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

Recurrences 한양대학교 정보보호 및 알고리즘 연구실 이재준 담당교수님 : 박희진 교수님

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

Relationship between asymptotic notations B. Recurrence Recurrence solution methods The substitution method The recursion-tree method

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**Analogy f(n) = Θ(g(n)) ≈ f(n) = g(n). f(n) = O(g(n)) ≈ f(n) ≤ g(n).**

3. Relationship between this notations Analogy f(n) = Θ(g(n)) ≈ f(n) = g(n). f(n) = O(g(n)) ≈ f(n) ≤ g(n). f(n) = Ω(g(n)) ≈ f(n) ≥ g(n). f(n) = o(g(n)) ≈ f(n) < g(n). f(n) = ω(g(n)) ≈ f(n) > 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)) 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)).

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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**

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

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**Recurrences Recurrence Solution method to example**

1) The substitution method guess a bound and prove our guess correct. 2) The recursion-tree method convert the recurrence into a tree whose nodes represent the costs incurred at various level of recursion 3) 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**

1) assumption of integer The running time T(n) of algorithm is only defined when n is an integer 2) floors / ceilings We often omit floors and ceilings 3) 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 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 1. Guess the form of the solution 2. Use mathematical induction to find the constant and show that the solution works. - It can be used to establish either upper or lower bounds on a recurrence.

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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 to meet the requirement about assumption T(n) ≤ cn lg n ( c>0 ) We using the Mathematical induction to show that our assumption holds for the boundary condition.

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**2. Recurrence solution methods**

Substitution Method Solution to example Mathematical induction We can prove something stronger for a given value by assuming something stronger for smaller values. → T(1), T(2), T(3), … ….. T(⌊n/2⌋ ), T(n) k k+1 k means ⌊n/2⌋ k+1 means n

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**2. Recurrence solution methods**

Substitution Method Solution to example Recurrence : Guess : T(n)= O(n) Prove : T(n) ≤ cn k means k+1 means n -> Substitution 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 -(c+1)n ( It’s our assumed fact )

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**2. Recurrence solution methods**

Substitution Method Solution to example Assumed Fact : cn lg n –(c+1)n Prove : T(n) ≤ cn Assumed Fact ≤ Prove T(n) ≤ cn lg n – (c+1)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 Skill - Subtleties We can prove something stronger for given value by assuming something stronger for a smaller values. Overcome difficulty by subtracting a lower-order term from our previous guess

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**2. Recurrence solution methods**

Substitution Method Skill - Subtleties example Recurrence : Guess : T(n)= O(n) Prove : T(n) ≤ cn Substitution k means k+1 means n → →

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**2. Recurrence solution methods**

Substitution Method Skill - Subtleties example ( It’s Impossible. Then T(n)= O(n) is wrong? ) Subtracting a lower-order term from our previous guess T(n) ≤ cn So, we now prove T(n) ≤ cn-b

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**2. Recurrence solution methods**

Substitution Method Skill - Subtleties example Recurrence : Guess : T(n)= O(n) Prove : T(n) ≤ cn-b ( b≥1, c must be chosen large enough ) T(n) 기존의 증명할 부등식에 상수 b를 subtracting 마친 부등식에 대하여 증명을 하는 것입니다. Substitution을 마친 가정된 사실보다 증명하고자 하는 Big Oh의 anonymous function 이 더 클경우는 이 추측은 증명이 되는것입니다. 때문에, 상수 b가 1보다 크거나 같고, 상수 c는 충분히 크다는 조건하에 추측한 가정사항은 만족함을 증명할수 있습니다.

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**2. Recurrence solution methods**

Substitution Method Skill - Avoid pitfalls We need to prove the exact form of the inductive hypothesis T(n) ≤ cn T(n) ≤ (c+1)n

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**2. Recurrence solution methods**

Substitution Method Skill - Avoid pitfalls example Recurrence : Guess : T(n)= O(n) Prove : T(n) ≤ cn ( Wrong !!! exact form :T(n) ≤ cn)

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**2. Recurrence solution methods**

Substitution Method Skill - 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 Skill - 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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**2. Recurrence solution methods**

Substitution Method Skill - 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 make a good guess by using recursion tree !!

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**2. Recurrence solution methods**

Recursion-tree - Definition For come up with good guess, convert the recurrence into a tree that each node represents the cost of a single subproblem somewhere in the set of recursive function invocations. We sum all the per-level costs to determine the total cost of all levels of the recursion Recusion-tree는 좋은 추측을 하기 위햐여 사용하는데, Recurrence를 tree로 변형을 하는데, 이 tree는, recursive function 이 발생하는 부분의 subproblem에 대한 비용으로 node를 표현하는 tree입니다.

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**2. Recurrence solution methods**

Recursion-tree Example The recurrence is T (n) = 3T (⌊n/4⌋) + Θ(n2) = 3T (n/4) + cn2 Use the recursion tree to making a good guess for upper bound for the solution.

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**2. Recurrence solution methods**

Recursion-tree Solution to example Draw recursion tree 3T (n/4) + cn2 cn2 T(n)

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**2. Recurrence solution methods**

Recursion-tree Solution to example cn2 c(n/4)2 T(1) … log4n When depth i then subprogram size is n/4i = 1 so, i = log4 n and it means tree’s height and tree has log4 n + 1 levels. The number of nodes at depth i is 3i and finally

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**2. Recurrence solution methods**

Recursion-tree Solution to example Total Cost Sum of cost for root to i-1 depth Sum of cost for i depth

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**2. Recurrence solution methods**

Recursion-tree Solution to example Total Cost is… ( When 0<x<1 then, )

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**2. Recurrence solution methods**

Recursion-tree Solution to example Prove T (n) = O(n2) by the substitution method Recurrence : Guess : T (n) = O(n2) Prove : T (n) ≤ dn2 (for some d > 0 and for the same c > 0) T (n) = 3T (⌊n/4⌋) + Θ(n2) = 3T(⌊n/4⌋) + cn2

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**2. Recurrence solution methods**

Recursion-tree Solution to example Find constant d T (n) = 3T(⌊n/4⌋) + cn2 T(n) ≤ 3T(⌊n/4⌋) + cn2 ≤ 3d⌊n/4⌋2 + cn2 ≤ 3d(n/4)2 + cn2 = 3/16 dn2 + cn2 ≤ dn2 (where the last step holds as long as d ≥ (16/13)c )

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**2. Recurrence solution methods**

Recursion-tree Example The recurrence T (n) = 3T (n/3)+T(2n/3)+O(n) Use the recursion tree to making a good guess for upper bound for the solution.

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**2. Recurrence solution methods**

Recursion-tree Solution to example Draw recursion tree T (n) = 3T (n/3)+T(2n/3)+ cn cn log3/2n Total : O(n log n) This recursion tree is not a complete binary tree. Not all leaves contribute a cost of exactly cn

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**2. Recurrence solution methods**

Recursion-tree Solution to example Draw recursion tree T (n) = 3T (n/3)+T(2n/3)+ cn cn log3/2n The longest path from root to a leaf is n→(2/3)n → (2/3)2n → … → 1. Since, (2/3)kn = 1 when k=log3/2 n and it means tree’s height is log3/2 n . The total cost = Cost of each level x Height = cn x log3/2 n = O(cnlog3/2n) = O(n lg n)

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**2. Recurrence solution methods**

Recursion-tree Solution to example Prove T (n) = O(n lg n) by the substitution method Recurrence : Guess : T (n) = O(n lg n) Prove : T (n) ≤ d n lg n (for some d > 0 and for the same c > 0) T(n/3) + T(2n/3) + cn

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**2. Recurrence solution methods**

Recursion-tree Solution to example Find constant d T(n)=T(n/3) + T(2n/3) + cn T(n) ≤ T(n/3) + T(2n/3) + cn ≤ d(n/3)lg(n/3) + d(2n/3)lg(2n/3) + cn = (d(n/3)lgn - d(n/3)lg 3) + (d(2n/3)lgn – d(2n/3)lg(3/2)) + cn = dnlgn - d((n/3)lg3 + (2n/3)lg(3/2)) + cn = dnlgn - d((n/3)lg3 + (2n/3)lg3 - (2n/3)lg2) + cn = dnlgn - dn(lg3 - 2/3) + cn ≤ dnlgn (as long as d ≥ c/(lg 3 - (2/3)) )

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**Content of next week Ch4. Recurrence**

Recurrence solution methods The substitution method The recursion-tree method The master method

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Recurrences 2008. 1. 11 : 1 Chapter 3. Growth of function Chapter 4. Recurrences.

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