# Behavior of Iterated 3-DES

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)

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

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