Re: SHA-1 broken

On 17/02/2005, at 1:52 PM, Bill Cox wrote:

> I think I get it.  The Chinese are just saying that in 2^69 random
> blocks, typically there will be two that collide, and not that I can
> find a random block that matches a specific signature.
>
> Thanks for the clarification.

Slightly more than that: If you picked the blocks randomly, it'd take 
2^80 blocks to find a collision. They're claiming a method of choosing 
blocks that reduces it to 2^69. If it's anything like the MD5 break, it 
seems likely we'll see an improvement that lets you choose all but a 
few bytes of the messages yourself. In this case, someone could, for 
example, generate two distinct torrents with identical info_hash 
values. (Granted 2^69 operations is still high enough it'd take a major 
distributed project to do so!)

-Nick Johnson



 
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