Exercise 2.58
Suppose we want to modify the differentiation program so that it works with ordinary mathematical notation, in which + and * are infix rather than prefix operators. Since the differentiation program is defined
in terms of abstract data, we ca modify it to work with different representations of expressions solely by changing the predicates, selectors, and constructors that define the representation of the algebraic
expressions on which the differentiator is to operate.a. Show how to do this in order to differentiate algebraic expressions presented in infix form, such as (x + (3 * (x + (y + 2)))). To simplify the task, assume that + and * always take two arguments and that
expressions are fully parenthesized.
b. The problem becomes substantially harder if we allow standard algebraic notation, such as (x + 3 * (x + y + 2)), which drops unnecessary parentheses and assumes that multiplication is done before addition.
Can you design appropriate predicates, selectors, and constructors for this notation such that our derivative program still works?
這道題乍一看很棘手，但是仔細一想其實挺簡單的，只要把建構函式和選擇器裡 + 和 * 的位置改一下就行了，還有原來的字首表示法要多加一個 + 或 *，現在也可以直接省略掉了。
(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 (sum? x) (and (pair? x) (eq? (cadr x) '+)))
(define (addend s) (car s))
(define (augend s)
(let ((last (cddr s)))
(if (null? (cdr last))
(car last)
last)))
(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))))
(define (product? x) (and (pair? x) (eq? (cadr x) '*)))
(define (multiplier p) (car p))
(define (multiplicand p)
(let ((last (cddr p)))
(if (null? (cdr last))
(car last)
last)))
(deriv '(x + (3 * (x + (y + 2)))) 'x)
(deriv '(x + 3 * (x + y + 2)) 'x)
; 執行結果
4
4