How to prove that the Greek cross tiles the plane?

Question

I hope to better understand how to prove that a given tile can indeed tile the plane. I can often visualize the tiling and draw by hand larger and larger regions that I can tile, but I'm unsure how to argue more rigorously that a given tiling actually works. This question is based off the simplest example that I'm confident can tile the plane but that I'm not sure how to prove.

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The Greek cross is a simple tile made of five squares.

I can draw patches of the plane using these tiles.

For example, using four of these tiles:

Using nine of these tiles:

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I can keep putting down Greek crosses in the spiral patterns above, and I believe that the Greek cross can tile the plane. How can I rigorously prove this?

I'm happy to view the plane as already being tiled by the faces of the square lattice, and a putative tiling as coloring such faces according to the shape of the Greek cross. I'm imagining attempting to give an explicit mapping of which face of the square lattice is part of which Greek cross, but I'm wondering if there might be a less explicit way.

Answer 1

As pointed out by Jaap Scherphuis in the comments, the Greek cross can be viewed as a deformation of the square:

Note that opposite sides are deformed identically. As Jaap noted, any parallelogram that has opposite sides deformed identically can tile the plane, so the Greek cross can tile the plane.

Below I show a four-by-four patch that highlights the undeformed squares underlying the pattern.

Answer 2

A proof without words (adding words because of word limit):

Answer 3

Idea for a formalization of the visual proof of Yuzuriha Inori using modular arithmetic.

The centers of white crosses will have coordinates of the form: $$(10m,10n),(10m+1,10n+3),(10m+2,10n+6),(10m+3,10n+9),(10m+4,10n+2),(10m+5,10n+5),(10m+6,10n+8),(10m+7,10n+1),(10m+8,10n+4),(10m+9,10n+7)$$ The centers of black crosses will have coordinates of the form: $$(10m+2,10n+1),(10m+3,10n+4),\dots$$ ...

Each square has a unique representation of the form $(10m+p,10n+q)$ with $p,q\in\{0,1,2,3,4,5,6,7,8,9\}$.

Answer 4

Let $L=\langle(2,-1),(1,2)\rangle\subset\mathbb{Z}^2$. We will show that $\mathbb{Z}^2/L$ identifies naturally with the Greek cross.

Given $(a,b)\in\mathbb{Z}^2$, we have $$(a,b)+b(2,-1)=(2b+a,0)$$and $b(2,-1)$ is the only multiple of $(2,-1)$ which added to $(a,b)$ lands in $\mathbb{Z} \times\{0\}$. Thus $$\mathbb{Z}^2/\langle(2,-1)\rangle\cong \mathbb{Z} \times\{0\}$$

Now $$(1,2)+2(2,-1)=(5,0)\in \mathbb{Z} \times \{0\}$$

Thus $$\mathbb{Z}^2/L=\left(\mathbb{Z}^2/\langle(2,-1)\rangle\right)/\langle(1,2)\rangle\\=\mathbb{Z} \times \{0\}/\langle(5,0)\rangle\\ =\{(-2,0),(-1,0),(0,0),(1,0),(2,0)\}\\ =\{(0,-1),(-1,0),(0,0),(1,0),(0,1)\}$$ as a set of distinct coset representatives of $\mathbb{Z}^2/L$.

In other words, if we translate the Greek cross $$\{(0,-1),(-1,0),(0,0),(1,0),(0,1)\}$$ by the elements of $L$, we get every point of $\mathbb{Z}^2$ exactly once.

Answer 5

I can always exhibit larger and larger regions of the plane that are tiled [...], but I feel unsure about arguing that I can tile the entire plane.

This comment directly suggests an approach for your proof: Induction.

1. We can trivially place a single cross on the plane.
2. Given a tiling of part of the plane, explain exactly how to add to it to tile a larger part of the plane.
3. Q.E.D.

Part of this will be finding the exact property of the partial tiling that's needed for step 2 to work, and showing that step 2 preserves this property. Personally, I would probably do two separate passes, first extending the pattern "to infinity" vertically (or diagonally) and then repeating the process in the other direction.

IDK if this'll be easier or harder than the coordinates approach, but I've no doubt it'll work.