Ms_7
Consider the recurrence relation for recursive algorithm $T(n)$ given by:
T(n) = \begin{cases} \Theta(1) & \text{if } n

Question

Ms_7
Consider the recurrence relation for recursive algorithm $T(n)$ given by:
T(n) = \begin{cases} \Theta(1) & \text{if } n < 2 \9T\left(\frac{n}{3}\right) + \Theta(n) & \text{otherwise}\end{cases}
What is the run time complexity of this algorithm?
The Master Theorem is provided below. Use it as you see fit:

BlackTom AI · Accepted Answer

To tackle the recurrence, I’ll compare the non-recursive work f(n) with the growth coming from the recursive term.
Option 1: $\Theta(n^2)$ - This aligns with the Master Theorem when a = 9 and b = 3, since n^{log_b a} = n^{log_3 9} = n^2. Because f(n) = \Theta(n) grows slower than n^2, the case where f(n) = O(n^{log_b a - ε}) applies, giving T(n) = \Theta(n^2). This makes sense as the recursion dominates.
Option 2: $\Theta(2^n)$ - An exponential growth function would only arise if the recurrence had a form like T(n) = a T(n−1) + f(n) or similar with large a, but here we reduce the problem size by a factor of 3, not by 1, and the branching factor leads to polynomial growth, not exponential. Therefore this is incorrect.
Option 3: $\Theta(n)$ - This would occur if the recursive cost didn’t accumulate more than a linear amount overall; however, the recursion expands with a^log_b a = n^2 growth at the top level, so the overall complexity cannot be linear. Hence this is incorrect.
Option 4: $\Theta(\log n)$ - Logarithmic complexity would imply very shallow or minimal recursive growth, which contradicts the 9-way branching at each level. This is not supported by the Master Theorem analysis here, so it’s incorrect.
Option 5: $\Theta(n \log n)$ - While a common result in some divide-and-conquer cases, the comparison f(n) = \Theta(n) vs n^2 shows the dominant term is n^2, not n log n. Therefore this option misestimates the recursion’s growth.
In summary, the only option that matches the Master Theorem reasoning for a = 9, b = 3, and f(n) = \Theta(n) is $\Theta(n^2)$.

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