4 ms·
Adding md5 (or whichever other hashing function), I have a feeling (but no direct proof of) it's not just equivalent to adding bits to the sha hash. Anyone wit
by Beltiras 14y ago
Adding md5 (or whichever other hashing function), I have a feeling (but no direct proof of) it's not just equivalent to adding bits to the sha hash. Anyone with a better sense for hashing than me why my intuition is wrong?
- gelisam 14y agoWhether 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.