Thread: encryption / decryption program

I am following this book and an asignment but I just cant figure the decryption part out! can anyone help?

Code:

/*
(Cryptography) A company wants to transmit data over the telephone,
but is concerned that its phones could be tapped.
All of the data are transmitted as four-digit integers.
The company has asked you to write a program that encrypts
the data so that it can be transmitted more securely.
Your program should read a four-digit integer and encrypt it as follows:

Replace each digit by (the sum of that digit plus 7) modulus 10.
Then, swap the first digit with the third,
swap the second digit with the fourth and print the encrypted integer.

  Write a separate program that inputs an encrypted
  fourdigit integer and decrypts it to form the original number.

  */

#include <iostream>
int Encrypt (int value)
{

	// compute seperate digits.
	// 1234
	int First,Second,Third,Forth;

	First = value/1000; // 1


	Second = (value%1000) / 100; // 2
	Third = (value%100) / 10; // 3
	Forth =  value % 10; // 4
	
	First = (First+7) % 10; // 8
	Second = (Second+7) % 10; // 9
	Third = (Third+7) % 10; // 0
	Forth = (Forth+7) % 10; // 1

	if ( Third <= 0 ) { Third = 1; } // if third = zero we get a 3digit back
	int Encrypted_number = (Third*1000)+(Forth*100)+(First*10)+(Second); // 1189
	return (Encrypted_number);

}
int Decrypt ( int value )
{
	// 1189
	int First,Second,Third,Forth;
	First = value/1000; // 1
	Second = (value%1000) / 100; // 1 
	Third = (value%100) / 10; // 8
	Forth =  value % 10; // 9

	First = (First%7) % 10; //  1
	Second = (Second%7) % 10; // 
	Third = (Third%7) % 10; // 
	Forth = (Forth%7) % 10; // 
	if ( Third <= 0 ) { Third = 1; } 
	int Decrypted_number = ((First*1000)+(Second*100)+(Third*10)+(Forth));
	return (Decrypted_number);

}

void Printmenu()
{
	std::cout <<"[1] To encrypt a 4 digit number\n[2] To decrypt a 4 digit number" << std::endl;
}
int GetChoise ()
{
	int choise;
	std::cin >> choise;
	while (choise != 1 && choise != 2)
	{
		std::cout <<"Please enter a menu item!\n";
		Printmenu();
		std::cin >> choise;
	}
	return choise;

}
int main ()
{

	
	Printmenu();
	int Choise;
	Choise = GetChoise ();
	if (Choise == 1)
	{
		int to_be_encrypted,encrypted; 
		std::cout <<"Please enter a 4 digit integer number for encryption\n";
		std::cin >> to_be_encrypted;

		encrypted = Encrypt(to_be_encrypted);

		std::cout <<"The original number: "  << to_be_encrypted << std::endl;
		std::cout <<"The encrypted number: " << encrypted << std::endl;
		
	}
	else if (Choise == 2)
	{
		int to_be_decrypted,decrypted; 
		std::cout <<"Please enter a 4 digit integer number for Decryption\n";
		std::cin >> to_be_decrypted;

		decrypted = Decrypt (to_be_decrypted);

		std::cout <<"Decrypted: "  << decrypted << std::endl;
		


	}
	std::cin.get();
	std::cin.get();
	return 0;
}

The opposite of + is -, not %. And you're not unshuffling the numbers.

On a side note, this line

Code:

if ( Third <= 0 ) { Third = 1; } // if third = zero we get a 3digit back

makes it impossible to decrypt the number, since there are multiple inputs going to the same output.

When doing the opposite of a series of operations, you not only have to do the opposite operation (like - instead of +) but you have to go in the opposite order.

Unfortunately, there's no direct opposite of %. You have to think about your encryption and the possible values before and after the call to % during that encryption, then use that knowledge to figure out how to go back for that step.

oh man I cant get it figured out

There are only 10 possible digits (0-9). Encrypt each digit on paper and find out what the result would be just for the digit encryption (don't rearrange the order like in your program).

Then, figure out how to go backwards. It isn't a simple exercise, although it is just logic.

Once you've figured that out, go back to your program. Your encryption encrypts each digit, then rearranges them. To decrypt, you need to rearrange the digits back into the correct order, then decrypt each digit individually like you figured out on paper.

ok man tnx i will try