问题
Say I have functions
g :: a -> b, h :: a -> c
and
f :: b -> c -> d.
Is it possible to write the function
f' :: a -> a -> d
given by
f' x y = f (g x) (h y)
in point free style?.
One can write the function
f' a -> d, f' x = f (g x) (h x)
in point free style by setting
f' = (f <$> g) <*> h
but I couldn't figure out how to do the more general case.
回答1:
We have:
k x y = (f (g x)) (h y)
and we wish to write k in point-free style.
The first argument passed to k is x. What do we need to do with x? Well, first we need to call g on it, and then f, and then do something fancy to apply this to (h y).
k = fancy . f . g
What is this fancy? Well:
k x y = (fancy . f . g) x y
= fancy (f (g x)) y
= f (g x) (h y)
So we desire fancy z y = z (h y). Eta-reducing, we get fancy z = z . h, or fancy = (. h).
k = (. h) . f . g
A more natural way to think about it might be
┌───┐ ┌───┐
x ───│ g │─── g x ───│ │
/ └───┘ │ │
(x, y) │ f │─── f (g x) (h y)
\ ┌───┐ │ │
y ───│ h │─── h y ───│ │
└───┘ └───┘
└──────────────────────────────┘
k
Enter Control.Arrow:
k = curry ((g *** h) >>> uncurry f)
回答2:
Take a look at online converter
It converted
f' x y = f (g x) (h y)
into
f' = (. h) . f . g
with the flow of transformations
f' = id (fix (const (flip ((.) . f . g) h)))
f' = fix (const (flip ((.) . f . g) h))
f' = fix (const ((. h) . f . g))
f' = (. h) . f . g
回答3:
This is slightly longer, but a little easier to follow, than (. h) . f. g.
First, rewrite f' slightly to take a tuple instead of two arguments. (In otherwords, we're uncurrying your original f'.)
f' (x, y) = f (g x) (h y)
You can pull a tuple apart with fst and snd instead of pattern matching on it:
f' t = f (g (fst t)) (h (snd t))
Using function composition, the above becomes
f' t = f ((g . fst) t) ((h . snd) t)
which, hey, looks a lot like the version you could make point-free using applicative style:
f' = let g' = g . fst
h' = h . snd
in (f <$> g') <*> h'
The only problem left is that f' :: (a, a) -> d. You can fix this by explicitly currying it:
f' :: a -> a -> d
f' = let g' = g . fst
h' = h . snd
in curry $ (f <$> g') <*> h'
(This is very similar, by the way, to the Control.Arrow solution added by Lynn.)
回答4:
Using the "three rules of operator sections" as applied to the (.) function composition operator,
(.) f g = (f . g) = (f .) g = (. g) f -- the argument goes into the free slot
-- 1 2 3
this is derivable by a few straightforward mechanical steps:
k x y = (f (g x)) (h y) -- a (b c) === (a . b) c
= (f (g x) . h) y
= (. h) (f (g x)) y
= (. h) ((f . g) x) y
= ((. h) . (f . g)) x y
Lastly, (.) is associative, so the inner parens may be dropped.
The general procedure is to strive to reach the situation where eta-reduction can be performed, i.e. we can get rid of the arguments if they are in same order and are outside any parentheses:
k x y = (......) y
=>
k x = (......)
Lather, rinse, repeat.
Another trick is to turn two arguments into one, or vice versa, with the equation
curry f x y = f (x,y)
so, your
f (g x) (h y) = (f.g) x (h y) -- by B-combinator rule
= (f.g.fst) (x,y) ((h.snd) (x,y))
= (f.g.fst <*> h.snd) (x,y) -- by S-combinator rule
= curry (f.g.fst <*> h.snd) x y
This is the same as the answer by @chepner, but presented more concisely.
So, you see, your (f.g <*> h) x1 just becomes (f.g.fst <*> h.snd) (x,y). Same difference.
1(because, for functions, (<$>) = (.))
回答5:
Control.Compose
(g ~> h ~> id) f
Data.Function.Meld
f $* g $$ h *$ id
Data.Function.Tacit
lurryA @N2 (f <$> (g <$> _1) <*> (h <$> _2))
lurryA @N5 (_1 <*> (_2 <*> _4) <*> (_3 <*> _5)) f g h
Related articles
- Semantic Editor Combinators, Conal Elliott, 2008/11/24
- Pointless fun, Matt Hellige, 2008/12/03
来源：https://stackoverflow.com/questions/38743576/write-f-in-pointfree-style