Talk:Structure and Interpretation of Computer Programs

Discussion of exercises from Structure and Interpretation of Computer Programs assumes the use of the MIT Scheme interpreter.

Chapter 1

Exercise 1.1

10
;Value 10

(+ 5 3 4)
;Value 12

(- 9 1)
;Value 8

(/ 6 2)
;Value 3

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

(define a 3)
;Value "a --> 3"

(define b (+ a 1))
;Value "b --> 4"

(+ a b (* a b))
;Value 19

(= a b)
;Value #f

(if (and (> b a) (< b (* a b)))
   b
   a)
;Value 4

(cond ((= a 4) 6)
     ((= b 4) (+ 6 7 a))
     (else 25))
;Value 16

(+ 2 (if (> b a) b a))
;Value 6

(* (cond ((> a b) a)
        ((< a b) b)
        (else -1))
  (+ a 1))
;Value 16

Exercise 1.2

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

Exercise 1.3

(define (sum-larger-squares a b c)
 (cond 
  ((and (<= a b) (<= a c)) (+ (* b b) (* c c)))
  ((and (<= b a) (<= b c)) (+ (* a a) (* c c)))	
  (else (+ (* a a) (* b b)))))

Exercise 1.4

Determine the operation based on the value of b. When b is greater than zero, add b to a. Otherwise, subtract b from a.

Exercise 1.5

Normal order:

(test 0 (p))
; substitute the definition of test
(if (= 0 0) 0 (p))
; if forces the first argument to be evaluated
(if #t 0 (p))
; because the first argument is #t, the second argument
; is evaluated and the third argument is ignored
;Value 0

Applicative order:

(test 0 (p))
; substitute the definition of test and the arguments
; (p) is defined as (p), so the interpreter enters an
; infinite loop

Exercise 1.6

Because new-if is a user defined function and user defined functions evaluate arguments in applicative order, the call to sqrt-iter will cause infinite recursion.