SICP Exercise 1.1 to Exercise 1.6

Exercise 1.1 This exercise is very easy, just evaluate some expressions.

1 ]=> 10

;Value: 10

1 ]=> (+ 5 3 4)

;Value: 12

1 ]=> (- 9 1)

;Value: 8

1 ]=> (/ 6 2)

;Value: 3

1 ]=> (+ (* 2 4) (- 4 6))

;Value: 6

1 ]=> (define a 3)

;Value: a

1 ]=> (define b (+ a 1))

;Value: b

1 ]=> (+ a b (* a b))

;Value: 19

1 ]=> (= a b)

;Value: #f

1 ]=> (if (and (> b a) (< b (* a b)))
          b
          a)

;Value: 4

1 ]=> (cond ((= a 4) 6)
            ((= b 4) (+ 6 7 a))
            (else 25))

;Value: 16

Exercise 1.2 Define a procedure that takes three numbers as arguments and returns the sum of the squares of the two larger numbers.

Solution: First, find the smallest number of three numbers. So the result is the sum of the squares of three numbers subtract the square of the smallest number.

(define (sum-of-square-bigers x y z)
              (let ((minimum (min x y z)))
              (- (+ (square x)
                    (square y)
                    (square z))
                 (square minimum)))) 

Exercise 1.3 Ben Bitdiddle has invented a test to determine whether the interpreter he is faced with is using applicative-order evaluation or normal-order evaluation. He defines the following two procedures:

(define (p) (p))
(define (test x y)
  (if (= x 0)
      0
      y))

Then he evaluates the expression

(test 0 (p))

What behavior will Ben observe with an interpreter that uses applicative-order evaluation? What behavior will he observe with an interpreter that uses normal-order evaluation? Explain your answer.