Elliott Kember dot Com - Latest Comments in The Kember Identity : Elliott Kember dot Com

Re: The Kember Identity : Elliott Kember dot Com

willy — Sun, 24 Apr 2011 02:26:39 -0000

Game is already lost for md5. On my C2Q 3GHz, I can do 1.8 million hashes per second per core. This is 7.2 million hashes per second. It will thus take 1498650426500835304023684 years on such a machine to scan the key space. Assuming your challenge becomes very popular and you get 100 million testers participating to it, all with an equally powerful hardware, it will still take 14986504265008353 years to those 100 million testers to scan the key space. This is one million times the age of the universe. It would thus take a lot less time to mathematically attack the algorithm. It might be done in 20 years by one single person.

Re: The Kember Identity : Elliott Kember dot Com

Demi — Fri, 02 Oct 2009 10:54:31 -0000

I've just burnt a rented server, after 3 days of computations the CPU went on fire.
Greetz

Re: The Kember Identity : Elliott Kember dot Com

Dustin Fineout — Thu, 18 Jun 2009 11:58:32 -0000

I am a bit concerned about your choice of wording in a couple of ways. The md5 algorithm does NOT output a 32-character string. Its output is a 128-bit binary number, which is commonly represented in hexadecimal for human readability.
Consider the differences of the following cases:

md5(X) == X

For this to be true, X must be a 128-bit value.

This is not the same as the following:

bin2hex(md5('string')) == 'string'

Which is what the Kember Identity Search is actually seeking. It is clear you are working with 32-character strings, not with 128-bit numbers, as the input to the md5 function, and thus are not seeking md5(X) == X.

That is not to say it would not be interesting to find such a string! You might also want to consider that the actual input values you are working with are directly dependent upon your character encoding. Different encoding schema will produce entirely different result sets.