Question

Prove by induction that the solution to the Merge sort recurrence,given below, is O(nlogn). T(n)<=2T(n/2)+n, for n>1,T(1)=1.

Answer 1

To prove by induction that the solution to the Merge sort recurrence is O(nlogn), we can establish a base case and then assume the inequality holds for all values of k up to n. By applying the merge sort recurrence and simplifying the equation, we can show that the time complexity is bounded by O(nlogn).

Step-by-step explanation:

To prove by induction that the solution to the Merge sort recurrence is O(nlogn), we will first establish a base case. When n = 1, T(1) = 1. Now, let's assume that the inequality T(k) ≤ cklog(k) holds for all values of k up to n. We can then substitute n/2 for k in the equation to get T(n/2) ≤ c(n/2)log(n/2).

Now, applying the merge sort recurrence to the inequality T(n) ≤ 2T(n/2) + n, we have:

1. T(n) ≤ 2(c(n/2)log(n/2)) + n
2. T(n) ≤ clog(n/2) + n
3. T(n) ≤ c(log(n)-1) + n
4. T(n) ≤ clog(n) + c - 1 + n
5. T(n) ≤ clog(n) + (c-1+n)
6. T(n) ≤ clog(n) + (c+n)
7. T(n) ≤ clog(n) + dn, where d = c+n

So, we have proven that T(n) ≤ clog(n) + dn. This inequality represents the Big O notation, where O(nlogn) is an upper bound on the time complexity of the merge sort algorithm.