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How can we attack by using chosen plaintext attack (and which plaintext should we choose) if AES didnt have the ShiftRow and Mixcolumn layers. I know each byte would be independent in the 16 byte scheme but couldnt figure out a way to attack.

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  • $\begingroup$ Think of ECB... $\endgroup$
    – Biv
    Commented Nov 1, 2017 at 10:36
  • $\begingroup$ @Biv where each of the 16 bytes in a block have a different transformation. $\endgroup$
    – DannyNiu
    Commented Nov 1, 2017 at 10:47
  • $\begingroup$ @DannyNiu A different transformation sure, but this time you can enumerate all the possible mappings ... ;) $\endgroup$
    – Biv
    Commented Nov 1, 2017 at 10:50
  • $\begingroup$ related crypto.stackexchange.com/questions/34928/… $\endgroup$ Commented Nov 2, 2017 at 1:15

1 Answer 1

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After the key expansion, there would be 16 different mappings for each of the 16 bytes in a block.

Suppose we are attacking the block cipher itself (in ECB mode), with each byte having 256 different values, we can query 256 ciphertext blocks (all-bytes-0, all-bytes-1, ... all-bytes-255) to get the exact plaintext-to-ciphertext mapping for all 16 bytes's 256 values.

To decrypt future blocks, you just need to have a 16x256=4096 byte (1 memory page) table of the inverse map which you can easily create now.

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