Broken Circles

Here are two circles with an eighth removed. As we can see, they can move together. What about adding more such broken circles? How densely can we slide them together?

Here is an example with three circles, each one sixth missing:

This is maybe a little bit remarkable: If you take two such broken circles and rotate them by 60º against each other, you can slide them along each other so that one end point of one circle moves on the other circle, and vice versa. The first question (which must have an easy answer, of course) is: why does this work for these angles?

As shown above, this allows us to pack three of these broken circles together, creating a mild form of prettiness.

Now let’s use four broken circles, with a quarter removed each, rotated by 90º, and colored appropriately:

Again, one can slide these circles along each other. Can you do this with six broken circles? Does this work with other angles?

Hopefully continued.

Tripods III

After realizing that while choices allow for free will, too many choices make everything possible and create only an illusion of free will. So let’s allow very few choices in our tripod-universe:

Above are the tripods we are allowed to use, and below the first three generations of our expanding universe if we pick the first of the three above and place it at the center:

We see that at each step, there are only two choices that occur along precisely one branch (marked red). So while the number of possible universe histories grows exponentially, the overwhelming majority of its inhabitants (i.e. the leaves at the end of the trees) don’t have a choice, their future is predetermined and can’t even be affected by the single monarch who can only determine their own future. More choice is needed.

So let’s allow all six tripods that use three colors with just one color occurring twice as leaves. We choose one of them for the Big Bang at time 0. At time 1 we already have 27 different possible histories, because at each leaf there are always 3 choices that can be made:

In the next generation, we will have already 19683 different possibilities. This looks promising, so let’s see how much these tripods can control their future. Below are two universes at time 2 that use only the colors yellow/green and yellow/blue at the leaves. Can you find a universe where all leaves are yellow? Or blue?

Some more questions:

• Suppose you succeeded in making all leaves yellow. How many more generations does it take you to make all leaves blue?
• Can you have a universe where no two neighboring leaves have the same color?
• Below is a universe where in generations 1,2, and 3 each we have an equal number of leaves of each color. Eg, in the current generation 3, there are 8 green, blue, and yellow leaves. Can you continue like this? Is there a recipe for it?

Tripods II

Let’s change the paradigm. Here are the eight tripods that will inhabit our universe today. Note that each tripod uses exactly four different colors.

At the beginning, in year 0,  there was only one tripod. Feeling lonely, it sprouted three more, augmenting each leg with two more legs.

We notice that there are choices: Each existing leg uses two of the four colors, and there are two possibilities to place the remaining two onto a newly sprouted tripod. So, in year 1, there are already 8 different possible universes. Free will is a good thing.

But there is also symmetry. At each node of the tree, we can exchange the branches, giving us lots of commuting involutions. We call trees that are obtained by these exchanges isometric. In particular, except for the coloring of the first tripod, all subsequent choices lead to isometric trees. Free will was an illusion. This insight should be hint enough to answer last time’s questions.

It also helps to go backwards in time, to uncover our past. For instance, the color at the center of the tree occurs most often at the rim of the universe, at year 2. It is easy to find more laws that allow us to understand the entire universe if we just know what it looks like in the leaves of year two. Even if the available information is only partial, we can often say a lot.

For instance, the partial information on the left universe above allows for two different pasts, one of which is shown on the right. Find the other one.

Another year has passed. Are the same laws still valid? Can you reconstruct the history? Is there more than one? Time is a complicated thing…

Tripods I

With all the emerging wildflowers, I have been using my tripod a lot lately, and this has led to today’s puzzle. We are going to color perfectly height balanced trivalent trees, like so:

This is of course too easy as it stands, so we have to impose restrictions. Today, I will insist that all tripods in your coloring look like this:

So a tripod is a perfectly height balanced trivalent tree of height 2, technically speaking. A quick inspection shows that the example above is of this sort. There still are many many such colorings, given the symmetries of the tree, and we’ll need further constraints for today’s puzzle. Before we get there, I have a few questions:

• The coloring above uses one color 21 times and the other color 25 times. Is this always the case?
• Is there always a single colored path from leaf to leaf through the center of the tree?
• How many different colorings exist if you disregard symmetries?

Above is a simple version of today’s puzzle. On the left, you see a partially colored tree. On the right there is a completed coloring, following the rules that all tripods in the tree must be colored as the two tripods above. In this case the solution is unique, as also in the puzzle below:

Enjoy. More tripods puzzles next week…

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?

Fractal Maze

A maze usually consists of a few rooms with options to move to other rooms, and the goal is to find a way from one given room to another one. In a fractal maze, some rooms might be copies of another maze, that itself contains rooms that are copies of other mazes, again and again. Here is a simple example:

We have four mazes, each consisting of two regular rooms (in white) and two substitution rooms (colored). The thick black lines tell you where you may move. When you enter a colored room, you enter the maze with that color, like so:

You can think about it like a dungeon where you descend to a lower level each time you enter one of these rooms. When you leave a maze to the left or right, you accordingly ascend one level back up (unless you have been eaten by a grue). Here is a sequence of rooms you can visit. The roman numerals indicate the level you are on, and the room you are in is indicated by a number.

In this example, it took 6 steps to get from the left room in the yellow maze to the left room in the purple maze, starting and ending at level I. Today’s puzzle is to get from the left room in purple to the left room in green. Below is a very simple solution, but this requires to go from level I up one level.

But alas, level 0 hasn’t been built. Can you nevertheless find a way?

Grow, Contact, and Trinity (Cooperation Games VII)

Let’s bring back a bit of color. Today we get a trio of games, played with similar cards and similar mechanics. The simplest version I call Grow, and it’s a game for two players One and Two, to be played with triangular cards on a triangular board. The pieces are called (top to bottom) root, branch, and twig.

We will need a single root-piece, 12 twigs for player One, and 12 branches for player Two. We begin by placing the root somewhere on the triangular board. Then we take turns by adding twigs and branches so that adjacent edges match and we have a connected growing tree.

There are two more rules: We don’t allow loops, and the branches must branch away from the root. The layout above on the left is still correct, but in the middle we illegally have closed a loop, and on the right we closed a loop and are also branching in the wrong direction.

Above are two solutions. The players need to collaborate in order to fill the entire board. Can we find a solution for every position of the root-tile? Can we solve this on a triangle with edge length 7 instead of 5?

Contact uses tiles with two colors, and adds leaf tiles. We begin by placing two socially distanced roots, one of each color, inside a regular hexagon of edge length 6, like so, for instance.

Again this game is played with two players, Orange and Green, but this time a player controls all pieces of the same color. There is only one root tile of each color, and sufficiently many leaves, twigs, and branches. The goal is to tile the entire hexagon by taking turns, so that all twigs and branches end in leaves, as shown below. Note in particular that no loose ends of twigs or branches are allowed on the boundary of the hexagon.

However, this solution has a flaw: We have used 27 orange and only 20 green tiles. Can we do it with the same number of tiles for Orange and Green?

Finally, Trinity is played with pieces in three colors, and two chiral leaf tiles. Again, each of the three players controls one color. This time, multiple roots are allowed, but watch out that different trees of the same color don’t grow together.

At the beginning, each player places one root tile on the board, again socially distanced. Then we take turns either growing our own tree, placing a leaf tile that fits at least our own tree, or placing a root for a new tree. Below are two solutions on a hexagon of edge length 6. Enjoy.

Ninety-Six (Collaboration Games VI)

Today Alan Schoen is celebrating his 96th birthday (below you can see him at the youthful age of 93),

and this gives me the opportunity to congratulate him with a puzzle. Its mechanics is motivated by the genus 3 Riemann surface which is the double cover branched over the vertices of a cube, the Riemann surface that underlies the Schwarz P surface, the Diamond surface, and Alan’s Gyroid, his best known discovery. Not quite incidentally, this surface has 96 symmetries. Today’s puzzle is also motivated by Alan’s puzzle RotoTiler.

Above is the first version of the puzzle, consisting of 12 squares. You are allowed to put two of them next to each other if the adjacent edges have the same color and the same orientation of arrows, but opposite shading. It is also prohibited to place two tiles next to each other that use the same pair of edge colors. All that is forbidden is shown below (wrong edge colors, wrong edge direction, wrong shading, same edge pairs).

It’s not that hard to lay out all 12 squares in a single connected way. You will notice that it is impossible to place four squares around a single vertex.

As a simple puzzle: The above layout fits in a 4×7 rectangle. Can you find a solution that fits into a smaller rectangle?

And, if you have a partner to play: print say 8 sets of the tiles so that you have 96. One player gets all the dark squares, the other the light squares, and you take turns creating a connected layout of all tiles. It is quite easy to get stuck, so the players have to collaborate.

The two crosses above give a hint how this is related to cubes. They can each be wrapped around a cube, one from the inside, the other from the outside, so that the two tiles on each face differ only in shading.

Next we have a rhombic version of the same puzzle, now with 24 different tiles. This more truthfully can be used to represent a walk on the Schwarz P-surface. We have one additional rule: 60º vertices are not allowed to meet with 120º vertices. All that’s forbidden is listed below:

Below are a few partial layouts. None of them can be extended.

Can you lay out all 24 tiles in a single, connected figure? That you can’t has to do with the Schwarz P surface having congruent insides and outsides, and it being impossible to reach the inside from the outside.

More down to earth is the following explanation: You can lay out all 24 tiles in the two rings above, but no tile from the left ring can be legally placed next to a tile from the right ring. To remedy this, we can relax the rules a bit. If we allow that two tiles that use the same pair of colored edges can be placed next to each other, a single chain can be found:

This was not so easy. Can you find smaller displays? And again, you can play with two player, using four sets of tiles, and take turns trying to complete a single layout.

Happy Birthday, Alan!

Binomino (Cooperation Games V)

Here is yet another domino variation. Below are the eight 3-binominoes:

You place them sideways so that colors of adjacent squares match. This is, alas, impossible, unless you use two identical binominoes. Therefore we are generous and allow the binominoes to be shifted up or down by one square, so that they have contact only along two squares. A chain using all eight might look like this:

The connectivity graph is surprisingly complicated. I have drawn the directed version, the target of an edge connects to the source by sliding it down one square.

A simple game for two players divides the eight binominoes in two sets of four. Each player gets one set, and they take turns placing binominoes in a chain so they match along two squares. If a player can’t play (either because their binominoes don’t fit, or because they are out of them), they have to take one from either end of the chain. The goal is for both players to finish simultaneously.

Above is a circular version with 4-binominoes, ragged so that not only the colors have to match, but the notches as well. I hope this looks appealing. The same rules apply, but now the goal is to create a closed chain, like so:

Here this is even done so that the binominoes are always shifted the same way. In other words, this solution represents a Hamiltonian cycle in the directed connectivity graph. I believe this graph is always Hamiltonian, for n-binominoes with arbitrary n.

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.