Estimating proportions of elements in finite symmetric and classical - PowerPoint PPT Presentation

Estimating proportions of elements in finite symmetric and classical groups Alice Niemeyer UWA, RWTH Aachen Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 1 / 64

Motivation Proportions of Elements Theorem Let G be a group in a family of groups. Then there exists some function c ( N ) of the size N of the input of G such that the proportion of elements in G with a particular property is at least c ( N ) . Such a theorem is often hard to prove. Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 2 / 64

Motivation Efficiency of algorithms Question Will any lower bound do? Answer The lower bound affects two things: Number of searches until success on correct input. Number of searches until we “give up” on incorrect input. Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 3 / 64

Motivation Efficiency of algorithms Question Will any lower bound do? Answer The lower bound affects two things: Number of searches until success on correct input. Number of searches until we “give up” on incorrect input. Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 3 / 64

Introduction Motivation Proportions of elements in S n : algorithmic applications theoretical interest applications to proportions in matrix groups Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 4 / 64

Introduction Notation We write permutations in disjoint cycle notation. The number of cycles always refers to such a decomposition. n , m , k denote positive integers. Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 5 / 64

Previous Work Euler (1707-1783) Euler: Quaestio curiosa ex doctrina combinationis How many of the n ! orderings of the numbers 1 , . . . , n are such that no number remains in its natural place? How many derangements are there in S n ? Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 6 / 64

Previous Work Previous Results: k cycles Let g ( n , k ) denote the proportion of elements in S n with exactly k cycles. Sylvester (1861) g ( n , k ) = 1 � � s . n ! s ∈ S S ⊆{ 1 ,..., n − 1 } | S | = n − k n ! g ( n , k ) is the Stirling number of the first kind. Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 7 / 64

Previous Work Previous Results: k cycles Let g odd ( n ) denote the proportion of elements in S n all of whose lengths are odd. Sylvester (1861) ( 1 · 3 · 5 · · · ( n − 2 )) 2 n  g odd ( n ) = 1 n odd  n ! · ( 1 · 3 · 5 · · · ( n − 1 )) 2 n even  Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 8 / 64

Previous Work Previous Results: order of elements Landau 1909 log ( max g ∈ S n ( o ( g ))) √ n log ( n ) lim = 1 . n →∞ Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 9 / 64

Previous Work Previous Results: order of elements Erdös and Turán wrote a series of papers on statistical group theory. Erdös and Turán (1965) For ε, δ > 0 and n ≥ N 0 ( ε, δ ) |{ g ∈ S n | e ( 1 / 2 − ε ) log 2 ( n ) ≤ o ( g ) ≤ e ( 1 / 2 + ε ) log 2 ( n ) } ≥ 1 − δ. n ! Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 10 / 64

Generating Functions Generating Functions Given a sequence of numbers, ( a n ) n ∈ N , e.g., a n the number of certain elements in S n . Quote from Wilf’s Book A generating function is a clothesline on which we hang up a sequence of numbers for display. Suggested reading: Wilf’s book Generatingfunctionology [3] or Analytic Combinatorics by Flajolet and Sedgewick [4]. Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 11 / 64

Generating Functions Ordinary Generating Functions The Ordinary Generating Function for a n is � a n z n . A ( z ) := n ≥ 0 We denote the coefficient of z n by [ z n ] A ( z ) . Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 12 / 64

Generating Functions Exponential Generating Functions (egf) We define the Exponential Generating Function for a n is a n � n ! z n . A ( z ) := n ≥ 0 When do we use egf? When the coefficients grow very fast. E.g. in S n the number of permutations is n ! and we can hope that a proportion a n / n ! is manageably small. Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 13 / 64

Generating Functions Generating Functions We study generating functions as formal power series in the ring of formal power series. Analytic questions, convergence etc. do not concern us just yet. Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 14 / 64

Generating Functions Multiplication of Generating Functions � ∞ � ∞ � n ∞ � � � a n z n b n z n z n . � � � � · = a k b n − k n = 0 n = 0 n = 0 k = 0 Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 15 / 64

Generating Functions Example Let b ≥ 1 be fixed integer and let a n denote the number of permutations in S n all of whose cycles have length at most b . List permutations by cycles of length d containing the point 1. � n − 1 � points for cycle of length d on 1 d − 1 ( d − 1 )! different cycles on these a n − d permutations on the remaining n − d points Then a n = n ! for n ≤ b and min { b , n } a n n ! = 1 a n − d � ( n − d )! . n d = 1 Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 16 / 64

Generating Functions Example Hence we get   min { b , n } ∞ ∞ a n 1 a n − d  z n � n ! z n � � A ( z ) := = 1 +  n ( n − d )! n = 0 n = 1 d = 1 ∞ b 1 a n − d � � ( n − d )! z n = 1 + n d = 1 n = d ∞ b 1 a n n ! z n + d � � = 1 + n + d d = 1 n = 0 Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 17 / 64

Generating Functions Example Hence b ∞ a n A ′ ( z ) � � n ! z n + d − 1 = d = 1 n = 0 b ∞ a n � z d − 1 � n ! z n = d = 1 n = 0 b � z d − 1 A ( z ) = d = 1 Thus b A ′ ( z ) z d − 1 � A ( z ) = d = 1 Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 18 / 64

Generating Functions Example b A ′ ( z ) z d − 1 � A ( z ) = d = 1 and so b z d � log ( A ( z )) = d . d = 1 Therefore Generating Function b z d � A ( z ) = exp ( d ) . d = 1 Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 19 / 64

Generating Functions Summary Proportions of elements important for algorithms Generating functions useful description Generating functions can often be found Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 20 / 64

Generating Functions Coefficients Question Can generating functions tell us about the limiting behaviour of the coefficients? Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 21 / 64

Generating Functions Saddlepoint Analysis Based on results of W. Hayman. Theorem (See [4]) j = 1 a j z j have non-negative coefficients and Let P ( z ) = � n suppose gcd ( { j | a j � = 0 } ) = 1 . Let F ( z ) = exp ( P ( z )) . Then 1 exp ( P ( r )) [ z n ] F ( z ) ∼ √ , r n 2 πλ � 2 P ( r ) . � r r where r is defined as rP ′ ( r ) = n and λ = dr Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 22 / 64

Generating Functions Example Saddlepoint Analysis z d � b Recall that A ( z ) = exp ( d ) is the generating function for the d = 1 number of elements all of whose cycles have length at most b . z d � b d . Then gcd ( { d | 1 Let P ( z ) = d � = 0 } ) = 1 . d = 1 Find r d = 1 r d − 1 = � b d = 1 r d ≥ r b . n = rP ′ ( r ) = r � b Find λ � 2 P ( r ) = r � b d = 1 dr d − 1 = � b d = 1 dr d ≤ b � m d = 1 r d = bn . � r r λ = dr Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 23 / 64

Generating Functions Example Saddlepoint Analysis z d � b Recall that A ( z ) = exp ( d ) is the generating function for the d = 1 number of elements all of whose cycles have length at most b . z d Let P ( z ) = � b d . Then gcd ( { d | 1 d � = 0 } ) = 1 . d = 1 Use r ≤ n 1 / b and λ ≥ bn and P ( r ) = d = 1 r d = n r d � b d ≥ 1 � b d = 1 b b Hence � n / b 1 exp ( P ( r )) 1 � e [ z n ] A ( z ) ∼ √ ≥ √ r n n 2 πλ 2 π bn Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 24 / 64

Generating Functions However ... This can be difficult when cycle lengths grow with n : For m fixed let c ( n , m ) = 1 n ! |{ g ∈ S n | g m = 1 }| . Let ∞ � c ( n , m ) z n C m ( z ) = n = 0 be the corresponding generating function. Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 25 / 64

Generating Functions Previous Results: c ( n , m ) Chowla, Herstein, and Scott (1952)   z d  .  � C m ( z ) = exp d 1 ≤ d | m Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 26 / 64

Generating Functions Previous Results: c ( n , m ) Warlimont (1978) � � 1 n + 2 c n 2 ≤ c ( n , n ) ≤ 1 n + 2 c 1 n 2 + O , n 3 − o ( 1 )  0 n odd  where c = n even . 1  Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 27 / 64