Algoeithm Design, Analysis and Implementation Assignment

Algoeithm Design, Analysis and Implementation - Assignment ExampleThis is done by choosing a comparison element and placing all the elements that are less than the comparison element in the first group and the rest of the elements in the second group. This procedure is iterate recursively until the elements are sorted (a part consist of only one element). T(n) = (n-1) + ?1 ? i ? n ti As 1,2,....k-elements are already sorted, we can severalise that ti =0, where i = 1,2, 3... k. Then, the contribution of quick sort when early stopping is used can be attached by, T(n)=(n+1)( ?k ? i ? n ti + ?(1)) = (n+1)( n lg +?(1)) =2n lg +?(n) Thus, T(n) for quick sort =O(nlg(n/k)). Given that, insertion sort is done on a partially sorted array (unsorted k-elements). In general, running sentence of insertion sort is O(n2 ), where n is the length of the array (total number of elements). In order to provide a solution to this problem, the total array is divided into subarrays of k-elements for eac h one, much(prenominal) that k/2? n ? k, then n = O(k) and the running fourth dimension of insertion sort is O(k2). The total number of such subarrays (m) would then be n/k ? m ? 2n/k., which implies m = O(n/k). The total time spent on insertion sort would then be O(k2)* O(n/k) = O(nk). T(n) for insertion sort = O(nk). Therefore, the total time for this sorting algorithm is as follows T(n) = O(nk + nlg(n/k) ). ... Solution From the above problem (1), we find that quick sort sorts k-elements of an n-element array O(n log(n/k)) time. Quick sort sorts by partitioning the given array Ap...r into two sub-arrays Ap...q and Aq+1... r such that every element in Ap...q is less than, or fair to middling to, elements in Aq+1... r. This process is repeated until all the elements are sorted. algorithmic rule for quick sort is given by AP is the pivot key upon which the comparison is made. P is chosen as the median value of the array at each step. If the element is less than, or equal to, the pivot key value, it is moved left. Otherwise, it is moved right. Assuming the best case scenario where each step produces two equal partitions, then T(n)=T(n/2)+T(n/2)+?(n) =2T(n/2)+ ?(n) By Masters Theorem case 2, T(n) = O(n lg n) In other words, the depth of recursion is log n and at each level/step, the number of elements to be treated is n. If only k-elements are sorted, then the depth of recursion would be n/k and the number of elements would be n at each level, the time interpreted by this sorting algorithm is given by T(n) = O(n lg (n/k)). 2.2 Show that we can sort a k-well-sorted array of length n in O(n log k) time. As the array is already sorted for k-elements, the remaining steps required to complete the sort would be k (using the results from 1), then T(n) = O(n lg k). 3. Computing the k-th smallest element in the junction of the two lists m and n using O(lg m +lg n) time algorithm Approach 1 Merge sort can be used in this case. It splits the list into two halves, recu rsively sorts each half, and then merges the two sorted sub-lists. In the given problem, the lists are already sorted hence, the