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Combinatorics 09.10.74 Combinatorics Lectures of Mark Dukes Typing and layout by Líney Halla Kristinsdóttir University of Iceland Spring Term 2008 Contents 1 Enumeration1 1.1 Counting Arguments...............................2 1.2 Elementary Counting Coecients........................3 1.3 Compositions & Partitions of an Integer....................7 1.4 The Reection Principle.............................9 1.5 Stirling Numbers................................. 10 1.5.1 Stirling Numbers of the 2nd Kind: S(n; k) ............... 10 1.5.2 Signless Stirling Numbers of the 1st Kind............... 12 1.6 Multinomial Coecients............................. 14 1.7 Generating Functions I.............................. 15 1.8 Generating Functions II............................. 22 2 Permutations & Permutation Statistics 27 2.1 Descents & the Eulerian Polynomial...................... 27 2.2 Cycle structure & left-to-right maxima..................... 32 2.3 Excedances & weak excedances......................... 33 2.4 Inversions & the Major index.......................... 35 2.5 Multisets, Permutations & q-series....................... 39 2.6 Subspaces of a Vector Space.......................... 43 2.7 Permutations as Increasing Binary Trees.................... 45 2.8 Standard Young Tableaux............................ 46 2.9 Robinson-Schensted-Knuth (RSK) correspondence............... 48 i 3 Partially Ordered Sets (Posets) 53 3.1 Posets...................................... 53 3.2 Lattices..................................... 60 3.3 Modular Lattices................................. 62 3.4 Distributive Lattices............................... 64 3.5 Incidence Algebra of a Poset.......................... 64 3.6 The Zeta Function................................ 66 3.7 The Möbius Inversion Formula......................... 67 3.8 Applications of the Möbius Inversion Formula................. 69 3.9 The Möbius Function of Lattices........................ 73 3.10 Young's lattice.................................. 74 3.11 Linear extensions................................. 76 3.12 Rank-selection.................................. 77 4 Other Things & Some Applications 79 4.1 Unimodality, Log-concavity & Real-rootedness................. 79 4.2 Transfer Matrix Method-Theory......................... 82 4.3 Transfer Matrix Method Applications..................... 86 4.4 Transfer Matrix Method Application to Ising Model............. 89 ii Chapter 1 Enumeration (1.1) Notation N = f1; 2; 3;:::g Natural numbers N0 = f0; 1; 2; 3;:::g C Complex numbers Z = f:::; −2; −1; 0; 1; 2;:::g Integers R Real numbers Q Set of rational numbers, fractions (1.2) Denition Floor function: bxc = largest integer ≤ x e.g. b−1:5c = −2 ; b1:5c = 1 b−2c = −2 ; b2c = 2 Ceiling function: dxe = smallest integer ≥ x e.g. d−1:5e = −1 ; d1:5e = 2 d−2e = −2 ; d2e = 2 Let X be a set. We denote by jXj the number of elements in the set X (card(X) and #X mean the same). Kroenecker delta function: ( 1 if i = j δij = 0 if i 6= j 1 CHAPTER 1. ENUMERATION (1.3) Denition Let A and B be (nite) sets. We dene: n o A × B = fa; bg : a 2 A; b 2 B Map(A; B) = fall functions f : A ! Bg Sur(A; B) = ff : A ! B; f is surjective (onto)g Inj(A; B) = ff : A ! B; f is injective (1-1)g Bij(A; B) = ff : A ! B; f is bijectiveg 1.1 Counting Arguments Basic Problem Given a set A, we want to determine jAj. (1.4) Fundamental Theorem of Enumeration (Discovered independently by several anony- mous cavemen) X jAj = 1 a2A Generic enumeration problem We are given an innite sequence of sets A1;A2;:::;An;::: which contain objects satisfying a set of combinatorial/algebraic properties. Determine an := jAnj for general n. What is an answer? Some instances of enumeration problems: n An = subsets of [[1; n]]; an = 2 An = permutations of [[1; n]]; an = n! complete binary trees with leaves, (2n−2)! An = n an = n!(n−1)! An = 'words' on the alphabet f1; 2g that sum to n, 0 p !n+1 p !n+11 bn=2c 1 1 + 5 1 − 5 X n − k an = p @ − A = = Fn+1 5 2 2 k k=0 (Fibonacci numbers). An = labelled connected simple graphs on n vertices, " n an = n! × coecient of x in the power series expansion +1 !# X 2k(k−1)=2 of Log xk k! k=0 An = latin squares of size n × n (every row and column is a permutation of [[1; n]]), an = ? (unknown for general n). 2 CHAPTER 1. ENUMERATION (1.5) Semi-denition An answer is a polynomial time algorithm (in n) for computing an. [Algorithms that are equivalent to the caveman formula are useless.] (1.6) Methods Rule of sum: If A; B are sets with A \ B = ;, then jA [ Bj = jAj + jBj Rule of product: jA × Bj = jAj · jBj Refinement: If n ] An = Bn(k) (disjoint unit) k=0 then X jAnj = jBn(k)j k Generating functions: Let 2 n f(x) = a0 + a1x + a2x + ··· + anx + ··· Find some recursive relation between the ai's (e.g. ai = ai−1 + ai−2; a0 = a1 = 1) and by using it obtain a function equation (e.g. f(x) = 1 + xf(x) + x2f(x)) and solve for f(x). Now Taylor expand this expression for f(x). Bijections: Want to count An. We know the size of Bn. If there exists a f : An ! Bn (bijection) then jAnj = jBnj for all n. 1.2 Elementary Counting Coecients Let N and R be sets vith jNj = n and jRj = r. (1.7) Proposition The number of mappings f : N ! R is j Map(N; R)j = rn Proof. Every map f : N ! R can be viewed as a word of length n from the alphabet R (e.g. N = fa; b; cg;R = f1; 2g, then the word 212 represents the function f(a) = 2; f(b) = 1; f(c) = 2). As there are r possibilities for each of the n letters, there are n r · r ··· r = r such maps. | {z } n times (1.8) Example Consider the set of n × n matrices with entries taking values in {−1; 0; 1g. Any such matrix can be thought of as a map M : f1; 2; : : : ; ng2 ! {−1; 0; 1g n2 Hence the number of these is 3 . ♦ 3 CHAPTER 1. ENUMERATION (1.9) Proposition The number of injective maps f : N ! R is j Inj(N; R)j = r(r − 1) ··· (r − n + 1) Proof. Inj(N; R) is the set of words of length n which contain no two letters the same. If N = fx1; : : : ; xng, there are r choices for f(x1) r − 1 choices for f(x2) . r − (n − 1) choices for f(xn) The result follows. (1.10) Denition The expression r(r − 1) ··· (r − n + 1) is called a lower factorial or falling factorial of length n. We write (r)n = r(r − 1) ··· (r − n + 1) (r)0 = 1 for all r 2 N0 (1.11) Example Let N = f1; 2g and R = fa; b; c; dg, then j Inj(N; R)j = (4)2 = 4 · 3 = 12 Further, ab ba ca da ac bc cb db ad bd cd dc Setting r = n in the previous proposition, we have (1.12) Corollary The number of permutations of a set of size n is jSnj = n! = (n)n (1.13) Proposition The number of n-subsets of an r-set is r (r) r! = n = n n! n!(r − n)! Proof. The function ' : b1b2 : : : bn ! fb1; : : : ; bng maps the set of words on R of length n onto the set of all n-subsets of R. Each n-subset is the image of exactly n! injective maps, obtained by permuting the 's. The result follows from the previous proposition. bi (1.14) Denition The numbers r, choose , for , , are called n r n 0 ≤ n ≤ r n; r 2 N0 the binomial coefficients. 4 CHAPTER 1. ENUMERATION (1.15) Proposition n n = = 1 for all n 2 0 n N0 n n − 1 n − 1 = + for all n ≥ 2 and 0 ≤ k ≤ n k k − 1 k Proof (Combinatorial). n = jfA ⊆ f1; : : : ; ng : jAj = 0gj = jf;gj = 1 0 n = jfA ⊆ f1; : : : ; ng : jAj = ngj = jff1; : : : ; nggj = 1 n Let n . The set can be written as An(k) = fX ⊆ [[1; n]] : jXj = kg; k = jAn(k)j An(k) the disjoint union of two sets: those subsets that contain n = ffng [ X : X 2 An−1(k − 1)g those subsets that do not contain n = An−1(k) Since An(k) = ffng [ X : X 2 An−1(k − 1)g ] An−1(k), we have n = jA (k)j = jA (k − 1)j + jA (k)j k n n−1 n−1 n − 1 n − 1 = + k − 1 k (1.16) Notation 123 $ f1; 2; 3g (1.17) Example The 3-subsets of f1; 2; 3; 4; 5g are f123; 124; 134; 234; 125; 135; 145; 235; 245; 345g =f123; 124; 134; 234g ] f125; 135; 145; 235; 245; 345g $f123; 124; 134; 234g ] f12; 13; 14; 23; 24; 34g = 3-subsets of f1; 2; 3; 4g ] 2-subsets of f1; 2; 3; 4g ♦ 5 4 4 = + ) 3 3 2 (1.18) Proposition For all 0 ≤ k ≤ n n n = k n − k Proof (Combinatorial). Let Sn(k) be the set of k-sets of f1; : : : ; ng. For A 2 Sn(n − k), dene f : Sn(n − k) ! Sn(k) by f(A) = Sn n A. The map f is injective & surjective so f is a bijection. Therefore jSn(n − k)j = jSn(k)j, i.e. n n = n − k k 5 CHAPTER 1. ENUMERATION (1.19) Theorem (Binomial theorem) Let A be a commutative ring and n 2 N0. For all a; b 2 A, n X n (a + b)n = akbn−k k k=0 Proof. Write (a + b)n =( a + b)(a + b) ··· (a + b) | {z } n times The right hand side is the sum of all products akbn−k where akbn−k appears as many times as we can choose k a's and (n − k) b's from the n factors.
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