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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?

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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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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