Card-cyclic-to-random shuffling with relabeling

0.33 MB·English

by  Johan Jonasson

Card-cyclic-to-random shuffling with
relabeling
5
1
0
2 Johan Jonasson∗†

p
e
S September 29, 2015

8
2

]
R
Abstract
P
h. The card-cyclic-to-random shuffle is the card shuffle where the n cards
t arelabeled1,...,naccordingtotheirstartingpositions. Thenthecardsare
a
m mixedbyfirstpickingcard1fromthedeckandreinsertingitatauniformly
[ randomposition,thenrepeatingforcard2,thenforcard3andsoonuntilall

cardshavebeenreinsertedinthisway. Thentheprocedurestartsoveragain,
3
v byfirstpickingthecardwithlabel1andreinserting,andsoon. Morris,Ning
5 andPeres[3]recentlyshowedthattheorderofthenumberofshufflesneeded
8
to mix the deck in this way is nlogn. In the present paper, we consider
4
5 a variant of this shuffle with relabeling, i.e. a shuffle that differs from the
0
above in that after one round, i.e. after all cards have been reinserted once,
.
1 we relabel the cards according to the positions in the deck that they now
0
have. The relabeling is then repeated after each round of shuffling. It is
5
1 shownthateveninthiscase,thecorrectorderofmixingisnlogn.
:
v
i Shorttitle: Card-cyclic-to-randomshuffling
X
AMSSubjectclassification: 60J10
r
a Keywordsandphrases: mixingtime,stabilityofeigenvalues
∗ChalmersUniversityofTechnologyandUniversityofGothenburg
†ResearchsupportedbytheKnutandAliceWallenbergFoundation
1
1 Introduction
ThesubjectofmixingtimesforMarkovchainsanimportantandexcitingresearch
fieldthathasattractedalotofattentioninrecentdecades. Anoutstandingsubclass
of Markov chains that has been studied extensively is card shuffling, i.e. Markov
chains on the symmetric group S of permutations of n items that one can think
n
ofasthecardsofadeck.
One of the early card shuffles to be studied was the random transpositions
shuffle, where each step of the shuffle is made by picking two cards uniformly
and independently at random and then swapping them. It was shown by Diaconis
and Shahshahani [2] that the mixing time of this shuffle has a sharp threshold at
1nlognshuffles. Itiseasytoseethatatleastorderofnlognshufflesisrequired,
2
since, by the coupon collector’s problem, it takes this order of shuffles until most
cardshavebeentouchedatall. Closelyrelatedtotherandomtranspositionsshuffle
is the top-to-random shuffle where at each step the card presently in position one
is moved to a uniform random position. The sharp threshold for this shuffle is
nlognandagainitiseasytoseethatatleastorderofnlognstepsisrequiredfor
mixing,forsimilarreasons.
Inrecentyearssomemoresystematicvariantsoftheseshuffleshavebeenpro-
posed and analyzed. Mossel, Peres and Sinclair [4] and Saloff-Coste and Zuniga
[6] analyzed the cyclic-to-random shuffle, where at time t the card presently in
position t mod n is swapped with a uniformly random card. Clearly at least once
per n steps, each card will be touched and one of the interesting questions about
this shuffle was if O(n) shuffles is also sufficient to mix the whole deck. The
answer turns out to be negative; indeed the mixing time is still of order nlogn.
Pinsky[5]laterintroducedthecard-cyclic-to-randomtranspositionsshuffle(CCR
shuffle), where at time t the card with label t mod n (i.e. the card that started
out in position t mod n) is moved to a uniformly random position. Again it is
obvious that every card will be touched once every n steps and again one main
question was if this way of systematically randomizing the cards, suffices to mix
the whole deck in O(n), or at least o(nlogn), steps. Again the answer turns out
to be negative; Morris, Ning and Peres [3] prove that nlogn is still the correct
order. In this paper we investigate the card-cyclic-to-random shuffle with relabel-
ing (the CCRR shuffle for short). For k = 1,2,... let round k consist of steps
kn+1,kn+2,...,kn+n of shuffling. The CCRR shuffle is the shuffle that is
exactly as the card-cyclic-to-random shuffle for the first round. After that how-
ever, the cards are relabeled 1,...,n according to their positions after the first
round. Next a new round of CCR shuffling is carried out according to the new
2
labels. After that the cards are relabeled again and a new round of CCR is done,
and so on. The main result of this paper is that relabeling does not help to speed
upmixingeither,atleastnotmorethanbyaconstant.
Theorem1.1 The mixing time of the card-cyclic-to-random transpositions with
relabelingisofordernlogn.
Here,themixingtimeisgivenby
1
τ := min{t : (cid:107)P(X ∈ ·)−π(cid:107) ≤ }
mix t TV
4
where X ∈ S is the state of the deck of cards after t steps of shuffling, π is the
t n
uniformdistributiononS and(cid:107)·(cid:107) isthetotalvariationnorm,giveningeneral
n TV
by
1 (cid:88)
(cid:107)µ(cid:107) := |µ(x)| = max{µ(A) : A ⊂ S}
TV
2
x∈S
forasignedmeasureµonafinitespaceS.
2 Proof of the main result
For the upper bound on τ , it suffices to note that the proof in [3] for the CCR
mix
shuffle goes through exactly as it stands there. Hence we will focus entirely on
the lower bound. The idea of the proof of the lower bound draws on the idea
behindWilson’stechniqueintroducedin[8]and[9],namelytouseaneigenvector
of the transition matrix for the movement of a single card to build a test function.
However since estimating the variance of the test function will in fact be quite
simple here, we will not need Wilson’s Lemma explicitly. A rough outline of the
proofis
1. Show that the position of a given card after one round of CCRR is deter-
√
mined,uptoarandomtermoforder n,bywhereitwasreinserted.
2. In the light of 1, study the idealized motion of a single card which is a
deterministicfunctionofwhereitwasreinserted.
3. Showthatthetransitionmatrixforoneroundofidealizedsinglecardmotion
hasaspectralgapboundedawayfrom1.
3
4. Use the eigenvector corresponding to the second eigenvalue to construct a
teststatistic.
5. Estimate, using 1, the expectation and variance of the test statistic applied
to the CCRR shuffle and establish the lower bound using Chebyshev’s in-
equality.
Because of the cyclic structure of the shuffle, the movement of a single card
is not time-homogenous if we consider individual steps of the shuffle. However
in terms of rounds, the movement of a given card is indeed a time-homogenous
Markovchain. LetA = A(n)denotethetransitionmatrixofthischainonncards.
It turns out that when analyzing this Markov chain, it is convenient to denote the
possible positions a card can have in the deck as 1/n,2/n,3/n,...,1 (instead
of the usual 1,2,...,n). Write Q := [n]/n = {1/n,2/n,...,1} for the set of
n
positions. As usual, we will identify a card with its starting position, i.e. when
we speak of card a, a ∈ Q , we are considering the card that starts in position a.
n
Since this is the na’th card from the top in the starting order of the deck, we may
sometimesalsospeakofthiscardascardna.
Itisdifficulttocomeupwithaclosed-formexpressionforA,buttheactionof
Acanbeprobabilisticallydescribedasfollows. Consideracardthatstartsaround
in position a ∈ Q . Let us refer to the cards 1/n,...,a−1/n as white cards and
n
to the cards a+1/n,...,1 as black cards. Now in a first stage the na−1 white
cardsaresequentiallypickedoutandreinsertedatindependentuniformpositions.
During this stage a certain number of cards will be reinserted above card a in the
deckwhereastheotherswillbeuniformlyspreadoutamongtheblackcardsbelow
card a. The cards that in this stage end up above card a will form a well-mixed
layer of white cards. Note that during stage 1, card a will move gradually higher
up in the deck. (Here we say that if a < b, then position a is higher up than, or
above,positionb.)
Next, after stage 1, card a itself is picked out and reinserted at a uniformly
random position U = U ∈ Q ; this is stage 2. In the third and final stage, the
n n
black cards are picked out and reinserted. If card a was reinserted in the white
layer at the top, then card a will move gradually down the deck during the whole
ofthisstage,whereasifnot,thenstage3dividesintothetwosub-stageswherein
thefirstofthese,stage3a,theblackcardsabovecardaarereinsertedandamoves
upwards and in the second, stage 3b, the black cards below card a are reinserted
andamovesdownthedeck.
Even though we will not need the exact distribution of where card a ends up
underthisprocedure,wewillstillneedagoodapproximatecontrol. Thefollowing
4
twolemmaswillbeusefulforthat.
Lemma2.1 Let the sequence Y ,Y ,...,Y be recursively defined by Y =
0 1 n(1−a) 0
a ∈ Q and Y = Y +1/n with probability Y and Y = Y with probability
n t+1 t t t+1 t
1−Y (wheretheseeventsareconditionallyindependentofY ,Y ,...,Y given
t 0 1 t−1
Y ). Then
t
(cid:18) 1(cid:19)t
E[Y ] = 1+ a
t
n
and
a (cid:88)2t (cid:18) 1(cid:19)j a2 (cid:18) 1(cid:19)2t
Var(Y ) ≤ 1+ − t 1+ .
t+1 n2 n n2 n
j=t
Inparticular,forallt,
2
Var(Y ) < n−1.
t
5
Proof. ByconditioningonY wegetthat
t
(cid:20) (cid:18) (cid:19) (cid:21) (cid:20)(cid:18) (cid:19) (cid:21)
1 1
E[Y ] = E Y Y + +(1−Y )Y = E 1+ Y
t+1 t t t t t
n n
whichprovestheexpressionfortheexpectation. Forthevariancepart,writev :=
t
Var(Y ). Thenv = 0andrecursively
t 0
Var(Y ) = E[Var(Y |Y )]+Var(E[Y |Y ]).
t+1 t+1 t t+1 t
BydefinitionoftheY ’s,Var(Y |Y ) = Y (1−Y )/n2andbytheaboveE[Y |Y ] =
t t+1 t t t t+1 t
(1+1/n)Y . Forforfirsttermwehave
t
E[Y ] E[Y2]
E[Var(Y |Y )] = t − t
t+1 t n2 n2
(cid:32) (cid:33)
(1+1/n)ty 1 (cid:18) 1(cid:19)2t
= 0 − v +a2 1+ .
n2 n2 t n
Addingthesecondtermandwritingc := 1+1/ngives
(cid:18) 1 (cid:19) cta
(cid:0) (cid:1)
v = c2 − v + 1−cta (1)
t+1 n2 t n2
1 1
< c2v + cta− c2ta2.
t n2 n2
5
Thisrecursionisreadilysolvedandgives
a (cid:88)2t a2
v < cj − tc2t.
t+1 n2 n2
j=t
By(1),v isincreasingint,sowegetanupperboundonpluggingint = n(1−a)
t
ontherighthandsideandthenget
ea
(cid:0) (cid:1)
v < e1−2a −e−a −a(1−a)e1−2a
t+1
n
2
< n−1,
5
wherethesecondinequalityfromstandardoptimizationovera.
(cid:50)
Lemma2.2 LetX ∈ L2(R)bearandomvariableandf : R → Rbecontractive,
i.e.|f(x)−f(y)| ≤ |x−y|forallx,y ∈ R. Then
Var(f(X)) ≤ Var(X).
Proof. LetX andX betwoindependentcopiesofX. Then
1 2
1
Var(X) = Var(X −X )
1 2
2
1
= E[|X −X |2]
1 2
2
1
≥ E[|f(X )−f(X )|2]
1 2
2
= Var(f(X)).
(cid:50)
Let Z = Z be the position that card a = a ∈ Q ends up in after one round
n n n
ofshuffling. Foreachb ∈ [0,1],defineG : [0,1] → [0,1]as
b
(cid:26) e1−bu, u ≤ u (b) := 1−(1−b)eb
G (u) = 0 (2)
b ee−b(1−u) −(1−u)e1−b, u > u (b)
0
Note that G (u) is continuous in (b,u) and for each b, G is differentiable for
b b
u (cid:54)= u (b). SeeFigure1toseeaplotofG forafewdifferentb. ThefunctionsG
0 b b
6
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Figure 1: The function G (u) for b = 0.3,0.5,0.7,0.9. The smaller the b, the
b
largertheascentattheorigin.
playacentralroˆleinthefollowinglemma,whichgivescontrolovertheasymptotic
distribution, expectation and variance of Z given U. The limiting distribution
is known and due to Pinsky, see Theorem 4 of [5]. Since we will also need a
quantifiedboundonthevariance,wewillforselfcontainedness,reprovetheresult
below.
Lemma2.3 Forallu ∈ Q ,
n
(cid:18) (cid:19)
2
E[Z|U = u] ∈ 1± G (u). (3)
a
n
and
Var(Z|U = u) < 9n−1,
orequivalently,writingZ = E[Z|U]+D,
Var(D) < 9n−1. (4)
Inparticular,ifa → a ∈ [0,1],then
n
d
Z → G (V)
n a
whereV isuniformon[0,1].
7
Proof. Let W = W ∈ Q be such that nW is the number of white cards
n n
that go to the top layer of white cards in stage 1. We will start by estimating the
varianceofZ givenW = w andU = u. Ifu ≤ w,sothatstage2movescardato
thetopwhitelayer,thenbyLemma2.1
(cid:18) 1(cid:19)n(1−a)
E[Z|U = u,W = w] = 1+ u
n
and
2
Var(Z|U = u,W = w) < n−1.
5
Thecaseu > wtakessomemorework. Inordertonotoverlyburdenthenotation,
we will until further notice, point out the conditioning on U = u and/or W = w
bywritingindexesuand/orw attheconditionalexpectationsandvariances.
Let S ∈ Q be such that nS is the number of black cards in positions w +
n
1/n,...,u − 1/n; these are the black cards that get reinserted in stage 3a. Let
nN be the number of cards below card a after these nS black cards have been
reinserted. Notethatwhenthefirstpartofstage3starts,thencardaisinuandat
that point, the number of black cards below a is n(1−a−S); these are the ones
thatwillgetreinsertedinthestage3b. HencebyLemma2.1,
(cid:18) 1(cid:19)n(1−a−S)
E [Z|N,S] = 1+ (1−N)
u,w
n
and
(cid:18) 1(cid:19)nS
E [N|S] = 1+ (1−u).
u,w
n
Hence
(cid:18) 1(cid:19)n(1−a−S)
E [Z|S] = (1−E [N|S]) 1+
u,w u,w
n
(cid:18) 1(cid:19)n(1−a−S) (cid:18) 1(cid:19)n(1−a)
= 1+ − 1+ (1−u).
n n
Itfollowsthat
(cid:34) (cid:35)
(cid:18) 1(cid:19)n(1−a−S) (cid:18) 1(cid:19)n(1−a)
E [Z] = E 1+ − 1+ (1−u). (5)
u,w u,w
n n
8
Wealsogetthat
Var (Z|S) = E [Var (Z|N,S)|S]+Var (E [Z|N,S]|S)
u,w u,w u,w u,w u,w
(cid:32) (cid:33)
2 (cid:18) 1(cid:19)n(1−a−S) (cid:12)
< n−1 +Var 1+ (1−N)(cid:12)S
5 u,w n (cid:12)
2 (cid:18) 1(cid:19)2n(1−a−S) 2
= n−1 + 1+ n−1
5 n 5
2
< (1+e2)n−1.
5
Therefore
Var (Z) = E [Var (Z|S)]+Var (E [Z|S])
u,w u,w u,w u,w u,w
(cid:32) (cid:33)
2 (cid:18) 1(cid:19)n(1−a−S)
< (1+e2)n−1 +Var 1+
u,w
5 n
2
≤ (1+e2)n−1 +Var (eS)
u,w
5
(cid:18) (cid:19)
2 13
< + e2 n−1
5 20
byLemma2.2,sincethemapS → (1/e)(1+1/n)n(1−a−S) iscontractiveandnS,
givenU = uandW = w,ishypergeometricwithvarianceatmostn/4.
Now bring back the conditioning on W into ordinary notation. What we have
just shown is among other things, that Var (Z|W) < Cn with C := 2/5 +
u
13e2/20. ThusVar (Z) = E [Var (Z|W)]+Var (E [Z|W]) < Cn+Var (E [Z|W]).
u u u u u u u
However,by(5)
|E [Z|W = w]−E [Z|W = w−1/n]|
u u
(cid:12) (cid:34) (cid:35) (cid:34) (cid:35)(cid:12)
(cid:12) (cid:18) 1(cid:19)n(1−a−S)(cid:12) (cid:18) 1(cid:19)n(1−a−S)(cid:12) (cid:12)
= (cid:12)E 1+ (cid:12)W = w −E 1+ (cid:12)W = w−1/n (cid:12)
(cid:12) u n (cid:12) u n (cid:12) (cid:12)
(cid:12) (cid:12)
(cid:34) (cid:35)
(cid:18) 1(cid:19)n(1−a−S)(cid:18)(cid:18) 1(cid:19) (cid:19)(cid:12) e
≤ E 1+ 1+ −1 (cid:12)W = w ≤ ,
u n n (cid:12) n
wherethefirstinequalityusesthattheconditionaldistributionsofnS givenW =
w and W = w − 1/n respectively, can easily be coupled so that the realizations
9
donotdifferbymorethan1. Itnowfollowsthat
2
Var (E [Z|W]) ≤ e2Var (W) < e2n−1
u u u
5
wherethesecondinequalityfollowsfromLemma2.1. Hence
(cid:18)8+21e2(cid:19)
Var (Z) < n−1 < 9n−1.
u
20
ThisallowsustowriteZ = E[Z|U]+D,whereD = Z −E[Z|U]has
Var(D) = E[Var(D|U)] = E[Var(Z|U)] < 9n−1.
Thisfinishestheproofofthevariancepartofthelemma.
Next let n → ∞, considering for each n card a , where a → a ∈ [0,1].
n n
Then,byLemma2.1,E[1−W ] → (1−a)ea sosinceVar(W ) ≤ 2n−1/5 → 0,
n n
W converges in probability to u (a) = 1 − (1 − a)ea. Given U = u → u
n 0 n n
and W = w → u (a), we get by the above that for the case u < u (a), so
n n 0 0
that u < w eventually, that Z →P e1−au. For u > u (a), so that u > w
n n n 0 n n
eventually,nSishypergeometricandhasexpectationn(u −w )(1−a )/(1−w ).
n n n n
Plugging in the limit u (a) of w , it follows that S converges in probability to
0 n n
1 − a − e−a(1 − u). Plugging this into (5) together with (4) and the fact that
P
W → u (a),givesthatconditionallyonU = u → uwithu > W ,
n 0 n n n
Z →P ee−a(1−a) −(1−u)e1−a.
n
Summing up, we get that the position of a card starting from position a ∈ Q ,
n n
a → a, after one round of CCRR shuffling converges in distribution to that of
n
G (V) where V is uniform on [0,1], as desired. Also, taking µ := E [S], we
a u,w
have,sinceVar (S) ≤ n−1/4,
u,w
(cid:18) (cid:19)
1
E [e−S] = e−µE [e−(S−µ)] ∈ e−µ(1,1+Var(S)) ⊆ e−µ 1± .
u,w u,w
4n
Plugging this into (5) , letting n → ∞ and again using the convergence of W
n
gives(3). (cid:50)
Recall that we write A = A(n) for the transition matrix of the movement
of a card under one round of CCRR. Write B = B(n) = [b ] for the tran-
ij
sition matrix of a card that moves according to G (V). More precisely, let V
a
10