Comment by zeroonetwothree
1 day ago
Then 'n' means kind of different things for sorting vs. multiplication though. For example for sorting we assume constant time comparison, which doesn't make sense inputs of O(n) bits
1 day ago
Then 'n' means kind of different things for sorting vs. multiplication though. For example for sorting we assume constant time comparison, which doesn't make sense inputs of O(n) bits
If you sort n k-bit items for a total time of O(nk logn), that scales more poorly in n than multiplying n-word integers. Of course if k is constant you can do radix sort, but I genuinely don't know under what conditions radix sort is more/less galactic than this multiplication algorithm.
this alg is way more galactic than radix sort. radix sort often wins in the hundreds of elements. the nlogn multiplication requires numbers with more digits than atoms in the universe (although that could probably be brought down a lot)
Ah thanks for pointing this out, for some reason I had always equated radix sort and bucket sort (with 2^k buckets) in my head. But I learned today that this isn't true!