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Sorting Algorithms

Highly-organized data can be critical for many algorithms, and often you want your data ordered from least to greatest. The art of getting your data in order is trickier than you might think!

Level 3


A large group of dogs have their collars removed before they enter a dog facility. Except for their sizes, the collars are indistinguishable. When they exit the facility they will all need their collars back. Your task is to assign every dog to their corresponding collar. The catch is that you are not able to compare collar to collar or dog to dog. You can only compare a dog to a collar and see if it is too small, too big or if it fits. You want to minimize the comparisons you make. The best way of solving this problem requires a modification of one of the following sorting algorithms. Which one is it?

We know that the insertion sort algorithm has a worth-case running time of \(O(n^{2})\) in general case. What if we could bound the algorithm for specific instance of the problem. One way to do that is by checking the number of swap operations needed by insertion algorithm to sort a list.

In the permutation of \(a_{1}\cdots a_{n}\) of \(n\) distinct elements the list inversion are pair of elements \((a_{i},a_{j})\) such that \(i<j\) and \(a_{i} > a_{j}\). What is the worst-case running time if the inputs are restricted to permutations of \(1\cdots n\) with at most \(n\) inversions?

Consider the modification of a 4 way merge sort which instead of dividing an array into two subarrays, 4-way merge sort divides the array into four sub-arrays and sorts each individual array recursively.

In the 2-way merge sort we have an index for each of the two sorted sub-arrays and we compare the elements they are pointing to and in worst case we perform \(2k-1\) comparison where \(k\) is the length of each array. Similarly in a \(4\)-way merge sort each of size \(k\) we have and index for each of the four arrays. It takes \(3\) comparisons to determine the smallest of the four. In worst case we must do this until each list has one element left for a total of \(12(k-1)\) comparison. Finally we perform \(3+2+1\) comparisons to finish the remaining list, thus for a total of \(12k-6\) comparisons.

Based on the above merge procedure which of the following represents the correct running time for a \(4\) way merge sort?

Details and Assumptions

-Ignore constant terms.

Is it possible to come up with a better worst case running time? is it asymptotically better?


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