scheme - 在 Racket 中使用调度表

Question

我正在编写一个小程序来计算以列表形式表示的表达式的简单导数，即 2x^2 表示为 (* 2 (exp x 2)) 和我定义了以下函数:

;; derivative of a constant is 0
(define (diff-constant x E) 0)

;; computes derivative of E with respect to x where E can only be of the form
;; (+ x x ...)
(define (diff-sum x E)
  (cond ((or (not (list? E)) (null? E)) 0)
        ((eq? (car E) x) (+ 1 (diff-sum x (cdr E))))
        (else (diff-sum x (cdr E)))))

;; computes derivative of E with respect to x where E can only be of the form
;; (* x y)
(define (diff-product x E)
  (cond ((and (eq? x (cadr E)) (eq? x (caddr E))) (list '* 2 x))
        ((eq? x (cadr E)) (list (caddr E)))
        ((eq? x (caddr E)) (list (cadr E)))
        (else 0)))

;; computes derivative of E with respect to x where E can only be of the form
;; (expt x y) which is x^y
(define (diff-expt x E)
  (cond ( not (eq? x (cadr E)) 0)
        ((eq? 1 (caddr E)) 0)
        ((eq? 2 (caddr E)) (cadr E))
        (else (list '* (caddr E) (list 'expt x (- (caddr E) 1))))))

我还有一个调度表定义为:

;; Dispatch Table of supported operators.
 (define diff-dispatch
   (list (list '+ diff-sum)
         (list '* diff-product)
         (list 'expt diff-expt)
         ))

我正在尝试编写一个函数diff，它采用方程式E(以列表形式)并计算关于x 的导数> 并使用调度表调用预定义函数返回结果

这是我目前所知道的，但我想不出其余的

;; Differentiate expression E with respect to x.
(define (diff x E)
  (cond ((number? E) (diff-constant x E))
        ;; insert code here to handle the other base case - variables
        ...
        (else    ; insert code here to lookup the appropriate function in the
                 ; dispatch table based on the operator in the expression,
                 ; then call that function to differentiate the expression
                     (diff-func x E))))))

ex: (diff 'x '(+ x (* x x))) 应该计算为 (+ 1 (+ (* 1 (* x)) (* x 1) ))(即 1 + x + x)

Answer 1

在SICP book有一整节详细解释了如何构建用于执行符号微分的 Scheme 程序，请查看 section §2.3 .特别要注意，您遗漏了一种情况——如果要派生的表达式是一个变量，会发生什么情况？检查链接以确保您在正确的轨道上。

现在，回答这个问题:根据代码中使用的表表示，实现调度程序很简单。这些方面的一些东西将有助于获得应用正确的微分程序:

((cadr             ; obtain the differentiation procedure
  (assoc           ; look for the differentiation procedure
   (car E)         ; obtain the operator in the expression
   diff-dispatch)) ; dispatch table
 x E)              ; apply the procedure with parameters `x` and `E`

请注意，找到要应用的正确过程的“诀窍”在于表是作为关联列表实现的，而 assoc 过程正是为在这样的列表中查找数据而设计的列表。在 documentation 中阅读更多相关信息.