Alter plaintext through IV in block cipher AES in CBC mode

Question

IV: 69bb375aa919c72aa561dd7b7ffacf66

Ciphertext: "c28ea025d10186510855d083747d6435c808d9857075abd9bfc54b094e60eaf3"

Plaintext: "SEND ENCRYPTED DATA"

I want to be able to change the resulting Plaintext to "SEND THE DATA", which if I understand correctly would involve altering the IV in this case.

I have read about the bit flipping and how this attack works, I am just having troubles implemeting it in this scenario, please help.

below the encryption (at the server side- we were given the code), includes the padding used,

  def encrypt(plaintext):
            log.info("[%d] encrypt data command invoked" % (client_id))
            iv = Random.new().read(AES.block_size)


        # add padding
        if (len(plaintext) % 16 != 0):
            plaintext += b"_" * (16 - len(plaintext) % 16)

        cipher = AES.new(shared_key, AES.MODE_CBC, iv)
        ciphertext = cipher.encrypt(plaintext)

        return hexlify(iv) + b"," + hexlify(ciphertext) + b"\n"

   def no_tranform(plaintext):
        log.info("[%d] send data command invoked" % (client_id))
        return hexlify(plaintext) + b"\n"

    commands = {"SEND ENCRYPTED DATA": encrypt,
                "SEND THE DATA": no_tranform}

Also you can see that the only allowed commands are "SEND ENCRYPTED DATA" and "SEND THE DATA" and I would like the latter.

Answer 1

Each bit you flipped in the IV changes the same bit in the decrypted plaintext. So if you want to change the first E character of ENCRYPTED into a T of THE then you need to XOR the binary representation of E and T (probably ASCII) and XOR that with the byte in the IV. This will of course only work for the first block of 16 bytes. Fortunately "SEND THE DATA" is small enough for that (13 bytes if I count correctly).

That leaves the problem of the padding. The original block contains the padding in the second block because of the size (19 bytes). However, you just have to adjust the padding of the first block by performing the XOR as displayed above. Instead of XOR'ing the final 16 - 3 = 3 plaintext bytes with a letter you XOR them with the value 03 in hex and XOR that with the IV. Now you've also generated the required padding.

Finally, you need to leave out the second block. And that's all me wrote, happy programming.