Y combinator in Scheme(jao.io)

68 pointsby momo-reina12 anni fa12 comments

http://jao.io/blog/2014/08/06/spj-s-y-combinator-in-scheme

6 comments

I like this explanation, because it focuses in it's latter portion on the one thing that actually made Y click for me: expanding it.

I think it's why it's so short: the Y-combinator makes perfect sense if you explain it in the context of expansion rather than recursion. The best demonstrator of how it combinators work isn't a bunch of algebraic derivations and explanations relative to tail-recursion, it's seeing how they expand to resolve the function given. For me, it was stepping through a factorial function in Racket.

In the end I wound up including it in the Heresy standard library.

viksit12 anni fa

That was exactly my problem too - I've tried in a complementary post to talk about some practical applications of the YCombinator.

http://www.viksit.com/tags/clojure/practical-applications-y-...

jarcane12 anni fa

That's pretty nice. I ultimately only included Y in Heresy as an academic exercise (I see the main purpose of it, if any, as being a playground for learning FP and Lisp concepts); on Racket at least, even with it's very good tail-call optimizations, it wasn't really performant enough by itself to be useful.

drostie12 anni fa

Yeah, I agree. To explain where I see it most firmly, though, I have to explain how I personally got to Y.

For me it was more interactive. You start with (Python syntax)

    def factorial(n):
        return 1 if n <= 1 else n * factorial(n - 1)

and a task: to remove the second word "factorial" from inside the function, with the hint that it becomes a parameter. (In other words, we will hand you the factorial function of type `int -> int` and you have to make the factorial function out of it.) You thus write:

    def factmaker(recurse):
        def fact_try(n):
            return 1 if n <= 1 else n * recurse(n - 1)
        return fact_try

This works if you pass in the function factorial -- but the problem is that now you don't have access to factorial at the outermost scope, so that

    factorial = factmaker(factmaker) # ...?

fails because the types are wrong: when `factmaker` becomes `recurse` it receives an int as its argument, but `factmaker` needs a function as its argument.

The key here for me was to accept `factorial = factmaker(factmaker)` and poke at `recurse` instead: how do we make it take a function as its argument? The only way to move forward is really that there really is which you can do with it is to replace `recurse` with `recurse(recurse)`, so that `recurse` can take a function as its argument:

    def factmaker(recurse):
        def fact_try(n):
            return 1 if n <= 1 else n * recurse(recurse)(n - 1)
        return fact_try
    factorial = factmaker(factmaker)

And surprisingly, that works! As you've said, you can now kind of see why it works in terms of expansion: you return 1 if n is the right value, or otherwise you return factmaker(factmaker), which we said was factorial, of n - 1. (You can also see why infinite types get involved: this requires a duck-typing approach where you say "function" but don't peek inside any further; the actual type of factmaker is (itself) -> int -> int.)

With a bit of abstraction to separate out the logic, restore the original form of factmaker, and handle more arguments, you can eventually get:

    def Y(maker):
        def F(f):
            def recurse(*a, **kw):
                return f(f)(*a, **kw)
            return maker(recurse)
        return F(F)
    
    def factmaker(fact):
        def fact2(n):
            return 1 if n <= 1 else n * fact(n - 1)
        return fact2
    
    factorial = Y(factmaker)

That's one step away from the actual lambda-calculus version.

throwaway_9983712 anni fa

klibertp12 anni fa

I recently found this talk: http://www.confreaks.com/videos/1287-rubyconf2012-y-not-adve... which I think it's a good introductory material: it not only derives Y, but starts with introducing all the tools needed for the derivation and explains rationale for using given tool at every step of derivation. Nice, however rather long.

Personally I found "The Little Schemer" explanation the first that made it click for me. Its unique style of making you do the exercises even if you don't do them (it's hard to explain, just go and read the book) gives you an impression that you really understand what is happening.

dionidium12 anni fa

There are some good summaries here, too:

http://stackoverflow.com/questions/93526/what-is-a-y-combina...

Full disclosure: I ended up answering there as well, but I won't say which one is mine.

stevenspasbo12 anni fa

I really liked this article: http://mvanier.livejournal.com/2897.html

I have been working through SICP and I ended up taking a weekend break from it to really understand everything in the article.

[deleted]12 anni fa

pash12 anni fa

To understand the theory of recursive functions you must understand fixed points, but I think there's a simpler way to understand the magic of the Y combinator.

Here's how it works. Say we have a recursive function like map:

    map f [] = []  -- We'll ignore the base case since there's no recursion here
    map f (x:xs) = f x : map f xs  -- This is the interesting part

We want to write this more primitively, without explicit recursion. How can we do that? Well, if it's possible at all, we obviously need to get rid of the map on the right hand side of the second case.

How? Well, let's abstract over the recursive call, replacing it with a function that we'll add as parameter to our definition:

    map' _ f [] = []  -- Still boring
    map' g f (x:xs) = f x : g g f xs  -- Pay attention for later!

(Read map' as "map-prime", i.e., a variant of the definition of map.)

All we've done so far is replaced map' where it should be on the right-hand side of its own definition with g, and then added g as parameter of the function. We end up with a double g on the right-hand side because of that added parameter.

OK, now what? Well, we need somehow to make g be map', the thing we're defining. which means we need somehow to pass our definition of map' to itself, so that (in the second case) we'd end up with something that evaluates to:

    map' map' f (x:xs) = f x : map' map' f xs

Then the right hand side would obviously be right; it's what we were shooting for at the start (assuming we can get map' map' to equal map, which is what we're trying to do). The left hand side looks a little strange [0], but it's just saying that we need the first parameter of our definition to be the the thing we're defining.

So how do we make this happen? Well, what is map map f (x:xs)? It's map applied to itself, then to some other arguments we need. So the key, it seems, is to figure out how to apply map to itself. Well, in the lambda calculus that's pretty easy [1]:

    why f x y = f f x y

That is, why is just

    lambda f. lambda x. lambda y. (f f) x) y

OK, so fix takes a function and applies it to itself (after applying it first to arguments x and y). Great. That's what we needed. Now we can take our defintion of map' (the one marked Pay attention for later! above) and do this:

    map = why map'

And I've re-used the name map there because we can easily show that this definition is equivalent to our original definition of map by showing that it reduces to the same expression:

    map f (x:xs) = why map' f (x:xs)
      == (by def. of `why`)
    map' map' f (x:xs)
      == (by def. of `map'`
    f x : map' map' f xs

Which, if you keep expanding the expression map' map' f xs, you will see is indeed equal to the definition of map.

Voilà. So that wasn't too hard. It's basically two steps: (1) abstract over the recursive call, then (2) figure out how to pass the function you're defining to itself. If you look at the Y combinator, you'll see that that's exactly what's going on there.

If this didn't make a lot of sense to you, it's probably because I wrote it on my iPad at 4:30 in the morning. It really is just those two steps: abstract over the recursive call, then self-apply. ...

To get deeper into the recursive mindset (if you're not there yet) and to build up deliberately to the Y combinator, take a look at one of my favorite books on functional programming, The Little Schemer [2]. (And watch out for the jelly stains!)

------

0. And it's ill typed, but never mind that. It's not possible to define the Y combinator (or any other fixed-point combinator) in the simply typed-lambda calculus, so just imagine that my Haskell-style syntax is untyped, like the basic lambda calculus.

1. Although, again, there's no way to give fix a good type in the simply typed lambda calculus.

2. http://www.amazon.com/The-Little-Schemer-4th-Edition/dp/0262...

amelius12 anni fa

Regarding your note "0", could anybody here explain how the Y combinator be properly typed in a more advanced type system? Is there a (practical) language that would allow it to be typed?

Also, would the Y combinator be a useful abstraction-aid in a compiler (for a typed language), or does it exist merely as a curiosity?

klibertp12 anni fa

Take a look at these examples:

    http://rosettacode.org/wiki/Y_combinator#Haskell
    http://rosettacode.org/wiki/Y_combinator#OCaml
    http://rosettacode.org/wiki/Y_combinator#Scala
    http://rosettacode.org/wiki/Y_combinator#Standard_ML
    http://rosettacode.org/wiki/Y_combinator#Swift

The Mu/Roll trick seems to be a default way of typing Y, while some languages (like OCaml) provide other type-system features like recursive types or polymorphic variants.

klibertp12 anni fa

> just imagine that my Haskell-style syntax

I wouldn't otherwise, but after reading this comment I have to ask: why did you use such a syntax instead of Scheme, which is what TFA uses?

EDIT: also, I second "The Little Schemer" suggestion. I'm working (for the nth time) through "The Seasoned Schemer" in preparation for "The Reasoned Schemer" and it still is very enjoyable. It's a bit like SICP but with a sense of humour and unique, engaging style of presentation. Well worth the read!

Sirenos12 anni fa

That was a really nice explanation. I like they you write.