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Recr_Ms_1 For the best case : Identify the recurrence relation   merge_sort function described below.

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A.T(n) = \begin{cases} \Theta(1) & \text{if } n <2 \\T\left(n-1\right) + \Theta(1) & \text{otherwise}\end{cases}
B.T(n) = \begin{cases} \Theta(1) & \text{if } n <2 \\T\left(n-1\right) + \Theta(n) & \text{otherwise}\end{cases}
C.T(n) = \begin{cases} \Theta(1) & \text{if } n <2 \\2T\left(\frac{n}{2}\right) + \Theta(n) & \text{otherwise}\end{cases}
D.T(n) =\begin{cases}\Theta(1) & \text{if } n < 2 \\2T\left(\frac{n}{2}\right) + \Theta(n \log n) & \text{otherwise}\end{cases}
E.T(n) = \begin{cases} \Theta(1) & \text{if } n <2 \\2T\left(n-1\right) + \Theta(1) & \text{otherwise}\end{cases}
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Question restatement: We are asked to identify the recurrence relation that describes the best-case time complexity for the described merge_sort function. Option A: T(n) = { Θ(1) if n < 2 ; T(n-1) + Θ(1) otherwise }. - This would model a process that reduces n by 1 at each step with constant work per step. It yields O(n) time, which is not consistent with the typical divide-and-conquer behavior of merge sort. Therefore this option is not appropriate for the best-case recurrence of merge......Login to view full explanation

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