An easy way to solve this is to ask lambdabot. Log on to the Haskell
IRC channel:
lennart: @pl \ f l -> l ++ map f l
lambdabot: ap (++) . map
Notice how it's much shorter than the Hughes' solution. :)
-- Lennart
On Sep 1, 2006, at 13:11 , John Hughes wrote:
From: Julien Oster <[EMAIL PROTECTED]>
Subject: [Haskell-cafe] Exercise in point free-style
I was just doing Exercise 7.1 of Hal Daumé's very good "Yet Another
Haskell Tutorial". It consists of 5 short functions which are to be
converted into point-free style (if possible).
It's insightful and after some thinking I've been able to come up
with
solutions that make me understand things better.
But I'm having problems with one of the functions:
func3 f l = l ++ map f l
Looks pretty clear and simple. However, I can't come up with a
solution.
Is it even possible to remove one of the variables, f or l? If so,
how?
Thanks,
Julien
Oh, YES!!
Two ways to remove l:
func3a f = uncurry ((.map f).(++)) . pair
func3b f = uncurry (flip (++).map f) . pair
And just to make sure they're right:
propab new f l =
func3 f l == new f l
where types = f :: Int->Int
quickCheck (propab func3a)
quickCheck (propab func3b)
If you don't mind swapping the arguments, then
propc f l =
func3 f l == func3c l f
where types = f :: Int->Int
func3c l = (l++) . (`map` l)
With the arguments swapped, you can even remove both!
propd f l =
func3 f l == func3d l f
where types = f :: Int -> Int
func3d = uncurry ((.(flip map)) . (.) . (++)) . pair
MUCH clearer!
The trick is to observe that l is duplicated, so you need to use a
combinator that duplicates something. The only one available here
is pair, which you then have to combine with uncurry.
It would be nicer to have
(f &&& g) x = (f x,g x)
available. (&&& is one of the arrow combinators). Then you could
remove l by
func3e f = uncurry (++) . (id &&& map f)
which is sort of readable, and remove both by
func3f = (uncurry (++).) . (id &&&) . map
John
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