(load "util.scm")

(define (require p)
  (if (not p) (amb)))

(define (an-element-of items)
  (require (not (null? items)))
  (amb (car items) (an-element-of (cdr items))))

(define (an-integer-starting-from n)
  (amb n (an-integer-starting-from (+ n 1))))

(display "\nex-4.35 - an-integer-between\n")

(define (an-integer-between a b)
  (require (<= a b))
  (amb a (an-integer-between (+ a 1) b)))

(define (a-pythagorean-triple-between low high)
  (let ((i (an-integer-between low high)))
    (let ((j (an-integer-between i high)))
      (let ((k (an-integer-between j high)))
        (require (= (+ (* i i) (* j j)) (* k k)))
        (list i j k)))))

(display "[done]\n")

(display "\nex-4.36 - all-pythagorean-triples\n")

; If we replace an-integer-between with an-integer-starting-from the variables
; i and j will stay at their initial value 1 while k will increment endlessly.
; Hence, only triplets of the form (1 1 n) will be generated.

(define (all-pythagorean-triples)
  (let ((i (an-integer-starting-from 1)))
    (let ((j (an-integer-starting-from i)))
      (let ((k (an-integer-starting-from j)))
        (require (= (+ (* i i) (* j j)) (* k k)))
        (list i j k)))))

; Write a procedure that actually will accomplish this. (That is, write a
; procedure for which repeatedly typing try-again would in principle eventually
; generate all Pythagorean triples.) 

; (display "[done]\n")

(display "\nex-4.37\n")

;(display "\nex-4.38\n")