[Haskell-cafe] Memoization in Haskell?
mpm at alumni.caltech.edu
Fri Jul 9 00:59:03 EDT 2010
Thanks, okay the next question is: how does the memoization work? Each
call to memo seems to construct a new array, if the same f(n) is
encountered several times in the recursive branching, it would be
computed several times. Am I wrong?
Gregory Crosswhite wrote:
> 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
>>> 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.
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