SICP 2.56
記号微分。面白いテーマだ。こんなことができるなんてすごい。
まず、同じ関数を何度も書くのが面倒なので、2.3.2.scmというファイルに定義した。
(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)))) (define (variable? x) (symbol? x)) (define (same-variable? v1 v2) (and (variable? v1) (variable? v2) (eq? v1 v2))) (define (make-sum a1 a2) (list '+ a1 a2)) (define (make-product m1 m2) (list '* m1 m2)) (define (sum? x) (and (pair? x) (eq? (car x) '+))) (define (addend s) (cadr s)) (define (augend s) (caddr s)) (define (product? x) (and (pair? x) (eq? (car x) '*))) (define (multiplier p) (cadr p)) (define (multiplicand p) (caddr p)) (define (=number? exp num) (and (number? exp) (= exp num))) (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))))
これを使って2.56の問題を解く。
(load "./2.3.2.scm") (define (exponentiation? e) (and (pair? e) (eq? '** (car e)))) (define base cadr) (define exponent caddr) (define (make-exponentiation b e) (cond ((=number? e 0) 1) ((=number? e 1) b) (else (list '** b e)))) (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) (make-product (exponent exp) (make-exponentiation (base exp) (make-sum (exponent exp) -1)))) (else (error "unknown expression type -- DERIV" exp)))) ; test (deriv '(** x 3) 'x) (deriv '(** x 1) 'x) (deriv '(** x 0) 'x)