a. Derivation of the Average-Case Time Complexity Expression

The average-case time complexity of Quick Sort depends on the position of the pivot, $p$, which partitions the array into two subarrays: one with $p$ elements and the other with $n-p-1$ elements. At each recursive step, we perform comparisons and partition the array around the pivot.

The recurrence relation for the time complexity can be expressed as: $T(n, p) = T(p) + T(n - p - 1) + C(n),$ where:

• $T(p)$ is the time complexity for sorting the left partition.
• $T(n-p-1)$ is the time complexity for sorting the right partition.
• $C(n)$ represents the comparisons made at the current level, which is proportional to $n-1$ because the pivot is compared with every other element.

Expression for $C(n)$:

At each level, the pivot is compared with $n-1$ elements, so: $C(n) = n-1.$

Incorporating $C(n)$ into the recurrence: $T(n, p) = T(p) + T(n-p-1) + (n-1).$

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b. Simplification to Represent the Total Number of Comparisons

To calculate the average-case time complexity, we consider the average pivot position. Since the pivot is chosen randomly, the probability of selecting any element as the pivot is uniform. For a pivot $p$, we need to average the contributions of all possible pivot positions $p$ from $0$ to $n-1$.

Average-Case Recurrence:

The average-case recurrence is given by summing over all possible pivot positions and dividing by $n$: $T(n) = \frac{1}{n} \sum_{p=0}^{n-1} \left[ T(p) + T(n-p-1) \right] + (n-1).$

Simplify the summation: $T(n) = \frac{1}{n} \left( \sum_{p=0}^{n-1} T(p) + \sum_{p=0}^{n-1} T(n-p-1) \right) + (n-1).$

Notice that $\sum_{p=0}^{n-1} T(n-p-1)$ is equivalent to $\sum_{p=0}^{n-1} T(p)$, so: $T(n) = \frac{2}{n} \sum_{p=0}^{n-1} T(p) + (n-1).$

Solve the Recurrence:

Using substitution, we analyze $T(n)$ asymptotically. Assume $T(n) \propto n \log n$ (based on intuition from the Quick Sort algorithm's behavior): $T(n) = \frac{2}{n} \cdot \frac{n(n-1)}{2} + (n-1),$ where the summation of $T(p)$ approximates to $\frac{n(n-1)}{2}$.

Simplify: $T(n) \approx 2H(n) + O(n),$ where $H(n) = \log n$.

Final Simplified Expression:

$T(n) = O(n \log n).$

This reflects the average-case time complexity of Quick Sort, where the pivot position is chosen uniformly at random.

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Do you have any questions?

Relative Questions:

1. How does pivot selection affect the worst-case time complexity of Quick Sort?
2. What is the role of $C(n)$ in determining the number of comparisons in Quick Sort?
3. Can you explain the difference between best-case, average-case, and worst-case time complexity for Quick Sort?
4. How would the time complexity change if the pivot is always the first or last element?
5. Why does Quick Sort have an average-case complexity of $O(n \log n)$ despite the randomness of pivot selection?

Tip: Always use a randomized pivot selection or a median-of-three strategy to minimize the likelihood of encountering the worst-case scenario in Quick Sort.