Tuple addition in pointfree

Question

What is the shortest way we can express the function

f(a,b)(c,d)=(a+c,b+d)

in point-free notation?

pointfree.io gives us

uncurry (flip flip snd . (ap .) . flip flip fst . ((.) .) . (. (+)) . flip . (((.) . (,)) .) . (+))

which with a little bit of work can be shortened to

uncurry$(`flip`snd).((<*>).).(`flip`fst).((.).).(.(+)).flip.(((.).(,)).).(+)

for 76 bytes. But this still seems really long and complex for such a simple task. Is there any way we can express pairwise addition as a shorter point-free function?

To be clear by what I mean by point-free, a point-free declaration of a function involves taking existing functions and operators and applying them to each other in such a way that the desired function is created. Backticks, parentheses and literal values ([],0,[1..3], etc.) are allowed but keywords like where and let are not. This means:

• You may not assign any variables/functions

• You may not use lambdas

• You may not import

Here is the same question when it was a CMC

Answer 1

44 bytes

Got this from \x y -> (fst x + fst y, snd x + snd y)

(<*>).((,).).(.fst).(+).fst<*>(.snd).(+).snd

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Or, 42 bytes using do:

do a<-fst;((,).(a+).fst<*>).(.snd).(+).snd

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Answer 2

44 bytes

-8 bytes thanks to Ørjan Johansen. -3 bytes thanks to Bruce Forte.

(.).flip(.)<*>(zipWith(+).)$mapM id[fst,snd]

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Translates to:

f t1 t2 = zipWith (+) (mapM id [fst, snd] $ t1) (mapM id [fst, snd] $ t2)

67 bytes

-8 bytes thanks to Ørjan Johansen. -1 byte thanks to Bruce Forte.

If tuple output is required:

(((,).head<*>last).).((.).flip(.)<*>(zipWith(+).)$mapM id[fst,snd])

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Yup, me manually doing it doesn't produce ripe fruit. But I am happy with the [a] → (a, a) conversion.

listToPair ∷ [a] → (a, a)
listToPair = (,) . head <*> last
-- listToPair [a, b] = (a, b)

Now if there was a short function with m (a → b) → a → m b.

Answer 3

54 bytes

I honestly doubt that we'll beat @H.PWiz's 44 bytes solution, but nobody was using the fact that (,) implements the type class Functor, so here's another interesting one which isn't too bad:

((<*>snd).((,).).(.fst).(+).fst<*>).flip(fmap.(+).snd)

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Explanation

The implementation of the type class Functor for 2-Tuples are very similar to that of Either (from base-4.10.1.0):

instance Functor ((,) a) where
    fmap f (x,y) = (x, f y)

instance Functor (Either a) where
    fmap _ (Left x) = Left x
    fmap f (Right y) = Right (f y)

What this means for this challenge, is that the following function adds the second elements while keeping the first element of the second argument:

λ f = fmap.(+).snd :: Num a => (a, a) -> (a, a) -> (a, a)
λ f (1,-2) (3,-4)
(3,-6)

So if only we got some little helper helpPlz = \a b -> (fst a+fst b,snd b) we could do (helpPlz<*>).flip(fmap.(+).snd) and would be done. Luckily we have the tool pointfree which gives us:

helpPlz = (`ap` snd) . ((,) .) . (. fst) . (+) . fst

So by simply plugging that function back in we arrive at the above solution (note that (<*>) = ap which is in base).

Answer 4

60 bytes

I'm not seeing any uncurry love here, so I figured I'd pop in and fix that.

uncurry$(uncurry.).flip(.)(flip(.).(+)).(flip(.).((,).).(+))

I thought, with all of the fst and snd, that unpacking the arguments with uncurry might yield some results. Clearly, it was not as fruitful as I had hoped.