Face no more (?)
(An exhaustive search of the convex pentagons which tiles the plane)
Michaël Rao
École Normale Supérieure de Lyon / CNRS, LIP, MC2 team
CRM
April 5, 2017
Michaël Rao
1 / 30
Sketch
We present an exhaustive search of all convex pentagons which tile the
plane
Introduction
Michaël Rao
2 / 30
Sketch
We present an exhaustive search of all convex pentagons which tile the
plane
Let P be a convex pentagon which tiles the plane.
Part 1: There exist a tiling by P such that each vertex type has
positive density
The set of vertex type (i.e. conditions implied by angles) must be
“good”
Part 2: There are only 371 good sets to consider
Part 3: For each good set : we do an exhaustive search
Result: we found only the 15 known families (and some special cases).
Introduction
Michaël Rao
2 / 30
Part 1: positive density tiling and good sets
Let P be a convex pentagon
the vertices are s1 , . . . s5 , in clockwise order
the angles are respectively α1 × π, . . . , α5 × π
∀1 ≤ i ≤ 5, 0 < αi < 1
5
X
αi = (1, 1, 1, 1, 1) · α = 3
i=1
Part 1/3 : Positive density tiling and good sets
Michaël Rao
3 / 30
Part 1: positive density tiling and good sets
Let P be a convex pentagon
the vertices are s1 , . . . s5 , in clockwise order
the angles are respectively α1 × π, . . . , α5 × π
∀1 ≤ i ≤ 5, 0 < αi < 1
5
X
αi = (1, 1, 1, 1, 1) · α = 3
i=1
Let T be tiling of the plane by P (we allow rotation/transtation/mirror)
(Note: no hypothesis on periodicity / transitivity)
Part 1/3 : Positive density tiling and good sets
Michaël Rao
3 / 30
Part 1/3 : Positive density tiling and good sets
Michaël Rao
4 / 30
Vector type
Let s be a vertex of T (that is a vertex of one pentagon in T )
The vector type of s, denoted V (s), is the vector v ∈ N5 s.t. there are vi
angles si around s.
Part 1/3 : Positive density tiling and good sets
Michaël Rao
5 / 30
Vector type
Let s be a vertex of T (that is a vertex of one pentagon in T )
The vector type of s, denoted V (s), is the vector v ∈ N5 s.t. there are vi
angles si around s.
Two cases of vertices:
“Half” : s is in the border of a tile P, but not a vertex of P
“Full” : s is a vertex of every tile around s
Part 1/3 : Positive density tiling and good sets
Michaël Rao
5 / 30
Vector type
Let s be a vertex of T (that is a vertex of one pentagon in T )
The vector type of s, denoted V (s), is the vector v ∈ N5 s.t. there are vi
angles si around s.
Two cases of vertices:
“Half” : s is in the border of a tile P, but not a vertex of P
“Full” : s is a vertex of every tile around s
The corrected vector type of s, denoted V c (s), is:
V (s) if s is full, or 2 × V (s) if s is half.
Part 1/3 : Positive density tiling and good sets
Michaël Rao
5 / 30
Vector type
Let s be a vertex of T (that is a vertex of one pentagon in T )
The vector type of s, denoted V (s), is the vector v ∈ N5 s.t. there are vi
angles si around s.
Two cases of vertices:
“Half” : s is in the border of a tile P, but not a vertex of P
“Full” : s is a vertex of every tile around s
The corrected vector type of s, denoted V c (s), is:
V (s) if s is full, or 2 × V (s) if s is half.
For every vertex s, V c (s) · α = 2
Part 1/3 : Positive density tiling and good sets
Michaël Rao
5 / 30
Vector type
Let s be a vertex of T (that is a vertex of one pentagon in T )
The vector type of s, denoted V (s), is the vector v ∈ N5 s.t. there are vi
angles si around s.
Two cases of vertices:
“Half” : s is in the border of a tile P, but not a vertex of P
“Full” : s is a vertex of every tile around s
The corrected vector type of s, denoted V c (s), is:
V (s) if s is full, or 2 × V (s) if s is half.
For every vertex s, V c (s) · α = 2
W : set of vector types of vertices in T
W c : set of corrected vector types
Part 1/3 : Positive density tiling and good sets
Michaël Rao
5 / 30
Vector type
Let s be a vertex of T (that is a vertex of one pentagon in T )
The vector type of s, denoted V (s), is the vector v ∈ N5 s.t. there are vi
angles si around s.
Two cases of vertices:
“Half” : s is in the border of a tile P, but not a vertex of P
“Full” : s is a vertex of every tile around s
The corrected vector type of s, denoted V c (s), is:
V (s) if s is full, or 2 × V (s) if s is half.
For every vertex s, V c (s) · α = 2
W : set of vector types of vertices in T
W c : set of corrected vector types
Note: W and W c are finite
Part 1/3 : Positive density tiling and good sets
Michaël Rao
5 / 30
An informal toy problem
Suppose that the density of each corrected vector type is definite.
density(v ) =
number of vertices s with V c (s) = v
number of tiles
Part 1/3 : Positive density tiling and good sets
Michaël Rao
6 / 30
An informal toy problem
Suppose that the density of each corrected vector type is definite.
density(v ) =
number of vertices s with V c (s) = v
number of tiles
W c is the following:
v1 = (1, 1, 1, 0, 0)
v2 = (0, 0, 0, 2, 2)
v3 = (1, 1, 0, 1, 0)
Part 1/3 : Positive density tiling and good sets
Michaël Rao
6 / 30
An informal toy problem
Suppose that the density of each corrected vector type is definite.
density(v ) =
number of vertices s with V c (s) = v
number of tiles
W c is the following:
v1 = (1, 1, 1, 0, 0)
v2 = (0, 0, 0, 2, 2)
v3 = (1, 1, 0, 1, 0)
What are the densities of vi ’s ?
Part 1/3 : Positive density tiling and good sets
Michaël Rao
6 / 30
An informal toy problem
Suppose that the density of each corrected vector type is definite.
density(v ) =
number of vertices s with V c (s) = v
number of tiles
W c is the following:
v1 = (1, 1, 1, 0, 0)
v2 = (0, 0, 0, 2, 2)
v3 = (1, 1, 0, 1, 0)
What are the densities of vi ’s ?
Respectively 1, 21 , 0
Part 1/3 : Positive density tiling and good sets
Michaël Rao
6 / 30
An informal toy problem
Suppose that the density of each corrected vector type is definite.
density(v ) =
number of vertices s with V c (s) = v
number of tiles
W c is the following:
v1 = (1, 1, 1, 0, 0)
v2 = (0, 0, 0, 2, 2)
v3 = (1, 1, 0, 1, 0)
What are the densities of vi ’s ?
Respectively 1, 21 , 0
Is v3 mandatory to tile with P ?
Part 1/3 : Positive density tiling and good sets
Michaël Rao
6 / 30
Positive density tilings
Let o ∈ R2 and r ≥ 0
B(o, r ) is the disk of radius r and center o
To,r be the set of tiles P ∈ T such that P ∩ B(o, r ) 6= ∅
IV(T 0 ) set of “interior vertices” of a sub-tiling T 0
For a
vector type v , let :
f
o,r (v )
=
|{s ∈ IV(To,r ) : V (s) = v }|
|To,r |
Part 1/3 : Positive density tiling and good sets
Michaël Rao
7 / 30
Positive density tilings
Let o ∈ R2 and r ≥ 0
B(o, r ) is the disk of radius r and center o
To,r be the set of tiles P ∈ T such that P ∩ B(o, r ) 6= ∅
IV(T 0 ) set of “interior vertices” of a sub-tiling T 0
For a
vector type v , let :
f
o,r (v )
=
|{s ∈ IV(To,r ) : V (s) = v }|
|To,r |
Definition (Positive density tiling)
T has positive density if for every v ∈ W and o ∈ R2 we have
lim inf r →∞ fo,r (v ) > 0
Part 1/3 : Positive density tiling and good sets
Michaël Rao
7 / 30
Positive density tilings
Let o ∈ R2 and r ≥ 0
B(o, r ) is the disk of radius r and center o
To,r be the set of tiles P ∈ T such that P ∩ B(o, r ) 6= ∅
IV(T 0 ) set of “interior vertices” of a sub-tiling T 0
For a corrected vector type v , let :
f 00 o,r (v ) =
|{s ∈ IV(To,r ) : V c (s) = v }|
|To,r |
Definition (Positive density tiling)
T has positive density if for every v ∈ W and o ∈ R2 we have
lim inf r →∞ fo,r (v ) > 0
00 (v ) > 0
imply also that ∀v ∈ W c , lim inf r →∞ fo,r
Part 1/3 : Positive density tiling and good sets
Michaël Rao
7 / 30
Tiling imply positive density tiling
Lemma
If a tiling by P exists, then a tiling of positive density by P exists.
Part 1/3 : Positive density tiling and good sets
Michaël Rao
8 / 30
Tiling imply positive density tiling
Lemma
If a tiling by P exists, then a tiling of positive density by P exists.
Otherwise, suppose v ∈ W with lim inf r →∞ fo,r (v ) = 0
Part 1/3 : Positive density tiling and good sets
Michaël Rao
8 / 30
Tiling imply positive density tiling
Lemma
If a tiling by P exists, then a tiling of positive density by P exists.
Otherwise, suppose v ∈ W with lim inf r →∞ fo,r (v ) = 0
There are sub-tilings of an arbitrarily large disk without a vertex of
vector type v
Part 1/3 : Positive density tiling and good sets
Michaël Rao
8 / 30
Tiling imply positive density tiling
Lemma
If a tiling by P exists, then a tiling of positive density by P exists.
Otherwise, suppose v ∈ W with lim inf r →∞ fo,r (v ) = 0
There are sub-tilings of an arbitrarily large disk without a vertex of
vector type v
(take a grid of girth x ∈ R: if there is v in every cell, then one have a
contradiction)
Part 1/3 : Positive density tiling and good sets
Michaël Rao
8 / 30
Tiling imply positive density tiling
Lemma
If a tiling by P exists, then a tiling of positive density by P exists.
Otherwise, suppose v ∈ W with lim inf r →∞ fo,r (v ) = 0
There are sub-tilings of an arbitrarily large disk without a vertex of
vector type v
(take a grid of girth x ∈ R: if there is v in every cell, then one have a
contradiction)
By compactness, one can construct a tiling which does not have a
vertex of vector type v
(warning: be careful with “fracture lines”)
Part 1/3 : Positive density tiling and good sets
Michaël Rao
8 / 30
Good set
Definition (Good set)
X ⊆ N5 is good if ∀u ∈ R5 with
P
u = 0, either:
u · v = 0 for every v ∈ X , or
there are v , v 0 ∈ X such that u · v < 0 < u · v 0 .
Part 1/3 : Positive density tiling and good sets
Michaël Rao
9 / 30
Good set
Definition (Good set)
X ⊆ N5 is good if ∀u ∈ R5 with
P
u = 0, either:
u · v = 0 for every v ∈ X , or
there are v , v 0 ∈ X such that u · v < 0 < u · v 0 .
In the toy example:
v1 = (1, 1, 1, 0, 0)
v2 = (0, 0, 0, 2, 2)
v3 = (1, 1, 0, 1, 0)
{v1 , v2 , v3 } is
Part 1/3 : Positive density tiling and good sets
Michaël Rao
9 / 30
Good set
Definition (Good set)
X ⊆ N5 is good if ∀u ∈ R5 with
P
u = 0, either:
u · v = 0 for every v ∈ X , or
there are v , v 0 ∈ X such that u · v < 0 < u · v 0 .
In the toy example:
v1 = (1, 1, 1, 0, 0)
v2 = (0, 0, 0, 2, 2)
v3 = (1, 1, 0, 1, 0)
{v1 , v2 , v3 } is not good, with u = (1, 0, −1, 0, 0)
Part 1/3 : Positive density tiling and good sets
Michaël Rao
9 / 30
Good set
Definition (Good set)
X ⊆ N5 is good if ∀u ∈ R5 with
P
u = 0, either:
u · v = 0 for every v ∈ X , or
there are v , v 0 ∈ X such that u · v < 0 < u · v 0 .
In the toy example:
v1 = (1, 1, 1, 0, 0)
v2 = (0, 0, 0, 2, 2)
v3 = (1, 1, 0, 1, 0)
{v1 , v2 , v3 } is not good, with u = (1, 0, −1, 0, 0)
{v1 , v2 } is
Part 1/3 : Positive density tiling and good sets
Michaël Rao
9 / 30
Good set
Definition (Good set)
X ⊆ N5 is good if ∀u ∈ R5 with
P
u = 0, either:
u · v = 0 for every v ∈ X , or
there are v , v 0 ∈ X such that u · v < 0 < u · v 0 .
In the toy example:
v1 = (1, 1, 1, 0, 0)
v2 = (0, 0, 0, 2, 2)
v3 = (1, 1, 0, 1, 0)
{v1 , v2 , v3 } is not good, with u = (1, 0, −1, 0, 0)
{v1 , v2 } is good since 2 × u · v1 + u · v2 = 0
Part 1/3 : Positive density tiling and good sets
Michaël Rao
9 / 30
Positive density imply W c is good
Lemma
If T has positive density, then W c is good.
Part 1/3 : Positive density tiling and good sets
Michaël Rao
10 / 30
Positive density imply W c is good
Lemma
If T has positive density, then W c is good.
P
Otherwise, suppose u ∈ R5 such that 5i=1 ui = 0 s.t.:
∀v ∈ W c with u · v ≥ 0.
there is a v + ∈ W c with u · v + > 0.
Part 1/3 : Positive density tiling and good sets
Michaël Rao
10 / 30
Positive density imply W c is good
Lemma
If T has positive density, then W c is good.
P
Otherwise, suppose u ∈ R5 such that 5i=1 ui = 0 s.t.:
∀v ∈ W c with u · v ≥ 0.
there is a v + ∈ W c with u · v + > 0.
|To,r | and |IV(To,r )| in Θ(r 2 )
Part 1/3 : Positive density tiling and good sets
Michaël Rao
10 / 30
Positive density imply W c is good
Lemma
If T has positive density, then W c is good.
P
Otherwise, suppose u ∈ R5 such that 5i=1 ui = 0 s.t.:
∀v ∈ W c with u · v ≥ 0.
there is a v + ∈ W c with u · v + > 0.
|To,r | and |IV(To,r )| in Θ(r 2 )
things on the “border” in O(r )
Part 1/3 : Positive density tiling and good sets
Michaël Rao
10 / 30
Positive density imply W c is good
Lemma
If T has positive density, then W c is good.
P
Otherwise, suppose u ∈ R5 such that 5i=1 ui = 0 s.t.:
∀v ∈ W c with u · v ≥ 0.
there is a v + ∈ W c with u · v + > 0.
|To,r | and |IV(To,r )| in Θ(r 2 )
things on the “border” in O(r )
We count the number of angles in To,r .
X
0
lim
v × fo,r
(v ) = (1, 1, 1, 1, 1)
r →∞
v ∈W c
Part 1/3 : Positive density tiling and good sets
Michaël Rao
10 / 30
Positive density imply W c is good
Lemma
If T has positive density, then W c is good.
P
Otherwise, suppose u ∈ R5 such that 5i=1 ui = 0 s.t.:
∀v ∈ W c with u · v ≥ 0.
there is a v + ∈ W c with u · v + > 0.
|To,r | and |IV(To,r )| in Θ(r 2 )
things on the “border” in O(r )
We count the number of angles in To,r .
X
0
lim
v × fo,r
(v ) = (1, 1, 1, 1, 1)
r →∞
v ∈W c
lim
r →∞
X
0
(u · v ) × fo,r
(v ) = 0
v ∈W c
Part 1/3 : Positive density tiling and good sets
Michaël Rao
10 / 30
Positive density imply W c is good
Lemma
If T has positive density, then W c is good.
P
Otherwise, suppose u ∈ R5 such that 5i=1 ui = 0 s.t.:
∀v ∈ W c with u · v ≥ 0.
there is a v + ∈ W c with u · v + > 0.
|To,r | and |IV(To,r )| in Θ(r 2 )
things on the “border” in O(r )
We count the number of angles in To,r .
X
0
lim
v × fo,r
(v ) = (1, 1, 1, 1, 1)
r →∞
v ∈W c
lim
r →∞
X
0
(u · v ) × fo,r
(v ) = 0
v ∈W c
Contradiction since:
X
0
0
lim
(u · v ) × fo,r
(v ) ≥ (u · v + ) × lim inf fo,r
(v + ) > 0
r →∞
v ∈W c
Part 1/3 : Positive density tiling and good sets
r →∞
Michaël Rao
10 / 30
PX
Definition (PX )
Given a subset X ⊆ N5 , we define by PX the subset of R5 such that
α = (α1 , . . . α5 ) ∈ PX if and only if
for every i ∈ {1, . . . , 5}, 0 ≤ αi ≤ 1,
P5
i=1 αi = 3 and
for every v ∈ X , α · v = 2.
Part 1/3 : Positive density tiling and good sets
Michaël Rao
11 / 30
PX
Definition (PX )
Given a subset X ⊆ N5 , we define by PX the subset of R5 such that
α = (α1 , . . . α5 ) ∈ PX if and only if
for every i ∈ {1, . . . , 5}, 0 ≤ αi ≤ 1,
P5
i=1 αi = 3 and
for every v ∈ X , α · v = 2.
PX is a convex polytope
Part 1/3 : Positive density tiling and good sets
Michaël Rao
11 / 30
PX
Definition (PX )
Given a subset X ⊆ N5 , we define by PX the subset of R5 such that
α = (α1 , . . . α5 ) ∈ PX if and only if
for every i ∈ {1, . . . , 5}, 0 ≤ αi ≤ 1,
P5
i=1 αi = 3 and
for every v ∈ X , α · v = 2.
PX is a convex polytope
In a tiling by a convex pentagon: α ∈ PW c , thus PW c ∩]0, 1[5 6= ∅
Part 1/3 : Positive density tiling and good sets
Michaël Rao
11 / 30
PX
Definition (PX )
Given a subset X ⊆ N5 , we define by PX the subset of R5 such that
α = (α1 , . . . α5 ) ∈ PX if and only if
for every i ∈ {1, . . . , 5}, 0 ≤ αi ≤ 1,
P5
i=1 αi = 3 and
for every v ∈ X , α · v = 2.
PX is a convex polytope
In a tiling by a convex pentagon: α ∈ PW c , thus PW c ∩]0, 1[5 6= ∅
What are the good sets X such that PX ∩]0, 1[5 6= ∅ ?
Part 1/3 : Positive density tiling and good sets
Michaël Rao
11 / 30
Compatible vectors, maximal set
span(X ): vectors which are linear combination of vectors in X
Compat(X ) = {w ∈ N5 :
(w , 2) ∈ span({(1, 1, 1, 1, 1, 3)} ∪ {(v , 2) : v ∈ X })}
Part 1/3 : Positive density tiling and good sets
Michaël Rao
12 / 30
Compatible vectors, maximal set
span(X ): vectors which are linear combination of vectors in X
Compat(X ) = {w ∈ N5 :
(w , 2) ∈ span({(1, 1, 1, 1, 1, 3)} ∪ {(v , 2) : v ∈ X })}
One have PX = PCompat(X ) .
Part 1/3 : Positive density tiling and good sets
Michaël Rao
12 / 30
Compatible vectors, maximal set
span(X ): vectors which are linear combination of vectors in X
Compat(X ) = {w ∈ N5 :
(w , 2) ∈ span({(1, 1, 1, 1, 1, 3)} ∪ {(v , 2) : v ∈ X })}
One have PX = PCompat(X ) .
If X is good then Compat(X ) is also good
Part 1/3 : Positive density tiling and good sets
Michaël Rao
12 / 30
Compatible vectors, maximal set
span(X ): vectors which are linear combination of vectors in X
Compat(X ) = {w ∈ N5 :
(w , 2) ∈ span({(1, 1, 1, 1, 1, 3)} ∪ {(v , 2) : v ∈ X })}
One have PX = PCompat(X ) .
If X is good then Compat(X ) is also good
A set X is maximal if X = Compat(X ).
W.l.o.g., we can work on maximal good sets
What are the maximal good sets X such that PX ∩]0, 1[5 6= ∅ ?
Part 1/3 : Positive density tiling and good sets
Michaël Rao
12 / 30
Part 2: Computation of all good sets
In this part, the order of the angles is not important.
(i.e. W.l.o.g. 1 > α1 ≥ α2 ≥ α3 ≥ α4 ≥ α5 > 0)
Part 2/3: Computing all good sets
Michaël Rao
13 / 30
Part 2: Computation of all good sets
In this part, the order of the angles is not important.
(i.e. W.l.o.g. 1 > α1 ≥ α2 ≥ α3 ≥ α4 ≥ α5 > 0)
P≥
X
Let P≥
X be the set of vectors (α1 , . . . α5 ) such that
1 ≥ α1 ≥ α2 ≥ α3 ≥ α4 ≥ α5 ≥ 0,
P
i αi = 3 and
for every v ∈ X , v · α = 2.
Part 2/3: Computing all good sets
Michaël Rao
13 / 30
Part 2: Computation of all good sets
In this part, the order of the angles is not important.
(i.e. W.l.o.g. 1 > α1 ≥ α2 ≥ α3 ≥ α4 ≥ α5 > 0)
P≥
X
Let P≥
X be the set of vectors (α1 , . . . α5 ) such that
1 ≥ α1 ≥ α2 ≥ α3 ≥ α4 ≥ α5 ≥ 0,
P
i αi = 3 and
for every v ∈ X , v · α = 2.
5
What are the good sets X such that P≥
X ∩]0, 1[ 6= ∅ ?
Part 2/3: Computing all good sets
Michaël Rao
13 / 30
Part 2: Computation of all good sets
In this part, the order of the angles is not important.
(i.e. W.l.o.g. 1 > α1 ≥ α2 ≥ α3 ≥ α4 ≥ α5 > 0)
P≥
X
Let P≥
X be the set of vectors (α1 , . . . α5 ) such that
1 ≥ α1 ≥ α2 ≥ α3 ≥ α4 ≥ α5 ≥ 0,
P
i αi = 3 and
for every v ∈ X , v · α = 2.
5
What are the good sets X such that P≥
X ∩]0, 1[ 6= ∅ ?
5
5
If P≥
X ∩ [0, 1] is non empty, let mX ∈ [0, 1] be such that
≥
(mX )i = min{αi : α ∈ PX }.
Part 2/3: Computing all good sets
Michaël Rao
13 / 30
Part 2: Computation of all good sets
In this part, the order of the angles is not important.
(i.e. W.l.o.g. 1 > α1 ≥ α2 ≥ α3 ≥ α4 ≥ α5 > 0)
P≥
X
Let P≥
X be the set of vectors (α1 , . . . α5 ) such that
1 ≥ α1 ≥ α2 ≥ α3 ≥ α4 ≥ α5 ≥ 0,
P
i αi = 3 and
for every v ∈ X , v · α = 2.
5
What are the good sets X such that P≥
X ∩]0, 1[ 6= ∅ ?
5
5
If P≥
X ∩ [0, 1] is non empty, let mX ∈ [0, 1] be such that
≥
(mX )i = min{αi : α ∈ PX }.
Note: (mX )1 ≥ 35 , (mX )2 ≥ 12 , (mX )3 ≥ 31 , and (mX )i ≥ (mX )i+1
Part 2/3: Computing all good sets
Michaël Rao
13 / 30
1: procedure Recurse(X )
2:
X ← Compat(X )
5
3:
if P≥
X ∩]0, 1[ = ∅ then return end if
4:
if X is good then
5:
Add X to the list of good sets
6:
end if
7:
Let u ∈ R5 such that:
8:
9:
10:
11:
12:
u · (1, 1, 1, 1, 1) = 0
∀v ∈ X , u · v = 0 and
∀i ∈ {4, 5}, (mX )i = 0 ⇒ ui < 0
V ← {v ∈ N5 : v · u ≥ 0 and v · mX ≤ 2}
for every w ∈ V \ X do
Recurse(X ∪ {w })
end for
end procedure
Part 2/3: Computing all good sets
Michaël Rao
14 / 30
1: procedure Recurse(X )
2:
X ← Compat(X )
5
3:
if P≥
X ∩]0, 1[ = ∅ then return end if
4:
if X is good then
5:
Add X to the list of good sets
6:
end if
7:
Let u ∈ R5 such that:
8:
9:
10:
11:
12:
u · (1, 1, 1, 1, 1) = 0
∀v ∈ X , u · v = 0 and
∀i ∈ {4, 5}, (mX )i = 0 ⇒ ui < 0
V ← {v ∈ N5 : v · u ≥ 0 and v · mX ≤ 2}
for every w ∈ V \ X do
Recurse(X ∪ {w })
end for
end procedure
5
Recurse(X ) computes all max good sets Y ⊇ X with P≥
Y ∩]0, 1[ 6= ∅
Part 2/3: Computing all good sets
Michaël Rao
14 / 30
1: procedure Recurse(X )
2:
X ← Compat(X )
5
3:
if P≥
X ∩]0, 1[ = ∅ then return end if
4:
if X is good then
5:
Add X to the list of good sets
6:
end if
7:
Let u ∈ R5 such that:
8:
9:
10:
11:
12:
u · (1, 1, 1, 1, 1) = 0
∀v ∈ X , u · v = 0 and
∀i ∈ {4, 5}, (mX )i = 0 ⇒ ui < 0
V ← {v ∈ N5 : v · u ≥ 0 and v · mX ≤ 2}
for every w ∈ V \ X do
Recurse(X ∪ {w })
end for
end procedure
5
Recurse(X ) computes all max good sets Y ⊇ X with P≥
Y ∩]0, 1[ 6= ∅
Line 7: such a u always exists
Part 2/3: Computing all good sets
Michaël Rao
14 / 30
1: procedure Recurse(X )
2:
X ← Compat(X )
5
3:
if P≥
X ∩]0, 1[ = ∅ then return end if
4:
if X is good then
5:
Add X to the list of good sets
6:
end if
7:
Let u ∈ R5 such that:
8:
9:
10:
11:
12:
u · (1, 1, 1, 1, 1) = 0
∀v ∈ X , u · v = 0 and
∀i ∈ {4, 5}, (mX )i = 0 ⇒ ui < 0
V ← {v ∈ N5 : v · u ≥ 0 and v · mX ≤ 2}
for every w ∈ V \ X do
Recurse(X ∪ {w })
end for
end procedure
5
Recurse(X ) computes all max good sets Y ⊇ X with P≥
Y ∩]0, 1[ 6= ∅
Line 7: such a u always exists
Line 8: V is finite
Part 2/3: Computing all good sets
Michaël Rao
14 / 30
1: procedure Recurse(X )
2:
X ← Compat(X )
5
3:
if P≥
X ∩]0, 1[ = ∅ then return end if
4:
if X is good then
5:
Add X to the list of good sets
6:
end if
7:
Let u ∈ R5 such that:
8:
9:
10:
11:
12:
u · (1, 1, 1, 1, 1) = 0
∀v ∈ X , u · v = 0 and
∀i ∈ {4, 5}, (mX )i = 0 ⇒ ui < 0
V ← {v ∈ N5 : v · u ≥ 0 and v · mX ≤ 2}
for every w ∈ V \ X do
Recurse(X ∪ {w })
end for
end procedure
5
Recurse(X ) computes all max good sets Y ⊇ X with P≥
Y ∩]0, 1[ 6= ∅
Line 8: V is finite
Line 7: such a u always exists
5
Recurse always terminates: finitely many good sets with P≥
X ∩]0, 1[ 6= ∅.
Part 2/3: Computing all good sets
Michaël Rao
14 / 30
Good sets: results
We execute Recurse(∅) and it finds 193 non-empty sets.
Part 2/3: Computing all good sets
Michaël Rao
15 / 30
Good sets: results
We execute Recurse(∅) and it finds 193 non-empty sets.
5
193 non-empty maximal good sets X with P≥
X ∩]0, 1[ 6= ∅
Part 2/3: Computing all good sets
Michaël Rao
15 / 30
Good sets: results
We execute Recurse(∅) and it finds 193 non-empty sets.
5
193 non-empty maximal good sets X with P≥
X ∩]0, 1[ 6= ∅
Take all permutations: 3495 non-empty maximal good sets X with
PX ∩]0, 1[5 6= ∅
Part 2/3: Computing all good sets
Michaël Rao
15 / 30
Good sets: results
We execute Recurse(∅) and it finds 193 non-empty sets.
5
193 non-empty maximal good sets X with P≥
X ∩]0, 1[ 6= ∅
Take all permutations: 3495 non-empty maximal good sets X with
PX ∩]0, 1[5 6= ∅
Keep only one represents for each class up to rotation/mirror, one
have the 371 sets.
Part 2/3: Computing all good sets
Michaël Rao
15 / 30
Good sets: results
371 sets to consider:
2 s.t. PX has dimension 3
26 s.t. PX has dimension 2
92 s.t. PX has dimension 1
251 s.t. PX has dimension 0
Part 2/3: Computing all good sets
Michaël Rao
16 / 30
Good sets: results
371 sets to consider:
2 s.t. PX has dimension 3
26 s.t. PX has dimension 2
92 s.t. PX has dimension 1
251 s.t. PX has dimension 0
90 of “Type 1” (that is αi + αi+1 = 1)
Part 2/3: Computing all good sets
Michaël Rao
16 / 30
Part 3 : Testing a family corresponding to a good set
For each family in the 371 families of maximal good sets, we do an
exhaustive search of a tiling.
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
17 / 30
Part 3 : Testing a family corresponding to a good set
For each family in the 371 families of maximal good sets, we do an
exhaustive search of a tiling.
We chose maximal good set X . Let P = PX .
We do an exhaustive search of all tilings, allowing only corrected vector
types in X .
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
17 / 30
Part 3 : Testing a family corresponding to a good set
For each family in the 371 families of maximal good sets, we do an
exhaustive search of a tiling.
We chose maximal good set X . Let P = PX .
We do an exhaustive search of all tilings, allowing only corrected vector
types in X .
We backtrack if the conditions (angles and lengths) imply
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
17 / 30
Part 3 : Testing a family corresponding to a good set
For each family in the 371 families of maximal good sets, we do an
exhaustive search of a tiling.
We chose maximal good set X . Let P = PX .
We do an exhaustive search of all tilings, allowing only corrected vector
types in X .
We backtrack if the conditions (angles and lengths) imply
we are in a known case: known family (Types 1 to 15 in Table 1),
or a special case of a known family (Types 16 to 19)
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
17 / 30
Part 3 : Testing a family corresponding to a good set
For each family in the 371 families of maximal good sets, we do an
exhaustive search of a tiling.
We chose maximal good set X . Let P = PX .
We do an exhaustive search of all tilings, allowing only corrected vector
types in X .
We backtrack if the conditions (angles and lengths) imply
we are in a known case: known family (Types 1 to 15 in Table 1),
or a special case of a known family (Types 16 to 19)
or no convex pentagon exists with these conditions
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
17 / 30
Backtracking: general idea
The object on which we work and backtrack is a pair:
a planar graph which represent the partial tiling (“Tiling graph”)
a set of conditions we know on the lengths of the pentagon: a linear
program (LP) Q on `1 . . . `5
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
18 / 30
Tiling graph
Tiling graph : planar graph with labels on angles and edges
Two types of faces: normal and special
normal: corresponds to a pentagon in the tiling. The degree is 5, and
the angles are marked from 1 to 5 (in CW or CCW)
special: corresponds to frontier between tiles, or an unknown area of
the plane. Angles are marked with ∅, π or ?
A special face is complete if there no “?”
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
19 / 30
Tiling graph: example
1
2
3
w
w’ 3
5 s
1
s’
5
2
z
4
∅
∅
4 3
∅
5
t ∅ u
∅ y 1
u’
∅ 4
1 ∅
5
5
4
1
r
4
r’
2
∅
2 x 2
3
3
Example of a tiling graph (Type 15). Unmarked angles are labeled “?”
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
20 / 30
Invariants on the tiling graph
The (forthcoming) operations on the tiling graph keep the following
conditions:
the graph is planar
there is at most one ? angle around a vertex
there is exactly one non-complete special face
if a special face is complete, it has exactly two ∅ angles
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
21 / 30
Completing vertices and staying in X
For a vertex s, let vs ∈ N5 s.t. there are vi angles labeled i adjacent to s
For every vertex s:
If there is no v ∈ X such that vs ≤ v , then we backtrack
If vs ∈ X , then we complete the vertex s : if there is an angle ?
adjacent to s, then we relabel it into ∅
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
22 / 30
Length suppositions and completing faces
A run on a special face is a succession of consecutive ∅ and π angles.
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
23 / 30
Length suppositions and completing faces
A run on a special face is a succession of consecutive ∅ and π angles.
Each run corresponds to aligned points in the tiling.
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
23 / 30
Length suppositions and completing faces
A run on a special face is a succession of consecutive ∅ and π angles.
Each run corresponds to aligned points in the tiling.
If there is a pair of vertices (s, s 0 ) on a same run, and Q implies that s and
s 0 are the same point in the tiling, then we merge s and s 0
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
23 / 30
Length suppositions and completing faces
A run on a special face is a succession of consecutive ∅ and π angles.
Each run corresponds to aligned points in the tiling.
If there is a pair of vertices (s, s 0 ) on a same run, and Q implies that s and
s 0 are the same point in the tiling, then we merge s and s 0
If Q does not permit to decide among the following 3 possibilities :
s and s 0 are the same point in the tiling,
s is on the right of s 0 ,
s is on the left of s,
then we branch on the 3 possibilities: we add the corresponding condition
in Q and recurse
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
23 / 30
Branching on length suppositions: example
1
2
3
w
w’ 3
5 s
1
s’
5
2
z
∅
4
∅
4 3
∅
5
y
∅
t
u
∅
1
u’
∅ 4 5
1 ∅
5
4
1
r
4
r’
2
∅
2 x 2
3
3
Q:
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
2
3
w
w’ 3
5 s
1
s’
5
2
z
∅
4
∅
4 3
∅
5
y
∅
t
u
∅
1
u’
∅ 4 5
1 ∅
5
4
1
r
4
r’
2
∅
2 x 2
3
3
Q : l4 − l5 = 0
(y,z) is a complete face. So we (already) have l4 = l5 in the LP Q
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
2
3
w
w’ 3
5 s
1
s’
5
2
z
∅
4
∅
4 3
∅
5
y
∅
t
u
∅
1
u’
∅ 4 5
1 ∅
5
4
1
r
4
r’
2
∅
2 x 2
3
3
Q : l4 − l5 = 0
(y,z) is a complete face. So we (already) have l4 = l5 in the LP Q
(w,t,w’) is a run. the length wt is `3 , and the length tw 0 is `3 . so we
merge w and w 0 ,
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
2
3
w 3
5 s
1
s’
5
2
z
∅
4
∅
4 3
∅
5
y
∅
t
u
∅
1
u’
∅ 4 5
1 ∅
5
4
1
r
4
r’
2
∅
2 x 2
3
3
Q : l4 − l5 = 0
(y,z) is a complete face. So we (already) have l4 = l5 in the LP Q
(w,t,w’) is a run. the length wt is `3 , and the length tw 0 is `3 . so we
merge w and w 0 ,
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
2
3
w 3
5 s
1
s’
5
2
z
∅
4
∅
4 3
∅
5
y
∅
t
u
∅
1
u’
∅ 4 5
1 ∅
5
4
1
r
4
r’
2
∅
2 x 2
3
3
Q : l4 − l5 = 0
(y,z) is a complete face. So we (already) have l4 = l5 in the LP Q
(w,t,w’) is a run. the length wt is `3 , and the length tw 0 is `3 . so we
merge w and w 0 , and mark angle (t, w , t) as ∅
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
2
3
∅
w 3
5 s
1
s’
5
2
z
∅
4
∅
4 3
∅
5
y
∅
t
u
∅
1
u’
∅ 4 5
1 ∅
5
4
1
r
4
r’
2
∅
2 x 2
3
3
Q : l4 − l5 = 0
(y,z) is a complete face. So we (already) have l4 = l5 in the LP Q
(w,t,w’) is a run. the length wt is `3 , and the length tw 0 is `3 . so we
merge w and w 0 , and mark angle (t, w , t) as ∅
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
2
3
∅
w 3
5 s
1
s’
5
2
z
∅
4
∅
4 3
∅
5
y
∅
t
u
∅
1
u’
∅ 4 5
1 ∅
5
4
1
r
4
r’
2
∅
2 x 2
3
3
Q : l4 − l5 = 0
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
2
3
∅
w 3
5 s
1
s’
5
2
z
∅
4
∅
4 3
∅
5
y
∅
t
u
∅
1
u’
∅ 4 5
1 ∅
5
4
1
r
4
r’
2
∅
2 x 2
3
3
Q : l4 − l5 = 0
(u, t, y , u 0 ) is also a run.
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
2
3
∅
w 3
5 s
1
s’
5
2
z
∅
4
∅
4 3
∅
5
y
∅
t
u
∅
1
u’
∅ 4 5
1 ∅
5
4
1
r
4
r’
2
∅
2 x 2
3
3
Q : l4 − l5 = 0
(u, t, y , u 0 ) is also a run.
Is u and u 0 the same vertex ?
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
2
3
∅
w 3
5 s
1
s’
5
2
z
∅
4
∅
4 3
∅
5
y
∅
t
u
∅
1
u’
∅ 4 5
1 ∅
5
4
1
r
4
r’
2
∅
2 x 2
3
3
Q : l4 − l5 = 0
(u, t, y , u 0 ) is also a run.
Is u and u 0 the same vertex ? Is `3 = `4 + `5 ?
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
2
3
∅
w 3
5 s
1
s’
5
2
z
∅
4
∅
4 3
∅
5
y
∅
t
u
∅
1
u’
∅ 4 5
1 ∅
5
4
1
r
4
r’
2
∅
2 x 2
3
3
Q : l4 − l5 = 0
(u, t, y , u 0 ) is also a run.
Is u and u 0 the same vertex ? Is `3 = `4 + `5 ?
We don’t know. We branch.
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
2
3
∅
w 3
5 s
1
s’
5
2
z
∅
4
∅
4 3
∅
5
y
∅
t
u
∅
1
u’
∅ 4 5
1 ∅
5
4
1
r
4
r’
2
∅
2 x 2
3
3
Q : l4 − l5 = 0
(u, t, y , u 0 ) is also a run.
Is u and u 0 the same vertex ? Is `3 = `4 + `5 ?
We don’t know. We branch.
first case : add `3 > `4 + `5 to Q and branch
second case : add `3 = `4 + `5 to Q and branch
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
2
3
∅
w 3
5 s
1
s’
5
2
z
∅
4
∅
4 3
∅
5
y
∅
t
u
∅
1
u’
∅ 4 5
1 ∅
5
4
1
r
4
r’
2
∅
2 x 2
3
3
(case 2) Q : l4 − l5 = 0, `3 − `4 − `5 = 0
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
2
3
∅
w 3
5 s
1
s’
5
2
z
∅
4
∅
4 3
∅
5
y
∅
t
u
∅
1
u’
∅ 4 5
1 ∅
5
4
1
r
4
r’
2
∅
2 x 2
3
3
(case 2) Q : l4 − l5 = 0, `3 − `4 − `5 = 0
(u, t, y , u 0 ) is a run, and we know that u and u 0 have the same position:
we merge u and u 0 ,
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
5 s
1
s’
5
z
4
∅
∅
5
y
∅
1
1 ∅
2
4 ∅3
t ∅
∅ 4
2
3
∅
w 3
u
5
1
2
r
5
r’
4
4
∅
2 x 2
3
3
(case 2) Q : l4 − l5 = 0, `3 − `4 − `5 = 0
(u, t, y , u 0 ) is a run, and we know that u and u 0 have the same position:
we merge u and u 0 ,
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
5 s
1
s’
5
z
4
∅
∅
5
y
∅
1
1 ∅
2
4 ∅3
t ∅
∅ 4
2
3
∅
w 3
π
u
5
1
2
r
5
r’
4
4
∅
2 x 2
3
3
(case 2) Q : l4 − l5 = 0, `3 − `4 − `5 = 0
(u, t, y , u 0 ) is a run, and we know that u and u 0 have the same position:
we merge u and u 0 , and the angle (t, u, y ) is labeled π.
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
5 s
1
s’
5
z
4
∅
∅
5
y
∅
1
1 ∅
2
4 ∅3
t ∅
∅ 4
2
3
∅
w 3
π
u
1
2
5∅ 5
r
r’
4
4
∅
2 x 2
3
3
(case 2) Q : l4 − l5 = 0, `3 − `4 − `5 = 0
(u, t, y , u 0 ) is a run, and we know that u and u 0 have the same position:
we merge u and u 0 , and the angle (t, u, y ) is labeled π.
u is now complete: the angle r , u, r 0 is labeled ∅
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on length suppositions: example
1
5 s
1
s’
2
4 ∅3
t ∅
∅ 4
2
3
∅
w 3
2
π
u
5∅ 5
1 ∅ 4
r
5
z
4
∅
∅
5
y
∅
1
1 ∅
4
∅
2 x 2
3
3
(case 2) Q : l4 − l5 = 0, `3 − `4 − `5 = 0
(u, t, y , u 0 ) is a run, and we know that u and u 0 have the same position:
we merge u and u 0 , and the angle (t, u, y ) is labeled π.
u is now complete: the angle r , u, r 0 is labeled ∅
in the run (r , u, r 0 ), r and r 0 have the same position: we merge...
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
24 / 30
Branching on a new tile
If we are not in any case of completing or branching on length supposition,
then we add a new normal face to the tiling graph.
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
25 / 30
Branching on a new tile
If we are not in any case of completing or branching on length supposition,
then we add a new normal face to the tiling graph.
We take a non-complete vertex w in the graph. We known that, if the
tiling graph corresponds to a sub-tiling T 0 of a tiling T by P, there is a
tile P ∈ T \ T 0 such that w is a vertex of P, and P shares a line segment
with T 0 .
Then we branch on on all theses possibilities of face addition.
(warning : be careful with “half” vertices)
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
25 / 30
Existence of the pentagon
Given the LP Q, we denote by Q the set of solutions
of Q with
P`i−1
Let s(α) be the vector such that s(α)i = (i − 1) − j=1 αi .
One have:
X
√
`i exp(s(α)i × π × −1) = 0.
P
`=1
(1)
i
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
26 / 30
Existence of the pentagon
Given the LP Q, we denote by Q the set of solutions
of Q with
P`i−1
Let s(α) be the vector such that s(α)i = (i − 1) − j=1 αi .
One have:
X
√
`i exp(s(α)i × π × −1) = 0.
P
`=1
(1)
i
We backtrack if there is no convex pentagon exists with the properties,
that is if the following condition is not fulfilled:
∃` ∈ Q∩]0, 1[5 , ∃α ∈ P∩]0, 1[5 ,
X
`i exp(s(α)i × π ×
√
−1) = 0
(2)
Michaël Rao
26 / 30
i
Part 3/3: Testing a family corresponding to a good set
Existence of the pentagon
Given the LP Q, we denote by Q the set of solutions
of Q with
P`i−1
Let s(α) be the vector such that s(α)i = (i − 1) − j=1 αi .
One have:
X
√
`i exp(s(α)i × π × −1) = 0.
P
`=1
(1)
i
We backtrack if there is no convex pentagon exists with the properties,
that is if the following condition is not fulfilled:
∃` ∈ Q∩]0, 1[5 , ∃α ∈ P∩]0, 1[5 ,
X
`i exp(s(α)i × π ×
√
−1) = 0
(2)
i
If dim(P) = 0 then α ∈ Q5 , and easy to decide: we compute on
Q[cos (π/q)] for a q ∈ N.
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
26 / 30
Existence of the pentagon
Given the LP Q, we denote by Q the set of solutions
of Q with
P`i−1
Let s(α) be the vector such that s(α)i = (i − 1) − j=1 αi .
One have:
X
√
`i exp(s(α)i × π × −1) = 0.
P
`=1
(1)
i
We backtrack if there is no convex pentagon exists with the properties,
that is if the following condition is not fulfilled:
∃` ∈ Q∩]0, 1[5 , ∃α ∈ P∩]0, 1[5 ,
X
`i exp(s(α)i × π ×
√
−1) = 0
(2)
i
If dim(P) = 0 then α ∈ Q5 , and easy to decide: we compute on
Q[cos (π/q)] for a q ∈ N.
If dim(P) > 0 : we backtrack if we have a certificate (computations in Q)
that there a no solution. Problem: this cannot detect “degenerate case”.
So we manually add some degenerate case. (Types 20 to 24 in Table 1).
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
26 / 30
Conditions for which we backtrack (Table 1)
Type 1
(i=1)
Type 3
(i=31)
Type 5
(i=4)
Type 7
(i=17)
Type 9
(i=15)
Type 11
(i=67)
Type 13
(i=63)
Type 15
(i=303)
Type 17
(i=25)
⊂ T2
Type 19
(i=23)
⊂ T1
Type 21
(i=12)
degen.
Type 23
(i=64)
degen.
a + b + c = 2π
Type 2
(i=2)
Type 4
(i=6)
a + b + d = 2π
C =E
a + b + d = 2π
2e = π
D =E
B =C
Type 6
(i=13)
Type 8
(i=14)
Type 10
(i=69)
d + 2e = 2π
a + c + d = 2π
d + 2e = 2π
2b + c = 2π
2c + d = 2π
b + c + e = 2π
a + 2b = 2π
c + 2d = 2π
b + d + e = 2π
a + 2b = 2π
c + 2d = 2π
b + d + e = 2π
a + 2b = 2π
b + c + e = 2π
2b + d = 2π
a + 2c = 2π
C =D =E
A=B
A=B =C =D
d + 2e = 2π
c + 2e = 2π
b + d + e = 2π
a + b + d = 2π
D =E
A=B
3e = 2π
d + 2e = 2π
b + 2e = 2π
3e = 2π
a + b + d = 2π
d + 2e = 2π
a + 2c = 2π
d + 2e = 2π
2a + c = 2π
C +E =D
A=B
c + 2d = 2π
b + d + e = 2π
a + 2b = 2π
b + 2d = 2π
a + b + d = 2π
2e = π
c + 2d = 2π
2b + e = 2π
2a + d = 2π
2e = π
c + 2e = 2π
2b + d = 2π
A = B = C + 2E
Type 12
(i=67)
A = 2B = 2C
Type 14
(i=67)
B =D =E
C = 2B
Type 16
(i=72)
⊂ T10
A=B =C =D =E
c + 2e = 2π
b + 2d = 2π
A=B =C =D
d + 2e = 2π
2a + b = 2π
A=B
C =D
2b + d = 2π
a + b + d = 2π
2e = π
A = 2C = 2D
Type 18
(i=73)
⊂ T2
Type 20
(i=2)
degen.
Type 22
(i=27)
degen.
Type 24
(i=69)
degen.
D =E
B =C
A=C =D =E
A=B =C =D
Part 3/3: Testing a family corresponding to a good set
A+C =D =E
A + C = B = 2E
A = B = 2C = 2E
2A = D = E
A=C
A=C +D
B =E
c + 2e = 2π
a + 2d = 2π
A=B =C =E
2c + d = 2π
b + c + e = 2π
a + 2b = 2π
2D = A + C
2E = A + C
Michaël Rao
27 / 30
Part 3: Results
For every family, the exhaustive search is finite
That is: if a pentagon does not respect condition of Type i for a
i ∈ {1, . . . 24}, then it cannot tile the plane.
Types 1 to 15 are the already known families
Types 16 to 19 are special cases of known families
Types 20 to 24 are “degenerate” (dim(P) > 0): there are no convex
pentagon which respects these conditions
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
28 / 30
Future / TODO
Recheck the proof
Recheck the code (∼5000 lines in C++)
And/or reproduce the exhaustive search
Formal proof ? (Coq or similar software)
Simplifications:
More direct proof for the positive density ?
Direct proof for the finiteness of goods sets ?
Part 3/3: Testing a family corresponding to a good set
Michaël Rao
29 / 30
Thanks !