Re: sorting problem

Bruno Wolff III <bruno@wolff.to> writes:

> Using an index to do an order by is an order N operation.

No, using an index to do an order by is actually still n*log(n). You have to
traverse all the parent pages in the binary tree of the index as well.

This only goes to show how small the log(n) component of the order is. It's
easily dwarfed by large constants such as the difference in i/o efficiency
from non-contiguous reads.

--
greg