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Convergence rate of i-cycles after an m-shelf shuffle

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Convergence rate of i-cycles after an m-shelf shuffle

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Content CONVERGENCE RATE OF I-CYCLES AFTER AN M-SHELF SHUFFLE by Haining Ren A Dissertation Presented to the FACULTY OF THE USC GRADUATE SCHOOL UNIVERSITY OF SOUTHERN CALIFORNIA In Partial Fulfillment of the Requirements for the Degree DOCTOR OF PHILOSOPHY (MATHEMATICS) August 2015 Copyright 2015 Haining Ren To mom and dad. ii Acknowledgements First I’d like to express my deepest and sincerest gratitude toward my advisors Jason Fulman and Richard Arratia. This work would not have been possible without them. Professor Fulman introduced me to the problem and provided unwavering encouragements and guidance along the way. Professor Arratia has beensoapproachableandgenerouswithhistimeandhisinsights,answeringagreat manyquestionsandteachingmeinvaluablethingsaboutlogarithmiccombinatorial analysis. Thanks also goes to them for their feedback on the thesis draft. Thanks to Professor Lototsky. Beyond his innumerable advice, he has been holding a wonderful eclectic probability seminar every semester. It was always fresh and interesting to come to class and I have garnered many useful Ph.D. student skills from these meetings. Thanks to Peter Ralph, Larry Goldstein for being on my oral and thesis com- mittee. ThanksalsotoAmyYungwhoalwaystakesgoodcareofthegraduatestudents. IhavemanyfondmemoriesofUSCmathematicsdepartmentanditsnurturing, friendly environment. I’m grateful for all the friendship I’ve received during my studies. Thanks especially to John Pike, Adam Ericksen and Brian Ryals, who inspired lots of mathematical discussions and listened to me patiently when I had doubts. iii Last but not least I’d like to thank my fiancé Alex Rozinov for the infinite love, encouragement, support and comic relief. iv Table of Contents Dedication ii Acknowledgements iii Abstract vii Chapter 1: Introduction 1 Chapter 2: Unimodal permutations 4 2.1 Gannon’s cycle structure decomposition for unimodal permutations 5 2.2 Analytical combinatorial structure of( ?) .............. 11 2.2.1 A bijection of( ?) c and Pset( ? )⇥ Mset( ? )........ 14 2.2.2 Useful counting result . . . . . . . . . . . . . . . . . . . . . 19 Chapter 3: Expected number of cycles of unimodal permutations 21 3.1 The number of components in combinatorial structures . . . . . . . 21 3.2 Expected number of cycles of unimodal permutations . . . . . . . . 23 Chapter 4: Convergence Rate of cycles of unimodal permutations 28 4.1 Set-up of the problem . . . . . . . . . . . . . . . . . . . . . . . . . . 28 4.2 Detailed calculations of D TV (L(C),L( ˜ C)) .............. 30 4.2.1 Term by term differences . . . . . . . . . . . . . . . . . . . . 36 4.2.2 Difference brought on by infinite tails . . . . . . . . . . . . . 38 4.2.2.1 Bounded size bias coupling . . . . . . . . . . . . . 39 4.2.2.2 Tail of ˜ W ....................... 41 Chapter 5: Convergence rate of i-cycles in m-shelf Shuffling 44 5.1 Introduction of m-shelf shuffles . . . . . . . . . . . . . . . . . . . . 44 5.2 M-shelf shuffles, self reciprocal polynomials and necklaces . . . . . . 46 5.2.1 Counting polynomials . . . . . . . . . . . . . . . . . . . . . 47 5.2.2 Necklace complexes and p non-prime . . . . . . . . . . . . . 52 5.3 Computations for q=2m ....................... 57 5.3.1 Bernougeo distributions . . . . . . . . . . . . . . . . . . . . 57 v 5.3.2 More computations and details . . . . . . . . . . . . . . . . 58 Chapter 6: Appendix 64 Bibliography 68 vi Abstract Casino Shelf shufflers with m shelves induce permutations on n cards. In the case where m =1, the permutations induced are unimodal permutations which are the discrete analog of unimodal maps from dynamical systems. We re-derive the generating functions for the cycle structures of uniformly randomly chosen uni- modal permutations and explore their expected number of cycles, parallel to the works of Arratia, Barbour and Tavaré on random permutations and other logarith- mic structures. As a generalization we will also show that the joint convergence rate of small cycles of the permutations induced by the shelf shuffle machine to Bin((2m) i ,m i )+Negbin( 1 (2m) i +1 ,m i ) is much better than o(1/n). vii Chapter 1 Introduction Sophisticated m-shelf shuffling machines were built by a manufacturer of casino equipment. A machine consists of an opaque box with ten shelves. A deck of cards is dropped at the top of the machine and an internal elevator moves the deck up down. Cards are dealt from the bottom of the deck onto the shelves chosen uni- formly at random. Each card is placed above or below the current pile on the shelf withprobabilityhalf. Inaquesttoanalyzetheshufflemachine’sefficacyatdeliver- ing perfectly shuffled decks, Diaconis, Fulman and Holmes derived several classical statistics in their paper [8]. One of such statistics is the i-cycle distribution. The statistics of i-cycle distribution for a uniform randomly chosen permutation from S n is a famous problem and needs little introduction. Let C n i denote the number ofi-cycles in a uniform randomly picked permutation. Asn!1 ,thedistribution of any finite number of the random variables {C n i } approaches n ˜ C i o , which are independent and Poisson distributed with ˜ C i = Poi( 1 i ).Incontrast,keepingthe same notation, in the limiting case ofm-shelf shuffle, the ˜ C i ’s are still independent, but they distributed as the convolution of a Binomial( 1 (2m) i +1 ,m i ) and Negative Binomial((2m) i ,m i ), where m i = 1 2i X d|n d odd µ(d)(2m) i/d . 1 In the case where m=1,thepermutationsaftertheshelfshufflearealways unimodal. The unimodal permutations are interesting in their own right - they have applications in dynamical systems and voting theory. Gannon discovered amonoidaloperationontheseunimodalpermutationsthatrespectstheircyclic structure in [13]. Since Gannon’s paper Thibon and Fulman gave the generating function for the number of unimodal permutations with a given cycle structure around the same time. Here the unimodal permutations serve as our prototypes to the general inquiry oftheconvergencerateoftheC n i ’sbym-shelfshufflers. Wegiveaconvergencerate forfinitelymanyC n i ’sbothofunimodalpermutationsandm-shelfshuffles. Wewill see that unimodal permutations occurs from the simplest 1-shelf shuffle. Along the way,weproveadecompositiontheoremoftheunimodaldistributions,henceoffera new, more elementary proof for the generating function of its cycle-structure using its analytical combinatorial structure, as well as derive new statistics counting the total number of cycles. We will prove that unimodal permutations can be regarded somewhat as a natural example of Mset⇥ Psets over the set of cyclic unimodal permutations, which we will make precise in Chapter 2. Chapter2beginswithadiscussionofunimodalpermutationsfromanalgebraic perspective. We make extensive use of Gannon’s operation on the cycle-types to arrive at a complete decomposition of the permutations, thereby achieving an understanding of their combinatorial structure. In the process we suggest a canonical way of representing the unimodal permutations. In Chapter 3 we calculate the expected number of cycles in unimodal permutations, the theorem can be generalized to the case of m-shelf shuffles. In Chapter 4 we calculate the convergence rate of i-cycles for unimodal permutations. Finally, we extend all the results to arbitrary m-shelf shuffles in Chapter 5. The techniques we use 2 come from analytical combinatorics; the paper that inspires the bigger structure of the arguments is [6] by Arratia, Barbour and Tavaré. These authors have also written extensively on the logarithmic combinatorial class and proven very general theorems that may be applied to our problem. We derive a bound that’s better than the general bound from their book [2]. 3 Chapter 2 Unimodal permutations Unimodal permutations are the discrete analogues of tent functions or continuous functions with one spike. To give the definition, let S n denote the permutation group of size n.Iffor ⇡ 2 S n there exists an ˆ m such that, ⇡ (i)<⇡ (i+1) when i< ˆ m and ⇡ (i)>⇡ (i+1) fori> ˆ m,then ⇡ is unimodal. The inequality is trivially satisfied by the identity permutation and the involution ⇡ (i)=(n+1 i). Let n be the set of unimodal permutations of size n.Weknowfromaneasy computation that | n | = P n i=1 n 1 i 1 =2 n 1 . The idea behind the computation is that we choose the position i where the peak of the permutation occurs and the numbers a 1 ...a i 1 that 1, 2,..., i 1 are being mapped to. Or, we can generate aunimodalpermutationusingthetwolinenotationbygoingbackwardsalongthe sequence {n,n 1,...1}: write down n,thenchoose n 1 to be either on the left or right of n,thenchoose n 2 to be on the left or right of the existing sequence etc. In either case, it’s clear that there are 2 n 1 unimodal permutations. Assume that we pick a unimodal permutation uniform randomly, the position where the peak occurs has expected value n+1 2 . The reason is that every time there is a unimodal permutation with peak at k,ifweconjugateitbyinvolution(i.e reflecting across the middle), the new permutation is still unimodal and the peak of the new unimodal permutation is at n+1 k.Hencetheexpectedpositionfor n is exactly in the middle. 4 2.1 Gannon’s cycle structure decomposition for unimodal permutations It is frequently interesting in analytical combinatorics to break down a combina- torial object into smaller and more ‘atomic’ structures. Arratia and Tavaré have provided many naturally occurring examples of multisets, sets and assemblies in the paper [5]. These include, but are not limited to, random mapping functions, partitions of a set as assemblies, mapping patterns, polynomials over finite fields as multisets, square-free polynomials over finite fields as selection (powerset). In our case, the crucial information is that when 2 S n is unimodal, so are each of its cycles. This implies the ‘unimodality’ can be carried over to smaller parts. The natural question to ask is if we can reverse this process: given a list of unimodal permutations that can be written in one cycle, in what sense can we combine them to arrive at a unimodal permutation while keeping the cycle structure? It turns out that the answer is elegant but the technique involved is not very simple. Before the technical discussion, we will define an important concept - transitive permutation. Definition 2.1.1. A transitive permutation ⌧ 2 S n is one such that any i 2 {1,...,n} can be sent to any j2{1,...,n} by ⌧ k (i)= j for a positive integer k. In the cycle notation, a transitive permutation can be written in one cycle, hence it is also called a unicycle. Let k denote the set of all transitive unimodal permutation of size k. We also would like to provide a notation table for future use, the notation follows Gannon’s original in [13]. 5 I n the set {1,2,..n} ( n) unimodal permutations of size n n transitive unimodal permutations of size n ( ?) [ n ( n) ? [ n n unimodal permutation m m( ) is the maximum of ˆ m 1 (m) is the number that comes before m in a subcycle It all starts with mappings between a set of two transitive unimodal cycles, one in j andonein k ,andoneunimodalpermutationin( j+k)withsubcyclesofsize j andk.Intuitivelyspeaking,weareableto‘nestle’twounimodaltransitivecycles together by re-indexing the numbers in both cycles while respecting the relative ordering within each cycle. The only decision we have to make is to decide which of the two peaks come first. The following map will tell us one way of mixing together the indices from two permutations. Definition 2.1.2. Pick any J ⇢ I k+l , |J| = k.Let ⌦ J : I k ! J and ¯ ⌦ J : I l ! I k+l \J be the order preserving maps. By the operation 1 J 2 we mean the operation: ( 1 J 2 )(i):= 8 > > < > > : ⌦ 1 ⌦ 1 (i) if i2 J ¯ ⌦ 2 ¯ ⌦ 1 (i) if i2 I j+k \J One obvious property is that the maximum ˆ m := ( 1 J 2 ) 1 (k + l) 2 ⌦(ˆm 1 ), ¯ ⌦(ˆm 2 ) , where ˆ m 1 = 1 1 (k), ˆ m 2 = 1 (l). 6 Let’s see an example. Consider (12) J (1234),let J = {4,6},sothat I 6 \J = {1,2,3,5}.Bydefinition, 1,2 are raised to 4,6 respectively. The complement of {4,6} in I 6 are {1,2,3,5} and the numbers in (1234) are mapped to {1,2,3,5} in the order preserving way, hence (12) J (1234) = (46)(1235). It’s clear that if 1 has size k and 2 has l then 1 J 2 2 S n , where n =l+k, and that ( 1 J 2 ) has one cycle of sizek and another of sizel.What’sinteresting is that there is a unique set J such that the permutation defined above will be unimodal satisfying⌦(ˆm 1 )< ¯ ⌦(ˆm 2 ) . Theorem 2.1.3 (Gannon). Given 1 2 k and 2 2 l ,thereexistsexactlyone set J with 1 J 2 2 ( k +j),satisfying⌦(ˆm 1 )< ¯ ⌦(ˆm 2 ). Interested readers are referred to the original paper [13] on the procedures of finding such an J. The process is very technical and omitted here. By the result of this theorem from [13], it makes sense to short-hand J to since such a set is unique if we require⌦(ˆm 1 )< ¯ ⌦(ˆm 2 ) to be always satisfied. Definition 2.1.4. If 1 2 k and 2 2 l ,let 1 2 be the unique unimodal sum 1 J 2 obeying⌦(ˆm 1 )< ¯ ⌦(ˆm 2 ). Corollary 2.1.5. If 1 6= 2 ,then 1 2 6= 2 1 .Thereisonlyoneuniqueway of combining two copies of the same transitive unimodal cycle so that the result is still unimodal. Definition 2.1.6 (Gannon). s 2 k is called acute if k = 1 s (k)mod2,other- wise call it grave. Definition 2.1.7. (Gannon) The cycle-type of (i 1 ,...,i n ) is (⌧ (i 1 ),...,⌧ (i n )), where ⌧ is the unique map that maps {i 1 ,...,i n } to {1,...n} and preserves the order of {i 1 ,...,i n }. The cycle-type of ⌧ 2 k is also referred to as the shape of ⌧ . 7 Definition2.1.8 (Gannon). Let 2 ( n),letJ bethesetofnumbersthatappear in a subcycle of so that (J)=J.Denote | J as this subcycle. A( ):= { s : s is an acute shape of a subcycle | J of } Let m( | J ) be the maximum of | J . A > ( ):= { s : s is the acute shape of | J , and m( | J )>m( s )} In other words, A > ( ) contains all acute shapes of subcycles such that the maximum of the subcycle is larger than the maximum of its shape. Similarly, we may defineA < ( ),andthedefinitionsforG,G < andG > are obvious. Now we are ready to state the most important theorem by Gannon that deals with the existence of unimodal sums, this will be useful later: Theorem 2.1.9 (Gannon). Let i 2 ( n i ),for i=1,2.Define ˆ m i = 1 i (n i ), J i = n i , 1 i (n i )= ˆ m i , 2 i (n i ),... ,and ˆ i =⌦ 1 J i i ⌦ J i so that ˆ i is the shape of subcycle of i containing the maxium. Then: iifeitherA > ( 1 )\ A > ( 2 ) orG < ( 1 )\ G < ( 2 ) are nonempty, then there are no unimodal sums of the form 1 2 or 2 1 ii if instead ˆ 1 2 G < ( 2 ) or ˆ 2 2 A > ( 1 ),thenthereisnounimodalsum 1 2 . iii otherwise, there is exactly one unimodal sum 1 2 .(⌦(ˆm 1 )< ¯ ⌦(ˆm 2 )) This theorem generalizes Theorem 2.1.3 because we no longer require i ’s be transitive. Theorem 2.1.10. Let i 2 ( ?). 8 iIfboth 1 2 and 2 1 exist, then they will be equal iff ˆ 1 = ˆ 2 ,usingthe notation of Theorem 2.1.9. ii Associativity holds when is defined. Corollary 2.1.11. Let 2 i .Thenthereisauniqueunimodalpermutationthat corresponds to ( ) k = ... . Proof. We will use Theorem 2.1.9 and Theorem 2.1.10 to argue for the existence and uniqueness of ( ) k . In the case where is acute, we will use induction. We know that always exists. Assume ( ) k 1 exists, consider the sum ( ) k 1 . A > ( )=G < ( )=; since there is only one shape in and s = so m( )= m( s ). Hence Theorem 2.1.9 i does not apply. Moving on to Theorem 2.1.9 ii,sinceG < ( )=; ,it’simpossiblethat ˆ k 1 2 G < ( ).Ontheotherhand, ˆ = 2 A > ( k 1 ) since all the shapes of k 1 are but for at least some J, m( k 1 J )>m( ).Hence ( ) k 1 does not exist. However, if we instead write k 1 , then Theorem 2.1.9 i does not apply for the same reason as we stated earlier, and neither does Theorem 2.1.9 ii since nowA > ( )=; andG < ( k 1 )=; . So given ( ) k 1 ,thereisonlyonewayof‘adding’ to the right. Hence there is one unique sum for ( ) k = ... (k times). On the other hand, if is grave, thenA < ( ) is empty and furthermore since is its own shape, G < ( ) is empty. Theorem 2.1.9 does not apply. ˆ = /2 A > (( ) k )=; . NowapplyTheorem2.1.10i. Clearly ˆ ( k 1 )= ˆ sothat ( ) k 1 = ( ) k 1 . In order to have a canonical representation of any permutation using only its cycle-types, it’s useful to have a total ordering of the cycle-types. It matters little which specific ordering we are using for what comes next. For simplicity’s sake we 9 will use the following lexicographical order: write all cycles i 2 ? beginning with (1). s < t if the length of s is smaller than that of t . For cycle-types with equal lengths, let i < j if the following happens: for the first s such that i (s)6= j (s), i (s)< j (s). It follows from definition that (1) < (12) < (1234) < (1324)...,for example. By Corollary 2.1.11,thereisauniqueunimodaldistributionthatcorresponds arepeated k sum of any cycle type, hence it makes sense to talk about a cycle type along with its multiplicity, we will call it a cycle power. Definition 2.1.12.Arepeated j Gannon sum of a cycle-type is referred to as a cycle-power and notated as j Theorem 2.1.13. Fixing the order of the list of distinct cycle powers and adding them recursively using the Gannon operation gives all the possible unimodal permu- tations with the same cycle-type decomposition. In other words, there is a canonical way of writing down a unimodal permutation as a Gannon sum of its cycle-types. Proof. Let 2 ( n),andtheorderedcyclepowersare a 1 1 , a 2 2 ,.... a k k where 1 < 2 <...< k . Take the largest cycle type with its multiplicity, call it a k k and decompose into its unique Gannon sum of either = L a k k or = a k k L , depending on whether⌦(ˆm( L ))<⌦(ˆm( an k )). This sum is defined since there are no identical shapes in L and a k k so we may apply Theorem 2.1.9. Now we continue the break-down: we decompose L into L = L 1 a k 1 k 1 or L = a k 1 k 1 L 1 . Keep applying this procedure on the piece that has more than one cycle-type, here denoted as L i , we will eventually have L k 2 = a 1 1 a 2 2 or L k 2 = a 2 1 a 1 2 , where a i i is a cycle-power. Now, going backwards, we are able to write down as non-commutative sums of the k cycle powers: Starting with a 1 1 ,add a 2 2 to the left or right according 10 to the ‘breakdown’, after having added a i i ,choosetoadd a i+1 i+1 to the left of the existing sum, or to the right of the existing sum. This operation stops after we’ve added all k parts. Since there are k cycle powers (cycle types with multiplicity), it takes us k 1 recursive Gannon sums to reach the end result. At each step of operation we have a choice of either adding the next largest cycle-power to the left or to the right of the existing sum. There are no repeated sums in this way, since if at any i step if we choose left instead of right, there will be a difference in the ordering of the⌦(ˆm( L k i )) and⌦(ˆm( a i i )),andhencetheresultingunimodalpermutations cannot be the same. There are altogether 2 k 1 different sums we can have with exactly the same k cycle-powers. Let’s illustrate Theorem 2.1.13 with some examples. The cycle types of (1249)(358)(67) are (1234), (123), (12). (1249)(358)(67) = (1234) (125)(34) = (1234) ((125) (12)). The first equality is true because ˆ m((1249)) < ˆ m((358)(67)).Secondqualityistruebecause ˆ m(125)< ˆ m(34). The cycle types of (129)(38)(47)(56) are (123) and (12) with a multiplicity of 3. Since2= ˆ m((129)) <(ˆ m(38)(47)(56)) = 3. We can write (129)(38)(47)(56) = (123) (12) 3 . 2.2 Analytical combinatorial structure of( ?) The previous section focuses on the technical elements that make the Gannon operation work without providing too much motivation. In this section we fill in the missing piece of the puzzle. In summary, we need Gannon’s operation to find out the combinatorial structure of the unimodal permutations. 11 Quite often, combinatorial classes decompose naturally into their components. For an object from the combinatorial class of total weight n,if C n i is the number of smaller components of weight i,wehave C n 1 +2C n 2 +···+nC n n =n. For example, a permutation has cycles as its components; a polynomial has irreducible polynomials as its components; a positive integer has prime factors as its components. They all satisfy the equation above. Given a class of structuresm that has m i different objects of weight i. There are typically 4 kinds of construction from this class: sequence, cycle, multiset and powerset. For a succinct introduction on the concepts of these constructions, [15] is a good place. Another source with an encyclopediac amount of examples is [10]. An integer can be regarded as a multiset of prime numbers. If we are given the list of primes along with their multiplicity it’s easy to multiply them and arrive at the positive integer. The multiset of prime numbers uniquely determines the positive integer. We do not need to worry about the order of the multiplication, since positive integers commute. In the unimodal permutation case, the compo- nents are cycle-types. When we are ‘combining’ the cycle types together, we are concerned with order, since the operation is non-commutative. We need to decide which ordering will end up in a different permutation, and which are the same. Hence it is worthwhile to come up with a unique decomposition of any unimodal sum into a non-commutative sum of its cycle-types as it will allow us to see the combinatorial structure. The corollaries from last Chapter suggest a way of representing any unimodal permutation by its cycle-types via the operation. The cycle powers j j of a 12 unimodal permutation is analogous to p j j in a factored integer. Now it almost seems as if the unimodal sum should be a multiset of cycle-types, but in fact this is not the case. Firstly the operation is non-commutative, secondly, if any cycle-type appears multiple times, they should appear together in our choice of presentation ((123) 2 (12) (123) isn’t defined). Our representation of unimodal permutations says that if we fix a particular ordering of the cycle-types, when we are adding the cycle-powers along the list, we can choose to add the next cycle power either to the left, or to the right of the existingnon-commutativesum. It’smuchliketravelingdownabinarytree, ateach node, we can choose to take the branch representing adding the next cycle-type with powers to the left, or take the other branch and add the next cycle-type with powers to the right. Each of the path represents a different unimodal permutation. And there are altogether 2 k 1 different unimodal permutations with the same k cycle-powers. a 1 1 a 1 1 2 a 2 ( a 1 1 a 2 2 ) a 3 3 a 3 3 ( a 1 1 a 2 2 ) a 2 2 a 1 1 ( a 2 2 a 1 1 ) a 3 3 a 3 3 ( a 2 2 a 1 1 ) ... The bracketing and keeping track of addition on the left or right is nice but a littletedious. Wecanincorporatethetreeintoastringofcycle-powersprovidedwe make note of which branch we take at each node. This should allow us to see the 13 structure of the unimodal permutations. So let’s write subscript L with a cycle- power if we are adding it to the left of the existing sum, so that 1 2 means 1 2 and 1 2L means 2 1 .Forexample, (12)(123) 3 L (1234) 2 is the representation for ((123) 3 (12)) (1234) 2 ,and (123) 5 (1234) 2 (1324) 3 L is the representation for (1324) 3 ((123) 5 (1234) 2 ).Letthese‘string’typeofrepresentationsbecalled unimodalstringrepresentations,andthesetofallstringsderivedassuchfrom the unimodal permutations be denoted ⇤ (n). Every element from ⇤ (n) starts with a cycle-power that has no subscript. This cycle-power is followed by other cycle-powers that may or may not have L as a subscript. For symmetries’ sake, we ‘complete’ ⇤ (n) by adding a mirror image L(w) for every representation w by writing an L as a subscript of the first cycle-power of w 2 ⇤ (n). Call the completed set c (n)= {w :w2 ⇤ (n)}[{ L(w),w2 ⇤ (n)} and we have proven: Lemma 1. | c (n)|=2| ⇤ (n)|=2|( n)|. Definition 2.2.1. Let( ?) c =[ i 1 c (i). 2.2.1 A bijection of ( ?) c and Pset( ? )⇥ Mset( ? ) Many enumeration problems in combinatorics could be approached symbolically using the generating functions. There is quite often a direct translation between operations over generating functions and set-theoretic constructions. In other words, there is a dictionary that translates the operations of union, cartesian prod- uct, set, multiset, and assembly into the language of generating functions. In this way, theproblemofenumeratingaclassofcombinatorialstructuresreducestofind- ing a proper specification expressed in terms of the basic constructions. A through introduction on combinatorial structures can be found in Chapter I of [10]. A combinatorial class is a finite or denumerable set on which a size function is defined, satisfying the following conditions: 14 1. the size of an element if a non-negative integer; 2. the number of elements of any given size is finite. Clearly, both the set of unimodal permutations ( ?) and the set of string expressions( ?) c arecombinatorialclasses. Wewouldliketofirstseethecombina- torial structure of( ?) c over ? ,thecombinatorialclassoftransitivecycle-types. Recall that[ i ( i )= ? . In this work we are primarily concerned of two examples of combinatorial con- structions, multisets and powersets, which will be defined below: Definition 2.2.2. Multiset construction. Multisets are like finite sets which ignores the order between elements, but arbitrary repetitions of elements are allowed. The notation is A = Mset(B) where A is obtained by forming all finite multisets of elements from B. Definition 2.2.3. Powerset construction. The powerset class A = Pset(B) is defined as the class consisting of all finite subsets of class B,orequivalently,as the class Pset(B)⇢ Mset(B) formed of multisets that involves no reptitions. Definition 2.2.4. Two combinatorial classes A and B are said to be (combinato- rially) isomorphic, which is writtenA ⇠ =B,iffthereexistsabijectionfromA to B that preserves size, and also one says that A and B are bijectively equivalent. Wewillnowconstructabijectivemapfrom( ?) c toPset( ? )⇥ Mset( ? ). The basic motivation behind the bijective map is to separate the cycle-types marked withLfromcycle-typesmarkedwithoutLwhileavoidingmultiplewaystoseparate the powers of each cycle type. The necessity of the last requirement will become more clear later on. Theorem 2.2.5. There is a bijection from( ?) c toPset( ? )⇥ Multiset( ? ). 15 Proof. Let a:( ?) c ! Pset( ⇤ ) and b:( ?) c ! Mset( ⇤ ). The rule is the following: 1. a( c ) picks out the distinct cycle-types marked with L,butignorestheir multiplicities. 2. b( c ) picks out the unmarked cycle types with its multiplicity (power) in its entirety and also cycle-powers with subscript L but with their multiplicities subtracted by 1. For example, a((123) 2 L (1234) 5 (12345) 7 L )) = {(123),(12345)} and b((123) 2 L (1234) 5 (12345) 7 L )) = {(123),(1234) 5 ,(12345) 6 } The map (a,b):( ?) c ! Pset( ⇤ ) is well-defined: assume (a( c ),b( c )) 6= (a( 0 c ),b( 0 c )),then a( c ) 6= a( 0 c ) or b( c ) 6= b( 0 c ). Hence the following scenarios can occur: 1) a( c )6= a( 0 c ), at least one cycle marked with left doesn’t appear in both c and 0 c so that c 6= 0 c ,or2) b( c )6= b( 0 c ),multiplicitiesoftheleftcycles or unmarked cycles in c is different from 0 c ,sothat c 6= 0 c . Thismapisone-to-one, if c 6= 0 c ,inthecasethat c and 0 c havedifferentcycle- types (subscripts ignored), it’s clear that either a( c )6= a( 0 c ) or b( c )6= b( 0 c ).In the case where c and 0 c have the same cycle-types with powers, and the only difference comes from the L-markings. In this case, a( c ) 6= a( 0 c ),since a keeps track of those cycles that are marked with L. We can also specify an inverse map g :Pset( ? )⇥ Mset( ? )! ( ?) c . 16 The mapg is defined in the following way: take all the cycle-types that appear in A[ B arrange them in order, combining the multiplicities if a cycle-type occurs in bothA andB. Mark the cycle-types with power with subscriptL,ifthatcycle- type makes an appearance inA, and do not mark otherwise. This means whenever there is an common cycle-type between A and B the resulting cycle-type power takes onL in the subscript as well. The fact thatg(a(w),b(w)) =w is obvious. Now we are ready for examples. Let A = {(12),(1234))}, B = {(123) 2 ,(1234)})},hencetheorderforallthecycle-typesthatappearin A[ B is (12) < (123) < (1234). The multiplicity for (12) is 1+0 = 1,themul- tiplicity for (123) is 0+2 = 2,themultiplicityfor (1234) is 1+1 = 2. Since (12), (1234) make appearance in A,thefinalresultis (12) L (123) 2 (1234) L . g({(12),(1234))},{(123) 2 ,(1234)})! (12) L (123)(1234) 2 L Since ( ?) c ⇠ = Pset( ? ) ⇥ Mset( ? ), we may write down the generating function for (C n 0 1 ,C n 0 2 ,...C n 0 n ) where C n 0 i counts the number of components of weight i in a weight-n element from ( ?) c .Fromthisgeneratingfunctionwe arrive easily at the generating function of (C n 1 ,C n 2 ,...C n n ) where C n i counts the number of i cycles in an unimodal permutation w. This theorem was first proved by Fulman in [12] and Thibon in [18]. Here we take a different approach - the proof here relies on the bijection in Theorem 2.2.5. Theorem 2.2.6. The cycle enumerator of unimodal permutations is 1+ X n 1 u n p a Y i x a i i = 1 2 1+ Y i 1 (1+u i x i ) M i Y i 1 (1 u i x i ) M i ! , where p a = p {a 1 ,a 2 ,...an} is the number of unimodal permutations with a i many i- cycles for i n. 17 Proof. It’soneoftheelementarytechniquesofgeneratingfunctionsthatthegener- atingfunctionofacrossproductofcombinatorialclassesisthemultiplicationofthe twogeneratingfunctionsofthecombinatorialclasses. Inmoreconcreteterms,since we have established an isomorphism in Theorem2.2.5, if 1 (~ x), ~ x=(x 1 ,x 2 ...x n ) is a generating function for Pset( ? ) such that the coefficient in front of x a i i x a j j is the number of elements in Pset( ? ) with a i cycle-types of size i and a j cycle- types of size j, a i =1 or 0 for all i,and 2 (~ x) is the generating function for Mset( ? ), where similarly, the coefficient in front of x a i i x a j j is the number of ele- ments in Mset( ? ) with a i cycle-types of size i and a j cycle-types of size j,then the generating function for( ?) c is simply 1 (~ x) 2 (~ x). The generating function ofPset( ? ) is Q i (1+x i ) m i and the generating func- tion of Mset( ? ) is Q i 1 (1 x i ) m i , where m i is the number of transitive unimodal permutations of size i: m i = 1 2i X d|i dodd µ(d)2 i d . The generating function for the number of size-i components in c (?) is hence: Y i (1+u i x i ) m i (1 u i x i ) m i . The coefficient ofu n Q k i x a i i is the number of unimodal permutations of sizen with a i cycles of size i,for 1 i k.Weknowfromtheconstructionof c (?) that for everyw2 ( ?),bothw andL(w)2 c (?)andtheyhaveidenticalcyclestructures. Hence, 1+ X n 1 u n p a Y i x a i i = 1 2 1+ Y i 1 (1+u i x i ) M i Y i 1 (1 u i x i ) M i ! . We add the extra constant term to make sure the coefficient ofu 0 is equal to 1. 18 2.2.2 Useful counting result Lastly, we would like to describe a very important counting result that makes appearance in the later chapters. We explain this result using Gannon’s operation. Let be a cycle-type of size j.Let Q n (N( ) 1) be the number of string representations of size n that have at least one copy of (with or without ‘L’ as its subscript) in its expression. Then the following recursive relation holds: Q n (N( ) 1) = 2 2 n j Q n j (N( ) 1) +Q n j (N( ) 1) (2.2.1) where Q l (N( ) 1) = 0 ifl<j. If 2 ( ?) c has a cycle-type j ,thenitmusthaveeitherexactly 1 copy, or more than 1 copy of j . That is, Q n (N( ) 1) =Q n (N( )> 1)+Q n (N( )=1). In the case where there is more than one copy of ,aftersettingasideonecopy of , by earlier results from Theorem 2.1.13, the remaining expression is of length n j and must contain at least 1 copy of with a fixed subscriptL orR,bysimply raising the exponent by 1 we will have arrived at an expression of length n that has strictly more than one copy of . Thus Q n (N( ) > 1) = Q n j (N( ) 1).In the case that there is exactly one cycle of size j,theremainingexpressionofsize n j must not contain any copies of at all. Since has the option of having either L or R as its subscript, we double the quantity 2 n j Q n j (N( ) 1) and Q n (N( )=1)=2 n j+1 2Q n j (N( ) 1). We can now in fact solve for Q n by applying this recursion until n kj <j : 19 Q n (N( ) 1) = 2 n j+1 Q n j (N( ) 1) =2 n j+1 2 n 2j+1 +Q n 2j =2 n+1 2 j (1 2 2j +2 3j ...+( 1) [ n j ]j 2 [ n j ]j ) =2 n+1 2 j ( 1+( 2) [ n j ]j 1+2 j ) =2 n+1 1+( 2) [ n j ]j 1+2 j (2.2.2) (2.2.1)inspires anotherimportantidentityforChapter4, whenwearecounting the occurrences of multiple cycle-types simultaenously. Let ¯ be a vector of cycle-types of lengthm> 1,weareinterestedincalcu- lating Q n (N( ¯ ) (c 1, ...c m )),and N( ¯ ) are the random variables which count the appearances of cycle-types ¯ .Againasetofsimilarrecursionholds: Let L have length m,andlet i have size s(i). Q n (N( 1 ... m ) (1,...,1)) = 2Q n s(m) (N( 1 ... m 1 ) (1,...,1)) Q n s(m) (( 1 ,..., m ) (1,...,1)) (2.2.3) 20 Chapter 3 Expected number of cycles of unimodal permutations 3.1 The number of components in combinatorial structures We begin the section by reviewing some of the key ideas in [11] which prove the number of components for certain ‘decomposable’ combinatorial structures tend to a limiting Gaussian distribution. The proofs are constructed by the continuity theorem for characteristic functions and singularity analysis techniques based on Hankel contours. The technical methods are described in detail in [9]. Letm be a class of combinatorial structures, a classP is said to be decompos- able over m if each of its elements may be uniquely decomposed into a multiset of elements ofm. We differentiate the cases where the structures are labelled or unlabelled. For the current discussion we are mainly interested in the unlabelled structures, where elements ofP are obtained by taking arbitrary sets with repe- tition allowed of elements ofm,andwemakenodistinctionsamongtheelements (examples of this are polynomials having irreducible polynomials as components). It translates into the classical relation for ordinary generating functions: P(z)= X n 0 P n z n =exp ✓ M(z) 1 + M(z 2 ) 2 + M(z 3 ) 3 ... ◆ , 21 where M(z)= P i m i z i is the ordinary generating function forM(z). m i is the number of distinct elements inm of sizei. P n is the total number of objects of size n.Weshallconsistentlyusethesamedefinitionforallclassesofstructures. We also need the power-set construction, where no component appears more than once. If Q is formed fromm in this way, then: Q(z)= X n 0 Q n z n =exp ✓ M(z) 1 M(z 2 ) 2 + M(z 3 ) 3 ... ◆ , where Q n has the obvious definition as in the multiset case. Our approach reliesoncontourintegrationofHankel-typepathsusedinconnectionwithCauchy’s coefficientformula. Classically,HankelcontoursservetoexpressGammafunctions. The contours come very close to the singularities then steer away. Interested readers can read [10] Chapter VI for reference and graphical representations. We first define a suitable domain: Definition 3.1.1. [10] Given two numbers , R, withR> 1 and 0<<⇡/ 2, the open domain( ,R ) is defined as ( ,R )= {z||z|<R, z6=1, |arg(z 1)|> }. Adomainisa -domain at 1 if it is a ( ,R ) for some R and .Foracomplex number⇣ 6=0,a -domainat⇣ istheimagebythemappingz7! ⇣z ofa -domain at 1.Afunctionis -analytic if it is analytic in some -domain. We are mainly interested in the generating functions which are satisfies the following definition of being logarithmic: 22 Definition 3.1.2. [10] A function M(z) analytic at 0,havingnon-negativecoef- ficients and finite radius of convergence ⇢ is said to be of (, )-logarithmic type, where  6=0, if the following conditions hold: 1. the number ⇢ is the unique singularity of G(z) on |z| = ⇢ ; 2. M(z) is continuable to a -domain at ⇢ ; 3. M(z) satisfies M(z)=  log 1 1 z/⇢ + +O ✓ 1 (log(1 z/⇢ )) 2 ◆ , as z! ⇢ in 3.2 Expected number of cycles of unimodal per- mutations The method we use to calculate the expected number of cycles as n!1 comes from Flajolet’s paper [11] and strongly mirrors the example of polynomials over finite fields. An improved version of this theorem appears in [10] Page 462, which alsosynthesizesresultsfrom[16]. FlajoletandSoriahadoriginallydefinedadecom- posable combinatorial class to be strictly a multiset (either labelled or unlabelled) over a smaller, atomic set and the theorems from both sources were intended to apply only to labelled assemblies or unlabelledP as multiset and powerset decom- posable overm.Despitethislimitationandthefactthatwehaveacrossproduct of two structures each decomposable overm, i.e Mset(m)⇥ pset(m),wefindthe assumption can be weakened and the method is equally effective. 23 We will use the following generating function to calculate the expected number of cycles as n!1 and give reasons in the proof. The function is a little modified from the one in last Chapter to keep track of the total number of cycles. The coefficient in front of u k z n is the number of unimodal string expressions with total weight n and k components: P(z,u)= Y k 1 (1+uz k ) m k · Y k 1 (1 uz k ) m k Theorem 3.2.1. Let K be the random variable that represent the total number of cycles of a randomly chosen unimodal permutation of size n. E[K]= µ n = logn 1 2 + +R 1 2 /2+O(n 3/2 ) where R(z)= P i>3, odd M (z i ) and ( )= d ds ( s) | s= . Furthermore, Var[K]= 2 n =logn + O(1).Inaddition, P(a< K µn n <b)! 1 p 2⇡ R b a e t 2 /2 dt Proof. Following the notation in [11], letP n,k be the number of structures of sizen that have k components, P n the number of total structures of size n.Let P n (u)= P k P n,k u k .If ⌦ n is the random variable representing the number of cycles in a uniformly chosen random unimodal permutation. ThenP(⌦ n =k)= P n,k Pn .Wesee that the mean and variance for⌦ n are µ n = P 0 n (1) P n (1) ; 2 n = P 00 n (1) P n (1) P 0 2 n (1) P 2 n (1) + P 0 n (1) P n (1) (3.2.1) Instead of the generating function listed in Theorem 2.2.6, we can use the generatingfunctionforthecomponentsof c (n)toprovethetheoremforunimodal permutations. This is because there are two string representations corresponding to one unimodal permutation, the constants cancel in (3.2.1). LetM(z)= P m i z i where m i is defined to be the number of unimodal transitive permutations of size 24 i,andintuitively M(z) can be viewed as a generating function of m i ’s; p(z,u) can be written: p(z,u)= X n,k 0 P n,k u k z n = Y k 1 (1+uz k ) m k · Y k 1 (1 uz k ) m k =exp( u 1 (M(z)+ u 2 2 M(z 2 )+ u 3 3 M(z 3 )+...) ·exp( u 1 (M(z) u 2 2 M(z 2 )+ u 3 3 M(z 3 ) u 4 4 M(z 4 )... =exp(2( X i=odd u i i M(z i )). Substituting u=1,let P(z):= p(1,z)=exp(2( P i 1,odd M(z i ) i )).Let L(z)= log(P(z)) = 2( P i=odd M(z i ) i ),considerthesetofalloddnumbers O = {1,3,5...}. Let L ⇤ (i)= L(z i ) and M ⇤ (i)=2M(z i ) both be defined on O. Since the set of all odd numbers is partially ordered and locally finite, we may apply the generalized version of the Möbius inversion using the divisor lattice, see page 194 of [1]. M(z)= 1 2 M ⇤ (1) 1 = 1 2 L ⇤ (1)+ 1 2 X i=odd L(z i )µ(i) = 1 2 log(P(z))+ 1 2 X i 3,odd L(z i )µ(i) (3.2.2) We know that there’s altogether 2 n string representations of total weight n. Hence: P(z)=p(z,1) = X n 0 ( X k 0 P n,k )z n = X n 0 2 n z n = 1 1 2z 25 Hence (3.2.2) becomes: M(z)= 1 2 " log ✓ 1 1 2z ◆ + X i 3,odd log ✓ 1 1 2z i ◆ µ(i) # (3.2.3) Since L(z i ) is analytic in the region |z| < 2 1/3 for i 3,odd,and |L(z i )| C st |z i |whichistruesincetheleadingtermintheexpansionof 1 1 2z i is 2z i ,thesum 1 2 P i 3,odd L(z i )µ(i) is analytic for |z|<r where 2 1 <r< 2 1/3 .Hence M(z) has an isolated singularity of logarithmic type at z=1/2< 1.  = 1 2 , = 1 2 X i 3,odd log ✓ 1 1 2 1 i ◆ µ(i) In fact our M(x) satisfies more than the logarithmic condition since using Taylor expansion: M(z)= 1 2 log ✓ 1 1 2z ◆ + 1 2 X i 3,odd log ✓ 1 1 2·2 i ◆ µ(i)+o(z 1 2 ), as z! 1 2 @p @u (z,1) = 2 X i 1,iodd M(z i ) ! p(z,1) so that @p @u (z,1) p(z,1) =log 1 1 2z +2 +R(z)+O(z 1 2 ) where R(z):= P i 3,odd log 1 1 2·z i µ(i) If F(z)=exp(M(z)),andlet ⇢ be the singularity, ⇢ = 1 2 here, then: F(z)= e +R(⇢ )/2 (1 z/⇢ )  [1+O(z 1 2 )] 26 so that by transfer theorem, see page 388 of [10]: [z n ]F(z)= e +R(⇢ )/2 (  ) ⇢ n ·(n  1 +O(n 3/2 )) Similarly, let f(z)= @p @u (z,1),then f(z)= e +R(⇢ )/2 (1 z/⇢ )  (log 1 1 2z +2 )[1+O(z 1 2 )] [z n ]f(z)= e +R/2 (  ) n  1 ⇢ n logn ( )+ +R(⇢ )/2+O(n 3/2 ) where ( )= d ds ( s) | s= . The expected number of cycles of a unimodal permutation can be found as a ratio, µ n = [z n ]f(z) [z n ]F(z) =logn ✓ 1 2 ◆ + +R ✓ 1 2 ◆ /2+O(n 3/2 ). In order to calculate the variance term, we can follow the same argument out- lined in [11] Proposition 1, taking coefficients of z n ,usingextensionofDarboux’s theorem to arrive at 2 n =logn+O(1). Using the same Hankel contour as outlined in [11], we may also conclude that once we normalize⌦ n , it will converge weakly to a limiting Gaussian distribution P(a< ⌦ n µn n <b)! 1 p 2⇡ R b a e t 2 /2 dt. 27 Chapter 4 Convergence Rate of cycles of unimodal permutations 4.1 Set-up of the problem Fulman, Diaconis and Holmes have shown in [8] that as n!1 ,thelimiting distribution of joint distributions of (C n 1 ,...,C n k ) approaches ( ˜ C 1 ,..., ˜ C k ) where each ˜ C i is distributed as the sum of two independent random variables Negative Binomial( 1 2 i ,m i )+Binomial( 1 2 i +1 ,m i ), and m i = 1 2 X d|n d odd µ(d)2 i/d . In this chapter we derive a rate for this convergence, and we will see that the method can be generalized to provide the rate for the convergence rate of i-cycles for the m-shelf shuffles in the following chapter. We observe that Negative Binomial( 1 2 i ,m i )= m i X 1 Ge( 1 2 i ) 28 and similarly Binomial( 1 2 i +1 ,m i )= m i X 1 Ber( 1 2 i +1 ) where Ge(✓ ) is a Geometric random variable with the following probability mass function supported on {0,1,2,...}. P(Ge(✓ )=k)=(✓ ) k (1 ✓ ), and Ber(✓ 1 ) is a Bernoulli random variable with p.m.f P(Ber(✓ 1 )=1)= ✓ 1 and P(Ber(✓ 1 )=0)=1 ✓ 1 . The idea of writing down component weight-i counting as a sum of distinct component weight-i counting is called ‘refining’. It is described in generality in book [2] page 45, for the case of polynomials see [6]. Let ˆ X j = Ge(2 j ), ˆ Y j =Be( 1 1+2 j ),and ˜ W j = ˆ X j + ˆ Y j .shorthanding 2 j = p 1 and 1 2 j +1 =p,fork> 0: P( ˆ X j + ˆ Y j =k)=p k 1 1 (1 p 1 )p+p k 1 (1 p 1 )(1 p) =p k 1 1 (1 p 1 )(p+p 1 (1 p)) =p k 1 1 (1 p 1 )(p+(p 1 pp 1 )) =2 j(k 1) ( 2 j 1 2 j )( 1 2 j +1 + 1 2 j +1 ) =2 jk+1 2 j 1 2 j +1 (4.1.1) 29 When k=0, P( ˆ X j + ˆ Y j =0)=(1 p 1 )(1 p)= 2 j 1 2 j 2 j 1+2 j this implies fork> 1, P( ˆ X j + ˆ Y j k)=2 j(k 1)+1 1 2 j +1 . 4.2 Detailed calculations of D TV (L(C),L( ˜ C)) Since the cycle types can be completely ordered, we can let ij denote the j’th cycle-type of size i,then | ij | = i.Let W n ij be associated with ij and count the number of times ij occurs in a unimodal permutation of size n.Let Q n (W n ij l) denote the number of unimodal permutations of size n in which the cycle-type ij occurs at least l times, in other words Q n is the counting measure. Clearly, Q n (W n ij l)=a for all 1 j m i . We know the following recursion holds from (2.2.1): Q n (W n ij 1) = 2(2 n i 1 ) Q n i (W n i ij 1). Letb·c be the floor function, after solving we get Q n (W n ij 1) = 2 n 1+( 2) b n i ci 1+2 i . 30 We note that after normalization which is achieved by dividing the counting measure with the number of total unimodal permutations, the result is fairly close to P( ˆ X j + ˆ Y j 1),infact: P(W n ij 1) =P( ˆ X j + ˆ Y j 1)+o(2 n ). Next, we are interested in counting the number of unimodal permutations of size n that has at least one copy of each W n i 1 j 1 ,...,W n imjm . Q n ((W n i 1 j 1 ...W n imjm ) (1,...,1)) = 2Q n im ((W n im i 1 j 1 ...W n im i m 1 j m 1 ) (1,...,1)) Q n im ((W n im i 1 j 1 ,...,W n im imjm ) (1,...,1)) This recursion is useful because it reduces the number of random variables in the joint distribution. Forn<i m ,thesecondtermontheRHSoftheequationis 0. Solving the case where m=2 may give us some insights: 31 Q n (W n j 1 s 1 1,W n j 2 s 2 1) = 2 b n j 2 j 1 c X k=1 ( 1) k+1 Q n kj 1 (W n kj 1 j 2 s 2 1) =2 b n j 2 j 1 c X k=1 ( 1) k+1 2 n j 1 k 0 @ 1 ( 1) b n j 1 k j 2 c 1 2 j 2 b n kj 1 j 2 c 1+2 j 2 1 A =2 n+1 1 1+2 j 2 ( 1 ( 2 j 1 ) b n j 2 j 1 c 1+2 j 1 + b n j 2 j 1 c X 1 ( 1) k+1+b n j 1 k j 2 c ·2 j 2 ·b n kj 1 j 2 c kj 1 ) =2 n+1 1 2 j 1 +1 1 2 j 2 +1 +E 1 (n,j 2 ,j 1 )+E 2 (n,j 2 ,j 1 ) =2 n 1 P( ˜ W j 2 1, ˜ W j 1 1)+E 1 (n,j 2 ,j 1 )+E 2 (n,j 2 ,j 1 ) E 2 =2 n+1 1 1+2 j 2 ( 2 j 1 ) b n j 2 j 1 c 1+2 j 1 and E 1 =2 n+1 1 1+2 j 2 b n j 2 j 1 c X k=1 ( 1) k+1+b n j 1 k j 2 c 2 j 2 ·b n kj 1 j 2 c kj 1  2 n+1 1 1+2 j 2 b n j 2 j 1 c X k=1 2 j 2 ·b n kj 1 j 2 c kj 1 b n j 2 j 1 c2 n+1 1 1+2 j 2 1 2 n j 2 < 2b n j 2 j 1 c (4.2.1) 32 The second to last inequality comes from the fact that n lb n l cl. Remark. We have achieved this bound by assuming the terms under the summation are all positive or negative. This may seem like a gross over-estimate but when j 1 = j 2 , the terms indeed have the same signs, in which case we will ‘practically’ have the equality E 1 =( 1) b n j c+1 b n j 1 c 1 Remark. We are also particularly interested in the case where j 1 j 2 =1,ifthis is true then |E 1 | j 1 We can solve for the joint distribution ofk differentW n ij ’sbysuccessivelyreduc- ing the number ofW n ij ’s one at a time via the recursion provided in (4.2.1). It’s not hard to realize that the error terms compound when we use the recursion k times for a joint distribution ofk cycle-types. Every time a new cycle-type is introduced, we are also creating a new error term that didn’t exist before the new recursion. LetE k i denote thei-th error term that occurs in an expression that has evolved k times. They can be thought of as the deviation from the Bernoulli and Geo- metric convolution before normalization. We reverse the normalization by letting ˜ Q n ( ˜ W i l)=2 n 1 Q n ( ˜ W i l) (similarly defined when there are more than one ˜ W i for easy tracking. Then we can see from solving the 2 -variable joint distribution that: Q 2 n = ˜ Q 2 n +E 2 1 +E 1 2 and in general: Q k n = ˜ Q k n + k X i=1 E k+1 i i E k 1 is the leading term of error, let’s see how it evolves when we add more cycle-types into considration. We list the first 4 E i 1 terms for i 4 and then write ageneralexpression. 33 E 1 1 = 2( 2) b n j 1 cj 1 1+2 j 1 =E 1 (n) E 2 1 =2 b n j 1 j 2 c X k 2 =1 ( 1) k 2 +1 E 1 (n k 2 j 2 )< 2b n j 1 j 2 c E 3 1 (n)=2 b n l j j 3 c X k 3 =1 ( 1) k 3 +1 E 1 (n k 3 j 3 ) 2 2 · b n l j j 3 c X k 3 =1 b n k 3 j 3 j 1 j 2 c E 4 1 (n)=2 b n l j j 3 j 4 c X k 4 =1 ( 1) k 4 +1 E 3 1 (n k 4 j 4 ) 2 3 · b n j j 3 l j 4 c X k 4 =1 b n k 4 j 4 l j j 3 c X k 3 =1 b n k 3 j 3 k 4 j 4 j 1 j 2 c In general, E r 1 (n)= 2 r 1 Br X kr=1 B r 1 X k r 1 =1 B 1 X k 3 =1 b n P r i=3 k i j i j 1 j 2 c Where B r =b n P r 1 i=3 j i j 1 j 2 j r c B r 1 =b n k r j r P r 2 i=1 j i j r 1 c B 3 =b n j 1 j 2 P k i=4 k i j i j 3 c To get an idea of how big E n 1 runs, we make the following simplification and then transform the sum into integral: since in the limiting case k is fixed and n!1 ,ourstrategymakessense. 34 B r  B 0 r = n j r B k  B 0 k = n P r i=k+1 k i j i j k E r 1 (n)< 2 r 1 Z Br 0 Z B r 1 0 ... Z B 3 0 n P r i=3 k i j i j 2 dk 3 dk 4 ...dk r =2 (2n) r 2 (r 2)! Q r k=2 j k Remark. The integral is easy to evaluate despite its appearance. For example, when integratingk 3 that term becomes k 2 3 j 3 j 2 ,theothertermsare k i j i k 3 j 2 ,andnk 3 /j 2 . So when j 3 =B 3 , a nice factorization along with cancellation happens. In general, E r i (n)< 2 r 1 n r i 1 (r i 1)! Q r k=1+i j k for i r 1 E r r (n)=1 Our analysis requires us to have bounds onQ n (W I >c I ),W I =(W i 1 ,...,W i k ), c I =(c i 1 ,...,c i k ) where not allc i ’s are 1. Fortunately the following identity relates nicely Q n (W I >c I ) to Q n (W I > (1,...,1)):if c i 1 for all c i ,then Q n ((W n i 1 j 1 ...W n imjm ) (c 1 ,...,c m )) =Q n P i (c i 1)j i ((W n P i (c i 1)j i i 1 j 1 ...W n P i (c i 1)j i imjm ) (1,...,1)) (4.2.2) 35 4.2.1 Term by term differences Let J k = n 1,2,··· P k i=1 m i o be the index set of (W n q k j k ,j k  k,1  q k  m k ) after rearrangements. By abuse of notation letW(J k )=(W n 1 ,...,W n k·m k ),thenby inclusion-exclusion principle, P(W(J k )=c)=P(W(J k ) c)+ |J k | X t=1 ( 1) t X B2 Dt P(W(J k ) c+ X s2 B e s ) where D r is the set of all r-subsets of J k , e s denotes the sth coordinate vector and c=(c 1 ,...,c km k ) is a constant vector. Let D n (W(J k )=c)= P(W(J k )=c) P( ˜ W(J k )=c) and D n (W(J k ) c) be defined correspondingly, then: D n (W(J k )=c) D n (W(J k ) c)+ X B2 Dt D n (W(J k ) c+ X s2 B e s ) This means D n (W(J k )=(1,...,1)) can be bounded using equation (4.2.2) by D n (W(J k )=1) 2 r 1 ( n r 2 (r 2)! Q j k + X i2 J k (n j i ) r 2 (r 2)! Q j k + X {i,q}2 J k (n j i j q ) r 2 (r 2)! Q j k +...) < 2 r 1 2 r n r 2 (r 2)! Q j k where r = |J k | and j k is the size of the k-th cycle-type. 36 Remark. the constant 2 r comes from the fact that there are a total of 2 r subsets from a set of size r. Since when c6=1 we can use the translation technique established by (4.2.2), we can calculate: X c D n (W(J k )=c) C n b n jr c X kr=0 b n jr j r 1 c X k r 1 =0 ··· b n k 2 j 2 k 3 j 3 ··· krjr j 1 c X k 1 =0 (n j 1 k 1 j 2 k 2 ··· j r k r ) r 2 where C n =2 n 4 r 2(r 2)! Q j k . Using integral estimation, [ n jr ] X kr=0 [ n jr j r 1 ] X k r 1 =0 ... [ n k 2 j 2 k 3 j 3 ... krjr j 1 ] X k 1 =0 (n j 1 k 1 j 2 k 2 ... j r k r ) r 2 =n r 2 [ n jr ] X kr=0 [ n jr j r 1 ] X k r 1 =0 ... [ n k 2 j 2 k 3 j 3 ... krjr j 1 ] X k 1 =0 (1 k 1 j 1 n k 2 j 2 n ...k r j r n ) r 2  n r 2 Z n jr 0 Z n jr j r 1 0 ... Z n k 2 j 2 k 3 j 3 ... krjr j 1 0 (1 k 1 j 1 n k 2 j 2 n ...k r j r n ) r 2 dk 1 ...dk r 1 dk r =n 2(r 2) 1 (2(r 2))! (r 2)! Q j k Remark. Vectorcrunsthroughallthepossiblenon-zerovectorssuchthatc·j n. We are over-estimating because if c has i zero values in its entries, then, D n (c) is really the magnitude of n r i .Butsuchcalculationsaretediousandmakeslittle difference in the magnitude of P D n (c). 37 4.2.2 Difference brought on by infinite tails Let P c k j k = l,since ˜ W ij = ˜ X i + ˜ Y i = Ge(2 i )+NegBin( 1 1+2 i ) have infinite tail, P( ˜ W =c+ P e s ) will be non-zero even when c+ P e s >n.Hencewemustalso account for their contributions. Ifc havew non-zero terms ands zero terms, then: E tail (W =c) |J k | X r=1 X B2 Dr 1 {l+ P i2 B j i >n} 2 w 2 l 2 P i2 B j i Y c i 6=0 (1 1 2 j i +1 ) Y j k 2 B 2(1 1 2 j k +1 ) The relative error brought by these ‘false zero’ tails is: O 0 @ k |J k | X r=1 X B2 Dr 1 {l+ P i2 B j i >n} 2 P i2 B j i Y j k 2 B 2 ✓ 1 1 2 j k +1 ◆ 1 A (4.2.3) For P j k =t,wemaketheobservationthat: X P j k =t Y j k 2 B 2 P i2 B j i 2 ✓ 1 1 2 j k +1 ◆  exp 3 2 l X i=1 2 d i ! l Y j=1 ( 1 t P i 1 j=1 d j ) (4.2.4) For now we state the result, but we will go into a detailed derivation in Section 5.3.2. The right hand side of this inequality is bounded by c 2 ·P(⇢ t 2 H 0 k ) where ⇢ t is a uniform random permutation of t objects and H 0 k is the set of permutations with all cycles of length no greater than k. The reason is that Q l j=1 ⇢ 1 t P i 1 j=1 d j is the probability of ‘age-cycles’ of permutations through Feller-coupling. This is remarkably similar to the polynomial case in paper [6] where no details were given on the bound. Here we bound P(⇢ t 2 H 0 k ) by a method outlined in [3] and discuss it in detail. 38 4.2.2.1 Bounded size bias coupling To bound P(⇢ t 2 H 0 k ),weusethecalculationsoutlinedinbook[2],andatheorem from [3]. P(⇢ t 2 H 0 k )=P( P k i=1 iC (t) i =t) whereC (t) i is the random variable that counts the number of i-cycles in a uniformly random chosen permutation from S t .Let Z i be poisson random variables with parameter 1 i . T 0k = P k i=1 iZ i ,then P( k X i=1 iC (t) i =t)=P(T 0k =t|T 0t =t). We may re-write the right side into P(T 0k =t|T 0n =t)= P(T 0k =t,X k ,...,X n =0) P(T 0n =t) . From [2], P(T 0n =t) c 1 n 1 for some c 1 > 0,and P(T 0k =t,X k ,...,X n =0)=P(T 0k =t)·e P n i=k+1 1/i , which implies that P(T 0k =t,X k ,...,X n =0) P(T 0k =t) e logn+logk c 1 n =c 2 P(T 0k =t) for some c 2 0. Now we have reduced the problem to bounding P(T 0k =t),andwedosobya clevertheoremfrom[3]usingaconceptcalledBoundedSizeBiasCoupling(BSBC). We will give the definition for size bias as well as BSBC. 39 Definition 1. [4] Let X be a non-negative random variable 0<E(X) <1 . X ⇤ has the X size biased distribution if and only if dF X ⇤ (x)= xdF(x) E(X) , in particular, when X is discrete the formula P(X ⇤ =k)= kP(X =k) E(X) applies. Bounded Size Bias Coupling 1. [3] X is a non-negative random variable with positive finite expected valueEX =a.andfor Y = d X ⇤ there exists a coupling in which Y  X +c for some c2 (0,1 ). It’s been well known, see [2] for instance, that T ⇤ 0k = T 0k + J k where J k is indepedent from T 0k and P(J k =j)= j · 1 j P k i=1 i·1/i = 1 k In other words, J k is uniformly distributed on {1,...,k}.Furthermore, ET 0k = P 1 i k i 1 i =k We may then use BSBC Arratia and Baxendale, this result can be found in their paper [3]. Theorem 4.2.1. Assume X satisfies Bounded Size Bias Coupling. Given x let k =b |x a| c c 40 so that k is a nonnegative integer, possibly zero. Then for x a, P(X x) Y 0 i k a x ic We let a = k, c = k since J k  k.Since P(T 0k = t) P(T 0k t),bythelast theorem P(T 0k n) k [ n k k ] n(n k)(n 2k)...(n [ n k k ]k)  1 [( n k ])! Hence E Tail from (4.2.3), via 4.2.4 is of order at most k X t>n l 1 [t/k]! =O ✓ k 2 [(n l)/k]! ◆ =O ⇣ k 2 exp n n 2k log n 2k + n 2k o⌘ 4.2.2.2 Tail of ˜ W Last but not least, we must consider the contribution to D TV from the tail of ˜ W. We know that (1 x) n  exp( nx) , (1 + x) n  exp(nx) , (1 p) n  exp{np+np 2 } for 0  p  3 5 . Furthermore, by (5.2.5), we have the following bound of the number of size-i cycle-types 2 i 1 /i 2 i/3 /i m i  2 i 1 /i. 41 By these estimations, E[z P i k i ˜ Y i ]=E[z P i k i P j M(i) ˜ W ij ] = Y i k E[z i ˜ W ij ] M i = Y i k  ( 2 i 2 i +1 + 1 2 i +1 z i ) 1 2 i 1 2 i z i M i = Y i k  (1+ 1 2 i +1 (z i 1)) M i  1 2 i 1 2 i z i M i  Y i k exp( z i 1 2 i +1 M i )exp( M i 2 i +M i (2 i z i +2 2i z 2i ) < Y i k exp ⇥ M i (2 i z i 2 i 2 i +2 i z i +2 2i z 2i ) ⇤ = Y i k exp ⇥ M i (2 i+1 z i 2 i+1 +2 2i z 2i ) ⇤  Y exp  1 i (2 i 1 )(2 i+1 z i 2 i+1 +2 2i z 2i )  exp " X i k 1 i (z i 1+2 i 1 z 2i ) # (4.2.5) for z 4 5 p q< 3 5 ·q and q=2 E[z P i k iX i ] exp( X i k 1 i (z i 1))exp( X i k 1 i 1 2 ( 4 5 ) i ) (4.2.6) < ( 5 2 )exp( X i k 1 i (z i 1)) (4.2.7) In fact, it is true that P i k z i 1 i  z k 1, and it can be shown quickly if we let f(z)= z k 1 P i k z i 1 i ,clearly f(1) = 0, f 0 (z)= kz k 1 P 1 i k z i 1 0 forz> 1. 42 Hence E[z P i k iX i ]< ( 5 2 )exp(z k 1) (4.2.8) Hence by Markov’s inequality, P( k X i i ˜ C i >n/2) 5 2 exp (n/2)logz +z k 1 z 4 5 q 1/2 so that P( k X i i ˜ C i >n/2) 5 2 exp ⇢ (n/2)log 4 5 p 2+( p 2 4 5 ) k 1 43 Chapter 5 Convergence rate of i-cycles in m-shelf Shuffling 5.1 Introduction of m-shelf shuffles In this chapter our main goal is to derive a bound for the cycle-types of ⇡ induced by a m-shelf shuffle. For a brief history on the mathematics of shuffling, [7] by Diaconis is a good source. The m-shuffles became of interest when the authors of [8] were contacted by a casino equipment manufacturer who were interested in theefficacyofaparticularkindofshufflingmachine. m-shelfshufflesareintricately connected to affine shuffles in [12]. A shelf-shuffle is defined in the following way. A deck of well-ordered cards is sequentially dealt from the bottom ontom shelves. For each card, a shelf is chosen uniformly randomly and the card can either be put above or below the current pile on the shelf with a probability of half. At the end the cards are collected from the m shelves from left to right (shelf 1 through m). Clearly this induces a permutation on the n cards. For example, if m=4,aftershuffleshelf 1 has cards {8,10,1},shelf 2 has cards {3,5},shelf 3 has cards {2,4,6},andshelf 4 has {9,7}, the permutation induced is 12345678910 81013524697. 44 An equivalent description is: label the back of n cards with numbers randomly chosen from 1 to 2m,thenfindallthecardslabeled 1 and put them in increasing order, find the cards labeled 2, put them under the current pile with their order reversed, then cards labeled 3 goes under the current pile with increasing order ... in the end all the odd numbered piles keep their order, and the even numbered piles have their orders reversed. It is not hard to see that there are (2m) n such configurations even though the permutations induced are not all distinct. We are interested in the convergence rate of small cycles of the permutations after the m-shelf shuffle. In the case where m=1,thepermutationsinducedby the shuffler are unimodal, and the measure is uniform. Our goal is the arrive at the convergence rate of the cycles of the m-shelf permutations when m 1 from generalizing the work we have done for unimodal permutations in the previous chapter. Wedosoprogressively, firstviaself-reciprocalpolynomialsofevendegree, thenthroughcomplexesofnecklaceswhichwewilldefine. Weendtheintroduction by the statement of the theorem we prove in this chapter. Theorem 5.1.1. Let C n i be the random variable that counts the number of i-cycles of ⇡ after an m-shelf shuffle of n cards, let ˜ C i := NegBin( 1 2 i ,m i )+Bin( 1 1+2 i ,m i ) D TV ((C n 1 ,...C n k ),( ˜ C 1 ... ˜ C k )) O 1 (2(r k 2))! 2 (r k 2)! ( Q r k i=1 i m i ) 2 (2m) n n 2(r k 2) 4 r k ! , (5.1.1) where m i = 1 2i P d|n d odd µ(d)(2m) i/d ,and r k = P k i=1 m i  2 k klog(2) . we have the more crude estimate of (5.1.1) in terms of k and n: D TV ((C n 1 ,...C n k ),( ˜ C 1 ... ˜ C k )) O n 2(2 k /klog(2) 2) (2m) n ! 45 5.2 M-shelf shuffles, self reciprocal polynomials and necklaces The connection of shelf-shuffle to signed permutations is the following: to each of the {1,2...n} assign uniformly and independently numbers 1,1, 2,2... m,m. Then a signed permutation is formed by starting with the cards assigned 1, arrange them in decreasing order then add negative signs, then continue with the numbers assigned 1 with increasing order with positive signs, then again pile the cards assigned to 2 with the decreasing order with negative signs and so on. An example is {3,5,6}! 1, {1,7}! 1, {2,4}! 2, {8}! +2. The permutation we get is in two-line notation 6 5 317 4 28. Now if we take the inverse of this signed permutation, ignored the signs, and conjugate by the involution i! n+1 i, we will arrive at a m shelf-shuffle. This allows us to study the problem as a type C affine shuffle [12]. The semisimple conjugacy classesc correspond to monic degree 2n polynomials (z) with non-zero constant term that are invariant under the involution sending (z) to ¯ (z):= z 2n (0) 1 (1/z).Wecallthesepolynomials self-reciprocal,andtheycanbe understood in a more qualitative sense; they are monic polynomials of degree 2n polynomials in which coefficient of z i is the same as that of z n i . Aself-reciprocalpolynomialsofdegree 2n factors uniquely as P(x)= Y i , ¯ i [ i ¯ ] r i ¯ i Y j : j = ¯ j s j j where i ’s are monic and irreducible. 46 The conjugacy classes of C n correspond to pairs of vectors (,µ ) where = ( 1 ,..., n ), µ=(µ 1 ,...,µ n ). i is the number of positive i-cycles of an element of C n ,viewedasasignedpermutation; µ i is the number of negative i-cycles of an element of C n ,viewedasasignedpermutation[12]. By this correspondence, in order to count the number of i-cycles, we are inter- ested in counting i = P i =i r i ¯ i andµ i = P | i |=2i s i and taking the sum i +µ i . 5.2.1 Counting polynomials If a polynomial is self-reciprocal of degree 2n,ithas 2n+1 coefficients. The coefficient of x 2n i should match that of x i ,andtheverymiddletermcanhave any coefficient. Since in addition the polynomial must be monic, we are then only choosing n coefficients. Therefore, there are p n self-reciprocal polynomials of degree 2n over F p . We want to draw readers to the attention of the fact that due to the corre- spondence we have established earlier, we are considering the polynomial pair i ¯ i as one object of weight i,andsimilarlyweletaselfreciprocalpolynomial 2i of degree 2i have weight i as well. When p is even, there are m p i number of weight i reciprocal pairs and self reciprocal polynomials. The fact that these quantities are equal is not so trivial. We provide a slightly more general result of this fact in the appendix where we also cover the case when p is odd. m p i = 1 2i X d|n d odd µ(d)p i/d . 47 Let i be an irreducible polynomial of degree i,thentheprobabilitythatthe pair i ¯ i occurring at least once in a self-reciprocal polynomial of degree 2n is P(r i ¯ i 1) =p n i /p n =p i (5.2.1) The explanation is very similar to the factorization of ordinary polynomials. If a self-reciprocal monic polynomial f can be written as f(x)= g(x)·s(x) with g(x) self-reciprocal, thens(x) must be self-reciprocal as well. Hence There arep n i polynomials having i ¯ i as one of its factors. P(s i =0) is the probability that i does not occur at all or an even number of times, and P(s i =1) is the probability that i occurs an odd number of times. Since there are m p i distinct many reciprocal pairs of weight i,and m p i distinct many self-reciprocal polynomials of weight i,wecanarrangetheminorder,pair up the reciprocal pair and the self-reciprocal polynomial according to their order, and assign one random variable that counts the occurrence of both. We make this ‘refinement’ process more precise in the following paragraphs. Let ij ¯ ij be the j-th conjugate pair of size 2i.and ij the j-th self-reciprocal polynomialofsize 2i.LetR n ij andS n ij berandomvariablesdefinedonself-reciprocal polynomials of degree 2n such that, for any f(x) that is self-reciprocal of degree 2n, R n ij (f(x)) = r ij ¯ ij and S n ij (f(x)) = s ij .Inotherwords, R n ij picks out the power r ij ¯ ij of ij ¯ ij in the factorization of f(x) and S n ij picks out the power s ij of ij . Clearly, S n ij =1 or S n ij =0.Weadd S n ij to R n ij and we would like to find P(S n ij +R n ij a i ). Since S n ij is binary, we have the following two scenarios, if S n ij +R n ij a i then either {S n ij =0,R n ij a i } or {S n ij =1,R n ij a i 1}.Let Q n be the counting 48 measure of polynomials of size 2n,throughthebasicfactorizationpropertywe mentioned earlier, Q n ({S n ij =0,R n ij a i })=Q n ia i (S n ia i ij =0), Q n ({S n ij =1,R n ij a i 1})) =Q n i(a i 1) (S n i(a i 1) ij =1)=Q n i(a i 1) i (S n i(a i 1) i ij =0). So there are equal number of polynomials satisfying both scenarios, and hence Q n (S n ij +R n ij a i )=2Q n ia i (S n ia i ij =0). Next we would like to calculate Q n (S n ij =0): Q n (S n ij =0)=Q n i (S n i ij =1)+Q n ( i doesn’t occur) The identity above is true because if S n ij =0,either ij doesn’t occur at all, or it occurs an even number of times, in which case setting aside one particular copy, the remaining polynomial of degree 2n 2i must have an odd number of ij occuring in the factorization. The recursion leads to: Q n (S n ij =0)=Q n i (S n i ij =1)+Q n ( i doesn’t occur) =Q n 2i (S n 2i ij =0)+Q n ( i doesn’t occur) =Q n 3i (S n 3i ij =1)+Q n 2i ( i doesn’t occcur)+Q n ( i doesn’t occur) =... 49 It’s easy to calculate Q n ( i doesn’t occur)=p n p n i . If n (2m+1)i<i,then Q n (2m+1)i (S n (2m+1)i ij =1)=0,andthefirstnon-zero term is Q n (2m)i ( ij =0)=p t wheret<i. Hence Q n (S n ij =0)= [ n i ] X k=0 ( 1) k p n ik So that Q n (S n ij +R n ij a i )=2Q n ia i (S n ia i ij =0)=2 [ n i ] X k=0 ( 1) (k+1) p n ik ia i Next we generalize this to multi-dimensions: We begin by introducing some notations to make counting easier: A n i = {S n ij = 0,R n ij a i },and B n i = {S n ij =1,R n ij a i 1}. Q n (S n i 1 j 1 +R n i 1 j 1 a 1 ,S n i 2 j 2 +R n i 2 j 2 a 2 ,...,S n injn +R n injn a n ) =Q n (A n i 1 ,B n i 2 ,A n i 3 ,B n i 4 ...)+Q n (B n i 1 ,A n i 2 ,A n i 3 ,B n i 4 )+... =2 n Q n (A n i 1 ....A n in ) =2 n Q n P n i=1 ia i (S n P n i=1 ia i i 1 =0,...S n P n i=1 ia i in =0) (5.2.2) Furthermore, another recursion holds. For clarity we suppress the superscript on S i k j k since it always matches the subscript of Q,andwealsoignore j k because only the weight information i k is important: Q n (S i 1 =0,...S in =0)=Q n in (S i 1 =0,...S in 1 =0) Q n in (S i 1 ,...S in =1) 50 So that Q n (S i 1 =0,...S in =0)= i=max X i=1 ( 1) k+1 Q n kin (S i 1 =0,...,S i n 1 =0) (5.2.3) Synthesizing both (5.2.2) and (5.2.3), we solve explicitly some simple cases as examples: Q n (S i +R i 1) = 2· [ n i ] X i=1 ( 1) k+1 p n ik =2p n 1 p [ n i ]i 1 p i Q n (S 1 +R 1 1,S 2 +R 2 1) = 2 2 [ n j 1 j 2 ] X k=0 ( 1) k+1 Q n kj 2 j 1 (S 1 =0) E 2 =2 2 p n 1 1+p j 1 ( p j 2 ) [ n j 1 j 2 ] 1+p j 2 E 1  2 2 p n  n j 1 j 2 1 1+p j 1 1 p n j 1 Now it’s clear that the recursions we had written done for unimodal permutations were simply a special case of these more general expressions when p=2.Wemay use the same error analysis as earlier. But before we do that, we would like to extend beyond the fact that p is prime. 51 5.2.2 Necklace complexes and p non-prime We are not restricted to even prime powers alone. In the more general case, we can consider the analogies of self-reciprocal polynomials when p=2m is not a prime power - these are self-reciprocal necklaces. Definition 5.2.1.Aprimitivenecklaceoflength n over p colors is an equivalence class of n beads over p colors, taking all rotations equivalent, and no two distinct rotations of a necklace within the class are equal. Definition 5.2.2.Aprimitiveselfreciprocalnecklaceof p colors is a primitive necklace made of p colors such that if we were to swap the i-th color with its complement (p i)-th color, the necklace remains unchanged. Definition 5.2.3. Let colors i and p i colors be complementary. A reciprocal of a necklace o is defined by the resulting necklace after replacing every bead with its complementary bead. Define ¯ o to be the reciprocal of o. Definition 5.2.4. If the two necklaces (P i , ¯ P i ) are reciprocals of each other, we call (P i , ¯ P i ) a reciprocal pair.Iffurthermore, P i is primitive, then (P i , ¯ P i ) is a a primitive reciprocal pair. Golomb was the first to realize a bijection between polynomials over field F p and necklaces in [14]. Miller proved that the same map also sends self-reciprocal polynomials to self-reciprocal necklaces in [17]. Even when p is not a prime, prim- itive self-reciprocal necklaces made of p colors still exist, and this fact allows us to introduce a new combinatorial class that’s of interest. The big idea is that we con- struct these combinatorial objects to have their weight generating function mimic the generating function of i-cycles after a m-shelf shuffle, and then we are able to carry out probability calculations using the discrete presentation and properties of the constructed objects. 52 Whenp is even, the total number of primitive weight-i self-reciprocal necklaces made of p colors is the same as the number of weight-i primitive reciprocal pairs. Let this number be M p i ,anditmakesfirstappearancein[17]: M p i = 1 2i X d|n d odd µ(d)p i/d . When p=2,thisispreciselythenumberofunimodalcycle-types. Wewould also like to point out that M p i is similar to N p i ,thenumberoflength i Lyndon words using p letters. N p i = 1 i X d, d|i µ(d)p i/d Next we describe a bound on M p i : Lemma 5.2.5. Let M p i be the number of equivalence classes(invariant under rotation) of self-reciprocal necklaces of weight-i made of p =2m colors, then p i p i/3 p p 1 < 2iN p i <p i if p 2. Proof. Let t be the smallest odd prime that divides i,then: 2i·N p i =p i p i/t + X d|n,dodd d>t µ(d)p i/d  p i p i/t + X d|n d>t p i/d  p i p i/t + i t 1 X k=0 p k =p i p i/t + p i/t 1 p 1 <p i (5.2.4) 53 Similarly, to achieve the lower bound: 2i·N p i =p i p i/t + X d|n,dodd d>t µ(d)p i/d p i p i/t X d|n d>t p i/d  p i p i/t i t 1 X k=0 p k =p i p i/t + p i/t 1 p 1 >p i p i/t p p 1 (5.2.5) M p i = 1 2 N p i when i is odd. In the case i is even, we can see that M p i is more than half of N p i from the following equation: X d|n d, odd µ(d)p n d = X d|n µ(d)p n d + X d| n 2 d, odd µ(d)p n 2 d . (5.2.6) If we again let i = P O:|O|=i r O andµ i = P O:|O|=2i s O . Then by the translation of analytical combinatorial structure into generating functions we have used in 2.2.5, the following generating function is true: 1+ X n 1 u n Y i 1 x i i y µ i i = Y i 1 ✓ 1+y i u i 1 x i u i ◆ m i . (5.2.7) The coefficient in front ofu n Q i 1 x i i y µ i i is the number of necklace complexes of total weightn having i primitive reciprocal pairs of weighti andµ i self-reciprocal necklaces of weight i. This function is the same one that appears in Theorem 7 of [12]: 54 Theorem 5.2.6 (Fulman). Let i (w) and µ i (w) be the number of positive and negative i-cycles of a signed permutation w in B n .If q is even, then 1+ X n 1 u n Y i 1 x i i y µ i i = Y i 1 ✓ 1+y i u i 1 x i u i ◆ m i where m i = 1 2i P d|n d odd µ(d)p i/d .Takingthecoefficientof u n Q i 1 x i i y µ i i on the left hand side of the equation and dividing by p n gives the probability that w chosen according to the probability shelf-shuffle measure has i positive i-cycles and µ i negative i-cycles. Going back to 5.2.7: letC n i bethetotalnumberofcomponentsofweighti from anecklacecomplexofsize n,then C n i = n i +µ n i .Weaddedthesuperscript n to remind us that these i and µ i are read from the term u n in the LHS of 5.2.7. If we wish to find the probability generating function of (C n 1 ,C n 2 ,...C n n ) when we pick anecklacecomplex o of weight n uniformly out of all the necklace complexes of weightn,wewouldhavetodividethisfunctionbyanormalizingconstant,i.e. the total number of necklaces in O(n).Bythewaywehaveconstructed O(n) earlier, this seems rather non-trivial. However, by Theorem 5.2.6 of [12], the normalizing constant is p n since we already know that p n is the number required to make the right side of Theorem 5.2.6 into a probability generating function. The two generating functions from 5.2.7 and Theorem 5.2.6 are identical therefore they must have the same normalizing constant. Again we emphasize that we no longer have the restriction that p is prime, but only need that p is even. Finally, we have transformed the problem of counting the cycles after a shelf-shuffle of n cards into counting the component weights of a necklace-complex after picking it uniformly random from O(n). We are ready to set up some recursion very similar to the previous section. 55 We can arrange (P ij , ¯ P ij )’s and P s (2i)j ’s among its kind, in order, separately, then after that we can pair them up. Let R n ij and S n ij be defined as before, i.e. R n ij is equal to the multiplicity of the pair (P ij , ¯ P ij ) in an object of the set of necklace complexes O(n). S n ij is either 0 or 1,andisequaltotheindexof P (2i)j .Weagain ignore the index j for clarity (implicitly we have matched up R ij with S ij ). Q n is the counting measure, and we rely on the subscriptn ofQ to remind us the sample space is O(n) of total weight n. Q n (R i +S i 1) =Q n (A i )+Q n (B i ) with A i and B i be the same as defined earlier. The complete notation with n says A n i = {S n ij =0,R n ij a i },and B n i = {S n ij =1,R n ij a i 1}. Now Q n (S i =0)=q n Q n (S i =1), the logic is pretty clear. Futhermore, Q n (S i =1)=Q n i (S i =0). If P s 2ij occurs once and we set this necklace aside, the remaining complex must have noP s 2ij ’s. Together the last two equations imply that both (5.2.2) and (5.2.3) holds and that q doesn’t need to be prime. 56 5.3 Computations for q=2m 5.3.1 Bernougeo distributions Let ˆ X j = Ge(q j ) be a geometric distribution with parameter q j and ˆ Y j = Be( 1 1+q j ) is a Bernoulli distribution with parameter 1 1+q j .Let ˜ W j = ˆ X j +Y j ,for lack of a better name we call ˜ W j a Bernougeo random variable with parameter q j .See(4.1)fortheirp.m.f.,then: P( ˜ W j =k)= p k 1 1 (1 p 1 )p+p k 1 (1 p 1 )(1 p) = p k 1 1 (1 p 1 )(p+p 1 (1 p)) = p k 1 1 (1 p 1 )(p+(p 1 pp 1 )) = q j(k 1) ( q j 1 q j )( 1 q j +1 + 1 q j +1 ) = q jk 2(q j 1) q j +1 for k 1. When k=0, P( ˜ W j =0)=(1 p 1 )(1 p)= q j 1 q j q j 1+q j = 1 q j 1+q j It is also an easy calculation to check that for k 1, P( ˜ W j k)=q jk 2q j q j +1 . It’s trivial that P( ˜ W j 0) = 1. 57 5.3.2 More computations and details We are able to generalize all the computations we have done in Section 4 here. Sincealltherecursionsarethesameapartfromachangeofparameter, it’sstraight forward to verify that the sum of the term by term difference is: X c D n (W(J k )=c) Cp n n 2(r k 2) 4 r k 1 (2(r k 2))! 2 (r k 2)! ( Q r k i=1 i m i ) 2 Next we prove an estimate we have used to establish E tail (W = c) in (4.2.3) earlier, with more generality. Let ˜ W j = ˆ X j + ˆ Y j like before, i.e. a Bernougeo random variable with parameter q j .Letallthenotationsfrom(4.2.1)bevalidhereaswell. Let c satisfy J k ·c =l, c have w non-zero terms and s zero terms, then P( ˜ W =c)=q l 2 w Y J k (1 2 1+q j k ) We knowP(W(J k )·J k >n)=0,howeverthesameisnottruefor ˜ Wsinceithas infinite tail. Hence we must account for this difference, call this termE tail (W =c) and E tail (W =c) |j k | X r=1 X B2 Dr 1 {l+ P i2 B j i >n} 2 w q l q P i2 B j i Y c i 6=0 (1 1 q j i +1 ) Y j k 2 B 2(1 1 q j k +1 ) And this term is of relative order O 0 @ k |j k | X r=1 X B2 Dr 1 {l+ P i2 B j i >n} q P i2 B j i Y j k 2 B 2 ✓ 1 1 q j k +1 ◆ 1 A 58 We design a sampling process of individual size i components that’s inspired by [6] and it will make the comparison to Feller’s coupling of random permutations apparent. In the case of necklaces, these individual j-th component of weight i is considered to be either the j-th reciprocal pair of necklaces of weight i or the j-th primitive self-reciprocal necklace of weight i.Inotherwords,weignorethe difference between these two previously matched up objects. We say that ij appears when either of the aforementioned objects appear in a factor. Let X ij denote the number of times ij occurs (in other words, either the j- th weight i reciprocal pair or the self reciprocal necklace occurs) in a uniformly chosen structure of size n. Consider a random sequence of 2-tuples constructed in the following way: choose o 2 O(n) uniformly at random, and then select its components one at a time, by sampling at random from those not already selected, with probabilities proportional to their size. If k 1 k 2 is selected at step p,thenset T p =(k 1 ,k 2 ). Then: P(T 1 =(k 1 ,t 1 )) = X ⇡ P(T 1 =(k 1 ,t 1 )|o)P(o) = X ⇡ X i ik 1 P(X k 1 t 1 =i |o) n P(o) = k 1 n E(X k 1 t 1 ) The general joint probability is then determined by the formula P(T 1 =(k 1 ,t 1 )....,T l =(k l ,t l )) = ( l Y i=1 k i n P i 1 j=1 k i ) E ( s Y j=1 (X k 0 j t 0 j ) m j ) In this notation, (k 0 j ,t 0 j )sarethedistinct2-tuplestakenby T j ’s with m j denoting the multiplicities in their occurrence. (x) r is the falling factorial 59 (x)(x 1)...(x r+1). P i k i  n. t j 1, P l i=1 k i  n Lemma 5.3.1. Let X j count the number of number of times a component jk of size j occurs as a factor in ⇡ ,thenforn> 10, E ( l Y j=1 (X j ) s j )  E ( l Y j=1 ( ˜ X j ) s j ) where ˜ X j ’s are independent random variables and defined on the positive inte- gers in the following way: P( ˜ X j m)=2 · q jm for m =1,2,3... and P( ˜ X j =0)=1 P( ˜ X j 1). Proof. We know that for c i 6=0, 1 i l, P(X 1 c 1 ,...X l c l )= l Y i=1 q j(c j 1) 2 q j +1 ±E(n) where E Cn l 2 q n Now, P( ˜ X 1 c 1 ,... ˜ X l c l )= l Y i=1 2 q jc j P(X 1 c 1 ,...X l c l ) for sufficiently large n since by Taylor series: l Y i=1 1 q j = l Y i=1 ✓ 1 q j +1 + x j q 2j ◆ l Y i=1 1 q j +1 + l X i=1 Y j6=i x i q j +1 +... where x j ’s are positive. Since the expansion doesn’t depend on n and E(n) ! 0 as n!1 ,wemay take n appropriately large for this estimation to be true. 60 For d 1 ,...d l not necessarily distinct by a counting result from lemma 5.3 [6]: X (k 1 t j ):k 1 =d 1 ... X (k l t l ):k l =d l E ( s Y j=1 (X k 0 j t 0 j ) m j )  E 8 < : t Y v=1 ( X k j =d 0 v ˜ X k j t j ) uv 9 = ;  t Y v=1 ✓ 2·q d 0 v M(d 0 v ) 1 q d 0 v ◆ uv = k Y j=1 ✓ 2·q d j M(d j ) 1 q d j ◆ If we let s(T i )=s(k i ,t i )=k i be the functional that picks up the weight of the object chosen at time T i ,then P (s(T 1 )=d 1 ...,s(T l )=d l ) l Y j=1 ✓ 2·q d j M(d j ) 1 q d j ◆ ( l Y j=1 d j n P i 1 j=1 d j )  l Y j=1 ✓ 1 1 q d j ◆ ( l Y j=1 1 n P i 1 j=1 d j )  exp 3 2 l X i=1 q d i ! l Y j=1 ( 1 n P i 1 j=1 d j ) For P j k =t, X P j k =t Y j k 2 B q P i2 B j i 2 ✓ 1 1 q j k +1 ◆  exp 3 2 l X i=1 q d i ! l Y j=1 ( 1 t P i 1 j=1 d j ) (5.3.1) So that E Tail =O ⇣ k 2 exp n n 2k log n 2k + n 2k o⌘ 61 Lastly, E[z P i k i ˜ Y i ]=E[z P i k i P j M(i) ˜ W ij ] = Y i k E[z i ˜ W ij ] M p i = Y i k  ( p i p i +1 + 1 p i +1 z i ) 1 p i 1 p i z i M p i = Y i k  (1+ 1 p i +1 (z i 1)) M p i  1 p i 1 p i z i M p i  Y i k exp( z i 1 p i +1 M p i )exp( M p i p i +M p i (p i z i +p 2i z 2i ) < Y i k exp ⇥ M p i (p i z i p i p i +p i z i +p 2i z 2i ) ⇤ = Y i k exp ⇥ M p i (p i+1 z i p i+1 +p 2i z 2i ) ⇤  Y exp  1 i (p i 1 )(p i+1 z i p i+1 +p 2i z 2i )  exp " X i k 1 i (z i 1+p i 1 z 2i ) # (5.3.2) Through the same reasoning as earlier by using Markov’s inequality, P( k X i i ˜ C i >n/2) 5 2 exp ⇢ (n/2)log 4 5 p 2+( p 2 4 5 ) k 1 Werealizethatthesetermsaresmallcomparedtothetermbytermdifferences, which means: D TV (L(W(J k )),L( ˜ W(J k ))) O p n n 2(r k 2) 4 r k 1 (2(r k 2))! 2 (r k 2)! ( Q r k i=1 i m i ) 2 ! (5.3.3) 62 Taking functionals never increases total variation distance, this fact is stated in the next lemma: Lemma 5.3.2 (Arratia and Tavaré). If h :S! T is a deterministic map between countable spaces, and X and Y are random elements of S,sothat h(X) and h(Y) are random elements of T,then d TV (h(X),h(Y)) d TV (X,Y). Hence, we have established (5.1.1). Finally, we present the parallel corollary to Corollary 3.2.1. Corollary 5.3.3. Let K be the random variable that represent the total number of cycles of a permutation of size n induced by a m-shelf shuffle. E[K]= µ n = logn 1 2m + + R 1 2m /2+ O(n 3/2 ) where R(z)= P i>3, odd M (z i ) and ( )= d ds ( s) | s= . Furthermore, Var[K]= 2 n =logn + O(1).Inaddition, P(a< K µn n <b)! 1 p 2⇡ R b a e t 2 /2 dt 63 Chapter 6 Appendix We begin with a general expression for the factorization of a self reciprocal poly- nomial of degree 2n over field F q ,callthispolynomial f . f(x)= Y , ¯ irred. ( i ¯ i ) r i Y ¯ = s i 2i Note s i 2{0,1} and we define i = P deg( )=i r and µ i = P deg( )=2i s . The generating function for i and µ i should look like (1 x i u i ) M 0 i (1+y i u i ) m 2i I will describe M 0 i and m 2i ,theshortversionisthattheyareequalapartfrom the case when i=1 in a field of odd size. In the case of even field size M 0 i =m 2i Let M i be the number of irreducible polynomials of degree i over F q .Weknow that for i 1, M i = 1 i X d|i µ(d)q i/d 64 and letm i be the number of self-reciprocal irreducible polynomials of degreei over F q .Weknowthat m k = 8 > > > > > > < > > > > > > : e if k=1 1 2i P d|i, dodd µ(d)(q 2i/2d +e 1) if k=2i,k > 0 0 if k=2i+1 Where e=2 when q is odd, and e=1 when q is even. The formula is telling us there are 2 self-reciprocal polynomials of degree 1 over an odd sized field, and only 1 self-reciprocal polynomial of degree 1 over an even sized field. Now I want to point out another expression of m 2i that will be useful [17]: m 2i = 1 2i X d|i, dodd µ(d)q 2i/2d +✏ 2i where ✏ 2i = 8 > > > > > < > > > > > : 1/2, if i=1 1 2i , if i=2 k 0 otherwise Let M 0 i equal to the number of distinct and allowable pairs (, ¯ ), we will see that degree 1 is a special case to this rule, in the sense that it is the only case where two polynomials can be paired together while not being reciprocals of each other, but we should include it in the factorization. i =1 is a special case, since when q is odd there are 2 self reciprocal polynomials of degree one, so there are three allowable pairs ( 1 , 2 ), ( 1 , 1 ) and ( 2 , 2 ). However when q is even, there’s only ( 1 , 1 ).Furthermore, M 1 = q 1 65 because we are not allowing x’s. Hence M 0 1 = q 1 e 2 +e+(e 1) = 1 2 µ(1)(q+1+e)+(e 1) =m 2 +(e 1) If i is larger than 1,wearecountingthedistinctpairs (, ¯ ) when is irre- ducible. Hence the following identity holds: M 0 i = M i m i 2 +m i When i is odd, M 0 i = M i m i 2 +m i = M i 2 . And M 0 i = 1 2i X d|i µ(d)q i/d = 1 2i X d|i, dodd µ(d)q 2i/2d =m 2i Note the second to last equality is true since ✏ 2i =0 as mentioned earlier in Miller’s counting expression. 66 If i is even, then M 0 i = M i 2 + m i 2 = 1 2i X d|i µ(d)q i/d + 1 2i X d|i, dodd µ(d)(q i/2d +e 1) = 1 2i X d|i µ(d)q i/d + 1 2i X d| i 2 , dodd µ(d)q i/2 d + 1 2i X d| i 2 , dodd µ(d)(e 1) = 1 2i X d|i, dodd µ(d)q i d + 1 2i X 2d|i, dodd µ(2d)q i 2d + X 2 2 d|i, dodd µ(2 2 d)q i 2 2 d +...+ 1 2i X d| i 2 , dodd µ(d)q i/2 d + 1 2i X d| i 2 , dodd µ(d)(e 1) = 1 2i X d|i, dodd µ(d)q i d + 1 2i X 2d|i, dodd µ(2d)q i 2d +0+ 1 2i X d| i 2 , dodd µ(d)q i/2 d + 1 2i X d| i 2 , dodd µ(d)(e 1) = 1 2i X d|i, dodd µ(d)q i d + 1 2i X d| i 2 , dodd µ(d)(e 1) =m 2i From (2) to (3) we have partitioned the divisors of i into odd divisors, divisors that look like 2d,divisorsthatlooklike 2 2 d etc. The step (5) to (6) is true since (2d)= (d) if d is odd, and (2 k d)=0, In conclusion, M 0 i = 8 > > < > > : m 2i = 1 2i P d|i, dodd µ(d)(q 2i/2d +e 1) ifi> 1 m 2 +(e 1) if i=1 and the generating function from Theorem 7 [12] is true. 67 Bibliography [1] Martin Aigner. AcourseinEnumeration.Dover,2007. [2] RichardArratia,A.D.Barbour,andSimonTavaré. Logarithmiccombinatorial structures: A probabilistic approach. EMS Monographs in Mathematics,1, 2003. [3] Richard Arratia and Peter Baxendale. Bounded size bias coupling: a gamma function bound, and universal dickman-function behavior. Probability Theory and Related Fields,pages1–19,July2014. [4] Richard Arratia and Larry Goldstein. Size bias, sampling, the waiting time paradox, and infinite divisibility: when is the increment independent? ArXiv e-prints,2010. [5] Richard Arratia and Simon Tavaré. Independent process approximations for random combinatorial structures. Adv. Maths,104,no.1:90–154,1994. [6] Richard Arratia , A. D. Barbour and Simon Tavaré. On random polynomials over finite fields. Mathematical Proceedings of the Cambridge Philosophical Society,114(02):347–368,1993. [7] PersiDiaconis.Mathematicaldevelopmentsfromtheanalysisofriffleshuffling. notes. [8] Persi Diaconis, Jason Fulman, and Susan Holmes. Analysis of casino shelf shuffling machines. Ann. Appl. Probab.,23(4):1692–1720,082013. [9] Philippe Flajolet and Andrew Odlyzko. Singularity analysis of generating funcitons. SIAM J. DISC. MATH.,3:216–240,1990. [10] Philippe Flajolet and Robert Sedgewick. Analytic Combinatorics. Cambridge Press, 2009. 68 [11] Philippe Flajolet and Michèle Soria. Gaussian limiting distributions for the number of components in combinatorial structures. Journal of Combinatorial Theory, Series A,53(2):165–182,1990. [12] Jason Fulman. Applications of the brauer complex: Card shuffling, permuta- tion statistics, and dynamical systems. Journal of Algebra,243(1):96–122,9 2001. [13] Terry Gannon. The cyclic structure of unimodal permutations. Discrete Math.,237(1-3):149–161,June2001. [14] S.W.Golomb.Irreduciblepolynomials,synchronizationcodes,primitiveneck- laces, and the cyclotomic algebra. Univ. of North Carolina Monograph Series in Probability and Statistics,no.4,1967. [15] Ian Goulden and David Jackson. Combinatorial Enumeration.Dover,1983. [16] Jennie C. Hansen. A functional central limit theorem for random mappings. Ann. Probab.,17,1989. [17] Robert L. Miller. Necklaces, symmetries and self-reciprocal polynomials. Dis- crete Mathematics,22(1):25–33,1978. [18] Jean-Yves Thibon. The cycle enumerator of unimodal permutations. Annals of Combinatorics,5(3-4):493–500,2001. 69

Abstract (if available)

Abstract Casino Shelf shufflers with m shelves induce permutations on n cards. In the case where m = 1, the permutations induced are unimodal permutations which are the discrete analog of unimodal maps from dynamical systems. We re-derive the generating functions for the cycle structures of uniformly randomly chosen unimodal permutations and explore their expected number of cycles, parallel to the works of Arratia, Barbour and Tavaré on random permutations and other logarithmic structures. As a generalization we will also show that the joint convergence rate of small cycles of the permutations induced by the shelf shuffle machine to Bin((2m)⁻ⁱ,mᵢ) + Negbin( 1 ,mᵢ) is much better than o(1/n).

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Creator Ren, Haining (author)

Core Title Convergence rate of i-cycles after an m-shelf shuffle

School College of Letters, Arts and Sciences

Degree Doctor of Philosophy

Degree Program Mathematics

Publication Date 07/29/2015

Defense Date 06/23/2015

Publisher University of Southern California (original), University of Southern California. Libraries (digital)

Tag analytic combinatorics,convergence rate,m-shelf shuffle,OAI-PMH Harvest,unimodal permutations

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Language English

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Advisor Fulman, Jason (committee chair), Arratia, Richard (committee member), Ralph, Peter L. (committee member)

Creator Email haining.ren@gmail.com,hren@usc.edu

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