David Malone wrote: > On Tue, Dec 27, 2005 at 03:26:59AM -0600, Travis H. wrote: >> On 12/26/05, Ben Laurie <[EMAIL PROTECTED]> wrote: >>> Surely if you do this, then there's a meet-in-the middle attack: for a >>> plaintext/ciphertext pair, P, C, I choose random keys to encrypt P and >>> decrypt C. If E_A(P)=D_B(C), then your key was A.B, which reduces the >>> strength of your cipher from 2^x to 2^(x/2)? > >> Almost true. The cardinality of the symmetric group S_(2^x) is >> (2^x)!, so it reduces it from (2^x)! to roughly sqrt((2^x)!). That's >> still a lot. > > I'm fairly sure knowing that E(P) = C reduces the key space from > (2^x)! to (2^x - 1)!, because you've just got to choose images for > the remaining 2^x - 1 possible blocks. > > I think a problem with Ben's arument is in assuming that knowing > E_A(P)=D_B(C) tells you that your key was A.B. For example, suppose > my key K is the permutation: > > 1 -> 2 > 2 -> 3 > 3 -> 4 > 4 -> 1 > > and my P = 2. Now we know E_K(P) = C = 3. Ben guesses A: > > 1 -> 1 > 2 -> 3 > 3 -> 2 > 4 -> 4 > > and B: > > 1 -> 1 > 2 -> 2 > 3 -> 3 > 4 -> 4 > > He sees that E_A(P) = E_A(2) = 3 = D_B(3), and so assumes that K = > A.B. But A.B = A != K. > > (In this example, imagine x = 2, and we label the blocks 00 = 1, > 01 = 2, 10 = 3, 11 = 4.)

If you don't have sufficient plain/ciphertext, then of course you can choose incorrect pairs. -- http://www.apache-ssl.org/ben.html http://www.thebunker.net/ "There is no limit to what a man can do or how far he can go if he doesn't mind who gets the credit." - Robert Woodruff --------------------------------------------------------------------- The Cryptography Mailing List Unsubscribe by sending "unsubscribe cryptography" to [EMAIL PROTECTED]