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Notes from 2015/16, with Minor Editing for 2016/17 DISCRETE AND ALGEBRAIC STRUCTURES Master Study Mathematics MAT.402 Winter Semester 2015/2016 & 2016/17 (Due to organisational reasons register for the course at TU and KFU) Parts I & II Contents I Combinatorics and Graph Theory 1 1 Combinatorics 2 1.1 Basics . .2 1.1.1 Binomial Coefficient and Binomial Theorem . .2 1.1.2 Landau- and Asymptotic Notations . .5 1.1.3 Useful Inequalities and Asymptotic Estimates . .6 1.2 Recursions . .7 1.2.1 Recursions . .7 1.2.2 Linear recurrences . .7 1.2.3 Non-linear recurrences . 12 1.3 Generating Functions . 13 1.3.1 Formal Power Series . 13 1.3.2 Ordinary and Exponential Generating Functions . 17 1.4 Symbolic method . 20 1.4.1 Unlabelled combinatorial objects . 20 1.4.2 Labelled combinatorial objects . 25 1.5 Analytic Methods . 28 1.5.1 Analytic functions . 28 1.5.2 Cauchy’s coefficient formula . 29 1.5.3 Lagrange Inversion Theorem . 29 1.5.4 Singularities . 33 1.5.5 Meromorphic functions . 33 1.5.6 Newton’s generalised binomial theorem . 34 1.5.7 Transfer theorem . 35 1.5.8 Multiple singularities . 36 2 Graph Theory 39 2.1 Matchings, Eulerian Tour and Hamiltorian Cycles . 40 2.1.1 Basic Terminologies . 40 2.1.2 Matchings . 42 2.1.3 Eulerian Tours and Hamiltorian Cycles . 44 2.2 Graph Decompositions . 46 2.2.1 Connected Graphs . 46 2.2.2 2-Connected Graphs . 46 i CONTENTS ii 2.2.3 3-Connected Graphs . 48 2.3 Planar Graphs and Trees . 49 2.3.1 Planar and Plane Graphs . 49 2.3.2 Trees . 51 3 Stochastic Aspects 55 3.1 Basics . 56 3.2 Ramsey Number . 59 3.3 Independence Number . 63 3.4 Subgraphs and Second Moment Method . 64 Part I Combinatorics and Graph Theory 1 Chapter 1 Standard Methods of Enumerative Combinatorics Lecture 1 1.1 Basics 1.1.1 Binomial Coefficient and Binomial Theorem Definition 1 (Set of integers). [n] := f1;2;:::;ng Definition 2 (Factorial). For any n 2 N [ f0g, n−1 n!:= n · (n − 1) · ::: · 2 · 1 = ∏ (n − i) i=0 0! := 1 Definition 3 (Falling factorial). For any n;k 2 N [ f0g satisfying k ≤ n, k−1 (n)k := n · (n − 1) · ::: · (n − k + 1) = ∏ (n − i) i=0 (n)0 := 1 Definition 4 (Binomial coefficient). For any n;k 2 N [ f0g, 8 n! = n·(n−1)·:::·(n−k+1) if k ≤ n n <> (n−k)!k! k! := k :> 0 if k > n 2 CHAPTER 1. COMBINATORICS 3 Definition 5 (Multinomial coefficient). For any n;k1;k2;:::;km 2 N [ f0g satisfying n = k1 + k2 + ::: + km, n n! := k1;k2;:::;km k1!k2! ::: km! Properties 1. For any n;k 2 N [ f0g satisfying k ≤ n, n n = k n − k n n − 1 n − 1 = + k k − 1 k For any k;m;n 2 N, n + m k n m = ∑ · k `=0 ` k − ` Theorem 1 (Binomial theorem). For any n 2 N, n n (1 + x)n = ∑ xk k=0 k where x is a formal variable or x 2 C. Theorem 2 (Multinomial theorem). For any n 2 N, n n k1 k2 k2 (x1 + x2 + ··· + xm) = ∑ x1 x2 ···xm k1;k2;:::;km where the sum is over all k1;k2;:::;km 2 N [ f0g satisfying n = k1 + k2 + ::: + km. Applying the binomial theorem with x = 1 and x = −1 we have n n 2n = ∑ i=0 k n n 0 = ∑ (−1)k i=0 k Example 1 (Subsets). The number of all subsets of [n] is equal to n n ∑ = 2n i=0 k CHAPTER 1. COMBINATORICS 4 Example 2 (Subsets of odd sizes). To calculate the number of subsets of [n] of odd sizes, we observe that n n n n 2n = ∑ + ∑ (−1)k i=0 k i=0 k n n n = 2 + + ··· + 0 2 2 · bn=2c This implies that the number of subsets of [n] of even sizes is 2n−1, and therefore the number of subsets of [n] of odd sizes is equal to 2n − 2n−1 = 2n−1 Definition 6 (Generalized binomial coefficient). For any a 2 C and k 2 N, a a · (a − 1) · ::: · (a − k + 1) := k k! (1) Note that the denominator has k terms. a (2) If a is in N and k > a, k = 0 as in Definition 4. Example 3. 1 1 · 1 − 1 · ::: · 1 − k + 1 2 := 2 2 2 k k! 1 · (1 − 2) · ::: · (1 − 2(k − 1)) = 2k k! (−1)k−1 2k = 4k (2k − 1) k Theorem 3 (Generalized binomial theorem). For any a 2 C and x 2 C with jxj < 1, ¥ a (1 + x)a = ∑ xk k=0 k Example 4. ¥ 1 1 k (1 − x) 2 = ∑ 2 (−x) k=0 k ¥ k−1 (−1) 2k k = ∑ k (−x) k=0 4 (2k − 1) k ¥ 1 2k k = − ∑ k x k=0 4 (2k − 1) k CHAPTER 1. COMBINATORICS 5 1.1.2 Landau- and Asymptotic Notations (1) O(g(n)) := f f (n) : 9c > 0;9n0 2 N s:t: 8n ≥ n0 j f (n)j ≤ c · jg(n)jg n j f (n)j o = f (n) : limsupn!¥ jg(n)j < ¥ f (n) = O(g(n)) ” f (n) is Big Oh of g(n)” ” f (n) grows at most as fast as g(n)”’ (2) W(g(n)) := f f (n) : 9c > 0;9n0 2 N s:t: 8n ≥ n0 j f (n)j ≥ c · jg(n)jg n j f (n)j o = f (n) : liminfn!¥ jg(n)j > 0 f (n) = W(g(n)) ” f (n) is Big Omega of g(n)” ” f (n) grows at least as fast as g(n)”’ (3) Q(g(n)) := O(g(n)) \ W(g(n)) f (n) = Q(g(n)) ” f (n) is Big Theta of g(n)” ” f (n) grows as fast as g(n)”’ (4) o(g(n)) := f f (n) : 8c > 0;9n0 2 N s:t: 8n ≥ n0 j f (n)j ≤ c · jg(n)jg n j f (n)j o = f (n) : limn!¥ jg(n)j = 0 f (n) = o(g(n)) or f (n) g(n) ” f (n) is Small Oh of g(n)” ” f (n) grows slower than g(n)”’ (5) w (g(n)) := f f (n) : 8c > 0;9n0 2 N s:t: 8n ≥ n0 j f (n)j ≥ c · jg(n)jg n j f (n)j o = f (n) : limn!¥ jg(n)j = ¥ f (n) = w (g(n)) or f (n) g(n) ” f (n) is omega of g(n)” ” f (n) grows faster than g(n)”’ CHAPTER 1. COMBINATORICS 6 (6) f (n) f (n) ∼ g(n) , lim = 1 n!¥ g(n) ” f (n) is of order of g(n)” ” f (n) is asymptotically equal to g(n)”’ Example 5. + 1 = + o( ) + 1 = + o( ) + 1 = + o( ) 1 n 1 1 vs 1 n2 1 1 vs 1 n2+1000 1 1 1.1.3 Useful Inequalities and Asymptotic Estimates + Theorem 4 (GM-AM-RMS inequality). For any x1;x2;:::;xn 2 R , 1 n ! n n s n 1 1 2 ∏xi ≤ ∑ xi ≤ ∑ xi i=1 n i=1 n i=1 In words, ”the geometric mean is smaller than the arithmetic mean, which is smaller than the root mean square”. Theorem 5 (Stirlings formula). For any n 2 N, p nn 1 1 n! = 2pn 1 + + O e 12n n2 p nn = (1 + o(1)) 2pn e Lemma 6 (Binomial coefficient approximation). For any n;k 2 N satisfying k ≤ n, nk n enk ≤ ≤ k k k n H n n+O( k) = 2 ( k ) log k where H is the Entropy function H (x) := −xlog2x − (1 − x)log2 (1 − x) CHAPTER 1. COMBINATORICS 7 Lecture 2 1.2 Recursions 1.2.1 Recursions Example 6. 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 sides 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 7. 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 7. 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 7. A first-order linear recurrence with constant coefficients of the form ( a0 = 0 an = cn · an−1 + dn; n ≥ 1; CHAPTER 1. COMBINATORICS 8 with ci 6= 0 has an explicit solution, n−1 an = dn + ∑ dici+1ci+2 ···cn: i=1 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 8.
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