# Similar Polyomino Constructions

As you most likely already know, a polyomino is a polygon that you get by joining unit squares together in such a way that each edge of a unit square touching another unit square touches exactly one other unit square. Here are some pictures of polyominoes: Those are all hexominoes. And before you ask, yes, that picture is from Wikipedia.

The concept that I am interested in is the construction of similar polyominoes. One would say that two polyominoes are similar if one could get from one to the other by replacing all unit squares in one with chunks of $4$ unit squares arranged in a $2$ by $2$ square. Here are two similar polyominoes: Let us call a polyomino self-similar if it can be assembled with other copies of itself to form a larger similar copy of itself. My example was self-similar, and here is the construction to prove it: My question is this: can anybody find a set of $3$ or more distinct polyominoes so that, for each polyomino in the set, a polyomino similar to it can be constructed using the other polyominoes in the set?

P.S. That last criterion is not essential - if you like, you could find a set in which, for each polyomino, a similar one can be constructed using all polyominoes in the set, including itself.

P.P.S. I have no idea if it's possible. If you suspect it isn't and you would like to shatter my beautiful problem by disproving the existence of such a set, go for it.

• I think you need to clarify your question. My set can be 3 squares of different size, and since all of them are similar, your criterion is satisfied. – greenturtle3141 Jun 30 '17 at 20:43
• Can you use multiple copies of each polyomino in the construction? – Deusovi Jun 30 '17 at 20:44
• @Deusovi No, you can't. – Franklin Pezzuti Dyer Jun 30 '17 at 21:30
• @greenturtle3141 No, you can't do that, because you can't assemble two or three distinct squares to get another square – Franklin Pezzuti Dyer Jun 30 '17 at 21:30
• It's not clear to me whether the similar polyomino must use only one copy each of the other polyominos. If so then the set must contain at least five polyominos for the simple reason that a linear factor of two corresponds to an areal factor of four. – Peter Taylor Jun 30 '17 at 22:28