Re: [Haskell-cafe] Fibonacci numbers generator in Haskell

16 Jun 2006


      Doug Quale  writes:
...
Mathew Mills  writes:
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Is there anything that can be done (easily) to reduce the rounding errors?
The hint that I gave before is one easy way.
...
fib :: Integer -> Integer
fib x = let phi = ( 1 + sqrt 5 ) / 2
         in truncate( ( 1 / sqrt 5 ) * ( phi ^ x + 0.5) )
You run out of precision eventually.  IEEE Double's give you about 15
decimal digits, so the results become approximate for x > 75.
Sorry, I suffered brain lock.  The correct expression is parenthesized
differently:
...
fib n = truncate(phi^n/sqrt 5 + 0.5)
   where phi = (1 + sqrt 5)/2