[Haskell-cafe] Haskell performance (again)!
hehx0sk02 at sneakemail.com
Mon Oct 9 02:31:45 EDT 2006
On 10/8/06, Udo Stenzel u.stenzel-at-web.de |haskell-cafe|
> 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) 
= foldl (+) (((0 + 1) + 2) + 3) 
= ((0 + 1) + 2) + 3
= (1 + 2) + 3
= 3 + 3
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 
= foldl' (+) 3 
= let u = 3 + 3 in u `seq` foldl' (+) u 
= foldl' (+) 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.
> Hast du zum Leben kein Motiv --
> steig mal vor, vielleicht geht's schief.
> -- aus einem Gipfelbuch
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