# How to golf a fork in Haskell?

I'm trying to make a pointfree function f that can act like APL's dyadic forks/Husk's §, so f g h i a b equals g (h a) (i b), where g is a binary function, h and i are unary functions, and a and b are integers (I'm only dealing with integers and not other types right now). I converted the function to pointfree in the following steps, but it's 26 bytes with a lot of parentheses and periods and even a flip, so I'd like to know how to golf it.

f g h i a b=g(h a)(i b)
f g h i=(.i).g.h
f g h=(.g.h).(.)
f g=(.(.)).(flip(.)).(g.)
f=((.(.)).).((flip(.)).).(.)


Try it online!

• Plugging into pointfree.io gives an 20-byter ((flip.((.).)).).(.). Aug 4 at 23:21
• @Bubbler ...Oh. Can you put that as an answer, with an explanation of how it works? It'd be very helpful.
– user
Aug 4 at 23:22
• The reason I wrote it in a comment is that I don't know how it works :P Btw, should the arguments be ordered exactly like that? Aug 4 at 23:22

# 18 bytes

(flip.).(.).((.).)


Try it online!

Since your function has no argument repetition and no argument deletion, it essentially becomes a BC calculus golf (in Haskell terms, a golf using just B=(.) and C=flip). I will use B and C combinators and convert to Haskell code later.

\g h i a b -> g (h a) (i b)
\g h i a -> B (g (h a)) i
\g h i -> C (\a -> B (g (h a))) i
\g h -> C (\a -> B (g (h a)))

-- choice 1
\g h -> C (B B (B g h))
-- choice 1.1
\g -> B C (B (B B) (B g))
B C.B (B B).B
(flip.).(((.).).).(.)  -- 21 bytes
-- choice 1.2
\g -> B (B C (B B)) (B g)
B (B C (B B)).B
((flip.((.).)).).(.)  -- 20 bytes; pointfree.io

-- choice 2
\g h -> C (B (B B g) h)
\g -> B C (B (B B g))
B C.B.B B
(flip.).(.).((.).)  -- 18 bytes; shortest


As a bonus, if you can swap i and a in the definition of f, it becomes an ordered LC and allows a very short form using just (.):

\g h a i b -> g (h a) (i b)
\g h a i -> B (g (h a)) i
\g h a -> B (g (h a))
\g h -> B (B B g) h
\g -> B (B B g)
B B (B B)
(.).((.).)  -- 10 bytes

• Excellent answer. It seems like a mechanical process of enumeration. Why is pointfree.io not able to find the shortest one? Is the original insight that the problem was equivalent to a BC calculus golf the stumbling block? Aug 5 at 0:27
• @Jonah The primary reason is that a.b.c can be both (a.b).c and a.(b.c), which give different results at the end. I'm pretty sure pointfree.io uses only one of the possible abstractions to get some correct result. Aug 5 at 0:34
• what does "BC calculus" mean? Where can I find more info to learn about it? Aug 5 at 17:23
• B and C are combinators, i.e. essentially functions that are used in a system with no explicit variables. Asking for a point-free definition of a function is more or less asking for a definition in terms of combinators. BC calculus is simply the system you get if you allow combinators B and C (+ applications) only.
– zale
Aug 5 at 20:36
• @NooneAtAll To add to zale's comment, BC calculus is equivalent to linear logic/linear lambda calculus where each argument is used exactly once in some order. Aug 6 at 0:02