Hashers and Crashers (Robbers)

Question

This is the robbers thread of this cops and robbers challenge. The cops thread is here.

Related, but hashes numbers instead of strings and uses a different scoring system.

Definitions

A hash collision is when two different strings produce the same hash.

Summary

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.

Format

A robber answer has to contain the hash collision, a link to the cracked answer, and the score.

Scoring

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.

The winner

With robbers and cops being ranked separately, the winner is the person with the best scoring answer.

Example

Cop:

Python

from hashlib import*
def singlecharhash(x):
  a = blake2b(digest_size=1)
  a.update(x)
  return a.hexdigest()

Robber:

4 Points, Cracks cop's answer

¯ and É, both 2 bytes.

Pinboard

I am thinking of changing the scoring system to make it the minimum of the two strings. Any thoughts?

Answer 1

Cracks Sisyphus's, 1 byte

b'' and b'\x00'

Try it online!

I'm not sure if this is the intended solution but it's definitely the simplest.

As @gsitcia has pointed out in the comments, for SHA-256 if the key is less than 64 bytes HMAC pads it with null bytes to length 64. So by adding a null byte to any key with less than 64 bytes we get the same result.

As @gsitcia also points out, for any key longer than 64 bytes, it is replaced with its hash. We can see this in action here. It is trivial to generate arbitrarily long collisions in this way.