Re:McGill will double your password if you don’t do it first
by dllthomas on Wednesday November 19, 2014 @07:40PM
(#8633740)
I don't think this follows. "Random output" is an idealized model that no actual, specific hash function achieves. I already said that having such an F would worry me, but "if F is computable, H is not secure" is a strong statement and I'm wondering whether it's backed up by math, not-quite-math-but-good-reasoning, or handwavy bluster. Please don't take this as an attack, though - I appreciate the attempt!
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Re:McGill will double your password if you don’t do it first
by bmm6o on Thursday November 20, 2014 @05:56PM
(#8638727)
I don't think I'm able to make a much more rigorous argument, but I will note that you seem to be applying two different criteria here. If you want to criticize modeling a hash function as a PRF as too idealized, then you aren't going to get a mathematical answer (since it will start with "let H be a PRF").
Re:McGill will double your password if you don’t do it first
by dllthomas on Thursday November 20, 2014 @06:43PM
(#8638928)
A maximally strong mathematical answer is showing that given that F and H(a), we can reconstruct too much about a, for any H and corresponding F. There are probably other similarly strong forms of argument - I'm not saying that's the only one - but you can see how it differs from "Well, it's just not random."