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Discrete Structures I – Combinatorics

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Discrete Structures I – Combinatorics Discrete Structures I – Combinatorics Lectures by Prof. Dr. Stefan Felsner Notes taken by Elisa Haubenreißer Summer term 2011 2 Contents 1 Introductory Examples 5 1.1 Derangements..................................... .. 5 1.2 TheOfficersProblem.................................. 9 2 Basic Counting 13 n 2.1 Models for k ...................................... 14 2.2 Extending binomial coefficients . ... 15 3 Fibonacci Numbers 17 3.1 Models for F -numbers .................................. 18 3.2 FormulasforFibonacciNumbers . .... 19 3.3 The -numberSystem.................................. 21 F 3.4 ContinuedFractionsandContinuants . ..... 22 4 The Twelvefold Way 25 5 Formal Power Series 33 5.1 Bernoullinumbersandsummation . ... 34 5.2 Compositionofseries................................. .. 35 5.3 RootsofFPS ....................................... 36 5.4 CatalanNumbers.................................... 37 6 Solvinglinearrecurrencesviageneratingfunctions 39 6.1 How to compute spanning trees of Gn ......................... 40 6.2 An example with exponential generating function . ....... 42 7 q-Enumeration 43 7.1 Another model for the q-binomial............................ 47 8 Finite Sets And Posets 49 8.1 Posets–Partiallyorderedsets. ..... 50 8.2 Thediagramofaposet ................................ 51 8.3 k-intersecting families . 52 8.4 Shadows – Another prooffor Sperner’s theorem . ......... 53 8.5 Erd¨os-Ko-RadofromKruskal-Katona. ........ 55 8.6 Symmetric chain decompositions . ... 57 8.7 An application: Dedekinds problem . ... 60 9 Duality Theorems 63 9.1 Thematchingpolytope ................................ 67 3 4 CONTENTS 10 P´olya Theory – Counting With Symmetries 71 10.1 Permutationgroupsandthecycleindex . ...... 71 10.2 LemmaofCauchy–Frobenius–Burnside . ..... 73 11 Linear Extensions And Dimension Of Posets 79 11.1 Dimension of n ..................................... 81 11.22-dimensionalposetsB ............................... .... 82 11.3 Algorithms to find antichain partitions/chain partitions . ........ 83 11.4 2-dimensional picture for 2-dimensional posets . .......... 83 11.5 The Robinson-Shensted-Correspondence . .......... 84 11.6Viennot’sskeletons................................. ... 85 12 Design Theory 91 12.1 Construction of an S6(3, 5, 10).............................. 93 12.2 Construction of S(3, 4, 10)................................ 93 12.3 Necessary conditions for the existence of designs . .......... 94 12.4Steinertriplesystems ............................... ... 95 12.5 Construction of STS(15) ................................ 96 12.6Projectiveplanes .................................. ... 97 12.6.1 Constructing projective planes . ... 97 12.7 Fisher’s inequality . 98 13 Inversions And Involutions 99 13.1M¨obiusinversion ................................... .. 99 13.2TheZeta-function ................................. 100 13.3 The M¨obius function of P ................................ 101 13.4Involutions ....................................... 104 13.5 LemmaofLindstr¨omGessel-Viennot . ...... 106 14 Catalan-Numbers 109 14.15morefamilies ...................................... 111 14.2 The reflection principle . 114 14.3Thecyclelemma ..................................... 115 14.4 A bijection for family (l) and a count via 13.5.1 . .. 115 14.5 BijectionbetweenNarayananumbers. ....... 116 14.6 Structure on Catalan families . .. 117 14.7Tamarilattice..................................... 119 A Practice Sheets 121 Chapter 1 Introductory Examples 1.1 Derangements Definition (permutation). A permutation is a bijection from a finite set X into itself. [n] := 1,...,n is the generic set of cardinality n. The set of permutations of the generic set of { } cardinality n is called Sn. Sn has some properties that we want to keep in mind: S has group structure • n S = n! • | n| two line notation for π S : • ∈ n 1 2 ... n π = π(1) π(2) ... π(n) Example. 1234567 π = 4352761 Definition (derangement). A derangement is a permutation without fixed points, i. e. π Sn with π(i) = i i [n]. The set of derangements in S is called D . ∈ 6 ∀ ∈ n n Problem (Montmort, 1708). How many derangements of n letters exist? Let d(n) be this number. We will list some possible answers for Montmorts problem and prove 5 6 CHAPTER 1. INTRODUCTORY EXAMPLES them later on. n 12345 6 7 (values) d(n) 0 1 2 9 44 265 1858 d(n) = (n 1) (d(n 1)+ d(n 2)) (recursion1) − − − n n n!= d(k) (recursion2) k k=0 nX n d(n)= ( 1)n+kk! (summation) k − kX=0 n! n! 1 d(n)= = + (explicit) e e 2 nn d(n) √2πn cn log n (asymptotic) ∼ en+1 ∼ d(n) e z zn = − (generating function) n! 1 z n 0 X≥ − For the table of values and more information have a look at the encyclopedia of integer sequences1. Proof. recursion 1 (proof by bijection) Let π D . ∈ n 1 2 3 ... x ... n 1 n π = n− y Now we construct a shorter permutation by shrinking column x: 1 2 3 ... x ... n 1 π = 0 y − If x = y then it follows that π0 Dn 1 else we remove the fixed point x and compactify the 6 ∈ − permutation to an even shorter permutation π00 and it therefore is π00 Dn 2. ∈ − Example. 1234 5 678910 π = 4713102896 5 123456789 π = 0 471352896 12345678 π00 = (compactification) 46132785 Claim 1.1.1. Every element of Dn 1 Dn 2 is exactly (n 1) times in the image of this map. − ∪ − − x Proof. Case 1: π0 Dn 1. Choose a column ((n 1) possibilities) and break it for n. We ∈ − y − ... x ... n obtain ... n ... y Case 2: π00 Dn 2. Choose b [n 1] and make a permutation π00 by increasing all elements ∈ − ∈ − b (this permutes 1,...,b 1 b +1,...,n 1 ). Add the fixed point b to get π0. Use b for ≥ { − }∪{ − } the column involving n. e 1http://oeis.org 1.1. DERANGEMENTS 7 Bijection: Dn [n 1] Dn 1 [n 1] Dn 2 ←→ − × − ∪ − × − (c, π ) c column π 0 ←→ (b, π00) b break element Proof of recursion 2. π S Fix(π)= i π(i)= i [n] ∈ n ⇒ { | }⊆ π ”is” (i. e. induces) a fixed point free permutation on [n] Fix(π). It follows: \ π00 Der ([n] Fix(π)) ∈ \ We can now define a bijection between Sn and A [n] Der(A) which adds a fixed point to any ⊆ element of Sn or deletes fixed points and restricts the derangement. S Example (S3). 1 2 3 1 2 3 π = Der( ) π = Der ( 1, 2 ) 1 1 2 3 ∈ ∅ 2 2 1 3 ∈ { } 1 2 3 1 2 3 π = Der ( 2, 3 ) π = Der ( 1, 2, 3 ) 3 1 3 2 ∈ { } 4 2 3 1 ∈ { } 1 2 3 1 2 3 π = Der ( 1, 2, 3 ) π = Der ( 1, 3 ) 5 3 1 2 ∈ { } 6 3 2 1 ∈ { } Now we can see the following equation. n!= S | n| = Der(A) | | A [n] X⊆ = d ( A ) | | A [n] X⊆ n n = d(k) k kX=0 To get an explicit formula for d(n) we need an inversion formula, which is a simplified version of the M¨obius inversion formula. Ê Proposition 1.1.2. Let f,g : Æ . The following equivalence holds: −→ n n g(n)= ( 1)kf(k) f(n)= ( 1)kg(k) k − ⇔ k − Xk Xk The fundamental lemma helps us proof this proposition. Lemma 1.1.3. A 0 B = ( 1)| | = 6 ∅ − 1 B = A B ∅ X⊆ Proof of 1.1.3. We know that A even contributes a +1 and an uneven A contributes a 1, so we show that there are as many even| | as odd subsets of B. Choose b B. Then we can form− pairs of subsets by removing b if b A or by adding b if b / A, we call the∈ new set A b. Then one of the sets A and A b is even and∈ the other one is odd∈ so the cardinalities add up⊕ to 0. ⊕ 8 CHAPTER 1. INTRODUCTORY EXAMPLES Proof of 1.1.2. For notational convenience let f(A) := f ( A ) and g(A) := g ( A ). The proposition is equivalent to | | | | A B g(B)= ( 1)| |f(A) f(C)= ( 1)| |g(B) − ⇔ − A B B C X⊆ X⊆ We proof this by B B A ( 1)| |g(B)= ( 1)| | ( 1)| |f(A) − − − B C B C A B X⊆ X⊆ X⊆ A B = ( 1)| |f(A) ( 1)| | − − A C B:A B C X⊆ X⊆ ⊆ A A X = ( 1)| |f(A)( 1)| | ( 1)| | − − − A C X C A X⊆ ⊆X\ =0 unless A=C C C = f(C)( 1)| |( 1)| | (1.1.3) − − | {z } = f(C) Proof of summation. We define g(k) := ( 1)kd(k) and f(k) := k!. So because of recursion 2 we get − n n f(n)= ( 1)kg(k) g(n)= ( 1)kf(k) k − ⇒ k − X n X n d(n) = ( 1)ng(n)= ( 1)kf(k)= ( 1)n+kk! ⇒ − k − k − X X To proof the explicit version, you have to use the following formulas n n! = k k!(n k)! − zk ez = k! k 0 X≥ n! 1 = n!e− e 1 Together with some estimates for the speed of convergence of the series of e− you get the result. √ n n For the asymptotic formulas you have to use the Stirling approximation of n! 2πn e together with the explicit formula. ∼ Proof of generating function. We define a new function and consider its derivative. d(n) n d(n) n 1 D(z) := z D0(z)= z − n! (n 1)! n 0 n 1 X≥ X≥ − 1.2. THE OFFICERS PROBLEM 9 From recursion 1 we obtain d(n + 1) d(n) d(n 1) = − n! − (n 1)! (n 1)! − − d(n) n 1 n (1 z)D0(z)= (z − z ) ⇒ − (n 1)! − n 1 X≥ − d(n + 1) d(n) = + zn − n! (n 1)! n 1 X≥ − d(n 1) recursion 1 = − zn − (n 1)! n 1 X≥ − = zD(z) (1 z)D0(z)= zD(z) − ⇒ − − z e− The solution D(z)= 1 z can be verified by computation. − 1.2 The Officers Problem Problem (Euler, 1782). 36 officers from 6 different countries (c) inherit 6 different ranks (r) such that every combination (c, r) appears exactly once. Can they be arranged in an 6 6 array such that each c and each r appears in every row and every column? × First of all we generalize the problem to n2 objects with two types of attributes, each with n variants.
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