I have been wondering how different standard Haskell functions could be implemented point-free. Currently, I am interested in uncurry and I feel this one is quite non-trivial.
The main problem is that we are unable (or as it seems to me) to group the arguments. If we had uncurry (in fact, uncurry ($) would suffice) in use, the solution would have been quite simple:
Make a tuple (f, (x, y)).
Apply assoc1 :: (a, (b, c)) -> ((a, b), c) to the tuple and get ((f, x), y).
Apply the uncurried ($) to the first element of the pair and get (f x, y).
Apply the uncurried ($) to the pair itself and get f x y.
Without the uncurried ($) we would have to extract both elements of the pair separately. E.g.:
uncurry f pair = f (fst pair) (snd pair)
I do not reckon this to be a smooth way to implement something point-free.
In fact, we have got this uncurried ($) at our behest: Control.Arrow.apply (other useful for the solution combinators could also be imported from Control.Arrow). Therefore:
import Control.Arrow ((>>>), (&&&), first, app)
myUncurry = let myAssoc1 = (fst &&& (fst . snd)) &&& (snd . snd)
in (,) >>> (>>> myAssoc1 >>> first app >>> app)
Yet, this feels a small bit like cheating.
Are there any other approaches towards this problem which do not require anything like app?
join on functions gives you (a -> a -> b) -> a -> b, so:
myUncurry f = join (\x y -> f (fst x) (snd y))
myUncurry f = join (\x -> f (fst x) . snd)
myUncurry f = join ((.snd) . f . fst)
myUncurry f = join ((.fst) ((.snd) . f))
myUncurry f = join ((.fst) ((.) (.snd) f))
myUncurry = join . (.fst) . \f -> (.) (.snd) f
myUncurry = join . (.fst) . ((.snd).)
join . (.fst) . ((.snd).) is very readable indeed
The artless, mechanical solution, by "pushing lambdas inward".
uncurry f (x,y) = f x y
uncurry f p = f (fst p) (snd p)
uncurry f = \p -> f (fst p) (snd p)
uncurry f = (<*>) (\p -> f (fst p)) (\p -> snd p)
uncurry f = (<*>) (f . fst) snd
uncurry = \f -> (<*>) (f . fst) snd
uncurry = flip (\f -> (<*>) (f . fst)) snd
uncurry = flip ((<*>) . (\f -> f . fst)) snd
uncurry = flip ((<*>) . (. fst)) snd
With Lambda Calculus' S combinator, Sabc = (a <*> b) c = a c $ b c,
uncurry f (x,y) = f (fst (x,y)) (snd (x,y))
= (f . fst <*> snd) (x,y)
uncurry f = (<*> snd) (f . fst)
= (<*> snd) . (. fst) $ f
hence,
uncurry :: (a -> b -> c) -> (a, b) -> c
uncurry = (<*> snd) . (. fst)
(edit:)
Still it's much more readable (and somewhat elucidating) with one explicit argument left there, as seen above:
uncurry f = f . fst <*> snd
But then this variant, shown by Jon Purdy in the comments,
uncurry f = liftA2 f fst snd
just might be the clearest.
This is because for functions, the monad and the applicative are equivalent in power,
(k =<< f) x = k (f x) x = flip k x (f x) = (flip k <*> f) x
-- i.e., uncurry f = flip (f . fst) =<< snd
and liftA2 f fst snd means, by definition,
= [ f a b | a <- fst ; b <- snd ]
=
do { a <- fst ;
b <- snd ;
return (f a b)
}
= \x -> let
{ a = fst x ;
b = snd x ;
}
in const (f a b) x
(the first one written with Monad Comprehensions). Thus,
uncurry f x = liftA2 f fst snd x
= let
{ a = fst x ;
b = snd x ;
}
in f a b
=
f (fst x) (snd x)
=
(f . fst <*> snd) x
=
(flip (f . fst) =<< snd) x
=
flip (f . fst) (snd x) x
=
(flip (f . fst) . snd) x x
=
join (flip (f . fst) . snd) x
=
join (flip (f . fst) <$> snd) x
following the well known equivalence, k =<< m = join (fmap k m) (and for functions, (<$>) = fmap = (.)).
So we've found yet another expression here,
uncurry f x = join (flip (f . fst) . snd)
= liftA2 f fst snd
= f . fst <*> snd
= flip (f . fst) =<< snd
The liftA2 one just might be the clearest and the least noisy.
Related
I tried for a long time to reduct this function in haskell, I want to express for example:
mySum x y = x + y
mySum x y = (+) x y
mySum x = (+) x
mySum = (+) -- it's Messi's goal!
My function it a little more complex, but I really can't do it, I was looking out here and there, and I know there are some techniques, like modify the right side, and use flip. I tried and I got stuck here:
zipWith' :: (a -> b -> c) -> [a] -> [b] -> [c]
zipWith' f x y = map (uncurry f) (zip x y)
Steps:
zipWith' f x y = map (uncurry f) (zip x y)
zipWith' f x y = flip map (zip x y) (uncurry f)
zipWith' f x y = flip map (zip x y) $ uncurry f
and then I don't know how to continue...
I'm looking for an answer that could explain step by step how to achieve the "Messi's goal", I know is a lot to ask, so I will add as soon as I can a bounty to thank the effort
zipWith' f x y = map (uncurry f) (zip x y)
Rewrite application to composition and eta-reduce:
-- \y -> let g = map (uncurry f); h = zip x in (g . h) y
-- let g = map (uncurry f); h = zip x in g . h
zipWith' f x = map (uncurry f) . zip x
Rewrite infix to prefix:
-- g . h = (.) g h
zipWith' f x = (.) (map (uncurry f)) (zip x)
Rewrite application to composition and eta-reduce:
-- \x -> let g = (.) (map (uncurry f)); h = zip in (g . h) x
-- let g = (.) (map (uncurry f)); h = zip in g . h
zipWith' f = (.) (map (uncurry f)) . zip
Rewrite infix to prefix:
-- g . h = (.) g h
zipWith' f = (.) ((.) (map (uncurry f))) zip
Use flip to move f to the right-hand side:
-- flip f x y = f y x
zipWith' f = flip (.) zip ((.) (map (uncurry f)))
Rewrite application to composition:
-- g (h (i x)) = (g . h . i) x
zipWith' f = flip (.) zip (((.) . map . uncurry) f)
Rewrite application to composition and eta-reduce:
-- \f -> let g = flip (.) zip; h = (.) . map . uncurry in (g . h) f
-- let g = flip (.) zip; h = (.) . map . uncurry in g . h
zipWith' = (flip (.) zip) . ((.) . map . uncurry)
Remove redundant parentheses:
zipWith' = flip (.) zip . (.) . map . uncurry
And simplify to infix if you like:
zipWith' = (. zip) . (.) . map . uncurry
This result isn’t very readable, though.
Often when writing fully point-free code, you want to take advantage of the -> applicative and arrow combinators from Control.Arrow. Rather than trying to write a function like \ f x y -> ..., you can start by grouping the arguments into tuples to make them easier to rearrange and pipe around. In this case I’ll use \ (f, (x, y)) -> ...
\ (f, (x, y)) -> map (uncurry f) (zip x y)
We can eliminate the unpacking of (x, y) by applying uncurry to zip:
\ (f, (x, y)) -> map (uncurry f) (uncurry zip (x, y))
\ (f, xy) -> map (uncurry f) (uncurry zip xy)
Now we have a simple case: applying two functions (uncurry and uncurry zip) to two arguments (f and xy), then combining the results (with map). For this we can use the *** combinator from Control.Arrow, of type:
(***) :: Arrow a => a b c -> a b' c' -> a (b, b') (c, c')
Specialised to functions, that’s:
(***) #(->) :: (b -> c) -> (b' -> c') -> (b, b') -> (c, c')
This just lets us apply a function to each element of a pair. Perfect!
uncurry *** uncurry zip
:: (a -> b -> c, ([x], [y])) -> ((a, b) -> c, [(x, y)])
You can think of uncurry f as combining the elements of a pair using the function f. So here we can combine the results using uncurry map:
uncurry map . (uncurry *** uncurry zip)
:: (a -> b -> c, ([a], [b])) -> [c]
And you can think of curry as turning a function on tuples into a multi-argument function. Here we have two levels of tuples, the outer (f, xy) and the inner (x, y). We can unpack the outer one with curry:
curry $ uncurry map . (uncurry *** uncurry zip)
:: (a -> b -> c) -> ([a], [b]) -> [c]
Now, you can think of fmap f in the -> applicative as “skipping over” the first argument:
fmap #((->) _) :: (a -> b) -> (t -> a) -> t -> b
So we can unpack the second tuple using fmap curry:
fmap curry $ curry $ uncurry map . (uncurry *** uncurry zip)
:: (a -> b -> c) -> [a] -> [b] -> [c]
And we’re done! Or not quite. When writing point-free code, it pays to break things out into many small reusable functions with clearer names, for example:
zipWith' = untuple2 $ combineWith map apply zipped
where
untuple2 = fmap curry . curry
combineWith f g h = uncurry f . (g *** h)
apply = uncurry
zipped = uncurry zip
However, while knowing these techniques is useful, all this is just unproductive trickery that’s easy to get lost in. Most of the time, you should only use point-free style in Haskell when it’s a clear win for readability, and neither of these results is clearer than the simple original version:
zipWith' f x y = map (uncurry f) (zip x y)
Or a partially point-free version:
zipWith' f = map (uncurry f) .: zip
where (.:) = (.) . (.)
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 <$> g) <*> h
but I couldn't figure out how to do the more general case.
We have:
k x y = (f (g x)) (h y)
and we wish to write k in point-free style.
The first argument passed to k is x. What do we need to do with x? Well, first we need to call g on it, and then f, and then do something fancy to apply this to (h y).
k = fancy . f . g
What is this fancy? Well:
k x y = (fancy . f . g) x y
= fancy (f (g x)) y
= f (g x) (h y)
So we desire fancy z y = z (h y). Eta-reducing, we get fancy z = z . h, or fancy = (. h).
k = (. h) . f . g
A more natural way to think about it might be
┌───┐ ┌───┐
x ───│ g │─── g x ───│ │
/ └───┘ │ │
(x, y) │ f │─── f (g x) (h y)
\ ┌───┐ │ │
y ───│ h │─── h y ───│ │
└───┘ └───┘
└──────────────────────────────┘
k
Enter Control.Arrow:
k = curry ((g *** h) >>> uncurry f)
Take a look at online converter
It converted
f' x y = f (g x) (h y)
into
f' = (. h) . f . g
with the flow of transformations
f' = id (fix (const (flip ((.) . f . g) h)))
f' = fix (const (flip ((.) . f . g) h))
f' = fix (const ((. h) . f . g))
f' = (. h) . f . g
This is slightly longer, but a little easier to follow, than (. h) . f. g.
First, rewrite f' slightly to take a tuple instead of two arguments. (In otherwords, we're uncurrying your original f'.)
f' (x, y) = f (g x) (h y)
You can pull a tuple apart with fst and snd instead of pattern matching on it:
f' t = f (g (fst t)) (h (snd t))
Using function composition, the above becomes
f' t = f ((g . fst) t) ((h . snd) t)
which, hey, looks a lot like the version you could make point-free using applicative style:
f' = let g' = g . fst
h' = h . snd
in (f <$> 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 <$> g') <*> h'
(This is very similar, by the way, to the Control.Arrow solution added by Lynn.)
Using the "three rules of operator sections" as applied to the (.) function composition operator,
(.) f g = (f . g) = (f .) g = (. g) f -- the argument goes into the free slot
-- 1 2 3
this is derivable by a few straightforward mechanical steps:
k x y = (f (g x)) (h y) -- a (b c) === (a . b) c
= (f (g x) . h) y
= (. h) (f (g x)) y
= (. h) ((f . g) x) y
= ((. h) . (f . g)) x y
Lastly, (.) is associative, so the inner parens may be dropped.
The general procedure is to strive to reach the situation where eta-reduction can be performed, i.e. we can get rid of the arguments if they are in same order and are outside any parentheses:
k x y = (......) y
=>
k x = (......)
Lather, rinse, repeat.
Another trick is to turn two arguments into one, or vice versa, with the equation
curry f x y = f (x,y)
so, your
f (g x) (h y) = (f.g) x (h y) -- by B-combinator rule
= (f.g.fst) (x,y) ((h.snd) (x,y))
= (f.g.fst <*> h.snd) (x,y) -- by S-combinator rule
= curry (f.g.fst <*> h.snd) x y
This is the same as the answer by #chepner, but presented more concisely.
So, you see, your (f.g <*> h) x1 just becomes (f.g.fst <*> h.snd) (x,y). Same difference.
1(because, for functions, (<$>) = (.))
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
I have seen a lot of functions being defined according to the pattern (f .) . g. For example:
countWhere = (length .) . filter
duplicate = (concat .) . replicate
concatMap = (concat .) . map
What does this mean?
The dot operator (i.e. (.)) is the function composition operator. It is defined as follows:
infixr 9 .
(.) :: (b -> c) -> (a -> b) -> a -> c
f . g = \x -> f (g x)
As you can see it takes a function of type b -> c and another function of type a -> b and returns a function of type a -> c (i.e. which applies the first function to the result of the second function).
The function composition operator is very useful. It allows you to pipe the output of one function into the input of another function. For example you could write a tac program in Haskell as follows:
main = interact (\x -> unlines (reverse (lines x)))
Not very readable. Using function composition however you could write it as follows:
main = interact (unlines . reverse . lines)
As you can see function composition is very useful but you can't use it everywhere. For example you can't pipe the output of filter into length using function composition:
countWhere = length . filter -- this is not allowed
The reason this is not allowed is because filter is of type (a -> Bool) -> [a] -> [a]. Comparing it with a -> b we find that a is of type (a -> Bool) and b is of type [a] -> [a]. This results in a type mismatch because Haskell expects length to be of type b -> c (i.e. ([a] -> [a]) -> c). However it's actually of type [a] -> Int.
The solution is pretty simple:
countWhere f = length . filter f
However some people don't like that extra dangling f. They prefer to write countWhere in pointfree style as follows:
countWhere = (length .) . filter
How do they get this? Consider:
countWhere f xs = length (filter f xs)
-- But `f x y` is `(f x) y`. Hence:
countWhere f xs = length ((filter f) xs)
-- But `\x -> f (g x)` is `f . g`. Hence:
countWhere f = length . (filter f)
-- But `f . g` is `(f .) g`. Hence:
countWhere f = (length .) (filter f)
-- But `\x -> f (g x)` is `f . g`. Hence:
countWhere = (length .) . filter
As you can see (f .) . g is simply \x y -> f (g x y). This concept can actually be iterated:
f . g --> \x -> f (g x)
(f .) . g --> \x y -> f (g x y)
((f .) .) . g --> \x y z -> f (g x y z)
(((f .) .) .) . g --> \w x y z -> f (g w x y z)
It's not pretty but it gets the job done. Given two functions you can also write your own function composition operators:
f .: g = (f .) . g
f .:: g = ((f .) .) . g
f .::: g = (((f .) .) .) . g
Using the (.:) operator you could write countWhere as follows instead:
countWhere = length .: filter
Interestingly though you could write (.:) in point free style as well:
f .: g = (f .) . g
-- But `f . g` is `(.) f g`. Hence:
f .: g = (.) (f .) g
-- But `\x -> f x` is `f`. Hence:
(f .:) = (.) (f .)
-- But `(f .)` is `((.) f)`. Hence:
(f .:) = (.) ((.) f)
-- But `\x -> f (g x)` is `f . g`. Hence:
(.:) = (.) . (.)
Similarly we get:
(.::) = (.) . (.) . (.)
(.:::) = (.) . (.) . (.) . (.)
As you can see (.:), (.::) and (.:::) are just powers of (.) (i.e. they are iterated functions of (.)). For numbers in Mathematics:
x ^ 0 = 1
x ^ n = x * x ^ (n - 1)
Similarly for functions in Mathematics:
f .^ 0 = id
f .^ n = f . (f .^ (n - 1))
If f is (.) then:
(.) .^ 1 = (.)
(.) .^ 2 = (.:)
(.) .^ 3 = (.::)
(.) .^ 4 = (.:::)
That brings us close to the end of this article. For a final challenge let's write the following function in pointfree style:
mf a b c = filter a (map b c)
mf a b c = filter a ((map b) c)
mf a b = filter a . (map b)
mf a b = (filter a .) (map b)
mf a = (filter a .) . map
mf a = (. map) (filter a .)
mf a = (. map) ((filter a) .)
mf a = (. map) ((.) (filter a))
mf a = ((. map) . (.)) (filter a)
mf = ((. map) . (.)) . filter
mf = (. map) . (.) . filter
We can further simplify this as follows:
compose f g = (. f) . (.) . g
compose f g = ((. f) . (.)) . g
compose f g = (.) ((. f) . (.)) g
compose f = (.) ((. f) . (.))
compose f = (.) ((. (.)) (. f))
compose f = ((.) . (. (.))) (. f)
compose f = ((.) . (. (.))) (flip (.) f)
compose f = ((.) . (. (.))) ((flip (.)) f)
compose = ((.) . (. (.))) . (flip (.))
Using compose you can now write mf as:
mf = compose map filter
Yes it is a bit ugly but it's also a really awesome mind-boggling concept. You can now write any function of the form \x y z -> f x (g y z) as compose f g and that is very neat.
This is a matter of taste, but I find such style to be unpleasant. First I'll describe what it means, and then I suggest an alternative that I prefer.
You need to know that (f . g) x = f (g x) and (f ?) x = f ? x for any operator ?. From this we can deduce that
countWhere p = ((length .) . filter) p
= (length .) (filter p)
= length . filter p
so
countWhere p xs = length (filter p xs)
I prefer to use a function called .:
(.:) :: (r -> z) -> (a -> b -> r) -> a -> b -> z
(f .: g) x y = f (g x y)
Then countWhere = length .: filter. Personally I find this a lot clearer.
(.: is defined in Data.Composition and probably other places too.)
When I don't grasp how an expression in Haskell works I often find it helps to decompose it into a more basic form.
Using the following definitions
sequenceA :: (Applicative f) => [f a] -> f [a]
sequenceA [] = pure []
sequenceA (x:xs) = (:) <$> x <*> sequenceA xs
instance Applicative ((->) r) where
pure x = (\_ -> x)
f <*> g = \x -> f x (g x)
I rewrote sequenceA [(+3),(+2)] 3 as
(\_ -> (:)) <*> (+3) <*> ((\_ -> (:)) <*> (+2) <*> (\_-> [])) $ 3
And then turned it into (please excuse the format; I'm not sure what the convention is for splitting lines)
(\d ->(\c->(\b -> (\a -> (\_ -> (:)) a (+3) a) b (\_ -> (:)) b) c (+2) c) d (\_ -> []) d) 3
This seems right when I work through it by hand, but I can't get GHCi to accept it. What have I done wrong here? My second question is how to convert from this form into functional composition. I've tried substituing dots in various combinations, but GHCi rejects all of them....
Being an idle goodfornothing, I thought I would make a computer do the expansion for me. So into GHCi, I typed
let pu x = "(\\_ -> " ++ x ++ ")"
let f >*< a = "(\\g -> " ++ f ++ " g (" ++ a ++ " g))"
So now I have funny versions of pure and <*> which map strings which look like expressions to string which look like more complicated expressions. I then defined, similarly, the analogue of sequenceA, replacing functions by strings.
let sqa [] = pu "[]" ; sqa (f : fs) = (pu "(:)" >*< f) >*< sqa fs
I was then able to generate the expanded form of the example as follows
putStrLn $ sqa ["(+3)","(+2)"] ++ " 3"
which duly printed
(\g -> (\g -> (\_ -> (:)) g ((+3) g)) g ((\g -> (\g -> (\_ -> (:)) g ((+2) g)) g ((\_ -> []) g)) g)) 3
This last, copied to the prompt, yielded
[6,5]
Comparing the output from my "metaprogram" with the attempt in the question shows a shorter initial prefix of lambdas, arising from a shallower nesting of <*> operations. Remember, it's
(pure (:) <*> (+3)) <*> ((pure (:) <*> (+2)) <*> pure [])
so the outer (:) should be only three lambdas deep. I suspect the proposed expansion may correspond to a differently bracketed version of the above, perhaps
pure (:) <*> (+3) <*> pure (:) <*> (+2) <*> pure []
Indeed, when I evaluate
putStrLn $ pu "(:)" >*< "(+3)" >*< pu "(:)" >*< "(+2)" >*< pu "[]" ++ " 3 "
I get
(\g -> (\g -> (\g -> (\g -> (\_ -> (:)) g ((+3) g)) g ((\_ -> (:)) g)) g ((+2) g)) g ((\_ -> []) g)) 3
which looks like it matches the (updated)
(\d -> (\c -> (\b -> (\a -> (\_ -> (:)) a ((+3) a)) b ((\_ -> (:)) b)) c ((+2) c)) d ((\_ -> []) d)) 3
I hope this machine-assisted investigation helps to clarify what's going on.
You rewrote (\_ -> (:)) <*> (+3) as \a -> (\_ -> (:)) a (+3) a, which is rewriting f <*> g as f x g x instead of f x (g x). I think you made that mistake for every <*>.
It might be easier to use combinators, e.g. _S and _K, symbolically, and not their definitions as lambda-expressions,
_S f g x = f x (g x)
_K x y = x
With functions, fmap is (.) and <*> is _S, as others already mentioned. So,
sequenceA [(+3),(+2)] 3 ==
( ((:) <$> (+3)) <*> sequenceA [(+2)] ) 3 ==
_S ((:).(+3)) ( ((:) <$> (+2)) <*> pure [] ) 3 ==
_S ((:).(+3)) ( _S ((:).(+2)) (_K []) ) 3 ==
((:).(+3)) 3 ( _S ((:).(+2)) (_K []) 3 ) ==
((:).(+3)) 3 ( ((:).(+2)) 3 (_K [] 3) ) ==
(6:) ( (5:) [] ) ==
[6,5]
So it might be easier to decompose expressions down to basic functions and combinators and stop there (i.e. not decomposing them to their lambda expressions), using their "re-write rules" in manipulating the expression to find its more comprehensible form.
If you wanted to, you could now write down for yourself a more abstract, informal re-write rule for sequenceA as
sequenceA [f,g,..., z] ==
_S ((:).f) . _S ((:).g) . _S ..... . _S ((:).z) . _K []
and so
sequenceA [f,g,..., z] a ==
((:).f) a $ ((:).g) a $ ..... $ ((:).z) a $ _K [] a ==
(f a:) $ (g a:) $ ..... $ (z a:) $ [] ==
[f a, g a, ..., z a]
and hence
sequenceA fs a == map ($ a) fs == flip (map . flip ($)) fs a
to wit,
Prelude Control.Applicative> flip (map . flip ($)) [(+3),(+2)] 3
[6,5]
import Prelude hiding (foldr)
import Control.Applicative
import Data.Foldable
import Data.Traversable
left, right :: (Applicative f, Traversable t) => (a -> b -> b) -> b -> t (f a) -> f b
left f z = fmap (foldr f z) . sequenceA
right f z = foldr (liftA2 f) (pure z)
I have a strong suspicion that the expressions left and right are equal, but how to prove it?
Here's a start at least:
\f z -> fmap (foldr f z) . sequenceA
== (definition of Foldable foldr)
\f z -> fmap (foldr f z . toList) . sequenceA
== (distributivity of fmap)
\f z -> fmap (foldr f z) . fmap toList . sequenceA
== (need to prove this step, but it seems intuitive to me)
\f z -> fmap (foldr f z) . sequenceA . toList
\f z -> foldr (liftA2 f) (pure z)
== (definition of Foldable foldr)
\f z -> foldr (liftA2 f) (pure z) . toList
If you can prove that fmap toList . sequenceA = sequenceA . toList, and that your original claim holds for t = [] you should be good to go.