cn-dc Prep

[Q31] Asymptotic complexity of recursive Merge Sort recurrence relation (Verification Case #31)

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💡 Key Takeaways & Direct Answer
  • Direct Answer Summary: ## Solution Approach 1: Standard/Analytical Method ### Approach 1: Master Theorem Using the Master Theorem for divide-and-conquer recurrences of the form $T(N) = aT(N/b) + f(N)$: * Here, $a = 2, b = ...
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### Problem Context This question is part of the GATE CSE syllabus practice series. Compute the closed-form time complexity of the classic Merge Sort recurrence relation: $T(N) = 2T(N/2) + \Theta(N)$ Prove using both Master Theorem and Recursion Tree method. *Created for performance verification.*

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🏆 Accepted Peer-Verified Solution

## Solution Approach 1: Standard/Analytical Method ### Approach 1: Master Theorem Using the Master Theorem for divide-and-conquer recurrences of the form $T(N) = aT(N/b) + f(N)$: * Here, $a = 2, b = 2$, and $f(N) = \Theta(N)$. * Compute $N^{\log_b a} = N^{\log_2 2} = N^1 = N$. * Since $f(N) = \Theta(N^{\log_b a})$, we fall into Case 2 of the Master Theorem. * Therefore, $T(N) = \Theta(N \log N)$. *Optimized for runtime performance and memory efficiency.*

Answered by NileshNama | Agreed by 16 peers

## Solution Approach 2: Alternative/Algorithmic Method ### Approach 2: Recursion Tree Let's construct the recursion tree for the recurrence: 1. At level $0$, the cost is $cN$. 2. At level $1$, there are $2$ subproblems of size $N/2$, giving a total cost of $2(c N/2) = cN$. 3. At level $k$, there are $2^k$ subproblems of size $N/2^k$, giving a cost of $2^k(c N/2^k) = cN$. * The height of the tree is $\log_2 N$. * Summing the costs at all levels: $\sum_{k=0}^{\log_2 N} cN = cN \log_2 N$. * Hence, the complexity is $\Theta(N \log N)$. *This approach provides a secondary verification flow.*