5 ms·

You can SVG it! <polygon points="89.3,146.9 8.5,194 35.3,240.5 143.3,240.5 170.2,193.6 250.8,240.7 331.7,193.9 305,146.9 251,146.9 251.7,53.4 170.1,6.5 143

You can SVG it! <polygon points="89.3,146.9 8.5,194 35.3,240.5 143.3,240.5 170.2,193.6 250.8,240.7 331.7,193.9 305,146.9 251,146.9 251.7,53.4 170.1,6.5 143.3,53.4 89.3,53.4 "/>

l0b0 3y ago
OK, follow-up puzzle: Can you specify the exact einstein polygon with only integer coordinates?
- javajosh 3y ago
  trivial. you can multiply by any n and then set the viewport attribute to compensate.
  - l0b0 3y ago
    No, I mean mathematically. I had a skim through the original paper[1], but unfortunately it's not easy to figure out what the coordinates are. If any of them are related as a factor of an irrational number, I don't think it's possible. [1] https://arxiv.org/abs/2303.10798 https://arxiv.org/abs/2303.10798
    - jameshart 3y ago
      The points of the Einstein tile all fall on a hex grid - see the illustration here: https://gametek.substack.com/p/the-einstein-tile https://gametek.substack.com/p/the-einstein-tile So in cartesian coordinates, there are √3 components that will always crop up. But there is always an integer number of √3's in each x and y coordinate, if that helps - every vertex can be written as (a + b√3, c + d√3) for integers a, b, c and d. Oh, and that goes for every vertex of every tile in the infinite tiling, not just for a single tile.
      - basil-rash 3y ago
        > Oh, and that goes for every vertex of every tile in the infinite tiling, not just for a single tile. Could it possibly not? i.e. is there any way to tile n-gons in the (A + B*sqrt(3))[] kernel that has vertices not in that kernel? What comes to mind is the counterexample of plain unit squares tiled with strips at irrational offsets from one another. But what about for n != 4? Does every tiling have some vertex-vertex co-location?
        jameshart 3y ago
        These aren't general n-gons on vertices in that set - the Einstein tiles have additional constraints: The angles of the Einstein tile are all ±90 or ±120° - so the edges are all parallel to one of the 12 30° meridians. And the edge lengths are all integer multiples of a common unit. And as you place these tiles edge-to-edge, and match vertices, the same applies to the edges that make up each subsequent tile. So each line takes integer steps along lines that are either parallel to the axes or are at a 30° angle from an axis. If we take 2 as the shortest length of an edge (so all the edges of the shape are either 2 or 4 units long), then: - if we move from a point 2 units along an edge parallel to an axis, we add or subtract 2 from one of our cartesian coordinates. - if we move 2 units along an edge at 30° from an axis, we must add or subtract 1 from one of our coordinates, and √3 from the other (it's the hypotenuse of a 30°/60° right triangle) Those are the only moves you can make, so in an infinite tiling of Einstein tiles, the vertices will all fall on those points.
        basil-rash 3y ago
        Right, but I’m trying to generalize the claim to beyond these tiles. The claim is that for every vertex-vertex tiling of n-gons with n not all equal to 4 and vertices in some kernel (A + B*sqrt(C)), for integer ABC, all verticies in the tiling will also be in that same kernel. Which is to say there’s nothing interesting about these Einstein tilings maintaining their NMsqet(3) kernel.
      - l0b0 3y ago
        > a + b√3 I assume there are no values a and b (including irrationals) which would make that sum an integer?
        kadoban 3y ago
        b = sqrt(3) gets you back into integers, sqrt(3)^2 is 3. Any factor of that with other integers also works. b=0 also works but that one seems rather cheap.
        jameshart 3y ago
        Bad assumption. It's an integer for all b=0, for a start. Also, among the multiples of √3, as you keep picking higher and higher values of b, you'll keep finding results that are closer and closer to (but never equal to) integers. And from them, you can choose any value of a to get a result equally close to any other integer. So you can also find an infinite number of combinations of an and nonzero b that get you arbitrarily close to any integer. The formulation for the Einstein tile grid points uses integer a and b, so those are the only cases of interest ('irrational' a or b doesn't matter) In the context of the Einstein grid, go look at the tiles and it makes sense. They have sides that are integer multiples of a unit length. They have 90 degree angles in them. So you can place them up against each other so you can walk along the edges taking only integer steps and 90 degree turns - staying on the integer cartesian grid. But once the tiling forces you to take one of the 120° angles you veer off at an angle and you're going to be an irrational distance from the grid. You might then find yourself turning back onto the grid and moving integer steps again - but those will not bring you back to the gridlines. To get back to the grid you have to take exactly the same number of irrational steps in the opposite direction.
        jlokier 3y ago
        Here are some positive a and b pairs. a = 2-√3, b = 1 a = 1, b = 1/√3 a = 1, b = √3 At least one √3 always appears somewhere in a or b (if they are non-zero), so they can't both be (non-zero) rationals. In some ways the √3 behaves like the imaginary unit i in complex numbers.
        quickthrower2 3y ago
    - DerekL 3y ago
      You can't. First, here's a lemma: Given three distinct lattice points (points with integer coordinates) A,B,C in the plane, either ⦟ABC is a right angle or tan ⦟ABC is rational. Proof: Assume that ⦟ABC is not a right angle. If either line AB or BC is vertical, rotate the plane by 90°. Then at least one line is horizontal, and the other can't be vertical, because ⦟ABC is not a right angle. Let x and y be the angles AB and BC make with a horizontal line. Then ⦟ABC = |x - y|. The tangents of x and y are rational, because the points have integer coordinates. Then |tan ⦟ABC| = |(tan x - tan y) / (1 + tan x tan y)|, which is rational. The hat has angles of 120°, and tan 120° = -√3, which is irrational, so you can't draw the hat with lattice points, no matter how you scale or rotate it.
      - l0b0 3y ago
        I'm not a mathematician, so I'm afraid I can't tell whether you or all the people saying to multiply by √3 are right. The fact that nobody has posted actual coordinates on the other side of the discussion is telling, though. SVG 2's bearing commands[1] might help, but I can't seem to find anything that supports it. [1] https://svgwg.org/specs/paths/#PathDataBearingCommands https://svgwg.org/specs/paths/#PathDataBearingCommands

You can SVG it! <polygon points="89.3,146.9 8.5,194 35.3,240.5 143.3,240.5 170.2,193.6 250.8,240.7 331.7,193.9 305,146.9 251,146.9 251.7,53.4 170.1,6.5 143.3,53.4 89.3,53.4 "/>

l0b0 3y ago

OK, follow-up puzzle: Can you specify the exact einstein polygon with only integer coordinates?

javajosh 3y ago
trivial. you can multiply by any n and then set the viewport attribute to compensate.
- l0b0 3y ago
  No, I mean mathematically. I had a skim through the original paper[1], but unfortunately it's not easy to figure out what the coordinates are. If any of them are related as a factor of an irrational number, I don't think it's possible. [1] https://arxiv.org/abs/2303.10798 https://arxiv.org/abs/2303.10798
  - jameshart 3y ago
    The points of the Einstein tile all fall on a hex grid - see the illustration here: https://gametek.substack.com/p/the-einstein-tile https://gametek.substack.com/p/the-einstein-tile So in cartesian coordinates, there are √3 components that will always crop up. But there is always an integer number of √3's in each x and y coordinate, if that helps - every vertex can be written as (a + b√3, c + d√3) for integers a, b, c and d. Oh, and that goes for every vertex of every tile in the infinite tiling, not just for a single tile.
    - basil-rash 3y ago
      > Oh, and that goes for every vertex of every tile in the infinite tiling, not just for a single tile. Could it possibly not? i.e. is there any way to tile n-gons in the (A + B*sqrt(3))[] kernel that has vertices not in that kernel? What comes to mind is the counterexample of plain unit squares tiled with strips at irrational offsets from one another. But what about for n != 4? Does every tiling have some vertex-vertex co-location?
      - jameshart 3y ago
        These aren't general n-gons on vertices in that set - the Einstein tiles have additional constraints: The angles of the Einstein tile are all ±90 or ±120° - so the edges are all parallel to one of the 12 30° meridians. And the edge lengths are all integer multiples of a common unit. And as you place these tiles edge-to-edge, and match vertices, the same applies to the edges that make up each subsequent tile. So each line takes integer steps along lines that are either parallel to the axes or are at a 30° angle from an axis. If we take 2 as the shortest length of an edge (so all the edges of the shape are either 2 or 4 units long), then: - if we move from a point 2 units along an edge parallel to an axis, we add or subtract 2 from one of our cartesian coordinates. - if we move 2 units along an edge at 30° from an axis, we must add or subtract 1 from one of our coordinates, and √3 from the other (it's the hypotenuse of a 30°/60° right triangle) Those are the only moves you can make, so in an infinite tiling of Einstein tiles, the vertices will all fall on those points.
        basil-rash 3y ago
        Right, but I’m trying to generalize the claim to beyond these tiles. The claim is that for every vertex-vertex tiling of n-gons with n not all equal to 4 and vertices in some kernel (A + B*sqrt(C)), for integer ABC, all verticies in the tiling will also be in that same kernel. Which is to say there’s nothing interesting about these Einstein tilings maintaining their NMsqet(3) kernel.
    - l0b0 3y ago
      > a + b√3 I assume there are no values a and b (including irrationals) which would make that sum an integer?
      - kadoban 3y ago
        b = sqrt(3) gets you back into integers, sqrt(3)^2 is 3. Any factor of that with other integers also works. b=0 also works but that one seems rather cheap.
      - jameshart 3y ago
        Bad assumption. It's an integer for all b=0, for a start. Also, among the multiples of √3, as you keep picking higher and higher values of b, you'll keep finding results that are closer and closer to (but never equal to) integers. And from them, you can choose any value of a to get a result equally close to any other integer. So you can also find an infinite number of combinations of an and nonzero b that get you arbitrarily close to any integer. The formulation for the Einstein tile grid points uses integer a and b, so those are the only cases of interest ('irrational' a or b doesn't matter) In the context of the Einstein grid, go look at the tiles and it makes sense. They have sides that are integer multiples of a unit length. They have 90 degree angles in them. So you can place them up against each other so you can walk along the edges taking only integer steps and 90 degree turns - staying on the integer cartesian grid. But once the tiling forces you to take one of the 120° angles you veer off at an angle and you're going to be an irrational distance from the grid. You might then find yourself turning back onto the grid and moving integer steps again - but those will not bring you back to the gridlines. To get back to the grid you have to take exactly the same number of irrational steps in the opposite direction.
      - jlokier 3y ago
        Here are some positive a and b pairs. a = 2-√3, b = 1 a = 1, b = 1/√3 a = 1, b = √3 At least one √3 always appears somewhere in a or b (if they are non-zero), so they can't both be (non-zero) rationals. In some ways the √3 behaves like the imaginary unit i in complex numbers.
        quickthrower2 3y ago
  - DerekL 3y ago
    You can't. First, here's a lemma: Given three distinct lattice points (points with integer coordinates) A,B,C in the plane, either ⦟ABC is a right angle or tan ⦟ABC is rational. Proof: Assume that ⦟ABC is not a right angle. If either line AB or BC is vertical, rotate the plane by 90°. Then at least one line is horizontal, and the other can't be vertical, because ⦟ABC is not a right angle. Let x and y be the angles AB and BC make with a horizontal line. Then ⦟ABC = |x - y|. The tangents of x and y are rational, because the points have integer coordinates. Then |tan ⦟ABC| = |(tan x - tan y) / (1 + tan x tan y)|, which is rational. The hat has angles of 120°, and tan 120° = -√3, which is irrational, so you can't draw the hat with lattice points, no matter how you scale or rotate it.
    - l0b0 3y ago
      I'm not a mathematician, so I'm afraid I can't tell whether you or all the people saying to multiply by √3 are right. The fact that nobody has posted actual coordinates on the other side of the discussion is telling, though. SVG 2's bearing commands[1] might help, but I can't seem to find anything that supports it. [1] https://svgwg.org/specs/paths/#PathDataBearingCommands https://svgwg.org/specs/paths/#PathDataBearingCommands