Where did the number of initial plaintexts required for impossible cryptanalysis on Mini-AES came from?

siba36

3/11/24, 7:52 PM

I'm implementing impossible differential cryptanalysis on AES and I've started with implementing it on mini-AES to fully understand the process using R.Phan's paper as a reference.

But I don't understand the initial pairs preperation step:

In the paper the author say to obtain $2^{13}$ plaintext $P$ and another $2^{13}$ plaintext $P^{'}$ which are equal in the second and third nibble and from those plaintexts we can obtain $2^{13}$ pairs.

But where did the $2^{13}$ come from if the input difference of the differential trail has two active nibbles and each nibble is $4\;bits$ so we have $2^8$ possible input differences for the pairs ?

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cryptanalysis

aes

differential-analysis

kodlu

3/14/24, 3:22 PM

In fact in the public (non paywall) version of the paper available on the author's page https://www.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.*

kodlu

3/14/24, 3:23 PM

so the author himself presents two inconsistent versions.

siba36

3/18/24, 3:47 PM

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

kodlu

3/18/24, 4:06 PM

That's exactly what's unclear to me, and it seems the author has two versions of the same claim.

siba36

3/18/24, 5:26 PM

well, regardless of the paper what do you think the number of the initial pairs should be? @kodlu

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