1. Compare the similarities and differences between traditional computing and the computing clouds launched in recent years. Also discuss possible convergence of the two computing paradigms in the future.
2. An increasing number of organizations in industry and business adopt cloud systems. Answer the following questions regarding cloud computing:
a. List and describe the main characteristics of cloud computing systems.
b. Discuss key enabling technologies in cloud computing systems.
c. Discuss different ways for cloud service providers to maximize their revenues.
d. Characterize the three cloud computing models using suitable examples.

Answer:

The actual cost of  (n) stack operations will be two ( charge for push and pop )

Explanation:

The cost of (n) stack operations involves two charges which are actual cost of operation of the stack which includes charge for pop ( poping an item into the stack) which is one and the charge for push ( pushing an item into the stack ) which is also one. and the charge for copy which is zero. therefore the maximum size of the stack operation can never exceed K because for every k operation there are k units left.

The amortized cost of the stack operation is constant 0 ( 1 ) and the cost of performing an operation on a stack made up of n elements will be assigned as 0 ( n )

The amortized cost of n stack operations, including backup copying after every k operations, remains O(n). This is achieved by assigning an amortized cost of 2 to each operation and distributing the copying cost evenly.

To analyze the cost of stack operations including backups, we consider the sequence of operations and calculate the total cost over time, assigning an amortized cost. Here’s the breakdown:

Since each operation has an amortized cost of 2, and every k operations involve an additional O(k) time for copying, the total cost for n operations is O(n) over time. This is because the cumulative effect of the every-k copying is distributed evenly across the individual operations, ensuring that the average cost per operation remains constant.

Therefore, the total cost comprises the individual stack operations plus the copying operations, yielding an overall cost of O(n).

Answer:

Check the explanation

Explanation:

Circle.java

public class Circle {

private double radius;

public Circle(double r) {

radius = r;

}

public double getArea() {

return Math.PI * radius * radius;

}

public double getRadius() {

return radius;

}

public String toString() {

String str;

str = "Radius: " + radius + "Area: " + getArea();

return str;

}

public boolean equals(Circle c) {

boolean status;

if (c.getRadius() == radius)

status = true;

else

status = false;

return status;

}

public boolean greaterThan(Circle c) {

boolean status;

if (c.getArea() > getArea())

status = true;

else

status = false;

return status;

}

}

I have created a class also to test this class below is the class: -

MainCircle.java

public class MainCircle {

public static void main(String args[]) {

Circle c = new Circle(2.5);

Circle c1 = new Circle(2.5);

Circle c2 = new Circle(4.5);

System.out.println(c);

System.out.println(c1);

System.out.println(c2);

System.out.println("Is c and c1 equal : "+c.equals(c1));

System.out.println("Is c2 greater than c1 : "+c1.greaterThan(c2));

}

}

Output: -

Radius: 2.5Area: 19.634954084936208

Radius: 2.5Area: 19.634954084936208

Radius: 4.5Area: 63.61725123519331

Is c and c1 equal : true

Is c2 greater than c1 : true

Answer:

floating point Decimal Number  = -78.1025390625

Explanation:

Hex Value =0xC29C3480

Convert this hex value into binary number, Use only C29C3480 as hexadecimal number. Each digit in hexadecimal will convert into 4 bits of binary. To convert this number into decimal we use hexadecimal to binary table. The conversion is as below:

1100 0010 1001 1100 0011 0100 1000 0000

Now this is a 32 bit number in binary, the bits are arranged as follows

1                        10000101                       00111000011010010000000

(Sign bit)            (Exponent bits)                      (Fraction bits)

In this 32 bit number, 32nd bit is sign bit,number is positive if the bit is 0 other wise number is negative. From 31st to 23rd bits 8 bits that are known as exponent bits, while from 22nd to 0 total 23 bits that are called fraction bit.

10000101 = 1x2⁷+0x2⁶+0x2⁵+0x2⁴+0x2³+1x2²+0x2¹+1x2⁰

                = 1x2⁷ + 1x2² + 1x2⁰

                = 128 + 4 + 1 =  133

      3. Now we calculate the exponent bias, that will be calculated through, the formula.

       exponent bias = 2ⁿ⁻¹ - 1

there n is total number of bits in exponent value of 32 bit number, Now in this question n=8, because there are total 8 bits in exponent.

  so,       exponent bias = 2⁸⁻¹ - 1 = 2⁷-1= 128-1= 127

          e = decimal value of exponent - exponent bias

             =     133 - 127 = 6

       4. Now, we calculate the decimal value of fraction part of the number, which is called mantissa.

         Mantissa=0x2⁻¹+0x2⁻²+1x2⁻³+1x2⁻⁴+1x2⁻⁵+0x2⁻⁶+0x2⁻⁷+0x2⁻⁸+0x2⁻⁹+1x2⁻¹⁰+1x2⁻¹¹+0x2⁻¹²+1x2⁻¹³+0x2⁻¹⁴+0x2⁻¹⁵+1x2⁻¹⁶+0x2⁻¹⁷+0x2⁻¹⁸+0x2⁻¹⁹+0x2⁻²⁰+0x2⁻²¹+0x2⁻²²+0x2⁻²³

               = 1x2⁻³+1x2⁻⁴+1x2⁻⁵+1x2⁻¹⁰+1x2⁻¹¹+1x2⁻¹³+1x2⁻¹⁶

               =  0.125 + 0.0625 + 0.03125 + 0.0009765625 + 0.00048828125 +                     0.00012207031 + 0.00001525878

Mantissa = 0.22035217284

Now the formula to calculate the decimal number is given below:

decimal number = (-1)^s x (1+Mantissa) x 2^e

there s is the sign bit which is 1. and e =6.

so,

decimal number = (-1)^1 x (1+0.22035217284 ) x 2⁶

                            = -1 x 1.22035217284 x 64

                            = -78.1025390625

Answer:

See explaination for the program code

Explanation:

//SocSecProcessor.java:

import java.util.Scanner;

public class SocSecProcessor

{

public static void main (String [] args)

{

Scanner keyboard = new Scanner (System.in);

String name;

String socSecNumber;

String response;

char answer = 'Y';

while (Character.toUpperCase(answer) == 'Y')

{

try

{

// Task #2 step 1 - To do:

// promote user to enter name and ssn number.

// save the input to variable name and SocSecNumber. Such as:

// System.out.print("Name? ");

// name = keyboard.nextLine();

// validate SSN number by calling isValid(socSecNumber).

// output the checking result.

System.out.print("Name?");

name = keyboard.nextLine();

System.out.print("SSN?");

socSecNumber = keyboard.nextLine();

if(isValid(socSecNumber)){

System.out.println(name + " " + socSecNumber + "is Valid");

}

}

catch (SocSecException e) // To do: catch the SocSecException

{

System.out.println(e.getMessage());

}

System.out.print("Continue? ");

response = keyboard.nextLine();

answer = response.charAt(0);

}

}

private static boolean isValid(String number)throws SocSecException

{

boolean goodSoFar = true;

int index = 0;

// To do: Task #2 step 2

// 1. check the SSN length. Throw SocSecException if it is not 11

if (number.length() != 11)

{

throw new SocSecException("wrong number of characters ");

}

for( index=0;index<number.length();++index){

if(index==3 || index==6){

if (number.charAt(index) != '-'){

throw new SocSecException("dashes at wrong positions");

}

}else if (!Character.isDigit(number.charAt(index))){

throw new SocSecException("contains a character that is not a digit");

}

}

// 2. check the two "-" are in right position. Throw SocSecException if it is not

// if (number.charAt(3) != '-')

// 3. check other position that should be digits. Throw SocSecException if it is not

// if (!Character.isDigit(number.charAt(index)))

return goodSoFar;

}

}

See attachment

Answer:

See explaination for program code

Explanation:

program code below:

#include<stdio.h>

int main()

{

//file pointer to read from the file

FILE *fptr;

//array to store the years

int years[50];

//array to store the population

float population[50];

int n = 0, j;

//variables for the linear interpolation formula

float x0,y0,x1,y1,xp,yp;

//opening the file

fptr = fopen("njpopulation.dat", "r");

if (fptr != NULL)

{

//reading data for the file

while(fscanf(fptr, "%d %f", &years[n], &population[n]) == 2)

n++;

//prompting the user

int year = -1;

printf("Enter the years for which the population is to be estimated. Enter 0 to stop.\n\n");

while(year != 0)

{

printf("Enter the year (1790 - 2010) : ");

scanf("%d", &year);

if (year >= 1790 && year <= 2010)

{

//calculating the estimation

xp = year;

for (j = 0; j < n; j++)

{

if (year == years[j])

{

//if the year is already in the table no nedd for calculation

yp = population[j];

}

else if (j != n - 1)

{

//finding the surrounding census years

if (year > years[j] && year < years[j + 1])

{

//point one

x0 = years[j];

y0 = population[j];

//point two

x1 = years[j + 1];

y1 = population[j + 1];

//interpolation formula

yp = y0 + ((y1-y0)/(x1-x0)) * (xp - x0);

}

}

}

printf("\nEstimated population for year %d : %.2f\n\n", year, yp);

}

else if (year != 0)

{

printf("\nInvalid chioce!!\n\n");

}

}

printf("\nAborting!!");

}

else

{

printf("File cannot be opened!!");

}

close(fptr);

return 0;

}

OUTPUT :

Enter the years for which the population is to be estimated. Enter 0 to stop.

Enter the year (1790 - 2010) : 1790

Estimated population for year 1790 : 184139.00

Enter the year (1790 - 2010) : 1855

Estimated population for year 1855 : 580795.00

Enter the year (1790 - 2010) : 2010

Estimated population for year 2010 : 8791894.00

Enter the year (1790 - 2010) : 4545

Invalid chioce!!

Enter the year (1790 - 2010) : 1992

Estimated population for year 1992 : 7867020.50

Enter the year (1790 - 2010) : 0

Aborting!!

Answer:

def student_minmax(filename):  

  """

      Description: Reads student names and finds min and max score

      Input: Name of the file that has the student data

      Output: Dictionary with name as key and value is a list with min and max values

  """

  # Dictionary that holds the results

  studentDict = {}

  # Opening and reading file

  with open(filename, "r") as fp:

      # Reading data line by line

      for line in fp:

          # Stripping new line

          line = line.strip()

          # Splitting into fields

          fields = line.split(' ')

          # Getting name and scores

          name = fields[0]

          scores = []

          scoresStr = fields[1:]

          # Converting to integer

          for score in scoresStr:

              scores.append(int(score))

          # Storing in dictionary

          studentDict[name] = []

          studentDict[name].append(min(scores))

          studentDict[name].append(max(scores))

  # Returning dictionary

  return studentDict  

# Testing function

resDict = student_minmax("d:\\Python\\studentMarks.txt")

# Printing dict

print(resDict)

Explanation:

Answer:

The solution is in C++

Explanation:

// C++ programs to search a word in a 2D grid  

#include<bits/stdc++.h>  

using namespace std;  

// Rows and columns in given grid  

#define R 3  

#define C 14  

// For searching in all 8 direction  

int x[] = { -1, -1, -1, 0, 0, 1, 1, 1 };  

int y[] = { -1, 0, 1, -1, 1, -1, 0, 1 };  

// This function searches in all 8-direction from point  

// (row, col) in grid[][]  

bool search2D(char grid[R][C], int row, int col, string word)  

{  

// If first character of word doesn't match with  

// given starting point in grid.  

if (grid[row][col] != word[0])  

return false;  

int len = word.length();  

// Search word in all 8 directions starting from (row,col)  

for (int dir = 0; dir < 8; dir++)  

{  

 // Initialize starting point for current direction  

 int k, rd = row + x[dir], cd = col + y[dir];  

 // First character is already checked, match remaining  

 // characters  

 for (k = 1; k < len; k++)  

 {  

  // If out of bound break  

  if (rd >= R || rd < 0 || cd >= C || cd < 0)  

   break;  

  // If not matched, break  

  if (grid[rd][cd] != word[k])  

   break;  

  // Moving in particular direction  

  rd += x[dir], cd += y[dir];  

 }  

 // If all character matched, then value of must  

 // be equal to length of word  

 if (k == len)  

  return true;  

}  

return false;  

}  

// Searches given word in a given matrix in all 8 directions  

void patternSearch(char grid[R][C], string word)  

{  

// Consider every point as starting point and search  

// given word  

for (int row = 0; row < R; row++)  

for (int col = 0; col < C; col++)  

 if (search2D(grid, row, col, word))  

  cout << "pattern found at " << row << ", "

   << col << endl;  

}  

// Driver program  

int main()  

{  

char grid[R][C] = {"GEEKSFORGEEKS",  

    "GEEKSQUIZGEEK",  

    "IDEQAPRACTICE"

    };  

patternSearch(grid, "GEEKS");  

cout << endl;  

patternSearch(grid, "EEE");  

return 0;  

}  

This response outlines the creation of a word search program including six key methods: horizontal, vertical, and diagonal search methods, along with their respective test methods to ensure accuracy.

In this task, we aim to develop a program that reads a word search from a file and includes six main methods. Below are the steps and methods you need to implement:

1. Horizontal Search Method

Create a method to search for a word placed horizontally in the word search. You can iterate through each row and check for the presence of the target word within that row.

2. Vertical Search Method

Create a method to search for a word placed vertically in the word search. In this method, you'll iterate over each column, constructing potential matches vertically.

3. Diagonal Search Method

Create a method to search for a word placed diagonally in the word search. Handle both diagonals from top-left to bottom-right and top-right to bottom-left.

4. Test Methods

Develop test methods for horizontal, vertical, and diagonal search methods. Ensure you have test cases that cover various scenarios to validate the correctness of your searches.

The code for the following programs are:

class WordSearchSolver:

   def __init__(self, filename):

       self.grid = self.read_word_search(filename)

   def read_word_search(self, filename):

       # Method to read word search from file and return grid

       pass

   def horizontal_search(self, word):

       # Method to search for a word horizontally in the grid

       pass

   def vertical_search(self, word):

       # Method to search for a word vertically in the grid

       pass

   def diagonal_search(self, word):

       # Method to search for a word diagonally in the grid

       pass

   def test_methods(self):

       # Test methods for horizontal, vertical, and diagonal search

       pass

# Test the class

if __name__ == "__main__":

   filename = "word_search.txt"  # Example filename

   solver = WordSearchSolver(filename)

   solver.test_methods()

By implementing these methods, the program can effectively search for words in different orientations within the word search grid.

Answer:

// This program is written in C++ programming language

// Comments are used for explanatory purpose

// Program starts here

#include<iostream>

using namespace std;

int main ()

{

// Declare and initialize variables

float connecfee = 1.99;

float first3 = 2.00;

float addmin = 0.45; float cost;

int minutes;

// Prompt user for minutes talked

cout<<"Enter Number of Minutes: ";

cin>>minutes;

// Calculate cost;

if(minutes < 1)

{

cout<<"Enter a number greater than 0";

}

else if(minutes <= 3)

{

// Calculating cost for minutes less than or equal to 3

cost = connecfee + first3 * minutes;

cout<<"Talk time of "<<minutes<<" minutes costs "<<cost;

}

else

{

cost = connecfee + first3 * 3 + addmin * (minutes - 3);

cout<<"Talk time of "<<minutes<<" minutes costs "<<cost;

}

return 0;

}

// End of Program

Answer:

Explanation:

The solution is written in Python 3.

To calculate the average cost shared by the guests, we create a function that takes two input, totalCost and numGuest (Line 1). Next apply the calculation formula to calculate the final cost after adding the 15% tips (Line 2). Next calculate the average cost shared by each guest by dividing the final cost by number of guest (Line 3). At last, use a for loop to print out the cost borne by each guest (Line 4-5).  

Answer:

Check the explanation

Explanation:

Given below is the code for the question. PLEASE MAKE SURE INDENTATION IS EXACTLY AS SHOWN IN IMAGE.

import csv

def read_data(csv_r, keys):

  data = []

  for row in csv_r:

      mydict = {}

      for i in range(len(keys)):

          mydict[keys[i]] = row[i]

      data.append(mydict)

  return data

SCREENSHOT:

Answer:

The answer is in the explanation and check the attached file for the output

Explanation:

GCD.py:

def gcd(a, b):

  if b > a:

       a, b = b, a

  while b != 0:

       t = b

       b = a % t

       a = t

   return a

def main():

   m = eval(input("Enter m: "))

   n = eval(input("Enter n: "))

   print("GCD(",m,",",n,") = ",gcd(m,n))

main()

Output: check the output in the attached file

Answer:

I am writing a C++ program.

#include <iostream> // for input output functions

using namespace std; // to identify objects like cin cout

int gcd(int m, int n) {

//function to compute greatest common divisor of two value m and n

  if (n == 0) //if value of n is equal to 0 (base condition)

  return m; // returns the value of m

  return gcd(n, m % n); }

// calls gcd function recursively until base condition is reached  

void get_value(int c, int d){ // function to obtain two values

  cout<<"Enter the two values: "<<endl;

//prompts user to enter values of c and d

  cin>>c>>d; //reads the input values of c and d

  while(c <= 0 || d<= 0){ // the loop continues to ask user to enter positive //values until user enters values greater than 0

      cout<<"Enter a positive value";

      cin>>c>>d;   } //reads values from user

    cout<<"GCD of "<< c <<" and "<< d <<" is "<< gcd(c, d); }

//calls gcd funtion to compute greatest common divisor of two values    

int main() { // main() function body

   int a,b; //two integer variables are declared

   get_value(a,b); // calls this method to take values of a and b from user

   char ch; // used to enter a choice by user to perform gcd again

   while(true)      {

//asks the user if he wants to continue to compute gcd

       cout<<"Would you like to do another computation or not?(Y/N)\n"<<endl;

       cin >> ch; //reads the choice user enters

/*if user enters Y or y then it calls get_value function to take values from user and computer gcd, if user types N or n then the program exits */

       if(ch == 'Y'|| ch =='y'){

           get_value(a,b);

       }else if(ch =='N'||ch =='n'){

           break;         }    }}

Explanation:

The program has three methods. gcd method that computes greatest common divisor (GCD) of two values can be computed using Euclid's algorithm. get_value method takes two positive integers from user to find gcd of these values. It keeps asking user to enter positive values until user enters positive values. Third is the main() function which calls get_value function and after computing gcd of two values, asks user to if he wants to find gcd again? If user types Y or y which indicates yes, it moves the program control to get_value function and if user enters n, then the program ends.

Final answer:

The minimum bandwidth required to transmit a signal generated at a rate of 1,000,000 ASCII characters per second, with an added parity bit for single error detection, is 4 MHz. This calculation is based on the bit rate of 8 million bits per second and assumes a binary signal requiring a minimum bandwidth that is half of the bit rate.

Explanation:

The question pertains to digital data transmission and requires calculating the minimum bandwidth necessary for a computer that generates data at a rate of 1,000,000 characters per second, using ASCII which has 7-bit binary-coded characters and includes an additional parity bit for error detection, making it 8 bits per character in total.

Bandwidth, in this context, refers to the range of frequencies necessary to transmit the digital signal. The data rate of the signal is given as 1,000,000 characters per second, and since each character is represented by 8 bits (7 bits for ASCII character and 1 for parity), the bit rate is 8,000,000 bits per second, or 8 Mbps (Megabits per second).

To calculate the minimum bandwidth using the Nyquist formula, we assume the signal is binary and requires a minimum bandwidth that is half of the bit rate. Hence, the minimum bandwidth required is 4 MHz (megahertz). This is because the maximum rate of change of a binary signal would be if it alternated between 0 and 1 for every bit, so the minimum bandwidth is also known as the Nyquist bandwidth.

Final answer:

The minimum bandwidth required to transmit ASCII characters with single error detection at 1,000,000 characters per second, assuming a parity bit is added to each character, is 8 Mbps. This figure is based on 8-bit as characters and could vary depending on the modulation technique used.

Explanation:

The minimum bandwidth required to transmit a signal with single error detection for ASCII characters at a rate of 1,000,000 characters per second can be calculated considering the addition of a parity bit to each character. ASCII uses 7 bits per character, making it 8 bits with the parity bit. To find the bandwidth, we multiply the bit rate per character by the number of characters per second. Since the computer generates data at 1,000,000 characters per second, and with the parity bit, we have an 8-bit as character, the data rate becomes 8,000,000 bits per second (or 8 Mbps).

However, the bandwidth can also be affected by various factors such as the modulation technique used. For instance, if a simple binary modulation scheme is used (such as non-return-to-zero, NRZ), where one bit is transmitted per signal change, the bandwidth would be 8 Mbps. But if a more complex modulation scheme is used, which transmits more than one bit per signal change, the required bandwidth could be lower.

Answer:

See explaination

Explanation:

The code here:

import java.io.*;

public class CISP401V11A5 {

public static void main(String[] args) throws Exception {

try {

CISP401Method();

} catch (Exception e) {

System.out.println("Exception thrown in " + e.getStackTrace()[0].getMethodName() + "\n");

StringWriter sw = new StringWriter();

e.printStackTrace(new PrintWriter(sw));

String exceptionAsString = sw.toString();

String bagBegin = exceptionAsString.substring(0, 19);

String bagEnd = exceptionAsString.substring(19);

System.out.println(bagBegin + ": Exception thrown in " + e.getStackTrace()[0].getMethodName() + bagEnd);

}

}

public static void CISP401Method() throws Exception {

try {

CISP401Method2();

} catch (Exception e) {

throw e;

}

}

public static void CISP401Method2() throws Exception {

throw new Exception();

}

}

See attachment for sample output

Answer:

Check the explanation

Explanation:

The loop runs for when i=0 to i=n-1 ie <n which consequently means that there are n iterations of the for loop, that is equal to array size n, therefore growth function will be

F(n) = n

Also big o notation is the upper bound of the growth function of any algorithm which consequently means that the big oh notation of this code's complexity O(n)

Answer:

See explaination

Explanation:

// Defination of MoneyMarketAccount CLASS

MoneyMarketAccount::MoneyMarketAccount(string name, double balance)

{

acctName = name;

acctBalance = balance;

}

int MoneyMarketAccount::getNumWithdrawals() const

{

return numWithdrawals;

}

int MoneyMarketAccount::withdraw(double amt)

{

if (amt < 0)

{

cout << "Attempt to withdraw negative amount - program terminated."<< endl;

exit(1);

}

if((numWithdrawals < FREE_WITHDRAWALS) && (amt <= acctBalance)){

acctBalance = acctBalance - amt;

return OK;

}

if((numWithdrawals >= FREE_WITHDRAWALS) &&((amt + WITHDRAWAL_FEE) <= acctBalance)){

acctBalance = acctBalance - (amt + WITHDRAWAL_FEE);

return OK;

}

return INSUFFICIENT_FUNDS;

}

// Defination of CDAccount CLASS

CDAccount::CDAccount(string name, double balance, double rate)

{

acctName = name;

acctBalance = balance;

interestRate = rate/100.0;

}

int CDAccount::withdraw(double amt)

{

if (amt < 0)

{

cout << "Attempt to withdraw negative amount - program terminated."<< endl;

exit(1);

}

double penalty = (PENALTY*interestRate*acctBalance);

if((amt + penalty) <= acctBalance){

acctBalance = acctBalance - (amt + penalty);

return OK:

}

else

return INSUFFICIENT_FUNDS;

}

void transfer (double amt, BankAccount& fromAcct, BankAccount& toAcct)

{

if (amt < 0 || fromAcct.acctBalance < 0 || toAcct < 0)

{

cout << "Attempt to transfer negative amount - program terminated."<< endl;

exit(1);

}

if(fromAcct.withdraw(amt) == OK){ // if withdraw function on fromAcct return OK, then me can withdraw amt from fromAcct

// and deposit in toAcct

toAcct.acctBalance = toAcct.acctBalance + amt;

}

}

Answer:

see explaination

Explanation:

#include <stdio.h>

// user defined functions

int min(int arr[], int n)

{

int m=arr[0];

for(int i=1;i<n;i++)

if(arr[i]<m)

m=arr[i];

// return minimum

return m;

}

int max(int arr[], int n)

{

int m=arr[0];

for(int i=1;i<n;i++)

if(arr[i]>m)

m=arr[i];

// return maximum

return m;

}

int main() {

int n;

// read N

printf("Enter number of elements: ");

scanf("%d",&n);

// read n values into list

int arr[n];

printf("Enter elements: ");

for(int i=0;i<n;i++)

scanf("%d",&arr[i]);

// find max and min

printf("The List is: ");

for(int i=0;i<n;i++)

printf("%d ",arr[i]);

printf("\nThe minimum value is: %d\n",min(arr,n));

printf("The maximum value is: %d\n",max(arr,n));

return 0;

}

see attachment for screenshot and output

Answer:

print("enter starting time hours")

st_hours=int(input()) #starting time hours

print("minutes")

st_minutes=int(input()) #starting time minutes

print("enter ending time hours")

et_hours=int(input()) #ending time hours

print("minutes")

et_minutes=int(input()) #ending time minutes

cost=2.50

if (st_hours<21 and et_hours>21): #if starting time is less than 21 and ending time is greater than 21

a_hours=et_hours-21

a_minutes=et_minutes #taking time after 21 from ending time ending time-21

time_minutes=60-st_minutes #converting all the time into minutes

time_hours=21-(st_hours+1)

time_hours=time_hours*60

time=time_hours+time_minutes

cost=(cost/60)*time

time=(a_hours*60)+a_minutes

cost=cost+(1.75/60)*time

print("cost=$",cost)

elif(st_hours>=21): #if starting time is greater than 21

time_hours=et_hours-(st_hours+1)

time_minutes=(60-st_minutes)+et_minutes #converting time into minutes

time=(time_hours*60)+time_minutes

cost=(1.75/60)*(time)

print("cost=$",cost)

elif(et_hours<=21): #if ending time is less than 21

time_hours=et_hours-(st_hours+1)

time_minutes=(60-st_minutes)+et_minutes

time=(time_hours*60)+time_minutes

cost=(2.50/60)*time

print("cost=$",cost)

Explanation:

This is a conditional program, the conditions applied are if and else if.

The resultant output was gotten using this three steps:

(1) If starting time is less than 21 and ending time is greater than 21

Then I converted the time into minutes before 21

ex:- 8:30

21-9= 12*60= 720 + 30 = 750

Then I calculated the bill.

750 * (2.50 / 60)

Then I converted the time after 21 and then calculated the bill and printed it.

(2)If starting time is greater than 21

Converted time into minutes and multiplied it with (1.75 / 60) to get the bill.

(3) If ending time is less than 21

Converted time into minutes and multiplied it with (2.50 / 60) to get the bill.

Please check attachment for program code screenshot.

Final answer:

In summary, the student's homework entails creating a program to calculate the total babysitting bill based on varying hourly rates before and after 9:00 PM, within a 24-hour period.

Explanation:

The question pertains to creating a program that calculates the total babysitting bill based on hourly rates that change after 9:00 PM. The program must accept a starting time and an ending time in hours and minutes and, with the given rates of $2.50 before 9:00 PM and $1.75 after 9:00 PM, compute the total cost considering partial hours.

An example of how the calculations may work is given: If you hired 5 people at a cost of $15 per hour each and they worked for 2 hours, the total cost would be 5 people \\* $15/hour \\* 2 hours = $150.

To apply this logic to the babysitting scenario, the program would need to calculate the number of hours worked at each rate, then multiply the number of hours by the corresponding rate and sum these amounts to arrive at the final bill.

Answer:

Relative greedy algorithm is not optimal

Explanation:

Proving that Relative’s greedy algorithm is not optimal.

This can be further proved by the following example.

Let us assumethe friends to be invited be Ali, Bill, David, Dennis, Grace, Eemi, and Sam.

The enemy list of each friend is shown below:

• Ali: Bill. David, Dennis

• Bill: Ali, Grace, Eemi, Sam

• David: Ali, Grace, Eemi, Sam

• Dennis: Ali, Grace, Eemi, Sam

• Grace: Bill. David, Dennis. Eemi . Sam

• Eemi: Bill, David, Dennis, Grace, Sam

• Sam: Bill, David, Dennis, Eemi, Grace

From the enemy list, the one with fewer enemies is Ali.

If we invite Ali, then Bill, David, and Dennis cannot be invited.

Next person that can be invited is one from Grace, Eemi, and Sam

If we choose any one of them we cannot add any other person.

For example, if we choose Grace, all other members are enemies of Ali and Grace. So only Ali

and Grace can only be invited.

So we get a list of two members using relative’s greedy algorithm.

This is not the optimal solution.

The optimal solution is a list of 3 members

• Bill, David, and Dennis can be invited to the party.

Hence, it is proved that relative’s greedy algorithm is not optimal, where the maximum number of friends is not invited to the party.

The Greedy algorithm is not optimal for this case because it cannot invite all the friends.

This refers to a modest approach that is taken to solve a problem by selecting the best option available to solve optimization problems.

Hence, we can see that if we use the greedy algorithm, we would have to make optimal selections to get the complete optimal system, and in this case, it is not optimal.

This is because, the optimal solution is a list of 3 members, and not all the friends can be invited.

Read more about greedy algorithm here:
https://brainly.com/question/13197481

Answer:

def string_with_most_the(string_list):

   index = 0

   count = []

   max_length = 0

   for i in range(len(string_list)):

       s1 = string_list[i].lower()

       count.append(s1.count("the"))

       if count[i] > max_length:

           max_length = count[i]

           index = i

   return string_list[index]

Explanation:

Create a function called string_with_most_the that takes one parameter, string_list

Inside the function:

Initialize the variables

Initialize a for loop that iterates through string_list

Set the lower case version of the strings in string list to s1

Count how many "the" each string in string_list contain and put the counts in the count list

If any string has more "the" than the previous one, set its count as new maximum, and update its index

When the loop is done, return the string that has maximum number of "the", string_list[index]

Answer:

Answer is in the provided screenshot!

Explanation:

Steps required:

check if pay is greater than 40, if so then handle logic for pay over 40, else treat normally.

Answer:

num1 = float(input("Enter the first number: "))

while True:

   print("Add, Subtract, Multiply, Divide, or Quit ?")

   choice = input("Your choice is: ")

   if choice == "Quit":

       break

   num2 = float(input("Enter the second number: "))

   if choice == "Add":

       num1 += + num2

       print("The subtotal is: " + str(num1))

   elif choice == "Subtract":

       num1 -= num2

       print("The subtotal is: " + str(num1))

   elif choice == "Multiply":

       num1 *= num2

       print("The subtotal is: " + str(num1))

   elif choice == "Divide":

       num1 /= num2

       print("The subtotal is: " + str(num1))

print("The final result is: " + str(num1))

Explanation:

Ask the user for the first number

Create a while loop that iterates until specified condition occurs inside the loop

Ask the user for the operation or quitting

If the user chooses quitting, stop the loop. Otherwise, get the second number, perform the operation and print the subtotal

When the loop is done, the user enters Quit, print the final value result

Answer:

Before running this program, make sure you have installed pandas module for python.

pip install pandas

import pandas as pd

import numpy as np

csv_file = input("Enter CSV File Name : ")

df = pd.read_csv(csv_file)

count_row = df.shape[0]

print("There were %d Film Permits in Total." %(count_row))

borough_count = df['Borough'].value_counts()

print(borough_count)

location_count = df['ParkingHeld'].value_counts().head(5)

print(location_count)

Explanation:

Answer:

pip install pandas

import pandas as pd

import numpy as np

csv_file = input("Enter CSV File Name : ")

df = pd.read_csv(csv_file)

count_row = df.shape[0]

print("There were %d Film Permits in Total." %(count_row))

borough_count = df['Borough'].value_counts()

print(borough_count)

location_count = df['ParkingHeld'].value_counts().head(5)

print(location_count)

Explanation:

Before running this program, make sure you already previously installed pandas module for python.

Import panda and numpy

Put csv file

count row

then tell to print

Answer:

i) The time taken for 1500 records = 15 seconds.

ii) The time taken for 1500 records = 50 seconds.

Explanation:

A is an O(n) algorithm.

An algorithm with O(n) efficiency is described as a linearly increasing algorithm, this mean that the rate at which the input increases is linear to the time needed to compute that algorithm with n inputs.

From the question, it is given that for 1000 records, the time required is: 10 seconds.

Algorithm time taken is O(n)

Hence,

1) For 1,500 records

=> 10/1000 = x/1500

=> 10x1500/1000 = x

x = 15 seconds

Thus, the time taken for 1500 records = 15 seconds.

2) For 5,000 records

=> 10/1000 = x/5000

=> 10x5000/1000 = x

x = 50 seconds

Thus, the time taken for 1500 records = 50 seconds.

Answer:

To look up functions and their purposes

Explanation:

Edg

Answer:

a on edge

Explanation:

just took the question

Answer:

Therefore, it is important to find opportunities to integrate security controls into day-to-day routines.

Do you believe this to be true- Yes.

In general, implementing security policies occurs in isolation from the business perspectives and organizational values that define the organization’s culture. Is this correct or incorrect? - Incorrect

Explanation:

Truly, it is important to find opportunities to integrate security controls into day-to-day routines, this is in order to minimize future security threats by formulating company-wide security policies and educating employees on daily risk prevention in their work routines. In the operational risk controls, vigilant monitoring of employees must be implemented in order to confirm that policies are followed and to deter insider threats from developing.

Flexing and developing policies as resources and priorities change is the key to operational risk controls.

These risk controls implementation in organizational security is not a one-time practice. Rather, it is a regular discipline that the best organizations continue to set and refine.

For better preparation of an organization towards mitigating security threats and adaptation to evolving organizational security needs, there must be a proactive integration of physical information and personnel security while keeping these risk controls in mind.

12. In general, implementing security policies occurs in isolation from the business perspectives and organizational values that define the organization’s culture - Incorrect.

When security policies are designed, the business perspectives and the organizational values that define the organization’s culture must be kept in mind.

An information security and risk management (ISRM) strategy provides an organization with a road map for information and information infrastructure protection with goals and objectives that ensure capabilities provided are adjusted to business goals and the organization’s risk profile. Traditionally, ISRM has been considered as an IT function and included in an organization’s IT strategic planning. As ISRM has emerged into a more critical element of business support activities, it now needs its own independent strategy to ensure its ability to appropriately support business goals and to mature and evolve effectively.

Hence, it is observed that both the Business Perspective and Organisational goals are taken into consideration while designing Security policies.

Answer:

import java.util.Scanner;

public class num10 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.println("Enter a word");

       String word = in.nextLine();

       String reversed = "";

       //Get the length of the string entered

       int length = word.length();

       //Loop through the string and reverse th string

       for ( int i = length - 1; i >= 0; i-- )

           reversed = reversed + word.charAt(i);

       //Compare the two strings "word" and "reversed"

       if (word.equals(reversed))

           System.out.println(word+" is a palindrome");

       else

           System.out.println(word+" is not a palindrome");

   }

}

Explanation:

Since we know that a palindrome word is a word that reads the same forward and backward, The Idea here is:

Final answer:

The code reads lines from two files and multiplies the integers from each line together, adding the products to a variable called scalar_product.

Explanation:

The task requires reading lines from two files and multiplying the corresponding integers from each line together. The products are then added to a variable called scalar_product. The code should stop when it reaches the end of either file. This can be achieved by using a loop to read lines from both files simultaneously and perform the multiplication and addition operations:

scalar_product = 0
with open('numbers1.txt') as file1, open('numbers2.txt') as file2:
   for line1, line2 in zip(file1, file2):
       num1 = int(line1.strip())
       num2 = int(line2.strip())
       scalar_product += (num1 * num2)
print(scalar_product)

This example demonstrates a simple but effective way to calculate the scalar product of two sets of integers from text files in Python.

Final answer:

The question asks for code that calculates the scalar product of integers read from two files. The process involves multiplying integers from corresponding lines and summing the products. The code in Python was provided with error handling for missing files or non-integer values.

Explanation:

The task given involves writing code that reads from two files containing lines of positive integers, multiplies corresponding numbers from each line of these files together, and adds the product to a variable named scalar_product, which has been initialized to zero. This process is reminiscent of computing the scalar product  or dot product in mathematics, though in this case, the operation is performed with integers from files instead of vector components. It is important that the code stops processing once it reaches the end of either file.

Below is an example of how the code could be implemented in Python:

scalar_product = 0
try:
   with open('numbers1.txt', 'r') as file1, open('numbers2.txt', 'r') as file2:
       for num1, num2 in zip(file1, file2):
           scalar_product += int(num1) * int(num2)
except FileNotFoundError:
   print("One of the files was not found.")
except ValueError:
   print("Found non-integer line.")
This code initializes the scalar_product, opens both files concurrently, and uses a zip function to read corresponding lines. It then multiplies these integers and adds the result to scalar_product. Additionally, the code includes error handling to deal with potential issues like missing files and invalid content.

Answer:

see explaination

Explanation:

/*

* To change this license header, choose License Headers in Project Properties.

* To change this template file, choose Tools | Templates

* and open the template in the editor.

*/

package knapsack;

import java.util.*;

/**

*

* atauthor Administrator

*/

public class MyKnapsack {

static int max(int a, int b) { return (a > b)? a : b; }

static int calculate(int W, int wt[], int val[], int n)

{

if (n == 0 || W == 0)

return 0;

if (wt[n-1] > W)

return calculate(W, wt, val, n-1);

else return max( val[n-1] + calculate(W-wt[n-1], wt, val, n-1),

calculate(W, wt, val, n-1)

);

}

public static void main(String[] args)

{

Scanner sc=new Scanner(System.in);

System.out.printf("|Enter value of capacity");

int Capacity =sc.nextInt();

System.out.printf("|Enter value of number of objects");

int n=sc.nextInt();

int profits[] = new int[n];

int weight[] = new int[n];

System.out.printf("|Enter values for profits");

for(int i=0;i<n;i++)

{

profits[i]=sc.nextInt();

}

System.out.printf("|Enter values for weights");

for(int j=0;j<n;j++)

{

weight[j]=sc.nextInt();

}

System.out.println(calculate(Capacity,weight, profits, n));

}

Note: change the at with the sign

}

Answer:

Preimage attack and collision attacks.

Explanation:

When SHA-2 hashing is used with only 128 bit hash, It usually makes the system vulnerable to attack.

Main disadvantage in case of truncated SHA is that there exist no reduction proofs to the original hash function. It implies that the truncated version is thus is very much resistant to preimage attack and collision attacks.

Collision resistant system does not always implies to collision free. If 128 bit hash is used for say any file greater than 16 bytes, then atleast two collision occurs. It can occur anytime in the beginning in starting one or two documents at any later stage. It will lead to a situation where one will not be able to write second or will ultimately overwrite the first one.

Final answer:

Reducing a SHA-2 hash to 128 bits makes it susceptible to a birthday attack due to the collision probability. The effectiveness of hash security relies on collision resistance, not just the length of the hash.

Explanation:

The Vulnerability of a 128-bit Hash

Using only the first 128 bits of a SHA-2 hash can make it vulnerable to a birthday attack. This type of attack exploits the mathematical Birthday Paradox, where the probability of two people sharing the same birthday rises disproportionately with the number of individuals in the group. In cryptographic terms, it means that with a hash of 128 bits in length, an attacker only needs roughly 264 attempts to find two different inputs that produce the same hash value, which is referred to as a collision. This is feasible with modern computing power. Therefore, reducing the length of the hash from its original size compromises its collision resistance.

Comparing hashing and encryption directly is not entirely appropriate. While a 128-bit encryption key may be considered secure because it requires 2128 attempts to crack, this doesn't translate the same way for hashing. Hashes are not encrypted; they are fixed-size outputs for inputs of arbitrary size, and their security is largely reliant on their collision resistance and unpredictability.

Final answer:

To find the satisfying assignment that minimizes the number of variables set to TRUE using the satisfiable num program, follow the provided pseudo-code.

Explanation:

To use the Boolean formula satisfiability program satisfiable num to find a satisfying assignment that minimizes the number of variables set to TRUE, you can follow the pseudo-code: