A new simple family of non-periodic tilings with square tiles

(a) Conditions (1) and (2) for $T^{\prime}$ follow from conditions 1 and 2 for $T$.

Condition (3): every crossing in $T^{\prime}$ either coincides with the corresponding crossing in $T$ and hence is of the form $C_{8}$, or is the crossing in the center of a tile in $T^{\prime}$ and hence is of the form $C_{4}$.

Condition (4) is verified by hand.

(b) Since in a correct tiling of the plane by triangles there can only be a crossing $C_{4}$ at the vertex of any right angle, its tiles can be grouped into quadruples, each of which forms a square tile from our set. Obviously, this partition is unique.

Since the local rules for tilings by squares essentially coincide with the local rules for tilings by triangles, the tiling $T$ is correct. ∎

This is proved by induction. The induction base is trivial, since any tile constitutes a correct tiling.

For the inductive step, we have to show that the decomposition of a correct tiling $T$ is again correct. But there is a problem here. This is true for tilings of the plane, but not for tilings of its parts. For example, a tiling consisting of two triangles with a common leg in Fig. 10,

is correct (the third condition is satisfied because there are no internal vertices), but its decomposition is not. To make the inductive step, it is necessary to impose a restriction on the crossings at the boundary points. To prove the lemma, it will suffice to require that the crossings with the center at any vertex of the boundary, except for the extreme ones, are halves of the legal crossings, that is, they have one of the two forms in Fig. 11.

Now we can make the inductive step. Assume that $T$ is a supertile and is correct. We have to show that so is its decomposition $\sigma T$. It is obvious that the first condition (tiles border side by side and colors and orientations of shared sides match) is inherited during decomposition.

Let us verify the second condition (adjacent triangles have different colors) for $\sigma T$. Let $F$ be a tile in $\sigma T$ and let $G$ be its sibling. W.l.o.g. assume that $F$ is a right, and hence green, tile (Fig. 12).

We need to prove that all neighbors of $F$ are red.

The tile $G$ is red by definition of the substitution. Besides $G$, the tile $F$ can have other neighbors. Namely, if the hypotenuse of $I$ does not lie on the edge of $T$, then the tiling $T$ contains the tile $E$ bordering $I$ along the hypotenuse. As a result of cutting $E$, two tiles will appear, of which it is the left, and therefore red, tile that borders $F$ along the leg.

Moreover, if the leg of the tile $I$ does not lie on the boundary of the supertile, then the tiling $T$ contains the tile $H$, which is the reflection of the tile $I$ with respect to this leg. Indeed, otherwise the crossing centered at the vertex $A$ would be illegal (right and acute angles cannot converge at a legal crossing). When the tile $H$ is decomposed, it is the left, and therefore red, tile that is adjacent to $F$.

Let us check the third condition for the tiling $\sigma T$ (that all crossings are legal). To do this, we need to check that during decomposition, a legal crossing is converted into a legal one, and the same holds for half crossings in Fig. 11. This is done by a direct verification. In addition, we need to check that all the crossings at the new vertices are legal as well.

New vertices are the centers of hypotenuses of tiles from $T$. The hypotenuse itself becomes the axis of the new crossing, and the new crossing is $C_{4}$ by the definition of the substitution.

Finally, we check the fourth condition. The crossings of the form $C_{4}$ are only the crossings at the centers of the hypotenuses of the tiles in $T$, and by the definition of substitution, all the left triangles in them are red and the right ones are green.

The third and fourth conditions for crossings centered on the boundary are verified in a similar way. ∎

Consider any tile of a given correct tiling $T$ and the vertex of its right angle. The crossing at this vertex is $C_{4}$. At this crossing, each tile has a sibling. Let us combine them into one, erasing the common leg, and leave all the colors and orientations as they were. The hypotenuse of the resulting triangle is formed by connecting two arrows on the axis of the crossing. Since they are equally colored and oriented, the hypotenuse will be correctly oriented and correctly colored. The resulting tiling is the composition of the original one. Let us prove that it is correct.

We are given that $T$ is correct and need to prove that $\sigma^{-1}T$ is correct. First we verify condition (1), that is, prove that $\sigma^{-1}T$ is side-to-side and shared sides have matching colors and orientations. This is obvious for the legs of the tiles from $\sigma^{-1}T$, since they are the hypotenuses of the tiles from $T$, and by the condition (1) for $T$ the hypotenuses are adjacent to the hypotenuses of the same color and orientation.

Now let us prove the same for the hypotenuses of tiles from $\sigma^{-1}T$. Consider a tile $F$ from $\sigma^{-1}T$ (Fig. 13(a)). W.l.o.g. assume that it is red. We need to show that the tile $G$ shown in Fig. 13(a) belongs to $\sigma^{-1}T$. To this end, consider the midpoint of the hypotenuse, denoted by the letter $A$. The crossing centered on $A$ in the tiling $T$ can only be $C_{4}$, so there are two tiles in $T$ under the hypotenuse $A$, as shown in Fig. 13(b). These two tiles, when composed, give the sought tile $G$.

As a bi-product of this argument, we see that the colors of the tiles $F$ and $G$ are different because the colors of the vertical arrows at the crossing $C_{4}$ are different. Therefore, condition (2) is satisfied for triangles $\sigma^{-1}T$ with a common hypotenuse. Let us verify condition (2) for tiles that share a leg. Let the tiles $F$ and $G$ have a common leg (Fig. 13(c)). Then their colors are equal to the colors of two arrows in the tiling $T$ (Fig. 13(d)), connecting the point $A$, the common vertex of right angles of $F$ and $G$, with the centers of their hypotenuses. These two arrows form a right angle and at their common origin (point $A$) there can only be the crossing $C_{8}$ in the tiling $T$. By a routine check, we can see that any two orthogonal arrows with a common origin at the center of the crossing $C_{8}$ have different colors.

Let us verify condition (3). In the course of composition, the crossings $C_{4}$ disappear. We need to show that crossings of the form $C_{8}$ are transformed into themselves or into $C_{4}$. In all crossings $C_{8}$ the sides of the tiles adjacent to the center of the crossing alternate — hypotenuse, leg, hypotenuse, and so on. And the colors of the tiles also alternate. Therefore, 4 options arise, depending on whether the alternation of colors begins with red or green, and the alternation of sides — from the leg or hypotenuse. These four options are shown in Fig. 14.

Let us first prove that the first option is impossible. Indeed, the crossings at the lower and upper blue points cannot be legal. Indeed, the axes of these crossings must be directed vertically, since otherwise an arrow would go from its center in the direction orthogonal to the axis. For the upper crossing, the axis should be directed downwards, and for the lower one, upwards. But then the colors of both triangles with a vertex at the blue points should be opposite to the existing colors.

In the next two crossings, the siblings to all tiles are not adjacent to the crossing. For this reason, when composed, the crossings do not change. Finally, in the fourth case, the sibling for each tile is also adjacent to the center. When decomposed, the eight tiles produce four tiles, and we get the crossing $C_{4}$. In this case, condition (4) is satisfied for the resulting crossing $C_{4}$. Indeed, in any $C_{8}$ crossing, the red arrow goes to the left of the arrow entering its center, and the green arrow goes to the right. Hence a red triangle is to the left of the arrow entering the center of the obtained $C_{4}$ crossing, and a green one to the right. ∎

Let $T$ be a correct tiling of the plane by triangles and $\sigma^{-1}T$ be its composition. Assume that $a\neq 0$ is its period, that is $T=T+a$. Then the vector $a/\sqrt{2}$ is the period of $\sigma^{-1}T$. Indeed,

It is proved similarly that $a/2$ is the period of the double composition of $T$. Repeating this argument many times, we get a tiling whose period is much less than the size of the tiles, which is impossible. ∎

Consider any fragment $P$ of a correct tiling $T$ of the plane. We have to show that $P$ occurs in a supertile. To this end add in $P$ a finite number of tiles from $T$ so that $P$ becomes an inner part of the resulting fragment $Q$ of $T$.

By Theorem 2 for any $k$ we can compose the given tiling $k$ times and the resulting tiling $\sigma^{-k}T$ is correct. Call a crown in $\sigma^{-k}T$ centered at a vertex $A$ of a tile in $\sigma^{-k}T$ the set of all tiles from $\sigma^{-k}T$ that include $A$. Since $\sigma^{-k}T$ is a correct tiling, all its crowns have a form shown on Fig. 15 (see the proof of Th. 2).

Consider the sets of the form $\sigma^{k}C$ where $C$ is a crown within the tiling $\sigma^{-k}T$. As $k$ increases, these sets increase as well. If $k$ is large enough, then the set $Q$ is covered by a single such set, say by $\sigma^{k}C$, that is, $Q\subset\sigma^{k}C$. As $\sigma^{-k}T$ is a correct tiling, it can have only crowns shown on Fig. 15.

Let us show that all crowns from Fig. 15 appear in supertiles provided we ignore colors and orientations of its outer sides (shown in black color on Fig. 15). For crowns of the form (a) it can be verified by hand: all the four such crowns occur within supertiles $S_{5}$ shown on Fig. 16.

Again it is easy to verify by hand that the substitution transforms the crowns as follows: (a)$\to$(b)$\to$(c)$\to$(d)$\to$(c). Hence all the four crowns (b) occur within supertiles $S_{6}$, the crowns (c) within supertiles $S_{7}$, and the crowns (d) within supertile $S_{8}$.

It follows that the crown $C$, whose $k$-fold decomposition covers $Q$, appears in a supertile, say in $S_{n}$. Therefore the tiling $\sigma^{k}C$ appears in the supertile $S_{n+k}$. Hence the patch $Q$ appears in that supertile provided we ignore colors and orientations of its outer sides. Since $P\subset Q$ and no side of the patch $P$ is an outer side of $Q$, we are done. ∎