1.2

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

 

1.3

(define (f x y z)
    (cond ((and (<= x y) (<= x z)) (+ (square y) (square z)))
          ((and (<= y x) (<= y z)) (+ (square x) (square z)))
          (and (<= z x) (<= z y)) (+ (square x) (square y))
          )
    )

1.4

 

1.5

Normal order 를 가정했을 때

(test 0 (p))

(if (= 0 0) 0 (p)))

0 가 리턴되고 끝

 

Applicative order 를 가정했을 때

(test 0 (p))

evaluate the arguments

(test 0 (p))

evaluate 작업을 무한히 반복하게 됨.