Secure Drive insecure? NOT
bryner@atlas.chem.utah.edu "Roger Bryner" wrote:
Again, the only way that MD5 can keep the entropy of a string is for every single 128 bit string to map itself onto a unique 128 bit string, for if two 128 bit strings produce the same output, then you loose entropy.
The question is, when md5 is restricted to 128 bit values, does it loose entropy, and if so how much? As much as a random mapping? if so, the 1024 bit itteration in secure drive HARMS security.
BZZZT! Read the code... MD5Init(&md5buf); MD5Update(&md5buf,pass1,strlen(pass1)); MD5Final(key,&md5buf); for(k=0;k<PASS_ITER;k++) { MD5Init(&md5buf); MD5Update(&md5buf,key,16); | Hmmm?!? MD5Update(&md5buf,pass1,strlen(pass1)); <----------| I wonder what MD5Final(key,&md5buf); | that line does? } Notice that the passphrase is included in each of the hashings. How can MD5(passphrase+16-byte hash) have lower entropy than MD5(passphrase)? The iteration just makes it slower to crack pass- phrases. Version 1.0, which used a single MD5, could be attacked at the rate of 1000 per second on a PC. --- Mike 1994: Wiretapping is privacy Secrecy is openness Obscurity is security
On Wed, 6 Jul 1994, Mike Ingle wrote:
BZZZT! Read the code... MD5Update(&md5buf,pass1,strlen(pass1)); <----------| I wonder what Gee, thanks for leting me play, do I get a consolation prize?:-)
Let me think, I still don't see any proof that this does not loose entropy, and it could, as if the two parts are not independent of each other. As soon as you start making a feedback machene, you have no guarentee that this is a maximal unless there is no state that is imediatly preceeded by two other states. Simply throwing a lot of stuff at a password is no substitute for a proof that the transformation does not loose entropy, which is available if you use an encryption algorithim for the last 1023 transformations, and a hash function only for the first one. Sorry to be a pain. Roger.
participants (2)
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Mike Ingle -
Roger Bryner