Example of SELECT (k, S) algorithm S = < 1, 10, 8, 5, 6, 3, 5, 4, 2, 4, 7, 5, 8, 8, 9, 24, 6, 16, 20, 11, 4, 3, 14, 19, 2 > Divide into subsequences of 5 and sort each: 1 2 5 6 2 5 3 7 11 3 6 4 8 16 4 M, the sequence of medians 8 4 8 10 10 5 9 220 14 24 19 Find the median m of M, and rearrange the columns: 2 2 1 5 6 3 3 5 7 11 4 4 6 8 16 m, the median of M 4 14 8 8 5 19 10 9 220 20 24 S1 = < 1, 5, 3, 5, 4, 2, 4, 5, 4, 3, 2> S2 = < 6,6> S3 = <10, 8, 7, 8, 8, 9, 24, 16, 20, 11, 14, 19 > SELECT (5, S) SELECT (5, S1) SELECT (12, S) SELECT (13, S) SELECT (17, S) SELECT (20, S) answer is 6 answer is 6 SELECT (4, S3) SELECT (7, S3) etc. |S1|= 11 | S2 | = 2 |s3| = 12 2 2 IA IɅ N≤3 ≤ + 4 IA 3 N≤M≤ < 4 IɅ VI ΙΛ 15516) IɅ 5 7 8 00 6 11 16 elements known to be less than or equal to m 4 14 8 8 20 VI 5 19 10 9 24 16 elements known to be greater than or equal to m 20 2 2 1 5 6 3 3 5 7 11 4 4 6 VI 8 VI IA ΙΛ 4 14 8 8 ΙΛ IɅ 5 19 10 9 6≤2≤2 24 Those elements in the green box are known to be less than or equal to m, the number of such elements is at least (3/5) x (1/2) x n = (3/10)n. Contrapositively, the number of elements greater than m is at most n - (3/10) n = (7/10) n. Therefore, the recursive call at Line 12, SELECT (k - |S1| - |S2, S3) takes time at most T(7n/10). Similarly, the elements in the blue box are guaranteed to be greater than or equal to m, and the number of such elements is at least (3/10)n. So the number of elements less than m is at most (7/10)n, and the recursive call at Line 10, SELECT (k, S1) also takes time at most T(7n/10). Note that depend on the choice for the length of the subsequences (in the algorithm discussed in class and the example above, the subsequences are of length 5) and n (number of elements in S), the two boxes may not be of equal size, in which case the recursive calls at Line 12 and Line 10 (observe that only one of the two may be executed) will have different complexities, and the larger one will dominate the expression for T(n).

Python please compare_sorts.py import random #Import module for generating random numbersimport time #Import module for getting the current timemaxvalue = 1000def merge(left, right):result = []left_idx, right_idx = 0, 0while left_idx < len(left) and right_idx < len(right):# to change direction of sort, change direction of comparisonif left[left_idx] <= right[right_idx]:result.append(left[left_idx])left_idx += 1else:result.append(right[right_idx])right_idx += 1if left:result.extend(left[left_idx:])if right:result.extend(right[right_idx:])return resultdef merge_sort(m):if len(m) <= 1:return mmiddle = len(m) // 2left = m[:middle]right = m[middle:]left = merge_sort(left)right = merge_sort(right)return list(merge(left, right))def insertion_sort(array):for slot in range(1, len(array)):value = array[slot]test_slot = slot - 1while test_slot > -1 and array[test_slot] > value:array[test_slot + 1] = array[test_slot]test_slot = test_slot - 1array[test_slot + 1] = valuereturn…

The dotted-decimal notation for the IP address 10010000 10101010 01011011 00101000 The dotted-decimal notation for the IP address 10000001 10100101 01010011 00110011 ​2 15​ 19​ 30​ 41​ 59​ 77​ 81​ 85​ 100​   Search for element 45 in the list above using a binary search algorithm. Clearly, show the first, middle, and last indexes as well as the target in every iteration. Show all the steps taken. Use the table below as a guideline with the provided format.   Instructions: Start your index at 0 Show all steps Mark the first row as header(The text will be in bold). Use the index number to show the first, last, and mid Use the floor function to find the middle element If struggling to insert a table, you may work this from a Word document and upload your work using the plus sign on the editing options.   ​First last​ mid​ 0​ 1​ 2​ 3​ 4​ 5​ 6​ 7​ 8​ 9​ target​ = 45

Q1Apply selection sort for the following list to sort it in ascending order. Record your answer for the first two passes. 80 Pass 1 29 Pass 2 73 43 97 How many comparisons and exchanges are done in each pass of the selection sort algorithm? Comparisons Passes Exchanges 52 Original Pass 1 Pass 2

Merge sort is an efficient sorting algorithm with a time complexity of O(n log n). This means that as the number of elements (chocolates or students) increases significantly, the efficiency of merge sort remains relatively stable compared to other sorting algorithms. Merge sort achieves this efficiency by recursively dividing the input array into smaller sub-arrays, sorting them individually, and then merging them back together. The efficiency of merge sort is primarily determined by its time complexity, which is , where n is the number of elements in the array. This time complexity indicates that the time taken by merge sort grows logarithmically with the size of the input array. Therefore, even as the number of chocolates or students increases significantly, merge sort maintains its relatively efficient performance. Regarding the distribution of a given set of x to y using iterative and recursive functions, the complexity analysis depends on the specific implementation of each…

Array x is [a, b, c, . . . , y] has 25 integers. To find median of medians we form five sub-arrays: [d, c, b, e, a], [i, f, h, j, g], [1, m, k, n, o], [p, s, q, r, t], [w, x, v, u, y]. We find median from these sub-arrays and generate new sorted array [k, h, q, b, v]. Median of new sorted array is q, so median of medians is q. List all elements guaranteed to be greater or equal to q.

Question Consider the following definition of a MysteryBox object: public class MysteryBox { private final int x0, x1, x2, x3; private final double y0; private final boolean z0; private final long[] a = new long [248]; } Using the 64-bit memory cost model from lecture, how many bytes does each object of type MysteryBox consume? Include all memory referenced by the object (deep memory). Answer Your answer should be a positive integer that is a multiple of 8.

I am learning the Algorithm for Order Statistics with the Select Recursive Procedure as shown in the pictures. In the algorithm, we separate out n into separate sequences of 5 elements each or n/5. I was wondering if the same algorithm would work if you did n/6, n/7, or n/10 instead and why. Thank you.

Below is an array of elements that can be traversed via Linear and Binary search. You will be given five elements for each searching algorithms. You need to list down the elements traversed until it reaches to the element we are looking for: 33 54 74 124 148 248 274 203 443 445 452 457 470 512 520 505 2 10 11 12 13 14 15 570 500 634 730 783 803 810 886 030 10 17 18 19 20 21 22 23 24 25 20 27 28 29 30 31 A. Linear Search 1. 763 2. 148 3. 476 4. 526 5. 74 B. Binary Search 1. 985 2. 590 3. 476 4. 887 5. 897

In the following Quick Sort algorithm, the last element in the array is selected as Pivot. We are going to select the Median of the data in the array as Pivot.  Make the appropriate changes in the following algorithms. Indicate which line (line number)  in each algorithm is to be modified and what are the changes.  In the original algorithm, the complexity of Quick Sort is "n Log (n)". What would be the complexity with Median as Pivot? Why?