SICP 习题2.10 区间除法 被除区间横跨0的问题

横跨0会导致除法的结果是错的，因为$\frac{x}{0}$是没有结果的

(define (add-interval x y)
  (make-interval (+ (lower-bound x) (lower-bound y))
         (+ (upper-bound x) (upper-bound y))))

(define (mul-interval x y)
  (let ((p1 (* (lower-bound x) (lower-bound y)))
    (p2 (* (lower-bound x) (upper-bound y)))
    (p3 (* (upper-bound x) (lower-bound y)))
    (p4 (* (upper-bound x) (upper-bound y))))
    (make-interval (min p1 p2 p3 p4)
           (max p1 p2 p3 p4))))

(define (out-error)
  (newline)
  (display "ERROR"))

(define (div-interval x y)
  (if (and (< (lower-bound y) 0) (> (upper-bound y) 0))
    (out-error)
    (mul-interval x 
        (make-interval (/ 1.0 (lower-bound y))
                   (/ 1.0 (upper-bound y))))))

(define (sub-interval x y)
  (make-interval (- (lower-bound x) (upper-bound y))
         (- (upper-bound x) (lower-bound y))))

(define (width-interval z)
  (/ (- (upper-bound z) (lower-bound z)) 2.0))

(define make-interval cons)
(define lower-bound car)
(define upper-bound cdr)

(define print-interval
  (lambda (z)
    (newline)
    (display (lower-bound z))
    (display " ~ ")
    (display (upper-bound z))))

(define a (make-interval 1 2))
(define b (make-interval 3 4))
(newline)
(display "a : ")
(display (width-interval a))
(newline)
(display "b : ")
(display (width-interval b))

(print-interval (add-interval a b))
(newline)
(display "add : ")
(display (width-interval (add-interval a b)))

(print-interval (mul-interval a b))
(newline)
(display "mul : ")
(display (width-interval (mul-interval a b)))

(print-interval (div-interval a b))
(newline)
(display "div : ")
(display (width-interval (div-interval a b)))

(print-interval (sub-interval a b))
(newline)
(display "sub : ")
(display (width-interval (sub-interval a b)))

(define c (make-interval -1 2))
(div-interval a c)