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nlexa [21]
3 years ago
7

Compare the time spent on each of the above operations, and describe your observations. If an operation is too fast, you may wan

t to repeat it for many times, i.e., 5000 times, and then take an average rsa aes
Computers and Technology
1 answer:
Artist 52 [7]3 years ago
3 0

Here's the complete question below that clarifies what you need to do

<u>Explanation</u>:

"In this task, we will study the performance of public-key algorithms. Please prepare a file ( message.txt) that contains a 16-byte message. Please also generate an 1024-bit RSA public/private key pair. Then, do the following:

1)Encrypt message.txt using the public key; save the the output in message_enc.txt.

2)Decrypt message_enc.txt using the private key.

3)Encrypt message.txt using a 128-bit AES key.

<em><u>Compare the time spent on each of the above operations, and describe your observations. If an operation is too fast, you may want to repeat it for many times, i.e., 5000 times, and then take an average.</u></em>

<em><u> After you finish the above exercise, you can now use OpenSSL's speed command to do such a benchmarking. Please describe whether your observations are similar to those from the outputs of the speed command?</u></em>

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In this chapter, you saw an example of how to write an algorithm that determines whether a number is even or odd. Write a progra
vagabundo [1.1K]

Answer:

// Program is written in Java Programming Language

// Comments are used for explanatory purpose

// Program starts here

public class RandomOddEve {

/** Main Method */

public static void main(String[] args) {

int[] nums = new int[100]; // Declare an array of 100 integers

// Store the counts of 100 random numbers

for (int i = 1; i <= 100; i++) {

nums[(int)(Math.random() * 10)]++;

}

int odd = 0, even = 0; // declare even and odd variables to 0, respectively

// Both variables will serve a counters

// Check for odd and even numbers

for(int I = 0; I<100; I++)

{

if (nums[I]%2 == 0) {// Even number.

even++;

}

else // Odd number.

{

odd++;

}

}

//.Print Results

System.out.print("Odd number = "+odd);

System.out.print("Even number = "+even);

}

8 0
3 years ago
Read 2 more answers
Write a program that asks the user for the speed of a vehicle (in miles per hour) and how many hours it has traveled. It should
HACTEHA [7]

Answer:

speed = float(input("Enter the speed: "))

hours = int(input("Enter the hours: "))

distance = 0

for i in range(hours):

   distance += speed * 1

   print("The distance after " + str(i+1) + ". hour(s): " + str(distance))

Explanation:

*The code is in Python.

Ask the user to enter the speed and the hours

Initialize the distance as 0

Create a for loop that iterates hours times. Inside the loop, calculate the cumulative distance traveled at the end of each hour and print it (Note that the distance = speed x hour)

5 0
3 years ago
What happens if i unplug my alarm system?
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6 0
3 years ago
Which option describes wearable technology?
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My opinion is the answers A and D. I would choose A if you can only choose one, though.
7 0
3 years ago
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Write a java program called allDigitsOdd that returns whether every digit of a positive integer is odd. Return true if the numbe
Vlada [557]

Answer:

public class Digits

{

   public static boolean allDigitsOdd(int num)

   {

       boolean flag=true;

       int rem;

       while(num>0)

       {

           rem=num%10;

           num=num/10;

           if(rem%2==0)    // if a even digit found immediately breaks out of loop

           {

               flag=false;

               break;

           }

       }

       return flag;     //returns result

   }

   public static void main(String args[])

   {

       System.out.println(allDigitsOdd(1375));    //returns true as all are odd digits

   }

}

OUTPUT :

true

Explanation:

Above program has 2 static methods inside a class Digits. Logic behind above function is that a number is divided by 10 until it is less than 0. Each time its remainder by 0 is checked if even immediately breaks out of the loop.

4 0
3 years ago
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