This is the robbers thread of this cops and robbers challenge. The cops thread is here.
A hash collision is when two different strings produce the same hash.
In this challenge, the cops will make a hash function, with a collision they know of. Then, the robbers will try to crack it by finding a hash collision in the cop's hash algorithm. To crack an answer, the hash collision the robber finds does not have to be the intended solution.
If a cop submission is not cracked for 7 days, it is safe, so mark it as such and reveal the intended crack and the score.
If a cop submission is cracked, mark it as such and edit in a link to the crack.
A robber answer has to contain the hash collision, a link to the cracked answer, and the score.
The score of a robber answer is the sum of the number of bytes of the two strings with the hash collision, with the higher score being better.
With robbers and cops being ranked separately, the winner is the person with the best scoring answer.
from hashlib import* def singlecharhash(x): a = blake2b(digest_size=1) a.update(x) return a.hexdigest()
4 Points, Cracks cop's answer
É, both 2 bytes.
I am thinking of changing the scoring system to make it the minimum of the two strings. Any thoughts?