Analysis of Algorithms -Quicksort-

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1 Aalysis of Algorithms -- Adreas Ermedahl MRTC (Mälardales Real-Time Research Ceter) Autum 2004 Proposed by C.A.R. Hoare i 962 Worst- case ruig time: Θ( 2 ) Expected ruig time: Θ( lg ) Costats hidde i Θ( lg ) are small Very efficiet i practice Usually beats HEAPSORT Sorts i place A divide- ad- coquer algorithm C.A.R Hoare To sort the subarray A[p..r]: Divide: Partitio A[p..r], ito two subarrays A[p..q-] ad A[q+..r] such that: Each elemet i first subarray A[p..q-] is A[q] Each elemet i secod subarray A[q+..r] is >A[q] p q- q q+ r A A[q] A[q] > A[q] All elemets pivot Pivot elemet All elemets pivot Coquer: Sort the two subarrays by recursive calls to QUICKSORT Combie: No work is eeded to combie the subarrays, because they are sorted i place (cot.) Iitial call: QUICKSORT(A,, ) Base case coditio = legth[a] Divide Coquer No Combie A call to QUICKSORT(A, p, r) sorts A[p..r] i-place Destructively updates A as a side effect of the call Sice we always sort distict segmets of A this is OK Divide step made by call to procedure PARTITION Returs q which marks positio separatig the subarrays Partitioig subroutie PARTITION(A, p, r): Selects A[r] as pivot elemet seds all elemets A[r] to the left seds all elemets > A[r]to the right returs the ew positio of rightmost elemet < A[r]: Example of partitioig j Iput array p A All elemets pivot i j x > x All elemets > pivot Not yet sorted r? x Pivot elemet All elemets pivot p i A x > x All elemets > pivot Not yet sorted j r? x Pivot elemet Resultig partitioed array

2 Aalysis of partitio Iput size = elemets i subarray A[p..r] Executes oce: Θ() Executes r-p+= times: Θ() Executes - times: Θ() Executes oce: Θ() Executes a most - times fl PARTITION is Θ() Valid each time PARTITION routie is called Aalysis of Ruig time T() of depeds o how successful the partitioigs are: Balaced partitioigs fl ca ru as fast as Mergesort Ubalaced partitioigs Θ( 2 ) fl ca ru as slowly as Isertiosort Base case: Θ() Θ( lg ) Partitio, rus i liear time: Θ() Time for recursive calls depeds o value of q Aalysis of Example array: ? Pivot elemet If pivot is 5 we get a balaced partitioig: ~ /2 elemets pivot ~ /2 elemets > pivot If pivot is 0 we get a ubalaced partitioig: No elemets > pivot Worst-Case of Occurs whe iput already sorted (or i reverse) Always produces a ubalaced partitioig Always partitio aroud max elemet Oe of the subarrays always becomes empty Time to sort elemets Problematic iput array Time to sort - elemets Time to sort elemet = Θ() Time to partitio elemets - elemets pivot Best case of Whe Partitio always split (sub)array evely Always perfectly balaced partitioigs Each subarray has /2 elemets Time to sort elemets T ( ) = 2T ( / 2) + Θ( ) Always geerate two ew subproblems of /2 size Cost for partitioig elemets Master- theorem (case 2) directly gives us: T ( ) = Θ( lg ) Balaced partitioig s average ruig time is much closer to the best case tha to the worst case Imagie that PARTITION always produces a 9-to- split Result i recurrece: 9 T ( ) = T ( ) + T ( ) + Θ( ) 0 0 Asymptotic boud for this recurrece? T() = Θ( log ) Ca be uderstood by recurrece tree method

3 log 0/9 9-to- Split Recursio-tree 9-to- partitioig gives ubalaced tree log 0 c/0 c 9c/0 c/00 9c/00 9c/00 8c/00 Shortest recursio brach peters out after log 0 levels Deepest recursio brach peters out after log 0/9 levels cost c c c 8c/ c/000 c Total cost: cost per level * umber of levels 9 T ( ) = T( ) + T( ) + c 0 0 c c Cost per full level = c Cost per o full level c c * log 0/9 Costat split aalysis We got that the total cost: T() c log 0/9 This gives that: T ( ) = O( log0 / 9 ) = O( lg ) Eve i a seemigly ubalaced case! Actually, ay split of costat proportioality yields: Recursio tree of depth Θ(lg ) With a O() cost of each level fl Ruig time is O( lg ) wheever the split has costat proportioality Average Case Aalysis Ituitio for average case: Not always costat splits i recursio tree There will usually be a mix of good ad bad splits throughout the recursio tree We expect some splits to be reasoably balaced ad some splits fairly ubalaced How does this affect the asymptotic ruig time of? Alteratig Splits Imagie that tree levels alterate betwee best-case ad worst-case splits 0 - Θ() (-) / 2 - (-) / 2 Alteratig best- ad worst-case splits Oly best-case splits Extra level oly adds to the hidde costat i Θ-otatio Same umber of subarrays to sort: oly twice as much work was doe to get to that poit. fl Both figures result i O(lg) time: larger costat hidde i the left figure (-) / 2 (-) / 2 Θ() Radomized Observatio: Not always true that all iput permutatios are equally likely How does this affect the average case ruig time? Idea: Use radom samplig to select a pivot i A[p..r] Do t always use A[r] as the pivot Partitio aroud a radom elemet i A[p..r] Several iterestig properties of resultig algorithm: Ruig time is idepedet of the iput order No assumptios required for the iput distributio No specific iput triggers the worst-case behavior Worst case determied oly by radom-umber geerator Radomized New algorithm calls radomized partitioig Radomly select a pivot elemet from subarray A[p..r] Equal chace that pivot is ay of the r-p+ elemets

4 Aalysis of Radomizatio stops ay specific type of array from causig worst case behavior Radomly selectig the pivot elemet will, o average, cause the split to be reasoably well balaced A already-sorted array o loger worst case Radomized- will help us uderstad origial I particular, we will aalyze: The worst-case ruig time of ad Radomized- (the same) The expected (average-case) ruig time of Radomized- Provig the Worst- Case O( 2 ) Boud What if we experiece a worst-case split at every level? Gives recurrece: Take max whe q rages from 0 to - We use substitutio method to show that T()=O( 2 ) Substitutig ito recurrece yields: We guess that T(k) ck 2 for some costat c ad all k < Maximum is obtaied whe q=0 or q=- We obtai: We use q=0 Givig that T() = O( 2 ) Average case aalysis The domiat cost of QUICKSORT is the partitioig Upper boud o the umber of partitio calls? Each call to PARTITION results i a selected pivot elemet This elemet ca ever be icluded i future recursive calls Thus, we ca have at most calls to PARTITION Sice QUICKSORT recurse o partioig we ca make at most recursive calls Partitioig routie ca called at most times Avarage case aalysis How may comparisos i total? Upper boud o the umber of comparisos? Each call to PARTITION takes costat time + a time proportioal to umber of pivot comparasios i the for loop A total upper boud for umber of comparasios gives a upper boud for the ruig time of QUICKSORT Let X = total umber of comparisos i all calls to PARTITION fl Total work doe over the etire executio is O( + X) We would like to boud X Boudig the Comparsios To boud X we must uderstad whe two elemets are compared ad whe they are ot We reame elemets i A as z,z 2,..., z where z i is the i:th smallest elemet i array A Let Z ij = {z i, z,..., z j } The set of elemets betwee z i ad z j (iclusive) The total umber of comparasios, X, are: X = (# times is compared to i= j= The i:th smallest elemet z i z j ) X = The total umber of comparasios The j:th smallest elemet Example Assume array A cotais elemets,2...0 The particular iput order is ot of iterest The Z 39 = {3,4,5,6,7,8,9} The set of elemets betwee 3 ad 9 (iclusive) Total umber of comparasios made are the: 0 0 X = (# times z i is compared to z j ) i= j= If two elemets are compared oe must be a pivot The pivot will ot be used later i the aalysis Each z i ad z j ca be compared at most oce

5 Average case aalysis Expected (or average) value of X could be determid by probability of comparig differet elemets: Avarage case time = expected value of X E( X ) = i= j= Pr{ z i is compared to z j } fl We have to fid probability that two elemets are compared Ivestigatig Z ij = {z i, z,..., z j } i more detail we ote If first pivot elemet selected from Z ij is x such that z i < x < z j the z i ad z j wili ever be compared If first pivot elemet selected from Z ij is z i or z j the z i ad z j will be compared Probability of comparig z i ad z j Z 2 9 = {2,3,4,5,6,7,8,9} Example Assume 4 is first elemet i Z 2 9 selected as pivot Set Z 2 3 = {2,3} will be sorted i isolatio from Z 5 9 = {5,6,7,8,9} Elemet 2 ad 9 will ever be compared Same if 4, 5,..., 8 is selected as pivot Assume 2 is first elemet i Z 2 9 selected as pivot Elemet 2 will be compared to elemet 9 Same if 9 is first elemet selected as pivot fl Elemet 2 ad 9 will oly be compared if either 2 or 9 is first elemet selected as pivot i the Z 2 9 set Average case aalysis This gives us that: Pr{z i is compared to z j } = Pr{z i or z j is first pivot elemet chose from Z ij } Set Z ij cotais j i + elemets Each elemet is chose radomly ad idepedetly Probability to choose a give elemet is therefore: / (j i + ) This gives us that: Pr{z i is compared to z j } = 2 * ( / (j i + )) = 2 /(j i + ) Z 2 9 = {2,3,4,5,6,7,8,9} Example Probability for a particular elemet i Z 2 9 is: / (9 2 + ) = / 8 We therefore get: Pr{elemet 2 is compared to elemet 9} = Pr{elemet 2 or elemet 9 is first selected from Z 2 9 } = 2 * ( / 8) = 2 / 8 = / 4 Expected value of X Average case aalysis Set variable k = j - i A.7: Harmoic series rule E( X ) = = = < = i= j= i= j= i i= k = i= k = i= O(lg ) = O( lg ) Pr{ z is compared to z } i 2 By previous slides j i + 2 k k = l + O() k k = j Upper boud o the umber of comparisos i the average case Ruig Time Summary Obtaied ruig time bouds: Best Case: Θ( log ) Average Case: Θ( log ) Worst Case: Θ( 2 ) Average case aalysis quite complicated Ivolves probabilities Not really i the course cotets Just skim this sectio of the course book!

6 i Practice A great geeral- purpose sortig algorithm Typically over twice as fast as Mergesort Behaves well eve with cachig ad virtual memory We ca make smaller modificatios to ehace performace further... Modificatios of Note: Costly to perform o small iputs Idea: Take advatage of the fact of fast ruig time of Isertiosort for almost sorted iput Modificatios: Whe is called with fewer tha k elemets, let it retur without sortig subarray After top-level call to returs, ru Isertiosort o etire array to fiish sortig Resultig algorithm rus i O(k + lg(/k)) See Exercise Modificatios of Note: behavior is rather sesitive to ubalaced partitioigs Idea: Somehow reduce probability that a bad pivot elemet is selected Modificatios: (Radomly) select three differet elemets from subarray istead of just oe elemet Let the media of the three elemets be the pivot Gets the same asymptotic bouds as origial, but with smaller costats Modificatios of Note: cotais two recursive calls to itself Forces us to do a lot of costly fuctio calls Forces us to save a lot of fuctio cotexts o the stack Idea: Rewrite algorithm to be tail-recursive Use a iterative cotrol structure istead of recursio QUICKSORT-TR(A, p, r). while p < r do 2. q PARTITION(A, p, r) 3. QUICKSORT-TR(A, p, q-) 4. p q + See 7-4 for more iformatio o tail-recursive versio The Ed!

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