## Tiling a rectangle: Another surprisingly tricky problem

For the discussion of math. Duh.

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Torn Apart By Dingos
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### Re: Tiling a rectangle: Another surprisingly tricky problem

Qaanol wrote:
Spoiler:
then rectangles can be made arbitrarily small in both dimensions simultaneously, and can thus tile any arbitrary rectangle.

Why should this be true? You can't tile a square with disks (no disk, no matter how small, can ever cover a corner), so why should these rectangles be able to do the job?

Talith
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### Re: Tiling a rectangle: Another surprisingly tricky problem

Let's take the rectangle 1.5*1.5. It's [imath]\mathbf{Z}[\sqrt{2}][/imath]- improper. However, an infinite tiling of proper rectangles exists. First use the tile 1.5*1, then the tile 1.5*(3-2 sqrt{2}). etc... given a distance of x that we have left to cover on the horizontal edge, we can always find some m and n in the integers so that m+n sqrt{2} < x. Hence, we can find a cauchy sequence in the set of proper rectangles who's tiling tends to an improper rectangle by taking arbitrarily close appoximations of one of the edge lengths.

 I'm a little concerned that this doesn't work because we never actually cover the edge opposite the one we fully cover at the start of the tiling, we can cover an arbitrarily high fraction of the improper rectangle in terms of area, but the set {<1.5, x> |x \in [0,1.5]} can never be covered. If we work, instead, with open rectangles, we have the equally troubling problem of tiles always having a set that isn't covered at their meeting edges.[/edit]

Torn Apart By Dingos
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### Re: Tiling a rectangle: Another surprisingly tricky problem

Alright, that (almost) works. I'd be interested in if an actual tiling can be made, but your construction works if we massage the definitions a little:

In my opinion, a tiling should be a partition of a set. So no overlapping of edges. Thus, we should ask if we can tile the "half-open" improper rectangle [0,1.5)*[0,1.5) with "half-open" proper rectangles (which contain their left and bottom sides, but not their top and right sides). Your construction does exactly this, with no gaps or overlapping.

Nitrodon
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### Re: Tiling a rectangle: Another surprisingly tricky problem

Spoiler:
A finite tiling of a [imath](2\sqrt{2}-1) \times (2\sqrt{2}-1)[/imath] square with "semi-proper" rectangles:

Talith
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### Re: Tiling a rectangle: Another surprisingly tricky problem

But the large rectangle that you've tiled is also semi-proper (in the the updated definition that aleph_one gave). Just to be clear, aleph_one's definition is that an m-by-n rectangle is semi-proper if either m, or n (or both) is of the form a+b*sqrt{2} where a and b are not both zero and are in the integers (ie, in [imath]\mathbf{Z}[\sqrt{2}][/imath]).

Qaanol
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### Re: Tiling a rectangle: Another surprisingly tricky problem

Talith wrote:Let's take the rectangle 1.5*1.5. It's [imath]\mathbf{Z}[\sqrt{2}][/imath]- improper. However, an infinite tiling of proper rectangles exists. First use the tile 1.5*1, then the tile 1.5*(3-2 sqrt{2}). etc... given a distance of x that we have left to cover on the horizontal edge, we can always find some m and n in the integers so that m+n sqrt{2} < x. Hence, we can find a cauchy sequence in the set of proper rectangles who's tiling tends to an improper rectangle by taking arbitrarily close appoximations of one of the edge lengths.

 I'm a little concerned that this doesn't work because we never actually cover the edge opposite the one we fully cover at the start of the tiling, we can cover an arbitrarily high fraction of the improper rectangle in terms of area, but the set {<1.5, x> |x \in [0,1.5]} can never be covered. If we work, instead, with open rectangles, we have the equally troubling problem of tiles always having a set that isn't covered at their meeting edges.[/edit]

Start by tiling around the perimeter with, say, 1×0.5 closed rectangles. Now we have a square 0.5×0.5 hole in the center to fill. Then infinitely tile that, making sure the limit of small rectangles occurs at an edge (or corner) of the 0.5×0.5 square, meaning the original rectangles already cover the limit points.
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