Discrete Mathematics Recurrences

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1 Discrete Mathematics Recurreces Saad Meimeh 1 What is a recurrece? It ofte happes that, i studyig a sequece of umbers a, a coectio betwee a ad a 1, or betwee a ad several of the previous a i, i <, is obtaied. This coectio is called a recurrece relatio. I spirit, a recurrece is similar to iductio, but while iductio is a proof techique, recurrece is more like a defiitio method. 2 Cuttig the plae Cosider lies i the plae such that: o two lies are parallel o three lies pass through the same poit We refer to such a cofiguratio as lies i geeral positio. The lies divide the plae ito a umber of regios. How may regios do we get? At the first glace, oe might thik that addig a lie doubles the umber of regios. For istace, oe lie divides the plae ito two regios; two lies ito four. However, it is easy to verify whe = 3 that this tred does ot cotiue; otherwise, we ed up with 2 regios. At this jucture, it might be reasoable to start experimetig with = 3 lies. The you will ivariably get 7 regios. With = 4 lies, you will always get 11 regios. It looks like after addig the th lie, the umber of regios icreases by. Let s prove this fact: The th lie itersects the other 1 lies i exactly 1 distict poits because the lies are i geeral positio. Tracig the th lie as it passes through these 1 poits, we get to a ew regio every time we cross a poit. So the umber of regios the th lie creates is oe larger tha the umber of crossig poits, thus. So if R is the umber of regios we get with lies i geeral positio, the R = R 1 +

2 ad we start with R 1 = 2. We ca easily guess relate this to the triagular umbers that + 1 R = which ca be proved by iductio. 3 Towers of Haoi The problem of the towers of Haoi is based o a system of discs of differet sizes that fit over three fixed pegs. At the start we have all the discs arraged o peg oe i size order smallest o top. The object is to trasport this whole pile to peg three by a series of moves. A move cosists of trasportig the top disc from oe peg to the top of a pile o either peg subject to the coditio that o disc is ever placed o a smaller disc. Let a be the miimum umber of moves required to complete the process. How big is a? It should be oted that the easiest way of provig that this process ca be performed at all is to calculate a. Evetually it will be ecessary to move the bottom largest disc from peg oe to peg three. The tur before this is doe all the 1 other discs are arraged o peg two. The miimum umber of moves required to shift these to peg two is a 1. The after the bottom disc is moved, the 1 smaller discs must be reassembled o top of it. This agai requires at least a 1 moves. This proves the followig recurrece: a = 2a from which a ca be determied. This recurrece, alog with the iitial coditio a 1 = 1, eables us to fid a. We have a 2 = = 3, a 3 = = 7, a 4 = = 15. It appears that a = 2 1. Let s cofirm this by a iductio proof. Our base case is a 1 = 1. For every > 1, assume that the property holds up to 1, ad prove that it remais true for. a = 2a = = = 2 1 Aother way to obtaiig this result is by repeatedly evaluatig a usig the recurrece: a = 2a = 22a = 222a =...

3 Cotiuig this way, we get a geometric series the followig equality ca be also proved by iductio 4 Fiboacci = = 2 1 Cosider a staircase with steps. We ca walk up by either takig oe or two steps at a time. I how may ways ca we go up the stairs? Let a deote this umber.clearly a 0 = 1 ad a 1 = 1. For > 1, we must start by oe of the two optios show below top view: I the first case we ca cotiue i a 1 ways, ad i the secod case we ca cotiue i a 2 ways. Sice we start differetly, by the additio priciple: a = a 1 + a 2 Therefore, we obtai a 2 = = 2, a 3 = = 3, a 4 = = 5, a 5 = = 8, a 6 = = 13, etc... The sequece a resembles the well kow Fiboacci sequece F a Clearly, a = F +1. We will later obtai a formula for F. 5 Flags A flag cosists of horizotal stripes, where each stripe ca be ay oe of blue, red, ad white, o two adjacet stripes havig the same color. Uder these coditios, the top stripe ca be ay of three colors, the secod has two possibilities, the third has two, ad so oe each stripe avoidig the color of the oe above it. By the multiplicatio rule there are possible flags. Suppose ow that, i order to avoid flyig the flag upside dow, the top ad bottom stripes should be of differet colors. How may such flags with stripes are possible? Let a deote that umber. If we let b be the umber of flags with matchig top ad bottom, the a + b = Here s a iterestig observatio: there is a oe-to-oe correspodece betwee flags of stripes with matchig top ad bottom, ad flags of 1 stripes with differet top ad bottom. Here s why: Give a flag of stripes with matchig top ad bottom, we ca remove the bottom stripe to uiquely obtai a flag of 1

4 stripes with differet top ad bottom. Similarly, give a flag of 1 stripes with differet top ad bottom, we ca duplicate the top stripe at the bottom to uiquely obtai a flag of stripes with matchig top ad bottom. By the oe-to-oe correspodece, b = a 1. Therefore, Similarly, a + a 1 = a 1 + a 2 = By multiplyig the secod equatio by 2 ad subtractig it from the first we get: a = a 1 + 2a 2 Kowig that a 1 = 0 why? ad a 2 = 6, we should be able to solve for a. It is ow time to show how to solve this type of recurrece. 6 Solutio for a = Aa 1 + Ba 2 Give the recurrece a = Aa 1 + Ba 2 ad two iitial coditios, say a 1 ad a 2, costruct the followig auxiliary equatio: x 2 = Ax + B Let p ad q be the solutios of the equatio. Case 1: p q The a = c 1 p + c 2 q satisfies the recurrece: Similarly, Therefore, after addig both sides: p 2 = Ap + B p = Ap 1 + Bp 2 c 1 p = c 1 Ap 1 + c 1 Bp 2 c 2 q = c 2 Aq 1 + c 2 Bq 2 c 1 p + c 2 q = Ac 1 p 1 + c 2 q 1 + Bc 1 p 2 + c 2 q 2 a = Aa 1 + Ba 2

5 Case 2: p = q First ote that if this is the case the A = 2p ad B = p 2 because x 2 = Ax+B becomes equivalet to x p 2 = x 2 2px + p 2 = 0. The a = c 1 p + c 2 p satisfies the recurrece: p 2 = Ap + B Similarly, p = Ap 1 + Bp 2 c 1 p = c 1 Ap 1 + c 1 Bp 2 c 2 p = c 2 Ap 1 + c 2 Bp 2 c 2 p = c 2 A 1p 1 + c 2 B 2p 2 + c 2 Ap 1 + 2c 2 Bp 2 }{{} The isolated above quatity is zero. Therefore, after addig both sides: c 1 p + c 2 p = A[c 1 p 1 + c 2 1p 1 ] + B[c 1 p 2 + c 2 2p 2 ] a = Aa 1 + Ba 2 I both cases, c 1 ad c 2 ca be obtaied from the two iitial coditios by equatig the expressios for a 1 ad a 2 to the values give. We ed up with a solutio for a that satisfies the recurrece ad the two iitial coditios. A similar techique ca be used to solve recurreces of the form a = A 1 a 1 + A 2 a A k a k by formig the auxiliary equatio x k = A 1 x k A k 1 x + A k, but we will ot cosider it here. 7 Let s capture that rabbit! The famous Fiboacci recurrece was associated by its ivetor with a hypothetical rabbit problem. Rabbits obey the followig law: every pair of rabbits, at least two moths old, will produce oe pair of rabbits every moth. Startig with oe pair of ewly bor rabbits, the th Fiboacci umber, F, is the umber of pairs at the begiig of the th moth by covetio F 0 = 0. Usig the above techique, let s solve for F. Kowig that we form the auxiliary equatio: F = F 1 + F 2 x 2 = x + 1 which has two solutios φ = 1 + 5/2 ad 1 φ. Therefore, our solutio for F has the form: c 1 φ + c 2 1 φ Let s determie c 1 ad c 2 from two iitial coditios. F 0 = c 1 φ 0 + c 2 1 φ 0 = c 1 + c 2 = 0

6 F 1 = c 1 φ 1 + c 2 1 φ 1 = c 1 φ + c 2 1 φ = 1 We obtai c 1 = c 2 = 1/2φ 1 = 1/ 5. Therefore, F = 1 5 [φ 1 φ ] Observe that 1 φ is approximately zero whe is large; therefore, F ca be obtaied for large by fidig the iteger value closest to the followig: F 1 5 φ 8 Let s fly the flag We previously determied that a = a 1 + 2a 2 is the umber of good flags. Let s solve for a. We form the auxiliary equatio: x 2 = x + 2 which has two solutios 2 ad 1. Therefore, our solutio for a has the form: c c 2 1 Let s determie c 1 ad c 2 from two iitial coditios. a 1 = c c = 2c 1 c 2 = 0 a 2 = c c = 4c 1 + c 2 = 6 We obtai c 1 = 1 ad c 2 = 2. Therefore, 9 Aother example a = Let s solve the recurrece a = 4a 1 4a 2 with a 0 = 0 ad a 1 = 2. We form the auxiliary equatio: x 2 = 4x 4 which has oe solutio 2 techically two solutios that are equal. Therefore, our solutio for a has the form: c c 2 2 Let s determie c 1 ad c 2 from the two iitial coditios. a 0 = c c = c 1 = 0 a 1 = c c = 2c 1 + 2c 2 = 2 We obtai c 1 = 0 ad c 2 = 1. Therefore, a = 2

7 10 Geeratig fuctios Goig back to the flag problem, the first recurrece that we established was a = a which does ot have the form a = Aa 1 + Ba 2. Geeratig fuctios represet aother method for solvig recurreces that we will ot fully explore. The geeratig fuctio of a sequece a 0, a 1, a 2... is defied as fx = a i x i = a 0 + a 1 x + a 2 x 2 + a 3 x i=0 For example, the geeratig fuctio of the Fiboacci sequece is x + x 2 + 2x 3 + 3x 4 + 5x Sometimes, give a recurrece, it is possible to fid the geeratig fuctio of the sequece ad the fid a by readig off the coefficiet of x. I will illustrate this for the flag problem. fx = a 1 x + a 2 x 2 + a 3 x = a 1 x a 1 x a 2 x = a 1 x + 32x x a 1 x 2 + a 2 x = 0 + 6x x x xfx We ca always assume that 2x < 1 so the geometric series above will coverge. fx = fx = 0 + 6x x xfx 6x x1 2x = 2x x x Expadig back ito geometric series, we get: fx = 4x x x x 2 1 x + x 2... Readig off the coefficiet of x gives a = =

8 11 Sortig Cosider the task of sortig a list of umbers from smallest to largest. Oe possible algorithm is the followig: idetify the smallest umber i the list ad swap it with the first umber. This correctly places the smallest umber at the begiig of the list. The process ca be ow repeated o the remaiig umbers by cosiderig them as a list of 1 umbers. To idetify the smallest umber i a list of umbers oe eeds to make 1 comparisos because each compariso ca elimiate oe umber as beig ot the smallest. Therefore, the total umber of comparisos is = This algorithm is called selectio sort. We say that selectio sort has a order 2 ruig time, deoted O 2. A better way to sort is usig the idea of mergig sorted lists. Give a list of umbers, split the list i two equal halves or almost equal if is ot eve. If we ca maage to sort these two halves separately, the we ca obtai a sorted list by mergig them as follows. Compare the first smallest umbers i the lists, ad take the smaller as the first umber of a ew list, crossig it out of its list. Repeat the process to fid the secod smallest, ad so o util oe of the two lists is exhausted. Apped the other list to the ed of the ew list. The umber of comparisos is at most 1 because each time we make a compariso we cross out a umber. However, sortig the two halves of the origial list usig selectio sort is ot goig to buy us ay gai, because the umber of comparisos would be approximately /2 2 + / = 2 / This is still a O 2 algorithm. But what happes if we recursively sort the two halves usig the same method? For simplicity of illustratio, let us assume that is a power of 2, so we ca always divide by 2 util we get 1. Let t be the umber of comparisos performed o a list of umbers. The Let a = t 2, the t = 2t /2 + 1 a = 2a with a 0 = 0 why?. Iteratig this few times reveals that a = which ca the be proved by iductio try. This meas t = a log2 = 2 log 2 log = log Aother possibility is to sacrifice accuracy ad seek a upper boud by sayig that comparisos are eeded as opposed to 1 to merge two lists of size /2. The

9 Similarly, t = 2t /2 + a = 2a a 1 = 2a By multiplyig the secod equatio by 2 ad subtractig it from the first we get: a = 4a 1 4a 2 with a 0 = 0 ad a 1 = 2 why?. a = 2. Therefore, We already foud the solutio for this as t = a log2 = log 2 2 log 2 = log 2 This algorithm is called merge sort ad it makes O log comparisos, a improvemet over the O 2 selectio sort algorithm. 12 Travelig i Mahatta ad the Catala umbers Imagie travelig i Mahatta. How may shortest paths ca you take by walkig o lies goig from poit A to poit B show below? Note that o a B A shortest path you ca oly move to the right or up. If we ecode each path by a strig of zeros right moves ad oes up moves, we will observe that every strig has legth 8 because every path must cosist of 8 moves. Moreover, 5 of those moves must be to the right ad 3 must be up. Therefore, our biary strig of legth 8 will cotai exactly 5 zeros ad 3 oes. It is ot hard to see that there is a oe-to-oe correspodece betwee shortest paths ad biary strigs of 5 zeros ad 3 oes. Every shortest path ca uiquely be ecoded by such a strig, ad coversely every such a strig ca uiquely idetify a path. Therefore, the umber of shortest paths is exactly the umber of biary strigs with 5 zeros ad 3 oes. How may are there? Well, we have to choose 5 zeros from 8 bits ad the remaiig 3 bits will be 1. Equivaletly, we ca choose 3 oes from 8 bits ad the remaiig 5 bits will

10 be 0. Hopefully this should ot be a surprise because you kow by ow that =. The umber of shortest paths is k k 8 8 = = I geeral, if we have a m grid, the umber of shortest paths will be m + m + = m ad if the grid is a square grid, i.e. m =, the that umber will be 2 Suppose we ow have a square grid, ad ask for the umber C of shortest paths from poit A to poit B which ever go above the diagoal AB. Let s call these the good paths. For the case of = 3 show below, there are 5 such paths; ecoded as biary strigs they are: , , , , Therefore, C 3 = 5. B A Ay good path from A to B must touch the diagoal at some poit prior to reachig B, eve if this happes oly at A. Let Ci, i be the last such poit, where 0 i < refer to the figure below. There are C i good paths from A to C. The path must the proceed from C to D ad evetually to E, but it must ever go above the diagoal DE, sice otherwise this would ivalidate the choice of C. But D ad E defie a square grid of size i 1, so there are C i 1 good paths from D to E. Therefore, by the multiplicatio priciple, there are C i C i 1 good paths with i, i as the last cotact with the diagoal AB. Sice i ca take ay value from 0 to 1, it follows from the additio priciple that C = C 0 C 1 + C 1 C 2 + C 2 C C 1 C 0 with a iitial coditio C 0 = 1 because i a grid of size 0, where A ad B coicide, there is oly oe way to go from A to B o a shortest path without goig above the diagoal; simply stay there!

11 B, E -i-1 Ci,i D A0,0 The sequece C defies the Catala umbers: but ca we fid a better expressio for C? Oe could possibly solve the recurrece usig the method of geeratig fuctios. But there is a easier way usig a oe-to-oe correspodece. Observe that the umber of good paths from A to B must be equal to the total umber of paths mius the umber 2 of shortest paths which do go above the diagoal AB let s call these the bad paths. Cosider ay bad path. There will be a first poit o that path above the diagoal AB, say Ri, i + 1. If we replace the part of the path from A to R by its mirror image with respect to GF refer to the figure below, the we get a shortest path from H 1, 1 to B,. Coversely, ay shortest path from H to B must cross GF somewhere, ad defies precisely oe bad path from A to B by mirrorig. Therefore, the umber of bad paths from A to B is equal to the umber of shortest paths from H to B, which is = Fially, C = = = Catala umbers appear i may situatios. Oe immediate iterpretatio is that C is the umber of biary strigs of legth 2 cotaiig exactly zeros ad oes, such that the umber of 1s up to a poit ever exceeds the

12 F B, Ri,i+1 H-1,1 G A0,0 umber of 0s. If we replace 0 by ad 1 by, C gives us the umber of valid parethesized expressios of legth 2 with pairs of parethesis. I how may ways ca we carry out the sum a 1 + a a? Let s take the example of = 4, so we have a 1 + a 2 + a 3 + a 4. We ca do the followig: a 1 + a 2 + a 3 + a 4 a 1 + a 2 + a 3 + a 4 a 1 + a 2 + a 3 + a 4 a 1 + a 2 + a 3 + a 4 a 1 + a 2 + a 3 + a 4 If S is the umber of ways of carryig out the sum a 1 + a a, it is ot hard to see that S 1 = 1 ad S 2 = 1 ad that S = S 1 S 1 + S 2 S S 1 S 1 S behaves exactly like C with oe idex shift. Therefore, S = C Stirlig umbers Cosider partitioig a set of size ito k subsets. This is similar to a problem that we have see before, amely x 1 + x x k = where x i 0 is a iteger. But there is differece. I the equatio above, the partitios are labeled. What is beig partitioed is ot. Now we have the opposite. The partitios are ot labeled, ad what is beig partitioed is. For example, while 4 ca be partitioed ito two labeled parts as follows 5 ways:

13 the S = {1, 2, 3, 4} ca be partitioed ito two subsets i the followig ways: {1} {2, 3, 4}, {2} {1, 3, 4}, {3} {1, 2, 4}, {4} {1, 2, 3}, {1, 2} {3, 4}, {1, 3} {2, 4}, {1, 4} {2, 3}. Let S, k deote the umber of ways we ca partitio a set of size ito k subsets. The, S4, 2 = 7. It is trivial that S, 1 = S, = 1. For other values of k, there is a ice recurrece for S, k. S, k = S 1, k 1 + ks 1, k where 1 < k <. Why is the above recurrece true? I ay partitio of S, the first elemet ca appear i a subset by itself or ca occur i a larger subset. If it appears by itself, the the remaiig 1 elemets are partitioed ito k 1 subsets, ad there are S 1, k 1 ways i which this ca be doe. O the other had, if the first elemet occurs i a subset of size at least two, the remaiig 1 elemets are partitioed ito k subsets - this ca be doe i S 1, k ways - ad the first elemet is itroduced ito o of these k subsets - ad this ca be doe i k ways. So by the additio ad multiplicatio priciples, we have S, k = S 1, k 1 + ks 1, k. It is ot hard to prove the followig iterestig equality defie Sm, 0 = 0 for m > 0 ad S0, 0 = 1: m = k=0 Sm, k k k! We will do this by a combiatorial argumet. Cosider placig m labeled balls ito labeled bis. There are m ways of doig this why?. Now imagie that we do it by first choosig a subset of bis of size k ca be doe i k ways, the partitioig the balls ito k subsets ca be doe i Sm, k ways, ad fially assigig the k subsets to the k bis ca be doe i k! ways. By the multiplicatio priciple, Sm, k k! is the umber of ways of placig k m balls ito exactly k of the bis. By the additio priciple, summig up from k = 1 to k = should give us the total m. As a special case whe k =, Sm,! is the umber of ways of placig m labeled balls ito exactly bis. It should be clear that this is also the umber

14 of oto fuctios f : X Y whe X = m ad Y =. It is possible to show that Sm,! = 1 k k m k k=0 which is sort of a iverted form of the above formula for m. Sice Sm,! is the umber of oto fuctios from X to Y, it must be equal to m the total umber of fuctios mius the umber of fuctios that leave at least oe elemet of Y umapped. Observe that the umber of fuctios that leave a give k elemets of Y umapped is k m. Therefore, by the iclusioexclusio priciple: [ Sm,! = m 1 m 2 m m] 1 2 }{{} Sm,! = 1 k k k=0 umber of fuctios that leave some elemets of Y umapped k m = 1 k k=0 k k m = 1 k k Note that the above quatity is zero whe m <. Example: cosider the task of rakig movies. You may put movies i the same rak ad you may exclude movies from beig raked at all. I how may ways ca we perform this task? Puttig i movies ito j raks ca be doe by first partitioig the i movies ito j sets i Si, j ways ad the orderig the j sets i j! ways, leadig to Si, jj! ways. Therefore, the total umber of ways of rakig movies is: i Si, jj! = i i=0 j=0 14 Balls ad bis i i=0 j=0 k=0 j i 1 j k j k k i I this sectio we cosider the differet variatios of placig balls i k bis. The results are summarized i the table below: Balls Bis Ay Bi Number Labeled Labeled Empty of Ways 1 No Yes No k + k 1 No Yes Yes Yes No No S, k Yes No Yes S, 1 + S, S, k Yes Yes No S, kk! Yes Yes Yes k k=0 k m

15 The secod etry refers to the problem x 1 + x x k = where x i 0 are itegers. The first etry refers to the problem x 1 + x x k = where x i 1 are itegers, which is equivalet to: x 1 + x x k = k where x i 0 are itegers why?. The case where both balls ad bis are ulabled is related to uordered iteger partitios that we do t cover here. For istace, 5 ca be partitioed i the followig ways:

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