With that analysis, the important conclusion is that we only need to attain overall coverage of just under 1.5% to get a match. That's almost two orders of magnitude less than a full search!
Chris
· 6 months ago
Seeing that there is a 37% chance that there will be no such code, I don't see how you can claim that only searching 1.5% would yield the desired result. Do you mind giving me the derivation?
Barney
· 4 months ago
Yeah, I totally jacked my "math". I realized this soon after, but didn't come back here and correct my statement. Though now that I'm looking, I don't think I can anyway.
ThorErik
· 4 months ago
I doubt it exsists, considering, which guy making such a algorithm would not make sure that would never happen? I would at least try to prevent it, since it kind of breaks the purpose of the algorithm(a one-way algorithm)
I'm still trying though, got 3 computers working on it with different ways of attacking(among one using the cuda c++ version) and a couple of versions I've made myself one in bash and one in php, plus trying a few of the pre-made ones(with some modification)
Ograll
· 4 months ago
If the designers knew that it didn't exist it would break one of the fundamental precepts of a Hash, that is given an output it is infeasible to construct the input. You also look at the Kember Identity as a negative. Its existence, or more importantly the inability to have a certainty about its existence is a strength of the hash since its one less thing you know about the Hash. When reversing a hash knowledge is an important part of the game, but if you can make certain statements, then the hash loses some of its strength.
All Comments
Is that just the longest common substring?
md5(a) = b
md5(b) = c
md5(c) = a
and so on. Nobody's found one yet, and there isn't any code which checks - but it's a good reason to print out your current hash every so often.
Not exactly a complete match but I'm done wasting cycles..
md5('deadbeefdc84955dfb53442f741f4ec9') == 1ebb63c68f0e40e6902e0deadfeefbfa
In[4]:= Limit[((n-1)/n)^n, {n -> Infinity}] //N
Out[4]= {0.367879}
http://ograll.blogspot.com/2009/06/mathematics-...
I would at least try to prevent it, since it kind of breaks the purpose of the algorithm(a one-way algorithm)
I'm still trying though, got 3 computers working on it with different ways of attacking(among one using the cuda c++ version) and a couple of versions I've made myself one in bash and one in php, plus trying a few of the pre-made ones(with some modification)