What does this combinator do: s (s k)

Question

I now understand the type signature of s (s k):

s (s k)  :: ((t1 -> t2) -> t1) -> (t1 -> t2) -> t1

And I can create examples that work without error in the Haskell WinGHCi tool:

Example:

s (s k) (\g -> 2) (\x -> 3)

returns 2.

Example:

s (s k) (\g -> g 3) successor

returns 4.

where successor is defined as so:

successor = (\x -> x + 1)

Nonetheless, I still don't have an intuitive feel for what s (s k) does.

The combinator s (s k) takes any two functions f and g. What does s (s k) do with f and g? Would you give me the big picture on what s (s k) does please?

Wilhelmina

The defination for S (S K) is missing. Is this the same s and k in #9592691 ? – Duologue

Btw, what is intuitive? Do you found en.wikipedia.org/wiki/Ouroboros intuitive? Can you imagine a snake eats itself and vanish? Or a robot that builds itself from itself? You need better sense on something acting on itself. – Duologue

Answer 1

Alright, let's look at what S (S K) means. I'm going to use these definitions:

S = \x y z -> x z (y z)
K = \x y   -> x

S (S K) = (\x y z -> x z (y z)) ((\x y z -> x z (y z)) (\a b -> a)) -- rename bound variables in K
        = (\x y z -> x z (y z)) (\y z -> (\a b -> a) z (y z)) -- apply S to K
        = (\x y z -> x z (y z)) (\y z -> (\b -> z) (y z)) -- apply K to z
        = (\x y z -> x z (y z)) (\y z -> z) -- apply (\_ -> z) to (y z)
        = (\x y z -> x z (y z)) (\a b -> b) -- rename bound variables
        = (\y z -> (\a b -> b) z (y z)) -- apply S to (\a b -> b)
        = (\y z -> (\b -> b) (y z)) -- apply (\a b -> b) to z
        = (\y z -> y z) -- apply id to (y z)

As you can see, it's just ($) with more specific type.

Halcomb

Another way to see this: the type is ((t1 -> t2) -> t1) -> (t1 -> t2) -> t1. Adding parentheses, we get ((t1 -> t2) -> t1) -> ((t1 -> t2) -> t1). Letting the type α stand for (t1 -> t2) -> t1, this is just α -> α, and so, by parametricity, s (s k) is the identity function with a more specific type. (And of course, ($) :: (a -> b) -> a -> b is also just the identity function with a more specific type.) – Helmsman

Indeed, if we η-reduce \y -> \z -> y z, we get \y -> y. – Halcomb

In combinators, S K y z = K z (y z) = z. Then S (S K) y z = S K z (y z) = K (y z) (z (y z)) = y z. – Murphree

@AntalS-Z, it's a bit cheeky to appeal to parametricity on a type you generalised yourself :P it's true that the type here pretty much demands that the function is a type-restricted identity, but e.g. (a -> a) -> (a -> a) is another type that is α -> α for some α, yet has plenty of non-identity values. – Bramwell

@benmachine: This is what I get for playing fast and loose with free theorems. Duly noted :-) – Helmsman