are the q-versions of n, n! and . The falling factorial is (x) k = x(x 1)(x 2)... (x k + 1).

Transcription

1 Lecture A jacques@ucsd.edu Notation: N, R, Z, F, C naturals, reals, integers, a field, complex numbers. p(n), S n,, b(n), s n, partition numbers, Stirling of the second ind, Bell numbers, Stirling of the firstind. [n] q, [n] q!, [ ( n ] are the q-versions of n, n! and n ). The falling factorial is (x) = x(x 1)(x 2)... (x + 1). 1 Introduction This is a first course in combinatorial enumeration for graduate students. I expect that you will have seen a smattering of the material already, such as using bijections to prove various combinatorial identities or proving some statements by induction. For instance, you might have seen many ways to count the number of positive integer sequences of length summing to a positive integer n, or some basic recurrence equations such as the Fibonacci equation, and you would be aware of some applications such as the use of the pigeonhole principle, or inclusion-exclusion to count permutations without fixed points. We will first review these basic results. The aim of this course is, as broadly as possible, to unify these approaches using general results from algebra and analysis, in particular, generating functions. On the algebraic side, we employ various basic group-theoretic approaches, such as Burnside s Lemma, the Pólya Redfield Theorem, Inversion Principles, and representation theory of S n to enumerate patterns, for instance, how many ways can one place pairwise non-attacing roos on an n by n chessboard? There are two essential approaches to proving combinatorial identities. The one involves a combinatorial representation of both sides of the equation, and the other more automatic approach in treating each side as coefficients in a formal power series and then showing that those formal power series are equal. Each approach has its advantages. The first gives a fundamental and perhaps deepest understanding of why the identity holds, whereas the second allows us to use the full power of techniques from algebra and analysis to derive various statistics related to the identity. Some fundamental examples are as follows: ( ) n = 2 n = the number of subsets of [n] =0 ( 1) n! = the number of derangements of [n]! =0 ( ) 1 2n = the number of binary trees with n nodes n + 1 n ( ) = 1 (φ n φ n ) φ = n =0 5 2 ( ) n 1 = the number of compositions of n 1 [x n 1 ] 1 x = the number of partitions of n =1 1

2 We are going to review these identities which come from basic principles of enumeration. While combinatorial identities are often beautiful, it is generally the case that a closed form expression for a sum cannot be obtained, and there are many natural counting problems where this occurs. Gosper s Theorem shows in many cases how to decide whether a sum can be computed in closed form. We are going to develop analytic study of formal power series, using tools from complex analysis, to deal with these examples. Some instances are the number of partitions of n, the number of non-isomorphic trees on n vertices, the number of primes up to n, and counting various graphs and subgraphs of lattices (for instances the number of spanning trees in a connected graph), which arise in statistical physics. In some of these situations, probability theory and its generating functions enter in a very natural way. Various general approaches such as the Exponential Formula, Perron-Mellin Transform, Fourier Analysis, the Residue Theorem, and a number of inversion formulae are powerful techniques to deal with these complicated situations. For these problems, we introduce various types of generating functions, including exponential generating functions, Dirichlet Series, tree generating functions, Euler products, and suitable counting polynomials such as the Tutte and chromatic polynomial, Combinatorial Nullstellensatz, and orthogonal series. You are required to be familiar with some basics of analysis, algebra and probability in this course. It is recommended that you go through the warm-up exercise sheet. 2 Basic Principles Here we will not go into any great detail, as it is assumed that these principles are familiar to you. However, we will describe some of the classical applications as a reminder. In combinatorics, we are primarily interested in enumerating sets and sequences, the difference being that sets are not ordered and sequences are. One of the fundamental principles is the following: Principle 1 (Bijection Principle) There is a bijection f between two finite sets A and B if and only if A = B. In a combinatorial identity, we might interpret each side of the identity as enumerating two sets and then find a bijection between those sets. This could also be referred to as counting in two different ways. A standard example is to show that the sum of the degrees of the vertices in a graph is twice the number of edges in the graph for the degree d(v) counts pairs (v, e) such that edge e contains v, but edge e is also counted by the pair (u, e) if u e. Unfortunately, in general, there is no easy way to find a bijection. Here are two basic counting principles: Principle 2 (Summation Principle) Let A 1, A 2,..., A n be finite disjoint sets. Then n A i = A j. 2

3 A basic course in mathematics confirms A B = A + B A B for finite sets A and B. This is a special instance of the inclusion exclusion formula, or combinatorial sieve. Here [n] denotes the set {1, 2,..., n}. Principle 3 (Inclusion-Exclusion) Let A 1, A 2,..., A n be finite sets. Then n A i = S [n] S ( 1) S +1. A i i S The proof is left as an exercise. Let us recall how this can be used to count derangements of n: that is permutations σ = (σ 1, σ 2,..., σ n ) such that σ i i for all i [n]. Set A i to be the set of σ for which σ i = i. Then the number of derangemenhts is n! A i = n! + S [n] S ( 1) S. A i i S The ey is that the size of the intersection is exactly (n S )!. By the summation principle, the number of derangements is n! + ( ) n ( 1) (n )! = n! + n! =1 ( 1) = n!! =1 ( 1).! Using the Taylor expansion of e 1, we see that in fact this is the nearest integer to n!/e. A second classical example is the number of integers in [n] which are relatively prime to n: this is the Euler φ-function, φ(n). Clearly this depends on the prime factorization of n, so let us suppose p 1, p 2,..., p are the distinct prime factors of n. The ey fact this time is that the number of integers in [n] which are divisible by primes p i : i S [] is exactly n/ i S p i. By inclusion-exclusion, letting A i be the set of integers in [n] divisible by p i, =0 ϕ(n) = ( ) n ( 1) S +1 i i S p i S = n (1 1 ). p i In the last step, we made a useful observation that if x 1, x 2,..., x are reals then x i = S [] i S (1 + x i ) applied with x i = 1/p i. In fact, this gives a simple way to prove the binomial theorem for (1 + x) with N. Recall the Cartesian product of sets A 1, A 2,..., A n is the set of sequences σ = (a 1, a 2,..., a n ) such that a i A i for i [n], and is denoted A 1 A 2 A n. 3

4 Principle 4 (Multiplication Principle) The number of sequences (x 1, x 2,..., x ) such that there are a i choices for x i after having chosen x 1, x 2,..., x i 1 for each i = 1, 2,..., n is exactly a 1 a 2... a n. In particular, for finite sets A 1, A 2,..., A n, n A 1 A 2 A n = A i. For instance, a wal of length on a graph G is a sequence (v 1, e 1, v 2, e 2,... v, e, v +1 ) where v i is a vertex and e i = {v i, v i+1 } is an edge of the graph. If every vertex of the graph is in d edges (the graph is said to be d-regular), then by the Multiplication Principle the number of wals of length is nd. We will return in detail to counting various types of wals in graphs at the ends of the course. For the meantime, you might determine how many wals of length in a d-regular n-vertex graph never reverse on an edge in consecutive steps. Also, refer to the warmup sheet for further exercises. Principle 5 (The Pigeonhole Principle) In a sequence of n+1 numbers with only n distinct entries, two of the entries are equal. This simple sounding principle, which we could call the principle of averaging, has some remarable consequences. For example, amongst any set S of n + 1 numbers in [2n], we claim that there is one number that divides another. To see this, we write each m S as m = 2 x y where y is odd. Then there are only n possible values of y, so we can find m = 2 x y and l = 2 z y with m, n S by the pigeonhole principle. Now clearly m l or l m, as required. Here is another application with arithmetic flavor, nown as Dirichlet s Box Principle. We claim that given any positive integer N, and any real number α, there exists a positive integer q N and a positive integer p such that α p 1 q qn. To see this, consider the fractional parts of the integers αi where i = 0, 1, 2,..., N. These are all real numbers in [0, 1], and therefore by the Pigeonhole Principle two of them are at most 1/N apart. In other words, the fractional parts of αi and αj differ by at most 1/N, for some i < j N. Letting q = j i, we see that for some integer p, qα p 1/N as required. This is crucial for diophantine approximation, equidistribution theory, and estimates on exponential sums and discrete fourier series. Here is a principle which should be familiar to you from early undergraduate courses. Principle 6 (The Principle of Induction) For each n N, let P (n) be a true-false statement. Suppose that P (1) is true and P (i) implies P (i + 1) for each i N. Then P (n) is true for every n N. We will be using this very frequently in proofs throughout the course, already see below for some examples, such as proving the Pascal and q-pascal identities. Let F be any field and let F[X] denote the ring of all polynomials in X with coefficients in F. Here is a fundamental fact from algebra: 4

5 Principle 7 (The Polynomial Principle) Let F be a field of characteristic zero, and let P, Q F[X] have degree, and suppose that P (x) = Q(x) for more than distinct values in x F. Then P and Q are identical. This follows immediately from the fact that a polynomial of degree has at most zeroes. 3 The Binomial Theorem Recall the definition of the generalized binomial coefficient ( ) a := a(a 1)...(a +1) for a Q.! In particular, if a = n is a positive integer, then this is the number of -element subsets of an n-element set. Here is the general form of the Binomial Theorem. There are many proofs (you are ased to give one when a is a positive integer using the technique in the last section). Theorem 1 (Binomial Theorem) For any a Q and x < 1, (1 + x) a = =0 ( ) a x. Proof (Setch) If f(x) and g(x) are the sides, then we see f and g satisfy (1 + x)f = af and f(0) = g(0) = 1. Furthermore, the radius ρ of convergence of g is ρ = 1. Later on we will see why this theorem is true in the ring of formal power series (without any restrictions on x). The Binomial Theorem supplies a large number of combinatorial identities automatically. For instance, how many pairs of sets (A, B) in [n] are there such that A B? By the summation principle, we can add up over the number of (A, B) such that B =. For then the number of choices of A is 2 and so the answer is ( ) n 2 = 3 n. =0 For an exercise, try to come up with a bijective proof of this identity, by finding a bijection from the pairs (A, B) to the set of ternary sequences of length n (sequences of 0s, 1s and 2s). The warmup sheet contains some other examples of identities from the Binomial Theorem. In particular, you should be able to prove Pascal s Triangle Identity in the general form: ( ) z = ( ) z 1 + ( ) z 1 1 where N and z K, a field of characteristic zero. The Binomial Theorem is a special case of the Multinomial Theorem stating that for any a Q, (x 1 + x x r ) a = 1, 2,..., r =0 5 ( a 1, 2,..., r ) r x i i.

6 The Binomial Theorem is important in probability: if an experiment is performed independently in n trials, and p is the probability of success on each trial, then the probability of exactly successes is ( ) n p (1 p) n. By the Binomial Theorem, this is a probability mass function since its sum over is (p+(1 p)) n = 1. The probabilistic aspect of identities is often very useful for enumeration, and we shall encounter it again. 3.1 The q-binomial Theorem There is another interesting generalization of the Binomial Theorem called the q-binomial theorem. Let us define, for a positive integer q, the q-integer n: The q-factorial is [n] q = 1 + q + q q n 1 = qn 1 q 1. [n] q! = n [] q. =1 The q-binomial coefficient or Gaussian coefficients are [ ] n [n] q! := [] q![n ] q!. q When q is a prime power, this is the number of -dimensional subspaces of an n-dimensional vector space over F q. One of the ey properties of [ ] n is the q-pascal Identity: q [ ] [ ] [ ] n n 1 n 1 = + q. q 1 q q This can be proved in a number of ways (see warmup sheet). The q-quantities match the ordinary factorial and binomial coefficients in the case q = 1. From now on, we omit the subscript q since q is considered fixed in what follows. The q-quantities also satisfy similar identities, [ ] n = 0 [ ] n = 1 n [ ] [ ] n n =. n They also lead to the q-binomial Theorem. This theorem has many proofs, we give a direct inductive proof here and discuss proofs using partitions later. Theorem 2 (The q-binomial Theorem) For any n N and q C, n (1 + q i x) = =0 6 [ ] n q (+1 2 ) x.

7 Proof For n = 1 the statement P (n) of the q-binomial Theorem is clear. Suppose n > 1 and P (n 1) is true. To prove P (n) it is sufficient to show n 1 [ ] n 1 [ ] (1 + q n x) q (+1 2 ) n x = q (+1 2 ) x. =0 For N, the coefficient of x on the left is [ ] [ ] n 1 q (+1 2 ) + q ( 2)+n n 1 = q (+1 2 ) ([ ] n 1 1 [ ] = q (+1 2 ) n by a suitable application of the q-pascal Identity. =0 [ ] n 1 ) + q n 1 The q-binomial coefficients also arise naturally in the following identity: for n N and q C, [ ] 1 n x n = 1 + (x q i ). =1 The proof of this fact can be deduced by induction from the q-pascal Identity. These proofs by induction do not reveal natural combinatorial objects being enumerated by each side of the identities above. We will be interpreting these quantities according to partitions and inversion numbers of permutations. 4 Partitions : First loo In this section we review some basic facts about various types of partitions as a basis for deeper study later. The main quantities we will be interested in are the partition numbers p(n), the Stirling Numbers s n, and S n, of the first and second inds, and the Bell Numbers b(n) = n =0 S n,. The latter three quantities are the subject of intense interest in combinatorics, and we will develop a number of methods for understanding them and their asymptotic behaviour. The partition numbers are important in many areas of mathematics, and we will be developing a lot of analytic approaches to study them. In this section, we tae a quic loo at some elementary facts about the partition numbers, by starting with a loo at compositions of integers. i=0 4.1 Compositions Perhaps the easiest type of partition is the composition: a composition of an integer n is a sequence (x 1, x 2,..., x ) of positive integers summing to n. Each x i is called a part of the composition and is the total number of parts. It is well-nown that the number of compositions of n N into parts is ( ) n

8 We imagine a composition of n as a selecting of integer intervals cutting up the interval [1, n], which is equivalent to selecting 1 numbers in [n 1] namely the beginning of the second interval, third interval, fourth interval, and so on. By the Binomial Theorem, there are then 2 n 1 compositions of n (with unrestricted number of parts). One of the popular enumeration is to count compositions with certain relations between the parts and certain restrictions on the parts. This is more-or-less automatic with the introduction of generating functions. 4.2 Partitions of integers A partition of n is a multiset (that is a set with repeated elements allowed) of positive integers whose sum is n, for example {1, 1, 1,..., 1} and {n} and {1, 1, n 2} are all partitions of n. Then p(n) is number of partitions of n. So partition numbers are the unordered version of compositions, just as multisets are the unordered version of sequences. A nice way to view partitions is via Euler products. Consider the product =1 1 1 x = (1 + x + x )(1 + x 2 + x )(1 + x 3 + x ) using the geometric series formula. The coefficient of x n is obtained by counting the number of ways to pic a power of x from each of the bracets so that the sum of the powers piced is n. If we pic say x m from the th bracet, then this is stating that the partition of n should have m repetitions of. So the coefficient of x n is precisely the number of partitions of n. Starting with this idea, we can determine all sorts of other partitions of n, for example the partitions of n into distinct parts corresponds to the coefficient of x n in the Euler product (1 + x ) and the partitions of n into parts which come from a set T is =1 T 1 1 x. There is no explicit formula for p(n) for all n. However, p(n) satisfies many beautiful identities, and one of the aims of this course is to obtain the remarable Hardy-Ramanujan asymptotic formula for p(n), namely p(n) eπ 2n 3. We are going to spend a lot of time on different representations of permutations, in particular on diagrams. 8

9 4.3 Stirling and Bell numbers Stirling numbers of the first and second ind refer to decompositions of permutations and partitions of sets. Throughout, we recall the falling factorial notation: for z C and N let (z) denote the polynomial z(z 1)... (z + 1). Traditionally, for N and n N {0}, S n, denotes the number of partitions of [n] into non-empty sets, and is the Stirling number of the second ind. Note there are! ways to order the sets in each partition, and this relates compositions of sets to partitions of sets (unlie in the case of integers there can be less than! ways to order the elements of a multiset of size ). A recurrence satisfied by S n, is quicly determined: for, n N: S n, = S n 1, 1 + S n 1, where S 0, = 0 for all N, S n,0 = 0 for all n N, and we define S 0,0 = 1. To see this, note that in a partition of [n], either {n} is a set in the partition or not. If it is, then removing {n} gives a partition of n 1 into 1 parts, and otherwise, given a partition of [n 1] into parts, we could add n to any of the parts to get a partition of [n] in which {n} is not a set in the partition. An equivalent interpretation of S n, is! times the number of surjections f : [n] []: for in this case the preimages f ({i}) form a partition of [n] into parts. Since the total number of maps from [n] to [] is then for n N and n : ( ) n = j!s n,j = () j S n,j. j j=1 Now consider the polynomials P (x) = x n and Q(x) = n j=0 (x) js n,j where x C. They both have degree n and they agree on N C. Clearly then they are identical, and so we have for x C by the polynomial principle: x n = (x) j S n,j. j=0 Using this one can quicly derive the recurrence for S n, given above. A similar method, replacing x with x + 1 in the above identity, or using a combinatorial argument shows ( ) n S n+1,+1 = S j,. j j=0 Many other identities satisfied by Stirling Numbers of the second ind are available. The Bell number b(n) is the total number of partitions of [n], hence b(n) = S n,. =0 The Bell numbers satisfy b(n) = ( ) n = 0 n b() and this is left as an exercise. Later on we will determine a formal power series relation for Bell numbers. 9 j=0

10 4.4 Stirling numbers of the first ind We represent a permutation of [n] as a sequence or word of n distinct elements of [n] and consider operations between permutations such as inverses and multiplications etc... in the symmetric group S n. Each permutation σ can be decomposed into cycles σ = σ 1 σ 2... σ and the decomposition is unique if we represent σ i as a word starting with the smallest element of σ i. Then the Stirling Number of the first ind, denoted s n,, is the number of permutations of [n] with exactly cycles, where, n N. The convention here is s 0,0 = 1 and s 0, = 0 for N. We observe the recurrence (by considering a cycle containing the element 1 as a fixed point or not) for, n N: s n, = s n 1, 1 + (n 1)s n 1,. This leads for x C to (x + n 1) n = s n, x. =0 10