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
Thank you all for your valuable answers.
On 22/04/2011 00:33, Dave Thompson wrote:
*Accidental* (birthday) collision is about 264 for MD5
and about 280 for SHA-1.
> SHA-256 should be much stronger, would this be sufficient
> for your needs? Or
> From: owner-openssl-us...@openssl.org On Behalf Of Steffen DETTMER
> Sent: Wednesday, 20 April, 2011 12:25
> * Luc Perthuis:
> > Is there any theoretical proof for a "good" selection of 2
> > HASH (computing the results of two different algorithms on
> > the same data) that would annihilate t
* Luc Perthuis:
> Hi all,
>
> I'm specially interested on finding a way to uniquely
> identify rather small data chunks (less than or equal to
> 128*1024 bytes in size) without using a byte per byte compare.
>
> Is there any theoretical proof for a "good" selection of 2
> HASH (computing the re
On 4/20/2011 1:18 AM, Luc Perthuis wrote:
Hi all,
I'm specially interested on finding a way to uniquely identify rather
small data chunks (less than or equal to 128*1024 bytes in size) without
using a byte per byte compare.
Is there any theoretical proof for a "good" selection of 2 HASH
(comput
Hi all,
I'm specially interested on finding a way to uniquely identify rather
small data chunks (less than or equal to 128*1024 bytes in size) without
using a byte per byte compare.
Is there any theoretical proof for a "good" selection of 2 HASH
(computing the results of two different algori
