[Haskell-cafe] turning an imperative loop to Haskell

[Axel Gerstenberger]

Thanks to all of you. The suggestions work like a charm. Very nice.

I still need to digest the advices, but have already one further 
question: How would I compute the new value based on the 2 (or even 
more) last values instead of only the last one?

[ 2, 3 , f 3 2, f((f 3 2) 3), f ( f((f 3 2) 3)  f 3 2)), ...]

(background: I am doing explicit time stepping for some physical 
problem, where higher order time integration schemes are interesting. 
You advance in time by extrapolating based on the old time step values.)

I guess I just wrote the definition and define iterate2 as

iterate2 history =
      case history of
          []   -> error "no start values"
          x1:x2:xs   -> iterate2 ([f x1 x2] ++ xs)
or

iterate2 :: [Double] -> [Double]
iterate2 history =
      case history of
          []   -> error "two start values needed"
          x1:[]   -> error "one more start values"
          x1:x2:xs   -> iterate2 (history ++ ([f a b]))
             where [a,b] = take 2 $ reverse history

however,I don't get it this to work. Is it possible to see the 
definition of the iterate function? The online help just shows it's usage...

Again thanks a lot for your ideas and the links. I knew there was a 
one-liner for my problem, but I couldn't find it for days.

Axel

Dougal Stanton wrote:
> On 06/09/07, Axel Gerstenberger <axel.gerstenberger at gmx.de> wrote:
> 
>> module Main where
>>
>> import System.IO
>> import Text.Printf
>>
>> main :: IO ()
>> main = do
>>      let all_results1 = take 20000 $ step [1]
>>      --print $ length all_results1 -- BTW: if not commented out,
>>                                    --      all values of all_results
>>                                    --      are already
>>                                    --      calculated here
>>      loop [1..50] $ \i -> do
>>          let x = all_results1!!i
>>          putStrLn $ show i ++ "  " ++ show x
>>
>> -- create an infinite list with values u_{n+1} ++ [u_n,u_{n-1},...,u_1]
>> -- where u_{n+1} = f (u_n)
>> step history =
>>      case history of
>>          []   -> error "no start values"
>>          xs   -> xs ++ (step [ f (head $ reverse (xs) )])
> 
> To create an infinite list where each f(u) depends on the previous u,
> with a single seed value, use 'iterate':
> 
> Prelude> let us = iterate f 3
> 
> That produces your infinite list of values, starting with [f 3, f(f3),
> f(f(f 3)), ...]. Pretty neat.
> 
> Then all you really need is
> 
> main = mapM_ (uncurry (printf "%d %f\n")) (zip [1..50] (iterate f 3))
> 
> You can probably shorten this a bit more with arrows but I've got a
> cold at the moment and not really thinking straight.
> 
> Cheers,
> 
> D.
>