3 ms·
You cannot put 51 pigeons in 50 pigeonholes. So the key length of your hash keys is m >= logn, which means the time to compare keys is also logn, which means ha
by danbruc 3y ago
You cannot put 51 pigeons in 50 pigeonholes. So the key length of your hash keys is m >= logn, which means the time to compare keys is also logn, which means hash tables are never actually O(1) access or insert time.
I am not sure I am following this argument. You are not going to have more than 2^64 pigeons and pigeonholes on any system soon and I almost dare to say you will never ever get to 2^128. And for 64 or 128 bit keys comparisons and many other operations are for all practical purposes constant time. I guess you could argue that this is sweeping a factor of log(n) under the rug because of things like carry chains which could be faster for smaller bit sizes but I am not sure that this is really useful on common hardware, an addition will take one clock cycle independent of the operand values.
- hinkley 3y agoAre your hash keys 8 bytes long? Mine aren't. Mine aren't even close.
- danbruc 3y agoSure, they can be and are often longer, but not because you were forced to use long keys, it just happened that the thing you want to store in a hash table is a long string. The way you worded it made it sound to me like you were saying that one has to use long keys, not that in practice one often has to deal with long keys. But even then I am not convinced that this should give an additional factor in the complexity analyses, I think I would argue, at least in some situations, that calculating hashes of long keys should still be considered constant time, that for the longest keys. But I can also imagine to take this into account if the key length is not only big but also highly variable.