Exercise 2.7
#lang racket(define (make-interval a b) (cons a b))(define (upper-bound interval) (cdr interval))(define (lower-bound interval) (car interval))
Exercise 2.8
#lang racket(define (sub-interval x y) (make-interval (- (lower-bound x) (upper-bound y)) (- (upper-bound x) (lower-bound y))))
Exercise 2.9
#lang racket;something like that(a b) + (c d) = (a+c b+d)width(ab) = (b-a)/2width(cd) = (d-c)/2width(add) = [(b+d)-(a+c)]/2 = [(b-a)+(d-c)]/2width(add) = width(ab) + width(cd);multiplication or division is not true;interval arithmethic is complicated and uncertain;especially div
Exercise 2.10
#lang racket(define (div-interval x y) (if (and (>= (upper-bound y) 0) (<= (lower-bound y) 0)) (error "Denominator spans zero" y) (mul-interval x (make-interval (/1.0 (upper-bound y)) (/1.0 (lower-bound y))))))
Exercise 2.11
#lang racket(define (mul-interval x y) (let ((upper-x (upper-bound x)) (lower-x (lower-bound x)) (upper-y (upper-bound y)) (lower-y (lower-bound y))) (cond ((and (>= upper-x 0) (>= lower-x 0) (>= upper-y 0) (>= lower-y 0)) (make-interval (* lower-x lower-y) (* upper-x upper-y)));x++,y++ ((and (>= upper-x 0) (>= lower-x 0) (>= upper-y 0) (< lower-y 0)) (make-interval (* upper-x lower-y) (* upper-x upper-y)));x++,y+- ((and (>= upper-x 0) (>= lower-x 0) (< upper-y 0) (< lower-y 0)) (make-interval (* upper-x lower-y) (* lower-x upper-y)));x++,y-- ((and (>= upper-x 0) (< lower-x 0) (>= upper-y 0) (>= lower-y 0)) (make-interval (* upper-y lower-x) (* upper-x upper-y)));x+-,y++ ((and (>= upper-x 0) (< lower-x 0) (>= upper-y 0) (< lower-y 0)) ;x+-,y+- need to do some special (let ((p1 (* lower-x lower-y)) (p2 (* lower-x upper-y)) (p3 (* upper-x upper-y)) (p4 (* upper-x lower-y))) (make-interval (min p1 p2 p3 p4) (max p1 p2 p3 p4)))) ((and (>= upper-x 0) (< lower-x 0) (< upper-y 0) (< lower-y 0)) (make-interval (* upper-x lower-y) (* lower-x lower-y)));x+-,y-- ((and (< upper-x 0) (< lower-x 0) (>= upper-y 0) (>= lower-y 0)) (make-interval (* upper-y lower-x) (* lower-y upper-x)));x--,y++ ((and (< upper-x 0) (< lower-x 0) (>= upper-y 0) (< lower-y 0)) (make-interval (* upper-y lower-x) (* lower-x lower-y)));x--,y+- ((and (< upper-x 0) (< lower-x 0) (< upper-y 0) (< lower-y 0)) (make-interval (* upper-x upper-y) (* lower-x lower-y)));x--,y-- (else (error "no this condition")))))
Exercise 2.12
#lang racket;transfer center width to lower and upper (define (make-center-width c w) (make-interval (- c w) (+ c w)));selector width(define (width i) (/ (- (upper-bound i) (lower-bound i)) 2));selector center(define (center i) (/ (+ (lower-bound i) (upper-bound i)) 2));transfer percent to width,then call make-center-width(define (make-center-percent c p) (let ((w (* c (/ p 100)))) (make-interval (- c w) (+ c w))));selector percent(define (percent i) (* 100 (/ (width i) (center i))))
Exercise 2.13-2.14
#lang racket(define (par1 r1 r2) (div-interval (mul-interval r1 r2) (add-interval r1 r2)))(define (par2 r1 r2) (let ((one (add-interval (div-interval one r1) (div-interval one r2))))));because the interval arithmetic r1/r1 is not 1.;so they can't be written in two algebraically equivalent ways:;(1/(1/R1)+(1/R2))*R1R2 = R1R2/(R2*R1/R1+R1*R2/R2)<>R1R2/(R1+R2)
Exercise 2.15-2.16
#lang racket;I think Eva Lu Ator is right,;by reducing the repeatability can make the tighter error bounds.;but i can't give the precise answer,just feel
Summarize
- 区间运算很复杂，一方面在于其边界条件考虑起来比较麻烦，另一个原因就是它的不确定性，没法保证区间除法产生固定的区间值。
- Don't repeat yourself,充分利用已有的方法，在已有的方法上堆砌新的方法。
- 题目很变态，累了就歇会~
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