performance problem in python 2.2

[Jeff Davis]

Paul Rubin wrote:

> Fernando Perez <fperez528 at yahoo.com> writes:
>> #!/usr/bin/perl
>> 
>> $p = 2**64;
>> $c = $ARGV[0];
>> 
>> $n = 1;
>> foreach $i (1..$c) {
>>         $n = ($n * ($p-$i)) / $p
>> }
>> print 1-$n, "\n";
>> 
>> But again, there's NO WAY you are going to write perl code to do this at
>> C speed (short of a trick like inlining C in perl).
> 
> Does
> 
>   $c = 0.0 + $ARGV[0];
> 
> making $c a float instead of a string help?
> 
> 
> Also:
> 
>  $p = 2**64;
>  $c = $ARGV[0];
>  $n = exp(-($c*$c) / (2*$p));
>  print 1-$n, "\n";
> 
> should give a pretty close approximation faster than any of the
> iterative methods listed.

Very nice! unfortunately, when I started to solve the problem I didn't 
understand the relationship to e with my algorithm. Upon further 
mathematical analysis (after seeing a little pattern develop), I realized 
the relationship. Of course, that still left me with my python questions 
:)

I learned to never underestimate the usefulness of mathematics in 
programming. Lucky for me, I had a teacher to ask and he gave me a lot of 
good info as well. After giving me some advice (including CS related tips 
that mostly boiled down to "Use assembly, or maybe C if you can about 
getting an answer"), he pointed me to Sterling's Theorem, which seems 
similar to what you did. 

Thanks!

Regards,
        Jeff