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**Master Theorem Chen Dan Dong Feb. 19, 2013**

the master theorem is extremely useful for time analysis in asymptotic terms (using Big O notation) forrecurrence relations. It can be applied to most of recurrence relationship. Chen Dan Dong Feb. 19, 2013

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**Outline Review of asymptotic notations Understand the Master Theorem**

Prove the theorem Examples and applications

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**Review of Asymptotic Notation**

Θ notation: asymptotic tight bound Θ(g(n)) = { f(n): there exist positive constants c1, c2, and n0 such that 0 ≤ c1g(n) ≤ f(n) ≤ c2g(n) for all n ≥ n0}. O notation: asymptotic upper bound O(g(n)) = { f(n): there exist positive constants c, and n0 such that 0≤ f(n) ≤ cg(n) for all n ≥ n0}. Ω notation: asymptotic lower bound Ω(g(n)) = { f(n): there exist positive constants c, and n0 such that 0≤ cg(n) ≤ f(n) for all n ≥ n0}. In mathematics, big O notation is used to describe the limiting behavior of a function when the argument tends towards a particular value or infinity, usually in terms of simpler functions. A description of a function in terms of big O notation usually only provides an upper bound on the growth rate of the function. For a given function g(n), we denote by big theta g(n) is the set of functions of f(n)

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**Review of Asymptotic Notation (Con.)**

Asymptotic notation in equations Theorem: For any two functions f(n) and g(n), we have f(n) = Θ(g(n)) if and only if f(n) = O(g(n)) and f(n) = Ω(g(n)).

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Master Theorem Theorem: Let a ≥ 1 and b > 1 be constants, let f(n) be a function and let T(n) be defined on the nonnegative integers by recurrence T(n) = aT(n/b) + f(n), Then T(n) has the following asymptotic bounds. Case 1. If for some constant ϵ > 0, then Case 2. If then Case 3. If for some constant ϵ > 0, and if a f(n/b) ≤ c f(n) for some constant c < 1 and all sufficiently large n, then T(n) = Θ(f(n)).

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**Proof of Master Theorem**

The proof consists of two parts The first part analyzes the recurrence under the simplifying assumption that T(n) is defined only on exact powers of b, for n = 1, b, b2, …. The second part extends the analysis to all positive integers n with handling floors and ceilings. Due to time limit, I will only show the proof for the first part for n as exact powers of b.

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Proof – Lemma 1 Lemma 1. Let a ≥ 1 and b > 1 be constants, let f(n) be a nonnegative function defined on exact powers of b. Define T(n) exact powers of b by the recurrence where j is a positive integer. Then

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Proof – Lemma 2 Lemma 2. Let a ≥ 1 and b > 1 be constants, let f(n) be a nonnegative function defined on exact powers of b. A function g(n) defined over exact power of b by has the following asymptotic bounds: Case 1. If for some constant ϵ > 0, then Case 2. If then Case 3. if a f(n/b) ≤ c f(n) for some constant c < 1 and all sufficiently large n, then g(n) = Θ(f(n)).

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**Combining Lemma 1 and Lemma 2**

Lemma 3. Let a ≥ 1 and b > 1 be constants, let f(n) be a nonnegative function defined on exact powers of b. Define T(n) exact powers of b by the recurrence Then T(n) has the following asymptotic bounds: Case 1. If for some constant ϵ > 0, then Case 2. If then Case 3. if for some constant ϵ > 0, and if a f(n/b) ≤ c f(n) for some constant c < 1 and all sufficiently large n, then g(n) = Θ(f(n)).

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**If n is not exact powers of b…**

If n is not exact powers of b, then n/b is an not integer. We need to obtain a lower bound on T(n) = a T(⌈n/b⌉) + f(n) and an upper bound on T(n) = a T(⌊n/b⌋) + f(n). If intersted, check out here.

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Master Theorem Theorem: Let a ≥ 1 and b > 1 be constants, let f(n) be a function and let T(n) be defined on the nonnegative integers by recurrence T(n) = aT(n/b) + f(n), Then T(n) has the following asymptotic bounds. Case 1. If for some constant ϵ > 0, then Case 2. If then Case 3. If for some constant ϵ > 0, and if a f(n/b) ≤ c f(n) for some constant c < 1 and all sufficiently large n, then T(n) = Θ(f(n)).

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**Some examples T(n) = 9 T(n/3) + n Case 1. T(n) = T(2n/3) +1 Case 2.**

T(n) = 2T(n/2) + n lgn Can’t apply Master Theorem.

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Binary Search Binary search finds the position of a specified value within a sorted array. Finding the recurrence relationship Applying Master Theorem. Case 2. T(n) = O(lg n).

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