Final answer:
The solution to the merge sort recurrence was proven to be O(n log n) through mathematical induction, starting with a base case for n=1, assuming the induction hypothesis holds for all halves of n, and then applying the induction hypothesis to show that it holds for n.
Step-by-step explanation:
The question asks to prove by induction that the solution to the merge sort recurrence is O(n log n).
Merge sort is an efficient, comparison-based, divide and conquer sorting algorithm that divides the input list into two halves, recursively sorts both halves, and then merges the sorted halves.
The recurrence relation for merge sort is T(n) = 2T(n/2) + n when n > 1 and T(n) = 0 when n = 1, where T(n) represents the time complexity to sort an array of n elements.
Induction Proof:
Base Case: For n = 1, the statement T(n) = n log n holds true as T(1) = 0 and 1 log 1 = 0. Thus, the base case is satisfied.
Induction Hypothesis: Assume T(k) = k log k for all n/2 ≤ k < n holds true.
Inductive Step: For n elements, the recurrence is T(n) = 2T(n/2) + n. By the induction hypothesis, this becomes
, which simplifies to T(n) = n(log n - 1) + n.
After simplification, we get T(n) = n log n, proving that the solution to the merge sort recurrence is indeed O(n log n).
Mathematical Induction - Merge sort: T(n) =2T(n/2) + O(n), T(2)=2 - Guess T(n) is O(nlgn) - Verify the guess by induction Merge sort: T(n) =2T(n/2) + n Use Mathematical induction to show that when n is exact power of 2, the solution of the recurrence is e(nlgn) Thus, we want to show that T(n) <c nlgn (it is sufficient for us to show it is Big-oh. The same analysis can be applied to $2()). Base Step: T(2) =2 • T(2) Sc 2lg2 2 Sc(2) ► 021 • Thus, let c =1 • Inductive Hypothesis • Assume T(n) Scnlgn for all n=2k, k>0 • Inductive step