CODEWITHSUNDEEP
javaencryption
Christopher Baldwin
Christopher Baldwin

Reputation: 347

Creating a simple encryption program

I'd like to use java to make a cipher of sorts, but im not sure how to go about it.

Basically, I'd want the machine to accept a string of text, say "Abcd"

and then a key, say '4532'

The program should move the characters forward in the alphabet if the number matching the place of the letter is even, and backward if it's odd.

If there is no number, the key should loop around until it's out of characters in the string to change.

the program would then print the key. Ideally, if im pseudocoding this correctly, deciphering the string would be a reverse process only applicable with the key.

I'm guessing i'd use a combination of an array and if/else statements.

I'm not sure where to start.

Example & edit String: 'hello' Key: '12'

a b c d e f g h i j k l m n o p q r s t u v w x y z

Because the corresponding key value is 1, h will travel backwards that many spaces.

h = g

because e has a 2, it'll move forward that many spaces.

e = g

the first l then becomes k, while the second becomes n. The Key is repeated because the string is out of numbers to compare. o turns into n because it's matched with 1.

hello would become ggknn with the key 42.

Upvotes: 0

Views: 30595

Answers (3)

Kick
Kick

Reputation: 4923

Find below the class for encyption

public class App 
{
    public static void main(String arg[]) 
    {

        String keys = "12";
        String codes = "hello";
        StringBuilder result = new StringBuilder();

        char[] codeList = codes.toCharArray();
        char[] keyList  = keys.toCharArray();

        int maxCount = keys.length();
        System.out.println("The length is "+maxCount);
        int i = 0;

         for (Character code : codeList) {

             int key = Character.getNumericValue(keyList[i]);

                 if(key % 2 == 0)
                    {
                      int res = code+key;
                      result.append((char)res);
                    }
                    else
                    {
                       int res = code-key;
                      result.append((char)res);
                    }
                 i++;
                 if(i==maxCount)
                 {
                    i = 0;
                 }
             }
         System.out.println("The result is "+result.toString());
        }
}

Upvotes: 0

schmidt73
schmidt73

Reputation: 191

There are many methods of Encryption, but if you want to learn about Encryption you should start with the study of XOR encryption. XOR Encryption uses a key and XORs the binary code of every character with the key. If the key is longer than the encrypted code it creates a One-Time Pad that is impossible to decrypt.

XOR - Exclusive OR - Unlike OR both values can not be true at the same time.

Simple Explanation:

  1. Pretend you want to encrypt the string "hello world" with the key 'c'.
  2. For every character in the string XOR it with the key c.

  3. Pretend the binary value of h is 1100011 and the binary value of c is 0010110 (these are made up and will not work) then you XOR every corresponding binary value.

    1100011
XOR
0010110
-------
1110101

1110101 is the XORed binary value.

  1. You then cast the binary value back into character and you do this for every step of the encrypted string.

Problems:

Insecure for short keys. but very powerful for long keys and creates a one time pad.

Example code:

http://www.ecestudents.ul.ie/Course_Pages/Btech_ITT/Modules/ET4263/More%20Samples/CEncrypt.java.html

Upvotes: 0

Martin Dinov
Martin Dinov

Reputation: 8815

Here are possible steps you can take to do this. This is not an exact and working solution, but it will hopefully get you started.

  1. Start by reading input from the console (via Scanner or a BufferedReader for example).
  2. Split your input on spaces perhaps, so that you have a String[] of words.
  3. Loop through the String[] of words, and loop again for which word. You can have a counter that is incremented in each iteration of an inner loop and gets reset at the end of an inner loop. You can use that counter variable to get a position into the key (key[counter%lengthOfKey]) in each iteration of the inner loop. If the (counter%lengthOfKey)%2 == 0, you have the even number case for the key, else the odd numbered case. Do whatever encryption at that point (simple substitution cipher for example).

Upvotes: 1

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