fix opt/8634

[Gerald Pfeifer]

On Wed, 9 Apr 2003, Andrew Pinski wrote:
> But really the math is K*n*O(n) = O(n^2) for any K>0 aka you call the
> function K*n times, and the function takes O(n) time. So really it is
> still O(n^2).

I was not talking about the specific algorithm here (and I agree with
what you write above), I just disputed

>>> If you take a linear operation (that is, O(n)), and then you perform
>>> it K*n times for some fixed constant K, then the program is O(n*n),
>>> which is quadratic.

which is not correct (as far as big-O is concerned).

Gerald
-- 
Gerald "Jerry"   pfeifer@dbai.tuwien.ac.at   http://www.pfeifer.com/gerald/