Yes , it’s true. In a known-plaintext attack (kpa), the cryptanalyst can only view a small portion of encrypted data, and he or she has no control over what that data might be.
The attacker also has access to one or more pairs of plaintext/ciphertext in a Known Plaintext Attack (KPA). Specifically, consider the scenario where key and plaintext were used to derive the ciphertext (either of which the attacker is trying to find). The attacker is also aware of what are the locations of the output from key encrypting. That is, the assailant is aware of a pair. They might be familiar with further pairings (obtained with the same key).
A straightforward illustration would be if the unencrypted messages had a set expiration date after which they would become publicly available. such as the location of a planned public event. The coordinates are encrypted and kept secret prior to the event. But when the incident occurs, the attacker has discovered the value of the coordinates /plaintext while the coordinates were decrypted (without knowing the key).
In general, a cipher is easier to break the more plaintext/ciphertext pairs that are known.
To learn more about Plaintext Attack click here:
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B, Parallel ports are faster than serial ports.
A parallel port can move a set of 8 bits at a time on eight different wires, it uses a 25 pin connector, called a DB-25 connector, whereas a serial port only has a DB-9 connector.
Answer:
The program is as follows:
#include <iostream>
#include <iomanip>
using namespace std;
int main(){
int num1, num2, num3;
cin>>num1>>num2>>num3;
cout << fixed << setprecision(2);
cout<<(num1 + num2 + num3)/3<<" ";
cout<<num1 * num2 * num3<<" ";
return 0;
}
Explanation:
This declares three integer variables
int num1, num2, num3;
This gets input for the three integers
cin>>num1>>num2>>num3;
This is used to set the precision to 2
cout << fixed << setprecision(2);
This prints the average
cout<<(num1 + num2 + num3)/3<<" ";
This prints the product
cout<<num1 * num2 * num3<<" ";
Idek but hope u figure it out!
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