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Improved mixing bounds for the anti-ferromagnetic Potts model on Z 2 - - PowerPoint PPT Presentation
Improved mixing bounds for the anti-ferromagnetic Potts model on Z 2 - - PowerPoint PPT Presentation
Improved mixing bounds for the anti-ferromagnetic Potts model on Z 2 Markus Jalsenius markus@dcs.warwick.ac.uk Department of Computer Science University of Warwick Joint work with Leslie Ann Goldberg, Russell Martin and Mike Paterson British
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Anti-ferromagnetic Potts model
Two parameters
◮ q, number of colours ◮ 0 ≤ λ ≤ 1
Weight of colouring = λnumber of monochromatic edges
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Anti-ferromagnetic Potts model
Two parameters
◮ q, number of colours ◮ 0 ≤ λ ≤ 1
Weight of colouring = λnumber of monochromatic edges 9 monochromatic edges, weight of colouring = λ9
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Distributions of colourings
σ = colouring Prob(σ) = weight(σ) Z Z =
- colourings σ
weight(σ)
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Distributions of colourings
Uniform distribution of proper colourings all colourings λ = 0 λ = 1 weight(σ) = 1 or 0 weight(σ) = 1
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Goal
◮ Want to sample from the distribution of colourings.
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Goal
◮ Want to sample from the distribution of colourings. ◮ For what values of q and λ can we sample efficiently?
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Goal
◮ Want to sample from the distribution of colourings. ◮ For what values of q and λ can we sample efficiently? ◮ Efficiently means polynomial time, in size of the region.
Method: Markov chains
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Markov chains
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Markov chains
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Markov chains
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Markov chains
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Markov chains
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Markov chains
Stationary distribution
The stationary distribution of the states is identical to the distribution we want to sample colourings from.
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Markov chains
Stationary distribution
The stationary distribution of the states is identical to the distribution we want to sample colourings from.
Question
How many steps does it take to get close to the stationary distribution?
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Markov chains
Stationary distribution
The stationary distribution of the states is identical to the distribution we want to sample colourings from.
Question
How many steps does it take to get close to the stationary distribution?
Total variation distance
dtv(D1, D2) < ǫ, where ǫ > 0
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Markov chains
Stationary distribution
The stationary distribution of the states is identical to the distribution we want to sample colourings from.
Question
How many steps does it take to get close to the stationary distribution?
Total variation distance
dtv(D1, D2) < ǫ, where ǫ > 0
Polynomial number of steps (rapidly mixing)
Want number of steps be polynomial in n and log 1
ǫ
- , where n
is the size of the region.
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Previous work
Theorem
[L.A. Goldberg, R. Martin and M. Paterson, 2005]
For any triangle-free graph with maximum degree ∆ ≥ 3 we have rapid mixing if q 1.76∆ − 0.47 and λ = 0.
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Previous work
Theorem
[L.A. Goldberg, R. Martin and M. Paterson, 2005]
For any triangle-free graph with maximum degree ∆ ≥ 3 we have rapid mixing if q 1.76∆ − 0.47 and λ = 0.
The lattize Z 2
∆ = 4 so the theorem above gives rapid mixing for q ≥ 7 and λ = 0. This result has now been improved [L.A. Goldberg, M. Jalsenius,
- R. Martin and M. Paterson, 2006].
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Path coupling
[R. Bubley and M. Dyer, 1997]
A B B’ A’
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Path coupling
[R. Bubley and M. Dyer, 1997]
Hamming distance(A, B) = 1
1 < 1 A B B’ A’
Want the expected Hamming distance(A′, B′) < 1
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Path coupling
Hamming distance 1 Three scenarios can happen when applying the ball.
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Path coupling
Hamming distance 1 Scenario 1. The discrepancy is outside of the ball. Hamming distance does not change.
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Path coupling
Hamming distance 1 Scenario 2. The discrepancy is inside the ball. Hamming distance drops to 0.
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Path coupling
Hamming distance 1 Scenario 3. The discrepancy is on the boundary of the ball. Hamming distance can increase. How much?
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Spatial mixing
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Spatial mixing
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Spatial mixing
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Strong spatial mixing
r r
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Strong spatial mixing
r r
◮ Probability of a different colour at distance r decreases
exponentially with r.
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Strong spatial mixing
r r
◮ Probability of a different colour at distance r decreases
exponentially with r.
◮ Expected total number of introduced discrepancies in
shaded region is bounded by a constant.
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Path coupling
d −1 d +k
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Path coupling
d −1 d +k
New expected Hamming distance = = 1 − 1 × |Ball volume| |Region| + k × |Ball boundary| |Region|
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Path coupling
d −1 d +k
New expected Hamming distance = = 1 − 1 × |Ball volume| |Region| + k × |Ball boundary| |Region| |Ball boundary| |Ball volume| ∈ Θ(d) Θ(d2) = Θ( 1 d )
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Path coupling
d −1 d +k
New expected Hamming distance = = 1 − 1 × |Ball volume| |Region| + k × |Ball boundary| |Region| < 1 |Ball boundary| |Ball volume| ∈ Θ(d) Θ(d2) = Θ( 1 d )
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