SICP - Solution: Exercise 1.6

# SICP - Solution: Exercise 1.6

## 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