Question

Apply the Master Theorem to the following: T() = 4T ( / 2 ) + ² log n

Answer 1

To solve the given recurrence relation, we can apply the Master Theorem. Using the formula and the provided values, we can determine the time complexity of the algorithm. In this case, the time complexity is O(n^2 log n).

Step-by-step explanation:

The given recurrence relation can be solved using the Master Theorem. The Master Theorem is a formula that allows us to determine the time complexity of divide-and-conquer algorithms. It has the following form:

T(n) = aT(n/b) + f(n)

Here, T(n) represents the time complexity of the algorithm, a represents the number of subproblems, n/b represents the size of each subproblem, and f(n) represents the time complexity of the work done outside of the recursive calls.

In this case, we have a = 4 (four recursive calls), b = 2 (each subproblem is half the size of the original problem), and f(n) = n^2 log n (work done outside of the recursive calls).

According to the Master Theorem, the time complexity can be determined by comparing the value of f(n) with n^logba.

1. If f(n) is smaller, then the time complexity is dominated by the recursive calls, resulting in a time complexity of O(n^logba).
2. If f(n) is equal, then the time complexity is O(n^logba log n).
3. If f(n) is larger, then the time complexity is dominated by f(n), resulting in a time complexity of O(f(n)).

In this case, n^logba is n^log24 = n^2. Since f(n) = n^2 log n is larger, the time complexity of the algorithm is O(n^2 log n).