Shuffling Cards via One-sided Transpositions Stephen Connor Joint - PowerPoint PPT Presentation

Shuffling Cards via One-sided Transpositions Stephen Connor Joint work with Oliver Matheau-Raven and Michael Bate SAMBa Conference July 2020

Introduction Consider the following method for shuffling a pack of n cards: • Right hand chooses a card uniformly at random ; • Left hand chooses a card uniformly from below the Right hand; • The two chosen cards are transposed . n Natural question j How many shuffles does it take to “randomize” the deck? i (What is this shuffle’s mixing time ?) 1

Card shuffles as random walks Most interesting card shuffles can be viewed as random walks on the symmetric group, S n , with uniform stationary distribution π n : • top-to-random • random k -cycles • riffle shuffle • adjacent transpositions • transpose top and random • semi-random transpositions Measure distance from equilibrium using the total variation metric : P t � � d n ( t ) = sup n ( B ) − π n ( B ) ∈ [0 , 1] . B ⊂ S n Define the ε -mixing time to be t mix ( ε ) = min { t : d n ( t ) ≤ ε } . n

The cutoff phenomenon Many shuffles exhibit somewhat surprising convergence behaviour... Definition The sequence of shuffles generated by { P n } n ∈ N exhibits a cutoff at time { t n } if  1 if c < 1  n →∞ d n ( ct n ) = lim 0 if c > 1 .  1.0 0.8 Cutoff implies that 0.6 t mix ( ε ) ∼ t n for all ε > 0. 0.4 n 0.2 0.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0

The one-sided transposition shuffle Our shuffle transposes cards in positions ( i , j ) with probability P n ( i , j ) = 1 jn , for all 1 ≤ i ≤ j ≤ n . This differs significantly from previously studied shuffles which have been analysed using group representation theory: • dependence between Left and Right hands n • generating set is entire conjugacy j class of transpositions, but P n is far from uniform on this set i 1

Our main results Theorem The one-sided transposition shuffle exhibits a cutoff at t n = n log n . Diaconis & Shahshahani (1981) showed that the standard transposition shuffle exhibits a cutoff at time n 2 log n . By biasing the Right hand, we can recover this result as a special case of the following: Theorem Suppose that the Right hand chooses card j with probability proportional to j α . Then we see a cutoff at time t n : α ( −∞ , − 1) − 1 ( − 1 , 1] (1 , ∞ ) ζ ( − α ) n − α log n n (log n ) 2 1 α t n 1+ α n log n 1+ α n log n

Upper bound We use the classical ℓ 2 bound on total variation distance. Lemma Let the eigenvalues of P n be 1 = β 1 > β 2 ≥ · · · ≥ β m > − 1. Then d n ( t ) 2 ≤ 1 � β 2 t . i 4 i � =1 Our analysis is inspired by recent work of Dieker & Saliola (2018) and Bernstein & Nestoridi (2019) on the Random-to-Random shuffle. To get a handle on the eigenvalues of P n we need to introduce the concept of Young tableaux .

Young tableaux Definition A partition of n is a decreasing tuple λ = ( λ 1 , λ 2 , . . . , λ l ) such that � i λ i = n and λ 1 ≥ · · · ≥ λ l . We denote this by λ ⊢ n . We may represent a partition using a Young diagram , e.g. (3 , 2) = (2 , 2 , 1) = (5) = A standard Young tableau (SYT) is an allocation of 1 , . . . , n to a Young diagram, such that rows and columns are increasing, e.g. 1 2 3 1 2 4 1 2 5 1 3 4 1 3 5 4 5 3 5 3 4 2 5 2 4 The dimension of λ , d λ , is the number of tableaux of shape λ .

Link to eigenvalues Theorem The eigenvalues of P n are labelled by standard Young tableaux of size n , and the eigenvalue represented by a tableau of shape λ appears d λ times. Lemma The eigenvalue corresponding to a tableau T is given by eig( T ) = 1 j − i + 1 � T ( i , j ) . n boxes ( i , j ) ∈ T 1 2 3 � 1 4 5 then eig( T ) = 1 1 + 2 2 + 3 3 + 0 4 + 1 � Example: if T = . 5 5

Main idea: we may lift the eigenvalues of P n to obtain those of P n +1 by following paths through Young’s lattice . 1 1 2 1 2 3 1 2 3 4

Upper bound on the mixing time Combining these results we obtain the bound: d n ( t ) 2 ≤ 1 = 1 � � � β 2 t d λ eig( T ) 2 t i 4 4 i � =1 λ ⊢ n T ∈ SYT( λ ) λ � =( n ) To establish how large t must be to make this small, we need to understand how the dimensions and eigenvalues behave for large n . Theorem For any c > 0, √ 2 e − c . lim sup n →∞ d n ( n log n + cn ) ≤ Analysis: exploit partial ordering of partitions, monotonicity of eigenvalues, and deal with large and small partitions separately.

Insight: the largest eigenvalue is given by the tableau 1 2 3 . . . n − 2 n − 1 n which has eigenvalue 1 − 1 n and dimension ( n − 1). This makes a contribution to the upper bound of at most � 2 t � 1 − 1 ( n − 1) 2 , n which at time t = n log n + cn is no greater than e − 2 c .

Lower bound Theorem For any c > 2, π 2 lim inf n →∞ d n ( n log n − n log log n − cn ) ≥ 1 − 6( c − 2) 2 . Sketch proof For any set of permutations B , d n ( t ) ≥ P t n ( B ) − π n ( B ) . Focus on cards near the top of the deck, since intuitively these should take longer to mix.

Let B n = { ρ ∈ S n : ρ has ≥ 1 fixed point in top n / log n cards } . Then • π n ( B n ) ≤ 1 / log n • P t n ( B n ) ≥ P (not all top n / log n cards touched by time t ) Now estimate how many shuffles it takes for all top n / log n cards to be touched, by coupling with a counting process. This is similar to the standard coupon-collector problem, but: • the Right and Left hands don’t “collect” cards independently • the counting process can increment by either one or two.

Final remarks • Our analysis yields an exact formula for all of the eigenvalues of the one-sided transposition shuffle • The results give both the cutoff time and a bound on the size of the cutoff window • Weighting the distribution of the Right hand is possible, and shows that the fastest mixing time is obtained when Right and Left hands are independent

References M Bernstein and E Nestoridi. Cutoff for random to random card shuffle. Ann. Probab. , 47(5):3303–3320, 2019. P Diaconis and M Shahshahani. Generating a random permutation with random transpositions. Z. Wahrscheinlichkeit , 57(2):159–179, 1981. AB Dieker and FV Saliola. Spectral analysis of random-to-random markov chains. Adv. Math. , 323:427–485, 2018.