When reading ''Algebraic Cryptanalysis'' by Gregory V. Bard, Springer Editions, I got confused by this sentence:

Suppose there were three naïve Cryptography students, who choose to use 3-DES iterated roughly one million times, because they are told that this will slow down a brute force attacker by a factor of one million.

I just wanted to make sure I understood this sentence fully. Does this mean that, for a given message m, we just compute 3DES(3DES(...(3DES(m))...)) one million times? (Assuming we keep the same keys)

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    $\begingroup$ IMHO. Bard is an expert on fixed points. His Keeloq attack was brilliant. $\endgroup$ – kelalaka Oct 23 '19 at 14:07

Yes, It is what you understood. You use the same key for each encryption.

If you think that they used different keys for each TDES (3DES) they need to store one million keys. This is not practical for the attacker but also for them.

It can also be written as $c = \operatorname{TDES}^{(1000000)}_k(m)$

The point in the section distinguishing these iterated from a random permutation and possible attacks that can be faster than brute-force.

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  • $\begingroup$ That's what I assumed but the result of this consideration is so ''brutal'' with the fixed points idea that I was a bit confused. Thank you for this quick answer. $\endgroup$ – Binou Oct 23 '19 at 14:10
  • $\begingroup$ Enjoy his book. As I said in comment, he is brilliant on fixed points. $\endgroup$ – kelalaka Oct 23 '19 at 14:11

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