Say that we have a polyomino $P$ which tiles the plane. In may cases, it can do so by forming a two-tile "patch" which tiles the plane. For instance, with the T pentomino:

enter image description here

Is there always a way to take the disjoint union of two congruent copies of $P$ to form a polyomino $Q$ which also tiles the plane?

This feels intuitively false to me - it doesn't feel like there should be any reason for it to hold, even in the case of periodic tilings (once the period is large enough, anyway). However, it holds for every polyomino on up to $9$ cells, so I've struggled to locate a counterexample.

Can anyone provide a counterexample to this claim (or, if my intuition is completely off, a proof)?

Note that one obstacle is the Conway criterion; any tile which satisfies it will generate such a $Q$.

This should be a much easier-to-disprove question than this one, which would require the discovery of a connected aperiodic prototile at a minimum.


Here is a counter example:

enter image description here

It does not take long to see this tiling is unique. (The red dots mark places where it is easy to analyze tiling possibilities.)

A periodic patch is 6 tiles, by looking at pairs of these, the remaining patch cannot be dissected into two shapes congruent to the pair, so no pair can tile the plane as a unit.

If you want to know how I arrived at this guess: I thought that a counterexample must tile the plane in a slightly complicated way, so I picked a 3-anisohedral tile and checked if it worked, and so it did. I did not thoroughly check other examples, but a few quick checks makes me think $k$-anisohedral tiles may be fertile ground for counter-examples. Here is a site that gives the sets of $k$-anisohedral polyomines.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.