Meet in the middle on 2DES uses $2^{56}$ memory.
Given the fact that the attacker has only $2^{45}$ memory.
How can the attacker adjust the attack so even with this memory limit, it will still be more efficient than looking after all the existing keys?
I guess you can separate this to a few look...
A 2DES like cipher $c=E^{(2)}_{K_2}(E^{(1)}_{K_1}(p))$ where both halves have an $n$ bit key is vulnerable to a meet-in-the-middle attack.
Meet-in-the-middle using a big table
Create a table containing $E^{(1)}_{K_1}(p)$ for all possible $K_1$ and computing $D^{(2)}_{K_2}(c)$ for all values of ...
@all As an aside: I just stumbled upon a paper called “Romantic Cryptography” (PDF). If you don´t know it yet, read it. It might make you smile… (that is, until someone asks us if it´s also quantum-secure)
@e-sushi I actually found them (well, at least the first and ignored the second) but none of them asks "How can I break 2DES?" but rather ask about specifics of the meat-in-the-middle attack ("how does run-time change if I reduce memory?" / "How to use MitM using cycle finding?").
This means we've got our very first "How do I break 2DES?" question today (after 3 years!).
This is a question I had in my exam today, and I'll be glad if someone can help me to find the answer.
A student built an encryption algorithm (something between DES and 3DES), in which the encryption is based on 2 keys, K1 and K2, and calculated this way:
Cipher = E_K1 (E_K2 (plaintext)
wher...
Guess so… or maybe not, because MITM is currently the best-known way to attack it (as the answer there describes). Oh, and btw: it´s called “meet-in-the-middle attack”, not “meat-in-the-middle attack” (meat = steak for example) – but something tells me that that mistake has become a bit of a trend lately. ;)
One thing´s for sure… it´s time for dinner at my side of the planet and all this meat-talk just decided I´m gonna grill myself a big steak tonight. Thanks for the inspiration. Talk to you later!