I am a little confused by the complexity proof of Radix Sort.
For counting sort, the complexity reported is $O(n+R)$, where $n$ is the number of items and $R$ is the range.
But this is not entirely correct, right? To do binary arithmetic on numbers as big as $R$, I need $\log_2 R$ operations for any deterministic hashing function. So the complexity of counting shot should be $O(\log_2 R \cdot (n+R))$.
Then the complexity of radix sort with base $b$ should be $O(\log_b R \cdot \log_2b \cdot (n+b))$, which is equivalent to $O(\log_2R \cdot n)$. For range $R = \Omega(n)$, this is $\Omega(n \log n)$ and thus not $O(n)$.
Can someone point out where I am going wrong in my proof? As far as I know, radix sort is $O(n)$ for $R = O(n^c)$.