開発環境
- OS X Lion - Apple(OS)
- Emacs、BBEdit - Bare Bones Software, Inc. (Text Editor)
- プログラミング言語: MIT/GNU Scheme
計算機プログラムの構造と解釈(Gerald Jay Sussman(原著)、Julie Sussman(原著)、Harold Abelson(原著)、和田 英一(翻訳)、ピアソンエデュケーション)の2(データによる抽象の構築)、2.3(記号データ)、2.3.2(例: 記号微分)の問題 2.57を解いてみる。
その他参考書籍
問題 2.57
コード
sample.scm
(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))))
((exponentiation? exp)
(let ((n (exponent exp))
(u (base exp)))
(make-product n
(make-product (make-exponentiation u
(make-sum n -1))
(deriv u var)))))
(else (error "unkown expression type -- DERIV" exp))))
(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 (addend s) (cadr s))
; 任意個の項の和のための修正箇所
(define (augend s)
(if (null? (cdddr s))
(caddr s)
(cons '+ (cddr s))))
(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 (multiplier p) (cadr p))
; 任意個の項の積のための修正箇所
(define (multiplicand p)
(if (null? (cdddr p))
(caddr p)
(cons '* (cddr p))))
(define (variable? x) (symbol? x))
(define (same-variable? v1 v2)
(and (variable? v1) (variable? v2) (eq? v1 v2)))
(define (sum? x)
(and (pair? x) (eq? (car x) '+)))
(define (product? x)
(and (pair? x) (eq? (car x) '*)))
(define (exponentiation? x)
(and (pair? x) (eq? (car x) '**)))
(define (base x) (cadr x))
(define (exponent x) (caddr x))
(define (make-exponentiation a b)
(cond ((=number? b 0) 1)
((=number? b 1) a)
(else (list '** a b))))
(define (=number? exp num)
(and (number? exp) (= exp num)))
入出力結果(Terminal, REPL(Read, Eval, Print, Loop))
1 ]=> (deriv '(* (* x y) (+ x 3)) 'x) ;Value 2: (+ (* x y) (* y (+ x 3))) 1 ]=> (deriv '(* x y (+ x 3)) 'x) ;Value 3: (+ (* x y) (* y (+ x 3)))
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