2. COSC 3101Winter, 2004 COSC 3101 Design and Analysis of Algorithms Lecture 2. Relevant Mathematics: –The time complexity of an algorithm –Adding made easy –Recurrence relations

3. COSC 3101Winter, 2004 Question from last class: recurrence relations T(1) = k for some constant k T(n) = 4 T(n/2) + k’ n + k’’ for some constants k’ and k’’ MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d RETURN MULT(a,c) 2 n + (MULT(a,d) + MULT(b,c)) 2 n/2 + MULT(b,d) Question from last class: how does Mult know the length of X and Y (i.e. n)? Answer: data type of X, Y, a, b, c, d must include length field.

4. COSC 3101Winter, 2004 The Time Complexity of an Algorithm Specifies how the running time depends on the size of the input.

5. COSC 3101Winter, 2004 Purpose?

6. COSC 3101Winter, 2004 Purpose To estimate how long a program will run. To estimate the largest input that can reasonably be given to the program. To compare the efficiency of different algorithms. To help focus on the parts of code that are executed the largest number of times. To choose an algorithm for an application.

7. COSC 3101Winter, 2004 Time Complexity Is a Function Specifies how the running time depends on the size of the input. A function mapping “size” of input  “time” T(n) executed.

8. COSC 3101Winter, 2004 Definition of Time?

9. COSC 3101Winter, 2004 Definition of Time # of seconds (machine dependent). # lines of code executed. # of times a specific operation is performed (e.g., addition). Which?

10. COSC 3101Winter, 2004 Definition of Time # of seconds (machine dependent). # lines of code executed. # of times a specific operation is performed (e.g., addition). These are all reasonable definitions of time, because they are within a constant factor of each other.

11. COSC 3101Winter, 2004 Size of Input Instance? 83920 Suppose input to the algorithm is the integer 83920.

12. COSC 3101Winter, 2004 Size of Input Instance Size of paper: # of bits: # of digits: Value: n = 2 in 2 n = 17 bits n = 5 digits n = 83920 2’’ 83920 5 1’’ Which are reasonable?

13. COSC 3101Winter, 2004 Size of Input Instance Intuitive 2’’ 83920 5 1’’ Size of paper: # of bits: # of digits: Value: n = 2 in 2 n = 17 bits n = 5 digits n = 83920

14. COSC 3101Winter, 2004 Size of Input Instance Size of paper: # of bits: # of digits: Value: n = 2 in 2 n = 17 bits n = 5 digits n = 83920 Intuitive Formal 2’’ 83920 5 1’’

15. COSC 3101Winter, 2004 Size of Input Instance Size of paper: # of bits: # of digits: Value: n = 2 in 2 n = 17 bits n = 5 digits n = 83920 Intuitive Formal Reasonable # of bits = 3.32 * # of digits 2’’ 83920 5 1’’

16. COSC 3101Winter, 2004 Size of Input Instance Size of paper # of bits # of digits Value - n = 2 in 2 - n = 17 bits - n = 5 digits - n = 83920 Intuitive Formal Reasonable Unreasonable # of bits = log 2 (Value) Value = 2 # of bits 2’’ 83920 5 1’’

17. COSC 3101Winter, 2004 Two Example Algorithms Sum N array entries: A(1) + A(2) + A(3) + … Factor value N: Input N=5917 & Output N=97*61. Algorithm: N/2, N/3, N/4, …. ? Time?

18. COSC 3101Winter, 2004 Two Example Algorithms Sum N array entries: A(1) + A(2) + A(3) + … Factor value N: Input N=5917 & Output N=97*61. Algorithm N/2, N/3, N/4, …. ? Time = N

19. COSC 3101Winter, 2004 Two Example Algorithms Sum N array entries: A(1) + A(2) + A(3) + … Factor value N: Input N=5917 & Output N=97*61. Algorithm N/2, N/3, N/4, …. ? Is this reasonable? Time = N

20. COSC 3101Winter, 2004 Two Example Algorithms Sum N array entries: A(1) + A(2) + A(3) + … Factor value N: Input N=5917 & Output N=97*61. Algorithm N/2, N/3, N/4, …. ? No! Time = N One is considered fast and the other slow!

21. COSC 3101Winter, 2004 Two Example Algorithms Sum N array entries: A(1) + A(2) + A(3) + … Factor value N: Input N=5917 & Output N=97*61. Algorithm N/2, N/3, N/4, …. ? Size of Input Instance?

22. COSC 3101Winter, 2004 Two Example Algorithms Sum N array entries: A(1) + A(2) + A(3) + … Factor value N: Input N=5917 & Output N=97*61. Algorithm N/2, N/3, N/4, …. ? size n = N size n = log N

23. COSC 3101Winter, 2004 Two Example Algorithms Sum N array entries: A(1) + A(2) + A(3) + … Factor value N: Input N=5917 & Output N=97*61. Algorithm N/2, N/3, N/4, …. ? Time as function of input size? size n = N size n = log N

24. COSC 3101Winter, 2004 Two Example Algorithms Sum N array entries: A(1) + A(2) + A(3) + … Factor value N: Input N=5917 & Output N=97*61. Algorithm N/2, N/3, N/4, …. ? Time = N = n Time = N = 2 n size n = N size n = log N

25. COSC 3101Winter, 2004 Two Example Algorithms Sum N array entries: A(1) + A(2) + A(3) + … Factor value N: Input N=5917 & Output N=97*61. Algorithm N/2, N/3, N/4, …. ? Linear vs Exponential Time! Time = N = n Time = N = 2 n size n = N size n = log N

26. COSC 3101Winter, 2004 Size of Input Instance? 14,23,25,30,31,52,62,79,88,98

27. COSC 3101Winter, 2004 Size of Input Instance # of elements: n = 10 elements 14,23,25,30,31,52,62,79,88,98 10

28. COSC 3101Winter, 2004 Size of Input Instance # of elements: n = 10 elements 14,23,25,30,31,52,62,79,88,98 10 Is this reasonable?

29. COSC 3101Winter, 2004 Size of Input Instance # of elements: n = 10 elements ~ Reasonable 14,23,25,30,31,52,62,79,88,98 10 If each element has size c # of bits = c * # of elements

30. COSC 3101Winter, 2004 Size of Input Instance # of elements: n = 10 elements Reasonable 6, 3, 8, 1, 2, 10, 4, 9, 7, 5 10 If each element is in [1..n] each element has size log n # of bits = n log n ≈ n

31. COSC 3101Winter, 2004 Time Complexity Is a Function Specifies how the running time depends on the size of the input. A function mapping: –“size” of input  “time” T(n) executed. Or more precisely: –# of bits n needed to represent the input  # of operations T(n) executed.

32. COSC 3101Winter, 2004 Which Input of size n? There are 2 n inputs of size n. Which do we consider for the time T(n)?

33. COSC 3101Winter, 2004 Which Input of size n? Provides an upper bound for all possible inputs. Easiest to compute Worst Case For what distribution?Average Case But what is typical?Typical Input

34. COSC 3101Winter, 2004 Time Complexity of an Algorithm O(n 2 ): Prove that for every input of size n, the algorithm takes no more than cn 2 time. Ω(n 2 ): Find one input of size n, for which the algorithm takes at least this much time. θ (n 2 ): Do both. The time complexity of an algorithm is the largest time required on any input of size n.

35. COSC 3101Winter, 2004 What is the height of tallest person in the class? Bigger than this? Need to look at only one person Need to look at every person Smaller than this?

36. COSC 3101Winter, 2004 Time Complexity of a Problem O(n 2 ): Provide an algorithm that solves the problem in no more than this time. Ω(n 2 ): Prove that no algorithm can solve it faster. θ (n 2 ): Do both. The time complexity of a problem is the time complexity of the fastest algorithm that solves the problem.

37. COSC 3101Winter, 2004 Adding: The Classic Techniques Evaluating ∑ i=1 f(i). n

38. COSC 3101Winter, 2004 ∑ i=1..n i = 1 + 2 + 3 +... + n = ? Arithmetic Sum

39. COSC 3101Winter, 2004 1+2+3+...+ n-1+n=S n+n-1+n-2+... +2+1=S (n+1) + (n+1) + (n+1) +... + (n+1) + (n+1) = 2S n (n+1) = 2S

40. COSC 3101Winter, 2004 1+2+3+...+ n-1+n=S n+n-1+n-2+... +2+1=S (n+1) + (n+1) + (n+1) +... + (n+1) + (n+1) = 2S n (n+1) = 2S

41. COSC 3101Winter, 2004 1+2+3+...+ n-1+n=S n+n-1+n-2+... +2+1=S (n+1) + (n+1) + (n+1) +... + (n+1) + (n+1) = 2S n (n+1) = 2S

42. COSC 3101Winter, 2004 1+2+3+...+ n-1+n=S n+n-1+n-2+... +2+1=S (n+1) + (n+1) + (n+1) +... + (n+1) + (n+1) = 2S n (n+1) = 2S

43. COSC 3101Winter, 2004 Let’s restate this argument using a geometric representation Algebraic argument

44. COSC 3101Winter, 2004 1 2........ n = number of white dots.

45. COSC 3101Winter, 2004 1 2........ n = number of white dots = number of yellow dots n....... 2 1

46. COSC 3101Winter, 2004 n+1 n+1 n+1 n+1 n+1 = number of white dots = number of yellow dots n n n n n n There are n(n+1) dots in the grid

47. COSC 3101Winter, 2004 n+1 n+1 n+1 n+1 n+1 = number of white dots = number of yellow dots n n n n n n Note =  (# of terms · last term))

48. COSC 3101Winter, 2004 ∑ i=1..n i = 1 + 2 + 3 +... + n =  (# of terms · last term) Arithmetic Sum True whenever terms increase slowly

49. COSC 3101Winter, 2004 ∑ i=0..n 2 i = 1 + 2 + 4 + 8 +... + 2 n = ? Geometric Sum

50. COSC 3101Winter, 2004 Geometric Sum

51. COSC 3101Winter, 2004 ∑ i=0..n 2 i = 1 + 2 + 4 + 8 +... + 2 n = 2 · last term - 1 Geometric Sum

52. COSC 3101Winter, 2004 ∑ i=0..n r i = r 0 + r 1 + r 2 +... + r n = ? Geometric Sum

53. COSC 3101Winter, 2004 S Srrrr...rr S 1r S r1 r1 23nn1 n1 n1  1rrr... 23 r n          Geometric Sum

54. COSC 3101Winter, 2004 ∑ i=0..n r i r1 r1 n1     Geometric Sum When r>1?

55. COSC 3101Winter, 2004 ∑ i=0..n r i r1 r1 n1     Geometric Sum θ(rn)θ(rn) Biggest Term When r>1

56. COSC 3101Winter, 2004 ∑ i=0..n r i = r 0 + r 1 + r 2 +... + r n =  (biggest term) Geometric Increasing True whenever terms increase quickly

57. COSC 3101Winter, 2004 ∑ i=0..n r i  Geometric Sum When r<1? 1 r n1   

58. COSC 3101Winter, 2004 ∑ i=0..n r i  Geometric Sum 1 r n1     θ(1) When r<1 Biggest Term

59. COSC 3101Winter, 2004 ∑ i=0..n r i = r 0 + r 1 + r 2 +... + r n =  (1) Bounded Tail True whenever terms decrease quickly

60. COSC 3101Winter, 2004 ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + …+ 1 / n = ? Harmonic Sum

61. COSC 3101Winter, 2004 f(i) = 1 ∑ i=1..n f(i) = n n Sum of Shrinking Function

62. COSC 3101Winter, 2004 f(i) = ? ∑ i=1..n f(i) = n 1/2 n Sum of Shrinking Function

63. COSC 3101Winter, 2004 f(i) = 1/2 i ∑ i=1..n f(i) = 2  Sum of Shrinking Function

64. COSC 3101Winter, 2004 f(i) = 1/i ∑ i=1..n f(i) = ? n Sum of Shrinking Function

65. COSC 3101Winter, 2004 Harmonic Sum NB: Error in Jeff’s notes, p.30, bottom:

66. COSC 3101Winter, 2004 ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + …+ 1 / n =  (log(n)) Harmonic Sum

67. COSC 3101Winter, 2004 Approximating Sum by Integrating The area under the curve approximates the sum ∑ i=1..n f(i) ≈ ∫ x=1..n f(x) dx

68. COSC 3101Winter, 2004 Approximating Sums by Integrating: Arithmetic Sums

69. COSC 3101Winter, 2004 Approximating Sums by Integrating: Geometric Sums

70. COSC 3101Winter, 2004 Harmonic Sum Approximating Sum by Integrating

71. COSC 3101Winter, 2004 Approximating Sum by Integrating Problem: Integrating may be hard too.

72. COSC 3101Winter, 2004 Adding Made Easy We will now classify (most) functions f(i) into four classes: –Geometric Like –Arithmetic Like –Harmonic –Bounded Tail For each class, we will give an easy rule for approximating its sum θ( ∑ i=1..n f(i) )

73. COSC 3101Winter, 2004 Classifying Animals Vertebrates Birds Mammals Reptiles Fish Dogs Giraffe

74. COSC 3101Winter, 2004 Classifying Functions Functions Poly. Exp. 2 θ(n)  ∑ i=1..n f(i) = θ(f(n)) 8 · 2 n / n 100 + n 3 θ(2 n / n 100 ) n θ(1) 2 θ(n) θ(2n)θ(2n) f(n) = 8 · 2 n / n 100 + n 3 θ(2 n / n 100 )  ∑ i=1..n f(i) = θ(2 n / n 100 ) 2 0.5n << 2 n 8 · 2 n / n 100 + n 3 Significant Less significant Irrelevant

75. COSC 3101Winter, 2004 Adding Made Easy 11 1 2 3 4 Four Classes of Functions

76. COSC 3101Winter, 2004 Adding Made Easy

77. COSC 3101Winter, 2004 If the terms f(i) grow sufficiently quickly, then the sum will be dominated by the largest term. f(n)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like: Classic example: ∑ i=1..n 2 i = 2 n+1 -1 ≈ 2 f(n)

78. COSC 3101Winter, 2004 If the terms f(i) grow sufficiently quickly, then the sum will be dominated by the largest term. f(n)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like: For which functions f(i) is this true? How fast and how slow can it grow?

79. COSC 3101Winter, 2004 r1 r1 n1      θ( ) r n Last Term when r>1. ∑ i=1..n r i 1rrr... 23 r n    θ(f(n)) ∑ i=1..n (1.0001) i ≈ 10,000 (1.0001) n = 10,000 f(n) Lower Extreme: Upper Extreme: ∑ i=1..n (1000) i ≈ 1.001(1000) n = 1.001 f(n) When f(n) = r n = 2 θ(n) Geometric Like:

80. COSC 3101Winter, 2004 Because the constant is sooo big, the statement is just barely true. ∑ i=1..n (1.0001) i ≈ 10,000 (1.0001) n = 10,000 f(n) Lower Extreme: Upper Extreme: ∑ i=1..n (1000) i ≈ 1.001(1000) n = 1.001 f(n) f(n)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

81. COSC 3101Winter, 2004 ∑ i=1..n (1.0001) i ≈ 10,000 (1.0001) n = 10,000 f(n) Lower Extreme: Upper Extreme: ∑ i=1..n (1000) i ≈ 1.001(1000) n = 1.001 f(n) Even bigger? f(n)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

82. COSC 3101Winter, 2004 2n2n 2i2i ∑ i=1..n 2 2 ≈ 2 2 = 1f(n) No Upper Extreme: Even bigger! ∑ i=1..n (1.0001) i ≈ 10,000 (1.0001) n = 10,000 f(n) Lower Extreme: f(n)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

83. COSC 3101Winter, 2004 2n2n 2i2i ∑ i=1..n 2 2 ≈ 2 2 = 1f(n) No Upper Extreme: Functions in between? ∑ i=1..n (1.0001) i ≈ 10,000 (1.0001) n = 10,000 f(n) Lower Extreme: f(n)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

84. COSC 3101Winter, 2004 f(n)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

85. COSC 3101Winter, 2004 f(n)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like: f(n) = 2 n 8·8· n 100 ∑ i=1..n f(i) f(n)  c · f(n) f(i)  · f(n) f(i)  · f(i +1 ) EasyHard (i +1 ) 100 i 100 2 =  (1 + 1 / i ) 100 1 2 1.9 = 8 · 2 (i+1) (i +1 ) 100 f(i +1 ) f(i) 8·2i8·2i i 100 =  1/.51  · f(i +1 ) f(i)

86. COSC 3101Winter, 2004

87. COSC 3101Winter, 2004 In general, if f(n) = c b an n d log e (n) c  ? a  ? b  ? d  ? e  ? f(n) = Ω, θ ? Then ∑ i=1..n f(i) = θ(f(n)).  2 Ω(n) > 0  0  1 (- ,  ) f(n)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

88. COSC 3101Winter, 2004 f(n)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like: Examples:

89. COSC 3101Winter, 2004 Do All functions in 2 Ω(n) have this property? Maybe not. f(n)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

90. COSC 3101Winter, 2004 Functions that oscillate with exponentially increasing amplitude do not have this property. f(n)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

91. COSC 3101Winter, 2004 Functions expressed with +, -, , , exp, log do not oscillate continually. They are well behaved for sufficiently large n. These do have this property. f(n)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

92. COSC 3101Winter, 2004 Adding Made Easy Done

93. COSC 3101Winter, 2004 Adding Made Easy

94. COSC 3101Winter, 2004 If the terms f(i) are increasing or decreasing relatively slowly, then the sum is roughly the number of terms, n, times the final value. f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

95. COSC 3101Winter, 2004 If the terms f(i) are increasing or decreasing relatively slowly, then the sum is roughly the number of terms, n, times the final value. Simple example: ∑ i=1..n 1 = n · 1 f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

96. COSC 3101Winter, 2004 If the terms f(i) are increasing or decreasing relatively slowly, then the sum is roughly the number of terms, n, times the final value. Is the statement true for this function? ∑ i=1..n i = 1 + 2 + 3 +... + n f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

97. COSC 3101Winter, 2004 Half the terms are roughly the same (within a multiplicative constant) ∑ i=1..n i = 1 +... + n/2 +... + n f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

98. COSC 3101Winter, 2004 But half the terms are roughly the same and the sum is roughly the number terms, n, times this value ∑ i=1..n i = 1 +... + n/2 +... + n ∑ i=1..n i = θ(n · n) f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

99. COSC 3101Winter, 2004 Is the statement true for this function? ∑ i=1..n i 2 = 1 2 + 2 2 + 3 2 +... + n 2 f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

100. COSC 3101Winter, 2004 Again half the terms are roughly the same. ∑ i=1..n i = 1 2 +... + (n/2) 2 +... + n 2 1 / 4 n 2 f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

101. COSC 3101Winter, 2004 Again half the terms are roughly the same. ∑ i=1..n i = 1 2 +... + (n/2) 2 +... + n 2 1 / 4 n 2 ∑ i=1..n i 2 = θ(n · n 2 ) f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

102. COSC 3101Winter, 2004 ∑ i=1..n f(i) ≈ area under curve f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like: Consider f(i) = n θ(1) (i.e. f(i) increasing)

103. COSC 3101Winter, 2004 area of small square  ∑ i=1..n f(i) ≈ area under curve  area of big square f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

104. COSC 3101Winter, 2004 n / 2 · f( n / 2 ) = area of small square  ∑ i=1..n f(i) ≈ area under curve  area of big square = n · f(n) f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

105. COSC 3101Winter, 2004 θ(n · f(n)) = n / 2 · f( n / 2 ) = area of small square  ∑ i=1..n f(i) ≈ area under curve  area of big square = n · f(n) f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

106. COSC 3101Winter, 2004 θ(n · f(n)) = n / 2 · f( n / 2 ) = area of small square  ∑ i=1..n f(i) ≈ area under curve  area of big square = n · f(n) ? f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

107. COSC 3101Winter, 2004 ∑ i=1..n i 2 = 1 2 + 2 2 + 3 2 +... + n 2 f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like: f(n) = n 2 f( n / 2 ) = θ(f(n)) The key property is = ?

108. COSC 3101Winter, 2004 The key property is f( n / 2 ) = (n/2) 2 = 1 / 4 n 2 = θ( n 2 ) = θ(f(n)) ∑ i=1..n i 2 = 1 2 + 2 2 + 3 2 +... + n 2 f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like: f(n) = n 2  = θ(n·f(n)) = θ(n · n 2 )

109. COSC 3101Winter, 2004 ∑ i=1..n f(i) = ? f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like: f(n) = n θ(1)

110. COSC 3101Winter, 2004 ∑ i=1..n i r = 1 r + 2 r + 3 r +... + n r f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like: f(n) = n θ(1) f( n / 2 ) = θ(f(n)) The key property is

111. COSC 3101Winter, 2004 The key property is f( n / 2 ) = (n/2) r = 1 / 2 r n r = θ( n r ) = θ(f(n)) ∑ i=1..n i r = 1 r + 2 r + 3 r +... + n r f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:  = θ(n·f(n)) = θ(n · n r ) f(n) = n θ(1)

112. COSC 3101Winter, 2004 ∑ i=1..n 2 i = 2 1 + 2 2 + 2 3 +... + 2 n f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like: f(n) = n θ(1) f( n / 2 ) = θ(f(n)) The key property is

113. COSC 3101Winter, 2004 The key property is ∑ i=1..n 2 i = 2 1 + 2 2 + 2 3 +... + 2 n f( n / 2 ) = 2 (n/2) = θ( 2 n ) = θ(f(n)) f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like: f(n) = n θ(1)  = θ(n·f(n)) = θ(n · 2 n )

114. COSC 3101Winter, 2004 Upper Extreme: All functions in between. f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

115. COSC 3101Winter, 2004 Adding Made Easy Half done

116. COSC 3101Winter, 2004 f(i) = 1 ∑ i=1..n f(i) = n n Sum of Shrinking Function

117. COSC 3101Winter, 2004 f(i) = ? ∑ i=1..n f(i) = n 1/2 n Sum of Shrinking Function 1/i 1/2

118. COSC 3101Winter, 2004 f(i) = 1/i ∑ i=1..n f(i) = log n n Sum of Shrinking Function

119. COSC 3101Winter, 2004 f(i) = 1/2 i ∑ i=1..n f(i) = 2 Sum of Shrinking Function 

120. COSC 3101Winter, 2004 If most of the terms f(i) have roughly the same value, then the sum is roughly the number of terms, n, times this value. Does the statement hold for functions f(i) that shrink? f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

121. COSC 3101Winter, 2004 If most of the terms f(i) have roughly the same value, then the sum is roughly the number of terms, n, times this value. ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + … + 1 / n Does the statement hold for the Harmonic Sum? f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

122. COSC 3101Winter, 2004 Does the statement hold for the Harmonic Sum? ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + … + 1 / n ∑ i=1..n f(i) = ? θ(n · f(n)) f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

123. COSC 3101Winter, 2004 Does the statement hold for the Harmonic Sum? ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + … + 1 / n ∑ i=1..n f(i) θ(n · f(n)) = ∑ i=1..n 1 / i = ? f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

124. COSC 3101Winter, 2004 Does the statement hold for the Harmonic Sum? ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + … + 1 / n ∑ i=1..n f(i) θ(n · f(n)) = ∑ i=1..n 1 / i =θ(log n) f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

125. COSC 3101Winter, 2004 Does the statement hold for the Harmonic Sum? ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + … + 1 / n ∑ i=1..n f(i) = θ(n · f(n)) = ∑ i=1..n 1 / i =θ(log n) ? f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

126. COSC 3101Winter, 2004 Does the statement hold for the Harmonic Sum? ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + … + 1 / n ∑ i=1..n f(i) = θ(n · f(n)) = ∑ i=1..n 1 / i =θ(log n) θ(1) = θ(n · 1 / n ) f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

127. COSC 3101Winter, 2004 Does the statement hold for the Harmonic Sum? ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + … + 1 / n ∑ i=1..n f(i) = θ(n · f(n)) = ∑ i=1..n 1 / i =θ(log n) θ(1) = θ(n · 1 / n ) ≠ No the statement does not hold! f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

128. COSC 3101Winter, 2004 Adding Made Easy not included

129. COSC 3101Winter, 2004 Does the statement hold for the almost Harmonic Sum? ∑ i=1..n 1 / i 0.9999 Shrinks slightly slower than harmonic. f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

130. COSC 3101Winter, 2004 Does the statement hold for the almost Harmonic Sum? ∑ i=1..n 1 / i 0.9999 ∑ i=1..n f(i) = θ(n · f(n)) = ∑ i=1..n 1 / i 0.9999 = ? f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

131. COSC 3101Winter, 2004 Does the statement hold for the almost Harmonic Sum? ∑ i=1..n 1 / i 0.9999 ∑ i=1..n f(i) = θ(n · f(n)) = ∑ i=1..n 1 / i 0.9999 =θ(n 0.0001 ) f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

132. COSC 3101Winter, 2004 Does the statement hold for the almost Harmonic Sum? ∑ i=1..n 1 / i 0.9999 ∑ i=1..n f(i) = θ(n · f(n)) = ∑ i=1..n 1 / i 0.9999 =θ(n 0.0001 ) ? f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

133. COSC 3101Winter, 2004 Does the statement hold for the almost Harmonic Sum? ∑ i=1..n 1 / i 0.9999 ∑ i=1..n f(i) = θ(n · f(n)) = ∑ i=1..n 1 / i 0.9999 =θ(n 0.0001 ) θ(n 0.0001 ) = θ(n · 1 / n 0.9999 ) f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

134. COSC 3101Winter, 2004 Does the statement hold for the almost Harmonic Sum? ∑ i=1..n 1 / i 0.9999 ∑ i=1..n f(i) = θ(n · f(n)) = ∑ i=1..n 1 / i 0.9999 =θ(n 0.0001 ) θ(n 0.0001 ) = θ(n · 1 / n 0.9999 ) = The statement does hold! f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

135. COSC 3101Winter, 2004 ∑ i=1..n 1 = n · 1 = n · f(n) Intermediate Case: Lower Extreme: ∑ i=1..n 1 / i 0.9999 = θ(n 0.0001 ) = θ(n · f(n)) All functions in between. f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

136. COSC 3101Winter, 2004 ∑ i=1..n 1 = n · 1 = n · f(n) Intermediate Case: Lower Extreme: ∑ i=1..n 1 / i 0.9999 = θ(n 0.0001 ) = θ(n · f(n)) Upper Extreme: ∑ i=1..n i 1000 = 1 / 1001 n 1001 = 1 / 1001 n · f(n) f(n) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

137. COSC 3101Winter, 2004 Arithmetic Like If f(n) = c b an n d log e (n) c  ? a  ? b  ? d  ? e  ? f(n) = Ω, θ ? then ∑ i=1..n f(i) = θ(n · f(n)). n -1+θ(1) > 0  0 or b  1 > -1 (- ,  ) (For +, -, , , exp, log functions f(n)) Conclusion

138. COSC 3101Winter, 2004 Adding Made Easy Done

139. COSC 3101Winter, 2004 Adding Made Easy Harmonic

140. COSC 3101Winter, 2004 Harmonic Sum ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + …+ 1 / n =  (log(n))

141. COSC 3101Winter, 2004 Adding Made Easy

142. COSC 3101Winter, 2004 If the terms f(i) decrease towards zero sufficiently quickly, then the sum will be a constant. The classic example ∑ i=0..n 1 / 2 i = 1 + 1 / 2 + 1 / 4 + 1 / 8 + … = 2. f(n)  n -1-Ω(1)  ∑ i=1..n f(i) = θ(1) Bounded Tail:

143. COSC 3101Winter, 2004 Upper Extreme: ∑ i=1..n 1 / i 1.0001 = θ(1) f(n)  n -1-Ω(1)  ∑ i=1..n f(i) = θ(1) Bounded Tail:

144. COSC 3101Winter, 2004 Upper Extreme: ∑ i=1..n 1 / i 1.0001 = θ(1) No Lower Extreme: 2i2i ∑ i=1..n 2 2 = θ(1). 1 All functions in between. f(n)  n -1-Ω(1)  ∑ i=1..n f(i) = θ(1) Bounded Tail:

145. COSC 3101Winter, 2004 Bounded Tail Conclusion

146. COSC 3101Winter, 2004 Adding Made Easy Done

147. COSC 3101Winter, 2004 Adding Made Easy Missing Functions 1 / nlogn logn / n

148. COSC 3101Winter, 2004 Adding Made Easy Geometric Like: If f(n)  2 Ω(n), then ∑ i=1..n f(i) = θ(f(n)). Arithmetic Like: If f(n) = n θ(1)-1, then ∑ i=1..n f(i) = θ(n · f(n)). Harmonic: If f(n) = 1 / n, then ∑ i=1..n f(i) = log e n + θ(1). Bounded Tail: If f(n)  n -1-Ω(1), then ∑ i=1..n f(i) = θ(1). ( For +, -, , , exp, log functions f(n) ) This may seem confusing, but it is really not. It should help you compute most sums easily.

149. COSC 3101Winter, 2004 Recurrence Relations T(1) = 1 T(n) = a T(n/b) + f(n)

150. COSC 3101Winter, 2004 Recurrence Relations  Time of Recursive Program procedure Eg(int n) if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a Eg(n/b) Recurrence relations arise from the timing of recursive programs. Let T(n) be the # of “Hi”s on an input of “size” n.

151. COSC 3101Winter, 2004 procedure Eg(int n) if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a Eg(n/b) Given size 1, the program outputs T(1)=1 Hi’s. Recurrence Relations  Time of Recursive Program

152. COSC 3101Winter, 2004 procedure Eg(int n) if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a Eg(n/b) Given size n, the stackframe outputs f(n) Hi’s. Recurrence Relations  Time of Recursive Program

153. COSC 3101Winter, 2004 procedure Eg(int n) if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a Eg(n/b) Recursing on a instances of size n/b generates T(n/b) “Hi”s. Recurrence Relations  Time of Recursive Program

154. COSC 3101Winter, 2004 procedure Eg(int n) if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a Eg(n/b) Recursing a times generates a·T(n/b) “Hi”s. Recurrence Relations  Time of Recursive Program

155. COSC 3101Winter, 2004 procedure Eg(int n) if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a Eg(n/b) For a total of T(1) = 1 T(n) = a·T(n/b) + f(n) “Hi”s. Recurrence Relations  Time of Recursive Program

156. COSC 3101Winter, 2004 Solving Technique 1 Guess and Verify Recurrence Relation: T(1) = 1 & T(n) = 4T(n/2) + n Guess: G(n) = 2n 2 – n Verify: Left Hand SideRight Hand Side T(1) = 2(1) 2 – 1 T(n) = 2n 2 – n 1 4T(n/2) + n = 4 [2( n / 2 ) 2 – ( n / 2 )] + n = 2n 2 – n

157. COSC 3101Winter, 2004 Recurrence Relation: T(1) = 1 & T(n) = 4T(n/2) + n Guess: G(n) = an 2 + bn + c Verify: Left Hand SideRight Hand Side T(1) = a+b+c T(n) = an 2 +bn+c 1 4T(n/2) + n = 4 [a ( n / 2 ) 2 + b ( n / 2 ) +c] + n = an 2 +(2b+1)n + 4c Solving Technique 2 Guess Form and Calculate Coefficients

158. COSC 3101Winter, 2004 Recurrence Relation: T(1) = 1 & T(n) = 4T(n/2) + n Guess: G(n) = an 2 + bn + 0 Verify: Left Hand SideRight Hand Side T(1) = a+b+c T(n) = an 2 +bn+c 1 4T(n/2) + n = 4 [a ( n / 2 ) 2 + b ( n / 2 ) +c] + n = an 2 +(2b+1)n + 4c Solving Technique 2 Guess Form and Calculate Coefficients c=4c c=0

159. COSC 3101Winter, 2004 Recurrence Relation: T(1) = 1 & T(n) = 4T(n/2) + n Guess: G(n) = an 2 - 1n + 0 Verify: Left Hand SideRight Hand Side T(1) = a+b+c T(n) = an 2 +bn+c 1 4T(n/2) + n = 4 [a ( n / 2 ) 2 + b ( n / 2 ) +c] + n = an 2 +(2b+1)n + 4c Solving Technique 2 Guess Form and Calculate Coefficients b=2b+1 b=-1

160. COSC 3101Winter, 2004 Recurrence Relation: T(1) = 1 & T(n) = 4T(n/2) + n Guess: G(n) = an 2 - 1n + 0 Verify: Left Hand SideRight Hand Side T(1) = a+b+c T(n) = an 2 +bn+c 1 4T(n/2) + n = 4 [a ( n / 2 ) 2 + b ( n / 2 ) +c] + n = an 2 +(2b+1)n + 4c Solving Technique 2 Guess Form and Calculate Coefficients a=a

161. COSC 3101Winter, 2004 Recurrence Relation: T(1) = 1 & T(n) = 4T(n/2) + n Guess: G(n) = 2n 2 - 1n + 0 Verify: Left Hand SideRight Hand Side T(1) = a+b+c T(n) = an 2 +bn+c 1 4T(n/2) + n = 4 [a ( n / 2 ) 2 + b ( n / 2 ) +c] + n = an 2 +(2b+1)n + 4c Solving Technique 2 Guess Form and Calculate Coefficients a+b+c=1 a-1+0=1 a=2

162. COSC 3101Winter, 2004 Recurrence Relation: T(1) = 1 & T(n) = aT(n/b) + f(n) Solving Technique 3 Approximate Form and Calculate Exponent which is bigger? Guess

163. COSC 3101Winter, 2004 Recurrence Relation: T(1) = 1 & T(n) = aT(n/b) + f(n) Guess: aT(n/b) << f(n) Simplify: T(n)  f(n) Solving Technique 3 Calculate Exponent In this case, the answer is easy. T(n) =  (f(n))

164. COSC 3101Winter, 2004 Recurrence Relation: T(1) = 1 & T(n) = aT(n/b) + f(n) Guess: aT(n/b) >> f(n) Simplify: T(n)  aT(n/b) Solving Technique 3 Calculate Exponent In this case, the answer is harder.

165. COSC 3101Winter, 2004 Recurrence Relation: T(1) = 1 & T(n) = aT(n/b) Guess: G(n) = cn  Verify: Left Hand SideRight Hand Side T(n) = cn  aT(n/b) = a [c ( n / b )  ] = c a b  n  Solving Technique 3 Calculate Exponent ( log a / log b ) = cn 1 = a b -  b  = a  log b = log a  = log a / log b

166. COSC 3101Winter, 2004 Recurrence Relation: T(1) = 1 & T(n) = 4T(n/2) Guess: G(n) = cn  Verify: Left Hand SideRight Hand Side T(n) = cn  aT(n/b) = c [a ( n / b )  ] = c a b  n  Solving Technique 3 Calculate Exponent 1 = a b -  b  = a  log b = log a  = log a / log b ( log a / log b ) = cn ( log 4 / log 2 ) = cn 2

167. COSC 3101Winter, 2004 Recurrence Relation: T(1) = 1 & T(n) = aT(n/b) + f(n) Solving Technique 3 Calculate Exponent If bigger then T(n) =  (f(n)) If bigger then ( log a / log b ) T(n) =  (n ) And if aT(n/b)  f(n) what is T(n) then?

168. COSC 3101Winter, 2004 Technique 4: Decorate The Tree T(n) = a T(n/b) + f(n) f(n) T(n/b) T(n) = T(1) T(1) = 1 1 = a

169. COSC 3101Winter, 2004 f(n) T(n/b) T(n) = a

170. COSC 3101Winter, 2004 f(n) T(n/b) T(n) = a f(n/b) T(n/b 2 ) a

171. COSC 3101Winter, 2004 f(n) T(n) = a f(n/b) T(n/b 2 ) a f(n/b) T(n/b 2 ) a f(n/b) T(n/b 2 ) a f(n/b) T(n/b 2 ) a

172. COSC 3101Winter, 2004 11111111111111111111111111111111...... 111111111111111111111111111111111 f(n) T(n) = a f(n/b) T(n/b 2 ) a f(n/b) T(n/b 2 ) a f(n/b) T(n/b 2 ) a f(n/b) T(n/b 2 ) a

173. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level 0 1 2 i h

174. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0 1 2 i h

175. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0n 1 2 i h

176. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0n 1 n/b 2 i h

177. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0n 1 n/b 2 n/b 2 i h

178. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0n 1 n/b 2 n/b 2 i n/b i h n/b h

179. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0n 1 n/b 2 n/b 2 i n/b i h n/b h

180. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0n 1 n/b 2 n/b 2 i n/b i h n/b h = 1 base case

181. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0n 1 n/b 2 n/b 2 i n/b i h n/b h = 1

182. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0n 1 n/b 2 n/b 2 i n/b i h = log n / log b n/b h = 1 b h = n h log b = log n h = log n / log b

183. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame 0n 1 n/b 2 n/b 2 i n/b i h = log n / log b 1

184. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame 0n f(n) 1 n/bf(n/b) 2 n/b 2 f(n/b 2 ) i n/b i f(n/b i ) h = log n / log b 1T(1)

185. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame # stack frames 0n f(n) 1 n/bf(n/b) 2 n/b 2 f(n/b 2 ) i n/b i f(n/b i ) h = log n / log b n/b h T(1)

186. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame # stack frames 0n f(n) 1 1 n/bf(n/b) a 2 n/b 2 f(n/b 2 ) a2a2 i n/b i f(n/b i ) aiai h = log n / log b n/b h T(1) ahah

187. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame # stack frames 0n f(n) 1 1 n/bf(n/b) a 2 n/b 2 f(n/b 2 ) a2a2 i n/b i f(n/b i ) aiai h = log n / log b n/b h T(1) ahah

188. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame # stack frames 0n f(n) 1 1 n/bf(n/b) a 2 n/b 2 f(n/b 2 ) a2a2 i n/b i f(n/b i ) aiai h = log n / log b n/b h T(1) ahah a h = a = n log n / log b log a / log b

189. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame # stack frames Work in Level 0n f(n) 1 1 n/bf(n/b) a 2 n/b 2 f(n/b 2 ) a2a2 i n/b i f(n/b i ) aiai h = log n / log b n/b h T(1) n log a / log b

190. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame # stack frames Work in Level 0n f(n) 1 1 · f(n) 1 n/bf(n/b) a a · f(n/b) 2 n/b 2 f(n/b 2 ) a2a2 a 2 · f(n/b 2 ) i n/b i f(n/b i ) aiai a i · f(n/b i ) h = log n / log b n/b h T(1) n log a / log b n · T(1) log a / log b Total Work T(n) = ∑ i=0..h a i  f(n/b i )

191. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) = ∑ i=0..h a i  f(n/b i ) If a Geometric Sum ∑ i=0..n x i  θ(max(first term, last term))

192. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame # stack frames Work in Level 0n f(n) 1 1 · f(n) 1 n/bf(n/b) a a · f(n/b) 2 n/b 2 f(n/b 2 ) a2a2 a 2 · f(n/b 2 ) i n/b i f(n/b i ) aiai a i · f(n/b i ) h = log n / log b n/b h T(1) n log a / log b n · T(1) log a / log b Dominated by Top Level or Base Cases

193. COSC 3101Winter, 2004

194. COSC 3101Winter, 2004

195. COSC 3101Winter, 2004

196. COSC 3101Winter, 2004

197. COSC 3101Winter, 2004

198. COSC 3101Winter, 2004

199. COSC 3101Winter, 2004 Sufficiently close to: T(n) = 4T(n/2)+ n 3 = θ(n 3 ). T 2 (n) = ? Evaluating: T 2 (n) = 4 T 2 (n/2+n 1/2 )+ n 3 θ(n 3 ).

200. COSC 3101Winter, 2004 Evaluating: T(n) = aT(n-b)+f(n) h = ? n-hb n-ib n-2b n-b T(0) f(n-ib) f(n-2b) f(n-b) f(n) n Work in Level # stack frames Work in stack frame Instance size a aiai a2a2 a 1 n/bn/b a · T(0) a i · f(n-ib) a 2 · f(n-2b) a · f(n-b) 1 · f(n) n/bn/b |base case| = 0 = n-hb h = n / b i 2 1 0 Level Likely dominated by base cases Exponential

201. COSC 3101Winter, 2004 Evaluating: T(n) = 1T(n-b)+f(n) h = ? Work in Level # stack frames Work in stack frame Instance size 1 1 1 1 1 f(0) f(n-ib) f(n-2b) f(n-b) f(n) n-b n n-hb n-ib n-2b T(0) f(n-ib) f(n-2b) f(n-b) f(n) h = n / b i 2 1 0 Level Total Work T(n) = ∑ i=0..h f(b·i) = θ(f(n)) or θ(n · f(n))

202. COSC 3101Winter, 2004 End