SICP - Solution: Exercise 1.6

September 28, 2018

Exercise 1.6 #

Alyssa P. Hacker doesn’t see why if needs to be provided as a special form. “Why can’t I just define it as an ordinary procedure in terms of cond?” she asks. Alyssa’s friend Eva Lu Ator claims this can indeed be done, and she defines a new version of if:

(define (new-if predicate
                then-clause
                else-clause)
  (cond (predicate then-clause)
        (else else-clause)))

Eva demonstrates the program for Alyssa:

(new-if (= 2 3) 0 5)
5

(new-if (= 1 1) 0 5)
0

Delighted, Alyssa uses new-if to rewrite the square-root program:

(define (sqrt-iter guess x)
  (new-if (good-enough? guess x)
          guess
          (sqrt-iter (improve guess x) x)))

What happens when Alyssa attempts to use this to compute square roots? Explain.

Solution #

Since new-if is a function, and not a special form, each parameter subexpression will be evaluated before the procedure is applied. It means that when executing the sqrt-iter function, (good-enough? guess x), guess and (sqrt-iter (improve guess x) x) will always be evaluated before new-if is applied, whatever (good-enough? guess x) evaluates to.

Since the second alternative (sqrt-iter (improve guess x) x) is calling the function recursively, the function will be stuck in an infinite loop. In this case, it is interesting to note that new-if will never be executed.

You can check by executing:

(define (new-if predicate then-clause else-clause)
  (cond (predicate then-clause)
        (else else-clause)))

(define (average x y) (/ (+ x y) 2))

(define (square x) (* x x))

(define (improve guess x) (average guess (/ x guess)))

(define (good-enough? guess x)
(< (abs (- (square guess) x)) 0.001))


(define (sqrt-iter guess x)
  (new-if (good-enough? guess x)
          guess
          (sqrt-iter (improve guess x) x)))


(sqrt-iter 1.0 4) ; Infinite loop