Show that the maximum number of comparisons performed by the Insertion Sort algorithm (Algorithm 7.1 ) is achieved when the keys are inputted in nonincreasing order.

Short Answer

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When the keys are inputted in nonincreasing order, each new key is smaller than all previously sorted keys. This forces the Insertion sort to make the maximum number of comparisons, totaling \((n-1)(n)/2\) comparisons for n elements. Therefore, the worst-case time complexity of the Insertion Sort is \(O(n^2)\).

Step by step solution

01

Understanding Insertion Sort Algorithm

Insertion Sort is a simple sorting algorithm that works by building a sorted sequence one element at a time. The algorithm starts from the second element and compares it with the first, and if the second element is smaller, they are swapped. Then we move to the third element, compare it with the second and the first, and place it in the right position in the sorted subsequence, so the sequence becomes sorted up to the third element. This process is repeated until all elements are sorted.
02

Nonincreasing Order as Worst-Case Scenario

When the elements are in nonincreasing order, each new element is smaller than all the elements in the already sorted subsequence. Therefore, every time we insert a new element, we need to compare it with all the elements already sorted. For example, for inserting the nth element, it needs to be compared with n-1 elements. This means we have to do a maximum number of comparisons.
03

Calculating the Maximum Number of Comparisons

The maximum number of comparisons that Insertion Sort can make is the sum of the first n-1 natural numbers, because for the 2nd element we make 1 comparison, for the 3rd two comparisons, and so on, up to n-1 comparisons for the nth element. The sum of the first n-1 natural numbers is given by the formula \((n-1)(n)/2\), which results in a time complexity of \(O(n^2)\), which is the worst-case time complexity of Insertion Sort.

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