Xor may be crack the NoPadding/CFB-AES encryption-

Hello
Looks like a password can be discovered based on the numerical value of an old known password (using the same key).

For example
In case I have an obvious password: AAA
and I have the password enrypted form of bytes: crypt (AAA).

I can decrypt any password based on the result of the Xor of the clear and encrypted password: (AAA xor crypt (AAA))
BBB = crypt (BBB) xor (xor crypt (AAA) AAA)
It looks like a security problem.
Is this a bug of java or mine?
BTW, if you use AES/CFB8/NoPadding, this is not the case.

Here is the Code example:

import javax.crypto.Cipher;
import javax.crypto.KeyGenerator;
import javax.crypto.SecretKey;
import javax.crypto.spec.IvParameterSpec;

public class MyTest
{
     static SecretKey m_secretKey = null;

     public static void main(String[] args)
     {
          try
          {
               MyTest test = new MyTest();
               test.init();
               test.encrypttionBreak();
          }
          catch (Exception e)
          {
               e.printStackTrace();
          }

     }

     private void init() throws Exception
     {
          if (m_secretKey == null)
               m_secretKey = KeyGenerator.getInstance("AES").generateKey();
     }



     private byte[] aesEncrypt(byte[] data, int mode) throws Exception
     {
          byte[] m_iv = new byte[16];

          for (int i = 0; i < m_iv.length; i++)
          {
               m_iv[i] = (byte) i;
          }

          IvParameterSpec iv = new IvParameterSpec(m_iv);
          Cipher cipher = Cipher.getInstance("AES/CFB/NoPadding", "SunJCE");
          cipher.init(mode, m_secretKey, iv);

          byte[] cipherInput = cipher.doFinal(data);

          return cipherInput;
     }

     public void encrypttionBreak() throws Exception
     {
          String clearText = "aaaaaaaaaaaa";
          byte[] cipherBytes = aesEncrypt(clearText.getBytes(),
               Cipher.ENCRYPT_MODE);

          byte clearBytes[] = clearText.getBytes();
          byte xorKey[] = new byte[clearBytes.length];
          for (int i = 0; i < xorKey.length; i++)
          {
               xorKey[i] = (byte) (clearBytes[i] ^ cipherBytes);

          }



          String password = "mypass";



          byte[] passwordEncrypted = aesEncrypt(password.getBytes(),

               Cipher.ENCRYPT_MODE);



          byte breakFinal[] = new byte[passwordEncrypted.length];

          for (int i = 0; i < breakFinal.length; i++)

          {

               breakFinal[i] = (byte) (passwordEncrypted[i] ^ xorKey[i]);

          }



          System.out.println("break final " + new String(breakFinal));



     }



}





Edited by: sabre150 on 19-Aug-2011 02:37



Moderator: added [ code] tags to make the code readable.

This is your bug, because you randomly pick a mode and a method without really understanding them. Just look at the definition of [https://secure.wikimedia.org/wikipedia/en/wiki/Block_cipher_modes_of_operation#Cipher_feedback_.28CFB.29]

Now, look at what your function "crypt". You use a fixed iv for everything, we just label the IV. So crypt ('AAA') = AES (key, iv) xor "AAA". That's why crypt ('AAA') xor 'AAA' = AES (key, iv). You should be able to understand what crypt('BBB") xor xor crypt ('AAA')"AAA ".

Tags: Java

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