sicp每日一题[2.7]

2.7

Alyssa’s program is incomplete because she has not specified the implementation of the interval abstraction. Here is a definition of the interval constructor:
(define (make-interval a b) (cons a b))
Define selector supper-bound and lower-bound to complete the implementation.

这道题目难度不大，比起前面的选择函数，就是多了一个取最小值为下界，最大值为上界的步骤，直接利用现成的 min, max 即可。

; x 表示某个区间
(define (lower-bound x) (min (car x) (cdr x)))

(define (upper-bound x) (max (car x) (cdr x)))

(define helper (make-interval 1.0 1.0))
(define r1 (make-interval 6.12 7.48))
(define r2 (make-interval 4.465 4.935))

; 计算并联电路的电阻，注意这里不能化简成 r1*r2/(r1+r2) 的形式
(define (parallel-resistance r1 r2)
  (div-interval helper (add-interval (div-interval helper r1)
                                     (div-interval helper r2))))
                                     
(display (parallel-resistance r1 r2))

; 执行结果
[2.581558809636278, 2.97332259363673]