# What is Case 2 of Master Theorem?

## What is Case 2 of Master Theorem?

Case 2. If f ( n ) = Θ ( n log b a ) f(n) = \Theta\left(n^{\log_b{a}}\right) f(n)=Θ(nlogba), then T ( n ) = Θ ( n log b a log n ) T(n) = \Theta\left(n^{\log_b{a}} \log{n} \right) T(n)=Θ(nlogbalogn).

## What are the cases of Master Theorem?

There are 3 cases for the master theorem:

- Case 1: d < log(a) [base b] => Time Complexity = O(n ^ log(a) [base b])
- Case 2: d = log(a) [base b] => Time Complexity = O((n ^ d) * log(n) )
- Case 3: d > log(a) [base b] => Time Complexity = O((n ^ d))

**What is K in Master Theorem?**

Note k = logb(n). The recurrence for the running time is: T(n) = aT(n/b) + f(n), T(1) = d . Here f(n) represents the divide and combine time (i.e., the non- recursive time).

### How do you prove the Master Theorem?

Page 1

- Proof of the Master Method.
- Theorem (Master Method) Consider the recurrence T(n) = aT(n/b) + f(n), (1) where a, b are constants. Then (A) If f(n) = O(nlogb a − ε) for some constant ε > 0, then T(n) = O(nlogb a).
- logb n.
- ∑
- i=0.
- aif(n/bi) + O(nlogb a). (2)
- f(n) f(n/b)

### What is A and B in master theorem?

T(n) = aT(n/b) + f(n), where, n = size of input a = number of subproblems in the recursion n/b = size of each subproblem. All subproblems are assumed to have the same size.

**What is the purpose of master theorem?**

Master Theorem is used to determine running time of algorithms (divide and conquer algorithms) in terms of asymptotic notations. Consider a problem that be solved using recursion.

#### What is the use of master theorem?

#### What is master theorem used for?

**What is the limitation of master theorem?**

Master Theorem Limitations The master theorem cannot be used if: T(n) is not monotone. eg. T(n) = sin n.

## How do you do master method?

The master method is a formula for solving recurrence relations of the form: T(n) = aT(n/b) + f(n), where, n = size of input a = number of subproblems in the recursion n/b = size of each subproblem. All subproblems are assumed to have the same size.

## Why is Masters theorem used?

Masters Theorem for divide and conquer is an analysis theorem that can be used to determine a big-0 value for recursive relation algorithms. It is used to find the time required by the algorithm and represent it in asymptotic notation form.

**What is the use of master theorem in algorithm?**

In the analysis of algorithms, the master theorem for divide-and-conquer recurrences provides an asymptotic analysis (using Big O notation) for recurrence relations of types that occur in the analysis of many divide and conquer algorithms.

### What are the cases of the master theorem?

There are 3 cases for the master theorem: Case 1: d < log(a) [base b] => Time Complexity = O(n ^ log(a) [base b]) Case 2: d = log(a) [base b] => Time Complexity = O((n ^ d) * log(n) ) Case 3: d > log(a) [base b] => Time Complexity = O((n ^ d)) Proof Of Master Theorem

### Can you make a connection to the three cases of the theorem?

Presumably you can make the connection to the three cases in the statement of the theorem yourself (they should be presented in the same order).$\\endgroup$– Yuval FilmusJun 8 ’15 at 0:43 Add a comment | 1 Answer 1 ActiveOldestVotes

**Is the master theorem a solution to recurrence relations?**

The master theorem provides a solution to recurrence relations of the form. T(n)=aT(nb)+f(n), T(n) = a T\\left(\\frac nb\\right) + f(n), T(n)=aT(bn)+f(n), for constants a≥1 a \\geq 1a≥1 and b>1b > 1 b>1 with f f f asymptotically positive. Such recurrences occur frequently in the runtime analysis of many commonly encountered algorithms.

#### How is the Black Box solved in the master theorem?

Master Theorem “Black Box” for solving recurrences Assumesall subproblems are of equal size (most algorithms do this) •The same amount of data is given to each recursive call An algorithm that splits the subproblems into 1/3 and 2/3 (or an algorithm that splits data randomly) must be solvedin a different manner. We’ll look at other methods later