On aperiodic monotile bySocolar-Taylor
Miyuki Komai(Niigata Univ)

[Main Proposition]
Ｉｆ Socolar-Taylor’s tiles cover the plane, then it is not periodic.
First, we switch to (b) having two shapes for the convenience.
(a)Both tiles of (b) have the samecolor arrangement except one
diagonal.
Left-handed flags→blue stripe
Let us call this diagonal having
Right-handed flags→red stripe different colors “red-blue(b)diagonal”.

Next, we prove Lemma 1,2 and five Propositions.
Later, using them, we prove the Main Proposition.

Socolar-Taylor’s tile has two matching rules.
1. Black stripes must be continuous across shared edges.
2. The colored (red or blue) segments that meet a given edge atopposite endpoints and are collinear with that edge must not be
the same color (as shown below).
(a)

(b)

[Lemma 1]
If the tiles cover the plane, then a small black ring like below must beformed. And it is formed at the opposite vertex, as well.
<Proof>First, we prove that a small ring must be formed.
Assuming that the direction of Tile A is like this, there are two cases.

A small ring is formed at vertex αby matching rule of black stripe.

A small ring is formed at vertex βby matching rule of black stripe.

Therefore, a small ring must be formed.

[Lemma 1]
If the tiles cover the plane, then a small black ring like below must beformed. And it is formed at the opposite vertex, as well.
<Proof: continued>
Next, we prove that a small ring is formed at the opposite vertex.

Assume that a small ring isformed at vertex γ.

By color matching rule, there aretwo ways to configure A as in thefigure above, but by matchingrule of black stripe, the former iscorrect.

Therefore, vertex δ has a small ring.

□The following lattice of holes is formed by Lemma 1.
There are three ways to fill a hole.

Hexagonal lattice of holes.

Putting a tile in the hole.

Let us call a tile in the hole in the lattice “vertex tile” and draw “V” in
it to fix the orientation.

[Lemma 2]
A black triangle using three tiles for an edge (Level 1 triangle) mustbe formed. And it is formed at the opposite side, as well.
<Proof>First, we prove that Level 1 triangle must be formed. As shown below,according to the lattice, there are two directions at A by matchingrule of black stripes.

The direction of tile A, B, C isdecided, and Level 1 triangle isformed here.

The direction of tile A, B, C, D, E, F
is decided, and Level 1 triangle isformed below.

[Lemma 2]
A black triangle using three tiles for an edge (Level 1 triangle) mustbe formed. And it is formed at the opposite side, as well.
<Proof: continued>
Next, we prove that if Level 1 triangle is formed, it is formed at theopposite side, too.

The directions of tile A and B
force the left color configurations,therefore, Level 1 triangle isformed at the opposite side, too.
□Two triangles at the opposite
sides share a “vertex tile”.

[Definition]
 If number of tiles of one side is Nn = 2n + 1 , then the triangle iscalled Level n triangle.
We do not count a small ring as a triangle.
 Tiles whose red-blue diagonals are coｌlinear are called “ tiles in thesame column”.
 If consecutive tiles from tile A to B contain a straight line, thendenote their length by NAB .

[Proposition 1]
If there is a straight line of m tiles which can’t be extended any more,then the line produce a triangle having m tiles at each side.

<Proof>When the line turns at two ends,
then the direction of turns mustcoincide by matching rule of black
stripe. Therefore, the other twosides starting from two ends
must be of length m as in thefigure, by matching rule of blackstripe.
□

[Proposition 2]
If there are two straight lines starting from a vertex tile, whose length
is longer than Nn + 1 , then we observe lined up triangles whosevertex tiles are in the same column in the middle of the two straightlines. They appear in the increasing order like, the vertex tile, a smallring, Level 1 triangle and Level 2 triangle ・・・ and form a Sierpinski’s
triangle shape.
<Proof>By induction on n.
A small ring and Level 1 triangle
are formed, as shown on the left,
by matching rule of black stripe.
(n=1 is satisfied because ofN1 = 3 = 21 + 1 )[Proposition 2]
If there are two straight lines starting from a vertex tile, whose length
is longer than Nn + 1 , then we observe lined up triangles whosevertex tiles are in the same column in the middle of the two straightlines. They appear in the increasing order like, the vertex tile, a smallring, Level 1 triangle and Level 2 triangle ・・・ and form a Sierpinski’s
triangle shape.
<Proof: continued>

n－3
NVA = 2 + ∑(Ni－1) + Nn－2
i=1= Nn－1= NAB

[Proposition 2]
If there are two straight lines starting from a vertex tile, whose length
is longer than Nn + 1 , then we observe lined up triangles whosevertex tiles are in the same column in the middle of the two straightlines. They appear in the increasing order like, the vertex tile, a smallring, Level 1 triangle and Level 2 triangle ・・・ and form a Sierpinski’s
triangle shape.
<Proof:continued>

Formation of triangles isrepeated, because tile V and A
are the same role and fromthe left figure.

[Proposition 2]
If there are two straight lines starting from a vertex tile, whose length
is longer than Nn + 1 , then we observe lined up triangles whosevertex tiles are in the same column in the middle of the two straightlines. They appear in the increasing order like, the vertex tile, a smallring, Level 1 triangle and Level 2 triangle ・・・ and form a Sierpinski’s
triangle shape.
<Proof:continued>

From the left figure,n－1 + 1－1
NDB = 2Nn－1－1 = 2 2= 2n + 1 = Nn
Therefore, the straight line whichconnects tile F to B form Level ntriangle by proposition 1.

()

[Proposition 2]
If there are two straight lines starting from a vertex tile, whose length
is longer than Nn + 1 , then we observe lined up triangles whosevertex tiles are in the same column in the middle of the two straightlines. They appear in the increasing order like, the vertex tile, a smallring, Level 1 triangle and Level 2 triangle ・・・ and form a Sierpinski’s
triangle shape.
<Proof:end>
Therefore, as in the figure,triangles form a Sierpinski’s
triangle shape.
□[Proposition 3]
The direction of red-blue diagonal of the vertex tile of Level n triangle
and that of the middle tile in the base line being opposite of thevertex tile, are the same.
<Proof>By induction.

First, the direction of colors ofFrom assumption, direction of
red-blue diagonal of the vertex tile, colors of red-blue diagonal of tile
tile A and B are the same as
B and A is the same. Iterating theshown.
same method, that of tiles fromtile V to A is the same.
□[Proposition 4]
If Level n triangle is formed, it must be formed at the opposite, too.
<Proof>By induction. Assume that for n-1 (Existence of Level 1 triangle wasalready shown.) the triangle is formed.

※Tile A, B and C is in the samecolumn.
And clearly NAV = NVC = Nn.
From proposition 2
From assumption ofinduction

[Proposition 4]
If Level n triangle is formed, it must be formed at the opposite, too.
<Proof:continued>

Color information is transmitted as A→D→B→C by proposition 3.
Direction of tile C is determined by matching rule of black stripe.
Therefore, proposition 4 is shown by proposition 1.
□[Proposition 5]
For all n, Level n triangle is formed.
<Proof>From proposition 2
From proposition 4

Assuming the existence of a Leveln-1 triangle, triangles whose level
is less than n-1 are formed. And,
consecutive tiles from tile A to B,from tile B to C and from tile C to
A contain a straight line.
Then we have,
NAB = NBC = NCA = Nn .

[Proposition 5]
For all n, Level n triangle is formed.
<Proof:continued>
By induction. Assume it for n-1. (Existence of Level 1 triangle wasalready shown.)
(a)

(b)

(c)

There are three directions of tile
A. But I prove only (a). Because
rotating figure (b) and (c), theyand (a) are same.

[Proposition 5]
For all n, Level n triangle is formed.
<Proof:end>
Look at the direction of tile B.
1. If tile B doesn’t extend astraight line.
Proposition 5 is shown byproposition 1.

2. If B extends a straight line. Tile
B and C must be as in figure.
Therefore, proposition 5 isshown by proposition 2.
□If the tiling has a period p , length of sides of triangles can notexceed p . This is inconsistent with lemma 5.
Therefore, the tiling is not periodic.
■