On 10/8/06, Udo Stenzel u.stenzel-at-web.de |haskell-cafe|
<...> wrote:
Yang wrote:
> type Poly = [(Int,Int)]
>
> addPoly1 :: Poly -> Poly -> Poly
> addPoly1 p1@(p1h@(p1c,p1d):p1t) p2@(p2h@(p2c,p2d):p2t)
> | p1d == p2d = (p1c + p2c, p1d) : addPoly1 p1t p2t
> | p1d < p2d = p1h : addPoly1 p1t p2
> | p1d > p2d = p2h : addPoly1 p1 p2t
> addPoly1 p1 [] = p1
> addPoly1 [] p2 = p2
> addPoly1 [] [] = []
>
> But this doesn't use tail recursion/accumulation
Indeed it doesn't. Now remind me, why is that supposed to be a Bad
Thing? The above code exhibits a maximum of lazyness and runs with no
useless space overhead. Apart from the expression (p1c + p2c), which
you probably want to evaluate eagerly, it is close to perfect.
> so I rewrote it: [...]
>
> But laziness will cause this to occupy Theta(n)-space of cons-ing
> thunks.
No, it doesn't. Insisting on accumulator recursion does. Actually,
using reverse does. Think about it, a strict reverse cannot use less
than O(n) space, either.
Well, in general, the problem you run into is this, where we use
linear space for the thunks:
foldl (+) 0 [1,2,3]
= foldl (+) (0 + 1) [2,3]
= foldl (+) ((0 + 1) + 2) [3]
= foldl (+) (((0 + 1) + 2) + 3) []
= ((0 + 1) + 2) + 3
= (1 + 2) + 3
= 3 + 3
= 6
whereas with strictness, you use constant space:
foldl' f z [] = z
foldl' f z (x:xs) = let u = f z x in u `seq` foldl' f u xs
foldl' (+) 0 [1,2,3]
= let u = 0 + 1 in u `seq` foldl' (+) u [2,3]
= foldl' (+) 1 [2,3]
= let u = 1 + 2 in u `seq` foldl' (+) u [3]
= foldl' (+) 3 [3]
= let u = 3 + 3 in u `seq` foldl' (+) u []
= foldl' (+) 6 []
= 6
> I was
> hoping for more in-depth insights on how to take advantage of laziness
> to write cleaner AND more efficient code.
Try to explain why your first iteration was bad. You'll achieve
enlightenment at the point where your explanation fails.
Udo.
--
Hast du zum Leben kein Motiv --
steig mal vor, vielleicht geht's schief.
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