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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