ECB decryption (Simple)

June 27, 2021

In this post we will investigate the byte-at-a-time ECB decryption exercise from Cryptopals. I highly recommend attempting the previous exercises yourself as they do a great job ramping up your knowledge on the subject.


The goal of this exercise is to figure out the contents of an unknown message that we know that will be encrypted under ECB.

The only thing we control in this exercise is the fact that we can add a prefix to this unknown buffer before encryption.

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

  1. Get our username (we control that)
  2. Concatenate it with something that is under the server control, let’s say ;admin=false;
  3. After steps 1 and 2 the server will encrypt something like: our_provided_name;admin=false;
  4. The server surfaces the encrypted value as a cookie

Our goal is discover that the server is appending ;admin=false; to our provided input.

This could be translated in Ruby as:

# Generating a random 16-bytes key to encrypt our unknown message with. # We can think of this key as something the web server holds. RANDOM_KEY = { rand(0..255) } # The ;admin=false; in our example above UNKNOWN_BUFFER ='unknown_buffer').strip def encryption_oracle(controlled_buffer) # Ignore this PKCS7_PAD function, all it does for our purposes in guarantee # that our message is a multiple of 16 bytes. We will understand the reason behind # this in the "ECB Encryption" section. padded_buffer = pkcs7_pad( controlled_buffer + UNKNOWN_BUFFER, 16 ) # Our ECB encryption using our RANDOM KEY # We will implement this method below. aes_ecb_encrypt(padded_buffer, RANDOM_KEY) end

ECB Encryption

Let’s see what ECB encryption actually is visually and talk about the reasons why encrypting things with ECB is not considered secure.

ECB Encryption by WhiteTimberwolf

According to the image above we have a plaintext (message) that we want to encrypt, and what the algorithm does is:

  1. Split our plaintext in blocks of 8 or 16 bytes, we will be using 16 bytes for our example
  2. Encrypt each of these blocks using a 16-byte key (the same size as our blocks)

This is considered insecure since encrypting the same block will always yield the same ciphertext (encoded block), and due to this characteristic we can observe the following characteristics from observing ciphertexts encrypted under the same key:

  1. Detect if plaintexts are equal or not
  2. Detect if plaintexts share a common prefix
  3. Detect if plaintexts share common substrings (depends on block alignment)
  4. Detect repetitive data in a plaintext

What happens when a plaintext is not a multiple of 16 bytes?

In this case we can use a padding algorithm, a commonly used one is called pkcs7.

Example in Ruby

Let’s use blocks of 16 bytes (128 bit encryption), meaning our message will be split into blocks of 16 bytes each.

We will create two plaintexts that share a common prefix of 16 bytes (the size of a block). What we want to prove is that the first block of each encrypted message will also be the same.

require 'openssl' KEY = 'supersecretkey?!' BLOCK_SIZE = 16 def aes_ecb_encrypt(buffer, key) raise 'Buffer must be composed of 16-byte chunks' unless (buffer.size % 16).zero? cipher ='AES-128-ECB') cipher.encrypt cipher.key = key.pack('C*') cipher.padding = 0 # we don't want padding in our example result = cipher.update(buffer.pack('C*')) + result.unpack('C*') end # The only difference between this method and the one above is the "cipher.decrypt" line. def aes_ecb_decrypt(buffer, key) raise 'Buffer must be composed of 16-byte chunks' unless (buffer.size % 16).zero? cipher ='AES-128-ECB') cipher.decrypt cipher.key = key.pack('C*') cipher.padding = 0 # we don't want padding in our example result = cipher.update(buffer.pack('C*')) + result.unpack('C*') end # The first 16 bytes are the same for both plaintexts, which is: "This message has" plaintext1 = 'This message has lots of content' plaintext2 = 'This message has amazing content' encrypted_plaintext1 = aes_ecb_encrypt( plaintext1.bytes, KEY.bytes ) # Sanity test so we can be sure our encryption/decryption works if aes_ecb_decrypt(encrypted_plaintext1, KEY.bytes) puts "Decrypting our encryption works!" end ciphertext1_first_block = aes_ecb_encrypt( plaintext1.bytes, KEY.bytes ).slice(0, BLOCK_SIZE) ciphertext2_first_block = aes_ecb_encrypt( plaintext2.bytes, KEY.bytes ).slice(0, BLOCK_SIZE) if ciphertext1_first_block == ciphertext2_first_block puts "Encrypting the same plaintext content yields the same ciphertext!" end # We can also see that the same plaintext encrypted twice will result in the same ciphertext first_encryption = aes_ecb_encrypt( plaintext1.bytes, KEY.bytes ) second_encryption = aes_ecb_encrypt( plaintext1.bytes, KEY.bytes ) if first_encryption == second_encryption puts "The same message will always yield the same encrypted result in ECB." end

Byte-at-a-time ECB decryption

With the knowledge of the algorithm in place, how would we attempt to reverse engineer our unknown message encrypted with ECB?

Let’s recap what we know so far:

  1. Our unknown message will be split into blocks of 16 bytes and each of these blocks will be encrypted under an unknown key
  2. We control the prefix, so we also control whatever our initial blocks will be
    • Take a moment to truly understand this statement

It’s time to work with a contrived example. :)

Contrived Example

Assume that:

  1. Our block size is 4 bytes
  2. Our buffer is [0,1,2,3,4,5,6,7] (the message that the server appends and we do not know)

We control our prefix, so let’s add add 3 ‘A’s (block size minus one) to our message. The final string that will be encrypted ends up being:
[A, A, A, 0, 1, 2, 3, 4, 5, 6, 7]

Notice that our first block is [A, A, A, 0] (4 bytes) and we know the first three bytes (our prefix), but we do not know the last byte of this block.

We let the server encrypt this value and take note of it.

With the knowledge that the same block encrypted under the same key produces the same ciphertext we can start our enumeration process. We want to feed to the server the same first block that it encrypted when we sent our three ‘A’s. To do that we need to send our three ‘A’s yet again plus the byte we want to enumerate.

  • [A, A, A, 0], [A, A, A, 1], [A, A, A, 2] and so on.
    • We now store every attempt and compare it with the original encrypted block

By comparing the encrypted value of [A, A, A, 0] (original encryption) with our dictionary we find that the first byte is 0, so it’s time to find the second unknown byte. To do this we need to send two ‘A’s this time instead of three! This is the case since we already know one byte of the buffer, and we want to figure out the second unknown byte. Our output after sending two ‘A’s is:
[A, A, 0, 1, 2, 3, 4, 5, 6, 7]

This means our first block is [A, A, 0, 1] and we again know the first three bytes, so we can enumerate only the last one.

  • [A, A, 0, 0], [A, A, 0, 1], [A, A, 0, 2] and so on.
    • We now store every attempt and compare it with the original encrypted block

By comparing the encrypted value of [A, A, 0, 1] (original encryption) with our dictionary we find that the second byte is 1.

We repeat the process always sending one less ‘A’ until we figure out the first full block of our unknown message, which is composed of the bytes
[0, 1, 2, 3].

It’s time to decode the second block, so let’s send again 3 ‘A’s, ending up with:
[A, A, A, 0, 1, 2, 3, 4, 5, 6, 7].

We are interested in our second block, which is [1, 2, 3, 4] and we know the first three bytes! So we can start our enumeration steps all over again.

We just need to repeat this same process until we have the full string.

Formal algorithm

  1. Craft an input block that is exactly 1 byte short (for instance, if the block size is 8 bytes, make “AAAAAAA”)
  2. Make a dictionary of every possible last byte by feeding different strings to the encryption function; for instance, “AAAAAAAA”, “AAAAAAAB”, “AAAAAAAC”, remembering the first block of each invocation.
  3. Match the output of the one-byte-short input to one of the entries in your dictionary. You’ve now discovered the first byte of unknown-string. Repeat for the next byte.


KEY_SIZE = 16 RANDOM_KEY = { rand(0..255) } UNKNOWN_BUFFER = base64_decode('12.txt').strip) BYTE = 'A'.ord def decrypt_byte(target, controlled_prefix, current_block) (0..255).each do |byte| # Our encryption function from the beginning of our post encryption = encryption_oracle(controlled_prefix + [byte]) return byte if encryption.slice(current_block * KEY_SIZE, KEY_SIZE) == target end raise "This shouldn't happen!" end def decrypt_aes_ecb known = [] UNKNOWN_BUFFER.size.times do current_block = known.size / KEY_SIZE prefix_size = (KEY_SIZE - known.size - 1) % KEY_SIZE prefix =, BYTE) controlled_encryption = encryption_oracle(prefix) known << decrypt_byte( controlled_encryption.slice(current_block * KEY_SIZE, KEY_SIZE), prefix + known, current_block ) end known end puts "Message found!\n\n#{decrypt_aes_ecb.pack('C*')}"

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