# Who drew on my program? 
![alt text](images/title.png)
## Challenge Description
![alt text](images/challengedescription.png)
The goal of the challenge is to calculate the IV. The challenge shows how keeping the IV  a secret does not protect against attacks. As it's easy to calculate using a known plaintext attack. 
The challenge gives us 14 bytes out of a 16 byte key and 29 bytes out of a 32 byte plaintext.
## The attack![alt text](images/wiki_CBC_decrypt.png)
By looking at the AES CBC decryption routine we see that the IV is used in the block-round when XORing the first block of the ciphertext with the first block of the plaintext. 
The basic premise of the attack is to get the first block of the ciphertext by XORing the decrypted second block of the ciphertext with the second block of the plaintext. As we don't have the full key and only parts of the first ciphertext we have to bruteforce the last two bytes of the key until we find a match with the known ciphertext bytes, either through encryption or decryption.  
When the full key is found we also have the first block of the ciphertext. To get the original IV, we simply XOR the first block of the plaintext first block of the ciphertext. 
The code for the python script which solved the challenge is found below.
``` python#!/usr/bin/python
from Crypto.Cipher import AESimport binascii, sys
# Key = "9aF738g9AkI112"+XX# IV = ?
# The last 16 bytes of the blocksizepartly_msg = binascii.unhexlify("808e200a54806b0e94fb9633db9d67f0")
plaintext = "The message is protected by AES!"
def encrypt(message, passphrase):    aes = AES.new(passphrase, AES.MODE_CBC, IV)    return aes.encrypt(message)
def decrypt(message, key):    aes = AES.new(key)    return aes.decrypt(message)
def xor(arr1, arr2):    dst = []    for i in range(len(arr1)):        dst.append(chr(ord(str(arr1[i])) ^ ord(str(arr2[i]))))    return ''.join(dst)
# First ciphertext block (IV1) = decrypt(ciphertext[16:]) ^ plaintext[16:]
# Creating all possible keyskey = "9aF738g9AkI112"possible_key_list = [] for i in range(127):    letter = chr(i)    for j in range(127):        letter2 = chr(j)        possible_key_list.append(key+letter+letter2)
CT1 = partly_msgPT1 = plaintext[16:]
key = ""IV1 = ""# Bruteforcing the key until we find the correct ciphertextfor i in possible_key_list:    iv1 = xor(decrypt(CT1, i), PT1)    if binascii.hexlify(iv1[0]) == "9e":        if binascii.hexlify(iv1[-1]) == "6a":            IV1 = iv1            key = i
#IV = decrypt(IV1) ^ plaintext[:16]IV = xor(decrypt(IV1, key), plaintext[:16])print IV
# RITSEC{b4dcbc#g}
```