3 ms·
No, 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
by Beltiras 14y ago
No, 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.