lazy-ycombination2.rkt

#lang lazy


(define g
    (lambda (f)
      (lambda (n)
        (if (= n 0)
            1
            (* n (f (- n 1))))))) 
(define f
  (lambda (n)
        (if (= n 0)
            1
            (* n (f (- n 1))))))


(define Y1
  (lambda (f)
    (f (Y1 f))))


(define (Y2 self)
  (lambda (f)
    (f ((self self) f)))) ;; 这里没有 f 

(((Y2 Y2) g) 5)

;; self self to let
;; but here we are lazy lang we can just use (self self)

(define (Y3 self)
  (lambda (f)
   (let ((g (self self))) (f (g f)))))
;; My Mistake

(((Y3 Y3) g) 5)

;; ;;  (let ((x <expr1>)) <expr2>)
;;  ==> ((lambda (x) <expr2>) <expr1>)

(define (Y4 self)
  ((lambda (g) (lambda (f) (f (g f))))(self self)))

;; My mistake 

(((Y4 Y4) g) 5)


(define Y5 (Y4 Y4))
((Y5 g) 5)

(define Y6
  (let ((Y4-local (lambda (self) ((lambda (g) (lambda (f) (f (g f))))(self self)))))
    (Y4-local Y4-local)))

((Y6 g) 5)


(define Y7
  (let ((h (lambda (self) ((lambda (g) (lambda (f) (f (g f))))(self self)))))
    (h h)))
((Y7 g) 5)

(define Y8
  ((lambda (h) (h h))
   (lambda (self) ((lambda (g) (lambda (f) (f (g f))))(self self)))))

((Y8 g) 5)


;;(lambda (k) (lambda (f) (lambda (g) (f (g f))) (k k))) to h

(define Y9
  ((lambda (self) ((lambda (g) (lambda (f) (f (g f))))(self self)))
   (lambda (self) ((lambda (g) (lambda (f) (f (g f))))(self self)))))

;; shot 
(define Y10
  ((lambda (h) ((lambda (g) (lambda (f) (f (g f)))) (h h)))
   (lambda (h) ((lambda (g) (lambda (f) (f (g f)))) (h h)))))

;; here we can apply (h h) to (lambda (f) (f (g f)))
;; at the end is

(define Y11
  ((lambda (h) (lambda (f) (f ((h h) f))))
   (lambda (h) (lambda (f) (f ((h h) f))))))

;; or and we can like fact make (lambda (g) (lambda (f) (f (g f)))) as big G
;; because (lambda (g) (lambda (f) (f (g f)))) = G has no relation with (h h)

(define Y12
    (lambda (G)
      ((lambda (h) (G (h h)))
       (lambda (h) (G (h h))))))