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The probability of a hash collision
(kevingal.com)
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this post was submitted on 05 Jul 2025
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Programming
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The assumption is there though.
Wouldn't multiplying the hash simply relabel the hash sites, as hashes non divisible by the factor simply be not accessible/not exist?
The hashes not being there isn't particularly relevant within a hash function outputting a specific size. If your hash function is always 64 bits for example, the fact that you have 3/4th of them not exist means you should be operating as if its a 16 bit hash, not a 64 bit hash. If you still do this math based on the 64 bits outputted (2^64 boxes) you'd arrive at very inaccurate numbers.
That would be 62 bit, not 16 bit.