## 1 Introduction

### Golden right triangles and tilings.

The altitude of any right triangle cuts it into two similar triangles. Those triangles are denoted by on Fig. 1(a).

If the angles of the original right triangle are chosen appropriately,
then the ratio of the (linear) size of the initial
triangle to the size of equals
the ratio of the size of to the size of .
More specifically, the ratio of the legs
of the initial triangle should be equal to the square root of
the golden ratio .
Such a triangle is shown on Fig. 1(b). The lengths of the sides
of triangles are shown on Fig. 1(с).
We will call triangles of this shape
*golden right triangles*.^{1}^{1}1The name ‘‘golden’’ in a similar context
was used to call isosceles triangles whose all angles are integer multiples of . To avoid confusion we add the attribute ‘‘right’’.

We will use triangles
and as prototiles and their isometric images are called *tiles*.
More specifically, isometric images of are called *large tiles*,
and isometric images of are called *small tiles*.
A *tiling* is a family of pair-wise non-overlapping tiles, which
means that the interiors of the tiles are disjoint.
On Fig. 2, we can see an example of a tiling.

We denote by the
union of all tiles from and say that
*tiles* , or that
*is a tiling of* .

### Decomposition and composition of tilings.

Consider the following operation on tilings.
Given a tiling we cut every large tile by its altitude into two triangles, and all small tiles remain intact.
We obtain a tiling of the same set by tiles of smaller size.
Then we apply to the resulting tiling some fixed pre-chosen homothety with the
coefficient . That homothety will be called *the reference homothety* in the sequel.
We call the resulting tiling
*the decomposition* of the initial tiling.
Each large tile produces a large and a small tile in
the decomposed tiling and each small tile becomes a large tile.
The decomposition of the tiling is denoted by .

The inverse operation is called *composition*.
More precisely, we call a tiling a composition of a tiling
if is the decomposition of .
There are tilings that have no composition, for instance,
the tiling consisting of a single small tile.
On the other hand, every tiling has at most one composition.
This property of our substitution is called *the unique composition property*.
The composition of a tiling (if exists) is denoted by .

It may happen that the composition of a tiling again has a composition.
In this case the initial tiling is called *doubly composable*. If
a tiling can be composed any number of times, we call it *infinitely composable*.

### Supertiles and substitution tilings.

*A supertile* is a tiling, which can be obtained from
a small or a large tile by applying decomposition several times.
Since the large tile is a decomposition of a small tile,
every supertile can be obtained from a small tile by
applying decomposition some times.
The number is called then the *level*
of the supertile. (In particular, the small tile
is a supertile of level .)
The supertile of level is denoted by . Fig. 2 shows supertiles of levels .

### Substitution tilings of the plane.

###### Definition 1.

*A patch* is a finite tiling.
A tiling is called a *substitution tiling*
if for each patch there is a
supertile including . For instance, all supertiles are substitution tilings.

There exist substitution tilings of the plane. This can be deduced by compactness arguments from the existence of substitution tilings of arbitrarily large parts of the plane. However, it is easier to prove this using the following argument. There are supertiles of levels 6 and 13, , such that and is included in the interior of . Indeed, on Fig. 3 we can see a supertile of level 6.

The interior of the triangle includes a small tile (shown in green color). Applying 7 decompositions to the supertiles and we get supertiles and , of levels 6 and 13, respectively. Since , we have . In this way we can construct a tower of supertiles

where each set extends the previous set in all directions. Therefore the tiling tiles the entire plane and is a substitution tiling by construction.

Substitution tilings associated with the above decomposition scheme were introduced in [5]. It is not hard to see that every substitution tiling of the plane has a composition, which is again a substitution tiling (Lemma 1 below). Thus every substitution tiling of the plane is infinitely composable. In particular, every substitution tiling of the plane contains supertiles of all levels. (To find a supertile of level in a substitution tiling of the plane, we can compose it times and then pick any large tile in the resulting tiling. The -fold decomposition of that tile is a supertile of level and is included in the original tiling .)

### Non-periodicity of substitution tilings of the plane.

The unique composition property implies that any infinitely composable (and hence any substitution) tiling of the plane is non-periodic. This folklore result is proved as follows. Assume that an infinitely composable tiling has a non-zero period , that is,

. Then the vector

is the period of . Indeed, the decomposition of the tiling is equal to the decomposition of shifted by , that is, to . By our assumption, we have . Thus both and are compositions of . By the unique composition property we then have . Repeating the argument we can conclude that the vector is a period of the tiling . Since is infinitely composable, in this way we can construct a tiling whose period is much smaller than the linear sizes of tiles, which is impossible.### Local rules.

A patch is called a
*fragment* of a tiling , if is a subset of .
If the diameter of a patch (the maximal distance between points of its tiles) is at most
, then we call that patch a *-patch*.
In a similar way we define -fragments.
A local rule is identified by a positive real
and by a division of all -patches
into *legal* and *illegal* ones.
A tiling *satisfies* the local rule,
if it does not include illegal
-patches.
The number is called the *diameter* of the local rule.

In the paper [5], it was shown that there is no local rule such that the family of tilings of the plane satisfying that local rule coincides with the family of substitution tilings. More specifically, it was proved that for any positive there is a periodic (and hence not substitution) tiling of the plane , whose all -fragments occur in supertiles. Hence for any local rule of diameter , either a -patch from is declared illegal by that rule and hence every substitution tiling does not satisfy that rule (recall that all substitution tilings include all supertiles and hence include ), or all -patches from are declared legal and hence the tiling satisfies that rule.

### New results.

In this paper we show that the family of substitution tilings of the plane is sofic. This means that we can color our prototiles in a finite number of colors and define a local rule for colored tiles so that the following holds: Every tiling of the plane satisfying the local rule is a substitution tiling (provided we ignore colors), and, conversely, every substitution tiling of the plane can be colored so that the resulting colored tiling satisfies the local rule (Theorem 3). In particular, all tilings of the plane satisfying the local rule are non-periodic.

Moreover, our substitution scheme can be naturally generalized to colored tiles. For the resulting substitution, we show that any tiling of the plane satisfying the local rule has a composition which again satisfies the local rule (Theorem 2). Then we prove that the family of tilings of the plane satisfying the local rule coincides with the family of substitution tilings associated with that substitution.

### Previous results.

The name *substitution tilings* was introduced in [3].
(Another name for the same notion, *self-affine tilings*,
was used in [9].) In general, a substitution is defined by a
finite family of polygons , a similarity ratio
and a way to cut every polygon from the family
into finite number of parts where each part is congruent to some
polygon from the family . Then in a similar way, as
for golden right triangles, we can define the notions of a decompostition, of a supertile and of a
substitution tiling.

Several families of substitution tilings appeared in the literature.
In the paper [3], Goodman-Strauss proved that
if a family of substitution tilings is side-to-side^{2}^{2}2This means that in every tiling from the family, no vertex
of its tile is in the interior of a side of another its tile., then it is sofic.
Although the family of substitution tilings considered in this paper is not side-to-side, one can add fictitious
vertices on the sides of large triangles so that the family becomes side-to-side.
In addition to the initial tiles, we obtain two quadrilaterals and one pentagon (see Fig. 4).

However, it is not clear how those quadrilaterals and the pentagon should be cut in parts so that the resulting substitution define the same family of tilings. Therefore, Goodman-Strauss theorem can not be applied to show that our family of substitution tilings is sofic. One could also try to use not the statement but the proof of Goodman-Strauss theorem. However, in that case we would not obtain such a simple local rule as the local rule defined below.

The common scheme to use substitutions is the following. One defines a substitution and a local rule so that the following properties hold. (1) For every tiling of the plane satisfying the local rule there is the unique tiling satisfying the local rule whose decomposition equals the initial tiling (the so called ‘‘unique composition property’’), (2) the decomposition of every tiling satisfying the local rule again satisfies the local rule (hence all supertiles satisfy the local rule). As explained above, the first property guarantees that all tilings of the plane satisfying the local rule are non-periodic. The second property is used to prove that there are such tilings. Usually people are not interested in the question whether the family of tilings satisfying the local rule coincides with the family of substitution tilings related to that substitution (the second property guarantees the inclusion in one direction only: every substitution tiling satisfies the local rule).

A classical example of this scheme is the proof of non-periodicity of Robinson tilings [8]. For the version of Robinson tilings from the paper [1], the proof follows exactly the above pattern. In the tiling of [1], there are prototiles, which are squares of the same size, and every tile is cut in four smaller squares. The local rule guarantees both above properties and it is easy to show that the family of substitution tilings for that substitution coincides with the family of tilings satisfying the local rule.

A simpler example is the Ammann A2 tilings, as defined in [4]. In Ammann A2 tilings there are only 2 hexagonal tiles (up to reflections and rotations) that are similar to each other. The substitution cuts every large tile into 2 smaller tiles that are similar to the initial tiles, and every small tile remains intact. The class of substitution tilings related to that substitution coincides with the class of tilings satisfying the local rule from [4]; this non-trivial result is proved in [2].

The proof on non-periodicity of Penrose tilings by golden isosceles triangles (see [7, 4]) also follows this pattern. However for Penrose tilings it is not clear whether every tiling satisfying the local rule is a substitution tiling.

We finish this section by the following lemma mentioned above.

###### Lemma 1.

Every substitution tiling of the plane has a composition, which is again a substitution tiling.

###### Proof.

If a small tile and a large tile are located, as shown on Fig. 1(a),
we say that and are *complements* of each other.
It is
easy to prove by induction that each supertile , except for , has the following property: the complement of every small tile from is also in .

We claim that this property holds for any substitution tiling of the plane as well. Indeed, otherwise an inner point of the large leg of a small tile belongs to a tile which is different from the complement of . Hence has a fragment , that does not belong to any supertile.

Thus every substitution tiling of the plane has the composition . It remains to show that every fragment of occurs in a supertile. Indeed, its decomposition is a fragment of . Add to the patch all the tiles from that share a point with a tile from (i.e., the neighbors of ). Let denote the resulting fragment of . By assumption is a fragment of a supertile . Then is a fragment of the supertile . Indeed, is the composition of and all tiles from are composed in in the same way, as they are composed in the tiling , as includes all neighbors of tiles from . ∎

## 2 Finite local complexity of substitution tilings

Substitution tilings of the plane have finite local complexity (FLC):

###### Theorem 1 ([5]).

For every there are only finitely many -fragments of substitution tilings, up to isometry (however their number depends on ).

Essentially, the theorem claims that there are only finitely many -fragments of supertiles. One can prove this as follows. Although substitution tilings are not side-to-side, we can add fictitious vertices on the sides of large tiles so that substitution tilings become side-to-side. In addition to initial triangles we obtain two quadrilaterals and one pentagon, see Fig. 4. (It is not hard to see that we have to increase the number of tiles: it is impossible to add finitely many vertices on the side of one large triangle to achieve the side-to-side property.) For every pair of the resulting 5 tiles there are only finitely many ways to attach to so that and share a side. Similarly, there are only finitely many ways to extend any patch by adding a new tile so that the resulting patch is side-to-side. Whatever way we chose, the area of the patch increases by at least some constant. Since the area of any -patch is at most , the number of patches obtained by adding tiles side-to-side is finite.

However, this is only a sketch of proof. To complete the proof, we still need
to show that there are indeed only three possible ways how the vertices of tiles
can divide edges of other tiles in supertiles. We will prove FLC property
in another, more constructive, way.
In the proof we will introduce the notion of a *crown*, which will play an important role in the sequel.

###### Definition 2.

Let be a tiling and a point from the set tiled by .
*A semi-crown centered at * is a fragment of
consisting of all tiles from that include the point (not necessarily as a vertex).
A semi-crown centered at is *an inner semi-crown of a tiling *, if is an inner point
of the set (that is, a neighborhood of is included in ).
For example, all semi-crowns of any tiling of the plane are inner semi-crowns.
*A crown in a tiling * is any its inner semi-crown whose center is a vertex of a tile
from .

###### Proof of Theorem 1.

We start with the following

###### Lemma 2.

Only finitely many crowns occur in substitution tilings of the plane.

###### Proof.

Every crown from a substitution tiling of the plane is
a crown of a supertile. So we have to show that
supertiles have only finitely many crowns (up to isometry).
Since the supertile of level is a disjunctive union
of supertiles of levels and , all crowns of the supertile of level
whose centers do not belong to the altitude of the triangle (let us call such semi-crowns *non-central*),
occur in supertiles of lower levels.
Thus it suffices
to show that supertiles have finitely many central semi-crowns. It turns out that, starting from the supertile ,
there are no new central semi-crowns.

To prove that, we first verify that all central semi-crowns of the tiling appear also in the supertile . This can be done by an exhaustive search (the supertiles are shown on the right on Fig. 5).

As we have explained, this implies that *all* inner semi-crowns of the
supertile appear also in
or in (and since , they appear in alone).
Let us deduce from this that the same holds for all inner semi-crowns of the
supertile .

Let be an inner semi-crowns in the
supertile centered at (on Fig. 5 this
semi-crown is highlighted in green on the left).
The supertile is obtained by the
decomposition of the supertile (shown to the right of ).
Consider the semi-crown of the supertile centered at the same point ,^{3}^{3}3More precisely,
we mean here the inverse image of the point under the reference homothety used in the definition of decomposition.
call this semi-crown by (the semi-crown is also highlighted in green). The semi-crown
can be obtained from by its decomposition followed by removing
all the obtained tiles that do not include the point .
We will call this operation *the reduced decomposition of a semi-crown*.
As all inner semi-crowns of the
supertile appear also in the supertile , the semi-crown
appears also in the supertile
. Let denote the center of this occurrence of in (that occurrence
is highlighted in green and an arrow from the point is directed to ). Consider now
the semi-crown centered at in the decomposition of the supertile (it is shown
on the left of ). This semi-crown is highlighted in green on the upper left supertile and
an arrow from the point is directed to . Since this semi-crown is obtained by the reduced
decomposition of the semi-crown , it coincides with . Hence
appears in the supertile (and hence in ).
∎

Now we have to generalise Lemma 2 to patches of bounded diameter. This can be done as follows.

###### Lemma 3.

If is large enough compared to , then for any substitution tiling of the plane for any its -fragment there is a crown in the tiling such that includes the entire patch .

###### Proof.

Consider supertiles of the form for , call them *-supertiles*.
These supertiles partition and hence is covered by a finite
number of -supertiles.
More specifically, is covered by those -supertiles which
intersect
. For small , for instance for ,
the respective tiles might not belong to a single crown of
the tiling . However, the sizes of -supertiles increase as increases,
and for a large enough
the respective tiles belong to a single crown of
the tiling .
Indeed, cover the set
by a disc of radius (centered at any point from ).
It suffices to show that if is large enough, then there is a crown
in such that covers .
In other words, covers , the inverse image of
under the th power of the reference homothety .
The radius of equals ,
therefore the claim follows from the following geometrical observation, whose proof
is moved to Appendix.

###### Lemma 4.

Let denote the minimal angle of the right golden triangle and the length of the altitude of the small right golden triangle. Let be a disc of diameter . If , then every tiling of the plain has a crown such that .

This lemma provides the relation between and we need. Assume that . Then any -fragment of the initial substitution tiling is covered by a disc of diameter and by Lemma 4 is included in for some crown from the tiling . ∎

Now we can finish the proof of the theorem. Given we first choose satisfying Lemma 3. By this lemma any -fragment of any substitution tiling is a subset of for a crown from another substitution tiling . By Lemma 2 there are only finitely many ways to choose . The number of patches in the -fold decomposition of any of those crowns depends only on . Hence there are only finitely many -fragments of substitution tilings. ∎

Сrowns that occur in substitution tilings are called *legal* in the sequel.
We will need a list of all legal crowns.
To make this list, it suffices to find
all the central crowns of supertiles .
There are no inner vertices on the altitudes of supertiles . For each of the remaining
supertiles there appears one new crown, and we obtain the following list of
legal crowns:

They are denoted by . The arrows indicate the reduced decomposition. The reduced decomposition of the crown equals the same crown rotated by clockwise. (This can be seen more easily on Fig. 9 on Page 9.)

## 3 Tilings by colored triangles

As we have said, there exists no local rule that defines the family of substitution tilings of the plane. However, there is such local rule provided tiles from substitution tilings are colored in a finite number of colors. In this section we explain how to do that.

We define first how we color the tiles. We first choose for every side of the tile its orientation (depicted by an arrow). Besides, every side is labeled by an integer number from 0 to 3. There is the following restriction for those labels: the hypotenuse and the small leg of the large triangle, and the large leg of the small triangle have even labels, and the remaining sides have odd labels. Tiles bearing orientation and digital labels on sides are called

*colored tiles*. Each of 2 prototiles produces colored prototiles.

### Decomposition and composition of colored tilings.

Now we generalize composition and decomposition to tilings by colored tiles. The decomposition of colored tilings consists of the following two steps: we first increment in a given tiling all digital labels by 1 modulo 4, and second we make substitution as described earlier. The newly appeared altitudes of large triangles are labeled by 0 and are oriented from the foot to the vertex. The axis of the altitude is divided into two segments, those segments keep their labels and orientations. It is not hard to verify that the requirement of evenness/oddness of labels is preserved and thus we obtain again a tiling by legally colored tiles.

On Fig. 6 we have shown a large colored tile, its decomposition, the decomposition of its decomposition and so on.

The digital labels are represented by colors: red is 0, yellow is 1, green is 2, blue is 3, orientations are shown by arrows. Long line segments of the same color represent identically oriented sides with the same digital label. This orientation is shown by an arrow at an end of the segment.

The inverse operation is called *the composition of colored tilings*. This is a partial operation
on colored tilings. If the composition of colored tilings exists, then it is unique.

*A colored supertile of level * is defined as the -fold decomposition of
a large colored tile.
Colored supertiles of levels 0,1,2,3,4
are shown on Fig. 6 and a colored supertile of level
10 is shown on Fig. 7.

A tiling of the plane or of its part (by colored tiles) is called a *substitution tiling* if all its fragments
appear in supertiles.
Our main goal is to define a local rule
such that the family of tilings of the plane obeying the local rule coincides with
the family of colored substitution tilings of the plane. To this end we need a notion of a
colored legal crown.

### Colored crowns and their decomposition.

###### Definition 3.

*A colored crown of a colored tiling centered at its vertex* is
defined similarly to the case of tilings by
non-colored tiles. However, there an important difference:
we ignore the digital labels and orientation of outer sides of the crown
(those sides of triangles that do not include the center of the crown).
The reduced decomposition of a colored crown is defined in exactly the same way
as for non-colored crowns. A colored crown is *legal*, if it appears in a colored
supertile.

An example if a legal colored crown is shown on Fig. 8.

Other examples of legal colored crowns can be seen on Fig. 7. The whole list of legal colored crowns is shown on Fig. 9.

The picture shows 7 crowns colored in 4 legal colors and also in black.
The black line segment is called
*the axis of a legal crown*. All the sides of the tiles from the crown that lie on the axis
must have matching labels and orientations. That label must have the same parity as the number of the crown
(that is, green or red for crowns and blue or yellow for the remaining crowns).
Although every side of a triangle that passes through the center of the crown
is shared by two triangles, it has only one orientation
and color, since the labels and orientations of segments of sides shared by different triangles match.

###### Lemma 5.

###### Proof.

(1) The first statement is proved by induction on the level of a supertile. Colored supertiles for have no crowns. Consider now a colored supertile for and consider its inner vertex . The supertile is defined as the decomposition of . Let us distinguish two cases.

*Case 1:* the vertex is a center of the crown
in . By definition of decomposition, all sides of that are absent in
, that is, altitudes of large triangles of ,
get 0 (red) label
and are directed from the foot to the vertex.
Hence the vertical line segment in
is oriented and labeled, as shown on Fig. 9. The horizontal line segment
(colored in black) is the hypotenuse
of the large tile from . Hence its label is even.
By definition of decomposition we first increment
that label (thus it becomes odd) and
both segments in which that hypotenuse gets divided
inherit the orientation and the label. Therefore
the requirement about the axis is fulfilled too.

Case 2: the vertex is the center of the crown in for some . We claim that is a vertex of a triangle in . Indeed, in all crowns – the center is a vertex of an acute angle of a small triangle and composition turns such vertices into vertices of large triangles. Hence the colored crown of centered at is a reduced decomposition of the colored crown of centered at the same vertex.

It is not hard to verify that the family of colored crowns from Fig. 9 is closed under reduced decomposition: see Fig. 10 where the action of reduced decomposition is shown by grey arrows.

By induction hypothesis the colored crown of centered at belongs to the list on Fig. 9. Hence the colored crown of centered at is in the list too.

(2) It suffices to prove the statement for colored crowns only. Indeed, for every the crown is obtained from by reduced decompositions. Thus, if occurs in , then occurs in . There are four crowns of the type : we have two ways to label the axis (yellow or blue) and two ways to choose orientation of the axis. The crowns with yellow axis of both orientations appear on the altitude of the supertile (Fig. 7). The crowns with blue axis appear on the altitude of the supertile on Fig. 6 (they appear also on Fig. 7, inside the supertile ). ∎

### Several remarks on legal colored crowns.

Every crown from Fig. 9 represents 4 crowns: there are two ways to choose the orientation of the axis and two ways to label it. Thus there are different legal colored crowns, up to an isometry. On Fig. 10, each of the crowns – is shown using two possible orientations of the axis. In the course of coloring, the number of legals crowns has increased by one—we had 6 legal crowns and now we have 7 ones. This is caused by the fact that in the colored supertiles crowns are colored in two different ways. It is more convenient to use different names for those different colored crowns: and .

There are one or two outgoing arrows from the center of any legal crown and
those arrows are orthogonal to the axis of the crown. All the remaining arrows are directed towards the center of the crown and form with the axis the acute angles and ,
called the *smaller* and the *larger* ones, respectively.
Let
denote the number of a legal crown. Then the digital labels of the arrows that go into or out of
the center of a legal crown are the following. On one side of the crown
the arrow that goes into the center of the crown and forms with the axis the smaller acute angle (if any) is labeled by , the arrow that goes into the center of
the crown and forms with the axis the larger acute angle
(if any) is labeled by ,
and the outgoing arrow (if any) is labeled by (addition modulo 4).
On the other side of the axis the digital labels are , and , respectively. There is a simple explanation of this rule: the altitude of the supertile of level
has the label . Its foot (that lies on the hypotenuse) belongs to the altitudes
of two supertiles of levels whose union is the initial supertile,
their altitudes have the labels , and form with the hypotenuse the larger and smaller
acute angles, respectively.

One can imagine that the center of a legal crown performs the following task on each side of the axis: it receives two signals, verifies their coherence and emits one signal. Besides, it transmits a signal emitted by another vertex using its axis.

### The local rule L.

###### Definition 4.

A tiling of the plane or its part by colored tiles satisfies the local rule L
if (1) any two sides of two different triangles that
share a common interval have matching orientations and digital labels
and (2) every its colored crown is legal.
Tilings satisfying the local rule L are called *L-tilings*.
A coloring of tiles in a tiling by ordinary (non-colored) tiles is called *correct*
if the resulting tiling is an L-tiling.

Actually, the requirement (1) follows from (2). Indeed, property (1) obviously holds for colored sides of any legal crown. Hence it holds for any L-tiling: assume that two sides of two different triangles share a common interval. Then we can move along these sides until we first meet an end of either of the sides. The crown centered at that vertex is legal and since it satisfies the property, the sides have the same orientation and digital label.

## 4 Results

The following two theorems are the main results of the paper.

###### Theorem 2.

Any -tiling of the plane has a composition, which is again an L-tiling.

###### Theorem 3.

(1) A tiling of the plane by colored tiles is a substitution tiling if and only if it is an L-tiling. (2) A tiling of the plane by non-colored tiles is a substitution tiling if and only if it has a correct coloring.

It follows from Theorem 2 that all L-tilings of the plane are non-periodic. Indeed, they are infinitely composable, and hence non-periodic, as explained above. We first derive Theorem 3 from Theorem 2 and then we prove the latter.

###### A derivation of Theorem 3 from Theorem 2.

(1) Every colored crown of every substitution tiling of the plane occurs in a colored supertile, hence is legal. Therefore every substitution tiling is an L-tiling.

To prove the converse, consider any fragment of an -tiling of the plane. We have to show that is legal. To this end add in a finite number of tiles from so that becomes an inner part of the resulting fragment . By Theorem 2 we can compose any number of times and the resulting tiling is an L-tiling. Consider the sets of the form where is a crown of the tiling . As increases, these sets increase as well. If is large enough, then the set is covered by a single such set, say by , that is, (Lemma 3). As is an L-tiling, all its crowns are legal. In particular, the crown is legal, that is, it appears in a supertile, say in . Therefore the tiling appears in the supertile . Hence the patch appears in a supertile provided we ignore labels and orientations of its outer sides. Since no side of the patch is an outer side of , we are done.

(2) Let be a substitution tiling of the plane (by uncolored tiles). Then every its fragment occurs in a supertile, and hence it has a correct coloring. However those colorings for different fragments may be inconsistent. Using compactness arguments, we will show that it is possible to choose consistent such colorings.

More specifically, enumerate all tiles from the tiling
and consider its
fragments of the form . For every such fragment
consider all its correct colorings. By compactness, there is a concentration
point of those colorings, which is a sought coloring of the initial tiling .^{4}^{4}4Here are more details. Consider the tree whose vertices are correct colorings of fragments
of the form . Edges connect a coloring of a fragment
to a coloring of the
fragment whenever the colorings are consistent. This
tree has arbitrary long branches. Any vertex of the tree has finitely many neighbors.
By König lemma [6], the tree has an infinite branch, which provides
a correct coloring of the entire tiling.

In the reverse direction: let be a tiling of the plane by non-colored tiles that has a correct coloring. By item (1) the resulting colored L-tiling is a substitution-tiling, and hence the initial tiling is a substitution tiling as well. ∎

###### Proof of Theorem 2.

Let be an L-tiling of the plane. We have to show that it has a composition and that its composition is again an L-tiling.

*Why has a composition?*
Let be any small tile from . Consider the crown of centered at the
vertex of the right angle of . We know that that crown is legal.
There are only two crowns in the list of legal crowns, whose center is a vertex of the right angle, the crowns and .

In both these crowns the small triangle has the complement labeled on the picture. Therefore the tiling has a composition which is obtained by removing the common legs of all complementary pairs of triangles and by decrementing all labels by 1 (and then applying , where is the reference homothety). Note that the new coloring is legal, that is, the hypotenuses and small legs of large tiles and large legs of small tiles have even colors and all other sides have odd colors.

*Why the composition of is an L-tiling?*
We have to show now that the resulting colored tiling is an -tiling.
Let be a vertex of a
triangle from .
We have to prove that the colored crown centered at in is legal.
First note that the crown centered at the same vertex in the initial tiling is different from
, as the centers of crowns
become inner points of sides in .
We claim that the composition transforms
the crowns by the inverse arrows from Fig. 10:

###### Lemma 6.

Assume that the crown centered at a vertex in an L-tiling is for
.
Consider the *non-colored*
crown centered at the same vertex in .
Then it is if ,
and is or if .

Unfortunately the proof of this lemma is very technical and consists of a large number of cases. Therefore we moved it to Appendix. We will finish the proof of the theorem assuming Lemma 6.

Note that Lemma 6 tells how the composition
transforms *non-colored* crowns. We claim that *colored* crowns are transformed by the inverse grey arrows from Fig. 10.
Indeed, in the course of composition, we decrement the labels and do not change
the orientation of sides. Hence legality of all crowns
in the tiling implies legality of all crowns in .
The theorem is proved.
∎

## 5 Acknowledgments

The author is grateful to Daria Pchelina and Alexander Kozachinskii for verifying the proofs and reading the preliminary version of the paper.

## References

9

## Appendix A Appendix: deferred proofs

###### Proof of Lemma 4.

We have to show that tiles intersecting the disc belong to a single crown. If there is a single such tile, then this is obvious. If there are exactly two such tiles, and , then they must share a part of a side and at least one end of these two sides belongs to both tiles. Then for the crown centered at that end we have . Finally, if there are three or more such tiles, then at least one of those tiles, call it , has common points with lying on two different sides of the tile . Let denote the common point of those sides and let denote the points from that belong to different sides of . The angle is one of the angles of the right golden triangle and the length of is at most . Hence . Therefore is at most . All the points from are at distance at most from and hence at distance at most from . That is, is covered by the disc with center and radius . That disc is covered by the crown centered at , provided the length of the altitude of small tiles is at least its radius . ∎

###### Proof of Lemma 6.

We first show that in any L-tiling every crown must have some fixed neighborhood,
called *the neighborhood of the crown*.
In this analysis we do not use labels and orientation of sides of triangles.
It is instructive, for reader’s convenience, to print out all the legal crowns from Fig. 11
and then to cut them out of paper. Matching tiles from
those paper crowns with the tiles from the figures below, it is easy to verify all the claims
that certain crowns do not fit in certain places.

### The neighborhoods of legal crowns.

In any L-tiling of the plane every crown has some fixed neighborhood
called *the neighborhood of the crown*.
The neighborhoods of the crowns
are shown on Fig. 12. They all are centrally symmetric.

The initial crown is colored in grey, the added tiles are colored in light-grey. These neighborhoods are obtained from each other by decomposition. One can verify that each of the first five crown indeed must have such neighborhood as follows.

*The crown *. Look at the blue vertex inside the grey crown
(Fig. 13). That vertex lies on the large leg of a large triangle.
One can easily verify that there is the unique legal crown whose center
lies on the large leg of a large triangle, namely the crown
. Hence the crown of centered at the blue vertex is again and we
get the sought neighborhood.

*The crown *.
The argument is similar to the previous one. The crown
has the following feature: it has a vertex (colored in blue)
that lies on the hypotenuse of its large
triangle. It is easy to verify that there is the unique crown
whose center lies on the hypotenuse of its large
triangle, namely the crown .

*The crown *.
The crown is the unique crown that has two
small triangles sharing the small leg. Hence the crown centered at the blue vertex is again
.

*The crown *.
The crown is the unique crown
that has two large triangles sharing the small leg. Hence the crown centered at the blue vertex is again
. However this crown does not complete the neighborhood:
the crowns centered at yellow vertices must be
and the crowns centered at red vertices again must be .

*The crown *.
The crown is the unique crown
that has two small triangles sharing the hypotenuse.
Hence the crown centered at the blue vertex again must be
. The crowns centered at yellow and green
vertices must be
. Furthermore, the crowns centered at red and white vertices must be
.

The neighborhoods of the crowns , are shown on Fig. 14.

One can verify in the following way that the crowns , indeed must have such neighborhoods.

*The crown .*
The crowns centered at yellow and green vertices
must be and the crowns centered at red and black vertices must
be (see Fig. 15).

Now we see that the crown centered at the brown vertex must be , which is added together with its neighborhood.

*The crown *.
Since the crown can be obtained from by rotation (ignoring labels and orientation),
the arguments are entirely similar to those for the crown .

We will see later that the neighborhood of the crown can be extended further. However, that extension needs a more complicated analysis that requires orientation of sides and their labels. Therefore we postpone that analysis until the place where we will really need it.

Now we can begin the proof of the lemma. Assume that the crown centered at a vertex in an L-tilling is where . We have to prove that the crown centered at in the composition of the tiling is for and is or for . We will treat all ’s separately.

### The crown of in is .

Consider the neighborhood of this crown (the vertex is colored in green and the initial crown is colored in dark-grey):

What is the crown of the green vertex in the composition of the tiling ? Two large triangles together with their complements form two large tiles. And the complements of the remaining two large triangles are missing, since their small legs lie on the hypotenuses of other large triangles. Hence they become small tiles. Therefore in the composition of the green vertex becomes the center of the crown .

### The crown of in is .

The vertex is colored green on Fig. 16(a) and its crown is colored in dark-grey. On Fig. 16(b) we show the neighborhood of that crown (added tiles are colored in light-grey). As we see, the fate of all tiles from the crown is completely determined except for the large triangle labeled by letter . If the tiling has a small triangle shown on Fig. 16(c), then the tile together with that small tile form a large tile in the composition of . In this case the crown of the green vertex in the composition of is illegal. Otherwise, if there is no such small triangle in the tiling , the tile becomes a small tile and the crown of the green vertex in the composition of is . Thus we have to prove that the tiling has no such small triangle.

For the sake of the contradiction
assume that there is an L-tiling that
includes the patch
shown on Fig. 16(c).
To derive a contradiction we will extend that patch by
adding certain crowns that must be centered at
certain vertices until no crown fits for the resulting
patch in some vertex.
We will
say that
*a crown fits for a given patch in a given its vertex*
if one can draw (an isometric copy of) centered at that vertex
so that each of its tiles either does not overlap all the tiles from that tiling, or belongs to that tiling.

Look at the vertex of the right angle of the triangle , which is colored black on Fig. 16(c). A quick look at the list of legal crowns reveals that only the crown fits for the patch from Fig. 16(c) in the black vertex. There is only one isometric image of that fits, the result of adding that image and its neighborhood is shown on Fig. 16(e).

Now look at the blue vertex on the bottom left on Fig. 16(e). In that vertex only the crowns and fit. Those crowns are shown on Fig. 16(d). The crown has non-matching orientation of the green arrow, hence cannot be the crown of in the blue vertex. Thus it is the crown . Adding the crown and its neighborhood we obtain the tiling shown on Fig. 17(a).

A small search reveals that only the crown fits for that patch in the brown vertex on the top left. Fig. 17(b) shows the tiling that is obtained by adding the crown and its neighborhood. Thus we conclude that the axis of the initial crown and its extension to the right must be directed from the right to the left (the yellow arrow on Fig. 17(c)). Now look at the beginning of the yellow arrow (the red point on the top right on Fig. 17(c)). Only the crowns shown on Fig. 17(d) fit for the resulting patch in the red vertex. However none of them can be there, since all they have non-matching orientation of the yellow arrow.

### The crown of the vertex is .

The vertex is shown by the green point on Fig. 18(a). And Fig. 18(b) shows the neighborhood of that crown. We have to prove that the complement of the large triangle is not in (in that case the crown of in the composition of is and we are done).

For the sake of contradiction assume that an L-tiling includes the patch shown on Fig. 18(c).

Only the crown fits for this patch in the blue vertex. Fig. 18(d) shows the patch that is obtained by adding the crown and its neighborhood. Look now at the yellow vertex on the bottom. Only the crowns , , , fit for the patch in that vertex, they are shown on Fig. 18(e). Note that the crowns (from the left column) have non-matching orientation of the blue arrow, which must direct downwards, hence cannot be there. We will consider the remaining two cases separately.

*Case 1: the crown is in the yellow vertex on Fig. 18(d).*
In this case we are able to derive a contradiction
quite easily. Adding the crown
and its neighborhood we obtain the patch shown on Fig. 19(a)

Only the crown fits for in the blue vertex. Adding that crown and its neighborhood, we obtain the patch shown on Fig. 19(b). Only the crowns (shown on на Fig. 19(c)) fit for the resulting patch in the green vertex on the right. However the crown has non-matching orientation of the green arrow, hence the crown centered at the green vertex is . Adding it and its neighborhood, we obtain the patch on Fig. 20(a).

Only the crowns (shown on Fig. 20(b)) fit for the patch in the brown vertex on the top right. However all those crowns have non-matching orientation of the yellow arrow in the lower half of the crown. Thus we have derived a contradiction in the first case.

*Case 2: the crown centered at yellow vertex on Fig. 18(d)
is .*
In this case we need a more involved analysis. Our plan is the following.
We first show that, in addition to all tiles in the patch
Fig. 18(d), the tiling must contain
all the tiles shown on Fig. 21.

Then we will show that the patch from Fig. 21 cannot be extended to an L-tiling.

Let us go back to Fig. 18(d) and add the crown and its neighborhood in the yellow vertex. We obtain the patch shown on Fig. 22(b).

Which crowns fit for the patch in the leftmost vertex (colored in red)? These are the crowns , which are shown on Fig. 22(a). In the case of we obtain the patch shown on Fig. 23(a).

In the case of we get a patch that differs from this one in orientation and labels of some sides. This difference does not matter and therefore we will consider only the case of .

Adding the neighborhood of the crown centered at the red point, we get the patch from Fig. 23(b). Look at the vertex colored black (on the top left). Only the crown fits for the patch in that vertex.

(On Fig. 24 we have drawn the crowns on the left side of the patch. The crowns ‘‘almost fit’’ for the patch, however their rightmost triangles overlap the initial crown.)

Adding and its neighborhood in the black vertex we obtain the patch shown on Fig. 25(a).

We have shown our tiling includes all the tiles from Fig. 21 except for both triangles labeled by letter and one triangle labeled by letter . We have also shown that the tiling includes the image of the initial crown under the inversion through the white point. Via central symmetrical arguments we can prove that includes also the other triangle labeled by letter . It remains to show that includes both triangles labeled by .

To this end look at the blue vertex on the right on Fig. 25(a). Only the crowns and , shown on Fig. 25(b), fit for the patch in that vertex. If the crown centered at the blue vertex is , we obtain the patch shown on Fig. 25(c), and no legal crown fits for it in the green vertex on the right.

In the remaining case the crown centered at the blue vertex is , and we get the patch shown on Fig. 26(a).

Adding its neighborhood, we get the patch that includes the sought triangle (see Fig. 26(b)). Via central symmetrical arguments we can prove that includes also the other triangle labeled by letter .

We have reached our first goal: we have proved that the tiling includes the patch shown on Fig. 21. That patch is copied on Fig. 27(a).

It is easy to verify that only the crowns fit for the patch in the rightmost brown vertex (Fig. 27(b)). In all three cases the axis of the initial crown (the horizontal black line) must be directed rightwards. However, similar arguments applied to the leftmost brown vertex show that that axis must be directed leftwards, which is a contradiction.

### The crown of the vertex in is .

Fig. 28(a,b) show the crown and its neighborhood. We need to show that the complement the large triangle is not in (in that case the crown produces the crown under the composition). For the sake of contradiction assume that the tiling includes the patch shown on Fig. 28(c).

Only the crown fits for this patch in the yellow vertex. Adding to the patch that crown and its neighborhood, we obtain the patch shown on Fig. 28(d). Now, by a simple search, we can verify that no legal crown fits for this patch in the blue vertex.

### The crown of in is .

Fig. 29(a,b) show the crown and its neighborhood. We can easily verify that the fate of all large triangles (under composition) is completely determined except the triangle .

If the complement of this triangle is in , then the initial crown produces the crown under the composition, and otherwise it produces .

### The crown of in is .

Fig. 30(a,b) show the crown and its neighborhood. We can easily verify that the fate of all large triangles (under composition) is completely determined except for the triangle .

We have to prove that the complement of the triangle is in and hence the crown of in the composition of is . (Otherwise the crown of in the composition of would be with the wrong axis—its axis would be orthogonal to the axis of the legal crown .)

Only the crowns fit for the patch in the blue vertex, they are shown on Fig. 30(c). If the crown in the blue vertex is , we are done, as in this case the complement of triangle is in . It remains to show that neither of the crowns can stand in the blue vertex.

It is easy to show that cannot be there. Indeed, adding that crown to the patch, we obtain the patch shown on Fig. 30(d). No legal crown fits for that patch in the yellow vertex.

We have to prove now that the crown centered at the blue vertex cannot be either. Assume the contrary, then we obtain the patch shown on Fig. 31(a).

First add the neighborhood of , we obtain Fig. 31(b). Only the crowns and , shown on Fig. 31(c), fit for the resulting patch in the yellow vertex. However the crown cannot be there, since its green arrow has the non-matching orientation. Hence the crown in the yellow vertex is . Fig. 31(d) shows the patch which is obtained by adding that crown and its neighborhood. Now look at the red vertex on the right. Only the crowns (Fig. 31(e)) fit for the patch in that vertex. In all the three cases there is a horizontal blue arrow that goes into the red vertex. That arrow lies on the axis of the initial crown.

Now we know the color and orientation of that axis (see Fig. 32(b)).

We are now very close to the end. We can find now the crowns in the leftmost and the bottommost vertices (both are colored blue). Indeed, only the crowns fit for the patch in the bottommost vertex, and two latter crowns fit in two ways (see Fig. 32(c)). In four cases the vertical arrow has the red (and not green) color. Hence those cases are impossible and only the lower crown can stand in the bottommost blue vertex.

A similar situation occurs in the leftmost blue vertex: five crowns fit for the patch there (two of them fit in two ways, see Fig. 32(a)). However only one crown, the lower , can have the matching color (blue) of the horizontal arrow.

Adding to the patch the crowns in the blue vertices and adding then their neighborhoods, we obtain the patch shown on Fig. 33(b).

Only the crown fits for that patch in the yellow vertex, that crown is shown on Fig. 33(a). Now it is obvious that neither of the legal crowns can stand in the adjacent black vertex, since the yellow arrow that goes out the yellow vertex into the black one cannot change its color to blue in the black vertex.

We have considered all the cases. The lemma is proved. ∎