Double Parity (From the Pillowbook VII)

Here are the 36 pillowminoes introduced last time, arranged by their imbalance, i.e. according to how many more convex than concave edges they have. Isn’t that a pretty bell curve?

Pillowminoes weight

This time we will focus on the pillowminoes near the border of existence, namely the six ones that have all but one edge either bulging in or out. They have an imbalance of +4 or -4. Gathered and recolored, here are the marginal pillowminoes:


Let’s tile some curvy shapes with these. A curvy rectangle has odd dimensions, so cannot be entirely tiled by pillowminoes. If we decide to leave a round hole in order to fix that, the entire curvy shape will be balanced. This means that we will need the same number of brownish pillowminoes (with imbalance +4) as bluish ones (with imbalance -4). In particular, we will need an even number of these pillowminoes, so the total area of our shape needs to be divisible by 4. That’s our double parity argument.

The simplest example is that of a 5×5 square with a center hole (it’s easy to see that skinny rectangles with one edge of length 3 are not tilable with marginal pillowminoes).


The example to the left is the only one I could find, up to the obvious symmetries. To the right you see how one can inflate it to make frames, proving:

Theorem: If you can tile an axb holy rectangle with marginal pillows, then you can also tile a holy (a+4)x(b+4) rectangle with marginal pillows.

We have seen this trick before, talking about ragged rectangles.

The next interesting case are 7×7 squares. Here is one example that also teaches us another trick:

Theorem: If you can tile an axb holy rectangle with marginal pillows, then you can also tile a holy ax(b+4) rectangle with marginal pillows.


This second trick decenters the holes, however.

Finally, two examples that employ all six different marginals. First a 5×7 rectangle with center hole, then another 7×7 square that uses four marginals of each kind, nice and symmetrically.



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