17 Dec
2006
17 Dec
'06
9:50 a.m.
I suppose that algorithm is O(n) if you take as O(1) examining a number with log n bits.
yes, I am thinking this is what Fred had in mind, because he was talking about an O(n log n) algorithm which "essentially sorts the permutation". And sorting integers takes O(n log n) comparisons and swaps of integers (not bits). But I might be completely wrong, Fred? Christoph