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]
