Lecture 1: Generating Functions

Transcription

1 Advaced Topics i Combiatorics Summer 08 Lecturer: Kavish Gadhi Lecture : Geeratig Fuctios Itroductio The geeratig fuctio is a very powerful tool i combiatorics, ad is the first oe we will be itroducig i this class, mostly because it has applicatios throughout combiatorics ad will be very useful i provig results i Ramsey theory, partitio theory, code theory, ad more It is also worth gettig to ow early, because at first it may be difficult to gai ituitio as to why we use them, ad just how powerful a techique they are Defiitio A (ordiary geeratig fuctio for a series a 0, a, a, is the power series a 0 + a x + a x + + a x + = a x As a example, the geeratig fuctio for the sequece,,, 0, 0, 0, 0, is + + x = ( + x, ad that for the sequece,,,,,, is + x + x + x 3 + = x Why is such a cocept useful? I may cases, eve if there are ifiitely may ozero etries to the series, we ca reduce this ifiite power series to a closed, fiite form; ad, as it turs out, we ca add ad multiply geeratig fuctios for differet power series i a meaigful, atural way (as well as divide, scale, differetiate, ad ay other meaigful operatios o polyomials This becomes extremely useful, as we will show with examples later i the lecture A particular simple example of where you ca see geeratig fuctios pop up is i the biomial theorem, which I m sure most of you ow ( Theorem (Biomial Theorem For a iteger ad variables x, y, (x + y = x y Now, if y is a variable, ote that this actually gives a geeratig fuctio for some series; it is ideed fiite, but a geeratig fuctio oetheless Now, at this poit, it may seem that the coectio to geeratig fuctios is rather fuzzy ad ot useful, but it becomes more clear whe you cosider a geeralized versio of the biomial theorem, i which the expoet is replaced by ay real umber The details of why it is defied this way are ot worth gettig ito, but either way, the series becomes ifiite ad you ca see why geeratig fuctios are a atural way to express such a sum, ad give a reaso to examie the uderlyig sequece Note that a biomial ( j for a oiteger j ad iteger is defied as j(j (j (j +! Theorem 3 (Newto s geeralized biomial theorem For ay real umber j ad variables x, y, we have ( j that (x + y j = x j y =0 These are examples where you ca see geeratig fuctios is everyday theorems, but why are geeratig fuctios themselves a powerful tool i solvig problems? To give some ituitio behid this, I will give a rather elogated problem i which geeratig fuctios tur out to be extremely useful: derivig a explicit form for the Catala umbers This techique will prove ivaluable i geeral for solvig recurreces =0 =0 -

2 - Lecture : Geeratig Fuctios Catala umbers: bac to the quadratic formula? Formally, the Catala umbers are a sequece C 0, C, C,, C which ca be described by the followig recurrece C 0 =, C = C i C i Now, this is a rather dry formalism ad may seem lie a cotrived recurrece that comes out of owhere, but the Catala umbers tur out to have a umber of very iterestig combiatorial applicatios For example, the Catala umbers are exactly the umber of ways to travel from (0, 0 to (,, oly movig uit distaces up or to the right, ad stayig below the lie y = x Alteratively, the Catala umbers are exactly the umber of ways to write a series of ope ad close paretheses that mae sese; ie, there ca ever be more close paratheses tha ope paretheses that have appeared before them (Why is this the same as the previous defiitio? The Catala umbers tur out to appear i a lot of places As oe might imagie, because of the ubiquity of the Catala umbers, it would be useful to derive a explicit formula for them, rather tha the rather cumbersome recurrece above How are we goig to do this? Geeratig fuctios, of course! What is this geeratig fuctio? We ow that i geeral, from our defiitio, the geeratig fuctio for the Catala umbers is of the form C x However, the utility of geeratig fuctios comes from writig this is a more compressed form, ad to this ed we will use the recurrece relatio of the Catala umbers Claim 4 The geeratig fuctio C(x for the Catala umbers satisfies x C(x = C (x Proof: We re goig to use our recurrece relatio to rewrite the geeratig fuctio To this ed, cosider a slight alteratio to our traditioal geeratig fuctio, the sum Usig our recurrece, ( is equivalet to C + x ( ( C i C i x But ote that this is actually just aother way of writig C(x, sice each term of this sequece is exactly equivalet to those you would get if you squared our origial geeratig fuctio However, we also ow that (, if we multiply by x, is just our origial geeratig fuctio, except without the C 0 x 0 term: i other words, it is x (C(x C 0 = x (C(x Thus, sice both of our derivatios must be equal sice they are from the same origial equatio, we have that x (C(x = C(x, as desired Now comes the weird part We ow how to solve this: use the quadratic formula! I particular, multiplyig by x, we have that xc(x C(x + = 0, so, usig our favorite formula, C(x = ± 4x Now, which solutio is it? To figure this out, as with may recurreces, we use the base case: C 0 = 0 Now, ote that if we plug i x = 0 ito our geeratig fuctio C x, we should just get the C 0 term, sice all others go to 0 However, if we plug i 0 ito the positive square root, we get 0 =, which is clearly impossible (with the egative square root, we get 0 0, which is idetermiate, ad still could be the correct solutio Thus, the solutio we are looig for is the egative square root: our geeratig fuctio C(x = 4x For those iterested, Richard Staley, a famous combiatorist at MIT, wrote up a compedium of hudreds of iterpretatios/ways to derive the Catala umbers which ca be foud here:

3 Lecture : Geeratig Fuctios -3 From here, we re goig to use a theorem I metioed earlier, Newto s geeralized biomial theorem (it was t just a digressio!, to derive a explicit form geeratig fuctio for the Catala umbers Propositio 5 (Explicit form of Catala Numbers For all 0, C = +( ( 05 Proof: Usig the geeralized biomial theorem, we ca write ( 4x / as ( 4 x The first step to simplifyig this is expadig ( 05 as 05(05 (05 +! Multiplyig this by ( 4, we get ( 4 ( ( ((6 (4 6 = = ((3 ( 3!!! Thus, ( 4x / = = ( x = = (! ((3 ( 3!! 4 + ( ( + + (! =!! ( = x = = ( ( x This gives us that C(x = x At this poit, we are doe, sice C = 4+ ( + +, although we ca, as it turs out, further simplify this (i a o-iterestig way to +(, the more commo explicit formula The, we are doe! 3 Fiboacci: where did the 5 come from? Now, the methods that we used to fid a explicit form for the Catala umbers are ot uique! They ca geerally be applied to solve may recurreces; multiply your give geeratig fuctio by x, plug i the recurrece, ad derive a equatio for the geeratig fuctio i terms of it ad x Usig the quadratic equatio or similar techiques, solve for the geeratig fuctio as a fractio i terms of x, ad use series expasios (the biomial theorem, Taylor expasios that you ll use i calculus, or others, to derive a explicit form for the geeratig fuctio ad thus the sequece as a whole A example of this ca be see with the Fiboacci sequece, which satisfies F 0 = 0, F =, F = F + F, for which we ll write the geeratig fuctio as F(x = F x The, usig our recurrece, x F(x = F x = F x + F x = F(x + xf(x + F Thus, we have that x F(x = = = F(x+xF(x+ Multiplyig by x ad solvig for F(x, we get that F(x = = x We wo t do the x x series expasio explicitly here, but it turs out that this series expasio gives that F = (+ 5 ( 5!! This is quite extraordiary 3, ad may beg the questio: where does such a result come from, ituitively? The mai explaatio I ca give is a shay oe; the roots of the polyomial i the deomiator of the geeratig fuctio, x x are exactly + 5 ad 5, ad it turs out that we do a partial fractio decompositio based o these roots to derive the explicit form geeratig fuctio This proof is adapted from that of Mie Spivey 3 At first it might ot be etirely obvious that this gives a result that is a iteger, util you realize that oly the 5 terms remai i the expasio of the umerator, ad dividig by the 5 i the deomiator gives at least a ratioal aswer; showig that this is a iteger is aother story 5

4 -4 Lecture : Geeratig Fuctios Either way, from these two results, you ca see that usig geeratig fuctios to solve recurreces is a extremely powerful tool, ad coupled with characteristic polyomials, which we may tal about i a further lecture, ca solve most recurreces that you ll ecouter 4 Geeratig Fuctios meet Cooies A commo problem ivolvig geeratig fuctios ivolves the fact that their product is meaigful i the cotext of a coutig problem I particular, let s say that p 0, p, p, p 3, are the umber of ways of choosig elemets from some set P, ad q 0, q, q, q 3, are the umber of ways of choosig elemets from some set Q Clearly, we ca write these sequeces as P(x = p x ad Q(x = b x Now, let r be the umber of ways of choosig elemets from either P or Q; clearly r is just p i q i, which gives that R(x = Q(xP(x This idetity shows that, i coutig problems, we ca multiply geeratig fuctios associated with choosig objects from particular sets to get the combied solutio This fact turs out to be extremely powerful For example, cosider the followig problem ivolvig differet types of cooies: oreos, milaos, chips ahoy, ad thi mits (my favorite! Problem 4 Let p( be the umber of ways to choose oreos, milaos, chips ahoy, ad thi mits, subject to the followig costraits: the umber of oreos we tae must be a multiple of 7, we ca oly tae 6 milaos, we ca tae oly chips ahoy, ad the umber of thi mits must be odd What is p( for all? To solve this problem, we will fid a geeratig fuctio for all of the differet quatities we are choosig, ad the multiply them Amazigly, this will give us a simple aswer to what seems to be a pretty ugly coutig problem geeratig fuctios elimiates all of the tediousess Solutio: First, let the umber of the oreos, milaos, chips ahoy, ad thi mits, respectively, be a, b, c, d The, our costraits are that a 0 mod 7, b 6, c, d mod We wat to write geeratig fuctios for all of these Cosider the oreos first; sice a 0 mod 7, the associated possibility sequece is 0, 7, 4,, 8 If we write our geeratig fuctio as taig a if a quatity is possible ad 0 otherwise, this gives us a geeratig fuctio of A(x = x 7 = Similarly, for milaos, the possibilities x7 6 are 0,,, 3, 4, 5, 6, givig a geeratig fuctio of B(x = x = x7 Derivig a similar geeratig x fuctio for chips ahoy ad thi mits gives C(x = + x ad D(x = x + x = Now, we use x the fact that we ca multiply; the total possibilities geeratig fuctio T (x is just A(xB(xC(xD(x = x x7 7 x + x x x T (x = x ( x To express this i typical form for a geeratig fuctio, I will tae a small digressio ito calculus, usig the fact that x has expasio + x + x + ad that the derivative of x is ( x Usig the fact that derivatives apply equal well to power series (a fact that I have ot prove, this gives that ( x ca be writte as ( + x, ad thus our desired geeratig fuctio is x ( x = x, which gives us that the umber of ways to choose cooies amog the four = types is exactly! Surprisig, huh?

5 Lecture : Geeratig Fuctios -5 I geeral, this id of method is very powerful whe we are ased to choose items with respect to some requiremet, ad ca mae our lives a whole lot easier whe doig may types of combiatorics problems! 5 Expoetial Geeratig Fuctios ad Bell umbers This sectio will itroduce a ew cocept, called the expoetial geeratig fuctio The mai purpose of this sort of fuctio is to solve a coutig problem i which our goal is to cout the umber of ways the - elemet set ca tae some structure; for ordiary geeratig fuctios, our objects were idistiguishable, i some sese (although multiplicatio sometimes allowed us to chage this, while with expoetial geeratig fuctios our objects are ecessarily ordered This is a rather vague prerogative, but it will become more clear as we wor through examples Formally, a expoetial geeratig fuctio is defied as follows: Defiitio 6 A expoetial geeratig fuctio for a series a 0, a, a, is the power series a 0 + a x + a + a 3x = a x It helps to wor through some trivial examples; if the series we are cosiderig is,,,,, the the x expoetial geeratig fuctio is just! = ex, by defiitio I may cases, however, it helps to thi of our series rather as a set; the first example described the geeral set, the secod the trivial 0-elemet set If we wat a eve-sized set, or a series, 0,, 0,,, the the expoetial geeratig fuctio is ex +e x, which is the defiitio of hyperbolic cosie You ca see that we re gettig some pretty weird fuctios, ad this helps i may cases i which recurreces are strage or the sequeces/structures we are studyig require us to impose structure o a set The additive rule we used frequetly for ordiary geeratig fuctios still applies to expoetial geeratig fuctios, but this is just about the oly rule that still applies Multiplicatio is ow differet; with ordiary geeratig fuctios, we used that r = p i q i However, if we multiply two expoe- ( tial geeratig fuctios, we istead get that r = p i q i ; so, for future referece, whe we say i that R(x = P(xQ(x for expoetial geeratig fuctios, this is the implicatio about the uderlyig sequeces 4 Usig these defiitios, I wat to prove a famous formula for the Bell umbers B, which represet the umber of partitios of a set of size For example, for the three elemet set {x, y, z}, the possible partitios are {{x}, {y}, {z}}, {{x, y}, {z}}, {{x}, {y, z}}, {{x, z}, {y}}, {{x, y, z}}, so B 3 = 5 It turs out that the Bell umbers satisfy the followig explicit formula: Theorem 7 (Dobisi s Formula The th Bell umber satisfies B = e The fact that we have a ifiite sum ad a e is a pretty good idicator i geeral that we d be usig expoetial geeratig fuctios, but we ll start from scratch 4 For further referece, shiftig to the left or right i the uderlyig sequece ca o loger be achieved by substitutig x ; istead differetiatig ad itegratig are the correspodig operatios We ll try to avoid calculus as much as possible, however! =0!

6 -6 Lecture : Geeratig Fuctios Proof: First, let s thi about fidig a recurrece relatio for B + To do this, thi about what happes oce we fix the subset that the first elemet of our set is i; it ca be of ay size + j, for which there are ( j ways of choosig the remaiig j elemets Summig over all such j, we get the recurrece + B + = j= ( j B + j, or as it is more typically writte =0 ( B, if we switch aroud some idices From here, we are goig to prove the followig lemma, which will help us derive the desired result Lemma 8 The expoetial geeratig fuctio B(x for the Bell umbers is e ex ( Proof: The first step is a bit tricy, ad ivolves the recurrece B + = B that we just derived =0 Remember that we ca actually write the multiplicatio of two expoetial geeratig fuctios p ad q as ( defiig a ew sequece such that r = p q I the above, if we let p = B ad q =, ad =0 the expoetial geeratig fuctio for the Bell umbers be B(x, we get that e x B(x = db dx The ex comes from the fact that this is the geeratig fuctio for the sequece of all s, which is the q i this case, ad the B (x comes from the fact that our multiplicatio shifts the values i the series oe over, which I briefly metioed was equivalet to differetiatig (this is ot so difficult to show, but we will ot get ito it right ow Either way, from here, movig terms, we get that db B = ex dx, which, itegratig, gives that l(b = e x + C B(x = e ex +C Usig the base case that B(0 = B 0 =, sice this gives us the first term of the series, we have that C =, so we get that the expoetial geeratig fuctio for the Bell umbers is e ex, as desired With this i had, we ow eed to just expad this expoetial geeratig fuctio to get our desired explicit formulatio I particular, B(x = e ; let s first expad e eex ex We ll have to use the so-called Maclauri series (which you ll lear i calculus for e x = x Just pluggig this i, this reduces our! =0 geeratig fuctio (disregardig the factor of e for ow to (e x From here, pluggig i the Maclauri! =0 expasio for each idividual term i the sum, we get that e x x =, so the geeratig fuctio ca! be expaded out as ( x e!! = x e!! =0 B x Sice we of course also ow that B(x =, this gives that B =! e =0 =0, as desired! 6 Let s solve some HMMT Problems! I thought we d fiish with some HMMT problems that are easily solvable usig techiques we ve addressed up to this poit We ve oly begu to scratch the surface of geeratig fuctios ad their utility, however, ad I d recommed that, if iterested, you delve more deeply ito the additive, multiplicative, ad differetiable properties I briefly metioed i this lecture, ad read some resources that will help you develop a eve more rigorous theory of geeratig fuctios Regardless, here goes!

7 Lecture : Geeratig Fuctios -7 Problem 6 (HMMT 007, Problem 9, Combiatorics Let S deote the set of all triples (i, j, of positive itegers where i + j + = 7 Compute ij (i,j, S Solutio: This problem may loo itimidatig, but we re goig to solve for the geeral case, ot just i + j + = 7! Let s = s x But ote that ij We wat to fid the geeratig fuctio i+j+= 0 this geeratig fuctio simply correspods to cubig the geeratig fuctio x, because cubig this 0 fuctio causes the coefficiet of x to be exactly the sum of the products of umbers that add up to Now, let S = x Multiplyig by x ad subtractig this from S, we get that S xs = x = x, by the x 0 ( 3 sum of a geometric series This gives that S = x x ( x, ad thus our geeratig fuctio is ( x = x 3 ( x 6 ( 6 From here, usig the geeralized biomial theorem, we have that ( x 6 = x ( The, we =0 expad ( 6 = 6( 7 ( 6 +! ; multiplyig ( gives this to be (+5 (6! = ( +5 5 Thus, our geeratig ( + 5 ( + fuctio is x 3 x = x Thus, we have that s = ( , so our desired aswer is =0 =3 ( 9 = 68, which is correct! 5 A problem that also falls easily i a similar way to geeratig fuctios argumets (quite similar to problems that we ve discussed, also from the HMMT, is the followig See if you you ca solve it! Problem 6 (HMMT 008, Combiatorics, Problem 0 Determie the umber of 8-tuples of oegative itegers (a, a, a 3, a 4, b, b, b 3, b 4 satisfyig 0 a, for each =,, 3, 4, ad a + a + a 3 + a 4 + b + 3b + 4b 3 + 5b 4 = 9 Now, fially, we ed with a problem that does t immediately loo lie it ca be solved with geeratig fuctios, but turs out to fall immediately to a geeratig fuctio that we derived earlier i this lecture! Problem 63 (HMMT 008, Algebra, Problem 0 Evaluate the ifiite sum Solutio: + ( ( 5 Recall that we ow the geeratig fuctio for the Catala umbers, which has the form x ad 4x If we multiply the Catala umber s equatio by x ad tae a derivative, we get the geeratig fuctio G(x = ( x, which is what we are looig for (we simply wat to plug 5 ito this geeratig fuctio Doig the same to our other ow explicit form for the Catala geeratig fuctio, we differetiate x = ( 4x, which gives G(x = 4x Pluggig ( 4x i 5, this gives a aswer to the origial problem of 4 5 = 5, which is ideed correct!

8 -8 Lecture : Geeratig Fuctios 7 Practice ad Challege Problems 7 Basic Problems Problem 7 Fid the geeratig fuctio for the sequeces,,,, 0, 0, 0, 0, ad, 7,, 35, 35,, 7,, 0, 0, Simplify your aswers as much as possible Problem 7 I how may ways ca we mae $ from peies, icels, dimes, ad quarters? Problem 73 Prove that the geeratig fuctio ( x µ(/ = e x, for µ( the Mobius fuctio defied as µ( =, µ( = 0 if is divisible by the square of a iteger greater tha oe, ad µ( = ( r if is the product of r distict primes Hit: tae logs! This is a beautiful, surprisig result, which does require some calculus Problem 74 Solve the liear recurrece a 4a + 4a = 0,, a 0 =, a = 4, usig geeratig fuctios Problem 75 Usig domioes that we ca place horizotally or vertically, i how may ways ca we tile a strip? These umbers may loo familiar! As a ote, the followig theorem is true i the geeral case: Theorem 9 (Kasteley, 96 Let Q m, be a rectagular grid with m eve The the umber of 05m ( ( π jπ domio tiligs of Q m, is cos + cos m + + = j= Is t that a astoishig result! We oly as you to prove the case, but it s still a fasciatig problem ad oe very approachable usig geeratig fuctios (i fact, the proof of the above relies o costructig such a geeratig fuctio 7 Competitio Problems Problem 76 (Romaia 003 How may -digit umbers, whose digits are i the set {, 3, 7, 9}, are divisible by 3? Problem 77 (IMO 995 Let p be a odd prime How may p-elemet subsets A of {,,, p} are there, the sum of whose elemets is divisible by p? Problem 78 (IMO Shortlist 998 The sequece 0 a 0 < a < a < is such that every oegative iteger ca be uiquely expressed as a i + a j + 4a, where i, j, are ot ecessarily distict Fid a Research Problems Geeratig fuctios are ubiquitious i research, especially i eumerative combiatorics I give a example below of a problem that is actually related to a famous NP-complete problem that baffled researchers for decades, for which the smaller cases ca be solved usig techiques of geeratig fuctios See if you ca figure out how! 5 Problem 79 (Subset sum problem Give a list of itegers, is there a oempty subset of these itegers whose sum is a particular iteger? If so, fid it 5 Note that this is t supposed to be especially elegat, but is a way that you could do this problem for small cases Pretty cool that somethig you ca lear i high school is still used i moder algorithms problems!