[Haskell-cafe] Memoization in Haskell?
Gregory Crosswhite
gcross at phys.washington.edu
Fri Jul 9 00:42:39 EDT 2010
On 7/8/10 9:17 PM, Michael Mossey wrote:
>
>
> Daniel Fischer wrote:
>>
>> If f has the appropriate type and the base case is f 0 = 0,
>>
>> module Memo where
>>
>> import Data.Array
>>
>> f :: (Integral a, Ord a, Ix a) => a -> a
>> f n = memo ! n
>> where
>> memo = array (0,n) $ (0,0) : [(i, max i (memo!(i
>> `quot` 2) + memo!(i `quot` 3) + memo!(i `quot`
>> 4))) | i <- [1 .. n]]
>>
>> is wasteful regarding space, but it calculates only the needed values
>> and very simple.
>
> Can someone explain to a beginner like me why this calculates only the
> needed values? The list comprehension draws from 1..n so I don't
> understand why all those values wouldn't be computed.
>
The second pair of each element of the list will remain unevaluated
until demanded --- it's the beauty of being a lazy language. :-) Put
another way, although it might look like the list contains values (and
technically it does due to referential transparency), at a lower level
what it actually contains are pairs such that the second element is
represented not by number but rather by a function that can be called to
obtain its value.
Cheers,
Greg
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