A New Algorithm for Sampling Closed
Equilateral Random Walks
Jason Cantarella, Clayton Shonkwiler, and Erica Uehara
University of Georgia, Colorado State University, and Ochanomizu University
Applied Math Seminar
November 13, 2014
Random Walks (and Polymer Physics)
Physics Question
What is the average shape of a polymer in solution or in melt?
Protonated P2VP
Roiter/Minko
Clarkson University
Plasmid DNA
Alonso-Sarduy, Dietler Lab
EPF Lausanne
Random Walks
Let Arm(n) be the moduli space of random walks in R3
consisting of n unit-length steps up to translation.
Then Arm(n) ∼
= S 2 (1) × . . . × S 2 (1).
{z
}
|
n
Random Walks
Let Arm(n) be the moduli space of random walks in R3
consisting of n unit-length steps up to translation.
Then Arm(n) ∼
= S 2 (1) × . . . × S 2 (1).
{z
}
|
n
Let Pol(n) ⊂ Arm(n) be the submanifold of closed random
walks (or random polygons); i.e., those walks which satisfy
n
X
i=1
~ei = ~0.
A Random Walk with 3,500 Steps
A Closed Random Walk with 3,500 Steps
Main Problem
Main Question
How can we sample closed equilateral random walks according
to the codimension-3 Hausdorff measure on Pol(n) ⊂ Arm(n)?
Motivating Question
How do we understand Pol(n)? What is this space?
Point of Talk
New direct sampling algorithms backed by symplectic
geometry. Sample ensembles are guaranteed to be perfect.
O(n3 ) complexity, but reasonably fast in practice.
(Incomplete?) History of Sampling Algorithms
• Markov Chain Algorithms
• crankshaft (Vologoskii 1979, Klenin 1988)
• polygonal fold (Millett 1994)
• Direct Sampling Algorithms
• triangle method (Moore 2004)
• generalized hedgehog method (Varela 2009)
• sinc integral method (Moore 2005, Diao 2011)
(Incomplete?) History of Sampling Algorithms
• Markov Chain Algorithms
• crankshaft (Vologoskii et al. 1979, Klenin et al. 1988)
• convergence to correct distribution unproved
• polygonal fold (Millett 1994)
• convergence to correct distribution unproved
• Direct Sampling Algorithms
• triangle method (Moore et al. 2004)
• samples a subset of closed polygons
• generalized hedgehog method (Varela et al. 2009)
• unproved whether this is correct distribution
• sinc integral method (Moore et al. 2005, Diao et al. 2011)
• requires sampling complicated 1-d polynomial densities
Action-Angle Coordinates
Definition
A triangulation T of the n-gon has n − 3 nonintersecting chords.
The lengths of these chords obey triangle inequalities, so they
lie in a convex polytope in Rn−3 called the moment polytope P.
Definition
If T n−3 = (S 1 )n−3 , there are action-angle coordinates:
α : P × T n−3 → Pol(n)/ SO(3)
14
✓2
d1
d2
✓1
c ~1) using the action-angle map.
FIG. 2: This figure shows how to construct an equilateral pentagon in Pol(5;
First, we pick a point in the moment polytope shown in Figure 3 at center. We have now specified diagonals
d and d of the pentagon, so we may build the three triangles in the triangulation from their side lengths,
Main Theorem
Theorem (with Cantarella, arXiv:1310.5924)
α pushes the standard probability measure on P × T n−3
forward to the correct probability measure on Pol(n)/ SO(3).
Corollary
Sampling P and T n−3 independently is a perfect sampling
algorithm for Pol(n).
Main Theorem
Theorem (with Cantarella, arXiv:1310.5924)
α pushes the standard probability measure on P × T n−3
forward to the correct probability measure on Pol(n)/ SO(3).
Ingredients of the Proof.
Kapovich–Millson toric symplectic structure on polygon space +
Duistermaat–Heckman theorem + Hitchin’s theorem on
compatibility of Riemannian and symplectic volume on
symplectic reductions of Kähler manifolds +
Howard–Manon–Millson analysis of polygon space.
Corollary
Sampling P and T n−3 independently is a perfect sampling
algorithm for Pol(n).
Understanding the Triangulation Polytope
If the chord lengths are di then the triangle inequalities are
0 ≤ d1 ≤ 2
1 ≤ di + di+1
|di − di+1 | ≤ 1
0 ≤ dn−3 ≤ 2
and the moment polytope is
2
d1
1
d2
0
0
1
2
Understanding the Triangulation Polytope
If the chord lengths are di then the triangle inequalities are
0 ≤ d1 ≤ 2
1 ≤ di + di+1
|di − di+1 | ≤ 1
0 ≤ dn−3 ≤ 2
and the moment polytope is
d2 ≤ 2
2
d1
d2 ≤ d1 + 1
d1 ≤ 2
1
d2
d1 + d2 ≥ 1
0
0
d1 ≤ d2 + 1
1
2
Understanding the Triangulation Polytope
If the chord lengths are di then the triangle inequalities are
0 ≤ d1 ≤ 2
1 ≤ di + di+1
|di − di+1 | ≤ 1
0 ≤ dn−3 ≤ 2
and the moment polytope is
(2, 3, 2)
d1
d2
d3
(0, 0, 0)
(2, 1, 0)
Hit-and-Run
Theorem
For any convex polytope P, the hit-and-run Markov chain is
uniformly ergodic with respect to Lebesgue measure on P.
Hit-and-Run
Theorem
For any convex polytope P, the hit-and-run Markov chain is
uniformly ergodic with respect to Lebesgue measure on P.
Hit-and-Run
Theorem
For any convex polytope P, the hit-and-run Markov chain is
uniformly ergodic with respect to Lebesgue measure on P.
Hit-and-Run
Theorem
For any convex polytope P, the hit-and-run Markov chain is
uniformly ergodic with respect to Lebesgue measure on P.
Hit-and-Run
Theorem
For any convex polytope P, the hit-and-run Markov chain is
uniformly ergodic with respect to Lebesgue measure on P.
Hit-and-Run
Theorem
For any convex polytope P, the hit-and-run Markov chain is
uniformly ergodic with respect to Lebesgue measure on P.
A Markov Chain for Closed Random Walks
Definition (TSMCMC(β))
If xk = (~pk , θ~k ) ∈ P × T n−3 , define xk +1 by:
• With probability β, update ~
pk by a hit-and-run step on P.
~k with a new uniformly
• With probability 1 − β, replace θ
sampled point in T n−3 .
At each step, construct the corresponding closed random walk
α(xk ) using action-angle coordinates.
A Markov Chain for Closed Random Walks
Definition (TSMCMC(β))
If xk = (~pk , θ~k ) ∈ P × T n−3 , define xk +1 by:
• With probability β, update ~
pk by a hit-and-run step on P.
~k with a new uniformly
• With probability 1 − β, replace θ
sampled point in T n−3 .
At each step, construct the corresponding closed random walk
α(xk ) using action-angle coordinates.
Proposition (with Cantarella)
TSMCMC(β) is uniformly ergodic with respect to the standard
probability measure on Pol(n)/SO(3).
A change of coordinates
If we introduce a fake chordlength d0 = 1, and make the linear
transformation
ai = di − di+1 , for 0 ≤ i ≤ n − 4,
an−3 = dn−3 − d0
then our inequalities
0 ≤ d1 ≤ 2
1 ≤ di + di+1
|di − di+1 | ≤ 1
0 ≤ dn−3 ≤ 2
become
−1 ≤ ai ≤ 1,
|
{z
X
|di −di+1 |≤1
ai = 0,
}
2
|
i−1
X
j=0
aj + ai ≤ 1
{z
di +di+1 ≥1
}
A change of coordinates
If we introduce a fake chordlength d0 = 1, and make the linear
transformation
ai = di − di+1 , for 0 ≤ i ≤ n − 4,
an−3 = dn−3 − d0
then our inequalities
0 ≤ d1 ≤ 2
1 ≤ di + di+1
|di − di+1 | ≤ 1
0 ≤ dn−3 ≤ 2
become
X
−1 ≤ ai ≤ 1,
ai = 0,
|
{z
}
easy conditions
2
i−1
X
j=0
aj + ai ≤ 1
|
{z
}
hard conditions
Basic Idea
Definition
The n-dimensional cross-polytope C is the slice of the
hypercube [−1, 1]n+1 by the plane x1 + · · · + xn+1 = 0.
0
→
1
2
2
2
1
1
0
0
0
1
2
Idea
Sample points in the cross polytope, which all obey the “easy
conditions”, and reject any samples which fail to obey the “hard
conditions”.
Relative volumes
Theorem (Marichal-Mossinghoff)
The volume of the projection of the (n − 3)-dimensional
cross-polytope C is
Pb n−2
2 c
j=0
n−2
j
(−1)j (n − 2j − 2)n−3
(n − 3)!
Theorem (Khoi, Takakura, Mandini)
The volume of the (n − 3)-dimensional moment polytope for
n-edge equilateral polygons P is
−
Pb n2 c
j=0
(−1)j (n − 2j)n−3
2(n − 3)!
n
j
Runtime of algorithm depends on acceptance ratio
Acceptance ratio =
bounded below by
Vol(P)
Vol(C)
1
n.
is conjectured ∼
6
n+2 .
It is certainly
0.14
0.12
0.10
Vol(P)
Vol(C)
0.08
graph of
0.06
6
n+2
(in red)
0.04
0.02
100
200
300
400
number of edges in polygon
500
Sampling the Cross Polytope
Definition
The hypersimplex ∆k ,n is the slice of the cube [0, 1]n with
Pn
Pn−1
n−1 with k − 1 ≤
i=1 xi = k or the slab of [0, 1]
i=1 xi ≤ k .
Theorem (Stanley)
There is a unimodular triangulation1 of ∆k ,n in [0, 1]n−1 indexed
by permutations of (1, . . . , n − 1) with k − 1 descents.
ψ −1
−→
Standard triangulation
1
Stanley triangulation
a decomposition into disjoint simplices of equal volume
Sampling the Cross Polytope
Definition
The hypersimplex ∆k ,n is the slice of the cube [0, 1]n with
Pn
Pn−1
n−1 with k − 1 ≤
i=1 xi = k or the slab of [0, 1]
i=1 xi ≤ k .
Theorem (Stanley)
There is a unimodular triangulation1 of ∆k ,n in [0, 1]n−1 indexed
by permutations of (1, . . . , n − 1) with k − 1 descents.
ψ −1
−→
Standard triangulation
1
Stanley triangulation
a decomposition into disjoint simplices of equal volume
Sampling the Cross Polytope
Definition
The hypersimplex ∆k ,n is the slice of the cube [0, 1]n with
Pn
Pn−1
n−1 with k − 1 ≤
i=1 xi = k or the slab of [0, 1]
i=1 xi ≤ k .
Theorem (Stanley)
There is a unimodular triangulation1 of ∆k ,n in [0, 1]n−1 indexed
by permutations of (1, . . . , n − 1) with k − 1 descents.
ψ −1
−→
Standard triangulation
1
Stanley triangulation
a decomposition into disjoint simplices of equal volume
Sampling the Cross Polytope
Definition
The hypersimplex ∆k ,n is the slice of the cube [0, 1]n with
Pn
Pn−1
n−1 with k − 1 ≤
i=1 xi = k or the slab of [0, 1]
i=1 xi ≤ k .
Theorem (Stanley)
There is a unimodular triangulation1 of ∆k ,n in [0, 1]n−1 indexed
by permutations of (1, . . . , n − 1) with k − 1 descents.
ψ −1
−→
Standard triangulation
1
Stanley triangulation
a decomposition into disjoint simplices of equal volume
Sampling the Cross Polytope
Definition
The hypersimplex ∆k ,n is the slice of the cube [0, 1]n with
Pn
Pn−1
n−1 with k − 1 ≤
i=1 xi = k or the slab of [0, 1]
i=1 xi ≤ k .
Theorem (Stanley)
There is a unimodular triangulation1 of ∆k ,n in [0, 1]n−1 indexed
by permutations of (1, . . . , n − 1) with k − 1 descents.
ψ −1
−→
Standard triangulation
1
Stanley triangulation
a decomposition into disjoint simplices of equal volume
Sampling the Cross Polytope
Definition
The hypersimplex ∆k ,n is the slice of the cube [0, 1]n with
Pn
Pn−1
n−1 with k − 1 ≤
i=1 xi = k or the slab of [0, 1]
i=1 xi ≤ k .
Theorem (Stanley)
There is a unimodular triangulation1 of ∆k ,n in [0, 1]n−1 indexed
by permutations of (1, . . . , n − 1) with k − 1 descents.
ψ −1
−→
Standard triangulation
1
Stanley triangulation
a decomposition into disjoint simplices of equal volume
Parity Issues
Recall that we’re interested in the cross polytope determined by
a0 + a1 + . . . + an−3 = 0
in the cube [−1, 1]n−2 . After translation and scaling, this
corresponds to the slice
x0 + x1 + . . . + xn−3 =
n−2
2
in the cube [0, 1]n−2 .
For even n this is just the hypersimplex ∆(n−2)/2,n−2 and
Stanley’s triangulation applies.
For odd n this is not a hypersimplex.
The Algorithm (for n even)
Moment Polytope Sampling Algorithm (with Cantarella
and Uehara)
1
Generate a permutation of n − 3 numbers with n/2 − 2
descents. O(n2 ) time .
2
Generate a point in the corresponding simplex of the
Stanley triangulation for ∆(n−2)/2,n−2 .
3
Project up to the cross polytope in [0, 1]n−2 , over to the
cross polytope in [−1, 1]n−2 and down to a set of diagonals.
4
Test the proposed set of diagonals against the “hard”
conditions. acceptance ratio > 1/n
5
Generate dihedral angles from T n−3 .
6
Build sample polygon in action-angle coordinates.
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example MPSA 160-gon
Example Sampling Calculation
We ran 10,000 samples of 48-gons using MPSA and 10,000
samples of 48-gons using TSMCMC and computed total
curvature:
Algorithm
MPSA
TSMCMC(β)
TSMCMC(β, δ)
Exact Value
Result
76.5568
78.0419
76.6142945311
76.6014
95% confidence interval
(76.4763, 76.6572)
(76.4848, 79.545)
(76.5163, 76.7122)
TSMCMC(β) (no permutations) used lagged autocovariances
up to lag 2018 in computing the error estimate, while
TSMCMC(β, δ) (90% permutations) used up to lag 4. MPSA
doesn’t need to compute any lagged autocovariances and just
uses the sample variance to find the confidence interval.
A Combinatorial Mystery
Recall the volumes of the cross polytope and moment polytope:
Vol(C) =
Pb n−2
2 c
Vol(P) = −
j=0
Pb n2 c
j=0
(−1)j (n − 2j − 2)n−3
(n − 3)!
(−1)j (n − 2j)n−3
2(n − 3)!
n
j
n−2
j
A Combinatorial Mystery
Recall the volumes of the cross polytope and moment polytope:
Vol(C) =
=
m
Pb n−2
2 c
j=0
(−1)j (n − 2j − 2)n−3
2n−3
(n − 3)!
(n − 3)!
n−3
,
n/2 − 2
n−2
j
where k is the Eulerian number which gives the number of
permutations of {1, . . . , m} with exactly k descents. The
Eulerian numbers satisfy the recurrence relation
DmE
m−1
m−1
= (k + 1)
+ (m − k )
k
k
k −1
Eulerian Numbers
m
k counts the number of permutations of {1, . . . , m} with
exactly k descents and satisfies the recurrence relation
DmE
m−1
m−1
= (k + 1)
+ (m − k )
k
k
k −1
m
k
k
m
1
1
1
1
1
1
1
1
1
1
4
11
26
57
120
247
502
1
11
66
302
1191
4293
14608
1
26
302
2416
15619
88234
1
57
1191
15619
156190
1
120
1
4293 247
1
88234 14608 502 1
A Combinatorial Mystery
Recall the volumes of the cross polytope and moment polytope:
Vol(C) =
Pb n−2
2 c
Vol(P) = −
j=0
Pb n2 c
j=0
(−1)j (n − 2j − 2)n−3
(n − 3)!
(−1)j (n − 2j)n−3
2(n − 3)!
n
j
n−2
j
A Combinatorial Mystery
Recall the volumes of the cross polytope and moment polytope:
Vol(P) = −
Pb n2 c
'
conj.
j=0
√
(−1)j (n − 2j)n−3
2(n − 3)!
n
j
3 3
Cn/2
8
where Cm is the mth Catalan number
2m
Γ(2m + 1)
1
Cm =
=
.
m+1 m
Γ(m + 2)Γ(m + 1)
Catalan Numbers and Moment Polytope Volumes
The ratio of Vol(P) to Cn/2 :
1.
0.8
Vol(P)
Cn/2
0.6
√
3 3
8
0.4
(in red)
0.2
0
20
40
60
80
number of edges in polygon
100
A Recurrence Relation for Vol(P)
The normalized volume (n − 3)! Vol(P) of the moment polytope
is the k = 1 case of a two-parameter family V (n, k ) given by the
recurrence relation
V (n, k ) = (n − k − 1)V (n − 1, k − 1) + (n + k − 1)V (n − 1, k + 1)
subject to the boundary conditions
V (n, 0) = 0,
V (3, 1) = 1,
V (3, 2) =
1
.
2
A Recurrence Relation for Vol(P)
The normalized volume (n − 3)! Vol(P) of the moment polytope
is the k = 1 case of a two-parameter family V (n, k ) given by the
recurrence relation
V (n, k ) = (n − k − 1)V (n − 1, k − 1) + (n + k − 1)V (n − 1, k + 1)
0 1
0
0
0
0
0
0
0
1
2
A Recurrence Relation for Vol(P)
The normalized volume (n − 3)! Vol(P) of the moment polytope
is the k = 1 case of a two-parameter family V (n, k ) given by the
recurrence relation
V (n, k ) = (n − k − 1)V (n − 1, k − 1) + (n + k − 1)V (n − 1, k + 1)
0
0
0
0
0
0
0
0
1
2
5
24
154
1280
13005
156800
1
2
1
4
22
160
1445
15680
199066
1
8
75
800
9821
137088
1
16
236
3584
58478
1
32
1
721
64
1
15232 2178 128 1
cf. OEIS A012249 “Volume of a certain rational polytope. . . ”
Catalan Numbers and Eulerian Numbers
Conjecture
√
3 3
Vol(P) '
Cn/2 .
8
Catalan Numbers and Eulerian Numbers
Conjecture
√
3 3
Vol(P) '
Cn/2 .
8
Combining the conjecture with Vol(C) =
Stirling’s approximation, and
2n−3
(n−3)!
D
n−3
n/2−2
Theorem (Giladi–Keller ’94)
n−3
n/2 − 2
'
s
6
(n − 3)!
π(n − 2)
would suffice to prove
√
Vol(P)
6 n−2
'
.
Vol(C)
n3/2
E
,
√
6 n−2
n3/2
√
6 n−2
n3/2
6
n+2 are basically
for Vol(P)
Vol(C) :
and
estimates
indistinguishable as asymptotic
1.
0.8
graph of
6
n+2
(in blue)
Vol(P) 0.6
Vol(C)
0.4
graph of
√
6 n−2
n3/2
20
30
(in yellow)
0.2
10
vs.
40
number of edges in polygon
50
6
n+2
Future Directions
√
• Prove conjectured volume Vol(P) ' 3 8 3 Cn/2 , which
implies the volume ratio estimate.
• Extend technique to odd numbers of edges. (Requires a
new method for sampling the cross-polytope.)
• Improve runtime to O(n2 ) using a reordering method.
• Improve algorithm for generating random k -descent
permutations.
• Reference implementation and reference ensembles of
polygons.
Thank you!
Thank you for listening!
Supported by:
Simons Foundation
NSF DMS–1105699
Background
• Probability Theory of Random Polygons from the
Quaternionic Viewpoint
Jason Cantarella, Tetsuo Deguchi, and Clayton Shonkwiler
Communications on Pure and Applied Mathematics 67
(2014), no. 10, 1658–1699.
• The Expected Total Curvature of Random Polygons
Jason Cantarella, Alexander Y. Grosberg, Robert Kusner,
and Clayton Shonkwiler
arXiv:1210.6537.
To appear in American Journal of Mathematics.
• The Symplectic Geometry of Closed Equilateral Random
Walks in 3-space
Jason Cantarella and Clayton Shonkwiler
arXiv:1310.5924.