Discrete mathematics , Fall Instructor: prof. János Pach. 1 Counting problems and the inclusion-exclusion principle

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1 Discrete mathematics , Fall Istructor: prof Jáos Pach - covered material - Special thaks to Jaa Cslovjecsek, Kelly Fakhauser, ad Sloboda Krstic for sharig their lecture otes If you otice ay errors, please sed a to: adriavalculescu@epflch 1 Coutig problems ad the iclusio-exclusio priciple Recall the followig formulas: i = ( + 1), i = ( + 1)( + 1) 6 - the umber of permutatios of a set of elemets is! - the umber of ways i which oe ca choose k objects out of distict objects,! assumig the order of the elemets matters, is ( k)! - the umber of ways i which oe ca choose k objects out of distict objects,! assumig the order of the elemets does ot matter, is = ( ( k)!k! k) This is the same as the umber of subsets of k elemets of a -elemet set The umber of subsets of a -elemet set is, sice we have ( ) ( ) ( ) = The umber of subsets of a -elemet set havig odd cardiality is 1 The umber of subsets of a -elemet set havig eve cardiality is 1 The equalities above ca be obtaied usig the biomial theorem (1 + x) = ( ) 0 + ( 0 ) x + + ( ) x = For x = 1, respectively x = 1, we obtai ( ) ( ) ( ) = = = ( ) 0 ( ) ( ) + + ( 1) = 1 i=0 i=0 i=0 ( ), i ( ) x i i ( ) ( 1) i, i

2 Addig, respectively substractig the two relatios, ad dividig each by two, oe obtais ( ) ( ) 1 = ( ) ( ) 1 = + +, 1 3 which proves the statemets about the umber of eve/odd sets Assume we have idetical objects ad k differet persos The, the umber of ways i which oe ca distribute the objects amog the k persos equals ( ) ( ) + k 1 + k 1 = k 1 If k ad each persos receives at least 1 object, the the umber of possible ways to distribute is ( k ) The followig is called Pascal s triagle The followig idetities hold: 1 ( ) ( k + ) ( k+1 = +1 k+1) ( k 1) is the k-th elemet i the -th lie of Pascal s triagle 3 All the diagoals of Pascal s triagle are strictly icreasig Theorem 1 (Multiomial theorem) The followig holds: (x x ) k = i 1,i,,i 0 i 1 ++i =k k! i 1!i! xi 1 1 x i Usig the biomial theorem, oe obtais the followig: ( ) ( ) = i

3 Theorem (Stirlig s formula)! ( ) π, e where is used to idicate that the ratio of the two sides teds to 1 as goes to Defiitio 3 Let f, g : Z + R We say that f is big-oh of g ad we write f(x) = O(g(x)) if there exist 0 ad c costats such that for all > 0, we have f() < c g() Defiitio 4 Let f, g : Z + R We say that f is little-oh of g ad we write f(x) = o(g(x)) if f() lim g() = 0 Examples: = O(), = O( 3 ), 10 = O( ), = o( ), 10 = o(3 ), but is ot o() Theorem 5 (Iclusio-Exclusio priciple) Let A 1,, A be fiite sets The, the followig holds A i = A i A i A j + A i A j A k +( 1) 1 A 1 A A 1 i<j 1 i<j<k Number of permutatios without fixed poits We are lookig for permutatios of the set {1,,, } without fixed poits I order to cout these, we apply the iclusio-exclusio priciple Let A be the set of all permutatios ad A i be the set of permutatios of the set {1,,, } for which i is a fixed poit The umber of permutatios with o fixed poits is A\ A i = A A i We kow that A =!, so we eed to cout A i We do this usig the iclusio priciple Note that A i A j represets the set of all permutatios for which i ad j are fixed poits Oe ca see that A i = ( 1)! for all i, while A i A j = ( )! Usig the same idea, we obtai A i A j A k = ( 3)! ad so o Altogether, this gives A i = ( 1)! ( )! + ( 3)! = = ( ) ( 1)! 1 ( ) ( )! + 1 i<j ( ) 3 1 i<j<k ( 3)! =!( ( 1) k+1 1 k! )!(1 1 e ) Thus, the umber of permutatios without fixed poits is!!(1 1 ) =!/e, as e 3 k=0

4 The umber of surjective fuctios Let X ad Y be two fiite sets with X = ad Y = m We wat to determie the umber of oto (that is, surjective) mappigs f : X Y Oce agai, we apply the iclusio-exclusio priciple Let A = {f : X Y } ad A i = {f : X Y i is ot i the image of f}, for all 1 i The umber of surjective maps is m A\ m A i = A A i By iclusio-exclusio, we obtai that m m A i = A i A i A j + + ( 1) m+1 A 1 A m 1 i<j m We have A = m, ad for every 1 i, A i = (m 1) (the umber of fuctios defied o X with values i {1,, }\{i} Aalogous, we obtai that A i A j = (m ) ad so o Altogether, this gives that the umber of surjective maps is m ( m 1 ) (m 1) + ( m ) (m ) = m ( m ( 1) k (m k) k Euler s fuctio For every positive iteger we defie φ() as the umber of positive itegers that are relatively prime with Formally, oe writes k=0 φ() = #{m {1,,, } gcd(m, ) = 1}, where gcd(m, ) deotes the greatest commo divisor of m ad If = p α, with p prime, we have Otherwise, = m pα i i oe obtais that φ() = φ(p α ) = p α p α 1 with p i prime factors I this case, usig iclusio-exclusio, φ() = m (1 1pi ) Refer as well to the followig: [Lov] 1 Sets, 13 Number of subsets, 15 Sequeces, 16 Permutatios, 17 Number of The Number of Ordered Subsets, 18 The Number of Subsets of a Give Size, 1 Iductio, 3 Iclusio-Exclusio, 4 Pigeohole priciple, 31 The Biomial Theorem, 3 Distributig Presets, 35 Pascals Triagle, 36 Idetities i Pascals Triagle [Mat] 34 Estimates: a itroductio - startig from 34 - Big Oh, little oh, 355 Estimate! - secod proof oly, 37 Iclusio - Exclusio, 38 The hatcheck lady 4 )

5 Elemets of graph theory Defiitio 6 A graph G is a ordered pair (V, E), where V is a set of elemets called vertices ad E is a set of -elemet subsets of V called edges Defiitio 7 Let G = (V, E) be a graph We call a sequece of distict vertices v 0,, v r a path if (v i, v i+1 ) is a edge of G, for every 0 i r 1, Defiitio 8 We say that a graph G = (V, E) is coected if for every two vertices v, u V there exist a path i G betwee u ad v Defiitio 9 For every vertex of a graph, we defie its degree as the umber of edges adjacet to it The followig lemma gives us a relatio betwee the degrees of the vertices of a graph ad the total umber of edges: Lemma 10 For every graph G = (V, E), the followig holds: E = v V d(v), where d(v) represets the degree of the vertex v Proof The proof is based o a double coutig of the total umber of edges O the oe had, we kow that this is E O the other had, whe coutig the sum of degrees of all vertices, we cout every edge twice Defiitio 11 A cycle i a graph G = (V, E) is a sequece of distict vertices v 1,, v r V such that v r = v 0 ad {v i, v i+1 } E for all i from 0 to r 1 Defiitio 1 A tree is a coected graph without cycles Defiitio 13 A vertex of degree oe i a tree is called a leaf Lemma 14 Every tree o vertices has at least oe leaf Proof Let S be the set of all the paths i the tree T We kow that every path o r vertices cotais exactly r 1 edges Cosider ow a path v 1,, v l of maximum legth Oe ca always fid a path of maximum legth sice every path i the tree ca cotai at most vertices (otherwise it will be self-itersectig, that is it will cotai a cycle, which is impossible sice i a tree we caot have cycles) We prove that both v 1 ad v l (the edpoits of the path) are leafs Assume at least oe of them is ot, say v 1 That meas that, there is at least aother edge apart from v 1 v icidet to v 1 Observe that u caot coicide with ay of the vertices of the path v 1,, v l (otherwise it will close a cycle) Therefore, we ca add u to the path without formig ay cycle But this is a cotradictio to the maximality of the legth of the path v 1,, v l Thus, both v 1 ad v l must be leaves 5

6 Theorem 15 Every tree o vertices has exactly 1 edges Proof By iductio o the umber of vertices We check for = 1, = ad assume that every graph o vertices cotais 1 edges We wat to prove that the statemet is true for + 1, that is, every tree o + 1 vertices has exactly edges Let T +1 = (V, E) be a arbitrary tree o + 1 vertices By the lemma above, we kow that T cotais at least oe leaf v We remove v ad its uique icidet edges from the tree T +1 This leaves us with a tree T o vertices, which, by the iductio hypothesis has exactly 1 edges Therefore, T +1 cotais exactly = edges, which completes the proof For other properties of trees, refer to the problem sets Refer as well to the followig: [Lov] 81 How to Defie Trees? [Mat] 41 The otio of a graph; isomorphism - oly the defiitio of graphs, 431 Sum of the degrees, 43 Hadshakes lemma, 51 Defiitio ad characterizatios of trees 3 Cayley s theorem ad Kruskal s algorithm I what follows, we will preset a result due to Cayley Before statig the theorem, we eed the followig lemma: Lemma 16 Let T be a tree o labeled vertices ad let d 1,, d be the degrees of the vertices The d i = E(T ) = ( 1), where by E(T ) deotes the edge set of the tree Now we ca state Cayley s theorem Theorem 17 (Cayley) The umber of trees o labeled vertices is We give two proofs to this theorem Oe of them is by iductio, while the other, due to Prüfer, is algorithmic Proof 1 of Cayley s theorem The proof is based o the followig lemma Lemma 18 The umber of trees o vertices labeled with 1,,, with degrees d 1,, d equals ( )! (d 1 1)!(d 1)! 6

7 By the lemma, we obtai that the total umber of trees o labeled vertices is the sum of the umber of trees over all possible values of degrees, that is d 1,,d 1 d 1 ++d = ( )! (d 1 1)!(d 1)! Let ow r i = d i 1, for all 1 i By substitutio, we obtai that the umber of trees o labeled vertices is r 1,,r 1 r 1 ++r = which, by the multiomial theorem equals ( )! (r 1 )!(r )!, Proof of Cayley s theorem We give ow the secod proof, due to Prüfer Deote the vertices by {1,,, } We will defie a oe-to-oe correspodece betwee the set of all trees o labeled vertices ad the set of all sequeces of legth cosistig of umbers i {1,,, } Sice the cardiality of the latter is, we obtai the desired result The followig algorithm takes a tree as iput, ad yields a sequece of itegers: Step 1: Fid the leaf with the smallest label ad write dow the umber of its eighbor Step : Delete this leaf, together with the oly edge adjacet to it Step 3: Repeat util we are left with oly two vertices We preset a algorithm that recostructs the tree from the Prüfer code Step 1: Draw the odes, ad label them from 1 to Step : Make a list of all the itegers (1,,, ) This will be called the list Step 3: If there are two umbers left i the list, coect them with a edge ad the stop Otherwise, cotiue o to step 4 Step 4: Fid the smallest umber i the list which is ot i the sequece Take the first umber i the sequece Add a edge coectig the odes whose labels correspod to those umbers Step 5: Delete the smallest umber from the list ad the first umber i the sequece This gives a smaller list ad a shorter sequece The retur to step 3 Defiitio 19 A weighted graph is a graph i which each edge is give a umerical weight We defie the weight of a graph as the sum of the weights of all its edges We are iterested i the followig problem: fid a miimum weight spaig tree T for a give weighted coected graph G Before that, we have to defie the otio of a spaig tree for a graph G Defiitio 0 Let G = (V, E) be a graph We way that a tree T is a spaig tree of G if it cotais all the vertices of V ad is a subgraph of G, that is every edge i the tree belogs to the graph G 7

8 Example Above is a example of a spaig tree: Oe way to solve the problem of fidig a miimum spaig tree is usig Kruskal s algorithm This works as follows: Step 1 Start with a empty graph Step Take all the edges that have t bee selected ad that would ot create a cycle with the already selected edges ad select it uless it creates a cycle Add the oe with the smallest weight Step 3 Repeat util the graph is coected A proof of the correctess of Kruskal s algorithm was give i solutios to problem set 6 Refer as well to the followig: [Lov] 83 How to Cout trees? 84 How to Store trees? [Mat] 53 Spaig trees of a graph, 54 Miimum spaig tree problem, 81 The umber of spaig trees, 8 A proof via scores, 84 A proof usig the Prüfer codes 4 Partially ordered sets ad Dilworth s theorem Defiitio 1 A partially ordered set (or simply poset) is a pair (X, ), where X is a set ad is a biary relatio over X, which is reflexive, atisymmetric, ad trasitive, ie, which satisfies the followig relatios, for all a, b, ad c i X: a a a (reflexivity); b if a b ad b a the a = b (atisymmetry); c if a b ad b c the a c (trasitivity) We write a < b if a b ad a b Defiitio Let (X, ) be a partially ordered set A chai i X is a sequece x 1,, x t X such that x 1 < x < < x t A atichai i X is a subset {x 1,, x t } of X such that o two elemets are comparable 8

9 Recall, that for a set X, X deotes the power set of X, that is the set of all subsets of X Example Cosider the partially ordered set ( {1,,3}, ) The sequece {1} {1,, 3} is a chai {1, }, {1, 3} is a atichai Theorem 3 (Dilworth) Let (X, ) be a partially ordered set a If the maximum size of a atichai is k, the X ca be decomposed ito k chais b If the maximum size of a chai is k, the X ca be decomposed ito k atichais Proof The proof ca be foud i [Juk], page 108 Corollary 4 Let (X, ) be a partially ordered set The, the followig hold: a The maximum size of a chai is equal to the miimum umber of atichais that cover X b The maximum size of a atichai is equal to the miimum umber of chais that cover X Refer as well to the followig: [Mat] Orderigs: the defiitios of poset, chai, atichai, Hasse diagram [Juk] 81 Decompositio i chais ad atichais 5 Matchigs i graphs ad Köig-Hall theorem Defiitio 5 A bipartite graph (or bigraph) is a graph G whose vertices ca be divided ito two disjoit sets A ad B such that every edge of the graph coects a vertex i A to oe i B (i other words, there is o edge of the graph betwee two vertices of A or two vertices of B Lemma 6 A graph is bipartite if, ad oly if, it does ot cotai a odd cycle (that is, a cycle of odd legth) Proof The proof ca be foud i the solutios to problem 1 i problem set 8 For other properties of bipartite graphs, refer to the problem sets Defiitio 7 Let G = (V, E) be a graph A subset E E of pairwise disjoit edges (that is, edges which do ot share ay vertex) is called a matchig i G Defiitio 8 A perfect matchig is a matchig where every vertex of the graph is icidet to exactly oe edge of the matchig Remark 9 A perfect matchig is therefore a matchig of a graph cotaiig / edges (where is the umber of vertices) Thus, perfect matchigs are oly possible o graphs with a eve umber of vertices! Example The blue edges i the graph o the ext page form a matchig: The matchig is ot a perfect matchig though 9

10 Theorem 30 (Köig-Hall) Let G = (V, E) be a bipartite graph with bipartitio V = A B For every X A, let B(X) = {b B : x X such that (x, b) E} The, a perfect matchig i the graph exists if ad oly if B(X) X, for all X A Proof You ca fid the proof i [Lov], chapter 103 The Mai Theorem or [Juk], page 83 Also refer to the followig: [Lov] 103 The Mai Theorem 6 Sperer s theorem ad applicatios We state the LYM iequality, Sperer s theorem ad preset the Littlewood-Offord problem as a applicatio of Sperer s theorem Theorem 31 (LYM iequality) Let X = {1,,, }, ad let F be a family of subsets A 1,, A m X such that A j A i, for all i j Let m k = {A F : A = k}, that is the umber of sets i F cotaiig k elemets(oe ca see that m m = m) The, the followig holds m ( i 1 i) Proof The proof ca be foud i [Mat] pages 7-8 i=0 Theorem 3 (Sperer) Let X = {1,,, } ad A 1,, A m X, with A j A i, for all i j The m ( / ) Proof Two proofs of Sperer s theorem (oe of them usig LYM iequality) ca be foud i [Mat], pages 7-9 Littlewood-Offord problem 10

11 Let a 1,, a 1 be fixed real umbers ad { } A = ɛ i a i, ɛ i = 0 or 1 We are iterested i fidig how may of the sums of A lie iside a give iterval I of legth 1 We will prove that, o matter how oe chooses I, we have ( ) A I / We do this usig Sperer s theorem We costruct a bijectio as follows: for every sum i=0 ɛ ia i we assig a uique characteristic set, that is {i ɛ i = 1} {1,,, } Oe ca easily see that this is a bijectio, so it is eough to boud the umber of characteristic sets We prove that the characteristic sets of sums cotaied i the same iterval satisfy the coditio i Sperer s theorem, that is o oe is cotaied i aother We assume the cotrary, that is, there exist two characteristic sets B, B with B B, B B such that the correspodig sums lie i the iterval I ɛ i a i I ad ɛ i a i I i B i B Sice the legth of I is less tha 1, the their differece has to be less tha oe O the other had, sice B B ad B B, we obtai that ɛ i a i < 1, i B \B where the sum is a sum of o-egative umbers, cotaiig at least oe o-zero term O the other had, sice every ɛ i is either 0 or 1 ad every a i 1, we obtai that the sum must be at least oe, which is a cotradictio Therefore, the coditio of Sperer s theorem is satisfied, so we obtai that the umber of sums iside the iterval I caot exceed ( / ), which completes the proof Also refer to the followig: [Bol] 3 Sperer systems, 4 The Littlewood-Offord problem [Mat] 7 Sperer s theorem o idepedet systems: Sperer theorem ad proof of Theorem 71 7 Erdős-Ko-Rado Theorem Defiitio 33 Let X be a set ad F be a family of subsets of X, that is F X We say that F is itersectig, if ay two members of F itersect 11

12 Lemma 34 Let X be a fiite set with X = ad F X a itersectig family The F 1 ad this boud is sharp (that is, there exists a itersectig family of size 1 ) Proof I total, X has subsets, which ca be arraged i pairs of sets that are complemets of each other We have 1 such pairs Out of each pair, F may cotai at most oe To see that the boud is sharp, take a arbitrary elemet of X, ad cosider all subsets of X cotaiig this elemet Theorem 35 (Erdős-Ko-Rado) Let k, X be a set with X = ad F a itersectig family of k-elemet subsets of X The ( ) 1 F k 1 Proof The proof ca be foud i [Juk], pages Remark The theorem above does ot hold i the case whe k > / Ideed, cosider the family of all k-elemet subsets of a -elemet set X This will be itersectig by the pigeohole priciple The cardiality of this family is ( ( k), which is larger tha 1 k 1) Defiitio 36 Let X be a set t 1 a iteger, ad F be a family of subsets of X We say that F is t-itersectig, if for every F 1, F F, we have F 1 F t We state the followig geeralizatio of the Erdős-Ko-Rado theorem proof of this i class We gave o Theorem 37 Let X be a set with X =, ad t a positive iteger Let F X be a t-itersectig family of k-elemet sets, such that (k t + 1)(t + 1) ad / k t The ( ) t F k t Theorem 38 Let X be a -elemet set, k / Let F be a itersectig family of subsets of X of cardiality k such that o two members of F cotai each other The ( ) 1 F k 1 Proof Our goal is to replace F by aother set family, which is of the same size as F, also itersectig, but has the additioal property that all of its members are k-elemet sets Clearly, from this, by Theorem 35, the result follows Assume that the miimum size of members of F is l < k We wat to add a elemet of X to each l-elemet member of F such that o two l-elemet subsets i F become the same (l + 1)-elemet subset If we ca do this, we ca repeat it as log as some sets are of size less tha k 1

13 We defie the followig bipartite graph Let A be the set of all l-elemet members of F Let B be the set of all (l + 1)-elemet subsets of X The vertex set of our graph is A B We coect a vertex F of A to a vertex G of B by a edge, if F G We claim that there is a matchig that covers A If it holds, the we are doe We check if Hall s coditio holds i the graph, so we cosider a set of t vertices from A Each vertex has l eighbors i B Each elemet i B has degree at most l + 1 Thus, the umber of eighbors of the set of t vertices is at least t( l) l + 1 Sice l < k /, this is at least t Thus, Hall s coditio holds, thus, there is a matchig coverig A This fiishes the proof Also refer to the followig: [Juk] 7 Erdős-Ko-Rado theorem 8 The probabilistic method Let X be a radom variable with the followig probability distributio where 0 p i 1 ad k p i = 1 The, the expected value of X is x 1, with probability p 1 x, with probability p X = x k, with probability p k E(X) = x 1 p x k p k, Some useful properties: a The probability of a uio of evets A 1,, A is at most the sum of the probabilities of the evets P (A 1 A ) P (A 1 ) + + P (A ) b If A 1,, A are idepedet evets, the P (A 1 A ) = P (A 1 ) P (A ) c The liearity of expectatio: If X 1,, X are radom variables ad a 1,, a are real umbers, the E(a 1 X a X ) = a 1 E[X 1 ] + + a E[X ] 13

14 A geeral framework for the probabilistic method is the followig: we are give a fiite set of objects A ad f : A R is a fuctio assigig to each object x A a real umber The goal is to show that there is at least oe elemet x A for which f(x) is at least a give value m For this, we defie a probability distributio P : A [0, 1] ad cosider the resultig probability space, where f becomes a radom variable Showig that the expected value of f is at least m is eough, sice, if this holds, the there exists at least oe value x A for which f(x) m Proof of Sperer s theorem usig the the probabilistic method Let F X be a family of subsets of X, where X =, such that A B, A, B F, A B We wat to prove that F ( / ) Recall that ( X, ) is a poset First, let us observe that the umber of maximal chais is! This is because a maximum legth chai has i it exactly oe k-elemet set for each 0 k Assume you start from the empty set There are ways to cotiue the chai to a 1-elemet set (thik of the umber of ways i which we ca add oe elemet) For each oe-elemet set, there are 1 ways to cotiue the chai to a -elemet set, ad so o We choose radomly with equal probability 1/! oe such maximum chai ad we defie a radom variable Y as follows Y = { 1 if F is i the chai Y F, where Y F = 0 otherwise F F F!( F )! We kow that E(Y F ) = P (Y F = 1) = = 1/ (! F ) I order to see this, we just cout the umber of ways ca we prolog a chai passig through a set F above ad below Below, there are F! possibilities, while above there are ( F )! possibilities O the other had, the total umber of chais is!, which gives the statemet above By the liearity of expectatio, we obtai E(Y ) = F F E(Y F ) = F F 1 ( ) F O the other had, a chai ca cotai at most oe elemet of F, so E(Y ) 1 Deotig by m i the umber of i-elemet subsets i F, we obtai that E(Y ) = i=0 m i ( i) 1 Therefore, 1 i=0 which implies F ( / ) m i ( i) i=0 m ( i ) = ( 1 / / ) m i = i=0 ( F ), / Schütte s problem 14

15 Defiitio 39 A touramet is a directed graph obtaied by assigig a directio for each edge i a udirected complete graph I a touramet, every pair of distict vertices is coected by a sigle directed edge Defiitio 40 We say that a touramet T = (V (T ), E(T )) has property S k if for ay k vertices v 1,, v k V (T ), there exist a vertex u V (T ) such that uv 1, uv,, uv k E(T ) Schütte s problem ca be formulated as follows: do such touramets exist, for every k fixed? The aswer is give by the followig theorem: Theorem 41 For every itegers k 1 ad k k (l() + o(1)), there exist a touramet o vertices havig property S k Proof The proof ca be foud i [Juk], pages Theorem 4 Every graph G = (V, E) cotais a bipartite subgraph with at least E / edges Proof The proof ca be foud i [Mat], page 307 Also refer to the followig: [Mat] 10 Probability ad probabilistic proofs [Juk] 3 Probabilistic Coutig 9 Geeratig fuctios ad Fiboacci sequeces Combiatorial applicatios of polyomials Example 1 Cosider the followig combiatorial problem: How may ways are there to pay the amout of 1 fracs if we have 6 oe-fracs cois, 5 two-fracs cois, ad 4 five-fracs cois? The required umber is i fact the umber of solutios of the equatio x 1 + x + x 3 = 1, with x 1 {0, 1,, 3, 4, 5, 6}, x {0,, 4, 6, 8, 10}, ad x 3 {0, 5, 10, 15, 0} I order to compute this, we associate to each variable x i above a polyomial p i as follows: p 1 (x) = 1+x+x +x 3 +x 4 +x 5 +x 6, p (x) = 1+x +x 4 +x 6 +x 8 +x 10, p 3 (x) = 1+x 5 +x 10 +x 15 +x 0 The umber of solutios of the equatio above will be the coefficiet of x 1 i the product p 1 (x) p (x) p 3 (x) Example We prove that k=0 k( k) = 1 By the biomial theorem, we kow that (1 + x) = k=0 xk( k) 15

16 Cosider the derivative of this polyomial This gives ( ) (1 + x) 1 = kx k 1 k Cosiderig x = 1 i the above, we obtai the desired idetity k=1 Example 3 We prove that k=0 ( ) = k ( ) We kow that (1 + x) (1 + x) = (1 + x) Cosider the coefficiet of x i this expressio O the oe had, it is ( ) O the other had, from the left, we obtai that this coefficiet is ( )( ( k=0 k k), ad sice ) ( k = k), we obtai the desired idetity Geeratig fuctio of a sequece Theorem 43 Let a 0, a 1, be a sequece of real umbers If a k c k for every k, where c is a positive real costat, the the series is coverget for all x with x < 1/c a 0 + a 1 x + a x + Defiitio 44 Let (a 0, a 1, ) be a sequece of real umbers The, its geeratig fuctio a(x) is a(x) = a 0 + a 1 x + a x + Example The geeratig fuctio of the sequece (0, 1, 1/, 1/3, ) is a(x) = 0 + x + 1 x x + = l(1 x) Theorem 45 (Geeralized biomial theorem) For every r R ad every iteger k 0, let ( ) r r(r 1)(r k + 1) = k k! The, the followig holds: for every x with x < 1 (1 + x) r = ( ) r + 0 ( ) r x + 1 ( ) r x + Fiboacci sequeces The Fiboacci sequece (F ) 0 is defied by the followig recursive formula: F 0 = 0, F 1 = 1, F = F 1 + F, 16

17 The sum of the first umbers of the Fiboacci sequece, is F k = F + 1 k=0 Aother way to iterpret the Fiboacci sequece is the followig: let F deote the umber of ways i which oe ca climb stairs if allowed to jump oe or two stairs at a time We wat to fid a explicit formula for the value of the -th Fiboacci umber We will preset several possible ways to do that Method 1 We will use the geeratig fuctios Let F (x) deote the geeratig fuctio of the Fiboacci sequece (F 0, F 1, ), that is F (x) = F 0 + F 1 x + F x + F 3 x 3 + Multiplyig F (x) by x, respectively x, we obtai that xf (x) = F 0 x + F 1 x + F x 3 + F 3 x 4 + x F (x) = F 0 x + F 1 x 3 + F x 4 + F 3 x 5 + Recall that for every, we have F = F 1 + F ad cosider F (x) xf (x) x F (x) Groupig together the coefficiets of x k for every k, oe obtais that F (x) xf (x) x F (x) = F 0 +x(f 1 F 0 )+x (F F 1 F 0 )+x 3 (F 3 F F 1 )++x k (F k F k 1 F k )+ This implies F (x) xf (x) x F (x) = x ad thus This meas, the geeral term is F (x) = x 1 x x F = F () (0),! where F () (0) is the value i 0 of the -th derivative of F (X) Method We factor 1 x x as (x x 1 )(x x ), where x 1, = 1± 1+4 This meas F (X) = From this we obtai that x 1 x x = A + x x 1 = 1± 5 B = A(x x ) + B(x x 1 ) x x (1 x x ) A + B = 1 ad Ax + Bx 1 = 0 17

18 This is a system of two equatios with A ad B as ukows, so we ca obtai exact values for A ad B Oe ca obtai that: A = A/x 1 = A ( ) x x = A x x 1 1 x/x 1 x 1 x 1 x +1 1 =0 B = B/x = B ( x x x 1 x/x x This implies that the geeral term F is =0 F = A x +1 1 x B x +1 ) = B =0 x x +1 =0 Usig F 0 = 0, F 1 = 1, we obtai the values of A ad B, we obtai (( ) F = 1 ( ) ) Method 3 Take the above formula ad prove it by iductio Method 4 We look first for a geometric series that satisfies F = F 1 + F, that is F = c α, This implies that cα = cα 1 +cα ad thus α α 1 = 0 Solvig this quadratic equatio, we get α 1, = 1± 5 Therefore, F = c 1 α 1 + c α = c 1 ( 1 + ) ( 5 1 ) 5 + c also satisfies F = F 1 + F for ay c 1, c R O the other had, we ca obtai the values of c 1 ad c from the followig 0 = F 0 = c 1 + c, 1 = F 1 = c Substitutig the values of c 1 ad c i ( F = c 1 ) + c ( 1 ) 5, c we obtai the formula for the geeral term of the Fiboacci sequece Refer as well to the followig: [Mat] 1 Geeratig fuctios [Lov] 4 Fiboacci Numbers 18

19 10 The liear algebra method The oddtow theorem Assume that i a tow there are ihabitats We wat to create m clubs such that every club has a odd umber of members, ad ay two clubs cotai a eve umber of commo members Our goal is to maximize the value of m We will prove that m Formally, we ca state this as the followig theorem: Theorem 46 Let A 1,, A m be distict subsets of {1,, }, so that the cardiality of each A i is odd, ad the cardiality of each A i A j is eve The, m Proof We assig to each set A i its characteristic vector v i, defied as { 1 if j A i v ij = 0 if j / A i We cosider these vectors i F, where F is the fiite field of the residues modulo The goal is to prove that the vectors v i, with 1 i m are liearly idepedet over the field F This will be eough, sice there are at most liearly idepedet vectors over a -dimesioal vector space, ad sice there are m such characteristic vectors, we obtai that m We prove that the vectors v i are liearly idepedet, by cotradictio We suppose the cotrary, so there is a liear combiatio m α iv i = 0 with ot all the coefficiets α i equalig zero Let us observe that, for every i, the value v i = v i, v i is the cardiality of the set A i modulo two, ad the value v i, v j is the cardiality of the itersectio A i A j modulo two for all i j Recall that we are workig over the field F, so sice the cardiality of each set A i is odd, ad the cardiality of every itersectio A i A j is eve, we obtai that v i = 1 ad v i, v j = 0 This meas that we have α i = m m α j v j, v i = α j v j, v i = 0 mod, j=1 j=1 for all 1 i m This cotradicts the assumptio that ot all α i are equal to zero, ad thus the iequality m follows Fisher s iequality Theorem 47 (Fisher s iequality) Let k be a fixed iteger ad A 1, A m {1, } such that A i A j = k, for all i, j {1,, m} with i j The m Proof We prove this theorem usig agai the liear algebra method As before, we prove that the characteristic vectors of the sets A i, for 1 i m are liearly idepedet over R, the field of real umbers We deote by v i the characteristic 19

20 vector of A i, ad we cosider it i R For the sake of cotradictio, we assume that these vectors are ot liearly idepedet, so there exist coefficiets α 1, α m, ot all zero such that m α iv i = 0 Sice the vector m α iv i is zero, its orm is zero, so we obtai that m m m m 0 = α i v i = α i v i, α i v i = αi v i + α i α j v i, v j 1 i j m The value v i equals the cardiality of A i, for each i, ad v i, v j equals the cardiality of the itersectio A i A j (so, it is k) Substitutig i the relatio above, we obtai that 0 = m αi v i + k α i α j = 1 i j m m αi A i + k α i α j = 1 i j m = m αi ( A i k) + k m αi + k 1 i j m α i α j = Now let us observe that 1 i,j m α iα j = ( m α i), so m αi ( A i k) + k 1 i,j m α i α j 0 = ( m m ) αi ( A i k) + k α i Clearly, A i k for all i ad A i = k for at most oe i, sice otherwise the itersectio coditio would ot be satisfied We distiguish two cases: if oe of the sets A i has cardiality k, the the first sum is zero if ad oly if all the α i s are zero, which is a cotradictio Otherwise, oe of the sets has cardiality k (we ca assume it is A 1 ), so all the coefficiets except α 1 are zero If α 1 is also zero, the we are doe, so we ca assume α 1 0 But the the right-had is greater tha 0 This is because the last sum vaishes oly if at least two of the coefficiets α i are ozero, which is agai a cotradictio Therefore, the vectors are liearly idepedet ad thus m, which completes the proof Refer as well to [Mat]: 13 Applicatios of liear algebra, ad [Juk] pages ad Part III - Liear algebra method Refereces [Lov] - Discrete Mathematics (L Lovasz, J Pelika, K Vesztergombi); [Bol] - Combiatorics: Set Systems, Hypergraphs, Families of Vectors ad Combiatorial Probability (B Bollobas); [Mat] - Ivitatio to Discrete Mathematics, (J Matousek, J Nesetril) [Juk] - Extremal combiatorics, (S Juka) 0