CBC or Cipher block chaining is another block cipher mode
The goal of CBC is to not be deterministic like ECB, ie two identical plaintext blocks should encrypt to different ciphertext blocks
How does it accomplish this?
CBC requires an extra initialization vector or IV for short, which is just random bytes the size of a block
For each plaintext block we want to xor our plaintext with something first, and then encrypt it with our cipher
For the first block we xor it with the IV before encrypting
For every other block we xor it with the last ciphertext block
Essentially “chaining” the different blocks together, so the resulting ciphertext of a block depends on the “sum” of all the previous blocks and the IV
https://en.wikipedia.org/wiki/Block_cipher_mode_of_operation#Cipher_block_chaining_(CBC) has some very useful diagrams
Implementing it is fairly straightforward once you understand whats going on
Decryption is just running this in reverse, block cipher decryption first and then xor
Note: ECB mode with one block is the same as just running the cipher, so you can reuse that as long as you’re encrypting/decrypting only one block
Some clarification on this challenge, I originally thought the random prefix would be generated each time, a new random prefix of random length each time you touch the oracle
I honestly don’t really know how one would go about solving that, so I and everyone else who has write ups for cryptopals instead assumed that the random prefix would be generated once and be reused for all later oracle calls
So now the trouble is really just one thing, how long is that random prefix? How can you figure that out?
Try to think about what happens in CBC decryption with the user data block and the block after it
How can we completely replace the second ciphertext block in a way to make the third block decrypt to what we want, namely “;admin=true;”? What happens to the second plaintext block in that case?
