Master MethodThe Master Method is used for solving the following types of recurrence T (n) = a T+ f (n) with a≥1 and b≥1 be constant & f(n) be a function and can be interpreted as Let T (n) is defined on nonnegative integers by the recurrence. T (n) = a T+ f (n) In the function to the analysis of a recursive algorithm, the constants and function take on the following significance:
Master Theorem:It is possible to complete an asymptotic tight bound in these three cases: Case1: If f (n) = for some constant ε >0, then it follows that: T (n) = Θ Example: T (n) = 8 T apply master theorem on it. Solution: Compare T (n) = 8 T with T (n) = a T a = 8, b=2, f (n) = 1000 n^{2}, log_{b}a = log_{2}8 = 3 Put all the values in: f (n) = 1000 n^{2} = O (n^{3ε} ) If we choose ε=1, we get: 1000 n^{2} = O (n^{31}) = O (n^{2}) Since this equation holds, the first case of the master theorem applies to the given recurrence relation, thus resulting in the conclusion: T (n) = Θ Therefore: T (n) = Θ (n^{3}) Case 2: If it is true, for some constant k ≥ 0 that: F (n) = Θ then it follows that: T (n) = Θ Example: T (n) = 2 , solve the recurrence by using the master method. As compare the given problem with T (n) = a T a = 2, b=2, k=0, f (n) = 10n, log_{b}a = log_{2}2 =1 Put all the values in f (n) =Θ , we will get 10n = Θ (n^{1}) = Θ (n) which is true. Therefore: T (n) = Θ = Θ (n log n) Case 3: If it is true f(n) = Ω for some constant ε >0 and it also true that: a f for some constant c<1 for large value of n ,then : Example: Solve the recurrence relation: T (n) = 2 Solution: Compare the given problem with T (n) = a T a= 2, b =2, f (n) = n^{2}, log_{b}a = log_{2}2 =1 Put all the values in f (n) = Ω ..... (Eq. 1) If we insert all the value in (Eq.1), we will get n^{2} = Ω(n^{1+ε}) put ε =1, then the equality will hold. n^{2} = Ω(n^{1+1}) = Ω(n^{2}) Now we will also check the second condition: 2 If we will choose c =1/2, it is true: ∀ n ≥1 So it follows: T (n) = Θ ((f (n)) T (n) = Θ(n^{2})
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