Lec2 Algorth

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Lec2 Algorth

  1. 1. Introduction to Algorithms 6.046J/18.401J LECTURE 2 Asymptotic Notation • O-, Ω-, and Θ-notation Recurrences • Substitution method • Iterating the recurrence • Recursion tree • Master method Prof. Erik Demaine September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.1
  2. 2. Asymptotic notation O-notation (upper bounds): We write f(n) = O(g(n)) if there We write f(n) = O(g(n)) if there exist constants c > 0, n00 > 0 such exist constants c > 0, n > 0 such that 0 ≤ f(n) ≤ cg(n) for all n ≥ n00.. that 0 ≤ f(n) ≤ cg(n) for all n ≥ n September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.2
  3. 3. Asymptotic notation O-notation (upper bounds): We write f(n) = O(g(n)) if there We write f(n) = O(g(n)) if there exist constants c > 0, n00 > 0 such exist constants c > 0, n > 0 such that 0 ≤ f(n) ≤ cg(n) for all n ≥ n00.. that 0 ≤ f(n) ≤ cg(n) for all n ≥ n EXAMPLE: 2n2 = O(n3) (c = 1, n0 = 2) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.3
  4. 4. Asymptotic notation O-notation (upper bounds): We write f(n) = O(g(n)) if there We write f(n) = O(g(n)) if there exist constants c > 0, n00 > 0 such exist constants c > 0, n > 0 such that 0 ≤ f(n) ≤ cg(n) for all n ≥ n00.. that 0 ≤ f(n) ≤ cg(n) for all n ≥ n EXAMPLE: 2n2 = O(n3) (c = 1, n0 = 2) functions, not values September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.4
  5. 5. Asymptotic notation O-notation (upper bounds): We write f(n) = O(g(n)) if there We write f(n) = O(g(n)) if there exist constants c > 0, n00 > 0 such exist constants c > 0, n > 0 such that 0 ≤ f(n) ≤ cg(n) for all n ≥ n00.. that 0 ≤ f(n) ≤ cg(n) for all n ≥ n EXAMPLE: 2n2 = O(n3) (c = 1, n0 = 2) funny, “one-way” functions, equality not values September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.5
  6. 6. Set definition of O-notation O(g(n)) = { f(n) :: there exist constants O(g(n)) = { f(n) there exist constants c > 0, n00 > 0 such c > 0, n > 0 such that 0 ≤ f(n) ≤ cg(n) that 0 ≤ f(n) ≤ cg(n) for all n ≥ n00 } for all n ≥ n } September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.6
  7. 7. Set definition of O-notation O(g(n)) = { f(n) :: there exist constants O(g(n)) = { f(n) there exist constants c > 0, n00 > 0 such c > 0, n > 0 such that 0 ≤ f(n) ≤ cg(n) that 0 ≤ f(n) ≤ cg(n) for all n ≥ n00 } for all n ≥ n } EXAMPLE: 2n2 ∈ O(n3) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.7
  8. 8. Set definition of O-notation O(g(n)) = { f(n) :: there exist constants O(g(n)) = { f(n) there exist constants c > 0, n00 > 0 such c > 0, n > 0 such that 0 ≤ f(n) ≤ cg(n) that 0 ≤ f(n) ≤ cg(n) for all n ≥ n00 } for all n ≥ n } EXAMPLE: 2n2 ∈ O(n3) (Logicians: λn.2n2 ∈ O(λn.n3), but it’s convenient to be sloppy, as long as we understand what’s really going on.) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.8
  9. 9. Macro substitution Convention: A set in a formula represents an anonymous function in the set. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.9
  10. 10. Macro substitution Convention: A set in a formula represents an anonymous function in the set. EXAMPLE: f(n) = n3 + O(n2) means f(n) = n3 + h(n) for some h(n) ∈ O(n2) . September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.10
  11. 11. Macro substitution Convention: A set in a formula represents an anonymous function in the set. EXAMPLE: n2 + O(n) = O(n2) means for any f(n) ∈ O(n): n2 + f(n) = h(n) for some h(n) ∈ O(n2) . September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.11
  12. 12. Ω-notation (lower bounds) O-notation is an upper-bound notation. It makes no sense to say f(n) is at least O(n2). September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.12
  13. 13. Ω-notation (lower bounds) O-notation is an upper-bound notation. It makes no sense to say f(n) is at least O(n2). Ω(g(n)) = { f(n) :: there exist constants Ω(g(n)) = { f(n) there exist constants c > 0, n00 > 0 such c > 0, n > 0 such that 0 ≤ cg(n) ≤ f(n) that 0 ≤ cg(n) ≤ f(n) for all n ≥ n00 } for all n ≥ n } September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.13
  14. 14. Ω-notation (lower bounds) O-notation is an upper-bound notation. It makes no sense to say f(n) is at least O(n2). Ω(g(n)) = { f(n) :: there exist constants Ω(g(n)) = { f(n) there exist constants c > 0, n00 > 0 such c > 0, n > 0 such that 0 ≤ cg(n) ≤ f(n) that 0 ≤ cg(n) ≤ f(n) for all n ≥ n00 } for all n ≥ n } EXAMPLE: n = Ω(lg n) (c = 1, n0 = 16) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.14
  15. 15. Θ-notation (tight bounds) Θ(g(n)) = O(g(n)) ∩ Ω(g(n)) Θ(g(n)) = O(g(n)) ∩ Ω(g(n)) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.15
  16. 16. Θ-notation (tight bounds) Θ(g(n)) = O(g(n)) ∩ Ω(g(n)) Θ(g(n)) = O(g(n)) ∩ Ω(g(n)) EXAMPLE: 1 2 n − 2 n = Θ( n ) 2 2 September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.16
  17. 17. ο-notation and ω-notation O-notation and Ω-notation are like ≤ and ≥. o-notation and ω-notation are like < and >. ο(g(n)) = { f(n) :: for any constant c > 0, ο(g(n)) = { f(n) for any constant c > 0, there is a constant n00 > 0 there is a constant n > 0 such that 0 ≤ f(n) < cg(n) such that 0 ≤ f(n) < cg(n) for all n ≥ n00 } for all n ≥ n } EXAMPLE: 2n2 = o(n3) (n0 = 2/c) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.17
  18. 18. ο-notation and ω-notation O-notation and Ω-notation are like ≤ and ≥. o-notation and ω-notation are like < and >. ω(g(n)) = { f(n) :: for any constant c > 0, ω(g(n)) = { f(n) for any constant c > 0, there is a constant n00 > 0 there is a constant n > 0 such that 0 ≤ cg(n) < f(n) such that 0 ≤ cg(n) < f(n) for all n ≥ n00 } for all n ≥ n } EXAMPLE: n = ω (lg n) (n0 = 1+1/c) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.18
  19. 19. Solving recurrences • The analysis of merge sort from Lecture 1 required us to solve a recurrence. • Recurrences are like solving integrals, differential equations, etc. o Learn a few tricks. • Lecture 3: Applications of recurrences to divide-and-conquer algorithms. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.19
  20. 20. Substitution method The most general method: 1. Guess the form of the solution. 2. Verify by induction. 3. Solve for constants. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.20
  21. 21. Substitution method The most general method: 1. Guess the form of the solution. 2. Verify by induction. 3. Solve for constants. EXAMPLE: T(n) = 4T(n/2) + n • [Assume that T(1) = Θ(1).] • Guess O(n3) . (Prove O and Ω separately.) • Assume that T(k) ≤ ck3 for k < n . • Prove T(n) ≤ cn3 by induction. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.21
  22. 22. Example of substitution T (n) = 4T (n / 2) + n ≤ 4c ( n / 2 ) 3 + n = ( c / 2) n 3 + n = cn3 − ((c / 2)n3 − n) desired – residual ≤ cn3 desired whenever (c/2)n3 – n ≥ 0, for example, if c ≥ 2 and n ≥ 1. residual September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.22
  23. 23. Example (continued) • We must also handle the initial conditions, that is, ground the induction with base cases. • Base: T(n) = Θ(1) for all n < n0, where n0 is a suitable constant. • For 1 ≤ n < n0, we have “Θ(1)” ≤ cn3, if we pick c big enough. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.23
  24. 24. Example (continued) • We must also handle the initial conditions, that is, ground the induction with base cases. • Base: T(n) = Θ(1) for all n < n0, where n0 is a suitable constant. • For 1 ≤ n < n0, we have “Θ(1)” ≤ cn3, if we pick c big enough. This bound is not tight! September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.24
  25. 25. A tighter upper bound? We shall prove that T(n) = O(n2). September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.25
  26. 26. A tighter upper bound? We shall prove that T(n) = O(n2). Assume that T(k) ≤ ck2 for k < n: T (n) = 4T (n / 2) + n ≤ 4 c ( n / 2) 2 + n = cn 2 + n = O(n 2 ) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.26
  27. 27. A tighter upper bound? We shall prove that T(n) = O(n2). Assume that T(k) ≤ ck2 for k < n: T (n) = 4T (n / 2) + n ≤ 4c ( n / 2) 2 + n = cn 2 + n = O(n 2 ) Wrong! We must prove the I.H. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.27
  28. 28. A tighter upper bound? We shall prove that T(n) = O(n2). Assume that T(k) ≤ ck2 for k < n: T (n) = 4T (n / 2) + n ≤ 4c ( n / 2) 2 + n = cn 2 + n = O(n 2 ) Wrong! We must prove the I.H. = cn 2 − (− n) [ desired – residual ] ≤ cn 2 for no choice of c > 0. Lose! September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.28
  29. 29. A tighter upper bound! IDEA: Strengthen the inductive hypothesis. • Subtract a low-order term. Inductive hypothesis: T(k) ≤ c1k2 – c2k for k < n. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.29
  30. 30. A tighter upper bound! IDEA: Strengthen the inductive hypothesis. • Subtract a low-order term. Inductive hypothesis: T(k) ≤ c1k2 – c2k for k < n. T(n) = 4T(n/2) + n = 4(c1(n/2)2 – c2(n/2)) + n = c1n2 – 2c2n + n = c1n2 – c2n – (c2n – n) ≤ c1n2 – c2n if c2 ≥ 1. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.30
  31. 31. A tighter upper bound! IDEA: Strengthen the inductive hypothesis. • Subtract a low-order term. Inductive hypothesis: T(k) ≤ c1k2 – c2k for k < n. T(n) = 4T(n/2) + n = 4(c1(n/2)2 – c2(n/2)) + n = c1n2 – 2c2n + n = c1n2 – c2n – (c2n – n) ≤ c1n2 – c2n if c2 ≥ 1. Pick c1 big enough to handle the initial conditions. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.31
  32. 32. Recursion-tree method • A recursion tree models the costs (time) of a recursive execution of an algorithm. • The recursion-tree method can be unreliable, just like any method that uses ellipses (…). • The recursion-tree method promotes intuition, however. • The recursion tree method is good for generating guesses for the substitution method. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.32
  33. 33. Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.33
  34. 34. Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: T(n) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.34
  35. 35. Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 T(n/4) T(n/2) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.35
  36. 36. Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 (n/4)2 (n/2)2 T(n/16) T(n/8) T(n/8) T(n/4) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.36
  37. 37. Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 (n/4)2 (n/2)2 (n/16)2 (n/8)2 (n/8)2 (n/4)2 … Θ(1) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.37
  38. 38. Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 n2 (n/4)2 (n/2)2 (n/16)2 (n/8)2 (n/8)2 (n/4)2 … Θ(1) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.38
  39. 39. Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 n2 5 n2 (n/4)2 (n/2)2 16 (n/16)2 (n/8)2 (n/8)2 (n/4)2 … Θ(1) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.39
  40. 40. Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 n2 5 n2 (n/4)2 (n/2)2 16 25 n 2 (n/16)2 (n/8)2 (n/8)2 (n/4)2 256 … … Θ(1) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.40
  41. 41. Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 n2 5 n2 (n/4)2 (n/2)2 16 25 n 2 (n/16)2 (n/8)2 (n/8)2 (n/4)2 256 … … Θ(1) Total = n 2 ( 5 + 5 2 1 + 16 16 ( ) +( ) +L 5 3 16 ) = Θ(n2) geometric series September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.41
  42. 42. The master method The master method applies to recurrences of the form T(n) = a T(n/b) + f (n) , where a ≥ 1, b > 1, and f is asymptotically positive. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.42
  43. 43. Three common cases Compare f (n) with nlogba: 1. f (n) = O(nlogba – ε) for some constant ε > 0. • f (n) grows polynomially slower than nlogba (by an nε factor). Solution: T(n) = Θ(nlogba) . September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.43
  44. 44. Three common cases Compare f (n) with nlogba: 1. f (n) = O(nlogba – ε) for some constant ε > 0. • f (n) grows polynomially slower than nlogba (by an nε factor). Solution: T(n) = Θ(nlogba) . 2. f (n) = Θ(nlogba lgkn) for some constant k ≥ 0. • f (n) and nlogba grow at similar rates. Solution: T(n) = Θ(nlogba lgk+1n) . September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.44
  45. 45. Three common cases (cont.) Compare f (n) with nlogba: 3. f (n) = Ω(nlogba + ε) for some constant ε > 0. • f (n) grows polynomially faster than nlogba (by an nε factor), and f (n) satisfies the regularity condition that a f (n/b) ≤ c f (n) for some constant c < 1. Solution: T(n) = Θ( f (n)) . September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.45
  46. 46. Examples EX. T(n) = 4T(n/2) + n a = 4, b = 2 ⇒ nlogba = n2; f (n) = n. CASE 1: f (n) = O(n2 – ε) for ε = 1. ∴ T(n) = Θ(n2). September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.46
  47. 47. Examples EX. T(n) = 4T(n/2) + n a = 4, b = 2 ⇒ nlogba = n2; f (n) = n. CASE 1: f (n) = O(n2 – ε) for ε = 1. ∴ T(n) = Θ(n2). EX. T(n) = 4T(n/2) + n2 a = 4, b = 2 ⇒ nlogba = n2; f (n) = n2. CASE 2: f (n) = Θ(n2lg0n), that is, k = 0. ∴ T(n) = Θ(n2lg n). September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.47
  48. 48. Examples EX. T(n) = 4T(n/2) + n3 a = 4, b = 2 ⇒ nlogba = n2; f (n) = n3. CASE 3: f (n) = Ω(n2 + ε) for ε = 1 and 4(n/2)3 ≤ cn3 (reg. cond.) for c = 1/2. ∴ T(n) = Θ(n3). September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.48
  49. 49. Examples EX. T(n) = 4T(n/2) + n3 a = 4, b = 2 ⇒ nlogba = n2; f (n) = n3. CASE 3: f (n) = Ω(n2 + ε) for ε = 1 and 4(n/2)3 ≤ cn3 (reg. cond.) for c = 1/2. ∴ T(n) = Θ(n3). EX. T(n) = 4T(n/2) + n2/lg n a = 4, b = 2 ⇒ nlogba = n2; f (n) = n2/lg n. Master method does not apply. In particular, for every constant ε > 0, we have nε = ω(lg n). September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.49
  50. 50. Idea of master theorem Recursion tree: a f (n) f (n/b) f (n/b) … f (n/b) a f (n/b2) f (n/b2) … f (n/b2) … Τ (1) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.50
  51. 51. Idea of master theorem Recursion tree: f (n) f (n) a f (n/b) f (n/b) … f (n/b) a f (n/b) a f (n/b2) f (n/b2) … f (n/b2) a2 f (n/b2) … … Τ (1) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.51
  52. 52. Idea of master theorem Recursion tree: f (n) f (n) a f (n/b) f (n/b) … f (n/b) a f (n/b) h = logbn a f (n/b2) f (n/b2) … f (n/b2) a2 f (n/b2) … … Τ (1) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.52
  53. 53. Idea of master theorem Recursion tree: f (n) f (n) a f (n/b) f (n/b) … f (n/b) a f (n/b) h = logbn a f (n/b2) f (n/b2) … f (n/b2) a2 f (n/b2) #leaves = ah … … = alogbn Τ (1) nlogbaΤ (1) = nlogba September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.53
  54. 54. Idea of master theorem Recursion tree: f (n) f (n) a f (n/b) f (n/b) … f (n/b) a f (n/b) h = logbn a f (n/b2) f (n/b2) … f (n/b2) a2 f (n/b2) … CASE 1: The weight increases … CASE 1: The weight increases geometrically from the root to the geometrically from the root to the Τ (1) leaves. The leaves hold aaconstant leaves. The leaves hold constant nlogbaΤ (1) fraction of the total weight. fraction of the total weight. Θ(nlogba) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.54
  55. 55. Idea of master theorem Recursion tree: f (n) f (n) a f (n/b) f (n/b) … f (n/b) a f (n/b) h = logbn a f (n/b2) f (n/b2) … f (n/b2) a2 f (n/b2) … … CASE 2: (k = 0) The weight CASE 2: (k = 0) The weight Τ (1) is approximately the same on is approximately the same on nlogbaΤ (1) each of the logbbn levels. each of the log n levels. Θ(nlogbalg n) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.55
  56. 56. Idea of master theorem Recursion tree: f (n) f (n) a f (n/b) f (n/b) … f (n/b) a f (n/b) h = logbn a f (n/b2) f (n/b2) … f (n/b2) a2 f (n/b2) … CASE 3: The weight decreases … CASE 3: The weight decreases geometrically from the root to the geometrically from the root to the Τ (1) leaves. The root holds aaconstant leaves. The root holds constant nlogbaΤ (1) fraction of the total weight. fraction of the total weight. Θ( f (n)) September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.56
  57. 57. Appendix: geometric series 2 n 1 − x n +1 1+ x + x +L+ x = for x ≠ 1 1− x 21 1+ x + x +L = for |x| < 1 1− x Return to last slide viewed. September 12, 2005 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson L2.57

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