Ok, I'll bite... The fibonacci sequence is often used in CS classes as "the poster child against recursion" because the recursive algorithm is just so horrific in performance.

Below I have coded both a non-recursive and a recursive algorithm in pretty much "run of the mill" implementations. They both produce the same results although the recursive algorithm is ridiculously slow.

Sorry I didn't crank up the iterations far enough to get a meaningful result for the non-iterative version because if I did that, I'd have to go take a long nap waiting for the recursive version to finish! However, the benchmark below shows just how silly the comparison between the two is. "Un-comment" the print statements and crank down the iterations to verify output is the same. 

#!/usr/bin/perl -w
use strict;
use Benchmark;

timethese (10,

 { Normal=> 
    q{
       foreach (20..25)
       {
          #print "number $_ : fibonacci ",fibonacci($_),"\n";
          my $f = fibonacci($_);
       }
     },
         
   Slow =>
    q{
        foreach (20..25)
        {
           #print "number $_ : fibonacci ",fibonacci_slow($_),"\n";
           my $f = fibonacci_slow($_);
        } 
     },
  } 
);

=prints:
Benchmark: timing 10 iterations of Normal, Slow...
    Normal:  0 wallclock secs ( 0.00 usr +  0.00 sys =  0.00 CPU)
            (warning: too few iterations for a reliable count)
      Slow:  7 wallclock secs ( 6.24 usr +  0.00 sys =  6.24 CPU) @  1
+.60/s (n=10)
=cut

sub fibonacci
{
  my $number  = shift;
  my $curnum  = 1;
  my $prevnum = 1;
  my $sum;
  
  if ($number ==1 || $number ==2){ return 1;}
  
  $number -= 2;
  while ($number--)
  {
     $sum = $curnum + $prevnum;
     $prevnum = $curnum;
     $curnum  = $sum;
  }
  
  return $sum;
}


sub fibonacci_slow
{
   my $number = shift;
   if ($number ==1 || $number ==2){ return 1;}
   
   return fibonacci_slow($number-1) + fibonacci_slow($number-2);
}

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In reply to Re: fibonacci numbers using subroutine? by Marshall
in thread fibonacci numbers using subroutine? by derpp

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