平方根を求めるsqrt関数 先日のもの（【scheme】1.1.7 例：Newton法による平方根＠SICP）
| (define avarage
(lambda (x y)
(/ (+ x y) 2)))
(avarage 10 20);;=> 15
(avarage 5.0 8.0);;=> 6.5 (define square
(lambda (x)
(* x x)))
(square 5);;=> 25
(square 12);;=> 144 (define improve
(lambda (guess x)
(avarage guess (/ x guess))))
(improve 10.0 5.0);;=> 5.25
(improve 1.5 2.5);;=> 1.5833333333333335 (define good-enough?
(lambda (guess x)
(< (abs (- (square guess) x)) 0.001))) (define sqrt-iter
(lambda (guess x)
(if (good-enough? guess x)
guess
(sqrt-iter (improve guess x) x)))) (define sqrt
(lambda (x)
(sqrt-iter 1.0 x))) (sqrt 9);;=> 3.00009155413138
(sqrt (+ 100 37));;=> 11.704699917758145
(sqrt (+ (sqrt 2)(sqrt 3)));;=> 1.7739279023207892
(square (sqrt 1000));;=> 1000.000369924366 |
各関数をsqrt定義内に納めたもの。
| ;; local scope difinition
(define sqrt
(lambda (x)
(define square
(lambda (x)
(* x x)))
(define avarage
(lambda (x y)
(/ (+ x y) 2)))
(define improve
(lambda (guess x)
(avarage guess (/ x guess))))
(define good-enough?
(lambda (guess x)
(< (abs (- (square guess) x)) 0.001)))
(define sqrt-iter
(lambda (guess x)
(if (good-enough? guess x)
guess
(sqrt-iter (improve guess x) x))))
(sqrt-iter 1.0 x)))
(sqrt 9);;=> 3.00009155413138 |
sqrtのパラメータxのブロック内スコープ（レキシカルスコープ）を利用したもの。 xを利用した関数を返すようにすればクロージャできそうでつね。
| (define square
(lambda (x)
(* x x)))
(define avarage
(lambda (x y)
(/ (+ x y) 2))) (define sqrt
(lambda (x)
(define improve
(lambda (guess)
(avarage guess (/ x guess))))
(define good-enough?
(lambda (guess)
(< (abs (- (square guess) x)) 0.001)))
(define sqrt-iter
(lambda (guess)
(if (good-enough? guess)
guess
(sqrt-iter (improve guess)))))
(sqrt-iter 1.0)))
(sqrt 9);;=> 3.00009155413138 |
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