ECBC algorithm

November 19, 2021

This is the second installment of our MAC series, read the previous posts to have a better understanding of what a MAC is and how it can be used to guarantee message integrity.

  1. Message Authentication Code (MAC)
  2. Encrypted CBC-MAC (ECBC)
  3. NMAC
  4. PMAC
  5. HMAC
  6. Poly1305-AES

In this post we will learn about MACs constructed using block ciphers by learning about the ECBC.

ECBC

ECBC, or encrypted CBC-MAC is a technique for constructing a message authentication code from a block cipher. Messages are encrypted with some block cipher algorithm in CBC mode to create a chain of blocks such that each block depends on the proper encryption of the previous block.

We have a previous posts explaining how CBC works, it will be helpful to have a grasp on the subject before moving forward. We will start by investigating how ECBC works through a diagram and we can discuss the implementation details afterwards.

ECBC by Tom Fitzhenry

Similarities between ECBC and CBC:

  1. Like CBC we have an initialization vector (IV) which is usually set to zero in this algorithm.
  2. Our message is still divided into blocks of 16 bytes assuming the encryption process is based on AES-128.
  3. Every encrypted block will be XORed with the next block before being encrypted under the same key.

Differences between ECBC and CBC:

  1. After the last block is encrypted we have a second key that will encrypt the final result of the chain.

It’s crucial for this second encryption to use a different key, without it we could forge a message from a previous message and tag. Before we attempt to exploit this vulnerability, let’s implement this algorithm.

Implementation

Most of the code here was already covered up in previous posts, the most relevant one for our understanding being the CBC Bitflipping Attack. Our example will assume that our messages are block aligned and no padding is required for now.

require 'openssl' BLOCK_SIZE = 16 KEY1 = BLOCK_SIZE.times.map { rand(0..255) } KEY2 = BLOCK_SIZE.times.map { rand(0..255) } ZERO_IV = Array.new(BLOCK_SIZE, 0) MESSAGE = 'ECBC is not as hard as I thought! I want a tag!!'.bytes def aes_ecb_encrypt(message, key) cipher = OpenSSL::Cipher.new('AES-128-ECB') cipher.encrypt cipher.key = key.pack('C*') cipher.padding = 0 # no padding result = cipher.update(message.pack('C*')) + cipher.final result.unpack('C*') end # Slightly modified from our previous implementation, now # we are just keeping the last encryption as opposed to the # encryption for every block. def aes_cbc_encrypt(message, key, iv) blocks = message.each_slice(key.size) previous_encryption = iv blocks.each do |block| xor = previous_encryption.zip(block).map { |a, b| a ^ b } previous_encryption = aes_ecb_encrypt(xor, key) end previous_encryption end def ecbc_sign(key1, key2, message) cbc_result = aes_cbc_encrypt(message, key1, ZERO_IV) aes_ecb_encrypt(cbc_result, key2).pack('C*') end def ecbc_verified?(key1, key2, message, tag) ecbc_sign(key1, key2, message) == tag end tag = ecbc_sign(KEY1, KEY2, MESSAGE) puts ecbc_verified?(KEY1, KEY2, MESSAGE, tag) # true puts ecbc_verified?(KEY1, KEY2, MESSAGE, 'wrong') # false

As we can see every step of the chain requires the computation of the previous step, making the algorithm impossible to parallelize. Let’s move on and explore the reason we need a second key in ECBC.

Example of an implementation mistake

Suppose we forgot to encrypt the final result of our CBC-MAC chain using a different key, how can this be exploited in order to forge a new message?

To make it clear we are changing this method:

def ecbc_sign(key1, key2, message) cbc_result = aes_cbc_encrypt(message, key1, ZERO_IV) aes_ecb_encrypt(cbc_result, key2).pack('C*') end


to:

def ecbc_sign(key, message) aes_cbc_encrypt(message, key1, ZERO_IV) end

Attack

Let’s imagine we have a message with a single block and we want to generate a tag for it.

The way the tag would be generated is:

Step 1: result = zero_iv XOR message
Step 2: tag = ENCRYPT(key, result)

How can we forge a new message with two blocks and the same tag based on this knowledge?

  1. Repeat the message for the first block
  2. Create a second block with value tag XOR message

This is all we need to generate the same tag for and entirely different message! Let’s deconstruct the operations to see why:

Step 1: result = zero_iv XOR message
Step 2: c0 = ENCRYPT(key, result)
Step 3: c0 XOR tag XOR message
Step 4: tag = ENCRYPT(key, c0 XOR tag XOR message)
Which really is:tag = ENCRYPT(key, message)

The trick here is that c0 and tag are exactly the same, so when we XOR both they will cancel each other and only the message remains!

Let’s implement this!

require 'openssl' BLOCK_SIZE = 16 KEY = BLOCK_SIZE.times.map { rand(0..255) } ZERO_IV = Array.new(BLOCK_SIZE, 0) MESSAGE = "I'm unforgeable!".bytes def aes_ecb_encrypt(message, key) cipher = OpenSSL::Cipher.new('AES-128-ECB') cipher.encrypt cipher.key = key.pack('C*') cipher.padding = 0 # no padding result = cipher.update(message.pack('C*')) + cipher.final result.unpack('C*') end def aes_cbc_encrypt(message, key, iv) blocks = message.each_slice(key.size) previous_encryption = iv blocks.each do |block| xor = previous_encryption.zip(block).map { |a, b| a ^ b } previous_encryption = aes_ecb_encrypt(xor, key) end previous_encryption end # This is now broken and cannot guarantee integrity anymore. def ecbc_sign(key, message) aes_cbc_encrypt(message, key, ZERO_IV) end def ecbc_verified?(key, message, tag) ecbc_sign(key, message) == tag end tag = ecbc_sign(KEY, MESSAGE) forged_message = MESSAGE + tag.zip(MESSAGE).map { |a, b| a ^ b } new_tag = ecbc_sign(KEY, forged_message) puts ecbc_verified?(KEY, MESSAGE, tag) # true puts ecbc_verified?(KEY, forged_message, new_tag) # true puts tag == new_tag # true

And that’s all we need to understand the encrypted CBC-MAC algorithm.

I will leave this post mentioning that if we are able to obtain tags for two one-block messages we can also generate a new message that is the concatenation of these messages. Reach out to me via email or Twitter if you would like to share or validate your solution. :)

Our next post will cover NMAC, which is the basis for other interesting constructs. See you there!


Bernardo de Araujo

Application Security Manager @Shopify.

© Bernardo de Araujo 2023