QUICKSORT(A, p, r)if p < rthen q + PARTITION(A, p, r)QUICKSORT(A, p, q - 1)QUICKSORT(A, q + 1, r)where the PARTITION procedure is as follows:PARTITION (A, p, r)x - A[r]i-p-1for j - p to r - 1do if A[j] ≤ xthen i + i + 1exchange A[i]exchange A[i+1] → A[r]return i + 1A[j]Draw the flowchart of the above algorithm.• Draw the corresponding graph and label the nodes as n1, n2, ... and edges as e1,e2, ...• Calculate the cyclomatic complexity of the above algorithm

QUICKSORT(A, p, r) if p < r then q - PARTITION(A, p, r) QUICKSORT(A, p. q - 1) QUICKSORT(A, q + 1, r) where the PARTITION procedure is as follows: PARTITION (A, p, r) x - A[r] i-p-1 for j - p to r - 1 do if A[j] ≤ x then i + i + 1 exchange A[i] → A[j] exchange A[i+1] → A[r] return i + 1 Draw the flowchart of the above algorithm. • Draw the corresponding graph and label the nodes as n1, n2, ... and edges as e1,e2, ... • Calculate the cyclomatic complexity of the above algorithm

QUICKSORT(A, p, r) if p < r then q - PARTITION(A, p, r) QUICKSORT(A, p. q - 1) QUICKSORT(A, q + 1, r) where the PARTITION procedure is as follows: PARTITION (A, p, r) x - A[r] i-p-1 for j - p to r - 1 do if A[j] ≤ x then i + i + 1 exchange A[i] → A[j] exchange A[i+1] → A[r] return i + 1 Draw the flowchart of the above algorithm. • Draw the corresponding graph and label the nodes as n1, n2, ... and edges as e1,e2, ... • Calculate the cyclomatic complexity of the above algorithm

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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QUICKSORT(A, p, r) if p < r then q ← PARTITION(A, p, r) QUICKSORT(A, p, q − 1)QUICKSORT(A, q + 1, r) where the PARTITION procedure is as follows:PARTITION(A, p, r) x ← A[r]i ← p − 1for j ← p to r − 1 do if A[j] ≤ x then i ← i + 1 exchange A[i] ↔ A[j] exchange A[i + 1] ↔ A[r]return i + 1 Draw the flowchart of the above algorithm. Draw the corresponding graph and label the nodes as n1, n2, … and edges as e1, e2, … Calculate the cyclomatic complexity of the above algorithm
Lomuto's partitioning algorithm Consider the implementation of Lomuto's partitioning algorithm shown below. def swap(a: Array[Int], i: Int, j: Int): Unit = val t = a(i); a(i) a(j); a(j) = t /* Partition the sub-array a(lo..hi). * Return the index of the pivot element in the result. */ def partition(a: Array[Int], lo: Int, hi: Int): Int = val pivot a(hi) var i lo var j = lo while j < hi do if a(j) <= pivot then swap(a, i, j) i += 1 j += 1 swap(a, i, hi) i = Assume that the array to be partitioned is a(0..6). The pivot element has already been moved to be the element a(6). Simulate the behaviour of the algorithm by showing which element swaps it performs. To perform a swap, click on the two elements in question. Remember to include possible swaps between element and itself, as well as the final swap moving the pivot in its correct place. 0 1 13 9 Swaps performed: 2 3 6 4 15 5 5 6 7 8 13 11 17 19
static E extends Comparable> int partition(E[] A, int l, int r, E pivot) { 27 28 29 30 31 32 33 34 35 36 37 38 39 40 do {// Move bounds inward until they meet while (A[++1].compareTo(pivot)0)); DSutil.swap(A, 1, r); } while (1 < r); DSutil.swap(A, 1, r); return 1; } // Swap out-of-place values // Stop when they cross // Reverse last, wasted swap // Return first position in right partition 2.3 (2.3.1)Produce a CFG of the partition function. Leverage the table based def-use pair approach in Question 1, and produce the table below for variable pivot. node i dcu(v,i) dpu(v,i)
import java.util.Scanner;import java.util.ArrayList; public class UserIDSorting {// TODO: Write the partitioning algorithm - pick the middle element as the // pivot, compare the values using two index variables l and h (low and high), // initialized to the left and right sides of the current elements being sorted,// and determine if a swap is necessarypublic static int partition(ArrayList<String> userIDs, int i, int k) {} // TODO: Write the quicksort algorithm that recursively sorts the low and // high partitionspublic static void quicksort(ArrayList<String> userIDs, int i, int k) {} public static void main(String[] args) {Scanner scnr = new Scanner(System.in); ArrayList<String> userIDList = new ArrayList<String>(); String userID; userID = scnr.next();while (!userID.equals("-1")) {userIDList.add(userID);userID = scnr.next();}// Initial call to quicksort quicksort(userIDList, 0, userIDList.size() - 1); for (int i = 0; i < userIDList.size(); ++i)…
A = [15, 12, 13, 19, 14, 10, 16, 20, 9, 18, 8, 7]B = [19, 14, 8, 16, 20, 9, 18, 15, 12, 13, 7, 10]vN = 0for i in range(len (A)):vN = A[i]for j in range(len (B)):if B[j] ==vN:print ('A[',i,'] with B[',j,']') implementation this code with JUST one for loop
in this code: def insertSort(A): for i in range(1,len(A)): j = i-1 while A[j]>A[j+1] and j>=0: A[j],A[j+1] = A[j+1],A[j] j -= 1 A = [3,1,11,9,10,2] insertSort(A) print(A) 1) What is the Hypothesis? 2) What is Your Induction
Professor JAson uses the following algorithm for merging k sorted lists, each having n/k elements. He takes the first list and merges it with the second list using a linear-time algorithm for merging two sorted lists, such as the merging algorithm used in mergesort. Then, he merges the resulting list of 2n/k elements with the third list, merges the list of 3n/k elements that results with the fourth list, and so forth, until he ends up with a single sorted list of all n elements. Analyze the worst-case running time of the professor’s algorithm in terms of n and k.
ALGORITHM Quicksort(A[l..r])//Sorts a subarray by quicksort//Input: Subarray of array A[0..n − 1], defined by its left and right// indices l and r//Output: Subarray A[l..r] sorted in nondecreasing order if l < r s = Partition(A[l..r]) //s is a split position Quicksort(A[l..s − 1]) Quicksort(A[s + 1..r]) The above code is the quicksort algorithm.Perform a time complexity analysis for quicksort. You need to explain 1) what makes a worst case, what makes a best case, 2) the time complexity for both the worst case and the best case, 3) how to avoid the worst case in practice.
Consider the following list of elements (M-2) 18, 40, 16, 82, 64, 67, 57, 50, 37, 47, 72,14, 17, 27, 35 Using quicksort algorithm, Use pivot as the median of the first, last, and middle elements of the list. a. What is the pivot? b. Give the resulting list after one call to the partition procedure.
Question 2 Choose the correct answer after considering the following declarations: map m; unordered_map um; O None of these answers. O m.find("X") runs in the same Big-Oh time as find(m.begin(),m.end(), "X") O m.find("X") runs in O(1) O um.find("X") runs in O(n log n) O find(m.begin(),m.end(), "X") runs in O(n log n)
please convert the code to C language //C++ program to check if two arrays //are equal or not #include <bits/stdc++.h>using namespace std; bool similar_array(vector<int> arr1, vector<int> arr2){ //create teo different hash table where for each key //the hash function is h(arr[i])=arr[i] //we will use stl map as hash table and //will keep frequency stored //so say two keys arr[0] and arr[5] are mapping to //the same location, then the location will have value 2 //instead of the keys itself //if two hash tables are exactly same then //we can say that our arrays are similar map<int, int> hash1; map<int, int> hash2; //for each number for (int i = 0, j = 0; i < arr1.size(); i++, j++) { hash1[arr1[i]]++; hash2[arr2[i]]++; } //now check whether hash tables are exactly same or not for (auto it = hash1.begin(), ij = hash2.begin(); it != hash1.end() && ij != hash2.end(); it++, ij++) {…
in h partition algorithm what is meaning of installation : repeat j ← j − 1 until A[j] ≤ pivot