According to RosettaCode, the Y Combinator in Scheme is implemented as
(define Y
(λ (h)
((λ (x) (x x))
(λ (g)
(h (λ args (apply (g g) args)))))))
Of course, the traditional Y Combinator is λf.(λx. f(x x))(λx. f(x x))
My question, then, is about h and args, which don't appear in the mathematical definition, and about apply, which seems like it should either be in both halves of the Combinator or in neither.
Can someone help me understand what is going on here?
Lets start off with the lambda calculus version traslated to Scheme:
(λ (f)
((λ (x) (f (x x)))
(λ (x) (f (x x)))))
I'd like to simplify this since I see (λ (x) (f x x)) is repeated twice. You can substitute the beginning there to this:
(λ (f)
((λ (b) (b b))
(λ (x) (f (x x)))))
Scheme is an eager language so it will go into an infinite loop. In order to avoid that we make a proxy.. Imagine you have + that takes two numbers, you can substitute it with (λ (a b) (+ a b)) without the result being changed. Lets do that with the code:
(λ (f)
((λ (b) (b b))
(λ (x) (f (λ (p) ((x x) p))))))
Actully this has its own name. It's called the Z combinator. (x x) is not done when f is applied only when the supplied proxy is applied. Delayed one step. It might look strange but I know (x x) becomes a function so this is exactly the same as my + substitution above.
In Lambda calculus all functions takes one argument. If you see f x y it's actually the same as ((f x) y) in Scheme. If you want it to work with functions of all arities your substitution needs to reflect that. In Scheme we have rest arguments and apply to do this.
(λ (f)
((λ (b) (b b))
(λ (x) (f (λ p (apply (x x) p))))))
This isn't neede if you only are going to use one arity functions as in lambda calculus.
Notice that in your code you use h instead of f. It doesn't really matter what you call the variables. This is the same code with different names. So this is the same:
(λ (rec-fun)
((λ (yfun) (yfun yfun))
(λ (self) (rec-fun (λ args (apply (self self) args))))))
Needless to say (yfun yfun) and (self self) does the same thing.
Related
While going through this article about Y-combinator (which I highly recommend), I stumbled over this transformation :
(define Y
(lambda (f)
((lambda (x) (x x))
(lambda (x) (f (x x))))))
Note that we can apply the inner lambda expression to its argument to get an equivalent version of Y:
(define Y
(lambda (f)
((lambda (x) (f (x x)))
(lambda (x) (f (x x))))))
Could someone please explain me how did we get to the second version of Y? Which steps did we follow to get there?
You are applying (lambda (x) (x x))(lambda (x) (f (x x)))
Do the application to get (lambda (x) (f (x x))(lambda (x) (f (x x))
Notice the left lambda creates 2 copies of its argument, which is the right lambda.
How can I change the code without the help of let that (g x) is just evaluated once?
(define f
(lambda (x)
(i (g x) (h (g x)))))
You can write the solution with the let, then transform it using the following equivalence between let and lambda:
(let ((name expression1))
expression2)
==
((lambda (name) expression2) expression1)
If let is not allowed, I would use define inside the define:
(define f
(lambda (x)
(define y (g x))
(i y (h y))))
However, I guess the other answer is the desired one. Because it uses the almighty lambda :) .
I am really new to scheme functional programming. I recently came across Y-combinator function in lambda calculus, something like this Y ≡ (λy.(λx.y(xx))(λx.y(xx))). I wanted to implement it in scheme, i searched alot but i didn't find any implementation which exactly matches the above given structure. Some of them i found are given below:
(define Y
(lambda (X)
((lambda (procedure)
(X (lambda (arg) ((procedure procedure) arg))))
(lambda (procedure)
(X (lambda (arg) ((procedure procedure) arg)))))))
and
(define Y
(lambda (r)
((lambda (f) (f f))
(lambda (y)
(r (lambda (x) ((y y) x)))))))
As you can see, they dont match with the structure of this Y ≡ (λy.(λx.y(xx))(λx.y(xx))) combinator function. How can I implement it in scheme in exactly same way?
In a lazy language like Lazy Racket you can use the normal order version, but not in any of the applicative order programming languages like Scheme. They will just go into an infinite loop.
The applicative version of Y is often called a Z combinator:
(define Z
(lambda (f)
((lambda (g) (g g))
(lambda (g)
(f (lambda args (apply (g g) args)))))))
Now the first thing that happens when this is applied is (g g) and since you can always substitute a whole application with the expansion of it's body the body of the function can get rewritten to:
(define Z
(lambda (f)
((lambda (g)
(f (lambda args (apply (g g) args))))
(lambda (g)
(f (lambda args (apply (g g) args)))))))
I haven't really changed anything. It's just a little more code that does exactly the same. Notice this version uses apply to support multiple argument functions. Imagine the Ackermann function:
(define ackermann
(lambda (m n)
(cond
((= m 0) (+ n 1))
((= n 0) (ackermann (- m 1) 1))
(else (ackermann (- m 1) (ackermann m (- n 1)))))))
(ackermann 3 6) ; ==> 509
This can be done with Z like this:
((Z (lambda (ackermann)
(lambda (m n)
(cond
((= m 0) (+ n 1))
((= n 0) (ackermann (- m 1) 1))
(else (ackermann (- m 1) (ackermann m (- n 1))))))))
3
6) ; ==> 509
Notice the implementations is exactly the same and the difference is how the reference to itself is handled.
EDIT
So you are asking how the evaluation gets delayed. Well the normal order version looks like this:
(define Y
(lambda (f)
((lambda (g) (g g))
(lambda (g) (f (g g))))))
If you look at how this would be applied with an argument you'll notice that Y never returns since before it can apply f in (f (g g)) it needs to evaluate (g g) which in turn evaluates (f (g g)) etc. To salvage that we don't apply (g g) right away. We know (g g) becomes a function so we just give f a function that when applied will generate the actual function and apply it. If you have a function add1 you can make a wrapper (lambda (x) (add1 x)) that you can use instead and it will work. In the same manner (lambda args (apply (g g) args)) is the same as (g g) and you can see that by just applying substitution rules. The clue here is that this effectively stops the computation at each step until it's actually put into use.
Here is the Y-combinator in Racket:
#lang lazy
(define Y (λ(f)((λ(x)(f (x x)))(λ(x)(f (x x))))))
(define Fact
(Y (λ(fact) (λ(n) (if (zero? n) 1 (* n (fact (- n 1))))))))
(define Fib
(Y (λ(fib) (λ(n) (if (<= n 1) n (+ (fib (- n 1)) (fib (- n 2))))))))
Here is the Y-combinator in Scheme:
(define Y
(lambda (f)
((lambda (x) (x x))
(lambda (g)
(f (lambda args (apply (g g) args)))))))
(define fac
(Y
(lambda (f)
(lambda (x)
(if (< x 2)
1
(* x (f (- x 1))))))))
(define fib
(Y
(lambda (f)
(lambda (x)
(if (< x 2)
x
(+ (f (- x 1)) (f (- x 2))))))))
(display (fac 6))
(newline)
(display (fib 6))
(newline)
My question is: Why does Scheme require the apply function but Racket does not?
Racket is very close to plain Scheme for most purposes, and for this example, they're the same. But the real difference between the two versions is the need for a delaying wrapper which is needed in a strict language (Scheme and Racket), but not in a lazy one (Lazy Racket, a different language).
That wrapper is put around the (x x) or (g g) -- what we know about this thing is that evaluating it will get you into an infinite loop, and we also know that it's going to be the resulting (recursive) function. Because it's a function, we can delay its evaluation with a lambda: instead of (x x) use (lambda (a) ((x x) a)). This works fine, but it has another assumption -- that the wrapped function takes a single argument. We could just as well wrap it with a function of two arguments: (lambda (a b) ((x x) a b)) but that won't work in other cases too. The solution is to use a rest argument (args) and use apply, therefore making the wrapper accept any number of arguments and pass them along to the recursive function. Strictly speaking, it's not required always, it's "only" required if you want to be able to produce recursive functions of any arity.
On the other hand, you have the Lazy Racket code, which is, as I said above, a different language -- one with call-by-need semantics. Since this language is lazy, there is no need to wrap the infinitely-looping (x x) expression, it's used as-is. And since no wrapper is required, there is no need to deal with the number of arguments, therefore no need for apply. In fact, the lazy version doesn't even need the assumption that you're generating a function value -- it can generate any value. For example, this:
(Y (lambda (ones) (cons 1 ones)))
works fine and returns an infinite list of 1s. To see this, try
(!! (take 20 (Y (lambda (ones) (cons 1 ones)))))
(Note that the !! is needed to "force" the resulting value recursively, since Lazy Racket doesn't evaluate recursively by default. Also, note the use of take -- without it, Racket will try to create that infinite list, which will not get anywhere.)
Scheme does not require apply function. you use apply to accept more than one argument.
in the factorial case, here is my implementation which does not require apply
;;2013/11/29
(define (Fact-maker f)
(lambda (n)
(cond ((= n 0) 1)
(else (* n (f (- n 1)))))))
(define (fib-maker f)
(lambda (n)
(cond ((or (= n 0) (= n 1)) 1)
(else
(+ (f (- n 1))
(f (- n 2)))))))
(define (Y F)
((lambda (procedure)
(F (lambda (x) ((procedure procedure) x))))
(lambda (procedure)
(F (lambda (x) ((procedure procedure) x))))))
How do I convert these procedures in Scheme to CPS form?
(lambda (x y)
((x x) y))
(lambda (x)
(lambda (f)
(f (lambda (y)
(((x x) f) y))))
((lambda (x) (x x)
(lambda (x) (x x))
*This is not any homework!
See Programming Languages, Application and Interpretation, starting around Chapter 15. Chapter 18 talks about how to do it automatically, but if you're not familiar with thinking about expressing a function that does "what to do next", you'll probably want to try the finger exercises first.
Don't have someone do it for you: you'll really want to understand the process and be able to do it by hand, independent of Scheme or otherwise. It comes up especially in Asynchronous JavaScript web programming, where you really have no choice but to do the transform.
In the CPS transform, all non-primitive functions need to now consume a function that represents "what-to-do-next". That includes all lambdas. Symmetrically, any application of a non-primitive function needs to provide a "what-to-do-next" function, and stuff the rest of the computation in that function.
So if we had a program to compute a triangle's hypothenuse:
(define (hypo a b)
(define (square x) (* x x))
(define (add x y) (+ x y))
(sqrt (add (square a)
(square b))))
and if we state that the only primitive applications here are *, +, and sqrt, then all the other function definitions and function calls need to be translated, like this:
(define (hypo/k a b k)
(define (square/k x k)
(k (* x x)))
(define (add/k x y k)
(k (+ x y)))
(square/k a
(lambda (a^2)
(square/k b
(lambda (b^2)
(add/k a^2 b^2
(lambda (a^2+b^2)
(k (sqrt a^2+b^2)))))))))
;; a small test of the function.
(hypo/k 2 3 (lambda (result) (display result) (newline)))
The last expression shows that you end up having to compute "inside-out", and that the transformation is pervasive: all lambdas in the original source program end up needing to take an additional argument, and all non-primitive applications need to stuff "what-to-do-next" as that argument.
Take a close look at section 17.2 of the cited book: it covers this, as well as 17.5, which talks about why you need to touch ALL the lambdas in the source program, so that the higher-order case works too.
As another example of the transform, applied for a higher-order case, let's say that we have:
(define (twice f)
(lambda (x)
(f (f x))))
Then the translation of something like this is:
(define (twice/k f k1)
(k1 (lambda ...)))
... because that lambda's just a value that can be passed to k1. But of course, the translation needs to run through the lambda as well.
We must first do the inner call to f with x (and remember that all non-primitive function applications need to pass an appropriate "what-to-do-next!"):
(define (twice/k f k1)
(k1 (lambda (x k2)
(f x (lambda (fx-val)
...)))))
... take that value and apply it again to f...
(define (twice/k f k1)
(k1 (lambda (x k2)
(f x (lambda (fx-val)
(f fx-val ...))))))
... and finally return that value to k2:
(define (twice/k f k1)
(k1 (lambda (x k2)
(f x (lambda (fx-val)
(f fx-val k2))))))
;; test. Essentially, ((twice square) 7)
(define (square/k x k) (k (* x x)))
(twice/k square/k
(lambda (squaresquare)
(squaresquare 7
(lambda (seven^4)
(display seven^4)
(newline)))))
You need to choose to what level you need/want to CPS-transform.
If you just want (lambda (x y) ((x x) y)) in continuation-passing(CP) style, then (lambda (k x y) (k ((x x) y))) will do fine.
If you want its arguments to be treated as being in CP style too, then you need a little more.
Suppose first that only the second argument (y) is in CP form and is thus really something like (lambda (k) (k y0)) and so needs to be called with some continuation to extract its value, then you would need:
(lambda (k x y)
(y (lambda (y0) (k ((x x) y0)) )) )
Finally assume that both x and y are in CP style. Then you would need something like:
(lambda (k x y)
(x (lambda (x0)
(x (lambda (x1)
(y (lambda (y0)
(k ((x0 x1) y0)) ))))
Here you have the freedom to reorder the calls to x and y. Or maybe you only need one call to x, because you know its value does not depend on the continuation it is called with. For example:
(lambda (k x y)
(y (lambda (y0)
(x (lambda (x0)
(k ((x0 x0) y0)) ))))
The other expressions you asked about can be transformed similarly.