# Are there polyominoes that can't tile the plane, but scaled copies can?

I'm wondering where there is a finite set $$\mathcal{T}$$ of polyominoes that are pairwise similar that can tile the plane, but a single element from the set cannot. (All orientations are allowed.)

To show what I mean, here is a tiling by two similar T-tetrominoes. This example is not interesting because T-tetrominoes of the same size already tile the plane.

The reason polyominoes cannot tile the plane is usually because of reasons that seem unlikely that the inclusion of scaled copies could solve, but showing this is the case generally seems difficult.

There are tilings of rectangles that are not possible with a single size, but can be done with multiple sizes, as this example shows. (This is also not an example of what I am looking for, since a single piece can in fact tile the plane).

Here are the small polyominoes that don't tile the plane; each of these (together with scaled copies), is a candidate set, although the ones I tried did not seem very promising.

• I am confused ... doesn't your last example show what you want to show? Oh wait, the single one is a tiler by itself .... – Bram28 Dec 7 '19 at 19:25
• @Jazzowner Sorry, I wasn't clear. I wasn't looking at all the little ones at the end, but the pair of little and big above it. But now that you mention that first one of all the little ones: wasn't the OP looking for a tiling of the plane? ... now I am even more confused what the OP is looking for ... – Bram28 Dec 7 '19 at 19:31
• I tried to clarify the question a bit. – Herman Tulleken Dec 7 '19 at 19:35
• @Jazzowner Ah, OK, I think I get it now. So that was an example of where the one shape by itself cannot tile a rectangle, but when you add a scaled copy, it can. So, likewise, the OP is looking for a one shape that cannot tile the plane by itself, but when you add a scaled copy, you can. And that same example doesn't do it, since the small one can tile the place all by itself. – Bram28 Dec 7 '19 at 19:37
• Koch snowflakes of two sizes tesselate the plane, but one size isn't enough! These are not polyominoes, though. – Ivan Neretin Dec 24 '19 at 14:51

We can achieve this using two polyominoes with one having double the dimensions of the other and the second copy rotated $$90^\circ$$ and reflected.

The basic element showing the two polyominoes is below. Its bounding box is $$165\times98$$ and the key to the tiling is that the longer vertical edges are both $$82$$ units long while the shorter vertical edges are both $$16$$ units long. It's clear that one polyomino can't tile the plane on its own; the second polyomino is needed for them to hook together.

And here's the tiling (click to enlarge):

The general approach is to hook two polyominoes together in the pattern below. Imagine this initially as a series of rectangles from bottom-left to top-right: $$2\times1$$ at bottom-left, then $$2\times4$$ just above that, then $$8\times4$$ and finally $$8\times16$$ in the large rectangle at top-right, with the other dimensions being fully determined from this.

A Python program was used to vary the dimensions of the smallest rectangle and redraw the resulting polyominoes. The data from the program was used to find the correct size of rectangle to generate the required matching edges. In the final polyomino the small rectangle is $$2\times16$$.

• Just in case the image of the full tiling doesn't enlarge, here's the direct Imgur link: i.stack.imgur.com/nU1yr.png – nickgard Dec 10 '19 at 16:39
• Truly amazing. I am struggling to follow your explanation. Starting with "both long edges"... which two are you referring to? (A labeled image would go a long way to clear things up.) – Herman Tulleken Dec 10 '19 at 17:26
• I figured it out now, but a diagram with edge lengths would still be helpful. (If you want I can add one.) – Herman Tulleken Dec 10 '19 at 18:13
• @HermanTulleken I've added a scale to the first image. I agree it helps the explanation. – nickgard Dec 10 '19 at 18:38
• Truly amazing! I think you deserve a bit more rep than that. – URL Jan 9 at 1:52

With an infinite set of tiles you can get the plane less one point with concentric rings. If you use four $$1 \times 2$$ rectangles you can make a hollow square that is $$3 \times 3$$ on the outside and $$1 \times 1$$ on the inside. Put notches on the rectangles to enforce this arrangement.

• Is the idea that the tiles that make the circles get bigger and bigger? (I specifically ruled out this case; only a finite number of versions of the tile is allowed). – Herman Tulleken Dec 8 '19 at 15:42
• Yes, that was the idea. I'll delete this if you want – Ross Millikan Dec 8 '19 at 16:19
• Well, the answer did get me thinking about notches, so maybe you can just add a line that says that it is an infinite tile set so does not answer the original question. (Some thoughts after playing with your example: It seems when there is a notch between a small and big tile, we need an infinite set. If notches are only between tiles of the same size, then the only way small and big tiles can touch is along smooth edges. These edges usually* make it so that the forced configurations that employ only either small or big tile can tile the plane. (Not sure what exactly *usually is here.) – Herman Tulleken Dec 8 '19 at 16:27
• I was prompted to think about notches by your original example that works in a rectangle. I think you are right that notches on scaled copies will force an infinite set and one that is infinite in both directions like the integers or my example. – Ross Millikan Dec 8 '19 at 16:35