Mail Archives: djgpp/1998/01/26/13:02:41
Charles Krug wrote:
>
> Heutchy wrote:
>
> > So the n! grows much faster. n^n would be worse however. I don't know
> > if there are any useful algorithms with O(n^n) though.
>
>
> n^n is the complexity of multiplication of hypermatrices of order n in n
> dimensions. Sure, I do it all the time. Yeah, right!
>
By the way, n! and n^n are not very far from each other: by Stirling
formula, n! is equivalent to sqrt(n) (n/e)^n. But again n! algorithm are
rare enough (I suppose it is equivalent to solving a travlling salesman
problem the hard way...).
Francois
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