Consider two sorting algorithms and a sequence A below and answer the followingquestions.A = {55,12,53,56,24,1,7,42}[ALGORITH I: INSERTION-SORT (A) ]1 for j= 2 to A.length2keyA[j]3// insert A[j]into the sorted sequence A[1 ... j- 1|i = j- 15while i > 0 and A[i] > keyA[i + 1] = A[i]7i = j – 1|8A[i + 1] = key[ALGORITHM II: SELECTION-SORT (A)]1 for i = 2 to A.length – 1- minlndex 3D і3for j= i+1 to A.length%3D4if A[j] < A[minIndex] and j = minIndexminlndex —і6.swap A[i] with A[minIndex]5.1. Find the computational complexity for ALGORITHM I? Justify your answer.5.2. Does ALGORITHM II have the same computational complexity? Justify your answer.5.3. Sort the sequence A using ALGORITHM I and show every change in A.5.4. Sort the sequence A using ALGORITHM II and show every change in A.

А Sоrting Аlgоrithm is used tо reаrrаnge а given аrrаy оr list elements ассоrding tо…

Answered: 5.1. Find the computational complexity…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Binary search Consider the binary search algorithm implementation below. def binarySearch[T](arr: Array[T], e: T) (using ord: Ordering[T]) = def inner (lo: Int, hi: Int): Boolean if lo <= hi then val mid = lo + ((hi lo) / 2) val cmp = ord.compare(e, arr(mid)) if cmp ==0 then true = // e arr(mid) else if cmp < 0 then inner(lo, mid-1) // e ‹ arr(mid) else inner(mid+1, hi) // e › arr(mid) else false inner(0, arr.length-1) Accessed indices: == Simulate its execution on the sorted array below when searching for the value -8. Click on the array elements in the same order the algorithm accesses them. Note: submission, reloading the page, and a session timeout will create a new instance! o 3 4 5 6 7 19 10 11 12 13 14 15
1)If we consider Recurrences and Running Time, what is the running time T(n) for size of an array n? BINARY-SEARCH (A, lo, hi, x) { if (lo > hi) return FALSE mid = (lo + hi)/2 if x = A[mid]Assignment 1 return TRUE if (x < A[mid]) BINARY-SEARCH (A, lo, mid-1, x) if (x > A[mid]) BINARY-SEARCH (A, mid+1, hi, x) } 2)6. Based on T(n) that you found on previous question, solve recurrence with iteration method.
Consider the following array of numbers A = {15, 4, 21, 12, 10, 0, 21, 12, 3, 14}. a. Show the result of the various passes needed to sort A : i. Using Insertion sort ii. Using Bubble sort iii. Using Selection sort iv. Using RadixSort (r = 5) b. Use a binary tree to show the different recursive calls resulting from using Merge Sort to sort A %3D
Consider a list L = {387, 690, 234, 435, 567, 123, 441}. Here, the number of elements n = 7, the number of digits l = 3 and radix r = 10. This means that radix sort would require 10 bins and would complete the sorting in 3 passes. Algorithm to illustrates the skeletal procedure for the LSD first radix sort.
Consider the following algorithm that uses a sorted list of n elements (alist). What is the worst case runtime of this algorithm? for each element in alist 1. ask the user for an input, call it value 2. search value in alist using binary search 3. if value exists in alist, print "successful" otherwise print "unsuccessful" Question options: a. O(log n) b. O(n) c. O(n log n) d. O(2^n) e. O(n^2) f. O(1)
In searching an element in an array, linear search can be used, even though simple to implement, but not efficient, with only O(n) time complexity. Assuming the array is already in sorted order, modify the search function below, using a better algorithm, so the average time complexity for the search function is O(log n). include <iostream> using namespace std; int search(int al), int s, int v) { 1/ Modify below codes. for (int i = 0; i <s; i++) { if (a[i] = v) return i; return -1; int main() { int intArray:10] = { 5, 7, 8, 9, 10, 12, 13, 15, 20, 34); // Search for element '12' in 10-elements integer array. cout << search(intArray, 10, 12); // '5' will be printed out. // Search for element '35' in 10-elements integer array. cout << search(intArray, 10, 35); // '-1' will be printed out. // Index '-l' means that the element is not found. return 0;
Create an algorithm that demonstrates the Fibonacci search procedure. The number of data elements n is chosen in such a way that: I Fk+1 > (n+1); and ii) Fk + m = (n +1) for some m 0, where Fk+1 and Fk are two consecutive Fibonacci numbers.
Linear Sort HereDATA is anarraywith Nelements. This algorithm sorts theelements. The algorithms sortthe elements in DATA. Algorithm A2: (Linear Sort)Linear(DATA,N) Repeatstep2and 3 forK=1 to N-1 Set PTR :=K+1 [Initialize pass pointerPTR] Repeat whilePTR£ N [Execute Pass](a).IfDATA[K]> DATA[PTR], then: Interchange DATA[K] andDATA [PTR][End ofIfstructure] (b)Set PTR: =PTR+1 [End ofinnerloop] [End ofstep 1 outer loop] Exit SelectionSort Selectionsortperformssortingbyrepeatedlyputtingthelargestelementintheunprocessedportion of thearrayto the end ofthe unprocessedportion until the whole arrayis sorted. Algorithm A2: (Selection Sort) SELECTION(A,N)This algorithm sorts thearrayA with Nelements. Initialize aloop: For(i =0;i <N; i++)SetMIN:=A[i] Initialize anotherloopFor(j =i+1 ; j<N ; j++) If (MIN >A[j]) SWAP(MIN,A[j]) Exit InsertionSort…
Linear Sort HereDATA is anarraywith Nelements. This algorithm sorts theelements. The algorithms sortthe elements in DATA. Algorithm A2: (Linear Sort)Linear(DATA,N) Repeatstep2and 3 forK=1 to N-1 Set PTR :=K+1 [Initialize pass pointerPTR] Repeat whilePTR£ N [Execute Pass](a).IfDATA[K]> DATA[PTR], then: Interchange DATA[K] andDATA [PTR][End ofIfstructure] (b)Set PTR: =PTR+1 [End ofinnerloop] [End ofstep 1 outer loop] Exit SelectionSort Selectionsortperformssortingbyrepeatedlyputtingthelargestelementintheunprocessedportion of thearrayto the end ofthe unprocessedportion until the whole arrayis sorted. Algorithm A2: (Selection Sort) SELECTION(A,N)This algorithm sorts thearrayA with Nelements. Initialize aloop: For(i =0;i <N; i++)SetMIN:=A[i] Initialize anotherloopFor(j =i+1 ; j<N ; j++) If (MIN >A[j]) SWAP(MIN,A[j]) Exit InsertionSort…
Given a sorted array of size N and an integer K, find the position at which K is present in the array using binary search. Example 1: Input: N = 5 arr[] = {1 2 3 4 5} K = 4 Output: 3 Explanation: 4 appears at index 3.
Given below is the Randomized Quick Sort Algorithm, where p and r represent lower and upper bounds of array A respectively. RANDOMIZED-PARTITION (A, p, r) 1 i = RANDOM(p,r) 2 exchange A[r] with A[i] 3 return PARTITION (A, p, r) RANDOMIZED-QUICKSORT if p
Python code for the following algorithms? Insertion Sort InsertionSort(A, n){for i  1 to ntemp  a[i]for j  i where j>0 and temp < a[j-1]a[j]  a[j-1] // shift elementsa[j]  temp // insert} Selection Sort SelectionSort(A, n) {for i  0 to n-1min  ifor j  i+1 to nif (A[j] < A[min])min  jswap(A[i], A[min])}swap( &lhs, &rhs ){tmp  lhslhs  rhsrhs  tmp}

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