My question is related to this question I posted in math forum:
https://math.stackexchange.com/questions/1512644/balls-and-bins-hash-table-a-concrete-example
but I could not get an answer that I need.
Please assume upper bound on set cardinality is $n$ (so the set has at most $n$ elements). A server wants to construct a hash table in away that each bucket would contain at most $d$ elements with a high probability. So for any client having set $S_i$ where $|S_i|<n$, any bucket of hash table would not overflow. In setting, any client can independently insert its elements in the hash table, given the hash table length and hash function; and send the hash table to the server.
I need to know how the server can compute the hash table length (or number of buckets in the hash table), given $n$ and $d$.
In [1,2] it is said $\frac{n}{k}+O(\sqrt[2]{(\frac{n}{k})\log k}+\log k)$ is a max load for a bucket with a high probability, where $n$ denotes number of elements and $k$ denotes number of buckets (alternatively we can re-write it as $\frac{n}{k}+C(\sqrt[2]{(\frac{n}{k})\log k}+\log k)$, where $C$ is a constant value). But, I need to know how this works in practice. More specifically how we can compute the probability and constant value, $C$.
