(define (tree->list-1 tree) (if (null? tree) '() (append (tree->list-1 (left-branch tree)) (cons (entry tree) (tree->list-1 (right-branch tree)))))) (define (tree->list-2 tree) (define (copy-to-list tree result-list) (if (null? tree) resul…

(define (union-set s t) (cond ((null? s) t) ((null? t) s) ((= (car s) (car t)) (cons (car s) (union-set (cdr s) (cdr t)))) ((< (car s) (car t)) (cons (car s) (union-set (cdr s) t))) ((< (car t) (car s)) (cons (car t) (union-set s (cdr t)))…

(define (adjoin-set x s) (cond ((null? s) (list x)) ((= x (car s)) s) ((< x (car s)) (cons x s)) (else (cons (car s) (adjoin-set x (cdr s)))))) ;; 同じ数字，またはxより大きい数字が出てきた時点で計算が終わるので順序付けられない表現に比べ半…

;; element-of-set? intersection-setはそのまま (define (adjoin-set x s) (cons x s)) (define (union-set s t) (append s t)) ;; element-of-set?やintersection-setについてはsetの中身が増えることで比較回数が増えて効率は下がる． ;; adjoin-set unio…

(define (element-of-set? x set) (cond ((null? set) false) ((equal? x (car set)) true) (else (element-of-set? x (cdr set))))) (define (unionset s t) (cond ((null? s) t) ((element-of-set? (car s) t) (unionset (cdr s) t)) (else (cons (car s) …

;; a (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)…

(define (augend s) (if (null? (cdddr s)) (caddr s) (cons '+ (cddr s)))) (define (multiplicand p) (if (null? (cdddr p)) (caddr p) (cons '* (cddr p)))) これ作るので精一杯でした． make-sumやmake-productを可変長引数に対応できるように変更するの…