旅の途中
你知道阿基米德原理吗?

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思路:

  1. 定义一个求导算法, 令其在抽象对象上执行求导操作。
    可以由以下规约规则完成:
    dc/dx=0
    dx/dx=1
    d(u+v)/dx=du/dx+dv/dx
    d(uv)/dx=u(dv/dx)+v(du/dx)
  2. 对抽象对象进行具体表示。

实现:

首先,让我们假定现在已经有了一些过程,可以实现对象的判断、构造以及选择。

(variable? x) ;是否为变量?
(same-variable? v1 v2) ;v1、v2是同一变量?
(=number? exp num) ;exp是否为数值且等于num?
(sum? x) ;x为加式?
(product? x) ;x为乘式?
(make-sum a1 a2) ;构造a1+a2 
(make-product m1 m2) ;构造a1*a2
(addend s) ;被加数
(augend s) ;加数
(multiplier s) ;被乘数
(multiplicand s) ;乘数

求导函数:

(define (deriv exp var)
  (cond ((number? exp) 0)
        ((variable? exp)
         (if (same-variable? exp var) 1 0))
        ((sum? exp)
         (make-sum (deriv (addend exp) var)
                   (deriv (augend exp) var)))
        ((product? exp)
         (make-sum
          (make-product (multiplier exp)
                        (deriv (multiplicand exp) var))
          (make-product (deriv (multiplier exp) var)
                        (multiplicand exp))))
        (else
         (error "unknown expression type -- DERIV" exp))))

判别函数,用来判断所求导的对象以及表达式类型。

(define (variable? x) (symbol? x)) 

(define (same-variable? v1 v2) 
  (and (variable? v1) (variable? v2) (eq? v1 v2)))

(define (=number? exp num) 
  (and (number? exp) (= exp num)))

(define (sum? x)
  (and (pair? x) (eq? (car x) '+)))

(define (product? x)
  (and (pair? x) (eq? (car x) '*)))

构造函数,构造和式和乘式:

(define (make-sum a1 a2)
  (cond ((=number? a1 0) a2)
        ((=number? a2 0) a1)
        ((and (number? a1) (number? a2)) (+ a1 a2))
        (else (list '+ a1 a2))))

(define (make-product m1 m2)
  (cond ((or (=number? m1 0) (=number? m2 0)) 0)
        ((=number? m1 1) m2)
        ((=number? m2 1) m1)
        ((and (number? m1) (number? m2)) (* m1 m2))
        (else (list '* m1 m2))))
;这里的加/乘式构造函数已经将式子进行了一定程度的化简

对加/乘数以及被加/乘数的选择函数。

(define (addend s) (cadr s)) 
(define (augend s) (caddr s))


(define (multiplier s) (cadr s)) 
(define (multiplicand s) (caddr s))
posted on 2017-05-04 14:58  CknightX  阅读(404)  评论(0编辑  收藏  举报