Mixing Times of Self-Organizing Lists Applications and Biased - PowerPoint PPT Presentation

Biased Card Shuffling - pick a pair of adjacent cards uniformly at random Card Shuffling - put j ahead of i with probability p j,i = 1- p i,j Applications 5 n 1 7 ... i j ... n-1 2 6 3 Previous Work New Results

Biased Card Shuffling - pick a pair of adjacent cards uniformly at random Card Shuffling - put j ahead of i with probability p j,i = 1- p i,j Applications 5 n 1 7 ... i j ... n-1 2 6 3 Previous Work New Results Positively biased: We make the assumption that p i,j ≥ 1/2 ∀ i < j

Biased Card Shuffling - pick a pair of adjacent cards uniformly at random Card Shuffling - put j ahead of i with probability p j,i = 1- p i,j Applications 5 n 1 7 ... i j ... n-1 2 6 3 Previous Work New Results Positively biased: We make the assumption that 1 3 5 p i,j ≥ 1/2 ∀ i < j

Biased Card Shuffling - pick a pair of adjacent cards uniformly at random Card Shuffling - put j ahead of i with probability p j,i = 1- p i,j Applications 5 n 1 7 ... i j ... n-1 2 6 3 Previous Work New Results

Biased Card Shuffling - pick a pair of adjacent cards uniformly at random Card Shuffling - put j ahead of i with probability p j,i = 1- p i,j Applications 5 n 1 7 ... i j ... n-1 2 6 3 Previous Work New Results If p i,j ≥ 1/2 ∀ i < j, then π favors increasing permutations

p ij p ji Biased Card Shuffling - pick a pair of adjacent cards uniformly at random Card Shuffling - put j ahead of i with probability p j,i = 1- p i,j Applications 5 n 1 7 ... i j ... n-1 2 6 3 Previous Work New Results Converges to: π(σ) = Π / Z i<j: σ(i)>σ(j) If p i,j ≥ 1/2 ∀ i < j, then π favors increasing permutations

Previous Work Conjecture: If the {p ij } are positively biased, then M is rapidly mixing. Card Shuffling Fill (2003): Gap problem Applications - Conjecture: Previous Work If {p ij } satisfies a “monotonicity” condition, then the spectral gap is max. when p ij =1/2 ∀ i,j New Results - proved this conjecture for n ≤ 4 Benjamini, Berger, Hoffman, Mossel (2005): If p i,j = p > 1/2 ∀ i < j , then θ ( n 2 ) steps

Previous Work Conjecture: If the {p ij } are positively biased, then M is rapidly mixing. Card Shuffling Fill (2003): Gap problem Applications - Conjecture: Previous Work If {p ij } satisfies a “monotonicity” condition, then the spectral gap is max. when p ij =1/2 ∀ i,j New Results - proved this conjecture for n ≤ 4 Benjamini, Berger, Hoffman, Mossel (2005): If p i,j = p > 1/2 ∀ i < j , then θ ( n 2 ) steps Bubley and Dyer (1998): If p i,j = 1/2 or 1 ∀ i<j , then O ( n 3 log n ) steps

Previous work: Biased sampling Simple Markov chain M NN : - pick a pair of adjacent elements (i,j) u.a.r. - swap them with prob. p if i<j, with prob. 1- p otherwise Card Shuffling Permutations : Applications 5 6 3 7 2 1 4 Previous Work 0 0 0 0 0 1 0 New Results 0 0 0 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 1 1 0 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1

Previous work: Biased sampling Simple Markov chain M NN : - pick a pair of adjacent elements (i,j) u.a.r. - swap them with prob. p if i<j, with prob. 1- p otherwise Card Shuffling Permutations : Lattice Paths: Applications 5 6 3 7 2 1 4 Previous Work 0 0 0 0 0 1 0 New Results 0 0 0 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 1 1 1 0 1 1 0 0 0 1 0 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 if p >1/2: τ = Θ (n 2 ) Benjamini et al.

Previous work: Biased sampling Simple Markov chain M NN : - pick a pair of adjacent elements (i,j) u.a.r. - swap them with prob. p if i<j, with prob. 1- p otherwise Card Shuffling Permutations : Lattice Paths: Applications 5 6 3 7 2 1 4 Previous Work 0 0 0 0 0 1 0 New Results 0 0 0 0 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 1 1 1 1 0 1 1 0 0 0 1 0 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 if p >1/2: if p >1/2: τ = Θ (n 2 ) Benjamini et al. τ = Θ (n 2 ) Benjamini et al.

Our Results Card Shuffling •Two classes - M rapidly mixing Applications 1. p i,j = r i ≥ 1/2 ∀ i < j Previous Work 2. {p i,j } have tree structure New Results •Thm: M is not always rapidly mixing. We identify {p i,j } that are positively biased but where M requires exponential time to mix!

Proof of Thm 1 Thm 1: If p i,j = r i ≥ 1/2 ∀ i < j , then M is rapidly mixing. Card Shuffling Applications Previous Work New Results

Proof of Thm 1 Thm 1: If p i,j = r i ≥ 1/2 ∀ i < j , then M is rapidly mixing. Card Shuffling Proof outline: Applications A. Define auxiliary Markov chain M’ B. Show M’ is rapidly mixing Previous Work C. Compare the mixing times of M and M’ New Results

Proof of Thm 1 Thm 1: If p i,j = r i ≥ 1/2 ∀ i < j , then M is rapidly mixing. Card Shuffling Proof outline: Applications A. Define auxiliary Markov chain M’ B. Show M’ is rapidly mixing Previous Work C. Compare the mixing times of M and M’ New Results M’ can swap pairs that are not nearest neighbors - maintain same stationary distribution - Define the probability of swapping i and j that are not nearest neighbors...

p ji p ij Proof of Thm 1 Thm 1: If p i,j = r i ≥ 1/2 ∀ i < j , then M is rapidly mixing. Card Shuffling M’ can swap pairs that are not nearest neighbors Applications same stationary π(σ) = Π / Z distribution: i<j: σ(i)>σ(j) Previous Work Permutation σ : Permutation τ : New Results 5 6 3 1 2 7 4 5 6 2 1 3 7 4

p ji p ij Proof of Thm 1 Thm 1: If p i,j = r i ≥ 1/2 ∀ i < j , then M is rapidly mixing. Card Shuffling M’ can swap pairs that are not nearest neighbors Applications same stationary π(σ) = Π / Z distribution: i<j: σ(i)>σ(j) Previous Work Permutation σ : Permutation τ : New Results 5 6 3 1 2 7 4 5 6 2 1 3 7 4 π ( τ ) P’( σ , τ ) p 2,1 p 2,3 p 1,3 = = p 1,2 p 3,2 p 3,1 π ( σ ) P’( τ , σ )

p ij p ji Proof of Thm 1 Thm 1: If p i,j = r i ≥ 1/2 ∀ i < j , then M is rapidly mixing. Card Shuffling M’ can swap pairs that are not nearest neighbors Applications same stationary π(σ) = Π / Z distribution: i<j: σ(i)>σ(j) Previous Work Permutation σ : Permutation τ : New Results 5 6 3 1 2 7 4 5 6 2 1 3 7 4 π ( τ ) P’( σ , τ ) p 2,1 p 2,3 p 1,3 = = p 1,2 p 3,2 p 3,1 π ( σ ) P’( τ , σ )

p ji p ij Proof of Thm 1 Thm 1: If p i,j = r i ≥ 1/2 ∀ i < j , then M is rapidly mixing. Card Shuffling M’ can swap pairs that are not nearest neighbors Applications same stationary π(σ) = Π / Z distribution: i<j: σ(i)>σ(j) Previous Work Permutation σ : Permutation τ : New Results 5 6 3 1 2 7 4 5 6 2 1 3 7 4 π ( τ ) p 2,3 P’( σ , τ ) p 2,1 p 2,3 p 1,3 = = = p 1,2 p 3,2 p 3,1 π ( σ ) p 3,2 P’( τ , σ )

p ji p ij Proof of Thm 1 Thm 1: If p i,j = r i ≥ 1/2 ∀ i < j , then M is rapidly mixing. Card Shuffling M’ can swap pairs that are not nearest neighbors Applications same stationary π(σ) = Π / Z distribution: i<j: σ(i)>σ(j) Previous Work Permutation σ : Permutation τ : New Results 5 6 3 1 2 7 4 5 6 2 1 3 7 4 π ( τ ) p 2,3 P’( σ , τ ) p 2,1 p 2,3 p 1,3 = = = p 1,2 p 3,2 p 3,1 π ( σ ) p 3,2 P’( τ , σ ) P’( σ , τ )= p 2,3

Proof of Thm 1 Thm 1: If p i,j = r i ≥ 1/2 ∀ i < j , then M is rapidly mixing. Card Shuffling Permutation σ : Permutation τ : Applications 5 6 3 1 2 7 4 5 6 2 1 3 7 4 Previous Work P’( σ , τ )= p 2,3 New Results - Can swap i and j across multiple smaller elements with probability p i,j Idea: Location of element i is independent of the location of all larger elements!

Inversion Tables Permutation σ : Card Shuffling 5 6 3 1 2 7 4 Applications Previous Work New Results

Inversion Tables Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications Previous Work New Results

Inversion Tables Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I σ (i) = # elements j > i appearing before i in σ Previous Work New Results

Inversion Tables Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I(2) I σ (i) = # elements j > i appearing before i in σ Previous Work New Results

Inversion Tables Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I(2) I σ (i) = # elements j > i appearing before i in σ Previous Work • 0 ≤ I σ (i) ≤ n - i New Results

Inversion Tables Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I(2) I σ (i) = # elements j > i appearing before i in σ Previous Work • 0 ≤ I σ (i) ≤ n - i New Results • I is a bijection from S n to T = { ( x 1 ,x 2 ,...,x n ): 0 ≤ x i ≤ n-i }

Inversion Tables Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I(2) I σ (i) = # elements j > i appearing before i in σ Previous Work • 0 ≤ I σ (i) ≤ n - i New Results • I is a bijection from S n to T = { ( x 1 ,x 2 ,...,x n ): 0 ≤ x i ≤ n-i } We will define M’ as a MC on the Inversion Tables.

Inversion Tables Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I(2) I σ (i) = # elements j > i appearing before i in σ Previous Work • 0 ≤ I σ (i) ≤ n - i New Results • I is a bijection from S n to T = { ( x 1 ,x 2 ,...,x n ): 0 ≤ x i ≤ n-i } What happens when you add 1 to x i ?

Inversion Tables Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I(2) I σ (i) = # elements j > i appearing before i in σ Previous Work • 0 ≤ I σ (i) ≤ n - i New Results • I is a bijection from S n to T = { ( x 1 ,x 2 ,...,x n ): 0 ≤ x i ≤ n-i } What happens when you add 1 to x i ? - swap element i with the first j>i to the right

Inversion Tables Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I(2) I σ (i) = # elements j > i appearing before i in σ Previous Work • 0 ≤ I σ (i) ≤ n - i New Results • I is a bijection from S n to T = { ( x 1 ,x 2 ,...,x n ): 0 ≤ x i ≤ n-i } What happens when you add 1 to x i ? - swap element i with the first j>i to the right - happens w.p. 1-r i

Inversion Tables Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I(2) I σ (i) = # elements j > i appearing before i in σ Previous Work • 0 ≤ I σ (i) ≤ n - i New Results • I is a bijection from S n to T = { ( x 1 ,x 2 ,...,x n ): 0 ≤ x i ≤ n-i } What happens when you subtract 1 from x i ? - swap element i with the first j>i to the left

Inversion Tables Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I(2) I σ (i) = # elements j > i appearing before i in σ Previous Work • 0 ≤ I σ (i) ≤ n - i New Results • I is a bijection from S n to T = { ( x 1 ,x 2 ,...,x n ): 0 ≤ x i ≤ n-i } What happens when you subtract 1 from x i ? - swap element i with the first j>i to the left - happens w.p. r i

Proof of Thm 1 Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I(2) Proof outline: Previous Work A. Define auxiliary Markov chain M’ B. Show M’ is rapidly mixing C. Compare the mixing times of M and M’ New Results M’ samples from {( x 1 ,x 2 ,...,x n ): 0 ≤ x i ≤ n-i }: - choose a column i uniformly - w.p. r i : subtract 1 from x i (if possible) - w.p. 1- r i : add 1 to x i (if possible)

Proof of Thm 1 Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I(2) Proof outline: Previous Work A. Define auxiliary Markov chain M’ B. Show M’ is rapidly mixing C. Compare the mixing times of M and M’ New Results M’ is just a cross-product of n independent, 1-dimensional random walks

Proof of Thm 1 Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I(2) Proof outline: Previous Work A. Define auxiliary Markov chain M’ B. Show M’ is rapidly mixing C. Compare the mixing times of M and M’ New Results Idea: Location of element i is independent of the M’ is just a cross-product of n location of all larger independent, 1-dimensional elements! random walks

Proof of Thm 1 Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I(2) Proof outline: Previous Work A. Define auxiliary Markov chain M’ B. Show M’ is rapidly mixing C. Compare the mixing times of M and M’ New Results M’ is just a cross-product of n independent, 1-dimensional random walks

Proof of Thm 1 Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I(2) Proof outline: Previous Work A. Define auxiliary Markov chain M’ B. Show M’ is rapidly mixing C. Compare the mixing times of M and M’ New Results M’ is just a cross-product of n independent, 1-dimensional M’ is rapidly mixing! random walks

Proof of Thm 1 Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I(2) Proof outline: Previous Work A. Define auxiliary Markov chain M’ B. Show M’ is rapidly mixing C. Compare the mixing times of M and M’ New Results

Proof of Thm 1 Permutation σ : Inversion Table I σ : Card Shuffling 5 6 3 1 2 7 4 3 3 2 3 0 0 0 Applications I(2) Proof outline: Previous Work A. Define auxiliary Markov chain M’ B. Show M’ is rapidly mixing C. Compare the mixing times of M and M’ New Results M’ rapidly mixing M is rapidly mixing

Our Results Card Shuffling •Two classes - M rapidly mixing Applications 1. p i,j = r i ≥ 1/2 ∀ i < j Previous Work 2. {p i,j } have tree structure New Results •Thm 2: M is not always rapidly mixing. We identify {p i,j } where {p i,j } are positively biased but M requires exponential time to mix!

Slow Mixing Results •Thm 2: M is not always rapidly mixing. Card Shuffling Applications Previous Work New Results

Slow Mixing Results •Thm 2: M is not always rapidly mixing. Card Shuffling Special Case: Bijection with staircase walks Applications Previous Work New Results

Slow Mixing Results •Thm 2: M is not always rapidly mixing. Card Shuffling Special Case: Bijection with staircase walks Applications n OR < i < j n 1 if i < j ≤ 2 always in order { 2 p ij = Previous Work n n 1 2 3 ... ... n +1 2 2 New Results Permutation σ : 1 2 6 7 3 8 9 4 5 10

Slow Mixing Results •Thm 2: M is not always rapidly mixing. Card Shuffling Special Case: Bijection with staircase walks Applications n OR < i < j n 1 if i < j ≤ 2 always in order { 2 p ij = Previous Work n n 1 2 3 ... ... n +1 2 2 New Results Permutation σ : 1 2 6 7 3 8 9 4 5 10 1 1 0 0 1 0 0 1 1 0

Slow Mixing Results •Thm 2: M is not always rapidly mixing. Card Shuffling Special Case: Bijection with staircase walks Applications n OR < i < j n 1 if i < j ≤ 2 always in order { 2 p ij = Previous Work n n 1 2 3 ... ... n +1 2 2 New Results Permutation σ : 1 2 6 7 3 8 9 4 5 10 1 1 0 0 1 0 0 1 1 0 n n 1’s and 0’s 2 2

Slow Mixing Results •Thm 2: M is not always rapidly mixing. Card Shuffling Special Case: Bijection with staircase walks Applications n OR < i < j n 1 if i < j ≤ 2 always in order { 2 p ij = Previous Work ?? else n n 1 2 3 ... ... n +1 2 2 New Results 1 2 Permutation σ : 6 7 1 2 6 7 3 8 9 4 5 10 3 8 1 1 0 0 1 0 0 1 1 0 9 4 5 10 n n 1’s and 0’s 2 2

Slow Mixing Results •Thm 2: M is not always rapidly mixing. Card Shuffling Special Case: Bijection with staircase walks Applications n OR < i < j n 1 if i < j ≤ 2 always in order { 2 p ij = Previous Work ?? else n n 1 2 3 ... ... n +1 2 2 New Results 1 2 Permutation σ : 6 7 1 2 6 7 3 8 9 4 5 10 3 8 1 1 0 0 1 0 0 1 1 0 9 4 5 10 p 37 n n 1’s and 0’s 2 2

Slow Mixing Results •Thm 2: M is not always rapidly mixing. Card Shuffling Special Case: Bijection with staircase walks Applications n OR < i < j n 1 if i < j ≤ 2 { 2 p ij = Previous Work ?? else New Results 1 2 Permutation σ : 6 7 1 2 6 7 3 8 9 4 5 10 3 8 1 1 0 0 1 0 0 1 1 0 9 4 5 10 p 37 n n 1’s and 0’s 2 2

Slow Mixing Results •Thm 2: M is not always rapidly mixing. Card Shuffling Special Case: Bijection with staircase walks Applications n OR < i < j n 1 if i < j ≤ 2 { 2 n p ij = So each choice of p ij where i ≤ < j 2 Previous Work ?? else determines the bias on square (i,n-j+1) New Results 1 2 Permutation σ : 6 7 1 2 6 7 3 8 9 4 5 10 3 8 1 1 0 0 1 0 0 1 1 0 9 4 5 10 p 37 n n 1’s and 0’s 2 2

Staircase Walks - each box has a different bias p x Card Shuffling x - M can add a box or remove a Applications box according to p x Previous Work New Results

Mixing Times of Self-Organizing Lists Applications and Biased - PowerPoint PPT Presentation

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