2
\$\begingroup\$

4.2 Machine Epsilon

Find the floating point number epsi that has the the following properties:

  1. 1.0+epsi is greater than 1.0 and
  2. Let m b e any number less than epsi. Then 1.0+m is equal to 1.0.

epsi is called machine epsilon. It is of great importance in understanding floating point numbers.

I have written the following program to try and find my machine epsilon value to a given search depth (10). Do you have any ideas how I could write this program better?

(defpackage :find-epsi (:use cl))
(in-package :find-epsi)

(defun smaller-scale (&OPTIONAL (epsi 1.0)) (if (> (+ 1.0 epsi) 1.0) (smaller-scale (/ epsi 10)) epsi))
(defun bigger (epsi inc-unit) (if (< (+ 1.0 epsi) 1.0) (bigger (+ epsi inc-unit) inc-unit) epsi))
(defun smaller (epsi dec-unit) (if (> (+ 1.0 epsi) 1.0) (smaller (+ epsi dec-unit) dec-unit) epsi))

(defun find-epsi (&OPTIONAL (search-depth 10) (epsi (smaller-scale)) (incdec-unit epsi))
  (if (= search-depth 0) epsi (find-epsi (1- search-depth) (bigger (smaller epsi incdec-unit) incdec-unit) incdec-unit)))

(format t "epsi: ~a ~%" (find-epsi))

It seems that it should be much simpler to find epsilon than I originally thought. What do you think about the following program?

(defpackage :find-epsi (:use cl))
(in-package :find-epsi)

(defun find-epsi (&OPTIONAL (epsi 1.0)) 
  (if (> (+ 1.0 epsi) 1.0)  ; if the variable epsi is still significant
    (find-epsi (/ epsi 2)) ; halve it and try again
    epsi)) ; otherwise, we have found epsilon

(format t "epsi: ~a ~%" (find-epsi))
\$\endgroup\$
1
  • \$\begingroup\$ As a general note you should get used to writing comments and (at least in this case) shorter lines. \$\endgroup\$
    – sepp2k
    Mar 17, 2011 at 22:55

1 Answer 1

3
\$\begingroup\$

If we assume that a float is represented in memory as a (sign, mantissa, exponent) tuple, and assume a radix of 2, then we can find the machine epsilon exactly. That is, if we can assume the machine stores floats using base-2 representations of the mantissa and exponent, then we know that:

  • The machine will store a value of 1 in floating point exactly - this would be stored as 1 for the mantissa, and 0 for the exponent, i.e. 1 * 2^0.
  • The machine will store all powers of two that it can represent using a single bit in the mantissa, and by varying the exponent. E.g. 1/4 could be represented as 1 * (2 ^ -2). Any representable power of two will be stored without losing information.
  • 1 + epsi will be the smallest value greater than 1 that can be stored in the mantissa of the floating-point number.

EDIT

The second version looks much better than the first, but I believe there's an off-by-one error in the number of times you recurse in find-epsi. I suggest that you create a test function, to see if your result is the machine epsilon:

(defun epsi-sig-p (epsi)
  (> (+ 1.0 epsi) 1.0))

You'll probably find that (is-sig (find-epsi)) is #f... This also suggests that you can refactor (under the DRY principle) find-epsi to use this test function in find-epsi:

(defun find-epsi (&OPTIONAL (epsi 1.0))
  (if (epsi-sig-p epsi)
    (find-epsi (/ epsi 2))
    epsi))

but we didn't change the behavior to fix the calculation, yet. For this, I'd suggest another routine, to check whether we should try the next possible epsi:

(defun next-epsi (epsi) (/ epsi 2))
(defun is-epsi-p (epsi)
  (and (epsi-sig-p epsi) (not (epsi-sig-p (next-epsi epsi)))))
(defun find-epsi (&OPTIONAL (epsi 1.0))
  (if (is-epsi-p epsi)
    epsi
    (find-epsi (next-epsi epsi))))

is-epsi-p should return #t, now.

\$\endgroup\$
1
  • 2
    \$\begingroup\$ I'm pretty sure jaresty is going through some collection of exercises voluntarily (or possibly in preparation for an exam), not as homework. It seems unlikely that a university would give out homework at the rate at which he's asking questions. \$\endgroup\$
    – sepp2k
    Mar 17, 2011 at 22:51

Your Answer

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

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