Am Donnerstag, 29. Januar 2009 20:29 schrieb Thomas Davie:

The reason that the second one is slower is that ghc makes a distinction that so called CAFs (constant applicative forms) are likely to be constants, and evaluates them once. Thus, your list (map fib [0..]) gets kept between runs. In the second form though, ghc sees a function, and evaluates it every time it gets called, which makes it into an exponential time algorithm.

However, if you compile it with -O2, the optimiser sees it's better to keep the list and it's again a linear time algorithm.

An aside: fibs !! 0 is usually defined to be 1.

I meet fibs !! 0 == 0 more often.

Here's another couple of definitions of fib for you to play with, and try and figure out the properties of: mfib :: Int -> Integer mfib = ((fibs 1 1) !!)

fibs :: Integer -> Integer -> [Integer] fibs n m = n : fibs m (n+m)

-- and fibs :: [Integer] fibs = 1 : 1 : zipWith (+) fibs (tail fibs)

Have fun

Not to forget the marvellous import Control.Monad.Fix fibs :: [Integer] fibs = fix ((0:) . scanl (+) 1)

Bob

Am Freitag, 30. Januar 2009 15:59 schrieb Ertugrul Soeylemez:

Daniel Fischer wrote:

...

Not to forget the marvellous

import Control.Monad.Fix

fibs :: [Integer] fibs = fix ((0:) . scanl (+) 1)

I don't like that one. My favorite is the following, because it's short, concise and still very comprehensible:

import Data.Function

fibs :: Num i => [i] fibs = fix (\r x y -> x : r y (x+y)) 0 1

But that's too easy to understand! What a waste of fix, sheesh ;-) My favourite is fibs = 0:1:zipWith (+) fibs (tail fibs) clear, elegant, accessible. But I also like the other one, precisely because, unless one is very experienced, one has to do a few rounds of "Wait, how on earth does this work?" before it clicks.

Replace 0 by 1, if you want to exclude it from the sequence. I prefer to include it.

Yep. Much more natural if the powers match the index.

Greets, Ertugrul.

Cheers, Daniel

On 30 Jan 2009, at 17:41, Patrick LeBoutillier wrote:

...

...

fibs = fix ((0:) . scanl (+) 1)

I don't like that one. My favorite is the following, because it's short, concise and still very comprehensible:

import Data.Function

fibs :: Num i => [i] fibs = fix (\r x y -> x : r y (x+y)) 0 1

Can someone please explain this one? I can't seem to figure out what 'fix' does (the definition in the documentation is a bit to mathematically inclined for me...).

It computes a fixed point for a function – that is a value which when passed to the function gives back the same answer. Some examples: fix id – any value is a fixed point for id, no matter what you give it, it always comes back the same! fix (+ 1) – no values, no matter what you give it, it'll always come out one bigger, so there's no fixed point here fix (1:) – the infinite list of 1s, if you put a 1 on the front of it, you get back the inifinite list of ones again. fix (\r x y -> x : r y (x+y)) – Lets try giving this function to itself: (\r x y -> x : (y : r (x+y) (y+(x+y)))), and again... (\r x y -

x : (y : ((x+y) : r (y+(x+y)) ((x+y)+y+(x+y))))... you can see where this is going, if we apply this to 0 and 1, we get 0 : 1 : (0+1) : (1 + (0 + 1)) : ... fibonacci numbers.

In general, we can write *any* recursive function using the fix function, we transform the function to accept another argument first, we replace all recursive calls with calls to this argument, and we wrap it in the fix function. Hope that helps. Bob

Patrick LeBoutillier wrote:

...

import Data.Function

fibs :: Num i => [i] fibs = fix (\r x y -> x : r y (x+y)) 0 1

Can someone please explain this one? I can't seem to figure out what 'fix' does (the definition in the documentation is a bit to mathematically inclined for me...).

Well, you can just try to replace fix by its definition and see what happens: fix (\r x y -> x : r y (x+y)) 0 1 = (let go = (\r x y -> x : r y (x+y)) go in go) 0 1 = let go = \x y -> x : go y (x+y) in go 0 1 = let go x y = x : go y (x+y) in go 0 1 In other words, fix can define a recursive function. -- http://apfelmus.nfshost.com