4 ms·

This is not the case because one could enumerate the entire 64bit space and perform a rainbow table attack on your scheme.

• jefftk 7y ago
  Simple brute force is not a rainbow table. Brute forcing 2^64 bits means calculating 2^63 MD5s in expectation. You can do ~100 GHash/sec, so ~2^37/s, so about 2^29s which is 17 gpu-years. So this is doable, but incredibly expensive.
  • thisacctforreal 7y ago
    Note that a it’s an entirely different story with a “real” kdf like scrypt or bcrypt. MD5 and SHA are specifically designed to be fast to compute, they shouldn’t be used for passphrases. Figured I’d bring it up in case there’s still PHP floating around with the once-typical practice of MySQL + MD5.
  • tialaramex 7y ago
    I'm glad my eyeballing the difficulty came off here. It was tempting to pick say 128-bits of randomness and SHA-512/256 where I'd stake actual money that it just cannot be done - but that's like twice as much die-rolling and typing. On the other hand if I do 32-bits (fewer rolls) and MD5 there's probably some loser out there who has already precomputed all of those for whatever reason and then somebody finds the answer with a Bing search and doesn't end up learning anything. echo -n 'F004672790DB5B1D' | md5sum f31fb042501c2a398974feca81afd8fa -
  • esotericn 7y ago
    Got me. I suppose I thought of 64bit as now being small. 48 would be doable.