DISCRETE AND ALGEBRAIC STRUCTURES

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1 DISCRETE AND ALGEBRAIC STRUCTURES Master Study Mathematics MAT.40 Winter Semester 015/016 & 016/17 (Due to organisational reasons register for the course at TU and KFU) Parts I & II

2 Contents I Combinatorics and Graph Theory 1 1 Combinatorics 1.1 Basics Binomial Coefficient and Binomial Theorem Landau- and Asymptotic Notations Useful Inequalities and Asymptotic Estimates Recursions Recursions Linear recurrences Non-linear recurrences Generating Functions Formal Power Series Ordinary and Exponential Generating Functions Symbolic method Unlabelled combinatorial objects Labelled combinatorial objects Analytic Methods Analytic functions Cauchy s coefficient formula Lagrange Inversion Theorem Singularities Meromorphic functions Newton s generalised binomial theorem Transfer theorem Multiple singularities Graph Theory 39.1 Matchings, Eulerian Tour and Hamiltorian Cycles Basic Terminologies Matchings Eulerian Tours and Hamiltorian Cycles Graph Decompositions Connected Graphs Connected Graphs i

3 CONTENTS ii..3 3-Connected Graphs Planar Graphs and Trees Planar and Plane Graphs Trees Stochastic Aspects Basics Ramsey Number Independence Number Subgraphs and Second Moment Method

4 Part I Combinatorics and Graph Theory 1

5 Chapter 1 Standard Methods of Enumerative Combinatorics Lecture Basics Binomial Coefficient and Binomial Theorem Definition 1 (Set of integers). [n] := {1,,...,n} Definition (Factorial). For any n N {0}, n! := n (n 1)... 1 = 0! := 1 n 1 (n i) i=0 Definition 3 (Falling factorial). For any n,k N {0} satisfying k n, (n) k := n (n 1)... (n k + 1) = (n) 0 := 1 Definition 4 (Binomial coefficient). For any n,k N {0}, ( ) n := k n! (n k)!k! k 1 (n i) i=0 = n (n 1)... (n k+1) k! if k n 0 if k > n

6 CHAPTER 1. COMBINATORICS 3 Definition 5 (Multinomial coefficient). For any n,k 1,k,...,k m N {0} satisfying n = k 1 + k k m, ( n ) k 1,k,...,k m := n! k 1!k!... k m! Properties 1. For any n,k N {0} satisfying k n, ( ) ( ) n n = k n k ( ) ( ) ( ) n n 1 n 1 = + k k 1 k For any k,m,n N, ( ) n + m k = k l=0 ( ) ( ) n m l k l Theorem 1 (Binomial theorem). For any n N, (1 + x) n = where x is a formal variable or x C. n k=0 ( n k ) x k Theorem (Multinomial theorem). For any n N, ( ) (x 1 + x + + x m ) n n = x k 1 1 k 1,k,...,k xk xk m m where the sum is over all k 1,k,...,k m N {0} satisfying n = k 1 + k k m. Applying the binomial theorem with x = 1 and x = 1 we have n n ( ) n = i=0 k n 0 = i=0( 1) k( ) n k Example 1 (Subsets). The number of all subsets of [n] is equal to n ( ) n = n k i=0

7 CHAPTER 1. COMBINATORICS 4 Example (Subsets of odd sizes). To calculate the number of subsets of [n] of odd sizes, we observe that n n ( ) n n = + i=0 k i=0( 1) k( ) n k [( ) ( ) ( )] n n n = n/ This implies that the number of subsets of [n] of even sizes is n 1, and therefore the number of subsets of [n] of odd sizes is equal to n n 1 = n 1 Definition 6 (Generalized binomial coefficient). For any α C and k N, ( ) α α (α 1)... (α k + 1) := k k! (1) Note that the denominator has k terms. () If α is in N and k > α, ( α k) = 0 as in Definition 4. Example 3. ( 1 ) 1 := ( 1 1 )... ( 1 k + 1 ) k k! 1 (1 )... (1 (k 1)) = k k! ) = ( 1)k 1 4 k (k 1) Theorem 3 (Generalized binomial theorem). For any α C and x C with x < 1, ( ) (1 + x) α α = x k k ( k k k=0 Example 4. (1 x) 1 = = k=0 k=0 = ( 1 ) ( x) k k ( 1) k 1 ( k 4 k (k 1) k k=0 1 4 k (k 1) ) ( x) k ( k k ) x k

8 CHAPTER 1. COMBINATORICS Landau- and Asymptotic Notations (1) O(g(n)) := { f (n) : c > 0, n 0 N s.t. n } n 0 f (n) c g(n) } f (n) = f (n) : limsup n g(n) < f (n) = O(g(n)) f (n) is Big Oh of g(n) f (n) grows at most as fast as g(n) () Ω(g(n)) := { f (n) : c > 0, n 0 N s.t. n} n 0 f (n) c g(n) } f (n) = f (n) : liminf n g(n) > 0 f (n) = Ω(g(n)) f (n) is Big Omega of g(n) f (n) grows at least as fast as g(n) (3) Θ(g(n)) := O(g(n)) Ω(g(n)) f (n) = Θ(g(n)) f (n) is Big Theta of g(n) f (n) grows as fast as g(n) (4) o(g(n)) := { f (n) : c > 0, n 0 N s.t. n } n 0 f (n) c g(n) } f (n) = f (n) : lim n g(n) = 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) ω (g(n)) := { f (n) : c > 0, n 0 N s.t. } n n 0 f (n) c g(n) } f (n) = f (n) : lim n g(n) = f (n) = ω (g(n)) or f (n) g(n) f (n) is omega of g(n) f (n) grows faster than g(n)

9 CHAPTER 1. COMBINATORICS 6 (6) f (n) f (n) g(n) lim n g(n) = 1 f (n) is of order of g(n) f (n) is asymptotically equal to g(n) Example n = 1 + o(1) vs n = 1 + o(1) vs n = 1 + o(1) Useful Inequalities and Asymptotic Estimates Theorem 4 (GM-AM-RMS inequality). For any x 1,x,...,x n R +, ( ) 1 n n x i i=1 1 n n x i i=1 1 n n xi 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 N, n! = ( n ) ( n πn ( )) 1 e 1n + O n = (1 + o(1)) ( n ) n πn e Lemma 6 (Binomial coefficient approximation). For any n,k N satisfying k n, where H is the Entropy function ( n ) ( ) k n ( en ) k k k k ( ) n = H( n k )n+o(logk) k H (x) := xlog x (1 x)log (1 x)

10 CHAPTER 1. COMBINATORICS 7 Lecture 1. Recursions 1..1 Recursions Example 6. Let a sequence (a n ) n N {0} of numbers be given by a recursion { a 0 = 0 a n+1 = a n + 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 a n = (a n + 1), n 0. Then we have a n + 1 = n (a 0 + 1) = n and therefore the sequence is explicitly given by a n = n 1, n 0. Example 7. Let a sequence (a n ) n N {0} of numbers be given by a recursion a 0 = 0 a 1 = 1 a n+ = a n+1 + a n, n 0. This sequence is called Fibonacci sequence and begins with 0, 1, 1,, 3, 5, 8, 13,... How can we derive the explicit form of the sequence? 1.. Linear recurrences Definition 7. We say that a sequence (a n ) n N {0} satisfies a k-th order recurrence (or recursion) if a n can be written as a n = f (a n 1,a n,...,a n k ), n k for a function f : C k C. First-order linear recurrences Theorem 7. A first-order linear recurrence with constant coefficients of the form { a 0 = 0 a n = c n a n 1 + d n, n 1,

11 CHAPTER 1. COMBINATORICS 8 with c i 0 has an explicit solution, n 1 a n = d n + d i c i+1 c i+ c n. i=1 Proof. Divide both sides by c n c n 1 c 1 : a n c n c n 1 c 1 = a n 1 c n 1 c n c 1 + d n c n c n 1 c 1. Change of variables: let We get the difference relation b n = a n c n c n 1 c 1. b n = b n 1 + d n c n c n 1 c 1, that is, Summing up, we get Thus, for a n we have b n b n 1 = b n = n i=1 a n = c n c n 1 c 1 d n c n c n 1 c 1. d i c i c i 1 c 1. ( n i=1 d i c i c i 1 c 1 n 1 = d n + d i c i+1 c i+ c n. i=1 ) Example 8. Returning to Example 6 we have { a 0 = 0 a n+1 = a n + 1, n 0, so taking c i =,d i = 1 for all i in Theorem 7 we have n 1 a n = 1 + i=1 n i = 1 + n (1 because n i=1 ri = r(1 rn ) 1 r for any r 1. ( ) ) 1 n 1 = n 1,

12 CHAPTER 1. COMBINATORICS 9 Higher order linear recurrences with constant coefficients Theorem 8. All solutions to the k-th order linear recurrence with constant coefficients c 1,c,...,c k a n = c 1 a n 1 + c a n c k a n k, n k, can be expressed as linear combinations of terms of the form α n,nα n,n α n,...,n m 1 α n where α is a root of order m of the characteristic polynomial q(z) = z k c 1 z k 1 c z k... c k. Remark 1. If q(z) = (z α) m p(z) for some polynomial p(z) where (z α) p(z), i.e. α is a root of multiplicity m, then q(α) = q (α) =... = q (m 1) (α) = 0. Proof of Theorem 8. Let α be a simple root of q(z) (i.e. the multiplicity is 1) and a n = α n. We want to check that a n is a solution to the recurrence, i.e. 0 =! α n c 1 α n 1 c α n... c k α n k ) = α n k (α k c 1 α k 1... c k = α n k q(α) = 0. Let α now be a root of multiplicity of m of q(z). Then, for 0 i m 1, a n = n i α n is a solution to the recurrence because 0! = a n c 1 a n 1 c a n... c k a n k = n i α n c 1 (n 1) i α n 1 c (n ) i α n...c k (n k) i α n k = α n k (n i α k c 1 (n 1) i α k 1... c k (n k) i) = α n k ((n k) i q(α) + α ( (n k) i (n k 1) i) q (α)+ α ( b 0 (n k) i + b 1 (n k 1) i + b (n k ) i) q (α) +... ) ( ( ) = α n k b j,l (n k l) )q i ( j) (α) = 0, 0 j i 0 l j because of the previous remark; the b j,l are constants. Furthermore, a linear combination of α n,n α n,...,n m 1 α n is also a solution to the recurrence. If q(z) has distinct roots α 1,...,α s with multiplicities m 1,...,m s where m m s = k, then a linear combination of all these n j α n i for 0 j m i 1, 1 i s, i.e. a n = s i=1 is also a solution to the recurrence. ( bi,0 + b i,1 n b i,mi 1n m i 1 ) α n i

13 CHAPTER 1. COMBINATORICS 10 We claim the opposite is also true. Let s be the set of sequences {a = (a n R) n 0 } and R be the set of solutions to the recurrence. Then R is closed under addition and scalar multiplication, and R /0. Therefore, R is a vector space. We claim that R has dimension k. Consider a map f from R to R k : f : R R k, a a 0 a 1. a k 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. Therefore, R is given (generated) by linear combinations of any k linearly independent solutions to the recurrence. Now it suffices to show that the set of k solutions {n j αi n 1 j mi 1,1 i s} to the recurrence is linearly independent. But this is true because these solutions have different orders of growth (in particular at ).. Example 9. Returning to Example 7 we consider the Fibonacci sequence { a n = a n 1 + a n, n a 0 = 0,a 1 = 1. (1.1) Its characteristic polynomial is q(z) = z z 1 = ( z and the solution to the recurrence for a n is { ( ) n ( a n = r r a 0 = 0,a 1 = 1. ) ( z 1 5 From the initial conditions, we get r 1 = 1 5, r = 1 5. The explicit solution to (1.1) is ( a n = ) n ( ) n 1 ) n 5. ), Example 10. Consider the second order linear recurrence with the initial conditions a 0 = 0,a 1 = 1. The characteristic polynomial is a n = 5a n 1 6a n, n q(z) = z 5z + 6 = (z )(z 3).

14 CHAPTER 1. COMBINATORICS 11 The solution will be of the form a n = r 1 n + r 3 n. Due to the initial conditions, r 1 = 1 and r = 1 and the solution is This can be solved in Maple as follows: a n = n + 3 n. rsolve({a(n) = 5 * a(n-1) - 6 * a(n-), a(0) = 0, a(1) = 1}, a(n)); In Mathematica: RSolve[{a[n] == 5 * a[n-1] - 6 * a[n-], a[0] == 0, a[1] == 1}, a[n], n] Exercise 1. Find initial conditions a 0,a 1,a for which the growth rate of the solution to the recurrence a n = a n 1 + a n a n 3, n 3 is (a) constant, (b) exponential, and (c) fluctuating in sign. Higher order linear recurrences with non-constant coefficients Example 11. Consider the recurrence { a n = na n 1 + n(n 1)a n, n a 0 = 0,a 1 = 1. Divide by n! to receive a n n! = a n 1 (n 1)! + a n (n )!. Change variables: let b n = a n n!. Then we get the recurrence { b n = b n 1 + b n, n b 0 = 0,b 1 = 1. for the Fibonacci sequence. From Example 9 we have ( ) n ( b n = ) n. This yields (( a n = n! ) n ( 5 1 ) n ) 5. Exercise. Solve the recurrence { n(n 1)a n = (n 1)a n 1 + a n, n a 0 = 0, a 1 = 1.

15 CHAPTER 1. COMBINATORICS Non-linear recurrences Example 1. Consider the second order non-linear recurrence { a n = a n 1 a n, n a 0 = 1, a 1 =. We use the logarithm function, since log a xy = log a x + log a y. Let b n = loga n. We get { b n = 1 (b n 1 + b n ), n b 0 = 0, b 1 = 1. This can be solved similar to the previous linear recurrences.

16 CHAPTER 1. COMBINATORICS 13 Lecture Generating Functions Formal Power Series Now let us return to Example 6. Example 13. Let a sequence (a n ) n N {0} of numbers be given by a recursion { a 0 = 0 a n+1 = a n + 1, n 0 (1.) and let z be a formal indeterminate variable. Multiplying both sides of (1.) by z n+1 and summing over n N {0}, we obtain a n+1 z n+1 = (a n + 1) z n+1. If it were true that (a n +1) z n+1 = z a nz n +z zn, we would obtain a n z n = z n=1 a n z n + z z n Defining A(z) := a n z n we have A(z) = za(z) + z z n equiv. (1 z)a(z) = z because a 0 = 0. From this we we would further obtain A(z)? =? = z? =? = = z 1 z n=1 n=1 z n? z = (z) n z (z) n ( n 1) z n ( n 1) z n, 1 z z n z n n=1 z n 1 1 z = z 1 1 z z 1 1 z

17 CHAPTER 1. COMBINATORICS 14 if the following were true: (1 z)a(z) = z z n? A(z) = z 1 z z n (1.3) (az) n? 1 = 1 az, a K (1.4) (a n + 1) z n+1? = z a n z n + z z n. (1.5) Summing up, we would have a n z n = (n 1) z n and therefore a n = n 1 for every n N {0}, by comparing the coefficients, if (1.3) (1.5) were true. When are they true? Definition 8 (Formal Power Series). Let K denote a commutative ring, usually we take K = C. Given a sequence (a n ) n N {0} with a n K and a formal variable z, we call an infinite sum of the form A(z) := a n z n = a n z n n 0 a formal power series. We call a n the coefficient of z n in A(z) and use the notation [z n ] A(z) := a n. The ring of formal power series is denoted by K[[z]] and endowed with the operations of addition and product: ( a n z n ) ( a n z n ) + ( b n z n ) ( b n z n ) := := (a n + b n ) z n (1.6) n ( k=0 a k b n k ) z n (1.7) We often skip for the product of two formal power series. As a special case of (1.7) we have, for any a K, a ( b n z n ) = a b n z n. (1.8) Given a formal power series A(z) = n 0 a n z n we also define the following algebraic operations: Differentiation: A (z) := n 1 na n z n 1 = n 0 (n + 1)a n+1 z n. Integration: z 0 A(t)dt := n 0 a n n+1 zn+1 = n 1 a n 1 n z n. A formal power series B(z) := b n z n is called a reciprocal of a formal power series A(z) := a n z n (and vice versa) if A(z)B(z) = B(z)A(z) = 1.

18 CHAPTER 1. COMBINATORICS 15 Propostion 1. A formal power series A(z) := a n z n has a reciprocal if and only if a 0 0. In that case, the reciprocal is unique and so we denote the reciprocal of A(z) by 1 A(z). Proof. Exercise! In Example 13 we wished to have (1 z)a(z) = 1 z n? = 1 1 z, (a n + 1) z n+1? = z? A(z) = 1 1 z a n z n + z From definition, equality (1 z)a(z) = 1 and Proposition 1, A(z) is the (unique) reciprocal of 1 z and vice versa, so z n. (1 z)a(z) = 1 A(z) = 1 1 z. For the second equality zn = 1 z 1, we shall show that the reciprocal of the formal power series A(z) := zn is B(z) := 1 z and vice versa. Letting A(z) := a n z n with a n = 1 for n 0 and B(z) := b n z n with b 0 = 1,b 1 = 1 and b n = 0 for n, we have A(z)B(z) (1.7) = n ) a k b n k ( k=0 z n = a 0 b 0 + n=1 (a n 1 b 1 + a n b 0 ) z n = 1. By Proposition 1, A(z) is the unique reciprocal of B(z) and so A(z) = 1 is, we have z n = 1 1 z B(z) = 1 1 z, that in the ring K[[z]]. Analogously one can show that for any a K, the reciprocal of the formal power series A(z) := (az)n is B(z) := 1 az and vice versa. In other words, for any a K, (az) n = 1 1 az (1.9) in the ring K[[z]]. For the third equality (a n +1) z n+1 = z a nz n +z zn, we use (1.6) and (1.7) to have n+1 (1.6) (a n + 1) z = a n z n+1 + n+1 (1.7) z = z a n z n + z z n.

19 CHAPTER 1. COMBINATORICS 16 That is, (a n + 1) z n+1 = z a nz n + z zn in the ring K[[z]]. Given a formal power series A(z) = n 0 a n z n and a constant β K we have the following properties: A(βz) = n 0 a n β n z n, so [z n ] A(βz) = a n β n = β n a n = β n [z n ] A(z) (scaling) (A(z) a 0 )/z = n 1 a n z n 1 = n 0 a n+1 z n z A(z) = n 0 a n z n+1 = n 1 a n 1 z n (right shift) A(z)/(1 z) = A(z) n 0 z n = n 0 ( 0 k n a k ) z n (left shift) (partial sum) Example 14. Let us study the recursion for the Fibonacci sequence a 0 = 0 a 1 = 1 a n+ = a n+1 + a n, n 0, using the generating function A(z) := a n z n. Multiplying both sides of the recursion by z n+ and summing over n N {0}, we obtain a n+ z n+ = A(z) = a n z n = z n= (a n+1 + a n ) z n+ n z n=1a n + z a n z n A(z) z = z A(z) + z A(z) (1 z z ) A(z) = z z 1 z z = 1 β 1 β (1.9) = 1 (1.7) = (1.6) = ( 1 1 β 1 z 1 ) 1 β z ) (β z) n ( 1 z) 5 (β n ( 1 β n 1 zn 5 where β 1 = 1+ 5 and β = 1 5. Therefore we have (( ) n ( a n = (β1 n β n ) zn, ) n ) ) 1 β n zn 5, n 0.

20 CHAPTER 1. COMBINATORICS Ordinary and Exponential Generating Functions Throughout the lecture we take K = C. Definition 9 (Generating Functions). The ordinary generating function, OGF, of a sequence (a n ) n N {0} is the formal power series A(z) := a n z n. The exponential generating function, EGF, of a sequence (a n ) n N {0} is the formal power series z n A(z) := a n n!. Definition 10 (Combinatorial Class, Counting Sequence and Generating Function). A combinatorial class A is a finite or denumerable set on which a size function is defined such that the size α of an element α is a non-negative integer and the number of elements of any given size is finite. An element of size 1 in a combinatorial class is an atom. Given a combinatorial class A we denote by A n the set of elements in A of size n for any n N {0}. The counting sequence of a combinatorial class A is the sequence (a n ) n N {0} where a n is the number of elements in A n, which is often denoted by A n (A n is also a combinatorial class). The ordinary generating function of a combinatorial class A is the ordinary generating function of the sequence of numbers a n = A n. Therefore we have Notation: [z n ]A(z) := a n. A(z) := a n z n = z α. α A The exponential generating function of a combinatorial class A is the exponential generating function of the sequence of numbers a n = A n. Therefore we have Notation: [z n ]A(z) := a n n!. A(z) := a n z n n! = α A z α α!. We say the variable z marks the size in the generating function A(z). Two combinatorial classes are said to be combinatorially equivalent if their counting sequences are identical.

21 CHAPTER 1. COMBINATORICS 18 Propostion. For the product of two exponential generating functions we have ( ) ( z ) n z n ( ) ) n z n a n b n = a k b n k n! n! k n!. Proof. Exercise! ( n k=0 Example 15 (Triangulations of Convex Polygon). Let T denote the class of all triangulation of convex polygons with one distinguished edge, in which the size of a triangulation in T is defined as the number of triangles it is composed of. Then T is a combinatorial class. For n 1 we let T n denote the class of all triangulation of convex polygons of size n and let t n := T n. Set t 0 := 1. That is, T n is the set of all triangulations of convex (n + )-gons (i.e. with n triangles) with one distinguished edge. The sequence (t n ) n N {0} begins with 1,1,,5,14,4,... By deleting the triangle incident to the distinguished edge we obtain the recursion t n = n 1 k=0 t k t n 1 k, n 1. Let T (z) := t n z n be the ordinary generating function of T. Multiplying both sides of t n = n 1 k=0 t k t n 1 k by z n and summing over n 1 we have n=1 t n z n = T (z) t 0 = z T (z) 1 = z n 1 ( n=1 k=0 n 1 ( n=1 k=0 n ( k=0 T (z) 1 (1.8) = z T (z). Therefore T (z) satisfies the quadratic equation among whose two solutions we choose t k t n 1 k ) z n z T (z) T (z) + 1 = 0, T (z) = 1 1 4z, z because the coefficients of T (z) are non-negative. From this, we get for n 1: t k t n 1 k ) z n 1 t k t n k ) z n t n = [z n ]T (z) = 1 [zn+1 ] 1 4z rescaling = 1 ( 4)n+1 [z n+1 ](1 + z) 1/.

22 CHAPTER 1. COMBINATORICS 19 Using the generalised binomial theorem ( 1 ) ( [z n+1 ](1 + z) 1/ ) (1 = = n ) n + 1 (n + 1)! we obtain t n = 1 ( 4)n+1 ( 1)n (n 1) n+1 (n + 1)! = (n 1) 4 6 (n) (n + 1)! n! Using Stirling s formula n! = ( 1 + O ( 1 n = 1 n + 1 )) πn ( n ) n, e = ( 1)n (n 1) n+1, (n + 1)! (n 1) n = (n + 1)! ( ) n. n we can derive the asymptotic number of triangulations of size n (i.e. of a convex n + - gon) t n = 1 ( ) n = 1 (n)! n + 1 n n + 1 n! n! 1 4 n n 3/. π

23 CHAPTER 1. COMBINATORICS 0 Lecture Symbolic method Unlabelled combinatorial objects For a given combinatorial class A, denote by A n the set of elements of size n in A, and let a n = A n. We say the OGF enumerates A. Basic constructions and OGF s A(z) = a n z n = z α n 0 α A (1) E is the neutral class that consists of a single element of size 0. The OGF of E is 1. () Z is the atomic class that consists of a single element of size 1. The OGF of Z is z. (3) The combinatorial sum (disjoint union) A + B of two combinatorial classes A and B with A B = /0 is the set of objects consisting of two disjoint copies of A and B, in which the size of an element α A + B is defined as α A if α A and α B if α B (i.e. the size of an element in α A +B is inherited from its size in its class of origin). In order to formalise A + B we introduce red marker to A and blue marker to B. The combinatorial sum A + B is a well-defined combinatorial class. Its OGF satisfies z α = z α + z α. α A +B α A α B (4) The Cartesian product A B of two combinatorial classes A and B is defined as A B = {(α,β) α A,β B}, in which the size of a pair (α,β) is defined as α + β. The Cartesian product A B is a well-defined combinatorial class. Its OGF satisfies (α,β) A B z (α,β) = (α,β) A B z α + β = z α + z β = A(z) B(z). α A β B For any n 1, we define A n = A n 1 A = A A recursively. (5) The sequence SEQ(A ) of a combinatorial class A with A 0 = /0 that consists of sequences of elements from A is the infinite sum E + A + A + +A

24 CHAPTER 1. COMBINATORICS 1 Note that the condition A 0 = /0 (equiv. a 0 = 0) guarantees the finiteness condition for sizes, and therefore the sequence construction SEQ(A ) is a well-defined combinatorial class. Its OGF satisfies 1 + A(z) + A(z) + A(z) = 1 A(z), where the latter equality is because [z 0 ]A(z) = 0. (6) The multiset construction MSET(A ) of a combinatorial class A with A 0 = /0 is the collection of all finite multisets (i.e. repetition allowed) of elements from A, more precisely, we define / MSET(A ) := SEQ(A ) R, where R is the equivalence class of sequences defined by (α 1,...,α n ) R (β 1,...,β n ) iff there is a permutation σ of [1...n] such that for all 1 i n, β i = α σ(i). We determine its generating sequence. Case A finite For a finite set A, we let α 1,α,... be a canonical listing of the elements of A. Then any multiset can be sorted in such a way that it can be viewed as formed by a sequence of repeated elements of α 1, followed by a sequence of repeated elements of α, and so on. It follows that MSET(A ) = SEQ(α). α A (a finite product). Therefore, the OGF of C = MSET(A ) satisfies ( ) 1 1 an C(z) = α A 1 z α = n 1 1 z n, where the latter equality holds because a 0 = 0. Consider the formal power series exp(z) := n 0 z n n!, ln(z) := n 1 Applying the exp-ln transformation we have ( ( )) C(z) = exp ln (1 z n ) a n = exp n 1 ( = exp n 1 a n k 1 (z n ) k ) ( 1 = exp k 1 ) k ( = exp A(z) + A(z ) + A(z3 ) ( 1) n 1 (z 1) n. n ( a n ln n 1 k n 1 ( 1 1 z n )) a n (z k ) n) ( 1 = exp k 1 k A(zk ) Infinite class A : The case of infinite class A follows by a limit argument.. )

25 CHAPTER 1. COMBINATORICS (7) The power set construction PSET(A ) of a combinatorial class A with A 0 = /0 is the collection of all finite subsets of A (without repetition). For a finite set A, we have PSET(A ) = (E + {α}), α A because the distributing the products in all possible ways forms all possible combinations of elements of A (i.e. the sets of elements from A without repetition). So, the OGF of C = PSET(A ) satisfies C(z) = α A = exp (1 + z α ) = (1 + z n ) a n = exp n 1 ( ) ( k 1 znk a n ( 1) = exp n 1 k 1 k k 1 ( = exp A(z) A(z ) + A(z3 ) ±... 3 ). ( ) a n ln(1 + z n ) n 1 ( 1) k 1 ) A(z k ) (8) A further construction is the substitution (see Chapter, Example 1 for an example with substitution). Let B,C be combinatorial classes. Then the substitution of C into B (or the composition of B and C ), note d B C or B[C ] is defined as B C = B[C] = B k SEQ k (C ) k 0 B[C ] means substitute elements of C for atoms of B. Applications Example 16. A binary tree is a combinatorial structure that is recursively defined such that It is either a single external node, or it consists of an internal node (the root ) and two binary trees attached to the root (left tree l and right tree r ), k or l r. Let B denote the class of all binary trees, in which the size of a binary tree is defined as the number of internal nodes. Then B is a combinatorial class. For n 0 we let B n denote the class of all binary trees of size n and let b n := B n. (Note that if a binary tree has n internal nodes, then it has n + 1 external nodes. Thus b n counts the number of binary trees with n + 1 external nodes.)

26 CHAPTER 1. COMBINATORICS 3 The sequence (b n ) n N {0} begins with 1,1,,5,14,4,... Let B(z) := b n z n be the ordinary generating function of the combinatorial class B. We have b 0 = 1, since the only (binary) tree is and for n 1,. l r Thus its OGF satisfies The solution for the quadratic equation is B(z) = 1 + zb(z). zb(z) B(z) + 1 = 0 B(z) = 1 ± 1 4z. z Since b n 0, B(z) increases along the real axis and therefore it increases. Thus the right solution of B(z) is B(z) = 1 1 4z. z As in Example 15 we get for n 1: b n = [z n ]B(z) = 1 [zn+1 ] 1 4z = 1 ( ) n. n + 1 n Theorem 9 (Number of binary trees). The number b n of binary trees with n internal nodes (equiv. n + 1 external nodes) is given by the so-called Catalan number b n = 1 ( ) n. n + 1 n Remark. Using Stirling s formula n! = ( 1 + O ( 1 n )) πn ( n ) n, e we can derive the asymptotic number of binary trees b n = 1 ( ) n 1 4 n n 3/. n + 1 n π Example 17. Let us return to triangulations of convex polygons we saw in Example 15. Let T n be the set of all triangulations of convex (n + )-gons (n triangles) with one edge distinguished. Using the basic constructions described above we have T = n 0T n = T n, T 0 = E, T 1 = Z. n 0

27 CHAPTER 1. COMBINATORICS 4 T 0 contains only one edge, T 1 one triangle. By removing one edge of the convex (n + )-gon, we end up with two separate convex triangulations (sharing one node), so T = E + T Z T among whose two solutions we choose T (z) = 1 + zt (z), T (z) = 1 1 4z, z because the coefficients of T (z) are non-negative. As in Example 15, t n = [z n ]T (z) = 1 [zn+1 ] 1 4z =... = 1 ( ) n. n + 1 n This suggests a bijection between binary trees and triangulations, where each node corresponds to one triangle (or alternatively, the root node of a binary tree B corresponds to the distinguished edge of a triangulation T, and each internal node of B to a diagonal edge of T, and each external node to the external edges of T except the distinguished edge of T. Example 18. Let S be the set of binary strings with no two consecutive 0 bits; for example, /0,0,1,01,10,11,010,011,... S = E + {0} + {1} S + {01} S. From (3) and (4) in Section we get and thus Exercise 3. Let S(z) be as above. S(z) = 1 + z + zs(z) + z S(z) S(z) = 1 + z 1 z z. Find the closed solution form of s n = S n. Find a recurrence of s n that leads to S(z).

28 CHAPTER 1. COMBINATORICS Labelled combinatorial objects A labelled combinatorial class A is a combinatorial class, where each atom carries an integer label in such a way that the labels of atoms occurring in an object are distinct, and the collection of the labels of atoms occurring in an object of size n is the complete integer interval [n]. Given a labelled combinatorial class A, we consider the EGF Basic constructions and EGFs z A(z) = n a n n 0 n! = α A z α α!. (1) Neutral class: E. The EGF of E is 1. () Atomic class: Z. The EGF of Z is z. (3) Disjoint union: A + B, for A and B labelled combinatorial classes. The EGF of A + B is A(z) + B(z). (4) Labelled product A B of two labelled combinatorial classes A and B is defined as A B := α β. α A,β B Given α A,β B, let α β denote the set of all pairs (α,β ) where the atoms of (α,β ) get distinct labels from [n] where n = α + β, such that the labelling preserves the relative order of labels of α,β. There are ( α + β ) α possibilities for such labelling (choose which labels go to the first substructure, then the previous relative order determines which label belongs to which element). In other words, if C = A B, C(z) = A(z) B(z). (5) Labelled sequence SEQ(A ) of a (labelled) combinatorial class A with A 0 = /0 is defined as SEQ(A ) = E + A + A A +... = SEQ k (A ), k 0 where SEQ k (A ) = A A is the labelled product of k copies of A. The EGF of SEQ(A ) satisfies 1 + A(z) + A(z) + A(z) = 1 1 A(z). (6) Set SET(A ) of a (labelled) combinatorial class A with A 0 = /0 is defined as / SET(A ) := SEQ(A ) R,

29 CHAPTER 1. COMBINATORICS 6 where R is an equivalence relation such that (α 1,...,α n ) R (β 1,...,β n ) if there exists a permutation σ of [n] such that for each 1 i n, β i = α σ(i). The EGF of SET(A ) satisfies 1 + A(z) + A(z)! + A(z)3 3! +... = exp(a(z)). (7) Cylces CYC(A ) of a (labelled) combinatorial class A with A 0 = /0 is defined as / CYC(A ) := SEQ(A ) S, where S is the equivalence relation such that (α 1,...,α n ) R (β 1,...,β n ) if there exists a cyclic permutation σ of [n] such that for each 1 i n, β i = α σ(i). The EGF of CYC(A ) satisfies 1 + A(z) + A(z) + A(z)3 3 Recall that a tree is a connected graph without cycles = log 1 A(z). Example 19 (Plane trees, ordered trees). Plane trees are trees embedded in the plane, so that subtrees attached to each vertex are ordered, say from left to right. There is a root vertex, implicitly defined; to a root vertex, subtrees are attached in a specified order. ordered So, The generating function is then P = Z SEQ(P). P(z) = z 1 1 P(z). This is a quadratic equation. As previously, since P(z) increases along the real axis, the negative solution is the correct one, P(z) = 1 1 4z = z B(z), where B(z) is the OGF of binary trees. Let p n be the number of plane trees on n vertices, and b n the number of binary trees on n + 1 external nodes, so p n+1 = b n. P(z) = zb(z) suggests that there is a combinatorial bijection between plane trees on n+1 vertices and binary trees on n+1 external nodes.

30 CHAPTER 1. COMBINATORICS 7 Exercise 4. Find such a bijection between plane trees and binary trees. Example 0 (-regular graphs). Let R be the set of all -regular labelled graphs, i.e. each vertex has exactly neighbours. Note that connected -regular graphs are undirected cycles of length 3. R = SET(connected -regular graphs) = SET(UCYC 3 (Z )), where UCYC is the undirected cyclic construction. Then, the generating function is R(z) = exp(ucyc 3 (z)) = e z/ z /4 1 z, since the generating function for UCYC 3 (Z ) is 1 How can we derive [z n ]R(z)? ( ) log( 1 z 1.

31 CHAPTER 1. COMBINATORICS 8 Lecture Analytic Methods Analytic functions Definition 11. Let Ω C be a region (i.e. open and connected). A function f : C C is called complex differentiable or holomorphic at z 0 Ω, if f (z) f (z 0 ) lim z z 0 z z 0 exists. It is called holomorphic in Ω if f is holomorphic for every point in Ω. Definition 1. Let Ω C be a region. A function f : C C is called analytic at z 0 Ω if ε > 0 such that z B ε (z 0 ), f (z) is representable by a convergent power series expansion around z 0 f (z) = c n (z z 0 ) n, for some c n C. n 0 It is called analytic in Ω if f is analytic for every point in Ω. Theorem 10. Let Ω C be a region. A function f : C C is holomorphic in Ω iff it is analytic in Ω. Definition 13. Let Ω C be a region and f : C C be holomorphic at z 0 Ω. Set f (0) (z 0 ) := f (z 0 ). Then the power series f (z) = f (z 0 ) + f (z 0 )(z z 0 ) + f (z 0 ) (z z 0 ) +... =! n 0 is called the Taylor series expansion of f (z) around z 0. f (n) (z 0 ) (z z 0 ) n n! Exercise 5. Find the maximum regions, in which the functions z e z, z 1 z 1, z (1+z) m (for fixed m N) are analytic. Show that their Taylor series expansions around the origin are given by e z = n!, 1 1 z = z n, n 0 z n n 0 Exercise 6. Find the Taylor series expansions of ( ) m (1 + z) m = z n. n 0 n 1 1 4z, sinz, cosz, z e z, lnz, 1 1 z ln 1 1 z, 1 ln 1 1 z 1 z around the origin and find the maximum regions where these power series converge.

32 CHAPTER 1. COMBINATORICS 9 Exercise 7. Prove Exercise 8. Prove n m ( ) n z n = m z m (1 z) m+1. H n z n = 1 n 0 1 z ln 1 1 z where H n = n k=1 1 k is the n-th harmonic number. Exercise 9. Find OGF for each of the following sequences: { n+1 } n 0, {n n+1 } n 0, {nh n } n 1, {n 3 } n Cauchy s coefficient formula Theorem 11 (Cauchy s coefficient formula, 1st version). Let Ω C be a region, f be analytic in Ω and γ be a simple loop in Ω. Then f = 0. γ Theorem 1 (Cauchy s coefficient formula, nd version). Let Ω C be a region and f be analytic in Ω. Let z 0 Ω and γ be a simple loop encircling z 0 such that the image of γ is contained in Ω. Then f (z 0 ) = 1 f (z) dz. πi γ z z 0 In general, for n 0 f (n) (z 0 ) n! Lagrange Inversion Theorem = 1 f (z) dz. πi γ (z z 0 ) n+1 Theorem 13 (Langrange Inversion, 1st version). Let φ(u) = k 0 φ k u k be a power series in C[[u]] with φ 0 0. Then the equation A = zφ(a) admits a unique solution in C[[z]] and the coefficients of are given by A(z) = a n z n n 0 a n = 1 n [un 1 ]φ(u) n. Furthermore, for any holomorphic function H, [z n ]H(A(z)) = 1 n [un 1 ] ( H (u)φ(u) n).

33 CHAPTER 1. COMBINATORICS 30 In particular, for k 1, [z n ]A(z) k = k n [un k ]φ(u) n. Theorem 14 (Langrange Inversion, nd version). Let the generating function satisfy the functional equation A(z) = a n z n n 0 z = ψ(a(z)), where ψ(0) = 0 but ψ (0) 0 (i.e. ψ is the compositional inverse of A). If ψ(a) = A φ(a), i.e. A = ψ(a) φ(a) = zφ(a), then a n = 1 n [zn 1 ]φ(z) n = 1 ( ) z n n [zn 1 ]. ψ(z) Application of Lagrange Inversion to Binary Trees Let B denote the class of all binary trees, in which the size of a binary tree is defined as the total number of all vertices (internal nodes and leaves alike). For n 1 we let B n denote the class of all binary trees of size n and let b n := B n and set b 0 = 0. Let B(z) be the ordinary generating function of the combinatorial class B. Then B(z) = z + z B(z) (1.10) Let B = B(z) and φ(u) := 1 + u. Then (1.10) can be rewritten as B = z φ( B) (1.11) Note that φ(u) := 1 + u = k=0 φ k u k is a formal power series in the ring C[[u]] with φ 0 = 1 0. Therefore by Lagrange Inversion Theorem, the equation (1.11) admits a unique solution B(z) := b n z n in the ring C[[z]] where the coefficients are given by b n = 1 n [un 1 ] φ(u) n = 1 n [un 1 ] (1 + u ) n = 1 n [un 1 ] ( n k=0 ( ) ) n u k k = { 0 if n is even ( n ) if n is odd. 1 n n 1

34 CHAPTER 1. COMBINATORICS 31 In other words, we have, for any n N, b n+1 = 1 ( n + 1 n + 1 n ) = 1 n + 1 ( ) n. n Remark 3. Recall that b n was defined in Theorem 9 as the number of binary trees of size n, in which the size of a binary tree is defined as the number of internal nodes. If a binary tree has n internal nodes, then it has n + 1 external nodes and therefore n + 1 vertices in total. Thus we have b n+1 = b n. Application of Lagrange Inversion to Labelled Trees Definition 14. A labelled tree (also called Cayley tree) is a tree whose vertices are distinctly labelled by numbers in N. The vertices of a Cayley tree on n vertices are labelled by distinct numbers from [n]. A rooted labelled tree on n vertices is a labelled tree on n vertices, in which one vertex is distinguished by a mark from the other vertices. Let C denote the class of all labelled trees, in which the size of a tree is defined as the number of vertices, and let C n denote the set set of all Cayley tree on n vertices. Then C is a (labelled) combinatorial class. (C n is also a (labelled) class, it is a subclass of C ). For n 1 we let c n := C n and set c 0 = 0. Let C(z) := c n zn n! be the exponential generating function of the combinatorial class C. The sequence (c n ) n 0 begins with 0,1,1,3,16,15... Theorem 15 (Cayley s formula (Cayley 1889)). c n = n n, n. Proof. Let T denote the (labelled combinatorial) class of all rooted labelled trees (i.e. the set of all Cayley trees in C, in which one vertex is distinguished by a mark from

35 CHAPTER 1. COMBINATORICS 3 the other vertices) and let T n denote the class of all rooted labelled tree on n vertices. For n 1 we let t n := T n and set t 0 = 0. Because there are n ways to choose a root vertex of a labelled tree on n vertices, the number of Cayley trees on n vertices is equal to t n = n c n. Let T (z) := t n zn n! be the exponential generating function of the combinatorial labelled class T. not ordered Its EGF T = Z SET(T ). T (z) = z exp(t (z)). (1.1) So T (z) is implicitly defined. u Let T = T (z) and φ(u) := exp(u) = k k 0 k!. Then (1.1) can be rewritten as T = z φ(t ) (1.13) Note that φ(u) := exp(u) = k=0 φ k u k is a formal power series in the ring C[[u]] with φ 0 = 1 0. Therefore by Lagrange Inversion Theorem, the equation (1.13) admits a unique solution in the ring C[[u]] whose coefficients are given by T (z) = t n z n, where t n = 1 n [un 1 ] φ(u) n = 1 n [un 1 ] (exp(u)) n = 1 n [un 1 ] exp(un) ) = 1 (un) k n [un 1 ] = 1 n n 1 k! n (n 1)! = nn 1 n!. ( k=0 In other words, for any n N, Therefore, we have t n = n n 1. c n = t n n = nn. Remark 4. There are several interesting proofs for this in The Book by M. Aigner and G. Ziegler. We shall see one proof in Section.3. (Chapter ).

36 CHAPTER 1. COMBINATORICS 33 Lecture Singularities Definition 15. Let f be an analytic function in a region Ω and z 0 be a point on the boundary of Ω. We say f is analytically continuable at z 0 if there is an analytic function g defined in a region Ω containing z 0 such that g(z) = f (z) in Ω Ω. Example 1. f (z) = 1 for z 0 = 1. 1 z is analytic for z < 1, and is analytically continuable except Definition 16. A function f is said to be singular at z 0 Ω or z 0 is called a singularity of f if f is not analytically continuable at z 0. Theorem 16 (Boundary singularities). Let f be analytic at the origin and let R be the finite radius of convergence of a power series expansion of f at the origin. Then f has necessarily a singularity on the boundary of the disc of convergence. Definition 17. If f is analytic at the origin with radius R of convergence then a dominant singularity is a singularity on the circle around 0 with radius R > 0. Theorem 17 (Pringsheim s Theorem). Let f be analytic at the origin and f (z) = n 0 f n z n be its convergence power series expansion at the origin with the radius of convergence R > 0. If f n 0, n 0, then the point z = R is a singularity of f ; in this case, z = R is a dominant singularity of f. The singularity z = R of Theorem 17 is often called the dominant singularity of f. Remark 5. The radius R of convergence of n 0 f n z n is 1 R =, limsup n f n 1 n so we have where f n = R n θ(n), lim sup θ(n) n 1 = 1. n Meromorphic functions Definition 18. A function f (z) is meromorphic at z 0 if z in a neighbourhood B ε (z 0 ) of z 0 with z z 0, it can be represented as f (z) = h(z) g(z),

37 CHAPTER 1. COMBINATORICS 34 where h(z) and g(z) are analytic at z 0 and g(z) 0. expansion of the form f (z) = f n (z z 0 ) n n M In this case f (z) admits an for z B ε (z 0 ), for some M Z. If f M 0, M 1, we say f (z) has a pole of order M at z 0. In this case, we have f (z) = f M (z z 0 ) M + O((z z 0 ) M+1 ) ( = f M ( z 0 ) M 1 z ) ( M ( + O 1 z ) ) M+1. z 0 z 0 Remark 6. Scaling rule: If f (z) analytic around 0 and singular at z 0, then g(z) = f (z 0 z) is singular at 1. If z 0 > 0 is a dominant singularity of f (z), then f (z) admits a convergent series expansion (around 0): f (z) = f n z n, z < z 0. n 0 Therefore, g(z) admits a convergent series expansion g(z) = g n z n, z < 1, g n = f n z n 0. n Newton s generalised binomial theorem For any α C, [z n ](1 z) α = Theorem 18. Let α C \ Z 0. where the Γ-function is defined as ( ) α? = n [z n ](1 z) α = nα 1 Γ(α) Γ(α) = 0 ( ) n + α 1 α 1 ( 1 + O e t t α 1 dt.? = Θ(n α 1 ). ( )) 1, n Properties of the Gamma function ( ) 1 Γ = e t t 1 dt = π. 0 { Γ(t + 1) = t Γ(t) Γ(1) = 1. For n N, Γ(n + 1) = nγ(n) =... = n!.

38 CHAPTER 1. COMBINATORICS Transfer theorem Transfer theorems allow to extract the asymptotit order of coefficients of error terms in singular expansions. For details see chapter VI in []. Definition 19. Given two numbers R,φ with R > 1 and 0 < φ < π, then the set = (R,φ) = {z z < R,z 1, arg(z 1) > φ} is called -domain at 1. A function f is called -analytic at 1 if it is analytic at some -domain at Theorem 19 (Transfer theorem). Let α C \ Z 0 and assume that f (z) is -analytic at 1. If f (z) = O((1 z) α ), then If f (z) = o((1 z) α ), then [z n ] f (z) = O(n α 1 ). [z n ] f (z) = o ( n α 1). Therefore, if f (z) = (1 z) α + o((1 z) α ), then ( ( )) [z n ] f (z) = nα O + o(n α 1 ). Γ(α) n Example (-regular graphs). Consider the EGF of -regular graphs, G = SET(UCYC 3 Z ). Then G(z) = exp ( ( ( ) )) 1 1 log z z = 1 e z z 4. 1 z 1 z 1 1 z is analytic in C \ [1,). G(z) is in particular -analytic at 1.

39 CHAPTER 1. COMBINATORICS 36 To derive the singular expansion of G (an expansion of the singularity of G) near 1, we use the Taylor expansion of e z z 4 at 1. h(z) = e z z 4 = h(1) + h (1)(z 1) + h (1) (z 1) +... = e e 3 4 (1 z) + e 3 4 (z 1) +... The singular expansion of G(z) near 1 is G(z) = e e 3 4 (1 z) 1 + e 3 4 (1 z) z ( ) = e 4 3 (1 z) 1 + o (1 z) 1. (z 1) Using the transfer theorem, we get [z n ]G(z) = e 4 3 n 1 ( ( )) 1 ( ) Γ ( ) 1 + O + o n 1 = e 3 4 (1 + o(1)). 1 n nπ The first term corresponds to h(1), the second is from the theorem Multiple singularities Theorem 0. Let f (z) be analytic in z < ρ. Suppose that f has a finite number of singularities on the circle z = ρ, at points ρ j = ρ e iθ j for j = 1,...,m. Assume there exists a -domain 0 at 1 such that f (z) is analytic in the region m D = ρ j 0, j=1 where ρ j 0 is the dilation (rotation) of the -domain 0 for each ρ j. ρ ρ m 0 D ρ 1 Let S = {(1 z) α α C \ Z 0 }.

40 CHAPTER 1. COMBINATORICS 37 Assume there exist m functions g 1 (z),...,g m (z), each a linear combination of the functions in S, and there exists a function h(z) = (1 z) α 0 such that ( ) ( ( )) z z f (z) = g j + O h as z ρ j in D. ρ j ρ j Then [z n ] f (z) = = m j=1 m j=1 ( ) ( ( zρ z ([z n ]g j + O [z n ]h j ρ j ( ) ρ n j [z n ]g j (z) + O ρ n j n α 0 1. ))) Remark 7. [z n ]g j (z) = [z n ] l k=1 Remark 8 (Recipe for singularity analysis). c jk (1 z) α j,k = l k=1 c jk n α j,k 1 Γ(α j,k ). Find the grammar that determines the combinatorial class and the corresponding generating function Locate the singularities Check the -analyticity of the generating function Do the singular expansion at each singularity separately Extract and sum the asymptotics using the Transfer Theorem and basic scaling. Example 3. Let P be the class of permutations with cycles of odd length. Then For the generating functions, this means z A(z) = k k=odd P(z) = e A(z) = exp P = SET(A ), A (Z ) = CYC odd (Z ). k = 1 log 1 + z 1 z, ( 1 log 1 + z 1 z ) 1 + z = 1 z. The dominating singularities of this function are {±1}. Let 0 be a -domain at 1. P(z) is analytic in the region D = 0 ( 1) 0 since 1 z is analytic in C\[1,) and 1 + z is analytic in C \ (, 1].

41 CHAPTER 1. COMBINATORICS 38 Singular expansions, with g(z) = 1 + z and writing P + (z) for P(z) as z +1 in D, P (z) for P(z) as z 1 in D: P + (z) = g(1) + g (1)(z 1) +... (as z +1 in D) 1 z 1 z = 3 ( ) 1 z + O (1 z) 3 1 z P (z) = 1 ) 1 + z + O ((1 + z) 3. (as z 1 in D) Extract asymptotics: [z n ]P(z) = 1 where Γ( 1 ) = π, Γ( 1 ) = π. Γ ( )n Γ ( Γ ( 1 )n 3 + O ( ) )n 3 + O n 5 ( ) n 5

42 Chapter Graph Theory 39

43 CHAPTER. GRAPH THEORY 40 Lecture 7 Suggested reading for Chapter : [1] and [4]..1 Matchings, Eulerian Tours, and Hamiltorian Cycles.1.1 Basic Terminologies Notation 1 (Subsets). Given a set X and k N, we denote by ( X k) the set of all k- element subsets of X. Definition 0 (Graphs). A graph is a pair (V,E) where E ( V ). The elements of V are called vertices (or nodes, points) and the elements of E are called edges. Two vertices v,w are called adjacent or neighbours if {v,w} E. A vertex v is said to be incident with an edge e E if v e. We call v an end of e. Note that all graphs considered will be simple (by the above definition) and finite (finitely many vertices, thus finitely many edges). If G is a graph we will write V (G) to denote the set of vertices of G and E(G) to denote the set of edges of G. Definition 1 (Walk, Paths, Cycles, Distance). Let G = (V,E) be a graph. (1) A walk in G is an alternating sequence W = v 0 e 0 v 1 e 1 v...v k 1 e k 1 v k of vertices and edges of G such that e i = {v i,v i+1 } for each 0 i k 1. It is said to be closed if v 0 = v k. () A walk v 0 e 0 v 1 e 1 v...v k 1 e k 1 v k in G is called a path if v 0,v 1,...,v k are all distinct. We write P = v 0 v 1 v...v k 1 v k and call it a v 0 -v k path. The length of P is k, i.e. the number of its edges. (3) A cycle in G is an alternating sequence C = v 0 e 0 v 1 e 1 v...v k 1 e k 1 v k e k v 0 of vertices and edges of G such that v 0,v 1,...,v k are distinct, e i = {v i,v i+1 } for 0 i k 1, and e k = {v k,v 0 }. (4) The distance of two vertices v,w in G is the length of a shortest v-w path (a shortest path starting at v and ending in w) in G. Definition (Subgraphs, induced subgraphs, spanning subgraphs). Given two graphs G = (V,E) and G = (V,E ), we set G G := (V V,E E ) G G := (V V,E E ).

44 CHAPTER. GRAPH THEORY 41 (1) If G G = /0, then we say that G and G are disjoint. () If V V and E E, then G is a subgraph of G. We write G G. (3) If G G and G G, then G is a proper subgraph of G. We write G G. (4) If G G and G contains all edges {v,w} E with v,w V, then G is called an induced subgraph of G. We write G = G[V ]. (5) If G G and V = V, then G is called a spanning subgraph of G. Definition 3 (Components, connectedness). Let G = (V,E) be a graph. (1) G is called connected if G contains a subgraph which is isomorphic to a path from v to w for any two vertices v,w V. () A maximal connected subgraph of G is called a component of G. (3) For k N, G is called k-connected if V > k and G[V \ X] is connected for every set X V with X < k. Definition 4 (Degrees). Let G = (V,E) be a graph. (1) For v V, d(v) := {w V : {v,w} E} is called the degree of the vertex v. A vertex of degree 0 is called isolated. () The number δ(g) := min {d(v) : v V } is called minimum degree of G, the number (G) := max {d(v) : v V } is called maximum degree of G, and the number is called average degree of G. d(g) := 1 V d(v) v V (3) If all the vertices have the same degree d, then G is called d-regular. A 3-regular graph is also called cubic. Remark 9. Clearly we have V d(g) = v V d(v) = E δ(g) d(g) (G)

45 CHAPTER. GRAPH THEORY 4 Definition 5 (Graph isomorphism). We say that two graphs G = (V,E) and G = (V,E ) are isomorphic and write G G, if there is a bijection φ : V V with {v,w} E {φ (v),φ (w)} E, v,w V. Definition 6 (Graph property). A class of graphs that is closed under isomorphism is called a graph property. An example of a graph property is the class of graphs containing a triangle. A graph property only depends on the abstract structure, not on a representation of a graph (like a drawing). Example 4. G G G G (1) Determine whether G,G,G are induced subgraphs of G or not () Determine whether G,G,G are spanning subgraphs of G or not.1. Matchings Definition 7 (Independent set, matchings, vertex-cover). Let G = (V,E) be a graph. (1) A set of vertices or a set of edges of G is called independent if no two of its elements are adjacent. An independent set is also called a stable set. () A set M of independent edges in G is called a matching of G. In other words, a matching is a set of pairwise non-adjacent edges. A vertex is called matched (or saturated) if it is an endpoint of an edge in the matching. (3) A d-regular spanning subgraph of G is called a d-factor of G. A 1-factor (indeed, its edge set) is called a perfect matching. (4) A set U V is called a vertex cover of G if every edge of G is incident with a vertex in U. Definition 8 (Independence number, clique number). Let G = (V,E) be a graph. (1) The independence number of G, denoted by, α (G), is the number of vertices in a maximum independent set (of vertices) in G. () The clique number of G, denoted by ω (G), is the number of vertices in a maximum clique (=complete graph) in G.

46 CHAPTER. GRAPH THEORY 43 Example 5. In Example 4, α (G) =?? and ω (G) =??. Remark 10 (Perfect matching). Let H G be a subgraph of G = (V,E). H G is a 1-factor E (H) is a perfect matching of G Every vertex in G is incident to exactly one edge of H Definition 9 (Bipartite graph). A graph G = (V, E) is called bipartite if V admits a partition into two sets, V = A B, such that every edge e E has its ends in different classes. Lemma 1. A graph G = (V,E) is bipartite if and only if it contains no odd cycle Proof. Exercise! Problem 1 (Maximum bipartite matching). Input: A bipartite graph G = (A B, E) Output: A matching in G with as many edges as possible Definition 30 (Alternating paths, augmenting paths). Let G = (A B,E) be a bipartite graph and M be an arbitrary matching in G. (1) A path which begins with an unmatched vertex and contains alternatingly edges from E \ M and from M is called an alternating path (with respect to M). () An alternating path that ends also at an unmatched vertex is called an augmenting path. Remark 11. An augmenting path P can be used to turn M into a larger matching M E(P) := (M \ E(P)) (E(P) \ M) It has one edge more than M, whose ends are the ends of P. Example 6. Let G = (A B,E) be a bipartite graph given below. a 1 b1 a 1 b1 a a 3 b a a 3 b a4 b 3 a4 b 3 a 5 b 4 a 5 b 4 A B A B