CSE 521: Design and Analysis of Algorithms I

Transcription

1 CSE 521: Design nd Anlysis of Algorithms I Divide nd Conquer Pul Beme 1

2 Algorithm Design Techniques Divide & Conquer Reduce prolem to one or more su-prolems of the sme type Typiclly, ech su-prolem is t most constnt frction of the size of the originl prolem e.g. Mergesort, Binry Serch, Strssen s Algorithm, Quicksort (kind of) 2

3 Fst exponentition Power(,n) Input: integer n nd numer Output: n Ovious lgorithm n-1 multiplictions Oservtion: if n is even, n=2m, then n = m m 3

4 Divide & Conquer Algorithm Power(,n) if n=0 then return(1) else if n=1 then return() else x Power(, n/2 ) if n is even then return(x x) else return( x x) 4

5 Anlysis Worst-cse recurrence T(n)=T( n/2 )2 for n 1 T(1)=0 Time T(n)=T( n/2 )2 T( n/4 )22 T(1)2 2 = 2 log 2 n More precise nlysis: log 2 n copies T(n)= log 2 n # of 1 s in n s inry representtion 5

6 A Prcticl Appliction- RSA Insted of n wnt n mod N ij mod N = (( i mod N) ( j mod N)) mod N sme lgorithm pplies with ech x y replced y ((x mod N) (y mod N)) mod N In RSA cryptosystem need n mod N where, n, N ech typiclly hve 1024 its Power: t most 2048 multiplies of 1024 it numers reltively esy for modern mchines Nive lgorithm: multiplies 6

7 Binry serch for roots (isection method) Given: continuous function f nd two points < with f() 0 nd f() > 0 Find: pproximtion to c s.t. f(c)=0 nd <c< 7

8 Bisection method Bisection(,, ε) if (-) < ε then return() else c ()/2 if f(c) 0 then return(bisection(c,,ε)) else return(bisection(,c,ε)) 8

9 Time Anlysis At ech step we hlved the size of the intervl It strted t size - It ended t size ε # of clls to f is log 2 ( (-)/ε) 9

10 Old fvorites Binry serch One suprolem of hlf size plus one comprison Recurrence T(n) = T( n/2 )1 for n 2 T(1) = 0 So T(n) is log 2 n 1 Mergesort Two suprolems of hlf size plus merge cost of n-1 comprisons Recurrence T(n) 2T( n/2 )n-1 for n 2 T(1) = 0 Roughly n comprisons t ech of log 2 n levels of recursion So T(n) is roughly 2n log 2 n 10

11 Eucliden Closest Pir Given set P of n points p 1,,p n with relvlued coordintes Find the pir of points p i,p j P such tht the Eucliden distnce d(p i,p j ) is minimized Θ(n 2 ) possile pirs In one dimension there is n esy O(n log n) lgorithm Sort the points Compre consecutive elements in the sorted list Wht out points in the plne? 11

12 Closest Pir in the Plne No single direction long which one cn sort points to gurntee success! 12

13 Closest Pir In the Plne: Divide nd Conquer Sort the points y their x coordintes Split the points into two sets of n/2 points L nd R y x coordinte Recursively compute closest pir of points in L, (p L,q L ) closest pir of points in R, (p R,q R ) Let δ=min{d(p L,q L ),d(p R,q R )} nd let (p,q) e the pir of points tht hs distnce δ This my not e enough! Closest pir of points my involve one point from L nd the other from R! 13

14 A clever geometric ide L R Any pir of points p L nd q R with d(p,q)<δ must lie in nd δ δ 14

15 A clever geometric ide L R Any pir of points p L nd q R with d(p,q)<δ must lie in nd δ/2 No two points cn e in the sme green ox δ/ 2 δ δ 15

16 A clever geometric ide L R Any pir of points p L nd q R with d(p,q)<δ must lie in nd δ/2 No two points cn e in the sme green ox δ δ Only need to check pirs of points up to 2 rows prt - constnt # of other points! 16

17 Closest Pir Recomining Sort points y y coordinte hed of time On recomintion only compre ech point in L R to the 12 points ove it in the y sorted order If ny of those distnces is etter thn δ replce (p,q) y the est of those pirs O(n log n) for x nd y sorting t strt Two recursive clls on prolems on hlf size O(n) recomintion Totl O(n log n) 17

18 Sometimes two su-prolems ren t enough More generl divide nd conquer You ve roken the prolem into different su-prolems Ech hs size t most n/ The cost of the rek-up nd recomining the su-prolem solutions is O(n k ) Recurrence T(n) T(n/)c n k 18

19 Mster Divide nd Conquer Recurrence If T(n) T(n/)c n k for n> then if > k log then T(n) is Θ(n ) if < k then T(n) is Θ(n k ) if = k then T(n) is Θ(n k log n) Works even if it is n/ insted of n/. 19

20 Proving Mster recurrence Prolem size n T(n)=T(n/)cn k # pros 1 n/ n/ T(1)=c d 20

21 Proving Mster recurrence Prolem size n T(n)= T(n/)c n k # pros 1 n/ n/ T(1)=c d 21

22 Proving Mster recurrence Prolem size n T(n)= T(n/)c n k # pros 1 cost cn k n/ c n k / k n/ 2 2 c 2 n k / 2k =c n k (/ k ) 2 1 T(1)=c d c n k (/ k ) d =c d 22

23 Geometric Series S = t tr tr 2... tr n-1 r S = tr tr 2... tr n-1 tr n (r-1)s =tr n - t so S=t (r n -1)/(r-1) if r 1. Simple rule If r 1 then S is constnt times lrgest term in series 23

24 Totl Cost Geometric series rtio / k d1=log n 1 terms first term cn k, lst term c d If / k =1 ll terms re equl T(n) is Θ(n k log n) If / k <1 first term is lrgest T(n) is Θ(n k ) If / k >1 lst term is lrgest T(n) is Θ( d )=Θ( (To see this tke log of oth sides) log n log ) ) =Θ(n 24

25 25 Multiplying Mtrices n 3 multiplictions, n 3 -n 2 dditions = o o o o

26 Multiplying Mtrices for i=1 to n for j=1 to n C[i,j] 0 for k=1 to n C[i,j]=C[i,j]A[i,k] B[k,j] endfor endfor endfor 26

27 27 Multiplying Mtrices = o o o o

28 28 Multiplying Mtrices = o o o o

29 Multiplying Mtrices 11 = A A A A B B B B A B A B A B A B o o o o A B A B A 21 B 12 A 22 B

30 Simple Divide nd Conquer A 11 A 12 B 11 B 12 A 21 A 22 B 21 B 22 = A 11 B 11 A 12 B 21 A 11 B 12 A 12 B 22 A 21 B 11 A 22 B 21 A 21 B 12 A 22 B 22 T(n)=8T(n/2)4(n/2) 2 =8T(n/2)n 2 8>2 2 so T(n) is Θ Θ Θ log log28 3 ( n ) = ( n ) = ( n ) 30

31 Strssen s Divide nd Conquer Algorithm Strssen s lgorithm Multiply 2x2 mtrices using 7 insted of 8 multiplictions (nd lots more thn 4 dditions) T(n)=7 T(n/2)cn 2 log 2 7 7>2 2 so T(n) is Θ(n ) which is O(n 2.81 ) Fstest lgorithms theoreticlly use O(n ) time not prcticl ut Strssen s is prcticl provided clcultions re exct nd we stop recursion when mtrix hs size out 32 nd use ordinry multipliction for suprolems 31

32 The lgorithm P 1 A 12 (B 11 B 21 ); P 2 A 21 (B 12 B 22 ) P 3 (A 11 - A 12 )B 11 ; P 4 (A 22 - A 21 )B 22 P 5 (A 22 - A 12 )(B 21 - B 22 ) P 6 (A 11 - A 21 )(B 12 - B 11 ) P 7 (A 21 - A 12 )(B 11 B 22 ) C 11 P 1 P 3 ; C 12 P 2 P 3 P 6 - P 7 C 21 P 1 P 4 P 5 P 7 ; C 22 P 2 P 4 32

33 Another Divide &Conquer Exmple: Multiplying Fster If you nlyze our usul elementry school lgorithm for multiplying numers Θ(n 2 ) time On rel mchines ech digit is, e.g., 32 its long ut still get Θ(n 2 ) running time with this lgorithm when run on n-it multipliction We cn do etter! We ll descrie the sic ides y multiplying polynomils rther thn integers Advntge is we don t get confused y worrying out crries 33

34 Polynomil Multipliction Given: Degree n-1 polynomils P nd Q P = 0 1 x 2 x 2 n-2 x n-2 n-1 x n-1 Q = 0 1 x 2 x 2 n-2 x n-2 n-1 x n-1 Compute: Degree 2n-2 Polynomil P Q P Q = 0 0 ( ) x ( ) x 2... ( n-2 n-1 n-1 n-2 ) x 2n-3 n-1 n-1 x 2n-2 Ovious Algorithm: Compute ll i j nd collect terms Θ (n 2 ) time 34

35 Nive Divide nd Conquer Assume n=2k P = ( 0 1 x 2 x 2... k-2 x k-2 k-1 x k-1 ) ( k k1 x n-2 x k-2 n-1 x k-1 ) x k = P 0 P 1 x k where P 0 nd P 1 re degree k-1 polynomils Similrly Q = Q 0 Q 1 x k P Q = (P 0 P 1 x k )(Q 0 Q 1 x k ) = P 0 Q 0 (P 1 Q 0 P 0 Q 1 )x k P 1 Q 1 x 2k 4 su-prolems of size k=n/2 plus liner comining T(n)=4 T(n/2)cn Solution T(n) = Θ(n 2 ) 35

36 Krtsu s Algorithm A etter wy to compute the terms Compute A P 0 Q 0 B P 1 Q 1 C (P 0 P 1 )(Q 0 Q 1 ) = P 0 Q 0 P 1 Q 0 P 0 Q 1 P 1 Q 1 Then P 0 Q 1 P 1 Q 0 = C-A-B So PQ=A(C-A-B)x k Bx 2k 3 su-prolems of size n/2 plus O(n) work T(n) = 3 T(n/2) cn T(n) = O(n α ) where α = log 2 3 =

37 Krtsu: Detils Mid B R 2n-1 n n/2 0 PolyMul(P, Q): // P, Q re length n =2k vectors, with P[i], Q[i] eing // the coefficient of x i in polynomils P, Q respectively. // Let Pzero e elements 0..k-1 of P; Pone e elements k..n-1 // Qzero, Qone : similr If n=1 then Return(P[0]*Q[0]) else A PolyMul(Pzero, Qzero); // result is (2k-1)-vector B PolyMul(Pone, Qone); // ditto Psum Pzero Pone; // dd corresponding elements Qsum Qzero Qone; // ditto C polymul(psum, Qsum); // nother (2k-1)-vector Mid C A B; // sutrct correspond elements R A Shift(Mid, n/2) Shift(B,n) // (2n-1)-vector Return( R); A 37

38 Multipliction Polynomils Nïve: Θ(n 2 ) Krtsu: Θ(n 1.59 ) Best known: Θ(n log n) "Fst Fourier Trnsform FFT widely used for signl processing, especilly s DCT (Discrete Cosine Trnsform) Integers Similr, ut some ugly detils re: crries, etc. due to Schonhge-Strssen in 1971 gives Θ(n log n loglog n), Improvement in 2007 due to Furer gives Θ(n log n 2 log* n ) Used in prctice in symolic mnipultion systems like Mple 38

39 Hints towrds FFT: Interpoltion Given set of vlues t 5 points 39

40 Hints towrds FFT: Interpoltion Given set of vlues t 5 points Cn find unique degree 4 polynomil going through these points 40

41 Multiplying Polynomils y Evlution & Interpoltion Any degree n-1 polynomil R(y) is determined y R(y 0 ),... R(y n-1 ) for ny n distinct y 0,...,y n-1 To compute PQ (ssume degree t most n-1) Evlute P(y 0 ),..., P(y n-1 ) Evlute Q(y 0 ),...,Q(y n-1 ) Multiply vlues P(y i )Q(y i ) for i=0,...,n-1 Interpolte to recover PQ 41

42 Interpoltion Given vlues of degree n-1 polynomil R t n distinct points y 1,,y n R(y 1 ),,R(y n ) Compute coefficients c 0,,c n-1 such tht R(x)=c 0 c 1 xc 2 x 2 c n-1 x n-1 System of liner equtions in c 0,,c n-1 c 0 c 1 y 1 c 2 y 12 c n-1 y n-1 1 =R(y 1 ) c 0 c 1 y 2 c 2 y 22 c n-1 y n-1 2 =R(y 2 ) c 0 c 1 y n c 2 y n2 c n-1 y n-1 n =R(y n ) known unknown 42

43 Interpoltion: n equtions in n unknowns Mtrix form of the liner system 1 y 1 y 12 y n-1 1 c 0 R(y 1 ) 1 y 2 y 22 y n-1 2 c 1 R(y 2 ) c 2 =... 1 y n y n2 y n-1 n c n-1 R(y n ) Fct: Determinnt of the mtrix is Π i<j (y i -y j ) which is not 0 since points re distinct System hs unique solution c 0,,c n-1 43

44 Hints towrds FFT: Evlution & Interpoltion P: 0, 1,..., n-1 Q: 0, 1,..., n-1 evlution t y 0,...,y 2n-1 O(?) ordinry polynomil multipliction Θ(n 2 ) c k i j= k i j R:c 0,c 1,...,c 2n-1 interpoltion from y 0,...,y 2n-1 O(?) P(y 0 ),Q(y 0 ) P(y 1 ),Q(y 1 )... P(y 2n-1 ),Q(y 2n-1 ) point-wise multipliction of numers O(n) R(y 0 ) P(y 0 ) Q(y 0 ) R(y 1 ) P(y 1 ) Q(y 1 )... R(y 2n-1 ) P(y 2n-1 ) Q(y 2n-1 ) 44

45 Krtsu s lgorithm nd evlution nd interpoltion Strssen gve wy of doing 2x2 mtrix multiplies with fewer multiplictions Krtsu s lgorithm cn e thought of s wy of multiplying degree 1 polynomils (which hve 2 coefficients) using fewer multiplictions PQ=(P 0 P 1 z)(q 0 Q 1 z) = P 0 Q 0 (P 1 Q 0 P 0 Q 1 )z P 1 Q 1 z 2 Alterntive Krtsu: evlute t 0,1,-1 (Could lso use other points) A = P(0)Q(0)= P 0 Q 0 C = P(1)Q(1)=(P 0 P 1 )(Q 0 Q 1 ) D = P(-1)Q(-1)=(P 0 -P 1 )(Q 0 -Q 1 ) Interpolting, P 1 Q 0 P 0 Q 1 =(C-D)/2 nd P 1 Q 1 =(CD)/2-A 45

46 Evlution t Specil Points Evlution of polynomil t 1 point tkes O(n) time So 2n points (nively) tkes O(n 2 ) no svings But the lgorithm works no mtter wht the points re So choose points tht re relted to ech other so tht evlution prolems cn shre suprolems 46

47 The key ide: Evlute t relted points P(x) = 0 1 x 2 x 2 3 x 3 4 x 4... n-1 x n-1 = 0 2 x 2 4 x 4... n-2 x n-2 1 x 3 x 3 5 x 5... n-1 x n-1 = P even (x 2 ) x P odd (x 2 ) P(-x)= 0-1 x 2 x 2-3 x 3 4 x n-1 x n-1 = 0 2 x 2 4 x 4... n-2 x n-2 - ( 1 x 3 x 3 5 x 5... n-1 x n-1 ) = P even (x 2 ) - x P odd (x 2 ) where P even (x) = 0 2 x 4 x 2... n-2 x n/2-1 nd P odd (x) = 1 3 x 5 x 2... n-1 x n/2-1 47

48 The key ide: Evlute t relted points So if we hve hlf the points s negtives of the other hlf i.e., y n/2 = -y 0, y n/21 = -y 1,,y n-1 = -y n/2-1 then we cn reduce the size n prolem of evluting degree n-1 polynomil P t n points to evluting 2 degree n/2-1 polynomils P even nd P odd t n/2 points y 02, y n/2-12 nd recomine nswers with O(1) extr work per point But to use this ide recursively we need hlf of y 02, y n/2-12 to e negtives of the other hlf If y n/4 2 = -y 02, sy, then (y n/4 /y 0 ) 2 = -1 Motivtes use of complex numers s evlution points 48

49 Complex Numers i 2 = -1 efi -1 cdi θϕ i ϕ θ i 1 To multiply complex numers: 1. dd ngles 2. multiply lengths (ll length 1 here) efi = (i)(cdi) e 2πi = 1 e π i = -1 -i i =cos θ i sin θ = e iθ cdi =cos ϕ i sin ϕ = e iϕ efi =cos (θϕ) i sin (θϕ) = e i(θϕ θϕ) 49

50 Primitive n th root of 1 /n ω=ω n = e i 2π/n ω 3 ω 2 =i ω Let ω = ω n = ei 2π /n = cos (2π/ π/n) i sin (2π/n) ω 4 =-1 ω 0 =1=ω 8 ω 5 ω 6 = -i ω 7 i 2 = -1 E 2π i = 1 50

51 Fcts out ω=e 2πi /n for even n ω n = 1 ω n/2 = -1 ω n/2k = - ω k for ll vlues of k ω 2 = e 2πi /m where m=n/2 ω k = cos(2kπ/n)i sin(2kπ/n) so cn compute with powers of ω ω k is root of x n -1= (x-1)(x n-1 x n-2 1) =0 ut for k 0, ω k 1 so ω k(n-1) ω k(n-2) 1=0 51

52 Gol: The recursive ide for n power of 2 Evlute P t 1,ω,ω 2,ω 3,...,ω n-1 Now P even nd P odd hve degree n/2-1 where P(ω k )=P even (ω 2k )ω k P odd (ω 2k ) P(-ω k )=P even (ω 2k )-ω k P odd (ω 2k ) Recursive Algorithm Evlute P even t 1,ω 2,ω 4,...,ω n-2 Evlute P odd t 1,ω 2,ω 4,...,ω n-2 Comine to compute P t 1,ω,ω 2,...,ω n/2-1 Comine to compute P t -1,-ω,-ω 2,...,-ω n/2-1 (i.e. t ω n/2, ω n/21, ω n/22,..., ω n-1 ) ω 2 is e 2πi / m where m=n/2 so prolems re of sme type ut smller size 52

53 Anlysis nd more Run-time T(n)=2 T(n/2)cn so T(n)=O(n log n) So much for evlution... wht out interpoltion? Given R(1), R(ω), R(ω 2 ),..., R(ω n-1 ) Compute c 0, c 1,...,c n-1 s.t. R(x)=c 0 c 1 x...c n-1 x n-1 53

54 Interpoltion: n equtions in n unknowns Mtrix form of the liner system c 0 R(1) 1 ω ω 2 ω n-1 c 1 R(ω) c 2 =... 1 ω n-1 ω 2n-2 ω (n-1)(n-1) c n-1 R(ω n-1 ) Let M e the interpoltion mtrix for these points Tht is: M ij = ω ij 54

55 The inverse of M Define mtrix N y N ij = ω -ij. Then (MN) ij =Σ k=0..n =0..n-1 ωik ω -kj = Σ k=0..n If i=j then this is Σ 1 = n k=0..n-1 If i j then this is k=0..n-1 ω k(i-j) 1ω i-j ω 2(i-j) ω (n-1)(i-j) =0 So MN is n times the identity mtrix; tht is M -1 =N/n 55

56 Interpoltion using FFT So C=M -1 R=N R/n; tht is R(1)/n c 0 1 ω -1 ω -2 ω -(n-1) R(ω)/n c 1. = c ω -(n-1) ω -(2n-2) ω -(n-1)(n-1) R(ω n-1 )/n c n-1 But mtrix N is just the mtrix for the evlution t points 1, ω -1, ω -2,, ω -(n-1)!!! So pply the sme FFT recursion for the interpoltion phse which is lso O(n log n) time. 56

57 Why this is clled the discrete Fourier trnsform Rel Fourier series Given rel vlued function f defined on [0,2π] the Fourier series for f is given y f(x)= 0 1 cos(x) 2 cos(2x)... m cos(mx)... where 2π 1 m = f(x) cos(mx) dx 2π 0 is the component of f of frequency m In signl processing nd dt compression one ignores ll ut the components with lrge m nd there ren t mny of these. 57

58 Why this is clled the discrete Fourier trnsform Complex Fourier series Given function f defined on [0,2π] the complex Fourier series for f is given y f(z)= 0 1 e i z 2 e 2i z... m e mi z... where m = 2π 1 π -mi z f(z) e dz 2 0 is the component of f of frequency m If we discretize this integrl using vlues t n eqully spced points etween 0 nd 2π we get 1 1 f e f n 1 n 1 iπ = -2km /n = km m k k ω n k= 0 n k= 0 just like interpoltion! where f k =f(2kπ/n) 2π/n prt 58

59 Beyond the Mster Theorem Divide nd conquer exmples Simple, rndomized medin lgorithm Expected O(n) time Not so simple, deterministic medin lgorithm Worst cse O(n) time Expected time nlysis for Rndomized QuickSort Expected O(n log n) time 59 59

60 Order prolems: Find the k th lrgest Runtime models Mchine Instructions Comprisons Mximum O(n) time n-1 comprisons 2 nd Lrgest O(n) time? comprisons 60

61 Medin Prolem k th lrgest for k = n/2 Esily done in O(n log n) time with sorting How cn the prolem e solved in O(n) time? Select(k, n) find the k-th lrgest from list of length n 61

62 Divide nd Conquer T(n) = n T(αn) for α < 1 Liner time solution Select lgorithm in liner time, reduce the prolem from selecting the k-th lrgest of n to the j-th lrgest of αn, for α < 1 62

63 Quick Select QSelect(k, S) Choose element x from S S L = {y in S y < x } S E = {y in S y = x } S G = {y in S y > x } if S L k return QSelect(k, S L ) else if S L S E k return y else return QSelect(k - S L - S E, S G ) 63

64 Implementing Choose n element x Idelly, we would choose n x in the middle, to reduce oth sets in hlf nd gurntee progress Method 1 Select n element t rndom Method 2 BFPRT Algorithm Select n element y complicted, ut liner time method tht gurntees good split 64

65 Rndom Selection Consider cll to QSelect(k, S), nd let S e the elements pssed to the recursive cll. With proility t lest ½, S < ¾ S d x good x good x elements of S listed in sorted order d x On verge only 2 recursive clls efore the size of S is t most 3n/4 65

66 Expected runtime is O(n) Given x, one pss over S to determine S L, S E, nd S G nd their sizes: cn time. Expect 2cn cost efore size of S drops to t most 3 S /4 Let T(n) e the expected running time T(n) T(3n/4) 2cn 2cn (¾) 2cn (¾) 2 2cn 2cn (1 (¾) (¾) 2 ) 66

67 Mking the lgorithm deterministic In O(n) time, find n element tht gurntees tht the lrger set in the split hs size t most ¾ n 67

68 Blum-Floyd-Prtt-Rivest-Trjn Algorithm Divide S into n/5 sets of size 5 Sort ech of these sets of size 5 Let M e the set of ll medins of the sets of size 5 Let x e the medin of M S L = {y in S y < x}, S G = {y in S y > x} Clim: S L < ¾ S, S G < ¾ S 68

69 BFPRT, Step 1: Construct sets of size 5, sort ech set 13, 15, 32, 14, 95, 5, 16, 45, 86, 65, 62, 41, 81, 52, 32, 32, 12, 73, 25, 81, 47, 8, 69, 9, 7, 81, 18, 25, 42, 91, 64, 98, 96, 91, 6, 51, 21, 12, 36, 11, 11, 9, 5, 17,

70 BFPRT, Step 2: Find medin of column medins

71 BFPRT Recurrence Sorting ll n/5 lists of size 5 c n time Finding medin of set M of medins Recursive computtion: T(n/5) Computing sets S L, S E, S G nd S c n time Solving selection prolem on S Recursive computtion: T(3n/4) since S ¾ n 71

72 T(n) cn T(n/5) T(3n/4) is O(n) Key property 3/4 1/5 < 1 (The sum is 19/20) Sum of prolem sizes decreses y 19/20 fctor per level of recursion Overhed per level is liner in the sum of the prolem sizes Overhed decreses y 19/20 fctor per level of recursion Totl overhed is liner (sum of geometric series with constnt rtio nd liner lrgest term) 72

73 Quick Sort QuickSort(S) if S is empty, return Choose element x from S pivot S L = {y in S y < x } S E = {y in S y = x } S G = {y in S y > x } return [QuickSort(S L ), S E, QuickSort(S G )] 73

74 QuickSort Pivot Selection Choose the medin T(n) = T(n/2) T(n/2) cn, O(n log n) Choose ritrry element Worst cse O(n 2 ) Averge cse O(n log n) Choose rndom pivot Expected time O(n log n) 74

75 Expected run time for QuickSort: Glol nlysis Count comprisons i, j elements in positions i nd j in the finl sorted list. p ij the proility tht i nd j re compred Expected numer of comprisons: Σ i<j p ij 75

76 Lemm: P ij 2/(j i 1) If i nd j re compred then it must e during the cll when they end up in different suprolems - Before tht, they ren t compred to ech other - After they ren t compred to ech other During this step they re only compred if one of them is the pivot Since ll elements etween i nd j re lso in the suprolem this is 2 out of t lest j-i1 choices 76

77 Averge runtime is 2nln n Σ i<j p ij Σ i<j 2/(j-i1) write j=ki = 2 n-1 n-i 1/(k 1) i=1 k=1 2 (n-1) (H n -1) where H n =11/21/31/4... = ln n O(1) 2n ln n O(n) 1.387nlog 2 n 77

78 Divide nd Conquer Summry Powerful technique, when pplicle Divide lrge prolem into few smller prolems of the sme type Smller prolems must e constnt fctor smller 78