# Master Method

The 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 non-negative 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:

• 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: 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 n2, logba = log28 = 3
Put all the values in: f (n) = 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 Therefore: T (n) = Θ (n3)
```

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, logba = log22 =1
Put all the values in f (n) =Θ , we will get
10n = Θ (n1) = 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) = n2, logba = log22 =1
Put all the values in f (n) = Ω ..... (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 If we will choose c =1/2, it is true: ∀ n ≥1
So it follows: T (n) = Θ ((f (n))
T (n) = Θ(n2)
```

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