Question

Show that the running time of the merge-sort algorithm on an \(n\) -element sequence is \(O(n \log n),\) even when \(n\) is not a power of 2 .

Short answer

\(O(n \log n)\).

Step-by-step solution

- Understand the Merge-Sort Algorithm

The merge-sort algorithm operates by dividing the sequence into two halves, sorting each half recursively, and then merging the two sorted halves back together.

- Basic Divide-and-Conquer Approach

In each step, the sequence of length n is divided into two approximately equal halves, resulting in subproblems of size n/2.

- Recurrence Relation

The recurrence relation for running time T(n) of merge sort is: \[T(n) = 2T(n/2) + O(n)\]

- Solve the Recurrence Relation

We use the Master Theorem to solve the recurrence relation. According to the Master Theorem, for the recurrence relation of the form: \[T(n) = aT(n/b) + f(n)\], where a = 2, b = 2, and f(n) = O(n), we have the case where \[f(n) = O(n^c)\] with c = 1.

- Apply the Master Theorem

Here, c = 1 and \[a/b^c = 2/(2^1) = 1\]. Since a and b^c are equal, it falls into the second case of Master Theorem, yielding: \[T(n) = O(n \text{ log } n)\]

- Conclusion for General n

This conclusion holds true even when n is not a power of 2, as the recurrence relation and the application of Master Theorem do not assume n being a power of 2.

Key concepts

Understanding the Merge-Sort Algorithm

The merge-sort algorithm is a popular sorting technique that uses the divide-and-conquer approach. It works by splitting an unsorted sequence into two halves. Each half is then sorted recursively. Finally, the two sorted halves are merged back together. This merging process ensures that the entire sequence is sorted.
Here’s how it works step-by-step:

• Divide: The sequence is divided into two equal or nearly equal halves.
• Conquer: Recursively sort each half.
• Combine: Merge the two sorted halves to create a sorted sequence.

Even if the sequence length isn’t a power of 2, the approach remains the same. The algorithm ensures that every element is compared and placed in the correct order.

Exploring Recurrence Relations in Merge-Sort

To analyze the time complexity of merge-sort, we need to understand its recurrence relation. A recurrence relation is an equation that defines a sequence based on its previous terms. For merge-sort, this relation is:
Combine step which takes O(n) Subproblems -T(n/2)-Merge Sort-T(n/2) -Final O(n) Therefore, we have:
T(n) = 2T(n/2) + O(n) This relation arises because:

• We split the sequence into two halves, leading to two subproblems each of size n/2.
• The merging process takes linear time, O(n), to combine these two halves.

Even when n is not a power of 2, the recurrence relation holds. The approach adapts and continues to divide the sequence until the base case (a sequence of length 1 or 0) is reached.

Applying the Master Theorem to Merge-Sort

The Master Theorem helps us solve recurrence relations of the form:
T(n) = aT(n/b) + f(n).
For merge-sort, we have:

• a = 2 - We break the sequence into 2 subproblems.
• b = 2 - Each subproblem is half the size of the original.
• f(n) = O(n) - The merging step takes linear time.

Using the Master Theorem, we compare f(n) with n^log_b(a) to determine the time complexity.
The Master Theorem provides three cases to consider:

• If f(n) = O(n^c) with c < log_b(a), then T(n) = O(n^log_b(a)).
• If f(n) = O(n^c) with c = log_b(a), then T(n) = O(n^log_b(a)log(n)).
• If f(n) = O(n^c) with c > log_b(a), then T(n) = O(f(n)).

For the merge-sort recurrence T(n) = 2T(n/2) + O(n), we find:

• a = 2
• b = 2
• c = 1

Since a/b^c = 1 and c = log_b(a), we fall into the second case:
T(n) = O(n log n).
This proves that the time complexity of merge-sort is O(n log n). This result holds irrespective of whether n is a power of 2 or not.