Exercise2-6 <---> Exercise2-8
Exercise 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 selectors upper-bound and lower-bound to complete the implementation.
Ru: Программа Лизы неполна, поскольку она не определила, как реализуется абстракция интервала. Вот определение конструктора интервала:
(define (make-interval a b) (cons a b))
Завершите реализацию, определив селекторы upper-bound и lower-bound.
Scheme solution:
It's a bit unclear what is smaller - a or b in this execrice, but we can figure out this from the text of chapter: a is smaller. Anyway, we can allow both variants using the following selectors:
(define (lower-bound interval)
(min (car interval) (cdr interval)))
(define (upper-bound interval)
(max (car interval) (cdr interval)))
But then it's better to implement correct order in constructor to simpluft selectors:
(define (make-interval a b)
(cons (min a b) (max a b)))
(define (lower-bound interval) (car interval))
(define (upper-bound interval) (cdr interval))
Exercise2-6 <---> Exercise2-8
Comments
| The answer above is incorrect. A and B are not end-points to an interval but each in itself is an interval. The add-interval function is adding two different intervals, each of which has a lower bound point and an upper bound point. The correct answer is: (define (upper-bound x) (cadr x)) (define (lower-bound x) (car x)) I might be wrong in my interpretation, would love to hear feedback. | ||||
| Posted by 76-14-65-64 at 2009-01-11 05:06:44 X | ||||
A and B are indeed intervals (each of them a pair with upper/lower bounds), but to extract the lower/upper bound, you definitely need min/max. Exercise didn't specify which of a or b is larger- (a & b comprises intervals A or | ||||
| Posted by 195 at 2009-02-18 08:04:39 X | ||||
Erm no, a and b are not intervals. They are end-points. We are taking two values like 1 and 2 and making an interval (1,2). You can see this is the case from the procedures given in the text. | ||||
| Posted by 204 at 2009-06-24 12:37:11 X | ||||
so the solution above is correct.