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**한양대학교 정보보호 및 알고리즘 연구실 2008. 2. 12 이재준 담당교수님 : 박희진 교수님**

Chapter 3. Growth of function Chapter 4. Recurrences 한양대학교 정보보호 및 알고리즘 연구실 이재준 담당교수님 : 박희진 교수님

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**Contents of Table A. Growth of function B. Recurrence**

2. Notation of asymptotically larger/smaller o-notation ω-notation 3. Relationship between this notations B. Recurrence 1. Technicalities for abbreviation 2. Recurrence solution methods The substitution method

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**f(n) is asymptotically smaller than g(n)**

2. Asymptotically larger/smaller notation o-notation Asymptotically smaller - Definition for all n, n ≥ n0 , any positive constant c, c>0, 0 ≤ f(n)< cg(n), the function f(n) is smaller to g(n) to within constant factor. f(n) is asymptotically smaller than g(n) f(n) o(g(n)) n0

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**o-notation 2. Asymptotically larger/smaller notation - Definition**

o(g(n)) = { f(n) : for any positive constant c>0, there exist a constant n0 >0 such that 0 ≤ f(n) < cg(n) for all n ≥ n0 } this limit shows the function f(n) becomes insignificant relative to g(n) as n approaches infinity

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**o-notation Example 1. Asymptotic notation**

Use the definition of o-notation to prove the following property. f(n) = o(n2)

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**o-notation Solution to example**

2. Notation of asymptotically larger/smaller o-notation Solution to example 2n = o(n2) If f(n) = o((g(n)) then

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**f(n) is asymptotically larger than g(n)**

2. Notation of asymptotically larger/smaller ω -notation Asymptotically larger - Definition for all n, n ≥ n0 , any positive constant c, c>0, 0 ≤ cg(n) < f(n), the function f(n) is larger to g(n) to within constant factor. f(n) is asymptotically larger than g(n) f(n) ω(g(n))

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**ω-notation 2. Notation of asymptotically larger/smaller - Definition**

ω(g(n)) = { f(n) : for any positive constant c>0, there exist a constant n0 >0 such that 0 ≤ cg(n) < f(n) for all n ≥ n0 } if the limit exists. That is, f(n) becomes arbitrarily large relative to g(n) as n approaches infinity.

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**ω -notation 2. Notation of asymptotically larger/smaller**

- We use ω-notation to denote a lower bound that is not asymptotically tight. - We use o-notation to denote an upper bound that is not asymptotically tight. - f(n) ∈ ω(g(n)) if and only if g(n) ∈ o(f(n)). Arbitrarily : 독단적으로, 제멋대로, 마음대로

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**ω -notation Example 2. Notation of asymptotically larger/smaller**

Use the definition of o-notation to prove the following property. f(n) = ω(n)

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**ω -notation Solution to example**

2. Notation of asymptotically larger/smaller ω -notation Solution to example =ω(n) If f(n) = ω((g(n)) then

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**3. Relationship between this notations**

f(n) O(g(n)) o(g(n)) ω(g(n)) Ω(g(n)) θ(g(n)) O(g(n)) Ω(g(n)) ω(g(n)) o(g(n))

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**Transitivity Transpose symmetry 3. Relationship between this notations**

f(n) = Θ(g(n)) and g(n) = Θ(h(n)) imply f(n) = Θ(h(n)) , f(n) = O(g(n)) and g(n) = O(h(n)) imply f(n) = O(h(n)) , f(n) = Ω(g(n)) and g(n) = Ω(h(n)) imply f(n) = Ω(h(n)) , f(n) = o(g(n)) and g(n) = o(h(n)) imply f(n) = o(h(n)) , f(n) = ω(g(n)) and g(n) = ω(h(n)) imply f(n) = ω(h(n)). f(n) O(g(n)) o(g(n)) ω(g(n)) Ω(g(n)) f(n) = O(g(n)) if and only if g(n) = Ω(f(n)), f(n) = o(g(n)) if and only if g(n) = ω(f(n)).

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**Reflexivity Symmetry f(n) = Θ(f(n)) f(n) = O(f(n)) f(n) = Ω(f(n))**

3. Relationship between this notations Reflexivity f(n) = Θ(f(n)) f(n) = O(f(n)) f(n) = Ω(f(n)) Symmetry f(n) = Θ(g(n)) if and only if g(n) = Θ(f(n)). θ(g(n)) O(g(n)) Ω(g(n))

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**When should we use the recurrence?**

Recurrences Recurrence When should we use the recurrence? When algorithm contains a recursive call to itself, Its running time can often be described by a recurrence.

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**Recurrences Recurrence - Definition**

Equation or inequality that describes a function in terms of its value on smaller inputs.

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**Recurrences Recurrence Example if n=1 if n>1**

Worst-case running time of T(n) could be described by the recurrence Merge-Sort procedure Show that the solution of T(n) = Θ(n log n) if n=1 if n>1 범위에 대한 가정을 하고 그 가정이 맞는지 수학적 귀납법을 이용하여 증명하는것입니다. Recurrence를 트리형태로 바꾸고 각 트리레벨에서의 cost를 구한후 이들의 총합을 구함으로써 전체 cost를 구하는 방법입니다. Recurrence 의 기본적인 Form에서 주어진 세가지 경우에 따라 값을 구하는 방법입니다.

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**Recurrences Recurrence Solution method to example**

- The substitution method guess a bound and prove our guess correct for small input. - The recursion-tree method convert the recurrence into a tree whose nodes represent the costs incurred at various level of recursion - The master method determining asymptotic bounds for many simple recurrences of the form 범위에 대한 가정을 하고 그 가정이 맞는지 수학적 귀납법을 이용하여 증명하는것입니다. Recurrence를 트리형태로 바꾸고 각 트리레벨에서의 cost를 구한후 이들의 총합을 구함으로써 전체 cost를 구하는 방법입니다. Recurrence 의 기본적인 Form에서 주어진 세가지 경우에 따라 값을 구하는 방법입니다. ( a ≥ 1, b ≥ 1, and f(n) is given function )

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**1. Technicalities for abbreviation**

- assumption of integer The running time T(n) of algorithm is only defined when n is an integer - floors / ceilings We often omit floors and ceilings - boundary condition The running time of an algorithm on a constant-sized input is a constant that we typically ignore.

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**1. Technicalities for abbreviation**

Technicalities Example The recurrence describing the worst-case running time of Merge-Sort if n=1 if n>1 Omit assumption of integer Omit floors and ceilings 생략되는 Technical details은 recurrence solution method를 보면서 보게 될 것이다. Floor : 같거나 더 큰 수치 중에서 가장 작은 정수 Ceiling : 같거나 더 작은 수치 중에서 가장 큰 정수 Omit boundary condition

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**2. Recurrence solution methods**

Substitution Method - Definition When recurrence is easy to guess form of the answer with the inductive hypothesis is applied to smaller values, substitution of the guessed answer for the function . 생략되는 Technical details은 recurrence solution method를 보면서 보게 될것이다. Floors와 ceilings의 경우 몇몇 특수한 경우를 제외하고는 걱정할 필요가 없습니다. - It can be used to establish either upper or lower bounds on a recurrence.

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**2. Recurrence solution methods**

Substitution Method Steps 1. Guess the form of the solution 2. Use mathematical induction to find the constant and show that the solution works.

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**2. Recurrence solution methods**

Substitution Method - Making a good guess prove loose upper and lower bounds on the recurrence and then reduce the range of uncertainty. - Guessing a solution takes : experience, occasionally, creativity - we can use recursion tree

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**2. Recurrence solution methods**

Substitution Method - Subtleties we can prove something stronger for given value by assuming something stronger for a smaller values. We guess that the solution is O(n), and show that T(n) ≤ cn Overcome difficulty by subtracting a lower-order term from our previous guess ( b≥1, c must be chosen large enough )

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**2. Recurrence solution methods**

Substitution Method - Avoid pitfalls In recurrence we can falsely prove T(n)= O(n) by guessing T(n) ≤ cn

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**2. Recurrence solution methods**

Substitution Method Example Let us determine an upper bound on the recurrence

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**2. Recurrence solution methods**

Substitution Method Solution to example 1. guess that the solution is 2. prove that ( c > 0 ) We need to find constant c and n0 - Assume that this bound holds for ⌊n/2⌋, that is, T (⌊n/2⌋) ≤ c ⌊n/2⌋ lg(⌊n/2⌋). T(n) ≤ 2(c ⌊n/2⌋lg(⌊n/2⌋)) + n ≤ cn lg(n/2) + n ( because, ⌊n/2⌋ < n/2 ) = cn lg n - cn lg 2 + n = cn lg n - cn + n ≤ cn lg n (as long as c ≥ 1)

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**2. Recurrence solution methods**

Substitution Method Solution to example - Find the constant n0 - we can take advantage of asymptotic notation, we can replace base case T(1) ≤ cn lg n T(1) = 1 but, c1 lg1 = 0 - So, we only have to prove T (n) = cn lg n for n ≥ n0 for n (n0 =2) Replace T(1) by T(2) and T(3) as the base cases by letting n0 =2.

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**2. Recurrence solution methods**

Substitution Method Solution to example - Find the constant c - choosing c large enough T (2) = 4 and T (3) = 5. T (2) = 4 ≤ c ( 2 lg 2 ) T (3) = 5 ≤ c ( 3 lg 3 ). Any choice of c ≥ 2 suffices for base case of n=2 and n =3

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**2. Recurrence solution methods**

Substitution Method Example Let us determine an upper bound on the recurrence

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**2. Recurrence solution methods**

Substitution Method Solution to example - Simplify this recurrence by changing variable Renaming m = lg n S(m) = T(2m) This new recurrence has same solution : S(m) = O(m log m)

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**2. Recurrence solution methods**

Substitution Method Solution to example Simplify this recurrence by changing variable This new recurrence has same solution : S(m) = O(m log m) Changing back from S(m) to T(n) T(n) = T(2m) = S(m) = O(m lg m) = O( lg n lg(lg n) )

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