# Solutions to Exercises Chapter 4: Recurrence relations and generating functions

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1 Solutios to Exercises Chapter 4: Recurrece relatios ad geeratig fuctios 1 (a) There are seatig positios arraged i a lie. Prove that the umber of ways of choosig a subset of these positios, with o two chose positios cosecutive, is F +1. (b) If the positios are arraged aroud a circle, show that the umber of choices is F + F 2 for 2. (a) Proof by iductio. If g() deotes this umber, the we have g(1) = 2 = F 2, g(2) = 3 = F 3. (For = 2, we caot occupy both positios; but all other choices are possible.) For > 2, we separate the seatig selectios ito those i which the last positio is uoccupied ad those i which it is occupied. There are g( 1) of the first kid. If the last positio is take, the the oe before it must be free, ad we have a arbitrary seatig pla o the first 2 positios; so there are g( 2) of these. Hece ad the iductive step is proved. g() = g( 1) + g( 2) = F + F 1 = F +1, (b) Cosider a particular positio o the circle. If it is uoccupied, we ca break the circle at that poit, ad obtai a lie with 1 positios, which ca be filled i g( 1) = F ways. If the positio is occupied, the its eighbours o either side are uoccupied, ad (if 3) we ca remove the positio ad its two eighbours, obtaiig a lie of 3 positios which ca be filled i g( 3) = F 2 ways. The same holds if = 2, sice there is just oe seatig pla with the give positio occupied, ad F 0 = 1. So we have F + F 2 seatig plas i all. 1

2 2 Prove the followig idetities: (a) F 2 F +1 F 1 = ( 1) for 1. (b) F i = F (c) F F 2 = F 2, F 1 F + F F +1 = F 2+1. (d) F = /2 ( i i ). (a) Iductio: the result holds for = 1. For 2, we have F 2 F +1 F 1 = F 2 (F +F 1 )F 1 = (F F 1 )F F 2 1 = (F 2 1 F F 2 ), so, if F 2 1 F F 2 = ( 1) 1 the F 2 F +1 F 1 = ( 1). (b) Iductio. The result is true for = 0 (the empty sum is zero). Assumig it for 1, we have F i = (F +1 1) + F = F (c) Agai, iductio. The result holds for = 1 by ispectio. Assume it for ; that is, F F 2 1 = F 2 2, F 2 F 1 + F 1 F = F 2 1. Addig these equatios ad usig the Fiboacci recurrece, we get F 2 F + F 1 F +1 = F 2. Usig (a) twice, this implies that F F2 = F 2. Now add this equatio to the secod displayed equatio usig the Fiboacci recurrece to get F 1 F + F F +1 = F 2+1. (d) Most easily, this follows from our origial iterpretatio of F as the umber of expressios for as a ordered sum of 1s ad 2s. Such a expressio with i 2s will have 2i 1s, hece i summads altogether; there are ( i) i ways to choose the positios of the 2s i the sum. Now summig over i gives the result. 2

5 9 (a) Solve the followig recurrece relatios. (i) f ( + 1) = f () 2, f (0) = 2. (ii) f ( + 1) = f () + f ( 1) + f ( 2), f (0) = f (1) = f (2) = 1. 1 (iii) f ( + 1) = 1 + f (i), f (0) = 1. (b) Show that the umber of ways of writig as a sum of positive itegers, where the order of the summads is sigificat, is 2 1 for 1. (a) (i) Let g() = log 2 f (). The g( + 1) = 2g(), g(0) = 1; so g() = 2 (Sectio 3.1), ad f () = 2 2. (ii) The characteristic equatio (see Sectio 4.3) is x 3 = x 2 + x + 1. This cubic has three distict real roots α, β, γ, ad so the geeral solutio is f () = aα + bβ + cγ, where a,b,c are determied by the three equatios a + b + c = 1, aα + bβ + cγ = 1, aα 2 + bβ 2 + cγ 2 = 1. (iii) We have f (1) = 1, sice the empty sum is zero. For 1, we have ( ) 1 2 f ( + 1) = 1 + f (i) = 1 + f (i) + f ( 1) = f () + f ( 1), ad so by iductio f () is the Fiboacci umber F. (b) There is oe expressio cosistig of the sigle umber. Ay other expressio eds with some positive umber j <, preceded by a expressio with sum j; so the umber f () of expressios satisfies 1 1 f () = 1 + f ( j) = 1 + f (i). j=1 i=1 Just as i (a)(iii), we fid that f ( + 1) = f () + f () = 2 f (). Sice f (1) = 1, we have f () = 2 1, as claimed. 5

7 for the coefficiet of t o the right is s( 1) + s( /2 ), ad the costat term o both sides is equal to 1. (b) The relatioship is u() u( 1) = s()/2 for 2. The proof is quite itricate, ad depeds o defiig aother sequece v() as follows: v() is the umber of sequeces (x 1,...,x k ) of positive itegers which satisfy x i > i 1 j=1 x j for all i k ad k i=1 x i =. Now such a sequece must have last term x k > /2, ad is obtaied by takig a sequece with sum x k ( 1)/2 ad appedig x k to it. So we have the recurrece v() = ( 1)/2 v(i). i=1 It follows from this that v(2) = v(2 1) = v(2 2) + v( 1) for all > 0. Now we claim that v(2) = s()/2 for all 1. This is proved by iductio, beig true for = 1. Assumig the result for 1, we have v(2) = v(2 2) + v( 1) = s( 1)/2 + s( /2 )/2 = s()/2, ad the iductio goes through. Now u() u( 1) is the umber of sequeces satisfyig the specificatio for u which have last term. These are obtaied by appedig to a sequece with sum strictly smaller tha ; so we have u() u( 1) = 1 v(i) = v(2) = s()/2. Remark: Sequeces satisfyig this coditio are called supericreasig; they arose i the first example of a public-key cryptosystem, the Merkle Hellma kapsack system. The two sequeces s ad u occur as umbers M1011 ad M1053 i the Ecyclopedia of Iteger Sequeces, where further refereces ca be foud, or you ca fid them i the olie versio here (for s) ad here (for u). 12 Let F(t) be a formal power series with costat term 1. By fidig a recurrece relatio for its coefficiets, show that there is a multiplicative iverse G(t) of F(t). Moreover, if the coefficiets of F are itegers, so are those of G. 7

8 Let F(t) = f t ad G(t) = g t, with f 0 = 1. The relatio F(t)G(t) = 1 is equivalet to g 0 = 1 ad the recurrece relatio g = i=1 f i g i for g i terms of earlier values. So the values g are determied, ad are itegers if the values f are. 13 A permutatio π of the set {1,...,} is called coected if there does ot exist a umber k with 1 k < such that π maps the subset {1,2,...,k} ito itself. Let c be the umber of coected permutatios. Prove that i=1 c i ( i)! =! Deduce that, if F(t) = 1!t ad G(t) = 1 c t are the geeratig fuctios of the sequeces (!) ad (c ) respectively, the 1 G(t) = (1 + F(t)) 1. For a permutatio π, let i(π) be the least positive iteger k such that π maps the set {1,...,k} ito itself. (Such a k exists, sice certaily π maps {1,...,} to itself.) The π is the composite of a coected permutatio o {1,...,i(π)} ad a arbitrary permutatio o {i(π)+1,...,} (this last set beig empty if i(π) = ). Summig over i, we obtai the stated recurrece. As i the precedig questio, this recurrece is equivalet to (1 + F(t))(1 G(t)) = Let (1 +t ) = a t. 1 0 Prove that a is the umber of ways of writig as the sum of distict positive itegers. (For example, a 6 = 4, sice 6 = = = ) A term i the ifiite product (1 + t ) is obtaied by selectig t m from the mth factor for a fiite umber of values of, ad 1 from all the other factors. So there is a cotributio of t for every expressio of as a sum of distict positive itegers. 8

10 of The radius of covergece of the power series 0 B t /! is the reciprocal lim sup ( ) 1/ B.! But the radius of covergece is ifiite, sice the series coverges everywhere. (I geeral, the radius of covergece is the distace from the origi to the earest sigularity i the complex plae; if the sum fuctio has o sigularities, the radius of covergece is ifiite.) So the limsup is zero, which meas (sice all the values are positive) that the limit is zero. 18 (a) Prove that the expoetial geeratig fuctio for the umber s() of ivolutios o {1,...,} (Sectio 4.4) is exp(t t2 ). (b) Prove that the expoetial geeratig fuctio for the umber d() of deragemets of {1,...,} is 1/((1 t)exp(t)). (a) The fuctio S(t) = exp(t t2 ) is the uique solutio of the differetial equatio S (t) = (1 +t)s(t) with the iitial coditio S(0) = 1. So it suffices to check that the e.g.f. of the umbers s() satisfies this differetial equatio. (Clearly it satisfies the iitial coditio.) Now, if we put S(t) = 0 s()t /!, the the coefficiet of t 1 /( 1)! i (1 + t)s(t) is s( 1) + ( 1)s( 2), whereas the coefficiet i S (t) is s() (sice differetiatio of the e.g.f. correspods to a left shift of the sequece). By the recurrece relatio i Sectio 4.4, these two expressios are equal. (b) We have by (4.4.1). Hece d()t 0! d() =! = = ( ) ( 1) i. i! ( ) ( ) ( t) i i 0 i! t k k 0 1 e t (1 t). (I the first lie, we put k = i; the i ad k ru idepedetly over the oegative itegers.) 10

11 19 The Beroulli umbers B() (ot to be cofused with the Bell umbers B ) are defied by the recurrece B(0) = 1 ad k=0 ( + 1 k ) B(k) = 0 for 1. Prove that the expoetial geeratig fuctio B()t f (t) = 0! is give by f (t) = t/(exp(t) 1). Show that f (t) t is a eve fuctio of t, ad deduce that B() = 0 for all odd 3. What is the solutio of the similar-lookig recurrece b(0) = 1 ad for 1? k=0 The recurrece ca be writte as ( k=1 k ( ) b(k) = 0 k ) B(k) = B( + 1) for 2. Multiplyig by t +1 /(+1)! ad summig over 2, the two sides are f (t)exp(t) ad f (t) with the costat ad liear terms omitted. Thus f (t)exp(t) 1 t t = f (t) t, whece f (t) = t/(exp(t) 1), as required. Now f (t)+ 2 1t = 1 2 t(exp( 1 2 t)+exp( 2 1t))/(exp( 1 2 t) exp( 1 2 t)) = 1 2 t coth 1 2 t, a eve fuctio. The last recurrece obviously has the solutio b(k) = ( 1) k. 11

12 20 For eve, let e be the umber of permutatios of {1,...,} with all cycles eve; o, the umber of permutatios with all cycles odd; ad p =! the total umber of permutatios. Let E(t), O(t) ad P(t) be the expoetial geeratig fuctios of these sequeces. Show that (a) P(t) = (1 t 2 ) 1 ; (b) E(t) = (1 t 2 ) 1/2 ; [HINT: Exercise 15 of Chapter 3] (c) E(t).O(t) = P(t); (d) e = o for all eve. Fid a bijective proof of the last equality. (a) Sice we are oly cosiderig eve vales of, we have ( ) (2m)! P(t) = t 2m = 1 m 0 (2m)! 1 t 2. (b) By Chapter 3, Exercise 16, the umber e 2m of permutatios of a (2m)-set with all cycles eve is ((2m)!!) 2. So we have ( ((2m)!!) 2 ) E(t) = t 2m = (1 t 2 ) 1/2, m 0 (2m)! the secod equality beig verified by expadig the last expressio usig the Biomial Theorem. (c) This is a istace of a very geeral coutig priciple. If F(t) ad G(t) are the e.g.f.s for labelled structures i two classes F ad G, with F(t) = f t /! ad G(t) = g t /!, the F(t)G(t) is the e.g.f. of structures cosistig of a partitio of the poit set with a F -structure o oe part ad a G-structure o the complemet. For, if H is the class of such composite structures, the the umber of H -structures o a -set is give by h = k=0 ( k ) f k g k, from which a simple calculatio shows that H(t) = h t /! = F(t)G(t). Now (c) follows o applyig this result, where F ad G deote the classes of permutatios with all cycles eve, resp., all cycles odd: give a arbitrary 12

14 This is a problem, but i practice ot too serious. If the lists to be sorted are truly radom, the the proportio of bad permutatios decreases faster tha expoetially. If we are likely to meet sorted or partially sorted lists quite ofte, we ca modify the algorithm by choosig a to be the middle item of the list, rather tha the first. 22 Let m be the miimum umber of comparisos required by QUICKSORT to sort a list of legth. Prove that, for each iteger k > 1, m is a liear fuctio of o the iterval from 2 k 1 1 to 2 k 1, with m 2 k 1 = (k 2)2k + 2. If = 2 k 1, what ca you say about the umber of orderigs requirig m comparisos? The miimum umber of comparisos will occur whe the sublists are as early equal as possible: both of legth ( 1)/2 if is odd, or of legths ( 2)/2 ad /2 if is eve. So we have the recurrece { 1 + 2m( 1)/2, if is odd, m = 1 + m ( 2)/2 + m /2, if is eve. Suppose that m = a + b for c d. The, for 2c + 1 2d + 1, we have m = 1 + 2(a( 1)/2 + b) = (a + 1) + (2b a 1). But m is liear (i fact idetically zero) for 0 1; by iductio o k, it is liear o each iterval 2 k k 1. If the value o this iterval is give by m = a k +b k, the we have a 0 = b 0 = 0, a k+1 = a k + 1, ad b k+1 = 2b k a k 1. By iductio, we fid that a k = k ad b k = 2 k+1 + k + 2. Settig = 2 k 1, we have as required. m 2 k 1 = k (2k 1) 2 k+1 + k + 2 = (k 2)2 k + 2, To cout the umber of orders which require the miimum umber of comparisos, ote that the first step i the algorithm splits the list ito sublists of legth i ad i 1, say; ow we require that each of these sublists ca be sorted with the miimum umber of comparisos, ad also that both i ad 1 i lie i a iterval i which m j is a liear fuctio of j, that is, oe of the form [2 d 1 1,2 d 1]. However, the two sublists ca be merged arbitrarily. So the umber f () of orders requirig the miimum umber of comparisos satisfies the recurrece ( ) 1 f () = f (i) f ( i 1), i 14

15 where the sum is over all i such that i ad i 1 lie i the same iterval of the form [2 d 1 1,2 d 1]. For example, if = 5 ad the list cotais 1,...,5 i some order, the it requires the miimum umber 6 of comparisos to sort if ad oly if oe of the followig holds: the first elemet is 3; the first elemet is 2, ad 4 precedes 3 ad 5; the first elemet is 4, ad 2 precedes 1 ad 3. There are 40 such lists, out of 120. As i the previous questio, the maximum umber of comparisos is 10, ad 16 lists require this umber. Check that the umbers requirig 7, 8, 9 comparisos are 32, 24, 8 respectively. Now check that the average agrees with the value calculated i the recurrece of Sectio

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