[Haskell-cafe] Sharing Subexpressions: Memoization of Fibonacci sequence

[SimonK77]

**Edit: formatting was bad**

Hallo everyone,

the last few weeks I was trying to get used to memoization in haskell. As I
found out in a previous post, memoization in haskell is, due to the graph
reduction strategy used in haskell, almost always implemented by sharing
subexpressions in an expression.

In examples as the following this is quite easy to see.

fibs = 0:1:zipWith (+) fibs (tail fibs)

But as I browsed through the search results from google for this topic, I
encountered the following implementation:

memoized_fib :: Int -> Integer
memoized_fib =
   let fib 0 = 0
       fib 1 = 1
       fib n = memoized_fib (n-2) + memoized_fib (n-1)
   in  (map fib [0 ..] !!)

You'll find it at Haskell.org. Here's the 
http://www.haskell.org/haskellwiki/Memoization link 

Let's assume we have the following implementation of the
higher-order-function map:

map :: (a -> b) -> [a] -> [b]  
map _ [] = []  
map f (x:xs) = f x : map f xs

The reduction sequence of memoized_fib 5 would start as follows:

//Reduction of memoized_fib 5

memoized_fib 5 = (map fib [0..] !!) 5
		   = (fib 0 : map fib [1..] !!) 5
		   = (0 : fib 1 : map fib [2..] !!) 5
		   = (0 : 1 : fib 2 : map fib [3..] !!) 5
		   = (0 : 1 : (memoized_fib 0 + memoized_fib 1) : fib 3 : map fib [4..]
!!) 5
		   = (0 : 1 : (map fib [0..] !! 0 + map fib [0..] !! 1) : (memoized_fib 1
+ memoized_fib 2) : map fib [4..] !!) 5
			.
			.
			.
			.
		   and so on!

I can't see where the sharing of subexpressions happens here. Any help is
highly appreciated.

Best regards

Simon
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