Andrew Rechnitzer Murray Elder Buks van Rensburg Thomas … · 2019. 9. 30. · Murray Elder Buks van Rensburg Thomas Wong Cameron Rogers Odense, August 2016 Rechnitzer. Counting

Counting From SAPs to groups Histograms Exact Results To do

Approximate enumeration of trivial words.

Andrew RechnitzerMurray Elder Buks van Rensburg Thomas Wong Cameron Rogers

Odense, August 2016

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COUNTING THINGS

EnumerationFind closed form expression for number of objects of size nor 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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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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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

• In 2d• Exponential growth known on hexagonal lattice

p1/nn →√

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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A HARD PROBLEM

• SAPs unsolved on any non-trivial lattice• In 2d

• Exponential growth known on hexagonal lattice

p1/nn →√

2 +√


• conformal invariance predictions for subdominant asymptotics

• In 3d = very open• series analysis — brute force + tricks• simulations — many different approaches

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A HARD PROBLEM

• SAPs unsolved on any non-trivial lattice• In 2d

• Exponential growth known on hexagonal lattice

p1/nn →√

2 +√


• 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][Aragão de Carvalho, Caracciolo & Frölich 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

• 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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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 `?• 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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• 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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• 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.

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〉

• 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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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

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

• If we know the counts then we can make uniform across size

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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(`) the # words with MSB `.

• 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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Make a new Markov chain• Let |x| be MSB of word x and c(`) the # words with MSB `.• Start at x = 0n



1,c(|x|)c(|y|)

}else keep x


Could fix our sampling long words issue. . .

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Make a new Markov chain• Let |x| be MSB of word x and c(`) the # words with MSB `.• Start at x = 0n



1,c(|x|)c(|y|)

}else keep x


Could fix our sampling long words issue. . .Rechnitzer


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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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|}


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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|}


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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)

— 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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TOWARDS A FIX


• As algorithm runs, improve g(n) — how?

Observation:


1,g(|x|)g(|y|)





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TOWARDS A FIX


• As algorithm runs, improve g(n) — how?Observation:


1,g(|x|)g(|y|)





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TOWARDS A FIX




1,g(|x|)g(|y|)



— 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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TOWARDS A FIX




1,g(|x|)g(|y|)





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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


1,g(|x|)g(|y|)

}, else keep x

• Increase g(n) 7→ g(n)× F

• Repeat until histogram is “flat” then• reduce F 7→

√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


1,g(|x|)g(|y|)

}, else keep x

• Increase g(n) 7→ g(n)× F• Repeat until histogram is “flat” then

• reduce F 7→√

F• reset histogram

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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


1,g(|x|)g(|y|)

}, else keep x

• Increase g(n) 7→ 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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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


1,g(|x|)g(|y|)

}, else keep x

• Increase g(n) 7→ g(n)× F

where now F = exp(t−ξ)• No flat-histogram check — flattens itself


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1,g(|x|)g(|y|)

}, else keep x

• Increase g(n) 7→ g(n)× F where now F = exp(t−ξ)

• No flat-histogram check — flattens itselfStochastic approximation Monte Carlo [Liang, Liu & Carroll 2007]

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1,g(|x|)g(|y|)

}, else keep x

• Increase g(n) 7→ g(n)× F where now F = exp(t−ξ)• No flat-histogram check — flattens itself


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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. . .

Need more test cases. . .

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0 250 50014.016

14.018

14.02

14.022

log( histogram )

Histogram is flat


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0 250 5000

200

400

600

log g(n)

n log 3

0 250 5000

0.25

0.5 g(n) ·n

3n

Consistent with g(n) ∼ 3n · n−1 X


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0 250 5000

200

400

600

log g(n)

n log 3

0 250 5000

0.25

0.5 g(n) ·n

3n

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

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COGROWTH SERIES

• Very few exact solutions• For Z2 a sneaky construction shows

c2n =

(2nn

)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: bk〈a〉

• 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

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 + q̄) G(z, q) + 2z2G(z; q)L(z; q)

L(z; q) = 1 + z (q + q̄) 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

• 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, b̄ moves to or from root.

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NOW DO BS(2, 2)


• 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, b̄ moves to or from root.

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NOW DO BS(2, 2)


• 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, b̄ moves to or from root.

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NOW DO BS(2, 2)


• 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, b̄ moves to or from root.

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NOW DO BS(2, 2)


• 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, b̄ moves to or from root.

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SCHEMATIC FACTORISATION

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

L(z; q) = 1 + z (q + q̄) 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

• 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 + q̄)L + z2L · [ΦN,M ◦ L + ΦM,N ◦ K]− z2 [ΦM,N ◦ K] · [ΦN,N ◦ L] ,

K = 1 + z(q + q̄)K + z2K · [ΦM,N ◦ K + ΦN,M ◦ L]− z2 [ΦN,M ◦ L] · [ΦM,M ◦ K] ,

G = 1 + z(q + q̄)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[z

n]G(z; q) grows exponentially with n.

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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. . .

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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. . .

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VERY SHORT RUNS USING SAMC

• Start with Kuksov’s exact results

In all cases relative error was 0.1 ∼ 1%

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〈a, b | a2, b3〉

0 250 5000

200

400

600exact

log g(n)


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〈a, b | a3, b3〉

0 250 5000

200

400

600exact

log g(n)


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〈a, b | a2, b2, c2〉

0 250 5000

250

500

750 exact

log g(n)


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• Braid3 and friends

Again, relative error was 0.1 ∼ 1%

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〈a, b | aba = bab〉

250 5000

200

400

600exact

log g(n)


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〈a, b | a2 = b3〉

250 5000

200

400

600exact

log g(n)


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〈a, b | a3 = b3〉

250 5000

200

400

600exact

log g(n)


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SLIGHTLY LONGER RUNS USING SAMC

• Baumslag Solitar groups with exact cogrowth


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〈a, b | ab = ba〉

250 5000

200

400

600exact

log g(n)


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〈a, b | aab = baa〉

250 5000

250

500exact

log g(n)


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〈a, b | a3b = ba3〉

250 5000

200

400 exact

log g(n)


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LONGER RUNS USING SAMC

• Baumslag Solitar groups, no exact solutions

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

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〈a, b | a2b = ba3〉

250 5000

200

400log g(n)

approxexact


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〈a, b | ab = ba2〉

250 5000

200

400

600

log g(n)

approxexact


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〈a, b | ab = ba2〉

250 5000

200

400

600

log g(n)

approxexact

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

— why?

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〈a, b | ab = ba2〉

250 5000

200

400

600

log g(n)

approxexact


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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 2000

100

200 log g(n)

exact

Thanks to [Elvey Price & Guttmann] for exact data

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NOW WE HAVE APPROXIMATE ENUMERATION DATA

What is the scaling of c(n)?

Consistent with c(n) ∼ 2.7n × 8.9−√

n

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100 2000

100

200 log g(n)

exact

Dominated by linear term


n

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100 2000.2

0.4

0.6

0.8

1n log g(n)

exact

1n log c(n) tending to a constant


n

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0 0.05 0.1

0.25

0.5

0.75

1

1n log g(n)

exact

log (3)

1n log c(n) against n

−1 — shows curvature


n

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0 0.1 0.2 0.3

0.25

0.5

0.75

1

1n log g(n)

exact

log (3)

1n log c(n) against n

−1/2 — appears straight


n

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0 0.1 0.2 0.3

0.25

0.5

0.75

1

1n log g(n)

exact

Rough line of best fit = 1n log c(n) ≈ 0.993− 2.19n−1/2


n

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MORE CAREFUL WEIGHTED LINEAR REGRESSION

• Run multiple independent simulations to get statistical errorbars

• Errorbars not visible

0 0.1 0.2 0.3

n−1/2

0.25

0.5

0.75

1

1ng(n)

0.995− 2.20x

• Most linear with n−1/2 correction• Line of best fit is 1n 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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MORE CAREFUL WEIGHTED LINEAR REGRESSION

• Run multiple independent simulations to get statistical errorbars• Errorbars not visible

0 0.1 0.2 0.3

n−1/2

0.25

0.5

0.75

1

1ng(n)

0.995− 2.20x

• Most linear with n−1/2 correction• Line of best fit is 1n 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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FOR COMPARISON — 〈a, b | ab = ba2〉

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

0 0.1 0.2 0.3

n−2/3

0.25

0.5

0.75

1

1ng(n)

y = 1.099− 2.44x

• Line of best fit is 1n log c(n) ≈ 1.099− 2.44n−1/2

• Consistent with c(n) ∼ 3n × 11.47−n1/3

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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?

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.

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WHAT NEXT?


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?


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WHAT NEXT?




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WHAT NEXT?




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CountingFrom SAPs to groupsHistogramsExactResultsTo do

Documents

Andrew Rechnitzer Murray Elder Buks van Rensburg Thomas … · 2019. 9. 30. · Murray Elder Buks van Rensburg Thomas Wong Cameron Rogers Odense, August 2016 Rechnitzer. Counting