Hi, thank you for clarification, Dave!
* Dave Thompson Friday, April 22, 2011 12:34 AM: > > so among 2^n+1 different messages, at least two of them > > must have the > > same 2^n bit hash (actually half because of birthday "attack"). > > To be exact: for an n-bit or 2^n-value hash, with 2^n + 1 > messages you are certain to have a collision. Because of the > birthday paradox, with about 2^(n/2) random messages you are > *likely* to have a collision. This is just a fact of nature; > "attack" is used for actions by an intelligent adversary. Just for sake of completeness I think it could be added (for the assumed intention of the OP's question) that there is no guarantee that two different messages have two different hashes, because the first two tested messages could have the same hash (of course with very very low probability). Could it even be possible that two messages shorter than n bit accidently have the same (strong n-bit) hash? Steffen About Ingenico: Ingenico is a leading provider of payment, transaction and business solutions, with over 15 million terminals deployed in more than 125 countries. Over 3,000 employees worldwide support merchants, banks and service providers to optimize and secure their electronic payments solutions, develop their offer of services and increase their point of sales revenue. http://www.ingenico.com/. This message may contain confidential and/or privileged information. If you are not the addressee or authorized to receive this for the addressee, you must not use, copy, disclose or take any action based on this message or any information herein. If you have received this message in error, please advise the sender immediately by reply e-mail and delete this message. Thank you for your cooperation. P Please consider the environment before printing this e-mail ______________________________________________________________________ OpenSSL Project http://www.openssl.org User Support Mailing List openssl-users@openssl.org Automated List Manager majord...@openssl.org
