ECB Cut and Paste Attack

July 11, 2021

In this post we will investigate the ECB cut and paste attack exercise from Cryptopals. I highly recommend attempting the previous exercises yourself as they do a great job ramping up your knowledge on the subject.

Let’s remember a key characteristic of ECB, which is:

Encrypting the same plaintext under the same key will always yield the same ciphertext

This characteristic will be the key to solving our exercise.


The goal of this exercise is to send to our server an input that once decrypted elevates our privileges to role=admin.

To make this exercise more relatable we can think of a web server performing an operation like:

  1. Read the user’s email
  2. Generates an output like:
  3. The server encrypts the value from the previous step
  4. The server surfaces the encrypted value as a cookie
  5. Once the user comes back to the site the server will read this cookie’s value and identify our user

Let’s see the code:

# The key used by the web server to encrypt and decrypt values KEY = { rand(0..255) } # Our helper method to encode a hash as a query string # Outputs something like: `` def encode_query_string(hash) { |k, v| "#{k}=#{v}" }.join('&') end # Our helper method to transform a query string into a Hash def decode_query_string(input) input.split('&').map { |kv| kv.split('=') }.to_h end # Our helper method to pad our input with PKCS7 and encrypt # it with ECB. # # See the post "Discovering ECB block size" for more details # on PKCS7. def encrypt_profile(input) aes_ecb_encrypt(pkcs7_pad(input, 16), KEY) end # Our helper method to decrypt a string with ECB, remove # its PKCS7 padding and calling the query string decoder. def decrypt_profile(input) decode_query_string( pkcs7_unpad(aes_ecb_decrypt(input, KEY)).pack('C*') ) end # The method that: # 1. Receives the user's email # 2. Strips dangerous characters like "&" and "=" # 3. Generates a hash representing our session # 3. Calls our encode_query_string method with the session def profile_for(email) encode_query_string({ 'email' =>'&=', ''), 'uid' => 10, 'role' => 'user' }) end

Let’s see the code in action:

input = "" ciphertext = encrypt_profile(profile_for(input).bytes) puts ciphertext.inspect # => [101, 60, 198, 183, 84, 111, 86, 117, 34, 98, 147, 118, 111, 130, 197, 153, 251, 158, 167, 73, 230, 124, 163, 111, 170, 87, 166, 18, 76, 188, 240, 247, 65, 25, 192, 127, 154, 7, 138, 219, 54, 209, 69, 244, 112, 148, 35, 17] profile = decrypt_profile(ciphertext) puts profile.inspect # => {"email"=>"", "uid"=>"10", "role"=>"user"}

Nothing fancy, we end up with a Hash with keys email, uid and role as expected.

Thought process

A few important things to keep in mind before we start:

  1. We know that the server encodes a value like the following:
  2. We know the block size, which is 16
  3. We know our string is padded with PKCS7 before encryption

I didn’t know anything about this attack before attempting to solve it, but the name of the exercise gives us a hint at how to solve our problem: Cut and Paste Attack

Maybe it is telling us that we need to cut blocks and paste them in a different order?

Plan of Attack

My plan of attack to solve this exercise was to figure out an email length that once encrypted will put the role= and user into different blocks, this way I would be able to maybe cut another block and replace it with the one that has the user value. Let’s see how we can do that:

  1. We know that email= length is 6, so to complete the first block we need 10 more characters.
  2. We know that &uid=10&role= is 13 characters, so we need 3 more characters to complete another block.

So if we send an email that is composed of 13 characters we end up with the following blocks:

Block 0: email=AAAAAAAAAA Block 1: AAA&uid=10&role= Block 2: user

The issue with this scenario is that there’s no block we can cut and paste in order to transform our role into admin.

My next attempt was to simply write 16 more characters to create yet another block in-between the email and role blocks, something like:

Block 0: email=AAAAAAAAAA Block 1: AAAAAAAAAAAAAAAA Block 2: AAA&uid=10&role= Block 3: user

This is much better! If we manage to make Block 1 represent “admin”, and we cut and paste it after the role block we can become admin:

Block 0: email=AAAAAAAAAA Block 2: AAA&uid=10&role= Block 1: AAAAAAAAAAAAAAAA

We know that Block 1 should yield admin in order to end up with role=admin, but how do we send only admin and nothing else?

  1. admin is 5 bytes long
  2. What do we do with the rest 11 bytes that cannot be used?

That’s when I remembered that we are padding our input with PKCS7, so we can likely exploit this by pretending our first block was padded.

If PKCS7 were to pad our first block we would end up with: admin\v\v\v\v\v\v\v\v\v\v\v. Don’t be confused by the \v character, it is just the visual representation of 11 in ASCII, which is the vertical tab.

Why 11? Because it is the number of bytes we need to pad our block with so it ends up with 16 bytes.

Let’s recap what we want to achieve:

  1. Send 10 characters in order for our first block to be email=<our_10_characters>
  2. Send admin\v\v\v\v\v\v\v\v\v\v\v so we can cut and paste this block
  3. Send 3 more characters so we have <our_3_characters>&uid=10&role=

Combining all of the above we end up with:

Block 0: email=AAAAAAAAAA Block 1: admin\v\v\v\v\v\v\v\v\v\v\v Block 2: AAA&uid=10&role= Block 3: user

After after cutting our Block 1 and pasting it on top of Block 3:

Block 0: email=AAAAAAAAAA Block 2: AAA&uid=10&role= Block 1: admin\v\v\v\v\v\v\v\v\v\v\v

Let’s see this in Ruby:

input = "AAAAAAAAAAadmin\v\v\v\v\v\v\v\v\v\v\vAAA" ciphertext = encrypt_profile(profile_for(input).bytes) block0 = ciphertext.slice(0, 16) block1 = ciphertext.slice(16, 16) block2 = ciphertext.slice(32, 16) ciphertext = block0 + block2 + block1 profile = decrypt_profile(ciphertext) puts profile # => {"email"=>"AAAAAAAAAAAAA", "uid"=>"10", "role"=>"admin"}

And we have reached the end of our exercise! Congratulations, take a moment to be proud of what you have achieved and I hope you are looking forward to the next posts as much as I am. :)

Bernardo de Araujo

Application Security Manager @Shopify.

© Bernardo de Araujo 2023