On the mixing time of the flip walk on
triangulations of the sphere
Thomas Budzinski
ENS Paris
Journées ALÉA
23 Mars 2017
Thomas Budzinski
Flips on triangulations of the sphere
Planar maps
Definitions
A planar map is a finite, connected graph embedded in the
sphere in such a way that no two edges cross (except at a
common endpoint), considered up to orientation-preserving
homeomorphism.
A planar map is a rooted type-I triangulation if all its faces
have degree 3 and it has a distinguished oriented edge. It may
contain multiple edges and loops.
Thomas Budzinski
Flips on triangulations of the sphere
Planar maps
Definitions
A planar map is a finite, connected graph embedded in the
sphere in such a way that no two edges cross (except at a
common endpoint), considered up to orientation-preserving
homeomorphism.
A planar map is a rooted type-I triangulation if all its faces
have degree 3 and it has a distinguished oriented edge. It may
contain multiple edges and loops.
=
Thomas Budzinski
Flips on triangulations of the sphere
Random planar maps in a nutshell
Let Tn be the set of rooted type-I triangulations of the sphere with
n vertices, and Tn (∞) be a uniform variable on Tn . Geometric
properties of Tn (∞) for n large ?
Thomas Budzinski
Flips on triangulations of the sphere
Random planar maps in a nutshell
Let Tn be the set of rooted type-I triangulations of the sphere with
n vertices, and Tn (∞) be a uniform variable on Tn . Geometric
properties of Tn (∞) for n large ?
Exact enumeration results [Tutte],
Thomas Budzinski
Flips on triangulations of the sphere
Random planar maps in a nutshell
Let Tn be the set of rooted type-I triangulations of the sphere with
n vertices, and Tn (∞) be a uniform variable on Tn . Geometric
properties of Tn (∞) for n large ?
Exact enumeration results [Tutte],
the distances in Tn (∞) are of order n1/4
[≈ Chassaing–Schaeffer],
Thomas Budzinski
Flips on triangulations of the sphere
Random planar maps in a nutshell
Let Tn be the set of rooted type-I triangulations of the sphere with
n vertices, and Tn (∞) be a uniform variable on Tn . Geometric
properties of Tn (∞) for n large ?
Exact enumeration results [Tutte],
the distances in Tn (∞) are of order n1/4
[≈ Chassaing–Schaeffer],
when the distances are renormalized, Tn (∞) to a continuum
random metric space called the Brownian map [Le Gall],
Thomas Budzinski
Flips on triangulations of the sphere
Random planar maps in a nutshell
Let Tn be the set of rooted type-I triangulations of the sphere with
n vertices, and Tn (∞) be a uniform variable on Tn . Geometric
properties of Tn (∞) for n large ?
Exact enumeration results [Tutte],
the distances in Tn (∞) are of order n1/4
[≈ Chassaing–Schaeffer],
when the distances are renormalized, Tn (∞) to a continuum
random metric space called the Brownian map [Le Gall],
the Brownian map is homeomorphic to the sphere
[Le Gall–Paulin].
Thomas Budzinski
Flips on triangulations of the sphere
A uniform triangulation of the sphere with 10 000 vertices
Thomas Budzinski
Flips on triangulations of the sphere
How to sample a large uniform triangulation ?
"Modern" tools : bijections with trees, peeling process.
Back in the 80’s : Monte Carlo methods : we look for a Markov
chain on Tn for which the uniform measure is stationary.
A simple local operation on triangulations : flips.
Thomas Budzinski
Flips on triangulations of the sphere
How to sample a large uniform triangulation ?
"Modern" tools : bijections with trees, peeling process.
Back in the 80’s : Monte Carlo methods : we look for a Markov
chain on Tn for which the uniform measure is stationary.
A simple local operation on triangulations : flips.
t
Thomas Budzinski
Flips on triangulations of the sphere
How to sample a large uniform triangulation ?
"Modern" tools : bijections with trees, peeling process.
Back in the 80’s : Monte Carlo methods : we look for a Markov
chain on Tn for which the uniform measure is stationary.
A simple local operation on triangulations : flips.
e1
t
Thomas Budzinski
Flips on triangulations of the sphere
How to sample a large uniform triangulation ?
"Modern" tools : bijections with trees, peeling process.
Back in the 80’s : Monte Carlo methods : we look for a Markov
chain on Tn for which the uniform measure is stationary.
A simple local operation on triangulations : flips.
Thomas Budzinski
Flips on triangulations of the sphere
How to sample a large uniform triangulation ?
"Modern" tools : bijections with trees, peeling process.
Back in the 80’s : Monte Carlo methods : we look for a Markov
chain on Tn for which the uniform measure is stationary.
A simple local operation on triangulations : flips.
flip(t, e1 )
Thomas Budzinski
Flips on triangulations of the sphere
How to sample a large uniform triangulation ?
"Modern" tools : bijections with trees, peeling process.
Back in the 80’s : Monte Carlo methods : we look for a Markov
chain on Tn for which the uniform measure is stationary.
A simple local operation on triangulations : flips.
e2
t
Thomas Budzinski
Flips on triangulations of the sphere
How to sample a large uniform triangulation ?
"Modern" tools : bijections with trees, peeling process.
Back in the 80’s : Monte Carlo methods : we look for a Markov
chain on Tn for which the uniform measure is stationary.
A simple local operation on triangulations : flips.
???
Thomas Budzinski
Flips on triangulations of the sphere
How to sample a large uniform triangulation ?
"Modern" tools : bijections with trees, peeling process.
Back in the 80’s : Monte Carlo methods : we look for a Markov
chain on Tn for which the uniform measure is stationary.
A simple local operation on triangulations : flips.
e2
flip(t, e2 ) = t
Thomas Budzinski
Flips on triangulations of the sphere
How to sample a large uniform triangulation ?
"Modern" tools : bijections with trees, peeling process.
Back in the 80’s : Monte Carlo methods : we look for a Markov
chain on Tn for which the uniform measure is stationary.
A simple local operation on triangulations : flips.
Thomas Budzinski
Flips on triangulations of the sphere
How to sample a large uniform triangulation ?
"Modern" tools : bijections with trees, peeling process.
Back in the 80’s : Monte Carlo methods : we look for a Markov
chain on Tn for which the uniform measure is stationary.
A simple local operation on triangulations : flips.
Thomas Budzinski
Flips on triangulations of the sphere
How to sample a large uniform triangulation ?
"Modern" tools : bijections with trees, peeling process.
Back in the 80’s : Monte Carlo methods : we look for a Markov
chain on Tn for which the uniform measure is stationary.
A simple local operation on triangulations : flips.
Thomas Budzinski
Flips on triangulations of the sphere
A Markov chain on Tn
We fix t0 ∈ Tn and take Tn (0) = t0 .
Conditionally on (Tn (k))0≤i≤k , let ek be a uniform edge of
Tn (k) and Tn (k + 1) = flip (Tn (k), ek ).
Thomas Budzinski
Flips on triangulations of the sphere
A Markov chain on Tn
We fix t0 ∈ Tn and take Tn (0) = t0 .
Conditionally on (Tn (k))0≤i≤k , let ek be a uniform edge of
Tn (k) and Tn (k + 1) = flip (Tn (k), ek ).
The uniform measure on Tn is reversible for Tn , thus
stationary.
Thomas Budzinski
Flips on triangulations of the sphere
A Markov chain on Tn
We fix t0 ∈ Tn and take Tn (0) = t0 .
Conditionally on (Tn (k))0≤i≤k , let ek be a uniform edge of
Tn (k) and Tn (k + 1) = flip (Tn (k), ek ).
The uniform measure on Tn is reversible for Tn , thus
stationary.
The chain Tn is irreducible (the flip graph is connected
[Wagner 36]) and aperiodic (non flippable edges), so it
converges to the uniform measure.
Thomas Budzinski
Flips on triangulations of the sphere
A Markov chain on Tn
We fix t0 ∈ Tn and take Tn (0) = t0 .
Conditionally on (Tn (k))0≤i≤k , let ek be a uniform edge of
Tn (k) and Tn (k + 1) = flip (Tn (k), ek ).
The uniform measure on Tn is reversible for Tn , thus
stationary.
The chain Tn is irreducible (the flip graph is connected
[Wagner 36]) and aperiodic (non flippable edges), so it
converges to the uniform measure.
Question : how quick is the convergence ?
Thomas Budzinski
Flips on triangulations of the sphere
Mixing time of Tn
For n ≥ 3 and 0 < ε < 1 we define the mixing time tmix (ε, n)
as the smallest k such that
max max |P (Tn (k) ∈ A) − P (Tn (∞) ∈ A)| ≤ ε,
t0 ∈Tn A⊂Tn
where we recall that Tn (∞) is uniform on Tn .
Thomas Budzinski
Flips on triangulations of the sphere
Mixing time of Tn
For n ≥ 3 and 0 < ε < 1 we define the mixing time tmix (ε, n)
as the smallest k such that
max max |P (Tn (k) ∈ A) − P (Tn (∞) ∈ A)| ≤ ε,
t0 ∈Tn A⊂Tn
where we recall that Tn (∞) is uniform on Tn .
Theorem (B., 2016)
For all 0 < ε < 1, there is a constant c > 0 such that
tmix (ε, n) ≥ cn5/4 .
Thomas Budzinski
Flips on triangulations of the sphere
Sketch of proof
We will be interested in the existence of small separating cycles.
Thomas Budzinski
Flips on triangulations of the sphere
Sketch of proof
We will be interested in the existence of small separating cycles.
Theorem (≈ Le Gall–Paulin, 2008)
Let `n = o(n1/4 ). Then, with probability going to 1 as n → +∞,
there is no cycle in Tn (∞) of length at most `n that separates
Tn (∞) in two parts, each of which contains at least n4 vertices.
Thomas Budzinski
Flips on triangulations of the sphere
Sketch of proof
We will be interested in the existence of small separating cycles.
Theorem (≈ Le Gall–Paulin, 2008)
Let `n = o(n1/4 ). Then, with probability going to 1 as n → +∞,
there is no cycle in Tn (∞) of length at most `n that separates
Tn (∞) in two parts, each of which contains at least n4 vertices.
Let Tn1 (0) and Tn2 (0) be two independent uniform triangulations of
a 1-gon with n2 inner vertices each, and Tn (0) the gluing of Tn1 (0)
and Tn2 (0) along their boundary.
Thomas Budzinski
Flips on triangulations of the sphere
Sketch of proof
We will be interested in the existence of small separating cycles.
Theorem (≈ Le Gall–Paulin, 2008)
Let `n = o(n1/4 ). Then, with probability going to 1 as n → +∞,
there is no cycle in Tn (∞) of length at most `n that separates
Tn (∞) in two parts, each of which contains at least n4 vertices.
Let Tn1 (0) and Tn2 (0) be two independent uniform triangulations of
a 1-gon with n2 inner vertices each, and Tn (0) the gluing of Tn1 (0)
and Tn2 (0) along their boundary. It is enough to prove
Proposition
Let kn = o(n5/4 ). There is a cycle γ in Tn (kn ) of length o(n1/4 ) in
probability that separates Tn (kn ) in two parts, each of which
contains at least n4 vertices.
Thomas Budzinski
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (0)
Perimeter :
Pen (0) = 1
Explored volume :
Ven (0) = 1
Tn2 (0)
Thomas Budzinski
exploration steps :
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (0)
Perimeter :
Pen (0) = 1
Explored volume :
Ven (0) = 1
e0
Tn2 (0)
Thomas Budzinski
exploration steps :
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (1)
Perimeter :
Pen (1) = 1
Explored volume :
Ven (1) = 1
Tn2 (1)
Thomas Budzinski
exploration steps :
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (1)
e1
Tn2 (1)
Thomas Budzinski
Perimeter :
Pen (1) = 1
Explored volume :
Ven (1) = 1
exploration steps :
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (1)
e1
Tn2 (1)
Thomas Budzinski
Perimeter :
Pen (1) = 1
Explored volume :
Ven (1) = 1
exploration steps :
1
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (2)
Perimeter :
Pen (2) = 2
Explored volume :
Ven (2) = 2
Tn2 (2)
Thomas Budzinski
exploration steps :
1
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (2)
Perimeter :
Pen (2) = 2
Explored volume :
Ven (2) = 2
Tn2 (2)
exploration steps :
1
e2
Thomas Budzinski
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (3)
Perimeter :
Pen (3) = 2
Explored volume :
Ven (3) = 2
Tn2 (3)
Thomas Budzinski
exploration steps :
1
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (3)
e3
Tn2 (3)
Thomas Budzinski
Perimeter :
Pen (3) = 2
Explored volume :
Ven (3) = 2
exploration steps :
1
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (3)
e3
Tn2 (3)
Thomas Budzinski
Perimeter :
Pen (3) = 2
Explored volume :
Ven (3) = 2
exploration steps :
1
3
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (4)
Perimeter :
Pen (4) = 3
Explored volume :
Ven (4) = 3
Tn2 (4)
Thomas Budzinski
exploration steps :
1
3
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (4)
e4
Tn2 (4)
Thomas Budzinski
Perimeter :
Pen (4) = 3
Explored volume :
Ven (4) = 3
exploration steps :
1
3
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (5)
Perimeter :
Pen (5) = 3
Explored volume :
Ven (5) = 3
Tn2 (5)
Thomas Budzinski
exploration steps :
1
3
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (5)
Perimeter :
Pen (5) = 3
Explored volume :
Ven (5) = 3
Tn2 (5)
exploration steps :
1
3
e5
Thomas Budzinski
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (5)
Perimeter :
Pen (5) = 3
Explored volume :
Ven (5) = 3
Tn2 (5)
exploration steps :
1
3
5
e5
Thomas Budzinski
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (6)
Perimeter :
Pen (6) = 4
Explored volume :
Ven (6) = 4
Tn2 (6)
Thomas Budzinski
exploration steps :
1
3
5
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (6)
Perimeter :
Pen (6) = 4
Explored volume :
Ven (6) = 4
Tn2 (6)
e6
Thomas Budzinski
exploration steps :
1
3
5
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (7)
Perimeter :
Pen (7) = 4
Explored volume :
Ven (7) = 4
Tn2 (7)
Thomas Budzinski
exploration steps :
1
3
5
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (7)
Perimeter :
Pen (7) = 4
Explored volume :
Ven (7) = 4
e7
Tn2 (7)
Thomas Budzinski
exploration steps :
1
3
5
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (7)
Perimeter :
Pen (7) = 4
Explored volume :
Ven (7) = 4
e7
Tn2 (7)
Thomas Budzinski
exploration steps :
1
3
5
7
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (8)
Perimeter :
Pen (8) = 5
Explored volume :
Ven (8) = 5
Tn2 (8)
Thomas Budzinski
exploration steps :
1
3
5
7
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (8)
Perimeter :
Pen (8) = 5
Explored volume :
Ven (8) = 5
Tn2 (8)
e8
Thomas Budzinski
exploration steps :
1
3
5
7
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (8)
Perimeter :
Pen (8) = 5
Explored volume :
Ven (8) = 5
Tn2 (8)
e8
Thomas Budzinski
exploration steps :
1
3
5
7
8
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (8)
Perimeter :
Pen (8) = 5
Explored volume :
Ven (8) = 5
Tn2 (8)
e8
Thomas Budzinski
exploration steps :
1
3
5
7
8
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (9)
Perimeter :
Pen (9) = 4
Explored volume :
Ven (9) = 6
Tn2 (9)
Thomas Budzinski
exploration steps :
1
3
5
7
8
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (9)
Perimeter :
Pen (9) = 4
Explored volume :
Ven (9) = 6
Tn2 (9)
e9
Thomas Budzinski
exploration steps :
1
3
5
7
8
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (10)
Perimeter :
Pen (10) = 4
Explored volume :
Ven (10) = 6
Tn2 (10)
Thomas Budzinski
e10
exploration steps :
1
3
5
7
8
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (11)
Perimeter :
Pen (11) = 4
Explored volume :
Ven (11) = 6
Tn2 (11)
Thomas Budzinski
exploration steps :
1
3
5
7
8
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (11)
Perimeter :
Pen (11) = 4
Explored volume :
Ven (11) = 6
Tn2 (11)
e11
Thomas Budzinski
exploration steps :
1
3
5
7
8
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (11)
Perimeter :
Pen (11) = 4
Explored volume :
Ven (11) = 6
Tn2 (11)
e11
Thomas Budzinski
exploration steps :
1
3
5
7
8
11
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (12)
Perimeter :
Pen (12) = 5
Explored volume :
Ven (12) = 7
Tn2 (12)
Thomas Budzinski
exploration steps :
1
3
5
7
8
11
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (12)
Perimeter :
Pen (12) = 5
Explored volume :
Ven (12) = 7
e12
Tn2 (12)
Thomas Budzinski
exploration steps :
1
3
5
7
8
11
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (12)
Perimeter :
Pen (12) = 5
Explored volume :
Ven (12) = 7
e12
Tn2 (12)
Thomas Budzinski
exploration steps :
1
3
5
7
8
11
12
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (12)
Perimeter :
Pen (12) = 5
Explored volume :
Ven (12) = 7
e12
Tn2 (12)
Thomas Budzinski
exploration steps :
1
3
5
7
8
11
12
Flips on triangulations of the sphere
Exploration of Tn (k)
Tn1 (13)
Perimeter :
Pen (13) = 3
Explored volume :
Ven (13) = 7
Tn2 (13)
Thomas Budzinski
exploration steps :
1
3
5
7
8
11
12
Flips on triangulations of the sphere
Peeling estimates
Claim
For all k ≥ 0, conditionally on (Tn1 (i))0≤i≤k , the triangulation
Tn2 (k) is a uniform triangulation with a boundary of length
|∂Tn1 (k)| and n − |Tn1 (k)| inner vertices.
Thomas Budzinski
Flips on triangulations of the sphere
Peeling estimates
Claim
For all k ≥ 0, conditionally on (Tn1 (i))0≤i≤k , the triangulation
Tn2 (k) is a uniform triangulation with a boundary of length
|∂Tn1 (k)| and n − |Tn1 (k)| inner vertices.
Consequence : the perimeter and volume of the red region
have the same transitions as for a fixed uniform triangulation.
Thomas Budzinski
Flips on triangulations of the sphere
Peeling estimates
Claim
For all k ≥ 0, conditionally on (Tn1 (i))0≤i≤k , the triangulation
Tn2 (k) is a uniform triangulation with a boundary of length
|∂Tn1 (k)| and n − |Tn1 (k)| inner vertices.
Consequence : the perimeter and volume of the red region
have the same transitions as for a fixed uniform triangulation.
We write τj for the times at which an exploration step is
performed. Let Pn (j) = Pen (τj ) and Vn (j) = Ven (τj ).
Thomas Budzinski
Flips on triangulations of the sphere
Peeling estimates
Claim
For all k ≥ 0, conditionally on (Tn1 (i))0≤i≤k , the triangulation
Tn2 (k) is a uniform triangulation with a boundary of length
|∂Tn1 (k)| and n − |Tn1 (k)| inner vertices.
Consequence : the perimeter and volume of the red region
have the same transitions as for a fixed uniform triangulation.
We write τj for the times at which an exploration step is
performed. Let Pn (j) = Pen (τj ) and Vn (j) = Ven (τj ).
We have Pn (j) ≈ j 2/3 and Vn (j) ≈ j 4/3 as long as j n3/4
[Curien–Le Gall].
Thomas Budzinski
Flips on triangulations of the sphere
Time change estimates
Conditionally on (Pn , Vn ), the τi+1 − τi are independent and
Pn (i)
geometric with parameters 3n−6
, so for ε > 0 small,
E [τεn3/4 |Pn ] =
3/4
εn
X
i=1
n × εn3/4
3n − 6
√
>
= εn5/4 ,
Pn (i)
n
so after kn = o(n5/4 ) flips, the number of exploration steps
performed is o(n3/4 ).
Thomas Budzinski
Flips on triangulations of the sphere
Time change estimates
Conditionally on (Pn , Vn ), the τi+1 − τi are independent and
Pn (i)
geometric with parameters 3n−6
, so for ε > 0 small,
E [τεn3/4 |Pn ] =
3/4
εn
X
i=1
n × εn3/4
3n − 6
√
>
= εn5/4 ,
Pn (i)
n
so after kn = o(n5/4 ) flips, the number of exploration steps
performed is o(n3/4 ). Hence,
√
Pen (kn ) = Pn (o(n3/4 )) = o( n),
Ven (kn ) = Vn (o(n3/4 )) = o(n).
Thomas Budzinski
Flips on triangulations of the sphere
Time change estimates
Conditionally on (Pn , Vn ), the τi+1 − τi are independent and
Pn (i)
geometric with parameters 3n−6
, so for ε > 0 small,
E [τεn3/4 |Pn ] =
3/4
εn
X
i=1
n × εn3/4
3n − 6
√
>
= εn5/4 ,
Pn (i)
n
so after kn = o(n5/4 ) flips, the number of exploration steps
performed is o(n3/4 ). Hence,
√
Pen (kn ) = Pn (o(n3/4 )) = o( n),
Ven (kn ) = Vn (o(n3/4 )) = o(n).
q
We can find a separating cycle of length Pen (kn ) = o(n1/4 ) in
Tn2 (kn ) [Krikun].
Thomas Budzinski
Flips on triangulations of the sphere
Is the lower bound sharp ?
Back-of-the-enveloppe computation :
in a typical triangulation, the distance between two typical
vertices x and y is ≈ n1/4 .
The probability that a flip hits a geodesic is ≈ n−3/4 .
The distance between x and y changes ≈ kn−3/4 times before
time k.
If d(x,
√ y ) evolves roughly like a random walk, it varies of
≈ kn−3/4 = n1/4 for k = n5/4 .
Thomas Budzinski
Flips on triangulations of the sphere
Is the lower bound sharp ?
Back-of-the-enveloppe computation :
in a typical triangulation, the distance between two typical
vertices x and y is ≈ n1/4 .
The probability that a flip hits a geodesic is ≈ n−3/4 .
The distance between x and y changes ≈ kn−3/4 times before
time k.
If d(x,
√ y ) evolves roughly like a random walk, it varies of
≈ kn−3/4 = n1/4 for k = n5/4 .
For triangulations of a convex polygon (no inner vertices), the
lower bound n3/2 is believed to be sharp but the best known
upper bound is n5 [McShine–Tetali].
Prove that the mixing time is polynomial ?
Thomas Budzinski
Flips on triangulations of the sphere
MERCI !
Thomas Budzinski
Flips on triangulations of the sphere