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Meet-in-the-middle attack on Double DES

us flag

This question is straight out of the book, I cant grasp my head around what it is stating, if someone can elaborate I would appreciate it. A meet in the middle attack on a double DES if the chosen plaintext is available, and the attacker recovers a 112-bit key, if the same work is needed for an exhaustive search to recover the 56-bit key which is about 2^55. (a) If we only have the known plaintext available, not the chosen plaintext, then what changes would we need to make to the double DES attack? (b) What is the work needed for the known plaintext in the Meet in the middle attack on the double DES?

For A I am assuming I would need to reconfigure the look up table every time the attack is done.

and B would the work factor be (2^56 ) + (2^55) since we have a known plaintext instead of the chosen?

fgrieu avatar
ng flag
It's hard to figure out what the question asks, because it's not told what the $2^{55}$ cost stated for chosen plaintext accounts for, so we have to figure out 1) if the unit is DES or 2DES-equivalent operations; 2) if that's max or average; 3) if that's making use of the DES complementation property or not; 4) if that accounts for precomputations that can be performed before the attack. As an aside the second sentence has a strange structure with two "if".
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