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icang [17]
4 years ago
5

Create an integer variable named ‘listSize’ and initialize the value to 1000.

Computers and Technology
1 answer:
Natalija [7]4 years ago
3 0

Answer:

The code is given in C++. The variable name is arraySize. It should be replaced with listSize

Explanation:

// header file

#include<iostream>

#include<fstream>

#include<cstdlib>

#include<chrono>

using namespace std;

using namespace std::chrono;

// bubble sort algorithm

void bubbleSort(int *a, int size) {

for (int i = 0; i<size - 1; i++) {

for (int j = 0; j<size - i - 1; j++) {

if (a[j]>a[j + 1]) {

//swap

int temp = a[j];

a[j] = a[j + 1];

a[j + 1] = temp;

}

}

}

}

// selection sort

void selectionSort(int *a, int size) {

for (int i = 0; i<size - 1; i++) {

int m_i = i;

for (int j = i + 1; j<size; j++) {

if (a[j]<a[m_i]) {

m_i = j;

}

}

//swap

int temp = a[i];

a[i] = a[m_i];

a[m_i] = a[i];

}

}

// insertion sort

void insertionSort(int *a, int size) {

for (int i = 1; i<size; ++i) {

int key = a[i];

int j=i-1;

for (; (j >= 0 && key < a[j]); j--) {

a[j + 1] = a[j];

}

a[j + 1] = key;

}

}

// make partition on array

int getPartition(int *a, int l, int h) {

int pvt = a[h];

int i = (l - 1);

for (int j = l; j <= h - 1; j++) {

if (a[j]<pvt) {

i++;

//swap

int tmp = a[i];

a[i] = a[j];

a[j] = tmp;

}

}

//swap

int tmp = a[i + 1];

a[i + 1] = a[h];

a[h] = tmp;

return(i + 1);

}

// quick sort

void quickSort(int *a, int l, int h) {

if (l<h) {

int p = getPartition(a, l, h);

quickSort(a, l, p - 1);

quickSort(a, p + 1, h);

}

}

// merge array

void merge(int *a, int l, int m, int r) {

int n1 = m - l + 1;

int n2 = r - m;

int *L = new int[n1];

int *R = new int[n2];

for (int i = 0; i<n1; i++) {

L[i] = a[i + l];

}

for (int j = 0; j<n2; j++) {

R[j] = a[m + j + 1];

}

int i = 0, j = 0, k = 1;

while (i<n1 && j<n2) {

if (L[i] <= R[j]) {

a[k] = L[i];

i++;

}

else {

a[k] = R[j];

j++;

}

k++;

}

while (i<n1) {

a[k] = L[i];

i++;

k++;

}

while (j<n2) {

a[k] = R[j];

j++;

k++;

}

}

// merge sort

void mergeSort(int *a, int l, int r) {

if (l<r) {

int m = l + (r - l) / 2;

mergeSort(a, l, m);

mergeSort(a, m + 1, r);

merge(a, l, m, r);

}

}

int* randomData(int size) {

int *ar = new int[size];

for (int i = 0; i<size; i++) {

ar[i] = rand();

}

return(ar);

}

void copyAry(int *a, int *b, int size) {

for (int i = 0; i < size; i++) {

a[i] = b[i];

}

}

int main() {

int arraySize = 1000;

int *arrLst = new int[arraySize];

int *arLst = new int[arraySize];

arrLst = randomData(arraySize);

//copy array

// because after sorting arry will be sorted

copyAry(arLst, arrLst, arraySize);

auto start1 = high_resolution_clock::now();

bubbleSort(arLst, arraySize);

auto end1 = high_resolution_clock::now();

auto duration1 = duration_cast<microseconds>(end1 - start1);

cout << duration1.count()<<endl;

// copy array for new sort

copyAry(arLst, arrLst, arraySize);

auto start2 = high_resolution_clock::now();

selectionSort(arLst, arraySize);

auto end2 = high_resolution_clock::now();

auto duration2 = duration_cast<microseconds>(end2 - start2);

cout << duration2.count() << endl;

// copy array for new sort

copyAry(arLst, arrLst, arraySize);

auto start3 = high_resolution_clock::now();

insertionSort(arLst, arraySize);

auto end3 = high_resolution_clock::now();

auto duration3 = duration_cast<microseconds>(end3 - start3);

cout << duration3.count() << endl;

// copy array for new sort

copyAry(arLst, arrLst, arraySize);

auto start4 = high_resolution_clock::now();

mergeSort(arLst, 0,arraySize-1);

auto end4 = high_resolution_clock::now();

auto duration4 = duration_cast<microseconds>(end4 - start4);

cout << duration4.count() << endl;

// copy array for new sort

copyAry(arLst, arrLst, arraySize);

auto start5 = high_resolution_clock::now();

quickSort(arLst, 0,arraySize-1);

auto end5 = high_resolution_clock::now();

auto duration5 = duration_cast<microseconds>(end5 - start5);

cout << duration5.count() << endl;

ofstream out("GroupAssignment2Results.csv");

out << "Array Size,Bubble Sort Time,Selection Sort Time,Insertion Sort Time,Quick Sort Time,Merge Sort Time\n";

out << arraySize << "," << duration1.count() << "," << duration2.count() << "," << duration3.count() << "," << duration5.count() << "," << duration4.count();

cout << "\nOutput in File.";

out.close();

cin.get();

return(0);

}

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Explanation:

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Which of the threats discussed in this chapter should receive Charlie’s attention early in his planning process?Ethical Decision
ratelena [41]

Answer:

Explanation:

1. Before the discussion at the start of this chapter, how do Fred, Gladys, and Charlie each perceive the scope and scale of the new information security effort? Did Fred’s perception change after that?

Answer:

Before the discussion, Fred, Gladys, and Charlie focused on other ends in regards to information security. Fred was more concerned with adding additional software to fix the malware issues when clearly there were easier steps that need to be taken

2. How should Fred measure success when he evaluates Gladys’ performance for this project? How should he evaluate Charlie’s performance?

Answer:

Gladys’s performance should be based on the new security measures and protocol that she has in place for the organization. This of course, is putting a lot of trust into Charlie’s performance as she was the one to introduce Charlie with his new plan on the organization’s new security. She practically had him nominated for CIO.

3. Which of the threats discussed in this chapter should receive Charlie’s attention early in his planning process?

Answer:

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Instead of Charlie being named CISO, suppose instead that Fred hired his son-in-law, an unemployed accountant, to fill the role. Assuming the person had no prior experience or preparation for a job in information security, did Fred make an ethical choice? Explain your answer.

Answer:

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CO! Oher...9
kondor19780726 [428]

Answer:

Altitude of the town(h1) = 1,196.8 m

Explanation:

Given:

Height of barometer(h) = 65 cm Hg = 0.65

Standard atmospheric pressure  = 76 cm Hg = 0.76

Density of mercury(Pm) = 13,600 kg/m³

Density of  air (Pa) = 1.25 kg/m³

Find:

Altitude of the town(h1)

Computation:

Pressure due column = Δp mercury column

(Pa)(h1) = (Pm)(h)

(1.25)(h1) = (13,600)(0.76-0.65)

(1.25)(h1) = 1,496

Altitude of the town(h1) = 1,196.8 m

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