SICP 题解集合

1.1（略）

1.2

 biwascheme>

 (/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 4 5)))))

    (* 3 (- 6 2) (- 2 7)))

 => -0.24666666666666667

1.3

 biwascheme>

 (define (BiggerSum x y z)

     (define s1 (+ x y))

     (define s2 (+ x z))

     (define s3 (+ y z))

     (cond ((and (> s1 s2) (> s1 s3)) s1)

           ((and (> s2 s1) (> s2 s3)) s2)

           (else s3)

     )

 )

 biwascheme> (BiggerSum 2 1 3)

 => 5

1.4 (略）

1.5

若解释器采用正则序求值，Ben可以看到解释器顺利返回0；

若解释器采用应用序，则程序会陷入无限循环。

1.6 (略）

1.7

 (define (square x)

  (* x x))

 (define (improve guess x)

  (average guess (/ x guess)))

 (define (average x y)

  (/ (+ x y) ))

 (define (good-enough? guess0 guess1)

  (< (abs (- (/ guess0 guess1) )) 0.001))

 (define (sqrt-iter guess x)

  (define new-guess (improve guess x))

  (if (good-enough? guess new-guess)

     guess

     (sqrt-iter new-guess x)))

 (define (sqrt x)

  (sqrt-iter 1.0 x))

1.8

 (define (square x)

  (* x x))

 (define (improve guess x)

  (/ (+ (/ x (square guess)) (*  guess))

     )

 )

 (define (good-enough? guess0 guess1)

  (< (abs (- (/ guess0 guess1) )) 0.001))

 (define (cube-root-iter guess x)

  (define new-guess (improve guess x))

  (if (good-enough? guess new-guess)

     guess

     (cube-root-iter new-guess x)))

 (define (cube-root x)

  (cube-root-iter 1.0 x))

巴特西

SICP 题解集合

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