Randomized Algorithms presentation

1. 2004 spring Randomized Algorithmspresentation B90902036許　平B90902050郭煜楓B90902057陳文祥

2. Randomized Selection in n+C+o(n)comparisons • Alexandros V. Gerbessiotis CS Department, New Jersey Institute of Technology, Newark • Constantinos J. Siniolakis The American College of Greece, 6 Gravias St., Aghia Paraskevi, Athens, 15342, Greece. Randomzied Algorithm

3. Part 1 Introduction

4. Introduction • With high probability for any constantρ > 1 requires n + C + o(n) comparisons to determine the C-th order statistic of n keys • Reduced by extending the algorithm of Floyd and Rivest and analyzing its resulting performance more carefully Randomzied Algorithm

5. Selection in O(n) : BFPRT algorithm • M. Blum • R. W. Floyd • V. R. Pratt • R. L. Rivest • R. E. Tarjan • 1973 • T(n) < c‧n Randomzied Algorithm

6. BFPRT algorithm • 1. Divide the items into floor(n/5) groups of5itemseach. Last group can be smaller. • 2. Find the median of each group(usingsorting). • 3. Use SELECT to recursively find themedian of the floor(n/5) group medians. Randomzied Algorithm

7. BFPRT algorithm • 4. Partition the input by using thismedian-of-median as pivot. • 5. Suppose low side of the partition has selements, and high side has n - s elements. • 6. If k ≤s, recursively call SELECT(k) onlow side; otherwise, recursively callSELECT(k - s) on high side. Randomzied Algorithm

8. BFPRT algorithm • Analysis : • x = pivot • At least half the medians are ≥ x • At least half of the floor(n/5) groups contributeat least3 items tothe high side. • |Item|≥ x are at least Randomzied Algorithm

9. BFPRT algorithm • Recursive call to SELECT is on sizeat most7n/10 + 6. • Step 3 has a recursive call T(n/5), and • Step 5 has a recursive call T(7n/10 + 6). Randomzied Algorithm

10. BFPRT algorithm • Inductively verify that T(n) ≤ cnfor scome constant c • T(n) ≤ c(n/5) + c(7n/10 + 6) + O(n) ≤ 9cn/10 + 6c + O(n) ≤ cn • In above, choose c so that c(n/10 - 6) beatsthe function O(n) for all n. Randomzied Algorithm

11. Part 1.5 The scenario of our random selection algorithm

12. Step 1. Random Sampling • 1. Let Y be randomly chosen subset of size2λ -1 < n/2 from X • 2. sort y in little time ! |X| = n |Y| = 2λ - 1 Randomzied Algorithm

13. Step 2. 左護法 右護法 • Assume : 2cλ > μ/2 for λ≧1, μ≧1 • L = y2cλ- μ/2 • R = y2cλ+ μ/2 • X0 = { x | x ∈ X and x < L } • X1 = { x | x ∈ X and L < x < R } • X2 = { x | x ∈ X and x > R } Randomzied Algorithm

14. Part 2 Claim

15. Sampling result X Y X0 X1 X2 Randomzied Algorithm

16. Probabilistic express • |X0| = n0 , |X2| = n2 • P0(j) : the probability that n0 = j • P2(j) : the probability that n2 = j Randomzied Algorithm

17. Claim 1 Randomzied Algorithm

18. That is … • P0(j) : 左護法 = j • P2(j) : 右護法 = N – J – 1 X0 左 X1 右 X2 Randomzied Algorithm

19. That is … |X| = n |Y| = 2λ - 1 Randomzied Algorithm

20. That is … L = y 2cλ- μ/2 X0 = j Select L-1 numbers into X0 from y Randomzied Algorithm

21. That is … = Randomzied Algorithm

22. Our goal • Establishing lower and upper bounds on the sizes of X0 and X2 , we guarantee that therequired statistic is located in set X1. • Proving that the probability of having such a set of size less than cn– f(n) for some functionf(n)of n is negligible Randomzied Algorithm

23. Proposition 1 • Proposition 1 • Let C ≤ N/2 , c = C/N . λ≥ 1, μ≥ 6, 2λ-1 < N/2 n ≥ N, 2cλ>μ/2, ρ>1 if f(N) = (Nμ) / ( 2λ) and μ= 2 ( 3ρ2cλlogn ) ^ 0.5 prove P( |X0| > cN + f(N) – 1 ) ≥ 1- n^(1- ρ) Randomzied Algorithm

24. Proposition 1 • Def Randomzied Algorithm

25. Proposition 1 • 證明 剩下證明 Randomzied Algorithm

26. Proposition 1 • 首先可以知道 接下來要用一個特殊的方法來算出上面那條等式的上界 Randomzied Algorithm

27. Proposition 1 • 觀察下列這道等式 可以看出 可以被視作為從 N 個物件裡, 取出 2λ-1個, 其中第 2cλ-μ/2個最大是 B 的機率 (…附註1) Randomzied Algorithm

28. 附註1 Concrete Math, p169, (5.26) l = N – 1, k = j, q = 0 m = 2(1-c)λ+μ/2 -1 n = 2cλ-μ/2 -1 Randomzied Algorithm

29. Proposition 1 • Let q = ( 2λ-1 ) / N . 在 N 的數中選出 2λ-1 的數, 每個數被選中的機率. 定義 p.s 前 j 個數中取 2cλ-μ/2-1 個數 取第 j+1 個數 , 一定是 2cλ-μ/2 th in sorted X 後 N-j-1 的數中取 2(1-c)λ+μ/2-1 個數 Randomzied Algorithm

30. Lemma1 Let Randomzied Algorithm

31. Lemma1-proof B.Bollobas. Random Graphs. Academic Press, New York, 1984 d = N, c = 2λ-1, a = q = (2λ-1)/N b = 1 - a Randomzied Algorithm

32. lemma2 我們要使用 generating function 的方法, 來證上面的式子 將上面那項當作是 多項式 ((1-q)+qx)^j 的 x^(2cλ-μ/2 -1) 係數 Randomzied Algorithm

33. lemma2-proof 那我們要求的方程式就會是 的 x^(2cλ-μ/2 -1) 係數 Randomzied Algorithm

34. lemma2-proof D. Angluin and L. G. Valiant. Fast probabilistic algorithms for hamiltonian circuits and matchings. Journal of Computer and System Sciences, 18:155-193,1979 Randomzied Algorithm Set 2cλ-μ/2 = ( 1+β)(B+1)q … β< 1 ? 見 lemma 3 可以得到hint

35. lemma3 Let f(N) = Nμ/(2λ) 因為c=C/N>0 可以省略 Randomzied Algorithm

36. Proposition 1 (…lemma1) (…lemma2 ) (…lemma1) (…lemma4) (後面兩項可以省略 因為都小於 1) 因為2λ-1<N/2 , 所以前面兩項相乘 < n Randomzied Algorithm

37. lemma3-review Let f(N) = Nμ/(2λ) 因為c=C/N>0 可以省略 Randomzied Algorithm

38. lemma4 Randomzied Algorithm

39. Lemma4-proof 剩下證 Randomzied Algorithm

40. Claim 2 • If Then • Proof : • Similarly inequation as Proposition 1, set f(n)=0, B = cn => Randomzied Algorithm

41. Claim 3 • If Then • Proof : • Symmetric as Claim 2 Randomzied Algorithm

42. Claim 4 • If Then • Proof : • Similarly inequation as Proposition 1 Randomzied Algorithm

43. Claim 5 • If Then • Proof : • Similarly inequation as Proposition 1 • c -> (1-c) • Same lower bound for μ: Randomzied Algorithm

44. Part 3 Algorithm

45. begin SELECT (X, C) letc = C/n, λ= , such that let select randomly a sample denote by if ( ) then let else let let fori = 1 to n do if ( ) then Line 1 ~ 11

46. let else if ( ) then let else let if ( ) then return SELECT_DET( ); else reexecute procedure SELECT; Line 12~20

47. Step 1: Sampling Original input : n numbers X A sample of 2λ+1 numbers S

48. Step 2 : sort S S

49. if ( ) let else let let Line 5~8 Randomzied Algorithm

50. Partition S using m and M m C M Rl Rr S m M Randomzied Algorithm