4 ms·
Whether you have two 128-bit hashes or one 256-bit hash, you're still trying to fit 2^n pigeons into (2^128)*(2^128) = 2^256 pigeonholes. Maybe your intuition
by gelisam 14y ago
Whether you have two 128-bit hashes or one 256-bit hash, you're still trying to fit 2^n pigeons into (2^128)*(2^128) = 2^256 pigeonholes.
Maybe your intuition is telling you that two hashes must be more secure because if one of the two hash functions turns out to be easy to break, then at least you will still be protected by the other?
- Beltiras 14y agoNo, what I was thinking was more like coverage. For any given hash, you do NOT have the same coverage for the other hash function. Let's make a bucket of all the originating bits of length n that will deliver a collision for one hash function. You are not guaranteed that the results of another hashing function has a uniform distribution. I still think I am wrong but you did not convince me.