post, but on 60 degree rhombi. I only use pieces with convex or concave edges, so there are seven different rhombic pillows up to symmetry, this time also not distinguishing between mirror symmetric pieces. The main diagonals of the original rhombi are marked white. For the purpose of the Alchemy game below, I call them elements.

These elements can be used to tile curvy shapes like the curvy hexagon below. Again, for the purpose of the game, I call such a tiled hexagon a Philosopher’s Stone.

I leave going through the brain yoga to discuss tileability questions to the dear reader. Instead, here is the game I designed these pieces for.

# Alchemy

### Purpose

To complete the Magnum Opus by crafting a Philosopher’s Stone.

### Material

• The seven elements above in seven colors, colored on both sides, at least 4 of each kind for each player;
• One transmutation card for each player;
• One Philosopher’s Stone outline for each player;
• Pencils and glue sticks.

Below is a template for the transmutation card. It shows a heptagon with the elements at its vertices, and all possible connections (transmutations, that is).

### Preparation

All elements are separated into resource piles according to color/shape. Each players takes a transfiguration card and an outline of the Philosopher’s Stone.

Above is an outline of the Philosophers stone, with little notches to indicate where the corners of the elements have to go. The elements are shown next to it to scale so that you get the elements in the right size.

### Completing the Magnum OpusGoals

The goal of the game is to accomplish the Opus Magnum by filling the outline of the Philosopher’s Stone with elements using as few transmutations as possible. Elements must be placed so that

• at least one corner matches a notch or a corner of another element that has already been placed;
• elements don’t overlap and don’t leave gaps;
• no two equal elements may share a curved edge (but they may share a vertex).

### Scoring

When a player has completed a Philosopher’s Stone, he or she determins the used transmutations:
A transmutation occurs in the Philosopher’s Stone when two elements share a curved edge.

The players record a transmutation on their transmutation card by drawing a straight red edge between two elements that share a curved edge in their completed Philosopher’s Stone.

The unused edges are then drawn black. The player with the largest number of black edges becomes the master alchemist.

Below is the completed transmutation card for the Philosopher’s Stone at the top. This was a pretty poor job, the player used all but three of all possible transmutations.

One can turn this game also into a puzzle. Can you tile the Philosopher’s Stone with the seven elements that its transmutation card is the one below?