Some observations
on a substitution rulewith singular vertex atlases
Kochi Univ.
Hiroko Hayashi

【1】 PreliminaryPenrose tiles

Penrose tiling

Divided Penrose tiles

The substitution rulesome of 21 expansions

Up-down generation

By erasing edges,
we have the Penrose tiling.

→Penrose tiling

5-fold rotational
symmmetryPrototiles of Danzer tilings

The substitution ruleWith 7-fold symmetry

Vertex atlas• Configurations(without gap andoverlapping)of tiles around avertex is called thevertex atlasin a tiling.

8 kinds of vertex atlases ofPenrose tiling【2】 Motivation

Motivation• Prof. Danzer said as a remark in the
appendix of Danzer's paper that
"29 kinds vertex atlases
appear in the Danzer tiling,
and that these vertex atlases
may serve as a matching rule."

How many vertex atlases ?
• 39 vertex atlases with arrows.
(our result)• 29 vertex atlases without arrows.
(remark of Prof. Danzer)

Up-down generation• Penrose tiling and Danzer tiling withrotational symmetry cannot be constructed
only by the up-down generation procedure.
• It is necessary to extend the tilings to thewhole plane by using reflection androtation.

Super tile

Scale-up procedure

All of Penrose tilings
with 5-fold symmetry

Scale-up procedure

The substitution rule

The center of 5-fold symmetry can not be found.

Vertex atlas(i)(ii)

The properties of (i) , (ii)• These do not exist in the list of 39 vertex
atlases.
• By substitution rule, the same typeappears at the center when thesesubdivide.
• Therefore, the plane can be covered by 39
kinds of vertex atlas and substitution.
• （There is no one of (i) and (ii) in thePenrose tiling. ）

(i)
(I)

• By substitution, we
have a tiling withvertex atlas (i).

• By trial and error, we can show that
configuration (I) is uniquely obtained from(i).
• If a tiling have just 39 vertex atlases andthe vertex atlas (i), then vertex atlas (i)appear only in one place.

【3】 Our observations

Our questions
Question 1• For which n, cantilings with n-foldsymmetry beconstructed only by
the up-down
generation procedure?

Our questions
Question 2• How can tilings beconstructed byattaching unbounded
configuration?

Primitive substitution• A substitution is primitive if the substitution
rule is a linear map that can berepresented by a primitive matrix.

Primitive matrix• An n × n matrix A is said to be primitive
if its entries are nonnegative integers andif there exists a positive integer k such thatk
all the entries of A are positive.

• If there is a substitution with n-fold tiling,
we can make a substitution that cangenerate an n-fold tiling from up-downgeneration.

Prototiles

Substitution

This is not primitive.

New question 1• For which n, can tilings with n-foldsymmetry be constructed only by the updown generation procedure of someprimitive substitution?

Forcing the border• A substitution is said to force the border
if there exists a positive integer n such thatany two level-n supertiles of the same typehave the same pattern of neighboring tiles.

Divided Penrose tiles

TL : level 0

TL : level 0

TL : level 1

TL : level 8

tL : level 8

red : the supertile
blue : the neighboring tiles
yellow : vertex atlas (i)Two supertiles of the same type• The way of attaching of Penrose tilings isunique.
• However, that of Danzer tiling is notunique.

New question 2• Nonexistence of singular vertex atlases .
⇔ Forcing the boder.
iff ?

• In particular, is the attachment ofunbouded configurations unique, if
singular vertex atlases do not exist?