SICP Exercise 1.1 ~ 1.10

Exercise 1.1 

10

(+ 5 3 4)

(- 9 1)

(/ 6 2)

(+ (* 2 4) (- 4 6))

(define a 3)

(define b (+ a 1))

(+ a b (* a b))

(= a b)

(if (and (> b a) (< b (* a b)))

    b

    a)

(cond ((= a 4) 6)

      ((= b 4) (+ 6 7 a))

      (else 25))

(+ 2 (if(> b a) b a))

(* (cond ((> a b) a)

         ((< a b) b)

         (else -1))

   (+ a 1))

Exercise 1.2

(/ (+ 5 4 (- 2 (- 3 (+ 6 4/5))))(* 3 (- 6 2)(- 2 7)))

Exercise 1.3

(define (square x)

  (* x x))

(define (sum-of-squares x y)

  (+ (square x) (square y)))

(define (sss x y z)

  (cond ((and (> x z) (> y z)) (sum-of-squares x y))

        ((and (> y x) (> z x)) (sum-of-squares y z))

        ((and (> x y) (> z y)) (sum-of-squares x z))

        ))

(sss 4 3 4)

Exercise 1.5

applicative-order에서는 (test 0 (p))가 절대 안끝난다. 왜냐면 (p)가 무한대로 불러지기 때문이다.

normal-order evaluation에서는 표현의 평가가 차례로 이뤄지므로 결과가 나온다.

Exercise 1.6

무한루프에 빠짐. Scheme은 applicative 방식이라 operand를 먼저 평가하기 때문임.

Exercise 1.7

(define (new-good-enough? guess x)

  (< (abs (- guess (improve guess x))) 0.0000001))

Exercise 1.8

(define (cube-improve guess x)

   (/ (+ (/ x (square guess)) (* 2 guess))

     3))

(cube-improve 3 27)

(define (cube-good-enough? guess x)

  (< (abs (- guess (cube-improve guess x))) 0.0000001))

(define (cube-iter guess x)

  (if (cube-good-enough? guess x)

      guess

      (cube-iter (cube-improve guess x)

                 x)))

(define (cube x)

  (cube-iter 1.0 x))

(cube 64.0)

Exercise 1.9

(+ 4 5)                        

(inc (+ (dec 4) 5))            

(inc (+ 3 5))                  

(inc (inc (+ 2 5)))            

(inc (inc (inc (+ 1 5))))      

(inc (inc (inc (inc (+ 0 5)))))

(inc (inc (inc (inc 5))))      

(inc (inc (inc 6)))            

(inc (inc 7))                  

(inc 8)                        

9                              

                               

(+ 4 5)                        

(+ (dec 4) (inc 5))            

(+ 3 6)                        

(+ (dec 3) (inc 6))            

(+ (dec 2) (inc 7))            

(+ (dec 1) (inc 8))            

(+ 0 9)                        

(9)                            

Exercise 1.10

(f n): 2n
(g n): 2^{n}
(h n): 2^2^… (n-1 times)