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Master Method

The Master Method is used for solving the following types of recurrence

T (n) = a TDAA Master Method+ f (n) with a≥1 and b≥1 be constant & f(n) be a function and DAA Master Methodcan be interpreted as

Let T (n) is defined on non-negative integers by the recurrence.

 T (n) = a TDAA Master Method+ f (n)

In the function to the analysis of a recursive algorithm, the constants and function take on the following significance:

  • n is the size of the problem.
  • a is the number of subproblems in the recursion.
  • n/b is the size of each subproblem. (Here it is assumed that all subproblems are essentially the same size.)
  • f (n) is the sum of the work done outside the recursive calls, which includes the sum of dividing the problem and the sum of combining the solutions to the subproblems.
  • It is not possible always bound the function according to the requirement, so we make three cases which will tell us what kind of bound we can apply on the function.

Master Theorem:

It is possible to complete an asymptotic tight bound in these three cases:

DAA Master Method

Case1: If f (n) = DAA Master Method for some constant ε >0, then it follows that:

T (n) = Θ DAA Master Method

Example:

T (n) = 8 T DAA Master Method apply master theorem on it.

Solution:

Compare T (n) = 8 T DAA Master Method with 
 T (n) = a T DAA Master Method
 a = 8, b=2, f (n) = 1000 n2, logba = log28 = 3
 Put all the values in: f (n) = DAA Master Method
     1000 n2 = O (n3-ε ) 
     If we choose ε=1, we get: 1000 n2 = O (n3-1) = O (n2)

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 DAA Master Method
   Therefore: T (n) = Θ (n3) 

Case 2: If it is true, for some constant k ≥ 0 that:

Example:

T (n) = 2 DAA Master Method, solve the recurrence by using the master method.
As compare the given problem with T (n) = a TDAA Master Method a = 2, b=2, k=0, f (n) = 10n, logba = log22 =1 
Put all the values in f (n) =Θ DAA Master Method, we will get 
	10n = Θ (n1) = Θ (n) which is true.
Therefore: T (n) = Θ DAA Master Method
      = Θ (n log n)

Case 3: If it is true f(n) = Ω DAA Master Method for some constant ε >0 and it also true that: a f DAA Master Method for some constant c<1 for large value of n ,then :

Example: Solve the recurrence relation:

T (n) = 2 DAA Master Method

Solution:

Compare the given problem with T (n) = a T DAA Master Method
a= 2, b =2, f (n) = n2, logba = log22 =1 
Put all the values in f (n) = Ω DAA Master Method ..... (Eq. 1)
If we insert all the value in (Eq.1), we will get 
  n2 = Ω(n1+ε) put ε =1, then the equality will hold.
  n2 = Ω(n1+1) = Ω(n2)
Now we will also check the second condition:
  2 DAA Master Method
If we will choose c =1/2, it is true:
  DAA Master Method  ∀ n ≥1 
So it follows: T (n) = Θ ((f (n))
    T (n) = Θ(n2)

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