11-12-2013, 02:10 AM
Well we know the ciphertext of 12345678, password, and adobe123. So that's enough needed. Also you wouldn't brute force this you would take all 2^56 keys and encrypt one of the plaintext blocks. Storing each value. Then start from the palintext's ciphertext and take all 2^112 key pairs and use one to decrypt the ciphertext then the other one to encrypt that value. Now just look up that inside of the 2^56 stored values. Basically ever 256 tries you will need to use another plaintext and ciphertext pair to verify if it is just a collision.
This would require 600 PB of ram and 2^111.0168 (((2^56)+(2^112+2^56)+(3/256*2^112)+(3/256/2^64*2^112))/2) DES operations on average. Or about 83 billion years if you can do a quadrillion DES operations/second... Good luck.
This would require 600 PB of ram and 2^111.0168 (((2^56)+(2^112+2^56)+(3/256*2^112)+(3/256/2^64*2^112))/2) DES operations on average. Or about 83 billion years if you can do a quadrillion DES operations/second... Good luck.