Hi, you may want to read yet another paper, where a different, faster method is introduced, than the chinese one (english):
http://cryptography.hyperlink.cz/md5/MD5_collisions.pdf Ferda > -----Original Message----- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Erwann ABALEA > Sent: Monday, March 14, 2005 6:27 PM > To: openssl-users@openssl.org > Subject: Re: [openssl-users] Re: The breaking of SHA1 > > Bonsoir, > > Hodie pr. Id. Mar. MMV est, Alicia da Conceicao scripsit: > > Of course, having a method in 2^69 calculations that find a second > > message that has the same SHA1 hash as a first message does > not mean > > that the second message would be of any use to an attacker/forger. > > This is not the result of the chinese team. What they did is > enhance the birthday paradox, from 2^80 to 2^69. But you have > to generate 2^69 random messages, and you then have 50% > chances that 2 of them have the same hash. This is completely > different that what you're saying: > finding a message that has the same hash as a choosed message. > > -- > Erwann ABALEA <[EMAIL PROTECTED]> > ______________________________________________________________________ > OpenSSL Project http://www.openssl.org > User Support Mailing List openssl-users@openssl.org > Automated List Manager [EMAIL PROTECTED] > ______________________________________________________________________ OpenSSL Project http://www.openssl.org User Support Mailing List openssl-users@openssl.org Automated List Manager [EMAIL PROTECTED]
