Analysis of Execution Costs for QuickSelect. Takéhiko Nakama

Transcription

1 Aalysis of Executio Costs for QuickSelect by Takéhiko Nakama A dissertatio submitted to The Johs Hopkis Uiversity i coformity with the requiremets for the degree of Doctor of Philosophy. Baltimore, Marylad August, 2009 c Takéhiko Nakama 2009 All rights reserved

2 Abstract QuickSelect is a search algorithm widely used for fidig a key of target rak i a file of keys. Whe the algorithm compares keys durig its executio, it must operate o the keys represetatios or iteral structures, which were igored by the previous studies that quatified the executio cost for the algorithm i terms of the umber of required key comparisos. I this dissertatio, we coduct aalyses of ruig costs for the algorithm that take ito accout ot oly the umber of key comparisos but also the cost of each key compariso. First we suppose that keys are idepedet ad uiformly distributed i the uit iterval (0, 1) ad are represeted as their biary expasios, ad we derive exact ad asymptotic expressios for the expected umber of bit comparisos required by QuickSelect. We also establish a closed formula for the expectatio that oly ivolves fiite summatio ad use it to compute the expected cost for various values of the target rak ad the umber of keys. The we ivestigate executio costs for the algorithm applied to keys that are represeted by more geeral sequeces of symbols, ad we idetify limitig distributios associated with the costs. Further, we derive itegral ad series expressios for the expectatios of the limitig distributios ad use them to recapture previously obtaied ii

3 ABSTRACT results o the umber of key comparisos required by QuickSelect. Primary Reader: Professor James Alle Fill Secodary Reader: Professor S. Rao Kosarau iii

4 Ackowledgemets It has truly bee a hoor ad a great pleasure workig with Professor James Alle Fill. I thak him for costatly ispirig ot oly with his exceptioal brilliace but also with his character. We are grateful for Professor Brigitte Vallée s collaboratio o the limitigdistributio aalysis described i Chapters 5 8 ad for Professor Rao Kosarau s helpful commets o a draft of this dissertatio. I am also wholeheartedly grateful to umerous people who have provided me with icalculable support; I would ot have bee able to coduct graduate studies without them. Fially, I thak Johs Hopkis. I have ever take for grated the privilege of pursuig my various itellectual iterests at this remarkable istitutio, which has provided me with woderful opportuities to lear. iv

5 Cotets Abstract ii Ackowledgemets iv List of Figures viii 1 Itroductio ad summary 1 2 Aalysis of µ(1, ) Prelimiaries Exact computatio of µ(1, ) Asymptotic aalysis of µ(1, ) Aalysis of µ( m, ) Exact computatio of µ( m, ) Exact computatio of µ 1 ( m, ) Exact computatio of µ 2 ( m, ) ad µ( m, ) v

6 CONTENTS 3.2 Asymptotic aalysis of µ( m, ) Derivatio of a closed formula for µ(m, ) 49 5 Backgroud ad prelimiaries for limitig-distributio aalysis Probabilistic source models for the keys Kow results for the umbers of key ad symbol comparisos QuickQuat ad QuickVal Aalysis of QuickVal Prelimiaries Covergece of S V / i L p for 1 p < Almost sure covergece of S V / Aalysis of QuickQuat Prelimiaries Covergece of S Q / i L p for 1 p < Aalysis of E S Computatio of E S: a itegral expressio Computatio of E S: a series expressio A Proof of (2.28) 95 B Tractable expressio for the measure ν 97 vi

7 CONTENTS Bibliography 106 Vita 110 vii

8 List of Figures 4.1 Expected umber of bit comparisos for Quickselect viii

9 Chapter 1 Itroductio ad summary QuickSelect, itroduced by Hoare [12] i 1961 ad also kow as Fid, is a search algorithm widely used for fidig a key of target rak i a file of keys. Suppose that there are distict keys (the keys are typically assumed to be distict, but the algorithm still works with a mior adustmet eve if they are ot distict) ad that the target rak is m, where 1 m. The algorithm fids the target key i a recursive ad radom fashio. First, it selects a pivot uiformly at radom from keys. Let k deote the rak of the pivot. If k = m, the the algorithm returs the pivot. If k > m, the the algorithm recursively operates o the set of keys smaller tha the pivot ad returs the rak-m key. Similarly, if k < m, the the algorithm recursively operates o the set of keys larger tha the pivot ad returs the (m k)-th smallest key from the subset. May studies have examied QuickSelect to quatify its executio costs (a oexhaustive list of refereces is Kuth [15], Mahmoud et al. [19], Prodiger [22], Grübel 1

10 CHAPTER 1. INTRODUCTION AND SUMMARY ad U. Rösler [11], Let ad Mahmoud [18], Grübel [10], Mahmoud ad Smythe [20], Devroye [4], Hwag ad Tsai [14], Fill ad Nakama [6], ad Vallée et al. [27]), ad all of them except for Fill ad Nakama [6] ad Vallée et al. [27] have coducted the quatificatio with regard to the umber of key comparisos required by the algorithm to achieve its task. As a result, most of the theoretical results o the complexity of QuickSelect are about expectatios or distributios for the umber of required key comparisos. (We will soo retur to [6] ad [27].) However, oe ca reasoably argue that aalyses of QuickSelect i terms of the umber of key comparisos caot fully quatify its complexity. For istace, if keys are represeted as biary strigs, the idividual bits of the strigs must be compared i order for QuickSelect to complete its task. Cosider applyig QuickSelect to fid the smallest key amog three keys k 1, k 2, ad k 3 whose biary represetatios are , , ad , respectively. If the algorithm selects k 3 as a pivot, the it compares each of k 1 ad k 2 to k 3 i order to determie the rak of k 3. Whe k 1 ad k 3 are compared, the algorithm requires 2 bit comparisos to determie that k 3 is smaller tha k 1 because the two keys have the same first digit ad differ at the secod digit. Similarly, whe k 2 ad k 3 are compared, the algorithm requires 4 bit comparisos to determie that k 3 is smaller tha k 2. After these comparisos, k 3 has bee idetified as smallest. Hece the search for the smallest key requires a total of 6 bit comparisos (resultig from the two key comparisos). This simple case illustrates that the umber of bit comparisos required by the algorithm more accurately reflects the actual executio cost. (We will co- 2

11 CHAPTER 1. INTRODUCTION AND SUMMARY sider bit comparisos as a example of symbol comparisos.) Whe QuickSelect (or ay other algorithm) compares keys durig its executio, it must operate o the keys represetatios or iteral structures, so these should ot be igored i fully characterizig the performace of the algorithm. Also, symbol-complexity aalysis allows us to compare key-based algorithms such as QuickSelect ad QuickSort with digital algorithms such as those utilizig digital search trees. Fill ad Jaso [5] pioeered symbol-complexity aalysis by aalyzig the expected umber of bit comparisos required by QuickSort. This well-kow ad popular sortig algorithm was also iveted by Hoare [13], ad, like QuickSelect, it uses a divide-adcoquer strategy. (I fact, QuickSelect is a oe-sided versio of QuickSort.) The algorithm picks a pivot key uiformly at radom from the give file of keys ad compares the other keys with the pivot to form two subfiles; oe of them cosists of keys larger tha the pivot, ad the other subfile cosists of keys smaller tha the pivot. QuickSort the recursively performs the same procedure o each of the resultig subfiles cotaiig more tha oe key. Fill ad Jaso assumed that QuickSort is applied to keys that are i.i.d. (idepedet ad idetically distributed) from the uiform distributio over (0, 1) ad represeted (via their biary expasios) as biary strigs, ad that the algorithm operates o idividual bits i order to sort the keys. They foud that the expected umber of bit comparisos required by QuickSort to sort keys is asymptotically equivalet to (l )(lg ) (where lg deotes biary logarithm), whereas the lead-order term of the expected umber of key comparisos is 2 l, smaller by a factor of order log. I their Sectio 6 they also 3

12 CHAPTER 1. INTRODUCTION AND SUMMARY cosidered i.i.d. keys draw from other distributios with desity o (0, 1). Their symbol-complexity aalysis was closely followed by Fill ad Nakama [6], who ivestigated the expected umber of bit comparisos required by QuickSelect. Chapters 2 4 of this dissertatio describe the results of [6] i detail. I these chapters, we assume that the algorithm is applied to keys that are i.i.d. uiform(0,1) ad that the keys are represeted as their biary expasios. We let µ(m, ) deote the expected umber of bit comparisos required to fid the rak-m key i a file of keys by QuickSelect. By symmetry, µ(m, ) = µ(+1 m, ). I this dissertatio, we will refer to QuickSelect for fidig the smallest key ad the largest key as QuickMi ad QuickMax, respectively. First, we develop exact ad asymptotic formulae for µ(1, ) = µ(, ), as summarized i the followig theorem. Theorem 1.1. The expected umber µ(1, ) of bit comparisos required by QuickMi has the followig exact ad asymptotic expressios: 1 µ(1, ) = 2(H 1) + 2 = c 1 l 2 (l )2 B + 1 ( ) ( 1)(1 2 ) ( 2 l (1.1) ) l + O(1), (1.2) where H ad B deote harmoic ad Beroulli umbers, respectively, ad 2 k c := k l = (1.3) 2 k k=0 =1 With χ k := 2πik l 2 ad γ := Euler s costat. = , the costat c ca alteratively be 4

13 CHAPTER 1. INTRODUCTION AND SUMMARY expressed as c = γ 9 l 2 4 l 2 k Z\{0} ζ(1 χ k )Γ(1 χ k ) Γ(4 χ k )(1 χ k ). (1.4) The asymptotic formula shows that the expected umber of bit comparisos is asymptotically liear i with lead-order coefficiet approximately equal to Hece the expected umber of bit comparisos is asymptotically differet from that of key comparisos required to fid the smallest key oly by a costat factor (the expectatio for key comparisos is asymptotically 2). Details of the derivatios of the formulae are described i Chapter 2. Complex-aalytical methods are utilized to obtai the asymptotic formula (1.2) with c i the form (1.4) ad seem to be idispesable for obtaiig asymptotics beyod the lead term. [We remark that, although it ivolves the imagiary umbers χ k, the expressio (1.4) is real because the terms with idices k ad k are complex cougates.] I Sectio 2.3 we agai use complex-aalytical methods to reexpress (1.4) i the form (1.3). Havig doe all this we suspected that there must be a purely real-aalytical way to obtai directly the lead-order asymptotics µ(1, ) c with c i the form (1.3). Ideed, there is: See Remark 2.3. I Chapter 3, we move o to derive exact ad asymptotic expressios for the expected umber of bit comparisos for the average case, where the target rak is chose uiformly radomly from {1, 2,..., }. We deote this expectatio by µ( m, ); hece µ( m, ) = 1 m=1 µ(m, ). QuickSelect for fidig a key of radom rak will be referred to as QuickRad. The derived asymptotic formula shows that µ( m, ) is also asymptotically 5

14 CHAPTER 1. INTRODUCTION AND SUMMARY liear i ; see (3.45). More detailed results for µ( m, ) are described i Chapter 3. I Chapter 4, we derive a exact expressio of µ(m, ) for each fixed m that is suited for computatios. Our prelimiary exact formula for µ(m, ) [show i (2.8)] etails ifiite summatio ad itegratio. As a result, it is ot a desirable form for umerically computig the expected umber of bit comparisos. Hece we establish aother exact formula that oly requires fiite summatio ad use it to compute µ(m, ) for m = 1,...,, = 2,..., 25. The computatio leads to the followig coectures: (i) for fixed, µ(m, ) icreases i m for m +1 ad is symmetric about +1 ; ad (ii) for fixed m, µ(m, ) 2 2 icreases i (asymptotically liearly). Vallée et al. [27] exteded the results of Fill ad Jaso [5] ad Fill ad Nakama [6] by ivestigatig QuickSort ad QuickSelect with respect to the umber of required comparisos of more geeral symbols. They cosidered keys represeted by sequeces of symbols geerated by ay of a wide variety of sources that iclude memoryless, Markov, ad other dyamical sources. (I both Fill ad Nakama [6] ad all but Sectio 6 of Fill ad Jaso [5], the symbols that compose each key represetatio are assumed to be radom uiform bits, geerated by a memoryless source.) We will provide details of their probabilistic source models i Sectio 5.1, sice we will use them i formulatig our limitigdistributio results for QuickSelect. Roughly summarized, Vallée et al. showed that the expected umber of symbol comparisos i processig a file of keys is of order log 2 for QuickSort ad of order for QuickSelect (applied to fid a key of target rak m with m / α for ay give α [0, 1]) if symbols are geerated by a suitably 6

15 CHAPTER 1. INTRODUCTION AND SUMMARY ice source. (For example, all memoryless sources are suitably ice.) For a more detailed discussio of sources ad the results of Vallée et al. [27] for QuickSelect, see Sectio 5.1. I this dissertatio, we further deepe the quatificatio of the performace of QuickSelect i terms of the umber of symbol comparisos used by the algorithm. I Chapters 5 8, we ivestigate limitig distributios for the umber of symbol comparisos required by QuickSelect. We believe that our study is the first to establish a limitig distributio for the umber of symbol comparisos required by ay key-based algorithm. Our elemetary approach allows us to hadle rather geeral kids of cost for comparig two keys, ad i particular to recover i a rather direct way kow results about the umber of key comparisos. There is o disadvatage to allowig geeral costs, sice our results rely o at most broad limitatios o the ature of the cost. I the last four chapters, we shall be cocered primarily with QuickQuat QuickQuat(, α), which is what we call the algorithm QuickSelect whe applied to fid the key of rak m i a file of size, where we are give 0 α 1 ad a sequece (m ) such that m / α. It turs out to be coveiet mathematically to aalyze a close cousi to QuickQuat itroduced by Vallée et al. [27], amely, QuickVal QuickVal(, α) (this algorithm is fully described i Sectio 5.3), ad the treat QuickQuat by compariso. For the same value of α, the operatio of QuickVal is quite close to that of QuickQuat, ad executio costs of the two algorithms are expected to be close; i fact, we will show that if S Q S Q (α) ad S V S V (α) deote the total costs of executig QuickQuat ad QuickVal, respectively, the S Q / ad S V / 7

16 CHAPTER 1. INTRODUCTION AND SUMMARY have the same limitig distributio uder proper coditios o the cost fuctio. I order to prove the covergece i distributio, we will defie all the radom variables S Q 1, S Q 2,... ad S1 V, S2 V,... o a commo probability space ad show that S Q / ad S V / both coverge i L p to a commo limit, call it S, for 1 p <. The limit S is defied at (6.7); see the ed of Sectio 6.1. (Uder a techical assumptio, we will also prove almost sure covergece for QuickVal i Sectio 6.3.) I order to establish limitig-distributio results for QuickVal ad QuickQuat, we first provide a careful descriptio of the probabilistic models used to gover the geeratio of keys i Sectio 5.1. The we review kow results about key ad symbol comparisos i Sectio 5.2 ad describe QuickVal i Sectio 5.3. I Chapter 6, we preset covergece results for QuickVal; see Theorems 6.1 ad 6.4. Fially we move o to QuickQuat i Chapter 7. Theorem 7.1 is the mai result of our limitig-distributio aalysis. We will also aalyze E S i Chapter 8 ad derive itegral ad series expressios for it. These expressios will be used to recover the previously obtaied results reviewed i Sectio

17 Chapter 2 Aalysis of µ(1, ) I this chapter, we aalyze the expected umber of bit comparisos required by QuickMi (QuickSelect for fidig the smallest key). By symmetry, the expectatio equals that for QuickMax (QuickSelect for fidig the largest key). As described i Chapter 1, we assume that QuickMi is applied to distict keys that are represeted as bit strigs ad that the algorithm operates o idividual bits i order to compare keys ad fid the target key. It is further assumed that the keys are uiformly ad idepedetly distributed i (0, 1). I Sectio 2.1, we describe the framework ad otatio for our bitcomplexity aalysis. I Sectio 2.2, we derive a exact expressio for µ(1, ). We coduct a asymptotic aalysis of this expressio i Sectio

18 CHAPTER 2. ANALYSIS OF µ(1, ) 2.1 Prelimiaries To ivestigate the bit complexity of Quickselect, we follow the geeral approach developed by Fill ad Jaso [5]. Let U 1,..., U deote the keys uiformly ad idepedetly distributed o (0, 1), ad let U (i) deote the rak-i key. The, for 1 i < (assume 2), P {U (i) ad U () are compared} = 2 m i m i + 1 if m i if i < m < if m. (2.1) To determie the first probability i (2.1), ote that U (m),..., U () remai i the same subset util the first time that oe of them is chose as a pivot. Therefore, U (i) ad U () are compared if ad oly if the first pivot chose from U (m),..., U () is either U (i) or U (). Aalogous argumets establish the other two cases. is give by For 0 < s < t < 1, it is well kow that the oit desity fuctio of U (i) ad U () f U(i),U () (s, t) := ( ) s i 1 (t s) i 1 (1 t). i 1, 1, i 1, 1, (2.2) 10

19 CHAPTER 2. ANALYSIS OF µ(1, ) Clearly, the evet that U (i) ad U () are compared is idepedet of the radom variables U (i) ad U (). Hece, defiig P 1 (s, t, m, ) = P 2 (s, t, m, ) = P 3 (s, t, m, ) = m i< 1 i<m< 1 i< m 2 m + 1 f U (i),u () (s, t), (2.3) 2 i + 1 f U (i),u () (s, t), (2.4) 2 m i + 1 f U (i),u () (s, t), (2.5) P (s, t, m, ) = P 1 (s, t, m, ) + P 2 (s, t, m, ) + P 3 (s, t, m, ) (2.6) [the sums i (2.3) (2.5) are double sums over i ad ], ad lettig β(s, t) deote the idex of the first bit at which the keys s ad t differ, we ca write the expectatio µ(m, ) of the umber of bit comparisos required to fid the rak-m key i a file of keys as µ(m, ) = = β(s, t)p (s, t, m, ) dt ds (2.7) s 2 k (l 1 2 )2 k l2 k k=0 l=1 (l 1)2 k (l 1 2 )2 k (k + 1)P (s, t, m, ) dt ds; (2.8) i this expressio, ote that k represets the last bit at which s ad t agree. 2.2 Exact computatio of µ(1, ) Sice the cotributio of P 2 (s, t, m, ) or P 3 (s, t, m, ) to P (s, t, m, ) is zero for m = 1, we have P (s, t, 1, ) = P 1 (s, t, 1, ) [see (2.4) through (2.6)]. Let x := s, y := 11

20 CHAPTER 2. ANALYSIS OF µ(1, ) t s, z := 1 t. The P 1 (s, t, 1, ) = z = 2z z 1 i< η 1 ( ) 2 x i 1 y i 1 z i 1, 1, i 1, 1, ( ) x i 1 y i 1 η dη i 1, 1, i 1, 1, 1 i< = 2z η 1 ( 1)(x + y + η) 2 dη z = 2z ( 1) z η 3 ( t η + 1 ) 2 dη. (2.9) Makig the chage of variables v = t + 1 ad itegratig, ad recallig z = 1 t, we fid, η after some calculatio, From (2.8) ad (2.10), µ(1, ) = = 2 = 2 = = (k + 1) k=0 (k + 1) k=0 (k + 1) k=0 (k + 1) k=0 P 1 (s, t, 1, ) = 2 2 k (l 1 2 )2 k l2 k ( ) ( 1) t 2. (2.10) l=1 (l 1)2 k 2 k (l 1 2 )2 k l2 k l=1 2 k l=1 2 k l=1 ( 1) ( ) 1 (l 1)2 k (l 1 2 )2 k P 1 (s, t, 1, ) dt ds (l 1 2 )2 k ( ) ( 1) t 2 dt ds ( ) l2 k ( 1) t 2 [(l 1 (l 1 2 )2 k (l 1)2 k ] dt 2 )2 k ( 1) ( ) 1 2 k {(l2 k ) 1 [(l 1 2 )2 k ] 1 } (k + 1)2 k=0 k 2k [l 1 (l 1 2 ) 1 ]. (2.11) l=1 12

21 CHAPTER 2. ANALYSIS OF µ(1, ) To further trasform (2.11), defie a,r = B r r ( ) 1 r 1 if r 2 1 if r = if r = 0, (2.12) where B r deotes the r-th Beroulli umber. Let S, := l=1 l 1. The S, = 1 r=0 a,r r (see Kuth [17]), ad 2 k l=1 [l 1 (l 1 2 ) 1 ] = S 2 k, 2 ( 1) 2k (2l 1) 1 = S 2 k, 2 ( 1) (S 2 k+1, 2 1 S 2 k,) = 2S 2 k, 2 ( 1) S 2 k+1, = 2 a,r 2 k( r) 2 ( 1) a,r 2 (k+1)( r) = 2 a,r 2 k( r) (1 2 r ). r=0 r=0 l=1 r=1 (2.13) From (2.11) ad (2.13), Here µ(1, ) = 2 ( 1) ( ) 1 1 (k + 1)2 k a,r 2 k( r) (1 2 r ). k=0 1 (k + 1)2 k a,r 2 k( r) (1 2 r ) = k=0 = r=1 r=1 r=1 1 (k + 1) a,r 2 kr (1 2 r ) k=0 1 r=1 1 a,r (1 2 r ) (k + 1)2 kr = a,r (1 2 r ) 1. k=0 r=1 13

22 CHAPTER 2. ANALYSIS OF µ(1, ) Hece µ(1, ) = 2 = 2 = 2 ( 1) ( ) 1 1 a,r (1 2 r ) 1 = 2 (1 2 r ) 1 1 r=1 r=1 ( 1) ( ) (1 2 r ) 1 B ( 1) ( )( 1 ) r r 1 1 r 1 r=2 =r+1 ( 1) ( ) [ (1 2 r ) 1 B ( 1) ( )( 1 ) r r 1 1 r 1 r=2 The sum ( 1) ( )( r 1) 1 =r ca be simplified as follows: 1 =r ( 1) 1 ( )( 1 r 1 ) = = 1 r 1 1 r 1 = ( 1)r r 1 =r =r+1 ( 1) ( ) 1 a,r ( 1)r( r r 1 )]. (2.14) ( )( ) 2 ( 1) r 2 ( )( ) 2 ( 1) r ( )( ) ( ) 1 r = ( 1)r + 1 r r r 1 r =r =r =0 = ( 1) r + 1 r. (2.15) r 1 Pluggig (2.15) ito (2.14) ad recallig B 2k+1 = 0 for k 1, we fially obtai µ(1, ) = 2 = 2 ( 1) ( ) r=2 ( 1) ( ) (1 2 r ) 1 B r r B [ ( 1) r ( r + 1) r 1 ) + 1 ( ( 1)(1 2 ) ( 1)r( r r 1 = 2(H 1) + 2t, (2.16) )] where H deotes the -th harmoic umber ad t := 1 B (1 2 ) 14 [ ( ) 1 1 ]. (2.17)

23 CHAPTER 2. ANALYSIS OF µ(1, ) The last equality i (2.16) follows from the easy idetity k=1 ( 1) k 1( ) k k = H. 15

24 CHAPTER 2. ANALYSIS OF µ(1, ) 2.3 Asymptotic aalysis of µ(1, ) I order to obtai a asymptotic expressio for µ(1, ), we aalyze t i (2.16) (2.17). The followig lemma provides a exact expressio for t that easily leads to a asymptotic expressio for µ(1, ): Lemma 2.1. For 2, let u := t +1 t (with t 2 = 0) ad v := u +1 u. Let γ deote Euler s costat (. = ), ad defie χ k := 2πik l 2. The (i) (ii) where where v = 1 H ( γ 1) l 2 l 2 2 Σ, ( + 1)( + 2) Σ := u = H + a a := 14 9 Σ := k Z\{0} k Z\{0} ζ(1 χ k )Γ( + 1)Γ(1 χ k ) ; (l 2)Γ( + 3 χ k ) ( H +1 γ 1 (l 2)( + 1) + l 2 1 ) Σ, γ 18 l 2 2 l 2 k Z\{0} ζ(1 χ k )Γ(1 χ k ) (l 2)(1 χ k ) ζ(1 χ k )Γ(1 χ k ) Γ(4 χ k )(1 χ k ), Γ( + 1) Γ( + 2 χ k ) ; (iii) t = (H 1) + a( 2) 1 2 l 2 ( γ 1 + l ) ( H 3 ) + b Σ, 2 [ H 2 + H (2) 7 ] 2 16

25 CHAPTER 2. ANALYSIS OF µ(1, ) where b := k Z\{0} Σ := k Z\{0} 2ζ(1 χ k )Γ( χ k ) (l 2)(1 χ k )Γ(3 χ k ), ζ(1 χ k )Γ( χ k )Γ( + 1) (l 2)(1 χ k )Γ( + 1 χ k ), ad H (2) deotes the -th Harmoic umber of order 2, i.e., H (2) := i=1 1. i 2 I this lemma, u ad v are derived i order to obtai the exact expressio for t i (iii). From (2.16), the exact expressio for t also provides a alterative exact expressio for µ(1, ). We kow Before provig Lemma 2.1, we complete the proof of Theorem 1.1 usig part (iii). H = l + γ O( 4 ), (2.18) H (2) = π O( 3 ). (2.19) Combiig (2.18) (2.19) with (2.16) ad Lemma 2.1(iii), we obtai a asymptotic expressio for µ(1, ): µ(1, ) = 2a 1 ( ) 2 l 2 (l )2 l l + O(1). (2.20) The term O(1) i (2.20) has fluctuatios of small magitude due to Σ, which is periodic i log with amplitude smaller tha Thus, as show i Theorem 1.1, the asymptotic slope i (2.20) is c = 2a = γ 9 l 2 4 l 2 k Z\{0} 17 ζ(1 χ k )Γ(1 χ k ) Γ(4 χ k )(1 χ k ). (2.21)

26 CHAPTER 2. ANALYSIS OF µ(1, ) Let S deote the sum i c: S := k Z\{0} ζ(1 χ k )Γ(1 χ k ) Γ(4 χ k )(1 χ k ) = k Z\{0} ζ(1 χ k ) (3 χ k )(2 χ k )(1 χ k ) 2, (2.22) where the formula Γ(1 + x) = xγ(x) is used to derive the secod expressio. Both expressios ivolve the imagiary umbers χ k, but S is a real umber. We ivestigate S ad express it usig oly real fuctios. We have the followig result: Theorem 2.2. Let S be the sum defied at (2.22). The where S l 2 = S ρ, (2.23) S := 2 k h(2 k ), h(m) := 1 2 (m l m l m! m) m 1, (2.24) k=0 ad 17 6γ ρ := 36 l Proof of Theorem 2.2. Choose ad fix 0 < θ < 1. We show that the itegral J := θ+i θ i ζ(1 s) (1 2 s )(3 s)(2 s)(1 s) 2 ds equals 2πi S o the oe had ad equals 2πi[ρ + (S/ l 2)] o the other had. Equatig these two expressios gives the desired result. To get the first expressio for J, we calculate J = k=0 θ+i θ i ζ(1 s)2 ks (3 s)(2 s)(1 s) ds = 1 θ+i 2 k ζ(t)2 kt 2 t 2 (1 + t)(2 + t) dt. k=0 1 θ i 18

27 CHAPTER 2. ANALYSIS OF µ(1, ) But, for ay positive iteger m ad ay α > 1, 1 θ+i 1 θ i ζ(t)m t t 2 (1 + t)(2 + t) dt = α+i α i ζ(t)m t t 2 (1 + t)(2 + t) dt 2πiRes t=1 [ ] ζ(t)m t, t 2 (1 + t)(2 + t) which follows from residue calculus, takig ito accout the cotributio of the simple pole of the itegrad at 1. Here ad Further, sice α+i α i we have by Melli iversio that equals [ ] ζ(t)m t Res t=1 = 1 t 2 (1 + t)(2 + t) 6 m ζ(t)m t t 2 (1 + t)(2 + t) dt = =1 α+i α i (/m) t t 2 (1 + t)(2 + t) dt. 1 t 2 (t + 1)(t + 2) = 3/4 + 1/2 + 1 t t 2 t + 1 1/4 t + 2, 1 α+i x t 2πi α i t 2 (1 + t)(2 + t) dt f(x) := l x + x 1 4 x2 for 0 x 1 ad equals 0 for x 1. (Note that this requires oly α > 0.) So α+i α i ζ(t)m t t 2 (1 + t)(2 + t) dt = 2πi ( ) f m [ 1 = 2πi 2 (m l m l m!) 1 3 m ] 24 m 1 = 2πi [h(m) + 16 ] m, 19

28 CHAPTER 2. ANALYSIS OF µ(1, ) where the sum is over 1 m (or 1 m 1), ad therefore 1 θ+i 1 θ i ζ(t)m t dt = 2πih(m). t 2 (1 + t)(2 + t) Thus we obtai our first expressio for J. Before we proceed, we examie the series expressio (2.24) for S. Applyig Stirlig s formula to l m!, we obtai l m! = m l m m l m l(2π) m 1 360m 3 + O(m 4 ). Thus h(m) = 1 4 l m + ( l(2π) ) 1 12m m 3 + O(m 4 ), ad the series S coverges geometrically rapidly. To obtai the secod expressio for J we move the horizotal (i.e., real) coordiate of the vertical lie of itegratio over from θ to C where C is large positive umber (C ). By residue calculus, we fid [ ] ζ(1 s) J = 2πi {Res s=0 (1 2 s )(3 s)(2 s)(1 s) 2 as desired. + S } ( = 2πi ρ + S ), l 2 l 2 Usig Theorem 2.2 it is straightforward to derive the alterative expressio 2 k c = k l (2.25) 2 k k=0 =1 for the liear coefficiet c i (2.21). Graber ad Prodiger [9] obtaied a earlier draft of this mauscript ad idepedetly coducted a similar aalysis of S leadig to (2.25). They also showed how to compute c efficietly to high precisio ad i particular computed c to 50 decimal places. 20

29 CHAPTER 2. ANALYSIS OF µ(1, ) Remark 2.3. The lead-order asymptotics µ(1, ) c with c i the form (2.25) ca also be obtaied simply usig real-aalytical argumets. Start with (2.7) with m = 1 ad recall that P (s, t, 1, ) = P 1 (s, t, 1, ) is give by (2.10) to see that µ(1, ) = 2 1 t 0 0 β(s, t)t 2 [(1 t) 1 + t] ds dt. A easy domiated-covergece argumet the shows that µ(1, ) c with c give i the itegral form Writig β(s, t) = c = 2 1 t 0 0 β(s, t)t 1 ds dt. 1(s ad t agree i their first k bits) k=0 ad breakig up the double itegral accordig to the first k bits of t leads to the summatio form (2.25) of c. Vallée et al. [27] followed ad further geeralized this approach. We omit the details. We do ot kow how to obtai asymptotics for µ(1, ) beyod the lead term by this sort of approach. Now we prove Lemma 2.1: Proof of Lemma 2.1. (i) Sice u = t +1 t = = B (1 2 ) [ ( ( + 1) +1 ) 1 B ( 1)(1 2 ) [( 1 1 ) ] ] 1, 1 B (1 2 ) [ ( ) 1 1 ] 21

30 CHAPTER 2. ANALYSIS OF µ(1, ) it follows that +1 v = u +1 u = B ( 1)(1 2 ) [( ) ] [( ) ] B + 1 ( 1)(1 2 ) 1 1 ( ) B k+2 = k (k + 2)(k + 1)[1 2 (k+2) ] k=0 1 ( ) = ( 1) k ζ( 1 k) (2.26) k (k + 1)[1 2 (k+2) ] k=0 = ( 1) ζ( 1 s)! ds, (2.27) 2πi (s + 1)[1 2 (s+2) ] s(s 1) (s ) C where C is a positively orieted closed curve that ecircles the itegers 0,..., 1 ad does ot iclude or ecircle ay of the followig poits: 2 + χ k (where χ k := 2πik l 2 ), k Z; 1; ad. Equality (2.26) follows from the fact that the Beroulli umbers are extrapolated by the Riema zeta fuctio take at oegative itegers: B k = kζ(1 k). [The coefficiets ( 1) k do ot cocer us sice the Beroulli umbers of odd idex greater tha 1 vaish.] Equality (2.27) follows from a direct applicatio of residue calculus, takig ito accout cotributios of the simple poles at the itegers 0,..., 1. Let φ(s) deote the itegrad i (2.27): φ(s) = ζ( 1 s)! (s + 1)[1 2 (s+2) ] s(s 1) (s ). We cosider a positively orieted rectagular cotour C l with horizotal sides Im(s) = λ l ad Im(s) = λ l, where λ l := (2l+1)π, l Z +, ad vertical sides Re(s) = θ ad l 2 22

31 CHAPTER 2. ANALYSIS OF µ(1, ) Re(s) = λ l, where 0 < θ < 1. By elemetary bouds o φ(s) alog C l ad the fact that θ+i θ i φ(s) ds = 0 (2.28) [this is implicit o page 113 of Flaolet ad Sedgewick [8] ad explicitly proved i Appedix A], oe ca show that lim φ(s) ds = 0. l C l Accoutig for residues due to the poles ecircled by C l, we obtai v = ( 1) +1 Res s= 1[φ(s)] + Res s= 2 [φ(s)] + Res s= 2+χk [φ(s)] where k Z\{0} = 1 H ( γ 1) l 2 l 2 2 Σ, (2.29) ( + 1)( + 2) Σ := k Z\{0} ζ(1 χ k )Γ( + 1)Γ(1 χ k ). (2.30) (l 2)Γ( + 3 χ k ) 23

32 CHAPTER 2. ANALYSIS OF µ(1, ) (ii) We have u 2 = t 3 t 2 = t 3 = 1. Hece, from (i), 9 1 u = u 2 + v = = v l 2 ( γ l = 1 9 (H H 2 ) l 2 ( γ l H +2 ( + 1)( + 2) ) ( + 1)( + 2) H +2 ( + 1)( + 2) ) ( ) = 14 9 γ ( γ 3 l 2 H + l 2 1 ) H +2 l 2 ( + 1)( + 2) 1 Σ. Σ Σ (2.31) Here 1 H +2 ( + 1)( + 2) = =3 H +1 = H =4 +1 =4 H H +1 H H (2.32) = H , (2.33) where we assume 3 for (2.32), but (2.33) holds also for = 2. I regard to 1 Σ, ote that Σ = k Z\{0} ζ(1 χ k )Γ(1 χ k ) (l 2)(1 χ k ) [ ] Γ( + 2) Γ( + 1), Γ( + 3 χ k ) Γ( + 2 χ k ) 24

33 CHAPTER 2. ANALYSIS OF µ(1, ) so that Defie 1 Σ = k Z\{0} Σ := ζ(1 χ k )Γ(1 χ k ) (l 2)(1 χ k ) k Z\{0} ζ(1 χ k )Γ(1 χ k ) (l 2)(1 χ k ) The, combiig (2.31), (2.33), ad (2.34), we obtai where u = H + a a := γ 18 l 2 2 l 2 [ Γ( + 1) Γ( + 2 χ k ) Γ(3) ]. (2.34) Γ(4 χ k ) Γ( + 1) Γ( + 2 χ k ). (2.35) ( H +1 γ 1 (l 2)( + 1) + l 2 1 ) Σ, k Z\{0} ζ(1 χ k )Γ(1 χ k ) Γ(4 χ k )(1 χ k ). (2.36) (iii) Closely followig the derivatio of u described above, we obtai (for 2) 1 1 t = t 2 + u = where u 1 = H + a( 2) 1 l 2 =3 = (H 1) + a( 2) 1 2 l 2 ( γ 1 + l ( H γ 1 + l [ H 2 + H (2) 7 2 ) ( H 3 ) ) ( H 3 ) + b Σ, (2.37) 2 b := k Z\{0} Σ := k Z\{0} ] 2ζ(1 χ k )Γ( χ k ) (l 2)(1 χ k )Γ(3 χ k ), (2.38) ζ(1 χ k )Γ( χ k )Γ( + 1) (l 2)(1 χ k )Γ( + 1 χ k ). (2.39) Σ 25

34 CHAPTER 2. ANALYSIS OF µ(1, ) Our aalysis shows that the expected umber of bit comparisos required by QuickMi is asymptotically liear i with the asymptotic slope approximately equal to Hece asymptotically it differs from the expected umber of key comparisos to achieve the same task oly by a costat factor. (The expectatio for key comparisos is asymptotically 2; see Kuth [15] ad Mahmoud et al. [19]). This result is rather cotrastive to the Quicksort case i which (see Fill ad Jaso [5]) the expected umber of bit comparisos is asymptotically (l )(lg ) whereas the expected umber of key comparisos is asymptotically 2 l. 26

35 Chapter 3 Aalysis of µ( m, ) I this chapter, we derive exact ad asymptotic expressios for the expected umber of bit comparisos required by QuickRad; Quickselect is applied to a file of keys ad fids a key of target rak that is chose uiformly radomly from {1, 2,..., }. We will cotiue to use the framework ad otatio established i Chapter Exact computatio of µ( m, ) We average µ(m, ) over m while the parameter is fixed. Usig the otatio defied i (2.3) through (2.7), we have µ( m, ) = 1 = µ(m, ) = 1 β(s, t)p (s, t, m, ) dt ds m=1 0 s 1 β(s, t) 1 P (s, t, m, ) dt ds = µ 1 ( m, ) + µ 2 ( m, ) + µ 3 ( m, ), m=1 1 0 s m=

36 CHAPTER 3. ANALYSIS OF µ( m, ) where, for l = 1, 2, 3, µ l ( m, ) = s β(s, t) 1 P l (s, t, m, ) dt ds. (3.1) m=1 Here µ 1 ( m, ) = µ 3 ( m, ), sice P 3 (1 t, 1 s, m + 1, ) = P 1 (s, t, m, ) by a easy symmetry argumet we omit, ad so µ 3 ( m, ) = = = s s = µ 1 ( m, ). β(s, t) 1 P 3 (s, t, m, ) dt ds m=1 β(1 t, 1 s ) 1 P 3 (1 t, 1 s, m + 1, ) dt ds s m =1 β(s, t ) 1 P 1 (s, t, m, ) dt ds m =1 Therefore µ( m, ) = 2µ 1 ( m, ) + µ 2 ( m, ), (3.2) ad we will compute µ 1 ( m, ) ad µ 2 ( m, ) exactly i Sectios Exact computatio of µ 1 ( m, ) We use the followig lemma i order to compute µ 1 ( m, ) exactly: 28

37 CHAPTER 3. ANALYSIS OF µ( m, ) Lemma β(s, t) 1 P 1 (s, t, m, ) dt ds 0 s m=2 1 ( 1) ( ) 1 = ( 1) 9 1 ( 1) ( ) 1 2 ( + 1)( 1)(1 2 ). ( 1) ( ) =3 B + 1 ( ) 1 1 ( 1)( 2)(1 2 ) Before provig the lemma, we complete the computatio of µ 1 ( m, ). Note that µ 1 ( m, ) = = s β(s, t) 1 = 1 µ(1, ) + 1 s P 1 (s, t, m, ) dt ds m=1 β(s, t) P 1 (s, t, 1, ) dt ds 0 1 s β(s, t) 1 Therefore, by (2.16) ad Lemma 3.1, we obtai s β(s, t) 1 P 1 (s, t, m, ) dt ds m=2 P 1 (s, t, m, ) dt ds. m=2 29

38 CHAPTER 3. ANALYSIS OF µ( m, ) µ 1 ( m, ) = =3 ( 1) ( ) ) ( 1) ( 1 ( 1) B B ( ) ( 1)(1 2 ) ( 1) ( ( 1 1 ( 1)( 2)(1 2 ) ) 1 ( 1) ( ) 1 2 ( + 1)( 1)(1 2 ) ( 1) ( ) 1 1 = 1 4 ( 1)( 2) ( ) B ( 1)(1 2 ) = ( 1) ( ) ( ) B 9 1 ( 1)( 2)(1 2 ) =3 1 ( 1) ( ) 1 2 ( + 1)( 1)(1 2 ), (3.3) where the secod equality holds sice 2 ( 1) ( ) ( 1) ( ) 1 ( 1) ( 1) ( 1)! = 2!( )!( 1) 2 ( 1) ( 1)! ( 1)!( )!( 1)( 2) =3 [ ( 1) ( 1)! 1 = ( 1)!( )!( 1) 1 ] 2 =3 ( 1) ( ) 1 1 = 1 4 ( 1)( 2). =3 I Sectio we combie the expressio for µ 1 ( m, ) i (3.3) with a similar expressio for µ 2 ( m, ) to obtai a exact expressio for µ( m, ). The remaider of this sectio is devoted to provig Lemma 3.1. For this, the followig expressio for P 1 (s, t, m, ) 30 )

39 CHAPTER 3. ANALYSIS OF µ( m, ) will prove useful: Lemma 3.2. Let m 2 ad let x := s, y := t s, z := 1 t. The the quatity P 1 (s, t, m, ) defied at (2.3) satisfies P 1 (s, t, m, ) = 2 x 0 1 (ξ + y) 2 [Υ 1(m,, ξ, x, y, z) Υ 2 (m,, ξ, x, y, z) + Υ 3 (m,, ξ, x, y, z)] dξ, (3.4) where Υ 1 (m,, ξ, x, y, z) := Υ 2 (m,, ξ, x, y, z) := Υ 3 (m,, ξ, x, y, z) := ( ) 1 (x ξ) m 2 ( m)(ξ + y + z) m+1, m 2 ( ) 1 (x ξ) m 2 ( m + 1)z(ξ + y + z) m, m 2 ( ) 1 (x ξ) m 2 z m+1. m 2 Proof of Lemma 3.2. By (2.2) (2.3), P 1 (s, t, m, ) = = = m i< m i< 2! ( m 1)! 2! m + 1 (i 1)!( i 1)!( )! xi 1 y i 1 z ( ) 2! m 1 m + 1 ( m 1)! i m, i 1, m i< 1 m + 1 ( (i m)! (i 1)! xi 1 y i 1 z ) m 1 i m, i 1, (i m)! (i 1)! xi 1 y i 1 z. (3.5) 31

40 CHAPTER 3. ANALYSIS OF µ( m, ) I order to compactly describe the derivatio of (3.4), we defie the followig idefiite itegratio operator T : T (f(x)) := x 0 f(ξ) dξ. We really should write (T f)(x) rather tha T (f(x)), but we would like to use shorthad such as T (x ) = x+1 +1 whe > 1. The operator T treats its argumet f as a fuctio of x; the other variables ivolved i f (amely, y ad z) are treated as costats. The otatio T l will deote the l-th iterate of T. I this otatio, for m < i, ad the sum i (3.5) equals Here so T m 1 ( m i< (i m)! (i 1)! xi 1 = T m 1 (x i m ), ( ) ) 1 m 1 x i m y i 1 z. m + 1 i m, i 1, 1 m + 1 z = z m+1 z η ( m+1) 1 dη, ( ( ) ) T m 1 1 m 1 x i m y i 1 z m + 1 i m, i 1, m i< ( [ ( ) ] ) = z m+1 T m 1 m 1 x i m y i 1 η +m 2 dη z i m, i 1, m i< ( ) = z m+1 T m 1 η +m 2 (x + y + η) m 1 dη z ( ( ) ) m 1 = z m+1 T m 1 t η 3 η + 1 dη (3.6) z 32

41 CHAPTER 3. ANALYSIS OF µ( m, ) (ote that x + y = t). Makig the chage of variables v = t + 1 ad itegratig, we obtai, η after some computatio, z η 3 ( t η + 1 ) m 1 dη = [ ( 1 ( m) 1 + t ) m+1 t 2 ( m + 1)( m) z ( ( m + 1) 1 + t ) ] m + 1. z (3.7) From (3.5) ad (3.6) (3.7), P 1 (s, t, m, ) = 2! ( m + 1)! T m 1 ( t 2 [ ( m)(z + t) m+1 ( m + 1)z(z + t) m + z m+1]). (3.8) Here t 2 [( m)(z + t) m+1 ( m + 1)z(z + t) m + z m+1 ] where = m+1 r=2 t r 2 Υ(m,, r, z), (3.9) ( ) ( ) m + 1 m Υ(m,, r, z) := ( m) z m+1 r ( m + 1) z m+1 r. r r (3.10) The, sice t = x + y, m+1 r=2 t r 2 Υ(m,, r, z) = m+1 r=2 r 2 ( r 2 Υ(m,, r, z) =0 ) x y r 2. (3.11) 33

42 CHAPTER 3. ANALYSIS OF µ( m, ) From (3.8) (3.11), P 1 (s, t, m, ) = = = 2! ( m + 1)! T m 1 m+1 2! ( m + 1)! r=2 m+1 2! ( m + 1)! r=2 ( m+1 r 2 ( ) r 2 Υ(m,, r, z) )x y r 2 r=2 =0 r 2 ( ) r 2 Υ(m,, r, z) y r 2 T m 1 (x ) =0 r 2 ( r 2 Υ(m,, r, z) =0 )y r 2 x +m 1 ( + 1) ( + m 1). (3.12) Because of the partial fractio expasio it follows that m 2 1 ( + 1) ( + m 1) = 1 (m 2)! l=0 ( 1) l( ) m 2 l + l + 1, r 2 ( r 2 )y r 2 x +m 1 ( + 1) ( + m 1) =0 r 2 ( m 2 r 2 r 2 x+m 1 ( 1) = )y l( ) m 2 l (m 2)! + l + 1 =0 l=0 m 2 1 ( m 2 x r 2 ( ) r 2 = ( 1) )x l m 2 l ξ l y r 2 ξ dξ (m 2)! l l=0 0 =0 m 2 1 ( m 2 x = ( 1) )x l m 2 l ξ l (ξ + y) r 2 dξ (m 2)! l = 1 (m 2)! l=0 x 0 0 (x ξ) m 2 (ξ + y) r 2 dξ. (3.13) 34

43 CHAPTER 3. ANALYSIS OF µ( m, ) From (3.12) (3.13), 2! P 1 (s, t, m, ) = ( m + 1)!(m 2)! ( ) 1 x = 2 m 2 0 r=2 ( ) 1 x = 2 m 2 Here, by (3.10), 0 m+1 m+1 r=2 (x ξ) m 2 (ξ + y) 2 Υ(m,, r, z) x 0 (x ξ) m 2 (ξ + y) r 2 dξ Υ(m,, r, z)(x ξ) m 2 (ξ + y) r 2 dξ m+1 r=2 Υ(m,, r, z)(ξ + y) r dξ. (3.14) m+1 r=2 = Υ(m,, r, z)(ξ + y) r m+1 r=2 = ( m) [ ( m) m+1 r=2 ( m + 1 r ( m + 1 r (ξ + y) r ) ( ] m z m+1 r ( m + 1) )z m+1 r r ) (ξ + y) r z m+1 r ( m + 1) m+1 r=2 ( m r ) (ξ + y) r z m+1 r = ( m)[(ξ + y + z) m+1 z m+1 ( m + 1)(ξ + y)z m ] ( m + 1)z[(ξ + y + z) m z m ( m)(ξ + y)z m 1 ] = ( m)(ξ + y + z) m+1 ( m + 1)z(ξ + y + z) m + z m+1. (3.15) Substitutio of (3.15) ito (3.14) gives the desired (3.4). Proof of Lemma 3.1. From Lemma 3.2, we have 1 m=2 x 1 P 1 (s, t, m, ) = 2 (ξ + y) 2 0 [Υ 1 (m,, ξ, x, y, z) m=2 Υ 2 (m,, ξ, x, y, z) + Υ 3 (m,, ξ, x, y, z)] dξ. (3.16) 35

44 CHAPTER 3. ANALYSIS OF µ( m, ) Here Υ 1 (m,, ξ, x, y, z) m=2 [ ( ) ] = (ξ + y + z) 2 d 1 (x ξ) m 2 w m dw m 2 m=2 w=ξ+y+z = (ξ + y + z) 2 d { w 1 [(x ξ + w) 1 (x ξ) 1 ] } dw w=ξ+y+z = (ξ + y + z) 2 { w 2 [(x ξ + w) 1 (x ξ) 1 ] = (x ξ) ( 1)(ξ + y + z) +w 1 ( 1)(x ξ + w) 2} w=ξ+y+z (ote that x + y + z = 1). Similarly, Υ 2 (m,, ξ, x, y, z) = z m=2 = z d dw [ d dw m=2 ( ) ] 1 (x ξ) m 2 w m+1 m 2 w=ξ+y+z [ (x ξ + w) 1 (x ξ) 1] w=ξ+y+z = z [ ( 1)(x ξ + w) 2] w=ξ+y+z = z( 1), ad Υ 3 (m,, ξ, x, y, z) = m=2 m=2 ( ) 1 (x ξ) m 2 z m+1 m 2 = (x ξ + z) 1 (x ξ) 1. Hece [Υ 1 (m,, ξ, x, y, z) Υ 2 (m,, ξ, x, y, z) + Υ 3 (m,, ξ, x, y, z)] m=2 = ( 1)(ξ + y) 1 + (x ξ + z) 1. (3.17) 36

45 CHAPTER 3. ANALYSIS OF µ( m, ) Therefore, from (3.16) ad (3.17), we obtai 1 P 1 (s, t, m, ) m=2 = 2 = 2 = 2 x 0 x 0 x 0 1 (ξ + y) [( 1)(ξ + y) 1 + (x ξ + 2 z) 1 ] dξ 1 (ξ + y) {( 1)(ξ + y) 1 + [1 (ξ + 2 y)] 1 } dξ 1 1 ( ) 1 ( 1) (ξ + y) dξ (ξ + y) 2 = 1 ( ) 1 x 2 ( 1) (ξ + y) 2 dξ 0 = 1 ( ) 1 (x + y) 2 ( 1) y 1 1 = 1 ( ) 1 t 2 ( 1) (t s) 1. 1 (3.18) We complete the proof by usig (3.18) to compute We have s = 2 = s β(s, t) s s β(s, t) 1 P 1 (s, t, m, ) ds dt. m=2 P 1 (s, t, m, ) ds dt m=2 β(s, t) β(s, t) s β(s, t) 1 ( ) 1 t ( 1) 1 (t s) 1 dt ds 1 1 ( ) 1 t ( 1) 1 dt ds 1 1 ( ) 1 (t s) ( 1) 1 dt ds. (3.19) 1 37

46 CHAPTER 3. ANALYSIS OF µ( m, ) Closely followig the derivatios show i (2.11) (2.16), oe ca show that = s 1 ( ) 1 t β(s, t) ( 1) 1 dt ds 1 ( 1) ( ) ( 1) ( ) 1 1 ( 1) =3 B + 1 ( ) 1 1 ( 1)( 2)(1 2 ). (3.20) Thus, i order to complete the proof, it remais to show that s β(s, t) 1 ( ) 1 (t s) ( 1) 1 1 dt ds = 1 ( 1) ( ) 1 ( + 1)( 1)(1 2 ). (3.21) Ideed, we have = = = s β(s, t) 1 ( ) 1 (t s) ( 1) (k+1) (k + 1)2 k k=0 2 (k+1) (k + 1)2 k k=0 k= k 1 2 (k+1) 2 k s 1 2 (k+1) s 1 ( 1 (k + 1)2 k ( 1) dt ds ( 1) ( 1 ) 2 k 0 ( 1) ( 1 v 1 1 ) (t s) 1 dt ds 1 ) v 1 dv ds 1 (2 k v) V 2 (k+1) [2 (k+1) v] W 0 ds dv. (3.22) Here (2 k v) V 2 (k+1) [2 (k+1) v] W 0 v if 0 v 2 (k+1) ds = 2 k v if 2 (k+1) < v 2 k. 38

47 CHAPTER 3. ANALYSIS OF µ( m, ) Thus 2 k 0 v 1 1 = 1 1 (2 k v) V 2 (k+1) [2 (k+1) v] W 0 ds dv [ 2 (k+1) v dv k 2 (k+1) v 1 (2 k v) dv = 2 k(+1) (1 2 ) ( + 1)( 1). (3.23) ] From (3.22) ad (3.23), we obtai = = = = ad (3.21) is proved. s β(s, t) k=0 1 ( ) 1 (t s) ( 1) 1 1 dt ds 1 ( ) 1 2 (k + 1)2 k ( 1) k(+1) (1 2 ) ( + 1)( 1) 1 ( 1 ( 1) ) 1 2 ( + 1)( 1) (k + 1)2 k 1 ( ) 1 ( 1) ( + 1)( 1) (1 2 ) 2 1 ( 1) ( ) 1 ( + 1)( 1)(1 2 ), (3.24) k= Exact computatio of µ 2 ( m, ) ad µ( m, ) The derivatios for obtaiig a computatioally preferable exact expressio for µ 2 ( m, ) are etirely aalogous to those for µ 1 ( m, ) described i the previous sectio (Sectio 3.1.1). Thus we omit details. As described i Sectio 2.2, P 2 (s, t, m, ) is zero 39

48 CHAPTER 3. ANALYSIS OF µ( m, ) for m = 1 ad for m =, so, from (3.1), µ 2 ( m, ) = s β(s, t) 1 1 m=2 P 2 (s, t, m, ) dt ds. (3.25) Therefore we first derive a computatioally desirable expressio for 1 Agai, let x := s, y := t s, z := 1 t. The 1 1 P 2 (s, t, m, ) m=2 = 1 1 = 1 m=2 1 i m< 1 m=2 ( 2 i + 1 S 1 (m,, x, y, z) 1 i 1, 1, i 1, 1, 1 m=2 S 2 (m,, x, y, z) 1 1 m=2 P 2(s, t, m, ). ) x i 1 y i 1 z 1 m=2 S 3 (m,, x, y, z), (3.26) where S 1 (m,, x, y, z) := S 2 (m,, x, y, z) := S 3 (m,, x, y, z) := 1 i< m i< 1 i< m ( ) 2 x i 1 y i 1 z, i + 1 i 1, 1, i 1, 1, ( ) 2 x i 1 y i 1 z, i + 1 i 1, 1, i 1, 1, 2 i + 1 ( i 1, 1, i 1, 1, ) x i 1 y i 1 z. Fill ad Jaso [5] showed that S 1 (m,, x, y, z) = 2 ( 1)( ) (t s) 2. Hece 1 1 S 1 (m,, x, y, z) = m=2 2( 2) ( ) ( 1) (t s) 2. (3.27) 40

49 CHAPTER 3. ANALYSIS OF µ( m, ) Followig the derivatios show i (3.5) through (3.18), oe ca show that 1 1 S 2 (m,, x, y, z) = 2y 2 x[(x + z) y( 1)] (3.28) m=2 To obtai a similar expressio for 1 = 2(t s) 2 s{[1 (t s)] (t s)( 1)} 1 ( ) 1 = 2s ( 1) (t s) 2. (3.29) + 1 m, i := + 1, := + 1 i, 1 m=2 S 3(m,, x, y, z), we ote that, lettig m := S 3 (m,, x, y, z) = m i < ( ) 2 i + 1, 1, i 1, 1, i 1 = S 2 ( + 1 m,, z, y, x). x y i 1 z i 1 Thus 1 1 S 3 (m,, x, y, z) = 1 m=2 Ispectig (3.28) (3.30), we fid 1 1 S 3 (m,, x, y, z) = 2(1 t) m=2 1 m=2 1 S 2 ( + 1 m,, z, y, x) = 1 S 2 (m,, z, y, x). (3.30) m=2 1 ( 1) ( 1 ) (t s) 2. (3.31) 41

50 CHAPTER 3. ANALYSIS OF µ( m, ) From (3.26), (3.27), (3.29), ad (3.31), 1 1 P 2 (s, t, m, ) m=1 2( 2) ( ) 1 ( ) 1 = ( 1) (t s) 2 2s ( 1) (t s) 2 1 ( ) 1 2(1 t) ( 1) (t s) 2 2( 2) ( ) 1 ( ) 1 = ( 1) (t s) 2 2 ( 1) (t s) 2 1 ( ) 1 +2 ( 1) (t s) 1. (3.32) Hece, from (3.25) ad (3.32), µ 2 ( m, ) = 2( 2) 2 +2 Fill ad Jaso [5] showed that s β(s, t) s s ( ) β(s, t) ( 1) (t s) 2 dt ds 0 s 1 ( ) 1 β(s, t) ( 1) (t s) 2 dt ds 1 ( ) 1 β(s, t) ( 1) (t s) 1 dt ds. (3.33) 1 1 ( ) ( 1) (t s) 2 dt ds = ( 1) ( ) ( 1)[1 2 ( 1) ]. (3.34) A careful term-by-term ispectio of the derivatios show i (3.22) (3.24) reveals that ( ) 1 1 ( 1) ( ) 1 β(s, t) ( 1) (t s) 2 dt ds = ( 1)[1 2 ( 1) ], 0 s s 1 β(s, t) ( 1) ( 1 ) (t s) 1 dt ds = 1 ( 1) ( ) 1 (3.35) ( + 1)(1 2 ). (3.36) 42

51 CHAPTER 3. ANALYSIS OF µ( m, ) Combiig (3.33) (3.36), we obtai µ 2 ( m, ) = 2( 2) 2 =0 ( 1) ( ) +2 ( + 1)( + 2)[1 2 (+1) ] 2 ( 1) ( ) ( 1)[1 2 ( 1) ] +2( 1) = 4 ( 1) ( ) + 2( 1). ( 1)[1 2 ( 1) ] (3.37) Fially, we complete the exact computatio of µ( m, ). From (3.2), (3.3), ad (3.37), we have µ( m, ) = 2µ 1 ( m, ) + µ 2 ( m, ) ( 1) ( ) 1 1 = 2( 1) 8 ( 1)( 2) ( ) B ( 1)(1 2 ) = ( 1) ( ) ( ) B 9 1 ( 1)( 2)(1 2 ) =3 1 ( 1) ( ) 1 4 ( + 1)( 1)(1 2 ) 4 ( 1) ( ) ( 1)[1 2 ( 1) ] +2( 1). (3.38) We rewrite or combie some of the terms i (3.38) for the asymptotic aalysis of µ( m, ) described i the ext sectio. We defie F 1 () := F 2 () := F 3 () := =3 1 1 ( 1) ( ) ( 1)( 2), B (1 2 ) ( 1) ( ) 1, 1 [ ( ) 1 1 ], 43

52 CHAPTER 3. ANALYSIS OF µ( m, ) F 4 () := F 5 () := 1 =3 B [ ( 1 1 ) 1 ( 1)(1 2 ) 2 ( 1) ( ) ( 1)( 2)[1 2 ( 1) ]. =3 1 ], The secod, third, fourth, ad fifth terms i (3.38) ca be writte as 8 F 1(), 4 F 2(), 4 F 9 3(), ad 4F 4 (), respectively. The last three terms i (3.38) ca be combied as follows: 1 ( 1) ( ) 1 4 ( + 1)( 1)(1 2 ) 4 ( 1) ( ) + 2( 1) ( 1)[1 2 ( 1) ] = 4 ( 1) ( ) ( 1)( 2)[1 2 ( 1) ] 4 ( 1) ( ) + 2( 1) ( 1)[1 2 ( 1) ] =3 = 8 ( 1) ( ) ( 1)( 2)[1 2 ( 1) ] = 8 F 5(). Therefore =3 µ( m, ) = 2( 1) 8 F 1() + 4 F 2() F 3() 4F 4 () + 8 F 5(). (3.39) 3.2 Asymptotic aalysis of µ( m, ) We derive a asymptotic expressio for µ( m, ) show i (3.39). The computatios described i this sectio are aalogous to those i Sectio 2.3. Hece we merely sketch details to derive the asymptotic expressio. First, we aalyze F 1 (). A routie complexaalytical argumet similar to (but much easier tha) the oe described i Sectio 2.3 shows 44

53 CHAPTER 3. ANALYSIS OF µ( m, ) that 2 [ ] F 1 () = ( 1) +1! Res s=k s(s 1) 2 (s 2) 2 (s 3) (s ) k=0 = ( 1) +1 [ ( 1) 2 + ( 1) H 1 + ( 1) ( 1) 2 ( H = 1 2 ( 1)H ( 1) H = 1 ( l + 4 γ ) 2 2 l + (γ + 1) + O(1). 2 2( 1) )] (3.40) Sice F 2 () is equal to t, which is defied at (2.17) ad aalyzed i Sectio 2.3, we already have a asymptotic expressio for F 2 (). Next we derive a asymptotic expressio for F 3 (): 1 { } F 3 () = ( 1) ( 1)! Res s=k s(s 1) 2 (s 2) [s ( 1)] k=0 = H 2 H = l + (γ 1) l + O(1). (3.41) To obtai a asymptotic expressio for F 4 (), we closely follow the approach of Sectio 2.3. Let ũ := F 4 ( + 1) F 4 (). The ũ = =3 B ( 1)( 2)(1 2 ) [( ) ] Let ṽ := ũ +1 ũ. The, by computatios similar to those performed for v i Sec- 45

54 CHAPTER 3. ANALYSIS OF µ( m, ) tio 2.3, 2 ( ) ṽ = ( 1) k ζ( 2 k) 1 (k + 2)(k + 1)[1 2 (k+3) ] k k=0 3 { } = ( 1) +1 ζ( 2 s) ( 1)! Res s= k (s + 2)(s + 1)[1 2 (s+3) ] s(s 1) [s ( 1)] k=1 { +( 1) +1 ζ( 2 s) Res s= 3+χk (s + 2)(s + 1)[1 2 (s+3) ] k Z\{0} } ( 1)! s(s 1) [s ( 1)] = 1 [ 9 1 ( + 1) 1 γ ( + 1)( + 2) l H ] +2 ξ, + 2 where Hece 1 ũ = ũ 2 + ṽ ξ := k Z\{0} = 1 9 H 1 + ã + ξ 1 2 l 2 ζ(1 χ k )Γ(1 χ k )Γ(). (l 2)Γ( + 3 χ k ) ( H H ) l 2 2γ l 2 1 ( + 1), where ã := ξ := 7 36 l γ 12 l 2 k Z\{0} k Z\{0} ζ(1 χ k )Γ(1 χ k )Γ() (l 2)(2 χ k )Γ( + 2 χ k ). ζ(1 χ k )Γ(1 χ k ) (l 2)(2 χ k )Γ(4 χ k ), 46

55 CHAPTER 3. ANALYSIS OF µ( m, ) Thus where Therefore 1 F 4 () = F 4 (2) + ũ = 1 9 H H 1 + ( ã 1 ) l l 2 2γ 8 l 2 2ã + b ξ + 1 H 3 + l 2 2γ l 2 4 l 2, (3.42) b := k Z\{0} ξ := k Z\{0} ζ(1 χ k )Γ(1 χ k ) (l 2)(2 χ k )(1 χ k )Γ(3 χ k ), ζ(1 χ k )Γ(1 χ k )Γ() (l 2)(2 χ k )(1 χ k )Γ( + 1 χ k ). F 4 () = 1 ( 9 l + ã γ 1 ) + 8 l + O(1). (3.43) 9 9 Fially, we aalyze F 5 (). Let G (s) :=! [1 2 (s 1) ]s 2 (s 1) 2 (s 2) 2 (s 3) (s ). The, by computatios that are etirely aalogous to those performed for F 1 (), F 2 (), 47

56 CHAPTER 3. ANALYSIS OF µ( m, ) ad F 4 (), 2 F 5 () = ( 1) +1 Res s=k G (s) + ( 1) +1 k Z\{0} k=0 k Z\{0} Res s=1+χk G (s) = 1 4 (2H ( 1) l 2) (H 2 l 2 3) 2 [ ( l 2 (H 1) ) H l 2 2 l 2 H(2) l 2 + l ] 2 + Γ( 1 χ k )Γ( + 1) (l 2)χ k (χ 2 k 1)Γ( χ k) = l l 2 γ l 2 (l )2 + ( 1 l 2 1 ) l + O(). 2 (3.44) Therefore, from (3.39) (3.41) ad (3.43) (3.44), we obtai the followig asymptotic formula for µ( m, ): µ( m, ) = 4(1 + l 2 ã) 4 ( ) 2 l 2 (l )2 + 4 l 2 1 l + O(1). (3.45) The asymptotic slope 4(1 + l 2 ã) is approximately Thus the expected umber of bit comparisos required by QuickRad remais asymptotically liear i. As i the QuickMi case, the expectatio is asymptotically differet from that for key comparisos oly by a costat factor. (The expected umber of key comparisos required by QuickRad is asymptotically 3; see Mahmoud et al. [19].) 48

57 Chapter 4 Derivatio of a closed formula for µ(m, ) The exact expressio for µ(m, ) obtaied i Sectio 2.1 [see (2.8)] ivolves ifiite summatio ad itegratio. Hece it is ot a preferable form for umerically computig the expectatio. I this chapter, we establish aother exact expressio for µ(m, ) that oly ivolves fiite summatio. We also use the formula to compute µ(m, ) for m = 1,...,, = 2,..., 20. As described i Sectio 2.1, it follows from equatios (2.6) (2.8) that µ(m, ) = µ 1 (m, ) + µ 2 (m, ) + µ 3 (m, ), (4.1) where, for q = 1, 2, 3, µ q (m, ) := k=0 2 k l=1 (l 1 2 )2 k (l 1)2 k l2 k (l 1 2 )2 k (k + 1)P q (s, t, m, ) dt ds. (4.2) 49

58 CHAPTER 4. CLOSED FORMULA FOR µ(m, ) The same techique ca be applied to elimiate the ifiite summatio ad itegratio from each µ q (m, ). We describe the techique for obtaiig a closed expressio of µ 1 (m, ) i detail. First, we trasform P 1 (s, t, m, ) show i (2.3) so that we ca elimiate the itegratio i µ 1 (m, ). Defie ( 2 C 1 (i, ) := 1(1 m i < ) m + 1 i 1, 1, i 1, 1, ), (4.3) where 1(1 m i < ) is a idicator fuctio that equals 1 if the evet i braces holds ad 0 otherwise. Sice s i 1 (t s) i 1 (1 t) i 1 = s i 1 it follows that u=0 ( i 1 ( )t u ( 1) i 1 u s i 1 u u v v=0 ) ( 1) v t v, where P 1 (s, t, m, ) = m i< = = m i< 2 f=m 1 C 1 (i, ) C 1 (i, ) f 2 h=0 f+1 C 2 (f, h) := i 1 u=0 2 ( )( i 1 u v v=0 f 2 f=i 1 h= f 2 ( )( i 1 s f t h f i + 1 ) s u 2 t v+u ( 1) i u v 1 h + f + 2 ) ( 1) h i +1 s f t h C 2 (f, h), (4.4) f+h+2 i=m =f+2 ( )( i 1 C 1 (i, ) f i h + f + 2 ) ( 1) h i +1.

59 CHAPTER 4. CLOSED FORMULA FOR µ(m, ) Thus, from (4.2) ad (4.4), we ca elimiate the itegratio i µ 1 (m, ) ad express it usig polyomials i l: µ 1 (m, ) = 2 f=m 1 f 2 h=0 C 3 (f, h) (k + 1) k=0 2 k l=1 2 k(f+h+2) [l h+1 (l 1 2 )h+1 ] [(l 1 2 )f+1 (l 1) f+1 ], (4.5) where C 3 (f, h) := 1 ( + 1)(f + 1) C 2(f, h). Note that Hece [ l h+1 ( l 1 2) h+1 = ( l 1 2) f+1 (l 1) f+1 = l h+1 = h ( ( h + 1 )l 1 h+1, 2) f ( [ f + 1 )l ( 1) f+1 1 =0 =0 ( l 1 ) ] [ h+1 ( l f h ( )( ) f + 1 h + 1 ( 1) f+h =0 =0 which ca be rearraged to where f+h+1 =1 C 4 (f, h, ) := ( 1) f+h +1 ( 1 2 ) f+1 (l 1) f+1 ] [ 1 ( ) ] f ( ) ] f+1 ( ) h l +, 2 2 C 4 (f, h, )l 1, (4.6) ) h +2 ( 1) V f [ 1 51 =0 W ( 1 h) ( )( f + 1 h + 1 ( ) ] f+1 ( ). 1 )

60 CHAPTER 4. CLOSED FORMULA FOR µ(m, ) Therefore, from (4.5) (4.6), we obtai µ 1 (m, ) = = 2 f=m 1 2 f=m 1 f 2 h=0 f 2 h=0 C 3 (f, h) f+h+1 =1 (k + 1) k=0 C 5 (f, h, ) 2 k l=1 (k + 1)2 k=0 f+h+1 2 k(f+h+2) =1 k(f+h+2) 2k C 4 (f, h, )l 1 l 1, l=1 where C 5 (f, h, ) := C 3 (f, h) C 4 (f, h, ). Here, as described i Sectio 2.2, 2 k 1 l 1 = a,r 2 k( r), l=1 r=0 where a,r is defied by (2.12). Now defie C 6 (f, h,, r) := a,r C 5 (f, h, ). The µ 1 (m, ) = = = 2 f=m 1 2 f=m 1 f 2 h=0 f 2 h=0 f+h+1 =1 f+h+1 =1 1 a,r C 5 (f, h, ) (k + 1)2 k(f+h+2+r ) r=0 k=0 1 C 6 (f, h,, r)[1 2 (f+h+2+r ) ] 2 r=0 1 C 7 (a)(1 2 a ) 2, (4.7) a=1 where C 7 (a) := 2 f=m 1 f 2 h=α f+h+1 =β C 6 (f, h,, a + (f + h + 2)), i which α := 0 (a f 1) ad β := 1 (f + h + 2 a). The procedure described above ca be applied to derive aalogous exact formulae 52

61 CHAPTER 4. CLOSED FORMULA FOR µ(m, ) for µ 2 (m, ) ad µ 3 (m, ). I order to derive the aalogous exact formula for µ 2 (m, ), oe eed oly start the derivatio by chagig the idicator fuctio i C 1 (i, ) [see (4.3)] to 1(1 i < m < ) ad follow each step of the procedure; for µ 3 (m, ), start the derivatio by chagig the idicator fuctio to 1(1 i < m ). Expectatio of bit comparisos μ(m,) m Figure 4.1: Expected umber of bit comparisos for Quickselect. The closed formulae for µ 1 (m, ), µ 2 (m, ), ad µ 3 (m, ) were used to compute µ(m, ) for = 1, 2,..., 20 ( represets the umber of keys) ad m = 1, 2,..., (m represets the rak of the target key). 53