問題1.9、問題1.10 – SICP(計算機プログラムの構造と解釈)その4
2008年11月10日
問題1.9
再帰的プロセスと反復的プロセス
次の手続きは再帰的プロセスを生成する。
(define (+ a b)
(if (= a 0)
b
(inc (+ (dec a) b))))
(+ 4 5)
(inc (+ (dec 4) 5))
(inc (inc (+ (dec 3) 5)))
(inc (inc (inc (+ (dec 2) 5))))
(inc (inc (inc (inc (+ (dec 1) 5)))))
(inc (inc (inc (inc (+ 0 5)))))
(inc (inc (inc (inc 5))))
(inc (inc (inc 6)))
(inc (inc 7))
(inc 8)
9
次の手続きは反復的プロセスを生成する。
(define (+ a b)
(if (= a 0)
b
(+ (dec a) (inc b))))
(+ 4 5)
(+ (dec 4) (inc 5))
(+ 3 6)
(+ (dec 3) (inc 6))
(+ 2 7)
(+ (dec 2) (inc 7))
(+ 1 8)
(+ (dec 1) (inc 8))
(+ 0 9)
9
問題1.10
(define (A x y)
(cond ((= y 0) 0)
((= x 0) (* 2 y))
((= y 1) 2)
(else (A (- x 1)
(A x (- y 1))))))
(A 1 10)
(A 0 (A 1 9))
(A 0 (A 0 (A 1 8)))
(A 0 (A 0 (A 0 (A 1 7))))
(A 0 (A 0 (A 0 (A 0 (A 1 6)))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 1 5))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 4)))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 3))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 2)))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 1))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 2)))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 4))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 8)))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 16))))))
(A 0 (A 0 (A 0 (A 0 (A 0 32)))))
(A 0 (A 0 (A 0 (A 0 64))))
(A 0 (A 0 (A 0 128)))
(A 0 (A 0 256))
(A 0 512)
1024
(A 2 4)
(A 1 (A 2 3))
(A 1 (A 1 (A 2 2)))
(A 1 (A 1 (A 1 (A 2 1))))
(A 1 (A 1 (A 1 2)))
(A 1 (A 1 (A 0 (A 1 1))))
(A 1 (A 1 (A 0 2)))
(A 1 (A 1 4))
(A 1 (A 0 (A 1 3)))
(A 1 (A 0 (A 0 (A 1 2))))
(A 1 (A 0 (A 0 (A 0 (A 1 1)))))
(A 1 (A 0 (A 0 (A 0 2))))
(A 1 (A 0 (A 0 4)))
(A 1 (A 0 8))
(A 1 16)
(A 0 (A 1 15))
(A 0 (A 0 (A 1 14)))
(A 0 (A 0 (A 0 (A 1 13))))
(A 0 (A 0 (A 0 (A 0 (A 1 12)))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 1 11))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 10)))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 9))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 8)))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 7))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 6)))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 5))))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 4)))))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 3))))))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 2)))))))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 1 1))))))))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 2)))))))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 4))))))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 8)))))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 16))))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 32)))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 64))))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 128)))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 256))))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 (A 0 512)))))))
(A 0 (A 0 (A 0 (A 0 (A 0 (A 0 1024))))))
(A 0 (A 0 (A 0 (A 0 (A 0 2048)))))
(A 0 (A 0 (A 0 (A 0 4096))))
(A 0 (A 0 (A 0 8192)))
(A 0 (A 0 16384))
(A 0 32768)
65536
(A 3 3)
(A 2 (A 3 2))
(A 2 (A 2 (A 3 1)))
(A 2 (A 2 2))
(A 2 (A 1 (A 2 1)))
(A 2 (A 1 2))
(A 2 (A 0 (A 1 1)))
(A 2 (A 0 2))
(A 2 4)
(A 1 (A 2 3))
(A 1 (A 1 (A 2 2)))
(A 1 (A 1 (A 1 (A 2 1))))
(A 1 (A 1 (A 1 2)))
(A 1 (A 1 (A 0 (A 1 1))))
(A 1 (A 1 (A 0 2)))
(A 1 (A 1 4))
(A 1 (A 0 (A 1 3)))
(A 1 (A 0 (A 0 (A 1 2))))
(A 1 (A 0 (A 0 (A 0 (A 1 1)))))
(A 1 (A 0 (A 0 (A 0 2))))
(A 1 (A 0 (A 0 4)))
(A 1 (A 0 8))
(A 1 16)
...
(A 0 32768)
65536
(define (f n) (A 0 n))
(A 0 n)
(* 2 n)
(f n)
は 2*n
を計算する。
(define (g n) (A 1 n))
(A 1 n)
(A 0 (A 1 (- n 1))) ;=> (* 2 (A 1 (- n 1)))
(A 0 (A 0 (A 1 (- n 2)))) ;=> (* 2 2 (A 1 (- n 2)))
(A 0 (A 0 (A 0 (A 1 (- n 3))))) ;=> (* 2 2 2 (A 1 (- n 3)))
(A 0 (A 0 (A 0 (A 0 (A 1 (- n 4)))))) ;=> (* 2 2 2 2 (A 1 (- n 4)))
(g n)
は 2^n
を計算する。
(define (h n) (A 2 n))
(A 2 n)
(A 1 (A 2 (- n 1))) ;=> 2^(h (- n 1))
(A 1 (A 1 (A 2 (- n 2)))) ;=> 2^2^(h (- n 2))
(A 1 (A 1 (A 1 (A 2 (- n 3))))) ;=> 2^2^2^(h (- n 3))
(A 1 (A 1 (A 1 (A 1 (A 2 (- n 4)))))) ;=> 2^2^2^2^(h (- n 4))
(h n)
は 2^2^2^...^2
(n個) を計算する。
計算機プログラムの構造と解釈
posted with amazlet at 08.11.07
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