Re^2: Fibonacci Numbers

If you stick with the original form of your answer (a recurrance-based one), try Memoizing it, viz.:

But the original form of his answer was not recursive:

sub fib{
($val0, $val1) =@_;
$sum= $val0 + $val1;
}
[download]

Comment on Re^2: Fibonacci Numbers Download Code

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Re^3: Fibonacci Numbers by dReKurCe (Scribe) on Feb 10, 2005 at 22:56 UTC
Honourable friar, Am I wrong to think that recurrsion is utilized by passing the value of $sum as computed by sub fib back into sub fib? Otherwise, the binet formula is extremely cool. Thanks.	[reply]
Re^4: Fibonacci Numbers by lidden (Curate) on Feb 10, 2005 at 23:33 UTC
This is a recursive way to do it but it gets slow quickly for larger values unless you memoze it. `print "$_ : ", fibo($_), "\n" for 0..23; sub fibo{ my $n = shift; if ( $n == 0){ return 0;} elsif( $n == 1){ return 1;} else { return fibo($n-1) + fibo($n-2);} }` [download]	[reply] [d/l]
Re^4: Fibonacci Numbers by blazar (Canon) on Feb 11, 2005 at 09:39 UTC
It is iterative. It is not recursive in that `fib()` does not call itself. FWIW I think that Binet's formula is cool too.	[reply] [d/l]