Questions

  • Homework - coming due Wednesday.
  • I deputize you to explain and to help others.

Y Combinator

Once again, if this was somewhat opaque to you, I highly recommend you read Vanier’s blog post (linked on Enrichment) to better understand what’s going on.

(lambda (x) 
  (lambda (y)
    (lambda (z) 
	  ((((x y) (z w)) ((a b) (p q)))))))

;; K (lambda (x) (lambda (y) x))
;; S (lambda (x) (lambda (y) (lambda (z) ((x z) (y z)))))

<your-fav-lambda-expression> = (((K K) S) (S (S K)))

(define Y ...) 


;; To get around the lack of define in our PL 
(define !
  (lambda (n)
    (if (zero? n)
	  1
	  (* n (! (sub1 n))))))

;; Turn your 1-arg function created w/define into another lambda binding the name.
;; Then pass that whole deal as an argument to Y.

(Y
  (lambda (!)
	(lambda (n)
	  (if (zero? n)
		1
		(* n (! (sub1 n))))))
	)

((Y
  (lambda (!)
	(lambda (n)
	  (if (zero? n)
		1
		(* n (! (sub1 n))))))
	)
 5)
120

;; To get to Y



;; ([Listof X] -> Nat) -> ([Listof X] -> Nat)
(lambda (len)
  (lambda (ls)
    (if (empty? ls)
       0
       (add1 (len (cdr ls))))))

;; (Nat -> Nat) -> (Nat -> Nat)
((lambda (!)
   (lambda (n)
     (if (zero? n)
	1
	(* n (! (sub1 n))))))
 <some real-honest-to-goodness-factorial-function>)

(lambda (!)
   (lambda (n)
     (if (zero? n)
	1
	(* n ((lambda (n)
                (if (zero? n)
                    1
                    (* n (! (sub1 n)))))
              (sub1 n))))))

   (lambda (n)
     (if (zero? n)
	1
	(* n (! (sub1 n)))))

   (lambda (n)
     (if (zero? n)
	1
	(* n (if (zero? (sub1 n))
                 1
                 (* (sub1 n) (! (sub1 (sub1 n))))))))

;; Omega 
((lambda (x) (x x)) (lambda (x) (x x)))


(lambda (f) ;; f, the "business logic" of the function we want to compute
  ((lambda (x) (f (x x)))
   (lambda (x) (f (x x)))))

;; add1 => (λ (n) (add1 n))


(lambda (f) ;; f, the "business logic" of the function we want to compute
  ((lambda (x) (f (λ (y) ((x x) y)))) ;; η (eta) expansion
   (lambda (x) (f (λ (y) ((x x) y))))))

(f (f (f (f 1))))

;; The "call by value Y combinator"

(lambda (f) 
  ((lambda (x) (f (λ (y) ((x x) y))))
   (lambda (x) (f (λ (y) ((x x) y))))))

History: The bad days

Dynamic Scope

How we might have (mis-)implemented our interpreter

#lang racket

(define (apply-env env y)
  (env y))

(define (extend-env x a env)
  (λ (y) (if (eqv? x y) a (apply-env env y))))

(define (valof exp env)
  (match exp
    [`(* ,n1 ,n2) (* (valof n1 env) (valof n2 env))]
    [`(sub1 ,n1) (sub1 (valof n1 env))]
    [`(zero? ,n1) (zero? (valof n1 env))]
    [`,y #:when (symbol? y) (env y)]
    [`,n #:when (number? n) n]
    [`(if ,n1 ,n2 ,n3) (if (valof n1 env) (valof n2 env) (valof n3 env))]
    [`(let ([,x ,e]) ,body) 
	  (let ([a (valof e env)])
        (valof body (extend-env x a env)))]
    [`(lambda (,x) ,body) (λ (a env^) (valof body (extend-env x a env^)))]
    [`(,rator ,rand) ((valof rator env) (valof rand env) env)]))

(define (empty-env)
  (λ (y) 
    (error 'empty-env "unbound identifier ~s\n" y)))

What happens as a consequence?

  1. Programmers are no longer free to choose their own variable names.

  2. A programmer can no longer, in general, determine the value of a given expression without evaluating the whole program.

  3. Programmers can no longer safely focus on understanding a small part of the program in isolation.

  4. Implementations suffer, as environment lookup becomes more expensive and we loose much of the “cactus stack” discipline of our lexically-scoped environments.

Syntax vs. Semantics

We have the same syntax, but different semantics. More pernicious yet, the two behaviors are often the same for most programs. Eek! The times when they don’t behave surely violate the principle of least surprise.

Examples

Is this a real thing? (Lurking example)

Emacs lisp

my vs. local in Perl

Updated: