What are the best known cryptanalytic attacks against AES-128 with 9 rounds?

I found many such attacks on AES-192 and AES-256 with 9 rounds, but not for AES-128 with 9 rounds.

Are there any cryptanalytic attacks that show that such a version of AES-128 would be weaker than the 10 rounds one?


1 Answer 1


The only academic paper for 9 rounds of AES-128 that I can find is Structural Evaluation of AES and Chosen-Key Distinguisher of 9-Round AES-128 at here.

I also find it weird that this hasn't been covered as much in academic literature, although it perplexes me why someone would choose 9-Rounds. As far as I'm aware, the CPB (Clocks Per Byte) on normal AES is already pretty low. Just seems like a poor optimization attempt.

If you have worries about AES-128 and this is why you are searching the literature for attacks, then don't worry, AES-128 is far from being broken, let alone 256.

  • 5
    $\begingroup$ Attacks against reduced-round versions of iterative ciphers isn't an attempt to see if a reduced-round version is secure enough to actually use. It's about determining the security margin of a cipher and trying to adapt attacks to more rounds. $\endgroup$
    – forest
    Commented Jan 9, 2023 at 2:04
  • $\begingroup$ @forest Thanks for the clarification! Have any reduced-round cryptanalysis attempts resulted in a successful (theoretical) attack? $\endgroup$
    – user104975
    Commented Jan 10, 2023 at 0:22
  • 2
    $\begingroup$ Any reduced-round attack is an attack against a reduced-round variant of an algorithm, not the full algorithm. It provides insight into the cipher's security, but an attack that works against N-1 rounds of a cipher with N rounds is not an attack against the full cipher. Also note that not all attacks are practical. There are attacks against the full 14 rounds of AES-256, but they're not much better than brute force. Likewise there are also "impossible differential attacks" which are performed against a cipher with a certain internal state that has a probability of 0 of actually existing. $\endgroup$
    – forest
    Commented Jan 10, 2023 at 0:33
  • $\begingroup$ @forest I see. Thank you. $\endgroup$
    – user104975
    Commented Jan 10, 2023 at 0:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.