On Tue, 6 Nov 2007, marnes wrote:
fib :: Integer - Integer
fib n = fibaux n 0 1 1
where
fibaux :: Integer - Integer - Integer - Integer - Integer
fibaux i a b c | i==0 = a
| i/=0 = fibaux (i-1) b c (b+c)
http://www.haskell.org/haskellwiki/Memoization
fib :: Integer - Integer
fib n = fibaux n 0 1 1
where
fibaux :: Integer - Integer - Integer - Integer - Integer
fibaux i a b c | i==0 = a
| i/=0 = fibaux (i-1) b c (b+c)
___
Haskell-Cafe mailing list
Throwing in a trace statement in fibaux tells me that fibaux i a b c is
not being memoized. If I do map fib [7..9], fibaux counts down to 0
afresh for each of 7, 8, and 9. Ideally, in map fib [0..n], fib (i-2)
and fib (i-1) should be memoized and fib i would be evaluated in
constant time. This
If people write any new variants, please add them to:
http://haskell.org/haskellwiki/The_Fibonacci_sequence
:)
westondan:
Throwing in a trace statement in fibaux tells me that fibaux i a b c is
not being memoized. If I do map fib [7..9], fibaux counts down to 0
afresh for each of 7, 8,
Thomas Hartman [EMAIL PROTECTED] writes:
- I just thought this was interesting, so I would share it.
- -- versus, try memoized_fibs !! 1
- memoized_fibs = map memoized_fib [1..]
- memoized_fib = ((map fib' [0 ..]) !!)
- where
- fib' 0 = 0
- fib' 1 = 1
- fib' n =
