Analyzing Recurrence Relations in Algorithms: T(n) = 3T(n/4) + Θ(n)

Math Problem Statement

Recurrence for the following algorithm is T(n) = { Θ(1) if n < 3, 3T(n/4) + Θ(n) otherwise}

The given algorithm follows a recurrence relation:

\begin{cases} \Theta(1), & \text{if } n < 3 \\ 3T(n/4) + \Theta(n), & \text{otherwise} \end{cases}$$ Let's analyze the correctness of this recurrence: 1. **Base Case:** If $$n < 3$$, the function returns a constant value (80), which is $$\Theta(1)$$. 2. **Recursive Case:** - The loop runs $$n$$ times, contributing $$\Theta(n)$$ complexity. - The function makes three recursive calls to $$recurse3(n/4)$$, leading to the recurrence: $$T(n) = 3T(n/4) + \Theta(n)$$ This matches the given recurrence equation, so the answer to whether the provided recurrence is correct is **True**. Would you like a more detailed breakdown of solving this recurrence using the Master Theorem? --- Here are five related questions to deepen your understanding: 1. How do we solve the recurrence $$T(n) = 3T(n/4) + \Theta(n)$$ using the Master Theorem? 2. What is the time complexity of the algorithm? 3. How does changing the number of recursive calls affect the recurrence relation? 4. What would happen if the for-loop had a different range, such as $$\text{range}(n/2)$$? 5. How does the recursion tree method help visualize the time complexity? **Tip:** The Master Theorem is a useful tool to solve recurrence relations of the form $$T(n) = aT(n/b) + f(n)$$. Always check if it applies to your recurrence.

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Recurrence Relations
Algorithm Analysis
Time Complexity

T(n) = 3T(n/4) + Θ(n)
T(n) = Θ(1) if n < 3

Master Theorem

College/University

Master Theorem: Solving T(n) = 2T(n/4) + n^2 Time Complexity

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