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by titfn 3933 days ago
instead of calculating multiple hashes every time, might as well use a better hash?
1 comments

We're way past my personal knowledge, but what I was thinking was that the probablistic combinations of two algorithms outweigh (in terms of probability of a collision) the likely hood of a single algo.

IE if you have some hashing function with a 1/2^5 chance of a collisions, using that in conjuntion with another hashing function with say 1/2^3 requires that both unlikely probabilities occur, resulting in 1/2^8, where a "stronger" hashing function might only have 1/2^6 (technically stronger than either one of them, but still not as strong as both)

I'd really love some correction on the logic above though, I am by no means well studied in probability or hashing functions, despite some effort, and would really appreciate any corrections.