[PPT] - Approximate enumeration of trivial words. Andrew Rechnitzer Murray PowerPoint Presentation

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Counting From SAPs to groups Histograms Exact Results To do

Approximate enumeration of trivial words.

Andrew Rechnitzer Murray Elder Buks van Rensburg Thomas Wong Cameron Rogers Odense, August 2016

Rechnitzer

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COUNTING THINGS

Enumeration

Find closed form expression for number of objects of size n

r generating function or recurrence or algorithm or . . .

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COUNTING THINGS

Enumeration

Find closed form expression for number of objects of size n

r generating function or recurrence or algorithm or . . .
Sometimes exactly, but very frequently we have to approximate.

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A FEW OF MY FAVOURITE THINGS

Self-avoiding walk

A path on a regular lattice that does not intersect itself
cn is # walks of n edges starting from origin.

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A FEW OF MY FAVOURITE THINGS

Self-avoiding polygon

An embedding of a simple closed curve into a regular lattice.
pn is # polygons of n vertices up to translations.

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A FEW OF MY FAVOURITE THINGS

Self-avoiding polygon

An embedding of a simple closed curve into a regular lattice.
pn is # polygons of n vertices up to translations.
pn(K) — polygons with knot-type K

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A HARD PROBLEM

SAPs unsolved on any non-trivial lattice

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A HARD PROBLEM

SAPs unsolved on any non-trivial lattice
In 2d
Exponential growth known on hexagonal lattice

p1/n

n

→

2 +

√ 2 [Duminil-Copin & Smirnov 2010]

conformal invariance predictions for subdominant asymptotics

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A HARD PROBLEM

SAPs unsolved on any non-trivial lattice
In 2d
Exponential growth known on hexagonal lattice

p1/n

n

→

2 +

√ 2 [Duminil-Copin & Smirnov 2010]

conformal invariance predictions for subdominant asymptotics
In 3d = very open
series analysis — brute force + tricks
simulations — many different approaches

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RANDOM SAMPLING OF SAPS

BFACF on Z2

Start with unit square, then

Pick a face adjacent to polygon
Flip edges around the face
Accept or reject with simple transition probability.

[Berg & Foerster 1981] [Arag˜ ao de Carvalho, Caracciolo & Fr¨

lich 1983]

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SAMPLING → COUNTING

BFACF samples from a Boltzmann distribution

— Pr(ϕ) ∝ β|ϕ| — samples at all lengths and uniform at each length.

Used to study “statistical topology” since moves preserve topology

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SAMPLING → COUNTING

BFACF samples from a Boltzmann distribution

— Pr(ϕ) ∝ β|ϕ| — samples at all lengths and uniform at each length.

Used to study “statistical topology” since moves preserve topology
To study knotting probabilities, extended BFACF → GAS

[Janse van Rensburg & R 2011]

Algorithm estimates ratios pn/pm

— approximate counting of loops on graphs.

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AT ABOUT THE SAME TIME. . .

Murray came to UBC and gave a talk on F.
How do we sample elements of geodesic length ℓ?
What is the growth series?
No unique construction — many geodesics for each element.

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AT ABOUT THE SAME TIME. . .

Murray came to UBC and gave a talk on F.
How do we sample elements of geodesic length ℓ?
What is the growth series?
No unique construction — many geodesics for each element.
One of the problems in developing SAP counting algorithm.

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AT ABOUT THE SAME TIME. . .

Murray came to UBC and gave a talk on F.
How do we sample elements of geodesic length ℓ?
What is the growth series?
No unique construction — many geodesics for each element.
One of the problems in developing SAP counting algorithm.
So we talked about sampling and counting, growth and cogrowth.. .

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BFACF ↔ ab = ba

We realised that BFACF moves are just insert-relation & cancel.

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BFACF ↔ ab = ba

We realised that BFACF moves are just insert-relation & cancel.

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BFACF ↔ ab = ba

We realised that BFACF moves are just insert-relation & cancel.

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BFACF ↔ ab = ba

We realised that BFACF moves are just insert-relation & cancel. So why not do BFACF on groups?

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BFACF ↔ ab = ba

We realised that BFACF moves are just insert-relation & cancel. So why not do BFACF on groups?

MC algorithm for trivial words in finitely presented groups

[Elder, JvR & R 2015]

See Murray + Cameron’s talks

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WHERE TO FROM HERE. . .

Mean length vs β for BS(1, 2) = a, b | ab = ba2

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WHERE TO FROM HERE. . .

Mean length vs β for BS(1, 2) = a, b | ab = ba2
Notice
mean length quite modest even for large β
exact data for comparison

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WHERE TO FROM HERE. . .

Mean length vs β for BS(1, 2) = a, b | ab = ba2
Notice
mean length quite modest even for large β
exact data for comparison
Two problems faced by our MC algorithm
difficult to sample long trivial words
need exact results for comparison

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TOWARDS LONGER WORDS — WARM-UP

Consider the following Markov chain that samples words in {0, 1}n

Start with 0n
Pick a bit uniformly at random and flip it 0 ↔ 1
Increment histogram bucket
Repeat

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TOWARDS LONGER WORDS — WARM-UP

Consider the following Markov chain that samples words in {0, 1}n

Start with 0n
Pick a bit uniformly at random and flip it 0 ↔ 1
Increment histogram bucket
Repeat

Converges to uniform distrubtion

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TOWARDS LONGER WORDS — WARM-UP

Consider the following Markov chain that samples words in {0, 1}n

Start with 0n
Pick a bit uniformly at random and flip it 0 ↔ 1
Increment histogram bucket
Repeat

Converges to uniform distrubtion

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GROUP BY SIZE

Let size be “most significant bit” in word

Start with x = 0n
Pick a bit uniformly at random and flip it 0 ↔ 1
Increment histogram bucket MSB(x)
Repeat

Samples fall in n + 1 buckets — not uniform, but.. .

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GROUP BY SIZE = MOST SIGNIFICANT BIT

#samples in each bucket ∝ counts = approximate enumeration scheme
but algorithm will spend all its time on large buckets,
and almost never sample smaller buckets

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GROUP BY SIZE = MOST SIGNIFICANT BIT

#samples in each bucket ∝ counts = approximate enumeration scheme
but algorithm will spend all its time on large buckets,
and almost never sample smaller buckets
If we know the counts then we can make uniform across size

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SET TRANSITIONS ACCORDING TO COUNTS

Make a new Markov chain

Let |x| be MSB of word x and c(ℓ) the # words with MSB ℓ.

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SET TRANSITIONS ACCORDING TO COUNTS

Make a new Markov chain

Let |x| be MSB of word x and c(ℓ) the # words with MSB ℓ.
Start at x = 0n
Create new word y by flipping a bit uniformly at random
Accept y with probability = min
1, c(|x|)

c(|y|)

else keep x

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SET TRANSITIONS ACCORDING TO COUNTS

Make a new Markov chain

Let |x| be MSB of word x and c(ℓ) the # words with MSB ℓ.
Start at x = 0n
Create new word y by flipping a bit uniformly at random
Accept y with probability = min
1, c(|x|)

c(|y|)

else keep x

Easy to check limiting distribution is uniform across size.

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SET TRANSITIONS ACCORDING TO COUNTS

Make a new Markov chain

Let |x| be MSB of word x and c(ℓ) the # words with MSB ℓ.
Start at x = 0n
Create new word y by flipping a bit uniformly at random
Accept y with probability = min
1, c(|x|)

c(|y|)

else keep x

Easy to check limiting distribution is uniform across size. Could fix our sampling long words issue.. .

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COUNTS UNKNOWN

Generally we do not know the counts — that is whole problem!

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COUNTS UNKNOWN

Generally we do not know the counts — that is whole problem!
but we do know c(n)1/n → µ
so approximate c(n) ∝ λn where λ ≈ µ.

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COUNTS UNKNOWN

Generally we do not know the counts — that is whole problem!
but we do know c(n)1/n → µ
so approximate c(n) ∝ λn where λ ≈ µ.
Then transition probability is then

Pr(x → y) = min

1, λ|x|−|y|

which is the transition probability in Murray’s talk

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WORKS WELL. . . BUT THERE CAN BE ISSUES

So if we approximate c(n) ∝ λn where λ ≈ µ, then

Need to choose λ carefully:
If λ > µ, difficult to grow large objects
If λ < µ, too easy to grow large objects — escape to ∞.

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WORKS WELL. . . BUT THERE CAN BE ISSUES

So if we approximate c(n) ∝ λn where λ ≈ µ, then

Need to choose λ carefully:
If λ > µ, difficult to grow large objects
If λ < µ, too easy to grow large objects — escape to ∞.
Even if we can set λ = µ, then there can be problems

— eg if c(n) ≪ µn then still difficult to grow large objects — the deficiency of [E, JvR & R] algorithm we’d like to fix.

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TOWARDS A FIX

Idea 1:

Improve single parameter approximation
Replace c(n) ≈ λn with c(n) ≈ g(n)
Initially set g(n) = λn
As algorithm runs, improve g(n)

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TOWARDS A FIX

Idea 1:

Improve single parameter approximation
Replace c(n) ≈ λn with c(n) ≈ g(n)
Initially set g(n) = λn
As algorithm runs, improve g(n) — how?

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TOWARDS A FIX

Idea 1:

Improve single parameter approximation
Replace c(n) ≈ λn with c(n) ≈ g(n)
Initially set g(n) = λn
As algorithm runs, improve g(n) — how?

Observation:

Transition probability from x → y is = min
1, g(|x|)

g(|y|)

if g(n) too big, then too few samples at size n
if g(n) too small, then too many samples at size n

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TOWARDS A FIX

Idea 1:

Improve single parameter approximation
Replace c(n) ≈ λn with c(n) ≈ g(n)
Initially set g(n) = λn
As algorithm runs, improve g(n) — how?

Observation:

Transition probability from x → y is = min
1, g(|x|)

g(|y|)

if g(n) too big, then too few samples at size n
if g(n) too small, then too many samples at size n

Idea 2:

use histogram to tune g(n) ≈ c(n)

— rather than setting g(n) ≈ c(n) to flatten histogram

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Counting From SAPs to groups Histograms Exact Results To do

TOWARDS A FIX

Idea 1:

Improve single parameter approximation
Replace c(n) ≈ λn with c(n) ≈ g(n)
Initially set g(n) = λn
As algorithm runs, improve g(n) — how?

Observation:

Transition probability from x → y is = min
1, g(|x|)

g(|y|)

if g(n) too big, then too few samples at size n
if g(n) too small, then too many samples at size n

Idea 2:

use histogram to tune g(n) ≈ c(n)

— rather than setting g(n) ≈ c(n) to flatten histogram

increase g(n) each time we sample an object of size n

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

TOWARDS A FIX

Idea 1:

Improve single parameter approximation
Replace c(n) ≈ λn with c(n) ≈ g(n)
Initially set g(n) = λn
As algorithm runs, improve g(n) — how?

Observation:

Transition probability from x → y is = min
1, g(|x|)

g(|y|)

if g(n) too big, then too few samples at size n
if g(n) too small, then too many samples at size n

Idea 2:

use histogram to tune g(n) ≈ c(n)

— rather than setting g(n) ≈ c(n) to flatten histogram

increase g(n) each time we sample an object of size n

— Wang-Landau algorithm [Wang & Landau 2001]

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WANG LANDAU ALGORITHM

Start at x = 0n, g(n) = λn and F = 2
Create new word y by flipping a bit uniformly at random
Accept y with probability = min
1, g(|x|)

g(|y|)

, else keep x
Increase g(n) → g(n) × F

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WANG LANDAU ALGORITHM

Start at x = 0n, g(n) = λn and F = 2
Create new word y by flipping a bit uniformly at random
Accept y with probability = min
1, g(|x|)

g(|y|)

, else keep x
Increase g(n) → g(n) × F
Repeat until histogram is “flat” then
reduce F →

√ F

reset histogram

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WANG LANDAU ALGORITHM

Start at x = 0n, g(n) = λn and F = 2
Create new word y by flipping a bit uniformly at random
Accept y with probability = min
1, g(|x|)

g(|y|)

, else keep x
Increase g(n) → g(n) × F
Repeat until histogram is “flat” then
reduce F →

√ F

reset histogram

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AVOID HISTOGRAM CHECKING

Similar idea but F decreases at each time step

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AVOID HISTOGRAM CHECKING

Similar idea but F decreases at each time step

Start at x = 0n, g(n) = λn and choose ξ ∈ (1/2, 1)
Create new word y by flipping a bit uniformly at random
Accept y with probability = min
1, g(|x|)

g(|y|)

, else keep x
Increase g(n) → g(n) × F

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AVOID HISTOGRAM CHECKING

Similar idea but F decreases at each time step

Start at x = 0n, g(n) = λn and choose ξ ∈ (1/2, 1)
Create new word y by flipping a bit uniformly at random
Accept y with probability = min
1, g(|x|)

g(|y|)

, else keep x
Increase g(n) → g(n) × F where now F = exp(t−ξ)

Rechnitzer

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AVOID HISTOGRAM CHECKING

Similar idea but F decreases at each time step

Start at x = 0n, g(n) = λn and choose ξ ∈ (1/2, 1)
Create new word y by flipping a bit uniformly at random
Accept y with probability = min
1, g(|x|)

g(|y|)

, else keep x
Increase g(n) → g(n) × F where now F = exp(t−ξ)
No flat-histogram check — flattens itself

Stochastic approximation Monte Carlo [Liang, Liu & Carroll 2007]

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AVOID HISTOGRAM CHECKING

Similar idea but F decreases at each time step

Start at x = 0n, g(n) = λn and choose ξ ∈ (1/2, 1)
Create new word y by flipping a bit uniformly at random
Accept y with probability = min
1, g(|x|)

g(|y|)

, else keep x
Increase g(n) → g(n) × F where now F = exp(t−ξ)
No flat-histogram check — flattens itself

Stochastic approximation Monte Carlo [Liang, Liu & Carroll 2007]

Rechnitzer

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HOW GOOD IS IT FOR COGROWTH?

Try on an easy example

Sample reduced trivial words in Z2 = a, b | ab = ba
Basic moves are conjugate + append relation to end

Let it run for half an hour. . .

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HOW GOOD IS IT FOR COGROWTH?

Try on an easy example

Sample reduced trivial words in Z2 = a, b | ab = ba
Basic moves are conjugate + append relation to end

Let it run for half an hour. . .

250 500 14.016 14.018 14.02 14.022 log( histogram )

Histogram is flat

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HOW GOOD IS IT FOR COGROWTH?

Try on an easy example

Sample reduced trivial words in Z2 = a, b | ab = ba
Basic moves are conjugate + append relation to end

Let it run for half an hour. . .

250 500 200 400 600 log g(n) n log 3 250 500 0.25 0.5 g(n) · n 3n

Consistent with g(n) ∼ 3n · n−1

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HOW GOOD IS IT FOR COGROWTH?

Try on an easy example

Sample reduced trivial words in Z2 = a, b | ab = ba
Basic moves are conjugate + append relation to end

Let it run for half an hour. . .

250 500 200 400 600 log g(n) n log 3 250 500 0.25 0.5 g(n) · n 3n

Consistent with g(n) ∼ 3n · n−1 Need more test cases.. .

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COGROWTH SERIES

Very few exact solutions
For Z2 a sneaky construction shows

c2n =

2n

n 2

Nice results by [Kouksov 1998]

a, b | a2, b3 a, b | a3, b3 a, b, c | a2, b2, c2

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COGROWTH SERIES

Very few exact solutions
For Z2 a sneaky construction shows

c2n =

2n

n 2

Nice results by [Kouksov 1998]

a, b | a2, b3 a, b | a3, b3 a, b, c | a2, b2, c2

So we went back and resolved Z2 by another method
Realised we could solve Baumslag-Solitar groups

BS(N, M) = a, b | aNb = baM

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COUNTING LOOPS IN BS(1, 1) THE HARDER WAY

Cut Z2 into cosets: bka
Horizontal steps move within coset
Vertical steps move between them.

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COUNTING LOOPS IN BS(1, 1)

Count all walks ending in a: G(z; q) =

k
w ≡ ak

z|w|qk

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COUNTING LOOPS IN BS(1, 1)

Count all walks ending in a: G(z; q) =

k
w ≡ ak

z|w|qk Use a standard factorisation for Catalan objects (eg Dyck paths, binary trees)

Cut walk into pieces at each visit to a

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SCHEMATIC FACTORISATION

G(z; q) = 1 + z (q + ¯ q) G(z, q) + 2z2G(z; q)L(z; q)

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SCHEMATIC FACTORISATION

G(z; q) = 1 + z (q + ¯ q) G(z, q) + 2z2G(z; q)L(z; q) L(z; q) = 1 + z (q + ¯ q) L(z; q) + z2L(z; q)L(z; q)

Solve for G(z; q) — algebraic function
Take constant term wrt q — D-finite function

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NOW DO BS(2, 2)

The Cayley graph is not so simple

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NOW DO BS(2, 2)

The Cayley graph is not so simple

It is not flat

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NOW DO BS(2, 2)

The Cayley graph is not so simple

It is not flat
Parity of x-ordinate decides if vertical step moves to a different sheet

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NOW DO BS(2, 2)

The Cayley graph is not so simple

It is not flat
Parity of x-ordinate decides if vertical step moves to a different sheet

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NOW DO BS(2, 2)

The Cayley graph is not so simple

It is not flat
Parity of x-ordinate decides if vertical step moves to a different sheet

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NOW DO BS(2, 2)

The Cayley graph is not so simple

It is not flat
Parity of x-ordinate decides if vertical step moves to a different sheet
Looked at from the side, cosets form a tree

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NOW DO BS(2, 2)

The Cayley graph is not so simple

It is not flat
Parity of x-ordinate decides if vertical step moves to a different sheet
Looked at from the side, cosets form a tree

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NOW DO BS(2, 2)

The Cayley graph is not so simple

It is not flat
Parity of x-ordinate decides if vertical step moves to a different sheet
Looked at from the side, cosets form a tree

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NOW DO BS(2, 2)

The Cayley graph is not so simple

It is not flat
Parity of x-ordinate decides if vertical step moves to a different sheet
Looked at from the side, cosets form a tree
Factor as before, but more care to decide if b,¯

b moves to or from root.

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SCHEMATIC FACTORISATION

G(z; q) = 1 + z (q + ¯ q) G(z, q) + 2z2G(z; q) [E ◦ L(z; q)]

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SCHEMATIC FACTORISATION

G(z; q) = 1 + z (q + ¯ q) G(z, q) + 2z2G(z; q) [E ◦ L(z; q)] L(z; q) = 1 + z (q + ¯ q) L(z; q) + z2L(z; q) [E ◦ L(z; q)] + z2 [O ◦ L(z; q)] [E ◦ L(z; q)]

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SCHEMATIC FACTORISATION

G(z; q) = 1 + z (q + ¯ q) G(z, q) + 2z2G(z; q) [E ◦ L(z; q)] L(z; q) = 1 + z (q + ¯ q) L(z; q) + z2L(z; q) [E ◦ L(z; q)] + z2 [O ◦ L(z; q)] [E ◦ L(z; q)]

Solve for G(z; q) — algebraic function
Take constant term wrt q — D-finite function

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MORE GENERALLY

D-finite solution for BS(N, N) = a, b | aNb = baN

Similar factorisation gives G(z, q) algebraic degree N + 1
Take constant term wrt q gives D-finite solution

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MORE GENERALLY

D-finite solution for BS(N, N) = a, b | aNb = baN

Similar factorisation gives G(z, q) algebraic degree N + 1
Take constant term wrt q gives D-finite solution
Growth rate of trivial words are algebraic numbers
The DE satisfied by the CT gets worse with N
BS(1, 1) — Write as elliptic integrals
BS(2, 2) — 6th order ODE, coeffs degree ≤ 47
BS(3, 3) — 8th order ODE, coeffs degree ≤ 105
BS(10, 10) — 22nd order ODE — 6 megabyte text file!
A big thanks to [Manuel Kauers] for help with this.

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MORE GENERALLY STILL — MESSIER

A similar factorisation gives

Functional equations for BS(N, M) = a, b | aNb = baM

L = 1 + z(q + ¯ q)L + z2L · [ΦN,M ◦ L + ΦM,N ◦ K] − z2 [ΦM,N ◦ K] · [ΦN,N ◦ L] , K = 1 + z(q + ¯ q)K + z2K · [ΦM,N ◦ K + ΦN,M ◦ L] − z2 [ΦN,M ◦ L] · [ΦM,M ◦ K] , G = 1 + z(q + ¯ q)G + z2G · [ΦN,M ◦ L + ΦM,N ◦ K] , where Φd,e ◦

k

cn,kqk =

j

cn,j·d qj·e [E, JvR, R & Wong 2014]

Unable to solve closed form — even for BS(1, 2).
Series generation hard since degq[zn]G(z; q) grows exponentially with n.

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Counting From SAPs to groups Histograms Exact Results To do

EXTENSION TO BRAIDS

We [E, Rogers & R] have extended this to

B3 = a, b | aba = bab = c, d | c3 = d2

Again, solution is constant term of algebraic function.
Extends to

c, d | cn = dm

But unfortunately doesn’t appear to work for Bn with n ≥ 4

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

EXTENSION TO BRAIDS

We [E, Rogers & R] have extended this to

B3 = a, b | aba = bab = c, d | c3 = d2

Again, solution is constant term of algebraic function.
Extends to

c, d | cn = dm

But unfortunately doesn’t appear to work for Bn with n ≥ 4
Now some results.. .

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

VERY SHORT RUNS USING SAMC

Start with Kuksov’s exact results

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

VERY SHORT RUNS USING SAMC

Start with Kuksov’s exact results

a, b | a2, b3

250 500 200 400 600 exact log g(n)

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

VERY SHORT RUNS USING SAMC

Start with Kuksov’s exact results

a, b | a3, b3

250 500 200 400 600 exact log g(n)

Rechnitzer

SLIDE 84

Counting From SAPs to groups Histograms Exact Results To do

VERY SHORT RUNS USING SAMC

Start with Kuksov’s exact results

a, b | a2, b2, c2

250 500 250 500 750 exact log g(n)

In all cases relative error was 0.1 ∼ 1%

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

VERY SHORT RUNS USING SAMC

Braid3 and friends

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

VERY SHORT RUNS USING SAMC

Braid3 and friends

a, b | aba = bab

250 500 200 400 600 exact log g(n)

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

VERY SHORT RUNS USING SAMC

Braid3 and friends

a, b | a2 = b3

250 500 200 400 600 exact log g(n)

Rechnitzer

SLIDE 88

Counting From SAPs to groups Histograms Exact Results To do

VERY SHORT RUNS USING SAMC

Braid3 and friends

a, b | a3 = b3

250 500 200 400 600 exact log g(n)

Again, relative error was 0.1 ∼ 1%

Rechnitzer

SLIDE 89

Counting From SAPs to groups Histograms Exact Results To do

SLIGHTLY LONGER RUNS USING SAMC

Baumslag Solitar groups with exact cogrowth

Rechnitzer

SLIDE 90

Counting From SAPs to groups Histograms Exact Results To do

SLIGHTLY LONGER RUNS USING SAMC

Baumslag Solitar groups with exact cogrowth

a, b | ab = ba

250 500 200 400 600 exact log g(n)

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

SLIGHTLY LONGER RUNS USING SAMC

Baumslag Solitar groups with exact cogrowth

a, b | aab = baa

250 500 250 500 exact log g(n)

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

SLIGHTLY LONGER RUNS USING SAMC

Baumslag Solitar groups with exact cogrowth

a, b | a3b = ba3

250 500 200 400 exact log g(n)

Again, relative error was 0.1 ∼ 1%

Rechnitzer

SLIDE 93

Counting From SAPs to groups Histograms Exact Results To do

LONGER RUNS USING SAMC

Baumslag Solitar groups, no exact solutions

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

LONGER RUNS USING SAMC

Baumslag Solitar groups, no exact solutions

a, b | a2b = ba3

250 500 200 400 log g(n) approx exact

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

LONGER RUNS USING SAMC

Baumslag Solitar groups, no exact solutions

a, b | ab = ba2

250 500 200 400 600 log g(n) approx exact

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

LONGER RUNS USING SAMC

Baumslag Solitar groups, no exact solutions

a, b | ab = ba2

250 500 200 400 600 log g(n) approx exact

BS23 well behaved, but BS12 and BS13 much slower to converge

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

LONGER RUNS USING SAMC

Baumslag Solitar groups, no exact solutions

a, b | ab = ba2

250 500 200 400 600 log g(n) approx exact

BS23 well behaved, but BS12 and BS13 much slower to converge — why?

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WHAT CAN WE SEE IN F?

Now look at F = a, b |
ab−1, a−1ba
,
ab−1, a−2ba2
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WHAT CAN WE SEE IN F?

Now look at F = a, b |
ab−1, a−1ba
,
ab−1, a−2ba2
100

200 100 200 log g(n) exact

Thanks to [Elvey Price & Guttmann] for exact data

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Counting From SAPs to groups Histograms Exact Results To do

NOW WE HAVE APPROXIMATE ENUMERATION DATA

What is the scaling of c(n)?

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Counting From SAPs to groups Histograms Exact Results To do

NOW WE HAVE APPROXIMATE ENUMERATION DATA

What is the scaling of c(n)?

100 200 100 200 log g(n) exact

Dominated by linear term

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NOW WE HAVE APPROXIMATE ENUMERATION DATA

What is the scaling of c(n)?

100 200 0.2 0.4 0.6 0.8

1 n log g(n)

exact

1 n log c(n) tending to a constant

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

NOW WE HAVE APPROXIMATE ENUMERATION DATA

What is the scaling of c(n)?

0.05 0.1 0.25 0.5 0.75 1

1 n log g(n)

exact log (3)

1 n log c(n) against n−1 — shows curvature

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

NOW WE HAVE APPROXIMATE ENUMERATION DATA

What is the scaling of c(n)?

0.1 0.2 0.3 0.25 0.5 0.75 1

1 n log g(n)

exact log (3)

1 n log c(n) against n−1/2 — appears straight

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

NOW WE HAVE APPROXIMATE ENUMERATION DATA

What is the scaling of c(n)?

0.1 0.2 0.3 0.25 0.5 0.75 1

1 n log g(n)

exact

Rough line of best fit = 1

n log c(n) ≈ 0.993 − 2.19n−1/2

Rechnitzer

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Counting From SAPs to groups Histograms Exact Results To do

NOW WE HAVE APPROXIMATE ENUMERATION DATA

What is the scaling of c(n)?

0.1 0.2 0.3 0.25 0.5 0.75 1

1 n log g(n)

exact

Rough line of best fit = 1

n log c(n) ≈ 0.993 − 2.19n−1/2

Consistent with c(n) ∼ 2.7n × 8.9−√n

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Counting From SAPs to groups Histograms Exact Results To do

MORE CAREFUL WEIGHTED LINEAR REGRESSION

Run multiple independent simulations to get statistical errorbars

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Counting From SAPs to groups Histograms Exact Results To do

MORE CAREFUL WEIGHTED LINEAR REGRESSION

Run multiple independent simulations to get statistical errorbars
Errorbars not visible

0.1 0.2 0.3 n−1/2 0.25 0.5 0.75 1

1 ng(n)

0.995 − 2.20x

Most linear with n−1/2 correction
Line of best fit is 1

n log c(n) ≈ 0.995 − 2.20n−1/2

Consistent with c(n) ∼ 2.705n × 9.0−√n

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Counting From SAPs to groups Histograms Exact Results To do

MORE CAREFUL WEIGHTED LINEAR REGRESSION

Run multiple independent simulations to get statistical errorbars
Errorbars not visible

0.1 0.2 0.3 n−1/2 0.25 0.5 0.75 1

1 ng(n)

0.995 − 2.20x

Most linear with n−1/2 correction
Line of best fit is 1

n log c(n) ≈ 0.995 − 2.20n−1/2

Consistent with c(n) ∼ 2.705n × 9.0−√n
Consistent with series analysis by [Elvey Price & Guttmann]

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Counting From SAPs to groups Histograms Exact Results To do

FOR COMPARISON — a, b | ab = ba2

Again errorbars not visible on plot
Most linear with n−2/3 correction

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Counting From SAPs to groups Histograms Exact Results To do

FOR COMPARISON — a, b | ab = ba2

Again errorbars not visible on plot
Most linear with n−2/3 correction

0.1 0.2 0.3 n−2/3 0.25 0.5 0.75 1

1 ng(n)

y = 1.099 − 2.44x

Line of best fit is 1

n log c(n) ≈ 1.099 − 2.44n−1/2

Consistent with c(n) ∼ 3n × 11.47−n1

/3 Rechnitzer

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WHAT NEXT?

Exact solution of cogrowth

Which presentations are amenable to this approach
If not closed forms, then polynomial time algorithms?
If not poly-time, then faster than brute-force?

Rechnitzer

SLIDE 113

Counting From SAPs to groups Histograms Exact Results To do

WHAT NEXT?

Exact solution of cogrowth

Which presentations are amenable to this approach
If not closed forms, then polynomial time algorithms?
If not poly-time, then faster than brute-force?

Random sampling and approximate enumeration

Understand convergence
Speed convergence to increase system sizes
Better asymptotics
Which statistics should we study?

Rechnitzer

SLIDE 114

Counting From SAPs to groups Histograms Exact Results To do

WHAT NEXT?

Exact solution of cogrowth

Which presentations are amenable to this approach
If not closed forms, then polynomial time algorithms?
If not poly-time, then faster than brute-force?

Random sampling and approximate enumeration

Understand convergence
Speed convergence to increase system sizes
Better asymptotics
Which statistics should we study?
What can we prove about the algorithm?

Rechnitzer

SLIDE 115

Counting From SAPs to groups Histograms Exact Results To do

WHAT NEXT?

Exact solution of cogrowth

Which presentations are amenable to this approach
If not closed forms, then polynomial time algorithms?
If not poly-time, then faster than brute-force?

Random sampling and approximate enumeration

Understand convergence
Speed convergence to increase system sizes
Better asymptotics
Which statistics should we study?
What can we prove about the algorithm?

Thanks for listening.

Rechnitzer