Re: SHA-1 broken

> Incorrect. In order to make this attack work, you need to break the
> hash's weakly collision-free property. The attack being mentioned is
> against the strongly collision-free property.
> 
> The above will still take 2^159 operations on average. Nick Johnson's
> explanation gets it right. Someone creating a torrent can make a torrent
> with a pair of blocks that both verify with 2^69 work (and note that
> neither of the pair can be chosen, you generate 2^69 of them and two of
> them match). Someone who isn't the creator still has to attack the weakly
> collision-free property and do 2^159 operations.
> 
> The strongly collision-free property is important for digital signatures,
> where the original content may not of been created by the signer. This
> isn't yet a problem for BT, just cause for concern.

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.

Bill




 
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