Combinatorics. Stephan Wagner

Transcription

1 Combiatorics Stepha Wager Jue 207

2 Chapter Elemetary eumeratio priciples Sequeces Theorem. There are differet sequeces of legth that ca be formed from elemets of a set X cosistig of elemets (elemets are allowed to occur several times i a sequece. Proof: For every elemet of the sequece, we have exactly choices. Therefore, there are } {{... } times differet possibilities. Example. Give a alphabet of letters, there are exactly -letter words. For istace, there are 8 three-digit words (ot ecessarily meaigful that ca be formed from the letters S ad O: SSS, SSO, SOS, OSS, SOO, OSO, OOS, OOO. The umber of 00-letter words over the alphabet A,C,G,T is 4 00, which is a 6-digit umber; DNA strigs as they occur i cells of livig orgaisms are much loger, of course... Permutatios Theorem.2 The umber of possibilities to arrage (distiguishable objects i a row (so-called permutatios is! ( Proof: There are obviously choices for the first positio, the remaiig choices for the secod positio (as opposed to the previous theorem, 2 for the third positio, etc. Therefore, oe obtais the stated formula.

3 CHAPTER. ELEMENTARY ENUMERATION PRINCIPLES 2 Example.2 There are 6 possibilities to arrage the letters A,E,T i a row: AET, ATE, EAT, ETA, TAE, TEA. Remar: By defiitio,! satisfies the equatio! (!, which remais true if oe defies 0! (iformally, there is exactly oe possibility to arrage 0 objects, ad that is to do othig at all. Example.3 I how may ways ca eight roos be placed o a 8 8-chessboard i such a way that o horizotal or vertical row cotais two roos? I order to solve this problem, let us assig coordiates (a-h ad -8 respectively to the squares of the chessboard. A possible cofiguratio would the be a3, b5, c, d8, e6, f2, g4, h7 (for istace. Geerally, there must be exactly oe roo o each vertical row (a-h, ad aalogously oe roo o each horizotal row (-8. Each permutatio of the umbers to 8 correspods to exactly oe feasible cofiguratio (i the above case, , ad so there are exactly 8! possibilities. Sequeces without repetitios Theorem.3 The umber of sequeces of legth whose elemets are tae from a set X comprisig elemets is ( ( 2... ( +! (!. Proof: The proof is essetially the same as for Theorem.2: for the first elemet, there are possible choices, the for the secod elemet, etc. For the last elemet, there are + choices left. Remar: The case of permutatios is clearly a special case of Theorem.3, correspodig to. is called a fallig factorial (read: to the fallig. Choosig a subset Theorem.4 The umber of possibilities to choose a subset of elemets from a set of elemets (the order beig irrelevat is (!!(!. Proof: Let x be the umber of possibilities that we are looig for. Oce the elemets have bee chose (for which there are x possible ways, oe has! possibilities (by Theorem.2 to arrage them i a sequece. Therefore, x! is exactly the umber of possible sequeces of distict elemets, for which we have the formula x!! (!

4 CHAPTER. ELEMENTARY ENUMERATION PRINCIPLES 3 by Theorem.3, so that x is obtaied immediately. Remar: The differece betwee Theorem.3 ad Theorem.4 lies i the fact that the order plays a role i the former, which is does ot i the latter. Each subset correspods to exactly! sequeces: for istace, the subset {A, E, T } of the set {A, B,..., Z} correspods to the sequeces AET, ATE, EAT, ETA, TAE, TEA. The formula for the biomial coefficiet oly maes sese if 0. This is also quite ituitive as o subset ca comprise more elemets tha the origial set. It is ofte useful to defie ( 0 if either < 0 or >. Later we will also give a more geeral defiitio of the biomial coefficiets. Example.4 The umber of six-elemet subsets of {, 2,..., 49} (Lotto is ( 49 49! 6 6! 43! Remar: A obvious property of ( is the idetity ( (, which follows immediately from the formula. However, it also has a combiatorial meaig: choosig elemets is equivalet to ot choosig elemets. The geeralisatio of this priciple leads us to the so-called multiomial coefficiet. Dividig a set ito groups Theorem.5 The umber of possibilities to divide a set X ito groups X, X 2,..., X r whose sizes are prescribed to be, 2,..., r respectively (where r is give by (!, 2,..., r! 2!... r!. Proof: By iductio o r; for r, the statemet is trivial. For r 2, oe has ( choices for the elemets of X, ad by the iductio hypothesis, (! 2! 3!... r! possible ways to divide the remaiig elemets. Therefore, we have exactly ( (! 2! 3!... r!!!(! (! 2! 3!... r!!! 2!... r! possibilities, as claimed.

5 CHAPTER. ELEMENTARY ENUMERATION PRINCIPLES 4 Choosig a multiset Theorem.6 The umber of ways to choose elemets from a set of elemets, repetitios allowed, is ( +. Proof: Let X {x, x 2,..., x } be the set. A choice is characterised by the umber of times that each of the elemets is selected. If l i deotes the multiplicity of x i i our collectio, the the problem is equivalet to determiig the umber of solutios of l + l l, where l, l 2,..., l have to be o-egative itegers. Equivaletly, we ca write m i l i + ad as for the umber of positive iteger solutios to the equatio m + m m +. (. Let us imagie + dots i a row. Each solutio to the equatio (. correspods to a way of separatig the dots by isertig bars at certai places (Figure.. Sice there are + positios for the bars, oe has ( ( + + possible ways to place the bars. Figure.: Dots ad bars. Remar: A fiite sequece m, m 2,..., m of positive itegers summig to (that is, m + m m is called a compositio of ; the above argumet ( dots ad bars shows that every positive iteger has exactly ( compositios ito summads.

6 Chapter 2 A combiatorial view of biomial coefficiets The idetity ( ( is oe of may iterestig properties of biomial coefficiets, may of which ca also be iterpreted combiatorially. I this chapter, some of these properties are cosidered. 2. The recursio Theorem 2. The biomial coefficiets satisfy the recursive formula ( ( ( + (0. Proof: The formula ca be verified easily by algebraic maipulatios: ( ( (! + (!(! + (!!(! (! ( (! +!(!!(! ( (!!!(!!(!. However, a (perhaps simpler way is to argue as follows: let x be a elemet of a - elemet set X. If oe wats to choose a subset of elemets, oe ca first decide whether to iclude x or ot. I the former case, there are ( possibilities to choose the remaiig elemets. I the latter, we have ( possible choices, which already completes the proof. Remar: Note that the equatio remais correct if oe defies ( to be 0 if either < 0 or >. A classical way to illustrate the recursio for the biomial coefficiets is Pascal s triagle: the -th lie cotais the biomial coefficiets ( (0. 5

7 CHAPTER 2. A COMBINATORIAL VIEW OF BINOMIAL COEFFICIENTS It is apparet that each umber is the sum of the two umbers above it. If oe defies the biomial coefficiet ( α for arbitrary real (or eve complex umbers α by ( α : α(α (α 2... (α +,! Theorem 2. remais correct, eve without the direct combiatorial iterpretatio. The followig lemma relates the case of egative α to the more familiar case that α is positive: Lemma 2.2 For ay real umber α ad ay o-egative iteger, oe has ( ( α α + (. Proof: ( α α( α ( α 2... ( α +! ( α(α + (α (α +! ( α + (. 2.2 The biomial theorem revisited The biomial theorem is certaily the most importat theorem that ivolves the biomial coefficiets; it ca be stated as follows: Theorem 2.3 (Biomial Theorem For ay iteger 0, oe has (x + y ( x y. Proof: The classical proof proceeds by iductio. However, it ca also be prove combiatorially: suppose that we expad the product (x + y (x + y(x + y... (x + y x x... x + y x... x +....

8 CHAPTER 2. A COMBINATORIAL VIEW OF BINOMIAL COEFFICIENTS 7 Each summad is obtaied by choosig either x or y for each of the factors. Therefore, we ed ( up with summads of the form x y, ad each of these summads occurs exactly times (which is the umber of ways to select out of factors from which a x is tae. This readily proves the theorem. Remar: It should be oted that oe usually defies 0 0 i this cotext, so that the biomial theorem remais correct eve if y 0 (ad/or x 0. The biomial theorem has a geeralisatio to the case that is ot ecessarily a positive iteger, which is ow as the biomial series: ( α ( + x α x. (2. Note that this is simply the Taylor series of ( + x α at x 0. If α 0 is a iteger, the this formula reduces to the biomial theorem, sice all terms i the ifiite sum that correspod to > are 0 by defiitio. It is possible to deduce several iterestig idetities directly from the biomial theorem, such as the followig: Theorem 2.4 Oe has the followig relatios: ( ( ( ( for ay iteger 0, ( ( for ay iteger > 0, for ay iteger > 0. eve ( 0 ( ( ( ( 0 odd ( 2 Proof: The first equatio is obtaied by settig x y i the biomial theorem, the secod equatio for x ad y. The third equatio is obtaied by addig (subtractig the first two equatios ad otig that holds. + ( { 2 eve 0 odd resp. ( { 0 eve 2 odd Remar: Let us iterpret these equatios combiatorially: the sum i the first equatio is exactly the umber of subsets of a -elemet set (sice ( couts subsets of size. Therefore, oe immediately eds up with the followig result:

9 CHAPTER 2. A COMBINATORIAL VIEW OF BINOMIAL COEFFICIENTS 8 Corollary 2.5 A -elemet set has precisely 2 distict subsets. Example 2. The set {A,E,T} has 8 subsets: {}, {A}, {E}, {T}, {A,E}, {A,T}, {E,T}, {A,E,T}. Of course, this corollary ca be obtaied without depedig o the biomial theorem: each elemet ca either be selected as a member of the subset or ot, so that oe has exactly possible optios. I this way, a subset of a -elemet set correspods directly to a 0--sequece of legth. O the other had, it follows from Theorem 2.4 that ay fiite o-empty set has exactly as may subsets of eve cardiality as of odd cardiality. This is somewhat obvious if is eve (ote the symmetry i Pascal s triagle!, but maybe a little more surprisig if is odd. However, there is agai a combiatorial proof: If we distiguish a elemet x of the set, the oe has a bijectio betwee subsets of eve cardiality ad subsets of odd cardiality: two subsets are associated to each other if their oly differece is that oe of them cotais x while the other oe does ot. I this way, we obtai 2 pairs, each of which cotais exactly oe subset of eve cardiality ad oe subset of odd cardiality. Istead of cosiderig eve or odd cardialities, let us see what happes if we restrict the cardiality of subsets to be divisible by 3: Example 2.2 We prove the followig equatio for all positive itegers : ( 3 ( (. 3 3 I order to achieve this, we would lie to use the same tric as before. Istead of pluggig i y ±, we cosider the roots of uity Oe has ζ 3 ζ 3 2 as well as ζ,2 ± i 3 2 e ±2πi/3. ζ + ζ 2 + 0, ζ 2 ζ 2 ad ζ 2 2 ζ. If we ow tae x ad y, y ζ ad y ζ 2 respectively i the biomial theorem ad add the resultig equatios, we obtai 3 l0 Now if l 3 is divisible by 3, the ( 3 ( + ζ l + ζ l l ( + ζ 3 + ( + ζ ζ l + ζ l ,

10 CHAPTER 2. A COMBINATORIAL VIEW OF BINOMIAL COEFFICIENTS 9 if l 3 +, the ad if l 3 + 2, the + ζ l + ζ l 2 + ζ + ζ 2 + ζ + ζ 2 0, + ζ l + ζ l 2 + ζ 2 + ζ + ζ 2 + ζ 0. Therefore, we ed up with ( ( + ζ 3 + ( + ζ ζ,2 ± 3i are sixth roots of uity: it is easy to chec that ( + ζ 2,2 3 (ad thus ( + ζ,2 6. So it fially follows that ( 3 ( (. 3 3 This shows that the proportio of subsets whose cardiality is a multiple of 3 is approximately (it teds to as. Let us ow have a loo at aother importat idetity 3 3 that ivolves the biomial coefficiets. 2.3 The Vadermode idetity Theorem 2.6 (Vadermode idetity For itegers N, M, 0, oe has ( ( ( N M N + M. Proof: Sums of the form a b ca be related to products of polyomials: ideed, if two polyomials A(x N a x ad B(x M l0 b lx l are give, the the product of the two is A(x B(x N M a b l x +l. The coefficiet of x is ow obtaied as the sum of those summads for which + l, or i other words l. Therefore this coefficiet is equal to a b, l0

11 CHAPTER 2. A COMBINATORIAL VIEW OF BINOMIAL COEFFICIENTS 0 where a is tae to be 0 if the degree N of the polyomial A is less tha (ad aalogously b l 0, if l > M. Similar ideas will be used extesively i Chapter 5. I this specific case, we apply the biomial theorem to the polyomial ( + x N+M that ca also be writte as the product of ( + x N ad ( + x M : ( N ( ( N M ( M ( + x N+M ( + x N ( + x M x x l l l0 N M ( ( N+M N M ( ( N M x +l x. l Comparig coefficiets with we obtai the desired idetity. l0 ( + x N+M N+M ( N + M x, Oce agai it is possible to provide a proof by coutig argumets as well: a set of N + M elemets is divided ito two groups cosistig of N ad M elemets respectively. Choosig elemets from the set is equivalet to choosig elemets from the first group ad the remaiig elemets from the secod group, where ca be ay iteger betwee 0 ad. The umber of possible choices for fixed is clearly ( ( N M. Summig over all, we obtai the total umber of choices, which is ( N+M. The followig special case of the Vadermode idetity is quite remarable: Corollary 2.7 ( 2 I a similar vei, we ca prove the followig: ( ( ( 2. Theorem 2.8 For a o-egative iteger, oe has ( { 2 ( ( ( /2 eve, /2 0 odd. Proof: We use the same type of argumet as before, comparig coefficiets i the equatio ( ( ( ( ( x ( x (+x ( x ( x 2 ( x 2.

12 CHAPTER 2. A COMBINATORIAL VIEW OF BINOMIAL COEFFICIENTS The coefficiet of x o the right had side is ( /2( /2 if is eve, ad otherwise 0. The coefficiet o the left had side obtaied by expadig the product is ( ( ( ( 2 (.

13 Chapter 3 The priciple of iclusio ad exclusio 3. A simple example A frequetly occurrig problem is to determie the size of the uio or itersectio of a umber of sets, as i the followig example: Example 3. All secod-year sciece studets may choose either mathematics, or physics, or both. The mathematics course is atteded by 50 studets, the physics course by 30 studets. 5 studets atted both courses. How may secod-year sciece studets are there? Let M be the set of studets taig mathematics ad P the set of all studets who tae physics. By our coditios, the set of all studets is the uio S M P. If we add the sizes of the two sets, all studets who atted both courses are couted twice. Therefore, we have to subtract the size of M P, which yields the formula S M P M + P M P, where X is the umber of elemets of a set X studets. Pluggig i, we fid that there are This priciple ca be geeralised to uios (or itersectios of a arbitrary umber of sets. Before we discuss the geeral formula, let us exted Example 3.: Example 3.2 Third-year sciece studets also have the opportuity to atted chemistry, but every studet has to tae at least oe of the three courses. Altogether, there are 40 studets i the mathematics class, 25 who atted physics, ad 20 who atted chemistry. Furthermore, we ow that 0 studets do both mathematics ad physics, 8 both mathematics ad chemistry, ad 7 physics ad chemistry. There are two particularly ee studets who atted all three courses. How may third-year sciece studets are there? 2

14 CHAPTER 3. THE PRINCIPLE OF INCLUSION AND EXCLUSION 3 Let M, P, C deote the respective sets of studets attedig mathematics, physics ad chemistry. Oce agai, we are looig for the size of the uio M P C. To this ed, we first add M, P ad C. Studets taig both mathematics ad physics are double-couted, so we have to subtract M P. The same applies to M C ad P C. However, those two studets that atted all the three courses are added three times ad subtracted three times as well. Hece we have to add M P C to mae up for this. Formally, we have M P C M + P + C M P M C P C + M P C. This meas that there are third-year studets. 3.2 The geeral formula The simple example preseted i the previous sectio ca be geeralized to a arbitrary umber of sets as follows: Theorem 3. (Iclusio-exclusio Let X, X 2,..., X be arbitrary fiite sets. For a set I {, 2,..., }, we deote the itersectio of all sets X i with i I by i I X i. The the followig formula holds: X i ( I + X i, (3. i I {,...,} I where the sum is tae over all o-empty subsets of {, 2,..., }. Remar: The cases 2 ad 3 correspod to our two examples; for istace, the formula X X 2 X 3 X + X 2 + X 3 X X 2 X X 3 X 2 X 3 + X X 2 X 3 is obtaied for 3. Proof: By iductio o. For, the formula reduces to the trivial idetity X X. The iductio step from to + maes use of the special case 2 that was discussed i our first example: ( + M i M i M + i i ( M i + M + M i M + i i M i + M + (M i M + i i i I

15 CHAPTER 3. THE PRINCIPLE OF INCLUSION AND EXCLUSION 4 I {,...,} I I {,...,} I I {,...,+} + I,I I {,...,+} I This completes the iductio. ( I + M i + M + i I ( I + M i + M + i I ( I + M i + i I ( I + M i. i I I {,...,+} + I I {,...,} I I {,...,} I ( I + (M i M + ( I + i I ( I + M i i I i I {+} Remar: The same formula holds true if uio ad itersectio are iterchaged (the proof beig completely aalogous: X i ( I + X i. (3.2 i I {,...,} I Remar: A commo iterpretatio of the iclusio-exclusio priciple ivolves probabilities: suppose that X, X 2,..., X are subsets of a set X of outcomes. The P (X i X i X is the probability that oe of the evets i X i occurs. Dividig formulas (3. ad (3.2 by X, we obtai ( P X i ( ( I + P X i ad i I {,...,} I ( P X i i I {,...,} I i I i I ( ( I + P X i. ( Note that P i X i is the probability that at least oe of the evets associated to ( X, X 2,... occurs, while P is the probability that all of them occur. i X i Remar: If we tae X X 2... X {x} i (3. or (3.2, the all itersectios ad uios have size, so that we obtai aother proof of the idetity ( ( ( ( ( ( 0, 0 i I M i

16 CHAPTER 3. THE PRINCIPLE OF INCLUSION AND EXCLUSION 5 see Theorem 2.4. Quite frequetly, the sets X i are subsets of some base set, ad oe is iterested i the umber of elemets that are cotaied i oe of the X i or ot i all of the X i. I the former case, oe has to determie the cardiality of X \ (X X 2... X, which is equal to X ( I + X i. (3.3 I {,...,},I The secod problem amouts to determiig the cardiality of X \ (X X 2... X, which is give by X ( I + X i. (3.4 I {,...,},I Naturally, the sizes of the sets i I X i are ot always explicitly give as i our first two examples. However, they are ofte easier to obtai tha those that oe is actually iterested i. I the followig sectio, some applicatios are discussed. i I i I 3.3 Applicatios Example 3.3 How may -digit umbers are there that do ot cotai the digits 0,, 2, but have to cotai the three digits 3, 4, 5? By the stated coditios, the digits have to be tae from the set {3, 4,..., 9}; let M (D be the umber of all -digit umbers whose digits are tae from the set D. The the set whose size we wat to determie cosists of all umbers i M ({3, 4,..., 9} that are ot cotaied i ay of M ({4, 5,..., 9}, M ({3, 5,..., 9} or M ({3, 4, 6,..., 9} (oe of the digits 3, 4, 5 may be missig. Note also that M (D M (D 2 M (D D 2. By Theorem., we have M (D D (if 0 D; otherwise, oe would have to exclude leadig zeros. Therefore, the iclusio-exclusio priciple yields M ({3, 4,..., 9} \ (M ({4, 5,..., 9} M ({3, 5,..., 9} M ({3, 4, 6,..., 9} M ({3, 4,..., 9} M ({4, 5,..., 9} M ({3, 5,..., 9} M ({3, 4, 6,..., 9} + M ({5, 6,..., 9} + M ({4, 6,..., 9} + M ({3, 6,..., 9} M ({6, 7,..., 9} Euler s totiet fuctio Example 3.4 A classical applicatio of the iclusio-exclusio priciple stems from umber theory: Euler s totiet fuctio ϕ( is defied as the umber of itegers x such that

17 CHAPTER 3. THE PRINCIPLE OF INCLUSION AND EXCLUSION 6 0 x < ad x ad do ot have a commo divisor (other tha. If p, p 2,..., p are the prime factors of, the this occurs precisely if x is ot divisible by ay of p, p 2,..., p. Note ow that the umber of itegers 0 x < that are divisible by a specific prime factor p i is exactly /p i. Liewise, if I is ay subset of {, 2,..., }, the the umber of itegers 0 x < that are divisible by all p i with i I (ad thus by their product i I p i is exactly / i I p i. Therefore, formula (3.3 yields ϕ( + I {,...,},I ( I + I {,...,},I ( I i I ( pi i i I p i The last step follows from the fact that the term i I p i occurs i the expasio of ( ( p p2... ( p p i with a coefficiet of ( I. Geerally, the product ( x i ( x ( x 2... ( x i expads to x i, I {,...,}( I i I where the product over the empty set is tae to be, while the product ( + x i ( + x ( + x 2... ( + x i is simply x i. I {,...,} i I Deragemets Example 3.5 A deragemet is a permutatio of {, 2,..., } without fixed poits; that is, the umber i does ot occur i positio i ( i. For istace, 3425 is a deragemet, but 3245 is ot (sice 2 occurs i secod positio. The umber of deragemets

18 CHAPTER 3. THE PRINCIPLE OF INCLUSION AND EXCLUSION 7 ca be determied by the iclusio-exclusio priciple: for a subset I of {, 2,..., }, let p (I be the umber of permutatios for which all elemets of I (ad possible others are fixed poits, i.e., they occur i their respective positios. The, we are looig for the umber of permutatios that belog to oe of p ({}, p ({2},.... The size of p (I is clearly ( I! (the I remaiig elemets ca be arraged i ay order, so that (3.3 yields the followig formula for the umber of deragemets:! ( I + p (I. I {,...,},I There are exactly ( subsets I of size I, ad so this reduces to (! + ( (!! + (!!! (!. Therefore, the probability that a radomly selected permutatio of {, 2,..., } is a deragemet is (! ±!. As, this value approaches the ifiite sum (! e

19 Chapter 4 Coutig by recursio May coutig problems ca be solved by settig up recursios ad solvig them. We have already ecoutered the recursive relatio that is satisfied by the biomial coefficiets. I the followig, we discuss three examples that exhibit the basic idea. The followig chapter provides a geeral method to solve recursios as they occur i the study of eumeratio problems. Example 4. We draw lies i the plae i such a way that there are o parallel lies ad o itersectios of three or more lies. These lies divide the plae ito several regios; how may such regios are there? Let us first discuss some trivial cases: if there is o lie ( 0, the there is precisely oe regio; for, there are two regios; for 2 ad 3 we obtai four ad seve regios, respectively. Oe might cojecture ow that the umber of regios icreases by + if we add a lie to existig lies. Ideed, this is the case, as the followig argumet shows. If we add a additioal lie to lies i the plae, the we obtai exactly ew poits of itersectio that divide the ew lie ito + segmets. Each of these segmets divides oe of the old regios ito two ew regios, while all other regios remai the same. Therefore, if a deotes the total umber of regios, we have the recursio a + a + ( +. A explicit form ca be deduced immediately: a a + a ( a ( + ( 2... a ( + ( a 0 + ( by the well-ow formula for the sum , which solves the problem. Example 4.2 Tom gets a allowace of R00 every moth, which he speds etirely o ice cream (R5, chocolate (R0 or cooies (R0. Every day, he buys exactly oe of these 8

20 CHAPTER 4. COUNTING BY RECURSION 9 util he rus out of moey. I how may possible ways ca Tom sped his moey? His older brother Phil gets a allowace of R50, which he also speds o ice cream, chocolate ad cooies oly. How may possible ways does he have? We study a more geeral problem: if the allowace is 5 ( ay positive iteger; ote that the prices are all divisible by R5, how may ways are there? Let this umber be deoted by a ; the it is easy to see that a ad a 2 3. Next we deduce a recursio for a : the first thig that Tom buys ca be either ice cream (so that he is left with 5( ad has a possibilities for the rest or chocolate or cooies (i each of these cases, the remaiig amout is 5( 2, so that he is left with a 2 possibilities. This shows that a a + 2a 2 holds. Note that this eve remais true if we set a 0 (without moey, there is oly oe optio, which is to buy othig at all. Oe obtais the followig sequece: A explicit formula is give by a a 3 ( (, which ca be prove by meas of iductio. A method to determie such a explicit formula from a recursio will be discussed i the followig chapter. Pluggig i 20 ad 30 respectively, we fid that Tom has optios, while Phil has differet ways to sped his moey. Our fial example leads to the famous Fiboacci umbers: Example 4.3 All the houses o oe side of a certai street are to be paited either yellow or red. I how may ways ca this be doe if there are houses ad there may ot be two red houses ext to each other? Let a be the umber that we wat to determie. We distiguish two cases: If the first house is paited yellow, the the remaiig houses ca be paited i ay of the feasible a ways, the first house ca be eglected. If the first house is paited red, the the secod house must be paited yellow. Usig the same argumet as before, we see that there are a 2 possibilities for the remaiig houses. Hece we have the recursio a a + a 2 with iitial values a 2 ad a 2 3. This yields the sequece 2, 3, 5, 8, 3, 2, 34, 55, 89,... of so-called Fiboacci umbers. They are usually defied by f 0 0, f ad f f + f 2 ; it is easy to see that a f +2 i our example. There is also a explicit formula for the Fiboacci umbers, see the followig chapter.

21 Chapter 5 Geeratig fuctios 5. Solvig recursios Geeratig fuctios provide a method to solve recursios, but they are actually much more versatile, as we will see later. We associate a power series A(x a x to ay sequece a 0, a,... of real (or complex umbers; i the cotext of combiatorics, the coefficiets a are typically itegers. The -th coefficiet of such a power series A(x is also writte as a [x ]A(x. For ow, we do ot care too much about covergece ad mostly regard geeratig fuctios as formal objects. A typical example of a geeratig fuctio is the geometric series qx q x, which is the geeratig fuctio of a geometric sequece, q, q 2, q 3,.... I particular, the geeratig fuctio of,,,... is. As a first example of the use of geeratig x fuctios, let us discuss the recursio a a + 2a 2 with iitial vales a 0 a that we ecoutered i the previous chapter. If A(x deotes the associated geeratig fuctio, the we obtai A(x a x a 0 + a x + + x + 2 (a + 2a 2 x 2 a x + 2 a 2 x 20 2

22 CHAPTER 5. GENERATING FUNCTIONS 2 Solvig for A(x, we fid which ca also be writte as + x + a x a x +2 + x + x(a(x a 0 + 2x 2 A(x. A(x x 2x 2, A(x 2/3 2x + /3 + x, maig use of partial fractios. We expad the two summads ito geometric series to obtai A(x 2 2 x + ( x ( x, so that we ca simply read off the coefficiets: a [x ]A(x 2+ +(, as claimed. I 3 the very same style, oe fids Biet s formula for the Fiboacci umbers, whose geeratig fuctio is x : x x 2 f 5 (( + ( (5. for ay 0. The method ca actually be geeralised to the etire class of liear recursios, whose geeral form is a c a + c 2 a c r a r. To see how such a recursio ca be solved for arbitrary coefficiets c, c 2,..., c r, we eed the followig lemma: Lemma 5. The power series associated to the fuctio ( qx ( qx ( + q x. Proof: Combie the biomial series (2. with Lemma 2.2 to obtai ( qx ( ( qx ( ( + ( qx is give by Note that for fixed, ( + is a polyomial (i of degree : ( + ( + ( ( +, (! ( + q x.

23 CHAPTER 5. GENERATING FUNCTIONS 22 for istace ( + +, ( ( + 2( ,... Now we are ready to prove the followig theorem: Theorem 5.2 The geeral solutio of a liear recursio of the form is give by a c a + c 2 a c r a r a s P i (qi, i where q, q 2,..., q s are the (possibly complex solutios of the characteristic equatio q r c q r + c 2 q r c r ad P (, P 2 (,..., P s ( are polyomials. The degree of P i is strictly less tha the multiplicity of q i as a solutio of the characteristic equatio (that is, the umber of times the factor q q i occurs i the factorisatio of the polyomial q r c q r c 2 q r 2... c r. Proof: Let A(x a x be the geeratig fuctio of the sequece a 0, a,.... The we have r A(x a x + r a x + r a x + r a x + r a x + r a x + a x r (c a + c 2 a c r a r x r r j r j c j c j r r j a j x a x +j ( r c j a x +j j r c j x j A(x j r j r j r j c j a x +j a x +j. Now write r N(x a x r j r j c j a x +j

24 CHAPTER 5. GENERATING FUNCTIONS 23 ad D(x r c j x j. Both N(x ad D(x are polyomials; solvig the equatio for A(x gives A(x N(x D(x, so that A(x is ivariably a ratioal fuctio. Now factor the deomiator: j D(x ( q x l ( q 2 x l 2... ( q s x ls Note that a factor ( q i x l i i the factorisatio occurs if ad oly if D( q i 0, which is equivalet to r r 0 c j q j i q (q r r c j q r j i, j i.e., q i is a solutio of the characteristic equatio. The multiplicity of q i as a solutio is precisely the expoet l i. Now expad the geeratig fuctio A(x ito partial fractios: j a x A(x s l i K ij ( q i x. j i j We ca apply Lemma 5. to each of the summads to obtai a x ad comparig coefficiets yields s l i i j K ij ( + j qi x, j a s i ( li j ( + j K ij qi j s P i (qi i for certai polyomials P i whose degree is less tha l i (the highest degree that occurs is that of ( +l i l i, which is li. This proves the theorem. Kowig the geeral form of a solutio, oe ca solve a recursio by meas of the method of udetermied coefficiets without traslatig it to the world of geeratig fuctios; this is exhibited i the followig example: Example 5. Suppose we wat to determie a explicit formula for the sequece that is defied by a 5a 3a 2 9a 3 with iitial values a 0, a ad a 2 5.

25 CHAPTER 5. GENERATING FUNCTIONS 24 The characteristic equatio is q 3 5q 2 + 3q + 9 (q 3 2 (q + 0, so that q 3 ad q 2, the multiplicities beig 2 ad respectively. Therefore, Theorem 5.2 shows that the solutio must have the form a (A + B3 + C(. Pluggig i 0,, 2 yields the system of equatios B + C, 3A + 3B C, 8A + 9B + C 5, which leads to the solutio A, B 3, C 5, so that fially a 4A (. 8 Theorem 5.2 actually oly treats the homogeeous case; the method ca be exteded to o-homogeeous recursios of the form a c a + c 2 a c r a r + b, the oly differece beig the o-homogeeous term b. geeratig fuctio of b, the oe obtais If B(x b x is the A(x N(x + B(x D(x as i the proof of Theorem 5.2. If the o-homogeous term is of the form Q(q for a polyomial Q, the oe ca deduce a explicit formula for a oce agai (the proof beig similar to that of Theorem 5.2: Theorem 5.3 Suppose that b Q(q, where Q is a polyomial of degree d. If q is ot a solutio of the characteristic equatio, the the solutio of the liear recursio is of the form a c a + c 2 a c r a r + b a s P i (qi + P (q, i where q, q 2,..., q s are the (possibly complex solutios of the characteristic equatio q r c q r + c 2 q r c r ad P (, P 2 (,..., P s (, P ( are polyomials. The degree of P i is strictly less tha the multiplicity of q i as a solutio of the characteristic equatio, ad the degree of P ( is d.

26 CHAPTER 5. GENERATING FUNCTIONS 25 If, o the other had, q is a solutio of the characteristic equatio (without loss of geerality, q q, the the solutio has the form a s P i (qi, i q, q 2,..., q s ad P (, P 2 (,..., P s ( as before, except for the degree of P, which is equal to the multiplicity of q plus the degree of Q. A sum of o-homogeeous terms of this form gives rise to a sum i the solutio; let us discuss a modified versio of Example 5. to exhibit the method: Example 5.2 Suppose we wat to determie a explicit formula for the sequece that is defied by a 5a 3a 2 9a with iitial values a 0, a ad a 2 5. The summad ca be iterpreted as (4 + 4 ad gives rise to a liear term A+B i the solutio; sice 3 is a double solutio of the characteristic equatio ad 4 3 occurs i the o-homogeeous term, we must have a summad (C 2 + D + E 3 i the solutio; fially, is a solutio of the characteristic equatio, so that we ed up with Pluggig ito the recursio yields a A + B + (C 2 + D + E 3 + F (. A + B + (C 2 + D + E 3 + F ( 5 ( A( + B + (C( 2 + D( + E 3 + F ( 3 ( A( 2 + B + (C( D( 2 + E F ( 2 9 ( A( 3 + B + (C( D( 3 + E F ( ad after collectig terms ( 8 3 C (8A 4 + ( 28A + 8B 4 0. Therefore, we have A, B 9 ad C 3. Fially, the iitial values ca be used to determie D, E, F, ad the solutio is foud to be a ( ( (.

27 CHAPTER 5. GENERATING FUNCTIONS 26 Remar: The parts that belog to a solutio of the homogeeous equatio ca be left out i the first step (i the example above, this meas that oe simply plugs i A+B +C 2 3 ito the recursio formula, sice the other parts have to cacel ayway. This priciple ca be stated as follows: the geeral solutio of a o-homogeeous recursio is the sum of a particular solutio ad the solutio to the homogeeous equatio (compare this to liear differetial equatios!. 5.2 Geeral rules for geeratig fuctios Before we tur to more advaced applicatios of geeratig fuctios, let us describe the effect that certai operatios o sequeces have o the geeratig fuctio level. Theorem 5.4 If {a } ad {b } ( 0 are sequeces ad A(x ad B(x their associated geeratig fuctios, the. the sequece c a + b has geeratig fuctio C(x A(x + B(x, 2. the sequece c a b has geeratig fuctio C(x A(x B(x, 3. the sequece c αa has geeratig fuctio C(x αa(x (α ay costat, 4. the sequece c { a m m, 0 otherwise, has geeratig fuctio C(x x m A(x, 5. the sequece c a +m has geeratig fuctio C(x A(x m a x x m, 6. the sequece c a has geeratig fuctio C(x xa (x, 7. the sequece c { a > 0, 0 otherwise, has geeratig fuctio C(x x 0 A(t a 0 t dt, 8. the sequece c a has geeratig fuctio C(x A(x x. Proof: Each of the statemets ca be obtaied by simple arithmetic:

28 CHAPTER 5. GENERATING FUNCTIONS C(x C(x C(x (a + b x C(x a b x a x + b x A(x + B(x. a b x a x b x ( ( a x b l x l a x b l x l A(x B(x. l0 C(x a m x m a +m x l0 αa x α a x αa(x. a x +m x m a x m x m m m C(x a x x a x d x a dx x x d dx C(x a x C(x x a t dt a x 0 a x x x 0 a x x m A(x. a x A(x m a x x m. a t dt a x a x a a x xa (x. x 0 x + x A(x x. A(x a 0 t dt.

29 CHAPTER 5. GENERATING FUNCTIONS 28 Remar: Geeratig fuctios ca be regarded as formal objects without cosiderig covergece. The sum, differece, product or quotiet of two power series is a power series agai (for the quotiet, oe has to assume that the deomiator has a o-zero costat coefficiet. For istace, oe ca formally multiply (a 0 +a x+a 2 x (b 0 +b x+b 2 x a 0 b 0 +(a b 0 +a 0 b x+(a 0 b 2 +a b +a 2 b 0 x or divide a 0 + a x + a 2 x b 0 + b x + b 2 x c 0 + c x + c 2 x , where the coefficiets c 0, c, c 2,... ca be foud by comparig coefficiets i the idetity (b 0 + b x + b 2 x (c 0 + c x + c 2 x a 0 + a x + a 2 x , so that c 0 a 0 b 0, c a b 0 a 0 b, etc. b 2 0 A applicatio of rule 2. for products of geeratig fuctios was give i Sectio 2.3 (Vadermode idetity; let us ow discuss a applicatio of the last rule for cumulative sums, which is a extesio of Example 4.2: Example 5.3 Tom decides that he might ot sped his etire allowace of R00 ad perhaps save some part of it istead. How may ways does he have ow to sped his moey? I this settig, Tom ca sped othig, or R5, R0,..., R00. Geerally, if his allowace is 5 ( ay o-egative iteger, the the amout that he speds is of the form 5, 0 ; if a deotes the umber of ways to sped exactly 5 Rad, the the umber of ways to sped at most 5 is exactly c a. Recall that the geeratig fuctio of a is A(x. Therefore, Theorem 5.4 shows x 2x 2 that the geeratig fuctio of c is C(x. Maig use of ( x( x x 2 ( x(+x( 2x partial fractios oce agai, we fid C(x 4 3( 2x 2( x + 6( + x ad thus c [x ]C(x (. I particular, if 20 (which correspods to a amout of R00, the this umber is 3980.

30 CHAPTER 5. GENERATING FUNCTIONS Noliear recursios While liear recursios are usually solved more quicly by meas of the method of udetermied coefficiets, geeratig fuctios are actually far more versatile ad ca be applied to may o-liear recursios as well (which the method of udetermied coefficiets ca ot. I this sectio, we treat two such examples. Example 5.4 We wat to cout the umber of stacs that ca be formed from cotiguous rows of cois i such a way that (from the secod row o every coi touches the two cois below it, where the umber of cois i the first row is (see Figure 5., which illustrates the case 3. Figure 5.: Differet stacs i the case 3. Let a deote the umber of cofiguratios with cois i the bottom row. Each such cofiguratio that is ot just a sigle row of cois is obtaied by placig a cofiguratio whose bottom row cosists of cois i ay of possible positios (there are available positios for cois altogether, so the first of the cois ca be i either the first, or the secod,..., or the ( -th of these positios, where ca be ay umber betwee ad. This results i the recursio a + ( a for > (with the iitial value a. For coveiece, we set a 0 0 ad rewrite this recursio as a + ( a. Let A(x be the geeratig fuctio of the sequece a ; ote that ( a is the -th coefficiet i the product of A(x ad x x d dx x x d dx x x ( x. 2

31 CHAPTER 5. GENERATING FUNCTIONS 30 Therefore, we obtai the equatio A(x x x + ( x A(x x 2 x + x ( x A(x. 2 Solvig for A(x, we fid x( x A(x 3x + x. 2 If oe computes the first few elemets of the sequece, oe fids a 2 2, a 3 5, a 4 3, a 6 34,.... Oe otices that all of these umbers are Fiboacci umbers. Ideed, oe ca show that a f 2, where f deotes the -th Fiboacci umber as i Example 4.3: f 0 0, f,.... Oe possibility to prove the idetity is to determie a explicit formula for a from the geeratig fuctio (by meas of partial fractios ad compare it to Biet s formula (5.. However, we will go aother way: if F (x f x x x x 2 deotes the geeratig fuctio for the Fiboacci umbers, the we fid the geeratig fuctio for the odd-idexed Fiboacci umbers by the same method that was also applied i the proof of Theorem 2.4: f x odd ( ( f x f x 2 2 f ( x which shows that f 2 x 2 ( x 2 x x x 2 + x x 2 thus ad fially f 2 x 2 x2 ( x 2 3x 2 + x 4 f 2 x x( x 3x + x 2 x( x2 3x 2 + x 4, F (x F ( x, 2 upo replacig x 2 by x, which is exactly the geeratig fuctio A(x that we foud before. The followig example itroduces the Catala umbers, which will be treated i more detail i Sectio 7.. Example 5.5 Cosider 2 poits o a circle. How may ways are there to coect them by lies such that there are o poits of itersectio (ad each poit is coected to exactly oe other poit? Figure 5.2 shows all cofiguratios i the case 3 (thi of 2 people shaig hads.

32 CHAPTER 5. GENERATING FUNCTIONS 3 Figure 5.2: Coectig six poits o a circle. Let the poits be umbered from to 2. If poit is coected to poit, this leaves two groups of sizes 2 ad that caot be coected ay more; therefore, 2l must be eve. Now each of the two groups ca be treated separately. If a deotes the umber of possibilities, the we obtai the recursio a a l a l, l the iitial value beig a 0. Oce agai, we wat to traslate this to a fuctioal equatio for the geeratig fuctio A(x. Note that a l a l l m0 a m a m is exactly the coefficiet of x i A(x 2, which is the coefficiet of x i xa(x 2. Therefore, A(x xa(x 2 +, where the last summad taes the iitial value ito accout. Solvig the quadratic equatio, we fid A(x 4x. 2x The egative sig has to be chose to mae sure that A(0 a 0, as it should be. How ca this be tured ito a formula for the coefficiets a? Recall the biomial series (2. that we ca apply ow (i the specific case α : 2 A(x 4x 2x 2x 2x ( 4x/2 2x 2x ( /2 ( 4x

33 CHAPTER 5. GENERATING FUNCTIONS 32 ( /2 ( 4x 2x 2 ( /2 ( 4 m+ x m, 2 m + m0 which shows that a [x ]A(x 2 follows: 2 ( /2 ( 4 x ( /2 + ( 4 + ; however, this ca be simplified as ( /2 ( 4 + (/2 ( /2 ( 3/2 (/2 ( ( +! 2 2 ( 3 (2 ( 4 + ( +! 2 3 (2 2 (2! ( +! ( +! (2 2 (2! ( +! 2 2 (2! ( +!! + ( 2. The umbers +( 2 are ow as Catala umbers; the first few elemets of the sequece are a, a 2 2, a 3 5, a 4 4.

34 Chapter 6 The symbolic method The detour via recursios is ofte ot ecessary if oe wats to obtai the geeratig fuctio for a certai family of combiatorial objects; the symbolic method that will be preseted i this chapter wors for may importat combiatorial structures, such as words, permutatios, compositios, trees, ad may others, that also play a frequet role i computer sciece ad other scieces. 6. Ulabelled structures We cosider combiatorial structures that are made up of certai atoms. The size of such a object is defied as the umber of its atoms. For istace, the atoms of a word over a give alphabet are its letters, ad the size is its legth. For a tree (a commo data structure, see Figure 6., a ode is a atom, ad the size is the umber of odes, ad there are may other examples. If there are a elemets of size i a certai family A of combiatorial objects, the the associated geeratig fuctio is A(x a x. Figure 6.: A simple biary tree. 33

35 CHAPTER 6. THE SYMBOLIC METHOD 34 Example 6. Let us start with a trivial example: piles of cois. The atoms are the cois, ad the size of a pile is its height (the umber of cois; if the cois are idistiguishable, the there is exactly oe pile of every o-egative iteger size (if we cosider the empty pile of size 0 as well, so that the family P of piles ca be described as P {,,,,...}, where deotes a atom (a sigle coi, ad the associated geeratig fuctio is P (x x x. There are certai atural trasformatios that ca be performed o families of combiatorial objects: Uios If A ad B are disjoit families of combiatorial objects, the their uio A B cosists of all objects that are elemets of either A or B, ad it is obvious that the associated geeratig fuctio is A(x + B(x. The empty family (that does ot cotai aythig at all has associated geeratig fuctio 0 ad acts as the eutral elemet i this regard. Pairs If A ad B are families of combiatorial objects (ot ecessarily disjoit, the we ca form pairs (A, B of objects A A ad B B. The family of all such pairs is typically deoted by A B (Cartesia product of sets. The size of a pair (A, B is the sum of the sizes of A ad B, so if a pair (A, B is to have size, the the sizes of A ad B have to be ad respectively, where is ay iteger betwee 0 ad. Therefore, if a (b deotes the umber of elemets i A (B, respectively whose size is, ad c deotes the umber of pairs of size, we fid c a b, which is exactly the coefficiet of x i the product of the associated geeratig fuctios. Therefore, the geeratig fuctio for the family A B is exactly A(x B(x. Of course, this priciple ca be geeralised to triples, quadruples, etc. The family E that oly cotais oe object ϵ (of size 0 has geeratig fuctio ad acts as the eutral elemet. Sequeces This builds o the ideas of the previous costructio; if A is ay family, the A A is the family of pairs, A A A the family of triples, etc. altogether, we obtai the set of all fiite sequeces of elemets of A. If we iclude the empty sequece ϵ as a elemet of size 0, the we obtai the specificatio for sequeces Seq(A as follows: B Seq(A {ϵ} A (A A (A A A...

36 CHAPTER 6. THE SYMBOLIC METHOD 35 This traslates to the world of geeratig fuctios as follows: B(x + A(x + A(x 2 + A(x A(x. Here, A itself must be assumed ot to cotai elemets of size 0 to avoid redudacies. Oe ca also cosider restricted sequeces: for istace, A(x is the geeratig fuctio for sequeces of legth (for which we write Seq, + A(x + A(x 2 + A(x A(x A(x+ A(x is the geeratig fuctio for sequeces of size at most (deoted Seq, ad A(x + A(x + + A(x A(x A(x is the geeratig fuctio for sequeces of size at least (deoted Seq. Example 6.2 Recall from Chapter that a compositio of is a fiite sequece of positive itegers addig up to, such as , which is a compositio of 9. The geeratig fuctio for positive itegers I is obviously x + x 2 + x x x, ad so the geeratig fuctio for compositios C Seq(I is C(x x x x 2x Now we ca mae use of the geometric series oce agai: x 2x ( + ( + 2 x 2 2x x, which shows that there are exactly 2 compositios of if, a result that ca also obtaied by the dots ad bars argumet. If we are iterested i compositios of legth, the we fid the geeratig fuctio to be ( x x x ( x x ( x ( + x + ( x by virtue of Lemma 2.2. Therefore, the umber of compositios of legth of a positive iteger is give by ( (, compare Theorem.6 ad the remar thereafter. Compositios ito restricted sets of itegers ca be cosidered as well: for istace, if oly summads ad 2 are allowed, the the associated family is Seq({, 2}, ad the geeratig fuctio is, which yields yet aother iterpretatio of the Fiboacci umbers. x x 2

37 CHAPTER 6. THE SYMBOLIC METHOD 36 Example 6.3 Cosider words over the simple alphabet A {a, b}. The we ca regard a ad b as our atoms, so that the geeratig fuctio is simply A(x 2x. The family W of words over A is ow specified by W Seq(A Seq({a, b}, so that the geeratig fuctio is W (x (i accordace with the fact that there 2x are 2 words of legth, see Theorem.. For a geeral fiite alphabet of size, the associated geeratig fuctio for words is. x Let us ow cosider a restricted family of words: oly a-b-words that do ot cotai two adjacet letters b are allowed. Such a word starts with a arbitrary (possibly empty sequece of a s, followed by a b, followed by a o-empty sequece of a s, followed by a b, followed by aother o-empty sequece of a s, etc. After the last sequece of a s, we may attach either oe more b or othig at all. This leads to the specificatio Seq({a} Seq({b} Seq ({a} {ϵ, b}, where ϵ deotes a empty word (of legth 0. This ca be traslated to the geeratig fuctio x ( + x + x x2 x x, 2 x ad we fid that the umber of such words is a Fiboacci umber (ote that this example is essetially equivalet to Example 4.3. Example 6.4 How may differet ways are there to put 50 teis balls ito 5 boxes, if the first box ca oly hold at most 9 balls ad the secod box at most 7 balls? While the iclusio-exclusio priciple could be used for this purpose, geeratig fuctios are somewhat faster. If stads for a sigle ball, the our situatio is equivalet to the specificatio Seq 9 ({ } Seq 7 ({ } Seq({ } Seq({ } Seq({ }, for which the geeratig fuctio is x 0 x ( 3 x8 x x8 x 0 + x 8 x ( x 5 we are iterested i the coefficiet of x 50, which is [x 50 ] x8 x 0 + x 8 [x 50 ] ( x 5 ( x x 8 5 [x50 ] ( x x 0 5 [x50 ] ( x + x 8 5 [x50 ] ( x 5 [x 50 ] ( x 5 [x42 ] ( x 5 [x40 ] ( x + 5 [x32 ] ( x 5

38 CHAPTER 6. THE SYMBOLIC METHOD 37 i view of Theorem 5.4. Now, Lemma 2.2 ca be applied to fid that this umber is ( ( ( ( ( ( ( ( ( ( ( ( The followig costructios are somewhat more advaced, but they ofte prove useful as well: Powersets The powerset PSet(A of A cosists of all subsets of A; if the size of a subset is the sum of the sizes of all its elemets (as i the case of sequeces, the we fid that PSet(A A A{A, ϵ}, sice every elemet A ca either be preset or ot; if a is the umber of elemets i A of size (assume that a 0 0, the we fid that the geeratig fuctio of B PSet(A is ( B(x ( + z a exp a log( + z ( ( ( m ( m exp a m zm exp a z m m m m ( ( ( m exp m A(zm exp A(x A(x2 + A(x3..., 2 3 m maig use of the Taylor series for the logarithm. Multisets Multisets MSet(A are closely related to the powerset costructio; we cosider collectios of elemets of A, where each elemet is allowed to occur more tha oce as well. This is equivalet to the formal expressio MSet(A A A Seq({A}, which leads to the followig geeratig fuctio for B MSet(A: ( B(x ( z a exp a log( z

39 CHAPTER 6. THE SYMBOLIC METHOD 38 exp exp ( a m zm m ( m m A(zm exp ( m m a z m ( exp A(x + A(x2 + A(x Example 6.5 Formally, a laguage is a collectio of distict words (this otio of a laguage plays a role i computer sciece. If a alphabet A of letters is give, how may laguages are there that cotai a total umber of letters? The accordig specificatio is L PSet(Seq (A, givig rise to the geeratig fuctio ( L(x exp m ( m m x m x m ( + x. While oe caot derive a closed formula from this geeratig fuctio, it ca be used to determie the first few coefficiets (the sequece starts l 2, l 2 5, l 3 6, l 4 42,... i the case 2, ad further iformatio about the coefficiets (i particular, their growth ca be extracted by meas of aalytic methods (which are beyod the scope of this course. 6.2 Labelled structures I may istaces, oe is iterested i structures where the atoms are labelled: if a structure cosists of atoms, the these atoms receive distict labels from to. The most typical example are permutatios: a permutatio of ca be regarded as a sequece of atoms bearig labels from to ; the 6 3! permutatios of {, 2, 3} (see Theorem.2 are thus { 2 3, 3 2, 2 3, 2 3, 3 2, 3 2 }. I the cotext of labelled structures, it is useful to wor with expoetial geeratig fuctios: if a is a sequece, the its expoetial geeratig fuctio is defied by A(x a! x. The expoetial geeratig fuctio associated to permutatios is thus!! x x x, sice there are! permutatios of {, 2,..., }. Note that, o the other had, the ordiary geeratig fuctio!x is ot a elemetary fuctio, ad it is eve diverget for

40 CHAPTER 6. THE SYMBOLIC METHOD 39 all x 0. The ame expoetial is due to the fact that the expoetial geeratig fuctio of the sequece,,... is e x (the associated structure is sometimes called urs : a ur is merely a collectio of atoms labelled to (thi of balls mared with these labels without additioal structure. Example 6.6 I how may ways ca oe form cycles of the umbers, 2,...,? Figure 6.2 shows all possibilities i the case 4; geerally, if oe starts at, the there are (! possible orders for the remaiig umbers. Therefore, the associated expoetial geeratig fuctio is (! x! x log( x Figure 6.2: All cycles of legth 4. For istace, oe ca thi of these cycles as eclaces made of four beads i differet colours. Agai, there are several useful trasformatios o families of labelled objects: Uios As i the case of ulabelled structures, oe ca defie the uio of two disjoit families A ad B, ad the associated geeratig fuctio is exactly the sum of the geeratig fuctios of A ad B. Pairs The situatio is slightly differet to the ulabelled case, sice a ordiary pair of labelled objects does ot costitute a proper labelled object agai (sice labels are duplicated. If

41 CHAPTER 6. THE SYMBOLIC METHOD 40 oe combies two labelled object A ad B of sizes ad respectively to form a ew object of size, the oe first has to distribute the labels amog the two objects, which ca be doe i ( ways (choose the labels that are give to A. Therefore, if two families A ad B with expoetial geeratig fuctios A(x a! ad B(x b! are give, the the umber of pairs of size that ca be formed i this way is c which ca be rewritte as ( a b c! a!!!(! a b, b (!. Note that this is exactly the coefficiet of x i the product A(x B(x, so that the geeratig fuctio of the family C A B of pairs is exactly A(x B(x. This fact is oe of the mai reasos why expoetial geeratig fuctios are used i this cotext. Sequeces As i the case of ulabelled structure, oe ca exted the above reasoig to triples, quadruples, etc. Geerally, sequeces of legth of elemets from a family A with expoetial geeratig fuctio A(x have expoetial geeratig fuctio A(x (we deote this family by Seq (A, ad the expoetial geeratig fuctios for sequeces Seq (A of legth at most ad sequeces Seq (A of legth at least are A(x + A(x ad A(x A(x respectively. Arbitrary sequeces Seq(A have expoetial geeratig fuctio A(x. Example 6.7 Words over a alphabet of size ca also be regarded as labelled objects, amely sequeces of urs (the i-th ur records the positios i the word where the i-th letter occurs of legth. Therefore, the expoetial geeratig fuctio is (e x e x! x, i agreemet with the fact that there are words of legth. Sets Sets of elemets from a family A of combiatorial objects are somewhat easier to hadle i the labelled case tha i the ulabelled oe. This is due to the fact that objects are ow distiguishable by their labels: to every set of objects (amog which the labels