# Building "oppositely bitten rectangles" with consecutive squares

We only consider rectangles with vertical and horizontal sides. Let us call a region $$R$$ in the plane an "oppositely bitten rectangle" if it is obtained from a rectangle by removing two smaller rectangles from two opposite corners. (The removed rectangles should be small enough, so that the remaining region is connected. We allow the possibility that one or both of the smaller removed rectangles be empty.) Here is an example of an oppositely bitten rectangle:

An oppositely bitten rectangle will be called $$n$$-squared (where $$n$$ is a positive integer) if it could be represented as the union of $$n$$ many non-overlapping squares with side lengths $$1,2...,n$$ respectively. Here is a picture of several $$n$$-squared, oppositely bitten rectangles for $$1\le n\le 9$$.

$n$-squared, oppositely bitten rectangles, $$1\le n\le 9$$" />
(Picture made with the help of Pattern designer for craft projects.)

Edit. I found a $$10$$-squared, oppositely bitten rectangle, here is a picture. (As seen, these are two slightly different versions, depending on where you choose to put the square with side length $$5$$.)

$10$-squared, oppositely bitten rectangle" />

Question 1. (a) Is there an $$11$$-squared, oppositely bitten rectangle? (b) For which $$n\ge11$$ is there an $$n$$-squared, oppositely bitten rectangle?

I got interested in this in relation to the following question by MandelBroccoli:
For which $$n$$ is it possible to find a region $$R$$ made of non-overlapping squares of side length $$1,2,...,n$$ which tiles the plane?
(See Tiling the plane with consecutive squares for more details.)

The rest of this question is a counterpart of that one.

The answer to MandelBroccoli's question is yes, for $$1\le n\le 9$$ (and I do not know the answer for $$n\ge10$$). Interestingly, for each $$n$$ with $$1\le n\le 9$$ one could find an $$n$$-squared, oppositely bitten rectangle $$R$$, which could be used (together with its image $$^-R$$ under $$180^o$$ rotation) to tile the plane. Apparently not every oppositely bitten rectangle could tile the plane (I need to look at that), but here are some cases when this works.

Trivially bitten rectangle, that is, when at most one corner is bitten. If $$R$$ is a trivially bitten rectangle, and $$^-R$$ is a copy that we obtain when we rotate $$R$$ by $$180^o$$, then we could put $$R$$ and $$^-R$$ together in an non-overlapping manner to obtain a symmetrically bitten rectangle (i.e. an oppositely bitten rectangle that is invariant under $$180^o$$ rotation with center the midpoint of two opposite non-bitten corners). Symmetrically bitten rectangles were introduced and used by Steven Stadnicki to resolve the $$n=3,4,5$$ cases of MandelBroccoli's question. Each symmetrically bitten rectangle tiles the plane (in a regular manner, by translations). Here is a picture for the $$n=3$$ case (that is, starting with a $$3$$-squared, trivially bitten rectangle).

For an oppositely bitten rectangle let us introduce the following terminology - remaining length, removed length, and opposite remaining length, as illustrated by the following picture. (Of course, there are in general four choices of the remaining length and corresponding removed length, and opposite remaining length that go with it.)

An evenly bitten rectangle is an oppositely bitten rectangle for which there is a choice of a remaining length and a corresponding removed length that are equal. Any evenly bitten rectangle $$R$$ could be put together with $$^-R$$ to form a symmetrically bitten rectangle, which could be used to tile the plane. Here is an illustration, starting with a $$6$$-squared, evenly bitten rectangle.

$6$-squared, evenly bitten rectangle" />

We introduce one more variation of an oppositely bitten rectangle, which we call a nicely bitten rectangle. This is the case (by definition) when a remaining length equals the sum of the corresponding removed length and opposite remaining length. In this case we put $$R$$ and $$^-R$$ together, and obtain a centrally symmetric shape which in general is not an oppositely bitten rectangle, but nevertheless could be used to tile the plane (in a regular manner, by translation). Here is an illustration, starting with a $$9$$-squared, nicely bitten rectangle.

$9$-squared, nicely bitten rectangle" />

So, if $$R$$ is either a trivially bitten rectangle, or an evenly bitten rectangle, or a nicely bitten rectangle, then $$R$$ (together with $$^-R$$) could be used to tile the plane. (I do not know if there are other types of oppositely bitten rectangles $$R$$ that together with $$^-R$$ could tile the plane, but I will think of that some other time.)

The example of a $$10$$-squared, oppositely bitten rectangle a picture of which is posted above (either of the two slightly different versions) is not a nicely bitten rectangle (nor is it evenly bitten, nor a trivially bitten rectangle). I do not seem to be able to use it to tile the plane.

Question 2. (a) Is there a $$10$$-squared, nicely bitten rectangle (or an evenly bitten, or perhaps a trivially bitten rectangle)? (b) For which $$n\ge10$$ is there an $$n$$-squared, nicely bitten rectangle (or an evenly bitten, or a trivially bitten rectangle)?

Here is an unsuccessful attempt at an $$11$$-squared, oppositely bitten rectangle. The result instead is what one might call an adjacently bitten rectangle. This $$R$$ (taken with its copy $$^-R$$ rotated $$180^o$$) does not seem to tile the plane.

$11$-squared, adjacently bitten rectangle" />

Here is another $$10$$-squared, oppositely bitten rectangle which is not nicely bitten (nor evenly bitten, nor trivially bitten). Moving the 10-square, one obtains an adjacently bitten rectangle.

$10$-squared, oppositely bitten rectangle" />

• As a heads-up, if you haven't heard of it you might want to look into the Conway Criterion, which I think subsumes the criteria for tiling with bitten rectangles... Sep 3, 2020 at 15:44
• @StevenStadnicki I had not heard of Conway Criterion and I am grateful for the reference. Certainly a "nicely" bitten rectangle satisfies this criterion. Oppositely bitten rectangles is just one of many possible shapes to tile the plane and I was curious about it. It seems it did help in cases $n\le9$, on the other hand it seems to become too stringent a requirement for larger $n$ (when perhaps tilings exist but not by oppositely bitten rectangles). I wouldn't rule out that as $n$ becomes much bigger,say $n>800$, then it might be easier to find $n$-squared, oppositely bitten rectangles again. Sep 4, 2020 at 3:48