Rick Harker wrote:
> > Someone wrote:
> > >>If N neurons have to meet up accurately with N locations, there are
> > >>N! (that's a factorial) ways of doing this. Given that N is very
>> I've haven't learned about factorials yet.
> But shouldn't this be ((n*n)-n)/2?
> (for the number of connections? It would eliminate counting a path both
No ... that's O(n^2), whereas factorial algorithms are O(x^n), for some
constant x ...
If you eliminate counting both ways, you are essentially dividing by a
constant factor, not changing from an exponential algorithm to a polynomial