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Cryptography

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  • $\begingroup$ In fact in the public (non paywall) version of the paper available on the author's page geocities.ws/dearphael/ImpMiniAES.pdf the same argument starts with obtain $2^{11}$ plaintexts P and P’ which are equal in the second and third nibble and differ in the other nibbles. Since each P and P’ forms a pair with passive second and third nibbles while the first and third nibbles are active, we then have $2^{11}$ such pairs. $\endgroup$ Commented Mar 14, 2023 at 15:22
  • $\begingroup$ so the author himself presents two inconsistent versions. $\endgroup$ Commented Mar 14, 2023 at 15:23
  • $\begingroup$ where does $2^{11}$ comes from? do we take variations of passive nibble or we just use a constant value and only change active varitaions ? @kodlu $\endgroup$ Commented Mar 18, 2023 at 15:47
  • $\begingroup$ That's exactly what's unclear to me, and it seems the author has two versions of the same claim. $\endgroup$ Commented Mar 18, 2023 at 16:06
  • $\begingroup$ well, regardless of the paper what do you think the number of the initial pairs should be? @kodlu $\endgroup$ Commented Mar 18, 2023 at 17:26