Question

It turns out that the recursive calls don’t have to overlap as much. Let R ∈ (0, 1) be some constant strictly between 0 and 1 in the following sorting algorithm: OVERLAPSORT2(A[1, . . . , n]): if n ≤ 10/(1 − R) INSERTIONSORT(A) else m := ⌈Rn⌉ OVERLAPSORT2(A[1, . . . ,m]) OVERLAPSORT2(A[n − m + 1, . . . , n]) OVERLAPSORT2(A[1, . . . ,m]) Don’t worry about the base case—it’s just to make sure that m < n.

(a) How small can you make R and still have a valid sorting algorithm? Give a brief informal justification as to why you can’t make it any smaller than your chosen value of R. (Hint: consider an array sorted in decreasing order)

Answer 1

The minimal value of R that still allows OVERLAPSORT2 to function as a valid sorting algorithm must be slightly over 0.5 to ensure all elements are compared and sorted, particularly in worst-case scenarios like inversely sorted arrays.

Step-by-step explanation:

The student is asking about the smallest possible value of R that can still result in a valid sorting algorithm in the context of a recursive sorting algorithm referred to as OVERLAPSORT2. For the algorithm to be valid, it must ensure that all elements are compared and sorted correctly. In the mentioned algorithm, the recursive calls involve sorting subarrays of different sizes, which means that the array is being divided into overlapping parts for sorting.

The smallest value of R that ensures a valid sorting algorithm must guarantee that the sorting includes all elements of the array. Imagine the worst-case scenario where the array is sorted in reverse order. Using a very small R-value, such as 0.5, will lead to a situation where the middle portion of the array might never get sorted. Therefore, the smallest R can be is slightly more than 0.5, ensuring that each recursive call has enough overlap to sort all elements encountered in the previous calls. Any smaller and we could end up with unsorted middle sections, hence invalidating the sorting algorithm.