if (++) were left associative ?
Konst Sushenko
konsu@microsoft.com
Sun, 7 Apr 2002 03:33:06 -0700
Thanks,
but you are assuming that foldl evaluates the intermediate result right
away. If it were, then I have no problem seeing why time is quadratic. I
am concerned about laziness though...
My understanding of foldl is thus:
foldl (++) [] [[1],[2],[3],[4]]
-> foldl (++) ([] ++ [1]) [[2],[3],[4]]
-> foldl (++) ([] ++ [1] ++ [2]) [[3],[4]]
-> foldl (++) ([] ++ [1] ++ [2] ++ [3]) [[4]]
-> foldl (++) ([] ++ [1] ++ [2] ++ [3] ++ [4]) []
-> [] ++ [1] ++ [2] ++ [3] ++ [4]
Which brings me back to my original question...
> -----Original Message-----
> From: David Feuer [mailto:dfeuer@cs.brown.edu]=20
> Sent: Sunday, April 07, 2002 3:21 AM
> To: haskell-cafe@haskell.org
> Subject: Re: if (++) were left associative ?
>=20
>=20
> On Sun, Apr 07, 2002, Konst Sushenko wrote:
> > Hello,
> >=20
> > this is probably embarrassing, but I just realised that I do not
> > understand why list concatenation operation takes quadratic=20
> time if it
> > is associated from left to right (see e.g. 'The Haskell School of
> > Expression' by Hudak, p. 335):
> >=20
> > cat1 =3D (++)
> > cat2 =3D (++)
> >=20
> > ( [1,2] `cat1` [3,4] ) `cat2` [5,6]
> >=20
> >=20
> > My understanding is that cat2 is invoked before cat1, and=20
> as soon as the
> > first element of cat1 is available, cat2 starts building=20
> its own result.
> > cat2 does not need to wait until the whole list is emitted=20
> by cat1, does
> > it? Where is quadratic time complexity?
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> Suppose we define
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> listseq [] x =3D x
> listseq (a:b) x =3D listseq b x
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> l =3D [1..n]
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> inter =3D foldl (++) [] (map (:[]) foo)
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> final =3D listseq inter inter
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>=20
> Then inter will be the result of concatenating [1],[2],[3],... from
> left to right. One meaningful question is: how long will it take to
> calculate "final"? This is (I believe) called the _shared cost_ of
> inter (while inter does not actually do any significant work, it
> produces a chain of suspensions that together do quite a=20
> bit.... I'm not
> sure if the way they are chained still allows this to be=20
> considered the
> shared cost, but I'm not sure).
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> 1. How long does it take to calculate final ?
> Well, let's try it for n =3D 4:
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> inter =3D foldl (++) [] [[1],[2],[3],[4]]
> =3D foldl (++) ([]++[1]) [[2],[3],[4]] =3D foldl (++) [1]=20
> [[2],[3],[4]]
> =3D foldl (++) ([1]++[2]) [[3],[4]] =3D foldl (++) [1,2] =
[[3],[4]]
> =3D foldl (++) ([1,2]++[3]) [[4]] =3D foldl (++) [1,2,3] [[4]]
> =3D foldl (++) ([1,2,3]++[4]) [] =3D foldl (++) [1,2,3,4] []
> =3D [1,2,3,4]
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> Note that in successive iterations, the following are calculated:
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> []++[1]
> [1]++[2]
> [1,2]++[3]
> [1,2,3]++[4]
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> Since l1++l2 takes order length(l1), each concatenation takes=20
> more time
> than the previous one (linearly), so the total time for all of them is
> quadratic.
>=20