I'm analysing how different components of AES influence its security. However, I cannot find a good explanation to what happens if the MixColumn step would use the identity matrix. Intuition tells me, that then AES won't be secure, but why?
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1$\begingroup$ mixcolumns mixes the column diffusively through matrix multiplication, if you use the identity matrix, that is equivalent to not applying any multiplication, so it would not... mix $\endgroup$– Richie FrameOct 25, 2018 at 3:24
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$\begingroup$ A Stick Figure Guide to the Advanced Encryption Standard (AES) and look at ShiftRows and MixColumn. This will help you to see poncho's answer. $\endgroup$– kelalakaOct 25, 2018 at 8:36
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2 Answers
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If you eliminate the MixColumn (or equivalently, replace it with an identity Matrix), then the resulting cipher will effectively be 4 independent 32 bit block ciphers. This happens because what happens in one 32 bit row no longer has any propagation to any of the other 3 32 bit rows.
This results in a much weaker cipher; not only would that bring up possible chosen plaintext/ciphertext attacks (where you modify one of the 32 bit blocks, but leave the other 3 the same), you would also have ciphertext only attacks with a birthday bound of circa 256k...
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$\begingroup$ I would like to add that; ShiftRows and MixXomun are together the permutations of the AES to form the SPN network. $\endgroup$– kelalakaOct 25, 2018 at 9:20
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1$\begingroup$ wouldnt it be 16 independent 8-bit block ciphers? related: crypto.stackexchange.com/questions/34928/… $\endgroup$ Oct 25, 2018 at 9:31
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$\begingroup$ @RichieFrame: hmmmm, good point; I had initially thought that the ShiftRows would do the intra-row propagation; obviously, it doesn't actually do that... $\endgroup$– ponchoOct 25, 2018 at 14:21
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The identity matrix has a branch number of 2 which means there is no mixing. one active cell in the input column will produce the same active cell in the output column.
