替代加密

Question

I'm trying to make a substitution encryption program in java, but I'm having trouble creating the code and I'm wondering if you have any tips. My idea was to make the alphabet into numbers and then compare it to the plainText and then make a new string to send back.

This is my code:

package ComputerSecurity;

public class Substitution {

    String str = "abcdefghijklmnopqrstuvwxyz";

    public String encrypt(int key, String plainText){
        StringBuilder temp = new StringBuilder();
        for(int i=0; i< plainText.length(); i++){
            int position = plainText.charAt(i);

            temp.append(str.charAt(plainText.indexOf(position)+key));
        }
        return temp.toString();
    }
}

key=3 but it's sent in by the main method plainText is "hello world".

Answer 1

I think you try to implement a Caesar Cipher.

plainText.charAt(i)

returns the actual current character, not a position. You can use the character to find the position in the alphabet:

char c = plainText.charAt(i)
int aPos = str.indexOf(c);

Then you can move this position and get the encrypted letter, but you need to wrap around. This is usually done with a modulo operation:

int encPos = aPos;
if (key >= 0) {
    encPos = (aPos + key) % str.length();
} else {
    // the key may be bigger than the alphabet length
    encPos = (str.length() + ((key + aPos) % str.length())) % str.length();
}
char ec = str.indexOf(encPos);

For every char where aPos is -1, you can skip the encryption. If you do this, then you can also encrypt spaces and other non-alphabet characters.

The problem with your idea to return a "number string" is that you don't know where one character number ends and where the next one begins. So, if you want that, you will have to add delimiters and this bloats your ciphertext unnecessarily.

Answer 2

There is 3 problems:

• You use str . It will limit the alphabet you can use. In your case uppercase H and W and the space between the 2 word will give the same output since they're not in str .
• Also, if you try to encrypt xyz with a key > 3 you'll end up trying to get a char in a position that doesn't exist in str .
• Finally, if you don't use a separator in our ouput it will be realy difficult to differentiate some case like 'ak' and 'ka' since both will output '111' with a key of 0.

My suggestion would be tu use the int value of the char . You can get it with plainText.charAt(i) .
For "abc" it will return 97, 98 and 99 since the code for a is 97. If you substract 96, you'll get 1 for a , 2 for b ...
You'll just have to add your key to it.

package ComputerSecurity;
public class Substitution {

    public String encrypt(int key, String plainText){
        StringBuilder temp = new StringBuilder();
        for(int i=0; i< plainText.length(); i++){
            int val = plainText.charAt(i) - 96;
            temp.append((val+key) + ";");
        }
        return temp.toString();
    }

encrypt(3, "Hello World") will output -21,8,15,15,18,-61,26,18,21,15,7, .
If you want to get 11,8,15,15,18,-61,26,18,21,15,7, put the plainText toLowerCase() .