28
votes
Accepted
Tiling the plane with pairwise non-congruent rational triangles
First answer: The plane can be tiled as requested. First, we tile the plane with equilateral triangles with side lengths $1$. Now each such triangle can be tiled into two rational-sided triangles in ...
- 20.8k
20
votes
Accepted
Aperiodic monotile without reflections?
The same authors have just released a preprint claiming a positive answer to this question.
EDIT: Here is a picture of the reflection-free aperiodic monotile:
More visualizations and other data are ...
- 109k
18
votes
Can the sphere be partitioned into small congruent cells?
Here is a partial result that says that if the sphere can be partitioned into small congruent polygons, then the polygons must be "thin" in some sense. Specifically:
Theorem Suppose a ...
- 109k
12
votes
Tiling the plane with pairwise non-congruent rational triangles
Yes, it is possible; in fact, we can do it entirely with $5-12-13$ right triangles at different scales.
First, note that we can three triangles at scales in the ratio $5:12:13$ to form a $5\times 12$ ...
- 2,298
10
votes
Decidability of completing Penrose tilings
Apparently it is decidable, as proved in theorem 27 here: https://people.maths.ox.ac.uk/ritter/masterclasses/ritter-lectures-on-penrose-tilings.pdf
- 449
9
votes
Tiling the plane with pairwise non-congruent rational triangles
This is overkill for your question, but in Carl Pomerance's paper, On a tiling problem of R. B. Eggleton (Discrete Mathematics 18 (1977), 63–70), he shows that the plane can be tiled using precisely ...
- 78.7k
9
votes
Tiling the plane with pairwise non-congruent rational triangles
The diagram below shows an alternative approach, using nested Pythagorean triangles whose right angles are all at the origin:
The nested Pythagorean triangles shown here are 5-12-13, 12-16-20 (3-4-5 ...
- 271
6
votes
How can one construct a four-coloring of a tiling of the plane with Smith, Myers, Kaplan, and Goodman-Strauss's aperiodic monotile?
The bijection between non-reflected tiles and a hexagonal tiling is the way to go. To construct it, we can associated to the hat tile a corresponding hexagonal tile, so that a tiling of hats gives a ...
- 61
6
votes
Can the sphere be partitioned into small congruent cells?
One possibility for a 120-way division with (slightly) smaller- diameter tiles uses a snub dodecahedron.
Start by inscribing the snub dodecahedron in the sphere and radially projecting the edges of ...
- 1,603
5
votes
Accepted
Does $\mathbb{Z}\times\mathbb{Z}$ have an aperiodic monotile?
Yes. A $2$-by-$2$ square $\{0,1\}^2$ can tile $\mathbb{Z}^2$ with just one period. So $\{0,2\}^2$ can tile $2\mathbb{Z}^2 \leq \mathbb{Z}^2$ with just one period. Break other periods in the other ...
- 6,337
4
votes
Accepted
Is there an L-system for aperiodic tilings of the plane with the "hat" monotile?
(a second answer because this one is an answer)
So, I misled myself staring at the H8 in Smith et al. The way to solve this is to look at the F-supertile. That tile has 5 edges, and 4 of them are F-...
- 196
4
votes
Is there an L-system for aperiodic tilings of the plane with the "hat" monotile?
I don't know the answer to the second part of your question (yet), about the self-avoiding fylfot fractal, but here's an L-system generating outlines of patches of monotiles, implied by the H7/H8 ...
- 196
3
votes
Tiling the plane with quadrilaterals that are mutually non-congruent and affine equivalent
I think the answer is trivially yes.
Tile the plane with squares of side 1.
Then split each square into two rectangles of sides $1, \alpha$ and $1, 1-\alpha$, of course with pairwise different $\alpha$...
- 1,990
2
votes
Examples of games developed purposely to analyze players' strategies for mathematics research
This meets your criterion of a "game developed purposely to analyze players' strategies for mathematics research," if I am allowed to consider a branch of mathematics, optimal control theory,...
- 178k
2
votes
To place copies of a planar convex region such that number of 'contacts' among them is maximized
The left shape below has $3$ contacts (circled)
"between pairs of units" and hull area $> 3$, while
the right shape has $2$ contacts and area $3$.
So minimizing the hull area does not ...
- 149k
2
votes
Accepted
Decidability of (restricted) periodicity of Wang tilings
This is undecidable. I refer to the proof of the periodic tiling problem which is Theorem 5.7 in the lecture notes of Jarkko Kari
https://users.utu.fi/jkari/wp-content/uploads/sites/1251/2021/10/part2....
- 6,337
2
votes
Tiling a rectangle with all simply connected polyominoes of fixed size
I believe $n = 17$ is also impossible for a similar reason as $n \geq 18$. According to a computer search, there are $219$ hole-less 17-ominoes that create a $4\times 3$ rectangular cavity, but only $...
2
votes
Tiling with ten-fold symmetry and (unoriented) Penrose tiles?
There are tiles with this star in the center. It also has inflation as you see in the image:
These are not Penrose tilings since they do not follow the same rules in its creation.
1
vote
How can one construct a four-coloring of a tiling of the plane with Smith, Myers, Kaplan, and Goodman-Strauss's aperiodic monotile?
Here are some computable colorings discovered by Simon Tatham. The algorithm is described at Appendix: four-colourings of the Hats and Spectre tilings.
Hat
Spectre
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1
vote
Does any set of dominoes tile some common figure?
What follows does not answer (the original question or any other question), just some thoughts.
Let $\mathcal C_n$ be the set of all dominoes contained in a $n$-square (i.e. $n$ by $n$ square...there ...
- 1,345
1
vote
What does the extension theorem for tilings state?
I first read about the extension theorem for tilings in a simpler form: if a finite protoset can tile an arbitrarily large disk, then it can the whole plane. My informal way of proving it is the ...
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