I have done exercise 1.7 in SICP (calculate square root precision when change in guesses is under a certain value), but I am calling the change-in-precision function twice in each iteration, which disturbs me. I wonder if there is a better way to implement this.

(define (average . ns) (/ (apply + ns) (length ns)))

(define (change-in-precision guess x)
  ( - (- guess (average guess (/ x guess)))))

(define (sqrt guess x)
  (if (< (abs (change-in-precision guess x)) (/ 0.00000001 guess))
  (+ guess (change-in-precision guess x))
  (sqrt (+ guess (change-in-precision guess x)) x)))

1 Answer 1


In change-in-precision, you can avoid the negation by swapping the operands for the subtraction.

In sqrt, if the guess is already close enough, why not just return guess?

To eliminate the repeated call to change-in-precision, use a let to define a variable delta.

The two branches of the if should be indented.

  • \$\begingroup\$ (/ 0.00000001 guess) is not the same as (/ guess 100000000); in the former, guess is in the denominator, whereas in the latter, guess is in the numerator. But (/ 1 guess 100000000) could work. Not sure if it's "more readable" at that stage, though. \$\endgroup\$ Commented Feb 20, 2016 at 18:04
  • \$\begingroup\$ @200_success Thank you for points 1,2 and 4. For point 3, I wanted to try and avoid a variable as we haven't done that in the book yet. \$\endgroup\$
    – leancz
    Commented Feb 21, 2016 at 9:37
  • \$\begingroup\$ @ChrisJester-Young Yes my way of doing it looks strange, but it helps me to think about what the precision implies. \$\endgroup\$
    – leancz
    Commented Feb 21, 2016 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.