It's well known that by injecting a byte fault during the 8th round of AES 128 just before the mix column the key can be recovered with $2^{32}$-key space. A brute force attack is then possible but it takes some time.

If instead a byte fault, it is just a bit flip does it reduce the key space for K10? And why?

In fact I was wondering if there is a kind of propagation of a 1 bit diff between the 8th round and 9th round that could reduce the key space for K10

  • $\begingroup$ If you do a search on the web you'll find a lot of papers describing such kind of attack. $\endgroup$ – Von k Jan 1 '19 at 16:58
  • $\begingroup$ For instance this one: eprint.iacr.org/2009/575.pdf $\endgroup$ – Von k Jan 1 '19 at 17:08
  • 4
    $\begingroup$ Welcome to crypto.se - here is some advice to help improve your question and get better answers: On stackexchange sites, it is expected for you to share your research and provide examples. If you hover over the upvote button, the text that appears is This question shows research effort; it is useful and clear. Leaving it to others to go out of their way to find examples does not make for a well researched question. $\endgroup$ – Ella Rose Jan 1 '19 at 17:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.