[Cryptography] blake2b 160

[jamesd at echeque.com]

On 31/12/2018 00:39, Alexander Kjeldaas wrote:
> 2^80 random blocks should give one collision, but for a given hash 2^160 
> blocks are needed to find a collision.

I know that, everyone knows that.  Does not seem to have the slightest 
relevance whatsoever to the question I asked.

Obviously the NSA can calculate 2^80 blocks without raising a sweat, but 
they cannot store and sort 2^80 blocks, which on the face of it they 
would need to do in order to find the collision that they might have 
generated.