Andy Latto wrote: On Fri, Jan 25, 2019 at 2:54 AM James Propp <jamespropp@gmail.com> wrote:
Here are some visualization challenges. See if you can answer the
questions
in your head without drawing anything (or if you must draw, draw in the air rather than on paper).
Picture the usual tiling of the plane by unit squares, and picture a shifted version of that tiling whose vertices all lie in the interiors of the squares of the first tiling. How many vertices of tiling #2 lie in each square cell of tiling #1?
Now try it with an infinite tiling of the plane by regular hexagons.
And now try it with an infinite tiling of the plane by equilateral triangles.
Bonus question: Is there an easy way to see, in each case, what the AVERAGE number of vertices-of-tiling-#2 per cell-of-tiling-#1 must be?
Sure. Each square has 4 vertices, and each vertex is a vertex of 4 squares. So in a large area, the total number of vertices = number of squares * 4 / 4, so one per square. With a hex tiling, each hex has 6 vertices, and each is a vertex of 3 hexes, so total number of vertices = number of hexes * 6 / 3 = twice the number of hexagons. For triangles, it's number of triangles * 3/6, so one per two triangles. For visualizing the square case, I found it easy to visualize moving the vertices in one grid to the centers of the squares in the other grid. For hexes, imagine the hexes having horizontal sides, and translate the grid vertically a small amount, and two of the vertices of each hex move into that hex. For triangles, if you translate along a direction that makes an angle of 30 degrees with the x-axis, every right-side-up triangle contains a vertex. Andy