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Discrete and Algebraic Structures DISCRETE AND ALGEBRAIC STRUCTURES Master Study Mathematics MAT.402 Winter Semester 2015/16 (Due to organisational reasons register for the course at TU and KFU) Edited by Mihyun Kang TU Graz Contents I Lectures by Mihyun Kang 0 1 Standard Methods of Enumerative Combinatorics 1 1.1 Basics . .1 1.2 Recursions . .2 1.2.1 Recursions . .2 1.2.2 Linear recurrences . .2 1.2.3 Non-linear recurrences . .6 1.3 Generating Functions . .7 1.3.1 Formal Power Series . .7 1.3.2 Ordinary and Exponential Generating Functions . 11 1.4 Symbolic method . 14 1.4.1 Unlabelled combinatorial objects . 14 1.4.2 Labelled combinatorial objects . 19 1.5 Analytic Methods . 22 1.5.1 Analytic functions . 22 1.5.2 Cauchy’s coefficient formula . 23 1.5.3 Lagrange Inversion Theorem . 23 1.5.4 Singularities . 26 1.5.5 Meromorphic functions . 27 1.5.6 Newton’s generalised binomial theorem . 28 1.5.7 Transfer theorem . 28 1.5.8 Multiple singularities . 30 2 Graph Theory 32 2.1 Matchings, Eulerian Tour and Hamiltorian Cycles . 33 2.1.1 Basic Terminologies . 33 2.1.2 Matchings . 33 2.1.3 Eulerian Tours . 33 2.1.4 Hamiltorian Cycles . 33 2.2 Graph Decompositions . 34 2.2.1 Connected Graphs . 34 2.2.2 2-Connected Graphs . 34 2.2.3 3-Connected Graphs . 34 2.3 Planar Graphs and Trees . 35 i CONTENTS ii 2.3.1 Embedding of Graphs in the Plane . 35 2.3.2 Euler’s Formula and its Applications . 35 2.3.3 Properties of Trees . 35 2.3.4 Prufer¨ Code . 35 2.4 Stochastic Aspects . 36 2.4.1 Binomial Random Graphs . 36 2.4.2 Useful Inequalities . 36 2.4.3 Ramsey Number . 36 2.4.4 Independence Number . 36 2.4.5 Subgraphs and Second Moment Method . 36 II Lectures by Karin Baur 37 3 Multilinear Algebra 38 4 Rings and Moduls 39 Part I Lectures by Mihyun Kang 0 Chapter 1 Standard Methods of Enumerative Combinatorics 1.1 Basics (hand-written notes to be typed in LaTex) 1 CHAPTER 1. STANDARD METHODS OF ENUMERATIVE COMBINATORICS 2 1.2 Recursions 1.2.1 Recursions Example 1. Let a sequence (an)n2N[f0g of numbers be given by a recursion ( a0 = 0 an+1 = 2an + 1; n ≥ 0 The sequence begins with 0;1;3;7;15;31;::: We add 1 in both side of the latter equation to obtain an+1 + 1 = 2(an + 1); n ≥ 0: n n Then we have an + 1 = 2 (a0 + 1) = 2 and therefore the sequence is explicitly given by n an = 2 − 1; n ≥ 0: Example 2. Let a sequence (an)n2N[f0g of numbers be given by a recursion 8 a = 0 <> 0 a1 = 1 > :an+2 = an+1 + an; n ≥ 0: This sequence is called Fibonacci sequence and begins with 0;1;1;2;3;5;8;13;:::. How can we derive the explicit form of the sequence? 1.2.2 Linear recurrences Definition 1. We say that a sequence (an)n2N[f0g satisfies a k-th order recurrence (or recursion) if an can be written as an = f (an−1;an−2;:::;an−k);; n ≥ k k for a function f : C ! C. First-order linear recurrences Theorem 1. A first-order linear recurrence with constant coefficients of the form ( a0 = 0 an = cn · an−1 + dn; n ≥ 1; with ci 6= 0 has an explicit solution, n−1 an = dn + ∑ dici+1ci+2 ···cn: i=1 CHAPTER 1. STANDARD METHODS OF ENUMERATIVE COMBINATORICS 3 Proof. Divide both sides by cncn−1 ···c1: a a d n = n−1 + n : cncn−1 ···c1 cn−1cn−2 ···c1 cncn−1 ···c1 Change of variables: let an bn = : cncn−1 ···c1 We get the difference relation dn bn = bn−1 + ; cncn−1 ···c1 that is, dn bn − bn−1 = : cncn−1 ···c1 Summing up, we get n di bn = ∑ : i=1 cici−1 ···c1 Thus, for an we have n ! di an = cncn−1 ···c1 · ∑ i=1 cici−1 ···c1 n−1 = dn + ∑ dici+1ci+2 ···cn: i=1 Example 3. Returning to Example 1 we have ( a0 = 0 an+1 = 2an + 1; n ≥ 0; so taking ci = 2;di = 1 in Theorem 1 we have n−1 n−1! n−i n 1 n an = 1 + ∑ 2 = 1 + 2 1 − = 2 − 1; i=1 2 n i r(1−rn) because ∑i=1 r = 1−r for any r 6= 1. Higher order linear recurrences with constant coefficients Theorem 2. All solutions to the k-th order linear recurrence with constant coefficients an = c1 · an−1 + c2 · an−2 + ::: + ck · an−k; n ≥ k; can be expressed as linear combinations of terms of the form an;nan;n2an;:::;nm−1an where a is a root of order m of the characteristic polynomial k k−1 k−2 q(z) = z − c1 · z − c2 · z − ::: − ck: CHAPTER 1. STANDARD METHODS OF ENUMERATIVE COMBINATORICS 4 m Remark 1. If q(z) = (z−a) · p(z) for some polynomial p(z) where (z−a) - p(z), i.e. a is a root of multiplicity m, then q(a) = q0(a) = ::: = q(m−1)(a) = 0. n Proof. Let a be a simple root of q(z) (i.e. the multiplicity is 1) and an = a . We want to check that an is a solution to the recurrence, i.e. ! n n−1 n−2 n−k 0 = a − c1a − c2a − ::: − cka n−k k k−1 = a · a − c1a − ::: − ck = an−k · q(a) = 0: i n Let a now be a root of multiplicity of m of q(z). Then, for 0 ≤ i ≤ m−1, an = n a is a solution to the recurrence because ! 0 = an − c1 · an−1 − c2 · an−2 − ::: − ckan−k i n i n−1 i n−2 i n−k = n a − c1(n − 1) a − c2(n − 2) a − :::ck(n − k) a n−k i k i k−1 i = a · n a − c1(n − 1) a − ::: − ck(n − k) = an−k · (n − k)iq(a) + a (n − i)i − (n − k − 1)iq0(a)+ 2 i i i 00 a b0(n − k) + b1(n − k − 1) + b2(n − k − 2) q (a) + ::: n−k i ( j) = a · ∑ · ∑ b j;0(n − k − `) q (a) = 0; 0≤ j≤i 0≤`≤ j because of the previous remark; the b j;` are constants. Furthermore, a linear combination of an;n·an;:::;nm−1an is also a solution to the recurrence. If q(z) has distinct roots a1;:::;as with multiplicities m1;:::;m j where m1 + ::: + j n ms = k, then a linear combination of all these n · ai for 0 ≤ j ≤ m − 1, 1 ≤ i ≤ s, i.e. s mi−1 n an = ∑ bi;0 + bi;1n + ::: + bi;m−1n · ai i=1 is also a solution to the recurrence. We claim the opposite is also true. Let s be the set of sequences fa = (an 2 R)n≥0g and R be the set of solutions to the recurrence. Then R is closed under addition and scalar multiplication, and R 6= /0.Therefore, R is a vector space. k We claim that R has dimension k. Consider a map f from R to R : 0 1 a0 B a1 C f : R ! k; a 7! B C: R B . C @ . A ak−1 This is a linear map and also an isomorphism, because any solution to the recurrence is uniquely determined by the k initial values. So, R has dimension k. CHAPTER 1. STANDARD METHODS OF ENUMERATIVE COMBINATORICS 5 Therefore, R is given (generated) by linear combinations of any k linearly indepen- dent solutions to the recurrence. Now it suffices to show that the set of k solutions j n fn ai 1 ≤ j ≤ mi − 1;1 ≤ i ≤ sg to the recurrence is linearly independent. But this is true because these solutions have different orders of growth (in particular at ¥). Example 4. Returning to Example 2 we consider the Fibonacci sequence ( an = an−1 + an−2; n ≥ 2 a0 = 0;a1 = 1: Its characteristic polynomial is p ! p ! 1 + 5 1 − 5 q(z) = z2 − z − 1 = z − · z − ; 2 2 and the solution to the recurrence for an is p n p n ( 1+ 5 1− 5 an = c1 2 + c2 2 a0 = 0;a1 = 1: From the initial conditions, we get c = p1 , c = − p1 . The explicit solution to (??) is 1 5 2 5 p !n p !n 1 1 + 5 1 1 − 5 an = p − p : 5 2 5 2 Example 5. Consider the second order linear recurrence an = 5an−1 − 6an−2; n ≥ 2 with the initial conditions a0 = 0;a1 = 1. The characteristic polynomial is q(z) = z2 − 5z + 6 = (z − 2)(z − 3): The solution will be of the form n n an = c12 + c23 : Due to the initial conditions, c1 = −1 and c2 = 1 and the solution is n n an = −2 + 3 : This can be solved in Maple as follows: rsolve({a(n) = 5 * a(n-1) - 6 * a(n-2), a(0) = 0, a(1) = 1}, a(n)); In Mathematica: RSolve[{a[n] == 5 * a[n-1] - 6 * a[n-2], a[0] == 0, a[1] == 1}, a[n], n] Exercise 1. Find initial conditions a0;a1;a2 for which the growth rate of the solution to the recurrence an = 2an−1 + an−2 − 2an−3; n ≥ 3 is (a) constant, (b) exponential, and (c) fluctuating in sign. CHAPTER 1. STANDARD METHODS OF ENUMERATIVE COMBINATORICS 6 Higher order linear recurrences with non-constant coefficients Example 6.
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