Recr_Q_6
What is the recurrence relation for the quick_sort function given below in the average case scenario as described below?

Question

Recr_Q_6
What is the recurrence relation for the quick_sort function given below  in the average case scenario as described below?

BlackTom AI · Accepted Answer

We start by restating the problem to ensure clarity: determine the average-case recurrence relation for the quick_sort function described.

Option 1: T(n) = { Θ(1) if n < 2; 2T(n/2) + Θ(n log n) otherwise }.
- Why this is unlikely: the non-base-case term Θ(n log n) suggests that combining or partitioning costs grow as n log n, which is not the standard cost for partitioning in quicksort. In typical average-case analyses, the dominant work outside the recursive calls is the partitioning step, which is linear in n, not n log n. Therefore this option overestimates the combination cost and does not align with the classic average-case recurrence for quicksort.

Option 2: T(n) = { Θ(1) if n < 2; 2T(n-1) + Θ(1) otherwise }.
- Why this is unlikely: this recurrence represents a linear-chain of subproblems of size n-1 with constant work per level, yielding Θ(n) total time. This would correspond to an algorithm that splits off a single element each time without reducing the problem size by a factor, which is not how quicksort partitions the array into two subarrays near n/2 on average. Therefore, this is not the average-case recurrence for quicksort.

Option 3: T(n) = { Θ(1) if n < 2; 2T(n/2) + Θ(n) otherwise }.
- Why this is correct: in the average case, quicksort partitions the array into two roughly equal halves, leading to two recursive calls on sizes about n/2. The partitioning step scans the array once, which costs Θ(n). This yields the standard recurrence T(n) = 2T(n/2) + Θ(n), with a base case T(1) = Θ(1). This recurrence solves to T(n) = Θ(n log n), matching the well-known average-case time complexity of quicksort.

Option 4: T(n) = { Θ(1) if n < 2; T(n-1) + Θ(1) otherwise }.
- Why this is unlikely: this recurrence implies a single recursive call of size n-1 with constant work per level, giving Θ(n) overall. This does not reflect quicksort’s two-way partitioning into subproblems, which is essential to its divide-and-conquer structure. Therefore, this option does not model the algorithm correctly.

Option 5: T(n) = { Θ(1) if n < 2; T(n-1) + Θ(n) otherwise }.
- Why this is unlikely: while the cost grows with n, the recurrence still only uses one subproblem of size n-1, which again ignores the two-way partitioning characteristic of quicksort. The added Θ(n) at each step would accumulate to a higher order than the actual average-case quicksort time, so this does not describe the standard average-case recurrence.

Option 6: T(n) = { Θ(1) if n < 2; T(n/2) + Θ(1) otherwise }.
- Why this is unlikely: this option shows a single recursive call on size n/2 with only constant additional work, which would lead to a much smaller overall complexity than quicksort’s average case. It ignores the two recursive branches and the linear cost of partitioning, so it does not capture the algorithm’s behavior.

In summary, among the provided possibilities, the one that aligns with the average-case analysis of quicksort is the form that has two recursive calls on roughly half-sized subproblems plus a linear partitioning cost. The remaining options either underestimate the combining cost, misrepresent the number of recursive calls, or fail to reflect the divide-and-conquer structure of quicksort.

Recr_Q_6 What is the recurrence relation for the quick_sort function given below in the average case scenario as described below? 单项选择题

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