Solution

The correct answer is : option 2.

Key Points

Explanation of the Recurrence Relation Solution

Given the recurrence relation: T(n) = T(2n/3) + 1

To solve this, we will use the Master Theorem for divide-and-conquer recurrences of the form:

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

In our case, a = 1, b = 3/2, and f(n) = 1.

Using the Master Theorem, we compare f(n) with n^log_b(a):

• a = 1
• b = 3/2
• log_b(a) = log(1) / log(3/2) = 0
• f(n) = 1

Since f(n) = 1 is Θ(1), it matches the case where f(n) = Θ(n^c) with c = 0. This corresponds to case 2 of the Master Theorem, where f(n) is polynomially smaller than n^log_b(a).

Therefore, the solution to the recurrence is:

T(n) = Θ(log n)

Thus, the correct answer is option 2.