SICP 习题2.8 区间减法

我的想法是区间减法是a区间的每一个数减去b区间的每一个数，这些相减的结果在一个区间上。

也就是x的下界-y的上界是最小的，x的上界-y的下界是最大的。其他所有结果都在这两个数之间。

(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 (div-interval x y)
  (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 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))

(print-interval (add-interval a b))
(print-interval (mul-interval a b))
(print-interval (div-interval a b))
(print-interval (sub-interval a b))