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MATH 895-4 Fall 2010 Course Schedule A DDITIONAL N OTES f ac ulty of sc ience depar tment of mathem atics Labelled combinatorial classes Labelled combinatorial classes Week Date Sections from FS2009 Part/ References Topic/Sections 1 Sept 7 I.1, I.2, I.3 Contents 1 Combinatorial Structures Part A.1, A.2 Labelling Atoms FS: Comtet74 3 21 II.1, II.2, II.3 1.1 Well-labellingHandout . . .#1. . (self study) 4 28 II.4, II.5, II.6 Symbolic methods 6 Multivariable GFs 2 14 I.4, I.5, I.6 Notes/Speaker Unlabelled structures . . . .Labelled . . . structures . . . .I. Labelled structures II 1.2 The Exponential Generating Function (egf) 1.3 Examples . . . . . . . . . . . . . . . . . . . . Combinatorial Combinatorial 5 Oct 5 III.1, III.2 parameters Parameters 1.4 Aside: comparing to unlabelled families . . FS A.III 12 IV.1, IV.2 (self-study) 2 7 Admissible constructions 19 IV.3, IV.4 Analytic Methods 2.1 Labelled products . . . . . FS: Part B: IV, V, VI 8 26 Appendix 2.2 Labelled product . .B4. . . . IV.5 V.1 Stanley 99: Ch. 6 9 2.3NovSequences 2 . Handout . . . .#1. . . . 2.49 SetsVI.1. . . . . (self-study) . . . . . . . . 10 2.512 Cycles . . . . . . . . . . . . A.3/ C 2.6 Admissibility Theorem . . 11 18 IX.1 3 Surjections 20 IX.2 . . . . . . Random Structures and Limit Laws FS: Part C (rotating presentations) Complex Analysis . . . . . . . . . .Singularity . . . .Analysis . . . . .Asymptotic . . . . methods . . . . . . . . . . . . . .Introduction . . . . to. Prob. . . . . . . . . . . . . Limit Laws and Comb . . . . . . . . . . . . . Asst . .#1. Due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2 2 2 3 . . . . . . . . . . . . . Asst . .#2. Due . . . Sophie . . . . . . Mariolys . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 4 5 5 5 5 6 Marni Discrete Limit Laws 6 Sophie Combinatorial IX.3 4 12Set23Partitions instances of discrete 4.125 Definition . . . . . . . . . . . . . . . . . . . . IX.4 Continuous Limit Laws 4.2 Set partitions as a labelled class . . . . . . Quasi-Powers and 134.330 Set IX.5 partitions as an unlabelled classlimit. laws . . . Gaussian 4.4 Wrapping up surjections and set partitions . Marni . . . . . . . . Sophie . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 7 7 8 9 5 Permutations 5.1 Sub-classes of permutations . . . . . . . . 5.1.1 Involutions . . . . . . . . . . . . . . 5.1.2 Derangements . . . . . . . . . . . . 5.1.3 Other variants . . . . . . . . . . . . 5.2 The number of cycles in a permutation . . 5.3 Application: 100 Prisoners . . . . . . . . . 5.3.1 The game . . . . . . . . . . . . . . . Dr. Marni MISHNA, of Mathematics, 5.3.2Department The strategy . SIMON . . . FRASER . . . .UNIVERSITY . . . . . Version of: 11-Dec-09 5.3.3 The analysis . . . . . . . . . . . . . 5.4 Computing Stirling cycle numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 10 10 10 11 11 11 11 12 13 13 14 Dec 10 Presentations . . . . . . . . . . Mariolys Asst #3 Due . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Trees maps and graphs 1 14 Labelling Atoms A great many objects in combinatorics like to be labelled — indeed in many cases counting labelled objects is significantly easier than counting their unlabelled siblings — graphs and trees particularly. So what do we mean by labelled? Since the objects we consider are constructed from atomic pieces, simply attatch a label (wlog a natural number) to each atomic piece. M ARNI M ISHNA , S PRING 2011 MATH 343: A PPLIED D ISCRETE M ATHEMATICS PAGE 1/15 MATH 895-4 Fall 2010 Course Schedule A DDITIONAL N OTES f ac ulty of sc ience depar tment of mathem atics Week Date Sections from FS2009 1 Sept 7 I.1, I.2, I.3 2 14 I.4, I.5, I.6 3 21 II.1, II.2, II.3 4 28 II.4, II.5, II.6 5 Oct 5 III.1, III.2 6 12 IV.1, IV.2 7 19 IV.3, IV.4 8 26 IV.5 V.1 9 Nov 2 The first unlabelled 9 VI.1 3. 10 1.1 11 12 Part/ References 1 Topic/Sections 2 Labelled combinatorial classes Notes/Speaker 2 1 3 2 4 2 4 1 4 3 1 1 2 4 3 Symbolic methods Combinatorial Structures FS: Part A.1, A.2 Comtet74 3 Handout #1 (self study) 4 Combinatorial parameters 1 FS A.III (self-study) 2Parameters 1 Unlabelled structures 3 Labelled structures I Labelled structures II Combinatorial Asst #1 Due 4 Multivariable GFs Complex Analysis Analytic Methods 2 3 V, VI 4 3 FS: Part B: IV, Singularity Analysis Appendix B4 Stanley 99: Ch. 6 Asst #2 Due Asymptotic methods Handoutcorresponds #1 graph to 4 different labelled (self-study) Sophie A.3/ C IX.1 Well-labelling 18 Introduction to Prob. Mariolys Limit Laws and Comb Marni graphs, while the second corresponds to Random Structures Discrete Limit Laws Sophie Labelling is a very intuitive notion, but we have a formal defintion anyhow: and Limit Laws FS: Part C Combinatorial 23 IX.3 Mariolys (rotating instances of discrete Definition. A labelled class of combinatorial objects is a combinatorial class such that every atomic 12 presentations) component (atom) is labelled by a distinct integer, andMarni the set of labels associated to an object of size n 25 IX.4 Continuous Limit Laws 20 IX.2 is the set {1, 2, . . . , n}. In this case weQuasi-Powers say thatand the objects are well-labelled. If the set of labels is not 13 30 IX.5 Sophie limit laws {1, 2, . . . , n}, but still distinct, then weGaussian say theobject is weakly labelled. 14 Dec 10 Presentations Asst #3 Due 4 Going back to the above graph example — there are 2(2) = 64 labelled graphs of 4 vertices (every edge is there or not), but only 11 unlabelled graphs. 1.2 The Exponential Generating Function (egf) We do not use ogfs to enumerate labelled classes (though one can), instead we use exponential generating functions. Definition. The exponential generating function (egf) of a sequence (An ) is the formal power series Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY Version of: 11-Dec-09 n X A(z) = An n≥0 z n! = X z |α| |α|! α∈A Again we need the neutral class that contains which has size zero and no label. Similarly the atomic class contains a single well-labelled object of size 1. So we have Z = {}. 1 The egfs are E(z) = 1 Z(z) = z, which coincides with their ogf, incidentally. 1.3 Examples We have already seen permutations. We now think of them as a labelled line of n vertices. Since Pn = n! the egf is P (z) = X n≥0 n! zn 1 = n! 1−z which is nice and simple. M ARNI M ISHNA , S PRING 2011 MATH 343: A PPLIED D ISCRETE M ATHEMATICS PAGE 2/15 MATH 895-4 Fall 2010 Course Schedule A DDITIONAL N OTES f ac ulty of sc ience depar tment of mathem atics Week Date Sections Part/ References Topic/Sections Labelled combinatorial classes Notes/Speaker from FS2009 Example. The class U of urns consists of completely disconnected labelled graph. Since there are no edges, there can only be a single labelling and thus Un = 1. Hence 1 Sept 7 I.1, I.2, I.3 Symbolic methods 2 14 I.4, I.5, I.6 3 21 II.1, II.2, II.3 4 28 II.4, II.5, II.6 Combinatorial Structures FS: Part A.1, A.2 Comtet74 Handout #1 (self study) X Unlabelled structures U (z) = Labelled structures I n≥0 zn = ez . n! Labelled structures II Example. Circular graphs, C, are oriented cycles (so 1 → 2 → 3 → 1 6≡ 1 → 3 → 2 → 1). Combinatorial Combinatorial 5 Oct 5 III.1, III.2 6 12 IV.1, IV.2 7 19 IV.3, IV.4 8 26 9 Nov 2 IV.5 V.1 Asst #1 Due parameters FS A.III (self-study) Parameters Analytic Methods FS: Part B: IV, V, VI Appendix B4 Stanley 99: Ch. 6 Handout #1 (self-study) Complex Analysis 1 2 1 3 1 4 Multivariable GFs 4 3 2 4 3 2 1 2 1 3 1 4 2 3 Singularity Analysis 4 3 4 Asymptotic methods 2 Asst #2 Due 9 VI.1 Sophieon either side of the One can simply decompose such cycles by cutting them 1 and unrolling them. 10 This gives aA.3/ permutation of 2, . . . , n and theretoare 1)! such objects. This is a bijective construction 12 C Introduction Prob.(n − Mariolys and thus CnIX.1 = (n − 1)!. 18 Limit Laws and Comb Marni 11 12 20 23 IX.2 IX.3 Random Structures and Limit Laws C(z) FS: Part C (rotating presentations) = X zn X Discrete Laws Sophie 1 (nLimit − 1)! zn = = log . Combinatorial n! n 1 − z Mariolys n≥1 instances of discrete n≥1 25 Limit Laws Marni Remark thatIX.4labelled cycles are much Continuous easier than unlabelled ones. 13 1.4 14 30 IX.5 Quasi-Powers and Gaussian limit laws Sophie Aside: comparing to unlabelled families Presentations Asst #3 Due Dec 10 More generally one can find examples of enumerations of labelled objects A and their unlabelled counterparts Â such that 1≤ An ≤ n! Aˆn To show both sides of this — consider the class of Urns. Here Un = Uˆn = 1. And similarly, consider permutations: Pn = n!, but one can consider them as labellings of a linear graph on n-vertices, so P̂ = 1. Note that one can also consider permutations to be sets of cycles, then the unlabelled version Dr.nMarni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY Version 11-Dec-09is a partition. of theof:object 2 Admissible constructions So the combinatorial sum still works just fine, but clearly we are going to have difficulties with cartesian product — the labels don’t work quite right. In particular, if we take β ∈ B and γ ∈ C and glue them together to get (β, γ) the label 1 appears twice, so the object is not well-labelled. Clearly we need to relabel things carefully to make this work. Proposition (Binomial convolution formula). Let A(z), B(z), C(z) be egfs of sequences An , Bn , Cn such that A(z) = B(z) · C(z) then n X n An = Bk Cn−k k k=0 M ARNI M ISHNA , S PRING 2011 MATH 343: A PPLIED D ISCRETE M ATHEMATICS PAGE 3/15 MATH 895-4 Fall 2010 Course Schedule A DDITIONAL N OTES f ac ulty of sc ience depar tment of mathem atics Week Date Sections 1 Sept 7 I.1, I.2, I.3 2 14 I.4, I.5, I.6 3 21 II.1, II.2, II.3 4 28 II.4, II.5, II.6 5 Oct 5 III.1, III.2 from FS2009 Proof. Expand: 6 12 IV.1, IV.2 7 19 IV.3, IV.4 8 26 9 Nov 2 10 11 12 2.1 IV.5 V.1 9 VI.1 12 A.3/ C 18 IX.1 20 IX.2 23 IX.3 Part/ References Combinatorial Structures A(z) = FS: Part A.1, A.2 Comtet74 Handout #1 (self study) Notes/Speaker Symbolic methods X zn zn X · Cn B(z)Unlabelled · C(z) structures = Bn n! n n! n Labelled structures I n X X Bk Cn−k Labelled structures II = k! (n − k)! n Combinatorial parameters FS A.III (self-study) Combinatorial Parameters ! zn k=0 Asst #1 Due X Multivariable= GFs n n X Bk Cn−k n! k! (n − k)! k=0 Analytic Methods FS: Part B: IV, V, VI Appendix B4 Stanley 99: Ch. 6 Handout #1 (self-study) Complex Analysis Random Structures and Limit Laws FS: Part C (rotating presentations) Discrete Limit Laws Sophie Combinatorial instances of discrete Mariolys Continuous Limit Laws Marni ! zn n! n XX zn n! Bk Cn−k k!(n − k)! n! n k=0Asst | #2 {z } Due Asymptotic methods n Sophie ( k ) n X XMariolys n zn Introduction to Prob. = Bk Cn−k . n! k Limit Laws and Comb Marni n Singularity Analysis = k=0 Labelled products IX.4 25 Topic/Sections Labelled combinatorial classes Quasi-Powers and labelled products. Gaussian limit laws Next how to generate InSophie the unlabelled case, a pair (β, γ) with β ∈ B and 13 we 30 describe IX.5 γ ∈ C gave rise to a single object of size |β| + |γ|. In the labelled case we will generate a set of objects. 14 Dec 10 #3 Due We describe how we get thisPresentations set. Essentially, we keepAsst the structure of the pair (β, γ) and we generate new labels that preserve the order relation among labels. We can do this formally using the following two operations: • reduction: For a weakly labelled structure size n, this reduces its labels to the standard interval [1, . . . , n] while keeping the relative order fixed. Eg h4, 6, 2, 9i becomes h2, 3, 1, 4i. Denote this operation by ρ(α). • expansion: roughly — the inverse of reduction. Let e : N → Z be any strictly increasing function. Then the expansion of a well-labelled object α by e is denoted e(α) and results in a weakly-labelled Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY object in which the label j is replaced by e(j). So h2, 3, 1, 4i may expand to h7, 10, 2, 88i or h3, 9, 1, 15i Version of: 11-Dec-09 etc etc. Now, given any two objects β ∈ B and γ ∈ C their labelled product (or product) is β ? γ. This is a set of well-labelled pairs (β 0 , γ 0 ) that reduce back to β, γ. β ? γ = {(β 0 , γ 0 ) s.t. it is well-labelled and ρ(β 0 ) = β, ρ(γ 0 ) = γ} So how big is this set? If |β| = n1 and |γ| = n2 then n1 + n2 n card(β ? γ) = = n1 , n2 n1 where n = n1 + n2 . You can see this by considering all the “new” labels and allocating n1 of them to the part of the object coming from β and the rest go to the part coming from γ. Then assign each block of labels so as to preserve the original orderings. Definition (Labelled product). The labelled product B ? C is obtained by forming all ordered pairs from B × C and computing all possible order-consistent relabellings. That is B ? C = ∪β∈B,γ∈C (β ? γ) Now we can look at the admissible constructions. M ARNI M ISHNA , S PRING 2011 MATH 343: A PPLIED D ISCRETE M ATHEMATICS PAGE 4/15 MATH 895-4 Fall 2010 Course Schedule A DDITIONAL N OTES f ac ulty of sc ience depar tment of mathem atics Week 2.2 Date Sections Part/ References Topic/Sections from FS2009 Labelled product Labelled combinatorial classes Notes/Speaker 1 Sept 7 I.1, I.2, I.3 Symbolic methods Combinatorial When A = B ? C the corresponding counting sequences satisfy Structures 2 14 I.4, I.5, I.6 Unlabelled structures FS: Part A.1, A.2 X n Comtet74 ALabelled Bn1 Cn2 3 21 II.1, II.2, II.3 n = structures I Handout #1 n1 , n2 n +n =n 4 28 (self study) II.4, II.5, II.6 1 2 Labelled structures II The product Bn1 Cn2 takes care of all Combinatorial possible pairings of objects and the binomial takes care of the Combinatorial 5 Oct 5 III.1, III.2 Asst #1 Due parameters Parameters convolution of the two sequences and thus possible relabellings. This is just the binomial 6 12 IV.1, IV.2 7 19 IV.3, IV.4 8 2.3 9 26 FS A.III (self-study) Multivariable GFs A=B?C ⇒ Analytic Methods FS: Part B: IV, V, VI Appendix B4 Stanley 99: Ch. 6 Handout #1 power of B is just (self-study) Sequences IV.5 V.1 Nov 2 The kth (labelled) 9 VI.1 The 10 sequence class is 12 11 18 A.3/ C A(z) = B(z) · C(z) Complex Analysis Singularity Analysis Asst #2 Due Asymptotic methods the k-fold labelled product of B with itself. It is denoted S EQk (B). Sophie Introduction to Prob. Mariolys S EQ(B) = E + B + B ? B + · · · = ∪k≥0 S EQk (B) IX.1 Limit Laws and Comb Marni These 20translate following operations on the Structures IX.2 to theRandom Discrete Limit Laws egf Sophie and Limit Laws 12 13 2.4 14 23 IX.3 A FS: Part C = S EQ k (B) 25 IX.4 A = S EQ(B) 30 (rotating presentations) ⇒ ⇒ Sets Continuous Limit Laws A(z)Marni = Quasi-Powers and Gaussian limit laws IX.5 Dec 10 Combinatorial A(z)Mariolys = B(z)k instances of discrete Presentations 1 1 − B(z) B0 = ∅ Sophie Asst #3 Due The class of sets of k-components of B is denoted S ETk (B). Formally it is defined as a k-sequence counted modulo permutation of the components. That is S ETk (B) = S EQk (B)/R where R is an equivalence relation that identifies two sequences if one is simply a permutation of the components of the other. The labelled set construction is then defined by S ET(B) = ∪k≥0 S ETk (B) If we take a labelled k-set, we can order its components in k! different ways to get k! labelled ksequences. Each component must be distinct since the k-set must be well-ordered. In reverse we Dr. Marni MISHNA, Department of Mathematics, FRASER UNIVERSITY of a k-sequnce (they must all be different) and then can consider the smallest labelSIMON in each component Version of: 11-Dec-09 reduce these labels to get a permutation of k. Hence we must have A = S ETk (B) A = S ET(B) 1 B(z)k k! A(z) = exp(B(z)) ⇒ A(z) = ⇒ Note that this is so much easier than the multiset construction for the unlabelled case since components must be distinct thanks to their labels. 2.5 Cycles The class of k-cycles of B is denoted C YCk (B). It is defined as a k-sequence counted modulo cyclic shifts of it components: C YCk (B) = S EQk (B)/S where S is an equivalence relation that identifies two sequences if the components of one is a cyclic shift of the components of the other. Since any k-sequence is well-labelled, its components are necesarrily distinct and thus there is a k-to-one mapping between k-sequences and k-cycles. Thus A = C YCk (B) A = C YC(B) ⇒ ⇒ 1 B(z)k k 1 A(z) = log 1 − B(z) A(z) = Again — this is much easier because the components are distinct thanks to their labelling. M ARNI M ISHNA , S PRING 2011 MATH 343: A PPLIED D ISCRETE M ATHEMATICS PAGE 5/15 MATH 895-4 Fall 2010 Course Schedule A DDITIONAL N OTES f ac ulty of sc ience depar tment of mathem atics Week 2.6 Date Sections Part/ References Topic/Sections from FS2009 Admissibility Theorem 1 Sept 7 I.1, I.2, I.3 Combinatorial Thus we have now proved 14 I.4, I.5, I.6 5 Oct 5 III.1, III.2 6 12 IV.1, IV.2 7 19 IV.3, IV.4 8 26 9 Nov 2 10 11 12 VI.1 12 A.3/ C 18 IX.1 20 IX.2 23 IX.3 25 IX.4 13 30 14 Dec 10 Combinatorial parameters FS A.III (self-study) Combinatorial Parameters A(z) = B(z) + C(z) Asst #1 Due Multivariable GFs A=B?C labelled product 9 labelled product, sequence, set and cycle all admis- A=B+C sum IV.5 V.1 Notes/Speaker Symbolic methods Structures Unlabelled structures FS: Part A.1, A.2 Theorem. The constructions of combinatorial sum, Comtet74 3 21 II.1, II.2, II.3 Labelled structures I Handout #1 sible. The operators are (self study) 4 28 II.4, II.5, II.6 Labelled structures II 2 Labelled combinatorial classes Analytic Methods FS: Part B: IV, V, VI Appendix B4 Stanley 99: Ch. 6 sequence Handout #1 (self-study) A(z) = B(z)C(z) Complex Analysis Singularity Analysis A = S EQ(B) Asst #2 Due A(z) = Asymptotic methods Sophie A(z) = B(z)k A = S EQk (B) set Random Structures and Limit Laws FS: Part C (rotating presentations) IX.5 Introduction to Prob. Mariolys Limit Laws and Comb Marni A = S ET(B) Sophie A(z) = exp(B(z)) 1 A(z) = B(z)k k! Discrete Limit Laws Combinatorial A = S ETk (B) Mariolys instances of discrete Continuous Limit Laws Marni Quasi-Powers and Sophie = laws C YC(B) GaussianA limit cycle Presentations 1 1 − B(z) A(z) = log Asst #3 Due A = C YCk (B) A(z) = 1 1 − B(z) 1 B(z)k k where for sequence, set and cycle it is assumed that B0 = ∅. Note how the cycle and set operations are nearly inverses of each other. 3 Surjections Consider the class of surjections: Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY Version of: 11-Dec-09 R = S EQ(S ET≥1 (Z)). http://www.nourishingdays.com/2009/02/make-yogurt-in-your-crock-pot/ A mapping from A to B is a (r) surjection if every element of B is mapped to be at least one element of A. Now let Rn denote the set (r) (r) of surjections from [1, n] onto [1, r]. Further let R = ∪n Rn . Consider the following surjection, φ from [1, 9] to [1, 4] 1→3 2→2 3→4 4→2 5→1 6→1 7→2 8→3 9→4 Now consider the preimage of each point φ−1 (1) = {5, 6} φ−1 (2) = {2, 4, 7} φ−1 (3) = {1, 8} φ−1 (4) = {3, 9} This is clearly an ordered sequence of non-empty sets. More generally one can decompose a surjection to [1, r] as an r-sequence of sets. Thus R(r) = S EQr (V) V = S ET≥1 (Z) V (z) = exp(z) − 1 M ARNI M ISHNA , S PRING 2011 R (r) MATH 343: A PPLIED D ISCRETE M ATHEMATICS (z) = (ez − 1)r PAGE 6/15 MATH 895-4 Fall 2010 Course Schedule A DDITIONAL N OTES f ac ulty of sc ience depar tment of mathem atics Week Date Sections Part/ References Topic/Sections Notes/Speaker ∼ fromV FS2009 Note that here = U − {} — non-empty urns (r) We can expand to get at Rn Symbolic methods 1 Sept 7 I.1, I.2, I.3 thisCombinatorial 2 14 I.4, I.5, I.6 3 21 II.1, II.2, II.3 4 28 II.4, II.5, II.6 5 Oct 5 III.1, III.2 6 12 Structures FS: Part A.1, A.2 Comtet74 Handout #1 (self study) Unlabelled structures r X (r) n j r RnLabelled = n![z ] I (−1) e(r−j)z structures j j=0 Labelled structures II r X j r n Combinatorial Combinatorial parameters FS A.III (self-study) IV.1, IV.2 = Parameters 4 26 9 Nov 2 4.1 9 VI.1 (r#1 −Due j) Asst , which we Complex Analysis r will learn about next. Singularity Analysis Set Partitions 10 j Multivariable GFs n FS: Part B: IV, V, VI Appendix B4 Stanley 99: Ch. 6 Handout #1 (self-study) IV.5 V.1 (−1) j=0 which is related numbers 7 19 IV.3, IV.4to stirling Analytic Methods 8 Labelled combinatorial classes Asymptotic methods Asst #2 Due Sophie Definition A.3/ C 12 Introduction to Prob. Mariolys IX.1 set-partitions. Consider Limit Laws and Comb [1,Marni Let us18look at the interval n]. We can partition this into r distinct subsets 11 Random Structures (called20blocks). [1, 2, 3, 4] can beLimit partitioned into IX.2 For example Discrete Laws Sophie • 123set =IX.3 [1, 2, 3, 4] 12 • 225sets =IX.4[1, 2, 3|4] 13 • 330sets =IX.5[1, 2|3|4] 14 10 • 4Dec sets = [1|2|3|4] and Limit Laws FS: Part C (rotating presentations) [1, 2, 4|3] [1, 3|2|4] Combinatorial instances of discrete Limit 3, Laws [1, 3,Continuous 4|2] [1|2, 4] Quasi-Powers and [1, 4|2|3] 3|4] Gaussian[1|2, limit laws Presentations Mariolys Marni [1, 2|3, 4] Sophie [1|2, 4|3] [1, 3|2, 4] [1, 4|2, 3] [1|2|3, 4] Asst #3 Due (r) So let Sn be the number of ways of partitioning an n-set into r-blocks, then we have (1) S4 (2) =1 S4 =7 (3) S4 =6 (4) S4 =1 (r) The numbers Sn are called the Stirling partition numbers (or Stirling numbers of the second kind) and are frequently denoted nr . Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY Version of: 11-Dec-09 4.2 Set partitions as a labelled class Now let S (r) denote the class of set partitions in r-blocks. Unlike the surjections we just studied, these blocks are not ordered. Thus we have a set of non-empty sets S (r) = S ETr (V) 1 S (r) (z) = (ez − 1)r r! V = S ET≥1 (Z) so we clearly have the relation Sn(r) = 1 (r) R r! n = r 1 X r (−1)j (r − j)n r! j=0 j Which is quite obvious, since we can take an r-partition and order the r-blocks in r! different ways to get surjections into [1, r]. M ARNI M ISHNA , S PRING 2011 MATH 343: A PPLIED D ISCRETE M ATHEMATICS PAGE 7/15 MATH 895-4 Fall 2010 Course Schedule A DDITIONAL N OTES f ac ulty of sc ience depar tment of mathem atics Week Date Sections Part/ References Topic/Sections Labelled combinatorial classes Notes/Speaker FS2009 It is the from work of a moment to extend these to get the total number of surjections from [1, n] and setpartitions of [1, 1 Sept 7 I.1, I.2, I.3 n] Symbolic methods 2 14 I.4, I.5, I.6 3 21 II.1, II.2, II.3 4 28 II.4, II.5, II.6 5 Oct 5 6 12 7 19 R= R= Combinatorial Structures (r) Unlabelled structures Part A.1, A.2 ∪FS: rR Comtet74 Labelled structures I SHandout EQ (S#1 ET ≥1 (Z)) (self study) Labelled 2 structures 3 II S = ∪r S (r) S = S ET(S ET≥1 (Z)) z 1 z = 1 + z + Combinatorial 3 + 13 + . . . Combinatorial z 2 − e 2! 3! III.1, III.2 Asst #1 Due parameters Parameters 3 2 FSz A.III z z e −1 IV.1,S(z) IV.2 = e(self-study) + 5 GFs + ... = 1 + z + 2Multivariable 2! 3! R(z) = IV.3, IV.4 Complex Analysis Analytic Methods FS: Part B: IV, V, VI 8 26 Singularity Analysis Appendix B4 IV.5 V.1 Stanley 99: Ch. 6 The Rn are called surjection numbers while the SDue n are called Bell numbers. 9 numbers Nov 2 Asst #2 Asymptotic methods #1 One can then quiteHandout easily compute the egfs of set partitions with any / odd / even number of blocks (self-study) 9 VI.1 Sophie of 10any / odd / even sizes, by making suitable restrictions of the set / sequence operators in the construcA.3/quite C tions. 12 In fact, generally we have Introduction to Prob. Mariolys 18 IX.1 23 IX.3 25 IX.4 Limit Laws and Comb Marni 11 Lemma. The class R(A,B) of surjections in which the cardinalities of the premiages lie in A ⊆ N and Random Structures 20 IX.2 Discrete Limit Laws Sophie the cardinality of the range and Limitlies Lawsin B ⊆ N is given by 12 13 30 14 Dec 10 FS: Part C (rotating R(A,B) (z) presentations) Combinatorial instances of discrete = β(α(z)) Limit Laws XContinuous za α(z) = Quasi-Powers and a! IX.5 a∈A Gaussian limit laws Presentations Mariolys Marni β(z) = Sophie X zb b∈B Asst #3 Due The class S (A,B) of set partitions with block sizes in A ⊆ N and number of blocks in in B ⊆ N is given by S (A,B) (z) = β(α(z)) X za α(z) = a! β(z) = a∈A 4.3 X zb ! b b∈B Set partitions as an unlabelled class Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY Version 11-Dec-09 Now, of:we can also consider set partitions using an unlabelled class. This shows that we need to keep (r) an open mind when deciding these properties. We encode set partitions as a family of words! Let Sn n (r) denote the set partitions of an n-set into r-blocks, so that r = card(Sn ). We can encode any such set partition as a word on B = {b1 , . . . , br }. In particular take each block and sort its elements smallest to largest. Then order the blocks according to the smallest element in each block. Label everything in the first block by b1 , the second block by b2 etc etc. Eg {1, 2, . . . , 9} = {6, 4} + {5, 1, 2} + {3, 8, 7} = {1, 2, 5} + {3, 7, 8} + {4, 6} | {z } | {z } | {z } b1 b2 b3 Then simply read off the label of each number in order ≡ [b1 , b1 , b2 , b3 , b1 , b3 , b2 , b2 ] Now this is not just any word in the letters bi there are some extra conditions • all r letters must occur • the first occurrance of b1 is before that of b2 which is before the first occurrance of b3 etc etc. So this means that our word looks like M ARNI M ISHNA , S PRING 2011 MATH 343: A PPLIED D ISCRETE M ATHEMATICS PAGE 8/15 MATH 895-4 Fall 2010 Course Schedule A DDITIONAL N OTES f ac ulty of sc ience depar tment of mathem atics Week Date Sections Part/ References Topic/Sections from FS2009 • a sequence of at least 1 b1 , then 1 2 3 Notes/Speaker I.1, I.2, I.3 Symbolic methods Combinatorial • aSept b2 7then a sequence of {b1 , b2 }, then Structures 14 I.4, I.5, I.6 21 II.1, II.2, II.3 Unlabelled structures FS: Part A.1, A.2 • a b3 then a sequence of {b1 , b2 , b3 }, then etc etc Comtet74 This gives 4 28 5 6 Labelled combinatorial classes Oct 5 II.4, II.5, II.6 III.1, III.2 IV.1, IV.2 12 A.3/ C Labelled structures II b1 S EQ (b1 )b2 S EQ(b1 + b2 )b3 S EQ(b1 + b2 + b3 ) . . . br S EQ(b1 + · · · + br ) Combinatorial Combinatorial Hence the ogf is 12 Labelled structures I Handout #1 (self study) parameters FS A.III (self-study) Asst #1 Due Parameters Multivariable GFs zr (r) S (z) = 7 19 IV.3, IV.4 Complex (1 Analysis − z)(1 − 2z) . . . (1 − rz) Analytic Methods FS: Part B: IV, V, VI 8 26we can re-writeAppendix Singularity Analysis Which using partial fraction decomposition, or other methods as: B4 IV.5 V.1 Stanley 99: Ch. 6 9 Nov 2 Asst #2 Due r Asymptotic methods Handout #1 1 X r (−1)r−j (self-study) S(z) = 9 VI.1 Sophie r! j=1 j 1 − jz 10 and so18 IX.1 20 IX.2 23 IX.3 25 IX.4 11 as12before. 4.4 13 Introduction to Prob. Random Structures and Limit Laws FS: Part C (rotating presentations) Mariolys Comb Marni LimitLaws andX r n 1 r−j r Discrete= Limit Laws (−1) Sophie jn, j r! j=1 r Combinatorial instances of discrete Mariolys Continuous Limit Laws Marni Quasi-Powers and Gaussian limit laws Wrapping up surjections and set partitions IX.5 Sophie 30 Proposition. The class R(A,B) of surjections in which the cardinalities of the premiages lie in A ⊆ N 14 Dec 10 Presentations Asst #3 Due and the cardinality of the range lies in B ⊆ N is given by R(A,B) (z) = β(α(z)) X za α(z) = a! β(z) = a∈A Proposition. The class S B ⊆ N is given by (A,B) X zb b∈B of set partitions with block sizes in A ⊆ N and number of blocks in in (A,B) S (z) = β(α(z)) Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY Version of: 11-Dec-09 α(z) = X za a! β(z) = a∈A 5 X zb ! b b∈B Permutations We just looked at sequences and sets of sets. We could do the same for cycles. Sequences of cycles are called alignments, but do not appear that frequently. Let us instead consider sets of cycles, also known as permutations: P = S ET(C YC(Z)) As suggested by this formula, every cycle can be naturally decomposed into a set of cycles. Consider the permutation σ = (10, 12, 2, 7, 1, 8, 6, 9, 4, 5, 3, 11) Let us track where 1 moves under the action of this permutation 1 → 10 → 5 → 1 2 → 12 → 11 → 3 → 2 4→7→6→8→9→4 So the above decomposes into 3 cycles of lengths 3,4,5; its cycle number is 3. Schematically this gives M ARNI M ISHNA , S PRING 2011 MATH 343: A PPLIED D ISCRETE M ATHEMATICS PAGE 9/15 MATH 895-4 Fall 2010 Course Schedule A DDITIONAL N OTES f ac ulty of sc ience depar tment of mathem atics Week Date 1 Sept 7 I.1, I.2, I.3 2 14 I.4, I.5, I.6 3 21 II.1, II.2, II.3 4 28 II.4, II.5, II.6 5 Oct 5 Using this 6 12 7 19 8 26 9 Nov 2 Sections from FS2009 Part/ References Topic/Sections 1 5 2 9 VI.1 18 IX.1 Analytic Methods FS: Part B: IV, V, VI Appendix B4 Stanley 99: Ch. 6 Handout #1 (self-study) 4 9 7 Labelled structures I 10 Combinatorial parameters FS A.III decomposition as a set IV.1, IV.2 (self-study) IV.5 V.1 12 Unlabelled structures 3 III.1, III.2 IV.3, IV.4 Notes/Speaker Symbolic methods Combinatorial Structures FS: Part A.1, A.2 Comtet74 Handout #1 (self study) Labelled combinatorial classes Labelled structures II 11 Combinatorial Parameters Asst #1 Due 8 6 of cycles we get Multivariable GFs Complex ∼ Analysis P = S ET(C YC(Z)) ∼ = S EQ(Z) 1 1 P (z) = exp log Asst #2 Due = Asymptotic methods 1 − z 1−z Singularity Analysis Sophie 10 fact that exp and log are inverses is reflected in cycle-decomposition of permutations. The 12 A.3/ C Introduction to Prob. Mariolys Now we can go further by playing with restrictions (as we did above) to get 11 Limit Laws and Comb Marni Proposition. P (A,B) of permutations with cycle Random Structures 20 IX.2 The class Discrete Limit Laws Sophie lengths in A ⊆ N and cycle number in B ⊆ N and Limit Laws is given by FS: Part C 12 23 IX.3 25 IX.4 13 30 IX.5 14 Dec 10 Combinatorial instances of discrete (rotating presentations) P (A,B) (z) Mariolys = β(α(z)) Continuous Limit Laws X za α(z) = Quasi-Powers and Gaussian a limit laws Marni β(z) = Sophie a∈A 5.1 5.1.1 X zb b! b∈B Presentations Asst #3 Due Sub-classes of permutations Involutions The permutation σ is an involution if σ ◦ σ = identity. This means that all cycles are length 1 or 2, but the number of cycles is unconstrained. In the language of the above theorem, A = {1, 2} and B = {0, 1, 2, . . . }. Hence I = S ET(C YC1,2 (Z)) Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY I(z) = exp z + z 2 /2 Version of: 11-Dec-09 = X zk k≥0 = k! (1 + z/2) = n/2 X j=0 k j+k XX k z k≥0 j=0 n/2 XX n−j n≥0 j=0 In = k j j 2j k! 1 zn 2j (n − j)! n! 2j (n − 2j)!j! One can of course extend this to consider permutations in which all cycles are of length ≤ r. 5.1.2 Derangements In the opposite direction, a derangement is a permutation in which no number stays put — hence all cycles must be 2 or longer. In the language of the above theorem, A = {2, 3, 4, 5, . . . } and B = M ARNI M ISHNA , S PRING 2011 MATH 343: A PPLIED D ISCRETE M ATHEMATICS PAGE 10/15 MATH 895-4 Fall 2010 Course Schedule A DDITIONAL N OTES f ac ulty of sc ience depar tment of mathem atics Week Date Sections 1 Sept 7 I.1, I.2, I.3 2 14 I.4, I.5, I.6 3 21 4 28 5 Oct 5 {0, 1, 2, . . . }. from FS2009 6 Part/ References Topic/Sections Labelled combinatorial classes Notes/Speaker Symbolic methods D = SCombinatorial ET (C YC >1 (Z)) Structures Unlabelled structures A.1, A.2 FS: Part 1 e−z Comtet74 − Labelled z = structures I II.1,D(z) II.2, II.3 = exp log Handout #1 1 − z 1−z (self study) II.4, II.5, II.6 Labelled structures II 2 Combinatorial (−1)n 1 Combinatorial + − · · · + D = n! 1 − III.1, III.2 n parameters1! 2! Parameters n! 12 IV.1, IV.2 FS A.III (self-study) 9 VI.1 Handout #1 (self-study) 12 A.3/ C 18 IX.1 20 IX.2 = X (−1)n n≥0 zn X n × z n! n≥0 Asst #1 Due Multivariable GFs Hence Dn /n! = probability that a permutation leaves nothing in place, is the truncation of the expansion IV.3, IV.4 Complex Analysis Analytic Methods of 7e−119 . Since this converges very quickly, the probability of a permutation being a derangment is FS: Part B: IV, V, VI 8 26 Singularity Analysis asymptotically 1/e ≈ 0.37. Appendix B4 IV.5 V.1 99: Ch. 6 to be length > r or longer one gets — 2if you requireStanley all cycles 9 So Nov Asst #2 Due Asymptotic methods 10 11 12 5.1.3 Sophie D(r) = S ET(C YC>r (Z)) Introduction to Prob. Mariolys z z2 zr X zk e− 1 − 2! −···− r! 1 Limit Laws and Comb Marni (r) = − D (z) = exp log 1−z k 1−z Random Structures Discrete Limit Laws Sophie k≤r and Limit Laws FS: Part C 23 IX.3 (rotating Other variants presentations) 25 IX.4 Combinatorial instances of discrete Mariolys Continuous Limit Laws Marni This approach will work well to consider other classes of permutations such as permutations with all Quasi-Powers and 13 cycles, 30 IX.5 Sophie even all odd cycles, cycles of even length, Gaussian limit laws all of odd length etc. It is also not hard to get all d permutations so that σ = identity for a given d. Give Asst these a shot. 14 Dec 10 Presentations #3 Due 5.2 The number of cycles in a permutation (r) Let P (r) be the class of permutations that decompose into r cycles. The number Pn Stirling cycle numbers (or Stirling numbers of the first kind), and their egf is ≡ n r are the P (r) = S ETr (C YC(Z)) r 1 1 (r) P (z) = log r! 1−z Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY Version of: 11-Dec-09 1 . So — the probability that a permutation has exactly r cycles is nr n! What do these numbers look like? Let pn,k be this probability, and consider n = 100. k pn,k 1 .01 2 .05 3 .12 4 .19 5 .21 6 .17 7 .11 8 .06 9 .03 10 .01 How do we interpret this? A random permutation of 100 has on average a few more than 5 cycles. It rarely has more than 10. What does this imply about the average length of a cycle? 5.3 Application: 100 Prisoners This example is a reformulation of an applied problem on probing and hashing. The wording is modified from the wikipedia page on Random permutation statistics. 5.3.1 The game A prison warden wants to make room in his prison and is considering liberating one hundred prisoners, thereby freeing one hundred cells. He assembles one hundred prisoners and asks them to play the following game: M ARNI M ISHNA , S PRING 2011 MATH 343: A PPLIED D ISCRETE M ATHEMATICS PAGE 11/15 MATH 895-4 Fall 2010 Course Schedule A DDITIONAL N OTES f ac ulty of sc ience depar tment of mathem atics Week Date Sections Part/ References Topic/Sections Labelled combinatorial classes Notes/Speaker from FS2009 1. he lines up one hundred urns in a row, each containing the name of one prisoner, where every prisoner’s name occurs exactly once; 1 Sept 7 I.1, I.2, I.3 Symbolic methods Combinatorial Structures isFS: allowed to lookUnlabelled insidestructures fifty urns. If he or she does not find his or Part A.1, A.2 Comtet74 the fifty will immediately be executed, otherwise the game II.1, II.2, II.3 urns, all prisoners Labelled structures I Handout #1 (self study) II.4, II.5, II.6 Labelled structures II 22. 14 I.4, I.5, I.6 every prisoner 3 one 21 of her name in continues. 4 28 The prisoners have a few moments to decide on a strategy, knowing that once the game has begun, they Combinatorial will not be able to communicate withCombinatorial each other, mark 5 Oct 5 III.1, III.2 Asst the #1 Dueurns in any way or move the urns or the parameters Parameters names inside them. Choosing urns at random, their chances of survival are almost zero, but there is a FS A.III 6 12 giving IV.1, IV.2 Multivariable GFs strategy them a(self-study) 30% chance of survival, assuming that the names are assigned to urns randomly what is 7 19 it? IV.3, IV.4 Complex Analysis Analytic Methods 99 100 FS: Part B: IV, V, VI (49) 1 8 First 26 of all, the survival Singularity Analysis = 2100 so this is definitely probability using random choices is Appendix B4 IV.5 V.1 (100 50 ) Stanley 99: Ch. 6 9 Nov 2 Asst #2 Due Asymptotic methods Handout #1 not a practical strategy. 10 5.3.2 9 (self-study) VI.1 Sophie 12The A.3/ C strategy Introduction to Prob. Mariolys 18 IX.1 Limit Laws and Comb Marni 1 BOX The 11 30% survival strategy is to consider the contents of the urns to be a permutation of the prisoners, Random Structures 20 IX.2 Discrete Limit Laws Sophie and traverse cycles. To the notation simple, assign a number to each prisoner, for example by and keep Limit Laws FS: Part C Combinatorial sorting23 theirIX.3names alphabetically. The urns may thereafter be considered to contain numbers rather Mariolys (rotating instances of discrete 12 names. Now clearly the contents than of the urns define a permutation. The first prisoner opens the presentations) 25 Continuous Limit Laws first urn. IfIX.4 he finds his name, he has finished and Marni survives. Otherwise he opens the urn with the number he found in the first urn. The processand repeats: the prisoner opens an urn and survives if he Quasi-Powers 13 30 IX.5 Sophie Gaussian limit laws finds his name, otherwise he opens the urn with the number just retrieved, up to a limit of fifty urns. 14 second Dec 10 prisoner starts Presentations #3 Duewith urn number three, and so on. This The with urn number two, theAsst third strategy is precisely equivalent to a traversal of the cycles of the permutation represented by the urns. Every prisoner starts with the urn bearing his number and keeps on traversing his cycle up to a limit of fifty urns. The number of the urn that contains his number is the pre-image of that number under the permutation. Hence the prisoners survive if all cycles of the permutation contain at most fifty elements. We have to show that this probability is at least 30%. Let us see how this works on a smaller example. Consider the permutation from the beginning of these notes: BOX 2 BOX 3 BOX 4 BOX 5 BOX 6 Dr. Marni Department of UNIVERSITY 10 MISHNA,12 2 Mathematics, 7 SIMON FRASER 1 8 Version of: 11-Dec-09 1 2 BOX 7 BOX 6 12 8 BOX 9 4 9 BOX 5 10 BOX 3 11 12 11 BOX 4 9 7 5 10 3 11 8 6 This simulates the situation where the warden has put a 10 in box 1, a 12 in box 2, a 2 in box 3 and in general σ(i) in box i. If the prisoners are following the strategy, they will each have 6 boxes to look in to, and since the maximum cycle size is 5, they will win. How will it go down? Prisoner one will open box 1. It will say 10. He then opens box 10, and it will say 5. He opens box 5, it says 1, and he breathes a sigh of relief. Prisoner two’s turn. He opens box 2: it says 12. He continues along the cycle and the fourth box he opens will have a 2 inside. Later, prisoner 4 is up. He follows this and he starts to sweat because it takes him 5 turns, but this is still less than 6, so he is okay. We know what this permutation looks like, and that every cycle is of length less than 6, so, as we already noted, if this were the scheme, the prisoners would win. Note that this assumes that the warden chooses the permutation randomly; if the warden anticipates this strategy, he can simply choose a permutation with a cycle of length 51. To overcome this, the prisoners may agree in advance on a random permutation of their names. M ARNI M ISHNA , S PRING 2011 MATH 343: A PPLIED D ISCRETE M ATHEMATICS PAGE 12/15 MATH 895-4 Fall 2010 Course Schedule A DDITIONAL N OTES f ac ulty of sc ience depar tment of mathem atics Week Date 5.3.3 Sections Part/ References from FS2009 The analysis Topic/Sections Labelled combinatorial classes Notes/Speaker 1 Sept 7 I.1, I.2, I.3 Symbolic methods Combinatorial First, let us consider the general problem with 2n prisoners choosing n urns. Let us calculate the egf Structures 2 14 I.4, I.5, I.6 Unlabelled that a random permutation (ofA.2any size) has astructures cycle of length n + 1, and then deduce the probability for FS: Part A.1, Comtet74 the3 case of length n, and these are 21 of permutations II.1, II.2, II.3 Labelled structures I the cases in which case this strategy will fail. Fix Handout #1 n and consider the class of all permutations which have a cycle of length at least n + 1. We take a sum (self study) 4 28 II.4, II.5, II.6 Labelled structures II over all permutations σ of the term z |σ| if it has no such cycle and z |σ| u if it does 5 Oct 5 III.1, III.2 Combinatorial Combinatorial Analytic Methods Complex Analysis Asst #1 Due parameters Parameters 1 (if σ has a cycle of length n);0 otherwise X z |σ| u 2z 2 zn z n+1 FS A.III =1+z+ + · · · + n! + (n!u + nn!) + ... g (z, u) = 6 12 IV.1, IV.2 Multivariable GFs (self-study) |σ|! 2! n! (n + 1)! (n) 7 19 σ∈P IV.3, IV.4 We first expand theFS:decomposition P = S ET(C YC(Z)) = S ET(C YC≤n (Z)+ C YC>n (Z)) which we then Part B: IV, V, VI 8 26 Singularity Analysis Appendix B4 expand as follows: IV.5 V.1 Stanley 99: Ch. 6 9 Nov 2 Asymptotic methods Asst #2 Due Handout #1 10 11 12 9 (self-study) VI.1 12 A.3/ C 18 IX.1 20 IX.2 23 IX.3 25 IX.4 g Sophie z n+1 z2 z3 zn z n+2 Introduction + to Prob. + · · · +Mariolys+ u +u + . . . (z, u) = exp z + 2 3{z n} n+1 n+2 | | {z } Limit Laws and Comb Marni C YC≤n (Z) Sophie C YC>n (Z) Random Structures Discrete Limit Laws n+1 and Limit Laws n+2 1 z z FS: Part C= Combinatorial exp (u − 1) + + ... . Mariolys (rotating 1 − z instances of discreten + 1 n + 2 (n) presentations) Continuous Limit Laws Marni The desired probability is Quasi-Powers and n+1 13 30 IX.5 Sophie Gaussian limit 1 laws z z n+2 2n (n) [z ] [u]g (z, u) = [z 2n ][u] 1 + (u − 1) + + ... , 1−z n+1 n+2 14 Dec 10 Presentations Asst #3 Due since exp(•) = 1 + • + •2 + . . . , and so once you square the argument to the exp, you will only create terms which are a power of z greater than 2n. This is equal to n+1 z n+2 1 z 2n (n) 2n + + ... [z ] [u]g (z, u) = [z ] 1−z n+1 n+2 ! 2n ` X X X 1 1 2n = [z ] z` = . k k ` Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY Theseof:are related to Harmonic numbers and have Version 11-Dec-09 we compute that [z 2n ] [u]g (n) (z, u) < log 2 =⇒ k=n+1 k=n+1 good estimates. Using one of these good estimates 1 − [z 2n ] [u]g (n) (z, u) > 1 − log 2 = 0.3068528 . . . That is, at least 30 %. 5.4 Computing Stirling cycle numbers We can follow a similar analysis to get formulas for Stirling cycle numbers nr which we recall are the number of permutations of n that decompose into r cycles. Start with the bivariate g.f. P (z, u) = ∞ X P (r) (z)ur r=0 ∞ X r 1 1 = u log 1−z r! r=0 1 = exp u log = (1 − z)−u 1−z X n −u = (−1) zn n n≥0 M ARNI M ISHNA , S PRING 2011 MATH 343: A PPLIED D ISCRETE M ATHEMATICS PAGE 13/15 MATH 895-4 Fall 2010 Course Schedule A DDITIONAL N OTES f ac ulty of sc ience depar tment of mathem atics Week Date Sections 1 Sept 7 I.1, I.2, I.3 2 14 I.4, I.5, I.6 3 21 II.1, II.2, II.3 Part/ References Topic/Sections from so the coeff of z nFS2009 is an expansion in u Combinatorial n Symbolic methods Structures X n r structures uUnlabelled = u(u + 1)(u FS: Part A.1, A.2 r Comtet74 r=0 Labelled structures I Handout #1 (self study) Labelled structures II Labelled combinatorial classes Notes/Speaker + 2) . . . (u + n − 1) 4 28 II.4, II.5, II.6 and setting u = 1 gives Differentiating this Combinatorial Combinatorial 5 Oct 5 III.1, III.2 Asst #1 Due n 1 X parameters Parameters r = 1 + 1/2 + 1/3 + . . . 1/n FS A.III n! Multivariable r 6 12 IV.1, IV.2 GFs (self-study) 7 is 19the expected IV.3, IV.4 Complex Analysis Analytic Methods This number of cycles in a permutation of length n and is approximately log n. 6 8 26 9 Nov 2 IV.5 V.1 FS: Part B: IV, V, VI Appendix B4 Stanley 99: Ch. 6 Handout #1 (self-study) Singularity Analysis Asymptotic methods Trees maps and graphs 10 9 VI.1 Asst #2 Due Sophie 12 A.3/ C to Prob. For trees, the labelled case is not soIntroduction different to theMariolys unlabelled case. We can consider planar and non-planar rooted trees. 18 IX.1 Limit Laws and Comb Marni 11 Random Structures 20 IX.2 A be the Discrete Limit Laws Sophie Example. Let class of rooted labelled planar trees whose vertex outdegrees must lie in the and Limit Laws set Ω. Then we have FS: Part C Combinatorial 12 23 IX.3 25 IX.4 (rotating presentations) A = Z S EQΩContinuous (A) Limit Laws 13 so30usingIX.5 and similar reasoning as the 14 Dec 10 instances of discrete Mariolys Marni Quasi-Powers and unlabelled case Sophie Gaussian limit laws Presentations Asst #3 Due A(z) = zφ(A(z)) φ(u) = X uω ω∈Ω and indeed this is identical to the ogf for the unlabelled version. Hence An = n!Aˆn — this is easy to prove by reading the vertices of a labelled tree in a canonical order (eg breadth first) to obtain a permutation and an unlabelled tree of the same shape. Using this idea or lagrange inversion we get that all labelled rooted planar trees are 1 2n − 2 = 2n−1 · 1 · 3 · · · (2n − 3) An = n! Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER n−1 n UNIVERSITY Version of: 11-Dec-09 So — to the non-planar case Example. Let T be the class of all non-planar rooted labelled trees. By deleting the root vertex one obtains a set of labelled trees. Thus T = Z ? S ET(T ) T (z) = zeT (z) Now φ(u) = eu and Lagrange inversion gives Tn = n![z n ]T (z) 1 n−1 u n [u ] (e ) = n! n nn = (n − 1)! = nn−1 n! Which is a famous result due to Cayley. These are usually called Cayley trees. A k-forest of Cayley trees gives a very similar result k n − 1 n−k (k) n T (z) Fn = n![z ] = n k! k−1 M ARNI M ISHNA , S PRING 2011 MATH 343: A PPLIED D ISCRETE M ATHEMATICS PAGE 14/15 f ac ulty of sc ience depar tment of mathem atics Week Date Sections 1 Sept 7 I.1, I.2, I.3 2 14 I.4, I.5, I.6 3 21 II.1, II.2, II.3 4 28 II.4, II.5, II.6 5 Oct 5 III.1, III.2 6 12 IV.1, IV.2 7 19 IV.3, IV.4 8 26 9 Nov 2 Part/ References MATH 895-4 Fall 2010 Course Schedule A DDITIONAL N OTES Topic/Sections Labelled combinatorial classes Notes/Speaker from FS2009 Similar games with restricted vertex degree lead to 10 11 12 IV.5 V.1 9 VI.1 12 A.3/ C 18 IX.1 20 IX.2 23 IX.3 25 IX.4 13 30 14 Dec 10 Symbolic methods Combinatorial T (Ω) = Z ? S ETΩ (T (Ω) ) Structures Unlabelled structures FS: Part A.1, A.2 (Ω) (Ω) Comtet74 T (z) = z φ̄(T Labelled (z)) structures I Handout #1 (self study) Labelled structures II Combinatorial parameters FS A.III (self-study) Combinatorial Parameters Analytic Methods FS: Part B: IV, V, VI Appendix B4 Stanley 99: Ch. 6 Handout #1 (self-study) Complex Analysis Random Structures and Limit Laws FS: Part C (rotating presentations) IX.5 φ̄(u) = X uω |ω|! ω∈Ω Asst #1 Due Multivariable GFs Singularity Analysis Asymptotic methods Asst #2 Due Sophie Introduction to Prob. Mariolys Limit Laws and Comb Marni Discrete Limit Laws Sophie Combinatorial instances of discrete Mariolys Continuous Limit Laws Marni Quasi-Powers and Gaussian limit laws Sophie Presentations Asst #3 Due Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY Version of: 11-Dec-09 M ARNI M ISHNA , S PRING 2011 MATH 343: A PPLIED D ISCRETE M ATHEMATICS PAGE 15/15