AES-CBC collision resistance of hash function

George

11/21/23, 10:41 AM

I'm very new to cryptography and am having some issues with the following question

A hash function H encrypts an n-block message m = (m1, m2 ... mn) in CBC mode:
- Block size is 128 bits
- Hash result is the last ciphertext
- IV is arbitrary 1 block long
- Encryption key is arbitrary number -> 128, 192 or 256 bits
- IV and Key are public

I need to find out if this function is collision resistant

If it is I need to provide proof
If it's not I need to provide a pair of collisions m and m' s.t -> H(m) = H(m')

Also discussing cases where block size varies over n=1,2,3

Any help would be greatly appreciated, thanks

117

0 + 5

aes

cbc

collision-resistance

Morrolan

11/21/23, 10:51 AM

Take a look at a figure showing how the CBC mode of operation works in the encryption direction. Try to answer the following: If you change a single bit of the message, in which way will this affect a) what comes out of AES for that block, b) what **goes in** to AES in the next block, and c) the final block of ciphertext (= the hash value). Can you use this knowledge to find two messages which collide?

George

11/21/23, 11:04 AM

@Morrolan thanks for your comment. I can already understand that for one block there will be no collisions and for anything >1 block there's a chance for collisions. Just having some issues actually proving that this is the case

Marc Ilunga

11/21/23, 11:12 AM

What is the origin of this question? There are a number of somewhat unrelated topics bundled together (e.g.: encryption and collision resistance). Another question is how does decryption work then?

fgrieu

11/21/23, 11:30 AM

Hint: how can you find the (unique, as you comment) 1-block message that has a given hash? Slightly adapt that technique to find several more-than-one block messages with a given hash.

Maarten Bodewes

11/21/23, 11:44 AM

Try this with a two block message just containing zero bits to make your life easier, then create another two block message that collides. Remember that XOR is associative, you can rearrange any formula, i.e. $P_1 \oplus C_0 = I_1$ means that $P_1 = \cdots$.

Elon Musk

I sit in a Tesla and translated this thread with Ai:

EN: AES-CBC collision resistance of hash function

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