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Advanced Algorithms (XIV)
Shanghai Jiao Tong University
Chihao Zhang
June 8, 2020
Advanced Algorithms (XIV) Shanghai Jiao Tong University Chihao - - PowerPoint PPT Presentation
Advanced Algorithms (XIV) Shanghai Jiao Tong University Chihao - - PowerPoint PPT Presentation
Advanced Algorithms (XIV) Shanghai Jiao Tong University Chihao Zhang June 8, 2020 Mixing Time via Coupling Mixing Time via Coupling The state space Mixing Time via Coupling The state space Transition matrix P Mixing Time
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Mixing Time via Coupling
The state space Ω
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Mixing Time via Coupling
The state space Ω Transition matrix P ∈ ℝΩ×Ω
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Mixing Time via Coupling
The state space Ω Two chains and
(X0, X1, …) (Y0, Y1, …)
Transition matrix P ∈ ℝΩ×Ω
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Mixing Time via Coupling
The state space Ω Two chains and
(X0, X1, …) (Y0, Y1, …)
Transition matrix P ∈ ℝΩ×Ω A distance d : Ω × Ω → ℝ≥0
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Mixing Time via Coupling
The state space Ω Two chains and
(X0, X1, …) (Y0, Y1, …)
Transition matrix P ∈ ℝΩ×Ω A distance d : Ω × Ω → ℝ≥0 Two chains are “coupled” so that:
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Mixing Time via Coupling
The state space Ω Two chains and
(X0, X1, …) (Y0, Y1, …)
Transition matrix P ∈ ℝΩ×Ω A distance d : Ω × Ω → ℝ≥0 Two chains are “coupled” so that: E[d(Xt+1, Yt+1) ∣ (Xt, Yt)] ≤ (1 − α) ⋅ d(Xt, Yt)
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In other words, is a super martingale
{d(Xt, Yt)}t≥0
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In other words, is a super martingale
{d(Xt, Yt)}t≥0
Recall the mixing time
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In other words, is a super martingale
{d(Xt, Yt)}t≥0
Recall the mixing time
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In other words, is a super martingale
{d(Xt, Yt)}t≥0
Recall the mixing time By coupling lemma
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In other words, is a super martingale
{d(Xt, Yt)}t≥0
Recall the mixing time By coupling lemma dTV(Xt, Yt) ≤ Pr[Xt ≠ Yt] = Pr[d(Xt, Yt) > 0]
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In other words, is a super martingale
{d(Xt, Yt)}t≥0
Recall the mixing time By coupling lemma dTV(Xt, Yt) ≤ Pr[Xt ≠ Yt] = Pr[d(Xt, Yt) > 0] For finite , we assume WLOG that
Ω min
x,y∈Ω:x≠y d(x, y) = 1
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In other words, is a super martingale
{d(Xt, Yt)}t≥0
Recall the mixing time By coupling lemma dTV(Xt, Yt) ≤ Pr[Xt ≠ Yt] = Pr[d(Xt, Yt) > 0] For finite , we assume WLOG that
Ω min
x,y∈Ω:x≠y d(x, y) = 1
Pr[d(Xt, Yt) > 0] = Pr[d(Xt, Yt) ≥ 1] ≤ E[d(Xt, Yt)] ≤ (1 − α)t ⋅ d(X0, Y0)
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Sampling Proper Colorings
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Sampling Proper Colorings
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Sampling Proper Colorings
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Sampling Proper Colorings
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Sampling Proper Colorings
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Sampling Proper Colorings
- the number of proper colorings
q
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Sampling Proper Colorings
- the number of proper colorings
q
- a graph of maximum degree
G Δ
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Sampling Proper Colorings
- the number of proper colorings
q
- a graph of maximum degree
G Δ
Is colorable using colors?
G q
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The problem is NP-hard in general
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The problem is NP-hard in general We consider the case when q > Δ
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The problem is NP-hard in general We consider the case when q > Δ Consider the chain obtained via the “Metropolis Rule”
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The problem is NP-hard in general We consider the case when q > Δ Consider the chain obtained via the “Metropolis Rule”
- Pick
and u.a.r.
- Recolor with if possible
v ∈ V c ∈ [q] v c
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The problem is NP-hard in general We consider the case when q > Δ Consider the chain obtained via the “Metropolis Rule”
- Pick
and u.a.r.
- Recolor with if possible
v ∈ V c ∈ [q] v c
The chain is irreducible when q ≥ Δ + 2
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The Coupling
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The Coupling
Two chains choose the same and
v c
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The Coupling
Two chains choose the same and
v c
v v c =
Good Move
Xt Yt
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The Coupling
Two chains choose the same and
v c
v v c =
Good Move
Xt Yt
d(Xt+1, Yt+1) = d(Xt, Yt) − 1
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The Coupling
Two chains choose the same and
v c
Pr[ ⋅ ] ≥ d(Xt, Yt) N ⋅ q − 2(Δ − 1) q
v v c =
Good Move
Xt Yt
d(Xt+1, Yt+1) = d(Xt, Yt) − 1
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The Coupling
Two chains choose the same and
v c
Pr[ ⋅ ] ≥ d(Xt, Yt) N ⋅ q − 2(Δ − 1) q
v v c =
Good Move
Xt Yt v v c =
Bad Move
Xt Yt
d(Xt+1, Yt+1) = d(Xt, Yt) − 1
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The Coupling
Two chains choose the same and
v c
Pr[ ⋅ ] ≥ d(Xt, Yt) N ⋅ q − 2(Δ − 1) q
v v c =
Good Move
Xt Yt v v c =
Bad Move
Xt Yt
d(Xt+1, Yt+1) = d(Xt, Yt) − 1 d(Xt+1, Yt+1) = d(Xt, Yt) + 1
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The Coupling
Two chains choose the same and
v c
Pr[ ⋅ ] ≥ d(Xt, Yt) N ⋅ q − 2(Δ − 1) q
v v c =
Good Move
Xt Yt v v c =
Bad Move
Xt Yt
Pr[ ⋅ ] ≤ 2d(Xt, Yt)Δ Nq d(Xt+1, Yt+1) = d(Xt, Yt) − 1 d(Xt+1, Yt+1) = d(Xt, Yt) + 1
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E[d(Xt+1, Yt+1) ∣ (Xt, Yt)] ≤ d(Xt, Yt) ⋅ (1 + 2Δ − (q − 2Δ + 2)) qN ) = d(Xt, Yt) ⋅ (1 − q − 4Δ + 2 qN )
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E[d(Xt+1, Yt+1) ∣ (Xt, Yt)] ≤ d(Xt, Yt) ⋅ (1 + 2Δ − (q − 2Δ + 2)) qN ) = d(Xt, Yt) ⋅ (1 − q − 4Δ + 2 qN )
So if , we have
q ≥ 4Δ − 1
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E[d(Xt+1, Yt+1) ∣ (Xt, Yt)] ≤ d(Xt, Yt) ⋅ (1 + 2Δ − (q − 2Δ + 2)) qN ) = d(Xt, Yt) ⋅ (1 − q − 4Δ + 2 qN )
So if , we have
q ≥ 4Δ − 1
E[d(Xt+1, Yt+1) ∣ (Xt, Yt)] ≤ (1 − 1 qN) d(Xt, Yt)
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E[d(Xt+1, Yt+1) ∣ (Xt, Yt)] ≤ d(Xt, Yt) ⋅ (1 + 2Δ − (q − 2Δ + 2)) qN ) = d(Xt, Yt) ⋅ (1 − q − 4Δ + 2 qN )
So if , we have
q ≥ 4Δ − 1
E[d(Xt+1, Yt+1) ∣ (Xt, Yt)] ≤ (1 − 1 qN) d(Xt, Yt)
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E[d(Xt+1, Yt+1) ∣ (Xt, Yt)] ≤ d(Xt, Yt) ⋅ (1 + 2Δ − (q − 2Δ + 2)) qN ) = d(Xt, Yt) ⋅ (1 − q − 4Δ + 2 qN )
So if , we have
q ≥ 4Δ − 1
E[d(Xt+1, Yt+1) ∣ (Xt, Yt)] ≤ (1 − 1 qN) d(Xt, Yt)
dTV(Xt, Yt) ≤ (1 − 1 qN )
t
⋅ N ≤ ε
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E[d(Xt+1, Yt+1) ∣ (Xt, Yt)] ≤ d(Xt, Yt) ⋅ (1 + 2Δ − (q − 2Δ + 2)) qN ) = d(Xt, Yt) ⋅ (1 − q − 4Δ + 2 qN )
So if , we have
q ≥ 4Δ − 1
E[d(Xt+1, Yt+1) ∣ (Xt, Yt)] ≤ (1 − 1 qN) d(Xt, Yt)
dTV(Xt, Yt) ≤ (1 − 1 qN )
t
⋅ N ≤ ε ⟹ τmix(ε) ≤ qN (log N + log ε−1)
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Geometric View of Mixing
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Geometric View of Mixing
A Markov chain is a random walk on the state space
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Geometric View of Mixing
A Markov chain is a random walk on the state space
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Geometric View of Mixing
A Markov chain is a random walk on the state space Which random walk mixes faster?
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Geometric View of Mixing
A Markov chain is a random walk on the state space Which random walk mixes faster? We will develop tools to formalize the intuition
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